问:
Delta method Taylor Expansion question

In the Taylor series expansion of the delta method, could someone help me why it leads to this equation?
I understand the first and second components, but the little-o function on the very right is confusing to me. How is this part of the taylor expansion? Is this function somehow equal to the rest of the expansion (remainder of the taylor formula)? Otherwise I have no idea how to make of the little-o function or understand how this can be part of the taylor expansion.

I would appreciate it if someone could explain where the little-o function came from and how the taylor expansion of f(Xn) at theta leads to the existence of this little-o function, thanks.

答:
Typically, the Delta methods is proven using the Mean value theorem. Write
$$
\tau_n(f(X_n)-f(\mu))=f'(\mu_n)\tau_n(X_n-\mu),
$$
where $\tau_n$ is a normalizing sequence (e.g., $\sqrt{n}$) and $\mu_n$ is a random variable satisfying $|\mu_n-\mu|\le |X_n-\mu|$. When $X_n$ converges in prob. to $\mu$ (i.e., $X_n-\mu=o_p(1)$), $\tau_n(X_n-\mu)=O_p(1)$, and $f'$ is continuous in the nbrhd of $\mu$,
\begin{align}
\tau_n(f(X_n)-f(\mu))&=f'(\mu)\tau_n(X_n-\mu)+(f'(\mu)-f'(\mu_n))\tau_n(X_n-\mu) \\
&=f'(\mu)\tau_n(X_n-\mu)+o_p(1)O_p(1),
\end{align}
and $o_p(1)O_p(1)=o_p(1)$.

Title: Delta method Taylor Expansion question


问:
Prove that $1+ \frac{5}{4}x ≤(1+x)^{\frac{5}{4}}≤1+ \frac{5}{4} x+\frac{5}{16} x^2$
How to prove this inequality:
$$1+ \frac54x \le (1+x)^{5/4} \le 1+ \frac54 x+\frac5{16} x^2$$

I tried applying binomial theorem to get this inequality, but I am not able to get it precisely. Please help.

答:
Using the Bernoulli's inequality, we have
$$(1+x)^{5/4} \geqslant 1 + \frac{5}{4} \cdot x,$$
and
$$(1+x)^{5/4} = (1+x)(1+x)^{1/4} \leqslant (1+x)\left(1+\frac{1}{4} \cdot x\right) $$
$$\leqslant (1+x)\left(1+\frac{1}{4} \cdot x\right) + \frac{1}{16}\cdot x^2 = 1+ \frac54 x+\frac5{16} x^2.$$
The proof is completed.

ContentLicense: CC BY-SA 4.0


问:
What is the expected value of the numbers of calls for a survey?
I was asked the following question by a medical doctor:

He is working on a survey. Assume that he called 120 persons and 24 persons agreed to do the survey. The second time, he called 57 persons among those who refused the first time and 10 persons agreed to do the survey. What is the expected value of the number of phone calls for people to agree to do the survey?

I first thought it would be

$$\frac{24\times 1+10\times 2}{120}.$$

Then I thought it would be

$$\frac{24}{120}\times 1+\frac{10}{57}\times 2.$$

I am confused by these two ideas. I prefer the second computation. Please enlighten me. Thanks a bunch.

答:
It really depends on what exactly is the question and what are the assumptions regarding the population and their willingness to participate in the survey. Here is my approach, which might not be exactly what OP wants.

I assume that each person has a fixed probability $p$ to agree, regardless of the number of phone calls. This is constant for all people and that the $57$ people were chosen at random. Then, the number of calls until someone agrees to participate has a Geometric distribution with parameter $p$, and the average number of calls needed is $1/p$.

It is left to evaluate $p$. Based on the data, $24$ people agreed on the first call, $10$ on the second call, $47$ didn't agree on the first two calls (so $X\geq 3$ for them) and $39$ people didn't agree on the first call (so $X\geq 2$ for them).
The likelihood function is therefore
$$L(p)=p^{24} \times [(1-p)p]^{10}\times {(1-p)^2}^{39}\times {(1-p)^3}^{47}$$
The maximum of this function for $p\in (0,1)$ is at $p=\tfrac{34}{263}$ so the expected number of calls for a person to agree is $\tfrac{263}{34}=7.7$

Sanity check: in the first round, the success rate was fifth, so the expectation based on that should be around $5$. The people that didn't agree twice pull the success rate down, increasing the expected number of phonecalls.


问:
How to write in $2^x=5$ in logarithmic form?
How do I write:
$$2^x = 5$$

In a logarithmic form?

I've looked for a solution for some time now, so I decided to try here.

答: Take $\log_e$ then you get $\log_e 2^x=\log_e 5 \implies x=\dfrac{\log_e 5}{\log_e 2}=\log_25$</p>

ContentLicense: CC BY-SA 3.0


问:
Iterative methods: What happens when the spectral radius of a matrix is exactly 1?
I know that an iterative method (I'm using Jacobi and Gauss-Seidel in this case) will converge iff the spectral radius (max absolute value of eigenvalues) of its iterative matrix is strictly less than 1.

I've encountered a situation where the spectral radius of the iterative matrix is exactly 1. I'm pretty sure that it will converge for certain initial "guesses", but not others. Why is this the case? Is there a way to determine which initial guesses result in convergence?

答:
Consider $A=M-N,$ where in Gauss-Seidel method $M=D-L,$ and $N=U,$ then the iteration matrix $B_{G}=$
$M^{-1} N=(D-L)^{-1} U .$ For the Gauss-Seidel method to converge, the spectral radius of $B_{G}$ should be less than $1,$ i.e.
$$
\rho\left(B_{G}\right)=\max _{\lambda \in \Lambda\left(B_{G}\right)}|\lambda|<1
$$
Let $(\lambda, v)$ be a pair of eigenvalue and eigenvector of $B_{G},\left(\lambda \in \Lambda\left(B_{G}\right)\right)$ then
$$
\begin{aligned}
B_{G} v &=(D-L)^{-1} U v=\lambda v \\
U v &=\lambda(D-L) v \\
-\sum_{j>i} A(i, j) * v(j) &=\lambda \sum_{j \leqslant i} A(i, j) * v(j)=\lambda A(i, i) * v(i)+\lambda \sum_{ji} A(i, j) * v(j)-\lambda \sum_{ji} A(i, j) * v(j)-\lambda \sum_{ji}|A(i, j)||v(j)|+|\lambda| \sum_{ji}|A(i, j)||v(j)|+\sum_{j<i}|A(i, j)||v(j)|\right) \quad\\
&=\max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i}^{n}|A(i, j)||v(j)|
\end{aligned}
$$
But $v(j) \leqslant\|v\|_{\infty},$ thus
$$
|\lambda||v(i)|\leqslant \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i}^{n}\left|\frac{A(i, j)}{A(i, i)} \right| \|v_{\infty}\|=\|v\|_{\infty} \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i}^{n}\left|\frac{A(i, j)}{A(i, i)}\right| \quad \text { for } i=1,2, \cdots, n
$$
Let $i_{0}$ be such that $\left|v\left(i_{0}\right)\right|=\|v\|_{\infty},$ then we have that
$$
\begin{aligned}
|\lambda|\left|v\left(i_{0}\right)\right| \leqslant &\|v\|_{\infty} \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i_{0}}^{n}\left|\frac{A\left(i_{0}, j\right)}{A\left(i_{0}, i_{0}\right)}\right| \\
\therefore & |\lambda|  \leqslant \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i_{0}}^{n}\left|\frac{A\left(i_{0}, j\right)}{A\left(i_{0}, i_{0}\right)}\right|
\;\llap{\mathrel{\boxed{\phantom{\therefore  |\lambda|  \leqslant \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i_{0}}^{n}\left|\frac{A\left(i_{0}, j\right)}{A\left(i_{0}, i_{0}\right)}\right|}}}}
\end{aligned}
$$
Considering $A$ to be strictly diagonally dominant, that is to say :
$$
|A(i, i)|>\sum_{j=1 \atop j \neq i}^{n}|A(i, j)|, \quad \forall i=1,2, \cdots, n
$$
We will have that :
$$
\begin{aligned}
|\lambda|& \leqslant \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i_{0}}^{n}\left|\frac{A\left(i_{0}, j\right)}{A\left(i_{0}, i_{0}\right)}\right| \\
&<\max \{1,|\lambda|\} \frac{1}{\left|A\left(i_{0}, i_{0}\right)\right|}\left|A\left(i_{0}, i_{0}\right)\right|=\max \{1,|\lambda|\}
\end{aligned}
\tag{1}
$$
\begin{aligned}
&\text { If }|\lambda| \color{red}{\geqslant} 1, \text { then this leads to a contradiction since equation (1) becomes }|\lambda|\color{red}{<}|\lambda| \text { . }\\
&\text { Thus, }|\lambda|<1 \text { and } :
\end{aligned}
$$
\boxed{\rho\left(B_{G}\right)=\max _{\lambda \in \Lambda\left(B_{G}\right)}|\lambda|<1}
$$

Body:
Consider $A=M-N,$ where in Gauss-Seidel method $M=D-L,$ and $N=U,$ then the iteration matrix $B_{G}=$
$M^{-1} N=(D-L)^{-1} U .$ For the Gauss-Seidel method to converge, the spectral radius of $B_{G}$ should be less than $1,$ i.e.
$$
\rho\left(B_{G}\right)=\max _{\lambda \in \Lambda\left(B_{G}\right)}|\lambda|<1
$$
Let $(\lambda, v)$ be a pair of eigenvalue and eigenvector of $B_{G},\left(\lambda \in \Lambda\left(B_{G}\right)\right)$ then
$$
\begin{aligned}
B_{G} v &=(D-L)^{-1} U v=\lambda v \\
U v &=\lambda(D-L) v \\
-\sum_{j>i} A(i, j) * v(j) &=\lambda \sum_{j \leqslant i} A(i, j) * v(j)=\lambda A(i, i) * v(i)+\lambda \sum_{j<i} A(i, j) * v(j) \quad \text { for } i=1,2, \cdots, n \\
\Longrightarrow \lambda A(i, i) * v(i) &=-\sum_{j>i} A(i, j) * v(j)-\lambda \sum_{j<i} A(i, j) * v(j) \quad \text { for } i=1,2, \cdots, n \\
\Longrightarrow|\lambda||A(i, i)||v(i)| &=\left|-\sum_{j>i} A(i, j) * v(j)-\lambda \sum_{j<i} A(i, j) * v(j)\right| \quad \text { for } i=1,2, \cdots, n \\
& \leqslant \sum_{j>i}|A(i, j)||v(j)|+|\lambda| \sum_{j<i}|A(i, j)||v(j)| \\
& \leqslant \max \{1,|\lambda|\}\left(\sum_{j>i}|A(i, j)||v(j)|+\sum_{j<i}|A(i, j)||v(j)|\right) \quad\\
&=\max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i}^{n}|A(i, j)||v(j)|
\end{aligned}
$$
But $v(j) \leqslant\|v\|_{\infty},$ thus
$$
|\lambda||v(i)|\leqslant \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i}^{n}\left|\frac{A(i, j)}{A(i, i)} \right| \|v_{\infty}\|=\|v\|_{\infty} \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i}^{n}\left|\frac{A(i, j)}{A(i, i)}\right| \quad \text { for } i=1,2, \cdots, n
$$
Let $i_{0}$ be such that $\left|v\left(i_{0}\right)\right|=\|v\|_{\infty},$ then we have that
$$
\begin{aligned}
|\lambda|\left|v\left(i_{0}\right)\right| \leqslant &\|v\|_{\infty} \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i_{0}}^{n}\left|\frac{A\left(i_{0}, j\right)}{A\left(i_{0}, i_{0}\right)}\right| \\
\therefore & |\lambda|  \leqslant \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i_{0}}^{n}\left|\frac{A\left(i_{0}, j\right)}{A\left(i_{0}, i_{0}\right)}\right|
\;\llap{\mathrel{\boxed{\phantom{\therefore  |\lambda|  \leqslant \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i_{0}}^{n}\left|\frac{A\left(i_{0}, j\right)}{A\left(i_{0}, i_{0}\right)}\right|}}}}
\end{aligned}
$$
Considering $A$ to be strictly diagonally dominant, that is to say :
$$
|A(i, i)|>\sum_{j=1 \atop j \neq i}^{n}|A(i, j)|, \quad \forall i=1,2, \cdots, n
$$
We will have that :
$$
\begin{aligned}
|\lambda|& \leqslant \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i_{0}}^{n}\left|\frac{A\left(i_{0}, j\right)}{A\left(i_{0}, i_{0}\right)}\right| \\
&<\max \{1,|\lambda|\} \frac{1}{\left|A\left(i_{0}, i_{0}\right)\right|}\left|A\left(i_{0}, i_{0}\right)\right|=\max \{1,|\lambda|\}
\end{aligned}
\tag{1}
$$
\begin{aligned}
&\text { If }|\lambda| \color{red}{\geqslant} 1, \text { then this leads to a contradiction since equation (1) becomes }|\lambda|\color{red}{<}|\lambda| \text { . }\\
&\text { Thus, }|\lambda|<1 \text { and } :
\end{aligned}
$$
\boxed{\rho\left(B_{G}\right)=\max _{\lambda \in \Lambda\left(B_{G}\right)}|\lambda|<1}
$$


问:
Help understanding a proof (Linear Transformation)

On a V vectorial space

we have that

$T(x+y) = Tx - Ty$

$T(\alpha x) = \alpha Tx$ , for $\alpha \geq 0$

Prove that T is linear.

I tried some stuff by myself and then I found an answer online, but I can't completely understand it.

It goes like this:

Suppose $\alpha < 0 $. We have $T[(\alpha-\alpha)x]= T(0x)=0Tx=0$

That part I get.

But then it goes,

so that $0=T( \alpha x- \alpha x) = T( \alpha x) + T(- \alpha x) = T(\alpha x) - \alpha T(  x) $

Thus, $T(\alpha x)= \alpha Tx$ if $\alpha < 0 $

The part I don't quite get is, how you get this equality:

$T( \alpha x- \alpha x) = T( \alpha x) + T(- \alpha x)  $

答:
If both conditions works for any vectors $x,y$ then $T(x)=0$ for all $x$.

First of all take $\alpha=0$ in the second condition, then $T(0)=0$.

Now take $x=0$ in the first condition. It give us $T(y)=-T(y)\to T(y)=0$, for all $y$.

By the way, your doubt is pertinent. The online solution is wrong. The correct expression should be, $0=T(\alpha x+(-\alpha x))=T(\alpha x)-T(-\alpha x)$.


问:
Showing a Vector is orthogonal to a Line
In order to show two vectors are orthogonal to each other, their dot product must be 0. How does this extend to lines? The question I have is as follows:

Show that the vector w is orthogonal to the line w$^T$x+$b=0$. Hint: Consider two points, x$_1$ and x$_2$ that lie on the line. What is the inner product of w$^T$(x$_1$ - x$_2$)?

Here is what I have so far. In order to for the vector w to be orthogonal to the line w$^T$x+$b=0$, the dot product between them must be $0$. Observe,

$$[w_1, w_2]^T[x_1, x_2] + b =0$$
$$w_1x_1 + w_2x_2 + b = 0$$

I feel like the answer is supposed to obvious, but I also do not know how the hint becomes useful.

