问:
Consequences of maximum modulus principle
Q. Let $f$ be holomorphic and satisfy $|f(z)| \leq A e^{|z|^{\alpha}},$ for all Re $z \geq 0,$ and where $0<\alpha<1$. If $|f(z)| \leq M$ on the imaginary axis, and $f(1)=0,$ prove that, for $x>0$
$$
|f(x+i y)| \leq M\left\{\frac{(1-x)^{2}+y^{2}}{(1+x)^{2}+y^{2}}\right\}^{1 / 2}
$$

Here, how to apply maximum modulus principle on the imaginary axis as the boundary for right half plane and to prove $|\frac{1+z}{1-z}f(z)|\leq M$?

答:
$T(z) = \frac{z-1}{z+1}$ is a Möbius transformation which maps the right halfplane onto the unit disk. In particular, $|T(z)| =1$ on the imaginary axis.

For $|z| \ge 2$ is
$$
 |T(z)| \ge \frac{|z|-1}{|z|+1} = 1 - \frac{2}{|z|+1} \ge \frac 13 \, .
$$

$f(1) = 0$ so that $g(z)=f(z)/T(z)$ is holomorphic in the right halfplane and continuous on its boundary. $|g(z)| \le M$ on the imaginary axis, and
$$
 |g(z)| \le 3 A e^{|z|^{\alpha}}
$$
if $\operatorname{Re}(z) \ge 0$ and $|z| \ge 2$. Then $|g(z)| \le B e^{|z|^{\alpha}}$ for some constant $B > 0$ and all $z$ with $\operatorname{Re}(z) \ge 0$.

Now we can apply the Phragmén–Lindelöf principle for a sector in the complex plane to $g$ and conclude that $|g(z)| \le M$ everywhere in the right halfplane, i.e.
$$
 |f(z)| \le M \cdot \left| \frac{z-1}{z+1} \right|
$$


问:
Prove $x^4-6x^2+4$ is irreducible
As title says, I'm asked to prove that the polynomial $x^4-6x^2+4$ is irreducible in $\mathbb{Q}[x]$.
This is what I tried so far:

• I've noticed it has no roots in $\mathbb{Q}$, since if it had then they would be of the form $a/b$ where $a|4$ and $b|1$ (and none of them satisfy the polynomial).

• I've also tried considering the polynomial (actually its image via ring morphism from $\mathbb{Z}[x]$ to $\mathbb{Z_p}[x]$) in $\mathbb{Z_2}$,$\mathbb{Z_3}$,$\mathbb{Z_5}$,$\mathbb{Z_7}$ but it turns out it is reducible over all of these rings. Also wolframalpha told me it is reducible in many other $\mathbb{Z_p}$.

• Finally I tried applying Eisenstein's criterion to the shifted polynomial $p(x+1), p(x+2), p(x+3)$ but had no success.

Seems to me that I've used all the tools that I've got. But maybe there are some consideration about its roots in a splitting field that can be done in order to prove its irreducibility, so I'm asking for your help. Thank you very much indeed.

答:
This polynomial is trickier to work with, because Eisenstein and mod $p$ thinking are both not helpful, as you say.

Instead, one can easily compute the roots of the given polynomial. Just putting $y=x^2$ , and solve $y^2-6y+4 = 0$ i.e. $y = 3 \pm \sqrt 5$. Thus, we get that the four roots of $x$ are $\pm \sqrt{3 \pm \sqrt 5}$.

Now, none of these roots are rational, so the factor theorem tells you that there can't be a degree one factor. However, if there is a degree two factor, say $x^2+\alpha x + \beta$, then two of the above numbers are its roots, and in particular the sum and product of those two numbers must be rational (using Vieta's formulas).

Therefore, the question reduces to :

Are there two numbers among $\pm\sqrt{3\pm \sqrt 5}$ such that both their product and sum are rational?

The answer to that is a clear no : for the product to be rational, we must have the pair to be either $\sqrt{3 \pm \sqrt 5}$ or $- \sqrt{3 \pm \sqrt 5}$ (easy to see) , but in either case the sum of the two said numbers is irrational.

For more on proving irreducibility when you can find the roots of the polynomial explicitly, read up on "proving irreducibility via Galois theory". Basically, if you want to prove that a polynomial is irreducible, then you find some elements of $\mathbb C$ such that all the roots of this polynomial are contained in the field containing $\mathbb Q$ and all those elements (for example, in this case $\mathbb Q$ and $\sqrt{3 + \sqrt 5}$ alone, would have worked). Now Galois theory will tell you that this field is an extension of $\mathbb Q$ of degree its minimal polynomial, and finding the degree can be done using the tower property of fields, for example.

ContentLicense: CC BY-SA 4.0

Title: Prove $x^4-6x^2+4$ is irreducible


问:
Prove that the sequence $b_n = \frac{n}{1 + n}$ converges to 1 (without using limits)
Let $(b_n)$ be a sequence on the Euclidean metric space $\mathbb{R}$ defined as $b_n = \frac{n}{n + 1}$. Prove that $b_n$ converges to 1 without using limits. I can't think of how to do it without using limits. I looked all over the internet but I couldn't find a formal proof example. All of them just state if it converges or diverges (which in this case is obvious).

答:
We can do it by definition. Pick $\varepsilon > 0$ and then pick $n_0 \in \Bbb{N}$ such that $\frac1{n_0} < \varepsilon$, for example take some $n_0 > \frac1\varepsilon$. For any $n \in \Bbb{N}$, $n \ge n_0$ we have

$$\left|1-\frac{n}{n+1}\right| = \left|\frac1{n+1}\right| = \frac1{n+1} \le \frac1{n_0+1} < \frac1{n_0} < \varepsilon$$

and therefore $\lim_{n\to\infty} \frac{n}{n+1}=1$.


问:
Solve using Bayes?
I'm trying to become more comfortable using Bayes or Conditional probability. I have the answer to this problem, but I'm wondering how to go about it in a more formal way. Care to help??

Do like you did before by considering both scenarios. One scenario is that you drew the biased coin and it came up heads $5$ times. That has probability $\frac15\cdot 1 = \frac15$. The other scenario is that you drew a fair coin, and it came up heads $5$ times. That has probability $\frac45\cdot\frac1{32} = \frac1{40}$. So biased is a $\frac15 : \frac1{40}$ or $8:1$ favorite over fair, so the probability that it is biased is $\frac89$, and the probability that it is fair is $\frac19$. The probability that it comes up heads on the next flip is

$$\frac89\cdot 1 +\frac19\cdot 0.5 = \frac{17}{18}\;.$$

Bayes theory is this right?

$$\Bbb P(A\mid B)=\frac{\Bbb P(B\mid A)\cdot\Bbb P(A)}{\Bbb P(B)}$$

答:
Calculating the probability that the coin drawn is biassed can indeed be done with Bayes’ theorem. Let $A$ be the event that you drew the biassed coin, and let $B$ be the event that the coin came up heads $5$ times; then you want $\Bbb P(A\mid B)$, the probability that it’s biassed given that it came up heads all $5$ times. Bayes’ theorem says that this is

$$\frac{\Bbb P(B\mid A)\cdot\Bbb P(A)}{\Bbb P(B)}\;.$$

It’s easy to compute $\Bbb P(B\mid A)$: if you drew the biassed coin, the probability of getting $5$ heads in $5$ tosses is $1$. You also know that $\Bbb P(A)=\frac15$. Calculating $\Bbb P(B)$ is a bit more work: it’s given by

$$\Bbb P(B)=\Bbb P(B\mid A)\cdot\Bbb P(A)+\Bbb P(B\mid\neg A)\cdot\Bbb P(\neg A)\;,$$

where $\neg A$ is not-$A$, the event that you drew the fair coin. This is

$$\Bbb P(B)=1\cdot\frac15+\left(\frac12\right)^5\cdot\frac45=\frac15+\frac1{40}=\frac9{40}\;,$$

so

$$\Bbb P(A\mid B)=\frac{1\cdot\frac15}{\frac9{40}}=\frac15\cdot\frac{40}9=\frac89\;.$$

As you can see, it’s pretty much what was done in the statement of the problem; I’ve just identified each piece with the appropriate element of Bayes’ theorem.

ContentLicense: CC BY-SA 3.0

Title: Solve using Bayes?


问:
Invertible map on a particular sequence space

For $a\in\mathbb R$, let $h_a$ be the Hilbert space of sequences defined by
$$ h_a=\left\{(x_n):\sum_{n\in\mathbb Z}(1+n^2)^a|x_n|^2<\infty\right\}$$
and inner product $\langle(x_n),(y_n)\rangle_a=\sum_{n\in\mathbb Z}(1+n^2)^ax_n\overline{y_n}$.

Define the function $f:h_{-a}\to(h_a)^*$ by
$$ f\big((x_n)\big)((y_n))=\sum_{n\in\mathbb Z}x_ny_n, $$
where $(x_n)\in h_{-a}$ and $(y_n)\in h_a$. Prove that

$f((a_n))$ is well-defined as a function on $(h_a)^*$.

$f$ is an invertible, continuous linear map and has a bounded inverse.

Attempt: I have shown already that the series $\sum x_ny_n$ is convergent: if $(x_n)\in h_{-a}$ and $(y_n)\in h_a$ then
$$
\sum(1+n^2)^{-a}|x_n|^2<\infty,\qquad\sum(1+n^2)^a|y_n|^2<\infty,
$$
so by Cauchy-Schwarz I have
$$
\begin{aligned}
\left|\sum|x_ny_n|\right|^2&=\left|\sum((1+n^2)^{-a/2}|x_n|)((1+n^2)^{a/2}|y_n|)\right|^2\\
&\leq\left(\sum\left|((1+n^2)^{-a/2}|x_n|)^2\right|\right)\left(\sum\left|((1+n^2)^{a/2}|y_n|)^2\right|\right)\\
&<\infty,
\end{aligned}
$$
which means $\sum x_ny_n$ is convergent.

To finish showing that $f$ is a function from $h_{-a}$ to $(h_a)^*$, I think that I need to show somehow that $f((x_n))$ is a continuous linear functional. That $f$ is linear is clear to me, but I don't know how to show it is continuous. I know that one way to show continuity is to show boundedness, but I also don't know how to proceed in that direction. As for Part (2), I am unsure how to begin as well; in particular, I'm having trouble visualizing what the inverse map from $(h_a)^*$ to $h_{-a}$ would look like.

Any help or hints on this problem would be greatly appreciated. Thank you in advance.

答:
You have proved that $|f((x_n)) ((y_n))| \leq \|(x_n)\|\|(y_n)\|$ where the norms are taken the appropriate spaces. This implies that $\|f((x_n))\|\leq \|(x_n))\|$ and hence $f$ is abounded operator with $\|f\|\leq 1$.

Hints for then second part: You know that $f$ is injective. To show that $f$is surjective use the fact that ant element of $h_a^{*}$ is given by an inner product with some element of $h_a$. This follows by Riesz Theorem. Completeness of the spaces $h_a$ can be proved exactly the way you prove completeness of $\ell^{p}$ spaces. Thus each $h_a$ is a Hilbert space and Riesz Theorem is applicable.


问:
Recover from the solution $y=\log (m \cos^{-1} x)$ the original equation $(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=ke^{-2y}$.
$$\frac{dy}{dx} = -\frac{1}{m\cos^{-1} \big(x \sqrt{1-x^2}\big)}$$
$$\frac{dy}{dx} = -\frac{1}{e^y\sqrt{1-x^2}}$$

So
$$\frac{d^2y}{dx^2} = \frac{1}{e^{2y}(1-x^2)}\left(-\frac{e^yx}{\sqrt{1-x^2}}+e^y \sqrt{1-x^2} \frac{dy}{dx}\right)$$

I am not able to rearrange this into the required from. How should I proceed?

答:
There is a mistake in your $\frac{dy}{dx}$. In fact $m$ should not be in the denominator.
$$\frac{dy}{dx}=-\frac{1}{\cos^{-1}(x)\:\sqrt{1-x^2}}$$
$$\frac{d^2y}{dx^2}=-\frac{x}{\cos^{-1}(x)\:(1-x^2)^{3/2}}-\frac{1}{(\cos^{-1}(x))^2\:(1-x^2)}$$
Puting both into the ODE and after simplification :
$$-\frac{m^2+k}{m^2(\cos^{-1}(x))^2}=0$$
Thus in general $y=\log (m \cos^{-1} x)$ is not solution of the ODE. It is solution of the ODE only in the particular case $k=-m^2$.


问:
Proof of non-zero set if $ \lambda $ is an eigenvalue of $A$
How would I go about proving that, for any real $ n \times n $ matrix, for the set

$ W = \lbrace \textbf{w} \in \mathbb{R}^n \mid (A-\lambda I)^2 \textbf{w} = \textbf{0} \rbrace $,

$ W \neq \lbrace  \textbf{0} \rbrace $ if and only if $ \lambda $ is an eigenvalue of $A$.

Honestly I have no idea where to start with this, any help would be appreciated.

答:
Hint

For the implication, if $\lambda $ is not an eigenvalue, then $A-\lambda I$ is invertible.

For the converse, notice that $$\ker(A-\lambda I)\subset \ker\big((A-\lambda I)^2\big).$$


问:
Calculating probability of getting 3 pairs in 13 cards(similar and related to another question I asked before)
Assuming there are 52 cards and dealt with 4 different players. Each of them will receive 13 cards. What are the possibilities of a player receive exactly 3 pairs?

*four-of-a-kind is considered 2 pairs.

This means we have  3 different pairs and 2 same pairs and one different pair.

For 3 different pairs,

${{13}\choose{2}}$ is for two ranks showing on the two pairs.

$ {{10}\choose{7}}$ is with 3 ranks being chosen and 10 ranks left, we choose from there.

$ {{4}\choose{2}}$ is the combination for the pair in the rank.

$ {{4}\choose{1}}$ is the suit for the non-matching cards.

For 2 same pairs and one different pair,

${{13}\choose{2}}$ is for two ranks showing on the two pairs.

$ {{10}\choose{7}}$ is with 2 ranks being chosen and 11 ranks left, we choose from there.

$ {{4}\choose{4}}$ is the combination for the 2 same pairs in the rank.

$ {{4}\choose{2}}$ is the combination for the pair in the rank.

$ {{4}\choose{1}}$ is the suit for the non-matching cards.

$$
P=\frac{\binom{13}{2}\cdot\binom{10}{7}\cdot\binom{4}{2}\cdot\binom{4}{2}\cdot\binom{4}{2}\cdot {{4}\choose{1}}^7}{\binom{52}{13}}+\frac{\binom{13}{2}\cdot\binom{11}{7}\cdot\binom{4}{4}\cdot\binom{4}{2}\cdot\binom{4}{1}^7}{\binom{52}{13}}=1.3707 \times{10}^{-5}
$$

Does this way of calculating probability include both situations or it doesn't?

