答:
Let me comment on the sentences you are saying you don't understand:

By the properties of the supremum, there exists some $a^{*}\in (c-\delta ,c]$ that is contained in $S$, and so $f(c)<f(a^{*})+\varepsilon \leq u+\varepsilon$.

As the supremum is the least upper bound of $S$, if there wasn't such $a^*$ contained in $S$, it would be then possible to find a smaller upper bound. Namely, $c-\delta$ would be a smaller upper bound of $S$, if you assume the whole interval $(c-\delta, c]$ is devoid of elements of $S$.

Now pick such $a^*\in S$, and you will have $f(a^*)\le u$, while simultaneously having $f(c)-\varepsilon\lt f(a^*)\lt f(c)+\varepsilon$ (which they assumed is not just valid for $a^*$ but for any value in $(c-\delta, c+\delta)$). Put those two inequalities together to get the desired inequality $f(c)<f(a^{*})+\varepsilon \leq u+\varepsilon$.

Picking $a^{**}\in (c,c+\delta )$, we know that $a^{**}\not \in S$ because $c$ is the supremum of $S$. This means that $f(c)>f(a^{**})-\varepsilon>u-\varepsilon$.

$a^{**}\not\in S$ because $c=\sup S$ is the upper bound of $S$, i.e. all elements of $S$ are "below" or equal to $c$, while $a^{**}$ has been chosen to be greater than $c$.

Not being in $S$ means that $f(a^{**})>u$, but with $f(c)-\epsilon\lt f(a^{**})\lt f(c)+\epsilon$ (which again is not just valid for $a^{**}$ but for any number in $(c-\delta, c+\delta)$) we can put together those two inequalities and get the desired inequality $f(c)>f(a^{**})-\varepsilon>u-\varepsilon$.

Title: I can't understand Intermediate Value Theorem Proof

Body:
Intermediate Value Theorem. If ƒ is a continuous function on a closed interval [a, b] and if u is any value between ƒ(a) and ƒ(b), then u = ƒ(c) for some c in [a, b].

I'm trying to understand this theorem from Wikipedia(https://en.wikipedia.org/wiki/Intermediate_value_theorem#Proof), it says that if $A:= \{x \in [a,b]:f(x)≤u\}$ and $s:=supA$ then $f(s)=u$.

I can't understand it. Please help me to prove this theorem.

Edited:I don't understand this part on Wikipedia:


问:
If $\lvert\varphi(t_0)\rvert=1$ holds for the characteristic function, then $P(X\in a+\frac{2\pi}{t_0} \mathbb{Z}) = 1$
I want to show the following: if for $t_0\ne 0$ we have $\lvert\varphi(t_0)\rvert = 1$, then there exists a constant $a$ such that $P(X\in a+\frac{2\pi}{t_0}\mathbb{Z}) = 1$.

I'm stuck on how to actually use the premise. Rewriting directly led me to
$$1=\lvert \varphi(t_0)\rvert  = \left\lvert\int e^{it_0 x}P_X(dx)\right\rvert \implies \int e^{it_0x}P_X(dx) = \cos(r) + i\sin(r)$$

But I don't see how this might help. I think I need to approach it from a different angle and then use the premise, but I don't know what that angle might be. Does anyone have any ideas?

答:
You have that $\mathbb{E}[e^{it_0 X}] = e^{it_0 a}$ for some $a \in \mathbb{R}$. It follows that $\mathbb{E}[e^{it_0 (X - a)}] = 1.$ In particular, we get for the real  part $\Re$ $$1 = \Re(\mathbb{E}[e^{it_0( X - a)}]) = \mathbb{E}[\cos(t_0(X -a))].$$
Note that $\cos(t_0(X -a)) < 1$ if $t_0(X -a) \notin 2\pi\mathbb{Z}$. It follows that $$\mathbb{P}\left(X \in a + \frac{2\pi}{t_0}\mathbb{Z}\right) = 1.$$

ContentLicense: CC BY-SA 4.0


问:
Proving limit using definition
I need to prove $$ \lim_{x\rightarrow\ 0}\frac{x^2-8}{{x-8}} =1 $$ using epsilon-delta definition.
I know I need to show that for every $\epsilon >0$ there exist a $\delta >0$ such that if $|x|<\delta$, then $| {\frac{x^2-8}{x-8}}-1|<\epsilon$

WHat I did:

$|{\frac{x^2-8}{x-8}}-1|=|{\frac{x^2-x}{x-8}}|$, but Im having troubles now deciding what my delta should be.

I would really appreciate if someone could give me an explanation on how should I choose delta and why.
Thanks.

答:
We have

$$\left|x^2-x\over x-8\right|=|x|\left|x-1\over x-8\right|\le|x|{|x|+1\over||x|-8|}$$

Now the main thing that causes a problem is the smallness of the denominator if $|x|\approx8$. But since we're interested in the limit as $x\to0$, it's easy to stay away from $8$ by requiring, say $|x|\le7$, in which case we have

$${|x|+1\over||x|-8|}\le{7+1\over|7-8|}=8$$

So taking $\delta=\min(7,\epsilon/8)$ we have

$$0\lt|x|\lt\delta\implies\left|x^2-x\over x-8\right|\le8|x|\lt8\delta\le\epsilon$$

as desired.

Remark: The key to limit write-ups of this type is to keep in mind that even though we're thinking of small epsilons, we don't want to be embarrassed by claiming an implication that isn't true in case someone decides to use a large $\epsilon$. E.g., if we just let $\delta=\epsilon/8$, we run the risk of the false implication $0\lt|x|\lt8\implies|(x^2-x)/(x-8)|\lt64$, which is falsified, for example, at $x=7.9$ (that is, $0\lt|7.9|\lt8$ but $|(7.9^2-7.9)/(7.9-8)|=545.1\gt64$). The role of the "min" is to avoid such false claims. As the other answers here show, there is no single prefered way to choose your delta; it's somewhat a matter of taste. My preference is to try to isolate what causes things to be uncontrollably large, usually a potential zero in a denominator, impose a restriction brings that term under control, and then look at what happens to everything else.

Title: Proving limit using definition


问:
How to apply the rule of absorption correctly to the following logical expressions?
How was the rule of absorption applied to the following logical expression?

$$ (A \land \neg B \land C) \lor (A \lor B \lor \neg C) \lor (\neg A \land B) \equiv (A \lor B \lor \neg C) \lor (\neg A \land B).$$

The step is not really clear to me, and I hope someone can explain it.

答:
Using associativity,
$$
(A \wedge \neg B \wedge C) \vee (A \vee B \vee \neg C) \vee (\neg A \wedge B)
= ((A \wedge \neg B \wedge C) \vee A) \vee (B \vee \neg C)\vee (\neg A \wedge B).
$$
Now, by absorption and associativity,
$$(A \wedge \neg B \wedge C) \vee A = (A \wedge (\neg B \wedge C)) \vee A = A,$$
and the expression becomes
$$A \vee (B \vee \neg C)\vee (\neg A \wedge B),$$
which is the desired result, using associativity again.


问:
Complex-valued Fourier integral: $ \int_{ - \infty }^{ + \infty } {\frac{{\cos (ax)}}{{{x^2} + 1}}{e^{ - ibx}}\,\mathrm dx} $
I'm working on the Fourier transform, but I don't know how to evaluate the integral:

$$I = \int_{ - \infty }^{ + \infty } {\frac{{\cos (ax)}}{{{x^2} + 1}}{e^{ - ibx}}\,\mathrm dx} $$

答:
Hint: Use the identity $\cos(ax)=\frac{e^{iax}+e^{-iax}}{2}$ and the fact that the Fourier transform of $\frac{1}{1+x^2}$ is

$$\int_{-\infty}^{\infty}\frac{1}{1+x^2} \rm e^{-ix w}dx=\pi\rm {e^{-|w|}}.$$

To evaluate the last integral, see the first example here. It is a worked out example.

ContentLicense: CC BY-SA 3.0


问:
Is $\sqrt[3]{-3} = -\sqrt[3]{3}$ true?
Well, Wolfram Alpha says no. Why?

答:
$$\sqrt[3]{-3} = -\sqrt[3]{3} \iff (\sqrt[3]{-3})^3 = (-1)^3(\sqrt[3]{3})^3 \iff -3=(-1)3 \iff -3=-3$$
So I would say it is true if we are talking about real numbers.

In fact in "Analisi Matematica" of Paolo Maurizio Sardi (CIttàStudi edizioni), we define roots in the following manner.

Theorem 1.7.1 Given an integer $n \ge 1$, and a real number $\alpha >0$, there exists one, and only one, real number $\beta>0$ such that
$$\beta^n=\alpha \mbox.$$

Definition 1.7.3 The number $\beta >0$  found in the Theorem above is called $n$-th root of $\alpha$ and is denoted by
$$\beta = \sqrt[n]\alpha \mbox.$$

Observation: $(\sqrt[n]\alpha )^m =\sqrt[n]{\alpha^m}$

Definition 1.7.4 Let $m\in \mathbb{Z}$, let $n\ge 1$ be an integer and let $\alpha >0$. We set
$$\alpha^{m/n}=(\sqrt[n]\alpha )^m $$

Observation: we now want to define all of this for negative real numbers. Let $-a<0$ and let $n \ge 1$ be an integer:

(1) if $n=2k$ is even, it is clear that there does not exists $\beta\in \mathbb{R}$ such that
$$\beta^{2k}=(\beta^2)^k=-\alpha \mbox,$$
so we cant' do this in the reals;

(2)if $n=2k+1$ is odd and let $\beta >0$ be such that $\beta^{2k+1}=\alpha$ ($>0$), then
$$(-\beta)^{2k+1}=-\alpha$$
so $-\beta$ is called the $n$-th root of $-\alpha$ an is denoted by
$$-\beta=\sqrt[n]{-\alpha} \mbox.$$

You could then continue with defining roots with real exponent (going back to $a>0$), showing that the following definition is well-defined:

Definition 1.7.5:

(1) if $\alpha \ge 1$ and $\beta >0$ we set
$$\alpha^{\beta}:= \sup\biggl{\{} \alpha^{m/n}:\frac{m}{n} \le \beta \biggr{\}}$$

(2) if $0<\alpha < 1$ and $\beta >0$ we set
$$\alpha^{\beta}:=\frac{1}{(1/\alpha)^{\beta}}$$

(3) if $\alpha >0$ and $\beta >0$ we set
$$\alpha^{-\beta}=\frac{1}{\alpha^{\beta}$$

(4) $\alpha^0=1$

(defined for every $\alpha>0$ and $\beta$ reals).

Title: Is $\sqrt[3]{-3} = -\sqrt[3]{3}$ true?

Body: Well, Wolfram Alpha says no. Why?


问:
eigenvalues and matrices kernel
Let B be an n × n matrix with eigenvalue λ. Note that Eλ(B) can be written as Ker(B −
λIn). Show that Ker(B − λIn) ⊆ Ker((B − λIn)^2). Please help.
Eλ(B)is the eigenspace associated with eigenvalue λ.
In is the identity matrix for nxn matrix</p>

答:
Let $v$ be an eigenvector with eigenvalue $\lambda$ then

$Bv=\lambda v$  and $v\in Ker(B-\lambda I)$

we want to prove that $Ker(B − λIn) ⊆ Ker((B − λIn)^2)$ which means

$v \in Ker(B − λIn)$ $\Rightarrow$ $v\in Ker((B − λIn)^2)$

$(B − λI)^2=B^2-2\lambda B+ \lambda^2I$

$(B − λI)^2v=B^2v-2\lambda Bv+ \lambda^2Iv=\lambda^2v-2\lambda^2v+\lambda^2v=0$ which means

$v\in  Ker((B − λIn)^2)$ so  $Ker(B − λIn) ⊆ Ker((B − λIn)^2)$

Title: eigenvalues and matrices kernel


问:
convergence sequence of xor bit operation
Given positive integer $m$, we consider the integers up to $m$ bits.

Let $mask = 2^m - 1$ and consider sequence, $a_0 = 0, a_1 = 1$,

For $i \ge 2$,

$$a_i = (a_{i-2}) \oplus (a_{i-1}) \oplus (a_{i-1}>>1) \oplus ((a_{i-1}<<1) \land mask)$$

$\oplus$: is bit XOR operation;

$>>$: is bit shift operation; $>>1$ is equivalent of dividing 2

$<<$: is bit shift operation; $<<1$ is equivalent of multipling 2

$\land$: is bit and operation; $\land mask$ operation makes sure the resulting number is up to $m$ bits.

Prove/disprove  that, for any positive integer $m$, there always exist $i>0$ such that $a_i = 0$.

Examples,

for $m=1$, ${\bf a} = [0, 1, 1, 0, 1, 1, 0, 1, 1, 0, ...]$

for $m=2$, ${\bf a} = [0, 1, 3, 1, 0, 1, 3, 1, 0, 1, ...]$

for $m=3$, ${\bf a} = [0, 1, 3, 5, 6, 4, 0, 4, 6, 5, ...]$

for $m=4$, ${\bf a} = [0, 1, 3, 5, 14, 0, 14, 5, 3, 1, ...]$

for $m=5$, ${\bf a} = [0, 1, 3, 5, 14, 16, 22, 1, 21, 20, 3, 16, 27, 16, 3, 20, 21, 1, 22, 16, 14, 5, 3, 1, 0, 1, ...]$

答:
I think the trick here is to recognise that you can play the sequence forwards as well as backwards. This is because:

$$a_{i-2} = a_i \oplus (a_{i-1}) \oplus (a_{i-1}>>1) \oplus ((a_{i-1}<<1) \land mask)$$

for $i\ge 2$ (i.e. you can interchange $a_i$ with $a_{i-2}$).

As the number of possible binary words of length $m$ is finite, which also means the number of pairs of those is finite, the sequence must at some point start repeating itself: let the first repetition (in lexicographic order) be $a_m=a_n, a_{m+1}=a_{n+1}$ (with $m<n$).

If $m=0$, we are done because $a_n=a_m=0$.

If $m>0$, then due to the property noted above we can prove that $a_{m-1}=a_{n-1}$ - which is a contradiction ($a_m=a_n$ was not the first repetition).


问:
Defining an equivalent measure starting from a continuous local Martingale
Suppose we have continuous local martingal $L$ given. We define $Z=\mathcal{E}(L)$, the stochastic exponential of $L$. I am interested in finding some condition such that $Z$ defines a density, i.e. I can define an equivalent measure $Q$ by setting $dQ:=Z_\infty dP$.

First of all: $Z$ is again a continuous local martingale, positive hence a supermartingale and $>0$ on $[0,\infty)$. Hence $Z_\infty$ exists. Now I need two assumption, such that everything works out nicely

$E[Z_\infty]=1$, which is equivalent to the property that $Z$ is a $P$-martingale on $[0,\infty]$, hence uniformly integrable.