答:
To prove that w is orthogonal to the line w$^T$x + $b$ = $0$, take 2 points on the line. Let these points be x$_1$ and x$_2$, since these 2 points lie on the line they will satisfy:

w$^T$x$_1$$+ \ b = 0$

w$^T$x$_2$$+ \ b = 0$

Subtracting these 2 equations will give you,

w$^T$(x$_1$-x$_2$)$= 0$

Giving you the result. I hope this helps.


问:
Finding a point on triangle.
I have a non-right triangle with known x,y coordinates. Lets say A,B,C is the points of triangle. I want to find a new point D which is perpendicular line from point B and lies on the line AC. Any simpled questions for D(x,y)?</p>

答:
First, given coordinates of $A$ and $C$, you can have the formula of the line $AC$.

Secondly, we know that line $BD$ and $AC$ are perpendicular, so we can get the slope of $BD$ (the product of slopes of perpendicular lines in $\mathbb R^2$ is $-1$). With coordinates of $B$, we can then get the formula of $BD$. And $D$ is just the point of intersection of $AC$ and $BD$.

Title: Finding a point on triangle.


问:
solution of $x^x = ax^b$
I am interested in the solution $x > 1$ of the transcendental equation $$x^x = ax^b$$ with $a > 1$, $b \in \mathbb{R}$. I am looking for both analytical and numerical methods, as well as approximations.

To see that the solution exists, write the equation as $\log x = A/(x-b)$ with $A = \log a$ and use the elementary bounds of the logarithm to obtain the bounds $$\frac{b+1}{2} + \frac{1}{2} \sqrt{(b-1)^2 + 4A} \leq x \leq \frac{b+1+A}{2} + \frac{1}{2} \sqrt{(b-1+A)^2+4 A}$$ for the solution. This is enough for general purpose numerical methods, but I would like to see if more specialised tools are available.

There is also a potential connection to the Lambert $W$-function: Write the equation as $x\log x = A + b \log x$ and set $y = x \log x$, by which $\log x = W(y)$, so that $$y = A + b \, W(y) \;.$$ This can be iterated with an initial guess based on the above bounds, but perhaps there are other useful results for the $W$-function of which I am not aware.

答:
From a formal point of view, there is probably a solution in terms of the generalized Lambert function (have a look at equation $(4)$).

This being said, consider that you look for the zero's of function
$$f(x)=(x-b)\log(x)-\log(a)$$
$$f'(x)=-\frac{b}{x}+\log (x)+1 \qquad \text{and} \qquad f''(x)=\frac{b}{x^2}+\frac{1}{x}$$ The first derivative cancels at
$$x_*=\frac{b}{W(e b)}$$ Good news : $f''(x_*) >0 \quad \forall b >0$.

Now, try a Taylor expansion to second order
$$f(x)=f(x_*)+\frac 12f''(x_*) (x-x_*)^2+O\left((x-x_*)^3\right)$$ This would give as a first estimate
$$x_0=x_*+\sqrt{-2\frac {f(x_*)}{f''(x_*)}}$$

Trying for $a=\pi$ and $b=e$, this would give $x_0=3.26318$ while the solution obtained using Newton method is $x=3.61002$. Using this estimate, the iterates of Newton method are
$$\{3.2631771,3.6338427,3.6101047,3.6100159\}$$
Trying for $a=e$ and $b=\pi$, this would give $x_0=3.51825$ while the solution obtained using Newton method is $x=3.87925$.

Try and let me know how this works for you (I have a few ideas for possible improvements).

Edit

From a series of numerical tests I performed, it seems that the left bound you proposed is good for the $x_0$ of Newton method. It can be improved at no cost using
$$\log(x) \geq 2\frac{x-1}{x+1}$$ the rhs being the simplest Padé approximant of $\log(x)$ built around $x=1$.

With $A=\log(a)$, this would give
$$x_0=\frac{1}{4} \left(A+2b+2+\sqrt{A^2+4 A (b+3)+4 (b-1)^2}\right)$$ which, for the worked examples, gives something quite close to the  geometric mean of your present lower and upper bound $(3.71264\text{ vs }3.69213)$ and $(3.97745\text{ vs }3.95909)$.

For $a=\pi$ and $b=e$, Newton iterates are
$$\{3.7126445,3.6115922,3.6100163,3.6100159\}$$
For $a=e$ and $b=\pi$, Newton iterates are
$$\{3.9774505,3.8806309,3.8792508,3.8792505\}$$

For these two cases, no overshoot during the path to solution.

Another possible approximation could be based on the fact that
$$\log(x) > \frac {e \log \left(\frac{e}{2}\right)+\left(x-\frac{e}{2}\right) \left(2+\log
   \left(\frac{e}{2}\right)\right)} {x+\frac{e}{2} }$$ as soon as $x > 1.36$.

For the worked examples, it could give $3.65464$ and $3.92416$

Update

Since $a >1$, as far as I can see, the solution looked for is $x>b$. So, considering
$$(x-b)\log(x)=\log(a)=A$$ approximate $\log(x)$ by its simplest Padé approximant around $x=b$ which is
$$\log(x)=\frac {2b\log(b)+ (2+\log (b))(x-b)}{x+b}$$ which leads to
$$x_0=\frac{A+4b+\sqrt{A (A+16 b)+4 b \log (b) (A+b \log (b))}}{2 (2+\log (b))}$$

For the worked examples, this gives $x_0=3.61112$ (the solution being $x=3.61002$) and $x_0=3.87962$ (the solution being $x=3.87925$).

Using $a=123456$ and $b=789$ this would give $x_0=790.75687946638$ while the solution is
$x=  790.75687946614$.

Using the same idea, we could expand $f(x)+A$ as a series around $x=b$
$$f(x)+A=(x-b)\log(b)+\sum_{n=2}^\infty (-1)^n\frac{ b^{1-n} }{n-1}(x-b)^n$$ and use series reversion to obtain
$$x=b+t\left(1+\sum_{n=1}^p c_n u^n\right)$$ where $$L=\log(b) \qquad\qquad t=\frac{A}{L}\qquad\qquad u=-\frac{t}{b L}$$ and the $c_n$'s form the sequence
$$\left\{1,\frac{L}{2}+2,\frac{L^2}{3}+\frac{5 L}{2}+5,\frac{L^3}{4}+\frac{11
   L^2}{4}+\frac{21 L}{2}+14,\cdots\right\} $$

Title: solution of $x^x = ax^b$


问:
Best way to integrate this?
$$ \int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{\sin^2(x)}{\sqrt{1+\cos(x)}}dx $$
What would be the best way to go about integrating and evaluating this integral?

答:
Substitute $\cos x =t \implies -\sin x dx =dt $: $$\int_{0}^{0.5} \frac{\sqrt{1-t^2}}{\sqrt{1+t}} dt =\int_0^{0.5} \sqrt{1-t}dt = \left[ -\frac 23 (1-t)^{3/2} \right]_0^{0.5} = \frac 23-\frac 23 \cdot \frac{1}{2\sqrt 2} = \frac 23 -\frac{1}{3\sqrt 2}$$

Second way: $$\frac{\sin^2x}{\sqrt{1+\cos x}}= \frac{4\sin^2(x/2)\cos^2 (x/2) }{\sqrt 2 \cos(x/2)} = 2\sqrt 2 \sin^2(x/2) \cos(x/2) $$ Then you have something of the form $u^2 du$.

Title: Best way to integrate this?


问:
Recovering $z_1,z_2\in\mathbb{C}$ from their modulus and $\mathrm{Re}(z_1z_2), \ \mathrm{Im}(z_1z_2)$
Let $z_1,z_2\in\mathbb{C}$ arbitrary but fixed. Now, consider $x_1,x_2,x_3,x_4\in\mathbb{R}$ given by $$
x_1:=\mathrm{Re}(z_1z_2), \quad x_2:=\mathrm{Im}(z_1z_2), \quad x_3:=\vert z_1\vert^2+\vert z_2\vert^2, \quad x_4:=\vert z_1\vert^2-\vert z_2\vert^2.
$$
My question is, if I know $x_1,...,x_4$, can I completely recover $z_1$ and $z_2$ as functions of $x_1,...,x_4$? I've been trying to do some algebraic manipulations but I have't suceed in completely identifying $z_1$ and $z_2$. Clearly, from $x_3\pm x_4$ we can recover the modulus of $z_1$ and $z_2$, but I am not sure how to obtain their argument (their phase in the complex plane). Does anyone know how to do it?

Body:
Hint: compute $x_1$, $x_2$, $x_3$, $x_4$ for the following two pairs $(z_1, z_2)$:
$$
z_1 = 1,\; z_2 = -1 \\
z_1 = z_2 = i \\
$$


问:
Compound interest formula $A= \frac{p((1+\frac rn)^{nt}-1)}{\frac rn} $
I am doing an MBA assignment and I have given this formula

$$A=\frac{p((1+\frac rn)^{nt}-1)}{\frac rn} $$

Is this formula correct? I have usually seen $A=p(1+\frac rn)^{nt}$

The specific problem is

"You make monthly payments of $500 into a retirement annuity that earns 4% compounded monthly. What is the value of your account after 10 years?  Round your answer to the nearest cent (two decimal places)."

答:
Note that $p(1+\frac rn)^{nt}$ is the compounded value of a single payment. The value in 10 years is the sum of all monthly payments, which is

$$A=\sum_{k=0}^{nt-1} p \left(1+\frac rn \right)^{k}=\frac{p((1+\frac rn)^{nt}-1)}{\frac rn}
$$


答:
$2^x = 5$

Now 'take log' on both sides:

$\log_2 2^x = \log_2 5$.

We can now use the property $\log a^b = b\times \log a$.

Thus,

$x \times \log_2 2 = \log_2 5$.

But $\log_2 2 = 1$, therefore,

$x = \log_2 5$.

Title: How to write in $2^x=5$ in logarithmic form?

Body:
How do I write:
$$2^x = 5$$

In a logarithmic form?

I've looked for a solution for some time now, so I decided to try here.


答: No, because $z_1,\,z_2\mapsto e^{i\theta}z_1,\,e^{-i\theta}z_2$ preserve the $x_j$ for all $\theta\in\Bbb R$. In polar coordinates, write $z_j=r_je^{it_j}$ so $r_1=\sqrt{\tfrac{x_3+x_4}{2}},\,r_2=\sqrt{\tfrac{x_3-x_4}{2}},\,r_1r_2\exp i(t_1+t_2)=x_1+ix_2$ to show this is the most general underdeterminarion of the solution.

Title: Recovering $z_1,z_2\in\mathbb{C}$ from their modulus and $\mathrm{Re}(z_1z_2), \ \mathrm{Im}(z_1z_2)$

Body:
Let $z_1,z_2\in\mathbb{C}$ arbitrary but fixed. Now, consider $x_1,x_2,x_3,x_4\in\mathbb{R}$ given by $$
x_1:=\mathrm{Re}(z_1z_2), \quad x_2:=\mathrm{Im}(z_1z_2), \quad x_3:=\vert z_1\vert^2+\vert z_2\vert^2, \quad x_4:=\vert z_1\vert^2-\vert z_2\vert^2.
$$
My question is, if I know $x_1,...,x_4$, can I completely recover $z_1$ and $z_2$ as functions of $x_1,...,x_4$? I've been trying to do some algebraic manipulations but I have't suceed in completely identifying $z_1$ and $z_2$. Clearly, from $x_3\pm x_4$ we can recover the modulus of $z_1$ and $z_2$, but I am not sure how to obtain their argument (their phase in the complex plane). Does anyone know how to do it?


问:
how to prove $\sin(kx)$ and $\cos(kx)$ are the basis of the Function space
At the beginning of learning the Fourier series. My teacher just told us the format but didn't prove the Completeness that why we can do it. So I'm confused.

I know $\langle\cos(nx),\cos(mx)\rangle= \pi\delta_{n,m}$ (that same with sin(nx)), which means the Orthogonality. But I can't think out why they are the basis.

Hope for someone to settle my doubts.

答:
In the context of the Fourier serie, we can use the following inner product $$\forall f, g \in \mathscr{C}([-\pi\pi]): \quad \langle f,g \rangle =\int_{-\pi}^{\pi}f(x)g(x)dx$$
and is well-know that if $f(x),g(x)\not\equiv 0$, so $$\langle f, g \rangle =0 \implies f,g \quad \text{and independent linearly}.$$
Now, note that $$\int_{-\pi}^{\pi}\cos(nx)\cos(mx)dx=0, \quad n\not=m$$
$$\int_{-\pi}^{\pi}\sin(nx)\sin(mx)dx=0, \quad n\not=m$$
$$ \int_{-\pi}^{\pi}\sin(nx)\cos(mx)dx=0, \quad n\not=m \quad \text{or} \quad n=m.$$
so, you have that the set $$S=\{\sin(nx),\cos(mx)\}_{n,m\in \mathbb{N}}$$is a system orthogonal. It's to say $$\beta_{\mathbf{T.S}}=\left\{1,\cos(x),\sin(x),\cos(2x),\sin(2x),\ldots,\cos(mx),\sin(mx),\ldots\right\}$$is orthogonal, so in particular $\beta_{\mathbf{T.S}}$ is independent linearly.

Now, we need to prove that $\{\sin(nx),\cos(mx): n,m \in \mathbb{N}\}$ is completness, so we need to prove that any continuous function on $[-\pi,\pi]$ can be approximated uniformly by trigonometric polynomials.

Note that the uniform convergence in $[-\pi,\pi]$ implies $L^{2}-$convergence and continuos functions are dense in $L^{2}([-\pi,\pi])$, it follows that trigonometric polynomials are dense in $L^{2}([-\pi,\pi])$, so  $\{\sin(nx),\cos(mx): n,m \in \mathbb{N}\}$ is a basis.

The idea behind the completness proof is to obtain a trigonometric polynomial approximation (Weierstrass approximation theorem) of a continuous function $f$ by taking the convolution of $f$ with an approximate identity that is a trigonometric polynomials.


问:
Proving inequality given some conditions
I want to show that $-x^2y^2-y^4+4y^2 \geq 0$, if $x^4+2y^2 \leq 4$ and $-x^2y^2-y^4+4y^2 \leq 0$, if $x^4+2y^2 \geq 32$. Is there a factoring trick that helps to prove this explicitly?

答:
For the first question you need to prove
$$
-2x^2-2y^2+8 \geq 0
$$
with the condition $-2y^2 \geq x^4 -4$. This condition implies $x^2 < 2$ since otherwise it can never be fulfilled. Combining the two gives the sharper $
-2x^2+x^4+4 \geq 0
$ or $(x^2-1)^2 +3\ge 0$, done.

For the second question you need to prove
$$
-2x^2-2y^2+8 \leq 0
$$
with the condition $-2y^2 \leq x^4 -32$. This condition needs to be applied only for $x^2 < \sqrt{32}$ since otherwise it is fulfilled anyway. So for this case, we have  $
-2x^2+x^4-24 \leq 0
$ or $(x^2-1)^2 \le 25$, or $x^2 \le 6$, but this is fulfilled with  $x^2 < \sqrt{32} < 6$.

