问:
Advection-style first order nonlinear PDE system
Given functions $a(t,x,y) > 0, d(t,x) >0$ and constants $b,c > 0$, I would like to find the three functions $u(t,x),v(t,x) , w(t,x,y)$ that solve the system:

\begin{align}
u_t - u_x &= -\left(\int_0^x a(t,x,y)\mathrm{d}y+b\right) u(t,x) + c v(t,x) + w_y(t,x,0) + d(t,x)\\
v_t - v_x &= b u(t,x) - c v(t,x) + b \int_0^x w(t,x,y) \mathrm{d} y\\
w_t - w_x - w_y &= a(t,x,y) u(t,x)  - b w(t,x,y),
\end{align}

and satisfy $u(t,0)=v(t,0)=w(t,x,0) = 0$.

Interpretation : the state of a job allocating platform

$u,w,v$ - is the number of workers $u$noccupied, $w$orking or on a sick lea$v$e.

$x$ - is the remaining duration of the working contract, by the end of which a worker leaves the system

$y$ - is the remaining duration of a job assignment

$a(t,x,y)$ - is the probability of being assigned a job of duration $y$ at date $t$

$d(t,x)$ - is the number of new workers joining the platform at date $t$ for the duration $x$

$b,c$ - is the chance of falling ill and that of recovering.

Once a job assignment is over ($y=0$), the busy worker joins the ranks of the unoccupied (the $w_y(t,x,0)$ term in the $u$-equation). Once the contract is over ($x=0$) the worker leaves the platform.

I kind of found the solution to a linear 1st order PDE system in the Andrei D. Polyanin's bestiary, but am absolutely at loss about what to do with the $w_y(t,x,0)$ and $\int w \mathrm{d}y $ terms.

Although not exactly a model of a physics phenomenon, this system looks like something that the physicists are most likely to be familiar with.
I would be immensely grateful for the solution and/or all the pointers to the sources treating this kind of problems (lectures, keywords, etc.).

Update

Noticing that for every worker the date of retirement, $T_x$,  and for every job the date of expiry, $T_y$, are constants, I can use the substitution $x = T_x - t$ and $y = T_y - t$ to transform equations 2 and 3 into ODEs, solve them and reverse the substitution to obtain $w,v$ in terms of $u$:

\begin{align*}
w(t,x,y) & = \int_0^t \int_{s}^{y+s} \mathrm{e}^{-b s} a_y(t-s,x+s,\xi) u(t-s,x+s) \mathrm{d}\xi \mathrm{d} s \\
v(t,x) & = b \int_0^t \int_{s}^{x + s} \mathrm{e}^{-cs} u_x(t-s,\xi) + W_x(t-s,\xi) \mathrm{d} \xi \mathrm{d} s.
\end{align*}

where

\begin{equation*}
    W(t,x) := \int_0^x w(t,x,y) \mathrm{d} y,
\end{equation*}

is the total size of the busy group across all job durations.

Plugging these expressions into the first equation one gets the following PDE in $u$:

\begin{align}
u_t =& u_x - \left(\int_0^x a\mathrm{d}y +b \right) u \\
     & + cb \int_0^t \int_{s}^{x + s} \mathrm{e}^{-cs} u_x(t-s,\xi)\mathrm{d} \xi \mathrm{d} s \\
     & + cb \int_0^t \int_{s}^{x + s} \mathrm{e}^{-cs} W_x(t-s,\xi) \mathrm{d} \xi \mathrm{d} s \\
     & + \int_0^t \mathrm{e}^{-bs} a_y(t-s,x+s,0) u(t-s, x+s) \mathrm{d}\xi \\
     & + d(t,x).
\end{align}

答:
I think this is a nice example of a system of equations solvable by the Laplace transform (I suppose that we are interested only in $x,y>0$ here):
$$
\bar{u}(p,t)=\int_0^{+\infty}dx u(x,t)e^{-px}, etc.
$$
This should reduce it to a system of linear ordinary differential equations in respect to time, which can be solved. Then one uses the Laplace transform table to try to obtain the solution in $(x,t)$ variables.

Update.
Another rather general approach that works particularly well for the first order PDEs is the method if characteristics (see the section "Two-dimensional theory" here).

OwnerDisplayName: Konstantin

OwnerDisplayName: Vadim


问:
Show that the solution of $P_n(X) = 0$ is $x=i\frac{e^{2i\theta}+1}{e^{2i\theta}-1}=\frac{1}{\tan(\theta)}$
I need to how that when $ k \neq 0$, we obtain a unique solution of $P_n(X)=\frac{(X+i)^{2n+1} - (X - i)^{2n+1}}{2i} = 0$ is $ x=i\frac{e^{2i\theta}+1}{e^{2i\theta}-1}=\frac{1}{\tan(\theta)} $ with $\theta=\frac{k\pi}{2n+1} \in ]-\frac{\pi}{2},\frac{\pi}{2}[ \backslash {0}$.

In the previous questions I've shown that there are no solutions when $k=0$, that if $P_n(x) = 0$, then $x$ verify $(\frac{x+i}{x-i})^{2n+1} = 1$, and that $P_n(x) = 0$ when $\frac{x+i}{x-i} = \exp(\frac{2ik\pi}{2n+1})$ with $k$ between $-n$ and $n$.

So my guess is that I need to replace $x$ in $\frac{x+i}{x-i}$ by $\frac{1}{tan(\theta)}$ and show that the equation $\frac{x+i}{x-i} = \exp(\frac{2ik\pi}{2n+1})$ with $k$ between $-n$ and $n$ is true? But I'm stuck and don't really know how to go further.

答:
Right, please see here and here and also here for a rather complete solution.

There, we show that $x=\frac{1}{\tan \theta_k} = \frac{\cos \theta}{\sin \theta} = i\frac{e^{i2\theta}+1}{e^{2i\theta}-1}$ where $k \in \{ \mp1,\mp2,\cdots,\mp n\}$ and $\theta_k = \frac{k\pi}{2n+1}$.

ContentLicense: CC BY-SA 4.0


问:
Verify that $z^{2n+1} = 1$ where the solutions are $\exp({\frac{2ik\pi}{2n+1}})$ where $k$ goes from $-n$ to $n$ doesn't have a solution when $k = 0$
I've shown that the solutions of $z^{2n+1} = 1$ are the $\exp({\frac{2ik\pi}{2n+1}})$ where $k$ goes from $-n$ to $n$ but now I need to show that it doesn't have a solution when $k = 0$.

So I tried to replace $k$ with $0$ in $\exp({\frac{2ik\pi}{2n+1}})$ and I obtain $\exp({\frac{0}{2n+1}}) = \exp(0) = 1$. So the equation $z^{2n+1} = 1$ still holds when $k = 0$. How do I show that there is no solution when $k=0$ ? Did I make any mistakes in my reasoning ?

In the very first question i've shown that if $P_{n}(x)=\frac{(x+i)^{(2n+1)} - (x-i)^{(2n+1)}}{2i} = 0$ then x verify $(\frac{x+i}{x-i})^{2n+1} = 1$

答:
Please see here and here and also here for a complete solution.

We show that $x=\frac{1}{\tan \theta_k} = \frac{\cos \theta}{\sin \theta} = i\frac{e^{i2\theta}+1}{e^{2i\theta}-1}$ where $k \in \{ \mp1,\mp2,\cdots,\mp n\}$ and $\theta_k = \frac{k\pi}{2n+1}$.

Please also see here.


问:
Compare the variances of restricted and unrestricted estimators?
Problem

Given a linear model $y_i = \beta_1 + \beta_2 x_i +\epsilon_i, \quad i = 1, \dots, n$

I need to compare the variance ordinary least squares estimator of $\beta_2$ without the restrictions and the variance of ordinary least squares estimator of $\beta_2$ under linear restriction $\beta_1= 0$ (i.e. $\Bbb Var(\beta_2^R)$ and $\Bbb Var(\beta_2^U) $).

Is $\beta_2^R$ unbiased and are there any violations of Gauss–Markov theorem?

My ideas

Intuitively, the restricted variance should be less than the unrestricted variance because the restrictions always damage the flexibility of a model hence reducing the variance.

However, I need a mathematical proof for the problem above.

With some help I have obtained that  $$ Var(\beta_2^{UR}) =  \frac{\sum(y_i - \beta_1- \beta_2x_i)^2/(n-2)}{\sum(x_i - \bar{x})^2}$$ and $$ Var(\beta_2^{R}) = \frac{\sum(y_i -  \beta_2x_i)^2/(n-2)}{\sum(x_i )^2}$$

答:
You can use this formula
$\mathbb{V}(\boldsymbol{\hat{\beta}}) = \sigma^2 (X^\text{T} X)^{-1}$

$\sigma^2$ is the variance of the disturbance term i.e. $\Bbb Var(\epsilon_i) = \sigma^2 $

If the regression has an intercept, then $X$ has the following form:
\begin{bmatrix}
1 & x_1  \\
1 & x_2 \\
\vdots & \vdots \\
1& x_n
\end{bmatrix}
If the regression has no intercept, then $X$ has the column form:

\begin{bmatrix}
x_1  \\
x_2 \\
\vdots \\
x_n
\end{bmatrix}

Therefore, using the formula above we obtain that

$$\Bbb {V}(\hat{\beta}_2^{UR}) = \frac{\sigma^2}{\sum x_i^2 - n \bar{x}^2}.$$

$$\Bbb {V}(\hat{\beta}_2^R) = \frac{\sigma^2}{\sum x_i^2}.$$

From these formulas the comparison is trivial.


问:
How many different 5 letters word can be possible by using the letters from "Cambridge" with all the vowels in it?
So I took the vowels (a,e,i) permute them 3p3 and then I choose 6c2.Here I get baffled and can't figure out what should I do next.

By the way multiply 3p3 and 6c2 comes up with 180 which is not the answer.The answer is 1800.

Please explain where I made the mistake.Or Am I missing something?

答:
You need to first choose the 5 letters in the word, then permute it.

You must choose all 3 vowels, and 2 consonants from the 6; there are $6C2 = 15$ choices.

Then you permute the 5 chosen letters; there are $5P5 = 5!$ choices.

Multiplying, we have $6C2 \times 5! = 15 \times 120 = 1800$ different words.


问:
Equivalency between a "mixed modular equation" of Gauss and a later theorem of Ramanujan.
In p. 476 of volume 3 of Gauss's collected works, appear several interesting identities on Jacobi theta functions which were used by the czech mathematician Karel Petr in an 1904 article "Bemerkung zur einer Gausschen Formel über die Thetafunktionen"  to derive relations giving the number of representations of a number $N$ by three quaternary quadratic forms: $x^2 + y^2 + 9z^2 + 9u^2$, $x^2+y^2+z^2+9u^2$, $x^2+9y^2+9z^2+9u^2$. The identities by Gauss are:

$$(\frac{3P^2-P^0\cdot P^0}{2})^2= p^4-4(\frac{pqr}{2})^{\frac{4}{3}}$$
$$(\frac{3Q^2-Q^0\cdot Q^0}{2})^2=q^4+4(\frac {pqr}{2})^{\frac{4}{3}}$$

where the relevant quantities are defined to be:

$$P(x^3,1)=P , P(x,1)=p, P^0=P(x^{\frac{1}{3}},1)$$
$$Q(x^3,1)=Q , Q(x,1)=q, Q^0 = Q(x^{\frac{1}{3}},1)$$
$$R(x^3,1)=R , R(x,1)= r$$

and the three functions $P(x,y),Q(x,y),R(x,y)$ are equivalent to Jacobi's theta functions $\vartheta_3,\vartheta_4,\vartheta_2$ and defined to be:

$$P(x,y)=1+x(y+\frac{1}{y})+x^4(y^2+\frac{1}{y^2})+x^9(y^3+\frac{1}{y^3})+...$$
$$Q(x,y)= 1-x(y+\frac{1}{y})+x^4(y^2+\frac{1}{y^2})-x^9(y^3+\frac{1}{y^3})+...$$
$$R(x,y)=x^{\frac{1}{4}}(y^{\frac{1}{2}}+y^{-\frac{1}{2}})+x^{\frac{9}{4}}(y^{\frac{3}{2}}+y^{-\frac{3}{2}})+x^{\frac{25}{4}}(y^{\frac{5}{2}}+y^{-\frac{5}{2}})+...$$

To see the equivalency between Gauss's notation and Jacobi's theta functions, look at the post Interpretation of a certain general theorem used by Gauss in his work on theta functions.. User Paramanand Singh helped me understand the meaning of Gauss's identities and remarked that they are essentialy a "mixed modular equation" which connects theta functions of $x^{\frac{1}{3}},x,x^3$ (or, equivalently, $\tau,\tau^3,\tau^9$), and said that Ramanujan also gave this modular equation.

According to Paramanand Singh's comments, the following identity (from p. 142 of the book "pi and the AGM") of Ramanujan is equivalent to Gauss's:

$$\frac{\theta_3(q)}{\theta_3(q^9)} - 1 = (\frac{\theta_3^4(q^3)}{\theta_3^4(q^9)}-1)^{\frac{1}{3}}$$

However, he didn't provide proof of equivalency. Therefore, it's not certain this is the desired identity of Ramanujan, and it's not even certain that Ramanujan stated an equivalent identity at all.

Therefore, my questions are:

Can anyone familiar with Ramanujan's results on mixed modular equations say which of Ramanujan's theorems is equivalent to Gauss's identities? and where can i find the relevant fragment of Ramanujan's writings on the internet?

Can someone also give a proof of equivalency between Ramanujan's theorem and Gauss's identities? this doesn't have to be a proof of correctness of Gauss's identities (which can be quite complicated), just a proof of equivalency to Ramanujan's theorem.

答:
The identity of Gauss has a typo and the correct form is $$\left(\frac{3P^2-P^0P^0}{2}\right)^2=p^4-4\left(\frac{pqr}{2}\right)^{4/3}\tag{1}$$ As in Somos's answer this translates to $$(3\phi^2(q^9)-\phi^2(q))^2=4\phi^4(q^3)-16qf^{4}(-q^6)\tag{2}$$ where $$\phi(q) =\vartheta_3(q)=\sum_{n\in\mathbb {Z}} q^{n^2},f(-q)=\prod_{n=1}^{\infty} (1-q^n)\tag{3}$$ We will use Ramanujan's notation and the proof borrows heavily from work of Dr. Bruce C. Berndt and his collaborators.

