问:
Inverse Fourier transform of $F(w)=\frac{1-\sum_{i=1}^n{a_ie^{-{r_k}{b_i}|w|}}}{|w|}$
I encountered this problem in my research. I don't know to calculate the inverse Fourier transform (IFT) of this function:
$$F(w)=\prod_{k=1}^m \frac{1-\sum_{i=1}^n{a_ie^{-{r_k}{b_i}|w|}}}{|w|}$$
where $a_i \in \mathbb{R}$, $b_i \in \mathbb{R^+} (i=1,...,n)$, $r_k \in \mathbb{R^+} (k=1,...,m)$ are known real parameters, $\sum_{i=1}^n{a_i} =1$ (so that $|F(0)|<+\infty$), and $m=1,2$ or $3$.

If the explicit expression can't be derived, then ideas for solving the problem will be great as well.

答:
I shall follow the convention that IFT of $F$ is  $\int_{\mathbb R}F(w)e^{2\pi i w t}dw$, take a scaling if necessary. First of all, a crucial lemma $$\int_0^{\infty } \frac{e^{-aw} \cos (b w)-e^{-cw} \cos (d w)}{w} \, dw=\frac{1}{2} \log \left(\frac{c^2+d^2}{a^2+b^2}\right)$$
Which is direct by applying Feynman's trick on either $a$ or $c$. Now, for case $m=1$ one have the IFT equals
$$I_1=\int_{\mathbb R} \frac{1-\sum_i a_i e^{-r_1 b_i |w|}}{|w|}e^{2\pi i w t}dw=2\int_{\mathbb R^+} \frac{1-\sum_i a_i e^{-r_1 b_i w}}{w}\cos(2\pi t w)dw=\sum_i a_i K_1(r_1 b_i)$$
Where we have took a reflection on part $\mathbb R^-$, applied Euler formula and rewritten $1$ as $\sum_i a_i$. Now the lemma gives
$$K_1(s)=\int_{\mathbb R^+}\frac{1-e^{-s w}}{w}\cos(2\pi t w)dw=\frac{1}{2} \left(\log \left(\frac{s^2}{4 t^2}+\pi ^2\right)-2 \log (\pi )\right)$$
Which completes the evaluation of this case. Note that this can be trivially generalized to cases that $m=2, 3$, namely elementary closed forms of IFT that $I_2=\sum_{i,j} a_i a_j K_2(r_1 b_i, r_2 b_j)$ and $I_3=\sum_{i,j,k} a_i a_j a_k K_3(r_1 b_i, r_2 b_j, r_3 b_k)$, where

$\small K_2(s,y)=\int_{\mathbb R^+}\frac{1-e^{-s w}}{w}\frac{1-e^{-y w}}{w}\cos(2\pi t w)dw=-\frac{1}{2} s \log \left(s^2+4 \pi ^2 t^2\right)+\frac{1}{2} s \log \left((s+y)^2+4 \pi ^2 t^2\right)+\frac{1}{2} y \log \left((s+y)^2+4 \pi ^2 t^2\right)-2 \pi  t \tan ^{-1}\left(\frac{2 \pi  t}{s+y}\right)+2 \pi  t \tan ^{-1}\left(\frac{2 \pi  t}{s}\right)-\frac{1}{2} y \log \left(4 \pi ^2 t^2+y^2\right)+2 \pi  t \tan ^{-1}\left(\frac{2 \pi  t}{y}\right)-\pi ^2 t$

$\scriptsize K_3(s,y,z)=\int_{\mathbb R^+}\frac{1-e^{-s w}}{w}\frac{1-e^{-y w}}{w}\frac{1-e^{-z w}}{w}\cos(2\pi t w)dw=\frac{1}{4} \left(2 \log (s) s^2+\log \left(\frac{4 \pi ^2 t^2}{s^2}+1\right) s^2-2 \log (s+y) s^2-\log \left(\frac{4 \pi ^2 t^2}{(s+y)^2}+1\right) s^2-2 \log (s+z) s^2+2 \log (s+y+z) s^2-\log \left(\frac{4 \pi ^2 t^2}{(s+z)^2}+1\right) s^2-8 \pi  t \tan ^{-1}\left(\frac{2 \pi  t}{s}\right) s+8 \pi  t \tan ^{-1}\left(\frac{2 \pi  t}{s+y}\right) s+8 \pi  t \tan ^{-1}\left(\frac{2 \pi  t}{s+z}\right) s-8 \pi  t \tan ^{-1}\left(\frac{2 \pi  t}{s+y+z}\right) s-4 y \log (s+y) s-2 y \log \left(\frac{4 \pi ^2 t^2}{(s+y)^2}+1\right) s-4 z \log (s+z) s+4 y \log (s+y+z) s+4 z \log (s+y+z) s-2 z \log \left(\frac{4 \pi ^2 t^2}{(s+z)^2}+1\right) s-8 \pi  t y \tan ^{-1}\left(\frac{2 \pi  t}{y}\right)+8 \pi  t y \tan ^{-1}\left(\frac{2 \pi  t}{s+y}\right)-8 \pi  t z \tan ^{-1}\left(\frac{2 \pi  t}{z}\right)+8 \pi  t z \tan ^{-1}\left(\frac{2 \pi  t}{s+z}\right)+8 \pi  t y \tan ^{-1}\left(\frac{2 \pi  t}{y+z}\right)+8 \pi  t z \tan ^{-1}\left(\frac{2 \pi  t}{y+z}\right)-8 \pi  t y \tan ^{-1}\left(\frac{2 \pi  t}{s+y+z}\right)-8 \pi  t z \tan ^{-1}\left(\frac{2 \pi  t}{s+y+z}\right)-8 \pi ^2 t^2 \log (s)-4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{s^2}+1\right)+8 \pi ^2 t^2 \log (2 \pi  t)-8 \pi ^2 t^2 \log (y)+2 y^2 \log (y)+8 \pi ^2 t^2 \log (s+y)-2 y^2 \log (s+y)-4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{y^2}+1\right)+y^2 \log \left(\frac{4 \pi ^2 t^2}{y^2}+1\right)+4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{(s+y)^2}+1\right)-y^2 \log \left(\frac{4 \pi ^2 t^2}{(s+y)^2}+1\right)-8 \pi ^2 t^2 \log (z)+2 z^2 \log (z)+8 \pi ^2 t^2 \log (s+z)-2 z^2 \log (s+z)+8 \pi ^2 t^2 \log (y+z)-2 y^2 \log (y+z)-2 z^2 \log (y+z)-4 y z \log (y+z)-8 \pi ^2 t^2 \log (s+y+z)+2 y^2 \log (s+y+z)+2 z^2 \log (s+y+z)+4 y z \log (s+y+z)-4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{z^2}+1\right)+z^2 \log \left(\frac{4 \pi ^2 t^2}{z^2}+1\right)+4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{(s+z)^2}+1\right)-z^2 \log \left(\frac{4 \pi ^2 t^2}{(s+z)^2}+1\right)+4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{(y+z)^2}+1\right)-y^2 \log \left(\frac{4 \pi ^2 t^2}{(y+z)^2}+1\right)-z^2 \log \left(\frac{4 \pi ^2 t^2}{(y+z)^2}+1\right)-2 y z \log \left(\frac{4 \pi ^2 t^2}{(y+z)^2}+1\right)+(s+2 \pi  t+y+z) (s-2 \pi  t+y+z) \log \left(\frac{4 \pi ^2 t^2}{(s+y+z)^2}+1\right)\right)$

There is no essential difficulty of calculating $K_2$ and $K_3$ in spite of their complicated forms (due to which I used Mathematica to generate the result). Take $K_2$ for example, just apply Feynman's trick again on $s$ and use the lemma again. For $K_3$, just differentiate $s$ twice, use the lemma, and integrate back twice (with good care of boundary values taken). The method described here should work for all $m>3$, since essentially we are integrating nothing but the forms $\int s^k \log(s^2+a^2) ds$ and $\int s^k \tan^{-1}(s)ds$, which will be expressible by $\log, \tan^{-1}$ again (no matter how harge $m$ is) without using polylog terms (like in, for instance, this question).

