问:
How can we show that $f_n$ diverges and $F_n$ converges
I just want to understand why the following statement are true :

Let $(f_n)$ be a sequence of continuous function , uniformly bounded over [$0,1$] and orthonormal with respect to scalar product $( \,,\, ) $ in $L^2[0,1]$

We define  : $F_n(x)=\int_{0}^{x}f_n(t)dt $ s.t $x\in [0,1]$

There is no subsequence of $(f_n)$ that converges in $L^2[0,1]$

We can find a subsequence of $F_n$ that converges in $L^2[0,1]$

I don't see why we can found for $f_n$ and not for $F_n$

Thanks in advance for your help .

答:
is nearly trivial, because $\|f_n-f_m\|^2=\|f_n\|^2-2\,(f_n,f_m)+\|f_n\|^2=2$ for $n\neq m$, and that's true for any subsequence as well, the contrary of a Cauchy sequence.

is a consequence of the celebrated Arzelà–Ascoli theorem (https://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem): since the $f_n$ are uniformly bounded, $F_n$ are equicontinuous, so $\{F_n\}$ is a (pre-)compact set in $C[0,1]$. Thus, there is a convergent subsequence in $C[0,1]$, and that's convergent in $L^2[0,1]$ as well.


问:
Solve $\int\frac{M\cos^2u-(M-1)\sec^2u}{\sqrt{1-M\sin^2u}}\,du$ without differentiating the solution
In this answer I said that "it is left as an exercise to show that"
$$\int\frac{M\cos^2u-(M-1)\sec^2u}{\sqrt{1-M\sin^2u}}\,du=(\tan u)\sqrt{1-M\sin^2u}+C\tag1$$
The original question asked for the derivation of a solution to an integral; the solution involves elliptic integrals. Because of the latter, I would ordinarily have the full power of elliptic integrals/functions at my disposal (i.e. Byrd and Friedman), but the OP said at the start

Yet again I might say I am very very rusty on my mathematics, but in this case I've never even heard of elliptic integrals up to the point I saw this paper.

And I did not want to scare the OP away by saying "oh, this is B&F XXX.XX, we have this and that". I wanted to be gentle.

However, I could not think of an easy way to justify the integral at the start of this question other than differentiating the solution I already knew. How could I solve $(1)$ from scratch in an elementary manner?

答:
Note

\begin{align}
I=&\int\frac{M\cos^2u-(M-1)\sec^2u}{\sqrt{1-M\sin^2u}}\,du\\
=& \int\frac{M\cos^2u-(M-1)\sec^2u}{\sqrt{(1-M)+M\cos^2u}}\,du
=\int\frac{M\cos^2u+(1-M)\sec^2u}{\sqrt{(1-M)+M\frac{\sin^2u}{\tan^2u}}}\,du\\
=& \int\frac{M\sin u \cos u+(1-M)\tan u\sec^2u}{\sqrt{(1-M)\tan^2 u+M\sin^2u}}\,du\\
\end{align}
Substitute $t= (1-M)\tan^2 u+M\sin^2u$. Then,
$$I= \frac12\int \frac{dt}{\sqrt t}= \sqrt t+C
$$

ContentLicense: CC BY-SA 4.0


问:
If $ 2^a - 5^b7^c =1$ ....
Find all possible values of positive integers a, b, c such that:
$$2^a - 5^b7^c =1$$
While solving this question I surmised that a must be a multiple of 4, since $5^a7^b$ always ends with a $5$. Add one to that and you get a six, which entails that $2^a$ has a 6 at the unit digit's spot, so a must be a multiple of $4$. After that, it's a dead end for me. Anyone have any idea how to proceed?

答:
Note that
$ 2^a\equiv 1\pmod 5$ implies $a\equiv 0\pmod4$,
so write $a=4m$.
Now
$$5^b7^c=2^a-1=(2^{2m}-1)(2^{2m}+1)$$
and the two factors are coprime.
Take it further,
$$5^b7^c=(2^{m}-1)(2^m+1)(2^{2m}+1)$$
and all three factors on the right are pairwise coprime. So at most one of them is divisible by $5$ and at most one is divisible by $7$, meaning that the smallest one must be $=1$, i.e., $m=1$. This gives us
$$5^b7^c=(2^1-1)(2^1+1)(2^2+1)=15$$
which is absurd.

Note that we used $b\ge 1$ to conclude $2^a\equiv 1\pmod 5$. If we were to allow $b=0$, we find additional solutions such as $a=1$, $b=c=0$ or $a=3$, $b=0$, $c=1$ and beyond that enter the realm of Catalan's conjecture.

Title: If $ 2^a - 5^b7^c =1$ ....


问:
Find the point on the graph of $y=x^2$ that is closest to the point $(0, 3)$
Here is Prob. 45, Sec. 2.3, in the book Calculus With Analytic Geometry by George F. Simmons, 2nd edition:

Find the point on the graph of $y = x^2$ that is closest to the point $(0, 3)$. ...

My Attempt:

Let $\left( a, a^2 \right)$ be the required point. Then the line through the points $(0, 3)$ and $\left( a, a^2 \right)$ is perpendicular to the line tangent to the curve $y = x^2$ at $\left( a, a^2 \right)$, and since the slope of the latter is $2a$, the slope of the former must satisfy
$$
\frac{ a^2 - 3}{ a - 0} = -\frac{1}{2a},
$$
provided of course that $a \neq 0$, which implies that
$$
2a^3 - 6a = -a,
$$
which implies that
$$
2a^3 = 5a,
$$
and as $a \neq 0$, so we must have
$$
a^2 = 5/2,
$$
and hence $a = \pm \sqrt{5/2}$.

Let us compute the distance between the points $(0, 3)$ and $(\pm \sqrt{5/2}, 5/2 )$.
$$
d \big( (0, 3), ( \pm \sqrt{5/2}, 5/2) \big) = \sqrt{ 5/2 + (5/2 - 3)^2 } = \sqrt{11}/2.
$$

For $a = 0$, we note that the distance between the point $(0, 3)$ and the point $(0, 0)$ on the graph of $y = x^2$ is $3$.

Since $\sqrt{11}/2 < \sqrt{36}/2 = 6/2 = 3$, we can conclude that the point on the graph of the parabola $y = x^2$ that is closest to the point $(0, 3)$ is either of the two points $( \pm \sqrt{5/2}, 5/2 )$.

Is my solution correct in each and every detail? Or, have I made any mistakes?

答:
its correct and rigorous  but can be simplified ...

we have to minimise $$f(a)=\sqrt{a^2+{(a^2-3)}^2}$$ or basically minimise $$g(a)=a^2+{(a^2-3)}^2$$ $$g(a)={\left(a^2-\frac{5}{2}\right)}^2+\frac{11}{4}$$ which happens when $$a^2=\frac{5}{2} \iff a=\pm \sqrt{\frac{5}{2}}$$ and we are done!


问:
Show asymptotic orthogonality of eigenvectors
Let $(A_n),(B_n)$ be sequences of $n\times n$ (growing size) matrices. Suppose also that

$$\lambda_1(A'_nA_n) \to c>0 \hspace{1cm} \text{as} \hspace{1cm} n\to \infty$$

$$\lambda_1(B'_nB_n) \to 0 \hspace{1cm} \text{as} \hspace{1cm} n\to \infty$$

where $\lambda_1$ denotes the largest eigenvalue. Let $Q_n$ and $R_n$ be the $n\times 1$ normalized eigenvectors associated to the largest eigenvalue of $A'_nA_n$ and $B'_nB_n$ respectively. Am trying to prove that

$$Q_n'R_n\to 0 \hspace{1cm} \text{as} \hspace{1cm} n\to \infty$$

Is this possible? Any ideas on how to proceed is greatly appreciated.

答:
I think you will need more qualifications for this to hold. Set
$$A_n = \text{diag}(1, 0, \cdots,0)$$
Then for each $A_n$, we have that $\lambda_1(A_n ' A_n) = 1$, and the corresponding eigenvector is $\hat{e}_1$. Now set
$$B_n = \text{diag}(1/n, 0, \cdots, 0)$$
Then for each $B_n$, we have that $\lambda_1(B_n' B_n) = 1/n^2 \to 0$ as $n \to \infty$. But the corresponding eigenvector is also $\hat{e}_1$, so it will always be the case that $Q_n' R_n' = 1$.


问:
permutation/combination problem
There are 3 doors to a lecture room. In how many ways can a lecturer enter the room from one door and leave from another door?

I have done like this: They way of entering is 3 and exiting is also
  3, therefore the total way will be 3*3=9.

答: For entering he can use each of 3 doors, for exit he can uses 2 other doors so total number of ways is $$3\cdot2=6$$</p>

ContentLicense: CC BY-SA 3.0

Title: permutation/combination problem


问:
Find total probability for Binomial pdfs

In a two-class classification problem with a single feature $X$ the pdfs are $\operatorname{Bin}(n,p)$ with $n=10$ for both classes and $p$-values $0.6$ and $0.4$ respectively. Find the decision boundary point for MAP classifier. Assume the two classes have equal prior probability.

What is the total probability of $X = x$?

I am struggling to find out the answer for no 2. Can anyone help me out with some hints?

Body:
Straightforward:

$P(X=x)=\frac{1}{2}\binom{10}{k}(0.6^k\times0.4^{10-k}+0.4^k\times0.6^{10-k})$


问:
Uniqueness of n and p in a discrete binomial distribution
For a discrete random variable with binomial distribution, if $E(X)=9.6$ and $\text{Var}(X)=1.92$, will $n$ and $p$ have the unique values of $12$ and $.8$ or are other pairs of values possible?

答:
If $p = 0$, then we will have $E[X] = \text{Var}[X] = 0$ so $n$ can be whatever we want.

However, if both the expectation and variance are nonzero, it is unique. Indeed, we know that $E[X] = np$ and $\text{Var}[X] = np(1-p)$. So we can immediately divide the two to get that $1 - p = \text{Var}[X] / E[X]$, and hence $p$ is uniquely determined. We can then substitute this value of $p$ into the first equation to conclude that $n$ is uniquely determined as well.


问:
Confusion on Uniform Integrability of Random Variables
We have the definition that a random process, $X_n$, is (1st power) uniformly integrable if

$$\lim_{M\to\infty}\sup_n\mathbb{E}(|X_n| ; |X_n|\geq M)=0$$.

My question is whether the following four statements are true and (intuitively) why this is or isn't the case:

"$|X_n|$ is bounded for all $n$ $\implies$ $X_n$ is 1st power uniformly integrable."

My guess: it's true since if it's bounded there must exist some $M\in \mathbb{R}$ such that $M>|X_n|$, and so $\lim_{M\to \infty}\mathbb{I}\{|X_n|\geq M\}=0 $

"$|X_n| < \infty$ $\implies$ $X_n$ is 1st power uniformly integrable."

My guess: intuitively to me it feels like this should be true by similar reasoning to (1); if $|X_n|$ is finite then as $M\to \infty$ we get $\mathbb{I}\{|X_n|\geq M\}=0$? And hence $\lim_{M\to\infty}\sup_n\mathbb{E}(|X_n| ; |X_n|\geq M)=0$? I think I'm wrong, however.

"$\mathbb{E}|X_n| < \infty$ $\implies$ $X_n$ is 1st power uniformly integrable."

My guess: I think this is wrong because $\mathbb{E}|X_n| < \infty$ doesn't imply that $|X_n|$ is finite and so $\lim_{M\to \infty}\mathbb{I}\{|X_n|\geq M\}$ is not necessarily $0$?

"If $\forall \epsilon>0, \exists\delta>0 $ such that $\mathbb{P}(|X_n|\geq M)<\delta $ implies  $\mathbb{E}(|X_n| ; |X_n|\geq M)<\epsilon$ $\implies$  $X_n$ is 1st power uniformly integrable."

My guess: I think this is not true since I think that this doesn't necessarily mean that we can find $\delta$ such that $\sup_n\mathbb{E}(|X_n| ; |X_n|\geq M)<\epsilon$? My question would also be - if we add in either the condition that $\mathbb{E}|X_n|<\infty$ or that $|X_n|<\infty$ then does this imply uniform integrability?

答:
It depends actually whether the bound is allowed to depend on $n$ or not. If $\sup_{n\geqslant 1}\lvert X_n\rvert$ is finite, then your arguement works. But if we only assume that $\lvert X_n(\omega)\rvert\leqslant c_n$ for almost every $\omega$, uniform integrability may fail: consider $X_n(\omega)=n$ if $0<\omega<1/n$ and $0$ otherwise on $(0,1)$ endowed with Lebesgue measure.

and 3. are covered by the previous counterexample.

For 4., the thing is that there are probability spaces such that $\mathbb P(A)$ is either $0$ or bigger than some $\delta_0$, like $\{a,b\}$ with $\mathbb P(\{a\})=\mathbb P(\{b\})=1/2$. In this case, the implication will be always satisfied with $\delta<1/2$, because the only way to have $\mathbb{P}(|X_n|\geq M)<\delta$ is that $\{|X_n|\geq M\}$ is empty. Hence a non-uniformly integrable sequence can also satisfy the implication.


问:
Show that $c$ is closed in $l^{\infty}$
Let $$c=\{ (a_i)_{i \in \mathbb{N}} \ ; \ \ a_i \in \mathbb{R}\ ,\ \forall i \in \mathbb{N} \ , \ \mbox{exist} \ \displaystyle \lim_{i \to \infty}(a_i)\}$$
 $$l^{\infty}=\{ (a_i)_{i \in \mathbb{N}} \ ; \ a_i \in \mathbb{R}\ ,\ \forall i \in \mathbb{N}; \  \displaystyle \sup_{i \in \mathbb{N}}|a_i|< \infty \}.$$ Show that $c$ is closed in $l^{\infty}$.

i am trying... taking a sequence of sequence in $c$ but i can't...i show that $c_0$ is closed in $l^{\infty}$.

答: Hint: a subset $A$ of a metric space $X$ is closed if and only if for any sequence $a_n\in A$ converging to an $x\in X$ you have $x\in A$.

Title: Show that $c$ is closed in $l^{\infty}$


问:
Tricky mortgage rate question
Can anybody suggest me how I might solve the following equation?

$$(1+x)^{300} -125x(1+x)^{300} = 1$$

where x is the unknown that I want to solve for.

The actual question in full is

Q. Assume that your are going to retire in 25 years time. You want a mortgage of £100,000 now to extend and renovate your house but want to have it paid in full before you retire.  The maximum repayment per month that your budge will allow id £800. What is the rate of interest you need from your bank to have the loan repaid in 300 monthly repayments (i.e 25 years).

