问:
Consecutive Percentage Increase-Decrease Problem
If the price of sugar increases by 10% on the first day and decreases by 20% on the second day and increases by 30% on the third day and decreases by 40% on the fourth day and like this it continues on. On which day the price of the sugar will become 1.2 times of the original price?

I came across this Arithmetic question which I was not able to solve mathematically i.e. I was not able to prove mathematically that this is not possible. However I deduced this logically that this is not possible as the percentage by which the price falls is always greater than the percentage by which the price of the sugar increases and hence there will be no such day the price of the sugar will become 1.2 times of the original price?

Can someone please tell me that if there is a mathematical way possible to prove that there could be no such day possible?

Also what will happen if the question is flipped like on the first day the price decreases by 10% and the second day the price increases by 20% and on the third day the price decreases by 30% and on the fourth day it increases by 40% and like this it continues. Can we find the day when the price could be 'x' times of the original price?

Thanks!

答:
On the tenth day the price of sugar decreases by $100\%$. So the sugar is worthless after day $10$, and thus you just need to check its value for the first nine days.

The prices are:

$$1 \to 1.1 \to 0.88 \to 1.144 \to 0.6864 \to 1.0296 \to 0.41184 \to 0.700128\to 0.1400256 \to 0.26604864 \to 0$$

so the price of $1.2$ is never reached.


问:
Proving that $\int_0^{2\pi}\frac{ 1-a\cos\theta}{1-2a\cos\theta + a²} \mathrm d\theta = 0$ for $|a|>1$

Let $|a|>1$. Show that $$\int_0^{2\pi}\frac{ 1-a\cos\theta}{1-2a\cos\theta + a²} \mathrm d\theta = 0$$

I'm also trying to solve this problem. I wasn't sure if I was supposed to start a new thread or contribute to previous asked questions about this integral.

I have some ideas that we could do something like this

$$ -i \oint _{|z|=1} \frac{1-az}{(z-2a) }dz = \int_{\theta=0}^{\theta=2\pi} \frac{1-a e^{i\theta}}{e^{i\theta}-2a } e^{i\theta}d\theta = \int_{\theta=0}^{\theta=2\pi} \frac{1-a e^{i\theta}}{1-2ae^{-i\theta} }d\theta \quad (2)$$

Then we have that the real part of $(2)$ is almost equal to the desired integral, except for the $a^2$ in the denominator.

An inequality to the right for the desired integral is then given from the real part of $(2)$. All we want to find is an inequality to the left for the desired integral. That hopefully and easily could be calculated with help of

$$ \oint _{|z-z_0|=r}(z-z_0)dz = \frac{0 \quad n \neq -1}{2\pi i \quad n = -1}  $$

Any suggestions?

答:
Edit: Conditions have been imposed on $a$ that remove the counterexamples below.

Nothing has been said about $a$. If $a$ is real and small in absolute value, then the integrand is positive, so the integral cannot be $0$.

ContentLicense: CC BY-SA 3.0


问:
Looking for different approach to find $\left(x_1+\frac1{x_1}\right)\left(x_2+\frac1{x_2}\right)\left(x_3+\frac1{x_3}\right)$

If $x_1,x_2,x_3$ be roots of $x^3+3x+5=0$, find the value of
$$\left(x_1+\frac1{x_1}\right)\left(x_2+\frac1{x_2}\right)\left(x_3+\frac1{x_3}\right)$$

To solve this problem I expanded the expression and used Vieta's formula and got the answer $-\frac{29}5$
. but I wonder is there any other approach to solve this problem?

I tried this way as an alternative ansewr:

Consider $x_1,x_2,x_3$ be roots polynomial $P(x)=x^3+3x+5$. then $\frac1{x_1},\frac1{x_2},\frac1{x_3}$ are roots of $P(\frac1x):$

$$P(\frac1x)=0\quad \rightarrow \quad P(\frac1x)=(\frac1x)^3+\frac3x+5=0\rightarrow 5x^3+3x^2+1=0$$
But I don't know whether it helps or not.

答:
Another approach: Let the roots of $$x^3+3x+5=0 ~~~~(1)$$  be $a,b,c$, then $a+b+c=0, ab+bc+ca=3, abc=-5$.
Let $$F=(a+1/a)(b+1/b)(c+1/c)=\frac{(a^2+1)(b^2+1)(c^2+1)}{abc}=\frac{y_1 y_2y_3}{abc}=\frac{y_1 y_2 y_3}{-5}$$
where $y_1=(a^2+1)$ etc. let us transform $x$ equation to $y$ equation as $y=x^2+1 \implies x= \sqrt{y-1}.$ Putting it in (1) and simplifying we get $$y^3+3y^2-29=0$$ Hence $y_1y_2y_3=29$ and finally, $$F=\frac{-29}{5}$$

ContentLicense: CC BY-SA 4.0


问:
Using the ratio test on $\sum_{n=1}^{\infty} \left(1+\frac{1}{n}\right)^{n^2}$
I already know that I should do the root test on this series due to the $n$th power, but I want to see if I can establish the result using the ratio test first. (Or would I always be stuck with only one kind of test)

I get something like this: $$\lim_{n \to \infty} \left|\frac{\left(1+\frac{1}{n+1}\right)^{\left(n+1\right)^2}}{\left(1+\frac{1}{n}\right)^{n^2}}\right|$$

But then I'm not sure how (or even if I could) simplify this. I've tried on Wolfram Alpha to make it simplify but the best I can get is numerical approximations that seem to converge to $e$

答:
$$a_n=\left(1+\frac{1}{n}\right)^{n^2}\implies \log(a_n)=n^2 \log\left(1+\frac{1}{n}\right)$$ Now, by Taylor
$$\log(a_n)=n-\frac{1}{2}+\frac{1}{3 n}-\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right)$$ Apply it a second time and continue with Taylor
$$\log(a_{n+1})-\log(a_n)=1-\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{a_{n+1} }{a_n }=e^{\log(a_{n+1})-\log(a_n)}=e-\frac{e}{3 n^2}+O\left(\frac{1}{n^3}\right)$$


答:
First, note that
$$
\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}
= 1-\frac{1}{(n+1)^2} \tag{1}
$$
We will use that to make the "right" exponent appear, since we want to use the known limit $\lim_{m\to\infty} (1+\frac{u}{m})^m = e^u$.

Now, since $n^2=(n+1)^2-(2n+1)$, we can rewrite
$$\begin{align*}
\frac{\left(1+\frac{1}{n+1}\right)^{\left(n+1\right)^2}}{\left(1+\frac{1}{n}\right)^{n^2}}
&= \left(1+\frac{1}{n}\right)^{2n+1}\cdot \frac{\left(1+\frac{1}{n+1}\right)^{(n+1)^2}}{\left(1+\frac{1}{n}\right)^{(n+1)^2}}
= \left(1+\frac{1}{n}\right)^{2n+1} \left(\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}\right)^{n^2} \\
&= \color{red}{\left(1+\frac{1}{n}\right)}\cdot \color{blue}{\left(1+\frac{2}{2n}\right)^{2n}}\cdot \color{green}{\left(1-\frac{1}{(n+1)^2}\right)^{(n+1)^2}} \tag{2}
\end{align*}$$
This is great! We know that
$$
\lim_{n\to\infty} \left(1+\frac{1}{n}\right) = \color{red}{1} \tag{3}
$$
and
$$
\lim_{n\to\infty} \left(1+\frac{2}{2n}\right)^{2n} = \color{blue}{e^{2}} \tag{4}
$$
and
$$
\lim_{n\to\infty} \left(1-\frac{1}{(n+1)^2}\right)^{(n+1)^2} = \color{green}{e^{-1}} \tag{5}
$$
so, combining (3), (4), and (5) into (2), we get
$$
\lim_{n\to\infty} \frac{\left(1+\frac{1}{n+1}\right)^{\left(n+1\right)^2}}{\left(1+\frac{1}{n}\right)^{n^2}}
= \color{red}{1} \cdot \color{blue}{e^{2}} \cdot \color{green}{e^{-1}} = \boxed{e}
$$

Title: Using the ratio test on $\sum_{n=1}^{\infty} \left(1+\frac{1}{n}\right)^{n^2}$

Body:
I already know that I should do the root test on this series due to the $n$th power, but I want to see if I can establish the result using the ratio test first. (Or would I always be stuck with only one kind of test)

I get something like this: $$\lim_{n \to \infty} \left|\frac{\left(1+\frac{1}{n+1}\right)^{\left(n+1\right)^2}}{\left(1+\frac{1}{n}\right)^{n^2}}\right|$$

But then I'm not sure how (or even if I could) simplify this. I've tried on Wolfram Alpha to make it simplify but the best I can get is numerical approximations that seem to converge to $e$


问:
Proofs inolving mapping from countably infinite product topology to another countably infinite product topology
Proposition: Let($X_i,\tau_i$) and ($Y_i,\tau^{‘}_i$), i∈N,be countably infinite families of topological spaces having product spaces( $\Pi X_i,\tau$)and( $\Pi Y_i, \tau_{‘}$), respectively.If the mapping
$h_i$: ($X_i,\tau_i$)$\mapsto$($Y_i,\tau^{′}_i$ ) is continuous for each i∈N, then so is the mapping h: (∏Xi,τ)$\mapsto$(∏Yi,τ′)given by h:($\Pi x_i$)=$\Pi h_i(x_i) $ that is,$h(x_1,x_2,...,x_n,…,)$=$〈h_1(x_1),h_2(x_2),...,h_n(x_n),...〉$

Note: we are working with the countable infinite product topology

If in the above proposition each mapping $h_i$ is also(a) one-to-one,(b) onto,(c) onto and open,(d) a homeomorphism, prove that h is respectively (a) one-to-one, (b) onto, (c) onto and open, (d) a homeomorphism

a)injective just means if f(a)=f(b) then a=b, so assume h(x)=h(y) where x and y are infinite tuples. So $(h_1(x_1),h_1(x_2),h_1(x_3)....)$=$(h_1(y_1),h_2(y_2),h_3(y_3)....)$. Now since each $h_i$ are one to one, we have $h_1(x_i)=h_1(y_1)$ so $x_1=y_1 $ and this is true for all $x_i $and $y_i$ so x=y. Thus h is injective.

b)onto means if y is in the codomain, there exist an x such that f(x)=y. If each $h_i$ is onto, then for all infinite n tuple ($y_1,y_2,y_3,y_4....$) there exist a infinite x tuple ($x_1,x_2,x_3,x_4)$ such that ($h_1(x_1),h_1(x_2),h_1(x_3)....)$
=($y_1,y_2,y_3,y_4... )$ By our definition of h, h is onto.

c) Now we assume each $h_i$ is onto and map open sets onto open sets. We have already shown that h is onto if each $h_i$ is onto. Take a basic open set in the product topology which is of the form $O_1\times O_2 \times O_3 \times R \times  R \times R $ with all but finitely many tuples are Rs. Now applying h we get, ($h_1(O_1),h_2(O_2),h_3(O_3),h_4(R)....).$ Since each $h_i$ are open mapping, each component in their respective codomain is open so h is open.

d)The definition of homeomorphism I will use is bijective, open, and continuous mapping and from the proposition above, continuity is already proven. Thus combining the four properties. if each hi is homeomorphic, then h is also homeomorphic.

答:
In fact, the cardinality of the index size of the product is irrelevant. It could be countable, uncountable or finite. The proofs are exactly the same.

You could just write $h: X:=\prod_{i \in I} X_i \to Y:=\prod_{i \in I} Y_i$ and prove the first continuity fact by noting $\pi'_i \circ h = h_i \circ \pi_i$ for all $i$ and so by the universal property of $(Y, \pi'_i, i \in I)$, $h$ is continuous. This makes for smoother notation (moreover one should use MathJax on this site for clarity's and readability's sake)

The injectivity and surjectivity facts are obvious as you noted.

$h(x)=h(y) \iff \forall i: \pi'_i(h(x)=\pi'_i(h(y) \iff \forall i: h_i(\pi_i(x)) = h_i(\pi_i(y))$ and as all $h_i$ are 1-1 we continue with $\iff \forall i: \pi_i(x)=\pi_i(y) \iff x=y$.

If $y\in Y$, for each $i$ we pick $x_i \in X_i$ so that $h_i(x_i) = \pi'_i(y)=y_i$  and then clearly $h(x)=y$ for $x= (x_i)_{i \in I}$.

It suffieces to check openness on basic open sets (a general proposition) and if $O = \prod_i O_i$ is basic open in $X$ (depending on some finite $F \subseteq I$, say), $h[O] = \prod_{i \in I} h_i[O_i]$ is of the same form, as $h_i[O_i]= h_i[X_i]=Y_i$ for all $ i \notin F$; this is where ontoness of the $h_i$ is used, so $h[O]$ is also (basic) open.

As a homeomorphism is 1-1 open and onto, the last follows immediately from the previous three.

This is all the proof comes down to, really.


问:
Proving that $\;\text{Im}\left[r\log\left(\frac{-1 + r}{r}\right)\right]= \pi r$

Assume $r \in (0,1)$, I want to prove that

$$ \text{Im}\left[r\log\left(\frac{-1 + r}{r}\right)\right]=\pi r$$

This is probably something trivial, but I don't have much experience with complex numbers. Are there any known identities for similar expressions?

As a side note, this logarithm cancels out a specific imaginary part of a Lerch Transcendent in another problem, so I can only guess that it's equal to $\pi i$.

答:
It seems to be that by $\;log\;$ you meant the complex log, something which isn't given...and thus:

$$\log\frac{-1+r}r=\ln\left|\frac{-1+r}r\right|+i\arg\frac{-1+r}r$$

Now, assuming the usual main branch of the complex logarithm, $\;-\pi<\arg\le\pi\;$ , we get that

$$0<r< 1\implies\frac{-1+r}r<0\implies \arg\frac{-1+r}r=\pi\implies\text{Im}\left(\log\frac{-1+r}r\right)=\pi$$

and multiplying by $\;r\;$ gives you the result.

