问:
Calculating interest rate for a payment plan
I really should know this from high school so I'm a bit ashamed to ask... :)
I've been offered several cars for lease and want to compare the offered interest rates.

E.g. I have:

```<code>Car: Ford Fiesta 1.0T
Lease period: 36 months
Price: 156619 SEK
Residual value: 37589 SEK (30%)
Monthly payment: 3159 SEK
</code>```

What's the formula to calculate the interest rate I'm being offered here?

答:
Price $P$. Residual value, or the amount for which you can successfully sell the car after $L$ months, $R$. Hence amount you are actually borrowing is effectively $B=P-R$.

Now you are paying a total of $T = L\times M$ where $M$ is the monthly payment. Therefore the percentage difference in the amount you pay versus the amount you borrow is
$$I=100\%\times( T-B)/B=100\%\times (LM/(P-R)  - 1)$$

If you want an equivalent monthly rate, then that can also be calculated fairly straightforwardly.


问:
A problem related to $\int_{0}^{1} f(x)(x-f(x))dx=1/12$

Consider  a differentiable  function satisfying $$\int_{0}^{1} f(x)(x-f(x))dx=1/12$$ Then find the nearest integer less than or equal to $\frac{1}{f'(1)}$.

Let $$F(x)=\int_{0}^{x}f(x)(x-f(x))$$ we notice $F(1)=1/12$ , $F(0)=0$ and $$F'(1)=f(1)-{f(1)}^2$$
What do i do next?
I have tried using LMVT and Rolle's but not getting anything nice.

答:
Thanks to @cosmo5  for inspiration we have in general $$ab\le \frac{{(a+b)}^2}{4}$$
$$\int_{0}^{1}f(x)((x-f(x))\le \int_{0}^1 \frac{{(f(x)+x-f(x))}^2}{4}=\int_{0}^{1}\frac{x^2}{4}=1/12$$
For equality to hold $$f(x)=x-f(x)\Rightarrow f(x)=\frac{x}{2}$$

ContentLicense: CC BY-SA 4.0


问:
How to prove $a x^{2a} -2 a x^a \geq x^{a+1}-(a+1)x$
For $x \in [0, 1]$ and $a \geq 1$, how to prove
$$a x^{2a} -2 a x^a \geq x^{a+1}-(a+1)x?$$

Because the inequality holds with equality at $x = 0$ and $x = 1$, we can not show that for each $a$, the function LHS - RHS is increasing / decreasing in $x$.

Because the inequality holds with equality at $a = 1$ for any $x$, this suggests that we can show, for each $x$, the LHS - RHS is increasing in $a$. But it is not easy to show this because the derivative is messy.

Is there any easy, elegant way to show this inequality?

Thanks

答:
The main idea is correct. Actually, the derivative is not messy with careful factoring.

Suppose $a$ is fixed. Define
$$f(x)=ax^{2a}-2ax^a-x^{a+1}+(a+1)x$$
Since you have found that $f(1)=0$ and $f(0)=0$, we may think we can do some thing with its monoticity! If we can prove $f(x)$ increases at first and decreases to $0$ when $x=1$,
$f(x)\ge 0$ will hold.

Thus its derivative can help us a lot:
\begin{align}
f'(x)&=2a^2x^{2a-1}-2a^2x^{a-1}-(a+1)x^a+a+1\\
&=2a^2x^{a-1}(x^a-1)-(a+1)(x^a-1)\\
&=(2a^2x^{a-1}-a-1)(x^a-1)
\end{align}
Define two new function $h(x)=2a^2x^{a-1}-a-1$ and $g(x)=x^a-1$, and then see how their value goes on $[0,1]$

Of course, $g(x)\le 0$ on $[0,1]$. While as to $h(x)$, it is an increasing fuction on $[0,1]$ and $h(0)=-a-1<0$, $h(1)=2a^2-a-1=(2a+1)(a-1)\ge0$. In this way, we should be able to find an $0<x_0<1$ such that $h(x_0)=0$. Therefore $h(x)\ge0$ on $[0,x_0)$ and $h(x)\le0$ on $(x_0,1]$.

Since $f'(x)=g(x)h(x)$, we can get that $f'(x)\ge0$ on $[0,x_0)$ and $f'(x)\le0$ on $(x_0,1]$. Thus, $f(x)$ will increase on $[0,x_0)$ and decrease on $(x_0,1]$, which fits the assumption we made at the beginning!

Factoring can help a lot in this kind of questions.


问:
Interpolation and approximation from Numerical Methods
I was reading about interpolation and approximation in Numerical Methods and came across this statement in my course material, "for n data points, there is one and only one polynomial of order (n − 1) that passes through all the points" for example, we have 3 data points on a straight line then how can a second order polynomial satisfy it?

答:
The statement could be rephrased with "of order at most $n-1$" or "with possibly zero coefficients". What matters is that these polynomials have exactly $n$ degrees of freedom, and more importantly, that Lagrangian interpolation is always possible and unique.

For completeness, the text might also specify $n$ data points "with different abscissas".


问:
Infinite sum of Wiener processes
Context

There are two different Wiener processes $W_t$ and $V_t$. It's known that they are independent. Additionally, we are given with third Wiener process $B_t$ that is given by the formula $$B_t = aW_t+bV_t, \quad \quad a^2 + b^2= 1.$$

Problem

Find the limit in $L^2$ of $$S_n = \sum_{i=1}^n\left[B_{it/n} - B_{(i-1)t/n}\right]\left[V_{it/n} - V_{(i-1)t/n}\right]$$

as $n$ tends to infinity.

My ideas

I assume that this is the type of task where we need to calculate the expected value and the variance. As the latter tends to $0$ (it should), we can say that the desired limit is the expected value. The issue is that it's very overextended work to calculate the $E(S_n)$.

Let $W_{it/n} = X_i$ and $V_{it/n} = Y_i$. We have

$$\Bbb E(S_n) = \Bbb E\sum_{i=1}^n [aX_i + bY_i - aX_{i-1} - bY_{i-1}][Y_{i} - Y_{i-1}]$$ which can be written as $$\sum \Bbb E\bigg( aX_iY_i + bY_i^2 - aX_{i-1}Y_i - bY_{i-1}Y_i   - aX_iY_{i-1} - bY_iY_{i-1} + aX_{i-1}Y_{i-1} + bY_{i-1}^2\bigg).$$ Next calculations confuse me (what is $\Bbb E(X_i Y_{i-1})$?) Is it zero? And the main question how to calculate the variance?

If I'm not mistaken, $\Bbb E(S_n) = nb \to \infty$, so we don't need variance. Am I right?

答: Let $P_i=W_{it/n}-W_{(i-1)t/n}$ and $Q_i=V_{it/n}-V_{(i-1)t/n}$. This means that $P_i,Q_i\sim{\sf N}(0,t/n)$, so\begin{align}\Bbb E[S_n]&=\sum_{i=1}^n\Bbb E\left[(aP_i+bQ_i)Q_i\right]=\sum_{i=1}^na\Bbb E[P_i]\Bbb E[Q_i]+b\Bbb E[Q_i^2]\\&=b\sum_{i=1}^n\Bbb V[Q_i]+\Bbb E[Q_i]^2=b\sum_{i=1}^n\frac tn\\&=bt.\end{align} Similarly, since $P_i$ is independent of $P_{i-1}$ and likewise for $Q_i$, we have \begin{align}\Bbb V[S_n]&=\sum_{i=1}^n\Bbb V\left[(aP_i+bQ_i)Q_i\right]\\&=\sum_{i=1}^n\Bbb E\left[(aP_i+bQ_i)^2Q_i^2\right]-\Bbb E[(aP_i+bQ_i)Q_i]^2\\&=\sum_{i=1}^na^2\Bbb E[P_i^2]\Bbb E[Q_i^2]+2ab\Bbb E[P_i]\Bbb E[Q_i^3]+b^2\Bbb E[Q_i^4]-\left(\frac{bt}n\right)^2\\&=\sum_{i=1}^na^2\left(\frac tn\right)^2+b^2\cdot3\left(\frac tn\right)^2-b^2\left(\frac tn\right)^2\\&=\frac{(a^2+2b^2)t^2}n\stackrel{n\to\infty}\to0.\end{align}

Title: Infinite sum of Wiener processes


问:
Object bounded by four planes
We have four surfaces : $z=0$, $y=0$, $z=1-x$, $y=1-x^2$. That is, the $xy$-plane, the $zx$-plane, a straight line contained in the $zx$-plane, and a parabola contained in the $xy$-plane. One might then determine the volume of this object with a triple integral. But isn't the volume of this object just zero? Meaning these planes do not contain within them any points.

I actually believe this set-up describes a half-cone, where the bottom half circle lies in the $xy$-plane, and its side is a triangle, which lies in the $zx$-plane. Is this the correct conclusion, and if so, how can I justify this from the given equations?

答:
You need to realize that, for instance, $y=1-x^2$ does not represent a line in the $xy$-plane. Since $z$ does not appear in the equation, it can take any value, and the equation actually represents a surface that corresponds to vertically moving the line that you mention. The same goes to $z=1-x$, which represents a plane.

If you want to picture it, draw the set $D = \{(x,y,0): -1 \leq x \leq 1, y\leq 1-x^2\}$ in the $xy$-plane and extend if perpencidularly to the $z$-axis making an infinite cylinder with a funny cross-section ($D$). Now just consider the portion of that cylinder that is above the plane $z=0$ (bottom) and the below the plane $z=1-x$ (top).

Title: Object bounded by four planes


问:
I need to compute the integral $\int _0^s\:x^2$ using Riemann sums.
I need to compute the integral $\int _0^s\:x^2$  using Riemann sums.

a) Consider the function $y=x^2$  on the interval [0, s] and divide it into n equal intervals. Each of these subintervals has length $\frac{s}{n}$.

b) Find the sum $U_n$ of all rectangles below the function $y=x^2$

$$U_n=\frac{s}{n}\left(0^3+\left(\frac{s}{n}\right)^3+\left(2\frac{s}{n}\right)^3+...+\left(n-1\right)\left(\frac{s}{n}\right)^3\right)= \frac{s^3}{n^3}\sum _{k=0}^{n-1}\:k^3$$

c) Now find the sum $O_n$ of all rectangles above the function $y=x^2$

$$O_n=\frac{s^3}{n^3}\left(1^3+2^3+3^3+...+n^3\right)=\frac{s^3}{n^3}\sum _{t=1}^nk^3$$

d) Take the limits to show that:
$$\lim _{n\to \infty }\left(U_n\right)=\lim \:_{n\to \:\infty \:}\left(O_n\right)$$

So My answer is:

$$U_n=\frac{s^3}{n^3}\sum _{k=0}^{n-1}\:k^3=\frac{s^3}{n^3}((\frac{n\left(n+1\right)}{6})-n^3$$

Therefore $U_n=\frac{s^3}{n^3}=\lim _{n\to \infty }\left(\frac{s^3}{3}+\frac{s^3}{2n}-\frac{s^3}{n}\right)=\frac{s^3}{3}$

Is it correct? I'd appreciate some guide and feedback if I am wrong.

答:
First a comment. If you assume that $x \mapsto x^2$ is Riemann integrable, you don't need to compute both upper and lower Riemann sums and verify that those are equal. This would be required to prove the Riemann integrability.

Now regarding the computation itself. You have

$$U_n=\frac{s}{n}\left(0^2+\left(\frac{s}{n}\right)^2+\left(2\frac{s}{n}\right)^2+...+\left(n-1\right)\left(\frac{s}{n}\right)^2\right)= \frac{s^3}{n^3}\sum_{k=0}^{n-1} k^2.$$

The power is $2$ and not $3$ as you're integrating $x \mapsto x^2$.

Using

$$\sum_{k=0}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6}$$ and taking the limit you get the right answer $$\int_0^s x^2 \ dx = \frac{s^3}{3}.$$


问:
Proving Morera's Theorem through the Cauchy-Goursat theorem for rectangles
In my book Morera's Theorem is stated as for if $f$ is a continuous function on an open set $U$, that has the property that around the boundary of any rectangle the closed integral is zero, then  $f$ is holomorphic.

I want to prove that this is equivalent to saying if $f$ is a continuous function on an open set $U$ that has the property $\oint f=0$ along any closed path in $U$, then $f$ his holomorphic.

A bit context: My book is Serge Lang's Introduction to Complex Analysis at an Undergraduate level. The book introduced the Cauchy-Goursat theorem for rectangles and then proved Cauchy's Integral Formula (for simply connected open sets)

Question:
Does anybody know how could I show that the conditions "around the boundary of any rectangle in $U$ the closed integral is zero" and "$\oint f$ along any closed path in $U$" are the same?

答:
The two conditions are not equivalent. Here is a counterexample: Let $U:=U_1(0)\backslash U_r(0)$, where $\frac{\sqrt2}{2}<r<1$. That's a thin annulus with the inner and outer radius chosen such that no rectangle can can be drawn around the center of the annulus. Essentially, if you draw any rectangle inside a circle of radius $1$, you can't fit a circle with radius larger than $\frac{\sqrt2}{2}$ inside the rectangle. So with the chosen radii, the inner boundary will never be inside any rectangle inscribed into the annulus. If any rectangle is inscribed into the annulus, then the inside of the rectangle is a subset of $U$.

Now take $f:U\to\mathbb C,~z\mapsto \frac{1}{z}$. According to the Goursat theorem, the integral $\oint_{\square}f(z)\mathrm dz$ along any rectangle $\square$ vanishes, since $f$ is holomorphic and the rectangle, including its interior, are included in $U$. But the integral $\oint_{\vert z\vert=\rho}f(z)\mathrm dz$ with $r<\rho<1$ is known to not vanish (it's $2\pi\mathrm i$).


问:
Collinearity of three points
In my book its given , three points A,B,C with position vectors a,b,c are collinear if and only if there exists scalars x,y,z not all zero simultaneously such that xa + yb + zc = 0, where x + y + z = 0.

Surprisingly, this is also the condition for coplanarity of three vectors. But All COPLANAR vectors are NOT collinear . So where is the the flaw in my argument ?

答:
Ok i think i've got it now.
For coplanarity the condition is that ax + by + cz = 0 for scalars x,y,z where not all x,y,z = 0 , but for collinearity there is an additional condition that x + y + z = 0   so that one point can be shown to to divide the other two points in some ratio .</p>

Title: Collinearity of three points


问:
Jordan Form of T and finding a basis relative to which T is in Jordan Form
Let $$T = \left(\begin{array}{ccc} 0&-1&-1\\-3&-1&-2\\7&5&6 \end{array}\right)$$
Find the Jordan Form of T, then a basis relative when T is in Jordan Form.

I've found that the characteristic polynomial $f(x)=$ minimal polynomial $p(x)=(x-1)(x-2)^2$, which gives me the eigenvalues of $\lambda=1,2$.

To find the Jordan Form, I looked at the dimensions of $Ker(T-I)$ and $Ker(T-2I)$ to figure out how many Jordan Blocks and their size. So I have determined that the Jordan Form is
$$J_T = \left(\begin{array}{ccc} 1&0&0\\0&2&0\\0&1&2 \end{array}\right)$$

Is this correct? If so, how do I continue with this to find the basis of T in jordan form?

答:
Apart from finding the dimensions of $\ker(T-I)$ and $\ker(T-2I)^2$ you need to find a basis for them: an eigenvector for the first, a chain for the second,  $(T-2I)v=w$, $(T-2I)w=0$. $$\ker(T-I)\mapsto \begin{pmatrix}0\\1\\-1\end{pmatrix}$$
$$\ker(T-2I)^2\mapsto\begin{pmatrix}1\\1\\-3\end{pmatrix}, \begin{pmatrix}-1\\0\\1\end{pmatrix}$$

So taking the change of basis matrix $P=\begin{pmatrix}0&-1&1\\1&0&1\\-1&1&-3\end{pmatrix}$ gives $$P^{-1}AP=J=\begin{pmatrix}1&0&0\\0&2&0\\0&1&2\end{pmatrix}$$


问:
Definition of regular point of a boundary with planar brownian motion
This is an exercise in G.Lawler's book Conformally invariant processes in the plane.

First he defined regular point of a boundary using brownian motion:

Suppose $D$ is a domain in $\mathbb{C}$ with boundary $\partial D$. A point $z\in\partial D$ is called regular if $\mathbb{P}^z[\tilde{\tau}_D=0]=1$,
where $\tilde{\tau}_D=\inf\{t>0:B_t\not\in D\}$.

The exercise is to show that $z\in\partial D$ is regular if and only if for every $\delta>0$,
$$
\lim_{\epsilon\to0+}\sup_{|w-z|<\epsilon}\mathbb{P}^w[|B_{\tau_D}-z|\ge\delta]=0,
$$
where $\tau_D=\inf\{t\ge0:B_t\not\in D\}$.

This seems intuitively to be right because if a brownian motion is started very near to $z$, it also exits the domain at a point near $z$. I know how to prove this when the boundary is a straight line. However here we don't know the exact form of the boundary. How can we use the stopping time information to show the equivalence?

答:
"$\Rightarrow$" Let $h>0$ arbritrary, then clearly

$$\mathbb{P}^w(|B_{\tau_D}-z| \geq \delta) = \underbrace{\mathbb{P}^w(|B_{\tau_D}-z| \geq \delta, \tau_D \leq h)}_{=:p_1} + \underbrace{\mathbb{P}^w(|B_{\tau_D}-z| \geq \delta, \tau_D > h)}_{=:p_2}$$

We have $$\begin{align*} p_1 &\leq \mathbb{P}^w \left( \sup_{t \leq h} |B_t-z| \geq \delta \right) \stackrel{\varepsilon \leq \frac{\delta}{2}}{\leq} \mathbb{P}^w \left( \sup_{t \leq h} |B_t-w| \geq \frac{\delta}{2} \right) \\ &= \mathbb{P}^0 \left( \sup_{t \leq h} |B_t| \geq \frac{\delta}{2} \right)  \leq \left( \frac{2}{\delta} \right)^2 \cdot \mathbb{E} \left( \sup_{t \leq h} |B_t|^2 \right) \\ &\leq 4 \cdot \left( \frac{2}{\delta} \right)^2 \cdot \mathbb{E}(B_h^2) = 4 \cdot \left( \frac{2}{\delta} \right)^2 \cdot d \cdot h \to 0 \qquad (h \to 0)\end{align*}$$ by applying Markov's inequality and Doob's maximal inequality.

For the second addend, we use that $$\forall h>0: \lim_{w \to z} \mathbb{P}^w(\tau_D > h) = 0$$ since $z$ is a regular point (see the theorem below).

"$\Leftarrow$" This is the easier direction, I leave it to you.

Theorem Let $z_0 \in \partial D$ a regular point. Then $$\lim_{w \to z_0} \mathbb{P}^w(\tau_D > h) = 0$$ for all $h>0$.

Proof: Let $w \in \mathbb{R}^d$, then $$\begin{align*} \mathbb{P}^w(\tau_D > h) &= \mathbb{P}^w(\forall s \in (0,h]: B_s \in D) \\ &= \inf_{n \in \mathbb{N}} \mathbb{P}^w(\forall s \in [1/n,h]: B_s \in D) \\ &= \inf_n \mathbb{E}^w \bigg( \mathbb{P}^{B_{1/n}}(\tau_D > h-1/n) \bigg) \\ &= \inf_n \mathbb{E}^w \big( \phi(B(1/n)) \big)\end{align*}$$ where $\phi(y) := \mathbb{P}^y(\tau_D>h-1/n)$ is bounded and borel-measurable. Hence $$w \mapsto \mathbb{E}^w(\phi(B(1/n)))$$ is continuous. This implies the assumption.

(cf. René L. Schilling/Lothar Partzsch, Brownian motion - An Introduction to Stochastic Processes, Theorem 8.16)

ContentLicense: CC BY-SA 3.0


问:
I need to solve an equation with both $\theta$ and $\sin\theta$ in it. How do I do it?
For example, if we have a circle as shown, with angle AOB=120 degrees, and radius 10 units. If we need to find area of the shaded segment, then we find area of sector and subtract area of triangle.

Here, area of sector would be $$\frac{120}{360}\pi\times10^2=\frac{100}{3}\pi$$
And the area of the triangle is $$25\sqrt3$$
So the area of segment is $$\frac{100}{3}\pi-25\sqrt3$$
Now, if the angle AOB was a variable $\theta$ and we were told that the shaded area is $\frac{100}{3}\pi-25\sqrt3$. Then we get an equation like $$\frac{\theta}{360}\pi\times100-50\sin\theta=\frac{100}{3}\pi-25\sqrt3$$
How would you solve this?
I know that I can equate the terms with $\pi$ in them and get the value of $\theta$, but what if we were given just values instead of something in terms of $\pi$ and $\sqrt3$? (The approximate value is 8.39468514934)

答:
The equation

$$y=x-\sin(x)$$ is not invertible analytically.

In the range $[0,\pi]$, you have the bracketing

$$\frac{x^3}{\pi^2}\le x-\sin(x)\le\frac{x^2}{\pi}$$ which can give you two gross approximations of $x$. You can extend this range by using symmetries. Then use a few Newton's iterations.


问:
Ambiguity on BODMAS, Multiplication and Division
$24\div4(8\div 4)$ - This is the equation. We need to solve. Lets do it then

$=> 24\div4(2)$

$=> 24\div8$

$=> 3$

Division is also the multiplication of the inverse. Therefore $24\div4(8\div 4)$ can be written as

$24*\frac{1}{4}(8*\frac{1}{4})$. Let us solve this

$=> 24 *\frac{1}{4}(2)$

$=> 24 *\frac{1}{2}$

$=> 12$

My question is which of the answer is correct? 3 or 12. I have followed BODMAS rule in both representation of the equation.

答:
The trouble with BOMDAS is that it's not usually implied.

