问:
Sketching a set of $ 2 < |z| \leq |z + 2| < 4 $ on the complex plane.
So I need to sketch $ 2 < |z| \leq |z + 2| < 4 $ on the complex plane. At first, it seamed pretty easy since I know that:

$2 < |z|$ is a circle with a center in point $0,0$ and radius = 2

|z + 2| < 4 is a circle with a center in point $-2,0$ and radius = 4

BUT here comes that part $ |z| \leq |z + 2| $ and I actually have no idea how to connect those two.

Edit - my proposed solution:

$ |z| \leq |z + 2| \implies z^2 \leq z^2 + 4z + 4 \implies 0 = \leq 4z + 4 \implies -1 \leq z $

Is that correct? Is it the missing condition?

答:
HINT

You cannot square both sides as if $z$ were real. Indeed, if we set $z = x + yi$, one has that
\begin{align*}
|z| \leq |z+2| & \Longleftrightarrow |z|^{2} \leq |z+2|^{2}\\\\
& \Longleftrightarrow z\overline{z} \leq (z+2)(\overline{z} + 2)\\\\
& \Longleftrightarrow z\overline{z} \leq z\overline{z} + 2(z+\overline{z}) + 4\\\\
& \Longleftrightarrow 0 \leq 4x + 4\\\\
& \Longleftrightarrow x\geq -1
\end{align*}


问:
Derivation of Quantifier Negation: $-(\forall x)(Px \rightarrow -Qx) \vdash (\exists x)(Px \wedge Qx)$
I am working on a series of derivations demonstrating the equivalence of negated quantifiers, etc.

I've worked through a bunch, but am stuck on one that seems to have Demorgan's law in it when I get my proof going.

$-(\forall x)(Px \rightarrow -Qx) \vdash (\exists x)(Px \wedge Qx)$

From there, I started with the premise, and then assumed:

$-(\exists x)(Px\wedge Qx)$

And then to:

$-(Pa\wedge Qa)$

From there, I dont really know what steps I should take. I think I would like to get to contradict on each parts of the conjunction and end with showing a contradiction between $-(\forall x)(Px \rightarrow -Qx) \wedge (\forall x)(Px \rightarrow -Qx)$, but without the availability to freely use Demorgan's law, I feel like it would be more work to try to insert a sub-proof to break it apart.

Does anyone have any suggestions or know if I am going about this the right way at all?

答:
Your basic strategy of proving this through a proof by Contradiction is correct. However, after assuming $\neg \exists x (Px \land Qx)$, I am not so sure you are allowed to infer $\neg (Pa \land Qa)$  .... it depends on how the rules are defined in your derivational system.

Here is what i would do: prove that $\forall x (Px \to \neg Qx)$, so that you get a contradiction with the very premise.

So, assume that $a$ is an arbitrary object, assume $Pa$, and show that $\neg Qa$ using another proof by contradiction: the assumption of $Qa$ leads to $Pa \land Qa$, and thus to $\exists x (Px \land Qx)$, contradicting you original assumption.

ContentLicense: CC BY-SA 4.0


问:
Is this condition equivalent to being differentiable?
Consider a function $f : U \subset \mathbb R^n \to \mathbb R$ and the following two assertions:

$f$ is differentiable at $p \in U$, i.e., there exists a linear map $df_p : \mathbb R^n \to \mathbb R$ such that

$$\lim_{v \to 0} \frac {f(p + v) - f(p) - df_p(v)} {\Vert v \Vert} = 0$$

$f$ is continuous at $p$, and there exists a linear map $df_p : \mathbb R^n \to \mathbb R$ such that

$$df_p(v) = \lim_{t \to 0} \frac {f(p + tv) - f(p)} t$$

Clearly, the first assertion implies the second. Does the converse also hold?

答:
The map $f : \mathbb R^2 \to \mathbb R$ defined by

$$f(x,y)=
\begin{cases}
\frac{x^2y}{x^4+y^2}\sqrt{x^2+y^2} & (x,y)\neq (0,0)\\
0 & (x,y) =(0,0)
\end{cases}$$

is a counterexample. It is Gâteaux differentiable, its Gâteaux derivative at the origin is the vanishing map and is therefore linear.

However $f$ is not Fréchet differentiable at $(0,0)$. If it was, its Fréchet derivative would be equal to zero at the origin. However

$$\lim\limits_{t \to 0} \frac{f(t,t^2)}{\sqrt{t^2+t^4}} \neq 0$$

A contradiction.


问:
Vector Valued Function Velocity and Acceleration
$$a(t)=2j-\frac{1}{t^2}k$$
particle movement is observed for t ≥ 1 second

v(1) = 2i+2j+k

r(1) = 3i-2j

I've encountered this kind of problem but all the problem I've encountered the t is $0$ so how can I can solve this problem ?

What I've tried is $$v(t) = \int 2j-\frac{1}{t^2}k$$
$$v(t) = (2t)j\space+\frac{1}{t}k\space+C$$
from what i know the C here is $v_0$ but what i have is v1 so how can i start to solve this problem ?

Thank you in advance

答: Your constant does not have to be the value at zero, it depends on where you integrate from. In this case, you start from $t=1$ and so that is your initial condition. Per your comment, notice that you are already given $v(1)$ in your problem. Compare it with your $v(t)$ obtained and deduce what $C$ should be.


问:
Row Rank and Column vectors of Matrix

For matrix $A \in M_{m×n}$, let the rank be k and $r_1 , . . . , r_k$ be a subset of rows of A forming a basis of the row-space. How can we show that there exists column vectors $c_j$ such that $A =\sum^k_{j=1} c_j r_j$?

I know that $c_j r_j$ is a mxn matrix and we are adding up a series of $c_j r_j$ together. But I have little idea on how to show there are column vectors $c_j$. Any help or hint is appreciated.

答:
HINT: Let $r_i$ be the ith row of $A$, and let $c_i^*$ be a column vector with $m$ entries consisting of a $1$ in the ith entry and a $0$ everywhere else. Can you show that this is true?

$$A = \sum_{i=0}^m c_i^{*} r_i$$

Once you see why this is true, can you figure out a way to use this fact to solve your problem?

Title: Row Rank and Column vectors of Matrix


问:
Intuitive understanding of why the sum of nth roots of unity is $0$
Wikipedia says that it is intuitively obvious that the sum of $n$th roots of unity is $0$.  To me it seems more obvious when considering the fact that $\displaystyle 1+x+x^2+...+x^{n-1}=\frac{x^n-1}{x-1}$ and that $\{\alpha \in \mathbb{C}: \alpha^n=1\}$ is a cyclic group.

For me, it is hard to, intuitively, see this in the case that $n$ is an odd composite natural number.

Here is my reasoning thus far:

Let $n=5$. (I understand this is simple when considering the minimal polynomial).

Since the conjugate of an $n$th root of unity is also an $n$th root of unity, it is easy to see that, when written in trigonometric form, the sum (possible reordering here)

$1+\alpha^1+\alpha^2+\alpha^3+\alpha^4$

$=1+( \cos (72)+i \sin (72))+( \cos (144) + i \sin (144))+( \cos (72)-i \sin (72))+( \cos (144)-i \sin (144))$

$=1+2 \cos (72)+2 \cos (144)$.

Somehow $2 \cos (72)+2 \cos (144)=-1$, but I'm not sure this is obvious, or perhaps I'm being silly here.

What do you think?  Silly?

答: The $n$ roots of unity of order $n$ (all of them, not just the primitive ones) are equally spaced around the unit circle. If their sum was nonzero, it would have an argument (angle relative to the $x$ axis) which would be a violation of symmetry. Therefore, by symmetry, the sum must be 0.</p>

ContentLicense: CC BY-SA 3.0


问:
$d\mid p\!-\!1\Rightarrow x^d-1\pmod{\!p^2}$ has exactly $d$ roots for odd prime $p$
I'm trying to figure out the number of solutions to the congruence equation $x^d \equiv1 \pmod{p^2}$ where $p$ is prime and $d\mid p-1$.

For the congruence equation ${x^d}\equiv1 \pmod p$ where $p$ is prime and $d\mid p-1$ I've shown that there are exactly $d$ solutions modulo $p$.

I'm trying to use the above result to extend it to the higher power. I'm aware of something called Hensel's Lemma which says if a polynomial has a simple root modulo a prime  $p$, then this root corresponds to a unique root of the same equation modulo any higher power of $p$. We can 'lift' the root iteratively up to higher powers.

I'm unsure exactly how this process works. All I have to start with is that I assume a solution of the form $m+np$ solves the 'base' congruence and I'm trying to somehow extend it to $p^2$.

Any help would be appreciated.
Thanks.

答:
Let $p$ be an odd prime, and work modulo $p^k$ (so $k$ does not have to be $2$). Then if $d$ divides $p-1$, the congruence $x^d\equiv 1\pmod{p^k}$ has exactly $d$ solutions.

To prove this, we use the fact that there is a primitive root $g$ of $p^k$, that is, a generator of the group of invertibles modulo $p^k$. Then for $1\le n\le (p-1)p^{k-1}$, we have $(g^n)^d\equiv 1\pmod{p^k}$ if and only if $nd$ is a multiple of $(p-1)p^{k-1}$. There are $d$ different such $n$, namely $i\frac{p-1}{d}p^{k-1}$ for $i=1,2,\dots, d$.


问:
Contour integration to evaluate a real-valued integral
I am evaluating this integral:

$$\int_{-\infty}^{\infty} \frac{\sin x}{x(x^2+1)^2}\,dx$$

with the formula

$$\int_{-\infty}^{\infty} f(x) \sin(sx) dx = 2\pi \sum\text{Re } \text{Res}[f(z) e^{isz}]$$

where the sum is over the residues in upper half plane.

So since the only two singularities that are inside the upper half plane are at $z = 0$ and $z=i$, I found that

$$\begin{align}
2\pi \sum\text{Re } \text{Res}\left(f(z) e^{isz}\right) &= 2 \pi \left(\text{Re }  \text{Res}_{z= 0}\left[\frac{1}{z (z^2+1)^2} e^{i z}\right]
+ \text{Re } \text{Res}_{z= i}\left[\frac{1}{z (z^2+1)^2} e^{i z}\right]\right) \\\\
  &=2 \pi \left(1 + \frac{-3}{4e}\right)
\end{align}$$

I am pretty sure that I calculated the two residues correctly, since in mathematica

```<code>Residue[E^(I z)/(z (z^2 + 1)^2), {z, 0}]
</code>```

is $1$ and

