答:
1) Trigonometric identities:
$$
\sin^2 x\cos^3x=(\sin x\cos x)^2\cos x=\left(\frac{\sin 2x}{2}\right)^2\cos x=\frac{1}{4}\sin^22x\cos x
$$
$$
=\frac{1}{4}\left(\frac{1-\cos 4x}{2}\right)\cos x=\frac{\cos x}{8}-\frac{\cos 4x\cos x}{8}
$$
$$
=\frac{\cos x}{8}-\frac{\cos 5x+\cos 3x}{16}
$$
$$
=\frac{\cos x}{8}-\frac{\cos 3x}{16}-\frac{\cos 5x}{16}
$$
2) Complex exponential:
$$
\sin^2x\cos^3x=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^2\left(\frac{e^{ix}+e^{-ix}}{2}\right)^3
$$
$$
=-\frac{1}{32}(e^{2ix}-2+e^{-2ix})(e^{3ix}+3e^{ix}+3e^{-ix}+e^{-3ix})
$$
$$
=-\frac{1}{32}(e^{5ix}+e^{3ix}-2e^{ix}-2e^{-ix}+e^{-3ix}+e^{-5ix})
$$
$$
=-\frac{1}{32}(2\cos 5x+2\cos 3x-4\cos x)
$$
$$
=\frac{1}{16}(2\cos x-\cos 3x-\cos 5x)
$$

Note: you made a mistake when you expanded $(e^{ix}-e^{-ix})^2$. I have no idea how you ended up with this $18$. You probably meant $16$.

Title: Finding the Fourier Series of $\sin(x)^2\cos(x)^3$

Body:
I'm currently struggling at calculation the Fourier series of the given function

$$\sin(x)^2 \cos(x)^3$$

Given Euler's identity, I thought that using the exponential approach would be the easiest way to do it.

What I found was: $$\frac{-1}{32}((\exp(2ix)-2\exp(2ix)+\exp(-2ix))(\exp(3ix)+3\exp(ix)+3\exp(-ix)+\exp(-3ix)))$$

Transforming it back, the result is:

$$ -\frac{1}{18}(\cos(5x)+\cos(3x)+2\cos(x))$$

(I've checked my calculations multiple times, I'm pretty sure it's correct.)

Considering the point $x = 0$ however, one can see that the series I found doesn't match the original function.

Could someone help me find my mistake?


问:
On Fourier-Bessel expansion
Expand $f(x)=2$, $0<x<1$, in a Fourier-Bessel series using the Bessel function of order zero that satisfies the boundary condition $J_0(α)=0.$

答:
enter image description here

one of these I think

ContentLicense: CC BY-SA 4.0

Title: On Fourier-Bessel expansion


问:
gammas division
Can anyone explain me how to calculate this expression?
$$\frac{\Gamma(n/2)}{\Gamma((n-1)/2)}$$
Thank you, I tried with doble factorial, but don´t know really how to continue.

答: It's already in the most helpful form. But its large-$n$ asymptotic behaviour is as $\sqrt{n/2}$, by linearizing $\ln\Gamma(z)$ viz.$$\begin{align}\ln\Gamma(n/2)-\ln\Gamma((n-1)/2)&\approx\frac{\ln\Gamma((n+1)/2)-\ln\Gamma((n-1)/2)}{2}\\&=\frac{\ln((n-1)/2)}{2}\\&=\ln\sqrt{(n-1)/2}.\end{align}$$

Title: gammas division


问:
How did this binomial $(x^2+3)^{-\frac 13}-\frac 23x^2(x^2+3)^{-\frac 43}$ get factored into these $(x^2+3)^{-\frac 43}(\frac 13x^2+3)$ factors?
This is an exercice in PRECALCULUS mathematics for calculus 7th, Section 1.3 Question 129.
And I just can't figure out how the book came up whith the answer.

I know how to factor this binomial:
$$(x^2+3)^{-\frac 13}-\frac 23x^2(x^2+3)^{-\frac 43}$$
... into these factors (it may be wrong):
$$(x^2+3)^{-\frac 43}  ((x^2+3)^{\frac 23}-\frac 23x^2)$$

But from there i have no idea how to continue, I'm following the basic strategy
of "pulling out" the smallest shared term $(x^2+3)^{-\frac 43}$

答:
The factorisation is wrong. The main idea is this: that when in the binomial $ab+ac,$ you factor out $a,$ then what you're in effect doing is rewriting the sum as $$a\left(\frac{ab}{a}+\frac{ac}{a}\right).$$

Hence by pulling out $(x^2+3)^{-4/3},$ you have to divide each summand by $(x^2+3)^{-4/3},$ or in other words multiply each summand by $(x^2+3)^{4/3},$ which gives you $$(x^2+3)^{-4/3}\left(x^2+3-\frac23 x^2\right),$$ from where you can continue.


答: If your question is connected to your previous one, you can leave the result as shown.</p>

Title: gammas division

Body:
Can anyone explain me how to calculate this expression?
$$\frac{\Gamma(n/2)}{\Gamma((n-1)/2)}$$
Thank you, I tried with doble factorial, but don´t know really how to continue.


问:
$m^*(E)=\infty$ in Caratheodory's criterion for Lebesgue measurability
I'm having trouble with the $m^*(E)=\infty$ case from this question. I've looked at the answers where $E$ is split into countably infinitely many bounded subsets, but I still cannot prove this case.

答:
Covering $\mathbb R$ with countably many boxes $B_n$ we see that for every elementary set $A$,
\begin{align}m^*(A\setminus (B_n\cap E))+m^*(A\cap(B_n\cap E))&\ge m^*(A)\tag{subadditivity}\\
&=m(A)\end{align}
and
\begin{align}m^*(A\setminus(B_n\cap E))+m^*(A\cap(B_n\cap E))
&=m^*(((A\cap B_n)\setminus E)\cup(A\setminus B_n))+m^*((A\cap B_n)\cap E))\\
&\le m^*((A\cap B_n)\setminus E)+m^*(A\setminus B_n)+m^*((A\cap B_n)\cap E))\tag{subadditivity}\\
&=m(A\cap B_n)+m(A\setminus B_n)\tag{$A\cap B_n$ and $A\setminus B_n$ are elementary}\\
&=m(A),\tag{additivity}
\end{align}
so $m^*(A\setminus (B_n\cap E))+m^*(A\cap(B_n\cap E))=m(A)$. By the finite case and the fact that countable unions of Lebesgue measurable sets are Lebesgue measurable, we conclude that $E=\bigcup_{n=1}^{\infty}(B_n\cap E)$ is Lebesgue measurable.


问:
What K value minimizes this function
I got a function $$ nK^{\frac{-1}{\beta}}\log{K} + ZK\log{n} $$ where $K$ and $n$ are integers and $Z > 0$.

I followed the typical way of differentiating the function and equating to zero.  But I couldn't find a closed form.  I got the following result

$$
\frac{Z\log{n}}{n} = \frac {\frac{\log{K}}{\beta}-1}{K^{\frac{1+\beta}{\beta}}}
$$

I am not sure how can I go from here.

答:
The last equation you wrote gave me an idea for simplifications.

For simplifications, let $$K=x^{\frac{\beta }{\beta +1}}$$ which makes the function to be
$$f(x)=x^{-\frac{1}{\beta +1}} \left(\frac{\beta  n}{\beta +1} \log (x)+x Z \log(n)\right)$$
$$f'(x)=\frac{\beta  x^{-\frac{\beta +2}{\beta +1}} ((\beta +1) (x Z \log (n)+n)-n \log(x))}{(\beta +1)^2}$$
So, after simplifications, we look for the zero of
$$g(x)=(\beta +1) (x Z \log (n)+n)-n \log (x)$$ The only analytical solution of this equation express in terms of Lambert function
$$x=-\frac{n\,\,W(t)}{(\beta +1) Z \log (n)} \quad \text{where} \qquad t=-\frac{e^{\beta +1} (\beta +1) Z \log (n)}{n}$$ Then $K$ from $x$.

It remains to prove that this solution is a minimum. So,
$$f''(x)=\frac{x^{-\frac{1}{\beta +1}-2} (\beta  (\beta +2) n \log (x)-\beta  (\beta +1)
   ((\beta +3) n+x Z \log (n)))}{(\beta +1)^3}$$ I have not been able to prove it.

Let us try using $n=56789$, $Z=\pi$  and $\beta=\log(10)$. It would give $K=10.4097$. But, this is a maximum.

So, if you look for a minimum value of the function other than $K=0$, we need to use the second branch of Lambert function, that is to say
$$x=-\frac{n\,\,W_{-1}(t)}{(\beta +1) Z \log (n)} \quad \text{where} \qquad t=-\frac{e^{\beta +1} (\beta +1) Z \log (n)}{n}$$ For the worked example, this will give $K=213.69$

For the evaluation of $W(t)$, assuming that $\beta >-1$ (wich makes $t<0$) you can use the series expansion
$$W(t)\approx L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(2L_2^2-9L_2+6)}{6L_1^3}+\cdots$$ where $L_1=\log(-t)$ and $L_2=\log(-L_1)$ .

If you just look for bounds, in the linked page, you will notice that in $2013$ was established the double inequality
$$-1 - \sqrt{2u} - u < W_{-1}\left(-e^{-(u+1)}\right) < -1 - \sqrt{2u} - \tfrac{2}{3}u $$

For the worked example this gives
$208.178 < K <229.578$

Title: What K value minimizes this function


问:
How to prove this inequality $2\cos{x} + \sec^{2}{x} - 3 \ge{0}$
How to prove this inequality in interval: $(0,\frac{\pi}{2})$
$
2\cos{x} + \sec^{2}{x} - 3 \ge{0}
$

I have to use it to show that the differentiation of a function is greater than or equal to 0 so that I can prove that the function is an increasing function.

Edit: Added the interval

答:
We have
$$2 \cos{x} + \sec^{2}{x} - 3 = 2 (\cos x - 1) + \left(\frac{1}{\cos^2 x} - 1\right) = \frac{(2 \cos x+1)(\cos x-1)^2}{\cos^2x}.$$
Thefore the inequality does not hold.


问:
Finding the Fourier Series of $\sin(x)^2\cos(x)^3$
I'm currently struggling at calculation the Fourier series of the given function

$$\sin(x)^2 \cos(x)^3$$

Given Euler's identity, I thought that using the exponential approach would be the easiest way to do it.

What I found was: $$\frac{-1}{32}((\exp(2ix)-2\exp(2ix)+\exp(-2ix))(\exp(3ix)+3\exp(ix)+3\exp(-ix)+\exp(-3ix)))$$

Transforming it back, the result is:

$$ -\frac{1}{18}(\cos(5x)+\cos(3x)+2\cos(x))$$

(I've checked my calculations multiple times, I'm pretty sure it's correct.)

Considering the point $x = 0$ however, one can see that the series I found doesn't match the original function.

Could someone help me find my mistake?

答:
You don't show enough for me to diagnose anything.  I can tell you by using the double-angle and add-multiply formulae, I get

$$\sin^2{x} \cos^3{x} = \frac{1}{16} (2 \cos{x}-\cos{3 x} - \cos{5 x})$$

The way I did this is to rewrite the LHS as

$$\begin{align} \cos^3{x}-\cos^5{x} &= \cos{x} (\cos^2{x}-\cos^4{x})\\&= \frac12 \cos{x} (1+\cos{2 x}) - \frac14 \cos{x} (1+\cos{2 x})^2 \\ &= \frac12 \cos{x}(1+\cos{2 x}) - \frac14 \cos{x}(1+2 \cos{2 x} + \cos^2{2 x}) \\ &=  \frac14 \cos{x} - \frac18 \cos{x} (1+\cos{4 x})\\ &= \frac18 \cos{x} - \frac{1}{16} \cos{3 x} - \frac{1}{16} \cos{5 x}\end{align}$$

The result follows.

ContentLicense: CC BY-SA 3.0


问:
Is a relatively weakly compact subset necessarily norm bounded (by the Banach-Steinhaus theorem)?
Let $E_i$ be a normed vector space (EDIT: Assume $E_2$ is complete), $\langle\;\cdot\;,\;\cdot\;\rangle$ be a duality pairing between $E_1$ and $E_2$ with $$\left|\langle x_1,x_2\rangle\right|\le\left\|x_1\right\|_{E_1}\left\|x_2\right\|_{E_2}\tag1,$$ $\sigma(E_2,E_1)$ denote the topology on $E_2$ generated by the seminorms $$p_{x_1}(x_2):=\left|\langle x_1,x_2\rangle\right|\;\;\;\text{for }(x_1,x_2)\in E_1\times E_2$$ and $\mathcal F\subseteq E_2$ be relatively $\sigma(E_2,E_1)$-compact.

Assume$^1$ $$\left\|x_2\right\|_{E_2}=\sup_{\left\|x_1\right\|_{E_1}\le1}p_{x_1}(x_2)\;\;\;\text{for all }x_2\in E_2\tag2.$$

Are we able to conclude that $\mathcal F$ is norm bounded, i.e. $\sup_{x_2\in\mathcal F}\left\|x_2\right\|_{E_2}<\infty$?

By $(2)$ and the Banach-Steinhaus theorem, it would be sufficient to show that $$\sup_{x_2\in\mathcal F}p_{x_1}(x_2)<\infty\tag3.$$ Can we show this?

$^1$ Do we actually explicitly need to assume this or is $(2)$ guaranteed to hold in the present setting?