```<code>It's like this
() brackets
close product (ie ab)
f  of   So cos ab = cos(a×b).
× and ÷  left to right
+ and -  left to right
</code>```

What actually happened in your calculation, is that when you switched the division into multiplication, then $4(8÷4)$ should had went to $\frac1{4(8÷4)}$.  This evaluates in both cases to 8, the division is thus $3$.

The fun really starts when you do something like $ad÷bc$ as $\frac{ad}{cd}$.  Because BOMDAS does not involve close division, then the equation ought be $\frac{acd}b$.

Normally, BOMDAS is designed to reduce brackets.  But the rule is faulty: The numeric appears at the front of any product, and any numerics further down are read some other way.  This allows, eg $2\cos 2\pi\theta$.

Body:
The trouble with BOMDAS is that it's not usually implied.

It's like this
() brackets
close product (ie ab)
f  of   So cos ab = cos(a×b).
× and ÷  left to right
+ and -  left to right

What actually happened in your calculation, is that when you switched the division into multiplication, then $4(8÷4)$ should had went to $\frac1{4(8÷4)}$.  This evaluates in both cases to 8, the division is thus $3$.

The fun really starts when you do something like $ad÷bc$ as $\frac{ad}{cd}$.  Because BOMDAS does not involve close division, then the equation ought be $\frac{acd}b$.

Normally, BOMDAS is designed to reduce brackets.  But the rule is faulty: The numeric appears at the front of any product, and any numerics further down are read some other way.  This allows, eg $2\cos 2\pi\theta$.


问:
Calculate the derivative of a complex norm
I'm stuck with a rather trivial looking question. How do you calculate the derivative of the norm of a complex number to it self? Like in

$$
\frac{d|a|^2}{da} = ?
$$

I think it would give rise to a real valued function but in stead I get

$$
\frac{d|a|^2}{da} = \bar{a}
$$

I can't understand this result.

Thanks in advance.

答:
Although this question is for previous years, I wanted to give a more complete and descriptive answer that might be useful to others.
$$
 \frac{d |a|^2}{da}=\frac{d (a a^{*})}{da}=\frac{1}{2}\left(\frac{d( a a^{*})}{d \Re{a}}-i \frac{d( a a^{*})}{\Im{a}}\right) \\
\frac{1}{2}((a+a^{*})-(a-a^{*}))=a^{*}
$$
$ a$ is complex number in $\mathbf{C}$.


问:
Nice lower bounds for $a^n + b^n$ for $a,b,n \geq 1$?
I know this isn't a very typical question, but i was wondering if anyone knows any good lower bounds for $a^n + b^n$. I'm looking for something akin to $a^n + b^n \leq (a + b)^n$ for $n \geq 1$.

The motivation is that I'm trying to find some nice upper bounds on $N$ for the minimum value of $N$ when $a_1^N + a_2^N + \dots + a_k^N \geq M$, for some given $1 \leq a_1, \dots, a_k \leq 2$ and $M > 0$. I don't need this to be particularly tight, but I was thinking if I could write this as some expression $a_1^N + a_2^N + \dots + a_k^N \geq f(a_1, a_2, \dots, a_k)^N \geq M$, then if $f$ is nice enough the $\log$ upper bound on $N$ should be good enough for my purposes.

I've tried using the AM-GM to get $a^n + b^n \geq nab$, but as that doesn't give a logarthmic bound it's not good enough for my purposes. I've scrolled through a bunch of other lists of inequalities and I can't find anything else that seems to work.

Does anyone have any ideas? Either for my original question or for the motivation? Thanks for the help!

答:
With respect to the original problem:

If all $a_j$ are equal to one then $a_1^N + a_2^N + \dots + a_k^N \ge M$ is equivalent to $k \ge M$, independently of $N$.

Otherwise Jensen's inequality for the convex function $t \to t^N$ (see for example Prove inequality using Jensen's inequality) gives the following estimate:
$$
 a_1^N + \ldots + a_k^N \ge k^{1-N}(a_1 + \ldots + a_k)^N
$$
and that is $\ge M$ if
$$
 \frac{(a_1 + \ldots + a_k)^N}{k^N} \ge \frac M k
$$
or
$$
 N \ge \frac{\log M - \log k}{\log(a_1 + \ldots + a_k) - \log k} \, .
$$


问:
Every Element of $\mathbb F_p$ is a Square in $\mathbb F_{p^2}$ if $p \equiv 3 \mod 4$
How can we show that every element of $\mathbb F_p$ is a square in $\mathbb F_{p^2}$ if $p \equiv 3 \mod 4$ is a prime?

This comes from the first paragraph of Koblitz's "Introduction to Elliptic Curves and Modular Forms" on page 61, self-study.

答:
Here is another proof which does not use the cyclicity of $F^\times$, nor the characterization of cyclic groups mentioned in Atticus' answer.

Let $F$ be a field with $q$ elements, where $q$ is odd. Then, $x\in F^\times $ is a square if and only if $x^{(q-1)/2}=1$.

Indeed, by Lagrange theorem, non zero squares are roots of $X^{(q-1)/2}-1\in F[X]$. Now, apply first isomorphism theorem to $x\in F^\times \mapsto x^2\in F^\times$ shows that there are $(q-1)/2$ nonzero squares, and we are done.

Now assume $q=p^2.$ It is known that $\mathbb{F}_p\subset \in F$, and that nonzero elements of $\mathbb{F}_p$ satisfy $x^{p-1}=1$ (Lagrange again, for example).

But $(q-1)/2=(p-1)(p+1)/2$. Thus, if $x$ lies in $\mathbb{F}_p^\times$, then $x^{(q-1)/2}=1$ and $x$ is a square in $F^\times$.


问:
$H_{1}([0,1]/A)$ can not be countable generated when $A=\{1/n\mid n\in\mathbf{Z},n>0\}\cup\{0\}$
I think this is another example when good pairs condition is not satisfied. Suppose $A=\{1/n\mid n\in\mathbf{Z},n>0\}\cup\{0\}$, from the exact sequence of homology group I can get that $H_1([0,1],A)$ is countably generated, but I don't know how to prove that $H_{1}([0,1]/A)$ is not countably generated. Can someone give me some ideas on how to prove that? Thanks!

(I tried to think of it as the wedge sum of countable many $S^{1}$, but its homology group can be countably generated...)

答:
Your space $A$ is homeomorphic to the Hawaiian earring (see my answer to Construct compact space with some homology group not finitely generated). This is not the wedge sum of countably many copies of $S^1$ In fact, this wedge sum is a non-compact CW-complex.

Now see Construct compact space with some homology group not finitely generated which answers your question.


