问:
Different approaches of binomial approximation using normal distribution yielding different results
I have the following question:

A process for manufacturing an electronic component yields items of
which $1$% are defective. A quality control plan is to select $100$
items from the process, and if none are defective, the process
continues. Use the normal approximation to the binomial to find the
probability that the process continues given the sampling plan
described.

Here we can define our random variable "X" as either the number of successes or number of failures, problem is that these should be equivalent in theory but are yielding different results in practice.

Approach A:

Defining X as number of failures, in other words we need $P(X=0)$
we have:

$n=100$,$p=0.01$,  $\mu=np=100*0.01=1$ and standard deviation=$\sqrt(npq)=0.99$

We need probability $P(X=0)$ which is $P(x<0.5)$ due to continuity correction.

Approximating this using the standard normal we get that this is equivalent to $P(Z<\frac{(0.5-1)}{0.99})$ which is equal to $P(Z<-0.503)$ that is equal to $0.3085$

Approach B:

Defining X as number of failures, in other words we need $P(X=1000)$
we have:

$n=100$,$p=0.99$,  $\mu=np=100*0.99=99$ and standard deviation=$\sqrt(npq)=0.99$

We need probability $P(X=100)$ which is $P(99.5<x<100.5)$ due to continuity correction.

Approximating this using the standard normal we get that this is equivalent to $P(\frac{(99.5-99)}{0.99}<Z<\frac{(100.5-99)}{0.99})$ which is equal to $P(0.505<Z<1.515$)

To get this we can do $P(Z<1.51)-P(Z<0.50)=0.93448-0.69146=0.24302$

Can someone explain why these results are different or what am I doing wrong? Any help would be appreciated.

答:
First of all, we should establish the exact probability.  For a single component selected at random, the probability it is defective should be $p = 0.01$.  Therefore, the probability that there are $X = 0$ defects in a batch of $n = 100$ components is simply $$\Pr[X = 0] = (1-p)^n = (0.99)^{100} \approx 0.366032.$$

Any approximation that does not yield an result that is reasonably close to this should be suspect.

For your Approach A, you seem to think that if the number of failures cannot be negative, then $\Pr[-0.5 < X < 0.5]$ should become $\Pr[X < 0.5]$.  But this is not consistent with your reasoning.  The probability $\Pr[X < 0.5]$ includes all negative values for $X$.  If you truly wanted to exclude negative values, you would compute $\Pr[0 \le X < 0.5] \approx 0.150212$.

Conversely, for Approach B, you don't argue that $X$ cannot exceed $100$, when this is equally true as your reasoning that $X$ cannot be negative.  This is the reason why your approaches do not match.

Another reason why you have discrepancies is because you are rounding your decimal precision far too early in your calculations.  You should never round to such low precision prior to computing a probability or quantile.

Finally, it should be noted that the argument for truncating the approximated distribution because the exact distribution has finite support, is invalid.  You shouldn't be doing it because you will lose half of the probability mass at the respective endpoint.  The correct calculation is as follows.

Let $X$ be the random number of defects.  The exact distribution is $$X \sim \operatorname{Binomial}(n = 100, p = 0.01),$$ but $X$ is approximately normal with mean $\mu = np = 1$ and variance $np(1-p) = 0.99$, therefore $$\Pr[X = 0] = \Pr[-0.5 \le X \le 0.5] = \Pr\left[\frac{-0.5 - 1}{\sqrt{0.99}} \le \frac{X - \mu}{\sigma} \le \frac{0.5 - 1}{\sqrt{0.99}}\right].$$  When employing continuity correction, we do .  We must include the entire probability mass, even at the endpoints of the support.  This gives us $$\Pr[X = 0] \approx 0.241817.$$

Note that this approximation is poor.


问:
Limit of two variables function by defintion
I have to prove by definition that
$$\lim_{(x,y)\to(3,\pi)}x^{2}\sin(y)=0$$

Since
$$|f(x,y)-0|=|x^{2}|\cdot|\sin(y)|$$
and
$$|\sin(y)|\leq1$$
then
$$|f(x,y)-0|\leq|x^{2}|$$

Can I say that since
$$|x|\leq\sqrt{x^{2}+y^{2}}$$
so
$$|x^{2}|=|x|^{2}\leq x^{2}+y^{2}$$
and therefore
$$|x|^{2}\leq(x-3)^{2}+(y-\pi)^{2}?$$

I'm not sure about the last inequality, but it's the only way I found to try to write $\delta$ using $\epsilon$.

EDIT

I tried to change the variables:
$$x=\bar{x}+3,\quad y=\bar{y}+\pi$$
so, I have the limit
$$\lim_{(\bar{x},\bar{y})\to(0,0)}(\bar{x}+3)^{2}\sin(\bar{y}+\pi)$$
and, still
$$|f(\bar{x},\bar{y})-0|\leq|(\bar{x}+3)^2|$$
but now I must to find $\delta$ such that
$$||(\bar{x},\bar{y})-(0,0)||<\delta$$
implies on
$$|f(\bar{x},\bar{y})-0|<\epsilon$$
Since
$$||(\bar{x},\bar{y})-(0,0)||=\sqrt{\bar{x}^{2}+\bar{y}^{2}}$$
then
$$\sqrt{\bar{x}^{2}+\bar{y}^{2}}<\delta\Rightarrow \bar{x}^{2}+\bar{y}^{2}<\delta^{2}$$
and, therefore
$$|\bar{x}|^{2}=\bar{x}^{2}\leq \bar{x}^{2}+\bar{y}^{2}<\delta^{2}\Rightarrow |\bar{x}|^{2}<\delta^{2}$$
and so
$$|\bar{x}|<\delta$$
Hence,
$$|(\bar{x}+3)^{2}|=|\bar{x}^{2}+6\bar{x}+9|\leq|\bar{x}^{2}|+6|\bar{x}|+9<\delta^{2}+6\delta+9=(\delta+3)^{2}$$
Then, if I take
$$\epsilon=(\delta+3)^{2}$$
the limit is proved.

I want to know if it's right.

答:
You are asked about the following limit: $\ \lim\limits_{(x,y)\to(3,\pi)}f(x,y)$

we have $x\to 3$ and $y\to \pi$ so we want to compare $f(x,y)$ to $f(3,\pi)=9\sin(\pi)=0$

Our mathematical toolbox works better with limits in $0$ or $\infty$ so when we have a limit in "random" points, we often shift the function to study it around $0$.

let set $\begin{cases}x=3+u &&u\to 0\\y=\pi+v&&v\to 0\end{cases}$

Notice $(u,v)\to(0,0)\implies (x,y)\to(3,\pi)$

You were on the right track but I feel that using $\bar x,\bar y$ just add some more confusion (also it is heavy to write), prefer the use of $(u,v)$ or $(h,k)$ or whatever seems familiar to you, using always the same letters will trigger math reflexes.

we now have to study $\bigg|f(x,y)-f(3,\pi)\bigg|=\bigg|(3+u)^2\sin(\pi+v)-0\bigg|=\bigg|(3+u)^2\sin(v)\bigg|$

Since $\sin(\pi+v)=-\sin(v)$

Now you recognize $\sin(v)\to 0$ when $v\to 0$, for that we can use the inequality $|\sin(v)|\le |v|$

On the other hand $(3+u)^2$ does not really interests us, let just say $u\to 0$ so it is small, without loss of generality we can assume $|u|<1$ and then $(3+u)^2<16$

Note: look at the end the choice of delta, $|u|<1$ is just convenient and not a limitation.

finally we arrived to $\bigg|f(x,y)-f(3,\pi)\bigg|\le 16|v|\to 0$

Now regarding the epsilon-delta definition of the limit, remember that norms in $\mathbb R^2$ are equivalent.

In particular using one of the $3$ standard ones $\begin{cases}\lVert(u,v)\lVert_1=|u|+|v|\\\\\lVert (u,v)\lVert_2=\sqrt{u^2+v^2}\\\\\lVert(u,v)\lVert_\infty=\max(|u|,|v|)\\\end{cases}$ is up to you.

Select the one which is more convenient for the problem at hand

since $\begin{cases}|u|\le |u|+|v|&&|u|\le\sqrt{u^2+v^2}\\|v|\le|u|+|v|&&|v|\le\sqrt{u^2+v^2}\end{cases}\implies \lVert(u,v)\lVert_\infty\le\lVert(u,v)\lVert_1\le2\lVert(u,v)\lVert_2$

also $\sqrt{u^2+v^2}\le\sqrt{u^2}+\sqrt{v^2}=|u|+|v|\implies \lVert(u,v)\lVert_2\le\lVert(u,v)\lVert_1\le2\lVert(u,v)\lVert_\infty$

In our case since we have to bound $|u|$ and $|v|$ then select $$\delta=\min(1,\frac{\epsilon}{16})\text{ and }\lVert(u,v)\lVert_\infty<\delta$$

We have our result since $\begin{cases}|u|\le\max(|u|,|v|)<\delta\le 1\\|v|\le\max(|u|,|v|)<\delta\le \frac{\epsilon}{16}\end{cases}$

Notice that we do not have to do particularly confusing operations with $x-3$ and $y-\pi$ and square roots. The fact that we used $u,v$ and norm equivalence leads to a natural solution for the problem.

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问:
Projection $E_i \to E$ in WOT is equivalent to converge in SOT when E is projection
I’m studying Conway’s Functional Calculus, and meet this question.

If $E_i, E$ are projections then $E_i \to E$ in WOT is equivalent to $E_i\to E$ in SOT.

I approached to this problem that
If it converge in SOT sense, then it trivially converge in WOT.

So we need to care about if WOT converge, then SOT converge.

I tried this formula.

$\langle E_i h, h \rangle \to \langle Eh, h \rangle$
And we know that $E_i $ is adjoint and $E_i^2 = E_i$.
So, $\langle E_i h, E_i h \rangle \to \langle Eh, Eh \rangle$

But it does not prove the relation $\|E_i h — Eh \| \to 0$

Can you help me?

答:
Note that:
$$\|(E-E_i)h\|^2 = \|Eh\|^2+\|E_ih\|^2 - \langle Eh, E_ih\rangle - \langle E_ih,Eh\rangle$$
As you have seen, $E_i\to E$ in WOT implies $\|E_ih\|^2\to \|Eh\|^2$, this is the only step that uses $E_i$ and $E$ being projections. From the definition WOT convergence it follows that $\langle Eh, E_ih\rangle\to\langle Eh,Eh\rangle = \|Eh\|^2$ and that $\langle E_ih,E\rangle \to \langle Eh, Eh,\rangle = \|Eh\|^2$. So in total you get:

$$\|(E-E_i)h\|^2 =  \|Eh\|^2+\|E_ih\|^2 - \langle Eh, E_ih\rangle - \langle E_ih,Eh\rangle \to \|Eh\|^2+\|Eh\|^2-\|Eh\|^2-\|Eh\|^2=0$$

for any $h$, which gives SOT convergence.


问:
Example of $f \in K[x]$ solvable by radicals but having a root that cannot be expressed by using only coefficients of $f$, $+,-,\cdot,\frac{..}{..}$
The definitions below are taken from Solvability by radicals implies a radical formula for its roots (question by Eparoh):

Definition 1: We say that a field extension $F/K$ is a radical extension if we can form a chain of fields
$$K=K_0 \leq K_1 \leq \cdots \leq K_n=F$$
where $K_{i+1}/K_i$ is a simple extension such that $K_{i+1}=K_i(a_i)$ and $a_i^{k_i} \in K_i$ for some positive integer $k_i$.

Definition 2: Let $K$ be a field and $f(x) \in K[x]$, we say that $f$ is solvable by radicals if there exists a radical extension $F/K$ such that $F$ contains a splitting field of $f$ over $K$.

This question has no answer, but it has a comment by reuns:

The radical formulas for the roots depend on constants of $K$, once the polynomial is fixed this is all we want (there are algorithms for the splitting field minimal polynomials and Galois group, if it is solvable we can unroll to find the radical formulas). What you are asking is if there are finitely many radical formulas $F_{d,l}$ of $d+1$ variables such that for every solvable polynomial $∑_{j=0}^{d} c_j x_j \in K[x]$ of degree $d$ its roots are given by $F_{d,l}(c_0,…,c_d)$ for some $l$. This is the problem of moduli space / parametrization of solvable polynomials of degree $d$.

Let $K$ be a field. Can you give an example of $f \in K[x]$ that is solvable by radicals but cannot be expressed by using only polynomial coefficients, $+, -, \cdot, \frac{...}{...}$ and proof of this fact?

Edit: the answer to this question is obvious and it is not what I actually wanted to ask. I forgot to specify $\sqrt[n]{...}$ as an operation that we can use in an expression. For this reason I created another question, Example of $f \in K[x]$ solvable by radicals but having a root inexpressible only by coefficients of $f$ and +, -, *, /, $\sqrt[n]{...}$ which asks the question I indeed wanted to ask.

答:
Any polynomial that is solvable with radicals but with no roots in $K$ would work. The four arithmetic operations cannot take you outside of $K$.

Take $K=\Bbb Q$ and $f(x)=x^2-2$, for instance.


问:
Does integral of L1 function vanishes on small intervals
I want to know if for $u \in L^1(\mathbb{R})$, we have $\int_0^\delta u \to 0 \text{ as } \delta \to 0$ ?

I know how to prove the result in $L^p(\Omega)$, $\Omega \subseteq \mathbb{R}^n$ and $p > 1$.

$ \left\vert \int_{B_q(\delta)} u \right\vert \leq \int_{B_q(\delta)} \vert u \vert$ where $B_q(\delta) \subseteq \Omega$ ($\delta$ small enough) is some ball in $q$-norm of radius $\delta$.

We have: $\int_{B_q(0, \delta)} \vert u \vert = \int_\Omega \vert u \vert \cdot 1_{B_q(0, \delta)}$, where $1_{B_q(0, \delta)}$ designates the indicator function on $B_q(0, \delta)$.

By Hölder inequality: $\int_\Omega \vert u \vert 1_{B_q(0, \delta)} \leq \Vert u \Vert_{L^p(\Omega)} \cdot \Vert 1_{B_q(0, \delta)} \Vert_{L^{p'}(\Omega)}$, with $p' < \infty$ the Hölder conjugate of $p$.

