问:
Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.
Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.

Attempt:
Notice that the $p-$ series $\sum_{n=1}^\infty \frac{1}{n^2}$ with $p=2$ is convergent. Then, we have
\begin{align*}
\lim\limits_{n \to \infty} \frac{\ln\left(\cos \frac{1}{n} \right)}{\frac{1}{n^2}} &= -\frac{1}{2} \cdot \lim\limits_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} \cdot \frac{1}{\cos \frac{1}{n}} \\
&= -\frac{1}{2}.
\end{align*}
Hence, the limit is equal to $-\frac{1}{2} \ne 0$. Since the $p-$ series is convergent with $p=2$, by the limit comparison test, the series is convergent.

Am I true?

答:
Just for the fun.

In the same spirit as @Alexander Conrad, we have the infinite series representation
$$\log \left(\cos \left(\frac{1}{n}\right)\right)=\sum_{k=1}^\infty (-1)^k\frac{ 2^{2 k-3} (E_{2 k-1}(1)-E_{2 k-1}(0)) }{k (2 k-1)!}\,n^{-2 k}$$ where appear Euler polynomials.

Truncated to any order, we can transform the Taylor series into a $[2k,2]$ Padé approximant $P_k$.

For  example
$$P_2=-\frac {3(10n^2-1) } {4n^2(15n^2-4) }=-\frac{3}{16 n^2}-\frac{75}{16 \left(15 n^2-4\right)}$$

$$\sum_{n=1}^\infty P_2=-\frac{1}{128} \left(75+4 \pi ^2-10 \sqrt{15} \pi  \cot \left(\frac{2 \pi}{\sqrt{15}}\right)\right)\sim -0.943375$$ Repeating for various values of $k$, we have more and more accurate values as shown below
$$\left(
\begin{array}{cc}
 1 & -0.9295780173 \\
 2 & -0.9433749128 \\
 3 & -0.9449845778 \\
 4 & -0.9452801699 \\
 5 & -0.9453463832 \\
 6 & -0.9453628819 \\
 7 & -0.9453672911 \\
 8 & -0.9453685316 \\
 9 & -0.9453688948 \\
 10 & -0.9453690046
\end{array}
\right)$$


问:
Prove that the difference between numbers with the same sum of digits is a multiple of $9$.
Prove that the difference between numbers with the same sum of digits
is a multiple of $9$.

答:
Ideally, you're supposed to show what you've tried, etc. on this website. However, since you're a new user, I do not want to be too harsh with regard to downvoting and closing, etc. I have answered (and corrected) your question below, but do bear in mind that you must show some effort while posting questions on MathSE in the future. Thank you, and welcome to the community!

Well, the difference is not exactly $9$, but it is a multiple of $9$. Consider two $n$-digit numbers $A = a_1a_2\ldots a_n$ and $B = b_1b_2\ldots b_n$ such that $$\sum_{i=1}^n a_i = \sum_{i=1}^n b_i \implies \sum_{i=1}^n (a_i-b_i) = 0$$
You are free to assume the first few digits of the number (or any other for that matter) to be zero, in order to produce numbers that have less than $n$ digits. The proof still goes through.

$$A - B = 10^{n-1}(a_1-b_1) + 10^{n-2}(a_2-b_2) + \ldots + (a_n-b_n)$$
Using $\sum_{i=1}^n (a_i-b_i) = 0$, we can simplify the above expression to get:
$$A - B = (10^{n-1}- 1)(a_1-b_1) + (10^{n-2}- 1)(a_2-b_2) + (10^{1}- 1)(a_{n-1}-b_{n-1})$$
Now, $10^k - 1$ is always a multiple of $9$. In addition, each of the terms in the expression of $A-B$ above, is a multiple of $10^k -1$ for some $k \ge 1, k\in\mathbb Z$. As a result, each term of the sum is a multiple of $9$, and therefore $A-B$ itself is also a multiple of $9$.

Why is $10^k-1$ a multiple of $9$ - you may ask? The binomial expansion suffices!
$$10^k - 1 = (9+1)^k - 1 = \text{sum of multiples of }9 = \text{multiple of }9$$
I leave the details of the above argument to you. Lastly, this works for all $n\in\mathbb N$, and so we have proved it for all numbers.

Summary: The difference between two integers with the same sum of digits, is divisible by $9$.

ContentLicense: CC BY-SA 4.0


问:
Expression for combining 3d and 2d matrices into resulting 2d matrix
I would like to know if there is a convenient mathematical expression for the following.

I want to construct a (n x n) matrix $\mathbf{C}$ for which i know that each column $\mathbf{C}_i$ is equal to $(\mathbf{A}_i \circ \mathbf{B})\mathbf{1}$. Matrix $\mathbf{B}$ is (n x n) whereas $\mathbf{1}$ is a (n x 1) vector of ones. Further, $\mathbf{A}$ is 3-dimensional (n x n x n) where i consider $\mathbf{A}_i$ to be the (n x n) submatrix at level $i$ of $\mathbf{A}$. Equivalently, if $\mathbf{A}$ would be viewed as a (1 x n) vector of 2d matrices, $\mathbf{A}_i$ would be the (n x n) matrix at position $(1,i)$ of that vector. To conclude, the $\circ$ operator indicates the Schur (element-wise) product.

Ideally, but i'm sure wrongfully, it would be nice to write $\mathbf{C} = (\mathbf{A} \circ \mathbf{B})\mathbf{U}$, with $\mathbf{U}$ being the (n x n) matrix of ones, but i think columns of $\mathbf{U}$ will not just correctly "hop over" to each level $i$ submatrix of $\mathbf{A}$ in constructing $\mathbf{C}$.

Is there a correct way for expressing $\mathbf{C}$, maybe via tensor algebra?

答:
I assume that "adheres to" means the same as "is equal to".

Let $\text{vec}$ denote the vectorization operator and $\text{diag}(M_1,\dots,M_k)$ the direct sum of matrices. We can write
$$
\operatorname{vec}(\mathbf C) = \operatorname{diag}(\mathbf A_1 \circ \mathbf B, \dots, \mathbf A_n \circ \mathbf B) \mathbf 1_{n^2 \times 1}
$$


问:
Solving linear differential equation with variable coefficients.
I have faced this task, when preparing to my university's "mathematical methods" exam, and I do not really understand how to solve it. The task content is as follows:

Given: $$(1-x^2)\cdot y''_{xx} -x\cdot y'_x +y = \frac{\sqrt{1-x^2}}{x}$$ Solve the equation by substituion of the independent variable: $x(t) = \cos(t), x\in(0,1).$

My attempt:

I have performed the suggested substitution:

$(1-\cos^{2}(t))\cdot \large\frac{\partial^2y}{\partial x(t)^2}\normalsize-\cos(t)\cdot\large\frac{\partial y}{\partial x(t)} \normalsize+ y = \large\frac{\sqrt{1-\cos^2(t)}}{cos(t)}.$

$\sin^2(t)\cdot \large\frac{\partial^2y}{\partial x(t)^2}\normalsize-\cos(t)\cdot\large\frac{\partial y / \partial t}{\partial x / \partial t} \normalsize+ y = \tan(t).$

$\sin^2(t)\cdot\frac{\large\partial\left(\frac{\dot{y}}{-\sin(t)}\right) / \partial t}{\large\partial x / \partial t} + \large\frac{\cos(t)}{\sin(t)}\normalsize\dot{y} + y = \tan(t).$

$-\ddot{y} + 2\dot{y}\cot(t) +y = \tan(t).$

That is now I have to solve: $-\ddot{y} + 2\dot{y}\cot(t) +y = \tan(t).$ And that is the point, where I am stuck, I have no idea how to proceed from here, and I did not manage to find the way to solve such a linear D.E. in my course book.

I would appreciate any help, hints or book references, thank you in advance!

Post Scriptum:

As correctly mentioned by @Aryadeva, I have made a mistake, when calculating $\large\frac{\partial^2 y}{\partial x^2}$, I simply lost "minus".

That is, after recalculating I had to solve: $$\ddot{y} + y = \tan(t), \text{ s.t. } x\in(0,1) \text{ and } x = \cos(t) \Leftrightarrow \forall x\in \mathbb{D}_x: \text{ }t = \arccos{x}.\tag{1}$$

My solution has involved appying vatiation of parameters method:

$\underline{\text{Step I}}$: find characteristic polynomial over $y(t)$ to get a general solution $(\large y_{\small G})$ for $(1)$:

$\lambda^2 + 1 = 0. \Rightarrow \lambda = \pm i.$

$\begin{bmatrix}
\lambda_1\\
\lambda_2\\
\end{bmatrix} = \begin{bmatrix}
i\\
-i\\
\end{bmatrix} \Rightarrow \large y_{\small G} \normalsize = C_1 e^{it}+C_2 e^{-it} = \dot{\tilde {C}}_1\cos(t) + \dot{\tilde{C}}_2 \sin(t).$

$\underline{\text{Step II}}$: Finding particular solution $(\large y_{\small P})$ for $(1)$, using Wronskian matrices:

I would omit my calculations here, would just note that for the given D.E.:

$\mathbb{W} = \begin{pmatrix}\cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{pmatrix}, \mathbb{W}_{\small \Delta} = \begin{vmatrix}\cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{vmatrix} = 1.$

$\mathbb{W}_{\small \Delta,\normalsize 1} = \begin{vmatrix}0 & \sin(t) \\ 1 & \cos(t) \end{vmatrix} = -\sin(t), \mathbb{W}_{\small \Delta,\normalsize 2} = \begin{vmatrix}\cos(t) & 0 \\ -\sin(t) & 1 \end{vmatrix} = \cos(t).$

Thus:
$$\dot{\tilde{C}}_1(t) = \int\frac{\tan(t)\cdot \mathbb{W}_{\small \Delta,\normalsize 1}}{\mathbb{W}_{\small \Delta}} \partial t = \sin(t) +\frac{1}{2}\ln\left(\frac{\sin(t)-1}{\sin(t)+1}\right) +\bar{C}_1,$$
$$\dot{\tilde{C}}_2(t) = \int\frac{\tan(t)\cdot \mathbb{W}_{\small \Delta,\normalsize 2}}{\mathbb{W}_{\small \Delta}} \partial t = -\cos(t) + \bar{C}_2.$$
Thereof:
$$\large y_{\small P} \normalsize = \left(\sin(t)+\frac{1}{2}\ln\left(\frac{\sin(t)-1}{\sin(t)+1}\right) +\bar{C}_1\right)\cos(t) +\left(-\cos(t) +\bar{C}_2\right)\sin(t).$$

$\underline{\text{Step III}}$: Now it just suffices to write down the whole solution and transfer from $t$ to $x$:

$\large y \normalsize (t) = \large y_{\small G} + y_{\small P} \normalsize = \left(\sin(t)+\frac{1}{2}\ln\left(\frac{\sin(t)-1}{\sin(t)+1}\right) +\tilde{C}_1\right)\cos(t) +\left(-\cos(t) +\tilde{C}_2\right)\sin(t).$

Now, as $x \in (0, 1): \cos(\arccos{x}) = x, \text{ }\sin(\arccos{x}) = \sqrt{1-x^2}:$
$$\text{Answer: } \begin{array}{|c|}
\hline
y(x) = \left(\sqrt{1-x^2} + \frac{1}{2}\ln\left(\frac{\sqrt{1-x^2} - 1}{\sqrt{1-x^2} +1}\right)+\tilde{C}_1 \right)x +\left(-x+\tilde{C}_2\right)\sqrt{1-x^2}.\\
\hline
\end{array}$$

答:
$$(1-x^2)\cdot y''_{xx} -x\cdot y'_x +y = \frac{\sqrt{1-x^2}}{x}$$
I don't know about the given hint but you can rewrite the DE as:
$$((1-x^2)\cdot y')'+(xy)' = \frac{\sqrt{1-x^2}}{x}$$
Integrate.

Note that we have:
$$x\dfrac {dy}{dx}=\cos t \dfrac {dy}{dt}\dfrac {dt}{dx}=-y'_t\dfrac {\cos t}{\sin t}$$
so that you have $$\dfrac {dy}{dx}=-\dfrac {dy}{dt}\dfrac 1 {\sin t}$$
Then you calculate $(1-x^2)y''_{xx}$
You should get:
$$y''_{tt}-y'_{t}\dfrac {\cos t }{\sin t}$$
This simplify your differential equation.
$$y''_{tt}+y=\tan t$$
It seems to me that you made a mistake when you calculated $y''_{tt}$.


问:
How to evaluate these types of definite integral?
$$\displaystyle\int\limits^{\cssId{upper-bound-mathjax}{1}}_{\cssId{lower-bound-mathjax}{0}} \sin^{2k}\left(\dfrac{{\pi}}{2}x\right)\,\cssId{int-var-mathjax}{\mathrm{d}x}$$.
I tried integration by parts but couldn't go far. The answer given in textbook is $$(2k)!/((2^
(2k))(k!)^2)$$

答:
Take $\dfrac{π}{2}x= u$
The integral becomes $$\dfrac{2}{π}\int_0^{\frac{π}{2}} \sin^{2k}(u)du $$
Using the definition of Beta function

$$\int_0^{\frac{π}{2}} \sin^{2m-1}(x) \cos^{2n-1}(x)dx= \dfrac{\Gamma{(m)} \Gamma{(n)}}{\Gamma{(m+n)}}$$

The integral is $$\dfrac{2}{π}\int_0^{\frac{π}{2}} \sin^{2(\frac{2k+1}{2})-1}(x)\cos^{2.(\frac{1}{2})-1}(x) dx= \dfrac{2}{π}\dfrac{\Gamma{(k+1/2)}\Gamma{(1/2)}}{\Gamma{(k+1)}}$$


问:
Is my proof of irrationality correct?
There is a lot of proof on this site about different numbers being irrational. This is just a generalized version.

THE STATEMENT:-

If $\forall c,k\in\mathbb{N},N\in\mathbb{N}:N\neq c^k$

then $\sqrt[k]{N}$ is irrational.

MY PROOF:-

Let $a,b\in\mathbb{Z}:\gcd(a,b)=1$

$\implies \gcd(a^k,b^k)=1$, call it $eq(1)$

Assume $\sqrt[k]{N}=\frac{a}{b}$

$\implies N=\frac{a^k}{b^k}$

$\implies a^k=Nb^k$

$\implies \gcd(a^k,b^k)=\gcd(Nb^k,b^k)=b^k$, call it $eq(2)$

Now, combining $eq(1)$ and $eq(2)$ would give

$b^k=1$

$\implies b=1$

$\implies \sqrt[k]{N}=a$

$\implies non-integer=integer$, which is a Contradiction

Which means our assumption was wrong and $\sqrt[k]{N}$ is irrational.$QED$

MY QUESTION:-

Is my proof entirely valid or is there a problem and if there is a problem then what is it?

答:
At the moment you say

Assume $\sqrt[k]{N}=\frac{a}{b}$

you should already mention that $b=1$ is ruled out, because you have $N\neq c^k$ in the initial conditions. This would let you finish the proof earlier, when you arrive at $b=1$ later, without having to use vague terminology like "non-integer = integer".