Consider the other case  $x^2 \ge \sqrt{32}$, then we see directly that $
-2x^2-2y^2+8 \leq -2x^2 +8 \leq - 2 \sqrt{32} +8 < 0
$


问:
Book recommendation for Pre-Calculus
So, I'm going to Pre-Calc honors next high school as a freshmen. I want to self-study right now so I won't get bad grades on my marks. Does anyone have a good book recommendation? Any tips for for getting good marks is also helpful.</p>

答:
I like the Art of Problem Solving series for self-study:

https://artofproblemsolving.com/store/item/precalculus

If you get the online version, there are embedded videos explaining various topics.  It's a different sort of text, that asks you first to wrestle with some questions, then it answers the questions, then it gives you some exercises.   The solution manual (and the solutions online) are very thorough.

Title: Book recommendation for Pre-Calculus


问:
Prove that $\{v_1+u_1,\cdots,v_n+u_n\}$ form a basis if $\{v_1,\cdots,v_n\}$ is an orthonormal basis and $\sum_{j=1}^n||u_j||^2<1$
Let $V$ be a finite dimensional inner product space with $\{v_1,\cdots,v_n\}$ as orthonormal basis vectors. Let $\{u_1,\cdots,u_n\}\in V$ be vectors such that $\sum_{j=1}^n||u_j||^2<1$, the prove that $\{v_1+u_1,\cdots,v_n+u_n\}$ form a basis for $V$, where $||u_j||$ is the norm of the vector $u_j$

I don't even see a clue of how the property $\sum_{j=1}^n||u_j||^2<1$ makes it a basis ?

答:
I assume your condition is $\sum_j||u_j||^2<1.$ If so, then we can prove the conclusion by contradiction. Suppose
$$\sum_ja_j(u_j+v_j)=0$$
for some $(a_1,\cdots,a_n)\neq 0.$ Then
$$\sum_ja_j^2=||\sum_ja_jv_j||^2=||\sum_ja_ju_j||^2\le \left(\sum_ja_j^2\right)\left(\sum_j||u_j||^2\right)<\sum_ja_j^2$$
since $\sum_ja_j^2>0.$ Thus the conclusion follows.


问:
Pullback of a morphism of schemes along the local scheme
This should be a very simple question but I can't seem to figure it out. If we have a morphism of schemes $f: X \rightarrow Y$, I am trying to understand what fiber is above some $\operatorname{spec} \kappa(f(p))$ for $p$ a point in $X$. Let's just work with affine schemes for now. If we have an $A$-algebra $\phi: A \rightarrow B$ which induces a morphism of schemes $f: \operatorname{spec}B \rightarrow \operatorname{spec}A$. Then firstly, if $p \in \operatorname{spec}B$ is a point, then what does the pullback along the morphism $\operatorname{spec} \mathcal{O}_{f(p)} \rightarrow \operatorname{spec}A$ look like? If $\mathfrak{p}$ is the prime ideal of $B$ corresponding to the point $p$, then obviously the resulting scheme is the tensor product $\operatorname{spec}(B \otimes_{A} A_{\phi^{-1}(\mathfrak{p})})$. But what does this scheme actually correspond to? Is there some simplification of this? I used to wrongly think that this was the scheme $\operatorname{spec}\mathcal{O}_{p}$ but I have realised there is no reason to think that is true.

So once I understand that, I would like to pullback further along the morphism $\operatorname{spec}\kappa(f(p)) \rightarrow \operatorname{spec}\mathcal{O}_{f(p)}$. I know this should be quite easy but I'm having trouble seeing it.

答:
The construction of the fiber product is an important construction in algebraic geometry hence this question may have general interest for students.

If $\phi: A \rightarrow B$ is a map of unital commutative rings with $X:=Spec(B), S:=Spec(A)$ and $f: X \rightarrow S$ the induced morphism of affine schemes, we may for any point $\mathfrak{p} \subseteq B$ consider the image under $f$ , $\mathfrak{q}:=\phi^{-1}(\mathfrak{p})$. We get an inclusion of affine schemes (the inclusion of the point $\mathfrak{q}$ in $S$):

I1. $i:Spec(\kappa(\mathfrak{q})) \rightarrow S$

and we may ask: What is the fiber of $f$ over the point $\mathfrak{q}$? The answer is given by an exercise in Atiyah-Macdonald (Exercise 4.21). It says that the fiber is given by the following formula:

I2. $f^{-1}(\mathfrak{q})\cong Spec(B\otimes_A \kappa(\mathfrak{q}))$.

Your question: "But what does this scheme actually correspond to? Is there some simplification of this?"

Answer: You must do Exercise 4.21 in AM. Here you prove that this is the scheme theoretic fiber of the map $f$ at $\mathfrak{q}$. The inverse image scheme is the schematic fiber product of $i$ and $f$, and this is given by the tensor product of the rings $\kappa(\mathfrak{q})$ and $B$ over $A$.

In Hartshorne Theorem II.3.3 they prove that the fiber product $X\times_S Y$ of two schemes $X/S,Y/S$ is unique up to unique isomorphism, and the fiber product is constructed for affine schemes using the tensor product. Hence if $S:=Spec(R), X:=Spec(U), Y:=Spec(V)$, it follows there is a canonical isomorphism

I3. $X\times_S Y \cong Spec(U\otimes_R V)$.

Hence the formula  I2

$f^{-1}(\mathfrak{q}):=Spec(B)\times_{Spec(A)} Spec(\kappa(\mathfrak{q}) \cong Spec(B\otimes_A \kappa(\mathfrak{q}))$

calulates the schematic inverse image of the point $\mathfrak{q}$.

More generally: If $\pi: X \rightarrow S$ is any morphism of schemes and if $s\in S$ is any point, let $i:=Spec(\kappa(s)) \rightarrow S$ be the inclusion of the pont $s$ into $S$. The fiber of $\pi$ at $s$ is by definition the fiber product $Spec(\kappa(s))\times_S X$.

Question: "I think I might be able to get the information I need from that answer, so thank you. Can I ask though, is it normal to take this long to grasp the basic foundations of algebraic geometry?"

Answer: If you do exercise 4.21 in AM and understand Theorem II.3.3 in Hartshorne you will get a better understanding of what the fiber product is.
For affine schemes it follows the universal property for the fiber product translates into a universal property for the tensor product of commutative unital rings. The exercise in AM establish a 1-1 correspondence between the prime ideals $\mathfrak{p} \subseteq B$ with $\mathfrak{p}\cap A= \mathfrak{q}$ and the set of prime ideals in $B\otimes_A \kappa(\mathfrak{q})$.


问:
Show that the completion of the Borel $\sigma$-algebra gives the collection of Lebesgue measurable subsets of $\Bbb R$
Show that the completion of the Borel $\sigma$-algebra gives the collection of
Lebesgue measurable subsets of $\Bbb R$ (with respect to Lebesgue measure)

I know that  $\mathcal B \subset \mathcal L$ and $\mathcal L$ is complete implies $\mathcal{\overline {B}} \subset \mathcal L$.
For the reverse inclusion, I chose some element of $\mathcal L$ and wanted to show it could be approximated on the outside by open sets and on the inside by closed sets and thus show it was an element of $\mathcal{\overline {B}}$ but got confused along the way.

答:
I suspect as i read your approach that your definition of completion is the following:

$$\tag{1}\overline{\mathcal{B}}=\{A\subseteq \mathbb{R}:\, \text{there exists}\ F,G\in \mathcal{B}\, \text{ with }\ F\subseteq A\subseteq G\ \text{and}\ \mu(G\setminus F)=0\} $$
Now, consider some $A\in \mathcal{L}$ with $\mu(A)<\infty$. We know, that the Lebesgue measure satisfies some normality conditions, i.e. as you said $A$ can be approximated above by open sets and below from closed sets. More formally, for $\epsilon>0$ there is an open set $G_\epsilon$ and a closed set $F_\epsilon$ such that $F_\epsilon\subseteq A\subseteq G_\epsilon$ and
$$\mu(G_\epsilon)<\mu(A)+\epsilon\\  \mu(F_\epsilon)>\mu(A)-\epsilon$$
Applying this for $\epsilon=1/n$ we may find a sequence of open sets $G_n$ and closed sets $F_n$ such that $F_n\subseteq A\subseteq G_n$ and
$$\tag{*}\mu(G_n)<\mu(A)+1/n,\\
\mu(F_n)>\mu(A)-1/n$$
for all $n$. Let $F=\bigcup_{n=1}^{\infty}F_n$ and $G=\bigcap_{n=1}^{\infty}G_n$. Observe that both $F,G$ belong to $\mathcal{B}$ ( $F$ is an $F_\sigma$ set and $G$ a $G_\delta )$. By $(*)$ we obtain that $\mu(G)\leq \mu(A)$ and $\mu(F)\geq \mu(A)$. Since, $F_n\subseteq A\subseteq G_n$ it follows that $F\subseteq A\subseteq G$. Since $\mu(A)<\infty$ we have that
\begin{align}
\mu(G\setminus F)&=\mu(A\cap G\setminus F)+\mu(A^c\cap G\setminus F)\\
&\leq \mu(A\setminus F)+\mu(G\setminus A)\\
&=\mu(A)-\mu(F)+\mu(G)-\mu(A)=0
\end{align}
From $(1)$ we get that $A\in \overline{\mathcal{B}}$. Now if $\mu(A)=\infty$ consider the sets $A_n=[-n,n]\cap A$. The $A_n's$ have finite measure, hence from the previous step they belong to $\overline{\mathcal{B}}$. Since, $$A=\bigcup_{n=1}^{\infty}A_n$$
it follows that $A\in \overline{\mathcal{B}}$ and the inclusion $\mathcal{L}\subseteq \overline{\mathcal{B}}$ is justified.


问:
interpretation for $f(x)=f(\frac{1}{x})$
Is there any interpretation geometric may be for a function that satisfies

$
f(x)= f(\frac{1}{x})\; \; \forall x \in \mathbb{R}^* $

Is it of kind $f(x)= f(g(x))$ ?

Thank you very much.

答:
Let us think of that property in group-theoretic terms. We have a function $f\colon \mathbb R_{>0}\to \mathbb R$ such that
$$\tag{1}
f(t)=f(t^{-1}), \qquad \forall t >0.$$
The domain $(\mathbb R_{>0}, \cdot)$ is a group, and (1) tells us that $f$ is invariant under the operation $t\mapsto t^{-1}$.

For the more familiar group $(\mathbb R, +)$, the corresponding operation would be $x\mapsto -x$. So, the property analogous to (1) for a function $g\colon \mathbb R\to \mathbb R$ would be
$$\tag{2}
g(x)=g(-x), \qquad \forall x \in \mathbb R,$$
which is a familiar one, as it expresses that $g$ is an even function.

Actually, (1) and (2) are equivalent. Indeed, the change of variable
$$t=e^x$$
is an isomorphism of the groups $(\mathbb R, +)$ and $(\mathbb R_{>0}, \cdot)$. Now, if $f$ satisfies (1), then denoting
$$
g(x):=f(e^x), $$
we see that $g$ satisfies (2). Conversely, if $g$ satisfies (2), then denoting
$$
f(t)=g(\log t) $$
we see that $f$ satisfies (1).

Conclusion. We can say that a function $f$ satisfying the property (1) is even in the group $(\mathbb R_{>0}, \cdot)$.

P.S. I hadn't read the comments before posting. I see that various users, such as Gae.S and Chrystomath, proposed this exact same idea.


问:
How to prove that this specific matrix has the following form when raised to a certain power?
Let $A = \sum{_{k=1}^{n-1}} kE_{k,k+1} $ be an $n \times n$ matrix where $E_{i,i+1}$ has zeroes everywhere except the $i,i+1$:th component where it has a one.

For example, if A was $3 \times 3$ it would be

$A =  \begin{matrix}
0 & 1 & 0\\
0 & 0 & 2\\
0 & 0 & 0
\end{matrix}$

It is clear that $A$ is nilpotent of order $n$, however I claim that $A^{n-1}$ has one entry in the $1, n$:th entry that is n! and zeros everywhere else. How would I go about to show this?

The reason I ask is that I need this for an induction proof. Thanks

答:
If $\;A=(a_{ij})\;$ , with $\;a_{ij}=0\;\;\forall\,i\neq i+1\;$, then

$$A^2=\left(\sum_{k=1}^n a_{ik}a_{kj}\right)\;,\;\;\text{where}\;\;a_{ik}a_{kj}=\begin{cases}0, &k\neq i+1\;\;or\;\;j\neq k+1\\{}\\
1,&k=i+1,\,j=k+1=i+2\end{cases}$$

Observe carefully the last line above...and there you go. Some small inductive argument and you shall be thru.


问:
An infinite hermitian matrix with each submatrix non singular
Let $A$ be an hermitian infinite matrix with the property that $\forall n \in \mathbb{N}: A_{n \times n}$, the submatrix of order $n$ (rows and columns from $1$ to $n$), is non-singular.

My question: is it true that $A$ is non-singular ?

Thanks.

答:
$A$ need not necessarily have an inverse. Let $$A=\begin{pmatrix}-1&1&1&1&\cdots\\1&1&0&\\1&0&1\\\vdots&&&\ddots\\1&&&&1&\\&&&&&\ddots\end{pmatrix}$$ Then every submatrix $A_{n\times n}$, with rows and columns from $1$ to $n$, is non-singular (with determinant $-n$).

If there were a matrix $B$ such that $BA=I$, then its first row would annihilate all the columns for $i>1$, so it would have to be of the form $a(-1,1,1,\ldots)$, yet multiplying out with the first column should give $1$, which is impossible.


问:
Proof of integral identity
I was looking for a proof of this identity:
\begin{equation}
\int_{\mathbb{R}^n}\phi \Delta\phi = - \int_{\mathbb{R}^n}|\nabla\phi|^2
\end{equation}
given $\phi \in C_0^{\infty}(\mathbb{R}^n)$.

答: The difference is a surface term,$$\int_{\Bbb R^n}(\phi\nabla^2\phi+\nabla\phi\cdot\nabla\phi)d^nx=\int_{\Bbb R^n}\nabla\cdot(\phi\nabla\phi)d^nx.$$

Title: Proof of integral identity


问:
Checking $\int_{0}^{\infty} {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}} dx}$ for convergence

Given $\displaystyle\int_{0}^{\infty} {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}}  dx}$, prove that it converges.

So of course, I said:

We have to calculate $\displaystyle \lim_{b \to \infty} {\int_{0}^{\infty} {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}}  dx}}$. And in order to do that we have to calculate the integral $\displaystyle \int {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}}  dx}$.

I got stuck calculating this integral, if anyone can give me a direction with this question I'll appreciate it!

答:
You're not meant to calculate the integral, you're meant to check it converges.

Since the integrand is finite and continuous everywhere, the only possible thing which can go wrong is that it does not converge at infinity.

The behaviour at large $x$ is asymptotically $\sin^2 x^{-7/3}$. Now this is bounded by $x^{-7/3}$ which has a finite integral $\propto x^{-4/3}$ which converges at infinity; the result is that the integral does indeed converge.


