问:
Does there exist a singular $3\times3$ stochastic matrix $A$ such that $a_{ii}>\max(a_{ij},a_{ji})$ for all $i\ne j$?
Yesterday, someone asked whether an $n\times n$ row-stochastic matrix $A$ that satisfies $a_{ii}>\max(a_{ij},a_{ji})$ for all $i\ne j$ is necessarily nonsingular. The answer is clearly positive when $n=1,2$. I have shown that the answer is negative for every $n\ge4$ and the counterexample $A$ can be chosen to be symmetric (hence doubly stochastic) and entrywise positive.

This leaves only the case $n=3$ open. I have a feeling that $A$ must be nonsingular but I cannot prove it. Any idea?

Here is one of my discoveries so far. Suppose $Av=0$ has a nontrivial solution $v=(x,y,z)^T$. By permuting the rows and columns of $A$ and by scaling $v$ (by a possibly negative factor) if necessary, we may assume that either $x=1\ge y\ge z>-1$ or $x=1\ge y\ge0>z=-1$. Note that none of $y$ and $z$ can be zero, otherwise some column of $A$ is a scalar multiple of another, but this is a contradiction to the assumption that $a_{ii}>\max(a_{ij},a_{ji})$. So, either $y>0>z$ or $y,z<0$. The former case is impossible, because it implies that
$$
a_{11}<a_{11}+a_{12}y=a_{13}|z|\le a_{13}.
$$
Hence we must have $y,z<0$. In turn, we obtain $x=1>0>y\ge z>-1$, but I cannot proceed any further.

答:
Lemma: If $x, y, a_1, a_2, b_1, b_2$ are non-negative terms such that $xa_1 + yb_1 \leq xa_2 + yb_2$, $x > 0$ and $a_1 > a_2$, then $b_2 > b_1$.

Proof: $0 \lt x (a_1 - a_2) \leq y (b_2 - b_1)$.

Note: If $ x \geq 0$, then $ b_2 \geq b_1$. The proof is similar.

Back to the problem Case: $ 1 = x \geq -y \geq -z > 0$.

$(x-y) a_{11}   + (z-y)a_{13} = -y = (x-y) a_{21}   + (z-y)a_{23}. $

Since the coefficients are non-negative and $a_{11} > a_{21}$, hence $a_{23} \geq a_{13}$.

On the other hand,

$ -y a_{22} - za_{23} = xa_{21} < xa_{11} = -ya_{12} -za_{13} $

Since the coefficients are positive and $a_{22} > a_{12}$, hence $a_{13} > a_{23}$.

Thus, we have a contradiction.


问:
How to calculate basis knowing only coordinates?
I am working on this exercise our teacher gave us:

So basically I am given nine coordinates in one base and nine in another and I should then figure out a matrix used to convert $e$ coordinates into $\tilde{e}$ coordinates? In my book there is this guide to solving problems like this I think.

Trying this, I got the change-of-basis matrix to be:

\begin{bmatrix}
13\quad & -7\quad & 37\\
-2\quad & 1\quad & -6\\
6\quad & -3\quad & 17
\end{bmatrix}

Am I thinking straight? There was another exercise similar to this that I used this exact method for, but the teacher said I was way off and I asked him why and he gave me an answer that I honestly did not understand nor didn't help me solve the problem and because of covid it would probably take at least a week, probably two weeks to get an answer to a question :( and I do not have that time. How do I solve these kind of problems and what is going on here?

答:
I suggested plotting on a comment to get a "feel" for what these matrices are doing, as well as the contravariant nature of the coefficients in a change of basis (as the basis vectors lengthen, the components of a given vector expressed in the new basis shrink). The answer by Chris Custer (+1) is the answer to the question, but to dispel the remaining questions I am going to look at one of the original vectors in standard Euclidean coordinates in terms of the new basis vectors, and use the matrix in Chris' answer as my translator to make sure that the order of matrix operations does make sense.

I chose the vector $\vec v=\small\begin{bmatrix}1\\3\\-1\end{bmatrix},$ just because I like it. This vector, in standard Euclidean coordinates is $\vec v= 1 \vec e_1 + 3 \vec e_2 -1 \vec e_3=\small 1\begin{bmatrix}1\\0\\0\end{bmatrix}+ 3 \begin{bmatrix}0\\1\\0\end{bmatrix}-1 \begin{bmatrix}0\\0\\1\end{bmatrix}.$ Nothing new. Here it is:

Now, this very same arrow, exactly as it is, can be looked at using the new basis vectors as calculated with the standard operation in Chris' answer:

$$\small\left\{\tilde {\vec e_1}=\begin{bmatrix}-1\\8\\5\end{bmatrix},\tilde {\vec e_2}=\begin{bmatrix}-2\\-1\\4\end{bmatrix}, \tilde {\vec e_3}=\begin{bmatrix}0\\-3\\-1\end{bmatrix} \right\}$$

The equation used in that answer is:

$$\mathbf C = \begin{bmatrix}\text{vectors}\\\text{in}\\ \text{old basis}\end{bmatrix}\; \begin{bmatrix}\text{vectors}\\\text{in}\\ \text{new basis}\end{bmatrix}^{-1}=\small\begin{bmatrix}-1&-2&0\\8&-1&-3\\5&4&-1\end{bmatrix}$$

where $\mathbf C$ is the matrix containing the new basis in its columns.

It follows that

$$\begin{bmatrix}\text{vectors}\\\text{in}\\ \text{old basis}\end{bmatrix} = \mathbf C  \begin{bmatrix}\text{vectors}\\\text{in}\\ \text{new basis}\end{bmatrix}$$

And since we are focusing on the second vector $\vec v,$ the operation will only involve the second vector in the new coordinate system, which is in the second column of the vectors expressed in the new basis: $\small \begin{bmatrix}-1&\color{red}1&1\\0&\color{red}{-1}&0\\-3&\color{red}2&2\end{bmatrix}$ as such:

$$\small \begin{bmatrix}1\\3\\-1\end{bmatrix}= \mathbf C\small \begin{bmatrix}1\\-1\\2\end{bmatrix}=\small\begin{bmatrix}-1&-2&0\\8&-1&-3\\5&4&-1\end{bmatrix}\small \begin{bmatrix}1\\-1\\2\end{bmatrix}$$

which is indeed what we get if we plot the equivalent expression:

$$\small \begin{bmatrix}1\\3\\-1\end{bmatrix}= 1\small \begin{bmatrix}-1\\8\\5\end{bmatrix} -1 \small \begin{bmatrix}-2\\-1\\4\end{bmatrix}+2 \small \begin{bmatrix}0\\-3\\-1\end{bmatrix}  =1\tilde {\vec e_1}-1\tilde {\vec e_2}+ 2 \tilde {\vec e_3}$$

as

And now closing the loop and transforming forward the coordinates from old to new, we can see why we need the inverse of the actual change of coordinates matrix $\mathbf C$:

$$ \begin{bmatrix}\text{vectors}\\\text{in}\\ \text{new basis}\end{bmatrix}= \mathbf C^{-1} \begin{bmatrix}\text{vectors}\\\text{in}\\ \text{old basis}\end{bmatrix}$$

So

$$\small \begin{bmatrix}1&-2&0\\8&-1&-3\\5&4&-1\end{bmatrix}^{-1}\big[\vec v\big]_\text{Std. Eucl.}=
\small \begin{bmatrix}13&-2&6\\-7&1&-3\\37&-6&17\end{bmatrix}\small \begin{bmatrix}1\\3\\-1\end{bmatrix}=\begin{bmatrix}1\\-1\\2\end{bmatrix}_{\text{New basis}}$$

ContentLicense: CC BY-SA 4.0


问:
Obtaining coefficients of a sum of exponential decays
Let $f:\mathbb R^+\to\mathbb R^+$ be such that
\begin{equation}
f\left(x\right)=\sum_{n=0}^\infty a_ne^{-4\pi nx}
\end{equation}
where $a_n$ are nonnegative integers. Is there some way of recovering $a_n$ from $f$? I have been trying to get them like coefficients of a Fourier series, and have tried taking a Laplace transform. Maybe something else works.

答:
Here's a way to recover the coefficients. Note that we can reduce $f$ to be a power series by "warping" its domain.
$$f\left(-\frac {\ln t} {4\pi}\right) = \sum_{n\geq 0}a_n t^n$$
Thus $$a_n=\frac 1 {n!} \left.\frac{d^n}{dt^n}f\left(-\frac {\ln t} {4\pi}\right)\right|_{t=0}\\$$
Looks great on paper. Numerically, not awesome though. Numerical differentiation is tricky.


问:
uncertain orthogonality of discrete Fourier transform on the ring of integers modulo some number
Update: I corresponded with one of the authors of CLRS, and he confirmed that the problem indeed should say "$n$ is a power of two", not "$n$ is even".

Original question:

CLRS Exercise 30.2-6 claims that for any positive even $n$ and positive integer $t$, a discrete Fourier transform can be defined on the ring of integers modulo $m = 2^{tn/2} + 1$, using $w = 2^t$ as the principle $n$th root of unity.

Wikipedia seems to say that if $w$ is a principal $n$th root of unity, then for $1 \leq k < n$, we must have

$$
T_k = \sum_{j=0}^{n-1} w^{jk} = 0 \mod m
$$

However, choosing $n = 6$ and $t = 3$, we have $m = 513$ and $w = 8$, and it's easy to see that $T_2 = 114 \mod m$. Isn't this supposed to be zero? If yes, where have I gone wrong above?

答:
Given any $m$ and $w\in \Bbb{Z/mZ}^\times$ of order $n$, the necessary and sufficient condition for $T_0 \in \Bbb{Z/mZ}^\times$ and $T_1=\ldots=T_{n-1}=0$ is that $\gcd(n,m)=1$ and for each $k\ne 0\bmod n, w^k-1$ is a unit ie. for each prime $p| m$,  $w\bmod p$ has order $n$.

If $\gcd(n,m)\ne 1$ then $T_0$ is not a unit.

If each $w^k-1$ is a unit then $w^{nk}-1=0$ implies $T_k=\frac{w^{kn}-1}{w^k-1}=0$.

If $w^k-1$ is not a unit then $w^k=1\bmod p$ for some $p|m$ thus $T_k = n\ne 0 \bmod p$

If for each prime $p| m$, $w\bmod p$ has order $n$ then each $w^k-1$ is a unit.

If for some prime $p| m$, $w\bmod p$ has order $k<n$ then $w^k-1$ is not a unit.

It is not obvious at all for which $n,t$ the condition is satisfied with $m=2^{nt/2}+1, w=2^t$ (except when $n$ is a power of $2$, see comment below)


问:
Theorem on lim sup of max of random variables.
Let $X_1,X_2,\ldots$ be iid random variables, then we are trying to prove that:
$$\limsup_{n\to\infty} \frac{1}{n} \max_{k\leq n} \lvert X_k \rvert = \begin{cases}0 &\textrm{if  } E\lvert X_1\rvert < \infty \\
\infty &\textrm{if  } E\lvert X_1\rvert = \infty \\
\end{cases}$$

In the proof given in class, we proved that for all $a>0$:
$$P(\lvert X_n \rvert > an \ i.o.) = \begin{cases}0 &\textrm{if  } E\lvert X_1\rvert < \infty \\
1 &\textrm{if  } E\lvert X_1\rvert = \infty \\
\end{cases}$$
Using the above result, if $E\lvert X_1\rvert = \infty$, then it is obvious that $\limsup \frac{1}{n} \max_{k\leq n} \lvert X_k \rvert > a$ for all $a> 0$ which gives us the conclusion. But, if $E\lvert X_1 \rvert <\infty$, i don't understand how the previous result guarantees that $\limsup \frac{1}{n} \max_{k\leq n} \lvert X_k \rvert < a$, since we're always taking the maximum over all elements.

答:
I think I have it figured out (can someone please verify this? or provide a simpler way if there is?):

If $E\lvert X_n\rvert < \infty$, then $P(\lvert X_n \rvert = \infty) = 0$. Consider the set $$\left(\bigcap_{n\geq 1} \{\lvert X_n \rvert < \infty\}\right) \cap \{ \lvert X_n \rvert > an \ i.o\}^c$$
This set has probability $1$ since it is the countable intersection of sets with probability $1$. Now, let $w$ be an element of the set above, then there exists $n_a$ such that $\forall n\geq n_a: \lvert X_n(w) \rvert \leq an$, and since $X_1(w),\ldots, X_{n_a-1}(w) < \infty$, there exists $n_0$ such that $$\forall n\geq n_0:
 \frac{1}{n}\max \{ X_1(w),\ldots, X_{n_0-1}(w) \} \leq a.$$
This gives us the required result by considering $n\geq \max\{n_0,n_a\}$.


答:
There is almost certainly an error in the CLRS claim, since $n$ being even is not sufficient, as demonstrated by the $n = 6$ counterexample.

On the other hand, $n$ being a power of two is sufficient, though it is still unclear to me whether it's necessary.

Proof (M. Pinson, private communication):

When $n = 2^\ell$, we have

$$
T_k = \frac{w^{nk} - 1}{w^k-1} = \prod_{J=0}^{\ell-1} \left( w^{2^J k} + 1\right).
$$

Without loss of generality, let $k = 2^R q$ where $q$ is odd. Since $0 < k < n$, we have $0 \leq R < \ell$, and thus we examine the $J = \ell - 1 - R$ factor and see that it becomes $w^{2^{\ell - 1}q} + 1$. Since $x^q + 1$ is always divisible by $x+1$ for any odd $q$ [which can be seen by computing $\sum_{i=0}^{q-1} (-x)^i $], we know that the factor $w^{2^{\ell - 1}q} + 1$ must be divisible by $m = w^{2^{\ell - 1}} + 1$, and we are done.

Update: I corresponded with one of the authors of CLRS, and he confirmed that the problem indeed should say "$n$ is a power of two", not "$n$ is even".

Title: uncertain orthogonality of discrete Fourier transform on the ring of integers modulo some number

Body:
Update: I corresponded with one of the authors of CLRS, and he confirmed that the problem indeed should say "$n$ is a power of two", not "$n$ is even".

Original question:

CLRS Exercise 30.2-6 claims that for any positive even $n$ and positive integer $t$, a discrete Fourier transform can be defined on the ring of integers modulo $m = 2^{tn/2} + 1$, using $w = 2^t$ as the principle $n$th root of unity.

