问:
why can only universal statements be vacuously true?
So I'm new to logic and im getting really confused on definitions namely that of universal and existential quantifiers and why vacuous truths only apply to statements with a universal quantifier $(\forall)$ and not for existential quantifier $(\exists)$.

I read that the 2 statements

(1) $\forall x\in A\,P(x)$

(2) $\forall x(x\in A\implies P(x))$

Are logically equivalent but why cant we say the same thing about the existential quantifier namely why cant we say that statements $(1)$,$(2)$ can be used.

(1) $\exists x\in A\,P(x)$

(2) $\exists x(x\in A\implies P(x))$

So why do we say that only universal or conditional statements are vacuously true?.

答:
The formula $\exists x \in A\;P(x)$ stands for $\exists x(x \in A \land P(x))$. This is because we want the following chain of equivalences to hold:

$$
\begin{align*}
\exists x \in A\;P(x) &\equiv \lnot \forall x \in A\;\lnot P(x) \\
 &\equiv \lnot \forall x(x \in A \implies \lnot P(x)) \\
 &\equiv \exists x(x \in A \land P(x))
\end{align*}
$$
where the first equivalence is the desired De Morgan law, the second follows from the definition as you give it and the third follows because to falsify $\phi \implies \psi$ you have to make $\phi$ true and $\psi$ false.

$\exists x(x \in A \implies P(x))$ would be trivially true (unless $A$ is the entire universe of discourse) since for $x \not\in A$, $x \in A \implies \phi$ is true for any $\phi$.

The analogue of the vacuously true universal: $\forall x \in \emptyset\;P(x)$ is the vacuously false existential: $\exists x \in \emptyset\;P(x)$.


问:
$ \sum_{i=1}^{10} x_i =5 $, and $ \sum_{i=1}^{10} x_i^2 =6.1 $ ; Find the largest value of these ten numbers

Ten non-negative numbers are such that  $ \sum_{i=1}^{10} x_i =5 $,
and $ \sum_{i=1}^{10} x_i^2 =6.1 $. What is the greatest value of the
largest of these numbers?

It presents a lot of confusion to me, but I think the largest number maybe a integer. Please help me with procedure.

Source : Moscow Institute of Physics and Technology admission assessment for computer Science

答:
Just apply Cauchy-Schwarz or AM-QM inequality to the nine smaller numbers, you get
(WLOG assume $x_1 \le x_2 \le \cdots \le x_{10}$)

$$
\left(\sum_{i=1}^9 x_i\right)^2 \le 9 \left( \sum_{i=1}^9 x_i^2\right)\\
\iff
(5-x_{10})^2 \le 9(6.1-x_{10}^2) \\
\iff 10x_{10}^2 - 10x_{10} - 29.9 \le 0 \\
\implies x_{10} \le 2.3.\blacksquare$$

ContentLicense: CC BY-SA 4.0


问:
Minimum Number of operations needed to break a number into numbers less than a target
Suppose we have two numbers $m$ and $n$, $m > n$. For a single operation, we are allowed to split $m$ into the sum of two smaller numbers each of which is greater or equal to one. We want to find the minimum number of operations to break $m$ down so that each component is less than or equal to $n$. The answer is the largest integer less than or equal to $\frac{m-1}{n}$.

For example, for $m = 4, n = 2$ you just need one operation to break $4$ into $2$ and $2$.

My thinking is that in the end the nicest situation is that you want each component to be at least $n$, so it is natural to look at $\frac{m}{n}$, but then you will have to use $\frac{m}{n} - 1$ operations because for each new component you need one operation and you originally start with one component. But this is certainly not correct if you try $m = 5, n = 2$.

答:
Any number $m$ can be  split up into into Int$(\frac{m}{n})$ copies of $n$ plus a non-negative remainder less than $n$, where Int$(x)$ represents the integer part of $x$.
This takes Int$(\frac{m}{n})$ operations unless the remainder is zero when one fewer operation is required.

Note that the restriction "each of which is greater than one" has to be removed. Otherwise you will not always be able to get to numbers less than or equal to $n$.


问:
Show that if $f(a) = 0$, then $a ^{−1}$ is a zero of the polynomial $g(x) = a_0x ^n + a_1x ^{n−1} + ... + a_n$
Let $F$ be a field, $f(x) = a_nx^n + ... + a_1x + a_0$ a polynomial in $F[x]$, and let $a$ be a nonzero element
of $F$.

Show that if $f(a) = 0$, then $a^{−1}$ is a zero of the polynomial $g(x) = a_0x^n + a_1x
^{n−1} + ... + a_n$.

My attempt:

We have $a.a^{-1}=1$ and thus $a^{-1} \not= 0$ since $a \not= 0$.

Now, if $f(a) = 0$, then $a_na^n +...+ a_1a + a_0 = 0$ and thus $a_n =...=a_1=a_0=0$ (since $a \not= 0$

$a^{-1}$ is a zero of $g(x)$ if $g(a^{-1})=0$.

Now $g(a^{-1}) = a_0(a^{-1})^n + a_1(a^{-1})^{n-1} +...+ a_n$

How to continue from here, any help please? Thank you.

答:
It is not true the every $a_k$ must be $0$. For instance, $a=2$ is root of $x-2$, but the coefficientis of $x-2$ are not all equal to $0$.

If $a_na^n+a_{n-1}a^{n-1}+\cdots+a_1a+a_0=0$, then$$\frac{a_na^n+a_{n-1}a^{n-1}+\cdots+a_1a+a_0}{a^n}=0,$$which is the same thing as asserting that$$a_0\left(a^{-1}\right)^n+a_1\left(a^{-1}\right)^{n-1}+\cdots a_{n-1}a^{-1}+a_n=0.$$


问:
Criteria of compactness of an operator
Suppose $K$ is a linear operator in a separable Hilbert space $H$ such that for any Hilbert basis $\{e_i\}$ of $H$ we have $\lim_{i,j \to \infty} (Ke_i,e_j) = 0$.

Is it true that $K$ is compact?

Thanks in advance for any help.

答:
Yes, $K$ is compact.

Note first that we can assume that $K$ is selfadjoint; indeed, if $K$ satisfies the hypothesis, so do $K^*$ and $K\pm K^*$, and if we prove that the real and the imaginary parts of $K$ are compact, then we have that $K$ is compact.

Using the Weyl-von Neumann-Berg Theorem (II.4.1 or II.4.2 in Davidson's C$^*$-algebras by example) we can write
$$
K=A+T,
$$
where $A$ is selfadjoint and diagonal in some basis; and $T$ is compact. Being compact, $T$ satisfies the hypothesis and so $A=K-T$ does too. Now, since $A$ is diagonal, the hypothesis implies that its diagonal tends to zero, and so it is compact. Then $K$ is compact, being a sum of compacts.

To finish, note that it is essential that the property of small entries of $K$ holds for any basis. If we fix a single basis, it is possible to have operators $K$ satisfying $\langle Ke_i,e_j\rangle\to0$ (in the sense of the question) but not compact. Indeed, let
$$
K=\bigoplus_n\begin{bmatrix}1/n&\cdots&1/n\\ \vdots&\ddots&\vdots\\ 1/n&\cdots&1/n\end{bmatrix}.
$$
This is an orthogonal sum of pairwise orthogonal projections, so it is an infinite-dimensional projection, thus not compact.

ContentLicense: CC BY-SA 3.0

Title: Criteria of compactness of an operator


问:
Prove $\ln(a \cdot b)=\ln(a)+\ln(b)$, $a,b>0 $ and $\lim_{a \to 0} \frac{\ln(1+a)}{a} =1$.But we shall not even know that the exponential function.
In Japanese high school education, the number $e$ is defined first. There is no disagreement that $e^q$ can be defined algebraically when the rational number $q$ is an exponent. However,  there is a sudden "deregulation" that $q$ can be a real number. Until now, I had no doubts about this. But now I'm confused about it.
(I am a user of college-level mathematics, and probably know the epsilon-delta argument. But I'm not familiar with rigorous mathematics.)

I found the way exponential functions were introduced in Japanese high school to be a bit mathematically imprecise. I overheard someone say, "That can be essentially eliminated with only high school knowledge by starting with defining the logarithmic function in terms of integrals. My question is based on that sense. Specifically;

We assume that we do not know the existence or properties of the $\exp(x)$.

We assume that we do not know that $\ln(x)$ is the logarithm.

But, we assume that we know general properties of differentiation and integration, such as integration by substitution.

We also assume that the function $\ln$ is defined as follows;
$$\ln(x):=\int_{t=1}^{t=x} \frac{1}{t}\ dt \tag{0-1}$$

We do not know any more about ln. For example, we do not know the following formulas;
$$\lim_{a \to 0}\ \frac{\ln(1+a)}{a} =1 \tag{0-2}$$

My Question:

Under the these assumptions, I would like to derive the following formulas. In particular, we want to pay attention to whether we are violating mathematical rules (e.g., improper exchange of integrals and limits).
$$\ln(a \cdot b)=\ln(a)+\ln(b) \ , a,b>0 \tag{1}$$
$$\lim_{a \to 0} \frac{\ln(1+a)}{a} =1 \ ,a>0 \tag{2}$$

Perhaps an outline of the solution would be as follows;

For (1)

$$\ln(a \cdot b):=\int_{t=1}^{t=a \cdot b} \frac{1}{t} dt$$
$$=\int_{t=1}^{t=a} \frac{1}{t}\ dt + \int_{t=a}^{t=a \cdot b} \frac{1}{t}\ dt$$
$$=\ln(a) + \int_{t=a}^{t=a \cdot b} \frac{1}{t}\ dt \tag{1-1}$$

Using $t=s*a$, second term on the right side is
$$\int_{t=a}^{t=a \cdot b} \frac{1}{t}\ dt
=\int_{s=1}^{s=b} \frac{1}{sa} \cdot a\ ds
=\int_{s=1}^{s=b} \frac{1}{s} \ ds
=\ln(b) \tag{1-2}$$

For (2)

$$\ln(1)=\int_{t=1}^{t=1} \frac{1}{t}\ dt=0 $$

Therefore,
$$\frac{\ln(1+a)}{a}
=\frac{\ln(1+a)+0}{a}$$
$$=\frac{\ln(1+a)-\ln(1)}{a}  + \frac{\ln(1)}{a}
=\frac{\ln(1+a)-\ln(1)}{a}  +0 $$
If we take the limits of both sides of the above equation, we can show the (2).
That is;
$${\lim}_{a \to 0} \frac{\ln(1+a)}{a}=
{\lim}_{a \to 0}\frac{\ln(1+a)-\ln(1)}{a}
=\ln'(1)=\frac{1}{1}=1$$
Again, I emphasize that my main concern is not the solution itself, but the  "violate mathematical rules". In other words, does the above solution do anything mathematically inappropriate? (e.g., use of circular argument, improper ordering of integrals and limits, etc.)

答:
Here's a different proof of the formula $\ln(ab)= \ln(a) + \ln(b)$.

Suppose that $a$ is a positive constant.

And suppose $f$ is the function defined over the positive reals by:

$f(x) = \ln(ax) - \ln(x) - \ln(a)$.

$f$ is differentiable over the positive reals. By the chain rule,

$f'(x) = a \frac{1}{ax} - \frac1{x} = 0$

So $f$ is constant.

As such, for all positive reals $x$, $f(x) = f(1) = \ln(a) - \ln(a) -\ln(1) = 0$.

Which concludes our proof.


问:
How to compute $1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$
How to compute $S = 1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$

Was thinking $\frac{S}{24} = {4\choose 4} + {6\choose 4} + {8\choose 4} + ... + {100\choose 4}$ but how do I sum this up?

Body:
Hint:

The $n$th term $T(n)$ is $$(2n-1)(2n)(2n+1)(2n+2) =16n^4+16n^3-4n^2-4n$$

WLOG $$T(n)=P(n+1)-P(n)$$ where $P(m)=\sum_{r=0}^ua_rm^r$

$$\sum_{r=0}^5a_r((n+1)^r-n^r)$$

$$=a_1+a_2(2n+1)+a_3(3n^3+3n+1)+a_4(4n^3+6n^2+4n+1)a+a_5(5n^4+10n^3+10n^2+5n+1)+\cdots$$

Clearly, $a_r=0$ for $r\ge6$

Now compare the  coefficients of $n^4,n^4,n^2,n$ and constants to find $a_j;0\le j\le5$

Comparing the coefficients of $n^4,$ $$5a_5=16$$


问:
proving that $\alpha_1+\alpha_2<-3$ which are the negative roots of $f(x)$
I am stuck on the following problem:

Given
\begin{align*}
f(x) &= x^4 + (-m - 2n + 6)x^3 + (mn - 5m - 8n + 10)x^2\\
&\quad + (3mn - 7m - 8n)x + mn-m-n-7,
\end{align*}
if $\alpha_1,\alpha_2$ are two negative roots of $f(x)=0$, show that $\alpha_1+\alpha_2< -3$ for $m,n\ge 4$.

I took some values of $m,n$ and found the following:

$m=n=4$, $\alpha_1=-2.476, \alpha_2=-0.621$;

$m=5,n=4$, $\alpha_1=-2.49, \alpha_2=-0.59$

I tried some other values of $m,n$,
In all the above cases, I am getting $\alpha_1+\alpha_2<-3$.

But I am not able to prove it. Can someone please help me out?

NOTE:

I found the following. I obtained $-1<\alpha_1<-k$ and $-3<\alpha_2<-3+k$
where $k$ is some function of $m$ and $n$.

I took $m=4,n=5$ and I got $\alpha_1=-2.492<-3+0.55, \alpha_2=-0.592<-0.55$.
I took some other values of $m,n$ and obtained similar stuff.
So I think my guess is correct.

Is there any way to find $k$ such that $k$ is a function of $m,n$?

Body:
We have
\begin{align}
f(-3) &= mn + 2m + 5n + 2 > 0, \\
f(-3/2) &= \frac{-20mn + 26m - 4n + 5}{16} < 0,\\
f(0) &= mn - m - n - 7 > 0, \\
f(m) &= -m^3 n+m^3-5 m^2 n+3 m^2-7 m n-m-n-7 < 0, \\
f(2m + 2n) &= 8 m^4+28 m^3 n+32 m^2 n^2+12 m n^3+28 m^3+78 m^2 n+66 m n^2\\
&\quad +16 n^3+26 m^2+51 m n+24 n^2-m-n-7\\
& > 0.
\end{align}
(Note: Simply letting $m = 4 + s, n = 4 + t$ for $s, t \ge 0$, all the inequalities are obvious.)

