问:
Calculating $\int_0^\infty \frac{\sin(x)}{x} \frac{\sin(x / 3)}{x / 3} \frac{\sin(x / 5)}{x / 5} \cdots \frac{\sin(x / 15)}{x / 15} \ dx$
I found the following result on this webpage:

$$\int_0^{\infty } \left(\prod _{k=0}^7 \frac{\sin \left(\frac{x}{2 k+1}\right)}{\frac{x}{2 k+1}}\right) \, \mathbb{d}x= \frac{\pi}{2} - \frac{6879714958723010531}{935615849440640907310521750000} \pi $$

However, I can't determine how to prove it.

答:
$\newcommand{\S}{\operatorname{sinc}}$

A 'brute force' solution to a beautiful problem. I don't claim this answer is as insightful as the others and the question is somewhat old, but I feel it is relevant and unique enough to merit posting.

As usual, we will let $\S(x) = \sin(x)/x$ with $\S(0)=1$.

The big idea: use angle-addition, Taylor series, and integration by parts to 'revert' the integral to a bunch- $128$, to be precise- of weighted integrals of the form $\int_0^{\infty}\sin(a x)/x\,dx$, $a>0$, each of which is $\pi/2$. Then we just add up the weights to produce the answer.

Angle-addition.

Recall that
\begin{align}
2\sin(\theta)\sin(\phi)=\cos(\theta-\phi)-\cos(\theta+\phi)\\
2\sin(\theta)\cos(\phi)=\sin(\theta-\phi)-\sin(\theta+\phi)\\
2\cos(\theta)\cos(\phi)=\cos(\theta-\phi)+\cos(\theta+\phi)\\
\end{align}

For instance, this tells us $\S(x)\S(x/3) = \frac{3}{2}x^{-2}(\cos(2x/3)-\cos(4x/3))$. You probably see what lies ahead, even if it's somewhat unappetizing. We have
$$
\S(x)\S(x/3)\cdots \S(x/15) =
$$
$$ \frac{1\cdot 3\cdots 15}{2^7 x^8} \sum_{e_i\in\{\pm 1\}} (-1)^{\#(e_i=-1)} \cos\left( x(1+e_1 \cdot 1/3 + e_2\cdot  1/5 +\cdots +e_7 \cdot 1/15)\right)
$$Thus, if we denote $W=(1\cdot 3\cdot \cdots \cdot 15)/2^7$, we have to evaluate $128$ integrals that look like $\displaystyle{W \int_0^{\infty}\frac{\cos(a x)}{x^8}\,dx}$, for some constant $a$. The trouble is that none of them are improperly integrable near $x=0$, which brings us to the second step in the process.

Taylor series.

The idea here is to add and subtract the same number of lower-order terms of the Taylor series of each $\cos(a x)$ to make the integrals converge. For example, $\displaystyle{\int_0^{\infty}\frac{\cos(x)}{x^8}\,dx}$ does not converge but $\displaystyle{\int_0^{\infty}\frac{\cos(x)-(1-x^2/2+x^4/24-x^6/720)}{x^8}\,dx}$ does. Since the sum of the arguments of the cosines is zero (check! A short explanation is that the $+$ and $-$ signs are equal in number) we will add and subtract up to the degree six terms to make the integrals convergent. Therefore, we have to evaluate
$$
\int_0^{\infty}\frac{\cos(a x)-(1-(ax)^2/2+(ax)^4/24-(ax)^6/720)}{x^8}\,dx
$$

Integration by parts.

We have
$$
\int_0^{\infty}\frac{\cos(a x)-(1-(ax)^2/2+(ax)^4/24-(ax)^6/720)}{x^8}\,dx
$$
\begin{align}
&=\left.-\frac{\cos(a x)-(1-(ax)^2/2+(ax)^4/24-(ax)^6/720)}{7x^7}\right|_0^{\infty}\\ &+\frac{1}{7}\int _0^{\infty} \frac{-a\sin(ax)+a^6 x^5/120 - a^4 x^3/6 + a^2 x }{x^7}\,dx
\end{align}The boundary terms vanish: at $\infty$, the denominator is a greater power and at $0$ the cosine series is $O(x^8)$. Continue this procedure until the denominator is just $x$:
$$
\cdots = \frac{a^7}{7!}\int_0^{\infty}\frac{\sin(ax)}{x}\,dx = \frac{a^7}{7!}\cdot\frac{\pi}{2}
$$

Add 'em up.

All that's left is to add up these values for all $128$ values of $a$ corresponding to the $2^7$ terms of the sum. This becomes
$$
W \cdot \frac{\pi}{2}\cdot \frac{1}{7!} \cdot \sum_{e_i\in\{\pm 1\}} (-1)^{\#(e_i=-1)} (1+e_1 \cdot 1/3 + e_2\cdot  1/5 +\cdots +e_7 \cdot 1/15)^7
$$A somewhat tedious but routine calculation gives the final answer of
$$
\frac{467807924713440738696537864469 \pi}{935615849440640907310521750000}
$$

For clarity, I'll do a smaller example (the three sinc case) that will evaluate to $\pi/2$.
$$
\int _0^{\infty} \S(x)\S(x/3)\S(x/5)\,dx = 15 \int_0^{\infty} \frac{\sin(x)\sin(x/3)\sin(x/5)}{x^3}\,dx
$$
$$
=\frac{15}{4} \int_0^{\infty} \frac{-\sin(7/15 x)+\sin(13/15 x)+\sin(17/15 x)-\sin(23/15 x)}{x^3}\,dx
$$
$$
=\frac{15}{4} \int_0^{\infty} \frac{-\sin(7/15 x)+\sin(13/15 x)+\sin(17/15 x)-\sin(23/15 x) }{x^3}+\frac{7/15 x-13/15 x-17/15 x+23/15 x}{x^3}\,dx
$$I'll just pick one of these, say the first one; the others are exactly the same.
$$
=\int_0^{\infty} \frac{-\sin(7/15 x)+7/15 x}{x^3}\,dx
$$
$$
= \left.\frac{-\sin(7/15 x)+7/15 x}{-2x^2}\right|_0^{\infty} + \int_0^{\infty} \frac{-7/15 \cos(7/15 x)+7/15}{2x^2}\,dx
$$
$$
=0+0 + \frac{(7/15)^2}{2} \int _0^{\infty}\frac{\sin(7/15 x)}{x}\,dx =\frac{(7/15)^2}{2}  \cdot \frac{\pi}{2}
$$So the total is
$$
\frac{15}{4}\cdot \frac{\pi}{2}\left(\frac{(7/15)^2}{2}-\frac{(13/15)^2}{2}-\frac{(17/15)^2}{2}+\frac{(23/15)^2}{2}\right)= \frac{\pi}{2}
$$

Of course, this method is merely computational: it has none of the theoretical trappings of the other answers. Nevertheless, the method could easily be generalized to other products and is quite amenable to reproduction on software. For instance, Mathematica thought long and hard about the original integral, but was able to perform the $128$ sub-integrals much quicker.

As a final remark, I'm reminded of the idea of quadrature from numerical analysis. It seems these integral-sinc-products, and more importantly their corresponding series, are 'exact' for the first seven until an error is introduced in the eighth. Perhaps I will investigate this further.


问:
What is the order of $\bar{2}$ in the multiplicative group $\mathbb Z_{289}^×$?
What is the order of $\bar{2}$ in the multiplicative group $\mathbb Z_{289}^×$?

I know that $289 = 17 \times 17$

so would it be $2^8\equiv 256\bmod17 =1$

and therefore the order of $\bar{2}$ is $8$? I'm not too sure about this

答:
$256 \equiv 1 \pmod {17}$ but $256\not \equiv 1 \pmod {289}$ which we need.

But not $289 = 17\times 17$ so $\phi (289) = 17\cdot16$ so $2^{17\cdot 16}\equiv 1\pmod {289}$ by Eulers theorem.

But the order might be something smaller that divides $17\cdot 16$.

We can figure that $2^8 = 17*15 + 1 \equiv 17*(-2) + 1\pmod{17^2}$ so

$2^{16} \equiv 17^2 *4 + 2*(-2)*17 + 1 \equiv -67 \pmod {289}$.

So the order of $2$ is not $16$ and thus not anything that divides $16$.  So the order of $2$ will be a multiple of $17$.  be a multiple of $17$ that divides $16*17$.

And $2^{17} \equiv -8*17+2$

$2^{2*17} \equiv (-8*17+2)^2 \equiv -32*17+ 4\equiv 2*17+4 \equiv 38\pmod{289}$.

$2^{4*17} \equiv 4^2*17^2 + 16*17 + 4^2 \equiv 16*17 +16\equiv 18*16\equiv 1*(-1)\equiv -1 \pmod {289}$.

And so $2^{8*17}\equiv (-1)^2 \equiv 1 \pmod {289}$.

So the order of $2$ is $8*17= 136$.

ContentLicense: CC BY-SA 4.0


问:
Prove Using L'Hopital's Rule And Mean Value Theorem.
how I can prove, for a function, $f(x)$ is differentiable such that $$\lim_{x\rightarrow \infty}f(x)=L\neq 0$$ and $$\lim_{x\rightarrow \infty}f'(x)$$ exists. Prove so that $\lim_{x\rightarrow\infty}f'(x)=0$. I have to use L'Hopital's Rule, and The Mean Value Theorem, do it 2 times.

How to prove using L'Hopital's Rule. I want to convert to indeterminate form. But I don't know the function, it is a random function. I converted this to $\frac{e^{x}f(x)}{e^x}$. What to do next?

How to prove using The Mean Value Theorem. I think I can use Mean Value Theorem to $f(x)$ on the interval (x, x+1), where x can be any number. And then calculate to $x\rightarrow\infty$. How to do that?

I have to prove this both ways.

答:
Sketch:
1.$$\lim_{x\to\infty}\frac{e^x f(x)}{e^x}=\lim_{x\to\infty}(f’(x)+f(x))\,,$$
which implies
$$L=\lim_{x\to\infty}f’(x)+L\,.$$

2.$$f(x+h)-f(x)=f’(\xi)h \,,$$
yields
$$L-L=\lim_{x\to\infty}f’(x) h\,.$$


答:
As shown in Baby Rudin we can apply L'Hopital's Rule as long as the denominator tends to $\infty$; it is not necessary that we have $\frac {\infty} {\infty}$ form. So apply the rule to $\lim \frac {f(x)} x$ to show that $f'(x) \to 0$.

Hint for proof using  MVT: If $f'(x) \ to l >0$ the $f(n+1)-f(n) \geq \frac l 2 $ for all $n$ sufficiently large. Show using this that $(f(n)$ is not even bounded.

Title: Prove Using L'Hopital's Rule And Mean Value Theorem.

Body:
how I can prove, for a function, $f(x)$ is differentiable such that $$\lim_{x\rightarrow \infty}f(x)=L\neq 0$$ and $$\lim_{x\rightarrow \infty}f'(x)$$ exists. Prove so that $\lim_{x\rightarrow\infty}f'(x)=0$. I have to use L'Hopital's Rule, and The Mean Value Theorem, do it 2 times.

How to prove using L'Hopital's Rule. I want to convert to indeterminate form. But I don't know the function, it is a random function. I converted this to $\frac{e^{x}f(x)}{e^x}$. What to do next?

How to prove using The Mean Value Theorem. I think I can use Mean Value Theorem to $f(x)$ on the interval (x, x+1), where x can be any number. And then calculate to $x\rightarrow\infty$. How to do that?

I have to prove this both ways.


问:
Family of circles passing through two points
Find the equation of family of circle center at (h,k) passing through the origin and (0,4).
So I obtained that the center would be at (h,2). Which has an equation of

(x-h)^2 + (y-2)^2 = r^2

But how can I restrict the above equation, so that all the circles will pass through the origin and (0,4)?

I also tried the S1 + LS2 = 0 and obtained
x^2 + y^2 - 4y - L4x = 0
I don't if it is the right equation to find the differential equation of the family of circles.

答:
So there are two conditions that help define this family of circles

The center of any circle lies on the perpendicular bisector of $(0,0)$ and $(0,4)$ - hence you get the following set of circles

$$(x-h)^2 + (y-2)^2 = r^2$$

The origin lies on the circle

$$h^2 + 4 = r^2$$

So now you can write the above family of circles in terms of parameter $h$, which represents the $x$ coordinate of it's centre (or location along the perpendicular bisector)

$$S_h : (x-h)^2 + (y-2)^2 = h^2 + 4$$

This is the equation in terms of $h$ - where each value of $h$ will give you a unique circle satisfying the condition


问:
Why does $\sqrt x = -a $ have no solution?
Algebraically, can't you solve this equation by squaring both sides?:
$$\sqrt x = -a $$
$$(\sqrt x)^2 = (-a)^2 $$
$$ x = a ^2$$

I can understand why $x^2 = -a$ has no solution, because a number multiplied by itself twice can't be a negative number. However, in the case above, $-4$ is technically still a square root of $16$.

答:
In the reals, if $a>0$ then $\sqrt x=-a$ has no solution since the square root in this context is usually understood to return a non-negative number.

In the complexes, $\sqrt x=-a$ has two solutions as expected: $x=\pm i\sqrt a$.


答:
Usually when we write $\sqrt{x}$ where $x$ is a real number, we usually mean the principal root. This means the positive square root. While $(-4)^2=16$, we don't say that $\sqrt{16}=-4$, because $\sqrt{16}=4$, and if $\sqrt{16}$ were both values, it wouldn't be a function. This is why when squaring an equation can sometimes lead to extraneous solutions.

When taking the square root of an equation, we attach a $\pm$. So if $a^2=16$, we do $a=\pm\sqrt{16}=4,-4$.

Title: Why does $\sqrt x = -a $ have no solution?

Body:
Algebraically, can't you solve this equation by squaring both sides?:
$$\sqrt x = -a $$
$$(\sqrt x)^2 = (-a)^2 $$
$$ x = a ^2$$

I can understand why $x^2 = -a$ has no solution, because a number multiplied by itself twice can't be a negative number. However, in the case above, $-4$ is technically still a square root of $16$.


问:
Help on theta region of polar double integral
I have a question about setting up the bounds on the theta integral of a polar double integral.

