问:
Calculate empty labeled boxes in labeled boxes-balls problem
The problem is to count the number of times that K boxes are empty when n labeled balls are distributed into m labeled boxes. So, if we have n balls and m boxes, we have $m^{n}$ possibilities. I would like to count the number of times that box 1 and/or box 2 for example are empty.

For example, how many times box 1 or box 4 (or both) are empty when we distribute 10 balls between 10 boxes?. There are $10^{10}$ possibilities. How can it be calculated?

答:
Hint: To ensure that some boxes are empty the balls have to be distributed between the remaining boxes.

For example:

The number of ways to distribute 10 balls so that the box 1 is empty is $9^{10}$.

The number of ways to distribute 10 balls so that the boxes 1 and 4 are empty is $8^{10}$.

Keep in mind that the above numbers count also the events with some other boxes being empty.


问:
How prove this $\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$
Prove that

$$\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$$

I have some question. Using this, find this integral is not converge, I'm wrong?

Thank you everyone

答:
First make the substitution $x=u^2$ to get:

$\displaystyle \int _{0}^{\infty }\!\sin \left( x \right) \sin \left( \sqrt {x}
 \right) {dx}=\int _{0}^{\infty }\!2\,\sin \left( {u}^{2} \right) \sin
 \left( u \right) u{du}$,

$\displaystyle=-\int _{0}^{\infty }\!u\cos \left( u \left( u+1 \right)  \right) {du}+
\int _{0}^{\infty }\!u\cos \left( u \left( u-1 \right)  \right) {du}$,

and changing variable again in the second integral on the R.H.S such that $u\rightarrow u+1$ this becomes:

$=\displaystyle\int _{0}^{\infty }\!-u\cos \left( u \left( u+1 \right)  \right) {du}+
\int _{-1}^{\infty }\!\left(u+1\right)\cos \left( u \left( u+1 \right)  \right) {du}
$,

$\displaystyle=\int _{0}^{\infty }\!\cos \left( u \left( u+1 \right)  \right) {du}+
\int _{-1}^{0}\! \left( u+1 \right) \cos \left( u \left( u+1 \right)
 \right) {du}
$.

Now we write $u=v-1/2$ and this becomes:

$\displaystyle\int _{1/2}^{\infty }\!\cos \left( {v}^{2}-1/4 \right) {dv}+\int _{-1/
2}^{1/2}\! \left( v+1/2 \right) \cos \left( {v}^{2}-1/4 \right) {dv}=$

$\displaystyle \left\{\int _{0}^{\infty }\!\cos \left( {v}^{2}-1/4 \right) {dv}\right\}$

$\displaystyle  +\left\{\int _{-1/2}
^{1/2}\!v\cos \left( {v}^{2}-1/4 \right) {dv}+\int _{-1/2}^{0}\!1/2\,
\cos \left( {v}^{2}-1/4 \right) {dv}-1/2\,\int _{0}^{1/2}\!\cos
 \left( {v}^{2}-1/4 \right) {dv}\right\},$

but the second curly bracket is zero by symmetry and so:

$\displaystyle \int _{0}^{\infty }\!\sin \left( x \right) \sin \left( \sqrt {x}
 \right) {dx}=\displaystyle \int _{0}^{\infty }\!\cos \left( {v}^{2}-1/4 \right) {dv}$,

$\displaystyle =\int _{0}^{
\infty }\!\cos \left( {v}^{2} \right) {dv}\cos \left( 1/4 \right) +
\int _{0}^{\infty }\!\sin \left( {v}^{2} \right) {dv}\sin \left( 1/4
 \right)
$.

We now quote the limit of Fresnel integrals:

$\displaystyle\int _{0}^{\infty }\!\cos \left( {v}^{2} \right) {dv}=\int _{0}^{
\infty }\!\sin \left( {v}^{2} \right) {dv}=\dfrac{\sqrt{2\pi}}{4}$,

to obtain:

$\displaystyle \int _{0}^{\infty }\!\sin \left( x \right) \sin \left( \sqrt {x}
 \right) {dx}=\dfrac{\sqrt{2\pi}}{4}\left(\cos\left(\dfrac{1}{4}\right)+\sin\left(\dfrac{1}{4}\right)\right)=\dfrac{\sqrt{\pi}}{2}\sin{\left(\dfrac{3\pi-1}{4}\right)}$.

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问:
If $\sum_{n=1}^{\infty} x_n$ $\lt$ $\infty$, $x_n$ must be $\lt$ $1$ for all $n$?
If $\sum_{n=1}^{\infty}$$x_n$ $\lt$ $\infty$,

does that mean that $x_n$ must be $\lt$ $1$ for all $n$?

答:
The other answers show that the claim is false.
However, it can be salvaged:

Since $x_n \to 0$, we have $x_n<1$ for all but finitely many $n$.


问:
Discrepancies in defining the curvature of a Cartesian function
I was thinking about how one might be able to define how curved a function was. I thought of two ways:

Method 1

The angle of a tangent of a function is $\arctan\left(\frac{dy}{dx}\right)$ (which we'll define as $\theta$). The rate that the angle of the tangent changes (or $\frac{d\theta}{dx}$) is, thus, $$\frac d{dx}[\theta]=\frac d{dx}\left[\arctan\left(\frac{dy}{dx}\right)\right]=\frac{d^2y}{dx^2}\arctan'\left(\frac{dy}{dx}\right)=\frac{\frac{d^2y}{dx^2}}{1+\left(\frac{dy}{dx}\right)^2}=\frac{f''(x)}{1+(f'(x))^2}$$
Hence the curvature of a function is $\frac{f''(x)}{1+(f'(x))^2}$.

Method 2

We can define the curvature of a circle $(x-a)^2+(y-b)^2=r^2$ as having curvature $\frac1r$. For each point on a function, we find $a,b,r$ that approximates the function well around that point, i.e. such that $(x-a)^2+(y-b)^2=r^2$ intersects at the point, and has the same first and second derivative as the function at that point, and calculate $\frac1r$ from there.

The first derivative of a circle is $\frac d{dx}\left[(x-a)^2+(y-b)^2\right]=\frac d{dx}r^2\Rightarrow \frac d{dx}\left[(x-a)^2\right]+\frac d{dx}\left[(y-b)^2\right]=0\Rightarrow 2\frac d{dx}\left[x-a\right](x-a)+2\frac d{dx}\left[(y-b)\right](y-b)=0\Rightarrow x-a+\frac {dy}{dx}(y-b)=\frac {dy}{dx}(y-b)=-(x-a)\Rightarrow\frac {dy}{dx}=-\frac{x-a}{y-b}$

The second derivative of a circle is $\frac d{dx}\left[\frac {dy}{dx}\right]=\frac d{dx}\left[\frac{a-x}{y-b}\right]\Rightarrow \frac{d^2y}{dx^2}=\frac{\frac d{dx}[a-x](y-b)-\frac d{dx}[y-b](a-x)}{(y-b)^2}=\frac{-(y-b)-\frac d{dx}[y-b](a-x)}{(y-b)^2}=-\frac{y-b+\frac{dy}{dx}(a-x)}{(y-b)^2}=-\frac{y-b+\frac{a-x}{y-b}(a-x)}{(y-b)^2}=-\frac{(y-b)^2+(a-x)^2}{(y-b)^3}=-\frac{r^2}{(y-b)^3}$

Thus, we have the following system of equations:

$$
\begin{cases}
(x-a)^2+(y-b)^2=r^2 \\
-\frac{x-a}{y-b}=y' \\
-\frac{r^2}{(y-b)^3}=y'' \\
\end{cases}
$$

To solve this equation, we first write $(y-b)^2=r^2-(x-a)^2$ and square both sides of the bottom two equations to get:

$$
\begin{cases}
(y-b)^2=r^2-(x-a)^2 \\
\frac{(x-a)^2}{(y-b)^2}=\left(y'\right)^2 \\
\frac{r^4}{(y-b)^6}=\left(y''\right)^2 \\
\end{cases}
$$

Now replace $(y-b)^2=r^2-(x-a)^2$ and multiply the bottom two equations by $(y-b)^2$ and $(y-b)^6$ respectively:

$$
\begin{cases}
(x-a)^2=\left(y'\right)^2\left(r^2-(x-a)^2\right) \\
r^4=\left(y''\right)^2\left(r^2-(x-a)^2\right)^3 \\
\end{cases}
$$

We can unpack the second equation:

$$(x-a)^2=\left(y'\right)^2\left(r^2-(x-a)^2\right)=\left(y'\right)^2r^2-\left(y'\right)^2(x-a)^2$$
$$\left(1+\left(y'\right)^2\right)(x-a)^2=\left(y'\right)^2r^2$$
$$(x-a)^2=\frac{\left(y'\right)^2}{1+\left(y'\right)^2}r^2$$

This leaves us with the equation $r^4=\left(y''\right)^2\left(r^2-(x-a)^2\right)^3$. Substituting $(x-a)^2=\frac{\left(y'\right)^2}{1+\left(y'\right)^2}r^2$ gives us

$$r^4=\left(y''\right)^2\left(r^2-\frac{\left(y'\right)^2}{1+\left(y'\right)^2}r^2\right)^3$$
$$r^4=\left(y''\right)^2\left(\frac1{1+\left(y'\right)^2}r^2\right)^3$$
$$r^4=\frac{\left(y''\right)^2}{\left(1+\left(y'\right)^2\right)^3}r^6$$
$$\frac{\left(y''\right)^2}{\left(1+\left(y'\right)^2\right)^3}r^6-r^4=0$$
$$r^4\left(\frac{\left(y''\right)^2}{\left(1+\left(y'\right)^2\right)^3}r^2-1\right)=0$$
(It seems that the quadruple root at $r=0$ is extraneous, so we can divide by $r^4$)
$$\frac{\left(y''\right)^2}{\left(1+\left(y'\right)^2\right)^3}r^2-1=0$$
$$\frac{\left(y''\right)^2}{\left(1+\left(y'\right)^2\right)^3}r^2=1$$
$$r^2=\left(\frac{\left(y''\right)^2}{\left(1+\left(y'\right)^2\right)^3}\right)^{-1}$$
$$\frac1r=\sqrt{\frac{\left(y''\right)^2}{\left(1+\left(y'\right)^2\right)^3}}=\frac{y''}{\left(1+\left(y'\right)^2\right)^\frac32}$$
Hence the curvature of a function is $\frac{f''(x)}{\left(1+\left(f'(x)\right)^2\right)^\frac32}$.

The two methods that I used to define curvature ended up giving different formulae for curvature, a result I was not expecting. According to the Wikipedia article for curvature the formula from Method 2 is the generally accepted formula for finding the curvature of . The definition seems to originally come from so-called "osculating circles", which seems remarkably similar to what I did in method 2. What I have yet to find, however, is why my first method gave a different result from my second result?

(Something I found note-worthy is that the integral of $\frac{f''(x)}{\left(1+\left(f'(x)\right)^2\right)^\frac32}$ is $\frac{f'(x)}{\sqrt{1+\left(f'(x)\right)^2}}$. Note that the denominator, $\sqrt{1+\left(f'(x)\right)^2}$, is the derivative of the arc length of $f(x)$. The second formula seems somehow related to arc length and the first one isn't. Not quite sure what to make of it.)

答:
Looking back at the question, I began to wonder why I defined the curvature of a circle as $\frac1r$ in the first place. I never really thought deep about it when thinking about curvature, and I wondered if the answer would lie there.

Intuitively the amount of "time" it would take for a circle to curve  is inversely proportional to its radius. Consider the equation $f(x)=\sqrt{r^2-x^2}$ (a semicircle). At $x=-r$ the tangent of the function is pointing $\frac\pi2$ degrees up, while at $x=-r$ the tangent is pointing $\frac\pi2$ degrees down. It took a change in $x$ of $2r$ to change the angle of the tangent (which we shall denote by $\theta$) by the same amount, $\pi$. However the (average) change in $\theta$ with respect to change in $x$ would then be $\frac\pi{2r}$, not $\frac1r$ as Method $2$ implies. Maybe I wasn't evaluating the change of $\theta$ with respect to $x$, but some other variable, to define curvature in Method $2$?

That's when it hit me. If I were to trace a path of length $l$ along a circle of radius $r$, by definition the angle change would be $\frac lr$. I was measuring the change in angle with respect to the change in arc length in method $2$, i.e. I was evaluating $\frac{d\theta}{dl}$ (where $l$ is arc length, and $dl=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$).  However, in method 1 I was evaluating $\frac{d\theta}{dx}$. Of course the two results would be different!

When looking at the results from this perspective, the two results actually agree with each other. We have:

$$\frac{d\theta}{dl}=\frac{d\theta}{dx}\cdot\frac{dx}{dl}=\frac{d\theta}{dx}\cdot\frac{dx}{\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx}=\frac{d\theta}{dx}\cdot\frac1{\sqrt{1+\left(f'(x)\right)^2}}$$

From method $1$ I found that $\frac{d\theta}{dx}=\frac1{1+\left(f'(x)\right)^2}$. Inserting this in, we have

$$\frac{d\theta}{dl}=\frac{d\theta}{dx}=\frac1{1+\left(f'(x)\right)^2}\cdot\frac1{\sqrt{1+\left(f'(x)\right)^2}}=\frac1{\left(1+\left(f'(x)\right)^2\right)^\frac32}$$

Which is the result I got in method $2$.

Looking back, I can understand why curvature isn't defined as $\frac{d\theta}{dx}$. $\theta$ is a co-ordinate found when using the polar co-ordinate system, while $x$ is a co-ordinate found when using the Cartesian co-ordinate system. It makes no sense to mix and match co-ordinates from different co-ordinate systems. On the other hand, arc length is a thing, both in the Cartesian and polar co-ordinate systems, and it does make more sense to talk about angles with respect to arc lengths.

As I was skimming though the Wikipedia page on curvature, I also noticed that vectors were used a lot. It is less nonsensical to talk about arc length (which in the case of vectors is just the norm), rather than talking about the length of a vector in some arbitrary direction.

In the end, this investigation into curvature has shown me that I should broaden my perspective when looking at mathematics, because sometimes the most obvious perspective may not end up being the most useful one. In this case, thinking about the change in the angle of the function in terms of $x$ was the most obvious perspective, but in reality it would have made more sense to think about the change in the angle of the function in terms of $l$.


