问:
Integration by parts twice problem with the answer
I am trying to integrate $\sin(x)e^{x(1-ik)}$ from $-\infty$ to $0$. The answer I am supposed to get is $\dfrac{k^2-2\mathrm{i}k-2}{k^4+4}$ but I get $\dfrac{-1}{1-(1-\mathrm{i}k)^2}$(I have checked and these are not equivalent). I looked at it many times but I cannot find the mistake I will put the working down below:

答: In your second integration by parts you did $ uv + \int{v du} $ when you meant $ uv - \int{v du} $.


问:
How to solve $z^3+z+1=0$?
I have troubles with solving an equation
$$z^3+z+1=0$$
in complex numbers. I know that in real numbers we can use Cardano's formula and it's clear to me. But what about the situation when $z \in \mathbb{C}$?

答:
Given the (real) cube roots $A,B$ in Cardano's mthod, so that the real root of your cubic is $A+B,$ the other two roots are complex conjugates, given by
$$  A \omega + B \omega^2 $$
$$  A \omega^2 + B \omega $$
where
$$ \omega = \frac{-1 + i\sqrt 3}{2}$$
and
$$ \omega^2 = \bar{\omega}  $$
and $$\omega^3 = 1 $$

ContentLicense: CC BY-SA 4.0

Title: How to solve $z^3+z+1=0$?


问:
How to prove that the sum $1!+2!+...+n! \equiv 0 \pmod 9$
I understand the divisibility rule, if the sum of a numbers digits are divisible by $9$, the number is divisible by $9$. I'm not sure how to integrate that, or if I even should.
Could I assume that for any $n>8$, $n!$ is divisible by $9$. And thus the sum of all $n!$ where $n>8$ would too be divisible by $9$?
Thanks in advance for any help.

答:
Hint 1

If $n\ge 6$ then

\begin{align}
n!
&=1\cdot 2\cdot \color{red}{3}\cdot 4\cdot 5\cdot \color{red}6\cdot \;\cdots\;\cdot n
= \color{red}9\cdot (1\cdot 2\cdot 4\cdot 5\cdot 2\cdot\;\cdots\; \cdot n).
\end{align}

Hint 2

$$1!+2!+3!+4!+5!=153=\color{red}9\cdot 17.$$


问:
Fourier transform as an integral on $L^2$
We often define Fourier transform $\mathcal F$ on $L^2$ by continuously extending it from $L^1$ or the Swartz space $\mathcal S$. There are of course other ways to do it.

Note: by "integral" I am referring to the Lebesgue integral.

But now, I am interested in the convergence of the integral
$$
\int_{\mathbb R^d} f(x) e^{-i\langle x, \xi\rangle} dx
$$
itself. Even if the function $f$ is not in $L^1$, this integral could converge, sometimes for almost every $\xi$. For example, if we pick a real-valued function $g(x)$ that is not $L^1$ but $L^2$, then the Fourier transform integral of $g(x)e^{i\langle x,\xi_0 \rangle}$ is undefined when $\xi=\xi_0$, but is defined for other values of $\xi$.

Does there exists a function $f\in L^2 (\mathbb R)$ such that the integral
$$
\int_{\mathbb R} f(x) e^{-ix \xi} dx
$$
diverges for $\xi$ on a set of non-zero measure?

答:
The inversion integral for $f\in L^2$ converges in $L^2(\mathbb{R})$. That is,
$$
           \lim_{R\rightarrow\infty} \left\|\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\hat{f}(s)e^{-isx}ds-f\right\|_{L^2(\mathbb{R})}=0.
$$
You can choose a subsequence that converges pointwise a.e. to $f$.


问:
Chebyshev's inequality assumptions
Why does Chebyshev's inequality demand that $\mathbb{E(}X^2) < \infty$?

答:
You can write  Chebyshev's inequality as $$\mathbb P(|X-\mu|\geq k\sigma) \leq \frac{1}{k^2}$$ or equivalently as $$\mathbb P(|X-\mu|\geq t) \leq \frac{\sigma ^2}{t^2}$$

with $k, t > 0$.

If $\mathbb E[X^2]=\infty$ then $\sigma^2=\mathbb E[X^2]-\mu^2=\infty$ and so you find $\mathbb P(|X-\mu|\geq t) \leq \infty$.

This would not be useful information, as you already know that $\mathbb P(|X-\mu|\geq t) \leq 1$ since it is a probability.

Title: Chebyshev's inequality assumptions


答:
But it's not true for $n= 1,2$

It is for $n=3$ by examination.  $1! +2! +3! = 1+2 +6=9$.

And $4! = 24$ is not divisible by $9$ but $1!+2!+3!$ is, then it can't be true for $n=4$ (and by examination for $n=4$ we git $(1!+2! +3!) + 4! = 9 + 24 = 33$).

As $4! \equiv 24 \equiv 6\pmod 9$ then $5!\equiv 6*5\equiv 30\equiv 3\pmod 9$ so $(1!+2!+3!)+4!+5! \equiv 0 + 6 + 3\equiv 0 \pmod 9$.

And for any $n\ge 6$ we have $3\times 6=18|n!$ so for all $n\ge 6$ we have $9|n!$ so
as $(1!+2! +3!) + (4! + 5!) + 6! +......... + n! \equiv (0) + (6 + 3) + 0 + ....... + 0\equiv 0 \pmod 9$.

but it's not true for $n=1,2,4$.

Title: How to prove that the sum $1!+2!+...+n! \equiv 0 \pmod 9$

Body:
I understand the divisibility rule, if the sum of a numbers digits are divisible by $9$, the number is divisible by $9$. I'm not sure how to integrate that, or if I even should.
Could I assume that for any $n>8$, $n!$ is divisible by $9$. And thus the sum of all $n!$ where $n>8$ would too be divisible by $9$?
Thanks in advance for any help.


问:
Is uniform convergence and pointwise convergence the same in function space C[X,R]?
In GF Simmon's book on Topology, I came across the definitions of
point wise convergence of a sequence of functions and uniform convergence of a sequence of functions.

For all future references,
$C[X,R]$ is the set of continuous bounded real functions defined on the metric space $X,$ and $B$ is the subset of all bounded real functions on $X.$

Let $\{f_n\}$ be a sequence of real functions defined on $X.$

Pointwise convergence is when if for each point $x\in X,$ the sequence of real functions $\{f_n(x)\}$ converges to a function
$f(x)$, i.e., at each $x,$ for a given $\varepsilon$, there exists a natural $N$ such that for all $n>N, $ $|f_n(x)-f(x)|<\varepsilon$.

Uniform Convergence is when the $\varepsilon$ doesn't depend on $x.$ Which means that a natural $N$ can be found for each $\varepsilon$ such that it works for all $x\in X.$

In terms of convergence of the sequence of functions to a function,
Pointwise means that for a given $\varepsilon$ you will find an $N$ for each $x\in X.$
Uniform means that for a given $\varepsilon$ you will find an $N$ which works for all $x\in X.$
Am I right?