```<code>Residue[E^(I z)/(z (z^2 + 1)^2), {z, I}]
</code>```

is $\frac{-3}{4e}$

But evaluating the integral

```<code>Integrate[Sin[x]/(x (x^2 + 1)^2), {x, -Infinity, Infinity}]
</code>```

mathematica gives $\pi - \frac{3 \pi}{2e}$.

I am wondering if this is because I did something wrong somewhere or if it is because mathematica gives the wrong answer.

Thank you!

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{\sin\pars{x} \over x\pars{x^{2} + 1}^{2}}\,\dd x} =
\Im\int_{-\infty}^{\infty}{\expo{\ic x} - 1\over x\pars{x^{2} + 1}^{2}}\,\dd x
\\[5mm] = &\
\Im\braces{2\pi\ic\,\lim_{x \to \ic}\,\totald{}{x}
\bracks{\pars{x - \ic}^{2}\,{\expo{\ic x} - 1\over x\pars{x^{2} + 1}^{2}}}}
\\[5mm] = &\
2\pi\,\Re\braces{\lim_{x \to \ic}\,\totald{}{x}
\bracks{{\expo{\ic x} - 1\over x\pars{x + \ic}^{2}}}}
\\[5mm] = &\
2\pi\,\Re\bracks{\lim_{x \to \ic}\
{\expo{\ic x}\pars{\ic x^{2} - 4x - \ic} + 3x + \ic \over
x^{2}\,\pars{x + \ic}^{3}}}
\\[5mm] = &\
\bbx{\pi - {3\pi \over 2\expo{}}} \approx 1.4080
\\ &
\end{align}


问:
Solution for a matrix commutation
I am new to linear algebra and having trouble with the matrix commutation logic below.

Given, $(\mu - \gamma)'\Sigma^{-1}1 = \lambda$ where $\lambda \in R$ and $\mu, 1, \gamma \in R^N$ and $\Sigma^{-1}$ is a NxN matrix.

I am trying to express $\gamma \text{ in terms of } \mu, \Sigma^{-1}, \lambda$.

Is this achievable and if yes, any advise on how I should proceed?

Thanks in advance.

答:
I think you can't. Here is what I tried and why I thought that it is impossible.

I assume $\gamma, \mu$ to be a 1$\times$n matrix, 1 is a n$\times$1 matrix whose entries are identically 1 and  $\Sigma^{−1}$ is an invertible matrix.

Now we can manipulate given formula as this :

$\mu\Sigma^{−1}1 - \gamma\Sigma^{−1}1=λ$
$\Rightarrow \gamma\Sigma^{−1}1 = \mu\Sigma^{−1}1 - λ$

And if I could find a 1$\times$n matrix $\alpha$ such that $\Sigma^{−1}1 \times \alpha = I$(here I put $\times$ to emphasize that it is a matrix multiplication, and $I$ represents n$\times$n identity matrix), then by right multiplying it both sides, we get
$\gamma = (\mu\Sigma^{−1}1 - λ)\alpha$.

But this is impossible.
Suppose there is such $\alpha$, that is, $\Sigma^{−1}1 \alpha = I$. Then since $\Sigma^{−1}$ is invertible matirx, $1 \alpha = \Sigma I = \Sigma$.
Hence $1 \alpha$ has to be an invertible matirx. But $rank(1 \alpha) = 1$ or $0$ for any 1$\times$n matrix $\alpha$. This leads to a contradiction unless you are dealing with 1$\times$1 matrices, which is just an ordinary real number arithmetic.

Title: Solution for a matrix commutation


问:
Prove that for any prime $p,$ there exists a positive integer $n$ such that $1^n+2^{n-1}+3^{n-2}+\dots+n^1\equiv 2020\pmod{p}.$
Prove that for any prime $p,$ there exists a positive integer $n$ such that
$1^n+2^{n-1}+3^{n-2}+\dots+n^1\equiv 2020\pmod{p}.$

Can I have hints for this problem? What I tried was finding $n$ for every prime, but I couldn't get any explicit formula.

答:
Sequence of hints, which mostly follows how I approached this problem based on wishful thinking that an olympiad problem has a nice solution, and also some backward deduction from the problem statement. You can reveal each part, though I encourage you to think about what the next part is first.

(Wishful thinking: $n$ is very likely to be a multiple of $p-1$, $p$, or $p+1$. This is a reasonable assumption, but not needed.)

Wishful thinking: 2020 is likely a red herring. It should work for all values $M$.

Wishful thinking: Knowing this suggests that we want to string multiple sequences together. To clarify what this means, we want $m$ consecutive sequences of $n$ terms to give us:

$$ \sum_{i=1}^{mn} i^{mn+1-i} \equiv m \times (\sum_{i=1}^n i ^{n+1 - i }) \pmod{p} \quad \quad (1)$$

Obviously this is very unique, so what values of $n$ could work?

 Since the bases repeat every $p$ times (mod $p$), and the powers repeat every $p-1$ times (by FLT), so we have $ i ^ {j+ p(p-1) } \equiv (i+p(p-1)) ^ j \pmod{p}$, and thus $(1)$ holds. This suggests $n = k p(p-1)$.

 Wishful thinking: If for some $k$ we have $ \sum_{i=1}^n i ^{n+1 - i } = N \neq 0 \pmod{p}$, then we just need to solve $mN \equiv M \pmod{p}$, or that $m = MN^{-1}$.

 Let's investigate $k=1$.

 Show that $$ \sum_{i=1}^{p(p-1)} i ^{n+1 - i }  \equiv \sum_{i=1}^{p-1} \sum_{j=0}^{p-2} i^j.$$

 This sum is much easier to deal with, e.g. using a geometric progression.

 As it turns out, the sum is $ \equiv -1 \neq 0$.

Hence we are done.


Body:
Prove that for any prime $p,$ there exists a positive integer $n$ such that
$1^n+2^{n-1}+3^{n-2}+\dots+n^1\equiv 2020\pmod{p}.$

Can I have hints for this problem? What I tried was finding $n$ for every prime, but I couldn't get any explicit formula.

Body:
Hopefully, I didn't make any stupid mistakes.

Let $$f(n)\equiv 1^n+2^{n-1}+3^{n-2}+\dots+n^1 \pmod{p}$$

Note that
$$f(n) \equiv (p+1)^n+(p+2)^{n-1}+(p+3)^{n-2}+\dots+(p+n)n^1 \pmod{p}
 $$
From where you can deduce that
$$f(n+p) \equiv 1^{n+p}+2^{n+p-1}+...+(p-1)^{n+2}+f(p) \pmod{p} \\
\equiv 1^{n+1}+2^{n}+...+(p-1)^{n-p+3}+f(p) \pmod{p}$$

Prove by induction on $k$ that
$$f(n+kp) \equiv 1^{n+p}+2^{n+p-1}+...+(p-1)^{n+2}+f(p) \pmod{p} \\
\equiv 1^{n+1}+1^{n+2}+...+1^{n+k}+2^{n}+2^{n+1}+...+2^{n+k-1}...+(p-1)^{n-p+3}+...+(p-1)^{n-p+k+2}+f(p) \pmod{p}$$

Next, note that for all $2 \leq k \leq p-1$ and all $n$ we have
$$k^m+k^{m+1}+...+k^{m+p-2}=(k-1)^{-1}k^m(1-k^{p-1})=0$$
since $k-1$ is invertible.

Moreover,
$$1^m+1^{m+1}+....+1^{m+p-2} \equiv -1 \pmod{p}$$

Deduce from here that
$$f(n+(p-1)p) \equiv f(n)-1 \pmod{p}$$
for all $n$. The conclusion follows immediatelly.


问:
How to transform quadratic terms to something like $u^2+v^2$?
I am told that by the substitution of $x = (a\cos \theta)u - (b\sin \theta) v$ and $y = (a\sin \theta)u + (b\cos \theta)v$, we can reduce $3x^2 + 2xy + 3y^2$ to $u^2 + v^2$.

I tried to plug in the substitutions, but I did not see how the quadratic term is reduce to $u^2 + v^2$. What I got was something complicated containing $a$, $b$, $\cos \theta$ and $\sin \theta$.

I am asked to reduce $3x^2 + 2xy + 3y^2 - x - 2y$ to $u^2 + v^2$ by similar substitutions. I do not know how to construct this substitution. Could you explain how do find this kind of substitutions in general?

答:
I haven't checked your algebra, but your second-to-last form looks promising. It is a quadratic expression involving only $u^2$, $v^2$, and $uv$, with coefficients in terms of $a$, $b$, and $\theta$.

Here's the idea: You are free to choose any value for the three parameters $a$, $b$, and $\theta$. So you can get rid of the $uv$ term by a suitable choice of $\theta$ that makes the coefficient zero. [It's better to not factor $\cos^2\theta-\sin^2\theta$ into $(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)$.] Having made your choice for $\theta$, you now get to choose $a$ to force the coefficient on $u^2$ to be $1$, and you can choose $b$ to force the coefficient on $v^2$ to be $1$.


问:
Easy Inequality Problem
Solve the given inequality by interpreting it as a statement about distances on the real line.

$$\lvert x+1\rvert  > \lvert x-3\rvert$$

I am confused on what this question is asking. Can anyone give me a quick explanation and answer?

Please show numerically and graphically.

Thanks

答:
They want you to find all values of $x$ for which $|x+1| > |x - 3|$.

The method used to solve the problem should involve interpreting $|x+1|$  as the distance from $x$ to $-1$ on the real number line, and interpreting $|x - 3|$ as the distance from $x$ to $3$ on the real number line.

Title: Easy Inequality Problem


问:
For a given transformation $T: V \rightarrow W$ why does $im(T) \cong V/ker(T)$
I was looking at this example on Wikipedia, but unfortunately I don't quite see the connection. I am mostly trying to understand it through this picture:

Does anyone have a better explanation? I think the main part that is throwing me off is the exclusion of ker(T).

答:
For any $w\in W$, any two elements of $T^{-1}(w)$ 'differ by' an element of the kernel.  That is, let $w=T(v_1)=T(v_2)$.  Then by linearity $T(v_2-v_1)=0$.  In other words, $v_2-v_1\in\rm{ker}T$.

Conversely, if you add any element of the kernel to a particular solution of $Tx=w$, you get another solution.

This may help with some intuition.

Incidentally, this one has a special name:  the first isomorphism theorem.


问:
Hartshorne's Exercise II.4.5(c). A third time.
It seems that I am the third one to pose this question. I think the proof in this question has a gap. He have not shown the lifting he constructed is compatible with $\mathrm{Spec}k(p) \rightarrow X$. And the proof given in wikiproof have not shown why Z satisfies the property as described in the exercise, which in my opinion is the most key part.

Any way to fill the gap is admired.

答:
Let me give an outline of proof, which is from my teacher. Indeed my knowledge is insufficient to fill the proof. I can just tell the geometry meaning of the proof.

First we can choose a completion of $X$, denoting by $\tilde{X}$. Then $K(X) = K(\tilde{X})$ (or they are similar. In fact, I don't know what exactly happens when taking completion. Maybe something like blowing up at $Z$ below. But I don't know blowing up neither.). Now we have a diagram
$$\require{AMScd}
\begin{CD}
\mathrm{Spec}(L) @>>> X;\\
@VVV @VVV \\
\mathrm{Spec}(S) @>>> \mathrm{Spec}(k).
\end{CD}
$$
The image of $\mathrm{Spec}(L)$ gives an irreducible closed subset $Z$ of $X$, whose generic point is $z$. We can pull back $S$ to be a valuation ring $S'$ of $k(z)$, and subsequently pull back $S'$ to be a ring in $\mathcal{O}_{X,z}\subset K(X)$, namely $S''$. Taking the maximum with respect to domination, we have a valuation ring $R$ of $K/k$, whose restriction to $\mathcal{O}_{X,z}$ is $S''$. Now $R$ dominates some $\mathcal{O}_{X,x}$ by hypothesis.

The problem is $x$ may not in $Z$. But for the commutative diagram
$$\require{AMScd}
\begin{CD}
\mathrm{Spec}(L) @>>> \tilde{X};\\
@VVV @VVV \\
\mathrm{Spec}(S) @>>> \mathrm{Spec}(k),
\end{CD}
$$
and because of the properness of $\tilde{X}$, we always have a lifting. By doing the same thing above, we have $\tilde{x} \in \tilde{Z}$, and $\mathcal{O}_{\tilde{X},\tilde{x}}$ is dominated by $R$. So as the counterpart corresponding to $\tilde{x}$, $x$ is in $Z$.


问:
Prove that if $f$ is continuous and $f$ is equal to its integral from 0 to $x$, then $f(x) = 0$ for all $x$
The question is basically we have a continuous function $f:R\to R$ in which we have $$f(x)=\int_0^xf$$ for all real x, and it asks us to show that $f(x) = 0$ for all x real.

The major drawback for me is that I only have theorems about derivatives and Taylor/power series (without the use of integrals), some integral propeties (like the ones which relates absolute values, sums, subtractions, I haven't got theorems about change of variable or integration by parts, for example) and the first statement from the Fundamental Theorem of Calculus to use: (no mention to differential equations nor the second statement of the FTC)

If $f:[a,b]\to R$ is a continuous function and $F:[a,b]\to R$ is defined as $$F(x) = \int_a^xf$$ for all $x\in[a,b]$. Then we have $F'(x) = f(x)$ for all $x\in (a,b)$.

All those integrals are defined with the Darboux definition which takes into account $sup$ and $inf$ of $f$ in each interval that we partition and then take the $sup$ and $inf$ of the lower and upper sums to calculate the integral. (So we're not using the Rienmann definition)

I thought that I could use the fact that the function is infinitely-derivable since $f'(x) = f(x)$ because of the first part of FTC, then $f^{(n)}(x) = f(x)$ and, since $f(0) = 0$ (because it's a degenerate interval for the integral), we'd have $f^{(n)}(0) = 0$ for all natural $n$, then I thought that I could use the Taylor series for the function $f$ around $x=0$ which would be simply $T(x) = 0$ for all real $x$, but then I'd need to proof that $f$ is analytic for all real $x$, which I don't know how to... And the only theorem that my book has about that idea of being analytic is this one:

Let $f:I\to R$ be an infinitely-derivable function in a interval $I$ which contains $0$ in its interior. Let $(-r,r)$ be the biggest open interval in that form such that it's contained in $I$ and such that for each $c$ with $0<c<r$ we have: $$\lim_{n\to\infty}\left[\frac{1}{n!}M_n(c)c^n \right] = 0,$$ where $M_n(c)$ is the maximum of the function $f^{(n)}(x)$ in the interval $[-c,c]$. Then the Taylor series of the function $f(x)$, relatively to $x=0$, converges to $f(x)$ in the interval $(-r,r)$.

But honestly, I don't exactly know how to proceed with my idea of proof using that theorem... any idea would be very welcome and appreciated!

答:
Hint: First show that it is zero on any interval $[0,b]$.

We have $f(0)=0$ and $f'(x)=f(x)$ so $|f'(x)|\leq |f(x)|$. Now this is an exercise in Rudin's Analysis book.

Let $M_0=\sup|f(x)|$ and $M_1=\sup|f'(x)|$ on $[0,b]$. Fix $x_0\in[0,b]$. For any $x\in[0,x_0]$,
\begin{align}
|f(x)|=|f(x)-f(0)|\stackrel{MVT}{=}|f'(c)|x\leq x_0|f(c)|\leq x_0M_0
\end{align}
Taking the supremum on the left, we have $M_0 \leq x_0 M_0$, which either implies $M_0=0$, if you ensure that $x_0<1$. You can repeat the proof, starting at $x_0$ until you fill up the whole interval. Since $b$ is arbitrary, the claim holds on $[0,\infty)$. A similar proof holds in the other direction.

EDIT: This assumes you don't know the function $e^x$. Obviously, The solution to $f'(x)=f(x)$ is $f(x)=Ce^x$ and $f(0)=0$ ensures $C=0$.


问:
Verify Functional Limit of $\lim_{h \to 0} \frac{\int_{1-h}^{1+h} f(x) \,dx}{h}$
I want to find $$\lim_{h \to 0} \dfrac{\displaystyle\int_{1-h}^{1+h} f(x) \,dx}{h}$$. Where $f$ is a continuous function defined on $\mathbb{R}$

What I have so far:

Since $$\lim_{h \to 0} \int_{1-h}^{1+h} f(x) \,dx = 0 \text{ and } \lim_{h \to 0} h = 0$$ we can use L'Hopital's rule.

$\dfrac{d}{dh}h =1$ and $\frac{d}{dh} \displaystyle\int_{1-h}^{1+h} f(x) \,dx = f(1+h)+f(1-h)$. (Is this part right?).

Then from that we apply L'Hopital's rule:

$$\lim_{h \to 0} \frac{\displaystyle \int_{1-h}^{1+h} f(x) \,dx}{h} = \lim_{h \to 0} f(1+h)+f(1-h) = f(1) +f(1)=2f(1)$$

Does this look right? Thanks!

答:
Checking the part you aren't sure: $$\frac{d}{dh} \int_{1-h}^{1+h} f(x) \,dx = \frac{d}{dh}\left(\int_{0}^{1+h} f(x) \ dx - \int_{0}^{1-h} f(x) \ dx\right) = \frac{d}{dh}\int_{0}^{1+h} f(x) \ dx  - \frac{d}{dh} \int_{0}^{1-h} f(x) \ dx = f(1+h) - (-f(1-h)).$$

Obtaining thus the result that you want. Then, using the L'hopital rule that you're mentioning, the result follows. Your answer is correct.


答:
The answer is $x=\frac{-1}{2}$, $b=-\frac{1}{\sqrt{2}}$, $\theta=\frac{\pi}{4}$ and
$$
x=(a\cos\theta)u-(b\sin\theta)v\textrm{, }y=(a\sin\theta)u+(b\cos\theta)v.\tag 1
$$
Setting this into $3x^2+2xy+3y^2$ you get $u^2+v^2$.

By demanding $kx^2+lxy+my^2$ to be $u^2+v^2$, you get after some calculations
$$
a=\frac{\pm1}{\sqrt{k\cos^2 \theta+l\cos \theta \sin \theta+m \cos^2\theta}}
$$
$$
b=\frac{\pm1}{\sqrt{m\cos^2 \theta-l\cos \theta \sin \theta+k\sin^2 \theta}}
$$
$$
l\cos 2\theta+(m-l)\sin 2\theta=0
$$

Title: How to transform quadratic terms to something like $u^2+v^2$?

Body:
I am told that by the substitution of $x = (a\cos \theta)u - (b\sin \theta) v$ and $y = (a\sin \theta)u + (b\cos \theta)v$, we can reduce $3x^2 + 2xy + 3y^2$ to $u^2 + v^2$.

I tried to plug in the substitutions, but I did not see how the quadratic term is reduce to $u^2 + v^2$. What I got was something complicated containing $a$, $b$, $\cos \theta$ and $\sin \theta$.

I am asked to reduce $3x^2 + 2xy + 3y^2 - x - 2y$ to $u^2 + v^2$ by similar substitutions. I do not know how to construct this substitution. Could you explain how do find this kind of substitutions in general?


答:
Ultimately, we want to solve the inequality. However, dealing with multiple absolute value signs is a headache to do algebraically, so the question statement is asking you to consider a geometric approach.

Remember that $|x|$ means the distance of $x$ from zero, and that $|x-A|$ can be interpreted as meaning the distance of $x$ away from $A$.

So, for example, $|x+1|>B$ means that $x$ is a distance of more than $B$ away from $-1$. In other words, to get from $-1$ to $x$, you have to travel at least $B$ units (maybe forward or maybe backward).

Can you use this to help settle your question?

Title: Easy Inequality Problem

Body:
Solve the given inequality by interpreting it as a statement about distances on the real line.

$$\lvert x+1\rvert  > \lvert x-3\rvert$$

I am confused on what this question is asking. Can anyone give me a quick explanation and answer?

Please show numerically and graphically.

Thanks


答:
Alternative solution

Let $g(x) = \displaystyle\int\limits_{0}^{x} f(t) \ dt$, then $\displaystyle\int\limits_{1-h}^{1+h} f(x) \ dx = g(1+h)-g(1-h)$. Therefore, $\lim\limits_{h\rightarrow 0}\frac{\int_{1-h}^{1+h}f(x)\ dx}{h}=\lim\limits_{h\rightarrow 0}\frac{g(1+h)-g(1-h)}{h}=\lim\limits_{h\rightarrow 0}\left(\frac{g(1+h)-g(1)}{h}+\frac{g(1)-g(1-h)}{h}\right)=2g'(1)=2f(1).$

I think Paresseux has given the most beautiful solution, though.

Title: Verify Functional Limit of $\lim_{h \to 0} \frac{\int_{1-h}^{1+h} f(x) \,dx}{h}$

Body:
I want to find $$\lim_{h \to 0} \dfrac{\displaystyle\int_{1-h}^{1+h} f(x) \,dx}{h}$$. Where $f$ is a continuous function defined on $\mathbb{R}$

What I have so far:

Since $$\lim_{h \to 0} \int_{1-h}^{1+h} f(x) \,dx = 0 \text{ and } \lim_{h \to 0} h = 0$$ we can use L'Hopital's rule.

$\dfrac{d}{dh}h =1$ and $\frac{d}{dh} \displaystyle\int_{1-h}^{1+h} f(x) \,dx = f(1+h)+f(1-h)$. (Is this part right?).

Then from that we apply L'Hopital's rule:

$$\lim_{h \to 0} \frac{\displaystyle \int_{1-h}^{1+h} f(x) \,dx}{h} = \lim_{h \to 0} f(1+h)+f(1-h) = f(1) +f(1)=2f(1)$$

Does this look right? Thanks!


问:
Integrals of powers of damped saw-tooth wave
Through a rather convoluted set of steps, I have been able to derive

$$\int_0^\infty \Big(\frac{\arcsin(\sin(t))}{\cosh(t)}\Big)^2dt = \frac{1}{2}\big(\frac{\pi^2}{3} - \pi\log{2} \big)$$

$$\int_0^\infty \Big(\frac{\arcsin(\sin(t))}{\cosh(t)}\Big)^4dt = \big(\frac{\pi^2}{6}-1\big)\big(\frac{\pi^2}{4} - \pi\log{2} \big) $$

I'm looking to extend the sequence to even powers of the integrand.  My current method cannot be continued because it relied on closed-form expressions for
$$ \sum_{m=1}^\infty \frac{(-1)^m}{m^{2k-1} \sinh{(\pi m x)} } \ , \ k=0, 1, 2 \ \text{ and } x=1. $$
As far as I'm aware, closed-form solutions for the previous equation with $k=3,4,...$ are not available. My method did not involve contour integration, which is something I infrequently use, so I'm hoping that more sophisticated methods might produce further examples.

