问:
Proof of Liouville's theorem and the construction of transcendental numbers - Hardy et al.
I've been trying to understand the proof of Liouville's theorem given in "Intro to Theory of Numbers"- G.H. Hardy et al. and not able to understand few steps. The proof snippet is attached.

Can someone please help me understand the following
a) How does the author conclude "there is a number $M(\epsilon)$ such that $|f^\prime(x)|< M \quad \epsilon -1 < x < \epsilon$"?

b) In the last line of the proof, he says that $\epsilon$ is not approximable to any order higher than n. But $p/q$ is an approximation of $\epsilon$. So with larger values of $n$, the lower bound should not hold. Maybe I'm not able to understand the conclusion this theorem wanted to prove in the first place.

答:
a) The function $f'$ is a polynomial function and therefore its restriction to any bounded interval is bounded. So, take $M>\sup_{x\in[\xi-1,\xi+1]}\bigl|f'(x)\bigr|$ and then $\xi-1<x<\xi+1\implies\bigl|f'(x)\bigr|<M$.

b) The goal is precisely to get a contradiction.


问:
Solve Logical Problem using contradiction rule and Truth Table
There are three containers, one container has diamond in it, other containers are empty. Each
container has label as clue whatever is inside the containers. Those labels are:

Container#1: The diamond is not present

Container #2: The diamond is not present

Container #3: The diamond is present in container#2

Only one statement is true and other are false. Which container has the diamond? Formalize this
problem into propositional statement and find a valid solution using truth table.

答:
Since the diamond can only be in one out of the two boxes, we can proceed with just one predicate
$$p=\mbox{Diamond is in box 1}$$
Scenario 1
$$\neg p\rightarrow \neg p\land p\equiv\mbox{false}$$
Scenario 2
$$p\rightarrow p\land p\equiv\mbox{true}$$
Scenario 3
$$\neg p\rightarrow p\land \neg p\equiv\mbox{false}$$
Solving it this way makes the solution very clear.

ContentLicense: CC BY-SA 4.0


问:
Series Summation: $\sum\limits_{k=1}^{N-1}\frac{1}{z-w_k}$ where $w_k=e^{\frac{2\pi i k}{N}}$
I have the series

$$\sum_{k=1}^{N-1}\frac{1}{1-w_k} $$ where $w_k=e^{\frac{2\pi i k}{N}}$, how can I find the summation of this , another question related to this sum

$$\sum_{k=1}^{N-1}\frac{1}{z-w_k}   $$

答:
The answer for the first sum is $$\sum_{k=1}^{N-1}\frac{1}{1-w_k}=\frac{N-1}{2}$$

And for the second:
$$\sum_{k=1}^{N-1}\frac{1}{z-w_k} = \sum_{k=0}^{N-1}\frac{1}{z-w_k}- \frac{1}{z-1}= \frac{N z^{N-1}}{z^N-1} -\frac{1}{z-1}$$

I believe series expansion in $w_k$, sum over $k$ and then summing back the series wll give you the above results.

ContentLicense: CC BY-SA 3.0


问:
Why $\log_b(x)$ = $-\log_{1/b}(x)$?
IN a textbook I am asked to consider the graphs of $f(x) = \log_{1/2}(x)$ and $-\log_2(x)$. They appear identical and then the textbooks tells the rule that $\log_b(x)$ = $-\log_{1/b}(x)$.

I tried to prove this to myself on paper.

I can write both equations in exponential form. Pretend $x=4$:

$$\log_{1/2}(4)=\frac{1}{2}^y=4$$

Then write the same for the other version:

$$-\log_2(4)=-2^y=4$$

If I know these are equivalent then the two exponent versions should be the same no?

$$4=\frac{1}{2}^y=-2^y$$

It's not clicking and I don't 'get it' from here. Is my last equation true? Why are they equivalent?

答:
$$log _{\frac 1 b} (x)=log _{\frac 1 b} (x)$$

$$x=(\frac 1b)^{log _{\frac 1 b} (x)}=b^{-log _{\frac 1 b} (x)}$$

We take log in base $\frac 1{\frac 1b}=b$ of both sides, we get:

$$log_b (x) =-log _{\frac 1 b} (x) log_b b=-log _{\frac 1 b} (x) $$

Title: Why $\log_b(x)$ = $-\log_{1/b}(x)$?


问:
Geometric interpretation of the non-uniqueness of the solution of a system of linear equations

Show that the equation $$\pmatrix{3&-7&0\\2&2&5\\1&3&4}\pmatrix{x\\y\\z} = \pmatrix{3\\2\\1}$$ does not have a unique solution, and give a geometrical interpretation.

Now, the first part of the question is fairly straightforward. I've shown that the determinant of the 3x3 matrix is 0; and thus the system of equation does not have a unique solution.

After that, I've rewritten the matrices into a set of linear equations.

$\begin{cases}3x-7y+0z=3\\2x+2y+5z=2\\
x+3y+4z=1\end{cases}$

Now this is where I'm slightly stumped, I've attempted this part of the  question but have no idea if I'm correct.

Through eliminating $x$ from pairs of equations, I've gotten $-4y-3z =0$ from 2 pairs and $-16y-22z=0$ from the other. What does this mean exactly?

答:
The three planes pass through the same line $r$ of equation
$$
r:\begin{cases}
3 x-7 y-3=0\\
4 y+3 z=0\\
\end{cases}
$$


问:
Index of submodule generated by multiplication by one element of free $\mathbb{Z}$-module
This question appears in an algebraic number theory notes with $M=\mathcal{O}_K$ for a number field $K$.

Let $M$ be a free $\mathbb{Z}$-module of rank $n$ which is a ring itself. Let $x\in M$, then multiplication by $x$, $\phi_x:y\mapsto xy$ is a $M$-endomorphism, hence can be represented by a $n\times n$ $\mathbb{Z}$-matrix C. Let $\{v_1\,\dots,v_n\}$ be a basis of $M$, and let $N$ be the submodule generated by $\{xv_1,\dots,xv_n\}$ over $\mathbb{Z}$. I want to show that $[M:N]=|\det(C)|$, where $[M:N]$ is the index of $N$ as a subgroup of $M$. Could anyone please help?

答:
Note that $N$ is just the image of $\phi_x$.

Let more generally $M$ be free abelian group of finite rang and let $\phi$ be an injective endomorphism of $M$.
Then
$$
  \lvert M / {\operatorname{im}(\phi)} \rvert
  =
  \lvert \det(\phi) \rvert.
$$
To show this, we may assume that $M = \mathbb{Z}^n$.
The endomorphism $\phi$ is then given by multiplication with some $\mathbb{Z}$-matrix $C$ of size $n \times n$.
The matrix $C$ has a Smith normal form.
There hence exist invertible $\mathbb{Z}$-matrices $S$ and $T$ of size $n \times n$ such that
$$
  C = S D T
$$
for some diagonal matrix
$$
  D
  =
  \begin{pmatrix}
    d_1 &        &     \\
        & \ddots &     \\
        &        & d_n
  \end{pmatrix},
$$
with $d_1, \dotsc, d_n > 0$.
(We may also assume that $d_i$ divides $d_{i+1}$ for all $i = 1, \dotsc, n-1$, but we won’t need this.)
Both $S$ and $T$ have determinant $\pm 1$ because they are invertible integer matrices. It follows that
\begin{align*}
  \lvert \det(\phi) \rvert
  &=
  \lvert \det(C) \rvert
  \\
  &=
  \lvert \det(SDT) \rvert
  \\
  &=
  \lvert \det(S) \det(D) \det(T) \rvert
  \\
  &=
  \lvert \det(S) \rvert
  \cdot
  \lvert \det(D) \rvert
  \cdot
  \lvert \det(T) \rvert
  \\
  &=
  d_1 \dotsm d_n .
\end{align*}
We find on the other hand that
$$
  M / \operatorname{im}(\phi)
  =
  M / C M
  =
  M / SDT M
  =
  M / SD M
$$
because $T$ is invertible and thus $TM = M$.
The isomorphism of abelian groups
$$
  M \to M,
  \quad
  x \mapsto Sx
$$
maps $DM$ onto $SDM$, and thus descends to an isomorphism of abelian groups between $M / DM$ and $M / SD M$.
We hence find that
\begin{align*}
  M / SD M
  &\cong
  M / DM
  \\
  &=
  \mathbb{Z}^n / D \mathbb{Z}^n
  \\
  &=
  ( \mathbb{Z} \oplus \dotsb \oplus \mathbb{Z} ) / (d_1 \mathbb{Z} \oplus \dotsb \oplus d_n \mathbb{Z})
  \\
  &\cong
  \mathbb{Z} / d_1 \mathbb{Z} \oplus \dotsb \mathbb{Z} / d_n \mathbb{Z}.
\end{align*}
This shows overall that
$$
  \lvert M / \operatorname{im}(\phi) \rvert
  =
  \lvert \mathbb{Z} / d_1 \mathbb{Z} \oplus \dotsb \oplus \mathbb{Z} / d_n \mathbb{Z} \rvert
  =
  d_1 \dotsm d_n.
$$


问:
USAJMO 2017 P4: triples $(a,b,c)$ such that $(a-2)(b-2)(c-2)+12$ is a prime number that divides the positive number $a^2+b^2+c^2+abc-2017$?

Are there any triples $(a,b,c)$ of positive integers such that $(a-2)(b-2)(c-2)+12$ is a prime number that properly divides the positive number $a^2+b^2+c^2+abc-2017$?

My progress: Let $x=a-2$, $y=b-2$, $z=c-2$, then we get $xyz+12 \mid \left( x+y+z+4 \right)^2 - 45^2 \implies xyz+12 \mid (x+y+z-41)(x+y+z+49)$. So $xyz+12 \mid (x+y+z-41)$ or $xyz+12|(x+y+z+49)$ .

Also $xyz+12 $ can't divide both or else then $xyz+12 |90$ ( which doesn't work ).

WLOG $x\ge y\ge z$, then clearly $x\ge 9$ , since $11^2\cdot 3+ 11^3\le 2017$ ( not satisfying properly divide property).

Also I got $x,y,z$ odd or not divisible by $1 \mod 3$, or else $(a-2)(b-2)(c-2)+12$ is not prime.

答:
If we continue from here: $\;\;\;xyz+12 \mid x+y+z-41\;\;\;$ or $\;\;\;xyz+12|x+y+z+49$.

We can assume $x\leq y\leq z$. We see that $x,y,z\notin \{0,2,3,4,6,8,9\}$ since $xyz+12$ is prime.

If $x>1$ then $x,y\geq 5$ and so $xyz+12\geq 25z+12$ so in

first case: $$25z+1\leq |x+y+z-41|\leq 3z+41\implies z\le 1$$ and thus no solution.

second case: $$25z+1\leq x+y+z+49\leq 3z+49\implies z\leq 2$$ so no solution again.

If $x=1$ then  $\;\;\;yz+12 \mid y+z-40\;\;\;$ or $\;\;\;yz+12|y+z+50$.

If $y\neq 1$ then $y\geq 5$. In first case we have $5z+12\leq 2z+40\implies z\leq 9$ so $z=7$ (and $y=5$ or $y=7$) or $z=5$ (and $y=5$). None works. In second case we have $5z+12\leq 2z+50\implies z\leq 12$. So $z\in\{11,7,5\}$ but none works.

If $y=1$ then $\;\;\;z+12 \mid z-39\;\;\;$ then $z+12\mid 51 \implies z=5$ or $\;\;\;z+12|z+51$, then $z+12\mid 39$ so $z=1$.

If $x=-1$ then we proccede similary like in a previous case...


问:
value of parameter a such that $\frac{1}{2}$ is a 3-periodic point
Let

\begin{cases}
T_a(x) = ax & \text{ if}  &  x \in  [0; \frac{1}{2}],\\
T_a(x) = a(1 − x) & \text{ if} &  x \in  [\frac{1}{2}; 1].
\end{cases}

Find a parameter value of $a$ for which $\frac{1}{2}$ is a $3$-periodic point.

I have tried to distinguish several cases.

If $0  \leq x \leq \frac{1}{2a}$.
Then $0 \leq ax \leq \frac{1}{2}$ and $T_a^2(x)= T(ax)=a^2x$.

If $\frac{1}{2a}  \leq x \leq \frac{1}{2}$.
Then $\frac{1}{2} \leq ax \leq \frac{a}{2}$ and $T_a^2(x)= T(a(1-x))=a(1-ax)$.

If $\frac{1}{2}  \leq x \leq \frac{a+1}{2}$.
Then $\frac{a(a-1)}{2} \leq a(1-x) \leq \frac{a}{2}$ and $T_a^2(x)= T(a(1-x))=a^2(1-x)$.

I am confused about the manner to mimic the response as shown on this post.
$$
f^{n}(x)=\left\{
\begin{array}{ccc}
a^nx & \text{if} & x\in\big[0,\tfrac{1}{2a^n}\big], \\
1-a^nx\big(x-\tfrac{1}{2a^n}\big) & \text{if} & x\in\big[\tfrac{1}{2a^n},\tfrac{2}{2a^n}\big], \\
a^n\big(x-\tfrac{2}{2a^n}\big) & \text{if} & x\in\big[\tfrac{2}{2a^n},\tfrac{3}{2a^n}\big], \\
& \vdots & \\
1-a^n\big(x-\tfrac{2a^n-1}{2a^n}\big) & \text{if} & x\in\big[\tfrac{2a^n-1}{2a^n},1\big],
\end{array}
\right.
$$

答:
(Fill in the gaps as needed. If you're stuck, explain what you've tried and why it doesn't seem to work.)

I'm assuming you're restricted to $ a \in [ 0, 2 ]$ for the function to make sense. In particular, we do not allow $a =-1.$

Hint: Deal with cases based on the supposed value of $ T_ a ( 0.5), T_a ( T_a (0.5) ) $.

IE Case 1: $T_ a ( 0.5) \leq 0.5 , T_a ( T_a (0.5) ) \leq 0.5$.

Show that in all cases, we require $ a = 1$.


问:
Is there a name for this given type of matrix?
Given a finite set of symbols, say $\Omega=\{1,\ldots,n\}$, is there a name for an $n\times m$ matrix $A$ such that every column of $A$ contains each elements of $\Omega$?

(The motivation for this question comes from looking at $p\times p$ matrices such that every column contains the elements $1,\ldots, p$).

答:
A sensible definition for this matrix would be a column-Latin rectangle, since the transpose is known as a row-Latin rectangle.  Example:

A. Drisko, Transversals in Row-Latin Rectangles, JCTA 81 (1998), 181-195.

The $m=n$ case is referred to as a column-Latin square in the literature (this is in widespread use).

I found one example of the use of column-Latin rectangle here (ref.; .ps file).


问:
Counterexample regarding basic properties of limits
Let $f, g$ be two functions defined on a deleted neighbourhood of
$a ∈ R$. If $g(x) \neq 0$ for every x on the deleted neighbourhood and $$\lim_{x\to a}{f(x)\over g(x)} = 1$$

then

$$\lim_{x\to a}{(f(x) - g(x))} = 0$$

I'm asked to prove or disprove this claim, along with others. My instinct in these type of questions is to start figuring out if there are any piecewise function that contradict the statements but I almost feel like I'm searching in the dark with no clear strategy.. is there an agreed upon process regarding searching for counter examples?