问:
Minkowski Theory: The embedding of fractional ideal is a full rank lattice.
I encountered this question when studying the proof towards Minkowski bound. Let $K/\mathbb{Q}$ be a number field of degree $n$. Let $r$ and $s$ be the number of real and complex embedding of $K$ respectively, then $n=r+2s$. Let $\sigma:K\to \mathbb{R}^r\times\mathbb{C}^s\cong \mathbb{R}^n$ be the usual embedding. The literature suggests that for any fractional ideal $I$ of $\mathcal{O}_K$, $\sigma(I)$ is a full rank lattice, i.e. a lattice of rank $n$. I have difficulty proving this fact.

I knew that $\mathcal{O}_K$ is a free $\mathbb{Z}$-module of rank $n$. Suppose $\{v_1.\dots,v_n\}$ is a $\mathbb{Z}$-basis of $\mathcal{O}_K$ and $I$ a non-zero ideal, then for a non-zero $a\in I$, $\{av_i\}\subset I$ is $\mathbb{Z}$-linearly independent. Furthermore, $\sigma(I)$ is discrete.

答:
$I$ contains $a O_K$ and $a O_K$ has finite index in $I$ so you already have the answer : discrete and full rank subgroup.

Otherwise

$K=  \Bbb{Q}[x]/(f)$, factorize $f=\prod_j f_j \in \Bbb{R}[x]$ so that $r_1=\# \{ f_j,\deg(f_j)=1\},r_2=\# \{ f_j,\deg(f_j)=2\}$.

Then
$$K\otimes_\Bbb{Q}\Bbb{R} = \Bbb{R}[x]/(f)\cong \prod_j \Bbb{R}[x]/(f_j) = \Bbb{R}^{r_1}\Bbb{C}^{r_2}$$
The isomorphism is given by the real/complex embeddings.

On the other hand $O_K= O_K\otimes_\Bbb{Z}\Bbb{Z}$ is a lattice in $K\otimes_\Bbb{R}\Bbb{R}$.

For a fractional ideal $I$ then $m I$ is a non-zero ideal for some integer, it has finite index $N(mI)$ in $O_K$ so $mI$ and $I$ are lattices too.


问:
Prove that $1+ \frac{5}{4}x ≤(1+x)^{\frac{5}{4}}≤1+ \frac{5}{4} x+\frac{5}{16} x^2$
How to prove this inequality:
$$1+ \frac54x \le (1+x)^{5/4} \le 1+ \frac54 x+\frac5{16} x^2$$

I tried applying binomial theorem to get this inequality, but I am not able to get it precisely. Please help.

Body:
Note
$$ (1+x)^{5/4}-(1+ \frac54x)=\frac54  \int_0^x (\sqrt[4]{1+t}-1)dt \ge 0
$$

and
\begin{align}
(1+ \frac54 x+\frac5{16} x^2 )- (1+x)^{5/4}
&\ge (1+x)[(1+ \frac14x )- (1+x)^{1/4}] \\
& =\frac14(1+x) \int_0^x (1- \frac1{(1+t)^{3/4} })dt \ge 0
\end{align}

Thus

$$1+ \frac54x \le (1+x)^{5/4} \le 1+ \frac54 x+\frac5{16} x^2$$


问:
Expressing $\sin(2x)$ as a polynomial of $\sin{x}$
Using trignometric identities ( double angle forumlas) one can see that $\cos{2x} = 2 \: \cos^{2}{x} - 1$ can be expressed as a polynomial of $\cos{x}$, where $p(\cos{x})=2 \: \cos^{2}{x}-1$. Then its natural to ask the same question for the sine function.

Can we express $\sin{2x}$ as a polynomial of $\sin{x}$?

答:
Just for the sake of completeness, here is a proof that you cannot express it that way.

(For a general proof, see the bottom of this answer).

Suppose we could express $\sin 2x$ as a polynomial in $\sin x$.

i.e.

$\sin x \cos x  = P(\sin x)$ for some polynomial $P$.

Since $\sin 0 = 0$, $x$ is a factor of $P(x)$. Hence we get

$\cos x = Q(\sin x)$ for a polynomial $Q$.

Put $x = \frac{\pi}{2}$ and $x=\frac{3\pi}{2}$ we see that $1,-1$ are roots of $Q(x)$.

Thus we see that for 0 < t < 1,

$\sqrt{1-t^2} = (t-1)(t+1)H(t)$ (because $Q(t) = (t-1)(t+1)H(t)$ for some polynomial $H$)

By squaring we see that the degree of the left hand side polynomial is $2$ while the degree of the right hand side polynomial is $>2$. This is not possible, as then their difference, which is not identically zero will have infinite roots.

Thus it is not possible to express $\sin 2x$ as a polynomial in $\sin x$.

General Case:

Here is a proof that in general $ \sin (2nx)$ cannot be written as a polynomial in $ \sin x$ based on the identity provided by J.M:

$$\frac{\sin(2nx)}{2\cos(x)}=(-1)^{\lfloor n-1/2\rfloor}\sum_{j=0}^{\lfloor n-1/2\rfloor}{(-1)^j \sin((2j+1)x)}$$

First notice that $\sin ((2m+1)x)$ can be written as a polynomial in $\sin x$ with integer coefficients (follows easily from the DeMoivre's identity).

Now suppose $ \sin (2nx)$ could be written as a polynomial $P$. Consider the $r^{th}$ derivative of that polynomial at $0$. This must be a rational number, as the $r^{th}$ derivative of $ \sin (2nx)$ at $0$ is rational. Thus the coefficients of $P$ must be rational.

Thus using the above identity we must have that $\cos x$ is of the form $\frac{P(\sin x)}{Q(\sin x)}$ where $P$ and $Q$ are polynomials with rational coefficients. Thus we must have that

$$\sqrt{1-t^2} = \frac{P(t)}{Q(t)}$$

$$ 0 < t < 1$$

Putting $t=\frac{1}{2}$ gives us a rational number on the RHS while an irrational number on the LHS.

Hence it is not possible to represent $ \sin (2nx)$ as a polynomial in $ \sin x$.

ContentLicense: CC BY-SA 2.5


问:
Any commutative associative operation can be extended to a function on nonempty finite sets
This is a fact we use very frequently in general mathematics when we write such notations as $1+2+3+4$: since we know that $+$ is commutative and associative, we can just "drop the parentheses" and not worry about order of operations. Of course I believe this, but how does one prove this in full generality? Even stating it is giving me trouble. Here's my attempt:

Assume an operation $\oplus:S\times S\to S$ is provided satisfying $x\oplus y=y\oplus x$ and $x\oplus(y\oplus z)=(x\oplus y)\oplus z$ for all $x,y,z\in S$.

Claim: Given any finite set $\emptyset\subset A\subseteq S$, there exists a unique $z\in S$ such that for any function $f:{\cal P}(A)\to{\cal P}(A)$ which satisfies $\emptyset \subset f(B)\subset B$ for all $B\subseteq A$ with $|B|\ge 2$ and any function $g:{\cal P}(A)\to S$ which satisfies $g(\{x\})=x$ for all $x\in S$ and $g(B)=g(f(B))\oplus g(B-f(B))$ for all $|B|\ge 2$, $g(A)=z$.

The operation $\oplus$ does not necessarily have an identity element, so we do not attempt to define an empty sum. Intuitively, this element $z$ represents the finite sum of the elements in $A$, so if $A=\{1,2,3\}$ and $f(\{1,2,3\})=\{1\}$ and $f(\{2,3\})=\{3\}$, then
$$z=g(\{1,2,3\})=g(\{1\})\oplus g(\{2,3\})=g(\{1\})\oplus (g(\{3\})\oplus g(\{2\}))=1\oplus(3\oplus 2).$$
There has got to be a better way to say that, but this is the only way I can think of to capture all the possibilities of parenthesization, and still be amenable to a formal proof. And now that I've stated it, how should I prove it? I suppose I should induct on something, but I've no idea what.

Edit: The goal here is to be able to define an operation $F$ such that $F(\{x_1,\dots,x_n\})=x_1\oplus\cdots\oplus x_n$ and be assured that the operation is well defined and satisfies $F(A\cup B)=F(A)\oplus F(B)$, when $A$ and $B$ are disjoint finite nonempty subsets of $S$.

答:
It turns out that the claim as originally stated cannot be proven by induction directly, since later steps may introduce repeated elements, which are not accounted for in the claim. Instead, we shall prove it for sequences and weaken it at the end.

Claim: Let $[n]=\{0,1,\dots,n-1\}$. Given any finite sequence $A=\langle x_0,x_2,\dots,x_{n-1}\rangle$ with $n\ge1$ and each $x_i\in S$, there exists a unique $z\in S$ (denoted $F(A)$) such that for any function $f:{\cal P}[n]\to{\cal P}[n]$ which satisfies $\emptyset \subset f(B)\subset B$ for all $B\subseteq [n]$ with $|B|\ge 2$ and any function $g:{\cal P}[n]\to S$ which satisfies $g(\{i\})=x_i$ for all $i\in [n]$ and $g(B)=g(f(B))\oplus g(B-f(B))$ for all $|B|\ge 2$, $g[n]=z=F(A)$.

Let's try to prove this by induction on $n$. In the base case, $A=\langle x_0\rangle$ for some $x_0\in S$. Then $g(\{0\})=g(\{x\})=x_0$ is unique, so we are done.

In the induction step, we assume that for all $|B|\le n$, there is a unique $F(B)$ as stated in the claim and we consider $|A|=n+1$. Then we claim that there exists some two-element set that is evaluated by $g$. (This step takes some justification; we were not precise about the "relevant" part of the domain of $g$, so this should say: $\exists k\in{\Bbb N}\ \exists h:[k]\to {\cal P}(n)$ such that $h(0)=[n+1]$ and $|h(k-1)|=2$ and $\forall m<k-1\,\big[h(m+1)=f(h(m))\vee h(m+1)=h(m)-f(h(m))\big]$. This is our way of specifying a "path" through the "binary tree" prescribed by $f$.) Suppose $h(k-1)=\{i,j\}$ is this two-element set. Then $g(\{i,j\})=x_i\oplus x_j$ or $g(\{i,j\})=x_j\oplus x_i$, and we know these are equal.

Now consider $B=\langle y_0,\dots,y_{n-1}\rangle=A-\langle x_i,x_j\rangle+\langle x_i\oplus x_j\rangle$, where we remove $x_i$ and $x_j$ from the sequence and append $x_i\oplus x_j$ to the end. There is an index mapping $\varphi$ which takes each element of $A$ to the corresponding element of $B$, that is, $y_{\varphi(m)}=x_m$, except that $y_{\varphi(i)}=y_{\varphi(j)}=x_i\oplus x_j$. If we define a function $f':{\cal P}[n]\to{\cal P}[n]$ by $f'(\varphi(X))=\varphi(f(X))$, it will satisfy the induction requirements (and a $g$ satisfying the requirements given $B$ and $f'$ always exists), so we can apply the induction hypothesis to $B$, to find that we can rearrange the sequence to
$$(\,((y_0\oplus y_1)\oplus y_2)\cdots\oplus y_{n-2})\oplus(x_i\oplus x_j)=((\,((y_0\oplus y_1)\oplus y_2)\cdots\oplus y_{n-2})\oplus x_i)\oplus x_j$$

by associativity. We're nearly done; we just need to put $x_{n+1}$ at the end instead of $x_j$. If $j=n+1$, great, otherwise, swap $x_{n+1}$ with $x_i$ in the left argument (which is of length $n$ and thus can be rearranged by hypothesis) and then swap $x_{n+1}$ with $x_j$ by regular associativity and commutativity. The left argument is now $\langle x_0,x_1,\dots,x_n\rangle$ in some order, so it adds to something unique by hypothesis; thus $F(A)=F(A-\langle x_{n+1}\rangle)\oplus x_{n+1}$ is independent of $f$ and $g$ and thus is unique.

There's my proof; as you can see I got tired of the $f$ and $g$ interpretation halfway through, but hopefully the attentive reader can fill in the details. Unfortunately, that's supposed to be me, so we'll see how the real proof goes.


问:
The limit of a series
I need help with this exercise!

$\lim\limits_{n \to \infty} \dfrac{1^4+2^4+...+n^4}{n^5}.$

I saw somewhere online that $1^4+2^4+...+n^4=\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$.

But I dont understand why!

So following that then,

$\lim\limits_{n \to \infty} \dfrac{1^4+2^4+...+n^4}{n^5}=$
$=\lim\limits_{n \to \infty} \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5}=$
$=\lim\limits_{n \to \infty} \dfrac{(n+1)(2n+1)(3n^2+3n-1)}{30n^4}=\dfrac15.$

And after that what!? Do I multiply and then use L'Hopital?

答:
This is what happens when you rush into learning calculus!

The quantity $S(n) = 1^4 + 2^4 + \cdots + n^4$ can be summed in a variety of ways. One such way is to guess that it grows like a fifth degree polynomial, then use Lagrange interpolation and induction. Another way might be to notice that

$$\sum_{i=0}^{n-1} (i+1)^5 - i^5 = n^5$$
is easy to evaluate. But alternatively,
$$(i+1)^5 - i^5 = 5i^4 + 10i^3 + 10i^2 + 5i + 1.$$
I leave it as an exercise to you on how to derive from here; now to your question on l'Hopital's.

You have $$\lim_{n\rightarrow\infty} \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5}.$$

The numerator is a fifth degree polynomial. The denominator is a fifth degree polynomial. So the limit will just be the ratio of the leading coefficients, $6/30 = 1/5.$ Why? Well, imagine we have a simpler case,
$$\lim_{n\rightarrow\infty} \frac{n^2 + 3n + 1}{n^2}.$$
Then
$$\frac{n^2+3n+1}{n^2} = 1 + \frac{3}{n} + \frac{1}{n^2}.$$
As $n$ goes to infinity, anything with an $n$ in the denominator will go to 0, and you're just left with 1. If you expanded out that fifth degree numerator, you'd quickly find the same trick applies and all but the leading term dies.

Title: The limit of a series


问:
How to fit an exponential curve to data?
I would like to fit a curve of the form $$\frac{1}{Y}=\frac{x^u}{c_1}+\frac{(1-x)^v}{c_2}$$ to a dataset with 7 data-points of (x,Y). How can I do this?
I already know the values of c1 and c2 and need to find out v.
I thought if I had some way to convert it into a linear combination of u,v and their powers, I could apply linear regression. But due to the nature of this research (in chemical engineering), I require an equation that will give me accurate values of Y for given x, even though I have very few data points.
Is there a way out?

答:
You have $n$ data points $(x_i,y_i)$ and you want to fit the model
$$\frac 1 y=\frac{x^u}{c_1}+\frac{(1-x)^v}{c_2}\tag 1$$ In fact, this is not correct since what is measured is $y$ and not $\frac 1 y$. So, even if we work with $(1)$ in a preliminary step, at the end, you will need to work with
$$y=\frac 1{ \frac{x^u}{c_1}+\frac{(1-x)^v}{c_2}}\tag 2$$ $(1)$ is a very highly nonlinear model and you need good estimates for  starting the full nonlinear regression.

