问:
Calculating the following integral $\int^{\infty}_{2}\frac{x-1}{x^3+4x+7}$
I'm trying to find the solution of the following integral, to check for convergence or divergence. Though, I'm finding the integral itself difficult to resolve. I welcome any feedback towards my approach.

$$\int^{\infty}_{2}\frac{x-1}{x^3+4x+7}$$

My approach:

$$\int^{\infty}_{2}\frac{x-1}{x^3+4x+7} \implies \frac{1}{2}\int^{\infty}_{2}\frac{2x-4}{x^3+4x+7}+\frac{1}{2}\int^{\infty}_{2}\frac{2}{x^3+4x+7}$$

Then to solve for the left-side first, I'm substiting u for $x^2-4x+7$, so the denominator looks like this: $xu$

Then to find x, such that: $x = \frac{u-7}{x-4}$

To get:

$$\frac{1}{2}\int^{\infty}_{2}\frac{du}{\frac{u-7}{x-4}\cdot u}$$

Though, I'm not sure if my apporach is valid, hence I would really enjoy some support towards my approach.

Body: Write $x^3+4x+7=(x-a)((x-b)^2+c^2)$ with $a\approx-1.25538,\,b\approx0.627692,\,c\approx2.2764$. Then$$\begin{align}\frac{x-1}{x^3+4x+7}&\equiv\frac{A}{x-a}+\frac{B(x-b)+C}{(x-b)^2+c^2}\\\iff x-b+b-1&\equiv A((x-b)^2+c^2)+(x-b+b-a)(B(x-b)+C)\\&=(A+B)(x-b)^2+((b-a)B+C)(x-b)+Ac^2+(b-a)C\\\iff A=-B&=\frac{a-1}{(b-a)^2},\\C&=\frac{b-1}{b-a}.\end{align}$$One antiderivative is $A\ln\tfrac{|x-a|}{\sqrt{(x-b)^2+c^2}}+\tfrac{C}{c}\arctan\tfrac{x-b}{c}$, so the integral is $A\ln\tfrac{\sqrt{(2-b)^2+c^2}}{|2-a|}+\tfrac{C}{c}\arctan\tfrac{c}{2-b}$.


问:
What is the relation?
The function $x^2 = y\quad$ limits two areas $A$ and $B$:

$A$ is further limited with the line $x= a$, $a\gt 0$.
$A$ rotates around the $x$-axis, which gives Volume $A = Va$.

$B$ is limited with the line $y=b$, $b\gt 0$.
$B$ rotates around the $y$-axis, which gives Volume $B = Vb$.

What are the relations between $a$ and $b$, when $Vb = Va$?

..................

I have come to the solution that:

$Vb = (\pi b^2 )/ 2$

$Va = (\pi a^5) / 5$

so the relation between them is:

$2.5b^2 = a^5$

Is that the final solution or is it more?

答:
If we take the region bounded by the $y$-axis, the $x$-axis, the line $x=a$ (with $a\gt 0$), and the parabola $y=x^2$, and rotate it about the $x$-axis, the volume of the resulting solid of revolution is easily computed (using, for example, discs perpendicular to the $x$-axis) to be
$$\text{Volume A} = \int_0^a \pi(x^2)^2\,dx = \frac{\pi}{5}x^5\Bigm|_0^a = \frac{\pi a^5}{5}.$$
If the region bounded by the $y$-axis, the $x$ axis, the line $y=b$ (with $b\gt 0$), and the parabola $y=x^2$ is revolved around the $y$-axis, then using discs perpendicular to the $y$-axis we obtain the volume to be:
$$\text{Volume B} = \int_0^b \pi (\sqrt{y})^2\,dy = \frac{\pi}{2}y^2\Bigm|_0^b = \frac{\pi b^2}{2}.$$
So your computations are correct there.

If the two volumes are the same, then we must have
$$\text{Volume A} = \frac{\pi a^5}{5} = \frac{\pi b^2}{2} = \text{Volume B};$$
there are many ways to express this: you can solve for one of $a$ or $b$ in terms of the other:
$$b = \sqrt{\frac{2a^5}{5}} = a^{5/2}\sqrt{\frac{2}{5}},$$
or, if you want to express $a$ in terms of $b$ instead,
$$ a = \sqrt[5]{\frac{5}{2}b^2} = b^{2/5}\sqrt[5]{\frac{5}{2}}.$$
Or you can simply express this relation by saying, say
$$2a^5 = 5b^2.$$

Note. If $a\lt 0$, then the volume of $A$ can be computed the same way, but the integral would go from $a$ to $0$, so that the volume would be $-\frac{\pi a^5}{5}$; to account for both possibilities, both $a\gt 0$ and $a\lt 0$, you can simply write that the volume is $\frac{\pi|a|^5}{5}$. For solid $B$, however, it makes no sense to talk about $b\lt 0$, because then we don't have a finite area "enclosed" by the curves in question.

ContentLicense: CC BY-SA 3.0

Title: What is the relation?


问:
Continuity Question, proving epsilon-delta
Question:

Suppose that $f$ is a continuous function at the point $x_0$ and that $f(x_0) \neq 0$.  Using the $\epsilon−\delta$ definition of the continuity of a function at $x_0$, show that there exists $\delta>0$ such that $$f(x_0) \neq 0 \ {whenever}\ |x-x_0| < \delta \ and \ x\in[a,b] $$

My thoughts are we use the defintion as stated in the question :

$|f(x) - f(x_0)| < \epsilon$ whenever $|x-x_0| < \delta$
and then we can use the triangle inequality "$|a-b|| \geq ||a| - |b||$" and then use the fact that $f(x_0) \neq 0 $ and then when we balance the inequality we get that $\epsilon - f(x_0) \neq 0 $ which we can say that it is $\neq 0$

If someone could guide me through how to complete this quesition to the end, it would be really appreciated, thank you!

答:
Suppose WLOG that $f(x_0)\lt 0$ so that $-f(x_0)\gt 0$

Since $f$ is continuous at $x=x_0$, it follows that for every $\epsilon\gt 0,\exists \delta_\epsilon\gt 0$ such that $|x-x_0|\lt \delta_\epsilon\implies |f(x)-f(x_0)|\lt \epsilon$. Choose $\epsilon = -f(x_0)$ and hence by continuity of $f$ at $x_0$, there exists a $\delta\gt 0$ such that

$\begin{align}|x-x_0|\lt \delta
&\implies |f(x)-f(x_0)|\lt -f(x_0)\\
&\implies f(x_0)\lt f(x)-f(x_0)\lt -f(x_0)\\
&\implies 2f(x_0)\lt f(x)\lt 0\\
&\implies \text{$f(x)\ne0$ for $x:|x-x_0|\lt \delta$}\end{align}$

Similarly, prove for $f(x_0)\gt 0$.

ContentLicense: CC BY-SA 4.0


问:
Infinitesimal limit of Poisson process
A Poisson process $N_t$ has a jump size $dN$ and a rate $\lambda$. Can a Poisson process be viewed as a diffusion process in the infinitesimal limit, i.e. as $dN \rightarrow 0$? Here is a related question, but my question is about scaling $dN$ in a finite time frame $t\in [0,T]$ in which case I don't think $\lambda$ would need to go to infinity.

The mean and variance of a Poisson process are both $\lambda$, so I was thinking the diffusion would be
$$dN_t = \lambda dt + \sqrt{\lambda} dB_t.$$

I don't know whether this form holds  or how to prove/disprove it. The Poisson process transition probability is
$$ P(N(t+dt) - N(t) = k) = \frac{(\lambda dt)^k e^{-\lambda dt}}{k!}$$
How can we extend this to the continuous case? Would it be, for $x\in \mathbb{R}$,
$$ P(N(t+dt) - N(t) = x) = \frac{(\lambda dt)^x e^{-\lambda dt}}{\Gamma(x)}?$$

答:
We have
$$
P(N(t+\Delta t)- N(t) = k) = \frac{(\lambda \Delta t)^k \mathrm{e}^{- \lambda \Delta t}}{k !}
$$
as you noted. Over large time intervals $\Delta t$ we thus have a convergence to a normal distribution
$$
\frac{N(t+\Delta t)- N(t) - \lambda \Delta t}{\sqrt{\lambda \Delta t}} \quad \xrightarrow[]{d} \quad \text{Normal}(0,1).
$$
as is standard for a Poisson distribution.

Now consider a random walk
$$
x(t+ d t) = \mu d t + \sigma d W_t
$$
for which we have
$$
\frac{x(t+\Delta t) - x(t) - \mu \Delta t}{\sigma \sqrt{\Delta t}} \sim \text{Normal(0,1)}
$$
Thus if one restricts to looking stroboscopically at time intervals $t \in \Delta t \mathbb{N}$ we see the two processes exhibit the same convergence in distribution if we set $\sigma = \sqrt{\lambda}$, and $\mu = \lambda$ in the limit of large $\Delta t$.

More coarsely, a Poisson process "looks" like a random walk as long as it is observed on timescales much larger than than the inverse rate constant
$$
\Delta t \gg \lambda^{-1}.
$$

Title: Infinitesimal limit of Poisson process


问:
How to define a Parse Tree, and the equality of parse tree?
I understand parse tree is used to describe a derive process of a context-free grammar $G=<V_N,V_T,P,S>$.

Thus in my point of view, it is a rooted tree $R=<V,E>$ with some kind of mapping of $f:V\to(V_N\cup V_T)$.

Furthermore, it also need to express the order of the produced string, so there must be an order relation $\prec$ on node of the tree $V$ .

Thus in my point of view, the parse tree need to be defined as $T=<V,E,f,\prec>$

But I would like to know how to precisely define each of the element within them. And furthermore, how to define the equality of 2 trees.

Because if the equality of tree is not formally defined, the ambiguity of a grammar cannot be defined.

Thanks a lot!

答:
There are some precise formulations of ordered rooted trees on Wikipedia. The first formalism proposed in that article defines

An ordered tree is a structure $(X, ≤_V, ≤_S)$ where $X$ is a non-empty set of nodes and $≤_V$ and $≤_S$ are relations on $X$ called vertical… order and sibling order, respectively.

For a parse tree, as you say, it is necessary to restrict $X$ to being a finite set, and to add a third relation $f: X\to (V_N \cup V_T)$ in order to label the nodes of a parse tree.

I've left out of the above definition a large number of restrictions on the relations which are required for the structure to be a well-formed parse tree. In addition to the restrictions provided in the linked Wikipedia article, there is a restriction of the labels in order that every node correspond to some production in the grammar. None of these restrictions is crucial to the definition of equality.

With that in place, we can say that two parse trees $(X, ≤_V, ≤_S, f)$ and $(X', ≤'_V, ≤'_S, f')$ are equivalent if there is an isomorphism between $X$ and $X'$ which preserves all of the relations.

But that's not the way that grammatical ambiguity is general defined. It is generally defined in terms of leftmost derivations.

Recall that a derivation of the sentence $\alpha \in V_T^*$ is an ordered sequence $\omega_0\to\omega_1\to\omega_2\to...\to\omega_n$ where each $\omega_i \in(V_N\cup V_T)^*, \omega_0 = S, \omega_n = \alpha$ and each $\omega_i$ is produced from $\omega_{i-1}$ by replacing some symbol $A$ in $V_N$ found in $\omega_{i-1}$ with the right-hand side of a production $A\to\nu\in P$. A derivation is a leftmost derivation if, at each step, the replaced non-terminal is the first non-terminal in the sentential form. (The point of this restriction is that every parse tree corresponds to precisely one leftmost derivation, although it might correspond to a number of different derivations.)

For completeness, two derivations are equal if they are equal symbol-for-symbol.

Then a grammar is unambiguous if there is no sentence which can be derived with two different leftmost derivations.


问:
Is existential instantiation related to "Let x be ..."
I have a question about existential instantiation.

So, I am familiar with statements like "Let $x$ be an integer" or "Let $x$ $\in$ $\mathbb{Z}$". These statements define $x$ to represent a particular, but arbitrarily chosen integer.

My question is, are we doing a similar thing when we do existential instantion? For example, lets start with the existential statement $P$ = "There exists an integer $x$ such that $x^2 = 4$". When we do existential instantiation on $P$, are we saying "Let $a$ be an element of the set of integers that make $P$ true?"

答:
Yes. If you already have a theorem $\exists x.P(x)$ then you can apply existential instantiation to derive $P(a)$, where $a$ is a new symbol. If you don't have the existence theorem, though, then introducing $P(a)$ constitutes making a new assumption that you need to keep track of.

In normal natural-language math exposition, we don't always clearly distinguish between these two situations at the point where $P(a)$ is introduced - the formal proof structure is something the reader reconstructs based on their understanding of what's being explained.


问:
Fourier series - unsure if I'm on the right track
I got

$$ f(x)=\begin{cases}-\dfrac{\pi}{2},& -\pi<x\le-\dfrac{\pi}{2}\\[1ex]
\phantom{-} x,&-\dfrac{\pi}{2}<x\le\dfrac{\pi}{2} \\[1ex]
\phantom{-}\dfrac\pi2,&\phantom{-}\dfrac{\pi}{2}<x\le\pi\end{cases}$$

Period is $2\pi$, which means $L$ is $\pi.$

Since this is an odd function, I need to find $b_n$, multiply that with $\sin\left(\frac{n\pi x}L\right)$ and that should be my series.

But I'm not sure if what I got is correct. I can't seem to be able to paste it into WolframAlpha with more than a few terms, either. The function itself is pretty straight forward, so you guys can probably see where I've gone wrong if I have!

Fourier series for an odd function should be $\sum\limits_{n=1}^\infty b_n \sin\left(\frac{n\pi x}L\right)$.

$$\sum_{n=1}^\infty \left(\frac{\sin\left(\frac{n\pi}2\right)}{n^2}-\frac{\pi\cos\left(\frac{n\pi}{2}\right)}{2n}+\frac{(-1)^n \pi}{2n}-\frac{\pi\cos\left(\frac{n\pi}{2}\right)}{2n} \right) \sin\left(\frac{n\pi x}\pi\right)$$

So, any fast math guys that can tell me where I should end up?

