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f58b1ac78a60d40365464118b4bde44060f2a937
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Q: A question about recurrence in a Newton's method problem? The equation $x^3-x=0$ has three roots,$ -1, 0, 1.$ We use the Newton's method to find the roots. And there are three cases
(i) If $x_0>1/\sqrt{3}$, the Newton's method will converge to $1$.
(ii) If $x_0<-1/\sqrt{3}$, the Newton's method will converge to $-1$.
(iii) If $-1/\sqrt{5}<x_0<1/\sqrt{5}$, the Newton's method will converge to $0$.
Here is the question I don't understand:
Consider $x_0$ is a little less than $1/\sqrt{3}$, there is an implicit recurrence relation that produces a decreasing sequence ${a_1=q/\sqrt{3}, a_2, a_3, ...}$, by means of which one you can easily find $\lim_{n\to\infty}x_n$ for any $x_0\in(1/\sqrt{5},1/\sqrt{3})$. Try to find the recurrence.
The answer provided is $a_i-f(a_i/f'(a_i))=-a_{i-1}$; $\lim_{n\to\infty}x_n=(-1)^i$ if $x_0\in(a_i,a_{i+1}); a_1=0.577, a_2=0.462, a_4 \approx a_i=1/\sqrt{5}=0.447$.
Could anyone kindly help explain what is $a_i$, what does this have to do with the Newton's method? Thanks so much!
PS: someone thinks my question is unclear, so I post the original question as the following. My question is 1.3(d).
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d4704c80ede3eb2f5f084e7aef32546263c3f82a
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Q: Need help creating a single formula to find the probability of successfully rolling multiple 6-sided dice with conditions. Example question: If you roll 7 dice, what is the probability of successfully rolling a 5 or higher, at least 4 times?
Let a=7 b=5 c=4.
My goal is to plug a,b,c into a long formula and have the probability of success for the example above.
A: Simulation. A simulation of a million 7-dice experiments in R is as follows. (At the end, the vector big contains the number of 5'5 or 6's seen on the seven dice rolled in each experiment.)
m = 10^6; n = 7; big = numeric(m)
for (i in 1:m) {
roll = sample(1:6, n, rep=T)
big[i] = sum(roll >= 5) }
mean(big >= 4) # proportion of expts with at least four 5's or 6's
## 0.173095 # Approx. answer
The following relative frequency histogram summarizes the one million results. (There were 463 results out of a million with seven big results, but they are relatively too few to show on this scale.)
Exact formula. Now for a formula that gives the exact answer, based on the
Comment by @AndreNicholas. Let $f(i, 7, 2/6) = {7 \choose i} (2/6)^i(4/6)^{7-i}$, for $i = 0, 1, \dots, 7,$ be the PDF
(or PMF, depending on text) of the Binomial distribution
based on $n = 7$ trials with success probability $p = 2/6$
on each trial. Also let $F(x, 7, 2/6) = \sum_{i=0}^x f(i, 7, 2/6)$
be the cumulative distribution function (CDF) of this binomial
distribution. You should match my notation with the notation
in your text.
In R, the function dbinom denotes the PDF and pbinom denotes the CDF.
Thus, your answer, $\sum_{i=4}^7 f(i, 7, 2/6) = 1 - F(3, 7, 2/6) = 0.1732968$
can be computed in R in two ways:
i = 4:7; sum(dbinom(i, 7, 2/6))
## 0.1732968 # exact answer summing PDF terms
1 - pbinom(3, 7, 2/6)
## 0.1732968 # exact answer using CDF
Notice that the simulation was accurate to three places.
The figure below shows a bargraph of the PDF of $Binom(7, 2/6),$
with the four probabilities that add to your answer (including the tiny probability at 7) highlighted as thick
blue bars.
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be44046db512bf84d0578fd42aa59b8ef3bccc16
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Q: Determine all positive rational numbers $r \neq 1$ such that $r^{\frac{1}{r-1}}$ is rational? Here's what I've got so far:
Let $r = \frac{a}{b}$, where $a$ and $b$ are integers. We then have $$r^{\frac{1}{r-1}} = \frac{a^{\frac{b}{a-b}}}{b^{\frac{b}{a-b}}}$$
Clearly, $a-b=1$ and $a-b=-1$ work. But it might be the case that $a-b \neq 1$ or $a-b \neq -1$, but $\frac{a^{\frac{b}{a-b}}}{b^{\frac{b}{a-b}}}$ still be rational, if it turns out that $a=p^{\frac{1}{a-b}}$ and $b=q^{\frac{1}{a-b}}$.
But how do I obtain a contradiction if $a-b > 1$, or $a-b<-1$? I let $a-b=n$, which means that $a=p^{n}$ and $b=q^{n}$, so $$a-b = p^n - q^n = n$$
This is strangely curious, because it's worked out well so far. I can see that the expression in the middle can be factored, but how will that help with the contradiction?
If you have your own way of approaching this problem, feel free to comment. Thanks.
A: It is much easier to start with $\frac{1}{1-r}=\frac{a}{b}$ so $r=\frac{a-b}{a}$. then you want $a-b$ and $a$ to be perfect $b$th power, or $b-a$ and $-a$ to be perfect $b$th powers. But $\left|(k+1)^b-k^b\right|$ is greater than $b$ when $b>1$ and $k\neq -1,0$, and the $k=-1,0$ cases are no good when $b>1$. So if $b>1$ then there are no solutions.
The case when $b=1$, however, gives a solution for any $a\neq 0$.
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c1775f39d6a018f7cf5b27d1d25421b3cabcc38d
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Q: Geodesic - How can we continue? I want to show that a geodesic with nowhere vanishing curvature is a plane curve if and only if it is a line of curvature.
I have done the following:
Let $\gamma$ be a geodesic with nowhere vanishing curvature and a plane curve.
Let $N$ be the unit normal of the surface.
We have that $\|N\|^2=1 \Rightarrow N\cdot N=1 \Rightarrow (N\cdot N)'=0\Rightarrow N'\cdot N=0 \Rightarrow N'\bot N$.
$\gamma'$ is the tangent vector of $\gamma$.
Since $N'$ is perpendicular to $N$, which is perpendicular to the curve $\gamma$ we have that $N'$ is parallel to $\gamma$, so it is also parallel to $\gamma'$.
So $N'=-\lambda \gamma'$, for some $\lambda$, which means that $\gamma$ is a line of curvature.
Is this correct?
$$$$
For the other direction:
Let $\gamma$ be a geodesic with noehere curvature and a line of curvature.
Without loss of generality, we suppose that $\gamma$ is unit-speed.
Let $a=N\times\gamma$. Differentiating this we get $a'=N'\times\gamma'+N\times\gamma''$.
Since $\gamma$ is a line of curvature we have that there is a $\lambda$ such that $N'= \lambda\gamma'$, i.e., $N'$ is paralle to $\gamma'$. So $N'\times\gamma'=0$. Since $\gamma$ is geodesic we have that $N\times\gamma''=0$.
How could we continue to conclude that the geodesic is a line of curvature.
A: Well, if $c\colon I\subset \mathbb{R}\longrightarrow S$ is a geodesic of a surface $S$ then we have : $$N(c(s))\parallel n(s),\ \forall s\in I,$$ where $N$ is the unit normal of $S$ and $n(s)$ the first normal of the curve $c$. Therefore, $$N(c(s))=\pm n(s),\ \forall s\in I.$$ By differentiating we obtain $$(N\circ c)'(s)=dN_{c(s)}(c'(s))=\pm n'(s),\ \forall s\in I.$$ By Frenet equations we have that $$n'(s)=-k(s)t(s)+\tau(s)b(s)$$ where $k(s)$ and $\tau(s)$ is the curvature and the torsion of $c$ respectively $(t(s):=c'(s)$ and $b(s)$ is the second normal of $c$. Thus, we obtain $$dN_{c(s)}(t(s))=-k(s)t(s)+\tau(s)b(s).$$ Now the result follows immediately:
*
*$(\Rightarrow)$ If $c$ is a plane curve then the torsion $\tau=0$. Thus, $$dN_{c(s)}(t(s))=-k(s)t(s)$$ which implies that $c$ is a line of curvature.
*$(\Leftarrow)$ If $c$ is a line of curvature then $dN_{c(s)}(t(s))=\lambda(s)t(s)$ and the equation above implies that $\tau=0$ and therefore $c$ is a plane curve.
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2ba80daf8a2ac57f6708a8131e39bd6bdfba662d
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Q: Prove that $aH\cap bH= \emptyset$ iff $b\notin aH$ and $a\in bH~ \text{iff}~ b\in aH$
Prove that
(i) $aH\cap bH= \emptyset$ iff $b\notin aH$ and
(ii) $a\in bH~ \text{iff}~ b\in aH$
where $H$ is a subgroup of a group $G$ and $a,b\in G$.
Attempt:
(i)
We know either $aH= bH$ or $aH\cap bH= \emptyset$. Then if $aH\cap bH= \emptyset$, then $aH\neq bH$. Then $b=be\in bH$ that implies $b\notin aH$ as $b\in bH$ & $aH\cap bH= \emptyset$.
I don't know whether the proof is correct or not. Also how to show the converse part of (i). Please help me to solve both the problems.
A: Part (i):
Let $aH\cap bH= \emptyset$. If $b\in aH$, then $b=ah$, for some $h\in H$. So, $a=bh^{-1}\in bH$ implies that $a\in aH\cap bH$, a contradiction.
Let $b\notin aH$. If $c\in aH \cap bH \neq \emptyset$, then $c=ah=bh'$ for some $h,h' \in H$, and so $b=ahh'^{-1} \in aH$, a contradiction.
Part (ii):
Let $a\in bH$. So, $a=bh$, for some $h\in H$, and we have $b=ah^{-1}\in aH$.
Let $b\in aH$. So, $b=ah$, for some $h\in H$, and we have $a=bh^{-1}\in bH$.
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e10d19bcca455d583537bc7ee132ea264c0bcd29
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Q: Determining a homomorphism between projective module groups In Milnor's Introduction to Algebraic K-Theory, he writes on page 7:
Suppose that $\Lambda$ can be mapped homomorphically into a field or skew-field $F$... Then we obtain a homomorphism $j_*: K_0 \Lambda \rightarrow K_0 F \cong \mathbb{Z}$. In the commutative case, this homomorphism is clearly determined by the kernel of $j$ [where $j: \Lambda \rightarrow F$ is a homomorphism], which is a prime ideal in $\Lambda$.
I'm having trouble seeing how and why $j_*$ is determined by the kernel of $j$. I agree that ker $j$ is a prime ideal in $\Lambda$, and I know that $j_*$ is a group homomorphism which must send generators to generators, and since $\mathbb{Z}$ is cyclic, I can believe that $j_*$ will be completely determined by what values it sends to 1, but where's the connection between ker $j$ and $j_*^{-1}(1)$?
A: Let $F_0$ be the field of fractions of $\Lambda/\ker j$. Then $j$ factors uniquely as a composite $\Lambda\to F_0\to F$, where the first map is canonical map. Since $F_0$ is a field, the induced map $K_0F_0\to K_0F$ can canonically be identified with the identity map $\mathbb{Z}\to\mathbb{Z}$. So $j_*$ can be identified with the map $K_0\Lambda\to K_0 F_0$, which depends only on $\ker j$.
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8b811c3f93ba5571b0ddd7b068f76768a77fd3e7
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Q: Probability of dealing pairs Texas Holdem I am in my first statistics/probability class and I am having trouble with this problem.
In two player Texas hold'em, what is the probaility that neither player is dealt a pair? What is the probability when you have three players?
I know you can solve it inelegantly by breaking it into a bunch of cases, but is there an easy general way to solve it?
A: I assume you can calculate the chance that the first player does not get a pair, which we can call $p$. An approximate approach is to say the chance the second player does not get a pair is independent of the first, so the chance neither player gets a pair is $p^2$ and for three players it is $p^3$. This is not exact. The fact that the first player did not get a pair has left a deck with $2$ ranks that have only $3$ copies and $11$ ranks that still have $4$. The right way is to condition on whether the first card drawn by the second player comes from a rank that has $3$ or $4$ cards at the start of his draw. This is not too bad. When you get to three players, it is a mess. Assuming the second player does not get a pair, there are three different deck compositions possible for the third player to draw from. Each one will give the third player a slightly different probability of getting a pair.
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fd9e30f86ff46277fe2dfc47a4a733861500943d
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Q: If a 10 digit number is formed using all the digits from 0 to 9 then find the following . A) Find the largest such number divisible by 11111 .
No matter what I try , I end up with atleast a digit repeating . Since the question says that the no. has all from 0 to 9 , therefore I cant solve it .
I think I need to find an algorithm for divisibility by 11111 to solve this . I cant find one .
B) How many no. of such numbers are possible that are divisible by 11111 ?
A: Our $10$-digit number $N$ is divisible by $11111$ if and only if it is divisible by $10^5-1$, for the digit sum of $N$ is divisible by $9$.
Let $a$ be the number whose decimal representation is given by the first $5$ digits of $N$, and $b$ the number whose decimal representation is given by the next $5$ digits of $N$. Then our $10$-digit number is divisible by $10^5-1$ if and only if $a+b$ is divisible by $10^5-1$. This is because $N=10^5a+b$, and $10^5\equiv 1\pmod{10^5-1}$.
Now that we know that $a+b$ must be divisible by $99999$, the rest should pose no difficulty.
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891c1734b0f405e6aa2ed2915bf09315c2ca8e4a
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Q: Regarding a Markov chain example state space $\{0,1\}$ I have trouble formulating a question.
The set up is $(X_n)$ is a Markov chain with the state space $\mathcal{S} = \{0,1\}$.
We know $X_0 =1$ and $X_2=1$ and the transition probability matrix, $p$.
We want to find $P(X_1=1)$.
My first guess is that it would be $p(1,1)$ but it seems reasonable that the knowledge that $X_2=1$ should have effect on the probability in question.
How should I approach this problem?
Thank you in advance!
A: My answer assumes you actually want to compute $P(X_1=1 \mid X_0=1, X_2=1)$. If you really wanted $P(X_1=1)$, see the answer by @probablyme.
By the definition of conditional probability,
$$P(X_1=1 \mid X_0=1, X_2=1) = \frac{P(X_0=1, X_1=1, X_2=1)}{P(X_0=1,X_2=1)}
= \frac{P(X_0=1, X_1=1, X_2=1)}{P(X_0=1,X_1=0,X_2=1)+P(X_0=1,X_1=1,X_2=1)}$$
By the Markov assumption,
$$P(X_0=1,X_1=x, X_2=1) = P(X_0=1) \cdot p(1,x) \cdot p(x,1).$$
Plugging this in above gives $\frac{p(1,1) \cdot p(1,1)}{p(1,0) \cdot p(0,1) + p(1,1) \cdot p(1,1)}$
A: $p(1,1)$ I'm assuming this notation means probability $P(X_1 = 1|X_0 = 1)$. So, not quite.
You're looking for
$$P(X_1=1) = P(X_0 = 0)P(X_1=1|X_0 = 0)+P(X_0= 1)P(X_1= 1|X_0=1),$$
by the law of total probability.
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82eb2e933654dd3ac69db14217f1dbbd8020e30f
|
Q: Differentiable Functions - Approximate Computation Write the formulas for approximate computation of the following values:
( $\sin \frac {\pi}{6} +\alpha$) for values of $\alpha$ near 0;
( $\sin 30^\circ +\alpha^\circ$) for values of $\alpha$ near 0;
( $\cos \frac {\pi}{4} +\alpha$) for values of $\alpha$ near 0;
( $\sin 45^\circ +\alpha^\circ$) for values of $\alpha$ near 0;
My friend gave me this hint: Use the formula y is approximately equal to $f'(x_0)$$(x-x_0)$+$f(x_0)$, in this case since you're searching for values near 0 you will have to set $x_0=0$, hence $f'(0)x+f(0)$.
But I'm having hard times to go further.
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30d52e5a2ca95009c0f7a7263524ee51b99246e9
|
Q: Why is $a \implies b$ is true when $a$ is false I understand that:
$True \implies True$, is true.
$True \implies False$, is False.
But why is it that
$False \implies True$, is True.
and
$False \implies False$, is True.
If $a$ is false I don't understand how we can say $a \implies b$ is true.
A: I was taught to think of implications like contracts. For example, say I told you "if you wash my car, I'll pay you ten dollars." Then the only way I could end up lying - the only way this statement could be false - is if I break my contract to you, and I stiff you after you wash my car. If you don't wash my car, I never lied to you, whether or not I end up paying you, and the contract always holds if the first part of the conditional is false.
A: What do you think of the statement "If 638,544 is divisible by 14, then it is divisible by 7"?
You probably already agreed that this statement was true before realizing that it was of the form $F \Rightarrow F$.
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9cd9f4f53e33fa20fe3b972b04ed4ffd55030108
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Q: How is it possible for a dot to have no dimension, while a line can have 1 dimension? In mathematics it is said that a "dot" has no dimension.
On a different context it is said that a line is made by joining different "dots".
Then how do line have only one dimension when its made up of something which has no dimension??
A: The first thing you need for this question to make sense is a definition of dimension. One convenient and commonly used notion is the Hausdorff dimension. The definiton may be too technical and beyond the level you seem to be asking at, but here it is anyway: the Hausdorff dimension of a metric space (or say, a subset of $\mathbb{R}^n$ for some $n$) is the smallest possible $d > 0$ such that your set can be covered with balls of radius $r_1, r_2, \ldots$, where the sum $\sum_{i} r_i^d$ is finite. (Actually, rather than "smallest possible", it's the infimum, in the case that there is no smallest $d$.)
Under this definition, it turns out that a point has Hausdorff dimension $0$ and a line has Hausdorff dimension $1$. How is this possible? Well, in your question, you seem to be assuming some principle like
The union of any collection of sets of dimension $d$ must have dimension $d$.
Assuming "dimension" = "Hausdorff dimension", this is true for a finite union. It is even true for a countably infinite union. But a line is composed of uncountably many points, and in the case of an uncountable collection the principle above is simply false.
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eb8b296e18fa23c42188ee7aa35bff4a184a3b71
|
Q: For each $x \in X$, $x/ \mathscr E=\{y \in X \mid y\mathscr Ex\}$? Shouldn't it be for each $y \in X$?
"Definition 6. Let $\mathscr E$ be an equivalence relation on a nonempty set $X$. For each $x \in X$, we define
$x/ \mathscr E=\{y \in X \mid y\mathscr Ex\}$
which is called the equivalence class determined by the element $x$.
The set of all such equivalence classes on $X$ is denoted by $X/\mathscr E$; that is, $X/\mathscr E=\{x/\mathscr E \mid x \in X\}$. The symbol $X/\mathscr E$ is read "$X$ modulo $\mathscr E$," or simply "$X$ mod $\mathscr E$".
Source: Set Theory by You-Feng Lin, Shwu Yeng T. Lin
In the definition, I don't understand "For each $x \in X$, we define
$x/ \mathscr E=\{y \in X \mid y\mathscr Ex\}$
which is called the equivalence class determined by the element x."
"$x/ \mathscr E=\{y \in X \mid y\mathscr Ex\}$" means x/$\mathscr E$ has y in X as its element and $y\mathscr Ex \Leftrightarrow (y, x) \in \mathscr E$. So shouldn't it be 'For each $y \in X$', rather than "For each $x \in X$'"?
A: He just means that for each $x \in X$ there is an equivalence class associated with that element, called $x/\mathscr{E}$. Within the definition of $x/\mathscr{E}$, yes you do look at for all $y \in X$, but that is not what he was talking about when he wrote "for each $x \in X$" earlier.
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7cb84b84cc408f01cf7679c59ee7a7968cfe2686
|
Q: Prove that $\|A\|=1$
Consider $C^0([0,1])$ with the $d_0(f,g)=\int_0^1|f(x)-g(x)| dx$. Let be $A: C^0([0,1])\to C^0([0,1])$ defined by $A(x)(s)=sx(s)$. Show that $\|A\|=1$.
Proof: The first implication
$$\|A(x)\|_0=\|sx(s)\|_0=|s|\|x\|_0\leq \|x\|_0, \text{for every $s\in[0,1].$}$$
Then
$$\|A\|=\sup_{x\in C^0} \frac{\|A(x)\|_0}{\|x\|_0}\leq 1.$$
How can I show that $\|A\|\geq 1?$
A: Your first argument has a mistake; bringing the $s$ out in front doesn't make sense as it's the variable of integration. What we get is
$$\|Af\| =\int_0^1|sf(s)|\, ds \le \int_0^1|f(s)|\,ds = \|f\|,$$
showing $\|A\|\le 1.$ To show $\|A\|\ge 1,$ consider the functions $f_n(s) = s^n, n \in \mathbb N.$
A: No. You are confuse. The correct is $\|Ax\|=\int_0^1|sx(s)|ds$ (which clearly is not $s|x(s)|$. Indeed, the latest has not a precisely meaning).
Now, as $s|x(s)|\le|x(s)|$ for all $s\in[0,1]$, then $\int_0^1|sx(s)|ds\le\int_0^1|x(s)|ds$. That is, $\|Ax\|\le\|x\|$ for all $x\in C[0,1]$. Then, $\|Ax\|\le 1$ for all $x\in C[0,1]$ such that $\|x\|=1$. Thus, $\|A\|\le 1$.
Now, set $\epsilon>0$. Take $a\in(0,1)$ such that $1-a/3<\epsilon$. Define $f\in C[0,1]$ by $f(s)=(-\frac{1}{a}s+1)1_{[0,a]}(s)$. Thus:
$\begin{eqnarray*}
\|f\|&=&\int_0^1(-\frac{1}{a}s+1)1_{[0,a]}(s)ds\\
&=&\int_0^a(-\frac{1}{a}s+1)ds\\
&=&\frac{a}{2}
\end{eqnarray*}$
and
$\begin{eqnarray*}
\|Af\|&=&\int_0^1s(-\frac{1}{a}s+1)1_{[0,a]}(s)ds\\
&=&\int_0^as(-\frac{1}{a}s+1)ds\\
&=&-\frac{1}{a}\int_0^as^2ds+\int_0^asds\\
&=&-\frac{1}{a}\frac{a^3}{3}+\frac{a^2}{2}\\
&=&\frac{a^2}{2}-\frac{a^2}{3}\\
&=&\frac{a^2}{6}
\end{eqnarray*}$
Thus, $\|Af\|/\|f\|=a/3>1-\epsilon$.
Conclusion: for all $\epsilon>0$ we can exhibit $f\in C[0,1]$ such that $1-\epsilon<\|Af\|/\|f\|<1$. Thus, $1=\sup\{\|Af\|/\|f\|\,\,:\,\,\|f\|\neq 0\}$. This proves $\|A\|=1$.
