What does $\mathbb{Q}(\sqrt{2},\sqrt{3})$ mean? What set is $\mathbb{Q}(\sqrt{2},\sqrt{3})$? Is it the set $X = \{a\sqrt{2}+b\sqrt{3}:a,b\in\mathbb{Q}\}$?

$\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the smallest field containing $\mathbb{Q}$, $\sqrt{2}$ and $\sqrt{3}$. One can say that $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the field generated by $\{\sqrt{2},\sqrt{3}\}$ over $\mathbb{Q}$. It an important fact that one has the following tower of extensions: $$\mathbb{Q}\hookrightarrow\mathbb{Q}(\sqrt{2})\hookrightarrow\mathbb{Q}(\sqrt{2},\sqrt{3}).$$
It is false to say that: $$\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}\sqrt{2}+\mathbb{Q}\sqrt{3}.$$
The latest set is not a field. Actually, one has: $$\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}+\mathbb{Q}\sqrt{2}+\mathbb{Q}\sqrt{3}+\mathbb{Q}\sqrt{6}.$$

Not quite.  It is $\{a\sqrt{2}+ b\sqrt{3}+ c : a,b,c \in Q\}$.
$Q(\sqrt{2}, \sqrt{3})$ is "the smallest field containing all rational numbers as well as $\sqrt{2}$, and $\sqrt{3}$" so must contain, for example, 1.


Probability mass function from the distribution I have the following distribution function:
\begin{equation*}
      F(x) = \begin{cases}
      0, x<-1 \\
      \frac{1}{2}, -1 \leq x < \frac{1}{2} \\
      \frac{3}{4}, \frac{1}{2} \leq x <2 \\
      1, x \geq 2
        \end{cases}
    \end{equation*}
How can i obtain the probability mass function? My attempt was to use the discontinuites and try to evaluate the probabilities, but i was no able to confirm anything. I know that the sums off all probabilities must be one.

If before -1 the distribution function equals 0 and after -1 it equals $\frac{1}{2}$, it means that there is a mass of probability $\frac{1}{2}$ in $x=-1$.
For the same reason, there is a mass of $\frac{3}{4} - \frac{1}{2} = \frac{1}{4}$ in $x=\frac{1}{2}$, and finally a mass of $1 - \frac{3}{4} = \frac{1}{4}$ in $x = 2$.
The sum of all probabilities is $\frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1$.

Take the differences from the DF, it will give you the pdf at the jump discontinuities.


About joint probability divided by the product of the probabilities Let $X$ and $Y$ be two events.

So $P(X)$ is the probability of $X$ happens, and $P(Y)$ is the probability of $Y$ happens.
So $P(X,Y)$ is probability of both $X$ and $Y$ happen.

So what is the meaning of the following function: $h(X,Y)=\frac{P(X,Y)}{P(X)P(Y)}?$

I know that when $h=1$, it means $X$ and $Y$ are independent. So what is the situation when

$h>1$ or $h<1$?

Since you say that $X$ and $Y$ are events, let us rename them $A$ and $B$, to avoid a confusion with random variables.

Then, at least in the environmental, medical and life sciences literature, $P(A\cap B)/(P(A)P(B))$ is called the observed to expected ratio (abbreviation o/e). The idea is that the numerator is the actual probability of $A\cap B$ while the denominator is what it would be if $A$ and $B$ were independent.

Obviously the o/e ratio is $1$ if $A$ and $B$ are independent, it is more than $1$ if $A$ is favored by $B$, or, equivalently, if $B$ is favored by $A$, and it is less than $1$ if the opposite holds.

In the statistical analysis of genomic sequences, the CpGo/e ratio is especially important, which represents the frequency of the word CG divided by the product of the frequencies of the letters C (cytosine) and G (guanine), see here for an example. The rough idea is that in non functional portions of the genome, CpGo/e is much less than $1$ due to some well-known biological and chemical processes (a methylation-deamination of the guanine when it is right next to a cytosine, if you want to know). By contrast, in  portions of the genome called CpG islands, CpGo/e is only slightly smaller than $1$ or even, greater than $1$, a fact which witnesses a repression of these processes and, as a consequence, may signal some functional regions.

Mathematical Statistics is the branch of math that is area of specialization dealing with topics such as this. I can explain it to some point only here. But; it is my second major and leave it to those whose amin ateş of expertise is Applied Mathematics.


Are any linear combination of normal random variables, normally distributed? It is easy to show that if we have n independent normally distributed random variables, then a linear combination fo them ar normally distributed.

It is also said that if (x1,x2,..,xn) is multivariate normally distributed, but not nececarrily independent, then any linear combination is also normally distributed. This is stated here: http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Definition

But does this mean that any linear combination of normally distributed random variables are normally distributed, even if they are not independent? This will follow from the definition if the joint distribution of set of normally distributed random variables(not nececarrily independent) are jointly multivarite distributed?

This will follow from the definition if the joint distribution of set of normally distributed random variables(not nececarrily independent) are jointly multivarite distributed?

Yes.

  But does this mean that any linear combination of normally distributed random variables are normally distributed, even if they are not independent?

No. A frequently mentioned counterexample is based on $X$ standard normal and $Y=SX$ with $S=\pm1$ symmetric and independent of $X$. Then $X$ and $Y$ are normal but $X+Y$ is not since $P(X+Y=0)=\frac12$, a property that no normal random variable satisfies.

Just thinking in the univariate case, note that if two RVs are iid, distributed as $N(\mu, \sigma^2)$, then their sum is distributed as $N(2\mu, 2\sigma^2)$.  If they are correlated with $\rho=1$, it is the same as $2X=X+X$, which is distributed as $N(2\mu, 4\sigma^2)$ - in other words, it is 'stretched out' twice as far in variance.  This makes sense, because uncorrelated would have a standard deviation that grows by $\sqrt 2$, whereas adding to itself must double the standard deviation (which is equivalent to $4\sigma^2$).

All other cases of correlation fall in-between.  Though, as people have pointed out, $pho=-1$ gives you zero as a degenerate point.


Find the remainder when $787^{777}$ is divided by $100$? Find the remainder when $787^{777}$ is divided by $100$?

MyApproach

$787^{20\times38+17}$=$787^{17}$=I will get the last digit of remainder as 7 but how to calculate tens digit in this question fast using this approach only.

  Similarly,Find the remainder when $948^{728}$ is divided by $100$.

On solving I get $948^8$=I will get the last digit of remainder as 7 but how to calculate tens digit in this question fast using this approach only.

Again here how to calculate the other digits fast.

Hint

$787^3 \equiv 03\pmod {100}$ and $3^{20} \equiv 01 \pmod {100}$

Now the problem becomes much simpler. The last two digits of $787^{780}$ are $01$. You can now easily work backwards.

Other Problem

You can tackle it similarly by observing

$948^2 \equiv 04\pmod {100}$ and $4^{16} \equiv 04 \pmod {100}$

(You are not going to be lucky because with an even number, you will never get a $01$)

Edit - Alternately for first problem

You can use Fermat's little theorem, knowing $\phi(100) = 40$.

So any number, relatively prime with $100$, raised to $40$ will give $01$ and hence the last two digits of $787^{780}$ are $01$


Ackermann function proof by Induction I'm currently studying discrete mathematics and i've been given an assignment to prove the following: $A(1, n) = n +2$ for all $ n \geq 0$ with induction. But i am somewhat unsure if i've done it correctly.

Ackermanns function is defined as:

$A(0, n) = n + 1, n \geq 0$
$A(m, 0) = A(m - 1,1) ,m > 0$
$A(m, n) = A(m - 1, A(m, n - 1)), m, n > 0 $

Here's what i've done so far:

Base Case, $n = 0$: $A(1,0) = A(1-1, 1) = A(0,1) = 1+1 = 2$

Induction Hypothesis, $n = k$: Assume true for $A(1, k) = k + 2$, when $k \geq 0$

Prove: $n = k + 1$:

I followed the third function because if we have $k + 1$ then $k > 0$

(1): $A(1, k + 1) = A[1-1, A(1, k+1-1)]$

Question: Am i allowed to set $k = 0$ when proving n = k + 1?

If $k = 0$

Calculate the inner function of (1): $A(1, k+1-1) = A(1,0)$

(2): $A(1, 0) = A(1-1,1)$

Calculate function from (2): $A(1-1,1) = A(0, 1)$

(3): $A(0, 1) = 1 + 1 = 2$

Replace the inner function of (1) with the result from (3)

(1): $A(1, 0 + 1) = A(0, 2) = 2 + 1 = 3$

The induction is on $n$.

Your first step is correct; for $n=0$ we have that:

  $A(1,0)=A(0,1)=1+1=2$,

that is $n+2$ for $n=0$.

For the induction step, we have to assume that the property holds for $k$ and prove it for $k+1$.

Thus, we assume the induction hypotheses: $A(1,k)=k+2$, and we have to compute:

  $A(1,k+1)=A(0,k+2)=k+3$,

which is $n+2$ for $n=k+1$.

Based on ackermanns function rules :
Show the following $A (1,n)=n+2$
For $n=0$

$A (1,0)=A (0,1)=1+1=2$ true

Assume $n=k$ then $A(1,k)=k+2$ its true

Proof that $n=k+1$ also true

$A(1,k+1)=A (0,A (1,(k+1)-1))$

$=A (0,A (1,k))$

$=A (0,k+2)$

$=(k+2)+1$

$=(k+1)+2$

Proved :))


How many subsets are there such that their pairwise intersection is empty? Let $A$ be a set of $n$ numbers, I need to find the max collection of non empty subset of $A$ such that intersection of any two pair is empty.

To me answer seem $n$, take the example $A = \{1,2,3\}$, there are only three non-empty subsets ({1},{2},{3}) which satisfy the required condition.

In general what is the maximum number of subsets in a collection of subsets such that intersection of any two pair is at most $k$?

No element of $A$ can be in two of the subsets, or it would be in the intersection of that pair.  The approach you hint at, taking all the singletons, is optimal.

I think the answer is $\sum_{i=1}^{k+1} \binom {n}{i}$ but a reasonable proof eludes me.
The sum counts all the nonempty subsets of size $ \le k+1$. The intersection of any two of them that are distinct can have no more than $k$ elements in common.  So the desired number is $\ge$ $\sum_{i=1}^{k+1} \binom {n}{i}$

Going the other way is harder. A collection larger than $\sum_{i=1}^{k+1} \binom {n}{i}$ must contain a subset of size $\ge k+2$. If a set of size $h: n\ge h \ge k+2$ is included then $ \binom{h}{k+1} $ sets of size $k+1$ must be excluded. Applying this process one should be able to show that any other collection of non-empty subsets with pairwise intersections of size $\le k$ must contain fewer subsets then $\sum_{i=1}^{k+1} \binom {n}{i}$.


How Does This Fourier Grapher Work? A friend sent me this link: http://toxicdump.org/stuff/FourierToy.swf.  I am not very versed in fourier series.  I know the basic definitions and some convergence stuff, what you'd learn in a basic real analysis course.  Would someone explain to me what this thing is doing?  What do the various numbers at the bottom represent?  Thanks.

The core concept of the Fourier series is that any wave form can be transformed into a series of (co)sin waves.  In general you can represent any wave form in the following way. $$ f(x) = \sum_{i=0}^{\infty}a_i \cos(ix) $$

All you have to do is pick the right values for $a_0, a_1, a_2, ...$

The boxes on that page each represent a harmonic of the wave.  The left most is the first harmonic or the fundamental.  The numbers along the bottom are the amplitudes of each of these harmonics; the $a_i$ values in the equation above.

You can play around with these values and you will see you can use it to generate other wave forms.  For example 1, 0.5, 0.25, 0.125 ... will get you a sawtooth looking wave.


Poisson summation formula clarification regarding Fejer kernel Define $$\mathbf{F}_R(t) =
  \begin{cases}
    R \left(\dfrac{\sin(\pi R t)}{\pi R t}\right)^2 &  t \neq 0\\[10pt]
   R &  t = 0
  \end{cases}
$$

A problem in Stein's Fourier Analysis asks us to prove that the periodization of $\mathbf{F}_R(t)$ is equal to the Fejer kernel on the circle

i.e.

    $$\sum_{n=-\infty}^{\infty}\mathbf{F}_N(x+n) = \sum_{n=-N}^{N}\left(1-\frac{|n|}{N}\right)e^{2 \pi i n x} = \frac{1}{N} \frac{\sin^2(N \pi x)}{\sin^2(\pi x)} $$

for $N \geq 1$ an integer

This strongly suggests an application of Poisson summation is needed, which would mean that we need to calculate $$\sum_{n=-\infty}^{\infty}\hat{\mathbf{F}}_N(n)e^{2 \pi i n x}$$

correct?

However, as I don't see how we can go from an infinite series to a finite series using Poisson, I assume we have to show that the above series converges to the closed form expression above?  I'm still a bit unclear as to how apply Poisson in this case: if this is correct, any hints as to how to tackle the integral
$$\int_{-\infty}^{\infty} \mathbf{F}_N(x)e^{-2 \pi i n x}dx$$ would be appreciated.

How about trying the following alternative approach. First, note that for integer $ N $, the numerator of $ F_N(t) $ factors out since:
$$
\sin^2(\pi N (t + n)) = \sin^2(\pi N t + \pi N n) = \sin^2(\pi N t)
$$
The remaining part is given by the identity:
$$
\sum_{n= -\infty}^{\infty} \frac{1}{\pi^2 (t+n)^2} = \frac{1}{\sin^2(\pi t)}
$$
I am not entirely sure how to prove this identity...but I hope this will help anyway.