So If the present value of each installment forms geometric sequence:

100,000 = 800/(1+i) + 800/(1+i)^2....

Where i is the monthly interest rate.

Then we can use the formula for Amortization - mortgages and loans:

A = Pi(1+i)^t/((1+i)^t -1)

Where :

A = repayment amount, P is the principle i is the term rate, and t is the term.

Then we get :

800 = 100,000i(1+i)^300/((1+i)^300-1)

Simplifying gives us:

$(1+i)^{300} -125i(1+i)^{300} = 1$

By the way the answer is 8.75% (yearly rate) which converts to a monthly rate of .70146% i.e a value of i of .0070146 which does indeed satisfy the equation above.

答:
You seem to be making a mistake in setting up the equation to be solved.  If the per-period discount rate is $v$, then we have $$100000=800\sum_{k=0}^{299}v^k=800\frac{v^{300}-1}{v-1}$$
or $$125(1-v)=1-v^{300},\tag1$$ assuming $v\neq1.$ If $x$ is the per-period interest rate, then $v = \frac 1{1+x}$,and we can rewrite $(1)$ as $$125x(1+x)^{299}=(1+x)^{300}-1$$ so it seems you made a slight error in setting up your equation, or it might be a difference in when the first payment is due.  In my experience the first payment on a mortage is due when the loan is received.  In any event, it will be easier to deal with in the form $(1)$. We rewrite $(1)$ as $$v=\frac{124+v^{300}}{125}\tag2$$  There's no way to solve $(2)$ explicitly, but we can solve it by iteration.  Write $$f(v)=\frac{124+v^{300}}125$$ and pick some starting value for $v$, say $v=0$.  Then $f(v)=f(0)=.992$.  Take this as the new guess for $v$ and compute $f(v)$ again.  $f(.992)=0.9927187727986595$.  Continue in this manner until we find a $v$ such that $f(v)=v$ to sufficiently many decimals.  We quickly arrive at $$v=0.9929610747842791$$ which gives $$x=0.007088822910052306$$  This is the monthly interest rate, so the nominal annual rate on the mortgage will be $12$ times that, or $$0.08506587492062767$$  I imagine that the expected answer is $8.5\%$.

Title: Tricky mortgage rate question


问:
How to find asymptotic expansion of $\frac{1}{\cos(z)}$

Find the asymptotic expantion of coefficients of the exponential generating function $f(z)=\frac{1}{\cos(z)}$ using all of its poles.

My work:

I know that:
$$
\frac{1}{\cos \left( z \right)}=\sum_{n\ge 0}{E_{2n}}\frac{z^{2n}}{\left( 2n \right) !}
$$

To find the poles we solve $\cos(z)=0$ where z is a complex number. Getting, $\chi_k=\pi/2+\pi k$
where $k \in Z$.
Now wrie $\frac{1}{\cos(z)} \sim  \frac{1}{z-\chi_k}= \frac{-1}{\chi_k} \frac{1}{1-z/\chi_k}$.(I'm not sure if this what I am supposed to do).
Then
$$
E_{2n}=-\left( 2n \right) !\sum_{k\in Z}{\frac{1}{\chi _k}}\frac{1}{\chi _{k}^{n}} \\
E_{2n}=-\left( 2n \right) !\left( \frac{1}{\left( \frac{\pi}{2} \right) ^{n+1}}+\sum_{k\ge 1}{\left( \frac{1}{\chi _{k}^{n+1}}+\frac{1}{\chi _{k_-}^{n+1}} \right)} \right)
$$

Then I can look at the internal sum in the last equation and try to simplify it.

Edited:

$f(z)=\frac{1}{cos(z)} \sim \frac{Res(f(z),\chi_k)}{z-\chi_k}$ where $\chi_k=\frac{\pi}{2}+2\pi k$ the
poles of $f(z)$ and  $Res(f(z),\chi_k)= lim_{z\to \chi_k} (z-\chi_k) f(z)$.

After calculations we get that:

$\chi_0= \pi /2$ with Res= -1, $\chi_{-1} = -\pi / 2$ with Res = 1 , $\chi_1=3\pi /2$ with Res = 1, $\chi_{-2}= -3\pi /2$ with Res = -1 etc.

So, $f(z)=\frac{1}{cos(z)} \sim \frac{Res(f(z),\chi_k)}{z-\chi_k} \sim \frac{-1}{z-1/2 \pi}  +\frac{1}{z+1/2 \pi} +\frac{1}{z-3/2 \pi} + \frac{-1}{z+3/2 \pi}+....$

Then if we expand these geometric coulomns and look at their coefficients we get:

$\frac{E_{2n}}{n!} \sim (2/ \pi)^{n+1} + (-2/ \pi)^{n+1} - (2/ 3\pi)^{n+1} +(2/ 3\pi)^{n+1} - (2/ 5\pi)^{n+1} -(-2/ 5\pi)^{n+1}-(2/ 7\pi)^{n+1} +(-2/7\pi)^{n+1}...)$

Thus, $E_{2n} \sim (2n)! (  (2/ \pi)^{n+1} + (-2/ \pi)^{n+1}  - (2/ 5\pi)^{n+1} -(-2/ 5\pi)^{n+1}-(2/ 7\pi)^{n+1} +(-2/7\pi)^{n+1}...)$

Thus $E_{2n} \sim (2n)! (  (2/ \pi)^{n+1} + (-2/ \pi)^{n+1} +O(2/ 5\pi)^{n})$

What do you think.
Any guide will be so appreciated.

Body:
You found that
$$
\frac{1}{{\cos z}} = \sum\limits_{k = 0}^\infty  {\left( {\frac{{( - 1)^k }}{{z + \left( {k + \frac{1}{2}} \right)\pi }} - \frac{{( - 1)^k }}{{z - \left( {k + \frac{1}{2}} \right)\pi }}} \right)}  = \sum\limits_{k = 0}^\infty  {( - 1)^k \frac{{(2k + 1)\pi }}{{\left( {k + \frac{1}{2}} \right)^2 \pi ^2  - z^2 }}} .
$$
This expansion is actually convergent for all $z \neq (k+1/2)\pi$, $k \in \mathbb{Z}$. If $|z|<\frac{\pi}{2}$, then, by the geometric series,
$$
\frac{1}{{\left( {k + \frac{1}{2}} \right)^2 \pi ^2  - z^2 }} = \frac{1}{{\left( {k + \frac{1}{2}} \right)^2 \pi ^2 }}\frac{1}{{1 - \left( {z/\left( {k + \frac{1}{2}} \right)\pi } \right)^2 }} = \sum\limits_{n = 0}^\infty  {\frac{{z^{2n} }}{{\left( {k + \frac{1}{2}} \right)^{2n + 2} \pi ^{2n + 2} }}} .
$$
Thus
\begin{align*}
\frac{1}{{\cos z}} & = \sum\limits_{k = 0}^\infty  {( - 1)^k (2k + 1)\pi \sum\limits_{n = 0}^\infty  {\frac{{z^{2n} }}{{\left( {k + \frac{1}{2}} \right)^{2n + 2} \pi ^{2n + 2} }}} }
\\ &
 = \sum\limits_{n = 0}^\infty  {( - 1)^n \left[ {( - 1)^n \left( {\frac{2}{\pi }} \right)^{2n + 1} 2(2n)!\sum\limits_{k = 0}^\infty  {\frac{{( - 1)^k }}{{(2k + 1)^{2n + 1} }}} } \right]\frac{{z^{2n} }}{{(2n)!}}} ,
\end{align*}
provided $|z|<\frac{\pi}{2}$. Consequently,
$$
E_{2n} = ( - 1)^n \left( {\frac{2}{\pi }} \right)^{2n + 1} 2(2n)!\sum\limits_{k = 0}^\infty  {\frac{{( - 1)^k }}{{(2k + 1)^{2n + 1} }}}
$$
for all $n\geq 0$.


问:
Projective objects in abelian categories having non-trivial morphisms
I'm trying to read an article about $p$-adic groups (based on the lectures of Joseph Bernstein), and I'm struggling to understand a certain argument regarding projectives objects in abelian categories wich goes like this:
let $X$ and $Y$ be two objects in an abelian category $M$, such that they are both projective objects and $X$ is a projective generator. Then $\hom(Y,X)$ is non-trivial.

If you could provide an explanation with as little categorical jargon as possible I would be grateful as my knowledge regarding category theory is quite limited.

答:
Presumably you want to assume that $Y$ is nonzero.

But even then, this is not true in a general abelian category.

For example, take the opposite category of the category of abelian groups. Take $X=\mathbb{Q}/\mathbb{Z}$ and $Y=\mathbb{Q}$. Both are injective in the category of abelian groups, and so projective in the opposite category. And $X$ is an injective cogenerator in the category of abelian groups, and so a projective generator in the opposite category.

But for $\operatorname{Hom}(Y,X)$ to be nontrivial in the opposite category, we need $\operatorname{Hom}(X,Y)$ to be nontrivial in the category of abelian groups. But there are no nontrivial group homomorphisms $\mathbb{Q}/\mathbb{Z}\to\mathbb{Q}$, since $\mathbb{Q}/\mathbb{Z}$ is a torsion group, and $\mathbb{Q}$ is torsion-free.

Maybe you are missing some assumption on the abelian category?


问:
How Can I Get This Quantification Deduction?
This is a question in my logic class.

Premises:

$(\exists x)(Px \land Lxa)$

$(y)(Py \supset Lay)$

$(x)(y)[(Lxa \land Lay) \supset Lxy]$

Deduce:

$(\exists x)[Px \land (y)(Py \supset Lxy)]$

So far what occurs to me is that EI 1,

$$Pu \land Lua$$ then UI 2,

$$Pu \supset Lau$$

Through MP I can get $$Lau$$
Since I have $Lau$ and $Lua$, So I can UI 3 and get $$Luv$$
Here's where I'm stuck at. Since $u$ showes up in a EI line, I can't use UG to get $$(y)(Py\supset Lxy)$$

Any idea? By the way, is there any general tips to these questions? Thanks!

答:
In yet a different proof system+format+notation, that of Feijen/Dijkstra/Gries/etc., we are given that

$\langle \exists x : P.x : L.x.a \rangle$

$\langle \forall y : P.y : L.a.y \rangle$

$\langle \forall x,y : L.x.a \land L.a.y : L.x.y \rangle$

and we are asked to prove

$\langle \exists x : P.x : \langle \forall y : P.y : L.x.y \rangle \rangle$

The proof is a simple calculation:
$$
\begin{align}
& \langle \exists x : P.x : \langle \forall y : P.y : L.x.y \rangle \rangle \\
\Leftarrow & \;\;\;\;\;\text{"weaken using 3 -- this is the only thing we know about $L.x.y$"} \\
& \langle \exists x : P.x : \langle \forall y : P.y : L.x.a \land L.a.y \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"logic: simplify by taking $L.x.a$, which does not contain $y$, outside of $\forall y$"} \\
& \langle \exists x : P.x : L.x.a \land \langle \forall y : P.y : L.a.y \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"logic: simplify by taking $\forall y$, which does not contain $x$, outside of $\forall x$"} \\
& \langle \exists x : P.x : L.x.a \rangle \land \langle \forall y : P.y : L.a.y \rangle \\
\equiv & \;\;\;\;\;\text{"using 1 and 2"} \\
& \textrm{true}
\end{align}
$$


问:
How to find $v_4$ such that $span\{v_1, v_2, v_3, v_4\}=\mathbb{R}^4$?
I am in trouble with these exercise: Consider the vectors $v_1=(1, 0, 0, 1), v_2=(2, h, h, 2), v_3=(1, 1+h, 2h +1, 1)$ with $h\in\mathbb{R}$. Determine the dimension of the subspace $S$ generated by $v_1, v_2, v_3, v_4$ by varying $h\in\mathbb{R}$. Thus, choosen $h\in\mathbb{R}$ such that $dim(S) =3$, find $v_4$ such that $span\{v_1, v_2, v_3, v_4\} = \mathbb{R}^4$.

In order to find the dimension of the generated subspace $S$ by varying $h\in\mathbb{R}$, I evaluate

$\begin{vmatrix}
1 &2&1\\
0&h &1+h\\
0& h&2h+1\\
1&2&1
\end{vmatrix}$

I observe that $rank(A) = 3$ for all $h\neq 0$ and if $h=0, rank(A) = 2$. So (I guess) the answer to the first question is that $dim(S) =3$ for all $h\neq 0$.

How to proceed to find $v_4$ such that $span\{v_1, v_2, v_3, v_4\}=\mathbb{R}^4$?
Could anyone please help me?

Thank you in advance!

答:
Let $W$ be linear space of finite dimension $n$ with $\{w_1, \dots , w_n\}$ a basis of $W$.

It is a theorem that for a linear subspace $V$ spanned by $\{v_1, \dots, v_m\}$ with $m \lt n$, you can pickup $w \in \{w_1, \dots , w_n\}$ such that $\{v_1, \dots, v_m, w\}$ is linearly independent.

So in your case, take the canonical basis $\{e_1, \dots e_4\}$. If $h \neq 0$, $\{v_1, v_2, v_3\}$ are linearly independent. You can pick up one of the vectors $e_i$ of the canonical basis in a way to get a basis $\{v_1,v_2,v_3,e_i\}$ of $\mathbb R^4$. Just test them one after the other... one will work!

In fact,

$$\begin{vmatrix}
1 & 1 &2&1\\
0 & 0&h &1+h\\
0 & 0& h&2h+1\\
0& 1&2&1
\end{vmatrix} = \begin{vmatrix}
0&h &1+h\\
0& h&2h+1\\
1&2&1
\end{vmatrix} = \begin{vmatrix}
h &1+h\\
h&2h+1
\end{vmatrix} = h^2 \neq 0$$

So you already win with $e_1$ and $\{v_1,v_2,v_3,e_1\}$ is a basis of $\mathbb R^4$.


问:
Apparently in this form the limit is "trivially" 1/2
I'll admit, I'm struggling to keep up with the material in my analysis classes, but it only demoralizes me more when the questions in my textbooks have explained solutions, but I don't even understand how you go from one step to another.

This is the final from, that's supposed to be close to the simplest form (so close that it's "trivial" for the student to find it):

$\lim _{n\to \infty }\left(\frac{\left(c+1\right)n^c-n^{c+1}+\left(n-1\right)^{c+1}}{\left(c+1\right)\left(n^c-\left(n-1\right)^c\right)}\right)=\frac{1}{2}$

Why is this true? I don't see it at all.

