问:
What is the sign of the integral $\int_{0}^{2\pi}e^{\sin(x)}\cos(nx)\,dx$?
Let $I_n = \int_{0}^{2\pi}e^{\sin(x)}\cos(nx)\,dx$ for a natural $n$.
I would like to prove that for $n$ odd this integral vanishes, while for $n = 4k$ for some natural $k$, the integral is always positive and for $n = 4k+2$ it is always negative.

I was able to do a few integrations by parts, and provided I didn't make mistakes, the result was the following rather ugly recurrence relation, valid for $n > 1$: $$I_{n+2} = -\frac1{n-1}\left[(4n^3-2n)I_n+(n+1)I_{n-2}\right].$$ Also we have $I_1 = I_3 = 0$ (the first is obvious from the integral, the second follows by integration by parts). And from here, clearly it is immediate to see that it vanishes for every $n$ odd, just by induction. However I'm not able to see why $I_{4k}$ is always positive or $I_{4k+2}$ is always negative from this relation. What I know is that $|I_n| \le I_0 \approxeq 7.95$.

I also tried to rewrite the integral using $e^x = 1 + x + \ldots$, moving the integral inside the summation, but still I couldn't come up with a solution. What do you suggest?

Note: this integral comes from the calculation of the Fourier coefficients of $e^{\sin(x)}$, and analogous properties hold for the coefficients of the sine terms. Also by a well-known theorem (Riemann-Lebesgue), $I_n$ goes to zero as $n\to\infty$, and hence it's not possible to find a fixed constant lower bound for $|I_n|$, making any estimation in the recurrence relation more difficult.

答:
Using the results listed here we can show that $I_n=2\pi i^n{\cal I}_n(1)$ where ${\cal I}_n$ is the modified Bessel function of the first kind.

Here we will perform the derivation from scratch. Applying the unit circle transformation we have \begin{align}I_n&=\Re\int_0^{2\pi}e^{\sin x}e^{inx}\,dx=\Re\oint_{|z|=1}z^n\exp\left(\frac{z-1/z}{2i}\right)\,\frac{dz}{iz}\end{align} which we can expand as \begin{align}I_n&=\Re\oint_{|z|=1}\sum_{r=0}^\infty\left(\frac i2\right)^r\frac{z^{n-r-1}(1-z^2)^r}{i\cdot r!}\,dz=\Re\left[2\pi\operatorname{Res}(f,0)\right]\end{align} where $$f(z)=\sum_{r=0}^\infty\left(\frac i2\right)^r\frac{z^{n-r-1}(1-z^2)^r}{r!}.$$ To evaluate this residue we seek the term independent of $z$ in $zf(z)$. This occurs when the power of $1-z^2$ is $(r-n)/2$ which means that $r=2s$ is even. Hence \begin{align}\operatorname{Res}(f,0)&=\sum_{s=n/2}^\infty\left(\frac i2\right)^{2m}\frac{(-1)^{(2s-n)/2}}{(2s)!}\binom{2s}{(2s-n)/2}\\&=\sum_{s=n/2}^\infty\frac{(-1)^s}{2^{2s}}\frac{(-1)^{s-n/2}}{(s-n/2)!(s+n/2)!}.\end{align} Substituting back into the integral gives  $$I_n=\sum_{s=n/2}^\infty\frac{(-1)^{n/2}\pi}{2^{2s-1}(s-n/2)!(s+n/2)!}=\sum_{t=0}^\infty\frac{(-1)^{n/2}\pi}{2^{2t+n-1}t!(t+n)!}=2\pi i^n{\cal I}_n(1)$$ and we can immediately conclude that $I_{4k}>0$ and $I_{4k+2}<0$.


问:
How to prove that $50^{44}+30! \cdot44^{50}-24 $ is divisible by $31$?
I am struggling with this example. This is a new topic for me. I would really appreciate some hints to understand it. I need part a) and b) to conclude the excercise.

I was thinking about Fermat, Euler or Wilson theorems but I am not sure if it is correct, of course I can use properties of congruences in number theory.

Part a): show that $50^{44}+30! \cdot 44^{50}-24 $ is divisible by $31$

Part b) : Conclude that $ (50^{44}+30!\cdot 44^{50})^{\varphi(31)} \equiv 1 \pmod{ 31}$

ContentLicense: CC BY-SA 4.0

Body:
\begin{align}
50^{44} + 30! \cdot 44^{50} &\equiv 50^{44} - 44^{50} \pmod{31}, \text{ By Wilson's Theorem } \\
&\equiv 2^{44} \cdot 5^{88} - 2^{100}\cdot 11^{50} \pmod{31} \\
&\equiv 2^{14}\cdot 5^{28} - 2^{10} \cdot 11^{20} \pmod{31}\text{, By Fermat's little Theorem } \\
&\equiv 2^{4} \cdot 5 - (-3)^{10} \pmod{31}, \\
&\equiv 2^4 \cdot 5 - 3^{10} \pmod{31} \\
&\equiv 2^4  \cdot 5 - 3^{3\times 3}\cdot 3 \pmod{31} \\
&\equiv 2^4 \cdot 5 - (-4)^3 \cdot 3 \pmod{31} \\
&\equiv 2^4 (5+12) \pmod{31} \\
&\equiv 2^4 (2^4+1) \pmod{31}
\\ &\equiv2^8  + 2^4 \pmod{31} \\
&\equiv 2^3 + 2^4 \pmod{31} \\
&\equiv 24 \pmod{31}
\end{align}


问:
Mathematical equivalent of Feynman's Lectures on Physics?
I'm slowly reading through Feynman's Lectures on Physics and I find myself wondering, is there an analogous book (or books) for math?  By this, I mean a good approach to mathematics given through sweeping motions, appeals to intuition and an emphasis on modern perspectives over historical development.

What strikes me about Lectures on Physics is its conceptual orientation. It is easy to read in that I don't need a pen and paper at hand to work through details, but I also find it necessary to regularly stop and contemplate his ideas and examples. Content-wise, it is not at the level of a calculus book. My impression is that it covers much of what would constitute an undergraduate education in physics. Of course, finding a single book to cover an entire undergraduate mathematics curriculum is likely impossible!  I'm looking for the same style, not necessarily the comprehensiveness.

答:
I.M Gelfand's books
on trigonometry,algebra,functions and graphs and calculus of variations(and much more) are comparable to Feynman Lectures. He has even stated his effort to write a book like feynman's in the book's preface. I strongly recommend the books. You can search the books in amazon for user reviews.

ContentLicense: CC BY-SA 3.0


问:
What function will be able to describe the distance between all perpendicular rays leaving a circle and hitting a parabola?
If I have a circle and parabola at the origin (very general)
$$
\text{p}\left(x\right):=a x^2 \\
\text{c}\left(x\right):=r-\sqrt{r^2-x^2}+b
$$

I want to know how I can apply Calculus techniques to determine the distance of all rays extended from the circle when they hit the parabola under it.

Will I be able to formulate a function that can describe what I am trying to explain?

Plot of what I mean:

Plot of what I am trying to explain

答:
Ok so I actually got the answer, it wasn't too bad, but please comment if you think I can "optimise" anything.

So firstly, any ray from the centre of the circle will be perpendicular to the circle. The tangent of the circle will always be:

$$\frac{x_1}{\sqrt{r^2-x_1^2}}$$
where $x_1$ is the x-value along the circle.

The line that is perpendicular to that tangent will be in the form:
$$ y=-\frac{\sqrt{r^2-x_1^2}}{x_1} x+c$$
Using the center of the circle as the "common point" (since we want a perpendicular ray) to solve for $c$ gives the following:

$$\text{d}\left(x\right)=-\frac{\sqrt{r^2-x_1^2}}{x_1} x+r+b$$
Note: the center of the circle is just $y=r + b$.

Then to find where the line intersects the parabola, just set both equations equal to each other.
$$
\text{p}\left(x\right)=\text{d}\left(x\right) \\
a x^2 = -\frac{\sqrt{r^2-x_1^2}}{x_1} x+r+b \\
\text{Solve polynomial (use positive value)} \\
x_2=\frac{-\sqrt{r^2-x_1^2}+\sqrt{4 a b x_1^2+4 a r x_1^2+r^2-x_1^2}}{2 a x_1}
$$
$x_2$ is the parabola intercept.

Now use the arc length formula and obtain equation in terms of $x_1$:

Using $x$ instead of $x_1$, we get:
$$\text{D}\left(x\right)=-\frac{\frac{r}{x}\left(2ax^{2}+\sqrt{r^{2}-x^{2}}-\sqrt{r^{2}+x^{2}\left(4a\left(b+r\right)-1\right)}\right)}{2ax}$$

Graph of the distance between the two curves at each x-value


问:
Quadratic surfaces and intersection curves
You are given a second order surface $Ax^2+By^2+Cz^2=α$ and a plane $x+z=β$. Determine the type of the surface and the type of the curve at the intersection of the surface and the plane.

Basically, I need to write a program, that will do task, getting as parameters $A, B, C, \alpha,\beta$.
So, the best way I found to make this, is to just check every equation by coefficients signs to determine quadratic surface. But what about curves of intersection? Is there a simple way to determine their type only by given parameters?

答:
Regarding the existence of intersections. Making the substitution of $x = \beta-z$ into $A x^2+B y^2+C z^2=\alpha$ we get

$$
B y^2 + C z^2+(\beta-z)^2 = \alpha
$$

so

$$
y = \pm\frac{\sqrt{\alpha-A (\beta -z)^2-C z^2}}{\sqrt{B}}
$$

and the intersection exists if $\Delta = \alpha-A (\beta -z)^2-C z^2\ge 0$ assuming $B\gt 0$

Now

$$
\Delta = \frac{(A+C) \alpha -A C  \beta ^2}{A+C}-(A+C) \left(z-\frac{A \beta }{A+C}\right)^2 \ge 0
$$

and assuming for instance that $A+C>0$ if

$$
(A+C) \alpha \lt A C  \beta ^2
$$

no intersection.


问:
Rankine-Hugoniot jump condition linear system
Ladies, gentlemen, I need your help with this one

Consider the linear system of equations

$u_t+Au_x=0$

with $u=(u_1,...,u_n)^T$ and $A\in n\times n$ matrix with constant entries.

Derive the Rankine-Hugoniot jump conditions for this system of linear hyperbolic equations

$A(u_l-u_r) =s(u_l-u_r) $

and show that $u_l-u_r$ is the eigenvector of $A$ and $s$ the associated eigenvalue.

I've tried to diag $A$ and then to decouple the system, but this doesn't work.

答:
Sure diagonalization works! Upon diagonalization, the proof is exactly the same as in the scalar case. The equality $A[\![u]\!]=s[\![u]\!]$ defines an eigenvalue problem, where $[\![u]\!]=u_l-u_r$ is the jump in $u$. Therefore, let us introduce the following eigendecomposition of the matrix $A$: $$
A = P \Lambda P^{-1} ,
$$
where the matrix $\Lambda = \text{diag}(\lambda_1 , \dots , \lambda_n)$ is diagonal and presumably real. Then, introducing $v = P^{-1} u$, we have:
$$
v_t + \Lambda v_x = 0 ,
$$
which is a decoupled system of $n$ linear advection equations $v_{i,t} + \lambda_i v_{i,x} = 0$. The Rankine-Hugoniot condition implies $\lambda_i [\![v_i]\!] = s[\![v_i]\!]$ component-wise, i.e. $\Lambda [\![v]\!] = s[\![v]\!]$. Then, upon left-multiplication of the previous equation by $P$, we get the desired result $$A [\![u]\!] = s[\![u]\!] \, ,$$
where we have used $[\![v]\!] = P^{-1} [\![u]\!]$.


问:
What is the intuitionist / contructivist view of the Fermat theorem proof?
Since Wiles proof is in essence proof by contradiction, it relies on the law of excluded middle. Which as I understand intuitionists / constructivists do not accept as an axiom. So what is their view of the Wiles proof? Do they still consider Fermat theorem not proved?</p>

答:
Wiles' proof is not, in essence, a proof by contradiction.

Wiles has shown that every elliptic curve over $\mathbb{Q}$ (well, a large enough subset of them) satisfies the following theorem. At the time, this was a conjecture, and a big one, and Wiles had begun his work on FLT after a connection was established between FLT and this conjecture.

A positive integer solution to FLT could be manipulated to construct an elliptic curve which isn't modular, but this kind of connection is only an indication that the modularity conjecture would imply FLT. So FLT would not be proven by assuming it's true and seeing it's impossible, but rather it was already established that if FLT is true, then something really strange happens, and Wiles had undertaken to prove that it was impossible.

Wiles, in fact, uses the word contradiction only 3 times throughout his amazing paper "Modular elliptic curves and Fermat's last theorem.". FLT isn't even stated directly after modularity is proven, since the connection was already known,

As for the proof itself, it was accepted as correct (not without some obstacles), and still is accepted, and will remain so. Wiles was awarded, among many other honors, a special plaque from the ICM (I heard someone refer to it as a "quantized" Fields medal), and in 2016 he was awarded the Abel prize.


问:
Existence of $\xi$ s.t. $f'''\left(\xi\right)=\frac{3f''\left(\xi\right)}{1-\xi}$

Given $f(x)\in C^3$, and $f(0)=f(1/2)=f(1)$. Prove that there exists at least one point $\xi$, such that $f'''\left(\xi\right)=\dfrac{3f''\left(\xi\right)}{1-\xi}$.

I tried to apply Mean Value Theorem to $f(x)$, but anyway I can't construct the denominator $1-\xi$. Can anyone help?

答:
by Rolles theorem there exist some $c_1,c_2$ such that $f'(c_1)=f'(c_2)=0$  where $c_1\in(0,1/2),c_2\in(1/2,1)$again by Rolle's there is some $c_3$ such that $f''(c_3)=0$, $c_3\in(c_1,c_2)$.

Now consider $$h(x)={(1-x)}^3 f''(x)$$ $$h(1)=h(c_3)=0$$ Then by Rolle's theorem there is some $\eta$: $$h'(\eta)=0\Rightarrow f'''(\eta)=\frac{3f''(\eta)}{1-\eta}$$


问:
Finding coordinates of maximum with unknown powers $f(x)=(x^m)*((1-x)^n)$ on a line segment [0;1]
So I need to find the coordinates of the maximum of a function on a line segment.So I approached this by taking the derivative and equated it with $0$. But due to $m$ and $n$ which are unknown, how do I find coordinates of the maximum?

答:
I suppose $m,n$ are positive integers.