Since $\Vert 1_{B_q(0, \delta)} \Vert_{L^{p'}(\Omega)} \leq \Vert 1_{B_\infty(0, \delta)} \Vert_{L^{p'}} = \delta^n$, we get $\left \vert \int_{B_q(0, \delta)} u \right\vert \to 0$ as $\delta \to 0$.

For $p = 1$, we cannot use the Hölder inequality to prove the statement as $\Vert 1_{B_\infty(0, \delta)} \Vert_{L^\infty(\Omega)} = 1$.

Is the statement true for $p = 1$ and how to prove it?

EDIT:

Yes, the statement is true in $L^1$:

By a density theorem (e.g. Theorem IV.3 in [1]) for $\epsilon > 0$, there is some $\phi \in C(\Omega)$ such that $\Vert u - \phi \Vert_{L^1(\Omega)} \leq \epsilon$.

Then we have $\int_{B_q(\delta)} \vert u \vert \leq \int_{B_q(\delta)} \vert u - \phi \vert + \int_{B_q(\delta)} \vert \phi \vert$.

We have: $\int_{B_q(\delta)} \vert u - \phi \vert \leq \int_\Omega \vert u - \phi \vert \leq \epsilon$

And: $\int_{B_q(\delta)} \vert \phi \vert \leq \delta^n \Vert \phi \Vert_{L^\infty(B_\infty(\delta))}$

Since $\phi$ continuous in $\Omega$, then $\phi$ continuous in $\overline{B_\infty(\delta)}$ (the closure of the ball) and $\Vert \phi \Vert_{L^\infty(B_\infty(\delta))} < \infty$.

Then the right hand side of the inequality can be made as small as we want by playing on $\delta$, proving the result in $L^1$.

[1] Brézis, Haïm, Functional analysis. Theory and applications, Collection Mathématiques Appliquées pour la Maîtrise. Paris: Masson. 248 p. (1994). ZBL1147.46300.)

答:
Yes, your solution is right. A more general result is this: let $(X,\mathcal A,\mu)$ a measure space and suppose that $\int_X |f|^pd\mu<\infty$ (for a measurable $f:X\to \Bbb R$ and some $p\in [1,\infty]$), then for every chosen $\epsilon>0$ there is a $\delta >0$ such that for any chosen $A\in\mathcal A$ it holds that $\mu(A)<\delta\implies\int_A |f|^pd\mu<\epsilon$.

Proof: the simple functions are dense in any $L^p$ space then there is a simple function $s$ such that $\|f-s\|_p<\epsilon/2$ for any chosen $\epsilon \in(0,1)$. Suppose that $p<\infty $, then as each simple function is bounded there is some $M\in(0,\infty )$ such that $|s|\leqslant M$, then by Minkowski's inequality we have that
$$
\left(\int_{A}|f|^pd\mu\right)^{1/p}\leqslant \left(\int_{A}|f-s|^pd\mu\right)^{1/p}+\left(\int_A|s|^pd\mu\right)^{1/p}\leqslant \frac\epsilon 2+M(\mu(A))^{1/p}
$$
Then choosing $\delta :=\left(\frac{\epsilon }{2M}\right)^p$ the statement is clear. For $p=\infty $ a shorter proof can be given.∎

Alternatively to Minkowski's inequality we can use the inequality
$$
|a+b|^p=2^p|\frac a2+\frac b2|^p\leqslant 2^p(\frac12|a|^p+\frac12| b|^p)=2^{p-1}(|a|^p+|b|^p)
$$
for $p\in[1,\infty )$.


问:
Weak convergence in Hilbert space L2 implies convergence in distribution?
Does weak convergence in $L^2$ (for $X_n, X \in L^2$ we say that $X_n$ converges weakly to $X$ ($X_n \rightarrow^w X$) if for every $Y\in L^2$ we have $\mathbb{E}X_nY \rightarrow \mathbb{E}XY$) implies convergence in distribution ($X_n \rightarrow^D X$ if $F_{X_n} \rightarrow F_{X}$ in every point of continuity of $F_X$)? Maybe we need some extra assumptions? If it is so, how to prove that fact?

Thanks a lot

答:
Weak convergence is not enough in order to give convergence in distribution. For example, take $\Omega:=[0,1]$, and $X_n\left(\omega\right):=\sin\left(\pi n\omega\right)$. Then $X_n\to 0$ weakly but not in distribution. Indeed, if a sequence $ X_n\to 0$ in distribution, then $X_n\to 0$ in probability and since $\left\lvert X_n\right\rvert\leqslant 1$, this implies that $\mathbb E\left\lvert X_n\right\rvert\to 0$, which is not the case (or more directly, in the definition of convergence in distribution given by $\mathbb Ef(X_n)\to\mathbb E f(X)$ for any continuous and bounded $f$, choose $f\colon x\mapsto \min\left\{1,\left\lvert x\right\rvert    \right\}$.               </p>

<p>But strong convergence in $L^2$ implies convergence in probability which implies convergence in distribution. </p>

ContentLicense: CC BY-SA 3.0

Body:
Weak convergence is not enough in order to give convergence in distribution. For example, take $\Omega:=[0,1]$, and $X_n\left(\omega\right):=\sin\left(\pi n\omega\right)$. Then $X_n\to 0$ weakly but not in distribution. Indeed, if a sequence $ X_n\to 0$ in distribution, then $X_n\to 0$ in probability and since $\left\lvert X_n\right\rvert\leqslant 1$, this implies that $\mathbb E\left\lvert X_n\right\rvert\to 0$, which is not the case (or more directly, in the definition of convergence in distribution given by $\mathbb Ef(X_n)\to\mathbb E f(X)$ for any continuous and bounded $f$, choose $f\colon x\mapsto \min\left\{1,\left\lvert x\right\rvert    \right\}$.

But strong convergence in $L^2$ implies convergence in probability which implies convergence in distribution.


问:
Group of order $63$ has an element of order $3$, without using Cauchy's or Sylow's theorems
I'm showing that any group of order $63$ has an element of order $3$, and can only use Lagrange's theorem not Cauchy's or Sylow's. I got it reduced to a case of having $62$ elements of order $7$ but I'm stuck now.

So let $|G|=63$, and consider an element $a \in G$, $a$ not equal to the identity. I'm allowed to use the theorem that if $|a|=n$, and $k$ divides $n$, then $|a^{n/k}|=k$.  Also, I know $G$ can only have elements of orders $1,3,7,9,21,63$.  If $|a|=63$, then $|a^{21}|=3$.  If $|a|=21$, then $|a^7|=3$.  If $|a|=9$, then $|a^3|=3$.  If $|a|=3$, then we are done. So $G$ must contain an element of order $3$ unless $G$ contains only the identity, and $62$ elements of order $7$. Now how do rule out this?

答:
|G|=63 By Lagrange theorem if a belongs to G and a is non identity element

Possible orders of  a= 1,3,6,7,9,21,63
If | a | =63 , order of a^21= 3
If  | a |= 21, order of a^7 =3
If | a | =9, order of a^3= 3
If | a | =6 ,  order of a^2=  3
If a has order 63, 21, 9, 6 and of course 3 then there is an element of order 3 in G
So if a does not have order 3 all non identity(62 ) elments of G must have order 7
But we know for |G |= 63 no of elements of order 7= 6k ( a multiple of 6)
But 62 is not a multiple of 6 hence there must be a non identity element of G whose order is 3.


问:
If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$
This is the $y=2^k$ case of this question.

Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?

Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that these two divisors are $2^k-1$ and $2^k+1$?

答:
This is just a hint, which may be useful, not a full answer.

$$(2^k-1)(2^k+1)=(2^k-x)(2^k+x)t$$

From this we see that $x$ should be odd. And easy to prove that the $$(2^k-x,2^k+x)=1 $$
As the $2^k+x > 2^k-1$ and $2^k+x > 2^k+1$ for the $x>1$ then $$(2^k+x,2^k-1)=a > 1$$ and $$(2^k+x,2^k+1)=b > 1$$ and $(a,b)=1$ and $ab=2^k+x$.

Let assume that $$(2^k-1,2^k-x)=c$$ and $$(2^k+1,2^k-x)=d$$ then obviously $(a,c)=1$, $(a,d)=1$, $(b,c)=1$, $(b,d)=1$, $(d,c)=1$ and $cd=2^k-x$.
And $a,b,c,d$ are odd.
$$ac | 2^k-1$$
$$bd | 2^k+1$$
$$ab+cd=2^{k+1}$$
$$ab-cd=2x$$

From this it seems that there should be some solution, but probably for the $k$-s for which the $2^k-1$ and $2^k+1$ has enough divisors.
In this scope it is also interesting to consider the Bang's theorem


问:
The stereographic projection is bijective
I want to prove that the stereographic projection:$f:\{\ (x,y):x^{2}+y^{2} = 1 \}\ \rightarrow \mathbb{R}$ where $f$ is defined as:
$$ f(x,y) = \frac{x}{1-y}$$
is bijective to the real number. I have already shown that $f$ is one-to-one, but I am struggling on showing this is onto. I also tried to find an inverse for $f$ but I failed. What is the easiest way to show $f$ is bijective?

答:
Let $\,S^1:=\{x^2+y^2=1\}\subseteq \Bbb R^2$. Consider $\,f :S^1-\{(0,1)\}\to \Bbb R\,$ defined by $\displaystyle \,f(x,y)=\frac x{1-y}$.

Given any $\,a\in\Bbb R$, we search, if exists, $\,(x,y)\in S^1-\{(0,1)\}\,$ such that

$$ f(x,y)=\frac x{1-y}=a. \tag{*}$$

From this we have, by squaring:

$$ \frac{x^2}{(1-y)^2}=\frac{1-y^2}{(1-y)^2}=\frac{1+y}{1-y}=a^2, $$

and hence

$$ y=\frac{a^2-1}{a^2+1}, \quad\text{ from which equation (*) gives }\quad x=\frac{2a}{a^2+1}.$$

As

$$ \left(\frac{2a}{a^2+1}\right)^2 + \left(\frac{a^2-1}{a^2+1}\right)^2=1 \qquad \text{and} \qquad \frac{a^2-1}{a^2+1}\neq 1, $$

and since you have already proved that $\,f\,$ is injective, it follows that there exists $\,f^{-1}\,$, given by

$$ f^{-1}(a)=\left(\frac{2a}{a^2+1},\; \frac{a^2-1}{a^2+1}\right). $$


问:
applying the chain rule for taking the derivative w.r.t a matrix
I have the following net $z=Wa$, where $z$ is a column vector 2X1, $a$ is a column vector of 3X1 and $W$ is a weight matrix with dimensions 2X3, and a loss function $l=L(z)$ where l is a scalar.

I'm now trying to calculate $\frac{dL}{dW}$ = the derivative of the loss w.r.t $W$, and I'm expecting the dimensions of the result to be 2X3. How do I do that using the generalized chain rule?

I calculated $\frac{dL}{dz}$, which is a 1X2 row vector.

I calculated $\frac{dz}{dW}$, which is a 3-d array, see below

My professor told us without proof that $\frac{dz}{dW}$ is simply $a$ as a 1x3 row vector!!!, so we can calculate the derivative using a simple outer product: $\frac{dL}{dz}^T\frac{dz}{dW}$, which indeed works out, and looks nice just like the chain rule - but why is that correct? how come my 3-d array below amount simply to $a$?

Looking for a formal explanation, along the lines of vector-derivatives

答:
$\def\p{\partial}$
First of all you are correct: the quantity $\left(\frac{\p z}{\p W}\right)$
is definitely a third-order tensor. The field of machine learning is full of folks (even professors) who use atrocious mathematics.

Second (as you have discovered), it is very difficult to apply the chain rule in matrix calculus because it requires intermediate quantities which are third and fourth order tensors. No one cares about these intermediate quantities, because the final result is always a scalar, vector, or matrix.

A simpler approach is to use differentials, because the differential of a matrix has the same dimensionality as the matrix, and it obeys all of the rules of matrix algebra.

One piece of notation that makes things less error prone (especially transpose errors) is the trace/Frobenius product (denoted by a colon) between two equally-sized matrices, i.e.
$$A:B = {\rm tr}(A^TB) = \sum_i\sum_j A_{ij}B_{ij}$$
The properties of the trace allow such products to be rearranged
in many equivalent ways, e.g.
$$\eqalign{
A:B &= B:A = B^T:A^T \\
CA:B &= A:C^TB = C:BA^T \\
}$$
This product also works for vectors (simply think of them as rectangular matrices)
$$a:b = {\rm tr}(a^Tb) = a^Tb \\$$
Assume we have a quadratic loss function
$$L = \tfrac 12 (z-\hat z):(z-\hat z)$$
and wish to calculate its gradient with respect to $W$.
Start by calculating the differential
$$\eqalign{
dL &= \tfrac 12d(z-\hat z):(z-\hat z)+\tfrac 12(z-\hat z):d(z-\hat z) \\
   &= (z-\hat z):d(z-\hat z) \\
   &= (z-\hat z):dz \\
   &= (z-\hat z):dW\,a \\
   &= (z-\hat z)a^T:dW \\
}$$
Now the gradient is easily identified as
$$\eqalign{
\frac{\p L}{\p W} &= (z-\hat z)a^T \\
}$$
But now you understand the origin of the $a^T$ term. It comes from rearranging a Frobenius product. It is definitely not equal to $\left(\frac{\p z}{\p W}\right)^T$

Notice that this gradient has the same dimensions as the $W$ matrix, exactly as you anticipated. This is due to the presence of the $dW$ factor on the RHS of the Frobenius product, which has the same dimensions as $W$.

The standard text for this stuff is Matrix Differential Calculus
by Magnus & Neudecker.


问:
$f(z) = 1$ for all $z \in \mathbb{C}$ with $|z| = 1 \implies f(z) = 1$ for all $z \in \mathbb{C}$?
I would like to know whether this statement is true or not:

If $f : \mathbb{C} \to \mathbb{C}$ is holomorphic such that $f(z) = 1$ for all $z \in \mathbb{C}$ and $|z| = 1$, then $f$ is constant with $f(z) = 1$ for all $z \in \mathbb{C}$.

My guess is that this statement holds. Still, I struggle to proof it, mainly because the set $M := \{z \in \mathbb{C} : |z| = 1\}$ is no domain. However, considering a domain (or at least an open set) is a basic assumption for almost every theorem that states something about constant holomorphic functions.

答: $1-f$ is $0$ over the non-discrete subset $S^1$, therefore it is constant $0$.


