Mixture and Alligation question: There is a Vessel holding 40 litres of milk. 4 litres of Milk is initially taken out from the Vessel and 4 litres of water is then poured in. After this 5 litres of mixtures of Mixture is replaced with six litres of water and finally six litres of Mixture is Replaced with the six litres of water. How Much Milk is there is in the Vessel?

I have tried:

Initally Vessel containing 40 litres of Milk :

4 litres out means  -> 36 litres

4 litres of water is poured in - > 4 litres

so Now total quantity is 40 litres

Mixture containing water and Milk in the ratio: 36:4 i.e 9:1

Again 5 litres of Mixture is replaced with the six litres of water

for that:

9x - 9/10*5 : x -1/10*5

Now the Ratio becomes:

90x - 45 : 10x -5 i.e 9x -9:2x -1

six litres of water is added

9x -9 :2x -5

again six litres of Mixture is replaced then

9x -9 - 9/10*6 : 2x -5 -9/10*6

that is

90x -144 :10x -84

after adding six litres of water again we got

90x -144 :10x -78

so Milk containing is

90x-144+10x -78 =41

100x -222 =41

100x = 263

x= 2.63

again substituting the value x=2.63 in 90x -144 then i am getting 92.7 milk

total itself 41 litres of Milk Please anyone guide me answer

what i am doing mistake please anyone guide me for the answer

Think in the terms of %. In the first iteration, $10$% is removed, in the second iteration, $12.5$% is removed and in the third iteration, $15$% is removed. In all these iterations, the same percentage of milk will be removed, because you are removing milk and not adding it back.

$40-10$%$=$$36$

$36-12.5$%$=$$31.5$

$31.5-15$%$=26.775$.

Therefore, after the third iteration, milk will be 26.775 liters.

Initial congifuration is $40$ litres of milk and no water, say $m=40$ and $w=0$.

Remove $4$ litres of milk and add $4$ litres of water: $m=36, w=4$.

The mixture is $36:4$ or $9:1$. $5$ litres of mixture contains $5\times \frac{9}{10}=\frac 92$ litres of milk and $5\times \frac{1}{10}=\frac 12$ litres of water.

Remove $5$ litres of mixture: $m=36-\frac 92=\frac{63}2, w=4-\frac 12 = \frac 72$.

Add $6$ litres water: $m=\frac{63}2, w=\frac 72+6=\frac{19}2$.

The mixture is now $63:19$. $6$ litres of mixture contains $6\times \frac{63}{82}=\frac{189}{41}$ litres of milk and $6\times \frac{19}{82}=\frac{57}{41}$ litres of water.

Remove $6$ litres of mixture: $m=\frac{63}2-\frac{189}{41}=\frac{2205}{82}, w=\frac{19}2-\frac{57}{41}=\frac{665}{82}$.

Add $6$ litres of water: $m=2205/82, w=\frac{665}{82}+6 = \frac{1157}{82}$.

Final configuration: $\frac{2205}{82}$ litres of milk (approx $26.9$) and $\frac{1157}{82}$ litres of water (approx $14.1$).


What is the inverse of the following function? What is the inverse function of $ y=x^3+x$? thanks

According to Wolfram Mathematica, the equation $x^3+x=y$ is solved by
$$
x=\frac{\sqrt[3]{\sqrt{3} \sqrt{27 y^2+4}+9 y}}{\sqrt[3]{2}
   3^{2/3}}-\frac{\sqrt[3]{\frac{2}{3}}}{\sqrt[3]{\sqrt{3} \sqrt{27 y^2+4}+9 y}}.
$$

I get $x= y-y^3+3y^5-12y^7+55y^9-273y^{11}+-...$ where the coefficients are given by http://oeis.org/A001764 with alternating signs.
This is of course convergent up to the point where the closed form solution has its poles. The reference shows that, as a generating function, the coefficients have nice closed forms in terms of binomial coefficients.


detail in proof of the mean value theorem for integrals I am trying to prove the following result

If $f$ is a continuous function on a closed interval $[a,b]$, then
there exists a $c \in (a,b)$ such that,
$$f(c)=\frac{1}{b-a}\int_a^bf(x)\;\mathrm{d}x$$

I have worked out the following proof
Because $f$ is continuous on the interval, it reaches a minimum $m$, and a maximum $M$ so that we have
$$\begin{aligned} m&\leq f(x)\leq M \\ m(b-a)\leq  & \int_a^bf(x)\;\mathrm{d}x\leq M(b-a) \\  m\leq &\frac{1}{b-a}\int_a^bf(x)\;\mathrm{d}x\leq M\end{aligned}$$
and the IVT guarantees the exxistence of $c$ as above.
How do I prove that $c \neq a\ \& \ c\neq b$?
Thanks for the help !

Assume $w<z, w,z\in [a,b]$ with $f(w) = m, f(z) = M$
If $m=M \implies f$ is a constant function then all $c\in(a,b)$ works.
Suppose $m<M$
We have that for $c\in [w,z]$, $f(c) = \frac{1}{b-a}\int_{a}^{b}f(x)dx$
Suppose $c = a$, then $a = w$, because $a\leq w \leq c \leq  z \leq b$
Then $f(a) = m$ is the  minimum of the function
$f(z) =M>m \implies$ there exist some $\delta >0$ such that $f(x)>m+\epsilon$ for some $\epsilon>0$ for all $x\in (z-\delta,z]$
Thus we have that $\begin{aligned}\int_{a}^{b}f(x)dx & = \underbrace{\int_{a}^{z-\delta}f(x)dx}_{\geq f(a)\cdot(z-\delta -a)} +\underbrace{\int_{z-\delta}^{z}f(x)dx}_{\geq (f(a) + \epsilon)\cdot\delta>f(a)\cdot \delta}+\underbrace{\int_{z}^{b}f(x)dx}_{\geq f(a)\cdot(b-z)}>f(a)\cdot(b-a)\\  &\underbrace{\frac{1}{b-a}\int_{a}^{b}f(x)dx}_{f(c)} > f(a)\end{aligned}$
that is a contradiction. Then $c \neq a$
I hope it help you,
I think the other cases will be analogous
(for example, apply this reasoning to $-f$)

Sometimes a figure is worth 1000 words:

What is the area in blue?
$$\int\limits_a^b f(x)\ dx$$
What is the area in orange?
$$f(c) \cdot (b-a)$$
QED.


Prove $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\gt1$ Prove that $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\gt1$, where a, b, c >0.
I tried,
$\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} = \frac{a}{\sqrt{a}\sqrt{a+b}}+\frac{b}{\sqrt{b}\sqrt{b+c}}+\frac{c}{\sqrt{c}\sqrt{c+a}} \ge \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\sqrt{a}\sqrt{a+b}+\sqrt{b}\sqrt{b+c}+\sqrt{c}\sqrt{c+a}}$.
Then $\sqrt{a}\sqrt{a+b} \le \sqrt{2}/2 (2a+(a+b))$.
This only gave me $\sqrt{2}/2 + \sqrt{2}(\frac{ab+bc+ca}{a+b+c})$.

Hint: $\dfrac{a}{a+b} > \dfrac{a}{a+b+c}$

We can use also your idea, but in another writing:
$$\sum_{cyc}\sqrt{\frac{a}{a+b}}=\sum_{cyc}\frac{\sqrt{a(a+b)}}{a+b}>\sum_{cyc}\frac{\sqrt{a\cdot a}}{a+b}>\sum_{cyc}\frac{a}{a+b+c}=1.$$


How to think of factorising $x^7+x^2+1$ to $(x^2+x+1)(x(x-1)(x^3+1)+1)$ (Thales 2016) I was just doing the following question:
Find a prime number which divides the number $A=14^7+14^2+1$.
I solved it by finding the result which is $A=105413504+196+1=105413701$ and then trying out all prime numbers till I found that 211 divides it. However, obviously this is extremely tedious. I hence looked at the solution which says that $x^7+x^2+1=(x^2+x+1)(x(x-1)(x^3+1)+1)$ and from here by saying that $x=14$ we get the solution. However I can't seem to think of how to intuitively turn  $x^7+x^2+1$ into $(x^2+x+1)(x(x-1)(x^3+1)+1)$. I realize that from the question it is obvious to go looking for factors of A and hence trying to factorize $14^7+14^2+1$, but I can't work out how to go about factorizing it, what are the steps which you need to take in order to factorize a given polynomial. Could you please explain to me how to go about factorizing such an expression and how to intuitively think of each step?

You can "see" that $\omega = \exp \left( \frac{2i\pi}{3}\right)$ and $\omega^2= \exp \left( \frac{-2i\pi}{3}\right)$ are roots of $x^7+x^2+1$, therefore $x^7+x^2+1$ factorizes by $x^2+x+1$.

Hint: $x^2+x+1$ always divides $x^a+x^b+x^c$ if $\{a, b, c) \equiv \{0, 1, 2\} \mod 3$.  Hence it is easy to note $14^7+14^2+1 \equiv 0 \pmod {14^2+14+1}$, and all that remains is checking primality of $211$.
In case you wish to find the other factor(s), in this case, it may be simpler to consider dividing by $x^3-1$, like so:
$$x^7+x^2+1 = x^4(x^3-1) + x(x^3-1)+(x^2+x+1)$$
$$\implies \frac{x^7+x^2+1}{x^2+x+1} = x^4(x-1)+x(x-1)+1=(x-1)(x^4+x)+1$$


A sequence of continuously differentiable functions with non-differentiable limit Working in the metric space $C[a,b]$, the set of continuous functions $f:[a,b]\rightarrow[a,b]$, with the supremum metric, I need to demonstrate that $C^1[a,b]$, the subset of continuously differentiable functions, is not closed.

So I believe that I need to construct a sequence of continuously differentiable functions which converge in the supremum metric to a continuous, non-differentiable limit.

It looks to me like converging with respect to the supremum metric is the same as uniform convergence. A function I know that is continuous but not differentiable is $|x|$ at $x=0$ so I can adapt this to put the non-differentiable bit in $[a,b]$ by translation.

So I want something that converges uniformly to $|x-\frac{a+b}{2}|=\sqrt{(x-\frac{a+b}{2})^2}$. My first thought is: $$f_n(x):=\sqrt{(x-\frac{a+b}{2})^2+\frac{1}{n}}$$

So now I need to work out what the value of the following is, and hope it tends to $0$ as $n \rightarrow \infty$:

$$ \sup_{x\in[a,b]}\sqrt{(x-\frac{a+b}{2})^2+\frac{1}{n}} $$

By differentiating, I see there's a minimum at $x=\frac{a+b}{2}$ and no other turning points, so the maximum is achieved at either $x=a$ or $x=b$ and it turns out that $f_n(a)=f_n(b)=\sqrt{(\frac{a+b}{2})^2+\frac{1}{n}}$. But this doesn't tend to $0$ as $n \rightarrow\infty$.

So do I need a different example, or is my reasoning mistaken?

Answering my own question. Per this answer to another question, we want not that $\sup_{x\in[a,b]}|f_n(x)|\rightarrow0$ as $n\rightarrow\infty$, but that:

$$\sup_{x\in[a,b]}|f_n(x)\color{red}{-f(x)}|\rightarrow0 \text{ as } n\rightarrow\infty$$

With the example $f_n(x):=\sqrt{(x-\frac{a+b}{2})^2+\frac{1}{n}}$, we have that

$$ \sup_{x\in[a,b]}|f_n(x)-f(x)|=|\sqrt{(\frac{a+b}{2})^2+\frac{1}{n}}-|\frac{a+b}{2}|| $$

and this tends to $0$ as $n\rightarrow\infty$. Hence the convergence is uniform.

This method also works for the example $\sqrt {x+\frac 1 {n^{2}}} \to \sqrt x$ given by another user.

Let $a=-1,b=1.$ By Weierstraß, there is a sequence of polynomials which converges uniformly to $|x|$ on $[-1,1]$, but $|x|$ is not differentiable at $x=0.$


Is $\log{(0.99\dots)}$ negative or it is $ 0$? It is known  that $0.99\dots =1$, but I'm afraid to say by substitution if it is allowed  that $\log{(0.99\dots)}=0$ , then is it negative or equal $0$ ?

$0.9999...$ is exactly one. Is not an approximation or some other construct; the symbol $0.999...$ and the symbol $1$ are the exact same object, and you can write one or the other in any context

Another approach:

$S_n:= 9(1/10)+$

$9(1/10)^2 +9(1/10)^3...+9(1/10)^n.$

$\lim_{n \rightarrow \infty}S_n =1$,

the limit of the infinite geometric series with $r=(1/10)$.

Since $\log$ is a continuous function :

$\lim_{n \rightarrow \infty} \log(S_n) =$

$\log(\lim_{n \rightarrow \infty}S_n)=0.$


Solve the Recurrence Relation T(n)=sum of T(i)T(n-i) $T(1)=1 \\
T(n) = \sum_{i=1}^{n-1}T(i)T(n-i) \\
$

I have been trying to get rid of the Summation using the T(n+1) - T(n) technique, but I can't figure out how to apply it to this.

I used recursion trees but got too messy.

Now I'm wondering if there is an elegant solution to this recurrence.


Characterization of Asymptotic Stability via KL-class functions Let us adopt the following definition of stability and asymptotic stability of a dynamical system of the form:

$$
\dot{x}=f(x)
$$

The trajectory of this system starting from the initial point $x_0$ at $t_0=0$ will be hereinafter denoted by $\varphi(t,x_0)$.

Definition 1 (Stability). Let $x^\star$ be an equilibrium point for $\dot{x}=f(x)$, i.e. $f(x^\star)=0$. This point is said to be stable if for every $\varepsilon>0$ there is a $\delta=\delta(\varepsilon)>0$ such that
$$
\|x-x^\star\|<\delta \implies \|\varphi(t,x_0)-x^\star\|<\varepsilon
$$
for all $t\geq 0$.