Given the nome $q$, the corresponding elliptic modulus $k$ and elliptic integral $K$ (or Ramanujan's $z$) are given in terms of theta functions $$\vartheta_2(q)=\sum_{n\in\mathbb {Z}} q^{(n+(1/2))^2},k=\frac{\vartheta_2^2(q)} {\vartheta_3^2(q)}, z=\frac{2K}{\pi}=\phi^2(q)\tag{4}$$ Let the parameter $k^2$ for nomes $q, q^3,q^9$ be denoted by $\alpha, \beta, \gamma $ respectively (so that for example $\beta=\vartheta_2^4(q^3)/\vartheta_3^4(q^3)$). Further let the parameter $z$ corresponding to these nomes be denoted by $z_1,z_3,z_9$ (so that $z_3=\phi^2(q^3)$).

The modular equations of degree 3 allow us to express $\alpha, \beta, \gamma $ in terms of multiplier $m$. Let us define $$m=\frac{z_1}{z_3},m'=\frac{z_3}{z_9}\tag{5}$$ Berndt gives us $$\alpha = \frac{(3 + m)^{3}(m - 1)}{16m^{3}}, \beta =\frac{(m - 1)^{3}(3 + m)}{16m}\tag{6}$$ and $$1-\alpha=\frac{(3 - m)^{3}(m + 1)}{16m^{3}}, 1-\beta =\frac{(m + 1)^{3}(3 - m)}{16m}\tag{7}$$ If we replace $m$ by $m'$ in above equations then $\alpha$ changes to $\beta$ and $\beta$ changes to $\gamma$. It should also be clear that if $0<q<1$ then we have $1<m,m'<3$.

Next Berndt uses another set of parameters $$u=\frac{m-1}{2}+\left(\frac{m-1}{2}\right)^2, v=\frac{m'-1}{2}+\left(\frac{m'-1}{2}\right)^2\tag{8}$$ (Berndt actually uses intermediate parameters $p, p'$ with $m=1+2p,m'=1+2p'$ and $q=p+p^2, q'=p'+p'^2$ instead of $u, v $, but this will conflict with the usage of $q$ for nome and hence we avoid it).

We have the obvious relations $$m=\sqrt {1+4u},m'=\sqrt{1+4v}\tag{9}$$ and the more difficult relations $$\alpha(1-\alpha)=u\left(\frac {2-u}{1+4u}\right)^3,\beta(1-\beta)=u^3\left(\frac{2-u}{1+4u}\right)\tag{10}$$ and $$\beta(1-\beta)=v\left(\frac{2-v}{1+4v}\right)^3,\gamma(1-\gamma)=v^3\left(\frac{2-v}{1+4v}\right)\tag{11}$$ Using the two expressions for $\beta(1-\beta)$ one gets a polynomial equation between $u, v$ of degree $4$ which Berndt successfully (!) solves using another parameter $t$ given by $v=2t^3$.

The algebra involved is tedious and I reproduce the necessary results directly from Berndt's Ramanujan's Notebooks Vol 3, page 354: $$m=\frac{(1+2t)^2}{\sqrt{1+8t^3}},m'=\sqrt{1+8t^3}\tag{12}$$ and $$\beta(1-\beta)=16t^3\left(\frac{1-t^3}{1+8t^3}\right)^3\tag{13}$$ The relation $(2)$ can be transcribed easily as $$\left(\frac{3}{m'}-m\right)^2=4-2^{8/3}(\beta(1-\beta))^{1/3}\tag{14}$$ which is easily verified using $(12),(13)$.

I am sure that Gauss would have provided a simpler proof using his relations between theta functions.


答:
You need to choose $2$ out of the $6$ letters. That is ${6 \choose 2}$ ways as you said.

Now you have $5$ letters and so just permute them. So your answer is ${6 \choose 2} \times 5!$

Title: How many different 5 letters word can be possible by using the letters from "Cambridge" with all the vowels in it?

Body:
So I took the vowels (a,e,i) permute them 3p3 and then I choose 6c2.Here I get baffled and can't figure out what should I do next.

By the way multiply 3p3 and 6c2 comes up with 180 which is not the answer.The answer is 1800.

Please explain where I made the mistake.Or Am I missing something?


问:
Determinant in calculation of orientation in Riemannian Geometry
Assuming not all of $y_1,..., y_n$ are zero. When $i\ne j$, define
$$
c_{ij}= \frac{2y_iy_j}{\left(\sum_{k=1}^n y_k^2\right)^2}
$$
and
$$
c_{ii}=\frac{-\left(\sum_{k=1}^n y_k^2\right)+2y_i^2}{\left(\sum_{k=1}^n y_k^2\right)^2}.
$$
How to show that
$$
\det(c_{ij})>0
$$
when $n$ is odd and
$$
\det(c_{ij})<0
$$
when $n$ is even?

答:
First use the fact that for any scalar $\alpha$ and $n\times n$ matrix $A$ we have $\det(\alpha A) = \alpha^n \det(A)$ to note that the sign of $\det(c_{ij})$ is the same as the sign of $\det(b_{ij})$ where

$$
b_{ij}= 2y_iy_j
$$

for $i \ne j$ and

$$
b_{ii}=2y_i^2-\sum_{k=1}^n y_k^2.
$$

Next, define $d = \sum_{k=1}^n y_k^2 = y^Ty \gt 0$ and note that

$$
\det(b_{ij}) = \det(2yy^T - d I) = (-d)^n \det\left(I - \frac{2yy^T}{d}\right).
$$

Now, by Weinstein–Aronszajn identity we have

$$
\det(b_{ij}) = (-d)^n \left(1 - \frac{2y^Ty}{d}\right) = (-1)^{n+1} d^n.
$$

Therefore,

$$
\det(b_{ij})>0
$$
when $n$ is odd and
$$
\det(b_{ij})<0
$$
when $n$ is even. Conclusion follows from the fact that $\det(c_{ij})$ has the same sign as $\det(b_{ij})$.


问:
Probability and the Symmetric Group
Let $p_{n,k}$ be the probability that a random permutation from $S_n$, the symmetric group of order $n!$, has exactly $k$ fixed points.  I am trying to compute $\lim _{n\to \infty}p_{n,k}$.  After playing with it a bit, I am relatively confident the limit is $\frac{1}{k!}$, although I have yet to come up with a proof.

Here's what I've tried so far.  If $a_{n,k}$ is the number of elements of $S_n$ with $k$ fixed points, then we have that

$$
a_{n,k}=\binom{n}{k}a_{n-k,0}
$$

That is, there are $\binom{n}{k}$ ways to choose $k$ fixed points, and after those are chosen, there are $a_{n-k,0}$ to permute the remaining $n-k$ elements with $0$ fixed points.  After computing $a_{n,0}$ by hand for small values of $n$, I was able to look up the recursion formula $a_n=na_{n-1}+(-1)^n$ from oeis.org.  Using this, I was able to come up with my "guess" of $\frac{1}{k!}$, but this direction doesn't seem to be leading me towards finding a proof.

Any ideas how to proceed?  Group theory is not my strong point, so I would not be surprised if there is a result that makes this problem almost trivial.  A pointer to any such results would be excellent.

答:
Here's a different take that continues along the lines you've already been going.  Use a different formula for $a_{n,0}$ from the OEIS; use $$a_{n,0} = n! \sum_{j=0}^n \frac{(-1)^j}{j!}.$$
This is a well-known formula for $a_{n,0}$ (the derangement numbers) and can be proved using inclusion-exclusion.  (Added: You might also want to look at the proof and general argument given by Qiaochu Yuan in his answer to a similar question.)

As $n \to \infty$, the sum in the expression for $a_{n,0}$ approaches $e^{-1}$.  With this and your expression $a_{n,k} = \binom{n}{k} a_{n-k,0}$, you're almost to the correct answer of $e^{-1}/k!$ as given by Yuval Filmus.

ContentLicense: CC BY-SA 3.0

Title: Probability and the Symmetric Group


问:
Does anyone know a closed-form expression for a bijection between $\mathbb{N}^k$ and $\mathbb{N}$?
I want to publish an article and one of its results is a simple closed-form expression for a natural bijection between $\mathbb{N}^k$ and $\mathbb{N}$. I wish to know whether it is already known.</p>

答:
Unfortunately, my manuscript was rejected by the American Mathematical Monthly. They said the following:

"The paper concentrates on the explicit construction of a bijection between $\bf N\times N$ and $\bf N$ and similar bijections for ${\bf N}^k$. However, for me it is quite unclear what is the intended audience and what is the main point of the paper. An explicit polynomial formula for the bijection indeed is nice, but it is definitely not new (I have seen it in logical textbooks). If the main point is a nice accessible explanation (for a wide audience) why this formula is true, it is possible, but the style of the paper with a lot of formulas is quite confusing and is not suitable for this goal.

So I am sorry to say that IMHO this paper does not look suitable for AMM (and, probably, for other journals I can think of)..."

So, as I have promised, I came back here to present you my formula:

$$\Psi_k(n_1, \dots, n_k) = \sum_{i=1}^{k}\binom{i-1+n_1+\dots +n_i}{i}$$

I will still try to present my work in some colloquium, before revealing how I got this formula.


问:
What is the distribution of $Z=X_1X_2$?
So my question went like this:

If $X_1\sim\text{Unif(0,a)}$ and $X_2\sim\text{Unif(0,a)}$ independently, then find the distribution of $Z=X_1X_2$.

I tried the following:

Let $Z_1=\frac{X_1}{a}\sim\text{Unif(0,1)}$ and $Z_2=\frac{X_2}{a}\sim\text{Unif(0,1)}$. Thus $Z_1$ and $Z_2$ are independent.

Also, we know that, $-2\log_eZ_1\sim\chi_2^2$ and $-2\log_eZ_2\sim\chi_2^2$.

This implies,

$-2\log_e{Z_1Z_2}\sim\chi_4^2\implies-2\log_e{\frac{X_1X_2}{a^2}}\sim\chi_4^2\implies-2\log_e{\frac{Z}{a^2}}\sim\chi_4^2$.

Let $Z^*=-2\log_e{\frac{Z}{a^2}}\sim\chi_4^2$.

Thus, applying the Jacobian Transformation, I found the pdf of $Z$ as:
$$f_Z(z)=\begin{cases}\frac{\left(-\log_e{z/a^2}\right)e^{-\log_e{z/a^2}}}{2z} & 0\leq z\leq a^2\\
0 & \text{otherwise}\end{cases}$$
Is what I did, right?

Actually, before proceeding like this, I tried using the direct method using the Jacobian transformation but I was getting something like this:

$$\begin{align}f_Z(z)&=\int_0^a{f_{X_1}\left(\frac{z}{x_2}\right)f_{X_2}\left(x_2\right)\frac{1}{x_2}}.dx_2\\
&=\frac{1}{a^2}\int_0^a\frac{1}{x_2}.dx_2\end{align}$$
But this integral is not defined.

So does this mean that the direct formula of the Jacobian Transformation is invalid in this case?

答:
Just to simplify the notation I set $Z=XY$.

To use the jacobian Method set

$$\begin{cases}
z=xy   \\
v=x
\end{cases}$$

The jacobian is $|J|=\frac{1}{v}$ thus

$$f_{UZ}(u,z)=\frac{1}{a^2v}\mathbb{1}_{(0;a^2)}(z)\cdot\mathbb{1}_{(\frac{z}{a};a)}(v)$$

Thus

$$f_Z(z)=\frac{1}{a^2}\int_{\frac{z}{a}}^a \frac{1}{v}dv=\frac{2\log a-\log z}{a^2}$$

Title: What is the distribution of $Z=X_1X_2$?


问:
How to check if the quadratic congruence has solution or not
How to check if the quadratic congruence has solution or not, for example check:
$$x^2\equiv 1141 \pmod{2020}$$

I use Euler's Criterion but it doesn't work:
$$1141^\dfrac{2020-1}{2} \equiv 1 \pmod{2020}$$ Because $\dfrac{2020-1}{2}$ is a decimal number. Any hints would be great!

答:
By the Chinese Remainder Theorem, since $2020=20\cdot 101$, it is enough to check whether the equation has roots modulo $20$ and $101$.

Modulo $20$: $x=1$ is a solution.

Modulo $101$: The equation is $x^2\equiv 30\pmod{101}$. Now, the quadratic residue is
\begin{align}
\left(\frac{30}{101}\right)&=\left(\frac{2}{101}\right)\left(\frac{3}{101}\right)\left(\frac{5}{101}\right)\\
&=(-1)^{\frac{101^2-1}8}\left(\frac{101}{3}\right)\left(\frac{101}{5}\right)\\
&=(-1)(-1)1\\
&=1.
\end{align}

Thus, there is a solution.

P.S.: $x=139$ is a solution to the original congruence.


问:
From $\{1,2,\ldots,2002\}$, remove numbers at positions congruent to $1\bmod 3$, repeating the process until one number remains.

We have a list of natural numbers $\{1,2,3,...,2002\}$. In each step, we remove the numbers whose positions are congruent to $1\bmod 3$. (The initial element is in position $1$.) Which is the last number remaining?

For example, in the list $\{1,2,3,\ldots,10\}$, the different steps are

$$\{\color{red}{1},2,3,\color{red}{4},5,6,\color{red}{7},8,9,\color{red}{10}\} \rightarrow
\{\color{red}{2},3,5,\color{red}{6},8,9\} \rightarrow
\{\color{red}{3},5,8,\color{red}{9}\} \rightarrow
\{\color{red}{5},8\} \rightarrow
\{8\}$$

So the last number remaining is $8$.

I prove that every element $n$, each step moves to position $n-\lceil n/3\rceil$. With Rstudio, I achieve the last number must be $1598$ and the number of steps must be $17$.

How can I prove it?

答:
Here's a sketch (omitting some details):  This process can be thought of as a map $f:\mathbb{N}\rightarrow \mathbb{N}$ sending $x\mapsto 0$ if $x\equiv 1$ mod 3 and $x\mapsto$ (new position) otherwise.  So e.g. $5\mapsto 3$, $3\mapsto 2$, $2\mapsto 1$, etc.

Now think of it in reverse.  Suppose it takes $k$ steps to get to just one number.  We know the last non-zero number (at step $k$) must be 1.  If we run the process in reverse, where does 1 go?  At step $k-1$, the numbers left are $\{1,2\}$, and 1 went to 2.  At the next step, 2 goes to 3, and then 3 goes to 5, etc.

In general, let's consider what happens at step $j$ to a list of numbers $\{1,2,\dots,n\}$.  To go backwards to step $j-1$, we insert a new number before the list and after every two successive numbers, giving something like this:
$$\{*,1,2,*,3,4,*,5,6,*,\dots\}$$
If $n$ is even, then there are $n/2$ new numbers inserted before it, so $n\mapsto 3n/2$ under this reverse process.  If $n$ is odd, then you can convince yourself that $n\mapsto (3n+1)/2$.  These two cases can be stated concisely as $$n\mapsto \lceil 3n/2\rceil$$
This gives a recursive formula for the last number standing.  The values are easy to compute: $$1,2,3,5,8,12,18,27,41,62,93,140,210,315,473,710,1065,1598,2397$$
The last number in this list that is less that 2002 is 1598, and it takes 17 steps for it to get to 1.