Observe though that the imaginary part of the logarithm is $\;\pi\;$ ...without the $\;i\;$ . Both the real and imaginary parts of a complex number are real


问:
Trigonometric definite integral $\int ^\frac{\pi}{2} _0 \cos^4(s)\mathrm{ds}-\int^\frac{\pi}{2}_0 \cos^6(s)\mathrm{ds} $
$$\int ^\frac{\pi}{2} _0 \cos^4(s)\mathrm{ds}-\int^\frac{\pi}{2}_0 \cos^6(s)\mathrm{ds}  $$

the answer is $\dfrac{3\cdot1}{4\cdot2}-\dfrac{5\cdot3\cdot1}{6\cdot4\cdot2}$

I really don't know why

There may be some tricks in this, but I have not seen it in the textbook

答:
It seems that one of the solution may be the formula for $\cos$ at even power:
$\cos^{2n}(s)=\frac{1}{2^{2n}}(e^{is}+e^{-is})^{2n}$:
$I(n)=\int ^\frac{\pi}{2} _0 \cos^{2n}(s)\mathrm{ds}=\frac{1}{4}\int ^{2\pi} _0 \cos^{2n}(s)\mathrm{ds}=\frac{1}{2^{2n}4}\int ^{2\pi} _0 (e^{is}+e^{-is})^{2n}\mathrm{ds}=$$\frac{1}{2^{2n}4}\int ^{2\pi} _0 (e^{2ins}+\binom{2n}{1}e^{is(2n-1)-is}+...+\binom{2n}{n}e^{is(2n-n)-isn}+...+e^{-2ins})\mathrm{ds}$

After integration the only surviving term is $\frac{2\pi}{2^{2n}4}\binom{2n}{n}=\frac{2\pi(2n)!}{2^{2n}4(n!)^2}$


问:
Normal subgroup if conjugate subgroup is subset
I find this explanation in Isaacs' Algebra:

Lemma. Let $H\subseteq G$ be a subgroup. Then $H$ is a normal subgroup if $H^g\subseteq H$ for all $g\in G$.

The reader should be warned that this lemma does not state that $H^g=H$ whenever $H^g\subseteq H$. Since the inner automorphism induced by the element $g$ is a bijection, it is certainly true that $|H^g|=|H|$, and if $H$ is finite, this equality of orders together with the containment $H^g\subseteq H$ certainly does imply that $H^g=H$. For infinite subgroups, however, this does not follow and is not generally true.

How does the lemma not state that? If $H^g\subseteq H$, then the lemma says that $H$ is a normal subgroup, which by definition means $H^g=H$. What am I missing?

答: Let $G$ be the group $\text{GL}_2(\mathbb{Q})$. This group has a subgroup $H \cong \mathbb{Z}$ consisting of all elements of the form $\left[ \begin{array}{cc} 1 & n \\ 0 & 1 \end{array} \right]$ where $n \in \mathbb{Z}$. Set $g = \left[ \begin{array}{cc} 2 & 0 \\ 0 & 1 \end{array} \right]$. Then $H^g = 2H$ is strictly contained in $H$. </p>


问:
Sum of a sequence of integral functions
For $n\in \mathbb N$, consider the real functions
$$f_n(x):=\int_2^x\dfrac{dt}{t^ne^t\log t}.$$
(1) Find the domain of the functions

(2) Show that $f(x)=\sum_{n=2}^{+\infty}f_n$ is $\mathcal C^1$ on $(1,+\infty).$

(3) Is $\underset{n\to+\infty}\lim f(x)$ finite?

Let $g_n(x)=\dfrac{1}{x^ne^x\log x}$ be the integrand function $\implies g_n(x)=\dfrac{1}{e^x\log x}\Big(\dfrac{1}{x}\Big)^n$ is defined in $(0,+\infty)-\{1\}$. If I now fix a real number $x_0\in \mathbb R$ with $n\to +\infty$, the functions $g_n(x)\not\to\infty\iff -1<\dfrac{1}{x}< 1\implies x>1$.

So the interval in which we have punctual and also uniform (I'd like to prove this more precisely)  convergence is $(1,+\infty)$ .

If I now want to study the differentiability of $f(x)$ I considered the derivative of the general term $\int_2^x\frac{dt}{t^ne^t\log t}$ in order to find a limit function $h(x)$ such that $f'_n(x)\to h(x)$ uniformly.
$$\dfrac{d}{dx}\int_2^x\dfrac{dt}{t^ne^t\log t}=\dfrac{1}{x^ne^x\log x}\underset{\text{uniformly}}\longrightarrow h(x)\equiv 0\text{ in } (1,+\infty)\implies$$
the series $\sum f_n$ is differentiable in $(1,+\infty)$.

For the third point I'm not sure about how to proceed...

答: Let $1<x<\infty$. We can write $f(x)= \sum\limits_{n=2}^{\infty}\int_2^{x} \frac 1 {t^{n}e^{t} \ln t}dt= \int_2^{x} \sum\limits_{n=2}^{\infty}\frac 1 {t^{n}e^{t} \ln t}dt$ since the interchange of sum and intetgral is justified by the non-negativity of the integrand. Using the formula for a geometric sum we get $f(x)=\int_2^{x} \frac 1 {t(t-1) e^{t}\ln t}dt$. By FTC this is differentiable on $(1,\infty)$ and $f'(x)=\frac 1 {x(x-1) e^{x}\ln x}$ and this derivative is continuous on that interval. Finally, the fact that $0< \frac 1 {t(t-1) e^{t}\ln t}<Ce^{-t}$ for some constant $C$ for lal $t>2$ shows that $\lim_{x \to \infty} f(x)$ exists and equals $\int_2^{\infty} \frac 1 {t(t-1) e^{t}\ln t}dt$.

Title: Sum of a sequence of integral functions


问:
Solve the equation: $\lfloor a/2\rfloor+\lfloor a/3\rfloor+\lfloor a/5\rfloor=a$, where $a> 0$.
Let $a\in\mathbb{Z}^+$, then $a$ we can write it as $a = 30k + i$, where $i\in\{0,1,2,\ldots, 29\}$ and $k\in\mathbb{Z}_0^+$. Thus substituting we have:
\begin{align*}
     \lfloor(30k + i)/2\rfloor + \lfloor(30k + i)/3\rfloor + \lfloor(30k + i)/5\rfloor &= 30k + i\\
     \lfloor 15k + (i/2)\rfloor + \lfloor 10k + (i/3)\rfloor + \lfloor 6k + (i/5)\rfloor &= 30k + i\\
     15k + \lfloor i/2\rfloor + 10k + \lfloor i/3\rfloor + 6k + \lfloor i/5\rfloor &= 30k + i\\
     31k + \lfloor i / 2 \rfloor + \lfloor i / 3 \rfloor + \lfloor i / 5 \rfloor & = 30k + i \\
     k &= i - (\lfloor i/2\rfloor + \lfloor i/3\rfloor + \lfloor i/5\rfloor)
\end{align*}
Thus we have for each $i$ we will have a value of $k$. Also note that if $i=0$,
\begin{align*}
     k &= 0 -(\lfloor 0/2\rfloor + \lfloor 0/3\rfloor + \lfloor 0/5 \rfloor) \Rightarrow k=0.
\end{align*}
So $a=0$, but $a>0$. So $i=0$ is not taken, for all others we have a solution.

I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.

答:
Yes, this is an excellent solution.

It would perhaps be a good idea, after obtaining
\begin{align*}
     k &= i -(\lfloor i/2\rfloor + \lfloor i/3\rfloor + \lfloor i/5 \rfloor)
\end{align*}
to show that $k$ is non-negative. This is not totally trivial and, since$$\frac{1}{2}+\frac{1}{3}+\frac{1}{5}>1,$$ it is only just true.


答: I'm not sure you stated things correctly, but I can tell you the best way to work with such an integrand. Basically,$$\frac{1-a\cos \theta}{1-2a\cos\theta + a^2}=\text{Re}\left(\frac{z}{z-a^2}\right)$$ when $z=a e^{i\theta}$, so you should choose the circle of radius $a$ centered at the origin as your contour to perform the integration. This should help you answer any further questions you have about these types of integrals.</p>

Title: Proving that $\int_0^{2\pi}\frac{ 1-a\cos\theta}{1-2a\cos\theta + a²} \mathrm d\theta = 0$ for $|a|>1$

Body:
Let $|a|>1$. Show that $$\int_0^{2\pi}\frac{ 1-a\cos\theta}{1-2a\cos\theta + a²} \mathrm d\theta = 0$$

I'm also trying to solve this problem. I wasn't sure if I was supposed to start a new thread or contribute to previous asked questions about this integral.

I have some ideas that we could do something like this

$$ -i \oint _{|z|=1} \frac{1-az}{(z-2a) }dz = \int_{\theta=0}^{\theta=2\pi} \frac{1-a e^{i\theta}}{e^{i\theta}-2a } e^{i\theta}d\theta = \int_{\theta=0}^{\theta=2\pi} \frac{1-a e^{i\theta}}{1-2ae^{-i\theta} }d\theta \quad (2)$$

Then we have that the real part of $(2)$ is almost equal to the desired integral, except for the $a^2$ in the denominator.

An inequality to the right for the desired integral is then given from the real part of $(2)$. All we want to find is an inequality to the left for the desired integral. That hopefully and easily could be calculated with help of

$$ \oint _{|z-z_0|=r}(z-z_0)dz = \frac{0 \quad n \neq -1}{2\pi i \quad n = -1}  $$

Any suggestions?


问:
Why is P(A |B) + P( A |B') ≠ P(A)?
The only explanation I could come up with is that when using Bayes Theorem, the denominators don't match up.

But is there a more intuitive explanation that can be verbalized?

答:
For example take 2 coin flips, the two events are $A=$'The first coin flip is heads' and $B=$'The second coin flip is heads'.

Then $P(A|B)=P(A|B^{C})=P(A)=0.5$, but clearly $0.5+0.5\neq0.5$.

There is no reason that these two probabilities should add up to $P(A)$, conditional probability is very different from the probability of an intersection.

Title: Why is P(A |B) + P( A |B') ≠ P(A)?


答:
The standard complex analysis approach is to make the substitution $z = e^{i\theta}$. This will map $[0,2\pi]$ onto the unit circle traversed counter-clockwise once. Using Euler's formulas,
$$ \cos \theta = \frac{e^{i\theta}+e^{-i\theta}}2 = \frac{z+z^{-1}}{2}.$$
We also have $dz = ie^{i\theta}\,d\theta$, i.e.
$$d\theta = \frac1{iz}dz.$$

Doing all this we get
\begin{align}
\int_0^{2\pi}\frac{ 1-a\cos\theta}{1-2a\cos\theta + a^2} d\theta &=
\int_{|z| = 1} \frac{ 1-a(z+z^{-1})/2}{1-a(z+z^{-1})+ a^2} \cdot \frac{1}{iz} dz \\
&= \frac1{2i} \int_{|z| = 1} \frac{2z- a(z^2+1)}{z(z-a(z^2+1)+ a^2z)} dz.
\end{align}

Note that $z(z-a(z^2+1)+ a^2z) = z(1-az)(z-a)$, so the integrand has simple poles at $z = 0$, $z = a$ and $z = 1/a$.

If $|a| < 1$, we get
\begin{align}
\int_0^{2\pi}\frac{ 1-a\cos\theta}{1-2a\cos\theta + a^2} d\theta
&= \pi \left( \operatorname{Res}\limits_{z=0} \frac{2z- a(z^2+1)}{z(1-az)(z-a)} +
 \operatorname{Res}\limits_{z=a} \frac{2z- a(z^2+1)}{z(1-az)(z-a)} \right) \\
&= \pi \left( \frac{-a}{-a} + \frac{2a- a(a^2+1)}{a(1-a^2)} \right) \\
&= \pi ( 1 + 1 ) = 2\pi.
\end{align}

I'll leave the case $|a| > 1$ to you.


问:
limit of a sequence of integral functions
I have to study punctual and uniform convergence of the following succession
$$f_n(x)=n\int_{x-\frac{1}{n}}^{x+\frac{1}{n}}\dfrac{1}{e^{t^2}}dt.$$
I showed that the functions $f_n$ are uniformly convergent in $\mathbb R$ but I'm not able to calculate explicitly the limit function $f(x):=\underset{n\to +\infty}{\lim}f_n(x)$.

My idea was to split the integral in half and use the characteristic function of the domain of integration but this seems to be inconclusive.

Thank you for your help and your time.

答:
For the updated question, the pointwise limit is $2e^{-x^2}$. Its easy to see that we just need to check
$$ 2n\int_{x-1/n}^{x+1/n}|e^{-t^2} - e^{-x^2}| dt \to 0$$
Since $e^{-t^2}$ is continuous at $t=x$, for each $\epsilon$ there is $\delta$ such that for all $|t-x| < \delta$,
$$  |e^{-t^2} - e^{-x^2}|\le \epsilon$$
Hence for all $n$ such that $1/n<\delta$,
$$2n\int_{x-1/n}^{x+1/n}|e^{-t^2} - e^{-x^2}| dt\le \epsilon $$
this shows exactly what we need. This basically proves a special case of the Lebesgue Differentiation Theorem
$$ f\in L^1_{loc}\implies f(x) = \lim_{r\to 0} \frac1{|B_r(x)|}\int_{B_r(x)} f(y)dy \quad \text{a.e.}$$
which is not trivial to prove in full generality, but much easier when $f$ is continuous.


问:
Showing that a subset of a Hilbert space is compact
My question is about this previous question by somebody.

Here, we know that a separable Hilbert space is isometrically isomorphic to $l^2,$ so it can be identified with $l^2.$ Also, it suffices to show that $K$ is sequentially compact, that is, any sequence in $K$ has a convergent subsequence. But I cannot follow the answer's last line: "Using the fact that $(x^{k,k})_k \subset K$ show that $x^{k,k}$ also converges in norm." Any elaboration would be appreciated.

答:
Let $(x^{k,k})_{k \in \mathbb{N}} \subseteq K$ be the sequence (of sequences in $\ell^2(\mathbb{R})$) as described in the linked question, and let $x = (x_j)_{j \in \mathbb{N}}$ be the constructed pointwise limit such that $x^{k,k}_j \to x_j$ as $j \to \infty$.

First, we can check that $(x_j)_{j \in \mathbb{N}}$ is in $K$ (e.g. since for each $k$, $\sum^\infty_{j=M_s} |x^{k,k}_j|^2 \leq \ell_s$, we can take the limit as $k \to \infty$ to deduce that $\sum^\infty_{j=M_s} |x_j|^2 \leq \ell_s$ for all $s$).

