问:
Complexified tangent vector, complex manifold
Consider a complex submanifold $M$ of a complex ambient vector space $X$. Suppose you have a base point $p \in M$ and a $C^1$ arc $\gamma(t)$ passing through $p$ and staying in $M$, with tangent vector at $p$ denoted by $u$ (the tangent vector is taken by considering $X$ as a real vector space, and $\gamma : [0,1] \rightarrow M$). Is there a complex analytic curve $\tilde{\gamma} : \Delta \rightarrow M$ ($\Delta$ being a complex disk) passing through $p$, whose image is in $M$, with tangent vector at $p$ equal to $u$ (as an element of $T_pM$) ? What if the curve $\gamma$ is only differentiable at $p$ ? Intuitively I think the answer is yes, but I can't construct such a curve $\tilde{\gamma}$. Thanks in advance for your help</p>

答: Just work in local coordinates, so you're starting with vector $v\in\mathbb C^k$, and consider the complex line spanned by $v$.</p>


问:
$|f|$ has a global minimum where $f:\mathbb{C} \to \mathbb{C}$ is any polynomial function
The following Lemma is used in a proof for the Fundamental Theorem of Algebra, but I cannot understand some steps.

Lemma: Let  $f:\mathbb{C} \to \mathbb{C}$  be any polynomial function.
Then there exists a point  $z_0 \in \mathbb{C}$  where the function
$|f|$  attains its minimum value in  $\mathbb{R}$.

The proof starts like this:

If  $f$  is a constant polynomial function, then the statement of the
Lemma is trivially true since  $|f|$  attains its minimum value at
every point in  $\mathbb{C}$ . So choose, e.g.,  $z_0=0$.

If  $f$  is not constant, then the degree of the polynomial defining
$f$  is at least one. In this case, we can denote  $f$  explicitly by

$f(z)=a_n z^n + ... + a_1 z + a_0$

with $a_n \neq 0$. Now, assume  $z \neq 0$ , and set
$A=max\{|a_0|,…,|a_{n−1}|\}$ . We can obtain a lower bound for
$|f(z)|$  as follows:

$f(z)=|a_n| |z|^n |1+\frac{a_{n-1}}{a_n} \frac{1}{z}+...+\frac{a_0}{a_n} \frac{1}{z^n}| \leq |a_n| |z|^n (1 - \frac{A}{|a_n|} \sum \limits_{k=1}^{\infty} \frac{1}{|z|^k})=|a_n| |z|^n (1 - \frac{A}{|a_n|} \frac{1}{|z|-1})$.

I don't see how these two steps follow. We can distribute the absolute value over all the factors, but how does one turn the second factor from $1+...$ into a $1-...$ using an infinite sum? After that it seems like an application of a geometric series, but how do we know that $|z|>1$?

I've been stuck for quite some while now with no notable progress. Am I missing anything obvious here?

Edit:

The next step of the proof is a follows (so it justifies the use of the formula for the geometric series):

For all $z \in \mathbb{C}$ such that $|z| \geq 2$, we can further
simplify this expression and obtain

$ (*) |f(z)| \geq |a_n| |z|^n (1 - \frac{2A}{|a_n| |z|})$.

It follows from this inequality that there is an $R > 0$ such that $|f(z)| > |f(0)|$, for all $z \in \mathbb{C}$ satisfying $|z| > R$.

Now this confuses me once again. We have $|f(0)|=a_0$, so how does this follow from the inequality given above? I think it ought to have something to do with the definiton of $A$, but I can't quite figure it out.

Could anyone please provide a hint or an explanation?

答:
We don't need geometric series, etc., in order to prove the Lemma. As you remark, we can assume
$$f(z)=a_nz^n+a_{n-1}z^{n-1}+\ldots+a_0,\qquad n\geq1,\qquad a_n\ne0\ .$$
This implies$${f(z)\over z^n}=a_n+{a_{n-1}\over z}+\ldots+{a_0\over z^n}\to a_n\quad\bigl(|z|\to\infty\bigr)\ ,$$
so that there is an $R>0$ with
$$\left|{f(z)\over z^n}\right|>{|a_n|\over2}\qquad\bigl(|z|>R\bigr)\ .$$
From this we can infer that there is an $R'\geq R$ with
$$\bigl|f(z)\bigr|>|z|^n{|a_n|\over2}>|a_0|\qquad\bigl(|z|>R'\bigr)\ .\tag{1}$$
On the compact set $\overline{D_{R'}}$ the continuous function $z\mapsto\bigl|f(z)\bigr|$ assumes a global minimum $\>\leq|f(0)|=|a_0|$. Due to $(1)$ this minimum value is not assumed when $|z|>R'$, hence is the global minimum of $|f|$ on ${\mathbb C}$.

ContentLicense: CC BY-SA 4.0


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: The equation $y=1$ is independent of $x,$ but your quotient of square roots is not. Working within the real numbers, the radical quotient makes sense only when $4-x > 0$ for two reasons: so you do not introduce complex numbers (if $4-x<0$); and so that the denominator is not $0$ (if $x=4$).


问:
$\int_{0}^{1} \sin x \cdot \ln x dx$
I would like to evaluate the following integral:

$$\int_{0}^{1} \sin x \cdot \ln x dx$$

I had a pretty decent idea to do it, and that is by turning $\sin x$ an $\ln x$ into their respective power series, turning it into a double summation, and integrating. This yields the following:

$$-\int_{0}^{1}\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{x^{2n+1}(-1)^n(1-x)^{m+1}}{(m+1)(2n+1)!} dx$$

Putting the integral inside the double summation and making use the beta function, we get:

$$ =-\sum_{n=0}^{\infty} \sum_{m=0}^{\infty}\frac{(-1)^n \beta(2n+1,m+1)}{(m+1)(2n+1)!}$$

A slight reindexing gives:

$$\sum_{n \space \text{odd}} \sum_{m \space\geq 1}\frac{\beta(n,m)}{n!m}$$

Notice the $-1$ in the front goes away by cancelling with the $-1$ from letting $n$ run through the odds.

Is this a valid strategy? If there is a closed form for this integral, I think this is a pretty cool summation identity. Can someone help out?

答:
Slightly different approach

Consider a parameter defined integral:

$$I(a)=\int_{0}^1x^a\sin x\;dx$$

We'll get back to the original integral after differentiating this wrt $a$ and setting $a=0$

Next, use the Taylor series of $\sin x$:

$$I(a)=\int_{0}^1\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1+a}}{(2k+1)!}\;dx$$

$$=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!(2k+2+a)}$$

Now, differentiate wrt $a$ and set $a=0$:

$$\int_{0}^1\ln x\sin x\;dx=-\frac{1}{4}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!(k+1)^2}$$

This infinite series converges very fast.

If we take only the first term of the series, $-\frac{1}{4}$, then there is an error only about $0.01$.

The sum of the first $2$ terms of the series is $-\frac{23}{96}$ and our deviation from the exact value is only about $0.00023$

Title: $\int_{0}^{1} \sin x \cdot \ln x dx$


问:
Is it true that a center of any group is contained in the center of any of its subgroups?
I'm reading the book "Abstract Algebra with Applications" by Audrey Terras and one of the exercises (2.4.11) has us find the centers of $D_5$ and $D_6$ (the dihedral groups).
It states then, as a hint: The center of any group is contained in the center of any of its subgroups.

Either I can't figure out exactly what this hint means, or I just don't see how it could be true.

Let $E$ be the identity symmetry element, $R$ be a rotation of 90 degrees, and $F$ a flip or mirroring about some axis. The center of $D_4$ (which is $\{E, R^2\}$) is certainly not contained in "any" of its subgroups, for example the subgroup $\{E, F\}$.

Am I misunderstanding something here?

答:
As has been noted, as stated the hint is very, very wrong. For example, $S_3\times C_2$ has center $\{e\}\times C_2$; the subgroup $S_3\times \{e\}$ has trivial center, so the center of the whole group is not contained in the center of this subgroup.

There are two possible corrections that might make it true.

As suggested by @ancientmathematician, they could have meant that the intersection of the center with any subgroup is contained in the center of the subgroup: $Z(G)\cap H\subseteq Z(H)$ for every subgroup $H$. This is certainly true. In the example above, it would tell you that the center does not contain any nontrivial elements of $S_3\times\{e\}$.

Alternatively, they could have been trying to say that the center of the group is contained in the centralizer of any subgroup. If $H$ is a subgroup of $G$, its centralizer is
$$C_G(H) = \{g\in G\mid gh=hg\text{ for all }h\in H\}.$$
And it is easy to verify that $Z(G)\subseteq C_G(H)$ for every subgroup $H$.

(I will note that it is also false that the center of a subgroup is contained in the center of the group; in $S_3$, every proper subgroup equals its center, but the whole group has trivial center.)


问:
Counting Number of Objects - When to Add One Back
I know this might be a very basic question. Sometimes to count objects, we just subtract. For example -- If there are 5 apples and I take away 1, then remaining are $5 - 1 = 4$ apples.

But other times, we have to add one back.
For example -- the number of days from the 1st to the 5th of any month = $5 - 1 + 1 = 5$. I added the last one back because I'm including the 1st.

What are these properties formally called? Are there more cases? Are there more formal explanations and properties of these cases in combinatorics?

答:
The number of days from the 1st to the 5th is only $5$ depending on how you define "from." How many days is it "from" 1st to the 2nd? Most people would say $1$ day, not $2$ days.

This is not a case of subtraction, but rather a case of counting a range - you are looking for the number of days between the 1st and 5th, inclusive.

I suppose one could say this is a difference between using cardinal numbers and ordinal numbers, but I think the real problem is your use of the word "from" as a synonym for subtraction, always. It is not, when it is dealing with a range.

ContentLicense: CC BY-SA 3.0


问:
X has finitely many connected components, can we get X locally connected?
say X has finitely many connected components, can we get X locally connected?

I'm thinking to prove this and then I can have every component is open.
Thanks!

答:
No: for instance, the space $X=\left\{(x,y)\in\Bbb R^2\,:\, x=0\lor y=\sin\frac1x\right\}$ with the topology inherited from $\Bbb R^2$ is connected, but not locally connected (all sufficiently small balls around $(0,t)$ for $t\in[0,1]$ contain infinitely many cuts of the topologist's sine curve but not a whole arc connecting them).

However, it is true that a topological space with finitely many connected components has all connected components open. This is because the connected components of a topological space are always closed, and therefore each connected component is $C_i=X\setminus\bigcup_{1\le j\le m\\ j\ne i} C_j$, which is in fact the complement of a closed set.


问:
A case in which the integral equals 0 implies that the function is zero
Let $x$,$y$ be real-valued functions defined and continuous on the closed interval $[a,b]$. Then,
$\int_{a}^{b} |x(t)-y(t)| \,dx = 0$ implies $|x(t)-y(t)| = 0$ for every $t \in [a,b]$. I know that continuity is a neccesary condition and I tried to used it with the definition of the Riemann integral (Rosenlicht, introduction to Analysis, chapter VI), but I didn't see a path.

Could you give me a hint?

答:
Let $f:[0,1]\to \mathbb R$ continuous and positive s.t. $$\int_0^1f=0.$$

Suppose $f\neq 0$. Therefore, since $f$ is continuous, there $t\in (0,1)$ s.t. $f(t)\neq 0$. By continuity of $f$, there is $\delta >0$ s.t. $(t-\delta ,t+\delta )\subset (0,1)$ and $f(s)\geq \frac{f(t)}{2}$ for all $s\in (t-\delta ,t+\delta ).$ Then $$\int_0^1f(s)\,\mathrm d s\geq \int_{t-\delta }^{t+\delta }f(s)\,\mathrm d s\geq f(t)\delta >0,$$
which is a contradiction with $\int_0^1 f=0$.


问:
Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\prod_{n=1}^{6} (g(a_n))$
Question:

Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\displaystyle \prod_{n=1}^{6} (g(a_n))$

My process:

I first thought of setting $z=a_i^2+2a_i+4$ and then putting the value back into $f(z)$ and then from there finding the product using Vieta's. But the process turned incredibly long and tedious and I found it to be impossible to do by hand.

So I just went to to the brute force method, expanding the whole expression out. Writing all the terms would be really tedious so I would like to abbreviate it. Let $P= a_1 \cdot a_2 \cdots \cdot a_6$ and let $e_n$ denote the sum of roots taken $n$ at a time.

Here is the expression I got:

$$\prod_{i=1}^{6} (a_i^2+2i+4)=P^2+2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P+2^5*4*e_5+2^4*4^2*e^4+2^3*4^3*e_3+2^2*4^4*e_2+2*4^5*e_1+4_6+4(\sum_{cyc}(a_1a_2a_3a_4a_5)^2))^+4^2(\sum_{cyc}(a_1a_2a_3a_4)^2)+4^3\sum_{cyc}(a_1a_2a_3)^2+4^4\sum_{cyc}(a_1a_2)^2+4^5\sum_{cyc}(a_1)^2$$

Which is a monster in its own right. However, I was extremely interested in the patterns that kept popping up. Like the $2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P$ where the power of 2 and the index of $e$ keeps increasing and decreasing with each other. Therefore I wondered if there could be any general formula for any $f(x)$ and $g(x)$.

And that is my question, is there any way to find the answer efficiently without all this unnecessary calculation; and also can this process be generalised to any polynomials $f(x)$ and $g(x)$. And if it cannot be generalised, then is there any sort of algorithm or methodical way that one can take while solving this type of problem?

答:
The leading coefficient of $f$ is equal to one, so $\prod_{n=1}^{6} (g(a_n))$ is the resultant of $f$ and $g$, and that can be computed as a determinant whose entries are the coefficients of the polynomials:

$$
\prod_{n=1}^{6} (g(a_n)) = \operatorname{res}(f, g) =
\begin{vmatrix}
-8 &   & 4 \\
0 & -8 & 2 & 4 \\
0 & 0 & 1 & 2 & 4 \\
-2 & 0 &  & 1 & 2 & 4 \\
0  & -2 &   &  & 1 & 2 & 4 \\
0 & 0 & & & & 1 & 2 & 4 \\
1 & 0 & & & & & 1 & 2  \\
 & 1 & & & & & & 1
\end{vmatrix}
= 1600
$$
where the  empty entries are zero.