Hint: You can make use of the function g in Exercise 2 of chapter 5 and the Fourier inversion formula to handle the inner integral you get as you apply the Poisson summation formula, i.e. the integral you are asking for.


Getting the sequence $\{1, 0, -1, 0, 1, 0, -1, 0, \ldots\}$ without trig? What's the simplest way to write a function that outputs the sequence:

{1, 0, -1, 0, 1, 0, -1, 0, ...}

... without using any trig functions?

I was able to come up with a very complex sequence involving -1 to some complicated formula, but I was hoping there is a more simple solution.

$n$ should start at 0 and go to infinity.

Update:

All the solutions you guys provided are great! I wasn't aware there were so many of them. I should have mentioned that I prefer a solution which doesn't use recursion; imaginary numbers; matrices; functions with if statements; or functions such as ceil, floor, or mod. I'm looking for something using basic algebra: addition/subtraction, multiplication/division, exponents, etc. However, I will accept anything since I didn't  include this clause originally.

This is what I came up with:

$$a_n=\frac{\left(-1\right)^n+1}{2}\cdot \left(-1\right)^{\left(\frac{n}{2}-\frac{\left(-1\right)^{n+1}+1}{4}\right)}$$

Is there a less complicated way (i.e. fewer terms) to get this same sequence?

Here is a solution using only 5 bitwise or arithmetic operations:

2 - (((x + 1) | (x + x)) & 3)


Reference request for $\sum_{i=1}^\infty\left\lfloor\frac{n}{p^i}\right\rfloor$ The highest power of a prime $p$ that divides $n!$ is give by $p^{\left(\sum_{i=1}^\infty\left\lfloor\frac{n}{p^i}\right\rfloor\right)}$

I am fully aware of how to prove this but I don't have space to prove it in my dissertation. Therefore I need to reference it but I can't find it anywhere. I thought it would be easy to find as it's such a well known fact. Can anyone help?

The numerical example below could be an acceptable proof without words.

$$13!=1\cdot2\cdot\underline3\cdot4\cdot5\cdot\underline6\cdot7\cdot8\cdot\underline9\cdot10\cdot11\cdot\underline{12}\cdot13\to3\cdot6\cdot9\cdot12=3^4\cdot 4!$$


Newton's method for multivariable function roots I know how to do Newton's method to find roots for a single variable function but then I got this problem and I am unsure of how to find the roots for multivariate functions using Newton's method:

Say you put this on an spreadsheet:

create a set (3 Rows 1 Column) of guesses for x, y, z; use 1 for each as specified in prob statement.

1

1

1

Create a 3x1 set of function values each evaluated with the current guesses for x, y, and z

f(x,y,z)

g(x,y,z)

h(x,y,z)

Create the 3x3 Jacobian Matrix.  1st row's three terms are the partial derivatives of the 1st equation with respect to x then y and then z.  Etc for g(x,y,z) and h(x,y,z) for rows 2 and 3.  These 9 terms are all evaluated using the current guessed values.
Use the MInverse function in Excel to find the inverse of the 3x3 Jacobian (a 3x3).  Call result J^-1.  Note: when entering a matrix function, select the 3x3, enter equation into active cell, hit Shift-Enter-Return to create the matrix results.  The equations will appear in braces; {.. }'s in the formula bar.
Use the MMult function to multiply the J^1 3x3 matrix times the 3x1 matrix array of function values created in #2.
The new values of x,y,z are the matrix of the x,y,z values created in #1 minus the result from #5.  You will not need a matrix function since each value may be subtracted piece wise: x2 = x1 - first term from #5, etc.
Repeat process starting at #1 but use new values for x,y,z calculated in #6 until convergence is satisfactory.

Note: non-linear systems can have multiple roots so the set you find depends on your initial guesses.

Saying it with mathematical succinctness, where X is an array of variables and f(X) an array of function values, repeating

X(i+1)=X(i)-J^-1 * f(X(i))

where J^-1 is the Jacobian of f(X(i))

will converge on X such that f(X) --> 0.


Does the inequality $0\leq a\leq b$ in a C*-algebra imply $\|a\|\leq\|b\|$? In relation to this question of mine: C* algebra inequalities

I am wondering if it is true that if $0\leq a \leq b$ in a C* algebra, does one have $||a||\leq||b||$?  If you need the C* algebra to be unital you can assume that since I could use the C* algebra unitization.  If it is not true, please state any reasonable conditions under which it is true because I see things like this used all the time, and it is possible that it merely applies in those special cases.

Yes it is true in general.  As you mention, the question isn't affected by going to the unitization, so assume that the algebra is unital.

Note that $b\leq \|b\|1$, because $\sigma(b)\subseteq[0,\|b\|]$, so that $\sigma(\|b\|1-b)\subseteq[0,\|b\|]$.  Thus $a\leq \|b\|1$, and since $\|a\|$ is in the spectrum of $a$, this implies that $\|a\|\leq \|b\|$.

You could use Gelfand theory here, because $a$ and $b$ both commute with $\|b\|1$, so you could consider $C^*(b,1)$ then $C^*(a,1)$.  However, all that is used above are the facts that $x\geq 0$ if and only if $x=x^*$ and $\sigma(x)\subseteq [0,\infty)$, and if $x\geq 0$, then the spectral radius of $x$ is $\|x\|$ (true for all normal elements, and quickly derived from the spectral radius formula and the C*-identity).  Regardless, the "trick" is to use $\|b\|1$ as an intermediary between the possibly noncommuting $a$ and $b$.

When $a$ and $b$ commute, it is true. You only need to look at the commutative C*-subalgebra generated by $a, b, 1$. But for the general case, I don't remember any thing right now.


Find the highest power of $4$ in $82! + 83!$ I'am only getting $4^{13}$ as answer, but the correct answer is $40$.
What am I missing?

We first find the maximum power of $2$ in $82!+83!=82!(1+83)=82!(84)$, since $83$ is odd we want the largest power of $2$ in $84!$.

Using the Legendre-Polignac formula this is:

$\sum\limits_{n=1}^{\infty}\lfloor\frac{84}{2^n}\rfloor=\lfloor\frac{84}{2}\rfloor+\lfloor\frac{84}{4}\rfloor+\lfloor\frac{84}{8}\rfloor+\lfloor\frac{84}{16}\rfloor+\lfloor\frac{84}{32}\rfloor+\lfloor\frac{84}{64}\rfloor=42+21+10+5+2+1=81$.

So the maximim power of $4$ that divides the number is $4^{40}$


When does pointwise convergence imply uniform convergence? On an exam question (Question 21H), it is claimed that if $K$ is compact and $f_n : K \to \mathbb{R}$ are continuous functions increasing pointwise to a continuous function $f : K \to \mathbb{R}$, then $f_n$ converges to $f$ uniformly. I have tried proving this claim for the better part of an hour but I keep coming short. I suspect a hypothesis on equicontinuity has been omitted — partly because the first half of the question is about the Arzelà–Ascoli theorem — but I don't have access to the errata for the exam so I can't be sure.

Here is my attempted proof: let $g_n = f - f_n$, so that $(g_n)$ is a sequence of continuous functions decreasing pointwise to $0$. Clearly, $0 \le \cdots \le \| g_n \| \le \| g_{n-1} \| \le \cdots \le \| g_1 \|$, so we must have $\| g_n \| \longrightarrow L$ for some constant $L$. $K$ is compact, so for each $g_n$, there is an $x_n \in K$ such that $g_n(x_n) = \| g_n \|$, and there is a convergent subsequence with $x_{n_k} \longrightarrow x$ for some $x \in K$. By hypothesis, $g_n(x) \longrightarrow 0$, and by construction $g_{n_k} (x_{n_k}) \longrightarrow L$. I'd like to conclude that $L = 0$, but to do this I would need to know that the two sequences have the same limit. This is true if, say, $\{ g_n \}$ is an equicontinuous family, but this isn't one of the hypotheses, so I'm stuck.

The result is true as stated and is called Dini's Theorem.  A proof can be found in Chapter III, Section 1.4 of these notes or indeed in this wikipedia article.  (My notes are for a sophomore-junior level undergraduate class, so it is stated in the case in which $K$ is a closed interval.  But it is clear that the argument works for any compact topological space.)

(For some reason this is one of those named theorems that tends to be assigned as an exercise or come up on exams...)

It would appear that there is a slight error in your argument – there exist compact topological spaces that are not sequentially compact, such as $[0,1]^{[0,1]}$ (to become more familiar with such pathological spaces, I recommend the book Counterexamples in Topology). Here is a corrected (and much shortened) argument:

Let $g_n=f-f_n$, and note that $\max g_n$ is decreasing, and converges to $0$ iff $g_n\to0$ uniformly. If $\max g_n\ge M>0$ for all $n$, then $\{g_n\ge M\}$ is a decreasing family of closed nonempty subsets of $K$, so by the finite intersection property, $\bigcap_n\{g_n\ge M\}$ is nonempty, which contradicts pointwise convergence.


Infimum over a class of holomorphic functions on the unit disk I am hoping the following is true but don't yet know.  Thoughts and ideas are appreciated.

  Let $k \ge 2$ be an integer. Let $0<\epsilon <1$. Then there is a holomorphic function $g: \mathbb{D} \to \mathbb{C}$ s.t. $$ \inf_{z \in \mathbb{D}} |1+z+z^kg(z)| > 1-\epsilon$$

Note: When $k=2$ the conclusion can be seen to follow using the Riemann Mapping Theorem.

How about this?

By the maximum/minimum principle, we know that $$\inf_{z \in \mathbb{D}} \left| 1+ z + z^k g(z) \right| = \min_{\left| z \right| =1} \left| 1 + z + z^k g(z) \right|.$$ Continuing this, it follows that \begin{eqnarray*}
\min_{\left| z \right| =1} \left| 1 + z + z^k g(z) \right| & > & \min_{\left| z \right| =1} \left| z + z^k g(z) \right| \\
&=& \min_{\left| z \right| =1} \left| z \right| \cdot \left| 1 + z^{k-1}g(z) \right| \\
&=& \min_{\left| z \right| =1} \left| 1 + z^{k-1}g(z) \right| \\
& > & \min_{\left| z \right| =1} \left( \left| 1 \right| - \left| z^{k-1}g(z) \right| \right) \\
&=& 1 - \min_{\left| z \right| =1} \left| z^{k-1}g(z) \right| \\
&=& 1 - \min_{\left| z \right| =1} \left| g(z) \right|,
\end{eqnarray*} choosing $g$ such that $\min_{\left| z \right| =1} \left| g(z) \right| = \varepsilon$ yields the result.


Natural Parameter Space of an Exponential Family is a convex set (Proof)? A natural parameter family is defined as follows
$$p(x|\eta) = h(x) \exp(\eta T(x) + A(\eta))$$
where
T: sufficient statistics
A: log partition function.

We want to prove that the natural parameter space $\mathcal{N}$ given by $$\mathcal{N} = \left\{ \eta: \int(\exp(A(\eta))) < \infty \right\}$$
is convex.

The proof rests on holder inequality and is given here.
I am attaching a picture for a quick reference
I have looked at the definition of holder inequality.
I am not really sure how the $1/\lambda$ and $1/1-\lambda$ are written in the denominator in eq 8.35 when applying the holder inequality in the proof given.

Also in Eq 8.36 how is the $e^{\lambda n^T T(x)}$ is discared for the integral.

Please help in explaining these things?

That looks like a mistake in print. Ideally, the $1/\lambda$ should be in the power for $\exp(\lambda\eta T(x))^{1/\lambda}$ where $p=1/\lambda$ in the Holder's inequality.

Also, it's not $\eta^T$ ($\eta$ raised to the power T), but $\eta^\mathrm{'}T$ which is just the multiplication of 2 matrices.


Is it possible to draw $85 ^{\circ} $ and $110 ^{\circ}$ angles by compass and straightedge construction? Is it possible to draw $85 ^{\circ} $ and $110 ^{\circ}$ angles by compass and straightedge construction?If yes then how?

I thought about it and the only thing coming to mind other than a protractor is to use the concept of trigonometry functions but that still needs calculator,and the other ides coming to my mind involves trisecting an angle which is impossible,any other feasible ideas?

We cannot construct these arcs by straightedge and compass.  For if we can draw an $85^\circ$ arc, we can draw a $5^\circ$ arc by subtraction from an easily constructible $90^\circ$ arc, and then construct a $20^\circ$ angle by repeated duplication, again easily done by straightedge and compass.  And if we can draw a $110^\circ$ arc, we can draw a $20^\circ$ angle by subtracting a $90^\circ$ arc.

However, it was proved by Wantzel, almost two centuries ago, that the $20^\circ$ angle cannot be drawn with straightedge and compass.  Equivalently, we cannot trisect the $60^\circ$ angle with straightedge and compass. For details, please see the Wikipedia article on Angle Trisection.

Comment: If we allow other tools than the classical Euclidean straightedge and compass, the answer changes. There are $3$-legged compass analogues (linkages) that can do trisection.  We can also trisect angles by Origami (the Japanese art of paper-folding).  From antiquity on, various correct trisection methods have been described. As they must, they all go beyond straightedge and compass.

Use the ruler and compass to draw two lines at 90 degrees. Then measure up the two lines the right lengths to form the two sides of a right triangle with an 85 degree angle on one corner and a 5 degree angle on the other. Joint the two measured lengths which gives you a line at 85 degrees and then draw the arc from the corner of the right triangle with the 85 degree angle.


Why is it considered that $(\mathrm d x)^2=0$? Why is it okay to consider that $(\mathrm d x)^n=0$ for any n greater than $1$?
I can understand that $\mathrm d x$ is infinitesimally small ( but greater than $0$ ) and hence its square or cube should be approximately equal to $0$ not exactly $0$ .