答:
The $n^{c+1}$ and $n^c$ powers in both halves of the fraction are zero by design. The highest non-zero power is $n^{c-1}$:
$$\frac{\binom{c+1}2n^{c-1}+\cdots}{(c+1)\binom c1n^{c-1}+\cdots}$$
As $n\to\infty$ these terms will dominate, so the limit is
$$\frac{\binom{c+1}2}{(c+1)\binom c1}=\frac12$$


问:
Is it possible to find an exact solution for $x$ in $\frac{x^2+9}{6 x}=\frac{\sin (\pi -1) x}{\cos \pi x}$?
I have this equation for $x>0$
$$\frac{x^2+9}{6 x}=\frac{\sin (\pi -1) x}{\cos \pi  x}$$
and I want to know if there is an xeact solution of $x$ in this equation.

答:
Here, I have a sort of "moral" argument for why there should be no poles. To begin, if there is a closed for for solutions to

$$\frac{(x^2+9)\cos(\pi x)}{x^2}=6\sin(\pi-1)$$

(which is equivalent to your problem) then we should expect that this neat formula will also give solutions to the equations

$$\frac{(x^2+9)\cos(\pi x)}{x^2}=c$$

for any constant $c$ since the constant $6\sin(\pi-1)$ is not nice at all and I cannot conceive of a formula that would solve this equation for only the value $c=6\sin(\pi-1)$. This means that we are looking for a "nice" function $f(c)$ such that

\begin{equation}
\frac{(f^2(c)+9)\cos(\pi f(c))}{f^2(c)}=c\tag{1}
\end{equation}

If $f(c)$ is a nice function, then surely a derivative of this function must exist (otherwise, I would not qualify $f(c)$ as a nice function). Thus,

$$\left(\frac{(f^2(c)+9)\cos(\pi f(c))}{f^2(c)}\right)^{'}=1$$

taking the derivative, we can then substitute in from (1) to write $\cos(\pi f(c))$
and $\sin(\pi f(c))$ in terms of a polynomial in $f(c),c$. Overall, this yields that

$$-18cf\left(c\right)^{-3}f'\left(c\right)=1+\frac{9}{f\left(c\right)^{2}}+\pi f'\left(c\right)\left(1+\frac{9}{f\left(c\right)^{2}}\right)^{2}\sqrt{1-\frac{c^{2}f\left(c\right)^{4}}{\left(f\left(c\right)^{2}+9\right)^{2}}}$$

Which, after expanding, is a polynomial in $f$ and $f'$. Apart from that, this differential equation has no special symmetry. It has been shown that general polynomials in $f$,$f'$, and $c$ have no closed form solution and so since we have no reason to expect that something special is going on here I would say that it is highly unlikely that your equation has a closed form. Not impossible, but definitely improbable.


问:
Graph of a morphism of affine schemes
Let $f:X\to Y$ be a morphism of schemes (here smooth, separated, of finite type over a field). Then the graph $\Gamma_f$ is defined as the image of the morphism $(id,f):X\to X\times Y$.

Claim: The structure sheaf of the graph is a coherent $\mathcal O_{X\times Y}$-module.

I tried to convince myself of this by considering the affine case first and got embarassingly confused:

So if $X=Spec\ (R)$ and $Y=Spec\ (S)$, then $X\times Y=Spec\ R\otimes S$. Let $f':S\to R$ be the corresponding ring homomorphism, then the graph should be just $\{s\otimes f'(s)\}$. And I guess I did something wrong here, because this is certainly no $S\otimes R$-module. The module generated by the set $\{s\otimes f'(s)\}$ on the other hand is the whole ring, because $f'(1)=1$.

It is easy to see that the graph is a closed subscheme, so it should be possible to write it as a quotient of $R\otimes S$.

答:
The graph $\Gamma_f\subset X\times Y$ is a closed subscheme of $X\times Y$ and as such corresponds to an ideal  of the ring $R\otimes_k S$, as you very correctly stated.

That ideal is the kernel $I$ of the ring morphism $$ R\otimes_k S\to R: r\otimes s\mapsto rf'(s) \quad (\star) $$The structure sheaf $\mathscr O_{\Gamma_f} $ of the graph is then the $\mathscr O_{X\times_kY}$- module $\widetilde {\frac {R\otimes_k S}{I}}$ corresponding to the $R\otimes_k S $- module $\frac {R\otimes_k S}{I}$  and is thus coherent.

 [Because $R\otimes_k S$ is noetherian. Beware that the structure sheaf of a scheme, even an affine one, needn't be coherent in the absence of noetherianness]

Note that the first projection $X\times_kY\to X$ restricts to an isomorphism $\Gamma_f \stackrel {\cong}{\to} X$ corresponding to the inverse of the isomorphism $\frac {R\otimes_k S}{I} \stackrel {\cong}{\to}  R $ deduced from $(\star) $.

Title: Graph of a morphism of affine schemes


答:
Compute the leading terms of binomial expansions in numerator and denominator and simplify

$$
\begin{align}
& \lim _{n\to \infty }\left(\frac{\left(c+1\right)n^c-n^{c+1}+\left(n-1\right)^{c+1}}{\left(c+1\right)\left(n^c-\left(n-1\right)^c\right)}\right)= \\
& = \lim _{n\to \infty }\left(\frac{\left(c+1\right)n^c-n^{c+1}+n^{c+1}-\binom{c+1}1 n^c + \binom{c+1}2 n^{c-1} + O(n^{c-2})}{\left(c+1\right)\left(n^c-n^c + \binom{c}1 n^{c-1}+O(n^{c-2})\right)}\right)\\
& = \lim _{n\to \infty }\left(\frac{\left(c+1\right)n^c-n^{c+1}+n^{c+1}-(c+1) n^c + \frac{c(c+1)}{2} n^{c-1} + O(n^{c-2})}{\left(c+1\right)\left(n^c-n^c + c n^{c-1}+O(n^{c-2})\right)}\right)\\
& = \lim _{n\to \infty }\left(\frac{\frac{c(c+1)}{2} n^{c-1} + O(n^{c-2})}{c\left(c+1\right)n^{c-1}+O(n^{c-2})}\right)\\
& = \frac{c(c+1)}{2}\frac{1}{c(c+1)} \\
& = \frac{1}{2}.
\end{align}
$$

Title: Apparently in this form the limit is "trivially" 1/2

Body:
I'll admit, I'm struggling to keep up with the material in my analysis classes, but it only demoralizes me more when the questions in my textbooks have explained solutions, but I don't even understand how you go from one step to another.

This is the final from, that's supposed to be close to the simplest form (so close that it's "trivial" for the student to find it):

$\lim _{n\to \infty }\left(\frac{\left(c+1\right)n^c-n^{c+1}+\left(n-1\right)^{c+1}}{\left(c+1\right)\left(n^c-\left(n-1\right)^c\right)}\right)=\frac{1}{2}$

Why is this true? I don't see it at all.


问:
Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$
Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?

$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$

Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell without touching our calculators?

答: $\sqrt x$ is a concave function, therefore $\alpha \sqrt x + (1-\alpha) \sqrt y < \sqrt {x+y }.$  Substitute $\alpha = \frac12, x=999, \text{and } y=1001$</p>


问:
How do you know if the graph starts from above or below?
I can't seem to find a way to know if a graph starts from below or above. By that I mean what is pictured below (where the graph starts from the top left).

For example, in this graph below where the equation is as in the picture, I know that the x values are 0, 1 and 3 but I'd have thought that the graph would start from the bottom left quadrant due to the fact that the x is positive.
Could someone please explain why in this case it starts from the top left and how to know for other similar cases.

Any help is appreciated, thank you.

答:
Let's get a few simple examples.

Notice that odd powers have one behaviour at the ends and even powers have the opposite behaviour.  This continues for higher powers.

You want to graph
$$  x^2(x-(\text{don't care}))(x-(\text{don't care})) \\
= x^4 + (\text{don't care})  \text{.}  $$
So we expect even power behaviour.

This is exactly what we do when the leading coefficient is positive.  When it is negative, flip the graphs upside-down.


问:
Show that $\mathbb{R}|_{\sim}$ is uncountable for $\sim (x, y) : y - x \in \mathbb{Q}$
I'm self-learning some measure theory. During the following lecture ( I had included the appropriate starting time inside the link, but timespan is 7:50-9:00):

https://youtu.be/llnNaRzuvd4?t=470

the Professor Claudio Landim states that if we take $\Lambda = \mathbb{R}|_{\sim}$ where $\sim (x, y)  :  y - x \in \mathbb{Q}$ then $\Lambda$ is uncountable. The lecturer explains that if indeed $\Lambda$ was countable, then every element in $\mathbb{R}$ could be represented by element of $\Lambda$ (some equivalence class) and by element of this equivalence class. Couldn't it? Why not? I cannot understand it.

答:
Your notations are not so clear. $$\mathbb R/_\sim=\{[x]\mid x\in \mathbb R\},$$
and $$[x]=\{y\in \mathbb R\mid y-x\in \mathbb Q\}.$$

So, what you want to prove is that $\Lambda :=\mathbb R/_\sim$ is uncountable. Suppose $\Lambda $ is countable, i.e. $\mathbb R/_\sim=\{[x_n]\}_{n\in\mathbb N}$ for some $x_n$. In one hand, if $x\in \mathbb R$, then $[x]=x+\mathbb Q$ is countable. In the other hand, if $\Lambda $ is countable, then $\mathbb R$ is a countable union of countable set, and thus $\mathbb R$ would be countable, which is a contradiction.


问:
How to compute the second derivatives?
Motivation:

In isogeometric analysis, state variables(e.g. displacement) are defined in the parametric domain, which can be mapped to the physical domain by $\boldsymbol{\xi}\mapsto \boldsymbol{x}$ as shown beneath. However the quantity related to displacment such as stress, strain are spacial derivatives of displacement. The following procedure is commonly used for solving those derivatives. $\blacksquare$

Know

\begin{equation}
u(\xi,\eta) = \sum_{i} c_iN^i(\xi,\eta)
\end{equation}
with
$$x(\xi,\eta) = \sum_{i} x_i N^i(\xi,\eta), \quad y(\xi,\eta) = \sum_{i} y_i N^i(\xi,\eta),$$
where $c_i,x_i,y_i$ are constants, with assumption that $(\xi,\eta)\mapsto(x,y)$ is bijective, i.e. inverse exists, $$J :=[\frac{\partial x_i}{\partial \xi_j}],\: |J| \neq 0\quad (\text{where }x_2 = y,\,\xi_2 = \eta).$$

By chain rule,
$$\frac{\partial u}{\partial \xi_j} = \frac{\partial u}{\partial x_i}\frac{\partial x_i}{\partial \xi_j}$$

or

$$
\begin{bmatrix}
\frac{\partial u}{\partial \xi}\\
\frac{\partial u}{\partial \eta}
\end{bmatrix}
=
\begin{bmatrix}
\frac{\partial x}{\partial \xi} &
\frac{\partial y}{\partial \xi}\\
\frac{\partial x}{\partial \eta} &
\frac{\partial y}{\partial \eta}
\end{bmatrix}
\begin{bmatrix}
\frac{\partial u}{\partial x}\\
\frac{\partial u}{\partial y}
\end{bmatrix}
 = J^T
\begin{bmatrix}
\frac{\partial u}{\partial x}\\
\frac{\partial u}{\partial y}
\end{bmatrix}.
$$

Hence,
$$
\begin{bmatrix}
\frac{\partial u}{\partial x}\\
\frac{\partial u}{\partial y}
\end{bmatrix}
=
(J^T)^{-1}
\begin{bmatrix}
\frac{\partial u}{\partial \xi}\\
\frac{\partial u}{\partial \eta}
\end{bmatrix}
$$

However, now I need to compute the second derivatives $$\frac{\partial^2 u}{\partial x_i\partial x_j},\quad i \text{ and }j \in \{1,2\}.$$

One may think displacement $\boldsymbol{u} = [u, v]^T$, then the goal is to compute
$$\nabla(\nabla\boldsymbol{u})$$ with information as above.

答:
I think total differential concept is very useful notation for your derivation. Let
$$u\big(x(\xi,\eta),x(\xi,\eta)\big) $$
Then it follows that
$$du=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \xi}d\xi+\frac{\partial u}{\partial x}\frac{\partial x}{\partial \eta}d\eta+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \xi}d\xi+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \eta}d\eta$$
$$\Rightarrow du=\bigg(\frac{\partial u}{\partial x}\frac{\partial x}{\partial \xi}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \xi}\bigg)d\xi+\bigg(\frac{\partial u}{\partial x}\frac{\partial x}{\partial \eta}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \eta}\bigg)d\eta$$
and you can use it for partial derivatives to construct your Jacobian
$$d\eta=0\Rightarrow\frac{\partial u}{\partial \xi}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \xi}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \xi}$$
$$d\xi=0\Rightarrow\frac{\partial u}{\partial \eta}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \eta}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \eta}$$
For second order
$$d^2u=d\bigg(\frac{\partial u}{\partial x}\frac{\partial x}{\partial \xi}\bigg) d\xi+d\bigg(\frac{\partial u}{\partial x}\frac{\partial x}{\partial \eta}\bigg)d\eta+d\bigg(\frac{\partial u}{\partial y}\frac{\partial y}{\partial \xi}\bigg)d\xi+d\bigg(\frac{\partial u}{\partial y}\frac{\partial y}{\partial \eta}\bigg)d\eta$$
where
$$d\bigg(\frac{\partial u}{\partial x}\frac{\partial x}{\partial \xi}\bigg)=d\bigg(\frac{\partial u}{\partial x}\bigg)\frac{\partial x}{\partial \xi}+\frac{\partial u}{\partial x}d\bigg(\frac{\partial x}{\partial \xi}\bigg)$$
$$=\frac{\partial^2 u}{\partial x^2}\bigg(\frac{\partial x}{\partial \xi}d\xi+\frac{\partial x}{\partial \eta}d\eta\bigg)\frac{\partial x}{\partial \xi}+\frac{\partial^2 u}{\partial x \partial y}\bigg(\frac{\partial y}{\partial \xi}d\xi+\frac{\partial y}{\partial \eta}d\eta\bigg)\frac{\partial x}{\partial \xi}+\frac{\partial u}{\partial x}\bigg(\frac{\partial^2 x}{\partial \xi^2}d\xi+\frac{\partial^2 x}{\partial \xi\partial\eta}d\eta\bigg)$$
$$=\bigg(\frac{\partial^2 u}{\partial x^2}\bigg(\frac{\partial x}{\partial \xi}\bigg)^2+\frac{\partial^2 u}{\partial x \partial y}\frac{\partial y}{\partial \xi}\frac{\partial x}{\partial \xi}+\frac{\partial u}{\partial x}\frac{\partial^2 x}{\partial \xi^2}\bigg)d\xi+\bigg(\frac{\partial^2 u}{\partial x^2}\frac{\partial x}{\partial \xi}\frac{\partial x}{\partial \eta}+\frac{\partial^2 u}{\partial x \partial y}\frac{\partial y}{\partial \eta}\frac{\partial x}{\partial \xi}+\frac{\partial u}{\partial x}\frac{\partial^2 x}{\partial \xi\partial \eta}\bigg)d\eta$$
You can calculate the other three differentials and find the total second order differential. Again by setting $d\eta$ and $d\xi$ to zero you can calculate second order partial differentials. Remember it's not an easy task...</p>

Title: How to compute the second derivatives?


