答:
we can write
$$f(x)=\frac{1}{\sqrt{x}} \cos\frac{1}{x}$$
Take $$x_n=\frac{1}{2n\pi}, x'_n= \frac{1}{(2n+1)\pi}$$
$f(x_n)=(2\pi n)^{1/2}\cos (2\pi n)>0$ and
$f(x'_n)=((2n+1)\pi)^{1/2} \cos((2n+1)\pi)<0$, these two values are essentially unequal. the $\lim_{x \to 0} f(x)$ does not exist.

Title: How to prove $ \lim_{x\rightarrow 0^+} \frac{1}{\sqrt{x}} \cos(1/x) $ does not exist

Body:
How do I prove that this limit does not exist?

$$
\lim_{x\rightarrow 0^+} \frac{1}{\sqrt{x}} \cos(1/x)
$$

When plugging in $0^+$ in $x$ we get $-\infty \times \cos(\infty)$ which is undefined. however I am trying to prove that it doesn't exist and am not capable? Do I use the squeeze rule or toast rule? I have tried a few different ways and haven't managed anything


问:
Why is ess sup $f$ not ess max $f$?
Consider a measure space $(X,\Sigma\,\mu)$. Given that one can easily prove that, $\mu$-a.e., $f \leq \text{ess} \sup_X f$, why is the notation not simply "$ \text{ess} \max_X f$"?

(Here $\text{ess} \sup_X f := \inf \{\xi \in \overline{\mathbb{R}}~|~f \leq \xi \text{ a.e.} \}$).

答: Usually "max" refers to a supremum that is attained, i.e., that is actually a value of the function.  Since the essential supremum need not be attained (let alone attained "essentially", which I suppose would mean attained on a set of positive measure), it's clearer to say "essential supremum".</p>

ContentLicense: CC BY-SA 3.0

Title: Why is ess sup $f$ not ess max $f$?


问:
Determinant of a Vandermonde-like matrix
Consider the determinant  $$ \Delta_n(k_1, x_1; ...; k_m, x_m) = \begin{pmatrix}
  M_{k_1}^n(x_1) \\
  M_{k_2}^n(x_2)\\
\vdots \\
  M_{k_m}^n(x_n)
\end{pmatrix}.
$$
where $ x_1 \cdots x_m$ are any variables; $ k_1 \cdots k_m$ are natural numbers such that $ k_1 + k_2 + \cdots + k_m = n$; $M_k^n(x)$ is a $k \times n $ matrix of the following form
$$ M_k^n(x) = \begin{pmatrix}
  1 & x & x^2 & \cdots & x^{n-1}\\
  0 & 1 & \binom{2}{1}x & \cdots & \binom{n-1}{1}x^{n-2}\\
  0 & 0 & 1 & \cdots & \binom{n-1}{2}x^{n-3}\\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
  0 & 0 & 0 & \cdots & \binom{n-1}{k - 1}x^{n-k}&
\end{pmatrix}.
$$
The problem is to prove that the determinant can be computed as $$
\Delta_n(k_1, x_1; ...; k_m, x_m) = \prod_{1 \le j < i \le m}(x_i-x_j)^{k_ik_j} $$
(Hint: when $k_1 = \cdots = k_m = 1, $ i.e. $m = n$, it's a Vandermonde determinant )

I tried to reduce this to Vandermonde determinant as was menthioned in hint, but with no success. Also it looks like each row of the $M_k^n(x)$ matrix is a linear combination of all the succesive rows, but I do not know how to use it.

答:
[Revised and completed.]

Let me slightly change your notation, and replace your $x_i$ by $X_i$.

Now start with the usual $n\times n$ Vandermonde determinant in the variables $x_1,x_2,\dots, x_n$.

Your determinant is obtained from this determinant by operating on it in the following way:

(step 1) differentiate w.r.t $x_2$ once, $x_3$ twice, ..., $x_{k_1}$ $(k_1-1)$ times; then differentiate w.r.t $x_{k_1+2} $ once, $x_{k_1+3}$ twice, ..., $x_{k_1+k_2}$ $(k_2-1)$ times, and so on;

(step 2) substitute $x_1=x_2=\dots=x_{k_1}=X_1$, $x_{k_1+1}=x_{k_1+2}=\dots=x_{k_1+k_2}=X_2$ , and so on;

(step 3) divide by $\prod_{i=1}^{m}\prod_{r=1}^{k_i} r!$.

It remains to see what happens when we apply these steps to $\prod_{i<j}(x_i-x_j)$.

Firstly note that if we apply $t$ differentiations to this product we get a sum/difference of subproducts each of which "omits" $t$ of the factors. Secondly note that when we substitute for the variables in a block, the only subproducts which don't evaluate to $0$ are those where the variables come from different blocks.

So consider the first block: we perform $k_1 \choose 2$ differentiations, and the only summand which will survive the substitution is the unique one which omits the factor $\prod_{i<j\leqslant k_1} (x_i -x_j)$.

Arguing in the same way about each block of variables we see that the only summand of the derivative (after we have carried out all the differentiations) which can survive the substitutions is the subproduct of $\prod_{i<j}(x_i-x_j)$ which omits $\prod_{r=1}^{m}\prod_{k_{r-1}<i<j\leqslant k_r}(x_i -x_j)$.

When we substitute we will clearly get a product of the form $\prod_{r<s}(X_r-X_s)^{h(r,s)}$. To determine $h(r,s)$ note that these factors arise from the factors $(x_i-x_j)$, where $x_i$ is one of the $k_r$ variables in the $r$-th block and $x_j$ is one of the $k_s$ variables in the $s$-th block. So $h(r,s)=k_r k_s$ as required.

We have therefore proved that the determinant in question is of the form $C\prod_{r<s}(X_r-X_s)^{k_r k_s}$, where the constant $C$ depends on the $k_i$. It remains to determine this constant.

I would prefer to give a combinatorial explanation of why $C=1$, but for present such an explanation eludes me. Instead let us determine on both sides the coefficient of the "highest" monomial $X_1^{n_1}X_2^{n_2}\dots X_m^{n_m}$, where highest means maximising $n_m$, then $n_{m-1}$, and so on.

As far as $\prod_{r<s}(X_r-X_s)^{k_r k_s}$ is concerned, the coefficient is clearly some $\epsilon$ where $\epsilon=\pm1$.

As far as the determinant is concerned, we maximise the power of $X_m$ by choosing the final entry of the $(k_1+k_2+\dots+k_{m-1}+1)$-th row, the next to last entry of the $(k_1+k_2+\dots+k_{m-1}+2)$-th row and so on: the antidiagonal of the bottom right-hand $k_m\times k_m$ block of the determinant. So far the coefficient is $\pm 1$: it is not hard to check that the parity is the same as the parity of our previous $\epsilon$.

Now this power of $X_m$ is must be  multiplied by the highest monomial in the determinant of the complementary block: inductively this is $1$ and we are done.

ContentLicense: CC BY-SA 4.0


问:
$\dfrac{O(1)}{x} = O(\dfrac{1}{x})$?
I just saw a lecture where the professor did this simplification in a proof but did not comment it further. How would you go about provning the statement in the title?</p>

答: Can this be considered a proof: let $h(x) = \dfrac{O(1)}{x} \Leftrightarrow x\cdot h(x) = O(1)$ meaning specifically $x\cdot h(x) \le  K\cdot 1\Leftrightarrow h(x) \le K\cdot \dfrac{1}{x}$ for some constant $K$. Thus $h(x) = O(1/x)$.

Title: $\dfrac{O(1)}{x} = O(\dfrac{1}{x})$?


问:
Compute $\int_0^\infty x^{n+1} J_n(kx) \exp(-a^2 x^2) dx$
I need advice on how to solve the following integral:

$$\int_0^\infty x^{n+1} J_n(kx) \exp(-a^2 x^2) dx$$

答:
First, make life simpler : let $x=\frac t k$ and $b=\frac{a^2}{k^2}$ which make
$$I_n=\frac 1{k^{n+2}}\int_0^\infty t^{n+1}\,J_n(t)\,e^{-b\,t^2}\,dt$$ and, from the link given by @Gary
$$I_n=\frac {\exp\left(-\frac 1{4b}\right) }{k^{n+2}(2b)^{n+1}}=\frac {k ^n}{2^{n+1}\, a^{2 (n+1)} } \exp\left(-\frac{k^2}{4 a^2}\right)$$


问:
Challenging Integrals for High School Students
I am now in my last year of high school. We have covered all the techniques useful for indefinite integration that are included in our Maths and Further Maths courses. This includes:

Integration by parts, inspection, substitution, partial fraction decomposition

Integration of regular and inverse trigonometric, regular and inverse hyperbolic, exponential, logarithmic, polynomial functions

I would like to have some challenging integrals to attack that are possible for me to solve at my current level of knowledge. By challenging, I mean integrals similar to the ones in this document. They were generally enjoyable and very satisfying to solve. If you have an integral that you think I could do that is more challenging than those in the aforementioned link, so much the better.

Thank you for your suggestions.

答:
In this question, $a$ is a positive constant.

(i)        Express $\cosh a$ in terms of exponentials.

             By using partial fractions, prove that $$\int_0^1 \frac{1}{x^2 + 2x\cosh a + 1} \mathrm dx = \frac{a}{2 \sinh a} .$$

(ii)       Find, expressing your answers in terms of hyperbolic functions,
$$\int_0^\infty \frac{1}{x^2 + 2x \sinh a -1} \mathrm dx$$
and
$$\int_0^ \infty \frac{1}{x^4 +2x^2 \cosh a + 1} \mathrm dx.$$

Find the area of the region between the curve $y = \frac {\ln x}{x} $
and the $x$-axis, for $1 \leq x  \leq a$. What happens to this area as $a$ tends to infinity?

Find the volume of the solid obtained when the region between the curve $y = \frac{\ln x}{x}$ and the $x$-axis, for $1 \leq x \leq a$, is rotated through $2π$ radians about the $x$-axis. What happens to this volume as $a$ tends to infinity?

For $n = 1, 2, 3, ..., $ let

$$I_n = \int_0^1 \frac{t^{n-1}}{(t+1)^n} \mathrm dt. $$

By considering the greatest value taken by $ \frac{t}{t+1} $ for $0 \leq  t \leq 1$ show that $I_n+1 < \frac{1}{2}I_n .$

Show also that $I_{n+1} = -\frac{1}{n2^n-1}.$

Deduce that $I_n < \frac{1}{n2^n-1}$

Prove that
$$ \ln 2 = \sum_{r=1}^n \frac{1}{r 2^r} + I_{n+1}$$

and hence how that $\frac{2}{3} < \ln 2 < \frac{17}{24}.$

The above questions are sourced from this website. All credits go to them.
If you need any more questions, please just drop a comment.


问:
How to prove $ \lim_{x\rightarrow 0^+} \frac{1}{\sqrt{x}} \cos(1/x) $ does not exist
How do I prove that this limit does not exist?

$$
\lim_{x\rightarrow 0^+} \frac{1}{\sqrt{x}} \cos(1/x)
$$

When plugging in $0^+$ in $x$ we get $-\infty \times \cos(\infty)$ which is undefined. however I am trying to prove that it doesn't exist and am not capable? Do I use the squeeze rule or toast rule? I have tried a few different ways and haven't managed anything

答:
$$\lim_{x\to 0^+} \frac{\cos(x^{-1})}{\sqrt{x}}=\lim_{n\to + \infty} \frac{\cos((\frac{1}{n\pi})^{-1})}{\sqrt{\frac{1}{n\pi}}}=\sqrt{\pi} \lim_{n \to +\infty} (-1)^n \sqrt{n}$$
And that limit clearly does not exist, so initial limit doesn't exist.


问:
prove $x \mapsto x^2$ is continuous
I am to show the continuity of this function with the help of $\epsilon$-$\delta$ argument.

The function is: $g: \Bbb{R} \rightarrow \Bbb{R}$, $x \mapsto x^2$.

Given the $\epsilon$-$\delta$ definition of limit, I tried as follows:

We must have: $|f(x)-f(x_0)|<\epsilon$, so then $|x^2-x_0^2|=|(x-x_0)(x+x_0)|=|(x-x_0)||(x+x_0)|$, so $|x-x_0|<\frac{\epsilon}{|x+x_0|}$. So now I will choose $\delta=\frac{\epsilon}{|x+x_0|}$.

Is it enough? I am writing this sentences like a machine, but I am not understanding intuitively.

答:
Assume we found
$\delta$,

$|x|\leq |x-x_0+x_0|\leq|x-x_0|+|x_0|<\delta+|x_0|.$

If this $\delta$ works, then any $\delta'\leq \delta$ also works, therefore we may assume $\delta<1$ as well.
Now we have an estimate of the form $|x|<1+|x_0|$. One can see that we just restricted $x$ in the open interval $(x_0-1,x_0+1)$.

Now, we have

$$|f(x)-f(x_0)|=|x^2-x_0^2|\leq (1+2|x_0|)|x-x_0|$$.

One can observe that $\delta<\frac{\epsilon}{(1+2|x_0|)}.$

Since we also want to estimate first, namely $|x|<1+|\delta|$.
Therefore we choose $\delta<\text{min}\{1,\frac{\epsilon}{(1+2|x_0|)}\}.$

Coming to the formal proof;

Fix $x_0\in\Bbb{R}$, given  $\epsilon>0$. Choose $\delta<\text{min}\{1,\frac{\epsilon}{(1+2|x_0|)}\}.$

Let $x\in\Bbb{R}$ be such that $|x-x_0|<\delta$. Then $|x-x_0|<1, |x|=|x-x_0+x_0|\leq |x-x_0|+|x_0|<1+|x_0|$.

Now,$|f(x)-f(x_0)|=|x^2-x_0^2|=|(x-x_0)(x+x_0)|=|x-x_0||x+x_0|\leq |x-x_0|(1+2|x_0|)<(1+2|x_0|)\delta<\epsilon.$

Title: prove $x \mapsto x^2$ is continuous


问:
Vector form of a plane

Express the following plane in vector form:

$\mathcal P_1\subseteq  \Bbb R^3$ with equation $4x-z=0$.