问:
Using accumulated value to match present value of future payments
Question:

Ms.Smith has two grand children, Adam and Evelyn. Adam will begin college on 9/1/03 and Evelyn will start college on 9/1/05. Ms.Smith wants both Adam and Evelyn to receive $1,000 at the beginning of each of their 4 years in college.

Ms.Smith will fund these payments by making five level annual deposits of P into an account earning an annual effective interest rate of 7%, with the first deposit on 9/1/1998.

Determine the value of P.

Okay so if she begins payment on 9/1/1998 and makes 5 annual payments, then the last payment is on 9/1/02, which is one year before Adam starts college. Thus, the account would accumulate by (1+i) after the last payment is made since it is earning interest for a whole year from 2002-2003.

And this value should be equal to the present value of all the payments that Adam and Evelyn would receive which is basically deferred annuities of 1000A-angle[6] + 1000A-angle[4] - 1000*A-angle[2]

Note: A-angle refers to annuity-immediate and the number that follows in brackets is the n-term.

So PV = 4766.54 + 3387.21 - 1808.02 = 6,345.73

Since this present value must match the accumulated value of Ms.Smith's account, we have s angle 5 (1+i)
where i=7% which gives:

6.15329074 * P = 6,345.73

P = 1031.27

However, my solution evaluates the accumulated value without the earned interest (1+i)

So s angle 5 = 5.75 thus,

5.75P = 6,345.73

P = 1,103.46

Can someone please explain to me why the textbook does not accumulate the fund by (1+i) even though it is sitting in the account for 1 year?

答:
I'm just going to do it the way I think it should be done.  I'm going to use the time of the last payment into the fund as the valuation time point; i.e., 01 Sep 2002.  Therefore, the payments are an annuity-immediate with accumulated value $$\require{enclose}
P s_{\enclose{actuarial}{5} i} = P(1+i)^4 + P(1+i)^3 + \cdots + P.$$
The fund is then withdrawn according to the following cash flow:
$$\begin{align}
1000v + 1000v^2 + 2000v^3 + 2000v^4 + 1000v^5 + 1000v^6
&= 1000(a_{\enclose{actuarial}{4}i} + v^2a_{\enclose{actuarial}{4}i}) \\
&= 1000(1+v^2)a_{\enclose{actuarial}{4}i}.\end{align}$$  Assuming the interest rate is constant at $i = 0.07$ and $v = 1/(1+i)$, the resulting equation of value yields $$P = \frac{1000(1+v^2)a_{\enclose{actuarial}{4}i}}{s_{\enclose{actuarial}{5}i}} = 1103.46. $$
As it is unclear which answer is yours and which is the textbook's, I will let you sort out the issue.

Note that because there is only one year between the last payment and the first withdrawal, no matter which time point you choose for the equation of value, the annuities involved in the equation must either be both annuities-immediate, or annuities-due.  Put another way, the equation should use $s$ and $a$, or $\ddot s$ and $\ddot a$.  You cannot mix the two.

To illustrate an alternative method of solution, we can choose as the valuation time point 01 Sep 1998, the first payment.  Then the equation of value looks like this:
$$P\ddot a_{\enclose{actuarial}{5} i} = 1000 v^6 \ddot a_{\enclose{actuarial}{4} i} + 1000 v^8 \ddot a_{\enclose{actuarial}{4} i}.$$  The LHS is the present value of the fund, and the RHS is the present value of the withdrawals, appropriately deferred.


问:
How many roots does the equation $a^x -x^n=0$ have?
I want to know if there is a simple method to be able to tell the number of roots of an equation in the form of : $ a^x - x^n = 0 $ (with a>0 and n>0)
If we take $3^x=x$ for example. I know that graphically for example, you can see that there are no roots. But is there another (simple) way to solve it?
What about $5^x=x^2$?

答:
Guess you are supposing $a>0$ and $n$ is a rational number which can be expressed as $\frac{p}{q}$ where $q$ is an odd here or the question will become pretty tricky.

Then consider what will happen when $x>0$.

We can move $x^n$ to the RHS and take logarithm functions on both sides and get $xlog(a)=nlog(x)$.

Hence we have $\frac{log(x)}{x}=\frac{log(a)}{n}$

For $\frac{d}{dx}\frac{log(x)}{x}=\frac{1-log(x)}{x^2}$ we can deduce that the function $\frac{log(x)}{x}$ has its maximum $\frac{1}{e}$ at $x=e$.

Hence, this function has 2 roots when $\frac{log(a)}{n}<\frac{1}{e}$, 1 root when $\frac{log(a)}{n}=\frac{1}{e}$ and no root when $\frac{log(a)}{n}>\frac{1}{e}$ if $x>0$.

Then consider $x<0$ (x cannot be equal to zero).

When $p$ is an odd, $x^n$ is definitely negative and $a^x$ is definitely positive. So $a^x-x^n=0$ has no root when $x<0.

When p is an even, we use the substitution $y=-x$. Hence the equation becomes $a^{-y}-(-y)^n=0$ which can be written as $(\frac{1}{a})^y-y^n=0$. Hence, this function has 2 roots when $-\frac{log(a)}{n}<\frac{1}{e}$, 1 root when $-\frac{log(a)}{n}=\frac{1}{e}$ and no root when $-\frac{log(a)}{n}>\frac{1}{e}$ if $x<0$. (Note that $log(\frac{1}{a})=-log(a)$)

According to the equation you give, $a=3$ and $n=1$ so $\frac{log(a)}{n}=log(3)$ which is greater than $\frac{1}{e}$. Since 3 is an odd, this equation has no root.


问:
Dirac Distribution in Polar Coordinates as a Measure
I have a question.

I was browsing different scripts on statistical mechanics and saw a definition of the "area of accessible phase space at energy $E$ " for a single harmonic oscillator. The author quickly corrected himself, stating, that this is more of a contour in phase space. So I calculated both, got different results and need someone explaining, where my mistake is or what is causing this discrepancy.

For simplicity let's define the Hamiltonian as:
$$
H(q,p) = p^2 + q^2
$$

And the curve $\mathcal{C}$ in phase space $\Gamma=\{(q,p)\in\mathbb{R}^2\}$ is the circle, where $H=E$ with the radius $R=\sqrt{E}$.

Now the odd defintion for the mentioned area / contour given by the mentioned textbook author, where $\delta$ is the Dirac Distribution:
$$
N(E)
=\int\mathrm{d}\Gamma~\delta(H(q,p)-E)
$$

If I remember correctly there is also a tool given by the Dirac $\delta$ to evaluate compositions with a function $f$ which will come in handy:

$$
\delta(f(x))
=\sum\limits_{i=1}^N\frac{\delta(x-x_i)}{|{f'(x_i)}|}
$$

Here $x_i$ denote the roots of $f(x)$.

My calculation would now read as follows:

\begin{align*}
N(E)
&=
\int\limits_{-\infty}^\infty
\mathrm{d}q
\int\limits_{-\infty}^\infty
\mathrm{d}p
~
\delta(p^2+q^2-R^2)
=
\int\limits_{0}^\infty
\mathrm{d}\rho
~
\rho
~
\delta(\rho^2-R^2)
\int\limits_{0}^{2\pi}
\mathrm{d}\varphi \\
&=
2\pi
\frac{1}{2R}
\int\limits_{0}^\infty
\mathrm{d}\rho
~
\rho
~
\delta(\rho-R)
=\pi
\end{align*}

Here I used the know parameterization $(q,p)^T=(\rho\cos\varphi,\rho\sin\varphi)^T$.
It's significant, that I get (if I'm not wrong) the area of a unit circle independent of the energy $E$.

That's completely against my intuition so I try it again with an line integral over $\mathcal{C}=\{(q,p)\in\mathbb{R}^2|q^2+p^2=R^2\}$.

$$
N(E)
=
\int\limits_\mathcal{C}
\mathrm{d}s
=
\int\limits_0^{2\pi}
\mathrm{d}\varphi
\left|
\frac{\partial \vec{s}}{\partial \varphi}
\right|
=
\int\limits_0^{2\pi}
\mathrm{d}\varphi
\left|
\begin{pmatrix}
-R\sin\varphi
\\
R\cos\varphi
\end{pmatrix}
\right|
=
2\pi R
=2\pi \sqrt{E}
$$
This is of course a very simple calculation in this case of a circle.

So now my questions: These two results are different right? If yes, why thinks the author of this textbook, he could define a contour in phase space with the help of this Dirac Distribution and if he didn't intend to do that, what is the use of an "area" defined like this (with a DD)?

答:
As I know little about the physical context that leads to the quantity $N(E)$, let me focus on the mathematical aspect. Note that

$$ \int_{\mathbb{R}^2} \mathrm{d}q\mathrm{d}p \, \delta(q^2+p^2-R^2)
= \left[ \frac{\partial}{\partial \epsilon} \int_{\mathbb{R}^2} \mathrm{d}q\mathrm{d}p \, \mathbf{1}_{\{ q^2 + p^2 - R^2 \leq \epsilon \}}  \right]_{\epsilon = 0}, $$

where $\mathbf{1}_{\{\cdots\}}$ is the indicator function notation, which takes value $1$ if $\cdots$ is true and value $0$ otherwise. On the other hand, the line integral in your second formulation corresponds to

$$ \int_{\mathcal{C}} \mathrm{d}s
= \int_{\mathbb{R}^2} \mathrm{d}q\mathrm{d}p \, \delta(\sqrt{\smash[b]{q^2+p^2}}-R)
= \left[ \frac{\partial}{\partial \epsilon} \int_{\mathbb{R}^2} \mathrm{d}q\mathrm{d}p \, \mathbf{1}_{\{ \sqrt{\smash[b]{q^2 + p^2}} - R \leq \epsilon \}}  \right]_{\epsilon = 0}. $$

In both cases, the integral is supported on the contour $H(q, p) = E$, but the surface measure is weighted in a different way, i.e., $ \mathrm{d}q\mathrm{d}p \, \delta(q^2+p^2-R^2) = \mathrm{d}q\mathrm{d}p \, \frac{1}{2R} \delta(\sqrt{\smash[b]{q^2 + p^2}}-R) $. Since this "energy shell" as a hypersurface in the phase space does not alone encode the information about the surface measure, it should be specified by other means. I guess that the definition

$$ N(E)
= \int \mathrm{d}\Gamma \, \delta(H(\Gamma) - E)
= \left[ \frac{\partial}{\partial \epsilon} \int \mathrm{d}\Gamma \, \mathbf{1}_{\{ H(\Gamma) \leq E + \epsilon \}}  \right]_{\epsilon = 0} $$

exactly serves this purpose, and perhaps the author's notion of "area" conforms to this choice.


问:
Linear Algebra determinant and rank relation
True or False?

If the determinant of a $4 \times 4$ matrix $A$ is $4$
   then its rank must be $4$.

Is it false or true?

My guess is true, because the matrix $A$ is invertible.

But there is any counter-example?

Please help me.

答:
You're absolutely correct. The point of mathematical proof is that you don't need to go looking for counterexamples once you've found the proof. Beforehand that's very reasonable, but once you're done you're done.

Determinant 4 is nonzero $\implies$ invertible $\implies$ full rank.

Each of these is a standard proposition in linear algebra.


问:
Hint for Spivak's Calculus Problem 2-4-a
(note: this is very similar to a related question but as I'm trying to solve it without looking at the answer yet, I hope the gods may humor me anyways)

I'm self-learning math, and an answer to an /r/math post about self-learning was finally enough to motivate me to try getting feedback on this site :) . After reading throughout the internet, I've decided to start with Spivak's Calculus. I'm loving the book thankfully, but I'm stuck on this problem and I don't want to look at the answer quite yet.

4 . (a) Prove that
$$\sum_{k=0}^l \binom{n}{k} \binom{m}{l-k} = \binom{n+m}{l}$$.
Hint: Apply the binomal theorem to $(1+x)^n(1+x)^m$ .

I've done all the prior problems, including Problem 3 (proving the Binomial Theorem) which is obviously closely tied to this, but even with the hint it feels like there's too much I don't know. I applied the hint and found:

$$\left(\sum_{i=0}^n \binom{n}{i} x^i\right)\left(\sum_{j=0}^m \binom{m}{j} x^j\right) = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k$$

And, setting $x=1$ got something even more interesting:

$$\left(\sum_{i=0}^n \binom{n}{i}\right)\left(\sum_{j=0}^m \binom{m}{j}\right) = \sum_{k=0}^{n+m} \binom{n+m}{k}$$

However, I don't know where to go after that...is there some property of multiplying sums that I need to prove first, to relate the multiplication of two sums of binomials with the sum of the multiplication of two binomials?

Thank you!

答:
Hint:

Think about the sum as the coefficient of $x^l$ in the expansion ${(1+x)}^n{(x+1)}^m$


问:
Proofs verification and question about equivalence relation v. partition
On page 3 / Proposition 2 of Dummit&Foote's Abstract Algebra (3rd ed.):

Let $A$ be a nonempty set.

(1) If $\sim$ defines an equivalence relation on $A$ then the set of equivalence classes of $\sim$ form a partition of $A$.

(2) If $\{A_i \mid i\in I\}$ is a partition of $A$ then there is an equivalence relation on $A$ whose equivalence classes are precisely the sets $A_i,i\in I$.

According to the book, those two statements show that the notions of an equivalence relation on $A$ and a partition on $A$ are the same. And the proofs are omitted by the book.

Below are my proofs and my questions related to this proposition:

Proof for (1):

$A$ is nonempty in the first place. Since equivalence classes are defined upon elements in the set, if we let $E=\{E_\alpha \mid \alpha\in A\}$ where $E_\alpha$ is the equivalence class of $\alpha\in A$, then $E$ shall catch all the equivalence classes of $\sim$ on $A$. And since every element in $E$ is a nonempty set (each contains its $\alpha$), all the equivalence classes are nonempty.

For now, let $I$ denote the injective-indexing set for the equivalence classes of $\sim$ on $A$ so that each equivalence class is uniquely denoted as $E_i$ for some $i\in I$.

If $A\neq \bigcup_{i\in I}E_i$, then there exists $a\in A$ such that $a\notin \bigcup_{i\in I}E_i$ or the other way around.

For the former, let's construct a set, called $E_c$, such that for each $x\in A$, $x\in E_c$ iff $x\sim a$. $E_c$ is then an equivalence class of $\sim$ on $A$. Then, $E_c=E_i$ for some $i\in I$. Thus, $a\in \bigcup_{i\in I}E_i$.