答:
If you assume that
$$\tag{1} ||x_2||_{E_2}=\sup_{||x_1||_{E_1}\leq 1}p_{x_1}(x_2)\ \text{ for all } x_2\in E_2$$
and $\mathcal{F}\subseteqq E_2$ is an $\sigma(E_2,E_1)-$compact subset of $E_2$ then indeed $\mathcal{F}$ is bounded with respect to the $||.||_2$ norm. This follows from the uniform boundedness theorem applied to the space $E_1$ ( here we need to assume that $E_1$ is a Banach space instead of only being a normed space ). Indeed, consider the linear functionals $f_{x_2}:E_1\to \mathbb{R}$ where
$$f_{x_2}(x_1)=\langle x_1,x_2 \rangle$$
for every $x_2 \in \mathcal{F}$. Then, by the assumption
$$\tag{2}|\langle x_1,x_2\rangle|\leq ||x_1||_{E_1}\cdot ||x_2||_{E_2}$$
$f_{x_2}$'s are continuous. Now, since $\mathcal{F}$ is $\sigma(E_2,E_1)-$compact it follows that for fixed $x_1\in E_1$ the set
$\{f_{x_2}(x_1):\, x_2\in \mathcal{F}\}$
is a bounded subset of $\mathbb{R}$. Hence, by the uniform boundedness theorem it follows that
$$\sup_{x_2\in \mathcal{F}}||f_{x_2}||<\infty$$
But, by $(1)$, $||f_{x_2}||=\sup_{||x_1||\leq 1}p_{x_1}(x_2)=||x_2||$. Hence, $\sup_{x_2\in \mathcal{F}}||x_2||<\infty$.

Now if you dont have that $(1)$ holds necessarily then the conclusion it is not true in general. An example is the following: Consider $E_1=E_2=\ell_2$ and for every $x,y\in \ell_2$ the duality pairing $\langle\;\cdot\;,\;\cdot\;\rangle_{\ell_2'} :\ell_2 \times \ell_2\to \mathbb{R}$ given by
$$\langle x,y\rangle_{\ell_2'}=\sum_{n=1}^{\infty}\frac{1}{n}x_n y_n$$
$(2)$ follows by the Cauchy-Schwarz inequality. Now, let $\mathcal{F}=\{ne_n:\, n\in \mathbb{N}\}\cup \{0\}$ where $e_n$ is the standard basis of $\ell_2$. Obviously, $\mathcal{F}$ is not bounded in $\ell_2$. Now for every $x\in \ell_2$ let $f_x:\ell_2\to \mathbb{R}$ given by
$$f_x(y)=\sum_{n=1}^{\infty}\frac{1}{n}x_ny_n$$
Then, the topology $\sigma(E_2,E_1)$ is the coarsest topology for which the functionals $f_x$'s are continuous. We claim that $\mathcal{F}$ is compact with respect to $\sigma(E_2,E_1)$. First observe that
$$\tag{3} f_x(\mathcal{F})=\{x_n:\, n\in \mathbb{N}\}\cup\{0\}$$
for every $x=(x_n)_{n=1}^{\infty}\in \ell_2$. Since, $x_n\to 0$ it follows that $f_x(\mathcal{F})$ is a compact subset of $\mathbb{R}$. Let $(W_i)_{i\in I}$ be an $\sigma(E_2,E_1)$-open cover of $\mathcal{F}$. Let $i_0$ such that $0\in W_{i_0}$. There is an $\epsilon>0$ and a finite set $\{x_1,...,x_m\}$ such that $0\in W'\subseteq W_{i_0}$ where
$$W'=\{y\in E_2:\, |f_{x_i}(y)|<\epsilon,\,\  i=1,...,m\}$$
By the description of $(3)$, if we fix an $N$ such that $|x_i(n)|<\epsilon$ for every $i=1,...,m$ and $n\geq N$ we obtain the inclusion
$$\{ne_n:\ n\geq N\}\cup\{0\}\subseteq W'$$
Since, the finite set $\{ne_n:\, 1\leq n\leq N\}$ can be covered by finitely many $W_i$'s it follows that $\mathcal{F}$ can be covered by finitely many $W_i$'s. Hence, $\mathcal{F}$ is an $\sigma(E_2,E_2)-$compact subset of $E_2$ which is unbounded in the $||.||_{\ell_2}$ norm.


问:
Wolframalpha and emathelp show different answers for limits
I have an infinite series of

$$
\sum_{n = 1}^{\infty}
\left[1-\cos\left(\frac{\pi}{n}\right)\right]
$$

to test if it is convergent or divergent.

My attempt:

I first split the infinite series into:

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

And then performed a divergence test on $\sum_{n=1}^\infty \left(1\right)$ which gives me:

$$\lim_{n\to \infty}1=1$$

Since the limit is $1$ $(\neq 0)$, hence $\sum_{n=1}^\infty \left(1-\cos\frac{\pi}{n}\right)$ is divergent too.

However, my TA told me that the infinite series is convergent. And when I used wolframalpha (here) to solve for the infinite series, it shows convergence. But when I used emathhelp (here), it shows divergent.

Question:

So is my TA and wolframalpha correct? Or am I or emathelp correct?

答: You cannot split the series into difference of two series since each of them is divergent. Use the fact that $0 \leq 1-\cos x \leq \frac {x^{2}} 2$ and use comparison test.


问:
Are these two expressions equal? Can this limit be exchanged for a derivative?
I'm trying to solve an excercise. If I prove the equation below, I will be able to finish the rest of the exercise pretty quickly.

We know that:
$$
  \lim_{x \to 0} f(x) = 0
$$
$$
  \lim_{x \to 0} \frac{f(2x)-f(x)}{x} = 0
$$

Can we then conclude (maybe from the definition of a derivative) that $f'(0)$ exists and is equal $0$?

答: Hint: If $f$ is differentiable at $0$, we need $f$ to be continuous at $0$, i.e. $\displaystyle\lim_{x \to 0}f(x) = f(0)$. However, we're only given $\displaystyle\lim_{x \to 0}f(x) = 0$. What happens if $f(0) \neq 0$? See if you can construct a function which satisfies $\displaystyle\lim_{x \to 0}f(x) = 0$ and $\displaystyle\lim_{x \to 0}\dfrac{f(2x)-f(x)}{x} = 0$, but $f(0) \neq 0$.


答:
By Taylor,

$$1-\cos\frac\pi n=\frac{\pi^2}{2n^2}-\frac{\pi^4}{4!n^4}+\cdots$$

and the individual terms form converging series. Then is not difficult to show that the global series is absolutely converging.

Title: Wolframalpha and emathelp show different answers for limits

Body:
I have an infinite series of

$$
\sum_{n = 1}^{\infty}
\left[1-\cos\left(\frac{\pi}{n}\right)\right]
$$

to test if it is convergent or divergent.

My attempt:

I first split the infinite series into:

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

And then performed a divergence test on $\sum_{n=1}^\infty \left(1\right)$ which gives me:

$$\lim_{n\to \infty}1=1$$

Since the limit is $1$ $(\neq 0)$, hence $\sum_{n=1}^\infty \left(1-\cos\frac{\pi}{n}\right)$ is divergent too.

However, my TA told me that the infinite series is convergent. And when I used wolframalpha (here) to solve for the infinite series, it shows convergence. But when I used emathhelp (here), it shows divergent.

Question:

So is my TA and wolframalpha correct? Or am I or emathelp correct?


问:
How can I solve this matrix?
I have a set of RGB colour values detected by a camera $C_{i_{RGB}}$ which are to be described by the following:

$C_{i_{RGB}} = X F_{i_{rgb}}$

where $F_{i_{rgb}}$ is the component incident light at the detector, where the components $rgb$ are fixed at narrow bands and $X$, an $(n\times 3)$ matrix, is a description of the colour band overlap (cross talk if you will) for specific frequencies.  In a simplified RGB scheme, $X$ would be described:

$\begin{align}
\mathsf X &= \begin{bmatrix}
  1 & R_{GR} & R_{BR} \\
  R_{RG} & 1 & R_{BG} \\
  R_{RB} & R_{G} & 1
\end{bmatrix}\end{align}$

That is, the diagonal entries of $X$ are $1$'s, and the off-diagonals are rational.

How can I solve for the values of $X$ only knowing $C_{i_{RGB}}$? Eventually, given $C_i$ I need to find the corresponding $F_i$ using the constant $X$.

Eventually added colour bands (band passes) will be added which will turn the above relation into something like this (and I fear the case where I have 5, 6, and 7 band passes to deal with):

$\begin{bmatrix}
  C_{i_R} \\
  C_{i_G} \\
  C_{i_B}
\end{bmatrix} = \begin{align}\begin{bmatrix}
  X_{aR} & X_{bR} & X_{cR} & X_{dR} & X_{eR} & X_{fR} \\
  X_{aG} & X_{bG} & X_{cG} & X_{dG} & X_{eG} & X_{fG} \\
  X_{aB} & X_{bB} & X_{cB} & X_{dB} & X_{eB} & X_{fB}
\end{bmatrix}\end{align}\begin{bmatrix}
  F_{i_a} \\
  F_{i_b} \\
  F_{i_c} \\
  F_{i_d} \\
  F_{i_e} \\
  F_{i_f}
\end{bmatrix}$

This is for a graphics programming application and I lament that my first year algebra was 20 years ago.

答:
Okay, as already mentioned the problem as posed is not tractable. However, reading between the lines, it looks to me a lot like Independent Components Analysis could work.

Basically, assuming you have a sequence of such observations from the same sources, and its reasonable to assume that the individual sources are independent of each other, then the original signals can be recovered up to scaling and order.

Here's a link to a nice tutorial paper by Aapo Hyvarinen:
http://www.cs.helsinki.fi/u/ahyvarin/papers/NN00new.pdf

ContentLicense: CC BY-SA 2.5

Title: How can I solve this matrix?


问:
Determine if function is little-o, little-omega or big-theta
Let $f(n) = n^3(5+2\cos(2n))$ and $g(n) = 3n^2+4n^3+5n$.
Given these two functions, I must determine the appropriate symbol where the underscore is:
$f(n) \in \_(g(n))$

So, first thing to do is take the limit: $$\lim_{n\to\infty}\frac{f(n)}{g(n)}=\lim_{n\to\infty}\frac{5n^3 + 2n^3\cos(2n)}{3n^2 + 4n^3 + 5n}$$
As n becomes very large, $3n^2 + 5n$ in the denominator become insignificant. So we get:
$$\lim_{n\to\infty}\frac{5n^3 + 2n^3\cos(2n)}{4n^3}$$
The $\cos$ function will oscillate between $-1$ and $1$. Things get a little sketchy here...I'm not sure if I can do this, but:
$$\lim_{n\to\infty}\frac{5n^3 ± 2n^3}{4n^3}$$
Which makes the limit as $n\to\infty$ in the range of $\frac{3}{4}$ to $\frac{7}{4}$ inclusive. Since the limit range is contained in $[0,\infty)$, then that means $f(n)\in \Theta(g(n))$.
Or, is $f(n) \notin \Theta(g(n))$ because the limit is not set to a single value?

答: Theta notation does not demand the limit will be a single value; it only demands that $C_1f(n)\le g(n)\le C_2 f(n)$ for large values of $n$. You may as well work with $\limsup$ and $\liminf$ here.</p>


问:
Root or zero...which to use when?
This may seem like a very basic question, but:

What exactly is the difference between a root of a polynomial, and a zero?  Of course I realise that they are technically exactly the same thing, but there seem to be subtle rules as to when to use each term, and a couple of times in the past I have been told I am using "root" where I should be using "zero".

Is it generally accepted that one should use "root" in an algebraic context,  and "zero" in a analytic context? If not, when should one use one or the other...and does it really matter?!

答: In general we will find zero of a function root of an equation</p>

Title: Root or zero...which to use when?


答:
Almost as @J.G. answered, for large values of $n$, using Stirlin approximation twice and continuing with Taylor series
$$\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma \left(\frac{n-1}{2}\right)}=\sqrt{\frac n 2}\left(1-\frac{3}{4 n}-\frac{7}{32 n^2}-\frac{9}{128 n^3}+\frac{59}{2048
   n^4}+O\left(\frac{1}{n^5}\right)\right)$$ which shows a relative error lower than $0.1$% as soon as $n>2$, lower than $0.01$% as soon as $n>3$, lower than $0.001$% as soon as $n>5$.

If you wish a very good approximation
$$\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma \left(\frac{n-1}{2}\right)}\sim\sqrt{\frac n 2}\frac{1-\frac{441823}{287784 n}+\frac{119909}{54816 n^2}-\frac{1473029}{877056
   n^3} } {1-\frac{225985}{287784 n}+\frac{697315}{383712 n^2}-\frac{7699031}{18418176
   n^3} }$$

Title: gammas division


问:
second order non linear differential equation
How do I solve the second order differential equation:

$ y'' - 9 y^2 = 0$ with $y(0) = y'(0) = 0 ?$

My attempt is the following:

Any kind of help is appreciated!

答:
$$y''=9y^2$$
Multiply by $2y'$ on both sides, to get
$$2y'y''=18y^2y' \implies \frac{d(y')^2}{dx}=18y^2 \frac{dy}{dx}$$
Integrate both sides w.r.t. $x$ to get
$$\int d(y'^2)= \int 18y^2 dy \implies y'^2=A+6y^3 \implies A=0$$
$$\implies y'=\pm \sqrt{6y^3} \implies \int \frac{dy}{\sqrt{6y^3} } dy=\pm \int dx.$$
$$\implies -\sqrt{\frac{2}{3}}\frac{1}{\sqrt{y}}=\pm x+B$$
the condition $y(0)=0$ is not compactible to give a finite solution.

OP's approach is correct.


问:
Is a Relationship Quadratic?
I have a relationship $y=f(x)$ for which I can obtain data through simulation.

I have good reason to suspect that this relationship is quadratic (rather than, say, exponential), and would like to provide evidence for this.

I was thinking of the following method, and I would like to ask if there is anything wrong with it:

a) Obtain $y$ data for a limited interval of $x$, say $x\in (0, x_a)$.

b) Fit a quadratic function to this data using least squares regression.

c) Obtain data for a larger value of $x$ (say $10x_a$) and check the percentage error that the model leads to when extrapolated to this value.

Does this prove anything?

答: A better method is to plot the points $(\log x,\log y)$ for all data points $(x,y)$. If the relationship is roughly quadratic, $y\approx Cx^2$, then $\log y\approx 2\log x+\log C$, which is a linear equation in $\log x$ and $\log y$. Straight lines are easier to detect in a point plot, and if you get a straight line with slope $2$, then you can be pretty sure your relationship is roughly quadratic.</p>

Title: Is a Relationship Quadratic?