问:
Indian National Olympiad $2000$ Problem $6$

For any natural numbers $n$, ( $n \geq 3$), let $f(n)$ denote the number of non-congruent integer-sided triangles with perimeter $n$. Show that
(i) $f(1999) > f (1996)$;
(ii) $f(2000) = f(1997)$.

My proof outline -

for second part -

Let $a,b,c$ be three integer sides for the triangle with perimeter $2000$ then $$a+b+c=2000$$ and by triangular inequality we have that $a+b>c,b+c>a,c+a>b$ so we get that $a,b,c<1000$

Now applying the formula of stars and bars we get that we have a total of ${1000\choose2}$ and since we wwant non congruent we divide this by $3$ ,similarly we do for $f(1997)$ but here answer is coming  ${999\choose2}$ .

So how $f(2000)=f(1997)$

Someone pls point out where is the mistake ?

thankyou

答:
You have misapplied stars and bars here. The quantity $\binom{1000}{2}$ partitions 1001 elements into three pieces. If you wanted to compute $f(2000)$,  you want to partition 2000 elements into three pieces, where each partition has at most 999 elements. Furthermore, to remove double-counting, you cannot simply divide by 3, since for example $(1,1,1)$ is unique for triangles of perimeter $3$.

I would suggest that a different approach be taken. Instead of attempting to compute $f(1997)$ or $f(2000)$ (and similarly for the first part of the exercise), find a way to map a triangle of the smaller perimeter to a triangle of the larger perimeter.

Hint: If $T_n$ denotes the set of sidelengths of such triangles with perimeter $n$, and $(a,b,c) \in T_n$, then $(a+1, b+1, c+1) \in T_{n+3}$. Now try to invert this correspondence. In particular, show why the inversion fails for $n = 1996$ but succeeds for $n = 1997$.


答:
You cannot rely on stars and bars. Stars and bars counts $\{a=600,b=650,c=750\}$ and $\{a=600,b=750,c=650\}$ as separate and in this case these values will permute in $3!=6$ ways, not $3$. On the other hand $\{600,700,700\}$ permutes in $3!/2!=3$ ways, not $6$. So it is clear that you can't obtain the answer by dividing by $3$ (or $6$).

We will denote a triangle by the unordered set containing its sides $\{a,b,c\}$. Let $A_n$ be the set of integer sided non-congruent triangles of perimeter $n$. Then $f(n)=|A_n|$. Define a map$$g:A_{1996}\to A_{1999}\\g(\{a,b,c\})=\{a+1,b+1,c+1\}$$Since $a+b>c\implies (a+1)+(b+1)>(c+1),g$ is well defined.

This map is injective, since$$g(\{a_1,b_1,c_1\})=g(\{a_2,b_2,c_2\})=\{p,q,r\}\\\implies \{a_1,b_1,c_1\}=\{a_2,b_2,c_2\}=\{p-1,q-1,r-1\}$$Thus $f(1999)\ge f(1996)$.

We now try to find a triangle in $A_{1999}$ that is not in $g(A_{1996})$. Let such a triangle be $\{a,b,c\}$, then$$a+b+c=1999\tag{1}$$and $\{a-1,b-1,c-1\}\not\in A_{1996}$. Since the perimeter of $\{a-1,b-1,c-1\}=1999-3=1996$, we require the triangular inequality to fail for $\{a-1,b-1,c-1\}$ i.e.$$(a-1)+(b-1)-(c-1)\le0\implies a+b-c\le 1$$Since $\{a,b,c\}\in A_{1999},a+b-c>0$. This gives$$a+b-c=1\tag{2}$$since $a,b,c$ are integral.

$c=999,a=500=b$ is a solution of $(1),(2)$. Hence we get $f(1999)>f(1996)$.

We can define a similar injection $g_1$ from $A_{1997}\to A_{2000}$. But we run into an issue when we try to find a triangle in $A_{2000}$ that is not in $g_1(A_{1997})$. Equations $(1),(2)$ get modified as$$a+b+c=2000\tag{1'}$$$$a+b-c=1\tag{2'}$$Adding these two we get $2(a+b)=2001$ which is not possible. Hence $f(2000)=f(1997)$.

Title: Indian National Olympiad $2000$ Problem $6$

Body:
For any natural numbers $n$, ( $n \geq 3$), let $f(n)$ denote the number of non-congruent integer-sided triangles with perimeter $n$. Show that
(i) $f(1999) > f (1996)$;
(ii) $f(2000) = f(1997)$.

My proof outline -

for second part -

Let $a,b,c$ be three integer sides for the triangle with perimeter $2000$ then $$a+b+c=2000$$ and by triangular inequality we have that $a+b>c,b+c>a,c+a>b$ so we get that $a,b,c<1000$

Now applying the formula of stars and bars we get that we have a total of ${1000\choose2}$ and since we wwant non congruent we divide this by $3$ ,similarly we do for $f(1997)$ but here answer is coming  ${999\choose2}$ .

So how $f(2000)=f(1997)$

Someone pls point out where is the mistake ?

thankyou


问:
Spivak chapter 5 exercise 36 - prove $\lim_{x \to 0^-} f(1/x) = \lim_{x \to -\infty} f(x)$
There're similar questions already, but I somewhat struggled to apply their reasoning to this particular statement.

I wanted to ask if my proof is correct (and clarify several things which I've seemingly figured out when writing it here). Here's the proof:

$$\mbox{(1) }\lim_{x \to 0^-} f\left(\frac1x\right)=l \mbox{ if } \forall\varepsilon>0(\exists\delta>0(\forall x(0<0-x<\delta \longrightarrow|f\left(\frac1x\right)-l|<\varepsilon$$

$$\mbox{(2) }\lim_{x \to -\infty} f(x)=m \mbox{ if } \forall\varepsilon>0(\exists N(\forall x(x<N \longrightarrow|f(x)-m|<\varepsilon$$

We need to show that $l=m$.

from (1) we have
$$-\delta>x>0\longrightarrow|f\left(\frac1x\right)-l|<\varepsilon$$

let $g(x)=f\left(\frac1x\right)$, then we have
$$-\delta>x>0\longrightarrow|g\left(x\right)-l|<\varepsilon$$

in (2) we suppose that $x<N$. Hence $\frac1x>\frac1N$. Let $x'=\frac1x$. Furthermore, since $-\delta$ is negative and we can assume that $N$ is negative. (If it's not, we can take $N$ to be -1 or any other negative number, because if the conclusion holds $\forall x<N$, it surely holds for a subset $(-\infty,-1)$). We can take $-\delta=\frac1N$, which gives us

$$\frac1N=-\delta<x'<0$$ and this is the hypothesis from (1). Hence we conclude that

$$|g\left(x'\right)-l|<\varepsilon$$ or $$|f\left(\frac1{\frac1x}\right)-l|<\varepsilon$$ or $$|f\left(x\right)-l|<\varepsilon$$

To summarise, $x<N\longrightarrow|f\left(x\right)-l|<\varepsilon$. By the definition of the limit $$\lim_{x \to -\infty}f(x)=l$$ but also $$\lim_{x \to -\infty}f(x)=m$$ Therefore $l=m$.

答:
The proof you give becomes a bit unwieldy and it does not make clear that $N$ must be negative.

I think it can be shortened as follows:

Assume $\lim_{x\to0^-} f(1/x)$ exists and equals $\ell$.  Then take any $\varepsilon > 0$.

Because the limit above exists, there exists $\delta > 0$ such that
$$\lvert f(1/x) - \ell\rvert \leqslant \varepsilon \text{ for all } -\delta \leqslant x < 0. \tag{1}\label{eq1}$$

Now take any $y \leqslant -1/\delta$, write $x=1/y$.  It follows $-\delta \leqslant x < 0$.  This satisifies the criterion in \eqref{eq1}, so $\lvert f(1/x)-\ell \rvert \leqslant \varepsilon$, which is the same as $\lvert f(y) - \ell\rvert \leqslant \varepsilon$.

Thus for every $\varepsilon > 0$ there exists $m$ (taking $m=1/\delta$) such that $\lvert f(y)-\ell\rvert \leqslant \varepsilon$ whenever $y \leqslant -m$. This provbes $\lim_{y\to-\infty}f(y)$ exists and equals $\ell$.


问:
Subobject Classifier of a Topos is Injective
Let $\mathcal{E}$ be a topos with subobject classifier $1\overset{t}{\rightarrow}\Omega$, then I want to show that $\Omega$ is injective.

So let $f:A\rightarrow\Omega$ and $g:A\rightarrow B$ be two maps in $\mathcal{E}$ with $g$ monic. Then I want to show that there exists a map $h:B\rightarrow\Omega$ such that $h\circ g = f$. Notice that by the property of a subobject classifier we have that there exists a unique map $\phi:B\rightarrow\Omega$ such that there is a pullback square
$\require{AMScd}$
\begin{CD}
A @>{\psi}>> 1\\
@VV{g}V @VV{t}V\\
B @>{\phi}>> \Omega
\end{CD}
Now I claim that $\phi$ does the job for $h$. But I don't see why $\phi\circ g= f$ (so maybe my claim is wrong). Any help would be appreciated!

答:
Eric's answer is spot on.

Here is an alternative method in which you might also be interested. Employing the internal language of the topos, we may reason in a naive element-based language similar to the ordinal mathematical language, as long as we take to restrict to intuitionistic logic (that is, not use the law of excluded middle). The question then becomes this:

Given an injective map $g : A \to B$ between sets and a map $f : A \to \Omega$, where $\Omega$ is the powerset of $1 = \{\heartsuit\}$, why is there an extension $g : B \to \Omega$?

And the answer is: We may set $g(y) := \{ p \in 1 \,|\, y \in A \Rightarrow 1 \in f(y) \}$. For $y \in A$, we have $g(y) = f(y)$, hence this map $g$ is indeed an extension.


问:
Compute $i^{i^i}$

Find $i^{i^i}$ Show that it does not coincide with $i^{i\cdot i} =
i^{−1}$

Attempt:
$$i^i = e^{i \log(i)} = e^{i[i(\pi/2 + 2n\pi)]} = e^{-\pi/2 - 2n\pi}$$

Thus,

$$i^{i^i}= e^{i^i \log (i)} = e^{ e^{-\pi/2 - 2n\pi}[i(\pi/2 + 2k\pi)]}$$

I am confused, does this have to be so much complicated? Also, how to show the second part exactly?

答: Your attempt on the first part looks good to me. For the second part, by Gelfond–Schneider theorem, if $a\ne0,1$ is an algebraic number and $b$ is not a real rational number, then $a^b$ is transcendental (regardless of the branch of logarithm one chooses when $a$ is complex). Apply the theorem with $a=b=i$, we see that $i^i$ is transcendental. Apply the theorem again with $a=i$ and $b=i^i$, we see that $i^{i^i}$ is transcendental. Hence it is not equal to $i^{-1}=-i$, which is algebraic.

Title: Compute $i^{i^i}$


问:
Let $D$ and $E$ be points on $AB$ and $BC$ of $\Delta ABC$ such that $AD = 7BD$ , $BE = 10CE$ . Let $AE$ and $BC$ meet at point $F$

Let $D$ and $E$ be points on segments of $AB$ and $BC$ of $\Delta ABC$ such that $AD = 7BD$ , $BE = 10CE$ . Let $AE$ and $BC$ meet at point $F$. If $AF = k \cdot FE$ , find $[k]$ .

What I Tried: Here is a picture :-

The first thing which I did was to try the process in Geogebra. There I got the value of $k$ to be approximately $7.74$ , so $[k] = 7,$ and I think that should be the answer. I also think that the initial value of $k$ varies, but $[k]$ is always $7$ . How will I be able to come up with a solution then?

Can anyone help me? Thank You.

答:
Use mass points method.

Place masses $70$ at $B$, $10$ at $A$, $7$ at $C$.

This gives $77$ at $E$ implying

$$ \dfrac{AF}{FE}= 7.7$$


问:
If $\operatorname{gcd}(a,b)=1$ then $\operatorname{gcd}(3a+b,a+3b)=$?

If $\operatorname{gcd}(a,b)=1$ then $\operatorname{gcd}(3a+b,a+3b)=$?

I try to solve it.

We are two cases :

If $a$ be even and $b$ is odd number.
Then $\operatorname{gcd}(3a+b,a+3b)=1$

$a$ and $b$ be odd numbers.then $\operatorname{gcd}(3a+b,a+3b)$ divided 8...

I guess  $\operatorname{gcd}(3a+b,a+3b)=2$.

答:
If $d$ divides  $3a+b, a+3b$

d must divide

$3(3a+b)-(a+3b)=8a$

$3(a+3b)-(3a+b)=8b$

$\implies d$ must divide $(8a,8b)=8(a,b)=8$

So, $g=(3a+b,a+3b)\in\{1,2,4,8\}$

$g=1$  if $3a+b$ is odd $\iff a+b$ is odd i.e., $a,b$ have opposite parities

Again   $g=8$  if $3a+b\equiv0\pmod8\iff b\equiv-3a\pmod8\iff a\equiv-3b\pmod8$

What about $g=2,4?$


问:
APR Calculation
I'm hoping someone can clarify this for me.

The model/example is this:

We lend an amount of 1498.50 (loan amount). Other fees total 39.95.
The term of the loan is for 12 months. There is no interest per month, per se, but a 10.00 fee is charged for each of the 12 months.

Using federally-approved software that calculates APR (to comply with Regulation Z, Appendix J), the calculation returns the APR to be 21.488%.

The monthly payment is calculated as follows:

(Loan Amount + (10.00 x 12 months)) / 12 = 134.87.

According to the amortization schedule, none of the interest for any particular month exceeds 14%.  How can the APR for the entire term be ~ 21% when none of the effective monthly interest rates even come close to 21%?

My intuition tells me that there is some averaging going on, but to achieve an average of 21%, some months would have to exceed 21% by quite a bit, while being offset by smaller percentages on the fringes of the curve.

If you are familiar with Reg Z, Appendix J, and the given example above, would you say that the formula being used is not appropriate?  (There are a few others that the feds provide; maybe we/the software applied the wrong one)

Additionally, playing around with the software, keeping all numbers the same, but changing the term (in months), I find that the APR also changes.  I find this confusing and counter-intuitive.

For example:
3 months, APR is 39.126%
6 months, APR is 27.393%
12 months, APR is 21.488%.