答:
$$\begin{align}
b_n&=\frac1\pi\int_{-\pi}^\pi f(x)\sin(nx)\,\mathrm dx\\[1ex]
&=-\frac12\int_{-\pi}^{-\frac\pi2}\sin(nx)\,\mathrm dx+\frac1\pi\int_{-\frac\pi2}^{\frac\pi2}x\sin(nx)\,\mathrm dx+\frac12\int_{\frac\pi2}^\pi\sin(nx)\,\mathrm dx\\[1ex]
&=\frac1\pi\int_{-\frac\pi2}^{\frac\pi2}x\sin(nx)\,\mathrm dx+\int_{\frac\pi2}^\pi\sin(nx)\,\mathrm dx\\[1ex]
&=\frac{2\sin\left(\frac{n\pi}2\right)-n\pi \cos\left(\frac{n\pi}2\right)}{n^2\pi}+\frac{\cos\left(\frac{n\pi}2\right)-\cos(n\pi)}n\\[1ex]
&=\frac{2\sin\left(\frac{n\pi}2\right)}{n^2\pi}-\frac{\cos(n\pi)}n\\[1ex]
&=\frac{2\sin\left(\frac{n\pi}2\right)}{n^2\pi}-\frac{(-1)^n}n\\[1ex]
&=\begin{cases}\dfrac{2\sin\left(\frac{2k\pi}2\right)}{(2k)^2\pi}-\dfrac{(-1)^{2k}}{2k}&n=2k\\[1ex]
\dfrac{2\sin\left(\frac{(2k-1)\pi}2\right)}{(2k-1)^2\pi}-\dfrac{(-1)^{2k-1}}{2k-1}&n=2k-1\end{cases}\\[1ex]
&=\begin{cases}-\dfrac1{2k}&n=2k\\[1ex]
\dfrac{2(-1)^{k+1}}{(2k-1)^2\pi}+\dfrac1{2k-1}&n=2k-1\end{cases}
\end{align}$$

where $k\ge1$ is a positive integer.


问:
independence of infinite collection of random variables
I am reading the book A First course in Probability and in the chapter of joint distribution, came through the text we say that an infinite collection of random variables is independent if every finite subcollection of them is independent.

How can this statement be proven and how can the finite subcollection be created from infinite collection of random variables since we need to be able to know all the elements in the collection to make all the possible subcollections.

答:
This statement is the definition of independence for an infinite collection. It does not need to be proved, we simply call a family of random variables independent if any finite subcollection is independent. That is, if $(X_i)_{i \in I}$ is any collection of random variables and it just so happens that for any $i_1,\ldots, i_n \in I$ we have that $(X_{i_1}, \ldots, X_{i_n})$ is independent in the sense that the joint probability distribution splits as a product of the marginal distributions, then we say that the family  $(X_{i})_{i \in I}$ is independent.

What would probably be your next question is that how could we possibly show that $(X_{i_1}, \ldots, X_{i_n})$ is independent for any finite subcollection when it seems to require checking an infinite number of conditions. Much like showing the square of any real number is non-negative, you do not need to have a separate proof for each real number. There is some extra structure that typically allows you to just take an arbitrary finite subcollection and directly show, regardless of which indices were selected, that the finite subcollection is independent.


问:
A rigid exponential question
This question was given by my friend

$$x^{x^{20}}=\sqrt[\sqrt 2]{2}$$

even after applying log this question dosen't simplifies

$x^{20}$ $ln$ $x$ $=$ $\frac{1}{\sqrt{2}}$ $ln$ $2$

if this question is put in wolfram alpha it gives and if I do a approx form it gives 1.067 how does this came

$x=e^{\frac{1}{20} W\left(10 \sqrt{2} \ln (2)\right)}$

How to solve this question

答:
Chances are this problem doesn't have an explicit solution.

Equations of the form

$$x\cdot e^x=c$$

$c\in\mathbb{R}\setminus\{0\}$

cannot be solved by simple algebra, but there is a function called Lambert W which can give us approximations.
It is used to solve such and similar equations (like $x\cdot\sin(x)=2$, $x\cdot\coth(x)=-4$).

Title: A rigid exponential question


问:
Prove or give a counterexample: $T$ is nilpotent $\iff$ $G(0,T)=V$ .
Suppose $V$ is a vector space and $T$ is a linear map on $V$. $\ G(\lambda,T) $ denotes the generalized eigenspace of $T$ corresponding to $\lambda $. In other words, $G(\lambda,T)=\{v\in V\mid {(T-\lambda I)}^{k}v=0\text{ for some integer }k>0\}$.

Prove or give a counterexample: $\ $ $T$ is nilpotent $\iff$ $G(0,T)=V $ .

BTW: It is true when V is finite-dimensional.$\;$This can be seen from 8.18 in Linear Algebra Done Right 3rd Edition.

Any insights are much appreciated.

答:
It does not follow that $T$ is nilpotent in the infinite-dimensional case. Take the vector space with countable basis
$$
\{ e_1, e_2, \ldots\}
$$
and let $T$ be the left-shift operator that annihilates $e_1$ and sends $e_{i+1}$ to $e_i$. Then the generalized eigenspace (with eigenvalue $0$) for $T$ is all of $V$ since for any basis vector, there is a power of $T$ that eliminates it, but $T$ is not nilpotent since there is no power of $T$ that eliminates all the basis vectors at once.


问:
a mesurable function and bounded a.e is of bounded variation?
Let $F:[0,1]\rightarrow \mathbb{R}$ be a measurable function, bounded a.e

Does this imply that $F$ is of bounded variation on $[0,1]$?

答: No, let $F$ be the indicator of the rational numbers. Then $F$ is bounded everywhere, measurable, and infinite variation.


答:
Not even true if we assume in addition that $F$ is continuous:
\begin{align}
F(x) &= \sin\frac{1}{x},\quad 0<x\le 1,
\\
F(0) &= 0
\end{align}

Title: a mesurable function and bounded a.e is of bounded variation?

Body:
Let $F:[0,1]\rightarrow \mathbb{R}$ be a measurable function, bounded a.e

Does this imply that $F$ is of bounded variation on $[0,1]$?


答:
Having rewritten the recurrence, we get an equation:

$$10R_{n-2}-7R_{n-1}+R_n=0$$

that can easily be solved.

The roots of a characteristical polymomial $10-7k+k^2=0$ are $k_1=2,k_2=5$.
So general solution is $R_n=c_12^{n}+c_25^{n}$, where $c_1$ and $c_2$ are arbitrary constants.

Title: $R_n = 3(2^n)-4(5^n)$, $n \geq0$, prove $R_n$ satisfies $R_n = 7R_{n-1}-10R_{n-2}$

Body:
So the question is:

$R_n=3(2^n)-4(5^n)$ for $n\ge 0$; prove that $R_n$ satisfies $R_n=7R_{n-1}-10R_{n-2}$.

I don't really know what to do from here. If I substitute

$$R_n = 3(2^n)-4(5^n)$$

into

$$Rn = 7R_{n-1}-10R_{n-2}$$

I end up getting

$$R_n = 7\Big(3(2^{n-1})-4(5^{n-1})\Big)-10\Big(3(2^{n-2})-4(5^{n-2})\Big)$$

Dont know what to do...

EDIT: Thanks to Zev, what I did was:

$$3(2^n)-4(5^n)=7\bigg[3(2^{n-1})-4(5^{n-1})\bigg]-10\bigg[3(2^{n-2})-4(5^{n-2})\bigg].$$
$$\begin{align*}
3(2^n)-4(5^n)&=21(2^{n-1})-28(5^{n-1})-30(2^{n-2})+40(5^{n-2})\\\\
3(2^n)-4(5^n)&=21(2^{n})(2^{-1})-28(5^{n})(5^{-1})-30(2^{n})(2^{-2})+40(5^{n})(5^{-2})\\\\
3(2^n)-4(5^n)&=21/2(2^{n})-28/5(5^{n})-30/2(2^{n})+40/5(5^{n})\\\\
3(2^n)-4(5^n)&=(2^{n})[21/2-30/4]+(5^{n})[40/25-28/25]\\\\
3(2^n)-4(5^n)&=(2^{n})[3]+(5^{n})[-4]\\\\
3(2^n)-4(5^n)&=3(2^{n})-4(5^{n})
\end{align*}$$


答:
Using by standard techniques, we have

$$\color{red}{\left(x^{x^{20}}\right)^{20}=\left(\sqrt[\sqrt 2]{2}\right)^{20}} \\\begin{align}&\implies x^{20 x^{20}}=\left(\sqrt[\sqrt 2]{2}\right)^{20} \\
&\implies \left({x^{20}}\right)^{ x^{20}}=\left(\sqrt[\sqrt 2]{2}\right)^{20}\end{align}$$

$$\color {blue}{ u(x)^{u(x)}=\left(\sqrt[\sqrt 2]{2}\right)^{20}}$$

$$\begin{align}&\implies \ln u(x)^{u(x)}=\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}\\
&\implies u(x) \ln u(x)=\ln \left(\sqrt[\sqrt 2]{2}\right)^{20} \\
&\implies \ln u(x) e^{\ln u(x)}=\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}\\
&\implies W \left(\ln u(x) e^{\ln u(x)}\right) =W\left(\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}\right) \\
&\implies \ln u(x)=W\left(\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}\right) \\
&\implies u(x)= e^{W\left(\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}\right)}\\
&\implies x^{20}=e^{W\left(\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}\right)}\end{align}$$

$$\begin{align}\color {gold}{\boxed {\color{black} {{x=e^{\frac {1}{20} W\left(\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}\right)}}}}}\end{align}$$

Note that ,

$$\left(\sqrt[\sqrt 2]{2}\right)^{20}=2^{10\sqrt 2}$$

$$\ln \left(\sqrt[\sqrt 2]{2}\right)^{20}=\ln 2^{10\sqrt 2}=10\sqrt 2 \ln 2.$$

Finally, the real root of our equation is equal to:

$$\large \color {gold}{\boxed {\color{black} {{{x=e^{1/20 ~W\left (10 \sqrt 2 \ln 2\right)}}}}}}$$

Title: A rigid exponential question

Body:
This question was given by my friend

$$x^{x^{20}}=\sqrt[\sqrt 2]{2}$$

even after applying log this question dosen't simplifies

$x^{20}$ $ln$ $x$ $=$ $\frac{1}{\sqrt{2}}$ $ln$ $2$

if this question is put in wolfram alpha it gives and if I do a approx form it gives 1.067 how does this came

$x=e^{\frac{1}{20} W\left(10 \sqrt{2} \ln (2)\right)}$

How to solve this question


问:
Is this a circular proof of Pythagorean Theorem? If not, are there better ones?
I found this proof of Pythagorean Theorem from  what 3Blue1Brown shows in his Lockdown math lecture  "Trigonometry Fundamentals":

In the following right triangle, project $\cos\alpha$ and $\sin\alpha$ back to the hypotenuse of length $1$, they become $\cos^2\alpha$ and $\sin^2\alpha$:
$\to\cos^2\alpha+\sin^2\alpha =1 $.

I thought this is what a simple, visual, and elegant proof!  My questions:

Is it a circular proof, why?

If not, are there better ones?

I want clarity about whether or not this proof is circular.  The argument only depends on the definition of the sine and cosine functions and nothing more (please correct me if I am wrong).  This is why I love it.  Many other geometry solutions involve more than three shapes.

答:
<img src="https://i.stack.imgur.com/w3aae.png" alt="Filling an $(a+b) \times (a+b) $ square" />

You can see above two squares of side $(a+b)$ filled with some squares and triangles. Can you spot the visual proof of Pythagorean Theorem behind these pictures?


问:
Does every convex planar set contain a centrally symmetric subset with at least $2/3$ its area?
Let $S$ be a bounded convex subset of the plane of unit area. Can we guarantee the existence of a centrally-symmetric subset $C⊆ S$ of area $2/3$?

If $S$ is any triangle, this bound is tight, attained by a hexagon whose vertices are $1/3$ of the way across each side:

I suspect the triangle is the unique worst case for this problem, but didn't see a great way to prove it - perhaps there is some other convex shape which only has a centrally symmetric subset of area at most $0.6$ or something, though I would be quite surprised.

It's easy to find a $C$ with at least half the area of $S$, because every convex set contains a rectangle of at least half its area (see e.g. this MO thread for references). How much can we improve this lower bound?

In the event of a positive resolution, I am curious whether one can find such a subset only using hexagonal $C$.

I imagine this question is discussed somewhere in the literature; if so, pointers to relevant papers or extensions to higher dimensions would be welcome.

答: Yes, this is the Kovner-Besikovich Theorem.</p>


答: I hope the author wasn't regarding that picture as a proof of Pythagoras. Since trigonometry depends on the Pythagorean theorem, the above "demonstration" employs circular reasoning and wouldn't count as a valid proof. So yes, there are better proofs.</p>

Title: Is this a circular proof of Pythagorean Theorem? If not, are there better ones?

Body:
I found this proof of Pythagorean Theorem from  what 3Blue1Brown shows in his Lockdown math lecture  "Trigonometry Fundamentals":

In the following right triangle, project $\cos\alpha$ and $\sin\alpha$ back to the hypotenuse of length $1$, they become $\cos^2\alpha$ and $\sin^2\alpha$:
$\to\cos^2\alpha+\sin^2\alpha =1 $.

I thought this is what a simple, visual, and elegant proof!  My questions:

Is it a circular proof, why?

If not, are there better ones?

I want clarity about whether or not this proof is circular.  The argument only depends on the definition of the sine and cosine functions and nothing more (please correct me if I am wrong).  This is why I love it.  Many other geometry solutions involve more than three shapes.


问:
infinite sum of analytic functions in this special case
Let $\mathbf{p}=\{p_{1},p_{2},\cdots, p_{k},\cdots\}$ be a probability sequence. Let $f_{k}(t)=p_{k}^t$ for $t\geq 1$. $f_{k}(t)$ is analytic on $[1,\infty)$. But is $$f(t)=\sum_{k\geq 1}f_{k}(t)$$ analytic on $[1,\infty)$? If not (which I suspect to be the case), what kind of conditions on $\mathbf{p}$ would make $f(t)$ to be analytic?