问:
$G$ is an abelian group and $x=a_1a_2...a_n$
Assume that $G$ is an abelian group of order $n$ and $a_1,a_2,...,a_n$ are its elements and we have $x=a_1a_2...a_n$.

If $G$ has more than one element such as $b\neq e$ and $b^2=e$, prove that $x=e$.

Answer:

Assume that $a\neq b$, $a^2=e$, $b^2=e$, $a\neq e$ and $b\neq e$.
$$(ab)^2=(ab)(ab)=a^2b^2=e$$
If $$ab=e\implies a=b^{-1}\implies a=b$$
But we assume that $a\neq b$. contradiction!

So, $ab \neq e$ and $(ab)^2=e$.

So, what should I do now?

STUCK!

答:
$\newcommand{\Set}[1]{\left\{ #1 \right\}}$The main point is what happens in the subgroup
$$
H = \Set { x \in G : x^{2} = e },
$$
which by assumption is a vector space of dimension $k$ over the field $F$ with $2$ elements.

Write the elements of $H$ with respect to a (multiplicative) basis. Looking at the exponents with respect to any given basis elements $b$, half of the elements of $H$ will have exponent $0$, and half will have exponent $1$. But half of $2^{k}$ is $2^{k-1}$, which is even as by assumption $k > 1$, so the sum $s$ of the exponents will be even, so that $b^{s} = e$.


问:
Importance and Intuition of Polynomial Rings
I know that a polynomial ring $R[x]$ is the ring with elements consisting of polynomials with coefficients in $R$. However, this definition leaves me confused when I try to really understand the concept of polynomial rings, rather than just accept that definition for what it is. Where did the polynomials come from? Why are polynomial rings important? Is there a way to “construct” the polynomials in the context of Abstract Algebra, so that we may see why studying rings made up of polynomials is important? Are polynomial rings actually just sets of polynomials, or are the polynomials just a concrete way to represent some more abstract idea? What does the variable $x$ mean in polynomial rings? I apologize for all the questions, I’m just having trouble seeing the importance or intuition behind them.

答:
Great question! Polynomial rings really get glossed over in my opinion, when they are actually quite complicated objects.

Super formally, we define the polynomial ring $R[x]$ as follows. Let $S$ be the set of all sequences $(r_0, r_1, r_2,\ldots)$ where the $r_i$ are elements of $R$ and only finitely many are nonzero. We explicitly define operations $+$ and $\cdot$ on this set by
$$
(a_0, a_1, a_2,\ldots) + (b_0, b_1, b_2, \ldots) = (a_0 + b_0, a_1 + b_1, a_2 + b_2, \ldots)
$$
and
$$
(a_0, a_1, a_2, \ldots) \cdot (b_0, b_1, b_2, \ldots) = (a_0b_0, a_1b_0 + a_0b_1, a_2b_0 + a_1b_1 + a_0b_2, \ldots).
$$
That is, if $a = (a_i)$ and $b = (b_i)$, then $(a+b)_i = a_i + b_i$ and $(ab)_i = \sum_{k=0}^i a_kb_{i-k}$.

It is easy to check that $(S, +, \cdot)$ is a ring with zero $(0, 0, 0, \ldots)$ and identity $(1, 0, 0, \ldots)$.

Furthermore, as an $R$-algebra, $S$ is generated by the element $x = (0, 1, 0, 0, \ldots)$ (if you don't know what an $R$-algebra is, the point is that we can "get to" every element of $S$ using just the element $x$ and the elements of $R$). To see this, note that for each $n$, the element $x^n$ is the sequence $(0, \ldots, 0, 1, 0, \ldots)$ with a $1$ in the $n^\text{th}$ entry and zeros elsewhere. The element $(a_0, a_1, \ldots)$ of $S$ is equal to
$$
a_0 + a_1x + a_2x^2 + \ldots.
$$
Thus we can "get to" $(a_0, a_1, \ldots)$ just by raising $x$ to various powers and multiplying by elements of $R$.

We denote this ring by $R[x]$. The intuition is that we are "adjoining" an "indeterminate" $x$ to the ring $R$, which means that we are adding some element that has no constraints on it. In some sense, the element $x$ is "free". In practice, this is how we always think about polynomials.

Polynomial rings are extremely important. Without them, mathematics would basically not be possible. The most ubiquitous example I can think of is in Linear Algebra, where the theory of polynomial rings allows us to prove results like the Cayley-Hamilton Theorem. Another example is in the theory of probability generating functions. Polynomials are also vital to number theory and geometry. Every field of mathematics I can think of is built at least implicitly on the theory of polynomials.

Edit I got a bit carried away with advanced topics in my examples. For a more basic one, we need the theory of polynomial rings to understand even basic things like factorising quadratics. To prove that factorisation exists and is unique, you need to understand the ring $\mathbb{R}[x]$. Factorising such polynomials is very useful, as any high school student will tell you.


问:
Limit: $\lim_{x \to 0}{[1+\ln{(1+x)}+\ln(1+2x)+\ldots+\ln(1+nx)]}^{1/x}.$
How can I find the following limit?

$$\lim_{x \to 0}{[1+\ln{(1+x)}+\ln(1+2x)+\ldots+\ln(1+nx)]}^{1/x}.$$

It's a limit of type $\displaystyle 1^{\infty}$ and if I note with $\displaystyle f(x)=\ln{(1+x)}+\ln(1+2x)+\ldots+\ln(1+nx)$ then I must do:

$$\displaystyle \lim_{x \to 0}{(1+f(x))}^{1/x}=\lim_{x \to 0}[(1+f(x))^{\frac{1}{f(x)}}]^{\frac{f(x)}{x}}. $$

Is ok? Thanks :)

答:
Write the expression above as

$$\lim_{x \to 0}\left [ 1 + \sum_{k=1}^n \log{(1+k x)}\right ]^{1/x}$$

Note that $n$ remains finite, so we can Taylor expand the logs to get

$$\lim_{x \to 0}\left [ 1 +\sum_{k=1}^n k x\right]^{1/x} = \lim_{x \to 0}\left(1+\frac12 n (n+1) x\right)^{1/x} = e^{n(n+1)/2}$$

Note that the error term in the Taylor expansion is $O(x^2)$, which in this limit produces an additional term as

$$(1+n(n+1)/2 x + c x^2)^{1/x} \sim  \left(1+\frac12 n (n+1) x\right)^{1/x} \left ( 1+\frac{c x^2}{1+n (n+1) x/2}\right )^{1/x} \sim\left(1+\frac12 n (n+1) x\right)^{1/x} e^{c x}$$

The error is then

$$(1+n(n+1)/2 x + c x^2)^{1/x} - (1+n(n+1)/2 x )^{1/x} = \left(1+\frac12 n (n+1) x\right)^{1/x} (e^{c x}-1)$$

As $x \to 0$, the error $\sim c x$.


问:
$101$-th power of three-digit integer $n$ ends with the same three digits of $n$
Prove that if a three-digit integer $n$ is relatively prime to $10$ then the $101$-th power of $n$ ends with the same three digits of $n$.

I tried using modular arithmetic but could only get

$$n^4 \equiv 1 \pmod{10}$$

答:
By Fermat's Little Theorem, if $n$ is relatively prime to $10$, then it is relatively prime to $125$, and
\begin{equation}
n^{\phi(125)} \equiv 1 \,(mod \, 125) \,.
\end{equation}
$\phi(125) = 100$, so we have $n^{101} \equiv n \,(mod \,125)$.
Likewise, we can look at $n^{\phi(8)} \equiv 1 \,(mod \,8)$ to get that $n^4 \equiv n^{100} \equiv 1 \,(mod \,8)$ and $n^{101} \equiv n \,(mod \, 8)$.

Together, they imply $n^{101} \equiv n \,(mod \,1000)$, which is what you want.


问:
$a < {\sqrt 3}b$ with certain conditions
Given two integers $a > b > 1$ such that $(a+b)|(ab+1)$ and $(a-b)|(ab-1)$. Prove that $a <$ $\sqrt 3$$b$. My approach was to firstly look for some inequalities but I couldn't come up with any useful. Then I tried to plug some numbers and I found out that if $a$ and $b$ differ by $2$ and both are $odd$ the conditions are satisfied, then I tried to look for some other $a$,$b$ that differ by $2n$ but I couldn't find any. So I think that it is the case in this problem, but I can't proceed. Any hints and was my approach correct so far? Any help appreciated.

答: First of all, we conclude that $a+b\mid ab+1\implies $gcd$(a,b)=1$. Also, observe that $$a+b\mid b\cdot (a+b)-(ab+1)=b^2-1\qquad \text{and} \qquad a-b\mid (ab-1)-b\cdot (a-b)=b^2-1$$ Besides, notice that, since gcd$(a,b)=1$, we also have gcd$(a+b, a-b)=\text{gcd}(2a, 2b)=1$ or $2$. This yields $$(a+b)(a-b)\mid 2\cdot (b^2-1)\implies a^2-b^2\leqslant 2(b^2-1)<2b^2\implies a<\sqrt3 b$$


问:
Is this set of matrices a vector space?
I was wondering if this set of matrices is a vector space:

$Q=\begin{bmatrix}u&v\\ -\overline v&\overline u\end{bmatrix} Such\space that\space u,v\in\mathbb{C}$

is it a vector space on field $\mathbb{R}$?

is it a vector space on field $\mathbb{C}$?

I have tried to prove it wrong but it didn't work, so I think they are both vector spaces, I have tried to prove that it's closed to addition and multiplying by scalar, and it seems fine but i'm not sure with it, because from what I see, no matter what I do (addition or multiplying by a scalar) from both fields, the vectors $u,v$ will stay in the complex field.

It would be really appreciated if someone can approve my thinking if its right, or tell me whats wrong if its wrong, thanks in advance.

答: Hint: If you multiply by $i$, will the diagonal elements still be complex conjugates?

Title: Is this set of matrices a vector space?


问:
Question on step of existing proof regarding area of disk
The following proof is from Apostol Calculus V1 Page 91.  I'm unsure how near the end of the proof:

$2r \int^1_{-1} \sqrt{r^2 - (rx)^2} dx = 2r^2 \int^1_{-1} \sqrt{1 - x^2} dx$.

How is $r * r$ on the LHS?

I'm using an image because I'm concerned the OCR may have caused on error.

答:
They factored the $r$ out of the square root, and then out of the integral because it is constant with respect to $x$:
\begin{align*}
2r \int^1_{-1} \sqrt{r^2 - (rx)^2} dx &= 2r\int_{-1}^{1}\sqrt{r^{2}(1 - x^2})\,dx\\
&=2r\int_{-1}^{1}\sqrt{r^2}\sqrt{1-x^2}\,dx\\
&=2r\int_{-1}^{1}r\sqrt{1-x^2}\,dx\\
&=2r^2\int_{-1}^{1}\sqrt{1-x^2}\,dx
\end{align*}


答: An $r$ has been factored out of  the integrand.  That is, $\sqrt{r^2-(rx)^2}=\sqrt{r^2(1-x^2)}=r\sqrt{1-x^2}$.

Title: Question on step of existing proof regarding area of disk

Body:
The following proof is from Apostol Calculus V1 Page 91.  I'm unsure how near the end of the proof:

$2r \int^1_{-1} \sqrt{r^2 - (rx)^2} dx = 2r^2 \int^1_{-1} \sqrt{1 - x^2} dx$.

How is $r * r$ on the LHS?

I'm using an image because I'm concerned the OCR may have caused on error.


问:
entire bijection of $\mathbb{C}$ with 2 fixed points
Besides the identity map, is there an entire function $f$ that is a bijection from $\mathbb{C}$ to $\mathbb{C}$ and has 2 fixed points? Thank you for the help.</p>

答:
Since invoking Picard seems like a bit of an overkill, here's a simple argument (essentially the same as Chris's answer, but replacing Picard by Casorati-Weierstrass and the open mapping theorem):

Claim. If $f: \mathbb{C} \to \mathbb{C}$ is entire and injective then $f$ is linear and non-constant, that is $f(z) = az + b$ for some $a \neq 0$.

It is easy to check that a linear function with two fixed points must be the identity, so we are left with proving the claim.

Proof of the claim. By hypothesis $f: \mathbb{C}^{\times} \to \mathbb{C}$ is injective where $\mathbb{C}^{\times} = \{z \neq 0\}$ is the set of non-zero complex numbers. Now $g(z) = f(1/z)$ is also injective $\mathbb{C}^{\times} \to \mathbb{C}$ and has a singularity at $0$. By injectivity, $g(\{|z| \gt 1\}) \cap g(\{0 \lt |z| \lt 1\}) = \emptyset$. On the other hand, both sets are open by the open mapping theorem, and by the Casorati-Weierstrass theorem $g(\{0 \lt |z| \lt 1\})$ would be dense if $g$ had an essential singularity at $0$. Therefore $g$ has a pole at $0$. But this implies that $f$ is a polynomial and hence it is linear and non-constant by injectivity.


问:
Show that $c$ is closed in $l^{\infty}$
Let $$c=\{ (a_i)_{i \in \mathbb{N}} \ ; \ \ a_i \in \mathbb{R}\ ,\ \forall i \in \mathbb{N} \ , \ \mbox{exist} \ \displaystyle \lim_{i \to \infty}(a_i)\}$$
 $$l^{\infty}=\{ (a_i)_{i \in \mathbb{N}} \ ; \ a_i \in \mathbb{R}\ ,\ \forall i \in \mathbb{N}; \  \displaystyle \sup_{i \in \mathbb{N}}|a_i|< \infty \}.$$ Show that $c$ is closed in $l^{\infty}$.

i am trying... taking a sequence of sequence in $c$ but i can't...i show that $c_0$ is closed in $l^{\infty}$.