Wikipedia seems to say that if $w$ is a principal $n$th root of unity, then for $1 \leq k < n$, we must have

$$
T_k = \sum_{j=0}^{n-1} w^{jk} = 0 \mod m
$$

However, choosing $n = 6$ and $t = 3$, we have $m = 513$ and $w = 8$, and it's easy to see that $T_2 = 114 \mod m$. Isn't this supposed to be zero? If yes, where have I gone wrong above?


问:
Counting the number of distinct groups with and without repeat items
Here's a simple question that popped into my head. I shouldn't be struggling with but am.

Suppose there are $n$ objects. We want to find the number of distinct groups.

First let's take the case where each of the $n$ objects is distinct. The number of distinct groups should be just $2^n$ right?

Here's where I get tripped up. Let's say all of the items are distinct except we have 1 item that is repeated. Let's say $\text{item}_1$ and $\text{item}_2$ are identical to each other. Well now the answer is not $2^n$ anymore because having $\text{item}_1$  and not $\text{item}_2$in my group is the same as having $\text{item}_2$ and not $\text{item}_1$. But having both and neither of the items should count as separate groups...

Does that make sense?

So let's break this into cases -

We have $\text{item}_1$ but not $\text{item}_2$. There are $2^{n-2}$ in this case because the number of distinct groups from the $n-2$ objects left is $2^{n-2}$

We have both $\text{item}_1$ and $\text{item}_2$. Again there are $2^{n-2}$ distinct groups in this case.

Now the case for neither of the items. Again there are $2^{n-2}$ distinct groups in this case.

We ignore the case where we have $\text{item}_2$ but not $\text{item}_1$ since we have already accounted for that in the first case and because the two items are identical.

How many distinct groups are there? There should be $3 (2^{n-2})$ since we just have to sum all three cases together.

Is this correct? How would I expand this for the general case where I have $k$ identical objects of one item in n objects. How about if I had $k_1, ..., k_r$ identical objects for items $1, ..., r$ amongst the n objects.

Going back to my example, I feel like back in highschool what I would do is just take the number of distinct n groups form before, $2^n$, and divide it by $2$ since I double counted one of the items. Why is this wrong or correct?

Also a general question - How can I tell when I've fully counted everything the correct number of times correctly. Any tips, resources, tricks and pointers are greatly appreciated. :)

答:
It would be better to call these "sets" or "multisets" because "group" has a technical meaning in mathematics.

Suppose you have $c_k$ copies of each item $k$, and let $x_k \in \{0,\dots,c_k\}$ be the number of copies you select.  The choices are independent, and you have $c_k+1$ choices for each item, so the total number of distinct multisets is $$\prod_{k=1}^n (c_k+1).$$

In the special case $c_k=1$ (all items are distinct), this product simplifies to $$\prod_{k=1}^n (c_k+1)=\prod_{k=1}^n (1+1) = 2^n.$$

In the special case $c_1=0$, $c_2=2$, and $c_k=1$ for $k>2$, this product simplifies to $$\prod_{k=1}^n (c_k+1)=(0+1)(2+1)\prod_{k=3}^n (1+1) = 3\cdot 2^{n-2}.$$

In the special case $c_1=n$ and $c_k=0$ for $k>1$ (all items are the same), this product simplifies to $$\prod_{k=1}^n (c_k+1)=(n+1)\prod_{k=2}^n (0+1) = n+1.$$

By the way, the general formula also counts the number of divisors of an integer whose prime factorization is of the form $$p_1^{c_1} p_2^{c_2} \dots p_n^{c_n}.$$  The "items" are the primes, and the numbers of copies selected are the numbers of times each prime appears in the divisor.   See https://oeis.org/A000005.


问:
A problem related to an integral equation
I am stuck on the following problem that is as follows:

The integral equation $\quad \varphi(x)-\lambda \displaystyle\int_{-1}^{1}\cos[\pi(x-t)]\varphi(t) dt= g(x)$  has

1.a unique solution for $ \lambda \ne 1$ when $g(x)=x$

2.no solution for $\lambda \ne 1$ when $g(x)=1$

3.no solution for $\lambda =1$ when $g(x)=x$

4.infinite number of solutions for $\lambda =1$ when $\quad g(x)=1.$

What I did:

$$\begin{align}\quad g(x)&= \varphi(x)-\lambda \displaystyle\int_{-1}^{1}\cos[\pi(x-t)]\varphi(t)dt \\   \implies g(x) &=\varphi(x)-\lambda \displaystyle \int_{-1}^{1}\cos[\pi x-\pi t]\varphi(t)dt\\ &= \varphi(x) -\lambda \displaystyle\int_{-1}^{1} [\cos \pi x \cos \pi t +\sin \pi x \sin \pi t]\varphi(t)dt \\&=\varphi(x)-  \lambda\cos \pi x\displaystyle\int_{-1}^{1} \cos \pi t \varphi(t)dt -\sin \pi x \displaystyle\int_{-1}^{1} \sin \pi t \varphi (t) dt \\&=....??\end{align}$$

Can someone explain with some details about how to tackle these types of problem.Thanks in advance for your time.

MY REATTEMPT TO THE PROBLEM:

( Using @Glen O 's suggestion)From the given problem we see,$$\begin{align}\quad \varphi(x)&=g(x)+\lambda \{A \cos (\pi x)+B \sin (\pi x)\} \end{align}$$,where $A=\displaystyle \int_{-1}^1 \cos(\pi t)\varphi(t)dt \longrightarrow \clubsuit$  and

$B=\displaystyle \int_{-1}^1 \sin(\pi t)\varphi(t)dt \longrightarrow \spadesuit$ .

After calculations,we get from  $ \clubsuit \space \text{and}\space \spadesuit \space ,  A(1-\lambda)=\displaystyle \int_{-1}^1 \cos(\pi t)g(t)dt  \space \text{and} \space B(1-\lambda)=\displaystyle \int_{-1}^1 \sin(\pi t)g(t)dt$ .

Now,(1) for $\lambda \ne 1,g(t)=t$,we get $A(1-\lambda)=0,B(1-\lambda)=\frac{2}{\pi}$  and so we can have a unique value of $A$ and $B$ and hence option $1$ is true.

(2)For $\lambda \ne 1,g(t)=1,$ we get $A(1-\lambda)=0,B(1-\lambda)=0$ and so we have $A=B=0$ and hence equation turns out to be $\varphi(x)=g(x)$ and hence no solution and so statement $2$ holds good.

(3)For $\lambda= 1,g(t) =t,$ we get $ 0 \times B=\frac{2}{\pi}$ which is impossible and hence statement $3$ does not hold.

(4)For $\lambda= 1,g(t) =t,$ we get $A(1-\lambda)=0,B(1-\lambda)=0$ and infinite number of values for $A,B$ is posssible. And hence statement $4$ holds true.

Am I missing something in my argument? Please feel free to comment.

答:
Hint: As you correctly note, you may write the integral equation in the form

$$
\varphi(x) - \lambda \cos(\pi x)\int_{-1}^1\cos(\pi t)\varphi(t)dt-\lambda \sin(\pi x)\int_{-1}^1\sin(\pi t)\varphi(t)dt=g(x)
$$
Notice that the integrals evaluate to simple numbers. So we have a solution of the form
$$
\varphi(x) = g(x)+\lambda\big(A\cos(\pi x)+B\sin(\pi x)\big)
$$
Now, $A=\int_{-1}^1 \cos(\pi t)\varphi(t)dt$. What do you get if you substitute our solution in for $\varphi(t)$, given $g(x)$? Similarly, what do you get for $B$?

ContentLicense: CC BY-SA 3.0


问:
Prove convergence/divergence of $ \sum_{k=0}^\infty ke^{-\sqrt k} $
Prove the convergence/divergence of the following series.
$$ \sum_{k=0}^\infty ke^{-\sqrt k} $$
The root and ratio tests don't work because r = 1. That leaves the integral and comparison tests, but I cannot for the life of me find an expression that works for the comparison test. And the integral seems too complicated for a homework problem from this book. (I tried.) Could someone just point me in the right direction? Thanks.

答: $\lim\limits_{k\rightarrow +\infty}k^3e^{-\sqrt{k}}=0$ and since $\sum\frac{1}{k^2}$ converges, $\sum ke^{-\sqrt{k}}$ converges aswell.


问:
How to simplify boolean expressions.
I'm struggling to understand what rules to apply when simplifying boolean expression.

For example:

$$ B+(A\cdot(C+B) \overline C) $$

I'm not sure how to simplify this expression. Here is my attempt.

$$
= B+AB\cdot(C+\overline C)
\\
= B+AB\cdot(1)
\\
= B(A+B)
\\
= B(B+A)
\\
= B
$$

答:
You can simplify this almost the same way you'd simplify an ordinary algebra expression:

$B+(A\cdot(C+B) \overline C)$

So just simplify it like an ordinary algebra expression first (technically we're using that AND is distributive over OR... in the same way multiplication is distributive over addition in ordinary algebra)

$=B+A(C\overline C+B\overline C)$. So here I'm just using $(C+B)\overline C = C\overline C + B\overline C$

Now I can use $C \overline C = 0$

$=B+A(0+B\overline C)$

$=B+A(B\overline C)$

$=B+B(A\overline C)$ (using commutativity and associativity of AND... just like in ordinary multiplication)

$=B(1+A\overline C)$ (factoring)

In boolean algebra 1+ anything is just 1.

$=B(1)$

$=B$

There are other tricks... but the initial main "tactic" is just to treat AND like multiplication and OR like addition.

Title: How to simplify boolean expressions.


问:
Express the integral of $|f|^{2}$ for a holomorphic function $f$ on a disc using its power series coefficients.
This post is related to this one: How to bound the $L^\infty$ norm of a holomorphic function by its $L^2$ norm on a larger domain?.

Suppose $f$ is a holomorphic function on $D_{R}(z_{0})$, then we know that it can be expressed as a power series $$f(z)=\sum_{n=1}^{\infty}a_{n}(z-z_{0})^{n},$$ where the power series converges in $D_{R}(z_{0})$, and converges absolutely and uniformly in $\overline{D_{r}(z_{0})}$ for all $0<r<R$.

The accpected answer of the above referred post seems suggesting that we can express the integral $$\int_{0}^{R}\int_{0}^{2\pi}f(z_{0}+re^{i\theta})\overline{f(z_{0}+re^{i\theta})}rdrd\theta$$ in terms of the coefficients of power series.

This is a pretty interesting fact, and I tried to prove it. At some stage I believe we need to interchange the summation and double integral, but I got stuck even at the beginning.  So $$f(z):=\sum_{n=1}^{\infty}a_{n}(z-z_{0})^{n}\ \ \text{and}\ \ \overline{f(z)}:=\sum_{n=1}^{\infty}\overline{a_{n}}\overline{(z-z_{0})}^{n},$$ this gives us
\begin{align*}
f(z)\overline{f(z)}&=\sum_{n=1}^{\infty}a_{n}(z-z_{0})^{n}\sum_{n=1}^{\infty}\overline{a_{n}}\overline{(z-z_{0})}^{n}\\
&=\sum_{n=1}^{\infty}|a_{n}|^{2}|z-z_{0}|^{2n}\sum_{k\neq m}a_{k}\overline{a}_{m}(z-z_{0})^{k}(\overline{z}-\overline{z}_{0})^{m}.
\end{align*}

Now we replace $z=z_{0}+re^{i\theta}$. The first summation then turned into $\sum_{n=1}^{\infty}|a_{n}|^{2}r^{2}$ and the second summation then turned into $$\sum_{k\neq m}a_{k}\overline{a}_{m}r^{k+m}e^{ik\theta-im\theta}.$$ What should I do to deal with the second sum? I am not sure what I should do now..

Thank you for any help!

答:
$$\int_0^{2\pi} (\sum_n a_n r^n e^{it n})(\sum_m \overline{a_m} r^m e^{-it m})dt$$
$$ = \sum_n\sum_m a_n \overline{a_m} r^{n+m} \int_0^{2\pi} e^{it (n-m)}dt =\sum_n a_n \overline{a_n} r^{2n} 2 \pi$$
This calculation is standard for (absolutely convergent) Fourier series


问:
find whether difference of square of two number is equal to cube of a given number
We are given a number b. We have to find is there any pair of integers exist which satisfy $a^2+b^3=c^2$.
Suppose b=3 then pair 3,6 exist. Is there any algorithm to check this existence?

答:
Every number not twice an odd number is a difference of two squares. No cube is twice an odd number, so every cube is a difference of two squares.

Given $n$, not twice an odd number, there is an easy way to find all ways to express $n$ as a difference of two squares (provided that you can factor $n$). I won't give the details, since surely this has been done already many times on this website. But for cubes it's particularly easy, since we have the factorization. $b^3=c^2-a^2$, where $c=(b^2+b)/2$ and $a=(b^2-b)/2$.


问:
Are there any finite examples of metachirality?
I learned about metachirality through an amazing Vihart series ([1] [2]), and the only examples I have encountered are infinitely long screws, and this infinite graph Wikipedia mentions.

Are there finite examples of metachirality? My intuition says no, because I can't even think of finite examples with glide symmetry or screw symmetry, both of which feel like a weaker thing to ask for. I also suspect that any example of metachirality must in some way rely on translation, and if I can somehow formalise and prove that notion, I think I'll be done. (Because I know that translational invariance implies that, at least in one direction, the object is infinite [3])

In the positive case, I am looking for examples of finite symmetry groups (so maybe the object to consider is infinite, but the symmetry group is somehow still finite), but examples with finite objects would be cool too.

In the negative case, are there any proofs that all examples of metachirality need to be finite? And do such proofs have anything to say about glide/screw symmetries?

答: In Conway & Smith's book "On Quaternions and Octonions", Section 4.6, it is stated that all the finite subgroups of O(3) coincide with their mirror images, i.e., are not metachiral. There is no proof given, but earlier in the chapter they enumerate all the finite subgroups of O(3), so you can just verify each one.</p>


问:
Let $P(x) = x^3 - 5 x^2 + 7 x + k = 0$. If P(x) has a double root in the set of integers, find the value of k using Vieta's Formulas.
I managed to derive Vieta's Formulas intuitively to find the values for each coefficient of $n$ degree of $x$ but didn't have success in finding the value of $k$. I figured out it was simultaneous equations however still couldn't find the value of $k$ being the constant in $P(x)$. Help would be great as questions like this will be in my year $10$ yearly exam.