The answer is $t(1,0,4)+s(0,1,0)$.
I don't understand how they got $(0,1,0)$ for the direction vector.

答:
The cartesian equation is $z=4x$

The parametrization is not unique. Anyway You need two linearly independent points on the plane.

One point can be $A=(1,0,4)$. The second one cannot be $(2,0,8)$ or other  points you can get from $A$ multiplying by a constant. Otherwise you get a line not a plane.

The second point cannot obviously be $(0,0,0$.

So we look for a point where the only nonzero coordinate is $y$, like $(0,1,0)$ or $(0,10,0)$

Finally the parametric equation of the plane is
$$(1,0,4)t+(0,1,0)s=0$$

Note that $(2,0,8)t+(0,10,0)s=0$ would be right.

Hope this helps

Title: Vector form of a plane


问:
System of two simple modular equations
$$x \equiv -7 \mod 13$$
$$x \equiv 39 \mod 15$$

I need to find the smallest x for which these equations can be solved. I've been always doing this using Chinese Reminder Theorem, but it seems that it doesn't work here, I'm not sure why though. Let's see:

$N = 13*15 = 195$

$N_{1} = \frac{195}{13} = 13$

$N_{2} = \frac{195}{15} = 15$

$GCD(13, 13) = 1*13 + 0*13$

$GCD(15,15) = 1*15 + 0*15$

$$x = 1 * (-7) * 13 + 1 * 39 * 15 = 480$$

The answer however is supposed to be $x = 84$...What's wrong?

答:
The way I usually go about these is to solve the two systems
$$
\mbox{$\begin{align}
x_{13}&\equiv1\pmod{13}\\
x_{13}&\equiv0\pmod{15}
\end{align}$}
\qquad\text{and}\qquad
\begin{align}
x_{15}&\equiv0\pmod{13}\\
x_{15}&\equiv1\pmod{15}
\end{align}
$$
Then combine these to get the solution to the equation given. These can both be solved at once using the Euclid-Wallis Algorithm:
$$
\begin{array}{r}
&&1&6&2\\\hline
1&0&1&-6&13\\
0&1&-1&7&-15\\
15&13&2&1&0\\
\end{array}
$$
This says that $(-6)15+(7)13=1$ which immediately gives $x_{13}\equiv(-6)15\pmod{195}$ and $x_{15}\equiv(7)13\pmod{195}$. The solution to the original question is
$$
\begin{align}
x
&\equiv-7x_{13}+39x_{15}\\
&\equiv84\pmod{195}
\end{align}
$$
Caveat: in the case of two equations as above, this only requires one application of the Euclidean algorithm. However, in the case of three or more equations, we need to apply the Euclidean algorithm once per modulus. For example, to solve
$$
\begin{align}
x&\equiv 1\pmod{3}\\
x&\equiv 2\pmod{5}\\
x&\equiv 4\pmod{7}
\end{align}
$$
we would need to solve each of
$$
\mbox{$\begin{align}
x_3&\equiv1\pmod{3}\\
x_3&\equiv0\pmod{35}
\end{align}$}
\quad\text{and}\quad
\mbox{$\begin{align}
x_5&\equiv1\pmod{5}\\
x_5&\equiv0\pmod{21}
\end{align}$}
\quad\text{and}\quad
\begin{align}
x_7&\equiv1\pmod{7}\\
x_7&\equiv0\pmod{15}
\end{align}
$$
using the Euclid-Wallis Algorithm.

Title: System of two simple modular equations


答:
Let $\pi\equiv 4x-z=0$ be your plane. We notice that the point $(0,0,0)\in\pi$ and the orthogonal vector to the plane is $\underline u=(4,0,-1)$.

The vector space $\text{dir}(\pi)$ has dimension $2$ and the vectors that span this space are orthogonal to the vector $\underline u$.

We can take for example the vectors $\underline v=(1,0,4)$ and $\underline w=(0,1,0)$, infact we have $$\langle\underline u,\underline v\rangle=1\cdot4+0+4\cdot(-1)=0\implies\text{ they are orthogonal}$$
$$\langle\underline u,\underline w\rangle=0\cdot4+1\cdot0+0\cdot(-1)=0\implies\text{ they are orthogonal}$$
Finally the equation of the plane $\pi$ is$$\begin{cases}x=0+\alpha\\y=0+\beta\\z=0+4\alpha \end{cases},\alpha,\beta\in\mathbb R.$$

Title: Vector form of a plane

Body:
Express the following plane in vector form:

$\mathcal P_1\subseteq  \Bbb R^3$ with equation $4x-z=0$.

The answer is $t(1,0,4)+s(0,1,0)$.
I don't understand how they got $(0,1,0)$ for the direction vector.


问:
I came across this problem in Binomial Theorem. I need help in proving this.
$$\sum_{k=1}^n\left(\frac{C_{2k-1}}{2k}\right)=\frac{2^n-1}{n+1}$$

I tried decoding that sigma on the LHS to identify the pattern. However, I wasn't able to proceed any further. I tried integrating the expansion of $(1-x)^n$ and its expansion, and then multiplying it with $(x+1)^n$ and its expansion; mainly to check if I am able to simplify anything. But no luck. I need help with this problem. Thanks in advance, for any and all help!

答:
Ok, I'm gonna assume

$$ C_{2k-1} = \binom{n}{2k-1}$$

Consider,

$$ f(x) = (1+x)^n = \sum_{i=0}^n \binom{n}{k} x^i$$

Notice, that odd coefficents are of interest here. We can extract out the odd part of function as:

$$  \frac{ f(x) - f(-x) }{2} = \sum_{k=1}^n x^{2k-1} \binom{n}{2k-1}$$

Now, to get the $ \frac{1}{k}$ part, integrate both sides:

$$ \int_{0}^1 \frac{ (1+x)^n - (1-x)^{n} }{2} dx = \sum_{k=1}^n \frac{1}{2k} \binom{n}{2k-1}$$

I'll leave it upto you to evaluate the LHS :^)


问:
Basis for Euclidean topology on $\mathbb{R}^2$
I am referring book Topology Without Tears

(i)In the book it is said,$\mathcal{B}$ collection of all 'open rectangles' $\{(x,y):(x,y)\in \mathbb{R}^2,a<x<b,c<y<d\}$ in the plane which have sides parallel to the $X$ axis or $Y$ axis is the basis for a topology on plane called Euclidean Topology.

(ii)It is further also proved : $\mathcal{B}$ set of all "open" equilateral traingles with the base parallel to the $X$ axis is a basis for the euclidean topology on $\mathbb{R}^2$

(iii)It is further stated " open rectangle " can be replaced by "open discs"

And there is a theorem for a collection to be basis of any topology:
Let $X$ be a non-empty set and $\mathcal{B}$ be a collection of subsets of $X$.Then $\mathcal{B}$ is a basis for a topology on $X$ if and only if $B$ has following properties:

(a)$X=\cup_{B\in \mathcal{B}}B $

(b)For any $B_1,B_2\in \mathcal{B}$ ,the set $B_1\cap B_2$ is a union of members of $\mathcal{B}$

From the above theorem the collection in (i), can truly be a basis for a topology as plane,$\mathbb{R}^2$ is the union of all of the open rectangle and intersection of two rectangle is a rectangle.

My question is how the collection of equilateral traingle with base parallel to $x$ axis can form basis according to the theorem above?
Because $\mathbb{R}^2$ may be seen as union of these open traingles but the intersection of two equilateral traingle need not be a equilateral traingle? or can it be written as union of equilateral traingle?

Same question in the case of disc,can intersection of two open disc can be written as union of open disc?

答:
By $B(x,r)$ I will denote an open ball around $x$ of radius $r$.

Intersection of two open discs need not be a disc. But it can be written as union of open discs. Take two open discs $D_1$ and $D_2$ and $x\in D_1\cap D_2$. Since $x\in D_1$ then there is $r_1>0$ such that $B(x,r_1)\subseteq D_1$. Analogously there is $r_2>0$ such that $B(x,r_2)\subseteq D_2$. Then for $r(x):=\min(r_1,r_2)$ we have $B(x,r(x))\subseteq D_1\cap D_2$. And so $D_1\cap D_2=\bigcup_{x\in D_1\cap D_2}B(x,r(x))$.

Analogously the intersection of equilateral triangles can be written as a union of equilateral triangles. Regardless of whether their sides are parallel to $x$, $y$ or whatever axis. For every point in an intersection $T_1\cap T_2$ you need to find a small enough equliateral triangle around $x$ contained in $T_1\cap T_2$. It is a bit harder compared to discs, but here's a sketch: first for $x\in T_1\cap T_2$ find an open disc $B(x,r)$ around $x$ fully contained in $T_1\cap T_2$. For that you need a well defined distance $r$ of $x$ from each side of $T_1$ and $T_2$. Then take an equilateral triangle around $x$ inside $B(x,r)$. The union of all such triangles around every point is of course $T_1\cap T_2$.


问:
Calculation of a multivariable integral
I read a paper which contains a tedious calculation and end up with the following integral:

$$
\int_{\mathbb{R^2}}\int_{\mathbb{R^2}}\frac{1}{1+\|x\|^2}\frac{1}{1+\|x+y\|^2}\frac{1}{1+\|y\|^2} \, dxdy
$$

Here, $\|\cdot\|$ denotes the Euclidean norm on $\mathbb{R^2}$. The author said that this integral is easily seen to be finite. My first attempt is by integrating in polar coordinates but I could not achieve the conclusion. Any hints? Thanks in advance

答:
I don't know about "easily." :) Don't attempt to do the integral. You merely want to give an estimate to be sure it converges. Ignoring the polar angles, if $r$ and $r'$ are the radial distances in the two respective $\mathbb R^2$'s, think about this: We can bound $\|x+y\|$ below by $\|y\|/2$ if
we know $\|y\|\ge 2\|x\|$. So our integral over $\{r'\ge 2r, r\ge R\}$ can be bounded above by
$$\lim_{R\to\infty}\int_R^\infty\int_{2r}^\infty \frac{rr'}{(r^2)(\frac12r')^2(r'^2)}\,dr'\,dr\,,$$
which is clearly finite. I believe that with a little thought, this will suffice to argue that the integral is convergent.

Title: Calculation of a multivariable integral


问:
Calculate the integral of a rational function using residues
I want to calculate integral :
$$\int_{0}^{\infty}\frac{x^2+16}{x^4+10x^2+9}$$
Which because its integrand is an even function of $x$, can be rewritten as $$\frac{1}{2}\int_{-\infty}^{\infty}\frac{x^2+16}{x^4+10x^2+9}$$
Consider the function $$f(z)=\frac{z^2+16}{z^4+10z^2+9}$$
The dominator has simple zeros at $z=\pm3i $ and $z=\pm i$.
Let's count
$Res[f(3i)] = \lim_{z \to 3i}\frac{z^1+16}{(z+3i)(z+i)(z-i)} = \frac{7}{48i}$
Similarly, $Res[f(-3i)] = \frac{7}{48i}$

$Res[f(i)] = \frac{15}{16i} = -Res[f(-i)]$.

So, $$\frac{1}{2}\int_{-\infty}^{\infty}\frac{x^2+16}{x^4+10x^2+9} = \frac{14\pi}{48i}$$
Is the idea good?
Of course, I am considering the semicircular contour $\gamma_R$, which starts at $R$, traces a semicircle in the upper half plane to $−R$ and then travels back to R along the real axis.
Why this integral is equal to zero if we integrate on semicircle in the upper half plane?

答:
You have the right idea, but the result is not correct. We have $\operatorname{res}_{3i} f=-\frac{7}{48i}$. So assuming the integral along the semicircle goes to zero we get: $$\frac{1}{2}\int_{-\infty}^\infty\frac{x^2+16}{x^4+10x^2+9}=\pi i\left(\operatorname{res}_{3i} f+\operatorname{res}_{i} f\right)=\pi i\left(\frac{-7}{48i}+\frac{15}{16i}\right)=\dots$$

We still need to argue why the integral over the semicircle goes to zero. Note for this that $\vert f(z)\vert\leq C\vert z\vert^{-2}$ for large enough $\vert z\vert$ and some constant $C$. So we can bound the integral over the semicircle $\gamma_R$ with radius $R$: $$\left\vert\int_{\gamma_R} f(z)dz\right\vert\leq CR^{-2}\operatorname{length}(\gamma_R)=CR^{-2}\pi R=\frac{C\pi}{R}\to0~~\text{as }R\to\infty $$


问:
Looking for a comprehensive text that compares the development of calculus using limits vs infinitesimals
I am looking for a book that covers the development of calculus using the ideas of limits working within the reals and also infinitesimals by extending the reals to hyperreals. I have seen people talk about how non-standard analysis provides more concise and intuitive proofs compare to the standard version using the idea of limits. I am looking for a self-contained text that covers the development of both simultaneously and provides a full comparison. Is there any such text available? Please help.</p>

答:
Having done my PhD in non-standard analysis, I've never seen such a book, and I doubt that you'll find one. The reason for that has nothing to do with the subject matter: mathematics books comparing multiple developments of the same subject in parallel just happen to be exceedingly rare, even in far more popular subfields.

You'd be hard-pressed to find a self-contained book-length treatment of sheaves on topological spaces that develops everything using both étalé spaces and functors, much less one that provides a deep comparison. You won't even find a book that combines and compares Linear Algebra Done Right with Linear Algebra Done Wrong in this manner!

There are books that teach (some) real analysis and (some) non-standard analysis at the same time [1] [2] [3], but they lack parallel developments of the same theorems. Instead, the authors will select what they consider the best approach to each theorem. Again, the same thing happens in other fields: a sheaf theory book might construct pullbacks using the étalé space perspective but pushforwards using the functorial perspective, and a linear algebra textbook might use determinants for some proofs but not others.