答:
This way of calculating is correct in the sense that it accounts for all possibilities and it does not double count some possibilities, but...
Apart from the typing mistakes which Alex already pointed out, there is also a conceptual error in the proposed solution.

When you have 3 pairs in different ranks, they can be treated as equal, so you have $\binom{13}{3}\cdot\binom{4}{2}^3$ combinations to take these pairs.
When you have 2 pairs in the same rank and 1 in another, the third pair can be distinguished from the first two, so there are more choices than $\binom{13}{2}$. It is easier to count these combinations as $13\cdot 12 \cdot \binom{4}{4}\cdot\binom{4}{2}$. This would give the following probability:

$$
P=\frac{\binom{13}{3}\cdot\binom{10}{7}\cdot\binom{4}{2}^3\cdot {{4}\choose{1}}^7}{\binom{52}{13}}+\frac{13\cdot12\cdot\binom{11}{7}\cdot\binom{4}{4}\cdot\binom{4}{2}\cdot\binom{4}{1}^7}{\binom{52}{13}} \approx 0.199
$$

Note also that this is the probability that one fixed player receives 3 pairs. The probability that at least one player gets 3 pairs, would be significantly harder to compute, since they cannot be considered independent. (This is how I interpreted the question at first.)


问:
Does this inequality hold for all odd $n$: $\sigma_{n-1}\sigma_{n+3} > \sigma_{n}\sigma_{n+2}$?
Let be $\sigma_n = \sum\limits_{d|n}d$ and let $n$ be an odd number, then the following inequality holds numerically up to $n = 10^7$.

$$
\begin{align}
\sigma_{n-1}\sigma_{n+3} > \sigma_{n}\sigma_{n+2}
\end{align}
$$

Question: Will this inequality hold for all odd $n$ or break at some point?

答:
Let $n= 470720125972210011720425021746900197623$. Then $\sigma_{n-1}\sigma_{n+3} < \sigma_{n}\sigma_{n+2}$. Here's the WolframAlpha input that substantiates my claim. But what is this monster? Checking prime factorizations:

\begin{align}n-1 &= 2 × 75083 × 3134665143722347347072073716732817\\n &= 659 × 5851 × 4680440291 × 6557737523 × 3977466362479\\n+2 &= 5^3 × 7^3 × 11 × 13 × 17 × 19 × 23 × 29 × 31 × 37 × 41 × 43 × 47 \\&\quad× 53 × 59 × 61 × 67 × 71 × 73 × 79 × 83 × 89 × 97\\n+3 &= 2 × 235360062986105005860212510873450098813\end{align}

Here I chose $n+2$ so it is a product of many distinct small primes. This increases the abundancy index of $n+2$($\approx 2.6859$). Since $\frac {n\pm1}2$ are prime/semiprime, their abundancy index is relatively small ($\approx 1.5$). Hence

$$\frac {\sigma_{n-1}\sigma_{n+3}}{n^2} \approx 1.5^2 < 2.6 \approx \frac {\sigma_{n+2}}{n} < \frac {\sigma_{n}\sigma_{n+2}}{n^2}$$

This $n$ is probably not be the smallest (to be frank, I just chose $\frac {5^27^2}6p_{25}\#$) but that invalidates your inequality. As the number of primes dividing $n$ increase, the abundancy index of $n$ can increase without bound, so it is intuitive to check near-primorials for counterexamples.

EDIT: $106992551791123$ is a much smaller counterexample, where $n+2=3^4 × 5^3 × 7^3 × 11 × 13 × 17 × 19 × 23 × 29$. This number has abundancy index $3.073$, which is greater than the original example since I realized $3$ being a divisor is not detrimental as long as the exponent $3^4$ is large.


答:
Assume that $A$ has eigenvalue $\lambda$. Then there exists an eigenvector $x \ne 0$ such that $(A-\lambda I )x = 0$ so in particular also $(A-\lambda I )^2x = 0$ so $x \in W$.

Conversely, if $W \ne \{0\}$ then there is some $x \ne 0$ such that $(A-\lambda I )^2x = 0$. In particular, the matrix $(A-\lambda I )^2$ is not invertible. Then $A-\lambda I$ also cannot be invertible since a product of two invertible matrices is invertible. Thus there is some $y \ne 0$ such that $(A-\lambda I)y = 0$, and hence $\lambda$ is an eigenvalue of $A$.

Title: Proof of non-zero set if $ \lambda $ is an eigenvalue of $A$

Body:
How would I go about proving that, for any real $ n \times n $ matrix, for the set

$ W = \lbrace \textbf{w} \in \mathbb{R}^n \mid (A-\lambda I)^2 \textbf{w} = \textbf{0} \rbrace $,

$ W \neq \lbrace  \textbf{0} \rbrace $ if and only if $ \lambda $ is an eigenvalue of $A$.

Honestly I have no idea where to start with this, any help would be appreciated.


问:
Why does $\cot\theta=\tan8\theta$ have two solutions?
The question is to find the general form of:
$$\cot\theta=\tan8\theta$$

By using $$x=n\pi+\alpha$$

I obtained
$$n\pi+\frac{\pi}{2}-\theta=8\theta$$
and
$$n\pi+8\theta=\frac{\pi}{2}-\theta$$

Using this formula interchangeably yields two different results.
But only the first one is given as the answer, so why can't I use the formula interchangeably? Have I done something wrong?

答: It's fine. Even though, the form of the answer is different, the set of solutions is the same. This is because $n$ can take negative values also.