P.S, please answer this question in reference to Vieta's Formulas and polynomial terminology, no calculus :) thanks.

答:
Let the roots be $r, r, s$. Vieta's formulas give the following system:
$$2r+s=5\\
r^2+2rs=7\\
r^2s=-k$$

Substituting the first equation into the second,
$$r^2+2r(5-2r)=7.$$

Can you take it from here?


问:
weak convergence and limit
This problem is in Real Analysis H.L. Royden 4th edition(pearson modern classic) No.25 at page 171.

Let X be normed linear space and for each $x \in X$, there is a bounded linear functional $T_{f} \in X^{*}$ such that $T_{f}(f)=||f||$ and $\|T_{f} \|_{*}=1$. Prove that if $\{f_{n}\}$ weakly converge to both $g_{1},g_{2}$, then $g_{1}=g_{2}$.

Here is my approach.

Since $\{f_{n}\}$ weakly converge to both $g_{1},g_{2}$, $T(f_{n}) \rightarrow T(g_{1})$ and $T(f_{n})\rightarrow T(g_{2})$ for every bounded linear functional on $X$. I know that $T(f_{n})$ is sequence on $\mathbb{R}$ and $\mathbb{R}$ is Hausdorff space, so I can conclude that $T(g_{1})=T(g_{2})$. So $T(g_{1}-g_{2})=0$ always.

Since $T($0$)=0$, I can conclude that $g_{1}=g_{2}$ if I can get a contradiction by assuming the contrary. I can guess that "$g_{1},g_{2}$ have same norm in $X$ thinking about assumptions in the problem, but still I can't prove that $g_{1}=g_{2}$. How can I approach to the next step?

答: I concluded. $T(g_{1})=T(g_{2})$ is equivalent to $T(g_{1}-g_{2})=0$, taking linear functional $T_{g}$ by assumption of the problem such that $T(g_{1}-g_{2})=\|g_{1}-g_{2}\|_{X}$, so I get $\|g_{1}-g_{2}\|_{X}=0$. Thus by the definition of norm, $g_{1}-g_{2}=$0$_{X}$, and $g_{1}=g_{2}$.

Title: weak convergence and limit


问:
Most compact representation of a plane
A common way to represent a plane is:

$$
a x + by + cz +d = 0
$$

Where (a, b, c) is the normal vector and corresponds to the orientation of the plane and $d$ is the distance from  the origin following the normal direction. This would define the plane using 4 scalars (a, b, c, d)

If (a, b, c) are unit vector, so... wouldn't be possible to represent a plane with only 3 numbers?

$$
(p_{x}, p_{y}, p_{z}) = (a, b, c)*d
$$

We could easily go back to the original form by doing:

$$
\frac{(p_{x}, p_{y}, p_{z})}{||(p_{x}, p_{y}, p_{z})||} = (a, b, c)
$$

$$
||(p_{x}, p_{y}, p_{z})|| = d
$$

We would still be representing both orientation and distance of the plane but with only 3 scalars $(p_{x}, p_{y}, p_{z})$ which give us advantages when thinking about memory performance on computer algorithms since we are fully defining the geometric information of a plane but with less variables.

Why isn't this representation more frequently used since it is more compact comparing to the hessian normal form? Is there something I'm missing?

This made me think: Is there a more compact representation of a plane in $\mathfrak{R}^3$ that uses less then 3 scalars?

答:
Sure, easiest way to see this is first starting in lower dimensions. If we have a point in $\mathbb{R}$, we can clearly parameterize this by a single value which satisfies the equation $x = c$. For a line in $\mathbb{R}^2$ we can again easily represent this with two real numbers, typically we write $y=mx+b$. Similarly in $\mathbb{R}^3$ we can parameterize a plane by $z= \alpha x  + \beta y + \gamma$ where if we are given a plane $ax + by + cz = d$, we find that the coefficients $\alpha = -a/c$, $\beta = -b/c$ and $\gamma = d/c$ will define the same plane, simply by solving for $z$.
$$$$
Note that we cannot parameterize a vertical line using the equation $y=mx+b$ for some $m,b\in\mathbb{R}$ but could instead use a polar parameterize where $(r,\theta)\mapsto \cos(\theta)x + \sin(\theta)y = r$. A similar problem occurs for the plane in three dimensions, it might be useful to work out a solution for yourself.

Title: Most compact representation of a plane


问:
Transforming a Weibull Distribution
X follows a Weibull distribution (k, $\lambda$), but I want to find the expected value of Y where
$$y = A e^{(-Bx)}$$
So I'm trying to solve:
$$E[Y] = {\int_0^\infty}A e^{(-Bx)}{f(x)}{dx}$$
where $f(x)$ is the pdf of the Weibull distribution.
$$f(x)=\frac {k}{\lambda} (\frac {x}{\lambda})^{k-1}e^{-(x/\lambda)}$$

This is as far as I've gotten:
$$E[Y] = A \frac {k}{\lambda^k} {\int_0^\infty} x^{k-1}(e^{-x})^{(\frac {1}{\lambda}+B)}{dx}$$

I can see the form of the Gamma function in there, but I'm not sure how to isolate it.

Note: I actually know the expected value of Y that I want (as well as the values of A, k, and $\lambda$). I'm hoping the solution here will be easy enough to invert and then solve for B.

答:
For the time being, let $a=B+\frac 1 \lambda$
$$\int_0^\infty x^{k-1}\,e^{-a x}\,dx=a^{-k} \Gamma (k)\quad \text{if} \quad \Re(k)>0\land \Re(a)>0$$ So, you want to solve for $a$ the equation
$$a^{-k} \Gamma (k)=\frac{\lambda ^k}{A k}\mathbb E[Y]\implies a=\frac 1\lambda\left(\frac{A k \Gamma (k)}{\mathbb E[Y]}\right)^{\frac{1}{k}}$$ then $B=a-\frac 1 \lambda$

Title: Transforming a Weibull Distribution


问:
Find the cosets in $\mathbb{Z}[\sqrt{-5}]/Q_1$
Find the cosets in $\mathbb{Z}[\sqrt{-5}]/Q_1$ where $Q_1=\{3a+(1+\sqrt{-5})b|a,b\in\mathbb{Z}[\sqrt{-5}]\}$. I am studying this and am having trouble understanding how to find cosets of quotient rings. I understand that it means the congruence classes modulo $Q_1$ and I know in this case there should be 3 distinct cosets. But how do I find the cosets? Please explain how you found them. Thank you.

答:
Otherwise
$$\Bbb{Z}+(1+\sqrt{-5})\Bbb{Z} =\bigcup_{a=0}^2 a+ 3\Bbb{Z}+(1+\sqrt{-5})\Bbb{Z}$$
(disjoint union)

then show that $Q_1=3\Bbb{Z}+(1+\sqrt{-5})\Bbb{Z}$


问:
How to find coordinates on a circle that are theta degrees away?
given:

cx,cy - coordinates of the center of a circle

x, y - coordinates on a circle

theta - number of degrees

How do you find cooridnates x',y' that are theta degrees away from x,y on the circle?

答:
Redefine $X = x - c_x$ and $Y = y - c_y$.

Calculate your starting angle as $\phi = \tan^{-1}(Y/X)$. (Watch the quadrant.)

Calculate the radius of the circle using $r = \sqrt{X^2 + Y^2}$.

Calculate your new $X'$ and $Y'$ as $X' = r \cos(\phi + \theta)$ and $Y' = r \sin(\phi + \theta)$.

Convert back: $x' = X' + c_x$ and $y' = Y' + c_y$.


问:
Intersection between sphere and cylinder
I have a sphere and a cylinder.

I have the center and the radius of each of them.

the sphere:

radius = $r_1$

center = $(x_1,y_1,z_1)$

the cylinder:

radius = $r_2$

height = $h_2$

center = $(x_2,y_2,z_2)$

how do I know if there is an intersection?

I read this one: http://en.wikipedia.org/wiki/Sphere%E2%80%93cylinder_intersection

but my center is not $(x_1,0,0)$ but $(x_1,y_1,z_1)$.

the radius in my sphere is 1, and the center point is $(x_1,y_1,z_1)$

the radius in my cylinder is 1, height is 10 and the center point is $(x_2,y_2,z_2)$.

I thought to convert my center point of $(x_1,y_1,z_1)$ into $(x_1,0,0)$ and then know if there is a intersection or not, according to the reference I linked..

any help appreciated!

答: You can translate the points to put the center of the sphere at the origin by making the center of the sphere $(0,0,0)$ and the center of the cylinder $(x_2-x_1,y_2-y_1,z_2-z_1)$  This is a translation of the space.  If you want your results in the original coordinate system, just add $(x_1,y_1,z_1)$ to them.</p>


问:
Counterexample of closed graph theorem when the image space is Hausdorff space but not compact?
Closed graph theorem on topological space requires the image space Y is compact Hausdorff. Some counter examples on Wikipedia are all non-Hausdorff which is considered more exotic examples.

What are the examples in which the map X->Y is continuous but not closed (as defined in http://en.wikipedia.org/wiki/Closed_graph_theorem, not the closed map in general topology), and the image space Y is Hausdorff but not compact?

答:
There is no such example. Aa long as $Y$ is Hausdorff the graph of $f$ is closed
: Let $(x_i,f(x_i))_{i \in I}$ be  a net in the graph converging to $(x,y)$ in the product topology. Then $x_i \to x$ and $f(x_i) \to y$. By continuity $f(x_i) \to f(x)$. Since limits are unique in a Hausdorff space  we get $y=f(x)$. Hence the graph is closed.


问:
find recurrence relation for strings of length n over {A,B,C}(none of the strings should have consecutive As)
Assume that an is number of strings of length n over A B C and none of strings have consecutive As

I have no idea how to find a relation for this

答:
For a recurrence relation, define $u_n$ to be the number of strings of length $n$ with letters $A,B,C$ such that there are no consecutive $A$s, and additionally, the string must end in $A$. Define $v_n$ to be the number of strings of length $n$ with letters $A,B,C$ such that there are no consecutive $A$s, and additionally, the string must NOT end in $A$. Hence, we are to find $u_n+v_n$. We provide a recursive relation:

To find $u_{n+1}$, we know that the last letter has to be $A$, and the first $n$ letters have to be forming a string of $A,B,C$ which does not end in $A$, and has no consecutive $A$s. Hence, $u_{n+1}=v_n$.

Next, to find $v_{n+1}$, we know that the first $n$ letters give us $u_n+v_n$ choices and the last letter can either be $B$ or $C$, giving us $2$ choices. Hence, $v_{n+1}=2(u_n+v_n)$.

Now, we can substitute $u_n=v_{n-1}$ in the recurrence for $v_{n+1}$ to yield $v_{n+1}=2(v_n+v_{n-1})$. After solving this recurrence, we can conclude that the answer is $u_n+v_n=v_n+v_{n-1}$.

This recurrence can be solved using the standard technique. The quadratic $x^2=2x+2$ has roots:
$$\frac{2 \pm \sqrt{(2^2+4 \cdot 2 \cdot 1)}}{2}=1 \pm \sqrt{3}$$
So, we can write $v_n=P(1+\sqrt{3})^n+Q(1-\sqrt{3})^n$. This has to hold good for $n=1,2$. Since $v_1=2$ amd $v_2=6$, we have:
$$\begin{cases}
(P+Q)+\sqrt{3}(P-Q)=2 \\
4(P+Q)+2\sqrt{3}(P-Q) = 6
\end{cases} \implies (P,Q)=\bigg(\frac{1+\sqrt{3}}{2},\frac{1-\sqrt{3}}{2} \bigg)$$
Hence, $v_n=\frac{1}{2}((1+\sqrt{3})^{n+1}+(1-\sqrt{3})^{n+1})$, and thus:
$$v_{n+1}+v_n=\frac{1}{2}((1+\sqrt{3})^{n+1}+(1-\sqrt{3})^{n+1})+\frac{1}{2}((1+\sqrt{3})^{n}+(1-\sqrt{3})^{n})$$


问:
How to graph $\ln(x^2+x+1)$
The question is to find the number of solutions of the equation $e^{\sin x+\cos x}=x^2+x+1$

To which I simplified to $\sin x+\cos x=\ln(x^2+x+1)$

I can graph $\sin x+\cos x$ quite easily. But I struggle in graphing $\ln(x^2+x+1)$

If a method could be given to graph such a function that would be highly appreciated.

答:
The red graph is that of $\ln(x^2+x+1)$, and the blue graph (easy to draw) is for $\sin x + \cos x$. How to draw $f(x) = \ln(x^2+x+1)$? Let's see.

I would start by noticing $f(0)=0$. Then I would find $$f'(x) = \frac{2x+1}{x^2+x+1}$$

Note that $x^2+x+1 = (x+0.5)^2 + 0.75 > 0$ for all $x\in\mathbb{R}$. So the sign of $f'(x)$ is determined by the sign of $2x+1$. $f'(x)>0$ for $x>-0.5$, $f'(x) = 0$ for $x=-0.5$ and negative otherwise. We deduce that $x=-0.5$ is a point of local minima for $f(x)$ and $f(x)$ strictly increases for $x>-0.5$ (and strictly decreases for $x<-0.5$). All that remains is to find $f(-0.5)$ which I'm sure you can do.

In the end, you get the graph below and exactly two solutions for the equation.

P.S.

It is worth noting that $g(x) = \ln x$ is an increasing function, so it would actually be enough to study the monotonicity of $h(x) = x^2+x+1$ to draw the graph of $f(x) = g(h(x))$. I hope you can see why.

Title: How to graph $\ln(x^2+x+1)$


问:
Probability, choose $2$ numbers from a set
Randomly choose two different numbers from $3,7,9,12,27$, one after the other.
If the first number chosen is $x$ and the second number chosen is $y$, what's the probability that $x^n = y$, where $n$ is an integer?

It might be a simply question but I can't get my answer match to the given answer which is $\frac{1}{10}$.

My approach as we can choose either $3$ for the first number and $9$ or $27$ for the second one.

Alternatively, we can have $9$ as first number and $27$ as second number.

