问:
How to find the value of series $\sum_{n=0}^\infty \frac{(-1)^n \pi^{2n+1}}{2^{2n+1}(2n+1)!}$
$$ \sum_{n=0}^{\infty} \frac{(-1)^n \pi^{2n+1}}{2^{2n+1}(2n+1)!} $$

Thanks to Wolfram Alpha, I know that the value is 1, but I have no clue how to get to that. Please help, I have been sitting on this for hours.

It looks like sinus or cosinus, so I could turn it into the form:

$$ \sum_{n=0}^{\infty} (-1)^n \cdot \dfrac{\pi^{2n} \cdot \pi}{2 \cdot 2^{2n} (2n+1) (2n)!}
= \sum_{n=0}^{\infty} (-1)^n \cdot \frac{\pi^{2n}}{(2n)!} \cdot \frac{\pi}{2^{2n+1} (2n+1)} $$

Now the first part of the equation looks like a $\cos(\pi)$, but what should I do with the second part? Something is telling me Cauchy product, but I'm not sure it that first here.

Another idea is of course

$$= \sum_{n=0}^{\infty} (-1)^n \cdot \frac{\pi^{2n+1}}{(2n+1)!} \cdot \frac{1}{2^{2n+1}} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{2n+1}} \cdot \frac{\pi^{2n+1}}{(2n+1)!} $$

But yeah, I pretty much get stuck here, I feel like I'm missing some information, but I don't know what I should be looking for.

答:
Write it as
$$\sum_{n=0}^{\infty} (-1)^n\frac{(\frac{\pi}{2})^{2n+1}}{(2n+1)!}=\sin\left(\frac{\pi}{2}\right)=1$$
using the Maclaurin series for $\sin$.


问:
A question on two polynomial ideals generated by the same polynomials have the same generator
Here is the statement I want to prove:

Let $K$ be a subfield of $F$, and $f,g\in{K[x]}$. Let $M_K=fK[x]+gK[x]$ and $M_F=fF[x]+gF[x]$. Show that $M_K$ and $M_F$ have the same monic generator.

My argument is the following: Suppose that $M_K=dK[x]$, then d is the unique and monic generator of $M_K$. Suppose $q\in{M_K}$, then there exists $q_1,q_2\in{K[x]}$ such that $q=fq_1+gq_2$ and there exists $q_3\in{K[x]}$ such that $q=dq_3$. So, $dq_3=fq_1+gq_2$. Also, since $q_1,q_2,q_3\in{K[x]}$, we have $q_1,q_2,q_3\in{F[x]}$. Thus, $fq_1+gq_2\in{M_F}$, i.e, $dq_3\in{M_F}$. I claim that since $q_3$ is an arbitrary polynomial in $F[x]$, it follows that $M_F=dF[x]$. Hence, d is also the unique and monic generator of $M_F$.

It seems logical but since I started with an assumption q belonging to $M_K$, I could not be sure. Is a converse implication needed, or is the idea completely wrong?

答: I think the point is that both ideals are generated by the monic greatest common divisor of $f$ and $g$. and this is the same over all fields containing the coefficients of $f$ and $g$.

ContentLicense: CC BY-SA 4.0


问:
Cardinality of an infinite tree paths
Suppose a tree with nodes located at levels $1,2,3...$. At each level the nodes branch into several nodes or do not branch.

Does the cardinality of the set of all infinite paths in this tree depend on the growth rate of the nodes by levels?

I mean, if the nodes do not branch at all, then the path is 1.

If the notes add a fixed amount of new nodes at each level, then the number of infinite paths seems to be countable. The same is if the number of nodes grows polynomially.

If the nodes branch at fixed rate at each level, e.g. each node gives 2 nodes, then the number of nodes grows exponentially and the set of all paths has cardinality of continuum.

What about intermediate branching rates?

答:
It’s possible to have $n+1$ nodes at level $n$ (counting the root level as level $0$) and still have $2^\omega$ branches. Here’s the start of an example:

```<code>                ---------x----------
                |                  |
          ------x------            °
          |           |            |
          °           °        ----x----
          |           |        |       |
      ----x----       °        °       °
      |       |       |        |       |
      °       °    ---x---     °       °
      |       |    |     |     |       |
      °       °    °     °  ---x---    °
      |       |    |     |  |     |    |
      °       °    °     °  °     °  --x--
      |       |    |     |  |     |  |   |
      x       °    °     °  °     °  °   °
</code>```

The nodes marked x are the branching nodes. If you collapse the path from each of them up to its nearest predecessor branching node to a single edge, you get the complete binary tree of height $\omega$. I’ve simply taken that tree and inserted nodes of degree $2$ so that there is only one branching node at each level.

This tree grows by $1$ node per level, but you could stretch it out even more and get an arbitrarily small growth rate and still have $2^\omega$ branches.

Title: Cardinality of an infinite tree paths

Body:
It’s possible to have $n+1$ nodes at level $n$ (counting the root level as level $0$) and still have $2^\omega$ branches. Here’s the start of an example:

                ---------x----------
                |                  |
          ------x------            °
          |           |            |
          °           °        ----x----
          |           |        |       |
      ----x----       °        °       °
      |       |       |        |       |
      °       °    ---x---     °       °
      |       |    |     |     |       |
      °       °    °     °  ---x---    °
      |       |    |     |  |     |    |
      °       °    °     °  °     °  --x--
      |       |    |     |  |     |  |   |
      x       °    °     °  °     °  °   °

The nodes marked x are the branching nodes. If you collapse the path from each of them up to its nearest predecessor branching node to a single edge, you get the complete binary tree of height $\omega$. I’ve simply taken that tree and inserted nodes of degree $2$ so that there is only one branching node at each level.

This tree grows by $1$ node per level, but you could stretch it out even more and get an arbitrarily small growth rate and still have $2^\omega$ branches.


答: Assuming each level of the tree is countable, the set of infinite paths is always either countable or has cardinality $\mathfrak{c}$.  To see this, let $T_n$ be the $n$th level of the tree.  The set of branches can then be considered as a subset $B$ of the product $\prod_n T_n$.  In fact, $B$ is a closed subset of $\prod_n T_n$ with respect to the product topology when you give each $T_n$ the discrete topology.  Since each $T_n$ is countable by hypothesis, the space $\prod_n T_n$ is a Polish space, and so $B$ is a Polish space as well.  In particular, this implies $B$ is either countable or has cardinality $\mathfrak{c}$ (as a consequence of the Cantor-Bendixson theorem).

Title: Cardinality of an infinite tree paths

Body:
Suppose a tree with nodes located at levels $1,2,3...$. At each level the nodes branch into several nodes or do not branch.

Does the cardinality of the set of all infinite paths in this tree depend on the growth rate of the nodes by levels?

I mean, if the nodes do not branch at all, then the path is 1.

If the notes add a fixed amount of new nodes at each level, then the number of infinite paths seems to be countable. The same is if the number of nodes grows polynomially.

If the nodes branch at fixed rate at each level, e.g. each node gives 2 nodes, then the number of nodes grows exponentially and the set of all paths has cardinality of continuum.

What about intermediate branching rates?


问:
Example of where the Mean Value Theorem of Integrals Fails
A circumstance where the Mean Value Theorem for Integrals fails is when the function is integrable but discontinuous.

The Mean Value Theorem for Integrals is,

$f(c)=\frac{1}{b-a}\int_{a}^{b}f(x)$

So if we want to prove it "fails" for a specific integral, then I assume we would want to prove that $f(c)$ takes on two different values. However, I really don't have any ideas for a discontinuous integral that would be simple to calculate yet suitable for the counterexample. A few suggestion would be really appreciated.

答:
I assume we would want to prove that $f(c)$ takes on two different values.

No, that's not how it fails. $f$ is a function, and therefore, by definition, no matter what $c$ is (as long as it's in the domain), $f(c)$ can only have a single value.

However, there is something else that can fail. Note what the mean value theorem says:

[...] there exists a $c\in [a, b]$ such that [...]

If $f$ is discontinuous, then it's possible that any candidate $c$ either makes $f(c)$ too large or too small. In that case there is no $c$ that makes $f(c)$ have the exact right value.

The easiest way to make sure that there is no such $c$ is to make $f$ take only two values, and make sure $\frac{1}{b-a}\int_a^bf(x)\,dx$ is not one of those two values.


问:
On the Pell-like $px^2-qy^2 = 1$ for prime $p,q$
Given any prime of form $p_n = u^2+nv^2$ for non-zero integers $u,v$. Consider,

\begin{aligned}
&p_2x^2-2y^2 = 1\\
&p_3x^2-3y^2 = 1\\
&p_7x^2-7y^2 = 1\\
&p_{11}x^2-11y^2 = 1\\
\end{aligned}

Question 1: Are these four families of Pell-like eqns always solvable in non-zero integers $x,y$?

Ex.

Let $n=11$, and since $53=u^2+11v^2$, then $53x^2-11y^2=1$ is solvable. (Other $p_{11}$ will do.)

Question 2: Is there another $n$?

Let $n=13$, and since $157=u^2+13v^2$, but $157x^2-13y^2=1$ is NOT solvable in the integers.

Have these questions been answered in the literature already?

答:
$$ 977 x^2 - 7 y^2 = 1$$
is not possible in integers. $$ 23^2 + 7 \cdot 8^2 = 977.  $$</p>

```<code>jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle
Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2
977 0 -7

  0  form            977           0          -7  delta    -11
  1  form             -7         154         130

           0          -1
           1         -11

To Return
         -11           1
          -1           0

0  form   -7 154 130   delta  1
1  form   130 106 -31   delta  -4
2  form   -31 142 58   delta  2
3  form   58 90 -83   delta  -1
4  form   -83 76 65   delta  1
5  form   65 54 -94   delta  -1
6  form   -94 134 25   delta  5
7  form   25 116 -139   delta  -1
8  form   -139 162 2   delta  81
9  form   2 162 -139   delta  -1
10  form   -139 116 25   delta  5
11  form   25 134 -94   delta  -1
12  form   -94 54 65   delta  1
13  form   65 76 -83   delta  -1
14  form   -83 90 58   delta  2
15  form   58 142 -31   delta  -4
16  form   -31 106 130   delta  1
17  form   130 154 -7   delta  -22
18  form   -7 154 130
</code>```

ContentLicense: CC BY-SA 3.0

Body:
$$ 977 x^2 - 7 y^2 = 1$$
is not possible in integers. $$ 23^2 + 7 \cdot 8^2 = 977.  $$

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle
Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2
977 0 -7

  0  form            977           0          -7  delta    -11
  1  form             -7         154         130

           0          -1
           1         -11

To Return
         -11           1
          -1           0

0  form   -7 154 130   delta  1
1  form   130 106 -31   delta  -4
2  form   -31 142 58   delta  2
3  form   58 90 -83   delta  -1
4  form   -83 76 65   delta  1
5  form   65 54 -94   delta  -1
6  form   -94 134 25   delta  5
7  form   25 116 -139   delta  -1
8  form   -139 162 2   delta  81
9  form   2 162 -139   delta  -1
10  form   -139 116 25   delta  5
11  form   25 134 -94   delta  -1
12  form   -94 54 65   delta  1
13  form   65 76 -83   delta  -1
14  form   -83 90 58   delta  2
15  form   58 142 -31   delta  -4
16  form   -31 106 130   delta  1
17  form   130 154 -7   delta  -22
18  form   -7 154 130


问:
Meaning of complex conjugate
I am reading about the Fourier series.

I understood the Fourier Series in terms of breaking down a function into a constant and a weighted series of sines and cosines.

I then read on in my text and it looked at the complex representation of a function.  I am having difficulty understanding the indexes of the complex coefficients.

Could someone just check my understanding of the following from my text?

The highlighted line is what I am uncertain about (C^-n = C^*n)$

I know what a complex conjugate is, e.g. The complex number a + ib has the complex conjugate a -ib.

Is the meaning of the -n subscript just referring to the symmetry of the function?  So if I have four elements in a Fourier series: -n2, -n, n, and n2, the complex value of term -n2 has to equal the complex conjugate of n2?

Thank you!

答: What that textbook is claiming is that, if$$f(x)=\sum_{n=-\infty}^\infty c_ne^{inx}$$and $f$ is a real function (that is, for each $x$, $f(x)\in\Bbb R$), then$$(\forall n\in\Bbb Z):c_{-n}=\overline{c_n}.$$That's so because, for each $x$\begin{align}\sum_{n=-\infty}^\infty c_ne^{inx}&=f(x)\\&=\overline{f(x)}\text{ (since $f(x)\in\Bbb R$)}\\&=\overline{\sum_{n=-\infty}^\infty c_ne^{inx}}\\&=\sum_{n=-\infty}^\infty\overline{c_ne^{inx}}\\&=\sum_{n=-\infty}^\infty\overline{c_n}e^{-inx}\\&=\sum_{n=-\infty}^\infty\overline{c_{-n}}e^{inx}.\end{align}

Title: Meaning of complex conjugate


问:
Cumulative distribution functions (cdfs) question
If I have two cumulative distribution functions $F(x)$ and $G(x)$, are $$F(x)G(x)\quad\text{ and }\quad [F(x)G(x)]^{0.5}$$ necessarily cdf? I feel a little weird, cause I am thinking to qualify to be a cdf, it has to qualify:

a. $0≤F(x)≤1$;

b. $F(x)$ and $G(x)$ are non-decreasing;

c. $F(x)$ and $G(x)$ are continuous from left to the right;

d. $\lim_{x\to-\infty} F(x)=0 $ and $ \lim_{x\to\infty}=1$

Right?
I think $F(x)G(x)$ qualifies to be a cdf ( not sure about b though), but I don't think $[F(x)G(x)]^{0.5}$ is cdf. Can any one give a counterexample maybe?