Therefore, for each $s$, $\sum^\infty_{j=M_s} |x^{k,k}_j - x_j|^2 \leq \sum^\infty_{j=M_s} |x^{k,k}_j|^2 + \sum^\infty_{j=M_s} |x_j|^2 \leq 2 \ell_s$. Hence,
$$
\lVert x^{k,k} - x \rVert_2^2 = \sum^\infty_{j=1} |x^{k,k}_j - x_j|^2 = \sum^{M_s-1}_{j=1} |x^{k,k}_j - x_j|^2 + \sum^\infty_{j=M_s} |x^{k,k}_j - x_j|^2 \leq \sum^{M_s-1}_{j=1} |x^{k,k}_j - x_j|^2 + 2\ell_s .
$$
Taking the limit as $k \to \infty$ shows that $\lim_{k \to \infty} \lVert x^{k,k} - x \rVert_2^2 \leq 2\ell_s$ by pointwise convergence of our constructed sequence. Since  this holds for all $s$, and $\ell_s$ tends to zero, this shows that $x^{k,k}$ also converges to $x$ in norm.


问:
For a countable infinite product, each (xi, Ti) is homeomorphic to a subspace of the product
"Let($X_i,\tau_i$),i∈N,be a countably infinite family of topological spaces. Prove that each ($X_i,\tau_i$) is homeomorphic to a subspace of $\Pi(X_i,\tau_i)$"

For each j $\ne$ i, fix an element $a_j$ in each j. For each i, define mapping $f_i(x)$: $(X_i,\tau_i$) $\mapsto$ $\Pi(X_i,\tau_i)$ by $f_i(x)$=($a,a_1,a_2$,$a_{i-1}$,x,$a_{i+1}$, $a_{i+2}$,...,$a_n$).

We claim that $f_i(x)$: ($X_i,\tau$’)$\mapsto$($f_i(X_i)$,τ′)is a homeomorphism, where τ′ is the topology subspace topology. Clearly, this mapping is bijective.  Let U∈$\tau_i$ Then$ f_i(U)$={$a_1$}×{$a_2$}×···×{$a_{i-1}$}×U×{$a_{i+1}$}×···×{$a_n$}=($X_1×X_2$×···×$X_{i−1}$×U×$X_{i+1}$×···×$X_n$)∩({$a_1$}×{$a_2$}×···×{$a_{i-1}$}×$X_i$×{$a_{i+1}$×···×{$a_n$})=($X_1×X_2$×···×$X_{i−1}$×
U×$X_{i+1}$×···×$X_n$)$\cap$ $f_i(X_i$))∈τ′ Thus $f_i(U)$ is open and in the subspace topology so we have shown the mapping is open.

Now we just need to show the mapping is continuous. Take any open set in the codomain which is of the form {($U_1×U_2×···×RxRxRXRxR)\cap$ $f_i(X_i)$. Since this is the infinite product topology, all basic open sets must have R for all but finite number of components. Now taking the preimage,$f^{-1}_i[(U_1×U_2$×...x X x X x X x X)∩$f_i(X_i)]$=$f^{-1}_i$($U_1×U_2$×...x X x X)$\cap$ $f^{-1}_i$=$f^{−1}_i$($U_1×U_2×...x X x X xX$)$\cap$($X_i$).

Now, if $a_j \notin$ $U_i$ then $U_i$ interection with $X_i$ is empty. Otherwise, the intersection is $U_i$ which is open in $T_i$ Either way, we get an open set, so the mapping is continuous.

One area of concern is whether it is possible to fix an element from each $X_i$ when we have an infinite set. I know it is possible for finite number of sets.

答:
You'll have to assume $X:=\prod_{i \in I} X_i$ is non-empty, or equivalently that all $X_i$ are non-empty. The Axiom of Choice then allows us me to pick a point $a \in \prod_{i \in I} X_i$ and I'll fix that point from now on.

Then for any $j \in I$ we define $e_j: X_j \to X$ by

$$\pi_i(e_j(x))= \begin{cases} x & i=j\\
                               a_i & i \neq j\\
\end{cases}$$

and note that $e_j$ is continuous as all $\pi_i \circ e_j$ are either a constant map (so continuous) or the identity on $X_j$ (also continuous). The universal property of products strikes again. No separate proof needed.

$e_j$ is 1-1: $x \neq x'$ implies $e_j(x)_i \neq e_j(x')_i$ so $e_j(x) \neq e_j(x')$.

To show its en embedding it is enough to it has a continuous inverse $e_j[X_j] \to X_j$ and this inverse is $\pi_j\restriction_{e_j[X_j]}$, which is clearly continuous and the required inverse. Don't overcomplicate things.

The idea of the map was OK though. There is no doubt in my mind that we can fix such a point. I believe in AC, as do most topologists. Not going to give up Tychonoff's theorem after all...


问:
How to represent a set containing sets following a certain pattern in mathematical notation?
I have a set that contains the following sets:

$$\{\{1\}, \{1, 2\}, \{1, 2, 3\}, ... \{1, 2, 3, ... , n\}\}$$

Is there an elegant way to represent the above using mathematical notation? I'm looking for something like

$$1 + 2 + 3 + ~~...~~ + n = \sum_{k=1}^n k$$

答:
A standard notation in combinatorics is to write $[n]$ for the set $\{1, \ldots, n\}$. With this notation, you could write your set as
$\bigl\{\ [k] \mid 1 \leqslant k \leqslant n \bigr\}$.


问:
Adding sets with forcing
Let $M$ be a c.t.m. of ZFC. Let $\mathbb{P}$ be the poset $Fn(\lambda \times \kappa, 2, \kappa) = \{p : |p| < \kappa \wedge p$ is a function $\wedge dom(p) \subseteq \lambda\times\kappa \wedge ran(p) \subseteq 2\}$, where $\lambda \leq \kappa$ and $\kappa$ regular in $M$. Let $G$ be a $\mathbb{P}$-generic over $M$. We know that this way in $M[G]$ we have $\lambda$ new subsets of $\kappa$. Do we know that everything of rank smaller than $\kappa$ remains unchanged?

I know that since $\kappa$ is regular, we have that $(\mathbb{P}$ is $\kappa$-closed$)^M$, but does this fact give us $$x \in M[G] \text{ and } (rank(x) < \kappa)^{M[G]} \Rightarrow x \in M.$$ If so, then do we also get that $(V_\kappa)^M = (V_\kappa)^{M[G]}$?

答:
This is certainly false if $\kappa$ is not inaccessible.

Consider this forcing when $\lambda=1$ and $\kappa=\aleph_1$. We add one new generic subset to $\omega_1$ without adding countable subsets to the universe.

Since $\aleph_1\leq2^{\aleph_0}$ we have that $|\omega_1|\leq|V_{\omega+1}|$ and therefore we can find an element of $V_{\omega+2}$ which encodes the generic set, but its rank is certainly countable.

Title: Adding sets with forcing


问:
Fundamental Theorem of Calculus for Lebesgue Integrable functions
I am trying to prove the following statement:

If $f$ is Lebesgue integrable on $[a,b]$ and if  $F(x)=\int_{a}^x f(t)dt$
then $′=()$ .. on $[,]$.

If $f$ is Riemann integrable it can be shown using the theorem that -"If $f$ is Riemann integrable, it is discontinuous in the set of measure of zero."

How to prove this statement for non-Riemann integrable functions?

答:
This theorem is definitely nontrivial and is at the basis of the theory that was published originally in Lebesgue's dissertation Intégrale, longueur, aire ("Integral, length, area") at the University of Nancy during 1902.

The main idea is to rephrase this statement in terms of measures and, with the help of measure-theoretic machinery, to solve the corresponding "paraphrased" problem with the help of Radon-Nikodym theorem and the notion of absolute continuity of measures. Finally, one goes back to the functional setting and gets the result.

A formal proof needs a bit of work, so I suggest that you consult books like Modern Real Analysis, by Ziemer or Real Analysis: Modern Techniques and Their Applications, by Folland.


问:
Why Is there no consistent factor in $A^2$ versus the other $A^X$ In the following sequence?
The following sequence is built of $((A^z)^x -1)/6$ where $z$ increases by $1$ and the conditions for $A$ must be that $(A - 1) \mod 6 = 0$ and for $x$ must be that $x$ is odd and $x > 1$:

$((A^1)^x -1)/6 $ , $((A^2)^x-1)/6$ , $((A^3)^x-1)/6$ , $((A^4)^x-1)/6$ ...

Examples:

$(7^3-1)/6 =57$ , $(49^3-1)/6 =19608$ , $(343^3-1)/6 = 6725601$....

$57, 19608, 6725601$ all have a factor of $19$

$(7^5-1)/6 =2801$ , $(49^5-1)/6 = 47079208$ , $(343^5-1)/6 = 791260261657$....

$2801, 47079208, 791260261657$ all have a factor of $2801$

$(13^3-1)/6 =366$ , $(169^3-1)/6 =1 804468$ , $(2197^3-1)/6 = 1767416562$....

$366, 804468, 1767416562$ all have a factor of $61$

Where as if $x$ is allowed to be $x=2$:

$(7^2-1)/6 = 8$ , $(49^2-1)/6 = 400$ , $(343^2-1)/6 = 19608$ , $(2401^2-1)/6 = 960800$....

$8$ has a factor of $2$

$400$ has factors of $2,5$

$19608$ has factors of $2,3,19,49$

$960800$ has factors  $2,3,5,1201$

As you can see there is no consistency in a common factor besides $2$.

$(11^2-1)/6 = 20$ , $(121^2-1)/6 = 2440$ , $(1331^2-1)/6 = 295260$....

$20$ has factors of $2,5$

$2440$ has factors of $2,5,61$

$295260$ has factors of $2,3,5,7, 19, 37$

As you can see there is no consistency in a common factor besides $2$ and $5$ ($5$ for obvious reasons related to the multiplications of 11).

The reason I allow myself to compare $x=2$ with other odd values where $x > 1$ is because $2$ is the only even prime.

From there on if $x$ is a multiplication of $2$ it can be rewritten in a way that $x$ is prime factor. Example $3^6=729$ can be written as $9^3$

The comparison is then reduced to a comparison between prime numbers with different $x$, and the result is that when $x=2$, there is no consistent common factor unlike when $x ≠ 2$

Why is there no consistent common factor when $x=2$?

Notes:

*I don’t take into account the common factors of 2,3 and 5

答:
The sequence that you are studying is the sequence of numbers $$\frac{A^{2z}-1}{6}, z=1,2,3,...$$
Since $$A^{2z}-1=(A^2-1)(1+A^2+A^4+ ... +A^{2z-2})$$ there is a consistent common factor of $$\frac{A^{2}-1}{6}.$$

In your example with $A=7$ this common factor is $8$.

The next $A$ would be $13$ and then the common factor would be $28$ which contains the prime $7$ and so is not in your excluded set of $\{2,3,5\}$.


答: I use the notation $$\mathbb{N}_n=\{k\in\mathbb{N}: 1\leq k\leq n\}$$ The set you have in your post would be $$\{\mathbb{N}_m: m\in \mathbb{N}_n\}$$ for some $n\in\mathbb{N}$.

Title: How to represent a set containing sets following a certain pattern in mathematical notation?

Body:
I have a set that contains the following sets:

$$\{\{1\}, \{1, 2\}, \{1, 2, 3\}, ... \{1, 2, 3, ... , n\}\}$$

Is there an elegant way to represent the above using mathematical notation? I'm looking for something like

$$1 + 2 + 3 + ~~...~~ + n = \sum_{k=1}^n k$$


问:
A proof of the conditions by which a union of two subspaces is a subspace
I came up with this proof which involves interpreting a truth table. I omitted the implication in the backwards direction because I'm more interested in reviewing the technique used in the first part. I wanted to post this here to get some feedback and to clarify any conceptual misunderstandings. If you know of any ways of improving or simplifying my proof I'd love to hear it. I know this question has been posted on here before but I think this is a pretty conclusive way of solving the problem in a way which actually shows you why the answer must be.

Let $W_1,W_2$ be subspaces of $V$. Prove the union $W_1 \cup W_2$ is also a subspace of $V$ if and only if $W_1 \subset W_2$, $W_2 \subset W_1$, or $W_1 = W_2$.

If $W_1 \cup W_2$ is a subspace of $V$, then $x,y \in W_1 \cup W_2 \iff x+y \in W_1 \cup W_2$.

By definition of union and closure under vector addition, $x+y \in W_1 \cup W_2 \equiv (x+y \in W_1) \lor (x+y \in W_2) \equiv (x \in W_1 \land x\in W_1)\lor (x \in W_2 \land x\in W_2).$

Clearly, this statement is only true when:

Both $x$ and $y$ are in both $W_1$ and $W_2$.

$x$ is in $W_1$ and $W_2$ but $y$ is only $W_1$ or only in $W_2$.

$y$ is in $W_1$ and $W_2$ but $x$ is only $W_1$ or only in $W_2$

$x$ and $y$ are only chosen from either $W_1$ or $W_2$ but not both.

These four cases can be true or false depending on the values of $x$ and $y$. However, only one of these cases may be true considering all possible $x,y \in W_1 \cup W_2$. Working backwards, we have $W_1 = W_2$ with (1), $W_1 \subset W_2$ or $W_2 \subset W_1$ with (2) and (3), and (4) can be discarded as it would necessitate that we cannot choose elements from one of the subspaces, which is to say one of the subspaces is empty, which is a contradiction.

We can also work backwards from here to show that if (1) or (2)/(3) are true, so it must be that $W_1 \cup W_2$ is a subspace of $V$ $\blacksquare$

答:
Welcome to MSE! There's a few misunderstandings here.

First, your criterion

If $W_1 \cup W_2$ is a subspace of $V$, then $x,y \in W_1 \cup W_2 \iff x+y \in W_1 \cup W_2$.

is not true. The forward direction is (that's more or less the definition of a subspace) but the backwards direction is not. For instance, if $v$ is any element not in $W_1 \cup W_2$, then $v + (- v) = 0 \in W_1 \cup W_2$, despite $v, -v \not \in W_1 \cup W_2$!