答:
What about

$$\begin{cases}
f(x) &= \frac{1}{(x-a)^2}+1\\
g(x) &= \frac{1}{(x-a)^2}
\end{cases}$$

The idea to find this counterexample is that if $\lim\limits_{x \to a} f(x)$ or $\lim\limits_{x \to a} g(x)$ exists and is finite, then the two limits are equal and therefore the limit of $f-g$ vanishes. So in the counterexample above, I looked at maps having non finite limits.


问:
$\lim_{x \to + \infty} \left \{ x-x^2 \ln (1+ \sin \frac{1}{x}) \right \}$
I've tried to solve this limit:
$\lim_{x \to + \infty} \left \{ x-x^2 \ln (1+ \sin \frac{1}{x}) \right \} $

$x-x^2 \ln (1+ \sin \frac{1}{x}) \sim x-x^2 \sin \frac{1}{x} \sim x-x = 0 $

but the final result should be $\frac{1}{2}$

答:
In the following proof, I won't be using L'hopital or series expansion.

Let $ x\in\mathbb{R}^{*} $, we have :

\begin{aligned}x-x^{2}\ln{\left(1+\sin{\left(\frac{1}{x}\right)}\right)}=\frac{\frac{1}{x}-\sin{\left(\frac{1}{x}\right)}}{\frac{1}{x^{2}}}+\left(x\sin{\left(\frac{1}{x}\right)}\right)^{2}\left(\frac{\sin{\left(\frac{1}{x}\right)}-\ln{\left(1+\sin{\left(\frac{1}{x}\right)}\right)}}{\sin^{2}{\left(\frac{1}{x}\right)}}\right)\end{aligned}

We ended with well known limits, that I'm sure have been computed many times on this site.

$ \textbf{Let's proof first that : $ \displaystyle\lim_{x\to 0}{\frac{x-\sin{x}}{x^{2}}}=0$ : } $

Note that for any $ x\in\mathbb{R}^{*} $, we have : $$ \frac{x-\sin{x}}{x^{2}}=\int_{0}^{1}{\left(1-y\right)\sin{\left(xy\right)}\,\mathrm{d}y} $$

Thus : \begin{aligned} \left|\frac{x-\sin{x}}{x^{2}}\right|&\leq\int_{0}^{1}{\left(1-y\right)\left|\sin{\left(xy\right)}\right|\mathrm{d}y}\\ &\leq\int_{0}^{1}{\left(1-y\right)\left|xy\right|\mathrm{d}y}\\ \left|\frac{x-\sin{x}}{x^{2}}\right|&\leq\frac{\left|x\right|}{6} \end{aligned}

Hence : $$ \lim_{x\to 0}{\frac{x-\sin{x}}{x^{2}}}=0 $$

$\textbf{Let's proof then that : $ \displaystyle\lim_{x\to 0}{\frac{x-\ln{\left(1+x\right)}}{x^{2}}}=\frac{1}{2} $ : }$

Note that for any $ x\in\left(-1,1\right)\setminus\left\lbrace 0\right\rbrace $, we have : $$ \frac{x-\ln{\left(1+x\right)}}{x^{2}}=\int_{0}^{1}{\frac{1-y}{\left(1+xy\right)^{2}}\,\mathrm{d}y}=\frac{1}{2}-x\int_{0}^{1}{\frac{y\left(2+xy\right)}{\left(1+xy\right)^{2}}\,\mathrm{d}y} $$

Since : \begin{aligned} \left|x\int_{0}^{1}{\frac{y\left(2+xy\right)}{\left(1+xy\right)^{2}}\,\mathrm{d}y}\right|&\leq\left|x\right|\int_{0}^{1}{\frac{y\left(2+\left|x\right|y\right)}{\left(1-\left|x\right|y\right)^{2}}\,\mathrm{d}y}\\ &\leq\frac{\left|x\right|\left(2+\left|x\right|\right)}{2\left(1-\left|x\right|\right)^{2}}\underset{x\to 0}{\longrightarrow}0 \end{aligned}

we get that : $$ \lim_{x\to 0}{\frac{x-\ln{\left(1+x\right)}}{x^{2}}}=\frac{1}{2} $$

$\textbf{Now let's apply this to our expression : }$

\begin{aligned}\lim_{x\to +\infty}{\left(x-x^{2}\ln{\left(1+\sin{\left(\frac{1}{x}\right)}\right)}\right)}&=\lim_{x\to +\infty}{\left(\frac{\frac{1}{x}-\sin{\left(\frac{1}{x}\right)}}{\frac{1}{x^{2}}}+\left(x\sin{\left(\frac{1}{x}\right)}\right)^{2}\left(\frac{\sin{\left(\frac{1}{x}\right)}-\ln{\left(1+\sin{\left(\frac{1}{x}\right)}\right)}}{\sin^{2}{\left(\frac{1}{x}\right)}}\right)\right)}\\ &=0 + 1^{2}\times\frac{1}{2}\\ &=\frac{1}{2}\end{aligned}


问:
Picard group of a ruled surface
Consider the ruled surface $F$ over $\Bbb P^1$ obtained by projectivizing $\mathcal{O}\oplus \mathcal{O}(-4)$. Let $s$ be denote the zero section $(0:1)$. Note that $\text{Pic}(F)=\Bbb Zs \oplus \Bbb Zf$ where $f$ is a fiber of $F$. Since $\mathcal{O}_F(1), \mathcal{O}_F(1)\otimes \mathcal{O}_{\Bbb P^1}(1) \in \text{Pic}(F)$, they are linear combinations of $s$ and $f$. But is there a way to determine the coefficients?

答:
Actually the basis you chose is not the one considered in Beauvilles book (he choses $\{f,h\}$ where $h$ is the class of the tautological bundle). Hence, I think it is sufficiently non-standard to write an answer. I will work with the vector bundle $E = \mathcal{O} \oplus \mathcal{O}(n)$ to make it more general

In Proposition 3.18 in Beauville's book here are two  equations concerning the class of the tautological bundle, which we call $h$, which essentially follow from the definition of ruled surface. Namely: $$1). h.f = 1 , $$
and $$2). h^2 = \deg(E) = n.$$

Now we consider the basis for the Picard group you, chose and try to write $h$ in terms of this. Unless I am mistaken you have chosen the section with negative self-intersection $-n$. So putting this together with the fact that the tautolical bundle restricts to $\mathcal{O}(1)$ on each fibre, gives that $s^2 = -n$,  $s.f=1$ and $f^2=0$. Writing $h = as+ b f$ we get from 1). above $$ a=1$$ and from 2.) $$ n = (s+bf)^2 = -n + 2b ,$$ hence $b=n$, i.e. $$h = s + nf.$$ Everything else can be worked out via direct calculation.

One final remark. If $s'$ is the other section with $s'^2=n$, then note that we have $s'-s = nf$, hence we have have $h = s'$ as classes in the Picard group.

Title: Picard group of a ruled surface


问:
Bound on the coordinates of torsion points on the elliptic curve: $y^2 = x^3 - k^2x + k^3$
Consider the elliptic curve $y^2 = x^3 - k^2x + k^3$ where $k$ is a non-zero integer.
I've come across a question that asks to show that if $(x,y)$ is a rational torsion point on the elliptic curve, then:
$$\vert{y}\vert \le 5\vert{k}\vert^3$$ and $$\vert{x}\vert \le 3\vert{k}\vert^2$$

Now by the Nagell-Lutz Theorem, $x$ and $y$ must be integers, and if $y$ is non-zero, then $y^2$ divides the discriminant of the curve, which I believe is $-4k^6 + 27k^6 = 23k^6$.

I therefore believe this should mean $y^2$ divides $k^6$ and so $\vert{y}\vert \le \vert{k}\vert^3$.

I'm therefore wondering why a factor of 5 has been included in the result of the question - have I missed something?

Further I'm not sure how to find the bound on $x$. I'm assuming I need to use the bound on $y$  but this seems difficult.

Any help much appreciated.

Body:
For $y$, $y^2|-23k^6$, then $|y|^2\leq |23k^6|\leq|25k^6|$, so $|y|\leq 5|k^3|$.

For $x$, we know $|x^3 - k^2x + k^3|\leq 25|k|^6$, thus $|x^3 - k^2x| \leq|k^3|+25|k|^6$, which is also $|x||x ^2- k^2|\leq|k^3|+25|k|^6$.

If $|x|>3k^2$, then $|x||x ^2- k^2|>3k^2(9k^4-k^2)$, so $3k^2(9k^4-k^2)<|k^3|+25|k|^6$, we now get $2k^6-3k^4<|k^3|$.

Now there are two possibilities.

If $k>0$, then $2k^3<3k+1$, so $k=1$. Now $y^2=x^3-x+1$ where $\Delta=-23$ and $y=1$ or $-1$, hence $x^3-x+1=1$, then we know $x=0,1,-1$ which doesn't satisfy $|x|>3|k|^2$.

If $k<0$, then $2k^3>3k+1$, this is always impossible too since $k$ is an integer.


问:
Prove that $\iiint _{f^{-1}[a,b]}f(x,y,z)dxdydz=\int _a ^b tF'(t)dt$
Suppose $f\in C^1(\mathbb{R}^3)$, for every $t\in \mathbb{R}$, $f^{-1}(t)$ is a simple(1) closed surface, and let the volume of the 3-D object surrounded by $f^{-1}(t)$ be $F(t)$. If $F:[0,+\infty] \to \mathbb{R}$ is continuously differentialble, prove that
$$\iiint _{f^{-1}[a,b]}f(x,y,z)dxdydz=\int _a ^b tF'(t)dt.$$
(1):To be clear, here is the definition of a simple domain:

My first thought was to use the coarea formula. Then
$$
\iiint _{f^{-1}[a,b]}f(x,y,z)dxdydz=\int_a^b dt \int _{f^{-1}(t)} \frac{f}{|\nabla f|} d\sigma = \int _a ^b tdt \int _{f^{-1}(t)} \frac{1}{|\nabla f|} d\sigma.
$$
Well, as you have seen the RHS is pretty close to $\int _a^b tF'(t)dt$. But from here I met 2 major difficulties:(1)the conditions required by the coarea formula were not all satisfied: we need  $f\in C^2(\mathbb{R})$.(2) even if the coarea formula is allowed, I still cannot proceed. By definition $F(t)=\iiint _D 1dxdydz$ where $\partial D=f^{-1}(t)$. But how should I get $F'(t)=\int _{f^{-1}(t)} \frac{1}{|\nabla f|} d\sigma$?

Thank you in advance.

Example: let $f(x,y,z)=x^2+y^2+z^2,[a,b]=[0,1]. F(t)=\frac{4\pi t^{3/2}}{3},F'(t)=2\pi \sqrt {t}.$ Hence,
$$
\iiint _{f^{-1}[0,1]}f(x,y,z)dxdydz=\iiint _{x^2+y^2+z^2 \leq 1}(x^2+y^2+z^2)dxdydz= \\ \int _{[0,2\pi] \times [0,\pi] \times [0,1]} r^2 \cdot r^2 \sin \phi d\theta d\phi dr = 4\pi/5.
$$
$$
\int _0 ^1 tF'(t)dt = \int _0 ^ 1 t \cdot 2\pi \sqrt {t} dt = 4\pi/5.
$$

答:
I won't go through the epsilontics, but I think your formula can be understood in terms of simple Riemann integrals. I write just ${\bf x}$ for your $(x,y,z)$.

The set $D:=f^{-1}\bigl([a,b]\bigr)$ can be viewed as an onion, foliated by the surfaces $f^{-1}\bigl(\{t\}\bigr)$, $a\leq t\leq b$.

Choose an $N\gg1$, and put
$$t_k:=a+{k\over N}(b-a)\qquad (0\leq k\leq N)\ ,$$  $$D_k:=\bigl\{{\bf x}\in D\bigm| t_{k-1}\leq f({\bf x})\leq t_k\bigr\}\qquad(1\leq k\leq N)\ .$$
We then have $\bigcup_{k=1}^N D_k=D$, and therefore
$$\int_Df({\bf x})\>{\rm d}({\bf x})=\sum_{k=1}^N \int_{D_k}f({\bf x})\>{\rm d}({\bf x})=\sum_{k=1}^Nf(\xi_k)\>{\rm vol}(D_k)\ .$$
By definition of $F$ we therefore have
$$\int_Df({\bf x})\>{\rm d}({\bf x})=\sum_{k=1}^Nf(\xi_k)\bigl(F(t_k)-F(t_{k-1})\bigr)=
\sum_{k=1}^Nf(\xi_k) F'(\tau_k)(t_k-t_{k-1})\ .$$
Since $\xi_k\in D_k$ we have $\tau_k^* :=f(\xi_k)\in[t_{k-1},t_k]$. The last sum therefore is a Riemann sum for the integral
$$\int_a^b t\>F'(t)\>dt\ .$$


问:
$\lim_{x \to + \frac{1}{2}} [ \tan(\pi x^2)+(2x-1)4^x+6x-4 ] \cot(x-\frac{1}{2})$
I've tried to solve this limit:
$\lim_{x \to + \frac{1}{2}}  \left [ \tan(\pi x^2)+(2x-1)4^x+6x-4 \right ] \cot(x-\frac{1}{2})$

the first parentesis should tend to $\frac{\sqrt{2}}{2}-1$ but I don't know how to procede.

The final result should be $2 \pi + 10$

答:
To make life easier, let $x=y+\frac 12$ to make the expression
$$\left(\left(4^{y+1}+6\right) y+\tan \left(\frac{1}{4} \pi  (2
   y+1)^2\right)-1\right) \cot (y)$$ and now, compose series one piece at the time
$$4^{y+1}=4 e^{y \log(4)}=4+4 y \log (4)+2 y^2 \log ^2(4)++O\left(y^3\right)$$

Now,   after expansion of the argument, and using $\tan(a+b)=\cdots$
$$\tan \left(\frac{1}{4} \pi  (2y+1)^2\right)=1+2 \pi  y+2 \left(\pi +\pi ^2\right) y^2+O\left(y^3\right)$$
$$\left(4^{y+1}+6\right) y+\tan \left(\frac{1}{4} \pi  (2
   y+1)^2\right)-1=(10+2 \pi ) y+y^2 \left(2 \left(\pi +\pi ^2\right)+4 \log(4)\right)+O\left(y^3\right)$$
$$\cot(y)=\frac{1}{y}-\frac{y}{3}+O\left(y^3\right)$$ Make the product and the expression is
$$(10+2 \pi )+y \left(2 \pi\left(\pi +1\right)+8 \log (2)\right)+O\left(y^2\right)$$ which shows the limit and also how it is approached.

