答:
Just start by bringing to a common denominator, then things quickly get simplified:
$$\begin{align}\frac{\dfrac{3}{\sqrt{1-x^2}}+1}{\dfrac{3}{\sqrt{1+x}}+\sqrt{1-x}}&= \frac{\cfrac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}}{\cfrac{3+\sqrt{1-x^2}}{\sqrt{1+x}}} \\[4mm]&= \cfrac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}\cdot\frac{\sqrt{1+x}}{3+\sqrt{1-x^2}}\\[4mm]&=\frac{\sqrt{1+x}}{\sqrt{(1-x)(1+x)}}\\[4mm]&=\frac{1}{\sqrt{1-x}}\end{align}  $$
Note: In the first line, I replaced your division sign ($:$) with the fraction sign ($\frac{*}{*}$).

Title: Simplify $\left(\frac{3}{\sqrt{1-x^2}}+1\right):\left(\frac{3}{\sqrt{1+x}}+\sqrt{1-x}\right)$

Body:
Simplify $\left(\dfrac{3}{\sqrt{1-x^2}}+1\right):\left(\dfrac{3}{\sqrt{1+x}}+\sqrt{1-x}\right)$.

I really don't have an idea how to start. Can you show me?


问:
Closed form for Chu–Vandermonde identity-like summations
I have been trying to analyze a randomized algorithm I cooked up and have found the need to compute some moments of a random variable modeling an aspect of the algorithm but I am unsure if there is a nice closed form. Suppose the random variable in question is $X$ and it can take a value $i \in \lbrace 0, 1, \cdots, k \rbrace$ with probability
$$\text{Pr}\lbrace X = i \rbrace = \binom{n}{k}^{-1}\binom{M}{i}\binom{n-M}{k-i}$$
where $n \geq M \geq 1$ and $n \geq k \geq 1$. Note that I assume that $\binom{a}{b} = 0$ if $a < b$ since there are $0$ ways to choose size $b$ sets out of a size $a$ set when this inequality holds. It is clear by the Chu-Vandermonde identity that indeed these values are probabilities because
$$\sum_{i=0}^{k} \text{Pr}\lbrace X = i \rbrace = \binom{n}{k}^{-1} \sum_{i=0}^k \binom{M}{i}\binom{n-M}{k-i} = 1$$
The first moment of $X$, namely its expectation, can be computed analytically as
\begin{align}
\mathbb{E}(X) &= \sum_{i=0}^k i \text{Pr}\lbrace X = i \rbrace \\
&= \binom{n}{k}^{-1} \sum_{i=1}^k i \binom{M}{i}\binom{n-M}{k-i} \\
&= M \binom{n}{k}^{-1} \sum_{i=1}^k \binom{M-1}{i-1}\binom{(n-1)-(M-1)}{(k-1)-(i-1)} \\
&= M \binom{n}{k}^{-1} \binom{n-1}{k-1} \\
&= \frac{Mk}{n}
\end{align}

I also need to compute $\mathbb{E}(X^2)$ so I can do a variance calculation but I am not sure how to go about it. One can find a loose upper bound of $Mk^2/n$ but at least empirically, it seems for the valid ranges of $k$, the upper bound can at least be made to be $Mk^{3/2}/n$. If I could find an exact expression to this moment, I would feel a lot better of where I might be able to go with this algorithm.

So does anyone recognize this probability distribution and have insight into the desired moment I want to compute?

答: I did some digging and found the probability distribution I was looking for was the Hypergeometric distribution. This gave me all the information I needed to finish the analysis of my randomized algorithm!</p>

ContentLicense: CC BY-SA 4.0


问:
What is the type of discontinuity at $x=0$ in $f(x)$ = {$\sin(1/x); x<0$ and $1/x ; x>0$}?
Left hand limit is oscillatory limit and Right hand limit is infinite limit then is it infinite discontinuity or oscillatory discontinuity?</p>

答: Here, $f(0^-)=0$ and $f(0^+)=\infty$, so the limit does not exist at $x=0$, hence there is essential (non-removable) discontinuity at $x=0$.


问:
If $\lim_{n\to\infty}\frac{a_n}{n^\alpha}=1,\alpha>0$, find $\lim_{n\to\infty}\frac{1}{n^{1+\alpha}}\sum_{k=1}^{n}a_k$ using definite integral.
So.. this looks pretty much like a reimann sum, $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}$$ something but it doesn't seem to be able to be written as a reimann sum.....is the final answer 1? I took a guess by grouping $n^\alpha$ to the $\sum_{k=1}^{n}a_k$ part, which would result in $n$ 1s and which would cancel out with the n at the denominator ....okay that's not rigorous I know.  Any leads? Thanks!

答:
Let $b_n=\displaystyle\sum_{k=1}^n k^\alpha$ and $c_n=\displaystyle\sum_{k=1}^n a_k$. Then (the "Riemann sum" argument) $$\lim_{n\to\infty}\frac{b_n}{n^{1+\alpha}}=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\left(\frac{k}{n}\right)^\alpha=\int_0^1 x^\alpha\,dx=\frac{1}{1+\alpha},$$ and we're given $\displaystyle\lim_{n\to\infty}\frac{c_n-c_{n-1}}{b_n-b_{n-1}}=\lim_{n\to\infty}\frac{a_n}{n^\alpha}=1$.

Hence $\displaystyle\lim_{n\to\infty}\frac{c_n}{b_n}=1$ by Stolz–Cesàro, and $\displaystyle\lim_{n\to\infty}\frac{c_n}{n^{1+\alpha}}=\frac{1}{1+\alpha}$ by the above.


问:
Consider $E_f = \{(x,(x,f(x)))| x \in \mathbb{R} \}$ for an arbitrary function, $[E_f]_x$ is a borel set
I have some sort of struggles solving this problem,
Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is an arbitrary function, and $E_f = \{(x,(x,f(x)))| x \in \mathbb{R}   \}$. I have to show that cross sections $[E_f]_x$ and $[E_f]^{(x,y)}$ are Borel sets. Also I have to show that $E_f$ is not necessarily a Borel subset of $\mathbb{R}^3$.

I think Both $[E_f]_x$ and $[E_f]^{(x,y)}$ are single points or an empty set, that is why they are Borel subsets of $\mathbb{R}^2$ and $\mathbb{R}$ respectively.

For the second part I need to know that what Borel subsets of $R^3$ look like, then I can give a counter example.

Do you have any idea?

答: Consider a non-measurable set $V$ like the Vitali set and take $f = \chi_V$ its characteristic function. If $E_{\chi_V}$ was a Borel set then $E_{\chi_V} \cap \{\mathbb R \times \{1\}\} = V \cap \{1\}$ would be Borel. Which is not the case as it not even measurable.


问:
Find the indefinite integral:$ \int \sqrt[3]{\frac{x+1}{x-1}} d x $
$$
\int \sqrt[3]{\frac{x+1}{x-1}} d x
$$
I tried a lot of ways, Wolfram gives me long ugly unswer, and always I came to a lot of partial fractions. I appreciate any help with this example

答:
Hint: If you substitute $$u=\sqrt[3]{\frac{x+1}{x-1}}$$ the integral becomes $$\int\frac{-6u^3}{(u^3-1)^2}du$$
Then, Ostrogradsky's integration method will help you.


问:
Cartesian arrow over an iso is an iso
Let $p:\mathcal{E}\to\mathcal{B}$ be a fibration. I'm working on showing that a Cartesian arrow $f:X\to Y$ over an isomorphism is also an isomorphism.  I see how to get a right inverse $u:Y\to X$ as below

but I'm having difficulty showing that $u$ is also a left inverse. Any assistance is appreciated.

EDIT: I was never quite able to fully work out the proof suggestion in the answer below, but it seems correct so I'm leaving it accepted.  For anyone who's curious, I ended up proving it as follows:

We have that $$f\circ 1_X=f=1_Y\circ f=f\circ u\circ f$$ and there is only one unique arrow $n:X\to X$ above $1_{p(X)}$ such that $f\circ n=f$ since $f$ is Cartesian, but $1_X:X\to X$ and $u\circ f:X\to X$ are both above $1_{p(X)}$ and satisfy this identity so $n=1_X=u\circ f$.

答: I would go at it in a different direction. Define $u$ as a Cartesian lift of $p(f)^{-1}$. Now since Cartesian morphism are closed under composition you still know $fu$ is an identity, but you also know $u$ is Cartesian. You can then show just as you have for $f$ that $u$ has a right inverse, and from $fu=1_Y$ and $uh=1_X$ we get $f=fuh=h$, so $f=u^{-1}$ as desired.

Title: Cartesian arrow over an iso is an iso


问:
Find polynomials such that $(x-16)p(2x)=16(x-1)p(x)$
Find all polynomials $p(x)$ such that for all $x$, we have $$(x-16)p(2x)=16(x-1)p(x)$$

I tried working out with replacing $x$ by $\frac{x}{2},\frac{x}{4},\cdots$, to have $p(2x) \to p(0)$ but then factor terms seems to create a problem.

Link: http://web.mit.edu/rwbarton/Public/func-eq.pdf

答:
Moron's solution and Dubuque's solution are essentilly the same, but I think their presentations can be made neater.

Clearly, $p$ is not a constant polynomial (unless it is zero).  So we may write $p(x) = C(x-z_1)(x-z_2)\ldots(x-z_n)$.  Then the given equality implies that

$(2x-z_1)(2x-z_2)\ldots(2x-z_n)(x-16) = 16(x-1)(x-z_1)(x-z_2)\ldots(x-z_n)$.

By comparing the leading coefficient on both sides, one sees $n=4$.  Thus,

$(x-\frac{z_1}{2})(x-\frac{z_2}{2})(x-\frac{z_3}{2})(x-\frac{z_4}{2})(x-16)=(x-1)(x-z_1)(x-z_2)(x-z_3)(x-z_n)$.

Hence $\lbrace \frac{z_1}{2}, \frac{z_2}{2}, \frac{z_3}{2}, \frac{z_4}{2}, 16\rbrace = \lbrace 1, z_1, z_2, z_3, z_4\rbrace$.  Now one easily sees that, up to a relabelling of indices, we must have $z_i = 2^i$.

OwnerDisplayName: anonymous

ContentLicense: CC BY-SA 2.5


问:
Norm of inverted matrix
Let be $ \|\cdot \| $ an induced matrix norm and $A, B\in \mathbb{R}^{n,n}$ where that $A$ is invertible. Further more it is $ \|B\| < \|A^{-1}\|^{-1} $.
Then $ A+B $ is invertible.

My idea was to show this: If $ (A+B)\cdot x=0 $ then it follows $ x=0\in \mathbb{R}^n $.

So I started with $ 0=\|(A+B)\cdot x\| $ and wanted to show that $ x=0 $ but I don't how I can use this estimation which mentioned above.

答: $(I+A^{-1}B)x=0$, $x \neq 0$ implies $\|x\|=\|A^{-1}Bx\|\leq \|A^{-1}\||B||\|X|| <\|x\|$ a contradiction. Hence $I+A^{-1}B$ is invertible. This implies $A+B=A(I+A^{-1}B)$ is invertible.

Title: Norm of inverted matrix


问:
Show that the binary Hamming Code is a perfect code.
Since the minimum distance of a binary hamming code is 3, we have 1 as the radius of the sphere. How is it being centered about a codeword in a binary hamming code to be called as a perfect code? Can anyone show an explanation for this? Thank you!</p>

答:
Hint: A perfect code $C$ of the field with $q$ elements with parameters $(n,k,d)$ fulfills the Hamming or Sphere packing bound with equality:

$q^n = q^k\cdot \sum_{i=0}^t {n\choose i} (q-1)^i$,

where $t=\lceil (d-1)/2\rceil$ is the number of correctable errors.

Please check for Hamming codes!

Interpretation: The spheres of radiums 1 around the codewords cover the whole ambient space.


问:
Matrix representation of graph to determine if two graphs are isomorphic
Based on the definition of Isomorphism i.e two graphs are isomorphic if there exists a Bijection between Vertices sets and Edge sets of the two graphs. Since a graph can be represented as a Matrix with rows and columns matrix representing the vertices set, can this matrix representation be used to prove if two graphs are Isomorphic ?

Thus if two matrices are equal then the graphs corresponding to those matrices are Isomorphic. Is this statement correct ? If not can you provide the correct statement.

答:
Thus if two matrices are equal then the graphs corresponding to those matrices are Isomorphic.

This statement is true. However, adjacency matrix and incidence matrix can be used in order to verify an isomorphism, rather than finding one. This is because when we write an adjacency matrix of two graphs that are isomorphic, if the ordering of vertices in matrix representations is not the same, then we probably get a different matrix. But, if we have an isomorphism between two graphs $G$ and $H$, say $\phi: G \to H$, with $V(G) = \{v_1,v_2,...,v_n\}$ and $V(H) = \{u_1, u_2,...,u_n\}$ then the ordering in adjacency matrices of these graphs should be like $(v_1,...,v_n)$ and $(\phi(v_1),...,\phi(v_n))$. Here, note that there might be more than one isomorphism and we may get the same matrices with different orderings.

In summary, if we can define a function $\phi: G \to H$ and suspecting that $\phi$ is an isomorphism, then we can verify it by comparing the matrices with orderings $(v_1,...,v_n)$ and $(\phi(v_1),...,\phi(v_n))$. If this is the case, matrices being the same proves that $\phi$ is an isomorphism.


问:
Is it true that for all $k>1$, $ x^k > x \log x $ for sufficiently large $x$?
I was doing some problems in Big-O notation, and I got intrigued by this question as it was never really addressed. If we plot $x^2$, it's obviously bigger than $x\log x$ for all $x > 0$. However, I started plotting graphs for smaller and smaller values of the exponent, and saw that the crossing point between $x^k$ and $x\log x$ got further and further away. I wondered if maybe there was some particular constant which was the smallest value for $k$ such that the two graphs would cross. But then I found out that even $x^{1.1} >x\log x$ when $x> 3.43 \times 10^{+15}$. This has lead me to believe that for any $k>1$, $x^k$ will eventually overtake $x\log x$, but I think it would be beyond my capabilities to prove it, and I don't know where to start. Does anyone know the answer to this, and is the proof understandable for an undergrad math student?

Edit: Also the functions cross at two points, and I noticed that the first crossing point converges to $e$ as $k$ goes to $1$. What's the explanation for this?