When you have a circle such as $(x-1)^2 + y^2 = 1$, the equation of the circle can be shown to be $2\cos(\theta)$. So $r$ will go from $0$ to $2\cos(\theta)$. Fair enough.

But the theta. I have watched one excellent online video (Prof. Leonard) who has the theta range of this integral as $0$ to $\pi$. On the other hand, the Stewart Calculus book has this same problem with a limit of $-\pi/2$ to $\pi/2$.

I know these may give the same answers, but which is the correct set up?

Thanks so much (and, yes, I've reviewed the other questions on the forum before posting this).

答:
As it pointed in  Jyrki Lahtonen's comment if we consider $(x-1)^2 + y^2 = 1$, then polar coordinates gives $0 \leqslant r \leqslant 2 \cos \theta$.
So we have region for $\theta$ from inequality $$0 \leqslant  \cos \theta$$
If initially we choose $ \theta \in (-\pi,  \pi]$, then we obtain solution $ \theta \in \left[-\frac{\pi}{2},  \frac{\pi}{2}\right]$.


Body:
What is the order of $\bar{2}$ in the multiplicative group $\mathbb Z_{289}^×$?

I know that $289 = 17 \times 17$

so would it be $2^8\equiv 256\bmod17 =1$

and therefore the order of $\bar{2}$ is $8$? I'm not too sure about this

Body:
This can be done very easy mentally using only trivial calculations.

$\!\bmod 17\!:\,\ 2^4\equiv -1\,\Rightarrow\, 2^8\equiv 1\Rightarrow 2\,$ has order $\,\color{#c00}{o(2) = 8}\,$ by the Order Test.

$\!\bmod 17^2\!:\ n\!:=\!o(2)\Rightarrow\,2^n\equiv 1\,$ thus $\bmod 17\!:\ 2^n\equiv 1\,$ thus $\, \color{#c00}8\mid n\,$ so $\,n = 8k$.

$\!\bmod 17\!:\ 2\equiv 6^2$ thus $\,2\,$ is a $\rm\color{#0a0}{square}\bmod 17^2\:\!$ too, so $\,o(2)=8k\mid \phi(17^2)/\color{#0a0}2 = 8\cdot 17$.

So $\,k\!=\!1$ or $17.\,$ But $\,k\!\neq\! 1\,$ by $\,2^8\!\equiv\! 256\!\not\equiv \!1\pmod{\!289}\,$ so $\,k\!=\!17,\,$ so $\,o(2)\! =\! 8(17)\!=\!136$.


答:
One more way to see why your argument is not valid,

Assuming $a$ is positive,
Define the function $f: [0,\infty) \to [0, \infty)$ by $f(x) = \sqrt x$ and generally when we use $\sqrt x$ we mean this function $f$.

So $\sqrt x$ is a unique,positive value. By definition of  $\sqrt x$, it cannot take negative values, hence the equation provided by you does not have a solution.


问:
Solve $(x^2 + 5)^2 - 15(x^2 + 5) + 54 = 0$
I got the square root of 14 and 11 but the answer book states that these answers are wrong. Can someone help me? I used this formula to find the individual roots </p>

<p>$x = -\frac{p}{2} \pm \sqrt{(\frac{p}{2})^2 - q}$ </p>

OwnerDisplayName: Kalina Kama

ContentLicense: CC BY-SA 3.0

Body:
Using middle term factor,  $$(x^2+5)^2-(9+6)(x^2+5)+6\cdot9=0$$

$$\implies (x^2+5)(x^2+5-9)-6(x^2+5-9)=0$$

$$\implies (x^2-4)(x^2-1)=0$$

$\implies x^2-1=0$ or $x^2-4=0$

Alternatively, using quadratic equation formula  for $x^2+5=\frac{15\pm\sqrt{15^2-4\cdot1\cdot 54}}{2\cdot1}=\frac{15\pm 3}2=9$ or $6$


问:
Looking for a fun topology to study
Sorry if this deviates from the norm a bit, but I need to come up with a topological space to write a report on in these last couple of weeks now that all of my other assignments are out of the way. I have such little understanding of the actual application of this topic outside of standard metric spaces and subspaces, so I was hoping someone might have an interesting idea related to my interests.

It's important to note that I initially was studying chemistry three years ago before switching to math because I absolutely loved linear algebra and elementary differential equations. I've taken the graduate level courses in both of those subjects (and PDEs) already... I'm not sure if that will be entirely helpful for what I need to do, but I'm not afraid to do something that might require more than a surface level understanding of those subjects. Also, I'm open to things that might apply to chemistry as well!

答:
Topology is a rather broad field, so there are very many interesting spaces to cover.

My personal favorite is the Sierpinski-space, which is the space $\{\bullet, \circ\}$, with precisely one nontrivial open set $\{\circ\}$. It is interesting out of the following reasons:

It is the smallest nontrivial topological space

It can be used to detect or classify open/closed subspaces of any topological space $X$

It is the Zariski-Topology of any arbitrary local integral domain

Possibly even more interesting facts

If you can pick a whole class of spaces I would suggest considering counterexamples for connectivity assumptions, by which I mean spaces like the Warshaw circle, Hawaiian earrings, topologists comb. They are interesting in the way how rather small modifications to a construction can fundamentally change topological properties.

These are examples rooted in abstract mathematics though and I would love to read about interesting spaces arising in applications...

Title: Looking for a fun topology to study


问:
Summation of the reciprocals of the product of consecutive integers
It is well known that there is a closed formula for:

$$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{(n)(n + 1)}$$

And likewise for:

$$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \cdots + \frac{1}{(n)(n + 1)(n+2)}$$

I am wondering if there is a closed formula for:

$$f(n, k) = \sum_{i=1}^n \frac{1}{\prod_{j=0}^k (i + j)}$$

Note that putting $k = 1$ and $k = 2$ in the above function yields the above two series.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\on{f}\pars{n, k} & \equiv
\bbox[5px,#ffd]{\sum_{i = 1}^{n}
{1 \over \prod_{j = 0}^{k}\,\pars{i + j}}} =
\sum_{i = 1}^{n}{1 \over i^{\overline{k + 1}}} =
\sum_{i = 1}^{n}{1 \over \Gamma\pars{i + k + 1}/\Gamma\pars{i}}
\\[5mm] & =
{1 \over k!}\sum_{i = 1}^{n}{\Gamma\pars{i}\Gamma\pars{k + 1} \over \Gamma\pars{i + k + 1}} =
{1 \over k!}\sum_{i = 1}^{n}
\int_{0}^{1}t^{i - 1}\,\pars{1 - t}^{k}\,\dd t
\\[5mm] & =
{1 \over k!}\int_{0}^{1}\pars{1 - t}^{k}
\sum_{i = 1}^{n}t^{i - 1}\,\,\dd t =
{1 \over k!}\int_{0}^{1}\pars{1 - t}^{k}\,\,
{t^{n} - 1 \over t - 1}\,\,\dd t
\\[5mm] = &\
{1 \over k!}\bracks{%
\int_{0}^{1}\pars{1 - t}^{k - 1}\,\,\dd t -
\int_{0}^{1}t^{n}\pars{1 - t}^{k - 1}\,\,\dd t}
\\[5mm] = &\
{1 \over k!}\bracks{%
{1 \over k} - {\Gamma\pars{n + 1}\Gamma\pars{k} \over \Gamma\pars{n + 1 + k}}} =
{1 \over k!}\bracks{%
{1 \over k} - {n!\pars{k - 1}! \over \pars{n + k}!}}
\\[5mm] = &\
\bbx{{1 \over k}\bracks{{1 \over k!} - {n! \over \pars{n + k}!}}} \\ &
\end{align}


问:
proving convergence of $a_{n+1}=1+\frac{1}{1+a_{n}}$

$a_1=1.$ $a_{n+1}=1+\frac{1}{1+a_{n}}$

Prove that the sequence is convergent.

I'm trying to prove the convergence of this sequence but having trouble. At first I thought this might be a monotone sequence since then I can try monotone convergence theorem to prove its convergence.

But after checking some terms, I realized it seemed the sequence is oscillating. So I'm not sure how to prove the convergence of this sequence.

Thanks.

答:
This sequence is a Cauchy sequence so it converges.

First you see $a_n>0, \forall n \in \mathbb{N}$ from recursive relation.  [$a_1=1$ and $a_{n+1}$ is defined to added positive terms]

Second since $a_n>0$ thus $a_{n+1} = 1 + \frac{1}{1+a_n} \leq 2 $

Now consider
\begin{align}
|a_{n+1} - a_n| = \left| \frac{1}{1+a_n} - \frac{1}{1+a_{n-1}} \right|
 = \frac{|a_n - a_{n-1}|}{(1+a_n)(1+a_{n-1})} \leq \frac{1}{4} | a_n - a_{n -1}|
\end{align}
and this is cauchy sequence. [Series with this form is called contractive and after repeatively apply the same procedure continuing to $|a_2-a_1|$, and by Squeeze theorem you can easily guess $a_n$ is a Cauchy sequence]

In $\mathbb{R}$ cauchy sequence implies convergences so it converges. Then by taking limits $\lim_{n\rightarrow \infty} a_n = \alpha$ we have $\alpha^2 = 2$ and from $a_n>0$, $\alpha = \sqrt{2}$.


答: You already proved $X_n \le \dfrac{3}{2}+\dfrac{1}{2n}$ for all integers $n \ge 1$. Now, note  that $\dfrac{1}{2n} \le \dfrac{1}{2}$ for all integers $n \ge 1$. Hence, $X_n \le \dfrac{3}{2}+\dfrac{1}{2n} \le \dfrac{3}{2}+\dfrac{1}{2} = 2$ for all integers $n \ge 1$.

Body:
$X_n=\frac{n+1}{n^2+1}+\frac{n+2}{n^2+2}+\frac{n+3}{n^2+3}+\dots+ \frac{n+n}{n^2+n} $ is bounded above?

I tried this:

$X_n=\frac{n+1}{n^2+1}+\frac{n+2}{n^2+2}+\frac{n+3}{n^2+3}+\dots+ \frac{n+n}{n^2+n} \leqslant \frac{n+1}{n^2}+\frac{n+2}{n^2}+\frac{n+3}{n^2}+\dots+ \frac{n+n}{n^2+n}$

$= \frac{1}{n^2}\sum_{k=1}^{n-1} k + \frac{n^2}{n^2}$

$=\frac{1}{n^2}\frac{(n+1)(n)}{2}+1$

$= \frac{n+1}{2n}+1= \frac{n}{2n}+\frac{1}{2n}+1=\frac{1}{2}+1+\frac{1}{2n}=\frac{3}{2}+\frac{1}{2n}$

I'm really not sure if I did it right, can someone help me? How can I conclude it?


问:
Convergence of $\sum_{n=1}^{\infty} \left(\frac{1}{n} - \sin\left(\frac{1}{n} \right) \right)^{\alpha}$
I am trying to seek the real values of $\alpha$ for convergence of the following series :
$\sum_{n=1}^{\infty} \left(\frac{1}{n} - \sin\left(\frac{1}{n} \right) \right)^{\alpha}$

My trial:
Since $n>1$, $0<\frac{1}{n}<1$, thus we can expand sin function as
\begin{align}
\sin\left(\frac{1}{n} \right) = \frac{1}{n}  - \frac{1}{3!} \frac{1}{n^3} + \cdots
\end{align}
Hence the series becomes
\begin{align}
 \sum_{n=1}^{\infty} \left(\frac{1}{n} - \sin\left(\frac{1}{n} \right) \right)^{\alpha}
= \sum_{n=1}^{\infty} \left(\frac{1}{3!}\frac{1}{n^3} - \cdots\right)^{\alpha} \leq \frac{1}{3!}\sum_{n=1}^{\infty} \left(\frac{1}{n}\right)^{3\alpha}
\end{align}
Thus my guess is from p-test $3\alpha >1$, i.e., $\alpha>\frac{1}{3}$.

Is this approach admissible?

答:
Your intuition is correct for using the Taylor expansion of $\sin x$ around $x=0$. Note that for large enough $n$:
$$
{1\over n}-{1\over 6n^3}\le\sin {1\over n}\le {1\over n}-{1\over 5n^3}
$$
You have already done the lower bound. The upper bound is just on its way!


问:
Show the convergence of $\sum_{n=0}^{\infty}\frac{1}{2}\cdot \frac{(2n)!}{n!(n+1)!}\cdot \left(\frac{1}{4}\right)^n$
As stated by the prompt, I'm looking to show the convergence of the series $$\sum_{n=0}^{\infty}\frac{1}{2}\cdot \frac{(2n)!}{n!(n+1)!}\cdot \left(\frac{1}{4}\right)^n$$
I've tried using the ratio, root, and Limit Comparison test (w/ a geometric series), and all have given me inconclusive answers.  If anyone has some insights for how I might be able to show this series converges, your help would be much appreciated.

答:
One possible solution using Raabe's Test
$$n\cdot\left(\frac{a_n}{a_{n+1}} -1\right) = n\cdot\frac{6n +6}{(2n+1)(2n+2)}\to \frac{3}{2}>1$$


问:
After hashing of how many keys will the probability that any new key hashed collides with an existing one exceed 0.5?
Consider a hash function that distributes keys uniformly. The hash table size is 20.

Here I am getting answer 11 considering the fact that probability will of collision will be 50% more only when we already have 10 entries in the hash table and we are about to map one more entry in the hash table .

Is this approach correct ?

答:
Let us first analyze the question a bit:

"After hashing of how many keys will the probability that any new key hashed collides with an existing one exceed $0.5$ ? "

I feel the question can be interpreted in two ways.

FIRST

After hashing of how many keys will the probability that any new key hashed NEXT collides with an existing one exceed $0.5$?

or

Find such $k$ for which after hashing $k$ keys the probability that just the next $(k+1)^{th}$ key (any new key hashed) collides any of the $k$ keys inserted till now exceed $0.5$ (without considering the chances of any collision which might have occurred while inserting the first $k$ keys)

The approach for this is as follows.