For OP's convenience, this command should generate the expression of $K_3$.

```<code>Integrate[(1 - Exp[-s w]) (1 - Exp[-y w]) (1 - Exp[-z w])/w^3*
Cos[2 Pi t w], {w, 0, Infinity},
Assumptions -> s > 0 && y > 0 && z > 0 && t > 0] // FullSimplify
</code>```

One may use them to write down explicit forms of IFT for certain $m$ and parameters. End of story.

Body:
I shall follow the convention that IFT of $F$ is  $\int_{\mathbb R}F(w)e^{2\pi i w t}dw$, take a scaling if necessary. First of all, a crucial lemma $$\int_0^{\infty } \frac{e^{-aw} \cos (b w)-e^{-cw} \cos (d w)}{w} \, dw=\frac{1}{2} \log \left(\frac{c^2+d^2}{a^2+b^2}\right)$$
Which is direct by applying Feynman's trick on either $a$ or $c$. Now, for case $m=1$ one have the IFT equals
$$I_1=\int_{\mathbb R} \frac{1-\sum_i a_i e^{-r_1 b_i |w|}}{|w|}e^{2\pi i w t}dw=2\int_{\mathbb R^+} \frac{1-\sum_i a_i e^{-r_1 b_i w}}{w}\cos(2\pi t w)dw=\sum_i a_i K_1(r_1 b_i)$$
Where we have took a reflection on part $\mathbb R^-$, applied Euler formula and rewritten $1$ as $\sum_i a_i$. Now the lemma gives
$$K_1(s)=\int_{\mathbb R^+}\frac{1-e^{-s w}}{w}\cos(2\pi t w)dw=\frac{1}{2} \left(\log \left(\frac{s^2}{4 t^2}+\pi ^2\right)-2 \log (\pi )\right)$$
Which completes the evaluation of this case. Note that this can be trivially generalized to cases that $m=2, 3$, namely elementary closed forms of IFT that $I_2=\sum_{i,j} a_i a_j K_2(r_1 b_i, r_2 b_j)$ and $I_3=\sum_{i,j,k} a_i a_j a_k K_3(r_1 b_i, r_2 b_j, r_3 b_k)$, where

$\small K_2(s,y)=\int_{\mathbb R^+}\frac{1-e^{-s w}}{w}\frac{1-e^{-y w}}{w}\cos(2\pi t w)dw=-\frac{1}{2} s \log \left(s^2+4 \pi ^2 t^2\right)+\frac{1}{2} s \log \left((s+y)^2+4 \pi ^2 t^2\right)+\frac{1}{2} y \log \left((s+y)^2+4 \pi ^2 t^2\right)-2 \pi  t \tan ^{-1}\left(\frac{2 \pi  t}{s+y}\right)+2 \pi  t \tan ^{-1}\left(\frac{2 \pi  t}{s}\right)-\frac{1}{2} y \log \left(4 \pi ^2 t^2+y^2\right)+2 \pi  t \tan ^{-1}\left(\frac{2 \pi  t}{y}\right)-\pi ^2 t$

$\scriptsize K_3(s,y,z)=\int_{\mathbb R^+}\frac{1-e^{-s w}}{w}\frac{1-e^{-y w}}{w}\frac{1-e^{-z w}}{w}\cos(2\pi t w)dw=\frac{1}{4} \left(2 \log (s) s^2+\log \left(\frac{4 \pi ^2 t^2}{s^2}+1\right) s^2-2 \log (s+y) s^2-\log \left(\frac{4 \pi ^2 t^2}{(s+y)^2}+1\right) s^2-2 \log (s+z) s^2+2 \log (s+y+z) s^2-\log \left(\frac{4 \pi ^2 t^2}{(s+z)^2}+1\right) s^2-8 \pi  t \tan ^{-1}\left(\frac{2 \pi  t}{s}\right) s+8 \pi  t \tan ^{-1}\left(\frac{2 \pi  t}{s+y}\right) s+8 \pi  t \tan ^{-1}\left(\frac{2 \pi  t}{s+z}\right) s-8 \pi  t \tan ^{-1}\left(\frac{2 \pi  t}{s+y+z}\right) s-4 y \log (s+y) s-2 y \log \left(\frac{4 \pi ^2 t^2}{(s+y)^2}+1\right) s-4 z \log (s+z) s+4 y \log (s+y+z) s+4 z \log (s+y+z) s-2 z \log \left(\frac{4 \pi ^2 t^2}{(s+z)^2}+1\right) s-8 \pi  t y \tan ^{-1}\left(\frac{2 \pi  t}{y}\right)+8 \pi  t y \tan ^{-1}\left(\frac{2 \pi  t}{s+y}\right)-8 \pi  t z \tan ^{-1}\left(\frac{2 \pi  t}{z}\right)+8 \pi  t z \tan ^{-1}\left(\frac{2 \pi  t}{s+z}\right)+8 \pi  t y \tan ^{-1}\left(\frac{2 \pi  t}{y+z}\right)+8 \pi  t z \tan ^{-1}\left(\frac{2 \pi  t}{y+z}\right)-8 \pi  t y \tan ^{-1}\left(\frac{2 \pi  t}{s+y+z}\right)-8 \pi  t z \tan ^{-1}\left(\frac{2 \pi  t}{s+y+z}\right)-8 \pi ^2 t^2 \log (s)-4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{s^2}+1\right)+8 \pi ^2 t^2 \log (2 \pi  t)-8 \pi ^2 t^2 \log (y)+2 y^2 \log (y)+8 \pi ^2 t^2 \log (s+y)-2 y^2 \log (s+y)-4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{y^2}+1\right)+y^2 \log \left(\frac{4 \pi ^2 t^2}{y^2}+1\right)+4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{(s+y)^2}+1\right)-y^2 \log \left(\frac{4 \pi ^2 t^2}{(s+y)^2}+1\right)-8 \pi ^2 t^2 \log (z)+2 z^2 \log (z)+8 \pi ^2 t^2 \log (s+z)-2 z^2 \log (s+z)+8 \pi ^2 t^2 \log (y+z)-2 y^2 \log (y+z)-2 z^2 \log (y+z)-4 y z \log (y+z)-8 \pi ^2 t^2 \log (s+y+z)+2 y^2 \log (s+y+z)+2 z^2 \log (s+y+z)+4 y z \log (s+y+z)-4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{z^2}+1\right)+z^2 \log \left(\frac{4 \pi ^2 t^2}{z^2}+1\right)+4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{(s+z)^2}+1\right)-z^2 \log \left(\frac{4 \pi ^2 t^2}{(s+z)^2}+1\right)+4 \pi ^2 t^2 \log \left(\frac{4 \pi ^2 t^2}{(y+z)^2}+1\right)-y^2 \log \left(\frac{4 \pi ^2 t^2}{(y+z)^2}+1\right)-z^2 \log \left(\frac{4 \pi ^2 t^2}{(y+z)^2}+1\right)-2 y z \log \left(\frac{4 \pi ^2 t^2}{(y+z)^2}+1\right)+(s+2 \pi  t+y+z) (s-2 \pi  t+y+z) \log \left(\frac{4 \pi ^2 t^2}{(s+y+z)^2}+1\right)\right)$

There is no essential difficulty of calculating $K_2$ and $K_3$ in spite of their complicated forms (due to which I used Mathematica to generate the result). Take $K_2$ for example, just apply Feynman's trick again on $s$ and use the lemma again. For $K_3$, just differentiate $s$ twice, use the lemma, and integrate back twice (with good care of boundary values taken). The method described here should work for all $m>3$, since essentially we are integrating nothing but the forms $\int s^k \log(s^2+a^2) ds$ and $\int s^k \tan^{-1}(s)ds$, which will be expressible by $\log, \tan^{-1}$ again (no matter how harge $m$ is) without using polylog terms (like in, for instance, this question).

For OP's convenience, this command should generate the expression of $K_3$.

Integrate[(1 - Exp[-s w]) (1 - Exp[-y w]) (1 - Exp[-z w])/w^3*
Cos[2 Pi t w], {w, 0, Infinity},
Assumptions -> s > 0 && y > 0 && z > 0 && t > 0] // FullSimplify

One may use them to write down explicit forms of IFT for certain $m$ and parameters. End of story.


问:
Asymptotic expansion of the inverse of $x\mapsto x+x^{\small\sqrt2}+x^2$ near zero
This is a follow-up to my previous question "Asymptotic expansion of the inverse of $x\mapsto x+x^\phi$ near zero".

Consider a continuous real-valued monotone increasing function $f:\mathbb R^+\to\mathbb R^+$ satisfying $f\big(x+x^{\small\sqrt2}+x^2\big)=x.$

I am interested in an asymptotic expansion of $f(z)$ for $z\to0^+$ in terms of powers of $z$. I was able to find a few initial terms by manually balancing coefficients:
$$f(z)=z-z^{\small\sqrt2}+\sqrt2\;z^{\small\unicode{x202f}2\unicode{x202f}\sqrt2-1}-z^2+\mathcal O\big(z^{\small\unicode{x202f}3\unicode{x202f}\sqrt2-2}\big), \quad z\to0^+.\tag{$\diamond$}$$
Computing next terms in an ad hoc fashion quickly becomes tedious, so I am looking for a more systematic approach that would allow to obtain a general formula for the terms of this series. I expect it to be a mix of integer powers of $z$ and irrational powers involving $\sqrt2$. Also, I would like to know the radius of convergence of that series.

More generally, I am looking for a uniform approach for inverting generalized polynomials of a single variable that may contain both rational and irrational powers.

答:
Some thoughts

Conjecture: When $y>0$ is sufficiently small, the unique positive real solution of $x + x^{\sqrt{2}} + x^2 = y$ is given by
\begin{align}
x &= y + \sum_{k=1}^\infty \frac{(-1)^k}{k!} \frac{\mathrm{d}^{k-1}}{\mathrm{d} y^{k-1}}\Big[(y^{\sqrt{2}} + y^2)^k\Big]\\
&= y - (y^{\sqrt{2}} + y^2) + (\sqrt{2}y^{2\sqrt{2}-1} + (2+\sqrt{2})y^{\sqrt{2}+1} + 2y^3) - \cdots \tag{1}
\end{align}

Motivation: The equation is written as $x = y + (-1)(x^{\sqrt{2}} + x^2)$. By the Lagrange reversion theorem (https://en.wikipedia.org/wiki/Lagrange_reversion_theorem),
we guess (1).

Numerical experiment using Maple: When $y = \frac{1}{100}$, the solution of $x + x^{\sqrt{2}} + x^2 = y$ is
$x_0 \approx 0.008704284007$,
while $y + \sum_{k=1}^{20} \frac{(-1)^k}{k!} \frac{\mathrm{d}^{k-1}}{\mathrm{d} y^{k-1}}\Big[(y^{\sqrt{2}} + y^2)^k\Big]
\approx 0.008704284006$.

However, when $y$ is larger, it may fail. For example, $y = 1/2$,
$y + \sum_{k=1}^{5} \frac{(-1)^k}{k!} \frac{\mathrm{d}^{k-1}}{\mathrm{d} y^{k-1}}\Big[(y^{\sqrt{2}} + y^2)^k\Big]\approx -19.46221804$.

So I guess we need sufficiently small $y$.

Further work: 1) What is the ROC of the series (1)? 2) Prove (or disprove) if it is the solution.

ContentLicense: CC BY-SA 4.0


问:
How do I sell out with abstract algebra?
My plan as an undergraduate was unequivocally to be a pure mathematician, working as an algebraist as a bigshot professor at a bigshot university.  I'm graduating this month, and I didn't get into where I expected to get into.  My letters were great and I'm published, but my GRE was bad and my grades were good but not perfect.  My current plan, I guess, is to start a PhD program at my backup school, then reapply to the better schools next year.

Reality is starting to hit, though, and I'm starting to think about "selling out."   I would still love to work in algebra, but I'm not as in love with the Ivory Tower as I was a few years ago, and I don't want to give up my entire life for it.  If the institution isn't going to let me do what I wanted to do, or if I'll never be as talented as I wanted to be, it isn't worth the sacrifice.  In other words, I'd rather be a well paid applied mathematician in industry than a poor, mediocre pure mathematician at a low end university.

The problem is it seems that most of the applied jobs out there are all about analysis / continuous mathematics, and I am firmly in the algebra / discrete camp. I really do not want to spend my life solving fluid flow PDEs. I always hear about cryptography as an "applied algebra" job, but I'm not particularly crazy about working for NSA or a telecom (plus crypto can't be the only option).

I read some of the answers from Can I use my powers for good? but it's not clear to me which of these suggestions value algebraic thinking.  Many seem very quantitative, rather than structural - is it possible to avoid this in industry?  Also, I have a lot of debt from a long undergraduate career across several majors, so "how much" is unfortunately also a concern.  I don't want to sell out cheap.

Are there applied math jobs in industry which focus on structural mathematics reminiscent of abstract algebra, earn an appreciably high salary, and aren't cryptography?

How would one best go about pursuing these jobs starting as a recent graduate / first year graduate student?

答:
NSA is not the only place for cryptographers. There are various research labs like HP   , Microsoft Research  labs where cryptographers and number theorists are hired.

Also, algebraic coding theory is another area where you can apply your skills of abstract algebra.

ContentLicense: CC BY-SA 3.0


问:
General Linear Group is a manifold
Already  seen a bunch of questions about proving that $(GL_{n}(\mathbb{R}),||-||)$ is a manifold, with the norm $||-||:M_{n}(\mathbb{R})\longrightarrow \mathbb{R}$ with $||(a_{i,j})||=\sqrt{A^tA}=\sqrt{\sum_{j=1}^{n}(\sum_{i=1}^{n}a_{i,j}^{2})}$, ($M_{n}(\mathbb{R})\cong\mathbb{R}^{n^2}$). None of them give any hints on how to prove it is a topological space ANII (or second countable) and that it is locally Euclidean.

For the first property, i have to prove that there exists a base which is countable, i.e. there exists a countable set $\{\mathcal{U}_{i}\}_{i\in I}$, $\mathcal{U}_{i}\subset GL_{n}(\mathbb{R})$ open, such that every open set in $GL_{n}(\mathbb{R})$ is written as $\mathcal{O}\subset GL_{n}(\mathbb{R}), \mathcal{O}=\cup_{i\in I}\mathcal{U}_{i}$. Is that true?

Also, i have no idea on how to see it is locally Euclidean.

I supose we have to use at any point the norm defined previously, although I have no idea how.

Any help will work, thanks.

答: $GL_n(\Bbb R)$ is an open subset of $M_n(\Bbb R)\cong \Bbb R^{n^2}$, namely the inverse image of $\Bbb R\setminus\{0\}$ under the continuous $\det$ map.

Title: General Linear Group is a manifold


答: Hint: Think of each element of $GL_n$, an $n \times n$ matrix, as a list of $n^2$ numbers (read left-to-right, top-to-bottom, for instance). Now you can see that $GL_n$ is just a (very large) subset of $X = \Bbb R^{n^2}$. That helps a lot with proving things. Because you can take "all balls in $X$ whose centers have rational-number coordinates, and whose radii are rational, and which contain no matrices of determinant zero" as your collection of open sets to use.

Title: General Linear Group is a manifold

Body:
Already  seen a bunch of questions about proving that $(GL_{n}(\mathbb{R}),||-||)$ is a manifold, with the norm $||-||:M_{n}(\mathbb{R})\longrightarrow \mathbb{R}$ with $||(a_{i,j})||=\sqrt{A^tA}=\sqrt{\sum_{j=1}^{n}(\sum_{i=1}^{n}a_{i,j}^{2})}$, ($M_{n}(\mathbb{R})\cong\mathbb{R}^{n^2}$). None of them give any hints on how to prove it is a topological space ANII (or second countable) and that it is locally Euclidean.

For the first property, i have to prove that there exists a base which is countable, i.e. there exists a countable set $\{\mathcal{U}_{i}\}_{i\in I}$, $\mathcal{U}_{i}\subset GL_{n}(\mathbb{R})$ open, such that every open set in $GL_{n}(\mathbb{R})$ is written as $\mathcal{O}\subset GL_{n}(\mathbb{R}), \mathcal{O}=\cup_{i\in I}\mathcal{U}_{i}$. Is that true?

Also, i have no idea on how to see it is locally Euclidean.

I supose we have to use at any point the norm defined previously, although I have no idea how.

Any help will work, thanks.


问:
How does this equation relate to a Gaussian distribution and what does the comma mean?
Before the following equation, the text says "The note transcription method takes as an input the
pitch track and outputs discrete notes on a continuous pitch
scale, based on Viterbi-decoding of a second, independent
hidden Markov model (HMM). [...] The likelihood of a non-silent state emitting a pitch track frame with pitch q is modelled as a Gaussian distribution centered at the note’s pitch p with a standard deviation of semitones, i.e.

enter image description here

where np is a state modelling the MIDI pitch p, z is
a normalising constant and the parameter 0 < τ < 1
controls how much the pitch estimate is trusted; we set
τ = 0:1. The probability of unvoiced states is set to
P(unvoiced|q) = (1 - v)=n, i.e. they sum to their combined
likelihood of (1 - v) and v = 0.5 is the prior likelihood
of a frame being voiced. The standard deviation
varies depending on the state: attack states have a larger
standard deviation (σ = 5 semitones) than stable parts (σ= 0.9).

I cannot see how this relates to these equation for Gaussian functions listed on Wikipedia:

enter image description here

enter image description here

Am I misunderstanding or can someone explain the relationship?

Moreover, does the comma in the equation mean, "or"?

答:
My guess is that they use just $\phi_{p,\sigma}(x)$ to mean:

$$\phi_{p,\sigma}(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-p)^2}{2\sigma^2}}$$

or something similar. The comma is there just to separate $p$ and $\sigma$, nothing else.

However, I feel you have not provided enough context, as there is further multiplication with $v$ (?!), division with $z$ (?!) and something denoted as $(\cdot)^{\tau}$ (?!). I am just wary that there might be more to this, hidden in this missing context.

Hope this helps.


问:
Calculating the value of $f'(x)$ if it exists.

Note that I am not looking for a solution, just a general advice on how to approach these questions and practical tips.

We've just started differentiability and I'm completely lost. I'm asked to calculate the value of $f'(x)$ at $x_0$ or prove it doesn't exist. But we haven't done much practice and only focused on theorems so far so I'm not sure how to approach these questions.

Are there any special tricks regarding calculating differential function limits or is it basically the same as ordinary limits?

$f\left(x\right)\ =\ \left|x^{2}-3x-4\right|,\ \ \ x_{0}=4$

$\lim _{x\to 4}\frac{\left(f\left(x\right)-f\left(4\right)\right)}{x-4} =\ \lim _{x\to 4}\frac{\left|x^2-3x-4\right|-0}{x-4}$

$\lim _{x\to 4^+}\frac{x^2-3x-4}{x-4} =\ \lim _{x\to 4^+}\frac{\left(x-4\right)\left(x-3\right)+4x-16}{x-4} =\ \lim _{x\to 4^+}\frac{\left(x-4\right)\left(x-3\right)+4\left(x-4\right)}{x-4} $

$=\ \lim _{x\to 4^+}\left(x-3\right)+4 =\ \lim _{x\to 4^+}\left(4-3\right)+4 = 5$

$\lim _{x\to 4^-}\frac{-x^2+3x+4}{x-4} =\ \lim _{x\to 4^-}\frac{\left(x-4\right)\left(3-x\right)-4x+16}{x-4} =\ \lim _{x\to 4^-}\frac{\left(x-4\right)\left(3-x\right)-4\left(x-4\right)}{x-4} $

$=\ \lim _{x\to 4^-}\left(3-x\right)-4 = \left(3-4\right)-4 = -5$

Therefore the limit doesn't exists, and f is not differentiable at 4.

b.

$f\left(x\right)\ =\ \sqrt[3]{x^{2}-\left|x\right|},\ \ x_{0}\ =0$

$\lim \:_{x\to \:0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0} =\ \lim \:_{x\to \:0}\frac{\left(\sqrt[3]{x^2-x}-0\right)}{x}$

$\lim _{x\to 0^+}\frac{\sqrt[3]{x^2-x}}{x}\cdot \frac{\sqrt[\frac{2}{3}]{x^2-x}}{\sqrt[\frac{2}{3}]{x^2-\:x}}=\lim \:_{x\to 0^+}\frac{\left(x^2-x\right)}{x\left(\sqrt[\frac{2}{3}]{x^2-\:x}\right)}=\lim \:_{x\to \:0^+}\frac{x-1}{\sqrt[\frac{2}{3}]{x^2-\:x}}=\frac{0-1}{\sqrt[\frac{2}{3}]{0^+}}=-∞$

A one sided limit doesn't exist, therefore f is not diffirintable at 0.

c.

$f\left(x\right)=
 \begin{cases}
      \frac{\sin x^{2}}{x} & if \ x ≠ 0\\
       0 & if \ x = 0 \\
   \end{cases}
\
, \ x_{0}\ =\ 0$

$\lim \:\:_{x\to \:\:0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0} =\ \lim \:\:_{x\to \:\:0^{ }}\frac{f\left(x\right)}{x}
$

$\lim _{x\to 0^+}\frac{\left(\frac{sin\left(x^2\right)}{x}\right)}{x}=\lim \:_{x\to 0^+}\left(\frac{sin\left(x^2\right)}{x^2}\right)=1$

$\lim _{x\to 0^-}\frac{\left(\frac{sin\left(x^2\right)}{x}\right)}{x}=\lim \:_{x\to 0^-}\left(\frac{sin\left(x^2\right)}{x^2}\right)=1$

Therefore the limit exists and f is differentiable at 0.

答:
In both cases you can analyze the difference quotient $\frac{f(x)-f(x_0)}{x-x_0}$ which is the basis of the definition of the derivative. As $f(x_0)=f(0)=0$ in both cases, this simplifies to
$$
-\frac{\sqrt[3]{|x|-x^2}}x=-\sqrt[3]{\frac1{x|x|}-\frac1x}
$$
which diverges at $x=0$ and
$$
\frac{\sin(x^2)}{x^2}
$$
which has a well-known limit at $x=0$.


问:
Find all positive integer solutions for the following equation:
Find all positive integer solutions for the following equation:

$(x^2+1)(y^2+1)+2(x-y)(1-xy)=4(1+xy)$

I've tried simplifying the equation and then refactoring but I can't find any solutions.

答:
The problem is that you are trying to solve this by simplifying instead of substituting for x and y. Hint to this problem is to first look at the graph of it on desmos perhaps. From this you can see the only integer solutions will ever be found on the axis'. Hence where x=0 or where y=0. This is the easiest method to do this of course. Doing this we can get:
When x=0 -> (y^2 +1)-2y =4  -> y^2-2y-3=0 use quadratic formula to solve and get:
y=-3 and y=1  two sols when x=0: (-3,0) and (1,0)

Now for the solutions when y=0
same process
y=0 -> (x^2+1)+2x =4 follow the same steps as earlier to get solutions where y=0: (0,3) and (0,-1)

Basically you had the right idea you simply needed to solve to place one variable as a zero first. Repeating the process with each variable. Hopefully this clears things up. If you want to know the reason why we can say there aren't any other integer solutions besides the intercept solutions would be a little more difficult/

Just remember that it asked for positive solutions. Therefor (0,3) and (1,0) are the only cases that work here.


问:
Finding the roots of $x^{2/3} + x ^{1 / 3} − 2$

If the radical expression given to us is $$x^{2/3}  + x ^{1 / 3} − 2$$ now the question is to find its roots.

I want to know: Can we multiply the whole equation in this way
$$x^2 + x - 8$$ I.e. each term by ^3
or I have to take its whole cube (which can be really lengthy).

Please do share if there is any other method I could use as well.

答:
Hint:

I presume the equation is $$x^{2/3}+x^{1/3}-2=0$$

How have  you reached to $x^2+x-8=0?$

Better use the following substitution:

$$x^{1/3}=y\implies x=y^3$$

$$x^{2/3}=(x^{1/3})^2=?$$  to form a quadratic equation $y$


问:
$f(z)$ has a pole if $\lim_{z\to a}\, f(z) = \infty$
here is the statement that I've found in 2 books:

If $\lim_{z\to a}f(z)=\infty$, the point $a$ is said to be a pole of $f(z)$, and we set $f(a) = \infty$

Intuitively I understand about pole. For example $\frac{1}{z-2}$ has a simple pole at the point $2$. But I don't know how do we get $\infty$ when we take the limit $z$ goes to $2$ since in the real plane this function has no limit cz the left and the right side of limit aren't the same. That's my point. So, how does it have $\infty$ as its limit if that function itself has a pole in the complex plane? So, What's the meaning of that statement actually?

It's been a year I didn't study about complex analysis. Hope you can help me to understand this.

