问:
Given particle undergoing Geometric Brownian Motion, want to find formula for probability that max-min > z after n days
Consider a particle undergoing geometric brownian motion with drift $\mu$ and volatility $\sigma$ e.g. as in here. Let $W_t$ denote this geometric brownian motion with drift at time $t$. I am looking for a formula to calculate:

$$
\mathbb{P}\big(\max_{0 \leq t \leq n} W_t - \min_{0\leq t \leq n} W_t > z\big)
$$
The inputs to the formula will be $\mu$, $\sigma$, $z$, and $n$.

答:
Let's start with the joint probability density of Brownian motion $W_t$ and its maximum $M_t = \max_{0 \le u \le t} W_u$

$\begin{equation}
f(m,w) = \frac{2(2m-w)}{T\sqrt{2 \pi T}} \exp \left\{ - \frac{(2m-w)^2}{2T} \right\}
\end{equation}$

We have to integrate

$f(\mu,\omega)$ over $\omega \in(-\infty,m), \mu \in (0,m), \mu \ge \omega$

to get the  distribution function of the maximum.

$P[M_t \le m] = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\frac{m}{\sqrt{t}}} e^{-\frac{-x^2}{2}} dx -\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{-\frac{m}{\sqrt{t}}} e^{-\frac{-x^2}{2}} dx $

Differentiating we'll obtain the pdf of the maximum $ f(m) = \frac{2}{\sqrt{2 \pi t}} e^{-\frac{m^2}{2t}} dm$

$m \in (0,\infty)$ is understandable, since $W_0 = M_0 =0$

By the reflection principle, the minimum $\min_{0 \le u \le t} W_u$ has the same pdf,  except that $m \in (-\infty,0)$

By the reflection principle we also know that

$\forall x \ge0: P[\min_{0 \le u \le t} W_u \le x] = P[\max_{0 \le u \le t} W_u \ge -x]$

therefore

$$
\begin{align}
P[\max_{0 \le u \le t} W_u - \min_{0 \le u \le t} W_u \ge x]
&= 2 P[\max_{0 \le u \le t} W_u \ge \frac{x}{2}]\\
&= 2 P[\min_{0 \le u \le t} W_u \le -\frac{x}{2}] \\
&= 2 \frac{2}{\sqrt{2 \pi}}\int_{-\infty}^{-\frac{x}{2}} e^{\frac{-m^2}{2}}{d}m\\
\end{align}
$$

$\forall x \in -\infty, 0)$

the only problem with this solution is that the probability adds together to 2 as opposed to 1. I guess the easiest way to solve this would be by eliminating the 2 in front, but is can't see any mathematical rationale behind doing so. Any suggestions?

ContentLicense: CC BY-SA 2.5


问:
Prove Or Disprove A Set Statement
So I saw this very trick question. Prove or disprove:

$$S=\left \{ \sum_{k=1}^{n}\frac{1}{k(k+1)}:n\in \mathbb{N} \right \}$$

$$T=\left \{\frac{m-1}{m}:m \in \mathbb{N} \right \}$$

Prove or disprove $S = T$. I tried to use different numbers and so far it is looking to be the same. Can someone help me with this?

答:
Following the @Kavi's comment, notice that $$\sum_{k=1}^{n}\frac{1}{k(k+1)}=\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right) $$ so

$$ \sum_{k=1}^{n}\frac{1}{k(k+1)}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}$$ therefore $$\sum_{k=1}^{n}\frac{1}{k(k+1)}=\frac{n}{n+1} $$

Well, indeed $T$ and $S$ are not the same, as the elements of $T$ are $$0,1,\frac{1}{2},\frac{2}{3},\frac{3}{4},\cdots  $$, and the elements of $S$ are $$1,\frac{1}{2},\frac{2}{3},\frac{3}{4},\cdots    $$ you couldn't get $0$ in $S$ as there is no way to sum positive values and get 0. To see why $1\in S$,  take the limit $$\lim_{n\to \infty} \sum_{k=1}^{n}\frac{1}{k(k+1)}=1$$ similarly $1\in T$ because $$ \lim_{m\to\infty} \frac{m-1}{m}=1$$ The sets are equal if $n\geq 2$.

ContentLicense: CC BY-SA 4.0

Title: Prove Or Disprove A Set Statement


问:
Law of large numbers of function of RV
We have that:
$$ \mu = E_{(\mathbf{x},y)\sim f(x,y)} \left[(Y - g(\mathbf{X}))^2 \right] $$

Where the function g(.) tries to estimate Y from $\mathbf{X}$. There is a true function that maps Y from $\mathbf{X}$. An example of such function is $Y = \mu$ and $g_{true}(\mathbf{X}) = \frac{1}{N}\sum_{k=1}^N X_k$, where $\mu$ is the mean of a Gaussian distribution and $X_i$ is a realization of this pdf (supposing $\sigma = 1$) and $f_M(\mu)=\mathcal{U}(0,2)$, i.e $f(x,y)= f(x|y)f(y) = \mathcal{N}(\mu,1)\frac{1}{2}$. So to benchmark this estimator function we use N realizations of $Y$ and $\mathbf{X}$:

$$ \frac{1}{N}\sum_{i=1}^N \left(Y_{i} -g(\mathbf{X}_i) \right)^2  \to E[(Y_i-g(\mathbf{X}_i))^2] $$

As $N \to \infty$ in probability from the law of large numbers. Is it true? If not, how can I assess an estimative of $E[(Y_i-g(\mathbf{X}_i))^2]$?

答:
In general, whenever you can guarantee apriori that the expectation $\mu$ is finite, you can use the Law of Large Numbers (LLN) to arrive at the desired result.

Here $X_i$'s are iid from Normal$(\mu, 1)$ where $\mu$ itself follows Uniform$(0, 2).$ We can set $Y = \mu$ and say that given $Y = y,$ $X_i$'s are iid from $N(y, 1),$ and so we estimate $y$ using the sample mean, $$g(\mathbf{X}) = \overline{X} = \frac{1}{n}\sum_{i=1}^n X_i,$$ and we have $$\mathbb{E}[(Y-g(\mathbf{X}))^2\mid Y] = \mathbb{E}(\overline{X} - y)^2 = \text{Var}(\overline{X}) = 1/n.$$
Now you can use the formula of total expectation to directly calculate $$\mathbb{E}(Y - g(\mathbf{X}))^2 =\mathbb{E}\left[ \mathbb{E}(Y - g(\mathbf{X}))^2\Big| Y\right] = \mathbb{E}\Big[\frac{1}{n}\Big] = \frac{1}{n}.$$

Title: Law of large numbers of function of RV


问:
Prove $(a^3+b^3+c^3)(a^2+b^2+c^2)\ge(a+b+c)(a^2b^2+b^2c^2+c^2a^2)$
Prove that $(a^3+b^3+c^3)(a^2+b^2+c^2)\ge(a+b+c)(a^2b^2+b^2c^2+c^2a^2)$, where a, b, c > 0.

I did LHS-RHS, broke parentheses and cancelled terms.

I think I might use $a^2+b^2+c^2\ge ab+bc+ca$. Then I only need to prove:

$a^5+b^5+c^5\gt abc(a^2+b^2+c^2)$.

答:
Hint:For your last inequality

Apply AM-GM like this $$3a^5+b^5+c^5\ge 5a^3bc$$


答:
Notice the last inequality is $a^5+b^5+c^5\geq abc(a^2+b^2+c^2)=a^3bc+ab^3c+abc^3$. One way to do it would be to quote Muirhead's Inequality directly, since $(5,0,0)\succ(3,1,1)$. Alternatively, using AM-GM, you obtain:

$$a^5+a^5+a^5+b^5+c^5\geq 5a^3bc$$
$$b^5+b^5+b^5+c^5+a^5\geq 5ab^3c$$
$$c^5+c^5+c^5+a^5+b^5\geq 5abc^3$$

I leave you to finish the proof using the above 3 inequalities if you do not want to use Muirhead's.

Title: Prove $(a^3+b^3+c^3)(a^2+b^2+c^2)\ge(a+b+c)(a^2b^2+b^2c^2+c^2a^2)$

Body:
Prove that $(a^3+b^3+c^3)(a^2+b^2+c^2)\ge(a+b+c)(a^2b^2+b^2c^2+c^2a^2)$, where a, b, c > 0.

I did LHS-RHS, broke parentheses and cancelled terms.

I think I might use $a^2+b^2+c^2\ge ab+bc+ca$. Then I only need to prove:

$a^5+b^5+c^5\gt abc(a^2+b^2+c^2)$.


问:
How many 4 digit number can be formed by using 1,2,3,4,5,6,7 if at least one digit is repeated
My approach for this was selecting 4 numbers from which one is being repeated

$$\boxed{A}\boxed{B}\boxed{C}\boxed{A} $$
$$\boxed{A}\boxed{B}\boxed{A}\boxed{A} $$
$$\boxed{A}\boxed{A}\boxed{A}\boxed{A} $$
and permitting them which gives 7$C_4$×4!/2! +7$C_4$×4!/3!+7$C_4$×4!/4! Since 2,3,4 no. Are being repeated numbers are being repeated

What they have done is
$7^4$-7$P_4$ subtracting cases where all 4 no. Are different from total cases. I agree but what is the flaw in my method

答:
The quick way

$$7^4 - \left[\binom{7}{4} \times 4!\right].$$

The hard way:

$$S_1 + S_2 + S_3 + S_4.$$

$S_1 = 7 =$ # ways all 4 numbers the same.

$S_2 = 7 \times 6 \times 4 =$ # ways 3 numbers the same.

$S_3 = \binom{7}{2}\times\binom{4}{2} =$ # ways of having two pairs of numbers.

$S_4 = 7 \times \binom{6}{2} \times 4 \times 3 =$ # of ways of having 2 of one number and two other numbers.


问:
Residue of a 1-form in a Riemann Surface does not depend of the chart
Let's suppose that $X$ is a Riemann Surface, $\omega$ is a meromorphic 1-form in $X$ and let $p$ be a pole of $\omega$ of order $M$. I want to show that the residue of $\omega$ at $p$, defined by
$$
Res_p(\omega )=\dfrac{1}{2\pi i}\int_\gamma\omega
$$
(here $\gamma$ is a closed $C^\infty$ path with $p\in Int(\gamma )$) does not depend of the chart over $X$. For this we will suppose that $\gamma$ is contained on the domain of two charts, $\phi_z:U_z\rightarrow V_z\subset\mathbb{C}$ and $\phi_w:U_w\rightarrow V_w\subset\mathbb{C}$, that is, $\gamma\subset U_z\cap U_w$. Here we'll soppuse tht both charts are centered at $p$ (that is, $\phi_z(p)=\phi_w(p)=0$.

(Please be careful to not be confused with the uses of $\omega$ and $w$!)

What I want is to show that the Residue is invariant under coordinate (or chart) changing.

This is my resolution for this (there is a proof for this in Otto's Lectures on Riemann Surfaces; it is proved as a corolary of the invariance of the integral of a 1-form over a path in Miranda's Algebraic Curves and Riemann Surfaces and in Farkas' Riemann Surfaces it is said that it can be done manipulating power series - which is the way I'm trying to do).

I can assume that in the coordinate $z$, $\omega$ is written as $\omega=f(z)dz$ (if not, then there is another local chart $\psi$ such that after coordinate changing $\omega$ has this form and then we lose no generality). We can then write $z=T(w)$ locally, where $T$ is holomorphic and $T(0)=0$.

Using Laurent Series, we have
$$
f(z)=\sum_{n=-M}^{\infty}c_nz^n
$$
near $p$. Note that $Res_p(\omega)=c_{-1}.$ On the $w$ coordinate, then
$$
\omega=f(T(w))T'(w)dw
$$
and also $$T(w)=\sum_{i=1}^{\infty}a_iw^i$$
therefore
$$
T'(w)=\sum_{i=1}^\infty a_iiw^{i-1}
$$

$\color{blue}{\text{This is the part where comes my dilemma}}$. I must obviously have $T(w)T(w)^{-1}=1$. Describing $T(w)^{-1}=\sum_{i=1}^\infty b_i w^{k_i}$, and working with the series, we have:
$$
\left(\sum_{i=1}^\infty a_i w^i\right)\left(\sum_{i=1}^\infty b_i w^{k_i}\right)=1
$$
$$
a_1b_1xx^{k_1}+[...]=1\Rightarrow k_1=-1\text{ and }b_1=a_1^{-1}
$$
and the rest of the terms will cancel themselves. Here $[...]$ are terms of higher/lower order.

Now, substituting the series in $\omega$,
$$
\omega=\left[\sum_{n=-M}^\infty c_n \left(\sum_{i=1}^\infty a_iw^i\right)^{-n}\right]\left[\sum_{n=1}^\infty na_nw^{n-1}\right]
$$
$$
=\left[ [...]+c_{-1}\left(\sum_{i=1}^\infty a_iw^i\right)^{-1}+[...]\right]\left[ a_1+[...]\right]
$$
$$
=\left[ [...]+c_{-1}(a_1^{-1}w^{-1}+[...])+[...]\right]\left[ a_1+[...]\right]
$$
$$
=[...]+c_1a_1^{-1}a_1w^{-1}+[...]
$$
Since $Res_p(\omega)=c_1$, the invariance is proved.

Can someone tell me if the series manipulation of the part in blue is correct? If not, a hint of how can I do it? Thanks!

答:
The flaw in this computation is that there are many cross-terms that could give you a $w^{-1}$ contribution. For example, what happened to $c_{-2}a_1^{-2}\cdot 2a_2$? The $[...]$ technique works with either higher or lower, but not both at once :)

What I guess you need, if you want to pursue this approach, is to check that each of the merormorphic $1$-forms $\dfrac{dw}{w^j}$, $j=2,\dots,M$, has $0$ residue. This can certainly be done by brute force.

ContentLicense: CC BY-SA 3.0


答:
You’ve miscounted all three of your types with repetitions. This is most obvious in the case of the $AAAA$ type: you counted $\binom74\cdot\frac{4!}{4!}=35$ of them, but there are clearly only $7$: $1111$, $2222$, $3333$, $4444$, $5555$, $6666$, and $7777$. For your $ABAA$ type you give a figure of $\binom74\cdot\frac{4!}{3!}=140$, but this is an underrestimate: there are $7$ ways to choose $A$, then $6$ ways to choose $B$, and finally $4$ possible places to put the $B$ digit, for a total of $7\cdot6\cdot4=168$ possibilities. For the $ABCA$ type you calculate $\binom74\cdot{4!}{2!}=420$; however, there are $7$ ways to choose the $A$ digit, then $\binom42=6$ ways to choose $2$ places for them, $6$ ways to choose the first of the $2$ remaining places, and $5$ ways to choose a digit for the one remaining place, for an actual total of $7\cdot6\cdot6\cdot5=1260$ numbers of this type.