I have to guarantee that $Z_\infty>0$ $P$-a.s.

For $2.$ we have $Z_\infty =e^{L_\infty -\frac{1}{2}\langle L\rangle_\infty}$. Therefore $L_\infty -\frac{1}{2}\langle L\rangle_\infty \not= -\infty$. Since the bracket process is increasing, we must have $\langle L\rangle_\infty <\infty$. The question is:

Can it happen that $\langle L\rangle_\infty <\infty$ but $L_\infty = -\infty$. If so, are there conditions which excludes $L_\infty=-\infty$ and $\langle L\rangle_\infty =\infty$?

答:
Let $L$ be a continuous local martingale with initial value zero. Define

$$
F = \{\omega\in\Omega\mid L_t \textrm{ converges as }t\to\infty\}\\
G = \{\omega\in\Omega\mid \langle L\rangle_t \textrm{ converges as }t\to\infty\}.
$$

It then holds that $F$ and $G$ are $P$-almost surely equal, in the sense that $P(F\setminus G)=P(G\setminus F)=0$. This implies that almost surely, it cannot happen that $\langle L\rangle_\infty<\infty$ but $L_\infty=-\infty.$

The proof of the above claim is not completely straightforward. You can find a detailed proof as the solution to Exercise 2.3 (in the appendix) in the following:

http://alexandersokol.dk/StochInt/ContinuousStochasticIntegration.pdf


问:
Sum of infinite series $\frac{k(k+1)3^k}{k!}$
Can someone suggest how to find the infinite series sum for

$$\frac{k(k+1)3^k}{k!}$$ where k goes from $1$ to infinity.

I know that $\sum_0\frac{3^k}{k!}=e^3$ but I'm not sure if that helps here.

Body: Hint:let $$f(x)=xe^x=\sum_{k=0}^{\infty}\frac{x^{k+1}}{k!}$$ what is $3f''(3)?$


答:
For simplicity $M=B-\lambda I_n$:
$$v \in Ker(M) \Rightarrow Mv=0 \Rightarrow MMv=M0 \Rightarrow M^2v=0 \Rightarrow v \in Ker(M^2)$$

Title: eigenvalues and matrices kernel

Body:
Let B be an n × n matrix with eigenvalue λ. Note that Eλ(B) can be written as Ker(B −
λIn). Show that Ker(B − λIn) ⊆ Ker((B − λIn)^2). Please help.
Eλ(B)is the eigenspace associated with eigenvalue λ.
In is the identity matrix for nxn matrix</p>


问:
Is $A \subset \overline{\mathrm{span} \{ f_n : n \in \mathbb{N} \}}$ if every element in $A$ is orthogonal to each $f_n$?
Let $H$ be a Hilbert space and $(f_n)_{n \in \mathbb{N}}$ a sequence in $H$. Further, let $A \subset H$ be a closed subspace of $H$. Suppose that the following property holds:

For every $a \in A$ the implication
$$
\langle f_n,a \rangle = 0 \ \ \forall n \in \mathbb{N} \implies a=0
$$
holds true. Can we conclude that $A \subset \overline{\mathrm{span} \{ f_n : n \in \mathbb{N} \}}$, i.e. $A$ is a subspace of the closure of the linear span of all $f_n$'s ?

答:
Unfortunately you cant conclude that $A\subseteq \overline{\text{span}\{f_n\,:\,n\in \mathbb{N}\}}$. For example take $H=\mathbb{R^3}$ with the Euclidean norm $||.||_2$. Take $f_1=(1,0,0)$ and $f_n=f_1$ for every $n$. For $A$ take $\text{span}\bigl((1,1,0)\bigr)$. Then,
$$\overline{\text{span}\{f_n\,:\,n\in \mathbb{N}\}}=\{(\lambda,0,0)\,:\, \lambda \in \mathbb{R}\}$$
and if  $(\mu,\mu,0)\cdot (1,0,0)=0$ then $\mu=0$. So the only vector of $A$ which is vertical to $\overline{\text{span}\{f_n\,:\,n\in \mathbb{N}\}}$ is the zero element. But of course $A\subsetneq \overline{\text{span}\{f_n\,:\,n\in \mathbb{N}\}}$. You can also construct an example where the sequence is non constant.

For example if $H=\ell_2$ with $||x||_2=\biggl(\sum_{k=1}^{\infty}|x_k|^2\biggr)^{1/2}$ where $x=(x_k)_{k=1}^{\infty}$. Then, take $f_n=e_{n+1}$ where $(e_n)_{n=1}^{\infty}$ is the standard basis of $\ell_2$. Let $A=\text{span}(e_1+e_2)=\{x\in \ell_2:\, x=(\lambda,\lambda,0,0,...),\, \lambda \in \mathbb{R}\}$. Then, you can check again that the only element which is perpendicular to $\overline{\text{span}\{f_n\,:\,n\in \mathbb{N}\}}$ is the zero element. But, $A\subsetneq \overline{\text{span}\{f_n\,:\,n\in \mathbb{N}\}}$ since
$$\overline{\text{span}\{f_n\,:\,n\in \mathbb{N}\}}=\{x\in \ell_2:\,x=(0,x_1,x_2,...),\, x_n\in \mathbb{R}\}$$
and $(1,1,0,0,0,...)$ belongs to $A$ but not to $\overline{\text{span}\{f_n\,:\,n\in \mathbb{N}\}}$.


问:
Integral and semi-local ring
Let $R$ be a ring and let $S$ be a subring of R. If $R$ is a semi-local ring and $R$ is integral over $S$, why $S$ is semi-local as well?</p>

答:
Because $R$ is integral over $S$, given any maximal ideal of $\mathfrak n$ of $S$, there is a maximal ideal $\mathfrak m$ of $R$ lying over $\mathfrak m$.  This shows
that the number of maximal ideals of $S$ is bounded above by the number of maximal ideals of $R$.  In particular, if $R$ has only finitely many such, then the same is true of $S$.</p>

Title: Integral and semi-local ring


问:
Calculate $\pi$ in an arbitrary base, to arbitrary precision
I need to calculate $\pi$ -- in base: 4, 12, 32, and 128 -- to an arbitrary number of digits. (It's for an artist friend).

I remember Taylor series and I've found miscellaneous "BBP" formulas, but so far, nothing that points to calculating the digits in arbitrary bases.

How can this be done in a practical manner?