The closest thing I can suggest: reading some articles that give nonstandard proofs of specific theorems, such as Monotone subsequence via ultrapower by Błaszczyk, Kanovei, Katz and Nowik. Articles of this type tend to include at least some kind of comparison with the more standard proofs. You can find some article-length treatments that compare standard and nonstandard proofs of common theorems along some philosophical angle as well, such as On the plausibility of nonstandard proofs in analysis by Farkas and Szabo. Katz's bibliography on the history, mathematics, and philosophy of infinitesimals is a great resource for finding articles of both kinds. Tao's blog post about A cheap version of non-standard analysis might also provide you with some intuition, explaining why certain types of non-standard arguments feel simpler than their ordinary counterparts (Beware, though: I've met many people who took away the wrong message from this post! Tao discusses one specific situation where non-standard analysis comes in handy, which is not the only, nor necessarily the most common such situation!). Needless to say, these resources are not self-contained, and most of them assume that you have some familiarity with both standard and nonstandard real analysis.

[1] R. F. Hoskins. "Standard and Nonstandard Analysis: Fundamental Theory, Techniques, and Applications",  Ellis Horwood Series on Mathematics, 1990.  Comment: this is a mention, not necessarily a recommendation for learning analysis.

[2]  H. J. Keisler. "Elementary Calculus: An Approach Using Infinitesimals", Prindle Weber & Schmidt, ISBN 978-0871509116, (online 3rd edition), 1976. Comment: The first, and still one of the few available, infinitesimals-first calculus textbooks; manages to teach a fair bit of standard first course real analysis material.

[3] N. Vakil: "Real Analysis through Modern Infinitesimals", Cambridge University Press, 2011. Comment: got a reminded of the existence of this book in personal correspondence about my answer; this one takes my preferred, Internal Set Theoretic approach, it does try to give self-contained $\varepsilon,\delta$ and infinitesimal characterizations of continuity etc. (again, without the parallel development of proofs that OP seeks)


问:
Show that the set $A := \{x \in Q : x < 0 \text{ or } x^{2} < 2\}$ has no maximum. Does it have a supremum? If yes, what is it?
Show that the set
$$A := \{x \in Q : x < 0  \text{ or }  x^{2} < 2\}$$

has no maximum. Does it have a supremum? If yes, what is it?

答: Since There is always some rational point as much as you want near to every number, in your question, $-\sqrt 2 \ \& \  0$, from both sides, So there is neither minimum nor maximum, respectively. But these number are infimum and supremum, respectively. Can you prove it now? Tell me how much you can continue from this hint...


问:
Struggeling with proof of an Equivalent transformation
I'm trying to proof the following equation:

$$(( ≥ ℎ ∧  = ) ∨ ( < ℎ ∧  = ))$$
$$⇕$$
$$( = ) ∨ ( ≥ ℎ)$$

It should be equivalent, but I'm struggeling proofing it.

I have got this so far:

$$( ≥ ℎ ∧  = ) ∨ ( < ℎ ∧  = )$$
$$⇕ Distributive law$$
$$(( < ℎ ∧  = ) ∨  = )∧$$
$$(( < ℎ ∧  = ) ∨  ≥ ℎ)$$
$$⇕ Distributive law$$
$$( < ℎ ∨  = ) ∧ ( =  ∨  = )∧$$
$$( < ℎ ∨  ≥ ℎ) ∧ ( =  ∨  ≥ ℎ)$$
$$⇕$$
$$( < ℎ ∨  = ) ∧ () ∧ () ∧ ( =  ∨  ≥ ℎ)$$
$$⇕ neutral$$
$$( < ℎ ∨  = ) ∧ ( =  ∨  ≥ ℎ)$$
$$⇕ Distributive law$$
$$(( =  ∨  ≥ ℎ) ∧  < ℎ)∨$$
$$(( =  ∨  ≥ ℎ) ∧  = )$$

Can someone please give me a hint. :)

答:
The equivalence is not true as written (If one is considering these as placeholders for sets, then one must consider the trivial cases where certain sets are empty).

In the $\Leftarrow$ direction consider $(found = true) \wedge (left \geq right)$ which satisfies the $RHS$, but does not satisfy either of the statements on the $LHS$.

If $(found = true) \wedge (left \geq right)$, then $(( ≥ ℎ ∧  = )$ is false since we can't have $found = true$ and $found = false$.

$ ( < ℎ ∧  = ))$ is false since we can't have $left \geq right$ and $left < right$.

Hence if $(found = true) \wedge (left \geq right)$ we can't have $(( ≥ ℎ ∧  = ) ∨ ( < ℎ ∧  = ))$.

So then $(found = true) \wedge (left \geq right) \nRightarrow $

$ (( ≥ ℎ ∧  = ) ∨ ( < ℎ ∧  = ))$.


问:
Two triangles of equal area

Claim. Given a convex quadrilateral $ABCD$ that isn't parallelogram. The Newton line intersects sides $AD$ and $BC$ of the quadrilateral at the points $H$ and $G$ , respectively . Then $Area(\triangle HGD)=Area(\triangle HBG)$ and $Area(\triangle HGC)=Area(\triangle HAG)$.

GeoGebra applet that demonstrates this claim can be found here.

Proof. Observe that diagonal $HG$ of the quadrilateral $HBGD$ lie on the Newton line of the quadrilateral $HBGD$. If we apply Anne's theorem on point $H$ and quadrilateral $HBGD$ we can write $Area(\triangle HGD)+0=Area(\triangle HBG)+0$ , hence $Area(\triangle HGD)=Area(\triangle HBG)$. Similarlly we can show that $Area(\triangle HGC)=Area(\triangle HAG)$.

Question. Is this proof acceptable? Can we use Anne's theorem in the case when the point lies on the side of the quadrilateral? Can you provide an alternative proof?

答:
$I$ is the midpoint of diagonal $BD$ by definition.

So $\triangle HBI$ and $\triangle HDI$ have the same area (same base length along $BD$ and same height from $H$ onto $BD$), and $\triangle GBI$ and $\triangle GDI$ have the same area.

Title: Two triangles of equal area


Title: Compute $\int_0^\infty x^{n+1} J_n(kx) \exp(-a^2 x^2) dx$

Body:
I need advice on how to solve the following integral:

$$\int_0^\infty x^{n+1} J_n(kx) \exp(-a^2 x^2) dx$$

Body:
A possible approach to get it is to substitute directly the power series for $J_n$:
\begin{align*}
\int_0^\infty x^{n+1}J_n(kx)e^{-a^2 x^2}\,dx
&=\int_0^\infty x^{n+1}e^{-a^2 x^2}\sum_{j=0}^\infty\frac{(-1)^j(kx/2)^{n+2j}}{j!\ \Gamma(n+j+1)}\,dx
\\&=\sum_{j=0}^\infty\frac{(-1)^j(k/2)^{n+2j}}{j!\ \Gamma(n+j+1)}\int_0^\infty x^{2(n+j)+1}e^{-a^2 x^2}\,dx
\\&=\sum_{j=0}^\infty\frac{(-1)^j(k/2)^{n+2j}}{j!\ \Gamma(n+j+1)}\frac{\Gamma(n+j+1)}{2a^{2(n+j+1)}}
\\&=\frac{k^n}{(2a^2)^{n+1}}\sum_{j=0}^\infty\frac{(-1)^j}{j!}\left(\frac{k}{2a}\right)^{2j}=\frac{k^n}{(2a^2)^{n+1}}e^{-(k/2a)^2}.
\end{align*}


问:
Topology, true or false?

a) $\forall j : R^2 \to R^2$  $C^1$ function, $\exists k : R^2 \to R$ : $\nabla k(x) = j(x)$

b) $R^n$ has only one open vector subspace

c) $R^n$ has several compact vector subspace

d) Let $h : R^2 \to R$ a continuous function and $A$ a compact of $R$, $h^{-1} (A)$ is a compact

I did not manage to answer these questions satisfactorily

for b) i said "true" because if there is an open vector subspace, this open vector subspace is $R^n$

c) A compact is a bounded closed space, i said "true" but i'm not sure, i said that there are several singleton

d) I said "true" because $h^{-1} : A \to R^2$  a continuous function so $h^{-1}(A)$ is a compact

I don't know how to answer for a)

Thank you

答:
For b), subspaces are closed, and not open, except $\Bbb R^n$ considered as a subspace of itself, which is both open and closed. Kevin's answer, now deleted, had an argument with hyperplanes, another almost equivalent one: subspaces are defined by linear equations. The preimage of $\{(0,0,\dots,0)\}$ by a continuous function is closed.

For c), all subspaces are unbounded, except the null subspace.

For d), use my comment for b) to find a counterexample: the preimage is closed, but not necessarily compact ($h^{−1}$ is not a function!)

For a), the curl of the gradient is always zero. So it's not always possible to find a potential $k$ for a given vector field $j$. However, if $\mathrm{curl}\; j=0$ (i.e., the vector field is conservative), then such a $k$ exists. See Why curl free field implies existence of potential function? In two dimensions you have to consider the scalar curl.

To find a counterexample, it's thus enough to find a vector field $j(x,y)$ such that $\dfrac{\partial j_1}{\partial y}\ne \dfrac{\partial j_2}{\partial x}$.

Let $j$ be the linear map:

$$j(x,y)=\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}\cdot \begin{pmatrix}
x \\
y
\end{pmatrix}$$

That is, $j_1(x,y)=y$ and $j_2(x,y)=-x$.

Then, a potential $k$ for this vector field must satisfy:

$$\begin{eqnarray}
\dfrac{\partial k}{\partial x}&=&j_1(x,y)&=&y\\
\dfrac{\partial k}{\partial y}&=&j_2(x,y)&=&-x
\end{eqnarray}$$

From the first equation, you get that $k(x,y)=xy+a(y)$, for some function $a$ of only $y$. But then

$$\dfrac{\partial k}{\partial y}=x+a'(y)$$

And since $a'(y)$ does not depend on $x$, there is no way this can be equal to $-x$ for all $x$.

Title: Topology, true or false?


Title: Calculate the integral of a rational function using residues

Body:
I want to calculate integral :
$$\int_{0}^{\infty}\frac{x^2+16}{x^4+10x^2+9}$$
Which because its integrand is an even function of $x$, can be rewritten as $$\frac{1}{2}\int_{-\infty}^{\infty}\frac{x^2+16}{x^4+10x^2+9}$$
Consider the function $$f(z)=\frac{z^2+16}{z^4+10z^2+9}$$
The dominator has simple zeros at $z=\pm3i $ and $z=\pm i$.
Let's count
$Res[f(3i)] = \lim_{z \to 3i}\frac{z^1+16}{(z+3i)(z+i)(z-i)} = \frac{7}{48i}$
Similarly, $Res[f(-3i)] = \frac{7}{48i}$

$Res[f(i)] = \frac{15}{16i} = -Res[f(-i)]$.

So, $$\frac{1}{2}\int_{-\infty}^{\infty}\frac{x^2+16}{x^4+10x^2+9} = \frac{14\pi}{48i}$$
Is the idea good?
Of course, I am considering the semicircular contour $\gamma_R$, which starts at $R$, traces a semicircle in the upper half plane to $−R$ and then travels back to R along the real axis.
Why this integral is equal to zero if we integrate on semicircle in the upper half plane?

Body:
$$I=\int_{0}^{\infty}\frac{x^2+16}{(x^4+10x^2+9)}dx=\int_{0}^{\infty}\frac{x^2+16}{(x^2+1)(x^2+9)}$$
$$I= \frac{15}{8} \int_{0}^{\infty}\frac{dx}{1+x^2}-\frac{7}{8}\int_{0}^{\infty} \frac{dx}{9+x^2}$$
$$I=\frac{15}{8} \frac{\pi}{2}-\frac{7}{8} \frac{\pi}{6}=\frac{19}{24}\pi$$


问:
The graph of the function $y=\log_bx$ passes through the point $(729,6)$. Determine $b$.
The graph of the function $y=\log_bx$ passes through the point $(729,6)$. Determine $b$.

Could someone show me a solution that is similar to mine if it is correct?

$\log_b 729 = 6$

$729 = b^6$

$b = 729^{1/6} = 3$

答: Your solution is fine. There are of course six different sixth roots of 729, but only one which is a positive real number.</p>


问:
Question about differentiability at an interval.

Let $f:(0,\infty)\to\mathbb R$ be a differentiable function for which
$f'\left(x\right)\cdot f'\left(\frac{1}{x}\right)<0$ for every
$0<x\ne1$. Then $f'(1)=0$

I have this question, and I was wondering how to switch the fact the f is differentiable for every positive x into differentiable in a chosen interval that will help me with my proof..
I tried to make quiet a general interval in $[\frac{1}{x},x]$ but I'm afraid that I can't acctualy do that, and that it's not accurate..

d. True

We know $f$ is differential in $(0,\infty)$, therefore $f$ is also differentiable in the interval $[\frac{1}{x},x]$ for any $x ∈ R$ such that $0<\frac{1}{x}<1<x$.

Since $f'\left(x\right)\cdot f'\left(\frac{1}{x}\right)<0$, by Darboux's theorem there exist a point $c ∈ R$ such that

$\frac{1}{x} < c < x$ and $f'(c)=0$.

Assume toward contradiction that c ≠ 1, then  we have $f'\left(c\right)\cdot f'\left(\frac{1}{c}\right)<0$.

But since $f'(c)=0$, we have $0 \cdot f'\left(\frac{1}{c}\right)<0 \ \ \ ➜ \ \ \ 0 < 0$. Contradiction!

Therefore $c = 1$ and $f'(1) = 0$.

答:
As @Proxava said, your proof is OK.

You can also show that $f$ is strictly monotone on $(0,1] \ \& \ [1,+\infty)$, in opposite directions.


问:
Investigate if the following polynomial in Q [x] is reducible or irreducible: $x^5+x+1$
Investigate if the following polynomial in Q [x] is reducible or irreducible: $x^5+x+1$

Attempt:

We can write $x^5+x+1 = (x^2+x+1)(x^3-x^2+1)$ which could be a hint to be reducible, if we look clously we see that $(x^2+x+1),(x^3-x^2+1)$ are both irreducible in $Q$, if we use rational root theorem, we see there is no root in $Q$, so its irreducible over $Q$?

答: Since it can be factored into two polynomials in $\mathbb Q[x]$, $x^5+x+1$ is reducible. That the two factors have no rational root is irrelevant.


问:
Let $f: S^1 \to T$ be an embedding into the torus. $X_f=T \setminus f(S^1)$. Show that $X_f$ is not compact, but it is locally compact.
Let $f: S^1 \to T$ be an embedding into the torus (from the circle). We define $X_f=T \setminus f(S^1)$.