What I would suggest is to fix one of the parameters $u$ or $v$ to an arbitrary value. Say that we fix $x$. So the model write
$$\frac 1 y-\frac{x^u}{c_1}=\frac{(1-x)^v}{c_2}\tag 3$$ that is to say
$$c_2\left(\frac 1 y-\frac{x^u}{c_1}\right)=(1-x)^v$$
$$\log\left( c_2\left(\frac 1 y-\frac{x^u}{c_1}\right) \right)=v\log(1-x)$$ which is more than simple (a linear regression with no intercept. So, for this given $u$, you know $v$; recompute the $\hat y_i$ and
$$\text{SSQ}(v)=\sum_{i=1}^n \left(\hat y_i -y_i\right)^2$$ Try a few values of $u$ and plot them until you see a minimum. Now you are ready.


问:
Range of a function
Find the range of

$$f(x)=\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}$$ where $a, b, c$ are distinct real numbers such $a\neq b\neq c\neq a$.

答:
This is an identity.Range of f(x) is 1.

This is because f(x)=1 is a quadratic equation.However f(a),f(b) and f(c)=1.

Since a quadratic equation has at most 2 distinct roots f(x)=1 must be an identity (i.e valid for all values of x).

Title: Range of a function


答:
Your condition should be $\sum_{i=1}^n \|u_i\|^2 < 1$. Assume
$$\sum_{i=1}^n \alpha_i(v_i+u_i) = 0$$
for some scalars $\alpha_i$ and consider the scalar product $\langle \cdot, v_j\rangle$. We get
$$0 = \sum_{i=1}^n \alpha_i \langle v_i+u_i, v_j\rangle = \alpha_j + \sum_{i=1}^n \alpha_i \langle u_i, v_j\rangle$$
so using Cauchy-Schwarz follows
$$|\alpha_j|^2 = \left|\sum_{i=1}^n \alpha_i \langle u_i, v_j\rangle\right|^2 \le \left(\sum_{i=1}^n |\alpha_i|^2\right)\left(\sum_{i=1}^n \left|\langle u_i,v_j\rangle\right|^2\right).$$
Summing this up over $j=1, \ldots, n$ and using Parseval we get
$$\sum_{j=1}^n |\alpha_j|^2 \le \left(\sum_{i=1}^n |\alpha_i|^2\right)\left(\sum_{i,j=1}^n \left|\langle u_i,v_j\rangle\right|^2\right) = \left(\sum_{i=1}^n |\alpha_i|^2\right)\underbrace{\left(\sum_{i=1}^n \|u_i\|^2\right)}_{<1}$$
which implies $\sum_{i=1}^n |\alpha_i|^2 = 0$ so all $\alpha_1 = \cdots = \alpha_n=0$.

Title: Prove that $\{v_1+u_1,\cdots,v_n+u_n\}$ form a basis if $\{v_1,\cdots,v_n\}$ is an orthonormal basis and $\sum_{j=1}^n||u_j||^2<1$

Body:
Let $V$ be a finite dimensional inner product space with $\{v_1,\cdots,v_n\}$ as orthonormal basis vectors. Let $\{u_1,\cdots,u_n\}\in V$ be vectors such that $\sum_{j=1}^n||u_j||^2<1$, the prove that $\{v_1+u_1,\cdots,v_n+u_n\}$ form a basis for $V$, where $||u_j||$ is the norm of the vector $u_j$

I don't even see a clue of how the property $\sum_{j=1}^n||u_j||^2<1$ makes it a basis ?


问:
If $Imf(z)\cdot Im(z)>0, z\in U\setminus\mathbb{R}\Rightarrow f'(z)\neq 0$
Let $f$ be a nonconstant analytic function in a neighborhood $U$ of the real axis $\mathbb{R}$. Assume that $\Im(f(z))\Im(z)>0, \forall z\in U\setminus\mathbb{R}.$ Then I want to prove that $f'(z)\neq 0, \forall z\in \mathbb{R}$.
Please tell me how should I proceed?

答:
Let me put an answer: first obviously the hypothesis needs to state $\Im(f(z))\Im(z)>0, \forall z\in U-\mathbb R$

Next assume there is $y \in \mathbb R, \Im f(y) \ne 0$ then if say $\Im f(y)=c>0$ then $\Im f(z) >c/2$ for $z$ near $y$ and that is a contradiction for $z=y-i\epsilon, \epsilon>0$; similarly we deal with $\Im f(y)=-c<0$ taking $z=y+i\epsilon$, so $f(\mathbb R) \subset {\mathbb R}$

Now since for $x \in \mathbb R$ one can compute all the derivatives of $f$ by considering the restriction of $f$ to $\mathbb R$, it follows by induction that $f^{(k)}(\mathbb R) \subset \mathbb (R)$ for any $k \ge 0$.

But if $f'(y)=0$ let $n \ge 2$ the first index for which $f^{(n)}(y)=b \ne 0, b \in \mathbb R$ from the above ($f$ is not constant by hypothesis), while $f(y)=a \in \mathbb R$

Writing the Taylor series of $f$ near $y$ we get $f(z)=a+b(z-y)^n+O(z-y)^{n+1}$. Since $n \ge 2$ there are roots of order $n$ of $i$ with imaginary parts strictly positive and negative respectively, so if say $b>0$ we can take $z=y+\omega \epsilon, \omega^n=i, \Im \omega <0$ and get a contradiction as $\epsilon>0$ small enough since $\Im f(z)=b\epsilon^n+O(\epsilon^{n+1})>0, \Im z<0$; for $b<0$ we use $\omega^n=i, \Im \omega>0$ hence we are done!


问:
Is $ f(x)=x^2 \sin x $ Uniformly Continuous on $\mathbb{R}$?
Hello and good afternoon. I've tried to smash $f$ on 2 other functions and prove that they are u.c. on $\mathbb{R}$ , but $x^2$ isn't, & I have seen that it is not necessary that the product of 2 functions be uniformly continuous in $\mathbb{R}$. It will help me to find the derivative and follow the theory of uniform continuity ( if it is u.c. on $\mathbb{R}$ ). And if it isn't, is it good to prove that it is not by the method of inopportune induction?

答:
Let $f(x)=x^{2}\sin(x)$ in $\mathbb{R}$. Assume that $f(x)$ is uniformly continuous in $\mathbb{R}$ and let $\varepsilon=1$, then there exists $\delta>0$ such that for $x,y\in\mathbb{R}$ with $|x-y|<\delta$ we have $|f(x)-f(y)|<1$. Let $y=x+\frac{\delta}{2}$ and $g(x)=x^{2}$. Thus, this means for $x\in\mathbb{R}$, we have :
\begin{align*}
\left|x^{2}\sin(x)-y^{2}\sin(y)\right|\leq&|x^{2}-y^{2}| \\
\implies &\left|g(x)-g\left(x+\frac{\delta}{2}\right)\right|<1 \\
\implies & \left|-\delta x -\frac{\delta^{2}}{4}\right|<1 \\
\implies & \delta x +\frac{\delta^{2}}{4}<1 \\
\implies &\text{$\forall x\in\mathbb{R}$ we have $x<\frac{1-\frac{\delta^{2}}{4}}{\delta}\implies$ Contradiction}
\end{align*}


问:
Angle between a surface and a curve.
Find the angle between the surface: $z=x^2+2y^2$ and a curve $$x=t-1, y=-\frac2t, z=t^2-1,$$
at the point $(1,-1,3)$.

My attempt:
I am first eliminating t from the given equation of curve. It is easy to see that
$$z=(t-1)(t+1)=x(t+1)=x(1-\frac2y)\implies xy-yz-2x-0.$$
Let us write $$\Phi_1(x, y, z)=x^2+2y^2-z=0, \ \Phi_2(x, y, z)=xy-yz-2x=0.$$
Then angle between them is given by $$\cos\theta =\frac{\nabla \Phi _{1}\cdot\nabla \Phi _{2}}{\left | \nabla \Phi _{1} \right |\cdot \left | \nabla \Phi _{2} \right |}$$
Please evaluate it and explain whether I am doing it right or not?

答:
I think $\Phi_2$ is no longer a curve but a surface, so I think it may probably not work (not sure). I think there is in fact no need to convert the curve into your $\Phi_2$.

To get the angle between the surface and the curve means to get the angle between the tangent plane of the surface and the tangent line of the the curve. The tangent line of the curve is just its derivative: $(1,\frac{2}{t^2},2t)$, so $(1,\frac{1}{2},4)$ at $t=2$. The tangent plane of the surface is harder to calculate but we can use the normal vector of the surface at $(1,-1,3)$ instead, which is $(2x,4y,-1)$ or $(2,-4,-1)$ at $(1,-1,3)$.

Hence what is left to do is to calculate the angle between $(1,\frac{1}{2},4)$ and $(2,-4,-1)$ and consider its complement angle.

Title: Angle between a surface and a curve.


答: Note $\sum_{k=1}^nk^4$ must be divisible by $n(n+1)$, so that the obvious extension to negative $n$ achieves $0-0^4=0$ at $n=-1$. The large-$n$ behaviour is asymptotic to $\int_0^nx^4dx=\tfrac15n^5$. We can determine coefficients in $\sum_{k=1}^nk^4=\tfrac15n(n+1)(n^3+An^2+Bn+C)$ from the sum at three values of $n$, say $n\in\{1,\,2,\,3\}$, thereby acquiring simultaneous equations. This gives $A=\tfrac32,\,B=\tfrac16,\,C=-\tfrac16$, after which the rational root theorem finds the factor $n+\tfrac12$.

Title: The limit of a series

Body:
I need help with this exercise!

$\lim\limits_{n \to \infty} \dfrac{1^4+2^4+...+n^4}{n^5}.$

I saw somewhere online that $1^4+2^4+...+n^4=\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$.

But I dont understand why!

So following that then,

$\lim\limits_{n \to \infty} \dfrac{1^4+2^4+...+n^4}{n^5}=$
$=\lim\limits_{n \to \infty} \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5}=$
$=\lim\limits_{n \to \infty} \dfrac{(n+1)(2n+1)(3n^2+3n-1)}{30n^4}=\dfrac15.$

And after that what!? Do I multiply and then use L'Hopital?


问:
Prove that/Explain how for independent random variables $X_i$, we have $f_i(X_i)$ are independent (in particular without measure theory)
I have seen a lot of posts that describe the case for just 2 random variables.

Independent random variables and function of them

Are functions of independent variables also independent?

If $X$ and $Y$ are independent then $f(X)$ and $g(Y)$ are also independent.

If $X$ and $Y$ are independent. How about $X^2$ and $Y$? And how about $f(X)$ and $g(Y)$?

Are squares of independent random variables independent?

Prove that if $X$ and $Y$ are independent, then $h(X)$ and $g(Y)$ are independent in BASIC probability -- can we use double integration? (oh I actually asked the 2 variable elementary case here, but there's no answer)

I have yet to see a post that describes the case for at least 3.

Please answer in 2 situations

1 - for advanced probability theory:

Let $X_i: \Omega \to \mathbb R$ be independent random variables in $(\Omega, \mathscr F, \mathbb P)$. Let $i \in I$ for any index set I think (or maybe has to be countable). Of course, assume $card(I) \ge 3$. Then show $f_i(X_i)$ are independent. Give conditions on $f_i$ such that $f_i(X_i)$ is independent. I read in above posts that the condition is 'measurable', which I guess means $\mathscr F$- measurable, but I could've sworn that I read before that the condition is supposed to be 'bounded and Borel-measurable', as in bounded and $\mathscr B(\mathbb R)$-measurable for $(\mathbb R, \mathscr B(\mathbb R), Lebesgue)$

2 - for elementary probability theory

Let $X_i: \Omega \to \mathbb R$ be independent random variables that have pdf's. Use the elementary probability definition of independence that is 'independent if the joint pdf splits up', or something. I guess the index set $I$ need not be finite, in which case I think the definition is that the joint pdf of any finite subset of is independent. Give conditions on $f_i$ such that $f_i(X_i)$ is independent. Of course we can't exactly say that $f_i$ is 'measurable'.

Context for the elementary case: I'm trying to justify the computation for the formula for the  moment-generating function for linear combination of independent random variables. See here: Proving inequality of probabilty to derive upper bound for moment-generating functions

Based on the application of Riemann–Stieltjes integral (or Lebesgue–Stieltjes integral) to probability, I think the condition is any $f_i$ such that $E[f_i(X_i)]$ exists (i.e. $E[|f_i(X_i)|]$ is finite).

This is the same condition in Larsen and Marx - Introduction to Mathematical Statistics and Its Applications.

I think $f$ bounded implies this but not conversely.

Update: Also related through another question If $g$ is a continuous and increasing function of $x$, prove that $g(X)$ is a random variable. --> More generally for what functions $g$ is $g(X)$ is a random variable? Of course in advanced probability just say $g$ is Borel-measurable or $\mathscr F$-measurable or whatever, but I think in elementary probability we say $g$ such that $E[g(X)]$ exists i.e. $E[|g(X)|] < \infty$, EVEN THOUGH this is, I believe, a stronger condition than that $g$ is 'measurable', whatever this means in elementary probability. But then again this is kind of weird since we don't even necessarily expect $E[X]$ to exist (i.e. $E[|X|] < \infty$) or well any higher moment $E[X^n]$ I guess.

答:
For $i\in I$ let $\sigma\left(X_{i}\right)\subseteq\mathscr{F}$ denote the $\sigma$-algebra
generated by random variable $X_{i}:\Omega\to\mathbb{R}$.

Then actually we have $\sigma\left(X_{i}\right)=X_{i}^{-1}\left(\mathscr{B}\left(\mathbb{R}\right)\right)=\left\{ X_{i}^{-1}\left(B\right)\mid B\in\mathscr{B}\left(\mathbb{R}\right)\right\} $.

The collection $(X_i)_{i\in I}$ of random variables is independent iff:

For every finite $J\subseteq I$ and every collection $\left\{ A_{i}\mid i\in J\right\} $
satisfying $\forall i\in J\left[A_{i}\in\sigma\left(X_{i}\right)\right]$ we have:

$$P\left(\bigcap_{i\in J}A_{i}\right)=\prod_{i\in J}P\left(A_{i}\right)\tag {1}$$

Now if $f_{i}:\mathbb{R}\to Y_{i}$ for $i\in I$ where $\left(Y_{i},\mathcal{A}_{i}\right)$
denotes a measurable space and where every $f_{i}$ is Borel-measurable
in the sense that $f_{i}^{-1}\left(\mathcal{A}_{i}\right)\subseteq\mathscr{B}\left(\mathbb{R}\right)$
then for checking independence we must look at the $\sigma$-algebras
$\sigma\left(f_{i}\left(X_{i}\right)\right)$.

But evidently: $$\sigma\left(f_{i}\left(X_{i}\right)\right)=\left(f_{i}\circ X_{i}\right)^{-1}\left(\mathcal{A}_{i}\right)=X_{i}^{-1}\left(f_{i}^{-1}\left(\mathcal{A}_{i}\right)\right)\subseteq X_{i}^{-1}\left(\mathscr{B}\left(\mathbb{R}\right)\right)=\sigma\left(X_{i}\right)$$
So if $\left(1\right)$ is satisfied for the $\sigma\left(X_{i}\right)$
then automatically it is satisfied for the smaller $\sigma\left(f_{i}\left(X_{i}\right)\right)$.