Title: Is my proof of irrationality correct?


问:
Functional equation about complex anlysis
Assuming $g: \mathbb C \rightarrow \mathbb C$ is a known function (assuming the definition
It is meaningful for or taking a value in $\infty$), use the function equation to find $f$:
$$
f\left(\frac{1}{1-z}\right)+f\left(\frac{z-1}{z}\right)=g(z)
$$

I have tried to find the function value of two points：
$$\frac{z-1}{z}=\frac{1}{1-z} \Rightarrow z^{2}-z+1=0 \quad z_{1,2}=\frac{1 \pm \sqrt{3} i}{2}$$
But i don't know what to do next.

答:
Sketch: formally by substituting $1-z \to 1/w$ so $z=\frac{w-1}{w}$ we get:

$f(w)+f(1/(1-w))=g(\frac{w-1}{w})$ or renaming $f(z)+f\left(\frac{1}{1-z}\right)=g(\frac{z-1}{z})$

Formally substituting $y=\frac{z-1}{z}$ so $z=\frac{1}{1-y}$ we get:

$f\left(\frac{y-1}{y}\right)+f(y)=g(\frac{1}{1-y})$ or renaming $f\left(\frac{z-1}{z}\right)+f(z)=g(\frac{1}{1-z})$

Now adding the three relations gives:

$2(f(z)+f\left(\frac{1}{1-z}\right)+f\left(\frac{z-1}{z}\right))=g(z)+g(\frac{z-1}{z})+g(\frac{1}{1-z})$ so finally one gets:

$2f(z)=g(\frac{z-1}{z})+g(\frac{1}{1-z})-g(z)$


问:
Different expressions of factorization
(This question might have incorrect tags)

I was playing with different expressions of factorization. Here's what I was doing. We all know that

$$\begin{align*}
a^2-b^2&=(a+b)(a-b)\\[5pt]
a-b&=(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})\\[5pt]
&=(\sqrt{a}+\sqrt{b})(\sqrt[4]{a}+\sqrt[4]{b})(\sqrt[4]{a}-\sqrt[4]{b})\\[5pt]
\end{align*}$$

Now if we continue this process infinitely then we get the formula:

$$a-b=\prod_{k=1}^{\infty}(a^{2^{-k}}+b^{2^{-k}})$$

and

$$\prod_{k=1}^{n}(a^{2^{-k}}+b^{2^{-k}})=\frac{a-b}{a^{2^{-n}}-b^{2^{-n}}}$$

But if I input this thing on wolfram alpha I get nothing.

And also if I input the second expression then I also get nothing.

So, are these formulas correct?

答:
The mistake in your expression of the product results from missing the difference factor.

$a-b=(a^{2^{-n}}-b^{2^{-n}})\prod_{k=1}^n {a^{2^{-k}}+b^{2^{-k}}}$

Title: Different expressions of factorization


问:
show $\lim_{x\rightarrow\infty} x^n\exp(-x^2)=0$
I want to show for all $n\in\mathbb N$
$$\lim_{x\rightarrow\infty}\frac{x^{n}}{\exp(x^2)}=0$$

I am pretty sure that I have to use L'Hospital. I've tried induction:

$n=1$: $$\lim_{x\rightarrow\infty}\frac x{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac1{2x\exp(x^2)}=0$$

And for $n\rightarrow n+1$:
$$\lim_{x\rightarrow\infty}\frac{x^{n+1}}{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac{(n+1)x^n}{2x\exp{x^2}}$$

And now I am stuck. The term $2x$ really annoys my for my induction hypothesis.

Any hints?

答:
For $ \ x > 1 \ , $

$$ 0 \ \le \ e^{-x^2} \ \le \ e^{-x} $$

and also $ \ 0 \ < \ x^n \ \text{for} \ n  \in \mathbb{N}$ , so

$$ 0 \ \le \ x^ne^{-x^2} \ \le \ x^ne^{-x} \ .$$

It is much easier to show $ \ \lim_{x \rightarrow \infty} \  x^ne^{-x} \ = \ 0  \ $ through l'Hopital's Rule.  Thence, the "Squeeze Theorem" leads to

$$ 0 \ \le \ \lim_{x \rightarrow \infty} x^ne^{-x^2} \ \le \ \lim_{x \rightarrow \infty} x^ne^{-x} \ \Rightarrow \ \lim_{x \rightarrow \infty} x^ne^{-x^2} \ = \ 0 \ . $$

ContentLicense: CC BY-SA 3.0


问:
Find numbers from GCD and LCM
two numbers gcd and lcm are respectively 6 and 600. What is the possible pairs of two numbers?</p>

答:
Let $m$ and $n$ be two positive integers whose gcd and lcm are $6$ and $600$ respectively.

Then, by dividing $m$ and $n$ by $6$, one reduces the problem to finding the number of unitary divisors of $100$.

The answer is $4$, since $100$ has exactly two distinct prime factors.

The four unitary divisors of $100$ ($1, 4, 25,$ and $100$) then correspond to the $(m,n)$ pairs $(6,600), (24,150), (150,24),$ and $(600,6)$ respectively.

Title: Find numbers from GCD and LCM


问:
Geometric view of isomorphism between sets
My studying of Commutative Algebra lead me to the following question (though not a strictly Commutative Algebra question) :

What is a (or the) geometric intuition of the isomorphism
$$\mathbb{R}[x,y]/(x^2-y,x-x^3+xy)\simeq\mathbb{R}$$
So $\mathbb{R}[x,y]$ is the ring of polynomials with variables $x,y$ and $(x^2-y,x-x^3+xy)$ is an ideal of $\mathbb{R}[x,y]$.

I would like to see an awnser for the spesific question for I have searched for 'geometric approaches' of isomorphisms in classics like the Atiyah MacDonald 'Introduction to Commutative Algebra' and in Algebraic Geometry material and I cannot figure out the idea. Really tough, I would like to know your mind considering geometric approaches of isomorphisms 'in general' (perhaps by giving an example that you have in mind, like the one above and of course, the more examples, the better). Also, any sources for further studying or links to previously related posts are welcome.

答:
I think you might want to get a better idea of algebraic quotients in general, here's an answer of mine where I try to go in depth for a beginner: Ideal generated by $v\otimes v - \Phi(v)1$

Most introductions to category theory give a good overview of quotients; I'd suggest Emily Riehl's Category Theory in Context.

The core geometric insight is that for any algebraic quotient, the kernel of the quotient homomorphism (the ideal) is reduced to $0$, and any affine space parallel to that ideal is also reduced to a single point (but nonzero).

If you really in need of a visualization, I'd suggest taking a 3D graphing calculator like GeoGebra, and graphing $f(x, y) = x^2 - y^2 + C$, making $C$ vary, and trying to understand how reducing these surfaces to a point (for each $C$) will draw a line/curve. Though this is not at all your example, it should give you an analogy of what's happening; your example is much simpler to understand algebraically than it is to visualize geometrically.

Edit: a previous version of this answer mistakenly claimed that $\mathbb{R}[x,y]/(x^2−y, x−x^3+xy)$ was a curve, when it is, in fact, a point. This has been corrected; and @Viktor Vaughn's answer accepted as correct. The points I make on algebraic quotients still hold, though, so I have left them unchanged.


问:
Find a value of $\;\lim\limits_{n\rightarrow\infty}\frac{a_{n}}{n}$

(If you're good at it, you can do it by heart.)

Let $a_n$ be the average value of lengths of all the diagonal lines of regular n-gon whose side is 1.

Find a value of
$$\lim_{n\rightarrow\infty}\frac{a_{n}}{n}$$
Source: Fujino_Yusui

After messing around with the cosine theorem and doing some calculations, I found the exact formula of $a_{n},$ so I did the rest of the calculations and the answer is
$$\lim_{n\rightarrow\infty}\frac{a_{n}}{n}= \lim_{n\rightarrow\infty}\frac{2\cos\frac{\pi}{n}- 1}{\left ( \left ( 1- \cos\frac{\pi}{n} \right )n \right )\left ( n- 3 \right )}= \frac{2}{\pi^{2}}$$
But I want to see a shortcut that leads $a_{n}\sim\frac{n}{2\pi}\cdot\frac{4}{\pi}\,{\rm as}\,n\rightarrow\infty$ without cosine theorem, I need to the help.

答:
Since you ask for a shortcut, here is what I have got--

Since we are talking about an infinite polygon, we may very well express it as a circle under limiting conditions. (A nice animation Here )

Hence, the average value of the length of diagonals becomes the average value of the length of line segments drawn from the circle to all of its points.

There are an infinite number of these lines, and hence we need to find an expression for their lengths. For that, we assume that in a cartesian plane, a circle of radius r is centred at the origin, with one of our endpoints along the y axis.
Consider any point P. the polar coordinates of P are $(rsin\theta,rcos\theta)$ while that of A are $(-r,0)$

AP=$\sqrt{(r+r cos\theta)^2+(r sin\theta)^2}=r\sqrt{2(1+cos\theta)}$

For finding the values of average, as continuous values are present, we integrate with respect to from $(0,\pi)$ and divide--

average value=$r . \frac{\int _0^{\pi }\left(\ \sqrt{2\left(1+\cos \theta \right)}\right)d\theta }{\int _0^{\pi }d\theta }=r . \frac{4}{\pi}$

Now the circumference is n. That is, $n=2\pi r$ we substitute value of r and get the desired answer

$a_{n}\sim\frac{n}{2\pi}\cdot\frac{4}{\pi}\,{\rm as}\,n\rightarrow\infty$

Hence proved................

Hope that my humble attempt provided you the 'shortcut' required....
If there are  more queries regarding this please do ask, and I would be happy to help :~)


答:
$$\lim_{x\rightarrow\infty}\frac{x^{n}}{\exp(x^2)}=\left( \lim_{x\rightarrow\infty}\frac{x}{\exp(x^2/n)}\right)^n=0$$

Now, you only need to apply L'H once.

Second solution

$$\lim_{x\rightarrow\infty}\frac{x^{n}}{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac{\exp(n \ln x)}{\exp(x^2)}=\lim_{x\rightarrow\infty}\exp(n \ln x-x^2)=0$$

Title: show $\lim_{x\rightarrow\infty} x^n\exp(-x^2)=0$

Body:
I want to show for all $n\in\mathbb N$
$$\lim_{x\rightarrow\infty}\frac{x^{n}}{\exp(x^2)}=0$$

I am pretty sure that I have to use L'Hospital. I've tried induction:

$n=1$: $$\lim_{x\rightarrow\infty}\frac x{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac1{2x\exp(x^2)}=0$$

And for $n\rightarrow n+1$:
$$\lim_{x\rightarrow\infty}\frac{x^{n+1}}{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac{(n+1)x^n}{2x\exp{x^2}}$$

And now I am stuck. The term $2x$ really annoys my for my induction hypothesis.

Any hints?


答:
I would try to avoid computations that look like high-rise buildings. By the suggested substitution, set $u(t)=y(\cos(t))$. Then $u'(t)=-y'(\cos(t))\sin(t)$ and
$$
u''(t)=\sin^2(t)y''(\cos(t))-y'(\cos(t))\cos(t)=\tan(t)-u(t)
$$
So that the task devolves to the standard exercise
$$
u''(t)+u(t)=\tan(t).
$$
One approach is variation of constants, one other is to observe that after multiplication with the cosine
$$
(\cos(t)u'(t))'+(\sin(t)u(t))'=\sin(t),
$$
etc.

Title: Solving linear differential equation with variable coefficients.

Body:
I have faced this task, when preparing to my university's "mathematical methods" exam, and I do not really understand how to solve it. The task content is as follows:

Given: $$(1-x^2)\cdot y''_{xx} -x\cdot y'_x +y = \frac{\sqrt{1-x^2}}{x}$$ Solve the equation by substituion of the independent variable: $x(t) = \cos(t), x\in(0,1).$

My attempt:

I have performed the suggested substitution:

$(1-\cos^{2}(t))\cdot \large\frac{\partial^2y}{\partial x(t)^2}\normalsize-\cos(t)\cdot\large\frac{\partial y}{\partial x(t)} \normalsize+ y = \large\frac{\sqrt{1-\cos^2(t)}}{cos(t)}.$

$\sin^2(t)\cdot \large\frac{\partial^2y}{\partial x(t)^2}\normalsize-\cos(t)\cdot\large\frac{\partial y / \partial t}{\partial x / \partial t} \normalsize+ y = \tan(t).$

$\sin^2(t)\cdot\frac{\large\partial\left(\frac{\dot{y}}{-\sin(t)}\right) / \partial t}{\large\partial x / \partial t} + \large\frac{\cos(t)}{\sin(t)}\normalsize\dot{y} + y = \tan(t).$

$-\ddot{y} + 2\dot{y}\cot(t) +y = \tan(t).$

That is now I have to solve: $-\ddot{y} + 2\dot{y}\cot(t) +y = \tan(t).$ And that is the point, where I am stuck, I have no idea how to proceed from here, and I did not manage to find the way to solve such a linear D.E. in my course book.

I would appreciate any help, hints or book references, thank you in advance!

Post Scriptum:

As correctly mentioned by @Aryadeva, I have made a mistake, when calculating $\large\frac{\partial^2 y}{\partial x^2}$, I simply lost "minus".