Definition 2 (Asymptotic Stability). Let $x^\star$ be an equilibrium point for $\dot{x}=f(x)$, i.e. $f(x^\star)=0$. This point is said to be asymptotically stable if it is stable and additionally there is an $\alpha>0$ such that
$$\lim_{t\to\infty}\|\varphi(t,x_0)-x^\star\|=0$$
for all $x$ with $\|x-x^\star\|<\alpha$.

Definition 3 (Classes of functions). We define the following four classes of functions:
A function $\alpha:\mathbb{R}^+\to\mathbb{R}^+$ is said to be

$\mathcal{K}$-class and we denote $\alpha\in\mathcal{K}$ if it is continuous, strictly increasing and $\alpha(0)=0$.
A function $\beta:\mathbb{R}^+\to\mathbb{R}^+$ is said to be $\mathcal{K}_\infty$-class and we denote $\beta\in\mathcal{K}_\infty$ if $\beta\in\mathcal{K}$ and it is unbounded.
A $\mu:\mathbb{R}^+\to\mathbb{R}^+$ is said to be $\mathcal{L}$-class and we denote $\mu\in\mathcal{L}$ if it is continuous, strictly decreasing and $\lim_{\tau\to\infty}\mu(\tau)=0$
A $\gamma:\mathbb{R}^+\times\mathbb{R}^+\to\mathbb{R}^+$ is said to be $\mathcal{KL}$-class and we denote $\gamma\in\mathcal{KL}$ if $\gamma(\cdot,t)\in\mathcal{K}$ for all $t\in\mathbb{R}^+$ and $\gamma(r,\cdot)\in\mathcal{L}$ for all $r\in\mathbb{R}^+$

Proposition 1. An equilibrium point $x^\star$ is asymptotically stable if and only if there is a $\gamma\in\mathcal{KL}$ and a constant $\kappa>0$ so that:
$$
 \|\varphi(t;x_0)-x^\star\|\leq \gamma(\|x_0-x_\star\|,t)
$$
for all $x_0$ such that $\|x_0-x^\star\|<\kappa$ and for all $t\geq 0$.

I have some problem with the proof!

Proof: 1. [The first part looks easy] Let us assume that $x^\star$ satisfies the inequality mentioned above. Let us choose an $\varepsilon>0$; $\gamma(\cdot,t)$ is $\mathcal{K}$-class, so we may choose a $\delta$ such that $0<\delta<\kappa$ and $\gamma(\delta,0)\leq \varepsilon$. Now for $\|x_0-x^\star\|<\delta$ we have:

$$
 \|\varphi(t;x_0)-x^\star\|\leq\gamma(\|x_0-x_\star\|,t)\leq\gamma(\|x_0-x_\star\|,0)\leq\gamma(\delta,0)\leq\varepsilon
$$

So if the inequality state above holds then $x^\star$ is stable and more:
$$
 0\leq\lim_{t\to\infty}\|\varphi(t;x_0)-x^\star\|\leq \lim_{t\to\infty}\gamma(\|x_0-x^\star\|,t)=0
$$
Therefore
$$
 \lim_{t\to\infty}\|\varphi(t;x_0)-x^\star\|=0\Leftrightarrow\lim_{t\to\infty}\varphi(t;x_0)=x^\star
$$

What about the converse? How can one prove it?

We need to prove that if $x^\star$ is a (locally) asymptotically stable equilibrium point, then there is a $\gamma\in\mathcal{KL}$ and a constant $\kappa>0$ so that:
$$
 \|\varphi(t;x_0)-x^\star\|\leq \gamma(\|x_0-x^\star\|,t)
$$
for all $x_0$ such that $\|x_0-x^\star\|<\kappa$ and for all $t\geq 0$.

I would appreciate a proof or at least a reference to some book or article where I can find it clearly explained.

A proof can be found in Khalil, Nonlinear Systems, 3rd edition, appendix C.6.


Prove or disprove: $n^2-n+11$ is a prime for every positive integer $n$ I'm having difficulty proving this statement.

  For every positive integer $n$, $n^2-n+11$ is a prime

So I know the obvious, that a prime can only be divided by $1$ and itself, but I'm not sure how to turn that into some generic equation to compare to the one mentioned in the question. Could someone give me a start?

Hint $\ f(11n)\, =\, 11\,(11n^2\!-n+1)\,$ always prime implies $\,11n^2\!-n+1 = 1 $ for all $\,n>0,\,$ contra a nonzero quadratic has at most $2$ roots.

Remark $\ $ I presented the proof this way because exactly the same proof shows that any nonconstant polynomials cannot produce only primes. You may find it much more instructive to prove this more general statement.

let $n=11$
$n^2$ is a multiple of $11$.
A multiple of $11$ minus $n$(a multiple of 11) is also a multiple of 11.
When a multiple of 11 is added to a multiple of 11, it is again a multiple of 11.
It means, 11 is a factor of the number -> the number has a proper factor 11.
Therefore $n$ is not prime.


$T:X\to Y$ a linear operator, $Y$ finite-dimensional, then $\ker T$ is closed iff $T$ is continuous I have to prove that if $T:X\to Y$ a linear operator on normed spaces, $Y$ finite-dimensional, then $\ker T$ is closed iff $T$ is continuous.

On a lot of places I see a proof that looks like this: it starts by assuming $Y=\mathbb{R}$ for simplicity. (I'm also aware of this that use a completely different construction). Anyway, I did not at all find it obvious that there was any kind of generalisation of the first linked proof to higher dimensions.

I will give my (quite long a tedious) generalisation here, but my question is: is there a more direct/easier generalisation of the construction given in the first linked proof? (I also wouldn't object to someone checking the actual correctness of this proof)

We will take $Y=\mathbb{R}^n$.

Take $x_n\in X$ s.t. $||x_n||=1$, $||Tx_n||\to \infty$. w.l.o.g. we may assume that $\frac{Tx_n}{||Tx_n||}\to v_1\in \overline{B(0,1)}\subset \mathbb{R^n}$, by the compactness of the closed unit ball in finite dimensions. We now take $y_n=x_j$ for some $j$ s.t. $||\frac{Tx_n}{||Tx_n||}-v_1||<\frac{1}{n}$. Then we adjust the $y_n$ a little bit: if $||Ty_n||>n$ we set $y_n=\frac{y_n}{||Ty_n||}n$. We now obtain a sequence $y_n$ s.t.
$$
\begin{align*}
||y_n||\leq 1\\
||Ty_n||\leq n\\
||Ty_n||\to \infty
\end{align*}
$$
In particular we have that $||Ty_n-||Ty_n||v_1||<1$ (this is the critical property that we needed).

We now take an orthonormal basis $\{v_i\}$ for $\mathcal{R}T$ containing $v_1$, and $v_i'$ s.t. $Tv_i'=v_i$. We conclude that for $j>1$ we must have $(v_j,Ty_n)<1$. We now define $z_n=y_n-\sum_{j>1}(v_j,Ty_n)v_j'$. Then
$$||z_n||=||y_n-\sum_{j>1}(v_j,Ty_n)v_j'||\leq ||y_n||+\sum_{j>1}(v_j,Ty_n)||v_j'||\leq 1+\sum_{j>1}||v_j'||\leq C$$
Hence these $z_n$ are bounded. Furthermore we clearly still have that
$$||Tz_n||\to\infty$$
because the norm of $T\sum_{j>1}(v_j,Ty_n)v_j'$ is bounded.

We can now finally define
$$w_n=v'_1-\frac{z_n}{||Tz_n||}$$
Then
$$Tw_n=v_1-\frac{1}{||Tz_n||}(Ty_n-\sum_{j>1}(v_j,Ty_n)v_j)=0$$
Since $Ty_n-\sum_{j>1}(v_j,Ty_n)v_j$ is by construction a vector pointing in the direction of $v_1$. But again $w_n\to v'_1$, which contradicts the closedness of the kernel.

For the intuition: We take the unbounded sequence in $X$. The the project is onto the unit ball, which is compact, hence the sequence a convergent subsequence. We take the sequence, with limit $v_1$. Then we 'limit the growth' of our unbounded sequence to make sure that if $\frac{Tx}{||Tx||}$ is close to $v_1$, then $Tx$ is close to the line spanned by $v_1$. Then we remove any orthogonal components from the $Tx$, and because $Tx$ is now close to $<v_1>$ we can remove the orthogonal components by subtracting a vector of a certain bounded length. This finally gives a bounded sequence in $X$ with unbounded images in $Y$ all lying on the lines $<v_1>$, and from here the previous argument goes through.

Thanks for reading.

If $T$ is continuous, then $\ker(T) = T^{-1}(\{0\})$ is clearly closed. Conversely, suppose $Z:= \ker(T)$ is closed, then the quotient map
$$
\pi : X\to X/Z
$$
is a well-defined continuous map between normed linear spaces. The first isomorphism theorem (from linear algebra) gives an injective map
$$
S : X/Z \to Y
$$
such that $T = S\circ \pi$. Now $S$ is an injective map, so $X/Z$ must be finite dimensional, and hence $S$ is continuous, and so it follows that $T$ must be continuous as well.


If there are $2$ linearly independent vectors $x,y \in X$ such that $||x+y||=||x||+||y||$, then the unit sphere $S(X)$ contains an interval Let $S(X)= \{x \in X: ||x||=1\}$ be the unit sphere in $X$. Assume that there are $x,y\in X$ linearly independent such that $||x+y||=||x||+||y||$. Prove that $S(X)$ contains the following set:$[x,y]=\{z\in X: z=tx+(1-t)y, t\in [0,1]\}$ for some $x,y$.

So it is obvious that I need to use the $x,y$ that are given to be linearly independent and form a $[x,y]$ in $S(X)$ but I don't know how to start.

This is similar to user1551's answer, but a little simpler.

For the given $x, y$, and all $\lambda,\mu \geqslant 0$, we have
as usual
$$
\|\lambda x + \mu y\| \leqslant \lambda\|x\| + \mu\|y\|,
$$
but also
\begin{align*}
(\lambda + \mu)(\|x\| + \|y\|) & = (\lambda + \mu)(\|x + y\|) \\
& = \|(\lambda + \mu)(x + y)\| \\
& = \|(\lambda x + \mu y) + \mu x + \lambda y\| \\
& \leqslant \|\lambda x + \mu y\| + \mu\|x\| + \lambda\|y\|,
\end{align*}
therefore
$$
\|\lambda x + \mu y\| \geqslant \lambda\|x\| + \mu\|y\|,
$$
therefore
$$
\boxed{\|\lambda x + \mu y\| = \lambda\|x\| + \mu\|y\|}
$$
Putting $\hat{x} = x/\|x\|$, $\hat{y} = y/\|y\|$, we have
$\hat{x} \ne \hat{y}$, $\|\hat{x}\| = \|\hat{y}\| = 1$, and if
$0 \leqslant t \leqslant 1$,
\begin{align*}
\|t\hat{x} + (1 - t)\hat{y}\| & =
\left\lVert\frac{t}{\|x\|}x + \frac{1 - t}{\|y\|}y\right\rVert \\
& = \frac{t}{\|x\|}\|x\| + \frac{1 - t}{\|y\|}\|y\| \\
& = 1.
\end{align*}


Joint density of two exponential random variables For independent random variables X ∼ Exp(1) and Y ∼ Exp(2), find the density of (Z, W) = (X-2Y, X).

My approach:

Since for any exponential distribution with parameter $\lambda$ the function is
$f(x) = \lambda e^{-\lambda x}$

$f_X(x) =  e^{-x}$

$f_Y(y) = 2e^{-2y}$

Therefore the joint density function is:
$$ f_{X, Y}(x, y) =f_X(x) f_Y(y) = \begin{cases} 2e^{-x-2y} \ & \mbox{ if } x \geq 0, y \geq 0;  \\ 0 \ & \mbox{ elsewhere}. \end{cases} $$

However I don't know how to use this to calculate $f_{Z, W}$

The map $g:(x,y) \mapsto (x-2y,x)$ is a differentiable and invertible function between $(0,\infty)\times (0,\infty)$ and $R=\{(z,w) | z< w \text{ and } w>0\}$, so first of all we get that the support for $(Z,W)=(X-2Y,X)$ must be $R$.

The transformation theorem for probability densities states that:

$$f_{Z,W}(z,w) = f_{X,Y}(g^{-1}(z,w)) |det(\frac{dg^{-1}}{d(z,w)}(z,w))|,$$
where $\frac{dg^{-1}}{d(z,w)}(z,w)$ is the jacobian of $g^{-1}$.

(see https://en.wikipedia.org/wiki/Probability_density_function#Vector_to_vector)

We first compute $g^{-1}(z,w)= (w,\frac{w-z}{2})$ and the jacobian
$$ \frac{dg^{-1}}{d(z,w)}(z,w) = \begin{pmatrix}0 & 1 \\ -\frac12 & \frac12 \end{pmatrix},$$
which has determinant $\frac12$ for all $z,w$. We now plugin, and get
$$ f_{Z,W}(z,w) = \frac12 f_{X,Y} ((w,\frac{w-z}{2})) = e^{-w}e^{-2\frac{w-z}{2}}=e^{z-2w}.$$
for all $(z,w) \in \{(z,w) | z< w \text{ and } w>0\}$. Just to verify, that this is in fact a valid density we compute
$$ \int_0^\infty \int_{-\infty}^w e^{z-2w} dzdw = 1$$

If I'm not wrong, you've formed the joint density function under the assumption that the random variables $X$ and $Y$ are independent.

In the formula for $f_{X,Y}(x,y)$, let us put $X-2Y$ in place of $X$, $X$ in place of $Y$, and let us also put $x-2y$ in place of $x$ and $x$ in place of $y$ to obtain
$$ f_{Z, W}(z, w) = 2e^{- (x-2y) - x } = 2 e^{2y-2x} $$
if $x-2y \geq 0$ and $x\geq 0$, that is, if $x \geq \max(2y, 0)$, and $0$ otherwise.

However, I'm not completely sure if my solution is correct.