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ef725140893d79604a91f36a7fcb9b53a5b565a4
|
Q: Finding the closest function to another in a Hilbert space.
Let H be the Hilbert space L$^2$([0,1)], and let $S$ be the subspace of functions f $\in$ H satisfying $\int^1_0(1+x)f(x)dx=0$.
Find the element of $S$ closest to the function $g\in H$ defined by $g(x)=1$. Prove that it is the closest element.
I think what I need to do is find the orthogonal projection of $g$ onto $S$. I'm not exactly sure how I would go about doing that. Any help would be appreciated.
A: Note that
$$S=(1+x)^\perp$$
Thus $S$ is closed and so we know, by the projection theorem, that there's a unique $f\in S$ such that $||f-g||=\min_{h\in S}\{||h-g||\}$. Furthermore, this $f$ is the unique element of $S$ satisfying $(f-g)\perp S$. But
$$S^\perp = ((1+x)^\perp)^\perp=\overline{(1+x)}=1+x$$
so we can solve for $f$.
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2a24a91462a00af2eb9d01e4bcb89bd86bd4a626
|
Q: Given a graph $G = (V, E)$, prove $e \leq \frac{n(n-1)}{2}$ for all $n$ I'm trying to figure out to solve this problem:
Given a graph $G = (V, E)$ prove $$e \leq \frac{n(n-1)}{2}$$ for all $n$, where $e$ is the number of edges and $n$ is the number of vertices.
I'm thinking that I should somehow be using math induction to figure out the correct answer and use $n = 1$ or $0$ for my hypothesis, but I'm getting a little stuck on what to do after -- if I assume $n = k$, then:
$$e \leq \frac{(k+1)k}{2}.$$
And if $n = k+1$ then $$e \leq \frac{k(k-1)}{2}.$$
As I understand it, each vertex has $n-1$ possible edges coming out, and there are n total vertices, which is where $n(n-1)$ comes from and dividing by $2$ gets rid of the repeats. But I am unsure how I am to prove this.
A: $\frac{n(n-1)}{2}$ is the number of edges in a simple, complete graph with $n$ vertices, for any $n \ge 2$. This is because, in a simple, complete graph with at least 2 vertices, every pair of vertices has an edge between them. So there are in total $\binom{n}{2}$ edges which is equivalent to $\frac{n(n-1)}{2}$. Assuming that your graph $G$ is also a simple graph, it will be either complete or not complete. In either case, $e \le \frac{n(n-1)}{2}$. For $n = 1$, check that the statement still holds.
Also, $\frac{n(n-1)}{2}$ is a value that you will probably encounter quite often in graph theory. It is quite helpful to remember that it represents the total number of edges in a simple, complete graph. Often it is quite a useful tool in graph theory inequality proofs.
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bddcd828eae81a878b0589fd48d48b13d6fd9d77
|
Q: Geometrical proof by induction
Given a segment $AB$ of length $1$, define the set $M$ of points in the following way: it contains the two points $A,B$ and also all points obtained from $A,B$ iterating the following rule: for every pair of points $X,Y$ in $M$, the set $M$ also contains the point $Z$ of the segment $XY$ for which $YZ = 3XZ$. Prove by induction that the set $M$ consists of points $X$ from the segment $AB$ for which the distance from the point $A$ is either $$AX = \dfrac{3k}{4^n} \hspace{3 mm} \text{or} \hspace{3 mm}AX = \dfrac{3k-2}{4^n}$$ where $n,k$ are nonnegative integers.
I am confused by this question since I am not used to doing induction geometrically. Seeing as how we have to iterate each time a new point, I find it hard to formalize an inductive argument to prove this.
A: Every point on the segment $AB$ has some distance from $A$.
So let's say you have two such points, $X$ and $Y$.
Let $x$ denote the distance of $X$ from $A$,
$y$ the distance of $Y$ from $A$.
Let $D$ be the set of distances from $A$ to each point in the
set $M$ that is constructed in the problem.
Recall the rule in the problem that says that if $X$ and $Y$ are members of
the set $M$, the point $Z$ of the segment $XY$ for which $YZ = 3XZ$ is too.
An equivalent rule is, if $x$ and $y$ are numbers in the set $D$,
then the number $z$ between $x$ and $y$ such that
$|z - y| = 3|z - x|$ also is in the set $D$.
The theorem you are to prove is that $D$ consists entirely of
numbers of the form $\dfrac{3k}{4^n}$ or of the form
$\dfrac{3k-2}{4^n}$.
There, now it's not a geometric induction any more. Just arithmetic.
There are still some complications, for example you can't just assume
$x < y$, you can't just assume that one of $x$ and $y$ has the form
$\dfrac{3k}{4^n}$ and the other has the form $\dfrac{3k-2}{4^n}$
(they might both have the same form),
and you can't just use one value of $k$ to write both $x$ and $y$.
You might have to consider several possible cases separately
for the induction step.
A: Using the same starting point as David K to transform this into something more arithmetic, his expression
$|z - y| = 3|z - x|$
is equivalent to $z=\dfrac {3x}4 + \dfrac{y}{4} \quad\forall x ,y \in D$.
If $y>z>x$,then $|z - y| = 3|z - x|$ is equivalent to
$y-z=3(z-x)$.
If, however, $x>z>y$, then $|z - y| = 3|z - x|$ is equivalent to
$z-y=3(x-z)$. Both are equivalent statements. Solving for $z$,
$z-y=3(x-z)$
$z-y=3x-3z$
$4z=3x+y$
$z=\dfrac{3x}4+\dfrac y4$
$z$ will always take the form of $y$ (either $\dfrac{3k}{4^n}$ or $\dfrac{3k-2}{4^n}$). Modular arithmetic may ease explanation. The numerator is either 0 (first case) or 1 (second case) mod 3. $3x=0\mod 3$ so we need only consider how $y$ affects the form of $z$.
If $x$ and $y$ have the same denominator, then the numerator of $z$ is simply $3x+y$ and $(3x+y)\mod\ 3=y\mod\ 3$.
If the denominator of $y$ is greater than that of $x$, the numerator of $z$ is $(3x)(4^n)+y$ and $(3x)(4^n)+y\mod\ 3=y\mod\ 3$.
If the denominator of $x$ is greater than that of $y$, the numerator of $z$ is $3x+y(4^n)$. In modular arithmetic $(ab)\mod c=[(a\mod c)(b\mod c)]\mod c$. $4^n=1 \mod c \quad\forall n \in \mathbb{Z}$.
Therefore, $3x+y(4^n) \mod\ 3=y\mod\ 3$.
|
24ca11004d5124231e678535fb73db44e646156e
|
Q: Question about a Symmetric random walk, Problem 4.1.1 in Durrett I am working on the following problem:
Let $X_1, X_2, \dots \in \mathbb{R}$ be i.i.d. with a distribution that is symmetric about $0$ and nondegenerate, i.e. $P(X_i=0)<1$. Show that $-\infty = \liminf S_n < \limsup S_n=\infty$. Here $S_n=X_1 + \dots +X_n$.
I know that my two events are exchangeable and so by a corollary of the Hewitt-Savage $0$-$1$ law one of the following events occurs with probability one:
$
i) \ S_n=0 \text{ for all } n,
$
$
ii) \ S_n\rightarrow \infty,
$
$
iii) \ S_n\rightarrow -\infty,
$
$
iv) -\infty = \liminf S_n < \limsup S_n=\infty.
$
Since $P(X_i=0)<1$, $(i)$ cannot occur. I am pretty sure that if $(ii)$ holds then $(iii)$ must hold by the symmetry of the distribution and vice versa, which would be a contradiction. However, I cannot think of a mathematically precise way to say this.
Any help is appreciated!
A: If $S_n\to\infty$ a.s. then for $\epsilon>0$ there is $N$ s.t. for $n>N$, $P[S_n>1]>1-\epsilon$, say. But you know for any finite $n$, $P[S_n<0]=P[S_n>0]$ (E.g. induction. Given symmetric independent RVs $U$ and $V$, $P[U+V<0]=\int_x F_U(-x)dF_V(x)=\int_x(1-F_U(x))dF_V(-x)=P[U+V>0]$, ignoring continuity issues.)
|
267acf786c4936189584131448dafced573d2097
|
Q: Given three arbitrary points in the euclidean plane, is there a way to homeomorphic map the plane so that the points become colinear? Suppose $p,q,r \in\mathbb{R}^2$. Is there a homeomorphism $h:\mathbb{R^2}\rightarrow \mathbb{R}^2$ such that $h(p),h(q),h(r)$ are colinear? I'm interested in whether such an $h$ be found for $n$ points in the plane, but perhaps I can figure that out from the $n=3$ case.
(Disclaimer: this question is relevant to a question on exam I had in a previous class, which ended weeks ago.)
A: I think when $n = 3$ the argument can be made quite a bit simpler, so because you want an argument that can be generalized, I'll go straight to $n$.
Pick a coordinate system in the plane so that the $y$-axis is not parallel to any of the lines through two of the points.
Let the points be $(x_1,y_1), \dots, (x_n,y_n)$. Let $f(x)$ be a continuous function whose graph passes through all $n$ points. Then $h \colon (x,y) \mapsto (x,y - f(x))$ is your required homeomorphism.
|
735484b640223244dcbf2d44562326f25887c39f
|
Q: how to visualise solenoidal vector field? How to visualise solenoidal vector field;I am thinking it as a solenoid going in a circular fashion with propagation in the z direction.
I was solving a problem to check that field is solenoidal or not ?
|
7aa14675e0173e54773c55b6b22200f5acc2c4ab
|
Q: An identity involving $[\sigma(n)]^2$ For a positive integer $n$, let $\sigma(n)$ denote the sum of the divisors of $n$. For example, $\sigma(1)=1$, $\sigma(2)=3$, $\sigma(4)=7$, etc.
I would like to prove the following identity:
For every positive integer $n$, $$ [\sigma(n)]^2=\sum_{d\mid n}
\frac{n}{d} \sigma(d^2)$$
where the sum ranges over all divisors $d$ of $n$.
Here is one attempt: We know $\sigma(n)$ is a multiplicative function, that is, $\sigma(ab)=\sigma(a)\sigma(b)$ for $\gcd(a, b)=1$. And I think one can show the right hand side is also multiplicative, so it suffices to check the equality when $n=p^k$ is a prime power. This last task boils down to showing that:
$$
(1+p+\cdots + p^k)^2 = \sum_{0\leq m\leq k} p^{k-m} (1+p+\cdots + p^{2m})
$$
Using the geometric series, the left hand side is $\left(\dfrac{p^{k+1}-1}{p-1}\right)^2$ and similarly the right hand side is
$$
\sum_{0\leq m\leq k} p^{k-m} \frac{p^{2m+1}-1}{p-1}
$$
So we just need to show:
$$
(p^{k+1}-1)^2 = (p-1)\sum_{0\leq m\leq k} p^{k-m}(p^{2m+1}-1) = (p-1) p^{k} \sum_{0\leq m\leq k} (p^{m+1}-p^{-m})
$$
and indeed, we can compute that:
$$
(p-1)p^{k} \sum_{0\leq m\leq k} (p^{m+1}-p^{-m}) = (p-1)p^{k}\left(\frac{p^{k+2}-p}{p-1} + \frac{p^{-k+1}-p}{p-1}\right) =
p^{2k+2} -2 p^{k+1} + 1 = (p^{k+1}-1)^2
$$
as desired. Is this proof correct? I am not really satisfied by it, because I don't seem to get any insight out of it. Does anyone have a more conceptual (perhaps combinatorial) proof?
A: Here's a direct-ish proof. at least, without using that the various functions are multiplicative.
First, let's prove a pair of counting lemmas.
Lemma 1: If $m\mid n^2$, then the number of ways to write $m=d_1d_2$ with $d_1\mid n$ and $d_2\mid n$ is equal to the number of divisors of $\gcd\left(m,\frac{n^2}{m}\right)$.
Proof: We need both that $d_1\mid m$ and $d_1\mid n$. So we need $d_1\mid \gcd(m,n)$. We also need $d_2=\frac{m}{d_1}\mid n$, so $\frac{m}{d_1}\mid\gcd(m,n)$ or $\frac{m}{\gcd(m,n)}\mid d_1 $. Therefore, we can choose any $d_1$ with:
$$\frac{m}{\gcd(m,n)}\mid d_1\mid \gcd(m,n)$$
If we let $d=\frac{d_1\gcd(m,n)}{m}$, we can pick any $d$ with $$d\mid\frac{\gcd(n,m)^2}{m}=\gcd\left(\frac{n^2}{m},m\right).$$
Lemma 2: If $m\mid n^2$, then the number of ways to write $m=\frac{n}{d_1}d_2$ with $d_1\mid n$ and $d_2\mid d_1^2$ is also equal to the number of divisors of $\gcd\left(\frac{n^2}{m},m\right)$.
Proof: Note that $\gcd\left(\frac{n^2}{m},m\right)\mid n$ (this follows from $\gcd(m^2,n^2)=\gcd(m,n)^2\mid mn$.)
So, if $d\mid \gcd\left(\frac{n^2}{m},m\right)$ then $d\mid n$ so let $d_1=\frac nd$ and let $d_2=\frac{md_1}{n}=\frac{m}{d}$, which is an integer.
Now, $\frac{d_1^2}{d_2} = \frac{n^2}{d^2}\frac{d}{m} = \frac{n^2}{md}$. But $d$ is a divisor of $\frac{n^2}m$ by definition, so this is an integer, and $d_2\mid d_1^2$.
On the other hand, if we have a pair $d_1,d_2$ with $\frac{n}{d_1}d_2=m$ and $d_1\mid n$ and $d_2\mid d_1^2$, then let $d=\frac{n}{d_1}$. We clearly have $d\mid n$.
Writing $\frac{n^2}{m}=\frac{n^2}{dd_2} = d\frac{d_1^2}{d_2}$ which is divisible by $d$.
Theorem: Let $\sigma_k(n)=\sum_{d\mid n} d^k$. Then:
$$\left[\sigma_k(n)\right]^2 = \sum_{d\mid n} \left(\frac{n}{d}\right)^k\sigma_k(d^2)$$
Proof: Both sides can be written as a sum of $k$th powers of divisors $m$ of $n^2$. The number of times $m^k$ occurs on the left side is counted in Lemma 1. The number of times $m$ occurs in the sum on the right side is counted by Lemma 2. These two values are the same.
It still feels almost accidental.
A: I don't know if this is what you're looking for but I believe there is a relatively simple way of seeing that
$$(1 + p + ... + p^k)^2 = \displaystyle\sum_{m=0}^k p^{k-m}(1 + p + ... + p^{2m})$$
Namely, let us consider the coefficient of $p^j$ on both sides of the above equation.
On the left hand side, this is clearly equal to the number of ways of writing $p^j = p^rp^s = p^{r+s}$ with $0 \leq r,s \leq k$, so it's the number of ways of writing $j = r+s$ with $0 \leq r,s \leq k$. Now if $j \leq k$, then this is possible for $r = 0,...,j$ and if $j \geq k$, it is possible for $r = j-k,...,k$.
On the right hand side, this coefficient is the number of $m$ such that the above sum contains $p^j$, which is clearly the number of $m$ such that $k-m \leq j \leq k+m$, with $0 \leq m \leq k$. Again, if $j \leq k$, then this is true
for $m = k-j,...,k$ and if $j \geq k$, it is true for $m = j-k,...,k$.
Thus the coefficient of $p^j$ is the same on both sides so they must be equal.
A: Another way to show the equality below that @jwsiegel proved:
$$(1 + p + \ldots + p^k)^2 = \sum_{m=0}^k p^{k-m}(1 + p + \ldots + p^{2m})$$
Group the terms in two different ways. (I think this is a nice direct visual proof.)
$$
\begin{array}{r}
\big(1 &+p &+p^2 &+\ldots &+p^k\big)\\
&+\big(p &+p^2 &+\ldots &+p^k &+p^{k+1}\big)\\
& &+\big(p^2 &+\ldots &+p^k &+p^{k+1} &+p^{k+2}\big)\\
& & & &\cdots\\
& & & &+\big(p^k &+p^{k+1} &+p^{k+2} &+\ldots &+p^{2k-1} &+p^{2k}\big)\\
\hline
& & & &=\\
& & & &p^k\\
& & &p^{k-1} &+p^k &+p^{k+1}\\
& & & &\cdots\\
& &p^2 &+\ldots+p^{k-1} &+p^k &+p^{k+1} &+p^{k+2} &+\ldots+p^{2k-2}\\
&p &+p^2 &+\ldots+p^{k-1} &+p^k &+p^{k+1} &+p^{k+2} &+\ldots+p^{2k-2} &+p^{2k-1}\\
1 &+p &+p^2 &+\ldots+p^{k-1} &+p^k &+p^{k+1} &+p^{k+2} &+\ldots+p^{2k-2} &+p^{2k-1} &+p^{2k}\\
\ \\
\end{array}
$$
|
72a3fe932cb8c23390e0f65d7e57411855b59337
|
Q: Weighted weak compositions of n into k parts I am new to combinatorics and am wondering about an extension of a weak composition of $n$ into $k$ parts where both $n \in \mathbb{N}$ and $k \in \mathbb{N}$. Classically, the (non-weighted) result is,
$$
\text{If, }A = \left\{ (b_1, b_2, \ldots, b_k) \in \mathbb{Z}_+^k : \sum_{i = 1}^k b_i = n \right\}, \quad \text{then } |A| = \binom{n + k - 1}{k - 1}
$$
where $|\cdot|$ denotes the cardinality of $A$ and $\mathbb{Z}_+$ denotes the field of non-negative integers.
My question is if it is possible to extend this to have a weighted weak composition of integers into $k$ parts? Namely, define a set of fixed weights $\alpha = (a_1, a_2, \ldots, a_k) \in \mathbb{R}_+^k\setminus\left\{ 0\right\}$ and
$$
A_\alpha = \left\{ (b_1, b_2, \ldots, b_k) \in \mathbb{Z}_+^k : \lfloor\sum_{i = 1}^k a_i b_i \rfloor = n \right\}
$$
Is it possible to compute $|A_\alpha|$? Or perhaps give bounds on $|A_\alpha|$?
A: As an example of how to do this with a recurrence, the number of ways $N_{s,i}$ of reaching a particular sum $s$ with $i$ weights $(a_1,a_2,\ldots,a_i)$ satisfies $$ N_{s,i} = N_{s,i-1}+N_{s-a_i,i}$$ starting with $N_{0,0}=1$ and $N_{s,0}=0$ for $s\not= 0$.
You then need to take $\displaystyle |A_\alpha|=\sum_{n \le s \lt n+1} N_{s,k}$ to get the result in your formulation.
For example suppose $n=2$, $k=5$ and the weights are $(0.7,0.5,0.9,0.8,0.6)$. The recurrence might then produce this table for $N_{s,i}$
0 1 1 1 1 1 1
0.1 0 0 0 0 0 0
0.2 0 0 0 0 0 0
0.3 0 0 0 0 0 0
0.4 0 0 0 0 0 0
0.5 0 0 1 1 1 1
0.6 0 0 0 0 0 1
0.7 0 1 1 1 1 1
0.8 0 0 0 0 1 1
0.9 0 0 0 1 1 1
1 0 0 1 1 1 1
1.1 0 0 0 0 0 1
1.2 0 0 1 1 1 2
1.3 0 0 0 0 1 2
1.4 0 1 1 2 2 3
1.5 0 0 1 1 2 3
1.6 0 0 0 1 2 3
1.7 0 0 1 1 2 3
1.8 0 0 0 1 2 4
1.9 0 0 1 2 2 4
2 0 0 1 1 2 5
2.1 0 1 1 2 3 6
2.2 0 0 1 1 3 6
2.3 0 0 0 2 4 7
2.4 0 0 1 2 4 8
2.5 0 0 1 2 4 8
2.6 0 0 1 2 4 9
2.7 0 0 1 2 4 10
2.8 0 1 1 3 5 11
2.9 0 0 1 2 5 12
giving the solution as the sum of the last ten numbers in the final column, i.e. $82$
|
477d740e83599157e9310e4eb61b8f754ca8ab55
|
Q: Common meeting point for 3 points to reach 4th point Problem statement: We are 3 friends at 3 different locations $A, B, C$ and want to reach a location $D$. Each person will take a separate cab to a common meeting point $E$, and then take a single cab from $E$ to $D$.
How can I find the point $E$ which minimizes the total distance traveled by all four cabs?
A: If $E$ is the common meeting point, you are trying to solve
$$\min_E \|A-E\| + \|B-E\| + \|C-E\| + \|D-E\|$$
The solution to this optimization problem is the geometric median $E$ of the three friends and the destination point $D$. In general computing the geometric median is not straightforward, but in the special case of four points there is an easy algorithm:
1) If one of the four points lies in the triangle formed by the others, take $E$ to be that point. In this case all friends travel directly to $D$ (if this is the inside point) or two friends first travel to the third and then all friends travel together to $D$.
2) Otherwise, the four points form a quadrilateral. Draw the diagonals and take $E$ to be the intersection point.
It's not so obvious why this algorithm correctly identifies the median; proofs are given here.
|
0a0be74aa4dd004e52bfd6a4d295dde9052d4082
|
Q: Path of completely bounded maps has uniformly bounded cb norm? If $\phi_t:A\rightarrow B$ is completely bounded for $t\in[0,1]$, and $t\mapsto\phi_t(a)$ is continuous for each $a\in A$, is $\sup_{t\in[0,1]}||\phi_t||_{cb}$ finite? Here, $A$ and $B$ are operator spaces or operator algebras.
In the setting of Banach spaces and bounded maps, it seems like the uniform boundedness principle gives a positive answer (or does it?).
|
9c7880b112e8b2d2d8341bc1963b58b2fc35182a
|
Q: Prove that $ (v_1,\ldots,v_n)$ is a basis for $F^n$ iff the matrix obtained by assembling the coordinate vectors of $v_i$ is invertible. Prove that a set $S = (v_1,\ldots,v_n)$ of vectors in $F^n$ is a basis $\Longleftrightarrow$ the matrix obtained by assembling the coordinate vectors of $v_i$ is invertible.
$(\Rightarrow):$ Suppose $S$ is basis of $F^n$. We can write every coordinate vector $[v_i]_S$ as $(0,0,\ldots,1,\ldots,0)$ where the $1$ is in the $i$th position. The matrix from assembling these vectors is $$M = \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{pmatrix},$$ which is an $n \times n$ identity matrix and so is clearly invertible.
Is this correct so far?
$(\Leftarrow):$ Now assume that the matrix $M$ obtained by assembling the coordinate vectors of $(v_1,\ldots,v_n)$ is invertible. Then there exists an $n \times n$ matrix $N$ such that $MN=NM = I$ where $I$ is an $n \times n$ identity matrix.