Edit for clarification about $\tan \left(\frac{1}{4} \pi  (2y+1)^2\right)$
$$\frac{1}{4} \pi  (2y+1)^2=\frac{\pi }{4}+\pi  y(y+1)=\frac{\pi }{4}+a$$
$$\tan \left(\frac{\pi }{4}+a\right)=\frac{1+\tan(a)}{1-\tan(a)}=\frac {1+a+\frac{a^3}{3}+\frac{2 a^5}{15}+O\left(a^7\right) } {1-a-\frac{a^3}{3}-\frac{2 a^5}{15}+O\left(a^7\right) }$$

Long division
$$\tan \left(\frac{\pi }{4}+a\right)=1+2 a+2 a^2+\frac{8 a^3}{3}+\frac{10 a^4}{3}+\frac{64 a^5}{15}+\frac{244
   a^6}{45}+O\left(a^7\right)$$ Now, replace $a=\pi  y(y+1)$

$$\tan \left(\frac{1}{4} \pi  (2y+1)^2\right)=1+2 \pi  y+2 \left(\pi +\pi ^2\right) y^2+O\left(y^3\right)$$


问:
I want to find maxima for a particular Euler product
I would like to obtain large numbers for a particular Euler product given by:

$\zeta(s)  = {\displaystyle \prod_{p} \left( 1-p^{-s} \right)^{-1} }$

for $s = 1 + ia$

$i$ being the imaginary unit.

In particular, I am interested in finding a large value of $\zeta(s)$ for large values - $a$ ~  $o(10^{33})$.

My current method is to loop through different values of $a$ in a brute force way. I hoped there might be some useful mathematical approach to make this search more efficient.

答: Fantastic question. Here is one attempt. Not clear if there is a better way. (I remark first in passing that the partial zeta sums here are almost periodic functions.) I would divide up the primes $p$ to those at or below $x$ and those above, taking $x$ to be as large as computationally possible, and divide up the sum of the $e^{-i a \log p}/p$ into the two corresponding parts. The high-$p$ part should act like noise and should largely cancel, but "cannot" be controlled in any case. On the low-$p$ part I would produce a record of many "early" values of $a$ for which the low-$p$ sum is large. As a first pass, each of these early values of $a$ could then be used to produce high-$a$ candidates by adding to it large multiples of the product of the early periodicities. In this almost-modular approach, the early zeta sum would remain large (indeed unchanged) and the later zeta sum would almost always be small except for the fact that the periodicities are not commensurate. What is desirable then is that the angles associated with the terms of the early zeta sum should keep "pointing" as much as possible in the same directions as before. Candidates for $a$ could then be sought among values for which the histogram of the angle differences was concentrated around $0$, and this histogram itself could be weighted towards the higher-impact lower values of $p$. Admittedly this still involves much computation. One possible upshot is that if one were indeed reduced to trial and error search, the searching could be substantially restricted to the early-$p$ sum


问:
what is ascending and descending induction?
On page 88 of Lang's "Topics in Cohomology of Groups", Lang mentions a technique he calls "ascending and descending induction".

Initially I felt a bit embarrassed that I did not know a sort of induction, but then when googling for a definition, I found very few mentions of it and none that helped me understand exactly what this sort of induction is.

Could someone explain to me what ascending and descending induction is, please?

A reference to its definition in the literature would also be very helpful.

答:
Let $(P_n)_{n \ge 1}$ be a set of statements. You want to prove that all $P_n$ are true.

In ascending induction, you prove (i) $P_1$ is true and (ii) $\forall n P_n \Rightarrow P_{n+1}$.

In descending induction, you prove (i) $P_n$ is true for infinitely many $n$ and (ii) $\forall n P_{n+1} \Rightarrow P_n$.

The best known example for descending induction is Cauchy's proof of the arithmetic-geometric mean inequality. Part (i) is done by proving that $P_n$ is true for all $n = 2^k$, which in turn is done by ascending induction over $k$. This is a "book" proof according to Erdos.


问:
Can someone help me with this other limit?
$$ \lim_{x \to 0} \frac{(\sin x - \tanh x)^2}{(e^{x}-1-\ln{(1+x)})^3} \overset{??}{=} \frac{1}{36} $$

If you consult the graph it is intuitive to think that when $ x \to 0 $ the function tends to 1/36. I know that you can apply L'Hospital's rule to solve it, but it's too long. The only solution I see is to apply infinitesimal equivalents, which I have not tried yet.

Thank you all!

Body:
You can use the Taylor expansion to evaluate. In fact, since
$$ \sin x=x-\frac16x^3+O(x^5),\tanh x=x-\frac13x^3+O(x^5),e^x-1=x+\frac12x^2+\frac16x^3+O(x^4), \ln(1+x)=x-\frac12x^2+\frac13x^3+O(x^4) $$
then you have
$$ \lim_{x \to 0} \frac{(\sin x - \tanh x)^2}{(e^{x}-1-\ln{(1+x)})^3}=\lim_{x \to 0} \frac{\bigg[(x-\frac16x^3)-(x-\frac13x^3)+O(x^5)\bigg]^2}{\bigg[(x+\frac12x^2+\frac16x^3)-(x-\frac12x^2+\frac13x^3)+O(x^4)\bigg]^3}=\frac{\frac{1}{6^2}}{
1}= \frac{1}{36}. $$


问:
Book recommendation on Discrete Geometry
I have started studying discrete geometry. And I need some good suggestions on discrete geometry books. I am looking for something rigorous with a lot of solved examples (if possible).
Some books that I am following right now are

Lectures on Discrete and Polyhedral Geometry by Igor Pak (found it a little bit advanced)

Lectures on Discrete Geometry Textbook by Jiří Matoušek.

Thanks in advance.

答: Try Matthias Beck, Sinai Robins: Computing the Continuous Discretely. This is an extremely well-written book for undergraduates.


问:
Significance of zero people out of 80 choosing something as their favourite from a selection of 42 things
80 people were asked to choose their favourite object from 42 objects. Not one person chose a particular object.

The criteria for people choosing objects might be random or there might be a pattern so that people like and dislike the same things.

I know that the chances that zero people choose a particular object are 14% - so this to me suggests there is something unfavourable about the object. That doesn't mean though that there is an 86% chance that this hasn't happened by random and there is a bias/pattern in the selections?

How do I determine the existence of whether there is a pattern or not and quantify the faith in which the fact that no person chose an object suggest it is unfavourable?

To me it seems pretty obvious that if no people out of 80 chose something as their favourite then we can say with reasonable certainty that it is not favoured. But somebody else is telling me that no it is not enough of a sample size.

I also have the results of all the 80 peoples selections if that helps.

答:
It seems to me that the relevant statistical question is "How likely is it that at least one of the 42 objects is not chosen, assuming that each item is equally likely to be chosen?".  If this probability is moderately high then it seems unfair to criticize an item which was not chosen, because we should expect some item(s) to be un-chosen just by chance.

So let's suppose that when a person chooses an item, each item is equally likely to be chosen, with a probability of $1/42$, and that each person makes his choice independently of other people's choices.  By inclusion/exclusion, the probability that at least one item is not chosen by anyone is
$$\sum_{m=1}^{42} (-1)^{m-1} \binom{42}{m} \left( 1 - \frac{m}{42} \right)^{80} = 0.99973$$
which is quite high, verging on certainty.  So we should not be surprised to find that some item is not chosen.


问:
If x is one of the K nearest neighbors of y in a dataset, is y one of the K nearest neighbors of x?
Suppose I have a set of vectors $X = \{x_i\}_{i=1}^n\subset \mathbb{R}^d$. Fix a vector $x\in X$ and assume that a different vector $y$ is one of the $K$ points in $X$ that is closest to $x$ in Euclidean distance. In other words, $y$ is one of the $K$ nearest neighbors of $x$. Does this mean that $x$ is one of the $K$ nearest neighbors of $y$ too?

Basically, is the statement "$x$ is one of the $K$ nearest neighbors for $y$" reflexive?

答:
That is not true.

Let $x=1$ and $y=0$ and let $K=1$.

Let $X=\{-\frac12, 0, 1\}$.

While $0$ is the closest neighbor of $1$, $1$ is not the closest neighbor of $0$.


Title: Can someone help me with this other limit?

Body:
$$ \lim_{x \to 0} \frac{(\sin x - \tanh x)^2}{(e^{x}-1-\ln{(1+x)})^3} \overset{??}{=} \frac{1}{36} $$

If you consult the graph it is intuitive to think that when $ x \to 0 $ the function tends to 1/36. I know that you can apply L'Hospital's rule to solve it, but it's too long. The only solution I see is to apply infinitesimal equivalents, which I have not tried yet.

Thank you all!

Body:
$$L=\lim_{x \to 0} \frac{(\sin x - \tanh x)^2}{(e^{x}-1-\ln{(1+x)})^3} \overset{??}{=} \frac{1}{36}$$ Expand the den. forsmall values of $x$ it is $$[1+x+x^2/2-1 -x+x^2/2+O(x^3)]=[x^2+O(x^3)]^3,$$ when $|x|<<1$.
So the Num. has to be expanded complete upto $x^3$ term using $\sin z= z-z^3/3!+.., \tanh z=z-z^3/3+..$, then $$L=\lim_{x\to 0} \frac{x^6/(36)}{x^6}=\frac{1}{36}.$$


问:
Flowing a vector along a vector field $X$ using the pushforward of the flow of $X$
On the three-sphere $S^3$, I'm given three vector fields $X$, $Y$ and $Z$, such that at each point $p\in S^3$, the tangent vectors $X_p$, $Y_p$ and $Z_p$ form an orthogonal basis of the tangent space $T_pS^3$.

I denote by $X^t:S^3\rightarrow S^3$ the flow of the vector field $X$, so $X^0(p)=p$ and $\frac{d}{dt}X^t(p)=X_{X^t(p)}$.

Its pushforward at the point $p$ is denoted $(X^t)_{*,p}$, so this is a map from $T_pS^3$ to $T_{X^t(p)}S^3$.

Now let $v$ be an arbitrary tangent vector in $T_pS^3$. We can write $v=v_xX_p+v_yY_p+v_zZ_p$. I want to prove the following:

$(X^t)_{*,p}(v)=v_x X_{X^t(p)}+...$.

Here the dots denote other terms in the span of $Y_{X^t(p)}$ and $Z_{X^t(p)}$, but it's only the $X_{X^t(p)}$ component I care about (plus I don't know the other components).

Is this maybe immediate from the definition of the flow? I tried using the defining equation $\frac{d}{dt}X^t(p)=X_{X^t(p)}$ of the flow of $X$, but I don't know how I can relate the time derivative with the pushforward...

Edit: so I think I can visualize my question as follows. Starting with a tangent vector $v$ at $p$, flowing this vector along $X$ (using the pushforward of the flow) does not change the component of $v$ along $X$. Sounds plausible...

答:
Just to set notation, let $X$ a vector field on $M$ and suppose the flow $\phi_t$ of $X$ exists for all $t$ (which is the case if $M$ is compact, like $S^3$ is). So, we are going to prove that $$D_p\phi_s(X_p)=X_{\phi_s(p)}.$$

We just need to use the "correct" definition of the pushforward: In general, for a map $f:M\rightarrow N$ we can compute $D_pf(u)$ by the following method. Let a curve $c:(-\epsilon,\epsilon)\rightarrow M$ such that $c(0)=p$ and $c'(0)=u$. Then $$D_pf(u)=(f\circ c)'(0).$$

In words, just pick a curve with velocity $u$, push forward the curve and compute the new velocity.

In our case now: To show our formula we have first to find a curve $c$ such that $c(0)=p$ and $c'(0)=X_p$. By definition, we can pick $c(t)=\phi_t(p)$! Therefore, to compute the right hand side of our formula all we have to compute is the velocity of $\sigma(t)=\phi_s(\phi_t(p))$. But, we have that $\phi_s(\phi_t(p))=\phi_t(\phi_s(p))$ therefore $$D_p\phi_s(X_p)=\sigma'(t)=\frac{d}{dt}\phi_t(\phi_s(p))=X_{\phi_s(p)}.$$

Let me just mention that this property holds because in fact $t\rightarrow \phi_t$ is a homomorphism from $\mathbb{R}\rightarrow Diff(M)$. If we have a $\textit{time-dependent}$ vector field $X_t$ then tthis property need not hold.


问:
f is uniformly continuous if for all $x$ and $y ∈ \mathbb{R}$, $|f(x) − f(y)| ≤ |x−y|^\frac{1}{2}$.
Let f : R → R be a given function. The following property will ensure that f is uniformly continuous:

for all $x$ and $y ∈ \mathbb{R}$, $|f(x) − f(y)| ≤ |x−y|^\frac{1}{2}$.

My attempt :
From definition a function $f$ is uniformly continuous if for a arbitrary $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that for any two points $x,y$ in $\mathbb{R}$,
$$ |x-y| \lt \delta \implies |f(x)-f(y)| \lt \epsilon
$$

Now let $\epsilon \gt 0$ be given. Then
$$
 |x−y|^\frac{1}{2} \lt \epsilon \\
\implies
 |x-y| \lt {\epsilon}^2 = \delta \\
\implies |f(x)-f(y)| \lt \epsilon
.$$

Therefore the function is uniformly continuous.

答:
You surely are on the right track. Given $\epsilon > 0$ you have to find a $\delta > 0$ such that the implication $|x-y| \lt \delta \implies |f(x)-f(y)| \lt \epsilon$ holds.

So the correct order of arguments would be: Let $\epsilon \gt 0$ be given. Set $\delta = \epsilon^2$. Then $\delta > 0$ and
$$
 |x−y| \lt \delta  \implies |f(x) − f(y)| ≤ |x−y|^{1/2} < \delta^{1/2} = \epsilon \, .
$$


问:
Why $\log_b(x)$ = $-\log_{1/b}(x)$?
IN a textbook I am asked to consider the graphs of $f(x) = \log_{1/2}(x)$ and $-\log_2(x)$. They appear identical and then the textbooks tells the rule that $\log_b(x)$ = $-\log_{1/b}(x)$.

I tried to prove this to myself on paper.

I can write both equations in exponential form. Pretend $x=4$:

$$\log_{1/2}(4)=\frac{1}{2}^y=4$$

Then write the same for the other version:

$$-\log_2(4)=-2^y=4$$

If I know these are equivalent then the two exponent versions should be the same no?

$$4=\frac{1}{2}^y=-2^y$$

It's not clicking and I don't 'get it' from here. Is my last equation true? Why are they equivalent?

答:
We will use the following two identity
$$\log_a(b)=\log_{a^x}b^x$$
$$log_a(b^n)=n \log_a(b)$$

We will start with the question
$$log_b(x)=-log_{\frac 1 b}(x)$$
Seeing $\frac{1}{b}$ as the base in the right hand side of the equation motivates us to use the second identity setting up $x = -1$.
$$log_b(x)=-log_{\frac 1 b}(x)=-log_b(\frac{1}{x})$$
Now we get rid of the minus sign using the second identity
$$log_b(x)=-log_b(\frac{1}{x})=log_b((\frac 1 x)^{-1})=log_b(x)$$

Title: Why $\log_b(x)$ = $-\log_{1/b}(x)$?


问:
very basic short exact sequence problem
Given a short exact sequence $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow0$ and $f:A \rightarrow B, g: B \rightarrow C$, why is $C$ isomorphic to $B/A$? All I can show is that $C$ is isomorphic to $B/ Im (A)$. I've look up a few books but I still don't understand why $Im (A) = A$.</p>

答: Since $f$ is injective (by exactness) we have $Im(A) \simeq A$, thus you're right, it's an acceptable abuse of language.</p>

Title: very basic short exact sequence problem


答:
You have the right idea, but the logic isn't quite there.

Fix $\epsilon > 0$. Now you need to show there exists a $\delta > 0$ such that if $| x - y| < \delta$, then $|f(x) - f(y) | < \epsilon$.