问:
Folland 5.1.12 e)

I'm working on part e) of the problem.

The quotient topology is defined as

I'm reading the proof in http://www-users.math.umn.edu/~bahra004/for-real/real2-hw4.pdf.

But I'm still not sure why it suffices to show $\pi: X \rightarrow X / U$ is open. Also, I'm not sure what the topology defined by the quotient norm is.

Also, is there an alternative way of doing part e)?

Thanks!

答:
The quotient norm is the norm $\|\cdot\|_{\mathcal X/\mathcal M}: \mathcal {X/M}\to\Bbb R, \ x+\mathcal M \mapsto \inf_{y\in\mathcal M}\|x+y\|$.

The way the solution you have given works is as follows:

Let $\pi: A\to B$ be a surjective map and consider two topologies $\tau_1, \tau_2$ on $B$ so that $\pi$ is open and continuous wrt both $\tau_1,\tau_2$. Then $\tau_1=\tau_2$. The argument showing this is as follows:

Let $U\in\tau_1$ be open, then by continuity $\pi^{-1}(U)$ is open in $A$. By surjectivity you have that $U=\pi(\pi^{-1}(U))$ and by openness this is open in $\tau_2$, hence $\tau_1\subseteq \tau_2$. You may repeat the argument with $\tau_1$ and $\tau_2$ swapped.

In your case the quotient map to the quotient space topology is continuous by definition, the map to the quotient norm topology is continuous by c). The map to the quotient space topology is open since for $U\subseteq X$ open you have that:
$$\pi^{-1}[\pi(U)] = U+\mathcal M = \bigcup_{x\in\mathcal M}(U+\{x\})$$
which is open as a union of open sets, hence $\pi(U)$ is open by definition of the quotient space topology. The map to the quotient norm topology is open by the argument described in the solution.

Title: Folland 5.1.12 e)


问:
Determine contribution quadratic in $B$ of $[\det (A+B) ]^{1/2}$
Let $A$ and $B$ be $n\times n$ matrices over $\mathbb R$, with $n\ge 4$. $A$ is a symmetric matrix with inverse $A^{-1}$ and $B$ is an antisymmetric matrix.

Consider
$$
f= \left[\det (A+B) \right]^{1/2}
$$
where $\det$ is the matrix determinant and we can assume that $\det (A+B)> 0$. I want to find an expression for the contributions to $f$ that are quadratic in the matrix elements $B_{ab}$ of $B$.

I believe it must be of the form $\alpha B_{ab} (A^{-1})^{ac}(A^{-1})^{bd}B_{cd}= -\alpha \, \mathrm{tr}\, (B A^{-1}B A^{-1})$ where tr is the standard matrix trace and $\alpha$ is some coefficient.

Unfortunately I have no idea how to show this in a simple way. Brute force calculation might help, but I don't know how I would have to handle the square root.

Any help on this will be much appreciated.

答:
Turns out the answer is pretty simple. The symmetry properties of $A$ and $B$ are irrelevant. Just use the expansion
$$
\det (\mathbb 1 + A) = 1 + \mathrm{Tr}\,{A} + \frac{1}{2} \left(\mathrm{Tr}\,{A}\right)^2 - \frac{1}{2} \mathrm{Tr}\,{A^2} + o(A^3)
$$
and then expand the square root. The terms with $\left(\mathrm{Tr}\,{A}\right)^2$ cancel out and the term $\mathrm{Tr}\,{A^2}$ has a factor $-1/4$.


问:
How to prove that this function is not 1-1
The following question was part of my analysis assignment and I couldn't solve this part out of 5parts.

Let $f: \mathbb{R}^2 \to \mathbb{R}^2 $ be a function $f(x,y)=( x^2 -y^2 , 2xy) \in \mathbb{R}^2$. Is  f 1-1 on $\mathbb{R}^2$ ?

I let $(x^2-y^2, 2xy)=(p^2-q^2, 2pq)$  but neither I am able to prove p=x and q=y nor able to find a contradictory exam ple.

Your guidance is required!

答: We have $f(x,y) = f(-x, -y)$ so it's not injective


答: Since $\Bbb C$ is connected, and since $S^1$ has non-isolated points (actually, none of its points is isolated), it follows from the identity theorem that$$\left(\forall z\in S^1\right):f(z)=1\implies(\forall z\in\Bbb C):f(z)=1.$$

Title: $f(z) = 1$ for all $z \in \mathbb{C}$ with $|z| = 1 \implies f(z) = 1$ for all $z \in \mathbb{C}$?

Body:
I would like to know whether this statement is true or not:

If $f : \mathbb{C} \to \mathbb{C}$ is holomorphic such that $f(z) = 1$ for all $z \in \mathbb{C}$ and $|z| = 1$, then $f$ is constant with $f(z) = 1$ for all $z \in \mathbb{C}$.

My guess is that this statement holds. Still, I struggle to proof it, mainly because the set $M := \{z \in \mathbb{C} : |z| = 1\}$ is no domain. However, considering a domain (or at least an open set) is a basic assumption for almost every theorem that states something about constant holomorphic functions.


问:
Computing a double integral involving $\sin(y^2)$

Let $T = \{(x,y) \in \mathbf{R}^2 \mid x\in [0,1] \land y \in[x, 1] \}$ Determine $$\int_D \sin(y^2)\ dy \ dx$$

I feel like I'm not interpreting the set $T$ correctly. Initially, I thought that it would represent the unit square, but I'm confused about $y\in [x,1]$. My inital approach was to evaluate $$\int_0^1 \int_0^1 \sin(y^2)\ dy \ dx$$

but even the integral of $\sin(y^2)$ is not trivial. I haven't taken complex analysis which seems to be needed to compute $\int_0^1 \sin(y^2)\ dy$. What might I be missing, this shouldn't need complex analysis since it was asked on analysis II.

Body:
The integral would be
$$ \int_{0}^{1} \int_{x}^{1} sin(y^2)\hspace{0.1cm}dy\hspace{0.1cm}dx$$
Or equivalently,
$$ \int_{0}^{1} \int_{0}^{y} sin(y^2)\hspace{0.1cm}dx\hspace{0.1cm}dy$$
$$ = \int_{0}^{1} y\hspace{0.1cm}sin(y^2)\hspace{0.1cm}dy$$
$$ = \frac{cos(y^2)}{2}\Bigg|_{y=1}^{y=0}$$
$$ = \frac{1-cos\hspace{0.05cm}1}{2}$$


问:
Prove that the vectors are linearly independent
Show that vectors $v_{1}, v_{2}, ..., v_{n} \in V$ are linearly independent only when $v_{1}, v_{1} + v_{2}, ..., \sum_{i \in [n]} v_{i}$ are linearly independent.
I know it must be true, because its the definition of linearly independent vectors, but I don't know how can I prove that.

答: We're given that if $\sum_{i=1}^nc_iv_i=0$ then all $c_i$ are $0$. You want to prove if$$\sum_{i=1}^nC_i\left(\sum_{j=1}^iv_j\right)=\sum_{i=1}^nc_iv_i=0,\,c_i:=\sum_{j=i}^nC_j$$then  all $C_i$ are $0$. Certainly all the $c_i$ are $0$. Then $C_i=c_i-c_{i+1}=0$ for $i<n$, while $C_n=c_n=0$.


问:
Why do we use normal approximation for sample proportions of cases involving a binomial distribution?
I'm in high school and am learning sample proportions. I have encountered a doubt that i cannot answer myself.

Why do we use normal approximation for sample proportions of cases involving a binomial distribution?

Why approximate? We have a binomial distribution, isn't it more accurate to just use this?

For example,

A company employs a sales team of 20 people, consisting of 12 men and 8 women. 5 sales people are to be selected at random to attend an important conference. Determine the probability that the proportion of men, in a random sample of 5 selected from the sales team, is greater than 0.7.

To solve this question, my teachers would say use normal distribution with $X$ being $N \sim (3, 0.048)$ to calculate the probabilty of $(X>0.7)$ in my calculator.

My problem here is: wouldn't it make more sense to calculate using binomial distribution with $X$ being $Bin \sim (5,0.6)$ then, as $0.7\times 5$ is $3.5$ round down to $3$ as you can't have $3.5$ men, calculating probability of $(X>3)$ in my calculator.

Can someone please clarify this? :D

Thanks

答:
In my opinion, the normal approximation is inappropriate for this problem because the sample size is simply too small, although, through sheer luck, it does happen to give a fairly good approximation to the exact probability, provided you choose the right variate value (namely $3.5$) to plug into it.

The exact probability that your random sample of $\ 5\ $ contains $\ m\ $ men and $\ 5-m\ $ women is
$$
p_m=\frac{{12\choose m}{8\choose5-m}}{20\choose5}\ .
$$
The first factor in the numerator of this fraction is the number of ways of choosing $\ m\ $ men from the $12$ available, the second factor is the number of ways of choosing $\ 5-m\ $ women from the $8$ available, and the denominator is the total number of ways of choosing
$5$ people from the $20$ available.  For the proportion of men in your sample of $\ 5\ $ to be greater than $\ 0.7\ $, $\ m\ $ must be $4$ or $5$, and the probability of this occurring is
$$
\frac{{12\choose4}{8\choose1}+{12\choose5}}{20\choose5}=\frac{99}{323}\approx0.3065\ .
$$
The binomial approximation, on the other hand, gives an estimate of
$$
{5\choose4}\times0.6^4\times0.4+{5\choose5}\times0.6^5\approx 0.337\ ,
$$
which is about $10\%$ too large.

The true standard deviation of the number of men in the sample is $\ \displaystyle\sqrt{\sum_{m=0}^5p_m(i-3)^2}\approx0.973\ $, while that of the approximating binomial distribution is $\ \sqrt{5\times0.6\times0.4}\approx1.095\ $, again about $10\%$ too large.  If $\ X\ $ is a normally-distributed random variable with mean $3$ and standard deviation $0.973$, then $\ P(X>0.7\times5)\approx0.3037\ $, which is a pretty good approximation to the exact probability.  However, $\ P(X>3)=0.5\ $ and $\ P(X\ge4)\approx0.152\ $, both very poor approximations.  Given that these three probabilities should be equal for an integer valued random variable, it becomes apparent that the closeness of $\ P(X>0.7\times5)\ $ to the exact probability is largely a matter of luck that the fraction chosen in the problem happened to have been  $0.7$ rather than some other fraction in the range $0.6$ to $0.8$.


问:
Integral question help me?
I am in the middle of solving a diff equation and I have  to solve $\int\dfrac{1}{\cos x}e^{\tan x}\,dx$? I am thinking about integration by parts but it is a very long way.is there another short way? Thank you :)

**Edit: The diff equation which is linear of first order that  I have to solve is (ysinx-1)dx +cosxdy=0

Edit again,I first find y=c*e^[-int(tgx)] then I find y' and then I replace both of them in y' +y*tgx=1/cosx..And here I have a c'=____ and then I find the integral of c' which is c and finally replace it in y=c*e^[-int(tgx)].The problem is that here I have this difficult integration..

答:
If the integral is $\displaystyle\int \dfrac{e^{\tan(x)}}{\cos^2(x)}dx$, then there is a nice form.

Set $\tan(x) = t$. This gives us $\sec^2(x) dx = dt \implies dx = \cos^2(x) dt$. Hence, we have we then have
$$\int \sec^2(x) e^{\tan(x)}dx = \int e^t dt = e^t + \text{ constant} = e^{\tan(x)} + \text{ constant}$$

However, if the integral is $\displaystyle\int \dfrac{e^{\tan(x)}}{\cos(x)}dx$, then there is no nice form as such, i.e.,
$$\int \sec(x) e^{\tan(x)}dx = \int \dfrac{e^t}{\sqrt{1+t^2}} dt$$
which cannot be expressed via a nice elementary function.

Title: Integral question help me?


问:
stopping times (definition)
I have the following question. If we want to verify that a random variable $T$ is a stopping time, according to the definition of a stopping time we have to show that $[T\le n]\in F_{n} $ for each $n\in\mathbb{N}$. I noticed that is enough to show only $[T= n]\in F_{n}$ for each $n\in\mathbb{N}$. Why is it enought?

答: It's enough because, $[T\leq n] = \cup_{k=1}^{n}[T=k]$ where $[T=k]\in\mathcal{F}_n\subset \mathcal{F}_n$ for all $1\leq k\leq n.$

Title: stopping times (definition)


答: It is assumed that $\mathcal F_n$'s are increasing. $[T=n]=[T=1] \cup [T=2]\cup...\cup[T=n]$ and $[T=i] \in F_i \subseteq F_n$ for each $i \in \{1,2...,n\}$. Hence $[T=n] \in F_n$

Title: stopping times (definition)

Body: I have the following question. If we want to verify that a random variable $T$ is a stopping time, according to the definition of a stopping time we have to show that $[T\le n]\in F_{n} $ for each $n\in\mathbb{N}$. I noticed that is enough to show only $[T= n]\in F_{n}$ for each $n\in\mathbb{N}$. Why is it enought?


问:
How to find the angle formed by an isosceles triangle next to another one?
The problem is as follows:

Using the figure from below: Find the unknown angle indicated as $x$.
Assume $AD=BC$ and $BD=DC$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&12^{\circ}\\
2.&10^{\circ}\\
3.&15^{\circ}\\
4.&16^{\circ}\\
4.&14^{\circ}\\
\end{array}$

What I attempted to do here was to add the angles in the isosceles which adds up to $4x$, this can be added to the $3x$ triangle but that's where I'm stuck.

In other words the only thing which I could spot was:

$\angle BDA = 2x+2x$

How exactly can it be used congruency to solve this problem?. Can someone help me here?. The intended approach is relying in euclidean geometry postulates, but I don't know exactly which sort of congruency of triangles identity should be used.

Please include a drawing in your answer because this part is difficult for me to spot with accuracy. Can you please use an explanation step-by-step.

答:
In the figure we have $$\dfrac{a}{\sin(7x)}=\dfrac{b}{\sin(3x)}\\\dfrac{a}{\sin(4x)}=\dfrac{b}{\sin(2x)}$$ It follows $$\sin(7x)=2\sin(3x)\cos(2x)$$
You have two options: solve this last equation or check which of the five answers given is correct. By both means (the second is preferable) you find $x = 10^{\circ}$(in boths sides you have $0.93969262$).