That is, after recalculating I had to solve: $$\ddot{y} + y = \tan(t), \text{ s.t. } x\in(0,1) \text{ and } x = \cos(t) \Leftrightarrow \forall x\in \mathbb{D}_x: \text{ }t = \arccos{x}.\tag{1}$$

My solution has involved appying vatiation of parameters method:

$\underline{\text{Step I}}$: find characteristic polynomial over $y(t)$ to get a general solution $(\large y_{\small G})$ for $(1)$:

$\lambda^2 + 1 = 0. \Rightarrow \lambda = \pm i.$

$\begin{bmatrix}
\lambda_1\\
\lambda_2\\
\end{bmatrix} = \begin{bmatrix}
i\\
-i\\
\end{bmatrix} \Rightarrow \large y_{\small G} \normalsize = C_1 e^{it}+C_2 e^{-it} = \dot{\tilde {C}}_1\cos(t) + \dot{\tilde{C}}_2 \sin(t).$

$\underline{\text{Step II}}$: Finding particular solution $(\large y_{\small P})$ for $(1)$, using Wronskian matrices:

I would omit my calculations here, would just note that for the given D.E.:

$\mathbb{W} = \begin{pmatrix}\cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{pmatrix}, \mathbb{W}_{\small \Delta} = \begin{vmatrix}\cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{vmatrix} = 1.$

$\mathbb{W}_{\small \Delta,\normalsize 1} = \begin{vmatrix}0 & \sin(t) \\ 1 & \cos(t) \end{vmatrix} = -\sin(t), \mathbb{W}_{\small \Delta,\normalsize 2} = \begin{vmatrix}\cos(t) & 0 \\ -\sin(t) & 1 \end{vmatrix} = \cos(t).$

Thus:
$$\dot{\tilde{C}}_1(t) = \int\frac{\tan(t)\cdot \mathbb{W}_{\small \Delta,\normalsize 1}}{\mathbb{W}_{\small \Delta}} \partial t = \sin(t) +\frac{1}{2}\ln\left(\frac{\sin(t)-1}{\sin(t)+1}\right) +\bar{C}_1,$$
$$\dot{\tilde{C}}_2(t) = \int\frac{\tan(t)\cdot \mathbb{W}_{\small \Delta,\normalsize 2}}{\mathbb{W}_{\small \Delta}} \partial t = -\cos(t) + \bar{C}_2.$$
Thereof:
$$\large y_{\small P} \normalsize = \left(\sin(t)+\frac{1}{2}\ln\left(\frac{\sin(t)-1}{\sin(t)+1}\right) +\bar{C}_1\right)\cos(t) +\left(-\cos(t) +\bar{C}_2\right)\sin(t).$$

$\underline{\text{Step III}}$: Now it just suffices to write down the whole solution and transfer from $t$ to $x$:

$\large y \normalsize (t) = \large y_{\small G} + y_{\small P} \normalsize = \left(\sin(t)+\frac{1}{2}\ln\left(\frac{\sin(t)-1}{\sin(t)+1}\right) +\tilde{C}_1\right)\cos(t) +\left(-\cos(t) +\tilde{C}_2\right)\sin(t).$

Now, as $x \in (0, 1): \cos(\arccos{x}) = x, \text{ }\sin(\arccos{x}) = \sqrt{1-x^2}:$
$$\text{Answer: } \begin{array}{|c|}
\hline
y(x) = \left(\sqrt{1-x^2} + \frac{1}{2}\ln\left(\frac{\sqrt{1-x^2} - 1}{\sqrt{1-x^2} +1}\right)+\tilde{C}_1 \right)x +\left(-x+\tilde{C}_2\right)\sqrt{1-x^2}.\\
\hline
\end{array}$$


问:
$f(x)$ is a bounded function such that $\int_a^b(f(x))dx=0$ then does it imply that $f(x)=0$
The function that I have chosen is $f(x)=1, a \le x \le \frac{a+b}{2},and f(x)=-1 , \frac{a+b}{2} < x \le b  $.

The function is bounded and with only one point of discontinuity so it is Riemann integrable

Now, if we choose our partition to be $\{a,\frac{a+b}{2},b\}$ then we see that $U(P,f)=0 $ and $L(P,f)=0$.We know by the property of Riemann integration that if we choose any partition $Q$ on $[a, b]$ then $U(Q,f)\ge L(P,f)$ )I  can't  really  conclude  after this.

答: Not necessarily. Take the function $$f(x)=sgn(x)$$ and take $a=-1 \ and \ b=1$. The integral is zero but the function is not itself zero over the range of integration.


问:
Find if this function is continuous or not
Find if this function is continuous or not. $$y= \lim_{\alpha\to +\infty}(1+x)\cdot \tanh(\alpha x)$$

Because $\tanh(\alpha x)=\frac{e^{2\alpha x}-1}{e^{2\alpha x}+1}$$\implies$that it is continuous.So $y$ is continuous.

But the answer is that this function has a first kind discontinuous point at $x_0=0$.Can you explain?

答:
You can first find what $y$ is given $x$ fixed. If $x > 0$, $y = 1 + x$; if $x < 0$, $y = -(1 + x)$. This actually shows that $$\lim_{x \rightarrow 0-} y(x) = -1 \neq 1 = \lim_{x\rightarrow 0+} y(x).$$
Also, this is the only discontinuity of $y$.

Hint: When $\alpha > 0$, express $\mathrm{tanh}(\alpha x)$ as $$\mathrm{tanh}(\alpha x) = \frac{1 - e^{-2ax}}{1 + e^{-2ax}}.$$

Edit: Basically $f(x;\alpha)$ is continuous in $x$ for each $\alpha$ does not imply $\lim_{\alpha\rightarrow \alpha_0} f(x; \alpha)$ is continuous in $x$. Think of $f_n(x) = x^n$ which is continuous on $[0, 1]$. However, $f(x) = \lim_{n\rightarrow \infty} x^n = \mathbb{I}_{x = 1}$ has a jump discontinuity at $1$.


问:
Subgroups of finite abelian group

Find all subgroups of order four in $\mathbb{Z}_4 \bigoplus \mathbb{Z}_4$.

I found a solution online from this link. The elements of order four are

(0,1) (0,3)

(1,0) (3,0)

(1,1) (3,3)

(1,2) (3,2)

(1,3) (3,1)

(2,1) (2,3)

Since there are 12 of them, there must be six cyclic groups of order 4.

The rest of the groups are of the form $\mathbb{Z}_2 \bigoplus \mathbb{Z}_2$

{0,0} {0,2},{2,0},{2,2}

I see that there are 12 order four elements, but I'm not sure how we can use that to conclude that there must be six cyclic groups of order 4? Furthermore, I'm not quite sure I understand the part about the non-cyclic groups or the notation being used?

I was also wondering in general if there's a systematic method that can be used to find the number of subgroups of a certain order $n$ in any finite abelian group given its decomposition into cyclic groups of prime powers?

Any help would be much appreciated!

答:
It's right that there are six cyclic subgroups of order  $4$, since there are $12$ elements of order  $4$ and $\varphi  (4)=2$ of them in each.

There are three elements of order  $2$.  Thus only one more subgroup of order  $4$, isomorphic the Klein four group.

Title: Subgroups of finite abelian group


问:
develop complex function to power series
I am trying to expand the function $$
f(z)=\frac{2 i}{2+i+z}
$$ to power series near $z_0=0$ and $z_0 =1$

I wanted to bring it to the form of "something" $\cdot$ $\frac{1}{1-az}$ and than get expression of geometric series.

\begin{align}
f(z) & =\frac{2 i}{2+i+z}=\frac{2i}{2+i}\cdot\frac{1}{1+\frac{z}{2+i}}  \\[8pt]
& =\frac{2i}{2+i}\cdot\frac{1}{1-\frac{z}{i(2i-1)}}
=\frac{2i}{2+i}\sum_{n=0}^{n=\infty} \left( \frac{1}{i(2i-1)} \right)^nz^n
\end{align}

I am not sure how should I continue from here, I am stuck with that constant before the sum, I need to bring in a some way into the sum so I will be able to determine my $a_n$.

I would also like for an help for $z_0=1$ if possible.

答:
What you've done is correct, but I think you could make it easier:
$$\begin{align}
f(z) & =\frac{2 i}{2+i+z}=\frac{2i}{2+i}\cdot\frac{1}{1+\frac{z}{2+i}}\\
&=\frac{2i}{2+i}\cdot\frac{1}{1+\frac{(2-i)z}5}\\
&=\frac{2i}{2+i}\sum_{n=0}^\infty(-1)^n\left(\frac{2-i}5\right)^nz^n\\
&=\sum_{n=0}^\infty(-1)^n\left(\frac{2i}{2+i}\right)\left(\frac{2-i}5\right)^nz^n
\end{align}$$
The radius of convergence is $$\limsup_{n\to\infty}\left\lvert\frac{a_{n+1}}{a_n}\right\rvert=\limsup_{n\to\infty}\left\lvert\frac5{2-i}\right\rvert=\sqrt5$$

To get the power series about $z=1$, just start by writing $$f(z)=\frac1{3+i+(z-1)},$$ and then take analogous steps.


问:
derivation of a recurrence relation
From this Stack post

Clever derivation of $\arcsin(x)$ Taylor series

I worked through the derivation of $c_{n+2}=\frac{n^2}{(2n+1)(2n+2)}c_n$ but couldn't figure out the derivation with $c_{2n+1}=\frac{(2n)!}{(2^nn!)^2(2n+1)}$.

any help would be greatly appreciated.

Title: derivation of a recurrence relation

Body:
You have

$$c_{n+2}=\frac{n}{n+2}\frac{n}{n+1}c_n$$

for each $n\ge 0$. Thus,

$$\begin{align*}
c_{2n+1}&=\frac{2n-1}{2n+1}\frac{2n-1}{2n}c_{2n-1}\\
&=\frac{2n-1}{2n+1}\frac{\color{red}{2n-1}}{2n}\frac{2n-3}{\color{red}{2n-1}}\frac{2n-3}{2n-2}c_{2n-3}\\
&=\frac{2n-1}{2n+1}\frac{\color{red}{2n-1}}{2n}\frac{2n-3}{\color{red}{2n-1}}\frac{\color{blue}{2n-3}}{2n-2}\frac{2n-5}{\color{blue}{2n-3}}\frac{2n-5}{2n-4}c_{2n-5}\\
&\;\;\vdots\\
&=\frac{\prod_{k=1}^n(2k-1)}{(2n+1)\prod_{k=1}^n2k}c_1\\
&=\frac{(2n-1)!!}{(2n+1)2^nn!}\,.
\end{align*}$$

Now

$$\begin{align*}
(2n-1)!!&=(2n-1)(2n-3)(2n-5)\ldots(3)(1)\\
&=\frac{(2n)!}{(2n)(2n-2)(2n-4)\ldots(4)(2)}\\
&=\frac{(2n)!}{2^nn!}\,,
\end{align*}$$

so

$$c_{2n+1}=\frac{(2n-1)!!}{(2n+1)2^nn!}=\frac{(2n)!}{(2n+1)(2^nn!)^2}\,.$$


问:
Why $t-7$ and $4$ are not coprime in $\mathbb Z[t]$?
I have to compute $(t-7,4)/(t-7)$ in $\mathbb Z[t]$. What I did is : since $t-7$ and $4$ are coprime, $(t-7,4)=\mathbb Z[t]$. Therefore, $$(t-7,4)/(t-7)\cong \mathbb Z[t]/(t-7)\cong \mathbb Z,$$
since $t-7$ is irreducible in $\mathbb Z[t]$. The Teacher assistant gave me a mark of $0$ out of $10$ and said that $(t-7,4)/(t-7)\cong 4\mathbb Z$. So, why $t-7$ and $4$ are not coprime, i.e. what would be the gcd ? $\color{red}{(solved)}$ And how to prove that $(t-7,4)/(t-7)\cong 4\mathbb Z$ ? this doesn't natural for me.

答:
$\gcd(t-7,4)=1$ and $(t-7,4)\ne\mathbb Z[t]$.

However $(t-7,4)/(t-7)\not\simeq\mathbb Z/4\mathbb Z$ (as groups, as rings, or whatever you want!) since the quotient is infinite. Instead it is an ideal of $$\mathbb Z[t]/(t-7)\simeq\mathbb Z\tag{*}.$$
Furthermore, the quotient $(t-7,4)/(t-7)$ is an ideal of $\mathbb Z[t]/(t-7)$ while $4\mathbb Z$ is an ideal of $\mathbb Z$. As I have already mentioned the rings are isomorphic, and the isomorphism is pretty obvious: define $\mathbb Z[t]\to\mathbb Z$ by $t\mapsto 7$. (Notice that the kernel is $(t-7)$.) Then $(t-7,4)/(t-7)$ corresponds to $4\mathbb Z$.


问:
Riemann's functional equation for $\zeta(s)$: what am I doing wrong?
Different sources appear to give different formulas for Riemann's famous functional equation for the zeta function. Specifically, Wikipedia (here) gives

$$\zeta(s) = 2^s \pi ^{s-1} 2 \sin \left(\frac{s \pi }{2}\right) \Gamma (1-s) \zeta (1-s)$$

whereas "Riemann's Zeta Function" by H.M. Edwards gives the formula

$$\zeta(s) = \Pi (-s) (2 \pi )^{s-1} 2 \sin \left(\frac{s \pi }{2}\right) \zeta (1-s)$$

His $\Pi$ notation is the original form of the gamma function, where $\Pi(s) = \Gamma(s-1)$. Thus $\Pi (-s) = \Gamma(-s-1)$, and so Edwards' formula is

$$
\begin{aligned}
\zeta(s) &= \Gamma (-s-1) (2 \pi )^{s-1} 2 \sin \left(\frac{s \pi }{2}\right) \zeta (1-s)
\\&= 2^s \pi ^{s-1} \sin \left(\frac{\pi  s}{2}\right) \Gamma (-s-1) \zeta (1-s)
\end{aligned}
$$

So, Wikipedia has $\Gamma (1-s)$ as a factor, where Edwards has $\Gamma (-s-1)$ - and only one of them can be right.

Trouble is, calculation using Mathematica suggests that neither of them are right! Using real values of $s$ in the range $s = [-1.5, 1.5]$ produces the following table:

What I have done here is use Mathematica's values for $\zeta(1-s)$, and plugged them into the two variants of the functional equation.

Clearly, I am doing something very badly wrong! There are two issues:

Why the disparity between Edwards and Wikipedia? Perhaps I am wrong to write $\Pi (-s) = \Gamma(-s-1)$?

Why does Mathematica disagree with the published versions on the functional equation?

It's clear to me that I must have made some very bad assumptions - but what are they?