Solve for $x, \;23+35 x ≡ 4$ (mod $ 13)$ I can simplify the equation, but I get stuck at the last step:   $\;x≡ 9^{−1} (4-10)$ (mod $13$).

My issue is that I am unsure how to compute this by hand.

You are looking for $9^{-1} \pmod {13}$.  You can find that using the Euclidean algorithm:

$$13 - \color{red}{9} = \color{violet}4$$
and
$$\color{red}9 - 2\times \color{violet}4=1.$$
Therefore,

$$\color{red}9 - 2 \times {(13 - 9)} = 1$$

i.e.,
$$3\times 9 -  2 \times 13  = 1.$$

Therefore $3 \times 9 \equiv 1 \pmod {13}.$

Addendum to answer comment:

$$\color{green}{9^{-1}}\times\color{orange}{(4-10)}\equiv\color{green}3\times\color{orange}{-6}\equiv\color{green}3\times\color{orange}7=21\equiv8 \pmod {13}.$$

We have 23 +35x =4(mod13)
It gives 35x =-19 =7(mod13)
It further gives 9x =7(mod13)
Implies 90x=70 (mod13) gives -x=5(mod13) implies x=8(mod13)


IMO Shortlist 2010 : N1 - Finding the sequence Find the least positive integer $n$ for which there exists a set
  $\{s_1, s_2, \ldots , s_n\}$ consisting of $n$ distinct positive
  integers such that $$ \left( 1 - \frac{1}{s_1} \right) \left( 1 -
 \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) =
 \frac{51}{2010}.$$

Here is my approach, WLOG $s_1<s_2<...<s_n$ and $s_i\ge i+1$, thus $\prod (1-\frac{1}{s_i})\ge \prod\frac{i}{i+1}=\frac{1}{n+1}$ thus $n\ge 39$.
But I can't find a sequence of length 39 to finish off.

Please show me how did you think of a sequence. There is a similar problem , but my question is basically "what's the motivation for finding such a sequence?". So show your scratch work(Not Literally) or thought process if possible.

There's an entire thread on AoPS about this problem.


What is the name of this method that can easily find $\lim_{x \to 0^+} \frac{x^x-1}{\ln(x)+x-1}$? I can easily find $\lim_{x \to 0^+} \frac{x^x-1}{\ln(x)+x-1}$ using these steps. However, I don't completely remember the rules my professor told me about it, and I want to know what it's name is so I can look it up.

$$\lim_{x \to 0^+} \frac{x^x-1}{\ln(x)+x-1} $$
$$= \lim_{x \to 0^+} (x^x-1)\cdot\lim_{x \to 0^+}(\frac{1}{\ln(x)+x-1}) $$
$$=0 \cdot 0$$
$$=0$$

It's confusing how $\lim_{x \to 0^+}(\frac{1}{\ln(x)+x-1}) = 0$ which puzzles me since $\ln(0)$ is undefined.

Given that
$$\lim_{x\to x_0} f(x)= f_0$$
$$\lim_{x\to x_0} g(x) = g_0$$
Where $f_0,g_0\in\mathbb{C}$ we can say that
$$\lim_{x\to x_0} \left(f(x)\cdot g(x)\right)=f_0\cdot g_0$$
A proof can be found here: http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx

The rule is if $\lim_{x\to a} f(x) = k$ and $k$ is finite and $\lim_{x\to a} g(x) = m$ and $m$ is finite then $\lim_{x\to a} f(x)g(x) = km$ and $\lim_{x\to a} (f(x)+g(x))=m+k$ and if $m \ne 0$ then $\lim_{x\to a} \frac {f(x)}{g(x}) = \frac km$.

As to we $\lim_{x\to 0}\frac {1}{\ln x +x - 1} = 0$.  There is a rule that if $\lim_{x\to a} f(x) = \pm \infty$ then $\lim_{x\to a} \frac 1{f(x)} = 0$.

We can extend these rules to cases wher $\lim_{x\to a} f(x) = \infty$ but we must be careful we actually know what we are talking about or we might end up talking gibberish.

If $\lim_{x\to a}f(x) = \infty$ and $\lim_{x\to a} g(x) = k$ finite.  Then $\lim_{x\to a} (f(x) \pm g(x))=\infty$.

If $k > 0$ then $\lim_{x\to a} f(x)g(x) = \infty$ and $\lim_{x\to a}\frac {f(x)}{g(x)} = \infty$.  If $k < 0$ then $\lim_{x\to a} f(x)g(x) = -\infty$ and $\lim_{x\to a}\frac {f(x)}{g(x)} = -\infty$. If $k=0$ then $\lim_{x\to a}\frac {g(x)}{f(x)} = 0$ but $\lim_{x\to a} f(x)g(x)$ is indeterminate.

As to youre Note:  In calculating a $\lim_{x\to 0^+} \frac 1{\ln x + x-1}$ we have $\lim_{x\to 0^+} \ln x = -\infty$ so $\lim (\ln x + x - 1)= -\infty$ and $\lim_{x\to 0^+} \frac 1{\ln x + x-1}=0$.

As to your comment that $\ln 0$ is undefined... Of course it is! That's the entire reason we are taking limits.  Notice $0^0$ is also undefined.

Note when we take limits $x \to a$, we are considering cases where $x$ is near $a$.  We are NEVER considering the the case where $x = a$.

$f(a)$ could be undefined, could be something utterly different, or could be completely irrelevent.  We are looking at $f(x)$ when $x$ is near $a$.  So nobody cares about what $f(a)$ does.


How to find $\int_{0}^{1}\dfrac{\ln^2{x}\ln^2{(1-x)}}{2-x}dx$ How to find

  $$
I=\int_{0}^{1}{\ln^{2}\left(x\right)\ln^{2}\left(1 - x\right) \over 2 - x}
\,{\rm d}x
$$

My idea:

Let $x=1-t$, then

  $$
I
=\int_{0}^{1}{\ln^{2}\left(1 - x\right)\ln^{2}\left(x\right) \over 1 + x}\,{\rm d}x
$$

I can't proceed any further. I can't remember that Math SE has this similar problem already posted.

this integral problem is from This Euler Sums  :
$$\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}} \left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{{k^2}}}} } \right) =  - \frac{{23}}{{1440}}{\pi ^4}\ln 2 + \frac{1}{{18}}{\pi ^2}{\ln ^3}2 - \frac{1}{{18}}{\ln ^5}2 - 4L{i_4}\left( {\frac{1}{2}} \right)\ln 2 - 8L{i_5}\left( {\frac{1}{2}} \right) + \frac{1}{{12}}{\pi ^2}\zeta \left( 3 \right) + \frac{{27}}{4}\zeta \left( 5 \right)$$

when I deal this series,then we must solve this integral .

Thank you.


Find the sum of series $\sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!}$ I wanted to know how can I start to find the sum of the series:

$$\sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!}=\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}\cdots$$

I am having no clue.

Thanks.

$$\begin{align*}
\sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!}
&=\sum_{n=0}^\infty\frac{1}{(4n+1)\times(4n+2)\times(4n+3)\times(4n+4)}
\\&=\frac{1}{8}\sum_{n=0}^\infty\frac{1}{32 n^4+80 n^3+70 n^2+25 n+3}
\end{align*}$$

Now that we have replaced the nasty factorials with a nice, clean, rational function, we can use a more general procedure from here.

Also, if all you need to do is prove convergence, the comparison test is the simplest conclusive test.

If you want some additional stuff, here are the Wolfram Alpha queries for the equations:
$\sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!}$, $\frac{1}{8}\sum_{n=0}^\infty\frac{1}{32 n^4+80 n^3+70 n^2+25 n+3}$


Graph Theory: Is this a correct statement? I'm asked to prove this statement in my assignment:

Suppose that $T$ is a tree. Prove that degree of every vertex of $T$ is odd if and only if for each $e \in E(t)$, size of both components of $T-e$ are odd.

Is this correct? Shouldn't size of each component be even?

Edit: Seems size of each components refers to number of vertices and not edges, therefore it's correct :)

No, the problem seems right as worded. Notice that it's talking about node counts, not edge counts. To take an example, look at a tree on four vertices such that one root and the other three are the root's children. This has odd degree for every vertex. Deleting any one edge gives two connected components: one vertex by itself and the other three together.

Yes. For instance, look at the tree with two vertices and one edge.


Inverse Matrix in Python My understanding is that I can use Python to initialize my matrix and then apply an inverse function to find the solution.

You can use matlab but:

1.) you will get a numerical approximation of the inverse,

2.) you won't learn anything from that.

Rather, try doing what the question asks you - work out what $z_1, z_2, z_3$ are in terms of $b_1, b_2, b_3$ (by writing out the equations and rewriting them a bit) or use any other technique to calculate an inverse of $3\times3$ matrix.

I'm not sure I understand your notation:

  deltaz = [-1 1 0][z1] = [z2 - z1] = [b1] = b [0 -1 1] [z2] [z3 - z2] [b2] [0 0 -1] [z3] [0 - z3] [b3]

Are you attempting Python pseudocode or are you defining a matrix? If the latter, please use MathJax for a more readable format.


Do addition and multiplication have arity? Many books classify the standard four arithmetical functions of addition, subtraction, multiplication, and division as binary (in terms of arity).  But, "sigma" and "product" notation often writes just one symbol at the front, and indexes those symbols which seemingly makes expressions like $+(2, 3, 4)=9$ meaningful.  Of course, we can't do something similar for division and subtraction, since they don't associate, but does the $+$ symbol in the above expression qualify as the same type of expression as when someone writes $2+4=6$?  Do addition and multiplication qualify as functions which don't necessarily have a fixed arity, or do they actually have a fixed arity, and thus instances of sigma and product notation should get taken as abbreviation of expressions involving binary functions?  Or is the above question merely a matter of perspective?  Do we get into any logical difficulties if we regard addition and multiplication as $n$-ary functions, or can we only avoid such difficulties if we regard addition and multiplication as binary?

Given any binary operation $(x,y) \mapsto F(x,y)$ on a set $X$ you can construct operations with "higher arity", in general in many different ways, e.g.

$g_{3,1}(x,y,z) = F(F(x,y),z)$,
$g_{3,2}(x,y,z) = F(x,F(y,z))$,
$g_{3,3}(x,y,z) = F(y,F(x,z))$,
$g_{3,4}(x,y,z) = F(x,F(y,y))$,

and so forth.  If the original binary operation is commutative and/or associative, then some of these induced operations will coincide.  In general, tracking down compatibilities between these sort of induced, higher arity operations is something that shows up a lot in category theory (and, especially, higher category theory).

So in particular you can absolutely define three-fold (and higher) subtraction operations; you'll just have more than one of them.  (Note also that in most number systems division is not a binary operation, since $a \div 0$ is not defined.)

That's the mathematics of it.  The rest of your question seems to be asking about "rules".  That depends on further context, which you have not provided.   For much of "mainstream" mathematics, the arity of an operation is not of explicit concern or interest.  If you are studying universal algebra, or model theory, or other branches of logic, or parts of computer science, it might matter, but there is no uniform set of rules.

Addition and multiplication have algebraic properties like commutativity, associativity, x x - +=0, etc.  Some of these properties only apply to binary operations, not operations of variable arity.  Regarding them as having variable arity would thus lead to a logical problem.  Therefore, addition and multiplication have a fixed arity of 2.


Counterexample? Other Alternative proofs? There exists real numbers $a$, and all real numbers $b$, such that $a-b^2<0$
I know you can use a counterexample of $a=5$, and $b=0$, but are there alternative proofs?
What about if there exists all natural numbers $a$?

Whenever $a<0$, then the statement is true. It is asking for a value $a$ that exists to prove this statement right for the very first part.
since $-b^2\leq0$, with $a<0$, we can always get a value that is lower than $0$.
For example let $a=-0.1$ and $b=0$, then for the first statement is true.
There exists an $a$, for all $b$s.
As for the second statement, it's obviously false as it asks for $a\in\mathbb{N}$, and we just proved that if $a<-b^2$, then the statement is true. But since it is for Natural numbers, $a>-b^2$, and thus it is false.


Two ships leaving a port at different times and different speeds. When do they meet? Can someone please show me the working out to this word problem I have the answer but have no clue how to do the working out.

At noon ship A leaves port steaming at 8 knots 2 hours later ship B leaves same port on the same course steaming at 12 knots.

How far from port will ship B overtake ship A? Answer(48 miles apart)

Show me the working out please.

First A goes 16 miles. Then B leaves, it takes some time to overtake A, let's say time $t$. In this time A travels $8t$ miles and B $12t$ miles. Given these we have that

$$12t = 16 + 8t$$
that is $t=4$. Hence the distance is $12t = 48$.

$S_1 - S_2 = S_3$ which is how many knots per hour so $S_3=t$.

$S_1 \times t = D$

$S_2 \times t = D$

$12 - 8 = 4$

$4 \times 12 = 48$ nm

$4 \times 8 = 48$ nm


show that $x^2e^x-\ln{x}-1>0$ For $x>0$ show that
  $$x^2e^x-\ln{x}>1$$

x^2e^x-lnx-1

if $x\ge 1$,we have
$$x^2e^x>2x^2>x\ge\ln{x}+1$$
But  for $0<x<1$ then It is hard.so How prove it?

Rewrite the inequality to be proved as

$$\left(1+u\over2\right)^2e^{(1+u)/2}-\ln\left(1+u\over2\right)-1\gt0$$

for $-1\lt u\lt1$.  This can be rewritten as

$$(1+u)^2\sqrt ee^{u/2}-4\ln(1+u)+4\ln2-4\gt0$$

Now $e^{u/2}\ge1+{u\over2}$, so

$$(1+u)^2\sqrt ee^{u/2}\ge(1+2u)\sqrt e\left(1+{1\over2}u\right)\ge\left(1+{5\over2}u\right)\sqrt e$$

Also, $\ln(1+u)\le u$, so

$$-4\ln(1+u)\ge-4u$$

Putting these together, we have, for $|u|\le1$,

$$\begin{align}
(1+u)^2\sqrt ee^{u/2}-4\ln(1+u)+4\ln2-4
&\ge\sqrt e+4\ln2-4+\left({5\over2}\sqrt e-4\right)u\\
&\ge\sqrt e+4\ln2-4-\left|{5\over2}\sqrt e-4\right|
\end{align}$$

It remains to compute $\sqrt e+4\ln2-4\approx0.4213$ while ${5\over2}\sqrt e-4\approx0.1218$.