答:
Write
$$x^k=x \log(x)\implies x^{k-1}=\log(x)\implies x^{k-1}=\frac 1{k-1}\log \left(x^{k-1}\right)$$
Let $y=x^{k-1}$ to make
$$y=\frac 1{k-1}\log(y)\implies y=\frac{W(1-k)}{1-k}$$ where $W(.)$ is Lambert function.

In the real domain, this function is defined as long as $1-k \geq -\frac 1e$ that is to say $k \leq 1+\frac 1e$.

Back to $x$, the  curves intersect at
$$x=\left(\frac{W(1-k)}{1-k}\right)^{\frac{1}{k-1}}$$ which is an increasing function.

When $k\to 1$, $x\to e$ and when $k\to  1+\frac 1e$, $x\to e^e$.

We can expand $x$ as series. For example, when $k$ is close to $1^+$, we have
$$x=e\left(1+(k-1)+2 (k-1)^2+\frac{13}{3} (k-1)^3+\frac{235}{24} (k-1)^4+O\left((k-1)^5\right) \right)$$ Try it for $k=\frac 54$; the approximation will give $x=\frac{3033 e}{2048}\approx 4.02566$ while the exact value is $4.17708$<.


问:
Notation about a randomized max cut algorithm.
http://users.cms.caltech.edu/~mccoy/publications/teatalk1.pdf

I'm trying to understand the lemma in this.

So we have

Lemma

Let $r$ be a random vector. For any unit vectors $u_{i}$ and $u_{j}$,

$\mathbb{P}\left(\operatorname{sign}(\langle u_{i},r\rangle) \neq \operatorname{sign}(\langle u_{j},r\rangle)\right) = \dfrac{\arccos(\langle u_{i},u_{j}\rangle)}{\pi}$

I'm a bit confused on notation. So what does $\mathbb{P}(\operatorname{sign}(\langle u_{i},r\rangle)$ mean. Also, I'm a bit confused on what a sign of a permutation is. I was under the impression that it's determined by the angle between the vectors.

答:
You missed a parenthesis (which I have now fixed). It's actually $\mathbb{P}\left(\operatorname{sign}(\langle u_{i},r\rangle) \neq \operatorname{sign}(\langle u_{j},r\rangle)\right)$. Is that what your confusion was about?

As for sign of a permutation, it is explained on Wikipedia and probably just about any introductory book which says anything about permutations or linear algebra. I don't see any signs of permutations mentioned in your link, though. The sign in this expression you (mis)quoted is probably just the sign of the inner product, which is determined by the angle, indeed.

ContentLicense: CC BY-SA 3.0


问:
In $\Delta ABC$ where $\angle A = 60^\circ$, $BP$ and $BE$ trisect $\angle ABC$ and $CP$ and $CE$ trisect $\angle ACB$ . Find $\angle BPE$ .

In $\Delta ABC$ where $\angle A = 60^\circ$, $BP$ and $BE$ trisect $\angle ABC$ and $CP$ and $CE$ trisect $\angle ACB$ . Find $\angle BPE$.

What I Tried: Here is a picture :-

It is not hard to realise that angle-chasing is useful here, and that's exactly the same thing I did. I got the required angles in the picture as shown, but I can't seem to understand how to get $\angle BPE$ once I join the line $PE$ , if there is no suitable idea other than angle-chasing which I can use here, what should I do?

Can anyone help me? Thank You.

Edit: I become a bit stupid sometimes, hence I didn't realise $E$ is the incenter which makes $PE$ the angle bisector, implying $\angle BPE = 50^\circ$.

答: HINT: Note that $BE$ and $CE$ are two of the angle bisectors of $\Delta BPC$, so where is the third one?


问:
Find at least one root of $\sin(t^2-1)=\sin(t)$
I have solved one exercise then I have to solve the trigonometric equation at last to finish the exercise.
The equation $\sin(t^2-1)=\sin(t)$

So I start to solve the following. Please give me a recommend or something to recover my solution:

$\begin{cases}
t^2-1=t+2k\pi\\
t^2-1=\pi-t+2k\pi
\end{cases} , k\in\mathbb{Z}$

Then I started to solve the quadratic equation using Discriminant.

Did I do this in correct way ? Or something give me a hint about this.

答:
Yes, you did this the correct way.

Rearranging the first equation to $t^2-t-2k\pi-1=0\\$ and then by the quadratic formula we have $$t=\frac{1\pm\sqrt{1+4(2k\pi+1)}}{2}=\frac{1}{2}\pm \frac{1}{2}\sqrt{5+8k\pi},\space\space k\in\mathbb Z$$
and similarly for the second quadratic $t^2+t-1-\pi-2k\pi=0$ we have $$t=\frac{-1\pm\sqrt{1+4(2k\pi+1+\pi)}}{2}=-\frac{1}{2}\pm \frac{1}{2}\sqrt{5+8k\pi+4\pi},\space\space k\in\mathbb Z$$


问:
Why the chain rule does not work for this question?

$f(x)=x^{14x}$

Find $f'(x)$

I used the chain rule and wrote it as $f(U)=U^{14x},U(x)=x $, and get an answer :$14x(x)^{14x-1}$

But it is wrong .The right answer should be make $y=x^{14x}$ then $\ln y=\ln x^{14x}$  then $\ln y=14x\ln x$ then differentiate each side with respect to $x$ .
Can anyone explain why my method is wrong ?

答:
There are two rules to use. The first $$(f^g)' = g\cdot f^{g-1} \cdot f'$$ if $g$ is constant and $$(f^g)' = \ln f \cdot f^g \cdot g'$$ if $f$ is constant. If neither are constant, the answer is the sum of both options:

$$(f^g)' = g\cdot f^{g-1} \cdot f' + \ln f \cdot f^g \cdot g'$$

This is also chain rule, but in a different form. You might have seen this pattern in product rule:

$$(fg)' = f'g+fg'$$

where you ferret out the dependence (derivative) in one function at a time. This is the intuition you can carry forward if you are careful about it.


答:
The statement is equivalent to
$$
x^\varepsilon   > \log x,
$$
for any $\varepsilon>0$ and sufficiently large $x$. From the power series expansion of the exponential function $e^x>x$ for all $x>0$, i.e., $x>\log x$ for all $x>0$. Thus,
$$
\log x = \frac{2}{\varepsilon }\log (x^{\varepsilon /2})  < \frac{2}{\varepsilon }x^{\varepsilon /2}  < x^{\varepsilon /2} x^{\varepsilon /2}  = x^\varepsilon  ,
$$
whenever $x^{\varepsilon /2}>2/\varepsilon \Leftrightarrow x > (2/\varepsilon )^{2/\varepsilon }$.

Title: Is it true that for all $k>1$, $ x^k > x \log x $ for sufficiently large $x$?

Body:
I was doing some problems in Big-O notation, and I got intrigued by this question as it was never really addressed. If we plot $x^2$, it's obviously bigger than $x\log x$ for all $x > 0$. However, I started plotting graphs for smaller and smaller values of the exponent, and saw that the crossing point between $x^k$ and $x\log x$ got further and further away. I wondered if maybe there was some particular constant which was the smallest value for $k$ such that the two graphs would cross. But then I found out that even $x^{1.1} >x\log x$ when $x> 3.43 \times 10^{+15}$. This has lead me to believe that for any $k>1$, $x^k$ will eventually overtake $x\log x$, but I think it would be beyond my capabilities to prove it, and I don't know where to start. Does anyone know the answer to this, and is the proof understandable for an undergrad math student?

Edit: Also the functions cross at two points, and I noticed that the first crossing point converges to $e$ as $k$ goes to $1$. What's the explanation for this?


问:
How to represent number that is the smallest possible value larger than another number?
How would I represent the number, i, that is one increment larger than $x$, where an increment is infinitely small.

For example, if $s \in (1,2)$, $i$ would be the smallest number $s$ can take on if $x=1$. I thought about making it $i= x+ (\lim_{n\to\infty} \frac{1}{n})$ but this evaluates to $x$. How would I represent this number $i$ for a given $x$?

I could represent it as $i=\text{min}(s)$ for $s \in (x,\infty)$ but this seems messy. I'm looking for a cleaner way to write this if possible

答: Based on the reading from Player3236's comment, I understand that this is not possible and more importantly why this is not possible. examples to understand why for any future readers</p>


问:
Reducing Boolean expressions
Just learning mathematical proof writing and came upon this interesting question Writing an expression using logic.

$$(P \land Q \land \lnot R) \lor (P \land \lnot Q \land \lnot R) \lor (\lnot P \land Q \land R)$$
Just to help myself prepare for these kind of questions (this question might have already been solved, but), how would you reduce this boolean sentence where as the question states, solving for (???) ?

答:
You can start by combining the first two terms:

$$\begin{align*}
(P\land Q\land\neg R)\lor(P\land\neg Q\land\neg R)&=\big((P\land\neg R)\land Q\big)\lor\big((P\land\neg R)\land\neg Q\big)\\
&=(P\land\neg R)\land(Q\lor\neg Q)\\
&=P\land\neg R\;.
\end{align*}$$

That leaves you with

$$(P\land\neg R)\lor(\neg P\land Q\land R)\;.$$

I'm not at all sure that you can do anything nicer with a disjunctive form.

Title: Reducing Boolean expressions


答: as :E is the incenter of $\Delta PEC$  therefore $PE$ bisects the angle

Title: In $\Delta ABC$ where $\angle A = 60^\circ$, $BP$ and $BE$ trisect $\angle ABC$ and $CP$ and $CE$ trisect $\angle ACB$ . Find $\angle BPE$ .

Body:
In $\Delta ABC$ where $\angle A = 60^\circ$, $BP$ and $BE$ trisect $\angle ABC$ and $CP$ and $CE$ trisect $\angle ACB$ . Find $\angle BPE$.

What I Tried: Here is a picture :-

It is not hard to realise that angle-chasing is useful here, and that's exactly the same thing I did. I got the required angles in the picture as shown, but I can't seem to understand how to get $\angle BPE$ once I join the line $PE$ , if there is no suitable idea other than angle-chasing which I can use here, what should I do?

Can anyone help me? Thank You.

Edit: I become a bit stupid sometimes, hence I didn't realise $E$ is the incenter which makes $PE$ the angle bisector, implying $\angle BPE = 50^\circ$.


问:
Finding a prime $p$ to solve a quadratic congruence $\pmod{p}$
I have a congruence of the form $$ax^2+bx \equiv -1 \pmod{p},$$
where $p$ is an odd prime and $a,b \in \mathbb{Z}$. Given $a$ and $b$, is there a general method to finding $p$ such that the above congruence holds?</p>

答:
The integers modulo a prime $p$, denoted ${\bf Z}/p{\bf Z}$, form a finite field (and to emphasize this it is also often denoted ${\bf F}_p$). Here, a quadratic $ax^2+bx+c$ has one root iff it has two roots (given one root, the other one can be computed as $b/a$ minus the first one).

The quadratic formula actually holds when $p\ne2$. Recall $\Delta=b^2-4ac$ is the discriminant.

$$au^2+bu+c=0\iff u=\frac{-b\pm\sqrt{\Delta}}{2a}.$$

Technically, $\sqrt{\Delta}$ by itself is ambiguous (it doesn't specify which square root), but with the $\pm$ there it should no longer matter. Thus, $ax^2+bx+c$ has a solution if and only if $\Delta$ is a square mod $p$, also known as a quadratic residue, which for $p\nmid\Delta$ can be expressed with the Legendre symbol

$$\left(\frac{\Delta}{p}\right)=1.$$

If you were going to do this for a specific $p$, you'd pick a residue of $\Delta$ mod $p$ and factor, then use multiplicativity. In general though you are correct that we need to factor $\Delta$ itself into a product of prime powers and $\pm1$. We can discard any squares that divide $\Delta$, so let $\Delta'$ be the radical of $\Delta$, and

$$\left(\frac{\Delta}{p}\right)=\left(\frac{\Delta' k^2}{p}\right)=\left(\frac{\Delta'}{p}\right)\left(\frac{k^2}{p}\right)=\left(\frac{\Delta'}{p}\right).$$

Thus it suffices to analyze $\Delta'$. In general, you need to look at the factorization of $\Delta'$ and invoke quadratic reciprocity, paying special care to $\Delta'$ being negative (I see this is what you were saying in the comments) and if it is even or odd.

Example: $\Delta'=-130$. For $p\ne 2,5,13$ we have $\displaystyle\left(\frac{\Delta}{p}\right)=1$ iff

$$1=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)\left(\frac{5}{p}\right)\left(\frac{13}{p}\right)\iff (-1)^{\Large \frac{p-1}{2}+\frac{p^2-1}{8}+\frac{5-1}{2}\frac{p-1}{2}+\frac{13-1}{2}\frac{p-1}{2}}=\left(\frac{p}{5}\right)\left(\frac{p}{13}\right).$$

The mod $8$ residues for which the power of $-1$ above yields $1$ are $p\equiv1,3$. The quadratic residues mod $5$ are $\pm1$, and the quadratic residues mod $13$ are $\pm1,\pm3,\pm4$. Therefore,

$$p\equiv\begin{cases} 1,3~(8) \\ \pm1~(5) \\ \pm1,\pm3,\pm4~(13)\end{cases}~{\rm or}~\begin{cases}1,3~(8) \\ \pm2~(5) \\ \pm 2,\pm5,\pm6~(13)\end{cases}~{\rm or}~\begin{cases}5,7~(8) \\ \pm1~(5) \\ \pm2,\pm5,\pm6~(13)\end{cases}~{\rm or}~\begin{cases}5,7~(8) \\ \pm2~(5) \\ \pm1,\pm3,\pm4~(13)\end{cases} $$

One may combine via Sun-Ze (aka Chinese Remainder Theorem) to find all $4\cdot2\cdot2\cdot6=96$ residues $r$ mod $2^3\cdot5\cdot13=520$ such that $p\equiv r~(520)$ implies $ax^2+bx+c$ has roots mod $p$, and the only exceptional cases not covered by these residues are $p=2,5,13$. ($\Delta\equiv0$ is a square mod $p$ so there is a root in those few cases.)


问:
Showing that the sequence is Cauchy if $|x_{n+1}-x_n|<a_n$

Let $a_n \ge 0$ for all $n\in \Bbb N$. Let $\{x_n\}$ be a sequence of real numbers such that $|x_{n+1}-x_n|<a_n$ for all $n\in \Bbb N$. If $\sum a_n$ is convergent then prove that $\{x_n\}$ is a Cauchy sequence in $\Bbb R$.