问:
Coefficient of $x^8$ in the expansion of $x^2 \cos(x^2)$ around $x=0$.
What is the Coefficient of $x^8$ in the expansion of $f(x)=x^2 \cos(x^2)$ around $x=0$.

Is there any compact way to solve this other than differentiating $f$ for $8$ times?

答: Write the power series for $\cos x$, substitute $x^2$ for $x$, multiply each term by $x^2$ and see the coefficient you want.


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:
This has confused me countless times. The example that always confuses me is this:$\require{AMScd}$
\begin{CD}0 @>>>\mathbb{Z} @>{\cdot 2}>> \mathbb{Z} @>{\pi} >> \mathbb{Z}/2 @>>> 0 \\\end{CD}

So $\mathbb{Z}/\mathbb{Z}=\mathbb{Z}/2$! But this only makes sense because of the identification $\mathbb{Z} \simeq 2\mathbb{Z}$.

Title: very basic short exact sequence problem


问:
Evaluate of $\int_{0}^{\infty}\frac{e^{x\pi}-1}{e^{x\pi}+1}\cdot \frac{x}{(1+x^2)(9+x^2)} dx$
Would like to know how would you evaluate $(1)$:
$$\int_{0}^{\infty}\frac{e^{x\pi}-1}{e^{x\pi}+1}\cdot \frac{x}{(1+x^2)(9+x^2)} dx=\frac{1}{8}?\tag1$$

Notices that, $$\tanh\left(\frac{x\pi}{2}\right)=\frac{e^{x\pi}-1}{e^{x\pi}+1}$$

$$\int_{0}^{\infty}\tanh\left(\frac{x\pi}{2}\right)\cdot \frac{x}{(1+x^2)(9+x^2)} dx\tag 2$$

$$8\int_{0}^{\infty}\tanh\left(\frac{x\pi}{2}\right)\cdot \frac{x}{1+x^2} dx-8\int_{0}^{\infty}\tanh\left(\frac{x\pi}{2}\right)\cdot \frac{x}{9+x^2} dx \tag3$$

答:
Basically, we note

$$
\int_{0}^{\infty} e^{-at}\sin(xt) \,\mathrm{d}t = \frac{x}{a^2+x^2}
$$

$$
\int_{0}^{\infty} {\frac{\cos(\alpha x)}{\cosh(\beta x)} \mathrm{d}x} = \frac{\pi}{2\beta} \operatorname{sech}\left(\frac{\pi\alpha}{2\beta}\right)
$$

then denote desired integral with $I(a,b)$ that is
$$
\begin{aligned}
I(a,b)
& = \int_{0}^{\infty} \frac{x}{(a^2+x^2)(b^2+x^2)}\tanh\left(\frac{\pi x}{2}\right)\mathrm{d}x\\
& = \frac1{b^2-a^2} \int_{0}^{\infty} \left(\frac{x}{a^2+x^2}-\frac{x}{b^2+x^2}\right)\tanh\left(\frac{\pi x}{2}\right)\mathrm{d}x\\
& = \frac1{b^2-a^2}\int_{0}^{\infty} \left(\int_{0}^{\infty} (e^{-at}-e^{-bt})\sin(xt) \,\mathrm{d}t\right)\tanh\left(\frac{\pi x}{2}\right)\mathrm{d}x\\
& = \frac1{b^2-a^2}\int_{0}^{\infty} (e^{-at}-e^{-bt}) \left( \int_{0}^{\infty} \sin(xt)\tanh\left(\frac{\pi x}{2}\right)\mathrm{d}x\right)\mathrm{d}t\\
& = \frac1{b^2-a^2}\int_{0}^{\infty} (e^{-at}-e^{-bt}) \left( -\frac{i}{2}\int_{0}^{\infty} \frac{\cos\left(t-\frac{i\pi}{2}\right)x-\cos\left(t+\frac{i\pi}{2}\right)x}{\cosh\left(\frac{\pi x}{2}\right)}\mathrm{d}x\right)\mathrm{d}t\\
& = \frac1{b^2-a^2}\int_{0}^{\infty} (e^{-at}-e^{-bt}) \left( -\frac{i}{2}\left(\operatorname{sech}\left(t-\frac{i\pi}{2}\right)-\operatorname{sech}\left(t+\frac{i\pi}{2}\right)\right)\right)\mathrm{d}t\\
& = \frac1{b^2-a^2}\int_{0}^{\infty} \frac{e^{-at}-e^{-bt}}{\sinh t} \mathrm{d}t = \frac1{b^2-a^2}\left(\psi\left(\frac{1+b}{2}\right) - \psi\left(\frac{1+a}{2}\right)\right)
\end{aligned}
$$
which easily reveals $I(1,3)=\frac1{8}$

Supplement for some identites
$$
\sinh\left(\frac{\pi x}{2}\right)=-i\sin\left(\frac{i\pi x}{2}\right), \quad\cosh\left(t\pm\frac{i\pi}{2}\right)=\pm i\sinh t
$$
and
Gauss's representation of digamma function for $\Re(z)>0$, namely
$$
\psi(z) = \int_{0}^{\infty} {\left( \frac{e^{-t}}{t} - \frac{e^{-zt}}{1-e^{-t}} \right) \mathrm{d}t}
$$

Edit for some necessary explanation, actually this integral
$$
\int_{0}^{\infty} \sin(xt)\tanh\left(\frac{\pi x}{2}\right)\mathrm{d}x
$$
does not converge under common meaning, there are some abuse of notation need to be justified, a quick insight is to recall
$$
\int_{0}^{\infty} {\frac{\sin(\alpha x)}{\sinh(\beta x)} \mathrm{d}x} = \frac{\pi}{2\beta} \tanh\left(\frac{\pi\alpha}{2\beta}\right)
$$
and we have
$$
\begin{aligned}
\int_{0}^{\infty} \sin(xt)\tanh\left(\frac{\pi x}{2}\right)\mathrm{d}x
& = \int_{0}^{\infty} \frac{\mathrm{d}w}{\sinh w} \left(\frac2{\pi}\int_{0}^{\infty} \sin(xt)\sin(xw)\mathrm{d}x\right)\\
& = \int_{0}^{\infty} \frac{\delta(t-w)-\delta(t+w)}{\sinh w} \mathrm{d}w = \frac1{\sinh t}
\end{aligned}
$$
by utilizing the Dirac's delta in such integral.

Or more directly, recalling $(w>0)$
$$
\int_{0}^{\infty} \frac{x\sin(wx)}{a^2+x^2}\mathrm{d}x = \frac{\pi}{2}e^{-aw}
$$
hence
$$
\begin{aligned}
I(a,b)
& = \int_{0}^{\infty} \frac{x}{(a^2+x^2)(b^2+x^2)}\tanh\left(\frac{\pi x}{2}\right)\mathrm{d}x\\
& = \frac1{b^2-a^2} \int_{0}^{\infty} \left(\frac{x}{a^2+x^2}-\frac{x}{b^2+x^2}\right) \left( \frac2{\pi}\int_{0}^{\infty} {\frac{\sin(wx)}{\sinh(w)} \mathrm{d}w} \right) \mathrm{d}x\\
& = \frac1{b^2-a^2}\int_{0}^{\infty} \frac{\mathrm{d}w}{\sinh w} \left(\frac{2}{\pi} \int_{0}^{\infty} \left(\frac{x\sin(wx)}{a^2+x^2} - \frac{x\sin(wx)}{b^2+x^2}\right) \mathrm{d}x\right)\\
& = \frac1{b^2-a^2}\int_{0}^{\infty} \frac{e^{-aw}-e^{-bw}}{\sinh w} \mathrm{d}w = \frac1{b^2-a^2}\left(\psi\left(\frac{1+b}{2}\right) - \psi\left(\frac{1+a}{2}\right)\right)
\end{aligned}
$$
which appears to be a better solution.


问:
$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$
I've tried to solve this limit:
$$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$

Here,
$$\frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$ $$\sim \frac{(x \arctan x-x^2)(\sqrt{1+2x^4}+1)}{2x^4}$$ $$\sim \frac{(x \arctan x-x^2)}{x^4}= \frac{ \arctan x-x}{x^3} \sim \frac{x-x}{x^3} \sim 0$$

But the final result should be $-\frac{4}{3}$.

Any help would be appreciated.

答:
$$L=\lim_{x \to 0}\frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$
You need to use Taylor series:
$$L=\lim_{x \to 0}\frac{\ln (1+x (x-\dfrac {x^3}3+...)-(1+x^2+\dfrac {x^4}{2}+o(x^4))+1}{x^4}$$
$$L=\lim_{x \to 0}\frac{ (x^2-\dfrac {x^4}3-\dfrac {x^4}2+o(x^4))-x^2-\dfrac {x^4}{2}+o(x^4)}{x^4}$$
$$L=\lim_{x \to 0}\frac{ -\dfrac {x^4}3-\dfrac {x^4}2-\dfrac {x^4}{2}+o(x^4)}{x^4}$$
$$L=\lim_{x \to 0}\frac{ -\dfrac {x^4}3- {x^4}+o(x^4)}{x^4}$$
$$L= -\dfrac {1}3-1=-\dfrac  43$$
Where :
$$\sqrt{1+2x^4}=1+x^4+o(x^4)$$
$$e^{x^2}=1+x^2+\dfrac {x^4}2+o(x^4)$$
$$\ln (1+x)=x-\dfrac {x^2}2 +...$$
$$\arctan x = x-\dfrac {x^3}3+o(x^4)$$

for the arctan and log term
$$S=\ln (1+x (x-\dfrac {x^3}3+...)$$
Remember that we have the Taylor serie for log:
$$\ln (1+z)=z-\dfrac {z^2}2$$
So that we have:
$$S=\ln (1+x (x-\dfrac {x^3}3+...)$$
$$S=\ln (1 +(x^2-\dfrac {x^4}3+...)$$
$$S=(x^2-\dfrac {x^4}3+...)-\dfrac 12(x^2-\dfrac {x^4}3+...)^2$$
We only keep terms with exponent less or equal to four...
$$S=(x^2-\dfrac {x^4}3+...)-\dfrac 12(x^4 \ \color {red}{ \text { + more terms with exponent> 4}})$$
We don't need the terms that have exponent greater than four. Because we take the limit at zero and all these terms will be zeros. They play no role. So that:
$$S=x^2-\dfrac {x^4}3-\dfrac {x^4}2+o(x^4)$$


答:
It's a little bit subtle.

When we think about apples we say that $5$ apples is four more apples than $1$ apple.  Because we're comparing the set of 5 whole apples to one whole apple.

When we think about days you're comparing the end of the first of June to the beginning of the fifth of June.
So the difference in days is only three.

So in fact it's a question of language not mathematics.  If your train is leaving Ambridge, and it stops at Brumley, Catford, and Dronby before arriving at Endton where you get off. You might say there are four stops until Endton (Brumley, Catford, Dronby and Endton) or you might say there are three (Brumley, Catford and Dronby).  It depends which country you come from.  In England we say four, in many European countries they say three. I haven't done an exhaustive survey. Most people are pretty convinced their way is right, but it's just convention.

So the only difference is whether the last item is included. Which differs by convention depending what the items are.  As mathematicians we think imprecision is bad so we don't talk about this sort of thing.

Title: Counting Number of Objects - When to Add One Back

Body:
I know this might be a very basic question. Sometimes to count objects, we just subtract. For example -- If there are 5 apples and I take away 1, then remaining are $5 - 1 = 4$ apples.

But other times, we have to add one back.
For example -- the number of days from the 1st to the 5th of any month = $5 - 1 + 1 = 5$. I added the last one back because I'm including the 1st.

What are these properties formally called? Are there more cases? Are there more formal explanations and properties of these cases in combinatorics?


问:
evaluate $\sum_1^\infty \frac{1}{(n \sqrt{n+1} + (n+1)\sqrt{n})}$
How can I evaluate the infinite sum $\sum_1^\infty \frac{1}{(n \sqrt{n+1} + (n+1)\sqrt{n})}$? I try to change that into the form $\sum_1^\infty \frac{1}{(\sqrt{n} + \frac{n}{n+1}\sqrt{n+1})}-\frac{1}{(\sqrt{n+1} + \frac{n+1}{n}\sqrt{n})}$ and try to reduce the terms. But it seems not valid. I'm thinking of using some fourier transform techniques, any suggestions?

答: Since $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{a}-\sqrt{b}}{a-b}$,$$\frac{1}{\sqrt{(n+1)^2n}+\sqrt{n^2(n+1)}}=\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)}=u_n-u_{n+1}$$with $u_n:=\frac{1}{\sqrt{n}}$, so the series telescopes to $u_1=1$.


问:
Schatten p Norm of adjoint matrix
Is it true that $\|A^*\|_p=\|A\|_p$?

Is there a simple argument?

$\| \cdot \|_p$ is the Schatten p norm defined with singular values.

Thanks in advance.

答:
The singular values are found by taking the square roots of the eigenvalues of the matrices $A^{*}A$ or $AA^{*}$ which are Hermitian positive semi-definite. These matrices only have real non-negative eigenvalues. So if you take $A^{*} = (U \Sigma V^{*})^{*}$ you end up with $V \Sigma^{*} U^{*}$ but for real values $*$ is just the transpose and the diagonal doesn't change. So $A^{*} = V \Sigma U^{*}$

Since the Schatten p-norm is defined as

$$\| \cdot\|_{p}  = \bigg( \sum_{i=1}^{n} \sigma_{i}^{p}(A) \bigg)^{\frac{1}{p}}$$

we have that $\| A\|_{p} =  \|A^{*}\|_{p}$

Title: Schatten p Norm of adjoint matrix


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 $2 \pi + 10$

Body:
Let $~f(x) = \left[\tan(\pi x^2)+(2x-1)4^x+6x-4 \right]~$ and

let $~g(x) = \tan[x - (1/2)].$

Note that $g(x)$ and each term of $f(x)$ are continuous functions.