问:
Composing inverse with a function vs manipulating to have inverse in function's expression
Consider a function $f(x)=2x$ with inverse $g(x)= \frac{x}{2}$, now notice that:

$$ f(x) = 2x = 4 \frac{x}{2} = 4g$$

Or,

$$ f(x) = 4g$$

Now it looks very tempting to say:

$$ f(g) =4g \tag{1}$$

The reason being that it's clear that $f$ is directly dependent on $g$, but.. this quantity is much different from composing $f$ with $g$ i.e:

$$ f(g) = x$$

I'm looking for an argument 'in words' as to why replacing the function dependencies as shown in (1) is not allowed.

Here is the source of confusion, consider the pic below

Usually composting is said as we have an input which we feed into a function, this function outputs and we feed that output into another function which gives the final output.

Now the structure of the above operations looks very similar to how the expression:

$$ f(x) = 4g$$

Behaves in the sense that you feed in an $x$, that goes into the $g$, you calculate the corresponding output by multiplying by that with definition of $f$

答:
The first case you provided a relation between the two functions you've defined:

However, if we are to saying 'composing functions' then that means to feed the input to a function, that function works on it and feeds it to another one.

Tl;dr equivalence between rules doesn't imply equivalence of composition of the rules.


问:
Show that $f$ has a removable singularity at $0$
Suppose that $f$ has an isolated singularity at $0$ and satisfies $|f(z)|\leq \frac{1}{\sqrt{|z|}}$ in some punctured neighborhood of $0$. Show that $f$ has a removable singularity at $0$.

I tried defining a function $g =f(z)\sqrt{|z|}$. From here we can see that $$0 \leq |g(z)|=|f(z)\sqrt{|z|}|\leq |f(z)|z|| \leq |\sqrt{|z|}| $$ and by squeeze theorem it is clear that $g(z)$ would go to $0$ and by Riemann $g(z)$ has a removable singularity at $0$. Since $g(z)$ has a removable singularity at $0$ so does $f$ and we are done, or am I missing something here?

答:
You cannot apply Riemann's theorem on removable singularities to $f(z)\sqrt{|z|}$ because that function is not holomorphic in a punctured neighbourhood of $0$.

But you can apply the theorem to $f$ directly: $\lim_{z \to 0} z f(z) = 0$ implies that $f$ has a removable singularity at $z=0$.


答:
What you could do is to switch variable and write
$$\frac {x''}{[x']^3}+9y^2=0$$ Reduction of order $p=x'$ gives
$$\frac {p'}{[p]^3}+9y^2=0\implies p=\pm \frac{1}{\sqrt{6 y^3- c_1}}$$ The condition $y'(0)=0$ makes $c_1=$ and
$$p=\pm \frac{1}{\sqrt{6 y^3}}$$ leads to @Z Ahmed's answer .

Title: second order non linear differential equation

Body:
How do I solve the second order differential equation:

$ y'' - 9 y^2 = 0$ with $y(0) = y'(0) = 0 ?$

My attempt is the following:

Any kind of help is appreciated!


问:
What is the limit of the multidimensional integral?
What is the limit of the integral $$\int_{[0,1]^n}\frac{x_1^5+x_2^5 + \cdots +x_n^5}{x_1^4+x_2^4 + \cdots +x_n^4} \, dx_1 \, dx_2 \cdots dx_n$$  as $n \to \infty ?$</p>

答:
Here is a heuristics: Let $U_i$ be i.i.d. random variables having the uniform distribution on $[0, 1]$. Then the integral is equal to

$$ \Bbb{E} \left[ \frac{U_{1}^{5} + \cdots + U_{n}^{5}}{U_{1}^{4} + \cdots + U_{n}^{4}} \right] $$

Now, from the strong law of large number we find that

$$ \frac{U_{1}^{5} + \cdots + U_{n}^{5}}{U_{1}^{4} + \cdots + U_{n}^{4}} \to \frac{\Bbb{E} U_{1}^{5}}{\Bbb{E} U_{1}^{4}} = \frac{5}{6} \quad \text{a.s.}  $$

Therefore the limit of the integral is $\frac{5}{6}$.

The only unjustified part of this argument is that we intercanged the order of the expectation and the limit. This can be justified by the dominated convergence theorem, together with the observation that

$$ \frac{U_{1}^{5} + \cdots + U_{n}^{5}}{U_{1}^{4} + \cdots + U_{n}^{4}} \leq 1. $$


问:
USAMO $2008/4$ Triangulation of regular n-gon

Let $ \mathcal{P}$ be a convex polygon with $ n$ sides, $ n\ge3$. Any set of $ n - 3$ diagonals of $ \mathcal{P}$ that do not intersect in the interior of the polygon determine a triangulation of $ \mathcal{P}$ into $ n - 2$ triangles. If $ \mathcal{P}$ is regular and there is a triangulation of $ \mathcal{P}$ consisting of only isosceles triangles, find all the possible values of $ n$.

Sol- In this solution ,page number $9$, I have some doubts which i mention below,

a) In the last claim when we are repeating
the argument with the triangles on each half why the number $n − 1$ must be a
power of $2$ ?

$n=17$ (for reference)

thankyou very much

答:
In the last claim when we are repeating
the argument with the triangles on each half why the number $n − 1$ must be a
power of $2$ ?

We shall take a look at small examples. For $n=7$:

For $n=9$:

As we can see from above, the inner lines that connect the outer points(which are paired) must be also paired. That is to say, the number of points $\dfrac{n-1}2$ should be always divisible by $2$ whenever doing $N\to N/2$. Thus, $\dfrac{n-1}{2}$ is a power of $2$ and so is $n-1$.


问:
Showing it is a joint probability density function
I have two random variables $X,Y$ with a joint density function $f_{X,Y}(x,y)=x+y$ if $(x,y)\in[0,1]\times [0,1]$ and otherwise $f_{X,Y}(x,y)=0$

I want to analyze this case in different cases, first of all, I want to show it is a probability density function. Well I think the best way is to calculate it $f_{X,Y}(x,y) = f_{Y\mid X}(y|x)f_X(x) = f_{X\mid Y}(x\mid y)f_Y(y)$ but how can we evaluate the condtional and marginal distributions ?

Another idea is to show $\int_x \int_y f_{X,Y}(x,y) \; dy \; dx= 1.$

In the next step I would like to determine the cumulative distributions $F_X(x), F_Y(y)$

答:
Yes, you should check that
$$\int_0^1\int_0^1(x+y)\,dydx=1\,.$$

By definition, $F_X(t)=P(X<t)$, and this can be expressed by the integral:
$$F_X(t)=\int_0^t\int_0^1 (x+y)\,dydx\,.$$


问:
Proving that $\cos(T) = \sum_{\lambda} \cos(\lambda) P_\lambda$ when $T$ is compact and self-adjoint

Let $H$ be a Hilbert space and let $T: H \to H$ be a compact and self-adjoint operator. Let $$T = \sum_{\lambda}\lambda P_\lambda $$ be its spectral decomposition (which converges in the operator norm), where $P_\lambda$ is the orthogonal projection on the $\lambda$-eigenspace.

Let us define $$\cos(T)x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}T^{2n}x. $$
How can we show that $$\cos(T)x = \sum_{\lambda}\cos(\lambda) P_\lambda x? $$

The proof seems pretty easy to me:
\begin{align}
\cos(T)x &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}T^{2n}x \\
&= \sum_{n=0}^\infty \sum_{\lambda} \frac{(-1)^n}{(2n)!} \lambda^{2n} P_\lambda  x\\
&= \sum_{\lambda} \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \lambda^{2n} P_\lambda x\\
&= \sum_{\lambda} \cos(\lambda) P_\lambda x,
\end{align}
or, denoting by $(\lambda_i)_i$ the sequence of non-zero eigenvalues of $T$, writing this in a more explicit manner:
\begin{align}
\cos(T)x &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}T^{2n}x \\
&= \lim_{k \to \infty} \sum_{n=0}^k \frac{(-1)^n}{(2n)!}T^{2n}x \\
&= \lim_{k \to \infty} \left( \sum_{n=0}^k \frac{(-1)^n}{(2n)!} \cdot \left( \lim_{l \to \infty} \sum_{i=1}^l \lambda_i^{2n} P_{\lambda_i} x \right) \right) \\
&= \lim_{k \to \infty} \left( \lim_{l \to \infty} \left( \sum_{n=0}^k \sum_{i=1}^l  \frac{(-1)^n}{(2n)!} \lambda_i^{2n} P_{\lambda_i} x \right) \right)
\end{align}
However, I run into the following problem: what is the reason we can interchange the two limits in the last equality (if we can even do that)? I believe this is because the power series of $\cos$ converges on every compact subset of $\mathbb{C}$, and $\sigma(T)$ is a compact subset, but I am not sure.

Body:
Observe that $\oplus_{\mu}P_{\mu}$ is the identity.

\begin{split}\sum_{n=0}^{\infty}\sum_{\lambda}\frac{(-1)^{n}}{(2n)!}\lambda^{2n}P_{\lambda}x&=\sum_{\mu}P_{\mu}(\sum_{n=0}^{\infty}\sum_{\lambda}\frac{{(-1)}^{n}}{(2n)!}\lambda^{2n}P_{\lambda}x)\\&=\sum_{\mu}\sum_{n=0}^{\infty}\sum_{\lambda}\frac{{(-1)}^{n}}{(2n)!}\lambda^{2n}P_{\mu}P_{\lambda}x\\&=\sum_{\mu}\sum_{n=0}^{\infty}\frac{{(-1)}^{n}}{(2n)!}\mu^{2n}P_{\mu}x. \end{split}


问:
Please help me with a single variable derivative applied question.
I'm Will. I have a problem with a question.