$f'(x)=m x^{m-1} (1-x)^n-n x^m (1-x)^{n-1}=x^{m-1} (1-x)^{n-1} [m (1-x)-n x]$

$$f'(x)=0\to x=0;\;x=1;\;m (1-x)-n x=0$$

$$m-mx-nx=0;\;(m+n)x=m\to x=\frac{m}{m+n}$$

$f'(x)=0\to x_1=0;\;x_2=1;\;x_3=\frac{m}{m+n}$

$$f''(x)=x^{m-2} (1-x)^{n-2} \left(m^2 (x-1)^2+m (x-1) ((2 n-1) x+1)+(n-1) n x^2\right)$$
Thus we are not sure if $x_1$ and $x_2$ are minima, maxima or inflection points. It depends on the actual values of $m$ and $n$.

$x_3\in[0,1]$ because $0\le \frac{m}{m+n}\le 1$. If this happens, then $x_3$ is a point of relative maximum for $f(x)$ because
$$f''(x_3)=-\frac{m^2 \left(\frac{m}{m+n}\right)^{m-3} \left(\frac{n}{m+n}\right)^n}{n}<0$$


答:
Notice we can use Fermat's little theorem to reduce the exponents using mod order reduction, based on these values:
$\  \color{#0a0}{2^5\equiv 1},\,$ $\, \color{#c00}{5^3\equiv 1},\,$ $\, \color{#90f}{11^3}\equiv 11(-3) \equiv \color{#90f}{-2},\,$ yielding

$$\begin{align}
   \bmod 31\!:\quad\     & 50^{44}\!+30! \cdot 44^{50}\\[,2em]
\equiv\ & 50^{14}\ \ \ \, -\,\ \ \ 44^{20}\qquad\ \ {\rm by}\ \ 30!\equiv -1\,\ {\rm by\ Wilson}\\[,2em]
\equiv\ &\color{#0a0}{2^{14}} \cdot \color{#c00}{5^{28}}-\color{#0a0}{2^{40}} \cdot \color{#90f}{11^{20}}\\[.3em]
\equiv\ &\ \ \ \ \ \frac{\color{#c00}{5}}{\color{#0a0}2}\,\ -\,\ \color{#0a0}1\cdot \color{#90f}{\frac{(-2)^{7}}{11} \ \ \ \text{by (*) below}}\\[.3em]
\equiv\ &\ \ \ \ \frac{\color{}{36}}{2}\ \ \ \ \ +\ \ \ \ \ \frac{\color{#90f}{2^{2}}}{11}\quad\ \ {\rm by}\ \ \color{#0a0}{2^5\equiv 1}\\[.3em]
\equiv\ &\ \ \ \ 18\ \ +\ \ \frac{\color{}{4+62}}{11}\\
\equiv\ &\ \ \ \ 24
\end{align}\!\!\!$$

$\color{#90f}{\text{(*)}}\,$ we used $\ \color{#90f}{11^3 = -2}\ \, \smash{\overset{\Large x^{7}\!}\Longrightarrow}\ 11^{21} \equiv (-2)^7\!\Rightarrow \color{#90f}{11^{20}\equiv (-2)^7/11},\ $ then in the next two lines we twiddle the fraction numerators by adding (or subtracting) small multiples of the modulus so to make the quotient exact.

Part b) is just FLT again.

Check here to see how modular fractions work.

Title: How to prove that $50^{44}+30! \cdot44^{50}-24 $ is divisible by $31$?

Body:
I am struggling with this example. This is a new topic for me. I would really appreciate some hints to understand it. I need part a) and b) to conclude the excercise.

I was thinking about Fermat, Euler or Wilson theorems but I am not sure if it is correct, of course I can use properties of congruences in number theory.

Part a): show that $50^{44}+30! \cdot 44^{50}-24 $ is divisible by $31$

Part b) : Conclude that $ (50^{44}+30!\cdot 44^{50})^{\varphi(31)} \equiv 1 \pmod{ 31}$


问:
Is the following map continuous?
Let $C(X,\mathbb{R})$ and $C(Y,\mathbb{R})$ be the space of continuous functions with $X$ is compact metric space and $Y$ be closed subset of $X$. The spaces are using the supremum norm. Is the map $R:C(X)\to C(Y)$ with $Rf=f|_Y$ a continuous map?
If it is how do I show it, thank you.

答:
If $f$ and $g$ are in $C(X,\mathbb{R})$ then
$$
  \|R(f)-R(g)\| = \sup_{y\in Y}|f(y)-g(y)| \leq $$$$\leq \sup_{x\in X}|f(x)-g(x)| =   \|f-g\|.
  $$
This shows that $R$ is Lipschitz, and hence continuous.

Title: Is the following map continuous?


问:
Determine whether $ [ (p \implies q) \land ( q \implies r)] \implies r $ logically implies the proposition $ (p \implies r) \implies r$ is True
Determine whether the proposition $ [ (p \implies   q) \land ( q \implies  r)] \implies r $ logically implies the proposition $ (p \implies r) \implies r.  $

Note that  $ [ (p \implies   q) \land ( q \implies  r)] \implies (p \implies r) $ is a  tautology, but this statement has some subtle difference.

Is there some laws/rules that may be of help here?

(It is indeed False, if we draw a Truth Table to verify)

答: The implication holds when $p \rightarrow r$ also implies $(p \rightarrow q) \land (q \rightarrow r)$. That is, $p \rightarrow r$ must imply $p \rightarrow q$. But as you know it is generally not true because we don't even know what $q$ will be. you can replace any statements only when they are equivalent.


答:
Assume that $(p \to r) \to r$ is False.

This means $r$ False and $p$ False.

Now consider the premise: $[(p \to q) ∧ (q \to r)] \to r$.

With the above values we have  $[(\bot \to q) ∧ (q \to \bot)] \to \bot$, that means: $[\top ∧ (q \to \bot)] \to \bot$.

If we choose True for $q$ we have that the premise will be: $[\top ∧ (\top \to \bot)] \to \bot$ that evaluates to True.

Conclusion: we have found that the truth assignment of False, True, False to $p,q,r$ respectively satisfies the premise and falsifies the conclusion, i.e. the conclusion is not logically implied by the premis.

Title: Determine whether $ [ (p \implies q) \land ( q \implies r)] \implies r $ logically implies the proposition $ (p \implies r) \implies r$ is True

Body:
Determine whether the proposition $ [ (p \implies   q) \land ( q \implies  r)] \implies r $ logically implies the proposition $ (p \implies r) \implies r.  $

Note that  $ [ (p \implies   q) \land ( q \implies  r)] \implies (p \implies r) $ is a  tautology, but this statement has some subtle difference.

Is there some laws/rules that may be of help here?

(It is indeed False, if we draw a Truth Table to verify)


问:
Linear combination of vectors in $\mathbb{R}^3$
Show that any linear combination of $\pmatrix{1\\\frac{3}{2}\\0}$ and $\pmatrix{0\\3\\6}$ is also a linear combination of $\pmatrix{2\\3\\0}$ and $\pmatrix{0\\1\\2}$

I'm not sure how to do this. I have a proof sketch for showing that any 2-dimensional vector is a linear combination of any other two nonparallel 2d vectors, but as far as I understand, this is not the case for 3d vectors. So how should I go about showing this?

Thanks.

答: Suppose $x=\alpha u_1 + \beta u_2$.  Then $x=(\frac{\alpha}{2})u_1'+(3\beta)u_2'$, where $u_1, u_2$ are your first two vectors and $u_1', u_2'$ are your second two vectors.</p>


问:
Evaluate $\int_{1}^{e} x^{n} \ln x \mathrm{d} x$
Evaluate the integral
$$
I=\int_{1}^{e} x^{n} \ln x \mathrm{~d} x, n \in \mathbb{Z}
$$

My Attempt: Substitute $\ln x$ by $t$, then $x=e^t$, and so we need to evaluate the following integral:
$$
I=\int_0^1 e^{tn} t\,\mathrm d (e^t)
$$
Integrate by parts, we get
$$
I=t^2e^{t\left( n+1 \right)}\mid_{0}^{1}-\int_0^1{t\left( e^{t\left( n+1 \right)}+t\left( n+1 \right) e^{t\left( n+1 \right)} \right) \text{d}t}
$$
Thus it seems to be more complicated!

So, my question is, how to evaluate this integral?

答: Call that integral $I_n$ instead so$$e^n=[x^n\ln x]_1^e=\int_1^e(x^{n-1}+nx^{n-1}\ln x)dx=\frac{e^n-1}{n}+nI_{n-1}$$and $I_{n-1}=\frac{(n-1)e^n+1}{n^2}$, i.e.$I_n=\frac{ne^{n+1}+1}{(n+1)^2}$. The only case this doesn't deal with is$$I_{-1}=\int_1^e\frac{\ln x}{x}dx=[\tfrac12\ln^2x]_1^2=\tfrac12.$$


答:
If $n \ne -1$, we use integration by parts:
$$\begin{align}\int x^{n} \ln x \mathrm{~d} x &= \frac{1}{n+1}x^{n+1}\ln x - \frac{1}{n+1}\int x^{n+1}\frac 1 xdx \\&= \frac{1}{n+1}x^{n+1}\ln x - \frac{1}{(n+1)^2}x^{n+1} + C \end{align}$$
Else, we use the substitution $u = \ln x$:
$$\int \frac{\ln x}{x}dx = \frac{1}{2}\ln^2 x + C$$

Title: Evaluate $\int_{1}^{e} x^{n} \ln x \mathrm{d} x$

Body:
Evaluate the integral
$$
I=\int_{1}^{e} x^{n} \ln x \mathrm{~d} x, n \in \mathbb{Z}
$$

My Attempt: Substitute $\ln x$ by $t$, then $x=e^t$, and so we need to evaluate the following integral:
$$
I=\int_0^1 e^{tn} t\,\mathrm d (e^t)
$$
Integrate by parts, we get
$$
I=t^2e^{t\left( n+1 \right)}\mid_{0}^{1}-\int_0^1{t\left( e^{t\left( n+1 \right)}+t\left( n+1 \right) e^{t\left( n+1 \right)} \right) \text{d}t}
$$
Thus it seems to be more complicated!

So, my question is, how to evaluate this integral?


问:
To find the power of the test against hypothesis test

This problem is all about probability density function and hypothesis which I am familiar with. But the problem is that I can't figure out how to approach the problem. A rigorous proof is needed. Thank you.

答:
In a very rigorous way it is enough  to apply the definitions.

(a) Find $c$ such that

$$P[X>c |\mu=0]=0.05$$

(b) Fixed $c$ found in (a), calculate

$$P[X>c|\mu=2]$$

These are the rigorous definitions of "size" and "power" . All you have to do is to do the calculations


问:
What is the real part of $(1-i)^{-i}$?

What is the real part of $(1-i)^{-i}$?

I am stuck in this question and not able to solve. Please help me. I am only able to reach to this conclusion.

The real part of complex number $a+ib$ is denoted by Re(z)=a and it's imaginary part Im(z) is b. There is no information (in my book) for $a^b$ where b is complex

$$(1-i)^{-i}=(e^{\frac{-i\pi}{4}})^{-i} \times (\sqrt{2})^{-i}$$

Disclaimer : I am a high school student.

答: Since $1-i=e^z$ iff $z=\tfrac12\ln2-\frac{\pi i}{4}+2n\pi i$ for some $n\in\Bbb Z$, there are multiple definitions of $(1-i)^{-i}$, namely those of the form $\exp(-\pi/4+2n\pi-i\tfrac12\ln 2)$, with respective real parts $e^{-\pi/4+2n\pi}\cos(\tfrac12\ln 2)$.

Title: What is the real part of $(1-i)^{-i}$?


Body:
If $n\ne -1$, simply use integration by part,
$$\begin{align}
\int_1^ex^n\ln(x)\;dx &= \frac{1}{n+1}\int_1^e\ln(x)\;dx^{n+1} \\
&= \frac{1}{n+1}\ln(x)x^{n+1}\bigg|_1^e-\frac{1}{n+1}\int_1^ex^{n+1}\;d\ln(x) \\
&= \frac{e^{n+1}}{n+1}-\frac{1}{n+1}\int_1^ex^n\;dx \\
&= \frac{e^{n+1}}{n+1}-\frac{e^{n+1}}{(n+1)^2}+\frac{1}{(n+1)^2} \\
&=\frac{ne^{n+1}+1}{(n+1)^2}
\end{align}$$
If $n=-1$, then
$$\int_1^e\frac{\ln(x)}{x}\;dx=\frac{1}{2}\ln^2(x)\bigg|_1^e=\frac{1}{2}$$
Actually, by L'Hopital's rule,
$$\lim_{n\to -1}\frac{ne^{n+1}+1}{(n+1)^2}=\frac{1}{2}$$
so the formula
$$\int_1^ex^n\ln(x)\;dx=\frac{ne^{n+1}+1}{(n+1)^2}$$
is valid even if $n=-1$.


答:
WARNING: I'm going to cheat and assume the "principal branch."

Rewrite $1 - i$ in polar form as $\sqrt{2}e^{-i\pi/4}$.

Then $(1 - i)^{-i} = (\sqrt{2}e^{-i\pi/4})^{-i} = \sqrt{2}^{-i}\left(e^{-i\pi/4}\right)^{-i}$.

This breaks down into two terms $\sqrt{2}^{-i}$ and $\left(e^{-i\pi/4}\right)^{-i}$.

For the first term rewrite $\sqrt{2}$ as $e^{\ln \sqrt{2}} = e^{(\ln 2)/2}$ so $\sqrt{2}^{-i} = \left(e^{(\ln 2)/2}\right)^{-i} = e^{-((\ln 2)/2)i}$ which is a number on the unit circle.

For the second term $\left(e^{-i\pi/4}\right)^{-i} = e^{(-i\pi/4)(-i)} = e^{-\pi/4} \approx 2.19$ which is a real number.

It is at this point that the cheat kicks in. We chose to write $1 - i$ with $\theta = -\pi/4$ but really we could have chosen $\theta = 7\pi/4$ or in general $\theta = 2 \pi n - (\pi/4)$. If we had done so we would have gotten $e^{i(2 \pi n - (\pi / 4))(-i)} = e^{2 \pi n - (\pi / 4)}$ for this term. The "principal branch" choice is that unique value of $\theta$ which satisfies $-\pi \lt \theta \leq \pi$, which in our case is $\theta = -\pi/4$.

Putting the two together we have $e^{-\pi/4} e^{-((\ln 2)/2)i}$. So in polar form we have $r = e^{-\pi/4}$ and $\theta = -(\ln 2)/2$.

The real part of this number is $r \cos \theta = e^{-\pi/4} \cos \left(-(\ln 2)/2\right) \approx 0.4288$.

Title: What is the real part of $(1-i)^{-i}$?

Body:
What is the real part of $(1-i)^{-i}$?

I am stuck in this question and not able to solve. Please help me. I am only able to reach to this conclusion.

The real part of complex number $a+ib$ is denoted by Re(z)=a and it's imaginary part Im(z) is b. There is no information (in my book) for $a^b$ where b is complex

$$(1-i)^{-i}=(e^{\frac{-i\pi}{4}})^{-i} \times (\sqrt{2})^{-i}$$

Disclaimer : I am a high school student.