But if this is so then how can we expect the results obtained from calculus to be exact and not just approximate ( like the slope or area under a curve )?

I have also noticed some anomalies, like $\sqrt{ (\mathrm d x)^2 + (\mathrm d y)^2 }$ is $0$ but $\mathrm d x\sqrt{1+ (\mathrm d y/\mathrm d x)^2 }$ is not $0$ when these two things are apparently the same . Moreover we can  claim that

$$(\mathrm d x)^2=(\mathrm d x)^3=(\mathrm d x)^4 = \cdots = 0$$

which is quite hard to believe.

Can you help me figure out the logic behind these things ?

It is not the square of $\mathrm{d}x$ that is $0$. It is $\mathrm{d}x\land\mathrm{d}x$ that is zero.

This comes into play in differential form and in changes of variables. Suppose that $u=x+y$ and $v=x-y$. Then
$$
\begin{align}
\iint f\,\mathrm{d}u\,\mathrm{d}v
&=\iint f\,\mathrm{d}(x+y)\,\mathrm{d}(x-y)\\
&=\iint f\,\mathrm{d}x\,\mathrm{d}x+\iint f\,\mathrm{d}x\,\mathrm{d}y-\iint f\,\mathrm{d}y\,\mathrm{d}x-\iint f\,\mathrm{d}y\,\mathrm{d}y\\
&=\iint f\,\mathrm{d}x\,\mathrm{d}y-\iint f\,\mathrm{d}y\,\mathrm{d}x\tag{1}
\end{align}
$$
Why do $\iint f\,\mathrm{d}x\,\mathrm{d}x=\iint f\,\mathrm{d}y\,\mathrm{d}y=0$? Well, inside the outer integral, $x$ is supposed to be held constant, so the inner $\mathrm{d}x$ will vanish. The same goes for the double $y$ integral.

Another consequence of this follows from
$$
\begin{align}
0
&=\iint f\,\mathrm{d}u\,\mathrm{d}u\\
&=\iint f\,\mathrm{d}(x+y)\,\mathrm{d}(x+y)\\
&=\iint f\,\mathrm{d}x\,\mathrm{d}x+\iint f\,\mathrm{d}x\,\mathrm{d}y+\iint f\,\mathrm{d}y\,\mathrm{d}x+\iint f\,\mathrm{d}y\,\mathrm{d}y\\
&=\iint f\,\mathrm{d}x\,\mathrm{d}y+\iint f\,\mathrm{d}y\,\mathrm{d}x\tag{2}
\end{align}
$$
That is, $\mathrm{d}y\land\mathrm{d}x=-\mathrm{d}x\land\mathrm{d}y$. Thus, the integral in $(1)$ is equal to
$$
2\iint f\,\mathrm{d}x\,\mathrm{d}y\tag{3}
$$

Note

It is not the case that $\sqrt{\mathrm{d}x^2+\mathrm{d}y^2}=0$. It is the same as $\mathrm{d}x\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}$ in whatever context they both make sense.

A little historical perspective: When mathematicians back in the 1500's - 1700's were solving problems using 'infinitesimals' - the backdrop of the invention of Calculus - their arguments were exactly what you are complaining about.  And people then complained about it too:

If 'h' isn't zero, you have no right to just drop the term as if it were; if 'h' is zero, you have no right to divide by it.  And don't give me that 'infinitesimal' nonsense, it has to be zero or nonzero.

To which the general reply was something on the order of:

Well, it works, so: get out of my face.

And the amazing thing was, it really did work.  But that was mostly because the early practitioners were basically geniuses - Newton, Leibniz, Bernoulli, Cauchy, Euler, etc.  It was only about a hundred years later, when people started trying to teach Calculus to us mere mortals, that the old 'hand-waving' arguments wouldn't do any more.  That's when you get Bolzano, Weirstrauss, etc., creating the notion of 'limit' and epsilon-delta arguments to make it all logically robust, partly so their students would 'buy' their arguments in class.  [that's not the only reason for the clean-up, but it was an important one]


Stability of homogeneous linear differential equation with variable coefficients I would like to know if a homogeneous linear differential equation, with variable coefficients which are periodic, is stable.

So the differential equation can be written as,

$$
\dot{y}(t)=A(t)y(t), \tag{1}
$$

$$
A(t+T)=A(t). \tag{2}
$$

I would suspect that the solution could be of the following form, similar to a linear time invariant system,

$$
y(t)=e^{B(t)\,t}y(0). \tag{3}
$$

This solution is stable if the eigenvalues of $e^{B(T)\,T}$ lie inside the unit disk.

Here $B(t)$ should satisfy the following differential equation,

$$
B(t)+t\dot{B}(t)=A(t), \tag{4}
$$

with the constraint that $B(t)\,t$ at $t=0$ is the zeros matrix.

In my case the matrix $A(t)$ is,

$$
A(t) = \begin{bmatrix}
0 & 1\\
a+\sin(b\,t) & -c
\end{bmatrix}. \tag{5}
$$

Solving for $B(t)$ yields,

$$
B(t)\,t=\begin{bmatrix}
0 & t\\
a\,t+\frac{1-\cos(b\,t)}{b} & -c\,t
\end{bmatrix}. \tag{6}
$$

However when I compare this solution to a numerical solution, they look very different. So where did I go wrong and how can I find the correct solution?

Another example, where the following $A(t)$ matrix is chosen,

$$
A(t) =
\begin{bmatrix}
0 & 1\\
\left(\sin(t) - 1\right)\left(1 - \sin(t) - 2t\cos(t)\right) & 2\left(\sin(t) - 1 + t\cos(t)\right)
\end{bmatrix}, \tag{7}
$$

such that the $B(t)$ matrix conveniently becomes,

$$
B(t) =
\begin{bmatrix}
0 & 1\\
-\left(\sin(t) - 1\right)^2 & 2\left(\sin(t) - 1\right)
\end{bmatrix}, \tag{8}
$$

which has the repeated eigenvalues $\lambda=\sin(t)-1$. The matrix exponential therefore becomes,

$$
e^{B(t)t} = e^{t\left(\sin(t) - 1\right)}
\begin{bmatrix}
1 + t - t\sin(t) & t\\
-t\left(\sin(t) - 1\right)^2 & 1 + t\sin(t) - t
\end{bmatrix}. \tag{9}
$$

However the original ODE is a rewritten form of a second order differential equation, so the second row of this exponential should be the derivative of the first row, however that is not the case. Namely taking the derivative of the first row gives,

$$
\frac{d}{dt}\left(e^{t\left(\sin(t) - 1\right)} \begin{bmatrix}1 + t - t\sin(t) & t\end{bmatrix}\right) = e^{t\left(\sin(t) - 1\right)}
\begin{bmatrix}-t\left(\sin(t) - 1\right)\left(\sin(t) - 1 + t\cos(t)\right) & 1+t\sin(t)-t+t^2\cos(t)\end{bmatrix}. \tag{10}
$$

So I suspect that in general the following is not true,

$$
\frac{d}{dt}e^{F(t)} = \left(\frac{d}{dt}F(t)\right)e^{F(t)}, \tag{11}
$$

where $F(t)$ is some square matrix as a function of time, $t$.

Let $\lambda_{max}(t)$ is the maximum eigenvalue of [$A^T(t)+A(t)$].

If there is a finite constant $\gamma$ such that
 $$ \int_\tau^t\lambda_{max}(\sigma)d\sigma\leq\gamma$$
$\forall t\geq\tau,$  then the system of variable coefficients ODE is uniformly stable.

The detailed proof of this claim can be found in corollary 8.3 p.p. 133 of linear system theory 2ed edition by Wilson J. Rugh.

The exponential of a matrix can be defined as

$$
e^{F(t)} = \sum_{n=0}^\infty \frac{\left(F(t)\right)^n}{n!} = I + F(t) + \frac{1}{2}\left(F(t)\right)^2 + \frac{1}{6}\left(F(t)\right)^3 + \cdots.
$$

When taking the derivative of this infinite sum you get,

$$
\frac{d}{dt}e^{F(t)} = \dot{F}(t) + \frac{1}{2}\left(\dot{F}(t)F(t)+F(t)\dot{F}(t)\right) + \frac{1}{6}\left(\dot{F}(t)F(t)F(t)+F(t)\dot{F}(t)F(t)+F(t)F(t)\dot{F}(t)\right) + \cdots.
$$

This is only equal to $\dot{F}(t)e^{F(t)}$ if $\dot{F}(t)F(t)=F(t)\dot{F}(t)$, however in general these matrices will not commute.


Intuitively, why does $\int_{-\infty}^{\infty}\sin(x)dx$ diverge? According to Wolfram Alpha, $\int_{-\infty}^{\infty}\sin(x)dx$ does not converge.

This makes no sense to me, intuitively, which I'll justify with a plot:

As we see, the positive and negative areas 'cancel out', so, for any $\alpha \in \mathbb{R}$, $\int_{-\alpha}^{\alpha}\sin(x)dx=0$ (I'm just thinking geometrically- in no way is this supposed to be a rigourous justification).

So, why is is that $\int_{-\infty}^{\infty}\sin(x)dx$ diverges?

A natural and intuitive reason, more than a rigourous one, would be best.

Thanks

The integral is
$$
\lim_{a \to \infty}\lim_{b \to \infty}\int_{-a}^b \sin x \, d x
$$
If you keep one fixed and vary the other limit of integration, there is no limit, the integral fluctuates between $-1$ and $1$. The value doesn't have to blow up for there not be a limit.

Since we know that $\sin x$ is a odd function. And we also know from the property of definite integral that if a function is odd then
limit from (-a) tends to (+a) integral f(x) dx = 0
and we know that sin x is a odd function
therefore
limit from (-infinite ) tends to (+infinite )
Integral sin x dx = 0


Outer Hausdorff measure/Infinite Let $\mathcal{H^s}$ be the s-dimensional outer Hausdorff measure.

For all $t \in (0,n)$ there is none countable cover $\lbrace F_i \rbrace^\infty_{i=1}$

of $\mathbb{R^n}=\bigcup_{i=1}^{\infty} F_{i}$ with $\mathcal{H^s}(F_i)<\infty$

I used the contradiction::

If it would be possible then

$n=$dim$_H(\mathbb{R^n})=$sup dim$_H(F_i)\leq s<n$

So there can't be such a cover.

Can it be proved like this or is there something to improve?

That looks fine, provided that you can prove $$\dim_H \left( \bigcup_i F_i \right) = \sup_i \dim_H(F_i).$$ It shouldn't be a difficult proof.


another follow up question: modeling with exponential distributions This a follow up question to the previous two:

modeling with exponential distributions
a follow up question about modeling with exponential distributions

I'm trying to do (c).

Denote the service time for John, Paul and Naomi as $X,Y,Z$ respectively and let $W=X-Y$. Let $T$ be the time Naomi spends in the post office. Then

$$
\begin{align*}
E(T)&=E(E(T|W))=E(T|W>0)P(W>0)+E(T|W<0)P(W<0)\\
&=E(Z+Y|W>0)P(W>0)+E(Z+X|W<0)P(W<0)\\
&=\left(\frac{1}{\lambda_2}+\frac{1}{\lambda_1+\lambda_2}\right)\frac{\lambda_2}{\lambda_1+\lambda_2}
+\left(\frac{1}{\lambda_1}+\frac{1}{\lambda_1+\lambda_2}\right)\frac{\lambda_1}{\lambda_1+\lambda_2}\\
&=\frac{3}{\lambda_1+\lambda_2}<
\max\left\{1/\lambda_1,1/\lambda_2\right\}
\end{align*}
$$

But I don't see how to get the answer $c=2+\sqrt{3}$.

(Thanks to Did's comment, one should not consider $E(Z)$ as the time Naomi spends in the offices.)

Rewrite the symbols $\lambda_1,\lambda_2$ as $a$ and $b$ respectively. The key point is that we are comparing Naomi's average waiting time to the average waiting time of John and Paul averaged together.

If one changes $\max\{1/a,1/b\}$ in OP to
$$
\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)
$$
then one would get
$$
\frac{3}{a+b}<\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)
$$
which would gives the correct answer $c=2+\sqrt{3}$.

Let N, J and P be the service times of Noami, John and Paul.  You have to find the expected time of event N< P-J

The probability distribution of N = P - J with parameter, $\lambda1$ and $\lambda2$ is a Laplace distribution  with pdf

$f_{N}(n) = \dfrac{e^{\frac{n}{\lambda_2}}}{\lambda_1 + \lambda_2}$ for n<0

$= \dfrac{e^{-\frac{n}{\lambda_1}}}{\lambda_1 + \lambda_2}$ for n>0

Find the expected value of this distribution and using the convention of the problem

$E(N) = \frac{(2\lambda_1 \lambda_2)}{\lambda_1+\lambda_2} $

Let $\lambda_2$>$\lambda_1$.. In other words, max($\lambda_1$,$\lambda_2$) = $\lambda_2$ and min($\lambda_1$,$\lambda_2$) = $\lambda_1$

We also know that $E(N)\lt max(E(P),E(J))$

Using the convention of the problem, where meantime is $\frac{1}{\lambda}$

This implies
$\frac{2\lambda_1\lambda_2}{\lambda_1+\lambda_2}<\frac{1}{\lambda_2}$

$\lambda_2 \gt 2\lambda_1$

In other words,
max($\lambda_1$,$\lambda_2$) > 2.min($\lambda_1$,$\lambda_2$)

Goes to prove that c = 2

I don't get the number $2+\sqrt{3}$


Finding out the area of a triangle if the coordinates of the three vertices are given What is the simplest way to find out the area of a triangle if the coordinates of the three vertices are given in $x$-$y$ plane?