Thus, $f(x) = 0$ has four real roots $x_1, x_2, x_3, x_4$ located in the following intervals respectively
$$x_1 \in (-3, -3/2), \quad x_2 \in (-3/2, 0), \quad x_3 \in (0, m), \quad x_4 \in (m, 2m+2n).$$

We need to prove that $x_1 + x_2 < -3$ or $x_2 < -3 - x_1$.

Since $-3 - x_1 \in (-3/2, 0)$, it suffices to prove that $f(-3-x_1) > 0$.

Since $f(x_1) = 0$, it suffices to prove that $f(-3-x_1) - f(x_1) > 0$ that is
$$(2x_1 + 3)\Big((m+2n)x_1^2 + 3(m+2n)x_1 + m+2n + 3\Big) > 0.$$

It suffices to prove that
$$x_1 > - \frac{3}{2} - \sqrt{\frac{5m + 10n - 12}{4m + 8n}}.$$

It suffices to prove that
$$f\left(- \frac{3}{2} - \sqrt{\frac{5m + 10n - 12}{4m + 8n}}\right) > 0$$
that is
$$\frac{m^3+2 m^2 n+2 m n^2+4 n^3-m^2-10 m n-16 n^2+3 m+6 n+9}{(m+2 n)^2} > 0$$
which is true. (Note: Simply letting $m = 4 + s, n = 4 + t$ for $s, t\ge 0$, this inequality is obvious.)

We are done.

$\phantom{2}$

For @Math_Freak: The Maple code for the last two equations is given by

f := x^4+(-m-2*n+6)*x^3+(m*n-5*m-8*n+10)*x^2+(3*m*n-7*m-8*n)*x+m*n-m-n-7
x := -3/2+Q
f1 := collect(expand(f), Q)
f2 := subs({Q^2 = (5*m+10*n-12)/(4*m+8*n), Q^3 = (5*m+10*n-12)*Q/(4*m+8*n), Q^4 = ((5*m+10*n-12)/(4*m+8*n))^2}, f1)
factor(f2)


问:
When does pointwise convergence imply uniform convergence?
On an exam question (Question 21H), it is claimed that if $K$ is compact and $f_n : K \to \mathbb{R}$ are continuous functions increasing pointwise to a continuous function $f : K \to \mathbb{R}$, then $f_n$ converges to $f$ uniformly. I have tried proving this claim for the better part of an hour but I keep coming short. I suspect a hypothesis on equicontinuity has been omitted — partly because the first half of the question is about the Arzelà–Ascoli theorem — but I don't have access to the errata for the exam so I can't be sure.

Here is my attempted proof: let $g_n = f - f_n$, so that $(g_n)$ is a sequence of continuous functions decreasing pointwise to $0$. Clearly, $0 \le \cdots \le \| g_n \| \le \| g_{n-1} \| \le \cdots \le \| g_1 \|$, so we must have $\| g_n \| \longrightarrow L$ for some constant $L$. $K$ is compact, so for each $g_n$, there is an $x_n \in K$ such that $g_n(x_n) = \| g_n \|$, and there is a convergent subsequence with $x_{n_k} \longrightarrow x$ for some $x \in K$. By hypothesis, $g_n(x) \longrightarrow 0$, and by construction $g_{n_k} (x_{n_k}) \longrightarrow L$. I'd like to conclude that $L = 0$, but to do this I would need to know that the two sequences have the same limit. This is true if, say, $\{ g_n \}$ is an equicontinuous family, but this isn't one of the hypotheses, so I'm stuck.

答:
The result is true as stated and is called Dini's Theorem.  A proof can be found in Chapter III, Section 1.4 of these notes or indeed in this wikipedia article.  (My notes are for a sophomore-junior level undergraduate class, so it is stated in the case in which $K$ is a closed interval.  But it is clear that the argument works for any compact topological space.)

(For some reason this is one of those named theorems that tends to be assigned as an exercise or come up on exams...)


问:
$f, g$ measurable, $\lVert g\rVert_1=1$ and $0 \le a \le f \le b$ so $\int_{\mathbb{R^n}} \frac{g}{f}dx \ge \frac{1}{\int_{\mathbb{R^n}} fg dx}$
Let $g \ge 0$ be measurable in $\mathbb{R^n}$ s.t. $\lVert g\rVert_1=1$ and $f$ measurable s.t. $a \le f \le b$, for $a, b \gt0$, show that: $$\int_{\mathbb{R^n}} \frac{g}{f}dx \ge \frac{1}{\int_{\mathbb{R^n}} fg dx}$$

My idea was:

$$1 = \lVert g\rVert_1 =\int_{\mathbb{R^n}} |g|dx = \int_{\mathbb{R^n}} \sqrt{fg}\sqrt\frac{g}{f}dx \le \biggl( \int_{\mathbb{R^n}} fg dx \biggr)^{\frac{1}{2}}\biggl(\int_{\mathbb{R^n}} \frac{g}{f}dx\biggr)^{\frac{1}{2}} $$

but, to use Hölder's inequality I have to be sure that $f \in L^1(\mathbb{R^n})$, how to see that (if my reasoning was right)?

答:
You cannot be sure that $f \in L^1(\mathbb{R}^n)$ because it is not True. For example, if $f = a$, you don't have $f \in L^1(\mathbb{R}^n)$.

However, you can "make" $f$ be in $L^1(\mathbb{R}^n)$ by considering $f\mathbb{1}_{||x||<n}$ for a natural number n. Appling the reasoning you used, you get $1 \leq (\int_{||x|| < n} fg)^{\frac{1}{2}} (\int_{||x|| < n} \frac{g}{f})^{\frac{1}{2}}$.

You can conclude by letting $n \to +\infty$, as $\frac{g}{a}$ and $gb$ are dominating functions for the dominated convergence theorem.


问:
Evaluating the indefinite integral $\int\log\!\left(x+\sqrt{x^2-1}\right)\!dx$
I came across the following integral, and I don't know how to solve it.
$$
\int\log\left(x+\sqrt{x^2-1}\right)dx
$$
I tried the "obvious" substitution of $x=\sec\theta$, which gives you:
$$
\int\tan\theta\sec\theta\log\left(\tan\theta+\sec\theta\right)d\theta
$$
However, this doesn't simplify it much, in the sense that I have no idea how to solve this now! Take out the common factor of $\sec$ from the log, or perhaps do $u=\tan\theta+\sec\theta$, but both of these lead to dead ends (at least for me).

In case you are wondering, Wolfram|Alpha claims the answer is:
$$x\log\left(x+\sqrt{x^2-1}\right)-\sqrt{x^2-1}+c$$

答:
Recognize $\log\left(x+\sqrt{x^2-1}\right)=\cosh^{-1}x$ and then integrate by parts

$$
\int\log\left(x+\sqrt{x^2-1}\right)dx=\int \cosh^{-1}x\>dx
= x \cosh^{-1}x-\int \frac x{\sqrt{x^2-1}}dx
$$
where the remaining integral is just $\sqrt{x^2-1}$.


答:
As I understand it, the implication $A\to B$ is said to be vacuously true if $A$ is false. We can see this from lines 3 and 4 of the truth table for implication:

$\neg A \to (A\to B)$ is also a tautology:

This can also be proven using a form of natural deduction:

Vacuous truth doesn't necessarily have anything to do with quantifiers or empty sets, though it is used, for example, to show that the empty set is unique, i.e. that all empty sets are equal. It is also used in proofs by cases when one or more of the cases being considered are proven or assumed to be false.

Title: why can only universal statements be vacuously true?

Body:
So I'm new to logic and im getting really confused on definitions namely that of universal and existential quantifiers and why vacuous truths only apply to statements with a universal quantifier $(\forall)$ and not for existential quantifier $(\exists)$.

I read that the 2 statements

(1) $\forall x\in A\,P(x)$

(2) $\forall x(x\in A\implies P(x))$

Are logically equivalent but why cant we say the same thing about the existential quantifier namely why cant we say that statements $(1)$,$(2)$ can be used.

(1) $\exists x\in A\,P(x)$

(2) $\exists x(x\in A\implies P(x))$

So why do we say that only universal or conditional statements are vacuously true?.


问:
The characteristics of a subfield of a field is same as that of the field.
How to show that the characteristics of a subfield of a field is same as that of the field??</p>

答:
If we suppose that $ char(F)=d $ and $char(K)=s $
Then   obviously $s<=d$
Now  if $ s<d $ then consider $ k= d-s$ ,a positive integer.
If we choose  an element  $b$ from the subfield $K$
Then it wil also belongs  to the field $F$
And $ k*b=0  $
Which imply that $ char( F)=k< d $
which is a contradiction  to our hypothesis that $ char(F)= d$
So we should have $s=d$


问:
$L^2$ convergence for variance of a Brownian motion
The Problem
Given $(B_s,s \in [0,t])$ a Brownian motion on $[0,t]$ and $0=t_0 < t_1 < ... < t_n \le  t$  a  partition of  $[0,t]$ such that  $\max_{j}|t_{j+1} - t_j| \to 0$ as $n \to \infty$. Show that the random   variable $$\sum_{j=0}^n (B_{t_j} - B_{t_{j-1}})^2$$ converges to $t$ in  $L^2$ as  $n \to  \infty$.

I am taking this class in stochastic calculus and I am very dry on probability theory. However, I think I understand key concepts of Brownian Motion, so I need some help putting together the pieces. I know I want to show that
$$\lim_{n \to \infty} \left\Vert \sum_{j=0}^n (B_{t_j} - B_{t_{j-1}})^2-  t \right\Vert_2 = \lim_{n \to \infty} \mathbb{E}\left[ \left| \sum_{j=0}^n (B_{t_j} - B_{t_{j-1}})^2-  t \right|^2 \right]=0$$

I also realize that $$\mathbb{E} \left[ \sum_{j=0}^n (B_{t_j} - B_{t_{j-1}})^2 \right ]= \sum_{j=0}^n \mathbb{E} \left[ (B_{t_j} - B_{t_{j-1}})^2 \right ]$$
$$=\sum_{j=0}^n t_j - t_{j-1}=t$$
by the property of the variance of Brownian motion and because the sum of the $t_i$ is $t$. I guess by biggest problem is I don't know how to mathematically break into this $L^2$ norm. Any help is greatly appreciated.

答:
Hint

In your partition, you took $t_n\leq t$, but you must take $t_n=t$. Then,

\begin{align*}
&\mathbb E\left[\left(\sum_{j=0}^{n-1}(B_{t_{j+1}}-B_{t_j})^2-t\right)^2\right]\\
&=\mathbb E\left[\left(\sum_{j=1}^{n-1}(B_{t_{j+1}}-B_{t_j})^2-(t_{j+1}-t_j)\right)^2\right]\\
&=\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}\mathbb E\left[\Big((B_{t_{j+1}}-B_{t_j})^2-(t_{j+1}-t_j)\Big)\Big((B_{t_{i+1}}-B_{t_i})^2-(t_{i+1}-t_i)\Big)\right].
\end{align*}

I let you continue.


问:
Transition maps
Let $S^1=\{(x,y) \in \mathbb R^2 : x^2+y^2=1\}$, $U=\{(\cos(a), \sin(a)): a\in]0,2\pi[\}, $ $V=\{(\cos(a), \sin(a)): a\in]-\pi,\pi[\}$.

$\phi: U\to \mathbb R$ and $\psi :V \to \mathbb R$ are defined by
$$\phi((\cos(a), \sin(a))=a $$ and $$\psi((\cos(a), \sin(a))=a. $$

I want to show that $\psi \circ \phi^{-1}$ is a transition maps.
I did find that $$\psi(U\cap V) =(]-\pi, 0[\, \cup\,]0,\pi[)\not \subset]0,2\pi[=\psi(U) $$  Also I couldn't find $\psi \circ \phi^{-1}$.

Any help is appreciated.

答:
Try to visualize how the transition maps are operating:
We know that $\psi \circ \phi^{-1}:(0, \pi)\cup(\pi, 2\pi) \rightarrow (-\pi, 0)\cup(0, \pi).$

But we know that cos and sin take the same values on $(\pi, 2\pi)$ and $(-\pi, 0)$, and it seems clear that the angle is not changed in the interval $(0, \pi)$.

Let's figure out explicitly $\psi \circ \phi^{-1}$ acts on $(0, \pi)$:

Let $\alpha \in (0, \pi) \ \Rightarrow \ \phi^{-1}: \alpha \mapsto (\text{cos }  \alpha, \text{sin }\alpha).$

Then $\psi: (\text{cos }  \alpha, \text{sin }\alpha) \mapsto \alpha$.

Thus the transition map acts as identity on $(0, \pi).$

Now let $\alpha \in (\pi, 2\pi) \ \Rightarrow \ \phi^{-1}: \alpha \mapsto (\text{cos }  \alpha, \text{sin }\alpha).$

Note that $\alpha \notin (-\pi, \pi)$, but $\alpha - 2\pi$ is! Then $\psi: (\text{cos }  \alpha, \text{sin }\alpha) \mapsto \alpha-2\pi$. To conclude we have:

$\psi \circ \phi^{-1}(\alpha)=\begin{cases}
\alpha & \text{if } \alpha \in (0, \pi) \\
\alpha - 2\pi & \text{if } \alpha \in (\pi, 2\pi)
 \end{cases} $

Note that since $\pi$ is not an element of the domain, this map is continous and smooth.

Title: Transition maps


问:
Uniqueness of 2nd order IVP
How do I show that the IVP

$x''(t)=\arctan{(x(t))}\arctan{(t)}$ with $x(0)=e$, $x'(0)=e^2$

is or isn't uniquely solvable and defined on $\mathbb{R}$?

答:
The solution is unique on $\mathbb{R}$.

The most popular way to verify whether a solution of dynamical system is unique is to show Lipschitzness condition as stated below.

[1, Theorem 3.2] (Global Existence and Uniqueness) Suppose that $f(t, x)$ is piecewise continuous in $t$ and satisfies $\lVert f(t, x) - f(t, y) \rVert \leq L \lVert x-y \rVert$
$\forall x, y \in \mathbb{R}^n$, $\forall t \in [t_0, t_1]$.
Then, the state equation $\dot{x} = f(t, x)$, with $x(t_0) = x_0$, has a unique solution over $[t_0, t_1]$.

In this case, $f(t, x) = [x_2 \ \arctan(t) \arctan(x_1)]^T$ where $x = [x_1 \  x_2]^T$.
Observe that $\lvert \arctan(t) \arctan(x_1) - \arctan(t) \arctan(y_1) \rvert \leq (\pi/2)\lvert x_1 - y_1 \rvert$ (Note that $g(x) := \arctan(x) \Rightarrow \lvert g'(x) \rvert=\frac{1}{1+x^2} \leq 1 \Rightarrow g$ is Lipschitz continuous with Lipschitz constant of $1$).