答:
A linear combination of $\pmatrix{1\\\frac{3}{2}\\0}$ and $\pmatrix{0\\3\\6}$ is a vector that can be expressed as
$$a\pmatrix{1\\\tfrac{3}{2}\\0}+b\pmatrix{0\\3\\6}$$
for some real numbers $a$ and $b$.

A linear combination of $\pmatrix{2\\3\\0}$ and $\pmatrix{0\\1\\2}$ is a vector that can be expressed as
$$c\pmatrix{2\\3\\0}+d\pmatrix{0\\1\\2}$$
for some real numbers $c$ and $d$.

But
$$2\pmatrix{1\\\tfrac{3}{2}\\0}=\pmatrix{2\\3\\0}\qquad \tfrac{1}{3}\pmatrix{0\\3\\6}=\pmatrix{0\\1\\2}.$$

Title: Linear combination of vectors in $\mathbb{R}^3$

Body:
Show that any linear combination of $\pmatrix{1\\\frac{3}{2}\\0}$ and $\pmatrix{0\\3\\6}$ is also a linear combination of $\pmatrix{2\\3\\0}$ and $\pmatrix{0\\1\\2}$

I'm not sure how to do this. I have a proof sketch for showing that any 2-dimensional vector is a linear combination of any other two nonparallel 2d vectors, but as far as I understand, this is not the case for 3d vectors. So how should I go about showing this?

Thanks.


问:
Necessary condition for selecting a New coordinate system (as tangent and normal vectors)
I wanted to understand intuitvily, how it is possible to make tangent and normal vectors as a coordinate system for a planar motion because my concern is that what I have read about coordinate system is that once it is fixed along a particular direction after that those directions will remain same regardless of  curve for example Cartesian coordinate

but

My concern with tangent and normal as coordinate system is that it keep changing for every new curve because tangent and normal for every other curve will be somewhat different which is quite unintuitive to what I have read earlier about coordinate system .

Can anyone clear my doubt ,what is basic criteria for selecting a coordinate system ?

答:
```<code>my concern is that what I have read about coordinate system is that once it is
fixed along a particular direction after that those directions will remain
same regardless of curve for example Cartesian coordinate
</code>```

That paragraph is where things go wrong, or at least where your confusion is revealed.

Let's think about a curve $\gamma$ in the plane. We have, for each $t$, the point
$$
\gamma(t) = (x(t), y(t))
$$

For instance, if $x(t) =  t$ and $y(t) = t^2$, the curve described by $\gamma$ is a parabola. It's made up of many different points, one for each possible $t$-value. The fact that $\gamma(0) = (0,0)$ doesn't prevent $\gamma(2)$ from being $(2, 4)$, right? If we think of $t$ as "time", then you could say that we have a "continuously changing point" moving along the curve, but in reality, we have many many points, each attached to a particular time-value. $\gamma(0) = (0,0)$, and it never changes, but $\gamma(t)$, as $t$ varies, does change.

The same thing is true for coordinate systems: the cartesian coordinate system is fixed once and for all. For our curve, the tangent vector at time $0$ is $[1, 0]$, and the normal is $[0, 1]$. That tangent-normal pair is fixed for all time. But at time $t = 1$, the tangent is $[1, 2]$ and the normal is something like $[-2c, 1c]$ for some value of $c$ that I'm not going to compute. We can use either of those to measure things like "the position of $\gamma(2)$".

The clever idea of the "moving frame" is that in measuring things associated with $\gamma(t)$, for some particular $t$, it may be useful not to use $T(0), N(0)$, or $T(1), N(1)$, but instead to express things in the coordinate system $T(t), N(t)$.

We do this all the time in real life. Imaging you're driving a car in the US; your friend Robin is in the passenger seat to the right of you. Your child is in a car-seat right behind you. You start out driving North, so Robin is to the East of you, but soon the road curves and you're driving East, and Robin is to the south of you. But in your mind, we ignore the global N/S/E/W coordinates, and you say "the whole time I was driving, Robin was to my right, and my child was behind me". In this case, you're using, at each moment, a coordinate system based on the car's orientate at that moment, i.e., a constantly moving frame-of-reference.

Body:
my concern is that what I have read about coordinate system is that once it is
fixed along a particular direction after that those directions will remain
same regardless of curve for example Cartesian coordinate

That paragraph is where things go wrong, or at least where your confusion is revealed.

Let's think about a curve $\gamma$ in the plane. We have, for each $t$, the point
$$
\gamma(t) = (x(t), y(t))
$$

For instance, if $x(t) =  t$ and $y(t) = t^2$, the curve described by $\gamma$ is a parabola. It's made up of many different points, one for each possible $t$-value. The fact that $\gamma(0) = (0,0)$ doesn't prevent $\gamma(2)$ from being $(2, 4)$, right? If we think of $t$ as "time", then you could say that we have a "continuously changing point" moving along the curve, but in reality, we have many many points, each attached to a particular time-value. $\gamma(0) = (0,0)$, and it never changes, but $\gamma(t)$, as $t$ varies, does change.

The same thing is true for coordinate systems: the cartesian coordinate system is fixed once and for all. For our curve, the tangent vector at time $0$ is $[1, 0]$, and the normal is $[0, 1]$. That tangent-normal pair is fixed for all time. But at time $t = 1$, the tangent is $[1, 2]$ and the normal is something like $[-2c, 1c]$ for some value of $c$ that I'm not going to compute. We can use either of those to measure things like "the position of $\gamma(2)$".

The clever idea of the "moving frame" is that in measuring things associated with $\gamma(t)$, for some particular $t$, it may be useful not to use $T(0), N(0)$, or $T(1), N(1)$, but instead to express things in the coordinate system $T(t), N(t)$.

We do this all the time in real life. Imaging you're driving a car in the US; your friend Robin is in the passenger seat to the right of you. Your child is in a car-seat right behind you. You start out driving North, so Robin is to the East of you, but soon the road curves and you're driving East, and Robin is to the south of you. But in your mind, we ignore the global N/S/E/W coordinates, and you say "the whole time I was driving, Robin was to my right, and my child was behind me". In this case, you're using, at each moment, a coordinate system based on the car's orientate at that moment, i.e., a constantly moving frame-of-reference.


问:
How to find the image of a linear map?
Lets say that $f:\mathbb{R}^2 \to \mathbb{R}^2`$ is a linear map with

$$ f(1,1)=(2,3) \\ f(0,1)=(1,-1) $$

How can we find $\text{Im}(f)$ ? What are the steps that we must follow?

答:
I'll try to expose you a general method to do the exercise.

The problem gives you two vectors $u,v \in \mathbb{R^2}$ and their images $f(u)$ and $f(v)$.

You know that we can take the canonical basis $\mathcal C=\{(1,0),(0,1)\}=\{e_1,e_2\}$ to write vectors in $\mathbb R^2$.

Note that $u=(1,1)=e_1+e_2$ and $v=(0,1)=e_2$ and since the map $f$ is linear we can see that $f(u)=f(e_1+e_2)=f(e_1)+f(e_2)=(2,3)$ and $f(v)=f(e_2)=(1,-1)\implies f(e_1)=(2,3)-(1,-1)=(1,4)$.

We can now write the representative matrix of the map $f$ respect to the canonical basis:

$$\mathcal M_{\mathcal C}^{\mathcal C}(f)=\begin{pmatrix}
1&&1\\
4&&-1
\end{pmatrix}$$
The vectors $(1,4)$ and $(1,-1)$ are linearly independent so we can conclude that $\text{dim}(f)=2$ and $Im(f)=\langle f(e_1),f(e_2)\rangle=\langle (1,4),(1,-1)\rangle$.

Title: How to find the image of a linear map?


问:
Getting from $A$ to $B$ (both constants) given $n$ iterations of $A*=x$. What's the ratio of $n/x$? How to express algorithmically?
For context, I created code of the form:

```<code>A = 0.125
n = 20
x = 1.1
for (int i = 0; i < 20; i++) {
  A = A * x
}
B = A
</code>```

But I don't necessarily always want to get to this particular $B$ value (0.8409378, unless I miscalculated) via 20 iterations. I may want to get to B via 30 iterations, with $x$ naturally needing modified. But I don't know how to modify $x$ given different values of $n$ such that I get from $A$ to $B$ every time.

How do I algorithmically find this relationship of $x:n$? What is the term for this kind of algorithm in mathematics?

答:
Solving the equation @the_candyman derived, yields:

$$B=Ax^n$$
$$\sqrt[n]{B}=\sqrt[n]{A}x$$
$$x=\sqrt[n]{\frac BA}$$

So as long as $A$, $B$ and $n$ are known, you can figure out what x is :)


问:
Variation of Lighthill-Whitham-Richards Traffic PDE: Finding Boundary Characteristics/Shocks
Reading a bit about  Lighthill-Whitham-Richards (LWR) Traffic PDEs, and I've realize I don't quite understand how to introduce shocks when a PDE has both an initial and boundary condition.

I propose a simple example:

Example. Define the following LWR with initial and boundary conditions
$$
\begin{align}
&u_t + 2uu_x = 0 \\
&u(x,0) = 1,\quad x > 0\\
&u(0,t) = 2,\quad 0 < t < 1\\
&u(0,t) = 1,\quad t > 1
\end{align}
$$

What I understand: We use the Method of Characteristics to solve, i.e.

$$\frac{dt}{1} = \frac{dx}{2u} = \frac{du}{0}.$$

From this it is easy to see that we get the implicit solution, $u(x,t) = F(x-tu)$. For the initial condition, we have characteristic curve, $x = tF(s) + s$. For $s > 0$, we get that $F(s) = 1$, and so

$$x = t + s \quad\Rightarrow\quad s = x - t \quad\Rightarrow\quad u(x,t) = 1,~ x > t.$$

Issue: So, now we look at the BP. Then using the same characteristic, $x = tF(s) + s$, we get
$$
\begin{align}
0 < t < 1 &\quad\Rightarrow\quad F(s) = 2 &\quad\Rightarrow\quad s = x - 2t &\quad\Rightarrow\quad u(x,t) = 2,~2t < x < 1 + 2t. \\
t > 1 &\quad\Rightarrow\quad F(s) = 1 &\quad\Rightarrow\quad s = x - t &\quad\Rightarrow\quad u(x,t) = 1,~x > 1 + t.
\end{align}
$$

I don't known how to visualize these in the $x-t$ plane (below line $x = t$), so that I can introduce shocks (two of them I think?). My picture so far would be:

答:
One needs to be careful when applying the initial/boundary conditions for the Lagrange-Charpit system $$
\frac{dt}{1} = \frac{dx}{2u} = \frac{du}{0} ,
$$
which gives $u = C_1$ and $x = 2ut+C_2$. The previous equations yield
$$u = C_1 = F(C_2) = F(x - 2u t) $$
or equivalently
$$u = C_1 = G\big({-\tfrac{C_2}{2C_1}}\big) = G(t-x/(2u)) \, . $$
We may rewrite the equation of characteristic curves starting from the initial condition as $x = 2u t + x_0$, along which $u = F(x_0)$ is uniformly equal to the value $u(x_0,0)$. The equation of characteristic curves starting from the boundary condition reads $t = x/(2u) + t_0$ where $u = G(t_0)$ is uniformly equal to the value $u(0,t_0)$.
Here is what the curves look like:

To determine the shock and rarefaction waves, you will need to use the conservative form obtained by substituting $2uu_x = (u^2)_x$. Note that the Lighthill-Witham-Richards traffic flow model may be recovered if we set $u = \frac12 - \rho$ where $0\leq \rho\leq 1$ is the car density, so that $$
u_t + 2u u_x = -[\rho_t + (1 - 2\rho) \rho_x] = 0 \, .
$$
Hence, the proposed values of $u$ in the problem statement aren't representative of a physical configuration.


问:
Which one can be a square number?
Which one can be a square number?

$1)\overline{ab\cdots18}$

$2)\overline{ab\cdots42}$

$3)\overline{ab\cdots46}$

$4)\overline{ab\cdots89}$

To solve this question I considered $17^2=289$ and last two digits of this perfect square is $89$ therefor I conclude the answer is $\overline{ab\cdots89}$. but I couldn't find a mathematically proof for that (I just tried a number and find out last digits can be $89$) . is there other approach to solve the qustion?

Title: Which one can be a square number?

Body:
Yes, there is a way. If you find the last two digits of every square from $0$ to $50$, you will find this pattern: $$00, 01, 04, 09, 16, 25, 36, 49, 64, 81, 00, 21, 44, 69, 96, 25, 56, 89, 24, 61, 00, 41, 84, 29, 76, 25, 76, 29, 84, 41, 00, 61, 24, 89, 56, 25, 96, 69, 44, 21, 00, 81, 64, 49, 36, 25, 16, 09, 04, 01, 00$$

This pattern of last two digits repeats every $50$ numbers. $89$ is your only two digits that appears in this set, so $89$ is the answer.


问:
Inverse of a function with absolute value function
For $\lambda > 0$ and $y \in \mathbb{R}$:

$$
f(y) = \frac{|y|^\lambda * sign(y) - 1}{\lambda}.
$$

What is the inverse function of $f$.

答:
A function $h$ has an inverse iff $h$ is bijective.

The domain of each function can be restricted so that a bijective restriction results.

You should split your functions into bijective restrictions.

The following holds for the partial inverses $f^{-1}$ of $f$:

$$f^{-1}(y)=\cases{?&$ \left( 1+{\it \lambda~}\,y \right) ^{{{\it \lambda}}^{-1}}<0$\cr 0&$ \left( 1+{\it \lambda}\,y \right) ^{{{\it \lambda}}^{-1}}=0$\cr \left( 1+{\it \lambda}\,y \right) ^{{{\it \lambda~}}^{-1}},- \left( 1+{\it \lambda}\,y \right) ^{{{\it \lambda}}^{-1}}&$\left( 1+{\it \lambda}\,y \right) ^{{{\it \lambda~}}^{-1}}>0$\cr}$$

Check the domains of the functions in the bottom line of the curly bracket.


问:
Is there a way to give a ring structure on the group of symmetric matrices?
The set of $n\times n$ symmetric matrices with entries in $\mathbb R$ is an abelian group under matrix addition. However, it is not closed under matrix multiplication so one can't give it a ring structure using matrix multiplication.

Is there any other way to give it a non commutative ring structure?

I tried entry-wise multiplication (Hadamard Product) and also $X\cdot Y:=XY+YX$ but both of these are commutative operations and not very interesting to me.

答:
One example I could think of is the following:
Define all operations just on the upper triangular part of the symmetric matrix and then just "define" the lower triangular part by mirroring the upper triangular part. This uses the idea that all information about a symmetric matrix is contained in its upper triangle</p>


答:
(a) $~~$At first, we need to find the distribution of $X = \max(0 , Z)$ . Clearly, if $x < 0$ , then $F_X(x) = 0$ , and if $x \geqslant 0$ , $F_X(x) = F_Z(x)$ .