As another point, I'm not sure how you're making step $\star$ in

$x+y \in W_1 \cup W_2
\equiv (x+y \in W_1) \lor (x+y \in W_2)
\overset{\star}{\equiv} (x \in W_1 \land x\in W_1) \lor (x \in W_2 \land x\in W_2).$

Judging by your truth table below, I think you meant to write

$(x \in W_1 \land y \in W_1) \lor (x \in W_2 \land y \in W_2)$

As we said earlier, though, knowing that $x+y \in W_1$ (resp. $W_2$) doesn't actually tell you that $x \in W_1 \land y \in W_1$ (resp. $W_2$)!

You then go on to make some other conclusions based on this faulty logic. These are also somewhat confusing, but I can't tell if they're just worded in a way I'm not used to or if there's another conceptual misunderstanding.

In particular, I'm not at all sure what you mean by

However, only one of these cases may be true considering all possible $x,y \in W_1 \cup W_2$.

There is, of course, a cleaner approach -- you should argue by contradiction
(Or by contrapositive if you prefer to think about it that way).

Say $W_1 \subseteq W_2$ and $W_2 \subseteq W_1$ are both false.

Then we can find an element $x_1 \in W_1 \setminus W_2$ and $x_2 \in W_2 \setminus W_1$.

Can you show that $x_1 + x_2 \not \in W_1 \cup W_2$? If you do, you'll have shown that $W_1 \cup W_2$ is not a subspace (which is what we want to show, to complete the contrapositive).

You'll need to use your (good!) idea from earlier that

$$x_1 + x_2 \in W_1 \cup W_2 \equiv (x_1 + x_2 \in W_1) \lor (x_1 + x_2 \in W_2)$$

you can then argue in a similar fashion to what you were doing earlier. It will be helpful to remember that $-x_1 \in W_1$ and $-x_2 \in W_2$, since they are both subspaces. Do you see how to take it from here?

I hope this helps ^_^


问:
Logic Puzzle of the age of three sons
There is a puzzle, it goes something like this:

Someone talks to a guy, and asks, Give me the age of my three sons,
The other guy asks for some clues:

The product of the age of the three sons (of someone) is equal to 36.

"I can't figure out their ages." says solver...

The sum of the ages of the three brothers is the same as the number of windows you can    see in this building (points to some building).

"I still can't figure out their
  ages." says solver...

The oldest has blue eyes.

"Now I know their ages." says solver!.

So, I do not have any clue of how to solve this logic puzzle...

How to link the number of windows in a building with the product of their ages?, or how would be the approach?...

Any idea?

答:
Well, first off, let's list all the possible combination of ages (and their sum):

$1,1,36; 38$

$1,2,18; 21$

$1,3,12; 16$

$1,4,9; 14$

$1,6,6; 13$

$2,2,9; 13$

$2,3,6; 11$

$3,3,4; 10$

I'm not sure what to make of the building one, but note the specific wording in the third clue: "older". The only reason you would say "older" when referring to THREE people (you would typically use "oldest") means that two of them must be twins. So, you now have three possibilities left:

$1,1,36; 38$

$2,2,9; 13$

$3,3,4; 10$

I don't know how to use the building clue to pare the choices down to one.

That help?

EDIT: Apparently, "older" should be "oldest". In that case, the solution could be any of them but one. In addition, the missing piece is that if the person solving the puzzle knows the number of windows in the building but still cant figure it out, then the two possibilities are:

$1,6,6; 13$

$2,2,9; 13$

At this point, the remark about "oldest" rules out the first one and leaves only $2,2,9$ as the correct answer.

Title: Logic Puzzle of the age of three sons


问:
If $X \sim N(μ,σ^2)$ how to find the PDF of $Y = 2X$
Given a random variable $X$ that follows the Gaussian Distribution (i.e  $X \sim N(μ,σ^2)$), I need to find the Probability Density Function (PDF) of $Y = 2X$.

It's been some years since I've done something similar so I am kinda stuck.
How do I approach this problem?

Of course, $f_X(x) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$

Obviously $Y \sim \frac{N(μ,σ^2)}{2} $

Therefore $f_Y(y) =f_{2X}(y) = ?$

Note: Please do not give me a specific answer, just a general guideline, as I would like to solve this by myself.

答:
Here are some guidelines:

The PDF is the derivative of the CDF.

$F_Y(z)=P(Y \leq z) = P(X \leq z/2)=F_X(z/2)$.


问:
Can any random vector $\mathbf{y}$ with covariance matrix $\Sigma$ be represented by $\mathbf{\Sigma}^{1/2}\mathbf{x}$?
Suppose that the components of the $n$-dimensional random vector $\mathbf{x}$ are independent with zero mean and variance $1$, that is,
$$
\mathsf{Cov}(\mathbf{x},\mathbf{x}) = \mathbf{I}.
$$
Let $\mathbf{y}=\mathbf{Ax}$, where $\mathbf{A}$ is a positive definite matrix, we know that
$$
\mathsf{Cov}(\mathbf{y},\mathbf{y}) = \mathsf{Cov}(\mathbf{Ax},\mathbf{Ax}) = \mathbf{A}\mathbf{A}^{\mathsf{T}}:=\Sigma.
$$

I wonder if the reverse holds or not. That is, if we know $\mathsf{Cov}(\mathbf{y},\mathbf{y})=\Sigma$, does there exist matrix $\mathbf{A}$ satisfying $\mathbf{A}\mathbf{A}^{\mathsf{T}}=\Sigma$ such that
$$
\mathbf{y}=\mathbf{Ax},
$$
where $\mathsf{Cov}(\mathbf{x},\mathbf{x})=\mathbf{I}$.

It is true if $\mathbf{y}$ follows a multivariate normal distribution. What about other distributions?

答:
For every symmetric positive-definite matrix $\Sigma$ (this is always satisfied for a  covariance matrix) there is a decomposition of the form $\Sigma=GG^T$. It is called the Cholesky decomposition. Then you have to solve the linear system
$$y=Gx.$$
But I think your solution $x$ might be not a random vector with independent components in general.


问:
Is there a special term for an array consisting only of ones?
Is there a special term for an array consisting only of ones?

Sorry for the rather elementary question. I am getting into MapReduce programming and am trying to frame my code to be nice and neat.

答:
I have used, and seen used, the term "all-ones vector". It is typically notated however other vectors are being notated in your writing. For instance, if $\vec{x}$ is a vector, then the all-ones vector would be $\vec{1}$. Likewise, if $\mathbf{x}$ is a vector, then you'd write $\mathbf{1}$.

As an example, consider the average of a list of numbers. If the list is in vector $\mathbf{x}$, then the average can be written:

$\mathrm{average}(\mathbf{x}) = \dfrac{\mathbf{x}\cdot\mathbf{1}}{\mathbf{1}\cdot\mathbf{1}}$

where "$\cdot$" represents the standard dot product.


问:
(convergence of series) Real Analysis

I've understood the part until the underline. That is if series a(n) is convergent then a(n) when n tends to infinity equals 0. But do not seem to understand the underlined part. Is it true that for every convergent sequence there would be a number when a(n) <= 1. How does the limit value equalling 0 further imply to this point? Thank you.

答:
That’s by the definition of limit. The statement
$$
\lim_{n\to\infty} a_n=0
$$

is equivalent to saying that for any $\varepsilon>0$, there exists a positive integer $K$ such that
$$
|a_n -0|=|a_n| \le \varepsilon
$$
whenever $n\ge K$. In the text, the author is just taking $\varepsilon=1$.

Title: (convergence of series) Real Analysis


问:
Is there any way to know the numbers which are NOT a term in a particular arithmetic progression??
For example say: 5n+2, 5=d, n = x-1 (for any x), a = 2. So, we can know that 3,5,8, etc. are definitely not terms in this progression, by hit and trial. But, is there any direct way by which can represent the terms which are NOT any term in the Arithmetic Progression?? The particular progression I'm looking for is:

3n+1 + (x-1)(2n+1). I want to know the numbers which are not a part of this progression. Thank you.

This is my first time posting here, please bare with me and spare me for my naivety.

答: One way to check if a progression of terms $T_n$ contains a particular term $T$ is to equate $T_n = T$ and solve for $n$. If you get a positive integer, you know the term $T$ will occur in the sequence $T_n$ in the usual (forward) direction. If you get $n$ as an integer such that $n \leq 0$, you know that extension of the sequence in the other (reverse) direction will yield that term. But if $n$ is not an integer at all, you know that the term won't occur in the sequence at all.


问:
Which of the statements is not necessarily true?
Let $A$ be a $3\times3$ matrix and $u, v, w$ be linearly independent vectors in $\mathbb{R}^3$ such that:

$Au = 2u, Av = 2v, Aw = 0$.

Which of the statements are NOT necessarily true?

Option 1: $w$ is an eigenvector of $A$.

Option 2: $\text{span}({u,v})$ is the eigenspace of $A$ associated with eigenvalue 2.

Option 3: $u+w$ is an eigenvector of $A$ associated with eigenvalue 2.

Option 4: $2u$ is an eigenvector of $A$ associated with eigenvalue 2.

Option 5: All the statements are necessarily true.

I am not sure about option 2 and 3, but I think option 1 and 4 are true. Please help. Thank you.

答:
Option 2 is certainly the most difficult. What do you know about the total sum of the geometric multiplicities of eigenvalues?

We know that $\text{span}(u, v)$ is at least a subset of the eigenspace of 2, so the geometric multiplicity of 2 is at least 2.

Assume that $\text{span}(u, v)$ is not all of the eigenspace of 2, then the geometric multiplicity of 2 is at least 3. But we know that $w$ is an eigenvector for 0, showing that the geometric multiplicity is at least 1. Is it possible for the geometric multiplicity of 2 to be at least 3 and that of 0 to be at least 1?

Option 3 is asking you whether $u+w$ is a nonzero vector that satisfies $A(u+w)=2(u+w)$. Is it?


问:
easy question about conditional expectations
I got a quite easy exercise I just don't get.

Let P be a probability measure, $\frac{dQ}{dP}=Z$, $Z>0$ a.s. and $E[Z]=1$, hence Q is an equivalent proba-measure to P. Then I shall prove that for a sub-sigma field G we have:
$E_{Q}[X|G](\omega)=\frac{E_{P}[XZ|G](\omega)}{E_{P}[Z|G](\omega)}=:h(\omega)$, what seems easy as pie, but when using the definition, I need to show that $E[X(\omega)*\mathbb{1}_A(\omega)]=E[h(\omega)*\mathbb{1}_A(\omega)]$ for all $A\in G$ what makes me confused. these double integrals and the fraction of integrals...

Thx for you help!!

答:
Let $A\in G$. We have
\begin{align*}
  \int_A \frac{E_P[XZ\mid G]}{E_P[Z\mid G]}\, dQ &= \int_A \frac{E_P[XZ \mid G]}{E_P[Z\mid G]}\,Z \, dP\\
  &= \int_A E_P\left[\frac{E_P[XZ \mid G]}{E_P[Z\mid G]}\,Z\Biggm | G \right]\, dP\\
  &= \int_A \frac{E_P[XZ \mid G]}{E_P[Z\mid G]}\,E_P[Z\mid G] \, dP & \text{the fraction is $G$-measurable}\\
  &= \int_A E_P[XZ\mid G]\,dP\\
  &= \int_A XZ\, dP\\
  &= \int_A X\, dQ.
\end{align*}
Hence
$$ E_Q[X \mid G] = \frac{E_P[XZ\mid G]}{E_P[Z\mid G]} $$
as wished.</p>


问:
Second Order Methods for conservation laws
Consider the scalar conservation law $u_t+f(u)_x=0.$
I understand that Lax-Wendroff scheme is second order accurate, because of the way it is derived using Taylor series. However since this scheme is not TVD there are various methods obtained from Lax-Wendroff scheme by introducing a correction term/Limiters. For example minmod, superbee, van Leer etc.

Now I have the following doubts:

Are these schemes second order accurate always? If so how to prove?

How to prove that these schemes converge to Kruzkov entropy solutions? (They are not monotone anymore!! So the standard monotone argument may not work)

答:
For the first part, these slope- or flux-limiter schemes are not second-order accurate anymore. In fact, their order of accuracy measured numerically is rather around 1.6-1.8. This becomes obvious after realising that they are a kind of compromise between first-order and second-order schemes, see e.g. the 2002 book by R.J. LeVeque.

For the second part, I may find some special cases where this property is proven in literature. To be updated.


问:
Why is the Transcendence degree of a point in $K^n$ equal to $0$?
Suppose $(x_1,...,x_n)\in K^n$ where $K$ is a field. Then we know that $\{(x_1,...,x_n)\}=Z(X_1-x_1,...X_n-x_n)$ i.e $\{(x_1,...,x_n)\}$ is a algebraic set (and variety). I want to show that the field of fractions of $K[X_1,...,X_n]/I(\{(x_1,...,x_n)\})$ has transcendence degree of $0$ over $K$. Does anyone know how to proceed. I think $I(\{(x_1,...,x_n)\})=(X_1-x_1,...X_n-x_n)$ but I do not know what $K[X_1,...,X_n]/(X_1-x_1,...X_n-x_n)$ is. Any Ideas?

答:
Let $u:P\in K[X_1,\ldots, X_n]\mapsto P(x_1,\ldots,x_n)\in K$.

Then $\ker(u)=I(\{x_1,\ldots,x_n\})$ and $u$ is obviously surjective. Then, first isomorphism theorem yields that your quotient is isomorphic to $K$, which has transcendence degree $0$ over $K$.

Moreover, you are right about $I(\{x_1,\ldots,x_n\})$, that I will denote $I$ for short. It is indeed $(X_1-x_1,\ldots, X_n-x_n)$.

Set $J=(X_1-x_1,\ldots, X_n-x_n)$, and note that $J\subset I$

To see that $I=J$ , note that $X_i=(X_i-x_i)+x_i$, so $X_i$ is congruent to a constant polynomial modulo  $J$. It follows that any polynomial $P$ is congruent to a constant modulo $J$, that is $P=c+Q,$ for some $Q\in J$.

Evaluation at $(x_1,\ldots,x_n)$ yields, using the fact that $J\subset I$ that $c=P(x_1,\ldots,x_n)$. Hence $P(x_1,\ldots,x_n)=0\iff P\in J$, that is $\ker(u)=J$, meaning $I=J$.