As you say, a good choice is $\delta = \epsilon^2 > 0$. Then
$$
\begin{align}
&\quad\quad\quad|x - y| < \delta
\\&\implies | x- y| < \epsilon^2
\\&\implies |x-y|^{\frac{1}{2}} < \epsilon
\\&\implies |f(x) - f(y)| \leq |x-y|^{\frac{1}{2}} < \epsilon
\\&\implies |f(x) - f(y)| < \epsilon
\end{align}
$$
This shows that there exists a $\delta > 0$ (namely $\epsilon^2$) such that if $|x - y| < \delta$, then $|f(x) - f(y)| < \epsilon$.

Title: f is uniformly continuous if for all $x$ and $y ∈ \mathbb{R}$, $|f(x) − f(y)| ≤ |x−y|^\frac{1}{2}$.

Body:
Let f : R → R be a given function. The following property will ensure that f is uniformly continuous:

for all $x$ and $y ∈ \mathbb{R}$, $|f(x) − f(y)| ≤ |x−y|^\frac{1}{2}$.

My attempt :
From definition a function $f$ is uniformly continuous if for a arbitrary $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that for any two points $x,y$ in $\mathbb{R}$,
$$ |x-y| \lt \delta \implies |f(x)-f(y)| \lt \epsilon
$$

Now let $\epsilon \gt 0$ be given. Then
$$
 |x−y|^\frac{1}{2} \lt \epsilon \\
\implies
 |x-y| \lt {\epsilon}^2 = \delta \\
\implies |f(x)-f(y)| \lt \epsilon
.$$

Therefore the function is uniformly continuous.


答:
The usual method for this is diagonalisation, but here there's a simpler way because $M$ is of a special form.

Let $J$ be the $5 \times 5$ matrix composed with ones only, so that $J^2=5J$, thus $J^n=5^{n-1}J$ for $n \geq 1$. Then $7M=2I+J$.

Therefore,

$$\begin{aligned} 7^n M^n &=\sum_{k=0}^n {\binom{n}{k} 2^k J^{n-k}}\\ &= 2^nI+\sum_{k=1}^n{\binom{n}{k}2^{n-k}5^{k-1}J}\\ &= 2^nI+\frac{1}{5}\left(\sum_{k=0}^n{\binom{n}{k}2^{n-k}5^k}-2^n\right)J\end{aligned}$$

So $7^nM^n=2^nI-\frac{2^n}{5} J+\frac{(2+5)^n}{5}J$. Thus, $M^n = \frac{J}{5}+\frac{2^n}{7^n}I-\frac{2^n}{5 \cdot 7^n}J$ and $M^n \rightarrow \frac{J}{5}$.

Title: Infinite power of doubly stochastic matrix without using diagonalization

Body:
Given
$$M := \frac17 \begin{bmatrix}
3 & 1 & 1 & 1 & 1 \\
1 & 3 & 1 & 1 & 1 \\
1 & 1 & 3 & 1 & 1 \\
1 & 1 & 1 & 3 & 1 \\
1 & 1 & 1 & 1 & 3
\end{bmatrix}$$ find $$\lim _{n \to \infty} M^n$$

We can express,

$$
M = \frac{1}{7}\left(\begin{matrix}
-1 & -1 & -1 & -1 & 1 \\
1 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & 1
\end{matrix}\right)\left(\begin{matrix}
2 & 0 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 2 & 0 \\
0 & 0 & 0 & 0 & 7
\end{matrix}\right)\left(\begin{matrix}
\frac{-1}{5} & \frac{4}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{-1}{5} & \frac{4}{5} & \frac{-1}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{4}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{4}{5} \\
\frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5}
\end{matrix}\right)
$$

And evaluate,
$$
\lim _{n \rightarrow \infty} M^{n} = \lim _{n \rightarrow \infty} P\left(\begin{matrix}
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1
\end{matrix}\right)P^{-1}
= \frac{1}{5}\left(\begin{matrix}
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1
\end{matrix}\right)
$$
where I used that,
$$
P = \left(\begin{matrix}
-1 & -1 & -1 & -1 & 1 \\
1 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & 1
\end{matrix}\right)
\,\,\,\,\,\&
\,\,\,\,\,
P^{-1} = \left(\begin{matrix}
\frac{-1}{5} & \frac{4}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{-1}{5} & \frac{4}{5} & \frac{-1}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{4}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{4}{5} \\
\frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5}
\end{matrix}\right)
$$

This traditional way of diagonalization involves not just finding eigenvalues but also eigenvectors (in other words find P).

Finding eigenvalues for this is easy.
I consider,
$$
A=\left[\begin{array}{lllll}
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1
\end{array}\right]
$$ which clearly has eigenvalues $\lambda_1 = 0$ (with algebraic multiplicity = $4$) and $\lambda_2 = 5$ (with algebraic multiplicity = $1$).

The eigenvalues for our first Matrix $M = \frac{1}{7}\left(A + 2I\right)$ are therefore, $\lambda_1 = \frac{2}{7}$ (with algebraic multiplicity = $4$) and $\lambda_2 = 1$ (with algebraic multiplicity = $1$).

Is there any way we can efficiently compute, $\lim _{n \rightarrow \infty} M^{n}$, without calculating eigenvectors i.e., without computing matrices $P$ and $P^{-1}$?


答:
Strictly speaking, $B/A$ doesn't make sense unless $A\subseteq B$. However, since the second arrow is a monomorphism, you can identify $Im(A)$ with $A$ (they are isomorphic).

So you're exactly right that $B/Im(A)\cong C$, and you go one step further by considering $A$ and its monomorphic image as the same thing.

Title: very basic short exact sequence problem

Body: Given a short exact sequence $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow0$ and $f:A \rightarrow B, g: B \rightarrow C$, why is $C$ isomorphic to $B/A$? All I can show is that $C$ is isomorphic to $B/ Im (A)$. I've look up a few books but I still don't understand why $Im (A) = A$.</p>


问:
Does the gambler's fallacy apply in this scenario?
An example of the gambler's fallacy:

Say I flip a coin ten times, and the first five times are all tails. People would think that in order to "even out" the probabilities, there will be a higher probability of heads compared to tails in the next five flips. This incorrect (and obviously dangerous) thinking has led many gamblers to lose their money, hence "the gambler's fallacy."

Now, consider this scenario:

I have an algorithm that can predict whether a coin will land on heads or tails, with 50% accuracy. I ask Bob to flip a coin ten times, and record the results. The results of the first five flips are then compared to what my algorithm predicted.

Let's say the algorithm got every single prediction wrong in the first five flips. Does this mean that there is a higher probability that the algorithm will be correct in the remaining five flips?

I'm not sure whether the gambler's fallacy applies here, so correct me if I'm wrong......

答:
50% probability does not mean that if you flip the coin 10 times, it will be heads five times and tails five times. It only means that the heads to tails  will approach the expected value the more you repeat the experiment.

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

In your case, I don't see why the probability would be higher in the next five flips.


问:
Existence of both right and left eigenvectors for the same eigenvalue
I started learning about Markov Chains recently and I've seen that for a transition matrix 1 is an eigenvalue and it has a right eigenvector and a left eigenvector for this value.

Is it always so that every eigenvalue has right and left eigenvectors? If not then is there any rule as to when this happens? Also is there any intuition behind what dictates this behavior?

Thanks

答: The key point is that $A$ and $A^T$ have the same eigenvalues, since they have the same characteristic polynomial.  So, if $\lambda$ is an eigenvalue of $A$ then there are vectors $X$ and $Y$ such that $AX=\lambda X$ and $A^TY=\lambda Y$.  Taking transposes in the last equation, $Y^TA=\lambda Y^T$, so that $A$ has both right and left eigenvectors associated with $\lambda$.


问:
derivative for the following matrix
I have a question about the following derivative. Let us have $X\in\mathbb{R}^{m\times n}, z\in\mathbb{R}^{n}$ and I would like to find the derivative $$\frac{\partial (z^{T}X^{T}Xz)}{\partial z}. $$ Any idea? It gives me a hard time. Thank you.

答:
Define a new vector
$$y=Xz$$
Write the function in terms of this vector. Then calculate the differential and gradient.
$$\eqalign{
\phi &= y:y \\
d\phi &= 2y:dy &= 2y:X\,dz &= 2X^Ty:dz \\
\frac{\partial \phi}{\partial z} &= 2X^Ty &= 2X^TXz \\
}$$

In the above, a colon is used to denote the trace/Frobenius product, i.e.
$$A:B = \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} = {\rm Tr}(A^TB)$$
This product is obviously commutative, i.e. $\,A\!:\!B=B\!:\!A$

In addition, the cyclic property of the trace permits the terms in such a product to be rearranged in a number of equivalent ways, e.g.
$$A:BC \;=\; AC^T:B \;=\; B^TA:C$$
Finally, the product is applicable to vectors by treating them as rectangular matrices (set $n=1$) in which case it becomes the familiar dot product.

Title: derivative for the following matrix


问:
Looking for an easy method of rearranging $W^2=\frac{T+W-1}{T}$
Rearrange the formula $(1)$ to make $W$ the subject

$$W^2=\frac{T+W-1}{T}\tag1$$
$$W^2T=T+W-1$$
$$W^2T-W=T-1$$
$$W^2-\frac{W}{T}=1-\frac{1}{T}$$
Using complete the square method
$$\left(W-\frac{1}{2T}\right)^2=1-\frac{1}{T}+\frac{1}{4T^2}$$
$$\left(W-\frac{1}{2T}\right)^2=\frac{4T^2-4T+1}{4T^2}$$
$$\left(W-\frac{1}{2T}\right)^2=\frac{(2T-1)^2}{4T^2}$$
$$W-\frac{1}{2T}=\pm\frac{(2T-1)}{2T}$$

$$W=\frac{1-T}{T}$$

Is there is an elegant method (or an easy way) of answering this question without doing completing the square method?

Body:
Note
$$W^2-\frac{T+W-1}{T}=W^2- 1-\frac1T(W-1)
= (W-1)(W+1-\frac1T)=0$$


问:
Evaluate $\int_0^\infty \frac{(\mathrm{e}^{-ax}-\mathrm{e}^{-bx})^2}{x^2}dx=\ln\frac{(2a)^{2a}(2b)^{2b}}{(a+b)^{2(a+b)}} $

(1) Prove that$$\int_0^\infty \frac{(\mathrm{e}^{-ax}-\mathrm{e}^{-bx})^2}{x^2}\mathrm{d}x=\ln\frac{(2a)^{2a}(2b)^{2b}}{(a+b)^{2(a+b)}}$$
(2) Prove that$$\int_0^\infty \frac{\mathrm{e}^{-ax^2}-\mathrm{e}^{-bx^2}}{x^2}\mathrm{d}x=\frac{1}{2}\ln\frac{b}{a} $$

I just learned improper integral with parameters and worked out a basic example below
$$
\int_0^\infty \frac{\mathrm{e}^{-ax}-\mathrm{e}^{-bx}}{x}\mathrm{d}x=\ln\frac{b}{a}
$$
But I do not know how to compute the two integrals above. I think the basic example may be of some help and tried integration by parts, inspired by $\int_0^\infty\frac{\sin x}{x}\mathrm{d}x$ and $\int_0^\infty\frac{\sin^2 x}{x^2}\mathrm{d}x$.

Any hint is appreciated.