Therefore, given that $\mu = 0$ , for $x \geqslant 0$ , $$F_X(x) = \int_{-\infty}^x \frac{1}{2}e^{-\vert z \vert} ~dz = \frac{1}{2} + \int_0^x \frac{1}{2}e^{-\vert z \vert}~dz = 1 - \frac{1}{2} e^{-x}$$ And, for $x < 0$ , we can see that $F_X(x) = 0 $ very clearly.

Since this is a one-tailed test, thus : $$F_X(c) = 0.95\quad\iff\quad e^{-c} = 0.1 \quad \iff\quad \boxed{~c = \log_e 10~}$$

(b) $~~$ Once (a) is done, (b) is straightforward. Power of the test will be : $$P(X > c~\vert~ \mu = 2) = F_X(c - 2) = 1 - \frac{1}{2} e^{-(\log_e 10 - 2)} = \boxed{~0.6305}$$

Hope this helps. Let me know if my answer is erroneous.

Title: To find the power of the test against hypothesis test

Body: This problem is all about probability density function and hypothesis which I am familiar with. But the problem is that I can't figure out how to approach the problem. A rigorous proof is needed. Thank you.


问:
Area between curves involving $e$
Find the area of the region bounded by the curve $ y= e^x$, the coordinate axes, and the line $x=2$

Can you tell me what is wrong with my solution?

$$\int_0^2 e^x \,dx =7.38\dots - 1 = 6.38\dots$$

Title: Area between curves involving $e$

Body:
You are right
$$A=\int_{0}^{2} e^{x} dx=e^2-1.$$


答:
When $n\ge2$, let $N=\frac12n(n+1)\,(\ge3)$ and let $P\subset\mathbb R[x]$ be the set of all polynomials of degree $\le N-2$. For any $p=\sum_{k=0}^{N-2}a_kx^k$ and $q=\sum_{k=0}^{N-2}b_kx^k$ in $P$, define $p\cdot q=a_0q=\sum_{k=0}^{N-2}a_0b_kx^k$. With this multiplication and the usual polynomial addition, $P$ becomes a non-commutative and non-unital ring ($1$ is not a unit element because $p\cdot1\ne p$ in general), and we may extend $P$ to a unital ring $P\times\mathbb R$ by defining an addition and a multiplication in a way akin to a Dorroh extension:
\begin{aligned}
(p,a)+(q,b)&=(p+q,a+b),\\
(p,a)\cdot(q,b)&=(p\cdot q+bp+aq,ab)
\end{aligned}
Now we can we pick a basis $\{S_0,S_1,\ldots,S_{N-1}\}$ of the vector space of all $n\times n$ real symmetric matrices and identify each symmetric matrix $\sum_{k=0}^{N-1}a_kS_k$ with the element $\left(\sum_{k=0}^{N-2}a_kx^k,\,a_{N-1}\right)$ in $P\times\mathbb R$.

Title: Is there a way to give a ring structure on the group of symmetric matrices?

Body:
The set of $n\times n$ symmetric matrices with entries in $\mathbb R$ is an abelian group under matrix addition. However, it is not closed under matrix multiplication so one can't give it a ring structure using matrix multiplication.

Is there any other way to give it a non commutative ring structure?

I tried entry-wise multiplication (Hadamard Product) and also $X\cdot Y:=XY+YX$ but both of these are commutative operations and not very interesting to me.


问:
Proving a statement using the definition of a limit
Let $f$ be a function that is defined on a deleted neighbourhood of $x_0 ∈ \Bbb R$,
such that $f(x) \ne 0$ for every $x$ on that deleted neighbourhood. Suppose that
$$\lim _{x\to x_0} f(x) = \lim_{x\to x_0} (f(x) + 1/|f(x)|) = 0 $$

(a) Prove that there exists a $δ > 0$ such that for every $0 < |x − x_0| < δ$, we have $f(x) < 0$.

(b) Set $g(x) = f(x)+ 1/|f(x)|$
for every $0 < |x−x_0| < δ$. Find an expression
for $f(x)$ as a function of $g(x)$ for every $0 < |x − x_0| < δ$.

(c) Conclude that $\lim_{ x→x_0}f(x) = −1$.

I'm having trouble with proving a, we are given an hint to prove first that  for every $y > 0$,  $y +
1/y ≥ 2$,
and I did, but I've been trying to manipulate the inequation in the definition of the limit without success.
How should I approach this?

EDIT -

Okay so I ended up proving the hint, manipulating it to look like this

if y>0 then y+1/|y|≥2 for every y in R

and then I used the contrapositive of this expression, which I found to be

if y+1/|y|<2 then 0≥y for every y in R.

and I used the contrapositive in the definition of the limit, chose epsilon = 2 and showed f(x)<0 in that deleted neighbourhood.

My question now is if my manipulation of the proof of the hint, and the contrapositive is legal..

答: If $f(x)>0$ for all $\delta$ such that $|x-x_0|<\delta$, you get $f(x)+\frac{1}{|f(x)|}=f(x)+\frac{1}{f(x)}\geq2$, and that is a contradiction, because $\lim\limits_{x\to x_0}f(x)+\frac{1}{f(x)}=0$. Then $f(x)<0$ for some $|x-x_0|<\delta$.


问:
Condition to union of connected sets be connected
Let $X$ and $Y$ be connected spaces in $\mathbb{R}^n$. If $\partial X \subset Y$ then $X \cup Y$ is connected.

$\partial X$ is the boundary of X.

i'm trying to prove this by showing somehow that $X \cap Y$ is non-empty and concluding that $X \cup Y$ is connected but i didn't have sucess

Any help would be appreciated!

答:
This result appears to hold in considerably more general topological settings. Consider a connected topological space $(X, \mathscr{T})$ and two subsets $S, T \subseteq X$ which form connected subspaces and such that $\partial S \subseteq T$. Then $S\cup T$ is also a connected subset of $X$.

We recall that in a connected space the only subsets of empty frontier are the extremal ones, and in either of the two cases $S=\varnothing$ or $S=X$ the claim is clearly true. Let us therefore assume that $\partial S \neq \varnothing$. In order to prove that $S \cup T$ is connected we will apply the following criterion: for any pair $(F, G)$ of separated subsets $F, G \subseteq X$ such that $S \cup T \subseteq F \cup G$ we have either $S \cup T \subseteq F$ or $S \cup T \subseteq G$. Recall that the property of $(F, G)$ being separated simply means that $\overline{F} \cap G=F \cap \overline{G}=\varnothing$.

Let us show that the conditions of the criterion are indeed met: for arbitrary separated pair $(F, G)$ such that $S \cup T \subseteq F \cup G$, since in particular $S, T \subseteq F \cup G$ and $S$, $T$ are both connected by hypothesis it follows that $(S \subseteq F \vee S \subseteq G) \wedge (T \subseteq F \vee T \subseteq G)$. Let us show that the "hybrid" situations $S \subseteq F \wedge T \subseteq G$ and the analogous one are impossible: assuming indeed that $S \subseteq F$ and $T \subseteq G$, we would infer that $\partial S \subseteq \overline{S} \subseteq \overline{F}$ and similarly $\partial S \subseteq T \subseteq G$, whence the contradiction $\partial S \subseteq \overline{F} \cap G=\varnothing$. The case $S \subseteq G \wedge T \subseteq F$ is ruled out by a similar reasoning and we are thus left with either $S, T \subseteq F$ or $S, T \subseteq G$ which entail either $S \cup T \subseteq F$ or $S \cup T \subseteq G$.

The criterion thus applies with the conclusion that $S \cup T$ is indeed connected.

As a final remark, the reasoning above can be adapted to establish the following related:

Proposition. Let $(X, \mathscr{T})$ be an arbitrary topological space and $S, T \subseteq X$ be two connected subsets which are not separated, in other words either $\overline{S} \cap T \neq \varnothing$ or $S \cap \overline{T}\neq \varnothing$. Then the union $S \cup T$ is also connected.

From this more general proposition it is easy to infer the following, as @mathcounterexamples.net points out (I am not however convinced by his argument for this claim but he is welcome to clarify the matter):

Corollary. Let $(X, \mathscr{T})$ be a topological space and $S, T \subseteq X$ two connected subsets such that $\partial S \cap T \neq \varnothing$. Then $S \cup T$ is also connected.

The corollary follows swiftly since $\partial S \cap T \subseteq \overline{S} \cap T$ for every subsets $S, T \subseteq X$.


答:
The hypothesis $\partial X \subseteq Y$ is even too strong. $\partial X \cap Y \neq \emptyset $ is sufficient. To prove it, suppose that $O_1, O_2 \subseteq \mathbb R^n$ are two open subsets such that $X \cup Y \subseteq O_1 \cup O_2$.

Without loss of generality, we can suppose that $X \subseteq O_1$ as $X$ is supposed to be connected. Let $a \in \partial X \cap Y$. By definition of the boundary $\partial X$, $a$ belongs to $\overline{X}$. If $a$ also belongs to $O_2$, then $O_2 \cap X \neq \emptyset$ in contradiction with $O_1 \cap O_2 \cap X= \emptyset$ as $X$ is supposed to be connected. Therefore $a \in Y \cap O_1$ and $Y \subseteq O_1$. And we get the desired conclusion.

Finally note that your willingness to prove that $X \cap Y$ is non-empty cannot work. Take for example $\mathbb R^2$, $X$ the open unit disk and $Y = \mathbb R^2 \setminus X$. $X,Y, X\cup Y$ are all connected however $X \cap Y = \emptyset$.

Title: Condition to union of connected sets be connected

Body:
Let $X$ and $Y$ be connected spaces in $\mathbb{R}^n$. If $\partial X \subset Y$ then $X \cup Y$ is connected.

$\partial X$ is the boundary of X.

i'm trying to prove this by showing somehow that $X \cap Y$ is non-empty and concluding that $X \cup Y$ is connected but i didn't have sucess

Any help would be appreciated!


问:
tower of separable extension
Let $K⊂L⊂M$ be a tower of fields.
Let $L/K$ and $M/L$ be separable, is it true that $M/K$ is separable?

I guess there are counterexamples, but I cannot point out them.
Thank you for your kind help.[I know if we substitute separable to normal, there are easy example.]

答: By the primitive element theorem, $L=K(\alpha),M=L(\beta)=K(\alpha)(\beta)$ for some $\alpha\in L,\beta \in M$. Then the minimal polynomial $m_{\beta, L} \mid m_{\beta, K}$, whence $m_{\beta, L}$ is separable. Suppose $m_{\beta, K}$ is inseparable, i.e. the characteristic of $K$ is $p\gt 0$ and $m_{\beta, K}=f(x^p)$ for some irreducible polynomial $f(x)$. Then $m_{\beta, L}=g(x^p)$ for some irreducible polynomial $g(x)$ and the characteristic of $L=K(\alpha)$ is also p, which implies that $m_{\beta, L}$ is inseparable, a contradiction. So $\beta$ is separable over $K$ and thus $M/K$ is separable.

Title: tower of separable extension


答:
Note that $dx^2\wedge dx^3\wedge dx^1 = - dx^2\wedge dx^1\wedge dx^3 = dx^1\wedge dx^2\wedge dx^3$. In general, if $\eta$ is a $k$-form and $\zeta$ is an $l$-form, then $\eta\wedge\zeta = (-1)^{kl}\zeta\wedge\eta$.

So the correct answer is $\alpha\wedge\beta = (a_1b_1 + a_2b_2 + a_3b_3)dx^1\wedge dx^2\wedge dx^3$.

Title: about Wedge product

Body:
Let $\alpha$ be a 1 -form and $\beta$ a 2 -form on $\mathbb{R}^{3}$. Then
$$
\begin{array}{l}
\alpha=a_{1} d x^{1}+a_{2} d x^{2}+a_{3} d x^{3} \\
\beta=b_{1} d x^{2} \wedge d x^{3}+b_{2} d x^{3} \wedge d x^{1}+b_{3} d x^{1} \wedge d x^{2}
\end{array}
$$
Compute $\alpha \wedge \beta$.

solution : $\alpha \wedge \beta=a_{1} d x^{1}\wedge(b_{1} d x^{2} \wedge d x^{3}+b_{2} d x^{3} \wedge d x^{1}+b_{3} d x^{1} \wedge d x^{2})+a_{2} d x^{2}\wedge(b_{1} d x^{2} \wedge d x^{3}+b_{2} d x^{3} \wedge d x^{1}+b_{3} d x^{1} \wedge d x^{2})+a_{3} d x^{3}\wedge(b_{1} d x^{2} \wedge d x^{3}+b_{2} d x^{3} \wedge d x^{1}+b_{3} d x^{1} \wedge d x^{2})=a_{1}b_{1} d x^{1}\wedge d x^{2} \wedge d x^{3}- a_{2}b_{2} d x^{1}\wedge d x^{2} \wedge d x^{3}+a_{3}b_{3} d x^{1}\wedge d x^{2} \wedge d x^{3}=(a_{1}b_{1}-a_{2}b_{2}+a_{3}b_{3})d x^{1}\wedge d x^{2} \wedge d x^{3}$

is this true ?

i think we have :

$a_{2}b_{2} d x^{2}\wedge d x^{3} \wedge d x^{1}= a_{2}b_{2} (d x^{2}\wedge d x^{3} )\wedge d x^{1}=-a_{2}b_{2} d x^{1}\wedge (d x^{2} \wedge d x^{3})$

is this true ?


问:
The Volterra-Lotka model (predator prey equations) -- Linearization ODE

Q: Find all fixed points of the equation, linearize the equation, substitute the origin point $(0, 0)$ into it and solve the linear version of Volterra-Lotka model.

The system looks like this (where $a,b,c,g,y,x_{0}$ are constants):
\begin{align}
\frac{dx}{dt}&=ax-gx^2-by\left(x-x_0\right)\\
\frac{dy}{dt}&=-cy+dy\left(x-x_0\right)
\end{align}

My take:

Critical point: $$
\begin{pmatrix}0\\ 0\end{pmatrix}
$$

Jacobian Matrix:
$$
\begin{pmatrix}a&bx_0\\ 0&-c-dx_0\end{pmatrix}
$$

$$
P(\lambda) = -(a-\lambda)(c+dx_0 + \lambda) =0  \implies \begin{cases}
\lambda_1 = a \\
\lambda_2 = -c-dx_0
\end{cases}
$$

I get two matrices:

For V1:
$$
\begin{pmatrix}0&bx_0\\ 0&-c-dx_0-a\end{pmatrix}
$$
For V2:
$$
\begin{pmatrix}a+c+dx_{0}&bx_{0}\\ \:\:0&0\end{pmatrix}
$$

Eigenvector 1 :
$$
\begin{pmatrix}1\\ 0\end{pmatrix}
$$
Eigenvector 2 :
$$
\begin{pmatrix}-bx_0\\ a+c+dx_0\end{pmatrix}
$$

So the solution is :
$$
C_1\cdot e^{at}\begin{pmatrix}0\\ 1\end{pmatrix}\:+\:C_2\cdot e^{-t\left(c+dx_0\right)}\cdot \begin{pmatrix}-bx_0\\ a+c+dx_0\end{pmatrix}
$$

I don't think that that's the solution - maybe I have an error with Eigenvector 2 ?
We didn't really went over this material so I'm having a hard time.

