问:
How do I prove $\mathbb{1}_{A_i} \rightarrow \mathbb{1}_{A}$ in a measure as $i \rightarrow \infty$ if and only if $\mu(A) <\infty$.?
Let $(X,M, \mu)$ be a measure space and let $A_1 \subset A_2 \subset ...$ be an increasing sequence of measurable sets such that $\mu(A_i)<\infty$ for every $I=1,2...$. Denote $A=\cup_{i=1} ^{\infty} A_i$

Prove that $\mathbb{1}_{A_i} \rightarrow \mathbb{1}_{A}$ in a measure as $i \rightarrow \infty$ if and only if $\mu(A) <\infty$.

Can anyone suggest some hints?

答:
The main task is to show that "$1_{A_i} \to 1_A$ in measure" in this case is equivalent to "$\mu (A \setminus A_i) \to 0$ as $i \to \infty$." If you do that, then you can handle the two cases fairly straightforwardly.

If $\mu(A) = \infty$, then $\mu(A \setminus A_i) = \mu(A) - \mu(A_i) = \infty$.

If $\mu(A) < \infty$, then use the fact that $\mu(A_i) \to \mu(A)$ to show that $\mu(A \setminus A_i) = \mu(A) - \mu(A_i) \to 0$.


问:
What is the largest value not attained by coins worth $\$0.25$ and $\$0.29$? (Frobenius Coin Problem)
I came across a math-related brain teaser recently that offered no solution, I present the gist of it and my reasoning so far.

Suppose I had an unlimited amount of two coin types that were worth $\$0.25$ and $\$0.29$ each. The question is: What is the maximum possible denomination that cannot be formed?

Now my reasoning:

Let z, a total producible denomination, be written as .25x + .29y = z.

Where x is the number of .25 coins, and y is the number of .29 coins and since we are counting the number of coins this limits x and y to the set of all positive integers.

By solving for y:  y = (4z-x)/1.16  I was able to come up with a "procedure" to determine if a particular denomination can be produced from a combination of the two coins.

It is as follows for a denomination d under consideration:

Set a to 4d

Divide a by 1.16, if fully divisible then the result is the number of .29 coins, and we are done if not go to step 3

Subtract 1 from a, the number of .25 coins is 1 more than before -- if a becomes < 0 then the amount is not producible from these coin combinations otherwise repeat step 2

Does this reasoning seem sound?  What I'm trying to do here is extrapolate out this process to prove that the maximum possible non-producible denomination is infinite.  (Such as a base case of 1.99 cannot work based on procedure above, and if you continued to add .25 coins to this amount to infinity there would never be a maximum possible non-producible denomination).

I'd appreciate any mathematical insight to approaching this kind of problem.

Brian's answer was helpful, now I know the name of the problem.  I see that $t = ma+nb$ can be the linear integral combination (from that PDF link) and t is representable only for solutions that have $m$ and $n$ with nonnegative integers, and that if you accept the inequality $0\le m \le (b-1)$  then the solution is $t = (b-1)a+(-1)b = (ab-a)-b = aa-a-b$.

Though, I have no idea where that inequality $0\le m \le (b-1)$ comes from, it seems to just be hand waved at.

So instead, I made a chart where (for my example problem) there's 29 rows and the chart is composed of all the counting numbers (one after the other).  This makes each column 29 apart.  Then it's elimination of all representable by finding the first number in each row divisible by 25 (in green).  It can then be seen each number to infinity for that row thereafter can be created by adding more 29's (so all green numbers and those crossed off are representable).  So in this chart, the red numbers are not representable, and the highest is 671 (which correlates to $ab-a-b = 25*29-25-29 = 671$).  Also, from this chart I can conceptually see that all numbers can be created after a certain number of unrepresentable, which was something I was originally having hard time seeing. (They are represented by the ... for infinity).

Anyways, back to the above: where does that inequality $0\le m \le (b-1)$ come from for the $ab-a-b$ solution?

答:
We'll begin by showing that for $n \ge 672$, we can solve

$\quad n = (x)\,25 + (y)\,29  \quad x,y \ge 0$

The two equations (see Bézout's identity),

$\tag 1 \quad \;(7)\,25 + (-6)\,29 = 1$

$\tag 2 (-22)\,25 + \;(19)\,29 = 1$

illuminate the problem and are relied on to compute all solutions in what follows.

To get the solution for $672$, multiply, say, $\text{(1)}$ by $672$

$\quad 672 = (7\cdot672)\,25 - (6\cdot672)\,29 = (4704) \,25 - (4032) \,29$

and divide,

$\quad 4704 = 29\cdot162+6$

and write

$\quad 672 = (4704) \,25 - (4032) \,25 = (6) \,25 - (4032 - 25\cdot162)\,29 = (6) \,25 + (18)\,29$

We can now write (after discovering some patterns on our scrap sheets),

$\quad 672 = (06)\,25 + (18)\,29$

$\quad 673 = (13)\,25 + (12)\,29$

$\quad 674 = (20)\,25 + (06)\,29$

$\quad 675 = (27)\,25 + (00)\,29$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \dots \text{skipping } 676,677,678$

$\quad 679 = (26)\,25 + (01)\,29$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \dots \text{skipping } 680,681,682$

$\quad 683 = (25)\,25 + (02)\,29$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad  \dots \text{skipping } 684,685,686$

$\quad 687 = (24)\,25 + (03)\,29$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad  \dots \text{skipping } 688,689,690$

$\quad 691 = (23)\,25 + (04)\,29$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad  \dots \text{skipping } 692,693,694$

$\quad 695 = (22)\,25 + (05)\,29$

$\quad 696 = (00)\,25 + (24)\,29$

To fill in the five blocks with $3$ missing entries each, we use $\text{(2)}$ for the first entry of each block and $\text{(1)}$ for the next two entries; there are clearly no obstacles to keeping both $x$ and $y$ non-negative.

Since we have a solid consecutive block of $25$ solutions, $672 \text{ thru } 696$,  solutions for $n \ge 672$ can be specified; see for example the work done here for a similar problem.

Our next step is to show that

$\tag 3 671 = (x)\,25 + (y)\,29  \quad x,y \ge 0$

can never be true.

Assuming $\text{(3)}$ is true, upon writing

$\quad 671 \equiv 29y \pmod{25} \; \text{ iff } \; 4y \equiv 21 \pmod{25} \; \text{ iff } \; y \equiv 24 \pmod{25}$

we substitute the smallest 'candidate' $y = 24$ into $\text{(3)}$, and observe that the rhs is greater than or equal to $24 \times 29$. We've already computed this number, $24 \times 29 = 696$, and, with a contradiction in hand, can now argue that  $671$ is the Frobenius number for this problem.

ContentLicense: CC BY-SA 4.0


问:
Evaluate $\lim_{x\rightarrow 0} \frac{\left( \cosh x \right) ^{\sin x}-1}{\sinh x\cos \left( \sin \left( x \right) -1 \right)}$
Evaluate the limit
$$
\lim_{x\rightarrow 0} \frac{\left( \cosh x \right) ^{\sin  x}-1}{\sinh x(\cos \left( \sin \left( x \right)  \right)-1)}
$$

My Attempt: I tried to use L'Hôpital's rule to evalute it, however I found that the $1$st and $2$nd derivative of the numerator is $0$ at $x=0$, and the $3$rd derivative is very complicated. And the method of Taylor's Series is too complicated here. So, my question is, is there any easier way to evaluate this limit?

The desired answer is $-1$.

答:
The first and second derivatives of both numerator and denominator are zero.  The third derivative of the numerator at $x=0$ is $3.$   The third derivative of the denominator at $x=0$ is $-3.$

The limit is $-1.$


问:
Complete the square for $f(x) = 2x^2 + 4x - 6$
I'm studying for a math test. This is the question:

$f(x) = 2x^2 + 4x - 6$. complete the square.

This is how much I get out of the question:
$$2x^2 + 4x - 6$$
$$2(x^2 + 2x - 3)$$
$$2(x^2 + 2x + 1^2 - 1^2 - 3)$$
$$2((x + 1)^2 - 4)$$

But I get stuck here. Can someone complete it and explain?

I'm sorry i forgot a important piece off the question
it says: Complete the square and derived to get the minimum value off $f(x)$

答:
Given:

$f(x) = 2x^2 + 4x - 6$. complete the square.

$f(x)=2x^2+4x-6$

$f(x)=2x^2+4x-6 $

$f(x)=2(x^2+2x)-6$

$f(x)=2[x^2+\dfrac{2x}{2}+(1)^2]-6$

$f(x)=[2(x+1)^2 -1]-6$

$f(x)=2(x+1)^2-8$  which is the same as: $f(x) = 2x^2 + 4x - 6$

This can be confirmed by using a graphing calculator such as: https://www.desmos.com/

ContentLicense: CC BY-SA 3.0


答:
And the method of Taylor's Series is too complicated here.

Not really, actually. Using the Taylor series approximations of $\cos$, $\sin$, $\ln$, $\exp$, and $\cosh$ (all standard) and composing them as we go along:
$$\begin{align*}
\frac{\left( \cosh x \right) ^{\sin  x}-1}{\sinh x(\cos \left( \sin \left( x \right)  \right)-1)}
&= \frac{e^{\sin  x \ln \cosh x}-1}{\sinh x(\cos \left( \sin \left( x \right)  \right)-1)}
= \frac{e^{(x+o(x)) \ln \left(1+\frac{x^2}{2}+o(x^2\right)}-1}{(x+o(x))(\cos \left( \left( x +o(x) \right)  \right)-1)}\\
&= \frac{e^{(x+o(x)) \left(\frac{x^2}{2}+o(x^2)\right)}-1}{(x+o(x))\left( -\frac{x^2}{2} +o(x^2) \right)}
= \frac{e^{\frac{x^3}{2}+o(x^3)}-1}{-\frac{x^3}{2} +o(x^3)}\\
&= \frac{\frac{x^3}{2}+o(x^3)}{-\frac{x^3}{2} +o(x^3)}
= \frac{1+o(1)}{-1+o(1)} \xrightarrow[x\to0]{} \boxed{-1}
\end{align*}$$

Title: Evaluate $\lim_{x\rightarrow 0} \frac{\left( \cosh x \right) ^{\sin x}-1}{\sinh x\cos \left( \sin \left( x \right) -1 \right)}$

Body:
Evaluate the limit
$$
\lim_{x\rightarrow 0} \frac{\left( \cosh x \right) ^{\sin  x}-1}{\sinh x(\cos \left( \sin \left( x \right)  \right)-1)}
$$

My Attempt: I tried to use L'Hôpital's rule to evalute it, however I found that the $1$st and $2$nd derivative of the numerator is $0$ at $x=0$, and the $3$rd derivative is very complicated. And the method of Taylor's Series is too complicated here. So, my question is, is there any easier way to evaluate this limit?

The desired answer is $-1$.


问:
How trouble with understanding what's being asked on my Taylor Series question?
I'm currently studying a practice exam for Calc II when I ran into a question that I don't understand. I'm not exactly looking for a solution; rather, a logical guide that I could use to understand what's being asked and some of the steps I can take to achieve the solution.

Here it is:

Suppose the Taylor series for a function f(x) is given by: $$ f(x) = 4 - x + 5x^2 +6x^3-x^4+x^6+...$$

Compute the first four terms of the Taylor Series for $f ''(x^2)$

Is it asking for me to convert $f(x)$ into a general series form and take its double derivative then plug in $x^2$ for $x$?

Here is the solution:
$$f''(x^2) = 10+36x^2-12x^4+30x^8$$

Thanks in advance

Body:
$f'(x)=-1+10x+18x^2-4x^3+6x^5\dots$

$f''(x)=10+36x-12x^2+30x^4\dots$

$f''(x^2)=10+36x^2-12x^4+30x^8\dots$


问:
Understanding entailment
I am trying to understand the concept of entailment when used within the context of inductive logic programming. Could somebody explain this to me?

In other words: if A entails B, this means that B is true when A is true. However how can I understand this in practice? For example, assuming the following:

```<code>append(c(U, V), W, c(U,X))) :- append(V, W, X) .     (let's call this A)
</code>```

and

```<code>append,(c(U1, c(V1,W1), X1, c(U1, c(V1,Y1)))) :- append(W1, X1, Y1) (let's call this B)
</code>```

could I say that A entails B? If so, how?

答:
The answer is yes, provided that you have access to the inference rule called Hypothetical Syllogism:

If $A$ entails $B$, and $B$ entails $C$, then $A$ entails $C$.

The "Hypothetical" part of the name results from the fact that it is not obvious from the conclusion (i.e., "$A$ entails $C$") that $B$ was involved in deriving it. So in proving that $A$ entails $C$, we may find use for an intermediary, "hypothetical" statement $B$.

So let us see where this Hypothetical Syllogism comes into play in your situation.

From the premise, i.e.:

$\mathsf{append}(\mathsf c(U_1, \mathsf c(V_1, W_1)), X_1, \mathsf c(U_1, \mathsf c(V_1, Y_1)))$

there's only one statement we can infer by using $A$, namely:

$\mathsf{append}(\mathsf c(V_1, W_1), X_1, \mathsf c(V_1, Y_1))$

Now we can apply our given rule $A$ to this expression again, and obtain:

$\mathsf{append}(W_1, X_1, Y_1)$

In conclusion, we have two valid applications of $A$:

$\mathsf{append}(\mathsf c(U_1, \mathsf c(V_1, W_1)), X_1, \mathsf c(U_1, \mathsf c(V_1, Y_1))) \vdash \mathsf{append}(\mathsf c(V_1, W_1), X_1, \mathsf c(V_1, Y_1))$

$\mathsf{append}(\mathsf c(V_1, W_1), X_1, \mathsf c(V_1, Y_1)) \vdash \mathsf{append}(W_1, X_1, Y_1)$

Using the Hypothetical syllogism, we can now infer $B$.

Title: Understanding entailment


问:
Topological spaces - Meager in itself
Hello someone know examples of topological spaces of first category in themselves (meager in itself), that are not countable? Note that the Cantor set is nowhere dense in $\mathbb{R}$ but it is not meager in itself, because it is Baire.