One approach is to find the length of each side from the coordinates given and then apply Heron's formula. Is this the best way possible?

Is it possible to compare the area of triangles with their coordinates provided without actually calculating side lengths?

What you are looking for is called the shoelace formula:

\begin{align*}
\text{Area}
&= \frac12 \big| (x_A - x_C) (y_B - y_A) - (x_A - x_B) (y_C - y_A) \big|\\
&= \frac12 \big| x_A y_B + x_B y_C + x_C y_A - x_A y_C  - x_C y_B - x_B y_A \big|\\
&= \frac12 \Big|\det \begin{bmatrix}
x_A & x_B & x_C \\
y_A & y_B & y_C \\
1 & 1 & 1
\end{bmatrix}\Big|
\end{align*}

The last version tells you how to generalize the formula to higher dimensions.

PS. Another generalization of the formula is obtained by noting that it follows from a discrete version of the Green's theorem:

$$ \text{Area} = \iint_{\text{domain}}dx\,dy = \frac12\oint_{\text{boundary}}x\,dy - y\,dx $$

Thus the signed (oriented) area of a polygon with $n$ vertexes $(x_i,y_i)$ is

$$ \frac12\sum_{i=0}^{n-1} x_i y_{i+1} - x_{i+1} y_i $$

where indexes are added modulo $n$.

The area $A$ of the triangle two of whose vertices lie on the axes, with coordinates $(a, 0)$, $(0, b)$, and a third vertex $(c, d)$ is obtained from previous formula by a mere horizontal axis shift of -a units as
                 $$A = \frac{|-ad + b(a - c)|}{2}$$


How would you explain why "e" is important? (And when it applies?) Possible Duplicate:
  Intuitive Understanding of the constant “e”

Let's say you want to explain this to your teenage son.  I understand the technical definition of $e$

$$ e=\lim_{n\to\infty}\left( 1 + \frac1n \right)^n $$

But, I don't want to get lost in technical babble.  While specific examples are welcome, I also want to understand the big picture.   I want to first know the general significance of why and when e is used.  Anyone got a link?    Is there a typical "pattern" that it models?  In what general sorts of situations does "e" arise?

I only know e from the classic continuous compounding example.  But, why does it appear in other applications of growth, science, etc?  Are all these examples just variations on this same limit that defines e?  This increasingly smaller interval when you apply an infinitely smaller percentage growth (1 + 1/n) but infinitely many times (to the power of n)

There is also a difference between an interest in understanding

the mathematical constant "$e$", which is an irrational, transcendental number, but real number, just as is $\pi$, but a number, nonetheless, which happens to be the value of your limit, and happens to be the value of the infinite sum $$e = \displaystyle \sum_{n=0}^\infty \frac{1}{n!}$$And more:
$$e^x = \sum_{n=0}^\infty{x^n \over n!} = 1 + x + {x^2\over2!}+{x^3\over3!}+{x^4\over4!}+\cdots$$
vs.
exponential functions (base $e$: (e.g., $f(x) = e^x$, or $e^{i\theta} = \cos \theta + i\sin \theta$), functions to which you seem to refer when you speak of its use in representing exponential growth, and other applications which crop up everywhere, so it seems. Certainly, functions involving $e$ tell us something about $e$, but they tell us so much more than that.

That is, the importance of $e$ itself isn't so much for its significance as a particular real number, but its significance as a base for exponential functions, and in terms of the ways that functions involving $e$ as a base appear in surprising ways, are powerful, and have applications in many domains.

So it's hard to know exactly what you are interested in: the number $e$, or the many functions involving $e$.

$e$ appears in a lot of places in calculus, but before calculus, it has little significance.


Fourier Transform of $ e^ {t/2} \operatorname{sech}(t) $ My question is what is the Fourier transform of the function

$f(t)=\exp(t/2)\text{sech}(t)$

Is there any closed form available? One possibility is that

$f(t)^2=\frac{1}{2} \text{sech}(t) + \frac{1}{4} (\dfrac{d}{dt}\operatorname{sech}t)$

and then the Fourier transform at frequency f is

$F(f(t)^2)=\frac{1}{2} F(\operatorname{sech}t)(1 + j\pi f )$

and $F(\operatorname{sech}t)=\pi \operatorname{sech}(\pi^2 f)$. But I am interested in $F(f(t))$ itself.

The best way to attack the problem is to use the definition of the hyperbolic secant, together with the Geometric Series.

$$\text{sech}(t) = \frac{2}{e^t + e^{-t}}$$

We then have:

$$\mathcal{F}\left(e^{t/2}\text{sech}(t)\right)(k) = \frac{2}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{t/2} e^{-i kt }\frac{\text{d}t}{e^t + e^{-t}}$$

Now we have to split the integral into the two ranges $(-\infty, 0]\cup[0, +\infty)$ in order to use geometric series.

Geometric series will be used for the denominator as follows:

$$\frac{1}{e^t + e^{-t}} = \frac{1}{e^t(1 + e^{-2t})} = \frac{e^{-t}}{1 + e^{-2t}} = e^{-t}\sum_{j = 0}^{+\infty}(-1)^j e^{-2jt} ~~~~~~~ \text{for} ~~ [0, +\infty)$$

$$\frac{1}{e^t + e^{-t}} = \frac{1}{e^{-t}(1 + e^{2t})} = \frac{e^{t}}{1 + e^{2t}} = e^{t}\sum_{j = 0}^{+\infty}(-1)^j e^{2jt} ~~~~~~~ \text{for} ~~ (-\infty, 0]$$

Hence the two integrals we need to calculate are:

I

$$\frac{2}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j\int_0^{+\infty} e^{t/2} e^{-ikt} e^{-t}e^{-2jt}\ \text{d}t = \frac{2}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j \frac{2}{1 + 4j + 2ik}$$

Provided that $4 \Re(j)+1>2 \Im(k)$ which should be the case in question.

The sum can be obtained by some knowledge of Special Functions, and it sums:

$$\frac{4}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j \frac{1}{1 + 4j + 2ik} = \frac{\Phi \left(-1,1,\frac{1}{4} (2 i k+1)\right)}{\sqrt{2 \pi }}$$

The Hurwitz function that obtained can be expanded into "more elementary" functions, like the Polygamma Function:

$$\frac{\Phi \left(-1,1,\frac{1}{4} (2 i k+1)\right)}{\sqrt{2 \pi }} \equiv \frac{\psi ^{(0)}\left(\frac{i k}{4}+\frac{5}{8}\right)-\psi ^{(0)}\left(\frac{i k}{4}+\frac{1}{8}\right)}{2 \sqrt{2 \pi }}$$

Here $\Phi$ denotes the Hurwitz function and $\Psi$ denotes the Polygamma.

II

Part two is similar to the first one, except for the calculations. In any case, similarly, we get:

$$\frac{2}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j\int_0^{+\infty} e^{t/2} e^{-ikt} e^{t}e^{2jt}\ \text{d}t = \frac{2}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j \frac{2}{4 j-2 i k+3}$$

Provided that $\Im(k)+2 \Re(j)>-\frac{3}{2}$ which again holds.

The sum can be evaluated in the same manner:

$$\frac{4}{\sqrt{2\pi}}\sum_{j = 0}^{+\infty}(-1)^j \frac{1}{4 j-2 i k+3} = \frac{\Phi \left(-1,1,\frac{1}{4} (3-2 i k)\right)}{\sqrt{2 \pi }}$$

And again:

$$\frac{\Phi \left(-1,1,\frac{1}{4} (3-2 i k)\right)}{\sqrt{2 \pi }} \equiv \frac{\psi ^{(0)}\left(\frac{7}{8}-\frac{i k}{4}\right)-\psi ^{(0)}\left(\frac{3}{8}-\frac{i k}{4}\right)}{2 \sqrt{2 \pi }}$$

The final result is thence the sum of the two previous results, which means:

Final Result

$$\mathcal{F}\left(e^{t/2}\text{sech}(t)\right)(k) = \frac{\psi ^{(0)}\left(\frac{i k}{4}+\frac{5}{8}\right)-\psi ^{(0)}\left(\frac{i k}{4}+\frac{1}{8}\right)}{2 \sqrt{2 \pi }} + \frac{\psi ^{(0)}\left(\frac{7}{8}-\frac{i k}{4}\right)-\psi ^{(0)}\left(\frac{3}{8}-\frac{i k}{4}\right)}{2 \sqrt{2 \pi }} = $$
$$ = \frac{-\psi ^{(0)}\left(\frac{i k}{4}+\frac{1}{8}\right)+\psi ^{(0)}\left(\frac{i k}{4}+\frac{5}{8}\right)-\psi ^{(0)}\left(\frac{3}{8}-\frac{i k}{4}\right)+\psi ^{(0)}\left(\frac{7}{8}-\frac{i k}{4}\right)}{2 \sqrt{2 \pi }}$$

FINAL BEAUTY

Through definitions of the Polygamma, and few algebra, we can transform that pretty ugly expression into something cute and simple. The "true" final result is then:

$$\color{red}{\frac{\pi  \tan \left(\frac{\pi  i k}{4}+\frac{\pi }{8}\right)+\pi  \cot \left(\frac{\pi  i k}{4}+\frac{\pi }{8}\right)}{2 \sqrt{2 \pi }}}$$

Which can be written also as

$$\color{blue}{\frac{\pi  \tan \left(\frac{1}{8} (2 \pi  i k+\pi )\right)+\pi  \cot \left(\frac{1}{8} (2 \pi  i k+\pi )\right)}{2 \sqrt{2 \pi }}}$$

If you already know that
$$\mathcal F[\operatorname{sech} t] = F(p) =
\int_{-\infty}^{\infty} e^{i p t} \operatorname{sech} t \,dt =
\pi \operatorname{sech} \frac {\pi p} 2,$$
then
$$\mathcal F[e ^{t/2} \operatorname{sech} t] =
\int_{-\infty}^{\infty} e^{i (p - i/2) t} \operatorname{sech} t \,dt = \\
F(p - i/2) = \pi \operatorname{sech} \frac {\pi (2p - i)} 4.$$


Find the last Digit of $237^{1002}$? I looked at alot of examples online and alot of videos on how to find the last digit  But the thing with their videos/examples was that the base wasn't a huge number. What I mean by that is you can actually do the calculations in your head. But let's say we are dealing with a $3$ digit base Number... then how would I find the last digit.

Q: $237^{1002}$

EDIT: UNIVERSITY LEVEL QUESTION.

It would be more appreciated if you can help answer in different ways.

Since the Last digit is 7 -->

$7^1 = 7$
$7^2 = 49 = 9$
$7^3 = 343 = 3$
$7^4 = 2401 = 1$

$.......$

$........$
$7^9 = 40353607 = 7$
$7^{10} = 282475249 = 9$

Notice the Pattern of the last digit. $7,9,3,1,7,9,3,1...$The last digit repeats in pattern that is 4 digits long.

Remainder is 1 --> 7
Remainder is 2 --> 9
Remainder is 3 --> 3
Remainder is 0 --> 1

So, $237/4 = 59$ with the remainder of $1$ which refers to $7$. So the last digit has to be $7$.

$$
237^{1002} = (23*10 + 7)^{1002} = \sum_{i=0}^{1001}23^{1002-i}10^{1002-i}\;7^i{1002 \choose i} + 7^{1002} =\\
[\textit{some huge honking multiple of }10] + 7^{1002} = \\
[\textit{some huge honking multiple of } 10] + 49^{501} = \\
[\textit{some huge honking multiple of }10] + (50 - 1)^{501} =\\
[\textit{some huge honking multiple of }10] + [\textit{some other gorfurshlugging multiple of }10] + (-1)^{501} =\\
[\textit{some huge honking multiple of }10] + [\textit{some other gorfurshlugging multiple of }10] - 1 = \\
[\textit{some huge honking multiple of }10] + [\textit{one less than some other gorfurshlugging multiple of }10] + 9
$$

So the last digit is $9$.  Thing is only the last digits matter, and the last digits will cycle between $1, 7, 9, 3, 1, 7, 9, 3$.  So you just need the last digit and the remainder of $1002$ divided by $4$.

=====

Crash course in modular arithmetic:

If you have some integer $N$ and we have two integers $a$ and $b$ so that $a = b \pm kN$ for some integer $k$ we say $a \equiv b \mod N$.  We are basically considering an arithmetic system where we consider numbers by how much more than a multiple of $N$ they are.

Ex:  If $4732895738927 \equiv 8647 \mod 10$ because $4732895738927 = 8647 + 10k$.  Basically if $a \equiv b \mod 10$ then $a$ and $b$ have same last digit as $a = b + 10k$ for some $k$.

Lemma:  if $a \equiv b \mod N$ and $c \equiv d \mod N$ then:

i) $a + c \equiv b+d \mod N$

ii) $ac \equiv bd \mod N$

iii) $a^n \equiv b^n \mod N$.

Pf:  i) $a = b + kN$, $c = d + jN$ so $a+c = b + d + (j+k)N$ so $a+c \equiv b+d \mod N$.

ii) $a = b + kN$, $c = d + jN$ so  $ac = (b+kN)(d+jN) = bd + (dk + bj)N + jkN^2 = bd + (dk + bj + jkN)N$.  so $ac \equiv bd \mod N$.

iii) by induction $a^1 \equiv b^1 \mod N$ and if $a^n \equiv b^n \mod N$ then $a^{n+1} = a^na \equiv b^nb \mod N \equiv b^{n+1} \mod N$.