答:
Suppose $p_k > 0$ and $\sum_{k=1}^\infty p_k$.  Consider the function of a complex variable
$$
F(t) = \sum_{k=1}^\infty p_k^t .
$$
Now $F(t)$ converges uniformly for $\operatorname{Re} t \le 1$ by comparison with $\sum p_k$ in the Weierstrass $M$ test.  Then note that $F$
is holomorphic in $\operatorname{Re} t > 1$ by Morera's theorem.

What does "analytic" mean at an endpoint $1$ ?  If $\sum p_k^t$ converges for some $t_0<1$, then the same reasoning shows that $F$ is holomorphic on $\operatorname{Re} t > t_0$.


问:
A function with the following properties is always bounded, uniformly continuous, and if $\exists \lim_{x\to \infty} f(x)$ then $f$ is constant.
Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function with the following property: Given $\epsilon > 0$, there exists a positive real number $r > 0$ and a set $D \subset \mathbb{R}$ such that

Any interval $I$ with length $r >0$, is such that $D \cap I \ne \emptyset$

$|f(x+t) - f(x)| < \epsilon$ for any $t \in D$ and $x\in \mathbb{R}$

Show that a function with those properties is always bounded, uniformly continuous, and if $\exists \lim_{x\to \infty} f(x)$ then $f$ is constant.

I've been stuck trying to prove this for a while. I think that any $f$ is a periodic function, like $\sin$, and that would imply that it is bounded and that if $\exists \lim_{x\to \infty} f(x)$ then $f$ is constant. But I don't know how to show that $f$ is periodic, nor if it is the best approach for this proof.

答:
This is a partial answer.

I prove only the last statement: if $\lim_{x \to \infty} f(x)$ exists, then $f$ is constant. The proof is by contradiction.

Suppose $f$ is not constant, and pick two distinct real numbers $x_0$ and $y_0$ such that
$$f(x_0) < f(y_0)$$
Call $3A = f(y_0) - f(x_0)$, which is a positive real number. Moreover, denote by $L= \lim_{x \to \infty} f(x)$.

By our assumption on $f$, for all $\varepsilon >0$ there exists $D( \varepsilon ) \subseteq \Bbb R$ and $r( \varepsilon) >0$ such that the two conditions are satisfied.

Consider for all $n \ge 0$ the sets
$$D_n = D( A 3^{-n} )$$ and the positive real numbers
$$r_n = r( A 3^{-n} )$$
Moreover, we construct a sequence $(x_n, y_n) \in \Bbb R^2$ recursively, starting from $(x_0,y_0)$ previously defined, and with the property that

$x_n, y_n \in D_n$

$x_{n+1}> x_n+1$ and $y_{n+1}>y_n+1$ (this is always possible, since $D_n$ is unbounded)

Note that $\lim_{n \to \infty} x_n = + \infty$ and $\lim_{n \to \infty} y_n = + \infty$.

Now we are ready to construct our contradiction. For all $n \ge 1$ you have
$$|f(x_0 + x_n) - f(x_0) | = \left| \sum_{k=1}^{n-1} f(x_0+x_{k+1}) - f(x_0 + x_k)  + (f(x_0 + x_1) - f(x_0))\right| \le$$
$$ \le \sum_{k=1}^{n-1} |f(x_0+x_{k+1}) - f(x_0 + x_k)|  + |f(x_0 + x_1) - f(x_0)| \le$$
$$ \le \sum_{k=1}^{n-1} |f(x_0+x_{k+1})- f(x_0)| + |f(x_0)- f(x_0 + x_k)|  + |f(x_0 + x_1) - f(x_0)| \le$$
$$ \le \sum_{k=1}^{n-1} A 3^{-(k+1)} + A 3^{-k}  + A 3^{-1} \le \sum_{k=1}^{\infty} A 3^{-(k+1)} + A 3^{-k}  + A 3^{-1} = A$$
With a similar method we can prove that
$$|f(y_0 + y_n) - f(y_0) | \le A$$
Taking the limit as $n \to \infty$ we get
$$|L - f(x_0)| \le A , \ |L - f(y_0)| \le A$$ thus
$$3A= |f(y_0) - f(x_0)| \le |L - f(x_0)| + |L - f(y_0)| \le 2A$$
A contradiction.

My argument relies on picking smaller and smaller $\varepsilon_n$, such that the series of all $\varepsilon_n$ is smaller than one fixed constant we need.
Since $f$ is not too far from $f(x_0)$ somewhere far from $x_0$ and the same holds for $y_0$, you can get a contradiction in this way.

I think that proving $f$ is bounded and uniformly continuous uses the same argument. Maybe one can show that $f$ is Lipschitz too.


问:
Closed unit ball is compact?
Let $X$ be a Banach space and let $\operatorname{Lip}_{0}(X)$ be the space of all real-valued Lipschitz functions which vanish at $0$. The space $\operatorname{Lip}_{0}(X)$ is a Banach space when it is equipped with the Lipschitz norm, defined by:

$$L(f)=\|f\|_{\operatorname{Lip}}=\sup\left\{\frac{f(x)-f(y)}{\|x-y\|}:\,x,y\in X,\,x\neq y\right\}$$

My goal is to show that the closed unit ball of the space $\operatorname{Lip}_{0}(X)$ is compact for the topology of pointwise convergence. I have try with no success to use Tychonoff theorem or Banach-Alaoglu theorem. I failed because the Banach-Alaoglu theorem concerned the closed unit ball of the dual space with the weak-star topology.

Thank for any help.

答:
Tychonoff's theorem does give it to you — the closed unit ball $B$ of $\def\Lip{\operatorname{Lip}}\Lip_0(X)$ with the topology of pointwise convergence is naturally realized as a subspace of the product

$Y = \prod_{x \in X} I_x$

where $I_x = [\hspace{.1 cm}-||x||, ||x||\hspace{.1 cm}]$. The intervals are chosen by considering the Lipschitz expression with $y = 0$. Now, the product space $Y$ is compact by Tychonoff's Theorem, so we must only show that $B$ is closed in this space, or that $Y \setminus B$ is open.

Suppose $f \in Y \setminus B$. Then there are some $z, w \in X$ such that $\cfrac{f(z) - f(w)}{||z - w||} = 1 + \varepsilon > 1$. Choose open subsets $O_z$ and $O_w$ of $I_z$ and $I_w$ respectively such that $f(z) \in O_z$, $f(w) \in O_w$, and if $a \in O_z, b \in O_w$, then $\cfrac{a - b}{||z - w||} > 1 + \frac{\varepsilon}{2}$. This allows us to form a neighborhood of $f$ disjoint from $B$, namely, $\{ g \in Y \mathrel: g(z) \in O_z \text{ and } g(w) \in O_w \}$.

Title: Closed unit ball is compact?


问:
Does this vector bundle have a natural description?
Work over $\mathbb{C}$. Consider a smooth divisor of bidegree $(2,1)$ in $\mathbb{P}^2 \times \mathbb{P}^2$, for example $$X = \{\sum_{i=1}^{2} x_{i} y_{i}^2 = 0 \} \subset \mathbb{P}^2 \times \mathbb{P}^2.$$

Then, if I am not mistaken, the projection to the $\mathbb{P}^2$ with coordinates $[y_0:y_1:y_2]$ is a $\mathbb{P}^1$-bundle in the strong sense, i.e. all fibres are isomorphic to $\mathbb{P}^1$. Hence, we may write $X = \mathbb{P}(E)$ where $E$ is a rank $2$ vector bundle over $\mathbb{CP}^2$ (since the Brauer group of $\mathbb{CP}^2$ vanishes). I did some calculations in the cohomology ring of $X$ which suggest that we can take $E$ with $c_{1}(E) = 2H$, $c_{2}(E) = 4H^2$, where $H$ in hyperplance class. In particular $E$ doesn't split. Although note that $E$ is only defined up to taking a tensor product with a line bundle.

Question: Is there a "natural description" of $E$?

More precisely I am asking if some representative up to the equivelance relation of tensoring with a line bundle has a natural description.

I am hoping for something analogous to the fact that a divisor $Y \subset \mathbb{P}^2 \times \mathbb{P}^2$ with bidegree $(1,1)$ satsifies $Y \cong \mathbb{P}(T\mathbb{P}^2)$.

答:
The description depends on the convention used for the projectivization. If $\mathbb{P}(E)$ is defined as the scheme parameterizing lines in fibers of $E$, then the bundle $E$ is defined by the exact sequence
$$
0 \to E \to \mathcal{O}^{\oplus 3} \stackrel{(y_0^2,y_1^2,y_2^2)}\to \mathcal{O}(2) \to 0.
$$
And if the opposite convention is used, you should replace the above bundle by its dual.


问:
Rank of a matrix and cross product between eigenvectors
I have two questions about linear algebra.
I was doing a exercise that says:
Let $A$ be a real symmetric matrix $3 \times 3$ and $\det A = 6  $. Suppose that $u =(4,8,-1)$ and $v=(1,0,4)$ are eigenvectors of $A$ and $1$ and $2$ the eigenvalues associated respectively.

My first conclusion about $\det A=6\neq 0$ is the matrix $A$ is full rank so its rows are L.I (Linear independent), I was trying to figure out a way to link this result with the number of eigenvalues of matrix $A$, because the first question says:

1) The eigenvalues of $A$ are only $1$ and $2$?

If my first conclusion is wrong(i'm not sure), how to use the information from the statement to conclude that is false.

2) The cross product $u \times v$ is necessarily a eigenvector ?

For this one, I've calculated the cross product that result $(32, -17, -8)$, but I don't how to follow from this and prove that.

Thanks for help!

答: Since $A$ is a real symmetric matrix, it is diagonalizable and there is an orthogonal basis of $\Bbb R^3$ which consists of eigenvalues of $A$. So, automatically you have that $(4,8,1)\times(1,0,-4)$ is an eigenvector of $A$. And the corresponding eigenvalue has to be $3$, since the product of the eigenvalues has to be $\det A$, which is $6$.


问:
In an if and only if proof, why do you have to prove sufficiency?
This my be a simple question and I am missing something fundamental. When doing if and only if proofs, doesn't necessity imply sufficiency, so why does the 'if' have to be proven if the 'only if' is true?</p>

答:
Necessity certainly does not imply sufficiency. As an example:

The planet Earth existing is necessary for a unicorn to exist and live a happy life, as they cannot survive in the vacuum of space.

The planet Earth existing is not sufficient for a unicorn to exist and live a happy life, as we can see, unicorns do not exist.


问:
Find the number of 3 element sets of positive integers $\{a,b,c\}$ such that $abc=2310$
Clearly the factors are $2310=2\times 3\times 5 \times 7 \times 11$

My approach was picking one number for $a$, this leaves 4 numbers for b and c. So b can take up 1 or 2 or 3 numbers out of those 4 numbers

So
$$\binom 51 (\binom 41 +\binom 42 +\binom 43)$$
$$=5(2^4-2)$$

Similar, 2 numbers can be picked for $a$, leaving 3 for b and c

Continuing this way gives
$$5(2^4-1) +10 (2^3-2) + 10 (2^2-2) $$which isn’t even close to 40, the answer.

What am I doing wrong?

I know this question already exists on MSE, but unfortunately it doesn’t solve my specific problem.

NOTE: I am using this problem as a reference for my answer The students $S_1, S_2,...S_{10}$ are divided into 3 groups A, B and C

答:
Knowing the prime factors makes things a lot easier. We basically just assign each factor to any of $a$, $b$, or $c$ and call it good. This means that there are a total of $3^5$ possible ways of assigning these factors, a total of $243$ ways.

This, however, overcounts several conditions. Firstly, if we ever end up in a situation where two of the numbers are the same, then we don't get a $3$ element set, so we need to exclude these. These conditions occur if ever the distribution of the prime factors could lead to two equal products. But, by the uniqueness of prime factorization, since all of our primes are distinct this can only happen if two of the numbers were both $1$, meaning that all of the factors ended up in a single number. This happens in exactly $3$ situations, so we remove these three.

We have still overcounted, however, since we do not care about the permutations of $a$, $b$, and $c$. We divide this new total by the total number of permutations that would lead to this collection of numbers, which is a total of $3!$ permutations. In all, we have

$$\frac{3^5-3}{3!}=\frac{243-3}{6}=\frac{240}{6}=40$$

Hope this helps!


问:
Is integration continuity-preserving?
I am curious about the following question: Suppose that $f(x;\theta)$ is a bounded function of $x$, where the domain of $x$ is $[a,b]\subseteq \mathbb{R}$, and $\theta$ is viewed as a parameter. If $f(x;\theta)$ is continuous in $\theta$, then under what condition(s) can we guarantee that the function
$$F(\theta)=\int_a^b f(x;\theta)\,\mathrm{d}x$$
is continuous in $\theta$ ? Can anyone give me some help? Thanks so much!</p>

答:
Since we are looking at continuity in $\theta$, we should instead interpret $f(x;\theta)$ as a family of continuous functions of $\theta$ parametrized by $x \in [a,b]$. Since

$$ |F(\theta + \delta) - F(\theta)| \le \int_a^b | f(x; \theta + \delta) - f(x; \theta)| \, dx ,$$

then a sufficient condition for $F(\theta)$ to be continuous is that the family $f(x; \theta)$ is equicontinuous, when viewed as parametrized by $x$.

Title: Is integration continuity-preserving?


问:
Find the distribution of $Y=\sqrt X$ when $X$~$N(0,1)$
PDF of $X$ is $f_x(t)=\frac{1}{\sqrt{2pi}}e^{\frac{-x^2}{2}}$.

I tried to found the CDF of $Y$: $F_Y(t)=P(Y \leq t)=P(\sqrt X \leq t)=P(X \leq t^2)=F_X(t^2)$

How do i get distrbution of Y from this? If I were to find the $P(X\leq t^2)$ I would get error function and i don't have the CDF of X to find $F_X(t^2)$

I tried to find the PDF of $Y$: $f_Y(t)=F'_Y(t)=F'_X(t^2)=2tf_X(t^2)$

Does PDF define the distribution of random variable? If not how to i find it?