问:
If $\mathbb E[X^nY^m]=\mathbb E[X^n]\mathbb E[Y^m]$ for all $n,m$, then $X$ and $Y$ are independent.
I had an exam this morning, and I had to prove that for $X$ and $Y$ bounded, if for all $n$ and all $m$, $$\mathbb E[X^nY^m]=\mathbb E[X^n]\mathbb E[Y^m],$$then $X$ and $Y$ are independents. Using Characteristic function I think that I could justify it using the fact that $$\cos(tX)=\sum_{k=0}^\infty \frac{(-1)^kt^{2k}X^{2k}}{(2k!)}\quad \text{and}\quad \sin(tX)=\sum_{k=0}^\infty \frac{(-1)^kt^{2k+1}X^{2k+1}}{(2k+1)!},$$
but how can I do without characteristic function, and out measure theory ? (it's a lecture of introduction of probability, so we don't have tools of measure theory as DCT or MCT, neither characteristic function).

My definition of the expectation is : for $X$ s.t. $\mathbb E|X|<\infty $, $$\mathbb E[X]=\lim_{n\to \infty }\mathbb E[X_n]=\sum_{k\in\mathbb Z}\frac{k}{2^n}\mathbb P\left(X_n=\frac{k}{2^n}\right),$$
where $X_n=2^{-n}\lfloor 2^nX\rfloor$.

答:
Are moment generating functions OK? Since $e^{sX}=\sum_{n=0}^\infty\frac{s^nX^n}{n!}$ and $e^{tY}=\sum_{m=0}^\infty\frac{t^m Y^m}{m!}$, by linearity
$$M_{X,Y}(s,t)=\mathbb E\left[e^{sX}\cdot e^{tY}\right]=\mathbb E\left[e^{sX}\right]\cdot\mathbb E\left[e^{tY}\right]=M_X(s)\cdot M_Y(t).$$
So $X,Y$ are independent.

This is probably similar to whatever method you used with characteristic functions, but I think normally MGFs are regarded as more elementary.


问:
Questions about intersection of algebraic subgroups
I have a question related to Proposition 1.49 of the book Algebraic Group of Milne. Here is the excerpt:

My question is the last line. I don't really understand the reasoning in that line. From what I understand, $H_j$ is a closed subscheme of $G$, which is affine, so if $G = Spec(R)$, then $H_j \subset G$, and $I(H_j) = \bigcap_{p \in H_j}p$, which is an ideal of $R$. So each $f \in I(H_j)$ is an element of $R$, which can be considered as a function on $Spec(R) = G$. How does this function $f$ act on some $g \in G(R) = Hom(SpecR, G)$? And why is that line true eventually?

If you don't understand any notion/symbol from the excerpt, please tell me. Thank you

答:
This notation is explained in 3.2 of the book. Let $G$ be an affine algebraic group.

For a $k$-algebra $R$, $G(R)\simeq\mathrm{Hom}_{k\text{-algebra}}(\mathcal{O}{}(G),R)\simeq
\mathrm{Hom}_{R\text{-algebra}}(\mathcal{O}{}(G)_{R},R)$.

An $f\in\mathcal{O}(G)$ defines an evaluation map
$f_{R}\colon G(R)\rightarrow R,\quad g\mapsto g(f)=f_{R}(g),$
which is natural in $R$.

If $H$ is the subgroup of an affine group scheme $G$ corresponding to an ideal $I$, then $H(R)=Hom(\mathcal{O}/I,R)
=\{g\in G(R)=Hom(\mathcal{O},R)|g(f)=0\text{ all }f\in I\}$


问:
Show that $\int_0^\pi \log( 1 - 2r\cos(t) + r^2)\, dt=0$
Show that for $r \in (-1,1)$

$$ \int_0^\pi \log( 1 - 2r\cos(t) + r^2)\, dt = 0$$

Here's what I did so far:

$$f(r,t) = \log(1 - 2r\cos(t) + r^2) = \log( (1-re^{it})(1-re^{-it}))$$
The Leibniz rule states that $$\dfrac{d}{dr} \int_0^\pi f(r,t)\ dt = \int_0^\pi \dfrac{\partial}{\partial r} f(r,t) \ dt$$

After calculating the right part I found $2\pi r$ which means that $\displaystyle\int_0^\pi f(r,t)\ dt = \pi r^2$ when it should be $0$.

Thanks in advance.

答:
By substitution $(z=e^{it})$ and computing residues, we have the well known integrals:

$$J_1 = \int_0^{2\pi} \frac{dt}{1-2r\cos t+ r^2} = \frac{2\pi}{1-r^2}, \quad (|r|<1),$$
$$J_2 = \int_0^{2\pi} \frac{\cos t\, dt }{1-2r\cos t+ r^2} = \frac{2\pi r}{1-r^2}, \quad (|r|<1).$$

Let $$J_0(r) = \int_0^{2\pi} \log \left( {1-2r\cos t+ r^2} \right) dt$$

$$\frac{dJ_0}{dr} = \int_0^{2\pi} \frac{\partial  \log \left( {1-2r\cos t+ r^2} \right)}{\partial r}dt=2 \int_0^{2\pi} \frac{r-\cos t}{ {1-2r\cos t+ r^2} }dt=rJ_1-J_2=0.$$
$$J_0(r) = \text{const}.$$
Since $J_0(0)=0,$ we have $$J_0(r) = 0,$$ but this is twice the integral we want, which is thus zero.


问:
Geometric sum with binomial coefficient
I'm looking for any kind of formula or insight concerning

$$
f(k,N)=\sum_{n=0}^N\binom{n}{k}\lambda^n
$$

There are loads of binomial identities, so maybe I was just missing something, but I couldn't find anything yet.

Addendum:
Further assumptions on $\lambda$, i.e. $\lambda=\exp(2\pi i\alpha)$ yield to the following problem: Gaussian like sum with binomial coefficients

答: To get a sum of about $k$ terms (opposed to $N$ originally), use $$\sum_{n=0}^N\binom{n}{k}\lambda^n=\frac{\lambda^k}{k!}\frac{d^k}{d\lambda^k}\sum_{n=0}^N\lambda^n=\frac{\lambda^k}{k!}\frac{d^k}{d\lambda^k}\frac{1-\lambda^{N+1}}{1-\lambda}$$ and apply Leibniz rule. This can also be viewed as a closed-form answer for a fixed $k$.

Title: Geometric sum with binomial coefficient


问:
Show that $1/x > \log(x)^2$, for all real $x < 1$
so i'm taking a Calc 1 class right now, and I am supposed to evaluate, whether there is a constant continuation of $f(x) = x^x$, where x approaches zero and is a positive real number.

I have a function $a(x) := e^x$, which is always higher than $f(x)$ for $x < e$ and $a(0) = 1$.

My other function is $b(x) := - \sqrt x +1$.  I know, that: $b(0) = 1$.

Right now, I want to prove, that: $a(x) > b(x), \forall  x \in (0, 1]$.

I already used the Bernoulli-Inequality, to get:

$-\sqrt x < \log(x^x) = x \log(x)$

Which should be equivalent to showing:

$1/x > \log(x)^2$

I already found out, that $\log(x)^2 = \log(1 / x)^2$ for $x < 1$,
but still, im a bit stuck here and don't know how to proceed.
Furhermore, I mustn't use differentiation, as we didn't introduce it yet in our course.

Any help would be appreciated, thanks!

答:
For $x\le0$, the inequality makes no sense .

For $x\in]0,1]$, set $y=\frac{1}{\sqrt x}$. Then $y\in[1,\infty[$. So $$\frac1x>\ln(x)^2\iff y^2>\ln\left(\frac1{y^2}\right)^2\iff y>-\ln\left(\frac1{y^2}\right)$$

The RHS is equal to $2\ln(y)$, so we have to prove $y>2\ln(y)$ for $y\in[1,\infty[$. Now,

$$y>2\ln(y)\iff \exp\left(\frac y2\right)>y,$$ which is true thanks to the classical inequality that for every $r\geq0$, $\exp(r)\geq 1+r+\frac{r^2}2$.

Indeed, setting $r=\frac y2$ gives us $$\exp\left(\frac y2\right)\geq\frac18 (y^2+4y+8)=\frac18 ((y-2)^2+8y+4)> y.$$

Remark. In fact, this is harder to prove, but the inequality is true for all $$x\in\left]0, \exp(2 W(1/2))\right[,$$ where $W(1/2)\approx 0.35$ is the Lambert-W-function applied to $\frac12$.


问:
How to prove this problem about uniform continuity
If $f$ Continuous bounded on $\mathbb{R}$，and
$$\lim_{h\to0}\sup_{x\in\mathbb{R}}\left|f(x+h)-2f(x)+f(x-h)\right|=0.$$
prove that $f$ uniformly continuous on $\mathbb{R}$.
I try to find a $\delta>0$ to make it arbitrary $x_1,x_2\in \mathbb{R}$and$\left|x_1-x_2\right|<\delta$have$$\left|f(x_1)-f(x_2)\right|<\varepsilon.$$but I don't know how to use this complex limit expression.I hope you can give me some tips, thank you!

答:
Let $M>0$ be an upper bound for $|f(x)|$, and $\epsilon > 0$ be given. From the given condition it follows that there is a $\delta > 0$ such that
$$
 |f(x+h)-2f(x)+f(x-h)| < \epsilon
$$
for all $x \in \Bbb R$ and all $h \in (0, \delta)$. We will now show that
$$ \tag{*}
 |f(x) - f(y) | \le 2 \sqrt{M \epsilon}
$$
for all $x, y\in \Bbb R$ with $|x-y| < \delta$, which implies that $f$ is uniformly continuous.

In order to prove $(*)$ we use the following result from Prove or disprove $\lim\limits_{n \to \infty}\Delta x_n=0.$ :

Let $(x_n)$ be a sequence of real numbers, and let $\Delta x_n = x_{n+1}-x_n$, $\Delta^2 x_n = \Delta x_{n+1}-\Delta x_n$ be the first and second differences.

If $|x_n| \le M$ and $|\Delta^2 x_n| \le K$ for all $n$ then $|\Delta x_n|^2 \le 4MK$ for all $n$.

For fixed $x, y \in \Bbb R$ with $0 < |x-y| < \delta$ we can apply this result to the sequence
$$
 f_n = f(x + n(y-x))
$$
which satisfies $|f_n| \le M$ and $|\Delta^2 f_n| \le \epsilon$. It follows that
$$
 |f(x) -f(y) | = |\Delta f_0| \le 2 \sqrt{M \epsilon}
$$
and this completes the proof.


问:
Tutte's 1947 proof and paper on a family of cubical graphs
It is known that if a graph is connected, cubic, simple and $t$-transitive, then $t \le 5$. A proof is given in [Biggs, Algebraic Graph Theory, Chapter 18], and this result is due to [Tutte, ``A family of cubical graphs,'' Proc. Cambridge Philosophical Society, 45, 459-474].  </p>

<p>My question is: Is the proof given in Biggs' text the same as the one in Tutte's paper?  I was unable to obtain Tutte's paper.  I would appreciate if someone could electronically post or mail to me his paper.  </p>

答: It is essentially the same. Richard Weiss produced shorter proofs later - there was a lot of work on $s$-arc regular and $s$-arc transitive graphs - but even these used the same basic strategy.</p>


问:
Conventions for defining inductive base case
I was reading a problem that stated:

There are a certain number of objects all of which have the same
weight except for one which has a different weight. In all other
aspects the objects are identical. Determine the unique one that has
the different weight using a pair of scales optimally. Show by induction on $n$ that at most $2*n$ comparisons are needed to identify the object when the total number of objects is $3^n$

My original thought was that the base case has to be $n = 1$ where we have a total of $3^1$ objects which we can figure out the one we are searching for with 2 comparisons.

But the solution uses as base case $n = 0$ considering in that case that the $3^0 = 1$ object is the one we are searching for. I originally rejected $n = 0 $ as base case because with $1$ object we can't do any comparison so we can't know if it is the one we search for or not.

So my question is: how do we determine in such cases how do we treat the base case? Could we also not have considered that for $n = 0$ the $1$ coin can not be the one we are searching for? Or even that as I originally thought that it is not meaningful to use $n = 0$ as base case?

Is there something in the problems we need to be cautious of, or some convention that exists that helps define the base case?

In this particular problem, it makes me think that there is some subtle implication that the specific unique object under search must be part of the set even if all the rest are missing which I did not pick up.

Is that so? Is there some convention for inductive approaches?

答:
You use whatever works for your base case, then carefully state what you have proved.  If you use $n=1$ as your base case, you have not proved the $n=0$ case.  You would have to state your result as "For all numbers of objects $n \ge 1$, it takes at most $2n$ comparisons ..."

If I were doing the proof, I would probably start with $n=1$.  Having done it, I would think about $n=0$ for completeness and prove the $n=0$ case separately.  I would wind up proving $n=0$ and $n=1$ directly, then using induction for everything above $1$.  The case of $n=0$ is often different and it can be easier to make the induction work starting from $n=1$.

The base case can be higher yet.  One we see here is to prove $n! \gt n^2$ for $n \ge 4$.  We can note that the statement is not true for $n=3$ as $3!=6 \not \gt 3^2=9$.  You use $4$ as the base case and the induction goes through nicely.


问:
How to show that a Group is not profinite?
I tried to show that $\mathbb{Z}$ is not profinite.
So I know that a topological group $G$ is profinite if $G\cong \underleftarrow{\lim}~ G/H$, with
$H$ all open normal subgroups of $G$. And I know
that there is no surjective map $\mathbb{Z}\longrightarrow\underleftarrow{\lim}~\mathbb{Z}/n\mathbb{Z}$.
So $\mathbb{Z}$ is not profinite with the profinite topology.
But how to show that there is no other topology that $\mathbb{Z}$ is profinite?

答:
An inverse limit of finite groups is either finite or uncountable. $\mathbb Z$ is neither of those.

For the proof, consider an inverse sequence of finite groups:
$$\cdots \xrightarrow{f_i} G_i \xrightarrow{f_{i-1}} \cdots \xrightarrow{f_2} G_2 \xrightarrow{f_1} G_1 \xrightarrow{f_0} G_0
$$

Case 1: Suppose there exists $I$ such that for all $i \ge I$ the map $f_i$ is an isomorphism. Then the inverse limit is isomorphic to $G_I$ and hence is finite.