The most obvious problem with your calculation is the binomial coefficient $\binom74$: that’s the number of ways to choose $4$ different digits from the set $\{0,1,2,3,4,5,6,7\}$, and for these types you’re choosing fewer than $4$ different digits. For the first type you’re actually choosing only $3$ different digits, and there are only $\binom73$ ways to do this. Then you have to choose which one of them is to appear twice; there are $3$ ways to do this. Then you can multiply by $\frac{4!}{2!}$ to get the correct figure of $1260$. Similarly, in your second type you’re choosing only $2$ different digits, something that can be done in $\binom72$ ways. Then you have $2$ ways to choose which one of those digits is used only once and $4$ ways to choose where to use it, for a total of $168$.

And as noted in the comments, you missed the $AABB$ type with $2$ distinct digits, each appearing twice. There are $7$ choices for the first digit, $3$ choices for the position of its mate, and $6$ choices for the other digit, for a total of $7\cdot3\cdot6=126$ possible numbers of this type. And sure enough, $7+168+1260+126=1561$.

Title: How many 4 digit number can be formed by using 1,2,3,4,5,6,7 if at least one digit is repeated

Body:
My approach for this was selecting 4 numbers from which one is being repeated

$$\boxed{A}\boxed{B}\boxed{C}\boxed{A} $$
$$\boxed{A}\boxed{B}\boxed{A}\boxed{A} $$
$$\boxed{A}\boxed{A}\boxed{A}\boxed{A} $$
and permitting them which gives 7$C_4$×4!/2! +7$C_4$×4!/3!+7$C_4$×4!/4! Since 2,3,4 no. Are being repeated numbers are being repeated

What they have done is
$7^4$-7$P_4$ subtracting cases where all 4 no. Are different from total cases. I agree but what is the flaw in my method


问:
Prove Set is disjoint
If $$n \in \mathbb{Z}, \geq3$$
$${A_{n}}=\left \{ {an+1: a \in \mathbb{Z}} \right \}$$

$${B_{n}}=\left \{ n|(b+1): b \in \mathbb{Z} \right \}$$

How do we prove that the set is disjoint? So This is easy to choose a number, an integer. Then I prove that An is not a subset of Bn and same time, Bn is not a subset of An. I think I need some help.

答: All elements in $A_n$ are congruent to $1$ modulo $n$ and all elements in $B_n$ are of the form $b=kn-1,k\in\Bbb Z$, hence congruent  to $-1$ modulo $n$. For $n\ge3$, is $1\equiv-1$?

Title: Prove Set is disjoint


答:
As the inequality is homogeneous, assume: $a+b+c = 3$.

We know
$$
(a^2 + b^2 + c^2)^2 \geqslant 3(a^2b^2 + b^2c^2 + c^2a^2) = (a+b+c)(a^2b^2 + b^2c^2 + c^2a^2)
$$
We need to prove now:
$$
a^3 + b^3 + c^3\geqslant a^2 + b^2 + c^2\quad \textrm{with}\quad a+b+c=3
$$
Which is
$$
3(a^3 + b^3 + c^3) \geqslant (a+b+c) (a^2 + b^2 + c^2)
$$Obviously true.

For proof, after expansion, you can add
$$
a^3 + b^3 + c^3 \geqslant a^2b + b^2c + c^2a\ \textrm{and}\ a^3 + b^3 + c^3 \geqslant ab^2 + bc^2 + ca^2
$$Proof of these are only by AM-GM.


问:
Prove that the equation $\tan (z)=z$ has only real roots.
Prove that the equation $\tan(z)=z$ has only real roots. How to do it?
The idea is that the increment of the argument need to look at the boundary of the square with a side of $\pi n$ and another that $\tan(z)-z$ has a pole at $0$. I do not know how to use it. </p>

答:
Suppose $\tan(x+iy)=x+iy$. Use the expansion $\tan(x+iy)=\frac{\sin2x}{\cos2x+\cosh2y}+i\frac{\sinh 2y}{\cos 2x+\cosh 2y}$ to get: $$\frac{1}{\cos2x+\cosh2y}(\sin2x+i \sinh2y)=x+iy .$$

This means that the 2 vectors $(\sin2x,\sinh2y)$ and $(x,y)$ are proportional, and therefore the determinant of the matrix $\begin{pmatrix}x & y \\ \sin2x &\sinh2y \end{pmatrix}=0$.

We get $x \sinh2y=y \sin2x$. Using the well-known inequalities $$|\sin t| \leq |t|,|\sinh t| \geq |t|$$ we see that we must have $x=0$ or $y=0$ (this is true because equality holds in the inequalities above iff $t=0$).

The case $y=0$ gives the wanted real solutions, and a small calculation shows that the case $x=0$ gives the one-and-only imaginary solution, namely $z=0$.


问:
The Functional Equation $ ( x + y ) \big( f ( x ) - f ( y ) \big) = ( x - y ) f ( x + y ) $, need solution have answer
Find all functions $ f: \mathbb R \to \mathbb R $ such that for all reals $ x $ and $ y $,
$$ ( x + y ) \big( f ( x ) - f ( y ) \big) = ( x - y ) f ( x + y ) \text . $$

I actually got the answer by guessing and checking, $ f ( x ) = a x ^ 2 + b x $, but I want to see the solution. My friend suggested surjectivity but I don't see how to continue.

$ f ( x ) = f \left( \frac x { f ( x ) ^ 2 } \right) $;
this is what I just got.

The answer is correct, just need a solution. Thanks!

答:
For any real numbers $ a $ and $ b $, if we define the function $ g : \mathbb R \to \mathbb R $ with $ g ( x ) = f ( x ) - a x ^ 2 - b x $, then using the original functional equation, we'll have
$$ ( x + y ) \big( g ( x ) - g ( y ) \big) = ( x - y ) g ( x + y ) \text . \tag 0 \label 0 $$
In particular, setting $ a = \frac { f ( 1 ) + f ( - 1 ) } 2 $ and $ b = \frac { f ( 1 ) - f ( - 1 ) } 2 $, we will also have $ g ( 1 ) = g ( - 1 ) = 0 $. Letting $ y = 1 $ in \eqref{0}, we get
$$ ( x + 1 ) g ( x ) = ( x - 1 ) g ( x + 1 ) \text , \tag 1 \label 1 $$
while substituting $ x + 1 $ for $ x $ and $ - 1 $ for $ y $ in \eqref{0} we have
$$ x g ( x + 1 ) = ( x + 2 ) g ( x ) \text . \tag 2 \label 2 $$
\eqref{1} and \eqref{2} together show that
$$ 2 g ( x ) = x ( x + 1 ) g ( x ) - ( x - 1 ) ( x + 2 ) g ( x ) = x ( x - 1 ) g ( x + 1 ) - ( x - 1 ) x g ( x + 1 ) = 0 \text . $$
Therefore $ g $ is the constant zero function, and hence $ f ( x ) = a x ^ 2 + b x $ for all $ x \in \mathbb R $. It's straightforward to see that any function of this form satisfies the original functional equation, and thus they form the class of all the solutions.


问:
Using Bayes theorem what is the probability of dropping the call after 50sec given call was not dropped during first 20 sec.
I have probabilities of dropping a call based on the duration of time customer has waited for before dropping.

Probability of dropping during  0-20sec: 0.23

Probability of dropping during 20-40sec: 0.11

Probability of dropping during 40-50sec: 0.04

Probability of dropping during 50-60sec: 0.61

Now I want to know the probability of dropping the call during 50-60sec when it is known that the call has not been dropped during 0-20sec.

How can we solve it using Bayes theorem ?

答:
Simply

$$\frac{0.61}{1-0.23}=\frac{61}{77}$$

This simple operation is exactly Bayes Theorem, or Conditional Probability formula...it's the same

Set:

$A$ as the event:  "dropping during 50-60sec"

$B$ the complement of the event "dropping during 0-20"

$$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}$$

being $A \subset B$


问:
Bhaskara approximation of cos(x)
In my knowledge the best aproximation of $\sin$ and $\cos$ is Bhaskara approximation which is :
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$
and
$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$

what is the best approximation of $\cos(x)$ for $ -\pi\leq x\leq 0$ (by a regular function ) ?

答:
In terms of approximation, you can do a lot of things depending on how many terms you want to include in the expression. Most of the time, this is done using Padé approximants.

For example, for your question, using the simple $[3,2]$  Padé approximant built around $x=-\frac \pi 2$, we get
$$\cos(x)=\left(x+\frac{\pi }{2}\right)\frac {1-\frac{7}{60} \left(x+\frac{\pi
   }{2}\right)^2 } {1+\frac{1}{20}\left(x+\frac{\pi
   }{2}\right)^2  }$$ which leads to an absolute error of $0.0042$ at the bounds.

For sure, you could go further and use the $[5,4]$  Padé approximant
$$\cos(x)=\left(x+\frac{\pi }{2}\right)\frac {1-\frac{53}{396} \left(x+\frac{\pi
   }{2}\right)^2+\frac{551 }{166320}\left(x+\frac{\pi
   }{2}\right)^4 } {1+\frac{13}{396} \left(x+\frac{\pi }{2}\right)^2+\frac{5 }{11088}\left(x+\frac{\pi
   }{2}\right)^4 }$$ which leads to an absolute error of $3.0\times 10^{-6}$ at the bounds. This is equivalent to the Taylor series to $O\left(\left(x+\frac{\pi   }{2}\right)^{11}\right)$.

To give you an idea of the possible improvements; consider that $P_n$ is the $[2n+1,2n]$ Padé approximant of $\cos(x)$ built around $x=-\frac{\pi
   }{2}$ and compute the norm
$$\Phi_n=\int_{-\pi}^0 \big[\cos(x)-P_n\big]^2\,dx$$

$$\left(
\begin{array}{cc}
 n & \Phi_n \\
 1 & 3.92 \times 10^{-6} \\
 2 & 1.26 \times 10^{-12}\\
 3 & 2.23 \times 10^{-20}
\end{array}
\right)$$

All of these have been done only based on the information (function and derivative values) at  $x=-\frac{\pi}{2}$.

But we can try to cover the whole range and consider for example
$$\cos(x) \sim \left(x+\frac{\pi }{2}\right)\frac {1+a \left(x+\frac{\pi
   }{2}\right)^2+b\left(x+\frac{\pi
   }{2}\right)^4 } {1+c \left(x+\frac{\pi }{2}\right)^2+d\left(x+\frac{\pi
   }{2}\right)^4 }$$ and obtain

$$a=\frac{-1952+672 \pi -192 \pi ^2+56 \pi ^3}{\pi ^3(\pi ^2-10)}$$
$$b=\frac{-3456+1056 \pi +768 \pi ^2-240 \pi ^3 } {\pi ^5 \left(\pi ^2-10\right) }$$
$$c=\frac{-192+72 \pi -104 \pi ^2+32 \pi ^3 } {\pi ^2 \left(\pi ^2-10\right) } $$
$$d=\frac{-4704+1504 \pi +400 \pi ^2-128 \pi ^3} {\pi ^4 \left(\pi ^2-10\right) }$$ which gives a maximum absolute error equal to $5.96\times 10^{-8}$  and a norm equal to $2.73\times 10^{-15}$ (which is $461$ times smaller than $\Phi_2$).

You could notice that the coefficients are extremely close to those of $P_2$ but these minor changes make a lot of difference.

All these approximations have been built for answering your question.

Title: Bhaskara approximation of cos(x)


问:
How to find the angle which is in a trapezoid formed by two triangles?
The problem is as follows:

The figure from below represents a quadrilateral $ABCE$. Using the
information, $BE=BC$,  $\angle\,BAC=\angle\,ADB=60^{\circ}$ and
$DC=10\,m\,AE=10\,m$. Find the angle $x$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&125^{\circ}\\
2.&140^{\circ}\\
3.&130^{\circ}\\
4.&120^{\circ}\\
\end{array}$

I couldn't really find a relationship other than spot an isosceles triangle on $\triangle BEC$. This would mean that $\angle BEC= \angle BCE$. I also could spot that the $\triangle ABD$ is an equilateral. But other than that I'm stuck. What sort of identity or construction here is required to solve this problem?.

Can somone help me?. Can this be solved relying only in Euclidean postulates?. Please an answer must include a drawing because in this figure I can't find exactly where to look for, but I think that it is related with congruence or maybe similarity.

答:
The following proportionately drawn figure shows how the problem has been set up. ( Labels as in given problem ):

Two equilateral triangles whose sides are scaled to a ratio

$$\dfrac{BE}{BA}=\dfrac{BC}{BD}$$

are rotated about fulcrum vertex $B$ from position $E$ to $C$ through  $60^{\circ}$.

The yellow triangle can be considered to be rigid hence congruent (like when made of plastic or cardboard).

So the external $\angle BAE $ in given figure is $120^{\circ} $ that gets carried forward. Option 4.


问:
Is $(0,0,0)$ saddle point?
$f(x,y,z)=x^3+y^3+z^3-3xy-3yz$

Is $(0,0,0)$ saddle point?

$\nabla f(0,0,0)=0$, so $(0,0,0)$ is one of the stationary point.

Also, because of my posture The reason why $f(0,0,0)$ is not a extreme value. ,
$f(0,0,0)$ is not a extreme value.

Therefore, I think $(0,0,0)$ is a saddle point.

However, according to wolfram alpha, saddle points of $f(x,y,z)=x^3+y^3+z^3-3xy-3yz$ don't exist.

答:
The Hessian at $(0,0,0)$ is:
\begin{pmatrix}
0 & -3 & 0 \\
-3 & 0 &-3 \\
0 & -3 & 0
\end{pmatrix}
Its npn-zero eigenvalues are $\pm3\sqrt2$. So by definition it is indefinite (not positive or negative definite) and so $(0,0,0)$ is a saddle point.

Title: Is $(0,0,0)$ saddle point?


问:
Showing two rings are not isomorphic
Consider $\mathbb{C}[x,y]/(y^2-x^3)$ and $\mathbb{C}[x,y]/(y^2-x^3-x^2)$.
Someone told me that these two rings are not isomorphic, but I don't know how to prove.
Please help me.

答:
As mentioned by Stahl in the comments, if these rings were isomorphic, then $\mathbb{C}[[x, y]] / (y^2 - x^3)$ and $\mathbb{C}[[x, y]]/(y^2 - x^3 - x^2)$ would be isomorphic.

But the former ring is an integral domain, and the latter is not.