Thus, we have $$\frac{1}{5} \times \frac{2}{4} + \frac{1}{5} \times \frac{1}{4} = \frac{3}{20}$$

I'm pretty sure I did something wrong here, but I can't figure it out at the moment. Any suggestion?

答: $9$ then $27$ does not satisfy $9^n=27$, so there are only $2$ of the $20$ choices that satisfy the requirment.


问:
If an open set $U$ in $\Bbb R^2$ contains an closed disk, then there is an larger closed disk containing the original one and contained by $U$

Let $U\subset \Bbb R^2$ be an open set such that the closed disk $\bar{B}(x,1)\subset U$. Then there is $\varepsilon >0$ such that $\bar{B}(x,1+\varepsilon)\subset U$.

It is obvious but I'm not sure how to prove it. If the boundary of $\bar{B}(x,1)$ intersects the boundary of $U$, then I think that the closed disk will not be contained by $U$. So, $\alpha=\inf\{\|t-y\| : t\in \bar{B}(x,1), y\in\partial U\}>0$, then if we choose $\varepsilon=\frac{\alpha}{2}$, it looks like $\bar{B}(x,1+\varepsilon)\subset U$.

But I not quite sure how to explain rigorously why $\alpha>0$. Hope someone could help me. Thanks for any suggestions.

答:
Hint:

If $\ U=\mathbb{R}^2\ $ then any $\ \epsilon $ will do. Otherwise, there exists a sequence $\ \big\{y_n\big\}_{n=1}^\infty\subseteq U^c\ $ such that $\ \lim_\limits{n\rightarrow\infty}\inf\big\{\big\|t-y_n\big\|: t\in\overline{B}(x,1)\big\}=\alpha\ $, and since it must be bounded, it must contain a convergent subsequence whose limit $\ y^*\ $ satisfies $\ \inf\big\{\big\|t-y^*\big\|: t\in\overline{B}(x,1)\big\}=\alpha\ $. Since $\ U^c\ $ is closed, then $\ y^*\in U^c\ $ and so $\ \inf\big\{\big\|t-y^*\big\|: t\in\overline{B}(x,1)\big\}\ne0\ $.

Reply to OP's query below

By the definition of infimum, for any positive integer $\ n\ $ there must exist $\ y_n\in\partial U\subseteq U^c\ $ and $\ t_n\in\overline{B}(x,1)\ $ such that $\ \big\|t_n-y_n\big\|<\alpha +\frac{1}{n}$. Then
$$
\alpha\le\inf\big\{\big\|t-y_n\big\|: t\in\overline{B}(x,1)\big\}\le\big\|t_n-y_n\big\|<\alpha +\frac{1}{n}\ ,
$$
so $\ \lim_\limits{n\rightarrow\infty}\inf\big\{\big\|t-y_n\big\|: t\in\overline{B}(x,1)\big\}=\alpha\ $ by the sandwich theorem.

Since  $\ \big\|y_n\big\|-\big\|x\big\|-1\le$$\big\|y_n\big\|-\big\|t\big\|\le$$\big\|t-y_n\big\|\ $ for all $\ t\in\overline{B}(x,1)\ $, then
$$
\big\|y_n\big\|\le\big\|x\big\|+1+\inf\big\{\big\|t-y_n\big\|: t\in\overline{B}(x,1)\big\}\ ,
$$
and since the sequence on the right of this inequality converges to $\ \|x\|+1+\alpha\ $, it must be bounded, and hence so must $\ y_n\ $.


问:
What does having a bar on a manifold mean?
If $M$ is a manifold then what is denoted as $\overline{M}$?

I am guessing that it means a reversal of orientation.

Related to the above I would like to understand the following construction of a $d$-dimensional Riemannian category that I recently came across:

Objects are $(d−1)$ Riemannian manifolds. Morphisms between two oriented $(d − 1)$-dimensional Riemannian manifolds $N_1$ and $N_2$ are oriented $d$-dimensional Riemannian manifolds $M$, such that $\partial M = N_1 ⊔ \overline{N_2}$. The orientation on all three manifolds should naturally agree, and the metric on $M$ agrees with the metric on $N_1$ and $N_2$ on a collar of the boundary. The composition is the gluing of such Riemannian cobordisms.

I would like to know more about this intuition of defining a morphism between two manifolds as another manifold itself! One is used to thinking of morphisms as maps.

答:
You've simply encountered the definition of a topological quantum field theory (in the formulation gave by M. Atiyah). I can provide you lots of documentation about that, but I'm a little hurried now... I'll edit this post during the day. For the moment I'm copypasting this:

Topological Quantum Field Theory, or TQFT for short, is a notion which originally arose from ideas by E. Witten in quantum physics (SUSY-Quantum Mechanics). Since then it has developed considerably in a number of directions, and in particular has had a pervasive influence in algebraic topology, by means of its capacity of computing algebraic invariants of a suitable class of manifolds, using so-called "cobordisms" between closed manifolds (smooth manifolds having them as boundaries).

Atiyah gave a first axiomatization of Witten's theory in his paper [1]; I'll try to focus on the illuminating analogy (duality) beween Atiyah's axioms  for TQFT and the Eilenberg-Steenrod axioms for homology as in [2].
Atiyah's approach is implicitly categorical; Jacob Lurie recovered this approach and considerably strengthened it in his paper [3]: by now we define an "n-dimensional, k-valued Topological Quantum Field Theory"as a monoidal functor
between the category of n-cobordisms and that of finite dimensional k-vector spaces.

If you have time and spirit to do so, I warmly recommend to give a glance to a "simplified n-categorical formulation" of TQFTs given by B. Toen
in his [4].

References.

[1] Atiyah, Michael (1988), "Topological quantum field theories", Publications Mathématiques de l'IHÉS 68 (68): 175–186

[2] Samuel Eilenberg, Norman E. Steenrod, Axiomatic approach to homology theory, Proc. Nat. Acad. Sci. U. S. A. 31, (1945). 117–120.

[3] Lurie, Jacob, On the Classification of Topological Field Theories

[4] Bertrand Toen, Higher categorical structures in TQFT. Lecture in MPI-Bonn, June 2000.


问:
Why is there missing solutions here

As we can see there is a missing portion in the purple graph, Why do we have missing solutions here?

Also happens in other graphing calculators.

I feel like we should get all the solutions but why are some parts missing?

答:
This has to do with numerical algorithms to evaluate the function and plot the curve.  One simple algorithm (that you may have learned in school) basically involves selecting various $x$-values and then calculating the corresponding $y$-values, then connecting adjacent points with a curve.  But a computer does this more rapidly than a human, so it can select many more points to evaluate.

But a computer doesn't handle points at which the evaluation of the function no longer yields a real number:  if $|x| > 1$, then $\pm \sqrt{1 - x^2}$ is an imaginary number.  So in the neighborhood of $x = 1$, the computer is happily plotting values like $x = 0.99351$, but then at the next step, it might choose $x = 1.00537$ and suddenly it encounters a result it cannot plot.  So there is no way for it to joint the previous point to the next one.  This is why the curve stops before $x = 1$, because the algorithm would somehow need to know that $x = 1$ is the largest value for which $y$ is defined.  But in practice, it is not generally possible for a computer to know this, because it involves solving for the inverse function.

What more sophisticated algorithms do instead is to leverage the available computing power, and in regions with discontinuities or irregular behavior, it will adaptively subsample the neighborhood--choosing finer increments for the $x$-values so that it can gather more information about the behavior of the function at that point.  But this comes at some computational cost, and even doing this may not adequately "close" the curve--if we zoom in further (as you show in your first picture), the computer also has to compensate by evaluating more points.

So, why does the red curve close when the purple and green ones do not?  This has to do with the choice of algorithm.  In the red curve example, you write $$y^2 = (1.1)^2 - x^2.$$  This way of writing the equation allows the use of a different algorithm to find the behavior of the relationship between $x$ and $y$.  For example, a crude method is to select values for both $x$ and $y$ and see how closely they satisfy the equation.  What I mean by this is, the computer might rewrite the equation as $$f(x,y) = x^2 + y^2 - (1.1)^2,$$ and try to detect where the function changes sign, since the curve lies where $f(x,y) = 0$.  For instance, the computer might select $(x,y) = (0.9, 0.2)$.  If the result is close to $0$, then it knows to search more in that area.  If it is off, then the computer knows to search elsewhere.  It does this for many ordered pairs in the viewing region, successively refining the guesses until the error is below a certain threshold.  Because the guesses don't result in imaginary numbers, and the sign of the guess also gives some indication as to which direction to search, this algorithm does not encounter the kind of problems that the first one does.  To be sure, it has other problems, but a detailed discussion is beyond the scope of this answer.

Title: Why is there missing solutions here


问:
Proving $\max(p) \geq \sum_d p_d^2$
I want to prove
$$\max(p) \geq \sum_d p_d^2,$$
where $p$ is a simplex, satisfying $\sum_d p_d = 1$.

It looks numerically true since I run the following code in R and got true:

library(gtools);
D<-5;
n<-100000;
sum(apply(rdirichlet(n,rep(1,D)), 1, function(x){max(x)-sum(x*x)})>=0)==n

Any suggestions are appreciated. Best,

答:
We have $p_d \leq \max(p)$ so assuming $p_d \geq 0,$ it follows that $p_d^2 \leq p_d \cdot \max(p).$

Then $$\sum_d p_d^2 \leq \sum_d p_d \cdot \max(p) = 1 \cdot \max(p)$$

Title: Proving $\max(p) \geq \sum_d p_d^2$


问:
Formula Manipulation?
I have a specific formula = $a$^2 + $b$^3 = $c$^2

and let's say I need to print out $a$ and $c$, but only $b$ is given to us.

Example (I could think of):

B = 3

3 ^ 2 + 3 ^ 3 => 9 + 27 = 36 (6 ^ 2)

So we would just output 3 and 6 since it works based on our formula. I feel like there's some sort of simple mathematical method for doing this in a way where it will always work. I eventually want to program this, but how can I manipulate this so it'll work everything (just by providing $b$)?

In short terms, I want to display the values of $a$ and $c$ where $b$ is given to us, and works with our formula: $a$^2 + $b$^3 = $c$^2

答:
Write the equation as $b^3=c^2-a^2=(c+a)(c-a)$.  You need to factor $b^3$ into two factors of the same parity as $c+a$ and $c-a$ are either both even or both odd.  For each factorization you can solve the simultaneous equations to get $c$ and $a$.

In your example we can factor $b^3=27=1\cdot 27=3 \cdot 9$  Taking the second
$$c+a=9\\c-a=3\\2c=12\\c=6\\a=3$$  Similarly for the first
$$c+a=27\\c-a=1\\2c=28\\c=14\\a=13$$
and we also have $13^2+3^3=14^2$

If $b$ is composite there will be more choices.

Title: Formula Manipulation?


问:
Prove convergence/divergence of $ \sum_{k=0}^\infty ke^{-\sqrt k} $
Prove the convergence/divergence of the following series.
$$ \sum_{k=0}^\infty ke^{-\sqrt k} $$
The root and ratio tests don't work because r = 1. That leaves the integral and comparison tests, but I cannot for the life of me find an expression that works for the comparison test. And the integral seems too complicated for a homework problem from this book. (I tried.) Could someone just point me in the right direction? Thanks.

答:
We can compare $k e^{-\sqrt{k}}$ and $\dfrac{1}{k(k+1)}$.

To do this, I assert that $\lim_{k\rightarrow \infty}\dfrac{k^{2}(k+1)}{e^{\sqrt{k}}}=0$.

Note that by L'hopital's rule again and again, we can calculate limit of $\dfrac{s^{4}(s^{2}+1)}{e^{s}}$ as follows.

$\lim_{s\rightarrow\infty}\dfrac{s^{4}(s^{2}+1)}{e^{s}}=\lim_{s\rightarrow\infty}\dfrac{s^{6}+s^{4}}{e^{s}}=\lim_{s\rightarrow\infty}\dfrac{6 s^{5}+4 s^{3}}{e^{s}}=\lim_{s\rightarrow\infty}\dfrac{30 s^{4}+12 s^{2}}{e^{s}}=\lim_{s\rightarrow\infty}\dfrac{120 s^{3}+24 s}{e^{s}}=\lim_{s\rightarrow\infty}\dfrac{360 s^{2}+24}{e^{s}}=\lim_{s\rightarrow\infty}\dfrac{720 s}{e^{s}}=\lim_{s\rightarrow\infty}\dfrac{720}{e^{s}}=0$

Since $\sqrt{k}$ goes $\infty$ as $k \rightarrow \infty$, we can conclude that $\lim_{k\rightarrow \infty}\dfrac{k^{2}(k+1)}{e^{\sqrt{k}}}=0$.

Thus, by definition of limit, we get there is $n\in\mathbb{N}$ such that $k \ge n$ gives $\dfrac{k^{2}(k+1)}{e^{\sqrt{k}}}<1$.

Dividing both side to $k(k+1)$, $k e^{-\sqrt{k}}<\dfrac{1}{k(k+1)}$.

So $\sum_{k=0}^{\infty} k e^{-\sqrt{k}} = \sum_{k=0}^{n} k e^{-\sqrt{k}}+\sum_{k=n+1}^{\infty} k e^{-\sqrt{k}} \le \sum_{k=0}^{n} k e^{-\sqrt{k}}+\sum_{k=n+1}^{\infty}\dfrac{1}{k(k+1)} <\infty$


问:
Calculate the limit $\frac{(2n-1)x_1+(2n-3)x_2+\dots +3x_{n-1}+x_n}{n^2}$ when $x_n\to x$.

Let $\{x_n\}$ be a sequence in $\Bbb R$ and $x_n\to x$ as $n\to \infty$.
Then $$\frac{(2n-1)x_1+(2n-3)x_2+\dots +3x_{n-1}+x_n}{n^2}\to x$$.

Do anyone know how to solve this kind of problem efficiently? I think I need to estimate $$\left|\frac{(2n-1)x_1+(2n-3)x_2+\dots +3x_{n-1}+x_n}{n^2}-x\right|$$ Is there some other ways to calculate the limit directly?

Thanks for any comments.

答:
(Fill in the gaps as needed. If you're stuck, explain what you've done and why you're stuck.)