答:
I assume the following approach is different from your approach since it's fairly straightforward.

Unfortunately it does not avoid having to evaluate alternating series involving the hyperbolic sine.

Taking the first integral as an example, we can expand $$s(x)= x^{2},  \ -\frac{\pi}{2} \le x \le \frac{\pi}{2} ,$$ in a Fourier series, replace $\left(\arcsin(\sin x)\right)^{2} $ with this series, switch the order of summation and integration, and then integrate.

Let's first find the coefficients for the Fourier series of $s(x)$.

By definition, $$a_{0} = \frac{2}{\pi} \int_{-\pi/2}^{\pi/2} x^{2} \,  \mathrm dx = \frac{\pi^{2}}{6},$$

$$a_{n}= \frac{2}{\pi} \int_{-\pi/2}^{\pi/2} x^{2} \cos(2nx) \, \mathrm dx = \frac{2}{\pi} \frac{\pi}{2n^{2}} \,  \cos(\pi n)= \frac{(-1)^{n}}{n^{2}}, $$

and $$b_{n}= \frac{2}{\pi}\int_{-\pi/2}^{\pi/2} x^{2} \sin(2nx) \, \mathrm dx =  0.$$

Therefore, $$s(x) = \frac{\pi^{2}}{12} + \sum_{n=1}^{\infty} \frac{(-1)^{n}\cos(2nx)}{n^{2}}.$$

Replacing $\left(\arcsin(\sin x) \right)^{2} $ with this Fourier series and then changing the order of summation and integration, we get $$\int_{0}^{\infty} \frac{\left(\arcsin(\sin x) \right)^{2}}{\cosh^{2}(x)} \, \mathrm dx = \left(\frac{\pi^{2}}{12} \int_{0}^{\infty} \frac{\mathrm dx}{\cosh^{2}(x)} + \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} \int_{0}^{\infty} \frac{\cos(2nx)}{\cosh^{2}(x)} \, \mathrm dx\right) . $$

The first integral is simple to evaluate since $\tanh(x)$ is an antiderivative of $\operatorname{sech}^{2}(x)$.

To evaluate the second integral, we can integrate the function $$f(z) = \frac{e^{2inx}}{\cosh^{2}(x)}$$ around a rectangular contour in the upper half of the complex plane of height $i \pi$.

We get $$\int_{-\infty}^{\infty} \frac{e^{2inx}}{\cosh^{2}(x)} \,  \mathrm dx - e^{-2 n \pi}\int_{-\infty}^{\infty} \frac{e^{2inx}}{\cosh^{2}(x)} \, \mathrm dx = 2 \pi i \operatorname{Res} \left[f(z), \frac{i \pi}{2}\right] = 2 \pi i   \left(2in   e^{- \pi n} \right).$$

Then equating the real parts on both sides of the equation, we get  $$\int_{-\infty}^{\infty}\frac{\cos(2nx)}{\cosh^{2}(x)} \, \mathrm dx = 2 \pi n \, \frac{2e^{- \pi n}}{1-e^{-2 \pi n}} = 2 \pi n \, \frac{2}{e^{\pi n}-e^{- \pi n}} = \frac{2 \pi n}{\sinh (\pi n)}. $$

Therefore, $$ \int_{0}^{\infty} \frac{\left(\arcsin(\sin x) \right)^{2}}{\cosh^{2}(x)} \, \mathrm dx = \frac{\pi^{2}}{12} + \pi  \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n \sinh(\pi n)}.$$

I asked about that particular series (and related series) here.


问:
Colouring $2n$ Points in $\mathbb{R}^2 $
Definition: A strip is the area in the plane enclosed between some two parallel lines. An axis-
parallel strip is a strip whose bounding lines are parallel to one of the axes.

Let $P$
be a set of $2n$ points in the plane (for some integer $n$). Prove that the points of $P$
can be colored with 'red' or `blue' such that in any axis-parallel strip the difference
between the number of red points and the number of blue points is at most 2.

The direction I was thinking about is definig monotone non-decreasing seq of the points, relatively to the $x$-axis, and another non-decreasing monotone sequence relatively to y-axis, from the given points. Then I would like taking a graph, such that the vertices are the points and the edges connect consecutive terms of both sequences. Im pretty stuck right now.

答:
Note: here is the hint that I used as a starting point.

We define two nondecreasing orderings of $P$ with $\leq_x$ and $\leq_y$ the component-wise comparisons. We write $P = \{p_1,\ldots,p_{2n}\} = \{\hat{p}_1,\ldots,\hat{p}_{2n}\}$ where $p_i \leq_x p_{i+1}$ and $\hat{p}_i \leq_y \hat{p}_{i+1}$.

Next, we define a graph on $P$ with the edge-set $E = \big\{\{p_{2k-1},p_{2k}\}\big\}_{k=1}^n \cup \big\{\{\hat{p}_{2k-1},\hat{p}_{2k}\}\big\}_{k=1}^n.$ Note that any vertex is a part of exactly 1 or 2 edges. If the vertex is part of 2 edges $e_1, e_2$ and $e_1$ is of the form $\{p_{2k-1},p_{2k}\}$, then $e_2$ is of the form $\{\hat{p}_{2k-1},\hat{p}_{2k}\}$.

We show that any connected component of $P$ has an even number of vertices. Suppose $v_1 v_2 \cdots v_{2k+1} \subsetneq P$ is a connected component with edges between $v_i$ and $v_{i+1}$. Without loss of generality, assume that $v_1 \leq_x v_2$. Since adjacent edges must be of differing types by our previous comment, we know that $v_{2k} \leq_x v_{2k+1}$. Then $v_{2k+1} = p_{2m}$ for some $m$, and we unravel the cahin to find that $v_1 = p_{2j}$ for some $j$. In particular this means that there exists $v_0 = p_{2j-1}$, and so $v_1$ is part of two edges of the same type. The same argument holds in the case where $v_1 \leq_y v_2$. So we know that any connected component of $P$ has an even number of edges.

Now that we have a decomposition of $P$ into components with an even number of edges, color the vertices of each component in an alternating pattern. This coloring satisfies the axis-parallel strip condition since you cannot expand a strip horizontally and pick up more than 2 vertices of a single color without picking up a vertex of the other color by construction, and similarly for expanding a strip vertically.


答:
We have $f'(x)=f(x)$. Consider the interval $[0,n]$ for some fixed but arbitrary $n\in \mathbb N.$ Set $M=\max_{0\le x\le 1/2}|f(x)|$. Then, for some $0<c<1/2,\ |f(x)|\le |f'(c)|x=|f(c)|x\le \frac{M}{2}$. Suping the left-hand side over $x\in [0,1/2],$ we get $M\le M/2$ which forces $M$ to zero. Now, repeat the argument on the interval $[1/2,1].$ Then, on $[1,3/2]$, etc. After $2n$ steps, we cover the entire interval $[0,n]$ and since $n$ is arbitrary, the result follows.

For another more intuitive proof, using the fact that we already know that we are dealing with the initial value problem $y=y':\ y(0)=0$, suppose there is an interval $(a,b)\subseteq [0,x]$ such that $f(a)=0$ but $f(y)\neq 0$ for $a<y<b.$ It is enough to show that this is impossible.

Without loss of generality, $f(y)>0$. This means that we can define $g(y)=\ln f(y)$ on this interval. Now note that $g'(y)=\frac{f'(y)}{f(y)}=1$ and that $g$ is increasing on this interval since $f'(y)=f(y)>0$ and $\ln $ is increasing.

Using this, together with the  Mean Value Theorem, we have for $a<y<\frac{a+b}{2},$

$g(\frac{a+b}{2})-g(y)=g'(c)(\frac{a+b}{2}-y)=\frac{a+b}{2}-y<b-a$ so $g(y)\ge g(\frac{a+b}{2})-(b-a).$ Letting $y\to a$ gives us a contradiction.

An identical argument works for $\frac{a+b}{2}<y<b.$

Title: Prove that if $f$ is continuous and $f$ is equal to its integral from 0 to $x$, then $f(x) = 0$ for all $x$

Body:
The question is basically we have a continuous function $f:R\to R$ in which we have $$f(x)=\int_0^xf$$ for all real x, and it asks us to show that $f(x) = 0$ for all x real.

The major drawback for me is that I only have theorems about derivatives and Taylor/power series (without the use of integrals), some integral propeties (like the ones which relates absolute values, sums, subtractions, I haven't got theorems about change of variable or integration by parts, for example) and the first statement from the Fundamental Theorem of Calculus to use: (no mention to differential equations nor the second statement of the FTC)

If $f:[a,b]\to R$ is a continuous function and $F:[a,b]\to R$ is defined as $$F(x) = \int_a^xf$$ for all $x\in[a,b]$. Then we have $F'(x) = f(x)$ for all $x\in (a,b)$.

All those integrals are defined with the Darboux definition which takes into account $sup$ and $inf$ of $f$ in each interval that we partition and then take the $sup$ and $inf$ of the lower and upper sums to calculate the integral. (So we're not using the Rienmann definition)

I thought that I could use the fact that the function is infinitely-derivable since $f'(x) = f(x)$ because of the first part of FTC, then $f^{(n)}(x) = f(x)$ and, since $f(0) = 0$ (because it's a degenerate interval for the integral), we'd have $f^{(n)}(0) = 0$ for all natural $n$, then I thought that I could use the Taylor series for the function $f$ around $x=0$ which would be simply $T(x) = 0$ for all real $x$, but then I'd need to proof that $f$ is analytic for all real $x$, which I don't know how to... And the only theorem that my book has about that idea of being analytic is this one:

Let $f:I\to R$ be an infinitely-derivable function in a interval $I$ which contains $0$ in its interior. Let $(-r,r)$ be the biggest open interval in that form such that it's contained in $I$ and such that for each $c$ with $0<c<r$ we have: $$\lim_{n\to\infty}\left[\frac{1}{n!}M_n(c)c^n \right] = 0,$$ where $M_n(c)$ is the maximum of the function $f^{(n)}(x)$ in the interval $[-c,c]$. Then the Taylor series of the function $f(x)$, relatively to $x=0$, converges to $f(x)$ in the interval $(-r,r)$.

But honestly, I don't exactly know how to proceed with my idea of proof using that theorem... any idea would be very welcome and appreciated!


问:
Prove Green formula
Let $(M^n,g)$ be an oriented Riemannian manifold with boundary $\partial M$.
The orientation on $Μ$ defines an orientation on $\partial M$. Locally, on the boundary, choose a positively oriented frame field $\{e\} ^{n}_{i=1} $ such that $e_1 =\nu$ is the
unit outward normal. Then the frame field $\{e\} ^{n}_{i=2} $ positively oriented on $\partial M$. Let $\{\omega ^i\} ^{n}_{i=1} $ denote the orthonormal coframe field dual to $\{e\} ^{n}_{i=1} $. The volume form of Μ is

$$d\mu=\omega^1 \wedge \cdots\wedge\omega^n $$

and the volume form of $\partial M$ is

$$d\sigma=\omega^2 \wedge \cdots\wedge\omega^n $$

By using divergence theorem prove that on a compact manifold,

$$\int _{M^n}(u\Delta v-v\Delta u)d\mu=\int_{\partial M^n} (u\frac{\partial v}{\partial \nu }-v\frac{\partial u}{\partial \nu })d\sigma .$$

Divergence theorem:Let $(M^n,g)$ be a compact oriented Riemannian manifola.
If $X$ is a vector fiela, then

$$\int_M div(X)d\mu=\int_{\partial M^n} \langle X, \nu \rangle d\sigma$$

where $div(X)= \nabla _i X^i.$

答:
Use the well known formula:
$f\Delta g + \langle \nabla f, \nabla g \rangle = div(f \nabla g) $ and the divergence theorem. </p>

Title: Prove Green formula


答:
Alternative solution 2

Since $f$ is continuous, for all $\epsilon>0$ there exists $\delta>0$ such that, if $1-\delta<x<1+\delta$, then $f(1)-\frac{\epsilon}{2}<f(x)<f(1)+\frac{\epsilon}{2}$. From this follows that $2h(f(1)-\frac{\epsilon}{2}) < {\displaystyle\int\limits_{1-h}^{1+h}f(x) \ dx} < 2h(f(1)+\frac{\epsilon}{2})$ for all $0<h<\delta$, which is equivalent to $2f(1)-\epsilon < \frac{\int_{1-h}^{1+h}f(x) \ dx}{h} < 2f(1)+\epsilon$ for all $0<h<\delta$ (it can easily be checked that the same holds when $-\delta<h<0$). Thus the limit follows directly from the definition.


问:
Why does multiplying a vector by $A^{-1}$ give the same result as reducing $A$ to RREF?
Given the linear operator  $T$ in $ℝ^2$ where

$$T:(x,y)↦(s,t)$$

defined by the equations

$$2x+y=s$$

$$6x+2y=t$$

Determine the image of the line $x+y=1$ under the given operator.

My question is why does method 1 give the same answer as method 2.

Method 1 (textbook answer):

Let $A = \begin{bmatrix}2&1\\6&2\end{bmatrix}$

$\begin{bmatrix}2&1\\6&2\end{bmatrix}$$\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}s\\t\end{bmatrix}$

$\begin{bmatrix}x\\y\end{bmatrix}=$ $A^{-1}$ $\begin{bmatrix}s\\t\end{bmatrix}$

Using Cramer's rule gives

$\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-s+\frac{1}{2}t\\3s-t\end{bmatrix}$

Thus the image of $x+y=1$ is $2s-\frac{1}{2}t=1$

Method 2 (The method I used):

Form the augmented matrix where column 1 represents x and column 2 represents y,

Let $A = \begin{bmatrix}2&1&s\\6&2&t\end{bmatrix}$

Reducing to REF gives (or RREF for lefthand side of matrix)

$\begin{bmatrix}1&0&-s+\frac{1}{2}t\\0&1&3s-t\end{bmatrix}$
Now see that $x=-s+\frac{1}{2}t$ and $y=3s-t$

Thus the image of $x+y=1$ is $2s-\frac{1}{2}t=1$

My question:  Why does method 1 give the same answer as method 2?

Method 2 kind of looks like the inversion algorithm
$$[A|I]$$
where reducing $A$ to $I$ undos $A$  and produces $A^{-1}$ on the right side, giving
$$[I|A^{-1}]$$

The problem with this is that the matrix in method 2

$$\begin{bmatrix}2&1&|s\\6&2&|t\end{bmatrix}$$

does not have the identity matrix on the righthand side of the matrix, so it does not have the form $[A|I]$

答: Every elementary row operation has the effect of multiplying the matrix by an elementary matrix $E_i$. Since the total effect of row reduction is to reduce the matrix to the identity matrix, that means that all told you've multiplied the matrix by its inverse. Since you're doing the same thing to the vector (that forms the last column in the augmented matrix), you're also multiplying it by the inverse of the matrix.


问:
How to simplify $\frac {\sin 3A - \cos 3A}{\sin A + \cos A} + 1$?
So I started by using $\sin 3A$ and $\cos 3A$ identities and then I added the lone $1$ to the trigonometric term. (Done in the picture below)

But after this I don't have any clue on how to proceed.

$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-\left(4 \cos ^{3} \theta-\sin \theta\right)}{\sin \theta+\cos \theta} + 1$$
$$=\frac{3 \sin \theta-4\sin^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta}{\sin \theta+\cos \theta} + 1$$
$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta+\sin \theta+\cos \theta}{\sin \theta+\cos \theta}$$
$$=\frac{4 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+4 \cos \theta}{\sin \theta+\cos \theta}$$
$$=\frac{4 \sin \theta+\cos \theta-\sin ^{3} \theta-\cos ^{3} \theta}
{\sin \theta+\cos \theta}
$$

Original image

答:
Recall the cube sum identity: $x^3+y^3 = (x+y)(x^2-xy+y^2)$

Continuing from where you left off:

\begin{align}&\quad4\frac {\sin\theta+\cos\theta-\sin^3\theta-\cos^3\theta}{\sin\theta+\cos\theta}\\
&=4-4\frac {\sin^3\theta+\cos^3\theta}{\sin\theta+\cos\theta}\\
&=4-4(\sin^2\theta - \sin\theta\cos\theta+\cos^2\theta)\\
&=4-4(1-\frac12\sin2\theta)\\
&=2\sin2\theta
\end{align}


答:
That's probably the worst diagram on the planet for trying to visualize the first isomorphism theorem for vector spaces.

Instead, think about ordinary 3-dimensional Euclidean space $\mathbb{R}^3$. The subspaces are: the origin (zero), lines through the origin, planes through the origin, and the whole space.

A linear map $L:\mathbb{R}^3\to\mathbb{R}^3$ always "kills off" a subspace (i.e. maps a subspace to the origin), which is called the kernel. Whatever it doesn't kill off is in the image subspace. For example, projection onto the $xy$-plane kills off the $z$-axis.

Taking a quotient by a subspace kills off that subspace. For example if $U$ is the $z$-axis, then $\mathbb{R}^3/U$ is (isomorphic to) the $xy$-plane.

This should hopefully help you understand the theorem.

Title: For a given transformation $T: V \rightarrow W$ why does $im(T) \cong V/ker(T)$

Body:
I was looking at this example on Wikipedia, but unfortunately I don't quite see the connection. I am mostly trying to understand it through this picture:

Does anyone have a better explanation? I think the main part that is throwing me off is the exclusion of ker(T).


答:
Just summarizing the comments discussion:

Notice that in Method 1, you have:
$\begin{bmatrix}2&1\\6&2\end{bmatrix}$$\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}s\\t\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}s\\t\end{bmatrix}$.

This is equivalent to the augmented system you have implicitly:
$\begin{bmatrix}2&1 & (x) &| & 1 & 0 & (s) \\6&2& (y) & | & 0 & 1 & (t) \end{bmatrix}$.

By conducting the elementary row operations, it works on all rows and you obtain your result, hope it helped!

Title: Why does multiplying a vector by $A^{-1}$ give the same result as reducing $A$ to RREF?

Body:
Given the linear operator  $T$ in $ℝ^2$ where

$$T:(x,y)↦(s,t)$$

defined by the equations

$$2x+y=s$$

$$6x+2y=t$$

Determine the image of the line $x+y=1$ under the given operator.

My question is why does method 1 give the same answer as method 2.

Method 1 (textbook answer):

Let $A = \begin{bmatrix}2&1\\6&2\end{bmatrix}$

$\begin{bmatrix}2&1\\6&2\end{bmatrix}$$\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}s\\t\end{bmatrix}$

$\begin{bmatrix}x\\y\end{bmatrix}=$ $A^{-1}$ $\begin{bmatrix}s\\t\end{bmatrix}$

Using Cramer's rule gives

$\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-s+\frac{1}{2}t\\3s-t\end{bmatrix}$

Thus the image of $x+y=1$ is $2s-\frac{1}{2}t=1$

Method 2 (The method I used):

Form the augmented matrix where column 1 represents x and column 2 represents y,

Let $A = \begin{bmatrix}2&1&s\\6&2&t\end{bmatrix}$

Reducing to REF gives (or RREF for lefthand side of matrix)

$\begin{bmatrix}1&0&-s+\frac{1}{2}t\\0&1&3s-t\end{bmatrix}$
Now see that $x=-s+\frac{1}{2}t$ and $y=3s-t$

Thus the image of $x+y=1$ is $2s-\frac{1}{2}t=1$

My question:  Why does method 1 give the same answer as method 2?

Method 2 kind of looks like the inversion algorithm
$$[A|I]$$
where reducing $A$ to $I$ undos $A$  and produces $A^{-1}$ on the right side, giving
$$[I|A^{-1}]$$

The problem with this is that the matrix in method 2

$$\begin{bmatrix}2&1&|s\\6&2&|t\end{bmatrix}$$

does not have the identity matrix on the righthand side of the matrix, so it does not have the form $[A|I]$


问:
Affine Plane of Order 4 Picture?
I am unable to construct an Affine Plane of Order 4, I can construct an Affine plane of Order 3, and 2. But am unable to find the construction of four anywhere,

It would be greatly appreciated if someone was able to supply a picture.

There was a similar question in the past, Affine plane of order 4?, but it was not very useful. (Plus one of the links are dead)

答: In this picture, opposite red points are the same.  For example, the vertical line appears to contain 5 points, but the 2 red points are the same, so it just contains 4 points.

Title: Affine Plane of Order 4 Picture?


答:
If $x$ and $y$ are real numbers, $|x-y|$ is the distance between the points $x$ and $y$ on the real line. Thus, $|x-3|$ is the distance between $x$ and $3$. The other absolute value expression in your problem is a little trickier, because it’s written as the absolute value of a sum rather than a difference, but you can rewrite it: $|x+1|=|x-(-1)|$ is the distance between $x$ and $-1$. To solve the inequality $|x+1|>|x-3|$, therefore, you’re looking for the real numbers $x$ that are farther from $-1$ than they are from $3$. It’s probably easier to think about if you turn that around: you’re looking for the real numbers that are closer to $3$ than they are to $-1$.