Here I get stuck. How do we prove $(v_1,\ldots,v_n)$ are independent and span $F^n$?
A: The first part of the proof is incorrect. Your $M$ is not the matrix obtained by assembling the vectors $v_i$.
I would say that what an acceptable proof of this fact is depends very much on the order in which theorems have been proved in your book. I would probably say that, letting $e_1,\dots,e_n$ be the standard basis, $M$ represents the linear mapping $\phi$ such that $\phi(e_1)=v_1, \dots, \phi(e_n) = v_n$. You may know that in these circumstances, since $v_1,\dots,v_n$ is a basis, $\phi$ is an automorphism of $F^{n}$. Let $N$ be the matrix of $\phi^{-1}$. Then $MN = NM = I$.
|
bf15c27310ecea7365ab48a35bf1d418732d1847
|
Q: Finding All Matrices Which Commute With a Given Matrix. I know this should be really simple, but for some reason I can;t figure it out.
I need to find all matrices which commute with the following 2x2 matrix A:
$$ B=\begin{bmatrix} 1 & -1\\ 5 & -4 \end{bmatrix} $$
I've tried using the definition of two commuting matrices, AB=BA and a generic matrix A where $ A=\begin{bmatrix} a & b\\ c & d \end{bmatrix} $ to generate a system of linear equations which I then solve to obtain a, b, c, and d such that A and B commute as follows:
$$ AB = \begin{bmatrix} a+5b & -a-4b\\ c+5d & -c-4d \end{bmatrix}=BA=\begin{bmatrix} a-c & b-d\\ 5a-4c & 5b-4d \end{bmatrix}$$
This gives the equations:
$$ c+ 5b=0 $$
$$ 5b-d+a=0 $$
$$ 5c-5a+5d=0 $$
$$5b+c=0 $$
This can then be converted to a matrix and reduced to row echelon form:
$$ \begin{bmatrix}
1 & 5 & 0 & -1 & 0 \\
0 & 5 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix} $$
Introducing a parameter s for the variable c and a parameter t for the variable d, it seems this system can be solved to give:
$ a=s+t $, $b =\frac {-s} {5} $, $ c=s $, and $ d=t $.
Substituting these values into AB or BA yields the following matrix which, by rights should represent all matrices which commute with c:
$$ \begin{bmatrix} t & \frac {-s} {5} -t \\ s+5t & -s-4t \end{bmatrix} $$
However, the given solution is:
$$ \begin{bmatrix} a & b \\ -5b & 5b+a \end{bmatrix} $$
Our solutions seem quite similar (except of course for different variable names), but I jsut can;t for the life of me figure out where I've gone wrong, any help would be greatly appreciated!
A: You've done everything just fine. You just need to perform the transformation $(a,b) \to (t,\frac{-s}{5}-t)$.
|
aed9e1571d1c84252e21792c4db3b47339a6d075
|
Q: If $p$ is prime and $p$ divides $a^n$ is it true that $p^n$ divides $a^n$? I'm stuck on this problem. A hint is to use the following corollary:
If $p$ is prime and $p$ divides $(a_1 \times a_2\times \dotsb\times a_n)$, then $p$ divides at least one of the $a_i$ .
So Assume $p$ is prime and $p$ divides $a^n$. Then, $p = k \times a^n$ for some integer $k$. Now, by the corollary, it is assumed that $p$ also divides on of the $a_i$, take $i = 1$. Then, $p = c \cdot a_1$. This is where I get stuck and keep going in circles. Any advice would be much appreciated.
A:
Theorem.
Let $c,b\in \Bbb Z$. If $p|cb$ then $p|c$ or $p|b$ (This is just a reformulation of yours).
Now, by induction, we want to prove that if $p|a^n$, $p|a$.
Base case is trivial, with $n=0,1$.
Now, if $p|a^n=a\cdot a^{n-1}$ then we have that (1) $p|a$ or (2) that $p|a^{n-1}$.
If it's (1), we're done. If it's (2), by inductive hypothesis we have that $p|a$.
A: Yes. Consider prime factorization of $a$,
$a= p{_1}^{k_1} p{_2}^{k_2} p{_3}^{k_3}.... p{_r}^{k_r}$, then
$a^n= p{_1}^{nk_1} p{_2}^{nk_2} p{_3}^{nk_3}.... p{_r}^{nk_r}$.
So if $p$ divides $a^n$, then $p$ divides one of the $p{_i}^{nk_i}$. Since $p$ and $p_{i}$ both are primes, thus it is only possible when $p = p_i$. Since $p{_i}^{nk_i}=p^{nk_i}$ divides $a^n$, hence $p^n$ must divide $a^n$
A: If $p|a$ then $a = p*c$ and $a^n = (p*c)^n = p^n*c^n$
Can you show that $p|a$?
A: Let $K(a)= \{ x \in \mathbf{N} : x \mid a, x \in \mathtt{PRIMES} \}$, then $K(a) = K(a^n)$.
We can see that $p \mid a$ if and only if $p \in K(a)$.
So $p \mid a^n \Leftrightarrow p \in K(a^n) \Leftrightarrow p \in K(a) \Leftrightarrow p \mid a \Rightarrow p^n \mid a^n$.
|
c4080faa38a40427558627b5624a9ec5a2dec3f1
|
Q: No. of parabolas possible for the given equation. Given, $f(x)=-x^2+qx+r$. $(q,r) \epsilon R$. $q,r$ are variables. A quadratic equation $f(x)=0$ has a maximum value $m$ ($m$ is a constant) and a root $x=a$. Does $f(x)$ correspond to a unique parabola? Why?
A:
Given, $f(x)=-x^2+qx+r$. $(q,r) \epsilon R$. A quadratic equation
$f(x)=0$ has a maximum value $m$
The function might have a maximum.
$$
f(x) = -(x^2 - qx - r) =
-\left( (x-q/2)^2 -(q^2/4) -r \right)
$$
The quadratic term is minimal and with the minus sign maximal, if $x=q/2$.
So $f$ has a maximum there. The value is
$$
m = (q^2/4) + r \quad (*)
$$
and a root $x=a$.
$$
0 = f(x) = -\left( (x-q/2)^2 -(q^2/4) -r \right) \iff \\
x = \frac{q \pm \sqrt{q^2+4r}}{2} = \frac{q \pm \sqrt{4m}}{2} \quad (**)
$$
The number and kind of roots are determined by the term $\Delta = q^2+4r$. If $\Delta = 0$, there is only one root $x=q/2$. If $\Delta > 0$ there are two real roots, if $\Delta <0$ there are no real valued roots (but two complex ones).
Does $f(x)$ correspond to a unique parabola? Why?
If that root $x=a$ is the only root, then the above gives that it is $a=q/2$ and that $q=2a$ from $0 =\Delta = q^2+4r$ we know that $r = -a^2$. Thus both coefficients $q, r$ are uniquely determined as the parabola is too.
If there is a second root or if we do not know the number of roots, and this seems to be the case for this task, then we need more information.
As additional information we got the value $m$ of the maximum of $f$, thus $m = f(x^*)$, for some unknown argument $x^*$.
We have two equations, for the maximum we have $(*)$ and for the root we have $(**)$. This leads to:
$$
q = 2a \mp \sqrt{4m} \quad (\#) \\
r = m - (q^2 / 4)
$$
If $m = 0$ then we have only one root, and the parabola is uniquely determined.
If $m > 0$, then we have two roots and we do not know which of the two possible signs for the square root in $(\#)$ applies, if the maximum is to the right ($a < x^*$) or to the left of $a$ ($x^* < a$).
So in this case we limit it down to two possible parabolas, with two different coefficients $q, r$, but we can not decide which of the two we are dealing with.
Here are graphs for $a = 2$ and $m = 0$ and $m = 1$:
A: What is given? What is required?
If $q,r$ are given, then the roots and extreme value are fixed automatically as
$ \dfrac{q \pm \sqrt{q^2+4r}}{2}, (r + q^2/4 ) $ respectively, obtained by standard method of finding roots and extreme values by differentiation.
The two constants $ q,r$ determine only one unique $ y = f(x) = -x^2 + q x + r $ parabola graph.
Do not assign new symbols to dependent quantities $ a, m$ as if they are new or independent!
However,
If $ a, m $ are given
Take the parabola equation in the form $ A y = -(x-a)(x-b) $
Differentiating, $ x_{max}= (a+b)/2 $
Maximum value of parabola is found as
$$ y_m = m = (b-a)^2/(4 A) $$
Find $b$ from above and plug into parabola equation to get one parabola for each value of $A$ chosen.
$$ A y = A f(x) = - (x-a)(x-a-2 \sqrt{m A} ) $$
So the parabolas are not unique like in the first case, they are infinitely many, as seen in the graph where I chose four values for the arbitrary constant $A$.
Treat the problems separately to avoid confusion.
In the example given $ m =1000, a= 1 , (0.05 < A < 0.2) $
EDIT 1:
We can drive home that $A$ is an arbitrary maximum curvature at all top points of same height $m$ for a particular boundary condition chosen. we can formulate a simple differential equation approach as well:
$$ y{''} = -1/A , y ' = -x/A + \tan \alpha , y = - x^2/ (2 A) + x \tan \alpha $$
( arbitrary constant set to $0$ for $x=0,y=0)$.
The $\alpha$ is the variable angle to ground fired by an imaginary gun for the projectile to reach same height $m$. It has two roots
$$ x=0, x = 2 A \tan \alpha ;$$
with maximum height for all $\alpha$
$$ y = y_{max} = m = A \tan {^2}\alpha /2 $$
for all $\alpha$ at $ x= A \tan \alpha.$ The variable $\alpha$ is given by:
$$ \tan \alpha = \sqrt{2 m/A} $$
[We can compare it to a dynamic projectile problem where the vertical component of velocity (proportional to maximum height reached) of a gun shot fired with velocity in various directions].
A: It need not be a unique parabola. Think geometrically of two possibilities consider a concave down parabola passing through the point $(a,0)$ with maximum value $m$. In one case the left branch passes through $(a,0)$ and in the other case the right branch passes through $(a,0)$. Hence not unique.
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8fa61e474a99571516f527639eb9841b52e39a1c
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Q: Discussion of $Z(A_4) = \{e\}$ I tried to answer the following question:
Why does the fact that the orders of the elements of $A_4$ are $1,2$
and $3$ imply that $|Z(A_4)|=1$?
My answer: Two cycles commute if and only if they are disjoint. But no combination of $2$-cycles and $3$-cycles in $A_4$ can ever be disjoint since there are only $4$ numbers. Hence the centre must be trivial.
Then I looked at the answer:
If there was $x$ in $Z(A_4)$ of order $2$ then a product with a $3$-cycle would have order $6$.
If there was $x$ in $Z(A_4)$ of order $3$ then a product with a $2$-cycle would have order $6$.
I have two questions about this question:
(1) Is my proof correct?
and:
(2) Is it a general principle that if the goal is to show the centre
of $S_n$ is trivial one assumes by contradiction it wasn't and then
produces an element with order too large to be in $S_n$?
|
ba9204b2d6b948928b277bd411f2b2783db8f832
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Q: Reference on properties of binary random vectors I am studying random vectors of length $n$ for which each element is an i.i.d Bernolli random variable with probability $\frac{1}{2}$. I think that many useful and interesting theorems can be proved for such random vectors and random vector spaces (like random graphs). Does anyone know a good reference which studies the statistical properties of such random vectors and vector spaces?
A: Random binary matrices have proven applications in Coding Theory, Information Theory and Cryptography.
In coding theory, Random Linear Coding is about considering such matrices and has strong connections with random graphs. One interesting report can be found here. Most of the modern coding theory, i.e., Low Density Parity Check (LDPC) codes and fountain codes are basically based on studying random binary matrices. MacKay has a very interesting paper about fountain codes that can be found here.
The most important theorems in Information Theory are based on considering random matrices with large (infinity in limit) dimensions. These can be found either in the seminal paper of Shannon entitled "A Mathematical Theory of Communications" or in the very nice book of Cover and Thomas named "Elements of Information Theory."
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1d2e12c1d4fa88ad1271c6fb714452bb09f632e3
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Q: Countable cartesian product and Axiom of Choice In the A taste of Topology book, when talking about Cartesian product $\prod\{S:S\in\mathcal{S}\}$, the author writes the following:
It is straightforward that $\prod\{S:S\in\mathcal{S}\}\neq\emptyset$
whenever $\mathcal{S}$ is finite and $S\neq\emptyset$ for any
$S\in\mathcal{S}$. The same can be shown if $\mathcal{S}$ is
countable, see Exercise 2.
The "Exercise 2" asks to prove that $\prod_n S_n$ of non-empty sets is non-empty, without invoking Zorn's lemma.
Is it indeed that straightforward to show the claim for countable $\mathcal{S}$? I was under the impression that while the full strength of AC is not needed, one still needs the weaker Axiom of Countable Choice to make that claim. Can it be proved in ZF alone?
A: Based on the information you provide this is a mistake. You are correct in that it is consistent that there is a countable family of finite sets whose product is empty (namely, they do not have a choice function).
For example in Cohen's second model for the failure of choice there is such family of sets of size $2$.
|
cd2e9e85ec7fdc7aeca5899fbe18a2d7171df534
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Q: Best Linear Prediction and variance decomposition Let $( X , Y )$ have moments of at least the second order ,and let $\hat{Y} = a +bX$ then we will choose the coefficients $a$ and $b$ such that $a=E(Y)-\sigma_{XY}/\sigma_{X}^2E(X)$ and $b=\sigma_{XY}/\sigma_{X}^2$ , so that $\hat{Y}$ is the best linear prediction of $Y$ , where best is taken to mean “minimum expected squared distance between outcomes of $Y$ and outcomes of $\hat{Y}$".
Now define $V= Y - \hat{Y}$ to represent the deviations between outcomes of $Y$ and outcomes of the best linear prediction of $Y$ .
How can I prove that $\mathbb E((Y-\mathbb E(Y))^2) =\sigma_{Y}^2\rho_{XY}^2 + \sigma_{Y}^2 (1-\rho _{XY}^2).$ ?
My attempt is :
If $\hat{Y}$ is the best prediction of $Y$ then $E(Y)=E(\hat{Y})$ that means $E(V)=0$
Then $E((Y-E(Y))^2)=E((\hat{Y}-E(\hat{Y})+V)^2)= \sigma_{\hat{Y}}^2 + \sigma_{V}^2 +\sigma_{\hat{Y}V}$ .
And $\sigma_{V}^2 =E(Y-\hat{Y})^2 =E((Y-(a+bX))^2)=E((Y-(E(Y)-\sigma_{XY}/\sigma_{X}^2E(X)+\sigma_{XY}/\sigma_{X}^2 X))^2) $
I could not do anything more
How can I prove that :
$$\sigma_{\hat{Y}}^2 + \sigma_{V}^2 +\sigma_{\hat{Y}V}=\sigma_{Y}^2\rho_{XY}^2 + \sigma_{Y}^2 (1-\rho _{XY}^2)$$
|
a30b92b6f7d0db6d300629fc9fb648290e9070d0
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Q: find coordinates from known angles and length in 3d Suppose I have 3 vectors with length a,b,and c. They are oriented in 3D space such that the angles between the three vectors are $\alpha$, $\beta$, and $\gamma$ (suppose all less than 90 degrees).
If I set the vectors with length a and b on the x-y plane with angel $\alpha$ between them (set the vector with length a to be along x-axis), I am able to find the coordinates of the first two vectors. But then I run into trouble when I try to find the coordinates of the third vector based on its length c, and it's angles with the other two vectors $\beta$, and $\gamma$.
Can someone help me out? Thank you.
A: Call the corresponding unit vectors $u, v, w$, and say $u = (1,0,0)$ and $v = (\cos \alpha,\sin\alpha,0)$. (That is, $u$ is on the positive $x$ axis and $v$ is in the half-plane $y \geq 0$ of the $xy$-plane.) Further assume that $w$ is the half-space $z \geq 0$.
Let $w = (x,y,z)$. We have
$$\cos \beta = u \cdot w = x, \qquad \cos \gamma = v \cdot w = x\cos \alpha+ y \sin\alpha.$$
Solving for $y$, we find $y = (\cos \gamma - \cos \alpha \cos \beta)/\sin \alpha$. To find $z$, we use the relation $x^2 + y^2 + z^2 = 1$. This yields
$$z = \frac{1}{\sin \alpha}(1-\cos^2 \alpha - \cos^2 \beta-\cos^2 \gamma + 2\cos\alpha\cos \beta\cos \gamma)^{1/2}.$$
Now your vectors are $au, bv, cw$.
A: Hint generate three equations . 2 with dot products with vectors with length $a,b$ as you know the one vector its magnitude and $\theta $ you will get two simultaneous equations with three variablesby assuming the third vectielrs as $xi+yj+zk$ and the third condition is $\sqrt{x^2+y^2+z^2}=c$ thats all solve them get values and thats the third vector.
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6e121b2cdd26b6eef8cc15b36f3c34826f8a78c1
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Q: How is this an Even function? How is this,
$$f(x) = \left\{\begin{array}{cc} |x + 1|, \quad &-2 \le x \le 0 \\
|x - 1|, \quad& 0 < x \le 2
\end{array}\right.$$
an even function?
A: Let $x>0$, then
\begin{align*}
f(-x) & = |-x+1| && (\because -x <0 \text{ we use the first part of the function})\\
& = |(x-1)(-1)|\\
&=|x-1|\\
&=f(x).
\end{align*}
Similarly prove it for the case when $x<0$.
A: For $-2< x < 0$, we have
$$f(x) = |x + 1| = |-(x+1)| = |-x-1| = f(-x).$$
|
1a83078dd0cf9fc9627330c1db5db0a825d4a686
|
Q: How to count the number of elments in a group generated by $2$ elements? Say we consider $a = (13)(24)$ and $b=(234)$ in $S_n$.
How can I find the number of elements in the group generated by $a$
and $b$?
I am looking for an intelligent way. So far I have multiplied together all possible pairs and got $b, b^2, a, ab, ba, ab^2, b^2a$ and they are all different. But unless I keep multiplying I cannot be sure I found them all.
This question came up when I was solving this exercise here:
Label the four locations of tires on an automobile with the labels $ 1,2,3 $ and $4$, clockwise. Let $a$ represent the operation of switching the tires in positions $1$ and $3$ and switching the tires in positions $2$ and $4$. Let $b$ represent the operation of rotating the tires in positions 2,3, and 4 clockwise and leaving the tire in position $1$ as is. Let $G$ be the group of all possible combinations of a and b. How many elements are in $G$?
Edit
This question comes before Lagrange's theorem is mentionend.
A: The permutations $a$ and $b$ are both even. Therefore the group generated by $a$ and $b$ has at most $12$ elements. The subgroup generated by $b$ has three elements. If you can find a subgroup of order $4$, then Lagrange's Theorem will tell you the answer is $12$. But $\{e,a,bab^{-1},b^2 a b^{-2}\}$ is such a subgroup. (Alternatively, you have already found seven distinct elements. The only divisor of 12 that is $\geq 7$ is 12 itself.)
Edit By request, here is an alternative proof not using Lagrange's Theorem. Let $H$ be the subgroup of $S_4$ generated by $a$ and $b$. Since $a, b \in A_4$, we have $H \subseteq A_4$. Let $K = \{e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)\}$. It is easy to see that $K$ is a subgroup of $A_4$. We have $K = \{e, a, bab^{-1}, b^2 a b^{-2}\} \subseteq H$. We will prove that the elements $b^i k$, where $k \in K$ and $i = 0, 1, 2$, are all distinct. There are twelve such elements, and they all belong to $H$, so this proves that $H = A_4$.
To prove the claim, assume that $b^{i_1} k_1 = b^{i_2} k_2$. Then $b^{i_2 - i_1} = k_2 k_1^{-1} \in K$. But the only power of $b$ that belongs to $K$ is $e$, so $b^{i_2 - i_1} = e$. Since $0 \leq i_1, i_2 \leq 2$ and $b, b^2 \ne e$, this implies $i_1 = i_2$.
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14b645fffea3f3f5ce173f151d54f74054c2ee25
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Q: whether the set is dense? $X=C[0,1]$ with sup norm,
$S =\lbrace f\in X: \int _0^1f(t)dt \neq 0 \rbrace $ then how to decide whether S dense, nowhere dense ?
The function which are outside S have integral $0$ so they are zero almost everywhere on $[0,1]$. Again all the positive$(f(x)>0$ $\forall x) $ and negative functions will be there in S. intuitively it is clear to me that this set is dense.
I have tried considering a function $f\to \int _0^1f$ so under this map $S$ is inverse image of $(-\infty,0) \cup (0,\infty)$. somehow i am not able to connect all this.. can anyone give some hint?
thank you.
A: First of:$\int_0 ^1 f = 0 $ does not implify $f \overset{\text{a.e.}}{=} 0$.
My attempt: Let $f\in C_\infty [0,1]$ if $\int_0 ^1 f \ne 0 $ then take the sequence $S \ni f_n = f$ obviously $f_n \rightarrow f$.
Otherwise, consider the sequence $f_n = f+ \frac{1}{n} $
Indeed $f\in C_\infty$ (as sum of such) and since $f \notin S$ we have:
$$\int _0 ^1 f_n = \int _0 ^{1} f +\int _0 ^1 \frac{1}{n}=0+1/n \ne 0$$
That implies $f_n \in f$. We also have $\|f_n - f\|_\infty =\|\frac{1}{n}\|_\infty \rightarrow 0$.
Followup question: I'm interested to see if someone can approach this via functional analysis. considering the functional $T \in C^\text{#}_\infty$:
$Tf = \int_0^1 f$. We can see $T$ is bounded since $|Tf| \leq |\int_0^1 f | \leq \|f\|_\infty$ which implies the kernel is closed. We also know that $\mathrm{Codim} \ker T =1$, but i can't get further with this.
A: A functional analysis aproach:
Lema: If $V$ is a vector subspace of a normed space $X$, then $V^\circ=\emptyset$ or $V=X$ (here, $V^\circ$ is interior of $V$).
Proof: Suppose $V\neq X$ and $V^\circ\neq\emptyset$. Set $x_0\in V^\circ$. How $T:X\to X$ given by $Tx=x-x_0$ is continuos, we can suppose $x_0=0$. Now, there is an $\epsilon>0$ such that $B(0,\epsilon)\subseteq V$. Take $v\in X\backslash V$. Then, $v\neq 0$ and there is a non zero multiple of $v$ that is in $B(0,\epsilon)$, say $tv$. Then, $tv\in V$. Thus, every multiple of $tv$ belongs to $V$. In particular $v$. This contradiction shows that $V=X$ or $V^\circ=\emptyset$.