So we can apply this to your problem:  $237 \equiv 7 \mod 10$ so $237^{1002} \equiv 7^{1002}$.

Notice:  If you consider $0, 1,.....,N -1, N, 1 + N, ......, 2N-1, 2N, 2N + 1....$  there are at most $0,1,.....,N-1$ distinct values that can be equivalent $\mod N$ so for all the $a^k$ there must only a finite number of distinct things for $a^k$ to be equivalent $\mod N$ so there must be some $a^k \equiv a^j \mod N$ where $k \ne j$.

And if $a^k \equiv 1 \mod N$ then $a^{nk} = (a^k)^n \equiv 1^n \mod N \equiv 1 \mod N$.

So for example $7^2 \equiv 49 \equiv 9 \mod 10$

$7^3 = 7^2*7 \equiv \mod 10 \equiv 9*7 \equiv 63 \equiv 3 \mod 10$

$7^4 = 7^3*7 \equiv 3*7 \equiv 21 \equiv 1 \mod 10$.

So $7^{1000} = (7^4)^{250} \equiv 1^250 \equiv 1 \mod 10$.

Putting this all together:

$237^{1002} \equiv 7^{1002} = 7^{1000}*7^2 \equiv (7^4)^{250}*49 \equiv 1^{250}*49 \equiv 1*49 \equiv 9 \mod 10$

So $237^{1002}$ and $9$ have the same last digit; $9$.

====

A theorem that is beyond this crash course is Euler's Thereom.  If $N$ and $a$ have not common factors, and if $\phi(N) = $ the number of numbers $1,2, ....,N$ that have no common factors with $n$... then $a^{\phi(N)} \equiv 1 \mod N$.

So in your problem $\phi(10) = 4$ because $1,3,7, 9$ have no factors in common with $10$ while $2,4,5,6,8,10$ do.  And $7$ and $10$ have no common factors... So $7^4 \equiv 1 \mod 10$.  (And we can test that and $7^4 = 49*49 = 40^2 + 2*9*40 + 9^2 \equiv 81 \equiv 1 \mod 10$.)

So $237^{1002} \equiv 7^{1002} \equiv (7^4)^{250}7^2 \equiv 1^{250}49 \equiv 9 \mod 10$.

You can solve this using Euler's theorem:

$\gcd(237,10)=1$
Therefore $237^{\phi(10)}\equiv1\pmod{10}$
Therefore $237^{\phi(2\cdot5)}\equiv1\pmod{10}$
Therefore $237^{(2-1)\cdot(5-1)}\equiv1\pmod{10}$
Therefore $237^{1\cdot4}\equiv1\pmod{10}$
Therefore $237^{4}\equiv1\pmod{10}$

Therefore
 $237^{1002}\equiv237^{4\cdot250+2}\equiv(\color\red{237^{4}})^{250}\cdot237^{2}\equiv\color\red{1}^{250}\cdot237^{2}\equiv56169\equiv9\pmod{10}$.


How to solve this exponential equation? I'm given this equation and i have to solve for the x.

$$ e^{2x} -(e^5 + e^2)e^x + e^7 = 0      $$

The results should be $x =2$ and $x = 5$.

At first i thought it would be an easy task, substituting $e^x$ with $y$, but I can't get rid of the coefficients. I'm sorry if this question is too simple for this site but I'd be really grateful if you could explain me. Thanks for your kindness.

You can make $e^{2x}-(e^2+e^5)e^x+e^7=(e^x-e^2)(e^x-e^5)=0$. So we have the result.


Show that every mapping $f: X \to Y$ is continuous where $(X,d)$ is a metric space with the discrete metric Let $(X,d)$ be a metric space with the discrete metric and let $(Y,d')$ be an arbitrary metric space. Show that every mapping $f: X\to Y$ is continuous.

I got a bit confused about this, initially, I thought that it would come quite directly from the definition, but I'm not sure how to proceed since I'm not given $f$ to work with. Perhaps there is some other definition for continuity that I'm not familiar? I've used the one that states $\forall \varepsilon >0 \ \exists \delta >0$ s.t. $d(x,y) < \delta \Longrightarrow |f(x)-f(y) | < \varepsilon.$

Hint: What happens if you set $\delta=\frac12$?


How to find center and radius of hand-drawn circle? You are given a set of points {(X1,Y1), (X2,Y2),...}  which represent a hand-drawn circle, so it's not perfect. You are asked to find the center and radius of this circle.

My intuition tells me this involves minimization, and I need to find the coefficients in a non-linear generalized least squares model. Am I not thinking of a simpler or more "proper" approach?

You can use a trick to make this a regular least-squares optimization, in an extension of the method for finding the center of a circle through three points. Draw a perpendicular bisector for every pair of points. If the points are all on a circle, they will all intersect at a common point; since it is an approximate circle, they will slightly miss each other. The squared distance to each line is a quadratic form, so the sum of the squared distances is also a quadratic form, and minimizing this is a least-squares problem. That's the center of the circle.

Given this, you can just find the radius as the mean of the distances of each point to the center.

For each point you are given, we assume that it satisfies the following relation:

$$
( x_i - x_0 )^2 + ( y_i - y_0)^2 = r^2.
$$

This can be written as

$$
\left[
\begin{array}{ccc}
1 & -2x_i & -2y_i
\end{array}
\right]
\left[
\begin{array}{c}
x_0^2 + y_0^2 -r^2 \\
x_0 \\
y_0
\end{array}
\right]
= -x_i^2 - y_i^2.
$$

Thus, using all of the $(x_i,y_i)$ pairs, we can set up an over-determined system $\mathbf{Ax}=\mathbf{b}$. The least squares solution $\bar{\mathbf{x}}$ to this system will give you your answer.


Linear equation using modulo The question is:  If $3X \equiv 2 \pmod 7$, what is $x$?

I used this article to solve, but I can't really get it. It's quite complicated....
I need guidance please

Hint: Use the Euclidean algorithm to find numbers $m$ and $n$ such that $3m+7n=1$. Then $3m\equiv 1\pmod 7$ (why?). The answer is then $X=2m$ (why?).

Basically mod 7 means you are operating on numbers in the ring $\Bbb{Z}_7$. This is the set $\{0, 1, 2, 3, 4, 5, 6\}$, composed of seven numbers. It's called a ring because you go "around in circles", as shown below.

We are given a number times 3 equals two. Basically, if it results in a number that doesn't appear in the set $\Bbb{Z}_7$, for example $9$, then we have $$9 \mod 7 = 9 + 7i = 2 + 7 + 7i = 2 + 7(1 + i) = 2 + 0 = 2.$$ We then say that $9$ is contained in the congruence class of $[2]_7$.

Thus, $2$ would equal $9$ in $\Bbb{Z}_7$. The equation then becomes $$3X = 9 \mod 7.$$ Does this help?


The error is in the step where you replace $g(x)-g(x_0)$ by its local linear approximation, $(x-x_0)g'(x_0)$.  You need to keep terms up to second order:
$$
g(x)-g(x_0)\approx(x-x_0)g'(x_0)+\frac{1}{2}(x-x_0)^2g''(x_0).
$$
In other words, you cannot assume that this second order term vanishes in the limit.

Adding to TheSimpliFire's note, I suspect you have confused yourself by using $x_0$ notation. Your definition of $f''$ treats the real variable as $x_0$, but later you subtly to switch to $x$.

After your definition of $f''$, you should have defined $g$ as $g(x_0) = f(2x_0 - x)$, not $g(x)$, and you won't get the contradiction.

In effect, you have defined a totally different curve $g$, and stated a point of intersection, and then falsely assumed that the local derivatives of the two curves $f$ and $g$ have to align.


Characterization of primes dividing $(x+y)(x+z)(y+z)$ in a [hypothetical] counterexample to Fermat's Last Theorem Assume $p$ is an odd prime, and $x,y,z$ are pairwise relatively prime nonzero integers, such that $x^p+y^p+z^p=0$.

In Ribenboim's Fermat's Last Theorem for Amateurs, he gives a proof (p. 101) that every prime divisor of $(x^p+y^p)/(x+y)$ is congruent to $1\!\!\pmod{\!2p^2}$; by symmetry the same is true also for $(x^p+z^p)/(x+z)$ and $(y^p+z^p)/(y+z)$.

QUESTION: Are there similar results about the size or shape of primes dividing $(x+y)(x+z)(y+z)$?

In particular, I'm interested in the following conjecture.

CONJECTURE: At least one of $x+y$, $x+z$, or $y+z$ must have a prime factor greater than $p$.

In the first case of Fermat's Theorem one can apply Sophie Germain's method to obtain factors. For example, if $2p+1$ is prime then $2p+1$ must be a factor of $(x+y)(y+z)(z+x)$. Similarly, one can use other primes of  the form $np+1$.

Here's an answer:

Assume $x^p+y^p+z^p=0$ for some integers $x,y,z$ and odd prime $p$. Then by Fermat's Last Theorem, we have a logical contradiction. It therefore follows that $1+1=2$, and also that $1+1=3$.

And also that $(x+y)(x+z)(y+z)$ is divisible by $47$. And that all divisors of $(x+y)(x+z)(y+z)$ are congruent to $1$ mod the order of the Monster group.

These are just some examples of  divisibility criteria on $(x+y)(x+z)(y+z)$ that follow from the assumption.


How do I find a partition of an equivalence relation? Say I have the function:
$$x\,R\,y \iff y = 3^k$$ for some $k \in \mathbb Z$ and the set is:

$$A = \{1,1/3,1/27,1/4,3,1/36 , 2,2/9,9/4, 5\}$$

So in this scenario, how do I find the partitions of the set?

Knowing a fairly example should allow me to know how to do it in the future.

An equivalence relation is always partition. In general you want to put it in the form of a set. Here, you can say,

$$A_{r}= \{(x,y) | y = \frac{x}{3^k}\}$$

Where $A_r$ denotes the equivalent partition corresponding to your relation. Also, $k$ is assumed to be an integer.

A partition of a set $A$ is a collection of $a_i \subset A$ such that each  $a_i$ is disjoint from the others and $\bigcup a_i = A$.

An equivalence relation always partitions a set. Given a set $A$ and some relation $R$, the set of equivalence classes $A/R$ is a partition of $A$. Each member of $A/R$ is a subset of $A$ such that each member is $R$ related.

Since your given set is a finite set, perhaps the simplest approach would be to start with an element and determine all the other elements that are $R$ related to it. The set of all these elements is one equivalence class in the partition. Then pick another element from $A$ that is not in this first equivalence class, and form another class. Once all the elements of $A$ have been grouped, you will have formed the partition of $A$ that you are looking for.


Solving the matrix equation and finding the matrices inverse Suppose that $E_1 \begin{bmatrix}12\\35\end{bmatrix} = \begin{bmatrix}48\\35\end{bmatrix}$ Find $E_1$ and $E_1^{-1}$

I set up the equation and saw that $E_1 * A = B$ so I know that:

$E_1 = A^{-1} * B$

where $I_m =$ Identity Matrix

So I attempted to find $A^{-1} = A*I_m = \begin{bmatrix}12&1\\35&0\end{bmatrix}$

After RREF I got: $A^{-1} =\begin{bmatrix}1&1/12\\0&35\end{bmatrix}$

So then I tried $E_1 = \begin{bmatrix}1&1/12\\0&35\end{bmatrix} * \begin{bmatrix}48\\35\end{bmatrix}$ but It is saying that I am incorrect? What am I doing wrong?

Apparently, $E_1$ is a $2\times 2$ matrix. Let, $E = \begin{bmatrix} a&b\\ c&d \end{bmatrix}$, thus we want to solve the linear system:
$$\begin{bmatrix} a&b\\ c&d \end{bmatrix}\cdot \begin{bmatrix}12\\35 \end{bmatrix}=\begin{bmatrix} 48\\35\end{bmatrix}
$$

Thus, we have the system of equations:
$\begin{cases}
12a+35b = 48\\
12c+35d = 35
\end{cases}\implies \begin{cases} b = \frac{48-12a}{35}\\[2ex] d = \frac{35-12c}{35}\end{cases}$

We can choose any values for $a,c$ (as long as $ad-bc\neq 0$ ), otherwise $E^{-1}$ will not be defined. Choosing $a = 1, c = 0\implies b=36/35, d = 1.$

So, we have the matrix $E = \begin{bmatrix} 1&\frac{36}{35}\\ 0&1\end{bmatrix}.$ You can indeed confirm that $\begin{bmatrix} 1&\frac{36}{35}\\ 0&1\end{bmatrix}\cdot \begin{bmatrix} 12\\35 \end{bmatrix} = \begin{bmatrix} 48\\35 \end{bmatrix}.$

Now, since $E$ is $2\times 2$, we can easily find the inverse. If $E = \begin{bmatrix} a&b\\c&d\end{bmatrix}$, then the inverse is given by the formula:
$$E^{-1} = \frac{1}{ad-bc}\cdot \begin{bmatrix} d&-b\\-c&a\end{bmatrix}.$$

there are errors:

even in the case defined $EA = B$ does not imply $E = A^{-1}B$,
for  a matrix A ,

the inverse matrix of $A$ (denoted $A^{-1}$)
exists iff the matrix $A$ is a square matrix and $detA \neg 0$.

you say " so I attempted  to find $A^{-1}=A*I_n$" there is no
reason in the text to look for this type of matrix


Is a curl-free vector field always a gradient? I tried to prove this problem using the Helmholtz decomposition theorem, but it seems the two are entirely contradictory--thus leaving me with empty hands. Does anyone know how to proceed?