Can someone please explain this?

*All numbers above are the input and output values of approved calculation software.

Huge thanks in advance!

PS: Apologies for the seemingly uninformative tag(s). I couldn't find one that was more meaningful.

Title: APR Calculation

Body:
I make the annual interest rate about 21.532%, though there may be some rounding issue.  Since $1.21532^{1/12}=1.01638$, this corresponds to a monthly interest rate of about 1.638%.

Initially the amount handed over less the fee is 1498.50-39.95 = 1458.55.

M   Inter-  Balance     Payment Balance
     est    before              after
            payment             payment

0           1498.500     39.950 1458.550
1   23.896  1482.446    134.875 1347.571
2   22.078  1369.649    134.875 1234.774
3   20.230  1255.004    134.875 1120.129
4   18.352  1138.481    134.875 1003.606
5   16.443  1020.048    134.875  885.173
6   14.502   899.676    134.875  764.801
7   12.530   777.331    134.875  642.456
8   10.526   652.982    134.875  518.107
9    8.488   526.595    134.875  391.720
10   6.418   398.138    134.875  263.263
11   4.313   267.576    134.875  132.701
12   2.174   134.875    134.875    0.000

To calculate the monthly interest rate $i$, you need to find the solution to

$$(L-f)(1+i)^m - p((1+i)^m-1)/i = 0$$

where $L$ is the loan, $f$ is the fee, $m$ is the number of months, and $p$ is the monthly payment.  Here, I get $i=0.01638$.

Then to get the annual interest rate

$$(1+i)^{12}-1$$

which I make $0.21532$.  Multiply by 100 to get percentages.


问:
Question about changing minimizer and impact on objective value
I have the a question about the following proof, page 5 lemma 2. Suppose we want for $m$ data samples $y_i$ to minimize $\sum_{i=1}^ml(y_i-\mu)$ where $l(x)=x\tau\mathbb{1}_{x\ge 0} -x\mathbb{1}_{x\le 0}(1-\tau) $. If $\mu^*$ is a minimizer of the optimization problem above and we denote with $m_+$ the number of points where $y_i>\mu^*$. If we slightly increase the $\mu^*$ by $\delta\mu^*$ why does the objective function change according to $[(1-m_+)(1-\tau)-m_+\tau]\delta\mu^*$?

I do see that the part $-m_+\tau\delta\mu^*$ is what we "gain". However, the first part is clearly from the fact that the other points below $\mu^*$ are getting a worse value. But it could well be that some change side. So how do they come up with that expression in particular the value $(1-m_+)$?

答:
I obtained that the function change $\delta f$ equals $(m(1-\tau)-m_+)\delta\mu^*$ for sufficiently small $\delta\mu^*$ as follows. We have $\delta f=\sum_{i=1}^m l(y_i-\mu^*-\delta\mu^*)- l(y_i-\mu^*)$. We pick $\delta\mu^*$ so small that $y_i-\mu^*-\delta\mu^*$ has the same sign as $y_i-\mu^*$ for each $i$ such that $y_i-\mu^*\ne 0$. Then

$$l(y_i-\mu^*-\delta\mu^*)- l(y_i-\mu^*)=$$
$$((y_i-\mu^*-\delta\mu^*)-(y_i-\mu^*))\tau\mathbb{1}_{ y_i-\mu^*\ge 0}-
((y_i-\mu^*-\delta\mu^*)-(y_i-\mu^*))(1-\tau)\mathbb{1}_{ y_i-\mu^*\le 0}=$$
$$-\delta\mu^*\tau\mathbb{1}_{ y_i-\mu^*\ge 0}+\delta\mu^* (1-\tau)\mathbb{1}_{ y_i-\mu^*\le 0}.$$
$$-\delta\mu^*\tau+\delta\mu^*\mathbb{1}_{ y_i-\mu^*\le 0}.$$

If $y_i=\mu^*$ then anyway

$$l(y_i-\mu^*-\delta\mu^*)- l(y_i-\mu^*)=l(-\delta\mu^*)-l(0)=\delta\mu^* (1-\tau)-0
=-\delta\mu^*\tau+\delta\mu^*\mathbb{1}_{ y_i-\mu^*\le 0}.$$

Thus

$$\delta f=\sum_{i=1}^m -\delta\mu^*\tau+\delta\mu^*\mathbb{1}_{ y_i-\mu^*\le 0}=(-m\tau+(m-m_+))\delta\mu^*.$$


问:
What is the distribution similar to the Poisson distribution with non-constant mean rate?
As the Poisson distribution expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event.

I have been asked the question that what is the distribution if the mean rate is non-constant?

I thought in this case it is about the sum of Poisson random variables which is also a Poisson random variable with a mean rate equal to the sum of the mean rates of each random variable.

Is it correct?

答:
If you have a Poisson process with constant rate $\lambda$, then the distribution of the number of events from time $t_1$ to time $t_2$ has a Poisson distribution with mean $(t_2-t_1)\lambda$

If instead the rate varies as a function of time so $\lambda(t)$ with events occurring  independently of the time since the last event (an inhomogeneous  Poisson process), then the distribution of the number of events from time $t_1$ to time $t_2$ has a Poisson distribution with mean $\int\limits_{t_1}^{t_2} \lambda(t)\, dt$


答:
Hint:Use Menelaus theorem in $\Delta ABE$ you will get result directly

$$\frac{BC}{CE}*\frac{EF}{FA}*\frac{AD}{DB}=1$$

About Menelaus https://artofproblemsolving.com/wiki/index.php/Menelaus%27_Theorem

Title: Let $D$ and $E$ be points on $AB$ and $BC$ of $\Delta ABC$ such that $AD = 7BD$ , $BE = 10CE$ . Let $AE$ and $BC$ meet at point $F$

Body:
Let $D$ and $E$ be points on segments of $AB$ and $BC$ of $\Delta ABC$ such that $AD = 7BD$ , $BE = 10CE$ . Let $AE$ and $BC$ meet at point $F$. If $AF = k \cdot FE$ , find $[k]$ .

What I Tried: Here is a picture :-

The first thing which I did was to try the process in Geogebra. There I got the value of $k$ to be approximately $7.74$ , so $[k] = 7,$ and I think that should be the answer. I also think that the initial value of $k$ varies, but $[k]$ is always $7$ . How will I be able to come up with a solution then?

Can anyone help me? Thank You.


问:
Continuity of a potential $(n+1$)-simplex from $n$-simplex: Rotman "Introduction to Algebraic Topology"
This is from Rotman's book Introduction to Algebraic Topology (p.72-73).

He starts with a convex subset $X$ of a Euclidean space, an $n$-simplex $\sigma\colon \Delta^n\to X$ and a point $b$ of $X$. He then defines $(n+1)$-simplex $b.\sigma\colon \Delta^{n+1}\to X$ as follows: $(b.\sigma)(x_0,x_1,\dots,x_n)$ is equal to $b$ if $(x_0,x_1,\dots,x_n) = (1,0,\dots,0)$ and to $x_0b + (1-x_0)\sigma\left(\dfrac{x_1}{1-x_0}, \dots, \dfrac{x_n}{1-x_0}\right)$ otherwise.

Rotman explains why this map is well-defined, and I can follow this explanation just fine. The problem lies in seeing why it is continuous.

The first idea is to use so-called "gluing lemma": given a map from a topological space which is a union of two open or two closed open subsets which agree on their intersection, we can check it's continuity by checking the continuity of respective restrictions. However, it doesn't seem to apply here: $\{(1,0,\dots,0)\}$ is closed in $\Delta^{n+1}$ since the latter is Hausdorff, so $\Delta^{n+1}\setminus\{(1,0,\dots,0)\}$ is open. Since $\Delta^{n+1}$ is path-connected, the only "clopen" subsets are $\varnothing$ and $\Delta^{n+1}$ itself.

The continuity of $b.\sigma$ at $(x_0,x_1,\dots,x_n) \neq (1,0,\dots,0)$ doesn't seem that hard at first. Indeed, the map $(x_0,x_1,\dots,x_n) \mapsto \left(\dfrac{x_1}{1-x_0}, \dots, \dfrac{x_n}{1-x_0}\right)$ is continuous as it is continuous coordinate-wise, $\sigma$ is continuous by assumption, and a multiplication and a sum of two continuous map is continuous. However, it would be two if the entire map $b.\sigma$ could be described like that, but it is defined point-wise. And, as I've said earlier, we can't separate the map into restrictions here.

Also, I don't see how to check continuity of $b.\sigma$ at $(1,0,\dots,0)$.

I have to confess that I don't know much of the theory of simplices, so I would prefer as elementary solution as possible. However, if some of it is needed, I would be extra grateful for a reference for the facts used.

答:
As you explained, the continuity on the open subset $\Delta^{n+1} \setminus
 \{(1,0\ldots,0)\}$ is easily verified. It remains to show continuity at $e_1 = (1,0,\ldots,0)$. The gluing lemma does not help us, we have to give a direct proof.

Write $x \in \Delta^{n+1}$ in the form $x = (x_0,\xi)$ with $x_0 \in \mathbb R$ and $\xi \in \mathbb R^n$. Then for $x \ne e_1$
$$(b.\sigma)(x) = x_0b + (1-x_0)\sigma(\frac{1}{1-x_0}\xi) .$$
Since $\Delta^n$ is compact, $\sigma(\Delta^n)$ is compact, thus bounded. That is, we find $R \ge 0$ such that $\lVert \sigma(y) \rVert \le R$ for all $y \in \Delta^n$. This shows that for $x \ne e_1$
$$\lVert (b.\sigma)(x) - e_1 \rVert = \lVert (x_0 -1) b + (1-x_0)\sigma(\frac{1}{1-x_0}\xi) \rVert = \lvert x_0 -1 \rvert \cdot \lVert  b - \sigma(\frac{1}{1-x_0}\xi) \rVert \\ \le \lvert x_0 -1 \rvert \cdot(\lVert  b \rVert + \lVert \sigma(\frac{1}{1-x_0}\xi) \rVert) \le \lvert x_0 -1 \rvert \cdot(\lVert  b \rVert + R) .$$
Sincbe $\lvert x_0 -1 \rvert \le \lVert x - e_1 \rVert$ we get
$$\lVert (b.\sigma)(x) - e_1 \rVert < (\lVert  b \rVert + R)\lVert x - e_1 \rVert $$
which proves continuity.


答:
Say $\angle BCD = \theta$.

Using Sine Law in $\triangle BCD$,

$\frac{\sin (\angle BDC)}{11y} = \frac{\sin \theta}{x}$ ...(i)

In $\triangle CEF$,

$\frac{\sin (\angle EFC)}{10y} = \frac{\sin \theta}{EF}$ ..(ii)

In $\triangle ADF$,

$\frac{\sin (\angle AFD)}{7x} = \frac{\sin \angle ADC}{AF}$ ..(iii)

$\angle ADC = 180^0 - \angle BDC$ (same sine value) and $\angle EFC = \angle AFD$

Substituting and solving, we get $\frac{AF}{FE} = \frac{77}{10}$


问:
Cross section is a regular hexagon.Is it a cube?
One of the cross sections in a rectangular box is a regular hexagon.Prove that the box is a cube

I tried to prove that certain lengths were equal by showing that certain triangles are congruent but that was of no use.Could someone help me with this one?

答:
Given $a$, $b$ and $c$ lengths of block(rectangular box) and $u$ length of edge of hexagon. Now we insert a line $u$ into a rectangle so that the side $a$ will divided to sections $x_a$ and $a-x_a$ and side $b$ will divided to $x_b$ and $b-x_b$. Inside this rectangle is a triangle for which the $x_a^2+x_b^2 = u^2$(see image)
, for rectangle $a\cdot c$ analogously $x_a^2+x_c^2=u^2$ and for rectangle $a\cdot c$ we get $x_a^2+x_c^2=u^2$.

Now

$x_a^2+x_c^2 = u^2 = x_b^2+x_c^2 \implies x_a=x_b$,

$x_a^2+x_b^2 = u^2 = x_b^2+x_c^2 \implies x_a=x_c$.

($x_a^2+x_b^2 = u^2 = x_a^2+x_c^2 \implies x_b=x_c$  is redundant).

All these equalities are valid for each the side and his opposite side, this mean for $a\cdot b$ and also for oposite side $a\cdot b$. And therefore, $x_a = a-x_a \implies x_a = \frac{a}{2}$, $x_b = b-x_b \implies x_b = \frac{b}{2}$ and $x_a = a-x_a$, $x_c = c-x_c \implies x_c = \frac{c}{2}$.

And from these and previous equalities follows $x_a = x_b = x_c$ and also $a = b = c \equiv v$, therefore, that hexagon can be entered only in the cube.

If length of edge of regular hexagon equals to $u$, the length of edge $v$ of this cube will $v = 2 \sqrt u$.


问:
Number of solutions of a polynomial
I have a polynomial with $n$ variables and degree $r$, of the form
$$f(x_1,\cdots,x_n)=c+\sum_{S\subset [n],|S|=r}c_s\prod_{i\in A}x_i$$
where $c$ is some constant and $c_s$ are coefficients that depend on the chosen subset $S\subset [n]$ (I believe this is called a multinomial but am not sure).
What is the maximal number of solutions that the equation $f(x_1,\cdots,x_n)=0$ can have? Is it the degree ($r$) as in the single variable case?

答: If $n>1$ then $f$ can have infinitely many solutions; consider the polynomial $f(x,y)=xy$.

Title: Number of solutions of a polynomial


问:
What is the probability that $\lfloor Z\rfloor$ is odd?
There are two IID standard uniform random variable ${X_1},{X_2}$. assume $Z = \frac{X_1}{X_2}$ ,what is the probability that $\lfloor Z\rfloor$ (floor function) is odd?

my solution is as follow:

${p_Z}(u) = \frac{1}{2}[\frac{{(H(u - 1))}}{{{u^2}}} + H(1 - u)]$

$\int {} {p_Z}(u)du = {P_Z}(u) = \frac{{{u^2} - {{(u - 1)}^2}H(u - 1)}}{{2u}}$

note that H is Heaviside step function.
we define ${P_{odd}}(u) = P(2u) - P(2u - 1) = \frac{{4{u^2} - {{(2u - 1)}^2}H(2u - 1)}}{{4u}} - \frac{{{{(2u - 1)}^2} - {{(2u - 2)}^2}H(2u - 2)}}{{2(2u - 1)}}$

we can show that ${P_{odd}}(1)$ is the probability of $Z$ being the first odd number and so on.
using wolfram alpha I found out

$\sum\limits_{i \in N} {{P_{odd}}(i)} = \frac{{\ln (4)}}{4}$

I tested this answer using Monte Carlo simulation and it seemed legit,I'm wondering is there any other simpler ways to answer this question? for example without using infinite sum.