Since there are $m=20$ slots in the hash table and there are $k$ slots already occupied in the table. So a collision shall occur for the $(k+1)^{th}$ key if it hashes into any of these $k$ slots.

The probability that this occurs is $\frac{k}{m}$. Now this value should be greater than $0.5$. So we have,

$$ \frac{k}{m} > 0.5 $$

$$\implies k> 10 \quad\quad\quad\text{as m=20}$$

Now as $k$ is an integer so , $k$ is at least $11$.

SECOND

After hashing of how many keys will the probability that any new key hashed TILL NOW collides with an existing one exceed $0.5$?

or

Find $k$ such that after hashing $k$ keys, the probability that all keys hashed till now (a key which is inserted for the first time is a new key and it can be any of the $k$ keys inserted till now) collide, which is the same as (1- probability that no two of the first $k$ keys hash to the same slot)$^{\dagger}$ exceeds $0.5$.

From $\dagger$ we clearly get the essence of the birthday problem.

Now,

probability that no two of the first $k$ keys hash to the same slot ($q$) = $\frac{m}{m}.\frac{m-1}{m}...\frac{m-k+1}{m}$

Now required probability $(p) = 1-q$.

Now,

$$p>0.5$$
$$\implies 1-\frac{m}{m}.\frac{m-1}{m}...\frac{m-k+1}{m} >0.5$$
$$\implies 1.\frac{19}{20}...\frac{20-k+1}{20} < 0.5 \quad\text { as m =20}$$

Let $\lambda(k) =1.\frac{19}{20}...\frac{20-k+1}{20}$

Now $\lambda(5) = 0.5814 \nless 0.5 $

$\lambda(6) = 0.43605 \lt 0.5 $

So $k = 6$


问:
Bound $\frac{\lVert x-y \rVert}{\lVert x \rVert}$ with perturbation
Firstly, I want to say I could not find a proper title sorry. I have a question.

Let $Ax =b$ and $(A+\Delta A)y = b + \Delta b $.

Suppose $\lVert \Delta A \rVert  \leq \epsilon \lVert E \rVert, \lVert \Delta b \rVert \leq \epsilon\lVert f\rVert $ and $\epsilon \lVert A^{-1} \rVert\lVert E\rVert < 1.$

Show that
$$
\frac{\|x-y\|}{\|x\|} \leq \frac{\epsilon}{1-\epsilon\left\|A^{-1}\right\|\|E\|}\left(\frac{\left\|A^{-1}\right\|\|f\|}{\|x\|}+\left\|A^{-1}\right\| \| E \|\right)
$$

I tried to apply some theorem from perturbation theory on this. But, I cannot do. Can you help me?

答:
First, let's obtain some closed forms for $x$ and $y$. For $x$ it is quite trivial: $x = A^{-1} b$.

For the $y$ we have
$$
y = (A + \Delta A)^{-1} (b + \delta b) = \left(A (I + A^{-1}\Delta A)\right)^{-1} (b + \Delta b).
$$
Under the given conditions the matrix $I + A^{-1} \Delta A$ is invertible:
$$
\left(I + A^{-1} \Delta A\right)^{-1} = I + \sum_{k=1}^{\infty} (-A^{-1} \Delta A)^k.
$$
This is known as Neumann series and is known to be converging when $\|A^{-1} \Delta A\| < 1$ (sufficient condition). And this is the case, since
$$
\|A^{-1} \Delta A\| \leq \|A^{-1}\| \|\Delta A\| < \frac{1}{\epsilon \|E\|}\epsilon \|E\| = 1.
$$
We can now write
$$
y = \left(I + \sum_{k=1}^\infty (-A^{-1} \Delta A)^k\right) A^{-1} (b + \Delta b).
$$
Let us introduce the auxiliary vector
$$
z = \left(I + \sum_{k=1}^\infty (-A^{-1} \Delta A)^k\right) A^{-1} b =
\left(I + \sum_{k=1}^\infty (-A^{-1} \Delta A)^k\right) x.
$$
Now apply the triangle inequality to get
$$
\|x - y\| \leq \|x - z\| + \|y - z\|. \tag{1}
$$
Estimating each term:
$$
\|x - z\| = \left\|\sum_{k=1}^\infty (-A^{-1}\Delta A)^k\right\| \|x\|\\
\|y - z\| = \left\|I + \sum_{k=1}^\infty (-A^{-1} \Delta A)^k\right\| \|A^{-1}\| \|\Delta b\|. \tag{2}
$$
Estimating the sums further by simple geometric series:
$$
\left\|\sum_{k=1}^\infty (-A^{-1}\Delta A)^k\right\|
\leq \sum_{k=1}^\infty \left\|A^{-1}\Delta A\right\|^k
\leq \sum_{k=1}^\infty \left\|A^{-1}\|^k \|\Delta A\right\|^k
= \frac{\left\|A^{-1}\|\|\Delta A\right\|}{1 - \left\|A^{-1}\|\|\Delta A\right\|} \tag{3}\\
\left\|I + \sum_{k=1}^\infty (-A^{-1}\Delta A)^k\right\|
\leq 1 + \sum_{k=1}^\infty \left\|A^{-1}\Delta A\right\|^k
\leq 1 + \sum_{k=1}^\infty \left\|A^{-1}\|^k\|\Delta A\right\|^k
= \frac{1}{1 - \left\|A^{-1}\|\|\Delta A\right\|}\\
$$
Note, that if we leave $\|A^{-1}\Delta A\|$ instead of $\|A^{-1}\|\|\Delta A\|$ we obtain a slightly tighter bound with $1 - \|A^{-1} \Delta A\|$ in the denominator. But it was not requested.

Plugging (3) to (2) and finally to (1) gives:
$$
\| x - y \| \leq \frac{1}{1 - \|A^{-1}\| \|\Delta A\|} \left(
\|A^{-1}\|\|\Delta A\| \|x\| + \|A^{-1}\| \|\Delta b\|
\right)
$$
Now divide both sides by $\|x\|$:
$$
\frac{\| x - y \|}{\|x\|} \leq \frac{1}{1 - \|A^{-1}\| \|\Delta A\|} \left(
\|A^{-1}\|\|\Delta A\| + \frac{\|A^{-1}\| \|\Delta b\|}{\|x\|}
\right)
$$
Putting $\|\Delta b\| = \epsilon \|f\|$ and $\|\Delta A\| \leq \epsilon \|E\|$ gives the desired estimation:
$$
\frac{\| x - y \|}{\|x\|} \leq \frac{\epsilon}{1 - \epsilon\|A^{-1}\| \|E\|} \left(
\|A^{-1}\| \|E\| + \frac{\|A^{-1}\| \|f\|}{\|x\|}
\right)
$$


问:
Doubt regarding a proof of intersection of the set of all neighbourhoods of a point $x\in\mathbb{R}$ is the singleton set $\{x\}$.
This is the exact same question as posted here. I was trying a different approach for solving this problem, but seem to have run into a dead end.

My approach: Let there be n such neighbourhoods of the point $x$ and let us denote them by $\left(a_1,b_1\right),\left(a_2,b_2\right),\left(a_3,b_3\right)...,\left(a_n,b_n\right)$. Thus $x$ is bounded above by $b_j$ and bounded below by $a_i$ $\forall i,j=1,2,...,n$.

Let us define a sequence $A_n:=\text{max}\{a_1,a_2,a_3,...,a_n\}$. Clearly, $A_n$ is a non-decreasing sequence bounded above by $x$.

Define another sequence $B_n:=\text{min}\{b_1,b_2,b_3,...,b_n\}$. Now this is a non-increasing sequence bounded below by $x$.

Thus, by Monotone Convergence Theorem, both $A_n$ and $B_n$ converges to some $A,B \left(\in \mathbb{R}\right)$ respectively. We can also say that $A\leq x\leq B$.

The only thing left to prove is: $A=B$. I seem to have run out of ideas here.

Can anyone please help me out?

答: Assume $A<x$ and observe that $(\frac {A+x}2,x+42)$ is a neighbourhood of $x$.


答:
The boundedness could also be shown quite directly:

$$X_n =\sum_{k=1}^n\frac{n+k}{n^2+k}\leq\sum_{k=1}^n\frac{2n}{n^2} = 2$$

Title: $X_n=\frac{n+1}{n^2+1}+\frac{n+2}{n^2+2}+\frac{n+3}{n^2+3}+\dots+ \frac{n+n}{n^2+n} $ is bounded above?


问:
If $\lim_{x\to c}f'(x)=A$ then $f'(c)=A$
I have problems with this exercise in the book Real Analysis of Miklós Laczkovich:

Let $f$ be continuous on $(a,b)$ and differentiable on $(a,b)\setminus \{c\}$, where $a<c<b$. Prove that if $\lim\limits_{x\to c}f'(x)=A$, where $A$ is finite. then $f$ is differentiable at $c$ and $f'(c)=A$

Since this problem is in the section of the mean value theorem, I have tried to attack it from that side, but I can not see what considerations to take to solve it, I would appreciate any help, thank you.

答:
First, let us look at the left-hand side. For all $x_1 \in (a,c)$, there exists $c_1 \in (x_1, c)$ such that $f(x_1) - f(c) = f'(c_1) (x_1 - c)$ by MVT. Note that this is so because $f$ is continuous on $[x_1, c]$. Therefore,
$$
\lim_{x_1 \nearrow c} \dfrac{f(x_1) - f(c)}{x_1 - c} = \lim_{c_1 \nearrow c} f'(c_1)
$$
since $x_1 \nearrow c$ implies $c_1 \nearrow c$ by the condition.
Hence, we have
$$
\lim_{x_1 \nearrow c} \dfrac{f(x_1) - f(c)}{x_1 - c} = A
$$
Likewise, we can do a similar process for the case that, say, $x_2 \in (c,b)$.


问:
Prove that there is a quadratic irreducible in $\mathbb Z_p[x]$ for $p$ prime.
I am kind of stuck at this problem. I know how to find irreducible polynomials in fields like $\mathbb Z_1,\mathbb Z_2$ or where the specific field is given.

I can't figure out a way to prove this. I am thinking to use theorems like:
If an irreducible polynomial $p(x)$ belongs to a field $F$, then $F[x]/\langle p(x)\rangle$ is a field.

Or should I use arguments like if $\mathbb Z_p$ is $\mathbb Z_7$, then polynomials like $x^2 + 3$ are irreducible. So, there exists a quadratic irreducible in $\mathbb Z_p[x]$?

Or is there any other method that I should follow? Any suggestions will be helpful.

答:
Just a comment on the very nice answer of @Robert Shore:

There are exactly $\frac{p(p-1)}{2}$ irreducible polynomials of degree $2$ in $\mathbb{Z}/p[x]$. They are of the form (assume $p$ odd)
$$(x-a)^2 - b$$
where $a\in \mathbb{Z}/p$ and $b$ is a quadratic non-residue $\mod p$.  There are $p$ choices for $a$ and $\frac{p-1}{2}$ choices for $b$.


问:
Study suggestions
I am a 16 year old high school student with too much free time. For the past two years I have been teaching myself university level maths in my free time. I have studied the basic first year subjects such as multivariable calculus, linear algebra and ordinary differential equations. My question is what should I study next?

I have tried more pure subjects such as real analysis, abstract algebra and number theory and additionally more applied subjects such as nonlinear dynamics, chaos theory and topology, but they either didn't spark my interest or were too difficult. I have even written a few papers that never really worked out.I posted one here I feel as if I need to regain my curiosity for mathematics and look for something beyond what academics present at seminars for children (referring to topology). Therefore, what is a subject that is seldom presented to an undergraduate but still is an valid field in mathematics?

答: I think you've already studied all the basic stuff common to science under-graduation courses. I think you should think about your interests now and from that move to more deeper and specific subjects. If you just want to have fun with mathematics, give another chance to set topology (did you really study set topology?), it would be your first rigorous math subject and then move to real analysis. Try different books, the fact is hard for you it means you can learn a lot from that (I don't think is too hard from your background, ask questions here).

Title: Study suggestions


答:
You can also have the explicit solution

$$x_n=\sum_{k=1}^n \frac{n+k}{n^2+k}=n+(n-1) n \left(H_{n^2}-H_{n^2+n}\right)$$

Using asymptotics of harmonic numbers and continuing with Taylor expansions
$$x_n=\frac{3}{2}-\frac{1}{3 n}-\frac{5}{12 n^2}+\frac{23}{60
   n^3}+O\left(\frac{1}{n^4}\right)$$ and you could generate very sharp upper and lower bounds.


问:
decidability of $\{x|W_x \text{is different from K in only finitely many elements}\}$
Is the following language decidable? Please explain your argument as I want to learn how such problems must be solved to do the rest on my own. </p>

<p>$$\{x \mid W_x \text{ is different from K in only finitely many
    elements}\},$$  where $W_x = \text{dom}(\Phi_x)$ and $K=\{x \mid \Phi_x(x) \downarrow\}$.</p>

答:
The usual way of proving something is not computable, not c.e., etc, it to reduce a problem known to be not computable, not c.e., etc to it. In this case, the details are as follows:

Let $A$ denote the set you are interested in.

Let $\bar{K}$ denote the complement of $K$. Define a partial computable function $\Psi$ as follows:

$\Psi(x,y) = \begin{cases}
0 & \quad \text{if }\Phi_{x,y}(x) \uparrow \wedge \Phi_y(y) \downarrow \\
\uparrow & \quad \text{otherwise}
\end{cases}$

where the function $\Phi_{x,y}(x)$ is defined to converges if and only if $\Phi_x(x)$ has converged by stage $y$. Intuitively, this algorithm runs $\Phi_x(x)$. For each $y$, if it has not converged yet at  stage $y$, then run $\Phi_y(y)$, outputting $0$ if $\Phi_y(y)$ converges and leaving it undefined if it never halts. This is clearly partial computable. By the s-m-n theorem, there is a computable function $f$ such that

$\Psi(x,y) = \Phi_{f(x)}(y)$

If $x \in \bar{K}$, then for all $y$, $\Phi_{x,y}(x) \uparrow$. Hence $\Phi_{f(x)}(y) = \Psi(x,y)$  halts if and only if $y \in K$. So $W_{f(x)} = K$ So $f(x) \in A$.