答:
I made a few edits to your question for things that looked like typos, because they didn't show up in your final answer. In particular, you had
$$\begin{pmatrix}-\color{red}{3}x_0\\ a+c+dx_0\end{pmatrix}$$
instead of
$$\begin{pmatrix}-\color{red}{b}x_0\\ a+c+dx_0\end{pmatrix}$$ and you had $$\begin{pmatrix}a+c+d\color{red}{g}&b\:\color{red}{g}\\ \:\:0&0\end{pmatrix}$$ instead of $$\begin{pmatrix}a+c+d\color{red}{x_{0}}&b\color{red}{x_{0}}\\ \:\:0&0\end{pmatrix}.$$

There's also a typo in your final answer, in particular you flipped your first eigenvector, so it should be
$$\begin{pmatrix}x\\y\end{pmatrix} = C_1e^{at}\color{red}{\begin{pmatrix}1\\ 0\end{pmatrix}}+C_2 e^{-t\left(c+dx_0\right)}\begin{pmatrix}-bx_0\\ a+c+dx_0\end{pmatrix}.$$

This is indeed the correct answer, and you can verify that it satisfies your linearized system of ODEs:
\begin{align}
\frac{dx}{dt} &= ax + bx_{0}y\\
\frac{dy}{dt} &= -(c+dx_{0})y.
\end{align}

Differentiating your solution
\begin{align}
\frac{dx}{dt} &=C_{1}ae^{at} + C_{2}bx_{0}(c+dx_{0})e^{-(c+dx_{0})t}\\
&= C_{1}ae^{at} + C_{2}bx_{0}(c+dx_{0})e^{-(c+dx_{0})t} + C_{2}bx_{0}ae^{-(c+dx_{0})t} - C_{2}bx_{0}ae^{-(c+dx_{0})t}\\
&=a\left(C_{1}e^{at} - C_{2}bx_{0}e^{-(c+dx_{0})t}\right) + bx_{0}C_{2}(a + c + dx_{0})e^{-(c+dx_{0})t}\\
&=ax + bx_{0}y\\
\frac{dy}{dt} &= -(c+dx_{0})C_{2}(a+c+dx_{0})e^{-(c+dx_{0})t}\\
&=-(c+dx_{0})y.
\end{align}


问:
Prove $f$ is measurable iff $f_0$ is measurable.
$f:\mathbb{R}^d\to\mathbb{R}$ called radial function if exists $f_0:\mathbb{R}_+\to\mathbb{R}$ such that for every $x\in\mathbb{R}^d$ $f(x)=f_0(||x||).$

Prove $f$ is measurable iff $f_0$ is measurable.

I've tried one way but I'm not sure about this.

Assuming that $f_0$ is measurable and defining the following sets

$f^{-1}(A)=\{x\in\mathbb{R}^d:||X||\in f_0^{-1}(A)\}=R_{f_0^{-1}(A)}$ and setting $B=f_0^{-1}(A)$

Assume that $B\subseteq\mathbb{R}_+$ is measurable and $g(x)=||x||$,$g$ is continuous function then for every open set $V(V\in \mathcal{B})$$\to$ $g^{-1}(V)\in \mathcal{B}$ is open, therefore $g$ is lebesgue measurable and then we get $R_B=\{x\in\mathbb{R}^d:||x||\in B\}=g^{-1}(B)$.

I have some doubts because $B$ is lebesgue and not borel and then $g^{-1}(B)$ is not borel measurable.

答:
The composition of measurable functions is always measurable. Since you are looking at measurability with respect to the Borel $\sigma$-algebra, the continuity of the norm implies its measurability. Hence, if $f_0$ is measurable, then $f$ is measurable as composition of measurable functions.

This should answer the implication you try to show. Since these more abstract results are sometimes easier to show, I recommend proving them and applying them to this special case.


问:
Stuck with this second derivative from first principles
It is given that

$a \in (0, \infty)$ and $f(a)=0$

$f(x)$ satisfies $x^2 f''(x) + xf'(x) -(x^2+1)f(x) =0$ for $x \in (0, \infty)$

The interval $I \subset \mathbb R$ contains $a$ and $f$ is differentiable once on $I$. It is also given that $f$ is twice differentiable at $a$. How do I prove that $$f''(a) = \lim_{h \to 0} \dfrac {f(a+h) - 2 f(a) + f(a-h)}{h^2}?$$

I tried using the "definition" of 2nd order derivative using limits, but could not proceed!

答:
What I'm going to show is a general result. It says :

Suppose $I \subseteq \mathbb{R}$ is an open interval containing $a$ . Let $f : I \to \mathbb{R}$ be a differentiable function on $I$ , and $f''(a)$ exists. Then : $$f''(a) = \lim_{h \to 0} \dfrac {f(a+h) - 2 f(a) + f(a-h)}{h^2}$$

Here's a proof :

\begin{align*}
&\lim_{h \to 0} \dfrac {f(a+h) - 2 f(a) + f(a-h)}{h^2}\\
&\stackrel{(\#)}{=} \lim_{h \to 0} \dfrac {f'(a+h) - f'(a-h)}{2h}\\
&= \frac{1}{2}\lim_{h \to 0} \dfrac {f'(a+h) - f'(a)}{h} + \frac{1}{2}\lim_{h \to 0} \dfrac {f'(a) - f'(a-h)}{h}\\
&= \frac{1}{2} f''(a) + \frac{1}{2} f''(a)\\
& = f''(a)\\
\end{align*}
Here, the equality marked by $(\#)$ is due to L'Hôpital's Rule.

Please let me know if this answers your query or not.


问:
Why is it okay to do this?
I am studying asymptotic recurrences for algorithms, and the book says:

$$T(n) = 2T(n/2) + \Theta (n)$$

is technically

$$T(n) = T(\lfloor n/2 \rfloor) + T(\lceil n/2 \rceil) + \Theta (n)$$

for an algorithm (e.g. Mergesort). However apparently it's okay to analyze the running time using the first recurrence. Why is it okay to approximate the second recurrence with the first one?

答: You are interested in the asymptotic behaviour; you can show that any $T$ satisfying the second relation is monotone (increasing), and then you can use the "sandwiching" theorem to relate solutions to the second to solutions to the first —up to some term which will, in hindsight (once you have the solution to the first relation) prove to be negligible.</p>

Title: Why is it okay to do this?


问:
number theoretical questions for division of numbers $1$ to $100$
Suppose we have a directed bipartite graph with $100$ vertices on both sides, and both consist of numbers $1$ to $100$. There is an arc $(i,b)$ from the left to the right side if and only if $i$ divides $b$. There are three questions:

Which pairs of vertices have at least $5$ neighbors in common (i.e. which pairs of vertices divide at least $5$ same numbers)?

How many arcs are there in total?

Take all possible combinations of vertices (single vertices, pairs, triples, etc) with at least $5$ neighbors in common. Find the maximal such sets.

Answers:

I guess here we have to count what each pair of numbers divides, so we have the number $\lfloor \frac{100}{\mathrm{lcm}(i,j)}\rfloor $. We first count how many numbers divide at least $5$ numbers on their own and that is $\lfloor \frac{100}{i}\rfloor \geq 5$. Here we see that $i\leq 20$. So we would have to look at ${20\choose 2}$ pairs. Another condition is that $\mathrm{lcm}(i,j) \leq 20$ but I do not know how to continue here to write the exact pairs? I guess each number and its multiples until $20$ is included, but there are some pairs such as $(2,3)$ which also appear.

I figured out the second question by counting how many divisors a number $b$ on the right has. From $1$ to $100$ we sum up $\lfloor \frac{100}{i}\rfloor $ and I get the total $482$.

I wrote a code to calculate these maximal sets, and I know the answer is
$$(1, 17), (1, 2, 3, 6, 9, 18), (1, 19), (1, 2, 4, 5, 10, 20), (1, 3, 5, 15), (1, 2, 4, 8, 16), (1, 13), (1, 2, 7, 14), (1, 2, 3, 4, 6, 12), (1, 11).$$
I am not sure how to prove it mathematically. It should be a generalization of the first question. I guess we would use the lcm again and for example the fact that some numbers are prime so they could not be contained in larger sets.

Any help is appreciated.

答:
Which pairs of vertices have at least 5 neighbors in common (i.e. which pairs of vertices divide at least 5 same numbers)?

The question in the parentheses only counts the pairs on the left side with five common neighbours. In other words, it only deals with out-neighbours. If you would like to find pairs with five common in-neighbours as well, you have to also find pairs that have at least five common divisors.

The former is, as you say, equivalent to finding pairs $(i,j), 1 \leq i < j \leq 100$ with $\text{lcm}(i,j) \leq 20$. There seem to be $56$ such pairs and they are easy to find by programming (and indeed easy, but tedious, to write out by hand).

The latter is equivalent to finding pairs $(i,j), 1 \leq i < j \leq 100$  such that $\text{gcd}(i,j)$ has at least five divisors. My computer says that there are $71$ such pairs.

This comes out to a grand total of $127$ pairs. But unless this was a programming assignment, I think the question was only aimed at giving a characterization of pairs with at least $5$ common neighbours (which we have done in terms of their $\text{lcm}$ and $\text{gcd}$).

How many arcs are there in total?

Your approach is sound and my computer says $482$ as well.

Take all possible combinations of vertices (single vertices, pairs, triples, etc) with at least 5 neighbors in common. Find the maximal such sets.

As in the first part, a set of numbers is valid if and only if the least common multiple $m$ of all the elements is at most $20$. Also notice that if $m \leq 20$, then we can add it to the set to get another valid set, and we can also add every divisor of $m$, since this doesn't increase the lowest common multiple of the members. This means that the maximal sets must be of the form $\{i : i \text{ divides } m\}$.

Finally, if this number $m$ was at most $10$, then we could also add $2m$ to the set to get a larger one. So in the end the maximal combinations are the sets of divisors of $m$ for $11 \leq m \leq 20$, and these are exactly the ones you gave in your answer.


问:
A parabola touches the bisectors of the angles formed by lines $x+2y+3=0$ and $2x+y+3=0$ at $(1,1)$ and $(0,-2)$. Find its focus and directrix.

A parabola touches the bisectors of the angles obtained by the lines $x+2y+3=0$ and $2x+y+3=0$ at the points $(1,1)$ and $(0,-2)$. Then find its focus and the equation of the directrix.

My approach is as follows:

The equation of bisector is
$$\frac{x+2y+3}{\sqrt{5}}= \pm \frac{2x+y+3}{\sqrt{5}}$$

We get the required bisectors as $x-y=0$ and $3x+3y+6=0$, or $x+y+2=0$.

$x-y=0$ is tangent to the parabola at $(1,1)$, whereas $x+y+2=0$ is tangent to the parabola at $(0,-2)$.

From here, how do I proceed?

答:
I have used the following properties derived from the basic equation of parabola.

(i)M is mid-point of AB. Line joining the point of intersection of tangents to the mid-point of the chord of contact is parallel to the axis of the parabola or perpendicular to the directrix.

(ii)Perpendicular tangents intersect on the directrix and the points of contact of tangents are the extremities of the focal chord.

(iii) The portion of tangent between the point of contact and the directrix always subtends a right angle at the focus.


问:
Maximal ideals and principal ideals domain
Let $R$ be a ring, and an ideal $I = ( p(x) )$ of $R[x]$.

Then consider the statement:
" $I$ is a maximal ideal if and only if $p(x)$ is irreducible "

Does that statement need the hypothesis "$R[x]$ is a principal ideals domain" to be true? Or is that statement true anyway?

答:
Tne statement is actually false even if $R$ is a PID.

E.g. consider the case $R=\Bbb Z$ and the chain of prime ideals with proper inclusions
$$
(0)\subset(x)\subset(2,x).
$$


问:
Fiber of $-1$ of norm sum in the ring of integers
Let $K$ be a real quadratic number field of discriminant $D>0$ with $\mathcal O_K$ being its ring of integers.

I'm interested in the map
$$ \varphi: \mathcal O_K \times \mathcal O_K \to \mathbb Z, \quad (\alpha,\beta) \mapsto N(\alpha)+N(\beta). $$
Is it surjective? Or can the fiber of $-1$ even be empty? If $\mathcal O_K$ has units of negative norm (which is for example the case when $D$ is prime) the latter is of course impossibe.

答:
Thm. The map $\varphi$ is surjective.

I will break the proof into three parts.

I. Preliminaries.

Write $K=\mathbb{Q}(\sqrt{d}),d>1$. One can assume that $d$ is squarefree.

First of all,  note that $\varphi(\alpha,\beta)=\det\pmatrix{\alpha & \beta \cr -\beta^* & \alpha^*}$
, where $*$ is the unique non trivial $\mathbb{Q}$-automorphism of $K$, so the image of $\varphi$ is a multiplicative monoid.

Note now that $0,=\varphi(0,0), 2=\varphi(1,1)$ and that $\varphi(m\alpha,m\beta)=m^2\varphi(\alpha,\beta)$ for all $m\in\mathbb{Z}$, so all in all, we are reduced to prove that the equation $$\varphi(\alpha,\beta)=a, \ \ \alpha,\beta\in\mathcal{O}_K$$ has at least a solution when $a$ is odd and squarefree.

Set $\omega=\sqrt{d}$ if $d\not\equiv  1 [4]$, and $\omega=\dfrac{-1+\sqrt{d}}{2}$ if $d=1+4k$, so that $\mathcal{O}_K=\mathbb{Z}[\omega]$.

If $\alpha=x+y\omega$ and $\beta=z+t\omega$, we have  $$\varphi(\alpha,\beta)=\left\lbrace\begin{array}{ccc}x^2+z^2-d(y^2+t^2)& if & d \not\equiv 1 [4]\cr  x^2-xy-ky^2+z^2-zt-kt^2 & if &  d=1+4k\end{array}\right.$$

Note that in the case $1+4k$, we also have $$\varphi(\alpha,\beta)=(x-\dfrac{y}{2})^2-dy^2+(z-\dfrac{t}{2})^2-dt^2  \ (*) $$
In particular, in both cases, $\varphi$ is an indefinite (since $d>0$) integral quadratic form in $4$ variables, which is non degenerate over $\mathbb{Q}$.

Now we have the following theorem.

Thm. (Eichler and al., 50's) Let $a\in\mathbb{Z}$, and let $q$ be an indefinite integral quadratic form in at least four variables which is non degenerate over $\mathbb{Q}$. Then $q$ represents $a$ over $\mathbb{Z}$ if and only if it represents $a$ over $\mathbb{Z}_p$ for all prime $p$.