答: If you consider the absolute value of $X$ instead (half-normal distribution), you should be able to derive the pdf of $Y:=\sqrt{|X|}$ via the density transformation formula.


答:
This is a great question. The original proof-as-I-read-it is valid, but some proofs can be "better" or worse in other respects. (By "as I read it", I mean with the implication that it calculates projections with similar triangles as I describe below, rather than stating their values without proof.) There are many criteria for saying one proof is "better" than another, so I'll focus on one germane to this question: making it obvious to all readers that there's no circularity (spoiler alert: there isn't).

Let the legs adjacent to and opposite $\alpha$ have respective lengths $a,\,b$,and let $c$ denote the hypotenuse's length. By similar triangles, your projections are $c(a/c)^2,\,c(b/c)^2$. Since they sum to the full hypotenuse $c$, $a^2+b^2=c^2$.

Since we define $\cos\alpha,\,\sin\alpha$ respectively as $a/c,\,b/c$, the above calculation can be restated as$$c\cos^2\alpha+c\sin^2\alpha=1\implies c(a/c)^2+c(b/c)^2=1\implies a^2+b^2=c^2,$$which is basically what your proof does. (It takes $c=1$ throughout, but that just changes the units to tidy the algebra; it's not an extra assumption as such, so it covers the general case.) But as I've shown above, there's no need to work with anything other than rational functions of $a,\,b,\,c$. The trigonometric approach switches from such rational functions to something more conceptually advanced, then "uninstalls" the trigonometry later.

One could argue this is a worse proof than the one I've presented, because in taking unnecessary detours it not only asks more of us, it also obscures the underlying reasoning. Well, whether it does is a psychological question, but people mistakenly thinking it's circular suggests as much.

Well, I want to be fair with an edit: as best I can tell from recent comments, those people either didn't realize which proof strategy is implicit in the OP's discussion of projections, or are making the actually salient point that the drawing itself doesn't necessarily communicate that as clearly as one would like in a proof without words. But the facts are these:

the proof can easily be spelled out in the aforementioned non-circular way,

it takes a bit more work to do that than is obtained in at least some readings of the picture and the text above it, and

none of these subtleties matter if we skip the trigonometry altogether, as aforesaid.


问:
Problem $13.7$ in Apostol's Mathematical Analysis.
I am looking at problem $13.7$ in Apostol's Mathematical Analysis.

Let $D=\{|z|<1\}$ and let $f=u+iv$ be such that

$(\rm i)$ $u,v\in{\mathcal C}^1(D)$

$(\rm ii)$ $f$ is continuous on $\overline D$

$(\rm iii)$ $f(z)=z$ on $\partial D$.

$(\rm iv)$ $J_f(z)>0$ if $z\in D$.

Show that

$(1)$ For every open $\Omega\subseteq D$, $f(\Omega)$ is open in $f(D)$.

$(2)$ $f(D)$ is an open ball of radius $1$.

$(3)$ For each $w\in f(D)$, $D\cap f^{-1}(w)$ is finite.

I have only proven $(1)$:

PROOF1 Since $u,v$ are of class $\mathcal C^1$ over $D$, so is $f$, and since the Jacobian doesn't vanish, the inverse function theorem says that for each $z\in D$ there exists an open nbhd $N_z$ such that $f\mid_{N_z}$ is a diffeomorphism of class $C_1$, thus $f\mid_{N_z}$ is a homeomorphism, so it is an open map. Let $\Omega\subseteq D$ be open, and choose $z\in\Omega$. Then there exists $N_z$ as above. Let $\widetilde N_z=N_z\cap \Omega$. This is open and lies inside $N_z$ so its image under $f\mid_{N_z}$ is open by the above. Then $f(\Omega)=f(\Omega)\cap f(D)=\bigcup_{z\in\Omega}f\left(N_z\right)\cap f(D)$ is the intersection of the union of open sets with $f(D)$, so it is open in $f(D)$. $\blacktriangle$.

PROOF2 Suppose that $f(D)\not\subseteq D$. Then certainly $f(\bar D)\not\subseteq \bar D$. Thus, there exists $q\in\bar D$ such that $|f(q)|>1$. But $|f|$ is continuous on the compact $\bar D$; thus it attains its maximum on some $p\in\bar D$. But we cannot have $p\in\partial D$ since we would have $|f(p)|=1$, so it must be the case $p\in D$. Let $K=\overline B$ where $B=B(0,r)\; ;\; r=|f(p)|$. Then $f(\bar D)\subseteq \bar B$, and $f(p)\in\partial B$. Let $N_p$ be an open nbhd of $p$ in $D$. Then we see $f(N_p)$ is not open in $f(D)$, contrary to what was proven in $(1)$. It follows that $f(D)\subseteq D$.

PROOF3 Suppose for the sake of a contradiction that there exists $w\in f(D)=D$such that $D \cap f^{-1}(w)$ is infinite. Since this is a subset of the compact $\bar D$, it must have an accumulation point in $\alpha \in \bar D$. Suppose first that $\alpha\in D$. Then for each $\epsilon >0$ the ball $B(\alpha;\epsilon)$ contains infinitely many points of $\{z\in D:f(z)=w\}$, in contradiction with the fact $f$ is a local diffeomorphism in $\alpha$. Suppose $\alpha\in \partial D$. Then $f(\alpha)=\alpha$, and by continuity, $f(\alpha)=w$, which is absurd since $D\cap\partial D=\varnothing$. $\blacktriangle$.

Chapter $13$ is "Implicit functions and extremum problems". I am not sure what theorems to use to prove $(2)$, and I haven't thought about $(3)$, so I will worry about $(3)$ later. I want to prove $(2)$ now. Any hints?

ADD If $A$ is any set, $\partial A$ denotes its boundary and $\overline A$ its closure.

答:
Part 2

We know that $f(\bar D)$ is compact subset of the closed unit ball, so it is closed, and its complement in the closed unit ball is open. This follows because $f$ is an open mapping, so it obeys a maximum principle: its maximum modulus on a compact region cannot be taken in the interior. So we must have $|f|<1$ for all $z\in D$.

We can write $f(\bar D)^c$ in $\bar D$ as $\bar D\cap U$ for some open set $U$ in $\mathbb R^2$. Then $f(\bar D)^c\cap D= U\cap D$ is an open set of $D$ in the subspace topology. So the set of points in $D$ omitted by $f(D)$ in is open in the subspace topology on $D$, as the boundary points of $\bar D$ are all mapped to the boundary of $D$.

But this set is also closed. We know $f(D)$ is open by part 1, and its complement in the plane is closed. So $f(D)^c\cap D$ is a closed set of $D$ in the subspace topology.

We see the omitted set must be empty by connectedness, as it is both open and closed.

Part 3

If the inverse image of some point $w$ were infinite, the set of inverse images would have a limit point in the closed unit ball by compactness. If $p$ is the limit point, then by continuity $f(p)=w$. If $p$ is in the open unit ball, then no neighborhood of the limit point is mapped diffeomorphically onto an open set, contradicting the third hypothesis and the inverse function theorem.

If the limit point is on the boundary, we get a contradiction, as $w$ must lie in the open unit ball and $f(p)$ on the boundary.


问:
Solve for $x$: $1024 = 2(3x+7)^{3/2}$
Could someone show me how to solve this problem step by step? I keep trying and get non-numerical answers. $x$ should be a whole number.

Title: Solve for $x$: $1024 = 2(3x+7)^{3/2}$

Body:
$$
1024 = 2 ( 3x+7) ^ {\frac{3}{2}} \\
\iff 512 = ( 3x+7) ^ {1.5}  \\
\iff 2^9= (3x+7) ^ {\frac{3}{2}}  \\
\iff 2^6 * 2^3 = (3x+7) ^ {1.5} = 64^{1.5}  \\
\iff 64 = 3x+7\\
\iff 57 = 3x \\
\iff 19 = x
$$

Can you follow these steps?


问:
Computing $\int_0^\infty \exp\left( -\frac{x^2}{A} -\frac{B}{x} -Cx \right){\rm d}x$
Exactly as the title suggests, I want to compute or approximate the integral of the form
$$\int_0^\infty \exp\left( -\frac{x^2}{A} -\frac{B}{x} -Cx \right){\rm d}x$$
where $A,B,C>0.$ I got the question from a friend who asked me how to evaluate it.

I was told that an asymptotic formula or a good approximation for large values of $t$ (if we integrate from $0$ to $t$ instead) is good enough, as long as it's written in terms of $A,B,C.$ I feel like contour integration might do the trick but I'm a bit too rusty on that. Thank you in advance.

答:
By rescaling $x$, we may set one constant to unity, so we can consider the integral

$$
I=\int\limits_0^\infty dx \ \exp \left( -ax^2-bx^{-1}-x\right)
$$

For large $a$

The integrand has a moveable maxima at $x_0(a,b)$. Let $x=ux_0$, then the maxima is fixed at $u=1$, and we have

$$
I=\int\limits_0^\infty du \ x_0 \exp \left( -ax_0^2u^2-bx_0^{-1}u^{-1}-x_0u \right)
$$

If you want to expand any of the exponentials, expanding this equation is a good idea. Expanding the whole exponent around $u=1$, we get a Gaussian integral, resulting in a complementary error function (for fixed $b$)

$$
I(a) \sim \frac{x_0}{2} \sqrt{\frac{\pi x_0}{b+ax_0^3}} e^{f(x_0)} \operatorname{erfc}\left[ g(x_0)\right] \quad ; \quad a \to \infty
$$

$x_0$, $f$, and $g$ are given at the end of the post. It turns out this is also good when $b \to \infty$ for fixed $a$. Here are plots together with the numerical integration:

Now consider the integral

$$
J(t)=\int\limits_t^\infty dx \ \exp \left( -ax^2-bx^{-1}-x\right)
$$

So that the integral in the comments is given by

$$
\int\limits_0^t dx \ \exp \left( -ax^2-bx^{-1}-x\right) = I-J(t)
$$

Where $a$ and $b$ are fixed. We develop the asymptotic series for $J$ by integration by parts.

$$
J(t)=\int\limits_t^\infty dx \ \frac{1}{-2ax+bx^{-2}-1} \frac{d}{dx} \left[ \exp \left( -ax^2-bx^{-1}-x\right)\right]
$$

$$
J(t)=\frac{\exp \left(-at^2-bt^{-1}-t \right)}{2at-bt^{-2}+1}-\int\limits_t^\infty dx \ \dots
$$

For the integral on the right, we have $\int dx \dots << J(t)$ for $t \to \infty$, so long as $a$ and $b$ are fixed. Neglecting this term, we have

$$
J(t) \sim \frac{\exp \left(-at^2-bt^{-1}-t \right)}{2at-bt^{-2}+1} \quad ; \quad t \to \infty
$$

Here is a plot with $a=b=3/2$:

Here are the functions (thanks Mathematica!)
$$
x_0(a,b)=\frac{1}{24a}\left[ -4 + \frac{2+2i\sqrt{3}}{(1-54a^2 b+6 \sqrt{3} \sqrt{a^2 b (27a^2 b-1})^{1/3}} +(2-2i\sqrt{3})(1-54a^2+6 \sqrt{3} \sqrt{a^2 b (27a^2 b-1)})^{1/3}\right]
$$

$$
f(x)=\frac{-3b^2+x^4-6bx^2(1+2ax)}{4x(b+ax^3)}
$$

$$
g(x)=\frac{x^2-3b}{2 \sqrt{bx+ax^4}}
$$

$x_0$ is real, despite the imaginary units floating around. Probably, it can be simplified if one was so inclined.


问:
Show $\int_0^1|x'(t)|^2 dt \leq M_2$ given that $\int_0^1|x'(t)|^2-|x| dt \leq M_1$ (using Hoelders inequality)
I am trying to prove the following statement:
$\int_0^1|x'(t)|^2-|x(t)| dt \leq M_1$ for some $M_1$, implies $\int_0^1|x'(t)|^2 dt \leq M_2$ for some $M_2$. Furthermore it is given that $x(0)=0$ and $x(1)=1$.

The only assumption I make is that $x(t)$ is absolutely continuous, so one can write $x(t)=\int_0^tx'(s)ds$.

What I have so far:

$M_1 \geq  \int_0^1|x'(t)|^2-|x(t)| dt $ =$ \int_0^1|x'(t)|^2-|\int_0^tx'(s)ds|dt \geq \int_0^1|x'(t)|^2-\int_0^t|x'(s)|dsdt$

Now using Hoelders inequality in the form $\int_0^t|x'(s)|ds\leq (\int_0^t|x'(s)|^2ds)^\frac{1}{2}$ we get to

$M_1 \geq  \int_0^1|x'(t)|^2 - (\int_0^t|x'(s)|^2ds)^\frac{1}{2}dt$.

One could then use $\sqrt(x) \leq 1+x$, to end up with $M_1+1 \geq  \int_0^1|x'(t)|^2 - (\int_0^t|x'(s)|^2ds)dt$

How could one proceed from here? I have tried now for several hours now but could not see any way to get to the conclusion. Or is there some completely different way?

Best

答:
Answer to original version, where the condition $x(0)=x(1)=0$ was left out:

Saying you want this for "arbitrary" constants makes little sense. You want to show that given $M_1$ there exists $M_2$ such that $\int(|x'|^2-|x|)\le M_2$ implies $\int|x'|^2\le M_1$.

This is more or less obviously false. Given $M_1$, choose $y$ so $\int|y'|^2>M_1$. For any $M_2$ there exists a constant $c$ so if $x=y+c$ then $\int(|x'|^2-|x|)<M_2$ although $\int|x'|^2=\int|y'|^2>M_1$.


问:
Derivative Question??
I am a little confused with how I am supposed to find $g'(4)$ when given this information. I just cannot seem to find a connection if I am not given the original $f(x)$ equation. Could anyone provide some insight:

You are given:
$$
g(x)= \sqrt{x}f(x) \\
f(4)= 3 \\
f’(4)= -5
$$

答:
Use the product-rule,

\begin{equation}
g'(x) = \frac{1}{2\sqrt{x}}f(x) + \sqrt{x}f'(x)
\end{equation}

Title: Derivative Question??