答:
Usually, ye calculate pi, and then convert it into the target base.

```<code>   3:16E8 E212,7796 7998,5967 5292,6847 6661,9725 5723         base 120
   3.141 592 653,589 793 238,462 643 383,279 502 884,197       base 10
</code>```

This is an implementation in rexx, that finds pi to so many digits in decimal, base 120, and any named base.

ContentLicense: CC BY-SA 2.5

Body:
Usually, ye calculate pi, and then convert it into the target base.

   3:16E8 E212,7796 7998,5967 5292,6847 6661,9725 5723         base 120
   3.141 592 653,589 793 238,462 643 383,279 502 884,197       base 10

This is an implementation in rexx, that finds pi to so many digits in decimal, base 120, and any named base.


问:
Number Theory from Polish maths olympiad
Because the date of sending tasks from this stage has ended I need to know answer of it. This is from the first stage of the Polish Maths Olympiad (task 11):

Let $p$ be a prime number. A natural number $n \in\mathbb N^+$ is beautiful only when the sum of the remainders of $n,\ n^2,\ n^3,\ n^4,\ \ldots,\ n^{p-1}$ after division by $p$ is equal to
$$\frac12p(p-1).$$
Show that the number of beautiful numbers in $\{1,2,3,4,\ldots,p-1\}$ is odd (uneven).

I was thinking about it quite a lot and thought maybe using the binominal theorem we can prove it but couldn't end proof ://

答:
Hint 1: If $n$ is beautiful then so is $n^{-1}$, the inverse of $n$ mod $p$.

If you are still stuck with this idea, feel free to ask for more details. Or click the spoiler below for the full idea of the proof:

 This means beautiful numbers come in pairs, except those for which $n\equiv n^{-1}\pmod{p}$. Show that there is precisely one beautiful number $n$ such that $n\equiv n^{-1}\pmod{p}$.


问:
Formal proof about ordered fields
Let $(\mathbb F,\mathbb F^{+})$ an ordered field.
Are the following statements true? Prove (or disprove).

(i) For every element $x \in \mathbb F$  there are two other elements in $\mathbb F$ which are smaller and larger than $x$.

(ii) For every $x,y \in \mathbb F$ there exists $z\in \mathbb F$ such that $x=0$ or $xz=y$.

Sorry that I'm completely lost. I think both statements are true. The first one is true since ordered fields can not be finite. The rational numbers are the smallest ordered field so my intuition is that the second statement is true as well. Can anyone help me to give a more formal proof of the two statements?

答:
(i) Your intuition is correct. You can simply take the numbers to be $x - 1$ and $x + 1$. These two are distinct from each other and from $x$ because $1 + 1 \neq 0 \neq 1$ in an ordered field (a consequence of $1 > 0$). $1 > 0$ also gives tells us that $x - 1 < x < x + 1$.

(ii) If $x = 0$, then you're done. Else, you may take $z = x^{-1}y.$ ($x^{-1}$ exists because every non-zero element in a field has a multiplicative inverse.)

Title: Formal proof about ordered fields


问:
A small doubt about the dominated convergence theorem

Theorem $\mathbf{A.2.11}$ (Dominated convergence). Let $f_n : X \to \mathbb R$ be a sequence of measurable functions and assume that there exists some integrable function $g : X \to \mathbb R$ such that $|f_n(x)| \leq |g(x)|$ for $\mu$-almost every $x$ in $X$. Assume moreover that the sequence $(f_n)_n$ converges at $\mu$-almost every point to some function $f : X \to \mathbb R$. Then $f$ is integrable and satisfies $$\lim_n \int f_n \, d\mu = \int f \, d\mu.$$

I wanted to know if in the hypothesis $|f_n(x)| \leq|g(x)|$ above, if I already know that each $f_n$ is integrable, besides convergent, the theorem remains valid? Without me having to find this $g$ integrable?

答:
As you guessed, you actually don't need a dominating function, but you need some condition in addition to the integrability and convergence of $f_{n}$.

Here's a simple example that shows that you don't need a dominating function. Let $X=(0,1]$ and $\mu$ be the Lebesgue measure. Let
$$
f_{n}(x) =
\begin{cases}
[1/n - 1/(n+1)]/n & \text{if $x \in (1/(n+1),1/n]$,}
\\ 0 & \text{otherwise.}
\end{cases}
$$
Then $\int f_{n} = 1/n$ and $\lim_{n} f_{n} = 0$. In this case, there's no dominating function that is integrable, but
$$
\lim_{n} \int f_{n} = \int \lim_{n} f_{n} = 0.
$$

There're various extensions of the dominating convergence theorem, and there's a recent paper that gives a necessary and sufficient condition for the first equality above to be valid. The exact condition is somewhat complicated, but it's strictly weaker than uniform integrability, which is also somewhat complicated and in turn weaker than the hypothesis of the dominated convergence theorem.

(This answer is almost identical to another answer of mine, but I'm a newbie here and I've just realized that a post that gives a link that answers the question can be deleted.)


问:
Area of a region bounded by a circle and vertical line using integration
What is the area of the right region bounded by $x^2 + y^2 =25$ and $x = -3$?

My attempt solution: I solved first for the area of the left region and subtracted it from the area of the circle, which is

$$ 25 \pi - \int_{-4}^4 -3 + \sqrt {25-y^2} \,dy $$
$$ 78.5 - 11.18... = 67.36 $$
So, the area of the right region is 67.36.

My other solution is:
$$ \int_{0}^5 \sqrt {25-y^2} \,dy + \int_{0}^4 -(-3)\,dy + \int_{4}^5 -(-\sqrt {25-y^2})= 67.36$$

Are the area (67.36) and the integration process all correct? Is there more precise integration for the area of the right bounded region, without using double or triple integration?

答:
I think both of your approaches are correct.

I went about it the following way:

$25\pi - 2 \cdot \int_{-5}^{-3} \sqrt{(-5)^2 - x^2} = 67.36$. When integrating I got $arcsin(3/5)$, which is where I rounded my values. I do not think, there is a way to avoid this other than leaving arcsin(3/5) in your reslut, if you want it to be more precise.


问:
Is it true that all senary numbers ending in 1 and 5 are primes?
I was reading the Wikipedia article on senary numbers (base 6), which states that:

all primes, when expressed in
  base-six, other than 2 and 3 have 1 or
  5 as the final digit

Unless I am converting to senary incorrectly, I find this not to be true. For example, the senary representation of the decimal number 2047 is '13251', which would be a prime according to the stated rule, but is not (2047 = 89 * 23).

Is my conversion correct? Is the stated rule incorrect?

答:
The relationship in the quote is a subset.  All primes in base 6, other than 2 and 3, is a subset of all numbers ending in 1, 5.  There are composite numbers that end in the same digits, since all numbers coprime to six end in 1,5 base 6.

41 = dec 25 = 5*5 , and 205 = 11 * 15 = dec 77 evidently are not primes, but are co-prime to 6.


问:
A partial converse to the stability of separatedness under base change
Let $S$ be a base scheme, and let $X/S$ be an $S$-scheme. Let $T/S$ be another scheme. Denote the base change of $X$ to $T$ by $X_T := X \times_S T$.

Now separatedness (resp. properness) are stable under base change, so if $X/S$ is separated then so is $X_T/T$.

I am searching for a (partial) converse to this. In its most general form

Suppose $X$ is a scheme over a base scheme $S$, and suppose that $T/S$ is a scheme. If $X_T$ is separated (resp. proper) over $T$, is $X$ separated (resp. proper) over $S$?

In this generality the claim is false: (and this may be an overkill example) consider $G$ the affine line with doubled origin over $S = \operatorname{Spec} k[t]$ (with $S$-group scheme structure), and its base change $G_T$ to $T = \operatorname{Spec} k(t)$. The first is not separated, but the second is a group scheme over a field, hence separated.

But can we impose sufficiently strong conditions (on $X$, $S$, and $T$) under which this holds?

An example of when this would be useful is if $k$ is a field of characteristic zero, $S = \operatorname{Spec} k$, $T = \operatorname{Spec} \mathbb{C}$, and $X$ is something reasonable (since in this case we can check hausdorfness and compactness in the analytification by GAGA).

答:
The general name for this problem is "descent". The answer to your question is affirmative: if $T\to S$ is fpqc (faithfully flat and quasi-compact) then separatedness descends along $T\to S$, so $X_T/T$ separated implies $X/S$ separated. For a source, see EGA IV2, 2.7.1(i), or these notes of Matt Emerton. (For quick lookups, the best place to check whether some property is stable under descent is the table at the end of Poonen's "Rational Points on Varieties", a copy of which is available at the author's website.)

Here's a sketch of the proof: consider the (cartesian) diagram

$$\require{AMScd}
\begin{CD}
X_T @>{\Delta}>> X_T\times_T X_T\\
@VVV @VVV \\
X @>{\Delta}>> X\times_S X
\end{CD}$$

where the horizontal maps are the diagonal morphisms. The top row is actually the base change of the bottom row along the map $T\to S$, so it's fpqc as that's preserved by arbitrary base change. The top row is also a closed immersion by assumption, so it suffices to show that closed immersions descend along fpqc maps. Checking this claim is local and amounts to the statement that if $A\to B$ is a surjective map of rings and $A\to A'$ is a faithfully flat map of rings, then $A\otimes_A A'\to B\otimes_A A'$ is also surjective. So $X\to X\times_S X$ is a closed immersion, and thus $X/S$ is separated.


问:
Reflexivity of finite dimensional space
I was given the statement in exam(for which I was supposed to mark it as correct or incorrect):

Every Finite dimensional space is reflexive. (I marked it as incorrect).

My attempt: I learnt the following theorem which says: Every finite dimensional normed space is reflexive. But since there was no use of word "normed", I thought it may happen that the definition of reflexivity may also be applicable to some other spaces where I could find a counterexample and so I marked it as incorrect.

My professor said, it is correct. So my question is : Is there any finite dimensional space which is not reflexive? Please throw some light.

答:
Hints: (Prove them as easy exercises, if you haven't already)

Any finite dimensional vector space can be given a norm.

On a finite dimensional normed space, any two norms are equivalent.

Bingo!

Title: Reflexivity of finite dimensional space


问:
Using Gram-Schmidt to find an orthonormal basis.
I am studying Linear Algebra and have been introduced to Gram-Schimdt algorithm and I am struggling on how to approach this question.

For the beginning inner product I'm unsure how you get the answer:

$u_{1}w_{1}+2u_{2}w_{2}+3u_{3}w_{3}$

as from my linear algebra notes it only gives me the information to obtain:

$u_{1}w_{1}+u_{2}w_{2}+u_{3}w_{3}$  and I dont know how to get the missing coefficients, but assume this is due to these vectors being in $R^{3}$.

Also next for the Gram-Schmidt algorithm, I believe that I have to use the inner product.

I have found from a similar post that I should get:

$\|(1,1,1)\|=\sqrt6$  and that the first vector we should get when applying Gram-Schmidt is $e_1=\frac1{\sqrt6}(1,1,1)$

And then we let:

$a_2=(1,0,1)-\langle(1,0,1),e_1\rangle e_1=\left(\frac13,-\frac23,\frac13\right)$

Its norm is $\sqrt{\frac43}$ and so we take $e_2=\frac1{2\sqrt3}(1,-2,1)$

And finally I'm supposed to take $a_3=(0,1,2)-\langle(0,1,2),e_1\rangle e_1+\langle(0,1,2),e_2\rangle e_2=\left(-\frac32,0,\frac12\right)$

Its norm is $\sqrt3$ and so im supposed to take $e_3=\frac1{\sqrt3}\left(-\frac32,0,\frac12\right)$

But I cant understand what is going on here and was wondering if anyone could explain what is happening here and how to obtain the orthonomal basis from this.

Thanks in advance.

答:
The algorithm of Gram-Schmidt is valid in any inner product space. If ${v_1,...,v_n}$ are the vectors that you want to orthogonalize (they need to be linearly independent otherwise the algorithm fails) then:
$$w_1=v_1$$
$$w_2=v_2-\frac{\langle v_2,w_1 \rangle}{\langle w_1,w_1 \rangle}w_1$$
$$w_3=v_3-\frac{\langle v_3,w_1 \rangle}{\langle w_1,w_1 \rangle}w_1 -\frac{\langle v_3,w_2 \rangle}{\langle w_2,w_2 \rangle}w_2$$
$$\\.\\.\\.\\$$
$$w_j=v_j-\sum_{i=1}^{j-1} \frac{\langle v_j,w_i \rangle} {\langle w_i,w_i \rangle}w_i$$
$$\\.\\.\\.\\$$
$$w_n=v_n-\sum_{i=1}^{n-1} \frac{\langle v_n,w_i \rangle} {\langle w_i,w_i \rangle}w_i$$

Simply apply the algorithm. The intuition behind the algorithm is that to make the vector $v_j$ orthogonal to $w_1,...,w_{j-1}$ you have to make sure that its projections on $w_1,...,w_{j-1}$ are $0$(a vector is orthogonal to another if the projection of the former on the latter is $0$). So you subtract from $v_j$ its projections on $w_1,...,w_{j-1}$.


问:
How to apply the rule of absorption correctly to the following logical expressions?
How was the rule of absorption applied to the following logical expression?

$$ (A \land \neg B \land C) \lor (A \lor B \lor \neg C) \lor (\neg A \land B) \equiv (A \lor B \lor \neg C) \lor (\neg A \land B).$$

The step is not really clear to me, and I hope someone can explain it.

答:
Given associativity of $\lor$, you can just drop parentheses:

$$ (A \land \neg B \land C) \lor (A \lor B \lor \neg C) \lor (\neg A \land B) $$

$$=(A \land \neg B \land C) \lor A \lor B \lor \neg C \lor (\neg A \land B)$$

So now the $A$ terms absorps the $(A \land \neg B \land C)$ term:

$$=A \lor B \lor \neg C \lor (\neg A \land B) $$

And note that the $B$ term will absorp the $(\neg A \land B) $ term:

$$=A \lor B \lor \neg C $$

When you drop unnecessary parentheses, these kinds of simplifications become a lot easier to see!


问:
Proof: $\mathbb{R}^n $ is separable, show that every open cover of $\mathbb{R}^n$ has a countable subcover
Using the fact that $\mathbb{R}^n $ is separable, show that every open cover of $\mathbb{R}^n$ has a countable subcover.

I was given a hint to prove the following claim and if needed, incorporate it into the original proof: ''A set is open if and only if it is equal to an arbitrary collection of open balls", which I have been able to prove both ways. However, I don't see how I can use that to go about the original proof.

Otherwise, I thought of the following statements

$(i)$ $\mathbb{R}^n$ is separable

$(ii)$ $\mathbb{R}^n$ is compact

$(iii)$ Any open cover of $\mathbb{R}^n$ has a countable subcover

It would have been great if I could have proved $(i)\Rightarrow(ii)\Rightarrow(iii)$ but the problems that arise are

1. Firstly, $\mathbb{R}^n$ is not compact. Also, ''Every compact metric space is separable", which means $(ii)\Rightarrow(i)$.

2. ''For a compact space, every open cover has a finite subcover", which is not equivalent to countability.

How should I prove this? How are separability and claimed fact requirements to the proof?

Edit. The attempted second method is obviously wrong because $\mathbb{R}^n$ is not compact.

答:
Cover $\Bbb R^n$ with any countable collection of compact sets (e.g. all closed balls of radius $\sqrt n$ centred on points with integral coordinates). Now just cover each of these sets separately with a finite subcover. A countable union of finite covers is countable.

I don't see where I've used the separability of $\Bbb R^n$, but I'm sure it's hidden somewhere in there...


问:
Invariant superoperators under unitary action