We have for any $p=1,2,\cdots$ that $|x_{n+p}-x_n|\le a_n+a_{n+1}+\cdots +a_{n+p-1}\le \sum_{k=1}^{n+p-1} a_k$. From here how can I show that $\{x_n\}$ is Cauchy ?

Any hint please?

Body: Hint: bound $a_n+\cdots+a_{n+p-1}\leq\sum_{k=n}^{\infty}a_k=\sum_{k=1}^{\infty}a_k-\sum_{k=1}^na_n$.


问:
Confusion when applying the Empty Set to the Axiom Schema of Replacement
In Kenneth Kunen's "The Foundations of Mathematics", one of the ZFC Axioms (The Axiom Schema of Replacement) reads as follows:

$$\forall A \Big ( \forall x \in A \  \exists !y \varphi(x,y) \rightarrow \exists B \ \forall x \in A \ \exists y \in B \ \varphi (x,y) \Big)$$

which can be reformulated more strictly as:

$$\forall A \Big ( \forall x \big(x\in A \rightarrow \exists!y \varphi(x,y)\big) \rightarrow \exists B \ \forall x \big(x \in A \rightarrow \exists y ( y \in B \land \varphi (x,y))\big)\Big)$$.

Consider when $A = \emptyset$, which has the property that $\forall x (x \notin \emptyset)$.

Looking at the more strict version of the axiom, we have:

$$\forall x \big(\color{red}{x\in \emptyset \rightarrow \exists!y \varphi(x,y)}\big) \rightarrow \exists B \ \forall x \big(x \in \emptyset \rightarrow \exists y ( y \in B \land \varphi (x,y))\big)$$

The "embedded" implication $\color{red}{x\in \emptyset \rightarrow \exists!y \varphi(x,y)}$ is vacuously true for any $x$. Therefore, the antecedent of the axiom when $A = \emptyset$ is true. Because we now have "access" to the consequent, this means that there exists a set of the following form:

$$\exists B \ \forall x \big(x \in \emptyset \rightarrow \exists y ( y \in B \land \varphi (x,y))\big)$$

Now, my question is, what exactly does $\exists B \ \forall x \big(x \in \emptyset \rightarrow \exists y ( y \in B \land \varphi (x,y))\big)$ mean?

Suppose $B'$ is a set that satisfies the formula: $\forall x \Big (x \in \emptyset \rightarrow \exists y \big (y \in B' \land \varphi(x,y) \big ) \Big )
$

Choose an arbitrary $x'$.

We know that $x' \in \emptyset \rightarrow \exists y \big ( y \in B' \land \varphi(x',y) \big )$ is a true statement.

However, because the antecedent of the implication is false, the truth of the implication can occur in two situations involving the consequent:

$\exists y \big ( y \in B' \land \varphi(x',y) \big )$ is false

$\exists y \big ( y \in B' \land \varphi(x',y) \big )$ is true

Situation 1. can be false in two (meaningfully) different ways:

1A. $\neg \exists y (y \in B')$. This would mean, by definition, that $B'$ is the empty set.

1B. $\neg \exists y \big (\varphi(x',y) \big )$ but $\exists y (y \in B')$. This would mean that $B'$ is not empty...but we cannot claim that any $y$ in the universe of all sets has a relationship with $x'$ as described by $\varphi$.

Situation 2. can be true in only one way:

2A. where $B'$ is non-empty and at least one $y$ has the property that $\varphi(x',y)$

Now, realize that the above comments were only made with respect to a particular $x$... (which we called $x'$)... in our domain of discourse (which is the universe of all sets). Obviously, because of the universal quantifier, claims will also be made for an x'', an x''', etc.

So, it seems as though the sets we can prove exist from this axiom when $A = \emptyset$ are the empty set and non-empty sets that may or may be related to an $x$ in the universe of all sets through $\varphi$.

Is this the correct interpretation?

If so, could some please help me understand what it means to prove the existence of a non-empty set that "may or may not" have particular properties? It's almost as if I am saying, "Here. I proved that an ambiguous object that I cannot describe with any clarity exists." I am having difficulties grasping what this means.

答:
You are right: $(x \in \emptyset \to \exists ! x \varphi(x,y))$ being vacuously true, the Axiom holds also for $\emptyset$.

But what does the axiom means? It is a conditional that is universally quantified: thus, it is true for every set $A$.

When $A=\emptyset$ the antecedent of $(x \in A \to \exists ! x \varphi(x,y))$ is False and thus the formula is True.

Thus, we can apply Modus Ponens to derive the consequent of the axiom: $∃B  ∀x(x∈\emptyset → ∃y(y∈B ∧ φ(x,y)))$.

What does it mean this formula? It means that the universally quantified conditional: $∀x(x∈\emptyset → ∃y(y∈B ∧ φ(x,y)))$ holds for some set $B$.

But again: the antecedent $x ∈ \emptyset$ is False for every $x$.

Thus, formula $∀x(x∈\emptyset → ∃y(y∈B ∧ φ(x,y)))$ holds for an arbitraty set $B$.

Conclusion: the empty set is not a counter-example to the Axiom.

The annoying case $A=\emptyset$ may be easily avoided re-writing Kunen's Axiom (call it: $\text {KA}$) as follows:

$(\text {KA}_1) \ ∀A[A \ne \emptyset \to (∀x∈A ∃!yφ(x,y) → ∃B ∀x∈A ∃y∈B φ(x,y))]$

It is interesting to note that the two are equivalent:

(i) $\text {KA} \to \text {KA}_1$: easily done, using tautology: $P \to (Q \to P)$.

(ii) $\text {KA}_1 \to \text {KA}$.

We know that: $\lnot (x \in \emptyset)$. Thus, by tautology: $\lnot P \to (P \to Q)$ we have: $x \in \emptyset → (∃y \in B) \varphi(x,y)$.

By Generalization: $\exists B\forall x (x \in \emptyset → (∃y \in B) \varphi(x,y))$.

Finally: $∀x∈ \emptyset ∃!yφ(x,y) → \exists B\forall x (x \in \emptyset → (∃y \in B) \varphi(x,y))$, and thus, discharging the assumption $A=\emptyset$:

$A=\emptyset \to [∀x∈A ∃!yφ(x,y) → \exists B\forall x (x \in A → (∃y \in B) \varphi(x,y))].$

Now we apply Proof by Cases to the last formula and $\text {KA}_1$ to conclude with $\text {KA}$.

Comment: what troubles you in Kunen's formulation of the Axiom?

The gist of Replacement is to provide a new tool to manufacture "big" sets whose existence is not licensed by other axioms: pair, union, power-set, infinity, Specification schema.

The "procedure" provided by Replacement is quite simple: for every existing set $A$ and every "functional" formula $\varphi(x,y)$, as long as $x$ spans over $A$ we "collect" the values $y$ such that $\varphi(x,y)$ into a new set $B$.

What happens when $x$ spans over $\emptyset$? That no new set $B$ is produced, exactly because every existing $B$ satisfies the "procedure" described above.

Thus, again, the empty set is not a counter-example to Replacement. But to apply the axiom to it is simply pointless.

Having said that, a different formulation is more transparent:

$∀x∃!y \varphi(x,y) → ∀A∃B∀y[y∈B ↔ ∃x(x∈A ∧ \varphi(x,y))]$.

And different versions are equivalent.


问:
Is there a limit to rooting?
I assume that there will always be limits on mathematical equations, but I never had a thought on it when it comes to square rooting.

I actually mean repetitive square rooting (which might be unnecessary):

$$ \sqrt{\sqrt{\sqrt{... \sqrt{n}}}} $$

Basically, I'm encapsulating an encapsulated square root that's already encapsulating another square root that is also encapsulating a square root that also... (you get the deal).

I'm not sure if there are limits to this, but if possible, how far can I go?

Example of repetitive square rooting:
$$ \sqrt{\sqrt{\sqrt{256}}} = \sqrt{\sqrt{16}} = \sqrt{4} = 2 $$

It probably could go well with puzzles.

答:
You can square root as much as you want. The only inefficiency is unexpectedly hitting the decimal values.

I can do multiple square roots and get a result of 5:

$$ \sqrt{\sqrt{\sqrt{\sqrt{152,587,890,625}}}} = \sqrt{\sqrt{\sqrt{390,625}}} = \sqrt{\sqrt{625}} = \sqrt{25} = 5 $$

You can try $ \sqrt{\infty} $ but it's probably not helping you.

Title: Is there a limit to rooting?


问:
Is it possible to construct circle $(DEC)$ with ruler and compass alone
I'm trying to draw this figure:

Given:

$\omega - \odot(O,OX=R)$

$\ell \perp OX$

$\gamma = (ABE)$ is given, it has radius $r$ and it is tangent to $\ell$ at $A$ and $\omega$ at $B$.

I'm trying to find point $D$ such that $(DEC)$ is also tangent to $\omega$ (in $C$) and $\ell$ (in $D$) and $D$ is colinear to $E = (DEC) \cap (AEB)$ and $B$.

My attempt was with analytic geometry after getting stuck with the basic results you can see above.

It didn't get pretty.

Let's call $Z = \ell \cap OX$ and define $x_0 = ZO > 0$.

$AZ = d_r$, so $d_r^2 = (R+x_0)(R-x_0-2r)$

If $\rho$ is the radius of $(DEC)$ and $d_{\rho}^2 = (R+x_0)(R-x_0-2\rho)$

Let's finally call $\square = (R+x_0)(1-\frac rR) + \frac{d_r-d_{\rho}}{\rho-r}(\frac rR d_{\rho} - d_r)$

The equation that I've found that solves $\rho  = f(R,r,x_0)$ is:

$(\square \cdot r \cdot (\rho-r))^2 = ((R+x_0)(d_r-d_{\rho})^2 + \square \cdot (R+x_0+r)(\rho - r))^2 + (\rho-r)^2((R+x_0)(d_r-d_{\rho}) + d_r \cdot \square)^2$

I kinda can't expand it with my notebook tho, is there any smart way to construct this image?

答:
OP's diagram correctly indicates that $\overleftrightarrow{AB}$, $\overleftrightarrow{CD}$, and the radical axis of the smaller circles all pass through $X'$. (The first two collinearities follow from dilating the semicircle on $\overline{XX'}$ with respect to $B$ and $C$. The third is less obvious, but since OP is aware of it we'll take it as "known".)

Altering notation considerably, consider a circle with diameter $\overline{PQ}$, take $\overline{OA}\perp\overline{PQ}$ (where $O$ is not necessarily the center of the circle), and $B$ on $\overline{OA}$. Extending $\overline{PA}$ and $\overline{PB}$ gives points of tangency $A'$ and $B'$.

Define $\alpha := \angle APQ$ and $\beta := \angle BPQ$, and also
$$p := |OP| \qquad q := |OQ| \qquad a := |OA|=p \tan\alpha \qquad b := |OB|=p\tan\beta \tag1$$
Coordinatizing with $O$ at the origin, we have
$$P = (-p,0) \qquad Q = ( q, 0) \qquad A = (0,a)=(0,p\tan\alpha) \qquad B = (0,b)=(0,p\tan\beta) \tag2$$
Also, since $|PA'|=(p+q)\cos\alpha$ and $|PB'|=(p+q)\cos\beta$, we can write
$$\begin{align}
A' &= P + (p+q)\cos\alpha\;(\cos\alpha,\sin\alpha) = \frac{p}{a^2+p^2}\;(pq-a^2,a(p+q))\\[4pt]
B' &= P + (p+q)\cos\beta\; (\cos\beta, \sin\beta)  = \frac{p}{b^2+p^2}\;(pq-b^2,b(p+q))
\end{align} \tag3$$

Let $E$ be the intersection of $\overline{A'B}$ and the radical axis of the two circles. Conveniently, since the radical axis must pass through the midpoint of tangent segment $\overline{AB}$, we don't need to know much else about the circles (centers, radii, etc) to know the axis' equation. In fact, we don't even assume that the circles meet or that $E$ lies on either circle.

Without too much trouble, we find that
$$E = \left(\frac{p(p q-a^2)}{a^2+2p^2+pq}, \frac{p (p + q)(a + b)}{a^2+2p^2+pq}\right) \tag4$$
Conveniently, the $x$-coordinate is independent of $B$. That is, all possible $E$-points lie on the line
$$x = \frac{p(p q-a^2)}{a^2+2p^2+pq} \tag5$$

For OP's specific needs —where the circles do meet and $E$ does lie on both— we take $E$ to be the intersection of this auxiliary line and the $A$-circle. So, "all we need to" is provide a construction for that line; this is actually pretty easy: we exploit the extreme positions of point $B$.

Extend $\overline{OA}$ to meet the circle at $S$ and $T$. Let $M$ and $N$ be the midpoints of $\overline{AS}$ and $\overline{AT}$, and extend $\overline{PM}$ and $\overline{PN}$ to meet $\overline{A'S}$ and $\overline{A'T}$ at $M'$ and $N'$. Then $\overleftrightarrow{M'N'}$ is the auxiliary line whose intersections with the $A$-circle are candidates for OP's point $E$. (Either point will do.) From there, although this isn't shown in the figure, we can extend $\overline{A'E}$ to meet $\overleftrightarrow{OA}$ at $B$, which defines the $B$-circle. Done.


问:
Elementary Ways to Solve System of Exponential Equation
Is there any elementary way (or using Lambert-W maybe) to solve this system of the exponential equation:
$$ \begin{cases}
3^{x+y}+2^{y-1}=23, \\
3^{2x-1}+2^{y+1}=43.
\end{cases} $$

I have tried to eliminate the exponent of 2 but it gets me
$$ 12 \cdot 3^{x + y} + 3^{2x} = 405 $$
which is more complicated.

I have also tried to substitute $ 3^x = u $ and $ 2^y = v $ but there is still $ 3^y $.

Any advice is welcome (it's okay to use non-elementary method). Thanks :)

答:
Hint:

$$(3^x)^2+(12\cdot3^y)3^x-405=0$$

The discriminant is $$(12\cdot3^y)^2+4\cdot405=16\cdot3^{2y+2}+3^4\cdot20=4\cdot3^2(4\cdot9^y+45)$$

For rational $3^x,$  we need $$(2\cdot3^y)^2+45$$ to be perfect square $=d^2, d\ge0$(say)

$$\implies45=d^2-(2\cdot3^y)^2=(d+2\cdot3^y)(d-2\cdot3^y)\le(d+2\cdot3^y)^2$$

$$\implies d+2\cdot3^y\ge\sqrt{45}>6$$

Again, $d+2\cdot3^y$ must divide $45,$ hence can be one of $$\{9,15,45\}$$

From here we can find $3^y$ and $d$ and hence $3^x$


问:
Finding integrating factor (IF) when integrating factor IF will be a function of both $x$ and $y$
I'm not finding any resource or description or systemic methodology to find integrating factors when the integrating factor will be a function of both x and y.