You want $\lim_{x\to (1/2)} \frac{f(x)}{g(x)}.$

Before applying L'Hopital's rule, first check $f(1/2)$ and $g(1/2)$.

$$f(1/2) = \tan(\pi/4) + (0)4^{(1/2)} + 3 - 4 = 1 + 3 - 4 = 0.$$

$$g(1/2) = \tan(0) = 0.$$

Applying L'Hopital's rule:

$$f'(x)|_{x=(1/2)}
~=~\frac{(2\pi)x}{\cos^2(\pi x^2)} + (2)(4^x) + (2x - 1)\log(4)[4^x] + 6 ~|_{x = (1/2)}
$$

$$= \frac{\pi}{(1/\sqrt{2})^2} + (2)(2) + (0)\log(4)(2) + 6
~=~ 2\pi + 4 + 0 + 6 ~=~ 2\pi + 10.$$

and

$$g'(x) ~|_{x = (1/2)}
~=~ \frac{1}{\cos^2(x - 1/2)} ~|_{x = (1/2)} = 1.$$

Thus, $$\frac{f'(1/2)}{g'(1/2)} = \frac{2\pi + 10}{1}.$$


问:
prove that a subset of the real line is connected if and only if it is an interval.
Do I need to use the Sufficient Condition, too?

Necessary Condition

Suppose $I$ is not an interval of $\mathbb{R}$.

Then $\exists$ $x,y\in I$ and $z\in\mathbb{R}-\mbox{I}$ such that $x<z<y.$

Consider the sets $I∩(-\infty..z)$ and $I∩(z..+\infty).$

Then $I∩(-\infty..z)$ and $I∩(z..\infty)$ are open by definition of the subspace topology on I.

Neither is empty because they contain $x$ and $y$ respectively.

They are disjoint, and their union is $I$, since $z∉I$.

Therefore $I∩(-\infty..z)∣I∩(z..+\infty)$ is a separation of $I$.

It follows by definition that $I$ is disconnected.

$\square$

答:
This question has been asked many times in this forum. Make a search for  "interval connected":

However, frequently the relevant arguments are spread over both question and answer, or only deal with a special type of interval (for example an open interval).

Let us a give a proof for any interval I which is open open / half-open / closed and bounded / unbounded.

Assume that $I$ is not connected. Then there exists a partition $(U_1,U_2)$ of $I$, i.e. a pair of nonempty open subsets $U_1, U_2 \subset I$ such that $U_1 \cup U_2 = I$ and $U_1 \cap U_2 = \emptyset$. We emphasize that these sets are open in the subspace $I$, it is possible that they are not open in $\mathbb R$. Note that the $U_i$ are clopen in $I$, i.e. are closed and open in $I$. Thus they have the form $U_i = I \cap C_i$ with closed $C_i \subset \mathbb R$.

Pick $x_i \in U_i$. W.l.o.g. we may assume $x_1 < x_2$ (otherwise we consider the partition $U'_1 = U_2, U'_2 = U_1$ and $x'_1 = x_2, x'_2 = x_1$). Define
$$y^* = \sup \{ y \in U_1 \mid y < x_2\} .$$
From this definition we get

$x_1 \le y^* \le x_2$. Thus $y^* \in I$ because $x_1, x_2 \in I$.

$y^*$ is the limit of an increasing sequence $(y_n)$ in $U_1 \subset C_1$. Since $C_1$ is closed in $\mathbb R$, we see that $y^* \in C_1$. Thus $y^* \in I \cap C_1 = U_1$ and therefore $y^* < x_2$ because $x_2 \notin U_1$.

The half open interval $(y^*,x_2]$ cannot contain points of $U_1$. But $(y^*,x_2] \subset I$ because $y^*, x_2 \in I$, thus $(y^*,x_2] \subset U_2$. Therefore $y^*$ is the limit of a decreasing sequence $(y'_n)$ in $U_2 \subset C_2$. Since $C_2$ is closed in $\mathbb R$, we see that $y^* \in C_2$. Thus $y^* \in I \cap C_2 = U_2$. This is a contradiction because $U_1 \cap U_2 = \emptyset$.


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:
Set $N_0=100$ and take $N_k$ to be the number of coins that remain after the $k^\text{th}$ trial in this process. So we can say something like $$P(N_1=81|N_0=100)={100 \choose 19}\Big(\frac{1}{2}\Big)^{100}$$

Now for $i\in \{0,1,\ldots, 100\}$ and $j\in \{0,1,\ldots ,i\}$ we have $$P(N_{k+1}=j|N_{k}=i)={i \choose j-i}\Big(\frac{1}{2}\Big)^i$$ Notice $\{N_k\}_{k=0}^{\infty}$ is an absorbing Markov chain with $0$ as an absorbing state. You're looking to compute the expected number of trials in this random process before being absorbed in state $0$ starting from state $100$. There are many ways to compute this expected value, the most efficient is probably by making use of the fundamental matrix which you can learn about here


问:
Result on idempotent matrices

Let $A$ be a $n\times n$ matrix with $A^2=A$ and $(A-A^T)^2 =0$. Then $AA^T$ is idempotent.

If $A$ is idempotent then $A^T$ is also idempotent. So, $(A^T)^2 =A^T$. Now, given that $(A-A^T)^2=0$ implies $A^2-A(A^T)-(A^T)A+(A^T)^2=0$, we have that $A-A(A^T)-(A^T)A+A^T=0$.

I cannot proceed further to prove the result. Thank you in advance

答:
$0=A(A-AA^T-A^TA+A^T)=A^2-A^2A^T-AA^TA+AA^T$

$=A-AA^T-AA^TA+AA^T=A-AA^TA=0$ implies that $0=(A-AA^TA)A^T=AA^T-AA^TAA^T=AA^T-(AA^T)^2$.

Title: Result on idempotent matrices


问:
how to show that $f'(x)=x/g(x)$?
Given that $$g(x)=1+\frac x{\sqrt{x^2+1}}$$

and that $$f(x)=\frac x{g(x)}$$

show that $$f'(x)=g(x)$$

Can you please provide some hints before giving out the full answer, thank you.

答: Have you tried using plain and simple differentiation? It's important to use chain rule and quotient rule here.</p>

Title: how to show that $f'(x)=x/g(x)$?


问:
Another basic short exact sequence problem
In the following commutative diagram of R-modules, all of the rows and columns are exact. Prove that $K$ is isomorphic to $L$.

\begin{array}{ccccccccccc} &&&&&&&&0 &&\\
&&&&&&&&\downarrow &&\\ &&&&&& 0 & & L &&\\ &&&&&& \downarrow && \downarrow &&\\ &&&& M^{\prime\prime} & \rightarrow& N^{\prime\prime} & \rightarrow & P^{\prime\prime} & \rightarrow& 0\\ &&&& \downarrow &  &  \downarrow   &&  \downarrow &&\\ && 0 & \rightarrow & M & \rightarrow & N & \rightarrow & P & \rightarrow & 0\\ &&&& \downarrow  &  &  \downarrow   &&  \downarrow &&\\ 0 & \rightarrow  &  K & \rightarrow  & M^\prime & \rightarrow  &  N^\prime& \rightarrow  & P^\prime & \rightarrow  & 0\\ &&&& \downarrow  &  &  \downarrow   &&  \downarrow &&\\
&&&& 0 &  & 0   && 0 && \end{array}

Again I have no idea how to do, it seems so complicated, please helps.

答:
HINT The snake lemma tells us that there is an exact sequence $$0 \to L \to M^\prime.$$

HINT 2 Kernels of $R$-modules have a universal property.

Added: The universal property of a kernel is this. A kernel of a morphism $f:M \to N$ is a morphism $i:K \to M$ such that $f \circ i = 0$ and such that is universal with respect to this property, meaning that if $i^\prime:K^\prime \to M$ is any other morphism with $f \circ i^\prime = 0$, then there exists a unique morphism $u:K^\prime \to K$  such that $i^\prime = i \circ u$.

By standard category theoretic considerations, the universal property implies that any two kernels are uniquely isomorphic. For more on this, see the Wikipedia article on kernels.


答:
Hint

that can reduce significantly the computation effort... But still some will be required!

Notice that the roots of $f$ are the cubic roots of $4$ and $-2$.