Thanks.

答: It is like the one-point compactification of $\mathbb C$. The topology on $\mathbb C\cup \{\infty\}$ is $\{X:X\subseteq \mathbb C\text { and $X$ is open, or $X^c$ is compact}\}$. Then you can verify that $\lim_{z\to a} f(z)=\infty$ iff $\lim_{z\to a}|f(z)|=+\infty$.


问:
How to find the factorial of a fraction?
From what I know, the factorial function is defined as follows:

$$n! = n(n-1)(n-2) \cdots(3)(2)(1)$$

And $0! = 1$. However, this page seems to be saying that you can take the factorial of a fraction, like, for instance, $\frac{1}{2}!$, which they claim is equal to $\frac{1}{2}\sqrt\pi$ due to something called the gamma function. Moreover, they start getting the factorial of negative numbers, like $-\frac{1}{2}! = \sqrt{\pi}$

How is this possible? What is the definition of the factorial of a fraction? What about negative numbers?

I tried researching it on Wikipedia and such, but there doesn't seem to be a clear-cut answer.

答:
The gamma function, shown with a Greek capital gamma $\Gamma$, is a function that extends the factorial function to all real numbers, except to the negative integers and zero, for which it is not defined. $\Gamma(x)$ is related to the factorial in that it is equal to $(x-1)!$. The function is defined as

$$\Gamma(z) = \frac{1}{z} \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}$$

Simply use this to compute factorials for any number. A handy way of calculating for real fractions with even denominators is:

$$\Gamma(\tfrac12 + n) = {(2n)! \over 4^n n!} \sqrt{\pi}$$

Where n is an integer. But keep in mind that the gamma function is actually the factorial of 1 less than the number than it evaluates, so if you want $\frac{3}{2}!$ use n = 2 instead of 1.

Or, you could just put the fraction into Google Calculator, which uses the gamma function to evaluate factorials of any number.

For some more examples of the gamma function's values, see here.

(If you don't understand this, don't worry, because I don't either, and the Wikipedia article on the function seems to lack a clear-cut definition of it or how it relates to $\sqrt{\pi}$.)


问:
Convex hull of points
We're studying a course and we're dabbling a bit in convex optimization.
We were told that the convex hull of a set of points x1,..,xn is a polytope and we were asked to show that any face of the convex hull must contain at least one of these points.

We defined a face to be the intersection of the polytope with a supporting hyperplane. So according to the definitions in this course the empty set is not considered a face.

I also understand that a subset of x1,..,xn must be the vertices but we aren't meant to use this property. any suggestions on how to go about proving this?

答: This is an intuition-jogger, so if at any point reading this you get the picture of the proof, feel free to finish it yourself. Every point is a convex combination of your vertices. So take an arbitrary point in a given face and write it as such. Visualise if you have to, to make a conjecture about some vertices of the convex hull having to be in the face. As you've written, a supporting hyperplane corresponding to this face is defined in terms of the inner product: for some $a$ in your vector space and $b\in\mathbb{R}$, it is the set of all points $x$ with $\langle a,x\rangle=b$. Substitute your point in the face into this expression. Use the linearity properties of the inner product in such a way that they play nicely with the convex combination. Use the fact that for all points $p$ in the convex hull $\langle a,p\rangle\leq b$ to check your conjecture.

Title: Convex hull of points


问:
Combinatorics problem about divisors
There is a combinatorics problem that need some Number Theory background:

How many numbers there are such that is divisible by 30 and have exactly 30 divisors?

I need some idea how to solve this problem.

答:
$30$ can factorize : $30=2*3*5$
so the number with mentioned conditions should have prime factors with powers $1,2,4$. since $3$ prime factors are obviously $2,3,5$. there are $ 3!=6 $ such numbers.

Title: Combinatorics problem about divisors


问:
Ito-Doeblin Formula (a step involved in the derivation)
This is an abbreviated version of Exercise 4.14 from Stocastic Calculus for Finance II by Steven Shreve but the context of the question should be sufficient to answer the question.

Define $Z_j = f''(W(t_j))[(W(t_{j+1}) - W(t_j))^2 - (t_{j+1} - t_{j})]$

where $W(t)$ is a Brownian motion , $t \geq 0.$ Show that $\mathbb{E}\sum_{j=0}^{n-1}Z_j = 0.$

My attempt:

\begin{align}
\mathbb{E}\sum_{j=0}^{n-1}Z_j
&= \mathbb{E}\sum_{j=0}^{n-1}f''(W(t_j))[W(t_{j+1}) - W(t_j)]^2 - \mathbb{E}\sum_{j=0}^{n-1}f''(W(t_j))(t_{j+1}-t_j)\\
&= \sum_{j=0}^{n-1}\mathbb{E}f''(W(t_j))\mathbb{E}[W(t_{j+1}) - W(t_j)]^2 - \mathbb{E}\sum_{j=0}^{n-1}f''(W(t_j))(t_{j+1}-t_j)\\
&= \sum_{j=0}^{n-1}\mathbb{E}f''(W(t_j))(t_{j+1}-t_j)  - \mathbb{E}\sum_{j=0}^{n-1}f''(W(t_j))(t_{j+1}-t_j)\\
&= 0.
\end{align}
where in the second equality, I have used the fact that $f''(W(t_j))$ is $\mathcal{F}(t_j)$- measurable while $[W(t_{j+1}) - W(t_j)]^2$ is independent of $\mathcal{F}(t_j)$ and in order to get the third equality, I have simply plugged in the value of $\mathbb{E}[W(t_{j+1}) - W(t_j)]^2$ which is $(t_{j+1}-t_{j}).$ Please comment if I'm right.

The official solution goes as follows:

$Z_j$ is $\mathcal{F}(t_{j+1})$- measurable and $\mathbb{E}[Z_j|\mathcal{F}(t_j)] = 0.$ I'm okay with both these steps and have easily proved it. But then they do the following working which I'm not clear with:

$\mathbb{E}\big[\sum_{j=0}^{n-1}Z_j\big] = \mathbb{E}\big[\sum_{j=0}^{n-1}\mathbb{E}[Z_j|\mathcal{F}(t_j)]\big] = 0.$

My question regarding their solution is the following:

$Z_j$ is $\mathcal{F}(t_{j+1})$- measurable so $\mathbb{E}[Z_j|\mathcal{F}(t_{j+1})] = Z_j$ so how can they replace $Z_j$ with $\mathbb{E}[Z_j|\mathcal{F}(t_{j})]$ ?

I know that $\mathbb{E}[Z_j|\mathcal{F}(t_{j})]$ is $\mathcal{F}(t_{j})$- measurable but can't understand the logic used in their solution. Could be a fundamental misunderstanding so any help highly appreciated.

答:
To you question: "Shouldn't the iterated conditioning work this way : $E[E[Z_j|\mathcal F_{t_{j+1}}|\mathcal F_{t_j}]$ since $\mathcal F_{t_j} \subset \mathcal F_{t_{j+1}}$?"

Yes, you are right, but the relevant $\sigma$-algebras here are  $\mathcal F_0$ and $\mathcal F_{t_j},$(not $\mathcal F_{t_{j+1}}$). We have

$$\mathbb{E}\big[\sum_{j=0}^{n-1}Z_j\big] = \mathbb{E}\big[\sum_{j=0}^{n-1}Z_j\big | \mathcal F_0]= \mathbb{E}\big[\sum_{j=0}^{n-1}\underbrace{\mathbb{E}[Z_j|\mathcal{F}(t_j)]}_{=0}\big| \mathcal F_0] = 0,$$
since
$\mathcal F_0 \subset \mathcal F_{t_j}$ and $\mathcal F_0$ is the trivial $\sigma$-algebra.


答:
The gamma function is defined by the following integral, which converges for real $s>0$:
$$\Gamma(s)=\int_0^\infty t^{s-1}e^{-t}dt.$$

The function can also be extended into the complex plane, if you're familiar with that subject. I'll assume not and just let $s$ be real.

This function is like the factorial in the when $s$ is a positive integer, say $s=n$, it satisfies $\Gamma(n)=(n-1)!$. It generalizes the factorial in the sense that it is the factorial for positive integer arguments, and is also well-defined for positive rational (and even real) numbers. This is what it means to take a "rational factorial," but I would hesitate to call it that. Many functions have those two properties, and $\Gamma$ is chosen out of all of them because it is the most useful in other applications. Rather than the notation used in that article you refer to, it would be more accurate for you to say that "the gamma function takes these values for these arguments." Gamma is not a function that intends to generalize factorials; rather, generalizing factorials came along as something of an accident following the definition. Its true purpose is deeper.

As for why $\Gamma(1/2)=\sqrt{\pi}$, this comes out of an interesting property of the $\Gamma$ function: some of them are here http://en.wikipedia.org/wiki/Gamma_function#Properties. The property you are interested in is the reflection formula: $$\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi z)}.$$ Set $z=1/2$ in the formula to get the desired identity.

If you want to learn more about the gamma function, the hard way is to learn a lot more math, in particular real and complex analysis. An easier way is to read this excellent set of notes: http://www.sosmath.com/calculus/improper/gamma/gamma.html.

Title: How to find the factorial of a fraction?

Body:
From what I know, the factorial function is defined as follows:

$$n! = n(n-1)(n-2) \cdots(3)(2)(1)$$

And $0! = 1$. However, this page seems to be saying that you can take the factorial of a fraction, like, for instance, $\frac{1}{2}!$, which they claim is equal to $\frac{1}{2}\sqrt\pi$ due to something called the gamma function. Moreover, they start getting the factorial of negative numbers, like $-\frac{1}{2}! = \sqrt{\pi}$

How is this possible? What is the definition of the factorial of a fraction? What about negative numbers?

I tried researching it on Wikipedia and such, but there doesn't seem to be a clear-cut answer.


问:
Prove that $A\cap B\subseteq A$
I am unsure if I got it right, I proved it as follows:

$$A\cap B \longleftrightarrow x\in A \text{ and } x\in B \text{ is true }\forall x\in A \longleftrightarrow \subseteq A$$

Althought I kind of doubt the last part about it being true for all $x\in A$. Am I missing something?

I apologize in advance if this post is a duplicate.

答: You are done after the first $\longleftrightarrow$ . This means that $x \in A \cap B \implies x \in A$, which satisfies the result by the definition of a subset.

Title: Prove that $A\cap B\subseteq A$


答:
Definition: $x\in A\cap B$ if and only if $x\in A$ and $x\in B$.

If it is false that $A\cap B \subseteq A$, then $\exists x\in A\cap B: x \notin A$. But this contradicts the definition. So, it must be that $A\cap B \subseteq A$.

Title: Prove that $A\cap B\subseteq A$

Body:
I am unsure if I got it right, I proved it as follows:

$$A\cap B \longleftrightarrow x\in A \text{ and } x\in B \text{ is true }\forall x\in A \longleftrightarrow \subseteq A$$

Althought I kind of doubt the last part about it being true for all $x\in A$. Am I missing something?

I apologize in advance if this post is a duplicate.


问:
mixture problem
From Stewart, Precalculus, $5$th ed, p.$71$, q.$55$

The radiator in a car is filled with a solution of $60\%$ antifreeze and $40\%$ water. The manufacturer of the antifreeze suggests that, for summer driving, optimal cooling of the engine is obtained with only $50\%$ antifreeze. If the capacity of the radiator is $ 3.6$L, how much coolant should be drained and replaced with water to reduce the antifreeze concentration to the recommended level?

I don't think there is a need to set up a model, since the currently concentration of antifreeze is at $60\%$ and desired contraction is $50\%$, the difference is $10\%$ of the total capacity of the radiator.

$10\%\times3.6L = 0.36L$

$0.36$L to be drained from the original $60\%$ concentration of $\dfrac{2.16}{3.6}L$ and replaced with water would give $50\%$ concentration at $\dfrac{1.8}{3.6}L$

The answer at the back of the book says $0.6$L to be drained. I don't understand why this is so, if $0.6$L is drained the new concentration would be $43.333\%$.

答:
Another way of addressing the same problem:

Whatever you do, you need to wind up with 1.8 litres of antifreeze in the car ($50\%$ of $3.6$ litres).

You already have $2.16$ litres of antifreeze in the car ($60\%$ of $3.6$ litres).

So you need to remove $0.36  (2.16 - 1.8)$ litres of antifreeze  from the radiator.

The solution in the radiator is a $60\%$ solution: $0.6$ litres of antifreeze in $1$ litre of solution, So:

$0.6$ litres of antifreeze in $1.0$ litre of solution

$1.0$ litres of antifreeze in $\frac{1}{0.6}$ litres of solution

$0.36$ litres of antifreeze in $\frac{0.36}{0.6}$ litres of solution, or $0.6$ litres of solution

So, removing $0.6$ litres of solution and adding $0.6$ litres of water will give the desired result.

Title: mixture problem


问:
Illusion in maths?? (Area of sectors and segments)
I was practising for my maths Olympiad when I stumbled upon this question...

This question is more than just a figure. It actually isn't what it looks like. I could've simply seen the solution at the back and moved further but that question stuck to me.

Yes, you would think at first that $b>a$ but $b$ is actually equal to $a$. I just want to know that why do I perceive it like that while mathematically it's untrue. I did found out the solution myself.

答: @Paul is right. The semi circles aren't drawn to scale; they don't project far enough into the quadrant, exaggerating $b$ while shrinking $a$. (A diagram could easily do the opposite with bulging semi circles, but I suspect the effect wouldn't fool as many.) This may have been deliberate, to encourage algebra while teaching you a valuable lesson.


问:
Prove that $\rho=1/2+it$

Given, $$\dfrac{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^{1-\rho}dx}{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^\bar{\rho}dx} = \dfrac{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^{\rho}dx}{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^{1-\bar{\rho}}dx}$$
where $[x]$ is the greatest integer function and $\rho\in \mathbb{C}$ is a constant , $0<\Re(\rho)<1$.

Prove that $\Re(\rho)=1/2$

My try:

Let
$$ I = \dfrac{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^{1-\rho}dx}{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^\bar{\rho}dx}$$

and $$J= \dfrac{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^{\rho}dx}{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^{1-\bar{\rho}}dx}$$

$$J= \frac{1}{\overline{I}}$$

$$I=J$$ gives $$ I= \frac{1}{\overline{I}}$$

$$\mid I \mid^2 =1$$

$$\mid I\mid=1$$

$$\left|\dfrac{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^{1-\rho}dx}{\int_1^\infty\dfrac{1-x+[x]}{x^2}x^\bar{\rho}dx}\right|=1$$
Note that $\rho=1/2+it_0$ satisfies the above integral.

$$\mid \frac{\int_1^\infty\frac{1-x+[x]}{x^2}x^{1-\rho}dx}{\int_1^\infty\frac{1-x+[x]}{x^2}x^\bar{\rho}dx}\mid=1$$
$$\mid \frac{\int_1^\infty\frac{1-x+[x]}{x^2}x^{1-2{\Re({\rho})}}x^{\overline{\rho}}dx}{\int_1^\infty\frac{1-x+[x]}{x^2}x^\bar{\rho}dx}\mid=1$$
Can we apply mean value theorem for improper integrals to conclude that there exists some $c\in(1,\infty)$ such that
$$c^{1-2\Re(\rho)}=1$$

$$c>1$$ gives $$\Re(\rho)=1/2$$
How to prove $\Re(\rho)=1/2$.?

Title: Prove that $\rho=1/2+it$

Body:
One more time $$F(s)=\int_1^\infty (\lfloor x\rfloor-x+1)x^{-s-1}dx= \frac{\zeta(s)}{s}-\frac{1}{s-1}+\frac1{s}$$

And $$|F(s)|^2=|F(1-s)|^2\ne 0$$ has plenty of solutions away from $\Re(s)=1/2$.

  F(s)= zeta(s)/s+1/s-1/(s-1); t=var('t'); plot(abs(F(0.48+i*t))^2-abs(F(1-0.48-i*t))^2,(13,14.5))

https://cocalc.com/projects/d7fa91cb-9f60-4060-8df0-eeb0459e729d/files/2020-12-21-122308.sagews?session=default


答: These are not semicircles, so the picture is simply wrong

Title: Illusion in maths?? (Area of sectors and segments)

Body:
I was practising for my maths Olympiad when I stumbled upon this question...

This question is more than just a figure. It actually isn't what it looks like. I could've simply seen the solution at the back and moved further but that question stuck to me.

Yes, you would think at first that $b>a$ but $b$ is actually equal to $a$. I just want to know that why do I perceive it like that while mathematically it's untrue. I did found out the solution myself.


答:
By definition, $C \subseteq A$ iff $x \in C \implies x \in A$.

Apply this to $C = A \cap B$:

$x \in A \cap B \implies x \in A \text{ and } x \in B \implies x \in A$

Title: Prove that $A\cap B\subseteq A$


问:
How to Calculate Integral $\int_0^1 \theta^{-1} (1 - \theta)^{n-1} d\theta$
Can anyone give any hints on how to calculate

$$\begin{equation}
   \int_0^1 \theta^{-1} (1-\theta)^{n-1} d\theta
\end{equation}$$
?

Thank You so much!

答:
The integral as currently stated is divergent. If you meant to have different bounds, this is how you can express the indefinite integral:

$$\int \frac{(1-x)^{n-1}}{x} dx \\ =\int \frac 1x \sum_{k=0}^{n-1} {{n-1}\choose k} (-x)^k dx \\ =\int\frac{dx}{x} +\int\sum_{k=1}^{n-1} {{n-1}\choose k}(-1)^k x^k dx \\ =\ln x +\sum_{k=1}^{n-1}{{n-1}\choose k}(-1)^k\frac{x^{k+1}}{k+1}
$$ See how evaluating the term at $x=0$ paired with the minus sign would give $-\ln(0) =\infty$?


问:
ABCD is a square, M, N, E are the midpoint of AD, DC, DN respectively, J is the intersection point of BE and AN.
ABCD is a square, M, N, E are the midpoint of AD, DC, DN respectively, J is the intersection point of BE and AN.
Prove:

a) BM perpendicular AN

b) MJ perpendicular BE

a) is an easy one, I solved it already.

b) I can finish it if AB = BJ is proved.

Or ∆DJN is a right angle

Or DJ parallel BM

Or by calling F is midpoint of BC and $J\in DF$

But I can't prove any of them

Please help me one

答:
Note that $DM^2=DE\cdot DC$. So $DA$ is tangent to $(CEM)$. But we can also see that $DA$ is tangent to $(BMC)$. So we get that $C$, $M$ lie on the circle with diameter $\overline{BE}$.

So the reflection of $E$ over $BM$ lies on $AB$. Since $AJ\perp BM$, it follows that the reflection of $J$ over $BM$ is $A$. Hence $\angle MJB=\angle BAM=90^{\circ}$.


问:
Find: The expected number of urns that are empty
A total of $n$ balls, numbered $1$ through $n$, are put into $n$ urns, also numbered $1$ through $n$ in such a way that ball $i$ is equally likely to go into any of the
urns $1, 2, . . . , i$. Find the expected number of urns that are empty.

Can somebody help me? I don´t understand.

Thanks very much.

答:
We proceed by induction.  For the case $n=1$, it is clear that the expected value of empty urns is $0$.  Now, suppose that with $n$ urns our expected value, $E(n)$, is given by $(n-1)/2$.

For the case $n+1$, we can think of our procedure in two steps:

Place $n$ balls (according to the rule for $n$ balls) into the first $n$ urns.

Place one ball in any urn.

After step 1, there are on average $E(n)+1$ empty urns.  As the last ball will end up in any urn with equal probability, there is a probability of $(E(n)+1)/(n+1)$ that our final ball will land in an empty urn.  Subtracting, there remains a $(n-E(n))/(n+1)$ chance that the final ball will land in a previously-occupied urn.  That is, our expected value on empty urns is given by

$$E(n+1)=E(n) \left(\frac{E(n)+1}{n+1}\right)+(E(n)+1)\left(\frac{n-E(n)}{n+1}\right)=\frac{n-1}{2}\cdot \frac{1}{2} + \frac{n+1}{2}\cdot \frac{1}{2},$$
in which we've used the inductive hypothesis $E(n)=(n-1)/2$.  The right-hand side of the expression above is $n/2$, and this proves the $n+1$ case (so that we conclude by induction).

I'd like to point out that I would not have known how to formulate this induction without doing some cases by hand.  I noticed that this pattern held for $n \leq 4$ before attempting the proof you see above.


答: Let the side length of the square be $4x$, you can see $EC=3x$ and therefore $BE=5x$. However since ${BJ\over JE}={AB\over EN}={4\over 1}$, we now $BJ=4x=AB$.

Body:
ABCD is a square, M, N, E are the midpoint of AD, DC, DN respectively, J is the intersection point of BE and AN.
Prove:

a) BM perpendicular AN

b) MJ perpendicular BE

a) is an easy one, I solved it already.

b) I can finish it if AB = BJ is proved.

Or ∆DJN is a right angle

Or DJ parallel BM

Or by calling F is midpoint of BC and $J\in DF$

But I can't prove any of them

Please help me one


问:
Primes approximated by eigenvalues?
Consider the infinite matrix starting:

$$\displaystyle T = -\left(   \begin{array}{ccccccc}   +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array}   \right)$$

defined by either the recurrence:

$$\displaystyle T(n,1)=-1, T(1,k)=-1, n>=k:T(n,k) = -\sum\limits_{i=1}^{k-1} T(n-i,k), n<k:T(n,k) = -\sum\limits_{i=1}^{n-1} T(k-i,n)$$

or:

A191898=A051731*transpose(A143256)

or:

$$\displaystyle T(n,k) = -a(GCD(n,k))$$

where $a$ is the Dirichlet inverse of the Euler totient function.

Does the largest eigenvalue of the matrix T(n,k) approximate the previous prime number sequence?

The signs of the eigenvalues appear to agree with the Möbius function.

As a Mathematica program this is:

```<code>Clear[b, t, n, k, i, j]
t[n_, 1] = -1;
t[1, k_] = -1;
t[n_, k_] :=
  t[n, k] =
   If[n >= k, -Sum[t[n - i, k], {i, 1, k - 1}], -Sum[
      t[k - i, n], {i, 1, n - 1}]];
nn = 42;
b = Range[1, nn]*0;
Do[m = Table[Table[t[n, k], {k, 1, j}], {n, 1, j}];
 b[[j]] = RankedMax[Eigenvalues[m], 1], {j, 1, nn}]
Round[b]
Table[NextPrime[i, -1], {i, 2, 43}]
</code>```

with the largest eigenvalues, rounded:

{-1, 1, 3, 3, 5, 5, 7, 7, 7, 8, 11, 11, 13, 13, 13, 13, 17, 17, 19, 19, 19, 19, 23, 23, 23, 23, 23, 23, 29, 29, 31, 31, 31, 31, 31, 31, 37, 37, 37, 37, 41, 41}

and the previous prime sequence:

{-2, 2, 3, 3, 5, 5, 7, 7, 7, 7, 11, 11, 13, 13, 13, 13, 17, 17, 19, 19, 19, 19, 23, 23, 23, 23, 23, 23, 29, 29, 31, 31, 31, 31, 31, 31, 37, 37, 37, 37, 41, 41}

as output.

Also, do the largest eigenvalues of this matrix approximate infinitely long sequences of consecutive prime numbers as the size of the matrix goes to infinity?

Again as a Mathematica program for a $200$ times $200$ matrix this is:

```<code>Clear[b, t, n, k, i, j]
t[n_, 1] = -1;
t[1, k_] = -1;
t[n_, k_] :=
  t[n, k] =
   If[n >= k, -Sum[t[n - i, k], {i, 1, k - 1}], -Sum[
      t[k - i, n], {i, 1, n - 1}]];
nn = 200;
m = Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}];
N[Sort[Eigenvalues[m], Less]]
</code>```

which outputs a long list of eigenvalues of which the last few are:

{..., 157.4, 163.302, 167.281, 173.217, 179.157, 181.163, 191.074, 193.065, 197.038, 199.026}

Rounding these gives us:

{..., 157, 163, 167, 173, 179, 181, 191, 193, 197, 199}

which are equal to the 37:th to 46:th primes.