Suppose $x \notin \bar{K}$. That is $x \in K$. So $\Phi_x(x) \downarrow$. Let $s$ be the least stage such that $\Phi_{x,s}(x) \downarrow$. Then $\Phi_{f(x)}(y)$ converges only for those $y \in K$ and $y < s$. So $W_{f(x)}$ is actually finite. $K$ is clearly infinite, so $f(x) \notin A$.

So this $f$ is the many to one reduction of $\bar{K} \leq_m A$. $K$ is not computable so $\bar{K}$ is not computable. So $A$ is not computable.


问:
Prove that $\sum_{n = 1}^\infty \frac{(-1)^n}{\sqrt{n}} \sin(1+\frac{x}{n})$
I know I can't use Weierstrass M-test for uniform convergence because it's not all positive elements. What other criteria can I use? It is the $(-1)^n$ that gives me conflict.

Sorry if I ask obvious questions, but I'm learning everything from books, and the books write that this is obvious.

答:
Note that $\sin \left(1 + \frac{x}{n}\right) = \sin(1)\cos \left(\frac{x}{n}\right)+ \cos(1)\sin \left(\frac{x}{n}\right),$ and

$$\tag{*}\sum_{n=1}^m\frac{(-1)^n}{\sqrt{n}}\sin \left(1 + \frac{x}{n}\right) = \sin(1)\sum_{n=1}^m\frac{(-1)^n}{\sqrt{n}}\cos \left(\frac{x}{n}\right)+ \cos(1)\sum_{n=1}^m\frac{(-1)^n}{\sqrt{n}}\sin \left(\frac{x}{n}\right)$$

For the second sum on the RHS of (*), we have uniform convergence on any compact subset $D$ (where $|x| \leqslant M$) by the Weierstrass M-test since

$$\left|\frac{(-1)^n}{\sqrt{n}}\sin \left(\frac{x}{n}\right) \right| = \frac{\left|\sin \left(\frac{x}{n}\right) \right| }{\sqrt{n}} \leqslant\frac{\left|\frac{x}{n}\right| }{\sqrt{n}} \leqslant \frac{M}{n^{3/2}}  $$

For the first sum on the RHS of (*), note that  $\sum \frac{(-1)^n}{\sqrt{n}}$is a convergent alternating series and is uniformly convergent as there is no $x$-dependence. Furthermore the sequence $\cos\left(\frac{x}{n} \right)$ is uniformly bounded and eventually monotone since for all $n > \frac{2M}{\pi}$ we have

$$- \frac{\pi}{2}< -\frac{M}{n} \leqslant \frac{x}{n} \leqslant \frac{M}{n} < \frac{\pi}{2},$$

so that $\cos \left(\frac{x}{n}\right)$ is monotone increasing as $n \to \infty$. By Abels' test the sum converges uniformly on $D$.


问:
Matrix Multiply and Finding Scalar Multiplication is need with a fast method?
I want to find the optimal scalar multiply for following matrix:

Answer is $405$. it means this is not homework !

I see a nice link Here wrote "For the example below, there are four sides: A, B, C and the final result ABC. A is a 10×30 matrix, B is a 30×5 matrix, C is a 5×60 matrix, and the final result is a 10×60 matrix. The regular polygon for this example is a 4-gon, i.e. a square..."

My matrix is: $A(5*10), B(10*3), C(3*12), D(12*5)$

is there any way to without lot's of calculation on MM with Dynamic Programming and using above method find there is $405$ scalar matrix is here for Multiplying this matrix in efficient way?

I talked about polygon method here...or any other method that can answer this question quickly? how many scalar multiplication is need efficiently for multiplying this matrix?

答:
When you multiply two matrices, $\mathbf{A}$ with $r$ rows and $n$ columns, and $\mathbf{B}$ with $n$ rows and $c$ columns, the result is matrix $\mathbf{C}$ with $r$ rows and $c$ columns, via
$$C_{i j} = \sum_{k=1}^{n} A_{i k} B_{k j} \quad \text{ for } \quad i = 1 \dots r, \quad j = 1 \dots c$$
which involves $r n c$ scalar multiplications.

When you multiply more than two matrices, you can choose the order freely, but it can affect how many scalar multiplications you need to do overall.

If you have three matrices, you can do either multiplication first, so there are $1 \cdot 2 = 2$ possible orders of doing the multiplications.
If you have four matrices, you have $1 \cdot 2 \cdot 3 = 6$ possible orders of doing the multiplications.
(See Combinatorics for details as to why this is.)
Simply put, if you have $N$ matrices to multiply, you have $(N-1)!$ possible orders of multiplying the matrices.

Since there are only four matrices here, there are only six possible ways/orders of doing the multiplications, and we can explore them all.

We have:
$$\begin{array}{c|c|c}
\text{Matrix} & \text{Rows} & \text{Columns} \\
\hline
\mathbf{A} & 5 & 10 \\
\mathbf{B} & 10 & 3 \\
\mathbf{C} & 3 & 12 \\
\mathbf{D} & 12 & 5 \\
\end{array}$$

$(((\mathbf{A}\mathbf{B})\mathbf{C})\mathbf{D}$:

$\mathbf{A}\mathbf{B}$: $5 \cdot 10 \cdot 3 = 150$ multiplications, yields a $5$-row, $3$-column matrix

$(\mathbf{A}\mathbf{B})\mathbf{C}$: $5 \cdot 3 \cdot 12 = 180$ multiplications, yields a $5$-row, $12$-column matrix

$((\mathbf{A}\mathbf{B})\mathbf{C})\mathbf{D}$: $5 \cdot 12 \cdot 5 = 300$ multiplications, yields a $5$-row, $5$-column matrix

Total $150+180+300 = 630$ scalar multiplications using this order

$(\mathbf{A}\mathbf{B})(\mathbf{C}\mathbf{D})$:

$\mathbf{A}\mathbf{B}$: $5 \cdot 10 \cdot 3 = 150$ multiplications, yields a $5$-row, $3$-column matrix

$\mathbf{C}\mathbf{D}$: $3 \cdot 12 \cdot 5 = 180$ multiplications, yields a $3$-row, $5$-column matrix

$(\mathbf{A}\mathbf{B})(\mathbf{C}\mathbf{D})$: $5 \cdot 3 \cdot 5 = 75$ multiplications, yields a $5$-row, $5$-column matrix

Total $150 + 180 + 75 = 405$ scalar multiplications using this order

$(\mathbf{A}(\mathbf{B}\mathbf{C}))\mathbf{D}$:

$\mathbf{B}\mathbf{C}$: $10 \cdot 3 \cdot 12 = 360$ multiplications, yields a $10$-row, $12$-column matrix

$\mathbf{A}(\mathbf{B}\mathbf{C})$: $5 \cdot 10 \cdot 12 = 600$ multiplications, yields a $5$-row, $12$-column matrix

$(\mathbf{A}(\mathbf{B}\mathbf{C}))\mathbf{D}$: $5 \cdot 12 \cdot 5 = 300$ multiplications, yields a $5$-row, $5$-column matrix

Total $360 + 600 + 300 = 1260$ scalar multiplications using this order

$\mathbf{A}((\mathbf{B}\mathbf{C})\mathbf{D})$:

$\mathbf{B}\mathbf{C}$: $10 \cdot 3 \cdot 12 = 360$ multiplications, yields a $10$-row, $12$-column matrix

$(\mathbf{B}\mathbf{C})\mathbf{D}$: $10 \cdot 12 \cdot 5 = 600$ multiplications, yields a $10$-row, $5$-column matrix

$\mathbf{A}((\mathbf{B}\mathbf{C})\mathbf{D})$: $5 \cdot 10 \cdot 5 = 250$ multiplications, yields a $5$-row, $5$-column matrix

Total $360 + 600 + 250 = 1210$ scalar multiplications using this order

Same as 2., but with $\mathbf{C}\mathbf{D}$ first; that does does not change the number of scalar multiplications needed

$\mathbf{A}(\mathbf{B}(\mathbf{C}\mathbf{D}))$:

$\mathbf{C}\mathbf{D}$: $3 \cdot 12 \cdot 5 = 180$ multiplications, yields a $3$-row, $5$-column matrix

$\mathbf{B}(\mathbf{C}\mathbf{D})$: $10 \cdot 3 \cdot 5 = 150$ multiplications, yields a $10$-row, $5$-column matrix

$\mathbf{A}(\mathbf{B}(\mathbf{C}\mathbf{D}))$: $5 \cdot 10 \cdot 5 = 250$ multiplications, yields a $5$-row, $5$-column matrix

Total $180 + 150 + 250 = 580$ scalar multiplications using this order

Therefore, we can calculate the product using $405$, $580$, $630$, $1210$, or $1260$ multiplications, without affecting the result (assuming no loss of precision in a multiplication operation – this is not exactly true when using e.g. floating-point numbers, as then some small variation due to limited precision may occur).

The simple approach to find the minimum number of multiplications to calculate the matrix chain product is to calculate the number of operations for all possible multiplication orders as above.  For $N$ matrices there are $(N-1)!$ possible orders, and examining each order requires $2(N-1)$ multiplications, so this brute force optimum finding operation itself requires a total of $2(N!)-2(N-1)!$ multiplications. As a sequence of increasing $N$ starting at 3, this is $8$, $36$, $192$, $1200$, $8640$, $70560$, $645120$, $6531840$, $72576000$, $878169600$, $11496038400$, and so on; meaning that if you have say eleven or twelve or more matrices to multiply, even the brute force search to find the optimum order to calculate the chain product efficiently takes too long!

As you can see from the above list, many of those multiplications are the same, and unnecessarily repeated.  The first dynamic programming option is memoization, or remembering the already calculated products.
See e.g. the Wikipedia Matrix chain multiplication, which says this brings the time complexity down to $O(N^3)$, which is already a significant improvement.

Much better improvement, at $O(N \log N)$ time complexity, is the polygon-based approach published by Hu and Shing in 1982 (Part I, PDF) and 1984 (Part II, PDF). If you consider the full matrix chain product a polygon, each matrix-matrix multiplication has similar rules to ear clipping in polygon triangulation.

It turns out that we can use a simple polygon with the chain product on one side (called base) and each matrix multiplicand as the other sides, to represent the matrix chain product operation in terms of scalar multiplication operations needed.  Each triangulation of that polygon represents a different way of calculating the matrix chain product. Each triangle represents one matrix-matrix product, with each vertex denoting the number of rows or columns.  Then, the product of the three vertices for each triangle is equal to the number of scalar multiplication operations needed to calculate that matrix-matrix product.

Hu and Shing discovered a way of finding the triangulation (partitioning) of such a polygon that minimises the sums of those products – and therefore represents the order that calculates the matrix chain product with the fewest number of scalar multiplication operations – in $O(N \log N)$ time!  This is, simply put, amazing improvement; almost unbelievable, unless you realize they found a more efficient method of describing the problem, and from there, a way to find the optimum solution very efficiently.  Ingenious, really.

Therefore, when calculating the product of many matrices – say dozen or more – of different sizes, using the Hu & Shing algorithm to find the optimum order of performing the matrix-matrix multiplications, is very important for efficiency.

I am not familiar enough with the two articles, so I cannot help with implementing those articles as example code right now.  They seem to be using similar geometric reasoning I've seen in other articles regarding triangulation, so perhaps other members here more familiar with triangulation and the above-linked Hu & Shing articles can be of help.


答:
Let $a = (x^2 + 5)$. Then $$(a-9)(a-6)=0$$

$$\implies a = 9\; or \;a = 6$$
$$\implies x^2+5 = 9 \; or \;6$$
$$\implies x = \pm\sqrt4\; or \pm\sqrt1$$
$$\implies x = \pm2\; or \pm1$$

This kind of equation is called a biquadratic equation. Cheers!

Title: Solve $(x^2 + 5)^2 - 15(x^2 + 5) + 54 = 0$

Body:
I got the square root of 14 and 11 but the answer book states that these answers are wrong. Can someone help me? I used this formula to find the individual roots </p>

<p>$x = -\frac{p}{2} \pm \sqrt{(\frac{p}{2})^2 - q}$ </p>


问:
Find Laurent series at $z_0 = 1$

Find the Laurent series that converges for $0 < |z - z_0| < R$ and determine the precise region of convergence. Show details

$\frac{e^z}{(z-1)^2}$ around $z_0 = 1$

So if I'm going to compare this to another well-known Taylor series, what could I do? It looks like the geometric series but there's a power. Can someone show me the way?

答:
You can start with the Taylor series of $e^z$ at $z_0 = 1$:
$$
e^z = \sum_{n=0}^\infty \frac{e}{n!} (z-1)^n
= e + e(z-1) + \sum_{n=2}^\infty \frac{e}{n!} (z-1)^n
$$

Dividing this by $(z-1)^2$ leads to the Laurent series of $\frac{e^z}{(z-1)^2}$ at $z_0 = 1$:
$$
\frac{e^z}{(z-1)^2}
= \frac{e}{(z-1)^2} + \frac{e}{z-1} + \sum_{n=0}^\infty \frac{e}{(n+2)!} (z-1)^n
$$

To calculate the radius $R$, let $a_n := \frac{e}{(n+2)!}$. Then
$$
\frac 1 R = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{1}{n+3} = 0 \; ,
$$
so $R = \infty$.

Title: Find Laurent series at $z_0 = 1$


问:
How many strings (alphanumerical) of length 7 have exactly one number?
My reasoning seems to fail here : I say there's  $26^6\times 10$*, but my answer is wrong and I don't understand why.

EDIT 1 : Thanks to Carl I realise I need to use a combination because I'm not counting differently ordered strings as being different. So why isn't *$26^6\times {10 \choose 1}$ the right answer ?

答: The quantity $26^6\cdot 10$ tells you a) how to pick the $6$ letters in order, and b) how to pick the one digit, but it doesn't tell you where the digit goes in the string. For example, you're not classifying the strings "8aabaea" and "aa8baea" as different. What do you need to change?


答:
You pick

the first letter

the second letter

the third letter

the fourth letter

the fifth letter

the sixth letter

the one and only digit

the position of the one and only digit

Title: How many strings (alphanumerical) of length 7 have exactly one number?