Hence, $\lVert f(t, x) - f(t, y) \rVert \leq (\pi/2)\lVert [(x_2-y_2) \  (x_1-y_1)]^T \rVert = (\pi/2) \lVert x - y\rVert$ implies that the dynamical system has a unique solution on $\mathbb{R}$ (Hint: applying this to forward and backward cases).

Reference

[1] H. K. Khalil, Nonlinear Systems, 3rd ed. Upper Saddle River, NJ: Prentice Hall, 2002.

Title: Uniqueness of 2nd order IVP


问:
Finding nth term for a recursive/iterative/term to term sequence
I have the sequence: 1, 16, 166, 1666 . Im trying to find the nth term for this sequence but since it is not linear or geometric I'm not sure how to. I worked out that the recursive formula is:
\begin{gather}
a_1 = 1 \\
a_{n + 1} = 10a_{n} + 6
\end{gather}
I'm trying to write this as an nth term and I saw that one solution was:
\begin{gather}
\frac{1}{6}\left ( 10 ^{n}-4\right )
\end{gather}
But I don't know how to derive this. I have seen other answers on this site but the explanations arent often very clear or involve math that is beyond my ability. For reference I am a high schooler in my final year, if anyone is familiar with the British Education system I am studying further maths for a level so I don't know any math more complex than that.
Thank you

答:
The Online Encyclopedia of Integer Sequences A246057 shows$\quad a(n) = (5\times10^n - 2)/3$

It generates
$\\
16\quad
166\quad
1666\quad
16666\quad
166666\quad
1666666\quad
16666666\quad
166666666\quad
\cdots$

Working backwards, we see that all of these number resemble $x+\dfrac23$ so we can try
$$3\times 16=48\qquad 3\times 166=498\qquad 3\times1666=4998
\quad\cdots\quad $$

and we can see that each is $2$ short of $5\times10^n.\quad$ From here it should be easy to see how the formula was developed and how it works.


问:
Eigenvalues of a linear transformation $M_{2,2}\ \rightarrow M_{2,2}$
I took linear algebra years ago, and for various reasons I'm going back to re-learn it. In that process, I came across the following question in an old exam from the course I once took:

A linear operator g: $M_{2,2}\ \rightarrow M_{2,2}$ is given by:

$$ A\rightarrow  \pmatrix{0 & 1\\1 & 0} A - A^T \pmatrix{0 & 1\\1 & 0} $$

Determine the eigenvalues of g and the dimensions of the relative eigenspaces.

Looking over the textbook I used in that course, I didn't see any material on eigenvalues of general linear transformations, but I decided to give it a shot.

To do this, I set $$A=\pmatrix{a & b\\c & d} $$ and computed the right hand side, so:

$$ \pmatrix{a & b\\c & d}\rightarrow  \pmatrix{0 & 1\\1 & 0} \pmatrix{a & b\\c & d} - \pmatrix{a & c\\b & d} \pmatrix{0 & 1\\1 & 0}= \pmatrix{c & d\\a & b}-\pmatrix{c & a\\d & b}=\pmatrix{0 & d-a\\a-d & 0}$$

Or:

$g \pmatrix{a\\b\\c\\d}=\pmatrix{0\\d-a\\a-d\\0}$

So for g to have an eigenvalue, $\lambda$, we would need

$g \pmatrix{a\\b\\c\\d}= \lambda \pmatrix{0\\d-a\\a-d\\0}$

for some vector, $\pmatrix{a\\b\\c\\d}$, that is not the zero-vector. It's clear that this would require $a=0$ and $d=0$. However, any $\pmatrix{0\\b\\c\\0}$ is mapped into the zero-vector.

Therefore, $g \pmatrix{a\\b\\c\\d}= \lambda \pmatrix{0\\d-a\\a-d\\0}$  is satisfied by any vector $\pmatrix{0\\b\\c\\0}$ with eigenvalue $\lambda=0$, with two eigenvectors, $\pmatrix{0\\1\\0\\0}$ and $\pmatrix{0\\0\\1\\0}$, meaning it has a two dimensional eigenspace.

Is my thinking here correct? Am I way off? I'd appreciate any comments.

Note: I made an edit to my question, and I originally said it had no eigenvalue.

答: Your computations are fine (well, see the comment that you got). However, after having got the linear map$$g(a,b,c,d)=(0,d-a,a-d,0),$$you can stay that its matrix with respect to the standard basis is$$M=\begin{bmatrix}0&0&0&0\\-1&0&0&1\\1&0&0&-1\\0&0&0&0\end{bmatrix},$$whose characteristic polynomial is $\lambda^4$. So, the only eigenvalue of $M$ is $0$. Now, all that remains to be done is to solve the system$$\left\{\begin{array}{l}0=0\\-a+d=0\\a-d=0\\0=0.\end{array}\right.$$


问:
Are global sections of varieties determined by their local rings?
Let $(X,O_X)$ be the ringed space associated to a "classical projective" variety over an algebraically closed field $k$. (i.e. $O_X$ is sheaf of regular functions, in the sense of Chapter I. Hartshorne.  or Chapter 5,6 this notes ). Is it true that
$$O_X(X)\simeq \bigcap_{p \in X} O_{X,p}?$$
More generally, is it true that for any open

$$O_X(V)= \bigcap_{p\in V} O_{X,p} ?$$
(Is there an analogous characterization for ringed spaces induced from schemes?)

I could prove this in the case of affine varieties: this follows from that if $B$ is an integral domain, then $$B=\bigcap B_{m} \subset Frac(B)$$
where $m$ ranges all the maximal ideals.

Any references would be appreciated.

答:
First, you need to assume $X$ is irreducible (which is maybe part of your definition of "variety") for the question to even make sense: what does $\bigcap_{p \in X} O_{X,p}$ even mean?  When $X$ is irreducible, you can consider all of these local rings as subrings of the function field of $X$ and take their intersection there.  Without this assumption, though, there is no natural way to consider all the local rings as subrings of some other ring.

In any case, here is a more general statement:

Theorem: Let $X$ be an integral scheme in which every nonempty closed set contains a closed point (this is automatic if $X$ is Noetherian, for instance).  Then $O_X(X)=\bigcap_p O_{X,p}$ as subrings of the function field of $X$, where $p$ ranges over all closed points of $X$.

Proof: Let $K(X)$ denote the function field of $X$.  An element $f\in K(X)$ is in $O_X(X)$ if for every $p\in X$, there is a neighborhood $U$ of $p$ such that $f\in O_X(U)$.  Indeed, if this is the case, then these local sections in $O_X(U)$ can be glued together to give a global section of $O_X$.  But now if $f\in O_{X,p}$, that by definition means that there is a neighborhood $U$ of $p$ such that $f\in O_X(U)$.

So, we have $O_X(X)=\bigcap_{p\in X}O_{X,p}$.  All that remains is to reduce the intersection down to just the closed points.  Now note that if $q\in X$ is any point, then there is a closed point $p$ in $\overline{\{q\}}$.  Then $O_{X,p}\subseteq O_{X,q}$, since every neighborhood of $p$ is also a neighborhood of $q$.  So, $\bigcap_{p\in X}O_{X,p}$ is the same as the intersection where you only consider closed points.


问:
Norm on a integral
Let $E= C([a,b])=\{f:[a,b]\rightarrow \mathbb{R}; f \ \text{is continuous} \}$ and $|| \quad||_1: E \rightarrow \mathbb{R}$, defined as $$||f||_1=\int_a^b|f(x)|dx.$$

Show that there is no $c>0$ for which $$\mathop{max}_{x \in [a,b]} f(x)<c||f||_1.$$

I tried to do some examples first, but i think that there is some information missing in the question, if we use $f(x)=a$, $a$ is some constant, we can find an c. Any tips on how to solve it?

答:
In other words, it says that the quantity
$$
\frac{\max_{x\in [a, b]} f(x)}{||f||_{1}}
$$
is unbounded. To make this large, imagine the situation where $||f||_{1}$ is fixed but $\max_{x\in [a, b]}f(x)$ being larger. How about a function that looks like a spike? More precisely, consider a triangle-like function where the area is fixed but the graph gets steeper and steeper.

Title: Norm on a integral


答:
You are correct in your proof that $\lambda$ must be $0$.

However, you have then overlooked the eigenvector $\pmatrix{1\\0\\0\\1}$.

Title: Eigenvalues of a linear transformation $M_{2,2}\ \rightarrow M_{2,2}$

Body:
I took linear algebra years ago, and for various reasons I'm going back to re-learn it. In that process, I came across the following question in an old exam from the course I once took:

A linear operator g: $M_{2,2}\ \rightarrow M_{2,2}$ is given by:

$$ A\rightarrow  \pmatrix{0 & 1\\1 & 0} A - A^T \pmatrix{0 & 1\\1 & 0} $$

Determine the eigenvalues of g and the dimensions of the relative eigenspaces.

Looking over the textbook I used in that course, I didn't see any material on eigenvalues of general linear transformations, but I decided to give it a shot.

To do this, I set $$A=\pmatrix{a & b\\c & d} $$ and computed the right hand side, so:

$$ \pmatrix{a & b\\c & d}\rightarrow  \pmatrix{0 & 1\\1 & 0} \pmatrix{a & b\\c & d} - \pmatrix{a & c\\b & d} \pmatrix{0 & 1\\1 & 0}= \pmatrix{c & d\\a & b}-\pmatrix{c & a\\d & b}=\pmatrix{0 & d-a\\a-d & 0}$$

Or:

$g \pmatrix{a\\b\\c\\d}=\pmatrix{0\\d-a\\a-d\\0}$

So for g to have an eigenvalue, $\lambda$, we would need

$g \pmatrix{a\\b\\c\\d}= \lambda \pmatrix{0\\d-a\\a-d\\0}$

for some vector, $\pmatrix{a\\b\\c\\d}$, that is not the zero-vector. It's clear that this would require $a=0$ and $d=0$. However, any $\pmatrix{0\\b\\c\\0}$ is mapped into the zero-vector.

Therefore, $g \pmatrix{a\\b\\c\\d}= \lambda \pmatrix{0\\d-a\\a-d\\0}$  is satisfied by any vector $\pmatrix{0\\b\\c\\0}$ with eigenvalue $\lambda=0$, with two eigenvectors, $\pmatrix{0\\1\\0\\0}$ and $\pmatrix{0\\0\\1\\0}$, meaning it has a two dimensional eigenspace.

Is my thinking here correct? Am I way off? I'd appreciate any comments.

Note: I made an edit to my question, and I originally said it had no eigenvalue.


答:
Suppose WLOG that $[a,b]=[0,1]$. Take $(f_n)_{n\geq 2}$ defined by $$f_n(x)=\begin{cases}0&x\in [0,1-\frac{1}{n}]\\ nx+1-n&x\in (1-\frac{1}{n},1].\end{cases}
$$
Then, $f_n\to 0$ in $(E,\|\cdot \|_1)$. Now, suppose that there is $c>0$ s.t. for all $f\in E$, $$\max_{[a,b]}|f|\leq c\|f\|_1.$$

In particular, for all $n\in\mathbb N$, $$1=|f_n(1)|\leq \max_{[a,b]}|f_n|\leq c\|f_n\|_1\underset{n\to \infty }{\longrightarrow }0,$$
which is a contradiction. Therefore, such $c>0$ doesn't exist.

Title: Norm on a integral

Body:
Let $E= C([a,b])=\{f:[a,b]\rightarrow \mathbb{R}; f \ \text{is continuous} \}$ and $|| \quad||_1: E \rightarrow \mathbb{R}$, defined as $$||f||_1=\int_a^b|f(x)|dx.$$

Show that there is no $c>0$ for which $$\mathop{max}_{x \in [a,b]} f(x)<c||f||_1.$$

I tried to do some examples first, but i think that there is some information missing in the question, if we use $f(x)=a$, $a$ is some constant, we can find an c. Any tips on how to solve it?


答:
Expanding my comment, it is clearer to write out what we want explicitly. We desire $\lambda$ and $A=\begin{bmatrix} a&b\\c&d \end{bmatrix}\neq 0$ such that the following holds.
$$g(\begin{bmatrix} a&b\\c&d \end{bmatrix}) = \begin{bmatrix} 0&d-a \\ a-d&0 \end{bmatrix}=\lambda\begin{bmatrix} a&b\\c&d \end{bmatrix}.$$

We separate for clarity the cases $\lambda=0$ and $\lambda\neq 0$.

For $\lambda=0$, we see that $d=a$ is the only restriction, so that $b,c$ can be chosen freely. Hence, the eigenspace corresponding to 0 is three dimensional. It has basis
$$\begin{bmatrix} 0&1\\0&0 \end{bmatrix},\begin{bmatrix} 0&0\\1&0 \end{bmatrix},\begin{bmatrix} 1&0\\0&1 \end{bmatrix}.$$

For $\lambda\neq 0$, we see that $a=d=0$ since $\lambda a = \lambda d=0$.  This forces $b=c=0$ since $\lambda c = a-d = 0 = d-a = \lambda b$. Thus there are no eigenvalues that are nonzero.


问:
Hints? Convergence behaviour.

I tried many things, including ratio test, root test, comparison test, cauchy criterion, rewriting w.r.t e^n and doing all these tests again, but no luck in any direction. Any hints?

It seems most likely that I missed out on rewriting this stuff to resemble something that can be compared.

答: Hint: Rewrite the summand so that the numerator and denominator are written as exponentials (that is, $n^{\ln n}=\exp(\ln^2(n))$. Now you can combine the numerator’s and denominator’s exponents since they have the same base. You can show the resulting exponent is greater than $n$ (for example), so you can make a comparison to $\sum e^{-n}$.

Title: Hints? Convergence behaviour.


答:
Root test should work since
$$\sqrt[n]{\frac{n^{\ln n}}{(\ln n)^n}} = \frac{n^{\frac{\ln n}{n}}}{\ln n} = \frac{e^\frac{\ln^2n}{n}}{\ln n}$$

Title: Hints? Convergence behaviour.

Body:
I tried many things, including ratio test, root test, comparison test, cauchy criterion, rewriting w.r.t e^n and doing all these tests again, but no luck in any direction. Any hints?

It seems most likely that I missed out on rewriting this stuff to resemble something that can be compared.


问:
Liouville's Theorem. Example when the theorem does not work if $f$ is not entire function.
I'm struggling with understanding Liouville's theorem, more precisely when does the theorem not hold. I'm looking for a non-holomorphic function to show that theorem doesn't work.