Thanks

答:
If $\{X_i\}_{i \in I}$ is a family of topological spaces, let $X = \coprod_{i \in I} X_i$ be their coproduct (a.k.a. disjoint union).  Then $X$ is meager iff each $X_i$ is meager in itself (iff each $X_i$ is meager as a subset of $X$).  This is not hard to prove, using the observation that a subset $Y$ of $X$ is nowhere dense iff its intersection to each $X_i$ is nowhere dense.

It follows from this that for every infinite cardinal $\kappa$, the coproduct $\coprod_{i \in \kappa} \mathbb{Q}$ of $\kappa$ copies of the rational numbers is a meager topological space of cardinality $\kappa$.  These spaces are all metrizable, as a coproduct of metrizable spaces is metrizable.

Title: Topological spaces - Meager in itself


问:
Showing that $\sum^\infty_{n=1}\left(\sin\left(\frac{1}{n}\right)-\ln\left(1+\frac{1}{\sqrt n}\right)\right)$ diverges
Separately, I can show that both parts of the sum diverge, but that doesn't help me very much. I don't see any test which could come in handy. I know both sides of the sum have popular Taylor forms, so I can partially represent them as such around $x=0$, but my question then what do I do with the remainder?

答:
$$\begin{split}\sin\left(\frac 1 n\right)-\ln\left(1+\frac 1 {\sqrt n}\right)&=\left[\frac 1 n +\mathcal O\left(\frac 1 {n^3}\right)\right]-\left[\frac 1 {\sqrt n}-\frac 1 {2n}+\mathcal O\left(\frac 1 {n\sqrt n}\right)\right]\\
&=-\frac 1 {\sqrt n}+\frac 3 {2n}+\mathcal O\left(\frac 1 {n\sqrt n}\right)\\
&\sim -\frac 1 {\sqrt n}\end{split}$$


问:
Determinant and inverse matrix calculation of a special matrix
Is there any smart way of calculating the determinant of this kind matrix?

\begin{pmatrix}
1 & 2 & 3 & \cdots &  n \\
2 & 1 & 2 & \cdots & n-1 \\
3 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & \cdots & 1 \end{pmatrix}

I encountered this in a problem for the case $n=4$. I need to find the inverse matrix.

I doubt the idea of this problem is to calculate all the cofactors and then the inverse matrix in the usual way.

I see the pattern here and some recursive relations... but I am not sure if this helps for calculating the determinant of the existing matrix.

答:
If it’s the determinant you are looking for, then what if you subtract column $j+1$ from column $j$ to get the matrix

$$\begin{pmatrix}
-1 & -1 & -1& \cdots &  n \\
1& -1 & -1& \cdots & n-1 \\
1& 1& -1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1& 1& 1& \cdots & 1 \end{pmatrix}_{n\times n}$$

Then you could add the bottom row to the top row to get $[0,0,...,n+1]$ then the determinant will be
$$(-1)^{n+1}(n+1)\begin{vmatrix}
1& -1 & -1& \cdots & -1\\
1& 1& -1 & \cdots & -1\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1& 1& 1& \cdots & 1 \end{vmatrix}_{(n-1)\times(n-1)}$$

From which you can always add the bottom row to the top to get $[2,0,...,0]$, so you end up getting

$(-1)^{n+1}(n+1)2^{n-2}$?

It looks like for the $n \times n$ matrix $A$ given, the inverse has the general form

$$B=
\begin{pmatrix}
\frac{-n}{2(n+1)}& \frac{1}{2} & 0 & 0 &\cdots & 0 & \frac{1}{2(n+1)} \\
\frac{1}{2} & -1 & \frac{1}{2} & 0 & \cdots & 0& 0 \\
0 & \frac{1}{2} & -1 & \frac{1}{2} & \cdots & 0&0 \\
0 & 0 & \frac{1}{2} & -1 & \cdots & 0&0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots &\vdots\\
0 & 0 & 0 & 0 & \cdots &-1&\frac{1}{2}\\
\frac{1}{2(n+1)} & 0 & 0 & 0& \cdots &\frac{1}{2}& \frac{-n}{2(n+1)} \end{pmatrix}
$$

I don't really have proof of how I got this except plugging in the matrix to an inverse matrix calculator for various values of $n$ and noticing a pattern. But one can certainly check that it is indeed the inverse, multiplying on the left and getting $B\cdot A=I$ ensures it is the inverse.

To check this, when multiplying this matrix on the left, it’s not hard to check for the rows of the form $[0,...,\frac{1}{2},-1,\frac{1}{2},...,0]$ because either we have three consecutive integers $n,n+1,n+2$ or $n,n-1,n-2$ $$\frac{n}{2}-(n+1)+\frac{n+2}{2}=0$$ $$\frac{n}{2}-(n-1)+\frac{n-2}{2}=0$$ or $$2\cdot\frac{1}{2}+(-1)\cdot 1+ 2\cdot \frac{1}{2}=1$$ which always is a diagonal entry. Then considering the top row:

$$\frac{-n}{2(n+1)} + 2\cdot \frac{1}{2} + \frac{n}{2(n+1)}=1$$

$$(i+1)\cdot\frac{-n}{2(n+1)}+ i\cdot \frac{1}{2} + (n-i)\cdot  \frac{1}{2(n+1)}=0$$ for $i=1,2,...,n-1$.

The bottom row acts similarly. Thus we indeed get $I$. This probably yields to a nice method to find the inverse through $[A|I] \rightsquigarrow [I|B]$.

Hope this helps!


问:
Proof verification and understanding
I am using Topology Without Tears by Sydney Morris

I am self studying it.

Theorem

If  is an infinite set of cardinality ℵ
prove that there are at least ℵ
distinct topologies on it.

My thoughts on it.(Edited)

Let X be infinite.

P(X)=$2^{\aleph}$

Let A$\in P(X)$ .Define  $T_A$={$\emptyset$,A,X}

$T_A$ is a topology.

Clearly $A\ne B \Rightarrow  T_A\ne T_B$

So C $\in$ P(X), $T_C$= {$\emptyset$,C,X} is a topology

By A1.2.1 A~C
So U={$T_a$ :A$\in $P(X)}is uncountable

And for our bijection

f:X$\mapsto $ P(X)

If A$\ne C \Rightarrow T_A\ne T_C$
Then B$\in $P(X) and $T_B$={$\emptyset$,B,X} is uncountable.

By A1.2.1 A~B and U={$T_A$ :A$\in $P(X)}is uncountable

Any help to make me understand this
stuff a bit easier would be helpful

Left out first theorem on pg 39
cause answers found. I am wondering
if there are other ways to do it
inductively.

答:
You’ve actually constructed $2^\aleph$ distinct topologies on $X$, which is what you actually need: you slightly misquoted the problem. I will note, however, that many of them are homeomorphic to one another: $\langle X,T_A\rangle$ is homeomorphic to $\langle X,T_B\rangle$ if and only if $|A|=|B|$. If $\aleph=\aleph_1$, for instance, you actually have only $\aleph_0$ pairwise non-homeomorphic topologies, far fewer than $2^{\aleph_1}$. (I’ve used $T_A$ for your $T_a$, since there is no necessary connection between $a$ and $A$: they are distinct symbols, and you want the name of the topology to refer to the subset of $X$ used to define it.) However, it appears from his hint that you’ve done what Morris had in mind.

Nevertheless, some of what you’ve said is incorrect.

$T_A$ is not infinite: it has either $2$ or $3$ elements, depending on whether $A\in\{\varnothing,X\}$ or not.

$U$ is indeed uncountable, since $\aleph$ is infinite, but that isn’t what you need to prove: you need to show that $|U|\ge|X|$.

No matter what the set $A$ is, there is no bijection from $A$ to $\wp(A)$: it is a general theorem that $|A|<|\wp(A)|$ for all sets $A$. It is also not clear what any such function would have to do with the problem. The relevant bijection is from $\wp(X)$ to $U$ and is given by the map $\varphi(A)=T_A$.

Added: You’re really almost done once you define the topologies $T_A$. You could write it up like this, for instance:

For each $A\in\wp(X)\setminus\{\varnothing,X\}$ let $T_A=\{\varnothing,A,X\}$; clearly $T_A$ is a topology on $X$ with $3$ open sets. Moreover, if $A,B\in\wp(X)\setminus\{\varnothing,X\}$, and $A\ne B$, then $T_A\ne T_B$, since $A\in T_A\setminus T_B$. Thus, if $U=\big\{T_A:A\in\wp(X)\setminus\{\varnothing,X\}\big\}$, the map $\varphi:\wp(X)\setminus\{\varnothing,X\}\to U:A\mapsto T_A$ is a bijection. Thus, $|U|=|\wp(X)\setminus\{\varnothing,X\}|\overset{(*)}=|\wp(X)|=2^{|X|}=2^\aleph$, where $(*)$ holds because $\wp(X)$ is infinite.

Note that $T_\varnothing=T_X$, so you can’t keep both of them in $U$ if you want $\varphi$ to be injective; I chose simply to delete both of them. You could keep one of them, but then the proof that $T_A\ne T_B$ when $A\ne B$ has to be broken into two cases, since the topology $T_\varnothing$ (or $T_X$) has to be handled separately.

Title: Proof verification and understanding


问:
Is every orthogonal matrix orthogonally diagonalizable?
Is every orthogonal matrix orthogonally diagonalizable? If so, how do you prove it?
And is it true that the entries of every orthogonal matrix is real?

Since every unitary matrix is unitarily diagonalizable, so is true that every orthogonal matrix orthogonally diagonalizable?
Thank you.

答:
Remark 1: Yes, orthogonal matrix is a real square matrix $A$ such that $AA^T=I$.

Claim: Orthogonal matrix is orthogonally diagonalizable if and only if all its eigenvalues are real.

Proof: Every orthogonal matrix $A$ is normal (i.e. commutes with its conjugate transpose) and the set of normal matrices is precisely the set of matrices that are unitarily diagonalizable. Therefore, there exists a unitary matrix $U$ and a diagonal matrix $D$ such that

$$
A = U D U^H\tag1.
$$

Notice that the columns of $U$ are the eigenvectors of A. If $v$ is an eigenvector of $A$ associated to eigenvalue $\lambda$ then

$$
Av = \lambda v\\
A\overline{v} = \lambda \overline{v}
$$

where we used the fact that $\lambda$ and all entries in $A$ are real. We see that $\overline{v}$ is an eigenvector of $A$ associated to eigenvalue $\lambda$. Therefore, the real vector $v + \overline{v}$ is an eigenvector of $A$ associated to eigenvalue $\lambda$. Consequently, we can replace $U$ in $(1)$ with a unitary matrix with real entries. Such a matrix is orthogonal.

For the converse, assume $A$ with an eigenvalue $\lambda \notin \mathbb{R}$ is orthogonally diagonalizable and write $(1)$ as

$$
AU = UD\tag2
$$

with $U$ orthogonal (and therefore real). Note that the left hand side of $(2)$ contains only real entries. Also, note that the entries in the row on the right hand side which corresponds to $\lambda$ are not all real. The contradiction reveals that the assumption that $A$ is orthogonally diagonalizable is false.

Remark 2: Orthogonal matrices with real eigenvalues have a simple geometric characterization. Since these matrices preserve the 2-norm, their eigenvalues have absolute value $1$. The only real numbers that satisfy this condition are $1$ and $-1$. Therefore, the only orthogonally diagonalizable orthogonal matrices are reflections (including identity as a trivial case).


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

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

Thanks in advance

答:
The Cauchy Condensation Test will do it.

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

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


答:
The short answer is no.

Any orthogonally diagonalizable matrix must be symmetric. Indeed, if $A = UDU^T$ where $U$ is orthogonal and $D$ diagonal, then it is easy to see $A^T = A$. On the other hand, there are plenty of orthogonal matrices which aren't symmetric. For example,
$$A = \begin{bmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}$$
is such a matrix.

As for the question "must the entries of orthogonal matrices be real?"--yes and no. When people say "orthogonal matrix" they mean a real orthogonal matrix.

On the other hand, one could define a set
$$O(n,\Bbb C) = \{A\in M(n,\Bbb C): A^TA=AA^T = I\}$$
but there isn't a good reason to look at such matrices. They don't preserve the complex inner product, so they're not a natural generalization of real orthogonal matrices (the unitary matrices are though, since they do preserve the complex inner product).

Body:
Is every orthogonal matrix orthogonally diagonalizable? If so, how do you prove it?
And is it true that the entries of every orthogonal matrix is real?

Since every unitary matrix is unitarily diagonalizable, so is true that every orthogonal matrix orthogonally diagonalizable?
Thank you.


问:
volume above a rectangle $[0,2]\times[1,4]$ and under the paraboloid $z=x^2+4y^2$
I'm trying to find the volume above a rectangle $[0,2]\times [1,4]$ and under the paraboloid $z=x^2+4y^2$

The answer should be 17, but I found 176.

$$\int_1^4\int_0^2\int_0^{x^2+4y^2} dzdxdy$$

I think the $x$ limits isn't correct. I tried the cylinder coordinates, but I got some weird numbers.

$$\int_0^{\frac{\pi}{2}}\int_1^{\sqrt{20}}\int_0^{r^2\cos^2\theta+4r^2\sin^2\theta} rdzdrd\theta$$

$r = \sqrt{0^2+1^2} = 1, r = \sqrt{4^2+2^2} = \sqrt20$

$z = r^2 cos^2\theta + 4r^2 sin^2\theta$

then for $\theta$ the rectangle is only in the +xy plan

答:
$z=x^2+4y^2\geq 0$ for all $x,y$ because a sum of squares is always positive.