```<code>Table[Prime[i], {i, 37, 46}]
</code>```

{157, 163, 167, 173, 179, 181, 191, 193, 197, 199}

Edit 1.8.2013: A more precise approximation of the primes by eigenvalues is given by this Mathematica program:

```<code>Clear[nn, n, k, d, kolumn];
a[n_] := If[n < 1, 0, Sum[d MoebiusMu@d, {d, Divisors[n]}]]
Do[nn = j;
 A3 = Range[nn]*0;
 Do[
   kolumn = i;
   A1 = Table[Table[a[GCD[n, k]], {k, 1, nn}], {n, 1, nn}];
   MatrixForm[A1];
   A1[[All, kolumn]];
   MatrixForm[
    Table[Table[
      If[Mod[n, k] == 0, MoebiusMu[n/k]*A1[[All, kolumn]][[k]],
       0], {k, 1, nn}], {n, 1, nn}]];
   a1 = Table[
     Total[Table[
       If[Mod[n, k] == 0, MoebiusMu[n/k]*A1[[All, kolumn]][[k]],
        0], {k, 1, nn}]], {n, 1, nn}];
   a2 = Sign[a1]*Exp[Abs[a1]];
   A2 = Table[
     Table[If[Mod[n, k] == 0, a2[[n/k]], 0], {k, 1, nn}], {n, 1, nn}];
   MatrixForm[A2];
   a3 = Table[
     Total[Table[If[Mod[n, k] == 0, a2[[n/k]], 0], {k, 1, nn}]], {n,
      1, nn}];
   A3[[i]] = a3;
   , {i, 1, nn}]
  MatrixForm[A3];
 Print[N[Log[-Min[Eigenvalues[A3]]], 12]]
 (*Print[N[Sign[Eigenvalues[A3]]Log[Abs[Eigenvalues[A3]]],12]]*)
 , {j, 1, 24}]
</code>```

which outputs:

{1.00000000000+3.14159265359 I,
1.71614308398,
2.89852243027,
2.94293185770,
4.98292271242,
4.98305676486,
6.99755465240,
6.99756090303,
6.99756737456,
6.99756777041,
10.9999546191,
10.9999546211,
12.9999938562,
12.9999938562,
12.9999938562,
12.9999938562,
16.9999998874,
16.9999998874,
18.9999999847,
18.9999999847,
18.9999999847,
18.9999999847,
22.9999999997,
22.9999999997}

Rounding these values we get:

{1, 2, 3, 3, 5, 5, 7, 7, 7, 7, 11, 11, 13, 13, 13, 13, 17, 17, 19, 19, 19, 19, 23, 23}

The program above basically uses the Riemann zeta function product that converges to the Dirichlet inverse of the Euler totient function with exponentiated divisors, analogously to the Riemann zeta function product that converges to the von Mangoldt function. I will explain more later, bedtime here now.

This latter matrix starts:

$$A_3 = \left(
\begin{array}{cccccc}
 e & e & e & e & e & e \\
 e & e-e^2 & e & e-e^2 & e & e-e^2 \\
 e & e & e-e^3 & e & e & e-e^3 \\
 e & e-e^2 & e & e-e^2 & e & e-e^2 \\
 e & e & e & e & e-e^5 & e \\
 e & e-e^2 & e-e^3 & e-e^2 & e & e-e^2-e^3+e^6
\end{array}
\right)$$

and is defined by taking the Möbius transform of each column of the first matrix $T$, exponentiating the divisors, multiplying with the Möbius function and then taking the inverse Möbius transform.

$$\displaystyle T = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \end{bmatrix}$$

Example: For the 6-th column with the entries: {1,-1,-2,-1,1,2,...} by Möbius inversion we get: {1,-2,-3,0,0,6} which is equal to the divisors of $6$ times the elementwise multiplication of the Möbius function of the divisors of $6$: {1,-2,-3,0,0,6}={1,2,3,0,0,6} times {1,-1,-1,0,-1,1}.
By then exponentiating the divisors we have: {e^1,e^2,e^3,0,0,e^6} times {1,-1,-1,0,-1,1}. The Inverse Möbius transform (=summing over the divisors) then gives: {e-e^2, e-e^3, e-e^2, e, e-e^2-e^3+e^6}

The claim is that the most negative eigenvalue of $A_3$ approximates the previous prime number sequence.

Of course this could be explained more clearly by starting with the exponentiated divisors, its matrix inverse and the row sums thereof, I believe.

Using the same algorithm as above and exponentiating and correspondingly logarithmating the terms in the eigenvalue sequence arbitrary many times, it appear that they - the logarithms of the eigenvalues - converge to the Möbius function times the natural numbers.

Example, double exponentiating of divisors and double logarithms of eigenvalues of a 12 times 12 matrix:
{-11.0000000000,10.0000000000,-7.00000000000,6.00171666577,-5.00000000000,-3.03392765715,-2.13750865340,1.00344457743,0,0,0,0}

Example, double exponentiating of divisors and double logarithms of eigenvalues of a 13 times 13 matrix:
{-13.0000000000,-11.0000000000,10.0000000000,-7.00000000000,6.00171666577,-5.00000000000,-3.03392765715,-2.13750865340,1.00344457743,0,0,0,0}

which appears to converge to: {1, -2, -3, 0, -5, 6, -7, 0, 0, 10, -11, 0, -13, 14, 15, 0, -17, 0, -19, 0, 21, 22, -23, 0, 0, 26, 0, 0, -29, -30, -31, 0}

Or as a Mathematica line:

```<code>Range[32]*MoebiusMu[Range[32]]
</code>```

Link to Pastebin with Mathematica program

```<code>Clear[n, k, a1, A1, a2, nn]
nn = 8;
b1 = Expand[
   Table[Limit[
     Zeta[s]*Total[
       MoebiusMu[Divisors[n]]*Exp[Exp[Exp[Divisors[n]]]]^(s - 1)],
     s -> 1], {n, 1, nn}]];
b1[[1]] = Exp[Exp[Exp[0]]];
A1 = Table[Table[b1[[GCD[n, k]]], {k, 1, nn}], {n, 1, nn}];
MatrixForm[A1]
a2 = Eigenvalues[A1];
N[Table[Sign[a2[[i]]] If[a2[[i]] == 0, 0, Log[Log[Abs[a2[[i]]]]]], {i,
1, nn}], nn]
</code>```

Output:
{-7.0000000, 6.0000000, -5.0000000, -3.0000000, -2.1375087, 1.0034446, 0, 0}

which are:

$$\text{sign}(Eigenvalues(A_1))\log(\log(abs(Eigenvalues(A_1))))$$
of:
$$A_1(n,k) = b_1(GCD(n,k))$$
where:
$$b_1(n) = \lim_{s\to 1} \, \zeta (s) \sum\limits_{d|n}\mu (d(n)) (\exp(\exp (\exp (d(n)))))^{s-1}$$

So in terms of this limit with the zeta function, there is one logarithm less than the number of exponentials.

As already mentioned above, the sequence that begins to emerge is found in the oeis:

Link to oeis sequence

{1, -2, -3, 0, -5, 6, -7, 0, 0, 10, -11, 0, -13, 14, 15, 0, -17, 0, -19, 0, 21, 22, -23, 0, 0, 26, 0, 0, -29, -30, -31, 0}

I believe this tendency becomes clearer with infinitely many exponentials of the divisors and equally many, minus one, logarithms of the eigenvalues. So something like:

$$\text{sign}(Eigenvalues(A_1))\log(\log(\log(\log(\log(\log(\log(abs(Eigenvalues(A_1)))))))))$$
of:
$$A_1(n,k) = b_1(GCD(n,k))$$
where:
$$b_1(n) = \lim_{s\to 1} \, \zeta (s) \sum\limits_{d|n}\mu (d(n)) (\exp(\exp(\exp(\exp(\exp(\exp(\exp (\exp (d(n))))))))))^{s-1}$$

Edit 22.8.2013:
Simplifying further to a infinite matrix

$$A_1 = \left(
\begin{array}{cccccccccccc}
 e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} \\
 e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 \\
 e^{e^{e^e}} & e^{e^{e^e}} & 0 & e^{e^{e^e}} & e^{e^{e^e}} & 0 & e^{e^{e^e}} & e^{e^{e^e}} & 0 & e^{e^{e^e}} & e^{e^{e^e}} & 0 \\
 e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 \\
 e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & 0 & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & 0 & e^{e^{e^e}} & e^{e^{e^e}} \\
 e^{e^{e^e}} & 0 & 0 & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & 0 & 0 & e^{e^{e^e}} & 0 \\
 e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & 0 & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} \\
 e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 \\
 e^{e^{e^e}} & e^{e^{e^e}} & 0 & e^{e^{e^e}} & e^{e^{e^e}} & 0 & e^{e^{e^e}} & e^{e^{e^e}} & 0 & e^{e^{e^e}} & e^{e^{e^e}} & 0 \\
 e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & 0 & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 \\
 e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & 0 & e^{e^{e^e}} \\
 e^{e^{e^e}} & 0 & 0 & 0 & e^{e^{e^e}} & 0 & e^{e^{e^e}} & 0 & 0 & 0 & e^{e^{e^e}} & 0
\end{array}
\right)$$

equal to $$(\exp(\exp(\exp(\exp(\exp(\exp(\exp(\exp(1))))))))))$$ if GCD(n,k)=1 and 0 otherwise, and taking equally many logarithms of the eigenvalues:

$$\text{sign}(Eigenvalues(A_1))\log(\log(\log(\log(\log(\log(\log(\log(abs(Eigenvalues(A_1))))))))))$$

we get the (rounded) eigenvalues:

{1.00000,-1.00000,-1.00000,-1.00000,-1.00000,1.00000,-1.00000,1.00000,-1.00000,1.00000,-1.00000,-1.00000,-1.00000,-1.00000,1.00000,1.00000,1.00000,-1.00000,1.00000,-1.00000,0,0,0,0,0,0,0,0,0,0,0,0}

which appears to be a rearrangement of the Möbius function:

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

Mathematica:

```<code>Do[
 nn = ii;
 A1 = Table[
   Table[If[GCD[n, k] == 1, Exp[Exp[Exp[Exp[1]]]], 0], {k, 1,
     nn}], {n, 1, nn}];
 a2 = Eigenvalues[A1];
 Print[N[Table[
    Sign[a2[[i]]] If[a2[[i]] == 0, 0,
      Log[Log[Log[Log[Abs[a2[[i]]]]]]]], {i, 1, nn}], 6]], {ii, 1, 32}]
MatrixForm[A1]
</code>```

Mussardos paper:

http://lanl.arxiv.org/pdf/cond-mat/9712010.pdf

http://people.sissa.it/~mussardo/Professional_web/Quantum_Mechanics_and_Number_Theory.html

The eigenvalues can be arbitrary. Here the zeta zeros as input, and as output the zeta zeros as eigenvalues:

```<code>(*Mathematica 8 program start*)(*The Mobius function times "n" \
approximately as the eigenvalues of a matrix*)Clear[nn, n, k, d, \
kolumn]
a[n_] := If[n < 1, 0, Sum[Im[ZetaZero[d]], {d, Divisors[n]}]]
Do[nn = j;
 A3 = Range[nn]*0;
 Do[kolumn = i;
   A1 = Table[Table[a[GCD[n, k]], {k, 1, nn}], {n, 1, nn}];
   MatrixForm[A1];
   A1[[All, kolumn]];
   MatrixForm[
    Table[Table[
      If[Mod[n, k] == 0, MoebiusMu[n/k]*A1[[All, kolumn]][[k]],
       0], {k, 1, nn}], {n, 1, nn}]];
   a1 = Table[
     Total[Table[
       If[Mod[n, k] == 0, MoebiusMu[n/k]*A1[[All, kolumn]][[k]],
        0], {k, 1, nn}]], {n, 1, nn}];
   a2 = Sign[a1]*Exp[Abs[a1]];
   A2 = Table[
     Table[If[Mod[n, k] == 0, a2[[n/k]], 0], {k, 1, nn}], {n, 1, nn}];
   MatrixForm[A2];
   a3 = Table[
     Total[Table[If[Mod[n, k] == 0, a2[[n/k]], 0], {k, 1, nn}]], {n,
      1, nn}];
   A3[[i]] = a3;, {i, 1, nn}] MatrixForm[A3];
 (*Print[N[Log[Log[-Min[Eigenvalues[A3]]]],12]]*)
 Print[N[Sign[Eigenvalues[A3]] Log[Abs[Eigenvalues[A3]]], 12]], {j, 1,
   10}]
(*program end*)
</code>```

答:
This related matrix seems to give the partial sums of the Euler totient times a factor:

```<code>(*Mathematica*)
Clear[n, s, a, A];
mm = 60;
Monitor[eig1 = Table[
   nn = ii;
   a = Table[N[0], {n, 1, nn}];
   a[[1]] = ii;
   A = Table[Table[a[[GCD[n, k]]], {k, 1, nn}], {n, 1, nn}];
   Max[Eigenvalues[A]], {ii, 1, mm}], ii]
ListLinePlot[Differences[%]]
(*end*)
</code>```

Title: Primes approximated by eigenvalues?

Body:
This related matrix seems to give the partial sums of the Euler totient times a factor:

(*Mathematica*)
Clear[n, s, a, A];
mm = 60;
Monitor[eig1 = Table[
   nn = ii;
   a = Table[N[0], {n, 1, nn}];
   a[[1]] = ii;
   A = Table[Table[a[[GCD[n, k]]], {k, 1, nn}], {n, 1, nn}];
   Max[Eigenvalues[A]], {ii, 1, mm}], ii]
ListLinePlot[Differences[%]]
(*end*)


问:
Applications of the concept of homomorphism
What are some interesting applications of the concept of homomorphism?

Example: If there is a homorphism from a ring $R$ to a ring $r$ then a solution to a polynomial equation in $R$ gives rise to a solution in $r$. e.g. if $f:R \rightarrow r$ and $X^2+Y^2=0$ then $f(X^2+Y^2)=f(0), f(X^2)+f(Y^2)=0, f(X)^2+f(Y)^2=0, x^2+y^2=0$

答:
Here are some striking examples from looking at the Lebesgue space $L^1(\mathbb{R})$ under Fourier transform. $L^1(\mathbb{R})$ is an algebra under convolution, that is it is a vector space with a multiplication (here multiplication is the convolution). The Fourier transform $\mathcal{F}: L^1(\mathbb{R})\to C_0(\mathbb{R})$ is an algebra homomorphism from $L^1$ into a sub-algebra $\mathcal(L^1(\mathbb{R}))$ of the space $C_0$ (continuous functions that vanish at $\pm\infty$).

In particular
$$\mathcal{F}(f*g) = \mathcal{F}(f)\cdot\mathcal{F}(g)$$
and instead of studying convolution equations in $L^1$ we may study multiplicative equations in the Fourier image.

An other example is The Wiener Tauberian Theorem, that states that the translates of $f$ span a dense subspace of $L^1$ if and only if $\mathcal{F}(f)$ is non-zero.

ContentLicense: CC BY-SA 2.5


问:
bound on the leading coefficient of a polynomial
Given a polynomial with real coefficients, that satisfies $\forall x \in \left[ -1,1 \right]: \ |p(x)|\leqslant 1$, I have to show, that its leading coefficient, $a_m$ satisfies $a_m \leqslant 2^{m-1}$.

I have obtained some easy partial results: for example under the given conditions the sum of all even coefficients has to be in $\left[ -1,1 \right]$ (and the sum of the odd in $\left[ -2,2 \right]$), but that didn't help me much...

答: This is directly related to Chebyshev polynomials.</p>


问:
Question 10.17 (a) Apostol's Mathematical analysis
This question is from apostol  Mathematical analysis Chapter 10.

Assume that $f \in \mathbb{R}$ on $[a,b]$ for every $b>a>0$. Define $g$ by the equation $x g(x)=\int_{1}^{x} f(t) dt$ if $x>0$, assume that $\lim_{x\to \infty} g(x)$ exists and $a$ and $b$ are fixed positive integers. Prove that :

$$\int_a^{b} f(x)/x dx=  g(b) -g(a) +\int_{a}^b g(x)/xdx.$$

I tried using  $x g(x)=\int_{1}^{x} f(t) dt$ and trying to work with both LHS and RHS 1 at a time and trying to get other but failed.

I am not getting any other Idea.

So, How should I attempt this?

答:
Taking the derivative  with respect to $x$ for
$$x g(x)=\int_{1}^{x} f(t) dt,$$

one gets:
$$ g(x)+xg'(x)= f(x) $$
divide both sides by $x\neq0$:

$$ g(x)/x+g'(x)= f(x)/x $$

Integrating both sides from $a$ to $b$ and using $\int_a^bg'(x)dx=g(b) -g(a)$, completes the result.