问:
Logistic proof in TFL of a theorem

This is supposed to be a hard one, I'm not sure what my beginning assumption should be. Should I assume -A and -B as two subproofs? The only hint here is this one uses IP rules (indirect proof)
With where to start it will be really helpful if you can teach me the main structure here! Thank you in advance

答:
$\def\fitch#1#2{~~~~\begin{array}{|l}#1\\\hline#2\end{array}}$

Your aim is to derive a contradiction under the first assumption.   So far you have derived $\lnot(A\to B)$.   That is not a bad start.

Consider that if the assumption really is inconsistent, you should also be able to derive $A\to B$.   A conditional proof would be required for this.

Now there are two ways you could derive $B$ under assumption of $A$: either directly or through the principle of explosion (also called 'ex falso quodlibet', or 'contradiction elimination').   Well, you have nothing that would directly derive it, so you must try the second option.

$$\fitch{}{\fitch{~~1.~\lnot((A\to B)\lor(B\to A)}{\fitch{~~2.~A\to B}{~~3.~(A\to B)\lor(B\to A)\\~~4.~\bot}\\~~5.~\lnot(A\to B)\\\fitch{~~6.~A}{~~~~\vdots\\~~~~.~~\bot\\~~~~.~ B}\\~~~~.~A\to B\\~~~~.~\bot}\\~~~~.~(A\to B)\lor(B\to A)}$$

Once you have filled out the missing detail, it should become evident that you can rearrange things for a more elegant proof.

Hint:

 $A\to(B\to A)$ is a tautology.

Title: Logistic proof in TFL of a theorem


问:
Show circle through feet of two altitudes and midpoint of third side passes through center of another circle
This is a question that was asked on the site a few days back and before an answer, it was deleted by the OP. While I was able to solve the problem, I feel there is a more elegant way.

Question:

In $\triangle ABC$, $BF \perp AC$ and $AG \perp BC$. Also $E$ is the midpoint of $AB$.

Show that the circumcircle of $\triangle EFG$ passes through the circumcenter of $\triangle CFG$.

My solution:

Let $N$ be the intersection of the perpendicular bisector of $FG$ and the circumcircle of $\triangle EFG$. I assumed circumradius of $\triangle ABC = R$. Then I showed circumradius of $\triangle EFG = \frac{R}{2}$. Using similarity of $\triangle CGF$ and $\triangle CAB$, I showed $MN = k^2R$ where $FG:AB = CF:BC = CG:AC = k:1$ so circumradius of $\triangle CFG = kR$. Finally I showed $FN = NG = kR$ which proves $N$ is the circumcenter of $\triangle CFG$.

Would like to see other solutions.

答:
Probably the easiest:

Enough to show that $FNGE$ is cyclic $\iff$ $FNG=180-FEG$. Also $FEG=180-(FEA+GEB)$ and clearly because $FE$ and $GE$ are the medians of right triangles $AFB$,$AGB$ respectively, it's well-known that $FEA=180-2A$ and $GEB=180-2B$ so $FEG=180-(180-2A)-(180-2B)=180-2C$ and $FEG=2C$ is clear because $N$ is the circumcenter of $CFG$.


问:
Quadratic-exponential integral divided by $x^2$. How to solve?
Is it possible to solve the following integral?

$$\int_a^b \frac{e^{-x^2-c x}}{x^2} dx$$

(where $c$ is a positive number)

I tried both with Mathematica and looking on an integral table (Gradshtein's big book) and I wasn't able to find an answer.

EDIT:

I've been asked to add a bit of context to this question.
I'm trying to calculate the solution of the 1D diffusion equation for arbitrary initial condition and inhomogeneous term independent on time:

$$\rho_t=D\rho_{xx}+v\rho_x+\phi\rho+\epsilon(x)$$
with a change of variable you reduce it to
$$f_t=D f_{xx}+\eta(x)$$
where $\eta(x,t)=\epsilon(x) e^{-\phi t}$

if $\phi$ is set to zero, $\eta$ would be independent on time.

The integral I've presented you is the time-integral of the one in Eq. 1.17 of
http://www-m6.ma.tum.de/~kuttler/script_reaktdiff.pdf
the coefficient $c$ in the integral is proportional to $\phi$, and I need it to impose degradation of the particles of my random walk.

I hope now the frame of the problem is a bit more clear .

答:
Starting fro @Turing's answer, considering $$I=\int\dfrac{e^{-(x+p)^2}}{x^2}\, dx$$ We could expand the numerator as a Taylor series around $p=0$. This would give
$$e^{-(x+p)^2}=e^{-x^2} \sum_{n=0}^\infty \frac{ Q_n(x)}{n!}\, p^n$$ where $Q_n(x)$ are polynomials of degree $n$ in $x$ (all coefficients being integer numbers).

For example, summing only four terms and grouping, we should have as integrand

$$e^{-x^2}\,\, \big[\frac {a_0} {x^2}+\frac {a_1} {x}+ {a_2} +{a_3} x+a_4 {x^2}\big]$$ where

$$a_0=\frac{p^4-2 p^2+2}{2} \qquad \qquad a_1=2 p \left(p^2-1\right)$$
$$a_2=-2 p^2
   \left(p^2-1\right)\qquad \qquad a_3=-\frac{4 p^3 }{3}\qquad \qquad a_4=\frac{2 p^4}{3}$$ and now we face integrals
$$J_n=\int x^{n-2}\,e^{-x^2}\,dx=-\frac{1}{2} \Gamma \left(\frac{n-1}{2},x^2\right)$$

For a very modest test, let $p=1$, $a=1$ and $b=2$ using $12$ terms. The approximation gives $0.003028$ while numerical integration gives $0.003012$.


答:
Quadrilateral $ABGF$ is cyclic so

$\angle CFG = \angle ABC$ and $\angle CGF = \angle BAC$.

$GE$ is the median to the midpoint of hypotenuse in a right angled triangle so $GE = BE = EE, \angle BGE = \angle ABC$. Similarly $FE = AE = BE, \angle AFE = \angle BAC$.

So $\triangle EFG$ is an isosceles triangle with $FE = GE$. That leads to,

$\angle GFE = \angle EGF = 180^0 - (\angle ABC + \angle BAC) = \angle ACB$

$\angle FEG = 180^0 - 2 \angle ACB \ $.

$ \therefore \ \angle FNG = 2 \angle ACB$ and as $N$ is a point on one of the perpendicular bisectors, it has to be the circumcenter of $\triangle FCG$.

Title: Show circle through feet of two altitudes and midpoint of third side passes through center of another circle

Body:
This is a question that was asked on the site a few days back and before an answer, it was deleted by the OP. While I was able to solve the problem, I feel there is a more elegant way.

Question:

In $\triangle ABC$, $BF \perp AC$ and $AG \perp BC$. Also $E$ is the midpoint of $AB$.

Show that the circumcircle of $\triangle EFG$ passes through the circumcenter of $\triangle CFG$.

My solution:

Let $N$ be the intersection of the perpendicular bisector of $FG$ and the circumcircle of $\triangle EFG$. I assumed circumradius of $\triangle ABC = R$. Then I showed circumradius of $\triangle EFG = \frac{R}{2}$. Using similarity of $\triangle CGF$ and $\triangle CAB$, I showed $MN = k^2R$ where $FG:AB = CF:BC = CG:AC = k:1$ so circumradius of $\triangle CFG = kR$. Finally I showed $FN = NG = kR$ which proves $N$ is the circumcenter of $\triangle CFG$.

Would like to see other solutions.


答:
$\angle HFC=90^{\circ}=\angle HGC$. So $CFHG$ is cyclic. Circumcircle of $CFG$ is the same as circumcircle of quadrilateral $CFHG$. $CH$ is its diameter. Midpoint of $CH$, $N$, is its center.

Now $FE$ is the median of right $\triangle AFB$ and $GE$ is median of right $\triangle AGB$. So $AE=FE=GE=BE=a/2$.

In isosceles $\triangle BEG$, $\angle EGB=B$. Since $CH$ is part of altitude on $AB$, $\angle NCG=90-B$. In isosceles $\triangle CNG$, $\angle NGC=\angle NCG=90-B$. $\Rightarrow \angle NGE=90^{\circ}$. Similarly $\angle NFE=90^{\circ}$. Thus $NFEG$ is also cyclic.

The circumcircle of $EFG$ is same as circumcircle of quadrilateral $NFEG$ whose diameter is $NE$.

Update :

The above is how one would explain in detail. But to quickly identify that the statement holds true, recall that the nine-point circle is the unique circle that passes through the midpoints of three sides of a triangle, the feet of three altitudes and midpoints of the three segments joining each vertex to the orthocenter. $\odot(EFG)$ is indeed the nine-point circle (you also show its radius is $R/2$). Once we identify the center of $\odot(CFG)$, $N$, as midpoint of $CH$, we're done.


问:
Number of ways to choose elements to swap
I'm trying to work through a question in a textbook.

For 3 elements, there are 3! = 6 permutations. That I understand. But I don't get the below statement:

"For 3 elements, the number of ways to choose elements to swap is 3 to the power of 3 = 27". How is it 27? Could someone here explain?

Thanks,

答:
Consider a 3 elements
A B C

The first iteration swaps A with another element at random, i.e. either A, B or C, so the three equally likely configurations after the first iteration are:

no change   ->   A  B  C

swap with B  ->  B  A  C

swap with C  ->  C  B  A

The second iteration starts with one of the three arrangements generated at the first iteration and the swaps are:

Second position with first position gives one of:

A B C -> B A C

B A C -> A B C

C B A -> B C A

Second position with second position gives one of:

A B C -> A B C

B A C -> B A C

C B A -> C B A

and finally second position with third position gives one of:

A B C -> A C B

B A C -> B C A

C B A -> C A B

So now we have nine equally likely arrangements after the second shuffle. That is:

B A C

A B C

B C A

A B C

B A C

C B A

A C B

B C A

C A B

The final step is to randomly move the final element to a new random position and this has three possibilities.

The third element is moved to the first element.

The third element is moved to the second.

And finally, the third element is moved to the third position.

Thus, in total resulting to 27 possible outcomes.

Reference : How Not To Shuffle : Mike James


问:
Solve the Integral in the following set
I have the Set $D=\{ (x,y) \in\mathbb{R}: |x|\le2,|x|\le y \le \sqrt{4-x^2}\}$ and I have to calculate $\int_{D}(x^2y+xy^2) dydx$.

So I cannot explain how hard it is for me to find the borders of the integral. I've tried to understand it but It's very hard for meand every time I think i have understood it, i do it wrong in another excercise.

I have an exam tomorrow and of course I can't do wonders with understanding them. We often do them in the lectures that we turn them somehow to polar coordinates or so.

However, my question is: is there any way, that I can read it from the definition of the set D for example, with x and y(without having to turn them in some other coordinates) and what suggestions could you give me in order to be able to do that?

For example in this specific excercise, i thought that since $|x|\le2$ then $-2\le x\le 2 $. Then I just took $y$ between $2$ (since the absolute value of $x$ can't be more than $2$, and $\sqrt{4-x^2}$.
So I calculated
$\int_{-2}^{2}\int_{2}^{\sqrt{4-x^2}}x^2y$. My answer is not the same as the answer in the book, so I got completely lost.

I would be very thankful for some help.

Annalisa

答:
If you read more carefully the definition of the domain $D$, then you will find that the given integral can be written as
$$\iint_D(x^2y+xy^2)\,dxdy=\int_{x\in [-2,2], |x|\leq \sqrt{4-x^2}}\left(\int_{|x|}^{\sqrt{4-x^2}}(x^2y+xy^2)\,dy\right) \,dx$$
Next step: draw the graphs of the functions $|x|$ and $\sqrt{4-x^2}$ over the interval $[-2,2]$ and find explicitly the upper limit and the lower limit of the domain of integration with respect to $x$.
Can you take it from here?

P.S. Now a big part of YOUR work (sketch the graphs + symmetry) has been done by Math Lover. This means that the interesting part of your exercise is gone... Well, that's a pity!

Title: Solve the Integral in the following set


问:
Derivation of $\frac{du}{dt}$ from The Chemical Basis of Morphogenesis by A. Turing
I am currently reading The Chemical Basis of Morphogenesis
 by A. Turing. On page $56$ of his article, Turing uses the substitution
\begin{align}
\xi&=b(u+v), \\
\eta&=(p-a')u+(p'-a')v,
\end{align}
to transform
\begin{align}
\frac{d\xi}{dt}&=a'\xi+b\eta+R_1(t),\\
\frac{d\eta}{dt}&=c\xi+d'\eta+R_2(t),
\end{align}
into $$\frac{du}{dt}=pu+\frac{p'-a'}{(p'-p)b}R_1(t)-\frac{R_2(t)}{p'-p}+\xi\frac{d}{dt}\left(\frac{p'-a'}{(p'-p)b}\right)-\eta\frac{d}{dt}\left(\frac{1}{p'-p}\right).$$ Here, $p$ and $p'$ are the (real) roots of $(p-a')(p-d')=bc$. I am seeking advice/hints on how I can derive this final expression.