Case 2: Suppose for all $I$ there exists $i \ge I$ such that the map $f_i$ is not an isomorphism. Thus one can find a subsequence $i_1 < i_2 < i_3 < \ldots$ such that $f_{i_k}$ is $n_k$-to-one where $n_k \ge 2$. Now one uses a Cantor-type diagonalization argument to prove that the inverse limit is uncountable.

ADDED: As pointed out in the comment of @AlexKruckman, you can also get exact cardinality $2^{\aleph_0}$, but that's more than you asked and so is really another question.


问:
$M||\textbf{x}||$ $\leq$ $f(\textbf{x})$?
Given a function $f$: $(V, || \circ ||_{\infty})$ $\rightarrow$ $(\mathbb{R}, |\circ |)$

how can you then show that there exists  $M$ $>$ $0$ such that $M||\textbf{x}||$ $\leq$ $f(\textbf{x})$ for every $\textbf{x} \in \mathbb{R}$?

I thought maybe that it could be done by showing that $f$ is a bounded linear operator but i'm not sure that it is.

答:
$\newcommand{\R}{\mathbb{R}}$
$\newcommand{\norm}[1]{\left\Vert #1 \right\Vert}$
Homogenuity of a function gives $f(sx) = |s|f(x)$ for all $s \in \R$ and all $x\in \R^n$.
We have
$$
f(x) = \norm{x}f\left(\frac{x}{\norm{x}}\right)
$$
for all $x \in \R^n\setminus\{0\}$. Therefore, if we find $a > 0$ such that $f(x/\norm{x}) \geq a$ for all $x \in \R^n\setminus\{0\}$, we are done. But as we know,
$$
\norm{\frac{x}{\norm{x}}}
= \frac{1}{\norm{x}}\norm{x}
= 1
$$
This is equivalent that we have to find $a > 0$ such that $f(y) \geq a $ for all $\norm{y} = 1$, that is,
$$
a \geq c= \min_{\norm{y}=1}f(y) > 0
$$
Hence, the smallest such coefficient is $c$.


问:
Which one is correct? Don’t we do as necessary as rigorous math?
The mathematical statement $$\exists h\in \mathbb{R} \quad  \dfrac{1}{h}=5 $$ seems a mathematical statement that is true.

However, its negation $$\forall h\in \mathbb{R} \quad  \dfrac{1}{h}\neq 5$$ is not a mathematical statement, because it claims that $\displaystyle\frac10\ne5,$ which is nonsense since $0$ does not have a multiplicative inverse.

Consequently, the two statements either are not negations of each other or are mathematically nonsense. Which is it?

And how do I formalise “there exists a real number whose multiplicative inverse is $5$” ?

答:
Assuming we interpret division to mean the usual notion of division in the real numbers, you're correct that the statement $(\exists h \in \mathbb{R})(1/h = 5)$ doesn't quite make sense, for the reason you stated. A more formal way of putting it is that $1/h = 5$ is not, in fact, a predicate with domain $\mathbb{R}$, so we can't quantify it over $\mathbb{R}$.

There are a few ways to correct this:

Rewrite the statement as $(\exists h \in \mathbb{R} \setminus \{0\})(1/h = 5)$ instead, so the domain matches the domain of division of real numbers.

Rewrite the statement as $(\exists h \in \mathbb{R})(1 = 5h)$, noting that $1/h = 5$ is equivalent to $1 = 5h$ for all $h \neq 0$, but the latter also makes sense (though it is false) for $h = 0$. I'd say this is the most straightforward way to formalize "there exists a real number whose multiplicative inverse is $5$".

Interpret the division symbol to mean division in the real projective line $\hat{\mathbb{R}}$, in which $1/0$ is a meaningful expression and is equal to $\infty$, a "point at infinity" that we add to the standard real numbers. With the division symbol interpreted in this way, $1/h = 5$ now makes sense for all $h \in \mathbb{R}$ (in fact, for all $h \in \hat{\mathbb{R}}$).

In practice, issues like this are often somewhat informally glossed over, because the intended meaning of the statement is clear and there's no ambiguity, as all these ways of resolving the formal syntactical issue lead to the same conclusion. But it's good to understand how a slightly informal notation like that can be correctly rewritten/reinterpreted to resolve any formal syntactical problems.


问:
Absolutely Homogenous Functions on Normed Spaces
I was given the following problem and am lost for where to start.

Let $\mathbb{R}^{m}$ be the real vector space of $m$-tuples of real numbers and let $\Vert \cdot \Vert : \mathbb{R}^{m} \to [0,\infty)$ be a norm on $\mathbb{R}^{m}$. Let $f: (\mathbb{R}^{m}, \Vert \cdot \Vert) \to (\mathbb{R},|\cdot|)$ be a continuous function with $f(\mathbf{x})>0$ for all non zero $\mathbf{x}\in\mathbb{R}^{m}$. Show that if $f$ is absolutely homogenous (i.e. $f(t\mathbf{x})=|t|f(\mathbf{x})$ for every real $t$ and every $\mathbf{x}\in\mathbb{R}^{m}$) then there exists a positive constant $\alpha > 0$ such that $\alpha \Vert \mathbf{x} \Vert \leq f(\mathbf{x})$ for every  $\mathbf{x}\in\mathbb{R}^{m}$

As I said I have no idea even where to start for this. Any advice is much appreciated.

答:
All norms on a finite dimensional space are equivalent, which in particular implies that the set $C = \{x\in \mathbb{R}^m: \|x\|=1\}$ is compact. The continuous function $f$ must attain a minimum on $C$ which is strictly positive, say $0<c = \min_{x\in C}f(x)$.

Then if $x\in \mathbb{R}^m\setminus \{0\}$ is arbitrary we conclude that

$$c\leq f(x/\|x\|) = \frac{1}{\|x\|}f(x)$$
since $x/\|x\|\in C$.


问:
Integral with residue theorem
I have to resolve the following integral for a proof of theorem. It is used the residue theorem. However I am not in confidence with this argument. I would like to have more detailed step. The integral is
$$\int_{-\infty}^{\infty}\frac{{\rm d}k_{1}}{2\pi} \frac{e^{{\rm i}k_{1}\,x_{1}}}{{\rm i}k_{0} + k_{1}}$$
The integral can be solved in the comples plane in the half circle with radius $R\rightarrow \infty$ that rests on the real axis and lies in the upper half plane.

$$\int_{-\infty}^{\infty}\frac{dk_1}{2\pi} \frac{e^{ik_1 x_1}}{ik_0+k_1}=\int_{half\,circle}\frac{dk_1}{2\pi} \frac{e^{ik_1 x_1}}{ik_0+k_1}
$$
The integral on the circumnference goes to zero, infact, using the change of variable $k_1=Re^{i\theta}$, one has:
\begin{align}
&\int_{\smallfrown R}\frac{dk_1}{2\pi} \frac{e^{ik_1 x_1}}{ik_0+k_1}
\\[5mm] = &\
\int_{0}^{\pi} \frac{d\theta i Re^{i\theta}}{2\pi}\frac{e^{ix_1Rcos\theta}e^{-x_1Rsin\theta}}{ik_0+Re^{i\theta}}\rightarrow 0 \,\,\,\mbox{for}\,R\rightarrow\infty
\end{align}
And here the first question: how can it goes to zero?

At this point one can resolves using the residue theorem. The unique pole is in $k_1=-ik_0$ so to have a non zero result, one needs $k_0<0$. Why is the integral equal to zero in the case $k_0>0$?

So in the hypothesis that $x_1>0$ and $k_0<0$ one has:
$$\int_{-\infty}^{\infty}\frac{dk_1}{2\pi} \frac{e^{ik_1 x_1}}{ik_0+k_1}=ie^{k_0x_1}$$

答:
For the half circle, the integral is bounded, up to a constant, by
$$
\int_0^\pi e^{-x_1 R\sin \theta}\,d\theta=2\int_0^{\pi/2}e^{-x_1 R\sin\theta}\,d\theta
\leq 2\int_0^{\pi/2}e^{-x_1R\theta/2}\,d\theta=\frac{2(1-e^{-x_1R\pi/4})}{x_1 R}\to0.
$$
Or, even easier, you can apply Dominated Convergence.

When $k_0>0$, the pole lies outside your semicircle, so Cauchy's Theorem gives you that the integral on $[-R,R]$ is the same as the integral on the semicircle.

Title: Integral with residue theorem


问:
Formula for eigenvectors of 2x2 matrices
I came across http://people.math.harvard.edu/~knill/teaching/math21b2004/exhibits/2dmatrices/index.html and I know that for a $n \times n$ Matrix there are at most $n$ eigenvectors. So what happens when b and c are not zero? There can't be four eigenvectors so is it me or the website missing something?

答:
Your statement "an $n \times n$ matrix has at most $n$ eigenvectors" is incorrect. In particular, if a matrix $\mathbf{A} \in \mathbb{R}^{n \times n}$ has an eigenvector $\mathbf{v} \neq \mathbf{0}$ with corresponding eigenvalue $\lambda \in \mathbb{R}$, then $\mu \mathbf{v}$ is also an eigenvector for all $\mu \in \mathbb{R} \setminus \{0\}$. So if there is an eigenvector, then there are infinitely many of them.

Thus, there being four eigenvectors for a $2 \times 2$ matrix is not a contradiction.


问:
How to solve matric question containing exponents
I am stuck on this matrix question as it contains complex numbers as exponents and I don't know how to go about it if anyone can guide me or show a worked solution it will be a massive help. Thanks in advance.

Question

答:
I will explain some complex numbers.

Consider the imaginary plane to have an $x$-axis consisting of the real part of the imaginary number and the $y$-axis consisting of the imaginary part. ie $3+4i$ is just the point (3,4).

in light of that consider polar coordinates. The angle $\theta$ is measured anticlockwise from from the positive real axis and the radius is as expected, both of which can be calculated with simple trigonometry.

Hence a complex number may be written as $z=re^{i\theta}$ where $r$ is the radius, and $e^{i\theta}= cos(\theta)+isin(\theta)$. I will leave you to prove this yourself using trigonometry and various Taylor expansions.

Now onto your question.

For $AB$ note that $e^{i\theta}e^{-i\theta}=(cos(\theta)+isin(\theta))(cos(-\theta)+isin(-\theta))=(cos(\theta)+isin(\theta))(cos(\theta)-isin(\theta))=cos^2(\theta)+sin^2(\theta)=1$

now apply this to the $2\times 2$ matrix you have and you should be fine!


问:
Subfields of splitting field
I recently started learnig field and Galois theory and was given this problem:

Given polynomial over Q

$\displaystyle f=2x^{6} \ −\ 4x^{5} \ −\ 3x^{4} \ −\ x^{3} \ +\ 8x^{2} \ +\ 6x\ −\ 6\ \in \ Z[ x] \ $

find splitting field K of f, and determine degree [K : Q].

Then find all subfields M of K satisfying [K : M]=2.

I managed to find splitting field K as $\displaystyle \mathbb{Q}\left( \sqrt[3] 2,\zeta _{3,}\sqrt{3}\right) \ $
and [K : Q] = 12

where $\displaystyle \zeta _{3}$ is 3rd root of unity.

I also think that galois group  of f is non abellian.

However when I try to figure out to what group it is precisely I get stuck.
If I missunderstood something or made mistake, please let me know. If not, how should I proceed?

答:
Hint: We have
$$
f=(x^3 - 2)(2x^2 - 6x + 3)(x + 1).
$$

Further references:

Galois group of a reducible polynomial over $\mathbb {Q}$

Galois group of a reducible polynomial

Title: Subfields of splitting field


问:
How to cut a cube out of a tree stump, such that a pair of opposing vertices are in the center?
I saw this picture of a cube cut out of a tree stump.

I've been trying to craft the same thing out of a tree stump, but I found it hard to figure out how to do it.

One of the opposing vertices pair is on the center of the tree stump:

I've been struggling to find the numbers and angle needed to make the cuts.

Any kind of advice would be greatly appreciated. Thanks for taking your time to read this and I'm sorry for my bad english.

答:
I hope the image will help cause otherwise I am not sure how to explain this without a visual support.

In the upper part of the image we have a rhombus ABCD that, when viewed from a top view, looks like the image on bottom of the figure. The point that corresponds to the center of the log is C.

Now the hard part that requires visualizing the process. Imagine that we can lift the point A on a line that is orthogonal on the initial plane of the rhombus. The ABCD shape remains a rhombus if both points B and C are also lift accordingly.

Using this process, we have to lift point A such that the new shape described by the shifted points A'B'CD' (C remains fixed) becomes a square, which is equivalent to having A'C=B'D' (after the shift of the points!)

With this info we can compute the angle $\theta$ between the now square and the initial plane using

$$\cos(\theta) = \frac{AC}{A'C} = \frac{AC}{BD}.$$

This formula is applied in the resulting triangle ACA'.

Now we have to relate the formula for this angle with the geometry of the log. A top view of the log with the cube inserted in it would look something like the following image where the point C is on the top or bottom flat surface of the log.

From this perspective it should be clear that during the shifting of mentioned above it is true that BD=B'D'. But BD is the edge of an equilateral triangle, and DC is the radius $r$ of the cylinder. So from this point you can compute the radius by knowing the edge of the triangle which gives

$$r = DC = BD \frac{\sqrt{3}}{3}.$$

This links the radius of the log with the angle $\theta$ through BD.

The final step is to link the height of the log with the angle. This should follow easily by observing that the height should be

$$h = l_{cube} \sqrt{3}$$

where $l_{cube}$ is the length of the edge of the cube. From the previous formula

$$\cos(\theta) = \frac{AC}{BD} = \frac{r}{l_{cube}\sqrt{2}}$$

which gives

$$r = \frac{\sqrt{2}}{3}h.$$

Now to put everything in terms of the shape of the log, the angle at which the cuts have to be made is

$$\theta = \arccos(\frac{1}{\sqrt{3}}) \approx 54.73^o$$

and the goemetry of the log has to satisfy the condition

$$r = \frac{\sqrt{2}}{3}h.$$

I would assume that the directions along which the cuts have to be made is not a problem for you. For others however, one can mark the center of a flat surface of the log and then draw 3 segments that have a $120^o$ between them. Then the cuts have to be made at the above mentioned angle with the drawn line being orthogonal to the intersection of the cutting plane and the flat surface.

EDIT: Correction applied based on comment from Joshua Wang. Thank you!