问:
$a_0 + a_1T + \cdots + a_k T^k$ is nilpotent if and only if $a_0 = 0$, where $T$ is nilpotent
Question:

Let $T: V \rightarrow V$ be a nilpotent linear transformation of an F-vector space. Show that $H := \sum^k_{i=0} a_i T^i$ with $a_0, \ldots, a_k \in F$ is nilpotent if and only if $a_0 = 0$

My attempt:

I have so far shown the backwards direction.

If $p$ is the smallest positive integer such that $T^p = T_0$, then I assume that $k < p$. Otherwise, I can just find maximal $n < p$ and then $H = \sum^n_{i=0} a_i T^i$ (everything greater than n "dies"), and I can apply the same argument below.

Then, if $a_0 = 0$, then the smallest power of $T$ in $H^p = (a_1T + \cdots + a_kT^k)^p$ is $T^p$. And so every term in the expansion of $H^p$ is $T_0$, so $H^p = T_0$. Hence, $H$ is nilpotent.

But I'm having trouble showing the forwards direction. My first thought was via a contradiction. So first I suppose $a_0 \neq 0$. Then I said that $H^m = a_0^m + b_1T + \cdots +b_{km}T^{km}$, for any positive integer $m$, where $b_i \in F$

But looking at the above, it seems possible that when I apply this to some $v \in V$, then the $a_0^mv$ and $b_1T(v), \ldots, b_{p-1}T^{p-1}(v)$ will somehow cancel out, and still leave $H$ being nilpotent, which is not what I want as I want to show some contradiction.

I would appreciate any feedback on my approach so far, and how one would show the forwards direction. Thank you!

答:
First, you need at least one $a_i\neq 0$, otherwise the statement is false. Suppose $H$ is Nilpotent. Then, there is $p>0$ s.t. $H^p=0$. Moreover,
$$0=H^p=a_0^p+\sum_{i=1}^{kp}\beta _iT^i,$$
for some $\beta _i\in \mathbb R$. Notice that at least one $\beta _j\neq 0$ (since $a_i\neq 0$ for some $i$). Let $J=\min\{j\mid \beta _j\neq 0\}$. Then
$$0=a_0^p+T^J\underbrace{\left(\beta _j+\sum_{s=j+1}^{kp}\beta _jT^{s-j}\right)}_{=:q(T)\neq 0}=a_0^p+T^Jq(T).$$

Therefore $a_0=0$ and $T^J=0$ as wished.


答:
It helps to simply calculate $g'$ directly, even if it has to be expressed in terms of $f(x)$:
$$
g'(x) = \frac{1}{2}x^{-1/2}f(x)+\sqrt{x}f'(x) \, .
$$
Once you have done this, it becomes clear why knowing the value of $g'(4)$ does not require you to know 'the original $f(x)$ equation'—in fact, you only need to know the values of $f(x)$ and $f'(x)$ when $x=4$.

Title: Derivative Question??

Body:
I am a little confused with how I am supposed to find $g'(4)$ when given this information. I just cannot seem to find a connection if I am not given the original $f(x)$ equation. Could anyone provide some insight:

You are given:
$$
g(x)= \sqrt{x}f(x) \\
f(4)= 3 \\
f’(4)= -5
$$


问:
Number of NxN grids with only one "spot" allowed
I'm working on a question where a spot is defined as "A 1 × 1 blue square that shares an edge with a red square to its left or above it". The square being in some NxN grid with all squares being either blue or red.

The question is asking how many grids are there with one and only one spot in it. An earlier part of the question was to get how many grids have no spots which I just did by adding them up but I'm looking for the method for how to actually solve this second part, rather than trying to count them all up

EDIT: Formula for Mike  (2(N!))/(N!*N!) or 10C5

答:
Hint

First, count the number of grids with a single "horizontal spot," i.e. a blue square with a red square to its left. At the end, multiply by two.

Suppose the red square is in row $r$ and column $c$. An example when $n=7, r=3, c=5$ is below. Note that placing the spot at row $r$ and column $c$ forces a lot of other entries to be red and blue in order to prevent any further spots. However, there are still two subgrids left to be filled, which are smaller instances of the zero-spot column. This lets you solve the problem for any particular $r,c$, and then you can sum over all possible values of $r, c$. The summations will simplify nicely if you group the terms together correctly.

```<code>        v
B B B B B B .
B B B B B B .
. . . . R B . <-
. . . . R R R
. . . . R R R
. . . . R R R
. . . . R R R
</code>```

Body:
Hint

First, count the number of grids with a single "horizontal spot," i.e. a blue square with a red square to its left. At the end, multiply by two.

Suppose the red square is in row $r$ and column $c$. An example when $n=7, r=3, c=5$ is below. Note that placing the spot at row $r$ and column $c$ forces a lot of other entries to be red and blue in order to prevent any further spots. However, there are still two subgrids left to be filled, which are smaller instances of the zero-spot column. This lets you solve the problem for any particular $r,c$, and then you can sum over all possible values of $r, c$. The summations will simplify nicely if you group the terms together correctly.

        v
B B B B B B .
B B B B B B .
. . . . R B . <-
. . . . R R R
. . . . R R R
. . . . R R R
. . . . R R R


问:
An identity about the Gamma function
When I do some numerical test, I can get:

$$\frac{2^{2a+1}\Gamma(a+1)\Gamma(a+\frac32)}{\Gamma(2a+2)}=\sqrt{\pi}.$$

Here $a$ is to be taken such that the Gamma function well-defined.

My question is that is there is a way to prove this identity.

I tried to apply the identity $\Gamma(z+1)=z\Gamma(z)$ but no further result.

Thanks for any suggestions.

答:
This is known as the duplication formula for the Gamma function. It is commonly written as
$$
\Gamma(2z) = \pi^{-1/2}2^{2z-1}\Gamma(z) \Gamma(z+1/2)
$$
and appears as formula 5.5.5 in the NIST handbook.

Title: An identity about the Gamma function


问:
KVPY Question: Sequence and Series.

$S_n = \cfrac{7}{4 \cdot 1 \cdot 2} + \cfrac{10}{4^2 \cdot 2 \cdot 3} + \cfrac{13}{4^3 \cdot 3 \cdot 4} + ....$, then $S_{\infty}=$

I found the general term:

$$\sum_{n=1}^{n \rightarrow \infty} \cfrac{(3n+4)}{n(n+1)4^{n}} = $$

How to evaluate further? Please help.

I am at an elementary level, so this question might seem dumb, sorry.

答:
$$\frac{4n+3}{(n+1)4^n}=\frac 1{4^{n-1}\cdot n}-\frac 1{4^n\cdot (n+1)}$$

Now we give values $n=1 \rightarrow n$ and sum up and cancel similar terms :

$$\frac 1{4^0\cdot 1}-\frac 1{4\cdot 2}$$

$$\frac 1{4\cdot 2}-\frac 1{4^2\cdot 3}$$

$$\frac 1{4^2\cdot 3}-\frac 1{4^3\cdot 4}$$

$$\frac 1{4^3\cdot 4}-\frac 1{4^4\cdot 5}$$

$$\vdots$$

$$\frac 1{4^{n-2}\cdot(n-1) }-\frac 1{4^{n-1}\cdot n}$$

$$\frac 1{4^{n-1}\cdot n}-\frac 1{4^n\cdot (n+1)}$$

$$\therefore S_n=1-\frac 1{4^n(n+1)}$$

$$\Rightarrow \lim_{n\rightarrow \infty} S_n=1$$

Title: KVPY Question: Sequence and Series.


问:
Solving For Variables In Simultaneous Equations
I'm doing some work in linear algebra and these came up and I realized I don't know how to solve them as they have quadratics in them. I'm sure I've done this before but if someone could give me a crash course on how to find the values of the variables, it would be most appreciated. For the following questions, let $a,b,c,d \in \mathbb{R}$

1)

$\begin{cases}
a^2+b^2=a\\
(a+d)b=b\\
b^2+d^2=d
\end{cases}$

2)

$\begin{cases}
a^2+bc=a\\
(a+d)b=b\\
(a+d)c=c\\
bc+d^2=d
\end{cases}$

答:
1)

$\begin{cases}
a^2+b^2=a\\
(a+d)b=b\\
b^2+d^2=d
\end{cases}$

in equation $2$ divide both side by b
$\begin{cases}
a^2+b^2=a\\
a+d=1\\
b^2+d^2=d
\end{cases}$

from eqn $2$ and $3$

$$a=1-d$$
$$b^2=d-d^2$$

put these value in eqn $1$

$$(1-d)^2+d-d^2=1-d\implies d=0$$
it will give $a=1,b=0$

2)
$\begin{cases}
a^2+bc=a\\
(a+d)b=b\\
(a+d)c=c\\
bc+d^2=d
\end{cases}$

in eqn $2,3$ divide both side by b and c

$
a^2+bc=a\\
a+d=1\\
a+d=1\\
bc+d^2=d
$

$
a^2+bc=a\\
a+d=1\\
bc+d^2=d
$

from eqn $1$ and $2$
$$bc=a-a^2$$
$$d=1-a$$

put these value in $3$
$$a-a^2+(1-a)^2=1-a\implies a=0$$
that will give $d=1$ and either of b and c will 0.
$$$$


问:
If Ratio Test for an alternating series is Divergent, Must I run the Alternating series test?
If I conduct the ratio test for a series that is alternating in sign, and find that $\lim$ as $n$ goes to $\infty$ is $ >1$, is it mathematically correct to say that the entire series (including the alternating piece) is divergent?  Or, must I conduct the alternating series convergence test?

答: If the ratio test for the absolute values of the terms gives a limit $>1$, the terms cannot go to zero and therefore the original series diverges by the Divergence Test.


问:
Is There an Abel-plana-type Exact Formula for the Imaginary Part of the Oscillatory Integral over Exp(iπx)x^(1/x) Between 1 and Infinity?
This is a question that is an important part of my life-long hobby of researching the MRB constant.

CMRB

I have posted Mathematica code with all my formulas below.

Richard J. Mathar did in-depth research into its Oscillatory Integral (CMKB=M1).

On Feb 24, 2021, at 11:00 am EST I discovered the following Abel-Plana type integrals for the exact real value of M1 or ReMKB (Re[CMKB]),

```<code>f[x_] = (-1)^x (x^(1/x) - 1); ReMKB =
 Im[N[NIntegrate[(f[(1-I t)] - f[( 1+I t)])/(Exp[2 Pi t] + 1), {t,
      0, Infinity}, WorkingPrecision -> 100], 50]]
</code>```

(0.070776039311528803539528021830282001365754696203363)

and

```<code>g[x_] = x^(1/x); ReMKB =
N[I NIntegrate[(g[(1-t I)] - g[(1+t I)])/(2 Exp[Pi t] ), {t, 0, Infinity}, WorkingPrecision -> 100], 50]
</code>```

(* 0.070776039311528803539528021830282001365754696203363.*)

If you need an exact formula for the imaginary part, you can use

Enter

```<code>g[x_] = x^(1/x); ImMKB =
 N[NIntegrate[I Sin[Pi n] g[n], {n, 1, Infinity},
    WorkingPrecision -> 100] - I/Pi, 50]
</code>```

(*
0.10^-51 - 0.68400038943793212918274445999266112671099148265500 I.)

The imaginary part, along with the real part, can be computed to many digits through partial integration with formulas like the following.

```<code> f[x_]:=x^(1/x);  Print[DateString[]]; Print[T0 = SessionTime[]]; prec =10000;
  Timing[Print[
  a = N[Re[-(136584/Pi^10) - (34784*I)/Pi^9 +
       5670/Pi^8 + (786*I)/Pi^7 - 90/Pi^6 -
                 (4*I)/Pi^5 - 3/Pi^4 - (2*I)/Pi^3 +
       1/Pi^2 - (2*I)/Pi] -
             (1/Pi)^10*
      NIntegrate[Cos[Pi*x]*D[f[x], {x, 10}], {x, 1, Infinity},
       WorkingPrecision -> prec,
                 PrecisionGoal -> prec], prec]];
 Print[SessionTime[] - T0, " seconds"];
     Print[
  N[b = -Im[-(136584/Pi^10) - (34784*I)/Pi^9 +
        5670/Pi^8 + (786*I)/Pi^7 - 90/Pi^6 -
                   (4*I)/Pi^5 - 3/Pi^4 - (2*I)/Pi^3 +
        1/Pi^2 - (2*I)/Pi] +
             (1/Pi)^10*
      NIntegrate[Sin[Pi*x]*D[f[x], {x, 10}], {x, 1, Infinity},
       WorkingPrecision -> prec,
                 PrecisionGoal -> prec], prec]]]; Print[
 SessionTime[] - T0, " seconds"];
  Print[N[Sqrt[a^2 + b^2], prec]]; Print[DateString[]];
</code>```

But is there an Abel-Plana-type exact formula for the imaginary part?

P.s. when looking for one I made the following discovery at 1:15 am, March 6, 2021.

```<code> f[x_] = (-1)^x (x^(1/x) - 1);

  CMRB = N[NSum[f[n], {n, 1, Infinity}, Method -> \
"AlternatingSigns", WorkingPrecision -> 100], 50]

 (* 0.18785964246206712024851793405427323005590309490014*)

 ReMKB =
 Im[N[NIntegrate[(f[(1 - I t)] - f[( 1 + I t)])/(Exp[2 Pi t] + 1), {t,
           0, Infinity}, WorkingPrecision -> 100], 50]]

 (*0.070776039311528803539528021830282001365754696203363*)

 NoMKB =
 Im[N[NIntegrate[(f[(1 - I t)] - f[( 1 + I t)])/(Exp[2 Pi t] - 1), {t,
           0, Infinity}, WorkingPrecision -> 100], 50]]

 (* 0.117083603150538316708989912223991228690148398696776*)

 ReMKB + NoMKB - CMRB

 (* 0.*10^-51*)
</code>```

Body:
[Some corrections made March 10, 2021.]