问:
Interpreting odds ratios
Suppose there is a state with 224 settlmenets to whose representatives I send a message randomly from either group R or group NR. The incoming reponses are like in the following matrix:

```<code>         R   NR   Sum
0       31   22   53
1       72   99  171
Sum    103  121  224
</code>```

Because I had several variables of interest in the experiment, I started analysing the effect sizes in odds ratios, so they were comparable.

odds (R) = success ⁄ failure  = 0,7 ⁄ 0,3  = 2,33

odds (NR) = 0,82 ⁄ 0,18  = 4,56

OR (R ⁄ NR ) = 2,33 ⁄ 4,55  = 0,51

which means

[1] The odds of R to receive a response is about half (0,51) that of the odds of NR to receive a response.

But for most people odds are a difficult concept to understand, and therefore I'd like to interpret the result in an everyday language too. So I calculate the probabilities which are

p(R = 1) = 73 ⁄ 103  = 0,70

p(NR = 1) = 99 ⁄ 121  = 0,82

My questions is whether it is a correct interpretation of the data if I give an estimate that [2] about every 8th person in group R could experience differential treatment in comparision to group NR.

(I got this result by simply dividing the probability of response in each group by ten. The ratio between 82 and 70 is like the ratio between ~ 8 and 7.)

If yes, then how is this sort of simplification called, when one tries to express effects in terms of smallest units. I tried looking at very different sources, like mathsisfun.com and Agresti's Categorical Data Analysis (2002), but the closest concept I could find was the difference of proportions which is, of course, not what I am doing here.

I apologise if this is a very basic question in this forum, but I find the magnitude of my results, as interpreted in terms of [1] and [2] quite counter-intuitive, and I really want to understand what I am saying.

Thank you.

答:
A note about your calculations.  You are rounding your intermediate calculations and therefore losing precision or getting incorrect digits.  The odds of response for Group NR is $99/22 = 4.5$ exactly.  You instead calculated $99/121 \approx 0.82$, then $22/121 \approx 0.18$, but you have discarded too much precision in doin so, leading you to an imprecise result of $0.456$ when even a few extra digits would not have led you astray; e.g., $$\frac{99}{121} \approx 0.81818, \quad \frac{22}{121} \approx 0.18182, \quad O_{NR} \approx 4.49995.$$

Also note that a shortcut to compute the odds ratio is to write $$OR_{R/NR} = \frac{(72)(22)}{(31)(99)}.$$

As for the "non-intuitive" difference between the odds ratio of $0.516129$ and the response rates, think of a more extreme example where the response rates among both groups are very nearly $1$; e.g.,

$$\begin{array}{|c|c|c|c|}
\hline
& \text{R} & \text{NR} & \text{Total} \\
\hline
0 & 2 & 1 & 3 \\
\hline
1 & 101 & 120 & 221 \\
\hline
\text{Total} & 103 & 121 & 224 \\
\hline
\end{array}$$

Here I have chosen the cell values such that the group (column) totals are unchanged from your example.  Now if we calculate the odds ratio, we get $$OR_{R/NR} = \frac{(101)(1)}{(2)(120)} \approx 0.42083,$$ meaning that the odds of response for group R is less than half of the odds of response of group NR, but in terms of their individual response rates, they are almost equal:  $$\Pr[R = 1] = \frac{101}{103} \approx 0.980583, \\ \Pr[NR = 1] = \frac{120}{121} \approx 0.991736,$$ meaning that about every $1$ in $100$ persons will have different responses between the two groups.

Why does this happen?  Because the odds of an event can be any positive number, whereas a probability can only be between $0$ and $1$.  The odds ratio doesn't capture the absolute difference in responses.  It looks at the relative odds of response versus non-response.

Title: Interpreting odds ratios


问:
Derivative of a linear vector function by matrix - 3d-Jacobian or a vector?
If I differentiate
$$
Y =
\pmatrix{a & b \\ c & d}
\pmatrix {x \\ y}
$$
with respect to the matrix on the left side, what do I get?

On one hand, I think I should get $\pmatrix{x \\ y}$ because it's a linear function.

On the other hand, I think I should get a 3-dimensional Jacobian matrix (if it's a real thing) where the each 4 entries are 2-d vectors

$$
{\partial Y \over \partial a} = \pmatrix{x \\ 0},
{\partial Y \over \partial b} = \pmatrix{y \\ 0} \\
{\partial Y \over \partial c} = \pmatrix{0 \\ x},
{\partial Y \over \partial d} = \pmatrix{0 \\ y} \\
$$

Which one is right?

答:
You can look at your function as $Y : \Bbb{R}^4 \to \Bbb{R}^2$ given by
$$Y(a,b,c,d) = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix} ax+by \\ cx+dy\end{pmatrix}$$

so clearly the second interpretation is correct. Te resulting $2\times 4$ matrix is
$$\begin{pmatrix} \frac{\partial Y}{\partial a} & \frac{\partial Y}{\partial b} & \frac{\partial Y}{\partial c} & \frac{\partial Y}{\partial d} \end{pmatrix} = \begin{pmatrix} x & y & 0 & 0 \\ 0 & 0 & x & y\end{pmatrix}$$

as you calculated. Now, recall that the differential $DY(a,b,c, d)$ at a point $(a,b,c,d) \in \Bbb{R}^4$ is actually a linear map $\Bbb{R}^4\to\Bbb{R}^2$ whose matrix is precisely the above Jacobian.

Therefore it acts as

$$DY(a,b,c,d)\begin{pmatrix} a' \\ b' \\ c' \\ d'\end{pmatrix} = \begin{pmatrix} x & y & 0 & 0 \\ 0 & 0 & x & y\end{pmatrix}\begin{pmatrix} a' \\ b' \\ c' \\ d'\end{pmatrix} = \begin{pmatrix} a'x+b'y \\ c'x+d'y\end{pmatrix} = Y(a',b',c',d')$$

so the linear map $DY(a,b,c,d)$ is precisely $Y$ itself. This is because $Y$ is a linear map, as you noticed in your first interpretation.


问:
Is symmetric group on natural numbers countable?
I guess it is too difficult a question to ask about the cardinality of $S_{\mathbb{N}}$ so I would like to ask whether it is countable or not.

I tried to prove it is uncountable somewhat mimicking the Cantor's diagonal argument but failed.

答:
Here is a slightly analytic approach, that I think isn't too similar to what anyone else has written.

Since $\mathbb Q$ is countably infinite, it suffices to show that there are uncountably many bijections $\mathbb Q \to \mathbb Q$. We exhibit an injection $\mathbb R_{>0} \to S_{\mathbb Q}$.

Given a positive real $\alpha > 0$, define
\begin{align*}
 f_\alpha: \mathbb Q &\to \mathbb Q \\
 q &\mapsto
 \begin{cases}
  q & \text{if $\lvert q \rvert \le \alpha$} \\
  -q & \text{if $\lvert q \rvert > \alpha$}
 \end{cases}
\end{align*}
Then $f_\alpha$ is an involution, so certainly a bijection.

Moreover, it is clear that $\sup \{q \in \mathbb Q \mid f_\alpha(q) = q\} = \alpha$. This shows that $\alpha \mapsto f_\alpha$ is an injection. So indeed $S_{\mathbb Q}$ is uncountable.

I personally like this proof a lot. I find it much more intuitive when things end up being uncountable "because of $\mathbb Q$", and particularly when it's because of a Dedekind-cut-like density-related construction like this, than when it's because of a diagonal argument.

(fun fact: the set of all bijections in $S_{\mathbb N}$ having no fixed points is also uncountable!)


问:
The determinant of a triangular matrix is the product of its diagonal entries — proof using permutations

Prove that the determinant of a triangular matrix is the product of its diagonal entries.

My attempt:

Suppose $A$ is an $n\times n$ matrix, then:

$$\det(A) = \sum_{\sigma} (\operatorname{sgn} \sigma) A_{1\sigma_1} \cdots A_{n\sigma_n}$$

The product of diagonal entries is $A_1\sigma_1 \cdots A_n\sigma_n$ when $\sigma_i = i$ for $i = 1, \ldots, n$.

I claim that when the permutation is not an identity, there exist $i$ and $j$ s.t. $\sigma_i < i$ and $\sigma_j > j$.

Let $\sigma_j \neq j$, then $\sigma_k = j$ where either $k > j$ or $k < j$.

If $k > j$, let $\sigma_\ell = k$ where $\ell > k$ or $\ell < k$.

If $\ell < k$, done. Otherwise, $\ell > k$ and suppose for $\sigma_\ell, \ldots, \sigma_n$ that $n>\cdots>\ell$ and follow the above sequence of operations incrementally.

Then, $\sigma_n = n - 1$.

This implies that $\sigma_p = n$ where $p \leq j$.

If $k <j$, let $\sigma_h = k$ where $h > k$ or $h < k$.

Likewise, if $h > k$, done. Otherwise, $h < k$ and suppose for $\sigma_1, \ldots, \sigma_h$ that $1<\ldots<h$ and follow that above sequence of operations incrementally.

Then, $\sigma_1 = 2$.

This implies that $\sigma_q = 1$ where $q \geq j$.

Thus, there is at least one $i$ and $j$ s.t. $\sigma_i < i$ and $\sigma_j > j$.

By definition, if $A$ is upper-triangular, $A_{ij} = 0$ when $i > j$ and so, $A_{i\sigma_i} = 0$. Alternatively, if $A$ is lower triangular, $A_{ij} = 0$ when $i < j$ and $A_{j\sigma_j} = 0$.

So, the determinant of a triangular matrix is indeed the product of diagonal entries.

QED.

I'm not sure if my proof is correct and I wanted someone to go over it. Please let me know if I made any mistake(s) anywhere and if so what I can and should do to correct it.

答:
Let ${\rm U}_n$ be an invertible $n \times n$ upper triangular matrix. Let

$${\rm U}_{n+1} := \begin{bmatrix} {\rm U}_n & {\rm c}_{n+1}\\ {\rm 0}_n^\top & u_{n+1}\end{bmatrix}$$

Using the Schur complement,

$$\det \left( {\rm U}_{n+1} \right) = \det \begin{bmatrix} {\rm U}_n & {\rm c}_{n+1}\\ {\rm 0}_n^\top & u_{n+1}\end{bmatrix} = \left( u_{n+1} - {\rm 0}_n^\top {\rm U}_n^{-1} {\rm c}_{n+1} \right) \det \left( {\rm U}_{n} \right) = u_{n+1} \det \left( {\rm U}_{n} \right)$$

Let ${\rm U}_{1} =: u_1$. Hence,

$$\begin{aligned} \det \left( {\rm U}_{1} \right) &= u_1\\ \det \left( {\rm U}_{2} \right) &= u_2 \, u_1\\ &\vdots\\ \det \left( {\rm U}_{n} \right) &= \color{blue}{u_n \, u_{n-1}\cdots u_2 \, u_1} \end{aligned}$$

The case where ${\rm U}_n$ is non-invertible is left as an exercise for the reader.

matrices block-matrices schur-complement determinant


问:
Which complete bipartite graphs (exhaustive) decompose into 2 isomorphic subgraph
I'm trying to find some hint regarding to this subject as I only found a 14 page paper dated to 1979 by S Quinn on Google.

I'm thinking about the cycle for a vertex and odd/even vertex sets.

Any thoughts or suggestions?

答:
An obvious necessary condition for $K_{m,n}$ to be decomposed into two copies of a graph $G$, is that the number of edges in $K_{m,n}$ is twice the number of edges in $G$.  Hence $mn$ must be even.  Without loss of generality, assume $m$ is even.

A drawing of a decomposition of $K_{4,5}$ into two isomorphic graphs is given below.

In general, we can split the vertices in the part of size $m$ into two halves $A$ and $B$, and retain only the edges incident with $A$ in one graph, and retain only the edges incident with $B$ in the other graph.

The result is a decomposition of $K_{m,n}$ into two copies of the graph $K_{m/2,n}$ together with $m/2$ isolated vertices.


问:
Finding the sum of an infinite series.
Question:

If $$x = \frac{7}{4\times 1\times 2} + \frac{10}{4^2\times 2\times 3} + \frac{13}{4^3\times 3\times 4} +\dots $$
then find the value of $x$.

I managed to find the $n^\text{th}$ term as

$$t_n = \frac{\frac{4}{n} - \frac{1}{1+n}}{4^n}$$

I also tried to find some arithmetic or geometric series hidden in the question, but $1\times 2$, $2\times 3$,... in the denominator were quite problematic for me to deal with.

Please provide a solution which does not make use of calculus as I haven't taken it yet. Thank you.

答:
Hint:your general term is $$\frac{4+3r}{4^r(r)(r+1)}=\frac{1}{4^{r-1}r}-\frac{1}{4^r(r+1)}$$

Now telescope.....

Title: Finding the sum of an infinite series.


问:
Submanifold such that $M-A$ is simply connected then $M$ is simply connected
I am trying to prove the following statement

Let $M$ be a manifold and $A$ a submanifold of codimension greater or equal to $2$. If $M-A$ is Simply connected then $M$ is simply connected.

Now I was able to see that $M$ is path connected. For the part where $\pi_1(M)=\{0\}$ is where I don't know what to do . Does anyone have any suggestions? Thanks in advance.

答:
A formalization hint. Any loop $\gamma$ in $M$ can be approximated by a differential loop $\alpha$ transversal to $A$.
Now: (i) approximation implies homotopy, and (ii) since
$\dim(\alpha)+\dim(A)<\dim M$, transversality just means
$\alpha\cap A=\varnothing$.

Two edits for the asking.

(i) One can suppose $M\subset\mathbb R^p$. Then any two maps $\gamma,\alpha:I\to M$ are homotopic in $\mathbb R^p$ by linear interpolation. Then if $\alpha$ is close enough to $\gamma$, you get that interpolation inside a tubular neighborhood $W$ of $M$ in $\mathbb R^p$, hence can retract it into $M$.

(ii) Density of transversality can be formulated in many ways, here it is enough the one that so-called “parametrized”. I think a good reference is Abraham-Robbin’s Transversal Mappings and flows. The formal statement is just what needed: any mapping can be approximated by a differential mapping transversal to any given submanifold.


答:
Hint. One classically has
$$
\sum_{n=1}^\infty\frac{x^n}{n}=-\ln(1-x),\quad |x|<1,
$$$$
\sum_{n=1}^\infty\frac{x^n}{n+1}=-1-\frac{\ln(1-x)}{x},\quad |x|<1.
$$

Title: Finding the sum of an infinite series.

Body:
Question:

If $$x = \frac{7}{4\times 1\times 2} + \frac{10}{4^2\times 2\times 3} + \frac{13}{4^3\times 3\times 4} +\dots $$
then find the value of $x$.

I managed to find the $n^\text{th}$ term as

$$t_n = \frac{\frac{4}{n} - \frac{1}{1+n}}{4^n}$$

I also tried to find some arithmetic or geometric series hidden in the question, but $1\times 2$, $2\times 3$,... in the denominator were quite problematic for me to deal with.

Please provide a solution which does not make use of calculus as I haven't taken it yet. Thank you.


问:
Intuition behind the general form of the solution in Frobenius Method?
When we solve the equation $z^{2}u''+p(z)zu'+q(z)u=0$ using Frobenius method, we first find zeros of the indicial polynomial $r_1>r_2$ to get a solution $u_1$. In the case that the difference of roots $r_1-r_2$ is an integer, we might need to find a second linearly independent solution using a different method. It is given on the Wikipedia page (and many other places) that
$$
u_{2}=Cu_{1}\ln x+\sum _{{k=0}}^{\infty }B_{k}x^{{k+r_{2}}}
$$
gives the desired second solution.

My question is: how do we come up with this general form of solution? I try to derive it in various ways, but find it hard to achieve - I have tried to write $u_2 =vu_1$, which reduces the order of the DE and gives
$$
v'= u_1^{-2}\exp\left(- \int \frac{p(z)}{z} dz\right),
$$
but this looks completely different from the formula above.

I do see some vague intuition about the logarithmic term - we are basically saying: if multiplying by $x^{-r}$ does not make the solution analytic, then something stronger like $\ln x$ will. In other words, we create an essential singularity by adding $\ln x$.

I wish to see a more satisfying explanation of this.

答:
The Frobenius method tells you that you can find a solution of the form $u(z)=z^ra(z)$ where $a(z)$ is a normal power series (with a positive radius of convergence). Inserting gives
$$
z^2[z^ra''(z)+2rz^{r-1}a'(z)+r(r-1)z^{r-2}a(z)]+zp(z)[z^ra'(z)+rz^{r-1}a(z)]+q(z)z^ra(z)=0
$$
Now one can remove the common power $z^r$ and evaluate at $z=0$
$$
z^2a''(z)+z[2r+p(z)]a'(z)+[r(r-1)+rp(z)+q(z)]a(z)=0\\
[r(r-1)+rp(0)+q(0)]a(0)=0
$$
to obtain the indicial equation. For its solutions we know $1-p_0=r_1+r_2$ by the Viete formulas.

If you set $p(z)=p_0+z\tilde p(z)$ then the solution formula becomes
$$
v'=z^{-2r_1}a(z)^{-2}z^{-p_0}\exp(-\int \tilde p(z) ds)
$$
Now $n=p_0+2r_1=1+r_1-r_2$ is an integer according to the assumptions for this case, all the other factors are regular power series, so all in all one gets
$$
v'(z)=z^{-n}c(z)=c_0z^{-n}+...+c_{n-1}z^{-1}+c_n+c_{n+1}z+...
$$
as $n$ is positive, $r_1$ being the larger root. Integrating this gives a logarithmic term from $c_{n-1}z^{-1}$ and a product $z^{1-n}C(z)$ with a power series $C(z)$ from all the other terms. Now the second solution takes the form
$$
u_2=c_{n-1}u_1\ln(z)+z^{1-n+r_1}a(z)C(z)
$$
Now recall that $1-n+r_1=r_2$ and name the power series product $B(z)=a(z)C(z)$ to get the claimed form of the second solution.


答:
You have a map $Y:\mathbb R^{2\times2}\to\mathbb R^2$ given by
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\mapsto\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}.$$
This map is linear, and the derivative of a linear map at any arbitrary point is always the linear map itself. So you have $\mathrm D Y(A)=Y$ for all $A\in\mathbb R^{2\times2}$.

You can also find a matrix representation of said map if you choose a basis. Let's say we choose the standard basis for $\mathbb R^2$, and the basis consisting of
$$\begin{pmatrix}1&0\\0&0\end{pmatrix},\begin{pmatrix}0&1\\0&0\end{pmatrix},\begin{pmatrix}0&0\\1&0\end{pmatrix},\begin{pmatrix}0&0\\0&1\end{pmatrix}$$
for $\mathbb R^{2\times2}$. I will call these matrices $B_1,\dots,B_4$.

Now we can simply calculate
$$Y(B_1)=\begin{pmatrix}x\\0\end{pmatrix},~Y(B_2)=\begin{pmatrix}y\\0\end{pmatrix},~Y(B_3)=\begin{pmatrix}0\\x\end{pmatrix},~Y(B_4)=\begin{pmatrix}0\\y\end{pmatrix}.$$
So the matrix representation with respect to this basis is
$$\begin{pmatrix}x&y&0&0\\0&0&x&y\end{pmatrix}.$$
The dimensions of the matrix shouldn't be surprising, since $Y$ maps from a 4d space $\mathbb R^{2\times2}$ to a 2d space $\mathbb R^2$, so it needs 4 columns and 2 rows.

Title: Derivative of a linear vector function by matrix - 3d-Jacobian or a vector?

Body:
If I differentiate
$$
Y =
\pmatrix{a & b \\ c & d}
\pmatrix {x \\ y}
$$
with respect to the matrix on the left side, what do I get?

On one hand, I think I should get $\pmatrix{x \\ y}$ because it's a linear function.

On the other hand, I think I should get a 3-dimensional Jacobian matrix (if it's a real thing) where the each 4 entries are 2-d vectors

$$
{\partial Y \over \partial a} = \pmatrix{x \\ 0},
{\partial Y \over \partial b} = \pmatrix{y \\ 0} \\
{\partial Y \over \partial c} = \pmatrix{0 \\ x},
{\partial Y \over \partial d} = \pmatrix{0 \\ y} \\
$$

Which one is right?


问:
If A is dense in (0, 1), then A is dense in R
Is it true or false? Explain please.
Thanks.

Dense definition:
Let $∅ ≠ A, B ⊆ R$. We say that $A$ is dense in $B$, if $A ⊆ B$, and, in
addition, if for every $x, y ∈ B$ such that $x < y$ there exists an $a ∈ A$ such
that $x < a < y$

答:
Yes, $A$ is dense in $[0,1]$.

Suppose you are given $x,y\in[0,1]$ with $x<y$. You want to find $a\in A$ such that $x<a<y$. You can't immediately use the denseness of $A$ in $(0,1)$, because it may be that $x=0$ or $y=1$. But you just have to adjust $x$ and $y$ a little, so that they do fall in $(0,1)$: put
$$x'=x+\frac13(y-x)$$
$$y'=x+\frac23(y-x)$$
Now $x'$ and $y'$ are in $(0,1)$, so from the denseness of $A$ in $(0,1)$, there exists $a\in A$ such that $x'<a<y'$. And therefore $x<a<y$, which is what you want.


答:
Consider an arbitrary ordered set $(A, R)$ and define the following notion of relative denseness as a binary relation on subsets of $A$:
$$X \leqslant_{\mathrm{d}} Y \colon \equiv X \subseteq Y \wedge (\forall x, y)\left(x, y \in Y \wedge x<_R y \Rightarrow (x, y)_R \cap X \neq \varnothing\right)$$
Let us show that this binary relation is transitive, in other words given subsets $X$, $Y$ and $Z$ such that $X \leqslant_{\mathrm{d}} Y$ and $Y \leqslant_{\mathrm{d}} Z$ it follows that $X \leqslant_{\mathrm{d}} Z$. Indeed, by definition we have that $X \subseteq Y$ and $Y \subseteq Z$ so $X \subseteq Z$ follows by virtue of the transitivity of inclusion.

Let us now consider $u, v \in Z$ such that $u<_Rv$. As $Y$ is dense in $Z$, there exists $x \in Y \cap (u, v)_R$. Since $Y \subseteq Z$ we have in particular that $x \in Z$. Since $x, v \in Z$ and $x<_R v$, by the same token of $Y$ being dense in $Z$ we infer the existence of $y \in Y \cap (x, v)_R$. Since $x, y \in Y$ and $x<_R y$, we appeal now to the fact that $X$ is dense in $Y$ to infer that $X \cap (x, y)_R \neq \varnothing$. Let us note that by construction $u<_R x<_R y<_R v$, hence $(x, y)_R \subset (u, v)_R$ and subsequently $X \cap (u, v)_R \supseteq X \cap (x, y)_R \supset \varnothing$. This shows that $X$ is dense in $Z$.

In your particular case, the ambient ordered set is $(\mathbb{R}, \mathrm{O})$ (I have taken the liberty to denote the standard order on the reals by $\mathrm{O}$) and you are given that $A \leqslant_{\mathrm{d}} (0, 1)$. The fact that $(0, 1) \leqslant_{\mathrm{d}} [0, 1]$ is immediate so the above property of transitivity applies.

Remark: given an arbitrary topological space $(X, \mathscr{T})$, one can define a similar binary relation of topological denseness between the subsets of $X$ by:
$$M \subseteq_{\mathrm{d}} N \colon \equiv M \subseteq N \subseteq \overline{M},$$
which is equivalent to saying that $M$ is a dense subset of $N$ with respect to the subspace topology (on $N$, of course). This relation is also transitive, and in the particular case of a topological space induced by a total order, the two notions of relative denseness between subsets are (almost) equivalent.

Title: If A is dense in (0, 1), then A is dense in R

Body:
Is it true or false? Explain please.
Thanks.

Dense definition:
Let $∅ ≠ A, B ⊆ R$. We say that $A$ is dense in $B$, if $A ⊆ B$, and, in
addition, if for every $x, y ∈ B$ such that $x < y$ there exists an $a ∈ A$ such
that $x < a < y$


问:
Determine the amount to save each year to cover expenses
Imagine owning a big house that require a lot of maintenance. To know what you need to spend on maintenance in advance you take in a professional that creates a maintenance plan covering 50 years in the future. The plan contains how much you need to spend each year to cover expected maintenance. Some years might have zero expenses, and some might have very big ones.

To avoid a chock when the big ones come, you want to save an amount of money each year to cover future maintenance spending.

The question is, how much to save each year?

Some additional requirements:

The amount can very from year to year, but the amount each year should differ as little as possible between years.

After the 50 years is over, the amount in the maintenance saving account should be as close to zero as possible.

The maintenance saving account could already have a starting amount of money.

Me not working in math I tried to think of some different methods to arrive at the best solution.
One would be to start with zero in saving each year and then traverse each year deducting the expenses for that year if any. Once the maintenance account got below zero, (say it went to -1000), then you divide the 1000 on the years prior and add that amount to the saving plan. Then you start again traversing the years from the beginning but with the updated savings plan. Do this over and over until you never reach below zero.

However as you are getting close to the 50 year you will be splitting larger expenses on all years prior, so that might create a larger amount each year in the first years then declining, which would violate the requirement to keep the sum as static as possible.

Is there a better way to think about this?

答:
You seem to be assuming no interest on accumulative savings and no inflation. That makes the calculation easier. On your third point; if the account starts with a lump sum, then it can be included as a negative amount in the cumulative spending calculations below, at least for the initial pass.

Suppose you save $X_1,X_2,\ldots,X_n$ each year and spend $Y_1,Y_2,\ldots,Y_n$ each year.  Let's call cumulative saving to year $k$: $C_k=\sum\limits_{i=1}^k X_i$; and cumulative spending: $S_k=\sum\limits_{i=1}^k Y_i$ (with any initial lump sum in the account deducted from all the $S_k$).  You have the constraints

total saving at the end equals total spending $C_n = S_n$ and

at any stage cumulative spending is at least cumulative spending $C_k \ge S_k$ for $1 \le k \le n$

Now consider average spending up to year $k$: $A_k=\frac1k S_k$.  Find which $k$ maximises this (if there is more than one take the greatest) and call it $m$.  Up to year $m$ you could save a constant $X_i=A_m$ each year for $1 \le k \le m$ and you would satisfy the second condition, leaving the account empty at year $m$.

If $m=n$ then you are done and you have a completely flat profile.  If not, then a flat profile over the whole period would not lead to $C_n=S_m$, so start again with $n-m$ years (instead of $n$) and $X_{m+1},X_{m+2},\ldots,X_n$ and $Y_{m+1},Y_{m+2},\ldots,Y_n$ and repeat.  Since you are reducing the number of years each time, this is a finite method which must terminate.

You will end up with a pattern for $X_1,X_2,\ldots,X_n$ which is flat up to each point at which $C_m=S_m$ so you cannot do better in the early years, and then reduces to another flat profile and so on until you end with $C_n=S_n$.


问:
Proving an identity algebraically
Show that $$\sum_{i=0}^{n} \binom{n}{i} i (i-1)=n(n-1)2^{n-2}$$
knowing that $$\sum_{i=0}^{n}\binom{n}{i}i=n2^{n-1}$$

I ended up after a while with

$$\sum_{i=0}^{n}\frac{n-1}{2}\binom{n}{i}i=n(n-1)2^{n-2}$$

How can I transform of the $\frac{n-1}{2}$?

答:
You can simply pull the constant factor of $\frac{n-1}2$ outside the summation: that’s just the distributive law. Alternatively, you start at the beginning and use the identity

$$\binom{n}ii=n\binom{n-1}{i-1}\;,$$

which you can verify either combinatorially or by expanding into factorials.

Then

$$\binom{n}i(i-1)=n\binom{n-1}{i-1}(i-1)=n(n-1)\binom{n-2}{i-2}\;,$$

and you can pull the factor of $n(n-1)$ out of the summation. Then you need only know that $\sum_{k=0}^m\binom{m}k=2^m$.

Title: Proving an identity algebraically


答:
It is true. Let $A$ be dense in $(0,1)$. We have to show that for any two points $x,y \in [0,1]$ such that $x < y$ there is a point $a \in A$ such that $x < a < y$.

Notice that

$$0 \le x < \frac{2x+y}3 < \frac{x+2y}3 < y \le 1$$

and that $\frac{2x+y}3, \frac{x+2y}3 \in (0,1)$ since $x,y$ cannot both be $0$ or both be $1$.

Now since $A$ is dense in $(0,1)$, there exists $a \in A$ such that

$$\frac{2x+y}3 < a < \frac{x+2y}3 $$

so in particular $x < a < y$.


问:
Using definition of limit to prove limit
I am having trouble with this problem:

Use the definition of limit to prove that:
  $$\lim_{x\to \infty} \frac{\sin x}{x(\sin x)^2 +1} = \,?$$

I have concluded that the limit must be 0, but I am having trouble proving it.

Using the limit definition, I must show that

$$\left\lvert \frac{\sin x}{x(\sin x)^2 +1}- 0\right\rvert < \epsilon$$

答:
This is constructed a special way. That is key to the solution. It is an example of a function that can be shown to converge to zero as x approaches infinity by two separate sequences on the function +1/x and -1/x. This values are taken by the function under consideration is sine is 1 and therefore the square of sine if 1.

These are neither maxima nor minima of the function but relative ones. The combination of x and sine makes it hard to calculate the values. The function is close to the periodicity of sine but drops like 1/x, -1/x to zero.

and on a longer intervall

This does not imply that +/-1/x is minorant or majorant to the function. Both are steady and infinitely differentiable and therefore close related in the graphs.

The function is limit on the positive reals since the value at 0 is 0. There is a maximum at 0.816834.

There are crossing points to -/+1/x by the functions on the positive reals. These start with x bigger than 2  .

They take place arount the maxima and minima of the sine. With that bounds to the crossings it is assured that the function decreases to 0 for x-> ∞. All these are numerical accessible value like the first maximum they are close to a transcendental number /4.

The series of crossings starts with