►If you have difficulty to solve the equation note that $$2\sin(3x)\cos(2x)=\sin(5x)+\sin(x)$$ and$$\sin(7x)-\sin(5x)=2\cos(6x)\sin(x)$$
Thus $$2\cos(6x)\sin(x)=\sin(x)\Rightarrow \cos(6x)=\dfrac12\Rightarrow  x=10^{\circ}$$


答:
jjagmath is right. Tip: When even-power terms appear prominently like that, always try plugging the negatives. But if you aren't able to find a counterexample just by "looking at it", you can try this to be more systematic (espacially if it's more complicated).

Assume $f(x,y) = (u,v)$, and solve for $(x,y)$. Then
$$
y = \frac{v}{2x}
$$
so
$$
x^2 - \frac{v^2}{4x^2} = u
\iff x = \pm\sqrt{\frac{u+\sqrt{u^2+v^2}}{2}}
$$
so
$$
y = \pm\frac{v}{2\sqrt{\frac{u+\sqrt{u^2+v^2}}{2}}}
$$
So we clearly have two solutions for each $(u,v)$. (Actually the above assumes $v\ne 0$ for convenience).

Title: How to prove that this function is not 1-1

Body:
The following question was part of my analysis assignment and I couldn't solve this part out of 5parts.

Let $f: \mathbb{R}^2 \to \mathbb{R}^2 $ be a function $f(x,y)=( x^2 -y^2 , 2xy) \in \mathbb{R}^2$. Is  f 1-1 on $\mathbb{R}^2$ ?

I let $(x^2-y^2, 2xy)=(p^2-q^2, 2pq)$  but neither I am able to prove p=x and q=y nor able to find a contradictory exam ple.

Your guidance is required!


问:
Can an arbitary Jordan curve be approximated by a smooth Jordan curve?
Given a Jordan curve $\gamma_1 \colon [a,b] \to \mathbb{C}$ and an $\epsilon > 0$, can we find a continuously differentiable Jordan curve, $\gamma_2 \colon [a,b] \to \mathbb{C}$, such that $ |\gamma_1(t) - \gamma_2(t) | < \epsilon$ for all $ t \in [a,b]$?

答:
Up to reparametrization, yes:

By the Jordan curve theorem and Caratheodory's conformal mapping theorem, we can find a homeomorphism $\varphi$ from $\gamma_1\cup \text{int}(\gamma_1)$ to the closed unit disc such that its restriction to $\text{int}(\gamma_1)$ is a conformal map. Now, let $0<r<1,\hat\gamma_r(t)=re^{i2\pi t}$, and note that $\varphi^{-1}(\hat\gamma_1(t))$ is a reparametrization of $\gamma_1$ . Since $\varphi^{-1}$ is a continuous function on a compact set, it is uniformly continuous. Thus we can find $\delta(\varepsilon)$ such that $|z_1-z_2|<\delta\rightarrow |\varphi(z_1)-\varphi(z_2)|<\varepsilon$, and so if we take $r$ such that $1-r<\delta$, we get that
$$\forall_{t\in[0,1]}|\varphi^{-1}(\hat\gamma_1(t))-\varphi^{-1}(\hat\gamma_r(t))|<\varepsilon$$ Since $\varphi^{-1}$ is a conformal map on the open unit disc, $\tilde\gamma_2(t):=\varphi^{-1}(\hat\gamma_r(t))$ is a smooth jordan curve, approximating  the reparametrization of $\gamma_1$ as desired.

To get the result withouth reparametrization, note that the reparametrization $f$ is monotone continuous, and as such can be uniformly approximated by monotone bernstein polynomials $f_n$ (see this thread on MO) with fixed endpoints.
Since $\tilde\gamma_2(t)$ is uniformly continous, $\tilde\gamma_2(f_n(t))\to \tilde\gamma_2(f(t))$ uniformly.
It is easy to see that $\gamma_2(t):=\tilde\gamma_2(f_n(t)))$ is a smooth jordan curve satisfying our requirement.


问:
Evaluate $\int_{0}^{\pi/2}\frac{x\cdot \sin(2x)}{(1+\cos(4x^2))}\,\mathrm{d}x$
Do you have a tip to accomplish this integral?

$$\int_{0}^{\pi/2}\frac{x\cdot \sin(2x)}{(1+\cos(4x^2))}\,\mathrm{d}x$$
I already tried by substitution and by parts, neither gave me a light how to solve it.

This is a multiple choice question from an Olympiad. The options are A) $\pi/2$, B) $\pi^2/4$, C) $\pi/8$ and D) $\pi^2/16$.

答:
Gabriela. Welcome to our community!

At first, you understood the problem incorrect. They ask you to evaluate such integral:

$$\int_0^{\pi /2} \frac{x \sin (2x)}{1+\cos^2 (2x)}dx=\frac{1}{4}\int_0^{\pi} \frac{t \sin (t)}{1+\cos^2 (t)}dt.$$
It's easier to get rid of those $2x$ before going further.

Now consider such integral first:
$$\int_0^{\pi} \frac{ \sin (x)}{1+\cos^2 (x)}dx= [\text{  substitute } t=\cos (x)]=\int_{-1}^{1} \frac{1}{1+t^2}dt=\frac{\pi}{2}.$$
It's a common integral, hope you will get the main idea.

Now get back to your problem. The trick here is to substitute $z=\pi-t$:
$$\frac{1}{4}\int_0^{\pi} \frac{t \sin (t)}{1+\cos^2 (t)}dt = \frac{1}{4}\int_0^{\pi} \frac{(\pi-z) \sin (z)}{1+\cos^2 (z)}dz=\frac{\pi}{4}\int_0^{\pi} \frac{\sin (z)}{1+\cos^2 (z)}dz- \frac{1}{4}\int_0^{\pi} \frac{z \sin (z)}{1+\cos^2 (z)}dz.$$

Now you'll probably notice similarities. Donate your integral as $I$, then:

$$I=\frac{\pi}{4}\int_0^{\pi} \frac{\sin (z)}{1+\cos^2 (z)}dz-I\implies I=\frac{\pi}{8}\int_0^{\pi} \frac{\sin (z)}{1+\cos^2 (z)}dz=\frac{\pi}{8} \cdot \frac{\pi}{2}=\frac{\pi^2}{16}.$$

So your answer ir $D$.


问:
Need help in formal grammar for the $L = \{wcw : w \in \{a,b\}^\ast\}$
I can't create formal grammar for the language like

$wcw$ where $w$ is the word in $\{a,b\}^\ast$ and $c$ is just a letter

Thanks!

答:
A solution:
$$\begin{align*}
\Sigma & \to \mathtt{c} & \tag{1} \\
\Sigma & \to i \mathtt{c} i & \mathrm{for} \: i \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{2} \\
\Sigma & \to i C_{j} A_{i} S_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{3} \\
C_{i} A_{j} & \to D_{i} A_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{4} \\
D_{i} A_{j} & \to D_{i} B_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{5} \\
B_{i} A_{j} & \to B_{i} B_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{6} \\
B_{i} S_{j} & \to B_{i} X_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{7} \\
B_{i} S_{j} & \to B_{i} Y_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{8} \\
B_{i} X_{j} & \to T_{i} X_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{9} \\
B_{i} Y_{j} & \to F_{i} Y_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{10} \\
B_{i} T_{j} & \to T_{i} T_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{11} \\
B_{i} F_{j} & \to F_{i} F_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{12} \\
D_{i} T_{j} & \to \mathtt{a} C_{i} T_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{13} \\
D_{i} F_{j} & \to \mathtt{b} C_{i} F_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{14} \\
C_{i} T_{j} & \to C_{i} Z_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{15} \\
C_{i} F_{j} & \to C_{i} Z_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{16} \\
Z_{i} T_{j} & \to Z_{i} Z_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{17} \\
Z_{i} F_{j} & \to Z_{i} Z_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{18} \\
Z_{i} X_{j} & \to Z_{i} A_{\mathtt{a}} S_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{19} \\
Z_{i} Y_{j} & \to Z_{i} A_{\mathtt{b}} S_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{20} \\
Z_{i} A_{j} & \to A_{i} A_{j} & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{21} \\
S_{i} & \to i & \mathrm{for} \: i \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{22} \\
A_{i} j & \to i j & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{23} \\
C_{i} j & \to i \mathtt{c} j & \mathrm{for} \: i , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} \tag{24}
\end{align*}$$

The idea of this context-sensitive grammar is similar to that of 4-stroke engine:

the sentential form
$$i_{1} i_{2} \cdots i_{n} C_{j} A_{i_{1}} A_{i_{2}} \cdots A_{i_{n}} S_{j}$$
becomes either
$$ i_{1} i_{2} \cdots i_{n} D_{j} B_{i_{1}} B_{i_{2}} \cdots B_{i_{n}} S_{j} \tag{go to 2} $$
by $\left( 4 \right)$, $\left( 5 \right)$ and $\left( 6 \right)$ or
$$ i_{1} i_{2} \cdots i_{n} j \mathtt{c} i_{1} i_{2} \cdots i_{n} j \tag{end} $$
by $\left( 22 \right)$, $\left( 23 \right)$ and $\left( 24 \right)$;

the sentential form
$$i_{1} i_{2} \cdots i_{n} D_{j} B_{i_{1}} B_{i_{2}} \cdots B_{i_{n}} S_{j}$$
becomes either
$$ i_{1} i_{2} \cdots i_{n} D_{j} T_{i_{1}} T_{i_{2}} \cdots T_{i_{n}} X_{j} \tag{go to 3} $$
by $\left( 7 \right)$, $\left( 9 \right)$ and $\left( 11 \right)$ or
$$ i_{1} i_{2} \cdots i_{n} D_{j} F_{i_{1}} F_{i_{2}} \cdots F_{i_{n}} Y_{j} \tag{go to 4} $$
by $\left( 8 \right)$,  $\left( 10 \right)$ and $\left( 12 \right)$;

the sentential form
$$i_{1} i_{2} \cdots i_{n} D_{j} T_{i_{1}} T_{i_{2}} \cdots T_{i_{n}} X_{j}$$
becomes
$$ i_{1} i_{2} \cdots i_{n} \mathtt{a} C_{j} Z_{i_{1}} Z_{i_{2}} \cdots Z_{i_{n}} X_{j} \tag{go to 5} $$
by $\left( 13 \right)$, $\left( 15 \right)$ and $\left( 17 \right)$;

the sentential form
$$i_{1} i_{2} \cdots i_{n} D_{j} F_{i_{1}} F_{i_{2}} \cdots F_{i_{n}} Y_{j}$$
becomes
$$ i_{1} i_{2} \cdots i_{n} \mathtt{b} C_{j} Z_{i_{1}} Z_{i_{2}} \cdots Z_{i_{n}} Y_{j} \tag{go to 6} $$
by $\left( 14 \right)$, $\left( 16 \right)$ and $\left( 18 \right)$;

the sentential form
$$i_{1} i_{2} \cdots i_{n} \mathtt{a} C_{j} Z_{i_{1}} Z_{i_{2}} \cdots Z_{i_{n}} X_{j}$$
becomes
$$ i_{1} i_{2} \cdots i_{n} \mathtt{a} C_{j} A_{i_{1}} A_{i_{2}} \cdots A_{i_{n}} A_{\mathtt{a}} S_{j} \tag{go to 1} $$
by $\left( 19 \right)$ and $\left( 21 \right)$;

the sentential form
$$i_{1} i_{2} \cdots i_{n} \mathtt{b} C_{j} Z_{i_{1}} Z_{i_{2}} \cdots Z_{i_{n}} Y_{j}$$
becomes
$$ i_{1} i_{2} \cdots i_{n} \mathtt{b} C_{j} A_{i_{1}} A_{i_{2}} \cdots A_{i_{n}} A_{\mathtt{b}} S_{j} \tag{go to 1} $$
by $\left( 20 \right)$ and $\left( 21 \right)$;

where $ n \geq 1 $, $ i_{1} , i_{2} , \cdots i_{n} , j \in \left\{ \mathtt{a} , \mathtt{b} \right\} $.