问:
Can every continuous function between topological manifolds be turned into a differentiable map?
Let $M$ and $N$ be topological manifolds that admit differential structures and let $f:M\to N$ be continuous. Can $M$ and $N$ always be given differential structures to become differentiable manifolds $\widetilde M$ and $\widetilde N$ such that $f:\widetilde M\to\widetilde N$ is differentiable? What if we impose further restrictions, such as $\widetilde M$ and $\widetilde N$ being smooth manifolds, or $f$ becoming smooth? Are there certain differential structures which we can't do this with (i.e. if we want to make $f$ differentiable, we can never make $N$ diffeomorphic to $\overline N$ where $\overline N$ is some differential structure on $N$)? What if $M$ already has a fixed differential structure?

答:
There is not necessarily a way to make a map smooth.  For example, suppose $M=\mathbb{R}$, and suppose $N$ is any topological manifold of dimension $2$ or more.  Let $f:M\rightarrow N$ be any continuous function whose image contains a non-empty open subset of $N$ (e.g., take a space filling curve onto $\mathbb{R}^n$ and then think of this $\mathbb{R}^n$ as a chart).

Then there are no smooth structures on $M$ and $N$ which makes $f$ smooth.  In fact, you cannot even make $f$ continuously differentiable.  One way to see this is to use Sard's Theorem: if you could make $f$ continuously differentiable, then the set of regular values would be open and dense.  Because $M$ has a lower dimension than $N$, regular values are points of $N$ which are not in the image of $f$.  But then the rest that the image of $f$ contains an open subset of $N$ means the set of regular values of $f$ is not dense.


问:
How can I calculate the rating of multiple Nodes?
I have four nodes, that sends numbers between 1 and 10000 to the main-node.
The value of every node should be between certain numbers:

$$10 < node1 < 15$$
$$12 < node2 < 45$$
$$50 < node3 < 76$$
$$800 < node4 < 1200$$

Use case 1: Ideal set of nodes, would look like this:
$$node1 = 12$$
$$node1 = 15$$
$$node1 = 55$$
$$node1 = 850$$

Use case 2: Non-ideal set would be:
$$node1 = 20$$
$$node1 = 15$$
$$node1 = 55$$
$$node1 = 850$$

As you can see, every node, except node1, sends the right values.

Now I want to take the values of all nodes, put into a formula and get a number, that can tell me, if every node has the ideal value or that at least one node is over the range, like in use case 2.

$$IdealMinValue < f(node1, node2, node3, node4) < IdealMaxValue $$
or
$$f(node1, node2, node3, node4) \approx IdealValue$$

答: For each node, define a function that returns True if the node is in tolerance and False otherwise.  Then take the conjunction of the four.  A False for the conjunction means one is out of range.</p>


问:
$\mathbb{R}^{n}$ is a strongly locally homogeneous space.
I am new to the study of CDH topological spaces, I wanted to study basic examples of this type of spaces for example for the fact that $\mathbb{R}^{n}$ is CDH some authors introduce the notion of Strongly Locally Homogeneous spaces.

A space $X$ is SLH if it has a basis of open subsets $\mathcal{U}$ such that if $p$ and $q$ are two point of $U \in \mathcal{U}$ then there is an homeomorphism $h:X\to X$ such that $h(p) = q$ and such that $h(x) = x$ for every $x$ not in $U$.

Unfortunately I have also failed to demonstrate that $\mathbb{R}^{n}$ is strongly locally homogeneous, does anyone have any ideas?

Thanks a lot

答:
$\def\conv{\operatorname{conv}}$
$\def\inte{\operatorname{int}}$

Let $A$ be a finite subset of $\Bbb R^n$  and $$\Lambda_A=\{(\lambda_x)_{x\in A}\in \mathbb R^A: \sum_{x\in A} \lambda_x=1\mbox{ and }\lambda_x\ge 0\mbox{ for each  }x\in A\}.$$ Define a map $\pi_A$ from $\Lambda_A$ to the convex hull $\conv A$ of the set $A$ by putting $\pi_A((\lambda_x)_{x\in A})=\sum_{x\in A} \lambda_x x$ for each $(\lambda_x)_{x\in A}\in \Lambda_A $. Since the map $\pi_A$ is continuous and the set $\Lambda_A$ is compact, the set $\conv A=\pi_A(\Lambda_A)$ is compact too. Moreover, if the set $A$ is affinely independent, that is there are no nonzero map $\lambda: A\to \Bbb R$ such that $\sum_{x\in A}\lambda(x)=0$ and $\sum_{x\in A}\lambda(x)x=0$, then the map $\pi_A$ is injective, and so a homeomorphism. Moreover, if $|A|=n+1$, then the set $\conv A$ has nonempty interior $\inte\conv A$ in $\Bbb R^n$, for instance, by the domain invariance theorem. It easily follows that $$\mathcal B=\{\inte\conv A: A\subset\Bbb R^n\mbox{ is an affinely independent set of size }n+1\}$$ is a base of the space $\Bbb R^n$.

We claim that $\mathcal B$ witnesess that $\Bbb R^n$ is a strongly locally homogeneous space. Indeed, let $A\subset\Bbb R^n$ be any affinely independent set of size $n+1$ and $p, q$ be any points of $\inte\conv A$. Then we have $$\conv A=\bigcup_{x\in A} \conv (\{p\} \cup A\setminus\{x\})=\bigcup_{x\in A} \conv(\{q\} \cup A\setminus\{x\}).$$

Let $B$ be any proper subset of $A$. Let $i_B$ be a map from $\Bbb R^{\{p\} \cup B}$ to $\Bbb R^{\{q\} \cup B}$ such that for each point $t=(t_x)_{x\in \{p\} \cup B}$ we have $i_B(t)=(i_B(t)_x)_{x\in \{q\} \cup B}$, where $i_B(t)_x=t_x$ for each $x\in B$ and $i_B(t)_q=t_p$. Clearly, $i_B$ is a homeomorphism. Now define a map $h_B$ from $\conv (\{p\} \cup B)$ to $\conv (\{q\} \cup B)$ as follows. Let $y$ be any point of $\conv (\{p\} \cup B)$. Since the set $\{p\}\cup B$ is affinely independent, the point $y$ has a unique representation $y=\lambda_p p+\sum_{x\in B} \lambda_x x$ with $\lambda_p+\sum_{x\in B} \lambda_x=1$, $\lambda_p\ge 0$, and $\lambda_x\ge 0$ for each $x\in B$. Put $h_B(y)=\lambda_p q+\sum_{x\in B} \lambda_x x$. It is easy to see that $h_B=\pi_{\{q\} \cup B\setminus\{x\}}\circ i_B\circ \pi^{-1}_{\{p\} \cup B\setminus\{x\}}$, so the map $h_B$ is a composition of continuous injections, and therefore a continuous injection too. It follows that $h_B$ is a homeomorphism between compact sets $\conv (\{p\} \cup B)$ and $\conv (\{q\} \cup B)$.

It is easy to see that for each points $x,y\in A$ the maps $h_{A\setminus\{x\}}$ and $h_{A\setminus\{y\}}$ coincide with the map $h_{A\setminus\{x,y\}}$ on the intersection $\conv (\{p\} \cup A\setminus\{x,y\})$ of their domains $\conv (\{p\} \cup A\setminus\{x\})$ and $\conv (\{p\} \cup A\setminus\{y\})$. Therefore there exists a map $h^p_q$ from $\conv A$ to $\conv A$ such that $h^p_q|\conv (\{p\} \cup A\setminus\{x\})= h_{A\setminus\{x\}}$ for each $x\in A$. In particular, the restriction of $h^p_q$ at the boundary of $\conv A$ is the identity map and $h^p_q(p)=q$. Since $\conv A$ is a union of the finite family $\{\conv (\{p\} \cup A\setminus\{x\})\}_{x\in A}$ of its closed subsets, the map $h^p_q$ is continuous. Since $h^p_q$ has a continuous inverse $h^q_p$, it is a homeomorphism of $\conv A$. Then the extension of $h^p_q$ at $\Bbb R^n\setminus\conv A$ by the identity map provides the required homeomorphism $h$.


答:
Just a suggestion:-Will delete later if someone comes up with a better function.
We know that $(x-a)(x-b) <0$ when $a<x<b$ and $(x-a)(x-b)>0$ for $x$ not in $(a,b)$.Note that $(a,b)$ are min and max values for a particular node.
From this you can define a function for every node and have a sort of "boolean function" to check if all the functions return negative values . In this case it will have "True Value" ie all of them are negative otherwise False.

Title: How can I calculate the rating of multiple Nodes?

Body:
I have four nodes, that sends numbers between 1 and 10000 to the main-node.
The value of every node should be between certain numbers:

$$10 < node1 < 15$$
$$12 < node2 < 45$$
$$50 < node3 < 76$$
$$800 < node4 < 1200$$

Use case 1: Ideal set of nodes, would look like this:
$$node1 = 12$$
$$node1 = 15$$
$$node1 = 55$$
$$node1 = 850$$

Use case 2: Non-ideal set would be:
$$node1 = 20$$
$$node1 = 15$$
$$node1 = 55$$
$$node1 = 850$$

As you can see, every node, except node1, sends the right values.

Now I want to take the values of all nodes, put into a formula and get a number, that can tell me, if every node has the ideal value or that at least one node is over the range, like in use case 2.

$$IdealMinValue < f(node1, node2, node3, node4) < IdealMaxValue $$
or
$$f(node1, node2, node3, node4) \approx IdealValue$$


问:
Katetov-maps with same support
Let (X,d) be a metric space. f: X $\rightarrow \mathbb{R}$ ist called "Katetov map" iff $\forall x, y, \in X : |f(x)-f(y)| \leq d(x,y) \leq f(x) + f(y)$. The set of all Katetov-maps on X is denoted by E(X).

If f $\in$ E(X) and S $\subset$ X are such that  $f(x) = \inf \{f(s) + d(x,s) : s \in S\}$, we say that f is "supported by S".

Show that if f, g $\in$ E(X) are both supported by S $\subset$ X it holds that $d(f,g)= \sup_{s \in S} |f(s)-g(s)| $ where d denotes the sup-metric.

Apparently this is an easy fact, but I am failing to prove it. Can someone help me please?

答:
Since I only need the statement for finite sets S, I am going to show how I proved that:

By applying the inequality $|\min\limits_{1\leq s \le1 n} a_s - \min\limits_{1\leq s \le1 n} b_s| \leq \max\limits_{1\leq s \le1 n}|a_s - b_s|$ we see that $\forall x$ it holds: $|f(x)-g(x)| = |\min\limits_{s \in S} \{f(s) + d(x,s)\} - \min\limits_{s \in S} \{g(s) + d(x,s)\}| \leq \max\limits_{s \in S} |\{f(s)-g(s)\}|$ and therefore $d(f,g) = \sup\limits_{x \in X} |f(x)-g(x)| \leq \max\limits_{s \in S} |\{f(s)-g(s)\}|$. The other inequality is trivially true.

Title: Katetov-maps with same support


问:
Conditional Variance Exercise
This is a simple exercise on Conditional Variance that I'm trying to understand the proof of.

Let $X$ and $Y$ be two real-valued random variables such that $Y$ is square-integrable. We call the random variable $\mathbf{E}\left((Y-\mathbf{E}(Y \mid X))^{2} \mid X\right)$ the conditional variance of $Y$ given $X$, denoted by $\operatorname{Var}(Y \mid X)$. Show that for all Borel measurable $f: \mathbf{R} \rightarrow \mathbf{R}$ such that $f(X)$ is square-integrable,
$$
\mathbf{E}\left((Y-f(X))^{2}\right)=\mathbf{E}(\operatorname{Var}(Y \mid X))+\mathbf{E}\left((\mathbf{E}(Y \mid X)-f(X))^{2}\right).
$$

Here is the solution, where I seem to be missing something simple.

We have
$$
\begin{aligned}
\mathbf{E}\left((Y-f(X))^{2}\right) &=\mathbf{E}\left([(Y-\mathbf{E}(Y \mid X))+(\mathbf{E}(Y \mid X)-f(X))]^{2}\right) \\
&=\mathbf{E}\left((Y-\mathbf{E}(Y \mid X))^{2}\right)+2 \mathbf{E}((Y-\mathbf{E}(Y \mid X))(\mathbf{E}(Y \mid X)-f(X))) \\
&+\mathbf{E}\left((\mathbf{E}(Y \mid X)-f(X))^{2}\right)
\end{aligned}
$$
The first term in the last line equals $\mathbf{E}(\operatorname{Var}(Y \mid X)),$ while the second term vanishes since $\mathbf{E}(Y \mid X)-f(X)$ is $\sigma(X)$-measurable. The proof is complete.

Okay, but what does $\mathbf{E}(Y \mid X)-f(X)$ being $\sigma(X)$-measurable have to do with anything? Why does it imply that $2 \mathbf{E}((Y-\mathbf{E}(Y \mid X))(\mathbf{E}(Y \mid X)-f(X))) = 0$?