```<code>{{x -> 6.47523}, {x -> 7.27203}, {x -> 8.38635}, {x -> 9.30196}, {x ->
    12.653}, {x -> 13.7369}, {x -> 14.5253}, {x -> 15.6393}, {x ->
   18.9056}, {x -> 20.0948}, {x -> 20.7404}, {x -> 21.9433}, {x ->
   25.1742}, {x -> 26.4219}, {x -> 26.9821}, {x -> 28.2376}}
</code>```

and for each natural positive n with n +  there is another crossing. If it would not be there the equation 1/2 (2 - x + x Cos[2 x] + 2 x Sin[x])=0 would have no solution. That can not be because the trigonometric functions are periodic.

Body:
This is constructed a special way. That is key to the solution. It is an example of a function that can be shown to converge to zero as x approaches infinity by two separate sequences on the function +1/x and -1/x. This values are taken by the function under consideration is sine is 1 and therefore the square of sine if 1.

These are neither maxima nor minima of the function but relative ones. The combination of x and sine makes it hard to calculate the values. The function is close to the periodicity of sine but drops like 1/x, -1/x to zero.

and on a longer intervall

This does not imply that +/-1/x is minorant or majorant to the function. Both are steady and infinitely differentiable and therefore close related in the graphs.

The function is limit on the positive reals since the value at 0 is 0. There is a maximum at 0.816834.

There are crossing points to -/+1/x by the functions on the positive reals. These start with x bigger than 2  .

They take place arount the maxima and minima of the sine. With that bounds to the crossings it is assured that the function decreases to 0 for x-> ∞. All these are numerical accessible value like the first maximum they are close to a transcendental number /4.

The series of crossings starts with

{{x -> 6.47523}, {x -> 7.27203}, {x -> 8.38635}, {x -> 9.30196}, {x ->
    12.653}, {x -> 13.7369}, {x -> 14.5253}, {x -> 15.6393}, {x ->
   18.9056}, {x -> 20.0948}, {x -> 20.7404}, {x -> 21.9433}, {x ->
   25.1742}, {x -> 26.4219}, {x -> 26.9821}, {x -> 28.2376}}

and for each natural positive n with n +  there is another crossing. If it would not be there the equation 1/2 (2 - x + x Cos[2 x] + 2 x Sin[x])=0 would have no solution. That can not be because the trigonometric functions are periodic.


问:
A bijective function between a square and its side
Is there a simple geometric proof that there exists a continuous bijective function between a square and its side?

And is there some explicit continuous function or formula $f^1(z)\mapsto (x,y)$ and $f(x,y)\mapsto z$, with $(x,y) \in [0,1]\times[0,1]$ and $z \in [0,1]$?

And is there a constructive proof that two sets of the same cardinality have a bijective function?

答:
A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Since $I\times I$ is not homeomorphic to $\partial(I\times I)$, there are no maps like the ones you are looking for.

To address your last question, it is usually taken as a definition that two sets have the same cardinality iff there is a bijection between them.


答:
$\text{Consider}$ $$\sum_{i=0}^n \dbinom{n}i x^i = (1+x)^n \tag{$\star$}$$
$\text{Differentiating $(\star)$, we get that}$
$$\sum_{i=0}^n \dbinom{n}i i x^{i-1} = n(1+x)^{n-1} \tag{$\perp$}$$
$\text{Differentiating $(\perp)$, we get that}$
$$\boxed{\color{red}{\displaystyle \sum_{i=0}^n \dbinom{n}i i(i-1) x^{i-2} = n(n-1)(1+x)^{n-2}}}$$
$\text{Now, set $x=1$ to get the answer, i.e.,}$
$$\boxed{\color{blue}{\displaystyle \sum_{i=0}^n \dbinom{n}i i(i-1) = n(n-1)2^{n-2}}}$$</p>

Title: Proving an identity algebraically

Body:
Show that $$\sum_{i=0}^{n} \binom{n}{i} i (i-1)=n(n-1)2^{n-2}$$
knowing that $$\sum_{i=0}^{n}\binom{n}{i}i=n2^{n-1}$$

I ended up after a while with

$$\sum_{i=0}^{n}\frac{n-1}{2}\binom{n}{i}i=n(n-1)2^{n-2}$$

How can I transform of the $\frac{n-1}{2}$?


问:
Difference of positive semidefinite matrices where $A_{ij} \geq B_{ij}$?
Let $A, B \in \mathbb R^{n\times n}$ be two positive semidefinite matrices. Assume that $A_{ij}\geq B_{ij}$ for $i,j\in \{1,\dots,n \}$. Does it hold that $C = A - B$ is positive semidefinite?

One possible simplification that I'm also interested in is if $B$ is of rank 1.

答:
In general the answer is NO.

Example:

Let $A=\begin{pmatrix}3&2\\2&3\end{pmatrix}$ and $B=\begin{pmatrix}2&0\\0&2\end{pmatrix}$

$C= A-B$ has eigenvalues $-1,3$, and is thus not positive semidefinite, although $A,B$ satisfy the conditions as mentioned in the question above.


问:
How to find the angles in a bisected golden rectangle?
If you split a golden rectangle along the diagonal, you get a right triangle with sides of 1 and phi.  I’ve been trying to figure out what the angles in such a triangle would be.  Obviously, I can plug phi or 1/phi into the arctan function, but I’ve been trying to figure out if there is some clean multiple of pi that expresses those angles.

Any one have thoughts or suggestions on how to solve this?

答:
This is a comment that is too long to be fit in a comment.

This reddit thread: https://www.reddit.com/r/math/comments/9std3s/arctangent_of_two_and_the_golden_ratio/ contains a beautiful Proof Without Words of the result:

$$\arctan\left(\frac1\phi\right)=\frac {\arctan 2}2$$

By the result in [Is] ArcTan(2) a rational multiple of $\pi$? (where the answers actually extends the results and determine exactly when the arctan of a rational number is a rational multiple of $\pi$), we see that $\arctan 2$ is indeed NOT a rational multiple of $\pi$. Hence it follows that neither $\arctan (\phi)$ nor $\arctan (\phi^{-1})$ could be rational multiples of $\pi$. The may not be a full answer though, since "clean multiple" is subjective.


问:
Closed form for $\sum_{n=1}^\infty\frac{1}{2^n\left(1+\sqrt[2^n]{2}\right)}$
Here is another infinite sum I need you help with:
$$\sum_{n=1}^\infty\frac{1}{2^n\left(1+\sqrt[2^n]{2}\right)}.$$
I was told it could be represented in terms of elementary functions and integers.</p>

Body:
Notice that
$$
\frac1{2^n(\sqrt[2^n]{2}-1)}
-\frac1{2^n(\sqrt[2^n]{2}+1)}
=\frac1{2^{n-1}(\sqrt[2^{n-1}]{2}-1)}
$$
We can rearrange this to
$$
\left(\frac1{2^n(\sqrt[2^n]{2}-1)}-1\right)
=\frac1{2^n(\sqrt[2^n]{2}+1)}
+\left(\frac1{2^{n-1}(\sqrt[2^{n-1}]{2}-1)}-1\right)
$$
and for $n=1$,
$$
\frac1{2^{n-1}(\sqrt[2^{n-1}]{2}-1)}-1=0
$$
therefore, the partial sum is
$$
\sum_{n=1}^m\frac1{2^n(\sqrt[2^n]{2}+1)}
=\frac1{2^m(\sqrt[2^m]{2}-1)}-1
$$
Taking the limit as $m\to\infty$, we get
$$
\sum_{n=1}^\infty\frac1{2^n(\sqrt[2^n]{2}+1)}
=\frac1{\log(2)}-1
$$


答:
The answer is no, even if $B$ has rank $1$. Consider
$$A = \begin{bmatrix} 4 & 2 \\ 2 & 1\end{bmatrix} \ge 0, \quad B = \begin{bmatrix} 1 & 1 \\ 1 & 1\end{bmatrix} \ge 0, \quad A-B = \begin{bmatrix} 3 & 1 \\ 1 & 0\end{bmatrix} \not\ge 0.$$

Title: Difference of positive semidefinite matrices where $A_{ij} \geq B_{ij}$?