Hence, we are reduced to prove the following facts:

Let  $a\in\mathbb{Z}$ be an odd squarefree integer, and let $p$ be a prime number . Then:

(a) the equation $x^2+z^2-d(y^2+t^2)=a, x,y,z,t\in\mathbb{Z}_p$ has at least one solution

(b) the equation $x^2-xy-ky^2+z^2-zt-kt^2=a, x,y,z,t\in\mathbb{Z}_p$ has at least one solution.

II. $p$ is odd.

I will make free use of the following well-known fact:

Fact. Let $a\in\mathbb{Z}_p$. If $a$ is a  nonzero square mod $p$, $a$ is a square in $\mathbb{Z}_p$.

Lemma 1. Let $a\in\mathbb{Z}_p^\times$.  Then the equation $a=u^2+v^2, u,v\in\mathbb{Z}_p$ has a solution.

Proof. It is known that this equation has a solution modulo $p$ (the set of elements of the form $\bar{v}^2,\bar{v}\in\mathbb{F}_p$ has $\dfrac{p-1}{2}+1=\dfrac{p+1}{2}$ elements, and same for the set of elements of the form $\bar{a}-\bar{u}^2$, so they cannot be disjoint).

Let $u_0,v_0\in\mathbb{Z}$ such that $a\equiv u_0^2+v_0^2 \ [p]$.

Since $a$ is not divisible by $p$ (this is a unit of $\mathbb{Z}_p$ !) , one may assume that $u_0\not \equiv 0 \ [p]$.

Hence $a-v_0^2$ is a nonzero square modulo $p$, hence a square in $\mathbb{Z}_p$ and we are done.

Proposition 2. Equations (a) and (b) have solutions in $\mathbb{Z}_p$.

Proof. Note that $2$ is a unit of $\mathbb{Z}_p$, so equation (b) has a solution in $\mathbb{Z}_p$ as soon as equation (a) has, since the substitution $(x,y,z,t)\longleftrightarrow (x-\dfrac{y}{2},y, z-\dfrac{t}{2},t)$ is bijective.

Hence it is enough to consider equation (a).

First case : $p\nmid a$.

Apply Lemma 1 to get $u,v\in\mathbb{Z}_p$ such that $u^2+v^2=a$. Then setting $x=u, y=0, z=v, t=0$ yields the desired solution for equation (a).

Second case : $p\mid a$.

First subcase: $p\mid d$, so $d=pd'$ where $p\nmid d'$ (since $d$ is squarefree). Write $a=pa'$, where $p\nmid a'$ (since $a'$ is squarefree). Then $d',a'$ are units of $\mathbb{Z}_p$, and so $-(d')^{-1}a'\in\mathbb{Z}_p^\times$ By Lemma 1, there exists $u,v\in\mathbb{Z}_p$ such that $u^2+v^2=-(d')^{-1}a'$. Then $-d(u^2+v^2)=pa'=a$.

Now set $x=0,y=u,z=0,t=v$.

Second subcase: $p\nmid d$. This time, $d$ is a unit of $\mathbb{Z}_p$, and so is $-d^{-1}(a-1)$. By Lemma 1, there exists $u,v\in\mathbb{Z}_p$ such that $u^2+v^2=-d^{-1}(a-1)$.
Then set $x=1,y=u,z=0,t=v$.

III. $p=2$.

I will make free use of the following well-known fact:

Fact. Let $a\in\mathbb{Z}_2$. If $a\equiv 1 \ [8]$, then $a$ is a square in $\mathbb{Z}_2$.

Proposition 3. Equations (a) and (b) have solutions in $\mathbb{Z}_2$.

Proof. Since $a$ is odd, we have $a\equiv 1,3,5,7 \ [8]$.

Using a CAS, one may check that equations (a) and (b) have a solution modulo $8$ of the form $(1,y_0,z_0,t_0)$ for all possible values of $a, d$ or $k$ modulo $8$ (of course, we have to exclude $d\equiv 0,4 \ [8]$ since $d$ is squarefree and stick to the values of $a$ listed above).

Hence $a- ( z_0^2-d(y_0^2+t_0^2))$ in case of Equation (a), and $a-(-y_0-ky_0^2+z_0^2-z_0t_0-kt_0^2)$ in case of equation (b), are congruent to $1$ modulo $8$, hence are square in $\mathbb{Z}_2$, and we are done.


问:
Show that $-Z$ is also a standard normal random variable.
Show that $-Z$ is also a standard normal random variable; that is, show that $P[-Z < x] = P[Z < x] \,\forall x.$</p>

答: Hint: The standard normal is a symmetric distribution.</p>


问:
System of equations$ \frac{x-a}{\sqrt{(x-a)^2+(y-b)^2}} + \frac{x-c}{\sqrt{(x-c)^2+(y-d)^2}} = 0$
While solving a bigger problem, I stumbeled upon a system of parametric equations
$$
\left\{
\begin{array}{ll}
\dfrac{x-a}{\sqrt{\left(x-a\right)^2+\left(y-b\right)^2}} + \dfrac{x-c}{\sqrt{\left(x-c\right)^2+\left(y-d\right)^2}} = 0\\
\dfrac{y-b}{\sqrt{\left(x-a\right)^2+\left(y-b\right)^2}} + \dfrac{y-d}{\sqrt{\left(x-c\right)^2+\left(y-d\right)^2}} = 0
\end{array}
\right.
$$
I need to solve it. I don't actually need all of the solutions to this system, one is enough.

Thanks, in advance.

答:
Note that $d_1=\sqrt{\left(x-a\right)^2+\left(y-b\right)^2}$ and $d_2=\sqrt{\left(x-c\right)^2+\left(y-d\right)^2}$ are the distances from
the point $(x,y)$ to the points $(a,b)$ and $(c,d)$, respectively. Rewrite the given system as

$$\frac{x-a}{c-x}= \frac{y-b}{d-y}= \frac{d_1}{d_2}$$

So, geometrically,  $(x,y)$ represent all the points on the line connecting $(a,b)$ and $(c,d)$.


问:
Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel.

Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel, and a circle of radius $(2 - \sqrt{2})$ cm covered within these squares. Find the length of $AD$.

What I Tried: Here is a picture :-

I made an attempt solving this and found rather a weird conclusion.

We have that $A,P,C,R$ are collinear, and also $AP = CR$.
Let $DL = LS = BK = KQ = x$ . We have $AP = x\sqrt{2}$ . Since $PC$ = $(4 - 2\sqrt{2})$ , $AS = (4 - 2\sqrt{2} + 2x\sqrt{2})$ .

Now, $PF = FC$ . So by Pythagoras Theorem on $\Delta PFC$ , $PF = FC = \sqrt{12 - 8 \sqrt{2}}$.

$AL = LS = \sqrt{12 - 8 \sqrt{2}} + 2x$ and $AS = (4 - 2\sqrt{2} + 2x\sqrt{2})$ . So now by Pythagoras Theorem :-

$$2\bigg(\sqrt{12 - 8 \sqrt{2}} + 2x\bigg)^2 = (4 - 2\sqrt{2} + 2x\sqrt{2})$$
$$\rightarrow 2\bigg(12 - 8\sqrt{2} + 4\sqrt{12 - 8\sqrt{2}}x + 4x^2\bigg) = 8x^2 + 16\sqrt{2}x - 16x - 16\sqrt{2} + 24$$
$$\rightarrow 24 - 16\sqrt{2} + 8\sqrt{12 - 8\sqrt{2}}x + 8x^2 = 8x^2 + 16\sqrt{2}x - 16x - 16\sqrt{2} + 24$$
$$\rightarrow \sqrt{12 - 8\sqrt{2}}x = 2\sqrt{2}x - 16x$$
$$\rightarrow \sqrt{12 - 8\sqrt{2}} = 2\sqrt{2} - 16$$
But this is obviously false.

So I literally ended up in a completely false equation, might be making a mistake in this complicated calculations, and I cannot find it. So can anyone tell me where I made the mistake, or I am solving this in the correct way or not?

答:
Let radius of circle be $r$. So side of square $a=2r$.

Let the circle touch $KR$ at $X$. So $KX=a/2=r$

Let $KQ=x$. So $PF=a-2x=2(r-x)$

Use power of point of $Q$, $$PQ\cdot QF=QX^2$$


问:
For a (not necessarily affine) scheme $X = \bigcup_{i=1}^n X_{f_i}$, does $(f_1, \ldots, f_n) = (1)$ in $\mathcal{O}_X(X)$?
This is of course true in the affine case, so it seems like it should be true in general, because $\mathcal{O}_X(X)$ should be "smaller" for a non-affine scheme than for a similar affine scheme (e.g. all global sections over a projective scheme are constants).  Just to be clear on notation:

$X$ is a scheme, not necessarily affine.

$f_1, \ldots, f_n \in \mathcal{O}_X(X)$.

$X_f := \{ x \in X \; | \; f_x \not \in \mathfrak{m}_{X, x} \}$, where $\mathfrak{m}_{X, x}$ is the maximal ideal of the stalk $\mathcal{O}_{X, x}$

$X = X_{f_1} \cup \cdots \cup X_{f_n}$

Does it follow that $(f_1, \ldots, f_n) = (1)$ in $\mathcal{O}_X(X)$?

We have $(f_1|_U, \ldots, f_n|_U) = (1)$ in $\mathcal{O}_X(U)$ where $U \subset X$ is open affine, but I don't see how to extend this to all of $X$.

答:
No. Let $X = \mathbb{A}^2 \setminus (0,0)$. Then $\mathcal{O}(X) = k[x,y]$ and $X = X_x \cup X_y$. However, the ideal $\langle x,y \rangle$ in $k[x,y]$ is not $1$.

The plane with a point deleted should be in your standard toolkit to test things against.


问:
An exercise on conditional expectation of L^1 random variables
I'm facing an exercise on conditional expectation, asking to prove/disprove the following:

Let $ X,Y \in L^1(\Omega, \mathcal{H},\mathbb{P}) $.

If $ E[X|Y] = Y $ and $ E[Y|X] = X $, then $ X=Y $ almost surely.

It is clear to me that the statement is true when $X,Y \in L^2 $. Does this hold even when $X,Y \in L^1 \setminus L^2$?

答:
Yes, this is true. We have $\mathbb{E}[(X-Y)|X]=0$ almost surely. From the definition of conditional expectation it follows that for each $t\in\mathbb{R}$:

$\mathbb{E}[(X-Y)1_{\{{X\leq t\}}}]=\mathbb{E}[\mathbb{E}[(X-Y)|X]\ 1_{\{X\leq t\}}]=0$

So then the following follows:

$\mathbb{E}[(X-Y) 1_{\{X\leq t, Y\leq t\}}]=\mathbb{E}[(X-Y)\ 1_{\{X\leq t\}}]-\mathbb{E}[(X-Y)\ 1_{\{X\leq t,Y>t\}}]=$

$=0-\mathbb{E}[(X-Y)\ 1_{\{X\leq t,Y>t\}}]$

By interchanging the roles of $X$ and $Y$ we can similarly get the following equality:

$\mathbb{E}[(X-Y)\ 1_{\{X\leq t,Y\leq t\}}]=-\mathbb{E}[(X-Y)\ 1_{\{Y\leq t, X>t\}}]$

Combining the results, we get:

$\mathbb{E}[(X-Y)\ 1_{\{X\leq t,\ Y>t\}}]=\mathbb{E}[(X-Y)\ 1_{\{Y\leq t,\ X>t\}}]$

But note that the right hand side is non negative, while the left hand side must be non positive. Since they are equal it follows that both expectations are equal to $0$. So $\mathbb{E}[(X-Y)\ 1_{\{X>t,\ Y\leq t\}}]=0$. A nonnegative random variable with $0$ expectation is $0$ almost surely. Thus $\mathbb{P}(Y\leq t<X)=0$. This is true for every $t\in\mathbb{R}$, so by taking rational values of $t$ we can easily conclude that $\mathbb{P}(Y<X)=0$. And similarly $\mathbb{P}(X<Y)=0$. So the result $\mathbb{P}(X=Y)=1$ follows.


问:
Simple proof of König's theorem
I have been trying to understand the proof of the König's following theorem on bipartite graphs: "The maximum cardinality of a matching in a graph $G$ is equal to the minimum cardinality of a vertex cover.". While working on it, I have found a simpler interpretation of the proof than the one given in Graph Theory book by Reinhard Diestel. Is the proof correct, or am I missing something here?

We can form the minimum vertex cover candidate by selecting the elements of the matchings that are connected to an unmatched element if there is one in the matching, and by selecting the element from the left hand side (or the right hand side, but consistently)  otherwise. The selection made in the example graph can be shown to be the minimum vertex since:

All the edges in the matching are covered by it since we have constructed this using one element of every matching,

All the edges that are connected to the unmatched elements are covered since we have selected the minimum vertex cover according to that. The critical point to emphasize here is that only a single element of a pair can be connected to a one (or many) unmatched element(s), because otherwise we would have an augmenting path that could have been used to form an additional matching which would violate the idea of the maximum matching.

For the K3-3 graph, the algorithm randomly picks a side to select the vertices from the maximum matching to form the minimum vertex cover. This is due to the fact that all elements from both sides are matched. I believe the size of the maximum matching is 3 for K3-3, and it can be seen that minimum vertex cover is of size 3 too.

答:
Consider the example below given by Diestel. Your algorithm would conceivably choose the top right vertex and then the left vertices of the rest of the matching edges. But this is not a vertex cover: there is an edge going between the left and right vertices of the first and second matching edges, respectively, that is not covered.

Of course, in this case if we had been more lucky and chose all of the vertices from the right side, we would have gotten a vertex cover. But it is not difficult (and maybe instructive) to extend this to an example where neither choice gives you a vertex cover. (Hint: you only have to add one unmatched vertex of degree one to this example.)

Title: Simple proof of König's theorem


答:
This seems to be : find the extremum of function
$$f(x,y)=\sqrt{(a-x)^2+(b-y)^2}+\sqrt{(c-x)^2+(d-y)^2}$$ If you do not have any constraint, it could be any point along the line explained by @Quanto in his/her answer.

Title: System of equations$ \frac{x-a}{\sqrt{(x-a)^2+(y-b)^2}} + \frac{x-c}{\sqrt{(x-c)^2+(y-d)^2}} = 0$

Body:
While solving a bigger problem, I stumbeled upon a system of parametric equations
$$
\left\{
\begin{array}{ll}
\dfrac{x-a}{\sqrt{\left(x-a\right)^2+\left(y-b\right)^2}} + \dfrac{x-c}{\sqrt{\left(x-c\right)^2+\left(y-d\right)^2}} = 0\\
\dfrac{y-b}{\sqrt{\left(x-a\right)^2+\left(y-b\right)^2}} + \dfrac{y-d}{\sqrt{\left(x-c\right)^2+\left(y-d\right)^2}} = 0
\end{array}
\right.
$$
I need to solve it. I don't actually need all of the solutions to this system, one is enough.