Proof that $\mathbb{C}[[x, y]] / (y^2 - x^3)$ is an integral domain: map $\mathbb{C}[[x, y]] \to \mathbb{C}[[t]]$ by $x \mapsto t^2$ and $y \mapsto t^3$. Let's check that the kernel is exactly $(y^2 - x^3)$. Any power series
$$
f(x, y) = a_{00} + a_{10}x + a_{01}y + a_{20}x^2 + a_{11}xy + a_{02}y^2 + a_{30} x^3 + \ldots
$$
in the kernel of this map satisfies
$$
a_{00} + a_{10}t^2 + a_{01}t^3 + a_{20}t^4 + a_{11}t^5 + (a_{02} + a_{30})t^6 + \ldots = 0,
$$
so in particular all the coefficients below $a_{02}$ are zero, and for each $n$,
$$
\sum_{2i + 3j = n} a_{ij} = 0.
$$
We are reduced to showing that any polynomial of the form
$$
\sum_{2i + 3j = n} a_{ij}x^iy^j
$$
is divisible by $y^2 - x^3$ when the sum of its coefficients is zero. We can replace every copy of $y^2$ with $x^3$ in the polynomial to check this, so (depending on the parity of $n$) our expression is either $y \sum a_{ij} x^{(n-1)/2}$ or $\sum a_{ij} x^{n/2}$, which is obviously zero either way.

Proof that $\mathbb{C}[[x, y]]/(y^2 - x^3 - x^2)$ is not an integral domain: we have $x^3 + x^2 = x^2(1+x)$. Now $(1+x)$ is a square in $\mathbb{C}[[x, y]]$, using the Taylor expansion of the square root function, that is, there exists $z$ with $x^3 + x^2 = z^2x^2$. Then $(y+zx)(y-zx) = y^2 - x^3 - x^2 =  0$ in the quotient, but neither $y + zx$ nor $y - zx$ are (they are degree $1$ in $y$ and $zx$ doesn't have any $y$-terms).

(Technically this only proves there is no $\mathbb{C}$-algebra isomorphism; I am not sure if there is an exotic ring isomorphism that is non-linear on the $\mathbb{C}$ part, since you have a bunch of transcendental garbage in $\mathbb{C}$ to work with in that case.)

(EDIT: oh, it's fine. The first ring has the property that for any maximal ideal, the completion is a domain. The latter doesn't have this property. So they can't be isomorphic.)

Title: Showing two rings are not isomorphic


问:
Why did Munkres write "unless one of the spaces $X$ or $Y$ happens to be empty"? On p.87 in "Topology 2nd Edition".
I am reading "Topology 2nd Edition" by James R. Munkres.

On p.87, there is the following definition:

Definition. Let $\pi_1 : X \times Y \to X$ be defined by the equation $$\pi_1(x, y) = x;$$ let $\pi_2 : X \times Y \to Y$ be defined by the equation $$\pi_2(x,y)=y.$$
The maps $\pi_1$ and $\pi_2$ are called the projections of $X \times Y$ onto its first and second factors, respectively.

We use the word "onto" because $\pi_1$ and $\pi_2$ are surjective (unless one of the spaces $X$ or $Y$ happens to be empty, in which case $X \times Y$ is empty and our whole discussion is empty as well!).

For example, if $X = \emptyset$, then $X \times Y = \emptyset$.

So if $Y \ne \emptyset$, then $\pi_2$ is the empty mapping $\emptyset \to Y$.

So $\pi_2$ is not surjective.

But,

for example, if $X = \emptyset$, then $X \times Y = \emptyset$.

So $\pi_1$ is the empty mapping $\emptyset \to \emptyset$.

So $\pi_1$ is bijective.

So $\pi_1$ is surjective.

Why did Munkres write "unless one of the spaces $X$ or $Y$ happens to be empty"?

答:
If $X\ne\emptyset$ and $Y=\emptyset$, then $X\times Y=\emptyset$ and $\pi_1$ is the empty map, which is not surjective. Symmetrically if $Y\ne\emptyset$ and $X=\emptyset$.

There might be a nuance in the English interpretation of "unless", though I am certain Munkres, nor any English-speaking mathematician, would use "$A$ unless $B$" as synonymous with "$A$ if and only if not $B$". Therefore let's say that, if you want as precise a statement as it can be, it should be that $\pi_1$ and $\pi_2$ are both surjective if and only if either both spaces are non-empty or they both are empty. Or, even better, that the projection to one of the factors is surjective if and only if that factor is empty or the other one isn't.


问:
Finding all distinct primes $p$ and $q$ such that $pq^3 + p^3q$ has exactly 12 divisors

How do you find all possible distinct prime numbers $p$ and $q$ such that $pq^3 + p^3q$ has exactly $12$ divisors?

Any help would be appreciated. Not sure if factoring into $pq(q^2+p^2)$ does anything, but I am lost on this.

答:
No solution.

As already indicated by David Stork's comment, since
$pq$ will have 4 divisors, $p^2 + q^2$ must have 3 divisors.  This means that
you must have a pythagorean triplet $p^2 + q^2 = P^2$, where $P$ is a prime.

Researching https://en.wikipedia.org/wiki/Pythagorean_triple,
you can assume WLOG that $q = 2.$

This means that $P^2 - 4 = p^2 \implies (P-2)(P+2) = p^2.$

Since $p$ must be odd, $p$ can't divide $2$.

This means that $p$ can't divide both $(P-2)$ and $(P+2).$

This means that $P-2 = 1.$
That doesn't work.


问:
finding certain colourings

Let $a\geq 2$ and $n\geq a.$ Let $K_n = (V,E)$ be the complete graph on $n$ vertices. A colouring of $E$ is a function $f : E\to \{green, orange\}.$ For an a-element subset $S$ of $V$, say a colouring is poor for $S$ if all edges with both ends in $S$ are assigned the same colour. Show that if $n\leq \sqrt{2}^{a-1},$ there is a colouring of $E$ that is not poor for any $S$, by first finding the number of colourings for $E$ and an upper bound for the number of colours of $E$ that are poor for some $a$-element subset.

The number of colourings for $E$ should be $2^{{n\choose 2}}$. But isn't this also an upper bound for the number of colourings of $E$ that are poor for some $S$? I think I need to find a better upper bound and use the fact that ${n\choose a} < \dfrac{n^a}{a!} \leq \dfrac{n^a}2.$ If I could show that the upper bound for the number of colourings of $E$ that are poor for some $S$ is strictly less than $2^{{n\choose 2}},$ then clearly there must be some colouring of $E$ that isn't poor for any $S$.

答:
For a fixed $a$-element set $S$ that we want to be poor, there are $2$ ways to color the edges with both ends in $S$ and $2^{\binom n2 - \binom a2}$ ways to color the other edges. Multiply this by $\binom na$ and we get
$$
   \binom na 2^{\binom n2 - \binom a2 + 1}
$$
as an upper bound.

If we want this to be less than $2^{\binom n2}$, then we want $2\binom na < 2^{\binom a2}$. We don't have to try hard: assuming $a \ge 2$,

$2\binom na < \frac{2n^a}{a!} \le n^a$.

$2^{\binom a2} = (\sqrt2^{a-1})^a$.

So if $n \le \sqrt2^{a-1}$, then $n^a \le 2^{\binom a2}$, from which we conclude $2 \binom na < 2^{\binom a2}$; therefore a coloring exists that is not poor for any $S$.

Title: finding certain colourings


问:
Differentiable almost every where
I need help in this problem ,royden 4th edition .
Let $g$ be integrable over $[a,b]$ .Define the antiderivative of $g$ to be the function $f$ defined on $[a,b]$ by  $f(x)=\int_{a}^{x} g~~ for~all~ x\in [a,b]$. Show that $f$ is differentiable almost everywhere on $(a,b)$.

I know from Lebesgues Theorem if the function $f$ is monotone on the open intervals $(a,b)$ , then it is differentiable almost everywhere on $(a,b)$ But integrable not always monotone .

and I can not show that if $f$ is not differentiable at $x$ then $x \in E$ and $m(E)=0$ These are the only information I have about differentiable almost everywhere

I saw your comment but I can not add comment , how can I discusses your comment with you

答:
Based on the comments from @Daniel Fischer

Since $g = |{g}| - (|{g}|-g)$ and $g$ is integrable, the linearity of Lebesgue integral implies
\begin{align*}
  f(x) = \int_a^x g = \int_a^x |{g}| - (|{g}|-g) = \int_a^x |{g}| - \int_a^x(|{g}|-g) = f_1(x)-f_2(x),
 \end{align*}
where $f_1(x) = \int_a^x |{g}|$ and $f_2(x) =  \int_a^x(|{g}|-g)$. Since $|{g}|\geq 0$ and $|{g}|-g\geq0$, the functions $f_1$ and $f_2$ are increasing on $(a,b)$. Thus, $f_1$ and $f_2$ are differentiable almost every on $(a,b)$. Since addition preserves differentiability, $f$ is differentiable a.e. on $(a,b)$.

Title: Differentiable almost every where


答: I don't see any problem. The statement "$\pi_1\text{ and }\pi_2\text{ are surjective}$" is not true when either $X\text{ or }Y$ is empty, since then atleast one of $\pi_1,\pi_2$ are not surjective as you have seen. Note that the negation of "$\pi_1\text{ and }\pi_2\text{ are surjective}$" is "$\pi_1\text{ OR }\pi_2\text{ is not surjective}$".

Title: Why did Munkres write "unless one of the spaces $X$ or $Y$ happens to be empty"? On p.87 in "Topology 2nd Edition".

Body:
I am reading "Topology 2nd Edition" by James R. Munkres.

On p.87, there is the following definition:

Definition. Let $\pi_1 : X \times Y \to X$ be defined by the equation $$\pi_1(x, y) = x;$$ let $\pi_2 : X \times Y \to Y$ be defined by the equation $$\pi_2(x,y)=y.$$
The maps $\pi_1$ and $\pi_2$ are called the projections of $X \times Y$ onto its first and second factors, respectively.

We use the word "onto" because $\pi_1$ and $\pi_2$ are surjective (unless one of the spaces $X$ or $Y$ happens to be empty, in which case $X \times Y$ is empty and our whole discussion is empty as well!).

For example, if $X = \emptyset$, then $X \times Y = \emptyset$.

So if $Y \ne \emptyset$, then $\pi_2$ is the empty mapping $\emptyset \to Y$.

So $\pi_2$ is not surjective.

But,

for example, if $X = \emptyset$, then $X \times Y = \emptyset$.

So $\pi_1$ is the empty mapping $\emptyset \to \emptyset$.

So $\pi_1$ is bijective.

So $\pi_1$ is surjective.

Why did Munkres write "unless one of the spaces $X$ or $Y$ happens to be empty"?


问:
Find coefficient of $x^r$ in $(x^5+x^6+x^7+...)^8$
My work:

We can rewrite the generating function $(x^5+x^6+x^7+...)^8$ as $x^{40}(1+x+x^2+...)^8$

We are looking for $x^{r-40}$ coefficient in the new generating function  $(1+x+x^2+...)^8$

We can rewrite $(1+x+x^2+...)^8$ as $(\cfrac{1}{1-x})^8$ = $(1-x)^{-8}$

So coefficient can be found by ${-8 \choose r-40} = {r - 33 \choose r-40 }$

so final answer  is ${r - 33 \choose r-40 }$

Is this correct?

This is a concept question so I want to make sure I'm understanding the material properly.

答:
There is a very close association between stars and bars and generating functions: see https://brilliant.org/wiki/integer-equations-star-and-bars/

Alternative approach

you want the # of non-negative integer solutions to

$x_1 + x_2 + x_3 + \cdots + x_8 = (r-40).$

This is immediately seen to be

$$\binom{[r-40] + [8-1]}{[8-1]}.$$


问:
Understanding why a generalized inverse of a right-continuous function is right-continuous
I'm trying to prove that the generalized iniverse $g^- (s) =\inf_t \{t:a(t)>s\}$  of a bounded from below, nondecreasing and right-continuous function is right-continuous.

There is a proof here which I'm not able to follow since I'm not sure what $\{t:a(t)>s\} = \inf_{\epsilon>0} \{t:a(t)>s+\epsilon\} $ means - though I would guess it means something like $\{t:a(t)>s\} = \{t:a(t)>s+\epsilon\}$ where $\epsilon $ is the greates lower bound of the set of real numbers $\eta \ge 0 $ such that for every $t $ in the left hand side $a(t)>\eta $. But I'm not able do do the rest of the proof using this.

Instead I tried the following: For any $s$ there are two cases: either $a(t) $ atains the value $s $ at some point $t_0$ or it "jumps past it". In the later case there exists an $\epsilon>0 $ such that $f $ also "jumps past" $s+ \epsilon $ in which case we have $\{t:a(t)>s\}=\{t:a(t)>s +\epsilon \}$ meaning that the generalized inverse $g^- $ is constant on $[y,y+\epsilon) $ and thus certainly right-continuous at $y $.

Is there an intuitive way to reason from the following two pictures, the first of a function $a(t)$ and the second of it's generalized invers $g^-(s)$, that from the conditions on $a $ follows the right-continuity of $g^- $ - and of course supply the abstract solution.?

Thanks in advance!

答:
Prove $\{t:a(t)>s\}=\bigcup\limits_{\epsilon>0}\{t:a(t)>s+\epsilon\}$.

Take $\inf$ on both sides, we have $\inf\{t:a(t)>s\}=\inf\bigl[\bigcup\limits_{\epsilon>0}\{t:a(t)>s+\epsilon\}\bigr]$.

We can prove that $\inf\bigl(\bigcup\limits_{\epsilon>0}\ldots\bigr)=\inf\limits_{\epsilon>0}(\inf\ldots)$. That is, informally, inf of a union is inf of each union element's inf. A similar proof of this equality can be seen here.

So, we have $\inf\{t:a(t)>s\}=\inf\limits_{\epsilon>0}\bigl[\inf\{t:a(t)>s+\epsilon\}\bigr]$, which is exactly $g^-(s)=\inf\limits_{\epsilon>0}g^-(s+\epsilon)$.

Prove that $g^-(s)$ is increasing.

When a function is increasing, its right limit is just the $\inf$ of the function values on the right, i.e., informally, $\lim\limits_{\epsilon\to0^+}$ is $\inf\limits_{\epsilon>0}$. Applying this to the right side of 4., we have $g^-(s)=\lim\limits_{\epsilon\to0^+}g^-(s+\epsilon)$, which is just the right-continuity.


问:
$x\mapsto x^k$ is an automorphism on $C_n$ iff $\gcd(n,k)=1$
Let $G = \langle g\rangle$ be a cyclic group of order $n$ (with multiplication as its operation). Prove that $f(x) = x^k$ is an automorphism on $G$ if and only if $\gcd(n, k) = 1$.

答: Since $G$ is cyclic, let $g$ be a generator.  That is, suppose $G=\langle g\rangle$.  Then since the order of $g^k$ is $n/\rm{gcd}(n,k)$, we have that under the mapping $x\stackrel{f}\mapsto x^k$, $g$ goes to another generator if and only if $\rm{gcd}(n,k)=1$.  In these cases, and only these cases, $f$ is an automorphism.