Now, let $T:C[0,1]\to C[0,1]$ given by $Tf=\int_0^1f(x)dx$. Note that $S=X\backslash\mathrm{Ker}(T)$. Then $\overline{S}=X\backslash\mathrm{Ker}(T)^\circ=X$ cause $\mathrm{Ker}(T)$ is a subspace of $X$ and it isn't $X$.
From that, $S$ is dense in $X$. Also, $S$ is an open set. Furthermore, note that we don't use that $T$ is continous. Then, if $T\neq 0$ is any linear operator, $\overline{X\backslash \mathrm{Ker}(T)}=X\backslash \mathrm{Ker}(T)^\circ=X$. Thus, $X\backslash\mathrm{Ker}(T)$ always is dense if $T$ isn't the null operator.
A: Hint :
$\displaystyle S=\left\{f\in X:\int_0^1f(t)\,dt \not=0\right\}=\left\{f\in X:\int_0^1f(t)\,dt<0\right\}\cup \left\{f\in X:\int_0^1f(t)\,dt>0\right\}$.
What is $\bar{S}$ ?
$\bar{S}=C[0,1]=X.$ and so ........
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c2a7de21740788f376841eb66898703b0d9bd886
|
Q: Prove by induction a property of a tree graph Prove by induction (and without the use of cycle definition) that if to delete a leaf vertex from a tree graph it will stay as a tree graph.
I think Ive got it wrong but what I did is the following:
By the inductive defenition of a tree graph of 'n' vertices, its a graph that been constructed out of a tree of 'n-1' vertices and an attached leaf.
But, the same goes for a tree with 'n-1'.
So, can I tell that this is why if we eliminate 1 vertice of a tree graph it will still be a tree graph?
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d0e5178d145c03cc7d1fa07f154892db9a619831
|
Q: Using the Gradient theorem along a parabolic path in 3D Check the fundamental theorem for gradient using $T=x^{2}+4xy+2yz^{3}$ from the point a = $\left ( 0,0,0 \right )$ to point b$=\left ( 1,1,1 \right )$
along the parabolic path where $z=x^{2}\ and\ y = x$
By definition:
$\vec{\nabla}T=\left \langle 2x+4y,4+2z^{3}6yz^{2} \right \rangle$
Looking at the xy-plane, we break the triangular path into the vertical and horizontal component
Along the horizontal component:
$d\vec{l}=\left \langle dx,0,0 \right \rangle where y=0\ and\ z=0$
Then,
$\int_{x=0}^{x=1}\vec{\nabla}T.d\vec{l}=1$
Along the vertical component:
$d\vec{l}=\left \langle 0,dy,0 \right \rangle$
where x=1 and z=0
Then,
$\int_{y=0}^{y=1}\vec{\nabla}T.d\vec{l}=4$
Along the path where the path goes from the xy-plane to the z-plane for which z=1
$d\vec{l}=\left \langle 0,0,dz \right \rangle$
Here, as given by the question $y=x\ and\ z=x^{2}$
Also, note that at all point where the path extends from $\left \langle 1,1,0 \right \rangle to \left \langle 1,1,1 \right \rangle$
we have x=y=1
Integrating over the z-component
$\int_{z=0}^{z=1}\vec{\nabla}T.d\vec{l}=\int_{z=0}^{z=1}6yz^{2}=2yz^{3}=2y=2 (since\ y=1)$
Is this correct? I haven't utilised the given information $z=x^{2}$
|
3365d77305c4b8c862d3a9db23a06ec455bba378
|
Q: Is the $\alpha$-curve a topological submanifold of $\mathbb{R}^2$ Consider the following subset of $\mathbb{R}^2$ defined by $\mathcal{M}=\{ \ (x,y)\in\mathbb{R}^2\ |\ y^2=x^2(x+1)\ \}$. I'm supposed to decide whether or not $\mathcal{M}$ is a topological submanifold (TS) of $\mathbb{R}^2$. It looks like this:
If it's a TS, then it would have to be 1-dimensional. Here is the definition I am using:
$\mathcal{M}$ is a TS of $\mathbb{R}^2$ if for every point $p \in \mathcal{M}$ there exists an open set $V \subset \mathbb{R}^2$ with $p \in V$, an open set $U\subset \mathbb{R}$ and a homeomorphism $\psi:U \to V$.
I don't think its a manifold. I think that a problem occurs at the origin since the curve crosses over itself there. I believe that a good homeomorphism cannot exist at the origin for this reason. How would I prove this?
I thought about using a proof by contradiction, saying suppose that there does exist a homeomorphism $\psi$ mapping an open ball $V=B_{\epsilon}(0,0)$ to some open interval $U$ of $\mathbb{R}$. Then $\psi(0,0,0)$ would be well-defined, but maybe not $\psi^{-1}(0,0,0)$? I'm not sure.
I am new to differential geometry, can someone give me a hint on how to approach this problem?
A: Let $U$ be a sufficiently small open neighbourhood of the origin in $\mathbb{R}^2$, then $U\cap \mathcal{M}$ is a neighbourhood of $(0, 0)$ in $\mathcal{M}$. It looks something like this
$\hspace{77mm}$
where now I have coloured the axes red to distinguish them from the set $U\cap \mathcal{M}$ (which is still coloured black).
How many connected components does $(U\cap \mathcal{M})\setminus\{(0, 0)\}$ have? If $\mathcal{M}$ were a topological manifold $U\cap \mathcal{M}$ would be homeomorphic to an open interval $I$. How many connected components does $I$ have if a point is removed?
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f037ad3834dff833eb5e217fc0d305ca3da2089d
|
Q: Rotations of sphere $\mathbb S^2$
*
*In the picture bellow; How to prove that the result of rotation about $P$ through
angle $\theta$, followed by rotation about $Q$ through angle $\varphi$ is rotation about $R$ through some angle? ــ Can it be proven by the reflections of the great circles $\mathscr{L,M,N}$? How?
*How to find the axis of the product rotation?
A: The first rotation is the reflection through $L$ followed by reflection through $M$. For the second, it's $M$ followed by $N$. The $M$'s cancel, and you're left with a rotation with axis $OR$.
|
a3432dbfa76f089b001ce1ab0280d2c277f774af
|
Q: Why is this a group action - what is the significance of $g^{-1}$?
Let $G$ be a group acting on a variety $X$ such that every $g\in G$
defines a morphism $\phi_g:X\rightarrow X$ given by $\phi_g(x)=g\cdot x$. If $X=\operatorname{Spec}(R)$ is affine then $\phi_g$ comes from a
map $\phi_g^*:R\rightarrow R$. We define the induced action of $G$ on
$R$ by $$g\cdot f=\phi_{g^{-1}}^{*}(f)$$ for $f\in R$
In other words, $(g\cdot f)(x)=f(g^{-1}\cdot x)$ for all $x\in X$
I have two questions -
*
*Why do we need $g^{-1}$ to ensure that $G$ induces an action on $R$? We have - $$(g_1 g_2)\cdot f=\phi^*_{(g_1g_2)^{-1}}(f)=\phi^*_{g_2^{-1}g_1^{-1}}(f)=(\phi^*_{g_2^{-1}}\phi^*_{g_1^{-1}})(f)$$
At this point I am confused as $$(\phi^*_{g_2^{-1}}\phi^*_{g_1^{-1}})(f)=\phi^*_{g_2^{-1}}(\phi^*_{g_1^{-1}}(f))=\phi^*_{g_2^{-1}}(g_1\cdot f)=g_2\cdot(g_1\cdot f)$$ which is not what I want to show that it is an action. So what am I doing wrong?
*Can someone explain to be how to show the "in other words" part?
Thank you.
A: (1) Your error is that $\phi^*_{g_2^{-1}g_1^{-1}}$ is $\phi^*_{g_1^{-1}}\phi^*_{g_2^{-1}}$, not $\phi^*_{g_2^{-1}}\phi^*_{g_1^{-1}}$. We have $\phi_{gh}=\phi_g\phi_h$, but passing from $\phi$ to $\phi^*$ is a contravariant functor, so it reverses the order of composition.
(2) Suppose $\phi:X\to X$ is a morphism and $x$ is a $k$-point of $X$. If $f\in R$, write $f(x)$ for the image of $f$ under the homomorphism $R\to k$ corresponding to $x$. Equivalently, $f(x)=i_x^*(f)$, where $i_x:\operatorname{Spec} k\to X$ is the map corresponding to $x$. Then $i_{\phi(x)}=\phi i_x$, so $$f(\phi(x))=i_{\phi(x)}^*(f)=i_x^*\phi^*f=(\phi^*f)(x).$$
In particular, in the case $\phi=\phi_{g^{-1}}$, the left-hand side is $f(g^{-1}\cdot x)$ and the right-hand side is $(g\cdot f)(x)$.
|
4438651ad9aca1fcf964fdd76ccf89f4d69101a8
|
Q: $\exists\ n \gt 34131$ with more than $7$ odd divisors $d_i \gt 1$ such as when $d_i+1$ are accumulated in increasing order to $1$ the sums are prime? In the same style as a previous test, I did a little test today looking for all the numbers such as the odd divisors, ordered in increasing order excluding $1$, when they are accumulated one by one to $1$ the resulting sums are always prime numbers:
The first elements of the sequence are $\{3,5,6,10,11,12,17...\}$, it is not included at OEIS:
$n=3$, odd divisors not including $1$ are: $\{3\}$ and $1+(3+1)=5 \in \Bbb P$.
$n=5$, odd divisors not including $1$ are: $\{5\}$ and $1+(5+1)=7 \in \Bbb P$.
$n=6$, odd divisors not including $1$ are: $\{3\}$ and $1+(3+1)=5 \in \Bbb P$.
For instance $n=39$ has $3$ divisors:
$n=39$, odd divisors not including $1$ are: $\{3,13,39\}$ and $1+(3+1)=5 \in \Bbb P$, $5+(13+1)=19 \in \Bbb P$, $19+(39+1)=59 \in \Bbb P$.
I tried to look for the one with the longest list of divisors, but it seems that the longest possible length is $7$ elements, for $n=34131$, at least up to $n=1.6 \cdot 10^9$ and in all cases $n$ is an odd number (thanks to @Peter for the feedback in the comments!):
$n=34131$, odd divisors not including $1$ are: $\{3,31,93,367,1101,11377,34131\}$ and $1+(3+1)=5 \in \Bbb P$, $5+(31+1)=37 \in \Bbb P$, $37+(93+1)=131 \in \Bbb P$, $131+(367+1)=499\in \Bbb P$, $499+(1101+1)=1601 \in \Bbb P$, $1601+(11377+1)=12979\in \Bbb P$, $12979+(34131+1)=47111 \in \Bbb P$.
So far heuristically I can see that the list of $n$'s includes some $p$ primes and $2p$.
I would like to ask the following question:
Why does it seem impossible to find a $n \gt 34131$ such as all the odd divisors comply with the property and the length of the list of those divisors is longer? Is there a theoretical reason behind? Is there a counterexample? (maybe a very big number due to the conditions of the property?)
Thank you very much in advance!
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0b60e78fc3924f58db5f427fc464600209fe24bb
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Q: Full Strength of Arzela-Ascoli We define a family of functions $\mathcal{F}$ on a domain $\Omega$ to be equicontinuous if for each point $x \in \Omega$ and any $\epsilon > 0$, there is a $\delta_x > 0$ such that $|f(x) - f(y)| < \epsilon$ whenever $| x - y| < \delta_x$ for all $f \in \mathcal{F}$.
Define a family of functions $\mathcal{F}$ on a domain $\Omega$ to be uniformly equicontinuous if for any $\epsilon > 0$, there is a $\delta > 0$ such that $|f(x) - f(y)| < \epsilon$ whenever $| x - y| < \delta$ for all $f \in \mathcal{F}$ and for all $x \in \Omega$.
The Arzela-Ascoli theorem is sometimes stated for the second class of families, but in fact it is true for both. I am reading a book that proves it only for the second, and they make the comment that the full version is almost never needed in applications. Since this is a complex analysis book I can definitely believe that, but I am wondering if there are any important results that have needed the full strength.
Thanks!
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66cb7ba3cd1d56d869f4dffb10caad0efe3a9581
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Q: Can this integral be evaluated/approximated? I've been trying to evaluate this integral without much success:
$$
\int_{-\infty}^{\infty}\mathrm{e}^{\mathrm{i}ax}\,\,
\frac{1 - \mathrm{e}^{-c\sinh^{\,2}\,\left(bx\right)}}{\sinh^{2}\left(bx\right)}\,\,\mathrm{d}x
$$
I've tried contour integration. There are no poles, but I can't find a suitable contour such that all contributions other than that from the real line disappear.
If anyone can help with evaluating/ approximating this in the regime where $c$ is large, that would be greatly appreciated.
A: Let's begin by converting this to a simpler double integral:
$$\begin{align}b \int_{-\infty}^{\infty} dx \, e^{i a x/b} \int_0^c du \, e^{-u \sinh^2{x}} &= b \int_0^c du \int_{-\infty}^{\infty} dx \, e^{i a x/b} \, e^{-u \sinh^2{x}}\\ &=b \int_0^c du \, e^{u/2}\,\int_0^{\infty} dx \, \cos{\left (\frac{a x}{2 b} \right )} e^{-(u/2) \cosh{x}} \\ &= 2 b\int_0^{c/2} du \, e^{u} \, K_{i \frac{a}{2 b}} \left (u \right ) \end{align}$$
I turned to Mathematica for this integral, and it is a little ugly but could be worse as an exact representation of the integral:
$$-b \operatorname{Re}{\left [2^{i a/b} c^{1-i a/(2 b)} \Gamma \left (-1-i \frac{a}{2 b} \right ) \, _2F_2 \left (\begin{array} \\ \frac12+i \frac{a}{2 b} & 1+i \frac{a}{2 b}\\2+i \frac{a}{2 b}&1+i \frac{a}{b} \end{array} ; c\right ) \right ]} $$
I did a numerical verification of this using $a=2$, $b=1$, and plotted against $c \in [0,1]$ and got indistinguishable plots.
Of course, we can debate whether this is truly a useful closed form. Within Mathematica, the answer is clear - it sure is. Mathematica can compute the closed form in a small fraction of the time it took to compute a numerical approximation to the integral (and it had a hard time doing so).
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917a7a0ef1031fef7f5c75d9c81c10f8f5b18ea0
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Q: Can there be an example of infinitely many, non-repeating mutually adjoint functors? Since it is very much possible to have functors with $F_0 \dashv F_1 \dashv F_2$ or even $F_0 \dashv F_1 \dashv F_2 \dashv F_3$, it begs the question:
Can there be an example of infinitely many, non-repeating mutually adjoint
functors?
Obviously, "non-repeating" is important, since $\operatorname{id} \dashv \operatorname{id} \dashv \dots$ would otherwise of course be a boring example (and every other pair of pseudo-invers functors would be too).
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b4d2c8be8c433bcc7ceef19897fb531a20192070
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Q: find the possible range of values for k for circles not touching
Circle 1 C has equation ${(x + 1)^2 + (y - 1)^2}$ = 121 A circle 2 C
with equation ${x^2 + y^2 -4x + 6y + p = 0}$ is drawn inside 1 C . The
circles have no points of contact. What is the range of values of p?
From my understanding, circles touch internally if distance ${C_1C_2 = r_1 - r_2}$
Then I need to make sure that distance ${C_1C_2 > r_1 - r_2}$
Using the distance formula, the distance between ${C_1}$ and ${C_2}$ is 5.
${\sqrt {(2 —1)^2 + (-3 -1)^2}}$ = 5
The radius of ${C_1}$ is 11
The radius of ${C_2}$ is ${\sqrt {2^2 + 3^2 - p}}$
=> ${\sqrt {13 - p}}$
Developing this I get
${5 > {\sqrt {13 - p}} - 11}$
I am not sure how to develop this any further.
A: Note that $C_1$ and $C_2$ are tangents in $P$ if $r_1=r_2+d((-1,1),(2,-3))$. Then, $r_2=11-\sqrt{(2+1)^2+(-3-1)^2}=11-5=6$
Then $0\le 13-p<36$. Thus $-23<p\le 13$
Editing: The main idea
First, we prove that if $C_1$ and $C_2$ are tangent, say in $P$, then their centers and $P$ are collinear.
Let $O_1$ and $O_2$ the centers. Suppose $C_2$ inside $C_1$. Then $O_2$ is inside of $C_1$. If $O_1$, $O_2$ and $P$ aren't collinear $O_2$ is not in the segment $O_1P$. Then, $\triangle O_1O_2P$ is not degenerate. Now, cause trought $P$ passes a common tangent to $C_1$ and $C_2$, say $\ell$, taking points $A,B\in\ell$ , we have $O_1P\perp AB$ and $O_2P\perp AB$. But $\angle APB=180$, and $\angle APB=\angle APO_1+\angle O_1PO_2+\angle O_2PB=180+\angle O_1PO_2$. From this $\angle O_1PO_2=0$. But $\triangle O_1PO_2$ isn't degenerated. Thus $0<\angle O_1PO_2=0$. This contradiction shows us that $O_1$, $O_2$ and $P$ are collinear.
The case $C_2$ outside $C_1$ is analogous.
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31c1ab73a76b81ea065fc28cc49f51759695eaa8
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Q: An exercise on $L^1$ function Suppose $f \in L^1 \cap C^1(\mathbf R)$, and assume $f'(x)$ goes to $0$ as $x$ goes to $\infty$. What is the best way to prove that $f(x)$ also goes to $0$ as $x$ goes to $\infty$.
A: As $f^\prime\in C^1$ and $\lim_{x\to\infty}f^\prime(x)=0$, then $f^\prime$ is bounded in $[0,\infty]$. Say $|f^\prime(x)|\le M$ for $x\ge 0$. Now, it follows that $f$ is uniformly continuous in $[0,\infty)$
Now, a classic result implies that $\lim_{x\to\infty}|f(x)|=0$ (for example $f$ uniformly continuous and $\int_a^\infty f(x)\,dx$ converges imply $\lim_{x \to \infty} f(x) = 0$ )
|
cc81b843253b09edc8b060c4eca1fa98f80c57a0
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Q: Lower bound on a subgroup generated by two elements $\langle a,b \rangle$? Let $G$ be a group and $a,b \in G$ such that $|a|=2$ and $|b|=3$ and $a\neq b$.
It follows immediately that $G$ contains at least $4$ elements since $\langle a\rangle$ cannot be a subgroup of $\langle b \rangle$ since $2$ does not divide $3$ and $a$ cannot equal $b$ or $b^2$.
Is there anything more that can be said about the lower bound on
$|\langle a,b \rangle|$?
Could it be $5$ or even $6$?
Edit
I'm very sorry I forgot to mention it but in the book Lagrange's Theorem has not been mentioned. So I'm trying to do this without it.
A: It must be at least $6$ because of a little theorem named after Joseph-Louis Lagrange. $\langle a\rangle$ and $\langle b\rangle$ are both subgroups of $\langle a, b\rangle$, and therefore $2$ and $3$ must both divide $|\langle a, b\rangle|$.
Of course, without any more information, you cannot say any more about the size (if it is finite, it must be divisible by $6$). It might be just $6$, which is the case for instance for $\langle 2, 3\rangle \subseteq \Bbb Z/6\Bbb Z$. Or it could be infinite, which is the case in the group
$$
\{a, b \mid a^2 = b^3 = 1\}
$$
or it might be anything in-between.
(If you are unfamiliar with that notation, here is a short explanation: $\{a, b\}$ without any restrictions is called the free group generated by $a$ and $b$. Its elements are finite words using the four letters $a, b, a^{-1}$ and $b^{-1}$, and the group operation is putting one word after the other, and then cancelling any inverses that might appear next to one another.
With the restriction $a^2 = b^3 = 1$, this means that any time you see the sub-word $aa$, you might remove it, and the same with $bbb$. Incidentally, this makes $a^{-1} = a$ and $b^{-1} = b^2$ as well, so now you can make do with only $a$ and $b$ as the letters of your words. Note that $|ab| = \infty$, which was the whole point of making this example.)
Edit Why there must be at least $6$ elements without resorting to Lagrange's theorem.
The six elements that are guaranteed to be different from one another is $e, a, b, b^2, ab, ab^2$. In general, if $\gcd(|a|, |b|) = 1$, then every element $a^ib^j$ with $0\leq i<|a|$ and $0\leq j<|b|$ will be different from every other such element. This implies that $|\langle a, b\rangle| \geq |a|\cdot |b|$. Here is why:
Assume, for contradiction, that $a^{i}b^{j} = a^{k}b^{l}$ with $0\leq i, k<|a|$ and $0\leq j, l<|b|$, and that at least one of $i \neq k$ and $j \neq l$ is true. This gives:
$$
a^ib^j = a^kb^l\\
a^{-i}a^ib^j = a^{-i}a^kb^l\\
b^j = a^{k-j}b^l\\
b^jb^{-l} = a^{k-j}b^lb^{-l}\\
b^{j-l} = a^{k-j}
$$
It's important to note that neither $a^{k-i}$ nor $b^{j-l}$ are the identity. We know that one of them is not, because of our assumption that either $i \neq k$ or $j \neq l$. Technically, this means that only one of the above is not the identity, but since we have equality, we know that it goes for both of them.
Now, raise this last equality to the power of $|a|$, and we get
$$
b^{|a|(j-l)} = a^{|a|(k-i)} =(a^{|a|})^{k-i} = e
$$
This means that $|b|$ divides $|a|(k-i)$. Since $\gcd(|a|,|b|) = 1$, we get that $|b|$ divides $k-i$. Since $k-i \neq 0$, we must have $|b|\leq |k-i|$ (absolute value signs because $k-i$ might be negative). Because of the restraints on $i$ and $k$ (they're both non-negative, and both smaller than $|a|$), we must have $|k-i| <|a|$. This gives $|b|<|a|$.
On the other hand, raising the same relation to the power of $|b|$ instead, a completely analoguous argument gives $|a|\leq |j-l| <|b|$. Thus we have our contradiction, and therefore we must have $a^ib^j \neq a^kb^l$.
Note that if we limit ourselves to $|a| = 2$ and $|b| = 3$, this proof becomes a bit shorter and more manageable: We know that $e, a, b, b^2$ are all different. The last ones to check are $ab$ and $ab^2$. In the following, $i, j = 1, 2$.
*
*$ab^i = e$ cannot be true since this implies $b^i = a$
*$ab^i = a$ implies $b^i = e$, which is also false
*$ab^i = b^j$ implies $a = b^{j-i}$, which is either $e, b$ or $b^2$, depending on $i$ and $j$. Either way, it's false
*Lastly, $ab = ab^2$ implies $b = b^2$, which is also false
And therefore there are at least $6$ elements in $\langle a, b\rangle$.