```<code>      --------------|------|-----|-------------|--------|----|----------------
                    a     -1     b             c        3    d
</code>```

Take a look at the numbers $a,b,c$, and $d$ in the sketch above; which are closer to $3$ than they are to $-1$? Can you find a way to tell whether a real number $x$ will be closer to $3$ than to $-1$ just by comparing it with a single real numbers?

Title: Easy Inequality Problem

Body:
If $x$ and $y$ are real numbers, $|x-y|$ is the distance between the points $x$ and $y$ on the real line. Thus, $|x-3|$ is the distance between $x$ and $3$. The other absolute value expression in your problem is a little trickier, because it’s written as the absolute value of a sum rather than a difference, but you can rewrite it: $|x+1|=|x-(-1)|$ is the distance between $x$ and $-1$. To solve the inequality $|x+1|>|x-3|$, therefore, you’re looking for the real numbers $x$ that are farther from $-1$ than they are from $3$. It’s probably easier to think about if you turn that around: you’re looking for the real numbers that are closer to $3$ than they are to $-1$.

      --------------|------|-----|-------------|--------|----|----------------
                    a     -1     b             c        3    d

Take a look at the numbers $a,b,c$, and $d$ in the sketch above; which are closer to $3$ than they are to $-1$? Can you find a way to tell whether a real number $x$ will be closer to $3$ than to $-1$ just by comparing it with a single real numbers?


问:
$f \in L^1$ and $f$ uniformly continous on $\mathbb{R}$ implies $f$ has compact support?
Is it true that if $f \in L^1$ and $f$ is uniformly continuous on $\mathbb{R}$, then $f$ has compact support?

答: No. Any continuous function on $\mathbb R$ that goes to zero at $\pm\infty$ is uniformly continuous.


问:
Show that for every positive integer $ f_n=\frac{\left ( \frac{1+\sqrt5}{2} \right )^n-\left ( \frac{1-\sqrt5}{2} \right )^n}{\sqrt5}$
Define the Fibonacci sequence via $f_1 = f_2 = 1$, and if $n > 2$, then $f_n =
f_{n−1} + f_{n−2}$ and show that for every positive integer $ f_n=\frac{\left ( \frac{1+\sqrt5}{2} \right )^n-\left ( \frac{1-\sqrt5}{2} \right )^n}{\sqrt5}$

I have been solving this problem using strong induction. This is my attemp.

For $n=1$.

$f_1=1$ and,

$\displaystyle \frac{\left ( \frac{1+\sqrt5}{2} \right )-\left ( \frac{1-\sqrt5}{2} \right )}{\sqrt5}=1$

For $n=2$

$f_1=2$ and,

$\displaystyle \frac{\left ( \frac{1+\sqrt5}{2} \right )^2-\left ( \frac{1-\sqrt5}{2} \right )^2}{\sqrt5}=1$ as well.

the statement is also true for $n = 3$

Now, assume that the statement is true for some $n-1$ and $n-2$. We get

$\displaystyle f_n =f_{n-1}+f_{n-2}=\frac{\left ( \frac{1+\sqrt5}{2} \right )^{n-1}-\left ( \frac{1-\sqrt5}{2} \right )^{n-1}}{\sqrt5}+\frac{\left ( \frac{1+\sqrt5}{2} \right )^{n-2}-\left ( \frac{1-\sqrt5}{2} \right )^{n-2}}{\sqrt5}$

From here, my intention is to show that the statement for $f_n$ is true. But I couldn't move on.