An example:
$$\begin{align*}
\Sigma & \implies \mathtt{a} C_{\mathtt{b}} A_{\mathtt{a}} S_{\mathtt{b}} & \mathrm{by} \: \left( 3 \right) \\
& \implies \mathtt{a} D_{\mathtt{b}} A_{\mathtt{a}} S_{\mathtt{b}} & \mathrm{by} \: \left( 4 \right) \\
& \implies \mathtt{a} D_{\mathtt{b}} B_{\mathtt{a}} S_{\mathtt{b}} & \mathrm{by} \: \left( 5 \right) \\
& \implies \mathtt{a} D_{\mathtt{b}} B_{\mathtt{a}} X_{\mathtt{b}} & \mathrm{by} \: \left( 7 \right) \\
& \implies \mathtt{a} D_{\mathtt{b}} T_{\mathtt{a}} X_{\mathtt{b}} & \mathrm{by} \: \left( 9 \right) \\
& \implies \mathtt{a} \mathtt{a} C_{\mathtt{b}} T_{\mathtt{a}} X_{\mathtt{b}} & \mathrm{by} \: \left( 13 \right) \\
& \implies \mathtt{a} \mathtt{a} C_{\mathtt{b}} Z_{\mathtt{a}} X_{\mathtt{b}} & \mathrm{by} \: \left( 15 \right) \\
& \implies \mathtt{a} \mathtt{a} C_{\mathtt{b}} Z_{\mathtt{a}} A_{\mathtt{a}} S_{\mathtt{b}} & \mathrm{by} \: \left( 19 \right) \\
& \implies \mathtt{a} \mathtt{a} C_{\mathtt{b}} A_{\mathtt{a}} A_{\mathtt{a}} S_{\mathtt{b}} & \mathrm{by} \: \left( 21 \right) \\
& \implies \mathtt{a} \mathtt{a} D_{\mathtt{b}} A_{\mathtt{a}} A_{\mathtt{a}} S_{\mathtt{b}} & \mathrm{by} \: \left( 4 \right) \\
& \implies \mathtt{a} \mathtt{a} D_{\mathtt{b}} B_{\mathtt{a}} A_{\mathtt{a}} S_{\mathtt{b}} & \mathrm{by} \: \left( 5 \right) \\
& \implies \mathtt{a} \mathtt{a} D_{\mathtt{b}} B_{\mathtt{a}} B_{\mathtt{a}} S_{\mathtt{b}} & \mathrm{by} \: \left( 6 \right) \\
& \implies \mathtt{a} \mathtt{a} D_{\mathtt{b}} B_{\mathtt{a}} B_{\mathtt{a}} Y_{\mathtt{b}} & \mathrm{by} \: \left( 8 \right) \\
& \implies \mathtt{a} \mathtt{a} D_{\mathtt{b}} B_{\mathtt{a}} F_{\mathtt{a}} Y_{\mathtt{b}} & \mathrm{by} \: \left( 10 \right) \\
& \implies \mathtt{a} \mathtt{a} D_{\mathtt{b}} F_{\mathtt{a}} F_{\mathtt{a}} Y_{\mathtt{b}} & \mathrm{by} \: \left( 12 \right) \\
& \implies \mathtt{a} \mathtt{a} \mathtt{b} C_{\mathtt{b}} F_{\mathtt{a}} F_{\mathtt{a}} Y_{\mathtt{b}} & \mathrm{by} \: \left( 14 \right) \\
& \implies \mathtt{a} \mathtt{a} \mathtt{b} C_{\mathtt{b}} Z_{\mathtt{a}} F_{\mathtt{a}} Y_{\mathtt{b}} & \mathrm{by} \: \left( 16 \right) \\
& \implies \mathtt{a} \mathtt{a} \mathtt{b} C_{\mathtt{b}} Z_{\mathtt{a}} Z_{\mathtt{a}} Y_{\mathtt{b}} & \mathrm{by} \: \left( 18 \right) \\
& \implies \mathtt{a} \mathtt{a} \mathtt{b} C_{\mathtt{b}} Z_{\mathtt{a}} Z_{\mathtt{a}} A_{\mathtt{b}} S_{\mathtt{b}} & \mathrm{by} \: \left( 20 \right) \\
& \implies \mathtt{a} \mathtt{a} \mathtt{b} C_{\mathtt{b}} Z_{\mathtt{a}} A_{\mathtt{a}} A_{\mathtt{b}} S_{\mathtt{b}} & \mathrm{by} \: \left( 21 \right) \\
& \implies \mathtt{a} \mathtt{a} \mathtt{b} C_{\mathtt{b}} A_{\mathtt{a}} A_{\mathtt{a}} A_{\mathtt{b}} S_{\mathtt{b}} & \mathrm{by} \: \left( 21 \right) \\
& \implies \mathtt{a} \mathtt{a} \mathtt{b} C_{\mathtt{b}} A_{\mathtt{a}} A_{\mathtt{a}} A_{\mathtt{b}} \mathtt{b} & \mathrm{by} \: \left( 22 \right) \\
& \implies \mathtt{a} \mathtt{a} \mathtt{b} C_{\mathtt{b}} A_{\mathtt{a}} A_{\mathtt{a}} \mathtt{b} \mathtt{b} & \mathrm{by} \: \left( 23 \right) \\
& \implies \mathtt{a} \mathtt{a} \mathtt{b} C_{\mathtt{b}} A_{\mathtt{a}} \mathtt{a} \mathtt{b} \mathtt{b} & \mathrm{by} \: \left( 23 \right) \\
& \implies \mathtt{a} \mathtt{a} \mathtt{b} C_{\mathtt{b}} \mathtt{a} \mathtt{a} \mathtt{b} \mathtt{b} & \mathrm{by} \: \left( 23 \right) \\
& \implies \mathtt{a} \mathtt{a} \mathtt{b} \mathtt{b} \mathtt{c} \mathtt{a} \mathtt{a} \mathtt{b} \mathtt{b} & \mathrm{by} \: \left( 24 \right)
\end{align*}$$


问:
Compute $a+b+c+d$ where $(ab+c+d =13),\ (bc+d+a = 27),\ (cd+a+b = 30),\ (da+b+c = 17)$
Compute $(a+b+c+d)\ $ if $\ a,b,c,d\ $ satisfy the system of equations below:

$$\begin{cases}
ab+c+d = 13\\\\
bc+d+a = 27\\\\
cd+a+b = 30\\\\
da+b+c = 17
\end{cases}$$

.................................................................................My method:

I added all equations to get:

$(ab+bc+cd+da)+2(a+b+c+d)=87$

$(a+c)(b+d)+2(a+c)+2(b+d)=87$

$(a+c)(b+d+2)+2(b+d)=87$

$(a+c+2)(b+d+2)=91$

$(a+c+2)(b+d+2)=13\times 7$

So, $a+c=11$ and $b+d=5$

or, $a+c =5$ and $b+d =11$

Hence, $a+b+c+d = 16$

But I am not sure if this is the correct method since it was not stated that $a,b,c,d$ were positive integers.

I would be thankful if someone provided a solution to this question.

Body:
Let $u = a+b+c+d$, the set of equations can be rewritten as

$$
\begin{cases}
(a-1)(b-1) + a + b + c + d &= 14\\
(b-1)(c-1) + a + b + c + d &= 28\\
(c-1)(d-1) + a + b + c + d &= 31\\
(d-1)(a-1) + a + b + c + d &= 18
\end{cases}
\iff
\begin{cases}
(a-1)(b-1) &= 14- u\\
(b-1)(c-1) &= 28 -u\\
(c-1)(d-1) &= 31 -u\\
(d-1)(a-1) &= 18 -u
\end{cases}$$
This leads to

$$(14-u)(31-u) = (a-1)(b-1)(c-1)(d-1) = (28-u)(18-u)$$
which can be simplified to $u = 70$. So $a + b + c + d = 70$.


问:
Uniform convergence to its integral proof

Given a function $f:[0,2]\to \mathbb{R}$ that is continuous and bounded on $[0,2]$, show that the function $f_n(x):[0,1]\to \mathbb{R}$ where $$ f_n(x) = \frac 1n \sum_{k=0}^{n-1} f\left(x + \frac kn\right)$$ is uniformly convergent.

So far I believe that $\left(f_n(x)\right) \to\int_x^{x+1} f(x)\;\text{dx}$ and I've broken it down as
$$
\begin{align}
|f_n(x) - f(x)| & = \left|\frac 1n \sum_{k=0}^{n-1} f\left(x + \frac kn\right) - \int_x^{x+1}f(x) \; \text{dx}\right| \\
& = \left|\frac 1n \left(f(x) + f(x + 1/n) + ... + f\left(x+\frac{n-1}{n}\right)\right) - \int_{x}^{x+1/n} f(x) \; \text{dx} - \int_{x+1/n}^{x+2/n} f(x) \; \text{dx} - ... - \int_{x+(n-1)/n}^{x+1} f(x) \; \text{dx} \right|\\
& \leq \left|\frac{1}{n} f(x) -  \int_{x}^{x+1/n} f(x) \; \text{dx} \right|+\left|\frac{1}{n} f(x+1/n) -  \int_{x+1/n}^{x+2/n} f(x) \; \text{dx} \right|+...+\left|\frac{1}{n} f(x+\frac{n-1}{n}) -  \int_{x+(n-1)/n}^{x+1} f(x) \; \text{dx} \right|
\end{align}
$$
Now I believe its a common result that $\left(\int_{a}^{a+1/n}f(x) dx\right) \to f(a)$ however that means I'm left with
$$
|f_n(x) - f(x)| \leq \left|\frac{1}{n} f(x) -  f(x)  \right|+\left|\frac{1}{n} f(x+1/n) -   f(x) \right|+...+\left|\frac{1}{n} f(x+\frac{n-2}{n}) -   f(x) \right|+\left|\frac{1}{n} f(x+\frac{n-1}{n}) -  f(x) \right|.
$$
I'm not sure where to go from here. I'm also unsure whether or not if some of the integrals go to $f(x)$ or $f(x+1)$. I was also thinking proving absolute convergence since it seems like the sum can be broken into finite monotonic parts?

答: It is not true that $\int_a^{a+\frac  1n} f(t)dt \to f(a)$. Write $\frac  1 nf(x+\frac k n) -\int\limits_{x+\frac {k-1} n}^{x+\frac k n} f(t)dt$ as $\int\limits_{\frac {k-1} n}^{\frac k n} [f(x+\frac k n)-f(x+s)]ds$ (where I have used the substitution $s=t-x$) and use uniform continuity of $f$.


问:
Jacobian of non-square matrices
Let $m>n$ and $L\colon \mathbb{R}^m \to \mathbb{R}^n$ be a linear map(=matrix). The "$n$-dimensional Jacobian"
$$
J^n(L) = \sqrt{\det(LL^t)}
$$
is of geometric significance! Let's say it makes the co-area formula possible.

Suppose $C\colon \mathbb{R}^m \to \mathbb{R}^m$ is an invertible linear map (think of it as a change of coordinates). I am interested in the precise formula relating $J^n(L)$ and $J^n(L \circ C)$.

For simplicity assume $L$ is surjective, i.e. it has rank $n$. Denote by $C^{ker}$ the restriction of $C$ to the $(m-n)$-dimensional $(\ker L\circ C)$. So,  $C^{ker}\colon (\ker L\circ C) \to \ker L$ can be seen as a map $\mathbb{R}^{m-n} \to \mathbb{R}^{m-n}$. I think the answer should be:
$$
J^n(L) = \frac{|\det C^{ker}|}{|\det C|}\, \cdot J^n(L \circ C)\, ,
$$
Because for certain $C$ I do have a proof, albeit quite a tricky one.

Questions:

Is the claimed identity true?

If not, what is the right formula?

How does one prove it?

答:
Since $C^{\ker}$ is not an endomorphism (because its domain $\ker(L\circ C)$ and its codomain $\ker(L)$ are different), you need to be precise about the meaning of $|\det C^{\ker}|$. One way to interpret it is as follows. Let $\theta:\mathbb R^{m-n}\to\mathbb R^m$ be any linear isometry (with respect to the usual inner products on $\mathbb R^{m-n}$ and $\mathbb R^m$) whose image is $\ker(L\circ C)$. Let $\pi:\mathbb R^m\to\mathbb R^{m-n}$ be an orthogonal projection in the sense that its restriction on $\ker(L)$ is an isometry and its restriction on $(\ker L)^\perp$ is zero. Then we interpret $|\det C^{\ker}|$ as the absolute value of the determinant of the automorphism $\pi\circ C\circ\theta$. Since all other linear maps $\theta'$ and $\pi'$ with the same properties as $\theta$ and $\pi$ can be written as $\theta'=\theta\circ q_1$ and $\pi'=q_2\circ\pi$ for some isometries $q_1,q_2$ on $\mathbb R^{m-n}$, $|\det(\pi\circ C\circ\theta)|$ is well-defined. (Actually, $J^n(L)$ can also be interpreted in a similar way. Let $\widetilde\theta:\mathbb R^n\to\mathbb R^m$ be an isometry whose range is $(\ker L)^\perp$. Then $J^n(L)$ coincides with $|\det(L\circ\widetilde\theta)|$.)

With this interpretation, the identity you mentioned is true. By picking two appropriate orthonormal bases of $\mathbb R^n$ and $\mathbb R^m$, we may assume that the matrix representation of $L$ is
$$
L=\begin{bmatrix}L'&0_{n\times(m-n)}\end{bmatrix}
$$
for some nonsingular matrix $L'$. Let $Q\in O(m,\mathbb R)$ be any orthogonal matrix whose last $m-n$ rows are orthogonal to the first $n$ rows of the matrix representation of $C$ with respect to the previously chosen basis of $\mathbb R^m$. Then $C$ is in the form of
$$
C=\begin{bmatrix}X&0_{n\times(m-n)}\\ Y&Z\end{bmatrix}Q
$$
where $Z\in M_{m-n}(\mathbb R)$. It follows that
\begin{aligned}
\ker(L\circ C)
&=\ker\left(\begin{bmatrix}L'X&0_{n\times(m-n)}\end{bmatrix}Q\right)
=\left\{Q^T\begin{bmatrix}0_{n\times1}\\ u\end{bmatrix}\colon u\in\mathbb R^{m-n}\right\},\\
\ker(L)
&=\ker\left(\begin{bmatrix}L'&0_{n\times(m-n)}\end{bmatrix}\right)
=\left\{\begin{bmatrix}0_{n\times1}\\ v\end{bmatrix}\colon v\in\mathbb R^{m-n}\right\}.\\
\end{aligned}
For any $x=Q^T\begin{bmatrix}0_{n\times1}\\ u\end{bmatrix}\in\ker(L\circ C)$, we have $Cx=\begin{bmatrix}0_{n\times1}\\ Zu\end{bmatrix}$. Therefore, if we define
\begin{align*}
&\theta:\mathbb R^{m-n}\to\mathbb R^m,\quad u\mapsto Q^T\begin{bmatrix}0_{n\times1}\\ u\end{bmatrix},\\
&\pi:\mathbb R^m\to\mathbb R^{m-n},\quad\begin{bmatrix}w\\ v\end{bmatrix}\mapsto v
\end{align*}
for all $u,v\in\mathbb R^{m-n}$ and $w\in\mathbb R^n$, then $\theta$ is an isometry whose image is $\ker(L\circ C)$, the restriction of $\pi$ on $\ker(L)$ is an isometry, $\pi=0$ on $(\ker L)^\perp$, and $\left(\pi\circ C\circ\theta\right)u=Zu$. Therefore $|\det C^{\ker}|=|\det Z|$ and the identity
$$
J^n(L) = \frac{|\det C^{\ker}|}{|\det C|}\, J^n(L \circ C)
$$
holds because it simply says that
$$
|\det(L')| = \frac{|\det(Z)|}{|\det(X)\det(Z)|}\, |\det(L')\det(X)|.
$$

Title: Jacobian of non-square matrices


问:
How to prove uniqueness of a solution to a recurrence relation?
I would like to know if there is a general method to prove the uniqueness of a solution to a recurrence relation, if yes can you provide an example?

I have a simple recurrence relation in mind, the following $$a_n=2a_{n-1} -1$$ with $a_1=3$ whose solution is $$a_n=2^n +1$$ now this is a linear recurrence relation and techniques from linear algebra could be used to prove the uniqueness, but I am not looking for a way to prove uniqueness by using the linearity of the relation but rather a more general method that would apply to nonlinear relations as well.

So how can I show that $a_n=2^n +1$ is the only solution?