Your integral integral is set up okay, I think it’s easiest to stick to rectangular coordinates for this particular case, and the answer is

$$\int_1^4\int_0^2x^2+4y^2dxdy=\int_1^4\frac{x^3}{3}+4xy^2\Bigg|_0^2dy=\int_1^4 \frac{8}{3}+8y^2dy=\frac{8y}{3}+\frac{8y^3}{3}\Bigg|_1^4=8+8\cdot 21= 176$$


问:
Show the limit of the sequence is $\sqrt{\alpha}$
Given: $\sqrt{\alpha}<x_1, \alpha >0, x_{n+1}=\frac{1}{2}\left(x_n+\frac{\alpha}{x_n}\right)$. Show $\lim_{n\to \infty}x_n=\sqrt{\alpha}.$

I have already shown the this is a monotonic decreasing function. Intuitively, I see in the end I get $\frac{1}{2}\left(\sqrt{\alpha}+\sqrt{\alpha}\right)$ but I'm not sure if I should keep going using $\lim$ notation or if I should switch to the definition of converging sequence.

答:
The problem is well-known here of course and in most intro to analysis books for sure. You probably come across few proofs or stumbled upon a page in some book that talks about this or similar problem. I just wanna write up what I see as perhaps what you want to see:

-For all $n\ge 1, x_n \ge \sqrt{\alpha}$. This fact can be established by the AM-GM inequality.

-The sequence is monotonically decreasing: $x_{n+1} - x_n = \dfrac{1}{2}\left(x_n+\dfrac{\alpha}{x_n}\right)-x_n=\dfrac{1}{2x_n}\left(\alpha - x_n^2\right)\le 0$

The above information shows that the sequence is bounded below and decreasing, hence converges to a limit $L$. You can show the $L = \sqrt{\alpha}$ from the equation defining $x_n$.


问:
Examples of properties that hold almost everywhere, but such that explicit examples are not known
In measure theory one makes rigorous the concept of something holding "almost everywhere" or "almost surely", meaning the set on which the property fails has measure zero.

I think it is very interesting that there are some properties which hold almost surely for which it is either very difficult to construct an example, an example requires the axiom of choice, or for which no known examples exist.

I was looking to put together a list of some of these properties.
More precisely, I am looking for (interesting) examples of measure spaces $(X,\mathscr{A},\mu)$ and $S\subseteq X$ such that there exists $N \in \mathscr{A}$ with $S^c \subseteq N$ and $\mu(N)=0$, but that it is hard to find/construct an explicit example of an element of $S$.

A good place to start is of course $(\mathbb{R}^n,\mathcal{B}^n,\lambda)$.

Here's one to get us started:

Theorem. Take $X=(0,1]$ with Lebesgue measure. Writing $\frac{p_n}{q_n}$ to be the $n$th continued fraction approximate of $x\in(0,1]$ in reduced terms, we have
  $$
\lim_{n \to \infty} \frac{\log q_n}{n} = \frac{\pi^2}{12 \log 2}
$$
  almost surely.
  Note: The result fails for all rationals.

答:
One of my favorite examples is more number theoretic:  Artin's conjecture

A weak form of it, paraphrased from Wikipidia, is the following.  Consider the set $\mathbb{Z}^\square$, consisting of all integers except squares and $-1$.

Pick any integer $a\in\mathbb{Z}^\square$ and let $S(a)$ denote the collection of all primes for which $a$ generates the mulitplicative group of $\mathbb{Z}/p$ (with $0$ removed).  Then $S(a)$ is infinite.

Heath-Brown showed that Artin's conjecture is true except for possibly at most 2 counterexamples.  But his proof is not constructive, and there is no known examples of an $a$ for which $S(a)$ is infinite.

In other words, Artin's conjecture holds almost everywhere on $\mathbb{Z}^\square$ (and possibly everywhere), but we don't know of a single point in $\mathbb{Z}^\square$ for which it holds.


问:
Subgaussian Absolute Constant
I  am looking at the proofs of equivalent properties for subgaussian random variables. Two listed below:

$$P(|X|\ge t)\le 2\text{exp}(\frac{-t^2}{K_1^2})$$

$$(\mathbb{E}|X|^{p})^{\frac{1}{p}}\le K_2\sqrt{p}$$

I am able to follow the proofs but the constants $K_i$ are stated to differ only by an absolute constant.
According to the definition they propose that if $i$ implies $j$ then $K_j\le C K_i$ where $C$ is an absolute constant. Does this means that $C$ is the same for any $(i,j)$ pairing? Looking at the proofs that doesn't seem to be true. Can someone help clarify what they mean by absolute constant?

答: There are only finitely many pairs, so $C$ can be taken to be a number larger than any of the values that appear in each part of the proof of Prop. 2.5.2.

Title: Subgaussian Absolute Constant


问:
How to find asymptotic expansion of $\frac{1}{\cos(z)}$

Find the asymptotic expantion of coefficients of the exponential generating function $f(z)=\frac{1}{\cos(z)}$ using all of its poles.

My work:

I know that:
$$
\frac{1}{\cos \left( z \right)}=\sum_{n\ge 0}{E_{2n}}\frac{z^{2n}}{\left( 2n \right) !}
$$

To find the poles we solve $\cos(z)=0$ where z is a complex number. Getting, $\chi_k=\pi/2+\pi k$
where $k \in Z$.
Now wrie $\frac{1}{\cos(z)} \sim  \frac{1}{z-\chi_k}= \frac{-1}{\chi_k} \frac{1}{1-z/\chi_k}$.(I'm not sure if this what I am supposed to do).
Then
$$
E_{2n}=-\left( 2n \right) !\sum_{k\in Z}{\frac{1}{\chi _k}}\frac{1}{\chi _{k}^{n}} \\
E_{2n}=-\left( 2n \right) !\left( \frac{1}{\left( \frac{\pi}{2} \right) ^{n+1}}+\sum_{k\ge 1}{\left( \frac{1}{\chi _{k}^{n+1}}+\frac{1}{\chi _{k_-}^{n+1}} \right)} \right)
$$

Then I can look at the internal sum in the last equation and try to simplify it.

Edited:

$f(z)=\frac{1}{cos(z)} \sim \frac{Res(f(z),\chi_k)}{z-\chi_k}$ where $\chi_k=\frac{\pi}{2}+2\pi k$ the
poles of $f(z)$ and  $Res(f(z),\chi_k)= lim_{z\to \chi_k} (z-\chi_k) f(z)$.

After calculations we get that:

$\chi_0= \pi /2$ with Res= -1, $\chi_{-1} = -\pi / 2$ with Res = 1 , $\chi_1=3\pi /2$ with Res = 1, $\chi_{-2}= -3\pi /2$ with Res = -1 etc.

So, $f(z)=\frac{1}{cos(z)} \sim \frac{Res(f(z),\chi_k)}{z-\chi_k} \sim \frac{-1}{z-1/2 \pi}  +\frac{1}{z+1/2 \pi} +\frac{1}{z-3/2 \pi} + \frac{-1}{z+3/2 \pi}+....$

Then if we expand these geometric coulomns and look at their coefficients we get:

$\frac{E_{2n}}{n!} \sim (2/ \pi)^{n+1} + (-2/ \pi)^{n+1} - (2/ 3\pi)^{n+1} +(2/ 3\pi)^{n+1} - (2/ 5\pi)^{n+1} -(-2/ 5\pi)^{n+1}-(2/ 7\pi)^{n+1} +(-2/7\pi)^{n+1}...)$

Thus, $E_{2n} \sim (2n)! (  (2/ \pi)^{n+1} + (-2/ \pi)^{n+1}  - (2/ 5\pi)^{n+1} -(-2/ 5\pi)^{n+1}-(2/ 7\pi)^{n+1} +(-2/7\pi)^{n+1}...)$

Thus $E_{2n} \sim (2n)! (  (2/ \pi)^{n+1} + (-2/ \pi)^{n+1} +O(2/ 5\pi)^{n})$

What do you think.
Any guide will be so appreciated.

答:
Here are the two facts you need to solve the problem:

If $f(z)=\sum_{n\ge 0} a_nz^n$ is analytic at the origin with radius of convergence $R$, then $a_n\in O\big((1/R+\epsilon )^{n}\big)$ for all $\epsilon>0$. When $f$ is meromorphic, recall that $R$ is the infimum of the magnitudes of the poles of $f$.

If $f(z)$ has a pole of order $1$ at $a$, then $f(z)-\frac{\text{Res}(f,a)}{z-a}$ has a removable singularity at $a$, and the same poles as $f$ elsewhere. Here, $\text{Res}(f,a)=\lim_{z\to a}(z-a)f(z)$.

This means that if $f$ only has simple poles, then by subtracting out $\frac{\text{Res}(f,a)}{z-a}$ as $a$ ranges over the poles of $f$, then coefficients of what remains will go to zero at an arbitrarily fast rate.

For example, $\frac1{\cos z}$ has a pole at $\chi_0=\pi/2$, and the residue at that pole is $\text{Res}(f,\pi/2)=\lim_{z\to\pi/2}\frac{z-\pi/2}{\cos z}=-1$. Therefore, $f-\frac{-1}{z-\pi/2}$ has one fewer pole. This is akin to what you attempted, but you forgot to compute the residue.

Similarly, you can subtract the pole at $\chi_{-1}=-\pi/2$, whose residue is $1$. When you do this, the current state of affairs is
$$
\frac{1}{\cos z}+\frac1{z-\pi/2}-\frac1{z-(-\pi/2)}
$$
and the poles of what remains are only at $3\pi/2,5\pi/2,\dots$ and $-3\pi/2,-5\pi/2,\dots$. The infimum the magnitude of these values is $3\pi/2$, which proves that
$$
{E_{n}}/{n!}-(2/\pi)^{n+1}+(-2/\pi)^{n+1}\in O\big((2/3\pi+\epsilon)^n)
$$
You can then keep subtracting out poles, accumulating more terms on the left hand side, and decreasing the error on the right hand side.

In general, this method does not give an equality for the coefficients of $f(z)$ in terms of an infinite series of the subtracted poles. It happens to work for $f(z)=1/\cos z$. However, in general, even after subtracting out all the poles of $f$, there may still be an entire function left over which contributes to the coefficents of $f$.


问:
High school contest math question (number theory) - prove:
Reposting with Mathjax - sorry, first time!

Let $S = \{4,8,9,16,...\}$ be the set of integers of the form $m^k$  for integers $m, k \ge  2$. For a positive integer $n$, let $f(n)$ denote the number of ways to write $n$ as the sum of (one or more) distinct elements of $S$.
For example, $f(5) = 0$ since there are no ways to express 5 in this fashion, and $f(17) = 1$ since $17 = 8+9$ is the only way to express 17.

(a) Prove that $f(30) = 0$

(b) Show that $f(n) \ge 1$ for $n \ge 31$.

(c) Let $T$ be the set of integers for which $f(n) = 3$. Prove that $T$ is finite and non-empty, and find the largest element of $T$.

I think that part a) is relatively easy to just check since none of the values in the first few values of the set $S = \{4,8,9,16,25,27,32,64,...\}$ will add to get to 30.

I'm not sure where to start with part b and part c. For part b, I was working at finding sums for each number but figured this was not an intelligent way to proceed. For part c) I'm not sure where to start at all.

Thanks for any help you can give.

答:
For part b, note that all multiples of $4$ can be represented because you have all the powers of $2$ except $1,2$.  Express any multiple of $4$ in binary and read off the numbers to add to get it.  All numbers equivalent to $1 \bmod 4$ that are at least $9$ can be expressed because the number minus $9$ is a multiple of $4$ and therefore expressible.  All numbers equivalent to $2 \bmod 4$ that are $34$ or greater are expressible because $34=9+25$.  All numbers equivalent to $3 \bmod 4$ that are $27$ or greater because we have $27$ available.  Therefore the greatest number that cannot be expressed is $30$.

For c, numbers that are large enough will have too many representations.  We will do each residue class $\mod 4$ in turn.

For $0 \bmod 4$ we have $\emptyset,36, 9+27, 25+27$ as ways to express numbers without any of the $2^n$ terms.  We can therefore express any number $52$ or greater in $4$ or more ways.

For $1 \bmod 4$ we have $9, 25, 9+36, 49$ so we can express any number $49$ or greater in $4$ or more ways.

For $2 \bmod 4$ we have $9+25, 9+49, 9+36+49, 25+49$ so we can express any number $74$ or greater in $4$ or more ways.

For $3 \bmod 4$ we have $27, 27+36, 9+25+49, 9+25+36+49$ so we can express any number $119$ or greater in $4$ or more ways.

The greatest number in $T$ is $115$, which can be expressed as $64+27+16+8, 36+32+27+16+4, 49+32+25+9$ but in no other ways.


问:
Prove that there does not exist a continuous map surjective map from $S^1$ to $\mathbb{R}$
Prove that there does not exist an surjective and continuous map from $S^1= \{(x,y) \in \mathbb{R}^2:x^2+y^2=1 \}$ (provided with the metric induced by the usual $ \mathbb{R}^2$)  in $ \mathbb{R}$.

How can I extend my proof below to prove that there does not exist an surjective and continuous map from $S^1$ to $\mathbb{R}$?

Let us take $K=[0,2π]\subset \mathbb{R}\rightarrow S^1$, $L=S1=\{(x,y)\in \mathbb{R}^2:x^2+y^2=1\}$ and $Φ:[0,2π]\rightarrow \mathbb{R}^2$, given by

$$
Φ(t)=(\cos t,\sin t)
$$

Proof: We have that $[0,2π]$ and $S^1$ are compact and $Φ:[0,2π]\rightarrow S^1$ is continuous and surjective. Let $g:[0,2π]\rightarrow \mathbb{R}^n$ be continuous such that $g(0)=g(2π)$. Let’s define $f:S^1\rightarrow \mathbb{R}^n$ from $g$ as: $f(\cos t,\sin t)=g(t)$, as $g(0)=g(2π)$, $f$ is well defined. We note that $g=f(g(Φ))$ is continuous.

Thus, to define a continuous map in the circle $S^1$, it is sufficient to define it in $[0,2π]$ so that it is continuous in this interval and assumes equal values in the extremes $0$ and $2π$.

答:
Perhaps the method that you started could be finished, but here's a quick way. It will use a common result, that the image of a compact space under a continuous map is compact. I suspect one of the above comments was hinting at this.

Suppose there were a continuous surjective map $f:S^1\rightarrow\mathbb{R}$. We know $S^1$ is compact, so by the above stated result, $f(S^1)=\mathbb{R}$ is compact. This is a contradiction, however, since $\mathbb{R}$ is not compact.