Thanks, in advance.


问:
Prove that $x \gt y \gt 0 \implies x \gt x-y$
Prove that $x \gt y \gt 0 \implies x \gt x-y$

I have approached it as below.

$y \gt 0 \implies y=x+(-x)+y \gt 0 \implies x+((-x)+y)=x+(-(x-y)=x-(x-y)>0 \implies x\gt x-y$

What I observe in this proof is that is does not consider the fact that $x \gt y$ and $x \gt 0$.

Is this proof valid?

答:
Alternative approach

$0 < y \implies  0 = [(-1) \times 0] > (-1) \times y = -y \implies$

$x = x + 0 > x + (-y) = x - y.$


问:
Calculation on a differential manifold
I do not know how to calculate:
Let $\alpha$ be a differential 1-form on $\mathbb{R}^{2}$ and let $V$ and $W$ be vector fields on $\mathbb{R}^{2}$. Compute $d \alpha(V, W)$ at the point $(0,1),$ where
$$
V=x \frac{\partial}{\partial y}-y \frac{\partial}{\partial x}, W=y \frac{\partial}{\partial y},
$$
and
$$
\alpha(V)=2 x-y^{2}, \alpha(W)=2 y.
$$

答:
From
$$
\alpha(V)=2 x-y^{2}, \alpha(W)=2 y, V=x \frac{\partial}{\partial y}-y \frac{\partial}{\partial x}, W=y \frac{\partial}{\partial y},
$$
we have $\alpha=ydx+2dy$.
\begin{equation*}
\begin{aligned}
d \alpha(V, W)&=V\alpha(W)-W\alpha(V)-\alpha([V,W])=V\alpha(W)-W\alpha(V)-\alpha(VW-WV)\\
&=2Vy-W(2 x-y^{2})-\alpha(x \frac{\partial}{\partial y}+y \frac{\partial}{\partial x})\\
&=2x+2y^2-(2x+y^2)=y^2.
\end{aligned}
\end{equation*}

Title: Calculation on a differential manifold


问:
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?

x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).

If yes, will large number of solutions give moderate rank EC?

If one drops $-1$ i.e. $xy(x-y)=n$ the answer is "yes" via multiples of rational point(s) and then multiplying by a cube.

EDIT: Suppose it is an open question.

EDIT: I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.

EDIT: What I find intriguing is that all integral points in this model come from factorization/divisors only.

EDIT: Current record is n=179071200 with 22 solutions with positive x,y. Due to Matthew Conroy.
Current record is n=391287046550400 with 26 solutions with positive x,y. Due to  Aaron Meyerowitz
Current record is n=8659883232000 with 28 solutions with positive x,y. Found by Tapio Rajala.

Current record is n=2597882099904000 with 36 solutions with positive x,y. Found by Tapio Rajala.

EDIT: Is it possible some relation in the primes or primes of certain form to produce records? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.

答:
This answer is primarily intended to remove this question from the Unanswered queue.

While no conclusive theoretical answers were given, there was a more fruitful discussion at the crosspost on MathOverflow, together with some interesting numerical data.

Further, the crosspost are also contains a number of nice references; check it out if you haven't!

ContentLicense: CC BY-SA 2.5


问:
Limit of integral of $\sin\frac{1}{t}$
Determine whether or not the limit below exists.
$$\lim_{x\rightarrow0}\frac{1}{x}\int_{0}^{\sin x}\sin \frac{1}{t} \cos t^{2}\mathrm{d}t$$
I tried to use the Mean value theorem integrals to prove the limit exists, but it does not exist for $\lim\limits_{x\rightarrow0}\sin \frac{1}{x}$. So I guessed the limit does not exist and used the Cauchy principle to prove it, but I failed.

Any idea will be helpful.

答:
If I am not wrong, you can easily majorate the integral using Triangular Inequality. Then, the limit should appear.
But you must be careful when writing the proof, considering that sin(x) is not constantly positive over $[0,+\infty)$

Title: Limit of integral of $\sin\frac{1}{t}$


问:
quadratic in linear form

$y = mx + b$ is a linear equation and represents a straight line. (The direction $m$ is constant.)

$y = x^2 + b$ is quadratic and represents a parabola. (The direction of the tangent is not constant.)

But $y = x^2 = x\cdot x$ in linear form with the first $x$ being the direction and y intercept being $0$ in this case.

So you could say that indeed the direction depends upon $x$.

However, for $x = 1$, $m = 2$ and not $1$ and indeed the derivative of $x^2$ is $2x$.

Why? Where does the $2$ come from?

答:
Since both the terms in the product $\color{blue}{x}*x$ are changing with $x$, you must use the product rule:

$$\color{blue}{x}*x$$

For a small change in $x$'s :
$$ (\color{blue}{x+\Delta x})*(x+\Delta x)$$

Take the difference

$$(\color{blue}{x+\Delta x})*(x+\Delta x) - \color{blue}{x}*x$$
Simplifying gives
$$\color{blue}{x}*\Delta x + \color{blue}{\Delta x}*x + \color{blue}{\Delta x}\Delta x$$

Divide by $\Delta x$
$$\color{blue}{x} + x + \Delta x$$

Since $\Delta x$ goes to $0$ as the secant approaches the tangent... see where  the $2x$ comes from?

Title: quadratic in linear form


答:
The map $f : t \mapsto \sin \frac{1}{t} \cos t^{2}$ is bounded by $1$. As $\sin$ is also bounded, the integral is bounded too. Therefore the limit is equal to zero.

Side comment: for $y \in \mathbb R$, the integral $\int_0^y f(t) \ dy$ exists as $f$ is bounded on $\mathbb R$ and continuous on $\mathbb R \setminus \{0\}$.

Title: Limit of integral of $\sin\frac{1}{t}$

Body:
Determine whether or not the limit below exists.
$$\lim_{x\rightarrow0}\frac{1}{x}\int_{0}^{\sin x}\sin \frac{1}{t} \cos t^{2}\mathrm{d}t$$
I tried to use the Mean value theorem integrals to prove the limit exists, but it does not exist for $\lim\limits_{x\rightarrow0}\sin \frac{1}{x}$. So I guessed the limit does not exist and used the Cauchy principle to prove it, but I failed.

Any idea will be helpful.


问:
Is my proof of convergence for $\sum\limits^{\infty}_{n=1} \frac {2^n} {n^n}$ true?
I want to prove that $\sum\limits^{\infty}_{n=1} \frac {2^n} {n^n}$ is convergent.
Therefore I want to use that it will be convergent if $\lim\limits_{n \to \infty}|\frac {2^{n+1}} {n^n}| < 1$ is true.

But I am not sure if all of my steps are legal to do in order to do it. Here is what I came up with:

$$\lim_{n \to \infty}|\frac {2^{n+1}} {n^n}| < 1$$
$$\lim_{n \to \infty}|\frac {\sqrt[n]{2^{n}} \sqrt[n]{2}} {\sqrt[n]{n^n}}| < \sqrt[n]1$$
$$\lim_{n \to \infty} \frac {2 \sqrt[n]2} {n} < 1$$
$$0<1$$

$\Rightarrow$ It is true.

答:
Of course converge.
$$
(\frac{2}{n})^n<\frac{1}{2^n}$$ when $n>4$.


问:
Finding the limit of a particular function
Can not understand the following limit in a past paper.

$$\dfrac{1}{[n(e^{\frac{1}{n}}-1)]}\rightarrow 1$$

答: Hint: $e^{\frac 1 n} - 1 \operatorname*{=}_{n\to\infty} \frac{1}{n} + o\left(\frac{1}{n}\right)$.


问:
How to Find an Equation of a Plane Containing Point Q and Perpendicular to Plane P
Given point $Q(1, 0, -1)$ and plane $P: 2x + 3y - 2z = 6$, how do I find an equation of a plane that contains point $Q$ and is perpendicular to plane $P$?

答:
With $p = (x,y,z),\vec n_1 = (2,3,-2)$ we have

$$
P:2x+3y-2z=6\equiv p\cdot \vec n_1 = 6
$$

The orthogonal plane will accomplish

$$
\cases{
(p-Q)\cdot\vec n_2 = 0\\
\vec n_1\cdot \vec n_2 = 0\\
\|\vec n_2\| = 1
}
$$

so there are infinite planes ...


问:
Proving equivalent statements regarding two sets
Let $A$ and $B$ be arbitrary sets with universal set $U$. Prove that the following statements are
equivalent.

a) $A ∩ B = ∅$

b) $A-B = A$

c) $A ⊆ B'$

d) $A' ∪ B' = U$

I've managed to put something together for $a\Rightarrow b$ and $b\Rightarrow c$, but I'm struggling with $c\Rightarrow d$.

I have that for every $y∈U$,then if $y∈A$, then $y∈B'$, where $B'$ denotes the complement of $B$. It then follows that if $y∈A$, then $y∉B$, but this is where I am getting stuck? I'm not sure what this does for me. Any help would be appreciated. Thanks.

答:
Let $x\in U$, i.e. $x$ is any arbitrary element. We need to show that $x\in A' \cup B'$.

There are two cases: $x\in A$ and $x\notin A$.

If $x\in A$, then since $A\subseteq B'$, $x\in B'$ and so $x\in A'\cup B'$.

If $x\notin A$, then $x\in A'$ and so $x\in A'\cup B'$.

In each case, we conclude $x\in A'\cup B'$. Therefore, $$U\subseteq A'\cup B'$$
We already know,
$$A'\cup B'\subseteq U$$

Combining the two we get,
$$A'\cup B'= U$$

as desired.

Hope this helps :)


答:
As the question has been changed from the limit when $x \rightarrow \infty$ to $x \rightarrow 0$ here is a new hint :

You still have to think about the continuity of the map $ f : t \mapsto \sin \frac{1}{t} \cos t^{2}$ for $t \neq 0$. Deduce from the behaviour of f around $0$ what happens to the integral when $x \rightarrow 0$.

Title: Limit of integral of $\sin\frac{1}{t}$


问:
Summation using binomial coefficient
I am trying to calculate the following using binomial coefficients and summation, but my memory is standing in the way:

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

Thanks!

With great help I got to:

$$ (n - 1)* 2 ^{n} - 1 $$

Can you please confirm this?

Title: Summation using binomial coefficient

Body:
Here is another approach, using
$$
\sum_{k=0}^n\binom{n}{k}=2^n
$$
and
$$
\sum_{k=j}^n\binom{k}{j}=\binom{n+1}{j+1}
$$
to get
$$
\begin{align}
\sum_{k=1}^nk\,2^{k-1}
&=\sum_{k=1}^nk\,\sum_{j=0}^{k-1}\binom{k-1}{j}\\
&=\sum_{j=0}^{n-1}\sum_{k=j+1}^n(j+1)\binom{k}{j+1}\\
&=\sum_{j=0}^{n-1}(j+1)\binom{n+1}{j+2}\\
&=\sum_{j=1}^nj\binom{n+1}{j+1}\\
&=\sum_{j=1}^n(j+1)\binom{n+1}{j+1}-\binom{n+1}{j+1}\\
&=\sum_{j=1}^n(n+1)\binom{n}{j}-\binom{n+1}{j+1}\\
&=(n+1)(2^n-1)-(2^{n+1}-n-2)\\[12pt]
&=(n-1)2^n+1
\end{align}
$$

Verification

Of course, to verify, we can try induction.

The formula works for $n=1$: $1=1$.

Assume the formula works for $n-1$, then
$$
\begin{align}
\sum_{k=1}^nk\,2^{k-1}
&=n\,2^{n-1}+\sum_{k=1}^{n-1}k\,2^{k-1}\\
&=n\,2^{n-1}+(n-2)2^{n-1}+1\\[9pt]
&=(n-1)2^n+1
\end{align}
$$
Thus, it is true for all $n\ge1$.


问:
Equivalent notation for the basis of L(V,W)
I am getting myself confused as I have seen two definitions for the basis of L(V,W). Are both correct? And if so can someone help me see how they are equivalent?

答: They are the same bases. Just note that the $T_{kl}$ from the first definition is equal to the $T_{lk}$ of the second one.


答:
The proof is indeed valid. And there's a slightly shorter proof which goes like this:
$$y > 0 \implies y+(-y) > 0+(-y) \implies 0 > -y \implies x+0 > x+(-y) \implies x > x-y
$$

Body:
Prove that $x \gt y \gt 0 \implies x \gt x-y$

I have approached it as below.

$y \gt 0 \implies y=x+(-x)+y \gt 0 \implies x+((-x)+y)=x+(-(x-y)=x-(x-y)>0 \implies x\gt x-y$

What I observe in this proof is that is does not consider the fact that $x \gt y$ and $x \gt 0$.

Is this proof valid?


问:
Solving $\frac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=c$ for $\alpha$
I have been struggling with this equation. Is it unsolvable?

$$\frac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=c$$

$\beta$ and $c$ are constant. How can I solve for the angle $\alpha$?

PS: As you might imagine, I'm still new to this, so please be easy on me. Thanks for your help.

答:
Rewrite $$\frac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=c$$ as $$
(1-c)\cos\alpha\cos\beta + (1+c)\sin\alpha\sin\beta = 0$$
Taking $a = (1-c)\cos\beta ~$ and $~b = -(1+c)\sin\beta$, we get
$$a\cos\alpha = b\sin\alpha$$
Now, the case $~\cos\alpha = 0~$ is not interesting. Assume $\cos \alpha \ne 0$ and divide by it to get
$$b\tan \alpha  = a$$
Hence, if either of $a$ and $b$ is $0$, then so is another. Assume, neither is zero and divide by $b$ to get
$$\tan \alpha = \frac ab = \frac{c - 1}{c + 1} \cot \beta$$

Now, you can decide if it is solvable or not.


问:
$z=f(u,v)$, with $u=x^2-y^2$ and $v=2xy$. Show that $\dfrac{\partial^2z}{\partial x \partial y}=-4xyf_{uu}+4(x^2-y^2)f_{uv}+4xyf_{vv}+2f_v$
given $z=f(u,v)$, with $u=x^2-y^2$ and $v=2xy$. Show that $\dfrac{\partial^2z}{\partial x \partial y}=-4xyf_{uu}+4(x^2-y^2)f_{uv}+4xyf_{vv}+2f_v$.
I have tried doing the tree and expanding, but what I get is not similar at all to what I must probe. I have looked for examples, but don't really understand

答:
Start with $$\dfrac{\partial z}{\partial x}= \dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x} + \dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial x } = 2xf_u + 2yf_v.$$
Then

$$\dfrac{\partial}{\partial y}\left[2xf_u + 2yf_v\right] = 2x\dfrac{\partial f_u}{\partial y} + 2y\dfrac{\partial f_v}{\partial y} + 2f_v.$$

Can you take it home from here?