Liouville’s Theorem: Let $f : \mathbb{C} \rightarrow \mathbb{C}$ be an bounded entire function.
Then $f$ is constant.

My first guess is, I should take function $f(z)=\frac{1}{z}$ on set $|z|>1$, which is ofcourse bounded but not holomorphic in $z=0$.

I don't know how to move futher with the example or should I find another example of non-holomorphic function?

答: What about $f(z)=\cos(\operatorname{Re}z)+\sin(\operatorname{Im}z)i$? It is bounded, non-holomorphic and non-constant. And, as an extra, it is continuous.


答:
I don't have reputation for comment. so trying to answer here.
${S_n= (2n-1)(2n)(2n+1)(2n+2)}$, simplifying it.

${S_n = (2n) \cdot (4n^2-1) \cdot (2n+2)}$,

${S_1+S_2+S_3+S_4= 24+360+1680+5040=7104}$

Simple program gives sum till ${100 = 33211999680 }$

Title: How to compute $1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$

Body:
How to compute $S = 1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$

Was thinking $\frac{S}{24} = {4\choose 4} + {6\choose 4} + {8\choose 4} + ... + {100\choose 4}$ but how do I sum this up?


问:
Show that the area vectors for a general tetrahedron sum to zero
Using vector addition and multiplication, it is possible to show that the sum of the area vectors for a general closed tetrahedron in $\mathbb{R}^3$ (3-space) is zero.

Hint: start by writing down three vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$ and derive relationships for the other sides in terms of $\vec{a}$, $\vec{b}$, and $\vec{c}$.

答:
If $\vec{n}$ is outward unit normal to a face $F$ in then the (signed) area of the projection of $F$ onto a plane with unit normal $\vec{p}$ is $\vec{n}\cdot\vec{p}$ times the area of $F$. For a closed polyhedron, the projection of the faces pointing in one direction exactly match the projection of the faces pointing in the opposite direction.

Consider any polygonal prism with axis parallel to $\vec{p}$ whose intersection with a polyhedron are the red and green faces.

$\hspace{2cm}$

The area of the green face times $\vec{p}\cdot\vec{n}_g$ is the cross-sectional area of the prism. The area of the red face times $\vec{p}\cdot\vec{n}_r$ is the negative of the cross-sectional area of the prism. Their sum is $0$ and so is the sum over all prisms parallel to $\vec{p}$.

Since this is true for any $\vec{p}$, the sum of the area vectors for a general closed polyhedron is $0$.


问:
To show that the field $\Bbb Z_p$ has no proper subfield when $p$ is a prime number.
To show that the field $\Bbb Z_p$ has no proper subfield when $p$ is a prime number.Is the reason of having no primes due to non factorization of $p$??</p>

答:
First look at the subrings of the field   $ Z_p$.
since  $Z_p$  is cyclic group (since prime order) so only subgroup of it ${0},Z_p$.

Since subrings should be also a subgroup with respect to addition. So the possible subrings are ${0},Z_p$.
Since for  a subfield  all elements are unit( i.e no divisor of zero)so it will be only  $Z_p$
So no proper subfiled exist .


问:
Any memory efficient mapping function from a square 2d space to 1d space?
I do programming questions but many a times I come across some problem in which I think I should make an effort to reduce both time and space complexity.
For example, I want to convert given coordinates (0,0 to n,n). Consider a square matrix. Now I want my program to mark the coordinates that I have already visited. For that I have to save the x,y coordinates in a pair<int,int> object. I was wondering if I could save them in a single primitive type variable.
E.g. x,y ---(Function 1)----> z   and  z--(Function 2)--> x,y

Is this possible?
Here we should get a unique value of z for every different pair of x,y and for each value of z we should get unique coordinates x,y.
I want to implement these 2 functions, that do the forward and backward mapping.

答: If you are dealing with integers then $$z=x+(n+1)y$$ would work in one direction while rounding down with the floor function (or integer division) $$y = \lfloor z/(n+1)\rfloor \text{ and } x=z-(n+1)\lfloor z/(n+1)\rfloor$$ would work in the other.  Since $0 \le x \le n$ and $0 \le y \le n$ you have $0 \le z \le n^2+2n$, and this is in effect a bijection.


问:
Union and intersections of logical consequences
I have a confusion with regards to the Tarskian consequence operator $Cn$.
I have an example where I am looking to find the intersection of 3 logical consequences but I am not sure how to find it.

$Cn(x) \cap Cn(y) \cap Cn(x,y) =\quad?$

I would say that the result is $Cn(x \vee y)$ because if I weaken the input $x$ in $Cn(x)$ with $Cn(x \vee y)$ and I do the same for $Cn(y)$ to get $Cn(y \vee x)$ I can get that the intersection $Cn(x \vee y)$.

Can anyone point me in the right direction or help me find materials where I can find examples about the union and intersections using Cn operators?

答:
First of all, by monotonicity, $C_{n}(x)\subseteq C_{n}(x,y)$ and $C_{n}(y)\subseteq C_{n}(x,y)$, meaning $C_{n}(x)\cap C_{n}(y)\cap C_{n}(x,y)=C_{n}(x)\cap C_{n}(y)$.

Now, take $z\in C_{n}(x\vee y)$:

since $x\vee y\in C_{n}(x)$ (given $x\vDash x\vee y$), by monotonicity and idempotence one finds $z\in C_{n}(x)$;

analogously, $z\in C_{n}(y)$.

So
$$C_{n}(x\vee y)\subseteq C_{n}(x)\cap C_{n}(y).$$
Reciprocally, assume $z\in C_{n}(x)\cap C_{n}(y)$: by elimination of disjunction (given $x\vDash z$ and $y\vDash z$, $x\vee y\vDash z$), one gets $z\in C_{n}(x\vee y)$, and so
$$C_{n}(x\vee y)=C_{n}(x)\cap C_{n}(y)=C_{n}(x)\cap C_{n}(y)\cap C_{n}(x,y),$$
as you suspected!


问:
The interpretation of a graph that is constructed using $\sin(x!)$

I wanted to know the interpretation of a graph that is constructed using
$$f(x) = \sin(x!)$$

$\sin(x!)$ : the plot of graph at $(170.65,0.633)$ ends

答: The graph of $\sin(x!)$ has a root whenever $x!$ is an integer multiple of $\pi$. Because $x!$ is convex, the spacing between the roots of $\sin(x!)$ approaches $0$ as $x$ tends to infinity. This explains the oscillatory behaviour of the graph. Eventually, the computer can't handle it, not only because $x!$ becomes extremely large, but because the lines on the graph gets so close to each other that the thickness of the lines make it look as if there is a solid block of colour.


问:
Prove that $\frac{x^2+y^2}{z}+\frac{y^2+z^2}{x}+\frac{z^2+x^2}{y}\geq 2(x+y+z)$
$x,y,z \in \mathbb R^+$ different than $0$, Prove that :
$$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$

My attempt:
First, we should prove that :
$$2\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x} \right)\geq 2(x+y+z)$$
I will consider it as proven, we gonna start by :
$$(x-y)^2 \geq 0 \iff x^2+y^2 \geq 2xy $$

by the same method we have :
$x^2+z^2 \geq 2xz$ and $y^2+z^2 \geq 2yz$

so we can replace $2xy$ by $x^2+y^2$ in this inequality
$$2\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x} \right)\geq 2(x+y+z)$$
so :

$$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$

I'm new to AM-GM inequality, if it exists a better proof using it please post it.

答:
$$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$
$$\Leftrightarrow (x^2 + y^2 + z^2) \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \geq 3(x+y+z)$$
Note that
$$(x^2 + y^2 + z^2) \geq \frac{1}{3} (x+y+z)^2 $$ (Cauchy-Schwarz ineq., equality: x = y = z)
and
$$(x+y+z) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \geq (1+1+1)^2 = 9$$ (Cauchy-Schwarz ineq., equality: $x=y=z$).

Applying them implies the original inequality (equality holds when $x=y=z$).


问:
Big-O notation, prove the following: $\sum\limits_{k=3}^n(k^2 - 2k)$ is $O(n^3)$.
Use the definition of Big-O notation to prove the following:

$\sum\limits_{k=3}^n(k^2 - 2k)$  is $O(n^3)$.

Can someone please give me some hints on how to expand $\sum\limits_{k=3}^n(k^2 - 2k)$?

答:
The easiest way to conclude that it is $\mathcal{O}(n^3)$ is to note that
$$k^2 - 2k < k^2 \leq n^2, \,\,\,\,\, \forall k \in \{3,4,\ldots,n\}$$
You could in fact evaluate the sum exactly and then prove that it is $\mathcal{\Omega}(n^3)$. Make use of the fact that
$$\sum_{k=1}^n k^2 = \dfrac{n(n+1)(2n+1)}6 \text{ and } \sum_{k=1}^n k = \dfrac{n(n+1)}2$$

Move your mouse over the gray area below for the complete solution.

Note that$$\sum_{k=3}^n (k^2-2k) = \sum_{k=3}^n k^2 - 2\sum_{k=3}^nk = \dfrac{n(n+1)(2n+1)}6-1^2-2^2 - 2 \left(\dfrac{n(n+1)}2-1-2\right)$$The above simplifies to$$\dfrac{(n-2)(n-1)(2n+3)}6$$


问:
limit of $\cos^n(\frac{\pi}{n})$
Pretty self explanatory - what is a nice way to show that
$(\cos(\frac{\pi}{n}))^n\to 1$ as $n\to\infty$, where $n$ are integers?

答:
Since $$\cos (x) = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-... \tag{$x \to 0$}$$
we have
$$\cos \frac{\pi}{n} = 1 + O\left(\frac{1}{n^2}\right) \tag{$n\to \infty$}$$
which implies $$\cos^n\frac{\pi}{n} = \left(1 + O\left(\frac{1}{n^2}\right) \right)^n \to 1 \tag{$n\to\infty$}$$

Title: limit of $\cos^n(\frac{\pi}{n})$


答:
$$\lim_{n\to\infty}\cos^n\left(\frac\pi n\right)=\lim_{n\to\infty}\exp\left(\ln\left(\cos^n\left(\frac\pi n\right)\right)\right)=\exp\left(\lim_{n\to\infty}n\ln\left(\cos\frac\pi n\right)\right)$$

$$\lim_{n\to\infty}n\ln\left(\cos\frac\pi n\right)=\lim_{n\to\infty}\frac{\ln\left(\cos\frac\pi n\right)}{\frac1n}$$

Let $x=\frac1n$

$$\lim_{n\to\infty}\frac{\ln\left(\cos\frac\pi n\right)}{\frac1n}=\lim_{x\to0}\frac{\ln(\cos(\pi x))}x$$

Applying l'Hopital's rule,

$$\lim_{x\to0}\frac{\ln(\cos(\pi x))}x=\lim_{x\to0}\frac{\frac{-\pi\sin(\pi x)}{\cos(\pi x)}}1=\lim_{x\to0}-\pi\tan(\pi x)=-\pi\tan0=0$$

Thus,

$$\exp\left(\lim_{n\to\infty}n\ln\left(\cos\frac\pi n\right)\right)=\exp(0)=1$$

Title: limit of $\cos^n(\frac{\pi}{n})$

Body:
Pretty self explanatory - what is a nice way to show that
$(\cos(\frac{\pi}{n}))^n\to 1$ as $n\to\infty$, where $n$ are integers?


问:
How to compute $1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$
How to compute $S = 1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$

Was thinking $\frac{S}{24} = {4\choose 4} + {6\choose 4} + {8\choose 4} + ... + {100\choose 4}$ but how do I sum this up?

答:
General term is:

$$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$$

Now we give values 1, 3,... n :

$1\times2\times3\times4=1^4+6\times1^3+11\times 1^2+6\times 1$

$3\times4\times5\times6=3^4+6\times3^3+11\times 3^2+6\times 3$

.

.

.

$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$

Now we sum up:

$A=S_4+6S_3+11S_2+6S_1$

We use  well known formulas for $S_n$ for odd numbers, for example:

$S_3=1^3+3^3+ . . . +n^3=n^2(2n^2-1)$

Here $n= 96$, we plug this in and get the sum.


问:
Formalization of $17\mid 77a+1$
I was solving a math problem when I came across this Question:

When is $17 \mid 77a+1$ With $a\in \mathbb N^*$

It’s easy to see that there’s infinitely many of values of $a$, you can’t try a little bit with this but you will end up with these values :
$$a\in \{15,32,49,66,...\}=A$$
In this set any number it’s equal to the previous one $+17$ .
I’ve found a relation with this set and the set contains all numbers divisible by $17$, let’s call it $D_{17}$
$$D_{17} =\{17,34,51,68,...\}$$
You can see that if $n \in D_{17} \iff n-2\in A$ And if $n\in A \iff n+2 \in D_{17}$.

My question is why $2$ appeared here, does there exist an equation that when we solve it we will get this $2$?

答:
$$77a+1\equiv 17 \bmod 17$$
$$77a\equiv -1 \bmod 17$$
$$9a\equiv -1 \bmod 17$$
The modular inverse of $9\bmod 17$ is $2$ therefore
$$a\equiv -2 \bmod 17$$
that is $$a=17n-2;\;\forall n\in\mathbb{N^*}$$

Title: Formalization of $17\mid 77a+1$


答:
Probably the following is shorter and easier.

Firstly, note that for positive reals $a$ and $b$ the inequality
$$
\frac{a^2}{b}\ge 2a-b
$$
holds (since it's equivalent to $\frac{(a-b)^2}{b}\ge 0$). Applying this observation six times gives
$$
\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\ge(2x-z+2y-z)+(2x-y+2z-y)+(2y-x+2z-x)=
\\
=2(x+y+z),
$$
as desired.

Comment. One can generalize this trick, namely the inequality
$$
\frac{a^x}{b^y}\ge\frac{xa^{x-y}-yb^{x-y}}{x-y}
$$
holds for $a,b,x,y>0$, $x\neq y$. This is often helpful when we need to estimate some fractions using polynomial expressions.

Title: Prove that $\frac{x^2+y^2}{z}+\frac{y^2+z^2}{x}+\frac{z^2+x^2}{y}\geq 2(x+y+z)$

Body:
$x,y,z \in \mathbb R^+$ different than $0$, Prove that :
$$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$

My attempt:
First, we should prove that :
$$2\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x} \right)\geq 2(x+y+z)$$
I will consider it as proven, we gonna start by :
$$(x-y)^2 \geq 0 \iff x^2+y^2 \geq 2xy $$

by the same method we have :
$x^2+z^2 \geq 2xz$ and $y^2+z^2 \geq 2yz$

so we can replace $2xy$ by $x^2+y^2$ in this inequality
$$2\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x} \right)\geq 2(x+y+z)$$
so :

$$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$

I'm new to AM-GM inequality, if it exists a better proof using it please post it.