I'm on this problem,

$$
( y - xy^2 ) dx + (x + x^2y^2) dy = 0
$$

In its present form its not exact, but multiplying through by an integrating factor $ \frac{1}{x^2y^2} $ does make it exact.

Now I want to know how you find that, when the integrating factor is both a function of x and y.

The technique on this page for finding an integrating factor $u(x)$ fails.

Which means there's no function of purely x that will make this equation exact.

So I'm trying something like this:

1. Multiply the original equation by a function u(x,y):

$$
u(x,y)( y - xy^2 )dx + u(x,y)(x + x^2y^2)dy = 0
$$

2. Attempt to set $ \frac{\partial M(x,y)}{\partial y} = \frac{\partial N(x,y)}{\partial x} $

$$
\frac{\partial}{\partial y} u(x,y)( y - xy^2 ) = \frac{\partial}{\partial x} u(x,y)(x + x^2y^2)
$$

(The two sides must be equal for u(x,y) to make the original equation exact, since $ \frac{\partial M(x,y)}{\partial y} = \frac{\partial N(x,y)}{\partial x} $ for an exact equation)

3. Now you have:

$$
u_y(x,y)(y-xy^2) + u(x,y)(1-2xy) = u_x(x,y)(x+x^2y^2) + u(x,y)(1+2xy^2)
$$

I'm kind of stuck now. What is the next step, or is this an incorrect start?

Body:
$(y-xy^2)~dx+(x+x^2y^2)~dy=0$

$(xy^2-y)~dx=(x^2y^2+x)~dy$

$\dfrac{dx}{dy}=\dfrac{x^2y^2+x}{xy^2-y}$

Let $x=yu$ ,

Then $\dfrac{dx}{dy}=y\dfrac{du}{dy}+u$

$\therefore y\dfrac{du}{dy}+u=\dfrac{y^4u^2+yu}{y^3u-y}$

$y\dfrac{du}{dy}=\dfrac{y^3u^2+u}{y^2u-1}-u$

$y\dfrac{du}{dy}=\dfrac{y^3u^2+u-y^2u^2+u}{y^2u-1}$

$y\dfrac{du}{dy}=\dfrac{(y-1)y^2u^2+2u}{y^2u-1}$


问:
Show $\lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n} = \|f\|_{\infty}$ for $f \in L^{\infty}$
I have a question that I need help with getting started (possibly I would be back for more help).

I have a measure space $(X,A,\mu)$ that is finite, and $f \in L^{\infty}(\mu)$. Also, defined is $a_n = \int_X\,|f|^n\,d\mu$. I need to show that the limit is: $$\lim_{n\to\infty}\,\frac{a_{n+1}}{a_n} = \|f\|_{\infty} .$$

I am stuck on getting started, anybody have any suggestions?

thanks much

答:
There is an alternative proof, also relying on the fact $\|f\|_p \rightarrow \|f\|_\infty$ as $p\rightarrow \infty$ ($\star$).

The inequality $\frac{a_{n+1}}{a_n} \le \|f\|_\infty$ being evident, the other one follows then from the following:

It holds true that

\begin{equation}
\liminf \frac{a_{n+1}}{a_n} \ge \liminf \sqrt[n]{a_n}
\end{equation}
for any strictly positive sequence $(a_n) \in \mathbb R^\mathbb N$. The same equality holds for $\limsup$, but with reversed inequality sign. This fact is the essential reason why the root test implies the ratio test for numerical series.

And yes, $(\star)$ and the fact to be proven here are linked, as pointed out by Did above, and the proofs are similar. But, in my opinion, very much as in elementary analysis, though their proofs are similar, the ''root'' result implies the ''ratio'' result, but not inversely. That implication is exactly what I wanted to point out here.


问:
How can I solve this definite Integral? In Bayesian Analysis
This integral is part of an exercise of Bayesian analysis, but i don't know how to integrate
I think completing a normal distribution can be helpful

Thanks in advance

答:
Note that if you expand the exponent it is in the form $$x_1a^2 + x_2a + x_3$$

Also note that $$\int_{-\infty}^{\infty} e^{-ax^2\,+\,bx\,+\,c}\,dx = \sqrt{\frac{\pi}{a}}\,e^{\frac{b^2}{4a}\,+\,c}$$


答:
Hints:

After expansion of the exponent and completing the square, you reduce to

$$\int_{-\infty}^\infty e^{-p(a-q)^2+r}da=e^r\int_{-\infty}^\infty e^{-pa^2}da$$

which is a familiar Gaussian integral.

Title: How can I solve this definite Integral? In Bayesian Analysis

Body:
This integral is part of an exercise of Bayesian analysis, but i don't know how to integrate
I think completing a normal distribution can be helpful

Thanks in advance


问:
If $f_n$ converges uniformly to $f$, then $\int f_n\ d\mu \to \int f\ d\mu$ as $ n \to \infty $
Let ($\Omega, \mathcal{F}, \mu)$, with $\mu (\Omega) <\infty$ and suppose  $f_n , n \geq 1$ is a sequence of integrable functions that converges uniformly on $\Omega$ to $f$. Given that $f$ is integrable, show that  $$\int f_n\ d\mu  \to  \int f\ d\mu \quad\text{ as }\quad n \to \infty $$

I tried :

$\left|\int_\Omega f_n \, d\mu - \int_\Omega f\, d\mu\right| $

: (By linearity of the integral)

$= |\int_\Omega (f_n - f)\, d\mu|\leq \int_\Omega |f_n - f|\, d\mu$

How do we prove the inequality above?

I tried the following but  I am not sure if it's correct or sufficient mathematical proof.

Since $f_n$ integrable, so  $\int_\Omega f_n \ d\mu$ is finite which means it exists.

$f$ integrable, so$\int_\Omega f\ d\mu$ is finite,   which means it also exists.

Thus, since they both exist  $(f_n - f)$ exists   $= |\int_\Omega (f_n - f)\, d\mu|$ is finite.

Hence from basic property of integral since  $= |\int_\Omega (f_n - f)\, d\mu|$ exists

Finally  how do we apply the Uniform convergence on      $  \int_\Omega |f_n - f|\, d\mu$ to conclude that

$\int_\Omega f\ d\mu  \to  \int_\Omega f\ d\mu$ as $ n \to \infty $ ?

答:
Step 1.
$$
\left|\,\int_\Omega f\,d\mu \,\right|\le \int_\Omega |f|\,d\mu
$$
Proof. Set
$$
f_-=\max\{-f,0\}, \quad
f_+=\max\{f,0\}.
$$
Then $f_-,f_+$ are measurable, $f_-,f_+\ge 0$ and also
$f=f_+-f_-$ and $|f|=f_++f_-$. So
$$
\left|\,\int_\Omega f\,d\mu \,\right|=\left|\,\int_\Omega (f_--f_-)\,d\mu \,\right|
=\left|\,\int_\Omega f_+\,d\mu -\int_\Omega f_-\,d\mu \,\right| \le
\left|\,\int_\Omega f_+\,d\mu\,\right| +\left|\,\int_\Omega f_-\,d\mu \,\right|
\\=\int_\Omega f_+\,d\mu+\int_\Omega f_-\,d\mu =\int_\Omega (f_-+f_-)\,d\mu
=\int_\Omega |f|\,d\mu.
$$

Step 2. If $f_n\to f$ uniformly on $\Omega$, then for every $\varepsilon>0$, there exists a $N$, such that
$$
n\ge N \quad\Longrightarrow\quad |f_n(x)-f(x)|<\frac{\varepsilon}{\mu(\Omega)+1}
$$
for all $x\in\Omega$.

Hence, if $n\ge N$, then
$$
\left|\,\int_\Omega f_n\,d\mu-\int_\Omega f\,d\mu\,\right|=
\left|\,\int_\Omega (f_n-f)\,d\mu\,\right|\le
\int_\Omega |f_n-f|\,d\mu \le \int_\Omega \frac{\varepsilon}{\mu(\Omega)+1}\,d\mu=\frac{\varepsilon\, \mu(\Omega)}{\mu(\Omega)+1}<\varepsilon.
$$
and thus
$$
\lim_{n\to\infty}\int_\Omega f_n\,d\mu=\int_\Omega f\,d\mu.
$$


问:
$B=\{f\in\{0,1\}^\mathbb N :f^{-1}(\{0\}) \,\text{and}\, f^{-1}(\{1\}) \, \text{infinite}\, \}$
$$B=\{f\in\{0,1\}^\mathbb N :f^{-1}(\{0\}) \,\text{and}\,
f^{-1}(\{1\}) \, \text{infinite}\,
\}$$
Prove that B is not a countable set. I thought about creating a function for every $n\in\mathbb N$ so $\pi :\mathbb N\longrightarrow B$, and another function $g_n=\pi(n)$ and another one $f(n)=1-g_n(n)$
Showing that $f\neq g_n$ for all $n\in\mathbb N$. So $\pi$ won't be subjective. But I have no idea how to address the condition of the group B. TBH, it seems like a messy incomplete solution.
Any ideas for how to address the question?

答: Hint: $\{f: f ^{-1}(\{0\}) \, \text { is finite}\}$  is countable and so is $\{f: f ^{-1}(\{1\}) \, \text { is finite}\}$. Remove these from the uncountable set $\{0,1\}^{\mathbb N}$ and you get an uncountable set.


问:
Prove that $f(x)=\frac{x}{\tan x}$ is a decreasing function for $x\in (0,1]$
$$f’(x)=\frac{\tan x -x\sec^2x}{\tan^2 x}$$

For it to be a decreasing function
$$\tan x -x\sec^2 x<0$$
$$\sin x\cos x>x$$
$$\implies \sin 2x >\frac x2$$

How do I prove that it this inequality exists for $0<x\le 1$?

答:
The last line in your calculation is wrong, together with the direction of the inequality sign. You really get

$\Rightarrow \sin 2x< 2x$.

That is, you want $\sin x< x$, which is well-known and simple (for all positive $x$).


问:
Is $U = V$ in the SVD of symmetric matrices?
Consider the SVD of matrix $A$:

$$A = U \Sigma V^\top$$

If $A$ is a symmetric, real matrix, is there a guarantee that $U = V$?

There is a similar question here that also posits $A$ is positive semi-definite. But I'm wondering whether $U$ would be equal to $V$ if $A$ is symmetric?

答:
If $A=USU^T$ is a SVD, $A$ has to be positive semidefinite, as shown in another answer here.

The converse is not true. If $A$ is positive semidefinite and $A=USV^T$ is a SVD, $U$ is not necessarily equal to $V$. E.g. $A=U0V^T$ is a SVD for any two unitary matrices $U$ and $V$.

However, if $A$ is positive definite, we must have $U=V$ in its SVD, as shown in the answers to Is $U=V$ in the SVD of a symmetric positive semidefinite matrix?


问:
What's the meaning of oscI(fn-f)?

I haven't seen the form of osc$_I(f_n-f)$.I expect your explanation.

答:
It is the oscillation over the set $I$ (in this case an interval) of the function $f_n-f$.

This is defined as the diameter of the set $(f_n-f)(I) = \{f_n(x)-f(x) \mathrel| x \in I\}$. As the name suggests, it measures how much the function varies over the set.

Title: What's the meaning of oscI(fn-f)?


问:
For Rudin 1.15, why can't B or C be both $0$
Question: Under what conditions does equality hold in the Schwarz inequality?

Below is the section in the textbook that I'm working on. I understand that I want $Ba_j - Cb_j = 0$ in order to get equality and, in a geometric interpretation, I can see why $a$ and $b$ would be linearly dependent (since that above equation is equivalent to $Ba_j = Cb_j$) but I don't understand how to get a specific relationship of $B$ and $C$ or what the restrictions are.

答: Since we have$$\sum_{j=1}^n|Ba_j-Cb_j|^2=B\left(AB-|C|^2\right),$$we have the equality $AB=|C|^2$ if and only if $|Ba_j-Cb_j|=0$ for each $j\in\{1,2,\ldots,n\}$. But, since we are assuming that $B\ne0$, this means that $a_j=\frac CBb_j$ for each $j\in\{1,2,\ldots,n\}$. In other words,$$(a_1,a_2,\ldots,a_n)=\frac CB(b_1,b_2,\ldots,b_n).$$


问:
Where did I go wrong in showing $\|x\|_1>\sqrt n\|x\|_2$
To show $\|x\|_1>\sqrt n\|x\|_2$ where the norm $\|\cdot||_1,\|\cdot\|_2$ are defined over $\mathbb R^n$

Let $x=(x_1,x_2,\ldots,x_n),y=(1,1,\ldots,1)\in\mathbb R^n$

Using  Hölder's Inequality

$$\sum_{i=1}^n|x_i y_i|\le\left(\sum_{i=1}^n|x_i|^2\right)^{1/2} \left(\sum_{i=1}^n|y_i|^2\right)^{1/2} \\ \implies \sum_{i=1}^n|x_i|\le  \left(\sum_{i=1}^n|x_i|^2\right)^{1/2}\sqrt n\implies \|x\|_1\le\sqrt n\|x\|_2$$

Where did I go wrong?

答:
Your argument correctly shows that $\|x\|_1\le \sqrt{n}\|x\|_2$, and thus disproves the opposite inequality stated in the title. </p>

<p>Bonus content: in the opposite direction, $\|x\|_2\le \|x_1\|$ holds. Indeed, squaring both sides we get $\sum_i x_i^2$ on the left and $\sum_i x_i^2 +\sum_i\sum_j |x_i|\,|x_j|$ on the right.   </p>

Body:
Your argument correctly shows that $\|x\|_1\le \sqrt{n}\|x\|_2$, and thus disproves the opposite inequality stated in the title.

Bonus content: in the opposite direction, $\|x\|_2\le \|x_1\|$ holds. Indeed, squaring both sides we get $\sum_i x_i^2$ on the left and $\sum_i x_i^2 +\sum_i\sum_j |x_i|\,|x_j|$ on the right.