答:
Hint 1: Convergence in $\ell_\infty$ and as a consequence in $c$ is a uniform convergence on $\mathbb{N}$

Hint 2: If $\{a^{(n)}:n\in\mathbb{N}\}\subset c$ is a uniformly convergent sequence, then
$$
\lim\limits_{i\to\infty}\lim\limits_{n\to\infty}a_i^{(n)}=
\lim\limits_{n\to\infty}\lim\limits_{i\to\infty}a_i^{(n)}
$$

Title: Show that $c$ is closed in $l^{\infty}$


问:
Push Forwards and Change of Variable Confusion
Taking a look at the Push-Forward measure and change of variable formula https://en.wikipedia.org/wiki/Pushforward_measure I thought that for a measure $\gamma$ on $\mathbb{R}^d$ (with density also denoted $\gamma$) and a map $f:\mathbb{R}^d\to \mathbb{R}^d$, and $g:\mathbb{R}^d \to \mathbb{R}$

$$ \int_{\mathbb{R}^d} g(f_{\#}\gamma(x)) df_{\#}\gamma = \int_{\mathbb{R}^d} g(\mu(x)) d\gamma. $$

In particular denoting Bolztman entropy as $H(\gamma):=\int \log\gamma(x) d\gamma$ this would imply

$$H(f_{\#}\gamma)-H(\gamma)=0~?$$

答:
You really need to sort out your notation and typos.

You write:

$$ \int_{\mathbb{R}^d} g(f_{\#}\gamma(x)) df_{\#}\gamma = \int_{\mathbb{R}^d} g(\mu(x)) d\gamma. $$

I'm assuming that $\mu$ is $\gamma$ and that's just a typo.

With that reading the right hand side $\int_{\mathbb{R}^d} g(\gamma(x)) d\gamma $ could make sense if $\gamma$ is both density and measure (bad notation). Then $\gamma(x)$ is the value of the density function, and $\gamma(x)\in \mathbb{R}$. But what is $g(\gamma(x))$? Your $g$ is of a wrong type for this to make sense.
Do you want $g$ to be from $\mathbb{R}\to \mathbb{R}$? Or do you just want to write  $g(x)$? Presumably the later.

The left hand side $ \int_{\mathbb{R}^d} g(f_{\#}\gamma(x)) df_{\#}\gamma$ has even more problems. Here $f_{\#}$ appears twice. Once as a pushforward of measure $\gamma$, and once, presumably, as pushforward of function $\gamma$ (?). The problem is, functions don't push forward, they pull  back: for a function $h(y)$ we have  $f^*(h)(x)=h(f(x))$. The whole expression is ill-formed.

Compare with what is written in Wiki:

$$\int_{X_2} h \, d(f_* \mu) = \int_{X_1} h \circ f \, d\mu$$

Here $h:\mathbb{R^d} \to \mathbb{R}$. Let $g(x)=h(f(x))$ for some $h(y)$. Then we get:

$$\int_y h(y) d(f_* \mu)=\int_x g(x) d \mu$$

Now if $g(x)=\log p(x)$ where $p(x)$ is the density of $\mu$, then the right hand side is the entropy $H(\mu)$. The left hand side is

$$\int_y h(y) d(f_* \mu)$$.

Assuming that $f$ is invertible, $h(y)=g(f^{-1}(y))$. Then the left hand side is

$$\int_y h(y) d(f_* \mu)=\int_y g(f^{-1}(y)) d(f_* \mu)= $$

$$\int_y \log p(f^{-1}(y)) d(f_* \mu)$$

Now, IF density of $f_* \mu$ was $p(f^{-1}(y))$, this would be
the entropy of $f_* \mu$, BUT IT IS NOT.

In fact the density of $f_* \mu$ is
$$q(y)=p(f^{-1}(y))|Df^{-1}(y)|$$

Then
$$H(f_* \mu)=\int_y \log [p(f^{-1}(y))|Df^{-1}(y)|] d(f_* \mu)=$$

$$\int_y \log [p(f^{-1}(y))]+ \log|Df^{-1}(y)|] d(f_* \mu)=$$

$$\int_y \log [p(f^{-1}(y))] d(f_* \mu)+ \int_y \log|Df^{-1}(y)| d(f_* \mu)=$$

$$H(\mu)+E_{f_* \mu}[ \log|Df^{-1}(y)|]=H(\mu)+E_{\mu}[ \log (1/|Df(x)|)]= H(\mu)-E_{\mu}[ \log |Df(x)|]$$

in agreement with https://en.wikipedia.org/wiki/Differential_entropy#Properties_of_differential_entropy

Some remarks:

Denoting the measure and its density by the same letter is a bad idea in this setting, because they don't transform the same way. If your measure is $\gamma$ has density $p(x)$ aka $\gamma=p(x)dx$ then the pushforward $f_{\#}\gamma$ has density $p(f^{-1}(y))|Df^{-1}(y)|$ aka $f_{\#}\gamma=p(f^{-1}(y))|Df^{-1}(y)|dy$ -- mnemonically $p(x)dx=p(x(y)) |dx/dy| dy$. Thus when computing the entropy you will not be integrating the transformed function $\log p(x(y))$ but rather the transformed log-density  i.e. $\log (p(f^{-1}(y))|Df^{-1}(y)|)$, so the integrals will not be equal.

The underlying reason for this is that the entropy of a measure is computed relative to the "background" Lebesgue measure (all the density functions are computed with respect to the Lebesgue measure - as Radon-Nikodym derivatives, if that means anything to you), and the Lebesgue measure on the target is not the pushforward of the Lebesgue measure on the domain, which is what makes the entropy not invariant under the coordinate changes.


问:
The closure in $(L_2[0,1] \,, \,||\,||_{2})$ is compact
I want to prove if this statement is either true or false but I haven't found any counter-example to prove that it's false..

The closure in $(L_2[0,1]$$,||\,||_{2})$ of  equicontinuous family of continuous function on $[0,1]$ is a compact

If anyone could help it would be a lot appreciated. Thanks in advance

答:
It is false. The family of constant functions furnishes a counterexample.

I think it is worth noting that a bounded set, $F$,  in your space is pre-compact (the closure is compact) if and only if $\lim\limits_{x\rightarrow0}\int_0^1 |f(x+y)-f(y)|^2\,dy=0$ uniformly for $f\in F$. C.f. Dunford and Schwartz, Linear Operators, vol. 1, Theorem IV.8.20.


问:
Characterization for compact sets in $\mathbb{R} $ with the topology generated by rays of the form $\left(-\infty,a\right) $
I'm trying to find a sufficient and necessary condition for a subset to be compact in  $\mathbb{R} $  when the topology is generated by the basis $\left\{ \left(-\infty,a\right)\,|\, a\in\mathbb{R}\right\} $. I thought it might be something similar to the standard characterization for compact sets in $\mathbb{R}$ with the standard topology like perhaps "closed and upper bounded" but that hasn't really led anywhere.

Help would be appreciated.

答: HINT: Let $A\subseteq\Bbb R$. Suppose that $A$ has a largest element, say $a$; what happens when you’ve covered $a$ with an open set? Now suppose that $A$ does not have a largest element, and consider the cover $\{(\leftarrow,a):a\in A\}$.</p>


问:
How to derive covariance formula for two random variables
I am a beginner in statistics and I have a little background on what is mean and variance.
Coming to the topic of covariance between two random variables, I am very curious to know why the formula is this.

I searched a lot online but I could not get anything that helped me understand the intuition behind the formula and how it is related to the point that the variables are related or not related.
Any help is greatly appreciated.

答:
The definition of the covariance of two random variables $X$ and $Y$ is

$$\mathbb E \left[(X-\mathbb E[X])(Y-\mathbb E[Y])\right]$$

If you want to estimate this from a sample size $n$, then it is natural to estimate $\mathbb E[X]$ from $\bar x=\frac1n\sum x_i$ and $\mathbb E[Y]$ from $\bar y=\frac1n\sum y_i$.  It might then  seem natural to estimate the covariance as $\frac1n\sum (x_i-\bar x)(y_i- \bar y)$ but it turns out this is biased slightly downwards in that its expectation is $1-\frac1n$ times the actual covariance, and to remedy this you could instead use the unbiased estimator of covariance $$\frac1{n-1}\sum (x_i-\bar x)(y_i- \bar y)$$

which is the formula you are asking about.  A similar issue arises with the sample variance.

If the random variables are independent and their covariance is then  $\mathbb E \left[(X-\mathbb E[X])(Y-\mathbb E[Y])\right] = \mathbb E \left[(X-\mathbb E[X])\right]\mathbb E \left[(Y-\mathbb E[Y])\right]$ which is $0 \times 0=0$ so they would have zero covariance. So a non-zero covariance indicates some sort of dependence (for example the sign of the covariance indicates whether an simple linear regression line is upward or downward sloping). The reverse is not necessarily true: two random variables can have zero covariance without being independent.


问:
Negation of quantificational logic statement help
$$\forall x(x<5\Rightarrow \exists y(x+y > 5)) $$
I want to negate this equation, am I thinking correctly?
$$ \exists x(x\geq5 \land\lnot\forall y(x+y \leq5))$$

答:
Here's one way to do it
$$\neg\forall x(x\lt5\rightarrow \exists y(x+y \gt 5))$$
$$\equiv\neg\forall x(x\ge5\lor\exists y(x+y \gt 5))$$
$$\equiv\exists x(x\lt5\land\neg\exists y(x+y \gt 5))$$
$$\equiv\exists x(x\lt5\land\forall y(x+y \le 5))$$


问:
Do there exist, $\underset{x\to \infty }{\text{lim}}\frac{3 \left| x\right| +2 x}{7 \left| x\right| -5 x}$?
Here is why I think the limit does not exists.

link to image if mathjax doesn't work

$\lim \limits_{x \to ∞ } \frac{3|x|+2x}{7|x|-5x}$ == $\lim \limits_{1/h \to 0 } \frac{3|1/h|+2(1/h)}{7|1/h|-5(1/h)}$

$\lim \limits_{1/h \to 0^+ } \frac{3|1/h|+2(1/h)}{7|1/h|-5(1/h)}$ == $\lim \limits_{1/h \to 0^+ } \frac{3+2}{7-5}$ == $\frac{5}{2}$

$\lim \limits_{1/h \to 0^- } \frac{3|1/h|+2(1/h)}{7|1/h|-5(1/h)}$ == $\lim \limits_{1/h \to 0^- } \frac{-3+2}{-7-5}$ == $\frac{1}{12}$

left limit does not match the right limit. so i think the limit does not exist. Is this solution correct?

Edit: please note here the question is whether the limit to the given equation exists or not, if it does what is it.

答:
$$\underset{x\to +\infty }{\text{lim}}\frac{3 \left| x\right| +2 x}{7 \left| x\right| -5 x}=\underset{x\to +\infty }{\text{lim}}\frac{3 x +2 x}{7x -5 x}=\underset{x\to +\infty }{\text{lim}}\frac{5x}{2x}=\frac{5}{2}$$
$$\underset{x\to -\infty }{\text{lim}}\frac{3 \left| x\right| +2 x}{7 \left| x\right| -5 x}=\underset{x\to -\infty }{\text{lim}}\frac{-3 x +2 x}{-7x -5 x}=\underset{x\to -\infty }{\text{lim}}\frac{-x}{-12x}=\frac{1}{12}$$
The limits at $\infty$ are two and they both exist. The limit at $x=0$ does not exist.


问:
alternative way of proving $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$ converges to $\ln 2$ without using integrals
I found a lot of answered to this problem using Riemann - sums, I myself solved it rewriting the logarithm sum, but there is another way where you should use the following hints.

Show that

$$\exp \left( \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k+n} \right) =
 2$$

using the following identities

(1). There is a nonnegative sequence $x_k$, converging to zero, and

$$\exp \left(\frac{1}{k} \right) = \left(1+\frac{1}{k} \right)\times
 \exp\left(\frac{x_k}{k} \right)$$

(2). $$\lim_{x \to 0} \frac{\exp(x)-1}{x}=1$$

To prove $(1)$, I wanted to rewrite

$$\exp\left(\frac{1}{k} \right)\times \frac{1}{1+\frac{1}{k}}$$ and expand $$\frac{1}{1+\frac{1}{k}} = \frac{1}{1-\frac{-1}{k}}$$  as geometric sequence to find $x_k$, but I don’t get anything useful.

For $(2)$ I thought may one should “stretch”

$$\sum_{k=1}^n \left(\exp\left(\frac{1}{k+n}\right)-1 \right)$$

which has the same limit (I proved that) as $\sum_{k=1}^n \frac{1}{k+n}$ and rewrite it as

$$\sum_{k=1}^n \frac{\exp\left(\frac{1}{k+n}\right)-1}{\frac{1}{k+n}}\frac{1}{k+n}$$

I’m thankful for any tips or answers!

Body:
Using the inequality$$\frac{1}{(n+1)}<\ln(\frac{n+1}{n})<\frac{1}{n}$$( Since $\frac{1}{(n+1)}*1<\ \int_{n}^{n+1}\frac{1}{x} \,dx <\frac{1}{n}*1$ (think area graphically) ),we can write $\ln(\frac{n+1}{n})<\frac{1}{n}$$<$$\frac{1}{n-1}<ln(\frac{n-1}{n-2})$. Then sum from $n$ to $2n$ and take $\lim_{n\to\infty}$.

$ \sum_{i=0}^{n}ln(\frac{n+i+1}{n+i})$$<$ $\sum_{i=0}^{n} \frac{1}{n+i} $$<$$ \sum_{i=0}^{n} ln(\frac{n+i-1}{n+i-2})$

=$\ln(\frac{n+1}{n}*\frac{n+2}{n+1}*...*\frac{2n+1}{2n})<\frac{1}{n}+ \frac{1}{n+1}+...+\frac{1}{2n}<ln(\frac{n-1}{n-2}*\frac{n}{n-1}*...*\frac{2n-1}{2n-2})$

= $\ln(\frac{2n+1}{n})<\frac{1}{n}+ \frac{1}{n+1}+...+\frac{1}{2n}<ln(\frac{2n-1}{n-2})$

$\lim_{n\to\infty}$$\ln(\frac{2n+1}{n})<$$\lim_{n\to\infty}$$(\frac{1}{n}+ \frac{1}{n+1}+...+\frac{1}{2n})<$$\lim_{n\to\infty}$$ln(\frac{2n-1}{n-2})$

$\lim_{n\to\infty}$$\ln(2+\frac{1}{n})<$$\lim_{n\to\infty}$$(\frac{1}{n}+ \frac{1}{n+1}+...+\frac{1}{2n})<$$\lim_{n\to\infty}$$ln(\frac{2-\frac{1}{n}}{1-\frac{2}{n}})$

You will see by the sandwich theorem the value comes to be $\ln2$.