问:
Prove $\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$
I have to prove that :
$$\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$$

I already know that the Maclaurin series of $\ln(1+x)$ is $x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+\frac{1}{5}x^5+\ldots$ But how can I use it to prove the first claim? Should I just multiply $\frac{1}{1+x}$ to the Maclaurin series of $\ln(1+x)$? Still, I don't know how to do that.

I tried to use the derivatives of $\frac{\ln(1+x)}{1+x}$ then substitute that into the Maclaurin formula. But I don't think that's the right way to do this efficiently.

答:
Multiply the two series $\ln(1+x) = x - \frac{1}{2} x^2 + \frac{1}{3} x^3 - \cdots$ and $\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots$.

For instance, the coefficient of $x^2$ in the product is $(-\frac{1}{2} x^2) \cdot 1 + x \cdot (-x) = -(1 - \frac{1}{2}) x^2$.


问:
Prove $\iiint_{\mathbb{R}^3}\left ( \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z} \right) dxdydz=0$
Assume $f(x,y,z)$ is continuously differentiable on $\mathbb R^3$, and both the integral
$$
\iiint_{\mathbb{R}^3}{\left| f\left( x,y,z \right) \right|\text{d}}x\text{d}y\text{d}z
$$
and
$$
\iiint_{\mathbb{R}^3}{\left( \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z} \right) \text{d}}x\text{d}y\text{d}z
$$
exists. Prove that
$$\iiint_{\mathbb{R}^3}{\left( \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z} \right) \text{d}}x\text{d}y\text{d}z=0$$

I don't have any ideas about the problem. Can anyone help?

答:
Let $\mathcal{C}_r = [-r,r]^3$ denote the cube of side length $2r$ centered at $0$. Then

\begin{align*}
\left| \int_{\mathcal{C}_r} \frac{\partial f}{\partial x} \, \mathrm{d}x\mathrm{d}y\mathrm{d}z \right|
&= \left| \int_{[-r,r]^2} \left( f(r, y, z) - f(-r, y, z) \right) \, \mathrm{d}y\mathrm{d}z \right| \\
&\leq \int_{\partial\mathcal{C}_r \cap \{x=\pm r\}} \left| f(\mathbf{x}) \right| \, \sigma(\mathrm{d}\mathbf{x}),
\end{align*}

where $\sigma$ is the surface measure on $\partial\mathcal{C}_r$. Since similar results hold along the other coordinate directions, we have

\begin{align*}
\left| \int_{\mathcal{C}_r} \left(\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}\right) \, \mathrm{d}x\mathrm{d}y\mathrm{d}z \right|
&\leq f(r), \qquad\text{where} \quad f(r) := \int_{\partial\mathcal{C}_r} \left| f(\mathbf{x}) \right| \, \sigma(\mathrm{d}\mathbf{x}).
\end{align*}

However, since

$$ \int_{0}^{\infty} f(r) \, \mathrm{d}r
= \int_{\mathbb{R}^3} \left| f(\mathbf{x}) \right| \, \mathrm{d}x\mathrm{d}y\mathrm{d}z
< \infty, $$

we have $\liminf_{r\to\infty} f(r) = 0$ and hence there exists a sequence $(r_n)_{n\geq 1}$ with $r_n \to \infty$ such that $f(r_n) \to 0$ as $n\to\infty$. From this, we obtain

$$ \left| \int_{\mathbb{R}^3} \left(\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}\right) \, \mathrm{d}x\mathrm{d}y\mathrm{d}z \right|
= \lim_{n\to\infty} \left| \int_{\mathcal{C}_{r_n}} \left(\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}\right) \, \mathrm{d}x\mathrm{d}y\mathrm{d}z \right|
\leq 0 $$

and therefore the desired conclusion follows.


Title: Prove $\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$

Body:
I have to prove that :
$$\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$$

I already know that the Maclaurin series of $\ln(1+x)$ is $x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+\frac{1}{5}x^5+\ldots$ But how can I use it to prove the first claim? Should I just multiply $\frac{1}{1+x}$ to the Maclaurin series of $\ln(1+x)$? Still, I don't know how to do that.

I tried to use the derivatives of $\frac{\ln(1+x)}{1+x}$ then substitute that into the Maclaurin formula. But I don't think that's the right way to do this efficiently.

Body:
You must multiply the series
$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right)$$
$$\frac 1{1+x}=1-x+x^2-x^3+x^4-x^5+O\left(x^6\right)$$
$$\frac{\log(1+x)}{1+x}=\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right) \right) \times$$ $$\left(1-x+x^2-x^3+x^4-x^5+O\left(x^6\right) \right) $$
$$\frac{\log(1+x)}{1+x}=x-\frac{3 x^2}{2}+\frac{11 x^3}{6}-\frac{25 x^4}{12}+\frac{137
   x^5}{60}+O\left(x^6\right)$$

In fact, almost as you wrote in the post
$$\frac{\log(1+x)}{1+x}=\sum_{n=1}^\infty (-1)^{n+1} H_n \,x^n$$


答:
All the other answers to your question focus on how to compute the Maclaurin series of this function using the known series. I will sketch a strategy to compute it by directly computing Maclaurin coefficients. Let $$(1+x)y=\ln(1+x)$$ for all $x\gt -1. $ Then lets look at higher derivatives to see a possible pattern: $$(1+x)\dfrac{dy}{dx}+y=\dfrac{1}{1+x}\tag1$$
$$(1+x)\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}=-\dfrac{1}{(1+x)^2}\tag2$$
$$(1+x)\dfrac{d^3y}{dx^3}+3\dfrac{d^2y}{d^2x}=\dfrac{2}{(1+x)^3}\tag3.$$ By looking at these, it is not difficult to guess that
$$(1+x)\dfrac{d^{n+1}y}{dx^{n+1}}+(n+1)\dfrac{d^ny}{d^nx}=\dfrac{(-1)^{n}n!}{(1+x)^{n+1}}\qquad \forall n\in\mathbb{N}\cup\{0\}.$$ We can use mathematical induction to establishes our claim rigorously. Since coefficients of Maclaurin series are given by
$$a_n=\dfrac{1}{n!}\dfrac{d^ny}{d^nx}\Big\vert_{x=0},$$ we can show that $a_n=(-1)^{n+1}H_n$ by solving the recursive relation $$a_{n+1}+a_n=\dfrac{(-1)^n}{n+1},\qquad a_0=0.$$ (You can easily solve this recursion by multiplying whole equation by $(-1)^n$ and then telescoping). In fact, we can easily use it to compute Taylor coefficients at any point we want.


问:
Test for convergence for improper integral $1/x^x$
I am having trouble determining if this is convergence or divergence

$$\int^1_0 1/x^x dx$$

答:
It suffices to show that

$$\lim_{x \to 0^+} x^{-x} =1,$$
because then we could consider the integral

$$\int_0^1 x^{-x} dx$$

as an integral of a continuous function on a compact set (which has convergent value).

Taking logs, this is the equivalent to

$$\lim_{x \to 0^+} -x \log x =0.$$

This is a standard result, which can be shown using l'Hopital's rule.


答:
\begin{align*}
\frac{\ln(1+x)}{1+x}
    &= \ln(1- -x) \cdot \frac{1}{1- -x}  \\
    &= \left( \sum_{i=1}^\infty \frac{-(-x)^i}{i} \right) \left( \sum_{j=0}^\infty (-x)^j \right),  & |x| < 1\\
    &=_{\text{why?}} \sum_{i=1}^\infty \sum_{j=0}^\infty \frac{-(-x)^i}{i} (-x)^j  \\
    &= \sum_{i=1}^\infty \sum_{j=0}^\infty \frac{(-1)^{i+j+1}}{i} x^{i+j}
\end{align*}

Now change variables to $k = i+j$ and $\ell$.  On a diagram of the terms in this double series (a plot of points on the $(i,j)$-plane corresponding to terms in the series), the lines of constant $k$ have slope $-1$ and meet $k$ points, so we may take $\ell = i$, running from $1$ to $k$.  (Again, this is vastly easier to see in the diagram.)  Now make $k$ be the outer sum and $\ell$ be the inner sum and you will immediately have your result,
$$  \sum_{k=1}^\infty \left( -(-1)^k x^k \sum_{\ell =1}^k \frac{1}{\ell} \right)  \text{.}  $$


问:
The set of all self-inverting functions in $\mathbb{R}$
I had asked a question about self-induced functions a few days ago and would like to discuss again in the topic a bit. We found out that if $f(x) = f^{-1}(x)$, then that for an invertible function $g(x)$, the function
\begin{align}
f_g(x) = g^{-1}(f(g(x)))
\end{align}
is also a self-inverting function again. What I am wondering now is whether with the function
\begin{align}
f(x) = \frac{ax+b}{cx-a}
\end{align}
all other self-invertible functions in the space $\,\mathbb{R} \,$ follow, so the set of self-inverting functions is thus completely described.

答:
There are many, many more functions on $\Bbb R$ that are their own inverse.  Most of them cannot be described.  You can split $\Bbb R$ into sets of one or two elements any way you want.  Having done so, a self-inverse function will take all the elements in single element sets to themselves and take all the elements in two element sets to the other one.

This gives us $2^{\mathfrak c}$ self-inverse functions. Take any partition of $\Bbb R$ into three element sets, of which there are $\mathfrak c$.  We can form a distinct self-inverse function by choosing any collection of one element out of each set, which we can do in $3^{\mathfrak c}$ ways.


问:
Are midpoints always the best estimate for the area under the curve using a finite number of rectangles?
Are there any scenarios where midpoints would produce a result which is less accurate than left-endpoints or right endpoints?

If so, how do I recognize when each rule would apply the most accurate result?

答:
Suppose we are to estimate $\int_a^b \; f(t) \,\mathrm{d}t$ using a left-, mid-, or right- sum.

Let $M_1$ be the maximum of $|f'(x)|$ on $[a,b]$ and $M_2$ be the maximum of $|f''(x)|$ on $[a,b]$.  Then (as you should know), the maximum error (absolute value of the difference between the value of the sum and the true value of the integral) in using a left- or right- sum of $n$ intervals (of width $\frac{b-a}{n}$ is
\begin{align*}
\text{left- and right-}  &&  \text{error} &\leq \frac{M_1(b-a)^2}{2n}  \\
\text{mid- }  &&  \text{error} &\leq \frac{M_2(b-a)^3}{24n^2}  \\
\end{align*}
We can see that as $n$ gets large, eventually the mid-sum error should become less (and stay less) that the left- and right- sum errors.  Let's show an example of this.

Let $\displaystyle f(x) = \frac{1}{10^4 x^2  +1}$ and $[a,b]=[0,3]$.  Then, we can show $M_1 = \frac{75 \sqrt{3}}{2}$ and $M_2 = 20\,000$.  With these, we can plot the upper bounds on the errors and see that eventually, the worst possible errors with the mid-sums will be less than the worst possible errors with the left- and right-sums (in this case).

As we can see, for $n$ less than about $50$, the interval containing the value of the integral is narrower for left- and right- sums than for mid-sums.  This reverses for $n$ greater than about $50$.  But none of this reveals the detailed errors for particular choices for $n$, just conservative bounds for how far a sum can be from the true value of the integral.  So let's study that.

(The error bounds are shown in light gray.)  Happily, we see that the actual errors are much less than the worst case bounds.  Initially, left-sums are very inaccurate.  (Expected, because this integrand is decreasing, rapidly for small $x$.)  However, right- and mid-sums have the same initial accuracy, with mid-sums gradually improving faster than right-sums.  Starting from $n$ a little less than $10^4$ and continuing to the right, the actual errors in the left- and right-sums exceed the error bound for the mid-sums, so the mid sums is guaranteed to be more accurate estimate of the integral than the left- and right-sums for large $n$.

The above is fairly generic behaviour for integrands that have a few monotonic regions on the interval of integration.  What if this is not the case?

Consider $g(x) = sin(1/x)$ on $[10^{-4},1]$.  It's a bit more work, but we can show that $M_1$ is a little less than $10^8$ and $M_2$ is a little less than $10^{16}$.  (By a little less, I mean "less that one part per thousand".)  Here, $n$ has to rise to around $10^7$ for the error bounds to match.

Notice that the error is actually bounded.  (The integrand is upper bounded by $1$ and lower bounded by $-1$, so the maximal possible error from a sum that samples the function is $2 \cdot (1-10^{-4})$, which is a bit less than $2$.  So the error bounds are (very) pessimistic until $n$ is large-ish.

But the detailed behaviour is much more complicated.

Which method has least error changes almost every time we increment $n$.  (It may seem that the error hovers around $0.7$, but this is a coincidence among the small $n$ sums.  Once $n$ is around $10^6$, the error starts to decrease -- this spacing of sample points is finally fine enough to see the individual oscillations in the function.)

So, yes, there are "scenarios where midpoints would produce a result which is less accurate than left-endpoints or right endpoints".  In general, there is no way to "recognize when each rule would apply the most accurate result".

What can be done is this:  Find the crossing $n$ for the two error estimates.  Decide whether that value of $n$ is small enough to exceed or is too large for reasonable computation.  If it is small, then use a mid-sum with larger $n$ (and possibly for a few values of $n$ to catch coincidences of large error).  If it is not small, then use left- or right-sums (and possibly for a few values of $n$).


问:
Showing set of tensored states span a space
I have the four states
$$
\lvert1\rangle \lvert1\rangle - \lvert0\rangle \lvert0\rangle \\
i\lvert1\rangle \lvert1\rangle + i\lvert0\rangle\lvert0\rangle \\
\lvert0\rangle \lvert1\rangle + \lvert1\rangle \lvert0\rangle \\
\lvert0\rangle \lvert1\rangle - \lvert1\rangle \lvert0\rangle
$$

What is a quick and easy way to show they are an orthogonal basis in $\mathbb{C^2}\otimes \mathbb{C^2}$. I've got the simple orthogonal bit down which I guess proves independence, but how can I show the span of the set?

答:
$\Bbb C^2\otimes \Bbb C^2$ is four dimensional, and you have four linearly independent vectors. They have to span the whole space because of dimensionality.