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bde40a3062c37b698e16294e3f62f88bdfdb7aea
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Q: How to evaluate this Integral $\int { {\sqrt{5^2+K^2}}dK \over {\sqrt{10^2+K^2}K}} $ While working on an Exact Differential Equation, I encounter the following Integral.
$$\int { {\sqrt{5^2+K^2}} \over {K\sqrt{10^2+K^2}}} dK$$
I have tried substitution and all the other elementary methods, but the Integral simply refuses to yield to all my attempts. I reckon that this requires some concept that I've yet not studied. Maybe elliptical Integrals? But that's just speculation. Please Help....
Thank You.
A: This is not a trivial one $$I=\int\frac{\sqrt{x^2+25}}{x \sqrt{x^2+100}}\,dx$$ Let us try using $$\frac{\sqrt{x^2+25}}{ \sqrt{x^2+100}}=u^2 \implies x=\frac{5 \sqrt{1-4 u^4}}{\sqrt{u^4-1}}\implies dx= \frac{30 u^3}{\sqrt{1-4 u^4} \left(u^4-1\right)^{3/2}}du$$ So, $$I=-\int\frac{6 u^5}{4 u^8-5 u^4+1}\,du$$ Now, since the denominator shows pretty nice roots (it is a quadratic in $u^4$), partial fraction decomposition leads to $$\frac{-6 u^5}{4 u^8-5 u^4+1}=\frac{u}{u^2+1}+\frac{u}{2 u^2-1}-\frac{u}{2 u^2+1}-\frac{1}{2 (u-1)}-\frac{1}{2
(u+1)}$$ makes the problem much more pleasant.
The result of the integration is just a sum of logarithms $$I=\frac{1}{4} \log \left(1-2 u^2\right)-\frac{1}{2} \log
\left(1-u^2\right)+\frac{1}{2} \log \left(1+u^2\right)-\frac{1}{4} \log \left(1+2
u^2\right)$$ Recombining all of that simplifies again and $$I=\tanh ^{-1}\left(u^2\right)-\frac{1}{2} \tanh ^{-1}\left(2 u^2\right)$$
Edit
Applying the same method to
$$I=\int\frac{\sqrt{x^2+a^2}}{x \sqrt{x^2+b^2}}\,dx$$ would lead to $$I=\tanh ^{-1}\left(u^2\right)-\frac{a }{b}\tanh ^{-1}\left(\frac{b }{a}u^2\right)$$
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a0bfc19a0fd7fd1c4b2c7e204dae0e18a6b03166
|
Q: Convergence of $\int_0^{\infty}\sin (p(t))dt$ Using contour integration, for integer $n\ge 2$, I could obtain the result $$\int_0^{\infty}\sin t^n dt=\sin\bigg(\frac{\pi}{2n}\bigg)\Gamma\bigg(1+\frac{1}{n}\bigg).$$ With the definition $f(x) = \int_x^{x+1}\sin(e^t)dt$ and substitution $e^t=u$, $$e^xf(x) = \cos e^x-e^{-1}\cos e^{x+1}-e^x\int_{e^x}^{e^{x+1}}\frac{\cos u}{u^2}du.$$Because $|\cos u |\lt 1$, it follows that $|\int_{e^x}^{e^{x+1}}\frac{\cos u}{u^2}du|\lt e^{-x}-e^{-x-1}$. Using this in the above equality one has
$$e^x|f(x)|\le|\cos e^x| + e^{-1}|\cos e^{x+1}|+ e^x|\int_{e^x}^{e^{x+1}}\frac{\cos u}{u^2}du|$$
$$e^x|f(x)|\lt1+e^{-1}+1-e^{-1}$$ $$\implies e^x|f(x)| \lt 2.$$ Since $$\int_0^{\infty}\sin(e^t)dt=\Big[\sum_{k=0}^{N}f(k)\Big]_{N\rightarrow \infty}$$ and the series $2\sum_{k=0}^{\infty}e^{-k}$ is a majorant of the series in the RHS of above, it follows that $\int_0^{\infty}\sin(e^t)dt$ is convergent. Now consider an integral of the form $$\int_0^{\infty}\sin(p(t))dt$$ where $p(t)$ is a polynomial in $t$ of degree $\ge 2$. It seems likely that such an integral should also converge. My question is does this integral converge, and if it does, then is there a general argument that uses the above facts to show this? Also, what can be said if $p(t)$ were a rational function with no non-negative poles?
A: Consider the set $\{t>0: p(t)=k\pi \text{ for some integer } k\}$, which is able to be written as an increasing sequence $\{a_1,a_2,\dots,\}$. Now since $p$ has degree $\geq 2$, the integral of $\sin p(t)$ over these intervals $[a_n,a_{n+1}]$ becomes an alternating sequence (eventually, as the highest order term dominates), since we have
$$
\lim (a_{n+1}-a_n)=0
$$
i.e. the roots are becoming closer and closer.
Thanks for the idea of the following:
ncmathsadist (https://math.stackexchange.com/users/4154/ncmathsadist), Prove: $\int_{0}^{\infty} \sin (x^2) dx$ converges., URL (version: 2012-02-03): https://math.stackexchange.com/q/105167
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650e96de42fcd33f044314a4b2a432748545bbce
|
Q: How To Prove The following equation? The equation arised in the paper:Exact and asympototic representations of the sound field in a stratified ocean.That is the equation(3.12) for solving the problem
$$\Delta p+k^{2}p=-\delta(z-z_{0})\frac{\delta(r)}{2\pi r}$$
$$p|_{z=0}=0$$
$$p|_{z=-h}=0$$
where p is the acouotic press and $\delta$ is the Dirac mass at the point 0.They said:using the fact that
$$(\frac{\partial^2}{\partial r^{2}}+\frac{1}{r}\frac{\partial}{\partial r}+ka_{n}^{2})H^{(1)}_{0}(ka_{n}r)=\frac{4i\delta(r)}{2\pi r}$$
where $H_{0}^{(1)}(ka_{n}r)$is the first kind Hankle Function of order zero.
Probelm:I don't know this fact because I can't derive from my knowledge.Please help me prove this equation.
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f3f37c969675850c69ec6de88a9799e5db6a4123
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Q: Can anyone explain the mathematical, geometric or physical meaning of $(-1)^{1/3}$ Suddenly, I was very confused on the mathematical or geometric meaning about cube root of a minus real number, say
$(-1)^{1/3}$. In my opinion, I think $(-1)^{1/3}=-1$, because $(-1)^3=(-1).(-1).(-1)=-1$. But when I ask Mathematica with
N[(-1)^(1/3)]
I got
(*0.5 + 0.866025 I*)
If the number is somewhat more complicated, say $\left(-\frac{27}{40}\right)^{1/3}$, how do I consider its sign, could it be considered as a positive number?
Or there is no plus or minus characteristic for this kind of number? Can I regard the cube root of a minus number as a minus one?
Sorry, if my question is too silly, kindly teach me. Thanks!
A: Typically, we do algebra in the real numbers ( $\mathbb{R}$), but here you're looking at a different set. You are getting different answers because $\mathbb{C}$ is algebraically closed. Namely, the degree of a polynomial corresponds to its number of roots. Interestingly enough, this is not the case in the real numbers ( take $x^{2}+1=0$ for example, which is solving for the roots of a parabola that does not intersect the x-axis.) In other words:
$$(-1)^3=-1.$$
However, $e^{i \pi/3}$ and $e^{-i\pi/3}$ are also legitimate solutions in the complex numbers.
This is true because $e^{i \pi/3}=\cos(\pi/3)+i\cdot \sin(\pi/3)\approx .5+i\cdot.86603$
This is a solution because $(e^{i \pi/3})^{3}=e^{i \pi}=-1$.
Hence the famous equation: $e^{i \cdot \pi}+1=0$.
This is part of what makes mathematics in the complex numbers so well-behaved.
A: In complex numbers every number, not equal to zero, has n, n-th roots. All distinct. The three cube roots of -1 are: $-1, 1/2 + \sqrt{3}/2 i, 1/2 - \sqrt{3}/2 i$.
So there is room for ambiguity in which of these should be $(-1)^{1/3}$. In general for $b >0$ $b^{1/3}$ is defined to be the specific and unique positive root which always exist and of which there is always only one. But if $b$ is not a positive real (or zero) then there isn't any positive real root. There is some, but not universal, consensus as to which of the n-th roots will get the honor of being defined $b^{1/n}$. (Others will correct me if I am wrong.)
Now one thing about n-th complex roots of $b$. There are always n of them and they always are positioned on the complex plane so that they are inscribed in a circle around the origin, 0, with radius $\sqrt[n]{|w|}$ and each equal angles apart.
So if there is a convention. It is usually to be the first one, one encounters on the circle when looking in a counter clockwise direction starting from the real axis. That way if $b$ is positive real the first we'd encounter is the real root.
If we follow this convention then the first 3rd root of -1 that we encounter is $1/2 + \sqrt{3}/2 i$. So that is why mathematica gave you what it did. (0.5 + 0.866025 I) is one cube root and it is the first one, one comes to in a counter-clockwise direction.
==== more advanced (slightly) =====
If we look at a complex number $a + b i$ in the complex plane, we can view it as an ordered pair. It is the point that is $a$ over on the x (real) axis and $b$ up on the y (imaginary) axis.
But we can also not that if we view it as a ray or vector form the origin, 0, we can see it has a length $r = \sqrt{a^2 + b^2}$ and it has an angle $\theta$ measured from the real axis.
So we can alternatively view the complex number as $r(\cos \theta + i \sin \theta)$. It's the same number as $a = r\cos \theta$ and $b = r\sin \theta$. But we are viewing it, instead of a rectangular coordinated point, we are iewing it as polar coordinated point.
There is a surprising benefit to viewing it this way:
If we let $w = e^{\ln r}(cos \theta + i sin \theta)$ is one complex number and $z = e^{\ln t}(cos \phi + i sin \phi)$ is another then $wz = e^{\ln r + \ln t}([\cos \theta \cos \phi - sin \theta \sin phi] + i [ cos \theta \sin \phi + \cos \phi \sin \theta]) = e^{\ln r + \ln t}(\cos (\theta + \phi) + i \sin (\theta + \phi))$.
In other words we found a way to do multiplication but adding the angles and multiplying the lengths.
(We go a step further and define and write $e^{a + b i} = e^a(\cos b + i \sin b)$ but we won't today.)
So anyway. This means: To find the n-th roots of $a + bi$ you convert it to $r(\cos \theta + i \sin \theta)$. Then... wait for it... the n-th roots will be: ${r^{1/n}(\cos (\theta/n + 2k\pi/n) + i \sin (\theta/n + 2k\pi/n)} $ for $k = 0, 1, ..., n-1$. (as raising those to the $n$th power will result in $({r^{1/n}}^n(\cos (n\theta/n + 2kn\pi/n) + i \sin (n\theta/n + 2kn\pi/n)= r(\cos(\theta + 2k\pi) + i \sin (\theta + 2k\pi)) = r(\cos \theta + i \sin \theta)$.)
Which is why the convention is the define the default i-th root by the formula above with k = 0.
So if $-1 = 1(cos \pi + i \sin \pi)$ then $-1^{1/3} = 1^{1/3}(\cos \pi/3 + i \sin \pi/3) = 1/2 + \sqrt{3}/2 i$.
But the other two are $(\cos (\pi/3 + 2\pi/3) + i \sin (\pi/3 + 2\pi/3)) = \cos \pi + i \sin \pi = -1$
And $\cos (\pi/3 + 4\pi/3) + i \sin (\pi/3 + 4\pi/3) = \cos -\pi/3 + i \sin -\pi/3 = 1/2 - \sqrt{3}/2 i$.
A: You are right, $(-1)^3=-1$, hence $-1=(-1)^{1/3}$.
Anyway, there are two other solutions in complex numbers, $\dfrac{1\pm i\sqrt3}2$. Indeed,
$$\left(\frac{1\pm i\sqrt3}2\right)^3=\frac{1\pm i3\sqrt3-3\left(\sqrt3\right)^2\mp i\left(\sqrt3\right)^3}8=-1.$$
Mathematica prefers one of these.
Interpretation:
In the reals, multiplying by $-1$ is a mirroring around the origin. Three mirrorings amount to a single one.
In the complex, numbers are represented by points in a plane, and multiplying by a complex number is the combination of a rotation and a scaling, around the origin.
The two complex solutions correspond to rotations of $\pm60°$, with unit scaling. Hence three rotations are equivalent to a single one by $\pm180°$, i.e. multiplication by $-1$.
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ba4495cebc0b302368dff7ff6ecdfe90ec93853c
|
Q: Solving a Diophantine Equation with 2 variables This is my answer for the following question: Find all natural numbers $(a,b)$ for which $a^b-b^a=1$.
When $a$ or $b$ equals $1$, $(a,b)=(2,1)$ is trivial. If $a,b>1$, I generalized the problem to $b^a|a^b-1$
Let $p$ be the smallest prime divisor of $b$. Then $a^ {p-1} \equiv a^ {b} \equiv 1\pmod p$.
Since $p$ is the smallest prime divisor of $b$, $gcd(b,p-1)=1$. Therefore $gcd(a^b-1,a^ {p-1}-1)=a-1 \equiv 0\pmod p$. $\therefore a \equiv 1\pmod p$.
By L.T.E
If $p$ is not $2$, $a{ v }_{ p }(b)\le { v }_{ p }(a^{ b }-1)={ v }_{ p }(a-1)+v_{ p }(b)$. Thus, ${ v }_{ p }(a-1)\ge a-1$. A contradiction.
If $p=2$,
$I)$ $a \equiv 1\pmod 4$: This implies that $a{ v }_{ 2 }(b)\le v_{ 2 }(a-1)+v_{ 2 }(b)$. Thus ${ v }_{ 2 }(a-1)\ge a-1$. A contradiction.
$II)$ $a \equiv 3\pmod 4$: This implies that $a{ v }_{ 2 }(b)\le v_{ 2 }(a+1)+v_{ 2 }(b)$. Thus ${ v }_{ 2 }(a-1)\ge a+1$, implying $a=3$ and ${v}_{2}(b)=1$.
If $b=2B, B \equiv 1\pmod 2$, $B^3|3^ {2B}-1$. If the smallest prime divisor of $B$ is $q$, $q$ is not $2$. However, $q|3^ {2B}-1$ and $q|3^ {q-1}-1$. Therfore $q|3^2-1$, a contradiction. Therfore $B=1$.
Thus the only solution is $(3,2),(2,1)$.
However, this solution seemed a bit strange. I think that there must be other solutions to this problem without using Catalan`s Conjecture. What are other elementary ways to solve this equation?
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839cb6953d94059109f080297fd881af6954ea8e
|
Q: Lifting a real quadratic twist of an Elliptic Curve to the modular surface Let $E$ be an elliptic curve of conductor $N\cdot p^2$ over $\mathbb{Q}$, defined by the equation
$$y^2=x^3+p^2b\cdot x + p^3\cdot c$$
and parametrized by a map
$$X_{0}(N\cdot {p}^{2})\rightarrow E$$
and let $E_{\sqrt(p)}$ be the curve
$$y^2=x^3+b\cdot x+c$$
parametrized by a map
$$X_{0}(N)\rightarrow E_{\sqrt(p)}$$
Both become isomorphic over $\mathbb{Q}[\sqrt{p}]$. Can a lifting of this isomorphism to a morphism: $X_{0}(N\cdot {p}^{2})\rightarrow X_{0}(N)$ be described explicitly?
|
3e7b84bde207cdb58c530d436ef7079fe1996f40
|
Q: Initial and final sub-spaces of a partial isometry Let $H$ be a Hilbert space and assume $H_0$ and $K_0$ are two sub-spaces of $H$ with dim$H_0$=dim$K_0$.
Question: Is there any partial isometry $u$ whose initial projection is $H_0$ and final projection is $K_0$?
A: Yes, the obvious construction works:
Let $(e_i)_{i\in I}$ and $(f_i)_{i\in I}$ be orthonormal bases of $H_0$ and $K_0$ respectively, and complete them to ONBs $(e_i)_{i\in J}$ and $(f_i)_{i\in J}$ of $H$. Define $ve_i=f_i$ for $i\in I$ and $ve_i=0$ for $i\in J\setminus I$. It is easily seen that
$$v^\ast f_i=\begin{cases}e_i,&\text{for }i\in I,\\0,&\text{for }i\in J\setminus I.\end{cases}
$$
It follows that $v^\ast v$ is the projection onto $\overline{\mathrm{lin}}\{e_i\mid i\in I\}=H_0$ and $vv^\ast$ the projection onto $\overline{\mathrm{lin}}\{f_i\mid i\in I\}=K_0$.
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08688b3ffab321521dd52a342d5dec310225711c
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Q: If n is positive integer, prove that the prime factorization of $2^{2n}\times 3^n - 1$ contains $11$ as one of the prime factors I have: $2^{2n} \cdot 3^{n} - 1 = (2^2 \cdot 3)^n - 1 = 12^n - 1$.
I know every positive integer is a product of primes, so that,
$$12^n - 1 = p_1 \cdot p_2 \cdot \dots \cdot p_r. $$
Also, any idea how to use math notation on this website?
A: We have to prove $11|12^n-1$. This follows easily from $$12^n\equiv 1^n=1\ (\ mod\ 11\ )$$
A: First, show that this is true for $n=1$:
$2^{2}\cdot3^{1}-1=11$
Second, assume that this is true for $n$:
$2^{2n}\cdot3^{n}-1=11k$
Third, prove that this is true for $n+1$:
$2^{2(n+1)}\cdot3^{n+1}-1=$
$\color\red{2^{2n}\cdot3^{n}-1}+11\cdot2^{2n}\cdot3^{n}=$
$\color\red{11k}+11\cdot2^{2n}\cdot3^{n}=$
$11(k+2^{2n}\cdot3^{n})$
Please note that the assumption is used only in the part marked red.
A: By the Binomial theorem,
$$12^n-1=(11+1)^n-1\\
=11^n+n\,11^{n-1}+\frac{n(n-1)}211^{n-2}+\cdots+\frac{n(n-1)}211^2+n\,11+1-1,$$
which is a multiple of $11$.
|
628b0b331a39960b1bfd9299cdbb7a985623c12e
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Q: How to find a sequence that maximizes a ratio Given positive parameters $n$, $P$ and $Q$, what is a sequence $a_1,\dots,a_n$ such that for every $k$:
$$
\frac{1}{k}\leq a_k \leq 1
$$
which maximizes the ratio:
$$
R = \frac{P + \sum_{k=1}^n a_k}{Q + \sum_{k=1}^n k\cdot a_k}
$$
?
NOTE: one observation is that a maximizing sequence should be weakly decreasing (for all $k>l$, $a_k\leq a_l$). PROOF: If there exists $k>l$ with $a_k>a_l$, then switch $a_k$ with $a_l$. This strictly decreases the denominator but does not change the nominator.
A: Suppose we have the optimal sequence $\{a_i\}$. Then $a_1=1$. Also, if $\frac{1}{k}\lt a_k \lt 1$ for some $k$, then a small perturbation ϵ in $a_k$ reduces the required ratio $R$. Now write$$R_m=\frac{N}{M}$$
where $R_m$ is the value of the ratio for the optimal sequence $\{a_i\}$. Then $$R_\epsilon=\frac{N+\epsilon}{M+k\epsilon}.$$
From the condition $R_m−R_ϵ>0$, we have $$N k \epsilon>\epsilon M.$$
Since a perturbation is possible in both directions for $a_k$, the inequality is a contradiction.
Your observation that the sequence is weakly decreasing is crucial here. For, if suppose that $a_1 = a_2 = .. a_{k-1} =1$ and $a_k=\frac{1}{k}$, then $a_t = \frac{1}{t}$ for $t>k$ is forced by the above argument.
|
bcc52e82e06c759241ef201814c9b811d95bae23
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Q: Given the lattice, find all pairs of elements that generate $D_{8}$ Given the subgroup lattice of the dihedral group $D_{8}$, find find all pairs of elements that generate $D_{8}$. There should be 8 pairs.
The given lattice is the image below:
I came to following set of pairs:
$\left \langle r, s \right \rangle$,
$\left \langle r^2, s \right \rangle$,
$\left \langle rs, r \right \rangle$,
$\left \langle r^3s, r \right \rangle$,
$\left \langle s, rs \right \rangle$,
$\left \langle s, r^3s \right \rangle$,
$\left \langle r^2s, rs \right \rangle$,
$\left \langle r^2s, r^3s \right \rangle$
and
$\left \langle s, r^3 \right \rangle$,
$\left \langle rs, r^3 \right \rangle$,
$\left \langle r^3s, r^3 \right \rangle$,
$\left \langle r^3, r^2s \right \rangle$
But I am not sure if this is right.
Also, how do you analyse the lattice step by step to reach the solution? Finding out first 8 was simple, because I had to find single-element subgroups and combine them to form pairs. However, the rest 4, I derived from found ones by searching for equivalent subgroups.
This is exercise 2.5.4 in Dummit & Foote "Abstract Algebra" (3rd ed.).
A: It's $12$, really.
You have to choose first a pair of distinct maximal subgroups. And then one element from each of these maximal subgroup, neither of which elements lies in their intersection.
For instance, if you take the pair of maximal subgroups
$$
\langle s, r^2 \rangle,
\quad
\langle r \rangle
$$
you can choose either $s$ or $s r^2$ from the first subgroup, and either $r$ or $r^3$ from the second one, This makes for $4$ generating pairs.
As there are three pairs of maximal subgroups, and for distinct maximal subgroups $M, N$ you have $\lvert M \setminus M \cap N \rvert = 4 - 2 = 2$, you get $2 \cdot 2 \cdot 3 = 12$ (unordered) pairs.
|
0429e1e84710328b932673dab473472c039c7aa0
|
Q: Example of subgroup $H$ with the property $g H g^{-1}$ is properly contained in $H$. After learning the definition of normal subgroup,
I would like to find an example of a group $G$ which has a subgroup $H$ such that there exists an element $g \in G$ such that $gHg^{-1} \subsetneq H$, namely, $gHg^{-1}$ is a proper subgroup of $H$.
Can anyone provide an example? Thank you very much!
A: See Example 1.33 in J.S. Milne's lecture notes.
|
3d69f52adf228dcf4a9521cf17eb55ba73fabd92
|
Q: Lp spaces (Hölder, Minkovski) Let $1<p<\infty$ and $f\in L_p(0,\infty)$. Show that
$$\lim_{x\to\infty}\frac{1}{x^{1-\frac{1}{p}}}\int_0^x f(t)dt=0$$
by assuming that f is compactly supported.
Any idea so it can help me how to show it?