I found what I was looking for.

g[x_] = x^(1/ x); ImMKB = -(N[
     I NIntegrate[(g[(1 - t I)] + g[(1 + t I)])/(2 Exp[Pi t]), {t, 0,
        Infinity}, WorkingPrecision -> 100], 50] + I/Pi)

(*0.*10^-67 - 0.68400038943793212918274445999266112671099148265500 I. *)

In summary,

   In[149]:= f[x_] = (-1)^x (x^(1/x) - 1);

In[150]:= CMRB = N[NSum[f[n], {n, 1, Infinity}, Method -> \
"AlternatingSigns", WorkingPrecision -> 100], 50]

Out[150]= 0.18785964246206712024851793405427323005590309490014

In[151]:= ReMKB =
 Im[N[NIntegrate[(f[(1 - I t)] - f[( 1 + I t)])/(Exp[2 Pi t] + 1), {t,
           0, Infinity}, WorkingPrecision -> 100], 50]]

Out[151]= 0.070776039311528803539528021830282001365754696203363

In[261]:= ReMKB -
 Im[N[NIntegrate[(f[(1 - I t)] + f[( 1 + I t)])/(Exp[2 Pi t] - 1), {t,
           0, Infinity}, WorkingPrecision -> 100], 50]]

Out[261]= 0.*10^-51

In[152]:= NoMKB =
 Im[N[NIntegrate[(f[(1 - I t)] - f[( 1 + I t)])/(Exp[2 Pi t] - 1), {t,
           0, Infinity}, WorkingPrecision -> 100], 50]]

Out[152]= 0.117083603150538316708989912223991228690148398696776

In[153]:= ReMKB + NoMKB - CMRB

Out[153]= 0.*10^-51

In[171]:= g[x_] = x^(1/x);

In[195]:= MKB =
 N[NIntegrate[Exp[I Pi t] g[t], {t, 1, 1 Infinity},
    WorkingPrecision -> 100], 50] - I/Pi

Out[195]= 0.07077603931152880353952802183028200136575469620336 -
 0.68400038943793212918274445999266112671099148265500 I

In[172]:= ImMKB = -(N[
     I NIntegrate[(g[(1 - t I)] + g[(1 + t I)])/(2 Exp[Pi t]), {t, 0,
        Infinity}, WorkingPrecision -> 100], 50] + I/Pi)

Out[172]=
0.*10^-67 - 0.68400038943793212918274445999266112671099148265500 I

In[162]:= ReMKB =
 N[I NIntegrate[(g[(1 - t I)] - g[(1 + t I)])/(2 Exp[Pi t]), {t, 0,
     Infinity}, WorkingPrecision -> 100], 50]

Out[162]= 0.070776039311528803539528021830282001365754696203363

In[196]:= ImMKB + ReMKB - MKB

Out[196]= 0.*10^-51 + 0.*10^-51 I


答:
I tried to solve 1) in your question.
$$\begin{cases}
a^2+b^2=a ----(1)\\
(a+d)b=b  ----(2)\\
b^2+d^2=d ----(3)
\end{cases}$$
when $b=0$, the equations (1) and (3) boil down to $a^2=a$ and $d^2=d$. In this case, we have the solutions below
$$\{a=0,b=0,d=0\},\\
\{a=0,b=0,d=1\},\\
\{a=1,b=0,d=0\},\\
\{a=1,b=0,d=1 \}.$$

If $b\ne0$, then equations (1), (2) and (3) becomes
\begin{cases}
a^2-a +b^2=0 ----(4)\\
a+d =1  -------(5)\\
d^2-d +b^2= 0----(6)
\end{cases}
From (4),(5)and (6), we can see $a$ and $d$ are the solutions of the equation $x^2-x+b^2=0$, where $b$ can be any real number. So in this case, the solutions could be
$$\left\{a=\frac{1}{2} \left(1-\sqrt{1-4 b^2}\right),d=\frac{1}{2} \left(\sqrt{1-4 b^2}+1\right)\right\},\\
\left\{a=\frac{1}{2} \left(\sqrt{1-4 b^2}+1\right),d=\frac{1}{2} \left(1-\sqrt{1-4 b^2}\right)\right\}.
$$
Hint: your question 2) can be solved in a similar way.

Title: Solving For Variables In Simultaneous Equations

Body:
I'm doing some work in linear algebra and these came up and I realized I don't know how to solve them as they have quadratics in them. I'm sure I've done this before but if someone could give me a crash course on how to find the values of the variables, it would be most appreciated. For the following questions, let $a,b,c,d \in \mathbb{R}$

1)

$\begin{cases}
a^2+b^2=a\\
(a+d)b=b\\
b^2+d^2=d
\end{cases}$

2)

$\begin{cases}
a^2+bc=a\\
(a+d)b=b\\
(a+d)c=c\\
bc+d^2=d
\end{cases}$


问:
Why doesn't the lack of pointwise convergence of $f_n(x)=x^n$ contradict the fact that $C[0,1]$ is Banach?
On this answer, the function $f_n(x)=x^n$ in the interval $[0,1]$ is given as a pathologic example with pointwise convergence.

Can I say that this Cauchy sequence does not (pointwise) converge because the limit of the sequence is a function like this (not continuous):

without specifying any particular norm? I read that pointwise convergence, doesn't imply $d_\infty$ (uniform) convergence, and that uniform convergence implies pointwise convergence. But does lack of pointwise convergence negate uniform convergence?

Does this contradict in any way (or under certain norms) the fact that $C[a,b]$ with respect to $\Vert f \Vert_{\infty}$ is a Banach space? In other words, why is not an example of a Cauchy sequence that does not converge to some $f\in C[0,1]$?

答:
After some research on the topic, I think that the answer is no (as canonically answered by Elchanan), and the critical bit is his sentence, "Pointwise convergence does not correspond to any metric (this can be proven)."

But there remains the meta-questions of what is happening with that "pathologic" function, and what words would give the uninitiated some sense of direction.

The problem with the sequence $\{x^n\}_{n=1}^\infty$ in the interval $[0,1]$ is that it is pointwise convergent, but not uniformly convergent.

The sequence $f_n \to f$ pointwise on $[0,1]$ if $f_n(x) \to f(x)$ as $n \to \infty$ for every $x \in [0,1].$ There is no $\varepsilon!$

The limit of a pointwise convergent sequence of continuous functions does not have to be continuous, and does not generally preserve boundedness. In this case:

$$\lim_{n\to\infty} f_n(x) = f(x) =
    \begin{cases} 0 & (0\leq x\lt 1) \\ 1 & (x=1) \end{cases}$$

In uniform convergence, the uniform norm is introduced. Uniform convergence implies pointwise convergence, but not the other way around.

A sequence of functions converges uniformly if for every $\varepsilon>0$ there is an $N\in \mathbb N$ such that for all $n≥N$ and all $x \in [0,1],$ $d(f_n(x),f(x))<\varepsilon,$ where $d(f, g) =  \sup_{x\in [0,1]} |f_n(x) - f(x)|.$

The space $C[0,1]$ with the infinity metric is complete (Banach), but the sequence $x^n$ is not Cauchy.

The link to Cauchy sequences is that $$\small \text{uniform convergence}\iff \text{Cauchy under the }\Vert \cdot \Vert_\infty \text{ uniform or infinity norm}. $$

The lack of uniform convergence is explained in Elchanan's answer, as well as here by contradiction:

If $f_n(x)$ converges uniformly, then the limit function must be $f(x)=0$ for $x∈[0,1)$ and $f(1)=1.$ Uniform convergence implies that for any $ϵ>0$ there is an $Nϵ \mathbb N$ such that $|x^n−f(x)|<ϵ$ for all $n≥ \mathbb N$ and all $x∈[0,1].$ Assuming this is indeed true we may choose $ϵ,$ in particular, we can choose $ϵ=1/2.$ Then there is an $N∈ \mathbb N$ such that for all $n≥N$ we have $|x^n−f(x)|<1/2.$ We may choose $n$ and $x.$ Let us choose $n=N,$ and $x=(\frac 3 4)^{1/N}.$ Then we have $f(x)=0$ and thus

$$|f_N(x)−f(x)|=x^N−0=\frac3 4>\frac 1 2, $$

a contradiction.


答:
Lemma: Let $R$ be any ring, $a,r\in R$, where $a$ is a unit and $r$ is nilpotent. Furthermore suppose that $a^{-1}$ commutes with $r$. Then $a+r$ is a unit.

Proof: Let $s=-a^{-1}r$ and $k\in \Bbb N$ such that $r^k=0$. As $a^{-1}$ and $r$ commute we also have $s^k=0$. Then $$(1-s)b=1-s^{k}=1=b(1-s)$$
where $b=1+s+\dots+s^{k-1}$. Hence $1-s$ and therefore also $a+r=a(1-s)$ is a unit.

Now assume that $a:=a_0\ne0$. By what you already showed $r:=a_1T+\dots+a_kT$ is nilpotent. Then (letting $R=\operatorname{End}(V)$ or $F[T]$) the hypotheses of the lemma are satisfied, so we get that $H=a+r$ is a unit. In particular it cannot be nilpotent.

Title: $a_0 + a_1T + \cdots + a_k T^k$ is nilpotent if and only if $a_0 = 0$, where $T$ is nilpotent

Body:
Question:

Let $T: V \rightarrow V$ be a nilpotent linear transformation of an F-vector space. Show that $H := \sum^k_{i=0} a_i T^i$ with $a_0, \ldots, a_k \in F$ is nilpotent if and only if $a_0 = 0$

My attempt:

I have so far shown the backwards direction.

If $p$ is the smallest positive integer such that $T^p = T_0$, then I assume that $k < p$. Otherwise, I can just find maximal $n < p$ and then $H = \sum^n_{i=0} a_i T^i$ (everything greater than n "dies"), and I can apply the same argument below.

Then, if $a_0 = 0$, then the smallest power of $T$ in $H^p = (a_1T + \cdots + a_kT^k)^p$ is $T^p$. And so every term in the expansion of $H^p$ is $T_0$, so $H^p = T_0$. Hence, $H$ is nilpotent.

But I'm having trouble showing the forwards direction. My first thought was via a contradiction. So first I suppose $a_0 \neq 0$. Then I said that $H^m = a_0^m + b_1T + \cdots +b_{km}T^{km}$, for any positive integer $m$, where $b_i \in F$

But looking at the above, it seems possible that when I apply this to some $v \in V$, then the $a_0^mv$ and $b_1T(v), \ldots, b_{p-1}T^{p-1}(v)$ will somehow cancel out, and still leave $H$ being nilpotent, which is not what I want as I want to show some contradiction.

I would appreciate any feedback on my approach so far, and how one would show the forwards direction. Thank you!


问:
Find a functions such $\int_0^{1} |f(t) + g(t)| dt = 1$
I'm trying to find different continuous functions $f,g$ such that:

$\int_0^1|f(t)| dt = 1$

$\int_0^1|g(t)| dt = 1$

and

$\int_0^1|f(t) + g(t)| dt = 1$

I tried a lot of possiblities but I end up with nothing. I tried to firsly choose $\int_0^1|f(t) + g(t)| dt = 1$ and then try to find $f$ and $g$, but I wasn't able to do so. Do dsuch functions exist ?

答: I think $f(x)=2x-2$ and $g(x)=2x$ works.


问:
Numerically solving a system of linear ODEs with different time conditions?
I am interested in solving the following system of ODEs numerically
\begin{align}
x'=2x+3y+z\\
y'=x-3y+2z\\
z'=x+2y-4z
\end{align}
with conditions at distinct times, even at time at infinity.
\begin{align}
x(0)=1, y(0)=1, z(+\infty)=0
\end{align}

Can I use PYTHON or MATLAB to solve it? If yes, where I can find such tutorial? Many thanks for your help!

答:
This is a linear system, you can solve it via eigen-decomposition. This produces a positive and two negative eigenvalues. The solution has to lie in the eigenspaces of the negative eigenvalues. This gives the missing initial condition as
$$
z(0)=-5.024155419109322.
$$
Long-term numerical simulation with this initial condition will likely not converge to zero in all components, as numerical errors will build up some small coefficient for the component with a positive eigenvalue, which will then grow exponentially.

$$
(λ_1,λ_2,λ_3)=( 2.93491425,\, -2.40143568,\, -5.53347857),
\\
V=(v_1,v_2,v_3)=
\begin{bmatrix}
 0.95198356& 0.56546078&  0.16012843\\
 0.2289107& -0.66866328& -0.64979254\\
 0.2032909 & -0.48284938&  0.74305352
\end{bmatrix}\\
(c_1,c_2,c_3)=(0.0,\,3.11077114,\, -4.74006433)
\\
y(t)=\sum_{k=1}^3c_cv_ke^{λ_kt}
$$


问:
Flux across surface
Find A uniform fluid that flows vertically downward is described by the vector field F (x, y, z) = (0, 0, −1).

Find the flux through the cone z =
$z= \sqrt{x^2 + y^2}$,
$x^2 + y^2 \le 1$.

I attempted this question with spherical coordinates and i don't know why it didn't work out. I used that $\sin(\theta) = \frac{\pi}{4}$ and evaluated the cross product T$\phi$ x Tr

This came to $(r\sin(\theta)\cos(\theta)\cos(\phi),-r\sin(\theta)\cos(\theta)\sin(\phi),r\sin^2(\theta))$
I then integrated

$\int_0^{2\pi}\int_0^1(0,0,-1).(r\sin(\theta)\cos(\theta)\cos(\phi),-r\sin(\theta)\cos(\theta)\sin(\phi),r\sin^2(\theta))~dr~d\phi$

=$-\int_0^{2\pi}\int_0^1r\sin^2(\theta)~dr~d\phi$

=$-\frac{\pi}{2}$ as $\theta = \frac{\pi}{4}$

Because I didn't get the right answer I checked my lecture notes, and I thought 'hey there is a $\hat n$ at the end of this integral and i didn't divide by magnitude. But to no avail.

I ended up getting -2$\pi$ by dividing by the magnitude and the answer is $\pi$

I now know that I can do it better with cylindrical coordinates and parametrization. But this way i don't divide by the magnitude

Do I divide by the normals magnitude to get $\hat n$ or not?

答:
If you know Divergence Theorem, then this problem looks like a standard "applying the Divergence theorem" exercise.