Suppose we work in a (finite) $d$-dimensional Hilbert space $ \mathcal H $ and let $ \mathcal B (\mathcal H) $ denote the set of bounded linear operators of this space. Define a superoperator $ \Phi : \mathcal B (\mathcal H) \to \mathcal B (\mathcal H) $ as a linear map between such bounded linear operators and let $ \mathcal S (\mathcal H ) $ denote the set of all such superoperators. In the below, we use $ U \in \mathcal B (\mathcal H) $ to denote a unitary operator in our space.

For vectors $ v \in \mathcal H $ we have the natural action $ v \mapsto U v $, which in turn induces an action $ f \mapsto U f U^{\dagger} := \mathcal U (f) $ due to the isomorphism $ \mathcal B (\mathcal H ) \simeq \mathcal H \otimes \mathcal H^{*} $(we use a calligraphic $ \mathcal U $ to denote the conjugation superoperator by the corresponding unitary $ U $).

On the level of superoperators, this action lifts to $ \Phi ( - ) \mapsto U^{\dagger} \Phi (U - U^{\dagger} ) U = \mathcal U^{\dagger} \circ \Phi \circ \mathcal U  $ (as before, we can view this as  arising from the natural isomorphism $ S( \mathcal H ) \simeq \mathcal B ( \mathcal H ) \otimes \mathcal B (\mathcal H)^{*} \simeq (\mathcal H \otimes \mathcal H^*) \otimes (\mathcal H \otimes \mathcal H^*) $).

My question is then what superoperators $ \Phi $ are invariant under this action for all unitaries, i.e. $ \forall \mathcal U, \; \mathcal U^{\dagger} \circ \Phi \circ \mathcal U = \Phi $? I've tried a few simple cases and it seems like the identity mapping $ \Phi(f) = f $ and the constant map $ \Phi(f) = I $ (where $ I $ is the identity operator) work, and hence any linear combination of the above but I am struggling to either find any more or prove these are the only ones.

Another idea I've noticed is that we could use view $ \Phi $ as a $ d^2 \times d^2 $ matrix and use vectorisation to transform the question into finding the invariant subspaces of the second tensor power, but in general it appears this question is difficult even if we know the invariant subspaces of the original space (for those familiar with the theory, the context of the question is related to finding trivial/irreducible representations of $\mathcal U (d) \otimes \mathcal U (d)$ but this isn't required to state the problem).

答:
Let $\Phi\in\mathcal{S}(\mathcal{H})$ and suppose that for any unitary
conjugation operator $\mathcal{U}\in\mathcal{S}(\mathcal{H})$,
$\mathcal{U}\Phi=\Phi\mathcal{U}$. Let $\{e_1,e_2,\dots,e_d\}$ be an
orthonormal basis for $\mathcal{H}$, let
$e_{ij}\in\mathcal{B}(\mathcal{H})$ denote the rank one operator
$e_i\otimes e_j$, the $d\times d$ matrix with $1$ in the $i$th row
$j$th column and zeros elswhere. Let $E_{ij}=\Phi(e_{ij})$. Claim,
there exists $r,s\in\mathbb{C}$ such that,
\begin{equation}
  \label{eq:wtf}
  E_{ij}=\begin{cases} re_{ij} & \text{ if } i\neq j, \text{ and }\\
    %
    re_{ij}+sI & \text{ if } i= j,
  \end{cases}
\end{equation}
so that for $B\in\mathcal{B}(\mathcal{H})$,
$\Phi(B)=rB+\mathrm{tr}(B)sI$. Accordingly, fix $1\leq i,j\leq d$ with
$i\neq j$, let $K_{ij}$ denote the span of $\{e_i,e_j\}$, and let
$P_{ij}$ denote the orthogonal projection onto $K_{ij}$. Notate the
compressions to $K_{ij}$ by
\begin{gather*}\epsilon_{ii}=P_{ij}E_{ii}\Bigm|_{K_{ij}}=
  \begin{bmatrix} a_{ii} & b_{ii} \\
    c_{ii} & d_{ii}  \end{bmatrix}\qquad
  %
  \epsilon_{ij}=P_{ij}E_{ij}\Bigm|_{K_{ij}}=
  \begin{bmatrix} a_{ij} & b_{ij} \\
    c_{ij} & d_{ij}  \end{bmatrix}\\
  %
  \epsilon_{ji}=P_{ij}E_{ji}\Bigm|_{K_{ij}}=
  \begin{bmatrix} a_{ji} & b_{ji} \\
    c_{ji} & d_{ji}  \end{bmatrix}\qquad
  %
  \epsilon_{jj}=P_{ij}E_{jj}\Bigm|_{K_{ij}}=
  \begin{bmatrix} a_{jj} & b_{jj} \\
    c_{jj} & d_{jj}  \end{bmatrix}
  %
\end{gather*}

Let $U_{1}$, $U_{2}$, and $U_{3}$ be the unitary matrices which fix
the orthogonal complement of $K_{ij}$ with action on $K_{ij}$ given by
$u_{1}=\begin{bmatrix} i & 0 \\ 0 & 1\end{bmatrix}$,
$u_{2}=\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$, and
$u_{3}=\frac{1}{\sqrt 2}\begin{bmatrix} 1 & 1 \\
  -1 & 1\end{bmatrix}$ respectively. Note that for
$A=\begin{bmatrix} a & b \\ c &
  d\end{bmatrix}\in\mathbb{C}\times\mathbb{C},$

$$u_1Au_1^\dagger=\begin{bmatrix} a & -ib \\ ic &  d\end{bmatrix}\quad
u_2Au_2^\dagger=\begin{bmatrix} d & c \\ b &  a\end{bmatrix}\quad
u_3Au_3^\dagger= \frac12\begin{bmatrix}
 a+b+c+d & -a+b-c+d \\ -a-b+c+d &  a-b-c+d \end{bmatrix}$$

For $k=1,2,3$, $U_kP_{ij}=P_{ij}U_k$ so that
$\mathcal{U}_{k}\Phi= \Phi\mathcal{U}_{k}$ implies for
$\ell,m\in\{i,j\}$ ,
\begin{equation}
  \label{eq:compression}
  u_k\epsilon_{\ell m}u_k^\dagger=
P_{ij}\Phi(U_ke_{\ell m}U_k^\dagger)\Bigm|_{K_{ij}}
\end{equation}
With $k=1$, equation (2) shows that the off-diagonal entries of
$\epsilon_{ii}$ and $\epsilon_{jj}$ equal zero and that all entries of
$\epsilon_{ij}$ and $\epsilon_{ji}$ except for $b_{ij}$ and $c_{ji}$
must equal zero. Since $\mathcal{U}_2(e_{ii})=e_{jj}$, $a_{ii}=d_{jj}$
and $d_{ii}=a_{jj}$. Since $\mathcal{U}_2(e_{ij})=e_{ji}$,
$b_{ij}=c_{ji}$. With these identities, it follows that
$2u_3\epsilon_{ij}u_3^\dagger=
\begin{bmatrix}b_{ij} & b_{ij} \\ -b_{ij} &
  -b_{ij} \end{bmatrix}$. Further, since
$2U_3e_{ij}U_3^\dagger=e_{ii}+e_{ij}-e_{ii}-e_{ii}$,
$$\begin{bmatrix}b_{ij} & b_{ij} \\ -b_{ij} &
  -b_{ij} \end{bmatrix}=
\begin{bmatrix} a_{ii}-d_{ii} & b_{ij} \\
  -b_{ij} & d_{ii}-a_{ii}\end{bmatrix},$$
so that $a_{ii}-d_{ii} = b_{ij}$. Letting $r=b_{ij}$ and  $s=d_{ii}$ one has,
$$\epsilon_{ii}=\begin{bmatrix} s+r & 0 \\
    0 & s  \end{bmatrix}\quad
  %
\epsilon_{ij}=\begin{bmatrix} 0 & r \\
    0 & 0  \end{bmatrix}\qquad
  %
\epsilon_{ji}=\begin{bmatrix} 0 & 0 \\
    r & 0  \end{bmatrix}\qquad
  %
\epsilon_{jj}=\begin{bmatrix} s & 0\\
    0 & s+r  \end{bmatrix}\qquad
  %
$$
Letting $i,j$ run through all unequal pairs yields equation (1).

Remark. Using similar techniques one can show the following.
Let $\Phi\in\mathcal{S}(\mathcal{H})$ and suppose that for any
orthogonal conjugation operator
$\mathcal{O}\in\mathcal{S}(\mathcal{H})$,
$\mathcal{O}\Phi=\Phi\mathcal{O}$. There exists $r,s,t\in\mathbb{C}$
such that, for $B\in\mathcal{B}(\mathcal{H})$,
$\Phi(B)=rB+sB^\top+\mathrm{tr}(B)tI$.


答:
$\mathbb R^n = \bigcup_{x \in \mathbb Z^n} ([0,1]^n + x)$, write $A_x = [0,1]^n + x$. The $A_x$ cover $\mathbb R^n$. Let $\mathcal U$ be a cover of $\mathbb R^n$, then $\mathcal U$ covers $A_x$ and has a finite subcover $\mathcal U_x$. The set $\bigcup_{x \in \mathbb Z^n} \mathcal U_x$ is a countable cover.

Here are generalizations, as mentioned in a comment:

Every hemicompact space is countably compact.

Every separable metric space is countably compact.

Title: Proof: $\mathbb{R}^n $ is separable, show that every open cover of $\mathbb{R}^n$ has a countable subcover

Body:
Using the fact that $\mathbb{R}^n $ is separable, show that every open cover of $\mathbb{R}^n$ has a countable subcover.

I was given a hint to prove the following claim and if needed, incorporate it into the original proof: ''A set is open if and only if it is equal to an arbitrary collection of open balls", which I have been able to prove both ways. However, I don't see how I can use that to go about the original proof.

Otherwise, I thought of the following statements

$(i)$ $\mathbb{R}^n$ is separable

$(ii)$ $\mathbb{R}^n$ is compact

$(iii)$ Any open cover of $\mathbb{R}^n$ has a countable subcover

It would have been great if I could have proved $(i)\Rightarrow(ii)\Rightarrow(iii)$ but the problems that arise are

1. Firstly, $\mathbb{R}^n$ is not compact. Also, ''Every compact metric space is separable", which means $(ii)\Rightarrow(i)$.

2. ''For a compact space, every open cover has a finite subcover", which is not equivalent to countability.

How should I prove this? How are separability and claimed fact requirements to the proof?

Edit. The attempted second method is obviously wrong because $\mathbb{R}^n$ is not compact.


问:
Terminology: Continuous Surjection/Quotient with Continuous Right-Inverse
A continuous map with a continuous left-inverse is called an embedding.  However, what is the correct dual notion?  On one side (if we go from the universal property perspective) then we get the quotient map concept.

However, what is the name of the "dual" object in the sense: a continuous surjection with continuous right-inverse.  Are there any known nice properties of these objects which are analogous to embeddings?

答:
Let's clarify things. Assume $f:X\to Y$. Then if $g:Y\to X$ is such that $f\circ g=id_Y$ then we say $g$ is right inverse of $f$. If $g\circ f=id_X$ then we say $g$ is left inverse of $f$.

With that you are talking about (category theoretic) sections and retractions. A left inverse is retraction, a right inverse is section. On the other hand a map having a left inverse is a section and a map having a right inverse is a retraction.

In topology there are very important special cases of retractions and special cases of sections.

Also note that typically we define an embedding as a homeomorphism onto image. Meaning $f:X\to Y$ is an embedding if the (co)restriction $f:X\to f(X)$ is a homeomorphism. Such map does not have to admit neither left nor right inverse. Your definition is too restrictive. A counterexample is the famous $S^n\to D^{n+1}$, $x\mapsto x$ embedding where $S^n$ is the $n$-dimensional sphere and $D^{n+1}$ is the $n+1$-dimensional disk (as a consequence of the Brouwer fixed point theorem).


问:
Given $A^2 = AB$ and $B^2=I+BA$, prove that $A=0$.
$A,B$ are a $n\times n$ matrices. given that

$A^2 = AB$ and $B^2=I+BA$

Prove that $A=0$.

Is the following proof correct?

If we write $A^2=AB \rightarrow A^2-AB=0 \rightarrow A(A-B)=0$

We get that $A$ or $(A-B)$ equals $0$.

We then write $B^2 = I+BA \rightarrow B^2 -BA = I \rightarrow B(B-A)=I$
and we get that $B^{-1}=(B-A)$

Now, since $-B^{-1}=-(B-A)=(A-B)$

And that $B(-B^{-1})=-(BB^{-1})=-I$ we get that $(A-B) \neq 0$ which means $A=0$

答: For matrices, you cannot conclude that $ST = 0$ implies at least one of $S$ and $T$ is the zero matrix in general (cf the example by player3236 in the comments). So that part isn't quite right though the rest seems okay to me. That said, the last bit is a bit clunky. I'd write it as $A(A-B) = 0$ however $(A-B)$ is invertible per the above and multiplying on the right on both sides by its inverse leads to $A = 0$.


答:
Note that
$$\tag1A^2=AB$$
and $$\tag2B^2=I+BA$$
imply
$$A^3\stackrel{(1)}=A^2B\stackrel{(1)}=AB^2\stackrel{(2)}=A(I+BA)=A+ABA\stackrel{(1)}=A+A^3$$
so that $A=0$.
Apparently, the conclusion is valid in any ring,  not just for matrices.

Nevertheless, your proof is nicely savable, even in spit of the major blunder where you say that one of the factors has to be zero:
You show correctly that $B(B-A)=I$, hence $B-A$ is invertible, hence so is $A-B$, hence $0=A(A-B)$ does imply that $A=0$ (namely, by multiplying with $(A-B)^{-1}$ from the right; you even elaborate that $(A-B)^{-1}=-B$. I am however certain that the mentioned blunder will lead to severe point deduction.

Title: Given $A^2 = AB$ and $B^2=I+BA$, prove that $A=0$.

Body:
$A,B$ are a $n\times n$ matrices. given that

$A^2 = AB$ and $B^2=I+BA$

Prove that $A=0$.

Is the following proof correct?

If we write $A^2=AB \rightarrow A^2-AB=0 \rightarrow A(A-B)=0$

We get that $A$ or $(A-B)$ equals $0$.

We then write $B^2 = I+BA \rightarrow B^2 -BA = I \rightarrow B(B-A)=I$
and we get that $B^{-1}=(B-A)$

Now, since $-B^{-1}=-(B-A)=(A-B)$

And that $B(-B^{-1})=-(BB^{-1})=-I$ we get that $(A-B) \neq 0$ which means $A=0$


问:
Sum of infinite series $\frac{k(k+1)3^k}{k!}$
Can someone suggest how to find the infinite series sum for

$$\frac{k(k+1)3^k}{k!}$$ where k goes from $1$ to infinity.

I know that $\sum_0\frac{3^k}{k!}=e^3$ but I'm not sure if that helps here.

Body:
$$\frac{k(k+1)3^k}{k!}=\frac{k(k-1)3^k}{k!}+\frac{2k3^k}{k!}=\frac{3^k}{(k-2)!}+2\frac{3^k}{(k-1)!}\\
=9\frac{3^{k-2}}{(k-2)!}+6\frac{3^{k-1}}{(k-1)!}$$
Now change variables


问:
Conjecture regarding the function $f(x)=e^{x^n}$ and asymptotics behavior .
Hi it's a follow up of my false conjecture see Existence of an asymptote for $g(x)=\frac{f(x)f'(x)+f(1)f'(1)}{f'(x)+f'(1)}-f\left(\frac{xf'(x)+f'(1)}{f'(x)+f'(1)}\right)$ .

It's a particular example wich seems to work :

Let $f(x)=e^{x^n}$ where $x\geq 0$ and $1\leq n$ a natural number then define :

$$g(x)=\frac{f(x)f'(x)+f(1)f'(1)}{f'(x)+f'(1)}-f\left(\frac{xf'(x)+f'(1)}{f'(x)+f'(1)}\right)$$

Claim: $g(x)$ have an asymptote as $x\to \infty$

Geogebra have some problem with big numbers so there is a big part of luck .

For the case $n=1$ I have tried Laurent's series and see here .

My question :

Have you an counter-example or a proof if so ?

Thanks in advance !

答:
I got
\begin{align*}
g(x) & = \frac{{nx^n e^{x^n } e^{x^n }  + ne^2 }}{{nx^n e^{x^n }  + ne}} - \exp \left[ {\left( {\frac{{nx^{n + 1} e^{x^n }  + ne}}{{nx^n e^{x^n }  + ne}}} \right)^n } \right]
\\ &
 = e^{x^n }  + \mathcal{O}\!\left( {\frac{1}{{x^n e^{x^n } }}} \right) - \exp \left[ {x^n \left( {1 - \frac{{ne}}{{x^n e^{x^n } }} + \mathcal{O}\!\left( {\frac{1}{{x^{n + 1} e^{x^n } }}} \right)} \right)} \right]
\\ &
 = e^{x^n }  + \mathcal{O}\!\left( {\frac{1}{{x^n e^{x^n } }}} \right) - e^{x^n }  + ne + \mathcal{O}\!\left( {\frac{1}{x}} \right) = ne + \mathcal{O}\!\left( {\frac{1}{x}} \right)
\end{align*}
as $x\to +\infty$. Thus, the line $x=ne$ is an asymptote of $g(x)$ as $x\to +\infty$.


问:
Conguences with power
Determine all the solutions of the congruence $5x^{44} ≡ 9 \pmod{29}$, using the index in base $2$
module $29$.

I don't know how to proceed with the presence of this $5$ by multiplying the $x$ ... should I multiply by the inverse class to eliminate it?

答:
$$2^5\equiv3\pmod{29}\implies9\equiv(2^5)^2$$

$2^{14}=(2^5)^22^4\equiv3^22^4\equiv-1\pmod{29}$

So, $2$ is a primitive root $\pmod{29}$

$$\implies5\equiv-2^3\cdot3\equiv2^{14+3+5}$$

Using Discrete logarithm with index in base $2,$

$22+44$ind$_2x\equiv10\pmod{\phi(29)}$

$\iff44$ind$_2x\equiv10-22\pmod{\phi(29)}\equiv-12+28\pmod{28}$

$\iff11$ind$_2x\equiv4\pmod7$

$\iff4$ind$_2x\equiv4\pmod7$ as $(11\equiv4\pmod7)$

$\iff$ind$_2x\equiv1\pmod7$ as $(4,7)=1$

$\implies x\equiv2^{1+7k}\pmod{29}, 0\le1+7k\le28\iff0\le k\le3$

Title: Conguences with power


问:
upper bound for $\frac{ax}{x-2}$
I need an upper bound for
$$\frac{ax}{x-2}$$
I know that $1\leq a< 2$ and $x\geq 0$.

This upper bound can include just $a$ and constant numbers not $x$.

thanks a lot.

Title: upper bound for $\frac{ax}{x-2}$

Body: $$\lim_{x\rightarrow 2^+}\frac{ax}{x-2}=+\infty$$


问:
characteristic equation's solution
Consider the characteristic equation for $\lambda \in \mathbb{C}$  $$\lambda = -e^{-\tau\lambda}\tag1$$
a) For $\lambda$ a root of (1), write $\lambda = x+iy$ and consider the real and imaginary part of (1). Rewrite (1) as a system of two equations.

b) Show that if $\lambda$ is a solution to (1) then $\bar\lambda $ is also a solution to (1).

I did the first part:
\begin{cases} x= -e^{-\tau x}\cos(\tau y) \\ y= e^{-\tau x}\sin(\tau y) \end{cases}
I really need help for the second part.

答:
$${\lambda} = -e^{-\tau{\lambda}}$$
$$ {\lambda} = -e^{-\tau x}(\cos (y\tau)-i\sin (y\tau))$$
$$\bar {\lambda} = -e^{-\tau x}(\cos (y\tau)+i\sin (y\tau))$$
$$\bar {\lambda} = -e^{-\tau x}e^{iy \tau}$$
$$\bar {\lambda} = -e^{-\tau (x-iy)}$$
$$\bar {\lambda} = -e^{-\tau \bar {\lambda}}$$
If $\lambda $ is a solution, so is $\bar {\lambda }$. You can read this page too

Or more simply.