Thanks

Moehringer

Read an answer here:      Helmholtz decomposition contradictions PDF http://analysis3.com/Helmholtz-decomposition-contradictions-pdf-e117376.html


Gauss-Seidel linear system solver - cases when it is unstable I have a Gauss-Seidel linear system solver that has always been able to solve many kinds of linear systems. However, it struggles when solving this very simple system:
$$
22.7986\,x_1 + 4981.2903 \,x_2 = 25.48488 \\ 10047.4775\,x_1 -4981.2903\,x_2 = 6.05489
$$

My initial guess is $(0.01,0.01)$ but I have tried changing it to other values and it's still is not converging.

Basically the new guess that the algorithm computes in each iteration consistently becomes larger, and so is the error. I was checking the Burden book to see if there are special cases in which Gauss-Seidel does not converge but there is little on that.

I know that there exists a solution to the above system because I have solved it using Excel's functionality to invert matrices and then use matrix multiplication. Particularly, the solution to the system is $(0.003131967,0.005101786)$.

Any help would be welcome.

The problem can be solved by rewriting the equations in the following way
4981.2903X2+22.7986X1=25.48488

-4981.2903X2+10047.4775X1 = 6.05489

Step 1: Assume X1 = 0 in the first equation and calculate X2 = 0.005116

Step 2: Taking X2 = 0.005116, calculate X1 from the second equation X1 = 0.003136

Step 3: Taking X1 = 0.003136, calculate X2 from the first equation X2 = 0.0051018

Step 4: Taking X2 = 0.0051018, calculate X1 from the second equation X1 = 0.003131

So the final answer is X1 = 0.003131 and x2 = 0.0051018


Unconditional expectation vs conditional expectation in regressions - does it really matter? I refer here to a simple linear regression whose true representation is given by the equation: $y_i=x_i'\beta+u_i$,

where as usual $x_i$ is a $K\times1$ vector of independent explanatory variables, $\beta$ is a  $K\times1$ vector of parameters and $u_i$ the error term of the $i$th observation and by construction of the true representation $u_i$ is considered to be independent of $x_i$.

I have trouble to understand why the conditional expectation $E[u_i\mid x_i]=0$ is considered stronger (or more natural) than the unconditional expectation $E[u_i]=0$. Almost every textbook naturally uses the conditional version instead of the unconditional one.

The reason for my difficulties is the following argument:

Firstly, it is clear to me that by the Law of Total Iteration $E[E[u_i\mid x_i]] = E[u_i]=0$. Hence conditional expectation implies unconditional expectation.

But don't we expect the error term $u_i$ to be independent of $x_i$ by construction? If that is the case, would not $E[u_i]=0$ imply $E[u_i\mid x_i]=0$ ? Or in other words, we if assume the error term to be independent of the explanatory variable, does the distinction of unconditional vs. conditional expectation really matter here?

I think there is a similar argument for the unconditional vs conditional homoskedasticity. Textbooks use the conditional one, i.e. $V[u_i^2\mid x_i] =\sigma^2$, which is preferred to the conditional variance $V[u_i^2]=\sigma^2$. Their difference should not matter if the error term is independent of the explanatory variable, should it?

Given a linear regression, I cannot find an example or perhaps a graphical representation or even a good story how an error term can be unconditionally mean-zero but conditionally biased. If I have a theoretical scatter plot of error term with mean zero, how am I suppose to slice out the $X$ dependent part and say, look here, the error is indeed biased, when by definition the error term exists only because of the regression set up in the first place.

Maybe I have difficulties understanding the conditional expectation fully. If I $X$ is given, or in other words, if I have knowledge of $X$, how can this change my expectation of $u_i$ when both are considered independent? What does this conditional expectation really mean and how does it improve my understanding of the underlying regression and to what means in contrast to the unconditional one?

Sorry if this seems to be confusing and probably a stupid question - I have been pondering about these concepts for a while and cannot find an illustrative answer that I am happy with.

Indeed, if E[u|x]=0 then E[u]=0 because E[u]=E[E[u|x]]. In the other direction, E[u]=0 does not imply E[u|x]=0 as the case x=u shows since E[u|u]=u. But if x and u are independent then E[u|x]=E[u] hence the implication [ if E[u]=0 then E[u|x]=0] becomes true.

Here is another answer to your question. You've correctly noted that the condition E[u|x]=0 --- called "mean independence" of u and x --- is weaker than full-on independence of u and x. Bruce Hansen's notes on Econometrics give a particularly simple counterexample. (The link to the book is here: http://www.ssc.wisc.edu/~bhansen/econometrics/. See pages 17-18.) Suppose u=ex where x and e are independent standard normals, i.e. x~N(0,1) and e~N(0,1). Then conditional on x, u has the distribution N(0,x^2). Therefore E[u|x]=0, but u is obviously not independent of x.

This simple counterexample illustrates how, in practice, the error terms might be DEPENDENT on a regressor variable but still MEAN-independent of it. In particular, the conditional distribution u|x might depend on x even though E[u|x]=0. Your problem may be that you're only imagining uncorrelated NORMAL distributions. If x and y are jointly normal but uncorrelated, then the conditional distribution u|x (where u=y-E[y|x]) doesn't depend on x: it's always the marginal distribution of y. In that special case, u and x are actually independent.

The point is that, in general, running a regression amounts to using E[y|x] to predict y in an arbitrary joint distribution (the normality assumption is used only for inference), and all you need to "assume," in that context, is that u is mean-independent of x. (I put scare quotes around "assume" because it's not really an assumption: if you DEFINE the error term as u:=y-E[y|x], then it's a simple CONSEQUENCE, not an assumption, that E[u|x]=0 [and from this it also follows fairly straightforwardly that u is uncorrelated with any function of x, though of course zero correlation doesn't imply independence]. The real assumption you make in running a basic regression is that E[y|x] is linear.) You don't need to assume the stronger condition that u and x are actually independent: nothing in the theory of regression requires it, so why impose it, if it unreasonably restricts the class of problems you can consider?


How can I prove that $MN$ is parallel to $AC$? Let $ABC$ be a triangle.  The internal angular bisectors of $\angle BAC$, $\angle CBA$, and $\angle ACB$ meet the circumcircle of the triangle $ABC$ at the points $A_1$, $B_1$, and $C_1$, respectively.  Suppose that $B_1C_1$ meets $AB$ at $M$, and $A_1B_1$ meets $BC$ at $N$.  Prove that $MN$ is parallel to $AC$.

So far, I have managed to prove that $M,I,N$ are collinear, where $I$ is the incentre of the circle and $AA_1$ is perpendicular to $B_1C_1$. I have also attempted to prove the result using radical axis / Brianchon's theorem but to no avail. Can anyone help me with this? Any help is greatly appreciated!

Showing MN is parallel to AC can be done by proving that B1C1 is the perpendicular bisector of AI. It is well-known that B1A=B1I=B1C, and C1 satisfy a similar relation. By this property we conclude that B1C1 is the perpendicular bisector of AI. And now that you have proved M,I,N are collinear, it suffices to notice that MI=MA (M is on the perpendicular bisector of AI), and so MIA=MAI=IAC, which implies that MI is parallel to AC. And we're done.

In the standard notation by the law of sines we obtain:
$$\frac{BM}{BB_1}=\frac{\sin\frac{\gamma}{2}}{\sin\frac{\beta+\gamma}{2}}$$ or
$$BM=\frac{BB_1\sin\frac{\gamma}{2}}{\cos\frac{\alpha}{2}}.$$
By the same way:$$BN=\frac{BB_1\sin\frac{\alpha}{2}}{\cos\frac{\gamma}{2}}.$$
Id est, it's enough to prove that
$$\frac{BM}{c}=\frac{BN}{a}$$ or
$$\frac{\sin\frac{\gamma}{2}}{\cos\frac{\alpha}{2}\sin\gamma}=\frac{\sin\frac{\alpha}{2}}{\cos\frac{\gamma}{2}\sin\alpha},$$ which is obvious.


Stirling numbers of the second kind on Multiset Stirling numbers of the second kind $S(n, k)$ count the number of ways to partition a set of $n$ elements into $k$ nonempty subsets. What if there were duplicate elements in the set? That is, the set is a multiset?

There is no known formulation for a general multiset. However, a paper at JIS tackles the case where the element 1 occurs multiple times.

By definition, a set only has distinct elements.  If elements are duplicated, then yes, the set in question would be a multiset.


prove validity of following sequent How to prove validity of following  sequent using rules of conjunction, disjunction, implication, negation etc.
Premises: $ c \wedge n \Rightarrow t$ ,  $h \wedge \sim s$, $h \wedge \sim(s\vee c) \Rightarrow p $
Conclusion: $ n \wedge \sim t \Rightarrow p $
It should be proceeded as follows:

$ 1- c \wedge n \Rightarrow t$     Premise

$2- h \wedge \sim s$               Premise

$3- h \wedge \sim(s\vee c) \Rightarrow p $   Premise

$4- n \wedge \sim t  $     Assumption

$5- n $    Using rule and_elimination1 on line 4

$6- \sim t $    Using rule and_elimination2 on line 4

$7-  \sim (c \wedge n)  $ Using MT rule on lones 6 and 1

$8- h $    Using rule and_elimination1 on line 2

$9-  \sim s $   Using rule and_elimination2 on line 2

$8- p $   By following which rules, we can get this p ?

Note that, up to step 9 in your proof, you have not yet used the 3rd premise :

  $h \land \lnot (s \lor c) \Rightarrow p$,

and we need it to complete the proof.

We will start from step 9.

$10 − \lnot c \lor \lnot n$ by De Morgan from 7 : $\lnot (c \land n)$

$11 - \lnot c$ by disjunctive syllogism from 10 and 5 [Note : disjunctive syllogism is a particular case of $\lor$-elimination]

$12 - \lnot c \land \lnot s$ using $\land$-introduction on 11 and 9

$13 - \lnot (c \lor s)$ again by De Morgan

$14 - h \land (s \lor c)$ using $\land$-introduction on 13 and 8

$15 - p$ using MP on 13 and 3 (the 3rd premise)

$16 - (n \land \lnot t) \Rightarrow p$ using $\Rightarrow$-introduction on 4 and 15, "discharging" the "temporary" assumption 4.

Thus, we conclude with :

    $c∧n⇒t, h∧ \lnot s, h∧ \lnot (s∨c)⇒p \vdash (n \land \lnot t) \Rightarrow p$.

I'll add my own loose answer here, and point out that we don't know exactly what your negation rule says.  The only rule I will use is a resolution rule of inference R which can get described by saying "given two formulas $\phi$ and $\psi$ which only contain disjunction symbols and literals, if there exists a literal $\alpha$ in $\phi$ such that it is the negation of a literal $\alpha$' in $\psi$, then we can infer any formula which consists solely of all of the remaining literals in $\alpha$ and $\psi$ and disjunction symbols."  You can verify that this rule holds by showing that disjunction commutes, that it associates, and that {AN$\alpha$$\beta$, $\alpha$} $\vdash$ $\beta$.

So, first we need to find equivalent formulas to each of the relevant formulas here.  The formulas contain only disjunction symbols and literals (lower case letters or the negations of lower case letters).  I use Polish notation.

CKcnt is logically equivalent to ANKcnt, since Cxy is logically equivalent to ANxy.  ANKcnt is logically equivalent to "AANcNnt" by the Petrus Hispanus law which says that NKxy is logically equivalent to ANxNy.

KhNs gives us "h" and "Ns".

CKhNAscp is logically equivalent to ANKhNAscp which is logically equivalent to AANhNNAscp by laws previously mentioned by name.  By the double negation law which says that NNx is logically equivalent to x, we then have "AANhAscp".

KnNt gives us "n" and "Nt".  Thus, I'll precede as follows:

assumption 1 AANcNnt
assumption 2 h
assumption 3 Ns
assumption 4 AANhAscp
assumption 5 n
assumption 6 Nt
R 4, 2     7 AAscp ("Nh" and "h")
R 7, 3     8 Acp (Ns, s... I'll just list one letter hereafter)
R 1, 5     9 ANct  (n)
R 9, 6    10 Nc  (t)
R 8, 10   11 p.  (c)

Here's another proof:

assumption 1 AANcNnt
assumption 2 h
assumption 3 Ns
assumption 4 AANhAscp
assumption 5 n
assumption 6 Nt
R 4, 1     7 AAAANhsNntp (c)
R 7, 2     8 AAAsNntp (h)
R 8, 3     9 AANntp (s)
R 9, 5    10 Atp  (n)
R 10, 6   11 p.  (t)

There does exist a more general resolution rule S here, which would give us a shorter proof:

assumption 1 AANcNnt
assumption 2 h
assumption 3 Ns
assumption 4 AANhAscp
assumption 5 n
assumption 6 Nt
S 4, 2, 3  7 Acp (h, s)
S 1, 5, 6  8 Nc (n, t)
S 7, 8     9 p.  (c)


Find $\lim_{n\to\infty} \sqrt[n]{a_n}$ where $a_n=\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n-k}{k}$ Let $a_n=\sum_{k=0}^{[n/2]} C_{n-k}^k$. Find $\lim_{n\to\infty} \sqrt[n]{a_n}$. Here $[n/2]$ is the largest integer $\leq n/2$, $C_{n-k}^k=\frac{(n-k)!}{k!(n-2k)!}$.

It sounds $a_{2m+1}-a_{2m}=a_{2m-1}$. Then...