Body:
Let $A, B \in \mathbb R^{n\times n}$ be two positive semidefinite matrices. Assume that $A_{ij}\geq B_{ij}$ for $i,j\in \{1,\dots,n \}$. Does it hold that $C = A - B$ is positive semidefinite?

One possible simplification that I'm also interested in is if $B$ is of rank 1.


问:
Find the equation of price - supply ,price - demand
I have a question as homework
I solved it using  online calculators but didn't understated it fully, even form my teacher

Here the solution form Online calculator

https://www.emathhelp.net/calculators/algebra-1/line-calculator/?lt=p&px=16.5&py=11.6&qx=15.3&qy=10.9&s=&ppx=&ppy=&steps=on

My question is that when I calculated by hand I got it wrong why is that ?

here my hand  calculation

答:
If $p=mx+b$ then I would have thought the slope was the change in price divided by the change in quantity so in this case $$\dfrac{11.6-10.9}{165045-152975}=\dfrac{0.7}{12070}\approx 0.000058$$

You seem to have lost a factor of $10000$ and divided the change in quantity by the change in price


问:
In a quaternion, are j and k not just equal to i?
I have been listening to many videos and reading but I am very confused. Firstly, I read that quaternions exist in $\mathbb{R}^4$ which would seem to exclude imaginary numbers completely (I would have guessed). But secondly, if a quaternion is sort of like a complex number, why do we need $j$ and $k,$ why can't we just use $i$ throughout ($a+ib+jc+kd$) is different somehow than ($a+ib+ic+id$) and if so, in what way does $i$ differ from $j$ and $k$?

Also, if quaternions have something to do with rotation is 3D space, why is a quaternion in a hypersphere? If we want to do something with rotations in on the plane, is 3D space involved?

答:
Firstly, I read that quaternions exist in $\mathbb{R}^4$ which would seem to exclude imaginary numbers completely (I would have guessed).

The text didn't mean that quaternions are real, but probably that they are 4-dimensional over the reals: $t+xi+yj+zk$ has 4 real parameters $t,x,y,z$.

But secondly, if a quaternion is sort of like a complex number, why do we need $j$ and $k,$ why can't we just use $i$ throughout ($a+ib+jc+kd$) is different somehow than ($a+ib+ic+id$) and if so, in what way does $i$ differ from $j$ and $k$?

It is sort of like a complex number, but it has two extra imaginary dimensions. It is an extended type of complex number. Note that $ijk=-1$ while $iii=-i.$ We therefore can not just replace $j$ and $k$ with $i$.

Also, if quaternions have something to do with rotation is 3D space, why is a quaternion in a hypersphere?

A rotation an angle $\theta$ around a direction $n=(n_x,n_y,n_z),$ where $|n|:=\sqrt{n_x^2+n_y^2+n_z^2}=1,$ can be described using a quaternion $r=\cos\theta+n_x \sin\theta\,i+n_y\sin\theta\,j+n_z\sin\theta\,k$. The magnitude of this is
$$
|r|=\sqrt{\cos^2\theta+n_x^2 \sin^2\theta+n_y^2\sin^2\theta+n_z^2\sin^2\theta} = 1.
$$
This means that $r$ as a 4-tuple $(\cos\theta, n_x\sin\theta, n_y\sin\theta, n_z\sin\theta)$ lies on the hypersphere $S^3 \subset \mathbb{R}^4.$

If we want to do something with rotations in on the plane, is 3D space involved?

For rotations in a plane, we can do with the ordinary complex numbers. Let $(x,y)\in\mathbb{R}^2$ be a point in the plane and set $z=x+iy.$ To rotate the point the angle $\theta$ around origin, just multiply $z$ with $e^{i\theta} = \cos\theta + \sin\theta\,i$ and take the real and imaginary parts of the result.


问:
In how many ways can I distribute 5 distinguishable balls into 4 distinguishable boxes such that no box is left empty.
So what I think is the way to solve this question is to first count the total number of ways of putting all the balls into boxes such that the restriction isn't satisfied (i.e. the total number of ways of putting the balls into the boxes). Then using the inclusion-exclusion principle for situations where the boxes are empty.

I think the number of total ways that you can distribute them would be $4^5$ but I'm unsure how to do the second part of the question.

答:
One box will have two balls. Choose which box that will be, choose two balls to put in it, then put one ball in each of the remaining boxes.

$$\binom{4}{1}\binom{5}{2}\times 3!=4\times 10\times 6=240$$


问:
If $\mu_n(A) \to \mu(A)$ for every $A$ such that $\mu(\partial A) = 0$, why does $\mu_n([0,x]) \to \mu([0,x])$?
Let $X = [0,1]$, and suppose $\mu_n$ is a sequence of finite measures on $(X, \mathcal{B}(X))$. Let $\mu$ be another finite measure on this space, and suppose that for each Borel set $A \subset [0,1]$ such that $\mu(\partial A) = 0$ we have $\mu_n(A) \to \mu(A)$ as $n \to \infty$.

I am trying to show that $\mu_n([0,x]) \to \mu([0,x])$ for each $x$ such that $\mu(\{x\}) = 0$. If $\mu(\{0\}) = 0$ this is immediate. I also know that there can be at most countably many singleton sets $\{x\}$ of positive measure. How can I show that $\mu_n(\{0\}) \to \mu(\{0\})$?

Could someone please give me a hint?

答:
It doesn't seem true. Let $\mu_n = \delta_{\frac{1}{n}}$ (ie $\mu_n(\{\frac{1}{n}\}) = 1$ and $\mu_n\left([0, 1] \setminus \{\frac{1}{n}\}\right) = 0$ and $\mu_n = \delta_0$.

If $0 \notin \delta A$, then either $[0, x] \subset A$ for some $x > 0$ or $[0, x] \cap A = \varnothing$ for some $x > 0$. In the former case, $\mu(A) = 1$ and $\mu_n(A) = 1$ for $n > \frac{1}{x}$. In the later case, $\mu(A) = 0$ and $\mu_n(A) = 0$ for $n > \frac{1}{x}$. In both cases, $\mu_n(A) \to \mu(A)$.

However, $\mu(\{0\}) = 1$ and $\mu_n(\{0\}) = 0 \not \to 1$.


问:
$\lim_{x\to 0}\frac{1}{x}−\frac{2}{e^{2x}−1}$ doesn't get solved as expected
A question says $f:\Bbb R\setminus\{0\} \to \Bbb R$, $f(x)=\dfrac{1}{x}−\dfrac{2}{e^{2x}−1}$. The limit as the function approaches $x=0$ is logically $0$, as in;

$\lim \limits_{x \to 0}\dfrac{e^{2x}-1}{2x} = 1$ This function then becomes $\dfrac{1}{x}−\dfrac{2}{2x} =0$

But when I graphed it online, the function approached $1$.

答:
You did an approximation error. When $x\to0$ it is $\frac{2}{e^{2x}-1}=\frac{2}{1+2x+\frac{1}{2}(2x)^2+\text{o}(x^2)-1}=\frac{1}{x+x^2+\text{o}(x^2)}$, so
$$\frac{1}{x}-\frac{2}{e^{2x}-1}=\frac{1}{x}-\frac{1}{x+x^2+\text{o}(x^2)}=\frac{1}{x}\left(1-\frac{1}{1+x+\text{o}(x)}\right)$$
Using the fact that $\frac{1}{1+\varepsilon}=1-\varepsilon+\text{o}(\varepsilon)$ when $\varepsilon \to 0$, you have
$$\frac{1}{x}\left(1-\frac{1}{1+x+\text{o}(x)}\right)=\frac{1}{x}\left(1-1+x+\text{o}(x)\right)=1+\text{o}(1) \to 1 \ \text{when} \ x \to 0$$


问:
Is second order logic even a logic?
Second order logic is a language, but, is it a logic?

My understanding is that a logic (or "logical system") is an ordered pair; it is a formal system together with a semantics. However, the language of second-order logic is associated with a variety of inequivalent formal systems and a variety of semantics. So, is it a logic?

I think not. Supposing not, what do sentences like the following (copied from wikipedia) even mean?

Second-order logic, for example, does not have a completeness theorem
  for its standard semantics (but does have the completeness property
  for Henkin semantics).

答:
Second-order logic has a collection of deduction rules. From a collection of statements, we can apply the deduction rules to derive new statements. This is called "syntactic entailment", or more simply, "formal proof".

Standard semantics talks about set-theoretic interpretations with the property that power types are interpreted as power sets. e.g. if $X$ is the interpretation of the domain of objects, then $\mathcal{P}(X)$ (or equivalently, $\{T,F\}^X$) is the interpretation of the domain of unary predicates on objects.

Given a collection of statements, we can consider set-theoretic models of those statements; interpretations where those statements are true. Using the laws of ZFC, one can show that the interpretations of statements are also true in every model. This is called "semantic entailment".

The incompleteness theorem is that syntactic entailment and semantic entailment (with standard semantics) do not coincide for second-order logic. (They do coincide for first-order logic!)

Incidentally, first-order logic also has variations. e.g. on the syntactic side, one can generalize to "free logic" (which, in some settings is what people mean when they say "first-order logic"), and people often consider various restricted fragments of it such as "regular logic" or "algebraic theories". On the semantic side, we can consider interpretations in terms of sheaves or objects in categories rather than in sets.

Title: Is second order logic even a logic?


答: Choose which two balls are to be paired, $\binom{5}{2}=10$, and then arrange the four objects, $4!=24$, for a total of $240$.

Title: In how many ways can I distribute 5 distinguishable balls into 4 distinguishable boxes such that no box is left empty.

Body:
So what I think is the way to solve this question is to first count the total number of ways of putting all the balls into boxes such that the restriction isn't satisfied (i.e. the total number of ways of putting the balls into the boxes). Then using the inclusion-exclusion principle for situations where the boxes are empty.

I think the number of total ways that you can distribute them would be $4^5$ but I'm unsure how to do the second part of the question.


问:
How does $1-\rho^2$ in the bivariate case, turn into $\det(\boldsymbol\Sigma)$ in the multivariate?
Bivariate mutual information for two Gaussian variables $X$ and $Y$ is well-known:

$$I(X;Y) = -\frac{1}{2}\ln(1-\rho^2)$$
where $\rho$ is the bivariate correlation coefficient.

Multivariate mutual information, according to Doquire and Verleysen (2012), is therefore

$$I(X_1; X_2;\dots ; X_n) = -\frac{1}{2}\ln (\det(\boldsymbol\Sigma)) $$
where $\boldsymbol\Sigma$ is the covariance matrix, and $\det$ is the determinant.

What is the connection between correlation and the determinant of the covariance matrix here? How does $1-\rho^2$ in the bivariate case, turn into $\det(\boldsymbol\Sigma)$? Please show steps of the derivation.

答:
It appears to me, that the cited formula is wrong. Also, the authors of the paper refer to another paper, in which I didn't find the mentioned formula at all. If we define the mutual information as customary as the KL divergence between the joint distribution and the product distribution of the respective marginals then in the Gaussian case one gets
\begin{equation}
I(X_{1},\dots,X_{n})=-\frac{1}{2}\ln\frac{\det(\Sigma_{0})}{\det(\Sigma_{1})},
\end{equation}
with $\Sigma_{0}$ being the covariance matrix of the multivariate Gaussian in question and $\Sigma_{1}$ the covariance matrix of the marginal product measure. Thus, $\Sigma_{1}$ is just the diagonal matrix with the same diagonal entries as $\Sigma_{0}$. And indeed in the two-dimensional setting, we recover your first formula since
\begin{equation}
\det(\Sigma_{0})=\sigma_{11}\sigma_{22}-\sigma_{12}^{2}\quad\text{ and }\quad \det(\Sigma_{1})=\sigma_{11}\sigma_{22}.
\end{equation}
The quotient thus reads
\begin{equation}
1-\frac{\sigma_{12}^{2}}{\sigma_{11}\sigma_{22}}
\end{equation}
which is $1-\rho^{2}$.


答:
Note that $ \dbinom{n}{i} i(i-1) = \frac{n!}{(n-i)! (i - 2)!}= n(n-1) \dbinom{n-2}{i} $.

Because $ \dbinom{n}{-2} = \dbinom{n}{-1} = 0 $, $$\sum_{i = 0}^n n(n-1)\dbinom{n-2}{i-2} = n(n-1) \sum_{i = 0}^{n-2} \dbinom{n-2}{i} = n(n-1)2^{n-2} $$

This only requires the knowledge that $ \sum_{i = 0}^n \dbinom{n}{i} = 2^n $ which is immediately obvious from the Binomial Theorem applied to $ (x + y)^n $ when $ x = y = 1 $.

If you wanted to explicitly use the given sum, you would consider $ \displaystyle \sum_{i = 0}^n \dbinom{n}{i} i^2 $. From there, $ \dbinom{n}{i} i^2 = n\dbinom{n-1}{i - 1}i $. Then shift the index up by $ 1 $ and you will once again have a linear function of $ i $ and $ \dbinom{n}{i} $.

Title: Proving an identity algebraically


问:
Find the limit of Riemann's sum
I have a question about the limit of Riemann’s sum. My Riemann’s sum is $\displaystyle   \sum_{i=1}^{2n}  \left(3 + \left(\frac{i}{n}\right)^2\right)\frac{1}{n}$. I tried to calculate the values of this sum using Sage but it was increasing as n was 10, 100, 1000, and did not approach any number. However, Symbolab says that there is a limit actually and it is 26/3.

Can you please help me to write integral and solve it for this problem?

Thank you a lot!

答:
Hint: Note that
\begin{eqnarray*} \sum_{i=1}^{2n}  \left(3 + \left(\frac{i}{n}\right)^2\right)\frac{1}{n}&=&\sum_{1\leq i \leq 2n} \frac{3+\left(\frac{i}{n}\right)}{n}\\
&=&\sum_{1\leq i \leq 2n}\frac{3}{n}+\frac{i^{2}}{nn^{2}}\\
&=&\sum_{1\leq i \leq 2n}\frac{3}{n}+n^{-3}i^{2}\\
&=&\frac{1}{3n^{2}}+\frac{2}{n}+\frac{26}{3}.
\end{eqnarray*}
This's a hint, so I think you can complete each the details.