答:
The answer for integral (2) is not correct.
It should be :

```<code>     (Pi)^(1/2) [ (b)^(1/2) - (a)^(1/2) ]
</code>```

fjaclot

Body:
The answer for integral (2) is not correct.
It should be :

     (Pi)^(1/2) [ (b)^(1/2) - (a)^(1/2) ]

fjaclot


问:
Combinatorial proof of a nice expression for Stirling numbers of second kind
While answering a question I came across the following nice identity:
$$
\sum_{1\le k_1\le k_2\le\dots\le k_n\le m}\prod_{1\le i \le n}k_i={n+m \brace m},
$$
where ${n+m \brace m}$ refers to the Stirling number of second kind. The equation can be readily proved by using generating functions but I strongly suspect that there should be a simple combinatorial explanation of the expression.

Any hint is appreciated.

答:
Hint:  Consider the following algorithm:

You have a list of the $n+m$ numbers and you start with $\{1\}$ as the partition, you start also with a pointer starting at $2$ that is going to increase $n$ times and in each number you have the following decision to make: If the number of blocks is exactly $m,$ you have $m$ options per number. If not, you can decide if you want to append the number to a block or create a new block where the number is going to be the smallest element of the block. When this happens the number of blocks will increase.

As an example and for simplicity, assume $m=2$ and take $n=3.$ We start with $1$ and we place it in a block so $\{1\}$ is what we have. Then $2,$ if we decide to place $2$ in the other block, then we will end up having $2^3$ options for the rest of numbers. If we decide to place $3$ as the smallest element of the new block, we will have $2^2$ options and so on. So we will have
$$\sum _{i=0}^{3}2^i=2^4-1=LHS,$$
but of course every partition can be created in this way. There you can check that this is the correspondence
$$\underbrace{1^3}_{\{1,2,3,4\},\{5\}}+\underbrace{1^2\cdot 2}_{\{1,2,3,\cdots \},\{4,\cdots \}}+\underbrace{1^1\cdot 2^2}_{\{1,2,\cdots \},\{3,\cdots \}}+\underbrace{ 2^3}_{\{1,\cdots \},\{2,\cdots \}}$$

So, to generalize, consider where do the numbers change, so I would suggest you write the LHS as
$$\sum_{d_1+d_2+\cdots +d_m=n}1^{d_1}2^{d_2}\cdots m^{d_m}.$$

Here the $d_i$ represent the frequency of $i$ in the sequence $k$ i.e., $$d_i = |\{j:k_j=i\}|.$$

A way to understand this algorithm is to think about it backwards. Suppose I give you a partition $\pi \vdash [n+m]$ with $m$ blocks. Suppose the blocks are labelled $\pi =\{B_1,\cdots , B_m\}$ where
the blocks are ordered by their minimal elements $b_i\in B_i: 1=b_1<b_2<\cdots<b_m\le n+m$. Consider further the consecutive differences of these numbers $d_i=b_{i+1}-b_{i}-1\ (0<i<m)$, $d_m=n+m-b_m.$ Notice that $d_i\ge0$ and $d_1+d_2+\cdots +d_m=n+m-(m-1)-b_1=n$. Notice further that $d_i$ is essentially the number of elements which can be placed arbitrarily in any of the first $i$ blocks, so that there are $i^{d_i}$ ways to perform this.


问:
Is the following proof of continuity correct?
I want to prove that $\max_{t\in[0,1]}x(t)$ is a continuous functional in $C[0,1]$.

I have that $\left | t-s \right |\leq 1$, $\forall t,s \in [0,1]$: then
$$
0=\left | \max_{t\in[0,1]}x(t)-\max_{s\in[0,1]}x(s) \right |\leq \left | x(t)-x(s) \right |<\epsilon,
$$
and I assume that  $\left | x(t)-x(s) \right |<\epsilon$ because $x(t)\in C[0,1]$.

Are my assumptions correct?

答:
Note: I assume below that $E=\mathcal C([0,1], \mathbb R)$ is endowed with the $\sup$ norm $\Vert \cdot \Vert_\infty$.

Your proof is not correct as to prove the continuity of

$$\begin{array}{l|rcl}
\phi : & \mathcal C([0,1], \mathbb R) & \longrightarrow & \mathbb R \\
    & x & \longmapsto & \max_{t \in [0,1]}  x(t)\end{array} $$

you need to consider the difference of two maps $x,y \in E$.

So take $x, y \in E$. For $t \in [0,1]$ we have

$$x(t) - y(t) \le \Vert x-y \Vert_\infty$$ hence
$$x(t) \le \Vert x-y \Vert_\infty + y(t)$$ and therefore

$$x(t) \le \Vert x-y \Vert_\infty + \max_{t \in [0,1]}  y(t) = \Vert x-y \Vert_\infty + \phi(y)$$ which implies
$$\phi(x) = \max_{t \in [0,1]}  x(t) \le \Vert x-y \Vert_\infty + \max_{t \in [0,1]}  y(t) = \Vert x-y \Vert_\infty + \phi(y)$$ i.e.

$$\phi(x) - \phi(y) \le \Vert x-y \Vert_\infty$$

As this proof is symmetric in $x,y$ we also get
$$\phi(y) - \phi(x) \le \Vert x-y \Vert_\infty$$ and finally
$$\vert \phi(y) - \phi(x) \vert \le \Vert x-y \Vert_\infty$$

From there it is clear that $\phi$ is continuous as it is Lipschitz.


问:
Show, that there exists a sequence $\left(B_{n}\right)_{n \in \mathbb{N}}$ of disjoint sets from $\mathcal{E}$
I am currently trying to solve the following problem.

Let $\mu$ and $v$ be two $\sigma$ -finite measures on a measurable space $(X, \mathcal{E}) .$ Show, that there exists a sequence $\left(B_{n}\right)_{n \in \mathbb{N}}$ of disjoint sets from $\mathcal{E},$ such that
$$
\bigcup_{n \in \mathbb{N}} B_{n}=X, \quad \text { og } \quad \mu\left(B_{n}\right), v\left(B_{n}\right)<\infty \quad \text { for alle } n
$$

Since both $\mu$ and $v$ are $\sigma$-finite measures, I make the sequences $A_n$ where $\mu(A_n)\leq\infty$ and $C_n$ where $v(C_n)\leq\infty$. I then tried to look at the intersection of these, which then is finite with respect to both measures, but I can't be sure that the union is $X$.

If someone has a good idea or hint it would be greatly appreciated.

答: We know that $X$ is measurable. By $\sigma$-finiteness we know that there exists a partition of $X$ into disjoint sets $B^{\mu}_1, B^{\mu}_2 \ldots$ that have finite measure wrt. $\mu$. Similarly, let $B^{\nu}_1, B^{\nu}_2 \ldots$ be a partition of $X$ into disjoint sets that have finite measure wrt. $\nu$. Then $B^{\mu}_1 \cap B^{\nu}_1, B^{\mu}_1 \setminus B^{\nu}_1, B^{\nu}_1 \setminus B^{\mu}_1, B^{\mu}_2 \cap B^{\nu}_2, B^{\mu}_2 \setminus B^{\nu}_2 \ldots$ is a partition of $X$ into disjoint sets that have finite measure wrt. both $\mu$ and $\nu$.


问:
$T(n) = nT(n/2) ,T(1)=1$ How can I solve this with backward subsitituon?
$n$ can be power of 2 or we can assume $Integerpart(n/2)$ ==> int casting.

I find the recurrence relation of function but How can I solve this with using backward subsitituon ?

My recurrence relation solution

答:
As pointed out in my comment, you can use the next hint:

Observe that $S(k) = 2^k S(k-1)$ and the fact that we know the value of $S(0)$. Next use backward substitution to "unroll" $S(k)$ until you get to $S(0)$ on the RHS. Using your "unrolled" expression for $S(k)$ you should be able to find $T(n)$

:)


问:
Truncation and $\textsf{isProp}$
In type theory, in particular, homotopy type theory. We have a notion of $\textsf{isProp}$ defined as follows (where we let $A:\textsf{Type}$):
$$\textsf{isProp}(A):=\Pi x,y:A.x=_Ay.$$
In words, $A$ has the property $\textsf{isProp}$ iff $A$ is either a singleton type or empty. Then, we have the truncation operation $||$-$||$. Let $A:\textsf{Type}$, then $||A||:\textsf{Prop}$.

My question is, if $\Sigma x:A.B(x)$ is known to have at most one inhabitant, that is, that $\Sigma x:A.B(x)$ in fact satisifes $\textsf{isProp}$, can we have the following?

$$\textsf{isProp}(\Sigma x:A.B(x))\rightarrow(||\Sigma x:A.B(x)||\leftrightarrow\Sigma x:A.B(x))$$

In this case, since $||\Sigma x:A.B(x)||$ is equivalent to $\Sigma x:A.B(x)$, we can apply the projection rules $\pi_1$ and $\pi_2$ of $\Sigma$.

答:
First, this:

In words, $A$ has the property $\mathsf{isProp}$ iff $A$ is either a singleton type or empty.

is incorrect unless one is assuming excluded middle. Constructively, propositions cannot be assumed to be either a singleton or empty.

The answer to your first question is, 'yes'. Truncation has an inclusion function:

$$|-| : A → \Vert A \Vert$$

And the induction principle for truncation allows for:

$$\mathsf{out}_{\Vert A \Vert} : \Vert A \Vert → A$$

provided $\mathsf{isProp}\ A$. Since both types are propositions and imply one another, it is even the case that $\Vert A \Vert =_\mathcal{U} A$ via univalence.

I'm unclear what precisely your second question means. You cannot exactly apply $π_1$ to a value of $\Vert Σ x:A. B(x)\Vert$ because it is ill typed. However, if the sigma type is a proposition, you can either use $\mathsf{out}$ to get a value that you can apply $π_1$ to, or you can transport $π_1$ along the equality between the types to get something of the right type to apply to the truncated type, which will have the same behavior as using $\mathsf{out}$.

Title: Truncation and $\textsf{isProp}$


问:
Monotonicity of incomplete Gamma function
For given $x \geq 0$, a function $f(a)$ is defined as

$f(a) = \frac{\Gamma(a, x)}{\Gamma(a)},$

where $\Gamma(a, x)$ is the upper incomplete Gamma function. I need to prove $f(a)$ is a monotonic function with $a \in [0,1]$. I have tried to compute the first derivative of $f(a)$, but I could not prove this derivative is nonnegative. Could you please help me?

答:
Partial answer. For $x>0$,
$$
\frac{{\Gamma (a,x)}}{{\Gamma (a)}} = \frac{1}{{\Gamma (a)}}\int_x^{+\infty}  {e^{ - t} t^{a - 1} dt}  .
$$
For $t\geq x\geq 1$,
$$
a \mapsto t^{a-1} ,\quad a \mapsto \frac{1}{{\Gamma (a)}}
$$
are monotonically increasing functions of $a$ on $[0,1]$, whence $\frac{{\Gamma (a,x)}}{{\Gamma (a)}}$ is monotonically increasing on $[0,1]$ for any fixed $x\geq 1$.


问:
Another bizarre sum involving a binomial coefficient and inverse powers of integers.
In the attempt to answer Binomial identity involving Harmonic numbers we stumbled on the following problem.
Let $i\ge 0 $ and $k \ge i+2 $ and $p \ge 1$  be integers. Consider a following sum:

\begin{equation}
{\mathfrak S}^k_{i,p} := \sum\limits_{j=2}^{k-i} \binom{k}{i+j} \frac{(-1)^j}{(j-1)^p} \quad (i)
\end{equation}

By using the usual trick $1/j^p = \int\limits_0^\infty \frac{t^{p-1}}{(p-1)!} e^{-j t} dt$ and then by summing up the resulting binomial series and finally by using the Faa di Bruno formula  we have arrived at the following:

\begin{eqnarray}
{\mathfrak S}^k_{i,p} &=& \lim\limits_{\theta \rightarrow 0} \left[
\frac{(-1)^{p-1+i}}{(p-1)!} \frac{d^{p-1}}{d \theta^{p-1}} \left[
B(k+1,\theta-i-1)
\right] - \sum\limits_{j=0}^{i+1} \binom{k}{j} (-1)^{j+i} \frac{1}{(\theta-i+j-1)^p}
\right] \\
&=& \lim\limits_{\theta \rightarrow 0} \left[(-1)^{p-1+i} B(k+1,\theta-i-1)
\sum\limits_{\sum\limits_{\xi=1}^{p-1} \xi m_\xi =p-1}
\frac{\prod\limits_{\xi=1}^{p-1} \left( \psi^{(\xi-1)}(\theta-i-1) - \psi^{(\xi-1)}(\theta-i+k)\right)^{m_\xi}}{\prod\limits_{\xi=1}^{p-1} (\xi!)^{m_\xi} m_\xi!}
- \sum\limits_{j=0}^{i+1} \binom{k}{j} (-1)^{j+i} \frac{1}{(\theta-i+j-1)^p}
\right] \quad(ii)
\end{eqnarray}
In the second line $\psi^{(j)}$ is the polygamma function .

The code below verifies the result:

```<code>{i} = RandomInteger[{1, 5}, 1]; p = RandomInteger[{3, 6}];
l1 = Table[
  Sum[ Binomial[k, j + i] (-1)^j/(j - 1)^p, {j, 2, k - i}], {k, i + 2,
    10}]
N[l1]
th =.;
l2 = Table[
   Normal[Series[(-1)^(p - 1 + i)/(p - 1)! D[
       Beta[k + 1, th - i - 1] -
        Sum[ Binomial[k, j] (-1)^j 1/(th - i + j - 1), {j, 0,
          i + 1}], {th, p - 1}], {th, 0, 0}]], {k, i + 2, 10}] // N
l3 = Table[
   Normal[Series[
     Beta[k + 1, th - i - 1] Total[
        DeleteCases[
         Flatten[Table[
           With[{mm = (p - 1 - Sum[xi m[xi], {xi, 1, p - 2}])/(p -
                 1)}, If[
             Element[mm,
              Integers], (-1)^(p - 1 + i)/
               Product[
                m[xi]!, {xi, 1, p - 1}] Product[((
                 PolyGamma[xi - 1, th - i - 1] -
                  PolyGamma[xi - 1, th - i + k])/xi!)^
                m[xi], {xi, 1, p - 1}] /. m[p - 1] :> mm, {}]],
           Evaluate[
            Sequence @@
             Table[{m[q], 0,
               Floor[(p - 1 - Sum[xi m[xi], {xi, 1, q - 1}])/q]}, {q,
               1, p - 2}]]], p - 3], {}]] -
      Sum[ Binomial[k, j] (-1)^j (-1)^(i)/(th - i + j - 1)^p, {j, 0,
        i + 1}], {th, 0, 0}]], {k, i + 2, 10}];
FullSimplify[l3]
</code>```

In particular we have:
\begin{eqnarray}
{\mathfrak S}^{k}_{i,1} &=& \binom{k}{i+1} \frac{1}{1!} \left( H_{k-i-1} - H_{i+1} \right) - \sum\limits_{j=0}^i \binom{k}{j} \frac{(-1)^{j+i}}{(j-i-1)^1} \\
{\mathfrak S}^{k}_{i,2} &=& \binom{k}{i+1} \frac{1}{2!} \left( (H_{k-i-1}-H_{i+1})^2 + H_{i+1}^{(2)} + H_{k-i-1}^{(2)}  \right) - \sum\limits_{j=0}^i \binom{k}{j} \frac{(-1)^{j+i}}{(j-i-1)^2} \\
{\mathfrak S}^{k}_{i,3} &=& \binom{k}{i+1} \frac{1}{3!} \left(
(H_{k-i-1}-H_{i+1})^3 + 3 (H_{i+1}^{(2)} + H_{k-i-1}^{(2)}) \cdot (H_{k-i-1}-H_{i+1}) + 2 (H_{k-i-1}^{(3)} - H_{i+1}^{(3)})
\right)-  \sum\limits_{j=0}^i \binom{k}{j} \frac{(-1)^{j+i}}{(j-i-1)^3}
\end{eqnarray}

```<code>In[127]:= Clear[H];
H[i_] := HarmonicNumber[i];
H[i_, p_] := HarmonicNumber[i, p];

l1 = Table[
   Sum[ Binomial[k, j + i] (-1)^j/(j - 1)^1, {j, 2, k - i}], {i, 0,
    5}, {k, i + 2, 10}];
l2 = Table[
   Binomial[k,
      i + 1] (HarmonicNumber[k - i - 1] - HarmonicNumber[i + 1]) -
    Sum[ Binomial[k, j] (-1)^(j + i)/(j - i - 1), {j, 0, i}], {i, 0,
    5}, {k, i + 2, 10}];
l1 - l2

l1 = Table[
   Sum[ Binomial[k, j + i] (-1)^j/(j - 1)^2, {j, 2, k - i}], {i, 0,
    5}, {k, i + 2, 10}];
l2 = Table[
   Binomial[k, i + 1] 1/
      2 ((H[1 + i] - H[k - i - 1])^2 + H[1 + i, 2] +
       H[-1 - i + k, 2]) -
    Sum[ Binomial[k, j] (-1)^(j + i)/(j - i - 1)^2, {j, 0, i}], {i, 0,
     5}, {k, i + 2, 10}];
l1 - l2

l1 = Table[
   Sum[ Binomial[k, j + i] (-1)^j/(j - 1)^3, {j, 2, k - i}], {i, 0,
    5}, {k, i + 2, 10}];
l2 = Table[
   Binomial[k, i + 1] 1/
     6 ((-H[1 + i] + H[k - i - 1])^3 +
       3 (H[1 + i, 2] +  H[-1 - i + k, 2]) (-H[1 + i] +
          H[-1 - i + k]) + 2 (-H[1 + i, 3] + H[-1 - i + k, 3])) -
    Sum[ Binomial[k, j] (-1)^(j + i)/(j - i - 1)^3, {j, 0, i}], {i, 0,
     5}, {k, i + 2, 10}];
l1 - l2

Out[132]= {{0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0}, {0,
  0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0}, {0, 0, 0, 0}}

Out[135]= {{0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0}, {0,
  0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0}, {0, 0, 0, 0}}

Out[138]= {{0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0}, {0,
  0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0}, {0, 0, 0,
  0}}
</code>```

Unfortunately the result $(ii)$ is highly singular and it is hard to carry out the limit $\theta \rightarrow 0$ in full generality. My question would therefore be how to perform this limit and obtain a closed form result for the sum in question.

答:
Now let us assume that $p \ge 2$ is an arbitrary integer. We also define coefficients as below:

\begin{equation}
{\mathcal S}^{\pm}_p(i):= \frac{1}{p!} \left. \frac{d^p}{d x^p}  e^{\sum\limits_{q=1}^\infty (\mp 1)^{q-1} \frac{H^{(q)}_i}{q} x^q } \right|_{x=0}
\end{equation}

We compute the first term in the right hand side as follows:

\begin{eqnarray}
&&\frac{(-1)^{p-1+i}}{(p-1)!}  \frac{d^{p-1}}{d \theta^{p-1}} B(k+1,\theta-i-1) =\\
&&\frac{(-1)^{p-1+i}}{(p-1)!}  \frac{d^{p-1}}{d \theta^{p-1}} \left(\frac{1}{\prod\limits_{p=1}^{i+1} (\theta-p)} \cdot \frac{k!}{\theta}  \cdot \frac{1}{ \prod\limits_{p=1}^{k-i-1} (\theta+p)} \right) = \\
&&\frac{(-1)^{p}}{(p-1)!}  \binom{k}{i+1} \frac{d^{p-1}}{d \theta^{p-1}} \left(\frac{1}{\prod\limits_{p=1}^{i+1} (-\frac{\theta}{p}+1)} \cdot \frac{1}{\theta}  \cdot \frac{1}{ \prod\limits_{p=1}^{k-i-1} (\frac{\theta}{p}+1)} \right) = \\
&&\frac{(-1)^{p}}{(p-1)!}  \binom{k}{i+1} \frac{d^{p-1}}{d \theta^{p-1}}
\left(
\frac{1}{\theta} \cdot
\frac{1}{\left[1+ \sum\limits_{q_1=1}^\infty (-\theta)^{q_1} {\mathcal S}^{+}_{q_1}(i+1) \right]} \cdot
\frac{1}{\left[1+ \sum\limits_{q_2=1}^\infty (+\theta)^{q_2} {\mathcal S}^{+}_{q_2}(k-i-1) \right]}
\right) = \\
&&
\frac{(-1)^{p}}{(p-1)!}  \binom{k}{i+1} \frac{d^{p-1}}{d \theta^{p-1}}
\left(
\frac{1}{\theta} \cdot
\sum\limits_{q_1=0}^\infty (+\theta)^{q_1} {\mathcal S}^{-}_{q_1}(i+1) \cdot
\sum\limits_{q_2=0}^\infty (-\theta)^{q_2} {\mathcal S}^{-}_{q_2}(k-i-1)
\right) = \\
&&
\frac{(-1)^{p}}{(p-1)!}  \binom{k}{i+1} \frac{d^{p-1}}{d \theta^{p-1}}
\left(
\frac{1}{\theta} +
(-1)^q \sum\limits_{q=1}^\infty \theta^{q-1}
\sum\limits_{q_1+q_2=q} (-1)^{q_1} {\mathcal S}^{-}_{q_1}(i+1) \cdot  {\mathcal S}^{-}_{q_2}(k-i-1)
\right) = \\
&&
\binom{k}{i+1} \left[
\frac{(-1)^1}{\theta^p} +
\left(\sum\limits_{q_1+q_2=p} (-1)^{q_1} {\mathcal S}^{-}_{q_1}(i+1) \cdot  {\mathcal S}^{-}_{q_2}(k-i-1)
\right) + O\left(\theta^{p+1} \right)
\right]
\end{eqnarray}

Now we can see that the first term in the square brackets cancels the last term , i.e. the $j=i+1$ term in the last sum in $(ii)$ . The singularities cancel out nicely and now it is straightforward to take the limit . So the result reads:

\begin{eqnarray}
{\mathfrak S}^k_{i,p} &=& \binom{k}{i+1} \left[\sum\limits_{q_1=0}^p (-1)^{q_1}{\mathcal S}^{-}_{q_1}(i+1) \cdot  {\mathcal S}^{-}_{p-q_1}(k-i-1)\right]
-
\sum\limits_{j=0}^i \binom{k}{j} \frac{(-1)^{j+i}}{(j-i-1)^p} \\
&=& \binom{k}{i+1} \cdot \left(
\sum\limits_{\sum\limits_{j=1}^p j \cdot m_j = p}
\frac{1}{\prod\limits_{j=1}^p m_j!} \prod\limits_{j=1}^p \left[ \frac{(-1)^j H^{(j)}_{i+1} + H^{(j)}_{k-i-1}}{j} \right]^{m_j}
\right)
-
\sum\limits_{j=0}^i \binom{k}{j} \frac{(-1)^{j+i}}{(j-i-1)^p}
\end{eqnarray}