A man 6 ft tall wants to construct a greenhouse of length L and width 18 ft against the outer wall of his house by building a sloping glass roof of slant height y from the ground to the wall, as shown in the figure below. He considers space in the greenhouse to be usable if he can stand upright without bumping his head. If the cost of building the roof is proportional to y, find the slope of the roof that minimizes the cost per square foot of usable space. Hint: Notice that this amounts to minimizing y/x.

```<code>      /|
     / |
  y /  |
   /   |
  /----|
 / |   |6
--------
18    x
</code>```

Hmmm please don't mind my bad drawing. This is a cross section of the greenhouse.
I have got to the position:

Let h be the height of the cross section.
h/ 18= (h- 6)/ x, using similar triangles.
So, xh= 18(h- 6).
Differentiating both sides with respect to x while h= (y^2- 18^2)^(1/2) and h'= yy'/(y^2- 18^2)^(1/2).I get:
xh'+ h= 18h',

xyy'+ y^2- 18^2= 18yy'.   (1)

At the same time, because I should minimize (y/x), I differentiate it and get (xy'- y)/x^2, and it should be 0.
So y'= y/x.
I plug y'=y/x into equation (1) and I get

x= 9y^2/(y^2- 9). (2)

Finally I'm totally stuck here. I get the solution which is only a '1' from the textbook without any explanations.
When I plug 1 as the slope into the figure, I find that x should be 12.
So, y should be 18*2^(1/2). But this is in contradiction to equation (2).

I'll be so appreciated for your help!

答:
With the hint, we wish to minimize $$\frac{y}{x},$$ given the following constraint $$y^{2} = 18^{2} + (\frac{108}{18-x})^{2}.$$

We use implicit differentiation, namely $$(\frac{y}{x})' = \frac{y'x-y}{x^{2}}$$ and $$(y^{2} = 18^{2} + (\frac{108}{18-x})^{2})'$$ gives $yy' = \frac{108^{2}}{(18-x)^{3}}.$

Now, the denominator of $\frac{y'x-y}{x^{2}}$ is positive (there must be some usable space), so the term vanishes if and only if the numerator vanishes. In other words, we require $y'= \frac{y}{x}.$ With this finding, we make a substitution such that $\frac{108^{2}}{(18-x)^{3}} = y (\frac{y}{x}) = \frac{y^{2}}{x}.$

We substitute the constraint in place of $y^{2}$ on the right hand side such that $\frac{108^{2}}{(18-x)^{3}} = \frac{18^{2} + (\frac{108}{18-x})^{2}}{x}.$ Then, we rewrite the equation as follows $\frac{x}{18-x} = 1 + (\frac{18-x}{6})^{2},$ where some algebraic steps have been omitted.

At this point, we substitute $u = 18-x,$ where $u$ denotes the horizontal length in the unusable part of the greenhouse. Namely, $\frac{18-u}{u} = 1 + (\frac{u}{6})^{2},$ which leads to the following cubic equation $u^{3} + 72u - 648 = 0.$ The only real solution is $u=6,$ which is in accord with John's previously posted solution. Therefore, the slope is equal to $1.$


问:
Why is the answer contradictory here in this double integral?
When we integrate the double integral
$$\int_{0}^{\pi}\int_{0}^{x}x\sin y\; \mathrm{d}y\,\mathrm{d}x$$
we get $2+π^2/2$ as answer. (Please see image)

When we use triangle area formula = $1/2 \cdot \mathrm{base} \cdot \mathrm{height}$, we get $π^2/2$ as answer. Where did the extra 2 come from? as here

答: When we use the triangle area formula, what we get is the area of the triangle, which is equal to$$\int_0^\pi\int_0^x1\,\mathrm dy\,\mathrm dx.$$But what you are computing is$$\int_0^\pi\int_0^xx\sin y\,\mathrm dy\,\mathrm dx.$$


答:
Not only it does converge as explained in comments and answers but you could also have a good approximation of the infinite sum.

Since we now the values for the first angles, write
$$S=\sum _{n=1}^{\infty } \left(1-\cos \left(\frac{\pi }{n}\right)\right)=\sum _{n=1}^{6 } \left(1-\cos \left(\frac{\pi }{n}\right)\right)+\sum _{n=7}^{\infty } \left(1-\cos \left(\frac{\pi }{n}\right)\right)$$ that is to say
$$S=\frac{1}{4} \left(25-2 \sqrt{2}-2 \sqrt{3}-\sqrt{5}\right)+\sum _{n=7}^{\infty } \left(1-\cos \left(\frac{\pi }{n}\right)\right)$$

Now
$$a_n=1-\cos \left(\frac{\pi }{n}\right)=\frac{\pi ^2}{2 n^2}-\frac{\pi ^4}{24 n^4}+\frac{\pi ^6}{720 n^6}-\frac{\pi
   ^8}{40320 n^8}+O\left(\frac{1}{n^{10}}\right)$$

$$\sum_{n=7}^\infty n^{-2k}=\zeta (2 k)-1-2^{-2 k}-3^{-2 k}-4^{-2 k}-5^{-2 k}-6^{-2 k}$$
$$\sum_{n=7}^\infty a_n=-\frac{5369 \pi ^2}{7200}+\frac{39931361 \pi ^4}{311040000}-\frac{47464376609 \pi
   ^6}{33592320000000}-\frac{423805258161097 \pi ^8}{967458816000000000}+\frac{\pi
   ^{12}}{680400}-\frac{\pi ^{16}}{381024000}$$

So

$$S\approx 4.87071896177$$ while the value computed with unlimited precision is
$S=4.87071896189\cdots$

These numbers are not recognized by inverse symbolic calculators.


问:
Topology generated by $(X,d_1)$ is contained in Topology genereted by $(X,d_2)$. What can be deduced from this?
If $\tau_1 \subset \tau_2$ where $\tau_1$ and $\tau_2$ generated by metric $d_1$ and $d_2$ respectively. Then what can be said about the metric spaces.

My Observations are as follows:

As $\tau_1 \subset \tau_2$ $\implies$ Every open ball in $(X,d_1)$ contains an
open ball from $(X,d_2) \implies (x_n)$ convergent in $(X,d_2)$ is also convergent in $(X,d_1)$ $\implies$ $id:
   (X,d_2)\rightarrow (X,d_1)$ is continuous.

My Questions are:

I think  $id^{-1}:(X,d_1)\rightarrow(X,d_2)$ is not continuous. (i.e there are
sequences in $(X,d_1)$ which is not convergent in $(X,d_2)$.)But I cannot
think of any counterexample.

A few nice examples of metric spaces with the above property.

What other information can be drawn from this given assumption?

答:
Indeed, a sequence/net converging in $\tau_2$ converges in $\tau_1$ as well. That's a straight forward application of the definition of convergence.

Neither the identity nor its inverse has to be continuous. Consider any topological (metric) space $(X,\tau)$ and let $\tau'$ be the discrete topology (which is metrical as well). Then $id:(X,\tau')\to (X,\tau)$ is continuous but its inverse $id^{-1}$ is continuous if and only if $\tau$ is discrete. This should give you plenty examples, e.g. $\mathbb{R}$ with the Euclidean and discrete metrics.

Not much can be said beyond that. Some axioms of separation are preserved under $\tau_1\subseteq \tau_2$ condition, but you are dealing with metric spaces to begin with.


问:
Geometric intuition for Properties of the Orthogonal Projection

Properties of the Orthogonal Projection: (Firm) Nonexpansivness

Let $C$ be a nonempty closed convex set Then:

1. For any $v,w\in\mathbb{R}^n$:

$\begin{aligned}
  (P_C(v)-P_C(w))^T(v-w)\geq||P_C(v)-P_C(w)||^2
 \end{aligned}$

2.(Non-expansiveness) for any $v,w\in\mathbb{R}^n$:

$\begin{aligned}
  ||P_C(v)-P_C(w)||\leq||v-w||
 \end{aligned}$

I've learnt this theorem in convex analysis course and I would like to get some geometric intuition for both of them

答:
Orthogonal projection in ${\mathbb R}^n$ can be thought of as shining a light onto an object and looking at the shadow it casts on the far wall.  That oversimplifies a little too much, because the "far wall" is a plane so we're projecting down into at most two dimensions, but it's not a bad image to start with.

When we project we are reducing the number of dimensions we are considering (this is sometimes referred to as 'setting co-ordinates to zero') in order to get a feel for the macroscopic properties of the object.  Your second point expresses this: when we reduce the number of dimensions that we consider an object in, we are losing part of the object and so the end-result cannot grow, it can stay the same or shrink.  (Mathematically: if $v = (v_1, v_2, \ldots ,v_n)\in {\mathbb R}^n$ is our vector and $Pv = (w_1, w_2, \ldots ,w_n)$ is an orthogonal projection then we have $|w_n| \leq |v_n|$ for each $n \in {\mathbb N}$ so $v$ is "smaller" in some sense than $Pv$).

Your first point expresses the concept that the orthogonal projection preserves orientation: whatever direction the vector $v-w$ points (i.e. what sign it has), the projection $Pv-Pw$ points in the same direction (has the same sign).


问:
How to show that $\lambda(D)=\infty$?

I have a bit trouble showing that $\lambda(D)=\infty$.

I have been giving the following problem:

Given $D=\{(x,y)\in \mathbb{R}^2 | 0<x=y<1\}$ and $\mathcal{N}=\{N \in \mathcal{B} | m_I(N)=0\}$ (where $m_I$ is the Lebesgue measure on $(\mathbb{R}, \mathcal{B})$ restricted on the interval $(0,1)$) - and the measure $\lambda:\mathcal{B}\otimes \mathcal{B} \to [0,\infty]$, where
$$\lambda(G)=
    \begin{cases}
        0 & \text{ if }\, \exists N_1,N_2 \in \mathcal{N}, B_1,B_2 \in \mathcal{B}, \text{ then } G \subseteq (B_1 \times N_1) \cup (N_2 \times B_2)  \\
        \infty & \text{ or else }\\
    \end{cases}
$$
Show that $\lambda(D)=\infty$.

I get that I might have to show it by contradiction, but I have only gotten so far:

Assume by contradiction, that $\lambda(D)\neq \infty \Longrightarrow \lambda(D)=0$, then $\exists N_1, N_2 \in \mathcal{N}$ and $B_1,B_2 \in \mathcal{B}$, where $D\subseteq (B_1 \times N_1) \cup (N_2 \times B_2)$.

For $D\subseteq (N_2\times B_2)$, I have that for $N_2\in \mathcal{N}$ it applies that $m_I(N_2)=0$.

Can somebody give me a hint, on how to get moving? I assume that for $D \subseteq (B_1 \times N_1)$, it follows analogously. Thank you.

答:
Assume that $D \subseteq (B_1 \times N_1) \cup (N_2 \times B_2)$. We want to find a contradiction.

Note that you can write the elements of $D$ in this way
$$D= \{ (t,t) : t \in (0,1) \}$$

Define $$I_1 = \{ t \in (0,1) : (t,t) \in (B_1 \times N_1) \}$$
and analogously
$$I_2 = \{ t \in (0,1) : (t,t) \in (N_2 \times B_2) \}$$
By our assumption you have that $(0,1) \subseteq I_1 \cup I_2$.

On the other hand, if you look carefully at the definition of $I_1$ and $I_2$ you can see that
$$I_1 \subseteq N_1 \qquad I_2 \subseteq N_2$$
To show this, let $t \in I_1$. Then $(t,t) \in B_1 \times N_1$. This means that $t \in B_1$ and $t \in N_1$. In particular this implies $t \in N_1$. By arbitrarity of $t \in I_1$, we have $I_1 \subseteq N_1$ (actually we can say more: $I_1 = B_1 \cap N_1$). The proof of $I_2 \subseteq N_2$ is similar.

Hence
$$(0,1) \subseteq N_1 \cup N_2$$
This contradicts monotonicity of the Lebesgue measure, since $N_1 \cup N_2$ has measure zero.

Title: How to show that $\lambda(D)=\infty$?


问:
find $\int_{0}^{1} x^2 e^x$ using trapezoidal rule with error less than $10^{-3}$
The question is to find $$\int_{0}^{1} x^2 e^x$$ using trapezoidal rule with error less than $10^{-3}$

Which is another way to say :”Find h” in the first place.
So we know that :

$f(x)=x^2 e^x$

$f’(x)=2xe^x+x^2e^x$

$f’’(x)=2e^x+4xe^x+x^2e^x= e^x(2+4x+x^2)$

So :
$| f’’(x)|=| 2e^x+4xe^x+x^2e^x | \le 7e$
So the upper bound (M)would be 7e

Using $\frac{b-a}{12} h^2 M\le 10^3$
I have to find h and then use trapezoidal rule to find T(h)
But the problem is h would be really small and using trapezoidal rule would be too long to do without using mathematical programs

Any help ?

答:
You were on the right way: just use the error formula to get

$$h^2 \leq \sqrt{\frac{12 \text{Tol}}{7e}}$$

which yields for $\text{Tol} = 10^{-3}$:

$h \approx 0.025$ and hence $n \approx 40$. All in all, just set $n>40$ and you'll observe an absolute error less that $\text{Tol}$ in your simulation.

The following C++ snippet shows you this fact

```<code>#include <iostream>
#include <cmath>
#include <iomanip>      // std::setprecision

double trapz(double f(double),int n, int a, int b){

    double sum{f(a) + f(b)};
    double h = double(b-a)/double(n);
    for (int i=1; i<n; ++i) {
        sum+=2*f(a+i*h);
//        std::cout << a+i*h <<std::endl; bound checking
    }
    return (0.5*h)*sum;
}

double f(double x){

    return x*x*exp(x);
}

int main(){
    double exact{exp(1)- 2};
    int n;
    std::cout <<"Set n: " <<std::endl;
    std::cin >> n;

    double res{trapz(f,n,0,1)};
    std::cout << "Numerical integration with n= " << n << " yields: "<< std::setprecision(9) << res
    << " with an error of " << std::scientific <<  std::fabs(exact - res) << std::endl;

    return 0;

}
</code>```

If you're not familiar with C/C++, just go to your terminal (in MacOS and Linux it's the same) and type the following in order to compile your source code

```<code>g++ -o file_name -std=c++14 file_name.cpp
</code>```

Then run it with