答:
Well, strictly speaking, the quantifiers in your last paragraph are not okay. Actually, it should be: "Pointwise mean that for a $\varepsilon>0$, and for each $x\in X$, there exists an $N\in\mathbb N$ such that...". Can you see the difference? We talk first about the point $x\in X$. Your understanding of uniform convergence is correct.

With respect to the title of you question, these kind of convergence are not the same, even when $X$ is compact. Consider $X=[0,1]$, the sequence $f_n(x)=x^n$ converge pointwise to a function which is not even continuous,  despite of that, the sequence does not have a uniform limit (since the uniform limit of continuous functions is continuous).


问:
Show that $\sin\left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{3!~n^3}+O\left(\frac{1}{n^5}\right)$

Show that $\sin\left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{3!~n^3}+O\left(\frac{1}{n^5}\right)$.

In fact, this result is pretty obvious but when I did this in homework I basically got no points because I was to shallow in my reasoning. So here is my new attempt.

Proof:

We know that the Taylor series at point $0$ of $\sin(x)$ exists so the "tail" $\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}$ of it also exists.

Basically, we now want to show that there exist a constant $C>0$ and an index $n_0$ such that for all $n>n_0$ it holds: $\Big|\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}\Big|\leq C \cdot \frac{1}{n^5}$ which is equivalent to saying $\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}=O\left(\frac{1}{n^5}\right)$.

For this we show that $\lim\limits_{n\to\infty}n^5\cdot\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}$ converges.

Because $\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}$ converges we are allowed to perform the multiplication with a fixed $n$: $n^5\cdot \sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}=\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k-4}}$.

Now the question is what happens to $\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k-4}}$ if $n\to\infty$? Again, this seems very obvious as $n$ only appears in the denominators but my tutor said that this is not a rigorous argument. So I guess that we must find a closed form expression for $\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k-4}}$ or we must at least find an upper bound which attains a closed form expression. I would simply take a geometric series $\sum\limits_{k=1}^{\infty}\frac{1}{n^k}=\frac{1}{n-1}$ as $\frac{(-1)^k}{(2k+1)!~~n^{2k-4}}\leq\frac{1}{n^k}$ holds for all $n$ and $k$.

Hence, for all $n$ we have:
$$
0\leq n^5\cdot\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}=\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k-4}}\leq \frac{1}{n-1}\\\implies \lim\limits_{n\to\infty}n^5\cdot\sum\limits_{k=2}^{\infty}\frac{1}{(2k+1)!~~n^{2k+1}}=0.
$$
So for a constant $C>0$ we find an index $n_0$ which is large enough such that for all $n>n_0$:

$$\Big|n^5\cdot\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}\Big|<C \\\implies \Big|\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}\Big|<C \frac{1}{n^5}\\\implies
\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}=O\left(\frac{1}{n^5}\right)\\\implies
\sin\left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{3!~n^3}+O\left(\frac{1}{n^5}\right).
$$

Is this correct?

答:
Comment on the Approach in the Question

It would be more rigorous to cite Dominated Convergence as a reason that
$$
\lim_{n\to\infty}n^5\sum_{k=2}^\infty(-1)^k\frac1{(2k+1)!}\frac1{n^{2k+1}}=-\frac1{120}
$$
thereby bounding the remainder of the series.

One could also bound by a geometric series:
$$
\begin{align}
\left|\sum_{k=2}^\infty(-1)^k\frac1{(2k+1)!}\frac1{n^{2k+1}}\right|
&\le\sum_{k=2}^\infty\frac1{120n^{2k+1}}\\
&=\frac1{120\!\left(n^5-n^3\right)}
\end{align}
$$

However, both of these approaches require that you have either defined $\sin(x)$ as its Taylor series, or that you've shown that the Taylor series converges, which seems to be doing more than this question asks, so circularity is a concern.

Another Approach

Integrate by parts $4$ times:
$$
\begin{align}
\sin(x)
&=\int_0^x\cos(t)\,\mathrm{d}t\\
&=x-\int_0^x(x-t)\sin(t)\,\mathrm{d}t\\
&=x-\frac12\int_0^x(x-t)^2\cos(t)\,\mathrm{d}t\\
&=x-\frac16x^3+\frac16\int_0^x(x-t)^3\sin(t)\,\mathrm{d}t\\
&=x-\frac16x^3+\frac1{24}\int_0^x(x-t)^4\cos(t)\,\mathrm{d}t\\
&=x-\frac16x^3+O\!\left(x^5\right)
\end{align}
$$
Since
$$
\begin{align}
\left|\int_0^x(x-t)^4\cos(t)\,\mathrm{d}t\right|
&\le\left|\int_0^x(x-t)^4\,\mathrm{d}t\right|\\
&=\frac15|x|^5
\end{align}
$$


问:
Can I take any capping surface to apply Stokes' theorem to?
I've been asked the following.

Verify Stokes' theorem for the vector field $\vec{a}=\vec{r}\times\hat{k}$, where $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$, and $\lbrace\hat{i},\hat{j},\hat{k}\rbrace$ is the standard basis for $\mathbb{R}^3$, given the hemisphere $C$ of radius $c$ centered at $(0,0,0)$ in the upper half region of $\mathbb{R}^3$ (that is, $z\geq0$).