We have the following lemma which follows from a stronger inequality proven in this question:

  If $(a_n)$ is a sequence in $\mathbb{R}^+$ and $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$ exists then
  $$\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$$

For this sequence we have that
$$a_{2n+1}=a_{2n}+a_{2n-1}$$
$$\frac{a_{2n+1}}{a_{2n}}=1+\frac{a_{2n-1}}{a_{2n}}\tag{1}$$
Also, by taking $n\to\infty$, we have that
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{a_{2n+1}}{a_{2n}}=\lim_{n\to\infty}\frac{a_{2n}}{a_{2n-1}}\tag{2}$$
Hence if we denote the desired limit by
$$\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L$$
then, using $(1)$ and $(2)$, we get the equation
$$L=1+\frac1{L}$$
which we can solve for $L\gt0$ giving
$$\boxed{\lim_{n\to\infty}\sqrt[n]{a_n}=\frac{\sqrt{5}+1}2}$$

How about schematically:

$a_n =$ the sum of the first half of the nth row of pascal's triangle

= half of 2^n when n is odd and a bit more than that when n is even

= roughly $2^{(n-1)}$

Hence the nth root of this is $2^{(1-1/n)}$ which tends to 2 as n-> inf


Prove $\sup(f+g) \le \sup f + \sup g$ Suppose $D$ is a nonempty bounded subset of reals. Let $f:D \to \mathbb R$ and $g:D \to \mathbb R$. Define $(f+g)(x)=f(x)+g(x)$. Prove $\sup(f+g)(D) \le \sup f(D) + \sup g(D)$ (also prove that $\sup (f+g)$ exists).

I understand why this is the case, just not how to prove it. Left side is pretty much $\sup  (f(x)+g(x))$ and right side is $\sup (f(x) + g(y))$ for $x,\,y \in D$. Basically $f+g$ has to use the same variable and $f(D)+g(D)$ use different ones. But I don't know how to go about proving this.

  The second part of the question is to find a specific example where strict inequality holds.

Let $D=\{a,b\}$ and $f: a \to 1,\, b\to 0, \,g: a \to 0,\, b\to 1$. $\sup f(D) = 1,\, \sup g(D) = 1,\, \sup f(D) + \sup g(D) = 2.$ $\sup (f+g)(D) = 1$ (if we choose a, $f+g = 1+0,\, b,\, f+g=0+1$).

Consider $x\in D$. Then $f(x)\le \sup f$ and $g(x)\le \sup g$, hence $(f+g)(x)=f(x)+g(x)\le \sup f+\sup g$.
Therefore, $\sup f + \sup g$ is some upper bound, but the least upper bound may be smaller.

Your example for strictness is fine.

To show that $\sup(f+g)\leq \sup(f) +\sup(g)$, suppose that $\sup(f)=A$ and $\sup(g)=B$, where $A$ and $B$ are the highest point of function $f$ and $g$ respectively.

If we add $f$ and $g$, then only two posibility happen:

1: If $A$ and $B$ are in same x, then $f(x)=A$ and $g(x)=B$. Then, we have that $(f+g)(t)=A+B$. Hence, $$\sup(f+g)=\sup(f)+\sup(g).$$

2: If $A$ and $B$ are not in same $x$, then $f(p)=A$ and $\ g(q)=B$. Hence, $$\sup(f+g)\leq A+B$$

From 1 and 2, we have that $$\sup(f+g)\leq\sup(f)+\sup(g).$$


Radius of $\sum a_n b_n x^n$ via radii of $\sum a_n x^n$ and $\sum b_n x^n$ Series $\sum a_n x^n$ and $\sum b_n x^n$ have radii of convergence of
  1 and 2, respectively. Then radius of convergence R of $\sum a_n b_n x^n$ is

  2
  1
  $\geq 1$
  $ \leq 2$

My attempt was to apply ratio test:

$$\lim_{n \rightarrow \infty} \frac{a_{n} b_{n}}{a_{n+1} b_{n+1}} = \lim_{n \rightarrow \infty} \frac{a_{n}}{a_{n+1}} \cdot \lim_{n \rightarrow \infty} \frac{ b_{n}}{ b_{n+1}} = 1 \cdot 2 = 2$$ i.e. option (1).

But the answer is (3). Why?

You can't apply the limit test because you don't know the limits $|\frac{a_n}{a_{n+1}}|$ or $|\frac{b_n}{b_{n+1}}|$. If these limits exist, you can use them to find a RoC, but knowing the radius doesn't imply that the limits exist.*

To show the answer is $3$, recall that a power series converges absolutely (and uniformly) inside it's radius of convergence.

Take any $x < 1$ and let $S(x) = \sum |a_nx^n|$. For any $n$ $|a_nb_nx^n|\leq S(x)|b_n|$ and $\sum S(x)|b_n|$ converges (since $1$ is inside the radius of convergence for $\sum b_nx^n$), so by the comparison test, $\sum a_nb_nx^n$ converges. Thus the radius of convergence is at least $1$.

Technically, this only proves that $3)$ is a possible answer. You would have to show that $1)$, $2)$ and $4)$ don't hold, by finding examples of power series that exclude these possibilities. It would in fact be enough to find a series that has RoC larger than $2$, since this would simultaneously exclude everything other than $3)$ (and if you assume that one of the answers is correct, this would actually be enough to prove that $3$ is the correct answer.)**

*Take for example, any convergent power series and replace the even coefficients with $0$. Clearly this gives a convergent power series, with RoC at least as large as the original, but the ratio test fails since the limit isn't defined.

**For an example of this, take $a_n = 1$ if $n$ is even, $0$ if $n$ is odd and $b_n = \frac{1}{2}^n$ if $n$ is odd, $0$ if $n$ is even. These have radii of convergence $1$ and $2$ respectively, as required, but $a_nb_n$ is identically $0$, so has raidus of convergence $\infty$.

we are given that $\limsup\sqrt[n]{|a_n|}=1$ and $\limsup\sqrt[n]{|b_n|}=\frac12$. Show that this implies $\limsup\sqrt[n]{|a_nb_n|}\le \frac12$.


What is the probability of rolling a 6-sided die 5 times, and getting at least 3 in a row? I had been working on this problem, and ran into trouble because I couldn't easily use the "find the opposite probability and subtract from one" trick. So for example I think I can find the probability of getting at least two in a row rolling a die 5 times, by finding the probability of getting no duplicates at all and then subtracting from one to find the opposite: ( 1 - (5/6)^4 )

But how to find the probability of a sequence of at least three? The extension of the problem is to write a formula generalizable to the number of trials, number of possible outcomes per trial, and the length of the repeated sequence.

I like working on this on my own so if you can just give a little hint or guidance I'd appreciate it very much!

You can do this using inclusion-exclusion. The conditions you want are the three conditions that the first, second or third triple of rolls are all the same. Adding the three counts for each of these conditions being met is straightforward. Then you need to substract the counts for each pair of conditions being met. This is only slightly more involved; you need to take into account that in one case the pair covers all five rolls and in the other two cases it only covers four rolls. Then you need to add the count for all three conditions being met – this is again straightforward.

The probability of getting exactly 3 in a row is obtained by counting the ways to select a number for those three, two numbers for the others (from the remaining five numbers, duplicates allowed) and choosing the start position for the triple, out of the total ways to roll the dice.$$\frac{6\cdot 4^2\cdot 3}{6^5} = \frac{4^2\cdot 3}{6^4} = \frac{25}{423}$$

Then to obtain the probability of at least three in a row, we also consider exactly 4 in a row, and all the same.$$\frac{5^2\cdot 3+5\cdot 2+1}{6^4} = \frac{43}{648}$$


Final all $x$ such that $|\sqrt{x} - x| < 1$ (simple but can’t resolve) I’m a beginner so be kind with me, I tried to resolve this for 2 hours now but always give me the same wrong result. I think I’m missing something big...

Can someone help me with this?

I began by assuming $x\geq 0$ (root) and then I started to make 2 systems, one with the argument $(\sqrt{x} - x) \geq 0$ and the other with the argument $<0$.
Once the systems are resolved I find a different result...

The result given by the textbook is $0\leq x<\frac{3 + \sqrt5}{2}$

Hint: it must be $$x\geq 0$$ so we get two cases:

$$\sqrt{x}\geq x$$ or by squaring

$$x(x-1)\le 0$$ this is only possible for $$0\le x\le 1$$ and we get

$$\sqrt{x}<x<1$$ and by squaring
$$0<x^2+x+1$$ which is true.
So we get

$$0\le x\le 1$$
the second case is for you!

We can set $\sqrt x=t$, which implies $t\ge0$. Then

$$|\sqrt x-x|<1\iff (t-t^2)^2<1\iff(t^2-t-1)(t^2-t+1)<0.$$

The second factor has no real roots and remains positive, and the second is negative between the roots $\dfrac{1\pm\sqrt 5}2$.

Hence

$$0\le x<\left(\dfrac{1+\sqrt 5}2\right)^2.$$


Is this true for biconditional statement? Given
A = At least ten people are there
L = Lecture will be given
Does A $\iff L$ means that lecture will be given only when at least ten people are there, means there cannot be other reason for lecture to be given and nothing else. And if ten people are not there then lecture will not be given at any cost.
Is this true ?

Material biconditional (symbol $X \leftrightarrow Y$) contains no idea of possibility or impossibility. It only means that, as a matter of fact, propositions $X$ and $Y$ do not have different truth values, that is, that they are either both true or both false. So as soon as (for a reason or another) the lecture is given and (for a reason or another) there are at least $10$ people, it is true that  $L \leftrightarrow A$.
Note: both material implication and bi-implication are very poor statements as to their informational contents

Logical equivalence (symbol: $\iff$, or $\equiv$) is more demanding. It requires not only that $X$ and $Y$ have the same truth value, but also that $X$ and $Y$ cannot possibly have different truth values. Or, if you prefer, it requires that $X$ and $Y$ have necessarily the same truth value.
So $X \iff Y$ standardly means that the material biconditional $X \leftrightarrow Y$ is always true, in all (logically) possible cases.

Normally, the kind of possibility / necessity invoked here is a logical one. But, in the example you take, it seems the necessity is more of a "legal" sort. The case you are considering is: "due to the rules of the university,
(a) the fact that there are at lest 10 students is necessarily a sufficient condition for the lecture to be given
and
(b) the fact that there are at least 10 students is necessarily a necessary condition for the lecture to be given".

Reference: Seymour Lipschutz, Schaum's Outline Of Set Theory (the chapter on the "Algebra Of Propositions"), at archive.org.


Calculating fluxintegral out of the surface $1-x^2-y^2$ I am trying to calculate the flux integral of the vector field
$$
\vec{F} = (x,y,1+z)
$$
Out of the surface $z=1-x^2-y^2$, $z\geq 0$

Answer : $\frac{5\pi}{2}$

I begin by defining a vector that traces out the surface and calculate the cross product of its derivate to get normal vector.

$$
\vec{r} = (x,y,1-x^2-y^2)\\
\vec{r}_x\times \vec{r}_y = (2x,2y,1)
$$
Next, I calculate the corresponding double integral using polar coordinates:
$$
\iint \vec{F} \cdot \hat{n}dS = \iint_D\vec{F}\cdot\vec{n}dxdy = \iint_D2+x^2+y^2dxdy\\
=\int_0^{2\pi}\int_0^{1}(2+r^2)*rdrd\theta \neq \frac{5\pi}{2}
$$
Where am I going wrong?

You should consider that $ds$=$\frac{dxdy}{\sigma}$ and thus projecting the whole surface on the XOY plane where $\sigma$=$|\vec n . \vec k|$ and $\vec n$ is the unit vector that is in this case $\vec n$ = $\frac{\vec grad(S)}{||\vec grad(S)||}$ = $\frac{2x\vec i + 2y\vec j + 2z\vec k}{\sqrt{4x^2 +4y^2 +4z^2}}$=$\frac{x\vec i + y\vec j + z\vec k}{\sqrt{x^2 +y^2 +z^2}}$=$\frac{x\vec i + y\vec j + z\vec k}{1}$ where $x^2 +y^2 +z^2=1$ and so you can evaluate your integral.

$$
\iint \vec{F} \cdot \hat{n}dS = \iint_D\vec{F}\cdot\vec{n}\frac{dxdy}{\sigma} = \iint_Dx^2+y^2+z(1-z)\frac{dxdy}{z}=\iint_D \frac{x^2+y^2-(1-x^2-y^2)+\sqrt{1-x^2-y^2}dxdy}{\sqrt{1-x^2-y^2}}=2 \iint_D \frac{x^2+y^2}{\sqrt{1-x^2-y^2}}dxdy -\iint_D \frac{1}{\sqrt{1-x^2-y^2}}dxdy + \iint_D dxdy$$. With D= {${{ (x,y) \in R^2 \ ,  x^2+y^2=1}}$} Now considering $x=rcos(\theta)$, $y=rsin(\theta)$ with the Jaacobian $|J|$ = $r$ you can solve the integral that will surely lead to $\frac{5\pi}{2}$


Surface Integral over a rhombus Evaluate the integral

$$\int\int_{R}(x-y)^2 cos^2(x+y)dxdy$$ where $R$ is the rhombus with successive vertices as $(\pi,0), (2\pi,\pi), (\pi,2\pi), (0,\pi).$

My attempt- I tried doing this surface integral by performing integration by parts but the answer solution seems to be very lengthy, also am unable to get the final answer by this method. Just wanted to know if there is any other simpler method or can it be solved by using some theorem in vector calculus perhaps Green's theorem or something.