Show that $X_f$ is not compact, but it is locally compact.

Context

In question a) we proved that $S^1$ and $T$ are not homeomorphic, since removing two points from the circle leaves it disconnected into two parts, but after removing two points from the torus we see that it is still connected.

In question b) we proved that $f$ is never surjective. This is because an embedding is always a homeomorphism to $f(S^1)$. If $f$ is surjective we get $f(S^1)=T$ and we see that this would imply that $S^1$ and $T$ are homeomorphic. This is a contradiction to a).

So we know that $f: S^1 \to T$ is a non-surjective embedding, which means $X_f$ is non-empty, how does this help us with compactness? any pointers?

答:
First of all observe the $T$ is Hausdorff and compact (and so locally compact).

Now if $X_{f}$ were to be compact, it would be closed because compact sets of Hausdorff spaces are closed. So we would have that $T\backslash X_{f} = f(S^{1})$ is open as it is the complement of a closed set. Now $f(S^{1})$ is compact because $S^{1}$ is and $f$ is continuous, so $f(S^{1})$ must be closed too. We just proved that $f(S^{1})$ is both open and closed, but this is impossible in a connected space such as $T$. So $X_{f}$ is not compact.
Now to prove locally compactness: we already know (by the previous argument) that $f(S^{1})$ is closed, so $X_{f}\subset T$ is open, and an open subspace of a locally compact Hausdorff space ($T$ is one of those) is againg locally compact.

If anything of what I said is unclear, feel free to ask.


问:
Truncation and $\Pi$-types
Truncated types or bracket types are used to recover traditional logic within type theory. I have a question about truncating $\Pi$-types. The question is very basic. Let $A:\textsf{U}$ and $B:A\rightarrow\textsf{U}$, then what is the relationship between the following two types?

$$||\Pi x:A.B(x)|| \quad(*),\quad\quad\quad\quad\Pi x:A.||B(x)||\quad(**)$$

Does $(*)\rightarrow(**)$ or the other way around? Thanks a lot!

In the Homotopy Type Theory book, I find the following (pp.119-120):

Lemma 3.8.2 The axiom of choice is equivalent to the statement that for any set $X$ and any $Y:X\rightarrow\textsf{U}$ such that each
$Y(x)$ is a set, we have $$(\Pi x:X.||Y(x)||)\rightarrow||\Pi x:X.Y(x)||.\quad\quad\quad(3.8.3)$$

Then immediately, in a remark, it says the following:

Remark 3.8.4. It is easy to show that the right side of (3.8.3) always implies the left. ...

I got sort of confused. Does this say that the two types are equivalent?

答:
First, note that $\Pi x:A.||B(x)||$ is a mere proposition.

Indeed, suppose $f , g : \Pi x:A.||B(x)||$.
We must show $f = g$.
By extensionality, it is sufficient to show that for all $x : A$ we have $f(x) = g(x)$.
So let $x : A$. Since $f(x) , g(x) : ||B(x)||$, we have $f(x) = g(x)$ due to the fact that $||B(x)||$ is a mere proposition.

Second, note that $\lambda f. \lambda x. |f(x)| : (\Pi x:A.B(x)) \to \Pi x:A.||B(x)||$.
Thus, from the recursion principle of truncation, there is a term of the type $||\Pi x:A.B(x)|| \to \Pi x:A.||B(x)||$.

So we have shown $(*) \to (**)$.
The other way around, i.e. $(**) \to (*)$, is the axiom of choice.
Read carefully lemma 3.8.2 that you quoted.
The book does not claim that the axiom of choice is valid, only that there are two equivalent statements of it.

Title: Truncation and $\Pi$-types


问:
Calculate Point based on distance in 2D-Space
I have a Point P in unit circle (on or in it) with a radius of r. How can I calculate a Point Q with a fixed radius of x, which has the same angle like P</p>

答:
Suppose $P$ is at location $(a,b)$, with $a,b$ assumed positive.  You want point $P$ to be at $(c,d)$, with $c^2+d^2=x$, but also you want $\frac{d}{c}=\frac{b}{a}$.

We divide both sides by $c^2$ to get $1+\left(\frac{d}{c}\right)^2=\frac{x}{c^2}$, or $$1+\left(\frac{b}{a}\right)^2=\frac{x}{c^2}$$

We can now solve for $c$ via $$c=\sqrt{\frac{x}{1+(b/a)^2}}$$

Having solved for $c$, we may solve for $d$ via $$d=\sqrt{x-c^2}$$

I leave the cases where $a$ and/or $b$ is nonpositive for you to consider as variations.


答: Beside the expressions of @freakish, notice that since your triangles have parallel bases, and therefore other sides too, the intersection of those becomes equilateral one too, also its base is parallel to others.</p>

Title: Basis for Euclidean topology on $\mathbb{R}^2$

Body:
I am referring book Topology Without Tears

(i)In the book it is said,$\mathcal{B}$ collection of all 'open rectangles' $\{(x,y):(x,y)\in \mathbb{R}^2,a<x<b,c<y<d\}$ in the plane which have sides parallel to the $X$ axis or $Y$ axis is the basis for a topology on plane called Euclidean Topology.

(ii)It is further also proved : $\mathcal{B}$ set of all "open" equilateral traingles with the base parallel to the $X$ axis is a basis for the euclidean topology on $\mathbb{R}^2$

(iii)It is further stated " open rectangle " can be replaced by "open discs"

And there is a theorem for a collection to be basis of any topology:
Let $X$ be a non-empty set and $\mathcal{B}$ be a collection of subsets of $X$.Then $\mathcal{B}$ is a basis for a topology on $X$ if and only if $B$ has following properties:

(a)$X=\cup_{B\in \mathcal{B}}B $

(b)For any $B_1,B_2\in \mathcal{B}$ ,the set $B_1\cap B_2$ is a union of members of $\mathcal{B}$

From the above theorem the collection in (i), can truly be a basis for a topology as plane,$\mathbb{R}^2$ is the union of all of the open rectangle and intersection of two rectangle is a rectangle.

My question is how the collection of equilateral traingle with base parallel to $x$ axis can form basis according to the theorem above?
Because $\mathbb{R}^2$ may be seen as union of these open traingles but the intersection of two equilateral traingle need not be a equilateral traingle? or can it be written as union of equilateral traingle?

Same question in the case of disc,can intersection of two open disc can be written as union of open disc?


问:
How to find formula for generic summation n/(n + C)
Note:  Completely revised in response to questions/comments on December 23, 2020 at 16:35 UTC.

I'm working on a coding challenge where you have to sum up a series of fractions.  These fractions form a matrix.  For example:

$n=0:\ \ 0$

$n=1:\ \ \frac{1}{2}$

$n=4:$

$\frac{1}{2}\ +\ \frac{2}{3}\ +\ \frac{3}{4}\ +\ \frac{4}{5}$

$\frac{1}{3}\ +\ \frac{2}{4}\ +\ \frac{3}{5}\ +\ \frac{4}{6}$

$\frac{1}{4}\ +\ \frac{2}{5}\ +\ \frac{3}{6}\ +\ \frac{4}{7}$

$\frac{1}{5}\ +\ \frac{2}{6}\ +\ \frac{3}{7}\ +\ \frac{4}{8}$

I started with a brute force approach.  I know this isn't a programming forum, but I'm hoping some simple Python code will illustrate this:

```<code>from fractions import Fraction

def matrix_sum(n):
    total = Fraction()
    for row_denominator in range(1, n + 1):
        for col_numerator in range(1, n + 1):
            total += Fraction(col_numerator, col_numerator + row_denominator)
    return total

>>> matrix_sum(0)
Fraction(0, 1)

>>> matrix_sum(1)
Fraction(1, 2)

>>> matrix_sum(4)
Fraction(8, 1)

>>> matrix_sum(8)
Fraction(32, 1)
</code>```

However, this doesn't work because the challenge purposefully uses huge numbers and imposes strict time constraints.  The above solution is correct, but way too slow (it's $O(n)^2$ - pretty bad algorithmically).  I even tried converting this to C++ for raw speed.  That fixes the computational speed, but because the numbers are so big, it overflows a 64-bit integer (maximum size for normal C++ integer that I can use in this challenge).  I know there are libraries and compilers have their own 128-bit numbers, but the challenge doesn't allow these.

That led me to think there must be some kind of series/sequence algorithm.  I looked at some series - arithmetic, geometric - but they don't quite work.  Here's how I started approaching this:

$\sum_{i=1}^n {col_i \over (col_i\ +\ row)}$

So this summation is for each row of the matrix.  If I let $n=4$ then I get:

Row 1: $\ \sum_{i=1}^4 {col_i \over (col_i\ +\ 1)} = \frac{1}{2}\ +\ \frac{2}{3}\ +\ \frac{3}{4}\ +\ \frac{4}{5}$

Row 2: $\ \sum_{i=1}^4 {col_i \over (col_i\ +\ 2)} = \frac{1}{3}\ +\ \frac{2}{4}\ +\ \frac{3}{5}\ +\ \frac{4}{6}$

Row 3: $\ \sum_{i=1}^4 {col_i \over (col_i\ +\ 3)} = \frac{1}{4}\ +\ \frac{2}{5}\ +\ \frac{3}{6}\ +\ \frac{4}{7}$

Row 4: $\ \sum_{i=1}^4 {col_i \over (col_i\ +\ 4)} = \frac{1}{5}\ +\ \frac{2}{6}\ +\ \frac{3}{7}\ +\ \frac{4}{8}$

I can then sum up those 4 series.  Would it be better to do these as a double summation that represents the rows too?  I apologize, but I'm rusty with summations.

However, in the process of explaining all this - I see there's a clever solution.  If you look along the diagonals, you'll notice that the fractions add up to one.  So, it appears that the clever answer for this series is the number of squares in the matrix divided by 2.  For example, note that when $n=4$ (16 squares), the answer is 8.

I'm still curious if I was thinking through this the right way.  I do a lot of programming and I think I would be much better off with stronger problem solving using pure mathematics.  Any suggestions on refining the summation/series/sequence, approaching this with matrix math/linear algebra, or otherwise solving this mathematically would be greatly appreciated.

Please let me know if any of this isn't clear and thanks.

--Jim

答:
If I did not misunderstand what you are asking for, it seems you have to evaluate the following $n$ sums:
$$
\begin{equation}
\frac{1}{2} + \frac{2}{3} + \dots + \frac{n}{n+1} \\
\frac{1}{3} + \frac{2}{4} + \dots + \frac{n}{n+2} \\
\vdots \\
\frac{1}{n+1} + \frac{2}{n+2} + \dots + \frac{n}{n+n}
\end{equation}
$$

Fix $i$ between $1$ and $n$. Then you have:

$$ \frac{1}{i+1} + \frac{2}{i+2} + \dots + \frac{n}{i+n} = \sum_{j=1}^n \frac{j}{i+j}$$

Wolframalpha shows a "closed form" for this sum you can check at this link.


答:
Let $x=w,w^2$, where $w$ cube root of unity. The $f(x)=x^5+x+1 \implies f(w)=w^2+w+1=0, f(w^2)=w+w^2+1=0$. So $g(x)=(x-w)(x-w^2)=x^2+x+1$ is a factor of $f(x)$ Further $\frac{f(x)}{g(x)}=x^3-x^2+1.$
So $f(x)$ is reducible.

Body:
Investigate if the following polynomial in Q [x] is reducible or irreducible: $x^5+x+1$

Attempt:

We can write $x^5+x+1 = (x^2+x+1)(x^3-x^2+1)$ which could be a hint to be reducible, if we look clously we see that $(x^2+x+1),(x^3-x^2+1)$ are both irreducible in $Q$, if we use rational root theorem, we see there is no root in $Q$, so its irreducible over $Q$?


答:
Let $P=(x_1, y_1)$ and $Q=(x_2,y_2)$, for unknown $(x_2,y_2)$. We want the points to be in the same line connecting the origin $(0,0)$ and $P$. (Let's assumme $P$ is not the origin!) This line is given by
$$
x_1 y = y_1 x
$$
Note that I didn't write $y=\frac{y_1}{x_1}x$ because $x_1$ may be zero.

Now let's plug $(x_2,y_2)$ in it:
$$
x_1 y_2 = y_1 x_2
$$

The radius is the distance from the point to the origin. I'll call it $q$ instead of $x$ to avois confusion. This is given by
$$
{x_2}^2 + {y_2}^2 = q^2
$$

Now you have two equations in two unknowns. Substituting the second in the first:
$$
{x_1}^2 (q^2-{x_2}^2) = {y_1}^2 {x_2}^2
$$

which is a second degree equation in $x_2$. The corresponding $y_2$ is found with the first equation. It has two solutions, because there are two points in the same line $\bar{OP}$ at the same distance from the origin. If you want, you can discard the solution pointing in the opposite direction.

Title: Calculate Point based on distance in 2D-Space

Body: I have a Point P in unit circle (on or in it) with a radius of r. How can I calculate a Point Q with a fixed radius of x, which has the same angle like P</p>


问:
Prove that for every set there is a free module on that set
While studying the book "Module theory: an approach to linear algebra" by T. S. Blyth (the electronic edition of the book is free to download), I came across a proof that confused me. We need to prove that for every non-empty set $S$ and every unitary ring $R$ there is a free $R$-module on $S$. Here's the complete proof:

First part of the proof
second part of the proof

I couldn't grasp the next to last line, specifically, the following equality:

$$h'(v) = \sum_{s \in S}v(s)h'[f(s)]$$

Edit:

It turned out that the source of my confusion was this equality:

$$v =  \sum_{s \in S}v(s)f(s)$$

I thought that $f(s)$ is an element of $R$, when in fact $f(s)$ is a mapping (from $S$ to $R$), and therefore the sum will also be a mapping.

答:
It seems that it is just

$$h^\prime(v) = h^\prime\left(\sum_{s \in S}v(s)[f(s)]\right) = \sum_{s \in S}v(s)h'[f(s)]$$

as $h^\prime$ is supposed to be an $R$-morphism.


答:
The equality follows from $R$-linearity of the morphism $h'$, and from the previous equation
$$\theta = \sum_{s\in S} \theta(s)f(s)$$
Applying $h'$ to such a sum results in the desired equation.

Title: Prove that for every set there is a free module on that set

Body:
While studying the book "Module theory: an approach to linear algebra" by T. S. Blyth (the electronic edition of the book is free to download), I came across a proof that confused me. We need to prove that for every non-empty set $S$ and every unitary ring $R$ there is a free $R$-module on $S$. Here's the complete proof:

First part of the proof
second part of the proof

I couldn't grasp the next to last line, specifically, the following equality:

$$h'(v) = \sum_{s \in S}v(s)h'[f(s)]$$

Edit:

It turned out that the source of my confusion was this equality:

$$v =  \sum_{s \in S}v(s)f(s)$$

I thought that $f(s)$ is an element of $R$, when in fact $f(s)$ is a mapping (from $S$ to $R$), and therefore the sum will also be a mapping.


问:
Addition rule for mutually exclusive events
Why does the addition rule not work here?

$$P(\text{win this week}) = 0.5$$
$$P(\text{win next week}) = 0.5$$

They are mutually exclusive, and you can only win once (although it is possible that neither occurs). What is the probability that either occurs?

$P(A \cup B) = P(A) + P(B)$ clearly does not give the correct answer, unless you are guaranteed to win (but that is not the case here).

答:
I think you confuse Independency & Exclusiveness. These are seems to be Mutually independent, not exclusive. Or there is some other information about game or lottery as @miracle173 noticed.

By informations of comments:

So, Yes, the probability of wins in whole $2$ weeks is $1$. But if you have problem to be winner surely, you can suppose some scenario in which you don't:

Suppose you randomly pick a number in $[0,1]$ and if it's a irrational number in $(0,1/2)$ you become the winner of the first week, and if it's a irrational number in $(1/2,1)$ you would be the winner of the next week. So you win by probability of $1$, but not surely, you lose if you be a infinitely bad luck rational picker!


问:
Closed form for Exponential Conditional Expected Value & Variance
I am wondering if there is a closed form for finding the expected value or variance for a conditional exponential distribution.

For example:
$$ E(X|x > a) $$ where X is exponential with mean $\lambda$.

Same question for variance.

What about for a joint distribution of independent exponentials?

$$ E(X|y > a) $$ where X is exponential with mean $\lambda$, Y is exponential with mean $\theta$ and X & Y are independent.

A sample problem for the actuarial P/1 exam (#124 for those also studying) asks:

The joint probability for $f(x,y) = 2e^{-x-2y}, ~ x > 0, ~ y > 0$. Calculate the variance of Y given $x > 3, ~ y > 3$.

The solution goes like this: (Math on the right, reasoning on the left)

$Var (Y|x>3, y>3) =$

$Var (Y|x>3) = ~~~~~$Independence

$Var (Y + 3) = ~~~~~$Memoryless

$Var (Y) + Var (3) =~~~~~$Independence of Y and 3.

$Var (Y) = ~~~~~ $ Since $Var (3) = 0$.

$0.25 ~~~~~ $Exponential Variance, $\lambda = 2$.

So this says to me that  $Var (Y|x>3) = Var (Y)$.  Is that true?  If so, is it always true?  If not, then how does this solution work?

Could one also replace E(Y) for Steps 1 - 4, Use $E(a) = a$ and get $E(Y| y>a) = E(y) + a$?

Shortcuts like this are immensely valuable for a timed test.  (Not just faster, but less error prone).

答:
Let $X$ be an exponentially distributed random variable with parameter $\lambda$. We can think of $X$ as the lifetime of a specific Carbon $14$ atom. Given that $X\gt a$, the additional lifetime $Y$ of the atom is exponentially distributed with parameter $\lambda$. This is a key property of the exponential, usually called memorylessness.

Given that $X\gt a$, $X=a+Y$. It follows that $E[X|X\gt a]=a+E[Y]=a+E[X]$.

The second question of the same kind is much simpler.  If $X$ and $Y$ are any independent random variables, then $E[X|Y\gt a]=E[X]$.

Back to the exponential. Since the random variable $X|X\gt a$ is just $a+Y$, where $Y$ is exponential, we have $\text{Var}(X|X\gt a)=\text{Var}(a+Y)=\text{Var}(Y)=\text{Var}(X)$.

Since we know that an exponentially distributed random variable $X$ with parameter $\lambda$ has mean $\frac{1}{\lambda}$ and variance $\frac{1}{\lambda^2}$, we can write down with no trouble the mean and variance of the conditional version.


问:
Splitting field over $\Bbb F_2$ with degree $8$

How do I determine the splitting field of this  polynomial
$x^8+x^6+x^5+x^4+x^3+x^2+1$ over $\Bbb F_2$.

I had referred to other question answers on the website, but all of them are with a lower degree (degree $3$). I had tried factorizing the polynomial but not sure what to do next. Thanks

答:
Note that
$$
f(x)=(x^3 + x^2 + 1)(x^3 + x + 1)(x^2 + x + 1).
$$
Then compare with

Degree of splitting field of polynomial over a finite field

Galois group of polynomials over finite fields


答:
The memorylessness property of the exponential distribution states that for if $X$ is exponentially distributed then for any function $f$ we have $E(f(X)|X>\alpha) = E(f(X+\alpha))$.  Essentially this means that the distribution of an exponential conditioned to be greater than $\alpha$ is exactly the same as the distribution of an exponential with $\alpha$ added to it.

So your method works for both the mean and the variance.  It's also useful to know what happens to the mean and variance of a random variable when you add a constant. (You add the constant to the mean, but leave the variance alone.)

In a maths exam you would be required to prove that the distribution of $X$ and $Y$ are independent exponentials.  I don't know if that would be true in an actuary's exam.

Title: Closed form for Exponential Conditional Expected Value & Variance

Body:
I am wondering if there is a closed form for finding the expected value or variance for a conditional exponential distribution.

For example:
$$ E(X|x > a) $$ where X is exponential with mean $\lambda$.

Same question for variance.

What about for a joint distribution of independent exponentials?

$$ E(X|y > a) $$ where X is exponential with mean $\lambda$, Y is exponential with mean $\theta$ and X & Y are independent.

A sample problem for the actuarial P/1 exam (#124 for those also studying) asks:

The joint probability for $f(x,y) = 2e^{-x-2y}, ~ x > 0, ~ y > 0$. Calculate the variance of Y given $x > 3, ~ y > 3$.

The solution goes like this: (Math on the right, reasoning on the left)

$Var (Y|x>3, y>3) =$

$Var (Y|x>3) = ~~~~~$Independence

$Var (Y + 3) = ~~~~~$Memoryless

$Var (Y) + Var (3) =~~~~~$Independence of Y and 3.

$Var (Y) = ~~~~~ $ Since $Var (3) = 0$.

$0.25 ~~~~~ $Exponential Variance, $\lambda = 2$.

So this says to me that  $Var (Y|x>3) = Var (Y)$.  Is that true?  If so, is it always true?  If not, then how does this solution work?

Could one also replace E(Y) for Steps 1 - 4, Use $E(a) = a$ and get $E(Y| y>a) = E(y) + a$?

Shortcuts like this are immensely valuable for a timed test.  (Not just faster, but less error prone).


问:
integral e with two variables
how are you? I have a problem with an integral. I get to a point where I don't know how to continue. I know there are several methods but I would like to follow this method.
i know that ∫dx e^a* a’ = e^a

so
$$
\int_{-\infty}^{\infty}\, dx\, x e^{-\frac{a}{2}x^2}
= -\frac{1}{a}
\int_{-\infty}^{\infty}\, dx\, (-ax) e^{-\frac{a}{2}x^2}
$$
https://i.stack.imgur.com/6Fwkm.png

And now I don't know whether to value x at infinity and minus infinity, or if before I would have had to substitute all that exponent with a single variable

答:
So you have
$$F(x)=\int x e^{\frac{-ax^2}{2}  }\, dx=-\frac{e^{-\frac{1}{2} \left(a x^2\right)}}{a}+C$$

and
$$\int_{-\infty }^{\infty } x e^{\frac{-ax^2}{2}} \, dx=\underset{M\to \infty }{\text{lim}}(F(M)-F(-M))$$

Edit

The result can be got without any calculation because $f(x)=x e^{\frac{-ax^2}{2}  }$ is an odd function, that is $f(-x)=-f(x)$, thus the integral is zero.

Title: integral e with two variables


问:
Sufficient statistic for variance in a normal distribution with mean know

Let $Y_{1},Y_{n},\ldots,Y_{n}\sim N(\mu,\sigma^{2})$ with mean know. Prove that $ \displaystyle T(Y)=\frac{1}{n}\sum_{1\leq i \leq n}Y_{i}^{2}$ is sufficient to $\sigma^{2}$.

My approach: I know that the Neyman-factorization theorem is useful here, then

\begin{eqnarray*}
  f(y_{1},y_{2},\ldots,y_{n};\sigma^{2})=\left( \frac{1}{\sigma \sqrt{2\pi}}\right)^{n}\exp \left\{-\frac{1}{2\sigma^{2}}\sum_{i=1}^{n}y_{i}^{2} \right\}\cdot \mathbf{1}
    \end{eqnarray*}
but I should conclude that $\displaystyle T_{\sigma^{2}}(Y)=\sum_{1\leq i \leq n} Y_{i}^{2}$ is sufficient statistic, but is not what is requered to prove. How can I continue from here?

For the following question my post is not same of the suggested by a member of MSE. However thanks.

EDIT: If $\sum_{n}Y^{2}$ is sufficient statistic, so $\frac{1}{n} \sum_{n}Y^{2}$ is a sufficient statistic? If it's true, how can I prove this? I think that if it's true so, my approach is correct and I can conclude.

Thank you so much @tommik for your explication.

答:
Sorry but there is an evident problem in  your statement.

$$Y_i\sim N(\mu;\sigma^2)$$

with $\mu$ known parameter; $\mu \in \mathbb{R}$

Thus

$$f(\mathbf{y},\mu,\sigma^2)=(2\pi\sigma^2)^{-n/2}\exp\Bigg\{-\frac{1}{2\sigma^2}\Sigma_i(y_i-\mu)^2\Bigg\}=$$

$$=(2\pi\sigma^2)^{-n/2}\exp\Bigg\{-\frac{\Sigma_i y_i^2}{2\sigma^2}+\frac{\mu\Sigma_i y_i}{\sigma^2}-\frac{n\mu^2}{2\sigma^2}\Bigg\}$$

Now using the Factorization theorem you get

$$f(\mathbf{y},\mu,\sigma^2)=1\times\underbrace{(2\pi\sigma^2)^{-n/2}\exp\Bigg\{-\frac{\Sigma_i y_i^2}{2\sigma^2}+\frac{\mu\Sigma_i y_i}{\sigma^2}-\frac{n\mu^2}{2\sigma^2}\Bigg\}}_{g[t(\mathbf{y}),\sigma^2]}$$

Thus

$T=\Sigma_i Y_i^2$ IS NOT SUFFICIENT FOR $\sigma^2$ with $\mu$ knonw UNLESS $\mu=0$

Applying the same theorem, what is valid for $T$ is also valid for $T/n$

(the proof is trivial, just substitute the new estimator in the factorized expression)


问:
Cardinals proof |A| ≤ |B| iff |B| ≥ |A|
So i have a little bit problems proofing this.

```<code> |A| ≤ |B| iff |B| ≥ |A|
</code>```

i know, that if |A| ≤ |B| => f: A => B, must be injective

and likewise, that if |B| ≥ |A| => g: B => A, A = ø or it must be surjective.

Good, but how do i exactly do this proof? I am a little bit lost on how to write it down in a mathematical accurate way. Would anyone be able to help me out?

Thanks in advance

答:
I will prove the first implication and leave the other to you.

Assume $\vert A\vert\leq \vert B\vert$.

If $A=\emptyset$, there is nothing to prove, so we can assume $A\ne\emptyset$. Since $\vert A\vert\leq \vert B\vert$ there is an injective map $f:A\to B$. We prove that this has a left-inverse, i.e. there is a $g:B\to A$ such that $g\circ f=\operatorname{id}_A$. Pick $a_0\in A$ arbitrary (possible since $A\ne\emptyset$). Then we define $g:B\to A$ as follows:
\begin{align*}
g(b)=\begin{cases} a ~~~\text{if }a\in A,f(a)=b\\a_0~~~\text{otherwise, i.e. if }b\notin f(A)\end{cases}
\end{align*}
Note that this is well-defined since $f$ is injective, i.e. if there is som $a\in A$ with $f(a)=b$ then this $a$ is uniquely determined. Similarly we can check that $g\circ f = \operatorname{id}_A$. Now we claim that this $g$ has to be surjective. Let $a\in A$. Then $g(f(a))=\operatorname{id}_A(a)=a$ hence $f(a)$ is a preimage of $a$ for $g$, so $g$ is surjective. So we proved that there is a surjective map $B\to A$, hence $\vert B\vert\geq\vert A\vert$.

Now do something similar for the other direction, i.e. $\vert B\vert\geq\vert A\vert\implies \vert A\vert\leq\vert B\vert$.

(Here I assumed that the definition of $\vert B\vert\geq\vert A\vert$ is that there is a surjective map $B\to A$ and for $\vert A\vert\leq\vert B\vert$ that there is an injective map $A\to B$)

Title: Cardinals proof |A| ≤ |B| iff |B| ≥ |A|


问:
Appearance of Formal Derivative in Algebra
When studying polynomials, I know it is useful to introduce the concept of a formal derivative. For example, over a field, a polynomial has no repeated roots iff it and its formal derivative are coprime. My question is, should we be surprised to see the formal derivative here?

Is there some way we can make sense of the appearance of the derivative (which to me is an analytic object) in algebra? I suspect it might have something to do with the fact that the derivative is linear and satisfies the product rule, which makes it a useful object to consider. It would also be interesting to hear an explanation which explains this in the context of algebraic geometry.

Thanks!