答:
Let $g\colon X\to X$ be any function and $x_1\in X$.
Also assume that $f_1,f_2\colon \Bbb N\to X$ are functions with $f_1(1)=f_2(1)=x_1$ and with $f_1(n+1)=g(f_1(n))$ for all a $n$ and $f_2(n+1)=g(f_2(n))$ for all a $n$, respectively.
Then $f_1=f_2$.
Indeed, assume otherwise and let $m\in\Bbb N$ be minimal with $f_1(m)\ne f_2(m)$. Then certainly $m\ne 1$ as we know $f_1(1)=f_2(1)$. Hence $m=k+1$ for some $k\in\Bbb N$, and by minimality $f_1(k)=f_2(k)$. But then
$$f_1(m)=f_1(k+1)=g(f_1(k))=g(f_2(k))=f_2(k+1)=f_2(m), $$
contradiction.


问:
Coloring the 1 x n field
I have a field of 1 x n size. I need to color it using: red, orange, green, blue. Also, I can color red only even amount of blocks, and orange only odd amount of blocks.

Finally I need to find a generating function describing how many combinations of different colors I can use.

How can I do it?

答:
Brian M. Scott's answer gets ugly when asking for explicit coefficients...

You are classifying sequences into (subindex is length):

Even red, even orange ($a_n$, with $a_0 = 1$)

Even red, odd orange  ($b_n$, with $b_0 = 0$)

Odd red, even orange  ($c_n$, with $c_0 = 0$)

Odd red, odd orange   ($d_n$, with $d_0 = 0$)

You can set up recurrences, by noting that to get the $a_{n + 1}$ even/even secuences you can start with the $a_n$ even/even and add green or blue ($2 a_n$ ways) or start with even/odd and add orange ($b_n$ ways) or odd/even and add red ($c_n$ ways). Completing this:
$$
\begin{align*}
a_{n + 1} &= 2 a_n + b_n + c_n \\
b_{n + 1} &= a_n + 2 b_n + d_n \\
c_{n + 1} &= a_n + 2 c_n + d_n \\
d_{n + 1} &= b_n + c_n + 2 d_n
\end{align*}
$$
Define generating functions $A(z) = \sum_{n \ge 0} a_n z^n$ and so on; multiply the recurrences by $z^n$ and add over $n \ge 0$ to get:
$$
\begin{align*}
\frac{A(z) - a_0}{z} &= 2 A(z) + B(z) + C(z) \\
\frac{B(z) - b_0}{z} &= A(z) + 2 B(z) + D(z) \\
\frac{C(z) - c_0}{z} &= A(z) + 2 C(z) + D(z) \\
\frac{D(z) - d_0}{z} &= B(z) + C(z) + 2 D(z)
\end{align*}
$$
Solve this linear system of equations for $B(z)$:
$$
B(z) = \frac{z}{1 - 4 z}
     = \frac{1}{4} \cdot \frac{1}{1 - 4 z} - \frac{1}{4}
$$
You can read the coefficients from the geometric series (the $[n = 0]$ is Iverson's bracket, $1$ if $n = 0$ and 0 otherwise):
$$
b_n = 4^{n - 1} - \frac{1}{4} [n = 0]
$$
I.e, $b_n = \langle 0, 1, 4, 16, \ldots \rangle$

Title: Coloring the 1 x n field


问:
Is there a canonical definition of a line?
Normally the definition of a line is either via two points or its slope and y-intercept.  But both methods of defining a line have problems.  With the first one is the problem, that many points are passing through a single line so you can define a line using every pair of them, which mean that a definition method is not unique.  And the second one can't handle lines with slope infinity (like x=0).  I searched and thought about it a little bit, but could find anything.

答: You can unambiguously define a line as a vector space of dimension 1 (the vector you choose for its basis will help describe it uniquely). Of course, depending on the vector space you are working in, some ligns are simply NOT ligns if you go up a dimension ($x=5$ is a plane in $2D$).


问:
Roots of unity - solving $z^n = 1$ and $z^n = -1$
If we are solving $z^n = 1$, the unique solutions are given by $e^{i\theta}  $ with $\theta$ being each of the angles you get when you split up $2\pi$ into $n$ angles (sorry for the bad wording - I didn't know how else to phrase it).

In a solution to a problem, the author solves $z^4 + 1 = 0$, and it is deduced that $z = e^{i\pi/4}, e^{3i\pi/4}, e^{5i\pi/4}, e^{7i\pi/4}$.

So is solving $z^n + 1 = 0$ equivalent to solving $z^n = 1$? And if so, why?

答: Note that if $z^n+1=0$, then $z^{2n}=1$ but $z^n\neq1$. So, to take your example, you can solve $z^8=1$, then from those 8 solutions remove any that also appear as solutions to $z^4=1$. The four that remain are solutions to $z^4=-1$.


问:
Fourier analysis exercise
I need a hand with this question:

If $f\in{L_1(\mathbb{R})}$ and $g\in{L_2(\mathbb{R})}$, then prove that $\widehat{f*g}=\hat{f}\cdot \hat{g}$

As a tip, i have been told to prove that:

$L_2(\mathbb{R}^n)\rightarrow L_2(\mathbb{R}^n)$

$h\rightarrow f*h$

and

$L_2(\mathbb{R}^n)\rightarrow L_2(\mathbb{R}^n)$

$h\rightarrow \hat{f}\cdot h$

are continuous.

Thanks a lot for any help.

答:
As a consequence of Fubini's theorem, the result is true when $g\in L^1\cap L^2$. Indeed, when $n=1$, write

$$\widehat{f *g}(x)=\int_{\Bbb R}e^{-itx}\int_{\Bbb R}f(t-s)g(s)dsdt$$
and $e^{-itx}=e^{-i(t-s)x}e^{-isx}$. The case $n$ arbitrary can be done similarily.

The purpose of the hint is to jump from this case to the general case. Indeed, thanks to Plancherel's theorem and the density of $L^1\cap L^2$ in $L^2$ we will be able to conclude. Indeed, let $g\in L^2$ and $(g_k,k\geqslant 1)$ a sequence of function of $L^1\cap L^2$ such that $\lVert g-g_k\rVert_{L^2}\to 0$ and $g_k\to g$ almost everywhere. Then $\widehat{f*g_k}=\widehat f\cdot\widehat{g_k}$ for each $k$. Now take the limit $k\to \infty$.

Title: Fourier analysis exercise


问:
Riemann-integral and Stone-Weierstrauss Theorem
Question:
Let ƒ be a continuous real-valued function on [0, 1].

Show that

$\lim_{n->\infty }\frac{\int_{0}^{1}x^nf(x)dx}{\int_{0}^{1}x^ndx} = f(1)$

My approach: I thought the denominator and the numerator are both differentiable in the given interval, so that I could use the L'Hospital's Theorem to prove this statement, by then making the question very straightforward.

But how can I use the Stone-Weierstrauss Theorem to prove this statement? I don't see any possible relation of solving this problem with the Stone-Weierstrauss Theorem.

答:
A trivial verification shows that the result is true when $f(x)=cx^{k}$ for some non-negative integer $k$ and $c$ is a constant. Hence it holds when  $f$ is a polynomial. Now choose a polynomial $p$ such that $|f(x)-p(x)| <\epsilon$ for all $x$ (which is possible by Weierstrass Approximation Theorem). Note that $|\frac {\int x^{n}f(x)} {\int x^{n}dx}-\frac {\int x^{n}p(x)} {\int x^{n}dx}|<\epsilon$. Can you finish?

Stone-Weierstrass Theorem is much stronger than Weierstrass Approximation Theorem. For this problem you only need Weierstrass Approximation Theorem.


问:
Smooth function with specific features
Im looking for a function that's continuous everywhere and whose second derivative is continuous everywhere with the following properties:

Near the origin, the function is approximately constant: $$f\approx c$$ with $c \in \mathbb{R}$

Far from the origin (outside some scale), the function is approximately linear: $$f(x) \approx m*x+b$$ with $m \neq 0$

So the function would look like

.

I just defined this function piecewise in mathematica as an example, but the second derivative is not continuous, because the curvy bits are quadratic and the flat bits are linear, so the second derivative is a series of steps.

I will eventually define a vector field from this scalar field which describes a flow in a flat manifold. Moreover, there is an energy functional that is defined in terms of the second derivatives of this scalar field, hence the reason for continuity.

I tried using bump functions and integrating backwards to get the scalar field I want, but the integrals became messy quite fast, as you might expect. Similarly, I tried using hyperbolic tangents, but I wasn't able to smoothly attach the pieces together.

I am also trying to minimize the magnitude of the second derivative (because it relates to the energy functional). So I am hoping to find a function that approaches the origin in a linear way, then curves in such a way that the second derivative has constant sign, then approaches zero again, then has constant opposite sign. Here is an example of such a second derivative using bump functions.

I tried integrating this and using the integration constants to satisfy my required properties, but the integrals are quite messy!

I appreciate any help!

答:
I actually found a descent enough function. I started with the conditions on the second derivative I wanted to satisfy. So I construct the following function, which is just the sum of two gaussians, such that the entire function is odd about the origin
$$ \alpha(x) = - e^{-\frac{(x+a)^2}{\sigma}} + e^{-\frac{(x-a)^2}{\sigma}}$$
with $a,\sigma \in \mathbb{R}$. Now, this function is what I want the second derivative to look like. To get the original function, I integrated this function twice to get

$$ \phi(x) = \frac{1}{2}\Bigg( e^{-\frac{(x-a)^2}{\sigma}} - e^{-\frac{(x+a)^2}{\sigma}}\Bigg)\sigma + C_1 + C_2 x + \frac{\sqrt{\pi \sigma}}{2}\Bigg( (a-x)ERF\bigg(\frac{a-x}{\sqrt{\sigma}}\bigg) - (a+x)ERF\bigg(\frac{a+x}{\sqrt{\sigma}}\bigg)\Bigg)$$

where $C_1, C_2$ are integration constants and $ERF$ is the error function from integrating a gaussian. Now, to have a linear function at large values of x and a flat function near the origin, we just massage the parameters until the function has vanishing derivative at the origin (flat) and constant non-zero at large x (linear). For example, I find $C_1=0$, $C_2 = 1.5$, and $a \approx 1.0087$ gives me:

Now, differentiating this twice yields the original function:

Hope this helps someone else in the future if you ever look for a function like this!

Title: Smooth function with specific features


问:
How to prove $(a \setminus b) ∪ c= (a ∪ c) \setminus (b ∪ c)$?
How to prove $(a \setminus b) ∪ c= (a ∪ c)  \setminus (b ∪ c)$? I tried to do some properties, but I don't know what to do next.
1

答:
We show the equality of two sets $S_1$ and $S_2$ by showing that $S_1 \subseteq S_2$ and that $S_2 \subseteq S_1$.

So take an arbitrary element from the left-hand side and show that it is also an element of the set on the right-hand side. And after that, take an arbitrary element from the right-hand side and show that it is also an element of the set on the left-hand side.


问:
Does $\operatorname{Re}(a+bi)^{n}=\operatorname{Re}(a+bi)^{n+1}$ have at least one solution $a,b\in\mathbb{Q}$, for all $n\in\mathbb{N}$?

Does
$$\operatorname{Re}((a+bi)^{n})=\operatorname{Re}((a+bi)^{n+1})$$
have a solution $a,b\in\mathbb{Q},|a|\neq |b|\neq0$, for all $n\in\mathbb{N}$?

I was working on my previous question, $\operatorname{Re}(a+bi)^m=\operatorname{Re}(a+bi)^n$, then all of the solutions I could find on PC was the case of $n-m=1$ and trivial cases. Is there a reason why many solutions can be found for $n-m=1$, and are there solutions for all $n\in\mathbb{N}$?

答:
Let $w = x + y \, \mathrm i$ for $x,y \in \mathbb Q$ such that $w^n+\overline w^n \neq 0$ and $w^{n+1}+\overline w^{n+1} \neq 0$. Take $$z = w \cdot \frac{w^n+ \overline{w}^n}{w^{n+1}+\overline w^{n+1}}.$$ Then $z$ has rational real and imaginary parts and $$\operatorname{Re}(z^{n+1}) = \operatorname{Re}(z^n).$$
Note that $z$ only depends on the direction of $w$: multiplying $w$ by a non-zero rational factor results in the same $z$.


问:
Stirling Numbers Combinatorics
The signless Stirling number of the first kind $\text{c(n,k)}$ is the sum of all products of $\text{(n-k)}$ integers from the range $\{1,2,...,\text{n-1}\}$, of which there are
$ \binom{n-k} {k-1} $ such products.
Verify that $\text{c}(6,2) = (1\times2\times3\times4)+(1\times2\times3\times5)+(1\times2\times4\times5)+(1\times3\times4\times5)+(2\times3\times4\times5) = 274$

Have this as a possible exam question but cannot figure it out. Any help please

答:
If I understand you correctly, you simply need to verify that ${6\brack 2}=274$, where $6\brack 2$ is the (unsigned) Stirling number of the first kind. How you’re to do this depends on how you’ve defined the Stirling numbers of the first kind. The usual definition is that $n\brack k$ is the number of permutations of $[n]=\{1,\dots,n\}$ with exactly $k$ disjoint cycles, so that’s the one that I’ll use.

$6\brack 2$ is then the number of permutations of $[6]=\{1,\dots,6\}$ having exactly $2$ disjoint cycles. Suppose that the cycle containing $1$ is a $k$-cycle, where of course $1\le k\le 5$. There are $\binom5{k-1}$ ways to choose the other $k-1$ members of the cycle, and there are then $(k-1)!$ ways to permute them, so there are

$$\binom5{k-1}(k-1)!=\frac{5!}{(6-k)!}$$

ways to choose the cycle containing $1$. The $6-k$ members of the other cycle are then completely determined, and there are $(5-k)!$ cyclically distinct orders for them. Thus, the number of permutations of $[6]$ with exactly $2$ cycles in which the cycle containing $1$ is a $k$-cycle is

$$\frac{5!}{(6-k)!}\cdot(5-k)!=\frac{5!}{6-k}\;,$$

and

$${6\brack 2}=\sum_{k=1}^5\frac{5!}{6-k}=\sum_{k=1}^5\frac{5!}k\;.$$

You can simply evaluate the sum, or you can recognize that it actually is the sum that you calculated in the question.

Title: Stirling Numbers Combinatorics


问:
How to solve $(1+y^2)+(x-e^{-\arctan y})y' = 0 $?
$(1+y^2)+(x-e^{-\arctan y})y' = 0    $

I have to solve this differential equation. Normally  I would just put all the terms with y variable on one side and x on the other and integrate. However when I do that there is y' in both numerator and denominator. Or I will make it in a form of $y + Px =  Q$ and find integrating factor.  There is other method of using auxiliary equation but that does not work. None of these methods work.