答: For convenience, instead of $X_1, X_2$, let us use $(X,Y)$ and observe their joint distribution is uniform on the unit square.  Then $Y/X$ is the slope of a line through the origin, and the set of all $(X,Y)$ such that $\lfloor Y/X \rfloor$ is odd, is the union of the set of triangles $A_k$ with vertices $$\{(0,0), (\tfrac{1}{2k-1}, 1), (\tfrac{1}{2k}, 1)\}$$ excluding the edge joining $(0,0)$ and $(\tfrac{1}{2k}, 1)\}$.  So the total probability is simply the sum of the areas of each $A_k$, and since they all have common height $1$, the answer is simply $$\frac{1}{2} \sum_{k=1}^\infty \frac{1}{2k-1} - \frac{1}{2k} = \frac{\log 2}{2}.$$


问:
If $f$ is integrable on $[a,b]$ and $g$ is a function on $[a,b]$ . Show $g$ is integrable and $\int_a^b f(x) = \int_a^b g(x)$.
Problem:

If $f$ is integrable on $[a,b]$ and $g$ is a function on $[a,b]]$ so that $f(x) = g(x)$ except for finitely many $x \in [a,b]$. Show $g$ is integrable and $\int_a^b f(x) = \int_a^b g(x)$.

I have looked at the solution from this page.

Where they use induction to prove this, such as this:

However my question is how do they arrive at $$t_k - t_{k-1}= \frac{\epsilon}{12B}$$
and $$|U(g,p) - U(f,p)| \le 2[B-(-B)]*\max \{ t_k - t_{k-1}\}  < \frac{\epsilon}{3} $$

Thank you very much.

答:
For $|U(g,p) - U(f,p)| \le 2[B-(-B)]*\max \{ t_k - t_{k-1}\}  < \frac{\epsilon}{3} $:

Remember that $f=g$ except for exactly one point. The worst case is that precisely this point is one of the points of the partition $P$. In this case, by the definition of upper sum, $U(f,P)$ and $U(g,P)$ only differes on what hapepens in these two intervals, so $U(f,P)-U(g,P)$ calcels except on  AT MOST, two (consecutive) intervals. So

$$U(f,P)-U(g,P)=(M_k(f)-M_k(g))(t_k-t_{k-1})+(M_{k+1}(f)-M_{k+1}(g))(t_{k+1}-t_{k}),
$$ where $M_j(f)$ is the max of $f$ in the $j$-th interval (and the same for $g$).

Can you finish from here?


问:
Equality of Series due to Rearrangements
Given $a_{n}\geq0$, let $b_{n}$ be rearrangement of $a_{n}$. Show that $\sum b_{n}=\sum a_{n}$

My Proof : Let $A_{n}=a_{0}+a_{1}+\cdots+a_{n}$ and let $B_{n}=b_{0}+b_{1}+\cdots+b_{n}$ be a rearrangements of $a_{n}$. Now we note that the sequence of the partial sums $A_{n}$ and $B_{n}$ are either finite, or either $A_{n}$ or $B_{n}$ approach infinity. Thus, it may either be the case that $A_{n}\leq\displaystyle\lim_{n\to+\infty}B_{n}$ or $B_{n}\leq\displaystyle\lim_{n\to+\infty}A_{n}$. We have that $\displaystyle\lim_{n\to+\infty}B_{n}$ involves all the the sum over all $b_{n}$ and so it involves the sum over $a_{0},a_{1},...,a_{n}$ and same for $\displaystyle\lim_{n\to+\infty}A_{n}$ . Taking the limit $n\to+\infty$ we find that $\displaystyle\lim_{n\to+\infty}B_{n}\leq \displaystyle\lim_{n\to+\infty}A_{n}$ and $\displaystyle\lim_{n\to+\infty}A_{n}\leq\displaystyle\lim_{n\to+\infty}B_{n}$ and thus we have that $\displaystyle\lim_{n\to+\infty}A_{n}=\displaystyle\lim_{n\to+\infty}B_{n}$

Is this a complete proof or is my proof missing some details?

答:
$\sum_{n\in \mathbb{N}} a_n = \sup S$, where $S$ is the set of all partial sums of $(a_n)$.

Let $\sigma: \mathbb{N} \to \mathbb{N}$, the bijection, such that $b_n = a_{\sigma (n)}$, $n\in \mathbb{N}$.

Now fix an arbitrary $m\in \mathbb{N}$. Then there exists a $M\in \mathbb{N}$, depending on $m$, such that $\{\sigma (1) , \sigma (2),\ldots , \sigma(m)\} \subseteq \{1,2, \ldots ,M\}$.

$\big($ for example if $b_1= a_1$ and $b_2=a_{100}$, then (for $M=100$), $\{\sigma(1), \sigma (2) \} = \{1, 100\} \subset \{1,2, \ldots ,100 \}$. In general, just consider $M=\max \{\sigma (1) , \sigma (2),\ldots , \sigma(m)\} \ \big)$

This means that $$\sum_{k=1}^{m}b_k = \sum_{k=1}^{m}a_{\sigma(k)} \overset{a_n \geq 0}{\leq} \sum_{k=1}^{M}a_{k} \leq \sup S = \sum_{n\in \mathbb{N}} a_n$$

Thus $$\sum_{n=1}^{\infty}b_n \leq \sum_{n=1}^{\infty}a_n$$

With the symmetric of this argument you have that $$\sum_{n=1}^{\infty}a_n \leq \sum_{n=1}^{\infty}b_n$$
and that completes the proof


问:
$\cos(x)+\cos(x\sqrt{2})$ is not periodic
Show that the function $$f(x)=\cos(x)+\cos(x\sqrt{2})$$ is not periodic. I tried $x = a$ and $a\sqrt{2}$. I am guessing that the method of contradiction would be of some help over here. What else should I try?</p>

答:
Assume $ \cos(x) + \cos(\sqrt{2} x) = \cos(x+T) + \cos(\sqrt{2} x+ \sqrt{2} T) $.

Let $x=0$. Then

$$2=\cos T + \cos \sqrt{2}T$$

Since the cosine is at most one, this means that simultaneously $\cos T = 1$ and $\cos \sqrt{2} T = 1$. This is equivalent to:

$$T = 2\pi n, \ \mbox{for some } n \in \mathbb{Z}$$
$$\sqrt{2}T = 2\pi m, \ \mbox{for some } m \in \mathbb{Z}$$

Substitute $T$ from the first into the second:

$$\sqrt{2} (2\pi n)= 2\pi m$$
$$\sqrt{2}=\frac{m}{n}$$

So $\sqrt{2} \in \mathbb{Q}$ which is a contradiction, or $T=0$. Either way, you're done.


问:
Logical translation of the verb " to be " in " the cat is a mammal ". ( Inclusion or membership?)

Standardly, the sentence " the cat is a mammal " would be translated as

for all x , if x is a cat, then x is a mammal,

meaning that the set of cats is included in the set of mammals : $C\subseteq M$.

There is another interpretation: the species " cat" is a mammals species. Under this alternative interpretation ( admitting that a " species" is some sort of set or collection), we would have : $ C \in M $*

with C = the set of cats and M* = the set of species of mammals.

The distinction I'm referring to is analogous to the medieval distinction between personal supposition and simple supposition of a term. In " horses are animals" the term " horse" supposes personnaly while in " horse is a species" , the term supposes simply.

What are the drawbacks of the second interpretation ( where " is" is interpreted as membership relation holding between a species and a set of species)? Why has the first interpretation become standard?

答:
"For all x , if x is a cat, then x is a mammal" is:

$\forall x (\text {Cat}(x) \to \text {Mammal}(x))$

and this is exactly the definition of set inclusion:

$\text{Cat } \subseteq \text { Mammal}.$

Regarding the multiple uses of "is" into natural language, we can identify three different "contexts" :

"Plato is a philosopher"; this context is relative to an object (or individual) belonging to a set or class.

In symbols this is expressed with $\text {Plato} \in \text {Philospher}$, i.e. $\text {Philospher}(\text {Plato})$.

"a man is a male"; this context is relative to a concept subsumed under a more general" one.

In symbols it is expressed with inclusion (your example above).

"$2+2 \text { is } 4$"; this context is relative to identity, i.e. the relation between two names denoting the same object.

It is simply: $2+2 = 4$.


问:
How to visualize Nerve of stars of all vertices of a simplicial complex?
I came across a theorem which states that a simplicial complex is a geometric realization of the nerve of the stars of its vertices. Here, we consider open stars (set of open simplices of which the vertex is a face).
I feel that the collection of stars of all the vertices of the complex will give the complex itself, and the complex is then trivially the geometric realization of the nerve of itself.
Could someone help me with visualizing the collection of stars of all the vertices of a simplicial complex?
Thanks</p>

答:
A face of the nerve of the covering of open stars is (by definition) precisely a non-empty finite set of vertices, whose open stars intersect non-trivially.

That is $(v_1v_2\cdots v_r)$ is a face of the nerve of the stars if and only if there is some point in the interior of a simplex, whose vertices include $v_1,v_2,\cdots,v_r$.  This happens if and only if $(v_1v_2\cdots v_r)$ is a simplex in the original simplicial complex.

That is you end up with the same simplicial complex that you started with, so of course they have the same geometric realisation.


答:
The two interpretations refer to two distinct terms. $M$ is the set of mammals, where each element of $M$ is an individual (a mammal); while $M^*$ is the set of species that are mammals, where each element of $M^*$ is a set (a species) of individuals (the mammals).

The meaning of the verb "to be" is an inclusion or a membership relation depending on whether the "predicate" you want to consider is $M$ or $M^*$.
In the former, $C$ is a subset of $M$, in the latter $C$ is an element of $M^*$.

In set theory, you can say that $C \subseteq M$ is equivalent to $C \in \mathcal{P}(M)$, for any sets $C$ and $M$, where $\mathcal{P}(M)$ is the powerset of $M$ (the set of all subsets of $M$).

Here, you are doing something similar. Your $M^*$ is not $\mathcal{P}(M)$ but a subset of it ($M^*$ "lives in the same realm" as $\mathcal{P}(M)$, if you like this ontological way of thinking), more precisely $M^*$ is the set of subsets of $M$ that are species.

From a psychological standpoint, I find more natural to interpret the sentence "cats are mammals" as $C \subseteq M$ because it requires less mathematical structures than $C \in M^*$. Anyway, the two interpretations are equivalent, so no one is preferable to the other one from a logical point of view (as soon as you are entitled to use powersets).

In first-order logic, it is quite natural and standard to interpret sentences in a domain of individuals to which quantifiers refer, and then to gather together individuals by means of predicates. So, according to Fregeian tradition, the sentence "the cat is a mammal" is translated in first-order logic as

for all $x$ , if $x$ is a cat, then $x$ is a mammal,

i.e., (for $C(x) =$ "$x$ is a cat", and $M(x) =$ "$x$ is a mammal")

$\forall x (C(x) \to M(x))$

which corresponds exactly the set-theoretic definition of $C \subseteq M$.

If you prefer to interpret the sentence "the cat is a mammal" as $C \in M^*$ where $C$ and $M$ are first-class elements in the domain of discourse, then you should define a way to interpret sentences which would require more mathematical structures  and where it would be clumsier to talk about single individuals.
I guess the domain should have many sorts of elements with a hierarchical structure: individuals (as me, you, my cat Zoe) and classes of these individuals (cats, mammals, human beings). It is not impossible, but technically more elaborated and less intuitive. See for instance order-sorted logic.

Title: Logical translation of the verb " to be " in " the cat is a mammal ". ( Inclusion or membership?)

Body:
Standardly, the sentence " the cat is a mammal " would be translated as

for all x , if x is a cat, then x is a mammal,

meaning that the set of cats is included in the set of mammals : $C\subseteq M$.

There is another interpretation: the species " cat" is a mammals species. Under this alternative interpretation ( admitting that a " species" is some sort of set or collection), we would have : $ C \in M $*

with C = the set of cats and M* = the set of species of mammals.

The distinction I'm referring to is analogous to the medieval distinction between personal supposition and simple supposition of a term. In " horses are animals" the term " horse" supposes personnaly while in " horse is a species" , the term supposes simply.

What are the drawbacks of the second interpretation ( where " is" is interpreted as membership relation holding between a species and a set of species)? Why has the first interpretation become standard?


问:
types of solutions of the functional equation $f\left(x^2+y^2\right)=f(x)^2+f(y)^2$

Are functions $0$, $\frac{1}{2}$, $x$ and $|x|$ the only continuous solutions of $$f\left(x^2+y^2\right)=f(x)^2+f(y)^2\text?$$

Does the equation have discontinuous solutions?

[Edit: This one is solved by  @Gae.S. in a comment below.]

(own problem, partially inspired by Is there a continuous function $f$ satisfying $f^{2}(x) = f(x^{2})$, $f(0)=1$ and $ f(1)=0$?)

答:
It can be shown that the only functions $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \left( x ^ 2 + y ^ 2 \right) = f ( x ) ^ 2 + f ( y ) ^ 2 \tag 0 \label 0 $$
are those of the following form:

$ f ( x ) = 0 $;

$ f ( x ) = \frac 1 2 $;

$ f ( x ) = \begin {cases} x & x \notin A \\ - x & x \in A \end{cases} $ for some $ A \subseteq \mathbb R ^ - $.

Among these, the continuous ones are exactly those you've mentioned. It's straightforward to check that these are all solutions. We prove that they are the only ones.