答:
Assuming $p \neq  {p'}$ and $b \neq  0$, we have

\begin{equation}
\renewcommand{\arraystretch}{1.5}  \left\{\begin{array}{rcl}{\xi}&=&b \left(u+v\right)\\
{\eta}&=&\left(p-{a'}\right) u+\left({p'}-{a'}\right) v
\end{array}\right. \Longleftrightarrow  \left\{\begin{array}{rcl}u&=&\displaystyle  \frac{{p'}-{a'}}{\left({p'}-p\right) b} {\xi}-\frac{1}{{p'}-p} {\eta}\\
v&=&\displaystyle -\frac{p-{a'}}{\left({p'}-p\right) b} {\xi}+\frac{1}{{p'}-p} {\eta}
\end{array}\right.
\end{equation}

Hence

\begin{equation}
{u'} = \frac{d u}{d t} = \frac{d}{d t} \left(\frac{{p'}-{a'}}{\left({p'}-p\right) b}\right) {\xi}-\frac{d}{d t} \left(\frac{1}{{p'}-p}\right) {\eta}+\frac{{p'}-{a'}}{\left({p'}-p\right) b} \frac{d {\xi}}{d t}-\frac{1}{{p'}-p} \frac{d {\eta}}{d t}
\end{equation}

Hence

\begin{equation}
{u'}-p u = \frac{d}{d t} \left(\frac{{p'}-{a'}}{\left({p'}-p\right) b}\right) {\xi}-\frac{d}{d t} \left(\frac{1}{{p'}-p}\right) {\eta}+\frac{{p'}-{a'}}{\left({p'}-p\right) b} \left(\frac{d {\xi}}{d t}-p {\xi}\right)-\frac{1}{{p'}-p} \left(\frac{d {\eta}}{d t}-p {\eta}\right)
\end{equation}

As $p$ and $p'$ are the two roots of the equation $x^2-(a'+d')x + (a' d' - b c) = 0$, it follows that $p + p'=a'+d'$ and $p p'=a' d' - b c$ by Vieta's formulas, hence we have

\begin{equation}
\renewcommand{\arraystretch}{1.5}  \begin{array}{rcccl}\displaystyle  \frac{d {\xi}}{d t}-p {\xi}&=&\displaystyle  \left({a'}-p\right) {\xi}+b {\eta}+{R}_{1}&=&\left({p'}-{d'}\right) {\xi}+b {\eta}+{R}_{1}\\
\displaystyle  \frac{d {\eta}}{d t}-p {\eta}&=&\displaystyle  c {\xi}+\left({d'}-p\right) {\eta}+{R}_{2}&=&c {\xi}+\left({p'}-{a'}\right) {\eta} + R_2
\end{array}
\end{equation}

Turing's formula clearly follows from these relations because $(p'-a')(p'-d')-b c = 0$ and

\begin{equation}
\renewcommand{\arraystretch}{2}  \begin{array}{rcl}\displaystyle  \frac{{p'}-{a'}}{\left({p'}-p\right) b} \left(\frac{d {\xi}}{d t}-p {\xi}\right)-\frac{1}{{p'}-p} \left(\frac{d {\eta}}{d t}-p {\eta}\right)&=&\displaystyle  \frac{\left({p'}-{a'}\right) \left({p'}-{d'}\right)-b c}{\left({p'}-p\right) b} {\xi}\\
&&\displaystyle  \qquad +\frac{\left({p'}-{a'}\right) b-\left({p'}-{a'}\right) b}{\left({p'}-p\right) b} {\eta}\\
&&\displaystyle  \qquad +\frac{{p'}-{a'}}{\left({p'}-p\right) b} {R}_{1}-\frac{1}{{p'}-p} {R}_{2}\\
&=&\displaystyle  0+0+\frac{{p'}-{a'}}{\left({p'}-p\right) b} {R}_{1}+\frac{1}{{p'}-p} {R}_{2}
\end{array}
\end{equation}


问:
two reduced row echelon matrices have the same nullspace, prove they are identical
I am trying to prove if R and R' are two reduced row echelon matrices, and have the same nullspaces, they are identical. I have observed this when testing them, but I have trouble finding a formal proof.

I tried reading this question and extracting a general solution, but this question is focused on a particular 2 by 3 matrix whereas I am trying to find a formal proof for an m by n matrix.

I believe I should start by writing two general reduced row echelon matrices, writing Rx=R'x=0 and finding a general answer for each row, then proving they are equal, but I have trouble proving it generally and without presumptions.

Thank you in advance.

答:
Perhaps there is an easier proof. I'll write it informally.

Consider the two system of equations defining the null spaces
$$
R_Ax=0\quad\text{respectively}\quad R_Bx=0
$$
To have the same null space means that the solutions of the two systems are identical.

Now we start from the last row $m$ of the two rref. Suppose that the two rows differ, $r_{A,m}\ne r_{B_m}$ ($r_{A,m}$ is the $m$th row of $R_A$), then(*) there is $x$ such that
$$
r_{A,m}\cdot x=0
\quad\text{and}\quad
r_{B,m}\cdot x\ne 0
$$
(or the other way around) contradicting the equality of the two null spaces.
Thus the two rows are equal. We proceed from row $m$ to row $1$ proving that all rows are identical.

(*) How to find such an $x$? We go back to the rules defining a rref.
The two rows are different and cannot be both zero.
If only one is zero the we have the whole $\mathbb{R}^n$ versus a strict subset of it.
If both are not zero, the first element of the row is $1$; if the position of the $1$ is different, w.l.o.g $r_{A,m}$ has less zero on the left side of the $1$ positioned at $k$, then $x_k$ appears in $r_{A,m}\cdot x=0$ and not in $r_{B,m}\cdot x= 0$, thus $\cdots$


问:
Continuity of identity map between normed linear space.
Let $f\in C[0, 1]$ and define $$\|f\|_p=(\int|f|^p dt) ^{\frac{1}{p}}, p=1, 2$$ Let $V_p$ be the space $C[0, 1]$ with norm $\|•\|$, then which of the following options are correct?

$1.$ the identity map is continuous from $V_1$ to $V_2$.

$2.$ the identity map is continuous from $V_2$ to $V_1$.

$3.$ the identity map is an isomorphism between $V_1$ and $V_2$.

I am unable to find constant $K$ or $L$ for  $\int |f|\leq K(\int |f|^2 dt) ^{\frac{1}{2}}$ or $(\int |f|^2)^{\frac{1}{2}}\leq L(\int |f| dt)$ as constants are changing with function $f$ . Please help me to solve the problem.

答:
As you say in the comments, as hinted by Ian, we have that $$\|f\|_1=\int_0^1|f(x)|dx=\int_0^1|f(x)|\cdot1dx\leq\big(\int_0^1|f(x)|^2dx\big)^{1/2}\cdot\big(\int_0^11^2dx\big)^{1/2}=\|f\|_2 $$

So the identity map $V_2\to V_1$ is continuous. On the other hand, the identity map $V_1\to V_2$ is not continuous: consider the functions $f_n(x)$, where $f_n(x)=n^{3/2}$ for $x\in[0,1/n^2]$, then $f_n$ descends linearly to $0$ in the interval $[1/n^2, 2/n^2]$ (so $f_n(2/n^2)=0$) and then $f_n$ is identically $0$. Note that
$$\int_0^1|f_n(x)|dx=n^{3/2}\cdot\frac{1}{n^2}+n^{3/2}\cdot\frac{1}{2n^2}=\frac{3}{2n^{1/2}}\to0 $$
so $f_n\to0$ in $\|\cdot\|_1$ while on the other hand
$$\big(\int_0^1|f_n(x)|^2dx\big)^{1/2}\geq\big(\int_0^{1/n^2}|f_n(x)|^2dx\big)^{1/2}=(n^3\cdot\frac{1}{n^2})^{1/2}=n^{1/2}\to\infty$$
so $(f_n)$ does not converge in $\|\cdot\|_2$. Thus only option B is correct.

edit: to see that $\|f_n\|_1\to0$ just draw these, their integrals are the area of a rectangle plus the area of a right triangle.


问:
What is the probability that there more rabbits than chickens in each of these three cages?
Suppose we have $4$ chickens and $12$ rabbits, and put them into $3$ cages. What is the probability that there more rabbits than chickens in each of these three cages?

I was trying to solve it by using P(more rabbits than chicken) = 1-P(more chicken than rabbits). Since the number of chickens is far less than the number of rabbits and it would be easier to calculate.

Then, the cardinality of the sample space would be $3^{12+4}$. But I don't know how to continue.

答:
To Reader: As discussed in the comments, and as @Matthew Pilling explained in the comments, my solution probably have an issue (read comments provided by @Matthew Pilling). I cannot delete my answer since it has been already accepted.

==============================================================

Assumptions: Chickens are indistinguishable as well as Rabbits.

Note:  For the case of distinguishable items, look at the other answer provided by @John L.

Let's say $c_{i}$ and $r_{i}$ are, respectively, the number of chickens and rabits in cage $i-th$. Therefore:

$$  c_{1} + c_{2} + c_{3} = 4  $$
$$  r_{1} + r_{2} + r_{3} = 12  $$

However, we want the number of rabbits to be more than the number of chickens in each cage. So,

$$ r_{1} = c_{1} + a_{1} + 1 $$
$$ r_{2} = c_{2} + a_{2} + 1 $$
$$ r_{3} = c_{3} + a_{3} + 1 $$

Where, $a_{i}$ is the auxiliary variable: $ a_{i} \ge 0 $

Finally, by substitution, we'll have:

$$  c_{1} + c_{2} + c_{3} = 4  $$
$$  a_{1} + a_{2} + a_{3} = 5  $$

So, we can first distribute the chickens and put them in cages, and then for each case, we can choose a different set of $a_{1}$, $a_{2}$, and $a_{3}$.

The final answer is:

$$ {4+3-1 \choose 3-1}{5+3-1 \choose 3-1} = {6 \choose 2}{7 \choose 2}  = 315 $$

Now, you have the number of ways putting chickens and rabits to satisfy condition. Divide it by total cases and you will get the probability.

For total number of cases:

$$  c_{1} + c_{2} + c_{3} = 4  $$
$$  r_{1} + r_{2} + r_{3} = 12  $$

total number of cases (no condition) =
$$ {4+3-1 \choose 3-1}{12+3-1 \choose 3-1} = {6 \choose 2}{14 \choose 2} = 1365 $$

$$ probaility = \frac{315}{1365} $$


问:
Problem regarding a MST proof of 0.99999...=1
I came across an answer of this (which is the highest voted, and also awarded bounties worth 50 reputations).
To quote, this is what the answer said:-

"Suppose this was not the case, i.e. $0.9999... \neq 1$. Then $0.9999... < 1$ (I hope we agree on that). But between two distinct real numbers, there's always another one (say $x$) in between, hence $0.9999... < x < 1$.

The decimal representation of $x$ must have a digit somewhere that is not $9$ (otherwise $x = 0.9999...$). But that means it's actually smaller, $x < 0.9999...$, contradicting the definition of $x$.

Thus, the assumption that there's a number between $0.9999...$ and $1$ is false, hence they're equal."

I have a problem regarding this, it merely tells that there exists no $x$ such that $$0.999...<x<1$$ and thus reaches the conclusion that $$0.9999...=x$$ What am I missing?

答: There is a problem with this proof, but it's not where you're looking for it. Its logic is supposed to be: assume both numbers are different, then, there must be a number strictly between them ("...between two distinct real numbers, there's always another one..."), and that leads to a contradiction. The way the contradiction is reached is faulty, though: "The decimal representation of $x$ must have ..." is implicitly assuming that there is a unique decimal representation of $x$, while proving that $0.9999\ldots=1.0000\ldots$, i.e. that decimal representation is not always unique. A proof proving its own invalidity is a rare specimen.


问:
Is a function that is continuous and doesn't blow up always uniformly continuous?

A function $f: D \rightarrow \mathbb{R}$ is called uniformly continuous if and only if $
\forall \varepsilon>0 \exists \delta>0 \forall x, x_{0} \in D:\left|x-x_{0}\right|<\delta \Rightarrow\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon$.

I am having difficulty understanding what uniform continuity really means.

I think a continuous linear function is always uniformly continuous, right? But how can a function that is not linear be uniformly continuous?

I also looked at the Wikipedia article on uniform continuity, where they have an animation of what uniform continuity means visually. I think this animation tells us that whenever a function blows up to infinity, then it is not uniformly continuous - no matter how small we choose $\delta$ for some $x_0$ there is always an $x$ s.t. $|x_0-x|<\delta$ and $f(x)=\infty$ and so $|f(x)-f(x_0)|>\varepsilon$, right? Is it also true that a function is uniformly continuous if it is continuous and doesn't blow up?

My question may not be very specific, but in a sense I am just asking what uniform continuity really means.

答:
Not sure if this is what you are looking for, but the essence in the difference of continuity and uniform continuity is the change in the order of the quantifiers:

Continuity: for each $\varepsilon>0$ and for each $x\in D$ there exists $\delta>0$ so that... Note that $\delta$ depends on $\varepsilon$ and $x$.

Uniform continuity: for each $\varepsilon>0$, there exists $\delta>0$ so that for each $x\in D$... Note that $\delta$ depends only on $\varepsilon$ here.

What this tells us is that, when a function is continuous, when we want to go $\varepsilon$-near the value $f(x_0)$ then we can find a quantity $\delta>0$ so that when we are $\delta$-near $x$, after applying $f$ we get values that are $\varepsilon$-near $f(x_0)$.

Imagine that you want to be at the same time $\varepsilon$-near $f(x_n)$ for an infinite set of points $\{x_n\}\subset D$. Can you find a distance $\delta$ so that, for all $n\geq1$, if $x$ is $\delta$-near some $x_n$ then $f(x)$ will be $\varepsilon$-near $f(x_n)$? Well, since $f$ is continuous, we find a $\delta_n$ depending on $x_n$ so that when $x$ is $\delta_n$-near $x_n$ then sure, $f(x)$ is $\varepsilon$-near $f(x_n)$. So a reasonable person would consider taking $\delta:=\inf_{n}\delta_n$. The problem is that this could be zero! In other words, what if the distance that you have to lie in so that $f$ maps you $\varepsilon$-near $f(x_n)$ gets arbitrarily small as $n$ becomes large?

On the other hand, uniform continuity does not get in such trouble: you can find a $\delta>0$ so that when you are $\delta$-near the set $\{x_n\}$ then your values through $f$ are $\varepsilon$-near the set $\{f(x_n)\}$.

Also, it depends on what you mean "blow up". For example, if a function is $C^1$ and its derivative is bounded (so it does not really "blow up": the rate under which its values change cannot get very large), then it is uniformly continuous. Can you see why? (hint: use the mean value theorem). On the other hand, there do exist bounded, continuous but not uniformly continuous functions: see for example Prove that the function$\ f(x)=\sin(x^2)$ is not uniformly continuous on the domain $\mathbb{R}$.

I hope this helps.


问:
Is there a mathematical way of finding b, given a to the power of b and a?
For example, we have $a^b = 256$ and $a = 2$.

Can I find $b = 8$ without trying all possible values of $b$ ?

答: Yes! $b=\log_2256$ - a calculator will give this result.


问:
Show that at one of these points the tangent to the curve is parallel to the $x$ axis
Show that at one of these points $(4,18)$,$(4,-2)$  the tangent line to the curve $y^2-4xy-x^2=20$ is parallel to the $x$ axis.