答:
We need to solve $$(1+y^{2})dx+(x-e^{-\tan^{-1}(y)})dy=0$$

Let $P(x,y)=y^{2}+1$ and $Q(x,y)=-e^{-\tan^{-1}(y)}+x$ so, we can see that $P_{y}\not=Q_{x}$.

Now, we can find an integrating factor $\mu(y)$ such that $$\mu(y)P(x,y)+\mu(y)Q(x,y)y'=0$$ is exact equation.

This means, $$(\mu(y)P(x,y))_{y}=(\mu(y)Q(x,y))_{x}$$
So, we have $$(y^{2}+1)\mu'(y)+2y\mu(y)=\mu(y)$$
Solving, we have $$\mu(y)=\frac{e^{\tan^{-1}(y)}}{y^{2}+1}$$
Note that if $\tilde{P}(x,y)=e^{\tan^{-1}(y)}$ and $\tilde{Q}(x,y)=\frac{e^{\tan^{-1}(y)}x-1}{y^{2}+1}$ so $\tilde{P}_{y}=\tilde{Q}_{x}$
Solving by exact method equation  (if you need more details here, you can say me and I will write the details of this section) we have the general solution:
$$\boxed{e^{\tan^{-1}(y)}x-\tan^{-1}(y)=c}.$$


问:
What is the c.d.f of median?
Given the simple random samples $X_{1}, X_{2}, X_{3}\sim Unif(0, 1)$, let
$Y$  is the median of $X_{1}, X_{2}, X_{3}$.
What is the c.d.f of $Y$?

答:
Using the Order Statistics Theory you easy find that

$$f_{X_{(2)}}(y)=3!y(1-y)=6y(1-y)$$

$y \in (0;1)$

This is the PDF... If you are intersted in the CDF, just integrate it

FYK: $X_{(2)}\sim Beta(2;2)$

Title: What is the c.d.f of median?


问:
How to find the angle formed by two joined triangles when they are multiples?
The problem is as follows:

The figure from below shows two triangles. Assume $PR=QS$. Using this
information find $x$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&20^{\circ}\\
2.&18^{\circ}\\
3.&22^{\circ}\\
4.&24^{\circ}\\
4.&30^{\circ}\\
\end{array}$

What I attempted to do to solve this problem was to use the identity which relates the exterior sum of the two angles in a triangle. In other words:

$\angle PQS= 6x$

But that's it, I came stuck there. Since I found this problem in a section belonging to triangle congruency, I think it should be solved using such approach. Can someone help me here?. If so, what sort of the cases related with congruency of triangles is it?. Could it be that is side-angle-side?.

Please include a drawing in the solution as I dont know how to spot those in triangles.

答:
[![enter image description here][2]][2]

This figure satisfies the conditions. If $\angle QPR=\angle PQR=54^o$ then $\angle SQR=36^o$   then $ PR=QR=QS$. So correct option is $18^o$. Other options are wrong.

Moreover we can solve following equation:

$2x+4x+3x+\alpha=180$

Where $\alpha=\angle PQS$

Among numerous solutions only one set of solution; $(x, \alpha)=(18, 18)$ satisfies the condition


问:
Operator Norm of an augmented matrix vs unaugmented matrix
This problem comes from Linear Algebra and its Applications by David C Lay and Steven R Lay.

Show that $$||[A B]|| \geq ||[A]||$$

Where $[A B]$ is a a matrix $A$ augmented with another matrix $B$
So if A = $$A = \begin{bmatrix}
a & b \\
c & d\\
\end{bmatrix}$$ and $$B = \begin{bmatrix}
e & f \\
g & h\\
\end{bmatrix}$$

Then $$[A B] = \begin{bmatrix}
a & b & e & f \\
c & d & g & h\\
\end{bmatrix}$$

My attempt:

We know that the operator norm of a matrix is the square root of the largest eigenvalue of the matrix. What's really throwing me off here is that what if by augmenting the matrix $B$, the eigenvalues of $[A B]$ are less than that of just $A$.

I think the approach I should be going for starts with an SVD approach(?)
$$[AB] = U_{AB}\Sigma_{AB}V^T_{AB}$$
$$[A] = U_{A}\Sigma_{A}V^T_{A}$$

From there I need to prove that the largest singular value of $[AB]$ is greater than that of $[A]$. Is this correct? Can someone share some insight?

答:
Although an answer is already provided and what I have in mind seems the same, I am still going to write one in order to break the problem down a bit more.

The norms of operators over non-trivial (ie. not $\{\vec{0}\}$) normed linear spaces $T:(V,\|\cdot\|_v)\rightarrow(W,\|\cdot\|_w)$ is defined as

$$
\|T\|:=\sup\{\|T(v)\|_w:\|v\|_v=1\}
$$

And that applies to the spectrum norm over $\mathbb{R}^n\rightarrow\mathbb{R}^m$ in your question as well.

For finite-dimensional vector spaces, the supremum is attained, meaning there is actually a particular vector $v_0$ such that $\|T(v_0)\|_v=\sup\{\|T(v)\|_w:\|v\|_v=1\}$. In that case, supremum becomes just maximum.

Using that fact, let $v_A$ be the vector where $\|A\|_2$ is attained. Note that $\|Av_A\|_2=\|A\|_2$.

Subscript 2 means 'spectrum norm'. We are only taking about one type of norm. From here, I am going to omit the subscript when I am referring to operator norm and keep the subscript for vector norm. Hopefully that makes reading easier for you.

Also note that the following vector has norm 1.

$$
w=\begin{bmatrix}
v_A\\0
\end{bmatrix}
$$

Thus $w$ belongs to the set of $v$ over which we are scanning for $\max \|[AB]v\|_2$ and so $\|[AB]w\|_2$ is smaller than or equal to the maximum out of the set (of vectors with norm 1).

$$
\therefore\|[AB]\|\geq\|[AB]w\|_2
$$

Now expand the matrix-vector multiplications.

$$
[AB]w=\begin{bmatrix}
Av_A\\0
\end{bmatrix}\\
\therefore
\|[AB]w\|_2=\|Av_A\|_2\\
$$

Combine that with the earlier inequality, we get

$$
\|[AB]\|\geq\|A\|
$$


答:
Interesting that you tried to make sens of the equation as a whole. I would start from one end and try to come out from the other :
$ (A \setminus B) \cup C = \{x, (x \in A \wedge x\not\in B) \vee x \in C\}$
$=\{x,(x\in A \vee x\in C)\wedge (x\not\in B \vee x \in C)\}=(A\cup C) \cap (\overline{B} \cup C),$ whereas you can esaily see that $(A \cup C) ∖ (B\cup C)=(A \cup C) \cap (\overline{B \cup C})$. You can clearly see here that the two sets are not the same. Any counter examples ?

Body:
How to prove $(a \setminus b) ∪ c= (a ∪ c)  \setminus (b ∪ c)$? I tried to do some properties, but I don't know what to do next.
1


答: Plücker coordinates of a line are in my opinion one of the best ways to represent a line. In $\mathbb{P}^2$, a line is represented by $[l_{01}, l_{02}, l_{12}]$, where $(l_{01}, l_{02})$ represent its direction vector and $l_{12}$ is called the moment. Given any point on the line $(p_1,p_2)$, $l_{12}=l_{01}p_2-l_{02}p_1$. Thus, the line equation turns out to be $$l_{02}x-l_{01}y+l_{12}w=0$$.

Title: Is there a canonical definition of a line?

Body: Normally the definition of a line is either via two points or its slope and y-intercept.  But both methods of defining a line have problems.  With the first one is the problem, that many points are passing through a single line so you can define a line using every pair of them, which mean that a definition method is not unique.  And the second one can't handle lines with slope infinity (like x=0).  I searched and thought about it a little bit, but could find anything.


问:
Mathematical Symbol
In the following paper, what does the symbol $\Phi$ in equation $3.1$ (page $3$) represent? Does it represent the normal distribution?</p>

答:
$\Phi(x)$ typically (and which is what is also means in the article you have linked to) represents a suitably normalized error function, equivalently the cummulative distribution function of the normal distribution, i.e.,
$$\Phi(x) = \dfrac1{\sqrt{2 \pi}}\int_{-\infty}^{x} \exp(-t^2/2)dt$$</p>

Title: Mathematical Symbol


问:
Solve the inequality $\sqrt[3]{1+(3+x)\sqrt{x}+3x} - \sqrt[3]{1-(3+x)\sqrt{x}+3x} > x + a$
So am trying to solve this inequality, $\sqrt[3]{1+(3+x)\sqrt{x}+3x} - \sqrt[3]{1-(3+x)\sqrt{x}+3x} > x + a$, the problem is of course for values $0<a<1$, I tried working through it multiple times but the solutions i found, which they are $\sqrt{1-\sqrt{1-a}}<x<1 $ are the wrong ones when I try wolfram alpha, can anyone show me the right way to do it please?

答:
You obviously need $x > 0$ and further have
$$\sqrt[3]{(1+\sqrt x)^3} - \sqrt[3]{(1-\sqrt x)^3} > x+a$$

$\iff  1-a > (\sqrt x - 1)^2 \iff \sqrt{1-a} > |\sqrt x - 1|$

For $x \in (0, 1)$, we have this as
$ \sqrt{1-a} > 1-\sqrt x \implies x > (1-\sqrt{1-a})^2 $

For $x \geqslant 1$, we have this as
$ \sqrt{1-a} > \sqrt x - 1  \implies (1+\sqrt{1-a})^2 > x $

Thus the solution set is $x \in \left((1-\sqrt{1-a})^2, (1 + \sqrt{1-a})^2\right)$.


问:
If $\varphi$ is a positive linear functional then the following are equivalent.
$\begin{align} &\text{(1) } \varphi(x^*x)=\varphi(xx^*)\text{ for all }x\in A.\\&\text{(2) } \varphi(u^*au)=\varphi(a)\text{ for all }a\in A \text{ and unitary } u\in \tilde A.\end{align}$

$\tilde A$ is the unitalization of $A$ by adding a $1$ ($A$ may have its own $1_A$).

my trying:

Extend $\varphi$ by defining $\varphi(a+k)=\varphi(a)+k\|\varphi\|$. Then $(1)\Leftrightarrow\varphi((a+k)^*(a+k))=\varphi((a+k)(a+k)^*)$ and $(2)\Leftrightarrow\varphi(u^*(a+k)u)=\varphi(a+k)$ .

Thus we may assume that $A$ is unital and prove that

$\begin{align} &\text{(1) } \varphi(x^*x)=\varphi(xx^*)\text{ for all }x\in A.\\&\text{(2) } \varphi(u^*au)=\varphi(a)\text{ for all }a\in A \text{ and unitary } u\in A.\end{align}$

are equivalent.

Since every $a$ can be decomposed in to $a=\text{Re}(a)^+-\text{Re}(a)^-+i\text{ Im}(a)^+-i\text{ Im}(a)^-$, and $\varphi(u^*xu)=\varphi(\sqrt x u^*u\sqrt x)=\varphi(x)$ for positive $x$, that (1) implies (2) is trivial.

However, I have no idea on how to show (2) implies (1). If $|x|$ is invertible, that there is a unitary $u$ such that $x=u|x|$ follows from polar decomposition.  Then $ux^*xu^*=xx^*$ follows. Similarly, if $x$ can be approximated by invertible elements, $\varphi(x^*x)=\varphi(xx^*)$ still holds. But this is true only for certain kinds of C*-algebras. Do I need to use Gelfand-Naimark representation on this?

答:
Use the following statement:

If $A$ is a unital $C^*$-algebra and $x\in A$ there are unitaries $u_1,...,u_4$ and complex numbers $a_1,...,a_4$ with $x=\sum_i a_i u_i$.

So:

$$\varphi(x^*x)=\varphi\left(\sum_i(a_iu_i)^*\sum_j(a_ju_j)\right)= \sum_{ij} \overline{a_i}a_j\varphi(u_i^*u_j)$$

now note that $\varphi(u_i^*u_j)= \varphi\left(u_i(u_i^*u_j)u_i^*\right) = \varphi(u_ju_i^*)$ by (2). Hence:
$$\varphi(x^*x)=\sum_{ij}\overline{a_i}a_j\varphi(u_ju_i^*) =\varphi\left(\sum_j(a_ju_j)\sum_i(a_iu_i)^*\right)=\varphi(xx^*)$$


问:
Constructing contradiction with existence and uniqueness theorem
Let a solution $y\colon I\to \mathbb{R}$, where $I$ is a subset of the real numbers, be defined as follows:

$$y'=\frac{\cos(t)}{4-(\sin(t))^2}(4-y^2),\quad y(0)=1$$

Is it possible to have a solution $y$ of the given IVP such that $y(a)=2$ for some $a$ in $I$?

Question: Can someone explain me how this $y(a)=2$ for some $a\in I$ contradicts with the uniqueness and existence theorem?

答:
Set $y(a)=b$ as initial value

get the solution$$y(t)=\frac{\sin (a) (b \sin (t)+4)-4 (b+\sin (t))}{\sin (a) (b+\sin (t))-b \sin (t)-4}$$

look for the maximum
$$y'(t)=-\frac{2 \left(b^2-4\right) (\cos (2 a)+7) \cos (t)}{(-2 \sin (a) (b+\sin (t))+2 b \sin (t)+8)^2}$$
$$y'(t)=0\to t=\pm\frac{\pi}{2}+2k\pi$$
$$f(\pi/2)=\frac{(b+4) \sin (a)-4 (b+1)}{(b+1) \sin (a)-b-4}$$
$$f(\pi/2)=2\to \sin a=-2\to\text{ impossible in }\mathbb{R}$$
if the max is at $-\pi/2$ this leads to $\sin a=2$ which is impossible.

Therefore the function is strictly less than $2$ for any initial value.

Maybe there is a shortcut, tho.


问:
Functions that are uniform limits
For $n \in \mathbb{N}$ let's consider $f_{n}: \mathbb{R}_{+} \to \mathbb{R}, x \mapsto e^{-nx}$, and let $E = \text{span}(f_{n})_{n \in \mathbb{N}}$.

The question is: determine the functions that are uniform limits of sequences of $E$.