答:
$\displaystyle f_n =f_{n-1}+f_{n-2}=\frac{\left ( \frac{1+\sqrt5}{2} \right )^{n-1}-\left ( \frac{1-\sqrt5}{2} \right )^{n-1}}{\sqrt5}+\frac{\left ( \frac{1+\sqrt5}{2} \right )^{n-2}-\left ( \frac{1-\sqrt5}{2} \right )^{n-2}}{\sqrt5}$

Hint: Consider only the numerator part:

$$\left ( \frac{1+\sqrt5}{2} \right )^{n-1}-\left ( \frac{1-\sqrt5}{2} \right )^{n-1} + \left ( \frac{1+\sqrt5}{2} \right )^{n-2}-\left ( \frac{1-\sqrt5}{2} \right )^{n-2}$$

$$= \left ( \frac{1+\sqrt5}{2} \right )^{n-2} \left( \frac{1+\sqrt5}{2} +1\right) - \left ( \frac{1-\sqrt5}{2} \right )^{n-2} \left( \frac{1-\sqrt5}{2} +1\right)$$

$$= \left ( \frac{1+\sqrt5}{2} \right )^{n-2} \left( \frac{3+\sqrt5}{2} \right) - \left ( \frac{1-\sqrt5}{2} \right )^{n-2} \left( \frac{3-\sqrt5}{2} \right)$$


问:
Calculate the volume of the region bounded by the coordinate planes and the surface $z=4-x^2-y^2$
Calculate the volume of the region bounded by the coordinate planes and the surface $z=4-x^2-y^2$

Then I think its $x=0, y=0$ and $z=0$ and the surface $z=4-x^2-y^2$

I think the integral is:

$\int_0^{2}\int_0^{\sqrt{{-x^2}+4}}4-x^2-y^2dydx$

is right?

答:
Use cylindrical coordinate: $z=4-r^{2}\implies r^{2}=4-z$

$$
\begin{aligned}
V&=\frac{1}{4}\int_{0}^{4}\pi r^{2}dz\\
&=\frac{\pi}{4}\int_{0}^{4}(4-z)dz\\
&=2\pi
\end{aligned}
$$


问:
Tor and flat base change
There is an interesting result in Rotman's homological algebra book. Let $A$ be an $R$-module, $B$ be an $(R,S)$-bimodule and $C$ an $S$-module. (All rings commutative). Then Corollary 10.61 (in the 2009 version) states

If ${}_R B_S$ is flat on either side then $$\text{Tor}^S_n(A \otimes_R B,C) \simeq \text{Tor}^R_n(A,B \otimes_S C)$$

If particular if we let $B = S$ (assuming this is also an $R$-module but not necessarily $R$-flat) then $$\text{Tor}^S_n(A \otimes_R S,C) \simeq \text{Tor}^R_n(A,C)$$

Yet if we start with a projective resolution of $A$, $P^\bullet \to A$, $\text{Tor}^R_*(A,C)$ is the homology of complex $P^\bullet \otimes_R C$. But it doesn't appear automatic that $P^\bullet \otimes_R S \to A \otimes_R S$ will be a projective resolution unless $S$ is $R$-flat.

What am I missing?

Edit: Here is an outline argument that Rotman gives. If $A_R$ and ${}_RB_S$ satisfy $\text{Tor}^R_i(A,B\otimes_S,P) = 0$ for all $i \ge 0$ whenever ${}_S P$ is projective, then there is a spectral sequence
$$\text{Tor}_p^R(A,\text{Tor}_q^S(B,C))\Rightarrow \text{Tor}_n^S(A \otimes_RB,C)$$

Now if $B_S$ is flat and ${}_SP$ is projective (hence flat) then $B \otimes_S P$ is a flat $R$-module$^\ast$, then the hypothesis of the spectral sequence holds, and it collapses to the desired isomorphism.

($^\ast$Is this true?)

答:
This is true in general (assuming, as I'm sure was meant, that $B$ is flat "on either side" means on both sides), and I don't think you need the rings to be commutative.  If $P_* \to A$ is a free resolution, then $P_* \otimes_R B$ has homology groups $Tor^R(A,B)$, and so it's also exact above degree zero.  It's then a resolution too -- not by free or projective objects, but (since $P_*$ was free) by direct sums of copies of $B$.  This makes $P_* \otimes_R B$ into a resolution of $A \otimes_R B$ by flat right $S$-modules, and that's good enough to compute Tor.

This makes the homology of the complex $(P_* \otimes_R B) \otimes_S C$ into something that computes $Tor_S(A \otimes_R B, C)$.

On the other hand, it is isomorphic to the complex $P_* \otimes_R (B \otimes_S C)$, which more directly is something that computes $Tor_R(A,B \otimes_S C)$.

(This feels a little ad-hoc.  One perspective on Tor is that you're allowed to take a resolution of one side, or the other side, or of both sides and tensor together the resolutions; this is how you usually show that Tor over a commutative ring is symmetric.  For this problem, that means that you can take resolutions of $A$ and $C$ simultaneously, which from a certain perspective might clean this argument up.)

Title: Tor and flat base change


问:
Calculus 1 - Find the area of shaded region

I have been working at this for like 3 hours and can find nothing online like it and there are no examples in the text book. I must be missing a simple step somewhere.

The correct answer is given as $$\frac{3 \pi \sqrt 2}{2}$$ but I cannot figure out how it was obtained.

Thank you for any help.

答:
Hint $\#1$: (calculus way)

The area you are trying to find is between the graphs

$$f(x) = 3 \sqrt 2 \qquad g(x) = 3 \sec x \tan x$$

on the interval $(-\pi/4,  \pi/4)$.

(I don't know why they expressed the latter function in terms of $\theta$, but whatever.)

The area is then given by

$$\text{shaded area} = \int_{-\pi/4}^{\pi/4} \Big( f(x) - g(x) \Big) \, dx$$

(In case the thing regarding $\theta$ in lieu of $x$ seems potentially an issue, WolframAlpha does happen to give the correct answer, if you just note that $\sqrt 2/2 = 1/\sqrt 2$.)

Hint $\#2$: (geometric way)

You should be able to convince yourself that the blue area makes up half of the area of the rectangle the picture defines:

(The side lengths of this rectangle are $\pi/2$ and $6 \sqrt 2$.) So if you can somehow convince yourself that the blue area is half, you can avoid calculus altogether!

(This can be somewhat justified by noticing that $y(x) = 3 \sec x \tan x$ is an odd function, i.e. $y(-x) = -y(x)$, and we are over an interval symmetric about zero.)


问:
What is the inverse image when $f$ is a constant function?
If I have a function $f: A\rightarrow B$ , The inverse image of a set $H\subseteq B$ is defined by $f^{-1}(H) = \{x\in A\mid f(x)\in H\}$

I think I have a reasonable understanding on finding inverse images when given something like $f(x) = x^2-1$, but I am having trouble comprehending what happens when $f(x) = c\quad$ for some  $c\in\mathbb{R}$. If $H$ is some set $(a,b), [a,b), [a,b], etc\in\mathbb{R}$. What is $f^{-1}(H)$?

Does the size of the set make a difference say $(-\infty,\infty)$ versus $[-3,2]$?

Does inclusion make a difference say $[-3,2]$ versus $(-3,2]$?

I know the elementary fact that the inverse of a constant function would just be a vertical line of the form $x=n,$ $n\in\mathbb{R}$ Which is not a function thus every constant function has no inverse. But I know the inverse-image which is a set has nothing to do with the inverse which is a function. Any guidance or knowledge would be appreciated. Also I am fairly new to working with functions in this manner so I would appreciate a delicate explanation.

答:
If $f\colon A\to B$ is a constant function, i.e. $f(A) = b \in B$ then

$$f^{-1}(\{b\}) = A.$$


问:
Show that there is no pair s.t. $a^1+a^2+a^3+...+a^n=b^m$
Let $a,b,m$ and $n$ belong to integer. Is there any pair exist such that