For the latter, we have $a\in E_i$ for some $i\in I$. By definition we then have $a\in A$.

Hence, $A= \bigcup_{i\in I}E_i$ by contradiction.

If there exist $i,j\in I$ with $i\neq j$ such that $E_i\cap E_j\neq \emptyset$, let $E_i\cap E_j=B$, then there exists $b\in B$. Let a representative of $E_i$ be $x$ and a representative of $E_j$ be $y$, we then have $b\sim x$ and $b\sim y$. Then we have $x\sim b$ and thus $x\sim y$. By definition, it follows that all elements in $E_i$ are in $E_j$ and vice versa (i.e. $E_i=E_j$). By injection, $i=j$. Hence, for all $i,j\in I$ with $i\neq j$, we have $E_i\cap E_j=\emptyset$ by contradiction.

Thus, the equivalence classes of $\sim$ on $A$ form a partition of $A$.

Proof for (2):

For personal preference, I won't use the index notation in this proof.

Given a partition of $A$, say $P$, let's define a binary relation $\sim$ by constructing a subset $R$ of $A\times A$ as follows:

$(a,b)\in R$ iff $a,b\in S$ for some $S\in P$.

Since for all $a\in A$, $a\in S$ for some $S\in P$, let $b=a$, we then have $(a,a)\in R$. Thus, $\sim$ is reflexive.

For all $a,b\in A$, if $(a,b)\in R$, then $a,b\in S$ for some $S\in P$. Then $b,a\in S$. Then $(b,a)\in R$. Thus, $\sim$ is symmetric.

For all $a,b,c\in A$, if $(a,b),(b,c)\in R$, then $a,b,c\in S$ for some $S\in P$. In particular, $a,c\in S$. Then $(a,c)\in R$. Thus, $\sim$ is transitive.

Hence, $\sim$ is an equivalence relation.

For each $a\in A$, let $E_a$ be the equivalence class of $a$. Then $E_a$ precisely contains all the $x\in A$ such that $(x,a)\in R$. Let $(x_1,a),(x_2,a)\in R$, we then have $x_1,a\in S_1$ and $x_2,a\in S_2$ where $S_1,S_2\in P$. If $S_1\neq S_2$, then $S_1$ and $S_2$ are disjoint. Since $a$ is in both, we must have $S_1=S_2$ by contradiction. Thus, all elements in $E_a$ must be in the same $S\in P$. That is, $E_a\subseteq S$ for some $S\in P$. Now, Assume this $S$ contains an element $b$ such that $b\notin E_a$. Since $S$ is a subset of $A$, we have $b\in A$. Since $b,a\in S$, we have $(b,a)\in R$. Together we have $b\in E_a$, which contradicts. Thus, $E_a=S$. Hence, the set of equivalence classes is a subset of the partition.

For each $S\in P$, since $S$ is nonempty, we can always fix an element $t\in S$. Clearly $t\in A$, thus there exists a set $E_t$ which is the equivalence class of $t$. For any $r\in S$, since $r,t\in S$, we have $(r,t)\in R$ and thus $r\sim t$. Since $r\in A$, we have $r\in E_t$. Thus, every element in $S$ is in $E_t$. If $E_t$ contains an element $\omega$ such that $\omega\notin S$, we have $\omega\sim t$ and thus $\omega,t\in S_3\in P$. In particular, $t\in S_3$. Then we must have $S_3=S$. Then $\omega\in S$. By contradiction, $S=E_t$. Hence, the partition is a subset of the set of equivalence classes.

This concludes that the set of equivalence classes is the partition.

Could someones please review my proofs to see if there are errors in any kind? Many thanks!

My question:

I think by the two statements in proposition 2, it is not sufficient to show that the notions of an equivalence relation on $A$ and a partition on $A$ are the same.

First, let me define how I understand "two notions are the same": say I have two notions $A$ and $B$, as long as I get an instance of $A$, I can transform it to an instance of $B$. And if now I give you this instance of $B$ without showing you what my instance of $A$ is, then you should always know what's in my hand by transforming the instance of $B$ to an instance of $A$.

The thing is, I think proposition 2 does not guarantee that when you transform the instance of $B$ back to $A$, you would always get the exact same instance of $A$ as what in my hand. Say I have an equivalence relation, then I can transform it into a partition [(1)], and if I show you the partition you could find an equivalence relation [(2)] that transforms into the partition in the same manner. However, to me, this is like a mapping: (1) guarantees that each $\sim$ can be mapped to a partition, (2) guarantees that the mapping is surjective, but we still need to prove that such mapping is injective for the two notions to be the same.

That is, we still need to prove that no two different equivalence relations could give the same partition.

May I ask if my concern is necessary?

答:
Your proof of (1) isn’t wrong, but you’re making things much harder than necessary. There is no reason to index $E$ injectively. For each $\alpha\in A$ we have $\alpha\in E_\alpha\in E$, so $\alpha\in\bigcup E$, and since $\alpha$ was an arbitrary element of $A$, it follows that $A\subseteq\bigcup E$. On the other hand, $E_\alpha\subseteq A$ for each $\alpha\in A$, so $\bigcup E\subseteq A$, and therefore $\bigcup E=A$.

Now suppose that $E_\alpha,E_\beta\in E$ are such that $E_\alpha\cap E_\beta\ne\varnothing$. Let $a\in E_\alpha\cap E_\beta$; then $a\sim\alpha$ and $a\sim\beta$. Let $x\in E_\alpha$; then $x\sim\alpha\sim a\sim\beta$, so $x\sim\beta$, and therefore $x\in E_\beta$. This shows that $E_\alpha\subseteq E_\beta$, and a similar argument shows that $E_\beta\subseteq E_\alpha$ and hence that $E_\alpha=E_\beta$. Thus, $E$ is a partition of $A$.

The proof of (2) is better in that respect, but one needn’t work quite that hard to show that the $\sim$-equivalence classes are the parts of the original partition. Fix $a\in A$; $P$ is a partition of $A$, so there is a unique $S_a\in P$ such that $a\in S_a$. Now let $x\in A$: clearly $x\in E_a$ iff $x\sim a$ iff there is an $S\in P$ such that $\{x,a\}\subseteq S$ iff $x\in S_a$, since $S_a$ is the only member of $P$ containing $a$. But $x\in E_a$ iff $x\in S_a$ simply means that $E_a=S_a$. That is, for each $a\in A$ the $\sim$-class of $a$ is identical to the unique member of $P$ containing $a$, so the partition of $A$ into $\sim$-classes (which we know from (1) is a partition) is identical to the partition $P$.

Your understanding of the sense in which the notions of equivalence relation on $A$ and partition of $A$ are ‘the same’ is correct. The demonstration of equivalence isn’t actually complete, because (1) doesn’t go quite far enough: now that we have (2), we should show that if we start with an equivalence relation $\sim$ on $A$, produce the associated partition as in (1), and then use (2) to produce an associated equivalence relation, we get back the original equivalence relation $\sim$. However, this is trivial, since two equivalence relations that have the same equivalence classes are easily seen to be the same relation.

This old answer of mine is another discussion of these matters.


问:
Deducing a coefficient from a cubic polynomial?
I fully answered the question, and got that $k=-3$, but the answer says it's positive. Can anyone show me my mistake?

"Given that $x-2$ is a factor of the polynomial $x^3 - kx^2 - 24x + 28$, find $k$ and the roots of this polynomial."

Using factor theorem, I realised that $P(2)$ is equivalent to $0$, therefore $2^3 - 2^2k - 24(2) + 28 = 0$

I solved it algebraically and got that $k=-3$, but the answers say it was $k=3$. Did I make a simple error?

Any help would be appreciated.

答:
No error on your part:

If you've copied the problem correctly, then your solution is correct: $k = -3$.

I was very careful in calculating, as I'm sure you were, in double checking, so if $(x - 2)$ is a factor for your given polynomial, then $k$ must be $-3$.

Typo/misprint I suspect, in your text: a typo in the solution, or a misprint of the desired polynomial.

E.g. If the polynomial had been

$$x^3 \color{blue}{\large \bf +} 4k^2 - 24x + 28\quad \text{and}\;\; (x-2) \;\text{is a factor}$$

then $k = 3$.


问:
Can't find the inverse of Laplace Transform.
Find $\mathcal{L}^{-1}{(F(s))}$, if given $F(s)=\dfrac{2}{s(s^2+4)}$.

I have tried as below.

To find inverse of Laplace transform, I want to make partial fraction as below.
\begin{align*}
\dfrac{2}{s(s^2+4)}=\dfrac{A}{s}+\dfrac{Bs+C}{s^2+4}=\dfrac{(A+B)s^2+Cs+4A}{s(s^2+4)}.
\end{align*}
After that, we have system of linear equation
\begin{align*}
A+B&=0\\
C&=0\\
4A&=2.
\end{align*}
Thus we have $A=2$, $B=-2$, and $C=0$.
Now, substituting $A, B, C$ and we have
\begin{align*}
\dfrac{2}{s(s^2+4)}=\dfrac{2}{s}+\dfrac{-2s}{s^2+4}.
\end{align*}
But the fact is
\begin{align*}
\dfrac{2}{s(s^2+4)}\neq \dfrac{2}{s}+\dfrac{-2s}{s^2+4} = \dfrac{8}{s^2+4}.
\end{align*}

I'm stuck here. I can't make a partial fraction for $F(s)$ and I can't find inverse of Laplace transform for $F(s)$.

Anyone can give me hint to give me hint for this problem?

Body:
Note that $$\frac{2}{s(s^{2}+4)}=\frac{1}{2s}-\frac{s}{2(s^{2}+4)}$$
So, you have $$\mathcal{L}^{-1}\left\{\frac{2}{s(s^{2}+2)} \right\}=\frac{1}{2}\mathcal{L}^{-1}\left\{\frac{1}{s} \right\}-\frac{1}{2}\mathcal{L}^{-1}\left\{\frac{s}{s^{2}+4} \right\}$$
Finally, you can use $$\mathcal{L}^{-1}\left\{ \frac{s}{s^{2}+4}\right\}=\cos(2t)$$
and $$\mathcal{L}^{-1}\left\{ \frac{1}{s}\right\}=1$$


答:
You made a mistake because $A \neq 2$

$4A=2$ then $A=\frac{1}{2}$

Title: Can't find the inverse of Laplace Transform.

Body:
Find $\mathcal{L}^{-1}{(F(s))}$, if given $F(s)=\dfrac{2}{s(s^2+4)}$.

I have tried as below.

To find inverse of Laplace transform, I want to make partial fraction as below.
\begin{align*}
\dfrac{2}{s(s^2+4)}=\dfrac{A}{s}+\dfrac{Bs+C}{s^2+4}=\dfrac{(A+B)s^2+Cs+4A}{s(s^2+4)}.
\end{align*}
After that, we have system of linear equation
\begin{align*}
A+B&=0\\
C&=0\\
4A&=2.
\end{align*}
Thus we have $A=2$, $B=-2$, and $C=0$.
Now, substituting $A, B, C$ and we have
\begin{align*}
\dfrac{2}{s(s^2+4)}=\dfrac{2}{s}+\dfrac{-2s}{s^2+4}.
\end{align*}
But the fact is
\begin{align*}
\dfrac{2}{s(s^2+4)}\neq \dfrac{2}{s}+\dfrac{-2s}{s^2+4} = \dfrac{8}{s^2+4}.
\end{align*}

I'm stuck here. I can't make a partial fraction for $F(s)$ and I can't find inverse of Laplace transform for $F(s)$.

Anyone can give me hint to give me hint for this problem?


答:
Firstly Take 3 vowels (1 way) & take 2 consonants out of 6 consonants => 6C2 = 1 * 15 = 15 ways to select required 5 letters.

Secondly, the number of ways to arrange those 5 letters is 54321 = 120

The answer is 120 * 15 = 1800


答:
Instead of setting $x=1$ in
$$
\left(\sum_{i=0}^n \binom{n}{i} x^i\right)\left(\sum_{j=0}^m \binom{m}{j} x^j\right) = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k,
$$
expand the LHS and collect terms of degree $k$ together to obtain
$$
\sum_{k=0}^{n+m} \Bigg( \sum_{\substack{0\le i\le n\\0\le j\le m\\i+j=k}} \binom{n}{i} \binom{m}{j}\Bigg) x^k = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k.
$$
Since $i+j=k$ determines $j=k-i$ this yields
$$
\sum_{k=0}^{n+m} \Bigg( \sum_{i=0}^k \binom{n}{i} \binom{m}{k-i}\Bigg) x^k = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k.
$$
Comparing coefficients yields the desired identity.

Title: Hint for Spivak's Calculus Problem 2-4-a

Body:
(note: this is very similar to a related question but as I'm trying to solve it without looking at the answer yet, I hope the gods may humor me anyways)

I'm self-learning math, and an answer to an /r/math post about self-learning was finally enough to motivate me to try getting feedback on this site :) . After reading throughout the internet, I've decided to start with Spivak's Calculus. I'm loving the book thankfully, but I'm stuck on this problem and I don't want to look at the answer quite yet.

4 . (a) Prove that
$$\sum_{k=0}^l \binom{n}{k} \binom{m}{l-k} = \binom{n+m}{l}$$.
Hint: Apply the binomal theorem to $(1+x)^n(1+x)^m$ .

I've done all the prior problems, including Problem 3 (proving the Binomial Theorem) which is obviously closely tied to this, but even with the hint it feels like there's too much I don't know. I applied the hint and found:

$$\left(\sum_{i=0}^n \binom{n}{i} x^i\right)\left(\sum_{j=0}^m \binom{m}{j} x^j\right) = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k$$

And, setting $x=1$ got something even more interesting:

$$\left(\sum_{i=0}^n \binom{n}{i}\right)\left(\sum_{j=0}^m \binom{m}{j}\right) = \sum_{k=0}^{n+m} \binom{n+m}{k}$$

However, I don't know where to go after that...is there some property of multiplying sums that I need to prove first, to relate the multiplication of two sums of binomials with the sum of the multiplication of two binomials?

Thank you!


问:
What does $2|z+3i|=|z-i|$ represent?

$z$ is a complex number. What does $2|z+3i|=|z-i|$ represent?

Putting $z=x+iy$, I get a circle with radius $\frac83$. But the equation of circle is $|z-z_1|=r$.

How to intuitively see that $2|z+3i|=|z-i|$ represents a circle?

答:
That is the Circles of Apollonius. In fact, if $|z - a| = k |z - b|$, we have
$$
\left |z - \frac{a - b k^2}{1 - k^2}\right | = \frac{k|a - b|}{|1 - k^2|},
$$
Thus it is a circle. See this post.