2)

The concept independence of random variables has impact on PDF's and calculation of moments, but its definition stands completely loose from it. Based on e.g. a split up of PDF's it can be deduced that there is independence but things like that must not be promoted to the status of "definition of independence". In situations like that we can at most say that it is a sufficient (not necessary) condition for independence. If we wonder: "what is needed for the $f_i(X_i)$ to be independent?" then we must focus on the definition of independence (not sufficient conditions). Doing so we find that measurability of the $f_i$ is enough whenever the $X_i$ are independent already.

BCLC edit: (let drhab edit this part further): There's no 'measurable' in elementary probability, so we just say 'suitable' or 'well-behaved' in that whatever functions that students of elementary probability will encounter, we hope that they are suitable. Probably, some textbooks will use weaker conditions than 'measurable' that will be used as the definition of independence for that book.

Edit: Functions that are not measurable (or not suitable, if you like) are in usual context very rare. The axiom of choice is needed to prove the existence of such functions. In that sense you could say that constructible functions (no arbitrary choice function is needed) are suitable.


问:
How to find probability using Bayes' theorem?
A company has four production sections viz., S1 , S2 , S3 and S4 which contribute 30% , 20% , 28% and 22% of the total output. It was observed that those sections respectively produced 1% , 2% , 3% and 4% defective units. If a unit is selected at random and found to be defective, what is the probability that the unit so selected has come from either S1 or S4.</p>

答:
Hint: We know

$P(S_1)=0.30$, $P(S_2)=0.20$, $P(S_3)=0.28$, $P(S_4)=0.22$. If $D$ is the defect case than we know also $P(D|S_1)=0.01$, $P(D|S_2)=0.02$, $P(D|S_3)=0.03$,and $P(D|S_4)=0.04$.

Now, the question is asking you to determine $P(S_1|D) + P(S_4|D)$. Can you determine these probabilities with the Bayes' theorem?


问:
For non-standard inner product, express the adjoint operator with a matrix and a metric matrix.
Exercise: Let $V = R^n$ with a non-standard inner product and let L be the multiplication by a matrix A. Find the matrix of $L^A$ (where $L^A$ is the adjoint operator) in terms of the matrix A and the metric matrix G.

My attempt: $ = A^AGL(B) = (L(B)^AG^AA)^A$.

Now $A^AGL(B) = (L(B)^AG^AA)^A$ if and only if

$(A^AGL(B))^A = L(B)^AG^AA$ if and only if

$L(B)^A = (A^AGL(B))^A(G^AA)^-1$.

The problem is that on both sides of the equation L(B) is present, whereas I only want it on the left hand side. I actually tried to rewrite it such that $=<L(A)^A|B>$, but did not succeed. Any help, whether it is answers or hints, are appreciated.

答:
Let $\cdot^G$ denote the adjoint w.r.t. your metric $G$ and let $\cdot^*$ denote the adjoint w.r.t. standard inner product. We have
$$\langle x, A^*G^*y\rangle = \langle GAx,y\rangle = \langle Ax,y\rangle_G = \langle x,A^Gy\rangle_G = \langle Gx,A^Gy\rangle = \langle x, G^*A^Gy\rangle$$
for all $x,y \in \Bbb{R}^n$ so
$$G^*A^G = A^*G^* \implies A^G = (G^*)^{-1}A^*G^* = (G^{-1})^*A^*G^*.$$
Note that $G$ is indeed invertible since $\langle \cdot, \cdot\rangle_G$ is an inner product.


问:
Sequence of functions in $L^1(\Omega)$ that converge in the $L^1$ norm to $f$.
Assume $(f_n)_n$ is a sequence of functions with values in $L^1(\Omega)$ that converges in the $L^1$ norm to $f$, that is $(f_n)_n \in L^1(\Omega)^{\mathbb{N}}$, $f_n \rightarrow f$ in $L^1(\Omega)$ with $\Omega$ with finite measure. Then we can easily show that the limit is in $L^1(\Omega)$ in this way (norms are the $L^1(\Omega)$-norm):

$$|| f || = ||f + f_n - f_n|| \le ||f_n - f|| + ||f_n|| $$

now taking the limit as $n \rightarrow \infty$  we get $$ || f || \le lim_{n \rightarrow \infty} {||f_n||} $$
and since $||f_n|| < \infty$ for every $n$, we get our conclusion that $f \in L^1(\Omega)$.

My question is if my proof is correct and, mostly, what is the topological conclusion we can draw. The statement above reminds me the definition of closed set for metric spaces. So, what can we infer about the set of $L^1(\Omega)$ functions? That it is a closed subset of which space of functions endowed with the $L^1$ norm topology? Thanks!

答: Yes, your proof is correct, but you should say that $\sup_n \|f_n\| < \infty$ since $(f_n)_n$ is a convergent sequence.


问:
Computing real integrals using the Residue Theorem where singularities are on the real line
How would you compute, for $a>0$ the integral $$\int_0^\infty \frac{\sin x}{x(x^2 + a^2)} dx \, \, ?$$

I've computed the residues of the function $$f(z) = \frac{e^{iz}}{z(z^2 + a^2)} $$ which I got, for the poles at $z = 0, ia, -ia$ residues of $\frac{1}{a^2}, \frac{-e^{-a}}{2a^2}, \frac{-e^a}{2a^2}$ respectively. But I'm now stuck as they are all on the real line? How could I used the residue formula to compute this?

答:
The poles at $\pm\mathrm ia$ aren't on the real line. To avoid integrating over a pole at the origin, use

$$
\int_0^\infty\frac{\sin x}{x(x^2+a^2)}\mathrm dx=\frac12\Im\int_{-\infty}^\infty\frac{\mathrm e^{\mathrm ix}-1}{x(x^2+a^2)}\mathrm dx\;.
$$


问:
How to derive velocity in polar coordinates
In one of my aerodynamics classes i need to use the following derivation to convert the velocity components u and v to a polar coordinate system:

$$
v_r = u \cos(\theta) + v \sin(\theta) \\
v_t = v \cos(\theta) - u \sin(\theta)
$$

I'm trying to get to this derivation to understand it but I cant figure it out.
Can someone show me how they converted the velocity in Cartesian coordinate's to polar coordinates?

答:
Let us introduce $[x, y]^T = [r\cos\theta, r\sin\theta]^T$. The velocity is given by time derivative of this position vector.

$$
\begin{bmatrix}
\dot{x}\\
\dot{y}\\
\end{bmatrix} = \begin{bmatrix}
\dot{r}\cos\theta -r\sin \theta \dot{\theta}\\
\dot{r}\sin\theta + r\cos \theta \dot{\theta}\\
 \end{bmatrix}
$$

$$
\implies \begin{bmatrix}
\dot{x}\\
\dot{y}\\
\end{bmatrix} =\begin{bmatrix}
\cos\theta & -r\sin \theta\\
\sin\theta &+ r\cos \theta \\
 \end{bmatrix}\begin{bmatrix}\dot{r}\\ \dot{\theta}\end{bmatrix}
$$

Now, invert the matrix on the right hand side and you will arrive at the expression in your question.

$$
\implies \begin{bmatrix}
\dot{r}\\
\dot{\theta}\\
\end{bmatrix} =\dfrac{1}{r}\begin{bmatrix}
r\cos\theta & r\sin \theta\\
-\sin\theta &+ \cos \theta \\
 \end{bmatrix}\begin{bmatrix}\dot{x}\\ \dot{y}\end{bmatrix}
$$

$$
\implies \begin{bmatrix}
\dot{r}\\
\dot{\theta}\\
\end{bmatrix} =\begin{bmatrix}
\cos\theta \dot{x} + \sin \theta \dot{y}\\
-\dfrac{1}{r}\sin\theta \dot{x}+ \dfrac{1}{r}\cos \theta \dot{y} \\
 \end{bmatrix}
$$

$$
\implies \begin{bmatrix}
\dot{r}\\
r\dot{\theta}\\
\end{bmatrix} =\begin{bmatrix}
\cos\theta \dot{x} + \sin \theta \dot{y}\\
-\sin\theta \dot{x}+ \cos \theta \dot{y} \\
 \end{bmatrix}
$$

Note that $v_t=r\dot{\theta}$.


问:
How do I determine all complex solutions of $|z|z=-i(\bar{z})$, for all $z\in \mathbb{C}$

Let $z\in \mathbb{C}$. How do I determine all complex solutions of $|z|z=-i(\bar{z})$?

My approach: We can see that $$|x+yi|(x+yi)=y-ix$$ Then I came up with the real part $$-y^2+x^2+x^4-y^4$$ and the imaginary part $$2ixy+x^2(2iyx)+y^2(2xiy)$$

I don't know how to go on and I think that is wrong anyway.I would appreciate every attempt to help!

答: Beside $z=0$, polar form gives $r^2e^{i\theta}=re^{-i(\theta+\pi/2)}$ so $r^2=r\implies r=1$, and $2\theta+\pi/2=2n\pi$ for $n\in\Bbb Z$, i.e. $\theta=-\pi/4+n\pi$. There are two nonzero solutions, $\pm e^{-\pi i/4}$.


答:
Hint:

If $|z|\, z=-i\bar z$, the moduli of both sides are the same, so $|z|^2=|z|$, which implies that $|z|=0$ (so $z=0$) or $|z|=1$.

For the latter case, the  best is to use the exponential form: $z=\mathrm e^{i\theta}$. The equation can be rewritten, since the modulus is $1$, as
$$\mathrm e^{i\theta}=-i\mathrm e^{-i\theta}=\mathrm e^{-i\bigl(\theta+\tfrac\pi 2\bigr)}.$$
Can you take  it from there?

Title: How do I determine all complex solutions of $|z|z=-i(\bar{z})$, for all $z\in \mathbb{C}$

Body:
Let $z\in \mathbb{C}$. How do I determine all complex solutions of $|z|z=-i(\bar{z})$?

My approach: We can see that $$|x+yi|(x+yi)=y-ix$$ Then I came up with the real part $$-y^2+x^2+x^4-y^4$$ and the imaginary part $$2ixy+x^2(2iyx)+y^2(2xiy)$$

I don't know how to go on and I think that is wrong anyway.I would appreciate every attempt to help!


问:
Finding order of orbit
Let $G$ be a finite group of odd order $n$. Suppose that $G$ acts on a set $X$ and let $Y\subseteq X$ be an orbit. Show that if $|Y| < n$ then $|Y| \le \frac n3$.

I know that if $G$ is a finite group acting on a set $X$, then every orbit is a finite set and its cardinality divides the order $|G|$ of the group, is this what I'm meant to use to solve this question?

I'm just a little unsure and I cant find anything similar in my notes

答:
HINT: As you suggested, by Orbit Stabilizer Theorem, we have

$$|G| = |O_a|\cdot|G_a|,\ \text{for any }a \in X$$

Thus, for an orbit $Y \subseteq X$, we have $|Y|\ \big{|}\ n=|G|$. If $n$ is odd and $|Y| \ne n$, what can we say about $|Y|$? What is the largest integer of the form $\frac{n}{k}$ if $k \ge 2$ with $k \in \mathbb{Z}^+$ when $n$ is odd?

Title: Finding order of orbit


问:
Estimating a expectation value of a function of a random variable and a non-probability variable using Monte Carlo simulation.
I have a function $g(x,Z)$, where $Z$ has a normal distribution, and $x$ is NOT a random variable (I have a range of pre-specified values to evaluate), and I want to generate a plot of the expectation value $E[g(x)]$, or $E[g](x)$.

In order to use Monte Carlo simulation to generate this plot with $n=1000$ draws of $Z$, must I resample $Z$ anew for every value of $x$ that I plan to evaluate? Or can I rely on the same $n=1000$ sample of $Z$ values for every value of $x$?

Thank you!

答:
You can consider that $g(x,Z)$ is a set of trajectories, where each trajectory is the curve associated with the function $g(\cdot, z)$, where $z$ is a sample of $Z$. The expected value $E[g](x)$ is just the mean of all these trajectories when evaluated on $x$. Using this point of view, we can say that we have $n = 1000$ trajectories, and we want to take the average on our values of $x$. In this scenario, we can use the same $n$ simulations to compute the average for each $x$. This has the advantage of being really efficient compared to take $n=1000$ simulations for each value of $x$. However, if our simulations are somehow skewed, all our estimates will be skewed as well, which could be a reason not to use the same simulations for each $x$.

At the end of the day, I think that as long as you have enough simulations, there's no problem in reusing them (for this particular case).


问:
Moment of inertia of hollow cone

Let $S$ be the boundary of the solid cone $x^2 + y^2 < z^2<2$ in the upper half of the $xyz$-space. Assume that $\mathcal S$ has constant density equal to $3$. Determine the moment of inertia with respect to the $z$-axis of the surface $\mathcal S$.

My solution is as follows:

$$I_z = \iint_S 3r^2 dS = \int_0^{2\pi}\int_0^{\sqrt 2} 3 z^2 (\sqrt{2}z)\, \mathrm d z \, \mathrm d \theta=6 \sqrt 2 \pi $$

However on Wikipedia it says the moment of inertia of a hollow cone is $I_z = \frac 1 2 m r^2$ which would give $I_z = 2 \sqrt 2 \pi$

Which answer is right?

答:
Your first answer is the correct one. The surface area of a right cone (minus its base) of radius $r$ and altitude $h$ is $\pi r\sqrt{r^2+h^2}$. In this case, $r=h=\sqrt{2}$ and we get $S = 2\pi\sqrt{2}$, so $m=6\pi\sqrt{2}$ and thus $I_z = 6\pi\sqrt{2}$.

As for what went wrong: You probably found the mass of the corresponding solid cone in your second calculation. If $3$ were volumetric density instead of surface density, the formula $V = \pi r^2h/3$ gives $m=2\pi\sqrt{2}$.

Title: Moment of inertia of hollow cone


问:
Finding the only multiple zero of a quadratic function
I don't know how to figure this problem out. Can someone please help?

For what value of $\;b\;$ will $\;f(x)=x^2 + bx + 400\;$ have $\;-20\;$ as its only zero?