问:
Automorphism from $\Bbb Q$ to $\Bbb Q$
Any automorphism of the group $\Bbb Q$ under addition is of the form $x\to qx$ for some $q\in \Bbb Q$. </p>

<p>I don't know how to proceed in this. Even if i say that $1\to q$, I can't claim that $x\to qx$ since $\Bbb Q$ is not cyclic.  </p>

答:
The following answer is given in the spirit of overly-pedantic exposition, spelling out ever detail for novice student.  This is not the kind of presentation that I would generally encourage in written form, but I might go into this much detail in a lecture to an introductory algebra class.

Theorem: Every automorphism of the additive group $\mathbb{Q}$ is of the form $x \mapsto qx$ for some $q \in \mathbb{Q}$.

Proof:  Let $\varphi$ be an automorphism of the additive group $\mathbb{Q}$.  That is, suppose that $\varphi: \mathbb{Q} \to \mathbb{Q}$ has the following properties:

the map $\varphi$ is bijective (that is, it is injective and surjective), and

if $x,y\in \mathbb{Q}$, then $\varphi(x+y) = \varphi(x) + \varphi(y)$.

To show that $\varphi(x) = qx$ for some $q \in \mathbb{Q}$, we can "bootstrap" our way up.  First, we need to know where $1$ goes.  The two properties given above don't put any restriction on the image of $1$, so all we know is that there is some rational number $q$ such that $\varphi(1) = q$.  Fortunately, that's good enough.

Next, an induction argument can be used to show that if $m\in\mathbb{N}$, then $\varphi(m) = qm$.  The base case ($m=1$) is straight-forward:  $\varphi(1) = q = 1\cdot q$.  Now suppose that $\varphi(k) = qk$ for some natural number $k$.  Then
\begin{align}
\varphi(k+1) &= \varphi(k) + \varphi(1) && (\text{property (2), above}) \\
&= kq + q && (\text{induction hypothesis}) \\
&= (k+1) q,
\end{align}
which completes the induction argument.

Any homomorphism must send the identity to itself, and so for any $m\in\mathbb{N}$
$$0 = \varphi(0) = \varphi(m-m) = \varphi(m) + \varphi(-m) = qm + \varphi(-m), $$
which implies that $\varphi(-m) = -qm$.  Thus $\varphi(m) = qm$ for any integer $m$.

Another quick induction argument can be used to show that if $r \in \mathbb{Q}$ is arbitrary and $n$ is any natural number, then
$ \varphi(nr) = n \varphi(r). $
To finish this part of the argument, let $x = m/n \in \mathbb{Q}$ be arbitrary.  Then
$$ n \varphi(x) = \varphi(nx) = \varphi\left( n \frac{m}{n} \right) = \varphi(m) = qm \implies \varphi(x) = \frac{qm}{n} = qx. $$
That is, if $x \in \mathbb{Q}$, then $\varphi(x) = qx$, which is the desired result.

It might be worth noting that the proof above shows that if $\varphi: \mathbb{Q} \to \mathbb{Q}$ is any homomorphism, then there must be some $q\in\mathbb{Q}$ such that $\varphi(x) = qx$.  Since every automorphism is a homomorphism, the theorem is proved.  However, this argument does not immediately tell us that any automorphisms exist, nor does it imply that every map of the form $x \mapsto qx$ is an automorphism.  For this, it is necessary that such maps are bijective.

Before doing this, it is, perhaps, worth noting that $x \mapsto 0$ is a homomorphism of the additive group $\mathbb{Q}$ of the form $x \mapsto qx$ (here, $q=0$), but this map is not an automorphism (for obvious(?) reasons).  So we don't have to worry about zero—this doesn't give us an automorphism.

On the other hand, let $q$ be a nonzero rational number, and let $\varphi(x) = qx$.  To see that $\varphi$ is injective, fix $x,y \in \mathbb{Q}$ and suppose that $\varphi(x) = \varphi(y)$.  Then
$$ \varphi(x) = \varphi(y) \iff qx = qy \iff x = y. $$
Thus $\varphi$ is injective.

Now suppose that $x \in \mathbb{Q}$.  Then
$$ x = q \frac{x}{q} = \varphi\left( \frac{x}{q} \right), $$
where $x/q$ is a rational number.  Thus $\varphi(x)$ is surjective.

This proves a slightly stronger statement that the original:

Theorem: A map $\varphi$ from the additive group $\mathbb{Q}$ to itself is an automorphism if and only if $\varphi(x) = qx$ for some $q \in \mathbb{Q} \setminus \{0\}$.

Title: Automorphism from $\Bbb Q$ to $\Bbb Q$


问:
The Derivative as a group homomorphism
This is actually a rather simple question, and I must be missing something - any help is appreciated. Let's consider the groups $(C^1(\mathbb{R}), +)$ and $(C^0(\mathbb{R}), +)$ and define the homomorphism $D: C^1 \rightarrow C^0$ such that $D(f) = f' = \frac{d(f)}{dx}$.

What is the kernel of $D$? Well, if $D(f) = 0$, that means $f'(x) = 0$,  $\forall x \in \mathbb{R}$ which implies that $f$ is a constant function. So $\ker(D) = \mathcal{C}$, where $\mathcal{C} = \{f \in C^1(\mathbb{R}) | f(x) = c, \forall x \in \mathbb{R}, c \in \mathbb{R}\}$. By the first isomorphism theorem, this then implies $\operatorname{Im}(D) \simeq C^1(\mathbb{R})/\mathcal{C}$.

So far, this seems ok. But consider the function $f(x) = \begin{cases}-\frac{x^2}{2} & \mbox{if } x \leq 0 \\ \frac{x^2}{2} & \mbox{if } x > 0\end{cases}$. It is a $C^1(\mathbb{R})$ function, and its derivative is the absolute value of $x$, which is not differentiable. So $f \in C^0(\mathbb{R}) \setminus C^1(\mathbb{R})$. But, according to the above, $D(f)$ is an equivalence class of $C^1(\mathbb{R})/\mathcal{C}$. And I don't think this can be, because $D(f)$ is not even in $C^1(\mathbb{R})$.

Where did I make mistakes? And why?

Thanks in advance for any help!

答: The equivalence class you are looking for is the set of all functions of the form $f(x)+\text{const}$, and these are $C^1$. Also, it is $f'(x)$ that is in $C^0\setminus C^1$.

Title: The Derivative as a group homomorphism


问:
How do you respond to "I was always bad at math"?
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.

答:
I get a variant of this depressingly often when people find out I am an engineer. You might be surprised how many people think "engineering = math."

It's worse than you suggest. Being "bad at math" has actual consequences in life in ways being bad at music or history or poetry do not. Being bad at math means you likely make foolish decisions, especially when it comes to finances. This is a problem because people who say "bad at math" normally mean "I avoid math at all costs but wish I was better at it."

So what can you respond? Realize people mean something different. They are really saying, "I dislike math and/or have no interest in learning but wish I magically knew it" - not "I am bad/was always bad/could never do math."

You can approach this a couple ways.

"Why do you say that?" is generally how I address it as this leads to lots of followup questions

"So was I" can make for interesting responses

"It takes a lot of work/practice"

"Oh wow! how do you maintain your life?!"  (I don't recommend this)

Many of these people have been conditioning themselves for years or even decades to convince themselves they are "bad at math."

Just a note, this used to bother me or even make me feel guilty, too. There is an implicit, "aww, man, I wish I was as lucky as you were to be good at math" in that response.

But at the end of the day, I've had to realize a lot of people have simply spent years telling themselves they are bad at math. That's not my fault.

Don't feel guilty by people saying a variation on "I was bad at math."


问:
Why function is maximum at intersection of the level curve with its constrained curve?
What I'm asking is the core question of Lagrange multiplier. Say we have a function $f:\mathbb{R^2}\rightarrow \mathbb{R}$. Now we intend to calculate at what $(x,y)$, the function achieves its maxima given constraint $g:\mathbb{R^2}\rightarrow\mathbb{R}$. The Lagrange multiplier says that we will find the maxima of the function where the level curve intersects the constrained function at exactly one point. Why is that? I'm not able to understand that why the function will achieve its maxima at that particular point

答:
The following is a standard explanation.

Consider this specific example:
$$\max f(x, y) = xy$$
$$\text{s.t. }  x^2 + y^2 = 1$$

Let $L_c = \{(x, y)|f(x, y) = c\}$ be the level curve of $f$ at $c$.  This is how the picture looks for $c = 4.3$:

As we decrease the value of $c$, the hyperbola comes closer and closer to the origin (you can interact with the graph here.) If we shrink the hyperbola until it is tangent to the circle, we will have a feasible point whose $f$ value is pretty high. But if we decrease $c$ even more, there will be $4$ points of intersection, but the objective value will be unnecessarily low.


问:
Can I reduce fraction under integral?
Can I perform such kind of fraction reduction under integral?

$$g(x)=\int \frac{d f(x)} {dx} dx = \int d f(x) = f(x) +C $$

答:
Unless you want to work in nonstandar analysis, the Leibniz notation $\frac{df(x)}{dx}$ is not a fraction. It means the derivative of the function $f$; in Lagrange's notation, it is the same as $f'(x)$.

The identity $\int f'(x)dx=f(x)+C$ is due to the (trivial) fact that $f$ is an antiderivative of $f$ (on some interval).

The identity $\int f'(x)dx=\int df(x)$ could be understood as change of variables. If there is anything to justify, you must define what the right-hand side means first.

Title: Can I reduce fraction under integral?


答: Your outcome is correct but it's not a matter of reducing the fraction.  This is more a case where the notation is chosen to line up with the result, not the other way around  Remember that $$\frac{df(x)}{dx} = \text{ the derivative of $f$ with respect to $x$}$$ and this is not a fraction.  Also, $$\int h(x)\,dx = \text{the antiderivative of $h$ with respect to $x$}.$$  So, combining these: $$\int \frac{df(x)}{dx}\,dx = \text{the antiderivative of $\frac{df(x)}{dx}$ with respect to $x$},$$ and the antiderivative of a derivative is the original function plus $C$, so we have $$\int \frac{df(x)}{dx}\,dx = f(x) + C$$

Title: Can I reduce fraction under integral?

Body:
Can I perform such kind of fraction reduction under integral?

$$g(x)=\int \frac{d f(x)} {dx} dx = \int d f(x) = f(x) +C $$


问:
(X,X) independent and moments
I have a question regarding a random variable X being independent of itself.
I have to proof, knowing that X has no moment, that the following real number exists:

$$ t_{0} = \inf_{} \{t\in \mathbb{R} | F_X(t) = 1 \}$$

Please, if my question is ambiguous or unclear, do not hesitate to edit it !
Thank you so much !

答:
Intuitively, if any event regarding X is independent with itself, when it realises, the probability of it realising doesn't change, so either it never realises, either it was already sure that it was gonna realise. Let us put this formally.

Consider the events $X\leq t$ for $t\in \mathbb{R}$. Since X is indepedent with itself, we know that $P(X\leq t) = P(X\leq t \cap X\leq t)= P(X\leq t)P(X\leq t)$ so $P(X\leq t) = 1$ or $P(X\leq t) = 0$.
So $\forall t \in \mathbb{R}, F_X(t) = 0$ or $F_X(t) = 1$.

Now, can it always be 0? No it cannot, because otherwise X would always be greater than any real number, which is impossible. Formally:

$P(X \leq n) = P(\cap_{k \leq n}X \leq k) \to P(X \in \mathbb{R}) = 1$

From this, we know that $\exists t_0 \in \mathbb{R}, F_X(t_0) = 1$. So the considered set is not empty.

Also, with the same argument, we know that $\exists t_1 \in \mathbb{R}, F_X(t_1) = 0$, and since $F_X$ is increasing, $\forall t \leq t_1, F_X(t) = 0$. So the set is bounded from the bottom.

Therefore, it has an inf. This is what we wanted to show.

Title: (X,X) independent and moments


问:
Ratio of corresponding sides of similar triangles, given the areas.
The area of two similar triangles are 72 and 162. what is the ratio of their corresponding sides?</p>

答:
if $\Delta ABC $ and $\Delta MNO$ are similar triangle and AD and MP are perpendicular on BC and ON then

$$\dfrac {AB^2}{MN^2}=\dfrac {AC^2}{MO^2}=\dfrac {BC^2}{NO^2}=\dfrac {AD^2}{MP^2} =\dfrac {area \Delta ABC}{area\Delta MNO}$$ (This is a theorem)

so in question : $$\dfrac {area \Delta ABC}{area\Delta MNO}=\dfrac{72}{162}$$

so ratio of corresponding sides are $$ \dfrac {AB^2}{MN^2}=\dfrac {72}{162}$$

$$ \dfrac {AB}{MN}=\sqrt {\dfrac {72}{162}}$$
$$\dfrac {AB}{MN}=\dfrac23$$
so ratio will be $$2:3$$


问:
Proving $|z|(|z|^{n-1}+1)<1$ with $|z|<1$
I'm trying to prove that, given a specific z such that $|z|<1$, there exists an $N$ value such that when we have $n>N$ then $|z|(|z|^{n-1}+1)<1$ .
I've done several number examples and I know it can work I'm just having trouble generalizing it. I think I'm missing something tiny. Any help would be very appreciated.

Is it just because when we have large enough $n$ then $|z|^{n-1}$ goes to zero so we have $|z|(1)$ and since $|z|<1$ the product is less than one? Or do I need to prove more?

答:
It's not too tricky to compute such an $N$. We have
\begin{align*}
|z|(|z|^{n - 1} + 1) < 1 &\iff |z|^{n-1} + 1 < |z|^{-1} \\
&\iff |z|^{n-1} < |z|^{-1} - 1 \\
&\iff n - 1 > \log_{|z|}(|z|^{-1} - 1) \\
&\iff n > 1 + \log_{|z|}(|z|^{-1} - 1).
\end{align*}
Note that $|z| < 1$ implies $|z|^{-1} > 1$, and so $|z|^{-1} - 1 > 0$. Also, the inequality sign reversed, as $|z| < 1$ implies $\log_{|z|}$ is a (strictly) decreasing function. So, we can simply take
$$N = 1 + \log_{|z|}(|z|^{-1} - 1),$$
or its ceiling, if you insist on $N$ being an integer.


问:
How do I prove $\exp(1)=e$ if $\exp$ is defined in a series expansion?
Right now I'm on a journey to clear up my confusion about exponential functions.
Thanks to your help in this stack, I was able to derive the basic properties of the ln function from the "position of defining ln in terms of integrals".