$${\lambda} = -e^{-\tau{\lambda}}$$
Change $ \lambda  \to \bar {\lambda}$
$$\bar {\lambda} = -e^{-\tau \bar {\lambda}}$$

Title: characteristic equation's solution


问:
The number of solutions for $|x^2-5|x|+6|=|\cos x|$
I wish to find the number of solutions for $|x^2-5|x|+6|=|\cos x|$ graphically

I can graph $|\cos x|$ to some accuracy but struggle in graphing $|x^2-5|x|+6|$

Is this question solvable graphically? If so could a method to sketch the graph be given.

Thanks in advance.

答:
To plot $y=|x^2-5|x|+6|$ you first plot

$y=x^2-5x+6$ and $y=x^2+5x+6$, then take the arc in the first quadrant for the first and in the second for the second.

Then you take the symmetric wrt $x$-axis of the negative parts, where $y$ is negative I mean.

Both function are even. Solutions are
$\pm 1.8054,\pm 3.5757$

$$
$$


问:
Fair and biased coins
Let's suppose we have $25$ coins: $14$ fair (F) and $11$ biased (B) with heads on both sides. We draw one coin, throw it, and repeat it ten times (with replacement). Calculate proability, that we get only heads.

So, at first I thought I can simply list all the possible outcomes with respect to biased coins, so $0$ biased, $1$ biased, $2$ biased, ..., $10$ biased, but it takes some time, and I guess it's not the best approach. What's more, I should permute all of the outcomes, as I can have, for example, for two biased coins, sequences $BBFFFFFFFF$ or $FBFFFFFBFF$ or assume, that order doesn't matter.

答:
Probability of picking biased coin $ = \frac{11}{25}$

Probability of picking fair coin $ = \frac{14}{25}$

Probability of getting head in a flip $ = \frac{11}{25} + \frac{14}{25} \times \frac{1}{2} = \frac{18}{25}$

For $10$ heads in a row, the probability is $(\frac{18}{25})^{10}$.

Title: Fair and biased coins


问:
An equality related to direction of affine subspaces
Given a finite-dimensional affine space $(X,\vec X,\Theta)$, for any two affine subspaces $M$ and $N$, prove that $$\overrightarrow{\langle M\cup N\rangle}=\overrightarrow{M}+\overrightarrow{N}+K\vec{ab},$$
where $a$ is an arbitrary point in $M$, $b$ an arbitrary point in $N$, and $K$ the field of scalars. If $Y$ is any subset of $X$,  $\langle Y\rangle$ is the notation for spanned subspace (smallest subspace containing $Y$). If $Y$ is an affine subspace of $X$, $\overrightarrow{Y}$ denotes the direction of the affine subspace ($=\Theta_a(Y)$ for any $a\in Y$).

Since I have not arrived at barycenter, I can't express elements in the spanned subspace using linear combination with sum of coefficients being 1. But this proposition appears before the concept of barycenter is introduced, so I think the equality can be proved without assistance of barycenter. I have proved $\overrightarrow{\langle M\cup N\rangle}\supseteq\overrightarrow{M}+\overrightarrow{N}+K\vec{ab}$. However, I just have no idea how to start proving $\subseteq$ part because the spanned subspace and its direction are too hard for me to handle. All I have at hand is the following two facts I have proven:

$M\cap N\ne\emptyset$ iff $\overrightarrow{ab}\in\overrightarrow{M}+\overrightarrow{N}$ for any $a\in M$ and any $b\in N$.

$M\cap N$ consists of a single point iff $\overrightarrow{ab}\in\overrightarrow{M}+\overrightarrow{N}$ for some $a\in M$ and some $b\in N$, and $\overrightarrow{M}\cap\overrightarrow{N}=\{0\}$.

But I don't know how to use them to establish the equality. It appears in two textbooks but both of them think it is too easy to provide a proof, leaving me so frustrated even after trying hard. Any help or hint would be greatly appreciated (not using barycenter). Thank you.

答:
Got an idea while bathing, so I think I can prove the equality myself. Criticisms welcomed.

$\supseteq$:

Although we do not know the concrete form of elements in $\langle M\cup N\rangle$, we do know $M$ and $N$ are in it. Each element in $\overrightarrow{M}+\overrightarrow{N}+K\vec{ab}$ has the form $\overrightarrow{ax}+\overrightarrow{by}+\mu\overrightarrow{ab}$ where $x\in M,y\in N$ and $\mu\in K$. Taking $a$ as the origin, we map this element back to $X$, which is $\Theta_a(\overrightarrow{ax}+\overrightarrow{by}+\mu\overrightarrow{ab})=\Theta_a(\overrightarrow{ax})+\Theta_a(\overrightarrow{by})+\mu\Theta_a(\overrightarrow{ab})$, with the fact that $\Theta_a$ is a vector space isomorphism between $\vec X$ and vectorialization $X_a$. $\Theta_a(\overrightarrow{ax})=x\in M$, $\Theta_a(\overrightarrow{by})=\Theta_a(\overrightarrow{ba}+\overrightarrow{ay})=-b+y$. Note that the above operations are carried out in vector space $X_a$. Since $N$ is a vector subspace of $X_a$ which preserves values of all these operations, $-b+y\in N$. The third term $\mu\Theta_a(\overrightarrow{ab})=\mu b$ which is in $N$ too. As a result, these three terms are in either $M$ or $N$, thus in $\langle M\cup N\rangle$. Since $\langle M\cup N\rangle$ is a vector subspace of $X_a$, $x+(-b+y)+\mu b$ is still in this vector subspace, which leads to $\overrightarrow{ax}+\overrightarrow{by}+\mu\overrightarrow{ab}\in\Theta_a(\langle M\cup N\rangle)$ and in turn $\overrightarrow{\langle M\cup N\rangle}\supseteq\overrightarrow{M}+\overrightarrow{N}+K\vec{ab}$.

$\subseteq$:

Since $\langle M\cup N\rangle$ is the smallest affine subspace, i.e., smallest vector subspace in $X_a$, containing $M\cup N$, if we can prove that $\Theta_a^{-1}(\overrightarrow{M}+\overrightarrow{N}+K\vec{ab})$ which is a vector subspace already, contains $M\cup N$, we will have $\Theta_a^{-1}(\overrightarrow{M}+\overrightarrow{N}+K\vec{ab})$ contains $\langle M\cup N\rangle$ by definition of spanned subspace. From $\overrightarrow{M}\subseteq\overrightarrow{M}+\overrightarrow{N}+K\vec{ab}$, we have $M\subseteq\Theta_a^{-1}(\overrightarrow{M}+\overrightarrow{N}+K\vec{ab})$. Likewise we have $N\subseteq\Theta_b^{-1}(\overrightarrow{M}+\overrightarrow{N}+K\vec{ab})$. But $\Theta_a^{-1}(\overrightarrow{M}+\overrightarrow{N}+K\vec{ab})=\Theta_b^{-1}(\overrightarrow{M}+\overrightarrow{N}+K\vec{ab})$ because $a$ and $b$ are both in this inverse image. So, $\langle M\cup N\rangle\subseteq\Theta_a^{-1}(\overrightarrow{M}+\overrightarrow{N}+K\vec{ab})$, which completes the whole proof.


问:
For real Lie algebras, is any invariant bilinear form a scalar multiple of the Killing form?
I know that for a complex Lie algebra, Schur's lemma can be used to show that any invariant bilinear form on a simple Lie algebra is a scalar multiple of the killing form, but Schur's lemma does not hold for $\mathbb{R}$, so I guess this isn't true for real simple Lie algebras. Does someone know a example? I can't seem to find one.

答:
Actually for a Lie algebra $\mathfrak{g}$ over a field $K$ and a field extension $K\subset L$, we have (by a straightforward linear algebra argument), with hopefully self-explanatory notation $$\dim_K \mathrm{InvBil}_K(\mathfrak{g})=\dim_L \mathrm{InvBil}_L(\mathfrak{g}\otimes_KL).$$

Consequence: for $\mathfrak{g}$ semisimple and $K$ of characteristic zero, $\dim_K \mathrm{InvBil}_K(\mathfrak{g})$ is equal to the number of simple factors of the "complexification" $\mathfrak{g}\otimes_K\bar{K}$.

In particular, $\dim_K \mathrm{InvBil}_K(\mathfrak{g})=1$ if and only if $\mathfrak{g}$ is absolutely simple.

Hence, to get counterexamples, it is enough to exhibit simple Lie algebras that are not absolutely simple. If $L$ is a finite extension of $K$ of degree $\ge 2$ and $n\ge 2$, $\mathfrak{sl}_n(L)$, viewed as Lie algebra over $K$, is such a Lie algebra.


问:
State domain, range, vertical asymptote and $x,y$ intercepts for $\log_4(x-1)+1$
State domain, range, vertical asymptote and $x,y$ intercepts for $\log_4(x-1)+1$

I have the solution in the answers section of my book:

Domain: $(1,\infty)$; Range: $(−\infty,\infty)$; Vertical asymptote: $x=1$;

$x$-intercept: $(5/4,0)$; $y$-intercept: Does not exist

I am able to figure out everything except the $x$-intercept.

Domain:

$$x-1 > 0$$
$$x>1$$

Domain therefore is $(1,\infty)$ and the vertical asymptote is 1.
Range is $(-\infty, \infty)$ and there is no $y$ intercept because it's a log function.

For the parent function $\log_4(x)$ the x-intercept is $(1,0)$. How can I arrive at $(\frac{5}{4},0)$?

Granular, baby steps appreciated.

答:
Looks all fine. You can solve the equation straightforward.

$\log_4(x-1)+1=0$

$\log_4(x-1)=-1$

Writing both sides into the exponent with base $4$.

$4^{\log_4(x-1)}=4^{-1}$

$x-1=\frac14$

$\vdots$


问:
find all $p\in \mathbb{R}$ s.t : $\int ^{\frac{\pi}{2}}_{0}(\sin(x))^{p} \ln(\sin(x)) \ dx$ converges.
Find all $p\in \mathbb{R}$ s.t : $\displaystyle \int ^{\frac{\pi}{2}}_{0}(\sin(x))^{p}\ln(\sin(x)) \ dx$ converges.

What I shown so far is that for $p > 0$ the integral converges because $f(x) =\sin(x)^{p} \ln(\sin(x))$ is continuous and bounded on $(0,\pi /2]$ and that the limit as $x\rightarrow 0^{+}$ exists.

For $p < 0$ I am not sure how to deal with that, I think it diverges but I am not sure how to see that.

It will be great if someone could give me a clue.

答:
Make $\sin(x)=t$
$$\int ^{\frac{\pi}{2}}_{0} \big[\sin(x)\big]^{p}\log(\sin(x)) \ dx=\int_0^1 \frac{t^p }{\sqrt{1-t^2}}\,\log (t)\,dt$$ Close to $t=0$
$$\frac{t^p }{\sqrt{1-t^2}}=\left(1+\frac{t^2}{2}+O\left(t^4\right)\right) t^p\sim t^p$$ and, using integration by parts,
$$\int_0^1 t^p\,\log(t),dt=-\frac{1}{(p+1)^2} \qquad \text{if} \qquad \Re(p)>-1$$


问:
Complex-valued Fourier integral: $ \int_{ - \infty }^{ + \infty } {\frac{{\cos (ax)}}{{{x^2} + 1}}{e^{ - ibx}}\,\mathrm dx} $
I'm working on the Fourier transform, but I don't know how to evaluate the integral:

$$I = \int_{ - \infty }^{ + \infty } {\frac{{\cos (ax)}}{{{x^2} + 1}}{e^{ - ibx}}\,\mathrm dx} $$

答:
The integral may be rewritten as the real part of

$$\int_{-\infty}^{\infty} dx \frac{e^{i (a-b) x}}{1+x^2}$$

This may be evaluated using residue theory.  Consider the following contour integral in the complex plane:

$$\oint_C dz \frac{e^{i (a-b) z}}{1+z^2}$$

where $C$ is the semicircle of radius $R$ aligned with the real axis and extending in the upper half-plane.  The contour integral is equal to

$$\int_{-R}^R dx \frac{e^{i (a-b) x}}{1+x^2} + i R \int_0^{\pi} d\phi\, e^{i \phi} \frac{e^{i (a-b)R \cos{\phi}} e^{-(a-b) R \sin{\phi}}}{1+R^2 e^{i 2 \phi}}$$

As $R \to \infty$, the second integral is bounded by

$$\frac{1}{R} \left | \int_0^{\pi} d\phi \, e^{-(a-b) R \sin{\phi}}|\right |\le \frac{2}{R} \left | \int_0^{\pi/2} d\phi \, e^{-2(a-b) R\phi /\pi}| \right |\le \frac{\pi}{(a-b)R^2}$$

where we used the inequality $\sin{\phi} \ge 2 \phi/\pi$ for $\phi \in [0,\pi/2]$.  Note that this integral converges only when $a>b$.  We will consider the case $ab$.  When $a<b$, we use the semicircle in the lower half-plane.  To traverse $C$ in a positive sense requires going along the real line from right-to-left, so the result for $a<b$ is

$$\int_{-\infty}^{\infty} dx \frac{e^{i (a-b) x}}{1+x^2} = \pi e^{a-b}$$

Putting this all together, and taking the real part of the integral, we finally have

$$\int_{-\infty}^{\infty} dx \frac{\cos{a x}}{1+x^2} e^{i b x} = \pi e^{-|a-b|}$$

Body:
The integral may be rewritten as the real part of

$$\int_{-\infty}^{\infty} dx \frac{e^{i (a-b) x}}{1+x^2}$$

This may be evaluated using residue theory.  Consider the following contour integral in the complex plane:

$$\oint_C dz \frac{e^{i (a-b) z}}{1+z^2}$$

where $C$ is the semicircle of radius $R$ aligned with the real axis and extending in the upper half-plane.  The contour integral is equal to

$$\int_{-R}^R dx \frac{e^{i (a-b) x}}{1+x^2} + i R \int_0^{\pi} d\phi\, e^{i \phi} \frac{e^{i (a-b)R \cos{\phi}} e^{-(a-b) R \sin{\phi}}}{1+R^2 e^{i 2 \phi}}$$

As $R \to \infty$, the second integral is bounded by

$$\frac{1}{R} \left | \int_0^{\pi} d\phi \, e^{-(a-b) R \sin{\phi}}|\right |\le \frac{2}{R} \left | \int_0^{\pi/2} d\phi \, e^{-2(a-b) R\phi /\pi}| \right |\le \frac{\pi}{(a-b)R^2}$$

where we used the inequality $\sin{\phi} \ge 2 \phi/\pi$ for $\phi \in [0,\pi/2]$.  Note that this integral converges only when $a>b$.  We will consider the case $a<b$ shortly.  In any case, this integral vanishes as $R \to \infty$.

The contour integral is equal to $i 2 \pi$ times the sum of the residues of the poles inside $C$.  The only pole inside $C$ is at $z=i$; this pole has residue

$$\frac{e^{-(a-b)}}{2 i}$$

so that we now have

$$\int_{-\infty}^{\infty} dx \frac{e^{i (a-b) x}}{1+x^2} = \pi e^{-(a-b)}$$

when $a>b$.  When $a<b$, we use the semicircle in the lower half-plane.  To traverse $C$ in a positive sense requires going along the real line from right-to-left, so the result for $a<b$ is

$$\int_{-\infty}^{\infty} dx \frac{e^{i (a-b) x}}{1+x^2} = \pi e^{a-b}$$

Putting this all together, and taking the real part of the integral, we finally have

$$\int_{-\infty}^{\infty} dx \frac{\cos{a x}}{1+x^2} e^{i b x} = \pi e^{-|a-b|}$$


问:
Quotient isomorphic to $(\mathbb{Z}/2\mathbb{Z})^n$
Suppose that groups G, H are such that $H \lhd G$, $H \cong (\mathbb{Z}/2\mathbb{Z})$, and $G / H \cong (\mathbb{Z}/2\mathbb{Z})^n$ for some $n \geq 1$. What then can we say about $G$? I think the only possibilities are that either $G \cong (\mathbb{Z} / 2 \mathbb{Z})^{n+1}$ or $G \cong (\mathbb{Z}/4\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z})^{n-1}$, but I'm not sure how you could prove anything like this.

答:
Writing $C_2$ for $\Bbb Z/2\Bbb Z$ we have a short exact sequence of groups
$$
1\rightarrow C_2\rightarrow G\rightarrow C_2^n\rightarrow 1.
$$
So $G$ is a group extension of $C_2^n$ by $C_2$. This is classified up to equivalence by the second cohomology group $H^2(C_2^n,C_2)$.

The isomorphism classes are "easier" to see.
Indeed, we may have the (semi)direct product if the sequence splits or some other possibilities if it doesn't split. Note that $G$ need not be abelian. Take for example the quaternion group.

Reference: How to find all possible extensions of a finite group by $C_2$

Some more details are also given here, but with $C_2$ and $C_2^n$ interchanged. Nevertheless it is useful.


问:
Sum to infinity using exponential series.
Evaluate $S=\frac{1}{2} - \frac{1}{3\times 1!} +\frac{1}{4\times 2!} -\frac{1}{5\times 3!} +\ldots$ up to infinity.

I am not able to plug in the exponential series. Please help

Body: Strong hint:$$S = \sum_{n=0}^\infty (-1)^n\frac{1}{(n+2)n!}= \sum_{n=0}^\infty (-1)^n \frac{n+1}{(n+2)!} = \sum_{n=0}^\infty (-1)^n \left(\frac{1}{(n+1)!}-\frac{1}{(n+2)!}\right)$$


Title: Sum to infinity using exponential series.

Body:
Evaluate $S=\frac{1}{2} - \frac{1}{3\times 1!} +\frac{1}{4\times 2!} -\frac{1}{5\times 3!} +\ldots$ up to infinity.

I am not able to plug in the exponential series. Please help


问:
Get the ratio of a products uses and price per use.
Not sure I am going to ask this well or even if the title is correct.

say you have a dataset like:

59 uses - price per use £0.09p

43 uses - price per use £0.06p

59 uses - price per use £0.04p

66 uses - price per use £0.12p

62 uses - price per use £0.1p

Obviously visually I can see which one in this small example has the best price to use ratio.

How do I write a formula that will result in a ratio for each item in the dataset?

I would also ask you keep you answer as simple as possible as I am not great at Math.

thanks

Above was confusing and imcomplete so I will add more hopefully it will help - I may be after the perctage im not sure.
Also adding missing data.