Show that if $ x_i \rightarrow x$, then$ \frac{1}{n} \sum x_i \rightarrow x$.

Show that if $ x_i \rightarrow x$, then $ \frac{1}{n^2} \sum (2i-1) x_i \rightarrow x$.

Hence, conclude that $ \frac{2}{n} \sum x_i - \frac{1}{n^2} \sum (2i-1) x_i \rightarrow x $


问:
Functional derivative of a complex fourier sum
I'm trying to parse a derivation of a lagrange-multiplier fourier technique, and can't quite grasp an intermediate step. The author sets up

$$ e_x(z,t) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty}{(f(\omega) e^{- j (\omega/c_0)n(\omega)z})\ e^{j\omega t} d\omega} $$

Where n() is a complex function of $\omega$. c, z, are real, and $f(\omega)$ is the function of interest. Now they set up a functional to minimize,

$$\xi = -e_x(z,t) + \lambda W $$

and then take the gradient of e_x W.R.T $f(\omega)$.

$$ \nabla \xi = - \frac{1}{\sqrt{2 \pi}} \left( e^{- j (\omega/c_0)n(\omega)z} \right)^\star \  e^{-j\omega T} +\ ...$$

where * is the complex conjugate.

Both Mathematica's VariationalD and Matlab's functionalDerivative both seem to balk at taking the gradient of that improper integral for some reason.

This is probably super straightforward, but...where did that complex conjugate come from? Previous questions would seem to suggest that the gradient of F(f(omega)) should be f(omega).

答:
I still have no mathematical understanding of how this could possibly be the case, but (as discussed in Waveform Optimizations for Ultra-Wideband Radio Systems, David M. Pozar) this problem is exhaustively considered in L. E. Franks, Signal Theory, revised edition, Dowden & Culver, Stroudsberg, PA, 1981..

In that text, the gradient of a transfer function is indeed given as the transfer function multiplied by its complex conjugate.

Note also that this isn't obviously the same kind of variational problem that tools like functionalDerivative are set up to solve; they seem to assume that you want to minimize the functional -f(x) + lambda W integrated over some path with well-defined endpoints - see Feynman's Principle of Least Action, but it's not obvious to me that this construction makes any sense for this problem. I might be grossly misunderstanding, however.


问:
Given a topology $\tau$ on $X \times Y$, can we find $\tau_X$ on $X$ and $\tau_Y$ on $Y$ such that $\tau_X \times \tau_Y = \tau$
Suppose we have spaces $X, Y$ and a topology $\tau$ on $X \times Y$. Can we find a topology $\tau_X$ on $X$, $\tau_Y$ on $Y$ such that $\tau_X \times \tau_Y = \tau$ (the product is referring to the product topology).

I think not. What if $\tau$ is a separable complete metric space?

I could also get away with somemthing weaker, like $\tau \subseteq \tau_X \times \tau_Y$ provided $\tau_X$ and $\tau_Y$ have sufficiently neat property (i.e, they're not discrete topologies or something).

答:
The answer is no even when $X=Y=\Bbb R$, and $\tau$ is a separable metric topology on $X\times Y$.

Let $\varphi:\Bbb R^2\to\Bbb R$ be a bijection, let $\tau_e$ be the Euclidean topology on $\Bbb R$, and let $\tau=\{\varphi^{-1}[U]:U\in\tau_e\}$; let $X$ denote $\Bbb R^2$ equipped with the topology $\tau$. Then $X$ is homeomorphic to $\langle\Bbb R,\tau_e\rangle$, so $X$ is path connected, and $X\setminus\{p\}$ is not path connected for any $p\in X$.

Let $\tau_0$ and $\tau_1$ be topologies on $\Bbb R$. If $\tau$ is the the product topology on $\Bbb R^2$ generated by $\tau_0$ and $\tau_1$, then $\langle\Bbb R,\tau_i\rangle$ must be path connected for $i\in\{0,1\}$. But it’s easily seen that if $Y$ and $Z$ are path connected, and $p\in Y\times Z$, then $(Y\times Z)\setminus\{p\}$ is path connected. (E.g., you can use the idea that I used in this answer, or the one that I used in this answer.) Thus, the product topology on $\Bbb R^2$ generated by $\tau_0$ and $\tau_1$ cannot be $\tau$.

For a rather different argument, notice that if the product topology were $\tau$, then $\langle\Bbb R,\tau_1\rangle$ would have to be connected. But then $\big\{\{x\}\times\Bbb R:x\in\Bbb R\big\}$ would be a partition of $X$ into uncountably many non-trivial connected sets, which is impossible: $X$ is homeomorphic to $\Bbb R$, every non-trivial connected subset of $\Bbb R$ contains a non-empty open interval, and $\Bbb R$ does not contain any uncountable family of pairwise disjoint, non-empty open intervals.


问:
Evaluating a recurrence relation with non-contiguous initial conditions
Is there a general way of evaluating the points in a recurrence relation when the initial conditions are not contiguous? For instance:

$$T(i) = f\left(T(i-1), T(i-2)\right),$$
with $T(a) = A$ and $T(b) = B$, and $a+1 \leq i \leq b-1$ (and with whatever $f$ and $a < b+1$).

I am concerned with the (possibly numeric) evaluation of $T(i)$, not with the solution of the above equation (since it has no closed-form solution in the general case).

If it is not possible in general, what about particular cases (other than $b$ = $a+1$ or $f$ "solvable").

答:
In the general case your recurrence boils down to a system of equations in the $T(i)$. You might try to solve that one directly (if the range isn't too big).

Another approach would be to start with the (known) $T(a)$, guess a $T(a + 1)$ and compute forward to $T(b)$. Use some iterative method (e.g. Brent's) to get the correct $T(a + 1)$.

If the recurrence has a particular form, it might be possible to solve it exactly (using methods similar to the ones that work for non-linear differential equations). Or you might apply some change of variable that makes it better behaved for some of the above techniques. Or approximate it by a differential equation, and solve that one...


问:
The sum $(1+i)^x + (1-i)^x = 0$
I recently saw that $(1+i)^x + (1-i)^x = 0$  for $x = 2\pmod 4$, but can't figure out how its true. Any help/hints would be greatly appreciated!

答: If $x\equiv 2\pmod 4$ is an integer, then $i^x=-1$ and hence $$(1-i)^x=-i^x(1-i)^x=-(i(1-i))^x=-(i+1)^x.$$ The conclusion follows.

Title: The sum $(1+i)^x + (1-i)^x = 0$


问:
The ideal generated by $x_\beta, h_\beta - \lambda(h_\beta), \beta \in \Phi^+,$ is the same as the ideal generated by taking all $\beta \in \Delta$.
Let $L$ be a semisimple Lie algebra with root system $\Phi$, Cartan subalgebra $H$, positive roots $\Phi^+$ and simple roots $\Delta$. Denote by $U(L)$ its universal enveloping algebra.

Fix an arbitrary $\lambda \in H^*$ and a system of generators  $\{x_\gamma,y_\gamma,h_\gamma: \gamma \in \Phi\}$ of $L$ such that the subalgebra generated by $x_\gamma, y_\gamma, h_\gamma$ is isomorphic to $\mathfrak{sl}_2$.

Let $I(\lambda)$ be the ideal of $U(L)$ generated by  $x_\beta, h_\beta -\lambda(h_\beta).1$ for all $\beta \in \Phi^+$ and $\tilde I(\lambda)$ be the ideal of $U(L)$ generated by $x_\alpha, h_\alpha - \lambda(h_\alpha).1$ for all $\alpha \in \Delta$. Prove that $I(\lambda) = \tilde I(\lambda)$.

Since the simple roots $\Delta$ are positive, $I(\lambda)$ contains all the generators of $\tilde I(\lambda)$, so $\tilde I(\lambda) \subseteq I(\lambda)$.

For the other direction, let $\beta \in \Phi^+$. All we have to do is to prove that $x_\beta$ and $h_\beta-\lambda(h_\beta).1$ lie in $\tilde I(\lambda)$. By some previous result in our book, there are $\alpha_1,\cdots, \alpha_s \in \Delta$ such that $\beta = \alpha_1+\cdots+\alpha_s $ with each partial sum $\alpha_1+\cdots +\alpha_i \in \Phi^+, 1\leq i \leq s.$ We prove by induction on $s$ that $x_\beta \in \tilde I(\lambda)$, the case $s=1$ being clear. Since $L_\beta = [L_{\alpha_1+\cdots+\alpha_{s-1}}, L_{\alpha_s}],$ then $x_\beta = c. [x_{\alpha_1+\cdots + \alpha_{s-1}}~,  x_{\alpha_s}]$ for some $c\neq 0$. Now the induction hyphotesis applied to $x_{\alpha_1+\cdots+\alpha_{s-1}}$ and the inclusion $L\subseteq U(L)$  assures that $x_\beta \in \tilde I({\lambda})$.

I can't prove that $h_\beta-\lambda(h_\beta).1$ lies in this ideal. Since $h_\beta = [x_\beta,y_\beta]$ and $x_\beta$ had seen to be in $\tilde I(\lambda)$, we have that $h_\beta \in \tilde I(\lambda)$. But the sum of $-\lambda(h_\beta).1$  with $h_\beta$ is actually bothering me and I am not sure how to finish the argument.

Any help?

答:
Since $\Delta \subset \Phi $ is a base, $\Delta$ spans $H^*$. Recall that $H^*$ is identified with $H$ via $\lambda \mapsto  t_\lambda$, where $t_\lambda$ is the unique element of $H$ such that $\kappa(t_\lambda, h)=\lambda(h)$ for all $h \in H$. Here $\kappa$ is the Killing form. For any $\alpha \in \Phi$, we have $h_\alpha = 2t_\alpha / \kappa(t_\alpha, t_\alpha)$. Thus $$h_\alpha - \lambda(h_\alpha).1 = \frac{2}{\kappa(t_\alpha, t_\alpha)}(t_\alpha - \lambda(t_\alpha).1)$$
in $U(L)$.

Now write $\beta = \alpha_1 + \dots + \alpha_s$. Then $t_{\alpha_i} - \lambda(t_{\alpha_i}).1 \in \tilde{I}(\lambda)$ for all $i$ since they are scalar multiples of $h_{\alpha_i}-\lambda(h_{\alpha_i}).1 \in \tilde{I}(\lambda)$. Observe that $t_\beta = t_{\alpha_1} + \dots + t_{\alpha_s}$, which implies $t_\beta - \lambda(t_\beta).1 \in \tilde{I}(\lambda)$. It follows that $h_\beta-\lambda(h_\beta).1 \in \tilde{I}(\lambda)$.


问:
Twice continuously differentiable function such that $f’’-f<0, \forall x\in(0,1)$.
Let $f:[0,1]\to\Bbb R$ be twice continuously differentiable function such that  $f’’(x)-f(x)<0, \forall x\in(0,1)$ and $f(0)=f(1)=0$, then which of the following statements is/are true about $f$?

$1.$ $f$ has at least one zero in $(0,1).$

$2.$ $f $ has at least two zeros in $(0,1).$

$3.$ $f(x)>0, \forall x\in (0,1)$.

$4.$ $f(x)<0, \forall x\in (0,1)$.

If i consider the example $x(1-x)$ on $[0,1]$ , then only option $3$ is correct one , but i want to solve the problem theoretically without using example or counter examples.  It seems that last option is false because that $f$ can’t be both negative and concave .  Please suggest how to discard rest options. Thanks .

答:
Assume that $f$ attains its minimum at some point $c \in (0, 1)$. Then
$$
 f''(c) < f(c) \le f(0) = 0 \, ,
$$
contradicting the fact that the second derivative is $\ge 0$ at a minimum in the interior of the interval.

It follows that $f$ attains its minimum on $[0, 1]$ only at the boundary points $x=0$ and $x=1$.

So (3) is true, which implies that the other statements are wrong.


答:
One notable fact about the product topology is that it is the weakest topology that makes both projection maps continuous.   Also note,  if X and Y  have topologies  A and B so that the projection maps from $X\times Y$ are continuous, then $A\times B$  is contained inside  the topology on $X\times Y$, C.  So if C is to be written as a product of A and B,  then A and B should be chosen to be the strongest topologies that make the projection maps continuous.

As a counterexample, let X and Y both be the real line.  Endow $X\times Y$ with the topology generated by the standard topology and open intervals on the diagonal. This topology is strictly larger than the standard topology however if we endow both X and Y  with the strongest topologies that make the projections continuous they would have the standard topologies on the real line.

Title: Given a topology $\tau$ on $X \times Y$, can we find $\tau_X$ on $X$ and $\tau_Y$ on $Y$ such that $\tau_X \times \tau_Y = \tau$

Body:
Suppose we have spaces $X, Y$ and a topology $\tau$ on $X \times Y$. Can we find a topology $\tau_X$ on $X$, $\tau_Y$ on $Y$ such that $\tau_X \times \tau_Y = \tau$ (the product is referring to the product topology).

I think not. What if $\tau$ is a separable complete metric space?

I could also get away with somemthing weaker, like $\tau \subseteq \tau_X \times \tau_Y$ provided $\tau_X$ and $\tau_Y$ have sufficiently neat property (i.e, they're not discrete topologies or something).


问:
How do I prove that $((2-2\cos(2x))/(\cos x \sin 2x)) = ((1/(\csc x - 1)) + (1/(\csc x +1)))$?
Does anyone know how to continue this proof? I am quite stuck at one point. Please note that I only have to equate one side of the equation.

$$
2-2\cos(2x)/\cos x (\sin 2x) = 1/(\csc x - 1) + 1/(\csc x +1)
$$

Here's my solution (I am equating only the left-hand side to be equal to the right-hand side):

$$
2-2(1-2 \sin^2x)/2(\cos^2x)\sin x
$$

$$
2\sin x/\cos^2x
$$

$$
2\sin x/1-\sin^2x
$$

$$
2\sin x/(1-\sin x)(1+\sin x)
$$

I am stuck at the last part. Does anyone have some tips to continue this? Thank you to those who would answer.