Just think what would happen if they didn't span the space. You would pick another vector outside of their span, and that would have to be linearly independent with the first four vectors. But that would mean the dimension of the space was at least 5 (an absurdity.)

In general, any $n$ linearly independent vectors of an $n$ dimensional space must span the space. Conversely, any $n$ vectors that span an $n$ dimensional vector space must be linearly independent.


问:
How can I solve this limit? $\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$
$\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$
Where $a$ can be any number.

I started by using $e$ like this:

$\lim _{x \to a}e^{\ln\left(\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}\right)} \to \lim _{x \to a}e^{\ln\left(\frac{\sin(x)}{\sin(a)}\right)\cdot\frac{1}{x-a}} \to \lim _{x \to a}e^{\frac{\ln\left(\sin(x)\right)-\ln\left(\sin(a)\right)}{x-a}}$

The answer should be $e^{\cot{a}}$

答:
$$\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$$

Let the limit = $L$, hence

$$L = \lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$$

Take the natural logarithm of both sides, and utilizing the fact it's continuous you get

$$ln(L) = \lim _{x \to a} \frac{1}{x-a}ln\left(\frac{\sin(x)}{\sin(a)}\right)$$

$$ln(L) = \lim _{x \to a} \frac{ln\left(\frac{\sin(x)}{\sin(a)}\right)}{x-a}$$

$$ln(L) = \lim _{x \to a} \frac{ln(sin(x))-ln(sin(a))}{x-a}$$

Now using L'hopitals rule

$$ln(L) = \lim _{x \to a} \frac{cos(x)}{sin(x)}$$

$$ln(L) = cot(a)$$

$$L = e^{cot(a)}$$


答:
Rewrite $$ x_{n+1}=\frac{1}{2}\left(x_n+\frac{\alpha}{x_n}\right)$$
$$ x_{n+1}= x_{n}-\Big[x_n-\frac{1}{2}\left(x_n+\frac{\alpha}{x_n}\right)\Big]$$ that is to say
$$ x_{n+1}= x_{n}-\frac {x_n^2-\alpha}{2x_n}$$ and recognize the Newton iterative scheme for finding the zero of function $f(x)=x^2-\alpha$.

Title: Show the limit of the sequence is $\sqrt{\alpha}$

Body:
Given: $\sqrt{\alpha}<x_1, \alpha >0, x_{n+1}=\frac{1}{2}\left(x_n+\frac{\alpha}{x_n}\right)$. Show $\lim_{n\to \infty}x_n=\sqrt{\alpha}.$

I have already shown the this is a monotonic decreasing function. Intuitively, I see in the end I get $\frac{1}{2}\left(\sqrt{\alpha}+\sqrt{\alpha}\right)$ but I'm not sure if I should keep going using $\lim$ notation or if I should switch to the definition of converging sequence.


问:
Conditional Probability (Diseases)
A town with a population 10,000 has suffered an outbreak of dragon pox, with 3% of the population being infected. There is a test to diagnose the disease. If you have the dragon pox, the test will correctly register positive 97% of the time and will yield a false negative 3% of the time. If you do not have the dragon pox the test will register that correctly 97% of the time and will yield a false positive 3% of the time. Given that you tested positive for the disease, what is the probability that you actually have dragon pox?

I just do not get this question at all.

答:
Hint

Need Bayes Theorem.

Let $A =$ event that you have the disease.

Let $B =$ event that you tested positive.

Let $(AB) =$ event that both of the events $A$ and $B$ occurred.

Per wikipedia article

$$p(A|B) = \frac{p(AB)}{p(B)}.$$

$p(AB)$ can be calculated as chance of having disease $\times$ chance of getting a true positive.

$p(B)$ can be calculated as the above chance

plus the chance of not having disease $\times$ chance of getting false positive.

Title: Conditional Probability (Diseases)


问:
Proving an inequality involving three real numbers
For any numbers $a,b,c \in \mathbb{R}$, prove that $\frac{3}{2} (a^2+b^2+c^2+1) \geq a(b+1) +b(c+1)+ c(a+1)$.

Body:
Left minus right can be written as
$$
\frac {a^2}{2}-ab+\frac {b^2}{2}
+\frac {b^2}{2}-bc+\frac {c^2}{2}
+\frac {c^2}{2}-ac+\frac {a^2}{2}
+\frac {a^2}{2}-a+\frac {1}{2}
+\frac {b^2}{2}-b+\frac {1}{2}
+\frac {c^2}{2}-c+\frac {1}{2},
$$
which is just
$$
\frac 12(a-b)^2+\frac 12(b-c)^2+\frac 12(c-a)^2+\frac 12(a-1)^2+\frac 12(b-1)^2+\frac 12(c-1)^2\geq 0.
$$


问:
Is this a contraction mapping?
Is $T(x)=\frac{1}{20}e^{-x^2}\sin(wx)$ a contraction mapping? $w$ can be any real number.

答: Let $w = 100$.  $T([-0.005,0]) = [-0.022{\dots}, 0]$.  So, no.  For this choice of $w$, the derivative exceeds $4$ on this interval, so this noncontractivity is entirely detectable by the usual method.

Title: Is this a contraction mapping?


问:
Show that $Y_t = (W_t + t)\exp\{-W_t - \frac{1}{2}t\}$ is a martingale.
Let $W$ be standard (one-dimensional) Brownian motion and define the process $Y = \left\{ Y_t, \mathcal{F}_t; 0 \le t < \infty \right\}$ by
$$ Y_t = (W_t + t)\exp\left\{-W_t - \frac{1}{2}t \right\}. $$
I am trying to show that $Y$ is an $\left\{ \mathcal{F}_t \right\}$-martingale using Ito's Formula.

Here's my approach:

For each $t \ge 0$, put $G_t := W_t + t$, $M_t := -W_t - \frac{1}{2}t$, and $H_t := \exp\{M_t\}$. Then $Y_t = G_t H_t$, and $M$ is an Ito process with $(u_t, v_t) = (-1/2, -1)$. Now, let $f(t,w) = f(w) = e^w$. Then $f(M_t) = f'(M_t) = f''(M_t) = H_t$; and so, by Ito's Formula,
$$ d H_t = \left[0 - \frac{1}{2}f'(M_t) d M_t + \frac{1}{2}f''(M_t)\right]dt - f'(M_t) dW_t = 0 - f'(M_t) dW_t = - H_t dW_t. $$
This means that
$$ H_t = 1 - \int_0^t H_s \, dW_s, \qquad t \ge 0;$$
and so, the process $H := \left\{H_t, \mathcal{F}_t; 0 \le t < \infty \right\}$ is a martingale.

Similarly, $G$ is an Ito process with $(u_t, v_t) = (1,1)$; i.e.,
$$ d G_t = d t + d W_t. $$
Now, using the integration-by-parts formula,
\begin{align*}
d Y_t &= G_t \, dH_t + H_t \, dG_t + dG_t \, dH_t \\
 &= -G_t H_t d W_t + H_t\left(d t + d W_t \right) - \left(d t + d W_t \right)\left(H_t d W_t \right)  \\
 &= 0 + (H_t - G_t H_t)d W_t \\
 &= (1 - t - W_t)H_t d W_t.
\end{align*}
Thus,
\begin{align*}
Y_t &= \int_{0}^t (1 - s - W_s)H_s d W_s.
\end{align*}
So, if we put $Q_t := (1 - t - W_t)H_t$, for each $t \ge 0$, then it clear that the process $Q := \left\{Q_t, \mathcal{F}_t ; 0 \le t < \infty \right\}$ is measurable and $\left\{\mathcal{F}_t \right\}$-adapted. Now, we just need to show that
\begin{align*}
\mathbb{E}\left[\left(\int_{0}^T Q_s^2 \, ds\right)^{1/2} \right] &= \mathbb{E}\left[\left(\int_{0}^T (1 - s - W_s)^2H_s^2 d s \right)^{1/2}\right] \\
&= \mathbb{E}\left[\left(\int_0^T (s^2H_s^2 + 2sW_sH_s^2 - 2sH_s^2 + W_s^2H_s^2 - 2W_sH_s^2 + H_s^2) ds \right)^{1/2}\right]
\end{align*}
is finite for all $T > 0$. Naturally, this is where I am stuck. Given the ugliness of this integrand, I can only I imagine I've done something wrong here (or that I am overlooking some "slicker" approach)...

答:
It is a consequence of Girsanov's theorem as per what was said earlier. If you define a probability measure $\mathbb{Q}$ such that :
$$\mathbb{dQ}_{| F_t} = e^{-W_t - \frac{1}{2}t}\mathbb{dP}_{| F_t}$$

Then it follows from Girsanov's theorem that the process $(B_t)_{t>0}$ defined as :
$$B_t = W_t +t$$

is a Brownian motion (hence a martingale) under $\mathbb{Q}$. You can check that this also implies that $e^{-W_t - \frac{1}{2}t}(W_t+t)$ is a martingale under $\mathbb{P}$ using Bayes' conditional expectation formula as follows :
$$\mathbb{E}[e^{-W_t - \frac{1}{2}t}(W_t+t)|F_s] = e^{-W_s - \frac{1}{2}s}\mathbb{E_{Q}}[W_t+t|F_s] = e^{-W_s - \frac{1}{2}s}(W_s+s)$$


答: Hint:  Use the limit comparison test.</p>

Title: Test for convergence for improper integral $1/x^x$

Body:
I am having trouble determining if this is convergence or divergence

$$\int^1_0 1/x^x dx$$

Body: Hint:  Use the limit comparison test.


问:
Show the limit of the sequence is $\sqrt{\alpha}$
Given: $\sqrt{\alpha}<x_1, \alpha >0, x_{n+1}=\frac{1}{2}\left(x_n+\frac{\alpha}{x_n}\right)$. Show $\lim_{n\to \infty}x_n=\sqrt{\alpha}.$

I have already shown the this is a monotonic decreasing function. Intuitively, I see in the end I get $\frac{1}{2}\left(\sqrt{\alpha}+\sqrt{\alpha}\right)$ but I'm not sure if I should keep going using $\lim$ notation or if I should switch to the definition of converging sequence.

答:
Since this hasn't been closed as duplicate,
here's my answer again.

A more direct proof.

$s_{n+1}
=\dfrac12(s_n+\dfrac{a}{s_n})
$.

$\begin{array}\\
s_{n+1}^2
&=\dfrac14(a_n^2+2a+\dfrac{a^2}{s_n^2})\\
&=a+\dfrac14(s_n^2-2a+\dfrac{a^2}{s_n^2})\\
&=a+\dfrac14(s_n-\dfrac{a}{s_n})^2\\
&\gt a\\
\end{array}
$

so
$s_n > \sqrt{a}
$.

$\begin{array}\\
s_{n+1}-\sqrt{a}
&=\dfrac12(s_n-2\sqrt{a}+\dfrac{a}{s_n})\\
&=\dfrac12(\sqrt{s_n}-\dfrac{\sqrt{a}}{\sqrt{s_n}})^2\\
&=\dfrac1{2s_n}(s_n-\sqrt{a})^2\\
\dfrac{s_{n+1}-\sqrt{a}}{s_n-\sqrt{a}}
&=\dfrac1{2s_n}(s_n-\sqrt{a})\\
&\lt \dfrac12\\
\end{array}
$

so $s_n$ converges to
$\sqrt{a}$.


问:
Solving Recursive Equations: How to transform the domain in such cases?
I understand that a widely-used recursive equation for the Binary Search is as follows:
$$
\begin{align}
T(1) &= 1\\
T(n) &= T(\tfrac{n}{2}) + 1, \quad n>1
\end{align}
$$

In order to solve the recursive equation we would simply transform the domain by setting $n = 2^k$, write the transformed equation, go through the telescoping and all that good stuff...

Anyway, my professor said that sometimes there may be an odd number of elements in a sorted list, in which case the equation would look like:
$$
\begin{align}
T(1) &= 1\\
T(n) &= T(\tfrac{n-1}{2}) + 1,\quad n>1
\end{align}
$$

Then I wondered, how would I perform a domain transformation on such an equation?

Thank you very much!

答:
In the case that $n$ is an odd number, we can just use $n=2k+1$. You'll end up with
$$T(2k+1)=T\left(\frac{2k+1-1}{2}\right)+1$$
I'll let you take care of the rest.


问:
Cardinality of the set of points on a circle whose coordinates are constructable is $\aleph_0$
Let $C=${$(x,y)\in\mathbb R^{2}|(x-a)^{2}+(y-b)^{2}=r^{2}$} be a circle of radius $r>1$ and centre $(a,b)$ where $a,b,r$ are all constructible. We want to show that the cardinality of the set of points on $C$ whose coordinates are constructible is $\aleph_0$
I do not have a good idea on how to approach this question. Is it enough that I show any coordinate with constructable numbers on the 2d plane is $\aleph_0$? Or should I show a bijection between the natural numbers and the set of points on C who's coordinates are constructable? If so, what bijection should I use? And I also do not know what is the use of the information that $a,b,r$ are constructable and $r<1$

答: I would follow the first approach, to show that the set of constructible points is $\aleph_0$.  At any stage in a construction, there are only finitely many points that can be constructed.  The number of stages of a construction is also finite.  The set of constructible points is then a countable union of finite sets.


答:
$a^2+b^2+c^2-2a-2b-2c+3[=(a-1)^2+(b-1)^2+(c-1)^2]\geq 0.$
Also $2a^2+2b^2+2c^2- 2ab-2bc-2ca\geq 0.$ Adding the two expressions give the desired result.

Title: Proving an inequality involving three real numbers

Body: For any numbers $a,b,c \in \mathbb{R}$, prove that $\frac{3}{2} (a^2+b^2+c^2+1) \geq a(b+1) +b(c+1)+ c(a+1)$.


答:
By the Cauchy-Schwarz inequality twice for the right hand side:
$a(b+1)+b(c+1)+c(a+1)\le\sqrt{a^2+b^2+c^2}\cdot \sqrt{(b+1)^2+(c+1)^2+(a+1)^2}\le \sqrt{a^2+b^2+c^2}\cdot \sqrt{2(b^2+1)+2(c^2+1)+2(a^2+1)}= \sqrt{x}\cdot \sqrt{2x+6}, x = a^2+b^2+c^2.$.