My attmept is as follows. Note firstly that $\vec{a}=y\hat{i}-x\hat{j}+0\hat{k}$. Note also that, due to the statement of stokes theorem, we may be immediately interested in obtaining an expression for the curl of $\vec{a}$, which is given by $\vec{\nabla}\times\vec{a}=0\hat{i}+0\hat{j}-2\hat{k}$. From here, note that the formulation of the question suggests that $C$ is the capping surface we are to use. For this reason, note that we may take the circular contour of radius $c$ centered at $(0,0,0)$ in the plane $z=0$, oriented counter-clockwise as the boundary of $C$, which we'll of course denote $\partial C$. Notice from here that $\gamma(t)=c\cos(t)\hat{i}+c\sin(t)\hat{j}+0\hat{k}$, for $0\leq t\leq 2\pi$, parameterizes $\partial C$. For this reason, consider
$$
\oint_{\partial C}\vec{a}\cdot d\vec{r}=\int_0^{2\pi}\vec{a}(\gamma(t))\cdot\gamma'(t)~dt
$$
$$
=\int_0^{2\pi}(c\sin(t)\hat{i}-c\cos(t)\hat{j}+0\hat{k})\cdot(-c\sin(t)\hat{i}+c\cos(t)\hat{j}+0\hat{k})~dt
$$
$$
=-c^2\int_0^{2\pi}~dt=-2c^2\pi
$$
And hence, by Stokes' theorem, this quantity should equate to
$$
I=\iint_C\mathrm{curl}(\vec{a})\cdot\hat{n}_C~dS
$$
Where $\hat{n}_C$ denotes the unit normal to the surface of $C$ (possibly as a function of $x$, $y$, and $z$, given $(x,y,z)\in C$). To ensure the preservation of orientation, we employ the trusty right hand rule to arrive at the conclusion that the normal to $C$ is given by $\hat{n}_C=\vec{r}/|\vec{r}|$. Note, however, that on the surface of $C$, $|\vec{r}|=c$, and so $\hat{n}_C=\vec{r}/c$. For this reason, we have that
$$
\mathrm{curl}(\vec{a})\cdot\hat{n}_C=(\vec{\nabla}\times\vec{a})\cdot\hat{n}_C=-\frac{2z}{c}
$$
Since, on $C$, $z=\sqrt{c^2-x^2-y^2}$ (we omit consideration of the negative square root of $z^2$, since we are told that $z\geq0$), we have that
$$
\mathrm{curl}(\vec{a})\cdot\hat{n}_C=-\frac{2}{c}\sqrt{c^2-x^2-y^2}
$$
And, for this reason, $I$ becomes
$$
I=-\frac{2}{c}\iint_C\sqrt{c^2-x^2-y^2}~dS
$$
We may, from here, convert to polar coordinates to arrive at the conclusion that
$$
I=-\frac{2}{c}\int_0^{2\pi}\int_0^cr\sqrt{c^2-r^2}~dr~d\theta
$$
$$
=-\frac{2}{c}\int_0^{2\pi}-\frac{1}{3}\left[(c^2-r^2)^{3/2}\right]_0^c~d\theta
$$
$$
=-\frac{2}{c}\int_0^{2\pi}\frac{c^3}{3}~d\theta=-\frac{4}{3}c^2\pi
$$
This result, however, is obviously not equal to $-2c^2\pi$, meaning that
$$
\oint_{\partial C}\vec{a}\cdot d\vec{r}\neq\iint_C\mathrm{curl}(\vec{a})\cdot\hat{n}_C~dS
$$
which, as far as my understanding of Stokes' theorem is concerned, cannot be the case. Can someone perhaps help me identify the mistake that I've made?

答:
I think your mistake is that you treated $dS$ as if it were $dx \, dy$ when you changed to polar coordinates. But $dS$ is the area element of the surface: if the surface $S$ is the graph of $z=f(x,y)$, then

$$ dS = \sqrt{1+f_x^2 + f_y^2} \, dx \, dy $$

So in your case, since the hemisphere is $z= \sqrt{c^2-x^2-y^2}$, this is

$$ dS = \frac{c}{\sqrt{c^2-x^2-y^2}} \, dx \, dy $$

This cancels with the $\sqrt{c^2-x^2-y^2}$ that's already in your integral, so you get

$$ I = -2 \iint\limits_D dx \, dy $$

where $D$ is the disc of radius $c$ in the $x,y$-plane. This is $-2$ times the area of the disc, which is $-2\pi c^2$, which agrees with the line integral.


问:
If graph G is an n-nodes Hamiltonian path, why χ(G)=2 or χ(G)=3?
graph $G$ is a $n$-nodes Hamiltonian path, $χ(G)$ is the least number of needed colors to make sure any two adjacent nodes not to have the same colors, then:
$$
χ(G)=\begin{cases}
2,&\text{if $n$ is even}\\
3,&\text{if $n$ is odd}
\end{cases}
$$
how could I yield this conclusion?

答:
To get you started...

Given a graph $G(V,E)$ let $H(V,E)$ be a Hamiltonial cycle (not just a path).

This means $H$ it essentially a "circle" of nodes. Start at any node and color it red. Then move to the next node (in either direction). That node can't be red, so color it blue. You can keep going this way, red $\rightarrow$ blue $\rightarrow$ red, etc...

Now, think about what happens with even and odd cycles.


问:
Intersection of subspace which projector do not commute
I am a little bit confused by the following.

Consider two subspace $U$ and $V$ of a given vector space. I associate to them $\Pi_U$ and $\Pi_V$ their projectors.

In the case $[\Pi_U,\Pi_V]=0$; the projector on $U \cap V$ is simply $\Pi_{U \cap V} = \Pi_U \circ \Pi_V = \Pi_V \circ \Pi_U$

But when the two projector do not commute, $\Pi_U \circ \Pi_V$ might not even be a projector.

My question is:

Is there a physical meaning for intersection of subspaces which projector do not commute ? Does that mean that the intersection is necesseraly $\{0\}$ ? If so I couldn't manage to show it.

答:
As far as I know, if the projectors do not commute there is not necessarily a significance of $U \cap V$. We may have cases where the projectors do not commute where $U \cap V \neq \{0\}$. For example, let
$$U = \text{span}\left(\hat{e}_1, \frac{1}{\sqrt 2}(\hat{e}_2 + \hat{e}_3)\right), ~ V = \text{span}\left(\hat{e}_1, \hat{e}_2\right)$$
The projectors do not commute, but clearly $\hat{e}_1 \in U \cap V$.

If they do not commute, we can say that there cannot be a complete basis of simultaneous eigenvectors of $\Pi_U$ and $\Pi_V$. Namely, there must be some vector $\vec{v}$ in either $U$ or $U^\perp$ that is not in $V$ or $V^\perp$ (and vice-versa), but as far as I know we can't really say more.


问:
Any reference for induces exact sequence $0 \to Ker(f) \to Ker(gf) \to Ker(g) \xrightarrow{\delta} Coker(f) \to Coker(gf) \to Coker(g) \to 0.$
Let $\mathscr{C}$ be an abelian category and $f:A \to B$ and $g:B \to C$ morphisms in $\mathscr{C}$, then we have the following exact sequence

$$0 \to \operatorname{Ker}(f) \to \operatorname{Ker}(gf) \to \operatorname{Ker}(g) \xrightarrow{\delta} \operatorname{Coker}(f) \to \operatorname{Coker}(gf) \to \operatorname{Coker}(g) \to 0.$$

I need to use this result which I know its true since I almost got it, but instead of looking help ending the proof. I'm wondering if someone knows a book or text where I can reference this result? This one looks pretty much like the Snake Lemma but I cannot find them in Literature. Or if someone help me see this as an application of a the Snake Lemma it would also be helpful. Thanks

答:
Try applying the snake lemma to the diagram
$$
\require{AMScd}
\begin{CD}
0 @>>> A @>{\begin{bmatrix} id \\ f \end{bmatrix}}>> A \oplus B @>{\begin{bmatrix} -f & id \end{bmatrix}}>> B @>>> 0 \\
@. @VVfV @VV{\begin{bmatrix} 0 & id \\ gf & 0 \end{bmatrix}}V @VV{-g}V \\
0 @>>> B @>{\begin{bmatrix} id \\ g \end{bmatrix}}>> B \oplus C @>{\begin{bmatrix} -g & id \end{bmatrix}}>> C @>>> 0.
\end{CD}
$$


答:
The Cardano formula gives the real root
$$a=\sqrt[3]{\frac{\sqrt{93}-9}{18}} - \sqrt[3]{\frac{\sqrt{93}+9}{18}} <0$$
Then, factorize the equation as
$$x^3+x+1=(x-a)(x^2+ax-\frac1a)=0$$
where the quadratic factor gives the pair of complex roots
$$x_{1,2}=\frac12\left( -a\pm i \sqrt{1-\frac3a}\right)$$

Title: How to solve $z^3+z+1=0$?