There might be some slicker way to do this that avoids the messy numerics at the end.  I'd like to see it.

Added later:  Oh, here's a way to avoid the messy numerics.  Note first that $e\gt2.56=1.6^2=(8/5)^2$, so ${5\over2}\sqrt e-4$ is positive.  This simplifies things to showing $4\ln2\ge{3\over2}\sqrt e$.  If we now allow ourselves to know that $\ln2\gt{2\over3}$, it suffices to show that ${16\over9}\gt\sqrt e$, or ${256\over81}\gt e$, which is clear, since ${256\over81}\gt3\gt e$.

If you want a quick proof that $\ln2\gt{2\over3}$, note that that inequality is equivalent to $8\gt e^2$, which follows from $8\gt{196\over25}=\left(14\over5\right)^2=2.8^2\gt e^2$.

And just to leave no numerical stone unturned, here's why we can confidently say that ${14\over5}\gt e\gt{64\over25}$:

$$e\gt1+1+{1\over2}+{1\over6}=2+{2\over3}\gt2+{14\over25}={64\over25}$$

since $2\cdot25\gt3\cdot14$, and

$$e^{-1}\gt1-1+{1\over2}-{1\over6}+{1\over24}-{1\over120}={11\over30}\gt{5\over14}$$

since $11\cdot14\gt5\cdot30$.

There's an ad-hoc solution at least going something like this:

You can use taylor expansion to show that (or the convexity of the function). Consider the function $f(x) = x^2e^x-\ln x - 1$ and you have:

$$f'(x) = (x^2+2x)e^x - 1/x$$
$$f''(x) = (x^2+4x+4)e^x + 1/x^2 = (x+2)^2e^x + 1/x^2$$

We see immediately that $f''(x)\ge 0$. This means that the expansion around any point we will have:

$$f(a+h) = f(a) + f'(a)h + f''(\xi)h^2/2 \ge f(a) + f'(a)h$$

This means that it will be above every of it's tangent. Now one only needs to pick a tangent sufficiently close to the right of it's minimum such that either of them are positive which will prove that $f(x)$ is positive (since it's above both tangents).

The rest is ugly computations. You have that $f(1/2) > 1/10$ and that $0<f'(1/2)<1/10$ (you can find this by Taylor expansion - or if you're lazy by using calculator). This means that we have:

$$f(x) = f(1/2 + x-1/2) \ge f(1/2) + f'(1/2)(x-1/2)$$

Which means:

$$f(x)\begin{cases}
\ge 1/10 + 1/10 (x-1/2) = (1+2x)/10 \ge 0 & \text{when } x\le 1/2
\\ \ge 1/10 \ge 0 & \text{when } x \ge 1/2
\end{cases}$$


Is there much theory developed for analytic functions of quaternions or of octonions? The quaternions are associative, so nonnegative integer powers of quaternions are well-defined, and one can consider analytic functions on $\mathbb{H}$ (functions that are given locally by power series).  The octonions are not associative, but from what I have read, I think they are "associative enough" that nonnegative integer powers of octonions are defined unambiguously (please correct me if I'm wrong), and one can define analytic functions of octonions.

My question is, how much of this have people actually done?  Has a lot of theory been developed about analytic functions on $\mathbb{H}$ and $\mathbb{O}$?  I haven't heard of very much. If not, is it because their properties are very similar to those of analytic functions on $\mathbb{C}$?

I know that a lot of people in engineering, kinematics, and computer graphics use quaternions (perhaps more than mathematicians do) so it is not the case that no one is interested in quaternions.  I have heard that some physicists are interested in octonions, but I don't know much about that.

Here's some information on quaternion analysis. The list of references is probably the most useful part.

http://www.zipcon.net/~swhite/docs/math/quaternions/analysis.html

I am familiar with the theory of holomorphicc functions on H totally similar to the theory of holomorphicc functions on C. See http://vixra.org/abs/1609.0006
Bilberry


Is there any way to find a number N if it's Carmichael function is given. I know to find the Carmichael function [ C() ] of a given no. But I want to know if there is any method or shortcut to find a number N if it's C() is given.

PARI/GP examples:

n-Carmichael <=> B(n)=eulerphi(n),

where B(n)=PRODUCT(p: p-1|n-1, where p|n and p-Prime)

It's analog for:
 n-Prime<=> n-1=eulerphi(n)

Your method:
  n-Carmichael <=> C(n)|n-1

where:
  C(n)=znstar(n)[2][1]


Evaluate $\int_{0}^\infty \sin\left(\frac{1}{x}\right)\sin\left(x^2\right)\,dx$ I don't really have much a reason for giving this question other than curiosity. I am looking to evaluate

$$\int_{0}^\infty \sin\left(\frac{1}{x}\right)\sin\left(x^2\right)\,dx$$

Using the complex definition of $\sin(x)$ doesn't seem to work, nor can I give a reasonable differentiating under the integral sign. Also, it doesn't seem Wolfram Alpha has a closed answer, but I would be curious to see if there is one.

This is too long for a comment.

Just out of curiosity, I used a CAS to see what could be the result of
$$I_{p,q}=\int_{0}^\infty \sin(\frac{1}{x^p})\sin(x^q)dx$$ and observed (for limited ranges of $p$ and $q$) that, if $p\neq q$, the result is either a Meijer G-function (as Phoenix already commented) or a nasty linear combination of hypergoemetric functions.

To my surprise, the very first cases where $p=q$ leads to quite simple results
$$I_{1,1}=\frac{\pi  }{2}J_1(2)$$
$$I_{2,2}=\frac{1}{4} \sqrt{\frac{\pi }{2}} \left(\frac{1}{e^2}+\sin (2)-\cos (2)\right)$$
$$I_{3,3}=\frac{1}{6} \left(\pi  \left(J_{\frac{1}{3}}(2)-J_{-\frac{1}{3}}(2)\right)+\sqrt{3}
   K_{\frac{1}{3}}(2)\right)$$
$$I_{4,4}=-\frac{\sqrt{2+\sqrt{2}}}{16}  \left(\sqrt{2} \pi
   \left(J_{-\frac{1}{4}}(2)-J_{\frac{1}{4}}(2)\right)-2 K_{\frac{1}{4}}(2)\right)$$ where appear Bessel functions.

$\int_0^\infty\sin\dfrac{1}{x}\sin x^2~dx$

$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin x^2}{(2n+1)!x^{2n+1}}~dx$

$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin x}{(2n+1)!(\sqrt x)^{2n+1}}~d(\sqrt x)$

$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin x}{2(2n+1)!x^{n+1}}~dx$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(n+1)!\sin\dfrac{n\pi}{2}}{2(2n+1)!}$ (according to http://eqworld.ipmnet.ru/en/auxiliary/inttrans/FourSin2.pdf)

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(n+1)\sin\dfrac{n\pi}{2}}{2^{2n+1}\left(\dfrac{3}{2}\right)_n}$ (according to http://mathworld.wolfram.com/PochhammerSymbol.html)

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(2n+2)}{2^{4n+3}\left(\dfrac{3}{2}\right)_{2n+1}}$


What do I compare the series $\sum_{i=2}^n \frac{i^2+1}{i^3-1}$ to? I have tried everything I can think of. I know that the answer is divergent and that you can use a comparison test, I just don't know what to compare it to.

$$\sum_{i=2}^n \frac{i^2+1}{i^3-1}$$

I would love any help possible.

Making the numerator smaller and the denominator larger makes the terms smaller. If it still diverges after this change, then the original series must also diverge.

With that in mind, subtract 1 from the numerators and add 1 to the denominators.

How to figure this out: You have a numerator where the dominating term is $i^2$, and a denominator where the dominating term is $i^3$. So it would be nice if we could make those the only terms.

This time it worked out in that making this change made the fractions smaller, and we wanted to show divergence. But what if the $+$ and the $-$ had been the other way? Then you note that $i^2-1\geq \frac12i^2$ and $i^3+1\leq 2i^3$. It still works more or less the same way, but the transformation isn't quite as simple.

If you perform the long division, you should get
$$\frac{i^2+1}{i^3-1}=\frac 1{i}+\frac 1{i^2}+\frac 1{i^4}+\frac 1{i^6}+\cdots >\frac 1{i} $$


Order of convergence limit Let $(x_n)$ be a real valued sequence converging to $\overline{x}$, with order of convergence $k \in \mathbb{N}$. This means the limit $$ \lim_{n \to \infty}\frac{|x_{n+1} - \overline{x}|}{{|x_{n} - \overline{x}|}^k}$$
exists. What I have to prove is that this limit is also equal to the limit $$ \lim_{n \to \infty}\frac{|x_{n+1} - x_n|}{{|x_{n} - x_{n-1}|}^k}$$
that is, to prove $$ \lim_{n \to \infty}\frac{|x_{n+1} - \overline{x}|}{{|x_{n} - \overline{x}|}^k} = \lim_{n \to \infty}\frac{|x_{n+1} - x_n|}{{|x_{n} - x_{n-1}|}^k} $$
I have been trying algebraic manipulations for hours now, but to no avail. At first, I though this problem was going to use the fact that Cauchy sequences converge in $\mathbb{R}$, but this didn’t get me anywhere, so I then decided to try using algebra and absolute value inequalities (like the triangle inequality), but I couldn’t get the correct denominator to appear.
So I set $k=1$ to see if it would make it easier, but the problem wasn’t in the value of $k$ itself, but rather in the product that would show up when trying to go from one side to the other.
I have tried going from the left hand side to the right one, and vice-versa, and neither of these approaches got me anywhere...
Could anyone help me? What am I missing?
Thanks in advance!

\begin{array}{l}
If\ \lim _{n\rightarrow \infty } x_{n} =L\ \ for\ some\ L\in R,\ then\ we\ let\ g( n) =x_{n} -L\ hence\ \lim _{n\rightarrow \infty } \ g( n) =0\\
\\
\\
If\ \lim _{n\rightarrow \infty } x_{n} =L\ \ for\ some\ L\in R,\\
then\ we\ can\ conclude\ that\ x_{n} -L=g( n) \ for\ some\ g\ where\ \lim _{n\rightarrow \infty } g( n) =0\ \\
the\ first\ limit,\lim _{n\rightarrow \infty }\frac{|x_{n+1} -L|}{|x_{n} -L|^{k}} =A\ for\ some\ A\in R\\
then\ becomes\lim _{n\rightarrow \infty }\frac{|g( n+1) |}{|g( n) |^{k}} =A\ \\
we\ conclude\ that\ \frac{|g( n+1) |}{|g( n) |^{k}} -A=f( n) \ for\ some\ f\ where\ \lim _{n\rightarrow \infty } f( n) =0\ \\
hence\ g( n+1) =[ g( n)]^{k} *[ A+f( n)] *( -1)^{S( n)} \ where\ S\ is\ some\ function\ N\rightarrow N\\
We\ need\ the\ ( -1)^{S( n)} \ function\ because\ we\ dont\ know\ how\ the\ signs\ of\ these\ two\\
sides\ relate\ to\ one\ another\ but\ we\ know\ they\ are\ equal\ in\ magintude.\\
\\
So\ now\ to\ look\ at\ the\ second\ limit:\\
LIMIT=\lim _{n\rightarrow \infty }\frac{|x_{n+1} -x_{n} |}{|x_{n} -x_{n-1} |^{k}} =\lim _{n\rightarrow \infty }\frac{|g( n+1) -g( n) |}{|g( n) -g( n-1) |^{k}}\\
=\lim _{n\rightarrow \infty }\frac{|[ g( n)]^{k} *[ A+f( n)] *( -1)^{S( n)} -g( n) |}{|[ g( n-1)]^{k} *[ A+f( n-1)] *( -1)^{S( n-1)} -g( n-1) |^{k}}\\
\\
factoring\ out\ we\ get:\\
=\lim _{n\rightarrow \infty }\frac{|[ g( n)] |*|\ [ g( n)]^{k-1} *[ A+f( n)] *( -1)^{S( n)} -1\ |}{|[ g( n-1)]^{k} |*|\ [ g( n-1)]^{k-1} *[ A+f( n-1)] *( -1)^{S( n-1)} -1|^{k}}\\
\\
but\ we\ know\ \lim _{n\rightarrow \infty }\frac{|g( n+1) |}{|g( n) |^{k}} =A\ \ hence\ \lim _{n\rightarrow \infty }\frac{|g( n) |}{|g( n-1) |^{k}} =A\ and\ so\\
\\
LIMIT=A*\lim _{n\rightarrow \infty }\frac{|\ [ g( n)]^{k-1} *[ A+f( n)] *( -1)^{S( n)} -1\ |}{|\ [ g( n-1)]^{k-1} *[ A+f( n-1)] *( -1)^{S( n-1)} -1|^{k}}\\
\\
Now\ if\ k >1:\ we\ notice\ that\lim _{n\rightarrow \infty }[ g( n)]^{k-1} =0\ \\
hence\ both\lim _{n\rightarrow \infty } |\ [ g( n)]^{k-1} *[ A+f( n)] *( -1)^{S( n)} -1\ |=|-1|=1\\
and\ \lim _{n\rightarrow \infty } |\ [ g( n-1)]^{k-1} *[ A+f( n-1)] *( -1)^{S( n-1)} -1|^{k} =1\\
and\ so\ LIMT=A*\frac{1}{1} =A\ so\ we\ are\ done.\\
\\
If\ k=1:\\
Then\ it\ is\ false:\\
Take\\
x_{n} =g( n) =2^{-n} *( -1)^{S( n)}\\
where\\
S( n) =\ repeating\ sequence\ of\ digits:\\
\{0,0,1\} \ \ \ \ \\
which\ repeats\ \ n\bmod 3\\
\lim _{n\rightarrow \infty }\frac{|g( n+1) |}{|g( n) |} =1/2\\
\\
Limit=\lim _{n\rightarrow \infty }\frac{|g( n+1) -g( n) |}{|g( n) -g( n-1) |} =\lim _{n\rightarrow \infty }\frac{|2^{-n-1} *( -1)^{S( n+1)} -2^{-n} *( -1)^{S( n)} |}{|2^{-n} *( -1)^{S( n)} -2^{-n+1} *( -1)^{S( n-1)} |}\\
\\
=\lim _{n\rightarrow \infty }\frac{|( -1)^{S( n+1)} -2( -1)^{S( n)} |}{|4( -1)^{S( n-1)} -2( -1)^{S( n)} |}\\
and\ this\ alternates\ between\ \frac{3}{2} ,\frac{1}{2} ,\frac{1}{6} \ and\ does\ not\ even\ appraoch\ a\ limit....\\
Unless\ you\ ask\ that\ the\ limit\ \lim _{n\rightarrow \infty }\frac{|x_{n+1} -x_{n} |}{|x_{n} -x_{n-1} |^{k}} MUST\ EXIST...\\
\\
\end{array}


Showing that $\sum_{n=3}^\infty\frac1{n(\log n)(\log\log n)}$ diverges I know that the series diverge, I'm just having hard time showing it.

$$\sum_{n=3}^\infty\frac1{n(\log n)(\log\log n)}$$

Thanks in advance

The Cauchy Condensation Test will do it.