Body:
Hint:
$$\dfrac{1}{\csc x - 1} + \dfrac{1}{\csc x + 1} = \dfrac{1}{\dfrac{1}{\sin x} - 1} + \frac{1}{\dfrac{1}{\sin x} + 1} = \dfrac{\sin x}{1 - \sin x} + \dfrac{\sin x}{1 + \sin x}$$


答:
You're almost done. Write $\sin x = \frac{1}{\csc x}$ to get,

$\begin{align}\frac{2\sin x}{(1-\sin x)(1+\sin x)} &= \frac{\frac2{\csc x}}{(1-\frac1{\csc x})(1+\frac1{\csc x})} = \frac{2\csc x}{(\csc x-1)(\csc x+1)}\\
\\&=\frac{(\csc x+1)+(\csc x-1)}{(\csc x-1)(\csc x+1)} = \frac{1}{\csc x-1}+\frac{1}{\csc x+1}\end{align}$

Title: How do I prove that $((2-2\cos(2x))/(\cos x \sin 2x)) = ((1/(\csc x - 1)) + (1/(\csc x +1)))$?

Body:
Does anyone know how to continue this proof? I am quite stuck at one point. Please note that I only have to equate one side of the equation.

$$
2-2\cos(2x)/\cos x (\sin 2x) = 1/(\csc x - 1) + 1/(\csc x +1)
$$

Here's my solution (I am equating only the left-hand side to be equal to the right-hand side):

$$
2-2(1-2 \sin^2x)/2(\cos^2x)\sin x
$$

$$
2\sin x/\cos^2x
$$

$$
2\sin x/1-\sin^2x
$$

$$
2\sin x/(1-\sin x)(1+\sin x)
$$

I am stuck at the last part. Does anyone have some tips to continue this? Thank you to those who would answer.


问:
Prove that this function has derivatives of all orders in some neighborhood of zero: $\sqrt{\frac{t}{\log\frac{1}{2e^{-t}-1}}}$
I want to prove that the following real function has derivatives of all orders in some neighborhood of zero:
$$f(t)=\sqrt{\frac{t}{\log\frac{1}{2e^{-t}-1}}}.$$
Moreover, there exists a constant $C$ such that $|f^{(k)}(0)|\leq C$ for all $k\geq 0.$
I conjecture that $f$ is analytic in some open interval $(-a,a)$ but am not entirely sure.

Edits: If the question regarding $C$ does not have positive answer, then can we show that $|f^{(k)}(0)|\leq 2^{\mathcal{O}(k)}$ as $k\to\infty$?

答:
This is probably a stupid answer.

In the real domain, the function
$$f(t)=\sqrt{\frac{t}{\log \left(\frac{1}{2 e^{-t}-1}\right)}}$$ exists as long as
$$-t \log \left(2 e^{-t}-1\right)\geq 0\implies t \leq \log(2)$$ So, as you wrote in comments, the domain is $(-\infty, \log (2))$.

Composing Taylor series around $t=0$
$$f(t)=\frac{1}{\sqrt{2}}\Big[1-\sum_{n=1}^\infty \frac {a_n} {n!} \,t^n\Big]$$ where the $a_n$ make the sequence
$$\left\{\frac{1}{4},\frac{5}{16},\frac{47}{64},\frac{711}{256},\frac{44851}{3072},\frac
  {401661}{4096},\frac{13128607}{16384},\frac{1516267757}{196608},\cdots\right\}$$


问:
Is there any common method to evaluate $\lim_{x\to +\infty} \left( \frac{\ln^4 x}{4}-\int_0^x \frac{\ln^3 t}{1+t} \, dt \right)$?
So is there any common method to solve this kind of problem?

Calculate that $$\lim_{x\to +\infty} \left( \frac{\ln^4 x}{4}-\int_0^x \frac{\ln^3 t}{1+t} \, dt \right)$$

Before this I have seen a similar problem which was solved by using digamma function or beta function. So is that mean we can always use special functions to solve them?

Body:
Note that
$$\frac{\ln^4 x}{4}=\int_{1}^{x}\frac{\ln^3 t}{t}dt.$$
So
\begin{align*}
\frac{\ln^4 x}{4}-\int_0^x \frac{\ln^3 t}{1+t}dt
  &=\int_{1}^{x}\frac{\ln^3 t}{t}dt-\int_0^x \frac{\ln^3 t}{1+t}dt\\
  &=\int_{1}^{x}\frac{\ln^3 t}{t}dt-\int_0^1 \frac{\ln^3 t}{1+t}dt-\int_1^x \frac{\ln^3 t}
     {1+t}dt\\
  &=\int_{1}^{x}\frac{\ln^3 t}{t(1+t)}dt-\int_0^1 \frac{\ln^3 t}{1+t}dt.
\end{align*}
Then
$$\lim_{x\to +\infty} \left( \frac{\ln^4 x}{4}-\int_0^x \frac{\ln^3 t}{1+t}dt \right)
=\int_{1}^{+\infty}\frac{\ln^3 t}{t(1+t)}dt-\int_0^1 \frac{\ln^3 t}{1+t}dt.$$
By variable substitution $t=\dfrac{1}{s}$,
$$\int_{1}^{+\infty}\frac{\ln^3 t}{t(1+t)}dt=-\int_0^1 \frac{\ln^3 s}{1+s}ds.$$
Henceforce,
$$\lim_{x\to +\infty} \left( \frac{\ln^4 x}{4}-\int_0^x \frac{\ln^3 t}{1+t}dt \right)
=-2\int_0^1 \frac{\ln^3 t}{1+t}dt\ \left(=\frac{7\pi^4}{60}\right).$$

So our main goal is to calculate improper integral:
$$\color{blue}{\int_0^1 \frac{\ln^3 t}{1+t}dt\ \left(=-\frac{7\pi^4}{120}\right)}.$$

Here is a proof:
\begin{align*}
I&=\int_0^1\frac{-\ln^3t}{1+t}dt\cr
&=-\sum_{n=0}^\infty \int_{0}^1(-t)^n \ln^3 t\,dt\cr
&=-\sum_{k=0}^\infty \int_{0}^1(-t)^{2k} \ln^3 t\,dt-\sum_{k=0}^\infty \int_{0}^1(-t)^{2k+1} \ln^3 t\,dt\cr
&=\color{red}{-\sum_{k=0}^\infty \int_{0}^1t^{2k} \ln^3 t\,dt+\sum_{k=0}^\infty \int_{0}^1 t^{2k+1} \ln^3 t\,dt}\cr
&\color{red}{=\sum_{k=0}^\infty\frac{6}{(2k+1)^4}-\sum_{k=0}^\infty\frac{6}{(2k+2)^4}}\cr
&=\sum_{k=0}^\infty\frac{6}{(2k+1)^4}+\sum_{k=0}^\infty\frac{6}{(2k+2)^4}-2\sum_{k=0}^\infty\frac{6}{(2k+2)^4}\cr
&=6\zeta(4)-\frac{6}{8}\zeta(4)\cr
&=\frac{21}{4}\zeta(4)\cr
&=\frac{7\pi^4}{120}.
\end{align*}

Indeed, we use the following result (red part), the change of variables
$x=e^{-t}$ shows that
\begin{align*}\color{red}{
\int_0^1x^n\ln^p(1/x)\,dx}
&\color{red}{=\int_0^\infty e^{-(n+1)t}t^pdt}\\
&\color{red}{=\frac{1}{(n+1)^{p+1}}\int_0^\infty e^{-u}u^pdu}\\
&\color{red}{=\frac{\Gamma(p+1)}{(n+1)^{p+1}}.}
\end{align*}


答: Here’s a hint/general roadmap as to how I did it. I did not use your manipulation by the way, so focus on the original form of the problem you provided. Use the fact that $$\sin(x)=2\sin\bigl(\frac{x}{2}\bigr)\cos\bigl(\frac{x}{2}\bigr)$$. Now, keep rewriting the recurring sin term in this expression similar to how I just did, where I halved the original argument. This generalizes to the product representation $$\sin(x) = 2^n\cos\bigl(\frac{x}{2^n}\bigr)\sin\bigl(\frac{x}{2^n}\bigr)\prod_{k=1}^{n-1}\cos\bigl(\frac{x}{2^k}\bigr).$$ To use this you’re gonna need to reindex the sum to start at $n=2$. Rewrite the inside of the sum (namely the inside of the $\cos$ argument) to allow you to plug in a certain value of $x$. You’re gonna need to use this identity to turn the sum into a telescopic series that will eventually result in the answer. I got $2-\frac{\pi^2}{16}$.

Body:
$\sum_{n=1}^{\infty} {\frac{1}{4^n \cos^2 (\frac{\pi}{2^{n+2}})}}$

How can I calculate this? Since there are $4^n$ and $\cos^2x$, I tried:
$$\sum_{n=1}^{\infty} {\frac{1}{4^n \cos^2 (\frac{\pi}{2^{n+2}})}} = 4\sum_{n=1}^{\infty}{\frac{\sin^2{\frac{\pi}{4 \cdot 2^n}}}{4^{n}\sin^2{\frac{\pi}{4\cdot2^{n-1}}}}}$$
to use $2\sin x \cos x = \sin2x$


问:
Wang Tiling problem - If you can tile the first quadrant then you can tile the whole plane
I have seen in some articles (see here for instance) -   the following claim without a proof regarding Wang tiles:

The first quadrant can be tiled iff the whole plane can be tiled

Can anyone explain why is that true?

答:
The plane case obviously implies the quarter-plane case, as we can just take a subset of any planar tiling. So we are interested in the implication from the quarter-plane to the plane.

If one has an infinite collection of tiles, this is not true: arrange squares in a quarter-plane, and then use a unique color on every adjacent pair of squares (as well as every side of a square which forms part of the quarter-plane's boundary). This produces an infinite collection of tiles. But then from any one of these tiles, there is at most one tile that can be added in any given direction, so the continuation from that tile is uniquely the original coloring described above (which cannot be extended past the quarter-plane).

However, in the case of a finite set of tiles, this is true. As alluded to in the comments, if we can tile a quarter-plane then we can tile an arbitrary finite $N\times N$ region. It turns out this statement is equivalent (in the finite case) to tiling the plane.

Proof: Consider, for each tiling of an $N\times N$ square with $N$ odd, the central tile; this is an infinite sequence, since there are infinitely many odd numbers, so one of our finite tiles must occur infinitely often within it. Let the tiling consisting of only this tile be $T_1$.

Now, by assumption there are infinitely many odd $N$ where $T_1$ is in the center of an $N\times N$ tiling. Consider the set of all central $3\times 3$ regions in each such tiling, for $N\ge 3$. Again, there are only finitely many possible $3\times 3$ tilings that can beformed, so one of them must work for infinitely many $N$. Let this tiling be $T_3$.

Iterating this procedure, we get a series of finite tilings $T_1, T_3, T_5,\ldots$, each containing the previous in its center. So the infinite union of all these tilings is well-defined, and serves as a tiling of the plane.

Less formally, the fundamental idea here is to at each step add a tile which can be built off of in all directions in infinitely many ways; by keeping the number of possible expansions infinite at every step, we guarantee that we never run out of possibilities to add a new tile. (The finiteness of our set of tiles is necessary for the assumption that, of the infinitely many ways to expand, the subsets of them which involve adding a specific new tile are not all finite.)


问:
Why is $\lim_{x \to 0^+}(e^{\sin x\, \cdot\, \ln x})$ equal to $e^{\lim_{x \to 0^+}(\sin x\, \cdot\, \ln x)}$?
Can someone give me a simple explanation of why this works (and what it's called)?

$$\lim_{x \to 0^+}(e^{\sin x\, \cdot\, \ln x}) = e^{\lim_{x \to 0^+}(\sin x\, \cdot\, \ln x)}$$

I can't find an explanation anywhere online...

答:
Here the function $e^x$ is a continuous function over the range of $\ln x \cdot \sin x$ (which is just $\mathbb{R}$). By definition of continuity, we know that as $x \to k$, $e^x \to e^k$. So in particular, as $\ln x \cdot \sin x \to L = \lim_{x \to 0^+} \ln x \cdot \sin x$ (supposing this limit existed), it would have to follow that $e^{\ln x \cdot \sin x} \to e^L$.

It's important to note that in general, when you are evaluating the limit of $f(g(x))$ as $x \to k$, you need $f(x)$ to be continuous over the range of $g(x)$, or at least in a neighborhood of $L = \lim_{x \to k} g(x)$ (again, supposing it existed) to conclude that $f(L) = \lim_{x \to k} f(g(x))$. It does not suffice to have $f$ continuous in a region around $k$. For example, if instead the problem were to evaluate $\lim_{x \to 2} f(\sin(x))$
where $f(x)$ was continuous at $2$ but not around $\sin(2)$, you could not conclude that
$$f\left(\lim_{x \to 2} \sin x\right) = \lim_{x \to 2} f(\sin x)$$


问:
Simplifying fractions - Ending up with wrong sign
I've been trying to simplify this
$$
1-\frac{1}{n+2}+\frac{1}{(n+2) (n+3)}
$$
to get it to that
$$
1-\frac{(n+3)-1}{(n+2)(n+3)}
$$
but I always end up with this
$$
1-\frac{(n+3)+1}{(n+2)(n+3)}
$$
Any ideas of where I'm going wrong?
Wolfram Alpha gets it to correct form but it doesn't show me the steps (even in pro version)

Thanks

答:
You just have the problem that while $$x-y+z = x-(y-z)$$

you are instead writing:$$x-y+z=x-(y+z)$$


问:
Determining an equivalence class
Let $S = \mathbb{Z}$ and $a ∼ b ⇐⇒ 3|(a + 2b)$.

Show that $∼$ is an equivalence relation.

Determine all equivalence classes for $∼$.

Part $1$ was pretty simple to prove, however I can't figure out how to proceed with part $2$. I'm thinking of either doing classes by the coefficient $m$ such that $3m = a+2b$ (ex: $cl(1) = (1,1) , (-1,2), (-3,3)... $  for $(a,b)$ ), by the sum of $a+2b$ (ex: $cl(0) = (-2,1)$ because $-2+2(1) = 0$ ), or just a single class where this relation exists (ex: all the values of $(a,b)$ where $3|(a + 2b)$. The last one wouldn't partition the set $S$ obviously, because it would only take the values divisible by $3$.

As a follow up thought, all $3$ of my methods seem incorrect to me, as the set $S$ is only $\mathbb{Z}$, which implies that we are only looking for single integer values in each class, not the $(a,b)$ coordinates that I put in my classes. Any help is appreciated. Thank you!