问:
Solve the differential equation $\frac{dx}{dt}=v(x)$
What techniques do I need to use to solve solve the differential equation $\frac{dx}{dt}=v(x)$, with $x(0)=0$? And would this classify as an ODE or PDE?

For a couple of particular cases, we could use $v(x)=\exp(-x)$, $v(x) = \tanh(x)$, or $v(x)=v_0 + k H(x-X)$, where $H$ is the Heaviside function.

答:
Ask yourself: do you see partial derivatives in the equation? If yes, it's a PDE. If not, it's a ODE.

As for solving, rewrite it as $$\frac{{\rm d}x}{v(x)} = {\rm d}t,$$so that $$t = \int \frac{{\rm d}x}{v(x)} + c,$$where $c$ will be chosen as to make $x(0)=0$. I don't think you can do better than this in full generality, but the algorithm goes: solve the integral on the right, and then algebraically solve for $x$, if possible.


问:
Question about direct factors
this is a problem from hungerford:

"A normal subgroup $H$ of a group $G$ is said to be a direct factor if there exists a (normal) subgroup $K$ of $G$ such that $G = H \times K$.

if $H$ is a direct factor of $K$ and $K$ is a direct factor of $G$, then $H$ is normal in $G$.

If $H$ is a direct factor of $G$, then every homomorphism $H \to G$ may be extended to an endomorphism $G \to G$. However, a monomorphism $H \to G$ need not be extendible to an automorphism $G \to G$"

I have a few questions:
1) why is the normality of $K$ put in brackets? Is there something that suggests that $K$ must be normal if it exists?

2) how would I go about proving these? For the first one, I know that $G = K \times L$ for some $L$ normal in $G$ and I know $K = H \times J$ for some $J$ normal in $K$.

So I can conclude $G = H \times J \times L$ but I don't see how to conclude normality of $H$.

答: Hungerford uses $H\times K$ to denote the internal direct product of H and K, so by definition, K is normal

Title: Question about direct factors


问:
How to find a Lyapunov function
I have the following system: $$\dot{x_1} = x_2(x_3-2) \\ \dot{x_2} = x_1(x_3-2) \\ \dot{x_3}=-x_3^3 $$ and I want to determine its equilibrium points together with their stability. To find the equilibrium points, I wrote $$x_2=0 \text{ or } x_3=2 \\ x_1=0 \text{ or } x_3=2  \\ x_3 =0$$
Thus, only equilibrium point $x^*$ for the system is the origin $0$. I found $\det{Df(0)} = 0$ so I expect that the origin is a non-hyperbolic equilibrium point, probably asymptotically stable. To determine its type, I want to find a Liapunov function. Now, I assumed I can take $V(x)=c_1x_1^2 +c_2x_2^2+c_3x_3^3$ for some positive constants $c_i$. We have $V(0)=0$ and $V(x)>0$ for any $x \neq 0$. Then, $$\dot{V}(x)=\nabla V\cdot f(x) = 2c_1x_1x_2(x_3-2)+2c_2x_1x_2(x_3-2)-2c_3x_3^4.$$

$(\star )$ Now, let $c_1=c_2=1$ and $c_3 =2$ so that we get $$\dot{V}(x)=4x_1x_2(x_3-2)-4x_3^4.$$

However, I do not know how to determine the sign of this expression to conclude the result. I want to ask

Is it enough to find a Lyapunov function which is always positive or negative to say that $x^*$ is stable/unstable/asymptotically stable?

Similar to the first question but am I allowed to set arbitary values for $c_i$'s? (The $\star$ part)

How can I deduce the result?

答:
The equations in their current form, don't need a Lyapunov function for their stability analysis. After all, it seems the system is ill-defined. Here is why.

The third equation describing $x_3$ is totally independent from the other two, and can/should be eliminated:
$$\dot{x}_3=-x_3^3\implies x_3(t)=\pm\frac1{\sqrt{2t-c_3}}$$
where $c_3$ is a constant that has to be determined according to the initial conditions:
$$x_3(0)^2=\frac1{-c_3}\implies x_3(t)=\pm\frac1{\sqrt{2t+1/x_3^2(0)}}$$
Anyway, you still can't get rid of the $\pm$ sign. Because both answers satisfy the equation in its current form.

For the other two, you can simply divide them like this:
$$\begin{matrix}\dot{x}_1=x_2(x_3-2)\\ \dot{x}_2=x_1(x_3-2)\end{matrix}
\implies\frac{\dot{x}_1}{\dot{x}_2}=\frac{x_2}{x_1}$$
But $\frac{\dot{x}_1}{\dot{x}_2}=\frac{d x_1/dt}{d x_2/dt}=\frac{d x_1}{d x_2}$, which implies
$$x_1 d x_1=x_2 d x_2\overset{\int}{\implies}x_1^2=x_2^2+c_2\quad\text{i.e.}\;x_2=\sqrt{x_1^2+c}$$
As you see, the equations are not independent. So the stability of the system cannot be determined without further information. If you put the above result into the first equation and also replace $x_3$ by the function we obtained for it, we will have a single ODE for $x_1$. I omit the rest.

Title: How to find a Lyapunov function


问:
Injective Linear Mapping
Could someone help me determine if the linear transformation $S(\begin{bmatrix}a&b\\c&d\end{bmatrix})= a+(2b-12c)x+(3d-a)x^2 $ is injective or not?

答:
We know that a linear map is injective if and only if $KerS=\{0\}$

$S(\begin{bmatrix}a&b\\c&d\end{bmatrix})= a+(2b-12c)x+(3d-a)x^2=0$ $\Rightarrow$ $a=0$ , $2b-12c=0$,$3d-a=0$

$3d-a=0$ $\Rightarrow$ $d=0$

$2b-12c=0$ $\Rightarrow$ $b=6c$

so $KerS=\{\begin{bmatrix}0&6c\\c&0\end{bmatrix}\} \neq \{0\}$ which means it is not injective

Title: Injective Linear Mapping


问:
How can I get the negation of $\exists!$ (unique existential quantification)?
How can I get the negation of $\exists!$ (unique existential quantification)? if it's $\forall$, So if I wanna re-negate the last one, I'll get $\exists$ but it's not the same as what we started with! Did I do something wrong here?

For example: $$P : \exists! x\in \mathbb{R} \text{ such that } x^2 = 0$$ it means $\exists x \in \mathbb{R}\text{ such that } x^2 = 0\wedge x $ is unique, So the negation is $\forall x \in \mathbb{R}\space x^2 \ne 0 \vee x \text{ isn't unique}$ is this a false statement?

答: Using the simplest definition of $\exists!$ I know, which is $$\langle \exists! x :: P(x) \rangle \;\equiv\; \langle \exists y :: \langle \forall x :: P(x) \equiv x = y \rangle \rangle \;,$$ the negation you're looking for is $$\langle \forall y :: \langle \exists x :: P(x) \equiv x \not= y \rangle \rangle \;.$$</p>


答:
When I started studying calculus, there still were books that distinguished between $+\infty$, $-\infty$ and $\infty$ (unsigned infinity).

The definition of $\smash{\lim\limits_{x\to\infty}}\,f(x)=l$ at unsigned infinity was

for every $\varepsilon>0$, there exists $M>0$ such that, for all $|x|>M$, it holds that $|f(x)-l|<\varepsilon$.

I don't think this has ever been useful, and nowadays it's quite rare to find it. Usually $\infty$ means what in those times was denoted as $+\infty$. Why is it useless? Because that (useless) limit exists if and only if the limits at $+\infty$ and $-\infty$ exist and they're equal. Which happens for rational functions when the degree of the numerator doesn't exceed the degree of the denominator: not so a general context.

In your case, you have
$$
f(x)=\begin{cases} 5/2 & x>0 \\[6px] 1/12 & x<0 \end{cases}
$$
so there's no doubt what the limit at positive and negative infinity are. Unless your textbook or instructor uses the old-fashioned unsigned infinity.

The “unsigned infinity” has its uses, though: when doing with analytic functions, it is common to “compactify” the complex plane by adding a single point, namely $\infty$. And the notion of limit at $\infty$ is exactly the same as the one spelled above, but with the variable in the complex numbers.

Title: Do there exist, $\underset{x\to \infty }{\text{lim}}\frac{3 \left| x\right| +2 x}{7 \left| x\right| -5 x}$?

Body:
Here is why I think the limit does not exists.

link to image if mathjax doesn't work

$\lim \limits_{x \to ∞ } \frac{3|x|+2x}{7|x|-5x}$ == $\lim \limits_{1/h \to 0 } \frac{3|1/h|+2(1/h)}{7|1/h|-5(1/h)}$

$\lim \limits_{1/h \to 0^+ } \frac{3|1/h|+2(1/h)}{7|1/h|-5(1/h)}$ == $\lim \limits_{1/h \to 0^+ } \frac{3+2}{7-5}$ == $\frac{5}{2}$

$\lim \limits_{1/h \to 0^- } \frac{3|1/h|+2(1/h)}{7|1/h|-5(1/h)}$ == $\lim \limits_{1/h \to 0^- } \frac{-3+2}{-7-5}$ == $\frac{1}{12}$

left limit does not match the right limit. so i think the limit does not exist. Is this solution correct?

Edit: please note here the question is whether the limit to the given equation exists or not, if it does what is it.


问:
Splitting a square into equal areas
I am looking to split a square into 4 equal areas, not necessarily needing to be the same shapes or sizes . I need to find a way to use the least length of lines ( which could be straight or curved lines) inside the square to split and divide it up. What would be the optimal solution. I need to code an algorithm to do it too but I wanted to understand the mathematical concept behind it first. Also how would I use this for a rectangle say side height 1 and length 2.

答:
My calculations show that the obvious configuration is not quite optimal.
Consider the square  $[0,1] \times [0,1]$.  The division by straight horizontal and vertical lines has total length $2$.  Let $\alpha$ be the root of the polynomial $768 z^8 + 64 z^6 - 72 z^4 + 1$ approximately $0.4694760731$, and
$\beta = 1/(4\alpha)$.
Partition the square using the points $[\alpha,\alpha]$, $[1-\alpha,1-\alpha]$,
$[0,\beta]$, $[\beta,0]$, $[1,1-\beta]$, $[1-\beta,1]$, looking like this:

If I'm not mistaken, this has equal areas and total length $4 \sqrt{\alpha^2 + (\alpha - \beta)^2} + \sqrt{2} (1-2\alpha) \approx 1.981089021 < 2$.

I'm not claiming that this is optimal, either.  You should do better by replacing the four straight line segments touching the boundary with circular arcs.

Title: Splitting a square into equal areas


问:
Prove $a^{7}|s$ using induction
Definition: a positive integer $s$, is 7-free if $a \in \mathbb{N}$, with $a^{7}|s$, then $a=1$. For example, 1, 4 are 7-free. 384 is not7-free.

How can I prove for any integer $m>1$, there exists positive integers, $s$ and $t$ such that $m=st^{7}$ and s is 7-free.

I think Induction would work here? What if I strongly induct? Can I use the definition of 7-free and convert it into gcd? For example, $gcd(a^{7},s)=1$?

答:
Strong induction hint: if $m$ is 7-free, write $s = m$ and $t = 1$.

If $m$ is not 7-free, find some $a > 1$ such that $a^7 \mid m$. Use the induction hypothesis on $\dfrac m {a^7}$.

Title: Prove $a^{7}|s$ using induction


问:
How many friends for birthday party every weekend?
A friend claims that he is invited to a birthday party every weekend.

I know he needs to have at least 52 friends for that, but what is the "realistic" amount of friends that chances are 100% for beeing invited every weekend

答: There is on $100\%$ guarantee, but you can pick any lower chance and find a number of friends.  Say we want a $50\%$ chance that he gets invited every weekend of the year.  If you have a favorite number, you can follow this through.  We need the probability $p$ of getting invited any given weekend to satisfy $p^{52}=\frac 12$, which gives $p = \left(\frac 12 \right)^{\frac 1{52}}\approx 0.98676$.  Now with $n$ friends, the chance of getting an invitation on a given weekend is $1-(1-\frac 1{52})^n$, so we want $$1-\left(1-\frac 1{52}\right)^n= \left(\frac 12 \right)^{\frac 1{52}}\\1-\left(\frac 12 \right)^{\frac 1{52}}=\left(1-\frac 1{52}\right)^n\\ \frac{\log\left(1-\left(\frac 12 \right)^{\frac 1{52}}\right)}{\log\frac {51}{52}}=n\\ n \approx 223 $$  This seems reasonable.  On average he will get about  $4.288$ invitations per weekend.  Using a Poisson model, that says the chance of a miss on a given weekend is one in $\boxed{73}$</p>


问:
Is there a field $K$ for which the kernel of the map $\mathbb{Z}[x]\to K$ is NOT principal?
Consider $K$ a field and a ring homomorphism $\phi:\mathbb{Z}[x]\to K$.

I guess this is wrong, because I was asked to give an example of precisely a field $K$ for which the kernel of the map above is not principal, but this is what I've thought so far: as long as $K$ is field (I had $K=\mathbb{Q},\mathbb{R}, \mathbb{C}$ in mind), $\ker{\phi}$ must be principal. The reason is that you consider taking $\mathbb{Z}[x]$ to its ring of fractions $\mathbb{Q}[x]$ and now consider the homorphism $\bar{\phi}:\mathbb{Q}[x]\to K$. Since $\mathbb{Q}[x]$ is an Euclidean domain, it's a PID, and so $\ker{\bar{\phi}}$ must be principal, generated by a polynomial of minimal degree (if not zero) having an element of $K$ as root. Now $\ker{\phi} = \ker{\bar{\phi}}\cap\mathbb{Z}[x]$ so it keeps being principal (if not zero).

Is my reasoning correct or is there a field $K$ for which the kernel of the map $\phi:\mathbb{Z}[x]\to K$ is not principal? I feel like I'm missing something important. Thank you so much!

答:
What about $f:\mathbb{Z}[X] \rightarrow \mathbb{Z}/2\mathbb{Z}$ defined for all $P \in \mathbb{Z}[X]$ by
$$f(P)=P(0) \quad \mod \text{ }2 \quad \quad ?$$


答:
If the map $\mathbb{Z}[x] \to K$ happens to be surjective, then its kernel will be a maximal ideal. So we could first ask if there are any maximal ideals of $\mathbb{Z}[x]$ which are not principal. How do we make a maximal ideal of $\mathbb{Z}[x]$? We can just take a maximal ideal of $\mathbb{Z}$, e.g. $(2)$, and "lift" it to $\mathbb{Z}[x]$ by adding in the generator $x$. The quotient $\mathbb{Z}[x]/(2,x)$ is isomorphic to $\mathbb{Z}/(2)$, which is a field, so $(2,x)$ is a maximal ideal of $\mathbb{Z}[x]$. Why can't this ideal be principal?