In less that 1 second:

```<code>from sympy import resultant
from sympy.abc import x

R = resultant(x ** 6 - 2 * x ** 3 - 8, x ** 2 + 2 * x + 4)
print(R)
</code>```

$R = 1600$

Title: Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\prod_{n=1}^{6} (g(a_n))$

Body:
Question:

Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\displaystyle \prod_{n=1}^{6} (g(a_n))$

My process:

I first thought of setting $z=a_i^2+2a_i+4$ and then putting the value back into $f(z)$ and then from there finding the product using Vieta's. But the process turned incredibly long and tedious and I found it to be impossible to do by hand.

So I just went to to the brute force method, expanding the whole expression out. Writing all the terms would be really tedious so I would like to abbreviate it. Let $P= a_1 \cdot a_2 \cdots \cdot a_6$ and let $e_n$ denote the sum of roots taken $n$ at a time.

Here is the expression I got:

$$\prod_{i=1}^{6} (a_i^2+2i+4)=P^2+2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P+2^5*4*e_5+2^4*4^2*e^4+2^3*4^3*e_3+2^2*4^4*e_2+2*4^5*e_1+4_6+4(\sum_{cyc}(a_1a_2a_3a_4a_5)^2))^+4^2(\sum_{cyc}(a_1a_2a_3a_4)^2)+4^3\sum_{cyc}(a_1a_2a_3)^2+4^4\sum_{cyc}(a_1a_2)^2+4^5\sum_{cyc}(a_1)^2$$

Which is a monster in its own right. However, I was extremely interested in the patterns that kept popping up. Like the $2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P$ where the power of 2 and the index of $e$ keeps increasing and decreasing with each other. Therefore I wondered if there could be any general formula for any $f(x)$ and $g(x)$.

And that is my question, is there any way to find the answer efficiently without all this unnecessary calculation; and also can this process be generalised to any polynomials $f(x)$ and $g(x)$. And if it cannot be generalised, then is there any sort of algorithm or methodical way that one can take while solving this type of problem?

Body:
Hint

that can reduce significantly the computation effort... But still some will be required!

Notice that the roots of $f$ are the cubic roots of $4$ and $-2$.

In less that 1 second:

from sympy import resultant
from sympy.abc import x

R = resultant(x ** 6 - 2 * x ** 3 - 8, x ** 2 + 2 * x + 4)
print(R)

$R = 1600$


问:
Show, that if the integral of f is non-negative for all subsets, then f is non-negative for mu-almost every x in X
I've been struggling with part two of a problem for a while now, and is hoping someone can help me out.

The problem goes.

Let $(X, \mathcal{E}, \mu)$ be a measurable space and let $f$ be a function from $\mathcal{L}_{\mathbb{C}}^{1}(\mu)$

(a) Show, that if $\int_{A} f d \mu \in \mathbb{R}$ for all $A$ in $\mathcal{E},$ then $f(x) \in \mathbb{R}$ for $\mu$ -almost all $x$ in $X$

(b) Show, that if $\int_{A} f d \mu \in[0, \infty)$ for all $A$ in $\mathcal{E},$ then $f(x) \in[0, \infty)$ for $\mu$ -almost all $x$ in $X$

I have already shown part (a), but am stuck on part (b).

So far, I have used that $\int_A f d\mu$ = $\int_A f^+ d\mu + \int_A f^- d\mu \geq 0$. So I want to show that if
$$
\int_A f^+ d\mu \geq \int_A f^- d\mu \quad \forall A \in \mathcal{E}
$$
Then $f\geq0$ for $\mu$-almost every x in X.

But I haven't really been able to find a good way of showing this.

Thanks in advance

答:
Although the link provided in the comments answers perfectly the question in a more general context i'll give a direct (standard) argument adjusted to the OP's question.

Let $A=\{x\in X:\, f(x)<0\}=\bigcup_{n=1}^{\infty}A_n$ where $A_n=\{x\in X:\, f(x)<-1/n\}$. We claim that $\mu(A_n)=0$ for all $n$. Indeed, if this wasnt the case, then $\mu(A_{n_0})>0$ for some $n_0$. Then, $f(x)<-1/n_0$ for every $x\in A_{n_0}$ implies
$$\int_{A_{n_0}}f\,d\mu\leq -\frac{1}{n_0}\mu(A_{n_0})<0$$
But since $A_{n_0}\in \mathcal{E}$ this contradicts the fact that $\int_{A_{n_0}}f\,d\mu \geq 0$. Hence, $\mu(A_n)=0$ for all $n$. And now, $A_n$ being the union of $A$ we get $\mu(A)=0$. Hence, $f\geq 0$ for almost all $x\in X$.


问:
Showing collection of subsets of R is a topology
Show that the collection of subsets of $\Bbb R$

$\mathscr T =\{(a,+\infty)|a ∈ \Bbb R\} ∪ \{\Bbb R\}  ∪ \{\emptyset\}$
is a topology i.e. a runs over all real numbers, i.e. for every a we have $(a,+\infty) \in \mathscr T $.

Show that $( \Bbb R, \mathscr T )$ is not a Hausdorff topological space

So I think I should be basing my answer for first part like part c from Prove that the following collection of subsets of $\Bbb R$ is a topology? but i'm not really sure, can anyone give any hints to help?

答:
Let  $\mathscr T =\{(a,+∞)|a ∈ \Bbb R \} ∪ \{\Bbb R\}∪\{\emptyset\}$

$\emptyset$, $\Bbb R \in \mathscr T$ by definition

Let $\{S_i\}_{i \in I}$ with $S_i \in \mathscr T$. If $\exists i \ $ s.t. $S_i=\Bbb R$ then $\cup_{i \in I}S_i=\Bbb R \in \mathscr T$. If $S_i\neq \Bbb R \  \forall i$ then $S_i=(a_i, +\infty)$ and so $\cup_{i \in I}S_i= (\inf a_i, +\infty) \in \mathscr T$ ($\inf$ can be $-\infty$

Let $S_1, S_2 \in \mathscr T$. If $S_i=\mathbb R$ or  $S_i=\emptyset$ $S_1 \cap S_2 \in \mathscr T$ trivially. Else $S_i=(a_i, +\infty)$ and $S_1 \cap S_2=(\max a_i,+\infty) \in \mathscr T$

$\mathscr T$ is not $T_2$: let $x,y\in \Bbb R$ s.t. $x < y$. Let $A \in \mathscr T$ an open not empty set s.t. $x\in A$. If $A= \Bbb R$ then $y \in A$ if $A=(a, +\infty)$ then $y\in A$. So you can't find $A,B$ disjoint open neighborhoods of $x$ and $y$


问:
Maximum value of $\sin(A/2)+\sin(B/2)+\sin(C/2)$?
So I came across a question in my textbook:

In triangle ABC, if $A$,$B$,$C$ represent angles, then find the maximum value of $\sin(A/2)+\sin(B/2)+\sin(C/2)$?

So I have already tried and best and put my blood,sweat and tears into this question..But I'm not able to go solve further!

So here's my approach:
By Using $\sin(C)+\sin(D)$ and $A+B+C= \pi$;

$2\sin(\frac{A+B}{4})\cos(\frac{A-B}{4})+\cos(\frac{A+B}{2})$
Now,
Using $\cos(2A)$ formula i.e, $1-2\sin^2(A)
$

$2\sin(\frac{A+B}{4})\cos(\frac{A-B}{4})+1-2\sin^2(\frac{A+B}{4})$

So I got quadratic in variable $\sin(\frac{A+B}{4})$

$-2\sin^2(\frac{A+B}{4})+2\sin(\frac{A+B}{4})\cos(\frac{A-B}{4})+1$

But I dunno what to do After that

Can I solve this question using this method or I have to use a different approach!

BTW, the answer is 3/2

Edit:I have just finished my high school and preparing for entrance exam IIT-JEE,So please don't use hard terms to solve this question.

This solution is sent by my teacher, atleast make me understand this one
[https://i.stack.imgur.com/51pCB.png]

答:
Since $\sin x$ is concave on acute $x$, by Jensen's inequality the maximum is found at $A/2=B/2=C/2=\pi/6$, as $3\sin\pi/6=3/2$.

Edit: since the OP mentioned in a comment on @B.Goddard's answer that they know differentiation, here's another proof the equilateral case achieves a maximum:

Keep using $\frac{C}{2}=\frac{\pi}{2}-\frac{A+B}{2}$. To extemize $\sin\frac{A}{2}+\sin\frac{B}{2}+\cos\frac{A+B}{2}$ simultaneously solve$$\tfrac12\cos\tfrac{A}{2}-\tfrac12\cos\tfrac{C}{2}=0,\,\tfrac12\cos\tfrac{B}{2}-\tfrac12\cos\tfrac{C}{2}=0$$viz. $A=B=C$. I'll leave the reader to check it's a maximum by considering second derivatives.


问:
Proving that for any any $n$ distinct real number the number of multiplications needed to compute their product is $n - 1$
I am studying for one of my exams, and and this question appeared in the practice problems for the exam. I tried proving it but got stuck midway. What I tried was using strong induction to induct on the number of brackets in a given expression. However, I am not quite sure how to build my proof based on this. I was able to do the base case and inductive hypothesis, but when it came to the inductive case, I was unsure how to proceed. Would inducting on the number of brackets be helpful in this scenario?

Any sort of help will be immensely appreciated.

答:
Strong induction.

Suppose we know that any group of $k < n$ terms the product will be the same no matter which order we do the operations in.

So if we have any $(a_1.....a_k)(a_{k+1}.....a_n)$ where $k <n$ then in wont matter which order we do them within parenthesis.

So that mean every way of arranging $n$  terms will have the same result as any other way of arranging them provided the two largest subgroupings are broken into the same to groups of $k$ and $n-k$ terms.

[For example:  $(a_1a_2a_3)(a_4a5a6)$ will be the same value whether $a_1a_2a_3$ is done as $a_1(a_2a_3)$ or as $(a_1a_2)a_3$ and the same for how we arrange $a_4a_5a_6$. And we know that $(a_1a_2)(a_3a_4a_5a_6)$ will be the same no matter how we arrange that $a_3a_4a_5a_6$.  But we currently DON'T know if $(a_1a_2)(a_3a_4a_5a_6)$ must be the same as $(a_1a_2a_3)(a_4a5a6)$.]

[That will be the next thing we have to show.  That $(a_1.....a_j)(a_{j+1}....a_n) = (a_1.....a_k)(a_{k+1}.....a_n)$ for any $j$ and $k$.  Will prove that with induction.  Basic induction.]

Now as it doesn't matter how we arrange the $a_1...a_k$ or the $a_{k+1}...a_n$ terms we have

$(a_1....a_k)(a_{k+1}....a_n)=$

$(\color{red}{(a_1....a_{k-1})}\color{blue}{a_k})\color{green}{(a_{k+1}....a_n)}$ is a product with three terms so

$(\color{red}{(a_1....a_{k-1})}\color{blue}{a_k})\color{green}{(a_{k+1}....a_n)}=$

$\color{red}{(a_1....a_{k-1})}(\color{blue}{a_k}\color{green}{(a_{k+1}....a_n)})=$

$(a_1....a_{k-1})(a_k.... a_n)=(a_1....a_k)(a_{k+1}....a_n)$.

And likewise $(a_1....a_k)(a_{k+1}....a_n)=$

$(\color{red}{(a_1....a_{k})}(\color{blue}{a_{k+1}}\color{green}{(a_{k+2}....a_n)})$

$(\color{red}{(a_1....a_{k})}\color{blue}{a_{k+1}})\color{green}{(a_{k+2}....a_n)})=$

$(a_1....a_{k+1})(a_{k+2}.... a_n)=(a_1....a_k)(a_{k+1}....a_n)$.

And thus by induction every way of arranging $n$ terms regardless of how you split them into two groups will have the same result.


问:
An identity for $\binom{-1/2}{n}$
I would like to show the formula
$$\forall n \in \mathbb{N},\quad\sum_{p=0}^n \binom{n}{p} \frac{(-1)^p}{p!} \prod_{k=0}^{p-1} \left(n+\frac{1}{2} + k\right) =  \binom{-1/2}{n} \, , $$
which is true according to Mathematica. I tried to use the identity
$$\binom{-1/2}{n} = \frac{1}{n!} \sum_{k=0}^n s_{n,k}\left(-\frac{1}{2}\right)^k \, ,$$
where $s_{n,k}$ denotes the Stirling numbers of first kind, but without any success. We end up with the following identity to prove
$$ \frac{1}{n!} \sum_{p=k}^n \binom{n}{p}\frac{s_{p,k}}{p!} \left(n+\frac{1}{2}\right)^k = s_{n,k} \, , $$
which is not a piece of cake for me neither. Among all the formulas I could find in Concrete Mathematics book from Ronald Graham, Donald Knuth, and Oren Patashnik, none comes in a handy. I confess I don't have a large culture of this field.

I will be thankfull for any hint.

答:
More than a hint: Recall that the falling factorial is defined as $x^\underline{k}=\frac{x!}{(x-k)!}=x(x-1)\cdots (x-k+1),$ and the rising factorial is defined as $x^\overline{k}=\frac{(x+k-1)!}{(x-1)!}=x(x+1)\cdots (x+k-1).$

This is the binomial theorem
$$(x+y)^n=\sum _{k=0}^n\binom{n}{k}x^ky^{n-k}.$$
The beautiful part is that this keeps happening for
$$(x+y)^{\underline{n}}=\sum _{k=0}^n\binom{n}{k}x^{\underline{k}}y^{\underline{n-k}},$$ notice that if you take $x=n,y=-1/2-n,$ then
$$(-1/2)^{\underline{n}}=\sum _{p=0}^n\binom{n}{p}n^{\underline{n-p}}(-1/2-n)^{\underline{p}}.$$
Then you have to check that $(-x)^{\underline{k}}=(-1)^k(x)^{\overline{k}}.$

Title: An identity for $\binom{-1/2}{n}$


问:
Mind too small for quadratic equation problem
I was struggling for a long time with the following question, eventually unable to find it. I'm really frustrated with it because the answer is not difficult per se but to me requires a complete shift in approaching the problem, and I am just unable to make such a leap by myself. So, is it a hard question or am I just stupid?

Here goes: Two pipes together can fill a reservoir in 6 hr 40 min. Find the time each alone will take to fill the reservoir if one of the pipes can fill it in 3 hr less time than the other.

Edit: The solution is in my textbook so not quite asking for that. Once you formulate the quadratic equation it's straightforward but for me this step really takes a leap of the imagination to find it and I wonder if I'm alone in this. Can't really post the answer because then it'll be obvious.

So basically I'm asking if anyone can formulate the equation and tell me if they found it difficult.

Thanks for the willingness to answer already!

答:
We are looking at the percentage of the reservoir that is full as a function of time, call the function $p_i(t)$, $0 \leq p_i(t) \leq 100$, where $i=1,2$ standing for pipe 1 or pipe 2. Note that regardless of the combination of pipes, the reservoir starts empty, so $p_1(0)=p_2(0)=0$.

One of the pipes fills the reservoir to $100$ after a certain period of time, call it $T$:
$$ p_1(T) = 100  .$$
The other pipe takes 3 hours less:
$$ p_2(T-3) = 100 .$$