$a^1+a^2+a^3+...+a^n=b^m$ where $a,b,m,n\ge 2$

Source code Pari/GP

```<code>for(m=2,10,for(n=2,10,for(a=2,100,for(b=2,100,if(sum(i=1,n,a^i)==b^m,print([a,n,b,m]))))))
</code>```

Edit: more on observation

There always have at least one solution for $\sum_{i=1}^{n}a^i=b^m+k$ for every integer $k\ne 0$

Apologies for the short question and shows no effort towards solution. Thank you.

答:
(This is incomplete)

$$a ( 1+ a^2 + \ldots a^{n-1}) = b^m$$

Since $ \gcd(a, 1+ a^2 + \ldots a^{n-1}) = 1$, both of these terms must be powers of $m$. Let $ a = c^m$ and $b = cd$, which gives us

$$ d^ m = \frac{ c^{mn} -1 } { c^m - 1}  > \frac{ c^{mn}}{c^m}   \Rightarrow d > c^{n-1}. $$

If $ n < m$, then $ (c^{n-1} +1)^m (c^m-1) > c^{mn} -1 \Rightarrow  c^{n-1} + 1 > d$. Hence, there are no integer solutions for $d$.

So we must have $ n \geq m$.


问:
Monte-carlo simulation for inflation linked derivatives
I'm trying to evaluate an inflation linked derivative using the Jarrow-Yildirim model (2003).

It goes like that:

$dn(t) = [\theta_{n}(t) - a_{n}n(t)]dt + \sigma_{n}dW_{n}(t)$

$dr(t) = [\theta_{r}(t) - \rho_{r, i}\sigma_{r}\sigma_{i} - a_{r}r(t)]dt + \sigma_{r}dW_{r}(t)$

$di(t) = i(t)[n(t)-r(t)]dt + \sigma_{i}i(t)dW_{i}(t)$

where $n(t)$, $r(t)$ and $i(t)$ are respectively the nominal interest rate, the reel interest rate and the inflation index. $a_{n}$ and $a_{r}$ are known constants and $W_{n}(t)$, $W_{r}(t)$, $W_{i}(t)$ are standard Wiener processes. $\theta_{j}(t)$ is defined as follows:

$\theta_{j}(t) = \dfrac{\partial f_{j}(0, t)}{\partial T} + a_{j}f_{j}(0, t) + \dfrac{\sigma_{j}^{2}}{2a_{j}}(1- e^{-2a_{j}t}),   j \in \{n, r \}$

The form of $f(0, t)$ can be obtained by the Nelson-Siegel model.

The solutions to the PDEs above are:

$n(t) = n(s)e^{-a_{n}(t-s)}+ \int_{s}^{t}e^{-a_{n}(t-u)}\theta_{n}(u)du + \sigma_{n}e^{-a_{n}t}\int_{s}^{t}e^{a_{n}u}dW_{n}(u)$

$r(t) = r(s)e^{-a_{r}(t-s)}+ \int_{s}^{t}e^{-a_{r}(t-u)}(\theta_{r}(u)-\rho_{r, i}\sigma_{r}\sigma_{i})du + \sigma_{r}e^{-a_{r}t}\int_{s}^{t}e^{a_{r}u}dW_{r}(u)$

$i(t) = i(s)e^{\int_{s}^{t}[n(u)-r(u)]du - \dfrac{\sigma^{2}_{i}}{2}(t-s) + \sigma_{i}(W_{t}-W_{s})}$

My question is concerning integrals under $n(t)$ and $r(t)$. Since i'm doing a Monte-Carlo simulation, I only need to find what laws drive those terms so I can simulate them.

First, $\int_{s}^{t}e^{-a_{n}(t-u)}\theta_{n}(u)du$ and $\int_{s}^{t}e^{-a_{r}(t-u)}(\theta_{r}(u)-\rho_{r, i}\sigma_{r}\sigma_{i})du$ are deterministic integrals but somehow, i'm having trouble to solve them. Since $\theta_{n}$ and $\theta_{r}$ are functions of time, should I only integrate each component of these terms with respect to time?

Also, it seems like $\sigma_{n}e^{-a_{n}t}\int_{s}^{t}e^{a_{n}u}dW_{n}(u)$ and $\sigma_{r}e^{-a_{r}t}\int_{s}^{t}e^{a_{r}u}dW_{r}(u)$ are respectively equivalent in law to $\dfrac{\sigma_{n}e^{-a_{n}t}}{\sqrt{2a_{n}}} \sqrt{e^{2a_{n}t}-e^{2a_{n}s}} \epsilon_{1}$ and $\dfrac{\sigma_{r}e^{-a_{r}t}}{\sqrt{2a_{r}}} \sqrt{e^{2a_{r}t}-e^{2a_{r}s}} \epsilon_{2}$ where $\epsilon_{1}$ and $\epsilon_{2}$ ~ $N(0, 1)$. Can someone familiar with such terms validate my answer?

答:
These terms somehow look really messy. And it seems you also made many typos

I'll start with your last question. According to my calculation, your observation is right.

Regarding the first question, I think for the first integral, you have written wrongly $a_r$ instead of $a_n$ ( in the exponential's power). The second integral looks okay.

As for how to integrate them, I don't think it is that complicated because it seems that :

$ e^{a_it} \theta_i(t)= \frac{\partial(g_i)}{\partial t}(t)+\text{some explicit functions}$

where $g_i(t)= e^{a_it}f_i(0,t)$


问:
Euclidean distances determine the points uniquely
I saw a conclusion: If we fix an arbitrary set of $n+1$ affine independent points $\{p_0,p_1, \dots, p_n\}$ in $\Bbb R^n$, then the Euclidean distances $\{|x-p_j|\}_{j=0}^n$ determine the points $x$ in $\Bbb R^n$ uniquely.

I have no idea of proving the above conclusion.

答:
HINT:

For two distinct points $x$, $x'$, the set of points $p$ such that $|p-x|=|p-x'|$ forms a hyperplane (perpendicular to $xx'$, through the middle of it), so it cannot contain $n+1$ affine independent points $p_1$, $\ldots$, $p_{n+1}$.

$\bf{Added:}$ This will not be true for other normed spaces. For instance, take $\mathbb{R}^2$ with the norm $\|(x_1, x_2)\| = \sqrt[4]{x_1^4 + x_2^4}$. Consider the points $x=(-1,1)$, $x'=(1,2)$.  The equal distance curve is not a line, so we can find three non-collinear points on it. [


问:
Randomly removing edges from a graph so that each vertex has N edges
I am an engineer working on an algorithm that solves a system of equations, where each equation is associated with an edge on a graph. Currently I am running the algorithm on a graph which has an edge between every pair of vertices (a complete graph, I believe?), but I am hoping to study the error and convergence for cases when there are fewer edges. However, when any vertex has fewer than 4 edges, the system will not converge, so I want to avoid such cases.

Is there a method to remove edges from a complete graph, such that every vertex has N edges? If not, is there anything that gets close, giving each vertex, say, N +/-M edges?

Thanks for any guidance.

答: If the number of vertices $V$ is prime and $N$ is even, you can number the vertices from $0$ to $V-1$ and choose $\frac N2$ pairwise coprime numbers, no pair of which adds to $V$.  Add and subtract each of your $\frac N2$ numbers from each vertex number $\bmod V$, keep the edges indicated, and you will have a graph that meets your requirement.  All these graphs will have a similar structure of being $\frac N2$ cycles that are interlaced.  That might or might not be a problem for you.  If $V$ is not prime you can do the same if you can find $\frac N2$ numbers that are pairwise coprime and coprime to $V$ with no pair summing to $V$.  As long as $N$ is not too large this should be possible.


问:
Doubts about the general solution of $y'' +7y' = e^{-7x}$
I have some doubts about how to find the general integral of this ODE:

$$y'' +7y' = e^{-7x}$$

I know that first we find the solution to the homogeneous ODE and then the particular solution. When I was solving the homogeneous  I got to:

$$y_h= K_1 e^0 + K_2 e^{-7x}$$

My doubt is that is it correct to put $y_h= K_1 + K_2 e^{-7x}$? Or is the $e^0$ some kind of special case I need to know and do something else. Also, to find the particular solution I'm doing

$$ y_p = A+Be^{-7x}$$

and then doing the substitutions and equalling them to the LHS of the original ODE.

Thanks.

Body:
$$y'' +7y' = e^{-7x}$$
Rewrite it as:
$$(y' e^{7x})'=1$$
And integrate.


问:
Is every subgroup of $S_n$ cyclic?
I've just showed that every subgroup of $S_3$ is cyclic.

I'm not familiar with permutation groups yet and I'm curious to know whether every subgroup of $S_n$ is cyclic.

As a counterexample might be $S_4$ probably, but I'm not sure at all.

Thanks in advance for explanation.

答:
That is a property that cyclic groups possess.  And while there are examples, such as $S_3$, of non-cyclic groups which possess it also, $S_n, n\gt3$ are not among them.

As already remarked, for instance, $S_k\le S_n\,,\forall k\lt n$.

Title: Is every subgroup of $S_n$ cyclic?


问:
Prove that if $f$ is continuous and $f$ is equal to its integral from 0 to $x$, then $f(x) = 0$ for all $x$
The question is basically we have a continuous function $f:R\to R$ in which we have $$f(x)=\int_0^xf$$ for all real x, and it asks us to show that $f(x) = 0$ for all x real.

The major drawback for me is that I only have theorems about derivatives and Taylor/power series (without the use of integrals), some integral propeties (like the ones which relates absolute values, sums, subtractions, I haven't got theorems about change of variable or integration by parts, for example) and the first statement from the Fundamental Theorem of Calculus to use: (no mention to differential equations nor the second statement of the FTC)

If $f:[a,b]\to R$ is a continuous function and $F:[a,b]\to R$ is defined as $$F(x) = \int_a^xf$$ for all $x\in[a,b]$. Then we have $F'(x) = f(x)$ for all $x\in (a,b)$.

All those integrals are defined with the Darboux definition which takes into account $sup$ and $inf$ of $f$ in each interval that we partition and then take the $sup$ and $inf$ of the lower and upper sums to calculate the integral. (So we're not using the Rienmann definition)

I thought that I could use the fact that the function is infinitely-derivable since $f'(x) = f(x)$ because of the first part of FTC, then $f^{(n)}(x) = f(x)$ and, since $f(0) = 0$ (because it's a degenerate interval for the integral), we'd have $f^{(n)}(0) = 0$ for all natural $n$, then I thought that I could use the Taylor series for the function $f$ around $x=0$ which would be simply $T(x) = 0$ for all real $x$, but then I'd need to proof that $f$ is analytic for all real $x$, which I don't know how to... And the only theorem that my book has about that idea of being analytic is this one:

Let $f:I\to R$ be an infinitely-derivable function in a interval $I$ which contains $0$ in its interior. Let $(-r,r)$ be the biggest open interval in that form such that it's contained in $I$ and such that for each $c$ with $0<c<r$ we have: $$\lim_{n\to\infty}\left[\frac{1}{n!}M_n(c)c^n \right] = 0,$$ where $M_n(c)$ is the maximum of the function $f^{(n)}(x)$ in the interval $[-c,c]$. Then the Taylor series of the function $f(x)$, relatively to $x=0$, converges to $f(x)$ in the interval $(-r,r)$.

But honestly, I don't exactly know how to proceed with my idea of proof using that theorem... any idea would be very welcome and appreciated!

答:
But, isn't this Gronwall...?

Like, letting $G(x)= e^{-x} \int_{0}^x f(s)ds$

$G'(x)= e^{-x}( f(x)- \int_{0}^x f(s)ds)=0$

Done.


答: $$\frac{\sin3x-\cos3x}{\sin x+\cos x}+1=\frac{\sin3x+\sin x+\cos x-\cos3x}{\sin x+\cos x}=\frac{\sin(2x+x)+\sin (2x-x)+\cos (2x-x)-\cos(2x+x)}{\sin x+\cos x}=\frac{\sin2x\cos x+\sin x\cos2x+\sin2x\cos x-\sin x\cos 2x+\cos 2x\cos x+ \sin 2x\sin x-(\cos 2x\cos x- \sin 2x\sin x)}{\sin x+\cos x}=\frac{2 \sin2x\cos x +2 \sin 2x\sin x}{\sin x+\cos x}=\frac{2 \sin2x(\cos x +\sin x)}{\sin x+\cos x}=2\sin2x$$

Title: How to simplify $\frac {\sin 3A - \cos 3A}{\sin A + \cos A} + 1$?

Body:
So I started by using $\sin 3A$ and $\cos 3A$ identities and then I added the lone $1$ to the trigonometric term. (Done in the picture below)

But after this I don't have any clue on how to proceed.

$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-\left(4 \cos ^{3} \theta-\sin \theta\right)}{\sin \theta+\cos \theta} + 1$$
$$=\frac{3 \sin \theta-4\sin^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta}{\sin \theta+\cos \theta} + 1$$
$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta+\sin \theta+\cos \theta}{\sin \theta+\cos \theta}$$
$$=\frac{4 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+4 \cos \theta}{\sin \theta+\cos \theta}$$
$$=\frac{4 \sin \theta+\cos \theta-\sin ^{3} \theta-\cos ^{3} \theta}
{\sin \theta+\cos \theta}
$$

Original image


问:
Find the GS of the System of DE's $\begin{cases} x' = x-3y\\ y'=3x+7y \end{cases}$
Find the GS of the following system of DE's where the independent variable is $t$ and $x$ and $y$ are the dependent variables

\begin{cases}
x' = x-3y\\
y'=3x+7y
\end{cases}

I know using eigenvalues and eigenvectors or operators is one way to do this. But I wish to double check my answer using a substitution method.

So my work:

The second DE $y'=3x+7y$ can be rewritten as $x = \cfrac{y'}{3}-\cfrac 73y$

then $x' = \cfrac{y''}{3}-\cfrac73y'$

When we plug these values of $x$ and $x'$ into the first DE ($x' = x -3y)$, we get with some rearranging

$\cfrac{y''}{3}-\cfrac83y'+\cfrac{16}{3}y = 0$

Which has a characteristic equation of

$\cfrac{r^2}{3}-\cfrac83r+\cfrac{16}{3} = 0$

with roots $r_1=4$ and $r_2 = 4$

Then the solution for $y$ is $y$ = $C_1e^{4t}+C_2te^{4t}$

Then we back sub to solve for $x$ using $x = \cfrac{y'}{3}$$-\cfrac73y$ with the solution of y we just found.

We get $x =-C_1e^{4t}-C_2te^{4t} + \cfrac{C_2}{3}e^{4t} =-C_1e^{4t}-C_2te^{4t} + C_3e^{4t}$

so the GS to the homo system is

\begin{cases}
x = -C_1e^{4t}-C_2te^{4t}+C_3e^{4t}\\
y = C_1e^{4t}+C_2te^{4t}
\end{cases}

If this solution is right, then I'm confident that I understand how substitution method works for solving DE systems. (Also it would boost my confidence in using the operator method to solve this as I got the same answer as this using the operator method). I'm a little thrown off on the roots being the same but I still think my methodology is still sound. I would appreciate it if someone could tell me if I've got this right cause then I know I completely understand how to solve a system of DE's.

If more work is necessary to show please let me know.

答:
Your method is correct.

Another way is to write the system as $X'=MY$

Where $M$ is the matrix $$\begin{pmatrix} 1 & -3 \\ 3 & 7 \end{pmatrix}$$
its eigen values are $m_1,m_2=4$ (repeated roots), so the solution for $x$ can be written as
$x=(A t+B)e^{4t}$, putting it in first equation one can get $y$. So there will be two
undetermined constants: $A,B$.


问:
Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$
Is there a closed form for the following infinite product?
$$\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$$</p>

答:
This uses the hints by Raymond Manzoni and Cameron Williams. I am not sure if the answer is correct, but this should give an idea of how to proceed. I believe the answer by Raymond Manzoni is correct.

$$ \dfrac{\Gamma\left(2^n+\frac{1}{2}\right)}{\Gamma\left(2^n\right)} = \dfrac{2^n(2^{n+1})!\sqrt{\pi}}{4^{2^n}(2^n)!^2}$$
Hence,
\begin{align}
\log \left( \dfrac{\Gamma\left(2^n+\frac{1}{2}\right)}{\Gamma\left(2^n\right)}\right) & = n \log(2) + \dfrac{\log(\pi)}2 - 2^{n+1} \log(2) + \log \left(\dbinom{2^{n+1}}{2^n}\right)
\end{align}
\begin{align}
S_n = \dfrac{\log \left( \dfrac{\Gamma\left(2^n+\frac{1}{2}\right)}{\Gamma\left(2^n\right)}\right)}{2^n} & = \dfrac{n \log(2)}{2^n} + \dfrac{\log(\pi)}{2^{n+1}} - 2 \log(2) + \dfrac1{2^{n}}\log \left(\dbinom{2^{n+1}}{2^n}\right)
\end{align}
Hence,
$$\sum_{n=1}^{\infty}\dfrac{n \log(2)}{2^n} = 2 \log(2)$$
$$\sum_{n=1}^{\infty}\dfrac{\log(\pi)}{2^{n+1}} = \dfrac{\log(\pi)}2$$
Recall that
$$\log \left(\dbinom{2k}k\right) \sim 2k \log(2) - \dfrac{\log(\pi)}2 - \dfrac{\log(k)}2$$
This gives us
$$\log \left(\dbinom{2 \cdot 2^n}{2^n}\right) \sim 2^{n+1} \log(2) - \dfrac{\log(\pi)}2 - n \dfrac{\log(2)}2$$
$$\dfrac{\log \left(\dbinom{2 \cdot 2^n}{2^n}\right)}{2^n} \sim 2 \log(2) - \dfrac{\log(\pi)}{2^{n+1}} - n \dfrac{\log(2)}{2^{n+1}}$$
This gives us $S_n \sim \dfrac{n}{2^{n+1}} \log(2)$ and hence I would hope (this step needs more justification and is probably wrong in fact) $$\sum_{n=1}^{\infty} S_n  \approx \log(2)$$
Hence, the answer you are looking for is approximately $$e^{\log(2)} = 2$$


问:
How to find the values which would equate the division between two numbers of swapped digits without knowing anything further?
I'm having difficulty in assess this problem in a method without relying too much in too many trial and error. I have noticed there is a similar puzzle but in such it includes the value of one of the digits, but how do you solve this without knowing any other information?.

The problem is as follows:

Using the expression from below:

$$\frac{\overline{ASIR}}{\overline{RISA}}=4$$

Find $\overline{RI}-\overline{SA}$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{75}\\