Thanks

1) Translate rhombus to origin:
$$
x' = x - \pi \quad y' = y - \pi
$$
$$
\int\int_{R}(x-y)^2 cos^2(x+y)dxdy \rightarrow \int\int_{R'}(x'-y')^2 cos^2(x'+y' + 2\pi)dx'dy'
$$
2) Change coordinate system:
$$
x' - y' = u \quad x' + y' = v
$$
Now:
$$
-\pi \leq u \leq \pi \quad -\pi \leq v \leq \pi
$$
$$
\int\int_{R'}(x'-y')^2 cos^2(x'+y')dx'dy' \rightarrow \int\int_{R''}u^2 cos^2(v)J(u,v)dudv
$$
3) Compute Jacobian: $J = 1/2$

4) Evaluate Integral:
$$
\frac{1}{2}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}u^2 cos^2(v)dudv = \frac{1}{2}\int_{-\pi}^{\pi} u^2 du \int_{-\pi}^{\pi}cos^2(v)dv = \frac{1}{2}\frac{2\pi^3}{3} \cdot \pi =  \frac{\pi^4}{3}
$$

This isn’t a surface integral because you’re integrating over a planar region R. Find a transformation that maps the rhombus in the xy-plane to a square in the uv-plane, and compute it’s jacobian. The problem incentivizes you to use the change of coordinates u = x-y, v = x+y (changes in coordinates like these are valuable because they tend to preserve the linearity of the transformation, which makes the problem easier to compute). The rhombus is a closed curve so Green’s Theorem applies, though I’d apply the change of coordinates first and then use the theorem. Keep in mind that doing so would force you to evaluate four separate line integrals.


Probability of a team winning the series? Suppose that two sports teams were to play each other in a series, and the winning team is the one that wins two games. One of the teams, Team A, has a probability of 80% chance of winning any one game. What is the probability of Team A winning the series?

The answer that I reached is 64%, and my reasoning is that for Team A to win, there must be two conditions:

Team A must win its first game
Team A must win its second game

With that in mind, I multiplied 80% (the chance of Team A winning its first game) with 80% again (because it has 80% chance of winning its second game). Is this correct? I'm getting a bit tripped up because the probability of Team A winning its second game is not dependent on its probability of winning the first game, so that's why I'm sort of leaning on the total probability being 80% again (because if we can assume Team A wins its first game then everything else is irrelevant; the chances of them winning again is 80%).

i will solve a general case , say probability of $A$ winning is $p$ and for a team to win it needs to win $n$ matches. now say $k$ matches have been played , and $A$ won the series , it means $A$ won the last match , thus out of the rest $k-1$ matches $A$ won $n-1$ games. thus for $A$ to win the series in $k$ matches probability is $$\binom{k-1}{n-1}p^{n}(1-p)^{k-n}$$

now $k>=n$ and $k<=2n-1$ ($A$ needs to win before $B$) thus our total probability is $$\sum_{k=n}^{2n-1} \binom{k-1}{n-1}p^{n}(1-p)^{k-n}$$

I stumbled upon this when I hit the same dilemma - probability of winning game 1 and game 2 are independent , its like picking a red ball from a jar that has 3 red and 2 blue with replacement - a general example problem.

However, on thinking  more, even though the individual game result is independent, the overall series win is dependent of the result of game 1 and game 2.

Thus to get the probability of winning the series, the P(winning game 1) has to be multiplied with P(winning game 2), which is:

0.8 * 0.8 = 0.64.


Value of an unbounded definite integral Evaluate the integration :

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(2x^{2}+2xy+2y^{2})}dxdy$$

The function is even about $x$ & $y$. So we can write, $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(2x^{2}+2xy+2y^{2})}dxdy=2\int_{0}^{\infty}\int_{0}^{\infty}e^{-(2x^{2}+2xy+2y^{2})}dxdy$$.

But now , how I will proceed it?

Take the integral
$$
I=\int_R\int_R dxdy e^{-\frac{1}{2}\vec{x}^t M \vec{x}}
$$
where $M$ is a positiv definite and symmetric $2\times2$ Matrix. In this case we can find an orthogonal transformation $S$ that fullfills $S^{t}MS=D$ where D is a diagonal matrix. Using $S^tS=1$ we can rewrite the exponent as $\underbrace{\vec{x}^tS}_{\vec{\zeta}^t}\underbrace{S^tMS}_{D}\underbrace{S^t\vec{x}}_{\vec{\zeta}}$, where the entries of $D$ are the two eigenvalues of $M$. The measure transforms as $dxdy \rightarrow \underbrace{|\det(S)|}_{1}d\zeta_1d\zeta_2$ so the whole integral reads:
$$
I=\int_R\int_R d\zeta_1d\zeta_2 e^{-\frac{1}{2}\vec{\zeta}^t D \vec{\zeta}}= \int_R\int_R d\zeta_1d\zeta_2 e^{-\frac{1}{2}\zeta_1 \lambda_1 \zeta_1}e^{-\frac{1}{2}\zeta_2 \lambda_2 \zeta_2}
$$

Which is just the product of two independent Gaussian integrals. It follows that I is just given by
$$
I=\sqrt{\frac{2\pi}{\lambda_1}}\sqrt{\frac{2\pi}{\lambda_2}}=\frac{2\pi}{\sqrt{\det(M)}}
$$

Which is in your case $(\lambda_1=2,\lambda_1=6)$

$$
\pi \frac{1}{\sqrt{3}}
$$

This method is generalizable to non symmetric positive definite $n\times n$ matirx by using the decomposition $2M=(M+M^t)+(M-M^t)$.

Edit: Correct Eigenvalues used

Assuming $x$ and $y$ are real, Wolfram gives the indefinite integral as
\begin{equation}
\frac{1}{2} e^{-2y^{2}} \sqrt{e^{y^{2}}} \sqrt{\frac{\pi}{2}} erf\left(\frac{2x+y}{\sqrt{2}}\right)
\end{equation}


Is $\sin(\alpha)=\frac{\tan(\alpha)}{\sqrt{1+\tan^2(\alpha)}}$ a true statment? I'm being asked to prove the following equality

$$\sin(\alpha)=\frac{\tan(\alpha)}{\sqrt{1+\tan^2(\alpha)}}$$

and I support the idea that they are not equal (the rest of my class seems to desagree with me which made me doubt a little).

The "proof" is simple:
$$\frac{\tan(\alpha)}{\sqrt{1+\tan^2(\alpha)}}=\frac{\sin(\alpha)}{\cos(\alpha)}\cdot \frac{1}{\sqrt{\frac{\cos^2(\alpha)+\sin^2(\alpha)}{\cos^2(\alpha)}}}\\=\frac{\sin(\alpha)}{\cos(\alpha)}\cdot |\cos(\alpha)|=\sin(\alpha)\cdot\frac{|\cos(\alpha)|}{\cos(\alpha)}=\sin(\alpha)$$
Now, this doesn´t make sense to me since $\exists\alpha$ s.t. $\cos(\alpha)>0$ and also $\exists\alpha$ s.t. $\cos(\alpha)<0$.

Does the equality hold?

Not much context is given when it comes to frasing the problem, the first question of our first physics homework.

I'm aware that this is very simple but my classmates seem to disagree with me on the fact that the equality is just false, maybe they know something I don´t. Thanks in advance.

If the domain is $-\pi/2 < \alpha < \pi/2$, then the two functions are identical. However, if the domain contains any values for which $\cos(\alpha) < 0$, then as you've shown the functions differ by a factor of $-1$.

You can see it on this graph - the functions track together, then the second one suddenly flips, then flips back, and so on.

First,$$\tan(\alpha)=\frac{\sin(\alpha)}{\cos(\alpha)}$$
Pythagorean identity: $1+\tan^2(\alpha)=\sec^2(\alpha)$
$$ \therefore \sqrt{1+\tan^2(\alpha)}=\sec(\alpha)$$
Putting it all together,
$$\sin(\alpha)=\frac{\sin(\alpha)}{\cos(\alpha)}\times\frac{1}{\sec(\alpha)}$$
But, $\sec(\alpha)=\frac{1}{\cos(\alpha)}$
$$\therefore \sin(\alpha)=\frac{\sin(\alpha)}{\cos(\alpha)}\times \cos(\alpha)=\sin(\alpha) $$
The statement is true.


General term for series $0,2,12,70, 408...$ Given the curves $y=\sqrt{2x^2+1}$ and $x=\sqrt{2y^2+1}$, the first line (line with smallest y coordinate) which intersects the curves with integer coordinates is $x+y=1.$ This line intersects the curves at $(0,1)$ and $(1,0)$. Along with this it intersects the curves at $(-2,3)$ and $(3,-2)$. If we use the larger integer coordinates and construct another line $x+y = 2+3=5$ and intersect it with the curves we can find the next integer coordinates which are $(2,3)$ and $(3,2)$, but along with this we find another two points of intersection $(-12,17)$ and $(12,-17).$ If we repeat this process and find the intersection of the curves with $x+y=12+17=29$ we can find the next two integer coordinates: $70,99$. If we continue this process we can iteratively find the next integer coordinates.
So the question at hand is to find the $n^{th}$ largest integer intersecting points on these curves or in other words... the $n^{th}$ largest point of the curve $y=\sqrt{2x^2+1}$ whose both coordinates are integers. I believe by reiterating this line process we can find all integer coordinates, i.e. there are no integer coordinates that can't be found by reiterating this process. We won't skip any. (Feel free to disprove this, this is just my intuition.) Anyways, the main question to generate the $n^{th}$ term in this series, i.e. finding a general term, one the doesn't require redrawing the next line to find the next point.

You are looking at a subsequence of the Pell numbers - probably every other one, with some variant on the initial condition:
$$
0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, \ldots
$$
https://oeis.org/A000129
You can write a recursive formula but I am pretty sure there is no closed form solution for the $n$th term.
They come from an analysis of the Pell equation $x^2−2y^2=1$.
http://en.wikipedia.org/wiki/Pell%27s_equation

I noticed sequence you got there,
$$ 1, 5, 29 , 169 , 985, 5741, \dots \dots $$
The formula to describe their $n$th term is
$$f[n] = 1+ 2\cdot \sum_{k=0}^{n-1} { \sum_{q=0}^{k+1} { 2^q \binom{2k+2}{2q+1} } }$$
The function grows very fast, $f[0] =1 , f[\infty ] = \infty$
I compared how fast $f[n]$ was rising with $n^n$ and $n!$ though it appears to rise faster, but it soon slow down with large $n$


Is it true that $f_n(x)=\left(\frac{n}{n+x}\right)^n$ uniformly converges to $e^{-x}$ Is it true that: $$f_n(x)=\left(\frac{n}{n+x}\right)^n$$ uniformly converges to $e^{-x}$?  I have tried to prove it using definition with supremum, but have not succeed. Can you give me a hint, please?

Hint for uniform convergence on $[0,\infty)$: let $\epsilon >0$ and choose $T$ such that $e^{-x} <\epsilon$ for all $x \geq T$. Next choose $m$ such that $(1+\frac T n)^{n} >\frac 1 {\epsilon}$ for $n \geq m$. [This is possible because $(1+\frac T n)^{n} \to e^{T}>1/\epsilon)$. Now you can easily check that $|(\frac n {n+x})^{n}-e^{-x}| <2\epsilon$ for all $x \geq T$ for all $n \geq m$. For uniform convergence on $[0,T]$ use Dini's Theorem.

Do you know that $\lim_{n\to\infty}\left(1+\dfrac{x}{n}\right)^n=\mathrm{e}^x$?

You may then use that $\left(\dfrac{n}{n+x}\right)^n= \left(\dfrac{n+x}{n}\right)^{-n}= \left(\left(1+\dfrac{x}{n}\right)^n\right)^{-1}$. Continuity of the function $g(y) = y^{-1}$ for $y\neq 0$ then gives the result.


A proof for a (non-constant) polynomial can't take only primes as value I know a proof of this statement, see How to demonstrate that there is no all-prime generating polynomial with rational cofficents?

My question is that, in the book

Introduction to Modern Number Theory - Fundamental Problems, Ideas and Theories

by Manin, Yu. I., Panchishkin, Alexei A,

it says (p16)

I would like to know if this is correct. If I understand it correctly, then $x^2+1$ is a counterexample of this, by considering the Legendre symbol $(\frac{-1}{p}) = (-1)^{\frac{p-1}{2}}$.

suppose the polynomial P(x) takes prime value for each integer k.assume r is a root of P(x).so P(r)=0

now k -r|P(k) -P(r)$ \Rightarrow $k -r|P(k)=m=a prime. so clearly k -r=m or 1
so clearly k-n=a prime or 1 for all roots n of P(x)
consider P(x)=a(x-r)(x-c)(x-d)....................(x-v)[r,c,d,.........,v are roots]
now P(k)=a(k-r)(k-c)(k-d)................(k-v)=a(a prime or 1)(a prime or1)....................(a prime or 1)=a constant
but P(x) is non constant.so a contradiction!!!!!!


Trigonometric Property How can I show that the following property holds?

$2\cos(4a)+2\cos(2a)+1=\displaystyle\frac{\sin(5a)}{\sin(a)}$

I've been trying to derive it to no avail. What would be a way to approach similar problems like this? Thank you for the help.

Using the arc addition formula one gets

$$\sin{5a}=\cos{4a}\sin{a}+\sin{4a}\cos{a}$$

Now use $\sin{4a}=2\sin{2a}\cos{2a}=4\sin{a}\cos{a}\cos{2a}$

$${\sin{5a}\over \sin{a}}=\cos{4a}+4\cos^2{a}\cos{2a}$$

Now keep in mind that $\cos^2{a}={\cos{2a}+1\over 2}$ to get

$${\sin{5a}\over \sin{a}}=\cos{4a}+2\cos^2{2a}+2\cos{2a}$$

And finally use $2\cos^2{2a}=\cos{4a}+1$ to get to

$${\sin{5a}\over \sin{a}}=2\cos{4a}+2\cos{2a}+1$$