```<code>./file_name
</code>```

to see the output

With $n=40$ I get:

```<code>Numerical integration with n= 40 yields: 0.718706544 with an error of 4.247156195e-04
</code>```

Body:
You were on the right way: just use the error formula to get

$$h^2 \leq \sqrt{\frac{12 \text{Tol}}{7e}}$$

which yields for $\text{Tol} = 10^{-3}$:

$h \approx 0.025$ and hence $n \approx 40$. All in all, just set $n>40$ and you'll observe an absolute error less that $\text{Tol}$ in your simulation.

The following C++ snippet shows you this fact

#include <iostream>
#include <cmath>
#include <iomanip>      // std::setprecision

double trapz(double f(double),int n, int a, int b){

    double sum{f(a) + f(b)};
    double h = double(b-a)/double(n);
    for (int i=1; i<n; ++i) {
        sum+=2*f(a+i*h);
//        std::cout << a+i*h <<std::endl; bound checking
    }
    return (0.5*h)*sum;
}

double f(double x){

    return x*x*exp(x);
}

int main(){
    double exact{exp(1)- 2};
    int n;
    std::cout <<"Set n: " <<std::endl;
    std::cin >> n;

    double res{trapz(f,n,0,1)};
    std::cout << "Numerical integration with n= " << n << " yields: "<< std::setprecision(9) << res
    << " with an error of " << std::scientific <<  std::fabs(exact - res) << std::endl;

    return 0;

}

If you're not familiar with C/C++, just go to your terminal (in MacOS and Linux it's the same) and type the following in order to compile your source code

g++ -o file_name -std=c++14 file_name.cpp

Then run it with

./file_name

to see the output

With $n=40$ I get:

Numerical integration with n= 40 yields: 0.718706544 with an error of 4.247156195e-04


问:
Simplifying $\sum_{k=0}^b{n+k \choose n+k-a}$
Let $a,b,n$ positive integers such that $a \leq n$. Can we simplify this formula?

$\sum_{k=0}^b{n+k \choose n+k-a}={ n \choose n-a } + { n+1 \choose n+1-a } + \dots + {n+b \choose n+b-a}$

I was stuck at the beginning except for the thought to use the hockey-stick identity somehow like mentioned in the comments.

Body:
$$S=\sum_{k=0}^{b} {n+k \choose n+k-a}=\sum_{k=0}^{b} {n+k \choose a}=[x^a]\sum_{k=0}^{b} (1+x)^{n+k}$$
$$S=[x^a] (1+x)^n \frac{(1+x)^{b+1}-1}{1+x-1}=[x^{a+1}]~((1+x)^{n+b+1}-(1+x)^n).$$
$$S={n+b+1 \choose a+1}-{n \choose 1+a}.$$
Here $[x^k]g(x)$ means the coefficient of $x^k$ in $g(x).$


问:
Using Fourier series techniques to solve $x'' + 3x = 7$ with $x'(0) = x'(5) = 0$
$$x'' + 3x= 7$$

Given conditions $x'(0)=x'(5)=0$.

I checked the list and I went through three books. I am doing intro to differential equations. I just don't know how to get the extensions... I was told if there is a derivative use $\cos$, but I can't hack it. If there is any simple explanation. Please feel free to expound. This is to simply find the formal solution of the said ODE using Fourier Series.
What do I substitute at right hand side? Since I think left hand is simply summation $\cos n \pi/L$.

......I just know that $L=5$? And $7$ is $f(t)$ or $f(x)$? Maybe?

Thanks.

答:
I will use the idea of @Mr.G and consider the equation for $y$. The equation is
$$
y''+3y=0, y'(0)=y'(5)=0.
$$
Obviously $y=0$ is a solution. Thus $x=7/3$ is a solution of the original problem.

The idea behind all this is that you can look for solutions which are trigonometric polinomials
$$
y_k(t)=a_k \cos(2\pi kt/5)+b_k\sin(2\pi kt/5).
$$
If you insert this ansatz into the boundary conditions you get
$$
y'_k(t)=0\Rightarrow b_k=0
$$
Thus, Neumann BC gives you cosines. As the ODE is linear you can have linear combination of it
$$
y(t)=\sum a_k \cos(2\pi kt/5).
$$
$$
\sum \left(-\left(2\pi k/5\right)^2+3\right)a_k\cos(2\pi kt/5)=0
$$
Now you can multply by $a_k\cos(2\pi kt/5)$ and integrate. You have a series with nonnegative terms equals to zero. You conclude that $a_k$ should be zero.

The Theorem behind this is that $\{e^{i\lambda t}\}$ form a basis of the Hilbert space
$$
L^2([0,5])=\{f| \int_0^5 |f(x)|^2 dx<\infty\}
$$
You can read the great book by Prof. Walter Strauss 'An introduction to Partial Differential Equations'.

I hope that this helps you


问:
Assume $G \cong A \times B $ and also $G\cong A \times C$ ,such that $A,B,C$ are normal subgroups in $G$,show that $B\cong C$ .
Assume $G \cong A \times B $ and also $G\cong A \times C$ ,such that $A,B,C$ are normal subgroups in $G$,show that $B\cong C$ .

I think it's a famous theorem,however I could not find any proof of that ,it would be highly appreciated if someone gives me a source for the proof or prove that here.

答:
Are you asking about the cancellation property for the direct product of groups? I.e. when $A\oplus B\simeq A\oplus C$ implies $B\simeq C$? It is famously known to be false, e.g. for any nontrivial group $G$ we have $$\bigg(\bigoplus_{i=1}^\infty G\bigg)\oplus G\simeq\bigg(\bigoplus_{i=1}^\infty G\bigg)\oplus 0$$ even though $G\not\simeq 0$ by assumption.

However it does hold in some cases, e.g. when $A$ is finite. You may want to read this classical paper: R. Hirshon "On Cancellation in Groups".


问:
second order non linear differential equation
How do I solve the second order differential equation:

$ y'' - 9 y^2 = 0$ with $y(0) = y'(0) = 0 ?$

My attempt is the following:

Any kind of help is appreciated!

答:
You have an autonomous equation or system, if written as first order system, $$y'=v, ~~ v'=9y^2,$$ that starts in an equilibrium position $y(0)=0$ with zero velocity $v(0)=0$. Thus the constant function with the equilibrium as value $y(t)\equiv0$ is a solution.

As the equation or right side of the system is continuously differentiable, it is also locally Lipschitz, enforcing uniqueness of the solution. Thus there is no other solution than $$y(t)=0~~~\forall t\in\Bbb R.$$


问:
Finding primitive element of field extension.
I wish to find the primitive element of $\mathbb{Q}(\sqrt[3]2,e^{\frac{2\pi i}{3}})$ over $\mathbb{Q}$. According to the proof of the theorem of primitive element. I suspect it is $\sqrt[3]2+e^{\frac{2\pi i}{3}}$ but I do not know how to show $\sqrt[3]2\in\mathbb{Q}(\sqrt[3]2+e^{\frac{2\pi i}{3}})$

Is there any hint to this?

答: The field $K=\mathbf Q(\sqrt [3]2,j)$, where $j$ is a primitive 3rd root of $1$, is obviously the field of decomposition over $\mathbf Q$ of the polynomial $X^3 - 2$, so the extension $K/\mathbf Q$ is galois, say with Galois group $G$. As $\mathbf Q(\sqrt [3]2)/\mathbf Q$ is not normal, $G$ is non abelian. Since $K$ is the compositum of the linearly disjoint extensions $\mathbf Q(\sqrt [3]2)$ and $\mathbf Q(j)$, $G$ has order $6$, hence $G \cong S_3$, which is (up to isomorphism) the unique non abelian group of order $6$, consisting of Id,  the three transpositions fixing the three cubic subfields $\mathbf Q(\sqrt [3]2), \mathbf Q(j\sqrt [3]2), \mathbf Q(j^2\sqrt [3]2)$, and the two 3-cycles generating the cyclic subgroup of order $3$ which fixes the quadratic subfield $\mathbf Q(j)$. Summarizing, $K$ contains only five strict subextensions, and it is straightforward to check that $j+\sqrt [3]2$ does not belong to any of these. Consequently, $j+\sqrt [3]2$ is a primitive element of $K/\mathbf Q$.


问:
Elementary proof of existence of integral kernel
I know the following statement is true.

Let $T \in B(L^1(\mathbb{R}^d), L^\infty(\mathbb{R}^d))$, where $B(X, Y)$ denotes all bounded linear operators from $X$ to $Y$.
Then, $T$ has the integral kernel $k(x,y) \in L^\infty(\mathbb{R}^d \times \mathbb{R}^d)$,
that is, for all $f \in L^1(\mathbb{R}^d)$ and for almost all $x \in \mathbb{R}^d$,
\begin{equation}
Tf(x) = \int_{\mathbb{R}^d} k(x, y)f(y) dy.
\end{equation}
Furthermore, $\|T\|_{B(L^1, L^\infty)} = \|k\|_{L^\infty(\mathbb{R}^{2d})}$.

This is a corollary of Theorem 2.2.5 of the following article:

N. Dunford and B. J. Pettis "Linear operations on summable functions", Trans. Amer. Math. Soc. 47, (1940), 323–392.
https://www.jstor.org/stable/1989960?seq=1#metadata_info_tab_contents

However, this article is written in a very abstract style. I would like to know more elementary proof.

答:
Hint Consider
\begin{equation}
\phi(g\otimes f) = \int_{{\mathbb R}^d} g(x) (Tf)(x) dx\in {\mathbb R}
\end{equation}
then extend $\phi$ by continuity (and linearity) to a continuous linear functional on $L^1({{\mathbb R}^d}\times{{\mathbb R}^d})$. Then use that the dual of $L^1({{\mathbb R}^d}\times{{\mathbb R}^d})$ is $L^{\infty}({{\mathbb R}^d}\times{{\mathbb R}^d})$.


问:
How decomposes an $R[x]/(x^n)$-module which is free of finite rank over $R$?
Let $R$ be a commutative ring (with unit) and let $R'$ be the $R$-algebra $R[x]/(x^n)$. Let $r$ be a positive integer. Let $V$ be a finitely generated $R'$-module which is free of finite rank over $R$.

Does $V$ decomposes as a direct sum $\bigoplus_{i=1}^s(x^{e_i})/(x^n)$ for integers $0\leq e_0 \leq e_1\leq ...\leq e_s<n$?
If it exists, is this decomposition unique?

I believe that, if true, one shall use $x$ as a nilpotent $R$-linear morphism on $V$, but I am unable to complete the argument.

Many thanks!

答:
Even the modified question is not true. Here is a counterexample: let $k$ be a field and consider $R = k[\varepsilon]$ with the relation $\varepsilon^2=0$. Let $R' = R[x]/(x^2)$, and consider the $R'$-module $V = R$, where $x$ acts by multiplication with $\varepsilon$.

If $V\cong \bigoplus_{i=1}^s(x^{e_i})/(x^2)$, for $0\le e_i \le 1$, then necessarily $s=1$ and $e_1 = 1$, because $V$ has rank $1$ as an $R$-module. But $x$ acts trivially on $(x)/(x^2)$, whereas it acts non-trivially on $V$. Hence $V$ does not admit such a decomposition.


问:
How can you use a (fair) coin to draw straws among 3 people? (Information Theory)
The following nice riddle is a quote from the excellent, free-to-download book: Information Theory, Inference, and Learning Algorithms, written by David J.C. MacKay.

How can you use a (fair) coin to draw straws among 3 people?

答:
Let $p_1, p_2, \ldots, p_n$ be probabilities adding up to $1$. Each $p_k$ represents the desired probability for the $k$th straw to be drawn.

Let

$q_0 = 0$

$q_1 = p_1$

$q_2 = p_1 + p_2$

$q_3 = p_1 + p_2 + p_3$

$\cdots$

$q_n = 1$

Consider the intervals $I_1 = (0, q_1), I_2=(q_1, q_2), \ldots, I_n=(q_{n-1},q_n)$ of the real line. We are going to produce a uniformly random real number between $0$ and $1$ and see which interval it lies in. The probability that it lies in the $k$th interval will then be $p_k$.

We will construct our number by flipping a coin repeatedly to form the bits of a binary fraction. In practice, we will stop as soon as we can determine which interval it lies in, but for simplicity we will deal with the entire sequence at once.

Let $r_1,r_2,\ldots$ be the results of the coin flips, where 1 represents heads and 0 represents tails.

Let $r = 0.r_1r_2\ldots$, interpreted as a binary fraction.

We will assume without proof that this procedure produces a uniform random distribution.

We will consider the bits of $r$ one at a time (as the coin is flipped).

Looking at the first $j$ bits of $r$ will allow us to see that $r$ lies in a certain interval $K_j$ of length $2^{-j}$: each additional bit halves the length of the interval. For example, if the first bit is $1$, then we know that $r \in [0.1,1]$. If the first bit is $1$ and the second is $0$, then we know that $r \in [0.1,0.11]$.

Note that once $2^{-j} < \min \bigl\{p_k: 1 \le k \le n\bigr\}$, or, equivalently, $j > \log_2 \bigl(1 \big/ \min \{p_k: 1 \le k \le n\}\bigr)$, $K_j$ will intersect at most two intervals $I_m, I_{m+1}$.

At this point, the only remaining question is whether $r < q_m$, $r > q_m$, or $r = q_m$. Since the case of $r = q_m$ has probability $0$, we will assume that it does not occur.

Then by the time $2^{-j} < \big|r - q_m\big|$, $K_j$ will lie entirely to the left or entirely to the right of $q_m$, and $r$ will have been determined to lie in $I_m$ or $I_{m+1}$.

The closer $r$ is to $q_m$, the longer the procedure will take, but as long as they are not actually equal, it will eventually terminate.

A somewhat more practical description:

Represent $q_0, q_1 \ldots, q_n$ in binary:

$q_0 = 0.000\ldots$, $q_1 = 0.b_{11}b_{12}\ldots, q_2 = 0.b_{21}b_{22}\ldots$.

Flip a fair coin. If the coin comes up $0$, then our number is in the interval $(0, 0.1)$, so we discard from consideration any interval whose left endpoint has a first bit of $1$. That is, if $b_{k-1,1} = 1$, then we discard $I_k$. If the first coin comes up $1$, we discard from consideration any interval whose right endpoint has a first bit of $0$. That is, if $b_{k,1} = 0$, we discard $I_k$.

Flip the coin again, this time focusing on the second bit of each endpoint.

Keep flipping until only one interval is left.

Note: there is probably a more efficient/effective to describe this procedure, but I don't remember where I read of it.