Title: What does $2|z+3i|=|z-i|$ represent?


问:
Power when exponent is i
Suppose we compute $z=3^i$.
$$3^i=(3*1)^i=3^i*(e^{2\pi ni})^i=3^i*e^{-2\pi n}\Rightarrow 3^i*(1-e^{-2\pi n})=0\Rightarrow 3^i=0,\textrm{ if }n\neq 0$$

So is $z=0$ a solution too ?

答:
The correct way: Write $$3^{i}=e^{\ln 3^i}= e^{i \ln 3}= \cos(\ln 3)+i \sin (\ln 3).$$

In your step $3^{i}$ remains as it is, you have made no progress.

Title: Power when exponent is i


问:
Countour Integration
I had to evaluate this integral $ \int_{-\infty}^{\infty} \frac{x \sin(x)}{x^2 - b^2}dx  $, according to Wolfram this had the following result $$ \int_{-\infty}^{\infty} \frac{x \sin(x)}{x^2 - b^2}  dx= \pi e^{ib}$$

However, when I integrated it:

$$\oint_C f(z) dz = \oint_{\gamma_{1}} f(z) dz + \oint_{\gamma_{2}} f(z) dz + \oint_{\Gamma} f(z) dz $$

The last term goes to zero applying Jordan's Lemma and to get the result of the integral over $\gamma_{1}$ and $\gamma_{2}$, in which $\gamma_{1}$ is the contour over the first pole and $\gamma_{2}$ is the contour over the second pole.

So, I get  $$\oint_{\gamma_{1}} f(z) dz = i \pi  \lim_{z \rightarrow -b} \frac{z \sin(z)}{(z-b)(z+b)}(z+b) =  i \pi \frac{\sin(-b)}{-2} $$

and

$$\oint_{\gamma_{2}} f(z) dz = i \pi  \lim_{z \rightarrow b} \frac{z \sin(z)}{(z-b)(z+b)}(z-b) =  i \pi \frac{\sin(b)}{2} $$

The result that I'm getting is:

$$\oint_C f(z) dz = i \pi \left( \frac{\sin(b)}{2} - \frac{\sin(-b)}{-2} \right) = \frac{\pi}{2} \left( e^{ib} - e^{-ib} \right)$$

That according to Wolfram is not the correct result.

What am I doing wrong?

Edit:

In this case, I used $$f(z) = \frac{z \sin(z)}{z^2 - b^2}$$

Edit 2:

Edit 3:

Using @Ted Shifrin 's suggestion to change $f(z)$ to a more well-behaved function when $|z|$ is very large, I got:

$f(z) = \frac{z \exp(iz)}{z^2 - b^2}$

So, I get  $$\oint_{\gamma_{1}} f(z) dz = i \pi  \lim_{z \rightarrow -b} \frac{z \exp(iz)}{(z-b)(z+b)}(z+b) =  i \pi \frac{\exp(-ib)}{2} $$

and

$$\oint_{\gamma_{2}} f(z) dz = i \pi  \lim_{z \rightarrow b} \frac{z \exp(iz)}{(z-b)(z+b)}(z-b) =  i \pi \frac{\exp(ib)}{2} $$

The result that I'm getting is:

$$\oint_C f(z) dz = i \pi \left( \frac{\exp(ib)}{2} + \frac{\exp(-ib)}{2} \right) = i \pi \cos(b)$$

and in conclusion:

$$\int_{-\infty}^{\infty} \frac{x \sin(x)}{x^2 - b^2}dx = \Im(i \pi cos(b)) = \pi \cos(b) $$

I still don't understand, according to some of you this integral doesn't converge how do I prove that?

This result is equal to @FelixMarin 's result and he used another method of integration, but is different of Wolfram's result and assuming that $b \in \Re $ this integral is supposed to diverge. What am I missing?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\left.\mrm{P.V.}\int_{-\infty}^{\infty}
{x\sin\pars{x} \over x^{2} - b^{2}}\,\dd x
\,\right\vert_{\,b\ \in\ \mathbb{R}}}
\\[5mm] = &\
{1 \over 2}\,\mrm{P.V.}\int_{-\infty}^{\infty}
{\sin\pars{x} \over x + b}\,\dd x +
{1 \over 2}\,\mrm{P.V.}\int_{-\infty}^{\infty}
{\sin\pars{x} \over x - b}\,\dd x
\\[5mm] = &\
{1 \over 2}\,\mrm{P.V.}\int_{-\infty}^{\infty}
{\sin\pars{x - b} + \sin\pars{x + b} \over x}\,\dd x
\\[5mm] = &\
{1 \over 2}\,\mrm{P.V.}\int_{-\infty}^{\infty}
{2\sin\pars{x}\cos\pars{b} \over x}\,\dd x =
\bbx{\pi\cos\pars{b}} \\ &
\end{align}

Title: Countour Integration


问:
general solution to non linear ordinary differential equation
Hi all I am building a computer program which needs a general solution to the following differential equation.

$$1 - \frac{dz}{dy} = (z/y + 1)^r$$

In this equation $r$ is a positive constant, $z > 0$, and $y > 0$.

I have tried solving this as a homogenous differential equation where I set $u = z/y$ doing this results in the following equation
$$y = C * e^{-\int_1^{\frac{z}{y}}\limits{\frac{du}{(u+1)^r + u - 1}}}$$

I know that I can approximate a result though due to computational limits of my computer program this is impractical and a general solution would be much better. I have tried feeding this equation to python sympy, wolfram mathematica, as well as matlab and none of them were able to find a solution. If anyone has any insight into this problem I would really appreciate hearing your feedback.

答:
Starting from @Joshua Wang's answer (now deleted)
$$\frac{dz}{z} = \frac{du}{1 - u - (u + 1)^{r}}$$ let $u=v-1$ to make
$$\frac{dz}{z} =- \frac{dv}{v^r+v-2}$$ If $r$ is an integer (otherwise, I do not think about a possible solution), we can write
$$\frac{dz}{z} =- \frac{dv}{\prod_{k=1}^r(v-s_k)}$$ where the $s_k$'s are the roots of the polynomial $v^r+v-2=0$.

Partial fraction decomposition
$$\frac{dz}{z} =- \sum_{k=1}^r\frac{a_k}{v-s_k}\,dv$$

$$\log(|z|)+C=- \sum_{k=1}^r a_k \log(|v-s_k|)$$ and we shall probably end with a sum of logarithms and arctangents.

For $r=3$, the rhs would be
$$-\frac{3 }{4 \sqrt{7}}\tan ^{-1}\left(\frac{2 v+1}{\sqrt{7}}\right)+\frac 14\log \left(\frac{(v-1)^2}{v^2+v+2}\right)$$


问:
Are we able to "combine" two analytic continuation to get another?
Say, I want to find the analytic continuation of the function $f(z)$ defined on $D=\{a:-\pi/5<arg(a)<\pi/5\}$ to the slit plane $\mathbb{C}-(-\infty,0]$. I assume that $f(z)$ is real on the boundary of $D$ and has analytic continuation on $\bar D$. I can shift $f(z)$ to $f(ze^{-\pi/5})$ and do the Schwarz reflection principle. I can do this twice, shift it back, and restrict the domain to get the analytic continuation on the upper half plane. I can do the same thing on the lower half plane to get the analytic continuation on the lower half plane. But I don't know if I'm able to combine these two analytic continuations to get the analytic continuation on the whole slit plane. Can anybody give me any tips?..

答:
If $f$ is analytic for $arg(z)\in (a,b),z\ne 0$ continuous on $arg(z) \in [a,b],z\ne 0$ and real on $arg(z)=b,z\ne 0$ then the Schwartz reflection along $arg(z)=b$ gives that $f$ extends analytically to $arg(z)\in (a,b+(b-a))$, and it is continuous on $arg(z)\in [a,b+(b-a)]$ and real on $arg(z)=2b-a$. So we can repeat and extend more and more the domain.

If $2b-2a>2\pi$ then $f(e^{2i\pi}z)$ doesn't have to be the same as $f(z)$, our analytic continuation may be "multi-valued" (try with $f(z)=z^{\pi},a=0,b=2$) that's the point of your branch cut on $(-\infty,0]$.


问:
Pearson's correlation and common variance
Does a coefficent of determination $r^2$ for the Pearson correlation coefficient $r$ exist if we are only interested in the relationship between 2 variables but not in regresssion?

In case $r^2$ is only defined in the context of regression then we first have to define a regressor, check for equal variance of residuals, and normally distributed residuals?

My question was motivated by material from a statistical course that discusses $r$ and $r^2$ without referencing regression. Translation of the source:

The coefficient of determination can be calculated from the correlation by squaring:
Coefficient of determination = $r^2$

For the example this results in:
$r^2 = .628^2 = .394$
If this value is multiplied by 100, the result is a percentage. This indicates what proportion of the variance in both variables is determined by common variance sources. For the present example, the share of the common variance is 39.4%.

It is unclear for me how the common variance shall be derived without a regression line.

答:
Consider two random variables $Y$ and $X$. Assume their correlation is $\rho$. Of course $\rho^2$ is always defined. However, consider a regression

$$Y=a+bX+\epsilon,$$

where $\epsilon$ is white noise. Here you can also define the regression the other way setting $X$ as the dependent variable.

Assume you estimate the parameters $(\hat{a},\hat{b})$ using OLS. Then you obtain the projection

$$\hat{Y}=\hat{a}+\hat{b}X.$$

This gives the linear relationship that best (lowest mean square distance) describes the dependency of $Y$ on $X$. You can essentially always run this regression and it is well defined assuming the relevant moments exist. Note that this does not require normality of residuals or the Gauss-Markov assumptions, which only imply that the regression has some additional "nice" properties. Of course this regression might not generally be the optimal way to model the dependency between $X$ and $Y$ nor OLS might not be the optimal way to estimate it, but in this case neither is the correlation coefficient an optimal measure of this dependency.

The coefficient of determination is

$$R^2\equiv\frac{Var(\hat{Y})}{Var(Y)}.$$

Now it turns out that $R^2=\rho^2$.

$R^2$ is defined in the context of a regression/projection. This tells you how much of the variance of $Y$ the projection explains.  Applying the concept in some other context would be either misusing or extending the original meaning.


问:
Prove $1+\sum_{k=1}^{p} \frac{(-1)^k.n(n-1)(n-2)\cdots(n-2k+1)}{2^k.k!.(2n-1)(2n-3)\cdots(2n-2k+1)}=\frac{2^n(n!)^2}{(2n)!}$
Prove that $$
a_n\bigg[1-\frac{n(n-1)}{2(2n-1)}+\frac{n(n-1)(n-2)(n-3)}{2\cdot4\cdot(2n-1)(2n-3)}-\cdots+\frac{n(n-1)(n-2)\cdots(n-2k+1)}{2\cdot4\cdots 2k\cdot(2n-1)(2n-3)\cdots(2n-2k+1)}\bigg]=1\\
\implies a_n=\frac{1}{1+\sum_{k=1}^{p} \frac{(-1)^k.n(n-1)(n-2)\cdots(n-2k+1)}{2^k\cdot k!\cdot(2n-1)(2n-3)\cdots(2n-2k+1)}}=\frac{(2n-1)(2n-3)\cdots 3\cdot1}{n!}=\frac{(2n)!}{2^n(n!)^2}
$$
where $p=n/2$ if $n$ is even.

This actually comes in the derivation of the Legendre's polynomials as the solutions of the Legendre's differential equation, when we try to do the standardization.

We can verify it for each $n$ but is it possible to directly evaluate the series to the required form ?

答:
First let's rewrite everything using factorials. Since $1\cdot 3\cdots(2n-1)=(2n)!/(2^n n!)$ (including $n=0$ if we interpret the "product of zero terms" as $1$ on the LHS), the general term is $$\frac{(-1)^k}{2^k k!}\frac{n!}{(n-2k)!}\frac{2^n n!}{(2n)!}\frac{(2n-2k)!}{2^{n-k}(n-k)!}=\frac{n!^2}{(2n)!}(-1)^k\binom{n}{k}\binom{2n-2k}{n}$$ (again, the "$1+{}$" part of the summation may be viewed as including the term with $k=0$).

And the key to a solution is indeed related to the Legendre polynomials, defined here by $$P_n(x)=\frac{1}{2^n n!}\frac{d^n}{dx^n}(x^2-1)^n$$ (Rodrigues' formula). If we expand $(x^2-1)^n$ and take the derivative termwise, we get $$P_n(x)=\frac{1}{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}\binom{2n-2k}{n}x^{n-2k},$$ so that the sum we're computing is equal to $\frac{2^n n!^2}{(2n)!}P_n(1)$.

Finally, $P_n(1)=1$, which is shown using $(x^2-1)^n=(x-1)^n(x+1)^n$ and Leibniz rule.


问:
How to integrate e to the power to the power?
How should I integrate?

$\int_0^\infty e^{-x^{1/3}}dx$

I think this is a simple question for the experts.  But a bit hard to tell Google what I want.  So, thanks for your help!  :)

And this looks terrifying to me.

答: Hint: Try the change of variable $u=x^{\frac{1}{3}}$ (so $dx=3u^2du$).


问:
Construct cut off function in the product form $\rho_1\otimes \rho_2\otimes...\otimes \rho_m$
Let $A_i\subset \Bbb{R}^n$,for $1\le i\le m$ are closed subset $\Bbb{R}^n$.Let $A_i^\epsilon$ denote the set $A_i+\overline{B_\epsilon(0)}$ which is also closed.Let $K$ be some closed subset of $\Bbb{R}^n$,Further suppose $A_1^\epsilon\times....\times A_m^\epsilon\bigcap K$ and $A_1\times....\times A_m\bigcap K$ are compact.

Prove there exist some cut-off function in the form $\rho_1(x_1)\otimes \rho_2(x_2)\otimes...\otimes \rho_m(x_m) = \rho(x_1,...,x_m)$ where $x_i\in \Bbb{R}^n$ such that $\rho_i \in C_c^\infty(\Bbb{R})$ satisfy the condition:

$\rho$ is supported on $A_1^\epsilon\times....\times A_m^\epsilon\bigcap K$ this compact set and constant 1 on neiborhood of $A_1\times....\times A_m\bigcap K$

Moreover can we choose each $\rho_i$ that are constant 1 on neiborhood of $A_i$ for each $i$

It seems no matter which $K$ is choosen the cut off function is not possible?