答:
$$f(x)=\frac{x}{\tan x} \implies f'(x)=\frac{\tan x-x \sec^2 x}{\tan^2 x}=\frac{g(x)}{\tan^2 x}$$
$$g(x)=\tan x-x \sec^2 x \implies g'(x)=\sec^2 x-\sec^2 x-2x\sec x \tan x<0~~if~~ x \in (0,\pi/2)$$
Therefore, $g(x)$ is a decreasing function in $(0, \pi/2) \implies g(x)\le g(0) =0.$
Finally, in $(0, \pi/2) $ $f'(x)<0$, this means $f(x)$ is a decreasing function.

Title: Prove that $f(x)=\frac{x}{\tan x}$ is a decreasing function for $x\in (0,1]$

Body:
$$f’(x)=\frac{\tan x -x\sec^2x}{\tan^2 x}$$

For it to be a decreasing function
$$\tan x -x\sec^2 x<0$$
$$\sin x\cos x>x$$
$$\implies \sin 2x >\frac x2$$

How do I prove that it this inequality exists for $0<x\le 1$?


问:
Proof limit even degree polynomial
Hello I'm trying to prove the following fact

Let
$f(x) = x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ with $n \in \mathbb{N}$ and $a_{n-1},\ldots,a_0 \in \mathbb{R}$. If $n$ is even, then we have  $\displaystyle\lim_{x\to+\infty}f(x)=+\infty$ and $\displaystyle\lim_{x\to -\infty}f(x)=+\infty$.

This seems quite intuitive since the highest degree of the polynomial and any number squared is even. However this is obviously not a proof. So my second thought was that there could be a way to prove that $f(x)$ is not injective, meaning that some values could have more than one corresponding $x$ value. However, this also does not quite hold (I think) for any polynomial. However this is where I'm stuck.
How can I properly prove it ?

答: Hint: Use the  fact that $$f(x)=x^{n}\left[1+\frac {a_{n-1}} x+\frac {a_{n-2}} {x^{2}}+\cdots+\frac {a_0} {x^{n}}\right].$$

Title: Proof limit even degree polynomial


问:
indefinite integral of $\int \frac{dx}{\sqrt{(x)(x+1)(x+2)}}$
I tried integration by parts considering $dx=du$

$\frac{1}{\sqrt{(x)(x+1)(x+2)}}=v$ but im not getting the answer.

My attempt....

$uv=\sqrt{\frac{x}{(x+1)(x+2)}}$ and-$\int{dv u}$=$\frac{I}{2}$+$\int{\frac{\sqrt{x}}{2\sqrt{(x+1)}(x+2)^{\frac{3}{2}}}}$+$\int{\frac{\sqrt{x}}{2\sqrt{(x+2)}(x+1)^{\frac{3}{2}}}}$

How do we calculate $\int{\frac{\sqrt{x}}{2\sqrt{(x+1)}(x+2)^{\frac{3}{2}}}}$ and
$\int{\frac{\sqrt{x}}{2\sqrt{(x+2)}(x+1)^{\frac{3}{2}}}}$

How do we do it?

Please help!

答:
Let $x=\sinh^2t$. Then

$$\int\frac{dx}{\sqrt{x(x+1)(x+2)}}=\int\frac{2\sinh t\cosh t\,dt}{\sqrt{\sinh^2t\cosh^2t(\sinh^2t+2)}}=\sqrt2\int\frac{dt}{\sqrt{1+\frac12\sinh^2t}}
\\=\sqrt2\int\frac{dt}{\sqrt{1-\frac12\sin^2it}}.$$


问:
How to draw the following set?
Given $S \subseteq \mathbb{R}^2$

We get the following:
$$S_1 = [-1, 1] \times [-1, 1], \:
S_2 = \{ \epsilon \in S_1 : || (\epsilon_1, 3)||\leq 5\epsilon_2 \}, \text{and }S_3 = \{ \epsilon \in \mathbb{R}^2 : 1-3\epsilon_1t + \epsilon_2 t^2 \geq 0\text{ for all }0 \leq t \leq 1\}$$

How can I draw these sets? I do not know how to solve it with the epsilons and t.  Does somebody have a hint to tackle such a problem?

答:
Don't be afraid of the epsilons. You can write $S_2$ as well in the form
$$S_2:=\big\{(x,y)\in S_1\bigm|\|(x,3)\|\leq 5y\bigr\}\ ,$$
and similarly

$$S_3:=\bigl\{(x,y)\in{\mathbb R}^2\bigm|1-3x t+yt^2\geq0\ (0\leq t\leq1)\bigr\}\ .$$
The $t$ here is an auxiliary variable needed to describe the intended condition on $(x,y)$.

This is a very nice problem. Be aware that there is more work involved as for a standard exercise. But you can learn a lot from it.

Title: How to draw the following set?


问:
Calculate the values of the "trigonometric" functions for the angles $a+b$ and $a-b$
Calculate the values of the "trigonometric" functions for the angles $a+b$ and $a-b$ if
$\sin a =\frac{3}{5} \, y\, \sin b= \frac{2\sqrt{13}}{13}$

I did for $\sin(a+b) = \sin(a)\cos(b)+\cos(a)\sin(b)= \frac{3}{5}\cos(b)+\frac{2\sqrt{13}}{13}\cos(a)$

but I don't know if am I the correct way and how to know how much is $\cos(b) $ and $\cos(a)$ how can I calculate it? just with calculator?

答:
$$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$$
$\sin(a)=\dfrac 3 5$ implies a right triangle with hypotenuse $5$ and side opposite $a$ of $3$

$\therefore$ side adjacent to $a$ is $4$ $\implies \cos(a)=\dfrac 4 5$

Now, $\dfrac {2\sqrt{13}} {13}=\dfrac 2 {\sqrt{13}}$, implying a hypotenuse of $\sqrt{13}$, opposite of $2$, and by Pythagoras, adjacent of $3$.

$\therefore \cos(b)=\dfrac 3 {\sqrt{13}}$

Substitution into angle formula and simplifying gives $\sin(a+b)=\dfrac {17\sqrt{13}} {65} $

I'll leave $\sin(a-b)$ to you.


问:
Posterior distribution of exponential prior and likelihood?
I have the prior density function:

$$e^{-\theta} \text{ for } \theta > 0$$

and the likelihood function:  $e^{\theta -x}$ for $x \geq \theta$

I have gotten the following in my attempt to derive the posterior distribution:

$$L(\theta) = \prod e^{\theta - x_i} = e^{n\theta}e^{-\sum x_i} \mathbb{I}_{\min X_i}$$

$$\pi(\theta \mid x) = \frac{ e^{-\sum x_i} e^{\theta(n-1)} \mathbb{I}_{\min X_i}}{e^{-\sum x_i}\int_0^{\min(x)}e^{n\theta}e^{-\theta} \, d\theta}$$

$$= \frac{e^{\theta(n-1)}\mathbb{I_{\min X_i}}}{\frac{1}{n-1}e^{\min(x)(n-1)} - \frac{1}{n-1}}$$

Is this correct? Does this further simplify, or is the posterior simply non-standard?

答:
I will illustrate how working with the kernel with respect to $\theta$ is easier.

For a sample $\boldsymbol x = (x_1, x_2, \ldots, x_n)$ drawn from a distribution for which $X - \theta \mid \theta \sim \operatorname{Exponential}(1)$, i.e., $X \mid \theta$ is a shifted exponential distribution with location shift $\theta$ and mean $1+\theta$, where $\theta \sim \operatorname{Exponential}(1)$ is itself random, we have the likelihood for $\theta$
$$\mathcal L(\theta \mid \boldsymbol x) = e^{-\theta} \mathbb 1 (\theta > 0) \prod_{i=1}^n e^{-(x_i-\theta)} \mathbb 1(x_i > \theta) = e^{-\theta} e^{-n \bar x} e^{n \theta} \mathbb 1 (x_{(1)} > \theta > 0) \propto e^{(n-1)\theta} \mathbb (0 < \theta < x_{(1)}) $$ where $\bar x$ is the sample mean.

While this is not exactly difficult to see, we could also have recognized if we observe that for a single observation, the model density is $$\mathcal L(\theta \mid x) = e^\theta e^{-x} \mathbb 1(\theta < x) \propto e^\theta \mathbb 1(\theta < x),$$ consequently the joint likelihood for the whole sample is proportional to $e^{-\theta} e^{n\theta} \mathbb 1(0 < \theta < x_{(1)}).$  Note this is a likelihood with finite support:  $\theta$ cannot exceed the sample minimum $x_{(1)}$ nor can it be less than or equal to zero due to the prior being exponentially distributed.

On the interval $0 < \theta < x_{(1)}$, the posterior density is then proportional to $e^{(n-1)\theta}$ in such a way that the area under the curve equals $1$; i.e., we seek a constant $c$ such that $$c \int_{\theta=0}^{x_{(1)}} e^{(n-1)\theta} \, d\theta = 1.$$  As this leads to $$c = \frac{n-1}{e^{(n-1) x_{(1)}}-1},$$ we obtain the posterior density $$f(\theta \mid \boldsymbol x) = \frac{n-1}{e^{(n-1) x_{(1)}}-1} e^{(n-1)\theta} \mathbb 1 (0 < \theta < x_{(1)}).$$


答:
Sketch: Consider any element $F$ of that set and consider two sequences $a_0< a_1<a_2<\cdots$ and $b_0< b_1< b_2<\cdots$ such that $F(a_i)=0$ and $F(b_j)=1$ for all $i,j\in\Bbb N$. Now, for $S\subseteq \Bbb N$, call $$\tau_S(m)=\begin{cases}b_i&\text{if }i\in S\land m=a_i\\ a_i&\text{if }i\in S\land m=b_i\\ m&\text{if }\forall i\in S, (m\ne a_i\land m\ne b_i)\end{cases}$$

Now, the goal is to prove that the map $$\Xi:\mathcal P(\Bbb N)\to B\\ \Xi(S)=F\circ\tau_S$$ is an injective function...

Body:
$$B=\{f\in\{0,1\}^\mathbb N :f^{-1}(\{0\}) \,\text{and}\,
f^{-1}(\{1\}) \, \text{infinite}\,
\}$$
Prove that B is not a countable set. I thought about creating a function for every $n\in\mathbb N$ so $\pi :\mathbb N\longrightarrow B$, and another function $g_n=\pi(n)$ and another one $f(n)=1-g_n(n)$
Showing that $f\neq g_n$ for all $n\in\mathbb N$. So $\pi$ won't be subjective. But I have no idea how to address the condition of the group B. TBH, it seems like a messy incomplete solution.
Any ideas for how to address the question?


答:
One often defines the oscillation of a function on an interval to be the difference between its supremum on the interval and its infimum on the interval.

If $|f| \leq M$, then the oscillation of $f$ on $I$ is at most $2M$.  That's what's going on here.

Title: What's the meaning of oscI(fn-f)?

Body: I haven't seen the form of osc$_I(f_n-f)$.I expect your explanation.


问:
An upper bound on Conditional Variance
Given $\sigma$-algebras $G \subset F$ and a random variable $X \in L^2(\Omega,F,P)$, I aim to show the following inequality:
$$
\text{Var}(X | G) \leq E[(X - \zeta)^2 | G] \text{ almost surely,}
$$
for all $\zeta \in L^2(\Omega,G,P)$. Conditional variance is defined as Var$(X|G) := E[(X - E[X|G])^2 | G]$.

I have so far been able to show that Var$(X | G) = E[X^2|G] - E[X | G]^2$ and $E[(X-\zeta)^2 | G] = E(X^2| G) - 2\zeta E[X|G] + \zeta^2$ using definitions of conditional expectation. I am now trying to show that $E[X|G]^2 \geq 2\zeta E[X|G] - \zeta^2$ or some identical rearrangement, but I am stuck. Is this the right way to go about this? Is there some conditional expectation trickery which gets us closer to the answer? In addition, why is the fact that we are in $L^2$ relevant?

Conceptually I understand what is going on; essentially $E[X|G]$ is the "closest" $G$-measureable random variable to $X$, in that the variance is minimized. Any suggestions would be appreciated.

答:
It's relevant that $X$ is in $L^2$ since otherwise, its conditional variance might not be defined.

Now, as for showing that the conditional expectation is the $G$-measurable minimizer of the conditional second moments, simply note that, using your computations,

\begin{align}
\mathbb{E}[(X-\zeta)^2|G]-\mathbb{E}[(X-\mathbb{E}[X|G])^2|G] &=\zeta^2-\mathbb{E}[X|G]^2-2\mathbb{E}[X|G](\zeta-\mathbb{E}[X|G])\\
&=(\zeta-\mathbb{E}[X|G])(\zeta+\mathbb{E}[X|G]-2\mathbb{E}[X|G])\\
&= (\zeta-\mathbb{E}[X|G])^2\geq 0,
\end{align}
et voila.

Title: An upper bound on Conditional Variance


问:
A question on subgroups, more specifically, a finite cyclic group
I define a binary operation $\oplus$ on $\mathbb{Z} _n$ as follows: $$[a] \oplus [b] = [a + b],$$ for all $[a], [b] \in \mathbb{Z} _n$. Then $(\mathbb{Z} _n, \oplus )$ is a finite cyclic group with order n.

This proposition in my notes says that the proof involves showing that the value of $[a] \oplus [b]$ is independent of the choice of equivalence class representatives.

Could someone please clarify to me why this is the case?

Kindest regards.

答:
Independence of choice of a representative means(as $\oplus$ is commutative):

$$\textrm{if} \; [a_1] = [a_2] \in \mathbb{Z} _n \; \textrm{ then, } \; [a_1] \oplus [b] = [a_2] \oplus [b] \; \; \; \forall \; [b] \in \mathbb{Z} _n$$

Example: See $\mathbb{Z} _5$, then $[1]=[6]$. If you calculate $$[1] \oplus [2] = [1+2] = [3]$$ and $$[6] \oplus [2] = [6+2]= [8]$$ you see that $[3] = [8]$, as we are in $\mathbb{Z} _5$. This means the result is the same, no matter which representative of the class $[1]= \{ ... ,-4, 1,6,11,... \} $ you chose.

Of course this was only an example, but this is the same way a proof will work.

You prove this, because you want your binary operation to be "well-defined".


问:
Friedman's approach of proving Cauchy-Schwarz inequality
In a linear vector space that is the Euclidean space $\mathbb{E}_{\infty}$, we have the Cauchy-Schwarz inequality

$$ |\langle x,y \rangle| \leq |x| |y|,$$

where both $x,y \in\mathbb{E}_{\infty}$. Explicitly $x=(\xi_{1},\xi_{2},\cdots)$ and $y=(\eta_{1},\eta_{2},\cdots)$, and we assume here that both are finite in length (i.e. the series for $|x|^{2}$ and $|y|^{2}$ converge).