答:
Hint:

$$a\sim b\iff a\equiv b\pmod 3$$

In other words $3|a+2b$ iff $3|a-b$

Title: Determining an equivalence class


Title: $\sum_{n=1}^{\infty} {\frac{1}{4^n \cos^2 (\frac{\pi}{2^{n+2}})}}$

Body:
Notice
$$\begin{align}\frac{1}{\cos^2\frac{\theta}{2}}
&= \frac{2}{1+\cos\theta}
= 2\frac{1 - \cos\theta}{1-\cos^2\theta}
= \frac{4 - 2(1+\cos\theta)}{1-\cos^2\theta}\\
&= \frac{4}{\sin^2\theta} - \frac{2}{1-\cos\theta}
= \frac{4}{\sin^2\theta} - \frac{1}{\sin^2\frac{\theta}{2}}
\end{align}
$$
We have
$$\begin{align}
\sum_{n=1}^p \frac{1}{4^n\cos^2\frac{\pi}{2^{n+2}}}
&= \sum_{n=1}^p \left[
\frac{1}{4^{n-1}\sin^2\frac{\pi}{2^{n+1}}} -
\frac{1}{4^n\sin^2\frac{\pi}{2^{n+2}}}
\right]\\
&=\frac{1}{4^{1-1}\sin^2\frac{\pi}{2^{1+1}}} - \frac{1}{4^p\sin^2\frac{\pi}{2^{p+2}}}\\
&= \frac{1}{\sin^2\frac{\pi}{4}} - \frac{\frac{16}{\pi^2}}{\left(\frac{2^{p+2}}{\pi}\sin\frac{\pi}{2^{p+2}}\right)^2}
\end{align}
$$
Since $\lim\limits_{x\to 0} \frac{\sin x}{x} = 1$, the denominator in last term tends to $1$ as $p \to \infty$, As a result,

$$\sum_{n=1}^\infty \frac{1}{4^n\cos^2\frac{\pi}{2^{n+2}}}
= \lim_{p\to\infty}
\sum_{n=1}^p \frac{1}{4^n\cos^2\frac{\pi}{2^{n+2}}}
= 2 - \frac{16}{\pi^2}
$$


问:
Are there any finite examples of metachirality?
I learned about metachirality through an amazing Vihart series ([1] [2]), and the only examples I have encountered are infinitely long screws, and this infinite graph Wikipedia mentions.

Are there finite examples of metachirality? My intuition says no, because I can't even think of finite examples with glide symmetry or screw symmetry, both of which feel like a weaker thing to ask for. I also suspect that any example of metachirality must in some way rely on translation, and if I can somehow formalise and prove that notion, I think I'll be done. (Because I know that translational invariance implies that, at least in one direction, the object is infinite [3])

In the positive case, I am looking for examples of finite symmetry groups (so maybe the object to consider is infinite, but the symmetry group is somehow still finite), but examples with finite objects would be cool too.

In the negative case, are there any proofs that all examples of metachirality need to be finite? And do such proofs have anything to say about glide/screw symmetries?

答:
I can't even think of finite examples with glide symmetry or screw symmetry, both of which feel like a weaker thing to ask for.

Your intuition here is correct. In the case of either glide or screw symmetry, you can consider the infimum and supremum of any finite object along its projection onto an axis parallel to the glide or screw motion. Since the motion would translate these bounds, it can't leave the object unchanged. So any bounded set in $\mathbb{R}^3$ cannot be invariant under such an isometry of the plane. (The same is true of translation, by the way, which perhaps is a more intuitively obvious case.)

This sort of argument would be more complicated in higher dimensions, and might fail altogether, with the greater variety of isometries and rotations available; it's possible that you could find examples in spaces beyond $\mathbb{R}^3$.


问:
How to solve the Asymptotic Estimation
How to prove that
$$\sum\limits_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^x}=\frac{1}{2} + \frac{\ln\pi-\ln2}{2}x+o(x)(x\to0^+)$$

答:
$$\sum\limits_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^x}=(1-2^{1-x})\,\zeta(x)$$

When $x$ is small
$$2^{1-x}=e^{(1-x)\log(2)}=2-2 x \log (2)+O\left(x^2\right)$$ and
$$\zeta(x)=-\frac{1}{2}-\frac{1}{2} x \log (2 \pi )+O\left(x^2\right)$$ Combine all of the above and simplify.

Title: How to solve the Asymptotic Estimation


问:
How to find the point of symmetry for $f(x) = \frac{6}{1+3\cdot(0.4)^x}$ algebraically?
In analyzing the graph of the function $$f(x) = \frac{6}{1+3\cdot(0.4)^x}$$ the textbook lists that it has symmetry around the point (1.2, 3). Is there a way to find the coordinates of this point algebraically, without tracing the graph or using calculus to find the inflection point?

Thank you!

答:
The value of the function at $-\infty$ is $0$, and at $+\infty$ is $6$ ( horizontal asymptotes if one is precise). Now if the graph has a center of symmetry, it is at the point of height $\frac{6}{2}=3$. So let $x_0$ such that $f(x_0) = 3$. This is equivalent to $3 (0.4)^{x_0}=1$.

One should check that $f(x_0- h) + f(x_0+ h) = 2 f(x_0)$.
This is equivalent to
$$\frac{6}{1 + 0.4^{-h}} + \frac{6}{1 + 0.4^h} = 6$$

or, with $0.4^h = t$
$$\frac{1}{1+1/t} + \frac{1}{1+t} = 1$$
which is true.

Note: $f$ is a logistic function.


问:
Calculate the limit $\frac{(2n-1)x_1+(2n-3)x_2+\dots +3x_{n-1}+x_n}{n^2}$ when $x_n\to x$.

Let $\{x_n\}$ be a sequence in $\Bbb R$ and $x_n\to x$ as $n\to \infty$.
Then $$\frac{(2n-1)x_1+(2n-3)x_2+\dots +3x_{n-1}+x_n}{n^2}\to x$$.

Do anyone know how to solve this kind of problem efficiently? I think I need to estimate $$\left|\frac{(2n-1)x_1+(2n-3)x_2+\dots +3x_{n-1}+x_n}{n^2}-x\right|$$ Is there some other ways to calculate the limit directly?

Thanks for any comments.

答:
This limit problem looks very much like intentionally designed for the use of the Stolz-Cesàro theorem:

So, let
$$a_n =\sum_{k=1}^n(2n-(2k-1))x_k = \sum_{k=1}^n(2(n-k)+1)x_k$$
and $$b_n = n^2$$

So, $$a_{n+1}-a_n = \sum_{k=1}^{n+1}(2(n+1-k)+1)x_k - \sum_{k=1}^n(2(n-k)+1)x_k$$$$ = x_{n+1}+2\sum_{k=1}^n x_k$$

Hence,
$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}= \frac{x_{n+1}+2\sum_{k=1}^n x_k}{2n}=\underbrace{\frac{x_{n+1}}{2n}}_{\stackrel{n\to\infty}{\longrightarrow}0} + \underbrace{\frac{\sum_{k=1}^n x_k}{n}}_{\stackrel{n\to\infty}{\longrightarrow}x}$$

where the last limit is just another application of Stolz-Cesàro.


问:
Elementary results from Algebraic Number Theory
The purpose of this question is to motivate me to study algebraic number theory.

Let me explain. My motivation for studying number theory is to learn about beautiful results with simple, accessible statements. For example, the theorem that a prime can be written as the sum of two squares if and only if it is 1 mod 4. I have been reading the first few pages of both Neukirch's Algebraic Number Theory and Serre's Local Fields. I have looked ahead to see what I have to look forward to in terms of such results.

A prime is the sum of two squares if and only if it is 1 mod 4 (done in the first few pages by describing characterizing factorization in the Gaussian integers)

Pythagorean triples (an exercise in the first section)

Solving Pell's Equation

Quadratic Reciprocity (using a result about cyclotomic number fields)

Lagrange's four-square theorem

All of these appear relatively early in the book, and are provable with elementary number theory accessible to a motivated high school student. Looking at some of the later chapters, I wonder: What is the point of all this theory, besides being aesthetically pleasing?

I do not wish to denigrate the absolutely beautiful theory contained in these books. It is marvelous, and I want to learn it. However, I wonder how it can be applied to produce results on the standard integers, especially ones not provable by elementary methods. Fermat's Last Theorem is the obvious example, but surely there are others. A reference where I could find a collection of such results would be especially appreciated. In case it is not clear, I am looking for results provable with methods and theorems developed in the books I mentioned above (or other similar standard graduate texts).

Perhaps the problem is that I don't yet appreciate that results about algebraic integers and number fields can be as thrilling as those about integers. If this is the case, I'm interested in seeing a short list of remarkable theorems about such objects. Maybe I just need to refine my tastes.

To summarize: How can the powerful theorems of algebraic number theory (e.g. the version of Rieman-Roch found in Neukirch, Class Field Theory) be used to give interesting elementary results?

答:
I don't know if you're still looking for motivational problems, but I began investigating algebraic number theory through Galois theory and the quest to characterize the algebraic extensions of $\mathbb{Q}$.

I used both of these books to develop mathematical background for the Kronecker-Weber Theorem, which says that any abelian extension of $\mathbb{Q}$ must be cyclotomic.  This is a rather beautiful result extending Galois's work, which is a rather easy result from class field theory, although it can also be proved using only ramification theory and valuations (which is why I used Neukirch and Serre as primary references).


问:
$\log(z_1z_2) = \log(z_1) +\log(z_2) + i2n\pi $ where log is the analytic branch of complex log.
In a problem provided in my course (the assignment was due a while ago, this isn't a homework help request), we needed to show :
$$\log(z_1z_2) = \log(z_1) + \log(z_2) + i2n\pi $$

I seem to understand the analytic branch of complex log, defined:

$$\log(z) = \ln|z| + i\arg(\theta)$$ where ln is the real natural log and $\arg(\theta) \in (-\pi, \pi)$.

In my professor's worked solution, he claims that this equality holds because :
$$e^{\log(z_1z_2)} = z_1z_2 = e^{\log(z_1)}e^{\log(z_2)} =e^{\log(z_1)+\log(z_2)} $$ and because $e^{i2n\pi} = 1$ we get that the equality $\log(z_1z_2) = \log(z_1) + \log(z_2) + i2n\pi $ holds.

I proved the problem in a different way, but want to see how this method is valid. I understand you can always multiply by 1 on the rightmost equation by multiplying by $e^{i2n\pi}$, but why does that imply that the equality holds? To me, it only implies that $z_1z_2=z_1+z_2+i2n\pi$ by treating the analytic log as the inverse of the exponential function, but this is not necessarily correct. Any help in understanding this method would be much appreciated!

答:
We have $e^{\log(z_1z_2)} = e^{\log(z_1)+\log(z_2)}$.  That implies,
$$
e^{\log(z_1z_2)-\log(z_1)-\log(z_2)} = 1$$
and the solution of $e^{\zeta} = 1$ is simply $z = 2n\pi i$.  The result follows.  There is no further need to treat the $\log$ as an inverse.


问:
Given a topology $\tau$ on $X \times Y$, can we find $\tau_X$ on $X$ and $\tau_Y$ on $Y$ such that $\tau_X \times \tau_Y = \tau$
Suppose we have spaces $X, Y$ and a topology $\tau$ on $X \times Y$. Can we find a topology $\tau_X$ on $X$, $\tau_Y$ on $Y$ such that $\tau_X \times \tau_Y = \tau$ (the product is referring to the product topology).

I think not. What if $\tau$ is a separable complete metric space?

I could also get away with somemthing weaker, like $\tau \subseteq \tau_X \times \tau_Y$ provided $\tau_X$ and $\tau_Y$ have sufficiently neat property (i.e, they're not discrete topologies or something).

答:
Notation: $P(Z)$ (the power-set of $Z$) is the set of all subsets of a set $Z.$

If $\tau=\tau_x\times\tau_Y$ then  $\tau \subseteq P(X)\times P(Y),$ which is generally not the case. Example: Suppose $X=Y=\{1,2\}$ and $s=\{(1, 2),(2,1)\}\in \tau.$ If $A\in P( X)$ and $B\in P (Y)$ and $s\subseteq A\times B$ then $(1,2)\in s\implies 1\in A$ and $(2,1)\in s\implies 2\in A,$ so $A=X.$ Similarly $B=Y.$ So $A\times B=X\times Y\ne s.$

Notation: $f[S]=\{f(s):s\in S\}$ when $f$ is a function and $S\subseteq dom(f).$

Given a topology $\tau$ on $X\times Y$: For $(x,y)$ let $p_X(x,y)=x$ and $p_Y(x,y)=y$... ($p_X$ and $p_y$ are the projections of $X\times Y$ to its $X$ & $Y$ co-ordinates.)... Let $\tau_X=\{p_X[s]:s\in \tau \}$ and $\tau_Y=\{p_Y[s]: s\in \tau\}.$  Then $\tau_X, \, \tau_Y$ are the strongest topologies on $X,Y$ such that $p_X,\,p_Y$ are continuous. And $\tau_X\times\tau_Y$ is a base (basis) for $\tau.$


问:
How did wolfram simplify this integral?
I am computing a convolution $e^{-ax^2}*e^{-bx^2}$. The integral is of course $\int_{-\infty}^{\infty} e^{-a(x-w)^2 - bw^2}dw$, and I have simplified this down to
$$\sqrt{\pi}\frac{e^{-x^2\frac{ab}{a+b}}}{\sqrt{a+b}}\bigg{(} 1+\text{erf}\big{(} \frac{ax}{\sqrt{a+b}}\big{)}\bigg{)} $$
this calculation can be found here. Also, putting that integral into wolfram alpha gives the same answer.
BUT when I give wolfram alpha the command convolution$(e^{-ax^2}, e^{-bx^2})$ it returns
$$\sqrt{\pi}\frac{e^{-x^2\frac{ab}{a+b}}}{\sqrt{a+b}} $$
Where is this value coming from/ why does it differ from the integral?

答:
What you need is the Gaussian integral
$$\int_{-\infty}^{\infty}e^{-c(t+d)^2}\,dt=\sqrt{\frac{\pi}{c}} $$
and note that
$$-a(x-w)^2-bw^2=-(a+b)\left(w-\frac{ax}{a+b}\right)^2-\frac{abx^2}{a+b} $$

Title: How did wolfram simplify this integral?