2.&\textrm{-75}\\
3.&\textrm{57}\\
4.&\textrm{-57}\\
5.&\textrm{-48}\\
\end{array}$

How exactly should this problem be assesed without just guessing?. Can someone help me with this?.

So far the only thing which I could come up with was with the following:

$\overline{ASIR}=4\overline{RISA}$

$1000A+100S+10I+R=4000R+400I+40S+4A$

From this it can be obtained:

$996A+60S=3996R+390I$

But that's it. That's the place where I came stuck. From then on I think its just trial and error, but on looking on the equation all letters with the exception of R and A can take values between 1 to 9. So how to constrain those or to solve this thing in a more orderly manner?. Can someone help me with a step-by-step approach and the most detailed as possible?.

答: Write it as $\overline {ASIR} \times 4 = \overline {RISA}$.  Note that $A$ can only be $1$ or $2$ because otherwise multiplying by $4$ would carry.  It has to be $2$ because it is the ones digit of the product.  Now $R$ must be $3$ or $8$ to make the $A$ in the product, and it must be $8$ because it is at least $4$.  We are left with $\overline {SI} \times 4 +3=\overline {IS}$, so $S$ is odd and must be $1$ to avoid a carry.  Then $I$ must be $7$ and we have $2178 \times 4 = 8712$


答:
$f_n=f_{n-1}+f_{n-2}$ is a difference equation whiose solution is found by putting
$f_n=t^n$, then $t_{1,2}=\frac{1\pm \sqrt{5}}{2}$. The solution is
$$f_n=A t_1^n+ B t_2^n$$. By putting $f_1=f_2=1$, we get $$A=-B=\frac{1}{\sqrt{5}}.$$

Title: Show that for every positive integer $ f_n=\frac{\left ( \frac{1+\sqrt5}{2} \right )^n-\left ( \frac{1-\sqrt5}{2} \right )^n}{\sqrt5}$

Body:
Define the Fibonacci sequence via $f_1 = f_2 = 1$, and if $n > 2$, then $f_n =
f_{n−1} + f_{n−2}$ and show that for every positive integer $ f_n=\frac{\left ( \frac{1+\sqrt5}{2} \right )^n-\left ( \frac{1-\sqrt5}{2} \right )^n}{\sqrt5}$

I have been solving this problem using strong induction. This is my attemp.

For $n=1$.

$f_1=1$ and,

$\displaystyle \frac{\left ( \frac{1+\sqrt5}{2} \right )-\left ( \frac{1-\sqrt5}{2} \right )}{\sqrt5}=1$

For $n=2$

$f_1=2$ and,

$\displaystyle \frac{\left ( \frac{1+\sqrt5}{2} \right )^2-\left ( \frac{1-\sqrt5}{2} \right )^2}{\sqrt5}=1$ as well.

the statement is also true for $n = 3$

Now, assume that the statement is true for some $n-1$ and $n-2$. We get

$\displaystyle f_n =f_{n-1}+f_{n-2}=\frac{\left ( \frac{1+\sqrt5}{2} \right )^{n-1}-\left ( \frac{1-\sqrt5}{2} \right )^{n-1}}{\sqrt5}+\frac{\left ( \frac{1+\sqrt5}{2} \right )^{n-2}-\left ( \frac{1-\sqrt5}{2} \right )^{n-2}}{\sqrt5}$

From here, my intention is to show that the statement for $f_n$ is true. But I couldn't move on.


答:
$$\begin{cases}
x = -C_1e^{4t}-C_2te^{4t}+C_3e^{4t}\\
y = C_1e^{4t}+C_2te^{4t}
\end{cases}$$
You should only end with two constants not three.
$$y=C_1e^{4t}+C_2te^{4t}$$
$$y'=4C_1e^{4t}+C_2e^{4t}+4tC_2e^{4t}$$
$$y'=4C_1e^{4t}+C_2e^{4t}(1+4t)$$
And since:
$$x = \cfrac{y'}{3}-\dfrac 73 y$$
You should only have $C_1$ and $C_2$ constant. You applied the method of substitution correctly.

Title: Find the GS of the System of DE's $\begin{cases} x' = x-3y\\ y'=3x+7y \end{cases}$

Body:
Find the GS of the following system of DE's where the independent variable is $t$ and $x$ and $y$ are the dependent variables

\begin{cases}
x' = x-3y\\
y'=3x+7y
\end{cases}

I know using eigenvalues and eigenvectors or operators is one way to do this. But I wish to double check my answer using a substitution method.

So my work:

The second DE $y'=3x+7y$ can be rewritten as $x = \cfrac{y'}{3}-\cfrac 73y$

then $x' = \cfrac{y''}{3}-\cfrac73y'$

When we plug these values of $x$ and $x'$ into the first DE ($x' = x -3y)$, we get with some rearranging

$\cfrac{y''}{3}-\cfrac83y'+\cfrac{16}{3}y = 0$

Which has a characteristic equation of

$\cfrac{r^2}{3}-\cfrac83r+\cfrac{16}{3} = 0$

with roots $r_1=4$ and $r_2 = 4$

Then the solution for $y$ is $y$ = $C_1e^{4t}+C_2te^{4t}$

Then we back sub to solve for $x$ using $x = \cfrac{y'}{3}$$-\cfrac73y$ with the solution of y we just found.

We get $x =-C_1e^{4t}-C_2te^{4t} + \cfrac{C_2}{3}e^{4t} =-C_1e^{4t}-C_2te^{4t} + C_3e^{4t}$

so the GS to the homo system is

\begin{cases}
x = -C_1e^{4t}-C_2te^{4t}+C_3e^{4t}\\
y = C_1e^{4t}+C_2te^{4t}
\end{cases}

If this solution is right, then I'm confident that I understand how substitution method works for solving DE systems. (Also it would boost my confidence in using the operator method to solve this as I got the same answer as this using the operator method). I'm a little thrown off on the roots being the same but I still think my methodology is still sound. I would appreciate it if someone could tell me if I've got this right cause then I know I completely understand how to solve a system of DE's.

If more work is necessary to show please let me know.


问:
Dartboard paradox and understanding independence

By definition, events $A$ and $B$ are independent if $$P(A \cap
    B) = P(A)\:P(B).$$ Thus if an event $A$ happens almost never
$\left(P(A)=0\right),$ then
$A$ is independent of all events, including itself. So, hitting the
exact centre of a dartboard (happens almost never) and hitting within its inner ring are independent events.

On the other hand, it is standard to characterise
independence as follows:

Two events are independent if the occurrence of one does not
affect the probability of occurrence of the other.

Since hitting the exact centre of a dartboard guarantees hitting
within its inner ring (which is otherwise not guaranteed), the
two events are dependent.

Isn't this a contradiction?? If yes, then do we accept that the definition of independence is not meant to fully correspond to its verbal/intuitive characterisation?

Is there any semantic difference between the following two versions?

$(i)$ Two events are independent if the occurrence of one does not
affect the probability of occurrence of the other.

$(ii)$ Two events are independent if the occurrence of one does not
affect the occurrence of the other.

ADDENDUM

On further pondering, I have resolved both questions:

2.

Consider this experiment: flip two fair coins, letting $H_1$ be the event that the first coin lands on Heads, and $X$ be the event that the coins land on different sides. Then $$ P\left(H_1 \cap X\right)=\frac14=P(H_1)\:P(X);$$ i.e., $H_1$ and $X$ are independent events.

Knowledge that $H_1$ happens reduces the possible number of ways that $X$ can eventuate—from $2$ (outcomes HT and TH) to $1$ (outcome HT)—but does not change the probability $\left(\frac12\right)$ of $X.$

1.

The following revision characterises pairwise independence more clearly and accurately:

Let $P(A)\neq0.$

Events $A$ and $B$ are independent iff. the knowledge that $A$ happens doesn't change $B$'s probability.

In this informal characterisation, almost-never events are now excluded from being conditioned on. (What does it even mean to say that an almost-never event has happened: in what sense have I hit the exact centre of a dartboard?) It motivates the definition of pairwise independence, which does allow both events to be impossible.

答:
"On the other hand, it is standard to characterise pairwise independence as follows:

Two events are independent if the occurrence of one does not affect the probability of occurrence of the other."

Errr no, I disagree; the usual and informal notion of independence is slightly different:

Two events are said to be independent if knowledge that one of them has occurred does not cause you to assign a different value to the probability of the other.

That is, if $A$ and $B$ the two events in question and we know that $A$ had occurred, then we should use $P(B\mid A)$ are instead of $P(B)$ as the probability of $B$ in our calculations. Then, $A$ and $B$ are said to be independent events if $P(B\mid A)$ happens to be equal to $P(B)$. Thus, informally, $A$ and $B$ are said to be independent if $P(B\mid A) = P(B)$, or if one prefers, $P(A\mid B) = P(A)$. This is an informal definition because the conditional probabilities are not necessarily defined: the usual definition $\displaystyle P(B\mid A) = \frac{P(B \cap A)}{P(A)}$ assumes that $P(A) > 0$. It is much better to use the standard definition of (mutual) independence $P(A\cap B) = P(A)P(B)$ which shows the symmetry of the relationship in contrast to $P(B\mid A) = P(B)$ which is asymmetric and if we stick in $P(A\mid B) = P(A)$ also to show symmetry, then we commit the sin of prolixity. Worse yet, both $A$ and $B$ might be events of probability $0$ making both $P(B\mid A)$ and $P(A\mid B)$ be undefined. In contrast, the definition of independence as $P(A\cap B) = P(A)P(B)$  has no such issues and is vastly to be preferred. $P(B\mid A) = P(B)$ is fine for informal use and for intuition, but not as a definition.  See also this answer of mine over on stats.SE.


问:
Proving $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.

Prove the distributive property for sets:

$A \cup (B \cap C) =  (A \cup B) \cap (A \cup C)$

I'm not good with proofs but my understanding is that I have to prove 2 things:

(1)  $A \cup (B \cap C) \subset (A \cup B) \cap (A \cap C)$

(2)  $A \cap (B \cap C) \supset (A \cup B) \cap (A \cup C)$

This is what I have done so far:

Part (1)

If $x\in A$, then $x \in (A \cup B)$ and $x \in (A \cup C)$.

$\therefore x \in (A \cup B) \cap (A \cap C)$

If $x \in (B \cap C)$ then $x \in (A  \cup B)$ and $x \in (A \cup C)$ because $x \in B$ and $x \in C$.

$\therefore x \in (A \cup B) \cap (A \cap C)$

$\therefore A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C)$

Part (2)

Now we have to prove the reverse inequality:  $(A \cup B) \cap (A \cap C)$.  Then $x \in A \cup B$ and $x \in (A \cup C)$

If $x \in A$, then $x \in A \cup (B \cap C)$

This is where I am up to.  I wanted to know whether my approach is correct and if I did part (1) correctly.  I'm stuck on part (2) and don't know how to proceed.  I'd appreciate any help.

Thank you!!

答:
You must first prove 2 cases:

(1)  $A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$

(2)  $(A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)$

Note that in mathematics we use the following symbols:

$\cap=$ AND = $\land$

$\cup=$ OR = $\lor$

Case 1:  $A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$

Let $x \in A \cap (B \cup C) \implies x \in A \land x \in (B \cup C)$

$\implies x \in A \land \{ x \in B \lor x \in C \}$

$\implies \{ x \in A \land x \in B \} \lor\{ x \in A \land x \in C \} $

$\implies x \in (A \cap B) \lor x \in (A \cap C)$

$\implies x \in (A \cap B) \cup (A \cap C)$

$\therefore x \in A \cap (B \cup C) \implies x \in (A \cap B) \cup (A \cap C)$

$\therefore A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$

Case 2: $(A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)$

Let $x \in (A \cap B) \cup (A \cap C) \implies x \in (A \cap B) \lor x \in (A \cap C)$

$\implies \{x \in A \land x \in B \} \lor \{ x \in A \land x \in C \}$

$\implies x \in A \land \{ x \in B \lor x \in C\}$

$\implies x \in A \land \{B \cup C \}$

$\implies x \in A \cap (B \cup C)$

$\therefore x \in (A \cap B) \cup (A \cap C) \implies x \in A \cap (B \cup C)$

$\therefore (A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)$

$\therefore A \cup (B \cap C) =  (A \cup B) \cap (A \cup C)$


答:
No MVT, no derivatives , no exponential function etc. This proof uses only basic properties of continuous functions.

Let us first show by induction that $f \equiv 0$ on $[0,n]$ for each positive integer $n$.

Let $M$ be the maximum of $f$ on $[0,1]$. There exist $x_0 \in [0,1]$ such that $f(x_0)=M$. Now $M=f(x_0)=\int_0^{x_0} f(y)dx \leq \int_0^{x_0}  M dx=M x_0 \leq M$. Thus equality holds throughout and this implies that $f$ has the constant value $M$ on $[0,x_0]$. Since $f(0)=0$ it follows that $M=0$ and  $f \leq 0$ and since $-f$ also satisfies the hypothesis we get $f \geq 0$ too. Hence $f \equiv 0$ on $[0,1]$

Suppose we know that $f \equiv 0$ on $[0,n]$. Consider the maximum value $M_n$ of $f$ on $[n,n+1]$ and repeat the argument above. Here we get $M_n=f(x_n)=\int_n^{x_n} f(y)dx \leq \int_n^{x_n}  M_n dx=M_n (x_n-n) \leq M_n$ and we can conclude as before that $f\equiv 0$ on $[n, n+1]$. The induction proof is now complete.

Now consider $[-1,0]$. Let $M$ be the  maximum of $|f(x))|$ in this interval. Then there exists $y \in [-1,0]$ with $|f(y)|=M$. Now $M=|f(y)|=|\int_0^{y} f(t)dt| \leq M|y| \leq M$. This implies that $|f(t)|$ has the constant value $M$ on $[y,0]$. Since $f(0)=0$ this constant value muse be $0$ so $M=0$. This implies that $f\equiv 0$ on $[-1,0]$. I will leave the induction argument to you.


问:
Find $\sqrt{a}+\sqrt{b}+\sqrt{c}$ only in terms of $p$.
If $a^{2} x^{3}+b^{2} y^{3}+c^{2} z^{3}=p^{5}, a x^{2}=b y^{2}=c z^{2}$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{p},$ find $\sqrt{a}+\sqrt{b}+\sqrt{c}$
only in terms of $p$.

I tried simple factorization like shifting terms here and there but didnt get anything.

答:
We are given
$$a^2x^3+b^2y^3+c^2z^3=p^5\quad (1)$$
$$ax^2=by^2=cz^2=t\ (\mathrm{let})\quad (2)$$
$$\frac 1 x +\frac 1 y +\frac 1 z =\frac 1 p\quad (3)$$
Squaring $(2)$, we get
$$a^2x^4= b^2y^4= c^2=z^4=t^2$$
$$a^2x^3=\frac t x ; b^2y^3=\frac t y ; c^2z^3=\frac t z$$
Substituting these values in $(1)$, we get
$$\frac t x + \frac t y +\frac t z = p^5$$
$$t\left(\frac 1 x +\frac 1 y +\frac 1 z\right)=p^5$$
Using $(3)$, we have
$$\frac t p = p^5 \implies t=p^6$$.
Now, we take the square root of $(2)$.
$$\sqrt{a}x =\sqrt{b}y=\sqrt{c}z=\sqrt{t}=p^3$$
$$\sqrt{a}=\frac{p^3}{x}; \sqrt{b}=\frac{p^3}{y}; \sqrt{c}=\frac{p^3}{z}$$
Therefore, we have
$$\sqrt{a}+\sqrt{b}+\sqrt{c}=\frac{p^3}{x}+\frac{p^3}{y}+\frac{p^3}{z}$$
$$\sqrt{a}+\sqrt{b}+\sqrt{c}=p^3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$$
Finally, using $(3)$ again, we have
$$\sqrt{a}+\sqrt{b}+\sqrt{c}=p^3\cdot \frac 1 p = p^2$$
Hope this helps :)


问:
Derivative of $(\arcsin (x))$
I was trying to solve the following exercise:

Show that $(\arcsin (x))'=\frac{1}{\sqrt{1-x^2}}$ for all $x\in (-1,1)$.

So far I got that $(\arcsin (x))'=\frac{1}{\cos(\arcsin (x))}$, but i don't know how to go from $\frac{1}{\cos(\arcsin (x))}$ to $\frac{1}{\sqrt{1-x^2}}$. Any help is appreciated.

答:
Consider a right triangle $\triangle ABC$ with a right angle at $B$. Let $AB=x\le1,AC=1$, then $\angle C=\arcsin x=\arcsin\sqrt{1-x^2}$.

So we have $\arcsin(x)=\begin{cases}\arccos\sqrt{1-x^2},&0\le x\le1\\-\arccos\sqrt{1-x^2},&-1\le x<0\end{cases}$

where for $x\in[-1,0)$, we added a minus sign before $\arccos\sqrt{1-x^2}$ since $\arcsin(-x)=-\arcsin(x)$.

Title: Derivative of $(\arcsin (x))$


答:
Hint: Let $ax^3 = by^3 = cz^3 = k $.

Show that $k = ax^2 + by^2 + cz^2$ and that $\sqrt[3]{k} = \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c}$, hence the result follows.

 $k = \frac{k}{x} + \frac{k}{y} + \frac{k}{z} = ax^2 + by^2 + cz^2$

 $\sqrt[3]{k} = \frac{\sqrt[3]{k}}{ x} + \frac{\sqrt[3]{k}}{ y} + \frac{\sqrt[3]{k}}{ z}  = \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c}$

Title: If $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 ;$ prove that $\sqrt[3]{a x^{2}+b y^{2}+c z^{2}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c} .$

Body:
If $a x^{3}=b y^{3}=c z^{3}$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 ;$ prove that $\sqrt[3]{a x^{2}+b y^{2}+c z^{2}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c} .$

Thanks!


问:
Solving this nonlinear system (a localization problem) with gradient descent.
I have the following algorithm designed to find the global minimum of the simple function $y=(x+5)^2$.