If you have not seen it, here is what it says. Suppose that $f(n)$ is positive nonincreasing. Then $\sum_{a}^\infty f(n)$ converges if and only if $\sum_a^\infty 2^nf(2^n)$ converges.

In our case, we have $f(n)=\frac{1}{n\log n\log\log n}$. Thus $2^n f(2^n)=\frac{1}{(n\log 2)(\log n+\log\log 2)}$. The divergence of $\sum 2^n f(2^n)$ now follows from the possibly familiar fact that $\sum_2^\infty \frac{1}{n\log n}$ diverges. If that fact is not familiar, we can use Cauchy Condensation again.

Hint:  Use the comparison test with 1/log(n)


Conjugate locus and local diffeomorphism problem in do Carmo's book. I'm reading DoCarmo's book, Riemannian Geometry and i don't understand something. At page 149 first lemma

My question is why $\text{exp}_p$ is local diffeomorphism?

For $p\in M$ identify $T_{0}T_{p}M$ with $T_{p}M$  then :

$d(\exp_{p})_{0}(X_{p})=\frac{d}{dt}_{t=0}(\exp_{p}(tX_{p}))=X_{p}$  , $X_{p}\in T_{p}M$ .

the rest concluded by inverse function theorem.


Subsequence Convergence If $\{f_{n}\}_{n\geq 1}$ is a sequence of continuous functions on $\mathbb R$, with $|f_{n}(x)|\leq 1$ for all $x\in \mathbb{R}$, and all $n\geq 1$. Does there exist a subsequence which converges uniformly (or pointwise) to some continuous function $f$?

As I know the Arzelà–Ascoli theorem works for closed intervals $[a,b]$, I don't know if there is something in case of $\mathbb{R}$?

EDIT: If this assumption help we can consider it: the sequence $\{f'_{n}\}$ is uniformly bounded on $\mathbb R$.

Take $f_n(x) = x^n 1_{(0,1)}(x)+1_{[1,\infty)}(x)$. Let $x_n = \frac{1}{\sqrt[n]{2}}$, and note that $x_n \to 1$. Also note that $f_n(x_n) = \frac{1}{2}$, $f_n(1) = 1$, and for a fixed $x$, $f_n(x)$ is non-increasing, so for $m\geq n$, $f_m(x_n) \leq \frac{1}{2}$.

Consequently no subsequence can converge to a continuous function. To get a contradiction, suppose $f_n(x) \to f(x)$, where $f$ is continuous. We must have $f(1) = 1$, so choose $\delta>0$ such that if $|x-1|<\delta$, then $|f(x)-1|<\frac{1}{4}$. Now choose $n$ such that $|x_n-1| < \delta$. Then if $m \geq n$ we have $f_m(x_n) \leq \frac{1}{2}$, hence $f(x_n) \leq \frac{1}{2}$, which is a contradiction.

Just noticed that there was an update to the question.

Answer to modified question:

If the derivatives are uniformly bounded, then the functions are uniformly Lipschitz, hence equicontinous. Since the functions themselves are uniformly bounded, we can apply Arzela-Ascoli on nested closed intervals to get the desired result:

Let $I_1 = [-1,1]$, and apply Arzela-Ascoli to get a subsequence (ie, an infinite subset) $N_1 \subset \mathbb{N}$ along which $\{f_n\}_{n \in N_1}$ converges uniformly to $f:I_1 \to \mathbb{R}$. (Since the convergence is uniform, $f$ is continuous.)

Now suppose we have a subsequence $N_k \subset \mathbb{N}$, an interval $I_k=[-k,k]$, and a continuous limit function $f:I_k \to \mathbb{R}$. Consider the functions $\{f_n\}_{n \in N_k}$ on the interval $I_{k+1} = [-(k+1),k+1]$. Again, by Arzela-Ascoli, there is a subsequence $\mathbb{N}_{k+1} \subset \mathbb{N}_k$ along which the functions converge uniformly to a continuous function $\phi: I_{k+1} \to \mathbb{R}$. Since the functions converge to $f$ on the subsequence $\mathbb{N}_k$, it follows that $\phi(x) = f(x)$, $\forall x \in I_k$. Hence we may abuse notation slightly by dropping the symbol $\phi$ and using $f$ to denote the new limit function, ie, $f:I_{k+1} \to \mathbb{R}$.

Continuing this way, we define a continuous function $f:\mathbb{R} \to \mathbb{R}$. If $x \in \mathbb{R}$, then for sufficiently large $k$, we have $x \in I_k$, and $\{f_n(x)\}_{n \in N_k}$ converges to $f(x)$.

The convergence is uniform on any bounded set (since a bounded set will be contained in $I_k$ for some $k$). The convergence need not be uniform on all of $\mathbb{R}$, however. For example, take $f_n(x) = 1_{(-2 n \pi, 2 n \pi)}(x) + \cos(x) 1_{(-\infty,-2 n \pi] \cup [2 n \pi, \infty)}(x)$. Both $f_n$ and $f_n'$ are uniformly bounded. Clearly, $f_n$ converges pointwise to $f(x) = 1$, but $\max_{x \in \mathbb{R}} |f_n(x)-f(x)| = 2$, $\forall n$, and so the convergence in not uniform.

No, it doesn't exist. Something very bad could happen in the infinity.


Standard Basis of $SU(2)$--where does the 1/2 come from? The most common matrix representation of $SU(2)$ is given by
$$
\begin{pmatrix}
a & b\\
b^* & -a^*\\
\end{pmatrix}
$$
where $a,b\in\mathbb{C}$. If we denote real components by the subscript $r$, imaginary by $i$, one can clearly write this as

$$
\begin{pmatrix}
a & b\\
-b^* & a^*\\
\end{pmatrix}
 =
a_r\begin{pmatrix}
1 & 0\\
0 & 1\\
\end{pmatrix}
+b_r\begin{pmatrix}
0 & 1 \\
-1 & 0 \\
\end{pmatrix}
+b_i\begin{pmatrix}
0 & i \\
i & 0 \\
\end{pmatrix}
+a_i\begin{pmatrix}
i & 0 \\
0 & -i \\
\end{pmatrix},
$$
so the Lie algebra should be obtained by discarding the coefficient of the identity matrix and identifying the remaining traceless matrices as the generators of $\mathfrak{su}(2)$. We quickly realize these are $i\sigma_a$, where $a = 1,2,3$ denote the Pauli spin matrices. Many books however define the standard basis of $\mathfrak{su}(2)$ as $(i/2)\sigma_a$. While this would normally be simply a trivial convention, R. Gilmore, on page 107 of "Lie Groups, Physics, and Geometry", seems to use this to show the $SU(2)\rightarrow SO(3)$ cover is 2:1 by splitting $\mathfrak{su}(2)$ into the form $\mathfrak{su}(2)\ni X = S_1 + (ia_i/2)\sigma_3$ where $S_1 =  (ib_i/2)\sigma_1 + (ib_r/2)\sigma_2$ and exponentiating $S_1$ and $(ia_i/2)\sigma_3$ individually to give

$$SU(2) \ni U = \exp(S_1)\times\begin{pmatrix}
e^{ia_i/2} & 0\\
0 & e^{-ia_i/2}\\
\end{pmatrix}.$$

Using a similar splitting of $\mathfrak{so}(3)$ into two components (an $SO(3)/SO(2)$ quotient and $SO(2)$) which are individually exponentiated, he concludes the quotients $SO(3)/SO(2)$ and $SU(2)/U(1)$ are both topologically $S^2$, thus bijective, and the thus the cover must be 2:1 because putting $\theta \mapsto \theta + 2\pi$ is a complete rotation in $SO(2)$, while $a_i \mapsto a_i + 4\pi$ is a complete rotation in the given representation of $U(1)$. I've been wanting to emulate this argument but can't do so without the factor of $1/2$. My questions are thus as follows:

(1) where does the $1/2$ come from? Is it just convention or something else?

(2) Is there any easy way to argue the cover is $2:1$ without using the $1/2$?

(3) I thought that generally the whole Lie algebra had to be exponentiated to map back onto the Lie group. How does separating the Lie algebra and exponentiating the terms individually make sense in the context of the BCH formula? Does this have to do with irreducible representations?

Evidently this is due to the basis with the $1/2$ giving the "fundamental representation" of $\mathfrak{su}(2)$, e.g. according to

https://en.wikipedia.org/wiki/Pauli_matrices

and

https://en.wikipedia.org/wiki/Representation_theory_of_SU(2),

But I don't know enough about representation theory to make a good argument about how that guarantees the covering map is 2:1 for $SU(2) \rightarrow SO(3)$ regardless of the representation.


Colouring of $\mathbb{N}$ that avoids all non-constant infinite arithmetic progressions? Can you color every positive integer either black or white such that there are no entirely white or entirely black non-constant infinite arithmetic progressions?

How about switching color every power of 2?

1,4,5,6,7,16,...,31,64,...

2,3,8,9,10,11,12,13,14,15,32,...

Any attempt at an AP would eventually get cut off.

Color the first integer white, the next two integers black, the next three integers white, etc. We will call these sections of successive integers with the same colors, "blocks". Thus we have successively blocks of length 1, 2, 3, 4, etc.

Now consider an arithmetic progression $a_n = a_0 + nd$ for some positive integers $a_0, d$ ($n = 0, 1, 2, \dots$). Let $k$ be a positive integer such that the block of length $kd$ comes after $a_0$. There is at least one element from the progression that lies in this block. Suppose that $a_n$ is the last element from the progression that lies in this block; then $a_{n+1}$ lies in the next block, of length $kd+1$, and therefore has the opposite color. Thus any arithmetic progression contains successive elements $a_n, a_{n+1}$ of opposite color, as we needed to prove.

If I Consider the irrational number $z=\sum_{i=1}^{\infty} 10^{-i!}=0,110001000...$ as a representation of natural numbers, where 1 means white and 0 means black. Let $\{a_n\}$ be an arithmetic progression, where $a_{n+1}-a_n=k$: if a similar series existed, it would mean that, in the decimal represntation of z there is the same digit repeating each k digits, but this is impossible because z is irrational. Therefore, this progression doesn't exist.


How to solve quadratic recurrence relation How to solve recurrence relation of the following form:

$U_n = a \times U_{n-1}^2 + b \times U_{n-1} + c$

where: $-1 < a < 0$ , $b = 1 - a$ , $c > 0$

Edit

I found here more cases where a quadratic recurrence is solvable.

Hint:

According to https://en.wikipedia.org/wiki/Iterated_function#Examples,

$U_n$ has easier forms when $c=\dfrac{(1-a)^2-2(1-a)}{4a}$ and $c=\dfrac{(1-a)^2-2(1-a)-8}{4a}$ .

With a few exceptions, general solutions of quadratic (or higher) recurrences can't be expressed in closed form.  This one is not an exception.


High School Math: Lines and Angles I have a simple high school math questions about intersecting lines. However, I was unable to answer this question.

I thought the answer should be "D" since I really don't see the questions are giving the answer. Please help.

Since the lines meet to the right,

$$2x+3x>180^{\circ}\implies{}x>36^{\circ}.$$

Hence Quantity $A$>Quantity $B$, which is option A.

That is like two lines: $3x+2x=5x$. There are two angels on the left side. Just look only at them. Do not think about left angels. The angel $3x$ and the angel $2x$ are more than $180$ degrees.


What does it mean to say a function is differentiable with respect to lebesgue measure? What does it mean to say a function is differentiable with respect to the lebesgue measure almost everywhere. A definition would be helpfull.

Do I need to learn about the Radon Nikodym derivative to understand this.

Thanks

After some thought and since I didn't really get any satisfactory answers. This is what I make of the statement.

The definition of differentiation in the statement is taken in the usual sense (the standard defintion from calculus, for some reason I thought the statement was referring to some new definition of differentiation!). So the statemtnt simply says, The set of points at which the function is not differentiable is a null set w.r.t. the lebesgue measure.

So the statement would have been clearer I think if the author had written

$f$ is differentiable almost everywhere with respect to the lebesgue measure ... I think thats right!!

So no new definition of differentiation, phew! just some confusing terminology.

I wonder what part of your question is not covered by this page.


Show that lim inf Bn and lim sup Bn equals to a null set Suppose that ${B_n: n \geq 1}$ is a sequence of disjoint set.