问:
Why is the factorial function always nonzero?
The factorial function is defined as
$$n!=\prod_{k=1}^{n}k$$
For $n>0$, $n$ being an integer. This definition can be extended to the complex numbers with positive real part by the Gamma function:
$$\Gamma(n)=(n-1)!$$
This is the plot of the factorial function, in which the values of the factorial of non-integral numbers are evaluated by the gamma function:

It can be seen that for the positive numbers, the factorial is always nonzero. Why is it so?

According to mathguy's comment, the factorial function is positive for all the complex numbers. Is this true? If yes, then why is it so?

答:
The complex variable function $z! = \Gamma(z+1)$ has no zeros anywhere in the complex plane.  Why?  One way to see this is to use Euler's reflection formula,
$$
z!(-z)! = \frac{\pi z}{\sin \pi z}, \quad z \neq 0
$$

See ProofWiki's "Euler's Reflection Formula" entry for a proof.

If $z!$ were ever zero, then the above formula would also be zero, but that obviously is not the case.  In fact $1/(z!)$ is an entire function, that is it is analytic for all $z$, and in particular has no poles, implying $z!$ has no zeros.

EDIT:

This does not necessarily preclude the case where $z!$ has a pole (is infinite) and $(-z)!$ is zero.  However, the only poles of $z!$ are at the negative integers, $z=-1, -2, \cdots$, which can be seen from the product formula,
$$
z! = \Gamma(z+1) = \Pi_{n=1}^\infty \frac{n}{n+z}\Big( 1+\frac{1}{n}\Big)^z
$$
convergent for all $z \neq -1, -2, \cdots$. At these values $-z$ is a positive integers and then $(-z)!$ is known to be non zero.


问:
numerical integration $f\left( x\right) =1-4\left( x-\dfrac{1}{2}\right) ^{2}$
$$f\left( x\right) =1-4\left( x-\dfrac{1}{2}\right) ^{2}$$
I want to be integrated this.
If you calculate normally, it will be like this
$$\int ^{1}_{0}\left( 1-4\left( x-\dfrac{1}{2}\right) ^{2}\right) dx=\dfrac{2}{3}\approx 0.66667\ldots$$

But I must use numerical integration.
$$\sum ^{n-1}_{i=0}0.5\times\left( f\left( x_{i}\right) +f\left( x_{i+1}\right) \right) \times d$$
If I use this calculate, I can answer the approximation of integration.

I think.

Put $n=100$
$$\begin{aligned}\sum ^{99}_{i=0}0.5\times \left( f\left( \dfrac{i}{100}\right) +f\left( \dfrac{i+1}{100}\right) \right) \times \dfrac{1}{100}\\ =\sum ^{99}_{i=0}0.5\times \left( 1-4\left( \dfrac{i}{100}-\dfrac{1}{2}\right) +1-4\left( \dfrac{i+1}{100}-\dfrac{1}{2}\right) \right) \times \dfrac{1}{100}\end{aligned}$$
$$=-\dfrac{6767}{5000}=-1.3534$$

Put $n=1000$
$$\sum ^{999}_{i=0}0.5\times\left( f\left( \dfrac{i}{1000}\right) +f\left( \dfrac{i+1}{1000}\right) \right)\times\dfrac{1}{1000}$$
$$=\sum ^{999}_{i=0}0.5\times \left( 1-4\left( \dfrac{i}{1000}-\dfrac{1}{2}\right) +1-4\left( \dfrac{i+1}{1000}-\dfrac{1}{2}\right) \right)\times\dfrac{1}{1000}$$
$$=-\dfrac{667667}{500000}= -1.335334$$

Obviously the approximation is wrong.
Where I made a mistake, please tell me.

答:
As I wrote in a very early comment, it seems that you lost the squares and for $n=100$ the formula should be
$$I=\sum ^{99}_{i=0}0.5\times \left( 1-4\left( \dfrac{i}{100}-\dfrac{1}{2}\right)^{\color{red}{2}} +1-4\left( \dfrac{i+1}{100}-\dfrac{1}{2}\right)^{\color{red}{2}} \right) \times \dfrac{1}{100}$$
Making it for general $n$
$$I=\sum ^{n-1}_{i=0}\frac 12\times \left( 1-4\left( \dfrac{i}{n}-\dfrac{1}{2}\right)^{2} +1-4\left( \dfrac{i+1}{n}-\dfrac{1}{2}\right)^{2} \right) \times \dfrac{1}{n}$$ Expanding
$$I=\sum ^{n-1}_{i=0} \frac {2( n-1)+ 4( n-1)i-4 i^2 } {n^3}=\frac{2 \left(n^2-1\right)}{3 n^2}=\frac{2}{3}-\frac{2}{3 n^2}$$ So, for $n=100$, the error is $\frac{1}{15000}=0.000067$.


答:
For $p,q$ distinct primes, $pq$ always has exactly $4$ divisors. It must be that
$q^2+p^2$ has $3$ divisors which means that $q^2+p^2 = x^2 \equiv 0,1 \mod 3$ and working in $\mod 3$ we see that there is no solution when $p,q \neq 3$. If $p=3$ then we only have solutions for $9+q^2 = x^2$ with composite $q$.

Title: Finding all distinct primes $p$ and $q$ such that $pq^3 + p^3q$ has exactly 12 divisors

Body:
How do you find all possible distinct prime numbers $p$ and $q$ such that $pq^3 + p^3q$ has exactly $12$ divisors?

Any help would be appreciated. Not sure if factoring into $pq(q^2+p^2)$ does anything, but I am lost on this.


问:
$(a^4+b^2) \leq (a^2+b^2)^2$ for all $a, b \in \mathbb{R}_+$?
Let $a, b \in \mathbb{R}_+$.

Question. Is valid that
$$(a^4+b^2) \leq (a^2+b^2)^2?\tag{1}$$

I tried what follows:
$$(a^2+b^2)^2=a^4+2a^2b^2+b^4 \geq a^4+2a^2b^2.$$
For this, can I to conclude that $(1)$ holds?

Body:
If $b\ne 0$ then $b^2>0$ so $(b^2C\le b^2D\iff C\le D)$ for any $C,D.$

So with $C=1$ and $D=2a^2+b^2,$ if $b\ne 0$ then $$a^4+b^2\le (a^2+b^2)^2 \iff a^4+b^2\le a^4+2a^2b^2+b^4 \iff$$ $$\iff b^2\le 2a^2b^2+b^4\iff$$ $$\iff b^2(1)\le b^2(2a^2+b^2)\iff$$ $$\iff 1\le 2a^2+b^2.$$ The last inequality above cannot be true for every $a,b>0.$ E.g. if $a=b=1/1000000000000.$


问:
Show that this map is a submersion.
Let $U\subset \mathbb{R}^n$ an open set and consider $f\in C^{\infty}(U,\mathbb{R})$ i need to show that the map
$$
\begin{array}{cccc}
F \ : & \! U\times (\mathbb{R}^n)^* & \! \longrightarrow
& \! \mathbb{R} \\
& \! (x,A) & \! \longmapsto
& \! f(x)+Ax
\end{array}
$$

it's a submersion. Well I tried calculate $D_1F_{(x,A)}=df(x)+A$, so when $A=-df(x)$ this is are zero, it's true this question?

答: You have to show that the total derivative $DF(x,A):\Bbb R^n \times (\Bbb R^n)^*\to \Bbb R$ is surjective. So pretend that $x$ and $A$ depend on a parameter $t$ and take the derivative with respect to $t$ to get $$DF(x,A)(\dot{x},\dot{A}) = Df(x)(\dot{x}) + \dot{A}x + A\dot{x}.$$Now, freeze $x$ and $A$, take $c \in \Bbb R$, and ask yourself: is it possible to solve $$Df(x)(\dot{x}) + \dot{A}x + A\dot{x}=c$$for $\dot{x}$ and $\dot{A}$? If $f = 0$ and $(x,A) =(0,0)$ it is impossible, as $DF(x,A)$ is the zero map. This shows that in general you need more assumptions. Rewrite the equation as $$(Df(x)+A)\dot{x} + \dot{A}x=c.$$If $x\neq 0$, then you can define a linear map $\dot{A}:\Bbb R^n \to \Bbb R$ by $\dot{A}x=c$ and $\dot{A}|_{x^\perp}=0$, so that $DF(x,A)(0,\dot{A})=c$, as wanted. We conclude that if $0\not\in U$, then $F$ is a submersion.

Title: Show that this map is a submersion.


问:
Ratio & Proportion - Splitting the ratio
A picture is made up of just two colours, red and green. They are used in the ratio 4:3. In the picture, if the ratio of red and green in the lower half of the picture is 2:3, then what is the ratio of red and green in the upper half?

Solution which we have

Total parts = 7
Multiply 7 to ratio lower half. You will get 14:21. So, total number of parts is 35. Then the total number of parts in upper half also 35. If we sum both it is 70.

Then, actual value for the total amount is 40 and 30. If we subtract with lower half, then we will get 26 and 9

Doubts which we have

Why ratio is summed and multiplied with lower half ratio?

If we are given the upper half ratio 26 and 9, how this problem would be approached?

Whether this problem can be approached differently?

To which questions, we can use this approach?

答:
Both the upper and lower half of the image consists of the same no. pixels that is the same no. red +green (per say) in that particular half . Think of it like you have red and green balls and u fill two halves of the box with them but in different ratios in each of them .
considering this in mind ,let say they occur in upper half in ratio x:y , the ratio of red balls in upper half is x:x+y and in the lower half is 2:(2+5) . The average of these ratios is given to us , for red is 4:(4+3) . Solve for x,y using this :
x/x+y + 2/5 = 2*(4/7) [as the average of these ratios is 4:7 therefore 2*(4/7)]

Similarly for green :
y/x+y  + 3/5 = 2*(3/7)

solving u get x=26 , y=9 so x:y is 26:9 . The main point here is the total no of pixels are same and red and green are divided among those pixels in a certain ratio . Hope this solution makes sense xD


问:
Proving that if $n$ is odd and $\gcd(m, n) = 1$, then $\gcd(2m + n, 2n) = 1$
I've been trying to crack this one for the last little while. I've tried a few approaches, but none have bore any fruit.

Let $n > 0$ be an odd integer. Prove that if $\gcd(m, n) = 1$, then $\gcd(2m + n, 2n) = 1$.

Body:
Since we have that $(m,n)=1$, there are $x,y$ so that
$$
mx+ny=1\tag{1}
$$
Then,
$$
\begin{align}
(\color{#C00000}{2m+n})2x+\color{#C00000}{2n}(2y-x)&=\color{#00A000}{4}\\
(\color{#C00000}{2m+n})n-\color{#C00000}{2n}m&=\color{#00A000}{n^2}
\end{align}\tag{2}
$$
Since $n$ is odd, there are $w,z$ so that
$$
2w+nz=1\tag{3}
$$
Squaring $(3)$ gives
$$
\color{#00A000}{4}(w^2+wnz)+\color{#00A000}{n^2}z^2=\color{#0000FF}{1}\tag{4}
$$
Combining $(2)$ and $(4)$, we get an integral combination of $\color{#C00000}{2m+n}$ and $\color{#C00000}{2n}$ that equals $\color{#0000FF}{1}$:
$$
(\color{#C00000}{2m+n})\left(2x(w^2+wnz)+nz^2\right)+\color{#C00000}{2n}\left((2y-x)(w^2+wnz)-mz^2\right)=\color{#0000FF}{1}\tag{5}
$$
Therefore, $(\color{#C00000}{2m+n},\color{#C00000}{2n})=\color{#0000FF}{1}$.


问:
Prove $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\gt1$
Prove that $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\gt1$, where a, b, c >0.

I tried,
$\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} = \frac{a}{\sqrt{a}\sqrt{a+b}}+\frac{b}{\sqrt{b}\sqrt{b+c}}+\frac{c}{\sqrt{c}\sqrt{c+a}} \ge \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\sqrt{a}\sqrt{a+b}+\sqrt{b}\sqrt{b+c}+\sqrt{c}\sqrt{c+a}}$.

Then $\sqrt{a}\sqrt{a+b} \le \sqrt{2}/2 (2a+(a+b))$.

This only gave me $\sqrt{2}/2 + \sqrt{2}(\frac{ab+bc+ca}{a+b+c})$.

Body:
Another way.
$$\sum_{cyc}\sqrt{\frac{a}{a+b}}=\sum_{cyc}\frac{\sqrt{a}}{\sqrt{a+b}}>\sum_{cyc}\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}>\sum_{cyc}\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1.$$


答:
We can use also your idea, but in another writing:
$$\sum_{cyc}\sqrt{\frac{a}{a+b}}=\sum_{cyc}\frac{\sqrt{a(a+b)}}{a+b}>\sum_{cyc}\frac{\sqrt{a\cdot a}}{a+b}>\sum_{cyc}\frac{a}{a+b+c}=1.$$

Title: Prove $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\gt1$

Body:
Prove that $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\gt1$, where a, b, c >0.

I tried,
$\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} = \frac{a}{\sqrt{a}\sqrt{a+b}}+\frac{b}{\sqrt{b}\sqrt{b+c}}+\frac{c}{\sqrt{c}\sqrt{c+a}} \ge \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\sqrt{a}\sqrt{a+b}+\sqrt{b}\sqrt{b+c}+\sqrt{c}\sqrt{c+a}}$.

Then $\sqrt{a}\sqrt{a+b} \le \sqrt{2}/2 (2a+(a+b))$.

This only gave me $\sqrt{2}/2 + \sqrt{2}(\frac{ab+bc+ca}{a+b+c})$.


问:
A question in Field theory about maximum degree of extension
I am trying assignment questions in Field Theory course and couldn't solve completely this particular question.

If u, v and are algebraic over K of degree m and n respectively, then prove that $[K(u, v) K]\leq mn $. If (m, n) =1 , then prove that [K(u, v) :K]= mn.

Using tower rule and if u and v are Linearly independent max possibility of [K(u, v) :K] =mn which will be maximum possibility.

But I am not able to think on how I will proceed when (m, n) =1 . It is equivalent to prove that u and v are independent of each other.

Kindly help me with this case.

答:
We have
$$
[K(u,v),K]=[K(u,v),K(u)]\cdot[K(u),K]\\
=[K(u,v),K(u)]\cdot m
$$
Which gives us $m\mid [K(u,v),K]$. Similarly we get $n\mid [K(u,v),K]$. If $m$ and $n$ are coprime, this implies $mn\mid [K(u,v),K]$. But since we also have $1\leq [K(u,v),K]\leq mn$, the only possibility is $[K(u,v),K]=mn$.