Body:
I have troubles with solving an equation
$$z^3+z+1=0$$
in complex numbers. I know that in real numbers we can use Cardano's formula and it's clear to me. But what about the situation when $z \in \mathbb{C}$?


问:
Proof of a compound proposition being a tautology
The question states the following compound proposition is a tautology, which I have to prove.

$$(p \rightarrow q) \wedge (q \rightarrow r) \rightarrow (p \rightarrow r)$$

My attempt is as follows:

$$\equiv (\neg p \vee q) \wedge (\neg q \vee r) \rightarrow (\neg p \vee r)$$
$$\equiv \neg((\neg p \vee q) \wedge (\neg q \vee r)) \vee (\neg p \vee r)$$

$$\equiv \neg[((\neg p \vee q) \wedge \neg q) \vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Distribution of conjunction over disjunction]}$$
$$\equiv \neg[((\neg p \wedge \neg q) \vee (q \wedge \neg q) \vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Distribution of conjunction over disjunction]}$$
$$\equiv \neg[((\neg p \wedge \neg q) \vee F )\vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Idempotent law]}$$
$$\equiv \neg[(\neg p \wedge \neg q) \vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Idempotent law]}$$
$$\equiv [(p \vee q) \wedge ((p \wedge \neg q) \vee \neg r)] \vee (\neg p \vee r)\text{ [De Morgan's law]}$$
$$\equiv (p \vee q) \wedge ((p \wedge \neg q) \vee \neg r) \vee (\neg p \vee r)$$
$$\equiv (p \vee q) \wedge (\neg p \vee r) \vee ((p \wedge \neg q) \vee \neg r)\text{ [Commutative law]}$$
$$\equiv [((p \vee q) \wedge \neg p) \vee ((p \vee q) \wedge r)] \vee ((p \wedge \neg q) \vee \neg r)\text{ [Conjunction over disjunction]}$$
$$\equiv [((p \wedge \neg p) \vee (q \wedge \neg p)) \vee ((p \vee q) \wedge r)] \vee ((p \wedge \neg q) \vee \neg r)\text{ [Conjunction over disjunction]}$$
$$\equiv [(F \vee (q \wedge \neg p)) \vee ((p \vee q) \wedge r)] \vee ((p \wedge \neg q) \vee \neg r)\text{ [Negation law]}$$

This goes on for quite a while, but I don't get the desired $T$ result. I'm guessing I've done something horribly wrong in the above calculation. I'll be so grateful for your help here. Anything regarding this text, from calculation error to using some incorrect terms. Thank yoou.

答:
The step:
$$\equiv [(p \vee q) \wedge ((p \wedge \neg q) \vee \neg r)] \vee (\neg p \vee r)\text{ [De Morgan's law]}$$
$$\equiv (p \vee q) \wedge ((p \wedge \neg q) \vee \neg r) \vee (\neg p \vee r)$$
is incorrect. You need the distributive law here, so it should be
$$
\equiv [(p \vee q)\vee (\neg p \vee r)] \wedge [((p \wedge \neg q) \vee \neg r) \vee (\neg p \vee r)]
$$
Now by commutative law
$$
\equiv [(p \vee \neg p)\vee q \vee r] \wedge [((p \wedge \neg q) \vee (\neg r \vee r) \vee \neg p] = T\wedge T = T
$$
So it is a tautology.


问:
For which $c\in \mathbb R$ the function $\kappa(t) := ct^4(1-t)^4$ can be curvature of regular, closed, convex curve
I have the following problem.

Suppose $\alpha:[0,1]\rightarrow \mathbb R^2$ is regular, closed and convex curve.

For which $c\in \mathbb R$ can the following function:

$\kappa:[0,1]\rightarrow\mathbb R$ ; $\kappa(t)= ct^4(1-t)^4$  can be it's curvature (meaning signed curvature , because curvature is always positive by definition)?

My (attempt of) solution:

Because $\alpha$ is convex we know that $\alpha\prime\prime\geq0$ $ \forall t\in[0,1]$

Because $\alpha$ is closed then there exists $T>0$ such that $\alpha(t) = \alpha(t+T)$ $ \forall t\in[0,1]$.

$c\neq0$ because otherwise we have that $\kappa=0$ therefore, $\alpha$ is part of straight line and can't be closed. contradiction.

I've read that convex curves curvature is $\geq0$ but I can't find a proper proof for this.

Therefore, in conclusion I get that for any $c>0$ $\kappa(t)$ can be it's curvature.

Did I'm miss something?

答:
Ok. So I've managed to combine a formal answer for this question.

$\gamma$ is convex i.e. simple curve by definition therefore, by Hopf Umlaufsatz theorem taking in account that $\gamma$ is simple and closed we get that the value of $\int_\gamma \kappa(t) dt = +/- 2\pi$.

Therefore by simple computation (and taking in account that $\kappa > 0$ because it is convex) we get that $c=1260\pi$


问:
Terminology for an element of a partition?
Suppose I'm dividing some region $\Theta \in \mathbb{R}^n$ into subregions $\theta_i, i=1,2,3$ such that $\theta_i \cap \theta_j = \varnothing, i\ne j$ and $\bigcup_i \theta_i = \Theta$.  I might say (perhaps loosely, even technically incorrectly) that I am partitioning the region $\Theta$.

Thus, a "partition" would be a particular configuration of $\{\theta_1,\theta_2,\theta_3\}$ that satisfies the aforementioned conditions.  But what would an element of a partition be referred to as?

Since I am partitioning a "region", it makes sense to say that I am partitioning a region into "subregions", but at a higher level, what is a correct term for an element of a partition?

答:
The terms cells, or classes, or blocks, even parts of a partition are often used to describe the "sub-regions" of a given partition, depending on the nature and/or context of the partition.

See for example, Partition of a Set.

ContentLicense: CC BY-SA 3.0


问:
How to see if the convergence is uniform?
I have the following : $f_n(x)=\frac{nx}{1+nx^3} \quad n=1,2,\dots \quad$ and $f(x)=\lim_{n \to \infty} f_n(x) ,$

and I have done the following :  $$|f_n(x) -f(x)| = \biggl|\frac{nx}{1+nx^3}-\frac{1}{x^2}\biggr| = \biggl|\frac{1}{x^2(1+nx^3)}\biggr |$$ where $\frac{1}{x^2}$ is the convergence of the series of the function. Now I know that $\Bigl|\frac{1}{x^2(1+nx^3)}\Bigr|< \epsilon$ but now I don't know how to choose $N$ in order to fully prove that it is uniformly convergent or not. Can somebody help me proceed with the proof?

答: Hint: The $f_n$'s are bounded on $(0,\infty).$ If the convergence were uniform there, then $f$ would be bounded there.


问:
If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$
This is the $y=2^k$ case of this question.

Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?

Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that these two divisors are $2^k-1$ and $2^k+1$?