问:
Basis dependence of sectional curvatures of $\mathfrak{su}(2)$
I am a little confused about an explicit calculation of sectional curvatures. Namely, I wanted to compute the sectional curvatures of the $SU(2)$ group. The sectional curvatures of Lie groups are known to be completely described by their values in the tangent space at the identity, i.e. the Lie algebra $\mathfrak{su}(2)$ in this case. I am working with the equations in this famous paper by [Milnor][1]. There, we can see that the sectional curvatures are determined once we have given: 1) an orthonormal basis $\{e_1,e_2,\cdots,e_n\}$ for the Lie algebra, 2) the algebra itself $$[e_i,e_j]=\sum_k\alpha_{ijk}\,e_k$$ with the structure constants $\alpha_{ijk}$, and 3) a choice of inner product on the algebra $<\cdot,\cdot>$ (this is actually necessary to prove orthonormality of the basis of course). With these ingredients, the sectional curvature $\kappa(e_i,e_j)$ of the tangential 2-plane spanned by the vectors $e_i$ and $e_j$ is given by: $$\kappa(e_i,e_j)=\sum_k\left(\frac{1}{2}\alpha_{ijk}(-\alpha_{ijk}+\alpha_{jki}+\alpha_{kij})-\frac
{1}{4}(\alpha_{ijk}-\alpha_{jki}+\alpha_{kij})(\alpha_{ijk}+\alpha_{jki}-\alpha_{kij})-\alpha_{kii}\alpha_{kjj}\right)$$.

For this case, I can take the standard Pauli matrices in their Hermitian OR anti-Hermitian version (see Wikipedia for their definition) as a basis. They are both equally good orthonormal bases for the $\mathfrak{su}(2)$ algebra w.r.t. to the canonical inner product given by the scaled Killing form, $\langle A,B\rangle=\frac{1}{2}Tr(AB)$. They, however, differ in their structure constants. $\alpha_{ijk}=2i\,\epsilon_{ijk}$ for the Hermitian basis and $\alpha_{ijk}=\epsilon_{ijk}$ for the anti-Hermitian (note the extra factor of i).

My question is:
My understanding was that the sectional curvature is independent of the choice of basis in the tangent plane, and for $SU(2)$ I expect only positive sectional curvatures since it is topologically a 3-sphere (and because I am using the Killing form as an inner product). However, plugging in the above mentioned structure constants yields results with opposite signs, comming from that extra "i" in the structure constants. How can this be the case if both bases are valid?

Edit: The equation for $\kappa$ is given by Milnor only for the case $\kappa(e_1,e_2)$. I simply generalized it for all the indices, but I find it suspicious that he did not do that. Is there some detail that prevents the straighforward generalization?
[1]: https://core.ac.uk/download/pdf/82428733.pdf

答:
The sectional curvature of a Riemannian manifold at a point $p$ is a function $K(X,Y)$ of couples of linearly independent vectors in $T_pM$ which happens to be dependent only on the two dimensional subspace generated by them, so it's really a function of planes in $T_pM$. In particular it's an object defined intrinsically but which, if you calculate it in a basis, depends on which basis you calculate it.

The tangent space of $SU(2)$ at the identity consists of antihermitian matrices, so the sectional curvatures computed with them is the legitimate sectional curvature.

If you calculate the sectional curvature on the hermitian matrices, that are outside the tangent space (they belong to its complexification), so that is not a sectional curvature.


问:
Need to find sample distribution with population distribution p given and sample size n given
A random sample of $100$ engineering students was chosen from Univesiti Teknologi Malaysia City Campus, Kuala Lumpur. $40\%$ of the students in the university work part time. The university authority wishes to know the following:

i) What percentage of the engineering students in the sample can be expected to have part time jobs?

ii) The probability of observing a sample proportion that is greater than $0.5$.

The answers for i) $0.4$ & ii) $0.0162$

For question 1 I'm not sure how to get the answer as in my view the confidence integers are not give and can't really be done while for question 2 the answer i got is $0.0207$ which is different than the given answers. Anyone can help explain? I'd appreciate any help.

答:
For i) the given answer is correct and it is simply due to the fact that

$$\mathbb{E}[\overline{X}_n]=\mu=40\%$$

For ii), using the continuity correction factor, you get

$$\mathbb{P}[\Sigma_i X_i>50]=1-\Phi\Big[\frac{50.5-40}{\sqrt{0.4\cdot0.6\cdot100}}\Big]=1-\Phi[2.14]\approx1.62\%$$

Your result is not wrong; simply you did not applied the continuity correction factor when approximating your discrete rv with a Gaussian.

Both results are an approximation of the true result.


答:
Hint : Make  use of the simple Pascal's identity, i.e. : $~\dbinom{n}{k} + \dbinom{n}{k+1} = \dbinom{n+1}{k+1}$

\begin{align*}
\sum_{k=0}^b \dbinom{n+k}{n+k-a}  &= \sum_{k=0}^b \dbinom{n+k}{a}\\
&= \dbinom{n}{a+1} + \dbinom{n}{a} + \dbinom{n+1}{a} + \cdots + \dbinom{n+b}{a} - \dbinom{n}{a+1}\\
&= \dbinom{n+1}{a+1} + \dbinom{n+1}{a} + \dbinom{n+2}{a} + \cdots + \dbinom{n+b}{a} - \dbinom{n}{a+1}\\
&\hspace{1.4cm}\vdots\hspace{2.2cm}\vdots\\
&= \dbinom{n+b}{a+1} + \dbinom{n+b}{a} - \dbinom{n}{a+1}\\
&= \dbinom{n+b+1}{a+1} - \dbinom{n}{a+1}
\end{align*}

Hope it helps.

Title: Simplifying $\sum_{k=0}^b{n+k \choose n+k-a}$

Body:
Let $a,b,n$ positive integers such that $a \leq n$. Can we simplify this formula?

$\sum_{k=0}^b{n+k \choose n+k-a}={ n \choose n-a } + { n+1 \choose n+1-a } + \dots + {n+b \choose n+b-a}$

I was stuck at the beginning except for the thought to use the hockey-stick identity somehow like mentioned in the comments.


问:
What is the meaning of the sequence e with k superscript and n subscript vertically aligned? How does it look like given its definition?
I am stuck on an exercise in my linear algebra textbook--I will not provide the source as it is an Italian textbook used at my university. A solution to the problem is provided, but I am not able to understand it as I do not understand what a sequence like the one given should look like in the first place, and also what the writing means.

Here's the problem.

Consider the set $\{e^k\}_{k\in\Bbb N}$ where for each $k\in\Bbb N$ the sequence  $e^k$ is such that $$e_n^k=\{\begin{cases} 1\; \text{if}\; n=k\\ 0\; \text{if}\; n\neq k\end{cases}\quad \forall n\in\Bbb N$$

Required: show that $\{e^k\}_{k\in\Bbb N}$ is linearly independent and that it is a basis of the vector space $W$, where
$W=\{x\in\Bbb R^{\Bbb N}: x \;\text{is eventually}\; 0\}$

The thing is that the $k$ and the $n$ in the sequence are, in my textbook, vertically aligned (I  just don't know how to write that here). The problem is then that I do not understand what a writing like that should mean, and what a sequence defined in this way could look like. Only then can I think about understanding and solving this (easy) exercise.

Thank you for your attention.

答:
We begin with the set $X$ of all sequences $${\bf x}:\quad{\mathbb N}\to{\mathbb R},\qquad n\mapsto x_n\ .$$
It is common to express such a sequence datawise as ${\bf x}=(x_1,x_2,x_3,\ldots)$. This set $X$ is an infinite-dimensional vector space under coordinatewise operations.  In your problem we then take the subspace $W\subset X$, consisting of the sequences having only finitely many $x_n\ne0$.

Now we consider the special sequences $${\bf e}^k=(e^k_1,e^k_2,\ldots)\qquad(k\in{\mathbb N})$$
defined by
$${\bf e}^k:=(0,0,\ldots,0,1,0,0,\ldots)\ ,$$
where the $1$ is at place $k$ of the sequence. Putting the index $k$ in the upper position is just a typographical manoeuvre; no exponentiation is meant.

It is claimed that the infinite system $$\bigl({\bf e}^k\bigr)_{k\in{\mathbb N}}$$ is a basis of the subspace $W$. This means that for each vector ${\bf x}\in W$ you can find a finite $N$ and uniquely determined numbers $\xi_k\in{\mathbb R}$ such that
$${\bf x}=\sum_{k=1}^N\xi_k\,{\bf e}^k\ .$$


问:
Let $p \colon X \rightarrow Y$ be a perfect map, and let $Y$ be compact. Show that $X$ is compact.

Let $p \colon X \to Y$ be a closed continuous map such that $p^{-1} ( \{y \} )$ is compact, for each $y \in Y$. (Such a map is called a perfect map.) Show that if $Y$ is compact, then $X$ is compact. [Hint: If $U$ is an open set containing $p^{-1} ( \{ y \} )$, there is a neighborhood $W$ of $y$ such that $p^{-1} (W)$ is contained in $U$.

I know that this exact question has been asked on this site, but I would like some guidance and hints to approach to solve it on my own. Here's my approach so far.

Let $X = \cup_\alpha U_\alpha$, with $U_\alpha$ open. Now, for $p^{-1}(\{y\}) \subseteq U_\alpha$, we may find $W_y$ such that $p^{-1}(W_y)$ is contained in $U_\alpha$. Now, $Y = \cup_{y \in Y} W_y$. So we get a finite subcover $W_{y_1} \ldots W_{y_n}$. So get $U_{\alpha_i}$ that contains $W_{y_i}$. As $\cup_i p^{-1}(W_{y_i}) = \cup_i p^{-1}(\cup W_{y_i})$, $\cup_i U_{\alpha_i} = X$ as desired.

Now, this proof doesn't seem to use many of the assumptions, so it can't be right.

Any help would be appreciated, thanks!

答:
First, if the map is not surjective, how can you be sure to be able to even say $p^{-1}(\{y\}) \subseteq U_{\alpha}$? You are considering as a trivial step the fact that we can find $W_y$ such that $p^{-1}(W_y)$ is contained in $U_{\alpha}$, while here you are using all the needed assumptions.

Look at this: Perfect map: If $Y$ is compact, then $X$ is compact?


问:
Is a Relationship Quadratic?
I have a relationship $y=f(x)$ for which I can obtain data through simulation.

I have good reason to suspect that this relationship is quadratic (rather than, say, exponential), and would like to provide evidence for this.

I was thinking of the following method, and I would like to ask if there is anything wrong with it:

a) Obtain $y$ data for a limited interval of $x$, say $x\in (0, x_a)$.

b) Fit a quadratic function to this data using least squares regression.

c) Obtain data for a larger value of $x$ (say $10x_a$) and check the percentage error that the model leads to when extrapolated to this value.

Does this prove anything?

答:
This is a problem of model comparison. A standard approach would be to fit the two competing models with same data and compare their performance on a test set.

You could do least squares fit, or if you have better idea about the noise distribution, you can use maximum likelihood estimation.

Title: Is a Relationship Quadratic?


问:
Find the least value of $m+n$, where $1\le m<n$ such that the last three digits of $1978^n$ are the same as $1978^m$.
I am no so sure how to do this without a calculator...

My method is like this.

So, $k=1978^n-1978^m$ is divisible by $1000$ which means $k$ is divisible by $8$ and $125$.

Since $8$ does not divide into $1978=2\times7\times 127$, $8$ divides $1978^m$ implies that $m\ge 3$.

Then obviously $125$ does not divide into $1978$ as well, so I need to figure out what $r$ is such that $1978^r\equiv 0\mod 125$. I have no idea how to find $r$ without using a calculator...

Could anyone help please?

Body:
Not a 'real' answer, but it was too big for a comment. I know that you're looking for a solution without using a calculator or PC but maybe this gives some insight.

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
ParallelTable[
  If[TrueQ[1 <= m < n &&
     Last@IntegerDigits[1978^n] == Last@IntegerDigits[1978^m] &&
     Part[IntegerDigits[1978^n], 1 + Floor[Log10[1978^n]] - 1] ==
      Part[IntegerDigits[1978^m], 1 + Floor[Log10[1978^m]] - 1] &&
     Part[IntegerDigits[1978^n], 1 + Floor[Log10[1978^n]] - 2] ==
      Part[IntegerDigits[1978^m], 1 + Floor[Log10[1978^m]] - 2]], {m,
    n, m + n}, Nothing], {n, 1, 200}, {m, 1, 200}] //. {} -> Nothing

Running the code gives:

Out[1]={{{3, 103, 106}}, {{4, 104, 108}}, {{5, 105, 110}}, {{6, 106,
   112}}, {{7, 107, 114}}, {{8, 108, 116}}, {{9, 109, 118}}, {{10,
   110, 120}}, {{11, 111, 122}}, {{12, 112, 124}}, {{13, 113,
   126}}, {{14, 114, 128}}, {{15, 115, 130}}, {{16, 116, 132}}, {{17,
   117, 134}}, {{18, 118, 136}}, {{19, 119, 138}}, {{20, 120,
   140}}, {{21, 121, 142}}, {{22, 122, 144}}, {{23, 123, 146}}, {{24,
   124, 148}}, {{25, 125, 150}}, {{26, 126, 152}}, {{27, 127,
   154}}, {{28, 128, 156}}, {{29, 129, 158}}, {{30, 130, 160}}, {{31,
   131, 162}}, {{32, 132, 164}}, {{33, 133, 166}}, {{34, 134,
   168}}, {{35, 135, 170}}, {{36, 136, 172}}, {{37, 137, 174}}, {{38,
   138, 176}}, {{39, 139, 178}}, {{40, 140, 180}}, {{41, 141,
   182}}, {{42, 142, 184}}, {{43, 143, 186}}, {{44, 144, 188}}, {{45,
   145, 190}}, {{46, 146, 192}}, {{47, 147, 194}}, {{48, 148,
   196}}, {{49, 149, 198}}, {{50, 150, 200}}, {{51, 151, 202}}, {{52,
   152, 204}}, {{53, 153, 206}}, {{54, 154, 208}}, {{55, 155,