答:
The desired equation
$$
2\mathbb{E}\!\left[\left(Y-\mathbb{E}[Y \, | \, X]\right)\left(\mathbb{E}[Y \, | \, X] - f(x)\right)\right]
=
0
$$
may be rewritten as
$$
2\mathbb{E}\!\left[Y\left(\mathbb{E}[Y \, | \, X] - f(x)\right)-\mathbb{E}[Y\left(\mathbb{E}[Y \, | \, X] - f(x)\right) \, | \, X]\right]
=
0
$$
due to the stated measureabilility. This equation holds as a consequence of the law of iterated expectations. To see this, let $Z:=Y\left(\mathbb{E}[Y \, | \, X] - f(x)\right)$ and write
$$
\mathbb{E}\!\left[Z-\mathbb{E}[Z \, | \, X]\right] = \mathbb{E}[Z] - \mathbb{E}\!\left[\mathbb{E}[Z \, | \, X]\right] = \mathbb{E}[Z] - \mathbb{E}[Z] = 0.
$$

Title: Conditional Variance Exercise


问:
Given $x$, $y$, $k$, convert $x$ to $y$ by multiplication by $k$ or subtraction of $1$ in the fewest steps

Problem Statement:

We are given three positive integers $x$, $y$, and $k$, and our aim is to convert $x$ to $y$ in the minimum number of steps using one of the following two operations at each step:

Multiply $x$ by $k$

Subtract $1$ from $x$

Let convert($x$,$y$,$k$) be the minimum number of steps to convert $x$ to $y$ using $k$. Then find the values for the following:

convert(3,10,2)

convert(4,92,3)

convert(11,104250,2)

Note: Following are the official answers for the three parts mentioned above:

3

7

24

My Approach: I think it is easier to go the other way round, i.e., to convert $y$ to $x$ using the opposite operations, which is to divide by $k$ or add $1$ at each step. I think this is easier because it restricts the possible cases as division is possible only when the number is a multiple of k. And while doing this we can avoid repeated calculations in case we encounter the same number some time later. For example, for convert(3,10,2), we have this,