答:
You could go further! If $F_{1}(x),F_{2}(x),\ldots,F_{n}(x)$ are all CDFs and $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\in\mathbb{R}$ and $\forall_i. \alpha_i>0$, then

$$F(x)=\prod_{i=1}^{n}F_{i}^{\alpha_{i}}(x)$$

is also a CDF and

$$f(x)=\left(\prod_{i=1}^{n}F_{i}^{\alpha_{i}}(x)\right)\sum_{j=1}^{n}\alpha_{j}F_{j}^{-1}(x)\frac{\mathrm{d}F_{j}(x)}{\mathrm{d}x} $$

the corresponding PDF.


问:
Gaussian sets modulo $p$.
I have the following problem, given the definition:

Let $p$ be an odd prime, and let $S$ be a set of $(p-1)/2$ integers. We call $S$ gaussian set modulo $p$ if $S\cup-S=S\cup\{-s\mid s\in S\}$ Is a reduced system of residues modulo $p$.

Reading the proof of a theorem uses that result. But the result is not explained.

Let $p$ be an odd prime. Prove that $\{1,2,...,(p-1)/2\}$ is a Gaussian set modulo $p$.

答: Note that if $T$ is a reduced system of residue then $T'$ will also be where $T'$ arises from $T$ by adding $p$ to some elements. So $S\cup -S$ is a reduced system iff $S\cup (-S+p)$ is. Now we have $S\cup (-S+p)=\{1,2,\dots,p-1\}$. Since $p$ is prime the elements of this set are all coprime to $p$ and it contains $\phi(p)=p-1$ incongruent elements. Therefore it is a reduced system of residues mod $p$.

Title: Gaussian sets modulo $p$.


问:
Showing that $B=\{(x,y) \in \Bbb{R^2} \mid \ xy>0 \}$ is an open set

Show that $$B=\{(x,y) \in \Bbb{R^2} \mid \ xy>0 \}$$ is an open set in the Euclidean space $(\Bbb{R^2},d_2),$ where $d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$.

Defining $A= \{(x,y) \in \Bbb{R^2} \mid x>0,y>0 \}$ and $E=\{  (x,y) \in \Bbb{R^2} \mid x<0,y<0 \}$ one sees that $B= A\cup E$ and thus $B$ will be open if $A$ and $E$ are open.

How can I show that $A$ and $E$ are both open? It seems that I would need to find open balls with arbitrary radius $B((x,y), r)$ for both cases that are contained in the set I'm working with?

答:
Suppose $x>0, y>0,$ and suppose $\min\{x,y\}=y.$ [A very similar proof can be done if $\min\{x,y\}=x$ ].

I want to do better than mechanodroid and show that $B\left(\ (x,y),\ y\ \right) \subseteq A$ which amounts to:

If $\ d\left(\ (x,y),\ (x',y')\ \right) < y,\ $ then $x' > 0$ and $y' > 0.$

Since showing that $x'>0\ $ is easier than showing $y' > 0,$ I'll concentrate on $y'$ and leave $x'$ as an exercise for the reader.

If $y' \geq y,$ then we are done since $y>0.$ So suppose further that $y' < y.$

By supposition, \begin{align} 0 < \sqrt{(x-x')^2 + (y-y')^2} < y \\
\implies (x-x')^2 + (y-y')^2 < y^2 \\
\implies 0 \leq (x-x')^2 < 2yy' - (y')^2 = y'(2y-y').
\end{align}

If $y'=0$ then $(x-x')^2 < 0 \ \Rightarrow\Leftarrow.$

If $y' < 0,$ then we require $2y - y' < 0$, which is true $\iff 0< 2y<y'<y\ \Rightarrow\Leftarrow.$

Therefore $y'>0.$


问:
the second order continued fraction expansion of an irrational number and the distance to its closest integer
Let $u$ be a posive irrational number and let

$$u=a_0+ \frac{1}{a_1+\frac{1}{a_2  \cdots} }$$

be its continued fraction expansion.

Consider the second-order finite continued fraction expansion of $u$:

$\frac{p_2}{q_2}:=a_0+ \frac{1}{a_1+\frac{1}{a_2}}$ with $p_2$ and $q_2$ being positive coprime integers.

I wonder if we have the following inequality and how to prove it:

$$|p_2-q_2 u|<\|u\|,$$

where $\|u\|$ denotes the distance from $u$ to its nearest integer.

Please also note that I am not asking how to prove $|\frac{p_2}{q_2}-u|<\|u\|$, which is simpler.

If the above is true (I have verified it with many examples), please feel free to use any very basic properties of continued fractions to prove it.

But please do not use any general fact about the optimality of the rational approximation with continued fractions to prove this because I actually want to use the above fact as one of the steps to prove optimality of the rational approximation with continued fractions.

答:
Hint:

The Continued Fraction expansion (and approximation) is tied to the Stern-Brocot tree.

In turn, the properties of the Stern-Brocot tree and related Farey sequence, ensure that a truncation of the CF, that is a truncated path on the tree, represents the best rational approximation (with the denominator being less than a fixed threshold).

The above is for the general case of taking $n$- terms of the CF.

A more simple reply to your specific case is that, being
$$
u = \left\lfloor u \right\rfloor  + \left\{ u \right\}\quad \left| {\;0 \le \left\{ u \right\} < 1} \right.
$$
the splitting of $u$ into the integral and fractional part, then
$$
a_{\,0}  = \left\lfloor u \right\rfloor
$$
and
$$
\eqalign{
  & \left\{ u \right\} = {1 \over {{1 \over {\left\{ u \right\}}}}}
 = {1 \over {\left\lfloor {{1 \over {\left\{ u \right\}}}} \right\rfloor
  + \left\{ {{1 \over {\left\{ u \right\}}}} \right\}}}
 = {1 \over {a_{\,1}  + {1 \over {{1 \over {\left\{ {1/\left\{ u \right\}} \right\}}}}}}} =   \cr
  &  = {1 \over {a_{\,1}  + {1 \over {\left\lfloor {{1 \over {\left\{ {1/\left\{ u \right\}} \right\}}}}
 \right\rfloor  + \left\{ {{1 \over {\left\{ {1/\left\{ u \right\}} \right\}}}} \right\}}}}}
 = {1 \over {a_{\,1}  + {1 \over {a_{\,2}  + r_{\,2} }}}} \cr}
$$

So you can conclude by yourself.


问:
Probability generating function of the negative binomial distribution.
I am using the definition of the negative binomial distribution from here. This is the same definition that Matlab uses. For convenience,

$$P(k) =  {r + k -1 \choose k}p^r(1-p)^k ,$$

where $p$ is the probability of success. $P(k)$ is the probability of $k$ failures before $r$ successes. The probability generating function is supposed to be,

$$ g(x) = \left(\frac{p}{1-(1-p)x}\right)^r.$$

However, I am trying to prove this. Steps:

$$g(x) = \sum^{\infty}_{k=0} P(k)\, x^k$$

$$= \sum^{\infty}_{k=0} {r + k -1 \choose k}p^r(1-p)^k \, x^k $$

$$ = p^r \sum^{\infty}_{k=0} {r + k -1 \choose k}(x(1-p))^k. $$

I suppose the next step would be to show that,

$$ \sum^{\infty}_{k=0} {r + k -1 \choose k}(x(1-p))^k = \frac{1}{(x(1-p))^r}.$$

Is there a formula (or theorem) for infinite sums involving binomial coefficients that I can apply to get this?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With
$\ds{\on{P}\pars{k} =
{r + k - 1 \choose k}p^{r}\,\pars{1 - p}^{k}}$:
\begin{align}
\bbox[5px,#ffd]{\on{g}\pars{x}} & \equiv
\sum_{k = 0}^{\infty}\on{P}\pars{k}\,x^{k} =
p^{r}\sum_{k = 0}^{\infty}{r + k - 1 \choose k}
\bracks{\pars{1 - p}x}^{\,k}
\\[5mm] & =
p^{r}\sum_{k = 0}^{\infty}\braces{{\bracks{-r - k + 1} + k - 1 \choose k}\pars{-1}^{k}}
\bracks{\pars{1 - p}x}^{\,k}
\\[5mm] & =
p^{r}\sum_{k = 0}^{\infty}{-r \choose k}
\bracks{-\pars{1 - p}x}^{\,k} =
p^{r}\,\bracks{1 - \pars{1 - p}x}^{\,-r}
\\[5mm] & =
\bbx{\bracks{p \over 1 - \pars{1 - p}x}^{r}} \\ &
\end{align}


问:
Vector field derived from Group action on smooth manifolds
It is known that given a smooth action $G$ on $M$,
$$G\times M\to M$$
one can
associate to each element of $v\in T_eG$ a vector field on $X\in \mathfrak{X}(M)$. I want to see a less obvious and concrete example and know that is this fact just about Lie groups of positive dimension or it is meaningful for discrete groups action too?

答: Nothing goes wrong in the case of a discrete (i.e. $0$-dimensional) Lie group; it's just a rather trivial case: If $G$ is a discrete group, then the tangent space is a singleton $T_eG=\{0\}$, and $0\in T_eG$ corresponds to the zero vector field $0\in\mathfrak{X}M$.


问:
Expected decay time with uniformly distribution

A strange radioactive sample initially contains $N>1$ atoms. Each day, if the sample contains $m$ atoms, then it decays in such a way that at the end of the day, the number of atoms remaining is uniformly distributed over $\{1,\dots,m-1\}$. Let $X_k$ be the number of atoms left at the end of day $k$ (with $X_0=N$). Define $T=\min\{k\geq0\mid X_k=1\}$. What is $\mathbb E[T]$?

Clearly the $X_k$ form a Markov Chain. I can write down the transition probabilities as $p_{ij}=\frac{1}{i-1}$ if $1\leq j\leq i-1$, else $p_{ij}=0$. However, I'm not sure how to find $\mathbb E[T]$. Certainly, $T<N$. Further, since in expectation the number of atoms halves each day, we might expect intuitively that $\mathbb E[T]$ will be $O(\log N)$, but this of course is not at all a rigorous argument.

答:
Let $k_i=\mathbb E[T\mid X_0=i]$. Then for $n>2$, we have
\begin{align}
k_n &= 1+\frac{1}{n-1}(k_2+\dots+k_{n-1}) \\
k_{n-1} &= 1+\frac{1}{n-2}(k_2+\dots+k_{n-2}).
\end{align}
So from the second equation, $k_2+\dots+k_{n-2}=(n-2)(k_{n-1}-1)$. Sticking this into the first,
$$k_n=1+\frac{1}{n-1}[(n-1)k_{n-1}-(n-2)]=k_{n-1}+\frac{1}{n-1}.$$
Now with $k_2=1$ and $k_3=1+\frac{1}{2}$, it follows inductively that
$$\mathbb E[T]=k_N=\sum_{i=1}^{N-1}\frac{1}{i}.$$
As you suspected, this is indeed $O(\log N)$.


问:
Check existence of cycle in graph with only number of vertices and their degree
I know that we can check existence of cycle in graph by simply traversing the graph and check whether the vertex has been visited.

However, if there are only number of vertices and their corresponding degree known, can we check if the graph contains at least a cycle?
Provided that there is no self-loop and no cycle involving exactly only 2 vertices with 2 degrees each. So it doesn't matter how they are connected to each other.

Is that possible without first randomly connecting each vertex according to the their degrees and then traversing the graph?

Edit

Actually the problem is to find a maximum value from a set of vertices if the vertex connected to other vertex. So let say I'm provided with a set of number $M$ and $|M| = |V| - 1$ with $V$ being a set of vertices, $M_n$ being the value that a vertex can have if the vertex is connected to $n$ other vertices (degree).

Eg.

$$M = (4, 1, 1, 3, 2)$$
In that case the $|V|=6$ and let say $V=(V_1,....,V_6)$.
That means if any of those vertices is connected to a vertex the value of both vertices would be $4$ ($M_1=4$, degree = 1) but if connected to other 3 vertices the value of that particular vertex would be $1$ ($M_3=1$, degree = 3).

From that, I'd like to find which combination yields the most value without introducing a cycle in the graph.

At first I was thinking of trying each combination one-by-one and constructing a graph from each combination then look for a cycle. However that approach wouldn't be feasible if there are a lot of vertices. So now I'm thinking of sorting the degree ($n$) first by the value ($M_n$) then start from there and thinking if it'd be possible to detect a cycle without constructing a graph.

答:
if there are only number of vertices and their corresponding degree known, can we check if the graph contains at least a cycle?

Do you mean a complete cycle (Hamiltonian) or any cycle at all? Vertex degrees fail in some strong ways for the problem of a complete cycle, because degree-based tests are too fast to be able to solve NP-complete problems.  To test if there is any cycle is to test whether the graph is a union of trees, which is computationally simple, number of edges = $n$ - (number of connected components), but that is not a local condition that can be determined by degrees of vertices.

A set of vertex degrees forces Hamiltonicity if and only if it satisfies Chvatal's numerical condition (as in the Bondy-Chvatal theorem, http://books.google.com/books?id=zxSmHAoMiRUC&pg=PA198 ), and non-Hamiltonicity is forced only in trivial situations like not having enough edges to form a complete cycle.  Most sets of vertex degrees are consistent both with a Hamilton cycle existing, and with a cycle not existing, so that there is not enough information in the degrees to test for a Hamiltonian cycle.  Degree-based tests and other local methods that look only a finite distance away from each vertex have polynomial run-time, but testing Hamiltonicity is NP-complete, which makes local tests hopeless for general graphs.


问:
calculating norm of a linear functional
I was solving a problem regarding Hanh-Banach extension of a linear functional and I encountered the following problem as a subpart of the original one -

Consider the normed linear space ($ \mathbb C^2$, $ $|| ||$_\infty$). $f$ is a linear functional on the space given by, $f(x,y)=ax+by$, for some constant a,b. Calculate norm of the functional $f$.