问:
Canonical examples of inner product spaces that are not Hilbert spaces?
That is, what are some good examples of vector spaces which are inner product spaces but in which not every Cauchy sequence converges?</p>

答: Lots of natural subspaces of Hilbert spaces aren't Hilbert spaces (which is the same thing as saying they're not closed). Two examples that come to mind are the subspace of sequences in $\ell^2(\mathbb{Z})$ at most finitely many terms of which are nonzero (the compactly supported sequences) and the continuous functions $C([0, 1])$ under the $L^2$-norm. (Note that you don't have to quotient out by anything here because it's already true that a continuous function with zero $L^2$-norm is identically zero.) </p>


问:
Do the adjoint functor theorems usefully dualise?
The special and general adjoint functor theorems exist to construct left adjoints to particular functors given certain conditions on them. However, I've not been able to find much mention – at least, not in my lecture notes nor in a quick Google search – of using them or something like them to construct right adjoints.

I suppose that if $F \dashv G$ then $G^\mathrm{op} \dashv F^\mathrm{op}$, so I can just apply the adjoint functor theorems to the opposites of the categories I'm interested in. Perhaps something like:

General Adjoint Functor Cotheorem: suppose $\mathcal C$ is locally small and cocomplete. Then a functor $F : \mathcal C \to \mathcal D$ has a right adjoint if and only if it preserves small colimits and, for each object $A$ of $\mathcal C$, the comma category $G \downarrow A$ has a weakly terminal set.

Special Adjoint Functor Cotheorem: suppose $\mathcal C, \mathcal D$ both locally small, $\mathcal C$ cocomplete, co-well-powered1, and with a separating set. Then a functor $F : \mathcal C \to \mathcal D$ has a right adjoint iff $F$ preserves small colimits.

Is that right? Is it useful?

1 Not sure if this is standard terminology, but I hope it's clear what I mean: for each object $A$, there is a category of epimorphisms with domain $A$, and a category would be co-well-powered if all these epimorphism categories were equivalent to a partially ordered set (they are already preorders, since if $fg = h$ with $g$ and $h$ epimorphisms, then $f$ is unique)

答:
as a sort of meta answer: there are more left adjoints than right adjoint (this is of course nonsense!!). More to the point, quite often we are interested in finding left adjoints to functors (in order to create free objects (which include all of the uses you mentioned)). Also quite often, the functors that typically interest us do not have right adjoints at all. This is related to the fact of life that mathematics is not self-dual. For whatever reason, many of the naturally occurring functors we meet tend to have left adjoint but often they lack right adjoints.

So, while category theory is self-dual, practiced mathematics is not. A similar phenomenon is the common concept of a cofibrantly generated model category, while there are very very few examples of naturally occurring categories that support the dual notion of a fibrantly generated model category.

Notice that this lack of duality in practiced mathematics starts very early on. The concept of cartesian product of sets is much more important than the dual notion of disjoint union. And while noticing this, it is the cartesian product of set that is at the heart of the definition of category. So the bias towards one concept over its dual is very deeply rooted.


问:
Show that $ax^2+bx+c \equiv 0 (p^2)$ has $0, 1, 2 \text{ or } p$ solutions
$f(x):=ax^2+bx+c$, $p$ an odd prime

I know that $f(x) \equiv 0 (p)$ can have $0, 1 \text{ or } 2$ solutions depending on the value of $b^2 - 4ac$. So using this, I started developing this and couldn't find a case where $f(x) \equiv 0 (p^2)$ might have just one solution. I've found the following:

If $f(x) \equiv 0 (p)$ has $0$ solutions, $f(x) \equiv 0 (p^2)$ will have no solutions (if there is one, then it's also a solution for the first one)

If $f(x) \equiv 0 (p)$ has 1 solution, means $b^2-4ac\equiv 0(p)$ and from a theorem I've seen, if $u$ is a solution for $f(x) \equiv 0 (p)$ then $2au+b\equiv 0 (p)$. So what I did was see what happens with $f(u+tp)$ mod $p$. After some calculations, I've reached that $f(u+tp)\equiv p(q+pq't)(p^2)$ where $pq=f(u)$ and $pq' = 2au+b$, so $f(u+tp)\equiv 0 (p^2) \iff q \equiv0(p)$, so it has either $0$ solutions or $p$ solutions

If $f(x) \equiv 0 (p)$ has 2 solutions, then these are $u, -u$ for some $u$, since $p$ is odd these are not the same number. So, calculating again $f(u+tp)$, I reach that the condition is that either $q+(2au+b)t \equiv 0(p)$ or $q+(-2au+b)t \equiv 0(p)$, and since neither of $2au+b$ and $-2au+b$ are $0$ because these are the solutions of $x^2 \equiv b^2-4ac \not\equiv 0(p)$, then for each one exists one t such that these hold, so there are two different solutions.

Is the book wrong? Or am I missing something? Thanks

答: $x \equiv 0 (p^2)$ is a simple example of such an equation with precisely 1 root modulo $p^2$.


问:
Operator acting on ONB decomposition
Consider the following setting: Given

a real Hilbert space $X$ with orthonormal basis $(e_k)_{k\in\mathbb{N}}$,

a linear (unbounded) operator $T$ with dense domain $D\subset X$ and that

$\{e_k\colon k\in\mathbb{N}\}\subset D$.

Let $x=\sum_{k=0}^\infty\alpha_ke_k\in X$. Is it true that $Tx=\sum_{k=0}^\infty\alpha_kTe_k$? If not, what are suitable assumptions for it to be true?

If $T$ was bounded this would clearly be true.

I have thought about closedness: Letting $x_n=\sum_{k=0}^n\alpha_ke_k$ we have $x_n\to x$ and $x_n\in D$. To conclude $Tx_n\to Tx$ I however still need that $Tx_n$ is convergent in $X$, which I don't think needs to be true in general.

What would be a suitable assumption on the operator $T$?

答:
In general this is not true:

The case $x_n\rightarrow x$ and $Tx_n\rightarrow y$ does not imply that $Tx = y$ in general:

Let $Y\subseteq l^2$ be the subspace

$$Y = \{(\alpha_k)_k\in l^2: \lim_{k\rightarrow \infty}k\alpha_k\text{ exists and is finite.}\}$$

Define $T:Y\rightarrow l^2$ by $T(\sum \alpha_ke_k) = (\lim_{k\rightarrow \infty}k\alpha_k)e_1$ then $T$ is linear and unbounded and furthermore $Te_k = 0$ for every $k$.

However if we let $x = \sum \frac{1}{k}e_k$ then $Tx = e_1$ and hence

$$T(\sum_{k=1}^{\infty}\beta_ke_k) = \sum_{k=1}^{\infty}\alpha_ke_k$$
does not imply that $\alpha_ke_k = \beta_kTe_k$ in general even though $e_k$ are eigenvectors.

Closed does not imply that $T(\sum_1^n \alpha_ke_k)\rightarrow \sum_{1}^{\infty} \alpha_kTe_k$ even though $\sum_1^\infty \alpha_ke_k$ exists.

Let $T(\sum \alpha_k e_k) = \sum k\alpha_k e_k$, then as a multiplication operator $T$ is closed. However if we let $x_n = \sum_{k=1}^n \frac{1}{k}e_k$ we know that $x_n\rightarrow \sum_{k=1}^{\infty}\frac{1}{k}e_k$ in $l^2$ while $Tx_n = \sum_{k=1}^{n}e_k$ diverges.

Title: Operator acting on ONB decomposition


问:
The spectrum of an invertible element $x$ is $\sigma(x^{-1})=\{\lambda^{-1}: \lambda\in \sigma(x)\}$
Suppose $x$ is invertible in the unital Banach algebra $A$.

How can I prove that $\sigma(x^{-1})=\{\lambda^{-1} : \lambda\in \sigma(x)\}$

答: If $x$ is invertible, $x \cdot v = \lambda v$ is equivalent to $v= x^{-1}x \cdot v= \lambda x^{-1} \cdot v$, that is $x^{-1} \cdot v=\lambda^{-1} v$.</p>


问:
Calculating $\int_{-1}^{1} e^{\sin(\frac{\pi}{2}t+\frac{\pi}{2})}dx$ by using gauss method
Today I had an exam ,
I’m reviewing my answers so I calculated one of the questions with maple after my exam and the answer is not anything near my answer
And also i don’t know what I’m doing wrong,i cannot find the flaw in my handcalculated answer

The question is :
Find the approximate integral using gauss method on two points
$$\int_{0}^{\pi} e^{\sin x} dx$$

What I did on paper :

$x=\frac{\pi}{2}t+\frac{\pi}{2}$

$\int_{-1}^{1} e^{\sin(\frac{\pi}{2}t+\frac{\pi}{2})}dt = f(\frac{-\sqrt{3}}{3})+ f(\frac{\sqrt{3}}{3}) $

$e^{\sin(\frac{\pi}{2} \frac{-\sqrt{3}}{3} +\frac{\pi}{2})} + e^{\sin(\frac{\pi}{2}\frac{\sqrt{3}}{3} +\frac{\pi}{2})} $

Then by expanding $\sin(a+b)$ we get

$\int_{-1}^{1} e^{\sin(\frac{\pi}{2}t+\frac{\pi}{2})}dx = 2.e^{\cos(\frac{-\sqrt 3}{6}\pi)}=3.70....  $

And by calculating it with maple it have to be somwhere around 6.2 which is absolutely not

Any help?

答: The problem is that you changed the variable in the function to integrate, but still have $dx$. That's not right. You need to replace $$dx=\frac{\pi}2 dt$$Then the answer will be approximately $\frac\pi 23.70=5.81$


问:
Determining the corresponding local field
Let $p$ be an odd prime number. Then there are three quadratic extensions of $\mathbb{Q}_p$. Assume that $t$ is a nonresidue module $p$, then these three extensions can be obtained by adjoining $\sqrt{t}$, $\sqrt{p}$ and $\sqrt{pt}$. Let $K/\mathbb{Q}$ be a Galois extension, for which the ramification index of $p$ is equal to $e=1$, and the inertia degree is equal to $f=2$. Let $\mathfrak{p}$ be a prime above $p$, then $K_{\mathfrak{p}}/\mathbb{Q}_p$ is an extension of degree $2$. But I can not decide to which of those three extensions it corresponds to. How can I find the corresponding extension? Which of these three extensions $\mathbb{Q}_p(\sqrt{t}), \mathbb{Q}_p(\sqrt{p}), \mathbb{Q}_p(\sqrt{pt})$ is obtained?

Edit: The comment by reuns answered my question. Now I have another similar question. How can we determine the corresponding extension, when $e=2$ and $f=1$? (The new question in detail: I mean Let $K/\mathbb{Q}$ be a Galois extension, for which the ramification index of $p$ is equal to $e=2$, and the inertia degree is equal to $f=1$. Let $\mathfrak{p}$ be a prime above $p$, then $K_{\mathfrak{p}}/\mathbb{Q}_p$ is an extension of degree $2$. But I can not decide to which of those three extensions it corresponds to. How can I find the corresponding extension? Which of these three extensions $\mathbb{Q}_p(\sqrt{t}), \mathbb{Q}_p(\sqrt{p}), \mathbb{Q}_p(\sqrt{pt})$ is obtained?)

答:
@NeoTheComputer  For $p$ odd, $\mathbf Q_p^*/{\mathbf Q_p^*}^{2}\cong {C_2}^2$. This comes from the general canonical decomposition $\mathbf Q_p^* \cong p^{\mathbf Z} \times  \mathbf \mu_{p-1} \times (1+p\mathbf Z_p)$, which implies that $\mathbf Q_p^*/{\mathbf Q_p^*}^{2}$ is as announced, with representatives ${1, p, u, up}$, where $u$  is a unit s.t. $(u/p)=-1$ . See e.g. Serre's "A course in arithmetic", chap. I, §3.2. I misread completely your question.

NB: It seems to me that your question has been answered by Reuns. What does still bother you ?


答: If there are two distinct eigenvalues, there is one eigenvector (up to scalar multiplication) associated with each, for a total of two.  If there is only one eigenvalue, it can be that all vectors in the space are eigenvectors, which gives two independent eigenvectors.  An example is the identity matrix.  There are matrices that have only one eigenvalue and one eigenvector.  An example is $$\begin {pmatrix} 1&1\\0&1 \end {pmatrix}$$ a possibility that the website does not mention.  In any case one would not consider there to be four eigenvectors.  f $b$ and $c$ are not both zero, the pairs given will be proportional to each other.

Title: Formula for eigenvectors of 2x2 matrices

Body: I came across http://people.math.harvard.edu/~knill/teaching/math21b2004/exhibits/2dmatrices/index.html and I know that for a $n \times n$ Matrix there are at most $n$ eigenvectors. So what happens when b and c are not zero? There can't be four eigenvectors so is it me or the website missing something?


问:
What are the math processes to get from $1-a$ to $1 - \frac{.1a}{1-.9a}$?
What are the math processes to get the following equation?
$$\begin{align}
&\;1 - a \\[4pt]
=&\; 1 -.9a-.1a \\[4pt]
=&\; 1 - \frac{.1a}{1-.9a}
\end{align}$$
I understand the second line, splitting the $a$ into $.9a$ and $.1a$. I just have no idea how they managed to get the final form.

答:
Obviously the case $a=0$ holds. Since
$$1-a=1-\frac{0.1a}{1-0.9a} \Rightarrow 1=\frac{0.1}{1-0.9a} \Rightarrow 1-0.9a=0.1a \Rightarrow 1=a,$$
your equations hold iff $a \in \{0,1\}$.


问:
Approximation of the Gamma function for small value
It is well known that $\Gamma(u) \stackrel{u \to 0}{\sim} \frac{1}{u}$.

I am looking for more precise information on the behavior of $\Gamma(x)$ when $x$ is small, ie: $x\to 0$.

My question is then, are there accurate (for small value) inequalities for the Gamma function ?

Any other information on the behavior is also very welcome.

答:
Laurent Series

Here is one way to show how the coefficients in GEdgar's answer were derived.

