答:
Assume that they can be in A.P.

$\sqrt{2}+(n-1)d=\sqrt{3}$

$\sqrt{2}+(m-1)d=\sqrt{5}$

$m$, $n$ are natural numbers.

$\Longrightarrow$ $~$ $\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\dfrac{n-1}{m-1}$

LHS is irrational and RHS is rational, which is contradiction.

Hence, they cannot be in A.P.

Title: Question on AP(sequences and series)

Body: Prove that sqrt(2), sqrt(3) and sqrt(5) cannot be terms of an A.P.(not necessarily consecutive)!</p>


问:
probability of occurrence a faulty job in final output in a chained queue(2 queue) when each queue has probability of generating a faulty job (b)
Consider we have a chained queue:

-> q1 ->q2 -> final output

the probability of generating a faulty job in each queue is b

Now, What's the probability of a faulty job in final output?

答: Since the probability of a good job in each stage is $1-b$, the probability of a good job after two stages is $(1-b)^2$, assuming stages are independent.  Therefore the probability of a fault at the end is $1-(1-b)^2=2b-b^2$.

ContentLicense: CC BY-SA 4.0


问:
How to solve this differential equation via power series
Studying Quantum Mechanics, I encountered the following differential equation:

how to solve $$y''-x^2y=0$$
Griffiths, the author of the book, just said that its solutions are
$$y=Ae^{x^2/2}+Be^{-x^2/2}$$
I can immediatelly see that this is true, but I do not know how to get to this solution, and I cannot find it solved anywhere else.
I tried to use the power series method:
$$y=\sum_{n=0}^{\infty}c_n x^n$$
So the equation becomes:
$$\sum_{n=0}^{\infty}c_{n+2}x^n(n+1)(n+2)-\sum_{n=0}^{\infty}c_n x^{n+2}=0$$
Here I need to get both sums to the same power of n, so I removed the first two terms from the first sum, so that I can later change the index:
$$2c_2+6c_3x+\sum_{n=2}^{\infty}c_{n+2}x^n(n+1)(n+2)-\sum_{n=0}^{\infty}c_n x^{n+2}=0$$
now I change the index in the fist sum:
$$2c_2+6c_3x+\sum_{n=2}^{\infty}c_{n+4}x^{n+2}(n+4)n(n+3)-\sum_{n=0}^{\infty}c_n x^{n+2}=0$$
Now I factor out the sum:
$$2c_2+6c_3x+\sum_{n=2}^{\infty}x^{n+2}(c_{n+4}(n+4)(n+3)-c_n)=0$$
Now all three terms must be zero at the same time, so $c_2=c_3=0$, and:
$$c_{n+4}(n+4)(n+3)-c_n=0$$
$$c_{n+4}=\frac{c_n}{(n+4)(n+3)}$$
The first terms of the series are:
$c_4=\frac{c_0}{4\times 3}$, $c_8=\frac{c_4}{8\times 7}=\frac{c_0}{8\times 7\times 4\times 3}$
However, I don't really know how to proceed from here. I tried putting the coefficients into the sum for $y$, but I can't really come close to the answer. Could you please guide me as how to advance from here?

答:
Since $c_2=0$ and $c_{n+4}=\frac {c_n}{(n+3)(n+4)}$, it follows that $c_6, c_{10},c_{14} ...=0$ and similarly $c_7, c_{11},c_{15} ...=0$. You can separate the remaining terms into $c_0x^0+c_4x^4+c_8x^8...$ and $c_1x^1+c_5x^5+c_9x^9...$. By relabeling the coefficients, we can get $y=
\sum_{n=0}^{\infty}a_nx^{4n}+\sum_{n=0}^{\infty}b_nx^{4n+1}$. Each of these sums has only one degree of freedom; the sequence $a_n$ is defined recursively from $a_0$ and similarly for $b_n$. This shows that $y = c_0f_0(x)+c_1f_1(x)$ for some $f_0,f_1$ that can be calculated numerically, if not in closed form.


问:
can someone explain the difference in uniform convergence and pointwise convergence
In wiki I found this explanation:

for uniform convergence:
$\forall \varepsilon >0\ \underline {\exists N\in \mathbb{N} \ \forall x\in D_{f}}\ \forall n\geq N:\quad \left|f_{n}(x)-f(x)\right|<\varepsilon ,$

for pointwise convergence: $\forall \varepsilon >0\ \underline {\forall x\in D_{f}\ \exists N\in \mathbb{N} }\ \forall n\geq N:\quad \left|f_{n}(x)-f(x)\right|<\varepsilon ,$

I'm trying to understand what the difference in the order of the two underlined parts exactly mean. And why that implies that pointwise convergence follows from uniform convergence.

答:
They already told you the formal things. The point is you cannot have a good understanding of this concept without visualising it graphically.

For this purpose, I'll give you two sequences of functions $(f_n) _n$ and $(g_n) _n$. The first one will converge pointwise, the second will converge uniformly (and therefore also pointwise).

You should DRAW two graphs, one for the $f_n$ and one for the $g_n$. Ready?

$f_n: (0,1) \rightarrow \mathbb{R}$ such that $f_n(x)=x^n$

$g_n: (0,1) \rightarrow \mathbb{R}$ such that $g_n(x)=\frac{x}{n} $

for $n\ge 0$. Note that you should only consider the domain $(0,1)$ for our purpose.

Now draw the graphs for the first $n$'s. You see both sequences converge to the zero function $f: (0,1) \rightarrow \mathbb{R}$ such that $f(x)=0$

The point is, since $f_n(1)=1$ for all $n$, at every fixed $n$ the maximum distance between $f$ and $f_n$ remains constantly 1, thus the sequence does not UNIFORMLY converge.

On the other hand, the $g_n$'s get pushed towards zero everywhere (it does not happen anymore that one extreme remains fixed at the same height). Thus, they converge uniformly.

The limit function $f$ being flat in this simple case, you can visualize the uniform convergence condition as "as $n$ grows, I can squeeze the graphs of $g_n$'s to zero with a horizontal line", which I can't do with the other sequence. The height of the horizontal line is the $\epsilon$ of your (first) definition.

Once this concept becomes clear, if you try and write down the formal definitions, you end up with the two (ugly) lines that you wrote.

EDIT: I'm happy I've been chosen as the main answer, thanks! If anyone finds it useful please leave an upvote :)


问:
How many increasing subsequences of length k can be made from an increasing sequence of length n.
For example, say you are given the sequence 1,2,3,4 and you want sequences of length 3, the sequences would be (1,2,3), (1,3,4), (2,3,4).

Is there a way I could compute the number of sequences like this.

答: There are $n\choose k$ $k$-element subsets of $\{1,2,\dots,n\}$. To see why $n\choose k$ is also the number of ordered $k$-element subsets of $\{1,2,\dots,n\}$ we describe a bijection between the two. First, notice that for each $k$-element subset there is exactly one ordering from smallest to largest, and no two subsets give the same ordered set of size $k$ when ordered, so each subset gives one ordered subset. On the other hand, consider an ordered subset of length $k$, each one of these gives exactly one subset of length $k$ because a subset doesn't care which order you list the elements, they're all the same subset.


问:
Derivative of a curve's velocity vector
Why is that in the equation below the second derivative of the position vector has a minus in the middle instead of plus after the product rule? I didn't get why we can just replace - sign in front of a if that's the reason.

where;

答:
Just the chain rule.

Note that $\dot{r}(t)=(aC'(t),S'(t))^T$.

Also, $C(t)=\cos\big(\pi-(1-\cos\omega t)\big)$ and
$$
C'(t)=-\sin\big(\pi-(1-\cos\omega t)\big)\cdot \pi \sin(\omega t)\cdot\omega
=\omega \pi\sin(\omega t)\cdot (-S(t))
$$

Title: Derivative of a curve's velocity vector


问:
Parenthesization in Lambda Calculus
I am trying to understand how parenthesization works in $λ$-calculus. All of the resources state the following:

Applications are left associative;

Abstractions are right associative and extend as much as possible.

When it comes to applications from what I understand this means  that it is not associative, that is: $$(ab)c \neq a(bc)$$ and the right way to fully parenthesize the expression $abc$ would be:
$$((a b)c)$$

However I do not understand the part of abstraction, what does it mean to extend as much as possible?

答:
Your understanding of left associativity for applications is perfect.

Concerning abstractions, the fact that they "are right associative and extend as much as possible" means that:

a term $\lambda x . N_1 N_2 \dots N_n$ must be read as $\lambda x. (N_1 N_2 \dots N_n)$ ("abstractions extend as much as possible");
for instance, $\lambda x. M N P$ must be read as $\lambda x. (M N P)$, and not as $(\lambda x. (M N))P$;

a term $\lambda x_1 . \lambda x_2. \dots \lambda x_n. N$ must be read as $\lambda x_1. (\lambda x_2. \dots (\lambda x_n. N) \dots )$ ("abstractions are right associative");
for instance, $\lambda x. \lambda y. \lambda z. N$ must be read as $\lambda x. (\lambda y. (\lambda z. N))$; note that any other parenthesization, for instance $\lambda x. ((\lambda y.\lambda z). N)$, would not yield a term, because $(\lambda y. \lambda z.)$ is not a term.

Title: Parenthesization in Lambda Calculus


问:
How do we find all possible values of r
enter image description here

I need help with this question

答: $f'(x)=2rx^{2r-1}arctan(\frac{1}{x})-\frac{x^{2r}}{1+x^2}$  Since $arctan(\frac{1}{x})$ is finite, $f'(0)=0$ for all $r\gt \frac{1}{2}$.  I don't see the restriction to $Q^+$.

Title: How do we find all possible values of r


问:
Circular Sector inscribed inside a similar sector
I was just thinking about this question. When can a circular sector be inscribed inside a similar  sector (with an equal angle and of course a larger radius)?

I don't have a diagram for this, but I don't think it's too complicated. I tried connecting the smaller radius to the point of tangency, but got nowhere. Thanks for the help.

Sketch of Figure

Body:
$AB=BC$

$AC=CE=2BC=2DE$

$tan(CDE)=\dfrac{2}{1}$

$\angle EDF=2tan^{-1}(2)≈2×63.435≈126.87°$


问:
Understanding how the hint will prove injectivity.
Here is the question I want to solve:

Let $M$ be a finitely generated $R$-module. Show that if $f \in \mathrm{End}_R(M)$ is surjective then it is also injective.

And here is the hint I got for the solution:

Let $R[X] \cong R^{[1]}$ and make $M$ into an $R[X]$-module by setting $X.m = f(m)$ for $m \in M.$ Show that $XM = M$ and apply Nakayama's Lemma.

My statement of Nakayama's Lemma is as follows:

If $M$ is a finitely generated $R$-module and $JM = M$ then there exists $r \in \operatorname{ann}(M)$ with $r \equiv 1 \pmod J.$

My question is:

How the hint will prove the injectivity? Any clarification of this will be appreciated.

答:
Taking $J=\langle X\rangle$, let $r$ be as in the conclusion of Nakayama’s Lemma. Then $r=p(X)X + 1$ for some $p(X)\in R[X]$. If $m\in\mathrm{ker}(f)$, then $p(X)X\cdot m = 0$, so
$$ m = 1\cdot m + p(X)X\cdot m = (1+p(X)X)\cdot m = r\cdot m = 0.$$


问:
Integral of a floor function $\iint_D [x + y] \, dx\, dy$
Given a domain
$$D=[0,2]×[0,2]$$
How can I calculate the integral $$\bbox[5px,border:2px solid #C0A000]{\iint_D [x + y]  \, dx\, dy}\quad$$

答:
According to the question, we have$$[x+y]=k-1((x,y)\in D_k,k=1,2,3,4).$$
So$$I=\sum\limits_{k=1}^{4}\iint\limits_{D_k}[x+y]\mathrm{d}x\mathrm{d}y=\sum\limits_{k=1}^{4}(k-1)|D_k|=|D_2|+2|D_3|+3|D_4|.$$
and the $|D_k|$ is the area of $D_k$.Hence
$$I=3+\frac{3}{2}.$$