Any help would be appreciated

答:
Hint:

Calculate the slope by differentiating the curve at both of these points. Slope of any tangent parallel to x-axis should be zero.


问:
Limit of function with log
Does $\displaystyle\lim_{x \to 1^{+}} \ln(3x-3)$ exist in real numbers ( and it's $-\infty$) or is it a complex solution?

Title: Limit of function with log

Body:
We have
$$\ln(3x-3) = \ln(3(x-1)) = \ln(3) + \ln(x-1)$$
and
$$
\lim_{x\to 1^+} \ln(x-1) = \lim_{x\to 0^+} \ln(x) = -\infty.
$$


问:
Find what values of $n$ give $\varphi(n) = 10$

For what values of $n$ do we get $\varphi(n) = 10$?

Here, $\varphi$ is the Euler Totient Function. Now, just by looking at it, I can see that this happens when $n = 11$. Also, my friend told me that it happens when $n = 22$, but both of these are lucky guesses, or educated guesses. I haven't actually worked it out, and I don't know if there are any more.

How would I go about answering this question?

答:
the general method will be demonstrated for 12.

The factors of 12 are 2*2*3.  this is the product of totients of the prime powers (p-1)p^q.  So we divide this up into eg products to 12 that are numbers of this form.

```<code> 1*12    1,2 * 13      13, 26
 1*2*6   1,2 *3* 7     21, 42
 2*6     4 * 7, 9      28, 36
</code>```

Since 3 can not be the totient of a prime power, the factorisation of 12=3*4 is not considered.

The case for 10involves 11 and 2 having totient products of 10 and 1, which leads directly to 11 and 22.

Body:
the general method will be demonstrated for 12.

The factors of 12 are 2*2*3.  this is the product of totients of the prime powers (p-1)p^q.  So we divide this up into eg products to 12 that are numbers of this form.

 1*12    1,2 * 13      13, 26
 1*2*6   1,2 *3* 7     21, 42
 2*6     4 * 7, 9      28, 36

Since 3 can not be the totient of a prime power, the factorisation of 12=3*4 is not considered.

The case for 10involves 11 and 2 having totient products of 10 and 1, which leads directly to 11 and 22.


问:
For any conditionally convergent series $\sum _{n=1}^\infty a_n,\ \exists\ k\geq 2\ $ such that the subseries $\sum _{n=1}^\infty a_{nk}$ converges.
A subseries of the series $\displaystyle\sum _{n=1}^\infty a_n$ is defined to be a series of the form $\displaystyle\sum _{k=1}^\infty a_{n_k}$, for $n_k \subseteq \Bbb N$.

Prove or disprove: For any conditionally convergent series $\displaystyle\sum _{n=1}^\infty a_n,\ \exists\ k\geq 2\ $ such that the subseries $\displaystyle\sum _{n=1}^\infty a_{kn}$ converges.

Certainly, every conditionally convergent series has some subseries that converge (decreasing alternating terms), and some subseries that diverge (the subseries of all of the positive terms; also the subseries of all of the negative terms).

$$$$

My original "answer" which I now doubt is a counter-example to the proposition:

Hint: Find a subset $\ A \subset \mathbb{N}\ $ with the property that, for each $\ n \in \mathbb{N},\ $ the set $ \{kn\}_{k \in \mathbb{N} }$  has finitely many elements in common with $A.$

Answer: The prime numbers is the standard example of such a subset of $\mathbb{N}$ with the property in the hint. This point is to make the 2nd, 3rd, 5th, 7th, ... members of our sequence have one sign, and every other member of our sequence have another sign. For example, $$ a_1 =  \frac{1}{1},\ a_2 =  \frac{-1}{4},\ a_3 =  \frac{-1}{4},\ a_4 =  \frac{1}{3},\ a_5 =  \frac{-1}{4},\ a_6 =  \frac{1}{5},\ a_7 =  \frac{-1}{6},\ a_8 = a_9 = a_{10} =  \frac{1}{3} \times \frac{1}{7},\ a_{11} = \frac{-1}{8},\ ... $$ This is simply the alternating harmonic series split up in a way that (I thought) answers the question.

That was my answer, but now I'm not sure the series above is a valid counter-example. I constructed it so that every subseries of the form $\displaystyle\sum _{n=1}^\infty a_{kn}$ contains finitely many negative numbers. And then I'm not sure what my logic was. It was either that "every subseries of a monotonic divergent series diverges", but now I realise that this is not what the link says, and is also not true. Or my reasoning was that, for each $k \in \mathbb{N},\ $ every subseries $\displaystyle\sum _{n=1}^\infty \frac{1}{kn}$ of the harmonic series, diverges, which is true, but the subseries of my series defined above do not match these subseries of the harmonic series.

So the problem remains open...

答:
Your primes idea is a good one! It just takes a little bit of finagling to make it clearer that the idea works. In particular, it relies on some nontrivial number theory -- but this kind of interplay is part of the fun of mathematics!

Here's the plan: We'll use the harmonic series on the composite numbers. This normally diverges (of course) but since there are infinitely many primes, we have infinitely many chances to knock it back down to $0$. Then we'll show that this converges (it's obvious that it does not converge absolutely). Lastly, we'll use your idea about arithmetic progressions only intersecting the primes finitely many times to conclude.

To be clear, our sequence will be:

$$
\begin{array}{|c|c|}
\hline
\mathbf{n} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\mathbf{a_n} & 1 & -1 & 0 & \frac{1}{4} & -\frac{1}{4} & \frac{1}{6} & -\frac{1}{6} & \frac{1}{8} & \frac{1}{9} & \frac{1}{10} & -(\frac{1}{8} + \frac{1}{9} + \frac{1}{10}) & \frac{1}{12}\\
\hline
\mathbf{\sum_{i \leq n} a_i} & 1 & 0 & 0 & \frac{1}{4} & 0 & \frac{1}{6} & 0 & \frac{1}{8} & \frac{1}{8} + \frac{1}{9} & \frac{1}{8} + \frac{1}{9} + \frac{1}{10} & 0 & \frac{1}{12} \\
\hline
\end{array}
$$

and so on.

It's obvious that the sum tends to stay around $0$, but it's not obvious that it converges to $0$. After all, who's to say we won't have a long dry spell between primes where we move above $10^{-100}$? This doesn't need to happen often. It only needs to happen infinitely many times, and convergence is ruined.

Thankfully, this can't happen. It's a corollary of the prime number theorem that for any $\epsilon$ we like, the distance between primes is eventually $\epsilon$ proportionally small. You can read more about this on wikipedia, but I've transcribed the relevant fact here:

For all $\epsilon > 0$, for all large $k$ (where "large" depends on $\epsilon$), we have $p_{k+1} - p_k \leq \epsilon p_k$

Why does this guarantee convergence? Well, let $\epsilon > 0$. We want to show that, eventually, we stay within $\epsilon$ of $0$. But if $n$ is bigger than $p_k$ (as guaranteed by the above theorem), we see:

$$
\begin{aligned}
\sum_{i \leq n} a_i
&\leq \sum_{i < p_{l+1}} a_i \\
&\leq H(p_{l+1}) - H(p_l) \\
&\sim \log(p_{l+1}) - \log(p_l) \\
&= \log(\frac{p_{l+1}}{p_l}) \\
&\leq \log(\frac{p_l + p_l \epsilon}{p_l}) \\
&= \log(1 + \epsilon) \\
&\leq \epsilon
\end{aligned}
$$

Here $p_l$ is the prime immediately before $n$, and $H(m)$ is the $m$th harmonic number.

But then for $n$ large, $\sum_{i \leq n} a_i \leq \epsilon$, and our series converges (conditionally) to $0$.

Now we can finish up! As you noticed, any sequence $(kn)_{n \in \mathbb{N}}$ can only intersect the primes finitely many times (indeed, at most once). So for any sequence of that form, we see $\sum_n a_{kn}$ contains a tail of the harmonic series (I guess we're off by a factor of $\frac{1}{k}$, but what's a constant between friends?), and diverges.

Thanks for asking this -- it was a lot of fun!

I hope this helps ^_^


问:
Point on a archimedean spiral given distance from a known point.
I have an archimedean spiral, and know these points:

angle and distance from center of the beginning of the spiral,
angle and distance from center of the end of the spiral

I would need the angle and distance from center of a point that is X mm in distance along the line from the start...

all values are in mm.

for example (angle, distance to center):

P start: 2Pi , 20

P end: 6Pi, 60

Distance from P(start) to px: 110mm

how can this be caluclated? In my case semplicity is more important than precision, it is enough to have a approximation.

答:
This spiral has equation
$$r=10 \frac{t}{\pi}$$
Length of the spiral is
$$L(x)=\int_0^x\sqrt{r^2+{r'}^2}\,dt$$
In this case we must have
$$L(x)=\int_{2\pi}^x \frac{10 \sqrt{t^2+1}}{\pi } \, dt=110$$
which means
$$\frac{5 x \sqrt{x^2+1}}{\pi }+\frac{5 \log \left(\sqrt{x^2+1}+x\right)}{\pi }-10 \sqrt{1+4 \pi ^2}-\frac{5 \log \left(\sqrt{1+4 \pi ^2}+2 \pi \right)}{\pi }=110$$
approximated
$$x\approx 10.4\sim 360°+235.7°$$
Distance from origin is given by the spiral equation
$$r=10 \frac{10.4}{\pi}\approx 33.1$$

$$...$$


答:
$|x| \leq 2 \implies -2 \leq x \leq 2$

$|x| \leq y \leq \sqrt{4-x^2}$

$\implies -x \leq y \leq \sqrt{4-x^2} \ $ for $x \leq 0$ and $x \leq y \leq \sqrt{4-x^2} \ $ for $x \geq 0$.

So here is the sketch of the region (you are asked to integrate over the shaded region).

Please note that integral of $xy^2$ will cancel out over region for $x \geq 0$ and for $x \leq 0$ due to $x-$ symmetry.

Your integral is split into two as below -

$\displaystyle \int_{-\sqrt2}^{0} \int_{-x}^{\sqrt{4-x^2}} (x y^2 + yx^2) \ dy \ dx +
\int_0^{\sqrt2} \int_{x}^{\sqrt{4-x^2}} (x y^2 + y x^2) \ dy \ dx$

As mentioned you can avoid integrating $xy^2$ as it will cancel out and will not change the result of your final integral.

Title: Solve the Integral in the following set

Body:
I have the Set $D=\{ (x,y) \in\mathbb{R}: |x|\le2,|x|\le y \le \sqrt{4-x^2}\}$ and I have to calculate $\int_{D}(x^2y+xy^2) dydx$.

So I cannot explain how hard it is for me to find the borders of the integral. I've tried to understand it but It's very hard for meand every time I think i have understood it, i do it wrong in another excercise.

I have an exam tomorrow and of course I can't do wonders with understanding them. We often do them in the lectures that we turn them somehow to polar coordinates or so.

However, my question is: is there any way, that I can read it from the definition of the set D for example, with x and y(without having to turn them in some other coordinates) and what suggestions could you give me in order to be able to do that?

For example in this specific excercise, i thought that since $|x|\le2$ then $-2\le x\le 2 $. Then I just took $y$ between $2$ (since the absolute value of $x$ can't be more than $2$, and $\sqrt{4-x^2}$.
So I calculated
$\int_{-2}^{2}\int_{2}^{\sqrt{4-x^2}}x^2y$. My answer is not the same as the answer in the book, so I got completely lost.

I would be very thankful for some help.

Annalisa


问:
Finding the funtion through expression
There is the expression:
$$2f'(x)f(x)=2f(x)+2xf'(x)$$
and we are given that $f(0)=1$.
We are asked to prove that $f(x)=x+\sqrt{x^2+1}$.

I tried to rewrite the given expression as $$[f(x)^2]'=[2xf(x)]',$$but I was not able to reach the specific conclusion.
I would appreciate your guidance.
Thank you very much in advance.

答: If $(f(x)^2)'=(2xf(x))'$ then we have that $f(x)^2=2xf(x)+C$, where $C$ is some constant. Since $f(0)=1$, $C=1$ by substitution. Hence we have that $f(x)^2=2xf(x)+1$, and so $f(x)^2-2xf(x)-1=0$. The quadratic formula tells us that $f(x)=\frac{2x\pm\sqrt{4x^2+4}}{2}$, which reduces to the required result. Substituting back in shows that it is indeed a valid solution. Note that we take the nonnegative square root because the solution $x-\sqrt{x^2+1}<0$ does not hold as $f(0)>0$.

Title: Finding the funtion through expression


答:
Derivate your curve like the following:

$\frac{d}{dx}(y^2-4xy-x^2)=2y(y')-2x-4x(y')-4y=\frac{dy}{dx}(20)=0$

$\implies y'=\frac{x+2y}{y+2x}$

So the tangent line in point $P=(x,y)$ is parallel to the x-axis $\iff y'=\frac{x+2y}{y+2x}=0 \iff x+2y=0 \iff x=-2y$ so the point $(4,-2)$ has the property.

Title: Show that at one of these points the tangent to the curve is parallel to the $x$ axis

Body:
Show that at one of these points $(4,18)$,$(4,-2)$  the tangent line to the curve $y^2-4xy-x^2=20$ is parallel to the $x$ axis.

Any help would be appreciated


答:
I'll try to describe it without the confusing math notation. For both continuity and uniform continuity, the game starts by choosing some usually small $\varepsilon>0$.

Continuity of $f$ means that at any given point $x$, if we get within a small enough $\delta$-neighborhood of $x$ (an open set containing all points whose distance to $x$ is less than $\delta$), then $f$ only takes values within an $\varepsilon$-neighborhood of $f(x)$. Essentially, we can force the function arbitrarily close to a particular value $f(x)$ by going close enough to $x$. What "close enough" means may depend on $x$.

Uniform continuity means that if we choose an arbitrary amount of values for $x$ under the condition that none of them are farther away from each other than a suitably small $\delta$, then their corresponding outputs under $f$ are no further away from each other than $\varepsilon$. Essentially, all intervals of length $\delta$ are mapped to within intervals of length $\varepsilon$, no matter where those $\delta$-intervals are. Or put another way, if $x$ varies by less than a suitably small $\delta$, then $f$ varies by less than $\varepsilon$.