So we can equate the two equations to see
$$ p_1(T) = p_2(T-3) . \quad (*)$$

Now assuming that in each pipe the water flows at a linear rate, we have that for any point in time $0 \leq t \leq T$,
$$ p_1 (t) = \left( \frac{p_1(T) - p_1(0) }{T-0} \right) (t-0) - p_1(0) = \frac{100t}{T} $$
and similarly, in combination with $(*)$,
$$ p_2 (t) = \frac{100t}{T-3}. $$

Finally we use the information that both pipes together fill the reservoir in $t=6+(4/6)=40/6=20/3$ hours.  So
$$  100 = p_1 (20/3) + p_2 (20/3) = (2000/3) \left( \frac{1}{T} + \frac{1}{T-3} \right) $$
Now solving for $T$ gives you your quadratic and your answer


问:
How to prove that $T \mapsto \exp (T)$ is continuous?

Let $\mathcal L (E)$ denote the space of all bounded linear operators on a normed linear space $E.$ Show that the map $\varphi : \mathcal L (E) \longrightarrow \mathcal L (E)$ defined by $T \longmapsto \exp (T),\ $ $T \in \mathcal L (E)$ is continuous.

I tried in the following way $:$

$$\begin{align*} \|\exp (T) - \exp (T_0) \|_{\text {op}}  & = \left \|\sum\limits_{j=0}^{\infty} \frac {T^j} {j!} - \sum\limits_{j=0}^{\infty} \frac {T_0^j} {j!} \right \|_{\text {op}} \\ & = \left \| \sum\limits_{j=0}^{\infty} \frac {T^j - T_0^j} {j!} \right \|_{\text {op}} \\ & = \left \| \sum\limits_{j=0}^{\infty} \frac {(T - T_0) (T^{j-1} + T^{j-2} T_0 + \cdots + T_0^{j-1} )} {j!} \right \|_{\text {op}} \\ & \leq \|T - T_0\|_{\text {op}} \left \| \sum\limits_{j=0}^{\infty} \frac {(T^{j-1} + T^{j-2} T_0 + \cdots + T_0^{j-1} )} {j!} \right \|_{\text {op}}  \end{align*}$$

Now if we can make the quantity $\displaystyle \left \| \sum\limits_{j=0}^{\infty} \frac {(T^{j-1} + T^{j-2} T_0 + \cdots + T_0^{j-1} )} {j!} \right \|_{\text {op}}$ bounded then we are through. But how do I do that? Any help in this regard will be appreciated.

Thanks in advance.

答:
If $\|T\|$ and $\|T_0\|$ are both $<M$ then for the tail we have
$$\left\|\sum_{j=N}^\infty\frac {T^j}{j!}-\sum_{j=N}^\infty\frac {T_0^j}{j!}\right\|\le 2\sum_{j=N}^\infty\frac {M^j}{j!}$$
and can be made $<\frac \epsilon2$, say, if only you make $N$ big enough. What is left, is polynomial.


答:
You can do it with Lagrange multipliers.  Maximize $f=\sin x/2 + \sin y/2+\sin z/2$  under the constraint $g=x+y+z = \pi$.

Then

$$\nabla f = \langle \cos(x/2)/2, \cos(y/2)/2, \cos(z/2)/2
\rangle =\lambda\langle 1,1,1 \rangle = \nabla g.$$

This shows that $x=y=z$ and the maximal triangle is equilateral.

Title: Maximum value of $\sin(A/2)+\sin(B/2)+\sin(C/2)$?

Body:
So I came across a question in my textbook:

In triangle ABC, if $A$,$B$,$C$ represent angles, then find the maximum value of $\sin(A/2)+\sin(B/2)+\sin(C/2)$?

So I have already tried and best and put my blood,sweat and tears into this question..But I'm not able to go solve further!

So here's my approach:
By Using $\sin(C)+\sin(D)$ and $A+B+C= \pi$;

$2\sin(\frac{A+B}{4})\cos(\frac{A-B}{4})+\cos(\frac{A+B}{2})$
Now,
Using $\cos(2A)$ formula i.e, $1-2\sin^2(A)
$

$2\sin(\frac{A+B}{4})\cos(\frac{A-B}{4})+1-2\sin^2(\frac{A+B}{4})$

So I got quadratic in variable $\sin(\frac{A+B}{4})$

$-2\sin^2(\frac{A+B}{4})+2\sin(\frac{A+B}{4})\cos(\frac{A-B}{4})+1$

But I dunno what to do After that

Can I solve this question using this method or I have to use a different approach!

BTW, the answer is 3/2

Edit:I have just finished my high school and preparing for entrance exam IIT-JEE,So please don't use hard terms to solve this question.

This solution is sent by my teacher, atleast make me understand this one
[https://i.stack.imgur.com/51pCB.png]


答:
Searching OEIS with @saulspatz first values, we can find that:

$$E_n = \frac{a(n)}{b(n)}$$

where $a(n)$ is OEIS A158466 and $b(n)$ is OEIS A158467. At OEIS A158466 you can find also the following formulas:

$$E_n = -\sum_{k=1}^n (-1)^k \frac{{n \choose k}}{1-\frac{1}{2^k}}$$

$$E_n = \sum_{k=1}^{\infty} k \left(\left(1-\frac{1}{2^k}\right)^n - \left(1-\frac{1}{2^{k-1}}\right)^n\right)$$

and thus (see here):

$$E_{100} \approx 7.983801535$$

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:
Circulation and line integrals.
The following is the problem I'm working on.

If $\overrightarrow {F} = <x,y^2z,-xy^2>$, calculate the circulation of $\overrightarrow {F}$ over the surface $z=x^2+y^2$ bounded by $C$ using a line integral. $C$ is given by $$C_1 : \overrightarrow {r}=<t,t,0>  (0\le t \le 2\sqrt{2})$$ $$C_2 : \overrightarrow {r} = <4\cos(t),4\sin(t),0> ({\pi \over 4} \le t \le \pi)$$ $$C_3 : \overrightarrow {r} = <t-4,0,0> (0 \le t \le 4)$$

I tried to solve it using $\int_C \overrightarrow {F}・d\overrightarrow {r}$, but the integral seems to be too complex and I don't think I'm doing this right.

I don't think neither Green's theorem or the Fundamental theorem of line integrals are applicable either because I don't think the curve is closed or $\overrightarrow {F}$ not seeming to be conservative.

Can someone explain this problem ?
I don't need much of the calculation. It's more of a conceptual idea.

It would also be great if someone could explain what it physically represents.

答:
The curve $C$, as given, is a curve in the plane (it traces out the boundary of a 'wedge of pie'); the first thing to do is lift it to the surface using the equation $z = x^2 + y^2$.

The curve is given as three smooth pieces, so you will need to calculate three separate line integrals; these become ordinary integrals over $t$ via
$$
d\vec{r} = \left(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\right)dt
$$

(Edit: I'm guessing by your question that you know the following, but I'll add it for completeness.  By Stokes' theorem, the line integral will be equal to the surface integral
$$
\int\!\!\!\!\int_S (\nabla\times\vec{F})\cdot\vec{n}~ dA
$$
where $S$ is the part of the surface bounded by the curve, $\vec{n}$ is a unit normal vector field, and $dA$ is the area element.)

Title: Circulation and line integrals.


问:
Are ratios preserved when multiplied infinitely?
If I offered to give you \$10 but steal $5 from you, you should take that deal. You should take it once, and again, and again. You will always gain money no matter how many times you take it and the ratio of money gained to money lost is constant.

But, what if you took the deal an infinite number of times? You will have both gained and lost an infinite amount of money. It's not as if you can have gained 10∞ and lost 5∞. The infinities have to cancel out, right?

But, then, at each step, the ratio of money gained to money lost is unchanged but, after an infinite amount of time, it is changed. That's clearly impossible, so what actually happens then?

What am I thinking about wrong here?

答:
This is the Ross-Littlewood paradox. The resolution I favor is the third one in the Wikipedia article; if I take the deal an infinite number of times, the amount of money I have remaining depends on the details of exactly which $5$ dollars you choose to steal from me at each step. This is surprising because if I only take the deal finitely many times it doesn't matter. Depending on exactly which dollars you choose to steal from me, I can end up with any finite amount of money left (including zero) or an infinite amount of money left.

Edit: Among other things, this means calculating the ratio of money gained to money lost is not a good way to understand what actually happens after taking the deal an infinite number of times.

Here's an example. Let's say all the dollar bills involved have serial numbers, starting $1, 2, 3, \dots$, and

whenever the you give me dollar bills you give me the ones with the smallest serial numbers you haven't given me yet, and also

whenever you take dollar bills, you take from me the ones with the smallest serial numbers.

Then it's not hard to see that you eventually steal every dollar bill from me, which means at the end I end up with no money. This is despite the fact that at every finite step I end up with more and more money!

On the other hand, suppose I changed the second bullet point to "whenever you take dollar bills, you take from me the ones with the largest serial numbers." Then it's not hard to see that I end up with infinitely many dollars. I'll leave it as an exercise to figure out how to modify this so that I end up with any particular finite amount of dollars.

(This is not just a mathematical curiosity either; it can be turned into an example where a limit of integrals is not an integral of (pointwise) limits, which explains why the hypotheses of the dominated convergence theorem are necessary.)


问:
Is this function of a sum of indicator functions convex?
let $x=x_1,x_2,...,x_n$ for $x \ge 0$,

let $I(x) =max(x_i, 0) $ for $i \in [1,n]$

$f(x) = I(c-x)a^T+I(x-c)b^T$ where $a,b,c$ are constant vectors

Is this a convex function? How is this proven?

答:
Yes, $f$ is a convex function. You just need to know some some basic properties that imply convexity:

Linear and convex functions are convex.

The maximum of convex functions is convex.

The composition of a convex function with an affine map is convex.

Nonnegative linear combinations of convex functions are convex.

From that we have:

The functions $x_i$ and $0$ are convex.

The components of $I(x)$, namely $\max(x_i,0)$ are convex.

The components of $I(c-x)$ and $I(x-c)$ are convex.

The nonnegative combinations $I(c-x)a^T$ and $I(x-c)b^T$ are convex, provided $a,b\ge0$. Likewise $f$ itself is convex.


问:
Need help with taylor polynomial of n-th degree
Lets assume that we calculate Taylor polynomial of $n^{\mathrm{th}}$ degree around the point $a=0$ for the function $\sin(x)$. How do I find the smallest value of $n$ that I can use so that the residual $R_{n}f(x) \leq 0.01, \, \forall x \in [0,1/3]$? And what is that value as well?

答:
$$\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$
is an alternating series, therefore the error is less than the absolute value of the first term neglected. In other words you should find $n$ such that
$$\frac{\left(\frac13\right)^{2n-1}}{(2n-1)!}\le 0.01$$
we get that $n=2$ terms are enough to keep the error less than $0.01$. Actually with two terms we get $\sin \frac13\approx \frac{53}{162}\approx 0.32716$ with an error less than $0.0001$ which means $4$ decimal exact
$$
\begin{array}{l|r}
 n & \texttt{error}\\
\hline
 1 & 0.333333 \\
 2 & 0.00617284 \\
 3 & 0.0000342936 \\
 4 & 9.1\times 10^{-8}\\
\end{array}
$$


问:
Can someone help me with this limit?
$ \lim_{x \to 0}x^{|\tan x |} \overset{??}{=} 1 ??$

If you consult the graph it is intuitive to think that when $ x \to 0 $ the function tends to $1$. I have tried to lower the exponent with logarithms and apply L'Hospital's rule, but nothing. The only solution I see is to apply infinitesimal equivalents, which I have not tried yet.

Thank you all!

Title: Can someone help me with this limit?

Body:
$$L=\lim_{x \to 0}x^{|\tan x |} \overset{??}{=} 1 ??$$
Take log
$$\ln L =\lim_{x \to 0}  {\ln x} \,{\tan x}$$
$$\ln L =\lim_{x \to 0} {\ln (1-(x+1))}\, {\tan x}$$
$$\ln L =\lim_{x \to 0}  {-(x+1)} \,{ x}$$
$$\ln L= 0 \implies L=e^0=1$$


问:
prove $a c^{-1} \equiv b d^{-1} \bmod m$
if $a \equiv b \bmod m$ and $c \equiv d \bmod m$ and $gcd(c,m) = 1$ prove that $a c^{-1} \equiv b d^{-1} \bmod m$ .

答: If $\gcd(c,m)=1$, then there exists $c^{-1}$ such that $cc^{-1}=1\pmod{m}$. Since $c\equiv{d}\pmod{m}$, $\gcd(d,m)=1$ and $dc^{-1}=1\pmod{m}$, and it's clear that $c^{-1}=d^{-1}\pmod{m}$. Then, it's obvious that $ac^{-1}=bd^{-1}\pmod{m}$.


问:
Expectation and variance of (this) yet another stochastic process
(Personal note: there is an incredible amount of questions whose title consists in some permutation of the words I chose for the title, and I guess someone may have already answered. I swear I looked many of them up to see if someone did. If I missed it, I am sorry for repeating it.)

Be $X_t:=\int_0^t e^s\text d W_s$, and $Y_t:=\int_0^t X_s\text dW_s$. Find $\mathbb E\left[Y_t\right]$ and $\mathbb V\left[Y_t\right]$.

I see that $Y_t$ must be a martingale because it is just the martingale part of an Itô integral (even if the integrand is given as an Itô integral?), so I state $\mathbb E\left[Y_t\right]=0$.

That said,
$\mathbb V\left[Y_t\right]=
\mathbb E\left[\left(\int_0^t X_s\text dW_s\right)^2\right]=
\mathbb E\left[\int_0^t X_s^2\text d s\right]$ by Itô isometry.
But by Itô lemma,
$$\text d X_s = e^s\text d W_s\Rightarrow\text d (X_s^2)=2X_s\text d X_s+\frac 12 e^{2s}\text ds=2\int_0^s e^u\text d W_u e^s\text d W_s+\frac 12 e^{2s}\text ds.$$

i) is the general reasoning sound? To me, everything seems to make sense, but I am absolutely not sure.

ii) How do I go past the expression for the variance?

答:
I am assuming there is a final time horizon $T.$