Note: this is "the same" as @TheSilverDoe's answer.

Note 2: Any map $f : \mathbb{Z}[x] \to K$ gives a surjection $\mathbb{Z}[x] \to \operatorname{img}(f)$ with the same kernel, and $\operatorname{img}(f)$ is a domain. Conversely, every surjection $\mathbb{Z}[x] \to R$ with $R$ a domain gives a map $\mathbb{Z}[x] \to \operatorname{Frac}(R)$ with the same kernel, and $\operatorname{Frac}(R)$ is a field. Thus, it's equivalent to find a surjection from $\mathbb{Z}[x]$ to a domain with non-principal kernel, which is equivalent to finding a non-principal prime ideal of $\mathbb{Z}$. In particular, I focused on maximal ideals, but if this didn't work you could think about more general primes!

Title: Is there a field $K$ for which the kernel of the map $\mathbb{Z}[x]\to K$ is NOT principal?

Body:
Consider $K$ a field and a ring homomorphism $\phi:\mathbb{Z}[x]\to K$.

I guess this is wrong, because I was asked to give an example of precisely a field $K$ for which the kernel of the map above is not principal, but this is what I've thought so far: as long as $K$ is field (I had $K=\mathbb{Q},\mathbb{R}, \mathbb{C}$ in mind), $\ker{\phi}$ must be principal. The reason is that you consider taking $\mathbb{Z}[x]$ to its ring of fractions $\mathbb{Q}[x]$ and now consider the homorphism $\bar{\phi}:\mathbb{Q}[x]\to K$. Since $\mathbb{Q}[x]$ is an Euclidean domain, it's a PID, and so $\ker{\bar{\phi}}$ must be principal, generated by a polynomial of minimal degree (if not zero) having an element of $K$ as root. Now $\ker{\phi} = \ker{\bar{\phi}}\cap\mathbb{Z}[x]$ so it keeps being principal (if not zero).

Is my reasoning correct or is there a field $K$ for which the kernel of the map $\phi:\mathbb{Z}[x]\to K$ is not principal? I feel like I'm missing something important. Thank you so much!


问:
Prove this equality with mathematical induction. The volume of $\Omega$. Using multivariable calculus.
$\Omega_n : = \{ (x_1, \cdots , x_n) \in \mathbb{R}^n \ | \ |x_1|+ \cdots +|x_n| \leqq r \}$

$V_n$ : Volume of $\Omega_n$ $\left( \text{therefore}, V_n = \int_{\Omega} 1 \ d(x_1, \cdots , x_n) \right) $

Problem :
\begin{equation}
\text{Prove that }V_n
\left(=
\int_{\Omega} 1 \ d(x_1, \cdots , x_n) \right)
=
 \dfrac{(2r)^n}{n \ !} \cdots \ast_1.
\end{equation}

(I calculated $V_1$ and $V_2$, and I derived $V_1=2r, V_2=2r^2$. These machies with $\ast_1$.)

Probably, I might be able to prove $\ast_1$ with mathematical induction., but I don't know how I have to work.

I have to suppose $V_{n-1}=\dfrac{(2r)^{n-1}}{(n-1)\ !}$ holds, and calculate $V_n$.

I don't know how I have to use $V_{n-1}=\dfrac{(2r)^{n-1}}{(n-1)\ !}$

in calculating $V_n$.

I would like you to tell me some ideas.

答:
For the purposes of this proof, it might help to add the number $r$ to the notation (in case you want to use different values of $r$ at different times). I suggest writing
$$ \Omega_n(r) = \left\{(x_1,\dots,x_n) ~\middle|~ \sum_{i=1}^n |x_i| = r \right\}  $$

I assume you are familiar with the fact that you can compute volume integrals "by slicing". Meaning if you take the cross section of $\Omega$ in the hyperplane $x_n = a$, and if that cross section has $(n-1)$-dimensional volume $v(a)$, then
$$ V_n(r) = \int_{-r}^r v(a) \, da $$
In this case, the cross-sectional volume is $v(a) = V_{n-1}(r-a)$, since the cross-sectional slice just looks like the set $\Omega_{n-1}(r-a)$. Putting this all together, you have
$$ \begin {align*}
\int_{\Omega_n(r)} dx_1 \cdots dx_n &= \int_{-r}^r \left( \int_{\Omega_{n-1}(r-x_n)} dx_1 \cdots dx_{n-1} \right) \, dx_n
\end {align*}
$$

Now you can use induction to evaluate the inner integral in the parentheses, and the leftover integral in the single variable $x_n$ is pretty simple.


问:
Importance and Intuition of Polynomial Rings
I know that a polynomial ring $R[x]$ is the ring with elements consisting of polynomials with coefficients in $R$. However, this definition leaves me confused when I try to really understand the concept of polynomial rings, rather than just accept that definition for what it is. Where did the polynomials come from? Why are polynomial rings important? Is there a way to “construct” the polynomials in the context of Abstract Algebra, so that we may see why studying rings made up of polynomials is important? Are polynomial rings actually just sets of polynomials, or are the polynomials just a concrete way to represent some more abstract idea? What does the variable $x$ mean in polynomial rings? I apologize for all the questions, I’m just having trouble seeing the importance or intuition behind them.

答:
The important thing to understand about polynomials is that they are templates for maps between rings. If you know a bit about coding, you'll probably know the concept that a template is something that can take different types of objects as inputs in a way that you don't have to specify what the objects are beforehand. For a polynomial, this means that it's a template for maps which take objects for which the following operations are meaningful:

multiplication of two or more of the objects

multiplication of an object with an element of the underlying ring

addition of two or more objects

For instance, take the polynomial $f=2X^2+4X-1\in\mathbb R[X]$. This polynomial defines a map
$$\begin{align}f_{\mathbb R}:&\mathbb R\longrightarrow\mathbb R,\\&x\mapsto 2x^2+4x-1.
\end{align}$$
But it also defines a map
$$\begin{align}
f_\mathbb C:&\mathbb C\longrightarrow\mathbb C,\\
&z\mapsto 2z^2+4z-1
\end{align}$$
Or even more exotic, it defines a map
$$\begin{align}
f_{\operatorname{Mat}_{n\times n}(\mathbb R)}:&\operatorname{Mat}_{n\times n}(\mathbb R)\longrightarrow \operatorname{Mat}_{n\times n}(\mathbb R)\\
&A\mapsto2A^2+4A-1 E_n.
\end{align}$$
This last one is important to formulate the Cayley-Hamilton theorem, for instance, which says that if $f$ is the characteristic polynomial of the matrix $A$, then $f_{\operatorname{Mat}_{n\times n}(\mathbb R)}(A)$ is the zero matrix.

More formally, this means that if $R$ is a commutative ring and $R\subseteq S$ a commutative ring extension, then for every $s\in S$ there is a unique evaluation map $R[X]\longrightarrow S$ which maps $X\mapsto s$ and $r\mapsto r$ for all $r\in R$, and is a homomorphism. The map evaluates each polynomial at $s$ (it "plugs in" $s$ for $X$). This property of the polynomial ring even uniquely characterizes it (up to isomorphism), and some will take this property as the definition of the polynomial ring. The construction via finite sequences in $R$ can then be used as proof of existence of such a ring.

In our examples from before, $\mathbb R\subseteq\mathbb R$ is a trivial ring extension. $\mathbb R\subseteq\mathbb C$ is another obvious ring extension which also gives us an evaluation map. And $\operatorname{Mat}_{n\times n}(\mathbb R)$ can also be viewed as a ring extension of $\mathbb R$ by identifying $r\in R$ with $rE_n\in\operatorname{Mat}_{n\times n}(\mathbb R)$, the matrix whose diagonal contains only $r$. This gives us a way to evaluate polynomials at matrices.

To summarize: Polynomials are there as templates for maps between ring extensions of the underlying ring - but only those special maps which are the result of applying these basic operations to an element $x$: addition, multiplication by an element of $R$, or multiplication by itself.


问:
Does there exist a shortest path which satisfies certain boundary conditions.
For context, I saw a variant of this question posted a couple of days ago, however, the question seems to have been deleted. $$$$
Let $S$ be the set of $C^1$ functions $y$ where $y:\mathbb{R}\longrightarrow \mathbb{R}$ such that $y(0) = y(1) = 0$, $y'(0) = \alpha$ and $y'(1) = \beta$. $\alpha$ and $\beta$ are fixed and at least one of them is non-zero. Let $l:S\longrightarrow\mathbb{R}$, $l(f)=\int_0^1\sqrt{1+(f')^2} dx$. Does $L =\{l(f):f\in S \}$ have a minimum?  $$$$
It is clear that $\inf(L) \geq 1$, however it does not seem like there should be a minimum of this set. I think this could be shown by first showing that $y = 0$ is the only function such that $l(y) = 1$ and $y(0) = y(1) = 0$-but clearly is not in $S$--then finding a subset $S'$ of $S$ such that $\inf(l(S')) = 1$, however, I was not able to accomplish this. Additionally would the question be different if each $y\in S$ was $C^{\infty}$?

答:
A better result: Again let $\alpha =1,\beta =0.$ This time define $f_n(x) = x(1-x)^n.$ Then

$$f_n'(x)= (1-x)^n +xn(1-x)^{n-1}(-1).$$

The first summand is $\le 1$ in absolute value on $[0,1].$ So is the second summand, as a little work shows. Thus $|f_n'| \le 2$ for each $n.$ Furthermore, $f_n'(x)\to 0$ pointwise on $(0,1).$ By the DCT,

$$\int_0^1\sqrt {1+(f_n')^2} \to \int_0^1\sqrt {1+0} = 1.  $$

The $f_n$ are polynomials, so we have improved on the previous result.

Previous result: For convenience, let $\alpha =1,\beta =0.$ For $n=2,3,\dots $ define $f_n(x) = x(1-nx)^2,$ $x\in [0,1/n],$ $f_n=0$ on $[1/n,1].$ We have

$$f_n'(x) = (1-nx)^2 +x2(1-nx)(-n),\, x\in [0,1/n].$$

This shows $f_n'(1/n)=0,$ hence $f_n\in C^1$ for each $n.$ Note that $|f_n'|\le 3$ on $[0,1/n].$ Thus

$$\int_0^1\sqrt {1+(f_n')^2} \le \sqrt {3}\cdot (1/n) +
1-1/n.$$

The relevant infimum over $n$ is therefore $1,$ but the infimum cannot arise as $l(f)$ with the given constraints. Other values of $\alpha, \beta$ should be similar.


问:
Finite difference method stability
I have shown that a finite difference method satisfies

$$\underline{u}^{n+1}=((1+6\mu)\mathbb{I}-36\mu A^{-1})\underline{u}^n$$

I don't think that the rest of the question is necessary but it is a finite difference method for the diffusion equation on $t>0, 0<x<1$ with boundary conditions $u(0,t)=0=u(1,t)$ and $u(x,0)=u_0(x)$.

The last part of the question asks to show that the scheme is stable if $\mu\leq 1/6$. The $A$ matrix was a tridiagonal symmetrix matrix with the numbers $1,4,1$ down the diagonal. It is said that the eigenvalues for this matrix are

$\lambda_i=4+2\cos\left( \frac{i\pi}{M+1} \right)$ for $i=1,\cdots,M$ if it is an $M\times M$ matrix.

Attempt

If I can show that the matrix

$$\left[(1+6\mu)\mathbb{I}-36\mu A^{-1}\right]$$

Is bounded with respect to some norm then the finite difference method is stable.

So I think that the norm will be the spectral norm since we have been given eigenvalues

If the eigenvalues are as above then take the reciprocal. So the eigenvalues of $A^{-1}$ is $\frac{1}{4+2\cos(\cdots)}$ which has a maximum value of 1/2.

So now using this I will try to show that

$$||B||=||(1+6\mu)\mathbb{I}-36\mu A^{-1}||=\sqrt{\rho(B^TB)}\leq C$$

I get stuck in evaluating this. My guess would be that the matrix is equal to its transpose so it is just:

$$||B||=\sqrt{\rho({B^2})} $$

My linear algebra is not very good I don't know if you can do this last part. Maybe take

$$(1+6\mu)^2\mathbb{I}-(1+6\mu)A^{-1}+36^2\mu^2A^{-1}A^{-1}$$

and if it were to act on some eigenvector then the eigenvalue would be

$$(1+6\mu)^2-\frac{1}{2}(1+6\mu)+\frac{36^2 \mu^2}{4}=\frac{1}{2}+9 \mu +360 \mu ^2$$

This doesn't seem to shed any light on it.

On the other hand, if I just take

$$|\rho(B)|=|1+6\mu-36\mu\frac{1}{2}|=|1-12\mu|$$

Then I for $0<\mu \leq 1/6$

I have $||B|| \leq 1$?

答:
You're right that the eigenvalues of $A^{-1}$ are $\lambda^{-1}$ for each eigenvalue $\lambda$ of $A$. (The notation is suggestive for a reason.) Furthermore, the identity matrix commutes with every matrix; so the eigenvalues of

$qI-rA^{-1}$

are just $q-r/\lambda$ for each eigenvalue $\lambda$ of $A$. Using this, you can compute the eigenvalues of $(1+6\mu)I-36\mu A^{-1}$.

You also know that the matrix $A$ is symmetric, so its inverse is also symmetric. For a symmetric matrix, the 2-norm $\|B\|_2 = \sqrt{\rho(B^\top B)}$ is just the absolute value of the eigenvalue of largest magnitude:

$\|B\|_2 = \max\{|\lambda| : \lambda \textrm{ is an eigenvalue of }B\}$.

I think you've more or less done this at the end of your answer without knowing the justification for why you can do it. You may also want to look at the smallest eigenvalue of $A^{-1}$.

Edit: There are actually plenty of definitions of stability, but I think the one you're looking for is that the solution's norm is bounded as $n\to\infty$. For that, you need that $\|B\| \le 1$ in some norm, where $B$ is the matrix corresponding to the iteration. Provided that you have a consistent discretization of your PDE, the Lax-Richtmeyer Theorem guarantees that, if the discretization is also stable, it will converge to the true solution as the mesh spacing $h$ goes to 0.

Title: Finite difference method stability


问:
ResNet derivative issue wrt the Hamiltonian
I'm reading Deep Learning as OCP and I've got some doubts about a derivative.