A: Let $\frac{1}{p}+\frac{1}{q}=1$. Then
\begin{eqnarray*}
\varphi _{f}(x) &=&\frac{1}{x^{1-\frac{1}{p}}}\int_{0}^{x}dtf(t)=\frac{1}{x^{%
\frac{1}{q}}}\int_{0}^{x}dtf(t)=x^{-1/q}\int_{0}^{x}dtf(t) \\
&=&x^{-1/q}\int_{0}^{\infty }dt\theta (x-t)f(t)
\end{eqnarray*}
Next
\begin{eqnarray*}
|\int_{0}^{\infty }dt\theta (x-t)f(t)| &\leqslant &\left[ \int_{0}^{\infty
}dt\theta (x-t)^{q}\right] ^{1/q}\left[ \int_{0}^{\infty }dt|f(t)|^{p}\right]
^{1/p} \\
&=&\left[ \int_{0}^{\infty }dt\theta (x-t)^{q}\right] ^{1/q}\left[
\int_{0}^{\infty }dt|f(t)|^{p}\right] ^{1/p} \\
&=&\left[ \int_{0}^{\infty }dt\theta (x-t)\right] ^{1/q}\parallel f\parallel
==\left[ \int_{0}^{x}dt\right] ^{1/q}\parallel f\parallel =x^{1/q}\parallel
f\parallel
\end{eqnarray*}
so
\begin{equation*}
|\varphi _{f}(x)|\leqslant \parallel f\parallel
\end{equation*}
Let $\{f_{n}(x)\}$ be a sequence of compactly supported functions converging
to $f$. We have
\begin{equation*}
\varphi _{f_{n}}(x)\rightarrow 0
\end{equation*}
Now
\begin{equation*}
|\varphi _{f_{n}}(x)-\varphi _{f}(x)|\leqslant \parallel f_{n}-f\parallel
\end{equation*}
where the left hand side converges to $|\varphi _{f}(x)|$ and the right hand
side to $0$. Hence
\begin{equation*}
\varphi _{f}(x)\rightarrow 0
\end{equation*}
|
96962b0077a55f0e63e57ca4df4d5e67304134f2
|
Q: Without using prime factorization, show if $m\mid n^2$ then $\gcd(m,n^2/m)\mid n$ It's easy to use prime factorization to show:
If $m\mid n^2$ then $\gcd(m,n^2/m)\mid n$.
Can anybody find some other proof - perhaps a simple reduction of some sort? Maybe solving $m^2x + n^2y=mn$, for example?
I suppose the result would follow if you could prove that $\gcd(m^2,n^2)=\gcd(m,n)^2$, since $\gcd(m,n)^2\mid mn$ since $\gcd(m,n)\mid m$ and $\gcd(m,n)\mid n$. That result doesn't even need $m\mid n^2$.
A: Since $m \mid n^2$, we have $n^2 = dm$ for some integer $d$. Inserting into the equation $m^2 x + n^2 y = mn$ gives
$$
m^2x + dmy = mn\\
mx + dy = n\\
\gcd(m, d) \mid n
$$
But $\gcd(m, d)^2 \mid md$, and we had $md = n^2$, so we're done.
A: By here $\,\color{#0a0}{(m^2,n^2)} = (m,n)^2 = (m^2,\color{#c00}{mn},n^2)\Rightarrow \color{#0a0}{(m^2,n^2)}\mid \color{#c00}{mn} \overset{\!\div\,m}\Rightarrow (m,n^2/m)\mid n$
A: Oh, this all follows from the result that if $\gcd(a,b)=1$ then $\gcd(a^2,b^2)=1$. This is true since if $ax+by=1$ then $$1=(ax+by)^3 = a^2X+b^2Y.$$
So we deduce that $\gcd\left(\frac{n^2}{\gcd(n,m)^2}\,,\frac{m^2}{\gcd(n,m)^2}\right)=1$, and hence that $\gcd(m^2,n^2)=\gcd(m,n)^2$, and hence that $m^2x+n^2y=mn$ has a solution.
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61a354994837a994850a69639f9c58abbb53ed36
|
Q: What's the ratio $\displaystyle \frac{e^{2(\theta+\phi)}-1}{e^{2\theta}+e^{2\phi}}$ equal to? I'm asked what the ratio $\displaystyle \frac{e^{2(\theta+\phi)}-1}{e^{2\theta}+e^{2\phi}}$ is equal to.
The answer I'm given is $\displaystyle \frac{\sinh(\theta+\phi)}{\cosh(\theta-\phi)}$. I could work backwards from this, but how do I show it's equal to this directly?
A: Diving the N & D by $e^{x+y}$ $$\dfrac{e^{2(x+y)}-1}{e^{2x}+e^{2y}}=\dfrac{e^{x+y}-e^{-(x+y)}}{e^{x-y}+e^{-(x-y)}}$$
Now use $e^A=\cosh A+\sinh A $
|
5daefdfaab46f74999fdcd41154da264d3ada7c2
|
Q: Factorization of polynomial with factors with degree at most $2$ For every polynomial $p(x)$ of degree $n>3$, can we find a factorization of factors with degree at most $2$? (factorization in $\mathbb R$).
Example. $$x^5-1=(x-1)(x^4+x^3+x^2+x+1)=(x-1)(x^2+\frac{\sqrt{5}+1}{2}x+1)(x^2+\frac{-\sqrt{5}+1}{2}x+1)$$
In case $n$ even we need to talk:
If $n$ be odd and the coefficient of $x^n$ be positive (negative), we see $p(-\infty)=-\infty$ and $p(+\infty)=+\infty$ $\Big(p(-\infty)=+\infty$ and $p(+\infty)=-\infty\Big)$ then it has a root and we reduce degree to case $n-1$.
A: Yes, this follows from (and is essentially equivalent to) the fundamental theorem of algebra. By the fundamental theorem of algebra, $p(x)$ has a complex root $a+bi$ for some $a,b\in\mathbb{R}$. If $b=0$, this is a real root and we're done, so suppose $b\neq 0$. Since $p(x)$ has real coefficients, $\overline{p(x)}=p(\bar{x})$ for any $x\in\mathbb{C}$, where the bar denotes complex conjugation. It follows that $0=\overline{p(a+bi)}=p(a-bi)$, so $a-bi$ is also a root of $p$. Now note that $$q(x)=(x-(a+bi))(x-(a-bi))=x^2-2ax+(a^2+b^2)$$ has real coefficients, and divides $p(x)$ since $a+bi$ and $a-bi$ are both roots of $p(x)$. Thus $p(x)$ has a quadratic factor with real coefficients, and by induction on degree we're done.
|
41bf68339528227dce00779e25296d17fa151fd6
|
Q: How to list elements of $\Bbb Z_j=\{x \in \Bbb Z\,|\, x-j=km \text{ for some } k \in \Bbb Z\}$ from its set builder notation? Let $m$ be any fixed positive integer. For each integer $j$, $0\le j \lt m$, let $\Bbb Z_j=\{x \in \Bbb Z\,|\, x-j=km \text{ for some } k \in \Bbb Z\}$.
The set notations for $\Bbb Z_0$, $\Bbb Z_1$, $\Bbb Z_2$ would be as below. Then how do we list elements of $\Bbb Z_0$, $\Bbb Z_1$, $\Bbb Z_2$ from the set notations below?
$\Bbb Z_0= \{x \in \Bbb Z\,|\, x=km \text{ for some } k \in \Bbb Z\}$
$\Bbb Z_1= \{x \in \Bbb Z\,|\, x=km+1 \text{ for some } k \in \Bbb Z\}$
$\Bbb Z_2= \{x \in \Bbb Z\,|\, x=km+2 \text{ for some } k \in \Bbb Z\}$
FYI
"Definition 6 Let F be an arbitrary family of sets. The union of the sets in F, denoted by $\bigcup\mathscr F$, is the set of all elements that are in A for some $A\in\mathscr F$.
$\bigcup\limits_{A \in \mathscr F}A$={$x\in U$|$x \in A$ for some $A\in \mathscr F$}"
"Definition 7 Let F be an arbitrary family of sets. The intersection of sets in F, denoted by $\bigcap\limits_{A\in\mathscr F}A$ or $\bigcap\mathscr F$, is the set of all elements that are in A for all $A \in\mathscr F$.
"
$\bigcap\limits_{A\in\mathscr F}A$={$x\in U$| x∈A for all $A\in \mathscr F$}
Source: Set Theory by You-Feng Lin, Shwu-Yeng T. Lin.
A: Are you looking for something like this?
$$\Bbb Z_0 = \{0, \pm m, \pm2m, \pm3m, \ldots\}\\
\Bbb Z_1 = \{1, \pm m +1, \pm2m + 1, \ldots\}\\
\Bbb Z_2 = \{2, \pm m +2, \pm2m + 2, \ldots\}
$$
I'll do a concrete example, so you can see how it works. Say $m = 3$. Then
$$
\Bbb Z_0 = \{x \mid x = 3k\text{ for some }k\in \Bbb Z\}
$$
It is spelled out "$\Bbb Z_0$ is the set of all $x$ such that $x = 3k$ for some $k \in \Bbb Z$". This means that $15 \in \Bbb Z_0$, since $15 = 3\cdot 5$ (in that case, $k = 5$). However, $43 \notin \Bbb Z_0$, because there is no $k\in \Bbb Z$ so that $43 = 3k$. We get
$$
\Bbb Z_0 = \{0, \pm 3, \pm 6, \pm 9, \ldots\} = \{0, 3, -3, 6, -6, 9, -9,\ldots\}
$$
On the other hand, we have
$$
\Bbb Z_1 = \{x \mid x = 3k+1\text{ for some }k \in \Bbb Z\}
$$
which means that a number $x$ is an element of $\Bbb Z_1$ iff there is a $k\in \Bbb Z$ such that $x = 3k+1$. For instance, $43 \in \Bbb Z_1$, since there is a $k\in \Bbb Z$ that makes $43 = 3k + 1$ (it's $14$). However, $15 \notin \Bbb Z_1$, since there is no integer $k$ such that $15 = 3k + 1$. This gives
$$
\Bbb Z_1 = \{1, \pm3 + 1, \pm6 + 1, \ldots\} = \{1, 4, -2, 7, -5, \ldots\}
$$
Lastly, in exactly the same way, we get
$$
\Bbb Z_2 = \{2, \pm3 + 2, \pm 6 + 2, \ldots\} = \{2, 5, -1, 8, -4,\ldots\}
$$
A: Start with another notation: $r+m\mathbb Z:=\{r+mk\mid k\in\mathbb Z\}$.
This instead of $\mathbb Z_r$.
Now define $\mathbb Z/m\mathbb Z:=\{r+m\mathbb Z\mid r=0,1,\dots,m-1\}$.
Then $\mathbb Z/m\mathbb Z$ is a list of the sets you mention.
Note that - defined like this - $\mathbb Z/m\mathbb Z$ is a partition of $\mathbb Z$.
Another commonly used notation for $\mathbb Z/m\mathbb Z$ is $\mathbb Z_m$.
However, that notation is unfortunately allready in use in your question.
Following your notation (I dislike it) we have $\mathbb Z/m\mathbb Z:=\{\mathbb Z_0,\dots,\mathbb Z_{m-1}\}$.
|
c790999bcaed18ad0ba6bca60e9d8ac4254b0def
|
Q: Using induction for an easy proof for formal languages I am having trouble to understand the way of using a induction for the following example:
Let $\Sigma \overset{\Delta} = \{a, b\}$ and $S_1 \overset{\Delta} = \{a^n \mid n \in \Bbb N\}$. Prove using induction that $$\forall w \in \Sigma^* . w \in S_1 \to \big( w = \lambda \lor (\forall \ 1 \le i \le |w| . (w)_i = a) \big) .$$
How to do that?
Any help is upvoted.
A: Induction on the lenght $n$ of the string.
i) $n=1$ : then $|w|=1$. But also $w=a$, by definition of $S_1$, and thus $(w)_1=a$.
ii) $n \to n+1$ : then $|w|=n+1$. But also $w=w'a$, where $|w'|=n$.
By induction hypotheses : $(w)_i=(w')_i=a$, for $1 \le i \le n$ and also $(w)_{n+1}=a$, and thus :
$(w)_i=a$, for $1 \le i \le n+1=|w|$.
As already commented by others, the use of induction here is quite "excessive".
We rely on the fact that, by definition of $S_1= \{a^n \mid n \in \mathbb N \}$, $(w)_i=a$, for any $i \le |w|$.
Thus, we already have the result, i.e.
$w \ne λ \to ∀i \ ((1≤i≤|w|) \to (w)_i=a))$.
|
bc4694aadb12885686026c1e8912d27179b5970b
|
Q: Orthogonal projection. Let $H$ be the Hilbert space $L^2[0,1]$, and let $S$ be the subspace of functions $f\in H$ satisfying $\int^1_0(1+x)f(x)dx=0$.
Find the element of $S$ closest to the function $g(x)=1$.
I know that $S$ contains the functions orthogonal to $x \mapsto 1+x$ and what I want to do is find the orthogonal projection of $g$ onto S but I'm not too sure how to go about doing this.
A: Let $\phi\colon[0,1]\to\mathbb{R},\,x\mapsto 1+x$. By the Hilbert projection theorem, $g$ can uniquely be written as $g=u+v$ with $u\in S$, $v\in S^\perp$, and $v$ is the orthogonal projection onto $\mathrm{lin}\{\phi\}$.
One can easily see that $v=\frac{\langle g,\phi\rangle}{\|\phi\|^2}\phi$ (just check that $g-v\perp\phi$). Thus,
\begin{align*}
u(x)=g(x)-v(x)=1-\frac{\int_0^1 (1+x)\,dx}{\int_0^1 (1+x)^2\,dx}(1+x)=1-\frac{9}{14}(1+x)=\frac{1}{14}(5-9x).
\end{align*}
|
d3ba841b5740d19bf580d74dc4cfa3364ad87c8d
|
Q: Find the number of balls in a jar The question is this:
A jar contains only red and green balls. Ten red balls are added and then the green balls now constitute 20 percent of the total. In addition, ten green balls are added, making the percentage of green balls equal to 40 percent of the total. How many balls were originally in the jar?
I don't know to solve this. Please help me.
A: HINT:
Solve the following system of two equations in two variables:
*
*$\color\green{y}=\frac{20}{100}\cdot(\color\red{x}+10+\color\green{y})$
*$10+\color\green{y}=\frac{40}{100}\cdot(\color\red{x}+10+\color\green{y}+10)$
|
9fc11aebf60f7d0af57d9246cb57feaf750d4af4
|
Q: For which values of $p$ is the Markov chain recurrent? ( sort of a RW with two steps forward and one back) I have a process with indices in $\mathbb{Z}$ with the following transition probabilities: $P(i,i+2)=p$ and $P(i,i-1)=1-p$, with $p\in(0,1)$, i.e. with probability $p$ I take two steps forward and with probability $1-p$ I go one step backwards. I have to determine for which values of $p$ this Markov chain is recurrent.
My idea was to compute the powers of the transition matrix (it´s easy to notice that the chain has period 3, so one is interested in the diagonal of the matrices $P^{3n}$), then check with the criterion saying that the chain is recurrent iff $\sum_{n}P^n(i,i)=\infty$. However, the computations seem cumbersome and I was wondering if anyone has a better idea on how to do this. A reference (on this process) would also be extremely appreciated.
A: To see why for $p \neq 1/3$ the process is transient, just apply Law of Large Numbers.
Let $X_i \in \{ -1, 2 \}$ where $X_i = -1 $(with probability $1-p$) and $X_i = 2$ (with probability $p$). Then, we are talking about the process $(S_n)_n$ where $S_n = \sum_{i=1}^n X_i $, right?
Well, we have $\mathbb{E}[X_i] = 3p-1$, thus, using independence of $X_i$'s, Law of Large Numbers implies that
$$
\frac{S_n}{n} \longrightarrow 3p-1
$$
almost surely. This means that your process has a drift to the right or the left (depending on $p$). This also implies that if you start the chain at zero, for example, you will visit it only a finite number of times and since the chain is irreducible, you have that the only chance of being recurrent is if $p=1/3$.
For $p=1/3$ you can apply similar argument employed to prove that the symmetric random walk is recurrent.
Hope this can help!
|
be5bf93a80cd3e313f2490de3f52a563bd0e33d1
|
Q: How can I find the Nash-equilibrium of the following zero sum game? I want to find the Nash-equilibrium of the following zero sum game. $$A=\begin{bmatrix}0&2&-1\\-2&0&3\\1&-3&0\end{bmatrix}$$
I used the Minimax Theorem.
$$min_{x \in X} max_{y \in Y} E(x,y)= min_{(u,x)} \{u; u \ge \sum^{n}_{i=1}a_{ij}x_i, \sum^{n}_{i=1}x_i=1, x_i \ge 0 \} $$
$$max_{y \in Y} min_{x \in X} E(x,y)=max_{(w,y)} \{w; w \le \sum^{m}_{j=1}a_{ij}y_j, \sum^{m}_{j=1}y_j=1, y_j \ge 0 \} $$
$$E(x,y)=x^TAy$$
I can get two linear programs
$$min \, u$$
$$-u-2x_2+x_3 \le 0 $$
$$-u+2x_1-3x_3 \le 0 $$
$$-u-x_1+3x_2 \le 0 $$
$$ x_1+x_2+x_3 =1 $$
$$ x_1,x_2,x_3 \ge 0 $$
and
$$max \, w $$
$$ w-2y_2+y_3 \le 0 $$
$$w+2y_1-3y_3 \le 0 $$
$$w-y_1+3y_2 \le 0 $$
$$ y_1+y_2+y_3 =1 $$
$$ y_1,y_2,y_3 \ge 0 $$
Then, how can I find the Nash-equilibrium?
A: Let one player's mixed strategy be $(p,q,r)$. Using the indifference condition for mixed NE, the other player's payoffs from each pure strategy must satisfy
$$
\begin{aligned}
-2q+r&=2p-3r\\
2p-3r&=-p+3q\\
1-p&=q+r
\end{aligned}
$$
Line 1 implies
$$
2r=p+q
$$
Line 2 implies
$$
p=q+r
$$
Together with line 3, we can deduce
$$
p=\frac12,\quad q=\frac16,\quad r=\frac13.
$$
So this is the mixed NE with symmetric strategies.
|
95269ded6e6e07da487380da8bfa131cfc685981
|
Q: Doubts on distributional derivative: null derivative and non negative derivative suppose we are in $\mathbb R^n$ and fix $E\subset \mathbb R^n$ and let $\chi_E$ be the indicator function of the set $E$. I'm reading an article that says that a vector field $X$ on $\mathbb R^n$, seen as a differential operator on $C^\infty(\mathbb R^n)$, has:
*
*distributional derivative $X\chi_E\geq 0$ if and only if $$-\int_{\mathbb R^n}\chi_E \,X\phi \,dx \geq 0 $$ for all $\phi\in C^\infty_c(\mathbb R^n)$ with $\phi\geq 0;$
*distributional derivative $X\chi_E=0$ if and only if $$\int_{\mathbb R^n}\chi_E \,X\phi \,dx = 0 $$ for all $\phi\in C^\infty_c(\mathbb R^n)$ with $\phi\geq 0.$
Now, my first doubt is why I require in both points above that the test function $\phi$ is non negative. Indeed, I think I've already seen definition 2., but, if I remember correctly, I've seen it with a general test function $\phi$ without requiring the non negativity. However, I can write any test function $\phi$ as $\phi=\phi^+-\phi^-$, thus, if $X\chi_E=0$ in the sense of 2., it also holds that $$\int_{\mathbb R^n}\chi_E \,X\phi \,dx = 0 $$ for all $\phi\in C^\infty_c(\mathbb R^n),$ right?
But I can't do a similar reasoning for the point 1., so I think that for it I have to ask the non negativity of the test function $\phi$.
My second doubt is the following one: if $X\chi_E$ in the sense of distribution is a measure, then clearly $X\chi_E\geq 0$ in the sense of 2. Is the viceversa true? I mean, if I have that $X\chi_E\geq 0$ in the sense of 2., is the distributional derivative $X\chi_E$ a measure?
I hope I made myself clear.
A: Before starting my answer, let me remark that all your questions concern a general distribution $T$, and isn't really related to the fact that $T=X\chi_E$.
In point 2, the usual definition is indeed with any test function (it is a trivial definition: $T=0\Leftrightarrow(\forall\phi\in C^\infty_c, T(\phi)=0)$; then the specificity of $T=X\chi_E$ rely on the fact that $XS(\phi)=-S(X\phi)$ with $S=\chi_E$).
However, your proof of this fact is incorrect as $\phi_+,\phi_-$ are not $C^\infty$. Though I trust the result, but you should find a proof of the fact : if $T$ satisfies $\forall \phi\geq 0, T(\phi)=0$, then $T=0$.
For point 1, you should require a sign for $\phi$. More generally, the definition of $T\geq 0$ when $T$ is a distribution is
$$\forall \phi\in C^\infty_c, \phi\geq 0\Rightarrow T(\phi)\geq 0.$$
A simple reason for that is to notice that if $T(\phi)=\int f\phi$ for a locally integrable function, then the above definition is equivalent to $f\geq 0 \; a.e.$.
If you didn't require a sign on $\phi$, then using $\pm\phi$, you would be lead to point 1!
About your last question:
if $T$ is a positive (Radon) measure (depending on the setting you work in, you could consider non-positive measure), then it is a positive distribution.
The converse is true: if $T$ is a positive distribution, then it is of order $0$, and therefore, according to Riesz representation theorem, can be represented with of positive (Radon) measure. The proof of that is a classical and short exercise (start on a compact set so that you don't have issue with the compact support of test functions)
|
96e4d31df37c1294428d0958a7e9185089a20ace
|
Q: Probability Problem: Find the probability Roy loses his job Melchester Rovers are playing a football tournament of $5$ matches.
The probabilities that they win, draw or lose a match are $\frac12$, $\frac16$ and $\frac13$ respectively.
These probabilities remain constant; the result of a match is independent of the results of other matches.
At the end of the tournament their coach Roy loses his job if they lose three consecutive matches, otherwise he does not lose his job.
Find the probability that Roy loses his job.
My attempt:
Ways for Roy to lose his job:
*
*$L,L,L,\neg{L},\neg{L}$
*$\neg{L},L,L,L,\neg{L}$
*$\neg{L},\neg{L},L,L,L$
*$L,L,L,L,\neg{L}$
*$\neg{L},L,L,L,L$
*$L,L,L,L,L$
For the first three cases which I collectively call Case A:
$$P(A)=3\left(\left(\frac13\right)^3\left(\frac12\right)^2+2\left(\frac13\right)^2\left(\frac12\right)^1\left(\frac16\right)^1+\left(\frac13\right)^2\left(\frac16\right)^2\right)$$
For the next 2 cases which I collectively call Case B:
$$P(B)=2\left(\left(\frac13\right)^4\left(\frac12\right)^1+\left(\frac13\right)^4\left(\frac16\right)^1\right)$$
The last case; Case C:
$$P(C)=\left(\frac13\right)^5$$
Adding all of these:
$$P(A \cup B \cup C)=\frac{17}{234}$$
However, this is not the correct answer.