For those who wish to replicate the table, the code is

```<code>Grid[Prepend[Table[{NumberForm[N[s], 3], NumberForm[N[Zeta[s]], 3],
     NumberForm[
 N[Gamma[-s - 1]*(2*Pi)^(s - 1)*2*Sin[(s*Pi)/2]*Zeta[1 - s]],
 3],
     NumberForm[
 N[2^s*Pi^(s - 1)*2*Sin[(s*Pi)/2]*Gamma[1 - s]*Zeta[1 - s]],
 3]},
   {s, -(3/2), 3/2,
1/4}], {Column[{"", TraditionalForm[s], "", ""}],
   Column[{"", TraditionalForm[Zeta[s]], "(Mathematica)", ""}],
   Column[{"",
 TraditionalForm[
  HoldForm[2^s*Pi^(s - 1)*Sin[(s*Pi)/2]*Gamma[-s - 1]*
             Zeta[1 - s]]], "(Edwards)", ""}],
   Column[{"",
 TraditionalForm[
  HoldForm[2^s*Pi^(s - 1)*2*Sin[(s*Pi)/2]*Gamma[1 - s]*
             Zeta[1 - s]]], "(Wikipedia)", ""}]}],
Alignment -> Left, Spacings -> {1, 0.5},
Dividers -> {{Thick, Thin, Thin, Thin, Thick}, {Thick, True, Thin,
Thin, Thin, Thin, Thin,
     Thin, Thin, Thin, Thin, Thin, Thin, Thin, Thick}}]
</code>```

答:
Note that $\Pi(s) = \Gamma(s+1)$ (easy check $\Pi(n)=n!, \Gamma(n)=(n-1)!$) so the two expressions in Edwards and Wikipedia match.

As for the computations, the third column has an extra $2$ factor (it has both $2^s$ and $2$ and it's either $2^s(\pi)^{s-1}$ or $2(2\pi)^{s-1}$, last one to make a homogeneous constant raised to same power $s-1$), and then indeed clearly the first column and the third column would then match (the third column being twice the first column)


问:
Spivak Calculus chapter 7 theorem 9
I am working through Spivak Calculus chapter 7 theorem 9. There is one statement that I can't quite understand.

The theorem states:
If $n$ is odd, then any equation
$$
\ x^n+a_{n-1}x^{n-1} +\cdots+a^0
$$
has a root.

proof: we would like to prove that $f$ is sometimes positive and sometimes negative. The
intuitive idea is that for large $|x|$, the function is very much like $g(x) = x^n$ and,
since $n$ is odd, this function is positive for large positive $x$ and negative for large
negative $x$. A little algebra is all we need to make this intuitive idea work.

$$
f(x) = x^n+a_{n-1}x^{n-1} +\cdots+a^0 = x^n \left(1+\frac{a_{n-1}}{x}+\cdots+\frac{a_0}{x^n}\right)
$$

Note that
$$
\left|\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\cdots+\frac{a_0}{x^n} \right|\le \frac{|a_{n-1}|}{|x|}+\frac{|a_{n-2}|}{|x^2|}+\cdots+\frac{|a_{0}|}{|x^n|}
$$

Consequently if we choose $x$ satisfying
$$
|x|>1,2n|a_{n-1}|,\ldots,2n|a_0| \tag{*}
$$

I am not sure how he comes to $(*)$

Thanks in advance

答:
You need to keep reading in the same page. He wants to make
$$
\left|\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\cdots+\frac{a_{0}}{x^n} \right|\le \frac12,
$$
in order to guarantee that the term multiplying $x^n$ is positive (one plus a number of absolute value less than a half will be positive).

So he chooses $x$ such that $|a_{n-k}/|x^k|\leq 1/2n$ (for all $k$), and then the sum will be less than $1/2$. He needs $|x|\geq1$ so that $|x^k|\geq|x|$.

Title: Spivak Calculus chapter 7 theorem 9


问:
If $f$ is a polynomial of degree $k$, can we quickly compute $f(1),f(2),\dots,f(n).$
$f(x)=\left(\sum_{i=1}^k{a_ix^i}\right) \pmod P$

$P > n \gg k.$

Obviously calculating $f(1),f(2)..f(n)$ takes $O(nk)$.

So is there a faster way to speed up the process? $k$ is not very big. $k$ is only about $30.$

答:
We are given
$$ f(x)=\left(\sum_{i=1}^k{a_ix^i}\right) \pmod P. $$
Since $\,f(x)\,$ is a polynomial mod $\,P\,$, then it is periodic.
That is, $\,f(x+P) = f(x) \pmod P.\,$ Thus, all you need to do
is compute $\,f(0),f(1),\dots,f(P-1)\,$ and there is no need for
any further computation of $\,f(n).\,$ Even though $\,P\,$ may be
big, it is a constant $\,O(1)\,$ compared to $\,O(nk).$

Of course, this is assuming that $\,P\,$ is fixed and that $\,n\,$
can be arbitrarily large. If $\,P\,$ depends on $\,n\,$ then that
is a very different situation. In that case, $\,P\,$ doesn't even matter.
Use a difference table up to $k$th order differences. Each new value
of $\,f(n)\,$ takes $\,k\,$ additions for time complexity of $\,O(nk)\,$
additions with no multiplications required. This is very fast in my opinion.


问:
bounded double sequence
Let $(a_{n,m})_{n,m\in\mathbb N}$ be a real double sequence.
Suppose that
$$ \forall m\in\mathbb N\quad \exists \lim_{n\to\infty} a_{n,m} = \lambda_m \in \mathbb R \,,$$
$$ \exists \lim_{m\to\infty} \lambda_{m} = \lambda \in \mathbb R \,,$$
Can we conclude that $(a_{n,m})_{n,m\in\mathbb N}$ is bounded?

答:
No, not in general. Consider, for example:
$$a_{n,m} = \begin{cases} n & \text{if } 1 \le n \le m \\ 0 & \text{if }n > m.\end{cases}$$
Then for any $m$, the sequence $a_{n,m}$ is eventually constantly $0$, hence:
$$\lim_{n \to \infty} a_{n,m} = 0.$$
The limit of these limits is clearly $0$ too. But, $a_{n,m}$ is not bounded, as $a_{n,n} = n \to \infty$ as $n \to \infty$.

Title: bounded double sequence


答:
Assuming that I’ve interpreted the question correctly, let

$$a_{n,m}=\begin{cases}
m,&\text{if }n=0\\
0,&\text{if }n>0\,;
\end{cases}$$

then $\langle a_{n,m}:n,m\in\Bbb N\rangle$ is unbounded, but $\lim\limits_{n\to\infty}a_{n,m}=0$ for each $m\in\Bbb N$, so of course $\lim\limits_{m\to\infty}\lim\limits_{n\to\infty}a_{n,m}=0$.

You can’t even guarantee that the double sequence is bounded by further requiring that $\lim\limits_{m\to\infty}a_{n,m}$ exist for each $n\in\Bbb N$: let

$$a_{n,m}=\begin{cases}
n,&\text{if }n=m\\
0,&\text{otherwise.}
\end{cases}$$

Then $\lim\limits_{n\to\infty}a_{n,m}=0$ for each $m\in\Bbb N$ and $\lim\limits_{m\to\infty}a_{n,m}=0$ for each $n\in\Bbb N$, but the double sequence is unbounded.

Title: bounded double sequence

Body:
Let $(a_{n,m})_{n,m\in\mathbb N}$ be a real double sequence.
Suppose that
$$ \forall m\in\mathbb N\quad \exists \lim_{n\to\infty} a_{n,m} = \lambda_m \in \mathbb R \,,$$
$$ \exists \lim_{m\to\infty} \lambda_{m} = \lambda \in \mathbb R \,,$$
Can we conclude that $(a_{n,m})_{n,m\in\mathbb N}$ is bounded?


问:
System of 7 equations and 3 unknowns in probability
I have the following data:
$P(A\cap B^c \cap C^c)=0.14$, $P(A^c\cap B \cap C^c)=0.1$, $P(A^c\cap B^c \cap C)=0.15$, $P(A\cap B \cap C^c)=0.13$, $P(A\cap B^c \cap C)=0.12$, $P(A^c\cap B \cap C)=0.16$ and $P(A^c\cap B^c \cap C^c)=0.16$. I must find $P(A),P(B)$ and $P(C)$.

What I came up with was to do $P(A)=x$, $P(B)=y$ and $P(C)=z$, and considering that the events are independent form the system of equations

$x(1-y)(1-z)=0.14$

$(1-x)y(1-z)=0.1$

$(1-x)(1-y)(1-z)=0.09$

$xy(1-z)=0.13$

$x(1-y)z=0.12$

$(1-x)yz=0.16$

$(1-x)(1-y)z=0.15$

according to wolfram said system of seven equations has no solution,
is there something wrong with my approach?

答:
HINT...you have assumed incorrectly that all these events are independent which gives you wrong equations.
Why not start by drawing a Venn diagram?</p>


问:
Proving that $\mathbb{Z}$ is not a field.
This question is related to this answer.  I always thought that we can simply say since $2 \in \mathbb{Z}$, and $2^{-1} = 1/2 \notin \mathbb{Z}$, $\mathbb{Z}$ is not a field.

Can someone explain why this proof is wrong as the answer claims?

答: Two is not equal to zero. Two times an integer is an even integer. Since $1$ is odd, there is no integer $n$ with $2n = 1$.</p>


问:
If $a_{n_k}$ denotes a subsequence of $a_n$. Why is $n_k \geq k$ for all $k$.?
I am trying to prove that If a sequence converges, then every subsequence converges to the same limit. Now going in the backwards direction, I know it is true because the sequence is a subsequence of itself. But what about proving if a sequence converges then every subsequence converges to the same limit.

I do not understand the notation $a_{n_k}$, $a_n$, $n_k$ and $k$
I know that subsequence need to be in order just like in the original sequence. Can someone please clarify the notation and explain why  $n_k \geq k$ for all $k$.?

答: The $n_k$ ($k=1,2,3,\ldots$) are a subsequence of $\Bbb N$, i.e., $n_1\ge 1$ and for all $k$, $n_{k+1}>n_k$. Then $n_k\ge k$ follows by induction.


问:
Difference between showing there is at least one periodic solution and showing that all solutions are periodic
Suppose that a system of differential equations is given in the following form
$$\frac{dy}{dx} = Ay(x) + b(x)$$
with $A$ a matrix. If $y(x)$ is a general solution, what is the difference between showing  there is at least one periodic solution and showing that all solutions are periodic?

答:
Assume there exists a periodic solution $y_1(x)$ and a non-periodic solution $y_2(x)$. Then the first statement is true and the second is false.

Or assume there is no solution at all. Then the first statement is false and the second is true.


问:
Show that if all the normal lines to a surface pass through a fixed point, then the surface is (a portion of) a sphere.
Show that if all the normal lines to a surface pass through a fixed point, then the surface is (a portion of) a sphere.

I'm not sure how to parametrize my surface.

In the book there is an example that says we can define a regular parametrization

$x:U \to M$, where $U$ is an open set in $\mathbb{R^2}$ and $M$ is a surface

$x(u,v) = \alpha(u) + v\beta(u)$

From here if the parametrization is correct,

$x_u = \alpha'(u) + v\beta'(u)$

$x_v = \beta(u)$

<=>

$\alpha'(u) + v\beta'(u) - x_u = 0$

$\beta(u) - x_v = 0$

Then if we take the scalar product with $n$

$\alpha'(u) \cdot n + v\beta'(u)\cdot n - x_u\cdot n= 0$ <=> $\alpha'(u) \cdot n + v\beta'(u)\cdot n= 0$

$\beta(u) \cdot n - x_v  \cdot n= 0$ <=> = $\beta(u) \cdot n = 0$ <=> $\beta(u)=0$ <=> $\beta'(u)=0$

Therefore

$\alpha'(u) \cdot n + v\beta'(u)\cdot n= 0$ <=>  $\alpha'(u) \cdot n = 0$ <=> $\alpha'(u) = 0$

Since $\alpha'(u) = \beta'(u) =0$, then $x(u,v)$ is always a fixed distance from some point which means its the a part of the sphere.

Is this correct?

答:
The hypothesis says that $x(u,v)= f(u,v)  N (u,v)$, where $f$ is a scalar, and $N$ the unit normal vector (if we fix the origin to be this point). We have to prove that $f$ is constant.

${\partial x\over \partial u }= {\partial f\over \partial u }N+ f {\partial N\over \partial u }$.

As $N$ is a unit vector, ${\partial N\over \partial u }$ is a tangential vector (derive $<N,N>=1$, as well as ${\partial x\over \partial u }$.
So computing the scalar product with $n$ we find ${\partial f\over \partial u }=0$, and similarly   ${\partial f\over \partial v }=0$, whence the result.


问:
SMO - 2010: finding least or greatest number

Among the five real numbers below, which one is the smallest?

$\text{(A)}\ \ \sqrt[\leftroot{-2}\uproot{2}2009]{2010};\quad\text{(B)}\ \ \sqrt[\leftroot{-2}\uproot{2}2010]{2009};\quad\text{(C)}\ \ 2010;\quad\text{(D)}\ \ \frac{2010}{2009};\quad\text{(E)}\ \ \frac{2009}{2010}.$

Among the five integers below, which one is the largest?

$\text{(A)}\ \ 2009^{2010};\quad\text{(B)}\ \  20092010^2;\quad\text{(C)}\ \ 2010^{2009};\quad\text{(D)}\ \ {3}^{(3^{(3^{3})})};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \text{(E)}\ \ 2^{10}+4^{10}+\cdots+2010^{10}$

As you can see, the above are the questions from $SMO - 2010$, I want to know if there exists a uniform method to solve such kind of questions.

All the options look equally promising.

In the $1^{st}$ part, option $2010$, $\frac{2010}{2009}$ and $\frac{2009}{2010}$get eliminated. I have no idea about how to make comparison between $A$ and $B$.

Similarly in the $2^{nd}$ part, I am clueless.

Please tell a uniform method which applies for all questions. Thanks.

答:
As mentioned in the comments, there isn't a "uniform" method. Some ideas are

Spot a pattern,

Compare to constants/powers like $1, 10^k, 2^k$.

Be aware of which functions grow quickly / slowly.

Hint for part 1: Show that one term is $<1$ and all the other terms are $>1$.

I'm guessing options A and B were put there to "scare" people / make them think the question is harder than it actually is.

Hint for part 2: The idea here is that tower of exponents grow very quickly. You should be aware of this fact, and so that is the most likely guess without further computation.

To prove it, you could bound 3^3^3^3 $> 10^{8040} > \max ( 2010^{2009}, 2009^{2010} )$.

 If you're stuck, use $ 3^5 > 100$.

 3^3^27 > $3^{10000000000} > 100^{10000000000/5 } $.


问:
How useless can the Mayer-Vietoris sequence be in general?
In an algebraic topology course I'm taking we are often asked to compute the homology groups of a space $X = A \cup B$ using the Mayer-Vietoris sequence, and it happens in all of the examples I've seen so far that it's possible to do this without knowing anything about the connecting homomorphisms $\partial_{\ast}$ (say on the level of chains); we only end up needing $H_{\ast}(A), H_{\ast}(B), H_{\ast}(A \cap B)$ and possibly some of the inclusion maps. </p>

<p>My guess is that this is not a typical situation; is there a relatively simple example of a nice space $X$ and nice subspaces $A, B$ such that knowing $H_{\ast}(A), H_{\ast}(B), H_{\ast}(A \cap B)$ is not enough to compute $H_{\ast}(X)$ without knowing the specific form of the connecting homomorphisms? (For maximal relevance to the course $X, A, B, A \cap B$ should be finite simplicial complexes.) </p>