```<code>cur_x = 3                  # the algorithm starts at x=3
rate = 0.01                # learning rate
precision = 0.000001       # this tells us when to stop the algorithm
previous_step_size = 1
max_iters = 10000          # maximum number of iterations
iters = 0                  # iteration counter
df = lambda x: 2*(x+5)     # gradient of our function

while previous_step_size > precision and iters < max_iters:
    prev_x = cur_x                                # store current x value in prev_x
    cur_x = cur_x - rate * df(prev_x)             # grad descent
    previous_step_size = abs(cur_x - prev_x)      # change in x
    iters = iters+1                               # iteration count
    print("Iteration",iters,"\nX value is",cur_x) # print iterations

print("The local minimum occurs at", cur_x)
</code>```

I'd now like to apply this to a localization problem, the Time Difference of Arrival problem, in 3-dimensions. That is, given the velocity $v$ of some signal, the coordinates $[x_i,y_i]$ of four observers (or, in general, $n+1$ observers for an $n$ dimensional solution), and the time of arrival at each observer, I want to reconstruct the coordinates $[x,y]$ of the signal source.

I've accomplished this in two dimensions using a slight variation of the approximation search algorithm found here: https://stackoverflow.com/questions/63707712/how-to-localize-a-signal-given-the-location-of-three-receivers-and-the-times-at/64318443#64318443. I'd now like to try doing so with gradient descent, although I'm not entirely sure how to apply it here (although I know it has been done).

I also know that the 2-dimensional solution can be described by the following nonlinear system:

$\sqrt{(x-x_1)^2+(y-y_1)^2}+s(t_2-t_1) = \sqrt{(x-x_2)^2 + (y-y_2)^2}$

$\sqrt{(x-x_2)^2+(y-y_2)^2}+s(t_3-t_2) = \sqrt{(x-x_3)^2 + (y-y_3)^2}$

$\sqrt{(x-x_3)^2+(y-y_3)^2}+s(t_1-t_3) = \sqrt{(x-x_1)^2 + (y-y_1)^2}$

How, precisely, might gradient descent be used to solve the problem in 3-dimensions?

I've had a look in the usual places (e.g., Wikipedia: https://en.wikipedia.org/wiki/Gradient_descent#Solution_of_a_non-linear_system), however I'm used to thinking of this "computationally" and I'm not familiar with the terminology/symbolism used there.

答:
Having worked this problem, what I found is that it is much better to work with absolute times in order to decouple the equations.

In three dimensions, eash equation write as
$$f_i=\sqrt{(X-x_i)^2+(Y-y_i)^2+(Z-z_i)^2}-v(t_i-T)=0$$ and you need to minimize
$$\Phi(X,Y,Z,T)=\frac 12\sum_{i=1}^n f_i^2$$ which is extremely nonlinear; this means that you need "reasonable" estimates of the four variables $(X,Y,Z,T)$ before starting anything.

What I did is to consider in a preliminary step the equations
$$g_i=(X-x_i)^2+(Y-y_i)^2+(Z-z_i)^2-v^2(t_i-T)^2$$ and built the $\frac {n(n-1)}2$ equations $(g_j-g_i)$ ($i$ varying from $1$ to $(n-1)$ and $j$ from $(i+1)$ to $n$); they write
$$2 (x_j- x_i) X+2 (y_j- y_i) Y+2 (z_j- z_i) Z+2 v^2 (t_i-t_j)T=$$
$$(x_j^2+y_j^2+z_j^2-v^2 t_j^2)-(x_i^2+y_i^2+z_i^2-v^2 t_i^2)$$ This system is very easy to solve in the least-square sense using matrices. So, at this point, we have the estimates  for the four variables $(X,Y,Z,T)$.

Now, we need to minimize $\Phi(X,Y,Z,T)$. Writing the partial derivatives, we have to solve the four equations
$$\frac{\partial \Phi(X,Y,Z,T)} {\partial X}= \sum_{i=1}^n f_i \,\frac{\partial f_i} {\partial X}=0$$
$$\frac{\partial \Phi(X,Y,Z,T)} {\partial Y}= \sum_{i=1}^n f_i \,\frac{\partial f_i} {\partial Y}=0$$
$$\frac{\partial \Phi(X,Y,Z,T)} {\partial Z}= \sum_{i=1}^n f_i \,\frac{\partial f_i} {\partial Z}=0$$
$$\frac{\partial \Phi(X,Y,Z,T)} {\partial T}= \sum_{i=1}^n f_i \,\frac{\partial f_i} {\partial T}=0$$
with
$$\frac{\partial f_i} {\partial X}=\frac{X-x_i}{\sqrt{(X-x_i)^2+(Y-y_i)^2+(Z-z_i)^2}}$$
$$\frac{\partial f_i} {\partial Y}=\frac{Y-y_i}{\sqrt{(X-x_i)^2+(Y-y_i)^2+(Z-z_i)^2}}$$
$$\frac{\partial f_i} {\partial Z}=\frac{Z-z_i}{\sqrt{(X-x_i)^2+(Y-y_i)^2+(Z-z_i)^2}}$$
$$\frac{\partial f_i} {\partial T}=v$$ This system of equations is quite easy to solve with Newton-Raphson method.


答:
Setting $g(x)= \arcsin(x) \in [ \frac{-\pi}{2}, \frac{\pi}{2} ]$

$\sin( g(x))=x \longrightarrow g'(x) \cos( g(x))= 1$

On the other hand, as $ g(x) \in [ \frac{-\pi}{2}, \frac{\pi}{2} ]$, we have:

$\cos( g(x))= \sqrt{1- \sin^2(g(x))}=\sqrt{1-x^2} $.
 So we're done.

Title: Derivative of $(\arcsin (x))$

Body:
I was trying to solve the following exercise:

Show that $(\arcsin (x))'=\frac{1}{\sqrt{1-x^2}}$ for all $x\in (-1,1)$.

So far I got that $(\arcsin (x))'=\frac{1}{\cos(\arcsin (x))}$, but i don't know how to go from $\frac{1}{\cos(\arcsin (x))}$ to $\frac{1}{\sqrt{1-x^2}}$. Any help is appreciated.


问:
What's the intuition for strongly consistent estimator?
I'm working through a portion of my statistics notes. Weak and strong consistency of estimators is defined. While I can intuitively understand the meaning of weak consistency (wikipedia has a very nice graph at the top of this article: https://en.wikipedia.org/wiki/Consistent_estimator), I cannot picture anything when thinking about strong consistency. Is there any intuition for it?</p>

答: This is basically the difference between convergence in probability and convergence almost surely. Almost sure convergence means that the events on which the estimate does not converge to the true value have probability 0. On the other hand, convergence in probability just means that the probability of deviation from the true value goes to 0 in the limit.</p>


问:
Showing that Brownian motion is bounded with non-zero probability
How do you show, that for every bound $\epsilon$, there is a non-zero probability that the motion is bounded on a finite interval.
i.e. $$\mathbb{P} (\sup_{t\in[0,1]} |B(t)| < \epsilon) > 0$$

I tried using the reflection principle, I can show that if B* is the motion reflected on the hitting time with the bound, $\tau$, then I think that
$$P(\sup_{t\in[0,1]} |B(t)| > \epsilon) = P(|B(1)| > \epsilon) + P(|B*(1)| > \epsilon) - P(\tau < 1, |B(1)-B(\tau)| > 2\epsilon)$$
However, I have no idea how to bound the third term (in a useful way, which would show that this probability is non-one)

答:
Here's three different methods of showing that $\mathbb{P}(\sup_{t\in[0,1]}\vert B_t\vert < \epsilon)$ is nonzero.

A simple argument based on intuition. You can break the unit interval into a lot of small time steps and, by continuity, the Brownian motion will not move much across each of these steps. By independence of the increments, there is a positive (but small) probability that they largely cancel out, so $B$ stays within $\epsilon$ of the origin. To make this precise, choose a positive integer $n$ such that $q\equiv\mathbb{P}(\sup_{t\le1/n}\vert B_t\vert < \epsilon/2)$ is nonzero ($q$ can be made as close to 1 as you like, by taking $n$ large). By symmetry, the event $\{\sup_{t\le1/n}\vert B_t\vert < \epsilon/2,\ B_{1/n}>0\}$ has probability $q/2$. Note that, if $\sup_{t\in[k/n,(k+1)/n]}\vert B_t-B_{k/n}\vert < \epsilon/2$ and $B_{(k+1)/n}-B_{k/n}$ has the opposite sign to $B_{k/n}$ for each $k=0,1,\ldots,n-1$ then $\vert B_t\vert$ will be bounded by $\epsilon/2$ at the times $k/n$ and, therefore, $\sup_{t\le1}\vert B_t\vert$ will be less than $\epsilon$. So, $\mathbb{P}(\sup_{t\le1}\vert B_t\vert < \epsilon)\ge(q/2)^n$.

Use a cunning trick. If $X,Y$ are independent Brownian motions over the interval $[0,1]$, then $B=(X-Y)/\sqrt{2}$ is also a Brownian motion. The sample paths of $X,Y,B$ can be considered as lying in the (complete, separable) metric space $C([0,1])$ of continuous functions $[0,1]\to\mathbb{R}$ under the supremum norm. By separability, $C([0,1])$ can be covered by countably many open balls of radius $\epsilon/\sqrt{2}$. So, by countable additivity of the probability measure, there exists at least one such ball containing $X$ with probability $q > 0$. By independence, $X,Y$ are both contained in this ball with probability $q^2 > 0$, in which case $\Vert B\Vert_\infty=\Vert X- Y\Vert_\infty/\sqrt{2}<\epsilon$.

Exact calculation. You can calculate an exact expression for the probability, as an infinite sum, and verify that it is dominated by a single positive term as $\epsilon$ goes to zero. This is not as simple as the intuitive argument I gave above, but has the advantage that it also gives an accurate asymptotic expression for the probability, which goes to zero like $e^{-\pi^2/(8\epsilon^2)}$ as $\epsilon\to0$ (this is positive, but tends to zero very quickly).

The probability can be calculated using the reflection principal (also see my comments and Douglas Zare's answer to this question). Writing $p(x)=(2\pi)^{-1/2}e^{-x^2/2}$ for the probability density function of $B_1$ and $f(x)=\sum_{n=-\infty}^\infty(-1)^n1_{\{(2n-1)\epsilon < x < (2n+1)\epsilon\}}$ (which is a kind of square wave function),
$$
\mathbb{P}\left(\sup_{t\le1}\vert B_t\vert < \epsilon\right)=\mathbb{E}[f(B_1)]=\int_{-\infty}^\infty f(x)p(x)\,dx.\qquad{\rm(1)}
$$
This expression comes from the reflection principle, which says that reflecting $B$ after it first hits $\pm\epsilon$ gives another Brownian motion. That is, $\hat B_t\equiv B_t+1_{\{t\ge T\}}2(B_T-B_t)$ is a Brownian motion, where $T$ is the first time at which $\vert B_T\vert=\epsilon$. As $f$ is antisymmetric about both $\epsilon$ and $-\epsilon$, the sum $f(B_1)+f(\hat B_1)$ vanishes whenever $T\le1$. So, $1_{\{T > 1\}}=(f(B_1)+f(\hat B_1))/2$, and taking the expectation gives (1).

You can perform the integral in (1) directly to express the probability as an infinite sum over the cumulative normal distribution function, but this is not so good in the limit where $\epsilon$ is small, as you don't have a single dominant term. Alternatively, the integral in (1) can be written as $\int_{-\epsilon}^\epsilon\theta(x)\,dx$ where $\theta(x)=\sum_{n=-\infty}^\infty(-1)^np(x+2n\epsilon)$. As $\theta$ has period $4\epsilon$ you can write it as a Fourier series, and working out the coefficients gives
$$
\theta(x)=\epsilon^{-1}\sum_{\substack{n > 0,\\n{\rm\ odd}}}\cos\left(\frac{n\pi x}{2\epsilon}\right)\exp\left(-\frac{n^2\pi^2}{8\epsilon^2}\right).
$$
This is a very fast converging sum, especially for small $\epsilon$ (the terms vanish much faster then exponentially in $n$). Actually, $\theta$ is a theta function and the fourier transform is the same thing as the Jacobi identity. Integrating it term by term gives
$$
\mathbb{P}\left(\sup_{t\le1}\vert B_t\vert < \epsilon\right)=\sum_{\substack{n > 0,\\ n{\rm\ odd}}}\frac{4}{n\pi}(-1)^{(n-1)/2}\exp\left(-\frac{n^2\pi^2}{8\epsilon^2}\right)
$$
As the first term goes to zero much more slowly than the sum of the remaining terms (as $\epsilon\to0$) this gives the asymptotic expression
$$
\mathbb{P}\left(\sup_{t\le1}\vert B_t\vert < \epsilon\right)\sim\frac{4}{\pi}\exp\left(-\frac{\pi^2}{8\epsilon^2}\right).
$$


问:
finding fixed field of automorphism
Let $F$ be a field and let $g:F(x) \to F(x)$ be the automorphism which maps $x$ to $x+1$.
I need to find the fixed field of this automorphism.
So far I know $g$ fixes $F$. I want to use Galois applications but the group generated by $g$ is infinite.
Can I have a hint how to start?
Thanks.</p>

答:
Hint: a standard way to show that a rational function over a field is constant is to show it has either has infinitely many zeroes (in its algebraic closure), or that it has no zeroes and no poles (in its algebraic closure).

I'm assuming you're in a field of characteristic zero? Otherwise, your automorphism group is finite.

Title: finding fixed field of automorphism