```<code>[
    {
        "weight" : 215,
        "price": 5.24,
        "weightPerUse": 3.65
    },
    {
        "weight" : 130,
        "price": 2.54,
        "weightPerUse": 3.04
    },
    {
        "weight" : 500,
        "price": 2.17,
        "weightPerUse": 8.5
    },
    {
        "weight" : 120,
        "price": 7.72,
        "weightPerUse": 1.83
    },
    {
        "weight" : 500,
        "price": 5.99,
        "weightPerUse": 8.12
    }
]
</code>```

This dataset results in

weighting 215 - number of uses 59 with a price per use of 0.09

weighting 130 - number of uses 43 with a price per use 0.06

weighting 500 - number of uses 59 with a price per use 0.04

weighting 120 - number of uses 66 with a price per use 0.12

weighting 500 - number of uses 62 with a price per use 0.1

I would like to know based on the information provided which has the best price to weight.

really sorry I am unable to explain myself better.

答:
What I would do is create a heuristic that increases as number of uses given increases, and decreases as price increases. The simplest one is $\dfrac{uses}{price}$.

If you add exponents to the numerator and denominator, you can change the "weight" you give each value. E.g., $\dfrac{uses}{(price)^3}$ will say a choice is very unfavourable even if the price is a little high and $\dfrac{uses^3}{price}$ will say a choice is very favourable even if you can get only a few more uses.

There is another technique, that I will not explain (you might not understand it) in great detail. We create something called a "loss function" that only looks at factors unfavourable to the problem. The task then is finding the minimum of this loss expression, invoking some calculus and even linear algebra.

E.g., if all you had a number of entries all of which offered the same number of uses, but at different prices, the loss function is easy $loss = {weightage} \times {price}$ (loss increases as price increases). Finding the minimum price here doesn't need any calculus or linear algebra, one merely has to look for the lowest price. In certain cases, the loss function that we arrive at is a curvy line than something that steadily increases or decreases: we use higher mathematical ideas there.


答:
Let $x$-intercept (intersection of curve with $x$-axis) of $y(x)=\log_{4} (x-1)+1$ be $(a,0)$. This means to find $a$, one has to solve $$1+\log_{4} (a-1) =0$$

$$\Rightarrow (a-1) = 4^{-1}$$

There is no $y$-intercept because it's a log function.

This is false. $y$-intercept is a point $(0,b)$ on the curve. For example, $y(x)=log_{4}(x+1)$ has $y$-intercept at $(0,0)$.

One has to check if $x=0$ is in the domain of $y(x)=log_{4}(x-1)+1$. Now argue that it is not. Domain is $x > 1$.

Body:
State domain, range, vertical asymptote and $x,y$ intercepts for $\log_4(x-1)+1$

I have the solution in the answers section of my book:

Domain: $(1,\infty)$; Range: $(−\infty,\infty)$; Vertical asymptote: $x=1$;

$x$-intercept: $(5/4,0)$; $y$-intercept: Does not exist

I am able to figure out everything except the $x$-intercept.

Domain:

$$x-1 > 0$$
$$x>1$$

Domain therefore is $(1,\infty)$ and the vertical asymptote is 1.
Range is $(-\infty, \infty)$ and there is no $y$ intercept because it's a log function.

For the parent function $\log_4(x)$ the x-intercept is $(1,0)$. How can I arrive at $(\frac{5}{4},0)$?

Granular, baby steps appreciated.


答:
Recall, $e^{-x}=\sum\limits_{n=0}^\infty (-1)^n\frac{x^n}{n!}$

Hence, $xe^{-x}=x-x^2+\frac{x^3}{2!}-\frac{x^4}{3!}+\dots$

Now define, $f(x)=\int_{0}^xte^{-t}dt$

For $x=1$, we are working within the radius of convergence of the above power series and hence term by term integration is allowed! Note that $f(1)=\frac{1}{2}-\frac{1}{3 \times 1!}+\frac{1}{4 \times 2!}-\frac{1}{5 \times 3!}+\dots$ and also from above, $$f(x)=-xe^{-x}-e^{-x}+1$$ Thus, $\frac{1}{2}-\frac{1}{3 \times 1!}+\frac{1}{4 \times 2!}-\frac{1}{5 \times 3!}+\dots= f(1)=1 - \frac{2}{e}$


问:
True or False: The line y=x is always at 45 degrees to the x-axis?
I have a doubt.

My teacher has posted a question that Is it always/sometimes/never true that the line y=x is at 45 degrees to the x-axis?

I know that the slope of the line y=x is always 1, and the line must make 45 degrees with x-axis.

I am a bit confused when I am plotting the line y=x at different X-scales (see Figures 1 and 2 attached).
I do not visually see the line at 45 degrees to the x-axis in Figure 2.

What am I missing?

Which answer would be correct and why?
(a) It is always true
(b) It is sometimes true because when we change the scale of x or y axis, graph will distort and not be at 45 degrees.

See images attached.
Fig 1 with equal x and y axis
Fig 2 with different axis

答:
Technically speaking the euclidean plane is a 2D affine euclidean space equipped with the standard product. The two axis(with their respective unit of measures) are simply a geometric representation of an orthogonal affine basis $(O,u,v)$ that we fix for the plane). Now $\mathcal{R}:y=x$ is a cartesian representation of an affine subspace in the affine basis $(O,u,v)$ (Where clearly $y$ is the second affine coordinate and $x$ the first). Formally an angle between two vectors $a,b$ is:
$$\arccos(\frac{\langle a,b \rangle}{||a||||b||})$$
(Where with $a,b$ i mean two vectors and not two coordinate vectors, this would be the same if the basis was orthonormal).
The angle between $\mathcal{R}$ and the "x-axis", is simply the angle between the coordinate vector $(1,1)$(which corresponds to the vector $v+u$) and the vector $u$. So:

$$\arccos(\frac{\langle v+u,u \rangle}{||u||||v||})$$

Since $v,u$ are orthogonal this becomes:

$$\arccos(\frac{||u||}{||v||})$$

That clearly depends on $\frac{||u||}{||v||}$ which is the scale factor of the two axes. So in general NO, that straight line doesn't always form a $45°$ angle with x-axis.


问:
I need help with the following exercise: find where $g:[0,5]\to\Bbb R$ is continuous and differentiable.
$g(x)=\int_{2}^{x}{f(t)dt}$, where

$$f(x)=\begin{cases} x^2,\qquad\quad\,\, 0\le x\le 1 \\ 2, \qquad\qquad 1 < x \le 4 \\ x-2, \qquad 4 < x \le 5 \end{cases}.$$

I thought to calculate the integral function, so I wrote:

if $x\in [2,4]$, $g(x)=\int_{2}^{x}{2 dt}$

if $x\in ]4,5]$, $g(x)=\int_{2}^{4}{2 dt}+\int_{4}^{x}{(t-2) dt}$.

Is this correct? Can someone add the missing cases?

What changes with $k:[0,5]\to\Bbb R, k(x)=\begin{cases} f(x),\qquad\quad\,\, x \not = {1,3} \\ 1000, \qquad\qquad x\in{1,3} \end{cases}$?

Isn't' it integrable for the same reason f is?

答:
If you just want to conclude on the continuity and differentiability, you don't need to perform any calculations. Since $f$ is piecewise continuous, it is integrable, and $g(x) = \int_2^x f(t) dt$ is automatically continuous and differentiable everywhere in its domain. If you really want to explicitly compute $g(x)$, considerer these cases

If $1 \leq x \leq 4$
$$
g(x)= \int_2^x f(t) dt = \int_2^x 2 dt = [2t]_2^x = 2(x-2)
$$

If $0 \leq x < 1$
$$
\int_2^x f(t) dt = \int_2^1 f(t) dt + \int_1^x f(t) dt = 2+[t^3/3]_1^x = 2+\frac 13(x^3 -1)
$$

and so on...


问:
Antiderivative of $\frac{1-x^n}{1-x}$
Is there a simple antiderivative of $\frac{1-x^n}{1-x}, \quad n \in \mathbb{N}$?

This functions seems realy easy, but if I try the product rule, I get a worse fraction.

Title: Antiderivative of $\frac{1-x^n}{1-x}$

Body: Hint: what is $(1-x)(1+x+x^2+...+x^{n-1})$?


Title: Antiderivative of $\frac{1-x^n}{1-x}$

Body:
Is there a simple antiderivative of $\frac{1-x^n}{1-x}, \quad n \in \mathbb{N}$?

This functions seems realy easy, but if I try the product rule, I get a worse fraction.

Body:
Use the fact that
$$x^n-1 = (x-1)(1+x+x^2+...+x^{n-1})$$
So
$$\int \frac{1-x^n}{1-x} dx= \int (1+x+x^2+\ldots+x^{n-1})dx$$


问:
A Conway quadratic (form) equation in four variables
In the Sensual Quadratic Form page 44, J.H. Conway is discussing isospectral lattices of dimension 4 when he states the following:

''... we find all solutions of the equation(s) $x^2+7y^2+13z^2+19w^2=48$ ... to find possible vectors of norm 4 ... in $L^+(1,7,13,19)$ and $L^-(1,7,13,19)$.''

Here $L^+$ and $L^-$ are the isospectral lattices in question. I know that 'all solutions' in this case refers to integer solutions, but I wonder how one goes about finding these. The statement is written in such a way that it would seem easy to do, yet I have no idea where to start.

答:
It suffices to find all solutions with $x$, $y$, $z$ and $w$ nonnegative. Because the quadratic form is positive definite, we see that
$$y^2\leq\frac{48}{7},\qquad z^2\leq\frac{48}{13},\qquad w^2\leq\frac{48}{19},$$
and so $y\leq2$, $z\leq1$, $w\leq1$, and for each choice of $(y,z,w)$ there is at most one (nonnegative) value of $x$ that satisfies the equation. So this leaves just $12$ cases to check, which isn't too much work to do by hand even.


问:
Why do Mathematica and Wolfram|Alpha say $\Gamma(-\infty)=0$?
According to Mathematica and Wolfram|Alpha, $\lim_{x\to -\infty}\Gamma(x)$ is equal to zero. See e.g https://www.wolframalpha.com/input/?i=gamma+function (at the bottom of the page), or try Limit[Gamma[x], x -> -Infinity] on Mathematica.

This contradicts the fact that $|\Gamma(x)|$ can be arbitrarily large when x is close enough to any negative integer.

Is Mathematica and Wolfram|Alpha wrong on this, or is there any other way to interpret this result?

答:
If $x$ is a continuous variable, then $\lim_{x \to -\infty} \Gamma(x)  $ does not exist.

But if  $-\infty$ is approached in a discrete manner, then the limit might exist.

For example, let's replace $x$ with $N+ \frac{1}{2}$, where $N$ is an integer.

The reflection formula for the gamma function states that $$\Gamma(x) = \frac{\pi \csc(\pi x)}{\Gamma(1-x)} $$ for all $x \notin \mathbb{Z}. $

Therefore,  $$ \lim_{N \to - \infty}\Gamma \left(N + \frac{1}{2} \right) = \lim_{N \to -\infty} \frac{\pi \csc \left(\pi \left(N + \frac{1}{2} \right) \right)}{\Gamma \left(\frac{1}{2}-N \right)} = \lim_{N \to -\infty} \frac{\pi \sec \left(\pi N \right)}{\Gamma \left(\frac{1}{2}-N \right)} =0 $$ since $\sec (\pi N)$ is just bouncing between $1$ and $-1$, while $\Gamma \left(\frac{1}{2}-N \right)$ is going to infinity.

EDIT:

In fact, as long as you stay greater than some fixed distance away from the negative integers, the limit will seemingly be zero.


问:
Double integral does not exist
I am trying to prove that the integral $$\int_{\mathbb{R}^2}|\frac{xy}{(x^2+y^2)^2}|\exp(\frac{-(x^2+y^2)}2) \, dx \otimes \, dy$$ does not converge.

I transformed into polar coordinates, with $x^2+y^2=r^2$ and $x=r\cos(\phi)$, $y=r\sin(\phi)$ and it became

$$\int_{[0, 2\pi) \times (0,\infty)}|\frac{r^2\sin(\phi)\cos(\phi)}{r^4}|\exp\Big(\frac{-r^2}{2}\Big) \, d\phi \otimes \, dr=$$

$$\int_{[0, 2\pi) \times (0,\infty)}|\frac{\sin(\phi)\cos(\phi)}{r^2}|\exp\Big(\frac{-r^2}{2}\Big) \, d\phi \otimes \, dr=$$

Using double angle formula and then grouping variables together

$$\int_{[0, 2\pi) \times (0,\infty)}|\frac{\sin(2\phi)}{2r^2}|\exp\Big(\frac{-r^2}{2}\Big) \, d\phi \otimes \, dr=\int_{[0, 2\pi) \times (0,\infty)}\frac{\exp(\frac{-r^2}{2})}{2r^2}|\sin(2\phi)| \, d\phi \otimes \, dr=$$

Transforming $r^2$ by $r^2=t$ and then using $1+x \leq \exp(x)$

$$\int_{[0, 2\pi) \times (0,\infty)}\frac{\exp(\frac{-t}{2})}{2t}|\sin(2\phi)| \, d\phi \otimes \, dt \geq \int_{[0, 2\pi) \times (0,\infty)}\frac{1-\frac{t}{2}}{2t}|\sin(2\phi)| \, d\phi \otimes \, dt=$$

$$\int_{[0, 2\pi) \times (0,\infty)}\frac{2-t}{4t} \, |\sin(2\phi)| \, d\phi \otimes \, dt$$

I feel like this function is simpler than at the start, however I fail to see why this integral does not converge. Any hints are welcome.