答:
Knowing only $H(A)$, $H(B)$ and $H(A\cap B)$ is not enough, of course.
For example, taking $A=B=S^1\times D^2$ and gluing them by $S^1\times S^1=A\cap B$ one can get either $X_1=S^2\times S^1$ or $X_2=S^3$. This gives two Mayer-Vietoris sequences with identical $H(A)$, $H(B)$ and $H(A\cap B)$ but different H(X).

As for the situation where one also knows inclusion maps, see Ryan Budney's excellent answer.


问:
Generalized eigenspace decomposition of vector space
I was curious whether the direct sum of generalized eigenspaces in a finite dimensional vector space is the largest decomposition of this space invector spaces that are invariant under the endomorphism or are there counterexamples available?</p>

答:
No. If you take the sum of two generalized eigenspaces, it will still be an invariant subspace, since generalized eigenspaces correspond to the blocks in the Jordan decomposition.

Even in finite dimension, the number of invariant subspaces can be infinite. To see the most basic example, consider the identity endomorphism: it obviously has infinitely many invariant subspaces. Or consider
$$
V=\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}
$$
as an endomorphism of $\mathbb R^3$. Then, for each $t\in[0,1]$, the subspace
$$
X_t=\{(0,ct,c(1-t)):\ c\in\mathbb R\}
$$
is invariant for $V$.


问:
Correctly prove that $\log(+1) = \Theta(\log())$
What is the correct mathematical reasoning to omit the $+ 1$ in $\log(+1) = \Theta(\log())$ since it is a constant?

答:
If $f=\Theta(g)$, then by the limit definition

$$0 < \liminf_{x \to \infty} \left|\frac{f(x)}{g(x)}\right| \le \limsup_{x \to \infty} \left|\frac{f(x)}{g(x)}\right|< \infty.$$

Note that if we choose $f=\log(n+1)$ and $g=\log(n)$,
$$\lim_{n\to\infty}\left|\frac{\log(n+1)}{\log(n)}\right| = \liminf_{n \to \infty} \left|\frac{\log(n+1)}{\log(n)}\right| = \limsup_{n \to \infty} \left|\frac{\log(n+1)}{\log(n)}\right|.$$

The $+1$ does not change the growth of the function, so
$\log(n+1)/\log(n)=1$ as $n\to\infty.$

In conclusion we have $$0<1\le1<\infty,$$
which is true and so $\log(n+1)=\Theta(\log n)$.


答: I will assume that the indices of the original sequence start at $0$; the proof is almost identical if they start at $1$ (or indeed at any other integer). Clearly $n_0$, the index of the first term of the subsequence, must be at least $0$: $n_0\ge 0$. Now suppose that $n_k\ge k$ for some $k$. We must have $n_{k+1}>n_k$, and $n_{k+1}$ and $n_k$ are both integers, so $n_{k+1}\ge n_k+1$. Finally, $n_k\ge k$, so $n_{k+1}\ge n_k+1\ge k+1$. By induction we conclude that $n_k\ge k$ for all $k\in\Bbb N$. (For me $\Bbb N$ includes $0$.)

Title: If $a_{n_k}$ denotes a subsequence of $a_n$. Why is $n_k \geq k$ for all $k$.?

Body:
I am trying to prove that If a sequence converges, then every subsequence converges to the same limit. Now going in the backwards direction, I know it is true because the sequence is a subsequence of itself. But what about proving if a sequence converges then every subsequence converges to the same limit.

I do not understand the notation $a_{n_k}$, $a_n$, $n_k$ and $k$
I know that subsequence need to be in order just like in the original sequence. Can someone please clarify the notation and explain why  $n_k \geq k$ for all $k$.?


答: As $n+1<n^2$ for $n>2$, we have $\ln(n+1)<2\ln n$ and so $\ln(n+1)=O(\ln n)$. On the other hand, $\ln n=O(\ln(n+1))$ follows from $\ln n<\ln(n+1)$.

Title: Correctly prove that $\log(+1) = \Theta(\log())$

Body: What is the correct mathematical reasoning to omit the $+ 1$ in $\log(+1) = \Theta(\log())$ since it is a constant?


答:
As I said somewhere, I prefer the symmetric discretizations that have automatically a higher order of accuracy. So with some infinitesimal $δt$ set $t_k=t_0+\frac12k\,δt$. Then $a_0=\frac{v_1-v_{-1}}{δt}+O(δt^2)$, $v_1=\frac{r_2-r_0}{δt}+O(δt^2)$, $v_{-1}$ similarly, and
\begin{align}
\frac{r_1×v_1-r_{-1}×v_{-1}}{δt}
&=\frac{(r_1-r_{-1})×(v_1+v_{-1})}{2δt}+\frac{(r_1+r_{-1})×(v_1-v_{-1})}{2δt}
\\
&=\frac{(v_0+O(δt)^2)×(v_0+O(δt)^2)}{2δt}+\frac{(r_0+O(δt)^2)×(a_0+O(δt)^2)}{2}
\\
&=r_0×a_0+O(δt^2)
\end{align}
Now remember that as an infinitesimal $δt$ has a decimal representation with a mind-boggling number of zeros after the decimal dot. This transfers to the remainder $O(δt^2)$. So in all integrations over finite time spans this remainder contributes at most $O(δt)$, which has no appreciable influence.

Title: $ \vec r \times m \frac {\delta^2 \vec r} {\delta^2 t} \neq \frac {\delta \vec L}{\delta t} $?

Body:
$$\forall m \land \forall \vec r_{(t1)}(r_{1x},r_{1y},r_{1z}) \land \forall \vec r_{(t2)}(r_{2x},r_{2y},r_{2z}) \land \forall \vec r_{(t3)}(r_{3x},r_{3y},r_{3z}) $$
$$\vec v_{(t1)}=\frac {\vec r_{(t2)} - \vec r_{(t1)}} {\delta t} \tag {1.a} $$
$$\vec v_{(t2)}=\frac {\vec r_{(t3)} - \vec r_{(t2)}} {\delta t} \tag {1.b} $$
$$\vec a_{(t1)}=\frac {\vec v_{(t2)} - \vec v_{(t1)}} {\delta t} \tag {1.c} $$
$$ \vec r_{(t1)} \times  m \vec v_{(t1)}=m(r_{1y} v_{1z} -r_{1z} v_{1y},r_{1z} v_{1x} -r_{1x} v_{1z},r_{1x} v_{1y} -r_{1y} v_{1x})= \vec L_{(t1)} \tag {2.a}  $$
$$ \vec r_{(t2)} \times  m \vec v_{(t2)}=m(r_{2y} v_{2z} -r_{2z} v_{2y},r_{2z} v_{2x} -r_{2x} v_{2z},r_{2x} v_{2y} -r_{2y} v_{2x})= \vec L_{(t2)} \tag {2.b}  $$
$$\frac {\vec L_{(t2)}-\vec L_{(t1)}} {\delta t} =(\frac {\delta l_{x}}{\delta t}, \frac {\delta l_{y}}{\delta t},\frac {\delta l_{z}}{\delta t} ) \tag 3$$
$$\frac {\delta l_{x}}{\delta t}=m(r_{2y}v_{2z}-r_{1y}v_{1z}-r_{2z}v_{2y}+r_{1z}v_{1y}) \tag {3.a}$$
$$\frac {\delta l_{y}}{\delta t}=m(r_{2z}v_{2x}-r_{1z}v_{1x}-r_{2x}v_{2z}+r_{1x}v_{1z}) \tag {3.b}$$
$$\frac {\delta l_{z}}{\delta t}=m(r_{2x}v_{2y}-r_{1x}v_{1y}-r_{2y}v_{2x}+r_{1y}v_{1x}) \tag {3.c}$$

$$\vec r_{(t1)}\times m \vec a_{(t1)}=m(r_{1y}a_{1z} - r_{1z}a_{1y}, r_{1z}a_{1x}-r_{1x}a_{1z},r_{1x}a_{1y} -r_{1y}a_{1x}) \tag 4$$

$$r_{1y}a_{1z} - r_{1z}a_{1y}=r_{1y}v_{2z}-r_{1y}v_{1z} -r_{1z}v_{2y}+r_{1z}v_{1y}\tag {4.a}$$
$$r_{1z}a_{1x} - r_{1x}a_{1z}=r_{1z}v_{2x}-r_{1z}v_{1x} -r_{1x}v_{2z}+r_{1x}v_{1z}\tag {4.b}$$
$$r_{1x}a_{1y} - r_{1y}a_{1x}=r_{1x}v_{2y}-r_{1x}v_{1y} -r_{1y}v_{2x}+r_{1y}v_{1x}\tag {4.c}$$

$$(4) \neq (3)$$

Correct pattern
$$  \frac {\delta \vec L}{\delta t} = m( \frac {\delta \vec r_{(t1)}} {\delta t} \times  \frac {\delta \vec r_{(t2)}} {\delta t} +  \vec r_{(t1)} \times   \frac {\delta^2 \vec r_{(t1)}} {\delta^2 t})   \tag 5$$

I proved it here but this result $(4) \neq (3)$ is contrary to bookish knowledge and (5) changes the understanding of point rotation, since this formula allows for the existence of non-zero internal moments of force, lawful conservation of angular momentum.

So the question is: I am wrong in the calculations or maybe the present knowledge is wrong?


问:
Solving linear differential equation with variable coefficients.
I have faced this task, when preparing to my university's "mathematical methods" exam, and I do not really understand how to solve it. The task content is as follows:

Given: $$(1-x^2)\cdot y''_{xx} -x\cdot y'_x +y = \frac{\sqrt{1-x^2}}{x}$$ Solve the equation by substituion of the independent variable: $x(t) = \cos(t), x\in(0,1).$

My attempt:

I have performed the suggested substitution:

$(1-\cos^{2}(t))\cdot \large\frac{\partial^2y}{\partial x(t)^2}\normalsize-\cos(t)\cdot\large\frac{\partial y}{\partial x(t)} \normalsize+ y = \large\frac{\sqrt{1-\cos^2(t)}}{cos(t)}.$

$\sin^2(t)\cdot \large\frac{\partial^2y}{\partial x(t)^2}\normalsize-\cos(t)\cdot\large\frac{\partial y / \partial t}{\partial x / \partial t} \normalsize+ y = \tan(t).$

$\sin^2(t)\cdot\frac{\large\partial\left(\frac{\dot{y}}{-\sin(t)}\right) / \partial t}{\large\partial x / \partial t} + \large\frac{\cos(t)}{\sin(t)}\normalsize\dot{y} + y = \tan(t).$

$-\ddot{y} + 2\dot{y}\cot(t) +y = \tan(t).$

That is now I have to solve: $-\ddot{y} + 2\dot{y}\cot(t) +y = \tan(t).$ And that is the point, where I am stuck, I have no idea how to proceed from here, and I did not manage to find the way to solve such a linear D.E. in my course book.

I would appreciate any help, hints or book references, thank you in advance!

Post Scriptum:

As correctly mentioned by @Aryadeva, I have made a mistake, when calculating $\large\frac{\partial^2 y}{\partial x^2}$, I simply lost "minus".

That is, after recalculating I had to solve: $$\ddot{y} + y = \tan(t), \text{ s.t. } x\in(0,1) \text{ and } x = \cos(t) \Leftrightarrow \forall x\in \mathbb{D}_x: \text{ }t = \arccos{x}.\tag{1}$$

My solution has involved appying vatiation of parameters method:

$\underline{\text{Step I}}$: find characteristic polynomial over $y(t)$ to get a general solution $(\large y_{\small G})$ for $(1)$:

$\lambda^2 + 1 = 0. \Rightarrow \lambda = \pm i.$

$\begin{bmatrix}
\lambda_1\\
\lambda_2\\
\end{bmatrix} = \begin{bmatrix}
i\\
-i\\
\end{bmatrix} \Rightarrow \large y_{\small G} \normalsize = C_1 e^{it}+C_2 e^{-it} = \dot{\tilde {C}}_1\cos(t) + \dot{\tilde{C}}_2 \sin(t).$

$\underline{\text{Step II}}$: Finding particular solution $(\large y_{\small P})$ for $(1)$, using Wronskian matrices:

I would omit my calculations here, would just note that for the given D.E.:

$\mathbb{W} = \begin{pmatrix}\cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{pmatrix}, \mathbb{W}_{\small \Delta} = \begin{vmatrix}\cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{vmatrix} = 1.$

$\mathbb{W}_{\small \Delta,\normalsize 1} = \begin{vmatrix}0 & \sin(t) \\ 1 & \cos(t) \end{vmatrix} = -\sin(t), \mathbb{W}_{\small \Delta,\normalsize 2} = \begin{vmatrix}\cos(t) & 0 \\ -\sin(t) & 1 \end{vmatrix} = \cos(t).$

Thus:
$$\dot{\tilde{C}}_1(t) = \int\frac{\tan(t)\cdot \mathbb{W}_{\small \Delta,\normalsize 1}}{\mathbb{W}_{\small \Delta}} \partial t = \sin(t) +\frac{1}{2}\ln\left(\frac{\sin(t)-1}{\sin(t)+1}\right) +\bar{C}_1,$$
$$\dot{\tilde{C}}_2(t) = \int\frac{\tan(t)\cdot \mathbb{W}_{\small \Delta,\normalsize 2}}{\mathbb{W}_{\small \Delta}} \partial t = -\cos(t) + \bar{C}_2.$$
Thereof:
$$\large y_{\small P} \normalsize = \left(\sin(t)+\frac{1}{2}\ln\left(\frac{\sin(t)-1}{\sin(t)+1}\right) +\bar{C}_1\right)\cos(t) +\left(-\cos(t) +\bar{C}_2\right)\sin(t).$$

$\underline{\text{Step III}}$: Now it just suffices to write down the whole solution and transfer from $t$ to $x$:

$\large y \normalsize (t) = \large y_{\small G} + y_{\small P} \normalsize = \left(\sin(t)+\frac{1}{2}\ln\left(\frac{\sin(t)-1}{\sin(t)+1}\right) +\tilde{C}_1\right)\cos(t) +\left(-\cos(t) +\tilde{C}_2\right)\sin(t).$

Now, as $x \in (0, 1): \cos(\arccos{x}) = x, \text{ }\sin(\arccos{x}) = \sqrt{1-x^2}:$
$$\text{Answer: } \begin{array}{|c|}
\hline
y(x) = \left(\sqrt{1-x^2} + \frac{1}{2}\ln\left(\frac{\sqrt{1-x^2} - 1}{\sqrt{1-x^2} +1}\right)+\tilde{C}_1 \right)x +\left(-x+\tilde{C}_2\right)\sqrt{1-x^2}.\\
\hline
\end{array}$$

答:
We are given this ODE:

$$(1 - x^2) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - x \frac{\mathrm{d}y}{\mathrm{d}x} + y = \frac{\sqrt{1 - x^2}}{x}$$

To solve this, we simply make the substitution $x = \cos t$ and then write $\frac{\mathrm{d}y}{\mathrm{d}x}$ in term of $\frac{\mathrm{d}y}{\mathrm{d}t}$ and $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ in term of $\frac{\mathrm{d}^2y}{\mathrm{d}t^2}$. After that, things will magically cancel out :)

Substitute $x = \cos t$ into the ODE gives

$$(1 - x^2) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - x \frac{\mathrm{d}y}{\mathrm{d}x} + y = \frac{\sqrt{1 - x^2}}{x}$$
$$(1 - \cos^2) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - \cos t \frac{\mathrm{d}y}{\mathrm{d}x} + y = \frac{\sqrt{1 - \cos^2}}{\cos t}$$
$$\sin^2 t \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - \cos t \frac{\mathrm{d}y}{\mathrm{d}x} + y = \frac{\sin t}{\cos t} = \tan t$$

Differentiate $x = \cos t$ on both side w.r.t. $x$  gives
$$1 = - \sin t \cdot \frac{\mathrm{d}t}{\mathrm{d}x}$$
$$\frac{\mathrm{d}t}{\mathrm{d}x} = -\csc t$$

Now $$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t} \cdot \frac{\mathrm{d}t}{\mathrm{d}x} = -\csc t \cdot \frac{\mathrm{d}y}{\mathrm{d}t}$$

and hence $$\begin{align}
\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
&= \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right) \\
&= \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x} \right) \cdot \frac{\mathrm{d}t}{\mathrm{d}x} \\
&= \frac{\mathrm{d}}{\mathrm{d}t} \left(-\csc t \frac{\mathrm{d}y}{\mathrm{d}t}\right) \cdot (-\csc t) \\
&= \left(\csc t \cot t \frac{\mathrm{d}y}{\mathrm{d}t} - \csc t \frac{\mathrm{d}^2y}{\mathrm{d}t^2}\right) \cdot (-\csc t) \\
&= \csc^2 t \left(\frac{\mathrm{d}^2y}{\mathrm{d}t^2} - \cot t \frac{\mathrm{d}y}{\mathrm{d}t}\right)
\end{align}$$

Next, write $\frac{\mathrm{d}y}{\mathrm{d}x}$ in term of $\frac{\mathrm{d}y}{\mathrm{d}t}$ and $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ in term of $\frac{\mathrm{d}^2y}{\mathrm{d}t^2}$

$$\sin^2 t \cdot \csc^2 t \left(\frac{\mathrm{d}^2y}{\mathrm{d}t^2} - \cot t \frac{\mathrm{d}y}{\mathrm{d}t}\right) - \cos t \cdot (-\csc t) \cdot \frac{\mathrm{d}y}{\mathrm{d}t} + y = \tan t$$
$$\frac{\mathrm{d}^2y}{\mathrm{d}t^2} - \cot t \frac{\mathrm{d}y}{\mathrm{d}t} + \cot t \cdot \frac{\mathrm{d}y}{\mathrm{d}t} + y = \tan t$$