Show that

$$\begin{align}\limsup_{n\rightarrow \infty}B_n &= \emptyset \text{ and}\\
\liminf_{n \rightarrow \infty}B_n&= \emptyset \end{align}$$

Where $$\begin{align}\limsup_{n\rightarrow \infty}B_n &= \{x \mid  x\in B_n \text{ for infinitely many } n\} \text{ and}\\
\liminf_{n \rightarrow \infty}B_n&= \{x \mid x \notin B_n \text{ for at most finitely many } n\}. \end{align}$$

Can someone help me to solve this?
Since it's a disjoint set, means there is no intersection in the sequence.  So the union of intersection of Bn will be empty set and that's why lim inf  Bn is and empty set too?
Please help me, thanks..

As a given element $x$ cannot be in more than one $B_n$, in particular it can't be in infinitely many, giving both results as the lim sup requires $B_n$ to be in infinitely many, and the lim inf is even stronger requiring it to be in all but finitely many.

Sorry, that was expressed in a confusing way. Start again!

As you know already, the default ordering for sequences of sets is inclusion. So in your case $B_{n+1}\subset B_n$.

The problem is the definition of lim sup and lim inf for sequences of sets.

lim sup is defined as the set of all points which belong to infinitely many $B_n$, and lim inf is defined as the set of all points which belong to all but finitely many $B_n$.

Since $k$ belongs to just $k$ of the $B_n$, these two sets are both empty.


Proving that $\lim\limits_{(x, y)\to (0, 0)} \frac{x^3 y^3}{(x^4+y^6)\sqrt{x^2+y^2}}$ doesn't exist I want to prove that $\displaystyle \lim_{(x, y)\to (0, 0)} \frac{x^3 y^3}{(x^4+y^6)\sqrt{x^2+y^2}}$ doesn't exist
I denoted $f(x, y):=\frac{x^3 y^3}{(x^4+y^6)\sqrt{x^2+y^2}}$. I tried looking at $f(\frac{1}{n}, \frac{1}{n})$ and other similar sequences, but no result. No matter what sequence I chose I got that the limit is $0$. However, WA says that this doesn't exist. What sequence should I pick?

As Maxim pointed out the limit actually exists and is equal to 0. I give an alternative proof for that now. It suffices to show, that for |x|,|y| sufficiently small there holds
$$\left| \frac{x^3y^3}{(x^4+y^6)\sqrt{x^2+y^2}}\right| \le \sqrt{2}\max(\sqrt{|y|},|x|^{\frac{1}{3}}).$$
This we see at follows. First by Cauchy-Schwart and the triangle inequality there holds
$$\frac{1}{\sqrt{2}}(|x|+|y|)\le \sqrt{x^2+y^2}\le |x|+|y|.$$
We assume in the following w.o.l.g. that $1>>x,y>0$. Then,
$$\frac{x^3y^3}{(x^4+y^6)\sqrt{x^2+y^2}}\le \sqrt{2} \frac{x^3y^3}{(x^4+y^6)(x+y)}.$$
First assume, that $x\ge y$. Then
$$\frac{x^3y^3}{(x^4+y^6)\sqrt{x^2+y^2}}\le \sqrt{2} \frac{x^3y^3}{(x^4+y^6)(x+y)}\le \sqrt{2}\frac{x^2y^3}{x^4} =\le \sqrt{2} \frac{y^3}{x^2}\le \sqrt{2} x$$
and the above statement is fulfilled. We now assume in the following, that $x\le y$. Then
$$\frac{x^3y^3}{(x^4+y^6)\sqrt{x^2+y^2}}\le \sqrt{2} \frac{x^3y^2}{x^4+y^6}.$$
Now we consider the two cases $x^4\le y^6$ and $x^4\ge y^6$. In the first case,
$$\frac{x^3y^2}{x^4+y^6} \le \frac{x^3y^2}{y^6}\le \frac{x^3}{y^4} \le \frac{y^\frac{3*6}{4}}{y^4} =  \sqrt{y}.$$
In the remaining case,
$$\frac{x^3y^2}{x^4+y^6} \le \frac{x^3y^2}{x^4} = \frac{y^2}{x}\le x^\frac{1}{3}.$$
Therefore, the statement is proved.
The following reasoning does not yield the statement, as Maxim pointed out:
Set $y=y(x)=x^q$ with $q>0$. You can try to tune $q$, such that the numerator dominates the denominator. By the above, $q$ has to be smaller than 1. For the denominator there holds
$$(x^4+x^{6q}) \sqrt{x^2+x^{2q}} = x^{\min(4,6q)}(1+O(1)) x^{min(1,q)} (1+O(1))$$
for $x\to0$. E.g., for $q=4/6$ we have
$$x^{3+3q} = x^5 > x^{28/6}(1+o(1)) =  x^{7q}(1+o(1)).$$
Thus for that choice, the sequence will diverge (This last deduction is wrong).

In this type of exercises of "non-existence of limits" you might prove it by 'Reductio ad absurdum':

You suppose that the limit exists. Now you are going to calculate its value.
In order to calculate it, you aproach the point $(0,0)$ in the $\mathbb{R}^2$ plane by one way (by $x=y,\ x=y^2,\ x=\frac{1}{n}\ y=\frac{1}{m}$, whatever) and it comes that its value is anything, igamine $1$.
Now, you take another aproach, different than the previous onne. The limit also exists, but its value its something different, imagine $0$.
So, the limit 'exists' and have two values. By the uniqness of the limit in $\mathbb{R}^n$, this can't happen, so it is a contradiction. So, your initial asumption "The limit exists" is false.

Conclude that the limit doesn't exists.


How to solve Cauchy problem if equation is not linear and there is no $t$ argument? Given the ODE $x'=3x^{2/3}$ with the initial condition $x(0)=0$, I tried to use an integration multiplicator method, but my solution is incorrect, I assume it's related to equation not being linear. As there are no $t$ argument, I cannot transform equation into $t(x)$ form or can I? Or maybe there is altogether different method how to solve this problem?

Assuming you mean $x'-3x^{2/3} = 0 \iff x' = 3x^{2/3}$, in which case you integrate both sides:
$$
\frac{dx}{dt} = 3x^{2/3} \iff \int dt = \int \frac{dx}{3x^{2/3}}
$$
can you finish?


Is $L:\mathbb P_4\to \mathbb P_5$, with $L(p)(x)=\int_0^x p(s)ds$, a linear function? If we consider a similar question, $L(p)(x)=7+\int_0^x p(s)ds$,

We can definitively say no.

Let $p(x)=q(x)=0$ be the zero polynomial. So we also have $p(x)+q(x)=p(x)$. We compute $$L(p)(x)=7+\int_0^x 0 ds = 7$$ But since $7+7\neq 7$ we have $L(p)+L(q)\neq L(p+q)$ so the additivity rule is not satisfied.

But if we remove the 7, does it still remain not a linear function, or does it become a linear function?

Let $p,q$ be in $\mathbb{P}_4[\mathbb{R}]$ and $\alpha, \beta \in \mathbb{R}$, then:

$$L(\alpha p + \beta q) (x)= \int_0^x \alpha p + \beta q ds = \alpha \int_0^x p(s)ds + \beta \int_0^x q(s) ds = \alpha L(p) (x) + \beta L(q)(x) $$
So $L$ is linear.


How do you solve the equation $6g + 8 = 9g - 25$? $6g+8=9g-25$

Can you simply solve for $g$? I'm having trouble with the steps.

we have to solve $$6g+8=9g-25$$ adding $25$ on both sides we get
$$6g+33=9g$$ substracting $6g$ this gives
$$33=3g$$ and dividing by $3$ and you have the answer.

This is one of the toughest questions I have seen on this site, but here are some steps that might lead to the solution.

We start with the original equation $$6g + 8 = 9g - 25$$ We take the square on both sides to get $${(6g + 8)}^2 = {(9g - 25)}^2$$ Expanding both sides, we get $$36g^2 + 96g + 64 = 81g^2 - 450g + 625$$Subtracting $81g^2 - 450g + 625$ from both sides leaves $$-45g^2 + 546g - 561 = 0$$

This is a quadratic equation, so we can use the famous Quadratic formula to get the solutions: $$ g = \frac{-546 \pm \sqrt{{546}^2 - 4 \cdot -45 \cdot -561} }{2 \cdot -45} $$ which simplifies to $$ g = \frac{546 \mp \sqrt{197136} }{90} $$

Now, the square root of 197136 is 444, so this leaves us with $$ g = \frac{546 \mp 444 }{90} $$ so $g$ is either $\frac{102}{90}$ or $\frac{990}{90}$. When you plug both values into the original equation, you will see which one is the real solution.


Nice examples of groups which are not obviously groups I am searching for some groups, where it is not so obvious that they are groups.

In the lecture's script there are only examples like $\mathbb{Z}$ under addition and other things like that. I don't think that these examples are helpful to understand the real properties of a group, when only looking to such trivial examples. I am searching for some more exotic examples, like the power set of a set together with the symmetric difference, or an elliptic curve with its group law.

Homological algebra. Let $A,B$ be abelian groups (or more generally objects of an abelian category) and consider the set of isomorphism classes of abelian groups $C$ together with an exact sequence $0 \to B \to C \to A \to 0$ (extensions of $A$ by $B$). It turns out that this set has a canonical group structure (isn't that surprising?!), namely the Baer sum, and that this group is isomorphic to $\mathrm{Ext}^1(A,B)$. This is also quite helpful to classify extensions for specific $A$ and $B$, since $\mathrm{Ext}$ has two long exact sequences. For details, see Weibel's book on homological algebra, Chapter 3. Similarily many obstructions in deformation theories are encoded in certain abelian groups.

Combinatorial game theory. A two-person game is called combinatorial if no chance is involved and the ending condition holds, so that in each case one of the two players wins. Each player has a set of possible moves, each one resulting in a new game. There is a notion of equivalent combinatorial games. It turns out that the equivalence classes of combinatorial games can be made into a (large) group. The zero game $0$ is the game where no moves are available. A move in the sum $G+H$ of two games $G,H$ is just a move in exactly one of $G$ or $H$. The inverse $-G$ of a game $G$ is the one where the possibles moves for the two players are swapped. The equation $G+(-G)=0$ requires a proof. An important subgroup is the class of impartial games, where the same moves are available for both players (or equivalently $G=-G$). This extra structure already suffices to solve many basic combinatorial games, such as Nim. In fact, one the first results in combinatorial game theory is that the (large) group of impartial combinatorial games is isomorphic to the ordinal numbers $\mathbf{On}$ with a certain group law $\oplus$, called the Nim-sum (different from the usual ordinal addition). This identification is given by the nimber. This makes it possible to reduce complicated games to simpler ones, in fact in theory to a trivial one-pile Nim game. Even the restriction to finite ordinal numbers gives an interesting group law on the set of natural numbers $\mathbb{N}$ (see Jyrki's answer). All this can be found in the fantastic book Winning Ways ... by Conway, Berlekamp, Guy, and in Conway's On Numbers and Games. A more formal introduction can be found in this paper by Schleicher, Stoll. There you also learn that (certain) combinatorial games actually constitute a (large) totally ordered field, containing the real numbers as well as the ordinal numbers. You couldn't have guessed this rich structure from their definition, right?

Algebraic topology. If $X$ is a based space, the set of homotopy classes of pointed maps $S^n \to X$ has a group structure; this is the $n$th homotopy group $\pi_n(X)$ of $X$. For $n=1$ the group structure is quite obvious, since we can compose paths and go paths backwards. But at first sight it is not obvious that we can do something like that in higher dimensions. Essentially this comes down to the cogroup structure of $S^n$. There is a nice geometric proof that $\pi_n(X)$ is abelian for $n>1$.

A set of permutations closed under composition.

These were historically the first thing called groups, every group algebra is instantiated by some set of permutations closed under composition and most of the theorems you see in group theory books have been developed to understand permutations closed under composition better and their relationship with solving algebraic equations.


Deal or No Deal: Monty Hall? This question was inspired by another question posted today: Monty Hall Problem Extended.

So I thought that the comments an answers brought up a great point about increasing the doors to 100 or something much larger, and using that as a way to help visualize why switching is always the best choice when trying to explain the problem to others.

And then I was thinking about the game show, Deal or No Deal. For those unfamiliar with Deal or No Deal: there are 26 cases, each containing amounts of money ranging from \$0.01 to one million dollars. You choose one case, and it's "yours" and out-of-play (this is analogous to choosing the first door in the Monty Hall problem). Throughout the game you open 24 of the remaining cases, and you see how much money was in each case.

In the end, you are left with 2 cases: "your" case, that you chose in the beginning, and the only other case you didn't open. This is where it becomes Monty Hall: you can either choose to keep your case, or switch cases and get the other one.

So what I'm wondering is, does the Monty Hall logic of "always switch doors/cases" apply here? The differences:

1) It's not a case of there being simply 1 car and a bunch of goats. All the money values are different in each case. You aren't always going to end up with a choice between a million dollars or something small... The two remaining cases might end up being \$10,000 and \$250,000. Or it might be \$10 and a million dollars. Or \$10 and $100.

2) I think part of what makes Monty Hall work is that the car always remains in play. Your first choice is a 1/26 probability of selecting the car/million dollar case. But in Deal or No Deal, the car/million dollar case can be eliminated partway through the game. So I'm thinking that probably changes things.

My first vague thoughts are... If you make it to the end and the million dollar case still is in play, Monty Hall applies and you should switch cases. Because it's the same idea; I had a 1/26 shot at the million. 24 have been eliminated. It's much more likely that the other case has the million.

But if the million is eliminated while you're playing, what then? Can Monty Hall not help us, because you can't compare the probability of selecting the million dollar case because now it's zero? I'm trying to think of a way to figure out whether or not you should switch, in an attempt to get the case with the most money in it. We know that \$1,000,000 is no longer available. But is there anything we can do to decide which case is likely to be more valuable? Or is this outside Monty Hall's bounds?

The key is: Monty knows where the car is (and will never open that door). We don't know where the million dollar is so we MIGHT open that door. For an illustration, we look at how the tree diagram differs for the two cases.