Now, it can be easily shown that $\frac {|f(x,y)|} {||(x,y)||_\infty} \leq |a|+|b|$ $\Rightarrow$ $||f|| \leq |a|+|b|$ . I know that the norm of the functional is $|a|+|b|$. However I am not being able to establish the equality $||f|| = |a|+|b|$. Someone kindly help me in proving the opposite inequaity. I have seen many hints that were available online and still have not been able to figure it out. So please be a little elaborate.

Thank you for your time.

答:
We can put $\tilde{x}=\overline{a}/\vert a \vert$ and $\tilde{y}=\overline{b}/\vert b \vert$. Then we have $ \Vert (\tilde{x}, \tilde{y}) \Vert_\infty = 1 $ and $$f(\tilde{x}, \tilde{y})= a \overline{a}/\vert a \vert + b\overline{b}/b = \vert a \vert + \vert b \vert.$$
Hence, we get
$$ \Vert f \Vert = \sup_{\Vert (x,y) \Vert_\infty=1} \vert f(x,y) \vert \geq \vert f(\tilde{x}, \tilde{y}) \vert=\vert a \vert + \vert b \vert.$$
In fact this is the general strategy to get the norm of a functional. When we have an upper bound (and believe that is indeed the norm), then we try to construct vectors, we will yield better approximations from below. Here we are lucky and can directly construct one vector where the norm is attained.

Title: calculating norm of a linear functional


问:
How much algebraic geometry is there in complex geometry (for example, Demailly)?
I wonder how much of modern algebraic geometry (schemes, etc) is there in complex (algebro-analytic) geometry. What I mean is complex algebraic (analytic) geometry in the sense of Demailly, Lasarsfeld etc.

I wonder if scheme theory (as in Grothendieck's work) is actually useful there.

答:
I am not a complex geometer, but my understanding is that since complex geometry is a vast and old field of research, the blend of methods (analysis, differential and algebraic geometry, topology) varies considerably depending on the topic.

To complement Georges Elencwajg's answer, let me mention the following. As far as scheme theoretic ideas go, they appear in the following elementary foundational topics (judging from what I have read in the standard textbooks of Demailly, Voisin and Griffiths-Harris):

Chow groups and intersection theory on smooth projective varieties. Apart from studying intersection theory in a purely topological (cup products in cohomology) and differential topological ways (wedge products of differential forms), it is also important to understand the algebro-geometric Chow groups. Voisin in Chapter 9 of Volume II mainly follows Fulton's treatment, which is purely algebraic and heavily uses scheme-theoretic machinery (multiplicities, generic points, nilpotent, etc.)

Deformation theory. It appears that the algebraic approach to deformation theory of complex manifolds and vector bundles is also pretty commonly used in complex geometry. E.g. in Chapter 9 of Volume I, Voisin uses infinitesimal neighbourhoods and rings of dual numbers to explain the Kodaira-Spencer map.

At the research level, it appears that complex geometry is even more interconnected with algebraic geometry. It is hard to tell which way the flow of new ideas go, e.g. multiplier ideal sheaves come to mind, and recent research on the interaction of various positivity notions in its differential- and algebro- geometric incarnations (Demailly, Lazarsfeld).


答:
cannot comment but some clarification would be fitting;

What is meant by ||[blah]|| in this context (supposing uϵ[blah]), or more to the point Why/how would it be equal to n minus [blah] given $nϵ{N},0<|u|<1$?

Title: the second order continued fraction expansion of an irrational number and the distance to its closest integer

Body:
Let $u$ be a posive irrational number and let

$$u=a_0+ \frac{1}{a_1+\frac{1}{a_2  \cdots} }$$

be its continued fraction expansion.

Consider the second-order finite continued fraction expansion of $u$:

$\frac{p_2}{q_2}:=a_0+ \frac{1}{a_1+\frac{1}{a_2}}$ with $p_2$ and $q_2$ being positive coprime integers.

I wonder if we have the following inequality and how to prove it:

$$|p_2-q_2 u|<\|u\|,$$

where $\|u\|$ denotes the distance from $u$ to its nearest integer.

Please also note that I am not asking how to prove $|\frac{p_2}{q_2}-u|<\|u\|$, which is simpler.

If the above is true (I have verified it with many examples), please feel free to use any very basic properties of continued fractions to prove it.

But please do not use any general fact about the optimality of the rational approximation with continued fractions to prove this because I actually want to use the above fact as one of the steps to prove optimality of the rational approximation with continued fractions.


问:
Absolute differences can be split into two sets with equal sums
Assume we have $n$ distinct real numbers. We replace these numbers with the absolute difference of each two of them. Prove that if $n$ is odd, then the resulting numbers can be divided into two sets of numbers, such that the sum of these two sets are equal.

We can order the numbers in descending order without changing the problem and by that get rid of the absolute function. How should I proceed? Does this have anything to do with number of pairs that can be chosen?

答:
As seen in the comments, you are stuck with the induction step. Consider the following decomposition just for the last two elements (when you induct) in the numbers $x_1<\cdots <x_{2n-1}<\color{blue}{x_{2n}<x_{2n+1}}.$
$$A=\{x_{2n+1}-x_{2n}\}\cup \{x_{2n+1}-x_i:1\leq i\leq n-1 \}\cup \{x_{2n}-x_j:n\leq j\leq 2n-1\}.$$
Take the rest of the elements in another set $B$ and conclude that both of them have as a sum $$nx_{2n+1}+(n-1)x_{2n}-\sum _{i=1}^{2n-1}x_i.$$


问:
Integrating $\int_0^\infty \sin(1/x^2) \, \operatorname{d}\!x$
How would one compute the following improper integral:

$$\int_0^\infty \sin\left(\frac{1}{x^2}\right) \, \operatorname{d}\!x$$

without any knowledge of Fresnel equations?

I was thinking of using the comparison theorem since $\sin x$ is never greater than 1. However, I can't find a function always greater than 1 between 0 and infinity such that its integral from 0 to infinity converges.

Thanks.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\sin\pars{1 \over x^{2}}
\,\dd x}
\,\,\,\stackrel{x\ =\ t^{-1/4}}{=}\,\,\,\,\,\,
{1 \over 4}\int_{0}^{\infty}
t^{\color{red}{1/4} - 1}\,\,\,
{\sin\pars{\root{t}} \over \root{t}}\,\dd t
\end{align}
The last integral can be evaluated with
Ramanujan's Master Theorem because
$\ds{{\sin\pars{\root{t}} \over \root{t}} =
\sum_{k = 0}^{\infty}
{\pars{-1}^{k} \over \pars{2k + 1}!}\,t^{k} =
\sum_{k = 0}^{\infty}\
\color{red}{\Gamma\pars{1 + k} \over \Gamma\pars{2 + 2k}}\,{\pars{-t}^{k} \over k!}}$.

Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\sin\pars{1 \over x^{2}}
\,\dd x} =
{1 \over 4}\,\Gamma\pars{\color{red}{1 \over 4}}\,
{\Gamma\pars{1 - \color{red}{1/4}} \over
\Gamma\pars{2 - 2\bracks{\color{red}{1/4}}}}
\\[5mm] = &\
{1 \over 4}\,{\pi \over \sin\pars{\pi/4}}\,{1 \over \pars{1/2}\root{\pi}}
= \bbx{{\root{2} \over 2}\root{\pi}} \approx 1.2533 \\ &
\end{align}


问:
Likelihood of Uniform Distribution Indicator Function
It is well known that the likelihood function for the uniform distribution on $[0,\theta]$ is given by

$$\frac{1}{\theta^n} \mathbf{1}_{\max(x_1,\ldots,x_n)\leq \theta}$$

Where the reason for this indicator is that the likelihood will be equal to $0$ if one of our observations $x_i$ exceeds $\theta$. But why do we not impose a similar condition on an observation being less than $0$? That is, also including $\mathbf{1}_{\min{(x_1,\ldots,x_n)\geq0}}$?

Sorry if I've misunderstood anything, please feel free to correct me!

答: No matter what $\theta$ is, your actual observations $x_i$ will be nonnegative (under the assumption that it comes from a uniform distribution on $[0, \theta]$). So technically, yes, you could include an indicator for $x_i \ge 0$, but that inequality will always hold.


问:
How many integers between $1$ and $1000$ are divisible by $2, 4,$ or $7?$
How many integers between 1 and 1000 are divisible by $2, 4,$ or $7?$

$A \cup B = A+B- A \cap B=500+142-71=571$

Body:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bound: $1\le\text{n}\le10^3$.

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
\[Alpha] = 10^3;
ParallelTable[
  If[TrueQ[IntegerQ[n/2] || IntegerQ[n/4] || IntegerQ[n/7]], n,
   Nothing], {n, 1, \[Alpha]}] //. {} -> Nothing

Running the code gives:

Out[1]={2, 4, 6, 7, 8, 10, 12, 14, 16, 18, 20, 21, 22, 24, 26, 28, 30, 32,
34, 35, 36, 38, 40, 42, 44, 46, 48, 49, 50, 52, 54, 56, 58, 60, 62,
63, 64, 66, 68, 70, 72, 74, 76, 77, 78, 80, 82, 84, 86, 88, 90, 91,
92, 94, 96, 98, 100, 102, 104, 105, 106, 108, 110, 112, 114, 116,
118, 119, 120, 122, 124, 126, 128, 130, 132, 133, 134, 136, 138, 140,
142, 144, 146, 147, 148, 150, 152, 154, 156, 158, 160, 161, 162, 164,
166, 168, 170, 172, 174, 175, 176, 178, 180, 182, 184, 186, 188, 189,
190, 192, 194, 196, 198, 200, 202, 203, 204, 206, 208, 210, 212, 214,
216, 217, 218, 220, 222, 224, 226, 228, 230, 231, 232, 234, 236, 238,
240, 242, 244, 245, 246, 248, 250, 252, 254, 256, 258, 259, 260, 262,
264, 266, 268, 270, 272, 273, 274, 276, 278, 280, 282, 284, 286, 287,
288, 290, 292, 294, 296, 298, 300, 301, 302, 304, 306, 308, 310, 312,
314, 315, 316, 318, 320, 322, 324, 326, 328, 329, 330, 332, 334, 336,
338, 340, 342, 343, 344, 346, 348, 350, 352, 354, 356, 357, 358, 360,
362, 364, 366, 368, 370, 371, 372, 374, 376, 378, 380, 382, 384, 385,
386, 388, 390, 392, 394, 396, 398, 399, 400, 402, 404, 406, 408, 410,
412, 413, 414, 416, 418, 420, 422, 424, 426, 427, 428, 430, 432, 434,
436, 438, 440, 441, 442, 444, 446, 448, 450, 452, 454, 455, 456, 458,
460, 462, 464, 466, 468, 469, 470, 472, 474, 476, 478, 480, 482, 483,
484, 486, 488, 490, 492, 494, 496, 497, 498, 500, 502, 504, 506, 508,
510, 511, 512, 514, 516, 518, 520, 522, 524, 525, 526, 528, 530, 532,
534, 536, 538, 539, 540, 542, 544, 546, 548, 550, 552, 553, 554, 556,
558, 560, 562, 564, 566, 567, 568, 570, 572, 574, 576, 578, 580, 581,
582, 584, 586, 588, 590, 592, 594, 595, 596, 598, 600, 602, 604, 606,
608, 609, 610, 612, 614, 616, 618, 620, 622, 623, 624, 626, 628, 630,
632, 634, 636, 637, 638, 640, 642, 644, 646, 648, 650, 651, 652, 654,
656, 658, 660, 662, 664, 665, 666, 668, 670, 672, 674, 676, 678, 679,
680, 682, 684, 686, 688, 690, 692, 693, 694, 696, 698, 700, 702, 704,
706, 707, 708, 710, 712, 714, 716, 718, 720, 721, 722, 724, 726, 728,
730, 732, 734, 735, 736, 738, 740, 742, 744, 746, 748, 749, 750, 752,
754, 756, 758, 760, 762, 763, 764, 766, 768, 770, 772, 774, 776, 777,
778, 780, 782, 784, 786, 788, 790, 791, 792, 794, 796, 798, 800, 802,
804, 805, 806, 808, 810, 812, 814, 816, 818, 819, 820, 822, 824, 826,
828, 830, 832, 833, 834, 836, 838, 840, 842, 844, 846, 847, 848, 850,
852, 854, 856, 858, 860, 861, 862, 864, 866, 868, 870, 872, 874, 875,
876, 878, 880, 882, 884, 886, 888, 889, 890, 892, 894, 896, 898, 900,
902, 903, 904, 906, 908, 910, 912, 914, 916, 917, 918, 920, 922, 924,
926, 928, 930, 931, 932, 934, 936, 938, 940, 942, 944, 945, 946, 948,
950, 952, 954, 956, 958, 959, 960, 962, 964, 966, 968, 970, 972, 973,
974, 976, 978, 980, 982, 984, 986, 987, 988, 990, 992, 994, 996, 998,
1000}

So, there are $571$ solutions. I found that, using the following code:

In[2]:=Clear["Global`*"];
\[Alpha] = 10^3;
Length[ParallelTable[
   If[TrueQ[IntegerQ[n/2] || IntegerQ[n/4] || IntegerQ[n/7]], n,
    Nothing], {n, 1, \[Alpha]}] //. {} -> Nothing]

Out[2]=571

Extending the bound to $1\le\text{n}\le10^9$ gives $571428571$ solutions.


问:
Determining conditions on two circles inscribed in square so that a tangent rectangle has integer area
In the accompanying figure, all seven contact points are tangential. It can be verified that if the areas of the major and minor circles, noted $A$ and $a$, are respectively equal to $8\pi$ and $2\pi$ then the area of the blue rectangle is equal to $7$.