问:
How to compute the probability using Naive Bayes assumption?
I am struggling a bit with this question (and it's on a practice test -- not an actual test).

```<code>Consider that  Y = wealth, and X1 = Gender, X2 = Hours worked.

Let us consider the following data:

P(Y = rich, X1 = Female, X2 < 40.5) = 0.0245895
P(Y = poor, X1 = Female, X2 < 40.5) = 0.253122
P(Y = rich, X1 = Female, X2 >= 40.5) = 0.0116293
P(Y = poor, X1 = Female, X2 >= 40.5) = 0.0421768
P(Y = rich, X1 = male, X2 < 40.5) = 0.0971295
P(Y = poor, X1 = male, X2 < 40.5) = 0.331313
P(Y = rich, X1 = male, X2 >= 40.5) = 0.105933
P(Y = poor, X1 = male, X2 >= 40.5) = 0.134106
Compute P(Y = rich | X1 = Female, X2 < 40.5) using Naiive Bayes assumption

**P(X1 = Female, X2 < 40.5| Y = rich)**

**P(Y = rich)/ P(X1 = Female, X2 < 40.5)**
</code>```

Firstly, I don't even understand what those last 2 equations mean -- how do they relate to the question?

Here is how I went about it (which is obviously not correct since it's not one of the answers):

P(Y = rich | X1 = female, X2 < 40.5)

= (P(X1 = female | Y = rich) * P (X2 < 40.5 | Y = rich) * P(Y = rich)) / P(X1 = female, X2 < 40.5)

= ((0.0245895 + 0.0116293) * (0.0245895 + 0.0971295) * (0.0245895 + 0.0116293 + 0.0971295 + 0.105933)) / (0.0245895 + 0.253122) = 0.003798

Possible answers:

0.07, 0.09, 0.91, 0.38

Body:
Using Naive Baye's:

P(Y = rich | X1 = Female, X2 < 40.5)

= (P(X1 = female | Y = rich) * P(X2 < 40.5 | Y = rich) * P(Y = rich)) / P (X1 = female, X2 < 40.5)

= (((0.0245895 + 0.0116293) / (0.0245895 + 0.0116293 + 0.0971295 + 0.105933)) * (0.0245895  + 0.0971295) / (0.0245895 + 0.0116293 + 0.0971295 + 0.105933)) * (0.0245895 + 0.0116293 + 0.0971295 + 0.105933)) / (0.0245895 + 0.253122)

= 0.066342

≈ 0.07


问:
I don't understand how this concentration problem is modeled.
This is the problem:

Ignore the part about inverses. I got that. I am confused why C is modeled this way. I don't get where the 100 + n came from. When I look up concentration and mixture problems, I get problems that are not represented as rational functions, like this:

Can someone walk me through how this function was created from the information provided?

答: The concentration is the amount of solute divided by the total volume. We started out with a 20% solution and 100ml, so 20 of the 100ml was acid. We added $n$ ml, and that solution was 40% acid, so we added $0.4n$ ml of acid and $n$ ml of solute. In the end, the total volume was $100+n$ and the total solute was $20+0.4n$.


答:
This isn't true for each $n$. For example, when $x=10$ and $n=5$
$$\sum_{i=2}^5(-1)^{i}\frac{10^{2i}}{(2i)!}=-\frac{707500}{567}<0 $$
However, for a fixed $x$, it will be true for sufficiently large $n$. This is because $\sum_{i=2}^n(-1)^{i}\frac{x^{2i}}{(2i)!}$ converges to $\cos x-1+x^2/2$ which is positive for $x\neq 0$, and so there's an $N\in\mathbb{N}$ such that for $n\geq N$, we can say that $\sum_{i=2}^n(-1)^{i}\frac{x^{2i}}{(2i)!}>0$. Note the $N$ depends on the value of $x$, and I don't think there's one independent of $x$.

Title: Prove that $0<\sum_{i=2}^n\ (-1)^n \frac{x^{2n}}{2n!}$ for $x>0$

Body:
So far, I've tried using induction:

Let P(n): $0<\sum_{i=2}^n\ (-1)^i \frac{x^{2i}}{2i!}$

Let $n=2$. Thus, P(2): $\frac{x^4}{4!}>0$ for $x>0$. Then, let P(k) be true (Induction Hypothesis). Now, prove the validity of P for n=k+1:

$P(k+1) : \sum_{i=2}^{k+1}\ (-1)^i\frac{x^{2i}}{2i!} = \sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} + (-1)^{k+1}\frac{x^{2(k+1)}}{2(k+1)!} = \sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} - (-1)^{k}\frac{x^{2(k+1)}}{2(k+1)!}$

Let $k=2m-1$:

$\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} - (-1)^{k}\frac{x^{2(k+1)}}{2(k+1)!}=\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} + (-1)^{2m}\frac{x^{4m}}{4m!}$ which is $>0$ since $\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!}>0$ and $\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} + (-1)^{2m}\frac{x^{4m}}{4m!} > \sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!}$ for $x>0$ according to the Induction Hypothesis.

I'm lost for the case where $k=2m$, I'd like some help there, please.


问:
How do I integrate $\ln(x)/(x-1)$
How do I integrate $\ln(x)/(x-1)$?

\begin{align}
f(x)&=\int\frac{\ln(x)}{x-1} \,dx\\
&=\ln(x)\ln|x-1|-\int\frac{\ln|x-1|}{x}\, dx\\
&=\ln(x)\ln|x-1|-\left(\ln|x-1|\ln(x)-\int\frac{\ln(x)}{x-1}\, dx\right)\\
&=\ln(x)\ln|x-1|-(\ln|x-1|\ln(x)-f(x))\\
2f(x)&=0\\
f(x)&=0
\end{align}
I end up getting this but I know its incorrect. How do I do solve this

答:
To handle this integral, we need to use polylogarithms where $|z:

$$
\operatorname{Li_s}(z)=\sum_{k=1}^\infty{z^k\over k^s}
$$

which satisfies the property

$$
\operatorname{Li}_{s+1}(z)=\int_0^z{\operatorname{Li}_{s}(t)\over t}\mathrm dt
$$

If we set $s=1$, we can use the power series of logarithm to get

$$
\operatorname{Li}_1(z)=-\log(1-z)
$$

As a result, we get

$$
-\operatorname{Li}_2(1-z)=\int_0^{1-z}{\log(1-t)\over t}\mathrm dt=\int_1^z{\log t\over t-1}\mathrm dt
$$

Title: How do I integrate $\ln(x)/(x-1)$


问:
A problem about Construct auxiliary function？

problem: f has the second differential function on $[a,b]$, prove:$\exists \alpha∈(a,b)$ such that :
$f(a)-2f(\frac{a+b}{2})+f(b)=\frac{1}{4} (b-a)^2  f^{''}(\alpha)$

My attempt:we choose $λ$ satisfied:
$f(a)-2f(\frac{a+b}{2})+f(b)=\frac{1}{4} (b-a)^2  λ$

And consider Auxiliary function $F(t)=f(t)-2f(\frac{t+a}{2})+f(a)-\frac{1}{4} (t-a)^2  λ$,it is also has the second differential function. We note $F(a)=F(b)=0$,$F^{'}(a)=0$, by mean value theorem, $\exists \beta ∈(a,b)$ such that $F^{''}(\beta)=0$
Then $λ =\frac{f^{''}(\beta)+f^{"}(\frac{\beta +a}{2})}{2}$.
But I want to prove $\exists \omega∈(a,b)$ such that $λ=f^{"}(\omega)$. I think this way is ture,but How to find the $\omega$? I want to use Darboux theorem of differential function to prove$f^{''}(\omega)=\frac{f^{''}(\beta)+f^{''}(\frac{\beta +a}{2})}{2}$. But I didn't succeed.

答:
I hate denominators, so i will introduce $A,B$ with $a=2A$, $b=2B$, and $A<B$. We are searching an intermediary point $c\in[2A,2B]$ with
$$
f(2A) -2f(A+B)+f(2B) = (B-A)^2f''(c)\ .
$$
In case of a function $f\in \mathcal C^2$, we can write:
$$
\begin{aligned}
&f(2B) -2f(A+B)+f(2A)
\\
&\qquad=
\int_A^B(f'(B+u)-f'(A+u))\; du
\\
&\qquad=
\int_A^B\int_A^Bf''(t+u)\; dt\; du
\\
&\qquad=
\int_A^B(A-B)f''(\tau+u)\; du
\\
&\qquad=
(A-B)(A-B)f''(\tau+\nu)
\end{aligned}
$$
for some intermediate points $\tau,\nu$. The argument can be rewritten to work for a general $f$, using a help function of two variables, $h(t,u)=f(t,u)-f(t,A)-f(A,u)+f(A,A)=(f(t,u)-f(t,A))-(f(A,u)-f(A,A))$.


答:
ok so the concentration is given as:
$$conc=\frac{mass\,of\,salt}{total\,mass}$$
but what you have to remember is that when we add a mass of salt its add to both the top and bottom of the fraction. We have:
$$m_{salt}=20\times\frac{20}{100}$$
$$m_{total}=20$$
so if we add an amount of salt we want:
$$\frac{20\times0.2+y}{20+y}=0.25$$
now rearrange and solve for y

Title: I don't understand how this concentration problem is modeled.

Body:
This is the problem:

Ignore the part about inverses. I got that. I am confused why C is modeled this way. I don't get where the 100 + n came from. When I look up concentration and mixture problems, I get problems that are not represented as rational functions, like this:

Can someone walk me through how this function was created from the information provided?


问:
Looking for analytical proof that this function given as a power series is constant.
Answering a recent question I came across the following function ($t$ is a positive integer), defined for $0\le x\le 1$ as:
$$
P_t(x)=x^t\sum_{n=0}^\infty \frac1{n+1}\binom{t(n+1)}n\left[(1-x)x^{t-1}\right]^n.
$$

Is there an analytical way to prove that $P_t(x)$ is a continuous function on $x\in(0,1)$ and:

$P_t(x)=1$ for all $x:\ 0\le 1-x\le\frac1t$,

$P_t(x)<1$ for all $x:\ \frac1t< 1-x\le1$.

I proved this for trivial cases $t=1,2$ but did not find a way to deal with general $t$.

The behavior of the function for $t=1,2,3,4$ is demonstrated below:

答:
It's analytic about $x=1$. Doing the substitution $x \mapsto 1-x$, collecting coefficients, and using the binomial series quickly reduces it to the following identity. I've used $\tau := t-1$.

Thm: For all $k, \tau \in \mathbb{Z}_{\geq 0}$,
$$\sum_{n=0}^k \frac{(-1)^{k-n}}{n+1} \binom{(n+1)(\tau+1)}{n} \binom{n\tau}{k-n}  = \binom{\tau+k}{k}. \label{*}\tag{*}$$

Proof: This identity was new to me, though @Semiclassical pointed me in the right direction in the comments! As Knuth says in his note on Convolution Polynomials, p.5, we have
$$\sum_{k=0}^n \binom{x+t(n-k)}{n-k} \binom{y+tk}{k} \frac{y}{y+tk} = \binom{x+y+tn}{n}. \label{**}\tag{**}$$
Let $t \mapsto \tau+1$, $n\mapsto k$, $k\mapsto n$, $y\mapsto \tau+1$, and $x\mapsto (\tau+k)-(\tau+1)(k+1)$ in $\eqref{**}$ and applying $\binom{\tau n}{k - n} (-1)^{k-n} = \binom{-\tau n + k - n - 1}{k - n}$ gives $\eqref{*}$. $\Box$

Remark: Knuth notes that $\eqref{**}$ goes back to Rothe in 1793 (!) in the guise of what is apparently known as the Rothe--Hagen identity. Knuth references Gould and Kaucky who provide further references and discussion as well. A short elementary proof is provided by Chu.


问:
formula of pascal's triangle
I want to make a program for pascal's triangle,I was reading through the details and found something like this:

```<code>0:                                1
1:                             1     1
2:                           1    2    1
3:                         1   3     3   1
4:                       1   4    6    4   1
5:                      1  5   10   10   5   1
</code>```

by looking at row number 5 of the triangle, one can quickly read off that:

$$(x + y)^5 = 1\cdot x^5 + 5\cdot x^4y + 10\cdot x^3y^2 + 10\cdot x^2y^3 + 5\cdot x y^4 + 1\cdot y^5$$

but I haven't been able to understand what are these x and y and what shall be their values.and also how shall I proceed for further rows.
I shall be really thatnkful if someone explained this to me.

答: These $x$ and $y$ are your variables. They can be any number in your domain(if it's a polynomial over the real, then $x$ and $y$ are real numbers; if it's a polynomial over the complex, then $x$ and $y$ are complex numbers.). They are arbitrary. So don't worry about what value to put in. This binomial theorem gives you a way to deal with things in general.</p>

ContentLicense: CC BY-SA 3.0

Title: formula of pascal's triangle


问:
"Extending" monoid action on monomials to the entire polynomial ring
Let $X$ be a set of variables, and consider the polynomial ring $K[X]$ in the variables $X$ over a field $K$. Let $Mon(K[X])$ denote the set of monomials of $K[X]$. Let $M$ be some monoid acting on $Mon(K[X])$ such that $\rho1=1$ and $\rho(ab)=\rho(a)\rho(b)$ for all $a,b\in Mon(K[X])$ and $\rho\in M$. How do I formally "extend" this action to a monoid action on all of $K[X]$? I want to do it in a natural way, so that things like $\rho(f+g)=\rho(f)+\rho(g)$ and $\rho(af)=a\rho(f)$ hold for all $f,g\in K[X]$, all $a\in K$ and all $\rho\in M$. Essentially I am starting with a monoid action on monomials and I wish to "linearly extend" the action to the entire polynomial ring.

My question is, is there a well-known formal way of doing this? How do I formally define the action on $K[X]$?

I thought about doing something like defining
$$
\rho\left(\sum_ic_im_i\right)=\sum_ic_i\rho(m_i)
$$
where the sum is finite and $c_i\in K$, $m_i\in Mon(K[X])$ and $\rho\in M$.

答:
Your construction is the way to go.

One sidenote: ${\rm Mon}(K[X])\cup\{1\}$ is basically the free commutative monoid on the set $X$, so the original action $\rho$ is already determined by its restriction to single elements of $X$, and that can be an arbitrary function $X\to{\rm Mon}(K[X])$ (which all can be directly seen right away).


问:
Proving combinatorial formula by induction and combinatorially
I need to prove this equation by induction and by combinatorial explanation. $$\sum_{k=0}^{n} \left( \begin{array}{c} 4n \\ 4k \end{array} \right) = 2^{4n-2} + (-1)^n \cdot 2^{2n-1}$$

My induction proof:

Base case: p(1): $$\left( \begin{array}{c} 4 \\ 0 \end{array} \right) + \left( \begin{array}{c} 4 \\ 4 \end{array} \right) = 2^{4*1-2} + (-1)^1 \cdot 2^{2*1-1}\;\;\;\;\;\;\;\checkmark$$

Inductive step: $p(k) \Longrightarrow p(k+1)$: $$\sum_{k=0}^{n} \left( \begin{array}{c} 4n \\ 4k \end{array} \right) + \left( \begin{array}{c} 4(n+1) \\ 4k \end{array} \right) = 2^{4n-2} + (-1)^n \cdot 2^{2n-1} + \left( \begin{array}{c} 4(n+1) \\ 4k \end{array} \right) \\ \sum_{k=0}^{n} \left( \begin{array}{c} 4(n+1) \\ 4(k+1) \end{array} \right) = 2^{4n-2} + (-1)^n \cdot 2^{2n-1} + \left( \begin{array}{c} 4(n+1) \\ 4k \end{array} \right)$$

I got stuck in this part and I can't find a combinatorial explanation to prove this.

答:
The problem is that the $n+1$ case isn't actually obtained from the $n$ case by adding an extra term to the end, the way you're thinking of it.

Here are the first few cases of this formula:

For $n=1$, $\binom 40 + \binom 44 = 2^2 - 2^1$.

For $n=2$, $\binom 80 + \binom 84 + \binom 88 = 2^6 + 2^3$.

For $n=3$, $\binom{12}0 + \binom{12}{4} + \binom{12}{8} + \binom{12}{12} = 2^{10} - 2^5$.

For $n=4$, $\binom{16}{0} + \binom{16}{4} + \binom{16}{8} + \binom{16}{12} + \binom{16}{16} = 2^{14} + 2^7$.

To prove this by induction, you might use Pascal's formula: $\binom nk = \binom{n-1}{k-1} + \binom{n-1}{k}$. But then you have to look at some intermediate cases as well: sums where the tops and/or the bottoms of the binomial coefficients aren't multiples of $4$. It's possible to get there, but the binomial theorem argument suggested in the comments is easier.