```<code>In[772]:= p = RandomInteger[{2, 15}]; Clear[H]; k =.; i =.; M = 10;
l1 = Table[
   Sum[ Binomial[k, j + i] (-1)^j/(j - 1)^p, {j, 2, k - i}], {i, 0,
    M}, {k, i + 2, 2 M}];
ff[k_, i_] =
  Binomial[k,
     i + 1] (Sum[ (-1)^p1/p1! 1/(p - p1)! D[
         Exp[ Sum[ H[i + 1, q]/q x^q, {q, 1, M}]], {x, p1}] D[
         Exp[ Sum[ H[k - i - 1, q]/q x^q, {q, 1, M}]], {x,
          p - p1}], {p1, 0, p}] /. x :> 0) -
   Sum[ Binomial[k, j] (-1)^(j + i)/(j - i - 1)^p, {j, 0, i}];
H[i_] := HarmonicNumber[i];
H[i_, p_] := HarmonicNumber[i, p];
l2 = Table[ff[k, i], {i, 0, M}, {k, i + 2, 2 M}];
l1 - l2

Out[778]= {{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0,
  0, 0, 0, 0, 0, 0, 0, 0}}
</code>```

Update: Here we also partially evaluate the remaining sum.
Take $i \ge 0 $ and $\xi =0,\cdots, i$ be integers then let us firstly define some coefficients:

\begin{eqnarray}
{\mathcal A}^{(\xi)}_{i,p} &:=& \sum\limits_{j=0}^\xi \binom{i+1}{\xi-j}  \frac{1}{(\xi-j-i-1)^{p-1}} \binom{-i+\xi-1}{j} \\
&=& \left\{
\begin{array}{rrr}
1 & \quad \mbox{if $p=1$}\\
H_{i-\xi}- H_{i+1} & \quad \mbox{if $p=2$} \\
\frac{1}{2} \left( (H_{i+1}^{(2)} - H_{i-\xi}^{(2)}) + (H_{i+1} - H_{i-\xi})^2 \right) & \quad \mbox{if $p=3$} \\
\vdots
\end{array}
\right.
\end{eqnarray}

now our remaining sum reads:

\begin{eqnarray}
\sum\limits_{j=0}^i \binom{k}{j} \frac{(-1)^{j+i}}{(j-i-1)^p} &=&
-\binom{k}{i+1} \cdot \sum\limits_{\xi=0}^i \frac{1}{(k-\xi)}  \cdot {\mathcal A}^{(\xi)}_{i,p} \quad (i)
\end{eqnarray}

```<code>In[858]:= p = RandomInteger[{1, 5}]; k =.;
H[i_, p_] := HarmonicNumber[i, p]; M = 10;
AA[xi_, i_, p_] :=
  Which[p == 1, 1, p == 2,
   HarmonicNumber[i - xi] - HarmonicNumber[1 + i], p == 3,
   1/2 (H[1 + i, 2] -
      H[i - xi,
       2] + (HarmonicNumber[1 + i] - HarmonicNumber[i - xi])^2), True,
    Sum[ Binomial[i + 1,
      xi - j] 1/(xi - j - i - 1)^(p - 1) Binomial[-i + xi - 1, j], {j,
      0, xi}]];
l1 = Table[
   Sum[Binomial[k, j] (-1)^(j + i)/(j - i - 1)^p, {j, 0, i}], {p, 1,
    5}, {i, 0, M}];
l2 = Table[-Binomial[k, i + 1] Sum[
     1/(k - xi) AA[xi, i, p] , {xi, 0, i}], {p, 1, 5}, {i, 0, M}];
FullSimplify[l1 - l2]

Out[862]= {{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0,
   0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0,
  0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}}
</code>```

Equation $(i)$ comes from factoring out the the binomial factor  and then decomposing the rest into partial fractions in the variable $k$ and then all the similar terms together and simplifying.

Body:
Now let us assume that $p \ge 2$ is an arbitrary integer. We also define coefficients as below:

\begin{equation}
{\mathcal S}^{\pm}_p(i):= \frac{1}{p!} \left. \frac{d^p}{d x^p}  e^{\sum\limits_{q=1}^\infty (\mp 1)^{q-1} \frac{H^{(q)}_i}{q} x^q } \right|_{x=0}
\end{equation}

We compute the first term in the right hand side as follows:

\begin{eqnarray}
&&\frac{(-1)^{p-1+i}}{(p-1)!}  \frac{d^{p-1}}{d \theta^{p-1}} B(k+1,\theta-i-1) =\\
&&\frac{(-1)^{p-1+i}}{(p-1)!}  \frac{d^{p-1}}{d \theta^{p-1}} \left(\frac{1}{\prod\limits_{p=1}^{i+1} (\theta-p)} \cdot \frac{k!}{\theta}  \cdot \frac{1}{ \prod\limits_{p=1}^{k-i-1} (\theta+p)} \right) = \\
&&\frac{(-1)^{p}}{(p-1)!}  \binom{k}{i+1} \frac{d^{p-1}}{d \theta^{p-1}} \left(\frac{1}{\prod\limits_{p=1}^{i+1} (-\frac{\theta}{p}+1)} \cdot \frac{1}{\theta}  \cdot \frac{1}{ \prod\limits_{p=1}^{k-i-1} (\frac{\theta}{p}+1)} \right) = \\
&&\frac{(-1)^{p}}{(p-1)!}  \binom{k}{i+1} \frac{d^{p-1}}{d \theta^{p-1}}
\left(
\frac{1}{\theta} \cdot
\frac{1}{\left[1+ \sum\limits_{q_1=1}^\infty (-\theta)^{q_1} {\mathcal S}^{+}_{q_1}(i+1) \right]} \cdot
\frac{1}{\left[1+ \sum\limits_{q_2=1}^\infty (+\theta)^{q_2} {\mathcal S}^{+}_{q_2}(k-i-1) \right]}
\right) = \\
&&
\frac{(-1)^{p}}{(p-1)!}  \binom{k}{i+1} \frac{d^{p-1}}{d \theta^{p-1}}
\left(
\frac{1}{\theta} \cdot
\sum\limits_{q_1=0}^\infty (+\theta)^{q_1} {\mathcal S}^{-}_{q_1}(i+1) \cdot
\sum\limits_{q_2=0}^\infty (-\theta)^{q_2} {\mathcal S}^{-}_{q_2}(k-i-1)
\right) = \\
&&
\frac{(-1)^{p}}{(p-1)!}  \binom{k}{i+1} \frac{d^{p-1}}{d \theta^{p-1}}
\left(
\frac{1}{\theta} +
(-1)^q \sum\limits_{q=1}^\infty \theta^{q-1}
\sum\limits_{q_1+q_2=q} (-1)^{q_1} {\mathcal S}^{-}_{q_1}(i+1) \cdot  {\mathcal S}^{-}_{q_2}(k-i-1)
\right) = \\
&&
\binom{k}{i+1} \left[
\frac{(-1)^1}{\theta^p} +
\left(\sum\limits_{q_1+q_2=p} (-1)^{q_1} {\mathcal S}^{-}_{q_1}(i+1) \cdot  {\mathcal S}^{-}_{q_2}(k-i-1)
\right) + O\left(\theta^{p+1} \right)
\right]
\end{eqnarray}

Now we can see that the first term in the square brackets cancels the last term , i.e. the $j=i+1$ term in the last sum in $(ii)$ . The singularities cancel out nicely and now it is straightforward to take the limit . So the result reads:

\begin{eqnarray}
{\mathfrak S}^k_{i,p} &=& \binom{k}{i+1} \left[\sum\limits_{q_1=0}^p (-1)^{q_1}{\mathcal S}^{-}_{q_1}(i+1) \cdot  {\mathcal S}^{-}_{p-q_1}(k-i-1)\right]
-
\sum\limits_{j=0}^i \binom{k}{j} \frac{(-1)^{j+i}}{(j-i-1)^p} \\
&=& \binom{k}{i+1} \cdot \left(
\sum\limits_{\sum\limits_{j=1}^p j \cdot m_j = p}
\frac{1}{\prod\limits_{j=1}^p m_j!} \prod\limits_{j=1}^p \left[ \frac{(-1)^j H^{(j)}_{i+1} + H^{(j)}_{k-i-1}}{j} \right]^{m_j}
\right)
-
\sum\limits_{j=0}^i \binom{k}{j} \frac{(-1)^{j+i}}{(j-i-1)^p}
\end{eqnarray}

In[772]:= p = RandomInteger[{2, 15}]; Clear[H]; k =.; i =.; M = 10;
l1 = Table[
   Sum[ Binomial[k, j + i] (-1)^j/(j - 1)^p, {j, 2, k - i}], {i, 0,
    M}, {k, i + 2, 2 M}];
ff[k_, i_] =
  Binomial[k,
     i + 1] (Sum[ (-1)^p1/p1! 1/(p - p1)! D[
         Exp[ Sum[ H[i + 1, q]/q x^q, {q, 1, M}]], {x, p1}] D[
         Exp[ Sum[ H[k - i - 1, q]/q x^q, {q, 1, M}]], {x,
          p - p1}], {p1, 0, p}] /. x :> 0) -
   Sum[ Binomial[k, j] (-1)^(j + i)/(j - i - 1)^p, {j, 0, i}];
H[i_] := HarmonicNumber[i];
H[i_, p_] := HarmonicNumber[i, p];
l2 = Table[ff[k, i], {i, 0, M}, {k, i + 2, 2 M}];
l1 - l2

Out[778]= {{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0,
  0, 0, 0, 0, 0, 0, 0, 0}}

Update: Here we also partially evaluate the remaining sum.
Take $i \ge 0 $ and $\xi =0,\cdots, i$ be integers then let us firstly define some coefficients:

\begin{eqnarray}
{\mathcal A}^{(\xi)}_{i,p} &:=& \sum\limits_{j=0}^\xi \binom{i+1}{\xi-j}  \frac{1}{(\xi-j-i-1)^{p-1}} \binom{-i+\xi-1}{j} \\
&=& \left\{
\begin{array}{rrr}
1 & \quad \mbox{if $p=1$}\\
H_{i-\xi}- H_{i+1} & \quad \mbox{if $p=2$} \\
\frac{1}{2} \left( (H_{i+1}^{(2)} - H_{i-\xi}^{(2)}) + (H_{i+1} - H_{i-\xi})^2 \right) & \quad \mbox{if $p=3$} \\
\vdots
\end{array}
\right.
\end{eqnarray}

now our remaining sum reads:

\begin{eqnarray}
\sum\limits_{j=0}^i \binom{k}{j} \frac{(-1)^{j+i}}{(j-i-1)^p} &=&
-\binom{k}{i+1} \cdot \sum\limits_{\xi=0}^i \frac{1}{(k-\xi)}  \cdot {\mathcal A}^{(\xi)}_{i,p} \quad (i)
\end{eqnarray}

In[858]:= p = RandomInteger[{1, 5}]; k =.;
H[i_, p_] := HarmonicNumber[i, p]; M = 10;
AA[xi_, i_, p_] :=
  Which[p == 1, 1, p == 2,
   HarmonicNumber[i - xi] - HarmonicNumber[1 + i], p == 3,
   1/2 (H[1 + i, 2] -
      H[i - xi,
       2] + (HarmonicNumber[1 + i] - HarmonicNumber[i - xi])^2), True,
    Sum[ Binomial[i + 1,
      xi - j] 1/(xi - j - i - 1)^(p - 1) Binomial[-i + xi - 1, j], {j,
      0, xi}]];
l1 = Table[
   Sum[Binomial[k, j] (-1)^(j + i)/(j - i - 1)^p, {j, 0, i}], {p, 1,
    5}, {i, 0, M}];
l2 = Table[-Binomial[k, i + 1] Sum[
     1/(k - xi) AA[xi, i, p] , {xi, 0, i}], {p, 1, 5}, {i, 0, M}];
FullSimplify[l1 - l2]

Out[862]= {{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0,
   0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0,
  0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}}

Equation $(i)$ comes from factoring out the the binomial factor  and then decomposing the rest into partial fractions in the variable $k$ and then all the similar terms together and simplifying.


问:
Approximate solution to the volume of an elliptic cylinder with unlike ellipses integration
I am attempting to solve for the volume of an elliptic cylinder with two different size ellipses. This is not the same as an elliptic truncated cone volume. The ellipses are two completely different sizes with different ratios (a1/b1 not equal to a2/b2)

If we assume the volume can be approximated as a sum of discs we can use the equation v ≈ Σ πab * Δh

From my attached work I end up with the integration constant C. Is there anything that can be done to approximate this solution? I need to converge on an answer for a practical problem.

Any help would be appreciated.

答:
Leaving $\pi$ on the side,

$$\int_0^H(a+\frac hH\Delta a)(b+\frac hH\Delta b)\,dh=abH+(a\Delta b+b\Delta a)\frac H2+\Delta a\Delta b\frac H3.$$


问:
Calculating error for $\pi$ in an expansion
I have the following equation: $\quad\pi = 4\arctan(1/2) + 4\arctan(1/3).$

I want to calculate how many terms of the expansion ($n$) I have to calculate in order for the error to be less than  $10^{-10}$.

I have an idea for the method to be used:

Taking the power series of $\arctan(x)$ and using the remainder integral of the form $$R_n \;=\; \int\frac{{(-1)^n t^{2n}}}{1+t^2}\,dt.$$

So I want the sum of the two remainders from the expansion of each $\arctan(x)$ to be $<$ $\frac{10^{-10}}{4}$.

$$\int_0^{1/2}\frac{{(-1)^n t^{2n}}}{1+t^2}\,dt \;+\; \int_0^{1/3}\frac{{(-1)^n t^{2n}}}{1+t^2}\,dt\;<\; \frac{10^{-10}}{4}.$$

Now to solve this, I can't do without making approximations since $t^{2n} > \frac{{(-1)^n t^{2n}}}{1+t^2}$ on my intervals; I approximate using this, getting $$\int_0^{1/2}t^{2n}\,dt \;+\; \int_0^{1/3}t^{2n}\,dt.$$

Evaluating, I get $$\frac{(1/2)^{2n+1}}{2n+1} + \frac{(1/3)^{2n+1}}{2n+1} < \frac{10^{-10}}{4}.$$

I have no idea how to solve this. Would I make another approximation? Or is my method completely wrong to start with.

Edit:
After making some huge approximations i.e. getting rid of the $2n+1$'s on the denominator and setting $1/3$ to be $1/2$, I got that $2n+1 =36.$ However, I am not sure whether this is correct.

答:
As the Taylor expansion of $\arctan=\sum _{n=1}^{\infty } \frac{(-1)^{n+1} x^{2 n-1}}{2 n-1}$ is an alternating series, the error is less than the abs value of the first term neglected, that is equivalent to find an $n$ such that
$$\frac{4 \left(\frac{1}{2}\right)^{2 n-1}}{2 n-1}+\frac{4 \left(\frac{1}{3}\right)^{2 n-1}}{2 n-1}=\frac{1}{10^{10}}$$
Using Mathematica I got an approximated value of $n=16$. Indeed
$$4 \sum _{n=1}^{16} \frac{(-1)^{n+1} \left(\frac{1}{2}\right)^{2 n-1}}{2 n-1}+4 \sum _{n=1}^{16} \frac{(-1)^{n+1} \left(\frac{1}{3}\right)^{2 n-1}}{2 n-1}\approx 3.14159 265357$$
While $\pi\approx 3.14159265358$.

Error is $1.14\times10^{-11}<10^{-10}$


答:
We suppose that

the ellipse axes are aligned with each other

the axis lengths are linearly interpolated between the two ends

Then the solid's volume is
$$\pi h\int_0^1((1-t)a_1+ta_2)((1-t)b_1+tb_2)\,dt=\frac{\pi h(2a_1b_1+a_1b_2+a_2b_1+2a_2b_2)}6$$

Title: Approximate solution to the volume of an elliptic cylinder with unlike ellipses integration

Body:
I am attempting to solve for the volume of an elliptic cylinder with two different size ellipses. This is not the same as an elliptic truncated cone volume. The ellipses are two completely different sizes with different ratios (a1/b1 not equal to a2/b2)

If we assume the volume can be approximated as a sum of discs we can use the equation v ≈ Σ πab * Δh

From my attached work I end up with the integration constant C. Is there anything that can be done to approximate this solution? I need to converge on an answer for a practical problem.

Any help would be appreciated.


问:
Toss 100 fair coins and take away the tails; toss the remaining coins and take away the tails. Continue until no coins remain.
100 participants have a fair coin each, on a given round, the not already discarded participants flip their coins, those who flip a tail are discarded from the game, the remaining ones continue to play until nobody is left (everyone has been discarded).

What would be the average number of trials (where each trial consists of a tossing and removing the tails) one would expect from doing this experiment?

Does conditional expectation work for something like this?

I know that each individual coin follows a Geometric distribution, but I am trying to figure out the sum of them to determine the average number of trials for a game like this.

My Logic/Thought Process: I started out trying to think of the probability that a particular coin makes it to round $r$ which is $\frac{1}{2^m}$. I then realized that each coin outcome can be modeled by a Geometric random variables with $p = 0.5$. I am just now unsure how to take the leap from this single case to a case with 100 coins. I presume it has to do with summing the geometric random variables, but I am not sure.

答:
I doubt that there's a simple expression for the expectation.  Let $E_n$ be the expected number of trials when $n$ coins remain, so that we are asked to compute $E_{100}$.  We know that $E_0=0$ and that $E_1=2$.  Now $$E_2=1+\frac14E_2+\frac12E_1+\frac14E_0$$ because we have to make one trial, and with probability $\frac14$ we throw two heads and still have two coins, with probability $\frac12$ we throw a head and a tail, and with probability $\frac14$, we throw two tails, and the experiment ends.  This gives $E_2=\frac83$.

We can continue in this manner: $$E_3=1+\frac18E_3+\frac38E_2+\frac38E_1+\frac18E_0$$ which gives $E_3=\frac{22}7$ if I'm not mistaken.

One could easily write a computer program to work back to $E_{100}$, but it would be easier to proceed by simulation.

EDIT

I wrote the script I suggested.  The exact value if a fraction whose numerator has $894$ decimal digits and whose denominator has $893$.  The approximate value is $7.98380153515692$.


问:
Finding a closed form, if it exists, for an integral
I encountered the following integral in statistical work and wondered if it was possible to find a closed form for it:
$$
I(r,a,b)=\int_0^\infty \frac{(1-z^2)^r}{(1+z^2)^r(1+az)^b} \mathrm{d}z
$$
where $r$ is an integer, $a > 0$ and $b > 1$.

I'm still unsure about whether a closed form does indeed exist. The closest thing to this integral I can find online is 3.253 of http://fisica.ciens.ucv.ve/~svincenz/TISPISGIMR.pdf.

Any help on this would be greatly appreciated, and if a closed form is not possible, what would the integral look like in terms of integrals such as the hypergeometric function or as an infinite summation in its simplest form? I tried, unsuccessfully, to use Wolfram's Computational Intelligence to check the value of the integral, but this requires purchasing 'pro' computational time.

答:
Maple does this in terms of the Meijer G function.  I general, probably nothing simpler can be expected.

If $b$ is rational, then Maple seems to find something where there is a sum over the roots of a certain polynomial... Example

$$
\int_0^\infty \frac{(1-z^2)\;dz}{(1+z^2)(1+az)^{7/4}} =
\frac{4a^2-4}{3a(a^2+1)} +\frac{a}{a^2+1} \sum \frac{(\rho^4-2)\log(1-\rho)}{\rho^3(\rho^4-1)}
$$
where the $\Sigma$ is over all eight roots $\rho$ of the polynomial
$Z^8-2Z^4+(1+a^2)$.


问:
Series convergence of function
Let $f:[0,\frac{\pi}{2}]:\to\mathbb{R}$ be defined as:
$$f(x):=\sum_{n=1}^{\infty}\frac{1}{2^n}\sin\left(\frac{x}{n^2}\right)$$.
How do I go about showing $f$ converges? And how o show it is increasing $(0,\pi/2)$?

答:
You may probably know that $\sum^\infty_{n=1} \frac{1}{2^n}$ converges.  Now you use the fact, that $\sin(a) \leq 1$ for all $a$. This gives you a dominant convergent series, namely $\sum^\infty_{n=1} \frac{1}{2^n}$.

Also for $x < y \in (0,\frac{\pi}{2})$ you can show (or you just know), that for all $n$ as defined here $\sin(\frac{x}{n^2}) < \sin(\frac{y}{n^2})$. From this by pairwise comparison it follows, that the series $f(x) < f(y)$. If you have difficulties to understand this you might want to consider what happens when $x=0$ and $y = \frac{\pi}{2}$, and then ask yourself how the values change inside of the interval. Another point of view is just to see, that sine is monotone on the relevant interval.

Title: Series convergence of function


问:
NP-completeness of undirected planar graph problem
I want to know whether a certain graph problem is NP-complete or not. The problem is as follows.

Given an undirected planar graph with in every vertex a number. Can you give every edge a direction such that the outdegree of every vertex is equal to the corresponding number?

Does anyone have an idea whether this problem is NP-complete, or know a related problem?

答:
It can be solved in polynomial time by reducing it to maximum flow problem, so it's unlikely to be NP-complete.

It's clear that sum of all numbers should be equal to number of edges. It it's so, the reduction works as follow.

For each vertex $v$ and edge $e$ from original graph we will have vertex in network. Also add source $s$ and sink $t$.

Edges in network are as follow:

for vertex $v$ with number $n$ add edge $s \to v$ with capacity $n$

for edge $e = (v_1, v_2)$ add edges $v_1 \to e$, $v_2 \to e$, $e \to t$ with capacities $1$

Now, if we have integer flow with value equal to number of edges (so that there is flow from every edge), we can assign directions: for edge $e = (v_1, v_2)$ there is flow of value $1$ to $e$ from either $v_1$ or $v_2$ - direct edge from $v_1$ to $v_2$ in the former case, and from $v_2$ to $v_1$ in the later. This way every edge is directed, and each vertex has number of outgoing edges equal to it's number.

Similarly, if we have a direction of edges, we can create flow with value equal to number of edges.

Thus, original graph can be directed iff there is a flow with value equal to number of edges. As max flow problem can be solved in polynomial time, so can be original problem.


问:
Computing $\frac{1}{\pi}\int\int_\mathbb{D}|\phi_\alpha '|^2 dxdy$

If $\phi_\alpha(z)=(\alpha-z)/(1-\bar\alpha z)$ for $|\alpha|<1$,