Can you find without much work necessary and sufficient conditions between the areas $A$ and $a$ so that the area of the blue rectangle is an integer? Or would it be too difficult?

答:
After observing that the centers of the two circles are on the diagonal involved, everything comes out easy with the $45º$ angles in play. Thus we arrive at the relationship between the radii $R$ and $r$ which expresses the integer area $n$ of the blue rectangle.
$$\frac12(6rR-R^2-r^2)=n$$
Moving then to areas $A$ and $a$ of the problem we have the equation
$$6\sqrt{aA}-A-a=2n\pi$$ from this an easy partial solution is given by the mentioned conditions in Stinking Bishop's answer above.

However the solution is much more general. Expressing the areas as a function of the radii, the transcendent $\pi$ disappears and we return to the original expression, so we can do

$$R=3r\pm\sqrt{8r^2-2n}$$ This suggests that for each natural integer $n$ there exists an infinity of real pairs $(r, R)$ such that the corresponding blue rectangle has an area equal to $n$.

We end with an example for area $n=1$ (there is an infinity possible).

Example.-$(n,r,R)=(1,\sqrt3,\space 3\sqrt3+\sqrt{22})$.

In fact $$6\sqrt3(3\sqrt3+\sqrt{22})-(\sqrt3)^2-(3\sqrt3+\sqrt{22})^2=2\cdot1$$


问:
The sum $\sum_{k=1}^{n-1}{10k+5\over (k+1)!2^k}$
I'm solving a problem, and I'm getting the following sum as the solution. $$\displaystyle\sum_{k=1}^{n-1}{10k+5\over (k+1)!2^k}$$

Wolfram Alpha says this can be simplified. How do I arrive at the result? I believe I've calculated similar sums in the past, but now I'm completely stumped.

Is there an elementary method to do it? Wolfram Alpha's result doesn't involve any symbols that a high school student wouldn't underdstand so maybe it's doable by elementary means?

Added. I think I made a mistake in obtaining the sum, and it should be $2^{k+1}$ in the denominator instead of $2^k$. I don't know if it changes anything. People were saying something about changing $(k+1)!$ to $k!$ in the comments, so maybe it does, but I completely don't see it. The expression Wolfram Alpha gives now isn't any simpler.

Body:
\begin{align}
\sum_{k=1}^{n-1}{\frac{10k+5}{2^k(k+1)!}} & =\sum_{k=1}^{n-1}{\frac{10k+10}{2^k(k+1)!}}-\sum_{k=1}^{n-1}{\frac{5}{2^k(k+1)!}} \\
& =10\sum_{k=1}^{n-1}{\frac{1}{2^kk!}}-10\sum_{k=1}^{n-1}{\frac{1}{2^{k+1}(k+1)!}} \\
& =10\sum_{k=1}^{n-1}{\frac{1}{2^kk!}}-10\sum_{k=2}^{n}{\frac{1}{2^kk!}} \\
& =10(\frac{1}{2^11!}-\frac{1}{2^nn!}) \\
& =5-\frac{10}{2^nn!}
\end{align}


问:
Are these continued fractions of integrals known?
Motivated by Bowman and McLaughlin (2000) on polynomial continued fractions, I considered an extension to functional continued fractions, where the numerator of each fraction is the integral of the previous fraction.

That is, define the continued fraction integral transform (CFIT) as \begin{align}\{\mathcal If(t)\}(s)=\dfrac{f(s)}{1+\dfrac{\int_0^s f(x)\,dx}{1+\dfrac{\int_0^s\int_0^u f(x)\,dx\,du}{1+\cdots}}}=\dfrac{f(s)}{1+\dfrac{f^{(-1)}(s)}{1+\dfrac{f^{(-2)}(s)}{1+\cdots}}}={\large{\mathop{\mathrm{K}}_{n=0}^{\infty}}}\frac{f^{(-n)}(s)}1\end{align} where $f\in C^\infty(\Bbb R)$ and $f^{(-k)}$ is Lagrange's notation for the $k$th antiderivative of $f$.

Some elementary properties of the transform include

$\{\mathcal I(0)\}(s)=0$;

$\{\mathcal If\}(0)=f(0)$;

$\{\mathcal If\}(s)=\dfrac{f'(s)}{\{\mathcal If'\}(s)}-1$ for all non-constant $f$ such that $f(0)=0$.

Considering the simplest, non-trivial case $f(t)\equiv1$, we have $f^{(-k)}(t)=t^k/k!$ so that $$\{\mathcal I(1)\}(s)=\dfrac{1}{1+\dfrac s{1+\dfrac{s^2/2!}{1+\cdots}}}.$$

We can see that as $n$ increases, $\mathcal I(1)$ produces more waves. As @PeterForeman mentioned in the comments, the limit $\lim\limits_{s\to-\infty}\mathcal I(1)$ does not exist since it is zero for odd $n$ and $+1$ for even $n$. However, I believe there are more things that can be investigated.

Questions

Has the CFIT already been studied/discovered? If so, reference to literature would be appreciated.

Going back to the example of $\{\mathcal I(1)\}(s)$, is it true that the maximum value of each wave increases as $s$ increases up to the first wave? Are there an infinite number of such waves, and is there a closed form for $\max\mathcal I(1)$?

答:
Not an answer, but one more picture, with the functions for 100 and 150 terms superposed.

It is not clear what is happening on the left.

EDIT:

Here is the code (using sagemath):

```<code>def fun(x, N=150):
    L = [x**k/gamma(k + 1) for k in range(N, -1, -1)]
    resu = 1.0
    for t in L:
        resu = t / (1 + resu)
    return resu
P1 = plot(lambda x:fun(x, 100),-12, 3, aspect_ratio=1, color='red'); P1
</code>```

Body:
Not an answer, but one more picture, with the functions for 100 and 150 terms superposed.

It is not clear what is happening on the left.

EDIT:

Here is the code (using sagemath):

def fun(x, N=150):
    L = [x**k/gamma(k + 1) for k in range(N, -1, -1)]
    resu = 1.0
    for t in L:
        resu = t / (1 + resu)
    return resu
P1 = plot(lambda x:fun(x, 100),-12, 3, aspect_ratio=1, color='red'); P1


答:
The point is just that the family of distributions you are considering has different values of $\theta$ for different distributions within the family, but the lower bound of the interval remains the same for all of them.

(The likelihood function is a function of $\theta$ with $x_1,\ldots,x_n$ fixed, so it is
$$
L(\theta) = \begin{cases} 1/\theta^n & \text{for } \theta\ge\text{something}, \\ 0 & \text{otherwise.} \end{cases}
$$
Thus the m.l.e. of $\theta$ is the lowest value that $\theta$ can attain without $L(\theta)$ being $0.$)

Title: Likelihood of Uniform Distribution Indicator Function

Body:
It is well known that the likelihood function for the uniform distribution on $[0,\theta]$ is given by

$$\frac{1}{\theta^n} \mathbf{1}_{\max(x_1,\ldots,x_n)\leq \theta}$$

Where the reason for this indicator is that the likelihood will be equal to $0$ if one of our observations $x_i$ exceeds $\theta$. But why do we not impose a similar condition on an observation being less than $0$? That is, also including $\mathbf{1}_{\min{(x_1,\ldots,x_n)\geq0}}$?

Sorry if I've misunderstood anything, please feel free to correct me!


问:
Solid of revolution of $\frac{1}{x^a}$
Consider the solid of revolution $L_a$ that arises with $$f(x)= \frac{1}{x^a}$$ with $0< x\leq 1$ about the $x$-axis.

I need to find all $a > 0$ where the volume $L_a$ would be finite and infinite.

I tried the following:

$$\pi \int_0^1\frac{1}{x^{2a}} dx$$
$$= \pi (\frac{1^{-2a+1}}{-2a+1})$$

The only solution that is not possible in this case, is when $a = 1/2$. But in this case we wouldn't get infinite, we would just divide by $0$, which is not possible. So the volume would be finite for all $\mathbb{R}\setminus \{1/2\}$. Which steps did I do wrong in this process?

答:
There's a mistake on the step
$$
\int_0^1\frac{1}{x^{2a}} dx = \frac{1^{-2a+1}}{-2a+1}
$$
which is not always true. To see why we'll split the problem in cases.

If $\color{blue}{a<\frac{1}{2}}$ we have that $2a<1 \implies 1-2a>0$, and since $1-2a$ is positive we can correctly assert that $\color{purple}{0^{1-2a} = 0}$. Notice that this is not always the case, as if we were to have a negative exponent we would have a $0$ in the denominator. Because of the previous discussion, we get that

\begin{align*}
\int_0^1\frac{1}{x^{2a}} dx &= \frac{x^{1-2a}}{1-2a}\Biggr|_{0}^{1}\\
&= \frac{1^{1-2a}}{1-2a}-\frac{\color{purple}{0^{1-2a}}}{1-2a} \\
& = \frac{1}{1-2a}
\end{align*}
which is correctly defined since $1-2a>0 \implies 1-2a \neq 0$. Hence, for $a < \frac{1}{2}$ the volume is finite.

If $\color{blue}{a=\frac{1}{2}}$ we see that
\begin{align*}
\int_0^1\frac{1}{x^{2a}} dx &= \lim_{\varepsilon \to 0^+}\int_\varepsilon^1\frac{1}{x^{2a}} dx\\
&= \lim_{\varepsilon \to 0^+}\int_\varepsilon^1\frac{1}{x} dx\\
& = \ln(1) - \lim_{\varepsilon \to 0^+}\ln(\varepsilon)\\
& = - \lim_{\varepsilon \to 0^+}\ln(\varepsilon)
\end{align*}
but since the natural log diverges at $0$, the integral diverges, and hence, for $a=\frac{1}{2}$ the volume is infinte.

If $\color{blue}{a>\frac{1}{2}}$ we have the following. Recall that $x^n  \le x$ for $x\in(0,1]$ and $n>1$ since
$$ x \le1 \overset{\color{purple}{n-1 >0}}{\implies} x^{n-1} -1  \le 0 \overset{\color{purple}{x  > 0}}{\implies}  x \left(x^{n-1} -1\right) \le 0 \implies x^n  \le x$$
Since in our integral we're interested in values of $x$ precisely on the interval from $0$ to $1$ we can say
$$
\underbrace{x^{2a} \le x}_{\color{blue}{2a>1}} \implies \frac{1}{x} \le \frac{1}{x^{2a}} \implies\underbrace{\int_0^1 \frac{1}{x} \ dx}_{\color{red}{\infty}}\le \int_0^1\frac{1}{x^{2a}} \ dx
$$
recalling that we previously showed that the case of $a = \frac{1}{2}$ is divergent. Since in this case every volume has to be greater than a divergent integral, all these volumes are also infinite.

In conclusion, the possible values of $a>0$ for which the volume would be finite would correspond to the values of $a$ in the interval
$$
a \in \left(0, \frac{1}{2}\right)
$$
and the volume would be infinite for every other positive value.

Title: Solid of revolution of $\frac{1}{x^a}$


问:
Calculating $\lim_{(x,y)\rightarrow (0,0)}\frac{1 - \cos(\pi x y ) + \sin (\pi(x^2 + y^2))}{x^2 + y^2}$ if it exists
I'm trying to calculate the above limit. I ran a few paths and found out that the limit is $\pi$ (which I also confirmed through WolframAlpha), but to prove it I use polar coordinates to get an expression of the form $F(r) \cdot G(\theta)$ where $F(r)\rightarrow$ when $r\rightarrow 0$ and $G$ is blocked. So far I have:

$$\lim_{r\rightarrow 0} \left|f(r\cos \theta,r \sin \theta)\right| = \lim \left|\frac{1-\cos(\pi r^2 \cos\theta \sin \theta ) + \sin (\pi r^2 (\sin^2\theta  + \cos ^2 \theta ))}{r^2\cos ^2 \theta + r^2 \sin^2 \theta}\right|
\\ \underset{\sin ^2 \theta + \cos ^2 \theta = 1}{=} \lim \left|\frac{1 - \cos(\pi r^2 \cos\theta \sin\theta ) + \sin (\pi r^2 )}{r^2}\right|$$

However, I don't know how to separate $r$ from the rest of the expression at this stage. But it occurred to me that this only works if the limit of the function is zero. So, how do I confirm this is the correct limit in a case such as this?

答:
On the one hand, $$\lim_{r\to 0}{\frac{\sin(\pi r^2)}{r^2}}=\pi\cdot\lim_{r\to 0}{\frac{\sin(\pi r^2)}{\pi r^2}}\overset{t=\pi r^2}{=}\pi\cdot\lim_{t\to 0}{\frac{\sin t}{t}}=\pi\cdot 1=\pi.$$
On the other hand,
\begin{align*}
\lim_{r\to 0}{\frac{1-\cos(\pi r^2\cos\theta\sin\theta)}{r^2}}&\overset{t=\pi r^2\cos\theta\sin\theta}{=}\pi\cos\theta\sin\theta\cdot\lim_{t\to 0}{\frac{1-\cos t}{t}}=0,
\end{align*}
because $$\lim_{t\to 0}{\frac{1-\cos t}{t}}=\lim_{t\to 0}{\frac{2\sin^2(t/2)}{t}}\overset{s=t/2}{=}\lim_{s\to 0}{\frac{\sin^2s}{s}}=\lim_{s\to 0}{\frac{\sin^2 s}{s^2}\cdot s}=1^2\cdot 0=0.$$
Therefore, the limit is $\pi$.


问:
Geometric interpretation of cubic curve?
Lines and conics have clear geometric meanings that are coordinate-free, but cubics seem to rely entirely on cubic equations and coordinate systems. Are there ways to define cubic curves without cubic equations?

I thought about this question by trying to generalize tangent lines and osculating circles. Since a tangent line is defined by taking the limit of two points on the curve, and an osculating circle is defined by taking the limit of two tangent lines on the curve, I thought about taking the limit of two osculating circles on the curve, but after playing around with evolutes and involutes, I couldn't come up with anything.

答:
A similar question has been asked and answered on Mathoverflow:

It is a "classical" fact that any nonsingular plane cubic curve $C$
can be projectively generated by means of a pencil of lines and a
pencil of conics.

The starting point of the construction is the observation that the
lines defined by any $g_2^1$ on $C$ all pass through the same point
$p$ of $C$, that following Sylvester is called the coresidual point.

Now, take four points $q_1, \ldots, q_4$ on $C$, such that any three
of them are not  collinear. Let $p \in C$ be the coresidual point with
respect to the $g_2^1$ on $C$ cut by the pencils of conics through
$q_1, \ldots, q_4$. Therefore such a pencil of conics and the pencil
of lines through $p$ projectively generate $C$.

All of this is explained in the paper by N. Fraser Kötter's
synthetic geometry of algebraic curves, Proceedings of the
Edinburgh Mathematical Society 7, 46–61 (1888). However, the
language is rather old-fashioned so the article is not easy to read
nowadays.

A modern treatment  can be found in Dolgachev's book Classical
Algebraic Geometry, Section 3.3 (this is the googlebook link).

See also Bunch, On the Straight Line Construction of Unicursal Cubics.  Here the construction uses a 1-2 correspondence between two pencils of lines.

And for something more down to earth: Matheq, A General Construction for Circular Cubics

OwnerDisplayName: Kopa Leo


Title: Calculating $\lim_{(x,y)\rightarrow (0,0)}\frac{1 - \cos(\pi x y ) + \sin (\pi(x^2 + y^2))}{x^2 + y^2}$ if it exists

Body:
I'm trying to calculate the above limit. I ran a few paths and found out that the limit is $\pi$ (which I also confirmed through WolframAlpha), but to prove it I use polar coordinates to get an expression of the form $F(r) \cdot G(\theta)$ where $F(r)\rightarrow$ when $r\rightarrow 0$ and $G$ is blocked. So far I have:

$$\lim_{r\rightarrow 0} \left|f(r\cos \theta,r \sin \theta)\right| = \lim \left|\frac{1-\cos(\pi r^2 \cos\theta \sin \theta ) + \sin (\pi r^2 (\sin^2\theta  + \cos ^2 \theta ))}{r^2\cos ^2 \theta + r^2 \sin^2 \theta}\right|
\\ \underset{\sin ^2 \theta + \cos ^2 \theta = 1}{=} \lim \left|\frac{1 - \cos(\pi r^2 \cos\theta \sin\theta ) + \sin (\pi r^2 )}{r^2}\right|$$

However, I don't know how to separate $r$ from the rest of the expression at this stage. But it occurred to me that this only works if the limit of the function is zero. So, how do I confirm this is the correct limit in a case such as this?

Body:
hint

$$1-\cos(2A)=2\sin^2(A)$$
$$=2\sin^2(|A|)\le 2A^2$$

$$\lim_{r\to0}\frac{\sin(\pi.r^2)}{r^2}=\pi$$


Body:
$$\frac{1 - \cos(\pi x y ) + \sin (\pi(x^2 + y^2))}{x^2 + y^2} = \frac{1 - \cos(\pi x y )  }{x^2 + y^2}+\pi\frac{ \sin (\pi(x^2 + y^2))}{\pi(x^2 + y^2)}$$
For first
$$\left|\frac{1 - \cos(\pi x y )  }{x^2 + y^2}\right|\leqslant\left|\frac{2\sin^2(\pi \frac{x^2 + y^2}{2} )  }{x^2 + y^2}\right| \to 0$$
While second is $\pi$


问:
Trying to prove uniform continuity for a function
I have been trying to prove that the following function is not uniformly continuous:

Consider $A=(0,\infty), B=(-\infty,0),$ and $f: A \cup B \rightarrow \mathbb{R}$ given by
$$
f(x)=\left\{\begin{array}{ll}
x+1 & x>0 \\
x-1 & x<0
\end{array}\right.
$$

What have i tried so far?

To show that $f$ is not Uniformly continuos, I am going to show that there exists $\epsilon > 0$ such that, $\forall \delta > 0$ there exists $x, y \in A \cup B$ such that $$|x-y|<\delta$$ but $$|f(x)-f(y)| \geq \epsilon$$

Choose $\epsilon = 1$ and consider $\delta > 0$ given. By taking $y = x + \frac{\delta}{2}$ and $x > 0$ from which we get that: $$|x-y|<\delta$$

Now, $$|f(x) - f(y)| = |(x+1) - (x - \frac{\delta}{2})| $$
which equals $$1 + \frac{\delta}{2}$$

Now i am stuck. I want the last quantity to be greater than $\epsilon = 1$ but i cant assure that, since i have no control over the value of $\delta$.

Can someone help me? Something is just not right. I would appreciate feedback followed by a detailed solution, i am really struggling to prove a function is not uniformly continuous in general.

Thanks in advance, Lucas

答:
Your computation looks wrong. Your $x$ and $y$ are both positive, so $f(x)=x+1$, and $f(y)=(x+\frac{\delta}{2})+1$. Thus $|f(x)-f(y)|=|(x+1)-(x+1+\frac{\delta}{2})|=\frac{\delta}{2}$. If $\delta$ is small then this will not be bigger than $1$.

Anyway, your choice of $x$ and $y$ will not work, because the function is uniformly continuous in both $A$ and $B$. What ruins uniform continuity in the union is when $x\in A$ and $y\in B$ or vice versa. So let $\delta>0$ and choose $x=\frac{\delta}{3}, y=-\frac{\delta}{3}$. Then $|x-y|<\delta$. However:

$|f(x)-f(y)|=|(\frac{\delta}{3}+1)-(-\frac{\delta}{3}-1)|=2+\frac{2\delta}{3}\geq 1=\epsilon$


答:
Take $\delta>0$. Now, take numbers $x$ and $y$ such that

$-\frac\delta2<x<0$;

$f(x)<-\frac12$;

$0<y<\frac\delta2$;

$f(y)>\frac12$.

Then $|x-y|<\delta$, and, since $f(x)<-\frac12$ and $\frac12<f(y)$,  $\bigl|f(x)-f(y)\bigr|>1$. So, $f$ is not uniformly continuous: there is some $\varepsilon>0$ (namely, $\varepsilon=1$) such that, for every $\delta>0$, there are numbers $x$ and $y$ in the domain of $f$ such that$$|x-y|<\delta\quad\text{and}\quad\bigl|f(x)-f(y)\bigr|\geqslant1.$$

Title: Trying to prove uniform continuity for a function

Body:
I have been trying to prove that the following function is not uniformly continuous:

Consider $A=(0,\infty), B=(-\infty,0),$ and $f: A \cup B \rightarrow \mathbb{R}$ given by
$$
f(x)=\left\{\begin{array}{ll}
x+1 & x>0 \\
x-1 & x<0
\end{array}\right.
$$

What have i tried so far?

To show that $f$ is not Uniformly continuos, I am going to show that there exists $\epsilon > 0$ such that, $\forall \delta > 0$ there exists $x, y \in A \cup B$ such that $$|x-y|<\delta$$ but $$|f(x)-f(y)| \geq \epsilon$$

Choose $\epsilon = 1$ and consider $\delta > 0$ given. By taking $y = x + \frac{\delta}{2}$ and $x > 0$ from which we get that: $$|x-y|<\delta$$

Now, $$|f(x) - f(y)| = |(x+1) - (x - \frac{\delta}{2})| $$
which equals $$1 + \frac{\delta}{2}$$

Now i am stuck. I want the last quantity to be greater than $\epsilon = 1$ but i cant assure that, since i have no control over the value of $\delta$.