问:
Closed form for n-th anti-derivative of $\log x$
Is it possible to write a closed-form expression with free variables $x, n$ representing the n-th anti-derivative of $\log x$?</p>

答:
An intuitive way to get the formula from the answer of Vladimir Reshetnikov without induction:

Recall that $\displaystyle\ln x=\lim_{s\rightarrow 0}\frac{x^s-1}{s}$. But the $n$th antiderivative of $x^s-1$ is very easy to compute:
$$\frac{1}{s}\underbrace{\int\ldots\int}_{n\;\text{times}}(x^s-1)dx=\frac{x^{s+n}}{s(s+1)\ldots(s+n)}-\frac{x^n}{s\cdot n!}=_{s\rightarrow0}\frac{x^n}{n!}\left(\ln x-\sum_{k=1}^n\frac{1}{k}\right).$$


问:
A discrete probability distribution, but the population is continuous vs. discrete
Suppose we have some sort of discrete distribution as follows:

fruit
frequency

apple
0.3

orange
0.5

tomato
0.2

This is a statistical description of the actual population, which is

```<code>orange, apple, tomato, apple, orange, apple, orange, orange, orange, tomato
</code>```

That is, the actual population consists of discrete items belonging to each category. It has a meaningful "size" (10) and its members cannot be subdivided. Of course, the same distribution could arise from a population of 100 fruit. For the purposes of this question, the overall population is unspecified, but it is assumed to be "knowable" in the sense that we could go out and count how many fruit there are.

Compare the following, identical distribution:

road material
frequency

dirt
0.3

asphalt
0.5

gravel
0.2

This is a statistical description of the road that runs from Smallville to Bigville. The road is ten miles long, but it isn't organized in ten neat segments (dirt, dirt, dirt, asphalt, asphalt ...). Instead the road shifts arbitrarily between materials, so it is impossible subdivide it into segments such that each segment consists of only one material. Imagine something like this, but with fractal complexity:

What term can I use to differentiate between these two kinds of population structures?

It seems appropriate to refer to the road as a continuum, but I'm not sure what term to use for the fruit. I want to say discrete, but this invites confusion because both of the probability distributions shown above are discrete (taking one of only three values).

答: According to the English Wikipedia page for "Probability space", it seems that both examples would have a finite "event space", but the "sample space" is finite/discrete in the first case, and a continuum (in various senses) in the latter case.</p>


问:
Why doesn’t the equation "break" when we put a value $x$ while solving partial fractions?
Let’s suppose we want to decompose $\frac {9}{9-x^2}$ by partial fractions. I was taught that we proceed by writing
$$ \frac{9}{9-x^2} = \frac{a}{3+x} + \frac{b}{3-x}, $$
where $a$ and $b$ are unknown constants.  Clearing the fractions, we get
$$(3-x)a + (3+x)b = 9.$$
To find $a$, we assume that $x=-3$ and solve the resulting equation; and similarly, to find $b$, we assume that $x=3$ and solve.  But when we assume it we are diving by $0$ in the original fraction which isn't possible!

I am confused.  How does this procedure give the correct result?

答:
$$\frac{9}{9-x^2} = \frac{a}{3+x} + \frac{b}{3-x} \tag{A}$$

LHS and RHS are continuous except at the two points $x=3$ and $x=-3$.

Multiplying through by $9-x^2$ gives

$$9 = a(3-x) + b(3+x) \tag{B}$$

where both LHS and RHS are continuous for all real numbers and can be solved for the parameters $a$ and $b$.

Once  you have found the values of $a$ and $b$, you can fill in the values for $a$ and $b$ and then divide both sides of equation (B) by $9-x^2$. You will get equation (A) with the values of $a$ and $b$ filled in and equation (A) will be true for all $x$ except at $x=3$ and $x=-3$.


问:
relationship between cosets and commutators
Let $G$ be a normal subgroup of $H$, and let $x \in G$ and $y \in H$. Is it true that the following equation:

$$
xyG = yxG
$$

holds if and only if $[x,y] \in G$ ?, where $[x,y]$ is the commutator of $x$ and $y$, i.e. $[x,y]=x^{-1}y^{-1}xy$.

答: Yes. If $xyG=yxG$, then there is a $g\in G$ such that $xy=yxg$, and therefore $[x,y]=x^{-1}y^{-1}xy=g\in G$. And if $[x,y]=g\in G$, then $xy=yxg$, so $xyG=yxgG=yxG$.


问:
$\sum_i x_i^n = 0$ for all $n$ implies $x_i = 0$
Here is a statement that seems prima facie obvious, but when I try to prove it, I am lost.

Let $x_1 , x_2, \dots, x_k$ be complex numbers satisfying:

$$x_1 + x_2+ \dots + x_k = 0$$

$$x_1^2 + x_2^2+ \dots + x_k^2 = 0$$

$$x_1^3 + x_2^3+ \dots + x_k^3 = 0$$

$$\dots$$

Then $x_1 = x_2 = \dots = x_k = 0$.

The statement seems obvious because we have more than $k$ constraints (constraints that are in some sense, "independent") on $k$ variables, so they should determine the variables uniquely.  But my attempts so far of formalizing this intuition have failed.  So, how do you prove this statement?  Is there a generalization of my intuition?

答: Apply Newton's identities. This shows the power sums form a basis for the symmetric polynomials, since the elementary symmetric polynomials are a basis and the elementary symmetric polynomials can be represented as linear combinations of power sums. So all symmetric polynomials evaluate to $0$ at the $x_i$. Consider a polynomial with the $x_i$ as roots. The coefficients are the elementary symmetry polynomials in the $x_i$, so they are all zero. The polynomial is then $x^k$, which shows $x_i=0$ for all $i$.</p>


答: Multiply both sides of the equation on the left by $x^{-1}y^{-1}$ to see that your equation is equivalent to $[x, y]G=G$, which is true iff $[x, y] \in G$.

Title: relationship between cosets and commutators

Body:
Let $G$ be a normal subgroup of $H$, and let $x \in G$ and $y \in H$. Is it true that the following equation:

$$
xyG = yxG
$$

holds if and only if $[x,y] \in G$ ?, where $[x,y]$ is the commutator of $x$ and $y$, i.e. $[x,y]=x^{-1}y^{-1}xy$.


问:
simplifying equation with logs
I have the following equation:

I would like to solve this for Ze. I have found the same equation expressed in terms of Ze in another paper:

I can't get my head around how this works. This is my attempt:

However, there is something about the last step that doesn't seem right to me, even though the answers match.

Can anyone explain the correct process for simplifying the first equation in terms of Ze?

答:
You made a mistake in the first step. Here's how it should work:
$$\begin{align*}
\ln\left(\frac{z_0}{z_e}\right) &= b_1Re^{0.25}+b_2 \\
\exp\left(  \ln\left(\frac{z_0}{z_e}\right) \right) &= \exp\left( b_1Re^{0.25}+b_2 \right) \\
\frac{z_0}{z_e} &= \exp\left( b_1Re^{0.25} + b_2 \right) \\
\left(\frac{z_0}{z_e}\right)^{-1} &=\left( \exp\left( b_1Re^{0.25} + b_2 \right) \right)^{-1} \\
\frac{z_e}{z_0} &= \exp\left(-1\left( b_1Re^{0.25} + b_2 \right) \right) \\
\frac{z_e}{z_0} &= \exp\left(-b_1Re^{0.25} - b_2 \right) \\
z_e &= z_0\exp\left(-b_1Re^{0.25} - b_2 \right) \\
\end{align*}$$

Note that since $x^a \cdot x^b = x^{a+b}$ and $(x^a)^b=x^{ab}$, we have the following identities:

$\exp(a+b)=\exp(a) \cdot \exp(b)$

$\left(\exp(a)\right)^b=\exp(ab)$

Title: simplifying equation with logs


问:
How to Show that an $n$-finned Hyperplane is not Homeomorphic to $\mathbb{R}^m$
This question came up again in an algebraic topology course, so I'm placing a bounty on it.  The answers below use algebraic topology - I am asking for a proof that does not use algebraic topology, e.g. homology, higher homotopy groups, or the Jordan Curve/Separation Theorem.  No smooth structures, just point-set stuff.

Original question:

Suppose that $X_n$ is the union of $n$ half closed, straight line segments, all joined at a common point (their boundary point).  So $X_2$ is just an open interval, and otherwise is an $n$-pointed asterisk with open ends.  How can one show that for $n \geq 2$ all the spaces $X_n \times \mathbb{R}$ are topologically distinct? These spaces are like $n$ sheets attached along a copy of $\mathbb{R}$.

More generally if we take $n$ copies of the $k$-dimensional closed half-space $\mathbb{H}^k$ and identify them along a $(k-1)$-plane, how do we show that this space is not homeomorphic to $\mathbb{R}^k$ for $n > 2$?

Discussion:

In the more general form, the $n=1$ case is known to be equivalent to the Invariance of Dimension Theorem, but for $n > 2$ it seems more geometrical that the spaces aren't homeomorphic.

Be careful trying to use a fact like "a homeomorphic image of a hyperplane disconnects Euclidean space into two pieces."  This may be the key ingredient, but would need a point-set proof; I only know of algebraic ones.

There is a theorem of Moore, simplified by Bing, that there is no uncountable collection of copies of a simple triod (three arcs attached at a common vertix, i.e. a three-legged asterisk) contained in the plane.  There is a Cantor Set of them along the spine of our three-finned 'plane.'  However, the proofs of this theorem depend crucially on the Jordan Curve Theorem and its corollaries.  It has higher-dimensional analogues, but they also depend on the higher-dimensional form of the Jordan Separation Theorem.

Similarly, evaluating the simplicial complex structures could differentiate them, but that is a lot of theory.  If we look at neighborhoods on, say, $X_3 \times \mathbb{R}$ of the points along the 'spine' we will get connected neighborhoods whose boundaries are the theta curve.  It can be shown that no connected set has boundary of a theta curve in $\mathbb{R}^2 = X_2 \times \mathbb{R}$ and this method also generalizes inductively, but again the details are not trivial and we are depending on at least the polygonal version of the Jordan Curve Theorem.  Putting a circle into one of the legit-planes doesn't separate an $n$-finned plane; similarly (equivalently, as an aside), the complete bipartite graph $K_{3,3}$ embeds in it.

Yet another method would be counting the number of components that removing the spine creates, which could distinguish $X_2$ from the others but seems harder to generalize.  This is one of the classic separation theorems, that a hyperplane cuts Euclidean space into two components, but we can't assume it.

And generalizing any of these methods (except for analyzing as complexes) to higher dimensions seem to have even more difficulties, for example by attaching three copies of threespace to a plane.  There is also the possibility of using symmetry groups and relating those to the bundle structure, but again it would use a lot of theory.

What is the point-set way to prove this?  Or is it equivalent to Jordan Curve Theorem, which would be really interesting?

答:
You can use the Jordan Curve Theorem.

Note that a circle consisting of two halves on two different sheets of $X_n \times \mathbb{R}$ is a simple closed curve with a connected complement for $n>2$. This proves that $X_n \times \mathbb{R}$ can't be homeomorphic to $\mathbb{R}^2$.

This also works in higher dimensions, by applying the Jordan Brouwer separation theorem to a hypersphere constructed half on one sheet and half on another. The complement stays connected as long as there is a sheet other than the two containing the image of the hypersphere.

The theorems do a lot of heavy lifting, but it makes the rest of the proof trivial.


问:
Let $S\subset \mathbb{R^2}$ be defined by $S=\{(m+\frac{1}{4^{|p|}}, n+\frac{1}{4^{|q|}}):m,n,p,q \in \mathbb{Z}\}$.
Let $S\subset \mathbb{R^2}$ be defined by $S=\{(m+\frac{1}{4^{|p|}}, n+\frac{1}{4^{|q|}}):m,n,p,q \in \mathbb{Z}\}$.
. Then,

S is discrete in $\mathbb R^2$ .

The set of limit points of S is the set {(m,n);m,n $\in \mathbb Z$}.

i don't understand this

答:
Discrete set: Discrete set is a set of points of a topological space such that each point in the set is an isolated point, i.e. a point that has a neighborhood that contains no other points of the set.

Clearly, $(5/4,1)$ is a limit point of $S$ and $(5/4,1)\not\in \{(m,n);m,n ∈\mathbb Z\}$.

Also $(1,1)\in S$ and $(1,1)$ is limit point as $(1,1)=(0+1/4^0, 0+1/4^0)$.

Thus $S$ is not discrete set.
Hence both the two options are incorrect.


问:
Complex Integral proof explanation

Compute $$\int_{-\infty}^{\infty} \frac{\cos(αx)}{(x^2+1)(x^2+4)} \mathrm dx. $$

DonAntonio gave a well-constructed answer here but I have a few confusions on his answer as I am  new to complex integral:

How did we derive $\text{Res}_{z=i}(f)=\lim_{z\to i}(z-i)f(z)=\frac{e^{-\alpha}}{2i\cdot 3i\cdot (-i)}=-\frac{e^{-\alpha}}6i\;\;$? (the step when solving the limit)

What is the purpose of using Jordan's Lemma to get
$\lim_{R\to\infty}\int\limits_{\gamma_R}f(z)dz=0$? The result equalling to 0 does not seem to be applied to the next equation.

Should the final answer be $\frac16\left(2e^{-\alpha}-e^{-2\alpha}\right)\pi$ or without $\pi$?

Thanks in advance.

答:
$$J=\int_{-\infty}^\infty\frac{\cos(\alpha x)}{(x^2+1)(x^2+4)}dx\tag{1}$$

$$\frac{1}{(x^2+1)(x^2+4)}=\frac13\left(\frac1{x^2+1}-\frac1{x^2+4}\right)$$
lets make a general integral:
$$I(\alpha,\beta)=\int_{-\infty}^\infty\frac{\cos(\alpha x)}{x^2+\beta^2}dx$$
this defines your integral as:
$$J=\frac{I(\alpha,1)-I(\alpha,2)}{3}\tag{2}$$

Lets use the Feynman technique:
$$I_\alpha=-\int_{-\infty}^\infty\frac{x\sin(\alpha x)}{x^2+\beta^2}dx=-\int_{-\infty}^\infty\frac{(x^2+\beta^2-\beta^2)\sin(\alpha x)}{x(x^2+\beta^2)}dx$$
$$I_\alpha=-\int_{-\infty}^\infty\frac{\sin(\alpha x)}{x}dx+\beta^2\int_{-\infty}^\infty\frac{\sin(\alpha x)}{x(x^2+\beta^2)}dx$$
now notice that this first integral is just the dirichlet integral so we can say:
$$I_\alpha=-\pi+\beta^2\int_{-\infty}^\infty\frac{\sin(\alpha x)}{x(x^2+\beta^2)}dx$$
now what we are going to do is differentiate wrt $\alpha$ again:
$$I_{\alpha\alpha}=\beta^2\int_{-\infty}^\infty\frac{\cos(\alpha x)}{x^2+\beta^2}dx$$
but notice that this is just the same as saying:
$$\frac{\partial^2}{\partial\alpha^2}I(\alpha,\beta)=\beta^2I(\alpha,\beta)$$
now this is just a pretty common pde which we can set up as the following system:
$$I=I(\alpha,\beta)$$
$$I_{\alpha\alpha}-\beta^2I=0$$
$$I(0,\beta)=\int_{-\infty}^\infty\frac{dx}{x^2+\beta^2}=\frac{\pi}{\beta}$$
$$I_\alpha(0,\beta)=-\pi$$

I get:
$$I(\alpha,\beta)=\frac{\pi}{\beta}e^{-\alpha\beta}$$
and so:
$$J=\frac13\left(\pi e^{-\alpha}-\frac{\pi}{2}e^{-2\alpha}\right)=\frac{\pi(2e^{-\alpha}-e^{-2\alpha)}}{6}$$

So yes the $\pi$ should be there. Just a quick note to add: The equation works for $\beta\in\mathbb{R}^+/\{0\},\alpha\in\mathbb{R}^+$

Title: Complex Integral proof explanation


问:
Weak convergence vs weak* convergence of a measure
I am reading Optimal Transport for Applied mathematicians by Filippo Santambrogio and in the first chapter he describes weak convergence and weak-* convergence in terms of a Banach space and its dual. He then goes on to (by an abuse of notation, apparently) extend this concept to signed measures  on a separable, locally compact metric space $X$ when he states the Riesz representation theorem by saying:

Let $\mathcal{X} = C_0(X)$ be the space of continuous function on X vanishing at infinity, i.e. $f \in C_0(x) \iff f \in C(X)$, and for every $\epsilon > 0$, there exists a compact subset $K \subset X$ such that $|f| < \epsilon$ on $X$ \ $K$. Let us endow this space with the sup norm since $C_0(X) \subset C_b(X)$ (this last space being the space of bounded continuous functions on $X$). Note that $C_0(X)$ is a Banach space and thaat it is a closed subset of $C_b(X)$. Then every element of $\mathcal{X}'$ (the dual of $\mathcal{X}$) is represented in a unique way as an element of $\mathcal{M}(X)$: for all $\xi \in \mathcal{X}'$ there exists a unique $\lambda \in \mathcal{M}(X)$ such that $\langle \xi, \phi \rangle = \int \phi d \lambda$ for every $\phi \in \mathcal{X}$; moreover, $\mathcal{X}'$ is isomorphic to $\mathcal{M}(X)$ endowed with the norm $||\lambda|| := |\lambda|(X)$.

For signed measures of $\mathcal{M}(X)$, we should call weak-* convergence the convergence in the duality with $C_0(X)$. Yet another interesting notion of convergence is that in duality with $C_b(X)$. We will call it (by abuse of notation) weak convergence and denote it through the symbol $\mu_n \rightharpoonup \mu$ iff for every $\phi \in C_b(X)$ we have $\int \phi d\mu_n \rightarrow \int \phi d \mu$.

My question is as follows: is weak-* convergence (i.e. weak convergence in duality with $C_0(X)$) the same as weak convergence in duality with $C_b(X)$, except for the fact that in the former condition we require $\int \phi d\mu_n \rightarrow \int \phi d \mu$ for all $\phi \in C_0(X)$ as opposed to $C_b(X)$?

答:
Firstly, we remark that $\mathcal{M}(X)$, as a set, is the collection
of all finite signed Radon measures (assuming that we are working
with the scalar field $\mathbb{R}$). That is, $\mu\in\mathcal{M}(X)$
if $\mu$ is a finite signed measure, outer-regular in the sense that
for each Borel subset $A\subseteq X$, $\mu(A)=\inf\{\mu(U)\mid A\subseteq U\mbox{ and }U\mbox{ is open}\}$,
and inner-regular for open set in the sense that for each open set
$U\subseteq X$, $\mu(U)=\sup\{\mu(K)\mid K\subseteq U\mbox{ and }K\mbox{ is compact.}\}$.
Note that if $X$ is second countable, then every finite signed measure
is automatically Radon. (see Folland, Real Analysis, Theorem 7.8).

For your question, the two modes of convergence are not the same. Obviously, given a sequence $(\mu_{n})$ in $\mathcal{M}(X)$
and $\mu\in\mathcal{M}(X)$,  $\mu_{n}\rightharpoonup\mu\Rightarrow\mu_{n}\rightarrow\mu$
in weak*-topology. The converse does not hold in general. The following
is a counter-example.

Let $X=\mathbb{R}$, equipped with the usual topology. For each $n\in\mathbb{N}$,
let $\mu_{n}(A)=\lambda(A\cap[n,n+1])$, where $\lambda$ is the usual
Lebesgue measure, then $\mu_{n}\in\mathcal{M}(X)$. Let $\mu=0$.
Clearly $\mu_{n}\rightarrow\mu$ in weak*-topology. However, $\mu_{n}\not\rightharpoonup\mu$
because the constant function $f=1$ is an element in $C_{b}(X)$
and $\int fd\mu_{n}=1\neq\int fd\mu$.


问:
Continuous $f : \Bbb{R}^n \rightarrow \Bbb{R}$, can always find $\delta$ with $\delta < \epsilon$?
I'm still new to rigorous proofs in multivariable calculus, does my attempt at the following work?

Let $f : \Bbb{R}^n \rightarrow \Bbb{R}$ be a continuous function with $f(\boldsymbol{0}) = 0$ and $f(\boldsymbol{x}) > 0$ otherwise.
Let $\epsilon > 0$ be given. Let
$$m = \min_{{\Vert\boldsymbol{x}\Vert = \epsilon}} f(\boldsymbol{x})$$

Show that it is always possible to find $\delta$ with $0 < \delta < \epsilon$ such that when $\Vert \boldsymbol{x} \Vert < \delta$, we have $\Vert f(\boldsymbol{x})\Vert < m$.

My attempt:

The existence of a $\delta$ such that when $\Vert \boldsymbol{x} \Vert < \delta$, we have $\Vert f(\boldsymbol{x})\Vert < m$ follows trivially from the definition of continuity of $f$ at $0$. All that remains is to show that is possible to pick a $\delta$ with $\delta < \epsilon$.

Assume we are given such a $\delta_1$. We have $\delta_1 \leq \epsilon$, because $\Vert f(x)\Vert \geq m$ for $\Vert \boldsymbol{x} \Vert = \epsilon$.

If $\delta_1 < \epsilon$, let $\delta = \delta_1$ and we are done. Otherwise, $\delta_1 = \epsilon$, so choose $\delta = \frac{\delta_1}{2}$ and we are done.

Does that proof work? Is there a simpler way to express the "idea" I'm trying to get at (that we can always shrink our potential $\delta$ down until the strict inequality $\delta < \epsilon$ holds)?

答: If you want to preserve your idea, rather than using Ted's idea from the comments, then instead of writing out the "if...then" explicitly, you could set $\delta=\min(\delta_1,\epsilon/2)$.


问:
Proving throught induction
How do I prove the following inquation using induction?

$2^n>1+n\sqrt{2^{n-1}}$ ,
$n\geq 2$

I did the base case, but I'm stuck at the induction process.

The induction: $2^{k+1}>1+(k+1)\sqrt{2^{k}}$

Now I used the hipothesis to prove the induction:

$2*2^{k}>2*(1+k\sqrt{2^{k-1}})$

$2^{k+1}>2+2k\sqrt{2^{k-1}})$

I can´t do the rest

答: Shouldn't your induction hypothesis be $2^k > 1 + k\sqrt{2^{k-1}}$ for some integer $k \geq 2$? It looks like you assumed $2^k > 1 + k\sqrt{2^{k}}$ instead.

Title: Proving throught induction


问:
Solve an algebraic equation involving complex numbers
Please, solve for $x$ and $y$,
$$(x+i)(3-iy)=1+13i,
\quad x,y \in \text{Reals}$$

答:
$$(x+i)(3-iy) = 1+13i $$

$$  3x+y+i(3-xy) = 1+13i \implies \left\{
3x+y = 1 \atop
3-xy = 13
\right.$$

Now solve the system for x and y.


问:
How to find $\int \frac{x\sqrt{1+\tan ^2x}}{\cos x}dx$
How to perform integration by substitution of the following expression?

$$\int \frac{x\sqrt{1+\tan ^2x}}{\cos x}dx$$

I've already reached to this part of the solution, but I'm having a hard time, figuring out, what to do next.

Step 1:  $$=\int\frac{x\sqrt{\frac{\cos ^2x+\sin ^2x}{\cos ^2x}}}{\cos x}dx$$
Step 2: $$=\int \frac{x\sqrt{1/\cos ^2x}}{\cos x}dx$$
Step 3: $$=\int \frac{x\sec x}{\cos x}dx$$
Step 4: $$\int \frac{x}{\cos ^2x}dx$$
but what next?

答:
Its probably easiest to write the integral as:
$$I=\int x\sec^2 x\,dx$$
now do integration by parts: $u=x\Rightarrow u'=1,v'=\sec^2x\Rightarrow v=\tan x$ so:
$$I=x\tan x-\int\tan x\,dx$$
can you finish it from here?


问:
Ellipse: Most direct algebraic demonstration that $\left|x\right|$ is maximal when $y=0$
Yes, I know this is the equation of an ellipse.

This is the starting point of the problem:

$$
r_1+r_2=2a,
$$

where

$$r_1\equiv\sqrt{(x+c)^2+y^2}; r_2\equiv\sqrt{(x-c)^2+y^2}.$$

I want to establish certain properties prior to deriving a parametric expression for the ellipse, or even the "standard form",

$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.
$$

I can do all of this with a couple tacks, a string and a pencil, but
I want to show things algebraically. Preferably in a way that is directly applicable to all second-degree plane curves.

The first property I wish to establish is the range of $x$ values.
I know from experience that the answer is going to be that the extreme
values of $x$ will occur when $y=0$. The question that I am asking
in this post is: what is the most direct (algebraic) way to show that $\left|x\right|$
is greatest when $y=0$?

I had hoped to establish that fact, and then use another (absolute value)
method of determining the actual values of $x$ when $y=0$. But, the best
I cam up with is this very long-winded demonstration.

First some definitions: Let $a,c,x,y\in\mathbb{R}$ where $a>c>0$
are constants, and $x$ and $y$ are variables such that, for $r_1\equiv\sqrt{(x+c)^2+y^2}$
and $r_2\equiv\sqrt{(x-c)^2+y^2}$ we have the equation

$$
r_1+r_2=2a.
$$

The degenerate case has been omitted.

I will presume the greatest value to be $\left|x\right|=a$, and take
the case of $x=a$. We square our equation

$$
(r_1+r_2)^2=4a^2
$$

$$
=r_1^2+2r_1 r_2+r_2^2
$$

$$
=(a+c)^2+y^2+2r_1 r_2+(a-c)^2+y^2
$$

$$
=2(a^2+c^2+y^2)+2r_1 r_2.
$$

So

$$
r_1(2a-r_1)=a^2-c^2-y^2
$$

$$
=2a\sqrt{(a+c)^2+y^2}-((a+c)^2+y^2)=a^2-c^2-y^2.
$$

Transpose, cancel and simplify

$$
2a\sqrt{(a+c)^2+y^2}=2a^2+2ac,
$$

$$
r_1=\sqrt{(a+c)^2+y^2}=a+c.
$$

And, therefore

$$
r_2=2a-r_1=a-c.
$$

Now, by definition we have

$$
\sqrt{\left(a+c\right)^2}\equiv\left|a+c\right|=r_1
$$

and

$$
\sqrt{(a-c)^2}\equiv\left|a-c\right|=r_2.
$$

If $y\ne0$ while $x=a$, then

$$
r_1=\sqrt{(a+c)^2+y^2}=\left|a+c\right|+\delta_1,
$$

and

$$
r_{2}=\sqrt{(a-c)^2+y^2}=\left|a-c\right|+\delta_2,
$$

where both $\delta_1>0$ and $\delta_2>0.$

Since $a>c>0,$ we have $r_1+r_2=2a+\delta_1+\delta_2$, in
contradiction to the equation we seek to satisfy. So $x=a\implies y=0.$

If $x>a$, let $\delta_3=x-a>0.$

So
$$
r_1+r_2=a+c+\delta_3+a-c+\delta_3>2a;
$$

another contradiction.

Thus far we have dertermined that $x=a\implies y=0,$ and for any
$y$ that $x\le a$.

If we set $x=-a,$ then

$$
(r_1+r_2)^2=4a^2
$$

$$
=r_1^2+2r_1 r_2+r_2^2
$$

$$
=(-a+c)^2+y^2+2r_1 r_2+(-a-c)^2+y^2
$$

$$
=2(a^2+c^2+y^2)+2r_1 r_2
$$

$$
r_{1}\left(2a-r_{1}\right)=a^{2}-c^{2}-y^{2}
$$

$$
2a\sqrt{\left(-a+c\right)^{2}+y^{2}}-\left(\left(-a+c\right)^{2}+y^{2}\right)=a^{2}-c^{2}-y^{2}
$$

$$
2a\sqrt{\left(-a+c\right)^{2}+y^{2}}=2a^{2}-2ac
$$

$$
r_{1}=\sqrt{\left(a-c\right)^{2}+y^{2}}=a-c>0.
$$

$$
r_{2}=2a-r_{1}=a+c.
$$

Since we now have

$$
r_{1}=\sqrt{\left(a-c\right)^{2}+y^{2}}=a-c,
$$

and

$$
r_{2}=\sqrt{\left(a+c\right)^{2}+y^{2}}=a+c,
$$

the argument showing that $x=-a\implies y=0,$ is essentially the
same. Likewise for the argument that $x\ge-a.$

And that's just the $x$ range.

NB: I don't "need" any of this to fully characterize the ellipse.  My goal is to find multiple ways of examining the same construct.

答:
There is a very simple and short answer.  Using the definition of monotonicity given in  Fundamentals of Mathematics, Volume 1: Foundations of Mathematics: The Real Number System and Algebra Edited by H. Behnke, F. Bachmann, K. Fladt, W. Süss and H. Kunle, which is a monotonic function is such that