In this answer, it is shown that for $f(x)=\log(\Gamma(x))$, $f(1)=0$, $f'(1)=-\gamma$, and for $k\ge2$,
$$
f^{(k)}(1)=(-1)^k(k-1)!\,\zeta(k)\tag1
$$
From which it can be recursively derived, using the Leibniz Rule, that
$$
\begin{align}
\Gamma'(1)&=f'(1)\Gamma(1)&&=-\gamma\tag{2a}\\[12pt]
\Gamma''(1)&=f''(1)\Gamma(1)+f'(1)\Gamma'(1)&&=\frac{\pi^2}6+\gamma^2\tag{2b}\\
\Gamma'''(1)&=f'''(1)\Gamma(1)+2f''(1)\Gamma'(1)\\
&+f'(1)\Gamma''(1)&&=-2\zeta(3)-\frac{\pi^2}2\gamma-\gamma^3\tag{2c}\\
\Gamma^{(4)}(1)&=f^{(4)}\Gamma(1)+3f'''(1)\Gamma'(1)\\
&+3f''(1)\Gamma''(1)+f'(1)\Gamma'''(1)&&=8\gamma\zeta(3)+\frac{3\pi^4}{20}+\gamma^2\pi^2+\gamma^4\tag{2d}\\
&\ \,\vdots\\
\Gamma^{(n)}(1)&=-\gamma\,\Gamma^{(n-1)}(1)+\sum_{k=2}^n(-1)^k\frac{(n-1)!}{(n-k)!}\,\zeta(k)\,\Gamma^{(n-k)}(1)\hspace{-4cm}\tag{2e}
\end{align}
$$
Thus, Taylor's Theorem says,
$$
\begin{align}
\Gamma(1+x)
&=1-\gamma x+\left(\frac{\pi^2}6+\gamma^2\right)\frac{x^2}2-\left(2\zeta(3)+\frac{\pi^2}2\gamma+\gamma^3\right)\frac{x^3}6\\
&+\left(8\gamma\zeta(3)+\frac{3\pi^4}{20}+\gamma^2\pi^2+\gamma^4\right)\frac{x^4}{24}+\dots\tag3
\end{align}
$$
and therefore, since $\Gamma(x)=\frac1x\Gamma(1+x)$,
$$
\begin{align}
\Gamma(x)
&=\frac1x-\gamma+\left(\frac{\pi^2}6+\gamma^2\right)\frac{x}2-\left(2\zeta(3)+\frac{\pi^2}2\gamma+\gamma^3\right)\frac{x^2}6\\
&+\left(8\gamma\zeta(3)+\frac{3\pi^4}{20}+\gamma^2\pi^2+\gamma^4\right)\frac{x^3}{24}+\dots\tag4
\end{align}
$$
Here is a Mathematica implementation of $\text{(2e)}$ to compute $\Gamma^{(n)}(1)$:

```<code>G[0] = 1; G[n_] := G[n] =
  Sum[(-1)^k (n-1)!/(n-k)! Zeta[k] G[n-k], {k, 2, n}] - EulerGamma G[n-1]
</code>```

Simplify[G[4]] should give the value in $\text{(2d)}$.

A Decent Approximation

I have often used the slight modification of Stirling's formula
$$
\Gamma(x)\approx\sqrt{2\pi\left(x+\tfrac16\right)}\,\frac{x^{x-1}}{e^x}\tag5
$$
which gives very good results, even for $x$ close to $0$. Although the Laurent expansion, $(4)$, gives a better approximation for $0\lt x\lt\frac13$, approximation $(5)$ stays within $2\frac13\%$ of the correct answer all the way to $0$.

Although we computed to the third order in $(4)$, the graph only contains terms to the second order.

For large $x$, the relative error of the approximation in $(5)$ is $\sim-\frac1{144x^2}$, whereas Stirling has a relative error of $\sim-\frac1{12x}$.

Body:
Laurent Series

Here is one way to show how the coefficients in GEdgar's answer were derived.

In this answer, it is shown that for $f(x)=\log(\Gamma(x))$, $f(1)=0$, $f'(1)=-\gamma$, and for $k\ge2$,
$$
f^{(k)}(1)=(-1)^k(k-1)!\,\zeta(k)\tag1
$$
From which it can be recursively derived, using the Leibniz Rule, that
$$
\begin{align}
\Gamma'(1)&=f'(1)\Gamma(1)&&=-\gamma\tag{2a}\\[12pt]
\Gamma''(1)&=f''(1)\Gamma(1)+f'(1)\Gamma'(1)&&=\frac{\pi^2}6+\gamma^2\tag{2b}\\
\Gamma'''(1)&=f'''(1)\Gamma(1)+2f''(1)\Gamma'(1)\\
&+f'(1)\Gamma''(1)&&=-2\zeta(3)-\frac{\pi^2}2\gamma-\gamma^3\tag{2c}\\
\Gamma^{(4)}(1)&=f^{(4)}\Gamma(1)+3f'''(1)\Gamma'(1)\\
&+3f''(1)\Gamma''(1)+f'(1)\Gamma'''(1)&&=8\gamma\zeta(3)+\frac{3\pi^4}{20}+\gamma^2\pi^2+\gamma^4\tag{2d}\\
&\ \,\vdots\\
\Gamma^{(n)}(1)&=-\gamma\,\Gamma^{(n-1)}(1)+\sum_{k=2}^n(-1)^k\frac{(n-1)!}{(n-k)!}\,\zeta(k)\,\Gamma^{(n-k)}(1)\hspace{-4cm}\tag{2e}
\end{align}
$$
Thus, Taylor's Theorem says,
$$
\begin{align}
\Gamma(1+x)
&=1-\gamma x+\left(\frac{\pi^2}6+\gamma^2\right)\frac{x^2}2-\left(2\zeta(3)+\frac{\pi^2}2\gamma+\gamma^3\right)\frac{x^3}6\\
&+\left(8\gamma\zeta(3)+\frac{3\pi^4}{20}+\gamma^2\pi^2+\gamma^4\right)\frac{x^4}{24}+\dots\tag3
\end{align}
$$
and therefore, since $\Gamma(x)=\frac1x\Gamma(1+x)$,
$$
\begin{align}
\Gamma(x)
&=\frac1x-\gamma+\left(\frac{\pi^2}6+\gamma^2\right)\frac{x}2-\left(2\zeta(3)+\frac{\pi^2}2\gamma+\gamma^3\right)\frac{x^2}6\\
&+\left(8\gamma\zeta(3)+\frac{3\pi^4}{20}+\gamma^2\pi^2+\gamma^4\right)\frac{x^3}{24}+\dots\tag4
\end{align}
$$
Here is a Mathematica implementation of $\text{(2e)}$ to compute $\Gamma^{(n)}(1)$:

G[0] = 1; G[n_] := G[n] =
  Sum[(-1)^k (n-1)!/(n-k)! Zeta[k] G[n-k], {k, 2, n}] - EulerGamma G[n-1]

Simplify[G[4]] should give the value in $\text{(2d)}$.

A Decent Approximation

I have often used the slight modification of Stirling's formula
$$
\Gamma(x)\approx\sqrt{2\pi\left(x+\tfrac16\right)}\,\frac{x^{x-1}}{e^x}\tag5
$$
which gives very good results, even for $x$ close to $0$. Although the Laurent expansion, $(4)$, gives a better approximation for $0\lt x\lt\frac13$, approximation $(5)$ stays within $2\frac13\%$ of the correct answer all the way to $0$.

Although we computed to the third order in $(4)$, the graph only contains terms to the second order.

For large $x$, the relative error of the approximation in $(5)$ is $\sim-\frac1{144x^2}$, whereas Stirling has a relative error of $\sim-\frac1{12x}$.


问:
curl identity on $\mathcal{D}'\left(\mathbb{R}^2\right)$ for $v\in H^1\left(\mathbb{R}^2\right)^2$
I need to show that if $u\in H^1\left(\mathbb{R}^2\right)^2$ such that $\text{div }u=0$, then if
$$w=\text{curl }u=\partial_1 u_2-\partial_2 u_1$$
in $\mathcal{D}'\left(\mathbb{R}^2\right)$ (space of distributions),
we have
$$\text{curl }\left(\left(u\cdot\nabla\right)u\right)=u\cdot \nabla w$$
where $$\left(u\cdot\nabla\right)u=\left(\sum_j u_j\partial_j u_i\right)_i$$
I have shown this holds if $u\in\mathcal{C}^2\left(\mathbb{R}^2\right)^2$, how can I pass it to the Sobolev space ? I thought about using density of smooth functions with compact support, but I do not know how this works in details.

I cannot pass all the derivatives to the test functions because I have a dot product.

Thanks for your answers !

答:
I think I have found the proof.

If $v\in H^{1}\left(\mathbb{R}^{2}\right)^{2}$, let $\varphi\in\mathcal{D}\left(\mathbb{R}^{2}\right)$.

On one side, we have
\begin{equation*}
  \int_{\mathbb{R}^{2}}\left(v\cdot \omega\right)\varphi=\int_{\mathbb{R}^{2}}\text{div}\left(v w\right)\varphi=-\int_{\mathbb{R}^{2}}wv\cdot\nabla\varphi =-\int_{\mathbb{R}^{2}}\text{curl}\left(v\right)v\cdot\nabla\varphi
 \end{equation*}
as $\text{div}(v)=0$.
and on the other side
\begin{equation*}
  \int_{\mathbb{R}^{2}}\text{curl}\left(\left(v\cdot\nabla\right)v\right)\varphi=\int_{\mathbb{R}^{2}}\left(v\cdot\nabla\right)v\wedge\nabla\varphi=\int_{\mathbb{R}^{2}}\left(v\cdot\nabla\right)v\cdot\left(\nabla\varphi\right)^{\perp}
 \end{equation*}
where we noted
\begin{equation*}
  \left(\nabla\varphi\right)^{\perp}=
  \begin{pmatrix}
   \partial_{2}\varphi\\
   -\partial_{1}\varphi
  \end{pmatrix}
 \end{equation*}
As such

\begin{align*}
  \int_{\mathbb{R}^{2}}\left(v\cdot\nabla\right)v\cdot\left(\nabla\varphi\right)^{\perp}+\text{curl}\left(v\right)v\cdot\nabla\varphi
  &=\int_{\mathbb{R}^{2}}\left[-v_{1}\partial_{1}v_{2}-v_{2}\partial_{2}v_{2}+v_{1}\partial_{1}v_{2}-v_{1}\partial_{2}v_{1}\right]\partial_{1}\varphi\\
  &~~~~~~~~+\left[v_{1}\partial_{1}v_{1}+v_{2}\partial_{2}v_{1}+v_{2}\partial_{1}v_{2}-v_{2}\partial_{2}v_{1}\right]\partial_{2}\varphi\\
  &=\int_{\mathbb{R}^{2}}-\left[v_{2}\partial_{2}v_{2}+v_{1}\partial_{2}v_{1}\right]\partial_{1}\varphi\\
  &~~~~~~~~+\left[v_{1}\partial_{1}v_{1}+v_{2}\partial_{1}v_{2}\right]\partial_{2}\varphi\\
  &=\int_{\mathbb{R}^{2}}\left[\frac{v_{2}^{2}}{2}+\frac{v_{1}^{2}}{2}\right]\partial_{2}\partial_{1}\varphi-\left[\frac{v_{1}^{2}}{2}+\frac{v_{2}^{2}}{2}\right]\partial_{1}\partial_{2}\varphi\\
  &=0
 \end{align*}
Everything is well defined as $v_{i}\in H^{1}\left(\mathbb{R}^{2}\right)$ for $i=1,2$ (use Cauchy-Schwarz inequality and the fact that $\partial_{i}\varphi$ is bounded for the first two lines, or the fact that $\partial_{1}\partial_{2}\varphi$ is bounded for the third).


答:
The result you got $2 e^{\cos \left(\frac{\pi }{2 \sqrt{3}}\right)}\approx 3.7$
is fine since the Mathematica exact result is $3.95262$.

Furthermore in the image below it can be seen that the rectangle approximates pretty well the desired integral.

Title: Calculating $\int_{-1}^{1} e^{\sin(\frac{\pi}{2}t+\frac{\pi}{2})}dx$ by using gauss method

Body:
Today I had an exam ,
I’m reviewing my answers so I calculated one of the questions with maple after my exam and the answer is not anything near my answer
And also i don’t know what I’m doing wrong,i cannot find the flaw in my handcalculated answer

The question is :
Find the approximate integral using gauss method on two points
$$\int_{0}^{\pi} e^{\sin x} dx$$

What I did on paper :

$x=\frac{\pi}{2}t+\frac{\pi}{2}$

$\int_{-1}^{1} e^{\sin(\frac{\pi}{2}t+\frac{\pi}{2})}dt = f(\frac{-\sqrt{3}}{3})+ f(\frac{\sqrt{3}}{3}) $

$e^{\sin(\frac{\pi}{2} \frac{-\sqrt{3}}{3} +\frac{\pi}{2})} + e^{\sin(\frac{\pi}{2}\frac{\sqrt{3}}{3} +\frac{\pi}{2})} $

Then by expanding $\sin(a+b)$ we get

$\int_{-1}^{1} e^{\sin(\frac{\pi}{2}t+\frac{\pi}{2})}dx = 2.e^{\cos(\frac{-\sqrt 3}{6}\pi)}=3.70....  $

And by calculating it with maple it have to be somwhere around 6.2 which is absolutely not

Any help?


问:
$F(u)=u|u|^a$ is locally Lipschitz
In my PDE class we have solved the nonlinear Schrödinger equation $$i\partial_t u+\Delta u =F(u), u(t=0)=u_0$$ where $F(u)=u|u|^a,a>0$ using a fixed point argument (under certain conditions which is not important for my question).
This involves showing the following inequality for $u,v \in \mathbb{C}$ $$|F(u)-F(v)|\leq C_a |u-v|(|u|^a+|v|^a)$$ for some $C_a$ depending on $a$. I have some problem seeing this. One can factor as follows $$|F(u)-F(v)|=|(u-v)(|u|^a+|v|^a)+v|u|^a-u|v|^a|$$ and thus I wonder if $$|v|u|^a-u|v|^a|\leq C_a |u-v|(|u|^a+|v|^a)$$ and it's here I got stuck. It's easy to see that $$|v|u|^a-u|v|^a|\leq (|u|+|v|)(|u|^a+|v|^a)$$ but how can I get the factor $|u-v|$?

Related question: Is the map $u\mapsto |u|^2 u$ globally or locally Lipschitz continuous in the $H_0^1$ norm?

答:
Here is what I could get for $a$ an even integer to show that:
$$\big\lvert v\lvert u\rvert^a - u\lvert v\rvert^a\big\rvert \leq C\lvert u-v\rvert(\lvert u\rvert^a + \lvert v \rvert^a)$$
First notice that $\lvert u \rvert^a = uu^{\frac a2-1}\bar u^{\frac a2}$ (with $\bar u$ the complex conjugate):
$$ \big\lvert v\lvert u\rvert^a - u\lvert v\rvert^a\big\rvert = \lvert uv\rvert\cdot\lvert u^{\frac a2-1}\bar u^{\frac a2} - v^{\frac a2-1}\bar v^{\frac a2} \rvert=\lvert uv\rvert\cdot\big\lvert u\lvert u\rvert^{a-1} - v\lvert v\rvert^{a-1}\big\rvert$$
and repetitively using that $$\big\lvert u\lvert u\rvert^{\alpha} - v\lvert v\rvert^{\alpha}\big\rvert \leq \lvert u-v\rvert(\lvert u \rvert^\alpha + \lvert v \rvert^\alpha) + \big\lvert v\lvert u \rvert^\alpha - u\lvert v \rvert^\alpha\big\rvert $$
you get a bound:
$$
\big\lvert v\lvert u\rvert^a - u\lvert v\rvert^a\big\rvert \leq C\lvert u-v\rvert(\lvert u\rvert^a + \lvert v \rvert^a) + \lvert uv\rvert^\frac a2\cdot\lvert u-v\rvert.
$$
Conclude by using Young's inequality on the $uv$ term:
$$\lvert u\rvert^\frac a2 \lvert v\rvert^\frac a2 \leq \frac 12(\lvert u \rvert^a + \lvert v \rvert^a).$$

I could not find how to do it for more general $a$.

Title: $F(u)=u|u|^a$ is locally Lipschitz


问:
Word problem confusion
The floor of a room measuring 5.94m by 3.60m is to be laid with tiles . Find the least number of identical square tiles required to cover the floor .

I do not have the measurements for the identical square tiles , so how do we solve this question ? Thanks in advance !!

答:
For imagery, let the room be measured in meters so
$\quad 5.94 m\times 3.6 m \approx 20'\times 12'. \quad $