Now to counterexamples to your hypothesis: bounded continuous functions are uniformly continuous. This is not true. Consider for instance $f:(0,\infty)\to\mathbb R,~f(x)=\cos\left(\frac1x\right)$. This is clearly bounded and continuous, but looking at a graph (use desmos or wolframalpha or whatever else you prefer), we can see that as we're getting closer to $0$, the distances between peaks and valleys of the cosine get arbitrarily small. But this also means that no matter how small we choose $\delta$, there are a peak and a valley which are closer to each other than $\delta$, so within the interval containing both, $f$ will never vary by less than $2$ (height of the peak to depth of the valley).

The other way around is also not true: uniformly continuous functions need not be bounded. Consider $f:\mathbb R\to\mathbb R,~f(x)=x$ as a simple counterexample.

However, what is true, and is also a pretty damn important thing to keep in mind, is that if $f$ is defined on a compact set and is continuous, then it is also uniformly continuous. This result is used in proving the fundamental theorem of calculus, for instance. But also more advanced stuff like Cauchy's generalized integral formula in complex analysis make use of this result.

Title: Is a function that is continuous and doesn't blow up always uniformly continuous?

Body:
A function $f: D \rightarrow \mathbb{R}$ is called uniformly continuous if and only if $
\forall \varepsilon>0 \exists \delta>0 \forall x, x_{0} \in D:\left|x-x_{0}\right|<\delta \Rightarrow\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon$.

I am having difficulty understanding what uniform continuity really means.

I think a continuous linear function is always uniformly continuous, right? But how can a function that is not linear be uniformly continuous?

I also looked at the Wikipedia article on uniform continuity, where they have an animation of what uniform continuity means visually. I think this animation tells us that whenever a function blows up to infinity, then it is not uniformly continuous - no matter how small we choose $\delta$ for some $x_0$ there is always an $x$ s.t. $|x_0-x|<\delta$ and $f(x)=\infty$ and so $|f(x)-f(x_0)|>\varepsilon$, right? Is it also true that a function is uniformly continuous if it is continuous and doesn't blow up?

My question may not be very specific, but in a sense I am just asking what uniform continuity really means.


问:
Spectrum and self-adjointness
For bounded operator on a Hilbert space, if its spectrum is a subset of $\mathbb{R}$, then is this operator self adjoint? If not, what is a counterexample?

答:
There are plenty of counterexamples. For example, consider any nilpotent operator $T: H \to H$ that is not self-adjoint. Then $T$ has spectrum $\{0\}$.

For example, consider
$$M:=\begin{pmatrix} 0 & 1\\ 0 & 0
\end{pmatrix}$$
Then $M^2 = 0$ but $M$ is not self-adjoint. If you really want this as an operator, it is
$$M: \mathbb{C}^2 \to \mathbb{C}^2: \begin{pmatrix}x \\ y\end{pmatrix}\mapsto \begin{pmatrix} 0 & 1\\ 0 & 0
\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}= \begin{pmatrix}y \\ 0\end{pmatrix}$$

Title: Spectrum and self-adjointness


问:
Nonlinear recurrence relation over $\mathbb{C}$
A sequence in $\mathbb{C}$ is given: $(a_n)_{n=1}^{\infty}$ s.t.

$a_1 = i$ and $a_{n+1} = \frac{3}{2 + a_n}$.

Assume $\lim(a_n)$ exists.

Using the fact that $\lim(a_n) = \lim(a_{n+1})$ and solving the resulting quadratic gives that $\lim(a_n)\in \{-3,1\}$.

I was trying to figure out which one the limit is. I tried finding a relation on the norm or the argument (using the initial condition, since clearly that is what the limit is determined by) and I also tried looking at individual terms and by this, the real part seems to stay positive and hover around one while the imaginary part we know goes to zero in any case (i.e., I am thinking the limit is 1). I wanted to ask for verification that I did this last part correctly, since $a_2 = \frac{6-3i}{5}$ if we can show that:

$Sgn[Re(a_m)] = Sgn[Re(a_{m+1})]$ $\forall m\geq 2$

we are done and the limit is 1. Suppose that $a_m = x+yi$ ($m\geq 2$) and $x > 0$. Then

$a_{m + 1}= \frac{3}{2 + x + yi}\frac{(2+x - yi)}{(2+x-yi)} = \frac{3(2+x) - 3yi}{(2+x)^2 + y^2}$

meaning that

$Sgn[Re(a_{m+1})] = Sgn\bigg[\frac{6 + 3x}{(2+x)^2 + y^2}\bigg] = 1$ since $x > 0$ and we are done. So the limit is 1. Does that look about right?

答:
We have that $-3$ and $1$ are fixed point of $f(z)=\frac{3}{2+z}$. I know that there's a theorem from dynamic systems theory (although I don't know the precise hypothesis) that says that a fixed point $z_0$ of $f$ is an attractor or a repeller depending on whether $|f'(z_0)|$ is less than or greater than $1$.

In this case it would show that the sequence converges to $1$.

May be some expert on the field could confirm if the theorem apply here?


问:
If $4 \mid x^{2}+y^{2} \implies 4\mid x^{2}$and $4\mid y^{2}$
I have a proof by contradiction but I’m not Sure if it’s correct .

My proof:

$4 \mid x^{2}+y^{2} \iff x^{2}+y^{2} =4n.$ other words it’s a multiple of $4$.

By the way of contradiction: suppose that $4 \nmid x^2$ and $4 \nmid y^2$

$$\iff x^2 \neq 4a, y^2 \neq 4b$$
$$\iff x^{2}+y^{2} \neq 4(a+b)$$

This means that $x^{2}+y^{2}$ it’s not a multiple of $4$. Which is a contradiction.

$\implies 4\mid x^{2}$and $4\mid y^{2}$

答:
No, it is not correct.

If you negate conjunction then you get disjunction so $4\nmid x^2$ or $4\nmid y^2$.

So suppose $4\nmid x^2$ then $2\nmid x$ so $x$ is odd. But then $x^2$ is odd and so $y^2=4n-x^2$ is also odd. So $x=2a+1$ and $y=2b+1$. Now we have $$4a^2+4a+4b^2+4b+2 =4n$$ and if we divide this by two we get $$2a^2+2a+2b^2+2b+1 =2n$$
$$2(a^2+a+b^2+b) +1 =2n $$which is contradiction since on LHS we have odd and on RHS we have even number.


答:
Your proof is not correct. If it were, you could use the same method to prove

If $4\mid x+y$ then $4\mid x$ and $4\mid y$

which is but falsified by $x=3$, $y=5$.

So try again.

Title: If $4 \mid x^{2}+y^{2} \implies 4\mid x^{2}$and $4\mid y^{2}$

Body:
I have a proof by contradiction but I’m not Sure if it’s correct .

My proof:

$4 \mid x^{2}+y^{2} \iff x^{2}+y^{2} =4n.$ other words it’s a multiple of $4$.

By the way of contradiction: suppose that $4 \nmid x^2$ and $4 \nmid y^2$

$$\iff x^2 \neq 4a, y^2 \neq 4b$$
$$\iff x^{2}+y^{2} \neq 4(a+b)$$

This means that $x^{2}+y^{2}$ it’s not a multiple of $4$. Which is a contradiction.

$\implies 4\mid x^{2}$and $4\mid y^{2}$


答:
You cannot have a good fitting with the function
$$V(r) = D[e^{-2\alpha(r-r_{eq})}-2e^{-\alpha(r-r_{eq})}]$$
because a constant term is missing.

Better use the function :

$$V(r) = a+D[e^{-2\alpha(r-r_{eq})}-2e^{-\alpha(r-r_{eq})}] \tag 1$$
where $a$ is a constant.

The fitting for the parameters $D,\alpha,r_{eq},a$ can be done thanks to non-linear regression. In fact this is least squares method but iterative. The principle can be found in many papers on the web, for example https://mathworld.wolfram.com/NonlinearLeastSquaresFitting.html

Of course non-linear robust fitting is not easy to implement. In practice one uses available softwares.

Note that the above function is a particular case of the more general function made of two exponentials :
$$V(r)=a+b\,e^{p\,r}+c\,e^{q\,r} \tag 2$$

Eq.$(2)$ is related to $(1)$ in particular case of $\quad p=-2\alpha \quad,\quad q=-\alpha\quad,\quad b=De^{2\alpha r_{eq}} \quad,\quad c=-2De^{\alpha r_{eq}}$

Thus the above numerical values of the parameters convenient for Eq.$(2)$ are not valid for Eq.$(1)$. If we require $p=2q$ the fitting will be slightly different (See at end).

FOR INFORMATION.

The above numerical values where computed thanks to the method explained in : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales

This method transforms the non-linear equation to a linear equation (wrt the parameters) thanks to an integral equation. In the present case the non-iterative calculus is shown below :

CASE OF EQUATION $(1)$ which is equivalent to :

$$V(r)=a+b\,e^{-2\alpha\,r}+c\,e^{-\alpha\,r} \tag 2$$

$b=De^{2\alpha r_{eq}} \quad,\quad c=-2De^{\alpha r_{eq}}\quad\implies\quad D=\frac{b^2}{4c}\quad,\quad r_{eq}=-\frac{1}{\alpha}\ln(-\frac{b}{2c}).$

A non-linear least squares regression is necessary only due to one parameter $\alpha$. The iterative calculus is simple and very fast starting from a rather good initial value $\alpha_{init}=-\frac{2p+q}{4}\simeq 1.4\quad$ from the values of $p$ and $q$ already known from the above calculus. The result is :
$$\alpha\simeq 1.433$$
Finally the linear least squares regression for $a,b,c$ , then the calculus of $D$ and $r_{eq}$ gives :

Title: Parameters of the Morse Potential using Least Squares method

Body:
If I have the data points for the Morse Potential '$V(r)$' at given distance '$r$' of the Morse Potential described by:
$V(r) = D[exp(-2\alpha(r-r_{eq}))-2exp(-\alpha(r-r_{eq}))]$, is it possible to find the parameters D,$\alpha$ and $r_{eq}$, using the least squares method?

The sample data looks like this

$V(r)$=[-1360.12,-1368.53,-1374.21,-1378.09,-1380.65,-1382.22,-1383.09,-1383.48,-1383.56,-1383.45,-1383.22,-1382.93,-1382.61,-1382.27,-1381.94,-1381.62,-1381.32,-1381.04,-1380.77,-1380.53,-1380.31,-1380.1,-1379.91,-1379.73,-1379.58,-1379.43,-1379.3,-1379.18,-1379.08,-1378.98,-1378.90,-1378.83,-1378.77,-1378.71,-1378.67,-1378.63,-1378.6,-1378.57,-1378.55,-1378.54,-1378.54,-1378.54,-1378.55,-1378.57,-1378.6,-1378.63,-1378.67,-1378.71,-1378.77,-1378.83]

r = [1.0 ,1.1 ,1.2 ,1.3 ,1.4 ,1.5 ,1.6 ,1.7 ,1.8 ,1.9 ,2.0 ,2.1 ,2.2 ,2.3 ,2.4 ,2.5 ,2.6 ,2.7 ,2.8 ,2.9 ,3.0 ,3.1 ,3.2 ,3.3 ,3.4 ,3.5 ,3.6 ,3.7 ,3.8 ,3.9 ,4.0 ,4.1 ,4.2 ,4.3 ,4.4 ,4.5 ,4.6 ,4.7 ,4.8 ,4.9 ,5.0 ,5.1 ,5.2 ,5.3 ,5.4 ,5.5 ,5.6 ,5.7 ,5.8 ,5.9]


问:
If the countable infinite product topology is compact, connected, or locally connected, then so is each component
For each i∈N, let(Xi,τi) be a topological space. Prove that

(i) if ∏(Xi,τi) is compact, then each (Xi,τi) is compact
(ii) if ∏(Xi,τi) is connected, then each (Xi,τi) is connected

Are we allow to use the projection function pi for each component in the infinite product? If we are allowed to, then we are done since the projection mapping is continuous and connected and compactness are preserved.

(iii) if ∏(Xi,τi) is locally compact, then each (Xi,τi) is locally compact and all but a finite number of (Xi,τi) are compact.

The definition of locally compact is if each point in the set has an open neighborhood that is compact.
Take a point y=(y1,y2,y3,y4.....) There exist a compact open neighborhoud which is of the form (O1,O2,O3,O4....Xi,Xi,Xi,Xi..) because a basic open set is of this form and all but a finite number of i are Xi. Now we can use the projection mapping pi which is a continuous mapping and preserves compactness and we can see that y1 is contained in O1 which is compact. Since x was an arbitrary point all yi contains a compact set in Xi. Now we also note that (O1,O2,O3,O4....Xi,Xi,Xi,Xi..) is a compact set that only contains a finite number of Oi and the rest are Xi implies that all but a finite number of Xi are compact because of the projection mapping.

答:
Yes, the $\pi_i$ are continuous and onto, so $X$ compact implies $\pi_i[X]= X_i$ compact by continuity.

Likewise for connectedness.

For local compactness we also need the $\pi_i$ are open. But that should be well-known to you.

When $f:X \to Y$ is open continuous and onto, and $X$ is locally compact then so is $Y$. We can apply  this to each $\pi_i$ when $X$ is assumed to be locally compact to directly conclude that all $X_i$ must be too.

More precision needed in the netx step:

If $X$ is locally compact it has a compact set $C$ with a non-empty interior.

To say more we go into the product structure: this interior contains a basic open subset $\prod_{i \in I} O_i \subseteq C$. Then if $F \subseteq I$ is the finite subset on which this basic set depends, we have that $X_i = \pi_i[\prod_{i \in I} X_i] \subseteq \pi_i[C] \subseteq X_i$ for all $i \notin F$, so for all but finitely many $i$, $X_i=\pi_i[C]$ is compact by continuity of $\pi_i$.

The reverse is clear: let $F$ be the set of indices of the non-compact $X_i$, which  is finite. For each $x \in X$, we pick a compact neighbourhood $C_i$ of $x_i$ by local compactness and then $\prod_{i \in I} C_i$ is a compact neighbourhood of $x$ (where $C_i = X_i$ for $i \notin F$) by Tychonoff's theorem (and the fact that it by definition contains a basic open neighbourhood of $x$).

(That last paragraph is bonus, you only needed the forward direction).