答:
Note
$$\int_{[0, 2\pi) \times (0,\infty)}|\frac{\sin(2\phi)}{2r^2}|\exp\Big(\frac{-r^2}{2}\Big) \, d\phi \otimes \, dr=\int_{[0, 2\pi)}|\sin(2\phi)| \, d\phi \int_{(0,\infty)}\frac{\exp(\frac{-r^2}{2})}{2r^2}\, dr. \tag1$$
But
$$\int_{[0, 2\pi)}|\sin(2\phi)| \, d\phi=const. $$
and
$$\int_{(0,\infty)}\frac{\exp(\frac{-r^2}{2})}{2r^2}\, dr=\int_{(0,1)}\frac{\exp(\frac{-r^2}{2})}{2r^2}\, dr+\int_{(1,\infty)}\frac{\exp(\frac{-r^2}{2})}{2r^2}\, dr.$$
Note the second integral converges and the second one diverges and so (1) diverges. This is because
$$ \frac{\exp(\frac{-r^2}{2})}{2r^2} \ge e^{-\frac12} \frac{1}{2r^2}$$
and $\int_0^1\frac{1}{2r^2}dr$ diverges and hence
$$ \int_{(0,1)}\frac{\exp(\frac{-r^2}{2})}{2r^2}\,dr $$
diverges.

Title: Double integral does not exist


问:
Four operand terms for arithmetic
Sorry for such a stupid/simple question, but I was wondering about the most common name for the arithmetic operator for division. For example:

Addition: + "plus". Example: two plus two.

Subtraction: - "minus". Example: two minus two.

Multiplication: * "times". Example: two times two.

Division: ???

The only one I've heard in common spoken english is "divided by", but was wondering if there was a simpler (non-formal) term like there are for the other three.

Finally, just to re-emphasize I don't mean the "slash" or "obelisk", I mean actually that anyone past second grade would know immediately what that term means (the only one I can think of is "divided by").

答:
Usually, $\frac{a}{b}$ can be read as "a over b." This is usually used more with fractions than whole numbers, though.

People also say "$a$ out of $b$," usually when dealing with data and/or whole numbers. For example, a basketball player may be "3 out of 4" from the free-throw line.

Sometimes this can be simplified to "$a$ for $b$." With the basketball player example, that same player is also "$3$ for $4$" from the free-throw line.

However, if speaking in a more mathematical sense, people tend to say "divided by," just because it's easier to understand; the operation is in the word.

Title: Four operand terms for arithmetic


问:
Show complex function is bijective
I'm asked to show that the function $f:\mathbb{C}\setminus\{1\}\rightarrow \mathbb{C}\setminus\{i\}$ given by
$f(z)=i\frac{z+1}{z-1}$ is bijective.

I get stuck very early on - I think I get thrown off by the fact that we're in $\mathbb{C}$ and not good old $\mathbb{R}$.

My attempt:

injective: assuming $f(z_1)=f(z_2)$, I need to show $z_1=z_2$. Usually I would just set it up like this: $$i\frac{z_1+1}{z_1-1}=i\frac{z_2+1}{z_2-1} $$
And then manipulate the expressios to get $z_1=z_2$, but I can't seem to do it here.

Am I choosing a poor strategy here?

In regards to surjectivity, I am just lost.

Any help is much appreciated.

答:
Writing $f(z)$ as
$$
f(z) = i\left (\frac{2}{z-1} + 1\right)
$$
shows that $f$ is the composition of bijective mappings:

$\Bbb{C}\setminus\{1\} \to \Bbb{C}\setminus\{0\} , \, z \mapsto z - 1$,

$\Bbb{C}\setminus\{0\} \to \Bbb{C}\setminus\{0\} , \, z \mapsto \frac{1}{z}$,

$\Bbb{C}\setminus\{0\} \to \Bbb{C}\setminus\{0\} , \, z \mapsto 2 \cdot z $,

$\Bbb{C}\setminus\{0\} \to \Bbb{C}\setminus\{1\} , \, z \mapsto z + 1$,

$\Bbb{C}\setminus\{1\} \to \Bbb{C}\setminus\{i\} , \, z \mapsto i \cdot z$

and therefore bijective.

Generally, every Möbius transformation is a composition of simple transformations (translation, inversion/reflection, homothety/rotation), and therefore a bijective mapping from the extended complex plane $\hat{\Bbb C} = \Bbb C \cup \{ \infty \}$ onto itself.

Title: Show complex function is bijective


问:
Remainder of division when $p$ is not prime
Calculate the remainder of the division of $1160^{602}$ by $35$.

What I thought of doing right away is to factor out $602$ until I get to a point where I can apply Fermat's theorem. However, $35$ is not a prime number. How to proceed in this case?

Body:
$$1160\equiv5\pmod{35}\implies1160^{602}\equiv5^{602}$$

Like How to find last two digits of $2^{2016}$,

As $(5^n,35)=5$ for $n\ge1,$

let us find $5^{602-1}\pmod{35/5}$

As $(5,7)=1$ and $601\equiv1\pmod6$   $$5^{602-1}\equiv5^1\pmod7$$

Now multiply $5$ throughout

$$5^{602-1}\cdot5\equiv5^1\cdot5\pmod{7\cdot5}$$


问:
Find the value of K for which the given matrix has rank 2

I think I've got the right intuition here. Row reduce and then by the knowlege that the matrix has a rank of 2, I know the third row will be one of zeros. Therefore, for the third entry in that third row, I will likely have some constant added or subtracted from K, which can be set equal to zero.

I've row reduced down but I'm essentially at a point where I can't get a rational number for K. Hopefully this is purely a computational error but please let me know if I'm going wrong elsewhere.

答:
The above given problem can be solved using 2 methods :
1.By echelon form
2.By using determinant

process of solving

process of solving


问:
Stirling number of the 2nd kind with pigeonhole theorem

I have just learnt Stirling number of the 2nd kind and I was also given a hint that this involves the pigeonhole theorem, but I have no idea how to approach this question. Any help would be much appreciated!

答:
Hint: First remember that ${n\brace k}$ means the number of ways to partition $n$ elements into $k$ non empty blocks (without any order).

If you have $3$ out of the $n$ elements in a block, how many elements are left? How many blocks are left? So how many elements per block are there left? (notice that there are $n-3$ elements and $n-3$ blocks, so there should be one element per block. Otherwise, one block would be empty!) That gives you $\binom{n}{3}\cdot 1,$ because you just have to care who are the elements that are together in the block of $3$ elements, the rest of them are in individual singletons.

Now you do not want to put $3$ elements in one block (you have done it before!), but then the maximal number of elements in one block has to be $2$(Pidgeon principle again) and so the two remaining elements ($n-(n-2)=2$ from the ones that have to determine a block) have to be put in two different blocks.

You have, then, $2$ blocks in which there are two elements each. Choose the $2$ elements for the first block in $\binom{n}{2}$ ways and, from the remaining $n-2,$ choose the other $2$ in $\binom{n-2}{2}$ ways. Notice that we use the word first, so we are giving this $2$ blocks an order. We can not do this! Eliminate the order.

Example: Suppose $n=4,$ then $n-2=2$ notice that we can have the following distribution. Either there is one element alone, which forces to have $3$ elements in the other block like in $\{\{1\},\{2,3,4\}\}$ or there are two blocks containing $2$ elements like in $\{\{1,2\},\{3,4\}\}.$

Notice that to create the first way we can choose who are the elements that are in the block of $3$ elements. In how many ways? $\binom{4}{3}.$ Notice that the remaining objects is forced to be alone in another block (so we do not have to do anything to take this into consideration).

For the second way, we want to choose the two pair of elements that are together. We can do this as $\binom{4}{2}$ to choose the first two, and then, from the remainder $4-2=2$ we can choose the other two. We get, by multiplication principle, $\binom{4}{2}\binom{2}{2}$ ways. The problem, tho, is that you gave this blocks an order, to take out the two possible orders, you divide by $2$ giving you $\frac{1}{2}\binom{4}{2}\binom{2}{2}.$

Putting the two ways together, you get your result.


答:
we know that $35=5.7$ and

$1160^{602} \equiv 0 \pmod{5}$

$1160 \equiv 5 \pmod{7}$

$a^{p-1} \equiv 1 \pmod{p} \Rightarrow$ $1160^6 \equiv 1 \pmod{7}$

$1160^6 \equiv 1 \pmod{7}$ $\Rightarrow$ $1160^{602}\equiv 1160^2\equiv 4 \pmod{7}$

$1160^{602}\equiv 4 \pmod{7}$ $\Rightarrow$ $1160^{602}=7k+4$ for some integer $k$ also we know that

$7k+4 \equiv 0 \pmod{5}$ $\Rightarrow 7k \equiv 1 \pmod{5}$

$7k \equiv 1 \pmod{5} \Rightarrow k \equiv 3 \pmod{5}$ and that means there exists an integer $m$ such that $k=5m+3$

$k=5m+3$ $\Rightarrow 7k+4=35m+25$ so

$1160^{602} \equiv 25 \pmod{35}$

Title: Remainder of division when $p$ is not prime

Body:
Calculate the remainder of the division of $1160^{602}$ by $35$.

What I thought of doing right away is to factor out $602$ until I get to a point where I can apply Fermat's theorem. However, $35$ is not a prime number. How to proceed in this case?


问:
Collinear points theorem
I was droodling a bit and a given moment I drew the following construction:

It appears that the three blue intersections are collinear (red line), no matter how I draw the construction lines. If this is always true, I assume that this a know fact [otherwise I have my first theorem! :-) erm conjecture, since I can't prove it :-(  ].

What's the theorem called?

TIA

Steven

答:
Congratulations! Looks to me like you have rediscovered Pappus' Hexagon Theorem.

The image from the link:

Title: Collinear points theorem


问:
Prime asymptotics from Euler product
It is said that the Euler product $$\prod_p \frac{1}{1-p^{-s}}$$ diverges as $s \to 1^+$ proves we can't find constants $C$,$\theta$ with $\theta < 1$ such that $\pi(x) < C x^\theta$ because that would imply that the product converges for $s > \theta$.

I don't understand this deduction at all, how is the fact about the prime counting function concluded?

答:
Taking logs of the Euler product, expanding out the resulting series, and comparing all the higher-order terms (i.e. terms involving $p^{-ns}$ for $n > 2$) to the convergent series $\sum_{m = 1}^{\infty} m^{-2s}$, we find that
the divergence as $s \to 1^{+}$ is equivalent to the divergence of the series
$\sum_p p^{-s}$ as $s \to 1^{+}$.  Now if $\pi(x) < C x^{\theta}$, then
the $n$th prime would have size $\sim n^{1/\theta}$, and so
this series would behave like $\sum_{n = 1}^{\infty} n^{- s/\theta},$ which
converges as $s \to 1^+$ (because $1/\theta > 1$), a contradiction.

Title: Prime asymptotics from Euler product


问:
upper bound for $\frac{ax}{x-2}$
I need an upper bound for
$$\frac{ax}{x-2}$$
I know that $1\leq a< 2$ and $x\geq 0$.

This upper bound can include just $a$ and constant numbers not $x$.

thanks a lot.

答:
Hint:

For $2>x \ge 0$, you get a negative value, and for $x>2$ you will see that the value gradually reduces.  What happens when $x \to 2$? Consider LHL and RHL.

Title: upper bound for $\frac{ax}{x-2}$


问:
Describe $n$ x $n$ matrix $A$ , such that it follows $A+A^T = 2A^{-1}$?
$A+A^T = 2A^{-1}$ => $(A+A^T)A= 2I$
$$\sum_{k=1}^n a_{kj}(a_{ik}+a_{ki}) = 2I_{ij}$$
So if $i = j$:
$$\sum_{k=1}^na_{ki}(a_{ik}+a_{ki}) = 2$$
And if $i != j$:
$$\sum_{k=1}^n a_{kj}(a_{ik}+a_{ki}) = 0$$
I am now stuck at this stage and can't figure out how to simplify the conditions.

答: Any symmetric and involutory matrix $B$ will fulfill the identity $B+B^{T}=2B^{-1}$. For example, the first Pauli matrix: $ \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$.


答:
Suppose we found $A \in \Bbb {GL_n}$ such that $A+A^T = 2A^{-1}$. We see that
$$
A+A^T = 2{A^T}^{{-1}}=2A^{-1}
$$
so
$$
A=A^T.
$$
Using the first equation, we get
$$
A=A^{-1}
$$
so
$$
A^2=I_n.
$$
For the other way, if $A^2=I_n$ and $A=A^T$, we have $A+A^T=2A=2A^{-1}$.
Thus
$$
A+A^T=2A^{-1} \iff A^2=I_n \text{ and } A=A^T.
$$

Title: Describe $n$ x $n$ matrix $A$ , such that it follows $A+A^T = 2A^{-1}$?

Body:
$A+A^T = 2A^{-1}$ => $(A+A^T)A= 2I$
$$\sum_{k=1}^n a_{kj}(a_{ik}+a_{ki}) = 2I_{ij}$$
So if $i = j$:
$$\sum_{k=1}^na_{ki}(a_{ik}+a_{ki}) = 2$$
And if $i != j$:
$$\sum_{k=1}^n a_{kj}(a_{ik}+a_{ki}) = 0$$
I am now stuck at this stage and can't figure out how to simplify the conditions.


问:
$X$, $Y$ be metric spaces and $f: X \to Y$. If $X$ is Baire and $Y$ is separable then $f$ is continuous in a dense $G_{\delta}$ of $X$

Let $X$, $Y$ be metric spaces and $f: X \to Y$. Suppose $X$ is Baire and $Y$ is separable. If $f ^{− 1} (O)$ is $F_{\sigma}$ for every open $O \subset Y$, show that $f$ is continuous in a dense $G_{\delta}$ of $X$

In the Dissertationes Matematicae (Rozprawy Matematyczne) book they prove that if $X$ is a Baire space, $Y$ is a space that has a countable base and $f: X \to Y$, then there exists a dense subset $D$ such that $f |_{D}$ is continuous (Theorem 3.6). Also, in How to show that the set of points of continuity is a $G_{\delta}$ they show that the set of continuity points of a function is $G_{\delta}$. Therefore $D$ would be $G_\delta$. But I don't know how to use the hypothesis that  $f ^{− 1} (O)$ is $F_{\sigma}$ for every open $O \subset Y$. Any idea please?