Next we prove: $\sqrt{x(2x+6)} \le \dfrac{3(x+1)}{2}\iff 2x^2+6x \le \dfrac{9}{4}\cdot (x^2+2x+1)\iff 8x^2+24x \le 9x^2+18x+9\iff 0 \le x^2-6x+9 \iff 0 \le (x-3)^2$, which is clearly true.

Equality occurs when $a = b = c = 1$.


问:
Ideal spline type for software optimized well-defined interpolation over time domain
I'm working on a software program which currently utilizes cubic Bézier splines for generating continuous well-defined function output (only 1 output value per input) in the time domain with varying sample rates. Initially I utilized some root finding code using Cardano's method, in order to convert time domain (X axis) positions into Bezier curve t position values [0, 1]. However, this is proving to have been a poor choice of algorithms for my purpose due to numeric instability. Also the current solution I am using utilizes trigonometry functions, which can be costly on many computer platforms, especially embedded ones.

I am now looking into other algorithms, but it also got me thinking as to whether cubic Bézier curves are even the optimal solution, when the application is not constrained to any particular spline type.

Ideal spline solution properties:

Intuitive user interface control of curve segments

Very stable "well defined" interpolation sampled output

Software optimization friendly floating point algorithm for CPU and possibly FPGA or DSP embedded solutions

Cubic Bézier curve splines and their 2 non-intersecting control points are familiar in many software programs for animation and illustration. So ideally I would continue to use this type of spline (at least on the front end) and choose the optimal solution algorithm which meets these requirements. One solution I am currently working with utilizes the Newton–Raphson method in combination with a binary search to find a t [0, 1] value given an X position, such as with this project: https://github.com/gre/bezier-easing
However, this was designed for rendering curves to screens and so I'm not yet sure if it will meet my numeric stability requirements. It does seem that the algorithm utilizes fairly basic math operations though, but if there is a better spline type to use in this type of situation, without having to go through this iterative approach of finding the t value for a given X, that may be a better option. Any helpful pointers would be greatly appreciated.

答:
I ended up using the solution described at the end of my question, namely the Newton–Raphson method in combination with a binary search to find a t [0, 1] value given an X position. Code for this can be found with this project: https://github.com/gre/bezier-easing

I was also able to optimize it to the point where it is comparable in performance to the original Cardano method I was using, which was much more complex, difficult to work with, and quite numerically unstable.  The performance/accuracy trade off the Newton-Raphson method can easily be tuned as well, just by adding more iterations to the loops. One downside is that it is not as conducive to SIMD optimizations, because each iteration depends on results from the previous one.


问:
Transformation law for metric tensor
For my topology class I have to calculate the transformation law of the metric tensor $g = g_{ij} \, dx^i \otimes dx^j$ under a coordinate transformation $x \longmapsto y=(x)$.

My approach:

$$ g(y) \enspace = \enspace g_{ij}(y) \, \big( dy^i \otimes dy^j \big) \enspace = \enspace g_{ij}(y) \, \bigg( \frac{\partial y^i}{\partial x^k} dx^k \otimes \frac{\partial y^j}{\partial x^{\ell}} dx^{\ell} \bigg) $$

$$= \enspace g_{ij}(y) \, \frac{\partial y^i}{\partial x^k} \, \frac{\partial y^j}{\partial x^{\ell}} \, \big(  dx^k \otimes dx^{\ell}  \big) \quad .$$

${}$

$\large \textbf{1.)} \enspace$ How do I know how the components $g_{ij}(y)$ transform?

${}$

$\large \textbf{2.)} \enspace$ If I identify the terms $\dfrac{\partial y^i}{\partial x^k}$ with the Jacobian $J^i{}_j$, then I get

$$ g(y) \enspace = \enspace g_{ij}(y) \, J^i{}_k \, J^j{}_{\ell} \, \big(  dx^k \otimes dx^{\ell}  \big) \quad .$$

Is it then true that:

$$ g(y) \enspace = \enspace \big( \boldsymbol{\operatorname{J}}^T \cdot \boldsymbol{\operatorname{g}}(y) \cdot \boldsymbol{\operatorname{J}} \big)_{k\ell} \, \big(  dx^k \otimes dx^{\ell}  \big) \quad ? $$

${}$

$\large \textbf{3.)} \enspace$ If I were to calculate the determinant of $g(y)$, how would I handle the tensor product in a mathematical rigorous way? Assuming that the equation in 2.) is true, would it just be

$$ \det g(y) \enspace = \enspace \det \boldsymbol{\operatorname{J}}^T \cdot \det \boldsymbol{\operatorname{g}}(y) \cdot \det \boldsymbol{\operatorname{J}} \quad ? $$

答:
Keep in mind your goal is to write $g(y) = g(y(x)) = g_{kl}'(x)(dx^k\otimes dx^l)$ and find this $g_{kl}'$. So you're basically done with (1). All that's left to do is read off the coefficients and conclude the transformation law:
$$g_{kl}'(x) = g_{ij}(y(x))\frac{\partial y^i}{\partial x^k}\frac{\partial y^j}{\partial x^l}.$$

(2) Yes, if you put $J^i_{\ k} = \partial y^i/\partial x^k$, it is a basic exercise in indices to check that $g' = J^TgJ$. It might help to consider the following fact about $n\times n$ matrices: $(A^TB)_{ij} = (A^T)_{ik}B_{kj} = A_{ki}B_{kj}$.

With that in hand, we have
$$g_{ij} J^i_{\ k}J^j_{\ l} = J^i_{\ k}g_{ij}J^j_{\ l} = (J^Tg)_{kj}J^j_{\ l} = (J^TgJ)_{kl}.$$
Thus $g_{kl}' = (J^TgJ)_{kl}$, as claimed.

(3) You don't have to worry about tensor product, because when people say $\det(g)$ they're referring to the determinant of the matrix $g_{ij}$. Thus the above calculations show
$$\det(g') = \det(J^TgJ) = \det(J)^2\det(g),$$
where we have used the fact $\det(J^T) = \det(J)$.

Incidentally, this proves the change of variables formula for integration on Riemannian manifolds. The volume form on an $n$-manifold is $\omega = \sqrt{|\det(g)|}\, dx_1\wedge\dotsm\wedge dx_n$. Given a diffeomorphism $x\mapsto y(x)$, the volume form becomes $\omega' = |\det(J)|\sqrt{|\det(g)|}\,dx_1\wedge\dotsm\wedge dx_n$, i.e., we scale by the absolute value of the Jacobian-determinant.

Title: Transformation law for metric tensor


答: $T'(0) = w/20$, so it's not a contraction if $|w| > 20$.  It does seem to be a contraction if $|w| \le 20$.

Title: Is this a contraction mapping?

Body: Is $T(x)=\frac{1}{20}e^{-x^2}\sin(wx)$ a contraction mapping? $w$ can be any real number.


问:
I have issue in the odd number part
Query on a Solution to problem:
If a is odd, then gcd(3a,3a+2)=1.
I tried to put a=2k+1 to be an odd number. But it does not solved.</p>

答:
By familiar properties of $\gcd$,
$$\gcd(3a+2, 3a) = \gcd((3a + 2) - 3a, 3a) = \gcd(2, 3a)  \text{.}  $$
Since $a$ and $3$ are odd, $3a$ is odd.  $2$ is prime, so $\gcd(2, 3a) > 1$ only if $2$ divides $3a$, but $2$ divides no odd number.  Therefore, $\gcd(2,3a) = 1$.

Title: I have issue in the odd number part


答:
The crux is to know $\gcd(a,b)=\gcd(a,b-a)=\gcd(b,b-a)=\gcd(b,a-b)$, or more generally, the Euclidean algorithm: $\gcd(a,b)=\gcd(b,a\bmod b)$

So, we have $$\begin{align}\gcd(3a,3a+2)&=\gcd(3a,3a+2-3a)\\&=\gcd(3a,2)\\&=\gcd(2,3a)=\gcd(2,3a-2a)=\gcd(2,a)\end{align}$$

which equals 2 when $a$ is even and 1 when $a$ is odd.

Title: I have issue in the odd number part

Body:
Query on a Solution to problem:
If a is odd, then gcd(3a,3a+2)=1.
I tried to put a=2k+1 to be an odd number. But it does not solved.</p>


答:
Using the Euclidean Algorithm we have the following chain of pairs:
\begin{gather*}
(3a+2,3a)\\
(3a,2) \\
(2,1).
\end{gather*}

This implies that $\gcd(3a+2,3a) = 1$.

Title: I have issue in the odd number part


问:
How to prove that $\sum\pi_i = \sum\frac{1}{E_iT_i} = 1$ in an irreducible Markov chain with stationary distribution $\pi$?
In Durrett's book Chapter 5 Theorem 4.6, It said that if p is irreducible and has stationary distribution, then $\pi_i=\frac{1}{E_iT_i}$. Where $T_i$ is the first time the markov process returns back to state i. $E_iT_i$ is the expected number of times needed to return back to state i given the initial state is $i$.

Also in the proof of theorem 4.7 said, if p is ireducible and i is positive recurrent, then, $\pi_j=\frac{\sum_{n=0}^{\infty}P_i(X_n=j, T_i \gt n)}{E_iT_i}$ is a stationary distribution.

Does that mean $\frac{\sum_{n=0}^{\infty}P_i(X_n=j, T_i \gt n)}{E_iT_i} = \frac{1}{E_jT_j}$? How can we prove that?

And also how can we prove that $\sum\frac{1}{E_jT_j} = 1$ in an irreducible Markov chain with stationary distribution $\pi$?

答:
I've found the answer myself, it's actually quite easy to prove:

In Durret's book, it's been proved that if $P$ is irreducible and recurrent, then $\mu_a$ is unique up to a constant multiple, and in its prove of this theorem, we found that the constant is $\nu(a)$, i.e. $\nu(x)=\nu(a)\mu_a(x)$.

That means $\mu_i(j)\mu_j(k)=\mu_i(k)$, here $\mu_i(j)$ is the constant.

If we sum both sides over $k$,

$$
\sum_{k}\mu_i(j)\mu_j(k)=\sum_{k}\mu_i(k)
$$

We get

$$
\mu_i(j)E_j(T_j)=E_i(T_i)
$$

We are done. We get

$$
\frac{\mu_i(j)}{E_i(T_i)} = \frac{1}{E_j(T_j)}
$$

I explained it in my video today: https://youtu.be/lZrzjxQ3YF8


问:
Smooth maps on a manifold lie group
$$
\operatorname{GL}_n(\mathbb R) = \{ A \in M_{n\times n} | \det A \ne 0 \} \\
\begin{align}
&n = 1, \operatorname{GL}_n(\mathbb R) = \mathbb R - \{0\} \\
&n = 2, \operatorname{GL}_n(\mathbb R) = \left\{\begin{bmatrix}a&b\\c&d\end{bmatrix}\Bigg| ad-bc \ne 0\right\}
\end{align}
$$

$(\operatorname{GL}_n(\mathbb R),\cdot)$ is a group.

$AB$ is invertible if $A$ and $B$ are invertible.

$A(BC)=(AB)C$

$I=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

$A^{-1}$ is invertible if $A$ is invertible.

$$(\operatorname{GL}_n(\mathbb R) := \det{}^{-1}(\{0\}))$$
$\det{}^{-1}(\{0\})$ is open in $M_{n\times n}(\mathbb R)$.

$\det : M_{n\times n}(\mathbb R) \to \mathbb R$ is continuous, why?

$\dim \operatorname{GL}_n(\mathbb R) = n^2 - 1$, why?

$(\operatorname{GL}_n(\mathbb R),\cdot)$ is a Lie group if:

$$
\mu : G\times G \to G \\
\mu(A,B) = A\cdot B \:\text{ is smooth} \\
I(A) = A^{-1} \:\text{ is smooth}
$$

How can I show this?

I want to show that general special linear group is Lie group. I could not show the my last step. How can I show that $AB$ and $A^{-1}$ are smooth? please help me I want to learn this. Thanks

Here the my handwritten notes http://i.stack.imgur.com/tkoMy.jpg

答:
$\det:M_n(\mathbb R)\to\mathbb R$ is continuous, since it is a polynomial map, if you regard it as a map $\det:\mathbb R^n\to\mathbb R$:

$$\det A=\sum\limits_{i+j=n}(-1)^{i+j}a_{ij}A_{ij}$$

where $A=(a_{ij}),$ and $A_{ij}$ is the determinant of the $(n-1)\times(n-1)$ matrix if we remove row $i$ and column $j$ from $A$.

Now, $\dim GL_n(\mathbb R)=\dim M_n(\mathbb R)=\dim \mathbb R^n=n^2$.

This is because
$GL_n(\mathbb R)=\det ^{-1}(\mathbb R\setminus \{0\})$ is an open subset of $M_n(\mathbb R)$, since $\det$ is continuous and $\mathbb R\setminus \{0\}$ is open. Also, it is well-known that an open subset of an $n-$manifold is an $n-$manifold.

Regarding the multiplication:

$$A\cdot B=(c_{ij}), \text{ where } c_{ij}=\sum\limits_{k=1}^na_{ik}b_{kj}$$
It is obvious that matrix multiplication is a vector function, with polynomial coordinate functions , therefore it is a smooth function.

Finally, consider the map: $I:GL_n(\mathbb R)\to GL_n(\mathbb R)$ such that $I(A)=A^{-1}$. Then:
$$I(A)=A^{-1}=\dfrac{1}{\det A}\mathrm {Adj}(A)$$
where $\det A$ is polynomial, and $\mathrm {Adj}(A) $, the adjoint of A,  is easily seen as a vector function with polynomial coordinate functions. Therefore, $I$ is a vector function with rational coordinate functions, which implies that it is smooth.

Title: Smooth maps on a manifold lie group


问:
An example of a differentiable function but not Lipschitz continuous.
I want to find a example of a differentiable function that is not Lipschitz, and I think if a function has no bounded derivative, it is not Lipschitz continuous.