Consider an Hamiltonian built in the following way

$$\mathcal{H}(y,p,u) = \langle p,f(Ky+\beta)\rangle$$
where $K\in\mathbb{R}^{n\times n}$ and $y,\beta\in\mathbb{R}^n$. Moreover $u=(K,\beta)$ is the control and $f$ acts component-wise on $Ky+\beta$. So $f: \mathbb{R}^n \to \mathbb{R}^n $. I have problems to calculate the adjoint system. I compute

$$
\dot p = -\partial_y\mathcal{H} = -\langle p, \partial_yf(Ky+\beta)\rangle
$$

At this point I made some considerations: since $f$ acts compenent-wise, $\partial_yf$ must do the same so I ended up with $\partial_yf = Kf'(Ky+\beta)$ so also $\partial_yf:\mathbb{R}^n \to \mathbb{R}^n$.

Now
$$
\dot p = -\langle p, \partial_yf(Ky+\beta)\rangle = -\langle p,Kf'(Ky+\beta) \rangle = -[Kf'(Ky+\beta)]^\top p
$$
but I don't think this is correct because in the article it's written something like
$$
\dot p = -K\;\partial_yf\odot p
$$
where $\odot$ is the component-wise product, and it seems correct since $p$ must be a vector as $y$.
Someone could enlighten me?

答:
At this point I made some considerations: since $f$ acts compenent-wise, $\partial_yf$ must do the same so I ended up with $\partial_yf = Kf'(Ky+\beta)$ so also $\partial_yf:\mathbb{R}^n \to \mathbb{R}^n$.

This does not make sense to me. If anything, $f$ acts componentwise on the vector $x=Ky+\beta$, so you could maybe say something like this (though I'm not sure what, exactly) if you were differentiating with respect to $x$; but since $K$ mixes components of $y$ this does not work for differentiating in $y$.

You just need to own up to this and do the chain rule. In fact its good to write $F:\mathbb{R}^n \to \mathbb{R}^n$ with components being $F_j(x)=f(x_j)$. We consider $\beta$ to be a constant. Then the chain rule says $D F (Ky+\beta)=DF\cdot D(Ky+\beta)=DF\cdot K$.  We then have

$$DH=D(
)=
=
=p^T DF K$$

and $\partial_y H=DH^T=K^T DF^T p= K^T (DF^T p)$

The matrix $DF$ is diagonal with $f'|_{(Ky)_j+\beta_j}$ on the diagonal, so $DF^T p$ has components $ f'|_{(Ky)_j+\beta_j}p_j$. Perhaps this is denoted by $\partial_y f \odot p$. Then this result almost matches with what you have written, except $K$ is $K^T$:

$$\partial_y H= K^T \partial_y f \odot p$$

However it's difficult for me to see how the result with $K$ can be correct, since the problem makes sense when $y\in \mathbb{R}^m$ and $K$ maps from $\mathbb{R}^m$ to $\mathbb{R}^n$; but then the gradient is also in $\mathbb{R}^m$, consistent with applying $K^T$ to the vector $\partial_y f \odot p \in \mathbb{R}^n$. Perhaps its an issue of notational conventions, and $\partial_y H$ is the derivative instead of the gradient -- but then $K$ would be in front. I am not entirely sure.

Body:
At this point I made some considerations: since $f$ acts compenent-wise, $\partial_yf$ must do the same so I ended up with $\partial_yf = Kf'(Ky+\beta)$ so also $\partial_yf:\mathbb{R}^n \to \mathbb{R}^n$.

This does not make sense to me. If anything, $f$ acts componentwise on the vector $x=Ky+\beta$, so you could maybe say something like this (though I'm not sure what, exactly) if you were differentiating with respect to $x$; but since $K$ mixes components of $y$ this does not work for differentiating in $y$.

You just need to own up to this and do the chain rule. In fact its good to write $F:\mathbb{R}^n \to \mathbb{R}^n$ with components being $F_j(x)=f(x_j)$. We consider $\beta$ to be a constant. Then the chain rule says $D F (Ky+\beta)=DF\cdot D(Ky+\beta)=DF\cdot K$.  We then have

$$DH=D(<p, F(Ky+\beta)>)=<p, D(F(Ky+\beta)>=<p,DF\cdot K>=p^T DF K$$

and $\partial_y H=DH^T=K^T DF^T p= K^T (DF^T p)$

The matrix $DF$ is diagonal with $f'|_{(Ky)_j+\beta_j}$ on the diagonal, so $DF^T p$ has components $ f'|_{(Ky)_j+\beta_j}p_j$. Perhaps this is denoted by $\partial_y f \odot p$. Then this result almost matches with what you have written, except $K$ is $K^T$:

$$\partial_y H= K^T \partial_y f \odot p$$

However it's difficult for me to see how the result with $K$ can be correct, since the problem makes sense when $y\in \mathbb{R}^m$ and $K$ maps from $\mathbb{R}^m$ to $\mathbb{R}^n$; but then the gradient is also in $\mathbb{R}^m$, consistent with applying $K^T$ to the vector $\partial_y f \odot p \in \mathbb{R}^n$. Perhaps its an issue of notational conventions, and $\partial_y H$ is the derivative instead of the gradient -- but then $K$ would be in front. I am not entirely sure.


问:
Give me some hints to prove this
Let $p$ and $q$ be distinct odd primes. Show that

$\sum_{j=1}^{\frac{p-1}{2}} \lfloor \frac{qj}{p} \rfloor +\sum_{j=1}^{\frac{q-1}{2}} \lfloor \frac{pj}{q} \rfloor = \frac{(p-1)(q-1)}{4}$

答:
A geometric hint is that, consider a line segment $L:y = \frac pq x$ between the origin and $(q,p)$. The following image shows an example where $p=7$ and $q=19$:

The first summation $\sum_{y=1}^{\frac{p-1}2}\left\lfloor\frac qpy\right\rfloor$ would correspond to some kind of area to the left of line $L$, explicitly the region:

$$0 \le y \le \frac {p+1}2,\quad 0 \le x \le \left\lfloor\frac qp\lfloor y\rfloor\right\rfloor$$

(Shown as a dark green region to the left of the dashed line and inside the blue rectangle)

The second summation $\sum_{x=1}^{\frac{q-1}2}\left\lfloor\frac pqx\right\rfloor$ would correspond to some kind of area below of line $L$, explicitly the region:

$$0 \le x \le \frac {q+1}2,\quad 0 \le y \le \left\lfloor\frac pq\lfloor x\rfloor\right\rfloor$$

(Shown as a light red region below the dashed line and inside the blue rectangle)

Separating the two regions is a path of neighbouring integer cells between the origin $(0,0)$ and the point $(\frac{q+1}2,\frac{p+1}2)$. (Shown as white cells around the dashed line and inside the blue rectangle)

By counting the number of horizontal and vertical grid lines the line $L$ crosses, the path contains $\frac{p-1}2 + \frac{q-1}2 + 1$ squares. The condition that $p$ and $q$ are coprime is important here, that $L$ does not pass through any integer lattice point between the origin and $(q,p)$.

Counting the three kinds of cells that together form the rectangle between the origin and $(\frac{q+1}2, \frac{p+1}2)$,

$$\begin{align*}
\sum_{y=1}^{\frac{p-1}2}\left\lfloor\frac qp y\right\rfloor +
\sum_{x=1}^{\frac{q-1}2}\left\lfloor\frac pq x\right\rfloor + \left(\frac{p-1}2 + \frac{q-1}2 + 1\right)
&= \frac{p+1}2\cdot \frac{q+1}2\\
&= \left(\frac{p-1}2+1\right)\left(\frac{q-1}2+1\right)\\
\sum_{y=1}^{\frac{p-1}2}\left\lfloor\frac qp y\right\rfloor +
\sum_{x=1}^{\frac{q-1}2}\left\lfloor\frac pq x\right\rfloor &= \frac{p-1}2\cdot \frac{q-1}2
\end{align*}$$

Title: Give me some hints to prove this


Title: Show that $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{2^{n}-1}>\frac{n}{2}$

Body:
Show that $$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{n}-1}>\frac{n}{2}.$$

I used the fact that for all $k \ge 1:k <2^k$
$$\sum_{k=1}^{2^{n}-1}\frac{1}{k}>\sum_{k=1}^{2^{n}-1}\frac{1}{2^{k}}=\frac{1}{2}\left(\frac{1-\frac{1}{2^{2^{n}-1}}}{1-\frac{1}{2}}\right)=1-\frac{1}{2^{2^{n}-1}}.$$

But I don't know how to continue.

Body:
Let $ n\in\mathbb{N}^{*} $, denoting $ H_{n}=\sum\limits_{k=1}^{n}{\frac{1}{k}} $, we have : $$ H_{2n-1}-H_{n-1}=\sum_{k=n}^{2n-1}{\frac{1}{k}}\geq\sum_{k=n}^{2n-1}{\frac{1}{2n-1}}=\frac{n}{2n-1}> \frac{1}{2} $$

Thus, if $ k\in\mathbb{N} $, then : \begin{aligned}H_{2^{k+1}-1}-H_{2^{k}-1}&> \frac{1}{2}\\ \Longrightarrow\sum_{k=0}^{n-1}{\left(H_{2^{k+1}-1}-H_{2^{k}-1}\right)}&> \frac{n}{2}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ H_{2^{n}-1}&>\frac{n}{2}\end{aligned}


问:
Best way to improve max-flow on a DAG
I am considering a problem related to max-flow on a weighted DAG and I saw a number of posts with related questions but nothing that really helps me.

I have a weighted DAG D = (V, A, W), a source S, a target T, and a budget B of weight that I can add to the arcs to increase the flow between S and T. The goal is to maximize the flow between S and T using the budget in the best way possible.
Now, a simple solution would be to compute the min-cut, pick $a$ random arc a from the set of arcs identified in this way, add weight of B/alpha to $a$, and restart the procedure alpha times.

I have a few problems with this solutions:

since the weights are real, setting alpha is not that easy (but I guess that a large enough alpha will work well in practice)

this method is terribly slow

I don't have any kind of theoretical guarantee (which I guess can be found but I don't have it yet)

Are you aware of any previous work in this area providing principled and efficient algorithms for this problem?

答:
You can solve the problem via linear programming as follows.  Let $x_{i,j} \ge 0$ be the flow across arc $(i,j)$, and let $y_{i,j} \ge 0$ be the weight added to arc $(i,j)$.  The problem is to maximize $\sum_{(S,j)\in A} x_{S,j}$ subject to
\begin{align}
\sum_{(i,j)\in A} x_{i,j} - \sum_{(j,i)\in A} x_{j,i} &= 0 &&\text{for $i\in N \setminus\{S,T\}$} \\
x_{i,j} &\le c_{i,j} + y_{i,j} &&\text{for $(i,j)\in A$} \\
\sum_{(i,j)\in A} y_{i,j} &\le B
\end{align}

Title: Best way to improve max-flow on a DAG


问:
Why compactness of control space implies Value Function $>-\infty$
I went through the Sec I.4 of the book Controlled Markov Process and Viscosity Solutions by Fleming and Soner. In the model, $x(s)$ for $s\in[t,t_{1}]$ follows the ODE,
\begin{eqnarray}
x^{\prime}(s)&=&f(s,x(s),u(s)) \\
x(t)&=&x
\end{eqnarray}
where $f(s,x,u)$ is a continuous function on the set $[t_{0},t_{1}]\times\mathbb{R}^{m}\times U$, where $U\subseteq\mathbb{R}^{m}$ is an open set called the control space. $f$ also follows the property,
\begin{eqnarray}
\forall s\in[t,t_{1}], x,y\in\mathbb{R}^{n},|u|\leq\rho,~~|f(s,x,u)-f(s,y,u)|&\leq & K_{\rho}|x-y|
\end{eqnarray}
for some constant $K_{\rho}>0$. The control $u(\cdot)$ must be bounded and belong to the set,  $\mathcal{U}^{0}(t)=L^{\infty}([t,t_{1}]; U)$, so that the IVP satisfied by $x(s)$ has a unique solution.  The problem is to minimize the cost function,
\begin{eqnarray}
J(x,t;u)&=&\int_{t}^{t_{1}}L(s,x(s),u(s))ds+\psi(x(t_{1}))
\end{eqnarray}
over the set of all admissible controls $\mathcal{U}^{0}(t)$. This minimum is defined as the value function,
\begin{eqnarray}
V(t,x)&=&\inf_{u\in\mathcal{U}^{0}(t)}J(x,t;u).
\end{eqnarray}
The book states that if the control space is a compact set, $V(t,x)>-\infty$. Is it because, the $J(x,t;u)$ is somehow continuous wrt $u$ in the space $\mathcal{U}^{0}(t)$? But still, where is the compactness of $U$ getting used.

Thanks in advance.

答:
This addresses a question in the comments as (assuming $L,\psi$ are continuous) the answer follows from the fact that the solution $x_{x_0,u}$ to the ode. is a continuous function of $x_0,u$.

A relevant result is Theorem 3 from Section XVI.1 of Kantorovich & Akilov, "Functional Analysis", 2nd. Ed:
Theorem 3. If $P_y$ satisfies condition (2) for each $y \in Y_0$, with $\alpha$ independent of $y$, and if $P_y$ is continuous in $y$ at a point $y_0 \in Y_0$, then the solution of (9) depends continuously on $y$ at $y=y_0$.
(The (2) means that $P_y$ is an $\alpha$ contraction, (9) means $x=P_y(x)$ and continuity means sequential continuity.) The result is remarkably straightforward to prove, it is surprising to me that it does not have the same prominence as a result as the basic Picard theorem given its utility in many situations.

Note that the Lipschitz rank $K_\rho$ is uniform as long as $\|u\|_\infty \le \rho$, and in the following I am usingthe $L^\infty$ norm (on a relevant interval).

Let $P_{(x_0,u)}(x)(\tau) = x_0+\int_t^\tau f(s,x(s),u(s))ds$. The usual Picard theorem shows that for for some $T$ (in $(t,t_1]$), if $\tau \in [t,T]$ then $P_{(x_0,u)}$ is a contraction (specifically if $K_\rho(T-\tau)<1$)and hence defines a unique solution $x_{(x_0,u)}$ on $[\tau,T]$ and this is repeated on subsequent intervals to show that there is a unique solution defined on $[t,t_1]$.

For the moment restrict attention to the interval $[t,T]$, then the above
shows that if $(x_0,u)_k \to (x_0,u)$, then the solution to $x=P_{(x_0,u)}(x)$ satisfies $x_{(x_0,u)_k} \to x_{(x_0,u)}$. To extend to the whole interval, note that the initial condition on the next interval starting at $T$ is $x_{(x_0,u)}(T)$ which is continuously dependent on $(x_0,u)$ hence we see that the entire solution on $[t,t_1]$ depends continuously on $(x_0,u)$.