答:
I continue from Thomas Browning's (the author of the question) answer. We desire to show that

$$nx^2-4(n-1)y^2=1$$

has no solutions. Note that any solution must satisfy $\gcd(nx,y)=1$. We can rewrite the equation as

$$(nx)^2-4n(n-1)y^2=n,$$

so if

$$x^2-4n(n-1)y^2=n$$

has no solutions with $\gcd(x,y)=1$ then we're done. I'm going to prove that using the fact that

$$\frac xy\approx \sqrt{4n(n-1)}\approx 2n$$

and then squeezing the inequalities together and proving that they're too tight to hold. This corner of number theory is called Diophantine Approximation, and I happen to know about it. Start with

$$\sqrt{4n(n-1)}=[2(n-1);\overline{1,4(n-1)}]$$

This is easier to prove backwards. Let

$$t=2(n-1)+\frac 1{1+\frac 1{t+2(n-1)}}$$

and then it's easy to find that the positive solution is $t=\sqrt{4n(n-1)}$.

Also if

$$x^2-dy^2=n$$

then

$$\frac xy=\sqrt{d+\frac n{y^2}}=\sqrt{d}\sqrt{1+\frac n{dy^2}}$$

$$\frac xy-\sqrt{d}<\frac n{2\sqrt{d}y^2}$$

In our case $n>0$ and $d=4n(n-1)$ so

$$0<\frac xy-\sqrt{4n(n-1)}<\frac 1{4y^2\sqrt{1-1/n}}$$

Now from Hardy and Wright intro to number theory page 153:

Theorem 184. If

$$\left|\frac pq -x\right|<\frac 1{2q^2}$$

then $p/q$ is a convergent.

Note that when H&W say convergent they require it to be in lowest terms. Which is true of our previous expression, so $x/y$ is a convergent of $\sqrt{4n(n-1)}$. But the residues $x^2-dy^2$ left by a convergent $\frac xy$ to the continued fraction of $\sqrt d$ are periodic with the same period as the continued fraction itself. You can verify that when $d=4n(n-1)$ the residues are $1$ and $-4(n-1)$.

\begin{align*}
[2(n-1)]&=\frac{2(n-1)}1                   &(2(n-1))^2-4n(n-1)1^2&=-4(n-1)\\
[2(n-1);1]&=\frac{2n-1}1                   &(2n-1)^2-4n(n-1)1^2&=1\\
[2(n-1);1,4(n-1)]&=\frac{8n^2-10n+2}{4n-3} &(8n^2-10n+2)^2-4n(n-1)(4n-3)^2&=-4(n-1)\\
[2(n-1);1,4(n-1),1]&=\frac{8n^2-8n+1}{4n-2}&(8n^2-8n+1)^2-4n(n-1)(4n-2)^2&=1\\
[2(n-1);1,4(n-1),1,4(n-1)]&=\frac{32n^3-56n^2+26n-2}{16n^2-20n+5}&(\dots)^2-4n(n-1)(\dots)^2&=-4(n-1)
\end{align*}

So $n$ can never be a residue, therefore our equation has no solution.


问:
How to evaluate these integrals in hydraulic?
I am studying about hydraulics, there are integrals that I don't know how to evaluate. Here is the problem:

Equation of velocity is:  $$U_r=K(R^2-r^2)$$ Where $U_r$ is velocity
at radius $r$ and $R$ is radius of the pipe and $K$ is constant. Find
energy and momentum coefficients ($\alpha$ and $\beta$).

We have:  $\alpha=\cfrac{\int v^3dA}{V^3.A}$ and $\beta=\cfrac{\int v^2.dA}{V^2A}$ (where $v$ is velocity and $V$ is average velocity).

I draw this:

In this figure $(R^2-r^2)$ lies on the shaded area. so $A=\pi(R^2-r^2)$. but I don't know how to calculate these integrals to find $\alpha$ and $\beta$

EDIT:

For $\alpha$ I tried it:

$V=\frac{Q}{A}$ where $Q$ is flow rate:

$$\alpha=\cfrac{\int v^3dA}{V^3.A}=\cfrac{\int v^3dA}{\dfrac{Q^3}{A^2}}=\frac{A^2}{Q^3}\int v^3 dA=\frac{A^2}{Q^3}\int(K(R^2-r^2))^3dA=\cfrac{K^3 (\pi R^2)^2}{Q^3}\times\int (R^2-r^2)^3dA $$ Using polar system: $dA=rdrd\theta$ But I am not sure what can I write for $(R^2-r^2)^3$ and what is the integral bound? I don't know how to change it to polar system integral.

答:
$\displaystyle \int v^3dA = 2\pi K^3  \int_r^{R} (R^2-\rho^2)^3 \rho \, d\rho$

$\rho(R^2 - \rho^2)^3 = R^6 \rho - \rho^7 - 3\rho^3R^4 + 3 \rho^5R^2$

So your integral becomes

$ \displaystyle 2\pi K^3  \int_r^{R} (R^6 \rho - \rho^7 - 3\rho^3R^4 + 3 \rho^5R^2) d\rho$

Can you take it from here?


问:
Verify the given linear approximation and determine the values of x for which it is accurate within 0.1

Verify the given linear approximation at $a=0$. Then determine the values of x for which the linear approximation is accurate within 0.1: $$ln(1 + x) ≈ x $$

The official method that was used to calculate a question similar to this was to use this formula:

$$ |f(x) - L(x)| < 0.1 $$

for example, in this case, it would be:

$$|ln(1 + x) - x| < 0.1 $$

My first question is, how do you solve for x algebraically in the equation above?

My second question is why isn't this the equation?

$$ |f'(x) - L(x)| < 0.1 $$

Logically I think the equation above makes more sense because you want to compare the approximation of the tangent line (L(x)) with the actual tangent line (f'x).

答:
I believe the inequality you would be solving would be

$$|x-\ln(1+x)|<0.1$$

because you're trying to find the difference between your approximation $x$ and the function $\ln(1+x)$ itself.

WolframAlpha gives the following solution to six decimal places:

$$-0.383183<x<0.516221$$


问:
Is (F, G) → G ◦ F linear or bilinear?
We just started the concept of "bilinear" in class and I don't think i totally get it.
Lets say U, V, W are 3 vector spaces and C is a map where

C :
L(U, V ) × L(V, W) → L(U, W)
(F, G) → G ◦ F
Can anyone briefly explain if C is linear or bilinear?

答: I have read you comment in the post, but I hace to tell you that $C$ is bilinear. Since the domain of $C$ is $\mathcal L(U,V)\times\mathcal L(V,W)$, when you take an element of that domain, it is of the form $(f,g)$. Note that in each entry, you can sum and multiply by scalars, and, then, apply $C$. Indeed, given $f,g:U\to V$, $h:V\to W$ linear maps and $\lambda \in\mathbb K$, then $s:=\lambda f+g:U\to V$ is linear and we can apply $C$ to $h$ and $s$. We have: $C(s,h)=h\circ s=h\circ(\lambda f+g)$, evaluating that composition in a LM arbitrary point $p\in U$, one obtains $h\circ(\lambda f+g)(p)=\lambda h\circ f(p)+h\circ g(p)= (\lambda h\circ f+h\circ g)(p)$. That is, $h\circ(\lambda f+g)=h\circ\lambda f+h\circ g$. So, finally, $C$ is linear in the first entry. Analogously, you can get that it is linear in the second entry. Therefore, $C$ is bilinear.