If we think in centimeters, we get
$\quad 594 cm \times 360 cm.\quad$
We find that
$\quad GCD(594,360)=18 \quad$
and dividing the sides by 18, we get
$ 33 \times 20=60$ tiles.

Title: Word problem confusion


问:
Prove this functional inequality
Given $f: R \to (0,\infty]$, twice differentiable with $f(0) = 0, f'(0) = 1, 1 + f = 1/f''$

Prove that $f(1) < 3/2$

I found some useless (I believe) facts about $f$, but nothing that gives me a light to answer it.

$f''(x) > 0$

$f'(x) \ge 1$

$e^{((f')^2-1)/2} = f+1$ is another way to write the differential equation

....
?

答:
As you've noticed, this can be written as
$$\frac12(f'^2-1)=\ln(1+f).$$ If we want a parametric equation, we might choose $f'=t$ as a parameter, and then, we have (denoting the derivative with respect to $t$ by a point)
$$f(t)=e^{(t^2-1)/2}-1$$ and $$\dot x=\dot f/t=e^{(t^2-1)/2},$$ and since $x=0$ corresponds to $t=f'(0)=1$, $$x(t)=\int^t_1 e^{(u^2-1)/2}\,du.$$ The latter may not be expressible by "elementary" functions, but there are a lot of rather common special functions one can use for that (Dawson's integral, error function or incomplete gamma function at some imaginary argument). So we can numerically solve for the value of $t$ that corresponds to $x=1$, finding $t\approx1.6517245235869685402116665754535807279$, and there, $f\approx1.3728623070052740075133959835820792705$.

Title: Prove this functional inequality


问:
Solving the equation $(x^2-7x+11)^{(x^2-13x+42)} = 1$
I have the following equation. $$(x^2-7x+11)^{(x^2-13x+42)} = 1$$ The question is to find the number of positive integer values possible for $x$. The answer is $6$. But I am only able to find $4 ( 2,5,6,7)$. What are the other two possible values?

答: I will give only the two other solutions the you missed. It is trick. Look carefully for the equation. RHS equal 1. if $x^2-13x+42$ is even number. then $x^2-7x+11=-1$ is a solution. By factoring the last equation , we have $x=3$ and $x=4$. So, the set of solution is $\{2,3,4,5,6.7\}$


问:
Do zero derivatives imply multiple zeros
Let $W$ be a polynomial and $a\in\mathbb{R}$. If $W^{(i)}(a)=0$ for all $i\le n$ does that imply $(x-a)^{n+1} | W$? It obviously does for $n=0$.

答: Yes, that is correct. If $n\geq0$ we can write $W=(x-a)^k P$ for some other polynomial $P$ with $P(a)\ne0$ and $k>0$. We then have $W'=k(x-a)^{k-1}P+(x-a)^kP'$, hence $W'(a)=0$ iff $k>1$. That means that if $n\geq1$ we have $(x-a)^2\mid W$. Similar for the higher powers.


答:
Partial answer : $a^b = 1 \rightarrow e^{b ln(a)} = 1 \rightarrow b ln(a) = 2 \pi i n$ for sone $n \in \mathbb{Z}$, where $ln$ denotes the principal branch of the natural logarithm.

Since $a$ is always a real number, $Im(ln(a)) = \begin{cases}
0 \text{ if $a > 0$}\\
\pi \text{ if $a < 0$}
\end{cases}$

Since you want the real part of $b ln(a)$ to be 0, and $b$ is a real number, you want the real part of $ln(a)$ to be 0 as well. This happens when $|a| = 1$ As $a$ is real, that means $a = 1$ or $a = -1$.

Try to finish it from here.

Title: Solving the equation $(x^2-7x+11)^{(x^2-13x+42)} = 1$

Body: I have the following equation. $$(x^2-7x+11)^{(x^2-13x+42)} = 1$$ The question is to find the number of positive integer values possible for $x$. The answer is $6$. But I am only able to find $4 ( 2,5,6,7)$. What are the other two possible values?


问:
Let $\{v_1,\cdots,v_k\}$ be linearly independent. Then we have $f$ st. $f(v_i)=1$ for all $i$
Given a linearly independent set $\{v_1,\cdots,v_k\}$, do we always have an element $f$ st. $f(v_i)=1$ for all $i$? My textbook states this is the case, but I don't really see how we can construct such element.

答:
This is one of the key ideas of linear algebra, so it's worth being confused about.

For convenience, let's assume that $\{v_1,\cdots,v_k\}$ is a basis for the space.  (If it isn't, we can just add vectors as needed to make any linearly independent set of vectors into a basis.  This gets messier if the space is not finite-dimensional, but I won't go there.)

Then, for any vector $u$ in our space, there is exactly one $k$-tuple of real numbers $(c_{u_1},...,c_{u_k})$ such that $$u=\sum_{i=1}^kc_{u_i}v_i$$  Let's assume that this appropriate choice of "coordinates" (relative to the basis) has been made for each vector $u$ in the space.  Notice that the coordinates of $v_1$ are $(1,0,...0)$, the coordinates of $v_2$ are $(0,1,0,...,0)$, and so on.  Also notice that the coordinates of $v_1+v_2$ are $(1,1,0,...,0)$ and the coordinates of $3v_1$ are $(3,0,...,0)$, so this correspondence between our vector space and $\mathbb R^k$ turns out to be a linear isomorphism.  (Some people use this identification to argue that $\mathbb R^k$ are the only finite dimensional vector spaces worth studying.)

With all this out of the way, the function you are looking for is just $$f(u)=\sum_{i=1}^kc_{u_i}$$  It's just a bunch of one-line proofs to show that $f$ is a linear form with $f(v_i)=1$ for each member of the basis.


问:
Is it possible to solve this nonlinear first order ODE using elementary methods?
Is there a trick, such as substitution or otherwise to solve this ode using elementary methods?

$$
  y = \left(y'\right)^3 y^2 + 2 x y'
$$

Maple says this is 1st order with linear symmetries. Tracing it, is seems to be using Lie methods which I do not know too well yet or something it calls 1st order, parametric methods which I also did not study.

I wonder if someone can see a simpler method to solve this? This is my attempt.

First solved for $y'$. This gives 3 ode's. Two of them are complex, so I threw these away as I am looking for real solution $y(x)$. The third one turned out to be isobaric. So I solved it using substitution as normally is done. But this gives integral which can't be solved. So my final solution was not verified by Maple and it is too complicated.

Maple gives these solutions (I show only the real ones). The first three solutions below are singular solutions. So the last two ones below is what I am interested in finding for now.

\begin{align*}
y \! \left(x \right) &=
-\frac{2 2^{\frac{1}{4}} 3^{\frac{1}{4}} \left(-x^{3}\right)^{\frac{1}{4}}}{3}\\
 y \! \left(x \right) &=
\frac{2 2^{\frac{1}{4}} 3^{\frac{1}{4}} \left(-x^{3}\right)^{\frac{1}{4}}}{3}\\
y(x) &=0\\
y \! \left(x \right) &= \sqrt{c_1^{3}+2 c_1 x}\\
y \! \left(x \right) &= -\sqrt{c_1^{3}+2 c_1 x}
\end{align*}

Is it possible to solve this ode, without using Lie symmetry methods using something more basic. most likely some smart substitution?

Just looking for hint.

答:
Set $u=y^2$, then
$$
u=u'x+\frac{u'^3}8
$$
is a Clairaut equation with solution family
$$
u=Cx+\frac{C^3}8=2cx+c^3~~~~(C=2c)
$$
and a singular curve $$
0=x+\frac{3u'^2}8\implies u=u'x-\frac{u'x}3=\pm\frac{2x}{3}\sqrt{-\frac{8x}{3}}
$$
which is defined for $x\le0$.


答:
You forgot that $(-1)^{2n} = 1$ for positive $n$. (The exponent is even). You have different cases for $x^y = 1$:

$$x \not= 0, y = 0$$

$$x = 1, y = \text{any real number} $$

$$x = -1, y = \text{even number} $$

You missed the last case.


问:
Newton's method in higher dimensions explained
I'm studying about Newton's method and I get the single dimension case perfectly, but the multidimensional version makes me ask question...

In Wikipedia Newton's method in higher dimensions is defined as:

$$\textbf{x}_{n+1} = \textbf{x}_n - [Hf(\textbf{x}_n)]^{-1}\nabla f(\textbf{x}_n), \;\;\; n \geq 0.$$

Where $\textbf{x}_n$ is the $p$-dimensional vector at $n$th iteration, $[Hf(\textbf{x}_n)]^{-1}$ is the inverse of the Hessian matrix of the function $f(\textbf{x})$ at $\textbf{x}_n$ and $\nabla f(\textbf{x}_n)$ is the gradient of the function $f(\textbf{x})$ at $\textbf{x}_n$. That is:

$$\left( \begin{array}{c}
x_1^{(n+1)}  \\
x_2^{(n+1)}  \\
\vdots \\
x_p^{(n+1)}  \end{array} \right) = \left( \begin{array}{c}
x_1^{(n)}  \\
x_2^{(n)}  \\
\vdots \\
x_p^{(n)}  \end{array} \right) - \left( \begin{array}{cccc}
\frac{\partial^2f}{\partial x_1^2}(\textbf{x}_n) & \dots & \dots &\frac{\partial^2f}{\partial x_p\partial x_1}(\textbf{x}_n)\\
\frac{\partial^2f}{\partial x_1\partial x_2}(\textbf{x}_n)  & \ddots & \vdots & \vdots\\
\vdots & \vdots & \vdots & \vdots\\
\frac{\partial^2f}{\partial x_1\partial x_p}(\textbf{x}_n) & \dots & \dots & \frac{\partial^2f}{\partial x_p^2}(\textbf{x}_n)  \end{array} \right)^{-1}\left( \begin{array}{c}
\frac{\partial f}{\partial x_1}(\textbf{x}_n)  \\
\frac{\partial f}{\partial x_2}(\textbf{x}_n)   \\
\vdots \\
\frac{\partial f}{\partial x_p}(\textbf{x}_n)   \end{array} \right)$$

Now my question is: "What is the intuition behind this formula?" This resembles somehow the gradient descent algorithm, but the inverse of the Hessian is like it came from the magician's hat :S Can somebody give me a similar kind of proof as is given here on the one-dimensional case:

Why does Newton's method work?

Why the Hessian? Why its inverse?! :) Intuition of the formula?

Thank you for any help :) P.S. I here is the page I got the formula above:

http://en.wikipedia.org/wiki/Newton%27s_method_in_optimization#Higher_dimensions

Note also that in my notation the topscript in the $x_i$s doesn't mean exponent, it's just an iteration label...

答:
I wonder if part of your confusion comes from the fact that one-dimensional Newton is usually presented as a method for finding the zero of a function $f(x)$, whereas your higher-dimensional formula is about finding the minimum of a function $f(x)$. So you first need to observe that your formula is about finding the zero of $\nabla f$.

In other words: you only need a first derivative in both cases, but since you are getting your formula in an optimization context, the function you are finding the zero of is already a derivative, hence the need for a second derivative.


问:
Is an infinite intersection of topology over X a topology?
I know that intersection of any two topology is a topology and so is any finite arbitrary number of intersections. But what about an infinite intersection? I can prove the first two case so if we just do the same proof with the infinite case, would it still be true or would we run into some problems?</p>

答:
Yes, it is. Let $X$ be a nonempty set and $\forall i\in I \; \tau _i$ be a topology on $X$ for an arbitrary indexing set $I$. Then,

$(\forall i\in I) \; \; \emptyset \in \displaystyle \bigcap _{i\in I} \tau _i$ and  $X\in \displaystyle \bigcap _{i\in I} \tau _i$ as all of $\tau _i$’s are topology on $X$.

Let $U_1, U_2, ..., U_n\in \displaystyle \bigcap _{i\in I} \tau _i$. Then, for all $i\in I$ $\; U_1,...,U_n \in \tau_i$. Then, since all of $\tau_i$’s are topology on $X$, $(\forall i\in I) \; \; U_1\cap U_2 \cap...\cap U_n \in \tau_i $ and so  $U_1\cap U_2 \cap...\cap U_n \in \displaystyle \bigcap _{i\in I} \tau _i$.

Let $\{U_j\}_{j\in J}   $ be a collection of some elements of  $\displaystyle \bigcap _{i\in I} \tau _i$. Then, $(\forall i\in I)$ since the collection $ \{U_j\}_{j\in J} $ is in $\tau _i$ and $\tau_i$ is a topology on $X$,  $\; \displaystyle \bigcup _{j\in J}U_j\in \tau_i $. And so,  $\; \displaystyle \bigcup _{j\in J}U_j\in \displaystyle \bigcap _{i\in I} \tau _i$.

As a result, $\displaystyle \bigcap _{i\in I} \tau _i$ is a topology on $X$.


问:
Time complexity (easy)
I know that this platform is probably not the best place for this question, on stack overflow I didn't receive any answers. Consider the following code snippet:

```<code>for (int i = n; i >= 1; i = i / 2) {
    for (int j = 1; j <= i; j = 2 * j) {
        f();
    }
}
</code>```

We want to determine the number of times f() get's called. My solution:
$$
\sum_{i=1}^{\log _{2}(n)} \sum_{j=1}^{\log _{2}(i)} 1 = \sum_{i=1}^{\log _{2}(n)} \log _{2}(i)$$
Apparently the solution is $\Theta(\log^2(n))$, how can I reduce the above further?

答:
Your statement is correct but it is not in the form expected for an answer.  We usually expect answers like this to be in the form $n^a (\log n)^b$ which makes them easy to compare.  As your expression is given it is not clear how it compares with other expressions.  It is similar to a problem where the expected answer is $3$ and somebody gives $\frac 62$.

You were expected to notice that you are summing $\log_2 n$ terms, each of which is at most $\log_2 n$, so the total is less than $(\log_2 n)^2$

Title: Time complexity (easy)


问:
Prove that $\langle \{a+b\sqrt[3]2 \mid a, b \in \mathbb{Q}\}, +, \cdot \rangle$ is not a field
I got this question in my real analysis course:

Prove that $\langle \{a+b\sqrt{2} \mid a, b \in \mathbb{Q}\}, +, \cdot \rangle$ is a field, but $\langle \{a+b\sqrt[3]{2} \mid a, b \in \mathbb{Q}\}, +, \cdot \rangle$ is not.

Obviously, I checked for the field properties (closure, commutativity, associativity, identity, inverse and distributivity) and they all check out except for the inverse. The additive inverse is trivial in both cases, and in the first case I found that the following is an inverse of $a+b\sqrt{2}$:

$$\frac{a}{a^2-2b^2}+\frac{-b}{a^2-2b^2}\sqrt{2}$$

$\forall$ $a, b \in \mathbb{Q}$. I have not been able to reproduce a similar result for the second set, which leads me to believe that a proof that the inverse does not always exist is the crux of the argument.

Any hints on how I can go about proving this?

Note: I have not had a course on abstract/modern algebra yet, so I only know the definition of a field and not any properties it may possess.

答: As J.W. Tanner and leoli1 mentioned in their comments, the second set does not satisfy the closure property due to $(\sqrt[3]{2})^2$ not being an element. Hence it is not a field and we are done.


问:
Action of $S_7$ on the set of $3$-subsets of $\Omega$
Reviewing the great book in Permutation Groups by J.D.Dixon, I encountered the following problem:

Show that $S_7$ acting on the set of $3$-subsets of $\Omega=\{1,2,3,4,5,6,7\}$ has degree $35$ and rank $4$ with subdegrees $1,4,12,18$.

I wanted to probe the problem by making a good program in GAP's environment so I did:

```<code> gap> S7:=SymmetricGroup(7);;
 gap> O:=[1..7];;
 gap> 3s:=Combinations( O,3);;
 gap> Size(3s);
                                          35
 gap> G1:=Stabilizer(S7,3s[1],OnSets);
               Group([ (6,7), (5,7), (4,7), (2,3)(5,7), (1,2,3)(5,7) ])
 gap> D1:=[];;
 gap> for k in [1..35] do D1[k]:=Size(Orbit(G1,3s[k],OnSets)); od;;
 gap> List([1..35],k->D1[k]);

 [ 1, 12, 12, 12, 12, 12, 12, 12, 12, 18, 18, 18, 18, 18, 18, 12, 12, 12, 12, 18,
   18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 4, 4, 4, 4 ]
</code>```

However, I can read the results as above, they are not in an order $1,4,8,12$ and I couldn't make them be happened just for one time.

答:
Try