问:
Where does $\frac {\pi}{4}$ come from in this inverse sine graph? $y=-3\sin\frac {x\pi}{4})$
I understand that the vertical shift is $0$ that is why the graph starts at $(0,0)$. Also I understand that the amplitude is $3$ because the maximum y value is $3$ and the minimum y value is $-3$. Last but not least the graph is inverse sine but it seems to be a period and a half. I am not sure where the  $\frac {\pi}{4}$ comes from. I need help understanding that part. Photo of problem attached below. Thanks!

https://imgur.com/gallery/coL4IUw

答:
The equation of a sine curve is

$$ y = A \sin \dfrac{2 \pi x}{\lambda}$$

So in your particular case amplitude and wavelengths are

$$A=-3, \lambda= 8$$

Your graph is negative in start amplitude compared to standard sine of positive initial slope, but is not an inverse sine graph. Inverse graphs have $(x,y)$ axes interchanged.


问:
A related question to the Weierstrass approximation theorem
Suppose $f : [a,b] \to \mathbb{R}$ is a continuous function. Then, Weierstrass Approximation Theorem asserts that there is a polynomial on $[a,b]$ which is as close to $f$ as we want (in a uniformly approximation sense).

It is my question :

Let $S$ be a set $\big\{ \big| \frac{f(y) - f(x)}{y-x} \big| : a \leq x < y \leq b \big\}$,
which is a set of absolute values of slopes between every two distinct points of $f$.

Then, is the set $S$ bounded above?

答:
No. Let $f(x) = \sqrt{x}$ on, for example, $[0,1]$.  The average rate of change of $f$ is unbounded for pairs of points near $0$.

Consider also
$$  f(x) = \begin{cases}
    1 - \sqrt{1-x^2} ,& 0 \leq x < 1  \\
    1 - \sqrt{1 - (x-2)^2} ,& 1 \leq x \leq 2
\end{cases}
$$
Its graph is two quarter circles meeting at a vertical tangent.

The average rates of change are unbounded for sample points near (and on the same side of) $1$.


答: The set is unbounded for a generic continuous function $f$. Take for example $f(x)=\sqrt{x}$ on $[0,1]$. If your set is bounded, your function is called Lipschitz continuous.

Title: A related question to the Weierstrass approximation theorem

Body:
Suppose $f : [a,b] \to \mathbb{R}$ is a continuous function. Then, Weierstrass Approximation Theorem asserts that there is a polynomial on $[a,b]$ which is as close to $f$ as we want (in a uniformly approximation sense).

It is my question :

Let $S$ be a set $\big\{ \big| \frac{f(y) - f(x)}{y-x} \big| : a \leq x < y \leq b \big\}$,
which is a set of absolute values of slopes between every two distinct points of $f$.

Then, is the set $S$ bounded above?


答:
No.

$f(x)=\sqrt{x}$, let $[a,b]=[0,1]$. $f$ is continuous but $\forall n \in \mathbb{N}$ : $\exists x \in [0,1]$ such that $\frac{f(x)-f(0)}{x} \geqslant n$


问:
Basis of Field F over $\mathbb{Q}$
I am self studying Field theory and got struck on this problem.

If $\;F=\mathbb{Q} \left(\sqrt{2}, \sqrt{3}\right),\;$ find
$\;\left[F:\mathbb{Q}\right]\;$ and a basis of $F$ over $\mathbb{Q}$.

I have proved that $\left[F:\mathbb{Q}\right] =4$, but there is a problem in basis elements.

My basis set is $\left\{a, b\sqrt{2}, c\sqrt{3}\right\}$ such that $a, b, c$ belongs to $\mathbb{Q}$. But what should be fourth element and why ?

I am not able to see .

Kindly help.

Title: Basis of Field F over $\mathbb{Q}$

Body:
A basis of $F$ over $\mathbb{Q}$ is $\;\left\{1,\sqrt{2},\sqrt{3},\sqrt{6}\right\},$ indeed for any element $\;x\in F\;$ there exist $\;r_1,\;r_2,\;r_3,\;r_4,\;r_5,\;r_6,\;r_7,\;r_8\in\mathbb{Q}\;$ such that

\begin{align}
x&=\dfrac{r_1+r_2\sqrt{2}+r_3\sqrt{3}+r_4\sqrt{6}}{r_5+r_6\sqrt{2}+r_7\sqrt{3}+r_8\sqrt{6}}\\
&=\dfrac{r_1+r_2\sqrt{2}+r_3\sqrt{3}+r_4\sqrt{6}}{r_5+r_6\sqrt{2}+r_7\sqrt{3}+r_8\sqrt{6}}\cdot\dfrac{r_5+r_6\sqrt{2}-r_7\sqrt{3}-r_8\sqrt{6}}{r_5+r_6\sqrt{2}-r_7\sqrt{3}-r_8\sqrt{6}}\\
&=\dfrac{\left(r_1+r_2\sqrt{2}+r_3\sqrt{3}+r_4\sqrt{6}\right)\left(r_5+r_6\sqrt{2}-r_7\sqrt{3}-r_8\sqrt{6}\right)}{\left(r_5+r_6\sqrt{2}\right)^2-\left(r_7\sqrt{3}+r_8\sqrt{6}\right)^2}\\
&=\dfrac{s_1+s_2\sqrt{2}+s_3\sqrt{3}+s_4\sqrt{6}}{s_5+s_6\sqrt{2}}\\
&=\dfrac{s_1+s_2\sqrt{2}+s_3\sqrt{3}+s_4\sqrt{6}}{s_5+s_6\sqrt{2}}\cdot\dfrac{s_5-s_6\sqrt{2}}{s_5-s_6\sqrt{2}}\\
&=\dfrac{\left(s_1+s_2\sqrt{2}+s_3\sqrt{3}+s_4\sqrt{6}\right)\left(s_5-s_6\sqrt{2}\right)}{\left(s_5+s_6\sqrt{2}\right)\left(s_5-s_6\sqrt{2}\right)}\\
&=q_11+q_2\sqrt{2}+q_3\sqrt{3}+q_4\sqrt{6}
\end{align}

where $\;q_1,\;q_2,\;q_3,\;q_4\in\mathbb{Q}\;$.


问:
Prove $(a^3+b^3+c^3)(a^2+b^2+c^2)\ge(a+b+c)(a^2b^2+b^2c^2+c^2a^2)$
Prove that $(a^3+b^3+c^3)(a^2+b^2+c^2)\ge(a+b+c)(a^2b^2+b^2c^2+c^2a^2)$, where a, b, c > 0.

I did LHS-RHS, broke parentheses and cancelled terms.

I think I might use $a^2+b^2+c^2\ge ab+bc+ca$. Then I only need to prove:

$a^5+b^5+c^5\gt abc(a^2+b^2+c^2)$.


问:
How to identify which states cannot be observed from the observability matrix?
I'm trying to determine observability of the following system:

$$
\mathbf{A} = \begin{bmatrix}-10 & 0 & -10 & 0 \\0 & -1 & 9 & 0\\0&-1&-1&0\\1&0&0&0\end{bmatrix}
$$
$$
\mathbf{B} = \begin{bmatrix}20&3\\0&-3\\0&0\\0&0\end{bmatrix}
$$
$$
\mathbf{C} = \begin{bmatrix}1&0&0&0\end{bmatrix}; \mathbf{D} = \begin{bmatrix}0\end{bmatrix}
$$

I found the observability matrix to be:
$$
\begin{bmatrix}
1 & 0 & 0 & 0\\
-10 & 0 & -10 &0\\
100 & 10 & 110 & 0\\
-100 &-120&-1020&0
\end{bmatrix}
$$
Which has the RREF form of:
$$
\begin{bmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&0
\end{bmatrix}
$$

Obviously the system is not completely observable, and I was able to determine through the $\mathbf{A}$ and $\mathbf{C}$ matrices that it would be the fourth state variable that is not observable, but I was unclear as to whether this could be determined from the observability matrix. Is it the fourth column being all zeroes that would tell me this, or is it the fourth row which would tell me this? Would it be the same for a controllability matrix?

答: In general you can find the null space of the observability matrix, whose basis should also be eigenvectors of $A$. So, the eigenvalues that correspond to those eigenvectors are the unobservable states.


问:
What's the name for "negative length" in pseudometric space?
In a pseudometric space (a metric space with signature $(1,1)$, for example), the "distance" between two points can be negative. That happens (in this example) when $d^2 = x^2 - y^2$ is negative because $y > x$.

But people don't like when I talk about "negative length" because apparently, the words "length" and "distance" are defined as always being positive.

In the case of 4-dimensional spacetime (which is Riemannian), the word to use is "interval."  But what's the word to use for negative distance in the general case of any pseudometric space?

答:
[...] the "distance" between two points can be negative. That happens (in this example) when $d^2 = x^2 − y^2$ is negative because $y \gt x$.

If $d$ is supposed to denote the distance value under consideration then the example you gave is not applicable because: the square root of a negative real number, i.e. for all cases $x^2 - y^2 \lt 0$ is not a negative real number but of course imaginary.

(And whether there may be objections to calling and considering values of imaginary or of complex numbers as values of "distance" or of "length" is a separate issue.)

An applicable example, somewhat based on the above, is instead:

$$d : \mathbb R^2 \rightarrow \mathbb R, \qquad d \mapsto \text{Sgn}[ \, x^2 - y^2 \, ] \, \sqrt{ \text{Sgn}[ \, x^2 - y^2 \, ] \, (x^2 - y^2) },$$

where $\text{Sgn}$ denotes the sign function.

But people don't like when I talk about "negative length" because apparently, the words "length" and "distance" are defined as always being positive.

Clearly, the (applicable) example function $d$ above does not satisfy all defining properties of a (plain) metric space.

Still, at least it shares one property of (plain) metric spaces, namely "Indiscernability of the Identical": $\forall x : d[ \, x, x \, ] = 0$.

(This illustrates by the way that even in the plain notion of distance its values are not always defined as strictly positive.)

However, obviously there are generalizations of the "plain" notion of distance.

In a pseudometric space

Example function $d$ defined above is (obviously) not an instance of pseudometric distance either. Therefore the relevant "larger" questions would be, to which type of generalization belongs the above example function, or how to call such a generalization of "plain" metric space.

For lack of knowing of any suitable precedence (on this website, in Wikipedia, or elsewhere), I have already proposed here to call the particular type of generalization examplified by the function $d$ above as quasihypometric distance.

In the case of 4-dimensional spacetime (which is Riemannian),

... actually, pseudo-Riemannian and even more specifically Lorentzian ...

the word to use is "interval."

Yes values of spacetime interval, $s^2$ can have negative real values (as well as non-negative values). But (alledgedly), the notion of spacetime interval is not applicable to Lorentzian manifolds in general, but only to flat manifolds (such as Minkowski space). A more general notion would be Synge's world function, $\sigma$.

But what's the word to use for negative distance in the general case

A notion for referring to the separation between timelike (or lightlike) separated events in general is Lorentzian distance.

(Apparently, terminology for naming the quantities $\text{Sgn}[ \, \sigma \, ] \, \sqrt{ \text{Sgn}[ \, \sigma \, ] \, \sigma }$ still needs to be proposed.)


问:
Is the given discrete random vector independent?

A random vector $(X,Y)$ has PMF given by $p(1,-1) = p(-1,1) = 1/2$. Are $X,Y$ independent?

I think they are dependent because $p(X=1)=p(Y= -1)= 1/2$ and $p(X=1,Y= -1)\ne p(X=1)*p(Y= -1)$. Is this correct?

答:
They are not independent. $X$ and $Y$ take the values $+1$ and $-1$. Note that $p(1,1)+p(-1,-1)+p(1,-1)+p(-1,1)=1$. Since $p(1,-1)+p(-1,1)=\frac 1 2+\frac 1  2 =1$ we get $p(1,1)=p(-1,-1)=0$. Now consider $p(1,1)=0$ This is not equal to $P(X=1)P(Y=1)$. In fact $P(X=1)=p(1,1)+p(1,-1)=\frac 1 2$ and $P(Y=1)=p(1,1)+p(-1,1)=\frac 1 2$ so $P(X=1)P(Y=1)=\frac 1 4$.

$X^{2}$ and $Y^{2}$ are independent since they are both equal to $1$ almost surely.


问:
Showing that pencil of conics through four points in general position in $\mathbb{P}^2$ forms a line when it's considered in $\mathbb{P}^5$

I know that there is a correspondence between points in $\mathbb{P}^5$ and conics in $\mathbb{P}^2$.

How do you show that family of conics through four points (pencil of conics) in general position forms a line when it's considered in $\mathbb{P}^5$?

If I consider a general conic given by
$$ax^2+by^2+cxy+dx+ey+f=0$$
we can divide both sides by $a$ if $a$ is nonzero
$$x^2+b'y^2+c'xy+d'x+e'y+f'=0,$$
and if I'm given the coordinates of four points, I can solve this for four variables, where the other one is a free variable. That's where the linearity comes from.

答: You don't have to worry so much about whether $a$ is zero or nonzero: you can just write down what the conditions "passing through $p$" mean. Explicitly, if $p=[p_0,p_1,p_2]$, then a conic $ax^2+by^2+cxy+dxz+eyz+fz^2$ passes through $p$ iff $ap_0^2+bp_1^2+cp_0p_1+dp_0p_2+ep_1p_2+fp_2^2=0$, which is a linear equation in $a,b,c,d,e,f$. Assuming the points are in general position gives you $4$ independent linear equations, so all you need to do is to understand that $V(l_1,l_2,l_3,l_4)$ is a $\Bbb P^1\subset \Bbb P^5$. (I should note that your equation of a conic is an affine equation and I'm upgrading to a projective equation - you should too!)


问:
Why does Excel fail to properly evaluate $ \lim_{x \to 0} \frac{x(e^x+1)-2(e^x-1)}{x^3} $?
I have this limit:
$$
\lim_{x \to 0} \frac{x(e^x+1)-2(e^x-1)}{x^3}
$$
By using L'Hospital rule and Maclaurin expansion I both found that limit equals to $1/6$.

I want to simulate this to my student by Excel but I failed.
This is my computation in Excel.

I did not know why Excel doesn't work in this situation?

Please help me. Thanks alot.

答:
If I do with Mathematica the same thing that you did with Excel, then I get a similar answer. That is due to numerical errors that are a consequence of a catastrophic cancellation, as you were told in the comments. However, I can force Mathematica to work with greater numerical precision and if I tell it to compute those numbers with ten significant figures, then I get a nice answer:

$$\begin{array}{c|c}n&f(10^{-n})\\\hline0&0.2817181715\\1&0.1752556563\\2&0.1675025056\\3&0.1667500250\\4&0.1666750003\\5&0.1666675000\\6&0.1666667500\\7&0.1666666750\\8&0.1666666675\\9&0.1666666668\end{array}$$I don't know whether or not you can do the same thing with Excel, but this could be a good lesson for your students.