Title: Is (F, G) → G ◦ F linear or bilinear?


答:
Something goes wrong here:

$$\equiv [(p \vee q) \wedge ((p \wedge \neg q) \vee \neg r)] \vee (\neg p \vee r)\text{ [De Morgan's law]}$$
$$\equiv (p \vee q) \wedge ((p \wedge \neg q) \vee \neg r) \vee (\neg p \vee r)$$
$$\equiv (p \vee q) \wedge (\neg p \vee r) \vee ((p \wedge \neg q) \vee \neg r)\text{ [Commutative law]}$$

You're basically going from something of the form $[A \land B] \lor C$ to $A \land B \lor C$ to $A \land C \lor B$

Note that the last two statements are both ungrammatical: since $(A \land B) \lor C$ is not equivalent to $A \land (B \lor C)$, you cannot just drop those parentheses.

Title: Proof of a compound proposition being a tautology

Body:
The question states the following compound proposition is a tautology, which I have to prove.

$$(p \rightarrow q) \wedge (q \rightarrow r) \rightarrow (p \rightarrow r)$$

My attempt is as follows:

$$\equiv (\neg p \vee q) \wedge (\neg q \vee r) \rightarrow (\neg p \vee r)$$
$$\equiv \neg((\neg p \vee q) \wedge (\neg q \vee r)) \vee (\neg p \vee r)$$

$$\equiv \neg[((\neg p \vee q) \wedge \neg q) \vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Distribution of conjunction over disjunction]}$$
$$\equiv \neg[((\neg p \wedge \neg q) \vee (q \wedge \neg q) \vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Distribution of conjunction over disjunction]}$$
$$\equiv \neg[((\neg p \wedge \neg q) \vee F )\vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Idempotent law]}$$
$$\equiv \neg[(\neg p \wedge \neg q) \vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Idempotent law]}$$
$$\equiv [(p \vee q) \wedge ((p \wedge \neg q) \vee \neg r)] \vee (\neg p \vee r)\text{ [De Morgan's law]}$$
$$\equiv (p \vee q) \wedge ((p \wedge \neg q) \vee \neg r) \vee (\neg p \vee r)$$
$$\equiv (p \vee q) \wedge (\neg p \vee r) \vee ((p \wedge \neg q) \vee \neg r)\text{ [Commutative law]}$$
$$\equiv [((p \vee q) \wedge \neg p) \vee ((p \vee q) \wedge r)] \vee ((p \wedge \neg q) \vee \neg r)\text{ [Conjunction over disjunction]}$$
$$\equiv [((p \wedge \neg p) \vee (q \wedge \neg p)) \vee ((p \vee q) \wedge r)] \vee ((p \wedge \neg q) \vee \neg r)\text{ [Conjunction over disjunction]}$$
$$\equiv [(F \vee (q \wedge \neg p)) \vee ((p \vee q) \wedge r)] \vee ((p \wedge \neg q) \vee \neg r)\text{ [Negation law]}$$

This goes on for quite a while, but I don't get the desired $T$ result. I'm guessing I've done something horribly wrong in the above calculation. I'll be so grateful for your help here. Anything regarding this text, from calculation error to using some incorrect terms. Thank yoou.


问:
How to solve $\int_0^\pi{\frac{\cos{nx}}{5 + 4\cos{x}}}dx$?
How can I solve the following integral?
$$\int_0^\pi{\frac{\cos{nx}}{5 + 4\cos{x}}}dx, n \in \mathbb{N}$$</p>

Body:
$2 I(n+1) + 5I(n) + 2I(n-1) = 0$

d’ou $I(n) = (-1/2)^n . I(0) = (-1/2)^n . \pi/3$

fjaclot


答:
Assume that $f_n \to f$ uniformly on $\langle 0,1\rangle$. For $\varepsilon = 1$ there exists $n_0 \in \Bbb{N}$ such that
$$n \ge n_0 \implies |f(x)-f_n(x)| < 1.$$

Now take $n := n_0 + 1$ and consider
$$\left|f\left(\frac1n\right) - f_n\left(\frac1n\right)\right| = n^2 - \frac1{1+\frac1{n^2}} \ge n^2-1 \ge n_0^2 \ge 1$$
which is a contradiction. This implies that the convergence is not uniform on $\langle 0,1\rangle$.

Title: How to see if the convergence is uniform?

Body:
I have the following : $f_n(x)=\frac{nx}{1+nx^3} \quad n=1,2,\dots \quad$ and $f(x)=\lim_{n \to \infty} f_n(x) ,$

and I have done the following :  $$|f_n(x) -f(x)| = \biggl|\frac{nx}{1+nx^3}-\frac{1}{x^2}\biggr| = \biggl|\frac{1}{x^2(1+nx^3)}\biggr |$$ where $\frac{1}{x^2}$ is the convergence of the series of the function. Now I know that $\Bigl|\frac{1}{x^2(1+nx^3)}\Bigr|< \epsilon$ but now I don't know how to choose $N$ in order to fully prove that it is uniformly convergent or not. Can somebody help me proceed with the proof?


问:
How to solve this multivariable limit?
I want to know if that function is continuous at $(0,0)$. So I tried to evaluate the limit at this point but I don't know how to solve it.
\begin{equation}
\lim_{(x,y)\to(0,0)} \left(\frac{x\sqrt{y}}{x^4+y^2}\right)
\end{equation}
Thanks in advance

答:
Hint

Try the two paths $y=x\to 0$ and $y=0,x\to 0$ to show that the limit does not exist.

Title: How to solve this multivariable limit?


问:
A tree is a connected graph with maximum possible vertices if given the number of edges?
My question is based on information that you are asked to deduce for yourself to confidently answer question #10 of Ralph P. Grimaldi's Discrete and Combinatorial Mathematics An Applied Introduction (5th Edition).

The question is:

The connected undirected graph $G=(V,E)$ has 30 edges. What is the maximum value that $|V|$ can have?

From what is written in the text I can sort of deduce that Theorem 12.3:

In every tree $T=(V,E), |V|=|E|+1$

will have something to do with it. However I would need to deduce that $G$ is a tree first.

My idea is to use the fact that if $G$ is planar and connected, then $v-e+r=2$ where $|V|=v$ and $|E|=e$ and $r$ is the number of regions made by the planar graph. To maximize $v$, minimize $r$. The smallest $r$ can be is 1 (zero inner regions, there is always 1 outer region). Then $v=e+2-1=e+1.$

If G is connected and nonplanar, then remove as many edges as necessary to make G planar. $|E|=30-k$ for our particular example, however we can make $|E|$ arbitrary. Then again, $v=e-k+2-1=e-k+1,$ for $k\geq1$, thus $v$ is smaller.