Denote the cone surface as $S=\{(x,y,z):z = \sqrt{x^2+y^2},\;z\leq 1\}$, the region surrounded by this surface is $D = \{(x,y,z):\sqrt{x^2+y^2}< z< 1\}$.

This region has a "cap": $C = \{(x,y,z): x^2+y^2\leq 1, z= 1\}$ such that $\partial D = S\cup C$. Apply the divergence theorem for $F = (0,0,-1)$:
$$
\iiint_D \nabla\cdot F\,dxdydz = \iint_{\partial D} F\cdot \nu\,dS =  \iint_{S} F\cdot \nu\,dS +  \iint_{C} F\cdot \nu\,dS,
$$
where $\nu$ are the unit surface normal pointing outward. The divergence of $F$ is zero, what we wanna compute is the flux across the cone surface $S$:
$$
\iint_{S} F\cdot \nu\,dS = - \iint_{C} F\cdot \nu\,dS.\tag{1}
$$
Basically this identity from divergence theorem tells us that the amount of fluid flowing across the cone surface outward away from the cone, is the same with the amount of the fluid flowing into the cone from the cap. Flowing in/out yields the sign  difference.

Now back to evaluating (1): $\nu$ on the cap is $(0,0,1)$, thus $F\cdot \nu = -1$, the right side is just the area of the cap $C$, which is $\pi$. Therefore:
$$
\iint_{S} F\cdot \nu\,dS = \pi,
$$
and we do not have to deal with the integral on a parametrized surface which is significantly messier than this approach.

Title: Flux across surface


问:
Distributing indistinguishable coins into distinguishable children
Abe wants to distribute 50 (indistinguishable) coins among his 12 (distinguishable) grandchildren, in such a way that all three of his children, A, B, C gets at least 3 coins each. In how many ways can Abe do this?

My solution begins by distributing three coins to A, B, C, and then use starts and bars to distribute the rest:

Without loss of generality, let's begin distributing to A, B, and C in that order.

There are $\binom{50}{3}$ ways to distribute to A.

There are $\binom{47}{3}$ ways to distribute to B.

There are $\binom{44}{3}$ ways to distribute to C.

Then there are 41 coins to be distributed among 12 grandchildren, so using stars and bars with the formula: $\binom{n+r-1}{r-1}$, where $n$ denotes the number of indistinguishable items to distribute, and $r$ denotes the bins.

We have $\binom{41+12-1}{12-1} = \binom{52}{11}$ ways to distribute the rest of the coins.

In total, we have $\binom{50}{3}\binom{47}{3}\binom{44}{3}\binom{52}{11}$ ways to distribute.

However, I got the question wrong, and the notes says "Not accounting for "at least 3 coins each" condition correctly".

I am wondering what was I missing here?

答:
There are $50$ identical coins. So you do not have ${50 \choose 3}$ ways to choose coins for grandchild $A$, for example.

What you are finding is positive integer solutions (and including zero) to

$a_1 + a_2 + ...+ a_{12} = 50$ with the constraint $a_1, a_2, a_3 \geq 3$.

This is equivalent to finding solutions to

$a_1 + a_2 + ...+ a_{12} = 41$ without constraint.

$\big[$Think of it as giving $3$ identical coins each to $a_1, a_2, a_3$ first and there is just one way to do that; then distributing rest $41$ to all $12$ grandchildren without constraint$\big]$.


问:
Find the distribution of $Y=\sqrt X$ when $X$~$N(0,1)$
PDF of $X$ is $f_x(t)=\frac{1}{\sqrt{2pi}}e^{\frac{-x^2}{2}}$.

I tried to found the CDF of $Y$: $F_Y(t)=P(Y \leq t)=P(\sqrt X \leq t)=P(X \leq t^2)=F_X(t^2)$

How do i get distrbution of Y from this? If I were to find the $P(X\leq t^2)$ I would get error function and i don't have the CDF of X to find $F_X(t^2)$

I tried to find the PDF of $Y$: $f_Y(t)=F'_Y(t)=F'_X(t^2)=2tf_X(t^2)$

Does PDF define the distribution of random variable? If not how to i find it?

答:
Here is a solution to finding the distribution of $Y=\sqrt{X}$ conditional on $X>0$ where unconditionally $X\sim \mathcal{N}(0,1)$. In this case, $Y=\sqrt{X}$, given $X>0$ is properly defined, because the square-root is undefined for negative reals. The conditional distribution of $X$ is given by the basic definition of conditional probability conditional on events:
$$P(X\leq x| X>0)=P(0<X\leq x)/P(X>0)=\frac{1}{1-\Phi(0)}[\Phi(x)-\Phi(0)],$$
so that then
$$P(Y\leq y|X>0)=P(\sqrt{X} \leq y |X >0)$$
$$=P(X\leq y^2 |X>0)=\frac{1}{1-\Phi(0)}[\Phi(y^2)-\Phi(0)],$$
now differentiate with respect to $y$ to find that the conditional probability-density-function of $Y$, conditional on $X>0$, is given by
$$f_Y(y|x>0)=\frac{2y \cdot \phi(y^2)}{1-\Phi(0)}$$

Here is a visual/numerical verification in R (black is the empirical density, blue is the exact density derived above):

Please comment for further clarifications/questions/corrections.


问:
Poisson distribution and manipulation of random variables
Let $X$ be a random variable with $X\sim \text{Pois}(\lambda)$. Define $Y=X^2+k$ and obtain the probability mass function $P(Y=y)$. My question is: Can we just do $$P(Y=y)=P(X^2+k=y)=P(X=\sqrt{y-k})$$ or there is some condition that make this wrong?

答:
Yes, it's correct. Consider that you must define $Y$-support too.

$$\mathbb{P}[Y=y]=\frac{e^{-\lambda}\lambda^{\sqrt{y-k}}}{(\sqrt{y-k})!}$$

$$y=k,k+1,k+4,k+9,k+16,\dots$$


问:
$e^{\pi i} = -1$ is this a fact or an assumption?
I think it is a fact but can someone explain why is it true intuitively?

I heard a lot of videos on youtube assuming it is the "natural" way of revolving around 0, many other explanations that does not make sense to me.

I only need a direct way of explaining why the imaginary power causes the number to become 1 in its radius (magically).

e.g. 1^i = 1

e.g. 9999^i = 1 (with some rotation)

the general intuition for exponentials is that it gives different numbers, but using an imaginary power always gives a 1 (in the radius)

this can also be proven using the formula: z = r(cos(θ) + i sin(θ)) where θ contains all the information about the number then it gets converted to an angle only without any radius other than 1.

答:
If you understand Taylor expansions you can prove this property of the exponential quite easily.  Assuming it is complex differentiable (see https://mathworld.wolfram.com/ComplexDifferentiable.html) you can express it as a Taylor Series (https://en.wikipedia.org/wiki/Taylor_series)

You can see that

$e^{i \theta} = \sum_{n=0}^{\infty} (i \theta)^n / n!  $

Using that $i^2 = -1$ we can separate the real even terms and imaginary odd terms

$\sum_{n=0}^{\infty} (i \theta)^n / n! = \sum_{n, even}^{\infty} (-1)^{n/2} / n!  + i\sum_{n, odd}^{\infty} (-1)^{(n-1)/2} / n!  $

The former can be identified as the series expansion for $\cos(\theta)$ and the latter as $\sin(\theta)$.  Therefore $e^{i \theta} = \cos(\theta) + i\times \sin(\theta)$ is proven.

If you were to define exponentiation to complex powers such that $9999^i = 1$, the operation $a^z$ would no longer obey the complex differentiability condition. You can define a function that behaves like normal exponentiation for real(z) but also satisfies $9999^z = 1$ if $z= i$, but then have just defined a new function that is not Holomorphic and not generally relevant to most mathematics.

Edit: \inf -> \infty


问:
Can someone give me an example of two compatible atlas for the same differential structure?
I'm a student of physics and i just began studying this topic. I read the definition of "compatible atlas" but i can't make clear examples of it by myself. Can someone help me?</p>

答: Here's a simple example. Take the usual atlas for the unit circle consisting of the four charts by (open) semicircles (two projecting to the $x$-axis, two projecting to the $y$-axis). Call these maps $\phi_i$, $i=1,2,3,4$. Now take an atlas consisting of the two charts by stereographic projection from the north pole and south pole respectively. Call these maps $\psi_j$, $j=1,2$. The two atlases are compatible, because on overlaps the maps $\phi_i\circ\psi_j^{-1}$ and $\psi_j\circ\phi_i^{-1}$ are smooth maps from appropriate open subsets of $\Bbb R$ to $\Bbb R$.


问:
AP Calculus multiple choice question
The position of a particle along a line is given by $z(t) = 2t^3 -24t^2 + 90t +7$ for $ t \geq 0$. For what values of $t$ is the speed of the particle increasing?

a) $3 < t < 4$ only

b) $t > 4$ only

c) $t > 5$ only

d) $0 < t < 3$ and $t > 5$

e) $3 < t < 4$ and $t > 5$

For this problem I said d) based on my calculations:

$z'(t) = 6t^2 - 48t + 90 = 6(t-5)(t-3)$. I then did a chart and found out that $z'(0) > 0$ ,$z'(4) < 0$ and $z'(6) > 0$ leading me to conclude d) however the textbook says e). I thought that I did everything right, what is wrong with my conclusion?

答: The speed is increasing $\iff$ the acceleration and velocity have the same sign. The second derivative (acceleration) is given by $z''(t) = 12t - 48$, so $z''(t) > 0$ for $t > 4$ and $z''(t) < 0$ for $t < 4$.   Now it's easy to see why (e) is the answer.  </p>

Title: AP Calculus multiple choice question


问:
Calculating probability for forming a triangle
I am having trouble coming up with a solution for this problem:

There is a stick of unit length. We break it into two parts.
  Now, we pick the bigger one and break it into two parts.
  I want to calculate the probability that the three pieces form a triangle.

The problem is from "Introduction to Probability, Charles M. Grinstead", Chapter 2.2, Exercise 13

答:
Following is my approach which gives the conditional and joint probability densities explicitly, then shows the feasible region in which it is possible to form a triangle, and finally computes the probability of forming a triangle by integrating the joint PDF over the feasible region.

Explicit expression for the Joint PDF $f_{X,Y}(x,y)$:

Let $(X, Y)$ be the bivariate r.v. in which $X$ represents the position of the first break and $Y$ represents the position of the second break. Note that $X$ and $Y$ are not independent.

Since the stick is of of unit length, the distribution of $X$ is given by $X \sim \mathcal{U}(0,1)$. Hence the density function of $X$ is given by $f_X (x) = 1$ when $x \in (0,1)$, and $0$ elsewhere.

Now, let the first break occur at $X = x$. (Note that we are dealing with continuous r.v. So we will be using the formula described here.)

If $x < \frac{1}{2}$, the bigger half of the stick is the right half. Hence the second break $Y$ will be distributed uniformly between $x$ and $1$. i.e. $(Y | X = x) \sim \mathcal{U}(x,1)$ when $x < \frac{1}{2}$. Hence the conditional density function is given by $f_{Y | X} (y | X = x) = \frac{1}{1 - x}$ for all $y \in (x, 1)$, and $0$ everywhere else in the region $0 < x < \frac{1}{2}$. (This is the region $AEFGA$ shown in the following picture.)

Similarly, if $x > \frac{1}{2}$, the bigger half of the stick is the left half. Hence the second break $Y$ will be distributed uniformly between $0$ and $x$. i.e. $(Y | X = x) \sim \mathcal{U}(0,x)$ when $x > \frac{1}{2}$. Hence the conditional density function is given by $f_{Y | X} (y | X = x) = \frac{1}{x}$ for all $y \in (0, x)$, and $0$ everywhere else in the region $\frac{1}{2} < x < 1$. (This is the region $ABCDA$ shown in the following picture.)

Now, using the formula $f_{X,Y}(x,y) = f_{Y|X}(y | X = x) \cdot f_X (x)$, we get the following joint density function:

\begin{equation*}
f_{X,Y}(x,y) =
\begin{cases}
\frac{1}{1-x} & \text{when $(x,y)$ is in region $AEFGA$} \\
\frac{1}{x} & \text{when $(x,y)$ is in region $ABCDA$} \\
0 & \text{elsewhere}
\end{cases}
\end{equation*}
(You can check that this is a legitimate PDF which satisfies $\iint_{\mathbb{R}^2} f_{X,Y}(x,y) dx dy = 1$).

Feasible Region

In region $AEFGA$, we have $0 < x < y < 1$, and the triangle will be formed if and only if the following three triangle inequalities are satisfied:

\begin{eqnarray*}
(x - 0) + (y - x) > 1 - y &\iff& y > \frac{1}{2} \\
(y - x) + (1 - y) > (x - 0) &\iff& x < \frac{1}{2} \\
(1 - y) + (x - 0) > y - x &\iff& y < x + \frac{1}{2}
\end{eqnarray*}

These three inequalities form the region inside $\Delta AIG$.

Similarly, in the region $ABCDA$, we can form a triangle if and only if $y < \frac{1}{2}$, $x > \frac{1}{2}$, and $y > x - \frac{1}{2}$. This is the region inside $\Delta AHD$.

Probability of forming a triangle

The probability of forming a triangle is given by:
\begin{eqnarray*}
P(\text{forming a triangle}) &=&
\iint \limits_{\Delta AIG} f_{X,Y}(x,y) dx dy
+ \iint \limits_{\Delta AHD} f_{X,Y}(x,y) dx dy \\
&=& \int \limits_{x \;=\; 0}^{1/2} \;\;\; \int \limits_{y \;=\; 1/2}^{x + 1/2}
        \frac{1}{1-x} \;dx \;dy +
    \int \limits_{x \;=\; 1/2}^{1} \;\;\; \int \limits_{y \;=\; x - 1/2}^{1/2}
        \frac{1}{x} \;dx \;dy \\
&=& \big( \ln(2) - \frac{1}{2} \big) + \big( \ln(2) - \frac{1}{2} \big) \\
&=& 2\ln(2) - 1 \\
&\approx& 38.6294.. \%
\end{eqnarray*}


问:
Question on decidability w.r.t. tag systems
I've read that it is known that tag systems in TS(2,2), that is, with two symbols and deletion number $2$, are decidable.