Substituting $ - x $ for $ x $ in \eqref{0} and comparing to \eqref{0} itself, we have
$$ f ( - x ) ^ 2 = f ( x ) ^ 2 \text . \tag 1 \label 1 $$
Defining $ a = f ( 0 ) $ and setting $ y = 0 $ in \eqref{0} we have
$$ f \left( x ^ 2 \right) = f ( x ) ^ 2 + a ^ 2 \text . \tag 2 \label 2 $$
\eqref{2} can be used to rewrite the right-hand side of \eqref{0} and get
$$ f \left( x ^ 2 + y ^ 2 \right) = f \left( x ^ 2 \right) + f \left( y ^ 2 \right) - 2 a ^ 2 \text , $$
which by defining $ g : \mathbb R ^ { 0 + } \to \mathbb R $ with $ g ( x ) = f ( x ) - 2 a ^ 2 $, can be transformed to
$$ g ( x + y ) = g ( x ) + g ( y ) \tag 3 \label 3 $$
for all $ x , y \in \mathbb R ^ { 0 + } $. \eqref{2} shows that $ g $ is bounded below; in particular $ g ( x ) \ge - a ^ 2 $ for all $ x \in \mathbb R ^ { 0 + } $. Using this fact together with \eqref{3}, shows that $ g $ must be of the form
$$ g ( x ) = b x \tag 4 \label 4 $$
for some constant $ b \in \mathbb R ^ { 0 + } $ and all $ x \in \mathbb R ^ { 0 + } $ (see here). Letting $ x = 0 $ in \eqref{2}, we see that $ a \in \{ 0 , \frac 1 2 \} $.

Case $ f ( 0 ) = 0 $:

Putting $ x = 1 $ in \eqref{2} and using \eqref{4}, we see that $ b \in \{ 0 , 1 \} $. If $ b = 0 $, then using \eqref{1} we can conclude that $ f $ is constantly zero on all $ \mathbb R $, which gives one of the mentioned solutions. If $ b = 1 $, we have $ f ( x ) = x $ for all $ x \in \mathbb R ^ { 0 + } $, and using \eqref{1} we can conclude that for any $ x \in \mathbb R ^ - $ we have $ f ( x ) \in \{ - x , x \} $. As we checked before, this can be chosen arbitrarily for negative $ x $, and leads to another class of solutions mentioned above.

Case $ f ( 0 ) = \frac 1 2 $:

Putting $ x = 1 $ in \eqref{2} and using \eqref{4}, we see that $ b = 0 $, which gives the constant solution $ f ( x ) = \frac 1 2 $.


问:
Reordering vector product
If I have vectors $a, b, c \in \mathbb{R}^3$, and if we have e.g. $a = b\times c$, is there any way to express $b$ in terms of the other two? </p>

答: Write out the vectors as $a=(a_1,a_2,a_3)$, and so on, do the cross product, set the two sides equal, and you'll get a system of three linear equations in the three unknown components of $b$. </p>

Title: Reordering vector product


问:
Partitioning a set of cardinal $np$
I wish to compute the number of possible partitions $S$ of a set of cardinal $np$ into $n$ subsets of cardinal $p$.
It is easy to obtain the formula :
$\displaystyle S=\sum_{k=0}^n {{pk}\choose{p}}$.
However i am getting stucked here without recouring to algebra
. I have in fact used generating functions to compute the sum, but i am looking for a non-algebraic argument (for example double couting ?) that could help me to compute this sum directly as i am sure there's a way to.

答:
"I wish to compute the number of possible partitions S of a set of cardinal np into n subsets of cardinal p."

Hint: Stirling numbers of 2nd kind: $S(n,k)$ is the number of partitions of an $n$ element set into $k$ subsets.

"It is easy to obtain the formula : $S=\sum_{k=0}^n{np\choose p}$." Summand is independent of $k$.

Title: Partitioning a set of cardinal $np$


问:
Second order inhomogeneous equation: $y''-2xy'-11y=e^{-ax}$
My question relates to an a second order inhomogeneous equation:

$$y''-2xy'-11y=e^{-ax}$$

First I need to investigate the homogeneous equation:

$$y''-2xy'-11y=0$$

$$y''-2xy'=11y$$

Forms Hermite's Equation where $\lambda = 11$

So I need a general solution of the homogeneous equation
$Ly(x) = 0$.

To do this I need to linearly independent solutions, $y_1(x)$ and $y_2(x)$ say, and
then the general solution of $Ly(x) = 0$.

becomes:

$$Ay_1(x) + By_2(x)$$

where $A$ and $B$ are arbitrary constants.

I am struggling to find two independent solutions of the hermite equation above:

My attempt

If I take a solution to be if the form $$y=\sum_{n=0}^{\infty}a_{n}x^n$$
Then putting this into the ODE I get the following:

$$\sum_{n=2}^{\infty}n(n-1)a_{n}x^{n-2}-2\sum_{n=1}^{\infty}na_{n}x^{n}-11\sum_{n=0}^{\infty}a_{n}x^{n}=0$$

Which I can reduce to:

$$\sum_{n=0}^{\infty}[(n+2)(n+1)a_{n+2}-2(n+\frac{11}{2})a_{n}]x^{n}=0$$

And I have found that rearranging gives:

$$a_{n+2} = \frac{-2(n+\frac{11}{2})a_{n}}{(n+2)(n+1)}$$

This can be used to develop a recurrence relation...

How can I use this to find my two independent solutions $y_1(x)$ and $y_2(x)$ which I need in order to calculate the inhomogeneous solution?

Maybe there are more efficient ways of calculating this ODE that I am unaware of.

Edit

I displayed the equation incorrectly and have since edited it.

Should my solution be in the form of an infinite sum rather than a finite polynomial because I believe that the solution could only be constructed as a series solution which terminates if and only if $\lambda = −2n$ where $n \in \mathbb N$.

In my case $\lambda$ isn't of the above form hence an infinte series is required.

答:
Approach $1$: intrgral solution

First consider $y_c''-2xy_c'-11y_c=0$ :

Similar to Help on solving an apparently simple differential equation,

Let $y_c=\int_Ce^{xs}K(s)~ds$ ,

Then $(\int_Ce^{xs}K(s)~ds)''-2x(\int_Ce^{xs}K(s)~ds)'-11\int_Ce^{xs}K(s)~ds=0$

$\int_Cs^2e^{xs}K(s)~ds-2x\int_Cse^{xs}K(s)~ds-11\int_Ce^{xs}K(s)~ds=0$

$\int_C(s^2-11)e^{xs}K(s)~ds-\int_C2se^{xs}K(s)~d(xs)=0$

$\int_C(s^2-11)e^{xs}K(s)~ds-\int_C2sK(s)~d(e^{xs})=0$

$\int_C(s^2-11)e^{xs}K(s)~ds-[2se^{xs}K(s)]_C+\int_Ce^{xs}~d(2sK(s))=0$

$\int_C(s^2-11)e^{xs}K(s)~ds-[2se^{xs}K(s)]_C+\int_Ce^{xs}(2sK'(s)+2K(s))~ds=0$

$-~[2se^{xs}K(s)]_C+\int_C(2sK'(s)+(s^2-9)K(s))e^{xs}~ds=0$

$\therefore2sK'(s)+(s^2-9)K(s)=0$

$2sK'(s)=(9-s^2)K(s)$

$\dfrac{K'(s)}{K(s)}=\dfrac{9}{2s}-\dfrac{s}{2}$

$\int\dfrac{K'(s)}{K(s)}ds=\int\left(\dfrac{9}{2s}-\dfrac{s}{2}\right)ds$

$\ln K(s)=\dfrac{9\ln s}{2}-\dfrac{s^2}{4}+c_1$

$K(s)=cs^\frac{9}{2}e^{-\frac{s^2}{4}}$

$\therefore y_c=\int_Ccs^\frac{9}{2}e^{-\frac{s^2}{4}+xs}~ds$

But since the above procedure in fact suitable for any complex number $s$ ,

$\therefore y_{c,n}=\int_{a_n}^{b_n}c_n(m_nt)^\frac{9}{2}e^{-\frac{(m_nt)^2}{4}+xm_nt}~d(m_nt)={m_n}^\frac{9}{2}c_n\int_{a_n}^{b_n}t^\frac{9}{2}e^{-\frac{{m_n}^2t^2}{4}+m_nxt}~dt$

For some $x$-independent real number choices of $a_n$ and $b_n$ and $x$-independent complex number choices of $m_n$ such that:

$\lim\limits_{t\to a_n}t^\frac{11}{2}e^{-\frac{{m_n}^2t^2}{4}+m_nxt}=\lim\limits_{t\to b_n}t^\frac{11}{2}e^{-\frac{{m_n}^2t^2}{4}+m_nxt}$

$\int_{a_n}^{b_n}t^\frac{9}{2}e^{-\frac{{m_n}^2t^2}{4}+m_nxt}~dt$ converges

For $n=1$, the best choice is $a_1=0$ , $b_1=\infty$ , $m_1=\pm1$

$\therefore y_c=C_1\int_0^\infty t^\frac{9}{2}e^{-\frac{t^2}{4}}\cosh xt~dt$ or $C_1\int_0^\infty t^\frac{9}{2}e^{-\frac{t^2}{4}}\sinh xt~dt$

Hence $y_c=C_1\int_0^\infty t^\frac{9}{2}e^{-\frac{t^2}{4}}\sinh xt~dt+C_2\int_0^\infty t^\frac{9}{2}e^{-\frac{t^2}{4}}\cosh xt~dt$


问:
Do we really need polynomials (In contrast to polynomial functions)?
In the following I'm going to call

a polynomial expression an element of a suitable algebraic structure (for example a ring, since it has an addition and a multiplication)
that has the form $a_{n}x^{n}+\cdots+a_{1}x+a_{0}$, where $x$ is
some fixed element of said structure,

a polynomial function a function of the form $x\mapsto a_{n}x^{n}+\cdots+a_{1}x+a_{0}$ (where the same algebraic considerations as above apply)

a polynomial an element of the form $a_{n}X^{n}+\cdots+a_{1}X+a_{0}$ with indeterminates $X$ (these
can be formalized, if we are in a ring, with strings of $0$s and
a $1$).

Note that when we are very rigorous/formal, polynomial expressions
and polynomials are something different (although in daily life we
often use them synonymously). Polynomial functions and expressions
are also different from each other although in this case the relationship
is a closer one, since every polynomials expression can be interpreted
as a polynomial function evaluated at a certain point (thus "polynomial
functions" are something more general than "polynomial expressions").

My question is: Why do we use polynomials ? It seems to me that every
piece of mathematics I have encountered so far, one could replace
every occurrence of polynomials with polynomial expressions/functions
without any major change in the rest of the proof/theorem/definition
etc.

My thoughts:

The only reasons that I can see to use polynomials are the following
two:

After one makes the idea precise that one can plug "suitable"
elements into polynomials (which may lie a ring containing the ring
in which the coefficients live in), one can save time in certain setting,
by handling the "plugging of suitable elements into polynomials"
more elegantly: For example in case of the theorem of Cayley-Hamilton,
which in its "polynomial function" version would look like:

Let $A$ be an $n\times n$ matrix over $K$, whose characteristic
polynomial (function) is $x\mapsto a_{n}x^{n}+\cdots+a_{1}x+a_{0}$.
Then
$$
a_{n}A^{n}+\cdots+a_{1}A+a_{0}I=0.
$$

whereas the "polynomial" version looks more elegant:

Let $A$ be an $n\times n$ matrix over $K$, whose characteristic
polynomial is $p_{A}\in K\left[X\right]$. Then
$$
p_{A}\left(A\right)=0.
$$

The only thing that polynomials can "do", but algebraic expressions/functions can't, is to be different, when the algebraic expressions/functions are the same (i.e. there's a theorem that tells us that the mapping of polynomials to polynomials expressions/functions
isn't injective, if the field is finite). Maybe this small difference makes a big enough difference
to consider polynomials after all, but as I said, I haven't encountered
any situation in which this difference could manifest itself.

(I'm guessing that maybe cryptography or higher number theory really needs
polynomials and not just polynomial expressions/functions. Since I don't know
anything about these subjects, I would be very happy with an example
of a theorem (whose content isn't merely technical as it is the case with the theorem above) involving polynomials, where these absolutely cannot
be replaced by polynomial expressions/functions. Conversely I would also be happy with an authoritative statement from someone knowledgeable that indeed we could dispense of polynomials, if we wanted to.)

答:
You write that polynomial functions are "more general" than polynomial expressions. In fact the exact opposite is the case: multiple polynomial expressions may be the same function (on a finite field), so it's polynomial expressions that are more general.

Here are a few ways that polynomial expressions are useful where it could (or would) be awkward to use polynomial functions.

If $F$ is  a field of $p^r$ elements, for a prime $p$ and positive integer $r$, then $F$ is the splitting field over ${\mathbf F}_p$ of $x^{p^r} - x$. But for every $a \in F$, $a^{p^r} = a$, so $a^{p^r} - a = 0$. You don't want to say $F$ is the splitting field of, well, "zero". The distinction between the nonzero polynomial $x^{p^r} - x$ in $F[x]$ and the zero polynomial is essential for that.

In case you think "I don't care about finite fields", well, some properties of sequences of integers are proved by making them into the coefficients of a polynomial and then reducing the polynomial mod p, which places the polynomial into ${\mathbf F}_p[x]$, where there are convenient features that are not available in ${\mathbf Z}[x]$. One example of this is a slick proof of Lucas's congruence on binomial coefficients. It's crucial for this that we can recover the coefficients of a polynomial in ${\mathbf F}_p[x]$ even if its degree is greater than $p$, which is impossible if you think of polynomials in ${\mathbf F}_p[x]$ as functions ${\mathbf F}_p \rightarrow {\mathbf F}_p$.

Quotient rings of $A[x]$, where $A$ is a ring, may not look like functions from an elementary point of view. Do you feel it's easier to work with ${\mathbf Q}[x]/(x^5)$ using the viewpoint of polynomials as functions rather than expressions?

Would you like to define the ring of noncommutative polynomials in two variables as functions?

Generating functions are important objects in mathematics (not just in combinatorics), and they are formal power series rather than numerical series. Most of them have positive radius of convergence (assuming their coefficients are in a ring like ${\mathbf R}$ or ${\mathbf C}$ and thus have a region where it makes sense to substitute a scalar for the variable at all), so they can be treated as functions on the domain of convergence, but it is convenient that the theory of formal power series doesn't really need to rely on the radius of convergence to get off the ground. Sometimes a generating function may even have radius of convergence equal to 0 but still contain useful information that can be legally extracted because formal power series can be developed without needing a radius of convergence. Of course a lot of important intuition about the way formal power series are manipulated is inspired by the properties of classical numerical power series as functions.