For a suitable integrand $f$ (that make the integral well-defined), the Ito integral (as a process) $\int_0^t f(s)dW_s, 0 \leq t \leq T,$ is a local martingale, not necessarily a martingale. For the Ito integral to be a (square-integrable) martingale you need the integrand  to satisfy $E\int_0^T |f(s)|^2 ds < \infty.$ In your case this condition is obviously satisfied since $e^s$ is a continuous function. Thus the process $X$ is a square-integrable martingale. Therefore, you have $E[Y_t]=0$ as you said.

For the variance, note that since $e^s$ is a deterministic function, we have that $\int_0^t f(s)dW_s \sim N(0, \int_0^t e^{2s}ds).$ Thus
$$E[X^2_t]= \int_0^t e^{2s}ds = \frac{1}{2}(e^{2t} -1).$$
Also, by Ito isometry and Fubini's theorem
$$var[Y_t]=E\left[\left(\int_0^t X_s dW_s \right)^2\right] = E\left[\int_0^t X^2_s d_s \right]=\int_0^t EX^2_s d_s=...$$


Body:
In a triangle ABC, $A+B+C=\pi$
$$f(x)=\sin(x/2) \implies f''(x)=-\frac{1}{4}\sin(x/2)<0, x\in[0,2\pi].$$ So by Jemsen's inequality $$\frac{f(A/2)+f(B/2)+f(C/2)}{3} \le f(\frac{A+B+C}{6}).$$
$$\implies \frac{\sin (A/2)+\sin(B/2)+\sin{C/2}}{3} \le \sin\frac{\pi}{6}.$$
$$\implies \sin(A/2)+\sin(B/2)+\sin(C/2) \le \frac{3}{2}$$


问:
When is the sum of the inverses of a finite, nondecreasing sequence of nonnegative numbers $1$?
This question is a simple generalization of a math problem I saw somewhere.

Consider the family of sets
$$
A_n \equiv \{(a_1, \ldots, a_n)\in \mathbb{Z}_{\geq 1}^n \, : \,  a_1 \leq \cdots \leq a_n \text{ and }  \sum_{k = 1}^n \frac{1}{a_k} = 1\}
$$

I think there are many interesting questions that could be asked about these $A_n$. Here are the ones I'm curious about:

Is there a paper or standard reference on these $A_n$ I should know about?

How fast does $|A_n |$ grow?

What's the best way to calculate the elements of $A_n$?

If Math SE etiquette requires pinning me down to one question, I'm most interested in reading about this problem elsewhere, assuming it has already been written about (which I'm sure is a safe assumption).

So far, I've come up with an algorithm for calculating the $A_n$. Suppose $a_1, \ldots, a_k$ are given with $k < n$. Then
$$
\sum_{j = k + 1}^n \frac{1}{a_j} = 1 - \sum_{j = 1}^k \frac{1}{a_j} > 0
$$
(If the RHS is $\leq 0$, there is no choice of $a_{k + 1}$ that works, and we are done).

But
$$
\sum_{j = k + 1}^n \frac{1}{a_j} \leq \frac{n- k}{a_{k + 1}}
$$
since $1 \leq a_1 \leq \cdots \leq a_n$. Thus
$$
\frac{n-k}{a_{k + 1}} \geq 1 - \sum_{j = 1}^k \frac{1}{a_j}
$$
Therefore
$$
a_{k + 1} \leq \frac{n - k}{1 - \sum_{j = 1}^k \frac{1}{a_j}}
$$
That is
$$
a_{k+1} \in \left\{n \in \mathbb{Z} : a_k \leq n \leq \left\lfloor \frac{n - k}{1 - \sum_{j = 1}^k \frac{1}{a_j}}\right\rfloor\right\} \equiv B_{k + 1}
$$
where $B_{k + 1}$ is clearly finite (and depends on $a_1, \ldots, a_k$). We may then pick a value for $a_{k +1}$ from $B_{k + 1}$ and similarly constrain the values for $a_{k + 2}$. We continue in this way until we get $B_j = \emptyset$ for some $j$ or until we've checked every element of $B_n$. We then select a different value for $a_{k + 1}$ from $B_{k + 1}$ and repeat. Once we've exhausted all the possibilities in $B_{k + 1}$, we make a different choice for $a_k$ from $B_k$ and so on until we've checked every possibility. Note that we always have $a_1 \in \{1, \ldots, n\}$, so there are definitely only finitely many possibilities to check.

Computer science is not my specialty. I'm sure there's a more elegant and rigorous way to describe this algorithm, but hopefully you get the idea. In particular, I hope it's clear that $A_n$ is finite and can be calculated by performing a finite (though possibly very large per this algorithm) number of computations.

If you're curious, the first few $A_n$ are:
\begin{align}
A_1 &= \{(1)\}
\\A_2 &= \{(2,2)\}
\\A_3 &= \{(2, 3, 6), (2, 4, 4), (3,3,3)\}
\end{align}

答:
Regarding question $2$ see OEIS A002966. There are a lot of sequences at OEIS related to Egyptian fractions along with references to papers discussing algorithms to compute decompositions.

In case you are interested on the particular case where the $a_i$ are powers of $2$, see this question and this question.


问:
properties of recursively enumerable sets
$A \times B$ is an r.e.(recursively enumerable) set,  I want to show that $A$ (or $B$) is  r.e. ($A$ and $B$ are nonempty)

I need to find a formula. I've got an idea that I should use the symbolic definition of an r.e. set. That is, writing a formula for the function that specifies $A$ or $B$, assuming a formula exists for $A \times B$. I guess I must use Gödel number somewhere.

I should've mentioned that I asked this question over there (Computer Science) but since I am more interested in mathematical arguments and looking for formulas, I'd ask it here as well.

答:
If $A\times B$ is recursively enumerable then there exists a bounded formula $\varphi(x,y,z)$ such that $\langle a,b\rangle\in A\times B$ if and only if $\exists z\varphi(a,b,z)$ is true.

Then we have that $a\in A$ if and only if $\exists b\exists z\varphi(a,b,z)$.

Therefore $A$ is definable by a $\Sigma_1$ formula, and therefore recursively enumerable.


问:
Limit of difference of pseudo-arcsinh and arcsinh
$$\DeclareMathOperator{\erf}{erf}$$
$$\DeclareMathOperator{\arcsinh}{arcsinh}$$
Let $f$ a function defined by : $$f(x) = \int_0^x \frac{\erf(u)}{u}du$$Where $\erf$ is the error-function. This function look like $\arcsinh(x)$ :

In black there is $f(x)$ and in red this is $\arcsinh(x)$. Now, we can see (or at least conjecture) that $$\kappa := \lim_\limits{x \rightarrow +\infty} \left[ \int_0^x \frac{\erf(u)}{u}du - \arcsinh(x)\right]$$exists and is finite. That would imply that $f(x) \sim \kappa + \arcsinh(x)$ and more accuratly $f(x) \approx \kappa + \arcsinh(x)$. By using $\arcsinh(x) \sim \ln(2x)$ we would have $$f(x) \sim \kappa + \ln(2x)$$Now, heres the graph of $f(x)-\arcsinh(x)$ :

In black this is $f(x)-\arcsinh(x)$. We can easily see that it converges toward $\kappa$ (in red). Here, we have $\kappa \approx 0.288607829951$.

My question is, what is the exact value of $\kappa$ ?

答: $$\kappa=\underset{x\to \infty }{\text{lim}}\left(\int_0^x \frac{\text{erf}(u)}{u} \, du-\text{arcsinh } x\right)=\frac{\gamma }{2}\approx 0.2886078324507665$$


问:
Reference for specific values of incomplete elliptic integrals
In  Wolfram's website specific values of incomplete elliptic integral of first, second and third kind are given in terms of complete elliptic integrals of first, second and third kind.

For the incomplete integral of first kind $\text{F}(z, m)$, the site lists the following relations:

$$
\text{F}(\pi/2, m)= \text{K}(m)\\
\text{F}(\text{csc}^{-1}(\sqrt{m}), m)= \frac{1}{\sqrt{m}}\text{K}\left(\frac{1}{m}\right)
$$

Similarly for incomplete elliptic integral of second kind $\text{E}(z,m)$:

$$
\text{E}(\pi/2, m)= \text{E}(m)\\
\text{E}(\text{csc}^{-1}(\sqrt{m}), m)= \sqrt{m} \left( \text{E}\left(\frac{1}{m}\right)+ \left(\frac{1}{m}-1\right)\text{K}\left(\frac{1}{m}\right)\right)
$$

and for the third kind $\Pi(n,\, z,\,m)$:

$$
\Pi(n,\,\pi/2,\, m)=\Pi(n,\,m)\\
\Pi(n,\, \text{csc}^{-1}(\sqrt{m}),\,m)=\frac{1}{\sqrt{m}}\Pi\left(\frac{n}{m},\,\frac{1}{m}\right)
$$

I would like to use this identities in a calculation however the site does not list references. I would appreciate a comprehensive reference list for these relations. Thanks in advance.

答: A good list of relations involving elliptic integrals is to be found in the DLMF. Your examples are obtained by applying the reciprocal modulus/parameter transformation with the amplitude $\phi=\frac\pi2$, which makes the integral(s) on one side complete.


问:
A stochastic Process where $X_{T_A}\not \in A$. Must it always be in $\bar A$?
So, I was wondering how to define a stochastic process $X$, with continuous paths, such that $X_{T_A}\not \in A$, where $T_A=\inf\{t>0:X_t \in A\}$.

For this, I was thinking of $X$ mimicking the behaviour of $\sin(1/t)$...

Also, is it true that we always have $X_{T_A}\in \bar A$?

答:
Yes, we will always have $X_{T_A} \in \overline A$ because $X$ has continuous paths.  For a specific example, you could do something as simple as $X_t = t$ and $A = (1,2)$.

EDIT:  To prove $X_{T_A} \in \overline A$, from the definition of $T_A$ we have that there exist $(t_n)$ decreasing to $T_A$ such that $X_{t_n} \in A$.  Since $X$ has continuous paths, $(X_{t_n}) \rightarrow X_{T_A},$ and since $X_{T_A}$ is the limit of points in $A$ we have $X_{T_A} \in \overline A$.


答:
The ratio actually stays 2. This can be shown via the rule of L'Hospital. Being $x$ the number of "deals":$$\lim_{x \rightarrow \infty}\frac{10x}{5x} \stackrel{\text{L'H}}{=} \frac{10}{5}=2 $$
Rule of L'Hospital means here taking the first derivation of the numerator and denominator, yielding the ratio 2.

Title: Are ratios preserved when multiplied infinitely?

Body:
If I offered to give you \$10 but steal $5 from you, you should take that deal. You should take it once, and again, and again. You will always gain money no matter how many times you take it and the ratio of money gained to money lost is constant.

But, what if you took the deal an infinite number of times? You will have both gained and lost an infinite amount of money. It's not as if you can have gained 10∞ and lost 5∞. The infinities have to cancel out, right?

But, then, at each step, the ratio of money gained to money lost is unchanged but, after an infinite amount of time, it is changed. That's clearly impossible, so what actually happens then?

What am I thinking about wrong here?


问:
Probability of rolling a pair of non-6's and a 6 on 3 dice
So the problem is asking for the probability of rolling 3 dice and getting a pair of non-6s and a 6. This is what I have so far:

Total possibilities $= 6^3=216$

Total possible pairs  $={(1,1), (2,2), (3,3), (4,4), (5,5)}=5$

Total ways to arrange a pair and a 6 $= 3$

So the total probability I got was $(5*3)/216=15/216=5/72$

My question is whether the total possible pairs is 5 or 10? I know that if you roll a 2 and a 3, it is considered a different arrangement than rolling a 3 then a 2. Does this concept apply at all if you're rolling a pair? Thank you!

答:
Say you throw a red die and a blue die. If you get a 3 on the blue die and a 2 on the red die, that is different from getting a 2 on the blue die and a 3 on the red die.

If you get a 5 on the blue die and a 5 on the red die, is that different from getting a 5 on the blue die and a 5 on the red die?

The answer is no. There are only 5 pairs for you to consider, not 10.


问:
Find the area of $A = \{ \langle x,y\rangle \in \mathbb{R}^2 \mathrel| (x+y)^4<a x^2 y,\ x>0 \}$?
I can't really think of how to set the limits</p>

答:
there are two ways to find the area

use wolframalpha,you can see the picture at once.

let $x=rcos(\theta), y=rsin(\theta)$, then you get $r \le \dfrac{a*cos^2(\theta)sin(\theta)}{(cos(\theta)+sin(\theta))^4}$

since $x>0 \to -\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$,I think you can go further this way.

Edit:$ay>0$, if $a>0 \to y>0 \to 0< \theta < \dfrac{\pi}{2}$

the shape is here

EDIT: I add the process also for reference:

$S=\dfrac{1}{2} \int_{0}^ {\frac{\pi}{2}} r^2 (\theta)d \theta$  and

$ \int r^2 (\theta)d \theta=\int \left[\dfrac{a*cos^2(\theta)sin(\theta)}{(cos(\theta)+sin(\theta))^4} \right]^2 d \theta=a^2*\dfrac{-280 sin(\theta)-210 sin(3\theta)+14 sin(5\theta)-385 cos(\theta)+147 cos(3 \theta)+105 cos(5 \theta)+13 cos(7 \theta)}{6720 (sin(\theta)+cos(\theta))^7}+C$

$S=a^2*\dfrac{-280+210+14-(-385+147+105+13 )}{2*6720}=\dfrac{a^2}{210}$


问:
Show that for natural $n \ge 2$ :$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$
Show that for natural $n \ge 2$ the following does hold:

$$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$$

First solution:
$$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$$
$$\iff$$
$$\left(1-\frac{1}{n}\right)^{n}\left(1+\frac{1}{n}\right)^{n}>1-\frac{1}{n}$$
$$\left(1-\frac{1}{n}\right)^{n-1}\left(1+\frac{1}{n}\right)^{n}>1$$

By Bernoulli inequality:
$$\left(1-\frac{1}{n}\right)^{n-1}\left(1+\frac{1}{n}\right)^{n}\ge2\left(1-\frac{n-1}{n}\right)=\frac{2}{n}$$

Which is not useful.

Second solution:

$$\left(1-\frac{1}{n^{2}}\right)^{n}=\left(1-\frac{1}{n}\right)^{n}\left(1+\frac{1}{n}\right)^{n}\ge0\cdot2=0$$

Which is not useful.

Third solution:

$$\left(1-\frac{1}{n^{2}}\right)^{n}=\left(1-\frac{1}{n}\right)^{n}\left(1+\frac{1}{n}\right)^{n}\ge2\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{n-1}=\left(1+\left(-\frac{1}{2}\right)\right)^{n-1}$$$$\ge1-\frac{n-1}{2}=\frac{3-n}{2}$$

Final solution:

$$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$$
$$\iff$$
$$\left(1-\frac{1}{n^{2}}\right)^{n-1}\left(1+\frac{1}{n}\right)>1$$
By Bernoulli inequality:
$$\left(1-\frac{1}{n^{2}}\right)^{n-1}\left(1+\frac{1}{n}\right)>1-\frac{n-1}{n^{2}}=1-\frac{1}{n}+\frac{1}{n^{2}}\ge\frac{1}{2}+\frac{1}{n^{2}}$$