Title: Find the limit of Riemann's sum


Title: Find the limit of Riemann's sum

Body:
I have a question about the limit of Riemann’s sum. My Riemann’s sum is $\displaystyle   \sum_{i=1}^{2n}  \left(3 + \left(\frac{i}{n}\right)^2\right)\frac{1}{n}$. I tried to calculate the values of this sum using Sage but it was increasing as n was 10, 100, 1000, and did not approach any number. However, Symbolab says that there is a limit actually and it is 26/3.

Can you please help me to write integral and solve it for this problem?

Thank you a lot!

Body:
\begin{gather*}
\frac{1}{n}\sum ^{2n}_{i=1}\left( 3+\frac{i^{2}}{n^{2}}\right) =\frac{1}{n}\left( 6n+\frac{2n( 2n+1)( 4n+1)}{6n^{2}}\right)\\
\lim _{n\rightarrow \infty }\frac{1}{n}\left( 6n+\frac{2n( 2n+1)( 4n+1)}{6n^{2}}\right) =\frac{26}{3}\\
\\
\end{gather*}


问:
Dual space kernel equality
I don't know how to show this one:
given elements $f,g ∈ V^∗$ of the dual space. I should now show that if $g = λf$ ,with $λ ∈ K\setminus\{0\}$ and $f$ is not $0$ ,
it follows that $\ker(f) = \ker(g)$.

答:
You have to show the equality of two sets.

$\ker(f)=\ker(g)$

So we have to show $\subseteq$ and $\supseteq$.

To show $\ker(f)\subseteq\ker(g)$ we take a vector $v\in\ker(f)$.

Then $f(v)=0$ but $g=\lambda f$ so $f=\lambda^{-1}g$. Then $(\lambda^{-1}g)(v)=\lambda^{-1}g(v)=0\Leftrightarrow g(v)=0$. So $v\in\ker(g)$.

You can show $\supseteq$ exactly the same way. I leave that for you.

Title: Dual space kernel equality


问:
Prove $var ( \Theta) = var(\hat \Theta) + var(\tilde \Theta)$ using the law of total variance
I am watching a lecture on Bayesian inference where the professor says this formula $(*)$ can be derived using the law of total variance but the derivation is not done in the lecture.

https://youtu.be/XtNXQJkgkhI?t=1812

$$var ( \Theta) = var(\hat \Theta) + var(\tilde \Theta) \tag{*}$$

So how can that be done?

Definitions:

a) $\Theta$ is an unknown parameter (r.v.) which we're trying to estimate

b) $\hat \Theta$ is the optimal LMS estimation of that unknown parameter $\hat{\Theta} := \mathbb{E}[\Theta \: | \: X]$

c) $\tilde \Theta$ is the error in the estimation of that unknown parameter i.e. $\tilde \Theta := \hat \Theta - \Theta$

I am looking at the law of total variance and I just don't see (yet) how to prove this formula by using it i.e. I don't see how it's related to $(*)$

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

Any hints/ideas?

答:
It is mentioned in the lecture slides, that $Cov(\hat{\Theta},\tilde{\Theta})=0$, and from this it easily follows, that
$$Var(\Theta)=Var(\hat{\Theta}+\tilde{\Theta})=Var(\hat{\Theta})+Var(\tilde{\Theta}) + 2\underbrace{Cov(\hat{\Theta},\tilde{\Theta})}_{=0}.$$
But this doesn't really show the relation to the law of total variance. From the lecture slides $\hat{\Theta}$ is introduced as
$$\hat{\Theta} := \mathbb{E}[\Theta \: | \: X],$$
so
$$Var(\hat{\Theta}) = Var(\mathbb{E}[\Theta \: | \: X]),$$
and thus the missing piece in order to relate the formula to the law of total variance is to show, that
$$Var(\tilde{\Theta}) = \mathbb{E}(Var(\Theta \: | \: X)).$$
I will now argue that this is indeed the case:
\begin{align*}
Var(\tilde{\Theta})&=Var(-\tilde{\Theta}) \\
&= Var(\Theta - \mathbb{E}[\Theta \: | \: X]) \\
&= \mathbb{E}[(\Theta-\mathbb{E}[\Theta \: | \: X])^2] &\text{(since $\mathbb{E}[\tilde{\Theta}] = 0$)} \\
&= \mathbb{E}[\mathbb{E}[(\Theta-\mathbb{E}[\Theta \: | \: X])^2 \: | \: X]] & \text{(Law of total expectation)} \\
&= \mathbb{E}[Var(\Theta \: | \: X)] &\text{(by definition of conditional variance)}
\end{align*}


问:
Perfect square involving the exponential law
If $n$ is a natural number, and $2^{10} + 2^{13} + 2^n$ is a perfect square, what is the value of $n$?

I've attempted to factor out $2^{10}$ and got $2^{10}(1 + 2^3 + 2^{n-10})$. How can I move further?

Body:
Complete the square:

$2^{10}+2^{13}+2^n=(2^5)^2+2\cdot2^5\cdot2^{n/2}+2^n=(2^5+2^{n/2})^2. $

$13=1+5+n/2$.

Can you take it from here?


问:
Problem name of the location of a new point the farthest away from any other point
For a project, I need to research a math and computer science problem. However, I am not sure what is the name of this problem.

I have N dimensions, each dimension has an upper and a lower bound. In this space, I possess p points. Now I want to find a new location of a point that is the farther apart from all the other points but still is inside the space.

I looked Voronoi diagram, p-dispersion problem, the metric k-center, but none of them seems to fit my problem setting.

If you know the name of this problem and/or the name of the algorithm that can solve it, it would be very much appreciated!

答:
It looks like you are on the concept of "farthest-point" Voronoi diagram which in a certain sense is dual to the closest point Voronoi diagram. See the last 10 slides of this excellent presentation.

Here is (see figure) an example issued from the Ph.D. thesis of Michel GOEB (references at the bottom of this answer) involving this concept in the framework of the measure of the set of circles including a certain set of points and excluding another set of points (This figure is extracted from his thesis).

These concepts can be extended from 2D to nD.

Fig. 1: Voronoi cells associated to 5 points: the "ordinary Voronoi" ones in red and the "Farthest Voronoi" in blue, with on the right, a 3D lifting of the left figure, one of the results of this Ph.D. thesis making a connection with a polyhedron obtained from tangent planes to paraboloid $z=x^2+y^2$.

References: Michel GOEB, "Modèles géométriques et mesures d'ensembles de cercles contraints", defended on the 2 Dec. 2008, Université de Saint-Etienne, France (I was the director of this Ph.D. thesis)


Title: Perfect square involving the exponential law

Body:
If $n$ is a natural number, and $2^{10} + 2^{13} + 2^n$ is a perfect square, what is the value of $n$?

I've attempted to factor out $2^{10}$ and got $2^{10}(1 + 2^3 + 2^{n-10})$. How can I move further?

Body:
$$(a+b)^2=a^2+2ab+b^2$$

$$2^{10}+2^{13}+2^n=(2^5)^2+2\times2^{12}+(2^7)^2=(2^5)^2+2\times2^{5+7}+(2^7)^2=(2^5+2^7)^2$$

Hence $n=14$


问:
$P_{n,k} = \frac{S_{n,k}}{k!}.$ Partiton of integers and surjective functions.
Let $S_{n,k}$ denote the number of surjective functions from $[n] \to [k]$ and $P_{n,k}$ denote the number of partitions (integer) of $n$ into $k$ parts. To prove by combinatorial argument that $$P_{n,k} = \frac{S_{n,k}}{k!}.$$

My Argument:

Surjection is actually partitioning the set $[n]$ into $k$ parts, by defining where each element is going to. Since the map is a surjection none of the parts are empty. So it is a valid partition. But we can rearrange the elements and get a different surjective map but the cardinality of each preimage set is same. So we are diving a factor of $k!$.

Am I going wrong somewhere in the argument?

答: Yes, you are wrong. The division by $k!$ takes out the order of the blocks, not the order of the elements in the domain of the function. So, for example, the set partition $\{1,2\},\{3\}$ gives you the integer partition $2,1$ but also this is given by $\{1,3\},\{2\}.$


问:
Number of ways of splitting set into disjoint subsets
How many distinct ways are there of splitting a finite set $X$ with $|X| = n$ into $k$ disjoint subsets of sizes $n_1, n_2, \dots, n_k$ with $\sum_{i=1}^k n_i = n$?

I would think the answer should be:

$$
\prod_{i=1}^k {n - \sum_{j=1}^{i-1} n_i\choose n_i}
$$

Because this is like first picking $n_1$ elements from $X$ in any order, then $n_2$ elements from the remaining $n - n_1$ elements of $X$ and so on.

Am I correct? And can I somehow simplify my solution?

答:
Just the typo that your binomial is backwards. Then
$$\binom{n}{n_1}\binom{n-n_1}{n_2}\cdots \binom{n_k}{n_k}=\frac{n!}{n_1!\cdot n_2!\cdots n_{k-1}!\cdot n_k!}.$$
That you can verify yourself by the definition of binomials using factorials i.e., $\binom{n}{k}=\frac{n!}{k!\cdot (n-k)!}.$

Now, do you care about the order of the sets? Notice that doing this the sets are labeled by the index of the sizes.


问:
Why doesn’t this socle of the lower triangular matrix ring contain more elements?
We know that the socle of a module is the sum of all the simple submodules. I was checking this for the following example, but it doesn't seem to line up with the definitions since there are obviously simple modules of the form $(0,1,0)^T$ whereas the socle of the example is only the sum of $(0,0,1)^T$ Thank you!
Example:
Confusion over Socle

答:
I think you are referring to the subset $\begin{bmatrix}0&0&0\\ a&b&0\\0&0&0\end{bmatrix}$ based on your attempted description of $\begin{bmatrix}0&0&0\\ 0&0&0\\ a&b&c\end{bmatrix}$ as “a sum of column vectors.”

One way to refute that is, as the link says, is that the Jacobson radical does not annihilate elements of that set. They are not as “obviously” simple as you thought.

It's true that the left socle of the lower triangular matrix ring over a field is a direct sum of copies of the set of matrices $[0,0,a]^T$, which is a simple left module.  But if you think the set of matrices $[0,a,0]^T$ is a simple left $R$ module, you'll have to check again: it isn't even closed under multiplication by $R$ on the left.


问:
Unidentifiable negative sign (-) resulting from U-substitution
I am working with General Forced Response for vibrations and I am trying to establish an easy procedure for solving integrals such as the one below.

$$
\int_{0}^{t}\tau\,
\sin\left(\,{\omega_{n}\left[\,{t - \tau}\,\right]}\,\right)\,\mathrm{d}\tau
$$

A quick conclusion is that first utilizing u-substitution, such that $u = (t - τ)$, is a good place to start to make things simpler with partial integration and such later. However, upon doing so, the result proposed by Symbolab has me a bit perplexed:
$$\int_t^0 -(-u+t)sin((ω_n(u))\;du$$
I am currently trying to understand where the constant negative sign $(-)$ before $(-u + t)$ comes from, since, surely, $(t-u)=τ\;$ as per the initial u-substitution? Does $(-u+t)$ then somehow not satisfy the original expression of $τ\;$ on its own?

I suspect that it may involve the new positioning of the internals of the parenthesis but I would very much like to hear from others on this. Thanks in advance!

答: It comes from the differential: if $u = t-\tau$, then $du = -d\tau$, so $d\tau = -du$.


问:
Laurent Series Expansion
Give the Laurent series development of the function $f(z)=\frac{1}{z(z-1)(z-2)}$ in the three rings $A_1=\{z:|z|<1\}$, $A_2=\{z:1<|z|<2\}$ and $A_3=\{z:2<|z|\}$.

I have gotten the partial fraction form $f(z)=\frac{1}{2z}-\frac{1}{z-1}+\frac{1}{z-2}$

For $A_1$, I think it's the standard Taylor expansion about 0 and I can do this for the second and third terms:

$-\frac{1}{z-1}=(1-z)^{-1}=\sum^\infty_{n=0}z^n$ and $\frac{1}{2(z-2)}=\frac{1}{4}\sum^\infty_{n=0}(\frac{z}{2})^n$. How do I get the Taylor expansion of the first term?

I foresee similar problems if I work out the expansion for $A_2$ and $A_3$ as well, so a general method is appreciated. Thanks!

答:
For $\,A_1\,$:

$$|z|<1\implies\frac{|z|}2<1\implies$$

$$\frac1{2z}+\frac1{1-z}-\frac12\frac1{1-\frac z2}=\frac1{2z}+\left(1+z+z^2+\ldots\right)-\frac12\left(1+\frac z2+\frac{z^2}4+\ldots\right)=$$

$$=\frac1{2z}+\frac12+\frac{3z}4+\frac{7z^2}8+\ldots$$

Using that $\,\displaystyle{\frac1{1-w}=1+w+w^2+\ldots\;,\;\;\text{for}\;\;|w|<1}\,$ . I let you to write the above with a sum symbol, if you want.

For $\,A_2\,$ , again we get:

$$1<|z|<2\implies\frac1{|z|}<1\;,\;\;\frac{|z|}2<1 \;$$

so

$$\frac1{2z}-\frac1z\frac1{1-\frac1z}-\frac12\frac1{1-\frac z2}=\frac1{2z}-\frac1z\left(1+\frac1z+\frac1{z^2}+\ldots\right)+\frac12\left(1+\frac z2\frac{z^2}4+\ldots\right)=\ldots$$

Title: Laurent Series Expansion


问:
Derivative of trace-based linear scalar field
I'm just getting into matrix calculus, so this question might be really easy for you guys out there. I'm trying to understand some of the simpler derivations in The Matrix Cookbook. I've been looking at derivatives of the trace predominantly. The following formulas are given:

$$
\frac{\delta}{\delta X} {\rm Tr}(XA) = A^T
$$

$$
\frac{\delta}{\delta X} {\rm Tr}(AX^T) = A
$$

I can't completely reproduce these example though. When I try to take the derivative with a small example manually, I always get $A$.

\begin{align}
\frac{\delta}{\delta X}
{\rm Tr}\left(
\begin{bmatrix}
x_1 & x_2 \\
x_3 & x_4
\end{bmatrix}
\begin{bmatrix}
a_1 & a_2 \\
a_3 & a_4
\end{bmatrix}
\right) &=
\frac{\delta}{\delta X}
{\rm Tr}\left(
\begin{bmatrix}
a_1x_1 + a_3x_2 & a_2x_1 + a_4x_2 \\
a_1x_3 + a_3x_4 & a_2x_3 + a_4x_4
\end{bmatrix}\right)
\\
&=
{\rm Tr}\left(
\begin{bmatrix}
\begin{bmatrix}
a_1 & a_2 \\
0 & 0
\end{bmatrix} &
\begin{bmatrix}
a_3 & a_4 \\
0 & 0
\end{bmatrix} \\
\begin{bmatrix}
0 & 0\\
a_1 & a_2
\end{bmatrix} &
\begin{bmatrix}
0 & 0 \\
a_3 & a_4
\end{bmatrix}
\end{bmatrix}
\right)\\
&=
\begin{bmatrix}
a_1 & a_2 \\
a_3 & a_4
\end{bmatrix}
\end{align}

\begin{align}
\frac{\delta}{\delta X}
{\rm Tr}\left(
\begin{bmatrix}
a_1 & a_2 \\
a_3 & a_4
\end{bmatrix}
\begin{bmatrix}
x_1 & x_3 \\
x_2 & x_4
\end{bmatrix}
\right) &=
\frac{\delta}{\delta X}
{\rm Tr}\left(
\begin{bmatrix}
a_1x_1 + a_2x_2 & a_1x_3 + a_2x_4 \\
a_3x_1 + a_4x_2 & a_3x_3 + a_4x_4
\end{bmatrix}\right)
\\
&=
{\rm Tr}\left(
\begin{bmatrix}
\begin{bmatrix}
a_1 & 0 \\
a_3 & 0
\end{bmatrix} &
\begin{bmatrix}
0 & a_2 \\
0 & a_4
\end{bmatrix} \\
\begin{bmatrix}
a_1 & 0\\
a_3 & 0
\end{bmatrix} &
\begin{bmatrix}
0 & a_2 \\
0 & a_4
\end{bmatrix}
\end{bmatrix}
\right)\\
&=
\begin{bmatrix}
a_1 & a_2 \\
a_3 & a_4
\end{bmatrix}
\end{align}

I'm sorry if the notation is wrong and thanks for your help.