$$\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + y = \tan t$$


问:
Constructing a primary decomposition for principal ideals in a Krull domain
Let $A$ be Krull ring and $\left\{R_l\right\}_{l \in L}$ a defining family of DVRs. Let $a \in A, a \neq 0$. Then $a R_l \neq R_l$ only for a finite number of indices $l$ (otherwise the image of $a$ under infinitely many valuations would be non-zero, contradiction). Let $R_1,\dots, R_t$ be the DVRs such that $a R_i \neq R_i$. Define $\mathfrak q_i = aR_i\cap A$ and $\mathfrak p_i=rad(R_i) \cap A$. Then $\mathfrak q_i$ is primary belonging to $\mathfrak p_i$.

Question: How can we see that $aA=\mathfrak q_1\cap \cdots \cap \mathfrak q_t$?

Remark: Certainly, $aA \subseteq \mathfrak q_1 \cap \cdots \cap \mathfrak q_t$, since $aA \subset \mathfrak q_i$ for every $i$. How about the other direction?

答:
Prove that
$$aR_1\cap\cdots\cap aR_t\cap A=aA.$$
"$\supseteq$" is clear. </p>

<p>"$\subseteq$" it's easy: take $x\in aR_1\cap\cdots\cap aR_t\cap A$. Then $a^{-1}x\in R_1\cap\cdots\cap R_t$. Since $aR_l=R_l$ for all $l\neq 1,\dots,t$ it follows that $a^{-1}\in R_l$, so $a^{-1}x\in R_l$, and that's all.</p>

Body:
Prove that
$$aR_1\cap\cdots\cap aR_t\cap A=aA.$$
"$\supseteq$" is clear.

"$\subseteq$" it's easy: take $x\in aR_1\cap\cdots\cap aR_t\cap A$. Then $a^{-1}x\in R_1\cap\cdots\cap R_t$. Since $aR_l=R_l$ for all $l\neq 1,\dots,t$ it follows that $a^{-1}\in R_l$, so $a^{-1}x\in R_l$, and that's all.


问:
Existence of diffeomorphism of high-dimensional manifold with boundary $M$ and $M\#\Sigma$.
Let $M$ be an oriented smooth $n$-manifold with non-empty boundary $\partial M$ where $n\geq 5$, and let $\Sigma\in \Theta_n$ be a homotopy sphere of dimension $n$. Does there exist a diffeomophism $M \cong M\# \Sigma$ by "sliding the exotic disk out to the boundary". We can assume that $M$ is simply connected and compact if needed.

I have tried to use the $h$-cobordism theorem but with no success. Any help would be appreciated.

答:
Using the reduction to the problem of if $(\partial(M) \times [0,1]) \sharp \Sigma \cong \partial(M) \times [0,1]$, we see that this is true if, $(\partial(M) \times [0,1]) \sharp \Sigma$ is an s-cobordism. It is certainly an h-cobordism since connect summing with a sphere does not change the homotopy type. But a similar analysis shows this is an s-cobordism as well:

For $W$ an h-cobordism, the question of if $M' \rightarrow W$ is simple can be phrased purely topologically. Hence, if we have even a homeomorphism $W \cong M' \times [0,1]$ (commuting with the inclusions $M' \rightarrow W$ and $M' \rightarrow M' \times [0,1])$, then we conclude that $M' \rightarrow W$ is simple because the latter clearly is. So we have reduced the problem to if connect summing with $S^n$ results in a homeomorphic manifold, where the homeomorphism is the identity on the boundary. This is clearly true since it is easily seen to be true for a disk.

Hence, the map $\partial(M) \rightarrow (\partial(M) \times [0,1]) \sharp \Sigma$ is an s-cobordism, so we apply the s-cobordism theorem to deduce the diffeomorphism $(\partial(M) \times [0,1]) \sharp \Sigma \cong \partial(M) \times [0,1]$. Meaning that yes, if $M$ has nonempty boundary, connect summing with an exotic sphere does not affect diffeomorphism type (in dimensions where s-cobordism holds).

To explicitly spell out the rest of the argument: removing an open collar of the boundary results in a diffeomorphic manifold to the original, so by equipping our manifold with this new choice of open collar obtained from the diffeomorphism we just proved, we obtain a manifold diffeomorphic to our original $M$, since we have removed where the connect sum has happened, but also diffeomorphic to $M \sharp \Sigma$ by our comments just now.


问:
What will happen to $\theta$ when $\sin \theta =\frac{\sqrt {{2{\pm}\sqrt i}}}{2}$ and i >4?
Some follow up questions  related to Two mysterious missing angles in the sine values of acute angle list?
.After finding two missing angles,   we have a list of 9 sine values for special angles between 0$^\circ$ to $90^\circ$: $\sin \theta =\frac{\sqrt {{2{\pm}\sqrt i}}}{2}, i=0,1,2,3,4 $. (A side note:This formula is easy to remember: start with
$\sin 45^\circ=$$\frac{\sqrt2}{2}$, let $\pm \sqrt i$ jump under the radical sign and follow  behind 2. Thanks @some guy to remind this trick!)

Now, I have some curious thoughts:

Why is the formula symmetric about $\theta = 45^\circ$?

One would naturally want to continue this list with i beyond 4,..  in which case |$\sin \theta$| will not be bounded by [-1,1]. What does this mean? What are those angles? I expect we step into complex numbers, but do not have a clear picture. Can you please help?

\begin{align}\sin 0^\circ =\frac{\sqrt {\color{green}{2-\sqrt {4}}}}{2},\sin 15^\circ =\frac{\sqrt {\color{green}{2-\sqrt3}}}{2},\sin 22.5^\circ =\frac{\sqrt {\color{green}{2-\sqrt2}}}{2} \end{align}

\begin{align}\sin 30^\circ =\frac{\sqrt {\color{green}{2-\sqrt{1}}}}{2}
, \sin 45^\circ =\frac{\sqrt {\color{green}{2-\sqrt{0}}}}{2}
, \sin 60^\circ =\frac{\sqrt {\color{green}{2+\sqrt{1}}}}{2}\end{align}

\begin{align}\sin 67.5^\circ =\frac{\sqrt {\color{green}{2+\sqrt2}}}{2},\sin 75^\circ =\frac{\sqrt {\color{green}{2+\sqrt 3}}}{2}, \sin 90^\circ =\frac{\sqrt {\color{green}{2+\sqrt{4}}}}{2}\end{align}

答:
The sine function takes all complex values on the complex numbers.  The definition is $$\sin z = \frac{e^{iz}-e^{-iz}}{2i},$$ so if we want to solve $\sin z =w$ we have to solve $$2iw=e^{iz}-e^{-iz}=x-\frac1x$$ where $x=e^{iz}$ and multiplying both sides by $x$, we have a quadratic in $x$.  Note that $e^{iz}$ never takes the value $0$, so there is never a division-by-zero problem.

Does this answer your question?


问:
Does every Abelian group of order $45$ have an element of order $9$?
I found this exercise

Prove that any Abelian group of order $45$ has an element of order $15$.
Does every Abelian group of order $45$ have an element of order $9$?

In the answer to this last question I am told no. Since $\mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_5$, does not have an element of order $9$.

This is a bit confusing to me, since if $m$ divide order of a finite Abelian group $G$, then $G$ has a subgroup of order $m$ and in this case $9\mid 45$. What detail am I missing?

答:
There's a difference between having a subgroup of order $9$ and an element of order $9$. While it's true that every abelian group of order $n$ has a subgroup of order $m$ for any $m|n$, this doesn't necessarily guarantee the existence of an element $x$ of order $m$; it's possible that (as in your case) the subgroup of order $m$ isn't cyclic, and so has no element with the same order as the order of the subgroup.

For concreteness, in your example of $\mathbb Z_3\times\mathbb Z_3\times \mathbb Z_5$, the subgroup $\mathbb Z_3\times\mathbb Z_3$ consisting of the first two factors is of order $9$ (it has $9$ elements), but all of its non-identity elements have order $3$ (have $x^3$ equal to the identity).


问:
Definition of anticommuting Mean Field Theory and real Grassmann fields
These words come from this article https://arxiv.org/abs/2008.04361, page 6. My question is what is anticommuting Mean Field Theory which appears in the equation (1.1), and what is real Grassmann fields which appears above equation (1.2)? Why the anticommuting Mean Field Theory can be characterized by the equation (1.2)?</p>

答: You need some info about Grassmann variables and Berezin integration, otherwise the article will be too hard to read. A quick intro is in https://arxiv.org/abs/math/0306396 In particular, you will see there is an analogue of Gaussian measures for these funny anticommuting variables. This is what the authors refer to by anticommuting MFT. Moments of (centered) Gaussians are entirely and explicitly determined by second moments, i.e., the covariance. The same is true here and that is why eq. (1.2) is enough information for the reader.</p>


问:
How to prove this sum is positive definite
I have a sum that is the norm for $l^2$ (the set of bounded series so $f = (f_n)_{n =1)}^{\infty}
  | \sum f_i^2 < \infty $). In order to prove that my norm is positive definite I must prove that the following is true:

$$\sum_{i=1}^\infty f_i^2+\frac 1 {8} f_if_{i+1} > 0$$ for $f\neq0$ and the sum being equal to $0$ when $f = 0 $.

I tried making appear $(f_i + f_{i+1})^2$ in order to prove that the sum satisfied the inequality but was unable to

Body: $$\sum_{i=1}^\infty\left(f_i^2+\frac18 f_i f_{i+1}\right)=\frac1{16}f_1^2+\frac78\sum_{i=1}^\infty f_i^2+\frac1{16}\sum_{i=1}^\infty(f_i+f_{i+1})^2.$$


答: About your first question about why they are symmetric about $45^\circ$. Let's say that you have an angle that is $x^\circ$ from the $45^\circ$ angle to make an angle of $45-x^\circ$. Thus, the symmetric angle must be $45+x^\circ$. So, if you add these angles up, you get $90^\circ$, which means that they are complementary. You should know that cosine is equal to the sine of the complementary angle. So, if we take a pair of symmetric angles $45-x^\circ, 45+x^\circ$, their sines will be $sin(45-x),sin(45+x)$ or $sin(45-x),cos(45-x)$, if we use our identity. Now, we will use one final identity, and that is that $sin(\theta)^2+cos(\theta)^2=1$. So, we get that $sin(45-x)^2+cos(45-x)^2=1$. Now, we will prove that if $sin$ of one of the angles, $sin(45-x) = \frac{\sqrt{2 - \sqrt{m}}}2$, then the other angle should be $sin(45+x) = \frac{\sqrt{2 + \sqrt{m}}}2$. Using our cosine identity, we know that $sin(45+x) = cos(45-x) = \frac{\sqrt{2 + \sqrt{m}}}2$. If we plug in our values into $sin(45-x)^2+cos(45-x)^2=1$, we get $(\frac{\sqrt{2 - \sqrt{m}}}2)^2+(\frac{\sqrt{2 + \sqrt{m}}}2)=1$ and if you simplify, you can see that this is true. Thus, our original assumptions that if $sin(45-x) = \frac{\sqrt{2 - \sqrt{m}}}2$, then the other angle should be $sin(45+x) = \frac{\sqrt{2 + \sqrt{m}}}2$ is true. If there is any confusion, let me know.

Title: What will happen to $\theta$ when $\sin \theta =\frac{\sqrt {{2{\pm}\sqrt i}}}{2}$ and i >4?

Body:
Some follow up questions  related to Two mysterious missing angles in the sine values of acute angle list?
.After finding two missing angles,   we have a list of 9 sine values for special angles between 0$^\circ$ to $90^\circ$: $\sin \theta =\frac{\sqrt {{2{\pm}\sqrt i}}}{2}, i=0,1,2,3,4 $. (A side note:This formula is easy to remember: start with
$\sin 45^\circ=$$\frac{\sqrt2}{2}$, let $\pm \sqrt i$ jump under the radical sign and follow  behind 2. Thanks @some guy to remind this trick!)

Now, I have some curious thoughts:

Why is the formula symmetric about $\theta = 45^\circ$?

One would naturally want to continue this list with i beyond 4,..  in which case |$\sin \theta$| will not be bounded by [-1,1]. What does this mean? What are those angles? I expect we step into complex numbers, but do not have a clear picture. Can you please help?

\begin{align}\sin 0^\circ =\frac{\sqrt {\color{green}{2-\sqrt {4}}}}{2},\sin 15^\circ =\frac{\sqrt {\color{green}{2-\sqrt3}}}{2},\sin 22.5^\circ =\frac{\sqrt {\color{green}{2-\sqrt2}}}{2} \end{align}

\begin{align}\sin 30^\circ =\frac{\sqrt {\color{green}{2-\sqrt{1}}}}{2}
, \sin 45^\circ =\frac{\sqrt {\color{green}{2-\sqrt{0}}}}{2}
, \sin 60^\circ =\frac{\sqrt {\color{green}{2+\sqrt{1}}}}{2}\end{align}

\begin{align}\sin 67.5^\circ =\frac{\sqrt {\color{green}{2+\sqrt2}}}{2},\sin 75^\circ =\frac{\sqrt {\color{green}{2+\sqrt 3}}}{2}, \sin 90^\circ =\frac{\sqrt {\color{green}{2+\sqrt{4}}}}{2}\end{align}


问:
Find a value of $\;\lim\limits_{n\rightarrow\infty}\frac{a_{n}}{n}$

(If you're good at it, you can do it by heart.)

Let $a_n$ be the average value of lengths of all the diagonal lines of regular n-gon whose side is 1.

Find a value of
$$\lim_{n\rightarrow\infty}\frac{a_{n}}{n}$$
Source: Fujino_Yusui

After messing around with the cosine theorem and doing some calculations, I found the exact formula of $a_{n},$ so I did the rest of the calculations and the answer is
$$\lim_{n\rightarrow\infty}\frac{a_{n}}{n}= \lim_{n\rightarrow\infty}\frac{2\cos\frac{\pi}{n}- 1}{\left ( \left ( 1- \cos\frac{\pi}{n} \right )n \right )\left ( n- 3 \right )}= \frac{2}{\pi^{2}}$$
But I want to see a shortcut that leads $a_{n}\sim\frac{n}{2\pi}\cdot\frac{4}{\pi}\,{\rm as}\,n\rightarrow\infty$ without cosine theorem, I need to the help.

答:
Inscribe the n-gon in a circle. When calculating the length of a diagonal, instead of using the cosine theorem, you can bisect the central angle subtended on the diagonal. The bisector, the radia leading to the end points of the diagonal, and the diagonal will create two right triangles, from which you can get that the length of the diagonal is
$$ L = 2R \sin\frac{\alpha}{2}$$
where $R$ is the radius of the circle cicrumscribed on the polygon and $\alpha$ is the central angle subtended on the diagonal. You can also use this formula to calculate the length of a side of the n-gon:
$$ a = 2R \sin\frac{\pi}{n}$$
Since $a=1$, we get