Let $f(x)=\log x$ on $(0,\infty)$. By mean value theorem, $$\begin{align}\left| f(x)-f(0)\right|=\frac{1}{c}|x|  \text{ for some $c$ between $x$ and $0$} .\end{align}$$
Let $x\to 0$, then $f$ is not Lipschitz continuous.
Am I correct?

答: As Clement said on his comment, your function is not defined on $\mathbb{R}$. Take a look at $f(x) = x^2$ for example. Can you see why this function is differentiable but not Lipschitz continuous? Take $x$ sufficiently large, and the derivative will pass any bound you like.


答: Try $f(x) = x^2 \sin { 1\over x^2}$ for $x \neq 0$ and $f(0)=0.$ This is differentiable everywhere but the derivative is unbounded on any neighbourhood of $0$.

Title: An example of a differentiable function but not Lipschitz continuous.

Body:
I want to find a example of a differentiable function that is not Lipschitz, and I think if a function has no bounded derivative, it is not Lipschitz continuous.

Let $f(x)=\log x$ on $(0,\infty)$. By mean value theorem, $$\begin{align}\left| f(x)-f(0)\right|=\frac{1}{c}|x|  \text{ for some $c$ between $x$ and $0$} .\end{align}$$
Let $x\to 0$, then $f$ is not Lipschitz continuous.
Am I correct?


问:
About substitution In integrals
In integrals we were taught substitution, but we were not said an intuitive way of thinking about it.

In some questions we substitute x=tan(¢) or something like that but how can we know they r equal. They are both varying how can we say the rate of change of the LHS equals the RHS how can a linear function equals this whole mess of infinite series. I know a bit about Taylor series but don't know why do we substitute some wierd things in an integral. Can someone help me out.

答:
The idea of a substitution is to make the resulting integral easier than the one before. A classic one for trigonometric substitution you mention is
$$
\int \frac{dx}{1+x^2},
$$
which is not at all clear how to find the anti-derivative for. The substitution $x = \tan t$ makes it easier since $1 + x^2 = 1 + \tan^2 t = \sec^2 t$ and $dx = \sec^2 t dt$, so
$$
\int \frac{dx}{1+x^2}
 = \int \frac{\sec^2 t\ dt}{1+\tan^2 t}
 = \int \frac{\sec^2 t\ dt}{\sec^2 t}
 = \int dt
 = t + C
 = \arctan x + C,
$$
since the resulting integral became trivial.

The intuitive reason you can do this is that substitution for integrals is like the chain rule for derivatives. In other words, Chain Rule says
$$
\frac{d}{dx} [f(g(x))] = f'(g(x)) g'(x),
$$
or equivalently in integral form,
$$
\int d[f(g(x)] = \int f'(g(x))g'(x)dx
$$
and the analog of what we do is substitute $u = g(x), du = g'(x) dx$ into the RHS to get
$$
\int d[f(g(x)] = \int f'(g(x))g'(x)dx = \int f'(u) du = \int d[f(u)].
$$

So as you can see, substitution is just an intuitive way of looking the implications of Chain Rule...

Title: About substitution In integrals


问:
Can I simply convert an integral into an expectation by "declaring" that the variable is a random variable?
I know that

$\int_{a}^{b}g(x)dx = (b-a)\int_{a}^{b}g(x)\frac{1}{b-a}dx = (b-a)E[g(X)]$

where $X$ has a $Uniform(a,b)$ distribution.

However, I don't understand how the second step turns the integral into an expected value.  It's as if I can simply "declare" that $X$ is a random variable.  Somehow, it seems like I'm hand-waving some magic into the equation, and I wonder if I am missing a step.

答:
As mentioned in the comments, this works because it follows from the definition of expectation of a continuous random variable.

Consider a continuous random variable $X$ defined on some interval $I$ with some distribution with pdf $f_X(x)$. Then,

$$\Bbb E[g(X)]:=\int_I g(x)f_X(x)~\mathrm dx$$

The question here corresponds to $X\sim\mathrm{Unif}(a,b)$ for which $f_X(x)=\frac 1{b-a}$ for $x\in [a,b]$ and $0$ otherwise, so we have,

$$\Bbb E[X]:=\int_a^b g(x)\frac 1{b-a}~\mathrm dx\implies (b-a)\Bbb E[X]=\int_a^b g(x)~\mathrm dx$$


问:
Question on sylow application
Without using the generalized cayley theorem prove there is no simple group $G$, of order $112$.

Answer: If we can find a pair of distinct  sylow $2-$ subgroups $A$ and $B$ such that $|A\cap B|=8$, then $N(A \cap B)\geq AB$ so that $N(A \cap B)=G$.

Why does this first statement hold here $?$ Why does $A,B$ being sylow $2$ subgroups such that $|A \cap B|=8$ imply $N(A \cap B)$ contains the subgroup $AB$? $N$ is the normalizer of subgroup in $G$.

答:
Under the assumption that $A$ and $B$ are Sylow-2 subgroups and their intersection has order $8$, then $AB$ cannot be a subgroup: its size is
$$|AB| = \frac{|A||B|}{|A\cap B|} = \frac{(16)(16)}{8} = 32,$$
which does not divide $|G|$.

But in fact, the statement does not claim that $AB$ is a subgroup, and you do not need that. The normalizer contains both $A$ and $B$, because $A\cap B$ is normal in each (being of index $2$). And since the normalizer contains both $A$ and $B$, it contains the subset $AB = \{ab\mid a\in A, b\in B\}$. That means that the normalizer contains at least $32$ elements, which in turn implies it must be all of $G$, since its order must divide $|G|$.

Title: Question on sylow application


问:
Using existential instantiation (logic)
Am I using EI right on line 6? (Actually, I'm pretty sure the answer is 'no', and there's a few sketchy lines after that, too. So maybe you could also give a hint about how to do this).

Prove:

(∃x)(∀y)(Gxy → Hxy)     (Premise)