I found this question in F. G. Friedlander, M. Joshi  Introduction to the Theory of Distributions page 56 as below:

答:
Seems to be impossible without restrictions on $K$ (or a clarification what $\otimes$ is doing). Let
$$
K:= \{ (x_1,x_2) : \ |x_1+x_2| \le 1, \ |x_1| \le 1\}.
$$
Let now $\phi_1,\phi_2$ be smooth cut-off functions with values in $[0,1]$. In order that
$\phi(x_1,x_2) =\phi_1(x_1)\phi_2(x_2)=1$ for $(1,0)$, $(0,1)$ it is necessary that
$$
\phi_1(0)=\phi_1(1)=\phi_2(0)=\phi_2(1)=1,
$$
which implies $\phi(1,1)=1$. But $(1,1)\not\in K$. Hence every cut-off function of the product form that is equal to one on $K$ has support much larger than $K$.


问:
Finding the flaw in $-12=(-12)^1=(-12)^{6/6}=\cdots=12$
I was tutoring today and came across a question that I was incapable of explaining sufficiently. It stems from the following false string of equalities...

$$-12=(-12)^1=(-12)^\frac{6}{6} =((-12)^6)^\frac{1}{6} =((-1)^6 12^6)^\frac{1}{6}=(12^6)^\frac{1}{6}=12$$

If anyone can explain the false step (Im assuming the third “=“), why this string of equalities is wrong, and how to avoid making it that would be most helpful for when I need to explain it again.

答:
$-12=(-12)^1=(-12)^\frac{6}{6} =((-12)^6)^\frac{1}{6}$

Last step above is wrong, because of the arbitrary convention that a positive number (i.e. $[(-12)^6]$) when taken to the $(1/6)$ power is equal to a positive number, not a negative number.


问:
How many roots does the equation $a^x -x^n=0$ have?
I want to know if there is a simple method to be able to tell the number of roots of an equation in the form of : $ a^x - x^n = 0 $ (with a>0 and n>0)
If we take $3^x=x$ for example. I know that graphically for example, you can see that there are no roots. But is there another (simple) way to solve it?
What about $5^x=x^2$?

答:
$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$

I want to know if there is a simple method to be able to tell the
number of roots of an equation in the form of : \begin{align} a^x -
> x^n &= 0 \tag{1} \end{align}

This is exactly what
the Lambert $\W$ function
(also known as ProductLog)
was introduced for.

\begin{align}
a^x &= x^n
,\\
x\ln a &= n\ln x
,\\
\tfrac1x\ln x
&=
\tfrac1n\ln a
,\\
\tfrac1x\ln \tfrac1x
&=
-\tfrac1n\ln a
,\\
\ln \tfrac1x
\exp(\ln \tfrac1x)
&=
-\tfrac1n\ln a
,\\
\W(\ln \tfrac1x
\exp(\ln \tfrac1x))
&=
\W(-\tfrac1n\ln a)
,\\
\ln \tfrac1x
&=
\W(-\tfrac1n\ln a)
,\\
\tfrac1x
&=
\exp\Big(\W(-\tfrac1n\ln a)\Big)
,\\
\tfrac1x
&=
-\frac{\tfrac1n\ln a}{\W(-\tfrac1n\ln a)}
,\\
x
&=
-\frac{\W(-\tfrac1n\ln a)}{\tfrac1n\ln a}
\tag{2}
.
\end{align}

The number of real roots of (1) is defined by the value of
$z=-\tfrac1n\ln a$,
the argument of $\W$ in (2):

\begin{align}
\begin{cases}
z\ge0:\quad &\text{ one real solution, }
\quad
x=\frac{\Wp(z)}z
,\quad
x\ge0
,\\
z\in(-\tfrac1\e,0):
\quad &\text{ two real solutions, }
\quad
x_0=\frac{\Wp(z)}z
,\ x_1=\frac{\Wm(z)}z,\quad 0<x_0<x_1
,\\
z=-\tfrac1\e:
\quad &\text{ one real solution, }
\quad
x=
\frac{\Wp(-\tfrac1\e)}{-\tfrac1\e}
=
\frac{\Wm(-\tfrac1\e)}{-\tfrac1\e}
=\e
,\\
z<-\tfrac1\e:\quad &\text{ no real solutions}
,
\end{cases}
\end{align}

where $\Wp$ is the principal branch
and $\Wm$ is the other real branch
of the Lambert $\W$ function.

For example, for $a=3$, $n=1$,
the solution of
\begin{align}
3^x&=x
\end{align}

must be

\begin{align}
x
&=
-\frac{\W(-\ln 3)}{\ln 3}
,
\end{align}

the argument of $\W$ in this case is
\begin{align}
z&=-\ln 3\approx -1.0986
<-\tfrac1\e
,
\end{align}

hence, there is no real solutions.

Edit:

As @alex.jordan pointed out, the expression (2)
does not cover all possible cases.

It misses the case of $x<0$
when $n$ is either even or
rational number with even numerator,
so in such cases
we need also to include the solutions $x=-t$
of the equation

\begin{align}
a^{-t}&=t^n
,\\
a^t&=t^{-n}
\tag{3}\label{3}
,
\end{align}
for positive $t$,
which we already know must be

\begin{align}
t
&=
-\frac{\W(-\tfrac1{-n}\ln a)}{\tfrac1{-n}\ln a}
=
\frac{\W(\tfrac1{n}\ln a)}{\tfrac1{n}\ln a}
,
\end{align}
so missing solution(s) are

\begin{align}
x
&=
-\frac{\W(\tfrac1{n}\ln a)}{\tfrac1{n}\ln a}
\tag{4}
.
\end{align}

For example, when $a=2,\, n=2$,
$z=-\tfrac12\ln 2\approx-0.34657\in(-\tfrac1\e.0)$

(2) provides two positive roots,

\begin{align}
x_0=-\frac{\Wp(-\tfrac12\ln 2)}{\tfrac12\ln 2}
=2
,\ x_1=-\frac{\Wm(-\tfrac12\ln 2)}{\tfrac12\ln 2}
=4
,
\end{align}

and since
the argument of $\W$ in (4) is
$\tfrac1{2}\ln 2>0$,
expression (4) gives just one more real root:

\begin{align}
x_3
&=
-\frac{\Wp(\tfrac1{2}\ln 2)}{\tfrac1{2}\ln 2}
\approx -0.7666647
.
\end{align}

Note that the arguments of $\W$
in (2) and (4) are $z$ and $-z$,
the maximal number of real roots of (1)
is $3$.

$\endgroup$


问:
Can such a matrix be singular?
Let $A$ be an $n\times n$ matrix that satisfies

i. All diagonal entries of $A$ are positive, even integers,

ii. All non-diagonal entries of $A$ are positive, odd integers,

iii. $A$ is symmetric: $A_{ij}=A_{ji}$.

iv. If $i\neq j$, then $A_{ij}<A_{ii}$ and $A_{ij}<A_{jj}$.

Question: can such $A$ be singular (i.e. $|A|=0$)?

I can prove the answer is NO, when $n$ is even, and without using condition iv: reduce $A$ mod $2$ we get a matrix with zero diagonal and all other entries $1$; then we can compute directly that the determinant of such matrix is odd, in particular not zero. When $n$ is odd this argument says $|A|$ is even, I don't know if it can be zero, and it seems condition iv will be important here.

I encountered it when I was trying to solve a question in this post.

答:
Yes, $A$ can be singular. Let $n\ge3$ be odd. Consider
$$
A=\left[\begin{array}{cccc|c}a&b&\cdots&b&d\\ b&a&\ddots&\vdots&\vdots\\ \vdots&\ddots&\ddots&b&\vdots\\ b&\cdots&b&a&d\\ \hline d&\cdots&\cdots&d&c\end{array}\right]
=\begin{bmatrix}(a-b)I_{n-1}+bee^T&de\\ de^T&c\end{bmatrix}
$$
where $a,c$ are even, $b,d$ are odd, $\max(b,d)<\min(a,c)$ and $e=(1,1,\ldots,1)^T\in\mathbb R^{n-1}$. Using Schur complement, $A$ is congruent to $B=\left[(a-b)I_{n-1}+(b-\frac{d^2}{c})ee^T\right]\oplus c$. The eigenvalues of $B$ are $a+(n-2)b-\frac{d^2}{c}(n-1),\,a-b$ and $c$. The latter two are odd integers. Hence $B$ and $A$ are singular if and only if
$$
[a+(n-2)b]c=(n-1)d^2.
$$
When $n\ge5$ is odd, we can always pick $(b,c,d)=(1,n-1,n-2)$ and $a=d^2-d=(n-2)(n-3)$. Note that we do have $\max(b,d)=d<d^2-d=\min(a,c)$ in this case.

When $n=3$, the above choice of $(a,b,c,d)$ does not work (because $d$ becomes zero), but a brute-force search gives a solution $(a,b,c,d)=(16,9,18,15)$, i.e.
$$
A=\begin{bmatrix}16&9&15\\ 9&16&15\\ 15&15&18\end{bmatrix}.
$$

Title: Can such a matrix be singular?


问:
Height of a cube edge from the floor

A cube $ABCD.EFGH$ has side length $2a$ cm. Point $A$ is lifted $a$ cm from the floor, point $C$ is still on the floor, point $B$ and point $D$ are on the same height from the floor. What is the height of point $E$ from the floor?

I'm sorry for my bad English, but I try to illustrate it as follows.

The left cube is the original one while the right side is the cube after we lifted the point $A$. In my mind, to find the height of point $E$ from the floor is to find the length of line $EN$. We can use Pythagorean theorem $EN^2=ME^2-MN^2$. But how to find the length of line $ME$ and $MN$?

答:
As your pivot is point $C$ and points $B$ and $D$ are at the same height, the vertical edge AE will rotate by the same angle as the horizontal line  $CA$.

(another way to think is $\triangle CAE$ rotating by angle $\theta$ in the same original plane)

If $\angle ACA' = \theta, \angle CAA' = 90^0 - \theta$.

where $A'$ is the vertical projection of point $A$ on the original surface.

Also $\angle CAE = 90^0$.

So, $\angle E'AE = 180^0 - (\angle CAE + \angle CAA') = 180^0 - (90^0 + 90^0 - \theta) = \theta$

Now $CA' = \sqrt{CA^2 - AA'^2} = \sqrt{{(2\sqrt2a)^2} - a^2} = a \sqrt 7$

So $\cos \theta = \frac{CA'}{CA} = \frac{a\sqrt7}{2a\sqrt2} = \frac{\sqrt7}{2\sqrt2}$

Now perpendicular distance of $E$ from the original horizontal plane

$= a + 2a \cos \theta \,$ ($\triangle AE'E \sim \triangle CA'A$)

$= A'A + AE' = a + 2a \frac{\sqrt7}{2\sqrt2} = a(1 + \sqrt{\frac{7}{2}})$

Title: Height of a cube edge from the floor


问:
How to find operator norm $T(\bar{x}) = (3x_1 + 2x_2,x_3-4x_4,3x_5 + 2x_6, ...)$
$$T(\bar{x}) = (3x_1 + 2x_2,x_3-4x_4,3x_5 + 2x_6, ...)$$
Is an operator from $l_2$ to $l_2$. I have tried to find upperbounds and lowerbounds for the supremum but I don't end up with anything I can work with. Any hints will be appreciated.

答:
Use the inequality
$$
(a+b)^2 = a^2 + 2ab + b^2 \le 2(a^2 + b^2)
$$
for each of the entries of in the sum defining $\|T(x)\|_{l^2}$.


问:
Examples of when Newton's Method will fail?
I'm currently working on Newton's Method, and my instructor gave four instances where Newton's Method will fail.

(A) Newton's method converges to another solutions x=b such that f(b)=0 instead of converging to the desired solution x=a.

(B) Newton's method eventually gets into the never ending cycle, bouncing between the same two approximations $x_i$ and $x_{i+1}$.

(C) Eventually, each next approximation $x_{i+1}$  falls further from desired solution $x_a$ than the previous approximation $x_i$  determined by the Newton's method.

(D) Newton's method is not able to find the next approximation $x_{i+1}$  because f'($x_i$)=0 or f'($x_i$) Does Not Exist.

However, there aren't any examples of when this happens. Would anyone be willing to provide examples of these instances?

答:
(A) $f(x) = x^2-1$ and start near $x=-1$ when you 'want' $x=1$.

(B) $f(x) = \sqrt{1+x^2}$ and start at $x=1$.

(C) $f(x) = \arctan x$ and $x_0>2$ (needs a little work to show that the Newton iteration diverges).

(D) $f(x) = x^3-1$ with $x_0 = 0$.


答:
I assume $a$ is positive and $n$ is a natural number.

Note right away that for $a\neq1,n=0$, there is one solution, $x=0$. For $a=1,n>0$, there is one solution when $n$ is odd and two solutions when $n$ is even. And when $n=0,a=1$, it's a bit silly, but all real numbers are solutions.

So we continue assuming $n>0, a\neq1$. For $n>0$, $x=0$ is never a solution. Any solutions are either positive or negative.

If $x$ is a positive solution, then $$\frac{\ln(x)}{x}=\frac{\ln(a)}{n}$$

So we want to understand the function $f(x)=\frac{\ln(x)}{x}$. Its derivative is $\frac{1-\ln(x)}{x^2}$. So $f$ increases on $(0,e)$, reaching a peak at $x=e$ with value $\frac{1}{e}$. And then decreases on $(e,\infty)$, with a limit of $0$. Also as $x\to0^+$, the limit is $-\infty$.

So $f$:

has no positive solutions if $\frac{\ln(a)}{n}>\frac{1}{e}$.

has one positive solution if $\frac{\ln(a)}{n}=\frac{1}{e}$.

has two positive solutions if $0<\frac{\ln(a)}{n}<\frac{1}{e}$.

has one positive solution if $\frac{\ln(a)}{n}\leq0$.

Now could there be negative solutions? Only when $n=2k$ is even, since otherwise $x^n$ is negative for negative $x$, whereas $a^x$ will still be positive. In this case we similarly write $$\frac{\ln(-x)}{x}=\frac{\ln(a)}{2k}$$ and study $g(x)=\frac{\ln(-x)}{x}$ for negative values of $x$. Its derivative is $\frac{1-\ln(-x)}{x^2}$. So $f$ decreases on $(-\infty,e)$, reaching a low at $x=-e$ with value $-\frac{1}{e}$. And then increases on $(-e,0)$, with a limit of $\infty$. Also as $x\to-\infty$, the limit is $0$.

So provided $n$ is even, we can add a negative roots to the list. There is:

one negative solution if $\frac{\ln(a)}{2k}=-\frac{1}{e}$.

two negative solutions if $0>\frac{\ln(a)}{2k}>-\frac{1}{e}$.

one negative solution if $\frac{\ln(a)}{2k}\geq0$.

otherwise has no negative roots.