Where did I go wrong?
A: Let $A$ be the event that matches 1,2,3 are lost, let $B$ be the event that matches 2,3,4 are lost and let $C$ be the event that matches 3,4,5 are lost.
Then you are asked to find $P(A\cup B\cup C)$ wich can be done using inclusion/exclusion and symmetry:
$$P(A\cup B\cup C)=3P(A)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$$
Also realize that $A\cap C=A\cap B\cap C$ and $P(A\cap B)=P(B\cap C)$, leading to:$$P(A\cup B\cup C)=3P(A)-2P(A\cap B)$$
A: If Roy's run of $3$ consecutive losses materialises on the 4th or 5th game,
it must be immediately preceded by a "not lost", hence $Pr = (\frac13)^3[1 + 2(\frac23)] = \frac{7}{81}$
[ Other results don't matter ]
Further explanation
With $\bullet$ indicating results that don't matter (i.e. could be any of $W, L$ or $D$) and $L'=$ not lose ($W$ or $D$), sequences with Roy losing job are:
$LLL\bullet\bullet\quad L'LLL\bullet\quad \bullet L'LLL$
|
cb2e7df8f605c6bce8bdd96f17ff680936acd378
|
Q: How can an angle be negative? How can be angle be negative like sine(-60) , cosine(-50) ?
Which quadrant do they fall if we have the negative angles ? I dont see any negative angles in full circle of 360
A: Angles increase positively by increasing counter clockwise. So a negative angle is one that starts in a clockwise direction. 60 is the angle 60 degrees above the x-axis so -60 is the angle 60 degrees below the x-axis.
Angle measures are considered cyclic and any angle $x$ is equal to $x \pm 360$. So $-60$ is the same thing as $300$.
In particular 180 = -180. Also convenient are -90 = 270. And of course 0 = 360.
====
Believe it or not you will encounter situations where you need to consider angles of 720 degrees. These are simply 0.
Sometimes you'll have to do something like sin(x + y) or cos(x - y) and have things like x = 135 and y = 243. It's useful to know that 125 + 243 = 368 = 8 and 135 - 243 = -108 make sense. (Actually -108 = 108 clockwise is far more intuitive than 252 degrees is.)
A: Hint:
$$
\sin (330°)=\sin (-30°)
$$
In other words: the positive angles are counterclockwise, the negative angles are clockwise.
A: For a negative angle, you simply need to rotate in the other direction (i.e., clockwise instead of anti-clockwise).
This means that for an angle of $-40^\circ$, you end up at the same point (and with the same values of $\sin$ and $\cos$ as an angle of $320^\circ$.
A: Negative angles are connected to an orientation on the circle, just as negative abscissae are connected to a direction of travel on a straight line: once you've decided what is the positive direction of travel (from left to right, or from right to left), the straight line becomes an (oriented) axis, and points on this axis can have a positive or negative abscissa.
Similarly, once you've decided what is a positive rotation on a circle (clockwise or counterclockwise), it becomes an oriented (or trigonometric) circle. Remember an angle, in a sense is but a curvilinear abscissa on a circle with radius $1$.
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62396f3a0f983e01d802aced1c40b06d724d535e
|
Q: Probability a random walk hits zero at specified time set Let $X_n \in \lbrace -1, 0, 1 \rbrace$ be sequence of i.i.d random variables taking $-1$ or $1$ with equal probability, and $0$ some positive probability. $S_n = \sum_{i = 1}^{n} X_i$ is a random walk. It is easy to see that $S_n$ is null recurrent. Let Q be the set of indexes so that $S_n = 0$, namely the set of return to zero times. Let $W \subset \mathbb{N}$ be any infinite subset of $\mathbb{N}$. Is it true or false that
\begin{align*} Pr[W \cap Q = \emptyset] = 0 \end{align*}
|
ff888767e044d7f5b861d76bd72b410b4a7608f9
|
Q: Equation of a spherocylinder/capsule I want to know if a 3D spherocylinder/capsule, that looks like this:
Can be approximated by a known equation. I want a shape with volume.
Edit: In my program I already have a cylinder with two hemispheres on the ends to represent the capsule, I wonder if there is a better idea.
A: Assume the upper/lower caps are hemispheres centered at $(0,0,\pm a)$ with radius $r$ and the middle portion is a circular cylinder, the equation of the capsule is
$$x^2 + y^2 + \frac14\left(|z-a|+|z+a| - 2a\right)^2 = r^2$$
I will encourage you to prove this yourself, it is sort of fun.
A: The top half sphere can be represented by the equation $$x^2 + y^2 + (z-h)^2 = a^2$$
where $z\geq h$.
The bottom can be represented as $x^2+y^2+z^2=a^2$ where $z\leq 0$
The cylinder in between can be $$x^2+y^2=a^2$$
where $0\leq z\leq h$
Put these three together and you get an equation $F(x,y,z)=a^2$ where $F$ is a continuous function (defined using three equations).
A: If $l$ is the 'length' of the cylindrical section of the capsule and $r$ is the 'radius' of the two half spheres then the volume of the capsule would be given by $$\pi r^{2} \left( \frac{4}{3}r+l \right)$$
|
8821281187e2cf2ceb1ef5c96f500c074a82a491
|
Q: Proof of hockey stick like theorem for multinomial coefficients I discovered following identity via a programming exercise. Looking for the way how it prove it by induction or some other way.
$$\binom{a_1+a_2+\cdots+a_t}{a_1,a_2,\cdots,a_t}=1+\sum_{i=2}^t \sum_{j=1}^{i-1} \sum_{k=1}^{a_i} \binom{ a_1+a_2+\cdots+a_{i-1}+k }{a_1,a_2,\cdots,a_j-1,\cdots,a_{i-1},k }$$
|
4e6696c7d6c074bafa74ca80e9589045c83a57ed
|
Q: What are the differences between large deviations theory & extreme value theory? I need to study both for my Master's thesis in finance. (Probably, I'll have to apply them on the Value at Risk and Conditional Value at Risk estimation, so, on quantile estimation, loosely speaking; I know they are both used to this purpose). Precisely, what are the differences and the relationships (if any) of these two theories?
|
69d9c192b7c5acc5fd959ae1377412e46b4a50d8
|
Q: Expectation value of $\cos^2\theta$ with Associated Legendre Polynomials as basis I am trying to solve this integral
$$\int_0^\pi P^m_p(\cos\theta) ~~\cos^2\theta ~~P^m_q(\cos\theta)~~\sin\theta d\theta$$
where $P$ is the Associated Legendre polynomial
I tried to solve it by the fact that
$$\cos^2\theta = \frac{2}{3}P^0_2(\cos\theta)+\frac{1}{3}$$
but this gives an integral that consists three Associated Legendre polynomial and I am not sure if this integral has straightforward solution.
This should give three terms that relate $(p,p)$, $(p,p+2)$ and $(p,p-2)$.
|
8eb9e2980d0ab79929a58c042531f7e13943178b
|
Q: $\lim_{x\to 1/2} \left(\frac{\tan(\pi x)}{2x-1}+\frac{2}{\pi(2x-1)^2}\right)$ using L' Hopital My task is to find
$$\lim_{x\to 1/2} \left(\frac{\tan(\pi x)}{2x-1}+\frac{2}{\pi(2x-1)^2}\right)$$
using L' Hopital, so I first wrote it in a way that the numerator and denominator have the limit $0$ at $x=1/2$:
$$\lim_{x\to 1/2}\frac{\pi (2x-1)^2\tan(\pi x)+4x-2}{\pi(2x-1)^3}$$
But when I calculate the derivative I still get a $0$ in the denominator, while the numerator doesnt seem to be $0$.
Can you please point me to the right direction?
A: Using first the idea of Claude Leibovici, let be $x=t/\pi+1/2$:
$$
\lim_{x\to 1/2}\frac{\pi (2x-1)^2\tan(\pi x)+4x-2}{\pi(2x-1)^3} =
\lim_{t\to 0}\frac{\pi(1-t\cot t)}{2t^2}.
$$
And before using L'Hôpital:
$$
\lim_{t\to 0}\frac{\pi(1-t\cot t)}{2t^2} =
\lim_{t\to 0}\frac{\pi(\tan t-t)}{2t^2\tan t} =
\lim_{t\to 0}\frac{\pi(\tan t-t)}{2t^3}
$$
by the equivalence $\tan t\sim t$ when $t\to 0$.
Now, you can easily use L'Hôpital and again the equivalence:
$$
\lim_{t\to 0}\frac{\pi(\tan t-t)}{2t^3} =
\lim_{t\to 0}\frac{\pi\tan^2 t}{6t} = \frac\pi6
$$
|
8fb5fe3dbd951fef32202c0f76665f69ed55d1f2
|
Q: Does there exist a self-adjoint operator whose spectrum is just the continuous spectrum? Does there exist a self-adjoint operator whose spectrum is just the continuous spectrum, i.e. no point spectrum and no residue spectrum?
If not, please prove it.
|
937b7ca37388eb3082e1ea5535d7bff8aaf63f8e
|
Q: How much of radioactive material substance will remain after $5$ days? Initially there are $8$ grams of a radioactive material in a container.The half-life of the material is $2$ days.
How much of the radioactive substance will remain after $5$ days ?
By exponential decay I know that after $2$ days I have $4$ grams of the radiactive material and that $2$ days later I will have $\cfrac{2}{2}=1$ gram of the substance.
Now I don't know how to find how much radioactive I have after $5$ days ,i.e. in the "beginning",roughly speaking, of the $6^{th}$ day.
My book provides as a solution the following equation $\cfrac{2-x}{2}=\cfrac{x-1}{x}$ ($x$ is the amount of material after $5$ days) from which it follows that $x=\sqrt{2}$ but I don't know how this equation have been set up,what is meaning behind this.
If someone can help me understand how that equation is derived I am really grateful.
A: The remaining quantities at the beginning of the days are known to be
$$0\to8g\\2\to4g\\4\to2g\\5\to xg\\6\to1g.$$
Then you express that the relative quantity that disappears in a single day (the daily decay) is a constant. With the disintegrations from day $4$ to day $5$ and day $5$ to day $6$,
$$\frac{2-x}2=\text{Cst}=\frac{x-1}{x}.$$
It is simpler to express the fraction that remains, giving
$$\frac x2=\text{Cst}=\frac 1x$$ and you immediately get $x=\sqrt2g$.
A: The amount of material after time $t$ (let's forget units for now) is $$m(t) = m(0) \cdot 2^{-\frac t{\tau}}$$
where $\tau$ is the half-life of the material (you can see that plugging in $t=\tau$ gives you $m(\tau)=m(0)\cdot\frac12$ which is what you would expect).
Now, what you want to know is what $m(t)$ will be equal to when $t=5$, given that $\tau=2$ and $m(0)=8$. Simply plug in the numbers and voila:
$$m(5)=8g\cdot 2^{-\frac{5\text{ days}}{2\text{ days}}} = \frac{8g}{2^\frac{5}{2}} = \frac{8}{2\cdot2\cdot\sqrt2}g = \sqrt2g$$
|
7e9cb745d70d1005660148704a5ab64625a28352
|
Q: Is there a function on a compact interval that is differentiable but not Lipschitz continuous? Consider a function $f:[a,b]\rightarrow \mathbb{R}$, does there exist a differentiable function that is not Lipschitz continuous?
After discussing this with friends we have come to the conclusion that none exist. However there is every chance we are wrong. If it is true that none exist how could we go about proving that? It is true that if $f$ is continuously differentiable then $f$ is Lipschitz, but what if we don't assume the derivative is continuous?
A: The map $f : [0,1] \to \mathbb{R}$, $f(0) = 0$ and $f(x) = x^{3/2} \sin(1/x)$ is differentiable on $[0,1]$ (in particular $f'(0) = \lim_{x \to 0^+} f(x)/x = 0$), but it is not Lipschitz (the derivative $f'(x)$ is unbounded).
A: If you require $f'$ to be continious then there can be no $f$ since $f'([0,1])$ is compact and indeed bounded.
|
10e9538c826bd6afeb764e462950656cb5451cfe
|
Q: Labelling the edges of a cube with {1, 2, 3,....,12} I did the following problem:
a) Is it possible to label the edges of a cube by $1, 2, \cdots 12$ (using each number only once) so that at each vertex, the labels of the edges leaving that vertex have the same sum.
b) A suitable edge label is replaced by $13$. Now is the equality of the sums possible?
The first is clearly not possible as, if each sum is equal to $k$ then we have, $78=4k$.
For the second one, to be possible (necessary condition) we get,
$78+13-i=4k$.
So, the possible values of $i$ are $3, 7, 11$.
Now my question is,
1)How to prove that for part b) the above condition is sufficient as well, without giving a construction?
I was able to give a construction(edge labelling) by some trial and error(smart guessing) for each $i=3, 7, 11$.
Or
2) Or is there a systematic way to give such a construction?
A: If there exists an integer $i$ where $1 \le i \le 12$ such that $4 |( 91 - i)$ then equality of the sums is possible. $91$ is not divisible by $4$ but $92$ is. So perhaps we can rephase the above in the following manner to make things easier: let $j = i + 1$, if there exists an integer $j$ where $2 \le j \le 13$ such that $4 |( 92 - j)$, then equality of the sums is possible. Clearly $j$ must be a multiple of 4 so possible candidates are $4, 8, 12$. It then follows that $i$ can be $3, 7, 11$.
EDIT: This is in response to the second part of your question where you were asking whether there's a systematic way to construct possible values of $i$.
A: One can write a program that finds the solutions, and yes there are solutions for all these.
Example of a python program that finds the solutions:
import itertools
edges = [
(0, 1), # 0
(0, 2), # 1
(0, 4), # 2
(1, 3), # 3
(1, 5), # 4
(2, 3), # 5
(2, 6), # 6
(3, 7), # 7
(4, 5), # 8
(4, 6), # 9
(5, 7), # 10
(6, 7), # 11
]
labels = [1,2,3,4,5,6,7,8,9,10,11,12]
labels[10-8] = 13 # 10 is the 11'th element
sref = sum(labels)/4 # this is the sum at each corner
cnt = 479001600
for x in itertools.permutations( labels ):
cnt -= 1
if cnt & 0xFFFFF == 0:
print cnt
if x[0] + x[1] + x[2] != sref:
continue
if x[0] + x[3] + x[4] != sref:
continue
if x[1] + x[5] + x[6] != sref:
continue
if x[3] + x[5] + x[7] != sref:
continue
if x[2] + x[8] + x[9] != sref:
continue
if x[4] + x[8] + x[10] != sref:
continue
if x[6] + x[9] + x[11] != sref:
continue
if x[7] + x[10] + x[11] != sref:
continue
print x
What can be noted is that there's $144$ solutions in the cases $i=3$ and $i=11$, and $48$ solutions in the case $i=7$. Since there's $24$ rotations of the cube and mirroring is possible the solution is unique except for this in the case $i=7$ and there's only three solutions otherwise.
|
70b52373d4f9aef6cbf79907c8cf2adb28d5eac5
|
Q: converges in $L^p(R)$ Consider the sequence of functions
\begin{align}
f_n(x)=
\begin{cases}
\frac{1}{n}\quad if\quad x\in [0,n]\\
0 \quad \mbox{otherwise}
\end{cases}
\end{align}
(1)Prove that $f_n\rightarrow0$ strongly in every space $L^p(
\mathbf{R})$ with $1<p\leq\infty$.
(2)On the other hand, show that in the space $L^1(\mathbf{R})$ this sequence is not strongly convergent.
I really don't know how to prove for this question.
A: Note that $\|f_n\|_p=\left(\int_0^n\frac{1}{n^p}dx\right)^{1/p}=\frac{1}{n}n^{1/p}=\frac{1}{n^{1-1/p}}$
But
$\dfrac{1}{n^{1-1/p}}\to\left\{\begin{array}{ccc}0&if&1<p<\infty\\1&if&p=1\end{array}\right.$
Moreover, if $p=\infty$, then $\|f_n\|_\infty=\frac{1}{n}\to 0$. This proves (1).
Now, suppose let $n,m\in\mathbb{N}$, $n>m$. Then $\|f_n-f_m|\|_1=\int_{\mathbb{R}}|\frac{1}{m}1_{[1,m]}-\frac{1}{n}1_{[1,n]}|dx=\int_0^m\frac{1}{m}-\frac{1}{n}dx+\int_m^n\frac{1}{n}dx=2$. Then $\{f_n\}$ is not a Cauchy sequence. As $L^1$ is a complete metric space (Banach space), then $\{f_n\}$ is not convergent.
|
2ddf09c2ce3fcfde27054b5af32ae2de3f66324a
|
Q: Examples of adjoint order morphisms Given posets $(X, \leq)$, and $(Y, \leq)$, an order morphism is a map $f : X \to Y$ such that $x \leq y$ implies $f(x) \leq f(y)$.
Given two order morphism $f: X \to Y$ and $g: Y \to X$, we say that $f$ is a left adjoint of $g$ (or that $g$ is a right adjoint of $f$) if for all $x \in X, y \in Y$ we have
$$
x \leq g(y) \Leftrightarrow f(x) \leq y
$$
in which case we also call $(f, g)$ an adjoint pair.
Classical examples are:
*
*For an extension of fields $F \subseteq K$, let $\mathcal{F}$ be the poset of intermediate fields (ordered by inclusion) and $\mathcal{G}$ the poset of subgroups of the Galois group Gal$(K/F)$, ordered by inclusion. Then
$$
f: \mathcal{G} \to \mathcal{F}^{op} : H \mapsto \text{Fix}(H) \\
g: \mathcal{F}^{op} \to \mathcal{G} : L \mapsto \text{Gal}(K/L)
$$
forms an adjoint pair.
*Let $(X, \leq)$ be a poset and for $Y \subseteq X$ define ub$(Y) = \lbrace x \in X \vert \forall y \in Y: y \leq x \rbrace$, similarly define lb$(Y) = \lbrace x \in X \vert \forall y \in Y: x \leq y \rbrace$. Letting $\mathcal{P}(X)$ be the power set of $X$, ordered by inclusion, then
$$
\text{ub} : \mathcal{P}(X) \to \mathcal{P}(X)^{op} : Y \mapsto \text{ub}Y \\
\text{lb} : \mathcal{P}(X)^{op} \to \mathcal{P}(X) : Y \mapsto \text{lb}Y
$$
forms an adjoint pair.
*For a field $k$ and the $n$-dimensional affine space $\mathbb{A}^{n}(k)$, the maps
$$
V : \mathcal{P}(k[X_1, ..., X_n]) \to \mathcal{P}(\mathbb{A}^{n}(k))^{op} : I \mapsto \lbrace x \in \mathbb{A}^{n}(k) \vert \forall f \in I : f(x) = 0 \rbrace \\
I : \mathcal{P}(\mathbb{A}^{n}(k))^{op} \to \mathcal{P}(k[X_1, ..., X_n]) : V \mapsto \lbrace f \in k[X_1, ..., X_n] \vert \forall x \in V : f(x) = 0 \rbrace
$$
form an adjoint pair.
All these examples, however, rely on arbitrarily considering the opposite relation on one of the two posets (for a poset $(X, \leq)$, the poset $(X^{op}, \leq^{op})$ is the set $X$ with order $x \leq^{op} y \Leftrightarrow y \leq x$). In fact, all maps considered 'reverse' the natural order and these structures are also known als general Galois connections; more examples are given in this question. Are there interesting examples of adjoint pairs which respect the natural order on both sets? I'm looking for adjoint pairs of posets, not of categories.
|
f24373cd74197e45420822e9bc1cee61d84333e2
|
Q: Geometry of the zeros of a power series. This is probably a basic question that is easily googlable, but it seems that I dont have the right keywords. So my question is, having some power series
$$
f(z)=\sum_{k=0}^{\infty}C_{k}z^{k}, z\in\mathbb{C}, C_{k}\in \mathbb{R}\;\wedge\;C_{0}=1
$$
Is there some theorem relating the coeficients $C_{k}$ of the power series to the geometry of the zeros? More specifically: Is there some theorem implying that the only zeros of $f(z)$ are real zeros if the coefficients $C_{k}$ have some specific properties?
References are welcome.
Thanks in advance.
|
ec07f5b01251d43345da9cadd393f9f6b8ff86d5
|
Q: A polynomial with integer coefficients with prescribed set of roots modulo infinitely many primes Given an infinite set of distinct prime numbers $p_1, p_2,\ldots,p_n,\ldots$, and arbitrary integres $a_1, a_2,\ldots,a_n,\ldots$, is there a nonzero polynomial $f(x)$ with integer coefficients such that $f(a_i)\equiv 0\pmod {p_i}$ for all $i\geq1$?
For example, let $p_1, p_2,...,p_n,...$ be all prime integers which are congruent to $1$ mod $4$. By a theorem of Euler, for every $i\geq1$, there exists some integer $a_i$, such that $a_i^2+1\equiv 0$(mod $p_i$). Thus the polynomial $f(x)=x^2+1$ has the required property.
A: Not in general. There are only countably many candidate polynomials, but uncountably many possible choices of input sequences.
To elaborate: Let us enumerate all nonzero polynomials in $\Bbb Z[X]$ as $f_1,f_2,f_3,\ldots$.
For each $i\in\Bbb N$ pick $a_i\in\Bbb Z$ with $f_i(a_i)\ne 0$. Then there are infinitely many primes that do not divide $f(a_i)$. Let $p_i$ be the smallest such prime that does not occur among $p_1,\ldots, p_{i-1}$. Then we have a sequence $p_1,p_2,\ldots$ of distinct primes and a sequence $a_1,a_2,\ldots$ such that no polynomial works. For if $f$ is a nonzero polynomial with integer coefficients, then $f=f_i$ for some $i$ and then $f(a_i)\not\equiv 0\pmod{p_i}$.
A: Not in general. If the sequence of primes increases exponentially but the sequence of integers increases linearly, then since a polynomial has only finitely many roots there would exist a polynomial s.t. its absolute value increases (at least) exponentially, which is impossible.
|
c00a16b3c1c56c9caab8cefaef416143063ab110
|
Q: Minimum value of $\frac{(1 + x + x^2)(1 + y + y^2)}{xy}$
What is the minimum value of $$\frac{(1 + x + x^2)(1 + y + y^2)}{xy},~~(x \neq 0)$$
Should we find the minimum value of each quadratic?