Body:
My reasoning seems to fail here : I say there's  $26^6\times 10$*, but my answer is wrong and I don't understand why.

EDIT 1 : Thanks to Carl I realise I need to use a combination because I'm not counting differently ordered strings as being different. So why isn't *$26^6\times {10 \choose 1}$ the right answer ?


问:
Why is the derivative of $b$ W.R.T $x$ $0$ in this problem?
$$\lim_{x \to \infty}{\left(\frac{x^2+1}{x+1} - ax - b\right)} =0$$

And you need to find the values of $a$ and $b$.

Now what will be the derivative of $b$ W.R.T $x$.

I have seen people answering $0$. But I don't understand how is it $0$ because $b$ does depend on $x$. For example if $x$ was approaching 100, we would get a different value of $b$ now. So $b$ does depend on $x$, so why do we take the derivative of $b$ W.R.T $x$ as $0$

答:
You have $$\frac{x^2+1}{x+1} - ax - b = x \left(\frac{x+ \frac{1}{x} }{x+1}-a\right) -b$$

First, if $a \neq 1$, then the above parenthesis does not tend to $0$ and then, the wholde expression cannot tend to $0$. So necessarily, $a=1$. Secondly, $$\frac{x^2+1}{x+1} - x - b = x \left(\frac{ \frac{1}{x}-1 }{x+1}\right) -b$$

which tends to $-1-b$, so for this to be equal to $0$, you must have $b=-1$.

So the values are $(a,b)=(1,-1)$, and your question on the derivatives has not a lot of sense.


问:
The MLE of this weird density function
I'm trying to solve the first question of this problem set.
I'm a beginner in this subject and it's the very first question I'm trying to solve. I really need help.

The question is:

Let $X_1,\ldots,X_n$ be $n$ i.i.d. random variables with density
$f_{\theta}$ with respect to the Lebesgue measure. For each case below
find the MLE of $\theta$.

$$f_{\theta}(x)= \theta\tau^{\theta}x^{−(\theta+1)}\unicode{x1D7D9}_{x
 \ge\tau},\ \theta>0$$

where $\tau>0$ is a known constant.

I supposed the function is concave and I took the derivative of the log and I equaled to zero:

$$\hat\theta = \frac{n}{\log(\frac{x_1\cdots x_n}{\tau^n})}$$
provided that $x_1,\ldots, x_n$ are greater or equal than $\tau$. The MLE is zero otherwise.

am I right?

答:
It's ok.

With different notation I get

$$\hat{\theta}_\text{ML}=\frac{n}{\sum_i \log X_i-n\log \tau}=\frac{n}{\log \prod_i X_i-\log \tau^n}=\frac{n}{\log\frac{\prod_i X_i}{ \tau^n}}$$

Just note that $X_i$ are rv's, so capital letter is required. They become $x_i$ after seeing their realizations

Now, if you want, you can now calculate also $\hat{\tau}$ supposing it is not known

Title: The MLE of this weird density function


问:
How to determine the number of divisors of an integer through combinatorics
How would you formally derive the fact that integer $100$ has 9 divisors $(1,2,4,5,10,20,25,50,100)$, knowing in advance that, apart from 1, its prime divisors are 2 and 5, each recurring 2 times $(2 \cdot 2 \cdot\ 5 \cdot 5 = 100)$?

This should be achievable using counting rules in combinatorics.

答: Yes, it is possible. Suppose we have $n = p_1^{k_1}...p_m^{k_m}$ where $p_1,...,p_m$ are distinct primes and $k_1,...,k_m \ge 1$. Then, for each prime factor $p_i$, we have the options $\{0,1,...,k_i\}$ for its power $k_i$ since $p_i^a|p_i^{k_i}, \forall a \text{ such that } 0 \le a \le k_i$. So we have $k_i+1$ options for each. Considering the fact that the prime factors are independent from other prime factors (as they are prime numbers), can you get a general result from here?


答: You're essentially trying to find the slant asymptote of $(x^2+1)/(x+1)$ when $x\to +\infty$. A rational function has a slant asymptote if and only if the degree of the numerator is precisely one higher than the degree of the numerator. By long-division (for instance), it's easy to see that $$\frac{x^2+1}{x+1} = x-1 +\frac{2}{x+1}.$$Thus $a=1$ and $b=-1$. I do not know what you mean by taking the derivative of $b$ with respect to $x$: both $a$ and $b$ are determined constants. The parameter $b$ is not a function of $x$. The limit is zero and you don't have any equality between the rational function and the asymptote, just the limit relation (hence asymptote).

Title: Why is the derivative of $b$ W.R.T $x$ $0$ in this problem?

Body:
$$\lim_{x \to \infty}{\left(\frac{x^2+1}{x+1} - ax - b\right)} =0$$

And you need to find the values of $a$ and $b$.

Now what will be the derivative of $b$ W.R.T $x$.

I have seen people answering $0$. But I don't understand how is it $0$ because $b$ does depend on $x$. For example if $x$ was approaching 100, we would get a different value of $b$ now. So $b$ does depend on $x$, so why do we take the derivative of $b$ W.R.T $x$ as $0$


答:
The divisor counting function is usually noted with $\tau(n)$ (tau). In other words, $n$ has $\tau(n)$ divisors.

$$n=\prod_{i=1}^{m}p_m^{a_m}\Rightarrow\tau(n)=\prod_{i=1}^{m}(a_i+1)$$

The proof for this? Well observe that $p^k$ with prime $p$ has indeed $k+1$ divisors ($1,p,...,p^k$). Finally, observe that $$\tau\bigg(\prod_{i=1}^{m}p_m^{a_m}\bigg)=\prod_{i=1}^{m}\tau(p_i^{a_i})$$ and you are done.

Title: How to determine the number of divisors of an integer through combinatorics

Body:
How would you formally derive the fact that integer $100$ has 9 divisors $(1,2,4,5,10,20,25,50,100)$, knowing in advance that, apart from 1, its prime divisors are 2 and 5, each recurring 2 times $(2 \cdot 2 \cdot\ 5 \cdot 5 = 100)$?

This should be achievable using counting rules in combinatorics.


问:
How to solve a linear congruence $8n+9\equiv 0\pmod{\!1163}$
Problem :

Solve in $\mathbb N$ :

$8n+9\equiv 0\pmod{1163}$

The mathematics give me : $n=435+1163k$ , $k\in\mathbb N$

$1163$ is a prime number

$8n\equiv -9\pmod{1163}$

$8n\equiv 1154\pmod{1163}$

I don't know any ideas to complete my work ?

I have to see your hints

答:
If you are a one trick pony you can get the answer

using the (not always useful) shortcut found here:

$\; 8n \equiv -9\pmod{1163} \; \text{ iff }$

$\; 8n \equiv 1154\pmod{1163} \; \text{ iff }$

$\; 4n \equiv 577\pmod{1163}$

And since $4 \mid 1164$, the solution is

$\quad x = \Large(\frac{1164}{4}) \normalsize (577) = (291)(577) = 167907 \equiv 435 \pmod{1163}$

Of course a better trick is the one given by Laassila souhayl ($+1$).


问:
Trying to get an intuitive understanding of compositions of rotations in $\mathbb{R}^3$
Lately I've been looking at the special orthogonal group $SO_\mathbb{R}(3)=\{M\in O(3),\, \det M =1\}$, where $O_{\mathbb{R}}(3)$ is the group of real matrices such that $AA^\top=I$. I'm looking to get a intuitive understanding of a "simple fact".

To my understanding its elements can be seen as rotations around a line through the origin (ie a subspace of dimension 1). $ SO_\mathbb{R}(3)$ forms a group with respect to matrix multiplication, so the product of two elements is still an element of the group. That is to say, if one composes two rotations, around two possibly different lines (through the origin), the resulting automorphism of $\mathbb{R^3}$ is another rotation around some line through 0. Why should that be?  Algebraically it checks out, but intuitively I can't see why it should be the case that combining rotations around different axes would have to be another rotation. Why is there always one line left unmoved, fixed by the composition of rotations?
Anyway, if anyone has a good mental image or intuitive perspective they'd like to share, I'd appreciate.

答:
It feels intuitive to me that any orientation preserving isometry of the sphere has at least a fixed point. I mean think of pulling one point of the sphere to another in a (more or less) straight line. The points orthogonal to the plane spanned by the path will stay fixed.

In case you know something about algebraic topology (ie. $H_2(\Bbb S^2)=\Bbb Z$) we can give a lovely proof of this fact.

First note that the composition of two rotations induces an isometry $\Bbb S^2 \rightarrow \Bbb S^2$ and that it suffices to show that this map has a point satisfying $f(x_0)=\pm x_0$. Being an isometry we will then have $f(-x_0)=\mp x_0$, as well so the span of ${x_0}$ is an axis of rotation. Technically there could be an inflection involved as well, but with some work (an inflection would change the orientation given by the fundamental class $[\Bbb S^2] := 1 \in H_2(\Bbb S^2)$ to $(-1)^3=-1=-[\Bbb S^2]$, but rotations preserve it and so do two rotations) we can indeed show $f(x_0)=x_0$.

Now we are left to show that any map $f:\Bbb S^2 \rightarrow \Bbb S^2$ has a point $f(x_0)=\pm x_0$. Assuming otherwise we can state explicit homotopies $\operatorname{id} \sim f \sim -\operatorname{id}$. But then the action of $\operatorname{id},-\operatorname{id}$ on the fundamental class would result in $1=(-1)^3=-1$, a contradiction.

I feel like I will have to defend this answer, so let me say this: What I like of this proof is that it is really conceptual, if one believes in homology, which is basically just a framework allowing you to speak about orientation (among other things). In my opinion this proof better mimics the intuition than algebraic manipulation of matrices does.


问:
If $f(x)$ is increasing and non-negative, then is it true that $xf(x)$ is convex on $ [0,+\infty)$?
If $f(x)$ is increasing and non-negative, then is it true that $xf(x)$ is convex on $ [0,+\infty)$?

##########
Update:

If $f(x)$ is second-order differentiabl, strictly increasing and non-negative on its domain, then is it true that $xf(x)$ is convex on $ [0,+\infty)$?

Would there be a neat and intuitive counterexample?

@Lapin gave an constructed conterexample.

If we allow $f(x)$ to be non-decreasing, then the counterexample proposed by @KaviRamaMurthy is very neat:
$$ f(x)=\left\{
\begin{aligned}
0, x\le 0 \\
1-e^{-x^2}, x>0
\end{aligned}
\right.
$$

答:
For the edited question, $
g(x)=\begin{cases}
0,&x\le 0,\\
\exp(-x^{-2}),&x>0,
\end{cases}$, $h(x)=\dfrac\pi2+\arctan(x),
$ then$f(x)=g(x)+0.00001h(x)$
is still a counter-example. It is infinitely differentiable, strictly increasing and non-negative on $\mathbb R$.


问:
How to determine confidence interval using central limit theorem?
How to determine the confidence interval for the unknown theta parameter of the Uniform($[0, \theta]$)-distribution using the central limit theorem, considering that the significance level is given and $\theta > 0$?

When generating the sample, theta should be considered as known and afterwards should be checked whether or not it's in the interval.

I know it can somehow be solved by generating a uniform sample, calculating the value for z and then using a polynomial with coefficients $a, b$ and $c$, then generating a $\delta$ and the roots will be the interval's endpoints, but I don't understand how I can get those coefficients.
I have a sample of size $N$ and I calculated the mean and the value for z but I'm stuck at this point. How do I calculate those coefficients for the polynomial?

答:
CLt is a very bad choise to calculate your Confidence interval. I do not know if this is explicitly asked or it is your own choice. Anyway, and in any case, first you have to derive your pivot quantity that is

$$Y=\frac{\text{Max}(X_i)}{\theta}$$

Knowing that, a Confidence interval can be easily calculated basing it on $Y$, amd leading to the following confidence interval:

$$\Bigg[X_{(n)};\frac{X_{(n)}}{\sqrt[n]{\alpha}}\Bigg]$$

Where $\alpha$ is the given significance level and $X_{(n)}$ is the maximum of the $X_i$

To use CLT I think you can start with estimating $\theta$ with Method of moments and not with ML. Thus you get

$$\hat{\theta}_{MoM}=2\overline{X}_n$$

And now, assuming $n$ great enough, you can use CLT, as $\overline{X}_n$ is asimptotically Normal (but it is a very ugly road...)


问:
Find centroid of shape given by a circle and a cardoid
I'm stuck on finding the centroid of this shape (the shaded area):

With the sphere having a radius of $r=2$, and the cardioide being given by $r=1+\cos\phi$

I found the Area, which is
$A=\frac12(2\pi r^2-\frac32 \pi)=\frac{13}{4} \pi$

Now I don't know how to proceed further. Could anyone point me in the right direction?
Thank you :)

答:
There is a mistake in your area calculation -

$A = \frac12(\pi r^2-\frac32 \pi)=\frac{5}{4} \pi$ (you wrote $2\pi r^2 $ instead of $\pi r^2$ by mistake)

Now coming to finding centroid,

We know in polar coordinates, $x = r\cos \theta, y = r \sin \theta$ so the centroid ($\overline{x}, \overline {y}$) of the polar region is given by

$\displaystyle \overline{x} = \dfrac{2}{3} \frac {\int_{\alpha}^{\beta} r^3 \cos\theta \,  \mathrm{d} \theta}{\int_{\alpha}^{\beta} r^2 \,  \mathrm{d} \theta} \,$ (numerator is $r \cos \theta .(r^2 d\theta))$

Similarly, $\displaystyle \overline{y} = \dfrac{2}{3} \frac {\int_{\alpha}^{\beta} r^3 \sin\theta \,  \mathrm{d} \theta}{\int_{\alpha}^{\beta} r^2 \,  \mathrm{d} \theta}$

The denominator is nothing but $2A \,$ where $A$ is the area of the region which you have already found. Also consider the polar region as a collection of infinitely small triangles and that is how this formula.