Body:
The number $\;-20\;$ is a zero of the function $\;f(x)=x^2+bx+400\;$ if and only if $\;f(-20)=0\;.$

$(-20)^2+b(-20)+400=0\;,$

$400-20b+400=0\;,$

$-20b=-800\;,$

$20b=800\;,$

$b=\dfrac{800}{20}=40\;.$

Moreover, if $\;b=40\;,\;$ then it results that $\;f(x)=x^2+40x+400=(x+20)^2\;,$

consequently,

$f(x)=(x+20)^2=0\iff x+20=0\iff x=-20\;,$

so $\;-20\;$ is the only zero of $\;f(x)\;.$

Hence, the value of $\;b\;$ such that $\;f(x)=x^2+bx+400\;$ has $\;-20\;$ as its only zero, is

$b=40\;.$

Another way to solve the exercise:

$f(x)=x^2+bx+400\;.$

$x^2+bx+400=0\;,$

and, by applying the quadratic formula’s determinant,

$\Delta=b^2-4ac=b^2-4\cdot1\cdot400=b^2-1600\;.$

$f(x)=x^2+bx+400\;$ has only a zero if and only if $\;\Delta=0\;.$

$b^2-1600=0\;,$

$b^2=1600\;,$

$b=\pm\sqrt{1600}=\pm40\;.$

If $\;b=-40\;,\;x=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{40\pm0}{2}=20\;.$

If $\;b=40\;,\;x=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-40\pm0}{2}=-20\;.$

Hence, the value of $\;b\;$ such that $\;f(x)=x^2+bx+400\;$ has $\;-20\;$ as its only zero, is

$b=40\;.$


答:
It seems that you can use a path with goes from $-M$ to $-\frac{1}{M}$ along the real line (path $\gamma_1$), then along a circle with radius $\frac{1}{M}$ around $0$ to $\frac{1}{M}$ (path $\gamma_2$), continues along the real line to $M$ (path $\gamma_3$) and finally goes back to $-M$ (path $\gamma_4$) along a circle of radius $M$ around $0$.

If you can for example find a bound for $f$ along the small circle $\gamma_2$ independent of $M$ (for sufficiently large $M$ suffices of course), then you know that the integral along $\gamma_2$ goes to zero as $M \to \infty$. Since $\frac{sin(x)}{x} \to 1$ as $x \to 0$, such a bound should exist.

That leaves you with having to shows that the integral along $\gamma_4$ goes to zero too, e.g by showing that $f$ is bounded by $\frac{C}{|z|^2}$ for some $C$ (for sufficiently large $|z|$, of course). If that is the case, the integral is bounded by $\frac{C}{M^2}\pi M$, since $\pi M$ is the length of the arc with radius $M$ around 0, and  the integral thus goes to 0 as $M \to \infty$. The fact that the denominator is a polynomial in $x$ of degree 3 should provide you with such a bound.

Title: Computing real integrals using the Residue Theorem where singularities are on the real line

Body:
How would you compute, for $a>0$ the integral $$\int_0^\infty \frac{\sin x}{x(x^2 + a^2)} dx \, \, ?$$

I've computed the residues of the function $$f(z) = \frac{e^{iz}}{z(z^2 + a^2)} $$ which I got, for the poles at $z = 0, ia, -ia$ residues of $\frac{1}{a^2}, \frac{-e^{-a}}{2a^2}, \frac{-e^a}{2a^2}$ respectively. But I'm now stuck as they are all on the real line? How could I used the residue formula to compute this?


问:
Prove $\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{3n}}$ for all $n$.
Prove for all $n$: $\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{3n}}$.

Using induction, I tried the brain-dead method and went straight for $$\frac{2n+1}{2n+2}\cdot\frac{1}{\sqrt{3n}}<\frac{1}{\sqrt{3n+3}}$$ $$...$$ $$1<0.$$ After embarrassing myself, I looked around and I found this thread. Using induction, we can then easily prove $$\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}$$ $$\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}<\frac{1}{\sqrt{3n}}.$$ This gets me to the original problem. But in a problem solving standpoint, how do you think to use $\frac{1}{\sqrt{3n+1}}$? Is there some point in the first induction that leads to this idea? Or is there a better method than the above?

答:
A First Approach
$$
\begin{align}
n\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)^2
&=\frac14\prod_{k=2}^n\left(\frac{2k-1}{2k}\right)^2\frac{k}{k-1}\tag{1a}\\
&=\frac14\prod_{k=2}^n\frac{2k-1}{2k}\frac{2k-1}{2k-2}\tag{1b}\\
&=\frac14\prod_{k=2}^n\frac{\color{#C00}{k-1/2}}{\color{#090}{k}}\frac{\color{#75F}{k-1/2}}{\color{#C90}{k-1}}\tag{1c}\\
&=\frac14\color{#C00}{\frac{\Gamma(n+1/2)}{\Gamma(3/2)}}\color{#090}{\frac{\Gamma(2)}{\Gamma(n+1)}}\color{#75F}{\frac{\Gamma(n+1/2)}{\Gamma(3/2)}}\color{#C90}{\frac{\Gamma(1)}{\Gamma(n)}}\tag{1d}\\[3pt]
&=\frac1\pi\frac{\Gamma(n+1/2)^2}{\Gamma(n+1)\,\Gamma(n)}\tag{1e}\\[3pt]
&\le\frac1\pi\tag{1f}
\end{align}
$$
Explanation:

$\text{(1a)}$: pull the $k=1$ term out front and bring $n$ inside as a telescoping product

$\text{(1b)}$: rearrange terms

$\text{(1c)}$: divide numerator and denominator by $2$

$\text{(1d)}$: write the products as ratios of the Gamma function, using $\Gamma(x+1)=x\,\Gamma(x)$

$\text{(1e)}$: collect terms using $\Gamma(1)=\Gamma(2)=1$ and $\Gamma(3/2)=\sqrt\pi/2$

$\text{(1f)}$: $\Gamma(x)$ is log-convex

Thus, we get the stronger
$$
\prod_{k=1}^n\frac{2k-1}{2k}\le\frac1{\sqrt{\pi n}}\tag2
$$

A Slightly Simpler Approach with a Better Bound
$$
\begin{align}
\prod_{k=1}^n\frac{2k-1}{2k}
&=\prod_{k=1}^n\frac{(2k-1)2k}{4k^2}\tag{3a}\\
&=\frac1{4^n}\binom{2n}{n}\tag{3b}\\
&\le\frac1{\sqrt{\pi\!\left(n+\frac14\right)}}\tag{3c}
\end{align}
$$
Explanation:

$\text{(3a)}$: multiply numerator and denominator by $2k$

$\text{(3b)}$: $\prod\limits_{k=1}^n(2k-1)2k=(2n)!$ and $\prod\limits_{k=1}^n2k=2^nn!$

$\text{(3c)}$: inequality $(9)$ from this answer

In fact, using inequality $(9)$ from this answer, we get
$$
\frac1{\sqrt{\pi\!\left(n+\frac13\right)}}\le\prod_{k=1}^n\frac{2k-1}{2k}\le\frac1{\sqrt{\pi\!\left(n+\frac14\right)}}\tag4
$$


问:
Solve the equation $h(x) = f(x) + g(x) = 0$, $f(x)$ and $g(x)$ having roots that are negatives of each other.
Let $f(x) = x^2 +bx+ 9$ and let $g(x) = x^2 +ax+c, a, b, c ∈ R$. The roots of $f(x) = 0$ and
$g(x) = 0$ are negatives of each other. If $h(x) = f(x)+g(x)$, then solve the equation $h(x) = 0$.

I'm not sure how to solve this at all, maybe Vieta's formulas for quadratics can help.

答:
One can certainly use Vieta's formulas, but I'll give a self-contained solution. (My argument essentially reproves them though.)

Suppose $r_1, r_2$ are the two roots of $f$. By the Fundamental Theorem of Algebra (and using the fact that $f$ is monic--i.e., the leading coefficient of $x^2$ is $1$), we may write
$$f(x) = x^2+bx+9 = (x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2.$$
Since $-r_1$ and $-r_2$ are the roots of $g$ and $g$ is also monic, we have
$$g(x) = x^2+ax+c = (x+r_1)(x+r_2) = x^2+(r_1+r_2)x+r_1r_2.$$
Comparing these two equations, we see $c=9$ and $a=-b$. Thus
$$h(x) = f(x)+g(x) = (x^2+bx+9)+(x^2-bx+9) = 2x^2+18.$$
So $h$ has two imaginary roots: $\pm 3i$.


答:
$f(x) = x^2 +bx+ 9;\;g(x) = x^2 +ax+c$

The sum of the roots of $f(x)$ is $s=-b$ and their product is $p=9$

The sum the roots of $g(x)$ is $-a$ and their product is $c$

So we must have $a=-b;\;c=9$

$f(x)=x^2+bx+9;\;g(x)=x^2-bx+9$

$h(x)=f(x)+g(x)=2x^2+18=0\to x=\pm 3i$

Title: Solve the equation $h(x) = f(x) + g(x) = 0$, $f(x)$ and $g(x)$ having roots that are negatives of each other.

Body:
Let $f(x) = x^2 +bx+ 9$ and let $g(x) = x^2 +ax+c, a, b, c ∈ R$. The roots of $f(x) = 0$ and
$g(x) = 0$ are negatives of each other. If $h(x) = f(x)+g(x)$, then solve the equation $h(x) = 0$.

I'm not sure how to solve this at all, maybe Vieta's formulas for quadratics can help.


问:
What does $f([n]) = h([n]) = X$ mean in functions, if $h$ is bijective and $X$ is a countable set?
What is the meaning of the [] brackets?
It was given that bijection h : [n] → X</p>

答: It's probably from the definition of 'finiteness' of the set $X$, whence $[n]$ denotes the $n$ element set $\{0,1,2,\dots,n-1\}$.


问:
Help understanding proof for integers than can be expressed as $a^{3} + b^{3} + c^{3} - abc$
Given $a^{3} + b^{3} + c^{3} - abc$, determine all the possible values of the expression where $A$, $B$, and $C$ are non-negative integers.

Note that this has been solved here

But I need help understanding why the proof generalizes when we let $f(A,A, A+1)$, $f(A, A, A-1)$ and $f(A, A+1, A-1)$.

The proof doesn't say anything about the following $f(A, A+j, A+k)$ where $k,j \in \mathbb{N}$.

But obviously it probably does as the proof is correct, I just can't seem to understand where. Any help understanding this would be great, thanks!

答:
When it considers $f(a,a,a\pm 1)$ and $f(a+1,a,a-1)$ it wants to show that you can get all numbers not divisible by $3$ or divisible by $9$.  These examples are sufficient to get all those numbers.  There is no need in this phase to consider any more values of the arguments.

The second phase is to show that numbers divisible by $3$ but not divisible by $9$ cannot be expressed.  Here you have to consider other values for the arguments because (so far) it could be that there is some set that will express $3$, for example.  It proves this by considering the numbers $\bmod 3$.


问:
Solving a 2-D SDE
Need to solve the system of SDE
\begin{align*}
dX^1&=-X^2dt+dW^1\\
dX^2&=X^1dt+dW^2
\end{align*}

I wrote the first eq. in integral form as $$X_t^1=X_0^1-\int_0^tX^2(s)ds+W_t^1$$
Then I plugged it into the second eq. and took the integral form of that which turned out to be
$$X_t^2=X_0^2+X_0^1t-\int_0^t\left(\int_0^tX^2(s)ds\right)ds+\int_0^tW^1(s)ds+W_t^2$$
Is this going in the right direction? If so, how much more can it be simplified?

答:
Good question!

It may first be helpful to rewrite as

\begin{align*}
\begin{pmatrix} dX_{t}^{(1)} \\ dX_{t}^{(2)} \end{pmatrix}
=
\begin{pmatrix} 0 & - 1 \\ 1 & 0 \end{pmatrix}
\begin{pmatrix} X_{t}^{(1)} \\ X_{t}^{(2)} \end{pmatrix}\, dt
+ \begin{pmatrix} dW_{t}^{(1)} \\ dW_{t}^{(2)} \end{pmatrix}.
\end{align*}

This is similar to vector ODE: $\vec{X}'(t) = A\vec{X}(t) + \vec{b}(t)$.

First diagonalize

\begin{align*}
A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} :&= PDP^{-1} = \begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix} \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \begin{pmatrix} \frac{1}{2} & -\frac{i}{2} \\ \frac{1}{2} & \frac{i}{2} \end{pmatrix}.
\end{align*}

Next compute

\begin{align*}
e^{At} &= Pe^{Dt}P^{-1} \\
&= \begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix} \begin{pmatrix} e^{it} & 0 \\ 0 & e^{-it} \end{pmatrix} \begin{pmatrix} \frac{1}{2} & -\frac{i}{2} \\ \frac{1}{2} & \frac{i}{2} \end{pmatrix} \\
&= \begin{pmatrix} e^{it} & e^{-it} \\ ie^{it} & -ie^{-it} \end{pmatrix} \begin{pmatrix} \frac{1}{2} & -\frac{i}{2} \\ \frac{1}{2} & \frac{i}{2} \end{pmatrix} \\
&= \begin{pmatrix} \cos(t) & \sin(t) \\ -\sin(t) & \cos(t) \end{pmatrix}
\end{align*}

Solve using d'Alembert's principle.

\begin{align*}
X_{t} &= e^{-At}\vec{X}_{0} + \int_{0}^{t} e^{-A(t-s)} \begin{pmatrix} dW_{s}^{(1)} \\ dW_{s}^{(2)} \end{pmatrix}
\end{align*}

This yields

\begin{align*}
&X_{t}^{(1)} = \cos(t)X_{0}^{(1)} - \sin(t)X_{0}^{(2)} + \int_{0}^{t} \cos(t-s)\, dW_{t}^{(1)} - \int_{0}^{t} \sin(t-s)\, dW_{t}^{(2)} \\
&X_{t}^{(2)} = \sin(t)X_{0}^{(1)} + \cos(t)X_{0}^{(2)} + \int_{0}^{t} \sin(t-s)\, dW_{t}^{(1)} + \int_{0}^{t} \cos(t-s)\, dW_{t}^{(2)}.
\end{align*}

You may wish to check, for example, that

\begin{align*}
dX_{t}^{(1)} &= -\sin(t)X_{0}^{(1)}\, dt - \cos(t)X_{0}^{(2)}\, dt + \left (\int_{0}^{t} \partial_{t}(\cos(t-s)) \, dW_{s}^{(1)}\right )\, dt \\
&\quad+ \cos(t-t)\, dW_{t}^{(1)} - \left (\int_{0}^{t} \partial_{t}(\sin(t-s)) \, dW_{s}^{(2)} \right )\, dt - \sin(t-t)\, dW_{t}^{(2)} \\
&= -X_{t}^{(2)}\, dt + dW_{t}^{(1)}
\end{align*}
as desired.

Title: Solving a 2-D SDE


问:
Probability distribution in the subset version of the coupon collector's problem
The original coupon collector problem was answered here:

Probability distribution in the coupon collector's problem

The subset version of the coupon collector's problem is:

There are $m$ different kinds of coupons to be collected from boxes. Assuming each type of coupon is equally likely to be found per box, what's the expected amount of boxes $(N)$ one has to buy to collect $q$ types of coupons, where $q<m$?

My question is, what is the "closed form" (as a sum perhaps) of $P(N\leqslant n)$?

Note that the collector is seeking a specific given subset of the $m$ coupons, rather than any $q$ of them.

答: You can solve this exactly like the original coupon collector's problem.  At the beginning, the probability that a randomly-selected coupon is one of the desired ones is $\frac qn$, so the expected waiting time is $\frac nq$.  The expected waiting time for the second coupon is $\frac n{q-1}$ and so on, down to the last coupon, which has an expected waiting time of $n$.  The overall expected waiting time is $$\sum_{k=1}^q\frac nk = nH_q\approx n\log q$$


答:
If $|z|=1$, then $\overline{z} = 1/z$ and the equation becomes
$$
z = -i/z,
\\
z^2 = -i
$$
with two solutions
$$
z = \frac{-1+i}{\sqrt{2}},\qquad z=\frac{1-i}{\sqrt{2}}
$$


问:
Evaluate $\displaystyle\int_{0}^{\infty}f\left(x+\dfrac1x\right)\cdot\dfrac{\log x}{x}\;\mathrm{d}x$
The answer is supposed to be an integer, but I don’t understand how the function $f$ is to be solved. What should be the first step?