I even coded a Breadth First Search solution for this problem (thought it is not so useful as the question was asked in a pen and paper based exam):

```<code>#include <bits/stdc++.h>
using namespace std;

using ll = long long;
const ll INF = 1e16;

int main() {
    const int lim = 10000;
    int t;
    cin >> t;
    while(t--) {
        ll x, y, k;
        cin >> x >> y >> k;
        map<ll,bool> vis;
        map<ll,int> dist;
        queue<ll> q;
        q.push(y);
        vis[y] = true;
        bool found = false;
        while(!q.empty()) {
            int u = q.front();
            q.pop();
            if(u == x) {
                found = true;
                break;
            }
            if(dist[u] < lim) {
                if(u+1<INF && !vis[u+1]) {
                    vis[u+1] = true;
                    dist[u+1] = 1+dist[u];
                    q.push(u+1);
                }
                if(u%k == 0 && !vis[u/k]) {
                    vis[u/k] = true;
                    dist[u/k] = 1+dist[u];
                    q.push(u/k);
                }
            }
        }
        if(found) cout << dist[x] << "\n";
        else cout << "-1\n";
    }

    return 0;
}
</code>```

So my question is:

Can we do something better than to look through each possibility? Any observations that can help? I am looking for an answer that can help compute the values without the help of a computer.

答:
(Fill in the gaps as needed. You should have everything that you need listed out here.)

Observations:

Each valid sequence of steps can be written as $ k (k (k ( \ldots ( x - a_n) - a_{n-1} ) - \ldots - a_1) - a_0= y $. This takes $ n + \sum_{i=0}^n a_i$ steps. This is a bijection.

Rewrite this as $k^n x - y = \sum a_i k^n$.

In particular, if $ k^n x - y < 0$, then there is no solution using multiplication by $k$ exactly $n$ times. (This should be obvious from the start).

For a fixed $N$ and $n$, the minimal $\sum_{i=0}^n a_i$ such that $ N=\sum_{i=0}^n a_i k^n$ happens when $ a_i < k$ for $ i < n$. There is a unique representation corresponding to the base $k$ representation of $N$. Note that $a_n$ is as large as it needs to be.

Hence, given a fixed $n, x, y, k$, the calculation for the minimum is straight forward.

We just need to test a several values of $n$, starting with $ \lceil \log_k \frac{y}{x} \rceil $, until $n >$ the minimum number of steps we're calculated so far. Note that the minimum doesn't always occur for the minimum $n$.

For example, in the first case,

$ 3 \times 2^1 - 10 < 0$, so no solution here.

$ 3 \times 2^2 - 10  = 2 = ( 1 \times 2^1) $ will take $ 2 + 1 = 3 $ steps.

$ 3 \times 2^3 - 10  = 14 = ( 1 \times 2^3 + 1\times 2^2 + 1 \times 2 )$ will take $ 3 + 3 = 6 $ steps.

We stop here since any larger $n$ is greater than the minimal steps calculated at this point.

And for the last case,

Start with $\lceil \log_2 104250 / 11 \rceil = 14$.

$11 \times 2^{14} - 104250 = 75974 = 410100011000110_2$, which takes $ 14 + 10 = 24$ steps.

$ 11 \times 2^{15} - 104250 = 256198  = 7110100011000110_2$, which takes $ 15 + 14 = 29$ steps.

Try $ n = 16, 17, \ldots, 24$ and see how many steps those take.

Here is the process for $11\times 2^{14}$ to get to $104250$.


问:
There exist three integers $r,l,w$ such that $rx^{n}+ly^{m}+wz^{q}=1$
We say that three positive integers $x,y,z$ are coprime if there exist three integers $a,b,c$ such that $$ax+by+cz=1$$

How one can prove that $x^{n},y^{m},z^{q}$ are also coprime, i.e., there exist three integers $r,l,w$ such that $$rx^{n}+ly^{m}+wz^{q}=1$$ and what is the relation between $a,b,c$ and $r,l,w$. Here $n,m,q$ are arbitrary.

答:
We have $ax+by+cz = 1$, raise it to the power $m+n+q$ and use the multinomial expansion.
$$
\sum_{r+s+t = m+n+q} \binom{m+n+q}{r,s,t} a^rx^rb^sy^sc^tz^t
$$

Every term of this summation has either $r \geq n$ or $s \geq m$ or $t \geq q$ (if all three of these were false then $r+s+t < m+n+q$, not possible). So we divide the terms into three parts such that in the first part, $r \geq n$, in the second part ,$s \geq m$ and in the third part $t \geq q$.

The first part is a multiple of $x^n$, the second part is a multiple of $y^m$, the third part is a multiple of $z^q$.  The sum of all three parts is $1$, so we have basically got that a sum of a multiple of $x^n$, of $y^m$ and of $z^q$ is $1$, as desired.

I leave you to write out the fine details if you want it, but this is a valid proof for certain.


问:
Is continuous and measurable function possible in discrete metric space?
The function defined by $f:X{\rightarrow}Y$ is continuous. If $X$ is endowed with discrete metric.

Then, ball $B(x,\dfrac{1}{2})=\{x\}$ for all $x\in X $.

For any $\epsilon >0$,

$f(B(x,\dfrac{1}{2}) \subset B(f(x), \epsilon) $

Now,consider characteristic function

$\chi_{A}:X{\rightarrow}R$ defined by

$\mathcal{X}_E(x) =\begin{cases}1, if& x \in E\\
0, & if& x \not\in E \\  \end{cases}$

is discontinuous if $X=R$ endowed with the standard metric.

But, is continuous if $X$ is $R$ endowed with the discrete metric.

And Now,

let (X, M) is a measurable space ,

where sigma algebra $M =\{\phi, X, \{a\}, \{b\}\}$ and $X=\{a, b\}$.

Then , the characteristic function is measurable.

Let $E\subset R$,

$\mathcal{X}_A^{-1}(E) =\begin{cases}X, & 0,1 \in E \\A, & 1 \in E, 0 \notin E \\A^C, & 1 \notin E, 0 \in E \\\emptyset, & o.w.\end{cases}$

This gives our characteristic function is measurable and continuous.
And according to definition of simple function a complex measurable function s on a measurable space $X$ whose range is only finitely many points is called simple function.
Hence,the characteristic function consider above is continuous simple function.

Thus,giving us a continuous simple function. I am correct or wrong please check and correct my understanding .I have used Rudin book of Real and complex analysis and some online notes .

答: If a topological space $X$ is discrete then any its subset is open (and so Borel) so any function  from $X$ to a topological space is continuous, and any function from $X$ to a measurable space is Borel-measurable.


问:
Prove or disprove basic facts about a subvergent(invented definition) series
I am self-learning real analysis from Understanding Analysis by Stephen Abbot. I'd like to ask if I have deduced the correct conclusions for the below assertions about a subvergent (invented definition) series.

$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

Definition. Let's say that a series subverges if the sequence of partial sums contains a subsequence that converges.

Consider this (invented) definition for a moment, and then decide which of the following statements are valid propositions about subvergent series:

(a) If $(a_n)$ is bounded, then $\sum a_n$ subverges.

(b) All convergent series are subvergent.

(c) If $\sum \absval{a_n}$ subverges, then $\sum a_n$ subverges as well.

(d) If $\sum a_n$ subverges, then $(a_n)$ has a convergent subsequence.

Proof.
(a) This proposition is false. As a counterexample, consider the sequence $(a_n):=1$. The sequence of partial sums is $s_1 = 1, s_2 = 2, s_3 = 3, \ldots, s_n = n,\ldots$. No subsequence of $(s_n)$ converges. So, $\sum {a_n}$ is not subvergent.

(b) Since the series is convergent, the sequence of the partial sums converges and therefore any subsequence of partial sums also converges to the same limit. Thus, all convergent series are subvergent.

(c) I think this proposition is true. Let $(s_n)$ be the sequence of partial sums of the absolute values and $(t_n)$ be the sequence of partial sums of the series $\sum a_n$.

By definition of subvergence, there is some subsequence $(s_{f(n)})$ of $(s_n)$ that converges. Without loss of generality, assume $(s_{2n})$ is one such convergent subsequence. Then, there exists a $N \in \mathbf{N}$ such that,
\begin{align*}
 \absval{\absval{a_{2m+2}} + \absval{a_{2m + 4}} + \ldots + \absval{a_{2n}}} < \epsilon
\end{align*}

for all $n > m \ge N$.

Using this fact, we can write a nice inequality for the subsequence $(t_{2n})$.
\begin{align*}
 \absval{t_{2n} - t_{2m}} &=
 \absval{a_{2m+2} + a_{2m+4} + \ldots + a_{2n}}\\
 &\le \absval{a_{2m+2}} + \absval{a_{2m+4}} + \ldots + \absval{a_{2n}}\\
 &\le \absval{\absval{a_{2m+2}} + \absval{a_{2m+4}} + \ldots + \absval{a_{2n}}}\\
 &< \epsilon
\end{align*}

for all $n \ge N$.

As the above holds true for all subsequences $(s_{f(n)})$ where $f(n):\mathbf{N} \to \mathbf{N}$ is a bijection, $\sum a_n$ is subvergent.

(d)  I can't think of a counterexample for this.

答:
For a) your proof is ok

For b), ok too

For c), I would have written :

Let's $a_n^+=\max \{0, a_n\}$ and $a_n^- = \max \{0, -a_n\}$ for all $n$.

Then for all $n$, $|a_n|=a_n^+ + a_n^-$ and $a_n = a_n^+ - a_n^-$.

Since $\sum |a_n|$ is subvergent, and $0\leqslant a_n^+  \leqslant |a_n|$ and $0\leqslant a_n^- \leqslant |a_n|$, we have that $\sum a_n^+$ and $\sum a_n^-$ are subvergent, so the sum $\sum a_n$ is subvergent.

(The fact that if $\sum u_n$ converges with $(u_n)$ positive, then for all $(v_n)$ positive such that $\forall n,v_n\leqslant u_n$ subverges would deserve a proof, but it's not that difficult)

For d) I define $(a_n)$ such that for $n\geqslant 0$,

$a_{2n} = -n$ and $a_{2n+1} = n + \frac{1}{n^2}$.

Then $\sum a_n$ converges since (if we note $S_n = \sum\limits_{k=0}^n a_n$) $S_{2n+1} = \sum\limits_{k=1}^n \frac{1}{k^2}$ converges when $n\rightarrow +\infty$.

But we clearly don't have a subsequence that converges.


问:
Elements and their properties in a finite field
I need help proving the following.

If $\alpha \in (\mathbb{Z}/p\mathbb{Z})[x]/\langle f\rangle$ for some irreducible $f\in (\mathbb{Z}/p\mathbb{Z})[x]$ of degree $n$, then both $\alpha+\alpha^{p}+\alpha^{p^2}+\cdots+\alpha^{p^{n-1}}\in\mathbb{Z}/p\mathbb{Z}$ and $\alpha\alpha^{p}\alpha^{p^2}\cdots\alpha^{p^{n-1}}\in\mathbb{Z}/p\mathbb{Z}$.

I can show it for specific cases of $p$ and $n$, but I don't even know where to start in proving it generally.

答:
Show that both the sum and the product satisfy $x^p=x$. Use $(a+b)^p=a^p+b^p$, a property that holds in characteristic $p$ thanks to the binomial theorem, and the fact that every unit in a finite field with $p^n$ elements must be a $p^n-1$ root of unity, hence $x^{p^n}=x$ holds for all $x$ here.

Every element of ${\bf Z}/p{\bf Z}$, a total of $p$ elements, satisfies $x^p=x$ as well. Given that this is a field, is it possible for any more than $p$ elements to satisfy it?

More generally, since the Frobenius map $x\mapsto x^p$ permutes $\{\alpha,\alpha^p,\cdots,\alpha^{p^{n-1}}\}$, any symmetric polynomial $f$ in $n$ variables evaluated at $\alpha,\alpha^p,\cdots,\alpha^{p^n}$ yields a value in ${\bf Z}/p{\bf Z}$.


问:
How did Euler and Rayleigh calculate $\tan(x) = x$?
I wanted to solve $\tan(x) = x$ without Newton's method. Along I have found various questions here on Mathematics like :

Solution of tanx = x? (Answer of JJacquelin)

Derivation of asymptotic solution of $\tan(x) = x$. (Answer of Antonio Vargas, Lagrange inversion formula) - This way I understand, how they get the result.

and

Location of the extrema of the sinc function.

In the last one the K. B. Dave named some sources like :

https://soundandscience.de/text/theory-sound-volume-1 (Page 278 or PDF-Page 293) (Rayleigh 1877)

and

https://www.digi-hub.de/viewer/image/BV044241472/325/ (Page 319 or PDF-Page 325) (Euler 1748)

Euler for example does not explain how he solved this. (or is it me who does not get it?)
How did they do this?

$$s =  x -  \frac{1}{3}x^3 + \frac{1}{5}x^5 - \frac{1}{7}x^7 ... (1)$$
$$(2n+1)q= \frac{1}{x} +  x -  \frac{1}{3}x^3 + \frac{1}{5}x^5 - \frac{1}{7}x^7 ... (2) $$

$$ \frac{1}{x} = (2n+1)q - s = (2n+1)q - \frac{1}{(2n+1)q} -  \frac{2}{3(2n+1)^3q^3} -  \frac{13}{15(2n+1)^5q^5} -  \frac{146}{105(2n+1)^7q^7} etc. $$

How to get from (1) and (2) to the last equation?

Further helpful links:

https://groups.google.com/g/sci.math/c/1KOaZXdFAG4/m/Ehqn-G6pJY8J?pli=1

and

https://mathworld.wolfram.com/TancFunction.html

答:
Here is an explanation.
Suppose we let $\,y := \frac1{(2n+1)q}.\,$ Euler's equation is
$$ \frac1y = \frac1x + x - \frac13 x^3 + \frac15 x^5 + \cdots. \tag{1}$$
He wants $\,\frac1x\,$ as a Laurent series in $\,y.\,$ First take the
reciprocal of equation $(1)$ to get
$$ y = x - x^3 + \frac43 x^5 - \frac{28}{15}x^7 + \cdots. \tag{2}$$
Use series reversion to express $\,x\,$ as a power series in $\,X\,$ to get
$$ x = y + y^3 + \frac53 y^5 + \frac{16}5 y^7 + \cdots. \tag{3}$$
Take the reciprocal of $(3)$ to get
$$ \frac1x = \frac1y - y - \frac23 y^3 - \frac{13}{15} y^5 + \cdots.
 \tag{4}$$
Euler had long experience with power series and series reversion
so he did not include the intermediate steps from $(1)$ to $(4)$.

Note that OEIS sequence A079330
"Numerators of coefficients of odd powers of 1/q in the solution series for Tan[x]/x=1" is relevant.


问:
How to calculate the second derivative of this example?
I am trying to calculate the second derivative of the following equation:

$$f(x)=\frac{6}{(x^2+3)}$$

$$f'(x)=\frac{-12x}{(x^2+3)^2}$$

The correct answer given was $$f''(x) = \frac{36(x+1)(x-1)}{(x^2+3)^3}$$ but the second derivative that I have gotten was different from the answer.

I have shown the calculations below and would like to understand where is the mistake. Would anyone be able to point out my mistake?

$$f'(x) =\frac{-12x}{(x^2+3)^2} = (-12x)(x^2+3)^{-2}$$

$$= (-12x)\frac{d}{dx} [(x^2+3)^{-2}]+(x^2+3)^{-2}\frac{d}{dx}[-12x]$$

$$=         48x^2(x^2+3)^{-3} -12(x^2+3)^{-2}$$

Body:
Keep going...

$$48x^2(x^2+3)^{-3}-12(x^2+3)^{-2}$$

$$=\frac{48x^2}{(x^2+3)^3}-\frac{12}{(x^2+3)^2}$$

$$=\frac{48x^2-12(x^2+3)}{(x^2+3)^3}$$

$$=\frac{36x^2-36}{(x^2+3)^3}$$

Factorise the numerator to match the given answer.


问:
What is the difference between $x=\sqrt{x}$, $x^2=x$, and $x=\pm\sqrt{x}$?
$$\begin{align}
x &= \sqrt{x} \tag1 \\[4pt]
x^2 &= x \tag2 \\[4pt]
x &=  \pm\sqrt{x} \tag3
\end{align}$$

What is the difference between equations $1$ and $3$?

If I were asked to find the value of $x$, and if I only solved using equation $1$, I get only one value. But if I solve using the equation $3$, I get two values of $x$. How to choose then ?

$$\begin{align}
1 &= \sqrt{1} \tag1 \\[4pt]
1^2 &= 1 \tag2 \\[4pt]
1 &=  \pm\sqrt{1} \tag3
\end{align}$$

The first equation has two solutions.0 and 1.

For 1st equation x = 0 and 1.

For 2nd equation , same 0and 1 only.

For 3rd we have , 0 = +- square root of 0.So it is only 0. But for 1 , it 1 = +- square root of 1.That is -1 and 1 since square root of 1 is 1only.

So total solutions = 1,-1,0. Now where am I wrong ?

答:
All three are statement forms. The solution set of a statement form is the set of all objects that turns the statement form in a true statement if the statement form's placeholder is substituted by that object.  If two statement form share the same solution set they're called equivalent.

The first one reads: "A number equals its square root."

The second one reads: "A number equals its square."

The third reads:" A number is either its root or its negative root."  as $x=\pm\sqrt x$ is just an abbreviation for $x=+\sqrt x\lor x=-\sqrt x$.

The solution set clearly is $\{0,1\}$ in all cases.

All three statement forms are equivalent.

EDIT (for the third equation): If a number equals its square root, it must be $1$ or $0$.  If a number equals its negative root it must be $0$.  Hence the solution set again is $\{0,1\}$.

EDIT2: All solutions of $x=\sqrt x$ must be non-negative as the square root is only defined for non-negative numbers. For non-negative numbers $x$ we therefore know $x=\sqrt{x}^2$ by definition of the square root. Now
$$\begin{align}x=\sqrt x&\iff x-\sqrt x=0\iff \sqrt{x}^2-\sqrt x=0\\
&\iff\sqrt x(\sqrt x-1)=0\iff \sqrt x=0\lor \sqrt x=1\\
&\iff x=0\lor x=1.\end{align}$$
Treat the other equations similarly.


问:
Fast generation of Pareto-distributed randoms.
I'm developing a library of routines for generating random numbers for simulations (it's on GitHub). I've included fast routines for normally distributed and exponentially distributed randoms, using Marsaglia's nice fast Ziggurat algorithm (with some improvements). I surmise from things I've read that Pareto-distributed randoms would also be useful, and I don't see anywhere on the net that does this with the Ziggurat, so I'd like to do that. I'm more of a codemonkey than a mathematician; I managed to calculate the constants, build the tables and make the code for the exponentials and normals, but I have questions about the "alpha" parameter of the Pareto distribution that are a little beyond my mathematical chops.

The Ziggurat code tables are based on a particular curve; the cases I have done areexp(-x) and exp(-0.5 * x * x), with no other parameters. If you want numbers from a curve with different parameters, this can be done afterward with reasonably fast multiply or add operations, still keeping the expensive exp() outside the loop. But I'm not sure about Pareto's alpha. If I build a Ziggurat based on, say, alpha = 1 (which would make f(x) = 1 / (x + 1) I believe; is that right?), could the user then obtain values distributed with a different alpha with fast operations, or would he have to do calculations that would make the Ziggurat a waste of effort?  Or, alternatively, could Pareto-distributed numbers be calculated cheaply from the other distributions, making the Ziggurat unnecessary?  Thirdly, perhaps there is a value of alpha that is so commonly used that a Ziggurat for that value alone would be worth the effort, even if one had to calculate the others expensively?

答:
Suppose $U$ is an exponentially distributed random variable and $\Pr(U>u)=e^{-au}$ for $u>0$ (and $\Pr(U>u)=1$ for $u\le 0$).  Then
$$
\Pr(e^U > x) = \Pr(U > \log_e x) = e^{-a\log_e x} = x^{-a} \text{ for }x>1,
$$
so $e^U$ is a Pareto-distributed random variable.  If you want the cutoff to be a positive number other than $1$, then just multiply $e^U$ by whatever that number is.</p>


问:
Function space, Harmonic analysis
I have been just reading "An introduction to harmonic analysis" by Yitzhak Katznelson, and try to finish the problem in it. There is a problem in Chapter 1:

Let $B$ be a Banach space on $\mathbb{T}$( $\mathbb{T}$ denotes torus), satisfying $\|f_\tau\|=\|f\|$, where $f_\tau=f(t+\tau)$. Define $B_c$ the set of all $f\in B$ such that $\tau\mapsto f_\tau$ is a continuous $B$ valued function. Then $B_c$ is the closure of the set of trigonometric polynomials in B.

To think about this s problem, I have tried to think of why trigonometric polynomial is so important in Banach space on $\mathbb T$ with translation invariant property, is it possible to find out a Banach space with translation invariant that even not include the space of trigonometric polynomials?

答:
$\newcommand{\T}{\mathbb T}\newcommand{\Z}{\mathbb Z}$Given any $f$ in $B_c$, and any $n\in {\Z}$, define
$$
  \hat f(n) = \frac1{2\pi }\int_0^{2\pi } e^{-int} f_t\, dt.
  $$
Observe that   $\hat f(n)$ is a well defined element of $B$ because the integrand is a continuous, hence
Riemann integrable $B$-valued function.  One should therefore notice that $\hat f(n)$ is not a scalar and hence should
not  be confused with the Fourier coefficient of $f$ (more to follow).

For $s$ in $\T$
denote by $T_s$ the map
$$
  T_s:f\in B\mapsto f_s\in B,
  $$
and
notice that $T_s$ is an isometric linear map on $B$ by hypothesis, hence
$$
  T_s\big (\hat f(n)\big ) =
  \frac1{2\pi }\int_0^{2\pi } e^{-int} T_s(f_t)\, dt = $$$$ =
  \frac1{2\pi }\int_0^{2\pi } e^{-int} f_{s+t}\, dt =
  \frac1{2\pi }\int_0^{2\pi } e^{-in(t'-s)} f_{t'}\, dt' = $$$$ =
  \frac{e^{ins}}{2\pi }\int_0^{2\pi } e^{-int} f_{t}\, dt =
  e^{ins} \hat f(n).
  $$
It follows that, for every $s$ in $\T$,
$$
  \hat f(n)(s) =
  \hat f(n)(0+s) =
  T_s\big (\hat f(n)\big )(0) =
  e^{ins} \hat f(n)(0) =
  ce^{ins},
  $$
where the constant $c$ is given by $c= \hat f(n)(0)$.

Therefore we see that, although $\hat f(n)$ is not a scalar, as noted above, it turns out to be a scalar
multiple of the trigonometric polynomial $p(s)=e^{ins}$.  This also shows that $B_c$ does indeed contain at least some
trigonometric polynomials!

To prove  that the trigonometric polynomials are dense in $B_c$, it is enough to show that any continuous linear functional $\varphi $ on
$B_c$ vanishing on all trigonometric polynomials must itself vanish.   So let us  fix  such a  $\varphi $.

Given any $f$ in $B_c$, consider the continuous, scalar valued function  $g$ given by
$$
  g(t) = \varphi (f_t),\quad\forall t\in  \T.
  $$
Then the $n^{\text{th}}$ Fourier coefficient (sic) of $g$ is given by
$$
  \hat g(n) =
  \frac1{2\pi }\int_0^{2\pi } e^{-int} \varphi (f_t)\, dt = $$$$ =
  \varphi \left(\frac1{2\pi }\int_0^{2\pi } e^{-int} f_t\, dt\right) =
  \varphi \big (\hat f(n)\big ) = 0,
  $$
because $\varphi $ vanishes on the trigonometric polynomial $\hat f(n)$ by our assumption.

Any continuous function whose Fourier coefficients vanish must coincide with the zero function, so we deduce that
$$
  0=g(0) = \varphi (f_0) = \varphi (f),
  $$
whence $\varphi =0$,  as desired.

Remark: I am  following the OP's convention according to which
$f_\tau (t)$ is given by $f(t+\tau )$ (rather than $f(t-\tau )$, as in Katznelson's book).

Title: Function space, Harmonic analysis


问:
Proof of identity in the sense of distributions
I am trying to prove this:

$\bullet$ Prove that $\Delta(\varrho_\epsilon \star u) = \varrho_\epsilon \star f $ in the sense of distributions, if $\Delta u = f$ in the sense of distributions, $ u \in L^1_{loc}(\mathbb{R}^n)$.

Can anybody help me?
Thank you in advance and I take advantage of the situation to wish you a happy new year :D

答: The only rule you need except for linearity is $\partial_k (u*v) = u * (\partial_k v).$


答:
You are getting confused. The plus or minus sign in the third equation does not mean there are two solutions in that equation.

When you solve both equation you have

$$ x^2 = x $$

Which has two solutions, 0 and 1. If you plug them in to both equation s you get that the solutions 0 and 1 do work.

The reason why $-1$ isn't a solution is because you can't do $\sqrt{-1}$ without entering the realm of complex numbers.

Also, $(-1)^2 \not= -1$.

I still don't think you understand the meaning of ${\displaystyle \pm }$. It means that it is either $+$ or $-$.

The third equation really says either:

$$ x = \sqrt{x}$$
or
$$ x = -\sqrt{x}$$.

Now understand?

Title: What is the difference between $x=\sqrt{x}$, $x^2=x$, and $x=\pm\sqrt{x}$?

Body:
$$\begin{align}
x &= \sqrt{x} \tag1 \\[4pt]
x^2 &= x \tag2 \\[4pt]
x &=  \pm\sqrt{x} \tag3
\end{align}$$

What is the difference between equations $1$ and $3$?

If I were asked to find the value of $x$, and if I only solved using equation $1$, I get only one value. But if I solve using the equation $3$, I get two values of $x$. How to choose then ?

$$\begin{align}
1 &= \sqrt{1} \tag1 \\[4pt]
1^2 &= 1 \tag2 \\[4pt]
1 &=  \pm\sqrt{1} \tag3
\end{align}$$

The first equation has two solutions.0 and 1.

For 1st equation x = 0 and 1.

For 2nd equation , same 0and 1 only.

For 3rd we have , 0 = +- square root of 0.So it is only 0. But for 1 , it 1 = +- square root of 1.That is -1 and 1 since square root of 1 is 1only.

So total solutions = 1,-1,0. Now where am I wrong ?


问:
Is $z^w$ for $z,w \in \mathbb{C}$ well-defined if you don't take a branch cut?
If I understand correctly, $z^w$ for $z,w \in \mathbb{C}$ is a shorthand for $\exp(w\log(z))$. The problem is that every non-zero complex number has infinitely many logarithms. Say $\exp(y)=z$. Then, so is $\exp(y\pm2i\pi)$, $\exp(y\pm4i\pi)$, etc. Therefore, $z^w$ is a mutli-valued function. This is usually fixed by taking a branch cut, so that we require $\Im\left({\log(z)}\right) \in (-\pi,\pi]$, for instance. However, say we don't choose a branch of the logarithm. Is $z^w$ well-defined? Can I, for instance, write that
$$
i^i=\{x \mid x=e^{-2 \pi n - \pi/2},n\in\mathbb{Z}\} \, ?
$$
Another concern I have is about the notation $e^z$ being used in place of $\exp(z)$. I find this notation confusing, since $\exp$ denotes a single-valued function, whereas $e^z$ does not. If $z^w=\exp(w\log(z))$, then doesn't
$$
e^z=\exp(w\log(e)) \, ,
$$
meaning that $e^z$ has infinitely many values?

答:
You're right that every non-zero complex number has infinitely many logarithms, but it is possible that exponentiation will make all these values equal.
For instance, let $ z= r e^{i \theta},$ then the possible values of $\log(z)$ are $ \log(r) + i \theta + 2 \pi i k$ for $k \in \mathbb{Z}.$   Now you would like
$\exp(w\log z) $ to be a well-defined function on $\mathbb{C}$ i.e. you want all the possible values of the logarithm of $z$ to give the same value of $w^z.$ Writing this out, you get:

$$ \exp(w(\log r + i\theta + 2\pi i k  )) = \exp ( w \log r + i\theta) \text{ for all } k \iff $$
$$  \exp( w \cdot 2\pi i k) = 1 \text{ for all } k \iff w \cdot 2\pi i k \in 2 \pi i \mathbb{Z} \text{ for all }k \iff  w \in \mathbb{Z}.$$
Thus only exponentiating by integers is well defined without taking a branch cut, for example: $z^{1} $ is a well-defined analytic function.


问:
Find matrix with respect to basis
I need help with this problem, the problem is that I don't know where to start.

$T(a) = -5a + (-2)b + 2c $

$T(a+b) = a + 7b + c$

$T(a+c) = -13a + (-9)b + 18c$

I know that the first one is just $(-5,-2,2)$ but I don't know what to do when it's written like the other two. Should I try to "remove" $b$ from $T(2a + b)$ in some way?

Title: Find matrix with respect to basis

Body: You know that $T(a)=-2a-2b+7c$. You also know the\begin{align}T(b)&=T(b+2a-2a)\\&=T(b+2a)-T(2a)\\&=8a + 4b + c-2T(a)\\&=8a + 4b + c-2(-2a-2b+7c)\\&=12a+8b-13c\end{align}and that\begin{align}T(c)&=T(c-b+b)\\&=T(c-b)+T(b)\\&=-12a-8b+17c+12a+8b-13c\\&=4c.\end{align}So, the matrix that you're after is$$\begin{bmatrix}-2&12&0\\-2&8&0\\7&-13&4\end{bmatrix}.$$