Can someone help me? Something is just not right. I would appreciate feedback followed by a detailed solution, i am really struggling to prove a function is not uniformly continuous in general.

Thanks in advance, Lucas


问:
Proof that a boundary of a cuboid is a zero-volume set
Let $ I $ be a cuboid in $ \mathbb{R}^n $.

We say that a set is of volume-zero, if for any $ \varepsilon> 0 $ there exists a finite number of cuboids $ I_1,\ldots,I_k $ such that

$$ \sum_{j=1}^k \operatorname{vol}(I_j)<\varepsilon, \qquad \text{and} \qquad A\subseteq\bigcup_{j=1}^{k}\text{int}(I_{j}) $$

where $ \operatorname{int}(I_j)$ denotes the interior of $I_j$

Now how can I prove that the boundary of a cuboid $ I $ is a zero-volume set?

I already proved that a compact set is of measure zero if and only if the set is of volume zero, so a proof that a boundary of a cuboid is a set of measure-zero would also be helpful.
(The definition I know for measure-zero set is the same as for volume-zero set that I mentioned here except that we allow the cover to be countable and not just finite).

Thanks in advance.

答:
Let your cuboid be $\prod_{i=1}^n[a_i,b_i]$. Then, its boundary is covered by $2n$ cuboid interiors:

$$X_i=(a_i-\delta, a_i+\delta)\times\prod_{j=1,j\ne i}^n(a_j-\delta,b_j+\delta)$$
$$Y_i=(b_i-\delta, b_i+\delta)\times\prod_{j=1,j\ne i}^n(a_j-\delta,b_j+\delta)$$

for $i=1,\ldots,n$. The volume of each of those interiors converges to $0$ as $\delta\to 0$:

$$\operatorname{vol}(X_i)=\operatorname{vol}(Y_i)=2\delta\prod_{j=1,j\ne i}^n(b_j-a_j+2\delta)$$

Thus, the (finite) sum of all those volumes also converges to $0$ as $\delta\to 0$. This means that the sum on the left-hand side of your definition of "volume zero" we can (by choosing suitably small $\delta$) make arbitrarily small (smaller than any pre-determined $\varepsilon>0$).

The intuition here is: to prove the boundary of a rectangle is of area $0$, we cover each side of the rectangle with very thin rectangles (of thickness $2\delta$), which "just" cover the sides. Then we let $\delta\to 0$.


问:
Is there a function that is both exponential or linear at one end and then logarithmic at the other (joining two types of functions generally)
This is more a general big picture conceptual question. I observe in life some phenomena display exponential growth. Wealth accumulation or growth of a new business or success as an artist can be included in this bucket. For years, visible signs of success may be slow, and then boom, things exponentiate.

I know some growth is logarithmic. You just can't get past a plateau. There's a kind of natural limit to a level of success. It strikes me that all things that grow exponentially must hit a limit or ceiling. You could also take the case of a foreign species introduced in a new environment in which they thrive. Their growth will be exponential, until it hits the upper limit of the carrying capacity.

Is there a type of function that combines these two concepts? It strikes me that indefinite, infinite exponential functions don't or can't exist in nature, except in rare cases (expansion of universe maybe).

I was wondering if such a thing exists and what the name for such a hybrid function would be, and where I can read up on it more, and how might it be described symbolically. I have never thought of merging or joining two separate types of functions and now I'm really curious.

答:
Of interest can be digamma function $\psi(x)$.

On the left side it resembles $-\pi\cot \pi x$ while on the right side it resembles logarithm.


问:
Elementary proof (involving no series expansion) that $\tan x \approx x+\frac{x^3}{3}$ for small $x$
For small values of $x$, the following widely used approximations follow immediately by Taylor expansion:

$\sin x \approx x$

$\cos x \approx 1-\frac{x^2}{2}$

$\tan x \approx x +\frac{x^3}{3}$

I am looking for a justification of these approximations without the use of series expansions.

By purely geometric considerations, it is easy to see that for small values of $x$, we have

$$ \sin x < x < \tan x.$$

Division by $\cos x$ and an application of the squeeze lemma yield

$$\frac{\sin x}{x}\xrightarrow{x\to0}1$$

and hence the approximation (1.). Using the identity $\cos x = 1-2 \sin^2 \frac{x}{2}$, the approximation (2.) also follows.

Can one justify the approximation (3.) by a similarly elementary argument without using Taylor expansion?

I tried around using the angle addition theorems, but I did not really get anywhere, mainly because I could not make the factor $\frac13$ appear anywhere.

答: Famously, a sector with a small angle proves $\lim_{x\to0}\tfrac{\sin x}{x}=1$ and hence$$\lim_{x\to0}\tfrac{1-\cos x}{x^2}=\lim_{x\to0}\tfrac{\sin^2\tfrac{x}{2}}{2(x/2)^2}=\tfrac12.$$We can improve the former result: if you approximate a sector's arc as a parabola, you can prove$$\lim_{x\to0}\frac{1-\tfrac{\sin x}{x}}{x^2}=\tfrac16.$$Finally,$$\lim_{x\to0}\frac{\tfrac{\tan x}{x}-1}{x^2}=\lim_{x\to0}\frac{\tfrac{\sin x}{x}-\cos x}{x^2}=\lim_{x\to0}\frac{1-\cos x}{x^2}-\lim_{x\to0}\frac{1-\tfrac{\sin x}{x}}{x^2}=\tfrac12-\tfrac16=\tfrac13.$$


问:
Help to understand the ring of polynomials terminology in $n$ indeterminates
In Hungerford's Algebra, page 150, the author defines a ring of polynomials in "$n$" indeterminates in the following manner:

After the author defines the operations in this ring with a theorem:

saying that the author defines another definition to a ring of polynomials with $n$ indeterminates:

My problem is with the terminology in this theorem which comes after:

If I understood well, the author defines a function $x_i$ which when we applies to $\epsilon_i$, we have the identity in $R$. following this reasoning what is $x_i^k(k\epsilon_i)?$ is this the composition of $x_i$?
I'm very confused with these terminologies, I need help.

Thanks a lot!

答:
It's not the composition, but just the multiplication of $x_i$ with itself $k$ times (as usual). Note that Theorem 5.3 defines a multiplication operation on $R[x_1,\ldots,x_n]$, via
$$(fg)(u)=\sum_{\large\substack{v+w=u\\ v,w\in\mathbb{N}^n}}f(v)g(w).$$
So for example, writing $x_i^2$ for the product $x_ix_i$,
$$(x_i^2)(2\epsilon_i)=\sum_{\substack{v+w=2\epsilon_i\\ v,w\in\mathbb{N}^n}}x_i(v)x_i(w)=\bigg[x_i(0)x_i(2\epsilon_i)\bigg]+\bigg[x_i(\epsilon_i)x_i(\epsilon_i)\bigg]+\bigg[x_i(2\epsilon_i)x_i(0)\bigg]$$
$$=[0\cdot 0]+[1\cdot 1]+[0\cdot 0]=1$$
(where we have used the defining property $x_i(\epsilon_i)=1$ and $x_i(u)=0$ for any other $u$).

The correspondence with our usual conception of polynomials is as follows: for any  polynomial $F\in R[x_1,\ldots,x_n]$, the corresponding function $\mathbf{F}:\mathbb{N}^n\to R$ is defined by
$$\mathbf{F}(k_1\epsilon_1+\cdots+k_n\epsilon_n)=\text{the coefficient of }x_1^{k_1}\cdots x_n^{k_n}\text{ in the polynomial }F.$$
Let's do an example in $\mathbb{Z}[s,t]$ (I'm using $s$ and $t$ to avoid confusion with the functions $x_1,x_2$):
$$F=2+t+5s^2$$
corresponds to the function $\mathbf{F}:\mathbb{N}^2\to \mathbb{Z}$ defined by
$$\mathbf{F}(u)=\begin{cases}
2 & \text{if }u=(0,0)\\
1 & \text{if }u=(0,1),\\
5 & \text{if }u=(2,0),\\
0 & \text{otherwise},
\end{cases}$$
and we can see that (as functions from $\mathbb{N}^2$ to $\mathbb{Z}$),
$$\mathbf{F}=2+x_1+5x_2^2.$$


问:
$2$ is prime, but $2=(1+i)(1-i)$ is composite.
If $2$ is a prime number, there is a corollary that says that every prime integer is irreducible on the Gaussian integers. But this is a contraction, for $2$ is composed of $(1+i)(1-i)$, and a number that can be composed cannot be irreducible.

How can this be?

I am totally confused. Does anyone propose to assist me?

答: Notice that $5 = \sqrt{5}\sqrt{5}$, so the prime $5$ can be factored in the ring $\mathbb{Z}[\sqrt{5}].$   "Prime" in this context means "can't be factored in the integers." It doesn't mean you can't add some more numbers to your system and create more factors.


问:
why is $\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$ correct?
Im trying to understand how the equation below was generated. A step-by-step solution would be awesome.
$$\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$$

答: Recall that $$\frac{1}{x}$$ means something so that, when you multiply it by $x$, you get $1$.  So check that $\sqrt{n+1}-\sqrt{n}$ is, indeed, something so  that, when you multiply it by $\sqrt{n+1}+\sqrt{n}$, you get $1$.


答:
That corollary probably says that every prime $p$ satisfying $p\equiv 3\pmod4$ stays prime among the Gauss integers.

Actually, every other primes (so $2$ and primes with $p\equiv1\pmod4$) do factor nontrivially in $\Bbb Z[i]$.

Title: $2$ is prime, but $2=(1+i)(1-i)$ is composite.

Body:
If $2$ is a prime number, there is a corollary that says that every prime integer is irreducible on the Gaussian integers. But this is a contraction, for $2$ is composed of $(1+i)(1-i)$, and a number that can be composed cannot be irreducible.

How can this be?

I am totally confused. Does anyone propose to assist me?


问:
Understanding why a vector topology is completely determined by any local base.
What is shown below is a reference from the text Functional Analysis by Walter Rudin.

If $X$ s a vector space over the field $\Bbb F$ then  the following notation well be used
$$
x+A=\{x+a:a\in A\}\\
x-A:=\{x-a:a\in A\}\\
A+B:=\{a+b:a\in A\wedge b\in B\}\\
\lambda A:=\{\lambda a:a\in A\}
$$
for any $A,B\subseteq X$, for any $x\in\ X$ and for any $\lambda\in\Bbb F$.

Definition

If $\tau$ is a topology on a vector space $X$ such that

every point of $X$ is a closed set

the vector space operation are continuous with respect to $\tau$

we say that $X$ is a topological vector space equipped with the vector topology $\tau$.

So we note that the condition $1$ imply that any topological vector space will be $T_1$ but many authors omitted the above condition from the definition of a topological vector space.

Proposition

Let be $X$ a topological vector space over the field $\Bbb F$ and for some $v\in X$ and $\lambda\in\Bbb F$ we define the $v$-traslation and $\lambda$-multiplication functions through the condition
$$
T_v(x):=x+v\,\,\,\text{and}\,\,\,M_\lambda(x):=\lambda x
$$
for any $x\in X$. So the above two defined functions are homeomorphisms.

Proof. Omitted.

One consequence of the above proposition is that every vector topology $\tau$ is translation-invariant, that is a set $E\subseteq X$ is open if and only if each of its translates $v+E$ is open and thus $\tau$ is completely determined by any local base so that the open sets of $X$ are then precisely those that are unions of translates of memebrs of any local base.

So I like to discuss the last statement: indeed since a neighborhood is not union of basic neighborhoods then the image of local base undera traslation is not a base so what means that a vector topology is completely determined by any local base? Could someone help me to understand this statement, please?

答: If $\mathscr{B}_x$ is a local base at $x$, $\mathscr{B}_y=\{B+(y-x):B\in\mathscr{B}\}$ is a local base at $y$, and $\bigcup_{y\in X}\mathscr{B}_y$ is a base for the topology of $X$. In other words, we can get a base for the topology and therefore the whole topology directly from that one local base $\mathscr{B}_x$.


问:
Orthogonal and orthonormal for dot-product
Consider the vectors $
u_1=\begin{pmatrix}
1\\
1\\
-1
\end{pmatrix},u_2=\begin{pmatrix}
2\\
2\\
4
\end{pmatrix},u_3=\begin{pmatrix}
1\\
-1\\
0
\end{pmatrix} $ in $\mathbb{R}^3$

and the map

$f:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by $f(x)=u_1\left \langle u_1, x \right \rangle + u_2\left \langle u_2, x \right \rangle + u_3\left \langle u_3, x \right \rangle$

Show that the family of vectors $u=(u_1,u_2,u_3)$ is a orthogonal family with respect to the dot-product in $\mathbb{R}^3$ and that it is a basis. Is it a orthonormal basis?

I know that we say that 2 vectors are orthogonal if they are perpendicular to each other. i.e. the dot product of the two vectors is equal to zero. Would I just have to simply apply the inner product on my space pairwise and conclude that it is equal to 0?

I know that an orthonormal basis for an inner product space with finite dimension is a basis for the space whose vectors are orthonormal i.e they are all unit vectors and orthogonal to each other.

Even though I seem to get the concepts and the thinking behind I do not know how to apply it in this example.

答: You're right that orthogonality means $\langle u_1,\,u_2\rangle=\langle u_2,\,u_3\rangle=\langle u_3,\,u_1\rangle=0$. As @MinusOne-Twelfth explained, the thogonality of $n$ nonzero vectors in $\Bbb R^n$ ensures they form a basis. For orthonormality, check whether $\langle u_1,\,u_1\rangle=\langle u_2,\,u_2\rangle=\langle u_3,\,u_3\rangle=1$; you should find this fails.


问:
Find y in terms of $t$
So I am given this equation for the rate of a reaction:
$$\frac{dy}{dt}=y^2-8y+15$$ when $t=0, y=0$ [what does this mean?]

How can I proceed to find $y$ in terms of $t$?

I checked out this site but I still don't get how you're supposed to find $y$ in terms of $t$ from it. Some guidance would be appreciated.

答:
I don't know if everything has cleared up for you yet but i wrote the whole thing here if someone still needs it.

Original problem: $$ \dfrac{dy}{dt}=y^2-8y+15 \; , \; y=0 \; t=0 $$

The problem is asking for a particular solution of this ODE at $ y=0 \; t=0$

The dependent and and independent variable is shown like this: $y(t)$, where $y$ depends on $t$ so by plugging in the initial contitions $y=0$ and $t=0\;$ we get $y(0)=0$.