A: $$\frac{(1+x+x^2)(1+y+y^2)}{xy} = \frac{1+x+x^2}{x} \cdot \frac{1+y+y^2}{y}$$
By AM–GM inequality $(1+x+\frac1x)(1+y+\frac1y)≥3*3=9$
|
7fb9c122159bd53a9ba13f1876a23207fb740964
|
Q: Does $x\longmapsto Ax$ invertible where $A=\begin{pmatrix}1&0\\0&1\\0&0\end{pmatrix}$? I had an exam this morning and a question was:
Let $A=\begin{pmatrix}1&0\\0&1\\0&0\end{pmatrix}$ and consider $$f:\mathbb R^2\longrightarrow \text{Span}\big((1,0,0),(0,1,0)\big)$$
defined by $f(x)=Ax$. Does $f$ is invertible ?
I answered yes since the column are linearly independent (and thus $f$ is injective), and it's range is $\text{Span}((1,0,0),(0,1,0))$. Therefore it's bijective and thus invertible, no ?
But the answer was no and I don't understand.
Thanks for the explanation.
A: You are correct with the injectivity.
If we view $f$ as a map to $span\{(1,0,0)^T,(0,1,0)^T\}$, then $f$ is indeed surjective, and hence bijective. If we consider $f$ mapping to $\mathbb R^3$, then it is not surjective anymore (as e.g. $(0,0,1)^T\notin ran A$).
But as stated in the question, we can even write down the inverse:
$$f^{-1}((a,b,0)^T)=\begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}\begin{pmatrix}a\\b\\0\end{pmatrix}.$$
|
bbe29eb2e5b589a4749a0c8aa5ad83922165bc11
|
Q: When does $az+b \bar z +c =0$ have one solution?
When does $az+b \bar z +c =0$ have one solution? $a,b,c \in \mathbb C$
What I did:
I rewrote:
$a=a_1+ia_2$,
$b=b_1+ib_2$,
$c=c_1+ic_2$, and
$z=z_1+iz_2$
Putting everything together we are solving:
$(a_1+ia_2)\times (z_1+iz_2) + (b_1+ib_2) \times (z_1-iz_2) +(c_1+ic_2)=0+i0$
To have one solution I assumed that the $i$ terms will cancel out.
$i(a_2z_1+z_2a_1-b_1iz_2+iz_1b_2)+i(c_2)=0$
$iz_1(a_2+b_2)+iz_2(a_1-b_1)+i(c_2)=0$
So $a_1=b_1$ which means that the real part should be the same?
I'm not sure if I am working correctly.
Thanks.
A: Your approach works. Setting $a=a_1+a_2i$, $b=b_1+b_2i$, $c=c_1+c_2i$ and $z=z_1+z_2i$, the equation becomes $(a_1+a_2i)(z_1+z_2i)+(b_1+b_2i)(z_1-z_2i)+(c_1+c_2i)=0$. This we can also write as $(a_1z_1-a_2z_2+b_1z_1+b_2z_2+c_1)+(a_1z_2+a_2z_1-b_1z_2+b_2z_1+c_2)i=0$. We now have an equation of the form $v+wi=0$, where $v,w\in\mathbb{R}$. For a complex number to be $0$, both the real and imaginary part must be zero; we may conclude $$a_1z_1-a_2z_2+b_1z_1+b_2z_2+c_1=0$$ $$a_1z_2+a_2z_1-b_1z_2+b_2z_1+c_2=0$$ Now we have two linear equations in $z_1$ and $z_2$, and we can solve this! We'll first write the equations to $$(a_1+b_1)z_1+(-a_2+b_2)z_2=-c_1$$ $$(a_2+b_2)z_1+(a_1-b_1)z_2=-c_2$$ Making the coefficient for $z_1$ in both equations equal and subtracting, we get $$(b_2^2-a_2^2-a_1^2+b_1^2)z_2=-c_1(a_2+b_2)+c_2(a_1+b_1)$$ Now if $b_2^2-a_2^2-a_1^2+b_1^2=0$, we have either no or infinitely many solutions for $z_2$, neither of which we want. Thus, a property our system must have is $b_1^2+b_2^2\neq a_1^2+a_2^2$, in other words $|a|\neq |b|$ (using the norm/length of the complex numbers). We can now divide by $|b|^2-|a|^2=b_1^2+b_2^2-a_1^2-a_2^2\neq 0$, and we get $$z_1=\frac{c_1(a_1-b_1)-c_2(-a_2+b_2)}{|b|^2-|a|^2}$$ $$z_2=\frac{(a_1+b_1)c_2-(a_2+b_2)c_1}{|b|^2-|a|^2}$$
The conclusion is, the system has exactly one solution only if $|a|\neq |b|$.
Hope this helps you!
A: An approach that does not involve using the "expanded" form of a complex number.
Let $w$ be a different solution than $z$, such that:
$aw+b \bar w +c=0$ and
$az + b \bar z + c= 0$.
By subtracting we get: $a(z-w)+b(\bar z- \bar w)=0$ $⇒ a(z-w)=-b(\bar z- \bar w)$ (1)The conjucate of this expression must also equal zero, thus:
$\bar a(\bar z- \bar w)=-\bar b(z-w)$ (2)
By multyplying (1) and (2) we get:
$a \bar a(z-w)(\bar z- \bar w)=b\bar b (z-w)(\bar z- \bar w) ⇒ a \bar a=b\bar b $ and finally we get: $|a|=|b|$
A: It's easier than that. What you wrote is a linear system of two equations (collect the real and imaginary terms into separate equations):
$$\begin{align}
a_1 z_1-a_2 z_2 + b_1 z_1+b_2 z_2&=-c_1\\
a_2 z_1+a_1 z_2 + b_2 z_1-b_1 z_2&=-c_2
\end{align}$$
This has a single solution, if the rank of the matrix on the left is full (its determinant nonzero) - that's basics of linear algebra. If the determinant is zero, then you'll have infinite amount of solutions or no solutions at all. Collecting the terms into a matrix and computing the determinant gives you the condition
$$(a_1+b_1)(a_1-b_1)+(a_2-b_2)(a_2+b_2)\neq 0$$
This can be rewritten as
$$a_1^2-b_1^2+a_2^2-b_2^2\neq 0$$
or
$$|a|\neq |b|$$
|
8026cf537ad51be319d4f6bf26fed557094421b3
|
Q: Continuity of a Linear function Continuity of a function always is my problem, I know a theorem about that : Given two normed vector spaces V and W (over the same base field) a linear map $A:V \to W$ is continuous if and only if there exist a real number $C$ such that $\|Av\|\le C\|v\|$ for all $v\in V$.
But I do not know how apply this theorem, and I am not sure that I am assuming correct theorem. for example we have:
$f(x)\to f(0.5)$ as linear functional on $(\mathcal C[0,1] , \|.\|_\infty)$
how could I show its continuous.
Thanks in advance!
A: You must exhibit $C$ such that $\|Tf\|\le C\|f\|_{\infty}$ for all $f\in C[0,1]$, where $\|\cdot\|$ is the norm on the normed space in which $Tf$ lives.
In our case, $Tf=f(0.5)\in\mathbb{R}$. Then, you want $|f(0.5)|\le C\sup_{x\in[0,1]}|f(x)|$. But it is clear that $C=1$ is a possible bound. Indeed, $|f(0.5)|\le \sup_{x\in[0,1]}|f(x)|$ cause $|f(0.5)|$ is a member of the set where we are taking $\sup$.
So, we have $\|Tf\|\le C\|f\|_{\infty}$ for all $f$ taking $C=1$.
|
0656083fcd9b631bbac3b451a2bc2a279b74a60b
|
Q: Matrix diagonalization depending on the eigenvalues and eigenvectors I'm trying to do an app that can give the eigenvalues and vectors for a given square matrix. I have already found how to compute and display the eigenvalues and vectors, but I don't know in which case return that the matrix can't be diagonalize and that there isn't any eigenvalue or eigenvector.
So can you tell me how to determine if the matrix has eigenvalues or eigenvectors (is diagonalizable in fact?) depending on the eigenvalues and eigenvectors computed by the algorithm?
I explain you my algorithm:
For each REAL eigenvalue, it puts the related NON NULL eigenvectors in a column of an "eigenvectors matrix" containing as many columns as the number of eigenvalues. For the first real eigenvalue, it puts the non null eigenvectors in the first column of the "eigenvector matrix". For the second real eigenvalue, it puts the non null eigenvectors in the second column of the "eigenvector matrix". ...
If an eigenvalue isn't real, then no eigenvector is added int the column corresponding to the eigenvalue and it goes to the next eigenvalue.
|
58d2eef885bfc17b7626b5ea9bf0f86e20468f13
|
Q: I've shown $\lim_{n\to\infty}\sqrt[n]{n}=e$, but it should be $1$. Where's the mistake? $$\lim_{n\to \infty} \sqrt[n] n = \lim_{n\to \infty} n^{\frac{1}{n}} = \lim_{n \to \infty} \{(1+(n-1))^{\frac{1}{n-1}}\}{^{(n-1)\frac{1}{n}}} = \lim_{n\to \infty} e^{\frac{n-1}{n}} = e$$
But this clearly isn't true as the actual limit is $1$.
Where did I go wrong?
A: Notice, your answer is incorrect. your mistake $\lim_{n\to \infty}\left(1+(n-1)\right)^{1/(n-1)}\ne e$ . Follow the method as $$\lim_{n\to \infty}\sqrt[n]{n}=\lim_{n\to \infty}n^{1/n}=\exp\lim_{n\to \infty}\frac 1n \ln(n)$$
Applying L-Hospital's rule for $\frac {\infty}{\infty}$ form,
$$=\exp\lim_{n\to \infty}\frac{\frac{1}{n}}{1}$$
$$=\exp\lim_{n\to \infty}\frac{1}{n}$$$$=e^0=\color{red}{1}$$
A: You can do it using another way $$A=n^{\frac 1n}\implies \log(A)=\frac {\log(n)}n$$ and you know that $\log(n)$ is "slower" than $n$. So, $\log(A)\to 0$ and then $A\to 1$.
A: This is a wrong answer because you do $(1 + \infty)^{(1/ \infty)}$ is $e$ but Euler defined this limit by $(1+(\text{something that go to zero}))^{(1/\text{the something in the base that go to zero})} = e$, and that for why your answer wrong.
A: $$\lim_{m\to\infty}\left(1+\dfrac1m\right)^m=e$$
not
$$\lim_{u\to\infty}\left(1+u\right)^{1/u}= e$$
Let $A=\lim_{u\to\infty}\left(1+u\right)^{1/u}\implies\ln A=\lim_{u\to\infty}\dfrac{\ln(1+u)}u=\lim_{u\to\infty}\dfrac1{1+u}=0$ (Using L'Hôpital's Rule)
$\implies A=e^0$
|
3a4f2357ffcf2d5d379da20be8ca95543bd9dab0
|
Q: How to find the best rotation matrix between two Gaussian random variables? My question is really simple, given two paired sets of points $\{x_i\}$ and $\{y_i\}$ defined in an N-dimensional space $\{(x_1,y_1), (x_2,y_2), ..., (x_n,y_n)\} \in {\rm I\!R}^N \times {\rm I\!R}^N $, it is possible to find the best translation vector and rotation matrix that transforms $\{x_i\}$ to $\{y_i\}$ using the Kabsch algorithm. To my understanding, this algorithm is purely geometrical and does not suppose any distribution for the random variables $X$ and $Y$ generating respectively $\{x_i\}$ and $\{y_i\}$.
Now, given that condition ($\{x_i\}$ and $\{y_i\}$ both have a multivariate Gaussian distribution), is there another version of the Kabsch algorithm that takes such hypothesis into account ? Or is there another technique to find the best translation / rotation vector between data points generated by multivariate Gaussian random variables ?
Any pointer would be of big help. Thanks !
A: I think I found the answer, there is a paper called Empirical Bayes hierarchical models for regularizing maximum likelihood estimation in the matrix Gaussian Procrustes problem which reformulates the Kabsch algorithm (solving the Procrustes problem) using a ML estimator supposing the data are generated by a Gaussian distribution.
|
822718fb4f4af06c91a4d903b14ab5379351b9c9
|
Q: the number of the conjugacy class sizes of $S_n$ If $a$ has the cycle
type $(k_1,k_2,...,k_t)$ and $b$ has the type $(l_1,l_2,...,l_t)$ which is different with the cycle
type of $a$, then can the equality $|C_{S_n}(a)|=|C_{S_n}(b)|$ occur?
I hope the answer is no.
A: Unfortunately, the answer is yes: there is a bijection between the $n-2$ cycles and the permutations of type (n-2,2).
More generally, permutations of type $(k_1,k_2,\dots, k_t)$ are $\dbinom{n}{k_1,k_2,\dots,k_t}=\dfrac{n!}{k_1!\,k_2!\,\dotsm\, k_t!}$, if you do not take into account permutations inside each cycle. Hence the cardinal of a cojugation class is
$$\dfrac{\displaystyle n!\prod_{i=1}^t( k_i-1)}{\displaystyle\prod_{i=1}^t k_i!}.$$
|
26173f22c06a7274f93151202f2d2b9cdf05b3bf
|
Q: Finding a vector in a subspace of a Hilbert I'm stuck in the following problem:
Let $H$ be an infinite dimension separable Hilbert Space and $\{e_{n}\}$ a Hilbert basis for $H$. $\forall \epsilon >0$ Find explicitly a vector $x_{\epsilon} \in span\{e_{1}-e_{2},e_{2}-e_{3},...\}$ such that $\|x_{\epsilon}-e_{1}\|< \epsilon$
I've proved that we can find that vector since $\|x_{\epsilon}-e_{1}\|< \epsilon \iff e_{1}\in \overline{span\{e_{1}-e_{2},e_{2}-e_{3},...\}}$, and $\overline{span\{e_{1}-e_{2},e_{2}-e_{3},...\}}$ turns out to be $H$.
From there, I'm quite lost. I'd really appreciate any hint or approach.
Thanks in advance.
A: $x_N:=\sum_{n=1}^N \frac{N-n+1}{N} (e_n - e_{n+1})$ converges to $e_1$ since
$$||e_1 - x_N||^2 = \sum_{n=2}^N\left|\frac{N-n-N+n-1}{N}\right|^2+\left|\frac{N-N+1}{N}\right|^2=\frac{N-1}{N^2}+\frac{1}{N^2}=\frac{1}{N}$$
converges to zero as $N \to \infty$.
To get the formula note:
$$\sum_{n=1}^N a_n (e_n-e_{n+1}) = a_1 e_1 + \sum_{n=2}^N (-a_{n-1}+a_n)e_n - a_N e_{N+1}$$
|
5bddfef54e64af419a7d2c323a16da0fff7bf4e6
|
Q: Invariant ring for $S_5$ For the standard representation of $S_5$, the ring of invariants is generated by the elementary symmetric polynomials and hence it is a polynomial ring. Now if we take the tensor product of standard and sign representation, it is again irreducible of degree $4$ but the ring of invariants is not a polynomial ring anymore. Now my question is what are the generators and relations for this ring? Is it known for general $n$ ?
|
fa5daf13e1c29c63f90c9d1e2b3b3ea2f20a1224
|
Q: Prove that $5a^2 \equiv k \bmod 12$ where $k \in \{0,5,8,9\}$.
Prove that $5a^2 \equiv k \bmod {12}$ where $k \in \{0,5,8,9\}$.
Hence show that the equation $24x^7 + 5y^2 = 15$ has no integer solutions.
I think I need to evaluate each case with the $k$ values first but not sure how...
A: $a\equiv0,\pm1,\pm2,\pm3,\pm4,\pm5,6$
$$\implies a^2\equiv0,1,4,9$$
$$\implies k\equiv5a^2\equiv0,5,,20\equiv8,45\equiv9$$
$$24x^2+5y^2=15\implies5y^2\equiv15\pmod{12}\iff y^2\equiv3$$
A: Consider the following cases:
*
*$a\equiv 0\pmod{12} \implies 5a^2\equiv5\cdot 0^2\equiv 0\equiv0\pmod{12}$
*$a\equiv 1\pmod{12} \implies 5a^2\equiv5\cdot 1^2\equiv 5\equiv5\pmod{12}$
*$a\equiv 2\pmod{12} \implies 5a^2\equiv5\cdot 2^2\equiv 20\equiv8\pmod{12}$
*$a\equiv 3\pmod{12} \implies 5a^2\equiv5\cdot 3^2\equiv 45\equiv9\pmod{12}$
*$a\equiv 4\pmod{12} \implies 5a^2\equiv5\cdot 4^2\equiv 80\equiv8\pmod{12}$
*$a\equiv 5\pmod{12} \implies 5a^2\equiv5\cdot 5^2\equiv125\equiv5\pmod{12}$
*$a\equiv 6\pmod{12} \implies 5a^2\equiv5\cdot 6^2\equiv180\equiv0\pmod{12}$
*$a\equiv 7\pmod{12} \implies 5a^2\equiv5\cdot 7^2\equiv245\equiv5\pmod{12}$
*$a\equiv 8\pmod{12} \implies 5a^2\equiv5\cdot 8^2\equiv320\equiv8\pmod{12}$
*$a\equiv 9\pmod{12} \implies 5a^2\equiv5\cdot 9^2\equiv405\equiv9\pmod{12}$
*$a\equiv10\pmod{12} \implies 5a^2\equiv5\cdot10^2\equiv500\equiv8\pmod{12}$
*$a\equiv11\pmod{12} \implies 5a^2\equiv5\cdot11^2\equiv605\equiv5\pmod{12}$
|
932848f338712a941fb88486f5d5b3c28dbe2b15
|
Q: left and right jacobians (Not derivatives) of a Lie group Let $G < Gl(n)$ be a Lie group, and let $g:\mathbb{R}^n\rightarrow G$ be a smooth curve, parametrized as $g(q(t))$ with $t\in\mathbb{R}$. In that case it holds that
$$
\tag 1
\frac{dg}{dt} g^{-1} \in \mathfrak{g}
$$
Some computer vision papers, as well as some books (e.g. for reference, this) define the "left-jacobian" (which apparently differs from the concept of left derivatives on lie groups) of $G$ as
$$
\begin{equation}
J_l=\begin{pmatrix}
\left(\frac{\partial{g}}{\partial{q_1}}g^{-1}\right)^{\vee} & ... & \left(\frac{\partial{g}}{\partial{q_n}}g^{-1}\right)^{\vee} \\
\end{pmatrix}
\end{equation}
$$
where, supposing that we note $G_1,...,G_n$ the generators of $\mathfrak{g}$ , $\vee$ is defined as
$$
\vee : \mathfrak{g}\rightarrow \mathbb{R}^n
$$
$$
\left(\sum_{i=1}^nx_iG_i\right)^{\vee}=(x_1,...,x_n)^T
$$
and then they seem to deduce that
$$
\tag 2
\frac{dg}{dt}g^{-1}=\sum_{i=1}^nw_iG_i
$$
with $w_i$ the $i$-th component of $W=J_l\frac{dq}{dt}$.
I fail to see why equation (2) is correct (because of $W$). Assuming that I can start from (1) by using the chain rule, i.e. writing $\frac{dg}{dq}\frac{dq}{dt}g^{-1}$, I am unsure about the way $\frac{dg}{dt}$ and $g^{-1}$ interact. I assume they do not necessarily commute, which confuses me even more, considering that $\frac{dg}{dt}$ ends up at the right-side of the expression of $W$.
|
2167970dd1ef9e9e94b4d6fecbe7ed2367b4facb
|
Q: Coherent states - operator algebra problem with physics motivation Motivation: I have a mathematical problem motivated by quantum field theory in physics. It should be quite easy to prove, but for some reason I can't do it.
Intro:
Let there be operators $\hat{a_i}$ and $\hat{a_i}^{\dagger}$ for some integer $i$. Let there be the state (an element of vector space) $\left | 0 \right >$. A commutation relation holds such that
$$
\left [ \hat{a_i} , \hat{a_j}^{\dagger} \right ] = \left ( \hat{a_i} \hat{a_j}^{\dagger} - \hat{a_j}^{\dagger} \hat{a_i} \right ) = \delta_{ij}
$$
It holds that $a_i \left | 0 \right > = 0$ for all $i$.
Problem: Let there be a state (an element of vector space)
$$
\left | \psi \right > = e^{\sum_i \phi_i \hat{a_i}^{\dagger}} \left | 0 \right > \, ,
$$
where I define the exponential via its Taylor series as $e^{\hat{A}} = 1 + \hat{A} + \hat{A}\hat{A}/2! + \hat{A}\hat{A}\hat{A}/3! + \dots$ .
I want to show that
$$
\hat{a_x} \left | \psi \right > = \phi_x \left | \psi \right > \, .
$$
Thanks a lot!
A: First, we establish a simple result. Suppose we have two operators $A,B$ with $[A,B]=1$. Then one can easily prove via induction that more generally $[A,B^k]=k B^{k-1}$ for any natural $k$. Therefore the map $d_B$ defined as $d_B(\cdot)=[A,\cdot]$ acts as a derivative on all such terms $B^k$, and since this map is linear we furthermore have that this extends to any function $f(B)=\sum_k f_k B^k$. In particular, it applies to the exponential $\displaystyle e^{t B}=\sum_k \frac{t^k}{k!}B^k$ and so $d_B e^{t B}=[A,e^{t B}]= t e^{tB}$.
With this, the problem is straightforward. First, the coherent state can be written as $|\psi\rangle = \prod_i e^{\phi_i \hat{a}_i^\dagger}| 0\rangle$ since all the creation operators commute. But if $i\neq x$ then $\hat{a}_x$ also commutes with $\hat{a}_i^\dagger$ (and therefore $e^{\phi_i a_i^\dagger}$); otherwise, the formula shown above gives $\hat{a}_x e^{\phi_x \hat{a_x}^{\dagger}} = e^{\phi_x \hat{a_x}^{\dagger}}(\hat{a}_x+\phi_x).$ Consequently $$\hat{a}_x |\psi\rangle =a_x\left(\prod_i e^{\phi_i \hat{a}_i^\dagger}\right)| 0\rangle=\left(\prod_i e^{\phi_i \hat{a}_i^\dagger}\right)(\hat{a}_x+\phi_x)| 0\rangle=\phi_x |\psi\rangle$$ since $\hat{a}_x$ annihilates the ground state.
|
e8638b2143684550a511743671604dfeaaf527e2
|
Q: What is the standard : long or short scales I know about long and short scales of numbering systems, but I want to know, what is the standard naming of large numbers?!! is that back on the long scale or the short scale?!! so if someone in the world say 1 billion, so I've to take it as 10^9 or 10^12?
I came across this Wikipedia link which clarifies some countries/regions that use short scale and some others that use long scale, but this is already not a clear answer to my question.
A: You've answered your own question:
some countries/regions that use short scale and some others that use long scale
There is no standard. Unfortunately. Just as how English words and idioms are different depending on where you say it and to whom.
|
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