问:
Solving $e^{0.0002n} + e^{0.0003n} = 0.01$
I need to solve an exponent equation which is:

$$
        e^{0.0002n} + e^{0.0003n} = 0.01
$$

Please help me with this equation.

Thanks.

Body:
HINT

Let $e^{0.0001n}=t$

$$t^2+t^3 = 0.01$$


问:
Equivalent Definitions of Subharmonic Function
I want to prove the following are equivalent.

(a) For any $p \in U$ and $r>0$ with $\overline{D(p, r )} \subseteq U$,

if $h$ is harmonic on a neighborhood of $\overline{D(p, r)}$ satisfying $f \le h $ on $\partial D(p, r)$, then $f \le h$ on $D(p, r)$.

(b) For any $p \in U$ and $r>0$ s.t. $\overline{D(p, r )} \subseteq U$,

if $h$ is continuous on $\overline{D(p, r)}$ and harmonic on $D(p, r)$,satisfying $f \le h $ on $\partial D(p, r)$, then $f \le h$ on $D(p, r)$.

(a) $\Leftarrow $ (b) is obvious.

But I don't know how to prove the converse. The author says that I should use a 'limiting argument', but I'm not sure what 'limiting argument' refers to.

Any help will be appreciated. Thanks.

答:
You didn't say what is $f$ so I assume it is a continuous function on $U$.

For sake of simplicity, let assume that $r = 1$.
If $h$ is harmonic on $D(p, 1)$ then for any $t \in (0, 1)$, the function $h$ is harmonic on a neighbourhood of $\overline{D(p, t)}$. Since $h$ and $f$ are continuous on the compact set $\overline{D(p, 1)}$, then there are uniformly continuous on it. Thus, for any $\varepsilon > 0$ there exists $t_{\varepsilon} \in (0, 1)$ such that
$$ \forall t \in (t_{\varepsilon}, 1),\ \forall z \in \partial D(p, 1),\ |h(tz)-h(z)| \leqslant \varepsilon\ \mbox{ and }\ |f(tz)-f(z)| \leqslant \varepsilon. $$
In particular, we have $f - 2\varepsilon \leqslant h$ on $\partial D(p, t)$ for any $t \in (t_{\varepsilon}, 1)$. According to (a), we got $f-2\varepsilon \leqslant h$ on $D(p, t)$ for all $t \in (t_{\varepsilon}, 1)$. Thus $f-2\varepsilon \leqslant h$ on $D(p, 1)$. Let $\varepsilon$ tends towards zero to conclude.


答:
Adding on @DatBoi's answer, if $t=0.1$, then

$$t^2+t^3=0.011$$ and we have a good approximation. Now using a Newton's iteration,

$$t'=t-\frac{t^2+t^3-0.01}{2t+3t^3}=0.1-\frac{0.001}{0.23}=0.09565\cdots$$

You can use a few more iterations, or use the general formula for the cubic equation. $t$ gives $n$.

Title: Solving $e^{0.0002n} + e^{0.0003n} = 0.01$

Body:
I need to solve an exponent equation which is:

$$
        e^{0.0002n} + e^{0.0003n} = 0.01
$$

Please help me with this equation.

Thanks.


问:
Vector Differentiation Derivation for Linear Least Mean Squares Estimators
Linear Estimation by Kailath, Sayed and Hassibi (Section 3.2.1) provides a derivation that the linear least mean squares estimator (LLMSE) $\hat{y}=K_0x$ satisfy the equation $K_0 R_y = R_{xy}$ where $x$ and $y$ are complex vector-valued zero-mean random variables (column vectors), $*$ denotes the complex conjugate transpose, and $R_x=E [xx^*]$ and $R_{xy}=E [xy^*]$ are the covariance matrices.  The proof simply says to differentiate with respect to $aK$ (where $a$ is a row vector):
$$a E[(x-Ky)(x-Ky)^*]a^* = a(R_x - R_{xy} K^* - KR_{yx} + KR_yK^*)a^*$$
and set to zero at $K=K_0$ to arrive at $K_0 R_y = R_{xy}$.

The simplified scalar case is clear. Unfortunatly, I don't know how do this vector differentiation?  For example, what is $\frac{\partial}{\partial (aK)}$ of $aR_{xy}K^*a^*$?  Any help is greatly appreciated.

答:
I think there is some misunderstanding. There is another book from one of the authors of the book that you have mentioned: A. H. Sayed, Adaptive Filters, Wiley, NJ, 2008. Perhaps it is easier to read than the book you have mentioned. I am sure there are many books/notes for the derivation of LMMSE.

Anyway, the optimization problem for LMMSE weight should read as following
\begin{align}
\text{minimize}_{K} \operatorname{Tr}(R_x - R_{xy} K^H - KR_{yx} + KR_yK^H),
\end{align}
where $(\cdot)^H$ denotes complex conjugate transpose, whereas $(\cdot)^*$ corresponds to complex conjugate.

To this end, we need to find an optimal weight matrix $K$. We have to use Wirtinger calculus for the gradient computation, i.e., treat $K$ and $K^*$ independently.

Before finding the gradient (and setting it to zero), here are some nomenclature:

Trace and Frobenius product relation $$\left\langle A, B C\right\rangle={\rm Tr}(A^TBC) := A : B C$$

Cyclic properties of Trace/Frobenius product
\begin{align}
A : B C
 &= A^*: B^*C^* \\
 &= BC : A \\
 &= C^TB^T : A^T \\
 &= A C^T   :  B \\
 &= A^T: C^TB^T \\
 &= {\text{etc.}} \cr
\end{align}

Let $f := \operatorname{Tr}(R_x - R_{xy} K^H - KR_{yx} + KR_yK^H) \equiv I: \operatorname{Tr}(R_x - R_{xy} K^H - KR_{yx} + KR_yK^H)$.

For simplicity, we will find the differential $dK^*$ (and ignore $dK$).
\begin{align}
df
&= d\left(I:R_x - R_{xy} K^H - KR_{yx} + KR_yK^H \right) \\
&= \left(I:0 - R_{xy} dK^H - 0 + KR_ydK^H \right) \\
&= \left(I^T: - dK^*R_{xy}^T + dK^* (KR_y)^T \right) \\
&= \left(\left[-R_{xy} + KR_y \right]: dK^*  \right)
\end{align}

Thus, the gradient is
\begin{align}
\frac{\partial f}{\partial K^*} = -R_{xy} + KR_y.
\end{align}

Now, you set the gradient to zero and obtain $K$ (assuming that $R_y$ is invertible and set $K = K_0$):

$$K_0 = R_{xy} R_y^{-1} .$$


问:
Finding all distinct primes $p$ and $q$ such that $pq^3 + p^3q$ has exactly 12 divisors

How do you find all possible distinct prime numbers $p$ and $q$ such that $pq^3 + p^3q$ has exactly $12$ divisors?

Any help would be appreciated. Not sure if factoring into $pq(q^2+p^2)$ does anything, but I am lost on this.

答:
If $p$ and $q$ are distinct primes, then the three factors in the factorization
$$pq^3+p^3q=pq(p^2+q^2),$$
are pairwise coprime. It follows that if the number of divisors of each factor is $a$, $b$ and $c$, respectively, then the product has $a\times b\times c$ factors. Of course $a=b=2$ because $p$ and $q$ are prime, so to get $abc=12$ we must have $c=3$. It follows that
$$p^2+q^2=r^2,$$
for some prime number $r$. Reducing mod $2$ shows that either $p$ or $q$ is divisible by $2$, and reducing mod $3$ shows that either $p$ or $q$ is divisible by $3$. Then
$$r^2=p^2+q^2=2^2+3^2=13,$$
a contradiction. Hence no solution exists.


问:
Definite Integral by Bernoulli numbers
How we can show the following definite integral on $[0,1]$?

$\begin{eqnarray}
 \int _0^1\left(\ln \frac{x}{1-x}\right)^kdx=(2^k-2)\pi ^k|B_k|,
\end{eqnarray}$

where $B_k$ are the $k$-th Bernoulli numbers, $k=1,2,\ldots$, respectively.

Note that the Bernoulli numbers $B_k$'s are a sequence of signed rational numbers. For every odd $k$ other than $1$, $B_k=0$. For every even $k$ other than $0$, $B_k$ is negative if $k$ is divisible by $4$ and positive otherwise. The first few Bernoulli numbers for $k=0,1,2,3,4,5,6,7,8,9,10$ are

$B_k=1,-\frac{1}{2}, \frac{1}{6}, 0,-\frac{1}{30},0,\frac{1}{42},0,-\frac{1}{30},0,\frac{5}{66}$.

答:
Performing the change of integration variable $\frac{x}{{1 - x}} = e^{t}$ from $x$ to $t$ turns your integral into
$$
\int_{-\infty}^{ + \infty } {\frac{{t^k e^t }}{{(e^t  + 1)^2 }}dt} .
$$
If $k$ is positive and even, the integrand is even. Hence, by http://dlmf.nist.gov/25.5.E4 and http://dlmf.nist.gov/25.6.E2, we find
$$
\int_{-\infty}^{ + \infty } {\frac{{t^k e^t }}{{(e^t  + 1)^2 }}dt} = 2\int_{0}^{ + \infty } {\frac{{t^k e^t }}{{(e^t  + 1)^2 }}dt} = 2(1 - 2^{1 - k} )k!\zeta (k) = (2^k-2)\pi^k |B_k|.
$$
For positive odd $k$, the integrand is an odd function, whence the integral is $0$. Thus your claim is indeed true for all positive integer $k$.

Title: Definite Integral by Bernoulli numbers


问:
probable squares in a square cake
There is a probability density function defined on the square [0,1]x[0,1].

The pdf is finite, i.e., the cumulative density is positive only for pieces with positive area.

Now Alice and Bob play a game: Alice marks two disjoint squares, Bob chooses the square that contains the maximum probability, and Alice gets the other square. The goal of Alice is to maximize the probability in her square.

Obviously, in some cases Alice can assure herself a probability of 1/2, for example, if the pdf is uniform in [0,$1 \over 2$]x[0,1], she can cut the squares [0,$1 \over 2$]x[0,$1 \over 2$] and [0,$1 \over 2$]x[$1 \over 2$,1], both of which contain $1 \over 2$.

However, in other cases Alice can assure herself only $1 \over 4$, for example, if the pdf is uniform in [0,1]x[0,1].

Are there pdfs for which Alice cannot assure herself even $1 \over 4$ ?

What is the worst case for Alice?

答:
I think Alice can always assure herself at least $1 \over 4$ cdf, in the following way.

First, in each of the 4 corners, mark a square that contains $1 \over 4$ cdf. Since the pdf is finite, it is always possible to construct such a square, by starting from the corner and increasing the square gradually, until it contains exactly $1 \over 4$ cdf.

There is at least one corner, in which the side length of such a square will be at most $1 \over 2$ . Suppose this is the lower-left corner, and the side length is a, so square #1 is [0,a]x[0,a], with a <= $1 \over 2$.

Now, consider the following 3 squares:

To the right of square #1: [a,1]x[0,1-a]

On top of square #1: [0,1-a]x[a,1]

On the top-right of square #1: [a,1]x[a,1]

The union of these squares covers the entire remainder after we remove square #1. This remainder contains $3 \over 4$ cdf. So, the sum of cdf in all 3 squares is at least $3 \over 4$ (probably more, because the squares overlap).

Among those 3, select the one with the greatest cdf. It must contain at least $1 \over 3$ of $3 \over 4$, i.e., at least $1 \over 4$. This is square #2.

So, Alice can always cut two squares that contain at least $1 \over 4$ cdf.

Note that this procedure relies on the fact (that I mentioned in the original question) that the pdf is finite. Otherwise, it may not always be possible to construct a square with $1 \over 4$ cdf.

Title: probable squares in a square cake


问:
Proof for Elementary Row Operations
The Elementary Row Operations are stated as follows:

Interchange two rows.

Multiply a row by a nonzero constant.

Add a multiple of a row to another row.

Consider the matrix below:

$$A =\begin{bmatrix}5 & 2 & 0 & 0 & -2\\0 & 1 & 4 & 3 & 2\\0 & 0 & 2 & 6 & 3\\0 & 0 & 3 & 4 & 1\\0 & 0 & 0 & 0 & 2 \end{bmatrix}$$

I am concerned with the third operation which is stated as $r_i \rightarrow r_i+cr_j$. Why is it wrong to interpret it as $r_i \rightarrow cr_i+r_j$? Is there any proof for this? Thank you!

答:
There is nothing wrong with that interpretation. In that case you are simply taking the row $r_i$ and substituting it by the multiple $cr_i$ and then adding the row $r_j$.

Of course in this case you have to consider that $c \neq 0$, otherwise the matrix that you will obtain will not be equivalent to the previous one (i.e., all steps in row operations must be reversible).

Title: Proof for Elementary Row Operations


问:
$\mathbb E[X]$ and $Var[X]$ for $X=IB$
Determine $\mathbb E[X]$ and $Var[X]$ for $X=IB$ where $I\sim Ber(p), p>0$ and $B$ with $\mathbb P[B=b]=1$ for some $b>0$ are independent.

I was wondering why the author makes the assumption $b>0$ here.

We simply have $\mathbb E[X]=\alpha b$ and $\mathbb E[X^2]=\alpha b^2$ (and can use this to determine the variance) or am I missing something?

答:
perhaps it is only a typo but

$$\mathbb{E}[X]=pb$$

I do not know where $\alpha$ comes from

Same observation for the second simple moment

Title: $\mathbb E[X]$ and $Var[X]$ for $X=IB$


问:
How to formally define a bijection $f\colon (A/\sim)\to R$, where $R\subseteq A$ is a set of representatives?
Given a set $A$ and an equivalence relation $\sim$ on $A$, how can I formally define a bijection $f\colon (A/\sim)\to R$, where $R\subseteq A$ is a set of representatives? Clearly, $[a]_\sim\mapsto a$ is not well-defined, because $a\ne b\in [a]_\sim\Longrightarrow [b]_\sim=[a]_\sim$, but $f([b]_\sim)=b\ne a$.

答:
It's easier to define a bijection $g$ from $R$ to $A/{\sim}$ given by $g(a)=[a]_\sim$; then the bijection from $A/{\sim}$ to $R$ is simply $f=g^{-1}$.

If we want an explicit formula for $f$, perhaps $f([a])=[a]\cap R$ works.