In this case our shaded region is bound by $r = (1 + \cos \theta) \,$ below and $r = 2$ above. Please also note that both curves go between $0 \leq \theta \leq \pi$ as we are interested in shaded area above $X$-axis. So,

$\displaystyle \int_{\alpha}^{\beta} r^3 \cos\theta \,  \mathrm{d} \theta = \displaystyle \int_{0}^{\pi} [2^3 - (1 + \cos \theta)^3] \cos\theta \,  \mathrm{d} \theta = -\frac{15 \pi}{8}$

$\displaystyle \int_{\alpha}^{\beta} r^3 \sin\theta \,  \mathrm{d} \theta = \displaystyle \int_{0}^{\pi} [2^3 - (1 + \cos \theta)^3] \sin\theta \,  \mathrm{d} \theta = 12$

$\displaystyle \int_{\alpha}^{\beta} r^2  \,  \mathrm{d} \theta \,$ is nothing but $2A$ and as you already found the area, our denominator is $\frac{5 \pi}{2}$.

So $\overline{x} = \frac{2}{3} \times \frac{-15 \pi/8}{5\pi/2} = -\frac{1}{2}$

$\overline{y} = \frac{2}{3} \times \frac{12}{5\pi/2} = \frac{16}{5\pi}$


问:
Basic probability question about delivery watches
The question is stated as:

A shipment of watches is shipped in two boxes. One contains 150 watches of which
21 are broken. The second box contains 100 of which 4 are broken. A box is chosen
at random and then from the chosen box a watch is picked at random. It turns out to
be broken. What is the probability is was shipped in the first box?

My attempt:

Define the following events,

$S_1 := \text{The watch is shipped in the first box.}$

$ B_1 :=\text{The first box is choosen}$

$A := \text{The choosen watch is broken}$

The probability we want to calculate is then given by bayes theorem,

$P(S_1|A)=\frac{P(A|S_1)P(S_1)}{P(A)}$

My question: is $P(A) = \frac{25}{250}$ or do you need to use the law of total probability

$P(A) = P(A|B_1)P(B_1) + P(A|B_1^c)P(B_1^c)$ ? Or is the approach wrong in general?

答:
I'll just go ahead and post a full solution. I'll change the notation a bit to my own taste. Let

$A=$watch is broken

$B_1=$ box 1 is chosen

$B_2=$ box 2 is chosen

The following probabilities are given:

$P(A|B_1) = \frac{21}{150}$

$P(A|B_2) = \frac{4}{100}$

$P(B_1) = P(B_2) = \frac{1}{2}$

The probability we want to calculate is
$P(B_1|A)$ and indeed we can use Bayes' theorem for this:
$$
P(B_1 | A) = \frac{P(A|B_1)P(B_1)}{P(A)}
$$
The denominator is always the tricky part in these calculations. We can use the law of total probability:
$$
P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2)
     = \frac{21}{150} \frac{1}{2} + \frac{4}{100}\frac{1}{2}
     = \frac{9}{100}
$$
Now we know all the values required, and we can just plug in the values to the final expression:

$$ P(B_1 | A) = \frac{\frac{21}{150} \frac{1}{2}}{\frac{9}{100}} = \frac{7}{9} $$


问:
Show that the localization $k[t]_{(t)}$ is a valuation ring
I want to see that $B=k[t]_{(t)}$ is a valuation ring. To show this I need to show that for every $x\in \operatorname{Frac}(B)$ we have at least one of $x\in B$ or $x^{-1}\in B$.
Intuitively i think $\operatorname{Frac}(B)$ should be $k(t)$ but I can't think of succinct way of showing this.

Let $x\in k(t)$ then $x=f/g$ with $f,g\in k[t]$. An element $y\in B$ looks like $h/t^n$. So we need to show that $f/g=h/t^n$ or $g/f=h/t^n$. I think we I should try use that $k[t]$ is a UFD and get a nice relation between $f$ and $g$. We can assume $f,g$ to be coprime,but even them I'm not sure ho we can deal with something like this $x=\frac{t-1}{t-2}$. If we could show that that $t$ must divide $f$ or $g$ we would be done but I am not sure if this is true.

Questions:
Why is $\operatorname{Frac}(B)=k(t)$?
Why can we assume $t$ must divide $f$ or $g$?

答:
$k[t]_{(t)}$ is a localization of $k[t]$, so it is canonically a subring of $\operatorname{Frac} k[t]=k(t)$. If $F$ were any field containing $k[t]_{(t)}$ properly contained in $k(t)$, this would contradict the fact that $k[t]$ has fraction field $k(t)$.

Your second problem has a key misunderstanding: $k[t]_{(t)}$ is defined to be the localization at the set of elements not in $(t)$, not the localization at $\{1,t,t^2,t^3,\cdots\}$. So $x$ is already in $k[t]_{(t)}$, for instance. Try again with this new information - I've put a full explanation under a spoiler below for you.

 If $f/g\in k(t)$ is written in lowest terms, then $t$ divides at most one of $f$ and $g$. If $t$ does not divide $g$, then $g\notin (t)$ so $f/g$ is already a member of $k[t]_{(t)}$. If $t$ divides $g$, then $t$ doesn't divide $f$, so $g/f$ is in $k[t]_{(t)}$.


问:
Integration of a rational function from +/- infinity
I am trying to calculate the integral

$$\int_{-\infty}^{\infty}{\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}}dx$$

where $\{a, b, c\}\in \mathbb{R}$. I have looked in a table of integrals for rational functions, but with no luck. Is there a smart trick I can utilize?

答:
Hint: Use partial fraction to write the integral as

$$ {\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}} = \frac{A+Bx}{b^2 + (a+x)^2} + \frac{C+D x}{1+c(a-x)^2}, $$

then you can write it as

$$ \frac{A}{b^2 + (a+x)^2} + \frac{C}{1+c(a-x)^2} + \frac{Bx}{b^2 + (a+x)^2} + \frac{D x}{1+c(a-x)^2}, $$

and finally make the change of variables $x-a=t$ and $x+a=t$ and things will be easy. See here for similar techniques and read the comments under my answer.


问:
What does a preorder have to do with the notion of ordering?
Given a set $\{a,b,c,d\}$, does the set of relations $\{a\leq b, b\leq c,c\leq d,d\leq a\}$ give rise to a preorder? If so, what does a preorder have to do with the notion of ordering if $a$ can be less than $b$, $b$ less than $a$, and $a\neq b$?

答:
A preorder is like a mix of a partial order and an equivalence relation. The transitive closure of your relations is a preorder in which every pair of objects is equivalent (equivalent means $a \le b$ and $b \le a$ - you can check that this is always an equivalence relation for any preorder), so they're all "the same" as far as the preorder is concerned. In other words, a preorder allows "ties."

The sense in which a preorder is "pre" a partial order is that every preorder gives rise canonically to a partial order obtained by identifying equivalent objects. Formally this is the left adjoint of the inclusion of partial orders into preorders.

It's a lot like considering topological spaces that are not $T_0$, and in fact the two can be related precisely using the notion of the specialization preorder of a topological space, which gives an equivalence between finite topological spaces and finite preorders restricting to an equivalence between finite $T_0$ spaces and finite partial orders.


问:
Same character table implies two groups have the same cardinality?
Suppose $G_1,G_2$ are two finite groups and their character tables are the same. Can I prove or disprove that $|G_1|=|G_2|$?

I have just started on character theory and I have been thinking proving this using row orthogonality and maybe also column orthogonality.

I have developed some thoughts using row orthogonality but I have not really gone far. Suppose $\{g_1,\cdots g_n\}$ is the complete set of conjugacy classes of $G_1$ and $\{h_1,\cdots,h_n\}$ is the conjugacy class of $G_2$. WLOG assume $g_i$ and $h_i$ are arranged such that they are in increasing order of cardinality so $|g_1|\leq|g_2|$ etc. Then my instinct is telling me that to satisfy the row orthogonality we needed $|h_i|=k|g_i|$ for some $k\in\mathbb{N}.$ However I am not sure if this is true or if there is a way of proving so.

How could I proceed? Any hint is much appreciated!

答: Hint: use the formula: $\sum_{\chi \in Irr(G)}\chi(1)^2=|G|$ (which basically is an orthogonality relation)


问:
Duality pairing between finite signed measures and bounded continuous functions
Let $E$ be a metric space, $C_b(E)$ denote the set of real-valued bounded continuous functions on $E$ and $\mathcal M(E)$ denote the set of finite signed measures on $\mathcal B(E)$.

How can we show that $$\langle f,\mu\rangle:=\int f\:{\rm d}\mu\;\;\;\text{for }(f,\mu)\in C_b(E)\times\mathcal M(E)$$ is a duality pairing between $C_b(E)$ and $\mathcal M(E)$? Do we need impose further restrictions on $E$ (e.g. completeness and/or separability) and/or $\mathcal M(E)$ (e.g. regularity of the measures)?

We need to show that \begin{align}\forall f\in C_b(E)\setminus\{0\}&:\exists\mu\in\mathcal M(E)&:\langle f,\mu\rangle\ne0\tag1;\\\forall\mu\in\mathcal M(E)\setminus\{0\}&:\exists f\in C_b(E)&:\langle f,\mu\rangle\ne0\tag2.\end{align}

$(1)$ Should be easy. If $f\in C_b(E)\setminus\{0\}$, there is a $x\in E$ with $f(x)\ne 0$. Now we can choose $\mu$ to be the Dirac measure $\delta_x$ concentrated at $x$ and obtain $\langle f,\mu\rangle=f(x)\ne0$.

But how can we show $(2)$?

答: If $C$ is any closed set and $U_n=\{x: d(x,C) <\frac 1 n\}$ then there exist continuous functions $f_n : E \to [0,1]$ such that $f_n(x)=1$ for all $x \in C$ and $f_n(x)=0$ if $x \notin U_n$. Since $\int f_n d\mu=0$ for all $n$ we get $\mu (C)=0$. If we write $\mu$ as $\mu_1-\mu_2$ where $\mu_1$ and $\mu_2$ are positive finite measures then we get $\mu_1(C)=\mu_2(C)$ for any closed set $C$. You can use $\pi -\lambda$ Theorem  to show that $\mu_1=\mu_2$.


答:
The “pre-” here is to suggest that this concept is somewhat order-ish, but is missing some of the properties that one would want an order to satisfy.  It's more order-like than a relation, but not as order-like as a total order, or even a partial order.

What properties would one like a relation to satisfy to be considered an order?  The prototypical example is numeric $<$:

Given any two elements $a$ and $b$, exactly one of $a<b$, $a=b$, and $b<a$ is true.

If $a<b$ and $b<c$, then $a<c$.

When this happens we have a "total order".

But sometimes it's useful to consider a more general idea, a "partial order".  In a partial order, we still have property (2), but property (1) becomes weaker:

Given any two elements $a$ and $b$, at most one of $a<b$, $a=b$, and $b<a$ is true.

There is less that you can say about partial orders, but there are more examples.  One typical example: suppose you have a set of tasks.  Some tasks are prerequisites for others. Write $a<b$ if task $a$ must be finished before task $b$ is begun.  It might happen that $a$ and $b$ are independent: they can be done in either order, or at the same time.  Then none of $a<b$, $a=b$, $a>b$ is true.

Sometimes it's useful to consider a "preorder", which is even more general.  Property (1) disappears completely.  But we still have property (2).  Again, there is less that you can say about preorders, but there are more examples.  One typical example is a communications network with one-way communications links.  Write $a<b$ if entity $b$ can send messages to entity $a$.  We might have both $a<b$ and $b<a$, or neither.  But if $a<b$ and $b<c$ then $a<c$ because $c$ can ask $b$ to relay a message to $a$.

If we have both $a<b$ and $b<a$ then that simply means that $a$ and $b$ have two-way communication.

Title: What does a preorder have to do with the notion of ordering?

Body: Given a set $\{a,b,c,d\}$, does the set of relations $\{a\leq b, b\leq c,c\leq d,d\leq a\}$ give rise to a preorder? If so, what does a preorder have to do with the notion of ordering if $a$ can be less than $b$, $b$ less than $a$, and $a\neq b$?


问:
Recurring decimals to fraction

$.\overline{36}=\frac{36}{99}$

$2.1\overline{36}=2\frac3{22}$

The part I do not understand however, is "you could used 1) to speed up the working of 2)" which is written in the book.

How would I use 1) to help me work out 2)?

Thanks.

答:
$.\overline{36}=\frac{36}{99}$

$2.1\overline{36}=2\frac3{22}$

part 1.

suppose
$$\begin {equation}\dfrac pq=0.\overline{36}\end {equation}$$
this is eqn (1).

mulipltly this eqn with 100
$$\begin {equation}100\dfrac pq=36.\overline{36}\end {equation}$$
this is eqn (2).Now subtract (1) from (2).
$$\begin {equation}100\dfrac pq-\frac pq=36.\overline{36}-0.\overline{36}\end {equation}$$
$$99\dfrac pq=36.0$$
$$\dfrac pq=\frac {36}{99}$$

part 2

$2.1\overline{36}=2+0.1\overline{36}$
$$\dfrac pq=0.1\overline{36}\implies 10\dfrac pq=1.\overline{36}$$this is eqn (1).

multilpy with 100 in eqn(1)
$$1000\dfrac pq=136.\overline{36}$$ then subtract (1)  from (2).
$$990\dfrac pq=135.0$$
$$\dfrac pq=\dfrac {135}{990}\implies\dfrac {3}{22}$$

so  $2.1\overline{36}\implies 2+0.1\overline{36}\implies 2+\dfrac{3}{22}\implies 2\dfrac3{22}$

Title: Recurring decimals to fraction


问:
Showing that a function is an automorphism of $\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$
I'm trying to prove that the function $\phi$ given $\phi(\sqrt{(2+\sqrt{2})(3+\sqrt{3})})=\sqrt{(2-\sqrt{2})(3+\sqrt{3})}$ and $\phi(q)=q$ for all $q\in \mathbb{Q}$ is an automorphism of  the extension field $\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3})}))/\mathbb{Q}$ but I can't prove the surjectivity.
More precisely, I cannot find an element $x$ in the extension field such that $\phi(x)=\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$
I try this: Assuming $\phi$ is a homomorphism (which I already proved)
Let $\alpha=\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$ and $\beta=\sqrt{(2-\sqrt{2})(3+\sqrt{3})})$
then
$\alpha=\frac{\alpha \beta}{\beta}=\frac{\sqrt{2}(3+\sqrt{3})}{\phi(\alpha)}$ and i would like that ${\sqrt{2}(3+\sqrt{3})}=\phi(y)$ some $y\in\mathbb{Q}(\alpha)$ but I get circular reasoning ...