prove that

$\frac{1}{\pi}\int\int_\mathbb{D}|\phi_\alpha '|^2 dxdy=1$ and
$\frac{1}{\pi}\int\int_\mathbb{D}|\phi'_\alpha|dxdy=\frac{1-|\alpha|^2}{|\alpha|^2}log\frac{1}{1-|\alpha|^2}$ where in the case $\alpha=0$ the expression on the right is understood
as the limit as $|\alpha|\to 0$.

[Hint: The first integral can be evaluated without a calculation. For
the second, use polar coordinates, and for each fixed $r$ use contour
integration to evaluate the integral in $\theta$]

I did the first question, but I have trouble doing the second question. I don't know how to use polar coordinates method here.

So I computed $\phi'_\alpha(z)=\frac{-1+\alpha\bar\alpha}{(1-\alpha z)^2}$ and by using polar coordinate method I know, I get $\frac{1}{\pi}\int\int_\mathbb{D}|\phi'_\alpha|dxdy=\frac{1}{\pi}\int^{2\pi}_0\int^1_0|\frac{-1+\alpha\bar\alpha}{(1-\bar\alpha re^{i\theta)^2}}|rdrd\theta$, but it looks wrong. Can you give me any idea? Thank you.

答:
$\frac{1}{\pi}\int\int_\mathbb{D}|\phi'_\alpha|dxdy=\frac{1}{\pi}\int^{2\pi}_0\int^1_0\frac{|-1+\alpha\bar\alpha|}{|1-\bar\alpha re^{i\theta)^2}|}rdrd\theta=(1-|\alpha|^2)\int^1_0r(\frac{1}{\pi}\int^{2\pi}_0\frac{1}{|1-\bar\alpha re^{i\theta)^2}|}d\theta)dr$

Clearly the integral is invariant under a circle rotation $\alpha \to \alpha e^{it_0}$ so we can assume $\alpha=c=|\alpha|>0$

$\frac{1}{|1-cre^{i\theta}|^2}=\frac{P(cr,\theta)}{1-c^2r^2}$ so $\frac{1}{\pi}\int^{2\pi}_0\frac{1}{|1-\bar\alpha re^{i\theta)^2}|}d\theta=\frac{2}{1-c^2r^2}=\frac{2}{1-|\alpha|^2r^2}$ as the integral of the Poisson kernel (divided by $2\pi$) is $1$ hence we get:

$\frac{1}{\pi}\int\int_\mathbb{D}|\phi'_\alpha|dxdy=2(1-|\alpha|^2)\int_0^1\frac{r}{1-|\alpha|^2r^2}dr=\frac{1-|\alpha|^2}{|\alpha|^2}\log\frac{1}{1-|\alpha|^2}$ so we are done!


问:
Shouldn't the graph of $y=1$ be the same as $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ because $1=\frac{\sqrt{4-x}}{\sqrt{4-x}}$?
Shouldn't the graph of $y=1$ be the same as $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ because $1=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ ?

So why is it that my calculator said otherwise? And I know that you can't square root a negative number, because it will gave an imaginary number. Is it because the calculator assume that if it there is an undefined value in the equation, the final result will be undefined too? Or is there a proof that I aren't aware of?

Unnecessary comment below:

And if graph of $y=1$ aren't the same as $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ just like the calculator said. That would be great, because you don't have to clarify the domain outside the expression. You could just do $y=2x\cdot\frac{\sqrt{4-x}}{\sqrt{4-x}}$ instead of using a set notation such as $\{x\in| \mathbb{R},x>4\}$

答:
If you define $y=1$ on all real numbers, including $4$, they cannot be the same as you cannot define the other function on the same domain, and for two functions to be the same (formally) their domain has to be the same.

On $\mathbb{C}$ the same holds, as for $x = 4$, the second function is not defined.

If you write these functions down formally you would have to say what there domain of definition is. If you define the first function $y=1$ on the same domain as the second they are the same. So it all depends on your choice of domain, which shows how important it is to keep in mind, that when defining a function you always have to state from where to where it maps, otherwise the definition is uncomplete.
I hope this also clarifies the question in your comment.


答:
This is essentially equivalent to computing the expected value of the maximum of $n=100$ iid geometric random variables, for $p=\frac12$

(BTW: The linked question includes the recursion given by @saulspatz's answer)

There is no closed form solution, but this approximation for large $n$ (with bounds) is given:

$$E_n \approx \frac{1}{2} + \frac{1}{\lambda} H_n$$

where $\lambda = - \log(1-p)=0.69314718\cdots$ and $H_n$ is the harmonic number.

For example, for $n=3$ this gives $E_3 \approx 3.14494$ , very near the exact $E_3=22/7=3.14285$

For $n=100$ this gives $E_{100} \approx 7.98380382$.

More in "Yet another application of a binomial recurrence order statistics",
W. Szpankowski; V. Rego, Computing, 1990, 43, 4, 401-410.

Title: Toss 100 fair coins and take away the tails; toss the remaining coins and take away the tails. Continue until no coins remain.

Body:
100 participants have a fair coin each, on a given round, the not already discarded participants flip their coins, those who flip a tail are discarded from the game, the remaining ones continue to play until nobody is left (everyone has been discarded).

What would be the average number of trials (where each trial consists of a tossing and removing the tails) one would expect from doing this experiment?

Does conditional expectation work for something like this?

I know that each individual coin follows a Geometric distribution, but I am trying to figure out the sum of them to determine the average number of trials for a game like this.

My Logic/Thought Process: I started out trying to think of the probability that a particular coin makes it to round $r$ which is $\frac{1}{2^m}$. I then realized that each coin outcome can be modeled by a Geometric random variables with $p = 0.5$. I am just now unsure how to take the leap from this single case to a case with 100 coins. I presume it has to do with summing the geometric random variables, but I am not sure.


问:
Flowing a vector along a vector field $X$ using the pushforward of the flow of $X$
On the three-sphere $S^3$, I'm given three vector fields $X$, $Y$ and $Z$, such that at each point $p\in S^3$, the tangent vectors $X_p$, $Y_p$ and $Z_p$ form an orthogonal basis of the tangent space $T_pS^3$.

I denote by $X^t:S^3\rightarrow S^3$ the flow of the vector field $X$, so $X^0(p)=p$ and $\frac{d}{dt}X^t(p)=X_{X^t(p)}$.

Its pushforward at the point $p$ is denoted $(X^t)_{*,p}$, so this is a map from $T_pS^3$ to $T_{X^t(p)}S^3$.

Now let $v$ be an arbitrary tangent vector in $T_pS^3$. We can write $v=v_xX_p+v_yY_p+v_zZ_p$. I want to prove the following:

$(X^t)_{*,p}(v)=v_x X_{X^t(p)}+...$.

Here the dots denote other terms in the span of $Y_{X^t(p)}$ and $Z_{X^t(p)}$, but it's only the $X_{X^t(p)}$ component I care about (plus I don't know the other components).

Is this maybe immediate from the definition of the flow? I tried using the defining equation $\frac{d}{dt}X^t(p)=X_{X^t(p)}$ of the flow of $X$, but I don't know how I can relate the time derivative with the pushforward...

Edit: so I think I can visualize my question as follows. Starting with a tangent vector $v$ at $p$, flowing this vector along $X$ (using the pushforward of the flow) does not change the component of $v$ along $X$. Sounds plausible...

答:
It may be easier to see at first with derivations. The flow is a smooth  $\theta:\mathbb R\times S^3\to S^3$ that satisfies $\theta(0,p)=p$ and $X_{\theta(0,p)}=\frac{d\theta(t,p)}{dt}|_{t=0}.\ $ Fixing $p\in S^3$ and $t_0\in \mathbb R$,  we flow from time $t=0$ to time $t=t_0$. This means that we move from $p$ to $\theta(t_0,p).$ We want to prove that $X_{\theta(t_0,p)}=\frac{d\theta(t,p)}{dt}|_{t=t_0}.\ $

There is an induced map between $T_pS^3$ and $T_{\theta(t_0,p)}S^3$ that takes the derivation $X_p$ to a derivation $X_{\theta(t_0,p)}$ given by $X_{\theta(t_0,p)}(f)=X_p(f\circ \theta(t_0,-)).$ This is the pushforward.

Now, $\theta(-,\theta(t_0,p))$ is a curve from some interval $(-\epsilon,\epsilon)$ into $S^3$ such that $\theta(0,\theta(t_0,p))=\theta(t_0,p)$  so $X_{\theta(t_0,p)}(f)=\frac{d f\circ \theta(t,\theta(t_0,p))}{dt}|_{t=0}.$

But since $\theta$ is a flow, $\theta(t,\theta(t_0,p))=\theta(t+t_0, p)$ so in fact $X_{\theta(t_0,p)}(f)=\frac{d f\circ \theta(t+t_0, p)}{dt}|_{t=0}=\frac{d f\circ \theta(t, p)}{dt}|_{t=t_0}.$

We conclude that $X_{\theta(t_0, p)}=\frac{d\theta(t, p)}{dt}|_{t=t_0}$ as expected.


问:
Regularization Of Solutions From Transport-Diffusion Equation
Let $\epsilon > 0$ be Real. Let $u \in L^\infty_{loc}(\mathbb{R}_+,L^2(\mathbb{R^2})^2) \cap L^2_{loc}(\mathbb{R}_+, \dot{H}^1(\mathbb{R^2})^2)$ be a divergence free vector field, and $w_0 \in L^2(\mathbb{R^2})$. I have successfully proved the existence of a solution $w \in L^\infty(\mathbb{R}_+, L^2(\mathbb{R^2}) \cap L^2(\mathbb{R}_+, \dot{H}^1(\mathbb{R^2}) \cap C_w(\mathbb{R}_+, L^2(\mathbb{R^2}) $ of the following PDE :

$$\partial_tw + (u\cdot\nabla) w - \epsilon\Delta w = 0 \ \text{in} \ ]0,+\infty[ \times \mathbb{R^2}\\ w_{|t=0} = w_0 \in L^2(\mathbb{R^2})
$$

Where we define $w \in C_w(\mathbb{R}_+, L^2(\mathbb{R^2}) $ if and only if $w \in L^\infty(\mathbb{R}_+, L^2(\mathbb{R^2}) $ and for any $\phi \in C_c^\infty(\mathbb{R}_+ \times L^2(\mathbb{R^2})) $ the map :
$$ t \mapsto \int_\mathbb{R^2} w(t,x)\phi(t,x)dx $$
is continuous. So $C_w$ stands for weak continuity. Also, $\dot{H}$ denotes the homogeneous Sobolev space and $u \cdot \nabla$ is the following differential operator : $$ (u \cdot \nabla) w = \sum_{i=1}^2 u_i \partial_i w$$

As $u$ is fixed, this term is linear.

I am trying to prove uniqueness of the solutions. In order to do that, as the PDE is linear, I am taking the weak formulation of the PDE, for $w_0 = 0$ wich is, for any $T > 0$ :
$$\int_\mathbb{R^2} w(T,x)\phi(T,x) = \int_0^T \int_\mathbb{R^2} w(t,x)[\partial_t\phi + u \cdot\nabla\phi + \epsilon\Delta\phi](t,x) dxdt$$

I'm trying to prove that for any $T>0$ :
$$ \int_\mathbb{R^2} w(T,x)\psi(x)dx = 0$$
for any function $\psi \in C_c^\infty(\mathbb{R^2})$.

So I'm looking for a solution that I know exists of the following PDE :
$$  \partial_t\phi + u \cdot\nabla\phi + \epsilon\Delta\phi = 0 \\ \phi(T,x) = \psi(x) \in C_c^\infty(\mathbb{R^2}) \subset L^2(\mathbb{R^2})
$$

If the solution $\phi$ was regular enough and compact supported that would prove my point. Unfortunatly $\phi$ does not have the regularity required to be taken as a test function in the weak formulation.

In order to fix this problem, I can regularize $\phi$ by convolution. and we then denote :

$$\phi_\delta = \phi \ast \rho_\delta$$

We obtain :

$$  \partial_t\phi_\delta + (u \cdot\nabla\phi) \ast \rho_\delta + \epsilon\Delta\phi_\delta = 0 \\
$$

My question is :

How do we modify the term $ (u \cdot\nabla\phi) \ast \rho_\delta$, in order the get a similar PDE for $\phi_\delta$  ?

Thank you very much.

答:
I have found the solution :

We have : $\partial_t\phi_\delta + (u \cdot\nabla\phi) \ast \rho_\delta + \epsilon\Delta\phi_\delta = 0 $

So : $$\partial_t\phi_\delta + u \cdot\nabla\phi_\delta+ \epsilon\Delta\phi_\delta = r_\delta
$$
where $r_\delta = u \cdot\nabla\phi_\delta - (u \cdot\nabla\phi) \ast \rho_\delta$
One may show the following statement (wich is a variation of the Di Perna-Lions Lemma shown in the proof of uniqueness in the Di Perna-Lions Theorem) :
$$ r_\delta \rightarrow 0$$ in $L^1([0,T] \times \mathbb{R^2})$ and $L^2([0,T], H^{-1}(\mathbb{R^2}))$

Let $\chi \in C_c^1\mathbb(R^2)$ be a cut-off function wich value is $1$ in $B(0,1)$. We define $\chi_R= \chi(\frac{\cdot}{R})$. We take $\Psi = \phi_\delta\chi_R \in C_c^1(\mathbb{R^2})$ as a test function in the weak formulation.

We have :
\begin{align}
\int_\mathbb{R^2} w(T,x)(\psi \ast \phi_\delta)\chi_Rdx = \int_0^T \int_\mathbb{R^2} w(t,x)[\partial_t(\phi_\delta\chi_R) + u \cdot\nabla(\phi_\delta\chi_R) + \epsilon\Delta(\phi_\delta\chi_R)](t,x) dxdt
= \int_0^T \int_\mathbb{R^2} w(t,x)[\partial_t\phi_\delta + u \cdot\nabla\phi_\delta + \epsilon\Delta\phi_\delta]\chi_R(t,x) dxdt
 + \int_0^T \int_\mathbb{R^2} \phi_\delta \partial_t \chi_R + (ect...)
= \int_0^T \int_\mathbb{R^2} w(t,x)r_\delta\chi_R +
\int_0^T \int_\mathbb{R^2} \phi_\delta \partial_t \chi_R + (ect...)
\end{align}
So we finally have :

\begin{align}
\int_\mathbb{R^2} w(T,x)\psi(x) = \int_\mathbb{R^2} w(T,x)[\psi - (\psi \ast \phi_\delta)\chi_R]dx + \int_0^T \int_\mathbb{R^2} w(t,x)r_\delta\chi_R +
\int_0^T \int_\mathbb{R^2} \phi_\delta \partial_t \chi_R + (ect...)\\ =
\end{align}

One can easily majorate the term $| \int_\mathbb{R^2} w(T,x)\psi(x) |$ by a constant $C$ that depends on $\delta$ and $R$ :
$$| \int_\mathbb{R^2} w(T,x)\psi(x) | \leq C(||w|| ||r_\delta||_{L^1}, \delta, R)$$

Then, as $\delta \rightarrow 0$ and $R \rightarrow \infty$ we conclude that :
$$\int_\mathbb{R^2} w(T,x)\psi(x) = 0$$
for any $\psi \in C_c^1(\mathbb{R^2})$ and $w = 0$ for any $T > 0$.
That gives us the Uniqueness.


问:
f is uniformly continuous if for all $x$ and $y ∈ \mathbb{R}$, $|f(x) − f(y)| ≤ |x−y|^\frac{1}{2}$.
Let f : R → R be a given function. The following property will ensure that f is uniformly continuous:

for all $x$ and $y ∈ \mathbb{R}$, $|f(x) − f(y)| ≤ |x−y|^\frac{1}{2}$.

My attempt :
From definition a function $f$ is uniformly continuous if for a arbitrary $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that for any two points $x,y$ in $\mathbb{R}$,
$$ |x-y| \lt \delta \implies |f(x)-f(y)| \lt \epsilon
$$

Now let $\epsilon \gt 0$ be given. Then
$$
 |x−y|^\frac{1}{2} \lt \epsilon \\
\implies
 |x-y| \lt {\epsilon}^2 = \delta \\
\implies |f(x)-f(y)| \lt \epsilon
.$$

Therefore the function is uniformly continuous.

答:
From definition a function $f$ is uniformly continuous if for a arbitrary $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that for any two points $x,y$ in $\mathbb{R}$,
$$ |x-y| \lt \delta \implies |f(x)-f(y)| \lt \epsilon
$$

Now let $\epsilon \gt 0$ be given. Then let $$\delta = {\epsilon}^2.$$ Then
$$ |x-y| \lt {\epsilon}^2 = \delta \\
\implies |x−y|^\frac{1}{2} \lt \epsilon \\
\implies |f(x)-f(y)| \lt \epsilon
.$$

Therefore the function is uniformly continuous.


答:
No

Your logic is sound in thinking that since $\frac{\sqrt{4-x}}{\sqrt{4-x}} = 1$, that $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ could be equal  to $y=1$, but since your value of $x$ isn't defined, in this case, $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ is not equal to $y=1$

Title: Shouldn't the graph of $y=1$ be the same as $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ because $1=\frac{\sqrt{4-x}}{\sqrt{4-x}}$?

Body:
Shouldn't the graph of $y=1$ be the same as $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ because $1=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ ?

So why is it that my calculator said otherwise? And I know that you can't square root a negative number, because it will gave an imaginary number. Is it because the calculator assume that if it there is an undefined value in the equation, the final result will be undefined too? Or is there a proof that I aren't aware of?

Unnecessary comment below:

And if graph of $y=1$ aren't the same as $y=\frac{\sqrt{4-x}}{\sqrt{4-x}}$ just like the calculator said. That would be great, because you don't have to clarify the domain outside the expression. You could just do $y=2x\cdot\frac{\sqrt{4-x}}{\sqrt{4-x}}$ instead of using a set notation such as $\{x\in| \mathbb{R},x>4\}$


问:
Implicit Differentiation Help
I have to use implicit differentiation to find $\frac{dy}{dx}$ given:

$$x^2 \cos(y) + \sin(2y) = xy$$

I don't even know where to begin, I missed the class where we went over implicit differentiation, and because of that, I am completely stuck.

Thank you everyone.

Edit: I don't know how to make the equation look all nice and whatnot, so sorry about that

答:
An equation $F(x,y)=0$ defines a curve $\gamma$ in the $(x,y)$-plane. In most cases (as here) it is impossible to solve the equation for $y$ explicitly. But one can say the following:  Any point $(x_0,y_0)\in\gamma$ where $F_y(x_0,y_0)\ne 0$ is the center of a rectangular window $W$, such that $\gamma \cap W$ is the graph of a ("local") function $y=\phi (x)$. One may ask for the derivative $\phi'$, in particular for its value at $x_0$. It is shown in the calculus course you missed that this value can be calculated ${\it without\ formally\ solving\ the\ equation}$  $F(x,y)=0$. The result is
$$\phi'(x_0)=-{F_x(x_0,y_0)\over F_y(x_0,y_0)}\ .$$
Doing the calculation you will end up with an expression containing both variables $x$ and $y$. If $(x_0,y_0)\in\gamma$ is given in advance you may just plug these coordinates in. If not, your derivative $\phi'$ appears not as a function of $x$ alone but as a function of $x$ and $y\ (=\phi(x))$. In addition the resulting expression is not uniquely determined insofar as it may be simplified using the equation $F(x,y)=0$.</p>

Title: Implicit Differentiation Help