Your question mentions 3 types of polynomial objects: polynomial expressions, polynomial functions, and polynomials as strings of coefficients. I'd say the first and third are really synonymous (the third is really just a rigorous form of the first, not materially different in practice), but they're not the same as polynomial functions if you work over a finite field. If the coefficients are an infinite field then different polynomial expressions do define different functions on the field.


问:
Proving that if $(x_n) \to x$ then $p(x_n) \to p(x)$ in Real Analysis
I am self-learning Real Analysis from Stephen Abott's Understanding Analysis. Exercise 2.3.8 asks to prove that when a polynomial function $p$ is applied to values of a convergent sequence $(x_n)$, then if the input sequence converges to $x$, the resulting sequence converges to $p(x)$.

Questions.

(a) Do you guys think, my proof of part (a) is technically correct and rigorous?

(b) For part (b), should I just construct any function, such that the function value is different from the limiting value?

Let $(x_n) \to x$ and let $p(x)$ be a polynomial.

(a) Show $p(x_n) \to p(x)$.

(b) Find an example of a function $f(x)$ and a convergent sequence $(x_n) \to x$ where the sequence $f(x_n)$ converges, but not to $f(x)$.

Proof.

$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

(a) We are given that $(x_n) \to x$, so we can make the distance $\absval{x_n - x}$ as small as we like. Suppose $p(x) = x^{k}$ where $k$ is a non-negative integer.

Observe that,

\begin{align*}
 \absval{p(x_n) - p(x)} &= \absval{x_n^k - x^k}\\
 &= \absval{(x_n - x)(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})}\\
\end{align*}

Also,

\begin{align*}
\tiny
 \absval{(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})} &\le \tiny \absval{x_n^{k-1}} + \absval{x_n^{k-2}\cdot x}  + \ldots + \absval{x_n \cdot x^{k-2}} + \absval{x^{k-1}}\\
 \tiny\frac{1}{\absval{x_n^{k-1}} + \absval{x_n^{k-2}\cdot x}  + \ldots + \absval{x_n \cdot x^{k-2}} + \absval{x^{k-1}}} &\le \tiny\frac{1}{\absval{(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})}}
\end{align*}

Now, there exists $N_1$ such that $\absval{x_n - x} < \frac{\absval{x}}{2}$ for $n \ge N_1$. If the distance between $x_N$ and $x$ is smaller than $\frac{\absval{x}}{2}$, then clearly $\absval{x_n} > \frac{\absval{x}}{2}$.

Thus,
\begin{align*}
 \frac{1}{k\cdot \frac{\absval{x^k}}{2^{k-1}}} &< \frac{1}{\absval{(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})}}\\
 \implies \frac{2^{k-1}}{k \cdot \absval{x^k}}&< \frac{1}{\absval{(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})}}
\end{align*}

There exists $N_2$ such that $\absval{x_n - x} < \epsilon \cdot \frac{2^{k-1}}{k \cdot \absval{x^k}} $ for all $n \ge N_2$.

Let $N = \max \{N_1,N_2 \}$. To show that this $N$ indeed works, we prove that:

\begin{align*}
 \absval{p(x_n) - p(x)} &= \absval{(x_n - x)(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})}\\
 &< \epsilon \cdot \frac{2^{k-1}}{k \cdot \absval{x^k}} \cdot \absval{(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})} \\
 &< \epsilon \cdot \frac{1}{\absval{(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})}} \cdot \absval{(x_n^{k-1} + x_n^{k-2}\cdot x + \ldots + x_n \cdot x^{k-2} + x^{k-1})} \\
 &= \epsilon
\end{align*}

for all $n \ge N$. Thus, $p(x_n) \to p(x)$.

By algebra of limits, any linear combination of the powers of the sequence $(x_n)$ must be convergent and $\lim p(x_n) = p(x)$.

答:
If you are using the algebra of limits in the end of your proof, you might as well use it in the beginning. Considering $p(x)=a_0 + a_1 x + \cdots + a_k x^k$, you have that
$$
\lim p(x_n) = \lim(a_0 + a_1 x_n + \cdots + a_k x_n^k)= a_0 + a_1 \lim x_n + \cdots + a_k (\lim x_n)^k = p(x).
$$


问:
Suppose $x$ and $y$ are unequal real numbers

Suppose $x$ and $y$ are unequal real numbers. If
$$
\sqrt[3]{\frac{x+y}{x-y}} + \sqrt[3]{\frac{x-y}{x+y}} = x+y
\qquad \text{and} \qquad
\sqrt{xy}=1
$$
then find the value of
$$
(x-y)^5 + 5(x-y)^3 - 2(x-y)^2+ 4(x-y)
$$

For the above question, I got $x$ not equal to $-1,0,1$ and $y = 1/x$.
Then, I tried finding $x-y$ using $\big((x+y)^2-4xy)\big)^{\frac{1}{2}}$ and then solving for $x$ and $y$.
I think I over complicated it and there must be some easier way. Please help me out.

答:
Intuitively, one should let $a=x+y$ and $b=x-y$.

Then from the condition $\sqrt {xy}=1$ one obtains $a^2-b^2=4$.

The other condition becomes

$$a = \sqrt[3]{\frac ab} + \sqrt[3]{\frac ba}$$

Cubing both sides, we have

\begin{align}a^3 &= \frac ab + 3\left(\sqrt[3]{\frac ab}\right)^2\sqrt[3]{\frac ba}+ 3\sqrt[3]{\frac ab}\left(\sqrt[3]{\frac ba}\right)^2+ \frac ba
\\&=\frac ab + \frac ba+ 3\left(\sqrt[3]{\frac ab} + \sqrt[3]{\frac ba}\right)
\\&=a\left(\frac 1b + \frac b{a^2}\right)+ 3a\end{align}

Eliminating $a$ from both sides gives

$$a^2 = \frac 1b + \frac b{a^2} + 3$$

$$4+b^2=\frac1b + \frac b{4+b^2} + 3$$

$$1+b^2=\frac {4+b^2+b^2}{b(4+b^2)}$$

$$b(4+b^2)(1+b^2)-2b^2 =4$$

Expanding $b(4+b^2)(1+b^2)-2b^2$ gives $b^5+5b^3-2b^2+4b$, which seems very familiar...


答:
Tricky one.

Let $x+y = u$, $x-y=v$.

$$ xy=1 \Rightarrow u^2-v^2 =4$$

and cubing original equation :$$ \sqrt[3]{\dfrac{u}{v}} + \sqrt[3]{\dfrac{v}{u}} = u $$

gives

$$ \dfrac{u}{v} + \dfrac{v}{u} + 3u = u^3 $$

$$ \Rightarrow u^2 + v^2 + 3u^2v = u^4v$$

Substitute $u^2=4+v^2$ ,

$$(4+v^2) + v^2 + 3(4+v^2)v = (4+v^2)^2v $$

directly yields

$$ \boxed{v^5 + 5v^3 -2v^2 + 4v =4}$$

Body:
Suppose $x$ and $y$ are unequal real numbers. If
$$
\sqrt[3]{\frac{x+y}{x-y}} + \sqrt[3]{\frac{x-y}{x+y}} = x+y
\qquad \text{and} \qquad
\sqrt{xy}=1
$$
then find the value of
$$
(x-y)^5 + 5(x-y)^3 - 2(x-y)^2+ 4(x-y)
$$

For the above question, I got $x$ not equal to $-1,0,1$ and $y = 1/x$.
Then, I tried finding $x-y$ using $\big((x+y)^2-4xy)\big)^{\frac{1}{2}}$ and then solving for $x$ and $y$.
I think I over complicated it and there must be some easier way. Please help me out.


答: There's only one value of $x$ for which $f(x)=2$. Proof: Since $\cos\phi\le1$ this is only possible, if $\cos x = 1$ and $\cos(x\sqrt2) = 1$ hold true at the same time. The first statement is equivalent to $\frac{x}{\pi}\in\mathbb Z$, the second is equivalent to $\frac{x}{\pi}\sqrt2\in\mathbb Z$. The value $x=0$ fulfills the requirements. If a non-zero $x$ would fulfill both $\frac{x}{\pi}\in\mathbb Z$ and $\frac{x}{\pi}\sqrt2\in\mathbb Z$, the you could devide the second integer by the first. The result would be a rational number with value $\sqrt2$. This is a contradiction. Hence, only $x=0$ fulfills both requirements.</p>

Title: $\cos(x)+\cos(x\sqrt{2})$ is not periodic

Body: Show that the function $$f(x)=\cos(x)+\cos(x\sqrt{2})$$ is not periodic. I tried $x = a$ and $a\sqrt{2}$. I am guessing that the method of contradiction would be of some help over here. What else should I try?</p>


问:
Why is the following map $\mu : G_K \to M_d(\Bbb C_K) : g \mapsto a_{ij}(g)$, where $g(w_j) = \sum_{i} a_{ij}(g)w_i$, continuous?
Let $K$ be a p-adic field and let $\Bbb C_K$ be the completion of its algebraic closure. Let $G_K := \operatorname{Gal}(\overline{K}/K)$. Fixing a basis $\lbrace w_1, \dots, w_d\rbrace$ of a $\Bbb C_K$-vector space $W$ we have a map

$$\mu := \mu_{(w_j)} : G_K \to M_d(\Bbb C_K), g \mapsto a_{ij}(g)$$

where the $a_{ij}(g)$ are given by $g(w_j) = \sum_{i}a_{ij}(g)w_i$ for all $j$.

The claim is that this $\mu$ defines a continuous map, but I can't see why this is the case. I assume the topology on $M_d(\Bbb C_K)$ is the product topology (let's say after identifying $M_d(\Bbb C_K)$ with $\Bbb C_K^{d^2}$).

答: Scratch that: a $\Bbb C_K$-representation of $G_K$ is an action $G_K \times W \to W$, with $W$ a $\Bbb C_K$-vector space which is continuous and semilinear. An action of $G_K$ on a $\Bbb C_K$-vector space is just a continuous homomorphism $G_K \to \operatorname{End}(W) = M_d(\Bbb C_K)$ (after fixing a basis of $W$) so this gives the continuity.


问:
Ambiguity on BODMAS, Multiplication and Division
$24\div4(8\div 4)$ - This is the equation. We need to solve. Lets do it then

$=> 24\div4(2)$

$=> 24\div8$

$=> 3$

Division is also the multiplication of the inverse. Therefore $24\div4(8\div 4)$ can be written as

$24*\frac{1}{4}(8*\frac{1}{4})$. Let us solve this

$=> 24 *\frac{1}{4}(2)$

$=> 24 *\frac{1}{2}$

$=> 12$

My question is which of the answer is correct? 3 or 12. I have followed BODMAS rule in both representation of the equation.

答:
For my two cents, the reason for this ambiguity stems from mixing two methods of writing multiplications and divisions, which is never justified.

Method 1: Using $\times$ and $\div$

$$\begin{align}
24 \div 4 \times (8 \div 4) &= 24 \div 4 \times 2 \\
&= 6 \times 2 \\
&= 12
\end{align}$$

Method 2: Not showing multiplications, writing divisions as fractions

$$\frac{24}{4} \left( \frac{8}{4} \right) = 6 (2) = 12$$

Both calculations are unambiguous and yield the same, correct answer.

The two methods should simply never be mixed; whenever they are, there is inevitably the possibility of ambiguity.

Note that either method could yield the other answer with a little re-writing:

Method 1:

$$\begin{align}
24 \div \left[ 4 \times (8 \div 4) \right] &= 24 \div \left( 4 \times 2 \right) \\
&= 24 \div 8 \\
&= 3
\end{align}$$

Method 2:

$$\begin{align}
\frac{24}{4 \left( \frac{8}{4} \right) } &= \frac{24}{4(2)}\\
&= \frac{24}{8} \\
&= 3
\end{align}$$


问:
Why can't this be a coset?
Let $H$ be a subgroup of $G$ and H is not normal, there are left  cosets $aH$ and $bH$ whose product isn't a coset.

My attempt:

$ab H\subset aHbH$ and if H is not normal, if $ah_1bh_2=abh_3$ for all $h_1$, $h_2$, and $b$ then $h_1b=bh'$ for all $b$ and $H$ is normal a contradiction. So,$ab H\neq aHbH$ for some  $a,b$ and as cosets don't overlap $aHbH$ is not a coset of $H$. However, I don't know how to show that this isn't a coset of some other subgroup.

答:
Others have given you hints as to why normality of $H$ follows, if we want $(aH)(bH)$ to be a coset of $H$ for all $a,b\in G$. I am fairly sure that this is the intended meaning of whoever gave you this exercise. This is because otherwise the claim is false. I proffer the following counterexample.

Let $G$ be the dihedral group of 8 elements, so it is generated by an element $r$ (=rotation by 90 degrees) and an element $s$ (= a reflection). These are subject to the relations $r^4=s^2=1, sr^i=r^{-i}s$ (for all integers $i$). We observe that the element $r^2$ (=rotation by 180 degrees) is in the center. Let $H=\langle s\rangle=\{1,s\}$. This is clearly not normal, for $rH=\{r,rs\}\neq \{r,sr=r^3s\}=Hr$. Yet I claim that $(aH)(bH)$ is always a coset of some subgroup of $G$.

All the cosets of $H$ are of the form $r^iH$ for $i=0,1,2$ or $3$. We have
$$
(r^iH)(r^jH)=\{r^{i+j},r^{i-j}s,r^{i+j}s,r^{i-j}\}.
$$
If $j$ is even, then $r^{i+j}=r^{i-j}$, and this is the coset $r^{i+j}H$ of $H$. If $j$ is odd, then $r^{i+j}=r^{i-j}r^{2j}=r^{i-j}r^2$. As $r^2$ is central this means that
$$
(r^iH)(r^jH)=r^iK,
$$
where $K$ is the subgroup generated by $s$ and $r^2$.

So it stands to reason that the (surrounding) context implies that the question is only about cosets of the fixed subgroup $H$.

Title: Why can't this be a coset?


问:
Notation for graph adjacency
Which is the best manner to express symbolically the adjacency relation in a graph? Is there a common notation for this?