答:
Mainlines for an answer

Denote

$$\begin{cases}
\pi &= \sqrt{2+ \sqrt 2}\\
\rho &= \sqrt{3+ \sqrt 3}
\end{cases}$$

$\pi, \rho$ are respectively roots of the rational polynomials
$$\begin{cases}
p(x) &= x^4 - 4 x^2 +2\\
r(x) &= x^4 - 9 x^2 + 3
\end{cases}$$

$p,r$ are irreducible according to Eisenstein's criterion. The set of roots

of $p$ is $\{ \pi_1=\sqrt{2+ \sqrt 2}, \pi_2=\sqrt{2- \sqrt 2}, \pi_3=-\sqrt{2+ \sqrt 2},\pi_4=-\sqrt{2-\sqrt 2}\}$

the one of $r$ is $\{ \rho_1=\sqrt{3+ \sqrt 3}, \rho_2=\sqrt{3- \sqrt 3}, \rho_3=-\sqrt{3+ \sqrt 3},\rho_4=-\sqrt{3-\sqrt 3}\}$.

$\alpha, \beta$ belong to the finite field extension $\mathbb Q(\pi, \rho)/\mathbb Q$ and the field homomorphism $\phi$ satisfies $\phi(\alpha) = \beta$.

You suppose that $\beta$ belongs to $\mathbb Q(\alpha) / \mathbb Q $ which remains to be proved.

To do so, notice that $\alpha \notin \mathbb Q(\pi, \sqrt 3)$. If that would be the case, $\rho$ would also belong to $\mathbb Q(\pi, \sqrt 3)$. Which in turn implies the contradiction that $\{1, \sqrt 2, \sqrt 3, \sqrt 6\}$ would be linearly dependent over $\mathbb Q$. Similarly $\beta$ doesn't belong to $\mathbb Q(\pi, \sqrt 3)$.

Therefore $\mathbb Q(\alpha) = \mathbb Q(\beta) = \mathbb Q(\pi, \rho)$ and if $\phi(\alpha) = \beta$, $\phi$ can be extended into an homeomorphism of $\mathbb Q(\alpha)$ which is also an automorphism.


问:
Why the solution for $1+y^{\prime}(x)^{2}-y(x) y^{\prime \prime}(x)=0$ is $y(x)=C_{1} \cosh \frac{x-\mathrm{C}_{2}}{C_{1}}$
Given $$1+y^{\prime}(x)^{2}-y(x) y^{\prime \prime}(x)=0$$
I don't understand how this is the solution for this differential equation. where $C_{1,2}$ are constants of boundary condition.

I substitude $$y(x)=Ae^x + Be^-x$$

and than I got $$AB=\frac{1}{4}$$ and than I concluded that $$A=B=\frac{1}{2}$$ thus the solution is just $$y(x)=cosh(x)$$
what I did wrong that I have no constants, and also why the solution $$y(x)=Ae^x + Be^-x$$ is work?

答:
This equation is separable,
$$
\frac{2y'y''}{1+y'^2}=\frac{2y'}{y}
$$
After that first integration you should be able to continue to solve the resulting first-order autonomous equation.

With your approach, you fixed too much arbitrarily. At least you should get $A=\frac12e^c$, $B=\frac12e^{-c}$ to get the solutions $y(x)=\cosh(x+c)$ that are to be expected for an autonomous equation. Then you could have a negative sign in both of the coefficients.

But your trial solution is too restrictive, note that the given equation is not linear, so in general you can not expect to get exponential solutions.


问:
Limit of the ratio of two non-Riemann sums.
Let $\left[ {a,b} \right] \subset \mathbb{R}$  and $f,g:\left[ {a,b} \right] \to \mathbb{R}$  be two Riemann-integrable functions.

Let $a = {x_0} < {x_1} < {x_2}... < {x_n} = b$  be a partition of $\left[ {a,b} \right]$  and let $\Delta x = \mathop {\max }\limits_{i = 0}^{n - 1} \left( {{x_{i + 1}} - {x_i}} \right)$.

Let ${t_i} \in \left[ {{x_i},{x_{i + 1}}} \right],\;i = 0,n - 1$  and let  $k \in {\mathbb{N}^*}$.

I’d like to prove that

$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sum\limits_{i = 0}^{n - 1} {{{\left( {{x_{i + 1}} - {x_i}} \right)}^k}f\left( {{t_i}} \right)} }}{{\sum\limits_{i = 0}^{n - 1} {{{\left( {{x_{i + 1}} - {x_i}} \right)}^k}g\left( {{t_i}} \right)} }} = \frac{{\int\limits_a^b {f\left( x \right){\text{d}}x} }}{{\int\limits_a^b {g\left( x \right){\text{d}}x} }}$

It is obvious for equally spaced partitions  ${x_{i + 1}} - {x_i} \equiv \Delta x$

$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sum\limits_{i = 0}^{n - 1} {\Delta {x^k}f\left( {{t_i}} \right)} }}{{\sum\limits_{i = 0}^{n - 1} {\Delta {x^k}g\left( {{t_i}} \right)} }} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sum\limits_{i = 0}^{n - 1} {\Delta xf\left( {{t_i}} \right)} }}{{\sum\limits_{i = 0}^{n - 1} {\Delta xg\left( {{t_i}} \right)} }} = \frac{{\int\limits_a^b {f\left( x \right){\text{d}}x} }}{{\int\limits_a^b {g\left( x \right){\text{d}}x} }}$

But I don’t see how to do it in the general case?

答:
I believe this reduces to two copies of a problem which has never been successfully solved in the history of mathematics. The best you could do is conjecture, but you should be able to at least show that your conjecture is reasonable.

As a matter of fact, you can't even prove the result which you claimed was obvious. It's obvious because we accept this definition of the integral. That does not mean that it can be proven unfortunately...


问:
The MLE of this weird density function
I'm trying to solve the first question of this problem set.
I'm a beginner in this subject and it's the very first question I'm trying to solve. I really need help.

The question is:

Let $X_1,\ldots,X_n$ be $n$ i.i.d. random variables with density
$f_{\theta}$ with respect to the Lebesgue measure. For each case below
find the MLE of $\theta$.

$$f_{\theta}(x)= \theta\tau^{\theta}x^{−(\theta+1)}\unicode{x1D7D9}_{x
 \ge\tau},\ \theta>0$$

where $\tau>0$ is a known constant.

I supposed the function is concave and I took the derivative of the log and I equaled to zero:

$$\hat\theta = \frac{n}{\log(\frac{x_1\cdots x_n}{\tau^n})}$$
provided that $x_1,\ldots, x_n$ are greater or equal than $\tau$. The MLE is zero otherwise.

am I right?

答:
We can also perform this calculation by considering the transformed random variable $Y_i = \log (X_i/\tau)$.  Then the density of $Y$ becomes $$f_Y(y) = f_X(\tau e^y) \tau e^y = \theta \tau^\theta (\tau e^y)^{-(\theta+1)} \tau e^y = \theta e^{-\theta y},  \quad y \ge 0.$$  This of course is an exponential distribution with rate $\theta$.  Then the MLE for $\theta$ is easily found:  $$\mathcal L(\theta \mid \tau, \boldsymbol y) \propto \theta^n e^{-\theta n \bar y} \mathbb 1(y_{(1)} \ge 0),$$ hence $$\ell(\theta \mid \tau, \boldsymbol y) \propto \log \theta - \theta \bar y.$$  The derivative of the log-likelihood is $$\frac{\partial \ell}{\partial \theta} = \frac{1}{\theta} - \bar y,$$ with critical point (and maximum) $$\hat \theta \mid \tau = 1/\bar y.$$  Then by the invariance principle, $$\hat \theta \mid \tau = \frac{n}{\sum_{i=1}^n \log (x_i/\tau)}.$$  This assumes $\tau$ is known.

We could also do the MLE calculation directly on $X$, namely $$\mathcal L(\theta \mid \tau, \boldsymbol x) \propto \theta^n \tau^{n \theta} \left(\prod_{i=1}^n x_i \right)^{-\theta} \mathbb 1(x_{(1)} \ge \tau).$$  Note we discarded a constant factor of $\prod x_i$. Then $$\ell(\theta \mid \tau, \boldsymbol x) \propto \log \theta + \theta \log \tau - \frac{\theta}{n} \log \prod_{i=1}^n x_i.$$  Then $$\frac{\partial \ell}{\partial \theta} = \frac{1}{\theta} + \log \tau - \frac{1}{n} \sum_{i=1}^n \log x_i = \frac{1}{\theta} - \frac{1}{n} \sum_{i=1}^n \log (x_i/\tau)$$ and the critical point is again the same as above.

The density you call "weird" is known as a Pareto distribution and is quite common.  We can see it arises naturally from a suitable transformation of an exponential distribution.

Title: The MLE of this weird density function


问:
What is the meaning of this sentence? "The right side of $a\equiv b \pmod c$ is less than or equal to the left side."
Let us consider a congruence of the form

$$a\equiv b \pmod c$$

What is the meaning of the following sentence?
$$\text{The right side of the congruence is less than or equal to the left side.}$$

For context, the Jones, et.al, American Mathematical Monthly article "Diophantine Representation of the Set of Prime Numbers" (PDF link via umd.edu) has the following (Lemma 2.4, pg 452; emphasis added):

For any numbers $p$, $n$ and $a\geq 1$ we have the congruence
$$\chi_a(n) \equiv p^n+\psi_a(n)(a-p) \pmod{2ap-p^2-1}$$
Furthermore, when $0< p^n<a$, the right side of the congruence is less than or equal to the left side.

答: I am not an expert on the topic, but I think you can use the definition of congruence modulo which is that $\displaystyle a\equiv b\ (\bmod c) \ \Leftrightarrow a-b=ck$, so I would think you can replace the equality sign with a greater than or equal to. It would then be $\displaystyle a-b\geqslant ck$. I am not sure, however.


问:
Order of product of disjoint cycles
I know the order of the product of disjoint cycles is their lcm. If both cycles are the same length, is this the order of their product or is lcm(x,x) not equal to x?</p>

答: We do have $\operatorname{lcm}(x,x)=x$. And we have that the order of the product of two disjoint cycles, both of length $x$, is equal to $x$. This is exactly as you say it should be.

Title: Order of product of disjoint cycles


问:
Maximum of N iid random random variables with Gumbel distribution
I know that the maximum of iid random variables has Gumbel distribution. For example, this is true when iid random variables have Exponential distribution.

Does this hold when iid random variables have Gumbel distribution themselves?

Let's say there are $N$ iid random variables with distribution $X_n\sim Gumbel(\mu, \beta),~n=1,...,N$. What is the distribution of $\text{max}_n~X_n$? If it is also a Gumbel distribution, say $Gumbel(\mu_{\text{max}}, \beta_{\text{max}})$, how are parameters $\mu_{\text{max}}$ and $\beta_{\text{max}}$ related to $\mu$ and $\beta$?

I tried to find an answer in multiple textbooks, but could not find anything.

答:
Does this hold when iid random variables have Gumbel distribution themselves?

Yes, due to the condition "iid".

 Without the condition "iid", that's not true.  You may consult this question for details.

The CDF of $X_i \sim {\rm Gumbel}(\mu,\beta)$ is
$$ F_i(x) = \exp(-\exp((\mu - x)/\beta)).$$
Then the CDF for $Y_n = \max_{i = 1,\dots,n} X_i$ is
$$\begin{aligned}
F_Y(y) &= \prod_{i=1}^n P(X_i \le y) \\
&= \prod_{i=1}^n F_i(y) \\
&= \exp(-n\exp((\mu-y)/\beta)) \\
&= \exp(-\exp((\mu+\beta \ln n-y)/\beta))
\end{aligned}$$
Therefore, $Y_n \sim {\rm Gumbel}(\mu+\beta\ln n, \beta)$.


问:
A "Rigorous" proof for this seemingly obvious proposition.
In mendelson logic textbook , there is this proposition,(I will replace some letters to make it easier to read)

"If $P$ is a tautology containing as statement letters $A_1,A_2,\ldots,A_n$ and $Q$ arises from $T$ by substituting statement forms $B_1,B_2,\ldots,B_n$ for $A_1,A_2,\ldots,A_n,$ respectively, then $Q$ is a tautology;that is, substituting in a tautology yields a tautology"

This proposition seems way too obvious too me. Then I wondered, is there a rigorous proof of this proposition. There was a proof in the textbook:

"Assume that $P$ is a tautology. For any assignment of truth values to the statement letters in $Q$, the forms $B_1,B_2,\ldots,B_n$ have truth values $x_1,x_2,\ldots,x_n$ (where each $x_i$  is $T$ or $F$). If we assign the values $x_1,x_2,...,x_n$ to  $A_1,A_2,\ldots,A_n,$ respectively, then the resulting truth value of $P$ is the truth value of $Q$  for the given assignment of truth values.Since $P$ is a tautology, this truth value must be $T.$ Thus, $Q$ always takes the value $T$."

The proposition itself is perfectly understandable and it is a very useful proposition(Maybe even too obvious for me). But the proof seems to be not even slightly understandable to me for some reason. I didn't found any other proofs for this theorem anywhere. Is there a better proof for this theorem or I am missing something? (Maybe the proposition is so obvious that proving it is impossible?)

Edit:I just noticed that when I said "I will replace some letters to make it easier to read" then that means "this proposition will stay true nomatter if i replace some symbols". But the proposition itself is "this proposition will stay true no matter if i replace some symbols". So it is like a propositon inside itself.So in shorter words, I am applying this proposition without even proving it in the first place. This is by far the weirdest part about this proposition to me.

答:
Long comment

Maybe, a more lengthy approach can help, considering the case of a single atom $p$.

Let $F$ a formula and let $p$ a propositional symbol occurring in $F$. This means that the formula is a truth-function $F(p)$: for every truth value assigned to $p$, the truth table corresponding to $F$ will outputs a truth value.