$$ R = \frac{1}{2 \sin\frac{\pi}{n}} $$
$$ L = \frac{\sin\frac{\alpha}{2}}{\sin\frac{\pi}{n}}$$
$$ a_n = \frac{1}{n \sin\frac{\pi}{n}} \sum_{k=2}^{n-2} \sin\frac{\pi k}{n}$$
\begin{align} \lim_{n\to\infty} \frac{a_n}{n} &= \lim_{n\to\infty} \frac{1}{n^2 \sin\frac{\pi}{n}} \sum_{k=2}^{n-2} \sin\frac{\pi k}{n} = \\
&= \lim_{n\to\infty} \frac{1}{n \sin\frac{\pi}{n}} \cdot \lim_{n\to\infty} \frac{1}{n} \sum_{k=2}^{n-2} \sin\frac{\pi k}{n} = \\
&= \frac{1}{\pi} \cdot \int_0^1 \sin(\pi x) dx = \\
&= \frac{1}{\pi} \cdot \frac{-1}{\pi}\cos(\pi x)\big|_{x=0}^{x=1} =\\
&= \frac{1}{\pi}\cdot\frac{2}{\pi} = \frac{2}{\pi^2} \end{align}


问:
Is it true that any set which contains elements with contradictory properties is the empty set?
Here's what I've "discovered" but can't verify.

Let's take a set $A$ defined as "the set of all $x$ which don't belong to $A$".

Obviously this definition wouldn't hold in particular axiomatic systems, but it would in others.

Let's analyze this in NST. $X\in A$ iff $X\not\in A$.

This is a logical contradiction, but it still doesn't prove that $A$ doesn't exist. In fact, I would argue that $A$ is the empty set.

If put all of the elements of $A$ into the empty set, we get $(X∈A \text { iff } X∉A) \text { iff } X\in \emptyset$.

This is a tautology for whatever $X$ and $A$. Now, if we switch two of the atomic sentences, we get $X\not\in A \text { iff } (X\in A \text { iff } X\in\emptyset$). We know this to also always be true, which means that either $A$ doesn't exist or $A$ is the empty set.

Therefore, the empty set does contain every element belonging to any set it doesn't belong to.

Is there some error in my proof?

答:
Suppose $A$ is a set such that $x\in A$ if and only if $x\notin A$.

Now it's certainly true that from $x\in A$ we can deduce $x\in\varnothing$.

Also, vacuously, from $x\in\varnothing$ we can deduce $x\in A$.

This does not mean $A$ exists and is $\varnothing$. All we have deduced is that if $A$ exists then it must be $\varnothing$. But going back and substituting $A=\varnothing$ doesn't work: taking $x=\varnothing$ too we have $\varnothing\notin\varnothing$, which should imply $\varnothing\in\varnothing$, but the latter is false. The only explanation is that our original assumption - that such an $A$ exists - was wrong.

This is analogous to the following argument.
Suppose there is a real number $x$ such that $x^2+x+1=0$. Multiplying both sides by $x-1$, we get $x^3-1=0$, i.e. $x^3=1$, and the only real number satisfying this is $x=1$.

What we have just done is to deduce that if such a real number exists then it must be $1$ - but it doesn't, so it isn't.


问:
Show that an uncountable set can be separated into two disjoint uncountable subsets
This is a question from Steven Abbott's Understanding Analysis and I am certainly aware of how many duplicates and duplicates there are on this website as well as the internet. However, of all the solutions, there is one essential part that still troubles me. For sake of completeness, I would show

The full question

Common solutions

My question

Let A be an uncountable set and let B be the set of real number that divides A into two uncountable set; that is, $s\in$ B if both $\{x:x\in A\text{ and } x<s\}$ and $\{x:x\in A\text{ and } x>s\}$ are uncountable. Show B is nonempty and open.

The general approach to this question is

Prove there exists a bounded uncountable subset

Prove there exists a countable subset in this form $(-\infty,s)\cap A$

Prove B is nonempty

Prove B is open

Sorry for being so brief since I am stuck in point 1 and 2.Here's my confusion. Why does the following makes sense?

Since A is a bounded uncountable set, there must exists s such that $(-\infty,s)\cap A$ is uncountable.

Of course, if it is bounded and there doesn't exist a real number, including the supremum of this set, that make the intersection uncountable, then the entire set will be countable, contradicts A is uncountable.

My question, however, is, why bounded? What is the significance behind it?

答:
It isn’t actually necessary to reduce the problem to the case in which $A$ is bounded. We can instead argue as follows.

Let

$$L=\{x\in\Bbb R:(\leftarrow,x)\cap A\text{ is countable}\}$$

and

$$R=\{x\in\Bbb R:(x,\to)\cap A\text{ is countable}\}\,.$$

If $L\ne\varnothing\ne R$, it is straightforward to show that $B=(\sup L,\inf R)\ne\varnothing$; this is the same argument as in the bounded case.

If $L=\varnothing=R$, then clearly $B=\Bbb R$.

If $L=\varnothing\ne R$, then $B=(\leftarrow,\inf R)$.

If $L\ne\varnothing=R$, then $B=(\sup L,\to)$.


问:
Discriminant of a non separable extension
Why is the discriminant of a non separable finite extension of a field $0$? I proved for the case when the extension is primitive but can't prove it for the general case. Can trying to use induction help?

答:
Let $L/K$ be non-separable. By Stacks 030K, the set of elements in $L$ that are separable over $K$ form an intermediate extension $L^s/K$ that is separable, and $L/L^s$ is purely inseparable (Stacks 09HE).

It then suffices to address the case where $L/K$ is purely inseparable (by letting $K := L^s$), by this formula about discriminants and towers of fields.

By Stacks 09HI, one can factorize $L/K$ into a tower $L = E_n / \cdots / E_2 / E_1 / E_0 = K$ where each $E_{i+1}/E_i$ is degree $p$ and purely inseparable and is formed by adjoining the $p$th root of some element in $E_i$. By the same formula quoted above, it suffices to address the case where $L/K$ is one of these steps in the tower.

So let $L = K(a^{1/p})$ for some $a \in K \setminus K^p$. The minimal polynomial of $a^{1/p}$ over $K$ is $f(x) = x^p - a$, so the discriminant of $L/K$ is $f'(a^{1/p}) = p \cdot a^{(p-1)/p} = {\fbox{0}}$, as required. Note that this last step depends on your definition of discriminant.


问:
Show that two $R$-algebras are not isomorphic.

Consider the $\mathbb{R}$-algebras $A := \mathbb{R}\left[a,b,c,d\right]\left[\frac{1}{a^2+b^2+c^2+d^2}\right]$ and $B := \mathbb{R}\left[a,b,c,d\right]\left[\frac{1}{ad-bc}\right]$ (so we are adjoining the inverses of $a^2 + b^2 + c^2 + d^2$ and $ad-bc$, respectively).
I claim that these are not isomorphic as $\mathbb{R}$-algebras, but I'm not sure how to formalize this.

My intuition says that this is because $$a^2 + b^2 + c^2 + d^2 = 0 \iff a=b=c=d=0,$$ while $ad-bc=0$ has nonzero solutions. I tried to formalize this using evaluation homomorphisms, but to no avail. I also tried assuming they were isomorphic and obtaining a contradiction, but haven't been able to.

Does anyone have any pointers on how to approach this?

答:
Proposition. Let $R$ be a UFD and $p,q\in R$ prime elements. Let $\phi:R_p\to R_q$ be a ring isomorphism. Then $\phi(p)=uq$ or $\phi(p)=uq^{-1}$ with $u\in R$ invertible.

Proof. Since $p$ is invertible in $R_p$ its image by $\phi$ is also invertible, that is, $\phi(p)=uq^s$ with $u\in R$ invertible and $s\in\mathbb Z$. Conversely, $\phi^{-1}(q)=vp^t$ with $v\in R$ invertible and $t\in\mathbb Z$. Then $q=\phi(v)u^tq^{st}$ and since $q$ is prime we get $st=1$, so $s=\pm1$.

Now let $R=\mathbb R[a,b,c,d]$, $p=a^2+b^2+c^2+d^2$ and $q=ad-bc$. Suppose that $\phi(p)=uq$ with $u\in\mathbb R\setminus\{0\}$. Then $\phi(a)^2+\phi(b)^2+\phi(c)^2+\phi(d)^2=uq$. We can write $\phi(a)=a_1/q^i$ and so on. Then we get $a_1^2+b_1^2+c_1^2+d_1^2=uq^{2m+1}$ with $m$ a nonnegative integer. Now define a monomial order on $\mathbb R[a,b,c,d]$ such that $a>b>c>d$. Then the leading term of the right hand side is $(ad)^{2m+1}$, while the leading term of the left hand side is a sum at most four squares from $\mathbb R$ times the square of a monomial. Since that coefficient can not be zero (as it could happen if one works over $\mathbb C$, for instance!), we get a contradiction.


问:
Equation of 2nd order kernel of arc-cosine kernel
In the paper "Kernel method for deep learning" (Cho and Saul, NIPS 2009), a part $J_n(\theta)$ of definition of the arc-cosine kernel is given as
$$J_n(\theta) = (-1)^n (\sin \theta)^{2n+1} \left( \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \right)^n \left( \frac{\pi - \theta}{\sin \theta} \right).$$

For $n=2$, eq.(7) in the paper shows
$$J_2(\theta) = 3 \sin \theta \cos \theta + (\pi - \theta) (1 + 2 \cos^2 \theta).$$

How to derive this equation?

In my calculation, $J_2(\theta)$ becomes $J_2(\theta) = 2 \sin \theta \cos \theta + (\pi - \theta) (1 + \cos^2 \theta)$.

答:
Most likely you made a computational mistake somewhere.

\begin{align}
J_2(\theta) &= (-1)^2 (\sin\theta)^5 \left( \frac{1}{\sin \theta}\frac{\partial}{\partial \theta}\right)^2\left( \frac{\pi - \theta}{\sin \theta}\right)\\
&= (\sin^5 \theta) \left( \frac{1}{\sin \theta}\frac{\partial}{\partial \theta}\right)\frac1{\sin \theta} \frac{-\sin \theta -[\cos \theta] (\pi - \theta) }{\sin^2 \theta} \\
&= -\sin^4 \theta \frac{\partial }{\partial \theta } \left( \frac{\sin \theta + [\cos \theta] (\pi - \theta)}{\sin^3 \theta}\right)\\
&= - \sin^4 \theta \left(\frac{\sin^3 \theta (\cos \theta -\cos \theta -[\sin \theta] (\pi - \theta)) - (\sin \theta + [\cos \theta] (\pi - \theta))(3\sin^2 \theta \cos \theta)}{\sin^6 \theta} \right)\\
&= -\frac1{\sin^2 \theta} \left([ -\sin^4 \theta] (\pi-\theta)  - (\sin \theta + [\cos \theta] (\pi - \theta))(3\sin^2 \theta \cos \theta)  \right)\\
&=  [\sin^2 \theta] (\pi - \theta)  + (\sin \theta + [\cos \theta](\pi-\theta))3\cos \theta)\\
&= 3 \sin \theta \cos \theta  +  (2\cos ^2 \theta + 1)(\pi - \theta)
\end{align}


问:
topology generated by a metric
What is the procedure to prove that a particular given metric generates a particular given topology?

Actually, I am given with a topology $\rho$ and a distance expression $d$ in my homework. Question is

Show that the topology $\rho$ is metrizable by showing that the distance $d$ defines a metric on $\mathcal T$ that generates the topology $\rho$.

I proved that the given expression $d$ is metric on space. But how to proceed next? How to show that the metric generates the topology?

答:
The topology given by the metric is generated by the balls
$$
B_r(x)=\{y:\ d(x,y)<r\}.
$$
This means that every open set is a union of balls. So you have to show that every ball $B_r(x)$ is in $\rho$ , and that for every $V\in\rho$ and every $x\in V$, there exists $r>0$ such that $B_r(x)\subset V$.

Title: topology generated by a metric


问:
Why $t-7$ and $4$ are not coprime in $\mathbb Z[t]$?
I have to compute $(t-7,4)/(t-7)$ in $\mathbb Z[t]$. What I did is : since $t-7$ and $4$ are coprime, $(t-7,4)=\mathbb Z[t]$. Therefore, $$(t-7,4)/(t-7)\cong \mathbb Z[t]/(t-7)\cong \mathbb Z,$$
since $t-7$ is irreducible in $\mathbb Z[t]$. The Teacher assistant gave me a mark of $0$ out of $10$ and said that $(t-7,4)/(t-7)\cong 4\mathbb Z$. So, why $t-7$ and $4$ are not coprime, i.e. what would be the gcd ? $\color{red}{(solved)}$ And how to prove that $(t-7,4)/(t-7)\cong 4\mathbb Z$ ? this doesn't natural for me.

答: Your problem is that you've confused two notions: a GCD domain and a principal ideal domain. The GCD of two elements $a,b\in R$ with $R$ a ring may be defined as the smallest principal ideal $(c)$ containing $(a,b)$. In your case, the GCD of $4$ and $t-7$ is indeed $(1)$. But $\mathbb Z[t]$ is not a principal ideal domain: not every ideal is principal. Thus even though $(1)$ is the smallest principal ideal containing $(4,t-7)$, we cannot conclude that $(1)=(4,t-7)$, as indeed it does not.


问:
Give an $\delta-\epsilon$ proof that the function $f:\mathbb{R}^3\to\mathbb{R}$ defined by $f(x,y,z)=x^2y+2xz^2$ is continuous at $(1,1,1)$

Give an $\delta-\epsilon$ proof that the function $f:\mathbb{R}^3\to\mathbb{R}$ definded by $$f(x,y,z)=x^2y+2xz^2$$ is continuous at $(1,1,1)$.

Let $\epsilon<0$ then,

$$
\begin{equation}
\begin{split}
|x^2y+2xz^2-3| & = |(x^2y-y)+(y-1)+(2xz^2-2x)+(2x-2)| \\
 & \leq |x^2y-y|+|y-1|+|2xz^2-2x|+|2x-2| \\
& = |y||x+1||x-1|+|y-1|+2|x||z+1||z-1|+2|x-1|
\end{split}
\end{equation}
$$

For single variable I know how to handle $|x+1||x-1|$ $(\text{Using }|x+1||x-1|<\epsilon \implies\delta=\min\{1,\frac\epsilon3\})$, but for multivariable how to avoid any dependencies between variables?

It will be great help If anyone provide a legit way/hint/idea to deal this kind of problem(more general will be appreciated).

答:
Let shift the limit point in $(0,0,0)$ (i.e. $x=1+u$ with $u\to 0$ and so on).

$\begin{align}|f(x,y,z)-f(1,1,1)|
&=|f(1+u,1+v,1+w)-3|\\
&=|v+4u+2uv+u^2+u^2v+4w+2w^2+4uw+2uw^2|\\
&\le|v|+4|u|+2|u||v|+|u|^2+|u|^2|v|+4|w|+2|w|^2+4|u||w|+2|u||w|^2\end{align}$

In $\mathbb R^3$ all norms are equivalent, in the present case $||(x,y,z)||_\infty=\max(|x|,|y|,|z|)$ is by far the most convenient to use.

Notice that $||(u,v,w)||_\infty<\delta$ means that:

by a choice of $\delta=1$ we have $|u|<1\implies |u|^n<|u|<1$

by a choice of $\delta=\epsilon$ we have $|u|<\epsilon$

Same for the other letters, therefore for $\delta=\min(1,\epsilon)$ then any block $$|u|^m|v|^n|w|^p\le\epsilon\times 1\times 1\le \epsilon$$

We can continue like below:

$$|f(x,y,z)-f(1,1,1)|\le \epsilon+4\epsilon+2\epsilon+\epsilon+\epsilon+4\epsilon+2\epsilon+4\epsilon+2\epsilon\le 21\epsilon$$