问:
Show that $\bar{\varphi} \pi = \pi \varphi.$
Here is the question I want to prove:

Let $R$ be a ring and $I \subset R$ a two-sided ideal, with quotient homomorphism $\pi : R \rightarrow R/I.$ Let $\operatorname{End_{I}(R)}$ be the set of $\phi \in \operatorname{End(R)}$ such that $\varphi(I) \subset I,$ and let $\operatorname{Aut_{I}(R)} = \operatorname{End_{I}(R) \cap Aut(R)}.$

$(a)$ Given $\varphi \in \operatorname{End_{I}(R)},$ show that there exists $\bar{\phi} \in \operatorname{End(R/I)}$ such that $\bar{\varphi} \pi = \pi \varphi.$

$(b)$ Given $\varphi, \psi \in \operatorname{End_{I}(R)},$ show that $\overline{\varphi \psi} = \bar{\varphi} \bar{\psi}.$

My question is:

1- For letter $(a)$I know that if I used the first isomorphism theorem, I would get $\bar{\varphi} \pi = \varphi.$ But how can I get the result that I want above? could anyone help me in figuring this out please?

2- For letter $(b),$ I do not know exactly what I am proving, I feel like I am proving well defined of taking the bar operation. But what does this operation means? How can I prove the required?

Any help would be appreciated.

答:
The definition of the "bar" operator is given in the first part of the problem.
The only possibility for $\bar{\varphi}$ is $\bar{\varphi}(\pi(x))=\bar{\varphi}(x+I)=\varphi(x)+I = \pi(\varphi(x))$. Can you show that this is indeed an endomorphism of $R/I$ using the property that $\varphi(I)\subseteq I$?

For the second part you need to show that the bar you just defined respects the composition of endomorphisms. Perhaps now that you see the definition you can show that $\overline{\varphi\psi}=\bar{\varphi}\bar{\psi}$.


问:
Ambiguity in delta equal sign
I tried asking this yesterday but my wording was too bad. I will try again because it really confuses me.

Let's look at an identity like $$x^n\delta(x) = 0$$
Does the equality in this case mean it is right for every $x$ value? For $x\neq0$ it is trivial, but we can't plug in $x=0$.

What I see they do in textbooks is taking the integral of $x^n\delta(x)$ around $0$, and see it is indeed zero. But why does that mean the expression itself is zero?

I seem to be missing an important notion concerning delta "functions". Are they defined only under an integral in the singularity? I see nowhere such definition.

If I'm still unclear, please let me know.

答:
Distributions, like $\delta$, are defined as linear functionals over a space of smooth functions with compact support ($C_c^\infty$).

The action of a distribution $u$ on a test function $\varphi \in C_c^\infty$ is often denoted $\langle u, \varphi \rangle$ or with some abuse of notation, as an integral $\int_{-\infty}^{\infty} u(x) \, \varphi(x) \, dx.$

Multiplication of a distribution with a smooth function $f \in C^\infty$ is defined by
$$
\langle fu, \varphi \rangle = \langle u, f\varphi \rangle.
$$
In integral notation this is just the quite obvious
$$
\int_{-\infty}^{\infty} (f(x)\,u(x)) \, \varphi(x) \, dx
= \int_{-\infty}^{\infty} u(x) \, (f(x)\,\varphi(x)) \, dx.
$$

The distribution $\delta$ is defined by $\langle \delta, \varphi \rangle = \varphi(0).$

By what I have now defined, we get
$$
\langle x^n \, \delta(x), \varphi(x) \rangle
= \langle \delta(x), x^n \, \varphi(x) \rangle
= (x^n\,\varphi(x))[x:=0]
= 0^n\,\varphi(0)
= 0.
$$
This is the quite exact meaning of $x^n\,\delta(x)=0.$

Title: Ambiguity in delta equal sign


问:
Why the Poisson and Binomial distributions are approximately normal?
Why the Poisson distribution P(n) and Binomial distribution B(n, p) are approximately
normal if n is a large positive integer and p ∈ (0, 1) is fixed? I found the examples of normal approximations, but didn't find the explanation why does it work and what will be the parameters of these approximations respectively.</p>

答:
Let's start with the question on Binomial.

As you probabily know, the binomial distribution can be expressed as the sum of $n$ independent Bernoulli. Then you can apply the central limit theorem in the version of  De Moivre Laplace

As per the Poisson is concerned, it can be wiewed a limit Law of the binomial thus what explained in 1. is still valid (when $n$ is great enough)

Both statement in 1. and 2. can be easily proved with MGF's properties

Proof of 2.

Setting $p=\frac{\theta}{n}$ the MGF of a Binomial $B(n;p)$ is the following

$$\Big(1-\frac{\theta}{n}+\frac{\theta}{n}e^t\Big)^n=\Big[1+\frac{\theta(e^t-1)}{n}\Big]^n$$

It is evident that

$$\lim\limits_{n \to \infty}\Big[1+\frac{\theta(e^t-1)}{n}\Big]^n=e^{\theta(e^t-1)}$$

Which is exactly the MGF of a Poisson distribution.

Proof of 1.

simply multiply n times MGF of the $B(1;p)$ and see that the result is the MGF of the $B(n;p)$


问:
Why in proximal gradient descent, "proximal" is referred to as "Backward" and the "gradient" is "Forward"?
I am not into monotone operators theory (but hope someday I can get a hang of it).

Sorry for asking probably stupid question.

Why in proximal gradient descent, "proximal" is referred to as "Backward" and the "gradient" is referred to as "Forward"?

I can imagine why "gradient" is referred to as "Forward" perhaps because it looks for the "downhill" and move forward. But I am not sure. However, I have no clue why "proximal" is referred to as "Backward". Can someone enlighten me? Thank you.

答:
Let $t > 0$. To minimize $f+ g$ where $f$ is smooth and $g$ is closed and convex, we need to find a point $x$ that satisfies
\begin{align}
& 0 \in \nabla f(x) + \partial g(x) \\
\iff & x - t \nabla f(x) \in x + t \partial g(x) \\
\iff & (I + t \partial g)^{-1}(x - t \nabla f(x)) = x.
\end{align}
The forward-backward method uses the fixed point iteration
$$
x^+ =  (I + t \partial g)^{-1}(x - t \nabla f(x)).
$$
Computing $x - t \nabla f(x)$ is the "forward" step. Applying the operator $(I + t \partial g)^{-1}$ is called the "backward" step, I guess because we are inverting the operator $I + t \partial g$.

If you are able to recognize that  $(I + t \partial g)^{-1}$ is the proximal operator of $g$, then you see that the forward-backward method is the same thing as the proximal gradient method.


问:
How to evaluate $\int_0^r \int_0^{2\pi} \lvert f'(te^{i\theta})\rvert^2 t\, d\theta dt$
Let $f:B_1(0)\subset \mathbb{C} \to \mathbb{C}$ be a holomorphic function with its taylor representation
$$f(w)=\sum_{n=1}^{\infty} A_n w^n, A_n\in \mathbb{C}, \lvert w\rvert<1$$
$$f'(w)=\sum_{n=1}^{\infty} A_n\cdot n\cdot w^{n-1}$$
I want to verify the following:
$$\int_{\lvert w\lvert <r} \lvert f'\rvert^2 dudv=\int_0^r \int_0^{2\pi} \lvert f'(te^{i\theta})\rvert^2 t\, d\theta dt \overset{!!!}{=} \pi \sum_{n=1}^{\infty} n \lvert A_n\rvert^2 r^{2n}$$

I already know how to calculate
$$\int_{0}^{2\pi}\lvert f(re^{i\theta})\rvert^2\,d\theta$$
which is equal to
$$2\pi \sum_{n=1}^{\infty} \lvert A_n\rvert^2 r^{2n} $$
(cf. here: Prove that $\int_0^{2\pi} \lvert \sum_{n=0}^{\infty} A_n (re^{i\theta})^n \rvert^2=2\pi \sum_{n=0}^{\infty} \lvert A_n\rvert ^2 r^{2n}$)
It immediatly follows that
$$\int_0^{2\pi} \lvert f'(te^{i\theta})\rvert^2\cdot t\,d\theta =2\pi \sum_{n=1}^{\infty} n^2 \lvert A_n\rvert^2 t^{2n-2}\cdot t=2\pi \sum_{n=1}^{\infty} n^2 \lvert A_n\rvert^2 t^{2n-1},$$
since you can basically just put a $2\pi$ in front of the sum and square the series inside the sum. Now if you could swap the series with the last integral, you would get
$$2\pi \sum_{n=1}^{\infty} \int_0^r n^2 \lvert A_n\rvert^2 t^{2n-1}\,dt=2\pi \sum_{n=1}^{\infty} \left[\frac{n^2}{2n} \lvert A_n\rvert^2 t^{2n} \right]_0^r=\pi \sum_{n=1}^{\infty} n \lvert A_n\rvert^2 r^{2n},$$
which gives my desired result. However, I was not able to show that the series
$$\sum_{n=1}^{\infty} n^2 \lvert A_n\rvert^2 t^{2n-1}$$
converges uniformly on $[0,r]$ for all $r\in [0,1)$. I have tried to calculate the radius of convergence, then move out the $\sqrt[n]{n^2}$ factor, which in the limit will be equal to one, but I wasn't able to continue.

答: The radius of convergence of $\sum n^{2}|A_n|^{2}w^{n}$ is $\frac 1  {\lim \sup n^{2/n}|A_n|^{2/n}}$. Since $\sum A_nw^{n}$ has radius of convergence at least $1$ we get $\lim \sup |A_n|^{n} \leq 1$. Hence $\frac 1  {\lim \sup n^{2/n}|A_n|^{2/n}} \geq 1$. This implies uniform convergence of $\sum n^{2}|A_n|^{2}w^{n}$ for $|w| \leq r$ if $r<1$.


问:
Bijection from $\mathrm{Hom}(\mathbb{Z}_2,G)$ to $H:=\{g\in G:g^2=e\}$

Let $G$ be a group. Construct a bijection from $\mathrm{Hom}(\mathbb{Z}_2,G)$ to $H:=\{g\in G:g^2=e\},$ where $\mathrm{Hom}(\mathbb{Z}_2,G)$ stands for the set of homomorphisms from $\mathbb{Z}_2$ to $G.$

Attempt. Let $f:\mathbb{Z}_2\to G$ a group homomorphism and we would like to map $f$
to an element $g$ of $G$ such that $g^2=e$ in a way such that the map is one to one and onto. The elements we have are $f(0)$ and $f(1)$ (here: $0$=set of even integers, $1$=set of odd integers), so a choice would be $f\mapsto f(0)$ (indeed, $f(0)^2=f(0^2)=f(0)=e$, since $f$ is  homomorphism). But if $f(0)=h(0)$ for homomorphisms $f,\,h$, then I don't see how $f=h$ (also, if $g^2=e$, how do we get a homomorphism $f$ such that $f(0)=g$?). Maybe another map would do the work.

Thanks in advance.

答:
I'll try to gather all comments and your attempt to an answer just for completion.

Let $W:Hom(\mathbb{Z}_2,G)\to H=\{g\in G| \ g^2=e\}$ with $W(f)=f(1)$. Then:

$W$ is well defined: it is $f(1)\in H$ since $f^2(1)=f(1+1)=f(0)=e$.

$W$ is $1-1$: Let $f(1)=g(1)$ for some $f,g\in Hom(\mathbb{Z}_2,G)$. Then since $f(0)=g(0)$ for all $f,g\in Hom(\mathbb{Z}_2,G)$ it is $f=g$.

$W$ is onto: Let $g\in G$ s.t. $g^2=e$. We define $f\in Hom(\mathbb{Z}_2,G)$ with $f(0)=e$ and $f(1)=g$. Then $f$ is indeed in $ Hom(\mathbb{Z}_2,G)$ and $W(f)=f(1)=g$.$\checkmark$


问:
Convergence and the limit of $x_n=(1+\frac{1}{2})(1+\frac{1}{2^2})...(1+\frac{1}{2^{2^n}})=\prod_{i=0}^n (1+\frac{1}{2^{2^i}})$ for $n \ge 0$
It is easy to see that the sequence is bounded below by $1$ and increasing as we multiply the terms by a number larger than $1$ each time.

I was also able to show that $x_n \le a_n$ for all $n \ge 0$ where $a_n=\sum_{i=0}^n \frac{1}{2^i}$. Since we know $(a_n)_n$ is bounded above by $2$, so do $(x_n)_n$ and now we know it also converges.

However, I couldn't find its limit. Can you help me on finding its limit?

Thanks for your effort and time in advance.

Body:
$$x_n=\left(1+\frac 1 2\right)\left(1+\frac 1 {2^2}\right)\cdots\left(1+\frac 1 {2^{2^n}}\right)\\=\frac{\left(1-\frac 1 2\right)\left[\left(1+\frac 1 2\right)\left(1+\frac 1 {2^2}\right)\cdots\left(1+\frac 1 {2^{2^n}}\right)\right]}{\left(1-\frac 1 2\right)}\\=\frac{\left(1-\frac 1 {2^{2^{n+1}}}\right)}{\left(1-\frac 1 2\right)}.$$
Thus, $$\lim x_n=\frac 1{1/2}=2.$$

See $x_n=1+\frac 1 2+\frac 1 {2^2}+\frac 1 {2^3}+\frac 1 {2^4}+\cdots+\frac 1 {2^{1+2+\cdots +2^{2^n}}}>\sum_{i=0}^n \frac 1 {2^i}.$


问:
Discrete subgroups of $(\mathbb{R},+)$
I  have a simple question on discrete subgroups.

We say that a subset $C$ of $\mathbb{R}$ is discrete if every point of $C$ is isolated in the topology inherited from $\mathbb{R}$.

Is it true that every discrete subgroup of $(\mathbb{R},+)$ is of the form $r\mathbb{Z}$ with $r \in \mathbb{R}$? And if so, how can i prove it?

\medskip

My attempt: I don't really know how to prove it, i mean, i cannot think of an example of a discrete subset of $\mathbb{R}$ which is not, somehow, related to $\mathbb{Z}$...

Thanks in advance!

答:
Even though this is an old question, I really like the following solution:

Let $G\subset \mathbb{R}$ denote a discrete subgroup. Then by classical covering space theory, $\mathbb{R}\to\mathbb{R}/G$ is a covering and $\mathbb{R}/G$ is a connected $1$-manifold. Furthermore by the classification of $1$-manifolds, it has to be compact (since the only connected non-compact $1$-manifold $\mathbb{R}$ is simply-connected), hence again by the classification it is $S^1$. By elementary covering space theory $\pi_1(S^1)\cong \mathbb{Z}\cong G$.

Since a map from $\mathbb{Z}$ to $\mathbb{R}$ is specified by the value of $1$, it has the required form.

Title: Discrete subgroups of $(\mathbb{R},+)$


问:
Why does Excel fail to properly evaluate $ \lim_{x \to 0} \frac{x(e^x+1)-2(e^x-1)}{x^3} $?
I have this limit:
$$
\lim_{x \to 0} \frac{x(e^x+1)-2(e^x-1)}{x^3}
$$
By using L'Hospital rule and Maclaurin expansion I both found that limit equals to $1/6$.

I want to simulate this to my student by Excel but I failed.
This is my computation in Excel.