I am familiar with several proofs of this inequality. However, I am unable to follow the particular approach that Friedman indicates in one of the problems (Problem 1.2, p.6) in his book (also I noticed it's posted online here) where he says that we can prove the inequality by using the result

$$ |\alpha x + \beta y|^{2}=\langle \alpha x + \beta y, \alpha x + \beta y\rangle = \alpha^{2} \langle x,x \rangle +2 \alpha \beta \langle x,y \rangle + \beta^{2} \langle y,y \rangle,$$

which holds for any $\alpha,\beta$ scalars, and by putting

$$ x_{n}=(\xi_{1},\xi_{2},\cdots,\xi_{n},0,0,\cdots), $$
$$ \alpha = |y|^{2}, $$
$$ \beta=\langle x_{n},y\rangle, $$

to prove that $\langle x_{n},y\rangle \leq |x_{n}| |y|$.

How do we proceed using this specific approach (not other approaches) to explicitly reach the inequality?

答:
I expect it's a typo, and $\ \beta\ $ should be $\ -\big\langle x_n, y\big\rangle\ $. You'll then get (with $\ \alpha=|y|^2\ $)
\begin{align}
\big|\alpha x_n+\beta y|^2&=|y|^4|x_n|^2-2|y|^2\big\langle x_n, y\big\rangle^2+ |y|^2\big\langle x_n, y\big\rangle^2\\
&=|y|^4|x_n|^2-|y|^2\big\langle x_n, y\big\rangle^2\ ,
\end{align}
or
\begin{align}
\big\langle x_n, y\big\rangle^2&= |y|^2|x_n|^2-\frac{\big|\alpha x_n+\beta y|^2}{|y|^2}\\
&\le |y|^2|x_n|^2\ ,
\end{align}
from which the Cauchy-Schwarz inequality follows immediately.


问:
Prove a function is not in a Sobolev space
A way to check that a $L^2(\mathbb{R})$ function is in $H^{3/2}(\mathbb{R})$, is to check that $(1+|\xi|^2)^{3/4}{\mathscr{F}}(f)$ is in $L^2(\mathbb{R})$.

I am interested on problems on the half-line. To check that a function in $L^2([0,\infty))$ is in  $H^{3/2}([0,\infty))$ if there is an extension of that function that is in $H^{3/2}(\mathbb{R})$. What about proving that something is not in $H^{3/2}([0,\infty))$? Is there a criterion, or a way to determine if a given function is not in  $H^{3/2}([0,\infty))$ other then, for each case, proving there is no extension.

I am using $k=3/2$, $p=2$, and the half-line because that is what I am interested in. But my question can be applied to $H^{k,p}(I)$ for other values of $k$ and p, and other open intervals of $\mathbb{R}$. This question is related to my other question,  Sobolev space on the half-line.

答:
One criteria follows from the following.

Theorem: If for $u \in H^k([0,\infty))$ with $k \leq 2,$ if we define
$$ Eu(x) = \begin{cases}
    u(x) & \text{ if } x \geq 0, \\
    u(-x)+3u(-x/3)-3u(-2x/3) & \text{ if } x < 0,
  \end{cases}$$
then $Eu \in H^k(\Bbb R),$ with comparable norms.

It may be possible to use a simpler extension here, but I would need to check the details. This can be shown by showing these fractional spaces can be equivalently defined using the Gagliardo seminorm, and establishing a suitable extension theorem. The details are essentially covered in the following reference:

Nezza, Palatucci, Valdinoci. Hitchhiker's guide to the fractional Sobolev spaces. arXiv:1104.4345v3  [math.FA]

This gives an explicit means to check if an element lies in $H^k([0,\infty));$ you compute its extension, and check if that lies in $H^k(\Bbb R).$ If $Eu \not\in H^k(\Bbb R),$ then you can conclude that $u \in H^k([0,\infty)).$

Note: If you are trying to show explicit functions don't lie in this Sobolev space, there are many necessary conditions for $u \in H^k([0,\infty))$ that might be easier to verify; local $H^k$ regularity ($\eta u \in H^k(\Bbb R)$ for all $\eta \in C^{\infty}_c((0,\infty))$), whether $u$ has the correct Hölder regularity, etc.


问:
Is there a limit to rooting?
I assume that there will always be limits on mathematical equations, but I never had a thought on it when it comes to square rooting.

I actually mean repetitive square rooting (which might be unnecessary):

$$ \sqrt{\sqrt{\sqrt{... \sqrt{n}}}} $$

Basically, I'm encapsulating an encapsulated square root that's already encapsulating another square root that is also encapsulating a square root that also... (you get the deal).

I'm not sure if there are limits to this, but if possible, how far can I go?

Example of repetitive square rooting:
$$ \sqrt{\sqrt{\sqrt{256}}} = \sqrt{\sqrt{16}} = \sqrt{4} = 2 $$

It probably could go well with puzzles.

答:
Mathematics allows such repetition as many times as you like;  the concept is not limited by what operations can be practically executed, but limited only but what you can imagine whilst remaining strictly logical.  These concepts can be formalized into a strict definition of what is allowed and what is not, but for day to day purposes, there is no reason why you cannot imagine taking the square root again and again.

In this case the only thing that might stop you repeating a square root would be if at any point you hit a negative number, but that won't happen here (as long as you start with a positive number) because each square root is positive.

In fact you are defining a new operation, or function to use the mathematical term, which you can denote $f(n, x)$.  It is equal to the square root of $x$ repeated $n$ times.  In this construction, $x$ can be any number (whole or a decimal) with $ x \geqslant 0 $ and $n$ any whole number with $n \geqslant 1$.  Or you could extend it to allow the case $n = 0$ as well, by saying $f(0,x)$ is just equal to $x$.

You would write this
$$
f(n, x) = \left\{ \begin{array}
\sqrt {\sqrt {\cdots \sqrt x}}, & \text{square root repeated }n\text{ times,}, n \geqslant 1, x \geqslant 0 \\
x & n=0, x\geqslant 0
\end{array} \right.
$$
and mathematicians would understand exactly what you mean.  You can go on to derive some properties of $f$, for example, when $n \geqslant 1$
$$ f(n,x) = \sqrt{ f(n-1,x)}.$$

Another property, more difficult to prove, is that as $n$ increases $f(n,x)$ gets closer and closer to $1$ provided $x$ is strictly positive, that is $x > 0$.  Mathematically this is written,
$$f(n,x) \to 1 \text{ as }n \to \infty \text{ if } x> 0$$

Title: Is there a limit to rooting?


问:
Rotman's Introduction to to the theory of groups. Exercise 3.45.
Can you give me a hint on the first part of the exercise?

Let $p$ be a prime and let $X$ be a finite $G$-set, where $|G| = p^n$ and $|X|$ is not divisible by $p$. Prove that there exists $x \in X$ with $\tau x = x$ for all $\tau \in G$.

I'm trying to prove that there's only one orbit on the $G$-set but the only "special feature" that I see is that the order of $G$ is a power of a prime. However, I cannot figure out how to use that in order to solve prove that there's only one orbit.

答: Hint: Decompose $X$ into a union of orbits and use the orbit-stabilizer theorem (or prove it), and the fact that any subgroup of a $p$-group has as its index a power of $p$.


答:
Notice that

$$\sqrt{\sqrt n}=n^{\frac12\frac12}$$ and after $m$ nestings, $$n^{\frac1{2^m}}, $$ which tends to $$n^0=1$$ by continuity of exponentiation.

If you mean nested square roots that result in an integer value, consider

$$\left(n^{2^m}\right)^{\frac1{2^m}}=n.$$

E.g.
$$\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{32768}}}}}}}}}}}}}}}=2.$$

Title: Is there a limit to rooting?

Body:
I assume that there will always be limits on mathematical equations, but I never had a thought on it when it comes to square rooting.

I actually mean repetitive square rooting (which might be unnecessary):

$$ \sqrt{\sqrt{\sqrt{... \sqrt{n}}}} $$

Basically, I'm encapsulating an encapsulated square root that's already encapsulating another square root that is also encapsulating a square root that also... (you get the deal).

I'm not sure if there are limits to this, but if possible, how far can I go?

Example of repetitive square rooting:
$$ \sqrt{\sqrt{\sqrt{256}}} = \sqrt{\sqrt{16}} = \sqrt{4} = 2 $$

It probably could go well with puzzles.


问:
Proving an Inequality from Finite Element Analysis
There is a uniform triangulation $\tau_h$. $v_h(x)$ is a piecewise linear function such that $$v_h(x)=\sum_{i=1}^nv_i\phi_i(x)$$ where the $v_i$'s are the nodal values, $\phi_i(x)$'s are the hat functions, and $n$ is the number of nodes. Also $\textbf{v}$ is the vector of the nodal values. Have to show that $$c_1\|v_h\|_{L^2(\Omega)}\leq h\|\textbf{v}\|_2\leq c_2\|v_h\|_{L^2(\Omega)}$$ where $\|\cdot\|_2$ is the Euclidean norm, and $c_1,c_2$ are constants.

I think I have got the right portion of the inequality using Poincare-Friedrichs inequality and the Inverse Estimate. But cannot figure out a proof for the left portion of the inequality. First I thought to go through the routine to prove that the two norms are equivalent but then I saw that of what the norms are being taken are not the same. Also the presence of $h$ means it is not saying that the norms are equivalent.

答:
First prove an estimate on a fixed cell $K$ of the triangulation:
$$
\|v_h\|_{L^2(K)}^2 \le c \cdot |K| \cdot (v_{K,1}^2 + \dots + v_{K,d+1}^2),
$$
where $v_{K,1}... v_{K,d+1}$ are the nodes values of $v_h$ in $K\subset\mathbb R^d$. Use element mass matrix to prove this. Constant $c$ is independent of $K$.

Then use the fact that each node is element of at most $L$ cells $K$, $L$ indepedent of the actual triangulation (follows from uniformity of triangulation).
$$
\|v_h\|_{L^2(\Omega)}^2 =
\sum_K \|v_h\|_{L^2(K)}^2 \le \sum_K c \cdot |K| \cdot (v_{K,1}^2 + \dots + v_{K,d+1}^2)
\le L \ c \ h^d \|v\|_2^2,
$$
where I used $|K| \le c h^d$.


问:
Prove by induction that for every $n \geq 1$, every non-empty ternary tree of height n has at most $(3^n − 1)/2$ nodes.
A ternary tree is a tree where each node has at most three children. Prove
that for every $n \geq 1$, every non-empty ternary tree of height $n$ has
at most $\dfrac{(3^n − 1)}{2}$ nodes.

I am confused what to do here, can someone explain me the question and how to solve it by induction?

答:
The case where $n=1$ is clear. Suppose the hypothesis holds for some $n \geq 1$. Then for the case of a ternary tree of height $n+1$ we have the subtree $n$ with at most $\frac {(3^{n}-1)}{2} $ nodes. For the extension of the tree at step $n+1$ we see that each of previous nodes in step $n$ can have at most $3$ child nodes by the definition of a ternary tree. So for the $n+1$ case we have $\frac{3^{n}-1}{2} + 3($# of nodes at step $n$ of the tree$)  \leq  \frac{3^{n}-1}{2} + 3(3^{n-1})=\frac{3^{n}-1}{2} + 3^{n} = \frac{3^{n}-1}{2} +\frac{2(3^{n})}{2} = \frac{3(3^{n}) -1}{2}=\frac {3^{n+1}-1}{2}$.

Hence, the maximal number of nodes for a tree of height $n+1 \leq \frac{3^{n+1}-1}{2}$. The first inequality with $3^{n-1}$ comes from the fact that this problem can be transformed into one concerning a finite geometric series, more specifically the following series: $1+3+3^2+...+3^{n-1}+3^n$ for a tree of height $n+1$.


问:
$A\neq\emptyset$ is bounded above in $\mathbb{R}$ such that $A$ does not contain a greatest element and if $x,y\in\mathbb{R},\cdots$
$A\neq\emptyset$ is bounded above in $\mathbb{R}$ such that $A$ does not contain a greatest element and if $x,y\in\mathbb{R}, x<y,y\in A$ then $x\in A.$

Then I think by the least upper bound property, $A=(-\infty,\sup{A}).$

I was wondering if there is a way to show $\exists a\in\mathbb{R}$ such that $A=(-\infty,a)$ using elementary set theory.

答:
Let $B$ be the set of upper-bounds for $A.$ By def'n of $\sup A$ we have $\sup A=\min B\in B.$

(1). By def'n of upper-bound, if $b\in B$ then $A \subseteq (-\infty,b].$ And $\sup A\in B.$ Therefore $$ A\subseteq (-\infty, \sup A].$$

(2). If $\sup A\in A$ then by (1), $\; \sup A \in A\subseteq (-\infty, \sup A] \implies \sup A=\max A,$ contrary to the constraint on $A.$ Therefore $\sup A\not \in A.$ Therefore by (1), $$A\subseteq (-\infty,\sup A).$$

(3). If $x<\sup A$ then $x<\min B$ so $x\not\in B$ so by def'n of $B, $ $ \;x$ is not an upper-bound for $A$ so by def'n of upper-bound there exists $y\in A$ with $x<y$ so the constraint on $A$ implies $x\in A.$ Therefore  $$(-\infty, \sup A)\subseteq A.$$


问:
Elementary Ways to Solve System of Exponential Equation
Is there any elementary way (or using Lambert-W maybe) to solve this system of the exponential equation:
$$ \begin{cases}
3^{x+y}+2^{y-1}=23, \\
3^{2x-1}+2^{y+1}=43.
\end{cases} $$

I have tried to eliminate the exponent of 2 but it gets me
$$ 12 \cdot 3^{x + y} + 3^{2x} = 405 $$
which is more complicated.

I have also tried to substitute $ 3^x = u $ and $ 2^y = v $ but there is still $ 3^y $.

Any advice is welcome (it's okay to use non-elementary method). Thanks :)

答:
Comment:

Finding by plotting the equations, using Wolfram we get following figure:

$(x, y)≈ (2.2, 1), (-2, 4.4)$


答: The question asks you to show that there is an orbit of size $1$, not to show that there is only one orbit. Note that orbits having size not equal to 1 must have size divisible by $p$ by orbit-stabilizer theorem.</p>

Title: Rotman's Introduction to to the theory of groups. Exercise 3.45.

Body:
Can you give me a hint on the first part of the exercise?