More specifically I would like to avoid things as $\{v, w\} \in E(X)$, where $E(X)$ is "edges of the graph $X$". I've seen in some notes the notation $v \sim w$, but I don't like the idea of using a symbol like $\sim$ for a non reflexive, non transitive relation.

答:
$v\sim w$ is the standard notation here. I'm working from home so don't have access to textbooks as sources, but I have never seen anything else used instead. From wikipedia:

The edges of a directed simple graph permitting loops $G$ is a homogeneous relation $\sim$ on the vertices of $G$ that is called the adjacency relation of $G$. Specifically, for each edge $(x,y)$, its endpoints $x$ and $y$ are said to be adjacent to one another, which is denoted $x\sim y$.

(They mention the most general case of directed graphs perhaps with loops, but the same notation is certainly used for undirected graphs.)

Title: Notation for graph adjacency


问:
Two questions on tensor (wedge) product
I am new to exterior algebra, and am very confused about the new concepts. I would appreciate a little help.

I saw in the textbook that if $\alpha$ is a $k$-covector where $k$ is odd, then $\alpha \land \alpha = 0$. If I assume that $\beta$ is a $k$-tensor, then does the same result hold?

I tried
$$\alpha \land \alpha = \frac{1}{2k!}\sum_{\sigma \in S_{2k}} sgn(\sigma)\, \sigma \,\alpha \otimes \alpha$$
but without any assumption about symmetricity or alternating property, I can't really say anything about the evaluation of the product. (perhaps this is the reason that wedge product is defined only on alternating functions)

Also, I saw this post: Wedge product of $\beta \wedge dx$
$\alpha = dx + dy + dz$ and $\beta = 2dx -dy  + dz$. Then,

i) Is $\alpha$ even a tensor? It doesn't take any argument. If it is not a tensor, how can I do wedge product on $\alpha \land \beta$?

ii) If $\alpha$ is a tensor, then is it a 3-tensor or 1-tensor? $\alpha$ is a linear combintation of three 1-tensors, but it takes three arguments, namely $dx, dy, dz$. If it is a 3-tensor, then $\alpha$ consists of basis $\{\hat e^i \otimes \hat e^j \otimes \hat e^k \}$ but it is not clear to me if $\alpha$ can be expressed in terms of the basis elements.

ii) I do not think $\alpha$ nor $\beta$ is alternating; if they are not alternating, how can I even do wedge sum? (note that on the link above, it does: $(dx+dy)∧(2dx−dy)=dx∧(2dx−dy)+dy∧(2dx−dy)=−dx∧dy+2dy∧dx=−3dx∧dy.$ I am confused since wedge sum is defined only on alternating functions.

(Perhaps, I am mixed up between differential form and wedge sum in the second question...)

Thank you in advance.

答:
I think you misunderstood the use of the wedge product. It is useful for alternating tensors as it makes the graded vector space $\Lambda^*V = \sum_{k\geqslant 0}\Lambda^k V$ a graded algebra. Of course you can expand the wedge product to arbitrary tensors but you loose a lot of sense. Moreover, your definition of $\alpha \wedge \alpha$ is false. If $(e_1,\ldots,e_n)$ is a basis of $V$, then a basis of $\Lambda^k V$ is $\left(e_{i_1}\wedge\cdots\wedge e_{i_k}\right)_{i_1<\cdots < i_k}$, where
\begin{align}
e_{i_1}\wedge \cdots \wedge e_{i_k} =\sum_{\sigma \in \mathbb{\mathfrak{S}_k}} \varepsilon(\sigma) e_{\sigma(i_1)}\otimes\cdots\otimes e_{\sigma(i_k)}
\end{align}
This is a definition. Moreover, the wedge product of two alternating tensors is defined to be coherent with the fact that $\left(e_{i_1}\wedge\cdots\wedge e_{i_k}\right)\wedge \left(e_{j_1}\wedge \cdots \wedge e_{j_l} \right)$ will be equal to $\left(e_{i_1}\wedge\cdots\wedge e_{i_k}\wedge e_{j_1}\wedge \cdots \wedge e_{j_l}  \right)$ and to be linear, thus the definition on the general case. Using the definition on the basis and linearity, one can show that if $\alpha$ is a $k$-alternating tensor and $\beta$ is a $l$ alternating tensor, then $\alpha\wedge \beta$ is a $(k+l)$ alternating tensor and
\begin{align}
\alpha \wedge \beta = (-1)^{kl}\beta \wedge \alpha
\end{align}
(show it on the basis and linearity gives you the result). Thus, if $\alpha$ is a $(2k+1)$-alternating tensor
\begin{align}
\alpha\wedge \alpha = (-1)^{(2k+1)^2}\alpha\wedge \alpha = -\alpha\wedge \alpha
\end{align}
and it follows that $\alpha \wedge \alpha = 0$.

In $\mathbb{R}^n$ with canonical basis $(e_1,\ldots,e_n)$, one define its dual space $\Lambda^1 \mathbb{R}^n = \left(\mathbb{R}^n\right)^* = L\left(\mathbb{R}^n,\mathbb{R}\right)$ with the dual the basis $({e_1}^*,\ldots,{e_n}^*)$, which is defined by ${e_i}^*(e_j) = \delta_{i,j}$. We write this basis $(\mathrm{d}x^1,\ldots,\mathrm{d}x^n)$. This is a notation. A vector $\mathbb{R}^n$ is defined in coordinates by $V = V_1 e_1 + \cdots V_n e_n$. A $1$ tensor on $\mathbb{R}^n$ is of the form $\alpha = \sum_{i=1}^n \alpha_i \mathrm{d}x^i$ where $\alpha_i$ are scalars. By the very definition of the dual basis, we can define
\begin{align}
\alpha(V) = \sum_{i=1}^n \alpha_iV_i
\end{align}
it is an alternating $1$ tensor as it takes only one argument (the vector $V$).

In $\mathbb{R}^3$, we prefer to use $(x,y,z)$ as coordinates, $\partial_x,\partial_y,\partial_z$ as the canonical basis and $\mathrm{d}x,\mathrm{d}y$ and $\mathrm{d}z$ for the basis of alternating $1$ tensors.

i) $\alpha = \mathrm{d}x + \mathrm{d}y + \mathrm{d}z$ is an alternating $1$-tensor as a linear combination of the basis of alternating $1$-tensors. Same for $\beta = 2\mathrm{d}x - \mathrm{d}y + \mathrm{d}z$. If $V$ is a vector field over $\mathbb{R}^3$, say $V = V_x \partial_x + V_y\partial_y + V_z \partial_z$:
\begin{align}
\alpha(V) &= V_x + V_y + V_z\\
\beta(V) &= 2V_x - V_y + V_z
\end{align}
As they are alternating tensors, their wedge product is well defined and by linearity, one has
\begin{align}
\alpha\wedge\beta &= \left(\mathrm{d}x + \mathrm{d}y + \mathrm{d}z \right) \wedge \left(2\mathrm{d}x - \mathrm{d}y + \mathrm{d}z \right)\\
&= \mathrm{d}x \wedge (2\mathrm{d}x) + \mathrm{d}x \wedge (-\mathrm{d}y) + \mathrm{d}x \wedge \mathrm{d}z \\
&~~~+ \mathrm{d}y \wedge (2\mathrm{d}x) + \mathrm{d}y \wedge (-\mathrm{d}y) + \mathrm{d}y \wedge \mathrm{d}z \\
&~~~+\mathrm{d}z \wedge (2\mathrm{d}x) + \mathrm{d}z \wedge (-\mathrm{d}y) + \mathrm{d}z \wedge \mathrm{d}z \\
&= -3\mathrm{d}x\wedge\mathrm{d}y + 2 \mathrm{d}y \wedge \mathrm{d}z - \mathrm{d}x \wedge \mathrm{d}z
\end{align}
(recall that as we use $1$-tensors, $\mathrm{d}x\wedge\mathrm{d}x = 0$, $\mathrm{d}y \wedge \mathrm{d}x = - \mathrm{d}x \wedge \mathrm{d}y$, etc.)

For ii) and iii), I already answered by saying that every $1$-tensor is an alternating tensor. This is because there is only one permutation of $1$ index, the identity, with signature signature $1$!

Commentary In the case of a manifold $M$, we are not looking at a fixed vector space $V$ and its exterior algebra but at a vector bundle $\Lambda^*(T^*M)$. For each $p \in M$, the cotangent bundle $T_pM^*$ has a local frame $(\partial_{x^1},\ldots,\partial_{x^n})$, thus we can define its exetrior algebra pointwisely the way we did for vector spaces. We define the exterior bundle of $M$ to be their union. We are then considering sections of this vector bundle. A $k$ differential form is defined to be a section of $\Lambda^k (TM^*)$, that is a smooth function $\alpha : p  \mapsto \alpha_p \in \Lambda^k(T_pM^*)$. In a local coordinate system, every $k$ differential form can be written $\alpha(p) = \sum_{i_1< \cdots< i_k} \alpha_{i_1,\cdots,i_k}\mathrm{d}x^{i_1}\wedge\cdots\wedge\mathrm{d}x^{i_k}$, where $\alpha_{i_1,\cdots,i_k}$ is a smooth function of $M$. For a fixed $p$, $\alpha(p)$ is a $k$ alternating tensor (in the vector space $\Lambda^k(T_pM^*)$). The definition of the wedge product is to be understood pointwisely:
\begin{align}
\alpha\wedge\beta : p \mapsto \alpha(p)\wedge \beta(p)
\end{align}
and if $\alpha$ and $\beta$ are $k$ and $l$ differential forms, then $\alpha\wedge \beta$ is a $k+l$ differential form.

Title: Two questions on tensor (wedge) product


问:
Winding number is 0
The initial exercise is to ask me to show the winding number is always an integer. I can prove this part but there is a part asking me to show also that $n(\gamma,a) = 0$ if $a$ is in the unbounded connected component.

My idea is if $a$ is in the unbounded connected component, then $a\notin \operatorname{int}(\gamma)$ which means $$n(\gamma,a) = \frac{1}{2\pi i}\int_\gamma\frac{1}{z-a}dz=0$$

Am I correct with this idea?

答: Let $[\alpha,\beta]$ be the domain of $\gamma$. Since the winding number is always a number and since the function$$\begin{array}{ccc}\Bbb C\setminus\gamma\bigl([\alpha,\beta]\bigr)&\longrightarrow&\Bbb Z\\a&\mapsto&\displaystyle\frac1{2\pi i}\int_\gamma\frac{\mathrm dz}{z-a}\end{array}$$is continuous, it is constant on any connected subset of $\Bbb C\setminus\gamma\bigl([\alpha,\beta]\bigr)$. On the other hand,$$\lim_{a\to\infty}\left|\frac1{2\pi i}\int_\gamma\frac{\mathrm dz}{z-a}\right|=0$$and therefore $\left|\frac1{2\pi i}\int_\gamma\frac{\mathrm dz}{z-a}\right|<1$ if $|a|$ is large enough. So, $\left|\frac1{2\pi i}\int_\gamma\frac{\mathrm dz}{z-a}\right|=0$ if $|a|$ is large enough, and tharefore it is $0$ on the unbounded connected component of $\Bbb C\setminus\gamma\bigl([\alpha,\beta]\bigr)$.

Title: Winding number is 0


问:
Problem separating variables (PDE)
I have the following PDE:

$$\nabla^2 u=-k^2u$$
where $k>0$.

The geometry demands cylindrical coordinates, which yields:
$$\frac{u_r}{r}+u_{rr}+\frac{u_{\varphi \varphi}}{r^2}+u_{zz}=-k^2u$$
So we're looking for a function:
$$u(r,\varphi,z)=R(r)Y(\varphi,z)$$
There's a  homogeneous boundary condition for $R(r_o)=0$, so separation of variables should be possible.
$$\frac{YR'}{r}+YR''+\frac{RY_{\varphi \varphi}}{r^2}+RY_{zz}=-k^2 RY$$
Divide both sides by $RY$:
$$\frac{R'}{rR}+\frac{R''}{R}+\frac{Y_{\varphi \varphi}}{r^2 Y}+\frac{Y_{zz}}{Y}=-k^2$$
But because the $r^2$ factor in the third denominator this is not separable?

I tried also this:
$$Y(\varphi,z)=\Phi(\varphi)Z(z)$$
$$\frac{Y_{\varphi \varphi}}{r^2 Y}=\frac{\Phi''}{r^2 \Phi}$$
$$\frac{Y_{zz}}{Y}=\frac{Z''}{Z}$$

But the problem remains. Is this not separable or am I getting this wrong?

答:
$$\frac{u_r}{r}+u_{rr}+\frac{u_{\varphi \varphi}}{r^2}+u_{zz}=-k^2u$$
$$u(r,\varphi,z)=R(r)P(\varphi)Z(z)$$
$$\frac{R'}{r}PZ+R''PZ+\frac{R}{r^2}P''Z+RPZ''=-k^2RPZ$$
$$\frac{R'}{rR}+\frac{R''}{R}+\frac{1}{r^2}\frac{P''}{P}+\frac{Z''}{Z}=-k^2$$
This implies :
$$\begin{cases}
\frac{P''}{P}=\lambda=\text{constant} \\
\frac{Z''}{Z}=\mu=\text{constant} \\
\frac{R'}{rR}+\frac{R''}{R}+\frac{\lambda}{r^2}+\mu+k^2=0
\end{cases}$$
$P(\varphi)=c_1e^{\lambda\:\varphi}$

$Z(z)=c_2e^{\mu\:z}$

Solving for $R(r)$ the third equation $\quad R''+\frac{1}{r}R'+(\frac{\lambda}{r^2}+\mu+k^2)R=0\quad$
involves Bessel functions.

Title: Problem separating variables (PDE)


答:
No, it's not possible, since infinitely many values of $b$ lead to the same value of $a$. This is obvious for $c=0$. For $c\neq0$ both $b_1=b$ and $b_2=b+c$ give the same $a$:
$$ b_2\times c = (b+c)\times c = b\times c + c\times c = b\times c = b_1\times c. $$</p>

Title: Reordering vector product

Body: If I have vectors $a, b, c \in \mathbb{R}^3$, and if we have e.g. $a = b\times c$, is there any way to express $b$ in terms of the other two? </p>