Which is true to claim that $$\frac{1}{2}+\frac{1}{n^{2}}>1-\frac{1}{n}$$
Since for natural $n \ge 2$ we have that $$\frac{1}{n}\left(\frac{1}{n}+1\right)>\frac{1}{2}$$

答:
Show that for natural $n \ge 2$ the following does hold:
$$\left(1-\frac{1}{n^{2}}\right)^{n}>1-\frac{1}{n}$$

By using the Bernoulli’s inequality

$(1+x)^{n}>1+nx$

for any natural $\;n\ge2\;$ and for any $\;x>-1\;,$

with $\;x=-\dfrac1{n^2}>-1\;,\;$

we get that

$\left(1-\dfrac{1}{n^{2}}\right)^{n}>1+n\left(-\dfrac{1}{n^2}\right)=1-\dfrac1n\;,$

for any natural $\;n\ge2\;.$


问:
Integral, Measurable Functions
Assume $\{f_n\}$ is a sequence of measurable functions on $[0,
1]$ and $\lim\limits_{n\to\infty} \int\limits_0^1 |f_n|=0$. Assume there is an integrable function $g$ on $[0,1]$ such that $|f_n|^2\leq g$ for all $n$. Show that $\lim\limits_{n\to\infty} \int\limits_0^1 |f_n|^2=0$.

Since the limit of the first integral is $0$, after some $n_0$ the functions $f_n$'s must be integrable. Then i tried to use Dominated Convergence Theorem and put the limit of the second integral inside the integral but couldn't continue with it. Any help is appreciated.

答:
You need to have a.s convergence to apply the dominated convergence. Your condition
$$\lim_{n\to \infty}\int_{0}^{1}|f_n|=0$$
assures that there exists a subsequence $f_{n_k}$ of $f_n$ that converges a.s to $0$. Indeed, observe that the above integral convergence implies that $f_n$ converges to $0$ in $L_1([0,1])$. Hence, we know that there is a subsequence $f_{n_k}$ that converges almost surely to $0$. Now if
$$\lim_{n\to \infty}\int_{0}^{1}|f_n|^2 \neq 0$$
There is a subsequence say $f_{n_k}$ of $f$ and $\epsilon>0$ such that
$$\tag{1} \int_{0}^{1}|f_{n_k}|^2\geq \epsilon$$
for every $k$. From the discussion above, we can extract a further subsequence of $f_{n_k}$ that converges a.s to $0$. Denote the subsequence of $f_{n_k}$ that converges a.s to $0$ again by $f_{n_k}$ for simplicity. Then, $|f_{n_k}|^2\leq g$ and $|f_{n_k}|^2\to 0$ a.s combined with the dominated convergence implies
$$\lim_{k\to \infty}\int_{0}^1|f_{n_k}|^2=0$$
which of course contradicts $(1)$.

Title: Integral, Measurable Functions


答:
On the condition that $T_A<\infty$.

By definition of $T_A$, there is a sequence of times $\tau_1,...,\tau_n$ such that:

$X_{\tau_n} \in A$

$\tau_n \longrightarrow T_A$

By the (right) continuity of $X$, $X_{T_A} = \lim X_{\tau_n}$.

By definition of $\overline{A}$, $\lim X_{\tau_n} \in \overline{A}$

Hence the conclusion.

Title: A stochastic Process where $X_{T_A}\not \in A$. Must it always be in $\bar A$?

Body:
So, I was wondering how to define a stochastic process $X$, with continuous paths, such that $X_{T_A}\not \in A$, where $T_A=\inf\{t>0:X_t \in A\}$.

For this, I was thinking of $X$ mimicking the behaviour of $\sin(1/t)$...

Also, is it true that we always have $X_{T_A}\in \bar A$?


问:
I'm asked to differentiate this $\dfrac{C}{2} \sum^{m}_{j' = 1} \| W^{j'} \|^{2}_{2}$ but I barely understand the notation.
I'm asked to differentiate

$$\dfrac{C}{2} \sum^{m}_{j' = 1} \| W^{j'} \|^{2}_{2},$$

according to $w^{j'}_{k}$ which is the $k$th weight of the vector of weight of $j'$.

It seems that $\| W^{j'} \|^{2}_{2}$ stands for the the $L2$ norm squared. This seems to indicate that we have:
$$\| W^{j'} \|^{2}_{2} = w^{2}_1 + w^{2}_2 + ... + w^{2}_k$$

Then to my understanding $\sum^{m}_{j' = 1} \| W^{j'} \|^{2}_{2}$ seems to indicate that we have:
$$\sum^{m}_{j' = 1} \| W^{j'} \|^{2}_{2} = (w^{2}_1 + w^{2}_2 + ... + w^{2}_k)_{j'=1} + (w^{2}_1 + w^{2}_2 + ... + w^{2}_k)_{j'=2} + ... + (w^{2}_1 + w^{2}_2 + ... + w^{2}_k)_{j'=m}$$

I have absolutely no clue how to derivate this formula.
Do I use the sum rule to get this?
$$\dfrac{\partial }{\partial w^{j'}_{k}} =\dfrac{C}{2} (2w_k)_{j'=1} + (2w_k)_{j'=2}  + ... + (2w_k)_{j'=m}$$

Edit:
This is from my machine learning course. This formula represents the regularization term we add to the loss function of an SVM (Support vector machines) in order to minimize the objective function.

Here $W$ is a vector containing scalars $w$.

答:
Use the standard basis vectors $\{e_k\}$ to extract the columns of the matrix, i.e.
$$W^j = We_j$$
Let's also use a colon to denote the trace/Frobenius product, i.e.
$$A:B \;=\; \sum_{i=1}^m\sum_{j=1}^nA_{ij}B_{ij} \;=\; {\rm Tr}(A^TB)$$
Then the objective function can be expressed in a form which is friendlier to matrix algebra.
$$\eqalign{
\phi &= \dfrac{C}{2} \sum^{m}_{j' = 1} \| W^{j'} \|^{2}_{2} \\
 &= \tfrac 12C\left(\sum_{j=1}^m We_j:We_j\right) \\
 &= \tfrac 12C\left(\sum_{j=1}^m e_je_j^T:W^TW\right) \\
 &= \tfrac 12C\Big(I:W^TW\Big) \\
 &= \tfrac 12C\,W:W \\
}$$
Now calculating the gradient is easy.
$$\eqalign{
d\phi &= \tfrac 12C\,(W:dW+dW:W) \\
 &= CW:dW \\
\frac{\partial\phi}{\partial W} &= CW \\
}$$
If you wish to extract individual components, simply pre/post multiply
by the basis vectors.
$$\eqalign{
e_i^T\left(\frac{\partial\phi}{\partial W}\right)e_j &= Ce_i^TWe_j \\
\frac{\partial\phi}{\partial W_{ij}} &= CW_{ij} \\
}$$


问:
A good book for learning mathematical trickery
I've seen several question here on what book to read to learn writing and reading proofs. This question is not about that. I've been doing that for a while, and I'm quite comfortable with proofs. I am looking for resources (books, ideally) that can teach not the concept of proofs, but rather some of the specific mathematical tricks that are commonly employed in proofs: those that mostly include clever number manipulation, ad-hoc integration techniques, numerical methods and other thing you are likely never to learn in theory-oriented books. I come mainly from applied math and engineering, and when I look at proofs from Stochastic Processes, Digital Signal Processing, Non-Linear Systems and other applied subjects, I feel like I need to learn a new method to understand every proof I read. Is there any good literature on such mathematical tricks?

答: The Tricki ("Trick Wiki") is an attempt to catalogue such things, although it is somewhat less successful than was initially hoped. </p>


问:
Parity relation for Legendre polynomials
Can anybody please tell me the parity relation for $(-1)^{-n} P_n(x)$. Note that $P_n(x)$ is the $n$th Legendre polynomial.

答:
Legendre polynomial $P_n(x)$ is given explicitly by Rodrigues' formula

$$
P_n(x) = \frac{1}{2^n n!}\frac{d^n}{dx^n}\left((x^2 - 1)^n\right).
$$

Note that $(x^2 - 1)^n$ is an even function for every integer $n$. Also, recall that the derivative of an even function is odd and the derivative of an odd function is even (this follows easily from the chain rule). We conclude that $P_n(x)$ is even when $n$ is even and $P_n(x)$ is odd when $n$ is odd.

Also, note that constant factors such as $(-1)^{-n}$ do not influence parity.


问:
evaluate $\sum_1^\infty \frac{1}{(n \sqrt{n+1} + (n+1)\sqrt{n})}$
How can I evaluate the infinite sum $\sum_1^\infty \frac{1}{(n \sqrt{n+1} + (n+1)\sqrt{n})}$? I try to change that into the form $\sum_1^\infty \frac{1}{(\sqrt{n} + \frac{n}{n+1}\sqrt{n+1})}-\frac{1}{(\sqrt{n+1} + \frac{n+1}{n}\sqrt{n})}$ and try to reduce the terms. But it seems not valid. I'm thinking of using some fourier transform techniques, any suggestions?

答:
J.G. was too fast for me...

My approach: I assumed there is no easy solution, so I tried to get an approximate solution. The values summed are roughly n^(-3/2). Even if you can't find an exact telescopic series, if you find one that is close to n^(-3/2) that gives you something you can easily sum up, plus a remainder that is a lot smaller, and then you can repeat the same approach.

Obviously with f(n) = -n^(-1/2), we have f(n+1) - f(n) ≈ n^(-3/2). So a guess is that with f(n) = -(n+c)^(1/2) for the right constant c, f(n+1) - f(n) should be a good approximation. How to find c? I used a spreadsheet to calculate some values. To my surprise, c = 0 gave an exact solution!

Now this isn't very elegant, but it helps with slightly different sums. For example if the divisor was n sqrt(n+2) + (n+1) sqrt(n), f(n) = -(n + 1/6)^(-1/2) gives a very nice approximation to a telescopic series; f(n+1) - f(n) has an error around 8 / n^3.5 when n is large. (I do prefer solutions that are not useful for one single problem only, but can be used for similar problems as well).


问:
What Im doing wrong in this combinatoric problem?
The task is to pick a team of 5 out of 17 boys and 13 girls, and there needs to be at least 2 of each.

I calculated the correct answer by taking $\binom{17}2\binom{13}3+\binom{17}3\binom{13}2=91936$.

When I tried different ways of imagining the problem I tried with typing it this way instead:

$$\binom{17}2\binom{13}2\binom{15+11}1$$ i.e. picking two from each gender then one out of the remaining, though this provides a different answer from above.

Where is my reasoning wrong on the latter?

Thanks!

答: From the formulas, I am assuming that you want to pick $5$ people. The problem is that when you pick $1$ person from the $15+11$ pool, what if you, in another configuration, have already picked that person when you picked the $2$ from the boys and the $2$ girls? So you are essentially marking one of the people of the team as special(a capitan?). Notice then that if you divide by $3$ the new number, you get the right answer. This division is taking out the special property of the person.


问:
Are $M_2= \langle(\sqrt{2}^{\sqrt{2}},\infty),\cdot \rangle$ , $M_1 = \langle(2,\infty),\cdot\rangle$ isomorphic structures?
$M_2= \langle(\sqrt{2}^{\sqrt{2}},\infty),\cdot \rangle$ ,  $M_1 =  \langle(2,\infty),\cdot\rangle$

attempt:

for all sign of function $f$

1.$h(F^{M}(a_1,a_2,...a_n))=F^N(h(a_1),h(a_2),...(h(a_n))$

2.Injective function

3.Surjective function

if i try to prove isomorphic in B. between $M_2$ , $M_1$
i dont know how to prove them with in open segment

答:
Let $\alpha,\beta\in\mathbb{R}$ such that $\alpha,\beta>1$. Then we claim
$M_\alpha:=\langle (\alpha,\infty),\cdot\rangle$ and $M_\beta:=\langle (\beta,\infty),\cdot\rangle$ are isomorphic. First note that $M_\alpha\cong\langle (\ln\alpha,\infty),+\rangle$ and $M_\beta\cong\langle (\ln\beta,\infty),+\rangle$, both via the map $x\mapsto\ln x$. Now we define an isomorphism $$f:\langle (\ln\beta,\infty),+\rangle\rightarrow \langle (\ln\alpha,\infty),+\rangle$$ via $x\mapsto\frac{\ln \alpha}{\ln \beta}x$. Since $\ln\alpha,\ln\beta>0$, this is a well-defined bijection, and, letting $\lambda=\frac{\ln{\alpha}}{\ln\beta}$, we have $f(x+y)=\lambda(x+y)=\lambda x+\lambda y=f(x)+f(y)$, so that $f$ is a $\{+\}$-embedding and hence an isomorphism, as desired.

Now, applying this to the case $\alpha=2$ and $\beta=\sqrt{2}^\sqrt{2}$ gives the desired result.


问:
distance travelled after nth bounce
A ball is thrown vertically to a height of $625$ meters from ground. Each time it hits the ground it bounces $\frac{2}{5}$ of the height it fell in the previous stage. How much will the ball travel during the first $20$th bounces? How can we derive a formula for finding this?</p>

答: Formula for $n$ bounces for your question: $$\displaystyle 625\left(1+2\sum_{k=1}^{n} \left(\frac{2}{5}\right)^k\right)$$</p>

Title: distance travelled after nth bounce


问:
Prove property of function using induction.
Show by using induction that if the function $f(x)$ is a solution to the equation $f(2t)+f(\frac{t}{2})=t$, then $f(x)$ has the following property: $f(mx)=m\cdot f(x)$ for any real number $m$.

So I tried to consider the function $f(kt)+f(\frac{t}{k})=t$.

If we assume that the function $f$ has the property $f(mx)=m\cdot f(x)$, it means that.

$$kf(t)+\frac{1}{k}f(t)=t$$

Any tips for what should I do next?

答:
I don't have the reputation to comment yet, so I'll post this as an answer.

You actually only need to prove the case of $m=2$ and $\frac12$: using $f(2t)+f(\frac t2)=t$, it is easy to see $f$ is proportional to $t$, and the claim follows then. But this this shows your claim that $f(kt)+f(\frac kt)=t$ is not true, so you probably want to think otherwise.

Another thing is that I don't see a reason for induction: it is used to prove claims about integers, but (I assume) the domain of $f$ is $\mathbb R$. Maybe some sort of continuity will come into play.


问:
Game Theory Solution to Brexit Negotiations?
Edit: This post is meant for fun: I was initially interested in seeing a wider range of mathematical techniques applied to computing the likelihood of a Deal or No-Deal outcome of the negotiations. Specifically, I was interested in how mathematics can be applied to a current real-world example, wanting to see how different fields of mathematics might approach the problem (i.e. Bayesian probabilists vs. Game Theorists).