答:
Formal derivatives appear naturally when trying to rewrite a polynomial in $ x $ ( over a characteristic $ 0 $ field ) as a polynomial in $ (x-c) $ :

Let $ f(x) = a_n x^n + \ldots + a_1 x + a_0 \in K[x] $ ( with $ a_n \neq 0 $ ). Given any $ c \in K $, we can expand $ f(x) $ as a polynomial in $ (x-c) $ to get
$ \displaystyle f(x) = \sum_{j=0}^{n} a_j x^j = \sum_{j=0}^{n} a_j (x-c+c)^j $ $ \displaystyle = \sum_{j=0}^{n} a_j \bigg( \sum_{r = 0}^{j} \binom{j}{r} (x-c)^{r} c^{j-r} \bigg) $ $ \displaystyle = \sum_{0 \leq r \leq j \leq n} (x-c)^{r} a_j \binom{j}{r}c^{j-r} $ $ \displaystyle = \sum_{r=0}^{n} (x-c)^{r} \bigg( \sum_{j=r}^{n} a_j \binom{j}{r}c^{j-r} \bigg) $

$ \big[ $ Here "$ \binom{j}{r} $" should be understood as $ 1 + \ldots + 1 $ with $ \frac{j(j-1)..(j-(r-1))}{r!} $ many $ 1$s $ \big] $

Now assuming $ \text{char}(K) = 0 $ we can write $ \binom{j}{r} = (r!)^{-1} ( j(j-1)..(j-r+1) )$ [ where again each term "$ n $" should be understood as $ 1 + \ldots + 1 $ with $n$ many $1$s ], giving us

$ f(x) \displaystyle = \sum_{r=0}^{n} \frac{(x-c)^{r}}{r!} \bigg( \sum_{j=r}^{n} a_j \: j(j-1)\ldots(j-(r-1)) \: c^{j-r} \bigg) $.

But $ \displaystyle \sum_{j=r}^{n} a_j \: j(j-1)\ldots(j-(r-1)) \: c^{j-r} $ is just $ (D^{r} f)(c) $, where $ D : K[x] \rightarrow K[x] $ is a map sending $ c_0 + c_1 x + c_2 x^2 + \ldots + c_n x^n $ $ \mapsto c_1 + 2 c_2 x + \ldots + n c_n x^{n-1} $.

To summarise : Let $ f(x) = a_0 + \ldots + a_n x^n \in K[x] $, $ c \in K $, and $ \text{char}(K) = 0 $. Then $ f(x) = f(c) + (Df)(c) \: (x-c) $ $ + \frac{(D^2 f)(c)}{2!} \: (x-c)^2 + \ldots + \frac{(D^n f)(c)}{n!} \: (x-c)^n $, where $ D : K[x] \rightarrow K[x] $ sends $ c_0 + c_1 x + \ldots + c_n x^n $ $ \mapsto c_1 + 2 c_2 x + \ldots + n c_n x^{n-1} $ [ especially if $ \alpha \in K $ is a root of $ f $, it's multiplicity is the least $ j $ for which $ (D^j f)( \alpha ) \neq 0 $ ].

Edit: [This should work in general characteristic]

For every $f \in K[x]$ there is a unique $\psi _f \in K[x,y]$ such that $f(y) - f(x) = (y-x) \psi _f (x,y).$

Explicitly, taking $f(x) = \sum _{0} ^{n} a _i x ^i,$ we get $f(y) - f(x)$ ${= \sum _{0} ^{n} a _i (y ^i - x ^i)}$ ${= (y-x) \left( \sum _{1} ^{n} a _i (y ^{i-1} + y ^{i-2} x + \ldots + x ^{i-2} y + x ^{i-1}) \right).}$

Now $K[x] \longrightarrow K[x,y]$ sending $f(x) \mapsto \psi _f (x,y)$ is linear.

$f(y) - f(x) = (y-x) \psi _f,$ and $g(y) - g(x) = (y-x) \psi _g.$ Taking ${\alpha \cdot (\text{eq}1) + \beta \cdot (\text{eq}2)}$ gives $\psi _{\alpha f + \beta g} = \alpha \psi _f + \beta \psi _g.$

Also $\psi _{fg} = \psi _f g + f \psi _g + (y-x) \psi _f \psi _g$.

$(fg)(y)$ ${= (f(x) + (y-x) \psi _f)(g(x) + (y-x) \psi _g)}$ ${ = (fg)(x) + (y-x)(\psi _f  g+ f \psi _g + (y-x) \psi _f \psi _g)}$

So the map $K[x] \overset{D}{\longrightarrow} K[x]$ sending $f(x) \mapsto \psi _f (x,x)$ is linear, with "Leibnitz rule" $D(fg) = (Df)g + f(Dg).$ From the expression for $\psi _f$ we get the explicit description $D\left( \sum a _i x ^i \right) = \left(\sum i a _{i} x ^{i-1} \right).$


问:
Cayley graphs obtained for different generating elements with the same orders in a finite group
When we consider the finite group $\mathbb{Z}_p \times \mathbb{Z}_p$, where $p$ is a prime, $p>2$, the set with the pair of elements $\{(0,1), (1,0)\}$ can generate the group. Moreover, a set $\{(1,0),(1,2)\}$ can also generate the group. Hence, there are many pairs of elements which can act as generating elements.

Consider the group $(\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes_{\phi} \mathbb{Z}_q$. A set $\{t,s\}$, where $|t|=p$ and $|s|=q$ is a minimal generating element set for some $p,q$ and $\phi$ values (Let $p$ and $q$ be distinct primes, $p,q>2$).

Under a certain $\phi, p, q$, the $t,s$ elements were,

$t=((1,0),0),\,\, s=((0,0),1)$

These generating elements can be used to draw a Cayley graph of the group.

However, if I take $t=((3,0),0)$ and draw the graph (under the same $\phi$. Suppose $\phi$ is defined using a general formula like, $\phi(t^i)=t^{i+1}$. (This is only an example). Then I can apply $\phi(t)$ to any $t$ value), the graph is the same as earlier graph if I forget the vertex lables.

Then can I bring an argument here, that I can use any element of order $p$ as $t$ and any element of order $q$ as $s$ in the generating set?

Thanks a lot in advance.

答:
Summarized idea of the answer of above question:

For a group $G = (\mathbb{Z}_p \times  \mathbb{Z}_p) \rtimes_{\phi} \mathbb{Z}_q$,

$\phi: \mathbb{Z}_q \rightarrow Aut(\mathbb{Z}_p \times \mathbb{Z}_p) $

defines the semidirect product. Let $S=\{s,t\}$, where $|s|=q$ and $|t|=p$ denote an irredundant generating set of $G$ and let $\{t,u\}$, where $|u|=p$ be a basis of $\mathbb{Z}_p \times \mathbb{Z}_p$. ($\phi_s(t) = s^{-1}ts =u$).

For an element $\phi(s)=\phi_s \in Aut(\mathbb{Z}_p \times \mathbb{Z}_p)$, in defining the automorphism,

$\phi_s(t^{i} u^{j})$ using the generating elements $t,u \in \mathbb{Z}_p \times \mathbb{Z}_p$, we can choose different candidates of order $p$ as $t$ and $u$, chosen appropriately.

i.e. as an example,

$t=((0,1),0), u=((1,0),0)$ is one choice, but also $t=((0,3),0), u=((1,0),0)$ can be another choice. Likewise, there can be several choices for the generating elements.


问:
Find the limit of $(n^3+2n^2)^{1/3}-n$
Find the limit of sequence.I tried and get this
$$
\lim_{n\to \infty} \frac{((n^3+2(n)^2)^{1/3}-n)\cdot((n^3+2 n ^2)^{1/3}+n)}{(2(n)^2+n^3)^{1/3}+n}
$$ I am getting limit 0 but this is wrong answer can you help?

Title: Find the limit of $(n^3+2n^2)^{1/3}-n$

Body: $(1+x)^{1/3} = 1+ \frac x 3+o(x)$ as $x \to 0$. The given expression is  $n[(1+\frac2  {3n}) -1+o(1/n)]  \to \frac 2  3$


问:
Isomorphism between groups and subgroups.
Say I know that $G_1,G_2$ are isomorphic groups, both with normal subgroups $N_1,N_2$ respectively. Can I take an isomorphism $\psi$ between $G_1,G_2$ s.t $\psi(N_1)=N_2$? Meaning that $\psi$ also defines a isomorphism between $N_1$ and $N_2$?

The thm I'm trying to prove is if $G_1 \cong G2, N1\cong N2,N1\unlhd G1, N2 \unlhd G2$ then $G1/N1 \cong G2/N2$, and proving the above statement I would be able to use the first isomorphism theorem and that's it. Is it even true?

Any hint would be helpful on how to approach this problem.

答: Let $\psi\colon G_1\rightarrow G_2$ be a group isomorphism and $N_1$ be a normal subgroup of $G_1$. Then of course $N_2=\psi(N_1)$ is a normal subgroup of $G_2$. However, the normal subgroups of a group may have different cardinality in general, so that not always $\psi(N_1)=N_2$ for arbitrary normal subgroups $N_1$ and $N_2$ in $G_1$ respectively $G_2$.


答:
Well, no. For example, take $G_1=G_2=N_1=S_3$ and $N_2=\langle (123)\rangle$. No isomorphism will satisfy $\psi(N_1)=N_2$, because they don't have the same cardinality.

Anyway, the statement you want to prove is false. If we take $G_1=G_2=\mathbb{Z}$ and $N_1=2\mathbb{Z}, N_2=3\mathbb{Z}$ then $G_1\cong G_2$ and $N_1\cong N_2$. However, $G_1/N_1$ and $G_2/N_2$ can't be isomorphic, as they have different cardinalities.

Title: Isomorphism between groups and subgroups.

Body:
Say I know that $G_1,G_2$ are isomorphic groups, both with normal subgroups $N_1,N_2$ respectively. Can I take an isomorphism $\psi$ between $G_1,G_2$ s.t $\psi(N_1)=N_2$? Meaning that $\psi$ also defines a isomorphism between $N_1$ and $N_2$?

The thm I'm trying to prove is if $G_1 \cong G2, N1\cong N2,N1\unlhd G1, N2 \unlhd G2$ then $G1/N1 \cong G2/N2$, and proving the above statement I would be able to use the first isomorphism theorem and that's it. Is it even true?

Any hint would be helpful on how to approach this problem.


答: It is not true, consider $G_1=G_2=C_4\times C_2$. Let $N_1=0\times C_2$ and $N_2 =H\times 0$ where $H$ is the subgroup of order $2$ in $C_4$. Then $N_1\cong N_2$, but $C_4=G_1/N_1\not\cong G_2/N_2=C_2\times C_2$


答:
$$L=\lim_{n \to \infty}(n^3+2n^2)^{1/3}-n =\lim_{n \to \infty} n(1+2/n)^{1/3}-n$$
$$\implies L=\lim_{n \infty} n(1+2/(3n))-n=\lim_{n \to \infty} n+2/3-n=2/3$$
Her we have used binomial approximation that $(1+z)^k\to 1+kz$, if $|z|\ll 1.$

Title: Find the limit of $(n^3+2n^2)^{1/3}-n$

Body:
Find the limit of sequence.I tried and get this
$$
\lim_{n\to \infty} \frac{((n^3+2(n)^2)^{1/3}-n)\cdot((n^3+2 n ^2)^{1/3}+n)}{(2(n)^2+n^3)^{1/3}+n}
$$ I am getting limit 0 but this is wrong answer can you help?


问:
The graph of the function $y=\log_bx$ passes through the point $(729,6)$. Determine $b$.
The graph of the function $y=\log_bx$ passes through the point $(729,6)$. Determine $b$.

Could someone show me a solution that is similar to mine if it is correct?

$\log_b 729 = 6$

$729 = b^6$

$b = 729^{1/6} = 3$

答:
After solving $$b=729^{1/6}$$ you get $$b=\pm 3$$

but as $b>0$ , base log is always positive and not equal to 1 . So, we get only $b=3$ as solution,


问:
Property of continuous map from compact space onto Hausdorff space
I am self studying Topology, and I am having difficulty in proving the following statement.

Let $f$ be a continuous mapping defined on a compact space $X$ onto a Hausdorff space $Y$. Suppose $y\in Y$ and $U$ is an open subset of $X$ that contains $f^{-1}[y]$. Show that there exists an open neighborhood $V$ of $y$ such that $f^{-1}[V] \subset U$.

I know that

$Y=f[X]$ is compact, and thus, being Hausdorff, it is also normal;

$f$ is a closed map;

$f^{-1}[y]$ is closed and compact.

Can you give me some hints on how to use these facts (or other consequences of the hypothesis) to prove the theorem?

答:
Because $U$ is open $X\backslash U$ is closed and therefore compact. Then $f(X\backslash U)$ is closed because $Y$ is a Hausdorff space. Also $y\not\in f(X\backslash U)$ because $f^{-1}(y)\subset U\ \cap X\backslash U = \varnothing$.

Let denote $V=Y\backslash f(X\backslash U)$ it is open and $y\in V$. $$f^{-1}(V)=f^{-1}(Y)\backslash f^{-1}(f(X\backslash U)) = X\backslash f^{-1}(f(X\backslash U))=W$$

$W$ is open because $f$ is continuous and $f(X\backslash U)$ is closed.

$X\backslash U \subset f^{-1}(f(X\backslash U))$ and therefore $W = X\backslash f^{-1}(f(X\backslash U)) \subset X\backslash(X\backslash U) = U$

$W\subset U$ as we need.


问:
If $A\cap B= \emptyset \Rightarrow \bar{A}\cap\dot{B} = \emptyset$
$A\cap B= \emptyset \Rightarrow \bar{A}\cap\overset{\,\circ}{B} = \emptyset$

This should be a really simple question but I cannot seem to work it out. I suppose there is an element $x \in \bar{A}\cap\overset{\,\circ}{B}$ then it must also be $x \in \bar{A}$ and $x\in B$. So I now only have to show that it must be in $A$ also. I suppose it is not, then it must be in $\partial A$. But I cannot seem to arrive at a contradiction.

答: By definition of interior there must be an open set $U$ such that $x\in U\subseteq B$. Since $x\in\overline{A}$ we have $A\cap U\ne\emptyset$, i.e there is some element $y\in A\cap U$. But then $y\in A\cap B$ (since $U\subseteq B$), a contradiction.