I suppose I'm asking for clarification on what that means in this context. I assume it means that at the very least, for any given system in that class, one can effectively calculate whether or not it eventually halts.

But what about more specific questions? In particular, I believe the simplest non-trivial system in TS(2,2) is $$\{a \rightarrow aab, b\rightarrow a\}.$$

It is immediately clear that almost all starting values grow without bound, and beyond that, there are definite patterns you can make headway on. However, it is not at all clear to me that, say, there's any closed-form way to determine whether the $k$th symbol added will be an $a$ or a $b$, or what the parity is after some number of rounds, or anything like that. My understanding is that Rice's Theorem wouldn't apply since a decidable system can't be Turing complete, so that doesn't answer it. So we know that any result we want can only take finitely long to calculate, but that leaves my main question:

Is it possible for a computational system to be decidable, yet have computationally irreducible properties which have no shortcuts and thus require actually doing all the intervening computation to determine?

答:
It looks like it specifically says it’s not decidable and is Turing complete.

Also, apparently the particular case for having an alphabet with just two letters is not an exception, unless I am misunderstanding something else.

Where did you read it was decidable?


答:
Another one that could maybe work is $f(x) = (2+\sqrt{3})x-\frac{1+\sqrt{3}}{2}$ and
$g(x)= - (2+\sqrt{3})x + \frac{3+\sqrt{3}}{2}$.

Those would be symmetric around $\frac{1}{2}$ and have an integral of $\frac{3}{4}$ above the x-axis and $\frac{1}{4}$ below.

Title: Find a functions such $\int_0^{1} |f(t) + g(t)| dt = 1$

Body:
I'm trying to find different continuous functions $f,g$ such that:

$\int_0^1|f(t)| dt = 1$

$\int_0^1|g(t)| dt = 1$

and

$\int_0^1|f(t) + g(t)| dt = 1$

I tried a lot of possiblities but I end up with nothing. I tried to firsly choose $\int_0^1|f(t) + g(t)| dt = 1$ and then try to find $f$ and $g$, but I wasn't able to do so. Do dsuch functions exist ?


问:
If $a+b$ is even, then $24|ab(a^2-b^2)$
If $a+b$ is even, then $24|ab(a^2-b^2)$.

Since $a+b$ is even, then $a+b=2n$ for some integer $n$. This also implies that both $a$ and $b$ are either even or odd. So there are two cases to consider.

Case $1$: Both $a$ and $b$ are even, then $a=2p$ and $b=2q$ for integers $p$ and $q$. Also $24=2^33$. So $$ab(a^2-b^2)=2^32q(p-1q)(p-0q)(p+1q).$$

If $(p-1q)(p-0q)(p+1q)$ is divisible by $3$, then case $1$ is done. I know that one of three consecutive integers is always divisible by $3$ and what I have found with this problem is the same thing, but where the consecutive aspect was replaced by shifting each $p$ term $q$ units. I'm trying to see how to incorporate this and know with certainty that I'm right and then do this for case $2$ where both $a$ and $b$ are odd.

I'm also certain that there is an easier way to do this problem because what I'm doing is unlike anything in the textbook I'm using, where the problem came from. Any hints or solutions are greatly appreciated. I feel like I'm using a tank to kill a roach.

答:
Hint: You need to prove that $ab(a^2-b^2)$ is divisible by $3$ and by $8$. Now consider these:

$ab(a^2-b^2)=a^3b-ab^3 \equiv ab-ab \bmod 3$ by Fermat.

$(2n+1)^2=8\binom{n+1}{2}+1$

$(2n)^3 = 8 n^3$

Title: If $a+b$ is even, then $24|ab(a^2-b^2)$


问:
Group acting on a set

Let $A$ be a set, and let $G$ be any subgroup of $S_A$.  $G$ is a group of permutations of $A$; we say it is a group acting on the set $A$. Assume here that $G$ is a finite group. If $u \in A$, the orbit of $u$ (with respect to $G$) is the set
  $$O(u)=\{g(u): g \in G\}.$$
  Prove that $G_u = \{g \in G: g(u)=u\}$ is a subgroup of $G$.

Any advice? Thanks a lot.

答:
Let $H$ be a group, whose operation is denoted by $\star$, and whose identity element is $e$.

By definition, a subset $K\subseteq H$ is a subgroup of $H$ when

$e\in K$.

For any $a,b\in K$, we also have $(a\star b)\in K$.

For any $a\in K$, we also have $a^{-1}\in K$.

Prove that $G(u)$ satisfies these requirements. Remember, the group operation of your group $G$ is composition of permutations, so the identity element is the trivial permutation (i.e. the one that does nothing). Also, remember that (by the definition of a group action), for any $g_1,g_2\in G$ and any $u\in A$, we have that $g_1(g_2(a))=(g_1\circ g_2)(a)$.

Title: Group acting on a set


问:
Possible directions for directional derivative
In which directions at the point (2,0) does the function $f(x,y)=xy$ have a rate of change of -1?

$$D_a⃗f(2,0) = ∇f(2,0)⋅a⃗ $$
$$-1 = ||(0,2)|| ||a⃗||cos(θ)$$
$$-1 = (2)(1)cos(θ)$$
$$-1/2 = cos(θ)$$
$$θ = 2π/3,4π/3$$

Is this answer correct, in that there are two angles 2π/3 and 4π/3 radians in the counter-clockwise direction from the gradient vector at point (2,0) at which the function has a rate of change of -1?

Is this the case for all functions in R3, that they always have two angles for each derivative value?

答:
This has nothing to do with directional derivatives. As you've noted, it's an observation about the dot product. Fixing $\vec x$ and the length of $\vec a$, then $\vec x\cdot\vec a$ will be the same for two positions of $\vec a$: Once you find one, reflecting across the line spanned by $\vec x$ gives you another. And those are the only ones.

(The simplest case to see is dot product $0$. Then you get the two vectors — in your case of length $1$ — orthogonal to $\vec x$.)

This is all about two dimensions. In three dimensions, you get a cone when you rotate $\vec a$ around the axis spanned by $\vec x$, and the tip of that vector traces out a circle. And so on in higher dimensions.


问:
Linear Programming: Absolute Value in the Objective Function
I am trying to reformulate a linear programming problem with an absolute value in the objective function. I decided to use a proxy method mentioned in the 'Absolute values in the objective function' section of this article.

I didn't get why 'if the objective is a minimization problem of the form $-|f(x)| + g(y)$ or is a maximization problem of the form $|f(x)| + g(y)$, unfortunately, the model cannot be reformulated as a standard LP model'.

For example, if we look at a maximization problem of the form $|f(x)| + g(y)$, can we not reformulate this as:

$\max U + g(y)$

s.t. $U \le |f(x)|$, i.e. $U\le f(x)$ and $U\le-f(x)$

I don't see why this formulation wouldn't work?

答: It does not work for maximizing $|f(x)|+g(y)$ because $U \leq f(x)$ and $U \leq -f(x)$ together with the objective of maximizing $U$ means that at optimality $U = \min\{f(x), -f(x)\} = -|f(x)|$. So you are maximizing $-|f(x)|+g(y)$ instead of $|f(x)|+g(y)$.


问:
Probability of at least one person in a car
There are 7 people, and 4 cars (the cars have no limit to how many people they can fit). Each person randomly chooses a car, and their decision is independent of what the others choose.
I need to find the probability that there is at least one person in each car.

I'm thinking to do 1 - probability there is at least 1 car with 0 people in, but I'm not sure how to do this?

答:
There are in total $4^7$ ways 7 person can choose a car. From this number we shall subtract the cases where one car is empty, which is $\binom 41 3^7$. In doing so we however multiply subtract the cases with two or more empty cars and shouls correct this. This is the inclusion-exclusion principle so that the number of cases with no car empty is
$$
\sum_{i=0}^4(-1)^i\binom4i(4-i)^7=4!{7\brace4}.
$$
Thus the number reported in the comment is correct.


问:
For what coinage systems does a greedy algorithm not work in providing change?
For the United States coinage system, a greedy algorithm nicely allows for an algorithm that provides change in the least amount of coins.

However, for a coinage system with 12 cent coins, a greedy algorithm would not work. For instance, change for 15 cents would be a 12 cent coin and 3 pennies (4 coins total) whereas a dime and a nickel (2 coins) would be optimal.

In what types of coinage systems does the greedy algorithm not work?

答:
With thanks to Florian Ingels, here is a paper discussing this problem.

We can define a coinage system $C$ as being composed of an ascending series of coin denominations $\{c_i\}$ with $c_1=1$. If the greedy algorithm always works to deliver minimum number of any coins for every target value, it is canonical (in the paper's definition).

There are a number of useful rules of thumb that apply to most practically-encountered coinage systems.

Clearly any two-denomination system $\{1,c\}$ is canonical. Using as many high-denomination coins as possible gives the smallest coin count.

For a three denomination system $\{c_1=1,c_2,c_3\}$, any problem with the greedy algorithm has to occur between the two-coin solutions involving the biggest coin, $c_3+1$ and $c_3+c_2$. So if $c_2=2$ the greedy algorithm will always work.

We can separate the problem for a coinage system $C$ into distinct challenges if for a given denomination $c_k$ we have $c_k\mid c_j$ for all $j>k$; then the system is canonical if the two systems $C]_{c_k}=\{c_1,...,c_k\}$ and ${}_{c_k}[C = \{c_k/c_k=1, c_{k+1}/c_k,...,c_n/c_k\}$ are both canonical. So for the US coin system $U=\{1,5,10,25\}$ we can split into a cent-based coinage $U]_5 =\{1,5\}$ and a nickel-based coinage ${}_5[U = \{1,2,5\}$ both of which are canonical.

For the counterexample of $A=\{1,5,10,12,25\}$, no split is possible. The paper cited earlier gives that a non-canonical system will have some value that can be expressed in two coins that the greedy algorithm will use more coins for. Here as you identify, $5+10=15$ cents will have a wrong greedy solution of $12+1+1+1$.


问:
Intersecting a non-principal filter on $\Bbb Z$ with $\mathcal P(\Bbb N)$
Does there exist a non-principal filter $\mathscr F$ on $\Bbb Z$ such that $\mathscr F \cap \mathcal P(\Bbb N)$ is principal?</p>

答:
I’ll get you started.

Suppose that $\mathscr{F}$ is a filter on $\Bbb Z$ such that $\mathscr{F}\cap\wp(\Bbb N)$ is a principal filter on $\Bbb N$. Let $\mathscr{G}=\mathscr{F}\cap\wp(\Bbb N)$; then there is some non-empty $A\subseteq\Bbb N$ such that $\mathscr{G}=\{G\subseteq\Bbb N:A\subseteq F\}$. In particular, $A\in\mathscr{G}\subseteq\mathscr{F}$.

Let $F\in\mathscr{F}$ be arbitrary; then $\varnothing\ne F\cap A\in\mathscr{F}$. But $F\cap A\subseteq\Bbb N$, so $F\cap A\in\mathscr{F}\cap\wp(\Bbb N)=\mathscr{G}$, and therefore $A\subseteq F\cap A$, i.e., $F\cap A=A$. In other words, $F\supseteq A$ for each $F\in\mathscr{F}$. What can you now conclude about $\mathscr{F}$?


答:
To start with, let me answer your first question. Decidability - in any context - refers only to the "raw" notion of the existence of an algorithm determining membership (or value, or etc.) which always halts. There is no assumption of feasibility, or "closed-formness," or anything else.

As to your main question, this is a very reasonable thing to consider, but it turns out to be deceptively difficult. The remainder of this answer merely addresses this the issue of framing your main question correctly; I am not going to attempt to answer it.

The issue is that the very notion of

computationally irreducible properties which have no shortcuts and thus require actually doing all the intervening computation to determine

is not nearly as clear-cut as it may appear. Roughly speaking, it's very difficult to "look inside" a computational process in a meaningful way without further assumptions on the nature of that process, and in particular "nicely analyzable" computations will always form a very small subclass of general computations. This is of course a highly informal principle (appropriately enough), but it is nonetheless an important obstacle to eventually come to terms with.

The best we can do in my opinion is look at complexity-theoretic concerns. For example, we could ask whether given a particular tag system the question "What symbol is added at stage $k$?" is computable by a polynomial-time machine. Note, however, that this only pays attention to the "extensional" behavior of the relevant algorithms, so it should not be construed as a fully satisfying framework for addressing your informal question.

Title: Question on decidability w.r.t. tag systems

Body:
I've read that it is known that tag systems in TS(2,2), that is, with two symbols and deletion number $2$, are decidable.

I suppose I'm asking for clarification on what that means in this context. I assume it means that at the very least, for any given system in that class, one can effectively calculate whether or not it eventually halts.

But what about more specific questions? In particular, I believe the simplest non-trivial system in TS(2,2) is $$\{a \rightarrow aab, b\rightarrow a\}.$$

It is immediately clear that almost all starting values grow without bound, and beyond that, there are definite patterns you can make headway on. However, it is not at all clear to me that, say, there's any closed-form way to determine whether the $k$th symbol added will be an $a$ or a $b$, or what the parity is after some number of rounds, or anything like that. My understanding is that Rice's Theorem wouldn't apply since a decidable system can't be Turing complete, so that doesn't answer it. So we know that any result we want can only take finitely long to calculate, but that leaves my main question:

Is it possible for a computational system to be decidable, yet have computationally irreducible properties which have no shortcuts and thus require actually doing all the intervening computation to determine?


问:
Proof identity of differential equation
I would appreciate if somebody could help me with the following problem:

Q: $f''(x)$ continuous in $\mathbb{R}$ show that
$$ \lim_{h\to 0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}=f''(x)$$

答: Hint: what is the limit definition of a derivative? Try writing it out, first applying it to $f(x)$, and then applying it to $f'(x)$. Can you apply this definition several times to get what you are looking for? (Note that you must be careful in this argument, as it requires the continuity of $f''(x)$!)</p>

Title: Proof identity of differential equation