答:
Let me recall some definitions. Let $X$ be a topological space (say just a metric space). A nowhere dense set is a set $N\subseteq X$ which is not dense in any open set, i.e. for any non-empty open $U\subseteq X$ there is an non-empty open subset $V\subseteq U$ with $V\cap N=\emptyset$. A set is meager if it can be covered by a countable union of nowhere dense sets.

Topological space is Baire if it satisfies the Baire category theorem, that is, the complement of every meager set is dense. A set $B\subseteq X$ has the Baire property if it can be written as $U\triangle N$, where $U$ is open and $N$ is meager.

Suppose that $f:X\rightarrow Y$ is a map, where $X$ is Baire and $Y$ is second countable (i.e. it has a countable basis of open sets) such that for every open $U\subseteq Y$, the preimage $f^{-1}(U)$ has the Baire property. Then there is a comeager (complement of meager) subset $G\subseteq X$ such that $f\upharpoonright G$ is continuous. Such $G$ moreover contains a dense $G_\delta$ set. Indeed, let $(O_n)_n$ be a countable basis of topology of $Y$. Set $B_n:=f^{-1}(O_n)$. By the assumption, $B_n=U_n\triangle M_n$, for some open $U_n$ and meager $M_n$. Set $G:=X\setminus \bigcup_n M_n$. Clearly $G$ is comeager and for every $n$, $f^{-1}(O_n)\cap G$ is relatively open in $G$.

So to answer your question it remains to check that $F_\sigma$ sets have the Baire property. As I wrote in the comment, sets with the Baire property form a $\sigma$-algebra containing all open sets, so it contains all Borel sets. For the specific case of $F_\sigma$ sets, we first check that every closed set $F\subseteq X$ has the Baire property. Let $\mathrm{int}F$ denote the interior of $F$ (can be empty of course) and let $\partial F$ denote $F\setminus \mathrm{int}F$. It is easy to check that $\partial F$ is nowhere dense, in particular meager, so $F$ has the Baire property. So let now $F=\bigcup_n F_n$ be an $F_\sigma$ set, where each $F_n$ is closed. Then $F=(\bigcup_n \mathrm{int}F_n)\cup F\setminus \bigcup_n \mathrm{int}F_n$. We have $F\setminus\bigcup_n \mathrm{int}F_n\subseteq \bigcup_n \partial F_n$, so it is meager.


问:
Conguences with power
Determine all the solutions of the congruence $5x^{44} ≡ 9 \pmod{29}$, using the index in base $2$
module $29$.

I don't know how to proceed with the presence of this $5$ by multiplying the $x$ ... should I multiply by the inverse class to eliminate it?

Title: Conguences with power

Body:
Yes, $\bmod 29\!:\ \color{#0a0}{\dfrac{9}5}\equiv \dfrac{54}{30}\equiv \dfrac{(\color{#c00}{-1})4}1\equiv \color{#c00}{2^{\large 14}}2^{\large 2}\equiv \color{#0a0}{2^{\large 16}},\,$ so with $\,x = 2^{\large n}\,$ we have

$ x^{\large 44}\!\equiv \color{#0a0}{\dfrac{9}5}\!\!\iff\!\!2^{\large 44n}\!\equiv \color{#0a0}{2^{\large 16}}\!\!\!\iff\! \bmod 28\!:\, \underbrace{44n}_{\textstyle 16n}\equiv 16\!\!\!\underset{\large \div\ 4\!\!}\iff\! \bmod 7\!:\ 4n\equiv 4\!\!\iff\! n\equiv 1$


问:
“$B$ does not follow from $A$” seems different from $\lnot(A\to B)$
I am a beginner in logic. I have a basic question about implication. If "$B$ follows from $A$", we can write $A\rightarrow B$; but what about "$B$ does not follow from $A$"? Someone might suggest $\neg(A\rightarrow B)$, but that's not what "$B$ does not follow from $A$" means. Because $\neg(A\rightarrow B)$ is logically equal to $A\wedge\neg B$. But saying that "$B$ does not follow from $A$", it might be the case that $B$ is undecidable given $A$. So, I wonder how this sentence can be analyzed using logical tools. Thanks in advance!

答:
In a lot of cases of trying to say "it doesn't follow that...." the issue is that there is implicit quantification that needs to be negated.  For example, let $S$ be "it is (S)aturday" and $W$ be "it is the (W)eekend".  From $S$ follows $W$, so we can say $$S \to W$$
But just because it is the weekend doesn't mean it is Saturday.  But as you identified, $\lnot(W \to S)$ doesn't say what we want** because it is propositionally equivalent to $W \land \lnot S$. What is missing is that your full statement is "for every day, if it is the weekend then it is Saturday", which is correct to negate as "there is a day which is a weekend but is not Saturday":

$$\lnot \forall d. W(d) \to S(d)$$
$$\exists d. W(d) \land \lnot S(d)$$

** Except sometimes it does, because you could have meant for example "if today is the weekend then today is Saturday" because you have ruled out that it isn't Sunday.  In this case $\lnot (W \to S)$ is correct.


答:
Claim: $A$ must be symmetric.

Proof: We have
$$
A + A^T = 2A^{-1} \implies A^{-1} = \frac 12[A + A^T].
$$
We see that $A^{-1}$ is symmetric since
$$
[A^{-1}]^T = \left(\frac 12[A + A^T]\right)^T = \frac 12[A^T + A^{TT}] = \frac 12[A^T + A] = A^{-1}.
$$
It follows that
$$
A^T = [(A^{-1})^{-1}]^T = [(A^{-1})^{T}]^{-1} = (A^{-1})^{-1} = A. \quad \square
$$

From there, we see that
$$
A + A^T = 2A^{-1} \implies 2A = 2A^{-1} \implies A^2 = I.
$$
$A$ will satisfy your condition iff $A$ is symmetric with $A^2 = I$.


问:
Difference between objective function values of any feasible solution and optimal solution
Suppose that $x^*$ is an optimal solution to the following least squares problem:
$$ \min_{x\in R^n} f(x): = \frac{1}{2} \| Ax - b \|_2^2.$$

Clearly, we have $f(x) \ge f(x^*) \ \forall x \in R^n.$ Prove that the difference $[f(x) - f(x^*)]$ between the objective function values of any feasible solution and optimal solution is the following:
$$ f(x) - f(x^*) = \frac{1}{2} \| Ax - Ax^*\|^2_2, \ \forall \ x \in R^n.$$

答:
Let's expand the objective function
$$\
f(x) = \frac{1}{2}\|Ax-b\|^2=\frac{1}{2}x^TA^TAx - b^TAx + \frac{1}{2}\|b\|^2
$$
Assume $x^*$ to be the minimum,
$$\
\nabla f(x^*) = A^TAx^* - A^Tb = 0 \\
\implies A^Tb = A^TAx^*
$$

Just plug this is in the below eqn to replace $b^TA$ and obtain the desired relation.
$$\
f(x) - f(x^*) = \frac{1}{2}x^TA^TAx - b^TAx - \left( \frac{1}{2}{x}^{*T}A^TAx^* - b^TAx^* \right)
$$


问:
Improper triple integral in spherical coordinates

Compute the integral $$\int_{B^3(0,1)} (\sqrt{x^2+y^2+z^2})^{-p}$$ when $x \ne0$ and $p<3.$

So $B^3(0,1)$ is just a sphere centered at the origin of radius $1$. Also in spherical coordinates I've managed to get the integral to the form $$\int \int_{0}^{2\pi}\int_{t}^{1}(r^2)^{-\frac{p}{2}} \cdot r^2 \cos(\varphi) \ dr \ d\theta \ d\varphi$$ as it is an improper integral I have to limit $t \to 0$ right? I'm not able to find the limits for $\varphi$. I tried solving for $\varphi$ from $r=\sqrt{x^2+y^2+z^2}$ with the appropriate substituion $x=r\cos(\theta)\sin(\varphi), y=r\sin(\theta)\sin(\varphi)$ and $z=r\cos(\varphi)$, but this ended up very messy and I couldn't solve it. Any hints on how to proceed here?

答:
With usual shperical coordinates you would get it with $\sin(\varphi)$, so
$$\int_0^{\pi} \int_{0}^{2\pi}\int_{t}^{1}(r^2)^{-\frac{p}{2}} \cdot r^2 \sin(\varphi) \ dr \ d\theta \ d\varphi=$$
$$= 2\pi \biggl[ -\cos(\varphi) \biggr]_0^{\pi}\int_{t}^{1}r^{2-p} \ dr =$$
which converges if and only if $p-2<1$ which means $p<3$, an it is:
$$= 4\pi\int_{t}^{1}r^{2-p}= 4\pi \biggl[ \frac{r^{3-p}}{3-p} \biggr]_0^1=\frac{4\pi}{3-p}$$


问:
How to express parameter constraints for $(x-a)\left(x-\frac{1}{a}\right)<0$ if $-1<a<1$ and $a\neq 0$
How would one express the constraints on the parameter $x$ to satisfy:
\begin{equation}
\left(x-a\right)\left(x-\frac{1}{a}\right)<0 \quad \mbox{given} \quad -1<a<1\quad \mbox{and} \quad -1<x<1 \quad \mbox{with} \quad x\neq0 \quad \mbox{and} \quad a\neq 0
\end{equation}
It ic clear that the conditions arise from the first bracket, as the second bracket requires $x$ to be outside of the prescribed range.

I thought $x>a$ if $a>0$ and $x<a$ if $a<0$ but is there a more elegant and precise way to say this?

答:
Note that the solution to $(x-b)(x-c)<0$ is $\min(b,c) < x < \max(b,c)$ as long as $b \ne c$.

Now $a-\frac{1}{a}=\frac{a^2-1}{a}$ with the numerator always negative, so together with the other constraint $-1<x<1, x\ne 0$ we have:

$a>0 \implies a<\frac 1a \implies a<x<\frac 1a \implies x \in (a,1)$;

$a<0 \implies a>\frac 1a \implies \frac 1a<x\implies x \in (-1,a)$;

Body:
Note that the solution to $(x-b)(x-c)<0$ is $\min(b,c) < x < \max(b,c)$ as long as $b \ne c$.

Now $a-\frac{1}{a}=\frac{a^2-1}{a}$ with the numerator always negative, so together with the other constraint $-1<x<1, x\ne 0$ we have:

$a>0 \implies a<\frac 1a \implies a<x<\frac 1a \implies x \in (a,1)$;

$a<0 \implies a>\frac 1a \implies \frac 1a<x<a \implies x \in (-1,a)$;


问:
Arranging rectangles in a circle with equal distance
I've originally posted this on StackOverflow and didn't get a lot of responses, since it is more a math question than a programming question. I do hope it is on topic here.

I'm writing a program that renders mind maps. So far, I've succeeded in drawing the root node and the first level of child nodes around it, in a circle.

This is the code that renders the mind map nodes:

```<code>const root = data.find((node) => node.parent === undefined);
const level1 = data.filter((node) => node.parent === root.id);
root.x = 0;
root.y = 0;
root.level = 0;

await addMindMapNode(scene, root);
const radius = 2;
const slice = (2 * Math.PI) / level1.length;
for (let i = 0; i < level1.length; i++) {
    const level1node = level1[i];
    level1node.level = 1;
    const angle = slice * i;
    const x = root.x + radius * Math.cos(angle);
    const y = root.y + radius * Math.sin(angle);
    level1node.x = x;
    level1node.y = y;
    await addMindMapNode(scene, level1node);
}
</code>```

You can find the whole program on CodeSandbox.

The output looks like this (with different number of child nodes, from 1 child to 9 children):

As you can see, depending on the number of child nodes, the distribution of the child nodes around the root node in the center looks more or less uneven and not pleasing to the eye, especially with 3, 7 or 9 child nodes.

This is because of the rectangular format of the child node boxes. The center points of the boxes have the exact same distance between them, but the length of the distance (shown as red line segments below) is different, depending on position:

I have to find a way to make it so that the red circle sections are equal in size.

So my question is:

How can I calculate the angle at which to render each purple box so that the distance between each box looks to be the same (i.e. take into account the width and height of each box)?

答:
Comment:

I hope these figures help you. I think you have to consider the line connection the mid points of height of rectangles. In first figure the orientation of 8 books on  circle with diameter 184 and 10 books on circle with diameter 560  with constant width of rectangle is shown. In second figure the orientation of 9 books is shown.


答: And a picture for the graphical learners.

Title: upper bound for $\frac{ax}{x-2}$

Body:
I need an upper bound for
$$\frac{ax}{x-2}$$
I know that $1\leq a< 2$ and $x\geq 0$.

This upper bound can include just $a$ and constant numbers not $x$.

thanks a lot.


问:
How can I rewrite $\frac{1}{(1-z)^4}$ as a power series?
I'm supposed to rewrite $\frac{1}{(1-z)^4}$ as a power series, using the Cauchy-Product. It is given that $|z| < 1$.

So far I have
\begin{aligned}
\frac{1}{(1-z)^4} &= \left( \frac{1}{(1-z)^2} \right)^2\\
&= \left( \sum_{n=0}^{\infty} (n+1)z^n \right)^2\\
&=  \sum_{n=0}^{\infty} \sum_{k=0}^{n} (n-k+1)z^{n-k}kz^k  \\
&=  \sum_{n=0}^{\infty} \sum_{k=0}^{n} k(k+1)z^{n}  \\
\end{aligned}
First of all, I'm not sure whether every step was correctly so far. Additionally, I don't know how I can reduce that to a power series.

答:
It looks right so far, though you're probably supposed to to get from the first line to the second by taking the Cauchy product of the series for $\frac1{1-z}$ with itself.

As for the last part, $$\sum_{n=0}^{\infty} \sum_{k=0}^{n} k(k+1)z^{n}=\sum_{n=0}^{\infty}z^{n} \sum_{k=0}^{n} k(k+1)$$

and $$k(k+1)=\frac{k(k+1)(k+2)-(k-1)k(k+1)}3$$


问:
Show that the odd prime divisors of $n^2+n+1$ are of form $6k+1$ (exclude 3)
Show that the odd prime divisors of $n^2+n+1$ are of form $6k+1$ (exclude 3)

I have started like below:

$n^2+n+1\equiv 0 \pmod {p_i}$

$(n+1)^2\equiv n \pmod {p_i}$

Any hints/help on how to proceed from here ?

答:
Let $p$ be an odd prime divisor of $n^2+n+1$

Then   $n^2+n+1  \equiv 0 \pmod{p}
\implies n^3-1  \equiv 0 \pmod{p}
\implies n^3  \equiv 1 \pmod{p}$

So $ord_pn=3$ (obvious!) and so $3|\phi(p)=p-1$

Note that $p-1$ is even. So $2|p-1$.

Since $gcd(2, 3)=1$, so $6|p-1\implies p-1=6k, k\in\Bbb{Z}\implies p=6k+1$