But $F$ is a tautology: thus for every assignments of truth values to atoms, the truth table will produce the value TRUE.

Consider now the formula $F' := F[A/p]$, where $A$ is a formula whatever. Irrespective of the "form" of $A$, every assignment will output either TRUE or FALSE as truth value for the formula.

Thus, when $A$ is evaluated to TRUE, we have to consider the lines in the original truth table for $F$ where $p$ is evaluated to TRUE. Due to the fact that $A$ is a tautology, in those lines the formula $F$ has value TRUE; thus also $F'$ will have TRUE in those lines.

But also when in the original truth table for $F$ the atom $p$ is evaluated to FALSE, the corresponding lines for formula $F$ have value TRUE; thus also $F'$ will have TRUE in those lines.

The argument can be iterated, taking into account that a formula is an expression of finite length and thus only a finite number of atoms occurs into it.


问:
Evaluate the triple integral $\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$ using spherical coordinates
How to evaluate triple integral $$\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$$
when $E$ is bounded by $x^2+y^2+z^2-x=0$?

I know that spherical coordinates mean that $$x=r\sin\theta\cos\varphi,\quad y=r\sin\theta\sin\varphi,\quad z=r\cos\theta$$
and this function in spherical coordinates is
\begin{align*}
&\iiint\limits_E\frac{yzdxdydz}{x^2+y^2+z^2} = \iiint\limits_E\frac{r^2\sin\theta}{r^2\sin^2\theta\cos^2\varphi + r^2\sin^2\theta\sin^2\varphi + r^2\cos^2\theta}drd\theta d\varphi = \\ &\iiint\limits_E\frac{r^2\sin\theta}{r^2(\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta)}drd\theta d\varphi = \iiint\limits_E\frac{\sin\theta}{\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta}drd\theta d\varphi
\end{align*}
but I don't know how to write $E$ as set and convert it to spherical coordinates, and also what happens with this function after conversion. Triple integrals is now topic for me and I have never used spherical coordinates before, so I would be grateful if anyone can help me with this.

答:
Using the spherical coordinates (with the convention you used) we have, $$yz= r^2\sin\theta\cos\theta\sin\varphi,$$ $$x^2+y^2+z^2 = r^2$$ $$\,dx\,dy\,dz = r^2\sin\theta\,dr\,d\varphi\,d\theta.$$ Also, the region $E$ is bounded by $x^2+y^2+z^2-x = 0$, which is a sphere centered at $(\tfrac{1}{2},0,0)$ with radius $\tfrac{1}{2}$.

The inside of this sphere is given by $x^2+y^2+z^2-x \le 0$, i.e. $r^2-r\sin\theta\cos\varphi \le 0$. Since $r \ge 0$, this simplifies to $0 \le r \le \sin\theta \cos\varphi$ or $r = 0$ (which is just one point).

In order for $0 \le r \le \sin\theta\cos\varphi$ to be a nontrivial range, we need $\sin\theta\cos\varphi \ge 0$. Since $\sin\theta \ge 0$ for all $\theta \in [0,\pi]$, we only need to restrict the bounds of $\varphi$ to be over the range where $\cos\varphi \ge 0$. We can do this either by $\varphi \in [0,\tfrac{\pi}{2}] \cup [\tfrac{3\pi}{2},2\pi]$ or $\varphi \in [-\tfrac{\pi}{2},\tfrac{\pi}{2}]$. The second is simpler, so we will use that instead.

So the integral becomes $$\iiint\limits_{E}\dfrac{yz \,dx\,dy\,dz}{x^2+y^2+z^2} = \int_{-\pi/2}^{\pi/2}\int_{0}^{\pi}\int_{0}^{\sin\theta\cos\varphi}r^2\sin^2\theta\cos\theta\sin\varphi\,dr\,d\theta\,d\varphi.$$


问:
Evaluating the improper integral $ \int_{0}^{\infty}{\frac{x^2}{x^{10} + 1}\mathrm dx} $
I am trying to solve the following integral, but I don't have a solution, and the answer I am getting doesn't seem correct.

So I am trying to integrate this:

$$ \int_{0}^{\infty}{\frac{x^2}{x^{10} + 1}\,\mathrm dx} $$

To integrate this, I want to use a contour that looks like a pizza slice, out of a pie of radius R. One edge of this pizza slice is along the positive x-axis, if that makes sense. Since $ z^{10} + 1 $ has 10 zeroes, the slice should only be one tenth of a whole circle. So let's call this contour $ C $. Then:

$$ \int_{C}{\frac{z^2}{z^{10} + 1}\,\mathrm dz} = 2 \pi i\,\operatorname{Res}(\frac{x^2}{x^{10} + 1}, e^{i \pi/10}) $$ This is because this slice contains only one singularity. Furthermore:

$$ \int_{C}{\frac{z^2}{z^{10} + 1}\,\mathrm dz} = \int_0^R{\frac{z^2}{z^{10} + 1}\,\mathrm dz} + \int_\Gamma{\frac{z^2}{z^{10} + 1}\,\mathrm dz} $$

And then, by the M-L Formula, we can say that $ \int_\Gamma{\frac{z^2}{z^{10} + 1}\,\mathrm dz} $ goes to $0$ as $R$ goes to infinity. Evaluating $ 2 \pi i\ \operatorname{Res}(\frac{x^2}{x^{10} + 1}, e^{i \pi/10}) $ I get $ \dfrac{\pi}{e^{i \pi/5}} $. Since this answer isn't real, I don't think this could be correct. What did I do wrong?

答:
This answer, applied with $n=2$ and $m=10$, yields
$$
\int_0^\infty\frac{x^2}{1+x^{10}}\,\mathrm{d}x=\frac\pi{10}\csc\left(\frac{3\pi}{10}\right)
$$</p>


问:
Confusion in solving this recursion
I am trying to solve the following recursion
$$
\sum _{i=1}^{n-1} -\mu (n-i)+\sum _{i=1}^{n-1} 2 i \mu (n-i)=(n-1)^2 (\mu (n)-1) \tag{1}
$$

By increasing $n$ to $n+1$, we get
$$
\sum _{i=1}^n -\mu (-i+n+1)+\sum _{i=1}^n 2 i \mu (-i+n+1)=n^2 (\mu (n+1)-1) \tag{2}
$$
and then by shifting index, we get
$$
\sum _{i=1}^{n-1} -\mu (n-i)+\sum _{i=1}^{n-1} 2 (i+1) \mu (n-i)+\mu (n)=n^2 (\mu (n+1)-1)\tag{3}
$$
Taking the difference of (1) and (3), we arrive at
$$
-2 \sum _{i=1}^{n-1} \mu (n-i)-\mu (n)=(n-1)^2 (\mu (n)-1)-n^2 (\mu (n+1)-1)\tag{4}
$$

Replace $n$ by $n+1$ and shift index again, we get
$$
(n+1)^2 \mu (n+2)+1= \\2 \sum _{i=1}^{n-1} \mu (n-i)+n^2 \mu (n+1)+2 \mu (n)+\mu (n+1)+2 n+2 \tag{5}
$$
Then taking the difference between (4) and (5), we get
$$
(n+1)^2 \mu (n+2)+(n-2) n \mu (n)=\left(2 n^2+1\right) \mu (n+1)+2 \tag{6}$$
This has a solution
$$
\mu(n)=\frac{2}{3} H_{n-1}:=\frac{2}{3} \sum_{i=1}^{n-1} \frac{1}{i}
$$
What bothers me is that this solution does not fit in the original equation (1)! How is this possible? I don't think I have made a mistake in the computations.

答: If you look at the derivation from (1) to (6) carefully, you would see that the argument can only go one direction. Any solution of (1) must satisfy (6), but not the other way around.</p>

Title: Confusion in solving this recursion


问:
Do we ever study "mixed" categories?
Consider a category whose object class includes the class of all topological spaces and the class of all topological groups. Furthermore, let the hom-sets between any two objects be the usual hom-sets in the category of topological spaces, unless both objects are topological groups, in which case the hom-set between them will be the usual hom-set in the category of topological groups.

Are "mixed" categories such as this ever studied?

答: What you've described is not a category, since it is not closed under composition of morphisms (since if you consider group $\to$ space $\to$ group, you allow the two arrows to be any continuous maps, and then there is little chance that there composite will be a group homomorphism).</p>

Title: Do we ever study "mixed" categories?


问:
Choose 3 people from a group of 10
My problem is the following : how many ways are there to form a commitee of $3$ people with a group of $10$ men ? My answer : We choose $3$ people; there are $10\times 9 \times 8=P(10,3)$ ways of doing that. Why is my reasoning wrong ? This obsiously should be $10\choose 3$ but I have no idea why.

答:
The reason for this is because if you have say A,B,C,D and you were to choose 2 then you can do this by selecting first one letter. For example A, with a there are 3 different combinations: AB, AC, AD. After A we can ignore A and move onto B since every letter has already been taken out with A. With B we have: BC, BD.
Lastly, since we are ignoring A and B there is just one last combination which is CD. Summing all of these together gives us 6 different combinations which is precisely nCr(4,2).

If you were selecting 3 different items from a list of 5 such as A,B,C,D,E you would follow a similar approach:

ABC Start with the first letter A and the second letter B

ABD Add all the combinations with B and D

ABE

ACD

ACE

ADE

BCD Ignore A and repeat process

BCE

BDE

CDE Ignore A and B and finish off

The reason behind the formula is because you have A nCr(number of letters,how many you are selecting-1) number of times times, B nCr(number of letters-1,how many you are selecting-1) and C nCr(number of letters-2,how many you're selecting -1).

Title: Choose 3 people from a group of 10


答: What's wrong with $10 \times 9 \times 8$ is that you take the order of the people chosen into account.  Say you're counting $ABC$ and $ACB$ as different objects.  That's the number of permutations of picking three objects from ten objects.  To count the number of combinations $C_3^{10}$ instead of $P_3^{10}$, we have to divide the number of permutations $P_3^{10}$ by $P_3^3 = 3!$.

Title: Choose 3 people from a group of 10

Body: My problem is the following : how many ways are there to form a commitee of $3$ people with a group of $10$ men ? My answer : We choose $3$ people; there are $10\times 9 \times 8=P(10,3)$ ways of doing that. Why is my reasoning wrong ? This obsiously should be $10\choose 3$ but I have no idea why.


问:
How many bit string of length n has no consecutive ones except the last two bits
I'm looking for the number of an n-bits long bit string having no consecutive ones except the last two bits. Let me clarify the question by making some examples. Let say $n=5$, then:

$$\text{String 1: } \color{red}{00011} \text{  Good}$$
$$\text{String 2: } \color{red}{01011} \text{  Good}$$
$$\text{String 3: } \color{red}{11011} \text{  No (two consecutive ones before the end)}$$
$$\text{String 4: } \color{red}{10111} \text{  No (two consecutive ones before the end)}$$
$$\text{String 5: } \color{red}{10010} \text{  No (one of the last two bits isn't 1)}$$

I'm looking for the number of "Good" strings. Although finding the number of strings with all zero except the last two bits is easy, I'm stuck with dealing with the case of alternating one's and zero's in the strings. I appreciate any help or tips :D.

答:
Any such binary number is a $n-3$ digits binary numbers with no consecutive ones followed by $011$. Therefore you just need to find out how many $n-3$ digits binary numbers with no consecutive ones exist.

Any $m$ digits binary number with no consecutive ones is either a $m-1$ digits binary number with no consecutive ones followed by $0$ or a $m-2$ digits binary number with no consecutive ones followed by $01$. Therefore, if $k_{m}$ is the number of $m$ digits binary numbers with no consecutive ones, we have the following recursive relation

$$
k_{m}=k_{m-1}+k_{m-2}
$$

I suggest to read more about recursive relation and how to solve them. I cannot write the full answer because I am working :)

The solution is $k_{m}=\frac{3+\sqrt{5}}{2\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{m}+\frac{-3+\sqrt{5}}{2\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{m}$

Now the answer to your original question is $k_{n-3}$


问:
Study suggestions
I am a 16 year old high school student with too much free time. For the past two years I have been teaching myself university level maths in my free time. I have studied the basic first year subjects such as multivariable calculus, linear algebra and ordinary differential equations. My question is what should I study next?

I have tried more pure subjects such as real analysis, abstract algebra and number theory and additionally more applied subjects such as nonlinear dynamics, chaos theory and topology, but they either didn't spark my interest or were too difficult. I have even written a few papers that never really worked out.I posted one here I feel as if I need to regain my curiosity for mathematics and look for something beyond what academics present at seminars for children (referring to topology). Therefore, what is a subject that is seldom presented to an undergraduate but still is an valid field in mathematics?

答:
At the risk of swamping you with too much, you may want to look through past issues of the following.

Pi Mu Epsilon Journal

Quantum

Mathematical Spectrum.

Also, see the books in the New Mathematical Library series and the many translations of Russian "school level" books.

Title: Study suggestions


问:
proving convergence of $a_{n+1}=1+\frac{1}{1+a_{n}}$

$a_1=1.$ $a_{n+1}=1+\frac{1}{1+a_{n}}$

Prove that the sequence is convergent.

I'm trying to prove the convergence of this sequence but having trouble. At first I thought this might be a monotone sequence since then I can try monotone convergence theorem to prove its convergence.

But after checking some terms, I realized it seemed the sequence is oscillating. So I'm not sure how to prove the convergence of this sequence.

Thanks.

答:
A method often useful with oscillating sequences: Let $b_n=|(a_n)^2-2|.$ Then $$0\le b_{n+1}=\frac {b_n}{(1+a_n)^2}\le \frac {b_n}{4} $$ because $1+a_n\ge 2$ by induction on $n$.

So $b_n$ decreases to  $0$. So $(a_n)^2\to 2$ with each $a_n>0.$

The motivation for the "$2$" in the definition of $b_n$ is that IF $a_n$ converges to a limit $L$ then $L=\lim_{n\to \infty}a_{n+1}=\lim_{n\to \infty} 1+\frac {1}{1+a_n}=1+\frac {1}{1+L},$ implying $L^2=2.$