Since $G$ can be connected and either planar or nonplanar, the maximum number of $|V|$ is $|E|+1.$ And $G$ being connected and having $|V|=|E|+1$ is equivalent to $G$ being a tree.

Am I on the right track? Or are my assumptions about planar and nonplanar too swift?

答:
The planarity seems irrelevant for the problem, but here is one way to look at it:

Given $m$ "vertex-less" edges you can add at most $2m$ vertices (-> $m$ disjoint edges) but you want to maximize $|V|$ while staying connected. Look at it as an algorithm, a step by step process of adding vertices. For the first edge, you have no choice but to give it $2$ vertices. For the second edge you have to add at least one vertex to it (supposing multiple edges are forbidden), but adding $2$ would produce a disjoint edge. For all the subsequent edges, you can add between $0$ and $2$ vertices to each edge. Adding $2$ is illegal because it disconnects the graph, adding $1$ is possible at each step, so you do it (you want to maximize $|V|$).

This produces $2+(m-1)\cdot1=m+1$ vertices.


问:
Probability of same numbers in a 5 digits password

I am trying to solve this probability problem but the result I get is wrong, would like some guidelines and tips.

Creating a random password made from 5 digits (0-9), need to calculate the probability that two numbers (for example 6 and 7) appear the same amount of times (could appear zero times as well).

The sample's size is $10^5$ (since it's a 5 digits password), and the wanted probability is as follows:

$5!$ for organizing the 5 numbers

$8^5$ in-case both numbers don't appear

${10\choose1}$${9\choose1}$${5\choose2}$ in-case for example 6 and 7 appeared once, the ${5\choose2}$ is for using the 6 and 7

${10\choose2}$${8\choose2}$${5\choose4}$ in-case each of those numbers was chosen two times

So,

$$\frac{5!(8^5+{10\choose1}{9\choose1}{5\choose2}+{10\choose2}{8\choose2}{5\choose4})} {10^5}$$

But the answer I get is bigger than 1 which is incorrect for probabilities...

答:
The digits $6$ and $7$ appear twice each, choose the positions for the sixes, then for the sevens, then choose the other digit:

$$\binom{5}{2}\binom{3}{2}\times 8=240$$

The digits $6$ and $7$ appear once each, choose the position for the six, then for the seven, then choose the other digits:

$$\binom{5}{1}\binom{4}{1}\times 8^3=10240$$

The digits $6$ and $7$ do not appear:

$$8^5=32768$$

For a total of $43248$, giving the probability to be:

$$\frac{43248}{10^5}=0.43248$$


答: I have usually seen them referred to as parts or blocks of the partition. Partition of a Set</p>

Title: Terminology for an element of a partition?

Body:
Suppose I'm dividing some region $\Theta \in \mathbb{R}^n$ into subregions $\theta_i, i=1,2,3$ such that $\theta_i \cap \theta_j = \varnothing, i\ne j$ and $\bigcup_i \theta_i = \Theta$.  I might say (perhaps loosely, even technically incorrectly) that I am partitioning the region $\Theta$.

Thus, a "partition" would be a particular configuration of $\{\theta_1,\theta_2,\theta_3\}$ that satisfies the aforementioned conditions.  But what would an element of a partition be referred to as?

Since I am partitioning a "region", it makes sense to say that I am partitioning a region into "subregions", but at a higher level, what is a correct term for an element of a partition?


问:
Solve $\int_{\frac{\sqrt{3}a}{2}}^a 1-\frac{2 \sqrt{x^2-\frac{3}{4}a^2}}{a} dx$
As part of a probability question I am trying to solve this integral:

$$\int_{\frac{\sqrt{3}a}{2}}^a 1-\frac{2 \sqrt{x^2-\frac{3}{4}a^2}}{a} \mathrm dx$$

But am stuck. help would be appreciated.

答:
Hint

Use the substitution $x={\sqrt 3\over 2}a\cosh u$ for $a>0$ and $x=-{\sqrt 3\over 2}a\cosh u$ for $a<0$.


问:
Solving integrals of combination of $x^2$ and $\sin^2(x)$
How do I solve the integral:   $\frac{2}{L}\int^{L}_{0}x^2 (\sin(\frac{\pi x}{L}))^2dx$

The answer is given as $\frac{L^2}{3}-\frac{L^2}{2\pi ^2} $

I know one solve using integration by parts, but I was wondering if there formula one can use to simplify this .

答:
First it is easier to deal with this undimensionalized integral instead

$$I = \frac{2L^2}{\pi^3}\int_0^\pi x^2\sin^2x\:dx$$

By double angle identity we get

$$I = \frac{L^2}{\pi^3}\int_0^\pi x^2dx - \frac{L^2}{\pi^3}\int_0^\pi x^2\cos 2x \:dx$$

The first piece evaluates to $\frac{L^2}{3}$. For the next piece, replace $2$ with the variable $a$ and recognize that

$$\int_0^\pi x^2\cos ax \:dx = -\frac{d^2}{da^2}\int_0^\pi \cos ax \:dx$$

Which means the second piece evaluates to

$$ \frac{L^2}{\pi^3}\frac{d^2}{da^2}\frac{\sin \pi a}{a}\Biggr|_{a=2} = -\frac{L^2}{2\pi^2}$$

as advertised.


问:
Continuous functions that coincide in dense set are equal
Any corrections or feedback to the proof below will be highly appreciated.

Proposition. We have that $f\colon\mathbb{R}\to\mathbb{R}$ e $g\colon \mathbb{R}\to\mathbb{R}$ are continuous functions. Let $X_\alpha$ be a dense set in $\mathbb{R}$. If $f(x)=g(x)$ for all $x\in X_\alpha$, then $f=g$.

Proof. As $f$ and $g$ are continuous functions, we have for every point $x_0\in\mathbb{R}$ such that $\lim\limits_{n\to \infty}{f(x_n)}=f(x_0)$ for every sequence of points $(x_n)_n \in \Bbb R$ with $\lim\limits_{n\to\infty} x_n = x_0$, and analogous definition is valid for $g$.

Because $X_\alpha$ is dense, then $\overline{X_\alpha}=\mathbb{R}$, that is, we can find for every $z\in\mathbb{R}$ a sequence $(z_n)_n$ in $X_a$ such that $z_n\to z$.

Then for every $z\in\Bbb R$ we do exactly that, defining a sequence $(z_n)_n$ in $X_\alpha$ such that $z_n\to z$. And as $f$ and $g$ are both continuous, we say that we have $f(z_n)\to f(z)$ as well as $g(z_n)\to g(z)$.