In this stack, the definitions of exp and e are as follows;

$$\exp(x):= 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots,\ $$

$$e:=\lim_{n\to \infty} \left( 1+\frac{1}{n} \right)^{n} , n\in \mathbb{N}.$$

When defined in this way, the relationship between $\exp(x)$ and $e$ seems to be non-trivial .
We further assume that;

The theory on limit operations such as the scissors-out theorem and the theory on convergence conditions for power series are assumed to exist separately and independently.

For any integer $a>0$ where $n$ is an integer, $a^n$ is assumed to be defined in an inductive way.

My question.

Under these assumptions, how can we mathematically prove the following?
$$\exp(1)=e$$

There is a similar stack, but it does not seem to go into the definition of Napier-number.
If possible, we focus my questions on the following "How to evaluate from the lower side.

As far as I tried, I was probably able to suppress it from the upper side.
In other words, I believe I was able to prove the following.

$$e\le \exp(1)$$

Therefore, perhaps the question is how to find the series to suppress from below.

This means that the most desirable answer is to show
$$e\ge \exp(1)$$

Proof of $e\le exp(1)$

From the binomial theorem,

$$\left( 1+\frac{1}{n} \right)^{n}
=\sum_{k=0}^{n} {}_{n} \mathrm{C}_{k} \frac{1}{n^k}\\
=\sum_{k=0}^{n} \frac{1}{k!} \left( 1- \frac{1}{n} \right)\cdots \left( 1- \frac{1-k}{n} \right)\\
\le \sum_{k=0}^{n} \frac{1}{k!}
\le \exp(1)
$$

Body: Let $N\in\Bbb N$. Then, for $n\geqslant N$,\begin{align}\left(1+\frac1n\right)^n&=\sum_{k=0}^n\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right)\\&\geqslant\sum_{k=0}^N\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right).\end{align}Therefore,$$e\geqslant\sum_{k=0}^N\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right)$$and so$$e\geqslant\lim_{n\to\infty}\sum_{k=0}^N\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right)=\sum_{k=0}^N\frac1{k!}.$$Since this takes place for each $N\in\Bbb N$,$$e\geqslant\sum_{k=0}^\infty\frac1{k!}=\exp(1).$$


问:
Interchanging finite sum(over prime numbers) and integration
Here is an expression I am trying to prove:
$$ \sum_{p\leq x}\frac{\ln(p)}{p}\int_{p}^{\infty}\frac{du}{u(\ln(u))^2} = \int_{2}^{x} \sum_{p\leq u } \frac{\ln(p)}{p} \frac{du}{u(\ln(u))^2} + \frac{1}{\ln(x)} \sum_{p\leq x}\frac{\ln(p)}{p}$$
where summation is over p  (prime number) $\leq x$.

I tried to solve this in this way.

$$ \sum_{p\leq x}\frac{\ln(p)}{p}\int_{p}^{\infty}\frac{du}{u(\ln(u))^2} =\sum_{p\leq x} \frac{\ln(p)}{p} \int_{p}^{x}  \frac{du}{u(\ln(u))^2} +
\sum_{p\leq x} \frac{\ln(p)}{p} \int_{x}^{\infty}  \frac{du}{u(\ln(u))^2} $$
$$\hspace{3cm}= \sum_{p\leq x} \frac{\ln(p)}{p} \int_{p}^{x}  \frac{du}{u(\ln(u))^2} +
\frac{1}{\ln(x)} \sum_{p\leq x} \frac{\ln(p)}{p}$$

But I am not able to prove
$$ \sum_{p\leq x} \frac{\ln(p)}{p} \int_{p}^{x}  \frac{du}{u(\ln(u))^2}=  \int_{2}^{x} \sum_{p\leq u } \frac{\ln(p)}{p} \frac{du}{u(\ln(u))^2} $$

I would like to know the solution for this or some hint. Or is my approach not correct?

答:
Let
\begin{align}a_n=\begin{cases}\frac{\ln n}{n}~~&\text{if $n$ is prime}\\0&\text{else}\end{cases}\end{align}
and $$\phi(y)=\int_y^x\frac{du}{u(\ln(u))^2}$$
Then we have $\phi(x)=0$ and $\phi'(y)=-\frac{1}{y(\ln(y))^2}$. Hence Abel's summation formula gives us:\begin{align*}
\sum_{p\leq x}\frac{\ln(p)}{p}\int_p^x\frac{du}{u(\ln(u))^2}=\sum_{1.5<n\leq x} a_n\phi(n)&=A(x)\phi(x)-A(1.5)\phi(1.5)-\int_{1.5}^xA(u)\phi'(u)du\\
&=\int_2^x\sum_{p\leq u}\frac{\ln(p)}{p}\frac{du}{u(\ln(u))^2}
\end{align*}
where $A(u)=\sum_{0\leq n\leq u}a_n$. (The '$1.5$' was only necessary here so that the conditions in the link are satisfied, note that $A(u)=0$ for $u<2$)


答:
The answer is yes. In order to avoid confusing fractions and differentials everywhere, I will write this in purely operator form to emphasize how simple this is. We can denote differentiation as a map between sets of differentiable/integrable functions, i.e
$$\mathcal{D}:\mathscr{D}\to\mathscr{I}~~,~~\mathcal{D}:f\mapsto \mathcal{D}(f)$$
Integration can be written similarly,
$$\mathcal{I}:\mathscr{I}\to\mathscr{D}~~,~~\mathcal{I}:f\mapsto\mathcal{I}(f)$$
Integration is defined to be an operation such that
$$(\mathcal{I}\circ\mathcal{D})=(\mathcal{D}\circ\mathcal{I})=\operatorname{id}$$
$$(\mathcal{I}\circ\mathcal{D})(f)=(\mathcal{D}\circ\mathcal{I})(f)=f$$
Where equality is taken to be up to a constant difference. This means, in words, that taking the derivative of the primitive of a function or taking the primitive of the derivative of a function will give you back the function. If we use more classical notation, this is
$$\int\frac{\mathrm{d}f}{\mathrm{d}x}\mathrm{d}x=\frac{\mathrm{d}}{\mathrm{d}x}\int f~\mathrm{d}x=f$$

Title: Can I reduce fraction under integral?


答:
This appears as Theorem 8.1.6 in Strichartz's The Way of Analysis.  The idea is the following:

First use the series definition to develop some basic properties of $\exp$: it is strictly positive on $\mathbb{R}$, it is its own derivative, it is strictly increasing, it satisfies $(\exp x)^n = \exp(nx)$.

Use the inverse function theorem to show that $\exp$ has an inverse function, call it $\ln$, which is differentiable.   Use the chain rule to show $\frac{d}{dx} \ln x = \frac{1}{x}$. Also verify that  $\ln(1)=0$ and $\ln (x^n) = n \ln x$.

Write
$$ \lim_{n \to \infty} \ln \left(1 + \frac{1}{n}\right)^n = \lim_{n \to \infty} \frac{\ln\left(1 + \frac{1}{n}\right) - \ln(1)}{\frac{1}{n}} = \left.\frac{d}{dx}\right|_{x=1} \ln x = 1.$$

Take $\exp$ of both sides and use the continuity of $\exp$ to conclude.

Title: How do I prove $\exp(1)=e$ if $\exp$ is defined in a series expansion?

Body:
Right now I'm on a journey to clear up my confusion about exponential functions.
Thanks to your help in this stack, I was able to derive the basic properties of the ln function from the "position of defining ln in terms of integrals".

In this stack, the definitions of exp and e are as follows;

$$\exp(x):= 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots,\ $$

$$e:=\lim_{n\to \infty} \left( 1+\frac{1}{n} \right)^{n} , n\in \mathbb{N}.$$

When defined in this way, the relationship between $\exp(x)$ and $e$ seems to be non-trivial .
We further assume that;

The theory on limit operations such as the scissors-out theorem and the theory on convergence conditions for power series are assumed to exist separately and independently.

For any integer $a>0$ where $n$ is an integer, $a^n$ is assumed to be defined in an inductive way.

My question.

Under these assumptions, how can we mathematically prove the following?
$$\exp(1)=e$$

There is a similar stack, but it does not seem to go into the definition of Napier-number.
If possible, we focus my questions on the following "How to evaluate from the lower side.

As far as I tried, I was probably able to suppress it from the upper side.
In other words, I believe I was able to prove the following.

$$e\le \exp(1)$$

Therefore, perhaps the question is how to find the series to suppress from below.

This means that the most desirable answer is to show
$$e\ge \exp(1)$$

Proof of $e\le exp(1)$

From the binomial theorem,

$$\left( 1+\frac{1}{n} \right)^{n}
=\sum_{k=0}^{n} {}_{n} \mathrm{C}_{k} \frac{1}{n^k}\\
=\sum_{k=0}^{n} \frac{1}{k!} \left( 1- \frac{1}{n} \right)\cdots \left( 1- \frac{1-k}{n} \right)\\
\le \sum_{k=0}^{n} \frac{1}{k!}
\le \exp(1)
$$


问:
Sitting students on a bench

In a certain bench there are x seats: $seat_{\{0 \le i < x | i,x\in\mathbb{N}\}}$

There are y students in total: $student_{\{0 \le i < y | i,y\in\mathbb{N}\}}$

x students should be seated on the bench for
$numOfDays = \lceil{y / x}\rceil$ days, such that on first day, $student_{\{0 \le i < x\}}$ will sit, on the second day $student_{\{x \le i < 2x\}}$ will sit and so on...

On the last day we may have to seat students who have already sat the day before.

For example, for $x = 3$, $y = 5$, on the first day $student_0$, $student_1$, $student_2$ will sit, and on the second day $student_2$, $student_3$, $student_4$ will sit.

Let $i$ be the index of the current day ($\{0 \le i\in\mathbb{N} < numOfDays\}$). Calculate the formulas of:

$min\_i$: the index of the first student to sit on the bench on day i

$max\_i$: the index of the last student to sit on the bench on day i

I tried to play with the data, but I kept getting tangled up with the last day. I can't find a formula that is correct for all days. I would love to get directions for a solution

答:
I think there might be a gap in your statement, but I will give you the answer that I think is correct. To simplify notation, let $S_i$ represent the $i^\text{th}$ student and write $s_i$ for the $i^\text{th}$ seat on the bench. We want to compute the minimum $m_i$ and the maximum $M_i.$

As you have mentioned in your statement, the last day is the only one that may cause us problems. Note that because the number of days is $d = \lceil \frac{y}{x} \rceil,$ then according to the rule in the statement, on the last day we should sit students $S_{(d - 1)x}$ through $S_{d x - 1}.$ However, note that $d x - 1 = \lceil \frac{y}{x} \rceil x - 1 > y - 1.$ Thus we will indeed encounter overlaps with the students from the previous day. Also according to the same rule, on the next to last day, we should sit students $S_{(d - 2)x}$ through $S_{(d - 1)x - 1}.$ It is important to note that $(d - 1)x - 1 = \lfloor \frac{y}{x} \rfloor x - 1 \leq y - 1,$ so up until the penultimate day, we encounter no problems at all.

Thus, we have established that for all $1 \leq i < d,$ the students that will be seated on the bench on day $i$ are students $S_{(i - 1)x}$ through $S_{i x - 1}.$ Hence we also deduce that $m_i = (i - 1) x$ and $M_i = i x - 1$ for all $1 \leq i < d.$ Let us analyse the case $i = d.$ From what I managed to gather from your statement, we do not want to sit again the first students, so we are interested in sitting once again $x$ people on the bench, but also manage to sit the last student $S_{y - 1}$ and this has to be done in the natural order. We deduce that the last student that has to be seated on the last day is exactly $S_{y - 1}.$ Since we need to have $x$ people on the bench, the first one must be $S_{y - x}.$ Therefore we also get $m_d = y - x$ and $M_d = y - 1.$ Note that this also works if $y \geq x.$ If this is not the case, then $d = 1$ and we will sit all the students on the bench on the first day in the natural order, which ends this short case.

The $y = x$ case is particularly interesting and noteworthy. For example, if $x = 3 = y,$ then $d = 2$ and we will sit on both days all the students, $S_0, \, S_1$ and $S_2$ in this order exactly.

Title: Sitting students on a bench


答:
You don’t actually have to work very hard at all:

$$\sum_{k=3}^n\left(k^2-2k\right)\le\sum_{k=3}^nk^2\le\sum_{k=3}^nn^2=(n-2)n^2\le n^3\;.$$

In other words, don’t expand the sum: bound it instead. As you can see, even a pretty crude bound works fine here.

Title: Big-O notation, prove the following: $\sum\limits_{k=3}^n(k^2 - 2k)$ is $O(n^3)$.

Body:
Use the definition of Big-O notation to prove the following:

$\sum\limits_{k=3}^n(k^2 - 2k)$  is $O(n^3)$.

Can someone please give me some hints on how to expand $\sum\limits_{k=3}^n(k^2 - 2k)$?


答:
So you want to show that
$$
\lim_{n\to\infty}\left(1+\frac1n\right)^n=\sum_{k=0}^\infty\frac{1}{k!}=:\exp(1)
$$

Let $s_n$ be the partial sum of the series and $t_n=\left(1+\frac1n\right)^n$.

Then the binomial theorem implies that
\begin{aligned} t_{n}=1+1+\frac{1}{2 !}\left(1-\frac{1}{n}\right)+\frac{1}{3 !}\left(1-\frac{1}{n}\right) &\left(1-\frac{2}{n}\right)+\cdots \\ &+\frac{1}{n !}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) \cdots\left(1-\frac{n-1}{n}\right) \end{aligned}

Thus $t_n\le s_n$, so that
$$
\limsup t_n\le \exp(1)\tag{1}
$$

On the other hand, if $n\ge m$, then
$$
t_{n} \geq 1+1+\frac{1}{2 !}\left(1-\frac{1}{n}\right)+\cdots+\frac{1}{m !}\left(1-\frac{1}{n}\right) \cdots\left(1-\frac{m-1}{n}\right)
$$
Let $n\to\infty$, keeping $m$ fixed you get
$$
\liminf _{n \rightarrow \infty} t_{n} \geq 1+1+\frac{1}{2 !}+\cdots+\frac{1}{m !}=s_m\tag{2}
$$

Combining (1) and (2) you are done.

Notes. This is Theorem 3.31 in Rudin's Principles of Mathematical Analysis.


问:
Question regarding $\varepsilon$-$\delta$
Consider the following, I try to prove a limit, and get $\frac \varepsilon 3$=$\delta$. So, the length $\delta$ $<$ $\varepsilon$. My question is how will every $\varepsilon$ correspond with a single and unique $\delta$?