I did not know why Excel doesn't work in this situation?

Please help me. Thanks alot.

答:
In the evaluation of the numerator you can incur a floating point error of about the size $(2+|x|)e^x\mu$, $\mu=2^{-52}$ being the machine constant. For $x\approx0$ this is $2\mu\approx 4\cdot 10^{-15}$. In the quotient this makes an error of $\frac{2\mu}{x^3}$.

In the series expansion the next term after the constant one is $\frac{x}{12}$. The floating point error becomes larger than the linear term at about $\frac{x}{12}=\frac{2\mu}{x^3}$, that is, $x=\sqrt[4]{24\mu}=2.7\cdot10^{-4}$. This is the region of minimal error, and also the point where your results turn away from convergence.


问:
Number of Dispersion Lines in A Square Dot Matrix
I want to find a formula to calculate the number of dispersion lines I can draw from any one corner of a square dot matrix of size $n \times n$.

By dispersion lines, I mean the lines that connect a particular point to all other points in a dot diagram. Here's a picture of what I mean :

(That's an approximate diagram of what I mean; sorry for the crude diagram)

Here, the horizontal blue line on the top row (when considered a straight line) connects many dots to a single point (thus all those dots on the line are collinear), so when we count the number of dispersion lines, we have to count lines connecting many collinear points as one and we mustn't take the subunits into consideration.

So when we consider a square dot matrix, where the dots are arranged as a square, we can draw $3$ apparent dispersion lines, plus some more. What I would like to find is a formula to calculate the number of dispersion lines in square dot grids (or dot matrices).

What I have at hand is this one formula that I tried to formulate today:$$n^2 - (n + 2(n - 1)) + 3$$
The problem is that this formula isn't working for all values of $n$ (I mean, the number of dots in a column/row of the square dot matrix), and also I haven't been able to spend sufficient time to find the number of dispersion lines for $n > 5$. If asked for, I'll attach the numbers below.

Any help in formulating an accurate formula is appreciated. If such a formula exists, please do tell me.

Thanks in advance.

答:
For $n=5$, the slopes of lines are in ascending order

$${\color{red}{0},\color{blue}{\frac{1}{4},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{3}{4},}\color{red}{1},\color{green}{\frac{4}{3},\frac{3}{2},2,3,4,}\color{red}{\infty}}$$

where blue and green are reciprocal of each other (symmetry about principal diagonal), hence equal in number.

Observe green slopes can be grouped as
$$ \color{green}{\underbrace{\frac{4}{3},\frac{4}{1}},\underbrace{\frac{3}{2},\frac{3}{1}},\underbrace{\frac{2}{1}}}$$

where numerators run from $2$ to $4$ and denominators are respectively coprime to numerators.

In general, this series runs from $i=2$ to $i=n-1$ and counts for all integers less than $i$ and coprime to it. We have Euler's totient function to count this, it is simply $\varphi(i)$.

Hence desired number of dispersion lines is $$f(n) =
 \color{red}{3} + \color{green}{2\sum_{i=2}^{n-1}} \color{green}{\varphi(i)} $$

We get following values

$n=2$, $f(n) = 3$

$n=3$, $f(n) = 5$

$n=4$, $f(n) = 9$

$n=5$, $f(n) = 13$

$n=6$, $f(n) = 21$

$n=7$, $f(n) = 25$

$f(n)$ satisfies the recurrence relation $$f(n) = f(n-1) + 2\varphi(n-1)$$


问:
Probability of exactly two heads in four coin flips?

When you flip a coin four times, what is the probability that it will come up heads
  exactly twice?

My calculation:

we have $2$ results for one flip : up or down

so flip $4$ times, we have $4\cdot2 = 8$ results total

Thus the probability is: $2/ 8 = 0.25 $
but the correct answer is $0.375$. Can anyone explain why I'm wrong?

答:
The generalized answer to r heads in n flips can be calculated this way

The sample space for n space is $2^n$

Number of ways in which you get r heads in n flips is

$nCr = \frac { {n!}}{(n-r)!(r!))} $

So $ P = \frac {{nCr}}{2^n}$

With n = 4 & r = 2

P = 6/16 = 0.375


问:
(x,y) co-ordinates of n equally spaced points on the perimeter of a square
Where a square of known dimensions has n points equally spaced around its perimeter, is there a way to calculate the set of (x,y) coordinates of these points?

I have a solution in C# language where starting from (0,0) I can add increments of perimeter/n to the x coordinate up to the max allowed value for x (i.e the length of the side of the square), and then spill the remainder into y and so on, but I can't help thinking there must be a better way.

Update with an example:
For instance, with number of points n=9 and the square sides of length=2, the distance between each pair of points around the perimeter is 0.889 and the set of points to two decimal places would be:

```<code>(0, 0)
(0.89, 0)
(1.78, 0)
(2, 0.67)
(2, 1.56)
(1.56, 2)
(0.67, 2)
(0, 1.78)
(0, 0.89)
</code>```

答:
Assume that each side of the square is $s$ units and that you are going to have $n$ points.

First, pretend that instead of placing these points around the perimeter of a square, you are placing them on 1 line segment, of length $4s$.  Then the $n$ points would be located at

$$0, ~\left(1 \times \frac{4s}{n}\right),
 ~\left(2 \times \frac{4s}{n}\right),
 ~\left(3 \times \frac{4s}{n}\right),
\cdots,
 ~\left((n-1) \times \frac{4s}{n}\right).$$

These one dimensional locations must then be translated into points on the perimeter of your square.

Let $k \in \{0,1,2,\cdots\},$ where
$k = \lfloor \frac{n}{4}\rfloor.$

That is, $k$ is the largest non-negative integer less than or equal to $\frac{n}{4}.$

First, one point goes to $(0,0)$.
Then, you can distribute $k$ points along the bottom edge of the rectangle, at points

$$\left(1 \times \frac{4s}{n}, ~0\right),
 ~\left(2 \times \frac{4s}{n}, ~0\right),
 ~\left(3 \times \frac{4s}{n}, ~0\right),
\cdots,
 ~\left(k \times \frac{4s}{n}, ~0\right).
$$

Similarly, you can then distribute another $k$ points along the left edge at

$$\left(0, ~1 \times \frac{4s}{n}\right),
 ~\left(0, ~2 \times \frac{4s}{n}\right),
 ~\left(0, ~3 \times \frac{4s}{n}\right),
\cdots,
 ~\left(0, ~k \times \frac{4s}{n}\right).
$$

Now, you have to "turn the corners".

First, calculate the "margin left over" along the bottom edge as

$$M = \left[s - \left(k \times \frac{4s}{n}\right)\right].$$

Then, the $k$ points along the right edge will be located at

$$\left(s, ~-M + 1 \times \frac{4s}{n}\right),
 ~\left(s, ~-M + 2 \times \frac{4s}{n}\right),
 ~\left(s, ~-M + 3 \times \frac{4s}{n}\right),
\cdots,
 ~\left(s, ~-M + k \times \frac{4s}{n}\right).
$$

Similarly, the $k$ points along the top edge will go at

$$\left(-M + 1 \times \frac{4s}{n}, ~s\right),
 ~\left(-M + 2 \times \frac{4s}{n}, ~s\right),
 ~\left(-M + 3 \times \frac{4s}{n}, ~s\right),
\cdots,
 ~\left(-M + k \times \frac{4s}{n}, ~s\right).
$$

Note that when $n$ is a multiple of $4$, so that $4k$ exactly equals $n$, then the above algorithm will end up having you place a point on the upper right corner of the square twice.

There is an exceptional case to be considered.

If

$$s + M ~\geq~ \left(k+1\right)
\times \left(\frac{4s}{n}\right)$$

then an additional point will need to be placed along both the right edge, and the top edge.


问:
Infinite products $\prod_{n=1}^\infty\sin\frac1n$ and $\prod_{n=1}^\infty\cos\frac1n$
This is my first question.

What are the values of the infinite products $\prod_{n=1}^\infty\cos\frac1n$ and $\prod_{n=1}^\infty\sin\frac1n$?

Body:
For the sines, it is quite simple
$$\prod_{n=1}^\infty\sin \left(\frac{1}{n}\right)=0$$ For the cosines the result is just a number
$$\prod_{n=1}^\infty\cos \left(\frac{1}{n}\right)\sim0.388536153335175859184329576$$ which is not recognized by inverse symbolic calculators.

However, we can approximate it using logarithms since
$$\log \left(\cos \left(\frac{1}{n}\right)\right)=\sum_{p=1 }^\infty (-1)^p\frac{ 2^{2 p-3} (E_{2 p-1}(1)-E_{2 p-1}(0)) }{p (2 p-1)!\,n^{2 p}}$$ which makes
$$\sum_{n=1 }^\infty \log \left(\cos \left(\frac{1}{n}\right)\right)=\sum_{p=1 }^\infty \frac{(-1)^{p+1} 2^{2 p-3} (E_{2 p-1}(0)-E_{2 p-1}(1)) \zeta (2 p)}{p (2 p-1)!}$$ which converges quite fast
$$\left(
\begin{array}{cc}
 p & \text{result} \\
 10 & 0.388538882923887752891524656369 \\
 20 & 0.388536153510144960460336253991 \\
 30 & 0.388536153335190159575729146913 \\
 40 & 0.388536153335175860483365339921 \\
 50 & 0.388536153335175859184454814359 \\
 60 & 0.388536153335175859184329588232 \\
 70 & 0.388536153335175859184329575688 \\
 80 & 0.388536153335175859184329575687
\end{array}
\right)$$


问:
Question 1.22 Unit 5 Hungerford Algebra
I am self studying Field theory and was trying some problems from Thomas Hungerford. I am struck on this problem  on page 242 and need help.

F is an algebraic extension of K if and only if for every intermediate field E every monomorphism $\sigma : E \to E $ which is the identity on K is in fact an automorphism of E.

I tried both sides of proof but couldn't do any.

Why if F is an algebraic extension then every $\sigma$ must be onto?

And conversaly, how if such $\sigma$ is an automorphism then why $\sigma $ is algebraic.

I have studied textbook thoroughly but there have been many questions which I asked.so, I would like to work on this problem by myself. Just give hints on which result  should I use as ultimately exercises are for my understanding not any others.

Thanks!!

答:
Let's first assume $F/K$ is algebraic and let $\sigma: E \longrightarrow E$ fixes $K$ (with $F/E/K$). Now let's take some $a \in E$. We want to show that this is in the image of $\sigma$. As $a \in E \subseteq F$, $a$ is algebraic over $K$. Thus, there is some $f \in K[x]$ nonzero such that $f(a) = 0$. Let $S \subseteq E$ be the set of all roots of $f$ in $E$. Note that as $\sigma$ fixes $K$, we have that $\sigma(f(b)) = f(\sigma(b))$ for all $b \in E$. Hence, $\sigma[S] \subseteq S$. We can therefore conclude that the restriction of $\sigma$ maps $K(S)$ to $K(S)$. Furthermore, there are only finitely many elements in $S$. As each is algebraic over $K$, we have that $K(S) / K$ is a finite extension. Thus, $\sigma: K(S) \longrightarrow K(S)$ is an injective map of finite dimensional $K$-vector spaces and is therefore an isomorphism on $K(S)$. As $a \in S$, we conclude that $a \in im(\sigma)$. Since $a$ was arbitrary, $\sigma$ is onto.

Now let's do the converse. As suggested in the comments, we'll approach this by contraposition. Suppose $F/K$ is not algebraic. Then there is some element $a \in F$ that is not algebraic over $K$. Consider then the map $\sigma: K(a) \longrightarrow K(a)$ via $a \mapsto a^2$. We claim that this is not an isomorphism. In particular, we show that $a$ is not in the image of $\sigma$. If it was, then there would be some $b \in K(a)$ such that $\sigma(b) = a$. We write $b = \sum_{i = 0}^{n} b_i a^i$ with $b_i \in K$. Let us also insist that $b_n \neq 0$. Indeed, clearly $b = 0$ cannot work here so this assumption is valid. Then $\sigma(b) = \sum b_i a^{2i} = a$. However, this is a polynomial relation for $a$! We can rewrite this as $\sum b_i a^{2i} - a = 0$. As we assumed $b_n \neq 0$, this is not the zero polynomial. However, we assumed that $a$ was not algebraic over $K$, so $a \notin im(\sigma)$.

Title: Question 1.22 Unit 5 Hungerford Algebra


问:
Check my proof: What determinants are possible for integer matrices whose inverses are also integer matrices?
I claim this is an adequate proof, but would appreciate criticism.

Trivially, we can see that this would be true for the special linear group ($|det(A)|=1$).

For the inverse condition to be true, the determinant must divide each element of the cofactor matrix, where $c_{ij} = (-1)^{i+j}M_{ij}$. Thus $det(A)|M_{ij} \forall i,j $ .

As the $(n-1)$st minor is a diagonal element, the "multiplicative trace" of $A$ must divide each element in the cofactor matrix.

Thus any integer determinant for A is attainable.

答:
I appear to have made an incorrect move regarding the minor.

I made the incorrect assumption that the minor preserves the diagonal of $A$, removing diagonal elements only one at a time.

We may recursively take minors of any matrix until arriving at a 1x1 matrix of any arbitrary element in $A$ (not only diagonal elements, as I had earlier assumed).

As $det(A)|M_{ij}$, and $M_{ij}|M_{ik,jl}, k\ne i, l\ne j$, and so on up to each single element matrix, $det(A)|a_{ij}$, that is, the determinant divides every element of $A$.

Suppose $A:n\times n;$ $|det A | \ne 1$. Let $|c|\ne1$ be the largest such element dividing each element in $A$. We may multiply each row in A by $c^{-1}$ to get $A'$ with $\det A'=c^{-n}\det A=1$,with the latter given by the condition that $c$ is the largest possible element.

Possible determinants for $A(\mathbb F^{n\times n})$ are therefore $c^n$ $\forall c\in\mathbb F$. In the case of integer matrices, $c\in\mathbb Z$, but then $c$ does not divide the elements of  $A^{-1}$. Therefore $|c|=1$ and A belongs to the special linear group.


问:
Limit by polar coordinates
Hi I need some help with this limit:
$$
\lim_{(x,y) \to (0,0)}  \frac{x^4y}{x^2+(x+y)^2}$$
I used polar coordinates and got that
$$
\frac{x^4y}{x^2+(x+y)^2}= \frac{r^3\cdot(\cos φ)^4  \sin φ } {1+ (\cosφ)^2+\sin(2φ)}
$$
I didn't know how to proceed from this.

答: $(x,y) \mapsto (r cos\phi, r sin\phi)$. So , when $(x,y) \to (0,0)$ it is equivalent to saying that $r\to 0$. Hence, the limit would be zero since you have $r^3$ in the numerator.

Title: Limit by polar coordinates