So obviously such functions are continuous, but else ... I do not know ... Maybe the fact that $f_{n} = (f_{1})^{n}$ is useful, but I can't see how.

答:
They are precisely continuous functions $f$ such that $f(x) \to 0$ as $x \to \infty$.

Proof: Given any such function $f$ define $g: [0,1] \to \mathbb R$ by $g(x)=f(- \ln x)$ if $x >0$ and $g(0)=0$. You can check that $g$ is continuous. By Weierstrass Theorem there is a polynomial $p$ such that $|g(x)-p(x)| <\epsilon$ for all $x$. Note that $|p(0)| <\epsilon$ since $g(0)=0$. We may therfore modify $p$ to a polynomial without constant term. This translates to $|f(x)- \sum\limits_{k=1}^{n} c_ke^{-kx}|<\epsilon$.

Title: Functions that are uniform limits


问:
Does $c^n(n!+c^n)\lt (n+c^2)^n$ hold for all positive integers $n$ and $c\gt 0$?
I am  not sure whether the following inequality is true? Some small $n$ indicates it is true. </p>

<p>Let $n$ be a positive integer and $c\gt0$, then
$$c^n(n!+c^n)\lt(n+c^2)^n.$$ </p>

答:
For $n=1$, you have $c+c^2 < 1+c^2$, which is equivalent to $c < 1$, hence always false because you supposed $c > 0$.

For $n=2$, you have $2c^2 + c^4 < c^4 +4c^2 + 4$, which is equivalent to $2c^2 > -4$ (always true).

Aryabhatta's example ensures us that we cannot conjecture that the inequality will always be true for $n > 2$.


问:
Abelian categories with direct sums
Does any abelian category admits direct sums?

If not, categories admiting direct sums have a special name?

I'm asking this since I am writing a proof that requires direct sums but I only know that the category is abelian.

Thank you very much for your answer!

Edit. In the tag description is written that indeed they posses finite direct sums.

答: Yes, by the definition of an Abelian category, it contains all finite coproducts ("direct sums"). Just look back to your definition of an Abelian product to see all the extra things an Abelian category is assumed to have.</p>

Title: Abelian categories with direct sums


问:
What are the lengths of the major and minor axes of an ellipse inscribed in a rhombus of an isometric grid?
Like this:

where inner the angles of this rhombus are 60° and 120°, and the sides have length 1.

We can see that the sides are tangent to the inscribed ellipse at their midpoints.

I believe that the two intersections of the perpendiculars of the sides shown here are the foci of the ellipse… but I can find no confirmation of that anywhere. If we can assume that is the case, then I worked out that the ellipse’s minor axis has length $\sqrt{\frac{1 + \sqrt 5}{6}}$ and the major axis is $\frac{1 + \sqrt 5 }{2 \sqrt 3} $.

EDIT: Nope! Bad assumption.  See below.

So a related question would be, are these really the foci of this ellipse? If not, then how to construct them properly?

I’ve looked everywhere for these answers, to no avail. I have found many videos and websites explaining how to approximate this ellipse using a compass, a technique which makes use of the pair of points that I’m assuming are foci. It seems like the only people interesting in this are draftsmen and artists, not mathematicians. Quite frustrating and disappointing!

A bit more discussion here:
https://www.quora.com/What-are-the-lengths-of-the-major-and-minor-axes-of-an-ellipse-inscribed-in-a-rhombus-of-an-isometric-grid-i-e-with-opposite-angles-of-60-and-120

答:
@Blue's comment shows you the way to go. Here is what your diagram looks like, squeezed horizontally by a factor of $\frac{1}{\sqrt 3}$:

Your rhombus with unit side lengths has become a square of side $\frac{1}{\sqrt2}$, so your ellipse has become a circle with diameter $\frac{1}{\sqrt2}$. Now stretch it back out; the minor axis remains unchanged, and the major axis increases by a factor of $\sqrt3$.


问:
A bijection between between the orbits $G$ on $G \backslash B \times G \backslash B$ and orbits of $B$ on $G \backslash B$?
Let $G$ be a group and $B$ a subgroup. Can somebody show me how that orbits of $G$ on $G \backslash B \times G \backslash B$ are in bijection with the orbit of $B$ on $G \backslash B$?

Here is what I've come up with so far:

Let $(aB,cB) \in G \backslash B \times G \backslash B$.

Then $G.(aB,cB) = \{(gaB,gcB) : g \in G\}$

By setting $ga = h$ we can see that this is in bijection with:

$=\{(hB,ha^{-1}cB) : h \in G\}$

Now, if we have $(hB,ha^{-1}cB) = (h'B,h'a^{-1}cB)$ then $(h'^{-1}hB,h'^{-1}ha^{-1}cB)=(B,a^{-1}cB)$ and in particular $h'^{1}h \in B$. I felt like this might be going in the write direction to showing the desired bijection.

Help appreciated!

答:
Two step hint:

step 1

show that every orbit of $G$ on $G/B\times G/B$ contains an element of the form $(B,gB)$.

step 2

Consider $\{(B,gB):g\in G\}$. What is the subgroup of $G$ that fixes this? Notice that orbits of $G$ are in bijection with the orbits of this stabiliser and up to a few explanatory words, you're there.


问:
$6!\cdot 7!=10!$. Is there a natural bijection between $S_6\times S_7$ and $S_{10}$?
Aside from $1!\cdot n!=n!$ and $(n!-1)!\cdot n! = (n!)!$, the only product of factorials known is $6!\cdot 7!=10!$.

One might naturally associate these numbers with the permutations on $6, 7,$ and $10$ objects, respectively, and hope that this result has some kind of connection to a sporadic relation between such permutations - numerical "coincidences" often have deep math behind them, like how $1^2+2^2+\ldots+24^2=70^2$ can be viewed as an ingredient that makes the Leech lattice work.

The most natural thing to hope for would be a product structure on the groups $S_6$ and $S_7$ mapping to $S_{10}$, but as this MathOverflow thread shows, one cannot find disjoint copies of $S_6$ and $S_7$ living in $S_{10}$, so a product structure seems unlikely.

However, I'm holding out hope that some weaker kind of bijection can be found in a "natural" way. Obviously one can exhibit a bijection. For instance, identify the relative ordering of $1,2,\ldots 7$ in a permutation of size $10$, and then biject $_{10}P_{3}=720$ with $S_6$ in some way. But I'd like to know if there is a way to define such a bijection which arises naturally from the permutation structures on these sets, and makes it clear why the construction does not extend to other orders.

I tried doing something with orderings on polar axes of the dodecahedron ($10!$) and orderings on polar axes of the icosahedron ($6!$), in the hopes that the sporadic structure and symmetry of these Platonic solids would allow for interesting constructions that don't generalize, but ran into issues with the dodecahedron (sequences of dodecahedral axes aren't particularly nice objects) and the question of how to extract a permutation of length $7$.

I'm curious if someone can either devise a natural bijection between these sets or link to previous work on this question.

答:
It may be connected with, of all things, the $3-4-5$ right triangle!  This triangle and its multiples stand out as having the sides in arithmetic progression.  Such an arithmetic progression leads to factorial expressions when the sides are multiplied together.

As a preliminary step, consider a relatively unheralded property of right triangles: the diameter of the incircle plus the hypotenuse equals the sum of the other two sides. Suppose that the legs are $a$ and $b$, and the hypoteneuse is $c$ where $c^2=a^2+b^2$.  The diameter of the incircle is then $2ab/(a+b+c)$ while the Pythagorean relation implies $$(a+b+c)(a+b-c)=(a^2+2ab+b^2)-(a^2+b^2)=2ab$$
Thereby the diameter of the incircle reduces to $a+b-c$.  Should there be a right triangle whose sides are in arithmetic progression, then, the diameter of the incircle will join this progression, making it longer and thus perhaps generating a bigger factorial upon multiplication.

In this question it is shown that the product of the sides of any triangle is half the product of the diameter of the circumcircle (circumdiameter), the diameter of the incircle (indiameter), and the perimeter.  Let us see where that leads if we apply it to a right triangle having sides $3,4,5$.
Multiplying the sides together then gives

$3×4×5=\text{circumdiameter}×\text{indiameter}×\text{perimeter}/2$

We double the sides of the triangle to clear the fraction on the right side:

$6×8×10=\text{circumdiameter}×\text{indiameter}×\text{perimeter}×4$

The circumdiameter is the hypoteneuse of the $3-4-5$ triangle, thus $5$ -- which is in the aforementioned arithmetic progression.  The indiameter is $2$ from the above lemma, which precedes $3,4,5$ in the arithmetic progression.  And the perimeter of the triangle is three times the longer leg, again due to the arithmetic progression, thus $4×3$.  Substituting these results into the above product equality then gives

$6×8×10=5×2×(3×4)×4=5!×4$

And there is our factorial.  To make it cleaner we should multiply by $3/2$, absorbing the dangling factor $4$ into the factorial.  We then get three different three-term products on the left side, depending on which of the factors $6,8,10$ we increment:

$\color{blue}{8×9×10}=6×10×12=6×8×15=5×2×(3×4)×6=6!$

And from the three-term product shown in blue, we have

$6!=10!/7!$

Why is this uniquely chosen?  We see that the sides of a right triangle being in arithmetic progression lead to the factorial on the right in two ways, by making the perimeter a simple multiple of one leg and by incorporating the circum diameter into the arithmetic progression.  Only the $3-4-5$ right triangle has these properties, and it leads specifically to $6!$ also being a factorial ratio.


答: In an abelian category you have much more than direct sums(=coproducts); you have biproducts, so that the object part is the same for binary products and coproducts and the arrows satisfy some very nice relations..http://en.wikipedia.org/wiki/Biproduct</p>

Title: Abelian categories with direct sums

Body:
Does any abelian category admits direct sums?

If not, categories admiting direct sums have a special name?

I'm asking this since I am writing a proof that requires direct sums but I only know that the category is abelian.

Thank you very much for your answer!

Edit. In the tag description is written that indeed they posses finite direct sums.


问:
Finding the norm of $(x_n) \mapsto \sum x_{n}2^{-n}$ in $(\ell^\infty)^*$
The question says: for $x \in \ell^{\infty}$, let $f(x)=\sum_{n \in \mathbb{N}}x_{n}2^{-n}$. Determine the norm of $f$ in $({\ell^{\infty}})^{*}$.

I have tried to use the definition of the dual space norm. I have written down that

$$\|f\|= \sup\left\{\frac{|\sum_{n \in \mathbb{N}}x_{n}2^{-n}|}{\sup\{x_n\}}\right\},$$ where the outer supremum is taken over all $x$ in $\ell^{\infty} \setminus \{0\}$.

I just don't know where to go from here.

Body:
First of all, for all $x\in\ell^{\infty}$ we have:

$|f(x)|=|\sum_{n=1}^\infty x_n2^{-n}|\leq\sum_{n=1}^\infty |x_n|2^{-n}\leq\sum_{n=1}^\infty ||x||_{\infty}2^{-n}=||x||_{\infty}\sum_{n=1}^\infty 2^{-n}=||x||_{\infty}$

And so $||f||\leq 1$. Also, let $y=(1,1,1,1,...)$. Then $||y||_{\infty}=1$, and so:

$||f||\geq\frac{|f(y)|}{||y||_{\infty}}=|f(y)|=\sum_{n=1}^\infty 2^{-n}=1$

So the norm of $f$ is $1$.


问:
What expectation can we infer about the daily total collections of the leaf litter from the two patches?

Two forest patches have, respectively, $100$ and $200$ teak trees of the same age. In a given season, all trees shed some of their leaves at random. The daily total collections of the leaf litter from the two patches are expected to have

$(1)$ nearly equal means, standard deviations and coefficients of variation

$(2)$ different means, nearly equal standard deviations and coefficients of variation

$(3)$ different means, nearly equal standard deviation and different coefficients of variation

$(4)$ different means, and standard deviations but nearly equal coefficients of variation

What conclusion can we make from the above data? What I know is that the mean $\mu = \dfrac {\sum x} {n}, \ $ standard deviation $\sigma = \sqrt {\dfrac {\sum x^2} {n} - \left (\dfrac {\sum x} {n} \right )^2}$ and coefficients of variation $\nu = \dfrac {\sigma} {\mu},$  where $n =$ sample size (in this case the number of trees in each of the forest patch) and $\sum x$ represents the total number of leaves collected from each forest patch in a day. Now how do I proceed? Any help in this regard will be appreciated.

Thanks for your time.

答:
The point of the problem is that the number of leaves shed by each tree on a given day is to be considered a random variable, and all the variables are independent and identically distributed.  So in one case, we have the sum of $100$ i.i.d. random variables, and in the other case we have the sum of $200$ i.i.d. random variables.  Let the mean and variance of each be $\mu$ and $\sigma^2$ respectively.

By linearity of expectation, the mean of the first patch is $100\mu$ and that of the second patch is $200\mu$. Also, the variance of the sum of independent random variables is the sum of the variances, so the variance of the first patch is $100\sigma^2$ and the standard deviation is $10\sigma$; the standard deviation of the second patch is $10\sigma\sqrt2$.


问:
Representing $\mathbf{Tr}(A + BC^{-1}B^T)< K$ as an LMI
I am trying to implement a convex optimization and I have the constraint $\mathbf{Tr}(A + BC^{-1}B^T)< K$ where $A\succeq0,B,C\succ0$ are the decision variables. Can it be transformed into a linear constraint or a linear matrix inequality?

答:
The trace can also be written as

$$
\mathbf{Tr}(A + B\,C^{-1}B^\top) = \sum_{i=1}^n e_i^\top (A + B\,C^{-1}B^\top)\,e_i, \tag{1}
$$

with $n$ such that $A \in \mathbb{R}^{n\times n}$ and $e_i$ the $i$th column of an $n \times n$ identity matrix. By introducing intermediate scalar variables $\alpha_i$ one can write the initial inequality in an indirect way by using

\begin{align}
e_i^\top (A + B\,C^{-1}B^\top)\,e_i &< \alpha_i, \forall\,i = 1 \dots n, \tag{2} \\
\sum_{i=1}^n \alpha_i &< K. \tag{3}
\end{align}

The inequality from $(3)$ is already a linear inequality. The inequality from $(2)$ can be formulated as linear (matrix) inequality by using the Schur complement. It can be noted that applying the Schur complement to each inequality from $(2)$ does require the additional assumptions that $C$ is positive definite.