答:
If you have $\lim_{x \to a} f(x) = l$ then for every $\epsilon$ there is a $\delta$ such that for every $x \in D$ (where D is the domain of $f$) with $|x-a|\lt \delta
$, you have that $|f(x) - l|<\epsilon$ by definition. Take one of these $\epsilon$'s and one $\delta$ that does the trick for that $\epsilon$. Now, take any $0<\delta_{o}<\delta$. Then, for any $x \in D$ with $|x-a|<\delta_{0}<\delta$, we have by definition of $\delta$ that $|f(x)-l|<\epsilon$. With this we conclude that there is actually never uniqueness of a $\delta$ correspondent to some $\epsilon$, since in the real numbers, there is always a $0<\delta_{0}<\delta$. And for each of these $\delta_{0}$, the condition is also satisfied. So the possible $\delta$'s are always infinite for each $\epsilon$


答:
We can get much more than the limit $$a_n=\cos^n\left(\frac{\pi}{n}\right)\implies \log(a_n)=n\log \left(\cos \left(\frac{\pi }{n}\right)\right)$$
$$\cos \left(\frac{\pi }{n}\right)=1-\frac{\pi ^2}{2 n^2}+\frac{\pi ^4}{24
   n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log \left(\cos \left(\frac{\pi }{n}\right)\right)=-\frac{\pi ^2}{2 n^2}-\frac{\pi ^4}{12
   n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log(a_n)=-\frac{\pi ^2}{2 n}-\frac{\pi ^4}{12
   n^3}+O\left(\frac{1}{n^5}\right)$$
$$a_n=e^{\log(a_n)}=1-\frac{\pi ^2}{2 n}+\frac{\pi ^4}{8 n^2}-\frac{\pi ^4 \left(4+\pi ^2\right)}{48
   n^3}+\frac{\pi ^6 \left(16+\pi ^2\right)}{384
   n^4}+O\left(\frac{1}{n^5}\right)$$

Title: limit of $\cos^n(\frac{\pi}{n})$


问:
Equation for the line joining north pole and a point on the complex plane
If I have a complex number in the z-plane and I want to find the equation of the line joining the north pole $(0,0,1)$ of the unit sphere and the complex number $(z=x+iy)$ on the x-y plane should be $$r=<0,0,1>+t<x,y,-1>$$ using the parametric equation of a line.
But the professor gave us a different equation:

$t(0,0,1)+(1-t)(x,y,0)$

This doesn't seem right nor can I find a justification for it anywhere.

答:
They are the same up to a change of variables $t \mapsto 1-t$.

Combining the vectors in your equation, you get:
$$\langle 0,0,1\rangle + t\langle x,y,-1\rangle = \langle tx,ty,1-t\rangle.$$

Combining the vectors in your professor's equation:
$$t\langle 0,0,1\rangle + (1-t) \langle x,y,0\rangle = \langle (1-t)x,(1-t)y, t\rangle.$$

Notice that these are the same except $t$ has been replaced by $1-t$.

In general the line segment between $\bf x$ and $\bf y$ can be parametrized by ${\bf x} + t({\bf y}-{\bf x})$ or by $t{\bf x} + (1-t){\bf y}$ for $0 \leq t \leq 1$. For a fixed value of $t$, the first equation gives you the point "$t$ of the way" to $\bf y$ from $\bf x$, while the second gives you the point "$t$ of the way" to $\bf x$ from $\bf y$.* You can think of it as a describing a weighted average of the points $\bf x$ and $\bf y$, for every possible combination of weights between $0$ and $1$ that sum to $1$.

*Here when I say "$t$ of the way" I don't mean in terms of absolute distance, but as a fraction of the distance between $\bf x$ and $\bf y$.


答: The general equation for a line joining the points $P$ and $Q$ is $t P + (1 - t) Q,$ where $t \in \mathbb{R},$ hence your professor's equation. In fact, note that this is exactly the form you also got since $(x, y, -1) = (x, y, 0) - (0, 0, 1),$ so that $(0, 0, 1) + t(x, y, -1) = t(x, y, 0) + (1 - t)(0, 0, 1),$ which is of course the same thing as before. The main point of my answer would be that there are different ways to get the equation of a line: you can express it as the all the convex combinations of two points on it (as your professor did) or you could write it as the points you get by starting at a point and going in a certain direction. Hope this helps. :)

Title: Equation for the line joining north pole and a point on the complex plane

Body:
If I have a complex number in the z-plane and I want to find the equation of the line joining the north pole $(0,0,1)$ of the unit sphere and the complex number $(z=x+iy)$ on the x-y plane should be $$r=<0,0,1>+t<x,y,-1>$$ using the parametric equation of a line.
But the professor gave us a different equation:

$t(0,0,1)+(1-t)(x,y,0)$

This doesn't seem right nor can I find a justification for it anywhere.


问:
Equivalence of triangulated categories of perfect complexes
I'm reading a paper by Paul Balmer where he supposes to have a nice (let's say noetherian) scheme $X$, with an open $U\subseteq X$, and proves the existence of an equivalence of triangulated categories
$$(\mathrm{Pf}(X)/\mathrm{Pf}_Z(U))^{\sim}\longrightarrow \mathrm{Pf}(U)$$

where $\mathrm{Pf}(X)$ is the triangulated category of perfect complexes on $X$, $\mathrm{Pf}_Z(X)$ is the triangulated subcategory of perfect complexes with support contained in $Z$, and the $\sim$ stands for the idempotent completion ($\overset{\sim}{\textbf{C}}$ is the full subcategory of presheaves on $\textbf{C}$ which are retract of a representable ones). The morphism is induces by the restriction functor $\mathrm{Pf}(X)\to \mathrm{Pf}(U)$ and the universal property of the Verider quotient (this gives the functor from the quotient), as well as the fact that $\mathrm{Pf}(U)$ is idempotent complete (giving the functor from the idempotent completion of the quotient).

Now, he first proves that the functor defined on the quotient is fully faithful, and it seems that one can deduce the same for the functor lifted to the idempotent completion (first question, why?). Then , once defined a triangulated structure for the idempotent completion, he claims this triangulated functor to be essentially surjective, and in order to prove that observes that it suffices, given a perfect complex $F$ on $U$, to prove that $F$ is a direct summand of a complex which is the restriction of a perfect complex on $X$. So my question is, why is that true? How is the idempotent completion used to prove that?

In more general terms my questions could be:

(1) Does a fully faithful functor $\textbf{C}\to \textbf{D}$ into an idempotent complete category lift to a fully fraithful functor $\overset{\sim}{\textbf{C}}\to \textbf{D}$?

(2) If a triangulated functor of triangulated categories $G:\textbf{C}\to \textbf{D}$, with $\textbf{D}$ idempotent complete, is such that an object $F$ in $\textbf{D}$ is direct summand of an object $F\oplus W$ isomorphic to $G(X)$ for some $X$ in $\textbf{C}$, is it true that $F$ is isomorphic to $\overset{\sim}G(Y)$ for an object $Y$ in $\overset{\sim}{\textbf{C}}$?

答:
These statements are "obvious" if you understand how idempotent completion works.  Let me back up and describe that first.  Suppose you have an object $c$ in a category and a morphism $f:c\to c$ which satisfies $f^2=f$.  Then a splitting of $f$ is an object $d$ together with a pair of morphisms $i:d\to c$ and $p:c\to d$ such that $pi=1_d$ and $ip=f$.

Now, the key point is, when you have a splitting of an idempotent, that uniquely determines what the morphisms to and from $d$ can be, in terms of morphisms to and from $c$.  Specifically, for any $e$, morphisms $g:e\to d$ are in natural bijection with morphisms $h:e\to c$ such that $fh=h$ (the bijection sends $g$ to $ig$, and its inverse sends $h$ to $ph$).  Dually, morphisms $g:d\to e$ are in natural bijection with morphisms $h:c\to e$ such that $hf=h$.

This means that given a category $C$ with an idempotent $f:c\to c$, we can simply formally adjoin a splitting of it to get a category $C'$ with one more object $d$, where morphisms in and out of $d$ are defined as certain subsets of the morphisms in and out of $c$ (namely, morphisms into $d$ are morphisms into $c$ such that $fh=h$, and morphisms out of $d$ are morphisms out of $c$ such that $hf=h$).  This is the only possibility for the morphisms in and out of a splitting of $f$, so if $C$ was a fully faithfully embedded in a category $D$ in which $f$ splits, then $C'$ will be too.  Explicitly, suppose $G:C\to D$ is fully faithful and $i:d'\to G(c)$, $p:G(c)\to d'$ is a splitting of $G(f)$.  Then we can extend $G$ to $G':C'\to D$ by defining $G'(d)=d'$.  We know that morphisms $G(e)\to d'$ are in bijection with morphisms $h:G(e)\to G(c)$ such that $G(f)h=h$.  Since $G$ is fully faithful, these are in bijection with morphisms $h:e\to c$ such that $fh=h$.  But those are the same as morphisms $e\to d$, so we can extend $G$ to all the morphisms into $d$ and it will remain fully faithful on them.  Similarly for morphisms out of $d$.  (Of course there are some details I am skipping over here about compatibility with composition but they are straightforward to check.)

Now, idempotent completion is just doing this process for every idempotent in $C$ at once (or if you prefer, doing it iteratively on them one by one by transfinite recursion).  So if $D$ is idempotent complete, any fully faithful functor $C\to D$ extends to a fully faithful functor $\tilde{C}\to D$.  This is just because every object of $\tilde{C}$ is a formally adjoined splitting of some idempotent in $C$, so you can map it to a splitting of the corresponding idempotent in $D$.  This is fully faithful because the new morphisms you get when you split an idempotent are uniquely determined by the morphisms you had in and out of the original object.

Now for your second question, suppose $G:C\to D$ is a fully faithful functor of additive categories and $D$ is idempotent complete.  If an object $F$ of $D$ is a direct summand of an object $G(X)$ in the image of $X$, that just means that it splits an idempotent on $G(X)$.  (Namely, if $G(X)=F\oplus W$, then the inclusion $i:F\to G(X)$ and the projection $p:G(X)\to F$ split the idempotent $ip:G(X)\to G(X)$.)  Since $G$ is fully faithful, we can lift that idempotent up to an idempotent of $X$.  This idempotent is split by some object $Y$ of $\tilde{C}$, and then $\tilde{G}(Y)$ splits the same idempotent on $G(X)$ as $F$ does.  This implies $\tilde{G}(Y)$ and $F$ are isomorphic.  (You can build an isomorphism explicitly, or it follows by Yoneda: morphisms in and out of $\tilde{G}(Y)$ and $F$ are in natural bijection, since they are both in natural bijection with certain morphisms in and out of $G(X)$.)


问:
Find the gradient (with respect to a matrix) of an expression containing a Frobenius norm and a Hadamard product.
I'm struggling with taking the gradient with respect to (w.r.t) the matrices $H_R$ and $H_I$ in the following expression

$$\left\| Z - I\odot(H_R^TA + H_I^TB) \right\|_F^2 + \left\|W-\begin{pmatrix} H_R & -H_I\\H_I & H_R\end{pmatrix}P_dS^T \right\|_F^2$$

where $\|\cdot\|_F^2$ represents the Frobenius norm, $\odot$ represents the Hadamard product, $(\cdot)^T$ is the transpose operation, $I$ is the identity matrix, and all the terms in the above expression have the proper dimension.

It would be very helpful to find at least some alternative on how to calculate the gradient w.r.t one of the above-mentioned matrices, say $H_R$, as the same analysis could be then extended to $H_I$. Any help would be highly appreciated.

答:
$\def\R#1{{\mathbb R}^{#1}}\def\v{{\rm vec}}\def\M{{\rm Reshape}}\def\m#1{\left[\begin{array}{r}#1\end{array}\right]}\def\p#1#2{\frac{\partial #1}{\partial #2}}$For
ease of typing, replace the subscripted variables with single-letter names
$$\eqalign{
R = H_R \qquad Q = H_I \qquad P=P_d \\
}$$
and define the matrices
$$\eqalign{
I_2 &= \m{1&0\\0&1}\qquad J_2 = \m{0&-1\\1&0} \\
M &= I\odot(R^TA+Q^TB)- Z \\
N &= (I_2\otimes R+J_2\otimes Q)PS^T - W \\
D &= I\odot M \\
C &= NSP^T \\
}$$
and finally, we'll need the SVD of the last matrix
$$\eqalign{
C &= \sum_{k=1}^{rank(C)} \sigma_ku_kv_k^T \\
U_k &= \M(u_k,n,2) \quad&\iff\quad u_k = \v(U_k) \\
V_k &= \M(v_k,n,2) \quad&\iff\quad v_k = \v(V_k) \\
}$$
Let's also use a colon to denote the matrix inner product, i.e.
$$\eqalign{
A:B &= {\rm Tr}(A^TB) \\
A:A &= \big\|A\big\|_F^2 \\
}$$
Write the objective function using this new notation. Then calculate its differential.
$$\eqalign{
\phi &= M:M + N:N \\
d\phi &= 2M:dM + 2N:dN \\
  &= 2M:(I\odot (dR^TA+dQ^TB) + 2N:(I_2\otimes dR+J_2\otimes dQ)PS^T \\
  &= 2D:(A^TdR+B^TdQ) + 2C:(I_2\otimes dR+J_2\otimes dQ) \\
  &= 2AD:dR + 2C:(I_2\otimes dR) + 2BD:dQ - 2(J_2\otimes I_Q)C:(I_2\otimes dQ) \\
}$$
The terms containing Kronecker products are tricky. Here's where the SVD becomes useful
$$\eqalign{
C:(I_2\otimes dR)
 &= \sum_k\sigma_ku_kv_k^T:(I_2\otimes dR) \\
 &= \sum_k\sigma_ku_k:(I_2\otimes dR)v_k \\
 &= \sum_k\sigma_ku_k:\v(dR\,V_kI_2) \\
 &= \sum_k\sigma_kU_k:dR\,V_k \\
 &= \left(\sum_k\sigma_kU_kV_k^T\right):dR \\
 &= E:dR \\
}$$
Substitute the SVD result and set $dQ=0$ to obtain the gradient wrt $R$
$$\eqalign{
d\phi &= 2(AD+E):dR \\
\p{\phi}{R} &= 2(AD+E) \\
}$$
A similar calculation for the gradient wrt $Q$ is a bit tricker, but the result is
$$\eqalign{
F &= \sum_k\sigma_kU_kJ_2V_k^T \\
\p{\phi}{Q} &= 2(BD-F) \\
}$$

There are other ways (besides the SVD) to handle the Kronecker term, but the formulas are longer/messier.

Update

Here's one way to deal with the Kronecker terms which doesn't require the SVD.