Suppose we have 3 doors, A, B and C and our car/million is in door A. We further assume we will always switch. (Once we understand this, we can extend it to $n$ doors and see that the situation will be similar.)

Case 1: Monty Hall Problem

If we switch, $P($Win$) = \frac{2}{3}$.

Case 2: Deal or No Deal scenario

Notice our assumption in the question is we only look at the situation if the million has not been opened. So we are in essence calculating a conditional probability. If we switch,

$P($Win $|$ Million not opened$) = \displaystyle \frac{P(\textrm{Win}\cap \textrm{Million not opened})}{P(\textrm{Million not opened})} = \frac{\frac{1}{6}+\frac{1}{6}}{\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}}=\frac{1}{2}$.

Firsly, let me preface this with "I'm happy to be wrong".. I'm not a probability or stats guru.. but surely the previous arguments are incorrect due to the fact they are not considering the pick of the initial box?

At the start of each Deal_Or_No_Deal game, someone at the studio picks a random box from 26 boxes to be the contestant's box. That person has a 1/26 of selecting the million for the player.

After this all the player really does, in a very long-winded way, is pick a second box before deciding whether to swap (if we are looking at conditional based probability, where we are assuming he/she makes it that far with the million intact)..

Then that person picks from a selection of 25 boxes.
now, is it not fair to say that it would be better to pick from the wider set of boxes and stick with the box the other person chose first, than yours?

worded another way: if i put 100 bits of paper on the floor, and one had a coin underneath... and i ask you if you want to try and guess which one had the coin under it first or let a second person go first.. while you might think that your odds would be better if the first person went first and got it wrong (they would obviously).. is it not better to be the first person in this scenario statistically?

might be wrong. interesting thought anyway.


$\int_{-1}^1{x^4}dx$ by substitution My student asked me: While solving $\int_{-1}^1{x^4}dx$, why can't we substitute as $t=x^2$
$$x^2 =t, 2xdx = dt, x=\pm \sqrt t, \int_{-1}^{1}(x^2 \cdot x \cdot x) dx = \int_1^1{\pm t
\sqrt{t}dt} =0 \ne \frac{1}{5}$$
I think the problem comes from $x = \pm \sqrt t$, but I can't properly explain why this happens. How can I prove that this substitution is wrong?

You write in a comment that the formula
$$\int_{\varphi (a)}^{\varphi (b)}{g(x)}dx = \int_{a}^{b} {g(\varphi (x))\varphi '(x)dx}$$
doesn't require $\varphi$ to be injective. This is true. But if you try and use this formula to evaluate your integral with $\varphi(x)=x^2$, you need to choose $g$ so that
$$g(\varphi(x))=\frac12 x^3$$
on $[a,b]=[-1,1]$. But this would require that $g(\varphi(-1))=-\frac12$ and $g(\varphi(1))=\frac12$, which is impossible, because $\varphi(-1)=\varphi(1)$.
So $\varphi$ has to be invertible on $[a,b]$. Otherwise no suitable $g$ can exist.
Taken together with the requirement that $\varphi$ be continuously differentiable on $[a,b]$, this implies that $\varphi$ is $monotonic$ on $[a,b]$. And this is where your substitution breaks down. As others have explained, you can get round this by integrating separately over $[-1,0]$ and $[0,1]$, because $\varphi$ is monotonic on each of these intervals.
This monotonicity requirement seems to be ignored by the various web pages which state the appropriate theorem. True, it is a logical consequence of the theorem, but not, I would say, an obvious consequence.

Since for $r>0$, $\dfrac{\partial x^r}{\partial x}=r\times x^{r-1}$ and if $r-1+r=4$ then $r=\dfrac{5}{2}$.
Thus,
\begin{align}\int_{-1}^1 x^4\,dx&\overset{x\rightarrow x^4 \text{is even function}}=2\int_0^1 x^4 dx\\
&\overset{y=x^{\frac{5}{2}}}=2\times \frac{2}{5}\int_0^1 ydy\\
&=\frac{4}{5}\int_0^1 y dy\\
&=\frac{4}{5}\left[\frac{y^2}{2}\right]_0^1\\
&=\frac{4}{5}\times \frac{1}{2}\\
&=\boxed{\frac{2}{5}}
\end{align}


Difficulty in Mathematical Writing Lots of people (including myself) face lot of problems in tackling Mathematical Problems, which appear as if we can solve it, but then writing out a solution becomes difficult.

Let us consider some examples:

I was asked this question, some time back in an exam. Give an example of a continuous function on $(a,b)$ which is not uniformly continuous. Well, one's obvious choice is $$f(x) = \frac{1}{x-a} \quad \text{or} \ \frac{1}{x-b}$$ I knew this as soon as i saw the problem, and started proving it. One actually has to make an observation that as $x \to a$ then, $\frac{1}{x-a}$ will be larger. But i found that i couldn't actually formally prove it.
Similarly, to prove that $f(x)=x^{2}$ is not uniformly continuous on $\mathbb{R}$, one again has to play with the quantifiers, to get the contradiction part.

So, these are two instances, where i have found the problem, which to me appeared that i could solve it, but writing out a formal solution became difficult.

How can students improve upon this? Are there any instances, which happened to you like this!

You need practice in formulating your intutions.  This is a skill that can be learned through practice, but like all skills, you can't begin just by jumping into the deep-end: you need to work consistently at extending your comfort zone, one small step at a time; over time you will then make substantial improvements.

What form should your practice take?  Well, you have to practice writing careful and complete proofs.   (The word "writing" here is important; it is not enough just to think about it and convince yourself that you could write a correct answer --- rather, sit down with pen and paper and carefully write out the full and correct answer.)

As a practical suggestion of where to begin (in light of your specific problem): rather than beginning directly with a concept like (failure of) uniform continuity, which involves a rather large number of quantifiers, be sure you can completely prove some simpler statements, such as "$1/(x-a)$ is unbounded on $(a,b)$" or "$x^2$ is unbounded on $\mathbb R$".   Once you can comfortably prove such statements, you can build up to carefully proving analytic statements with more quantifiers.

One more thing: When trying to prove statements (using precise defintions)
which you feel that you have a good intuitive understanding for, try to self-conciously refelect on how your intuition accords with the formal definitions.  For example, ask yourself "why do I think that $x^2$ is unbounded on $\mathbb R$?" Try to analyze your own intuition and extract some statement out of it.  Then compare this with the formal defintion of unbounded and see if they are the same.  If so, great (!) ; you've got good intuition.  If they dont' quite agree, try to understand why; what was missing in your intuition?  what extra aspect of the situation is the formal definition trying to capture?  Again, reflecting on and analyzing your mathematical intuition is a skill that can be practiced and which you can improve --- but, like all skills, you have to practice it if you want to get better.

The problem isn't really about writing. It's about being able to do math.

One of the big problems in math is the P = NP problem. Most mathematicians believe that the know that the two aren't equal.
There's however no accepted proof. As a result a lot of smart people work on the problem. Writing proofs is the main activity of professional mathematicians.

If you 'or'-sign instead of the 'and'-sign because you don't remember with is turned with way, then you have a problem with writing stuff down in a formal matter.
If you however have no idea about what theorems you need to use to prove the solution then you misstate your problem by saying it's about writing stuff down in a formal matter.

Math centers around finding a way to prove your result. Faith isn't enough.


to compute $\int_{2}^{3}\sqrt{1+\frac{1}{x^{2}}} \: \: dx$ Help me please to compute: $\int_{2}^{3}\sqrt{1+\frac{1}{x^{2}}} \: \: dx$

Thanks a lot!

Hint: let $x=\tan{\theta}$, $dx=\sec^2{\theta} d \theta$:

$$\begin{align} \int_{2}^{3} dx \: \sqrt{1+\frac{1}{x^{2}}} &= \int_{\arctan{2}}^{\arctan{3}} \frac{d \theta}{\sin{\theta} \cos^2{\theta}} \\ &= \int_{\arctan{2}}^{\arctan{3}} \frac{d \theta \, \sin{\theta}}{\sin^2{\theta} \cos^2{\theta}} \end{align}  $$

Now let $y=\cos{\theta}$, $dy=-\sin{\theta} d \theta$:

$$\begin{align} \int_{2}^{3} dx \: \sqrt{1+\frac{1}{x^{2}}} &= \int_{\frac{1}{\sqrt{10}}}^{\frac{1}{\sqrt{5}}} \frac{dy}{y^2 (1-y^2)} \\ &= \int_{\frac{1}{\sqrt{10}}}^{\frac{1}{\sqrt{5}}} dy \: \left ( \frac{1}{y^2} + \frac{1}{1-y^2} \right )\\ &= \int_{\frac{1}{\sqrt{10}}}^{\frac{1}{\sqrt{5}}} dy \: \left [ \frac{1}{y^2} + \frac{1}{2} \left ( \frac{1}{1-y} + \frac{1}{1+y} \right ) \right ]  \\ \end{align} $$

I think you can take it from here.

Hint : set $x = f(y)$ where is $f$ is a very well-known function.


Why is the Fourier transform self-inverse? I've seen the standard proof that the Fourier transform is self-inverse (up to an overall factor determined by conventions), which is essentially equivalent to
$\int_{-\infty}^\infty e^{2 \pi i \omega x} dx = \delta(\omega)$.
That isn't too hard to see from definitions.

But intuitively, the Fourier transform of a function gives the amplitude of each component frequency in an obvious sense.  Is it intuitively obvious that this operation is self-inverse or is there a deep reason why it must be so?

This won't be the most rigorous answer, but just a representation my intuition with regards to the Fourier transform.

In nonrelativistic quantum mechanics there are two important operators which act on $L^2(\mathbb{R})$.

The position operator,

$$ (\hat{x} \psi)(x) = x\psi(x),$$

and the momentum operator,

$$ (\hat{p} \psi)(x) = -i \frac{\partial \psi}{\partial x} (x).$$

Now the "eigenvectors" of $\hat{x}$ are $a_{x'}(x) = \delta(x-x')$ and the "eigenvectors" of $\hat{p}$ are $b_p(x) = e^{ipx}$.

Now if you study linear algebra at all you know that the choice of basis is arbitrary. Also you would know that the eigenvectors of a Hermitian operator always form a complete basis. Although our operators our Hermitian, their eigen-functions certainly don't live in $L^2(\mathbb{R})$. I'm going to completely ignore that fact and pretend that I can use them as basis.

For instance the expansion of $\psi(x)$ in terms of  $a_x$'s will be,

$$ \psi(x) = \int \mathrm dx'\ c(x') a_{x'}(x),$$

we can use what we know about delta functions to conclude that the appropiate expansion coefficients, $c(x')$, are in fact $c(x')=\psi(x')$.

If we want to expand $\psi$ in the momentum basis (the $b_p$'s) then we write the superposition and try to determine the coefficients.

$$\boxed{ \psi(x) = \int \mathrm dp \ c(p) b_{p}(x)},$$

the coefficient function, $c(p)$, is a vector in $L^2(\mathbb{R})$ as well. In fact what we have writtend down here is an isomorphism between the x-basis and the p-basis. However the bahavior of the operators on functions expressed in this new bases is different.

$$ \hat{x} \psi(x)  = \int \mathrm dp \ c(p) x e^{ipx} = \int \mathrm dp \ c(p) \left(-i\frac{\partial}{\partial p} \right)e^{ipx} = \int \mathrm dp \ i \frac{\partial c(p)}{\partial p}  e^{ipx}   $$

$$ \boxed{ \hat{x} = i \frac{\partial}{\partial p}} $$

$$ \hat{p} \psi(x)  = \int \mathrm dp \ c(p) \left(-i \frac{\partial}{\partial x} \right) e^{ipx} = \int \mathrm dp \ c(p) p e^{ipx} = \int \mathrm dp \ i \frac{\partial c(p)}{\partial p}  e^{ipx}   $$

$$ \boxed{ \hat{p} = p} $$

This means that in the p-basis the eigenvalues of $\hat{x}$ are $a_x(p) = e^{-ipx}$. I can write $c(p)$ in terms of the x-basis then by the integral,

$$ c(p) = \int \mathrm dx \ \phi(x) e^{-ipx} $$

I've used $\phi(x)$ to represent the coefficients in this expansion, but in practice we actually know the vector in the x-basis which is isomorphic to $c$, that would be $\psi(x)$.

$$ \boxed{ c(p) = \int \mathrm dx \ \psi(x) e^{-ipx} }$$

One issue is the normalization of the transforms are off. This is because the don't have finite norms. For that you need to actually prove the Fourier inversion theorem.

$$
                     f=\sum_{n}(f,e_n)e_n,\;\;\;\; e_n(x)=\frac{1}{\sqrt{2\pi}}e^{inx} \\
                     g=\int_{s}(f,e_s)e_s,\;\;\;\; e_s(x)=\frac{1}{\sqrt{2\pi}}e^{isx}
$$
See any similarities? The first is the expansion of a function in terms of the Fourier series on $[-\pi,\pi]$. The second is the expansion for the Fourier integral. The discrete Fourier transform is $\hat{f}(n)=(f,e_n)$, which is the coefficient function required to reconstruct $f$ in terms of the orthonormal basis $\{e_n \}_{n=-\infty}^{\infty}$. The Fourier transform looks more symmetric because both the coefficient function $\hat{f}(s)=(f,e_{s})$ and the reconstruction using this coefficient function are both integrals. But notice there's a difference, and the sign change is not just superficial.

The roles of the forward transform and the reverse integral reconstruction are quite different, even though they end up being related in this case. It's more of a coincidence that the computation of the coefficient function $\hat{f}(s)=(f,e_s)$ and the inverse reconstruction $\int_{s} \hat{f}(s)e_{s}$ are related. The original context definitely distinguished between the two, and they are distinguished by a sign change, too. Similarity between the forward transform and the reconstruction integral is interesting, but it seems to me that one loses context in viewing it that way. There are many other such integral eigenfunction expansions where it is important to distinguish between the coefficient computation and the integral reconstruction.