问:
Solving a system of Diophantine equations
For a problem that I'm working on, I need to solve the following system of Diophantine equations:-
$a^3+40033=d$, $b^3+39312=d$, $ c^3+4104 = d$ where $a,b,c>0$ are all DISTINCT positive integers, and $a,b,c$ ∉ {$ 2, 9, 15, 16, 33, 34$}

How does one go about solving this? Is brute-force the only possible way? Or could there be a case that no integer solutions exist for this equation?

Also, are there any online computing engines, that allow me to set constraints, and solve Diophantine equations of this sort?

Any and all help is appreciated! Thanks!

答:
Here's a first step / hint to get you started.

Since the first two equations both equal $d$, we have that
$$
a^3 + 40033 = b^3 + 39312
$$

Subtract $a^3$ and $39312$ from both sides, and then factor each side.  What do you get?


问:
Solving $-x^3-2x^2-4x-5.4+K=0$
The equation is :
$$
    -x^3-2x^2-4x-5.4+K=0
$$
and I want my solutions in terms of $K$.

How do I simplify this polynomial?

答:
You face a cubic equation in $x$. So, just follow the steps given in the Wikipedia page.

The first is to compute
$$\Delta=-27 k^2+\frac{898 k}{5}-\frac{9363}{25}$$ which is always ngative since this quadratic in $k$ does not show real root.

So, use the hyperbolic method for one real root and get
$$x=-\frac{2}{3} \left(1+2 \sqrt{2} \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{449-135 k}{160 \sqrt{2}}\right)\right)\right)$$

Title: Solving $-x^3-2x^2-4x-5.4+K=0$


问:
Property of the $\zeta := \exp(2\pi i/k)$ function.
Working on something I arrived at a point where I can conclude if the following is true:

Let $\zeta := \exp(2\pi i/k)$ with a $k\ge 1$ integer (if it helps it can be $k\ge 2$), and let $\mu$, $\nu$ be integers such that $\mu \not =\nu$ (mod $k$); then
$$\sum_{l=0}^{k-1}\zeta^{(\mu - \nu)l} = 0$$

Observation: I think I found that if $\mu =\nu$ (mod $k$), then there exist $h\in \mathbb{Z}$ such that $\nu = \mu +hk$, so
$$\sum_{l=0}^{k-1}\zeta^{(\mu - \nu)l} = \sum_{l=0}^{k-1} \zeta^{-hkl}\zeta^{0} = \sum_{l=0}^{k-1} \mbox{ exp}(2\pi i)^{hl} = \sum_{l=0}^{k-1} (-1)^{2hl} = \sum_{l=0}^{k-1}1=k$$

Could somebody please help me with proving or disproving the claim? I really have been struggling with it for a while.

答:
Let $q=\zeta^{\mu-\nu}$. Since $\mu\not\equiv\nu\pmod k$ we have $q\ne1$, so we can apply the geometric sum formula to get:$$\sum_{l=0}^{k-1}q^l=\frac{q^k-1}{q-1}$$
Now we have $q^k=\zeta^{(\mu-\nu)k}=1$, hence the sum equals $0$.


问:
Proving the limit of a fraction at infinity by replacing the fraction by a larger fraction.
To prove $$\lim_{x \to -\infty} \frac{x^3}{5x^3 - 2x^2 - 4} = \frac{1}{5},$$ one can identify an $N$ such that $x < N$ implies that $$\left|\frac{x^3}{5x^3 - 2x^2 - 4} - \frac{1}{5}\right| < \epsilon$$ by working backwards. That is $$\left|\frac{x^3}{5x^3 - 2x^2 - 4} - \frac{1}{5}\right| = \left|\frac{2x^2 + 4}{5x^3 - 2x^2 - 4}\right|.$$ It would be nice to simplify this rather messy expression; something you can do as long as you do not introduce changes that prevent the final inequality from holding. In this case, the $4$ term in the numerator of $\frac{2x^2 + 4}{5x^3 - 2x^2 - 4}$ is an inconvenience, and it would be nice to remove it. Simply removing this positive term would make the absolute value of the fraction smaller when what is needed is to make the fraction larger. The same thing happens when the expression $-2x^2 - 4$ is removed from the denominator of our fraction. How should I replace this fraction with a simpler fraction that is clearly larger?

答: Hint: You want $|\frac {\frac 2 x+\frac4 {x^{3}}} {5-\frac 2 x+\frac 4 {x^{3}}}| <\epsilon$. Use the fact that $|5-\frac 2 x+\frac 4 {x^{3}}|>5-|\frac 2 x|-|\frac 4 {x^{3}}|$. I will leave the rest to you.


问:
Proof that golden angle successively divides the largest gap by the golden ratio?
The golden angle divides the circumference of a circle by the golden ratio.

"If radial spokes are placed successively into the circle, each spaced by a golden angle increment, then each additional spoke divides the currently largest existing azimuthal gap between two successive spokes."

I have read frequently about this property of the golden angle, and I can see this behaviour by plotting (or calculating) the first few spokes.

Does anybody know a proof for the above mentioned property of the golden angle (preferably citeable)?

Note:
The following answer proves that the golden angle is the irrational angle wich leads to spokes with a maximal distance to its neighbours, but it does not explicitly prove the above mentioned property: Does the golden angle produce maximally distant divisions of a circle?

答:
The property that the last point divides one of the largest gaps is true for all irrationals $\alpha$. Please refer to this paper by Tony Ravenstein, specifically proposition 4.2 and the first statement of its proof. (Also note that, the question's situation of dividing a circle's circumference is equivalent to dividing the interval [0,1] by fractional-part of multiples of $\alpha$).

Now, if $\alpha$ is the golden-ratio, it can be proved that the ratio of the new produced gaps is also in golden-ratio. In fact, all of the gaps (there are 3 distinct, due to Three-Gap Theorem) are always in golden-ratio. To prove this, we need to use the relation of the three gap lengths to the continued fraction of $\alpha$. For that, in addition to above paper, we may also refer this (theorem 3) and this (theorems 2 and 4; this one is written by me).

These references tell us about gap-lengths being related to leftmost/rightmost corner-points ordinal. These two corner points (their ordinal) themselves are related to the continued-fraction.

For golden-ratio, these corner points ordinal will come out to be some consecutive fibonacci-numbers (we need to work out from the details in these references). Once we have the gap-lengths expressed in terms of fibonacci-numbers, we can use fibonacci-number's properties to derive that their ratio is golden-ratio itself.


Title: Proving the limit of a fraction at infinity by replacing the fraction by a larger fraction.

Body: To prove $$\lim_{x \to -\infty} \frac{x^3}{5x^3 - 2x^2 - 4} = \frac{1}{5},$$ one can identify an $N$ such that $x < N$ implies that $$\left|\frac{x^3}{5x^3 - 2x^2 - 4} - \frac{1}{5}\right| < \epsilon$$ by working backwards. That is $$\left|\frac{x^3}{5x^3 - 2x^2 - 4} - \frac{1}{5}\right| = \left|\frac{2x^2 + 4}{5x^3 - 2x^2 - 4}\right|.$$ It would be nice to simplify this rather messy expression; something you can do as long as you do not introduce changes that prevent the final inequality from holding. In this case, the $4$ term in the numerator of $\frac{2x^2 + 4}{5x^3 - 2x^2 - 4}$ is an inconvenience, and it would be nice to remove it. Simply removing this positive term would make the absolute value of the fraction smaller when what is needed is to make the fraction larger. The same thing happens when the expression $-2x^2 - 4$ is removed from the denominator of our fraction. How should I replace this fraction with a simpler fraction that is clearly larger?

Body: Assume that $x<-2$, then $|2x^2+4|\le 2x^2+|x|^2= 3x^2$ and $|2x^2+4|\ge 2x^2-4\ge x^2$. Hence your fraction is less than $|3x^2/(5x^3+x^2)|=3/|5x+2|$ if $x<-2$.


问:
Show that $G=HN$ for $H$ subgroup and $N$ normal subgroup
I have a practice problem that states following:

Lad $G$ be a group and let N be a normal subgroup in $G$ with index $p$ which is a prime.

Let $H$ be a subgroup in $G$ which is not contained in $N$. Show that $G=HN$.

My answer:

I have for $x \in HN$ then clearly $x \in G$ and therefore $HN \subseteq G$

However for $x \in G$ I have $gNg^{-1} \in N$ and $gHg^{-1} \notin N$

But how do I continue from here to show $G=HN$?

答:
Hint: (For the case $G$ is finite.) What can be the order of the subgroup generated by $N\cup\{x\}$ where $x\in H\setminus N$?

(More generally.) Consider the quotient map $G\to G/N\cong\Bbb Z/p\Bbb Z$.


问:
Investigate if the following polynomial in Q[x] is reducible or irreducible: $x^3-5x^2+2x+1$

Investigate if the following polynomial in $\mathbb Q[x]$ is reducible or irreducible: $x^3-5x^2+2x+1$.

Attempt:

$x^3-5x^2+2x+1\bmod 2$ we get $x^3+x^2+1$.

We see that the polynomial is of degree 3, so it is reducible if it has a root in $\mathbb Q$, using rational root theorem in $\mathbb Q$, we see there is no root in $\mathbb Q$, so it's irreducible?

答: By the Rational Root Theorem the polynomial has no root over $\Bbb Q$ and hence is irreducible. This is probably the most direct way.


答:
As you have already calculated, the polynomial $\bmod 2$ is of the form $x^3+x^2+1$.

If this was reducible in $\mathbb{Q}$, it would also be reducible $\bmod 2$. As you have already noted if it is reducible, it must have a root in $\mathbb{Q}$. But more importantly, if it is reducible, it must also have a root $\bmod 2$. This is because the polynomial is of degree $3$.

So $\bmod 2$ we only have to try two values: $1$ and $0$. Both are no root, thus it is irreducible $\bmod 2$. Then from it follows that it's also irreducible in $\mathbb{Q}$.

Title: Investigate if the following polynomial in Q[x] is reducible or irreducible: $x^3-5x^2+2x+1$

Body:
Investigate if the following polynomial in $\mathbb Q[x]$ is reducible or irreducible: $x^3-5x^2+2x+1$.

Attempt:

$x^3-5x^2+2x+1\bmod 2$ we get $x^3+x^2+1$.

We see that the polynomial is of degree 3, so it is reducible if it has a root in $\mathbb Q$, using rational root theorem in $\mathbb Q$, we see there is no root in $\mathbb Q$, so it's irreducible?


答: First, since $N$ is a normal subgroup, we can conclude that $HN$ is a subgroup of $G$. Next, let $x \in {H - N}$. Then, we know that $x \in H \implies x \in HN$ but $x \not \in N$. Thus, $N < HN \leqslant G$. By Lagrange's theorem, the index of $HN$ in $G$ is a proper divisor of the index of $N$ in $G$, which is a prime $p$. Since the only proper divisor of a prime is $1$, the index of $HN$ in $G$ is $1$, and hence, $HN = G$.

Title: Show that $G=HN$ for $H$ subgroup and $N$ normal subgroup

Body:
I have a practice problem that states following:

Lad $G$ be a group and let N be a normal subgroup in $G$ with index $p$ which is a prime.

Let $H$ be a subgroup in $G$ which is not contained in $N$. Show that $G=HN$.

My answer:

I have for $x \in HN$ then clearly $x \in G$ and therefore $HN \subseteq G$

However for $x \in G$ I have $gNg^{-1} \in N$ and $gHg^{-1} \notin N$

But how do I continue from here to show $G=HN$?


答:
A reducible polynomial must be a product of two non-constant polynomials with lower degree.  In particular, a reducible cubic polynomial has a degree-one and a degree-two factor.  Having a degree-one factor is equivalent to having a root.
You've correctly applied the rational root theorem to conclude the absence of rational root for the cubic polynomial in the question, so it's irreducible over $\mathbb{Q}$.


问:
Is diffirentiability in finite dimensional space is equivalent to the existence of partial derivatives
Let $f$ be a continuous mapping, defined on $\mathbb{R}^2$. I want to prove that $f$ is differentiable at a point $(a,b)\in \mathbb{R}^2$ : Is it sufficient to prove that $f$ has first order derivative at $a$ and $b$ i.e $f_x(a,b)$ and $f_y(a,b)$ exists (since we have a finite dimensional space)?

Is there a reference book I could look at to find such result ?

答:
$f$ might even fail to be continuous at $(a,b)=(0,0)$ :
$$
f(x,y)=\left\{
\matrix{
0, & \text{ if } x=0 \text{ or } y=0,\hfill\cr
1, & \text{ if } x\neq0 \text{ and } y\neq0.
}
\right.
$$


问:
Approximating a Fourier transform
Suppose the Fourier transform $\hat{f}(k)$ (with $k \in \mathbb{R}^d$) is given, and one intends to get some information about its position-space counterpart $f(x)$.
When the analytical calculation of the inverse Fourier transform of $\hat{f}(k)$ is not possible, one may still be able to extract useful information by specializing to specific regions of $k$ space; for instance, in statistical physics, it is often customary to study the "macroscopic" properties of, e.g., correlation functions, by examining the $k\to 0$ limit of their Fourier transforms. It appears to me that such a process is somewhat analogous to looking at the Taylor series of a Fourier transform, i.e.,
\begin{equation}
\hat{f}(k) = \hat{f}\big\rvert_{k=0} + k \partial_k\hat{f}\big\rvert_{k=0} + \ldots
\end{equation}
If one truncates this series and then tries to perform on it the inverse Fourier transformation, $$
\int \frac{dk}{2\pi} e^{ikx} \hat{f}_{\rm trunc}(k),
$$
in some cases one might find that the result diverges as $k\to\infty$. However, in many theories, and especially in field theories, there is an upper cutoff for $k$ which determines the range of validity of that theory; such a cutoff often resolves the possible divergence of the inverse Fourier transform.

Question Does the position-space function that is obtained from the inverse transformation of the truncated Taylor series $\hat{f}_{\rm trunc}$, with some cutoff $\Lambda$, approximate the original function $f(x)$ in any sense? otherwise, is there a systematic way of obtaining such an approximate form from its Fourier transform $\hat{f}(k)$?