Given matrices of the following dimensions and a cost function
$$\eqalign{
&I_p\in\R{p\times p}
\qquad
X\in\R{m\times n}
\qquad  Y\in\R{pm\times pn} \\
&\psi = Y:\left(I_p\otimes X\right) \\
}$$
To take advantage of the block-diagonal structure of the RHS,
define block-matrix analogs of the standard $\{e_k\}$ basis vectors
(i.e. columns of the $I_p$ identity matrix)
$$\eqalign{
E_k &= (e_k\otimes I_m) &\in\R{pm\times m} \\
F_k &= (e_k\otimes I_n) &\in\R{pn\times n} \\
}$$
and note that
$$\eqalign{
E_j^T(I_p\otimes X)F_k &= \left(e_j^T\otimes I_m\right) \left(I_p\otimes X\right) \left(e_k\otimes I_n\right) \\
 &= \left(e_j^Te_k\right)X \\
 &= \delta_{jk}X \\
}$$
Evaluate the cost function using block-wise summation and calculate its gradient.
$$\eqalign{
\psi &= \sum_{j=1}^p\sum_{k=1}^p\;E_j^TYF_k:E_j^T(I_p\otimes X)F_k \\
  &= \sum_{j=1}^p\sum_{k=1}^p\;E_j^TYF_k:\delta_{jk}X \\
  &= \sum_{k=1}^p\;E_k^TYF_k:X \\
d\psi &= \sum_{k=1}^p\;E_k^TYF_k:dX \\
\p{\psi}{X}
 &= \sum_{k=1}^p\;E_k^T Y F_k \\
 &= \sum_{k=1}^p\;\left(e_k^T\otimes I_m\right)Y\big(e_k\otimes I_n\big) \\
}$$


问:
Related rates, calculus
Suppose that $k^{2} + h^{3} = 9$. Find $\frac{dh}{dt}$ when $k=1$ and $\frac{dk}{dt} = 3$ ans $= \frac{1}{2}$. I'm differentiating with respect to $t$ but I cannot get the answer if you could show me the steps, or how to approach this question, would be very helpful :)

答:
Use the fact that $$\frac{dh}{dk} = \frac{\frac{dh}{dt}}{\frac {dk}{dt}} \iff \frac{dh}{dt} = \frac{dh}{dk}\cdot \frac{dk}{dt}$$

Integrate the equation with respect to $k$, evaluate $\frac{dh}{dk}$ at $k = 1$, use that value, and use the given value for $\frac {dk}{dt}$, and solve for $$\frac{dh}{dt} = \frac{dh}{dk}\cdot \frac{dk}{dt}$$

Title: Related rates, calculus


问:
$\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$
$\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$

I used the ratio criterion for the calculation and I got to this, can I say now that it is zero or is it still an undefined expression?

$\frac{10+(10/\sqrt{(n+1)})}{\sqrt{(n+2)}+1}$

10/inf < 1 ----> lim an=0

答:
Hint: No, it's not undefined, since the numerator goes to 11.

Just explaining a bit about the above answer:

$$100^n \leq  n!$$

and because we can see that:

$$n \cdot log(100) < log(n!) = \sum_{j = 0 }^{n} log(j)$$

for some $j = 100$, from this point the sums on the right will be bigger and overcome the difference, this is, on the left you are summing each time $log(100)$, but at some moment a gap will appear as we said, and this can be proved formally.


问:
Can't understand how the authors got this equation?
Let us consider a two-qubit system (A and R) initially entangled as given by:
$$
\left| {{\Psi _{AR}}} \right\rangle  = \alpha \left| {{0_A}} \right\rangle  \otimes \left| {{1_R}} \right\rangle  + \beta \left| {{1_A}} \right\rangle  \otimes \left| {{0_R}} \right\rangle
$$
where ${\left| \alpha  \right|^2} + {\left| \beta  \right|^2} = 1$. By using the fact that Rob’s detector is the only one which interacts with the field, the initial state of the total system is:
$$\left| {\Psi _{ - \infty }^{AB\phi }} \right\rangle  = \left| {{\Psi _{AR}}} \right\rangle  \otimes \left| {{0_M}} \right\rangle $$
where $\left| {{0_M}} \right\rangle $ the Minkowski vacuum.  The final state of the qubit-field system can be described as:
$$\left| {\Psi _\infty ^{AB\phi }} \right\rangle  = \left( {I + a_{RI}^\dagger(\lambda )R - {a_{RI}}(\bar \lambda ){R^\dagger}} \right)\left| {\Psi _{ - \infty }^{AB\phi }} \right\rangle $$
where $I$ is the identity matrix with the same dimension of $\left| {\Psi _{ - \infty }^{AB\phi }} \right\rangle $. ${D^\dagger}\left| 1 \right\rangle  = D\left| 0 \right\rangle  = 0$, $D\left| 1 \right\rangle  = \left| 0 \right\rangle $, and ${D^\dagger}\left| 0 \right\rangle  = \left| 1 \right\rangle $, ($D \in A,R$) are the raising and lowering operators for the qubit.

Using the first two equations in the last one, the authors got:
$$\left| {\Psi _\infty ^{AB\phi }} \right\rangle  = \left| {\Psi _{ - \infty }^{AB\phi }} \right\rangle  + \alpha \left| {{0_A}} \right\rangle  \otimes \left| {{0_R}} \right\rangle  \otimes \left( {a_{RI}^\dagger(\lambda )\left| {{0_M}} \right\rangle } \right)\\
+ \beta \left| {{1_A}} \right\rangle  \otimes \left| {{1_R}} \right\rangle  \otimes \left( {{a_{RI}}(\bar \lambda )\left| {{0_M}} \right\rangle } \right)$$
Please, could someone explain to me how the authors got this equation?

The intended equation is Eq. (3.6) in 10.1103/PhysRevA.80.032315

答:
First,
$$
\left| {\Psi _\infty ^{AB\phi }} \right\rangle
= \left( {I + a_{RI}^\dagger(\lambda )R - {a_{RI}}(\bar \lambda ){R^\dagger}} \right)\left| {\Psi _{ - \infty }^{AB\phi }} \right\rangle
= \left( {I + a_{RI}^\dagger(\lambda )R - {a_{RI}}(\bar \lambda ){R^\dagger}} \right)
\left( \left| {{\Psi _{AR}}} \right\rangle  \otimes \left| {{0_M}} \right\rangle \right)
\\
= \left| {{\Psi _{AR}}} \right\rangle  \otimes \left| {{0_M}} \right\rangle
+ a_{RI}^\dagger(\lambda )R \left( \left| {{\Psi _{AR}}} \right\rangle  \otimes \left| {{0_M}} \right\rangle \right)
- {a_{RI}}(\bar \lambda ){R^\dagger}\left( \left| {{\Psi _{AR}}} \right\rangle  \otimes \left| {{0_M}} \right\rangle \right)
$$
Then, for the second term,
$
a_{RI}^\dagger(\lambda )R \left( \left| {{\Psi _{AR}}} \right\rangle  \otimes \left| {{0_M}} \right\rangle \right)
= R\left| {{\Psi _{AR}}} \right\rangle \otimes a_{RI}^\dagger(\lambda )\left| {{0_M}} \right\rangle,
$
where
$$
R\left| {{\Psi _{AR}}} \right\rangle
= R \left( \alpha \left| {{0_A}} \right\rangle  \otimes \left| {{1_R}} \right\rangle  + \beta \left| {{1_A}} \right\rangle  \otimes \left| {{0_R}} \right\rangle \right)
= \alpha \left| {{0_A}} \right\rangle  \otimes R\left| {{1_R}} \right\rangle  + \beta \left| {{1_A}} \right\rangle  \otimes R \left| {{0_R}} \right\rangle
\\
= \alpha \left| {{0_A}} \right\rangle  \otimes \left| {{0_R}} \right\rangle  + \beta \left| {{1_A}} \right\rangle  \otimes R 0
= \alpha \left| {{0_A}} \right\rangle  \otimes \left| {{0_R}} \right\rangle
.
$$
Likewise for the third term, $- {a_{RI}}(\bar \lambda ){R^\dagger}\left( \left| {{\Psi _{AR}}} \right\rangle  \otimes \left| {{0_M}} \right\rangle \right)$.


问:
limit of $\cos^n(\frac{\pi}{n})$
Pretty self explanatory - what is a nice way to show that
$(\cos(\frac{\pi}{n}))^n\to 1$ as $n\to\infty$, where $n$ are integers?

答:
Let $L=\lim_{n\to+\propto}\begin{pmatrix}
cos\begin{pmatrix}
\frac{\pi}{n}\\
\end{pmatrix}\\
\end{pmatrix}^n$. We obtain: $log(L)=\lim_{n\to+\propto}n\cdot log\begin{pmatrix}
cos\begin{pmatrix}
\frac{\pi}{n}\\
\end{pmatrix}\\
\end{pmatrix}$.

As $n \to +\propto$, we know that:

$cos\begin{pmatrix}\frac{\pi}{n}\\ \end{pmatrix}=1-\frac{\pi^2}{2n^2}+o\begin{pmatrix}\frac{1}{n^3}\\ \end{pmatrix}$. So, using Taylor expansion again:

$log\begin{pmatrix}
cos\begin{pmatrix}
\frac{\pi}{n}\\
\end{pmatrix}\\
\end{pmatrix}=log\begin{pmatrix}1-\frac{\pi^2}{2n^2}+o\begin{pmatrix}\frac{1}{n^3}\\ \end{pmatrix}
\end{pmatrix}=-\frac{\pi^2}{2n^2}+o\begin{pmatrix}\frac{1}{n^3}\\ \end{pmatrix}+o\begin{pmatrix}-\frac{\pi^2}{2n^2}+o\begin{pmatrix}\frac{1}{n^3}\\ \end{pmatrix} \\ \end{pmatrix}$
$=-\frac{\pi^2}{2n^2}+o\begin{pmatrix}\frac{1}{n^2}\\ \end{pmatrix}$

Thus we can finally calculate: $log(L)=\lim_{n\to+\propto}n\cdot log\begin{pmatrix}
cos\begin{pmatrix}
\frac{\pi}{n}\\
\end{pmatrix}\\
\end{pmatrix}=\lim_{n\to+\propto}n\cdot \begin{pmatrix}
-\frac{\pi^2}{2n^2}+o\begin{pmatrix}\frac{1}{n^2}\\ \end{pmatrix}\\\end{pmatrix}=0$.

So $L=\lim_{n\to+\propto}\begin{pmatrix}
cos\begin{pmatrix}
\frac{\pi}{n}\\
\end{pmatrix}\\
\end{pmatrix}^n=1$.

Title: limit of $\cos^n(\frac{\pi}{n})$


问:
first order linear non-homogenous equation
I have the following circuit:

I have solved the equation this far $$ I_C + I_{R1}=I , U=I_{R1}R_1 \implies I_{R1}=\frac{U}{R_1}, Q=CV \implies I_C=C\frac{dU}{dt}, \frac{dU}{dt}=\frac{I_C}{C}=\frac{I-I_{R1}}{C}=\frac{I}{C}-\frac{U}{R_1C}$$
this can be solved as
$$ U_i=\Delta t(\frac{-U_{i-1}}{R_1C}+\frac{I}{C})+U_{i-1}$$
but for first order linear non homogenous we get:
$$ U=A \exp {\frac{-t}{RC}}+IR_1$$
I need the derivation to get this answer instead:
$$U_{i+1}=U_i\exp {\frac{-\Delta t}{RC}} +IR_1*(1-\exp{\frac{-\Delta t}{RC}})$$

I is not constant and can be a function of time

答:
$$I=I_C + I_{R1}$$
$$ U=I_{R1}R_1 \implies I_{R1}=\frac{U}{R_1}$$
$$Q=CV \implies I_C=C\frac{dU}{dt}$$
$$\frac{dU}{dt}=\frac{I_C}{C}=\frac{I-I_{R1}}{C}=\frac{I}{C}-\frac{U}{R_1C}$$
$$R_1C\frac{dU}{dt}=R_1I-U$$
but we know that the 1st order linear non homogenous solution is:
$$ U=A \exp {\frac{-t}{RC}}+IR_1 $$

In time series analysis we know for first order equations of the form:
$$ \dot x(t)=ax(t)+bu(t) \implies x[k+1]=e^{a\Delta t}x[k]+\frac{1}{a}\left( e^{a \Delta t} -1\right) bu[k]$$
in our differential we have:
$$ a=-\frac{1}{R_1C}$$ $$ bu(t)=\frac{I}{C}$$ $$ \dot x=\frac{dU}{dt}$$
$$ U_{i+1}=e^{-\frac{\Delta t}{R_1C} }U_i-IR_1 \left( e^{-\frac{\Delta t}{R_1C} } -1\right)$$
$$ U_{i+1}=U_i\exp {\frac{-\Delta t}{R_1C}} +IR_1*(1-\exp{\frac{-\Delta t}{R_1C}})$$


问:
Differential equation of type $u_{tt}=u_{xx} + u$ with specific boundary conditions
I need to solve the following differential equations.

$u(x, t)\;;x\in(0,2)\;;t>0$

$u_{tt} = u_{xx} + u$

Boundary conditions:

$u(x,0) = 0$

$u_t(x,0) = 0$

$u(0, t) = 2t$

$u(2, t) = 0$

I have tried to separate variables: $u(x,t) = X(x)T(t)$.

$XT'' = X''T + XT$

$\frac{T''}{T} = \frac{X''}{X} + 1 = \lambda$

But whatever lambda I choose ($<0;=0;>0$), $T$ is equal to $0$ due to boundary conditions:

$T = ax + b$

$0 = T(0)X(x) = bX(x)$

$0 = T'(0)X(x) = aX(x)$

$a = b = 0$ if $\exists x\;X(x)\neq0$

$T = ae^{kt} + be^{-kt}$

$0 = T(0)X(x) = (a+b)bX(x)$

$0 = T'(0)X(x) = k(a-b)X(x)$

$a = b = 0$ if $\exists x\;X(x)\neq0$

$T = a\sin{kt} + b\cos{kt}$

$0 = T(0)X(x) = bX(x)$

$0 = T'(0)X(x) = kaX(x)$

$a = b = 0$ if $\exists x\;X(x)\neq0$

What is wrong? Is separation of variables not a valid way for solving equations of such type? How could it be solved then?

答:
Hint: you are going to have to solve the following two equations:

$T'' - T(\lambda -1) = 0$

and also:

$X'' - X(\lambda -1) = 0$

And you have to considerer solutions of the form $e^{rx}$.

Obs.: it's always good to note if the solution will or not be linearly independent.