```<code>SortParallel(D1, 3s);
</code>```

and then redo the steps

```<code>D := [];
for k in [1..35] do D1[k]:=Size(Orbit(G1,3s[k],OnSets)); od;;
D1;
</code>```

Is this that you want?

Body:
Try

SortParallel(D1, 3s);

and then redo the steps

D := [];
for k in [1..35] do D1[k]:=Size(Orbit(G1,3s[k],OnSets)); od;;
D1;

Is this that you want?


问:
Random walk following $N(\mu,\sigma^2)$
Consider a general random walk

$$ X_n=x_0+\sum_{k=1}^{n} J_k $$

where the jumps $J_k$ are IID RV's following $N(\mu,\sigma^2)$. Find the distribution of $X_n$ and:

$$\mathbb{P}(X_n \leq x, X_m \leq y) $$

where $n<m$. I don't know how to start,
I'm just learning what random walks are.

答:
The key lemma we need is:

Lemma. If $X\sim\mathcal N(\mu_1,\sigma_1^2)$ and $Y\sim\mathcal N(\mu_2,\sigma_2^2)$ are independent, then $$X+Y\sim\mathcal N(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2).$$

The proof is easy -- just look at the moment generating functions.

This lemma immediately tells us that $X_n\sim\mathcal N(x_0+n\mu,n\sigma^2)$.

For the other question, by the Markov property
$$f_{X_n,X_m}(x,y)=f_{X_m\mid X_n}(y\mid x)\cdot f_{X_n}(x).$$
Of course, we know the density of $X_n$. But conditional on $X_n=x$, the Markov property implies that $X_m\sim\mathcal N(x+(m-n)\mu,(m-n)\sigma^2)$. So we can find the joint density of $X_n$, $X_m$, and integrating over the appropriate region will give you $\mathbb P(X_n\leq x,X_m\leq y)$.

Title: Random walk following $N(\mu,\sigma^2)$


答:
If $2^{2/3} = a+b2^{1/3}$ then $2^{1/3}$ is a root of $X^2-bX-a$ which means that $2^{1/3} = \frac{b\pm \sqrt{b^2+4a}}{2}$ an obvious contradiction.

(if you prefer then $X^3-2\in \Bbb{Q}[x]$ is irreducible so $2^{1/3}$ can't be the root of a degree two polynomial in $\Bbb{Q}[x]$)

Title: Prove that $\langle \{a+b\sqrt[3]2 \mid a, b \in \mathbb{Q}\}, +, \cdot \rangle$ is not a field

Body:
I got this question in my real analysis course:

Prove that $\langle \{a+b\sqrt{2} \mid a, b \in \mathbb{Q}\}, +, \cdot \rangle$ is a field, but $\langle \{a+b\sqrt[3]{2} \mid a, b \in \mathbb{Q}\}, +, \cdot \rangle$ is not.

Obviously, I checked for the field properties (closure, commutativity, associativity, identity, inverse and distributivity) and they all check out except for the inverse. The additive inverse is trivial in both cases, and in the first case I found that the following is an inverse of $a+b\sqrt{2}$:

$$\frac{a}{a^2-2b^2}+\frac{-b}{a^2-2b^2}\sqrt{2}$$

$\forall$ $a, b \in \mathbb{Q}$. I have not been able to reproduce a similar result for the second set, which leads me to believe that a proof that the inverse does not always exist is the crux of the argument.

Any hints on how I can go about proving this?

Note: I have not had a course on abstract/modern algebra yet, so I only know the definition of a field and not any properties it may possess.


问:
How to understand the following derivation result
Known $$F(t)=\iint_{[0,t]^2}f(xy)dxdy$$
So$$F^{\prime}(t)=\int_{0}^{t}f(tx)dx+\int_{0}^{t}f(ty)dy$$
I try to use the product derivative formula to find the derivative. It's different from what I think. Can you explain it for me, thank you

答:
Let $$\psi(y,t):=\int_0^tf(xy)dx$$
Then $$F(t)=\int_0^t\psi(y,t)dy$$
So by the Leibniz integral rule we have: \begin{align*}
F'(t)&=\psi(t,t)+\int_0^t\frac{\partial\psi}{\partial t}(y,t)dy\\
&=\psi(t,t)+\int_0^t\left(\frac{\partial}{\partial t}\int_0^tf(xy)dx\right)dy\\
&=\int_0^tf(tx)dx+\int_0^tf(ty)dy
\end{align*}
(Of course, I assume that all the derivatives and integral exist)


问:
Why it is opposite for negative integers when it comes to bigger or smaller number?
Example :Why (-5) is smaller than (-2). And can you explain the reason behind it.</p>

答: $a$ is less than $b$ just when $b-a$ is positive. Geometrically, that means $a$ is to the left of $b$ on the usual number line.


问:
Integral of a function over a ball of radius tending to zero
If it is true, can anyone give me a rigorous proof for the second equality?

for $w$ any continuous function defined on a locally compact metric space $(X,d)$,

for $B[x,r]$ the ball whith center $x$ and radius $r$.

Thank you in advance.

$$\lim_{r \to 0} \int_{B(x,r)} w \, d\mu = \lim_{r \to 0} \int_{B(x,r)} w \chi(B(x,r)) \, d\mu =\\ \lim_{r \to 0} \int_{B(x,r)} w(x) \chi(B(x,r)) \, d\mu = \\ w(x)\lim_{r \to 0} \int_{B(x,r)}  \chi(B(x,r)) \, d\mu$$

答:
You need Lebesgue differentiation theorem:

Theorem. Let $f : \mathbb{R}^d \to \mathbb{C}$ be an $L^1$ function. Then, for almost every $x$, we have, $$ \lim_{r\to 0}
\cfrac{1}{\mu(B(x, r))} \int_{B(x, r)} f(y) dy = f(x). $$

See Tao's 'An Introduction to Measure Theory', Theorem 1.6.19.

(Actually you can relax the $L^1$ requirement to $L^1_{loc}$.)


问:
Finding the volume of the solid below $z=y$ and above the region in the $xy$-plane bounded by $y=x$ and $y=x^2$. (Seeking advice on bounds.)
So, I have this question which states:

Find the volume of the solid that lies below the surface,
$$z=y$$
and above the region $T$ in the $xy$-plane enclosed by the two curves $y=x$ and $y = x^2$.

As shown in the diagram.

And I realise this is through triple integrals and was wondering what the bounds would be.

My initial thoughts are let us use spherical coordinates: $$ 0 \leq \theta \leq 2 \pi$$
$$0 \leq \phi \leq \frac{\pi}{4}$$
and for $$\rho$$ I'm not too sure.

I was hoping someone could clarify what the bounds would be.
A simple (or complex) explanation would really help thank you

答:
Yes you have to do triple integral. You can just do it in cartesian coordinates.

$0 \leq z \leq y, x^2 \leq y \leq x, 0 \leq x \leq 1$

Order of integration will be $dz$ first, then $\ dy$ and $\ dx$ last. If you change the order of integration, the limits will have to change accordingly.


问:
Determining if a recursive sequence is converging with induction
Consider this sequence

$$ a_0 = 2,$$
$$ a_{n+1} = \frac 1 a_n $$

Now to prove for convergence I simply put in values ($a_1, a_2, a_3$) and got the the sequence has this behaviour: $2, 1/2, 2, 1/2,\dots$. Now since it isn't approaching any value it shouldn't be converging.

But I would like to prove this, so I looked up how to do it and you're supposed to prove monotony and limitations of the sequence. Now to prove monotony we would have to assume that the sequence is falling or raising. But how can I do that here? Since the sequence is changing its value from <1 to >1 in every iteration what should I assume? Is there a way to solve this without the "obvious" answer?

Thank you!

答:
Sequence $\{a_n\}$ is convergent if and only if
$$\limsup_{n\to\infty}a_n=\liminf_{n\to\infty}a_n$$
which does not hold in your case since $$\limsup_{n\to\infty}a_n = 2 \quad\ne\quad \frac{1}{2} = \liminf_{n\to\infty}a_n$$


问:
Why did James Stewart regard it "more elegant" if you define the logarithm as an integral, and the exponential function its inverse?
Stewart didn't explain why the "latter method" is "less intuitive", but I think this is because you can define the exponential function first, without integrals, that he does starting on p. 391. And integrals aren't intuitive. Correct?

Personally I like intuition. I just can't imagine how something can be "less intuitive but more elegant". If it's less intuitive, then it must less elegant!

I read What do we mean by an "Elegant Proof"?, but English is not my first language. Perhaps I don't understand the meaning of "elegant here".

There are two possible ways of treating the exponential and logarithmic functions and each
method has its passionate advocates. Because one often finds advocates of both approaches
teaching the same course, I include full treatments of both methods. In Sections 6.2, 6.3,
and 6.4 the exponential function is defined first, followed by the logarithmic function as
its inverse. (Students have seen these functions introduced this way since high school.) In
the alternative approach, presented in Sections 6.2*, 6.3*, and 6.4*, the logarithm is
defined as an integral and the exponential function is its inverse. This latter method is, of
course, less intuitive but more elegant. You can use whichever treatment you prefer.

James Stewart, Calculus 7th ed. 2011. Not the Early Transcendentals version. Page xiv from Preface.

答:
Intuition and Elegance

The question seems to assert that the more intuitive an argument is, the more elegant it is.  I don't think that this is the way in which most mathematicians understand the terms.  Indeed, I think that "elegance" and "intuition" are often opposites of each other.  To explain:

Intuition is about the way in which we understand how a theory is built.  An argument is intuitive if we can start from a solid foundation and then build up one step at a time, essentially "following our noses" through a line of reasoning.  Intuitive arguments are often easier to understand, but can get bogged down in computational detail.

Elegance is about an economy of ideas.  A proxy for elegance is brevity:  in general, shorter arguments are considered more elegant.  I think that there is also a a kind of "Ah ha!" associated with elegance—elegant arguments often pull in seemingly unrelated arguments in a surprising way, and evoke some awe in the reader.

From this point of view, intuition and elegance are often at war with each other:  an intuitive idea rarely evokes awe, and often requires tedious computations in order to make rigorous, while elegant proofs often pull rabbits out of hats.

Of course, what is "intuitive" or "elegant" is very much a matter of opinion.  What is intuitive to one person may be utterly opaque and mysterious to another, and what I consider elegant might be a horrendous mess to another.

Working with Exponentials and Logarithms

Both logarithms and exponentials are important classes of functions, which were understood and studied separately at different moments in history.  Both classes of functions appear in relatively intuitive ways:  exponential functions show up as the solutions to certain "nice" differential equations, while logarithms provide a useful computational tool (the logarithm is a homomorphism from the multiplicative group of positive real numbers to the additive group of all real numbers—tables of logarithms were extremely useful before the advent of modern computers).

Regarding the introduction of logarithms and exponentials, I can see three basic approaches:

Introduce exponential functions first, then define logarithms to be the inverses of exponentials.

Stewart calls this approach intuitive.  I am not sure that I agree, but I can, I think, make an argument in support of Stewart.  My recollection is that Stewart first defines exponential functions over the natural numbers: given $a > 0$ and $n \in \mathbb{N}$, define
$$
a^n = \prod_{j=1}^{n} a = \underbrace{a\cdot a\cdot \dotsb \cdot a}_{\text{$n$ times}}.
$$
That is, define exponentiation to be repeated multiplication.  This invokes very simple ideas which most American students learn in the second or third grade.

Using properties of this definition (e.g. $a^{m+n} = a^ma^n$, $(a^m)^n = a^{mn}$, and so on), it is possible to extend this definition to the rational numbers in a relatively straight-forward manner.  Then, because Stewart is writing a calculus text, he can extend this definition to the real numbers by continuity (it is necessary to show that such a continuous extension is possible, but this isn't too hard, and the underlying idea is central to calculus, so this is a reasonable next step).

Once this is done, we can note that the exponential function is strictly increasing, therefore one-to-one, therefore invertible.  Define the inverse to be the logarithm, and we're done.

The essential idea here is that we build up the theory in stages, one little bit at a time, using relatively intuitive ideas (properties of repeated multiplication, continuity, functional inverses, etc).  There is a "narrative arc" to the development of the theory, with each step following in a relatively intuitive way.  On the other hand, this approach may not be terribly elegant, as it requires a lot of little steps, each requiring some new mathematical tool.

Introduce logarithmic functions first, then define exponentials to be the inverses of logarithms.

Defining the logarithm to be an antiderivative of $t \mapsto 1/t$ is not an "intuitive" or "obvious" thing to do (although I think that there is a narrative which can make it obvious, starting with John Napier, but this may be too much of a diversion from the material that a calculus instructor may want to introduce), but it is a computationally and theoretically "elegant" approach, in the sense that it doesn't require nearly the amount of boot-strapping as starting with the exponential.

In this approach, we define the function $\log : (0,\infty) \to \mathbb{R}$ by
$$
\log(x) := \int_{1}^{x} \frac{\mathrm{d}t}{t}.
$$
That this gives a well-defined function follows immediately from the fact that continuous functions are integrable over closed intervals (a theorem which should be old hat by the time we are working with logarithms), and most of the properties of a logarithm very quickly fall out of this definition.

Once the logarithm is defined, taking the exponential function to be the inverse (by definition) is a simple next step, and most of the properties of the exponential function follow fairly quickly from the properties of logarithms and/or the inverse function theorem.

There is an economy of ideas here which is, from a certain point of view, very elegant.  We really only need to know a couple of things (not much more than basic properties of the Riemann integral), but get a large number of important results and properties almost automatically.  This is not necessarily an obvious approach, but it requires very little development.

Introduce exponential and logarithmic functions separately, then show that these functions are inverse to each other.

Personally, this is the approach that I prefer.  My feeling is that both exponential and logarithmic functions are natural objects to study, and so I like to study them separately, and then show that they are inverse to each other.  I typically introduce the exponential function via some variation on the approach mentioned above (essentially extend $\mathbb{Q} \to \mathbb{Q} : q \mapsto a^q$ by continuity to a function on $\mathbb{R}$, then consider the special value of $a$ such that $(a^x)' = a^x$).  This typically happens near the end of the discussion of derivatives—once we know a lot of things about differentiation, we can start talking about exponential functions.

Later, once the integral is fairly well understood, I introduce logarithms as described above.  I find it motivating to discuss this in the context of an "inverse power rule"—once the Fundamental Theorem of Calculus is proved, it is straightforward to show that if $q \ne -1$, then
$$ \int_{a}^{b} x^q \mathrm{d}x
= \frac{1}{q+1} x^{q+1}. $$
But this leaves a gap:  what happens when $q = -1$?  To investigate this, define a function $u : (0,\infty) \to \mathbb{R}$ by
$$ u(x) = \int_{1}^{x} t^{-1} \,\mathrm{d}t. $$
Again, we know that this defines a function, so explore the properties of this function.

Finally, having introduced the exponential and logarithm functions independently, it is not too difficult to show that they are inverse to each other.  I kind of like doing this, because it leads some students to say "Ah ha!" when the see the connection between these two, seemingly unrelated (families of) functions.


答:
The terminology is not that $-5$ is "smaller than" $-2$, instead it is that $-5$ is less than $-2$, which we express in notation as $-5 < -2$. And the reason for this notation and terminology is that the $<$ relation obeys some important and useful mathematical properties, in particular the trichotomy law, and the transitive law, and other laws like $a<b \implies a+c < b+c$.

When using inequalities, the proper notation is $a<b$ or $a>b$, which you may properly read out loud as "$a$ is less than $b$" or "$a$ is greater than $b$". But should avoid the "larger" and "smaller" terminology, which does not have a standard meaning when discussing inequalities.

Nonetheless, sometimes one informally uses "larger" and "smaller" terminology in the sense of absolute values, where one thinks of the absolute value as a measurement of how large the number is: perhaps one might informally say that $-5$ is "larger than" $-2$ because $|-5|=5$ is greater than $|-2|=2$.

Title: Why it is opposite for negative integers when it comes to bigger or smaller number?

Body: Example :Why (-5) is smaller than (-2). And can you explain the reason behind it.</p>


问:
Possessing derivative for a complex function of a complex variable vs Being differentiable for a corresponding $\Bbb R^2\to\Bbb R^2$ function
Assume $f(z)$ is a complex function of a complex variable $z$. The definition of $f$ possessing a derivative at $z$ is the existence of the limit $$\lim\limits_{h\to0}\frac{f(z+h)-f(z)}{h}.$$ See, e.g., $\S$1.1 of "Complex Analysis (3rd Ed.)" by Lars V. Ahlfors.

A multivariate vector-valued function $\mathbf{f}$ mapping from $\Bbb R^n\to\Bbb R^m$ is differentiable at $\bf{x}$ if there is a linear transformation $A$ of $\Bbb R^n$ into $\Bbb R^m$ such that $$\lim\limits_{{\bf{h}}\to\bf{0}}\frac{|{\bf{f}}({\bf{x}}+{\bf{h}})-{\bf{f}}({\bf{x}})-A{\bf{h}}|}{|{\bf{h}}|}=0,$$ which is equivalent to $$\lim\limits_{{\bf{h}}\to\bf{0}}\frac{|{\bf{r}}({\bf{h}})|}{|{\bf{h}}|}=0,$$ where ${\bf{r}}({\bf{h}})={\bf{f}}({\bf{x}}+{\bf{h}})-{\bf{f}}({\bf{x}})-A{\bf{h}}$. See, e.g., "DIFFERENTIATION" Section in Chapter 9 of Rudin's "Principles of Mathematical Analysis (3rd Ed.)".

I've been being tempted to think of complex functions of a complex variable $f(z)=u(z)+iv(z)$ as a $\Bbb R^2\to\Bbb R^2$ function $\mathbf{f}(\mathbf{x})$ where $\mathbf{x}=(x,y)$ corresponds to $z$ and $\mathbf{f}=(u,v)$, and subsequently the condition that $f(z)$ possesses derivative at $z$ (first paragraph) is equivalent to (i.e., can imply each other) $\mathbf{f}$ being differentiable at $\mathbf{x}$ with $n=m=2$ (second paragraph). But today, when I thought more about this, I found myself standing in front of an obvious contradiction that arises from the following theorem (P26 of Ahlfors' "Complex Analysis (3rd Ed.)"):

If $u(x,y)$ and $v(x,y)$ have continuous first-order partial derivatives which
satisfy the Cauchy-Riemann differential equations, then $f(z) = u(z) + iv(z)$
is analytic with continuous derivative $f'(z)$, and conversely.

If the above equivalence claim were correct, the continuity of first-order partial derivatives will guarantee differentiability of $\mathbf{f}$ (by, e.g., Theorem 9.21 on P219 of Rudin's text) and in turn that $f$ possesses derivative. Then the Cauchy-Riemann differential equations would be useless in the proof. Since the Ahlfors' text is so classical, I'm sure this condition is indispensable (I guess it is needed in line 11 and 12 of P26 in Ahlfors' text, right?). As a result, the equivalence claim in the above paragraph must be wrong. Then I tried to figure out why on earth they are not equivalent. Later I came to a point that (1) $\Rightarrow$ (2) while (2) $\not\Rightarrow$ (1), where (1) stands for $f$ possessing derivative and (2) for the corresponding $\Bbb R^2\to\Bbb R^2$ function $\mathbf{f}$ being differentiable. That is, (1) is stronger than (2). I guess the strongerness comes from the fact that $\Bbb C$ defines a multiplication operation, which is not available in $\Bbb R^2$. The contradiction can be solved by this relation: the continuity of first-order partial derivatives establishes (2), but we need additional conditions (the Cauchy-Riemann differential equations) to arrive at a stronger conclusion (1).

To show the above relation, First, I gave a proof of (1) $\Rightarrow$ (2): Assume the derivative is a complex number $\alpha$ and we define linear transformation according to complex number multiplication: if ${\bf{h}}=(x,y)$, $A{\bf{h}}\cong\alpha(x+iy)$ (an isometry between $\Bbb R^2$ and the complex plane). Since $\lim\limits_{z\to0}|f(z)|=0$ iff $\lim\limits_{z\to0}f(z)=0$, we have $$\begin{eqnarray}
\lim\limits_{{\bf{h}}\to\bf{0}}\frac{|{\bf{f}}({\bf{x}}+{\bf{h}})-{\bf{f}}({\bf{x}})-A{\bf{h}}|}{|{\bf{h}}|}&=&\lim\limits_{h\to0}\frac{|f(x+h)-f(x)-\alpha h|}{|h|}\\
&=&\lim\limits_{h\to0}\frac{f(x+h)-f(x)-\alpha h}{h}\\
&=&\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}-\alpha\\
&=&\alpha-\alpha=0
\end{eqnarray}$$
For (2) $\not\Rightarrow$ (1): From the equivalent formulation of differentiability, we have $f(z+h)-f(z)=A{\bf{h}}+r(h)$ where $r(h)=o(h)$. If we can write $A{\bf{h}}\cong\alpha h$ for some complex number $\alpha$ as in the last paragraph, then the derivative of $f$ is immediate. But the thing is, we can't (at least I can't). $A{\bf{h}}/h$ is not necessarily a constant complex number for a given general linear transformation $A$. That's why (2) does not imply (1).

Now comes my question: 1) Is it true that possessing derivative for a complex function of a complex variable is not equivalent to its corresponding $\Bbb R^2\to\Bbb R^2$ function being differentiable? 2) Is my relation that (1) $\Rightarrow$ (2) while (2) $\not\Rightarrow$ (1) a correct argument to explain this inequivalence? 3) Is there anything wrong (or not rigorous) throughout my argument? Thank you.

答:
Update of my longer answer:

$ $ 1) Yes, 2) Yes, 3) No.

Concerning "complexifying": This was a word invented by me describing a thought process that is common, and that you do all the time in your question.


答:
Don't confuse "smaller" and "larger" with "less than" and "greater than"! The words "smaller" and "larger" refer to magnitude, whereas "less than" and "greater than" refer to order. $-5$ is larger than $-2$ because it is further from $0$, that is,
$$|-5|>|-2|$$
However $-5$ is less than $-2$ because it is to the left of it, that is
$$-2-(-5)=3>0$$


答: The set ${\mathbb Z}$ of all integers is usually defined as the Grothendiek group constructed from the monoid $({\mathbb N},+)$ of natural numbers. In this construction, the ordering is defined by $[(a, b)]<[(c, d)]$ iff $a+d < b+c$. It is very easy to check that this definition implies $-5 < -2$ because $-5$ is the class $[(0, 5)]$ and $-2$ is the class $[(0, 2)]$.


答: $-5°$ is colder than $-2°$, while $2°$ is colder than $5°$.


问:
Can the values of the expressions $\frac{1}{\sqrt{2a+1}},\frac{1}{\sqrt{2a-1}},3(a^2-19)\sqrt{2a-1},18\sqrt{2a+1}$ be in G.P.?
Can the values of the expressions $\dfrac{1}{\sqrt{2a+1}},\dfrac{1}{\sqrt{2a-1}},3(a^2-19)\sqrt{2a-1},18\sqrt{2a+1}$ be in geometric progression (in the given order)?

I am confused by the fact that the expressions aren't defined for every value of $a$. Should I determine the range of $a$ or this won't be necessary in the solution? I mean we will have $$\begin{cases}2a+1>0\\2a-1>0\end{cases}\iff\begin{cases}a>-\dfrac12\\a>\dfrac12\end{cases}\iff a\in\left(\dfrac12;+\infty\right).$$ Can we use this further in the solution in some way? Thank you in advance!

答:
$\frac{\dfrac{1}{\sqrt{2a-1}}}{\dfrac{1}{\sqrt{2a+1}}}=\frac{18\sqrt{2a+1}}{3(a^2-19)\sqrt{2a-1}}\Rightarrow \frac{6}{a^2-19}=1$

$\frac{6}{a^2-19}=1 \Rightarrow a^2=25$ which means $a=5$ or $a=-5$

but $a=-5 \Rightarrow 2a+1<0$ so $a=5$

$\frac{3(a^2-19)\sqrt{2a-1}}{\dfrac{1}{\sqrt{2a-1}}}=3(a^2-19)(2a-1)=162 \neq 5$ which means it is a fact that values given above  cannot express a geometric progression