$$x<y\implies f\left(x\right)<f\left(y\right)$$
or such that
$$x<y\implies f\left(y\right)<f\left(x\right),$$

and using the results of the elementary theory of numbers developed in Chapter 1B-1 for exponents, the proof is straight forward.  That was the part I was missing.  The result is "intuitively obvious", but I did not have a theorem to apply.  In particular, I had no proof of the the blatantly obvious fact
$$\sqrt{(x-c)^2+y^2}\ge\sqrt{(x-c)^2}.$$


问:
List of interesting math podcasts?
mathfactor is one I listen to.  Does anyone else have a recommendation?</p>

答: More or Less is a BBC Radio 4 programme about maths and statistics in the news, and there is a free podcast. It's presented by Tim Harford, the Undercover Economist from the Financial Times.</p>

ContentLicense: CC BY-SA 2.5

Title: List of interesting math podcasts?


问:
What is a real number (also rational, decimal, integer, natural, cardinal, ordinal...)?
In mathematics, there seem to be a lot of different types of numbers. What exactly are:

Real numbers

Integers

Rational numbers

Decimals

Complex numbers

Natural numbers

Cardinals

Ordinals

And as workmad3 points out, some more advanced types of numbers (I'd never heard of)

Hyper-reals

Quaternions

Imaginary numbers

Are there any other types of classifications of a number I missed?

答:
Natural numbers

The "counting" numbers.  (That is, all integers, that are one or greater).

Whole numbers

The Natural numbers, and zero.

Integers

The Whole numbers, and the negatives of the Natural numbers.

Rational numbers

Any number that may be expressed by any integer A divided by any integer B, where B is not zero.

Irrational numbers

Any number that cannot be expressed as a rational number, but is not imaginary.  All irrational numbers have an infinite decimal representation.

Real numbers

All of the Rational and Irrational numbers.

Imaginary numbers

All Real numbers, multiplied by the square root of negative one.  Imaginary numbers are signified by the letter i.

Complex numbers

Numbers composed of the sum of a Real and an Imaginary number.  This includes all Real and all Imaginary numbers.


问:
What is larger -- the set of all positive even numbers, or the set of all positive integers?
We will call the set of all positive even numbers E and the set of all positive integers N.

At first glance, it seems obvious that E is smaller than N, because for E is basically N with half of its terms taken out.  The size of E is the size of N divided by two.

You could see this as, for every item in E, two items in N could be matched (the item x and x-1).  This implies that N is twice as large as E

On second glance though, it seems less obvious.  Each item in N could be mapped with one item in E (the item x*2).

Which is larger, then?  Or are they both equal in size?  Why?

(My background in Set theory is quite extremely scant)

答:
Each item in N could be mapped with one item in E  (the item x*2).

Yes. Both sets have cardinality aleph-0.


问:
Is a prime number still a prime when in a different base?
Is a prime number in the decimal system still a prime when converted to a different base?</p>

答:
Here is the same question which has been asked at Mathforum: Here is the link

Link

from the link

The fact of being prime or composite is
just a property of the number itself, regardless of the way you write
it. 15 and F and Roman numeral XV all mean the number, which is 3
times 5, so it is composite. That is the way it is for all numbers, in
the sense that if a base ten number N has factors, you can represent
those factors in Hex and their product will be the number N in Hex.

OwnerDisplayName: anonymous


问:
Proving that if $f'$ has at most $n-1$ zeros, then $f$ has at most $n$ zeros
Is this proof correct?

The problem is the following.

Let $n$ be a natural number. Suppose that the function $f:\mathbb{R}\to\mathbb{R}$ is differentiable and that the following equation has at most $n-1$ solutions:
  $$f'(x)=0, \quad x \in \mathbb{R}.$$
  Prove that the following equation has at most $n$ solutions:
  $$f(x)=0,\quad x \in \mathbb{R}.$$

My proof is:

Let $f'(x)=0$ have solutions $x_1$, $x_2,\ldots ,x_{n-1}$.

since
$$f'(x_1)=\lim\limits_{{x}\to{x_1}}\frac{f(x)-f(x_1)}{x-x_1} = 0$$
and doing so, $f(x)=f(x_{n-1})$
Therefore $f(x)=f(x_1)=f(x_2)=\cdots=f(x_{n-1})$.

Let $x_n$ be one of solutions of $f(x)=0$.
$$0=f(x_n)=f(x_1)=f(x_2)=\cdots=f(x_{n-1}),$$
so $f(x)=0$ has solutions like $x_1$, $x_2,\ldots, x_n$.

If there is wrong part, please let me know.

答:
You cannot simply assume that $f'(x)$ is a polynomial. That is not given in the problem, and there are plenty of other functions that can have a specific number of zeros.  (Also, the problem says that $f'(x)=0$ has at most $n-1$ solutions, not that it has exactly $n-1$ solutions, yet you assume it has this maximum).

Even if you know that $f'(x)$ is a polynomial, you do not know that the degree is at most $n-1$; it's possible for a polynomial to have degree strictly larger than $k$, and yet have only $k$ distinct real roots.

It's false that the fact that $f'(x_1)=0$ implies that $f(x_1)=0$.

The statement "and doing so, $f(x)=f(x_{n-1})$" makes no sense to me.

You never showed that there are at most $n$ solutions to $f(x)=0$.

You seem to believe that the same numbers that are zeros of the derivative are zeros of the original function. This is not true, even for polynomials. $f(x) = x^2 - 3x + 2$ has zeros at $x=1$ and $x=2$, but the zero of $f'(x) = 2x-3$ is at $x=\frac{3}{2}$.

So, I would say that pretty much all of this argument is incorrect.

Suggestion. Either use the Mean Value Theorem, or use Darboux's Theorem.

Here's how an argument using the latter would go:

Suppose the zeros of the derivative are $a_1\lt a_2 \lt\cdots \lt a_{n-1}$. By Darboux's Theorem, the derivative cannot change signs except at the points where it is equal to $0$. That means, for example, that it is either always positive or it is always negative on $(-\infty,a_1)$. That means that $f(x)$ is strictly monotone on $(-\infty,a_1)$: either strictly increasing, or strictly decreasing. How many zeros can a strictly monotone function have? So, how many zeros can $f(x)$ have on $(-\infty,a_1)$?

What about $(a_1,a_2)$? $(a_2,a_3)$? Etc.

But I suspect you'll want to use the Mean Value Theorem; Darboux's Theorem is not often covered in basic analysis courses.


答:
Be careful:

$e^{a+b}$ is not the same as $e^a+e^b$ (it's actually $e^a.e^b$).

When it comes to the final step, one only needs to note that $e^{-a}=\frac{1}{e^a}$.

The full derivation might look like this:
$$\ln\left(\frac{z_0}{z_e}\right) = b_1Re^{0.25}+b_2$$
Get rid of the logarithm:
$$\frac{z_0}{z_e} = \exp(b_1Re^{0.25}+b_2)$$
Multiply by $z_e$ and divide by the exponential:
$$z_e = z_0\frac{1}{\exp(b_1Re^{0.25}+b_2)}$$
Use the fact that $\exp{-a}=\frac{1}{\exp{a}}$:
$$z_e = z_0\exp(-(b_1Re^{0.25}+b_2))$$

Title: simplifying equation with logs

Body:
I have the following equation:

I would like to solve this for Ze. I have found the same equation expressed in terms of Ze in another paper:

I can't get my head around how this works. This is my attempt:

However, there is something about the last step that doesn't seem right to me, even though the answers match.

Can anyone explain the correct process for simplifying the first equation in terms of Ze?


Title: Proving throught induction

Body:
How do I prove the following inquation using induction?

$2^n>1+n\sqrt{2^{n-1}}$ ,
$n\geq 2$

I did the base case, but I'm stuck at the induction process.

The induction: $2^{k+1}>1+(k+1)\sqrt{2^{k}}$

Now I used the hipothesis to prove the induction:

$2*2^{k}>2*(1+k\sqrt{2^{k-1}})$

$2^{k+1}>2+2k\sqrt{2^{k-1}})$

I can´t do the rest

Body:
Hip: $2^n>1+n\sqrt{2^{n-1}}$.

Then,

$2 \cdot 2^n>2(1+n\sqrt{2^{n-1}})$

$2^{n+1}>2+2n\sqrt{2^{n-1}}=2+n\sqrt{2}\sqrt{2^{n}}>1+(n+1)\sqrt{2^{n}}$

since $n\sqrt{2}>n+1$, if $n>2$.


问:
solve the Bernoulli equation xy' - y = xy^2

Solve the Bernoulli equation $xy' - y = xy^2$.

I started with diving both sides by $x$, and ended up with $y' - \frac{y}{x} = y^2$. Then, I divided both sides by $y^2$ and got $\frac{y'}{y^2} - \frac{1}{xy} = 1$. Can someone help me finish this problem?

答:
Let $g(x)=\frac{1}{y(x)}$ then
$$g'(x)=\frac{-y'(x)}{(y(x))^2}$$
so we reduce the equation to
$$-g'(x)-\frac{g(x)}{x} = 1$$

Can you finish from here?


问:
2/3 log x = log 4.So in the equation,I just divide 4 and 2/3 and i get 6. So the remaining equation is log x = log 6. Do I just cancel the log?
I forgot about the log rules. Can someone help me by explaining it. And if we need to find the approximate value of f(x) increases from 12.0 to 12.4. Do I need to use the two value so I will get two answer. the f(x) = 3x^2/3</p>

答:
Two important properties of logarithms:

\begin{align*} \color{blue}{1)}& \qquad  a\log(x) = \log(x^{a})\\
\color{blue}{2)}& \qquad a^{\log_a(b)} = b
\end{align*}

In your problem, you apply them as follows
\begin{align*}
&\frac{2}{3} \log(x) = \log(4) \\
\implies \quad&\log(x)  = \frac{3}{2} \log(4)\\
\overset{\color{blue}{1)}}{\implies} \quad&\log(x)  =  \log\left(4^{\frac{3}{2}}\right)
\end{align*}
Now, assuming that by $\log(x)$ you mean $\log_e(x)$ we get
\begin{align*}
&\log(x)  =  \log\left(4^{\frac{3}{2}}\right)\\
\implies \quad& e^{\log_e(x)} = e^{\log_e\left(4^{\frac{3}{2}}\right)} \\
\overset{\color{blue}{2)}}{\implies} \quad& x = 4^{\frac{3}{2}}\\
\implies \quad& x = \sqrt{4^3}\\
\implies \quad & x = \sqrt{64} = \boxed{8}
\end{align*}

Indeed, if we have any two logarithms of the same base being equated, as was pointed out in the comments we are allowed to cancel out the logarithms because the logarithm function is bijective, and hence has an inverse function in the exponential.


答: Start over.  Divide through by $y^2$.  Express in terms of differentials (dy and dx).  The LHS is an exact differential.

Title: solve the Bernoulli equation xy' - y = xy^2

Body:
Solve the Bernoulli equation $xy' - y = xy^2$.

I started with diving both sides by $x$, and ended up with $y' - \frac{y}{x} = y^2$. Then, I divided both sides by $y^2$ and got $\frac{y'}{y^2} - \frac{1}{xy} = 1$. Can someone help me finish this problem?


问:
How do you make dependent Brownian motions independent?
Can someone explain to me how to take 2 correlated Brownian motions and make them independent?  I can't seem to grasp this process.

Just assuming $dB_1(t)dB_2(t) = \rho dt $

From what was explained to me, but I dont understand entirely:

You set the Brownian motions equal to the following functions of new uncorrelated brownian motions:

$B_1(t) = W_1(t)$

$B_2(t) = \rho W_1(t) + \sqrt{1-\rho ^2} W_2(t)$

Then if this is written in matrix form:

$dB(t) = \begin{bmatrix}1 & 0\\ \rho & \sqrt{1-\rho ^2}\\ \end{bmatrix} dW(t)$

So now that right hand side could be used as an indepedent Brownian motion?

答: $W$ is a pair of independent Brownian motions, but $B$ is not. The trick is just to assume that $B_1$ is a certain Brownian motion (e.g. $W_1$) and that $B_2$ is a linear combination of $B_1$ and a Brownian motion that is independent from $B_1$, e.g. $W_2$.  Coefficients of this linear combination are computed based on the correlation condition $\mathrm dB_1 \mathrm dB_2 = \rho\mathrm dt.$</p>


问:
Prove $\int_{0}^{1}{x\arcsin(x) dx} \leq \frac{\pi}{4}$
I know how you can show this geometrically, but is there any way to prove this algebraically?</p>

答:
Since $\arcsin x$ is bounded above by $\pi / 2$, by monotonicity of the integral we have
$$\int_0^1 x \arcsin x dx \leq \frac \pi 2 \int_0^1 x dx = \frac \pi 4$$


问:
Drunk man with a set of keys.
I found this problem in a contest of years ago, but I'm not very good at probability, so I prefer to see how you do it:

A man gets drunk half of the days of a month. To open his house, he has a set of keys with $5$ keys that are all very similar, and only one key lets him enter his house. Even when he arrives sober he doesn't know which key is the correct one, so he tries them one by one until he chooses the correct key. When he's drunk, he also tries the keys one by one, but he can't distinguish which keys he has tried before, so he may repeat the same key.

One day we saw that he opened the door on his third try.

What is the probability that he was drunk that day?

答:
I'll do it for the case where the man has $k$ keys and gets the correct key on the $j$th try. Essentially, we're trying to find

$$P(\text{Drunk|$j ^{th}$ attempt successful})$$
By Bayes' Theorem,

$$P(\text{Drunk|$j ^{th}$ attempt successful}) = \frac{P(\text{Drunk $\cap$ $j ^{th}$ attempt successful})}{P(\text{$j ^{th}$ attempt successful})}$$

Let's find the denominator first. We can do this with the Law of Total Probability
$$P(\text{$j ^{th}$ attempt successful}) = P(\text{$j ^{th}$ attempt successful|Drunk})P(\text{Drunk}) + P(\text{$j ^{th}$ attempt successful|Sober})P(\text{Sober})$$

We are told that he comes home drunk about half the time, so $P(\text{Drunk}) = \frac{1}{2}$ and $P(\text{Sober}) = \frac{1}{2}$.

We also know that when drunk he has the chance to repeat keys and when sober he tries the keys without repetition, so
$$P(\text{$j ^{th}$ attempt successful|Drunk}) = P(\text{$j$ attempts with repetition}) \\ = \left( \frac{k-1}{k}\right)^{j-1}\frac{1}{k}  \\ (\text{Because  we get the wrong key $j-1$ times and then the right key}) $$

$$P(\text{$j ^{th}$ attempt successful|Sober}) = P(\text{$j$ attempts without repetition}) = \frac{1}{k}$$

Putting this together:$$P(\text{$j ^{th}$ attempt successful}) = \left( \left( \frac{k-1}{k}\right)^{j-1}\frac{1}{k}   \right)\left( \frac{1}{2} \right) + \left( \frac{1}{k} \right) \left( \frac{1}{2} \right) \\ = \left( \frac{1}{2} \right)\left[ \left( \frac{k-1}{k}\right)^{j-1}\frac{1}{k}    + \left( \frac{1}{k} \right)\right]$$

Now all that's left is the numerator which we've honestly already done a lot of the work for $$P(\text{Drunk $\cap$ $j ^{th}$ attempt successful}) = P(\text{$j ^{th}$ attempt successful|Drunk})P(\text{Drunk}) \\ \text{(This is the first term in our law of total probability equation)} \\ = \left( \frac{1}{2} \right) \left( \frac{k-1}{k} \right)^{j-1}\frac{1}{k}  $$

Altogether now we have:

$$P(\text{Drunk|$j ^{th}$ attempt successful}) = \frac {{\left( \frac{1}{2} \right) \left( \frac{k-1}{k} \right)^{j-1}\frac{1}{k}}}{\left( \frac{1}{2} \right)\left[ \left( \frac{k-1}{k}\right)^{j-1}\frac{1}{k}    + \left( \frac{1}{k} \right)\right]} \\ = \frac{\left( \frac{k -1 }{k} \right)^{j-1}}{1 + \left( \frac{k -1 }{k} \right)^{j-1} }$$

If you substitute in the numbers you have, you should end up with $\frac{16}{41} = \overline{.39024}$

Title: Drunk man with a set of keys.


问:
Behaviour of the curve $(y^4 − 4y^2 + 3)+i(-y^3+2y)$ in the complex plane
I was reading an example of the application of the Argument Principle and I encountered the following reasoning, which I do not understand.

At a certain point in the example, we consider $f(iy)=(y^4 − 4y^2 + 3)+i(-y^3+2y)$ and the following asseveration is made: since $\lim_{y\to \infty}\frac{-y^3+2y}{y^4 − 4y^2 + 3}=0$, then $f(iy)$ reaches the positive real axis as $y\to\infty$.
(You can see the context of this reasoning on page 6 of this PDF http://legacy-www.math.harvard.edu/archive/115_fall_06/argument_principle.pdf).

I do not understand why this is true. I fact, I have ploted the curve (in this context $y$ takes values from $0$ to $\infty$), and it does not seem like it reaches the positive real axis as $y\to\infty$.

If anyone could clarify why this reasoning is valid it would be very helpful.

Thanks.

Body:
$$(y^4 − 4y^2 + 3)+i(-y^3+2y)=y^3[(y-4/y+3/y^2)-i(1-2/y^2)] $$

$$\arg f(iy)= \tan^{-1} \frac{2/y^2-1}{y-4/y+3/y^2} \approx -\tan^{-1} \frac{1}{y}\to 0.$$


问:
Is it possible to determine f(7)if f(x) = 3 for all x < 7 and f is right-continuous at x = 7? What if f is left-continuous?

I am having a hard time really understanding the question, but what I am gathering is if the question had said "f(x)=3 for all x GREATER THAN 7" then we would be able to determine f(7) because it is right continuous until it gets to 3? Which 3 would be our value for f(7) in the new condition I created?

Maybe I just need a rewording of the question.

答:
The problem tells you $y=3$ for $x<7$. This is the first picture. $\lim_{x\rightarrow7-}f(x)=3$ However you need some information about continuity to determine more about the problem.

Now, if y is continuous from the left, as you MENTIONED IN YOUR POST AS A CONJECTURE, you can conclude that $f(7)=\lim_{x\rightarrow7-}f(x)=3$.

But if you are told y is continuous on the right, AS STATED IN THE PROBLEM, it's not necessarily the case that $f(7)=3$. $\lim_{x\rightarrow 7+}f(x)=L=f(7)$. The limit COULD be 3 but doesn't have to. You wouldn’t be able to determine it as the solution says.


问:
Darth Vader Rule: what is the reason for its name, and a formal proof?
I often hear the term "Darth Vader Rule" when calculating the expected value using the survival function and taking the integral where it is defined.

I am not quite sure why it is called that (is it customary?) and I would also like to know a formal proof of it.  I tried to look around, but I have a feeling that the name of this rule is not official and I cannot seem to find it right away.

答:
On a formal proof:

There are many proofs of this rule on math.SE.  Duplicate requests for proof are redirected here.  Intuition on this rule can be found in this post. IMO the slickest proof argues as follows:

Claim: Let $X$ be a nonnegative random variable. Then
$$
E[X]=\int_0^\infty P(X>t).
$$

Proof: Write $X$ as the integral of the constant $1$ from $0$ to $X$:
$$
X = \int_0^X1\,dt=\int_0^\infty H(t)dt
$$
where
$$H(t) = \begin{cases}1&\text{if $t<X$}\\ 0&\text{otherwise}\end{cases}$$
To compute the expectation of $X$, interchange the order of expectation and integration (Fubini-Tonelli):
$$
E[X] = E\left[\int_0^\infty H(t)dt\right]\stackrel{\text{Fubini}}=\int_0^\infty E[H(t)]dt$$
But for each $t>0$, $H(t)$ is a zero-one random variable, so its expectation is the probability that it equals $1$:$$ E[H(t)] = P(H(t)=1) = P(t<X) = P(X>t).$$

The same argument proves the alternative form $E[X]=\int_0^\infty P(X\ge t)$.


问:
Calculation Of Integral Related To Sequence
Let's evaluate the following integral. Many trials but no success.
$$\int_{-\pi}^{\pi}\dfrac{\sin nx}{(1+\pi^{x})\sin x}dx$$</p>

答:
Hint:

Fix $x=-x \implies dx=-dx$

$$I=-\int_{\pi}^{-\pi}\dfrac{\sin n(-x)}{(1+(\pi)^{-x})\sin (-x)} dx$$

$$I= \int_{-\pi}^{\pi}\dfrac{\sin n(-x)}{(1+(\pi)^{-x})\sin (-x)} dx$$

$$I=\int_{-\pi}^{\pi}\dfrac{\pi^x \sin n(x)}{(\pi^x+1)\sin x} dx$$

Add this to $I=\int_{-\pi}^{\pi}\dfrac{\sin nx}{(1+\pi^{x})\sin x}dx$

$$2I= \int_{-\pi}^{\pi} \dfrac{\sin nx}{\sin x} \dfrac{\pi^x+1}{1+\pi^x}dx$$

If you have $\int_{-a}^a f(x) dx$ and $f(x) $ is even, the same integral can be written as $2\int_0^af(x)dx$. I believe you can do the further caculations.


问:
How can I proove this using integrals
I want to know how to prove this: Assume a function, $f:[0,\pi] \rightarrow \mathbb{R}$ is continuous. Then prove $$\int_{0}^{\pi}xf(\sin(x))dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin(x))dx$$ First time, I tried using the fundamental theorem and substitute $u= \pi - x$. How can I continue?

I can still use this to evalute $$\int_{0}^{\pi}\frac{x\sin(x)}{1+\cos^{2}x}dx$$Can I do that using the proof here?

答:
Assume that the integral is $\mathcal{I}$.

\begin{align}
\mathcal{I} &= \int_{0}^{\pi}x f(\sin(x)))dx \\
&\text{substitute} ~~ u=\pi-x \\
\mathcal{I} &= \int_{\pi}^{0}(\pi-u) f(\sin(\pi-u)))d(\pi-u) \\
 &= \int_{\pi}^{0}(\pi-u) f(\sin(u))(-du) \\
&= \int_{\pi}^{0}(\pi-u) f(\sin(u))(-du) \\
&= \int_{\pi}^{0}-(\pi-u) f(\sin(u))du \\
&= \int_{0}^{\pi}(\pi-u) f(\sin(u))du ~~~~~\text{; because} \int_{b}^{a}f(x)dx=-\int_{a}^{b}f(x)dx\\
&= \int_{0}^{\pi}\pi f(\sin(u))du - \int_{0}^{\pi}u f(\sin(u))du \\
& \text{Since 'u' is a dummy variable can be substituted by any other dummy variable 'x'}\\
&= \int_{0}^{\pi}\pi f(\sin(x))dx - \int_{0}^{\pi}x f(\sin(x))dx \\
&= \pi\int_{0}^{\pi}f(\sin(x))dx - \mathcal{I} \\
i.e,~~~ 2\mathcal{I} &= \pi\int_{0}^{\pi}f(\sin(x))dx \\
\mathcal{I} &= \frac{\pi}{2}\int_{0}^{\pi}f(\sin(x))dx \\
i.e, ~~ \int_{0}^{\pi}x f(\sin(x)))dx &= \frac{\pi}{2}\int_{0}^{\pi}f(\sin(x))dx \\
\end{align}

Title: How can I proove this using integrals


答:
You can use the following two known inequalities
$$x \leq |x|$$ and $$\left|\int_a^b f(x)dx\right|\leq \int_a^b|f(x)|dx.$$
We have
\begin{align*}
\int_0^1x\sin^{-1}xdx &\leq \left|\int_0^1x\sin^{-1}xdx\right|\\
&\leq \int_0^1x|\sin^{-1}x|dx\\
&\leq \frac{\pi}{2}\int_0^1xdx=\frac{\pi}{4}.
\end{align*}

Title: Prove $\int_{0}^{1}{x\arcsin(x) dx} \leq \frac{\pi}{4}$

Body: I know how you can show this geometrically, but is there any way to prove this algebraically?</p>


问:
Does the Jordan curve theorem apply to non-closed curves?
A Jordan curve is a continuous closed curve in $\Bbb R^2$ which is simple, i.e. has no self-intersections. The Jordan curve theorem states that the complement of any Jordan curve has two connected components, an interior and an exterior.

Now let's define an unbounded curve to be a continuous map $f: (-\infty,\infty)\to\Bbb R^2$ such that $f((-\infty,0))$ and $f((0,\infty))$ are both unbounded. My question is, does the complement of a simple unbounded curve always have two connected components? It seems intuitively true, since you'd expect the curve to have two sides, but considering how long it took to prove the Jordan curve theorem, things may not be as straightforward as they appear.

Any help would be greatly appreciated.

Thank You in Advance.

EDIT: As @dfeuer suggested, let's also require that the curve goes off to infinity in both directions.  To make this precise, let's say that there exists two lines $L_1$ and $L_2$, parametrized by $L_1(t) = (a_1 + b_1 t, c_1 + d_1 t)$ and $L_2(t) = (a_2 + b_2 t, c_2 + d_2 t)$, such that the limit of $d(f(t), L_1(t))$ as $t$ goes to $-\infty$ is $0$, and the limit of $d(f(t), L_2(t))$ as $t$ goes to $\infty$ is $0$.  Under that condition, does the complement of the curve have two connected components?

答:
You can use the single point compactification of $\mathbb{R}^2$, call this $S^2$ to formulate a sufficient criterion for a non-closed curve to divide the plane in two connected components.

If the curve stays unbounded at both ends, i.e. for each $M$ there is an $x_0$ such that if $|x|>x_0$ then $|f(x)| > M $, the curve will cut the plane in two connected halves.

This is equivalent to saying that $f$ can be completed in the one point compactification of ${\mathbb{R}^2}$ which is a sphere, as $f^*$. This is a closed curve.

Take a point not on the image of $f^*$, say $P$ and consider its complement in the sphere $S^2 \setminus P$, this space is homeomorphic to the plane. So, we constructed a simple closed curve in the plane, which has a bounded interior and an unbounded exterior. Because the interior is bounded, this space is disconnected from $P$. So the image of $f^*$ separates $S^2$ into two connected components. These two connected components are exactly the complement of the image of $f$ in $\mathbb{R}^2$, because the point added to $\mathbb{R}^2$ to create $S^2$ is in the image of $f^*$.

My explanation may be a bit chaotic, but there's a valid proof in there.

Answer to the EDIT: Yes that is enough, but the requirement can be weakened quite a bit.


问:
Validity of Limits? Proof that $\lim_{x \to 1} \frac{x^{\alpha}-1}{x-1} = \alpha$ when $\alpha \in \mathbb{R}$
Synopsis

In Tao's Analysis 1, we are asked the following:

Let $\alpha$ be a real number, and let $f:(0,\infty) \to \mathbb{R}$
be the function $f(x) = x^{\alpha}$. Show that  $\lim_{x \to 1} \frac{f(x)-f(1)}{x-1} = \alpha$ using Exercise 10.4.2 and the comparison principle; you may need to consider right and left limits separately. The fact that the rationals are dense in the reals might also help.

Now, Exercise 10.4.2 proves the result for $f(x) = x^q$ where $q \in \mathbb{Q}$ and the comparison principle is given as follows:

Lemma 6.4.13 (Comparison principle) - Suppose that $(a_n)_{n=m}^{\infty}$ and $(b_n)_{n=m}^{\infty}$ are two sequences of real numbers such that $a_n \leq b_n$ for all $n \geq m$. Then we have the inequalities $$\text{sup}(a_n)_{n=m}^{\infty} \leq \text{sup}(b_n)_{n=m}^{\infty}\\
\text{inf}(a_n)_{n=m}^{\infty} \leq \text{inf}(b_n)_{n=m}^{\infty} \\
\lim \text{sup}_{n \rightarrow \infty}a_n \leq \lim \text{sup}_{n \rightarrow \infty}b_n\\
\lim \text{inf}_{n \rightarrow \infty}a_n \leq \lim \text{inf}_{n \rightarrow \infty}b_n$$