答: When you truncate the Taylor expansion around $0$, you are saying that you are interested in modes with long wavelength. These are often the modes that are long-lived, so that for long times they will approximately describe your system. In spirit, it is like doing a coarse graining: you forget about the fast microscopic dynamics and retain only macroscopoic information. In a more rigorous sense, one has $|| \mathcal{F}^{-1} [\hat f_{trunc}](x) - f(x) ||_2 = || \hat f_{trunc}(k) - \hat f (k) ||_2$, so if the approximation of your fourier transform is good in the $L^2$ sense so it will be the approximation of the position space $f(x)$.

Title: Approximating a Fourier transform


问:
Question on Summation of a equation
Find $$\sum_{x= 1/2}^{25 / 2} x^3 + 3x + 2.$$ Are there any simpler methods to solve this summation ? I will be thankful to any helps.

Counting is in the format of ${1 \over2}, 1, {3 \over2},....   $

答:
Put $\frac y2 = x$.  Then the lower limit, which was $x=\frac12$, becomes $\frac y2=\frac12$, so $y=1$, and the upper limit, which was $x=\frac{25}2$, becomes $\frac y2=\frac{25}2$, so $y=25$.  We now want to sum $y$ from 1 to 25:

$$\sum_{y=1}^{25}\left(\frac18y^3 + \frac32y + 2\right) \\
= \frac18\sum_{y=1}^{25} y^3 + \frac32\sum_{y=1}^{25} y + \sum_{y=1}^{25}2
$$

Do you know how to evaluate the three sums separately?

(The important thing to check here it that we didn't somehow mess up the sums by changing the index from $x$ to $y$.  There are 25 terms in the sum now, and there should have been 25 in the sum before.  You said that is correct ($\frac12, 1, \frac32,\ldots \frac{25}2$ is 25 terms) so this substitution is correct. If the sum before had had only 13 terms ($\frac12,\frac32,\ldots \frac{25}2$ is only 13 terms) then new summation would not be the same.)

Title: Question on Summation of a equation


问:
A question related to when is F Galois over K and when F is splitting field
This question was part of my abstract algebra assignment and I was unable to solve 1 part among it. So, I am posting it here as I need help.

Suppose [F : K] is finite. Then the following conditions are equivalent :
(i) F is Galois over K ;
(ii) F is separable over K and a splitting field of a polynomial f $\in$ K[x] ;
(iii) F is a splitting field over K of a polynomial $f \in  K[x]$ whose irreducible
factors are separable.

I am having problem only in the part in which I ahve to assume (iii) and prove (i) holds.

Can you please help with that? I have to prove that $Aut_K F = K $ itself. But i don't see any reason why it should hold.

Do you mind giving some hints.

答:
I assume your definition of 'Galois-extension' in the finite case is $F^{Aut_KF}=K$. Hints:

First conclude from the given assumptions in (iii) that $F/K$ is normal and separable.

Then show that in general for a normal extension $F/K$ we have that $F^{Aut_KF}$ is purely inseparable over $K$ by showing that there is only one $K$-homomorphism $F^{Aut_KF}\to C$ where $C$ is some algebraic closure of $F$.

Conclude under the additional assumption $F/K$ separable that $F^{Aut_KF}=F$ holds.


问:
How to understand the following formula
Known
$$D:\{x^2+y^2\leqslant1\}$$
and $$\iint\limits_D\left(x\frac{\partial f}{\partial x}+
y\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y=\int_0^1\mathrm{d}r\int\limits_{x^2+y^2 =
r^2}\left(x\frac{\partial f}{\partial x}+ y\frac{\partial
f}{\partial
y}\right)\mathrm{d}s$$
I want to know what kind of Gauss formula this belongs to. I've looked up a lot of data, but I can't find it.

答:
An intuitional idea is:

You can tile $D$ by rectangles, left hand side,

Or by rings, right hand sides,

Or by circular sector, if you change the order of $s \ \& \ r$.

But the details is provided by what @KaviRamaMurthy referenced in comment.

Title: How to understand the following formula


问:
Prove that Dirichlet function is non monotonic.
We have Dirichlet function which is $1$ if $x$ is in $\mathbb Q$ and $0$ if $x$ is in $\mathbb I=\mathbb R\setminus\mathbb Q$.We need to prove that function is not monotonic and is invertible.

For monotonic I tried to get two points $x_1$ from $\mathbb Q$, $x_2$ from $\mathbb I$, and then same for $x_1$ from $\mathbb Q$, $x_2$ from $\mathbb Q$ but I couldn't mathematically prove that it is not monotonic.

For function to be invertible needs to be $x_1\neq x_2 \implies f(x_1)\neq f(x_2)$ and $\forall y \exists x, f(x)=y$. But I doubt I did it correctly. Can you help?

答:
The Dirichlet function $f$ is not monotonic since $f(0) = 1$, $f(\pi) = 0, f(4) = 1$ where $0 < \pi< 4$.

To find an inverse of $f$, we must find a function $f_{1}$ such that $f_{1}(f(x)) = x$ for all $x$.


问:
Energy functional in Poisson's equation: what physical interpretation?
Let's consider this boundary-value problem:

$$\begin{cases} -\Delta V = \rho & \rm{in}\ \Omega
\\
V=0 & \rm{on}\ \partial \Omega \end{cases}.$$

We know that this problem has a variational formulation: its solutions are critical points of the energy functional

$$I[u]=\int_{\Omega} \left( \frac{1}{2}\lvert \nabla u \rvert^2 - u\rho \right)\, dx,$$

(cfr. Evans Partial differential equations 2nd ed., §2.2.5).

Now let us look at one (of the many) possible physical interpretations of this problem, the electrostatic one. If units are chosen so that $\varepsilon_0=1$, this equation describes the electric potential $V$ in a region $\Omega$ filled with a charge distribution $\rho$ and whose boundary is grounded.

It would be nice if the energy functional introduced above coincided with the total energy of this physical system. Unfortunately, it seems to me not to be so. In fact, as I read in Feynman's Lectures on physics, vol.II, §8-5, the total energy of this system consists of two additive parts: one is

$$\frac{1}{2}\int_{\Omega} \lvert \nabla V \rvert^2\, dx$$

and is due to the electric field; the other is

$$\frac{1}{2}\int_{\Omega} V\rho\, dx$$

and is due to the charge distribution.

This induces me to think that the
  energy functional should be

$$U[u]=\int_{\Omega} \left(\frac{1}{2}\lvert \nabla u \rvert^2 + \frac{1}{2}u\rho \right)\, dx,$$
  instead of the correct
  $$I[u]=\int_{\Omega} \left( \frac{1}{2}\lvert \nabla u \rvert^2 - u\rho \right)\, dx.$$
  Why am I wrong?

答:
The energy functional has to agree with the total energy and there can't be any ambiguity or "freedom of conventions" that would add the factors of two, among other things. After all, the total energy/mass is a subject to conservation laws and it determines both inertial of the system as well as the strength of its gravitational field.

If you integrated Feynman's two $\int E^2/2$ and $\int \rho V/2$ contributions over the whole 3-space, you would surely be double-counting the energy. The correct energy is just one of them, as given by the Evans formula. In the localized case, the two expressions may be shown to be equal.

You may also use Feynman's formula but you must be careful to use the $\int E^2/2$ exclusively for the part of the electric fields that are "unrelated" to localized charged sources.


答:
Observe that $f(0) = 1$ and $f(1) = 1$, but $f(1/\sqrt2) = 0$ and $0 < 1/\sqrt2 < 1$.

In fact, the range of $f$ is $\{0, 1\}$, with $f^{-1}(\{0\}) = \mathbb{R}\setminus\mathbb{Q}$ and $f^{-1}(\{1\}) = \mathbb{Q}$.  Since both pre-images are not a singleton, so it would be impossible to invert $f$.

Title: Prove that Dirichlet function is non monotonic.

Body:
We have Dirichlet function which is $1$ if $x$ is in $\mathbb Q$ and $0$ if $x$ is in $\mathbb I=\mathbb R\setminus\mathbb Q$.We need to prove that function is not monotonic and is invertible.

For monotonic I tried to get two points $x_1$ from $\mathbb Q$, $x_2$ from $\mathbb I$, and then same for $x_1$ from $\mathbb Q$, $x_2$ from $\mathbb Q$ but I couldn't mathematically prove that it is not monotonic.

For function to be invertible needs to be $x_1\neq x_2 \implies f(x_1)\neq f(x_2)$ and $\forall y \exists x, f(x)=y$. But I doubt I did it correctly. Can you help?


问:
$ G/H \cong K $ and $ G/K \cong H\implies G \cong H \times K $
Let $ (G, *) $ be a group.

Say, $ H, K \lhd G $ such that $ G/H \cong K $ and $ G/K \cong H $

Example:$\ G = Z_{4}, \ H = \{ 0, 2\}, \ K = \{ 0, 2\}  \ $ satisfies the above condition, but $ G \not\cong \{(0,0),(0,2),(2,0)(2,2) \} = H \times K$,

Here we observe that $H \cap K \not= \{(0,0)\}$

Example: $ \ G = GL(n,R), H = SL(n,R), K = Z(G)$ also satisfies the above condition, and $G = H \times K$,

Here we observe that $H \cap K = \{I_{n \times n}\}$

So, intutively I can feel that
$ \ \ G \cong H \times K \iff H \cap K = \{ e \} $

Can anyone prove or counter it?

(In case the statement is wrong, please provide an iff condition for $ G \cong H \times K$ where $ H, K \lhd G $ such that $ G/H \cong K $ and $ G/K \cong H $)

答: In case $G$ is finite it is true when $H \cap K=1$: from $G/H \cong K$ it follows that $|G|=|H||K|=|HK||H \cap K|$, whence $G=HK$. From this it follows that $G \cong H \times K$.


问:
Is every graph "planar" on some geometric smooth object?
I don't know much about geometry and topology so forgive me for asking this like that, but is there a theorem that says that every simple graph is pLaNaR (clearly every graph is not planar, I dont't know what to call the property that I am thinking of, so I called it pLaNaR)?

And by pLaNaR I mean something that relates to planarity on a geometric object's surface.

For example if you consider a torus and you discretize it (create a mesh) say you're a computational person and you do so numerical calculations on a torus, you can apply the finite difference or finite element method. The graph resulting from this discretization is by definition pLaNaR with respect to the torus.

Hopefully this gives you an idea of what I mean.

So my question is: is every graph pLaNaR for some smooth geometric object?

Any kind of insight and reference is welcome.

答:
Every time you add an edge $vw$, you can add an extra handle to your surface to make that edge easy to draw without crossing other edges. Delete a disk from the surface somewhere close to $v$, another disk somewhere close to $w$, and join the two holes with a cylinder.

This is incredibly wasteful, but shows that any $n$-vertex graph can be drawn on a surface with genus $\binom n2$: a torus with $\binom n2$ holes. (If we're more careful, then even the complete graph can be drawn on a surface with genus $\lceil \frac16 \binom{n-3}{2}\rceil$, as the link to Wolfram in the comments points out.)


问:
Leaves of a tree made up of internal nodes with either 3 or 2 children
I'm currently struggling with solving the following problem: given a tree T made up of n internal nodes, where n-k nodes have 2 children and k nodes have 3 children, demonstrate that T has n+k+1 leaves.

Can anyone help me or point me to the right direction? Thank you in advance!

答: Try thinking of the tree as a family tree, and what a node with 2 children does to the number of nodes in the next generation... What does a node with 3 children do? Hope this helps</p>


问:
$|x^{k}\varphi^{(q)}(x)| \leqslant C_{q}A^{k}$ if $|x|>A$
given

$$|x^{k}\varphi^{(q)}(x)| \leqslant C_{q}A^{k}k^{k\alpha}$$

I am told in a textbook that if the infinitely differentiable function $\varphi(x)$ vanishes for $|x|>A$ then

$$|x^{k}\varphi^{(q)}(x)| \leqslant C_{q}A^{k}$$

I dont really see how this is the case though. Could anyone explain?

答:
Outside the support of $\phi$ , $\phi^{(q)}=0$ also.  So for $|x|>A$, $x^k \phi^{(q)}(x) = 0$. For $|x|\le A$,
$$ |x^k \phi^{(q)}(x)| \le A^k \| \phi^{(q)}\|_{L^\infty(B(0,A))}. $$
taking $C_q = \| \phi^{(q)}\|_{L^\infty(B(0,A))} < \infty$  gives the result.


问:
Efficiently invert threediagonal symmetric matrix
I need to invert a threediagonal symmetric $1000\times10000$ matrix (or of similar dimensions). Specifically I need to solve the following equation:
$$Ax=b$$
with
$$A=\begin{pmatrix}v_1&-\frac{dt}{2}&0&\cdots\\-\frac{dt}{2}&v_2&-\frac{dt}{2}&0&\dots\\0&-\frac{dt}{2}&v_3&-\frac{dt}{2}&0&\dots\\\vdots&0&\ddots&\ddots&\ddots\end{pmatrix}$$
To be even more precise:
$$A=\mathbf1-\frac{dt}{2}\cdot(D2+V)\qquad b=(\mathbf1 + \frac{dt}{2} * (D2 + V)) * \psi$$
with
$$D2=\begin{pmatrix}-2&1&0&\cdots\\1&-2&1&0&\cdots\\0&1&\ddots&\ddots&0&\cdots\\\vdots&\vdots&\ddots&\vdots&\vdots&\ddots\end{pmatrix}\qquad V=\begin{pmatrix}V_1&0&\cdots\\0&V_2&0&\cdots\\\vdots&\ddots&\ddots&\ddots\end{pmatrix}$$
If you're curious, I need this for solving the one dimensional Schrödinger equation via the Crank-Nicolson iteration.

Does an efficient algorithm exist for solving such a system (or for inverting the matrix)? So far the best I found is the Gaussian elimination, which works in $\mathcal O(n^3)$ - but this would yield a billion iteration steps.

答:
Just too long for a comment.

You don't need (and don't have to) invert a matrix: just solve the linear system. That said, for tridiagonal matrices you can use Thomas Algorithm, which is just Gaussian Elimination applied to a tridiagonal system. It's composed by two parts: the forward sweep and the backward one and it's easy to implement. You can find all the relevant information in the Wikipedia page