And we are done.

Ultimately you can select $\delta=\min(1,\frac{\epsilon}{21})$ to get to a bare epsilon in the end, but not really required.

The dependencies between variables is hidden mostly in the choice of a suitable norm, but also in ignoring terms of higher degree (instead of having $\epsilon^2,\epsilon^3$ you just have $<\epsilon$ by bounding the other variables by $1$).

This is why the schema $\delta=\min(1,\epsilon)$ is so frequently used.


问:
Lathe Bed Differential Wear Tool Offset Calculation
This is probably an odd one here,

I'm trying to calculate the linear offset that a tool needs to have to meet a circle of any radius r with an arbitrary offset y and arbitrary angle ϴ. The tool holder "height" from the bed surface is fixed and can be taken as a constant, but is also arbitrary and will vary lathe to lathe. I am trying to "map" this offset on my own, but would like to have a general formula to calculate the offset programmatically from measured values to incorporate into a LinuxCNC component so other users with this issue can measure and compensate their own machines.

I've linked an example with a green circle showing the distance between the circle and "tool" I'm trying to calculate, note this distance should be along theta and not along the origin axis. All of the dimensions are exaggerated for effect, in reality the difference will likely be within a few thousandths of an inch. Note that I have assumed that as the bed wears the left edge remains in the same plane. This is the edge of the lathe carriage that is "guided," the other is free to move laterally.

I've racked my brain and just can't come up with anything to describe how this gap interacts with the variables. CAD has been no help, either. I'd really appreciate it if someone could help me out with this.

答:
If the lathe tool passes through the center of the axis, no compensation is necessary, since the coordinate system is just rotated around the axis by $\theta$.

Consider the following diagram, with differential wear $y$, causing the tool carriage to tilt/turn by angle $\theta$,
$$\tan\theta = \frac{y}{L} \tag{1}\label{G1}$$

The vertical error $v$ at the tool tip is
$$v = \frac{L - w_5}{L} y = \frac{w_1 + w_2 - w_5}{w_1 + w_2} y \tag{2a}\label{G2a}$$
and if the axis is centered between the rails ("pivot points" of the rails), then the tool height error $\epsilon$ (i.e., how much below the center of the axis the tool height is, when it is directly below the center of the axis), is
$$\epsilon = \frac{w_1}{L} y = \frac{w_1}{w_1 + w_2} y = \frac{y}{2} \quad \text{if} \quad w_1 = w_2 \tag{3a}\label{G3a}$$

Let's use $\beta$ for the "balance factor",
$$\beta = \frac{w_1}{w_1 + w_2} \tag{4}\label{G4}$$
If the axis is exactly midway between the rails, then $\beta = 0.5$.  If the axis is closer to the tool edge, $\beta \gt 0.5$; if the axis was right over the tool edge pivot point, then $\beta = 1$.  If the axis is closer to the other edge, $\beta \lt 0.5$; if the axis was right over the other edge pivot point, then $\beta = 0$.  As far as I know, $\beta = 0.5$ for most lathes.

Then, the tool height error $\epsilon$ is
$$\epsilon = \beta y \quad \iff \quad y = \frac{\epsilon}{\beta} \tag{3b}\label{G3b}$$
and the vertical error at the tool $v$ is
$$v = \left(\frac{L - w_5}{L}\right)\left(\frac{\epsilon}{\beta}\right) = \left(1 - \frac{w_5}{L} \right)\left(\frac{\epsilon}{\beta}\right) \tag{2b}\label{G2b}$$

We really only need to know the tool height error $\epsilon$ to compensate for the misalignment, because the rotation/tilt angle $\theta$ does not affect accuracy at all.

Let's say $d$ is the distance from the tool to the center of the axis, the tool height error is $\epsilon$, and $x$ is the distance from the center of the axis measured along the tool carriage.  Then, because $\epsilon$ and $x$ are perpendicular, Pythagoras tells us
$$d^2 = x^2 + \epsilon^2 \tag{5a}\label{G5a}$$
i.e.
$$x = \sqrt{d^2 - \epsilon^2} \tag{5b}\label{G5b}$$
or equivalently,
$$d = \sqrt{x^2 + \epsilon^2} \tag{5c}\label{G5c}$$

Here, $d$ is the real world distance between the tool tip and the center of the axis, and $x$ is the DRO measurement, with $x = 0$ exactly below the center of the axis and $\epsilon$ the distance between the tool tip and the center of the axis at that point.


问:
Probability distributions named after Greek letters
There are some probability distributions named after Greek letters, such as Gamma, Beta, Chi, and Zeta distributions.

I wonder if there are some rules of why and how they are named after Greek letters?

Can we make use of the rules (if exist) to help us memorize these distributions?

Are there other such distributions?

Thanks and regards!

答:
I think that there are no such rules; there is only history.

Whether to generalize rules that don't exist is a very good question.  I think sometimes that happens.  In other words, no rule brought this about, but maybe by coincidence they fit some simple pattern that no one's noticed, and once we do notice, we can extrapolate, and that extrapolation may be valuable.  But simplicity is essential here.  Obviously there's such a rule if we don't require simplicity, namely: the naming conventions are precisely that we call this one the Beta distribution, and this one the Gamma distribution, etc.  And it's easy to extrapolate then!


问:
Find all positive $x,y\in\Bbb{Q}$ such that $x+\frac{1}{y}$ and $y+\frac{1}{x}$ are natural numbers.
What I did is, the product is also a natural number.

$xy+\frac{1}{xy}=m$ is also a natural number. Assuming $xy=z$, we get the equation $z^2-mz+1=0$

Hence, $z=\frac{m\pm \sqrt{m^2-4}}{2}$. Since z is rational $\sqrt{m^2-4}$ has to be rational. It is rational only when $m=2$ (not sure). So $z=xy=1$. So, $y=\frac{1}{x}$

Hence $2x$ and $2y$ are natural numbers.

I can't proceed further. Is my way correct? Any help please.

答:
Hint If $\sqrt{m^2-4}=\frac{p}{q}$ is rational with $\mbox{gcd}(p,q)=1$, then $\frac{p^2}{q^2} \in \mathbb Z$. Use $\mbox{gcd}(p,q)=1$ to conclude that $q=1$.

This gives that $\sqrt{m^2-4}=n$ for some $n \in \mathbb N$ and hence
$$
4=m^2-n^2=(m-n)(m+n)
$$
Look at all possible factorizations to conclude that $m=2$.

To complete the problem, observe that $2x=n, 2y=k$ and $xy=1$ gives
$$
kn=4
$$
There are only 3 possibilities (or two by symmetry), just check them.


问:
How can I solve $(y+2)^2 y' + (x + 2y + 3)^2 = 0 $?
I'm stuck trying to solve this ODE and I wonder if someone can help, thanks in advance. The ODE is the following:

$$(y+2)^2 y' + (x + 2y + 3)^2 = 0 $$

I first tried expanding the squared terms and trying to separate the variables but it was not very helpful. Then I noticed that the equation is homogenious but trying to make a substitution got me nowhere. Thanks for your help.

答: As given, it is not homogeneous, but you can make a substitution to make it homogeneous: $Y = y+2$, $X = x - 1$


答:
Hint :

Substitute $y=Y+k$, $x=X+h$ to do away with the constant terms. Then this will reduce to homogeneous form when the substitution $Y=VX\Rightarrow Y'=V+XV'$ works.

On first substituion, you will find $k=-2, h=1$.

Title: How can I solve $(y+2)^2 y' + (x + 2y + 3)^2 = 0 $?

Body:
I'm stuck trying to solve this ODE and I wonder if someone can help, thanks in advance. The ODE is the following:

$$(y+2)^2 y' + (x + 2y + 3)^2 = 0 $$

I first tried expanding the squared terms and trying to separate the variables but it was not very helpful. Then I noticed that the equation is homogenious but trying to make a substitution got me nowhere. Thanks for your help.


问:
Roots of minimal and characteristic polynomials
Why is it that for matrix $A \in M_n(\mathbb{C})$ the characteristic polynomial $\chi_A(t)$ and the minimal polynomial $\mu_A(t)$ have the same roots?

Since $\chi_A(t) = \mu_A(t) \cdot p(t)$ it should be easy to follow, that $\chi_A(t)$ has roots where $\mu_A(t)$ has roots.

But why can't $\chi_A(t)$ have roots where $\mu_A(t)$ hasn't?

答:
I remember of something like this : if $\mu_A(t)=\prod_{\lambda \in Sp(A)}(t-\lambda)$, then $\chi_A(t)=\prod_{\lambda \in Sp(A)}(t-\lambda)^{\nu_\lambda}$. The two polynomials have the same roots, the eigenvalues of $A$, and the order of multiplicity of $\lambda$ is $1$ in the minimal polynomial and $\nu_\lambda$ in the characteristic polynomial.

This is proved by Dennis Gulko's answer.

Further, the order of multiplicity $\nu_\lambda$ is the the dimension of the eigenspace $\nu_\lambda = \dim E_\lambda=\dim \ker (\lambda I - A)=\dim \{x|Ax=\lambda x\}$.

This is shown by looking at the restriction of $A$ to $E_\lambda$, where it acts as the homothety $x \rightarrow \lambda x$ and its matrix is the square diagonal matrix $\left( \matrix{\lambda&0&...&0 \\ 0&\lambda&...&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&...&\lambda} \right)$.


问:
The number of word groups possible by taking at least $4$ letters from $3$ words

The number of word groups by taking at least $4$ letters of each words length, stroke and number are:

$$\text{(A)}\ 221\quad\text{(B)}\ 1068\quad\text{(C)}\ 22\quad\text{(D)}\ 66$$

My Attempt:

Selecting 4 letters from each words: $^6C_4 \cdot ^6C_4 \cdot ^6C_4 = 15^3 = 3375 $

Selecting 5 letters from each words: $^6C_5 \cdot ^6C_5 \cdot ^6C_5 = 6^3 = 216 $

Selecting all 6 letters from each words: $^6C_6 \cdot ^6C_6 \cdot ^6C_6 = 1 $

Thus total words group are $33375 + 216 + 1 = \boxed{3592}$

I don't know what is wrong. Is there any mistake in understanding the question?

答:
Guessing the intent a little... I'm assuming we have to pick a number of different letters that are in the given words and fulfill the criterion of having at least four letters in common with each source word. Your answer, which I would say is also a valid interpretation, doesn't account for the potential of selecting different numbers of letters from different words (so your answer should be even higher).

I think the point might be to exploit the fact that each word has three unique letters, two that are respectively shared with only one of the other two words, and one letter shared by all three source words:

To fulfill the criterion, we can break down choices relative to what we pick from the shared letters ENTR:

All $4$, ENTR: we must choose at least one more from each word, and up to all three. So $(2^3-1)^3 = 7^3 = \fbox{343}$ options for this (meaning the only valid answer choice left would be $1068$, but we'll see).

E plus two of NTR: So we'll have at least one extra letter from one word and at least two from each of the others, so $\binom 32\left(7\cdot 4^2\right) = 3\cdot 112  = \fbox{336}$ options.

E plus one of NTR: Now we need at least two extra letters from two words and all three unique letters from the third so $\binom 31\left(4^2\cdot 1\right) = 3\cdot 16 = \fbox{48}$ options.

E only, none of NTR: This needs all three unique letters from all source words, $\fbox{1}$ option only.

for a total of $343+336+48+1 = \fbox{728}$, not an available option...

So, change of plan, let's limit it to getting exactly four letters from each word, in which case:

All $4$, ENTR:  one extra letter per word, $3^3 = \fbox{27}$ options here.

E plus two of NTR: So exactly one extra letter from one word and two from each of the others, so $\binom 32\left(\binom 31{\binom 32}^2\right) = 3^4 = \fbox{81}$ options.

E plus one of NTR: two extra letters from two words, all three from the third so $\binom 31\left({\binom 32}^2\binom 33\right) = 3^3 = \fbox{27}$ options.

E only, none of NTR: again $\fbox{1}$ option only.

Then this total - for choosing a set of letters that has exactly four letters in common with each source word - is also different to any offered, $27+81+27+1 = \fbox{136}$.

So in conclusion: Interesting, but I still don't know what the original answers related to.


问:
Maximum cliques of the transitive closure of a chordal DAG
Let $G=(V,A)$ be a directed acyclic graph, for which the underlying
undirected graph is chordal (so that every induced cycle in the
underlying undirected graph is a triangle).

It is known that in a chordal graph the number of maximum cliques
is linearly bounded in the number $|V|$ of vertices.

Let $G'=(V,A')$ be the transitive closure of $G$, so that for
every directed path $p= (v_i,\ldots,v_j)$ from $v_i$ to $v_j$
in $G$, there exists a directed edge $(v_i,v_j)$ in $A'$.

Question: Is there any polynomial/linear bound on the number
of maximum cliques in the undirected graph that underlies $G'$?

答:
Consider the following graph, with edges oriented from left to right :

This graph is chordal.

The transitive closure of this graph is the complete graph without every vertical edges.

A clique in the transitive closure is any subset of the vertices such that there isn't two vertices with the same $x$ coordinate.

A maximal clique is thus a maximal set of such vertices, so one for each different $x$ coordinate.

There is an exponential number of those : $2^{\frac {n-1} 3}$.

This construction isn't optimal, but I guess this lower bound gives the idea.