(∀x)(∃y)￢Hxy        (Premise)

```<code>    ∴ ￢(∀x)(∀y)Gxy
</code>```

(∀y)(Gay → Hay)     (1, Existential Instantiation)

Gab → Hab           (3, Universal Instantiation)

(∃y)￢Hay            (2, Universal Instantiation)

￢Hab                (5, Existential Instantiation)

￢Gab                (4, 6, Modus Tolens)

(∃y)￢Gay            (7, Existential Generalization)

￢(∀y)Gay            (8, Quantifier Negation)

(∃x)￢(∀y)Gxy        (9, Existential Generalization)

￢(∀x)(∀y)Gxy        (10, Quantifier Negation)

答:
The below is verbatim from a comment by Zhen Lin, which was requested by the OP to be posted as an answer, but that wasn't done.

The primary intent of this answer is to remove this question from the Unanswered queue.

Zhen Lin wrote:

(6) is no good, because you introduced b in (4). But you have essentially the right idea: just put (4) after (6)!

Title: Using existential instantiation (logic)


问:
Logistic differential population equations
Given a population growth $P$ milligrams at time $t$ hours, such that $\frac{dP}{dt} = 0.005P(120-P)$ If the initial population is $10$mg find $P$ as a function of $t$ in order to express how many hours it will take for the population to reach $110$mg. State the value of $lim P(t)$ as $t→∞$

Now, I believe I can arrange $P$ as a function of $t$ by separating variables

$\frac{dP}{P(120-P)} = 0.005dt$

$\frac{1}{P(120-P)} = \frac{A}{P} + \frac{B}{120-P}$

$1 = A(120-P) + BP = 120A + (B-A)P$   when $B-A = 0$, $120A = 1$ so $A = B = \frac{1}{120}$

$\frac{1}{P(120-P)} = \frac{1}{120}[\frac{1}{P} + \frac{1}{120-P}]$

$∫\frac{dP}{P(120-P)} = \frac{1}{120}∫[\frac{1}{P} + \frac{1}{120-P}]dP = ∫0.005dt$

$\frac{1}{120}[ln|P|-ln|120-P|] = 0.005t + c$

$ln|120-P| - ln|P| = -0.6t - 120c$

using log laws $ln|\frac{120-P}{P}| = -0.6t - 120c$

$|\frac{120-P}{P}| = e^{(-0.6t-120c)} = e^{-0.6t}e^{-120c}$

$\frac{120-P}{P} = Ae^{-0.6t}$ where $A = e^{-120c}$

$\frac{120-P}{P} = \frac{120}{P} - 1 = Ae^{-0.6t}$

$\frac{120-P}{P} = 1 + Ae^{-0.6t}$

$P = \frac{120}{1 + Ae^{-0.6t}}$

For the initial condition, $A = \frac{120-10}{10} = \frac{110}{10} = 11$

$P = \frac{120}{1 + 11e^{-0.6t}}$

Is my working out correct? If so, my assumption is that to find when the population is $110$ we find $t$ when

$110 = \frac{120}{1 + 11e^{-0.6t}}$

As for the limit, I'm not totally sure how to calculate that. I assumed that $t→∞$ would be when $P = 120$ but there must be some method that can calculate it correctly? Thanks.

答:
All your work is correct.

When you arrive at $$P = \frac{120}{1 + 11e^{-0.6t}}$$ when $t\to \infty$, $e^{-0.6t}\to 0^+$ and $P_\infty=120^-=120$.

What you missed is the solution of $t$ such that
$$110 = \frac{120}{1 + 11e^{-0.6t}}\implies 11e^{-0.6t}=\frac 1{11}\implies e^{-0.6t}=\frac 1{121}$$ that is to say
$$e^{0.6t}=121\implies 0.6t=\log(121)\implies t=\frac{\log(121)}{0.6} \approx 8$$


问:
Fourier Transform - Dirac Delta
How can I compute this Fourier transform?

$$\int \frac{d^3q}{(2\pi)^3}(\vec q \cdot \vec a)(\vec q \cdot \vec b)\frac{1}{q^2}\,e^{i \vec q \cdot \vec r }\,$$

My idea was to write it as

$$-( \vec a \cdot \vec \nabla)(\vec b \cdot \vec \nabla)\int \frac{d^3q}{(2\pi)^3}\frac{1}{q^2}\,e^{i \vec q \cdot \vec r }$$

and use the fact that $\int \frac{d^3q}{(2\pi)^3}\frac{1}{q^2}\,e^{i \vec q \cdot \vec r }=\frac{1}{4\pi r}$, together with $\Delta \frac{1}{r} = -4\,\pi \,\delta(\vec r)$ but is is not exactly the same.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
In order to deal with possible issues related to integrals convergence, I'll include the $\ds{\underline{\it parameter}\,\,\, \beta}$. Its relevance is discussed at the very end.
\begin{align}
&\bbox[5px,#ffd]{{1 \over \pars{2\pi}^{3}}
\iiint_{\mathbb{R}^{3}}\pars{\vec{q}\cdot\vec{a}}
\pars{\vec{q}\cdot\vec{b}}
{\expo{\ic\vec{q}\cdot\vec{r}} \over
q^{2 + \beta}\,}\,\,\dd^{3}\vec{q}}
\\[5mm] = &\
{1 \over \pars{2\pi}^{3}}\,\vec{a}\cdot\bracks{%
\iiint_{\mathbb{R}^{3}}\vec{q}\,\vec{q}\,
{\expo{\ic\vec{q}\cdot\vec{r}} \over q^{2}}{\dd^{3}\vec{q} \over q^{\beta}}}\cdot\vec{b}
\\[5mm] = &\
{1 \over \pars{2\pi}^{3}}\,\vec{a}\cdot\bracks{%
\int_{0}^{\infty}\pars{%
\int_{\Omega_{\vec{q}}}\hat{q}\,\hat{q}\,
\expo{\ic\vec{q}\cdot\vec{r}}\,\dd\Omega_{\vec{q}}}
q^{2 - \beta}\,\,\dd q}\cdot\vec{b}\label{1}\tag{1}
\end{align}
Lets perform the angular integration where, for simplicity, $\ds{\vec{r}}$ is chosen along the $\ds{\hat{z}}$-axis. Later on, the $\ds{\vec{r}}$-general direction can be suitable restored:
\begin{align}
&\int_{\Omega_{\vec{q}}}\hat{q}\,\hat{q}\,
\expo{\ic\vec{q}\cdot\vec{r}}\,\dd\Omega_{\vec{q}} =
\int_{0}^{2\pi}\int_{0}^{\pi}
\hat{q}\,\hat{q}\expo{\ic qr\cos\pars{\theta}}\,
\sin\pars{\theta}\,\dd\theta\,\dd\phi
\\[5mm] = &\
\int_{0}^{\pi}
\bracks{\sin^{2}\pars{\theta}\,\pi\,\hat{x}\,\hat{x} +
\sin^{2}\pars{\theta}\,\pi\,\hat{y}\,\hat{y} +
\cos^{2}\pars{\theta}\pars{2\pi}\hat{z}\,\hat{z}}\ \times
\\ &\ \phantom{\,\,\int_{0}^{\pi}}
\,\,\,\expo{\ic qr\cos\pars{\theta}}\,\,\sin\pars{\theta}
\,\dd\theta
\\ = &\
\pi\pars{\hat{x}\,\hat{x} + \hat{y}\,\hat{y}}
\int_{-1}^{1}\pars{1 - \xi^{2}}\expo{\ic qr\xi}\,\dd\xi +
2\pi\,\hat{z}\,\hat{z}
\int_{-1}^{1}\xi^{2}\expo{\ic qr\xi}\,\dd\xi
\\ = &\
4\pi\pars{\hat{x}\,\hat{x} + \hat{y}\,\hat{y}}
\on{f}\pars{qr} +
4\pi\,\hat{z}\,\hat{z}\,\on{g}\pars{qr}
\\[5mm] &\ \mbox{where}\quad
\left\{\begin{array}{rcl}
\ds{\on{f}\pars{\xi}} & \ds{\equiv} &
\ds{\sin\pars{\xi} - \xi\cos\pars{\xi} \over \xi^{3}}
\\[2mm]
\ds{\on{g}\pars{\xi}} & \ds{\equiv} &
\ds{\pars{\xi^{2} - 2}\sin\pars{\xi} + 2\xi\cos\pars{\xi} \over \xi^{3}}
\end{array}\right.
\end{align}

(\ref{1}) becomes
\begin{align}
&\bbox[5px,#ffd]{{1 \over \pars{2\pi}^{3}}
\iiint_{\mathbb{R}^{3}}\pars{\vec{q}\cdot\vec{a}}
\pars{\vec{q}\cdot\vec{b}}
{\expo{\ic\vec{q}\cdot\vec{r}} \over q^{2 + \beta}}\,\dd^{3}\vec{q}}
\\[5mm] = &\
{1 \over 2\pi^{2}}\,{1 \over r^{3 - \beta}\,}
\,\,\vec{a}\,\cdot
\\[2mm] &\
\bracks{\pars{\hat{x}\,\hat{x} + \hat{y}\,\hat{y}}
\int_{0}^{\infty}\on{f}\pars{\xi}\,\xi^{2 - \beta}\,
\,\dd\xi +
\hat{z}\,\hat{z}
\int_{0}^{\infty}\on{g}\pars{\xi}\,\xi^{2 - \beta}\,
\,\dd\xi}
\cdot\vec{b}
\end{align}
The integrations are given by
\begin{align}
&\int_{0}^{\infty}\on{f}\pars{\xi}\,\xi^{2 - \beta}\,
\,\dd\xi =
{\Gamma\pars{2 - \beta}\sin\pars{\pi\beta/2} \over \beta}
\,,\quad\,\,\,\,\,\, 0 < \Re\pars{\beta} < 1
\\[5mm] &\
\int_{0}^{\infty}\on{g}\pars{\xi}\,\xi^{2 - \beta}\,
\,\dd\xi =
-\,{\Gamma\pars{3 - \beta}\sin\pars{\pi\beta/2} \over \beta}
\,,\quad 1 < \Re\pars{\beta} < 3
\end{align}
The first integral $\ds{\to {\pi \over 2}}$ as $\ds{\beta \to 0^{+}}$ while such a limit is not allowed for the second integral -unless we assume an analytical continuation- which we left as a discussion for the OP.

Title: Fourier Transform - Dirac Delta


问:
Find formula for $\sum_{k = 1}^{n}(k^2 \cdot x^{k - 1})$.
It is almost obvious that we have to use derivative in order to find the answer. However, I couldn't see this one to be derivative of something.

It would be amazing if you share your thoughts on this problem. Also, I would be thankful if you would explain what was intuition behind your solution. Thanks!

Body:
An old trick !
$$S=\sum_{k = 1}^{n}k^2 \, x^{k - 1}=\sum_{k = 1}^{n}\big[k(k-1)+k \big]\, x^{k - 1}$$ that is to say
$$S=\sum_{k = 1}^{n}k(k-1) \, x^{k - 1}+\sum_{k = 1}^{n}k \, x^{k - 1}$$
$$S=x\sum_{k = 1}^{n}k(k-1) \, x^{k - 2}+\sum_{k = 1}^{n}k \, x^{k - 1}$$
$$S=x \left(\sum_{k = 1}^{n}x^k \right)''+\left(\sum_{k = 1}^{n}x^k \right)'$$


问:
Etymology of 'connex' and 'transitive' for binary relations
There is a great question here about the etymology behind the term 'reflexive' for binary relations. How about
'connex' and 'transitive'? Where do these names come from?

I expect that transitivity is related to the idea of transitive verbs in grammar based on the etymology for reflexivity, but I can't see the relation exactly.  Connex I have no idea about, it looks kind of like convex?

答:
The mathematical meaning of connex is a new use of an obsolete English adjective meaning ‘connected’, for which the OED has citations from 1653, 1677, and 1685. It is derived from Latin connexus, classical Latin cōnexus, the past participle of connectō. The idea is that a connex order is ‘connected’ in the sense that no two elements are incomparable.

The mathematical sense of transitive goes back at least to De Morgan; the OED has a citation from 1856 for him. The word derives from Latin transeō ‘to pass over’; the idea is that the relationship makes a transition through the intermediate term.


问:
Plotting $||x|-|y||=1$: How to verify the extra line segments aren't included in the plot?
I have to plot the graph of $||x|-|y||=1$, I did the following: As $|x|=\max(x,-x)$ I've obtained the following equations:

$$|x|-|y|=1 \hspace{2cm} |y|-|x|=1$$

From there, I've obtained the following equations:

$$x-y=1 \hspace{1cm} x+y=1\hspace{1cm}-x-y=1\hspace{1cm}-x+y=1$$

The plot of these equations is the first here, and the plot of $||x|-|y||=1$ is the second.

$ \hspace{4mm}$

So I am not so far from the answer. My trouble is the following: How do I verify that the extra line segments in the first plot are not included? I tried to verifiy some values, for example: $x=-1/2,y=1/2$ and notices that for them $||x|-|y||<1$ so I tried to use the triangle inequality expecting to obtain $||x|-|y||\leq |x-y|<1$ for the extra line segments but I failed to do so.

答:
For each of the four equations you obtained, the signs of $x$ and $y$ must be considered. We have:

$$x-y=1 \text{ for } x,y \ge 0 \text { (satisfying $|x|-|y|=1$) or } x,y \le 0 \text { (satisfying $|y|-|x|=1$) }$$

Notice that the part of the line $x-y=1$ in $x \ge 0, y \le 0$ (Quadrant IV) is omitted.

Similarly:

$$x+y=1 \text{ for } x \ge 0, y \le 0 \text{ or } x \le 0, y\ge 0$$

$$-x-y=1 \text{ for } x \ge 0, y \le 0 \text{ or } x \le 0, y\ge 0$$

$$-x+y=1 \text{ for } x, y \ge 0 \text{ or } x , y \le 0$$

These extra constraints conveniently corresponds to the four quadrants, omitting Quadrants I, III, II respectively. Plotting these lines according to these constraints will avoid the extra lines in the 'undesired' quadrants.


问:
Closed sets in a given topology
I came across a topology on $ \mathbb Z \times \mathbb Z $ whose basis is defined as follows:

$ B(m,n) = \lbrace (m,n) \rbrace $ if both m and n are odd

$ B(m,n) = \lbrace (m+a, n) | a = -1, 0 ,1 \rbrace $ when m is even and n is odd

$ B(m,n) = \lbrace (m, n+a) | a = -1, 0 ,1 \rbrace $ when m is odd and n is even

$ B(m,n) = \lbrace (m+a, n+b) | a,b = -1, 0 ,1 \rbrace $ when both m and n are even

I was trying to determine the smallest possible closed sets given a point, as the problem requires me to. I showed that when both the coordinates are even, its just the singleton set which is closed.

But in the remaining cases, I think it has to be the entire set itself. However I am lacking a rigorous proof to establish my claim. Can anyone guide me??

答:
HINT: Consider the point $\langle 0,1\rangle$, say. A point $p\in\Bbb Z\times\Bbb Z$ is in the closure of $\{\langle0,1\rangle\}$ if and only if every open nbhd of $p$ contains $\langle 0,1\rangle$. This is the case if $p$ is $\langle 0,2\rangle$ or $\langle 0,0\rangle$, but that’s it. To see this, let $p=\langle m,n\rangle$. If $m<-1$ or $m>1$, $p\notin B(m,n)$, and similarly if $n<0$ or $n>2$. Thus, we need only consider the nine points in the set $\{-1,0,1\}\times\{0,1,2\}$, and in all cases except the two already noted, $p\notin B(m,n)$.

This argument easily generalizes to show that the closure of $\{\langle 2k,2\ell+1\rangle\}$ is the three-point set $\{\langle 2k,2\ell\rangle,\langle 2k,2\ell+1\rangle,\langle 2k,2\ell+2\rangle\}$ for any integers $k$ and $\ell$. That is, if $p$ is an even-odd point, so to speak, the only points that can’t be separated from it are the two adjacent to it vertically.

Now try to analyze the remaining cases in the same general fashion.

Title: Closed sets in a given topology


问:
Expressing $\sum_{n=0}^{\infty} P(n) x^n$ as a rational function
For $x \in (-1, 1)$, the power series $f(x) = \sum_{n=0}^{\infty} P(n) x^n$, where $P(n)$ is a polynomial, converge. Within this interval of convergence, how to write $f$ as a rational function $\frac{p(x)}{q(x)}$, where $p$ and $q$ are polynomials?

答:
Note that for $i \in \mathbb{Z}^+$, we have
$$
\frac{1}{(1 - x)^i} = \sum_{n = 0}^{\infty} \binom{n + i - 1}{n} x^{n} \, .
$$
Thus a finite linear combination of $\{ (1-x)^{-i} \}_i$ can be written as
$$
\sum_{i = 1}^{d} \frac{a_i}{(1-x)^{i}} = \sum_{n = 0}^{\infty} \sum_{i = 1}^{d} a_i \binom{n + i - 1}{n} x^{n} \, ,
$$
where $d - 1$ is the degree of $P(n)$.

As such, we only need to make the following expressions equivalent:
$$
\sum_{i = 1}^{d} a_i \binom{n + i - 1}{n} = P(n) \, .
$$
This can be done in quite a few ways, one of them being solving the following system of equations:
$$
\sum_{j=0}^{i+1} \binom{i}{j} a_{j+1} = P(-i) \, .
$$

For example, solving this with $P(n) = n^2 - 4$ gives $a_1 = -3, a_2 = -3, a_3 = 2$. I.e.,
$$
\sum_{n=0}^{\infty} (n^2 - 4)x^n = \frac{-3}{1-x} + \frac{-3}{(1-x)^2} + \frac{2}{(1-x)^3} = \frac{-3x^2 + 7x - 4}{(1-x)^3} \, .
$$


问:
Structure of a bijection on a set of bitstrings to itself
Let $S$ be the set of bitstrings of length $n$. Let $f:S\to S$ be a bijection defined as follows. Let $x\in S$, then $f(x)=y$ where $y$ is constructed from $x$ by

Adding all the $n$ bits of $x$ modulo $2$ and appending it to $x$

Popping the first bit of $x$

For example $f(1010)=0100$.

Let $W(x)$ be the Hamming weight of $x\in S$ i.e. number of ones in $x$. It is clear that $f$ maps $x$, such that $W(x)=w$, to bitstrings whose weights are $w,w+1$ or $w-1$. I am trying to find for a given $n$ and $w$, how many of the bitstrings $x$, s.t. $W(x)=w$, are mapped to bitstrings of weight $w,w+1,w-1$ under $f$?

Looking at a few $n$, it looks like for small $w$, the mappings to weight class $w+1$ are more than that of weight class $w-1$ and for $w=n/2$, the mappings to weight class $w+1$ is equal to weight class $w-1$. But I couldn't prove it precisely.

答:
Let $x\in S$, and let $x_1$ be the first bit of $x$. It’s not hard to verify that

$W(f(x))=W(x)$ iff $W(x)\equiv x_1\pmod2$;

$W(f(x))=W(x)+1$ iff $W(x)$ is odd and $x_1=0$; and

$W(f(x))=W(x)-1$ iff $W(x)$ is even and $x_1=1$.

Assume that $n\ge 2$. Then there are $2^{n-2}$ bit strings of length $n-1$ whose Hamming weight is even and $2^{n-2}$ whose Hamming weight is odd, so there are altogether $2^{n-1}$ $w\in S$ such that $W(w)\equiv w_1\pmod2$. In other words, half of the $2^n$ strings in $S$ satisfy $W(f(w))=W(w)$. Half of the remainder, or $2^{n-2}$ bit strings, have $W(w)$ odd and $w_1=0$, and the other half, or $2^{n-2}$ bit strings, have $W(w)$ even and $w_1=1$.

For strings $x$ of a given weight $w$ matters are more complicated. If $w=0$, then $f(x)=x$. If $w=n$, then $W(f(x))=n$ if $n$ is odd, and $W(f(x))=n-1$ if $n$ is even. Suppose that $0<w<n$.

There are $\binom{n-1}w$ bit strings $x$ of weight $w$ with $x_1=0$; for these we have $W(f(x))=w$ if $w$ is even, and $W(f(x))=w-1$ if $w$ is odd.

There are $\binom{n-1}{w-1}$ bit strings $x$ of weight $w$ with $x_1=1$; for these we have $W(f(x))=w$ if $w$ is odd, and $W(f(x))=w+1$ if $w$ is even.


问:
Can someone help solve this discrete maths question?

I have an attempt done there and this is the last question I have to do. I'm not sure how to solve
this one. Can someone help me or at least give me pointers on how I can solve this question as
I've been trying for an hour and have absolutely no idea? Thank you very much :)

答:
Here are some facts that can help you figure this out. First find the total probability
$$P(E)=\sum_{k=1}^n P(E\ \vert A_k)P(A_k)$$
Once you find $P(E)$, then you can use it to find the other three answers. Just apply the following three times and you're done.
$$P(A_k\vert E)=\frac{P(E\ \vert A_k)P(A_k)}{P(E)}$$
Let me know if you have any questions.


问:
if $\mathbf x$ is sampled randomly from a hypercube on $R^n$, what is the probability density for $|\mathbf x| = d$
if the vector $\mathbf x$ is sampled randomly from a uniform distribution on $[0, 1]^d$, what is the probability density function for $|\mathbf x|$? Is it easy to scale for $[0, n]^d$?</p>

答:
There are many different probability densities of continuous distributions on $[0,1]^d$.  There are also many probability distributions on that set that do not have densities, including, but not limited to, discrete distributions and mixtures of discrete and continuous distributions.  A point could be randomly sampled from any of those distributions.

There is a somewhat uninformed usage that is commonplace among mathematicians who are not probabilists, according to which "randomly" means "uniformly distributed".  It that is what is meant, then the density is equal to $1$ everywhere within that cube.  On $[0,n]^d$ the density of the uniform distribution is $n^{-d}$ at every point.

Later edit: As noted in the comments, my answer ends before getting to the hard part.  Maybe I'll add more later if no one beats me to it.

The conditional density given the order in which the coordinates appear when sorted, does not depend on which order it is.  Therefore, we can condition on the event that $x_1<x_2<x_3<\cdots<x_n$ and we get the same distribution.