As we get our general solution these initial values should be plugged in and then we can solve our equation for the constant of integration.

By analyzing the specific initial values of our ODE we can find their unique particular solution which by the way can be graphically represented by an integral curve after writing our specific result of the constant of integration back to the general solution.

In some literature $\dfrac{dy}{dt}$ is often marked with $y'(t)$ or $\dot{y}$ .

This is a First Order Inhomogeneous ODE.

This can be solved by separation of variables so integrating both sides with respect to $dy$ and $dt$ respectfully.

$$ \int \dfrac{dy}{y^2-8y+15}= \int dt $$

This LHS integral can be simplified by Partial Fraction Decomposition.

$$\dfrac{1}{y^2-8y+15}= \dfrac{1}{(y-3)(y-5)}= \dfrac{A}{y-3} + \dfrac{B}{y-5}= \dfrac{A(y-5)+B(y-3)}{y^2-8y+15} \Rightarrow$$

where $A, B \in \mathbb{R} $

$$ 1= A(y-5)+B(y-3)= Ay-5A+By-3B$$
$$ y^0: \; 1 = -5A -3B $$
$$ y^1: \; 0 = A+B$$
Solving for $A$ and $B$ we get $A= -\frac{1}{2}$ and $B=\frac{1}{2}$

Substituting back our simplified problem becomes:

$$ \int \left(\dfrac{-\frac{1}{2}}{y-3}+\dfrac{\frac{1}{2}}{y-5}\right)\; dy = \int dt $$

which can be solved by u-substitution or by the $\int \frac{f'}{f}$ rule.

Note: There are 3 integrals actually here, as from the property of linearity we integrate the LHS as two integrals so we'd technically get 3 constants of integration, 2 from $dy$ and 1 from $dt$.

We get:

$$ -\dfrac{1}{2}\ln(|y-3|)+ \dfrac{1}{2}\ln(|y-5|)+C_1 = t+C_2$$

where $C_1, C_2 \in \mathbb{R}$, let $C_2-C_1=C$ which is a real number too.

$$ -\dfrac{1}{2}\ln(|y-3|)+ \dfrac{1}{2}\ln(|y-5|)= t+C$$
$$ -\ln(|y-3|)+ \ln(|y-5|)= 2(t+C)$$

By the quotient rule of logarithms:

$$  \ln\left(\left|\dfrac{y-5}{y-3}\right|\right)= 2(t+C)$$

After exponentiating both sides:

$$ \left|\dfrac{y-5}{y-3}\right| = e^{2(t+C)} $$

Ideally we'd want to solve the equation for $y$ explicitly but this solution is enough in our case.

Let's plug in $y(0)=0$, we can use any equation between $y$ and $t$ after the integrations.
Since we are trying to solve for $C$, let's choose $  \ln\left(\left|\frac{y-5}{y-3}\right|\right)= 2(t+C)$

We get

$$  \ln\left(\left|\dfrac{0-5}{0-3}\right|\right)= 2(0+C) $$

After a bit of cleaning up

$$ C = \dfrac{\ln\left(\dfrac{5}{3}\right)}{2} $$

Which is approximately $$ C \approx 0.25541281188 $$

Let's subtitute $C$ back into our general solution:

$$ \left|\dfrac{y-5}{y-3}\right| = e^{2(t+C)} $$

In an engineering class 2-3 decimal points are usually enough.

So our (approximated) solution is
$$ \boxed{ \left|\dfrac{y-5}{y-3}\right| = e^{2(t+0.256)} } $$

Whose graph looks like this.

For a rather pure/rigourous mathematical setting they'd expect you to give an exact(non-approximated) closed-form solution, which would be:

$$ \boxed{\left|\dfrac{y-5}{y-3}\right| = exp\left({2t+\ln\left(\dfrac{5}{3}\right)}\right)} $$

( exp( for better visibility)

Title: Find y in terms of $t$


答:
To show that the vectors are an orthogonal family, yes you just have to apply the dot product pairwise and show that each pair of vectors has a dot product of $0$.

To next argue that the vectors form a basis for $\Bbb{R}^3$, recall that a set of orthogonal vectors, none of which are the zero vector, is linearly independent. So $\{ u_1, u_2, u_3\}$ is a set of three linearly independent vectors in $\Bbb{R}^3$, and hence a basis for $\Bbb{R}^3$.

Finally, to check whether it is an orthonormal basis, you just have to check whether the norm of all the vectors is $1$ (equivalently whether $u_i\cdot u_i = 1$ for all $i=1,2,3$). Since $u_1\cdot u_1 = 3 \ne 1$, this is not the case. Hence the basis is not orthonormal.

Title: Orthogonal and orthonormal for dot-product

Body:
Consider the vectors $
u_1=\begin{pmatrix}
1\\
1\\
-1
\end{pmatrix},u_2=\begin{pmatrix}
2\\
2\\
4
\end{pmatrix},u_3=\begin{pmatrix}
1\\
-1\\
0
\end{pmatrix} $ in $\mathbb{R}^3$

and the map

$f:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by $f(x)=u_1\left \langle u_1, x \right \rangle + u_2\left \langle u_2, x \right \rangle + u_3\left \langle u_3, x \right \rangle$

Show that the family of vectors $u=(u_1,u_2,u_3)$ is a orthogonal family with respect to the dot-product in $\mathbb{R}^3$ and that it is a basis. Is it a orthonormal basis?

I know that we say that 2 vectors are orthogonal if they are perpendicular to each other. i.e. the dot product of the two vectors is equal to zero. Would I just have to simply apply the inner product on my space pairwise and conclude that it is equal to 0?

I know that an orthonormal basis for an inner product space with finite dimension is a basis for the space whose vectors are orthonormal i.e they are all unit vectors and orthogonal to each other.

Even though I seem to get the concepts and the thinking behind I do not know how to apply it in this example.


问:
Understanding the definition the norm of a principal ideal of a number field
Hi I'm a grad student and I was re reading the lecture notes my professor from a previous semester posted for algebraic number theory.

The lecture notes state: "If $O_K$ is a number field and $I$ is an ideal of $O_K$, then $N(I)$ is defined as $N(I) = |O_K : I|$."

I took this definition to mean, if $N(I) = n$ then there are $n$ elements $a_1, a_2, \cdots a_n \in O_K$ such that every element of $a \in O_k$ is of the form $i + a_j$ where $i \in I$, and $1 \leq j \leq n$.

Later on the lecture notes state:

"Let $\gamma \in O_K$, and $\{\beta_1, \cdots \beta_n \}$ be an integral basis of $O_K$. Let $$\gamma\beta_j = \sum_{k = 1}^n a_{j,k}\beta_k, \, \, \, \, \, j = 1, 2, \cdots, n,$$
then \begin{equation*} \begin{bmatrix}
\gamma \beta_1 \\
\vdots \\
\gamma \beta_n
\end{bmatrix} = A\begin{bmatrix}
 \beta_1 \\
\vdots \\
\beta_n
\end{bmatrix}, \, \, \, \, \, A = [a_{j,k}] \in \mathbb{Z}^{n \times n} \end{equation*}

Let $I=\langle \gamma \rangle$.
Now $N(I) = |O_K :I| = |det(A)|$."

So I'm having trouble seeing why $|O_K :I| = |det(A)|$?

答:
See Theorem 5.19 here.  The point is that if $M$ is a finite-free $\mathbf Z$-module and $N$ is a submodule of finite index, then $[M:N]$ can be computed as $|{\rm det}(A)|$ where $A$ is the matrix expressing an arbitrary basis of $N$ in terms of an arbitrary basis of $M$.

Here is the basic idea. There is a pair of aligned bases in $M$ and $N$: there's a basis $e_1,\ldots,e_n$ of $M$ and nonzero integers $a_1,\ldots,a_n$ such that $a_1e_1,\ldots,a_ne_n$ is a basis of $N$.  That means
$$
M/N \cong \prod_{i=1}^n \mathbf Z/(a_i),
$$
so $[M:N] = |a_1\cdots a_n|$. The matrix expressing $a_1e_1,\ldots,a_ne_n$ in terms of $e_1,\ldots,e_n$ is a diagonal matrix ${\rm diag}(a_1,\ldots,a_n)$, which has determinant $\prod a_i$, so the absolute value of this determinant is $[M:N]$.  What if we used other bases for $M$ and $N$?

The transition matrix between two bases of the same finite-free abelian group has determinant $\pm 1$, so the transition matrix $A$ between a new basis of $N$ and new basis of $M$ differs from the transition matrix between the aligned bases of $N$ and $M$ by matrices with determinant $\pm 1$, so $|\det A| = |\prod a_i| = [M:N]$.


答:
$(a+b)(a-b) = a^2 - b^2.$

Put $\ a = \sqrt{n+1},\quad b = \sqrt{n}\ .$

$\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right) = \left(\sqrt{n+1}\right)^2 - \left(\sqrt{n}\right)^2 = n+1 - n =1.$

Therefore, starting with the RHS and rationalising the denominator,

\begin{align}\frac{1}{\sqrt{n+1}+\sqrt{n}} =\frac{1}{\sqrt{n+1}+\sqrt{n}} \times 1\\
\\
=\frac{1}{\sqrt{n+1}+\sqrt{n}} \times \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}\\
\\
= \frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\\
\\
= \frac{\sqrt{n+1}-\sqrt{n}}{1}\\
\\
= \sqrt{n+1}-\sqrt{n}\ .\\
\end{align}

Title: why is $\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$ correct?

Body:
Im trying to understand how the equation below was generated. A step-by-step solution would be awesome.
$$\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$$


问:
How to find how many integer values does it exit in a segment in a triangle?
The problem is as follows:

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{7}\\
2.&\textrm{6}\\
3.&\textrm{5}\\
4.&\textrm{4}\\
\end{array}$

I'm totally clueless here. But from looking at the way how this question is presented. I'm feeling that it will require the use of triangle inequality. And it seems as well if the approach of euclidean geometry is intended then the answer will use triangle congruence however the given clue of $\angle \alpha < \angle \beta$ is somewhat confusing.

Can someone help me here?.

So far I have ran out of ideas. Please include a drawing in your answer. Can this be solved relying only in euclidean geomtry?

答:
This question seems a bit unclear.  Do they want the number of integer values as a single answer (i.e. there are 5 possible integers) or to identify which integers from that list are possible answers (i.e. AH could be 7 or 6, but not 4 or 5)?  My calculation is that there are 5 possible integer solutions: 6, 7, 8, 9 and 10.  I get these by looking at the two extremes.

If CMH is a right angle then \beta = \alpha.  CMH and AMH are congruent right angled triangles (by SAS).  AH therefore = 5, but this case is specifically excluded by the requirement that \beta > \alpha.
A=B limit image

Otherwise, we are very constrained by the question.  HC must be 5.  HM must be 3.  CM cannot be less than 4 otherwise \beta will not be less than \alpha.  Since \beta must be greater than \alpha, H must move closer to line AC, opening up that angle.  H will move in an arc towards A keeping 5cm from C.  But as it does so, M will have to move left in order to keep HM at 3cm - so A will also move left to keep AM and CM the same.
H moving down

At the opposite limit, H will reach CM.  At this point, CM=MH+HC=3+5=8.  AM therefore also equals 8.  AH = AM + MH = 8+3=11.  But at this point, triangle CMH has disappeared, so this also cannot be a solution.

Between these two extremes, then, AH can be any integer value between 5 and 11.  So 6,7,8,9, or 10.  There are 5 possible integer solutions - which I assume is what option 3 means in the question.


问:
Intuitive explanation of 'Normal' integral
For $f$ given as (standard normal density)
\begin{equation}
f(x) = \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{x^2}}}{2}}},
\end{equation}
the follwing holds
\begin{equation}
\int\limits_a^\infty  {xf(x)dx}  = f(a).
\end{equation}
This result isn't hard to derive. However, I'm having a difficulty to intuitively understand why 'expected' value of $X\sim N(0,1)$ over a region is equal to the density at the lower bound. I know it isn't a true expected value but still. Why a result of this integral is equal to $f(a)$?

答: A trick is that $f(x)=\frac1{\sqrt{2\pi}}e^{-x^2/2}$ satisfies the differential equation $f'(x)=-xf(x)$ and its limit as $x\to\infty$ is $0$ (this condition is already a consequence of being a solution of the differential equation, but there's no harm in mentioning it). Therefore $\int_x^\infty tf(t)\,dt=f(x)$.


问:
Evaluate $\lim_{x\to 0}\frac{x-\sin x}{x\sin x}$ without to use L'Hopital

Evaluate $$\lim_{x\to 0}\frac{x-\sin x}{x\sin x}$$ Without L'Hopital's Rule

$$\lim_{x\to 0}\frac{x-\sin x}{x\sin x}=\lim_{x\to 0}\frac{x(1-\frac{\sin x}{x})}{x\sin x}=\lim_{x\to 0}\frac{1-\frac{\sin x}{x}}{\sin x}$$

But I can not find a way to deal with $\sin x$ that does not result with a limit of the type $"\frac{0}{0}"$

答:
Given your steps, without derivates and Taylor,
$$\lim_{x\to 0}\frac{x-\sin x}{x\sin x}=\lim_{x\to 0}\frac{x(1-\frac{\sin x}{x})}{x\sin x}=\lim_{x\to 0}\frac{1-\frac{\sin x}{x}}{\sin x}$$

an knowing that
$$\lim_{x\to0}\frac{1-\tfrac{\sin x}{x}}{x^2}=\frac16$$

you have:

$$\lim_{x\to 0}\frac{1-\frac{\sin x}{x}}{\sin x}=\lim_{x\to 0}\frac{x^2}{x^2}\left(\frac{1-\frac{\sin x}{x}}{\sin x}\right)=\lim_{x\to 0}\left[\underbrace{\frac{1-\tfrac{\sin x}{x}}{x^2}}_{\to \frac 16}\left(\underbrace{\frac{x}{\sin x}}_{\to 1}\right)\right]\, x=0$$


问:
Is Einstein notation universally applicable?
It gets surprisingly monotonous to write the sum symbol repeatedly, so I'm wondering if one can just swap in Einstein notation at will to avoid the monotony, assuming that the symbol that's being summed over appears exactly twice?

For example, the wiki definition of $\nabla$ uses the sum symbol:

$\nabla = \sum_{i=1}^n \vec e_i {\partial \over \partial x_i} $

Could one simply write

$\nabla = \vec e_i {\partial \over \partial x_i} $

instead, and if so, would it need to be prefaced with "Using Einstein summation notation... " or is even that necessary?

If this question is more appropriate for physics.se, I'll post it there. Math.se has more hits on the search "Einstein notation," so I thought I'd ask here first.

Thank you for any help.

答:
I would just like to add few comments to the answers posted here.

First of all, I am afraid there is nothing universally accepted, especially when regarding notation. It is true however that Einstein notation alleys in almosy every scientific paper related to General Relativity and is also common in theoretical physics. In maths is more difficult to see it, but I do think most of the geometers are aware of its existence (although they might not like it).