问:
Partition a simple undirected graph with degree of every vertex at most 4 into two graphs with degree of every vertex at most 2
Suppose $G = (V, E)$ is a simple undirected graph (with no self-loop) such that the degree of every
vertex is at most $4$.
How to prove that it is possible to partition the edges $E$ into two sets $E_1$ and $E_2$ such that in
each of the graphs $G_1 = (V, E_1)$ and $G_2 = (V, E_2)$, the degree of every vertex is at most $2$?

答:
We have $G$ with $\Delta(G) \le 4$. Show that

There is some finite $4$-regular graph $R$ such that $G \subseteq R$.

The above statement generalizes nicely for any $\Delta(G)$. Then, we can apply Petersen's $2$-factor Theorem:

If a graph $G$ is $2k$-regular, then it has a $2$-factor. Consequently, $E(G)$ decomposes into $k$ many $2$-factors.

A $2$-factor is a $2$-regular spanning subgraph. There's a proof in the link, but you can also look at this MSE question to find an argument that generalizes nicely.

From these two, we can find a $4$-regular graph containing our graph, use the $2$-factor Theorem, and then delete vertices as required to obtain $G_1 = (V_1,E_1)$ and $G_2 = (V_2,E_2)$ as desired.

This is probably the way to think about it, but I wouldn't be surprised if there is another way using edge colorings. For example, the result follows immediately for Class $1$ graphs, but Class $2$ graphs may require a bit more thinking (See Edge Coloring.)


答: Shorter: $x = (x-y)+y > x-y+0=x-y.\blacksquare$

Title: Prove that $x \gt y \gt 0 \implies x \gt x-y$


问:
Prove general Euclid's Lemma in a UFD using prime factorization
I have seen plenty of proofs for this theorem:
In a UFD if $(a,b)=1$ and $a|bc$ then $a|c$.
They mostly use the gcd distributive law e.g. here.
Well I wanted to prove this just by relying on the properties that the UFD had.

My attempt:
Since $a|bc$ then for some $r$ we have $ar=bc$.
Now by the existence, since we know that any non unit element like $a$ can be rewritten as $t_1×....t_n$ where $t_i$ are irreducible, we can do this:

$$p_1^{α_1}...p_n^{α_n} g_1^{ε_1}...g_m^{ε_m} =q_1^{β_1}...q_k^{β_k}h_1^{ψ_1}...h_i^{ψ_i}$$
(Where $p_i$, $g_i$, $q_i$ and $h_i$ are primes.)
By uniqueness the set that is on right should be on left too, Am I right?
But since $(a,b)=1$ then $a$ and $b$ shouldn't share prime elements. Somehow it's like $A$ is a subset of $C$. I can't really manage this but it's becoming like a problem in set theory.

Can you please help me with my own approach??

答:
It has a natural proof by induction on the number $\:\!k\:\!$ of prime factors of $\,a,\,$ using as inductive step Euclid's Lemma (if a prime divides a product then it divides some factor). If $\,k=0\,$ then $\,a\,$ is a unit so $\,a\mid c.\,$ Else $\,a = p\bar a\,$ for a prime $\,p\,$ so $\,p\bar a\mid bc\,\Rightarrow\,p\mid b\,$ or $\,p\mid c,\,$ so $\,\color{#c00}{p\mid c}\,$ by $\,(p,b)=1\,$ by $\,(p\bar a,b)=1$. Cancelling $\,p\,$ from $\,p\bar a\mid bc\Rightarrow \bar a\mid b\,\color{#c00}{c/p},\,$ and $\,(\bar a,b)=1\,$ by $\,(p\bar a,b)=1.\,$ Notice $\,\bar a\,$ has fewer  prime factors than $\,a=p\bar a,\,$ thus $\,\bar a\mid \color{#c00}{c/p}\underset{\textstyle\times\, p}\Rightarrow p\bar a\mid c\ $ (i.e. $\,a\mid c),\,$ by induction.

Exercise $ $ Make explicit all  implicit uses of the existence and uniqueness of prime factorizations that are employed in the proof (necessary to be completely rigorous).


问:
Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel.

Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel, and a circle of radius $(2 - \sqrt{2})$ cm covered within these squares. Find the length of $AD$.

What I Tried: Here is a picture :-

I made an attempt solving this and found rather a weird conclusion.

We have that $A,P,C,R$ are collinear, and also $AP = CR$.
Let $DL = LS = BK = KQ = x$ . We have $AP = x\sqrt{2}$ . Since $PC$ = $(4 - 2\sqrt{2})$ , $AS = (4 - 2\sqrt{2} + 2x\sqrt{2})$ .

Now, $PF = FC$ . So by Pythagoras Theorem on $\Delta PFC$ , $PF = FC = \sqrt{12 - 8 \sqrt{2}}$.

$AL = LS = \sqrt{12 - 8 \sqrt{2}} + 2x$ and $AS = (4 - 2\sqrt{2} + 2x\sqrt{2})$ . So now by Pythagoras Theorem :-

$$2\bigg(\sqrt{12 - 8 \sqrt{2}} + 2x\bigg)^2 = (4 - 2\sqrt{2} + 2x\sqrt{2})$$
$$\rightarrow 2\bigg(12 - 8\sqrt{2} + 4\sqrt{12 - 8\sqrt{2}}x + 4x^2\bigg) = 8x^2 + 16\sqrt{2}x - 16x - 16\sqrt{2} + 24$$
$$\rightarrow 24 - 16\sqrt{2} + 8\sqrt{12 - 8\sqrt{2}}x + 8x^2 = 8x^2 + 16\sqrt{2}x - 16x - 16\sqrt{2} + 24$$
$$\rightarrow \sqrt{12 - 8\sqrt{2}}x = 2\sqrt{2}x - 16x$$
$$\rightarrow \sqrt{12 - 8\sqrt{2}} = 2\sqrt{2} - 16$$
But this is obviously false.

So I literally ended up in a completely false equation, might be making a mistake in this complicated calculations, and I cannot find it. So can anyone tell me where I made the mistake, or I am solving this in the correct way or not?

答:
Using the fact that the side of the outer square and diagonal of the inner square are equal to the diameter of the circle,

$AD = \frac{AL + PE}{2} = \frac{2r + r\sqrt2}{2} = \frac{(2 + \sqrt2)(2 - \sqrt2)}{2} = 1$


问:
How to sketch $\ln|\sin x|$
The question is to find the number of solutions of
$$\ln|\sin x|=-x^2+2x$$ but I just don't know how to sketch a logarithmic function like $\ln|\sin x|$.

A step by step method would be highly appreciated.
Thanks in advance.

答:
If you use inspection, you would notice that, if
$$f(x)=\log (|\sin (x)|)+x^2-2 x$$
$$f\left(\frac{2 \pi }{3}\right)=\log \left(\frac{\sqrt{3}}{2}\right)+\frac{4 \pi ^2}{9}-\frac{4 \pi }{3}$$ is very small $(\approx 0.054)$ while
$$f\left(\frac{ \pi }{2}\right)=\frac{1}{4} (\pi -4) \pi \approx -0.674$$
$$f\left(\frac{3 \pi }{4}\right)=\frac{1}{16} \left(-24 \pi +9 \pi ^2-8 \log (2)\right) \approx 0.493$$

On the other hand, there is a problem on each side of $x=n\pi$ because of the logarithm.

So, an infinite number of roots $(\sim 2, n\pi-\epsilon,n\pi+\epsilon)$

Title: How to sketch $\ln|\sin x|$


问:
Integral calculation question
Is my following calculation correct? For some smooth $u: \mathbb{R} \to[0,1]$, $$\int\limits_0^1 \big[\chi_{\{u(x)<t\}}(x)\chi_{\{u(y)\geq t\}}(y)+\chi_{\{u(x)\geq t\}}(x)\chi_{\{u(y)<t\}}(y)\big]\mathrm{d}t =|u(x)-u(y)|.$$

答:
The integral can be written in the following way
$$\int_0^1 \chi_{ \{ u(x) <t\le u(y) \}} (x,y) dt + \int_0^1 \chi_{ \{ u(y) <t\le u(x) \}} (x,y) dt$$
If $u(x) < u(y)$, then the second integral is zero, and you get
$$\int_0^1 \chi_{ \{ u(x) <t\le u(y) \}} (x,y) dt = \int_{u(x)}^{u(y)} 1 dt = u(y)-u(x)$$

Similarly, if $u(y) < u(x)$ the first integral is zero and you get $u(x)-u(y)$.

Title: Integral calculation question


问:
Finding the expected value of $\frac{1}{(x+1)(x+2)}$ where X is binomial
So i was trying to solve one of late questions from past midterms and had that:

Find E($\frac{1}{(x+1)(x+2)}$) where $X∼b(n,p)$

i know i can solve the sum of g(x)f(x) but i got stuck in solving that one

any solutions?

Body: The PGF of $X$ is $\Bbb Et^X=(q+pt)^n$ with $q:=1-p$. Integrating from $0$ to $t$ twice, $\Bbb E\frac{t^{X+1}}{X+1}=\frac{(q+pt)^{n+1}-q^{n+1}}{p(n+1)}$ and $\Bbb E\frac{t^{X+2}}{(X+1)(X+2)}=\frac{(q+pt)^{n+2}-q^{n+2}}{p^2(n+1)(n+2)}-\frac{q^{n+1}t}{p(n+1)}$. Setting $t=1$, $\Bbb E\frac{1}{(X+1)(X+2)}=\frac{1-q^{n+2}}{p^2(n+1)(n+2)}-\frac{q^{n+1}}{p(n+1)}$.


问:
What does having a bar on a manifold mean?
If $M$ is a manifold then what is denoted as $\overline{M}$?

I am guessing that it means a reversal of orientation.

Related to the above I would like to understand the following construction of a $d$-dimensional Riemannian category that I recently came across:

Objects are $(d−1)$ Riemannian manifolds. Morphisms between two oriented $(d − 1)$-dimensional Riemannian manifolds $N_1$ and $N_2$ are oriented $d$-dimensional Riemannian manifolds $M$, such that $\partial M = N_1 ⊔ \overline{N_2}$. The orientation on all three manifolds should naturally agree, and the metric on $M$ agrees with the metric on $N_1$ and $N_2$ on a collar of the boundary. The composition is the gluing of such Riemannian cobordisms.

I would like to know more about this intuition of defining a morphism between two manifolds as another manifold itself! One is used to thinking of morphisms as maps.

答:
I would like to know more about this intuition of defining a morphism between two manifolds as another manifold itself! One is used to thinking of morphisms as maps.

One intuition is from physics: one thinks of $n$-manifolds as "space," and of $n+1$-dimensional cobordisms between them as "time evolution." A nice introduction to these ideas is Baez's Physics, Topology, Logic and Computation: a Rosetta Stone.


问:
Sequence in $c_0$ (related with $\ell_1$).
I need some help with this cuestion:

Let b a sequence $a=\{a(n)\}$ in $\ell_1$. We define, for each $x \in c_0$, the sequence $$T_a: c_0 \longrightarrow c_0, \hspace{0.15cm} T_a(x)(n)=\sum_{k=n}^{\infty}a(k)x(k), n \in \mathbb{N}.$$ Give the properties of the sequences $a \in \ell_1$ for which $T_a$ reach its norm.

I think that I have to give an if-and-only-if property, but I don't know exactly how to do it. I'm trying to prove that maybe $\exists k \in \mathbb{N}$ such that $|a(k)|=||a||_{\infty}$, or something about $||a||_1$, but I don't arrive to any conclusion.

答:
Here is a proof that uses the fact that $\ell_1$ is isometrically isomorphic to $c_0^*$ via the map

$\tag1 a\mapsto f:x\mapsto \sum_{n=1}^\infty a_nx_n$

We have

$\tag2 \|T_a\|=\sup\{\|T_a(x)\|:\|x\|\le 1\}$ and

$\tag3 \|T_a(x)\|=\sup \{|T_a(x)(n)|\:n\in \mathbb N\}$

Now, if we regard $T_a(x)(n)=\sum _{k=n}^\infty a_kx_k$ as arising from the linear functional  $f_n:=T(\cdot)(n): x\mapsto \sum _{k=n}^\infty a_kx_k$ then $(1)$ implies that $\|f_n\|=\sum _{k=n}^\infty |a_k|.$ And  since $\sup \{\|f_n(x)\|\:\|x\|\le 1\}\le \sum _{k=1}^\infty |a_k|$ holds for all integers $n$, we have from $(3)$ that $ \|T_a(x)\|\le \sum _{k=1}^\infty |a_k|$, and finally from $(2)$ that

$\tag4 \|T_a\|\le \sum _{k=1}^\infty |a_k|=\|a\|.$

But for each integer $n$ if we define the sequence $x_n\in c_0;\ \|x_n\|=1$ by: $x_k=\operatorname {sgn}\ a_k:k\le n$ and $0:k>n$ then, $\|T_a(x_n)\|=\sum_{k=1}^n|a_k|$ and so $(2)$ implies that $\|T_a\|=\|a\|.$

This argument shows that if $a\in c_{00}$ then the norm is attained because in this case, $(x_n)$ is finite. On the other hand, if $x$ is any other norm one sequence, other than the one defined above, then there is an integer $j$ such that $|x_j||a_j|<|a_j|$  and so $\|T_a(x)(n)\|\le \sum_{k=1}^\infty |a_k||x_k|<\sum_{1=k\neq j}^\infty |a_k||x_k|+|a_j|\le \sum_{k=1}^\infty |a_k|$. It follows then that the norm is not achieved and therefore that if $a\notin c_{00}$ then $x\notin c_0.$ We conclude that the norm of this operator is achieved if and only if $a\in c_{00}.$


问:
Is $ \operatorname{lin}(\mathcal{V}) $ the same as $ GL(n,\mathbb{C} )$?
In an introductory linear algebra course, the notation " $\in $ lin$(\mathcal{V}) $ " is frequently used to mean the set of linear operators over a previously defined vector space, and I'm wondering if  the notation " $ \in GL(n,\mathbb{C})$ " would be loosely equivalent?

答: No. The set $GL(n,\Bbb C)$ is the set of all invertible $n\times n$ complex matrices. It can be seen as the set of all invertible linear maps from $\Bbb C^n$ into itself, but not as the set of all linear maps.


问:
Ask for two symbols
I find several symbols in Differentiable Manifold. I list them in the following,
$$
\int_{\Omega}*(V)^b.
$$
I wanna ask about the meaning of $V^b$ so that I can calculate it.

For $*$, I guess it means the dual space of $V^b$.

答: If $V$ is a vector space, $V^*$ is often used to denote the dual space, but in this case, the thing in the integral must be a differential form, so "$*$" probably means something like the Hodge-star operator. The superscript "b" doesn't mean anything to me except that maybe it means "apply hodge star b times," but that doesn't really make sense. More context is needed.

Title: Ask for two symbols