   210}}, {{56, 156, 212}}, {{57, 157, 214}}, {{58, 158, 216}}, {{59,
   159, 218}}, {{60, 160, 220}}, {{61, 161, 222}}, {{62, 162,
   224}}, {{63, 163, 226}}, {{64, 164, 228}}, {{65, 165, 230}}, {{66,
   166, 232}}, {{67, 167, 234}}, {{68, 168, 236}}, {{69, 169,
   238}}, {{70, 170, 240}}, {{71, 171, 242}}, {{72, 172, 244}}, {{73,
   173, 246}}, {{74, 174, 248}}, {{75, 175, 250}}, {{76, 176,
   252}}, {{77, 177, 254}}, {{78, 178, 256}}, {{79, 179, 258}}, {{80,
   180, 260}}, {{81, 181, 262}}, {{82, 182, 264}}, {{83, 183,
   266}}, {{84, 184, 268}}, {{85, 185, 270}}, {{86, 186, 272}}, {{87,
   187, 274}}, {{88, 188, 276}}, {{89, 189, 278}}, {{90, 190,
   280}}, {{91, 191, 282}}, {{92, 192, 284}}, {{93, 193, 286}}, {{94,
   194, 288}}, {{95, 195, 290}}, {{96, 196, 292}}, {{97, 197,
   294}}, {{98, 198, 296}}, {{99, 199, 298}}, {{100, 200, 300}}}

So, we can see that the lowest number it found is $\text{m}=3$ and $\text{n}=103$ such that $\text{m}+\text{n}=106$. This is true because:

In[2]:=1978^(3)

Out[2]=7738893352

In[3]:=1978^(103)

Out[3]=3245694113774954778728642579697330790826430286814377973924914348349172
1776912127907693002626855998102667117472974423953886955433471331120533
2387497964129735391536809964330889620953818913346685089195593781962547
1616597048762535229892714999678772716316673743259286605339715498711285
462611923097914966231058794697896139243678438050112682852352


问:
Is there a deeper reason why exponentiation is not associative?
Addition can be thought of as repeated counting; multiplication can be thought of as repeated addition; and exponentiation can be thought of as repeated multiplication. And yet, while the first three operations listed—counting, addition, and multiplication—are associative, exponentiation is not. Even its most primitive form, where $a^n$ simply means
$$
\underbrace{a \times a \times a \times \ldots \times a}_{n \text{ times}} \, ,
$$
exponentiation does not have this property:
$$
(a^b)^c \neq a^{(b^c)} \text{ in general.}
$$
Is there a deeper reason for this phenomenon, something that makes exponentiation fundamentally different to counting, addition, and multiplication?

答:
I've tried to omit most of the technical details and go for a more intuitive explanation, since I believe one should always try to explain elementary topics like this one in an accessible way. I also mostly used set-theoretic terminology for the sake of familiarity, but the underlying concepts fundamentally come from category/type theory.

The operations of addition, multiplication and exponentiation on natural numbers are important and useful not because they form part of an infinite sequence, but because they mimic natural constructions we can apply to collections of things (after all, natural numbers were originally used for counting). In particular, given two sets $A$ and $B$, we can construct:

The disjoint union $A\sqcup B$ containing labeled elements $(1,a)$ and $(2,b)$ where $a\in A, b\in B$. The labels $1$ and $2$ tell us which set the element came from, to avoid conflict in case $A$ and $B$ have nonempty intersection.

The Cartesian product $A\times B$, whose elements are all pairs $(a,b)$ where $a\in A, b\in B$.

The function set $A\to B$, whose elements are all functions $f$ from $A$ to $B$.

If we denote by $|X|$ the number of elements of a set $X$, we have $|A\sqcup B|=|A|+|B|$, $|A\times B| = |A|\cdot |B|$ and $|A\to B| = |B|^{|A|}$, so these are indeed the counterparts of addition, multiplication and exponentiation respectively. In fact function sets are frequently denoted by $B^A$, and sometimes disjoint union is denoted by $+$; I deliberately chose different notations here to distinguish between the arithmetic and set-theoretic settings. (I am excluding the operation of succession because it is fundamentally different from the others, as it is a unary operation instead of binary, so it doesn't make much sense to talk about its associativity or commutativity; however, it does have a counterpart in set theory, namely the disjoint union with a prescribed one-element set).

In this set-theoretic language, arithmetic identities such as $a^{b+c} = a^b \cdot a^c$ turn into isomorphisms $(B\sqcup C)\to A \simeq (B\to A) \times (C\to A)$ between the underlying sets.

There are two main phenomena at play:

Repetition

What does it mean for a set construction to be another construction "repeated"? The following is enough for the purposes of this answer: if we have two constructions $\star$ and $\odot$, we will say that $\star$ is repeated $\odot$ if it is possible to define an operator $\bigodot_{x\in A} B(x)$ taking as inputs a family of sets $B(x)$ indexed by elements of another set $A$, such that the two operations are limiting cases of it, that is,

$$\bigodot_{x \in \{1,2,\ldots,n\}} \: B(x) \simeq ((B(1) \odot B(2)) \odot \ldots) \odot B(n),$$

and

$$\bigodot_{x \in A} \: B \simeq A \star B$$

whenever $B$ does not depend on $x$. (Note that for the big operator to be defined independently of the ordering of $A$, $\odot$ already needs to be commutative and associative up to isomorphism, so addition and multiplication are the only operations in the hyperoperation sequence whose set-theoretic counterparts can be "repeated", at least under this definition of repetition).

In the first case we have an obvious candidate, the generalized disjoint union $\bigsqcup_{x \in A} \: B(x)$ whose elements are pairs $(a,b)$ such that $a \in A$ and $b \in B(a)$. We can check by induction that the two above isomorphisms hold, so we can say that $\times$ is repeated $\sqcup$.

For the second case, we need to use an equivalent$^{(*)}$ definition of the Cartesian product:

The Cartesian product $A\times B$ can be equivalently defined as the set of functions $f$ from $\{1, 2\}$ to $A \cup B$ such that $f(1) \in A$ and $f(2) \in B$.

We can think of $f$ as the pair $(f(1), f(2))$ encoded as a function of its two slots. Defining the generalized Cartesian product $\prod_{x \in A} \: B(x)$ whose elements are functions $f$ from $A$ to $\bigcup_{x\in A} B(x)$ such that $f(a) \in B(a)$, we can again check that the above properties hold, so $\to$ is repeated $\times$.

Duality

The phenomenon of duality is found everywhere in category theory; each concept has a corresponding opposite or co-concept. For example, the set-theoretic Cartesian product (under the second definition) is an example of what in category theory is called a product, and the disjoint union is the corresponding coproduct. Similarly, the function set corresponds to an exponential object, which is dual to the product in one of its arguments (categories having all three constructions are called bicartesian closed). Actually, to properly express the universal properties of exponential objects (and products under the first definition) would require something stronger than a category, called a Cartesian multicategory (see e.g. this PDF, or this recent MathOverflow answer). Category theory can even handle the big operators defined above, though their definition is more complicated.

To avoid introducing so much new terminology and state the dualities in the most obvious way possible, let me recapitulate what the elements on each construction do, in terms of the information they give you:

Labeled element: "I give you $1$ and I give you an element of $A$, or I give you $2$ and I give you an element of $B$".

Pair (definition I): "I give you an element of $A$ and I give you an element of $B$".

Pair (definition II): "You give me $1$, then I give you an element of $A$, or you give me $2$, then I give you an element of $B$".

Function: "You give me an element of $A$, then I give you an element of $B$".

Categorical duality in this case corresponds to replacing one or more of the outputs by inputs or viceversa, that is, swapping some "I give you... and" by "you give me... then". Hence there is an obvious duality between $A\sqcup B$ and $A\times B$ (under definition II), and another partial duality between $A\times B$ (under definition I) and $A\to B$, which affects only the first argument. I don't want to go into much detail, but this duality is manifested in category theory itself by interchanging source and target in the relevant morphisms (which in this context are just functions between sets, or multivariable functions in the case of multicategories).

Categorical duality is a powerful notion, and lets us translate properties of one construction to properties of the other. For example, to prove that $X\sqcup Y \simeq Y\sqcup X$, we note that if we swap the labels $1\leftrightarrow 2$ in the element of $A\sqcup B$ described by "You give me $1$, then I give you an $a$, or you give me $2$, then I give you a $b$", we get an element of $B\sqcup A$, and if we do it again we recover the original element, so we have an isomorphism. Now, applying duality, we immediately find an isomorphism $X\times Y \simeq Y\times X$. It's easy to see that the same argument allows us to translate any equational property of $\sqcup$ alone to a corresponding property of $\times$.

In the case of exponentiation, the partial duality means that not all morphisms are reversed, so in some sense we can only translate arithmetic properties that involve exclusively the exponents or exclusively the bases: defining $g_a(b,c)=(a^b)^c$, we have commutativity $g_a(b,c)=g_a(c,b)$ and something reminiscent of associativity, namely that $g_a(b,c,d,\ldots)$ exists and is well-defined for any number of arguments (but note that $g_a(g_a(b,c),d)\neq g_a(b,g_a(c,d))$).  Intuitively, in a multivariable function described by "You give me a $c$, then you give me a $b$, then I give you an $a$", the elements $b$ and $c$ can freely switch places without modifying the overall meaning, but we can't do the same with $a$.

The question remains, what goes wrong if we try to dualize $A\times B$ with respect to both arguments? Taking the cardinality of the resulting construction, we do indeed obtain an operation which has all desirable properties: it is commutative, associative and distributes over $\times$. What is this mysterious operation?

Sadly, the operation turns out to be just $(1^a)^b=1$. The issue is that the corresponding construction receives two inputs but gives no output, and any function that gives no output must be isomorphic to the empty function, the unique member of the singleton set $\mathbf{1}$.

$(*)$: The fact that there are two equivalent definitions of Cartesian product is important, because there are situations where these definitions are not equivalent!

For example, in settings such as intuitionistic linear type theory (whose category-theoretic counterpart would be something like monoidal closed categories, or rather multicategories, as I mentioned above), there are two versions of the product, the tensor product $\otimes$ (definition I) and Cartesian product $\&$ (definition II), such that $\otimes$ is repeated $\sqcup$ and $\to$ is repeated $\&$, but $\otimes \neq \&$. This is because in this setting, the ambient logic (called linear logic) only allows you to call a function exactly once, so given a pair in function form (an element of $\&$) you will never be able to recover both coordinates, only one of them. Thus $\otimes$ needs not always share all the properties of $\sqcup$, and can be, for example, noncommutative (and indeed there is a noncommutative version of linear logic).

This might seem like a strange setting, but it has found applications in fields such as quantum mechanics (because of the so-called no-cloning theorem) and computer science (to handle situations where the availability of resources matters).


问:
Homeomorphism between convex set and a sphere
For $C$ a compact convex set of $\mathbb{R}^n$ $(n\geq 1)$ whose interior is nonempty and whose boundary $\partial C$ is a smooth manifold, is $\partial C$ homeomorphic to the sphere $\mathbb{S}^{n-1}$?

答:
Yes. You can suppose that $B_1(0)\subset C$. Then in the halfline of positive scalar multiples of every unit vector $v$ there is only one point of $\partial C$, because if $\lambda>\mu>0$, $\mu v$ is in the convex envelope of $B_1(0)$ and $\lambda v$, which is open, and so it can't belong to $\partial C$.

So the continuous function $\rho:{\mathbf R}^n\setminus 0 \to {\mathbf S}^{n-1}$ restricts to a bijective continuous function $\partial C\to {\mathbf S}^{n-1}$. Since $\partial C$ is compact and ${\mathbf S}^{n-1}$ is Hausdorff's, by a famous lemma of topology it is a homeomorphism.

By extending it linearly on every ray, it can even been proved that $C$ is homeomorphic to $\overline{B_1}(0)$.


问:
Determine all n element N, so that (1-i)^n is natural
I've done following steps thus far:

```<code>define z:= 1-i; then z=sqrt(2) * e^((-pi/4)i)
(1-i)^n = z^n = (sqrt(2)*e^((-pi/4)i)^n = sqrt(2)^n * e^((-pi/4)in
That gives two conditions:
1) sqrt(2)^n needs to be element N -> n=2k, k element N
2) e^((-pi/4)in needs to be element N -> pi/4 * n = pi * k -> n=4k, k element N
</code>```

At this point I'm lost, though. I know the solution from straight thinking - n needs to be 8*k, but how do I get to that point from where I am with my steps so far? E.g. From the two conditions I have here, n could also be 4, but (1-i)^n is -4 which is not element N.

Cheers
Anton

答:
Alright, I just found my mistake...

of course coniditon 2) is wrong. It needs to be as followed

```<code>2) e^((-pi/4)in needs to be element N -> pi/4 * n = 2*pi * k -> n=8*k, k element N
</code>```

now 2) implies 1), therefore the only condition that needs to be accounted for is 2). Done.

Body:
Alright, I just found my mistake...

of course coniditon 2) is wrong. It needs to be as followed

2) e^((-pi/4)in needs to be element N -> pi/4 * n = 2*pi * k -> n=8*k, k element N

now 2) implies 1), therefore the only condition that needs to be accounted for is 2). Done.


问:
Rigorous graduate probability textbook for self-study.
I have checked several answers that request graduate-level probability theory textbooks, but I have wanted to receive advice that fits my particular needs.

My background is in mathematics and I am currently pursuing Ph.D. degree in statistics, but I am still more geared toward mathematics. In my first-year Ph.D. course for probability theory, we used Durrett's Probability Theory and Examples, but I found that the book is too terse for a first reading. I had no problem reading Terrace Tao's Introduction to Measure Theory and Folland's Real Analysis in my real analysis course, but the proofs in Durrett's book were often incomplete and most of the exercises were too hard to solve for someone who is studying the topic for the first time. Therefore, I am planning to use another textbook to supplement Durrett's, and I wanted to ask you advice on which books would be the best choice for me

I wish the book is suitable for self-study, that is, it should have rather complete proofs to show each step especially in the earlier chapters. Since I have background in mathematics, I am looking for a book that is still rigorous so that it does not omit significant amount of proofs.

Some of the alternatives I have found are Erhan Çinlar's Probability and Stochastics and Shiryaev's Probability 1, 2. I have asked to my professor about the textbook and it seems like he approves of it, but I wanted to see other options as well. I would love to hear your recommendations and your rationales behind recommending the books.

[Edited] Added Shiryaev's Probability series.

答:
I highly recommend Jean-Francois Le Gall's "Brownian Motion, Martingales and Stochastic Calculus", as I have used this book for self-study myself. As the title says, it focuses mainly on stochastic calculus, but also contains some general theory of martingales and Markov processes. I believe that these topics are usually covered in advanced probability courses.

I also recommend Kallenberg's "Foundations of modern probability", which covers all the basics, but also has quite some depth, perhaps even too much depth for self-study. If Kallenberg's book is too much and you're more interested in getting an comprehensive introduction to the basics of probability and measure theory, I recommend Billingsley's "Probability and Measure".


问:
Do I understand this proof for AoC $\implies$ well-ordering theorem correctly?
Am a real beginner in set-theory. I read that the Axiom of Choice (AoC) implies the Well-Ordering Theorem...

From what I understand, the gist of the proof is that given any set $X$, we can use AoC to well-order $X$ using induction. Namely, given any element $a_i \in X$ in the ordering, we can use the choice function $c$ (implied by AoC) to select $a_{i+1}=c(X-\bigcup_{j \leq i} a_j)$, and then extend the ordering to include $a_i \leq a_{i+1}$, and so on and so forth ...

But suppose that $X=\mathbb{R}$ ... if we try to well-order $\mathbb{R}$ using the proof above, up to what extent should we do the induction ?  Can we do the induction an uncountable number of times? --- up to say ... $2^{\omega}+1$ (where $\omega$ is the cardinal number of $\mathbb{N}$) i.e. to "exhaust" all of $\mathbb{R}$)

答:
Firstly, we're using recursion, rather than induction.

Secondly, the recursion can go as long as any ordinal $\alpha$ such that $|\alpha|=|\Bbb R|$, and there are a lot of these, at least $\aleph_2$, and in fact $(2^{\aleph_0})^+$ of them.

To see this in action, fix a well-ordering of $\Bbb N$ which is not of order type $\omega$, e.g., $\omega+1$. Then use this to define a choice function: $c(A)=\min A\setminus\{0\}$ or $0$ if $A=\{0\}$. If we use this as our choice function to well-order $\Bbb N$, we will end up with an order of type $\omega+1$. And the same happens with any other countably infinite ordinal.

So "how long"? Well, some ordinal of cardinality $2^{\aleph_0}$, but there's a lot of them, so it's not possible to pre-determine which one. But that is exactly what makes transfinite recursion trans-finite. The ability to do it over very long well-orders.

(And as a side note, especially when we're dealing with recursion over the ordinals, it's better to use $\aleph$ notation to discuss cardinals, so you should write $2^{\aleph_0}$ rather than $2^\omega$. This way there is no chance of mistaking $2^\omega$ for ordinal exponentiation, which is just $\omega$ again.)


问:
Does convergence in $L^p(U)$ and $L^q(U)$ implies same limit.
Let $1\le p<q<\infty$ such that $x_n\to x_p$ in $L^p(U)$ and $x_n\to x_q$ in $L^q(U)$ for some open set $U$.Does the limit $x_p = x_q$ a.e?

I have two approach to do it,if $U$ is bonded ,the result holds.since convergence in $L^q(U)$ implies convergence in $L^p(U)$.Hence the limit is unique in normed space.

For unbounded case,extract almost everywhere convergence subsequence then limit must coinside a.e.

Is my proof correct?

答:
You are right. (Formally) the uniqueness of the limit (in the single space you are considering) comes from the fact that normed spaces are $T_2$. But, the main argument in both cases is the subsequence that converges almost everywhere to the limit function. Both the bounded and unbounded cases work.

Look at this for more: Convergence of a sequence in two different $L_p$ spaces


答:
If $n>m$ are such that $1978^n - 1978^m$ is a multiple of $1000$, then we factorize as $1978^{m}(1978^{n-m} - 1)$ is a multiple of $1000$.

This is equivalent to $1978^{m}(1978^{n-m}-1)$ being a multiple of $8$ and $125$.

In the first case, note that $1978^{n-m} - 1$ is odd because $n \neq m$, hence coprime to $8$. $1978$ as a multiple of $2$ has multiplicity $1$ . So it follows that $m \geq 3$ is necessary and sufficient for divisibility by $8$ to hold.

In the second case, $1978^m$ is coprime to $125$ (by prime factorization) so we need the smallest value of $T = n-m$ such that $1978^T \equiv 1 \pmod{125}$.

Note that $1978 \equiv 103 \pmod{125}$. So we need to find $T$ such that $103^T \equiv 1 \pmod{125}$. Now, since $\phi(125) = 100$ it follows that $T$ is a divisor of $100$.

Start with the basic : if $103^T \equiv 1 \pmod{125}$ then in particular $103^T \equiv 1 \pmod{5}$, which comes to $3^T \equiv 1 \pmod{5}$, which is true only if $T$ is even. Thus, all odd values of $T$ are ruled out.

Next, we go modulo $25$ to see that $3^T \equiv 1 \pmod{25}$ must be true. We see that $\phi(25) =20$, so $T$ must be a factor of $20$. It is easy (because $3$ is small) to see that no number smaller than $20$ works. Thus, $T=20$ or $T = 100$.

Unfortunately, checking that $103^{20} \not \equiv 1 \pmod{125}$ is to be checked by hand. One can use repeated squaring to make this easier. (Also refer to the answer linked by J.W.Tanner below).

Thus, $T=100$, and $n = 3+100 = 103$ and $m=3,n=103$ form the smallest set of values with $m+n = 106$.

Title: Find the least value of $m+n$, where $1\le m<n$ such that the last three digits of $1978^n$ are the same as $1978^m$.

Body:
I am no so sure how to do this without a calculator...

My method is like this.

So, $k=1978^n-1978^m$ is divisible by $1000$ which means $k$ is divisible by $8$ and $125$.

Since $8$ does not divide into $1978=2\times7\times 127$, $8$ divides $1978^m$ implies that $m\ge 3$.

Then obviously $125$ does not divide into $1978$ as well, so I need to figure out what $r$ is such that $1978^r\equiv 0\mod 125$. I have no idea how to find $r$ without using a calculator...

Could anyone help please?


问:
Proofs using modular arithmetic.
In my textbook, it says,

The sum of the cubes of any three consecutive integers is divisible by
$9$

This statement is equivalent to proving,

$(n^3 + (n+1)^3 + (n+2)^3 ) \mod 9 =0$  is true for $n=0,1, \ldots, 8$

Now I understand how the modulo function works but I just don't see how the latter part is the same as the former. Can anyone walk me through it?

Note that, I'm not really looking for the proof, what I am looking for is how the first and the second statement are related.

Follow up question which begs to be asked,

Given some expression of $a \in \mathbb{Z}$, say $P(a)$, are the following 2 statements equivalent?

$n$ divides $P(a), \, \forall a \in \mathbb{Z}$ and for some fixed $n \in \mathbb{N}$

$P(a) \mod n = 0, \forall a \in \{0,1, \ldots, n-1\}$

Can anyone explain me whether or not the two statements are equivalent? Thanks!

答: If $n$ and $m$ are equivalent modulo $9$, then so are $n^3$ and $m^3$. Hence there is no need to try all $n$, but only those that yield different values modulo $9$.

Title: Proofs using modular arithmetic.


问:
Show linear independence of sum of vectors from a base
let $V$ be a vectorspace and $\{a_1,a_2,a_3,a_4\}$ a base of $V$;
Moreover $b_1:=2a_1-a_2,$ $b_2:=a_2+a_3+a_4,$ $b_3:=a_3-a_4$;
Show that $b_1,b_2,b_3$ are linear independent;

By definition of linear independence:

(1) $x\cdot b_1+y\cdot b_2+z\cdot b_3 = 0 \Rightarrow x=y=z=0$

so

(2) $x\cdot_(2a_1-a_2)+y\cdot (a_2+a_3+a_4)+z\cdot(a_3-a_4) = 0 \Rightarrow x=y=z=0$.

This is where I'm stuck already. I've tried resorting (2) so that I have $a_1\cdot(2x)+a_2\cdot(\dots$ but that doesn't get me anywhere.

Any help or pointers are much welcome!

Thanks
Anton

答: Suppose that $x(2a_1-a_2)+y(a_2+a_3+a_4)+z(a_3-a_4)=0$. This is the same thing as asserting that $2xa_1+(-x+y)a_2+(y+z)a_3+(y-z)a_4=0$. Since $\{a_1,a_2,a_3,a_4\}$ is a basis, then$$\left\{\begin{array}{l}2x=0\\-x+y=0\\y+z=0\\y-z=0.\end{array}\right.$$It is easy to see that the only solution of this system is $x=y=z=0$.


问:
Why M$\otimes$N can be considered as a domain for a Linear Map but M $\times$ N is a domain of Bilinear Maps?
In one of the lecture notes, I have read that:

Instead of bilinear maps
from M $\times$ N we can then always consider linear maps from the tensor product, and thus use all the machinery that we have developed so far for homomorphism.

I can't see why M$\otimes$N can be considered as a domain for a Linear Map but M $\times$ N is a domain of Bilinear Maps?

答:
First of all a bit of notation: a space cannot be a linear map.

A tensor product is a vector space, hence you can consider linear maps having it as the domain. Do not confuse maps having it as domain, and its elements.


答: $$ n | P(a) \ \forall \ a\in \mathbb Z \iff P(a) \mod n = 0 \ \forall \ a\in \{0,1,2\dots n-1\}$$ The key here is that if all elements $e$ in this set satisfy this property, then all numbers of the form $e+nk$ for some integer $k$ also satisfy this property, as the $kn$ term gets vanished mod $n$. For instance, $$ n^3 +(n+1)^3 +(n+2)^3 \equiv (n+9k)^3 +(n+9k+1)^3 +(n+9k+2)^3 \pmod 9 $$ because $$(n+9k)^3 = n^3 +9^3k^3 +3n^2(9k) +3(n)(9^2k^2) = n^3 +9(\text{stuff}) $$ and similarly for the other two terms.

Title: Proofs using modular arithmetic.

Body:
In my textbook, it says,

The sum of the cubes of any three consecutive integers is divisible by
$9$

This statement is equivalent to proving,

$(n^3 + (n+1)^3 + (n+2)^3 ) \mod 9 =0$  is true for $n=0,1, \ldots, 8$

Now I understand how the modulo function works but I just don't see how the latter part is the same as the former. Can anyone walk me through it?

Note that, I'm not really looking for the proof, what I am looking for is how the first and the second statement are related.

Follow up question which begs to be asked,

Given some expression of $a \in \mathbb{Z}$, say $P(a)$, are the following 2 statements equivalent?

$n$ divides $P(a), \, \forall a \in \mathbb{Z}$ and for some fixed $n \in \mathbb{N}$

$P(a) \mod n = 0, \forall a \in \{0,1, \ldots, n-1\}$

Can anyone explain me whether or not the two statements are equivalent? Thanks!


答:
Consider your equation $x(2a_1-a_2)+y*(a_2+a_3+a_4)+z*(a_3-a_4)=0$.

It gives

$2xa_1  + (-x+y) a_2+ (y+z)a_3 + (y-z)a_4=0$.

Since $\{a_1,\ldots,a_4\}$ is a basis, it follows that $2x=0$, $y-x=0$, $y+z=0$ and $y-z=0$.
Thus $x=0$ and then in order $y=0$ and $z=0$, as required.

Title: Show linear independence of sum of vectors from a base

Body:
let $V$ be a vectorspace and $\{a_1,a_2,a_3,a_4\}$ a base of $V$;
Moreover $b_1:=2a_1-a_2,$ $b_2:=a_2+a_3+a_4,$ $b_3:=a_3-a_4$;
Show that $b_1,b_2,b_3$ are linear independent;

By definition of linear independence:

(1) $x\cdot b_1+y\cdot b_2+z\cdot b_3 = 0 \Rightarrow x=y=z=0$

so

(2) $x\cdot_(2a_1-a_2)+y\cdot (a_2+a_3+a_4)+z\cdot(a_3-a_4) = 0 \Rightarrow x=y=z=0$.

This is where I'm stuck already. I've tried resorting (2) so that I have $a_1\cdot(2x)+a_2\cdot(\dots$ but that doesn't get me anywhere.

Any help or pointers are much welcome!

Thanks
Anton


问:
Origin of a problem in graph theory/planar geometry
Does anyone remember where does the following problem comes from:

Let $P_n$ be a set of $n$ points on the plane, and denote by $d$ the minimal distance between any two points of $P_n$ (i.e. $d=\min\{d(p_i,p_j)|p_i\neq p_j\in P_n\}$). Then there exists a subset $X \subset P_n$ such that for all $p_i,p_j\in X$, $d(p_i,p_j)>d$ and $|X|\geq\frac{n}{4}$

If I remember correctly, this was an open problem (which was solved quite easily using the four color theorem). I just can't remember who asked it (presumably Erdős, but i'm not sure) and when.

Thanks in advance.

答:
I think you'll enjoy Gyorgy Csizmadia, On the independence number of minimum distance graphs, freely available at http://www.math.nyu.edu/~csiz/cikkek/penz.ps

Csizmadia says Erdos asked for bounds in 1983. The only Erdos paper he cites is Some combinatorial and metric problems in Geometry, in Intuitive Geometry, Colloquia Mathematica Societatis Janos Bolyai 48 (1985) 167-177. The application of the FCT appears in R. Pollack, Increasing the minimum distance of a set of points, J Combinatorial Theory Series A 40 (1985) 450.


问:
The population of a certain bacteria can multiply threefold in 24 hours. If there are 500 bacteria now, how many will there be in 96 hours?
The population of a certain bacteria can multiply threefold in 24 hours. If there are 500 bacteria now, how many will there be in 96 hours?

I figured out this bacteria $=500(3)^{96/24}$

but then my mate told me it's bacteria $=500(1+3/1)^{96/24 }$

what do I do what is the next step.

答: It depends on the intended meaning of multiply threefold. You’re interpreting it as meaning increase by a factor of $3$, which is how I would interpret it. Your mate is interpreting it to mean that the amount by which the population increases is $3$ times the present population; I would call that a fourfold increase. However, I can imagine someone using the term multiply threefold in that way; the only way to be certain what’s intended is to ask the person who set the problem.


答:
You are on the right track but you've made a small error.  Let $a = 1978$.  Then as you noted, $1000 \mid a^n - a^m$, which leads to $a^m (a^{n-m} - 1) = 1000k$ for some integer $k$.  Since $a^r - 1$ is odd for all positive integer $r$, we must have $8 \mid a^m$ so $m \ge 3$, and $125 \mid a^{n-m} - 1$.  So instead of having a power of $a$ being congruent to $0$ modulo $125$, it leaves a remainder of $1$.  From Euler's theorem, for $\gcd(a, b) = 1$, we have $$a^{\varphi(b)} \equiv 1 \pmod b$$ where $\varphi(b)$ is the Euler totient function that counts the number of positive integers between $1$ and $b$ that are relatively prime to $b$.  One property of $\varphi$ is that if $b$ is the power of a prime, say $b = p^k$, then $$\varphi(p^k) = p^k - p^{k-1}.$$

Since $\gcd(1978,125) = 1$ and $125 = 5^3$, it follows that $\varphi(125) = 5^3 - 5^2 = 100$ and $$1978^{100} \equiv 1 \pmod {125}.$$  Thus $(m,n) = (3, 103)$ satisfies the given conditions.  All that is left is to show that such an $n$ is minimal; i.e., there is no smaller exponent $r^* < 100$ for which $a^{r^*} \equiv 1$.  This is not difficult to show and I leave as an exercise for the reader.


问:
Geometric series and Fourier transform
In my book solution example, I have this signal

discrete square wave

I don't understand how the right hand side is attained from the expression on the left
$$\sum_{m=p-M}^{p-1}e^{-imk\omega_0}=e^{iMk\omega_0}\sum_{m=0}^{M-1}e^{-imk\omega_0}$$

Where the last step is accomplished by changing variables and simplifying using the fact that
$$\omega_0 = \frac{2\pi}{p}$$

Is there some book can I read to understand things like that?

答:
Starting with the last expression
$$e^{iMk\omega_0}\sum_{m=0}^{M-1}e^{-imk\omega_0}$$
$$=\sum_{m=0}^{M-1}e^{iMk\omega_0}e^{-imk\omega_0}$$
$$=\sum_{m=0}^{M-1}e^{-i(m-M)k\omega_0}$$
Here we can shift the index by $p-M$ and now we have
$$\sum_{m=p-M}^{M-1+p-M}e^{-i(m-M-p+M)k\omega_0}$$
$$=\sum_{m=p-M}^{p-1}e^{-i(m-p)k\omega_0}$$
$$=\sum_{m=p-M}^{p-1}e^{-imk\omega_0}e^{ipk\omega_0}$$
Since $\omega_0=\frac{2\pi}{p}$, we have
$$\sum_{m=p-M}^{p-1}e^{-imk\omega_0}e^{2\pi ik}$$
$$=\sum_{m=p-M}^{p-1}e^{-imk\omega_0}$$
Therefore,
$$\sum_{m=p-M}^{p-1}e^{-imk\omega_0}=e^{iMk\omega_0}\sum_{m=0}^{M-1}e^{-imk\omega_0}$$

Title: Geometric series and Fourier transform