答:
$\def\p{\partial} \def\m#1{ \left[\begin{array}{r}#1\end{array}\right] }$
Before you start differentiating, you must reduce the trace to a scalar expression.

So here's how you should perform the calculate for the first example
$$\eqalign{
t &= {\rm Tr}(XA) \\&= a_1x_1+a_3x_2 + a_2x_3+a_4x_4 \\\\
\frac{\p t}{\p X}
 &= \m{
\frac{\p t}{\p x_1}&\frac{\p t}{\p x_2} \\
\frac{\p t}{\p x_3}&\frac{\p t}{\p x_4} }
 = \m{
a_1&a_3 \\
a_2&a_4 }
 = A^T
}$$


问:
Scope in $\forall x P(x) \rightarrow Q(x)$
For $ \exists x P(x) \wedge Q(\color{blue}x) $

The scope of the quantifier, is just the predicate $P(x)$, thus the occurrence of x in blue is free.

I am unsure exactly the scope of the quantifier in this formula:

$ \forall x P(x) \rightarrow Q(\color{red}x) $

Based off the previous formula, I think the occurrence of the x in red is free but I'm not sure.

Is the occurrence of $\color{red}{x}$ free or bound? Would this vary based off convention?

答:
You are right.

The "usual" convention is that quantifiers, like negation, apply as little as possible.

Thus, $∃xα∧β$ is $(∃xα∧β)$, and not $∃x(α∧β)$.

The same for $∀xP(x) → Q(x)$: without parentheses, the scope of $∀x$ is only $P(x)$. Thus, the occurrence of $x$ into $Q(x)$ is free.


问:
How to make sense of the Green's function of the 4+1D wave equation?
In the paper "Wakes and waves in N dimensions" by Harry Soodak and Martin S. Tiersten, equation $(36)$ gives the Green's function for the 4+1D wave equation in the following form:

$$G_4(r,t)=\frac1{4\pi^2c^3}\left(\frac{\delta(t-r/c)}{r(t^2-r^2/c^2)^{1/2}}-\frac{\eta(t-r/c)}{c(t^2-r^2/c^2)^{3/2}}\right),\tag{36}$$

where $$r=\sqrt{x^2+y^2+z^2+w^2}$$ is distance from origin, $c$ is wave propagation speed, $\delta(\cdot)$ is the Dirac delta, and $\eta(\cdot)$ is the Heaviside step function.

Trying to understand its meaning, I've stumbled upon the fact that the first term contains Dirac delta with singularity at $r=ct$, multiplied by another function, which is singular (has algebraic branch point) at the very same point. This means that not only is the multiplier infinite at the delta's singularity, but it doesn't even have a well-defined complex phase at this point.

How should this expression be interpreted? Does it even make sense? If not, does Green's function not exist at all in this case?

答:
The retarded Green's function$^1$

$$\begin{align} G_{\rm ret}(\vec{r},t)
~=~&i\theta(t)G_C(\vec{r},t)
~\stackrel{(B)}{=}~2\theta(t) \bar{G}(\vec{r},t)
~\stackrel{(C)}{=}~2\theta(t)~{\rm Re}\,G_F(\vec{r},t),\tag{A}\cr
iG_C(\vec{r},t)~=~&G_{\rm ret}(\vec{r},t)-G_{\rm adv}(\vec{r},t)
~=~2~{\rm sgn}(t)\bar{G}(\vec{r},t),\tag{B}\cr
\bar{G}(\vec{r},t)
~=~&\frac{G_{\rm ret}(\vec{r},t)+G_{\rm adv}(\vec{r},t)}{2}
~=~{\rm Re}\,G_F(\vec{r},t),\tag{C}\cr
G_F(\vec{r},t)
~=~&\lim_{\epsilon\searrow 0^+}iG_E(\vec{r},(i+\epsilon)t),\tag{D}\cr
{\rm Re}\,G_F(\vec{r},t)~\stackrel{(D)}{=}~&-\lim_{\epsilon\searrow 0^+}{\rm Im}\,G_E(\vec{r},(i+\epsilon)t)\tag{E}  \cr
~\stackrel{(H)}{=}~&\lim_{\epsilon\searrow 0^+} \left\{ \begin{array}{lcl}
\frac{1}{2}{\rm Im}\sqrt{-t^2+i\epsilon}  &{\rm for}& d= 1,\cr
\frac{1}{4\pi}{\rm Im} \ln(r^2-t^2+i\epsilon) &{\rm for}& d= 2,\cr
 -\frac{1}{(d-2){\rm Vol}(\mathbb{S}^{d-1})}{\rm Im} \frac{1}{(r^2-t^2+i\epsilon)^{d/2-1}} &{\rm for}& d\geq 3,
\end{array}\right.\tag{F}
\end{align}$$
for the d'Alembertian
$$ (\partial_t^2-\vec{\nabla}^2)G(\vec{r},t)
~=~\delta^{d-1}(\vec{r})\delta(t) \tag{G}$$
in $d$-dimensional Minkowski spacetime can be derived as a generalized function by analytic continuation/Wick rotation of the Euclidean Green's function
$$ G_E(\vec{r})~=~\left\{ \begin{array}{lcl}
-\frac{r}{2} &{\rm for}& d=1, \cr
-\frac{\ln(r)}{2\pi} &{\rm for}& d=2, \cr
\frac{r^{2-d}}{(d-2){\rm Vol}(\mathbb{S}^{d-1})} &{\rm for}& d\geq 3,\end{array}\right. \tag{H}$$
for the Laplacian $$ -\vec{\nabla}^2 G_E(\vec{r})~=~\delta^d(\vec{r}) \tag{I}$$
in $d$ dimensions.

OP's expression (36) is supposed to be the retarded Green's function $G_{\rm ret}(\vec{r},t)$ in 4+1D. It is a linear combination of a singular function times a distribution. OP is correct that eq. (36) is not well-defined in distribution theory. This is why we propose to use analytic continuation/Wick rotation (D) instead.

If one rewrites the retarded Green's functions $G_d(r^2,t)$ with a square radial argument $r^2$ (and regularization $\epsilon>0$) then one may check that the formulas (F) satisfy the recurrence relations (25) and (29) of Ref. 1 in the following form
$$ G_{d+2}(r^2,t)~=~-\frac{1}{\pi} \frac{\partial G_d(r^2,t)}{\partial (r^2)}\tag{29'} $$
and$^2$
$$ G_{d-1}(r^2,t)~=~2\int_{\mathbb{R}_+}\!\mathrm{d}s~G_d(s^2+r^2,t). \tag{25'} $$
Ref. 1 argues that dimensional reduction from $d$ to $d\!-\!p$ dimensions corresponds to inserting a $(d\!-\!p)$-dimensional delta function source with a $p$-dimensional transversal support in $d$ dimensions. If $p=1$, this yields eq. (25'). If $p=2$, one gets
$$ G_{d-2}(r^2,t)~=~\int_{\mathbb{R}_+}\!2\pi s\mathrm{d}s~G_d(s^2+r^2,t)
~=~\pi\int_{r^2}^{\infty}\!\mathrm{d}(s^2)~G_d(s^2,t). \tag{28'} $$
Differentiation of eq. (28') yields eq. (29').

More calculations:
$$\begin{align}{\rm Re}\,&G_F(\vec{r},t)
~\stackrel{(E)}{=}~ -{\rm Im}\,G_E(\vec{r},(i+\epsilon)t)\cr
~\stackrel{(F)}{=}~&\left\{ \begin{array}{lclcl}
\frac{1}{2}{\rm Im}\sqrt{-t^2+i\epsilon}
&=& \frac{|t|}{2} &{\rm for}& d=1, \cr
\frac{1}{4\pi}{\rm Im}\ln(r^2-t^2+i\epsilon)
&=& \frac{1}{4}\theta(t^2-r^2) &{\rm for}& d=2, \cr
-\frac{1}{4\pi}{\rm Im}\frac{1}{\sqrt{r^2-t^2+i\epsilon}}
&=&-\frac{1}{4\pi}{\rm Im}\sqrt{\frac{r^2-t^2-i\epsilon}{(r^2-t^2)^2+\epsilon^2}}\cr
&=&\frac{1}{4\pi}\sqrt{\frac{\sqrt{(r^2-t^2)^2+\epsilon^2}-(r^2-t^2)}{2((r^2-t^2)^2+\epsilon^2)}}\cr
&=&\frac{\theta(t^2-r^2)}{4\pi\sqrt{|t^2-r^2|}}+\text{sing. terms} &{\rm for}& d=3, \cr
-\frac{1}{4\pi^2}{\rm Im}\frac{1}{r^2-t^2+i\epsilon}
&=&-\frac{1}{4\pi^2}{\rm Im}\frac{r^2-t^2-i\epsilon}{(r^2-t^2)^2+\epsilon^2}\cr
&=&\frac{1}{4\pi^2}\frac{\epsilon}{(r^2-t^2)^2+\epsilon^2}\cr
&=&\frac{1}{4\pi}\delta(t^2-r^2) &{\rm for}& d=4, \cr
-\frac{1}{8\pi^2}{\rm Im}\frac{1}{(r^2-t^2+i\epsilon)^{3/2}}
&=&-\frac{1}{8\pi^2}{\rm Im}\sqrt{\frac{(r^2-t^2-i\epsilon)^3}{((r^2-t^2)^2+\epsilon^2)^3}} \cr
&=&\frac{1}{8\pi^2}{\rm sgn}(r^2-t^2-\epsilon^2/3)\cr
&\times&\sqrt{\frac{\sqrt{((r^2-t^2)^2+\epsilon^2)^3}-(r^2-t^2)^3+3\epsilon^2(r^2-t^2)}{2((r^2-t^2)^2+\epsilon^2)^3}} \cr
&=&-\frac{\theta(t^2-r^2)}{8\pi^2|t^2-r^2|^{3/2}}+\text{sing. terms} &{\rm for}& d=5. \cr
\end{array}\right.\end{align} \tag{J}$$
Here the singular terms have support on the light-cone $\{(\vec{r},t)\in\mathbb{R}^d | r^2=t^2\}$. Therefore
$$\begin{align}  G_{\rm ret}(\vec{r},t)
~\stackrel{(A)}{=}~& 2\theta(t)~{\rm Re}\,G_F(\vec{r},t)\cr
~\stackrel{(J)}{=}~&\left\{ \begin{array}{lcl}
t^+=\max(t,0) &{\rm for}& d=1, \cr
\frac{1}{2}\theta(t-r) &{\rm for}& d=2, \cr
\frac{\theta(t-r)}{2\pi\sqrt{t^2-r^2}} +\text{sing. terms}&{\rm for}& d=3, \cr
\frac{\delta(t-r)}{4\pi r} &{\rm for}& d= 4, \cr
-\frac{\theta(t-r)}{4\pi^2(t^2-r^2)^{3/2}}+\text{sing. terms} &{\rm for}& d=5.\end{array}\right.\end{align} \tag{K}$$
Eq. (K) makes contact to the ill-defined expressions (35) & (36) of Ref. 1.

References:

H. Soodak & M.S. Tiersten, Wakes and waves in $N$ dimensions, Am. J. Phys. 61 (1993) 395.

--

$^1$ In this answer, we put the speed of light $c=1$. The limit symbol $\lim_{\epsilon\searrow 0^+}$ is implicitly implied from now on. Here $${\rm Vol}(\mathbb{S}^{d-1})
~=~ 2\frac{\pi^{d/2}}{\Gamma(\frac{d}{2})}\tag{L}$$
is the volume of the $(d\!-\!1)$-dimensional sphere $\mathbb{S}^{d-1}$. So
the constant in eqs. (F) & (H) is
$$(d\!-\!2){\rm Vol}(\mathbb{S}^{d-1})
~=~ 4\frac{\pi^{d/2}}{\Gamma(\frac{d}{2}-1)}.\tag{M}$$

$^2$ Concretely, eq. (25') was checked for low $d$ (high enough that the integral (25') is convergent). Now use the recurrence relation (29') to extend eq. (25') to arbitrary high $d$. Eq. (29') does not hold if we remove the regularization $\epsilon>0$.


问:
Help in showing that if the number of upcrossings is finite then the left and right limits of a function always exist
Let $I\subset \mathbb{R}$ be bounded and $f:I\rightarrow \mathbb{R}$ be a real function. The number of upcrossings of $f$
across the interval $[a,b]$, denoted by $U_f[a,b]$ is defined as the supremum of the nonnegative integers such that there exists $s_k,t_k\in I$ satisfying
$s_1<t_1<s_2<t_2<\cdots<s_n<t_n$,
and for which $f(s_k)\le a<b\le f(t_k)$, i.e.

$$ \begin{aligned}
U_f[a,b]:= \sup\big\{n\in\mathbb{N}\ |\ &(\exists s_1,...,s_n\in I)(\exists t_1,...,t_n\in I)(s_1<t_1<...<s_n<t_n \\ &\& \ (\forall k\leq n)(f(s_k)\le a<b\le f(t_k)))\}
\end{aligned}\tag{1}\label{uc}$$

The following lemma is taken from La Gall's Brownian Motion, Martingales ad Stochastic Calculus(Lemma 3.16). No proof is given in the text and the domain of the function is taken to be a countable dense set $D\subset\mathbb{R}_+$, as in stochastic analysis a common trope is to prove results in the discrete case first then use a limiting argument to go to the continuous case, however in this particular case I don't think we need the domain to be a countable set, if I am wrong please let me know why? In the text the function is assumed to be bounded and the upcrossings finite on the sets $D\cap [0,T]$ for every $T\in D$. So I am not sure what is more general, by considering the entire domain, but only bounded sets or the countable subsets?