It should be noted that Terence Tao builds the reals through Cauchy sequences of rational numbers, and defines real exponentiations in the following manner:

Definition 6.7.2 (Exponentiation to a real exponent) - Let $x> 0$ be real, and let $\alpha$ be a real number. We define the quantity $x^\alpha$ by the formula $x^{\alpha} = \lim_{n \to \infty} x^{q_n}$, where $(q_n)_{n=1}^{\infty}$ is any sequence fo rational numbers converging to $\alpha$.

Note that this exercise is given when we have not been taught L'Hopital's rule yet, and when we have yet even to prove that the derivative of $x^{\alpha} = \alpha x^{\alpha -1}$ for real $\alpha$. The previous exercise proved the statement for rationals and the exercise before that proved it for integers. This is not a question of computation but of generalization. Please take this into account.

My Attempt

I immediately thought of using double limits. Let $(q_k)_{k=1}^{\infty}$ be a sequence of rationals converging to $\alpha$. Then  $$ \lim_{x \to 1} \frac{x^{\alpha}-1}{x-1}  = \lim_{x \to 1}  \frac{\lim_{k \to \infty}(x^{q_k})-1}{x-1} = \lim_{k \to \infty} \lim_{x \to 1}  \frac{x^{q_k}-1}{x-1} = \lim_{k \to \infty} q_k = \alpha.$$ But this solution doesn't utilize the comparison principle or left and right limits or anything else Tao gives as hints except for the previous exercise. Furthermore, I'm not even sure if the limits are valid. Can any of you show me what's wrong with my solution (if it is wrong), or what Tao may have intended?

Update

Because of a helpful comment, I've realized that I can't abuse limits like I did above. But I'm out of ideas. I can't seem to think of anything for some reason. It's been bothering me for a hour. Any help would be appreciated!

答:
If you break the problem into approaching from the right and the left of 1, then the inequalities will be more clear I think.

Note that for $x>1$ $\frac{x^{\alpha}-1}{x-1}$ is increasing in the $\alpha$ variable so for $q>\alpha>p$ (p and q rational) the inequality:

$$\frac{x^{p}-1}{x-1} < \frac{x^{\alpha}-1}{x-1} < \frac{x^{q}-1}{x-1}$$

Then you may want to take a sequence approaching 1 from the right $x_{n}$ and with the comparison:

$$p = \lim\inf \frac{x_{n}^{p}-1}{x_{n}-1} \leq \liminf \frac{x_{n}^{\alpha}-1}{x_{n}-1} \leq  \limsup \frac{x_{n}^{\alpha}-1}{x_{n}-1} \leq \limsup\frac{x_{n}^{q}-1}{x_{n}-1} = q$$

So then liminf and limsup of $\frac{x_{n}^{\alpha}-1}{x_{n}-1}$ are trapped between any such p and q rational. This is enough to show that the distance between the liminf and limsup is zero and the limit and $\alpha$ is zero, you can choose sequences $p_{k}$ and $q_{k}$ converging to alpha if you want to show this.

Approaching from the left the order of the inequalities will be reversed but it is the same idea.


问:
Solve for the column vector (matrices within a matrix)?
Given that the $A_{11}$ and $A_{22} \in \mathbb{R}^{3 x 3}$ are invertible, $A_{21} \in \mathbb{R}^{3 x 3}$, and $b_1, b_2, x_1, x_2 \in \mathbb{R}^{3}$, then solve for $x_1$ and $x_2$ from

$\begin{bmatrix}
A_{11} & 0\\
A_{21} & A_{22}\\
\end{bmatrix}$ $\begin{align}
    \begin{bmatrix}
           x_{1} \\
           x_{2} \\
         \end{bmatrix}
  \end{align} =$ $\begin{align}
    \begin{bmatrix}
           b_{1} \\
           b_{2} \\
         \end{bmatrix}
  \end{align}$

What are $x_1$ and $x_2$ in terms of $A_{11}, A_{21}, A_{22}, b_1, b_2$?

答:
Split it up into two equations

$$ \begin{aligned}
A_{11} x_1 & = b_1 \\
A_{21} x_1 + A_{22} x_2 & = b_2
\end{aligned}$$

Solve for $x_1$ and in the first equation and use it in the second

$$ \begin{aligned}
 x_1 & = A_{11}^{-1} b_1 \\
 A_{22} x_2  & = b_2 - A_{21} A_{11}^{-1} b_1 \\
\end{aligned} $$

Solve for $x_2$

$$ x_2 = A_{22}^{-1} \left(b_2 - A_{21} A_{11}^{-1} b_1 \right) $$

Re-combine the solution

$$ \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix}
A_{11}^{-1} & 0 \\ -A_{22}^{-1} A_{21} A_{11}^{-1} & A_{22}^{-1}
\end{bmatrix}  \begin{bmatrix} b_1 \\ b_2 \end{bmatrix} $$


问:
Brauer–Nesbitt theorem -- Does Finite-dimensionality matter?
The Brauer–Nesbitt theorem that many literatures cite is stated as follows:

(BS.1): Let $KG$ be a group algebra of a finite group $G$ over a field $K$. Let $M,N$ be $KG$-modules. Suppose [A condition that will be quoted later], then $M$ and $N$ have the same composition factors iff for each $g \in G$, the matrices $M(g), N(g)$ have the same characteristic roots (counted according to their multiplicities).

This is quoted from (30.16) of:

[1]  Charles W. Curtis, Irving Reiner - Representation Theory of Finite Groups and Associative Algebras -John Wiley & Sons Inc (1962)

However, there is another Brauer–Nesbitt theorem that I met:

(BS.2) Two Galois representations $\rho, \rho^{\prime}: G_K \rightarrow GL_n(L)$ have isomorphic semisimplifications if and only $\rho(g), \rho^{\prime}(g)$ have the same characteristic polynomials for each $g \in G_K$. [Other claims omitted.]

This is quoted from (2.5) of this note, whose author said that this is (BS.1).

[2]  http://wwwf.imperial.ac.uk/~tsg/Index_files/ArizonaWinterSchool2013.pdf

Question 1: Since the absolute Galois group is profinite (not necessarily finite), how can we directly apply (BS.1) to show (BS.2)?

My attempts: I have read the proof quite thoroughly and feel that the finite-dimensionality of the group algebra $KG$ is essential. The finiteness is used to show that the characters of simple $FG$-modules are linearly independent, when $FG$ is split over $F$. (I used the Wedderburn-Artin Theorem to prove the linear independence)

I have also found a "purely algebraic" result on this, which seems to be quite broad:

(BS.3): Let $k$ be any field and $A$ a $k$-algebra. Let $M, N$ be two $A$-modules
which are finite-dimensional as $k$-vector spaces. If for all $a \in A$, the characteristic polynomials on $M$ and $N$ are equal, then $M$ and $N$ have the same composition factors.

We see that (BS.2) is a direct corollary of (BS.3). This is quoted from this note:

[3] https://math.uni.lu/~wiese/notes/GalRep.pdf

Question 2: In [3], the proof is essentially the same as the one in [1]. But in [3], the author said that "For this question we may and do assume that $A$ is a semisimple finite dimensional k-algebra". It seems that this claim is a gaint leap for me! If $A$ were a finite dimensional $k$-algebra, I know that by quotient out the Jacobson radical, $A$ could indeed be assumed as semisimple. BUT again, why can we assume that $A$ is a finite dimensional $k$-algebra?

Sorry for such a lengthy post, and sincerely hope that someone may help me out! Thank you for commenting or answering!

P.S. The condition omitted in (BS.1) in the braket "[]":

there exists an extension field $L$ of $K$ such that $L$ is a splitting field for $G$, with the property that if $V$ is a completely reducible $KG$-module, then $V^L$ is a completely reducible $LG$-module. (We may take $L=K$ if $K$ is already a splitting field.)

答:
With the OK of the OP, I'm just going to turn my comment into an answer.

The question that he is after is answered in Theorem 5.40 of this.

.


问:
How to calculate the Rotation Matrix from other known values?
Dear Math Stackexchange community, how can I compute $R$ rotation matrix in this equation ($P=Rp+t$)? (Please see the figure. More detailed explanation is under the figure.)

The equation is for the transformation of $p=p_1,p_2,p_3$ points. $p$ is in Oxyz (local) reference frame.     $p$ consists of 3 points of $p_1,p_2,p_3$. And $p_1,p_2,p_3$ each has 3 coordinates of $x,y,z$. Therefore p is 3x3 matrix. $p=\begin{bmatrix} p_{1x} & p_{2x} & p_{3x} \\ p_{1y} & p_{2y} & p_{3y} \\ p_{1z} & p_{2z} & p_{3z} \end{bmatrix}$

$t$ is the distance between Oxyz(local) and OXYZ(global) reference frame.  $t=\begin{bmatrix} t_x \\ t_y \\ t_z \end{bmatrix}$. To be able to do matrix operations, we can take $t$ as $t=\begin{bmatrix} t_x&t_x&t_x \\ t_y&t_y&t_y \\ t_z&t_z&t_z \end{bmatrix}$

$P$ is the coordinates of p_1, p_2, and p_3 in OXYZ global reference frame. $P$ is a 3x3 matrix.
$P=\begin{bmatrix} P_{1x} & P_{2x} & P_{3x} \\ P_{1y} & P_{2y} & P_{3y} \\ P_{1z} & P_{2z} & P_{3z} \end{bmatrix}$

$R$ is the (orthogonal) rotation matrix.
$R=\begin{bmatrix} r_{11}&r_{12}&r_{13} \\ r_{21}&r_{22}&r_{23} \\ r_{31}&r_{32}&r_{33} \end{bmatrix}$

Therefore, all in all:
$P=Rp+t$

$\begin{bmatrix} P_{1x} & P_{2x} & P_{3x} \\ P_{1y} & P_{2y} & P_{3y} \\ P_{1z} & P_{2z} & P_{3z} \end{bmatrix} = \begin{bmatrix} r_{11}&r_{12}&r_{13} \\ r_{21}&r_{22}&r_{23} \\ r_{31}&r_{32}&r_{33} \end{bmatrix}.\begin{bmatrix} p_{1x} & p_{2x} & p_{3x} \\ p_{1y} & p_{2y} & p_{3y} \\ p_{1z} & p_{2z} & p_{3z} \end{bmatrix}+\begin{bmatrix} t_x&t_x&t_x \\ t_y&t_y&t_y \\ t_z&t_z&t_z \end{bmatrix}$

We can derive the following form of the equation of  $P=Rp+t$

$(P-t)=R.p$

$(P-t).p^{-1}=R$

$p$ is an arbitrary matrix, and not always be taken of its inverse. I can't take the inverse of $p$, if last raw of $p$ is zero.

Let's look at an example to make it clear:
$P=Rp+t$

$p_1=(1,1,0)$ , $p_2=(2,4,0)$ , $p_3=(5,10,0)$ is the coordinates of three points. So in matrix form:
$p=\begin{bmatrix} 1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0 \end{bmatrix}$

$t=\begin{bmatrix} 0 \\ 0 \\ 50 \end{bmatrix}$

$\alpha=10^\circ $ (rotation around the x axis)

$\beta=15^\circ $ (rotation around the y axis)

$\gamma=20^\circ $ (rotation around the z axis)

$R_x=\begin{bmatrix} 1 & 0 & 0 \\ 0 & cos(\alpha) & -sin(\alpha) \\ 0 & sin(\alpha) & cos(\alpha) \end{bmatrix}$

$R_y=\begin{bmatrix} cos(\beta) & 0 & sin(\beta) \\ 0 & 1 & 0 \\ -sin(\beta) & 0 & cos(\beta) \end{bmatrix}$

$R_x=\begin{bmatrix} cos(\gamma) & -sin(\gamma) & 0 \\ sin(\gamma) & cos(\gamma) & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$R=R_z.R_y.R_x$

Therefore, for $\alpha=10^\circ $, $\beta=15^\circ $, $\gamma=20^\circ $, the rotation matrix is :

$R=\begin{bmatrix} 0.9077 & -0.2946 & 0.2989 \\ 0.3304 & 0.9408 & -0.0760 \\ -0.2588 & 0.1677 & 0.9513 \end{bmatrix}$

$P=R.p+t$

$P=$$\begin{bmatrix} 0.9077 & -0.2946 & 0.2989 \\ 0.3304 & 0.9408 & -0.0760 \\ -0.2588 & 0.1677 & 0.9513 \end{bmatrix}$$.\begin{bmatrix}  1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0  \end{bmatrix}$  $+\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 50 & 50 & 50 \end{bmatrix}$

$P=\begin{bmatrix} 0.6131 & 0.6370 & 1.5925 \\ 1.2712 & 4.4239 & 11.0597 \\ 49.9089 & 50.1533 & 50.3832 \end{bmatrix}$

$p_1=(0.6131,1.2712,49.90)$,$p_2=(0.6370,4.4239,50.1533)$,$p_3=(1.5925,11.0597,50.3832)$.

Now let's take opposite example with the same values to find $R$, with the values of $P,p,t$ as known.

$P=R.p+t$

$\begin{bmatrix} 0.6131 & 0.6370 & 1.5925 \\ 1.2712 & 4.4239 & 11.0597 \\ 49.9089 & 50.1533 & 50.3832 \end{bmatrix}=R.$ $\begin{bmatrix} 1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0 \end{bmatrix}+$ $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 50 & 50 & 50 \end{bmatrix}$

Could you find $R$ here in the above equation?

To repeat the formulation:

$P=Rp+t$

$(P-t)=Rp$

$(P-t).p^{-1}=R.p.p^{-1}$

$(P-t).p^{-1}=R$

For the inverse of $p$ (where $p$ is $\begin{bmatrix}  1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0  \end{bmatrix}$ ), I can't take the inverse of $p$.

Here I am stuck. It is interesting for me, because for the condition that $R,p,t$ is known, $P$ can be calculated(9 unknown elements in 3x3 matrix of $P$). However, for the condition that $P,p,t$ is known, I am having problem to compute $R$ (9 unknown elements in 3x3 matrix of $R$). I should be able to compute the equivalent $R$ matrix for that inverse problem.

Do you have any idea about how to find $R$ rotation matrix? Any suggestion would be  appreciated.

Thanks in advance.

答:
One of the most robust ways of finding the rotation matrix would probably be by using Wahba's problem, which in your case has the following associated cost function

$$
J(R) = \frac{1}{2} \sum_{k=1}^3 a_k \| P_k - t_k - R\,p_k\|^2, \tag{1}
$$

with $a_k > 0$ weights (but if you don't have any uncertainty in $P_k,t_k,p_k\in\mathbb{R}^3$ you could just set those to one). This cost function will have one global minima with $R\in SO(3)$ if the vectors $P_k-t_k$ or $p_k$ span at least a plane (which would be the case for your example).

The minimum of the cost function from $(1)$ can be found using

$$
B = \sum_{k=1}^3 a_k (P_k - t_k)\,p_k^\top. \tag{2}
$$

Now find the singular value decomposition of $B$ such that

$$
B = U\,S\,V^\top, \tag{3}