Let $p$ be a prime and let $X$ be a finite $G$-set, where $|G| = p^n$ and $|X|$ is not divisible by $p$. Prove that there exists $x \in X$ with $\tau x = x$ for all $\tau \in G$.

I'm trying to prove that there's only one orbit on the $G$-set but the only "special feature" that I see is that the order of $G$ is a power of a prime. However, I cannot figure out how to use that in order to solve prove that there's only one orbit.


答:
It turns out that my Professor is making a mistake. The correct system of equations is
$$ \begin{cases}
3^{x + y} + 2^{y - 1} = 239, \\
3^{2x - 1} + 2^{y + 1} = 43.
\end{cases} $$

Answer:
We know that $ 239 = 243 - 4 = 3^5 - 2^2 $ and $ 43 = 27 + 16 = 3^3 + 2^4 $, so we have new system of equations that satisfy
$$ \begin{cases}
x + y = 5, \\
y - 1 = 2, \\
2x - 1 = 3, \\
y + 1 = 4.
\end{cases} $$
Which only satisfied for $ (x, y) = (2, 3) $.

And also, thanks for your answer. I really appreciate it.

Title: Elementary Ways to Solve System of Exponential Equation

Body:
Is there any elementary way (or using Lambert-W maybe) to solve this system of the exponential equation:
$$ \begin{cases}
3^{x+y}+2^{y-1}=23, \\
3^{2x-1}+2^{y+1}=43.
\end{cases} $$

I have tried to eliminate the exponent of 2 but it gets me
$$ 12 \cdot 3^{x + y} + 3^{2x} = 405 $$
which is more complicated.

I have also tried to substitute $ 3^x = u $ and $ 2^y = v $ but there is still $ 3^y $.

Any advice is welcome (it's okay to use non-elementary method). Thanks :)


问:
In how many ways can $kn$ people be arranged around $k$ tables such that each table has $n$ people?
Find a general formula to arrange $kn$ people around $k$ tables such that each table has $n$ people?

The way I dealt with it was to form $k$ groups of $n$ and got ans as (kn)!/(n!)^k which is wrong please tell the mistake

答:
The first part of your proposed solution is to count the number of distinct ways to partition $kn$ people into $k$ groups of $n$ people.  This is done by multinomial coefficient:  there are $$\binom{kn}{n,n,\ldots,n} = \frac{(kn)!}{(n!)^k}$$ such ways.  To see why, think of the people as being labeled $1, 2, \ldots, kn$.  For each person, we assign a table numbered $1, \ldots, k$ in the following manner:  first, we select $\binom{kn}{n}$ people for the first table.  Then out of the remaining $(k-1)n$ people, we select $\binom{(k-1)n}{n}$ people for the second table; and so on until the final table has only $\binom{n}{n}$ possible choices.  Thus the total number of arrangements is $$\binom{kn}{n}\binom{(k-1)n}{n} \cdots \binom{n}{n} = \prod_{i=1}^k \binom{in}{n} = \prod_{i=1}^k \frac{(in)!}{n!((i-1)n)!} = \frac{1}{(n!)^k} \prod_{i=1}^k \frac{(in)!}{((i-1)n)!} = \frac{(kn)!}{(n!)^k}.$$

The second part of the solution is to then consider the number of distinct circular permutations of $n$ people around a table.  An arrangement is considered equivalent to another arrangement if they are invariant with respect to a rotation of the table.  So the way to do this is to fix one person's position at the table and count the number of ways to arrange the others relative to that person; i.e., there are $(n-1)!$ circular permutations of $n$ people.  Since there are $k$ tables, for each partition of the whole group into tables, there are $((n-1)!)^k$ ways to seat them.  So the total number of seating arrangements is $$((n-1)!)^k \frac{(kn)!}{(n!)^k} = \frac{(kn)!}{n^k}.$$


问:
By using primitive roots how does one solve $x^2 \equiv -1 \pmod p$ for $x$, given prime $p$.
By using primitive roots how does one solve $x^2 \equiv -1 \pmod p$ for $x$, given prime $p$ (when solvable).

In particular I want to solve for very large $p$ efficiently.

答:
Unless you are enquiring on “how to compute” the primitive roots $\mod p$ themselves (or perhaps I’m misunderstanding your question?), then this is actually straightforward because once you already have a primitive root, say $r$ , modulo $p\equiv 1\pmod 4$, then we have $$(r^{(p-1)/4})^2\equiv-1\pmod p\,.$$
That is, $x= r^{(p-1)/4}$ is such a solution.

Added Later

In addition to VIVID’s comment below on finding a primitive root, it is perhaps worth remarking that there is no general formula for finding them; in general, for sufficiently large $p$, combing through $\{1,2,\ldots,\lfloor\sqrt{p}\rfloor\}$ is enough to find the least primitive root modulo $p$ (indeed, McGown and Trudgian (2020) have recently claimed that $p\ge 10^{56}$ suffices). It is also perhaps worth mentioning that you may also obtain a solution to your problem in $\{1,2,\ldots,p-1\}$ as $$x_0=r^{(p-1)/4}-\lfloor p^{-1}r^{(p-1)/4}\rfloor p\,.$$


问:
Using $(\neg \psi \to \neg \phi )\to (\phi \to \psi )$ as axiom3 of $L$, show $\vdash (((\psi \to \phi )\rightarrow \psi )\rightarrow \psi )$？
How to show $\vdash (((\psi \to \phi )\rightarrow \psi )\rightarrow \psi )$？Equivalently,$((\psi \to \phi )\rightarrow \psi )\vdash \psi $?Of course, by formal language $L$. The Axiom3 of $L$ is $(\neg \psi \to \neg \phi )\to (\phi \to \psi )$--not same to the answer already put on website.
This question came from Logic for mathematicians written by A. G. Hammilton, in the chapter 2 exercise3.c.

答:
No big change needed...

In A.G. Hammilton, Logic for mathematicians's proof system we have:

Example 2.7 (page 31): $\vdash (\phi\to \phi)$.

Prop.2.8 (Deduction Theorem, page 32).

Prop.2.11 (page 35): (a) $\vdash (\lnot \phi \to (\phi \to \psi))$
and (b) $\vdash ((\lnot \phi\to \phi) \to \phi)$.

Now for the main proof :

$(ϕ → ψ) → ϕ$ --- premise

$\lnot \phi$ --- premise

$\vdash \lnot \phi \to (\phi \to \psi)$ --- Prop.2.11 (a)

$\phi \to \psi$ --- from 2) and 3) by MP

$\phi$ --- from 1) and 4) by mp

$(ϕ → ψ) → ϕ \vdash \lnot \phi \to \phi$ --- from 2) and 5) by Deduction Th

$(ϕ → ψ) → ϕ \vdash ϕ$ --- Prop.2.11 (b) and MP

$\vdash ((ϕ → ψ ) → ϕ) → ϕ$ --- from t) by Deduction Th.


问:
Solving linear equations including semi-positive definite matrix
Suppose $A \in \mathbb{R}^{n \times n}$ is a positive semi-definite matrix. Can we solve the following for $x \in \mathbb{R}^n$?

\begin{equation}
\begin{split}
A(x-a) + \lambda I_{\bullet i}&=0\\
I_{\bullet i}^Tx&=0,\,\,\,\text{or }(x_i=0)
\end{split}
\end{equation}
where $\lambda \in \mathbb{R}$, $a \in \mathbb{R}^n$, $I_{\bullet i}$ is the i-th column of the identity matrix, and $i$ is a fixed and given index.
I did the following but cannot go further.

$x^TA(x-a) + \lambda x^TI_{\bullet i}=0$

$x^TA(x-a)=0$. Hence, $x- a \in N(A)$ or $x\in a+ N(A)$ and $\lambda=0$.

Also, $(x-a)^TAx=0$. Hence, $x \in N(A)$ and $Aa = \lambda I_{\bullet i}$ (I have no idea how to find $\lambda$ in this case).

Is there any other solutions for $x$ and consequently for $\lambda$?

答:
Assuming that $a,i$ are given quantities (but not $\lambda$), the problem has a solution when

$Ay=-e_i$  ($e_i:=I_{\bullet,i}$) has a solution $y_0$, i.e., whether $e_i\in\mathrm{im}A$, since the first condition is equivalent to $Ay=-\lambda e_i$ (for some $\lambda$);

For any $b\in N(A)$ such that $e_i^T(y_0+b)\ne0$, let  $\lambda:=-e_i^Ta/e_i^T(y_0+b)$, so that $e_i^T(a+\lambda y_0+\lambda b)=0$, then the solution can be checked to be $$x=a+\lambda(y_0+b)$$ since $A(x-a)=\lambda Ay_0=-\lambda e_i$, and $e_i^Tx=e_i^T(a+\lambda y_0+\lambda b)=0$

[If $e_i^Ta=0$ then the solutions are $x=a+b$ for any $b\in N(A)$, since $(x-a)^TA(x-a)=-\lambda(x-a)^Te_i=0$ so $x-a\in N(A)$.]

If $e_i\notin\mathrm{im}(A)$, then the equation $Ay=-\lambda e_i$ can only be satisfied if $\lambda=0$ and $Ay=0$. Thus $x=a+y$ for some $y\in N(A)$. The other condition $e_i^Tx=0$ is then satisfied iff there is a non-empty intersection $(a+N(A))\cap e_i^\perp\ne\emptyset$. This is equivalent to $N(A)$ not being a subspace of $e_i^\perp$, i.e., $e_i\notin N(A)^\perp$.


问:
Computing the flows $\theta, \Psi$ of $X$ and $Y,$ and verifying that the flows do not commute.
This is Problem 9-18 from Professor Lee's Intro to Smooth Manifolds (2nd edition). I believe I have most of the proof done, however, I am unsure how to finish it from where I currently am at. Can someone please help me? Thank you so much!

Define vector fields $X$ and $Y$ on the plane by $$X = x\frac{\partial}{\partial x} - y\frac{\partial}{\partial y}, \hspace{20 pt} Y = x\frac{\partial}{\partial y} + y \frac{\partial}{\partial x}.$$ Compute the flows $\theta, \Psi$ of $X$ and $Y,$ and verify that the flows do not commute by finding explicit open intervals $J$ and $K$ containing $0$ such that $\theta_s \circ \Psi_t$ and $\Psi_t \circ \theta_s$ are both defined for all $(s,t) \in J\times K,$ but they are unequal for some such $(s,t).$

$\textit{Proof.}$ Recall that the flow is a family of integral curves for a specific vector field.

Let $\gamma(t) = (x(t),y(t))$ be the candidate for the flow. We must solve $$X(\gamma(t)) = \gamma'(t).$$ We need to delineate $\gamma'(t).$ It must be an element of the tangent space at $\gamma(t).$ So, now we verify that. $$\gamma'(t) = x'(t)\partial_{x(t)} + y'(t)\partial_{y(t)}.$$ On the other side we have $X(\gamma(t)) = x(t)\partial_{x(t)} -y(t)\partial_{y(t)}.$ Once $X(\gamma(t)) = \gamma'(t)$ it assigns us $$x'(t) = x(t), y'(t) = -y(t).$$ These equations have the solutions $$x(t) = a\cos t = b\sin t, \hspace{20 pt} y(t) = a\sin t + b \cos t,$$ where $a,b$ are arbitrary constants.

答:
You have the right idea by computing the flows, but you did a mistake while solving the differential equation. The solutions of $x' = x$ are $t\mapsto ae^t$, and the solutions of $y'=-y$ are $t\mapsto be^{-t}$.

For the flow of $Y$, the solutions of $x'= y$ and $y' = x$ are $x(t) = x_0\cosh t + y_0 \sinh t$ and $y(t) = y_0\cosh t + x_0 \sinh t$.

Thus the flows of $X$ and $Y$ are
\begin{align}
\varphi_X(t,(x_0,y_0)) &= (x_0e^t,y_0e^{-t}) \\ \varphi_Y(s,(x_0,y_0)) &= (x_0\cosh s + y_0 \sinh s,y_0\cosh s + x_0 \sinh s)
\end{align}

Try now to find $t$ and $s$, and $x_0$ and $y_0$ such that $\varphi_X\left(t,\varphi_Y(s,(x_0,y_0)) \right)\neq \varphi_Y\left(s,\varphi_X(t,(x_0,y_0))\right)$.

I would add two things to help you understand your mistakes:

when you solved the differential equations, you had two parameters for each solution, that is uncoherent with the fact that you tried to solve a first order linear equation

if you know the formulation of Lie brackets, the flows of two vector fields $X$ and $Y$ commute if and only if $[X,Y]=0$. Consequently, if you compute $[X,Y]$, and it is non zero, you know that the flows do not commute. But you know more than that: if you identify the non-zero component of $[X,Y]$, you can identify how they do not commute, and that can help you find $s,t,x_0$ and $y_0$

Edit
Here is a precise computation.
\begin{align}
\varphi_X\left(t,\varphi_Y(s,(x_0,y_0)) \right) &= \varphi_X\left(t,(x_0\cosh s + y_0 \sinh s,y_0\cosh s + x_0 \sinh s)\right)\\
&=\left( x_0(\cosh s) e^{t} + y_0(\sinh s) e^t,y_0(\cosh s) e^{-t} + x_0(\sinh s)e^{-t}\right) \\
\varphi_Y(s,\varphi_X(t,x_0,y_0))) &= \varphi_Y\left(s,(x_0e^{t},y_0e^{-t} )\right) \\
&= \left(x_0e^t(\cosh s) +y_0e^{-t}(\sinh s), y_0e^{-t}(\cosh s) + x_0e^t (\sinh s) \right)
\end{align}


问:
Parametric Equation for Great Circle
So I've been doing a lot of searching and haven't found exactly what I'm looking for.  My math skills are a bit rusty, so I haven't had luck deriving this on my own.

What I'm looking for is an equation (or set of equations) where I can plug in starting and ending spherical coordinates, plus a percentage (ie [0,1]) and output spherical coordinates of some point in between (ie, progress along a great circle).

The idea is basically to chart the progress of a plane between two cities and draw it on a globe or map.

Inputs:

lat1,lon1

lat2,lon2

r (radius)

p (progress, from 0->1)

Output:

lat_x,lon_x (point in-between)

答:
Perhaps the most succinct and easiest answer is (as quoted from here)

You are given two points $u$ and $v$ on the unit sphere. Think of them as position vectors $\vec u$ and $\vec v$ . It is easy enough to calculate cross products. Calculate $\vec w = (\vec u \times \vec v)\times\vec u$. Then $\vec w$ and $\vec u$ are unit vectors perpendicular to each other and in the plane of the circle. So a parameterization of the circle is $$\vec R (t)=\vec u \cos t+\vec w \sin t.$$

Correction: $\vec w$ is only a unit vector if $\vec u$ and
$\vec v$ are perpendicular.