Lemma. If $f:I\rightarrow \mathbb{R}$ is bounded and for every $a<b$, $U_f[a,b]<\infty$ then $\lim\limits_{s\downarrow t}f(s)$ and $\lim\limits_{s\uparrow t}f(s)$ exist for all $t\in I$.

We can just consider the right limits as the left case dually follows, my argument is a proof by contradiction and is as follows:

Suppose that the right limit does not exist for a particular $x\in I$. Then for a decreasing sequence $\{x_n\}\subset I$ converging to $x\in I$ we have
$$l:=\liminf\limits_{n\rightarrow\infty}f(x_n)<\limsup\limits_{n\rightarrow\infty}f(x_n):=u.\tag{2}$$
Now take $a,b\in(l,u)$, with $a<b$, then for any $n\in\mathbb{N}$ there must be a $n'>n$ such that $f(x_{n'})>b$ (otherwise $u\le b$ an impossibility); and similarly there is a $n^*>n$ such that $f(x_{n^*})<a$. We can then define the following sequence:
$$ \begin{aligned}
t_0 &=\inf I \\
s_k &=\inf\{m\geq t_{k-1} \ | \ f(x_m)\le a\} &\\
t_k &=\inf\{m\geq s_{k} \ | \ f(x_m)\ge b\}  \ &\text{for}\ k\in\mathbb{N}.
\end{aligned}\tag{3}\label{ucs} $$

Where we see that $s_1<t_1<s_2<t_2\cdots$ and $(\forall k\in\mathbb{N})(f(s_k)\le a<b\le f(t_k)$, thus $U_f[a,b]=\infty$, and the supposition was wrong and the right limit exists for every $x\in I$.

Would appreciate any feedback and input in particular please let me know:

If the assumptions and the statement of the lemma is correct, and what is the most general case of the lemma?

If the proof of the lemma is correct; if there are errors can they be fixed or if the whole argument has the be scrapped? If the latter please let me know how to proceed?

If there is a better argument, a neater one than the one above please let me know?

Is the formula for $U_f[a,b]$ in (1) correct? Should the conjunction (&) be replaced with an implication, $(\Rightarrow)$? Is there a better way to write a formula for $U_f[a,b]$?

Thanks in advance, any feedback will be appreciated.

答:
Generally, a bounded sequence $(x_n)_n$ has a subsequence $(x_{n(k)})_k$ that decreases to $\limsup_{n \to \infty}x_n$. Similar result holds for $\liminf_{n \to \infty}x_n$.

Let $(t_n)_n$ be a sequence in $D$ decreasing to $t >0.$
If $l<u$, then we have two subsequences of $(f(t_n))_n$ that decreases to $l$ and increases to $u$, respectively. This implies $M^{f}_{ab}([0,t_0]\cap D)=\infty$.


问:
Biggest circle under a polynomial curve
What is the biggest circle that is contained in the region bounded by the graph of the polynomial $f(x) = x(1-x)(2x+1)$ and the x-axis interval $[0, 1]$?

(Here's the thing in Desmos)

Here's what I have tried

Let's denote the center of the circle by $(c_x, c_y)$ and the points of tangency by $(x_k, y_k)$, $k=1,2$. From distance to the x-axis, we know that the radius of the circle is then $c_y$. We can express the tangency by the dot product of the vectors $(x_k-c_x, y_k-c_y)$ and $(1, f'(x_k))$ being zero. So we get the group of equations (all for $k=1,2$)

$$\begin{cases}
(x_k-c_x)^2 + (y_k-c_y)^2 &=& c_y^2 \\
x_k-c_x + (y_k-c_y)f'(x_k) &=& 0 \\
f(x_k) &=& y_k
\end{cases}
$$

Then I tried to solve this group by using Gröbner basis and elimination ideals. I wrote this SageMath code:

```<code>f(x) = x*(1-x)(2*x+1)
fd = f.derivative()

R.<x1, y1, x2, y2, cx, cy> = PolynomialRing(QQ, order='lex')
polys1 = [(cx-xk)**2+(cy-yk)**2 - cy**2 for (xk, yk) in [(x1,y1), (x2, y2)]]
polys2 = [(xk-cx)+(yk-cy)*fd(xk)  for (xk, yk) in [(x1,y1), (x2, y2)]]
polys3 = [f(xk)-yk for (xk, yk) in [(x1,y1), (x2, y2)]]
I = R.ideal(polys1+polys2+polys3)
gb = I.groebner_basis()
for poly in gb:
    print (poly)
print (I.dimension())
</code>```

It gave me a long list of polynomials, the last one being

$$c_x^6 - 2c_x^4c_y^2 + \frac{5}{4}c_x^4c_y + \frac{1}{64}c_x^4 + c_x^2c_y^4 + \frac{3}{4}c_x^2c_y^3 + \frac{3}{16}c_x^2c_y^2 + \frac{1}{64}c_x^2c_y$$

and said the dimension is $1$. I now realize the reason we don't get dimension $0$ has probably to do with the fact that the points $(x_1, y_1)$ and $(x_2, y_2)$ could be the same point(?) So for a solution, the point $(c_x, c_y)$ should be a zero of that polynomial? But is even that correct, since the points could be collapsed and the other side go over the curve??

How to solve this even numerically? I tried with Sage by minimizing the square sum of all those polynomials that should be zero. I also tried adding constraints for the variables to be near the solution that can be seen from the picture but that didn't work either.

答:
When $(u,v)$ is a point on your curve $\gamma: \>y=f(x)$ then we have
$$v=f(u)=u+u^2-2u^3,\qquad m:=f'(u)=1+2u+6u^2\ .$$
Given $u$, $v$, $m$ there is exactly one circle touching $\gamma$ at $(u,v)$ as well as the $x$-axis at some $q$. This circle has a center $(q,r)$ and a radius $r$. In order to compute $q$ and $r$ we introduce the following quantities (see the figure):
$$\tan(2\alpha)=m,\quad l={v\over\sin(2\alpha)}\ .$$
Then
$$\eqalign{q(u)&=u+v\tan\alpha=u+v\>{\sqrt{1+m^2}-1\over m},\cr r(u)&=l\tan\alpha=v\>{1+m^2-\sqrt{1+m^2}\over m^2}\ .\cr}\tag{1}$$
This leads to the green curve
$$u\mapsto\bigl(q(u),r(u)\bigr)\qquad(0\leq u\leq1)$$
in the figure. Each point of this curve is the center of a circle that touches $\gamma$ and the $x$-axis. This curve has a point $S$ of selfintersection, coming from two values $u_1\approx0.4507$, $u_2\approx0.7940$. The circle centered at this point touches $\gamma$ in two points $(u_i,v_i)\in\gamma$ as well as the $x$-axis, and it seems that this circle is the solution to your problem. The point $S\approx(0.5962,0.2578)$ can be found numerically with arbitrary precision, using the equations coming from $(1)$.

Title: Biggest circle under a polynomial curve


问:
A function satisfying $f''(x)(1+f(x))=1+x$

let $f$ be defined for $x\ge 0$ and has a continous derivative and $f(0)=1$ and $f'(0)=0$ such that $$f''(x)(1+f(x))=1+x$$Then the range that $f(1)$ can take-

I think the key to the question is to find a function whose derivetive/double derivative is $f''(x)(1+f(x))$.So far i have thought of using,$H(x)=e^xf(x)$ but $H''(x)$ is coming to be quite messy.

Also integrating both sides;$$f'(x)=\int \frac{1+x}{1+f(x)} $$.

I think it has something to do with finding some bound on $f'(x)$ and then using LMVT?

答:
Assume there exists some $x_0 \ge 0$ such that $f(x_0)+1=0$. Then, from the given equality, we get $1 + x_0 = 0$. Hence, there does not exist such $x_0$.

Since $f$ is continuous and the function $f(x)+1$ is non-zero, for all $x$ in $[0, \infty)$ we have $1+f(x) > 0$. Therefore, $\forall x: ~f''(x) >0$. It means that $f'(x)$ is strictly increasing. Since $f'(0) = 1$, we get that $\forall x: ~f'(x) > 0$, as well. Hence, $f(x)$ is strictly increasing too.

Now, since $f(0)=1$, we must have (at least) $f(1) > 1$.


问:
Introduction to Abstract Harmonic Analysis for undergraduate background
I'm looking for a good starting book on the subject which only assumes standard undergraduate background.

In particular, I need to gain some confidence working with properties of Haar measures, so I can better understand the spaces $L^{p}(G)$ for a locally compact group $G$.

For some perspective on what I currently know, I can't yet solve this problem:  Verifying Convolution Identities

Something with plenty of exercises would be ideal.

答:
I suggest the short book by Robert, "Introduction to the Representation Theory of Compact and Locally Compact Groups" which is leisurely and has plenty of exercises. The only prerequisite of this book is some familiarity of finite dimensional representations.

A second book you should look at is Folland's "A Course in Abstract Harmonic Analysis", which is more advanced, and requires more experience with analysis (having seen Banach spaces is not a bad thing), but the advantage of this book is that it has very clearly written proofs, that are easily to follow (I do algebra mostly, and I find many analysis tracts a bit opaque in this regard). Unfortunately, this book does not have exercises, and should be approached once you have plenty of examples in mind.

Donald Cohn's "measure theory" has a large number of exercises on the basics of topological groups and Haar measure, but it doesn't do representation theory or much else on locally compact groups except an introduction.


问:
why $g(x) < \infty$ almost everywhere?
I have  some confusion in RCA book

My confusion is given below and marked in red box

My attempt  :  If $g(x) < \infty$  this  implies $x \notin A $ and $ \mu(E_k)=0$

But almost everywhere mean  that  property that doesn't hold  on set  measure $0$

so i think  $g(x) = \infty$ will converge  almost  everywhere

My doubt :why  $g(x) < \infty$  almost everywhere ?

答: As it was pointed out in the comments, given $g\in L^1(\mu)$ we must have $g(x)<\infty$ a.e., since otherwise $g(x)=\infty$ on a measurable set $E\subset X$ with positive measure $\mu(E)>0$, and by monotonicity of the Lebesgue integral that would imply $\int|g|_X\,\text{d}\mu\ge\int_E|g|\,\text{d}\mu=\int_Eg\,\text{d}\mu=\int_E\infty\,\text{d}\mu=\infty\mu(E)=\infty$, which is a contradiction since $g\in L^1(\mu)\Rightarrow\int_X|g|\,\text{d}\mu<\infty$.

Title: why $g(x) < \infty$ almost everywhere?


问:
A question about induction.
Are we allowed to perform induction only on set of natural numbers or can we also perform induction on say set of even integers?</p>

答:
In the strictest theoretic sense, no. Mathematical induction applies only to predicates taking an argument $n \in \Bbb{N}$ (i.e. a true/false statement that depends on an unknown $n \in \Bbb{N}$, e.g. "$n$ is even"). It states that, if we have a predicate $P(n)$, where $n \in \Bbb{N}$, such that $P(0)$ is true and $P(n) \implies P(n + 1)$ for all $n \in \Bbb{N}$, then $P(n)$ is true for all $n \in \Bbb{N}$.

In practice, this principle is far more flexible. If we want to prove $Q(n)$ for all $n \ge 7$, we can simply let $P(n)$ be $Q(n + 7)$ and apply induction as above. If we want to prove $Q(n)$ for positive even integers, that's easy too: let $P(n)$ be $Q(2n + 2)$, then use induction.

We could even induct over all the integers. If we wish to prove $Q(n)$ for all $n \in \Bbb{Z}$, we let $P(n)$ be $Q(n)$ and $P'(n)$ be $Q(-n)$ for $n \in \Bbb{N}$. We can then prove $P(n)$ and $P'(n)$ for all $n$ using induction, and that will prove $Q(n)$ for all $n \in \Bbb{Z}$. Induction over all even integers would work similarly.

What sets can't you use induction on? Essentially, our predicates need to be indexed by a well-ordered set, and preferably order-isomorphic to $\Bbb{N}$ (unless you use the more complex transfinite induction). Inducting over the real numbers wouldn't work so well, because there is no "next" real number, so the induction step is kinda meaningless.

I hope that helps.

Title: A question about induction.


问:
Integral of a binomial-like function through Integration by parts
I wonder how to calculate the following through integration by parts:
\begin{align*}
\int_{0}^{1}\binom{n}{i}u^{i}(1-u)^{n-i}\, du
\end{align*}
$i=0,1,2,\cdots,n$. I suppose Gamma function might be useful here?

Body:
Let
$$I:=\int_0^1 \binom{n}{i} u^i(1-u)^{n-i} \text{d}u$$
So
$$(n+1)I=\sum_{i=0}^{n}\int_0^1 \binom{n}{i} u^i(1-u)^{n-i} \text{d}u=\int_0^1 \sum_{i=0}^{n}\binom{n}{i} u^i(1-u)^{n-i} \text{d}u$$
$$=\int_0^11^n\text{d}u=1\implies I=\frac{1}{n+1}$$


问:
Homeomorphism between $R^1 \times S^1 $ & $ R^2 -(0,0)$
My question is in the context of topological spaces. I can show that if $X_1$ & $X_2$ are homeomorphism and $Y_1$ & $Y_2$ are homeomorphism, then $X_1 \times Y_1$ & $X_2 \times Y_2$ are homeomorphism too.
Can I use this property to show homeomorphism between
$R^1 \times S^1 $ & $ R^2 -(0,0)$ ?

If not, can we do this using only a function like $f(x,t)=x^t$?

答:
Generous hint; not a complete solution

You can't apply that theorem here, no, because the right-hand side is not a product in any obvious way.

I suggest you look at
$$
f(s, \theta) = (s \cos \theta, s \sin \theta).
$$

That would be a perfect solution if the left hand side were $\Bbb R^{+} \times S^1$. You should work through that carefully to see why.

And then you should modify it to work for all of $\Bbb R$ instead of just the positive part of the real line.