答:
We are going to prove that

if  $\;\displaystyle\int_{0}^{\infty}f\left(x+\dfrac1x\right)\cdot\dfrac{\log x}{x}\;\mathrm{d}x\;$ converges, then its value is $\;0\;.$

Proof:

By letting $\;t=\dfrac1x\;,$ we get that

$\;\displaystyle\int_{0}^{\infty}f\left(x+\dfrac1x\right)\cdot\dfrac{\log x}{x}\;\mathrm{d}x=\displaystyle\int_{\infty}^0f\left(\dfrac1t+t\right)\big(-t\log t\big)\left(-\dfrac1{t^2}\right)\mathrm{d}t=$

$=-\displaystyle\int_{0}^{\infty}f\left(t+\dfrac1t\right)\cdot\dfrac{\log t}{t}\;\mathrm{d}t=-\displaystyle\int_{0}^{\infty}f\left(x+\dfrac1x\right)\cdot\dfrac{\log x}{x}\;\mathrm{d}x\;.$

Hence,

$\displaystyle\int_{0}^{\infty}f\left(x+\dfrac1x\right)\cdot\dfrac{\log x}{x}\;\mathrm{d}x=-\displaystyle\int_{0}^{\infty}f\left(x+\dfrac1x\right)\cdot\dfrac{\log x}{x}\;\mathrm{d}x\;,$

consequently,

$\displaystyle\int_{0}^{\infty}f\left(x+\dfrac1x\right)\cdot\dfrac{\log x}{x}\;\mathrm{d}x=0\;.$


问:
Checking $\int_{0}^{\infty} {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}} dx}$ for convergence

Given $\displaystyle\int_{0}^{\infty} {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}}  dx}$, prove that it converges.

So of course, I said:

We have to calculate $\displaystyle \lim_{b \to \infty} {\int_{0}^{\infty} {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}}  dx}}$. And in order to do that we have to calculate the integral $\displaystyle \int {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}}  dx}$.

I got stuck calculating this integral, if anyone can give me a direction with this question I'll appreciate it!

答:
You have the inequalities

$$
0\le\frac{\sin^2x}{\sqrt[3]{x^7 + 1}}\le
\frac{1}{\sqrt[3]{x^7 + 1}}\le
\frac{1}{x^{7/3}}
$$

so your integral will converge if

$$
\int_1^{\infty}\frac{1}{x^{7/3}}\,dx
$$

does. The change of the integration bounds is irrelevant, because
$$
\int_0^1\frac{\sin^2x}{\sqrt[3]{x^7 + 1}}\,dx
$$
poses no problem.

Now

$$
\int_1^{a}\frac{1}{x^{7/3}}\,dx=\left[-\frac{3}{4}x^{-4/3}\right]_1^a
=-\frac{3}{4}+\frac{3}{4}a^{-4/3}
$$

and

$$
\lim_{a\to\infty}a^{-4/3}=0
$$

It's quite improbable that you can find the "exact" value of the proposed integral, but the question was only to check for convergence.


问:
Find the common PDF of $|X|,|Y|$

Consider $X,Y$ R.Vs which their common PDF is uniform on the triangle $(-1,-1),(-1,1),(1,-1)$

Find the common PDF of $|X|,|Y|$

I've found that $f_{X,Y}=1/2$ but didn't really know how to work with $|X|,|Y|$ any hint will be welcome :)

答:
Well, I'll do it for $X$.

Clearly $\mathbb{P}(|X|\leq1)=1$. Now, with $0\leq k \leq 1$,  $\mathbb{P}(|X|\leq k) = \frac12\int\int_T \chi_{|x|\leq k} dx dy = \frac12\int_{-k}^k\int_{-1}^{-x}dydx = \frac12\int_{-k}^k(1-x)dx = \frac12\big(2k\big) = k$.

So $|X|$ is uniformly distributed on $[0,1]$.

Title: Find the common PDF of $|X|,|Y|$


问:
Convergence of a complex series, root test
I got series
$\sum\limits_{n=1}^{\infty}\left(\frac{zn}{n+|z|}\right)^{n^2}$ and I'm trying to test its convergence.
I have tried root test

$C=\limsup\limits_{n\rightarrow \infty} \sqrt[n]{\left|\left(\frac{zn}{n+|z|}\right)^{n^2}\right|}=\limsup\limits_{n\rightarrow \infty}\left(\frac{|z|n}{n+|z|}\right)^{n}$

but I don't know how to calculate this limit and if it is the right way to solve this problem.

答:
$$\log C=\underset{n\to \infty }{\text{lim}}n\log\left(\frac{|z| n}{|z|+n}\right)$$
thus $C\to\infty$ for $|z|>1$ and $C\to 0$ for $|z|<1$

$C=\frac{1}{e}$ for $|z|=1$

So the series converges absolutely for $|z|\le 1$ and diverges for $|z|>1$


答:
We know that for $0 \le a < b$ and $0 \le c < d$

$$
\begin{align}
\mathbb P((|X|, |Y|) \in (a, b) \times (c, d))  =& \mathbb P((X, Y) \in (a, b) \times (c,d)) \\
&+ \mathbb P((X, Y) \in -(a, b) \times (c,d)) \\
&+ \mathbb P((X, Y) \in (a, b) \times -(c,d)) \\
&+ \mathbb P((X, Y) \in -(a, b) \times -(c,d)) .
\end{align}
$$

But we know what the summands on the right hand side are: Suppose that $f_{X, Y}$ equals the distribution function of $(X, Y)$. Then
$$
\begin{align}
\mathbb P((X, Y) \in (a, b) \times (c,d))& = \int_a^b \int_c^d f_{X, Y}(x, y) dx dy, \\
\mathbb P((X, Y) \in -(a, b) \times (c,d))& = \int_{-b}^{-a} \int_c^d f_{X, Y}(x, y) dx dy \\
& = \int_a^b \int_c^d f_{X, Y}(-x, y) dx dy, \\
\mathbb P((X, Y) \in (a, b) \times -(c,d)) &= \int_a^b \int_c^d f_{X, Y}(x, -y) dx dy, \text{ and} \\
\mathbb P((X, Y) \in -(a, b) \times -(c,d)) &= \int_a^b \int_c^d f_{X, Y}(-x, -y) dx dy.
\end{align}
$$
From this, we get that the probability distribution function on the upper right quadrant is
$$
f_{X, Y}(x, y) + f_{X, Y}(-x, y) + f_{X, Y}(x, -y) + f_{X, Y}(-x, -y).
$$
Since both components of $(|X|, |Y|)$ are non-negative, the probability distribution function is zero on the other quadrants.

Now you just note that
$$
f_{X, Y}(x, y) = \mathbf 1_{x + y \le 0 \wedge x \ge -1 \wedge y \ge -1}
$$
and substitute that into the above expression.

Title: Find the common PDF of $|X|,|Y|$

Body:
Consider $X,Y$ R.Vs which their common PDF is uniform on the triangle $(-1,-1),(-1,1),(1,-1)$

Find the common PDF of $|X|,|Y|$

I've found that $f_{X,Y}=1/2$ but didn't really know how to work with $|X|,|Y|$ any hint will be welcome :)


问:
Figuring out winding number of a curve using crossing rule of a point
In page-340 of Tristan Needham's Visual Complex Analysis, he introduces a theorem to find number of a loop. To motivate the theorem, he shows the situation of a point coming closer and closer to a segment of a curve:

For the right most picture, he writes a relation between winding numbers:

$$ v(K,r) = v(K,s)  + v(L,s)$$

Or,

$$ v(K,r) = v(K,s) -1 $$

Or,

$$ v(K,r) +1 =v(K,s) \tag{1}$$

Explanation given:

Start from outside L, where you know the winding number is zero, move from
region to region using crossing rule to add or subtract one at each
crossing of L

The equation (1) and the idea behind its derivation allows us to relate winding numbers between the interior and exterior of the sets in which the curve partitions the plane. However, how can this be used to figure out the winding numbers easily?

Note:

$v(k,s) $ means the winding number of the loop $k$ around the point $s$

答:
Let's think of $L$ as a one way road. The equation states that when crossing the road, your winding number increases by one if traffic is coming from your left and decreases by one if traffic is coming from your right.

Now look at the example: Starting at the exterior region $D_4$ we have $v_4=0$. To get into $D_3$ we have to cross the road with traffic coming from our left, so the winding number increases to $v_3=1$. From $D_3$ to $D_2$ we have to cross $K$ with traffic coming from our right, so $v_2=v_3-1=1$. From $D_3$ to $D_1$ we cross $K$ with traffic coming from our left, so $v_1=v_3+1=2$.


问:
Help with finding a function, $\lim \ne \inf$
I am looking for an example for a function where:

$$\lim\limits_{x \to a^+} f(x) \neq \inf\, \{f(x) \mid a\leq x \leq b\},\quad a<b,$$

$$f:[a,b]\to\mathbb{R}\;\;\text{ and }\;f\;\text{ is an increasing function.}$$

Can you guys give me a hint of the way of thinking to find such a function? (Notice that $x \to a^{+}$)

The function must be an increasing function.

Thanks!

答:
Actually, you can generate lots of functions satisfying this condition since the function is not supposed to be continuous.

For example, $$f(x) = \begin{cases} 0, \ \ x = 0 \\ \sqrt x + 1, \ \ x \in (0, 1] \ \ \end{cases}$$
Then, we have $$\lim_{x\to0^+} (\sqrt x + 1) = 1 \quad \ne \quad 0 = \inf\{f(x) \mid 0 \leq x \leq 1\}$$

$\quad\quad\quad\quad$


问:
How to show subset of $\mathbb R$ is not an element of $\mathbb R$
How do you show that for some $A\subset \mathbb R$, $A\not \in \mathbb R$.

Intuitively it makes sense, but how do you actually prove it?

答: It depends upon how the real numbers are defined.  For example, if a real number is an equivalence class of Cauchy sequences of rational numbers, then a set of real numbers is not a real number, because its elements are not equivalence classes of Cauchy sequences, but sets of such equivalence classes.</p>


问:
If $f^{-1}$(a) $\in$ $\chi$ for all a $\in$ $\mathbb{R}$ then f is a measurable function
$f$ : $X$ $\to$ $\mathbb{R}$ is a function on a measurable space ($X$, $\chi$).

I would like to show that this statement is $false$:

If $f^{-1}$(a) $\in$ $\chi$ for all a $\in$ $\mathbb{R}$ then f is a measurable function

I know that if $f$ is a measurable function then $f^{−1}$(a) ∈ $\chi$ for all a $\in$ $\mathbb{R}$

but I am having trouble showing the converse is false.

I can see that $f^{-1}$(B) $\in$ $\chi$, B being a Borel subset, but just a $\in$ $\mathbb{R}$?

any help would be appreciated thanks.

答: Easiest counterexample is probably to set $X = [0,1]$ with the Lebesgue $\sigma$-algebra, $V \subset X$ a non-measurable set, and $f(x) := x + 1_V(x)$.  Then $f$ is injective so $f^{-1}(a)$ is a singleton for all $a \in \mathbb{R}$ and hence measurable, but $f$ is clearly not measurable (if it were, then $f(x) - x$ would also be, but $f(x)-x = 1_V$ which was chosen to be non-measurable).


问:
Finding tangent vector for a composite function at a given point
I would want to insert the whole question but it's kinda complex/difficult to type it out :( sorry guysss
Basically the problem looks similar to

$T(x,y,z)=(u,v)$

$u=x+y+z^2$

$v=\cos(x)+\tan(y)-z+3$

vector $r(t)=(t\cdot e,(t^2+2)^t, e^{-s^2})$

At point $t=0$ what's the tangent vector of $T(r(t))$?

Anybody who can help me out with this problem?
Would appreciate sooooo much $x$.

答:
Its an application of the Chain rule. Note that $r(0)=(0,1,1)$ and $r'(0)=(e,\ln 2,0)$. By the chain rule
$$\tag{*}\frac{\partial}{\partial t}\biggl|_{t=0}T(r(t))=J(T)_{r(0)}\cdot r'(0)$$
where $J(T)_{r(0)}$ is the Jacobian of $T$ evaluated at $r(0)$. That is,
\begin{align}
J(T)_{r(0)}&=
\begin{pmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}& \frac{\partial u}{\partial z} \\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z}
\end{pmatrix}
_{\bigl|r(0)}\\
\\
&=\begin{pmatrix}
1 & 1 & 2z\\
-\sin x & \frac{1}{\cos ^2 y} & -1
\end{pmatrix}
_{(0,1,1)}\\
\\
&=\begin{pmatrix}
1 & 1 & 2\\
0& \frac{1}{\cos^2 1} & -1
\end{pmatrix}
\end{align}
Multiplying the above matrix with the vector $r'(0)=(e,\ln 2,0)$ and using $(*)$ we obtain
\begin{align}
\frac{\partial}{\partial t}\biggl|_{t=0}T(r(t))&=\begin{pmatrix}
1 & 1 & 2\\
0& \frac{1}{\cos^2 1} & -1
\end{pmatrix} \cdot
\begin{pmatrix}
e \\ \ln 2 \\ 0
\end{pmatrix}\\
&\\
&=(e+\ln 2,\  \frac{\ln 2}{\cos^2 1})
\end{align}
the desired tangent vector of $T(r(t))$ at $t=0$.


问:
Interesting thing about sum of squares of prime factors of $27$ and $16$.
Let
$$n=p_1×p_2×p_3×\dots×p_r$$
where $p_i$ are prime factors and
$f$  is the functions
$$f(n)=p_1^2+p_2^2+\dots+p_r^2$$
If we put $n=27,16$ and $27=3×3×3$, $16=2×2×2×2$ then
$$\begin{split}f(27)&=3^2+3^2+3^2=27\\f(16)&=2^2+2^2+2^2+2^2=16.\end{split}$$
I checked it upto $n=10000$, I did not find another number with this property $f(n)=n$.

Can we prove that other such numbers do not exist?

Some approaching values

$f(992)=981\\f(1058)=1062\\f(1922)=1926\\f(5396)=5410\\f(7198)=7206\\f(9506)=9511$

Sequence: A067666, Sum of squares of prime factors of n (counted with multiplicity).

Edit

We can show there are infinitely many $n$ s.t. $f(n)=n+4$

Proof: put $n=2\cdot p^2$ where $p$ prime number

gives $f(2\cdot p^2)=2^2+p^2+p^2=4+2\cdot p^2$.

答:
Giorgos Kalogeropoulos has found 3 such numbers, each having more than 100 digits.

You can find these numbers if you follow the links in the comments of OEIS A339062 &
A338093

or here https://www.primepuzzles.net/puzzles/puzz_1019.htm

So, such numbers exist! It is an open question if there are infinitely many of them...