One big summary:

If $n$ is odd:

no solutions if $\frac{\ln(a)}{n}>\frac{1}{e}$. (Alternatively written as $a>e^{n/e}$.)

one solution if $\frac{\ln(a)}{n}=\frac{1}{e}$. ($a=e^{n/e}$)

two solutions if $0<\frac{\ln(a)}{n}<\frac{1}{e}$. ($1<a<e^{n/e}$)

one solution if $\frac{\ln(a)}{n}\leq0$. ($a\leq1$.)

If $n\geq2$ is even:

one solution if $\frac{\ln(a)}{n}>\frac{1}{e}$. ($a>e^{n/e}$)

two solutions if $\frac{\ln(a)}{n}=\frac{1}{e}$. ($a=e^{n/e}$)

three solutions if $0<\frac{\ln(a)}{n}<\frac{1}{e}$. ($1<a<e^{n/e}$)

two solutions if $\frac{\ln(a)}{n}=0$. ($a=1$)

three solutions if $0>\frac{\ln(a)}{n}>-\frac{1}{e}$. ($e^{-n/e}<a<1$)

two solutions if $\frac{\ln(a)}{n}=-\frac{1}{e}$. ($a=e^{-n/e}$)

one solution if $\frac{\ln(a)}{n}<-\frac{1}{e}$. ($a<e^{-n/e}$)

If $n=0,a\neq1$:

one solution ($x=0$) no matter what $a$ is.

If $n=0,a=1$:

all real $x$ are solutions

Arranged an alternative way:

All real numbers are solutions when:

$n=0,a=1$

There are three solutions when:

$n$ is even and $e^{-n/e}<a<e^{n/e}$ but $a\neq1$

There are two solutions when:

$n$ is odd and $1<a<e^{n/e}$

$n$ is even, at least $2$, and either $a=e^{n/e}$, $a=1$, or $a=e^{-n/e}$

There is one solution when:

$n=0$

$n$ is odd and $a\leq1$ or $a=e^{n/e}$

$n$ is even, at least $2$, and either $a>e^{n/e}$ or $a<e^{-n/e}$

There are no solutions when:

$n$ is odd and $a>e^{n/e}$

Title: How many roots does the equation $a^x -x^n=0$ have?

Body:
I want to know if there is a simple method to be able to tell the number of roots of an equation in the form of : $ a^x - x^n = 0 $ (with a>0 and n>0)
If we take $3^x=x$ for example. I know that graphically for example, you can see that there are no roots. But is there another (simple) way to solve it?
What about $5^x=x^2$?


问:
Uniform convergence of $\sum \frac{nx}{1+n^2x^2}$
I know that $\frac{nx}{1+n^2x^2}$ isn't uniformly convergent.

However I am not sure about the series. None of the tests for series convergence, namely wierstrass m test, Abel test or Dirichlet test give anything about proving non convergence.

Does non-uniform convergence of individual terms imply non-uniform convergence of series?

答:
Let $f_n(x)=\frac{n x}{1+n^2 x}$, the series $\sum_{n=1}^{\infty} \frac{n x}{1+n^2 x}$ is not defined on $x=\frac{-1}{n^2},\,\, n=1,2,3,...$. and for all other values of $x$, we have $\frac{n x}{1+n^2 x} \to 0,\,\, as\,\, n \to 0$.

According to M-test theorem, the series convergent uniformly on any bounded interval $(a,b],\,\, b >a>0$ for

$$\frac{n x}{1+n^2 x} \le \frac{b} {a n^2} $$

and it is clear that
$$\frac{b}{a}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{b \pi^2}{6 a}.$$

The series diverges at $x= \frac{1}{1-n^2},\,\,\, n=1,2,3,...$ because the expression $\frac{n x}{1+n^2 x} =1$ at $x=\frac{1}{1-n^2}$ and the infinite summation for the last diverges.


答: Hint: You can use the Cauchy-Schawrz Inequality to show that $$(3a+2b)^2 = \langle(3,2),(a,b)\rangle^2 \le \|(3,2)\|^2\|(a,b)\|^2 = 13(a^2+b^2)$$ and similarly, $$(a-4b)^2 = \langle(1,-4),(a,b)\rangle^2 \le \|(1,-4)\|^2\|(a,b)\|^2 = 17(a^2+b^2).$$ Using these facts, try to show that $$\|T(\bar{x})\|_{\ell_2}^2 \le C\|\bar{x}\|_{\ell_2}^2,$$ i.e. $$\sum_{k = 0}^{\infty}(3x_{4k+1}+2x_{4k+2})^2 + \sum_{k = 0}^{\infty}(x_{4k+3}-4x_{4k+4})^2 \le C\sum_{n = 1}^{\infty}x_n^2.$$ This will establish that $\|T\| \le \sqrt{C}$. Then, all you need to show is that there is some non-zero sequence $\bar{x}$ such that $\|T(\bar{x})\|_{\ell_2} = C\|\bar{x}\|_{\ell_2}$.

Title: How to find operator norm $T(\bar{x}) = (3x_1 + 2x_2,x_3-4x_4,3x_5 + 2x_6, ...)$

Body:
$$T(\bar{x}) = (3x_1 + 2x_2,x_3-4x_4,3x_5 + 2x_6, ...)$$
Is an operator from $l_2$ to $l_2$. I have tried to find upperbounds and lowerbounds for the supremum but I don't end up with anything I can work with. Any hints will be appreciated.


问:
Contour question - Draw contours for In (xy) = c
Can anyone explain me how to draw the contour for surface z = in(x y) where its in(x y) = z , and z = 0, 1, 4.

The in function is defined as

$$ y(x) =e^x/x $$

I don't have a graphing calculator. I already understand the sections are z = in x and z = in y.

But I am completely lost in how to draw the contour.