Title: Parametric Equation for Great Circle


问:
Finding a Taylor expansion of $e^{-x}$ using $e^{x}$
I want a Taylor expansion about $x = - 4$ for the function $f(x)=e^{-x}$.

I instead find the Taylor expansion of $f(x)=e^{x}$ about $x=-4$ and then replace $x$ by$-x$ but this gives me a wrong answer.

This is what I did:

$f(x)=e^{x}$, expanding it about $x=-4$ gives,
$\sum_{n=0}^{\infty} \frac{e^{-4}}{n !}(x+4)^{n}=e^{x}$

Replacing $x$ by$-x$ I get
$e^{-x}=\sum_{n=0}^{\infty} \frac{e^{-4}}{n !}(-x+4)^{n}$

.Which is incorrect. Can anyone please show me what's wrong in my approach?

Body: $$e^{-x}= e^4e^{-(x+4)}=e^4 \sum_{n=0}^{\infty}(-1)^n\frac{(x+4)^n}{n!}.$$


问:
Derivative of $F$ with respect to $W$
Let $g$ be the derivative of a function $F$ with respect to $Z$ where $Z = WX$

(i.e. $\frac{\partial}{\partial Z}F = g$ where $Z = WX$ )

If I want to find the derivative of $F$ with respect to $W$. How do I find that?

(i.e. What is the value for $\frac{\partial}{\partial W}F$ ?)

Note: I know chain rule is involved somehow but I don't know how exactly is it involved

答:
Well, instead of the answer, you can try understading the theory. Here's a screenshot from Thomas' Calculus (13th ed) which might help you...

And another one:

If you can, read the explanation from the book as well. Try your question again.

Title: Derivative of $F$ with respect to $W$


问:
Show that $f$ is continuous if and only if for each $x \in X$ there is a neighborhood $U$ of $x$ such that $f|U$ is continuous.
Useful Definition: $f$ is continuous at $x_o\in X$ if for every $\epsilon >0, \exists \delta = \delta(\epsilon,x_o)<0$ such that if $x \in X$ and $\|x-x_o\|<\delta$ then $\|f(x)-f(x_o)\|<\epsilon$

$f$ is continuous if it is continuous for all $x\in X$.

Useful Theorem: $f$ is continuous (on all of $x$) $\iff$ for every open set $V$ in $R^m$, $f^{-1}(V)$ is open in $X$.

Exercise: Let $f:X\rightarrow Y$.  Show that $f$ is continuous if and only if for each $x \in X$ there is a neighborhood $U$ of $x$ such that $f|U$ is continuous.

My attempt:

proof.

($\implies$) Suppose that $f$ is continuous $\forall x\in X$.  Then there exists some open set in $V\subset Y$ such that $U \subset f^{-1}(V)$ where $f^{-1}(V)$ is open. Now since $U$ is a neighborhood it is open (I've proved that neighborhoods are open previously).  So by the Theorem above $f|U$ is continuous.

($\Longleftarrow$) Suppose that for each $x \in X$ there is a neighborhood $U$ of $x$ such that $f|U$ is continuous...

--
I'm not quite sure how to prove the rest of this or whether my first part of the proof was correct.  Any guidance would be appreciated.

答:
I guess this exercise is the exercise 5 in section 3 in chapter 1 on p.30 in "Analysis on Manifolds" by James R. Munkres.

On p.26 in this book, there is the definition of "neighborhood":

If $U$ is any open set containing $x_0$, we commonly refer to $U$ simply as a neighborhood of $x_0$.

My solution is here:

Assume that $f$ is continuous.

For each $x\in X$, we choose $U:=X$.

Then, $U$ is a neighborhood of $x$ and $f=f|U$ is continuous.

Conversely, assume that for each $x\in X$, there is a neighborhood $U$ of $x$ such that $f|U$ is continuous.

Let $x$ be any point in $X$.

Let $V$ be any open set in $Y$ which contains $f(x)$.

By assumption, there is a neighborhood $W$ of $x$ such that $f|W$ is continuous.

Since $(f|W)^{-1}(V)$ is an open set in $W$, there is an open set $U$ in $X$ such that $(f|W)^{-1}(V)=W\cap U$.

Since both of $W$ and $U$ are open sets in $X$, their intersection $(f|W)^{-1}(V)$ is also an open set in $X$ and $(f|W)^{-1}(V)$ contains $x$.

Obviously, $f((f|W)^{-1}(V))\subset V$ holds.

Therefore, $f$ is continuous.


问:
Finding a linear map when given a matrix
I was wondering if anyone could help me. I am currently studying linear algebra and have came across some questions about finding a linear map associated to a given matrix.

I have been working a lot with the opposition of this statement with finding an associated matrix given the linear map but do not know how to tackle this new situation, although I have a few ideas.

So if I have a matrix $$E=\begin{pmatrix} 3 & -6\\ 1 & -2\end{pmatrix}$$

would the the linear map for this associated matrix be:

$$T=\{3 + x, -6 - 2x\}$$

or am I wrong?

Thanks for the help in advance xx

答:
$$T[(x',y') \rightarrow (3x-6y, x-2y)]$$ Because
$$\begin{pmatrix}x' \\ y' \end{pmatrix}= \begin{pmatrix} 3 & -6 \\ 1 & -2\end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$


问:
Non-homogeneous linear recurrence relation
I have this recurrence relation to solve:

$$a_{n+1}=3a_n+2^{n-1}-1.$$

The homogeneous part's solution is obviously $a_n=k3^n.$ Now I don't know how to solve the original equation, but I do know what to do in the case of a seemingly analogous differential equation. Consider the equation $$y'=3y+2^{x-1}-1.$$ The homogenous part is solved by $y=Ce^{3x}.$ To find the solution of the non-homogeneous equation, I put $y=C(x)e^{3x}$. Then $y'=C'(x)e^{3x}+C(x)3e^{3x}$. But then, substituting, I get $$C'(x)e^{3x}+C(x)3e^{3x}=3C(x)e^{3x}+2^{x-1}-1,$$ or $$C'(x)e^{3x}=2^{x-1}-1,$$ which can be easily solved.

Now, I thought what I should do for my recurrence relation is just take $a_n=k(n)3^n$ like for the differential equation, but this doesn't seem to work. The problem is that now the equation doesn't simplify after the analgous substitution. Simply, if $a_n=k(n)3^n$, then $a_{n+1}=k(n+1)3^{n+1},$ and not $k(n+1)3^n+k(n)3^{n+1}.$

So what's the algorithm here? If possible, please make the algorithm easy to remember by analogy with the algorithm for differential equations given above (for an example equation, but I hope it's clear what I do in the general case). I'm looking more for a way to remember this than for a way to understand it fully. Sorry if this approach offends anyone, but I'm sure that if I know the answer and have an easy way to remember it, then I can figure out a way to understand it.

答:
Your approach will work fine for the difference equation. $$k(n+1)3^{n+1}=k(n)3^{n+1}+2^{n-1}-1$$

Hence $3^{n+1}(k(n+1)-k(n))=2^{n-1}-1$, which you can solve the same way as in the differential equation case.  In discrete-land $f(n+1)-f(n)$ serves the same role that $\frac{d}{dx}f(x)$ does in continuous-land.

Nb. In neither case would I choose this approach to use; it's usually easiest to use undetermined coefficients with a suitable guess, as @Qiaochu suggests.


问:
How to change the orger of integration of the integration in polar co-ordinate?
Change the order of integration of the following double integral $\int_{\theta=\pi/4}^{\pi/2}\int_{r=0}^{\csc(\theta)}r^2\cos \theta drd\theta.$

My attempt:-
I got the area of integration. I plotted it. $r=\csc(\theta) \implies \theta=\arcsin(1/r).$ hence variation of theta from $\theta=\pi/4 \to  \theta=\arcsin(1/r)  $ and $r$ from $r=\sqrt{2} \to r=1.$  But when I calculated using wolfram alpha I got the different answer. What is the wrong in the change of order?

答:
When you are changing order you are only integrating between $1 \leq r \leq \sqrt2$. You are missing to integrate over the circular sector of radius $1$ between $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$. So you can set it up as follows -

$I = \displaystyle \int_{0}^{1} \int_{\pi/4}^{\pi/2} r^2 \cos \theta d\theta dr + \int_{1}^{\sqrt2} \int_{\pi/4}^{\arcsin(1/r)} r^2 \cos \theta d\theta dr$


答:
By matrix multiplication we obtain

$$\begin{pmatrix} 3 & -6\\ 1 & -2\end{pmatrix}\begin{pmatrix} x\\ y\end{pmatrix}=\begin{pmatrix} 3x-6y\\ x-2y\end{pmatrix}=(3x-6y, x-2y)$$

Title: Finding a linear map when given a matrix

Body:
I was wondering if anyone could help me. I am currently studying linear algebra and have came across some questions about finding a linear map associated to a given matrix.

I have been working a lot with the opposition of this statement with finding an associated matrix given the linear map but do not know how to tackle this new situation, although I have a few ideas.

So if I have a matrix $$E=\begin{pmatrix} 3 & -6\\ 1 & -2\end{pmatrix}$$

would the the linear map for this associated matrix be:

$$T=\{3 + x, -6 - 2x\}$$

or am I wrong?

Thanks for the help in advance xx


问:
The professor of a large calculus class randomly selected 6 students
and asked the amount of time (in hours) they spend for his course per week. The data are given below.

10, 8, 9, 7, 11, 13

Estimate the standard error of the estimated mean time spent in a week for this course by the students who are taking this course.

Ok
my work:

I've found $\hat{\mu}=\bar{x}=\cfrac{29}{3}$

$\hat{\sigma}=s=2.16 $ approximately

so $\hat{\sigma_\hat{\mu}}=s_\hat{\mu}=\cfrac{\hat{\sigma}}{\sqrt{n-1}}=\cfrac{2.16}{\sqrt{5}} = .966$ approximately

so final answer should be $.966$ hours

But the book's answer says that the answer is $0.882$ hours

You get this answer via $\cfrac{2.16}{\sqrt{6}} = .882$ approximately

But this makes no sense to me. This would be population standard deviation for $\bar{x}$ NOT sample standard error/deviation for $\bar{x}$. Am I mistaken? Shouldn't it be $\sqrt{n-1}$ in the denominator not just $\sqrt{n}$ because it's a sample? If not why? Can someone explain this to me so that I understand the concept?

答:
The reason why there is difference between the variance of the sample data whether the mean is known or not could be explained with the degree of freedom. Assume we have $N$ independent data from a dataset, if the mean is unknown, it should be estimated as
$$
\overline x = \frac{1}{N}\sum_i^N x_i
$$
the expectation of discrepancy between sample variance and population variance is
$$
E \{ \sigma^2 - s^2 \} = E\left \{  \frac{1}{N}\sum_{i=1}^N (x_i-\mu)^2 - \frac{1}{N}\sum_{i=1}^N(x_i-\overline x)^2 \right \} \\
= E\left \{  \frac{1}{N}\sum_{i=1}^N \left ( ( x_i^2 - 2x_i\mu+\mu^2) - (x_i^2 - 2x_i\overline x + \overline x^2) \right) \right \}\\
= E\left \{  \frac{1}{N}\sum_{i=1}^N \left (  \mu^2 - \overline x^2 +  2x_i(\overline x - \mu) \right) \right \} = E\left \{ \mu^2 - \overline x^2  +  \frac{1}{N}\sum_{i=1}^N \left (2x_i(\overline x - \mu) \right) \right \} = E\left \{ (\overline x - \mu)^2 \right \} = var(\overline x) = \frac{\sigma^2}{N}
$$
So the expected value of sample variance have the following relation with the actual variance,
$$
E \{ s^2 \} = \sigma^2 - \frac{\sigma^2}{N} = \frac {N-1}{N} \sigma^2
$$
However, if our population only consists of the given numbers, then actual mean and estimated one would be identical and thus, there will be no difference between $\sigma$ and $s$.


问:
Show that $\mathbb{RP}^2$ has four "lines", no three of which have a common "point"
I'm going through Stillwells The Four Pillars of Geometry and this is one of the questions from an exercise. I tried searching for this both here and other places online but all of them contain language that I am unable to comprehend. I am a complete beginner to projective geometry and i'm finding it hard to wrap my head around these concepts.

Here's what i've done so far: I know that any "point" in  $\mathbb{RP}^2$ is a line in  $\mathbb{R^3}$, and any "line" in  $\mathbb{RP}^2$ is a plane in $\mathbb{R}^3$. This reduces the problem to one in $\mathbb{R}^3$:  given that I have 4 planes that pass through $\text{O}$, i need to show that no three of them intersect in a line. Is my approach correct? How do I go forward with this?

答:
The problem asks to

Show that $\Bbb{RP}^2$ has four lines such that no three of them has a common point.

You are right about the correspondence of lines in $\Bbb{RP}^2$ with planes in $\Bbb R^3$ going through the origin.

Thus, to solve the problem, it's enough to specify 4 planes in $\Bbb R^3$ such that the intersection of any 3 of them is trivial.

Can you find a 4th plane for the 3 coordinate planes $x=0,\ y=0,\ z=0$?


问:
Why the proof of existence of Riemannian metric does not apply to pseudo Riemannian metric?
I just looked on the proof in Wikipedia Existence of Riemannian metric on smooth manifold, but I don’t see why this proof failed for pseudo-Riemannian metric, could anyone point out why this is not applicable for pseudo-Riemannian metric?</p>

答:
It fails even at the pointwise level: in the Riemannian case,
$$\tag{1} \sum_\beta \tau_\beta g_\beta$$
is always positive definite since $g_\beta >0$, $\tau _\beta \ge 0$ and $\tau_{\beta_0} >0$ for some $\beta_0$. But if each $g_\beta$ is only non-degenerate, then (1) might give you a degenerate symmetric two tensor.

Related question about the existence of Puesdo Riemannian metric: here


问:
$\lim\limits_{n\to \infty}\sin(n) \neq 0.5$
I need to prove using definition,
I tried to prove that the limit isn't $0.5$ by adding and subtracting and finding some $n$ using floor function but it doesn't work.
Any help?
Thank you

答: $\pi>1$ so that in any period of the sine there are integers such that $\sin(n)<0$. You can conclude.