问:
let $f: R\to R$ be $f(x) = \begin{cases} 1,& x = 0 \\ 0, & \text{otherwise}\end{cases}$ is $\space\lim\limits_{x\to a}f\left(x\right) = 0\space$?
let $f: R\to R$ be  $f(x) = \begin{cases} 1,& x = 0 \\ 0, & \text{otherwise}\end{cases}$ defined on a deleted neighbourhood of $x_0 \in R$.

and I'm trying to prove that $\space\lim\limits_{x\to a}f\left(x\right) = 0\space$ for every $\space a\in R$

I have managed to prove that for $a = 0, \lim\limits_{x\to a}f\left(x\right) = 0 $ but I'm having troubles trying to prove that $\lim\limits_{x\to a}f\left(x\right) = 0 $ when $a\ne 0$

That is my prove so far:

for every $\epsilon > 0$ there exists a $ \delta$ such that for $0 < |x-a| < \delta $

$|f\left(x\right) - 0| < \epsilon$

and for $f(x) = 0$ it's obvious , but what about $f(x) = 1$ how do I choose a $\delta$ then? And isn't that mean that for $ 0<\epsilon<1\space $ $0$ is not the limit?

I'm new to calculus so I'm very sorry if my question is too easy and I'm just missing some definitions :(
but I would love to get an explanation so I can learn for next time.

答:
See this answer for Heine's definition of a limit of a function.

According to that, take
$$\forall (x_n): \lim x_n = 0, ~ x_n \ne 0$$
But, obviously (since $x_n \ne 0$), we have
$$\lim f(x_n) =\lim 0=0$$

Hence, the limit at $0$ is $0$.


问:
Has this function ever been discussed? Is it an elementary function?
$f^{(n)}(0)=\frac{1}{n!},\ \ n=1,2,\cdots\\$

With Taylor series, this function can also be written as

$f(z)=\sum_{n=0}^{\infty}\frac{x^n}{(n!)^2}$

Has this function ever been discussed? Is it an elementary function? Does it have anything to do with $\Gamma(s)$ (or other famous functions about factorials)?...

答: $$f(z)=\sum_{n=0}^{\infty}\frac{z^n}{(n!)^2}=I_0\left(2 \sqrt{z}\right)$$ where appears the modified Bessel function of the first kind.


答:
Everywhere there is a minus sign, replace it with plus a negative.

So with your original expression, try instead simplifying
$$
1+\frac{-1}{n+2}+\frac{1}{(n+2) (n+3)}
$$
and you should be much less prone to error.

Title: Simplifying fractions - Ending up with wrong sign

Body:
I've been trying to simplify this
$$
1-\frac{1}{n+2}+\frac{1}{(n+2) (n+3)}
$$
to get it to that
$$
1-\frac{(n+3)-1}{(n+2)(n+3)}
$$
but I always end up with this
$$
1-\frac{(n+3)+1}{(n+2)(n+3)}
$$
Any ideas of where I'm going wrong?
Wolfram Alpha gets it to correct form but it doesn't show me the steps (even in pro version)

Thanks


问:
Proving a function bounded between integrable functions is integrable?
Let $I=[a,b]$, $f,g,h$ bounded functions in $I$ with real values. Suppose $f(x)\leq g(x)\leq h(x)$ $\forall x\in R$. Prove that if $f$ and $h$ are integrable in $I$ and if $\int_a^b f =\int_a^b h$ then $g$ is also integrable in $I$ and $\int_a^b g =\int_a^b f$.

My attempt:
Given that f and g are integrable then we have that $L(f)=U(f)$ and $L(h)=U(h)$ and since the integrals of $f$ and $g$ in $I$ are equal then I guess that $L(f)=L(h)$ and $U(f)=U(h)$ ?

I think that f and h share the same upper and lower integral, and I can see the solution to this problem intuitively, I was thinking there must be a way to get to $L(g)=U(g)$ and that it's integral is equal to the integral of f using the fact that g is bounded in between f and h.

Any help would be appreciated. Thank you.

答:
Set $f^+=\max\{f,0\}$ and $f^-=\max\{-f,0\}$. A function is integrable if and only if the positive part $f^+$ and the negative part $f^-$, which are both nonnegative, are integrable meaning
$$0\leq\int f^+ dx<\infty  \text{ and } 0\leq\int f^- dx<\infty.$$

Now, check that by hypothesis, $g\leq h$ implies $g^+\leq h^+$ and so if $\int h^+ dx<\infty$, then $\int g^+ dx<\infty$ by comparison theorem. In the same way, $f\leq g$ implies $g^-\leq f^-$ and integrability of $f^-$ implies that of $g^-$.  So $g$ is integrable.

Now that the integrability is proved, you just take the integral of all sides and if the first integral equals the third integral, the second integral also must be equal to the other two (circle inequalities in algebra: $a\leq b\leq a$ implies $a=b$)


问:
If $E(X)=E(X^2)$ then X is Bernoulli
I read on my textbook that:

"The distribution of $X$ is concentrated on the points $0$ and $1$ if $E(X)=E(X^2)$.

Is this statement implies that always $X$ is going to be a Bernoulli random variable?, and if is it true, why?

答: No, it is not necessarily a Bernoulli distribution. For example, let $X \sim N\left(\frac{1}{2}, \frac{1}{4}\right)$. Then $E[X] = E\left[X^2\right] = \frac{1}{2}$. In fact it would hold for any distribution such that $0 < \mu < 1$ and $\text{Var}(X) = \mu-\mu^2$, where $\mu$ is the mean.

Title: If $E(X)=E(X^2)$ then X is Bernoulli


问:
Finding particular integral to $ (D^3 + 4D)y = \sin2x$
The question is to find particular integral of this differential equation

$$ (D^3 + 4D)y = \sin2x.$$

where $$ D = \frac{d}{dx} $$
Please looking for your quick tip guys.

答:
Given:- (D³+4D)y=sin2x
Condition : 1/f(D), if after equal to sinax or cosax given means you have to replace D² into -(a²)
Ans-  (1/D³+4D)*sin2x
[1/D(D²+4)]*sin2x    Here, D² is replaced by -(a²) that is

D²= -(2²)
[1/D(-4+4)]*sin2x  (important first square the number and put (-) symbol.)
[1/D(0)]*sin2x [1/0]*sinx ( condition if denomerator  is zero multiply  numerator by x and differentiate f(D)).
[x/3D²+4]*sin2x ( substitute D² value)
[x/3(-4)+4]*sin2x [x/(-12+4)]*sin2x
Therefore [-x/8]*sin2x.


问:
Convexity of abs of quadratic form
We know that $f(x) = x^T Q x$ is convex iff $Q \geq 0$. How about $g(x) = |x^T Q x|$ where $Q$ is an arbitrary symmetric matrix?

答:
Let $Q=\operatorname{diag}(1,-1)$.

Then $g(x) = |x_1^2 -x_2^2|$, and if we let $\phi(t) = g((t,1)) = |t^2-1|$ we see that $\phi(-1) = \phi(1) = 0$ but $\phi(0) = 1$, hence $g$ is not convex.

Title: Convexity of abs of quadratic form


问:
Taylor and geometric series
1) iF $f(x) = x^2 +x$, find the taylor series for f centered at a = 2.

2)what is the sum from 1 to infinity of $(.95)^n$

I got these questions wrong on my last test, and I'm not really sure how to go about solving it.

答:
The Taylor series of an infinitely differentiable function at a point is a representation of a function given by an infinite sum of terms of a specific form. Let $f(x)$ be the function and $a$ be the point, then the Taylor series is given by $$f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \cdots $$ where $f^{(n)}$ is the $n$-th derivative of the function.

In your case, we have $$f(x) = 6 + \frac{5}{1!}(x-2) + \frac{2}{2!}(x-2)^2 + \frac{0}{3!}(x-2)^3.$$ The last term given above is zero as we have a zero in the numerator. All subsequent terms are  zero as $f^{(n)}=0$ for $n\geq 3$.

For Two. I suppose that you want to find $$\sum_{n=1}^\infty (0.95)^n = 0.95 + 0.95^2 + 0.95^3 + \cdots $$

As this is a geometric series with $a=0.95$ and $r=0.95$ we have the sum given by $$\frac{a}{1-r} = \frac{0.95}{1-0.95}=19$$

Title: Taylor and geometric series


问:
Show that no non-trivial $T$ -invariant subspace has a complementary $T$ -invariant subspace
Let $T$ be a linear operator on a finite-dimensional vector space $V$. Suppose that
(a) the minimal polynomial for $T$ is a power of an irreducible polynomial;
(b) the minimal polynomial is equal to the characteristic polynomial.
Show that no non-trivial $T$ -invariant subspace has a complementary $T$ -invariant subspace.

Solution:
$T$ has a cyclic vector.

We first prove that every $T$ -invariant
subspace is $T$ -cyclic. Suppose, to get a contradiction, that the restriction of $T$ to a $T$ -invariant subspace of dimension $d$ was annihilated by a polynomial $p$ with degree less than $d$. Then the image $W$ of $p$ would be a $T$ -invariant subspace of dimension at most $dim (V) − d$ (by Rank-Nullity Theorem) and annihilated by some polynomial $q$ with degree at most $dim (W)$, such as the characteristic polynomial for $T$ restricted to $W$. Then $qp$ would annihilate $T$ because $p$ maps $V$ into $W$ which is annihilated by $q$.

But $deg (qp) = deg (q) + deg (p) < dim (W) + d ≤ dim (V) − d + d = dim (V) $, which
contradicts that the minimal polynomial for $T$ must equal its characteristic polynomial, which has degree $dim (V)$.

Hence, the minimal polynomial of the $T$ -invariant subspace has degree d. But that minimal polynomial also divides
the characteristic polynomial for the $T$ -invariant subspace and so must equal the characteristic polynomial. Hence, the $T$ -invariant subspace is $T$ -cyclic.

Now suppose, to get a contradiction, that $V = W_1 \oplus W_2$ where $W_1$, $W_2$ are non-trivial $T$ -invariant subspaces. By the previous paragraph, $W_1$, $W_2$ are also $T$ -cyclic, so the minimal polynomials for T restricted to $W1$, $W2$ have degrees
equal to $dim (W_1)$, $dim (W_2)$, which are each less than $dim (V)$ because $W_1$, $W_2$ are non-trivial. Let sm be the minimal
polynomial for $T$ where s is irreducible. Then because the minimal polynomials for $T$ restricted to $W_1$, $W_2$ divides
$s^m$, they are $s^j$ and $s^k$ where j and k are each less than m. The minimal polynomial for $T$ is the least common multiple of $s^j$ and $s^k$, which is s^max(j,k), but that contradicts that the minimal polynomial is $s^m$.
Thus, no non-trivial $T$ -invariant subspace has a complementary $T$ -invariant subspace.

This is the proof which I have written and it would be great to have some feedback on it.

答:
If $U = V\oplus W$ with $V,W$ invariant (and non-zero) then $charpoly(T)=charpoly(T|_V)charpoly(T|_W)$ and $minpoly(T)=lcm(minpoly(T|_V),minpoly(T|_W))$

So $charpoly(T)=minpoly(T)$ implies that $minpoly(T|_V),minpoly(T|_W)$ are coprime contradicting that $minpoly(T)$ is the power of an irreducible polynomial.


答:
As pointed out by other answers/comments, the claim is not true without extra assumptions. Here are some comments:

Given that the counterexamples mentioned involve both positive and negative values, one might suspect that adding the condition $X\geq 0$ might salvage the claim. However, this is still not enough since we have the following example:

If $X \sim \operatorname{Uniform}([\frac{1}{2},\frac{1+\sqrt{3}}{2}])$, then $ \mathbb{E}[X] = \mathbb{E}[X^2] = \frac{2+\sqrt{3}}{4}$.

If we impose the condition that $0 \leq X \leq 1$, then the claim is true by noting that $X(1-X)$ is a non-negative and $\mathbb{E}[X(1-X)] = 0$. This implies that $X(1-X) = 0$ a.s. and hence $X$ is Bernoulli.

In this posting, it is shown that the condition $\mathbb{E}[X^2]=\mathbb{E}[X^3]=\mathbb{E}[X^4]<\infty$ is enough to guarantee that $X$ is Bernoulli.

Title: If $E(X)=E(X^2)$ then X is Bernoulli

Body:
I read on my textbook that:

"The distribution of $X$ is concentrated on the points $0$ and $1$ if $E(X)=E(X^2)$.

Is this statement implies that always $X$ is going to be a Bernoulli random variable?, and if is it true, why?


问:
Computing the conditional density function

Let $(X,Y)$ have joint density $ f(x,y)=c\exp(-(1+x^2)(1+y^2)) $, $(x,y)\in\mathbb R^2$, $c$ is a constant such that $f$ is indeed a probability density function. Then what is the conditional density of $X$ given $\{Y=y\}$.

We compute the conditional distribution of $X$ given $\{Y=y\}$.
\begin{align}
\Pr(X<x_0|Y=y)&=\frac{\int\int_{-\infty}^{x_0}f(x,y)dxdy}{\int f(x,y) dx}.
\end{align}

Applying the Fubini-Tonelli theorem and taking the derivative in terms of $x_0$ gives us $$\frac{\int f(x_0,y)dy}{\int f(x,y) dx}.$$

Direct computation shows that
$$ \frac{\int f(x_0,y)dy}{\int f(x,y) dx}=\frac{\exp(-(1+x_0^2))\sqrt{1+y^2}}{\exp(-(1+y^2))\sqrt{1+x_0^2}} $$

But I think there should be a better way to compute the density other than taking the derivative of the distribution function. Can we do that?

答:
With direct computation I get (similar to yours but I calcualated also the normalizing constant)

$$f_Y(y)=\sqrt{\frac{\pi}{1+y^2}}e^{-(1+y^2)}$$

Thus the conditional density is, by definition,

$$f_{X|Y}(x|y)=\frac{f_{XY}(x,y)}{f_Y(y)}=\sqrt{\frac{1+y^2}{\pi}}e^{-x^2(1+y^2)}$$

Thus immediately we recognize that

$$(X|Y=y)\sim N\Bigg(0;\frac{1}{2(1+y^2)}\Bigg)$$