Show that $f(x)=x^2+x+4$ is irreducible over $\mathbb{Z}_{11}$ I know this can be done by evaluating $f$ at the points $0,1,...10$ to check if $f$ has a linear factor.
Is there any other shorter way?

We start with $$x^2+x+4=0$$ multiplying with $4$ we get $$4x^2+4x+16=0$$ This can be written as $$(2x+1)^2+15=0$$ modulo $11$ this is equivalent to $$(2x+1)^2-7=0$$

So we have to solve $$(2x+1)^2\equiv 7\mod 11$$

Using the quadratic reciprocity law we can calculate the legendre symbol

$$(\frac{7}{11})=-(\frac{11}{7})=-(\frac{4}{7})=-1$$

Hence $u^2\equiv 7\mod 11$ is not solveable because $7$ is not a quadratic residue modulo $11$. Since $x^2+x+4$ has no root in $\mathbb Z_{11}$, it follows that $x^2+x+4$ is irreducible in $\mathbb Z_{11}$

The discriminant of the quadratic form $u^2 + uv + 4 v^2$ is $-15.$
$$ (-15|11) = (-4|11) = (-1|11) (4|11) = (-1|11)=-1,  $$
the last part because $11 \equiv 3 \pmod 4.$

Therefore, if
$$ u^2 + uv + 4 v^2 \equiv 0 \pmod {11}, $$ it follows that both $u,v$ are divisible by $11.$ General theorem with proof at Prime divisors of $k^2+(k+1)^2$

So, with integer $x,$ $x^2 + x + 4$ is the same as $x^2 + xy + 4 y^2$ restricted to $y=1;$ that is, this $y$ is not divisible by $11.$ So $x^2 + x + 4$ cannot be divisible by $11$

I like this way of writing things, in quadratic forms this behavior is called "anisotropic." My experience is that existence of a solution that, however, violates some condition, is easier for the student to deal with than "infinite descent" arguments. All is the same, just a matter of emphasis.

Sigh. Same thing: $x^2 + 1$ is not divisible by any prime $q \equiv 3 \pmod 4.$ The discriminant of $u^2 + v^2$  is $-4.$ For such $q,$ if $q | (u^2 + v^2)$ then both $q|u$ and $q | v.$ However, it is not possible to have $q | 1,$ so $q$ cannot divide $x^2 + 1$ either.


Is there a closed form expression for the infinity symbol? I was looking for a closed form expression which plots the infinity symbol.

The Lemniscate of Bernoulli is a plane curve with a shape similar
to the infinity symbol. It can be plotted  using the parametric equation
$$
x = \frac{a \sqrt2 \cos(t)}{\sin^2(t)+1} \,, \quad y = \frac{a \sqrt2 \cos(t)\sin(t)}{\sin^2(t)+1}\, , \quad 0 \le t \le 2 \pi.
$$
$a$ is half of the distance between the "foci" of the lemniscate.

(From http://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli)

Here is another way you could put it in your calculator:
$$X(t)=2\sin (t)$$
$$Y(t)=\sin (2t)$$
the reason I say like this is because with my calculator I cannot write 2sin(t) as
2sin(t) in my calculator, so I just want to give this solution to put the multiplication sign in it. Also If you want a smoother infinity symbol, instead of 2sin(t), do 3sin(t)(3sin(t)), this will make it bigger but smoother.


Is the point of a shape with the greatest average ray length also the "centroid"? By a shape I mean one with a enclosed area. One example can be with $x^2+y^2+\sin{4x}+\sin{4y}=4$. I am dealing with implicit relations on a 2-d plane.

By the "ray length" I mean the length of the line segment from any chosen point inside the enclosed area to a point on the boundary of the shape.

Just in case you do not know how to find the average ray length you can start by converting an implicit relation into polar form. Just substitute $x=r\cos{\theta}$ and $y=r\sin{\theta}$ in $f(x,y)=0$.

Then to choose a point inside an implicit relation just convert $f(x+u,y+v)=0$ (because we want our chosen point at the pole) and with substitution get $f(r\cos\theta+u,r\sin\theta+v)$. From here you solve for $r$ In terms of $\theta$ $u$ and $v$ as $r=g(\theta)$ then take the area as $\frac{1}{2\pi}\int_{0}^{2{\pi}}g(\theta){d\theta}$ and find the $u$ and $v$ (u,v) with the greatest area. If this is not possible you could implicitly find the area under $f(r\cos\theta+u,r\sin\theta+v){d\theta}$ ,where $r>0$ and the area between $0<\theta<2\pi$. Then divide the area by $2\pi$ and once again find the $u$ and $v$ in point (u,v) with the greatest area.

(If you are not clear with how the area should be calculated treat the polar equation as cartesian implicit function. Replace $r=y$ and $\theta=x$ to get $f(y\cos{x}+u,y\sin{x}+v)$ and then proceed.

Here is an example with the circle https://files.acrobat.com/a/preview/38a501f2-5b63-4b76-b720-6cadb9c3e142

The centroid can be found online in https://en.wikipedia.org/wiki/Centroid.

When it came to a circle and ellipse I found the point of "highest average ray length" and "center of mass" to be the same. However I am not sure how to figure this out with more complicated implicit relations like $x^2+y^2+\sin{4x}+\sin{4y}=4$. Are there any approaches to finding the difference using calculus or numerical integration?

No, consider a C-shaped area, where the center of mass is obviously outside of the shape, and cannot be within the shape. Perhaps you might want to consider convex shapes but even with those I do not believe center of mass is the point of the highest average radius. I think your idea could only (perhaps) work if you only consider the boundary of the shape and its mass (but the area within has no mass).

Consider a line segment, say with a length of one. The average radius with respect to the midpoint is $1/4$, but the average radius with respect to either endpoint is $1/2$.

If you want a $2$-dimensional example, just consider a very thin rectangle where the same argument holds.


Solution to ODE with Dirac Delta satisfies ODE I am working on a problem where I have the following ODE. $$m\dot{v}+bv=\delta_I(t)$$
where
$$\delta_I(t)=\begin{cases}0, & \text{for}&t\ne0\\ I, & \text{for} &t=0\end{cases}.$$
The solution $v(t)$ was derived using Laplace transforms, the ODE in the Laplace domain is (with $0$ initial conditions)$$(ms+b)V(s)=I$$ giving $$v(t)=\frac{I}{m}e^{-bt/m}.$$
How does this solution satisfy the original ODE though? At $t\ne0$ everything is good,$$-\frac{Ib}{m}e^{-bt/m}+\frac{Ib}{m}e^{-bt/m}=0,$$ while at $t=0$, $$-\frac{Ib}{m}+\frac{Ib}{m}=I$$ the result seems to be saying $0=1$ which is obviously false. What am I missing here?

So I start by find the solution to the homogeneous eqn: $$my'+by=0 \to y_h=Ce^{-bt/m}$$

Then, we can determine the inhomogeneous part using a Fourier transform of the original eqn: $$my'+by=\delta(t) \to i\omega m \tilde{y}+b\tilde{y}=1 \to \tilde{y}=\frac{1}{m}\frac{1}{b/m+i\omega}$$

From a table of Fourier transforms $$\frac{1}{b/m+i\omega} \to \theta(t)e^{-bt/m}$$ where $\theta(t)$ is the heaviside step function.

So, $f_i= \theta(t)e^{-bt/m}/m$, which when combined with the homogeneous solution will give the general solution for this differential equation.

The reason why you got a wrong answer is because this ODE isn’t properly defined at t=0. The dirac delta function itself is singular at t=0. So is the ODE at t=0, holding no responsibility for t=0. The ODE only cares about the solution at t>0,. The only piece of precise info the dirac delta provides is that its time integral is 1.


What does this ensemble of symbols mean: $C^{2}(\bar{\Omega})$? For Context, I'll provid the sentence in which the symbols lie; it sets up the conditions for a proof of Green's Identity.

Let $\Omega \subset \Re^n$ be domain with a smooth boundary $\partial \Omega$. Let u,v $\in $ $C^{2}(\bar{\Omega})$, where $\bar{\Omega}$ denotes the closure of $\Omega$.

What does this ensemble of symbols mean: $C^{2}(\bar{\Omega})$ ?

From [Adams-Fournier, p. 80]:

$C^j(\bar{\Omega})$ denotes the closed subspace of $C^j_b(\Omega)$ consisting of functions having bounded, uniformly continuous derivatives up to order $j$ on $\Omega$ normed by
$$
\|\phi\|_{C^j(\bar{\Omega})} = \max_{0 \leq \alpha \leq j} \sup_{x\in \Omega} |D^{\alpha}\phi(x)|.
$$

Here $C^j_b(\Omega)$ is the space of functions having bounded, continuous derivatives up to order $j$ on $\Omega$.

$C^2 (\bar{\Omega})$ represents the $C^2$ functions on this domain i.e. functions whose second derivatives (both pure and mixed) exist and are continuous.

Edit: As Shifrin points out below, this is incorrect.


which rule or definition apply ${\displaystyle {\begin{aligned} \operatorname {E} [Y_{i}^{2}] = (\sigma ^{2}+\mu ^{2}) \quad (3.1) \end{aligned}}}$ I am learning the justification of Sample variance

$${\displaystyle {\begin{aligned}
\operatorname {E} [\sigma _{Y}^{2}]&=\operatorname {E} \left[{\frac {1}{n}}\sum _{i=1}^{n}\left(Y_{i}-{\frac {1}{n}}\sum _{j=1}^{n}Y_{j}\right)^{2}\right] \quad (1.1)\\[5pt]
&={\frac {1}{n}}\sum _{i=1}^{n}\operatorname {E} \left[Y_{i}^{2}-{\frac {2}{n}}Y_{i}\sum _{j=1}^{n}Y_{j}+{\frac {1}{n^{2}}}\sum _{j=1}^{n}Y_{j}\sum _{k=1}^{n}Y_{k}\right ] \quad (1.2)\\[5pt]
&={\frac {1}{n}}\sum _{i=1}^{n}\left[{\frac {n-2}{n}}\operatorname {E} [Y_{i}^{2}]-{\frac {2}{n}}\sum _{j\neq i}\operatorname {E} [Y_{i}Y_{j}]+{\frac {1}{n^{2}}}\sum _{j=1}^{n}\sum _{k\neq j}^{n}\operatorname {E} [Y_{j}Y_{k}]+{\frac {1}{n^{2}}}\sum _{j=1}^{n}\operatorname {E} [Y_{j}^{2}]\right] \quad (1.3)\\
&={\frac {1}{n}}\sum _{i=1}^{n}\left[{\frac {n-2}{n}}(\sigma ^{2}+\mu ^{2})-{\frac {2}{n}}(n-1)\mu ^{2}+{\frac {1}{n^{2}}}n(n-1)\mu ^{2}+{\frac {1}{n}}(\sigma ^{2}+\mu ^{2})\right] \quad (1.4)\\
&={\frac {n-1}{n}}\sigma ^{2}
\end{aligned}}}$$

that post is discussing how equation (1.2) would conduct to equation (1.3)

in this post, i am trying to understand how equation (1.3) would conduct to equation (1.4)

It seems that the procedure contains following equation

$${\displaystyle {\begin{aligned}
\operatorname {E} [Y_{i}^{2}] = (\sigma ^{2}+\mu ^{2}) \quad (3.1)
\end{aligned}}}$$

does equation (3.1) hold? if yes, which rule or definition apply here?

By definition $\sigma^{2}=EY_i^{2}-(EY_i)^{2}=EY_i^{2}-\mu^{2}$. Just add $\mu^{2}$ to both sides.

the Definition part of wiki you linked gives the answer


Number of solutions to $x^{3^{n+1}+1} = 1$ in a field of order $3^{2n+1}$ Let $F$ be a field such that $|F|=3^{2n+1}$ and $r=3^{n+1}$. I want to find the number of $x\in F$ that satisfies the equation $x^{r+1}=1$.

Say we want to count the solutions to $x^k=e$ in a cyclic group of order $m$. We may as well write everything additively and work in $\mathbb{Z}/m\mathbb{Z}$. Writing $\overline{k}=k/\gcd(m,k)$ and $\overline{m}=m/\gcd(m,k)$,

$$kx\equiv 0~\bmod m ~\iff~ \overline{k}x\equiv 0\bmod\overline{m} ~\iff~ x\equiv 0\bmod \overline{m}$$

The first $\Leftrightarrow$ follows from dividing/multiplying by $\gcd(m,k)$, and the second by multiplying by the inverse of $\overline{k}$ mod $\overline{m}$ (which exists since $\gcd(\overline{m},\overline{k})=1$). Therefore, the solutions $x$ mod $m$ are precisely $\ell\overline{m}$ for $0\le \ell<\gcd(m,k)$, of which there are $\gcd(m,k)$ many.

If $F$ is a field of order $3^{2n+1}$, then $|F^\times|=3^{2n+1}-1$ and with the Euclidean algorithm we may calculate the number of solutions to $x^{3^{n+1}+1}=1$ as

$$\begin{array}{ll} \gcd(3^{2n+1}-1,3^{n+1}+1) & =\gcd\big(3^{2n+1}-1-3^n(3^{n+1}+1)),3^{n+1}+1\big) \\ & = \gcd(-3^n-1,3^{n+1}+1) \\ & =\gcd(3^n+1,3^{n+1}+1) \\ & = \gcd\big(3^n+1,3^{n+1}+1-3(3^n+1)\big) \\ & = \gcd(3^n+1,-2) \\ & =2. \end{array}$$

Well, since the multiplicative field of $\;\Bbb F\;$ , namely $\;\Bbb F^*:=\Bbb F-\{0\}\;$ is cyclic of order $\;3^{2n+1}-1\;$, we get that forall

$$x\in\Bbb F^*\;,\;\;x^{3^{2n+1}-1}=1\;$$

You though want some $\;x\in\Bbb F^*\;$ s.t.

$$x^{3^{n+1}+1}=1\iff \left(3^{n+1}+1\right)\mid\left(3^{2n+1}-1\right)$$

Try to take it from here.