$$

where $U$ and $V$ are unitary matrices and $S$ a diagonal matrix. The rotation matrix $R$ that minimizes $(1)$ can be obtained using

$$
R = U
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & \det(U)\det(V)
\end{bmatrix} V^\top. \tag{4}
$$


问:
Factoring question on binomial
I'm trying to reduce the following expression.  The steps follow an online CAS and I have a question regarding a step.

$= (b-1)(\frac {1-b^n} {1-b})$

$= \frac {(1-b^n)(b-1)} {1-b}$

The next steps indicates cancellation.  Does anyone know what is being canceled?

$= -(-b^n + 1) = b^n - 1$

答:
I have put the cancelled parts in boldface so you can see them.

$$(b-1)\left(\frac {1-b^n} {1-b}\right)
=\frac{(b-1)(1-b^n)}{1-b}
=\frac{\mathbf{(b-1)}(1-b^n)}{-\mathbf{(b-1)}}=\frac{(1-b^n)}{-1}=(b^n-1)$$

Title: Factoring question on binomial


问:
How do we prove the proposed expression is nonnegative?
So the expression which I am interested in is given by
\begin{align*}
f(b,\theta) = \frac{2b-1}{b\sqrt{(1-\theta)(2b-1)^{2} + \theta}} + \frac{\sqrt{(1-\theta)(2b-1)^{2} + \theta} - 1}{2b^{2}(\theta - 1)}
\end{align*}
where $b\in(0,1]$, $\theta\in\mathbb{R}_{>0}$ and $\theta\neq 1$.

It is part of my research project and I would like to know a good way to approach it.

Any hint is appreciated.

EDIT

According to WA, the solution is given by $0 < b \leq 1$ when $\theta > 0$.

答:
$$
f(b,\theta)\geq 0
$$
\begin{align*}
 \implies\frac{2b-1}{b\sqrt{(1-\theta)(2b-1)^{2} + \theta}} + \frac{\sqrt{(1-\theta)(2b-1)^{2} + \theta} - 1}{2b^{2}(\theta - 1)}&\geq 0\\ \\
 \implies 2b-1+\frac{\left(b\sqrt{(1-\theta)(2b-1)^{2}+\theta}\right)\cdot\left(\sqrt{(1-\theta)(2b-1)^{2} + \theta} - 1\right)}{2b^{2}(\theta - 1)}&\geq 0\\\\
 \implies 2b-1+\frac{(1-\theta)(2b-1)^{2}+\theta-\sqrt{(1-\theta)(2b-1)^{2}+\theta}}{2b(\theta-1)}&\geq 0\\
 \implies 2b(\theta-1)(2b-1)+(1-\theta)(2b-1)^{2}+\theta-\sqrt{(1-\theta)(2b-1)^{2}+\theta}&\geq 0\\\\
\implies 2b\theta-2b+1-\sqrt{(1-\theta)(2b-1)^{2}+\theta}&\geq 0
\end{align*}
This implies the following :

First Case : $b=0$ which is not the case since $b\in(0,1]$.

Second Case : $0<b \leq 0.5$ and $\theta >\displaystyle \frac{4b^{2}-4 b+1 }{4  b^{2}-4  b}$

Third Case : $0.5<b\leq 1$ and $\theta> 0$

Hence it follows that $\forall b\in(0,1]$ and $\forall\theta\in\mathbb{R}_{>0}$ with $\theta\neq 1$ we have that $f$ is non-negative.


问:
How to express series summation notation in reverse?
Recently started learning about series summation.

For example

$\Large \frac{1}2+\frac{1}4+\frac{1}8+...=\sum_{n=1}^{\infty} (\frac{1}2)^n=1$

and can be visually proved with square area $1$

containing the infinite series summation from large to small areas:

However, is it possible to express a reverse series summation from small to large areas?

$\Large ...+ \frac{1}{2^n}+\frac{1}{2^{n-1}}+...+\frac{1}2=\sum_{n=\infty}^{1} (\frac{1}2)^n=1$

The idea is to convey meaning by notation. For example, emptying a square by half each step while the reverse is filling it.

Thanks.

答:
So you will have to start somewhere, with a first term, call it a.

Let's say the terms represent area as in the example, then they will be positive, and if they go from smaller to larger then you will expect all of the remaining terms to be bigger than the first term a.

This represents a problem since we expect that the sum of infinitely many terms  bigger than a lower bound a should diverge.

So for instance if you started with a square and kept doubling it, it would diverge.

All of that being said, the phrase 'express a reverse series' which you used is vague enough that it is hard to definitively answer no to the question.

If an expression such as the one you gave in the statement meets the criteria that you set out, then that would suggest that it can be done for any series, since you are just re-arranging a finite segment of the series to that the smallest are ordered first and having the usual trailing dots go from right to left instead of left to right and be located in the front.

We can surely rearrange how we write series so that the smallest terms are left most and the trailing dots go from right to left and are located in front just as you demonstrated there. The notation once it is defined and agreed upon is unambiguous.

The phrase 'small to large areas' indicates to me that the first interpretation is more along the lines of what you are thinking, in that case then no. If those areas are supposed to represent the partial sums of the sequence then they will move from small to large and the answer is yes.


问:
Prove that $\lfloor 5x\rfloor + \lfloor 5y\rfloor \ge \lfloor {3x+y}\rfloor + \lfloor {x+3y}\rfloor$
Prove that $\lfloor 5x\rfloor + \lfloor 5y\rfloor \ge \lfloor {3x+y}\rfloor + \lfloor {x+3y}\rfloor \,\,\forall x,y \ge 0$

My attempt so far:

Suppose that either $x$ or $y$ is greater than $0$.

Now $\lfloor 5x\rfloor + \lfloor 5y\rfloor \ge \lfloor 4x\rfloor + \lfloor x\rfloor +\lfloor 4y\rfloor + \lfloor y\rfloor \ge \lfloor 4x\rfloor + \lfloor 4y\rfloor +1\ge \lfloor 4x+4y\rfloor \ge \lfloor 3x+y\rfloor + \lfloor x+3y\rfloor$

For $x,y \in [0,1)$!

答:
Proof that
$\lfloor 5x\rfloor + \lfloor 5y\rfloor \ge \lfloor {3x+y}\rfloor + \lfloor {x+3y}\rfloor$
is FALSE.

Let $x=^-1$ and $y=0$ then the inequality above becomes $^-5 + 0 \ge ^-3 + ^-1$; which is FALSE.


问:
I have 16 features or terms and a maximum degree of 5 , how do I form unique polynomials , what is the number of unique polynomials
A number of possible combinations out of 16 variables and degree of polynomial being 5?

Combinations: If I have 16 variables and the max number of polynomial degrees is 5(i.e, 2, 3, 4 and 5 are included), how many possible unique combinations of polynomials can I form?

To begin with, I started with unique combinations of 16 variables which is 2^n = 65536, now when the polynomial degree is involved how will the number of unique combinations increase, i am out of ideas.

答:
You have 16 variables, call them $a,b,...,p$. Each can be degree 1,2,3,4, or 5. So these are the possible terms in your polynomial:

$a, a^2, a^3, a^4, a^5\\b, b^2, b^3, b^4, b^5\\...\\p, p^2,p^3,p^4,p^5$

of which there are $5*16=80$ of them.

Since each can be included or not included in the polynomial, you get $2^{80}=1.208926e+24$ possible polynomials. If you include the possibility of including a constant, you'd multiply this number by 2.


问:
Understanding how we can make $M$ into an $R[X]$-module by setting $Xm = f(m)$ for $m \in M.$
Here is the question I want to answer:

Let $M$ be a finite $R$-module. Show that if $f \in \operatorname {End_R(M)}$ is surjective then it is also injective.

Hint: Let $R[X] \cong R^{[1]}$ and make $M$ into an $R[X]$-module by setting $Xm = f(m)$ for $m \in M.$ Show that $XM = M$ and apply Nakayama's Lemma.

My question is:

1- I do not understand why setting $Xm = f(m)$ for $m \in M$ will make $M$ into an $R[X]$-module. Could anyone clarify this to me, please?

2- Also, I do not see how I will use surjectivity in the proof that $XM = M$. Could any one elaborate this to me please?

答:
Saying $M$ is an $R$-module is same as giving an abelian group $M$ along with a ring homomorphism $\psi: R\rightarrow \operatorname{End}_\mathbb Z (M)$. Since $f\in End_R(M)$, by the universal property of polynomial algebra $\exists! \ \Psi :R[X]\rightarrow \operatorname{End}_\mathbb Z (M)$ which extends $\psi$ and takes $X\mapsto f$. This makes $M$ an $R[X]$ module.

Note every $m\in M$ is of the form $f(m')=X.m'$ for some $m'\in M$.


Title: How to express series summation notation in reverse?

Body:
Recently started learning about series summation.

For example

$\Large \frac{1}2+\frac{1}4+\frac{1}8+...=\sum_{n=1}^{\infty} (\frac{1}2)^n=1$

and can be visually proved with square area $1$

containing the infinite series summation from large to small areas:

However, is it possible to express a reverse series summation from small to large areas?

$\Large ...+ \frac{1}{2^n}+\frac{1}{2^{n-1}}+...+\frac{1}2=\sum_{n=\infty}^{1} (\frac{1}2)^n=1$

The idea is to convey meaning by notation. For example, emptying a square by half each step while the reverse is filling it.

Thanks.

Body:
Geometric progression
$$\frac{a}{1-x}=a+ax+ ax^2+ax^3+....+..., ~~~|x|<1.$$
Here, $a=1/2$ and $x=1/2$
So $$\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....=\frac{1}{2} \frac{1}{1-1/2}=1=\sum_{n=1}^{\infty}\frac{1}{2^n}$$


问:
What is the definition of surjective morphism of schemes?
Let $f: X \to Y$ be a morphism of schemes, when talking about the surjectivity of $f$, there are at least several possibilities.

(1) $f$ is surjective at the level of sets, that is $\forall \  y \in Y$, there exist $x \in X$, such that $f(x)=y$.

(2) $f$ is surjective in the sense of category morphisms. This means for any scheme $Z$, and morphisms $g_1, g_2 : Y \to Z$ such that $g_1 \circ f=g_2 \circ f$ implies $g_1 =g_2$.

(3)$f$ is surjective at the level of schemes, i.e. $\overline{f(X)} = Y$. This seems unlikely to be correct definition of surjectivity, and I do not know how to make sense of the scheme structure of the closure of $f(X)$ (that is $\overline{f(X)}$). But if $Y$ is reduced, then there is no ambiguity.

My question is two folds:
(1) Which is "correct" definition of surjective morphism between schemes. (2) Since both definition 1 and 2 seems reasonable in some sense, are they equivalent? If $Y$ is reduced scheme, are three definitions equivalent?

答:
1) The correct definition of "surjective" for a morphism of schemes $f:X\to Y $ is that the underlying map of sets $|f|:|X|\to |Y|$ be surjective.

"Correct" means "as decreed by  Grothendieck" : Cf. EGAI, Chap. I, Prop 3.5.2 .

2) The  categorical  notion mentioned in your point (2) is called epimorphism.

3) If $\overline{f(X)} = Y$, we say that $f$ is dominant.

This is much weaker than surjectivity (even for reduced schemes), as witnessed by the inclusion of any  dense open strict subset  of a scheme, say $\mathbb A^1\setminus \{0\}\hookrightarrow \mathbb A^1$.

4) Surjective morphisms needn't be epimorphisms:

Let $k$ be a field and let $X=Spec(k)$, $Y=Spec(k[T]/(T^2))=k[\epsilon]$ be  respectively the simple point and the double point over $k$.

Let     $f:X\hookrightarrow Y$ be the closed immersion of the simple point into the double point corresponding to the $k$-algebra morphism $k[\epsilon]\to k$ sending $\epsilon$ to $0$.

The two $k$-algebra morphisms $k[\epsilon]\to k[\epsilon]$ sending $\epsilon$ to respectively $0$ and $\epsilon $ give rise to two morphisms $g_1, g_2:Y\to Y$ satisfying $g_1 \circ f=g_2 \circ f$ but  $g_1 \neq g_2$.

Hence the scheme morphism  $f:X\to Y$ is not an epimorphism although it is surjective,  as are all maps between  singleton sets.

5) Epimorphisms needn't be surjective (?):

Let $Y=\mathbb A^1_k$ (with $k$ an algebraically closed field) and $X$ be the discrete scheme obtained from the disjoint union (=coproduct) of all the closed points of $Y$.

The natural morphism $f:X\to Y$ has image $Y$ minus the generic point of $\mathbb A^1_k$ and is thus not surjective.

However it should be an epimorphism, but I haven't written out a proof of that.


问:
Prove that $ \lfloor 2x \rfloor \leq 2\lfloor x \rfloor + 1$

Theorem. For $x \in \mathbb R$, $$2 \lfloor x \rfloor \leq \lfloor 2x \rfloor \leq 2 \lfloor x \rfloor +1.$$

I tried to prove this theorem by first proving a few helper theorems. I have proved the following Lemma.

Lemma. For $x \in \mathbb R$ and $n \in \mathbb N$,
  $$n \leq x \Leftrightarrow n \leq \lfloor x \rfloor. $$

This allowed me to prove the first inequality, because
$$2 \lfloor x \rfloor = \lfloor x \rfloor  + \lfloor x \rfloor \leq x + x = 2x, $$
and because $2 \lfloor x \rfloor$ is integer, $2 \lfloor x \rfloor \leq \lfloor 2x \rfloor$ due to the Lemma above. However, I can not get a hold of the other inequality. Anyone any ideas? Note: I have seen approaches using the fact that $x = \lfloor x \rfloor + x_1$ with $x_1 \in [0,1)$, however I prefer to not take such an approach, and rather work with the properties described above.

答:
You will find a property here

https://en.wikipedia.org/wiki/Floor_and_ceiling_functions

that

$$\lfloor u \rfloor + \lfloor v \rfloor
  \le \lfloor u + v \rfloor
  \le \lfloor u \rfloor + \lfloor v \rfloor + 1$$

letting $u = v = x$, we get

$$ 2\lfloor x \rfloor
   \le \lfloor 2x \rfloor
   \le 2\lfloor x \rfloor + 1$$