Regarding how useful/useless it is, I would say it depends. It depends on the equations you need to write. If in your expressions repeated symbols are always adding then go ahead. You can use it even if that's not the case for all your expressions, you can just add a warning remark between brackets "(no sum)" or "(no sum over $i$)" before displaying   the exceptional ones.

For the other pathological case mentioned in a comment, I suggest two possible solutions: either the indices are clearly different (for instance of you are considering a tensor product of different spaces it is common to denote the first coordinates by $x_i$ for example and use $y_\alpha$ for the others and the reader knows the ranges of each indexing set) or in that particular case the sum is explicitly written.

Summarising, that would be the way I  suggest that you solve the possible issues:

Throughout the text, Einstein notation will be adopted when there will be no risk of confusion. In such cases, and if nothong else is stated, he reader is intended to interpret repeated indices affected by a sum symbol, ranging over the corresponding indexing sets. In the cases where there may be a possible misunderstanding, sums will be explicitly written.


问:
Residue of $p.v.\int_{-\infty}^{\infty}\frac{e^{2x}}{\cosh(\pi x)}dx=\text{sec}1$

Show that $$p.v.\int_{-\infty}^{\infty}\frac{e^{2x}}{\cosh(\pi
x)}dx=\text{sec}1$$ by integrating $\frac{e^{2z}}{\cosh(\pi z)}$ around
  rectangles with vertices at $z=\pm p,p+i,-p+i.$

I asked this similar question, and was recommended this link stackexchange, but the answer provided is confusing.

I am not sure how they got those vertices, but with those vertices I got the parameterization:

$\begin{cases} \Gamma_1 = t, & -p\leq t\leq p \\\\ \Gamma_2= p+it, &0\leq t \leq 1 \\\\ \Gamma_3 = i-t, & -p\leq t\leq p \\\\ \Gamma_4=i(1-t)-p, & 0\leq t\leq 1. \end{cases}$

Thus I have $\Gamma_p =\Gamma_1 +\Gamma_2+\Gamma_3+\Gamma_4  .$

$f(z)=\frac{e^{2z}}{\cosh(\pi z)} = \frac{2e^{z(2+i\pi)}}{e^{2i\pi z}+1}$, where the denominator has a $\text{pole}=(2n+1)\frac{1}{2}$.

And this is as far as I can get.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\large\mbox{A}\ \Re eal\mbox{-Approach}}$:
\begin{align}
&\bbox[5px,#ffd]
{\int_{-\infty}^{\infty}{\expo{2x} \over \cosh\pars{\pi x}}
\,\dd x} =
2\int_{0}^{\infty}{\cosh\pars{2x} \over \cosh\pars{\pi x}}
\,\dd x
\\[5mm] = &\
4\int_{0}^{\infty}{\sinh\pars{\pi x}\cosh\pars{2x} \over \sinh\pars{2\pi x}}\,\dd x =
{\tt J}\pars{\pi + 2} + {\tt J}\pars{\pi - 2}
\label{1}\tag{1}
\\[2mm] &\
\mbox{where}\ {\tt J}\pars{\alpha} \equiv
2\int_{0}^{\infty}{\sinh\pars{\alpha x} \over \sinh\pars{2\pi x}}
\,\dd x
\end{align}

\begin{align}
{\tt J}\pars{\alpha} & =
2\int_{0}^{\infty}
{\expo{-\pars{2\pi - \alpha}x} -
\expo{-\pars{2\pi + \alpha}x} \over 1 - \expo{-4\pi x}}\,\dd x
\\[5mm] \stackrel{4\pi x\ \mapsto\ x}{=}\,\,\, &\
{1 \over 2\pi}\int_{0}^{\infty}
{\expo{-\bracks{1/2 - \alpha/\pars{4\pi}}x}\,\,\, -
\expo{-\bracks{1/2 + \alpha/\pars{4\pi}}x} \over 1 - \expo{-x}}
\,\dd x
\\[5mm] = &\
{1 \over 2\pi}\int_{0}^{\infty}
{1 - \expo{-\bracks{1/2 + \alpha/\pars{4\pi}}x} \over
1 - \expo{-x}}\,\dd x
\\[2mm] - &\
{1 \over 2\pi}\int_{0}^{\infty}
{1 - \expo{-\bracks{1/2 - \alpha/\pars{4\pi}}x} \over
1 - \expo{-x}}\,\dd x
\\[5mm] = & {1 \over 2\pi}
\bracks{\Psi\pars{{1 \over 2} + {\alpha \over 4\pi}} -
\Psi\pars{{1 \over 2} - {\alpha \over 4\pi}}}
\\[5mm] = &\
{1 \over 2\pi}\,\pi
\cot\pars{\pi\bracks{{1 \over 2} - {\alpha \over 4\pi}}} =
{1 \over 2}\tan\pars{\alpha \over 4}\label{2}\tag{2}
\end{align}

With (\ref{1}) and (\ref{2}):
\begin{align}
&\bbox[5px,#ffd]
{\int_{-\infty}^{\infty}{\expo{2x} \over \cosh\pars{\pi x}}
\,\dd x} =
{1 \over 2}\tan\pars{\pi + 2 \over 4} +
{1 \over 2}\tan\pars{\pi - 2 \over 4}
\\[5mm] = &\
{1 \over 2}\,{1 + \tan\pars{1/2} \over 1 - \tan\pars{1/2}} +
{1 \over 2}\,{1 - \tan\pars{1/2} \over 1 + \tan\pars{1/2}} =
{1 + \tan^{2}\pars{1/2} \over 1 - \tan^{2}\pars{1/2}}
\\[5mm] = &\
{1 \over \cos^{2}\pars{1/2} - \sin^{2}\pars{1/2}} =
{1 \over \cos\pars{2\bracks{1/2}}}
\\[5mm] = & \bbx{\sec\pars{1}} \approx 1.8508 \\ &
\end{align}


问:
Given a collection of cuboids, How to find another collection that cover the first collection 2 times the volume
Assume we have a countable collection of cuboids $  \left\{ J_{i}\right\} _{i\in I} $

each cuboid is an n-dimenional cubould $ J_i\subset \mathbb{R}^n $

Such that $ \sum_{i\in I}\text{vol}\left(J_{i}\right)<\varepsilon $

Is it possible to manipulate each cuboid and get a new collection $ \{\tilde{J}_{i}\}_{i\in I} $ such that each new cuboid  $ \tilde{J}_{i} $ would satisfy:

$ J_{i}\subset\text{int}\left(\tilde{J}_{i}\right) $ and also

$ \sum_{i\in I}\text{vol}\left(\tilde{J}_{i}\right)<2\varepsilon $?

Im pretty sure that its possible but I cant see how to formally prove it, my intuition is that I want to make each one-dimesional segment a little bit bigger, such that the new volume would only increase by $ \frac{1}{2^{n}}\varepsilon $, so that $ \text{ vol}\left(\tilde{J}_{i}\right)=\left(\frac{1}{2^{n}\varepsilon }+\text{vol}J_{i}\right) $  and then since the index set $ I $ is countable, I can assume that the cardinality is $ \aleph_0 $ and the write:

$ \sum_{i=1}^{\infty}\text{vol}\tilde{J}_{i}=\sum_{i=1}^{\infty}vol\left(J_{i}\right)+\sum_{i=1}^{\infty}\left(\frac{1}{2^{n}}\varepsilon\right)<\varepsilon+\varepsilon=2\varepsilon $

But I could not find out how to do it precisly. Thanks in advance

答:
Let $u = 2^{\frac{1}{n+1}}$. Replace $J_i$ with a cuboid whose center is the same, but where each edge's length is multiplied by $u > 1$. Then you have a cuboid containing $J'_i$ containing $J_i$, and the volume of $J'_i$ is
$$
v' = u^n v
$$
where $v$ is the volume of $J_i$. But $u^n = 2^\frac{n}{n+1} < 2^1 = 2$, so $v' < 2v.$


问:
Construction of the Associated projective hyperplane.
Let $V$ be $K-$vector space of dimension $n$. Let $H$ be a hyperplane given by the kernel of the linear map
$l_H:V\longrightarrow K$ defined by
$$l_H(x)=a_1x_1+\cdots+a_nx_n$$
where $x_1,\cdots,x_n$ are the coordinates of $x$ in a chosen basis of $V$.

Let $P(V)$ be the projective space associated to $V$. Consider the map: $l_{H'}:P(V)\longrightarrow K$
defined by
$$l_{H'}([x_1:\cdots:x_n])=a_1x_1+\cdots+a_nx_n$$
Does the kernel of $l_{H'}$ defines a projective hyperplane in $P(V)$?

答: The answer is yes. A point $M \in P(V)$ belongs to the projective hyperplane $P(H)$ if and only if one of its representative, $\vec{M} \in H \iff l_H(\vec{M}) = 0$ (which is an equation of $P(H)$.


问:
Proving one function is greater than another
How can I prove $f(x)$ $>$ $g(x)$ for all $x > 0$ given $f(x) = (x+1)^{2}$ and $g(x) = 4qx$ where $q$ is a constant in $(0, 1)$?

My approach was to show that $(x+1)^2 > 4qx$ for the interval endpoints, e.g. $q=0$ and $q=1$. E.g. $(x+1)^2 \geq 4x$ for all $x$ and $(x+1)^2 > 0$ for all $x$. However, $q \neq 0,1$ so $f(x) > g(x)$ for all $x$. However, I'm looking for something more mathematically rigorous. Any suggestions?

答:
Since $q \in (0,1)$ and $x > 0$, we know $4xq < 4x$, so it suffices to prove the stronger claim $(x+1)^2 \geq 4x$ for all $x > 0$. The latter is equivalent to showing $(x+1)^2 - 4x \geq 0$. To that end, we have
\begin{align*}
(x+1)^2 - 4x &= x^2 + 2x + 1 - 4x\\
&= x^2 - 2x + 1\\
&= (x - 1)^2\\
&\geq 0.
\end{align*}</p>


问:
Proving a formula for the stationary distribution of a finite irreducible Markov chain
Let $\{X_n\}_{n=1}^{\infty}$ be an irreducible Markov chain on a finite state space $S = \{1,2,\ldots,N\}$.  Let $P$ be its transition matrix (i.e. $P_{ij} = P(X_{n+1} = j | X_n = i)$ for all $(i,j) \in S^2$). Show that the Markov chain has a (unique) stationary distribution given by
$$\pi = (1,1,\ldots,1) (I - P + Q)^{-1},$$
where $I$ is the $N \times N$ identity matrix and $Q$ is the $N \times N$ matrix whose entries are all 1.

My attempt:
I take it for granted (by a theorem in my text) that a finite irreducible Markov chain has a unique stationary distribution (i.e. there exists a probability vector $\pi$ such that $\pi P = \pi$), so I'm just trying to show that the above formula for $\pi$ is correct.

First I establish that $(I-P+Q)$ is invertible: Suppose $\vec{0} = (I-P+Q)\vec{x}$. This implies that
\begin{align*}
       \vec{0} &= \vec{x} - P \vec{x} + Q \vec{x} \\[2pt]
       &= \pi \vec{x} - \pi P \vec{x} + \pi Q \vec{x} && \text{(Multiplying by } \pi \text{ on the left}) \\[2pt]
       &= \pi \vec{x} - \pi \vec{x} + \pi Q \vec{x} && (\pi P = \pi) \\[2pt]
       &= \pi Q \vec{x}.
\end{align*}
So we have $\vec{0} = \pi Q \vec{x}$. By a theorem from my text, the entries of $\pi$ are strictly positive, so it follows that $\vec{0} = Q \vec{x}$. Substituting this into the first of the above equations gives $\vec{0} = \vec{x} - P \vec{x}$, which implies $P \vec{x} = \vec{x}$. Therefore, $Q \vec{x} - P \vec{x} = (Q-P)\vec{x} = -\vec{x}$. Now since $Q_{ij} = 1$ and $0 \leq P_{ij} \leq 1$ for all $(i,j)$, the entries of the matrix $Q-P$ are all nonnegative. Thus, the only way for $(Q-P) \vec{x} = -\vec{x}$ to be true is if $\vec{x} = 0$. And this shows that $(I-P+Q)\vec{x} = \vec{0} \implies \vec{x} = \vec{0}$, and so $(I-P+Q)$ is invertible.

At this point I got stuck. I don't see how to show the given formula for $\pi$. Of course the most direct approach would be to show that
$$(1,1,\ldots,1) (I-P+Q)^{-1} P = (1,1,\ldots,1) (I-P+Q)^{-1}, $$
but I have no idea how to proceed with this. Any help would be greatly appreciated!

答:
(This is a long comment.) In your proof of the invertibility of $I-P+Q$, you wrote:

the entries of the matrix $Q-P$ are all nonnegative. Thus, the only way for $(Q-P) \vec{x} = -\vec{x}$ to be true is if $\vec{x} = 0$.

This isn't quite correct. For instance, $-1$ is an eigenvalue of $Q-P=\pmatrix{0&1\\ 1&0}$ when $P=I_2$.

You need to use the condition that $P$ is irreducible. Suppose $(I-P+Q)x=0$. Then $\mathbf1^Tx=0$:
\begin{aligned}
0
&=\mathbf\pi^T(I-P+\mathbf 1\mathbf1^T)x\\
&=\mathbf\pi^Tx-(\mathbf\pi^TP)x+(\mathbf\pi^T\mathbf 1)(\mathbf1^Tx)\\
&=\mathbf\pi^Tx-\mathbf\pi^Tx+\mathbf1^Tx\\
&=\mathbf1^Tx.
\end{aligned}
It follows that $0=(I-P+\mathbf 1\mathbf1^T)x=x-Px$. That is, $Px=x$. Hence $x$ lies in the eigenspace of $P$ corresponding to the eigenvalue $1$. However, as $P$ is an irreducible row-stochastic matrix, $x$ must be a scalar multiple of $\mathbf1$, by Perron-Frobenius theorem. The equality $\mathbf1^Tx=0$ thus implies that $x=0$. Therefore $I-P+Q$ is invertible.


问:
Problem 13.7 of Additional Exercises by Stephen Boyd: Energy consumption optimization problem
Consider a unit mass electric vehicle with position $x(t)∈ℝ$, moving with velocity $v(t)$ and acceleration $a(t)$. The vehicle needs to be maneuvred from initial position of rest to final position $x(T)$ at time $t=T$ satisfying $x(T)≥x_f$. Requirements are $\dot v≤A$, $0≤v(t)≤V$, $v(T)≤v_f$.

Let $p(t),p_b(t),p_d(t)$ be power delivered by battery, power lost in braking, power lost due to drag respectively and $k(t)$ be kinetic energy of vehicle which is $v(t)^2/2$. Then, $$p(t)=\dot k(t)+p_b(t)+p_d(t)$$ where where, $p_d(t)=cv(t)^3$ and the power used for braking $p_b(t)$ can be controlled.

Take $T=nh$ and divide interval [$0,T]$ into $n$ subintervals of length $h$ to discretize the vehicle equations of motion, $\dot x=v,\dot v=a$. In discrete form we get following equations

$$x_{r+1}=x_r+h(v_r+v_{r+1})/2$$
$$v_{r+1}=v_r+ha_r$$
$$p_r=(k_{r+1}−kr)/h+p_{b,r}+p_{d,r}$$
$$k_r=v^2_r/2$$

Total energy consumed is $E=h \sum_{r=0}^{n}p_r$.

Formulate an optimization problem for minimum energy consumption. Write all variables and constraints on them. Indicate decision variables. Comment on the nature of the problem.

Solve the problem by taking $x_f=12 m, v_f=2 m/s, V=10 m/s, A=2 m/s^2, c=2, h=0.1, T=5 s$.

I am fairly new to convex optimization and solving these kinds of problems. I am not sure how to go around with this one. I think formulating the optimization problem is the tough part. I think I can approach solving the problem, once I know what exactly needs to be done. Thoughts and hints are appreciated.

Thanks

Edit: I recently found out that this problem is the same as problem 13.7 of the additional exercises in Stephen Boyd's book- 2020 edition.

答:
$E = h \sum_{r=0}^{n}p_r$

This seems to be an error because it sums $n+1$ intervals of width $h$.

The optimization problem is almost written out in full already:
\begin{align}
\min \quad & h \sum_{r=0}^{n-1}p_r \\
\text{s.t.} \quad & x_{r+1}=x_r+h(v_r+v_{r+1})/2 \quad \forall r \in \{0,1,\ldots,n-1\} \\
&v_{r+1}=v_r+ha_r \quad \forall r \in \{0,1,\ldots,n-1\} \\
&p_r=(k_{r+1}−k_r)/h+p_{b,r}+p_{d,r} \quad \forall r \in \{0,1,\ldots,n-1\} \\
&p_{d,r}=cv_r^3 \quad \forall r \in \{0,1,\ldots,n-1\} \\
&k_r=v^2_r/2 \quad \forall r \in \{0,1,\ldots,n\} \\
& x_0 = 0 \\
& x_n \geq x_f \\
& a_r \leq A \quad \forall r \in \{0,1,\ldots,n-1\} \\
& 0 \leq v_r \leq V \quad \forall r \in \{0,1,\ldots,n\} \\
& v_n \leq v_f \\
& a,p,p_b,p_d \in \mathbb{R}^{n-1}, \; x,v,k \in \mathbb{R}^n
\end{align}

The optimization variables and their dimensions are listed on the last line. The constraints $p_{d,r}=cv_r^3$ and $k_r=v^2_r/2$ are not convex, but they can be rephrased as inequalities: $p_{d,r} \geq cv_r^3$ and $k_r \geq v^2_r/2$, which are convex inequalities on the domain $v_r \geq 0$. At optimality, the inequalities will be tight.