答:
Mathematica  program has this function
ContourPlot3D

```<code> a = 1.8; in[u_] = Exp[u]/u; ContourPlot3D[z == in [x y], { x, -a, a},{y, -a, a}, {z, -a, a}]
</code>```

It shows intersections at not only for the level curves $z =$ const., but also for $x= $const., and $ y= $ const.

Body:
Mathematica  program has this function
ContourPlot3D

 a = 1.8; in[u_] = Exp[u]/u; ContourPlot3D[z == in [x y], { x, -a, a},{y, -a, a}, {z, -a, a}]

It shows intersections at not only for the level curves $z =$ const., but also for $x= $const., and $ y= $ const.


问:
When can you not use inclusion-exclusion in probability?
I came across this question:
Alice and Bob both try to climb a rope. Alice and Bob will get to the top with probability 1/3 and 1/4 respectively. given that exactly one person got to the top, what is the probability that the person is Alice.

The way the textbook solved it, they did $\frac13\times\frac34+\frac23\times\frac14$.

And I got that answer, but I was thinking, why can't we do inclusion-exclusion on the probability that exactly on person makes it to the top. ie, why cant we do $P[\text{Alice} ]+P[\text{Bob}]-P[\text{both}] = 1/3+1/4-1/12$ The calculation yields 2/3, not 5/12

So I'm confused -- when can we not use inclusion-exclusion?

答:
Actually, both answers are wrong.

Breaking up into mutually exclusive cases,

P(only A reaches top) $= \frac 1 3\cdot \frac 3 4 = \frac 3 {12}$

P(only B reaches top) $= \frac 1 4\cdot \frac 2 3 = \frac 2{12}$

P(both reach top) $= \frac 1 3\cdot \frac 1 4 = \frac 1 {12}$

P(neither reach top) $= \frac 2 3 \cdot \frac 3 4 = \frac6 {12}$

The book has computed P(exactly one has reached top), you have computed P(at least one has reached  top),
whereas  you are asked  to find out P(A has reached top |exactly one has reached top)

You should be able to proceed now to get the correct answer

The correct answer is

  P(A reaches top | exactly one reaches top) = $ \dfrac 3 {12}\over {\dfrac3 {12} + \dfrac 2 {12}}$


问:
Frobenius norm minimization with additions
I am trying to find minimizer X for which

$$ \arg \min_{X} {\left\|| A X + X B - Y \right\||}_{F} $$

when all matrices are square. I know to use the pseudoinverse of A for

$$ \arg \min_{X} {\left\|| A X - Y \right\||}_{F} $$

but don't understand how to rewrite the argument of the first equation to a solvable form.

答: Since $\left\|A X + X B - Y \right\|_F=\|(I\otimes A+B^T\otimes I)\operatorname{vec}(X)-\operatorname{vec}(Y)\|_2$ (see Wikipedia for the meaning of the $\operatorname{vec}$ operator), this is just an ordinary least-square problem and a global minimiser is given by $\operatorname{vec}(X)=(I\otimes A+B^T\otimes I)^+\operatorname{vec}(Y)$.


问:
Deducing a coefficient from a cubic polynomial?
I fully answered the question, and got that $k=-3$, but the answer says it's positive. Can anyone show me my mistake?

"Given that $x-2$ is a factor of the polynomial $x^3 - kx^2 - 24x + 28$, find $k$ and the roots of this polynomial."

Using factor theorem, I realised that $P(2)$ is equivalent to $0$, therefore $2^3 - 2^2k - 24(2) + 28 = 0$

I solved it algebraically and got that $k=-3$, but the answers say it was $k=3$. Did I make a simple error?

Any help would be appreciated.

Body:
I got k = -3 also. I think the answers may be wrong:

2 | 1     -k      -24           28
  |        2      2(-k + 2)     4(-k + 2)-48
  |____________________________________________
    1   (-k + 2)  2(-k + 2)-24  4(-k + 2)-48+28

-4k+8-48+28 = 0
-4k = 12
k = -3


问:
Prove that * is a binary operation
I have the set $S=\mathbb{R}-\{-1\}$ and the operation $a*b=a+b+ab$ and I have to prove that $*$ defines a binary operation on $S$ my question is if is the same prove that is a binary operation and prove $S$ and $*$ is a group?

答: No, because it is necessary for $\ast$ to be an operation in order for $(S,\ast)$ to be a group in the first place. Sometimes this is explicitly stated, sometimes not, but either way it is a necessary condition.

Title: Prove that * is a binary operation


答:
No, is not the same. The relation $*$ is a binary operation on $S$ if is a function from $S\times S$ to $S$, i.e., if for each $a,b\in S$, then $a*b$ is well-defined and belongs to $S$. It is clear that $a*b$ is well-defined in $\mathbb{R}$, so you have to check that if $a,b\neq -1$, then $a*b\neq -1$. This is easy, indeed $a*b=-1$ iff $a+b+ab+1=0$ iff $(a+1)(b+1)=0$ iff $a$ or $b$ is equal to $-1$.

So $*$ is really an operation on $S$, which means that $(S,*)$ is a magma (or grupoid). If you can also prove that $*$ is associative, has an identity element and that every element in $S$ is invertible, than you have that $(S,*)$ is a group.

Title: Prove that * is a binary operation

Body: I have the set $S=\mathbb{R}-\{-1\}$ and the operation $a*b=a+b+ab$ and I have to prove that $*$ defines a binary operation on $S$ my question is if is the same prove that is a binary operation and prove $S$ and $*$ is a group?


问:
Show $8\mid n^2-1$ if $n$ is an odd positive integer.
Show that $n^2-1$ is divisible by $8$, if $n$ is an odd positive integer.
Please help me to prove whether this statement is true or false.

答:
We know that any odd positive integer is of the form $4q+1$ or, $4q+3$ for some integer $q$.

So, we have the following cases:

Case 1 when $n=4q+1$: In this case, we have

$n^2 -1=(4q+1)^2 -1=16q^2 +8q+1-1=16q^2 +8q=8q(2q+1)$

$∴ n^2 -1$ is divisible by 8 [$∵ 8q(2q+1)$ is divisible by 8]

Case 2 when $n=4q+3$: In this case, we have

$n^2 -1=(4q+3)^2 -1= 16q^2 +24q+9-1=16q^2 +24q+8$

⇒ $n^2 -1=8(2q^2 +3q+1)=8(2q+1)(q+1)$

$∴ n^2 -1$ is divisible by 8 [$∵ 8(2q+1)(q+1)$  is divisible by 8]

Hence, $n^2 -1$ is divisible by 8.


问:
Does there exist a faithful representation of degree $n$?
Let $G$ be a group of order $n$ where field is real or complex. Does there exist a faithful $FG$-module of dimension $n$ (in other words a representation of degree $n$) other than the regular $FG$- module (or representation)?

I think it should be no, as every such representation will be isomorphic to regular one.

答: They can exist. For instance, consider the Klein four group: the regular rep decomposes as a direct sum of all four 1D irreps, but we can construct another faithful 4D rep by direct summing any four 1D reps at least two of which are distinct nontrivial. Indeed, you can take the direct sum of two distinct nontrivial permutation 2D reps.</p>


答:
You can use inclusion-exclusion to find $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.

But you need the probability that exactly one event happens, which is $P(A\cup B)-P(A\cap B)$. So calculating this with inclusion-exclusion gives $P(A)+P(B)-2P(A\cap B)=5/12$, the same as the other method.

(You also have an error in calculation: $1/3+1/4-1/12=1/2$, not $2/3$.)

Title: When can you not use inclusion-exclusion in probability?

Body:
I came across this question:
Alice and Bob both try to climb a rope. Alice and Bob will get to the top with probability 1/3 and 1/4 respectively. given that exactly one person got to the top, what is the probability that the person is Alice.

The way the textbook solved it, they did $\frac13\times\frac34+\frac23\times\frac14$.

And I got that answer, but I was thinking, why can't we do inclusion-exclusion on the probability that exactly on person makes it to the top. ie, why cant we do $P[\text{Alice} ]+P[\text{Bob}]-P[\text{both}] = 1/3+1/4-1/12$ The calculation yields 2/3, not 5/12

So I'm confused -- when can we not use inclusion-exclusion?


问:
Set partition refinement lattice, maximal chain length
Denote $\Pi_n$ the set of all set partitions of [n]. For set partitions $S=\{S_1,S_2,...,S_p\}$ and $T=\{T_1,T_2,...,T_q\}$ of [n], define $S \leq T$ iff for every i=1,2,...,p there exists a j=1,2,...,q such that $S_i \subseteq T_j$.

a)Show that $(\Pi_n,\leq)$ is a partial order

b)Find the length of a maximal chain in $(\Pi_n,\leq)$

To prove a relation is partial order we have to prove reflexivity, antisymmetry, and transitivity. It seems proving reflexivity and transitivity is easy. I'm just not sure about antisymmetry.

Here is one example. This is called the set partition refinement lattice. It looks like the max chain length is n. I don't know if that is correct.

答:
Well for antisymmetry it's also straight forward:

Assume $S\leq T\leq S$ then for every $i$ there exists a $j$ and a $k$ so that $S_i\subseteq T_j \subseteq S_k$ thus $S_i\subseteq S_k$ and since $S=\{S_1,\ldots,S_p\}$ is a partition $S_i\cap S_k \neq \emptyset$ implies $S_i = S_k$ and hence $S_i = T_j = S_k$ since this is valid for every $i$ and exchanging the roles of $S$ and $T$ this is valid for every $j$ we get that $T=S$ proving anti-symmetry.

As for the maximal chain length in $\Pi_n$ it necessarily is $\geq n$ necessairily since we can always consider the chain
$$ \{\{1\},\{2\},\ldots, \{n\}\}\leq \{\{1,2\},\{3\},\ldots, \{n\}\}\leq \cdots \leq \{\{1,\ldots, n\}\}$$

(note that $\{\{1\},\{2\},\ldots, \{n\}\}$ is minimal and $\{\{1,\ldots,n\}\}$ is maximal with respect to $\leq$)

Now to prove the maximal chain can use induction:

for $n=1$ the maximal chain has length $1$ (the minimal and maximal partitions coincide)

Now assume that for $n$ we have a maximal chain of length $m \geq n + 1$, of the form $S^1\leq S^2\leq \cdots \leq S^m$.

Note also that if a chain is maximal then only one of the $S^i_j$ is strictly contained in some $S^{i+1}_k$, for all the others equality holds otherwise the chain can be further extended contradicing maximality.

Now without loss of generality let $n\in S^i_1$ for all $i$. Since by hypothesis the chain is maximal, it starts on the minimal elemnt, thus $S^1_1 = \{n\}$ let $j$ be the minimum for which $S^1_1\subsetneq S^j_1$, then if we simply delete $S^i_1$ for all $i< j$ and remove $n$ from $S^i_1$ for all $i\geq j$ we get a sequence in $\Pi_{n-1}$ that is not strict only at step $j$, thus if we delete step $j$ we obtain a chain of length $n-1$ in $\Pi_{n-1}$ contradicting the induction hypothesis.


问:
Inclusion-exclusion of values of $\Omega$ (the arithmetic function).
For any number $n \in \Bbb{Z}$, define

$$
f(n) = \Omega(n) - \sum_{q \mid n} \Omega(n/q) + \sum_{q,q' \mid n}\Omega(n/(qq')) - \dots \pm \Omega(1),$$

where each sum is taken over only prime divisors and includes multiple copies of the same prime (multiplicity).

Will this sum always be zero?

Is there a way to view it using inclusion-exclusion?

答:
Let $n=q_1 q_2\dots q_r$ be a factorization into primes and consider the set of indices $Q=[r]=\{1,\dots,r\}$ so that $\Omega(n)=|Q|=r$.

For each $i\in[r]$ let $Q_i=Q\setminus \{i\}$ and note that $\Omega(n/q_i) = |Q_i|.$

When $r>1$, the inclusion-exclusion principle for $Q=Q_1\cup Q_2\cup\dots\cup Q_r$ yields
$$
|Q|\ - \sum_{1\le i_1\le r} |Q_{i_1}|\ + \sum_{1\le i_1<i_2\le r} |Q_{i_1}\cap Q_{i_2}|\ - \ \cdots\ \pm |Q_1\cap Q_2\cap \dots \cap Q_r| = 0.
$$
For $1\le i_1<\dots<i_n\le r$ we have
$$
n/\prod_{k=1}^n q_{i_k} = \prod_{i\in Q'} q_i
\qquad\text{for}\qquad
Q'=Q\setminus\{i_1,\dots,i_n\} = Q_{i_1}\cap\dots\cap Q_{i_n}.
$$
Hence,
$$
\Omega\left(n/\prod_{k=1}^n q_{i_k}\right) = \left|Q_{i_1}\cap\dots\cap Q_{i_n}\right|
$$
and the above identity is exactly $f(n)=0$.

When $r=1$, so that $n=p$ is a prime, we have
$$
f(p) = \Omega(p) - \Omega(1) = 1 - 0 = 1 \neq 0.
$$


问:
Ring Homomorphism (mapping of unit element) between a ring and an integral domain
Let R and S be two rings with unit elements. Let us assume that the unit element of R is $1_R$
and the unit element of S is $1_S$. Let $\phi$ : R → S be a ring homomorphism. Show that if S is
an integral domain, then $\phi$ ($1_R$) = $1_S$.

Note: $\phi$($1_R$) $\neq 0_S$ (Zero element in S)

I want to prove this via contradiction specifically and have proceeded as follows:

$\exists y \in S \ \ s.t.$  $ y \ \phi(1_R)  \neq \ y $

Then-

$ y \ \phi(1_R)- y  \neq  0_S $

I am stuck here and don't know how to proceed ahead with this proof using contradiction.

答:
If $\phi(1_R) \neq 1_S$, then $\phi(1_R)-1_S \neq 0_S$ and $\phi(1_R) \neq 0_S$ but
$$(\phi(1_R)-1_S)\phi(1_R) = \phi(1_R) - \phi(1_R) = 0_S,$$ that is, $\phi(1_R)-1_S$ and $\phi(1_R)$ are zero divisors in $S$.


问:
Why does the manipulation of expressions allow for L'Hôpital?
It's been some time since I've taken a maths course and I remember to have always had this question in the back of my mind and I thought I should finally ask.

When we have expressions that are not in a form for which we can use L'Hôpital's on, we manipulate it very easily and voila, we get $\frac{0}{0}$ once we take the limit and thus we can use L'Hôpital's. Why is this the case and why don't we have to use the simplest form of an expression?

Here is an example:

$$\lim_{x \to \infty} x\sin\left(\frac{1}{x}\right)$$
From the get go, this is not a form we can simply use L'Hôpital's on, so all we do is $x = \frac{1}{x^{-1}}$ and thus
$$\lim_{x \to \infty} \frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}$$

I just don't understand why this makes everything work all of a sudden. Why are we not obliged to simplify this. Is it because we can derive from $\sin\left(\frac{1}{x}\right)$ that $x$ cannot be zero and use this rule to play around with $x$?

I would appreciate some explanation, thanks in advance!

答:
Why is this the case and why don't we have to use the simplest form of an expression?

Because L'Hopital is concerned with a ratio of functions; it is not concerned with the ratio being in the simplest form, and nowhere mandates that. (Plus sometimes what is "simplest" is a subjective matter anyways.)

In the case of your example, we have

$$\frac{f(x)}{g(x)} = \frac{\sin(1/x)}{1/x}$$

as the ratio we're taking the limit of. That is, we assign $f(x) = \sin(1/x)$ and $g(x)=1/x$. Then, by L'Hopital,

$$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)}$$

if the necessary conditions hold.

Bear in mind that this fact is L'Hopital's rule (for limits $x \to \infty$ anyhow), and to your example we apply L'Hopital's rule with the definitions of $f,g$ are described earlier. The fact that $f/g$ simplifies your desired function allows us to claim the limit for your function in turn.

Is it because we can derive from $sin(\frac{1}{x})$ that $x$ cannot be zero and use this rule to play around with $x$?

That has nothing to do with it. Since we're taking the limit as $x \to \infty$, eventually $x$ will far exceed $0$, and the singularity becomes irrelevant in that sense.


答:
Let $p = \angle ACA'$, $q = \angle ECA$ (in the original figure), where $A,C,A'$ are the same as in Math Lover's answer.

Then $\tan p = \frac{1}{\sqrt 7}, p = \tan^{-1} \frac{1}{\sqrt 7}$ and $\tan q = \frac{2}{2 \sqrt 2} = \frac{1}{\sqrt 2}, q = \tan^{-1} \frac{1}{\sqrt2}$. Now using the arctangent addition formula:

$$p + q = \tan^{-1} \frac{1/\sqrt7 + 1/\sqrt2}{1 - (1/\sqrt7)(1/\sqrt2)} = \tan^{-1} \frac{\sqrt{2} + \sqrt{7}}{\sqrt{14}-1} = \tan^{-1} \frac{\sqrt{2\cdot2\cdot7}+\sqrt{2\cdot7\cdot7}+\sqrt{2}+\sqrt{7}}{13} = \frac{1}{13} \left(8 \sqrt2 + 3 \sqrt7 \right).$$

So $\sin(p+q) = EN / EC \Rightarrow EN = 2\sqrt{3}a \sin(p+q)$. Now since $\sin(\tan^{-1} x) = \frac{x}{\sqrt{x^2+1}}$ (draw a triangle), then we have the monstrous task of simplifying this expression:

$$\sin(p+q) = \frac{(1/13) \left(8 \sqrt2 + 3 \sqrt7 \right)}{\sqrt{(1/169)(191+48\sqrt{14})+1}}$$

and after incorporating the $1$, denesting the square root, and dividing, I can confirm that this eventually gives $EN = a(1 + \sqrt{7/2})$.

Title: Height of a cube edge from the floor

Body:
A cube $ABCD.EFGH$ has side length $2a$ cm. Point $A$ is lifted $a$ cm from the floor, point $C$ is still on the floor, point $B$ and point $D$ are on the same height from the floor. What is the height of point $E$ from the floor?

I'm sorry for my bad English, but I try to illustrate it as follows.

The left cube is the original one while the right side is the cube after we lifted the point $A$. In my mind, to find the height of point $E$ from the floor is to find the length of line $EN$. We can use Pythagorean theorem $EN^2=ME^2-MN^2$. But how to find the length of line $ME$ and $MN$?


问:
Why the induced orientation need the manifold is connected
Picture below is from 18th page of  Do Carmo's Riemannian Geometry. In my view, the induced orientation does not connected. Assume $(U_\alpha, x_\alpha)$ is orientation of $M_1$, then $(\varphi(U_\alpha), \varphi\circ x_\alpha)$ is a orientation of $M_2$. Seemly, the connected is not used in here.

答:
If $M$ has $n$ different connected components and if $M$ is orientable, each connected component is orientable and $M$ has $2^n$ different possible orientations.

Thus, if $f : M_1 \to M_2$ is a diffeomorphism between oriented manifolds with finite connected components, $f$ could be

orientation preserving ($1$ case)

non orientation preserving ($2^n -1$ cases)

In case there is just one component, being non orientation preserving leads to only one case. In that last case, we say that $f$ is orientation reversing and this is not ambiguous which orientation is induced on $M_2$.


问:
How to prove this inequality if $\|x_i\|=1$.
How to prove that for given $k\geq 1$, if $\|x_i\|=1$ for all $i=1,\dots,n$, $\alpha_1\neq0$ and $|\lambda_1|>|\lambda_2|\geq |\lambda_3|\geq\dots\geq |\lambda_n|$, then

\begin{equation}
\|\sum_{i=2}^n \frac{\alpha_i}{\alpha_1}\left(\frac{\lambda_i}{\lambda_1}\right)^k x_i \|_2 \leq \left\lvert\frac{\lambda_2}{\lambda_1}\right\rvert^k\left(\sum_{i=2}^n \left(\frac{\alpha_i}{\alpha_1}\right)^{2}\right)^{1/2}
\end{equation}

This is a convergence analysis of the power method.

答:
This text does not make much sense. One can always normalize the generalized eigenvectors.

$\|\sum c_ix_i\|$ will be close to $|\sum c_i|$ if the $x_i$ are almost equal, $\sqrt{\sum |c_i|^2}$ is usually much smaller.

It appears that the Euclidean norm is for the coordinates in the eigen-basis, then the question-critical inequality is the Pythagorean identity.

In general you can get
$$
\|\sum_{i=2..n}c_ix_i\|\le \sum_{i=2..n}|c_i|\|x_i\|\le\sqrt{n-1}\sqrt{\sum_{i=2..n} |c_i|^2},
$$
the last by the Cauchy-Schwarz inequality. This is also sufficient for the convergence proof.


答:
The operation of taking limit is very "distributive" it gets distributed obver, sum, product or even is exponents: $L(f(x)+g(x))=L(f(x))+L(g(x))$.
$L(f(x)g(x))=L(f(x)) L(g(x))$, $L(f(x)^{g(x)})=L(f(x))^{L(g(x))},$ All these are allowed iff the final limit is finite.

Next, $L(\frac{f(x)g(x)h(x)}{k(x)p(x)})=L\frac{f(x)}{k(x)} L\frac{g(x)}{p(x)} L(h(x))=L\frac{h(x)}{k(x)} L\frac{g(x)}{p(x)} L(f(x)),$ Provided, these results are finite.

Similarly, reshuffling of the expression is allowed if the final limit exists (it is finite).

Title: Why does the manipulation of expressions allow for L'Hôpital?

Body:
It's been some time since I've taken a maths course and I remember to have always had this question in the back of my mind and I thought I should finally ask.

When we have expressions that are not in a form for which we can use L'Hôpital's on, we manipulate it very easily and voila, we get $\frac{0}{0}$ once we take the limit and thus we can use L'Hôpital's. Why is this the case and why don't we have to use the simplest form of an expression?

Here is an example:

$$\lim_{x \to \infty} x\sin\left(\frac{1}{x}\right)$$
From the get go, this is not a form we can simply use L'Hôpital's on, so all we do is $x = \frac{1}{x^{-1}}$ and thus
$$\lim_{x \to \infty} \frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}$$

I just don't understand why this makes everything work all of a sudden. Why are we not obliged to simplify this. Is it because we can derive from $\sin\left(\frac{1}{x}\right)$ that $x$ cannot be zero and use this rule to play around with $x$?

I would appreciate some explanation, thanks in advance!


问:
On Landau notations
How common it is to write e.g. $1-o(1)$ for a function that eventually approaches $1$ from below (or eventually equals $1$)? Would a better notation be $1-|o(1)|$ or what is meant is already obvious from $1-o(1)$ ? Obviously, precisely defining everything will work eventually, but I was wondering about the general convention. Similarly, are notations like $|\omega(1)|$ common?

Edit: In other words, would you say "What the hell is this?" is you saw both $1+o(1)$ and $1-o(1)$ in the same sentence, or would you say "Hmm, ok I got it, these are all assumed to be positive"?

答:
The definition of little-Oh I recall is:
$$
f=o(g) \Leftrightarrow \exists\varepsilon, \varepsilon\to 0, f=\varepsilon g
$$
In particular, it doesn't really make sense to write "$-o(1)$"; according to this definition, if $f$ is $o(g)$, $(-f)$ is also $o(g)$. Another convention to circumvent that is to write $f\to 1^-$ or $f\to 1^+$ (if the limit is reached resp. from below or above). Or $\lim f= 1^-$ ($\lim f= 1^+$).

Rk: I've also seen teachers write $f\nearrow 1$ and $f\searrow 1$, but I'm not sure it is standard.

Title: On Landau notations