Since there have been no answers so far, I have decided to narrow the "question" specifically to Game Theory, as I would be most interested in seeing how this particular discipline could be applied here. If any game theorists are interested in responding, I would be very interested in seeing an example.

Original post: This post is meant for fun: given that we have two days of negotiations left, use any mathematical technique or reasoning to compute the probability of a No-Deal. Be creative: game theory, probabilistic theory, Monte Carlo: it's all allowed.

I will post my own answer in a few hours.

As per the request in the comments below, here are some details:

Negotiations must end by Sunday, whether a deal has been reached or not (this post is written on Fri evening, i.e. two days before the deadline)

EU wants guarantees on aligned regulation ("level playing field"), to make sure that UK can't change it's regulatory environment in key areas to "undercut the EU" and gain a competitive advantage

UK rejects the above, because it argues the point of Brexit is to be able to pass its own laws (i.e doesn't wanna be kept in the EU's "regulatory orbit")

Both sides have much to loose: EU exports many goods to the UK (trade between Germany and UK alone was worth $139bn in 2019), whilst the goods exported by the UK to the EU amounted to £294bn in 2019 (43% of all UK exports): no deal means trade tariffs governed by WTO (mathematically, treat no deal as economically damaging to both sides)

A deal would be economically beneficial to both sides, but EU demands regulatory alignment, which is rejected on political grounds by the UK (and there is an argument for economic benefits in the longer run if own laws can be passed in the future)

Mathematically, you can look at current odds (and their evolution over the past week and months), current market moves in the GBP currency, or use other techniques: use anything you like to come up with a mathematically-backed answer.

Try to stay apolitical.

答:
Initial super-basic model to get the thread started (I hope to see some Game theorist & Bayesian probabilist contributions!)

Event $A$: EU willing to offer trade deal without UK fully committing to regulatory alignment

Event $B$: UK willing to accept a trade deal whilst agreeing to commit to regulatory alignment

Event $C$: No-Deal Brexit

Event $C^{'}$: Some sort of Deal

Now:

$$\mathbb{P}(C')=\mathbb{P}(A)+\mathbb{P}(A'\cap B)=\\=\mathbb{P}(A)+\mathbb{P}(A')\mathbb{P}(B)=\\=\mathbb{P}(A)+(1-\mathbb{P}(A))\mathbb{P}(B)$$

My subjective guesstimate is that $\mathbb{P}(A)=0.15$, whilst $\mathbb{P}(B)=0.35$. Then clearly:

$$\mathbb{P}(C')=0.15+0.85*0.35=0.4475$$

So that would mean:

$$\mathbb{P}(C)=1-0.4475=55.25\%$$

Latest odds of "no deal" are now $8/11$, which implies: $\mathbb{P}(C)=\frac{8}{8+11}=42.1\%$.

Clearly my very basic model is off. I might want to do some Bayesian refinements and assume distributions of probabilities, instead of constants, and regress these on latest events (as proposed in the comments in the original question: might have a shot 2mrw!).

I am sure a game theory answer can beat my initial super-basic model!


问:
Given $y=\frac{\sqrt{\cos(x)-1/2}}{\sqrt{6+35x-6x^2}}$ find the domain of the function

Find the domain of the function
$$y=\frac{\sqrt{\cos(x)-1/2}}{\sqrt{6+35x-6x^2}}$$

I'm unable to find the values for which $\cos(x)\geq 1/2$.

PS: This is not a homework question.

答:
$\begin{cases}\cos(x)-1/2\ge 0\\6+35x-6x^2>0\end{cases}\iff\begin{cases}\cos(x)\ge1/2 \text{ (*)}\\6x^2-35x-6<0\text{ (**)}\end{cases}$

From $(*)$ we have to find the value $x\in \mathbb R$ such that $\cos(x)\ge1/2\implies-\dfrac{\pi}{3}+2k\pi\le x \le\dfrac{\pi}{3}+2k\pi$, $k\in \mathbb Z$.

Note: for $\alpha\in[0,2\pi]$, $\cos(\pi/3)=\cos(5\pi/3)=1/2 \implies$ if we want to solve the inequality $\cos(x)>1/2$ we get $\pi/3<\alpha<5\pi/3$. Since the $\cos(x)$ function is periodic you have to consider all the other values in $\mathbb R$ which satisfies the condition.

From $(**)$ we have to solve $6x^2-35x-6<0\iff \dfrac{35-37}{12}<x<\dfrac{35+37}{12}\iff-\dfrac{1}{6}<x<6$.

In conclusion $dom:=\Big[-\dfrac{\pi}{3}+2k\pi,\dfrac{\pi}{3}+2k\pi\Big]\cap\Big(-\dfrac{1}{6},6\Big)=\left(-\dfrac{1}{6}, \dfrac{\pi }{3}\right]\cup \left[\dfrac{5 \pi }{3},6\right)$


答:
$$
\begin{cases}
\cos x-\frac{1}{2}\ge 0\\
-6 x^2+35 x+6 > 0\\
\end{cases}
$$
The second inequality is quite simple. The solution is $-\frac{1}{6}<x<6$

For the first inequality $\cos x \ge \frac12$ see the graph below

$$-\frac{\pi}{3}\le x\le \frac{\pi}{3}\lor \frac{5\pi}{3}\le x\le 2\pi$$
The intersection of the two intervals is
$$-\frac{1}{6}<x\leq \frac{\pi }{3}\lor \frac{5 \pi }{3}\leq x<6$$
with interval notation
$$\left(-\frac{1}{6}, \frac{\pi }{3}\right]\bigcup \left[\frac{5 \pi }{3},6\right)$$

Title: Given $y=\frac{\sqrt{\cos(x)-1/2}}{\sqrt{6+35x-6x^2}}$ find the domain of the function

Body:
Find the domain of the function
$$y=\frac{\sqrt{\cos(x)-1/2}}{\sqrt{6+35x-6x^2}}$$

I'm unable to find the values for which $\cos(x)\geq 1/2$.

PS: This is not a homework question.


答:
The map $f : x \mapsto x^{\sqrt 2}$ is an homomorphism from $M_2= \langle(\sqrt{2}^{\sqrt{2}},\infty),\cdot \rangle$ onto $M_1 = \langle(2,\infty),\cdot\rangle$.

as for any $x,y \gt 0$ we have $(x \cdot y)^{\sqrt{2}} = x^{\sqrt{2}} \cdot y^{\sqrt{2}}$. Also

$$f\left( \sqrt 2^{\sqrt 2}\right)=\left(\sqrt 2^{\sqrt 2}\right)^{\sqrt 2} = (\sqrt 2)^2 = 2$$

and as $f$ is continuous strictly increasing it is bijective. Finally $f$ is an isomorphism between the two structures.

Title: Are $M_2= \langle(\sqrt{2}^{\sqrt{2}},\infty),\cdot \rangle$ , $M_1 = \langle(2,\infty),\cdot\rangle$ isomorphic structures?

Body:
$M_2= \langle(\sqrt{2}^{\sqrt{2}},\infty),\cdot \rangle$ ,  $M_1 =  \langle(2,\infty),\cdot\rangle$

attempt:

for all sign of function $f$

1.$h(F^{M}(a_1,a_2,...a_n))=F^N(h(a_1),h(a_2),...(h(a_n))$

2.Injective function

3.Surjective function

if i try to prove isomorphic in B. between $M_2$ , $M_1$
i dont know how to prove them with in open segment


问:
Number of possible states on a $6\times 6$ grid
You have a 6x6 square grid. You have 18 black and 18 white pieces. How many ways can you fill the board (one piece per square) such that there are 3 black and 3 white pieces per column, and such that there are 3 black and 3 white pieces per row?

I've tried to do it purely combinatorally , just looking row by row, but it gets confusing very fast. Is there perhaps some way to do it utilizing symmetries, like how a solution translated horizontally or vertically is a solution, or how the 90 degree rotation around the center of a solution is a solution?

答:
By considering the case of a $2 \times 2$ grid with $1$ black piece and $1$ white piece for each row and each column you can count $2$ ways (i.e. BW-WB and WB-BW). It is not too difficult to count patiently all cases for a $4 \times 4$ grid, totalling $90$ ways.

Searching OEIS with $2,90$, you can find OEIS sequence A058527 and, for the $6 \times 6$ grid the value $297200$.


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:
As fjaclot says, $(2)$ is wrong. Making the substitution $u = x\sqrt{a}$ yields $$\sqrt{a}\int_{0}^{\infty}\frac{e^{-u^{2}}-e^{-tu^{2}}}{u^{2}}du$$ where $t = \frac{b}{a}$. Let $f(t)$ be the integral. Then $$f'(t) = \int_0^{\infty} e^{-tu^2} du = \frac{\sqrt{\pi}}{2\sqrt{t}}$$

Since $f(1) = 0$, $(2)$ is $$\sqrt{a}\left( \sqrt{\pi t}-\sqrt{\pi} \right) = \sqrt{\pi}\left(\sqrt{b}-\sqrt{a}\right)$$

Similarly for $(1)$, make the substitution $u=ax$ and let $f(t) = \int_{0}^{\infty}\frac{\left(e^{-u}-e^{-tu}\right)^{2}}{u^{2}}du$. Then $$f'(t) = 2\int_{0}^{\infty}\frac{e^{-\left(1+t\right)u}-e^{-2tu}}{u}du$$

Then you can apply that formula you found so that $$f'(t) = 2 \ln\left( \frac{2t}{1+t} \right) = 2\ln\left(2t\right)-2\ln\left(1+t\right)$$

The integral is then $f(t) = 2t\ln\left( \frac{2t}{t+1} \right)-2\ln(t+1)+2\ln(2)$. Therefore the answer for $(1)$ is $af\left(\frac{b}{a}\right)$ which simplifies to $$\ln\left(\frac{(2a)^{2a}(2b)^{2b}}{(a+b)^{2(a+b)}}\right)$$


问:
Arbitrage pricing for financial options
Are the following arguments valid for why $C(K_1) \geq C(K_2)$?

There are several outcomes for the payoff.

If $S<K_1$ then the payoff of option is $0$

If $K_1<S<K_2$ then the payoff is $S−K_1$ which is positive

If $K_1<K_2<S$ the the payoff is $-(S−K_2)−(S−K_1)$ which is $K_2−K_1$. Now since K2>K1 then this is positive.

This is a Type B arbitrage and so $C(K_1) \geq C(K_2)$ must hold

Exercise 1 (Monotonicity of the price of a call)
Two call options on a stock are identical apart from the strike price. The value of a call option with strike price $K$ is denoted by $C(K)$. Use an arbitrage argument to show that
$$
K_{1}<K_{2} \Rightarrow C\left(K_{1}\right) \geq C\left(K_{2}\right) .
$$

答:
An arbitrage exists if there is a portfolio of options with initial value $V_0 \leqslant 0$  , that is with a nonpositive cost of construction, such that upon expiration at time $T$ we have

$$\tag{*}\mathbb{P}(V_T \geqslant 0) =1, \quad \mathbb{P}(V_T > 0) > 0. $$

If $C(K_1) < C(K_2)$, then a portfolio that leads to arbitrage is long one option with strike $K_1$ and short one option with strike $K_2$.  The initial cost here is $C(K_1) - C(K_2) < 0$.  Using the payoff scenarios that you correctly specified, show that the arbitrage condition (*) is met.  This requires that the probability of the stock price exceeding $K_1$ be greater than zero -- which is true under any reasonable stochastic model for the price of a common stock.

Note also that $C(K_1) = C(K_2)$ also leads to an arbitrage as the initial cost  of the long-short portfolio is zero and you can make money from nothing with non-zero probability.  To preclude arbitrage we must see $C(K_1) > C(K_2)$.

All of this assumes that you can transact at those prices, i.e., there is no bid-ask spread.

Title: Arbitrage pricing for financial options


问:
Understanding the effect of affine connection and exterior derivative on a Tensor field
If we have a differentiable manifold $M$ and a vector field $X$, with $d$ the exterior derivative and $\nabla$ an affine connection on $M$. If $f$ is a smooth function, what are the $(q,r)$ tensor field components for the following tensor fields:
$$ T = \nabla_X (df) $$
$$ U = d(L_X (df)) $$
Where $L_X$ is the Lie derivative.

I am struggling to understand how the affine connection and the exterior derivative effects the tensor field components.

Any help appreciated.

答:
If $M$ is a manifold with a connexion $\nabla$, and if $f$ is a function on $M$, $X$ and $Y$ are vector fields on $M$.

First:
$$
\left(\nabla_X\left(\mathrm{d}f\right)\right)Y = \nabla_X\left(\mathrm{d}f(Y)\right) - \mathrm{d}f\left(\nabla_XY\right) = \nabla_X\left(\nabla_Y f\right) - \nabla_XY \cdot f = X\cdot\left(Y\cdot f\right) - \nabla_XY\cdot f
$$
is the Hessian of $f$ with respect to $\nabla$. It is usually denoted $\nabla^2_{X,Y}f$. It is a bilinear form. In the case of a torsion free connexion, for example in the riemannian case, it is symmetric in $X$ and $Y$.

In local coordinates, say $\mathrm{d}f = \frac{\partial f}{\partial x^i}\otimes\mathrm{d}x^i$, then
$$
\nabla_X \mathrm{d}f = \nabla_X\left(\frac{\partial f}{\partial x^i} \right)\otimes \mathrm{d}x^i + \frac{\partial f}{\partial x^i}\otimes \nabla_X \mathrm{d}x^i
$$
and you can express $\nabla_X\mathrm{d}x^i$ thanks to the Christoffel symbols and the components of $X$.
Second:
$$
d\circ L_X\circ d = d\circ \left( d\circ i_X + i_X\circ d\right)\circ d = d^2 \circ i_X \circ d + d\circ i_X \circ d^2 = 0
$$
because of the Cartan formula ($L_X = d\circ i_X + i_X \circ d$) and because $d^2=0$. Hence, $d\left(L_X\left(df\right)\right)=0$.


问:
How many vertices of degree 1 in a tree?
How many vertices of degree 1 are there in a tree with no vertices of degree more than 4?
The only thing that I have right now is that the number of edges in a tree is n-1 where n is the number of vertices. Any help would be appreciated. Thanks in advance.

答:
Let $k$ be the number of leafs in the tree. Let $n$ be the number of vertices.

It is known that $k \geq 2$ with equality for $T=P_n$.

As for the upper bound, the sum of degrees is at most $k+4(n-k)$. Thus

$$2(n-1)\leq k+4(n-k)=4n-3k \,$$

Solving this equality we get

$$k \leq \frac{2n+2}{3} \,.$$

Since $k$ is integer, we must have

$$k \leq \lfloor \frac{2n+2}{3} \rfloor \,.$$

The upper bound can be reached. Let $r$ be the remainder when $2n+2$ is divided by $3$. Then (as the above proof suggests) we need to construct a tree with $\lfloor \frac{2n+2}{3} \rfloor$ leafs, $r$ vertices of degree $2$ and $n-r- \lfloor \frac{2n+2}{3} \rfloor$ vertices of degree $4$. You can prove that such a tree exists by induction on $n$, the $P(n) \Rightarrow P(n+3)$ is easy (replace a leaf by a vertex of degree $4$ and connect it to 3 new leafs.)