And if $\lim\limits_{n\to\infty}{f(z_n)}=f(z)$ e $\lim\limits_{n\to\infty}{g(z_n)}=g(z)$, then we have that

$$\forall \epsilon_1>0\ \exists n_f\in\mathbb{N}:\ n>n_f\implies \left|f(z_n)-f(z)\right|<\epsilon_1$$

and also

$$\forall \epsilon_2>0\ \exists n_g\in\mathbb{N}:\ n>n_g\implies \left|g(z_n)-g(z)\right|<\epsilon_2$$

We take $n_0=\min{(n_f,n_g)}$, and we have

$$n>n_0\implies \left|f(z_n)-f(z)\right|<\epsilon_1\ \text{e}\ \left|g(z_n)-g(z)\right|<\epsilon_2$$

We define $\varepsilon=\epsilon_1+\epsilon_2$, and therefore

$$\left|f(z_n)-f(z)\right| + \left| g(z_n)-g(z)\right|<\varepsilon$$

If $f(x)=g(x)$ for all $x\in X_\alpha$, then $f(z_n)=g(z_n)$ for all natural $n$, because $z_n\in X_\alpha$. And by triangular inequality we have

$$\left|f(z)-g(z)\right|\leq \left|f(z_n)-f(z)\right| + \left| g(z_n)-g(z)\right|$$

But $\left|f(z)-g(z)\right|$ is absolute value, therefore

$$0\leq \left|f(z)-g(z)\right|<\varepsilon;\ \text{for all $\varepsilon>0$}$$

We claim that $0= \left|f(z)-g(z)\right|$, because if we assume the opposite, we would have $\left|f(z)-g(z)\right|= a >0$, then we could define $\varepsilon=\frac{a}{2}$ and find $a=\left|f(z)-g(z)\right|<\frac{a}{2}$, contradiction.

Hence,

$$0= \left|f(z)-g(z)\right|\implies f(z)=g(z)\ \text{for all $z\in\mathbb{R}$}$$

Ergo $f=g$. $\tag*{$\blacksquare$}$

$\rule{10cm}{0.4pt}$

What if $X_\alpha$ is not a dense set?

If $X_\alpha$ is not dense set, that is, $\overline{X_\alpha} \subsetneq \mathbb{R}$ is a proper subset. Well, this means that there is some $z\in\mathbb{R}$ such that $z\not\in\overline{X_\alpha}$, or yet, $z$ is not a limit point of $X_\alpha$, then there is no sequence $(z_n)_n$ of $X_\alpha$ such that $z_n\to z$.

答:
Since you included the general-topology tag, I’ll point out that this is a special case of a much more general result that has a short, simple proof.

Theorem. Let $f$ and $g$ be continuous functions from a space $X$ to a Hausdorff space $Y$, and let $D$ be a dense subset of $X$ such that $f\upharpoonright D=g\upharpoonright D$; then $f=g$.

Proof. If not, there is an $x\in X$ such that $f(x)\ne g(x)$. $Y$ is Hausdorff, so there are disjoint open sets $U$ and $V$ in $Y$ such that $f(x)\in U$ and $g(x)\in V$. Let $W=f^{-1}[U]\cap g^{-1}[V]$; clearly $x\in W$, so $W\ne\varnothing$, and the continuity of $f$ and $g$ ensures that $W$ is open in $X$, so $W\cap D\ne\varnothing$. Let $y\in W\cap D$; then by hypothesis $f(y)=g(y)$, but $f(y)\in U$, $g(y)\in V$, and $U\cap V=\varnothing$, so $f(y)\ne g(y)$. This contradiction shows that $f=g$. $\dashv$

It’s not hard to adapt this proof to your specific setting.


问:
Example of $f \in K[x]$ solvable by radicals but having a root inexpressible only by coefficients of $f$ and +, -, *, /, $\sqrt[n]{...}$
Remark: at a very first glance it might seems as a duplicate of Example of $f \in K[x]$ solvable by radicals but having a root that cannot be expressed by using only coefficients of $f$, $+,-,\cdot,\frac{..}{..}$ but it is not. I made overlook in that question - I forgot to specify $\sqrt[n]{...}$ as an operation that we can use in an expression.

The definitions below are taken from Solvability by radicals implies a radical formula for its roots (question by Eparoh):

Definition 1: We say that a field extension $F/K$ is a radical extension if we can form a chain of fields
$$K=K_0 \leq K_1 \leq \cdots \leq K_n=F$$
where $K_{i+1}/K_i$ is a simple extension such that $K_{i+1}=K_i(a_i)$ and $a_i^{k_i} \in K_i$ for some positive integer $k_i$.

Definition 2: Let $K$ be a field and $f(x) \in K[x]$, we say that $f$ is solvable by radicals if there exists a radical extension $F/K$ such that $F$ contains a splitting field of $f$ over $K$.

This question has no answer, but it has a comment by reuns:

The radical formulas for the roots depend on constants of $K$, once the polynomial is fixed this is all we want (there are algorithms for the splitting field minimal polynomials and Galois group, if it is solvable we can unroll to find the radical formulas). What you are asking is if there are finitely many radical formulas $F_{d,l}$ of $d+1$ variables such that for every solvable polynomial $∑_{j=0}^{d} c_j x_j \in K[x]$ of degree $d$ its roots are given by $F_{d,l}(c_0,…,c_d)$ for some $l$. This is the problem of moduli space / parametrization of solvable polynomials of degree $d$.

Let $K$ be a field. Can you give an example of $f \in K[x]$ that is solvable by radicals but cannot be expressed by using only polynomial coefficients, $+, -, \cdot, \frac{...}{...}$ and operations of taking roots of natural ($N_+$) degrees and proof of this fact?

As I understand the comment I quoted above such polynomials and roots exist. I ask this question after a long quest for the answer to exactly the question asked in the linked post. I have no idea where to look for examples of such polynomials and roots in literature. To find the answer to the original question was itself hard. I have tried to google for "moduli space / parametrization of solvable polynomials" just having little hope that it will return information relevant to the posed problem, but no luck (as could be of course expected since it is not something directly related).

I must say that I do not exactly understand the quoted comment, but I think I will ask another question to resolve my doubts.

Edit 1: I have added this because it seems that the repliers make some implicit assumption which is exactly the point of my question. What I'm asking about is if there is an example of $f \in K[x]$ that is solvable by radicals but cannot be expressed by using ONLY polynomial coefficients, $+,−,\cdot,$ and operations of taking roots of natural ($N_+$) degrees. That is, it is not allowed to use these members of K that are not expressible in this form. The question from which I have taken definitions asks a very similar question, but it is not the same as it does not ask for such a specific example. I do not exactly understand the comment by reuns and I have some doubts if it is correct (please see the questions from which I took definitions, I think it will be beneficial for this discussion). But as I understand it, it says that if the assumptions I listed are satisfied then it is always possible to express roots using only members of K, $+, -, \cdot, \frac{...}{...}$ and operations of taking roots of natural ($N_+$) degrees, but not necessarily only polynomial coefficients, $+, -, \cdot, \frac{...}{...}$ and operations of taking roots of natural ($N_+$) degrees.

答:
I think this is more of a confusion of language and nothing else. If $f(x) \in K[x] $ is a specific polynomial then the coefficients of $f$ are nothing but specific members of $K$.

And then if you have a formula for roots of $f$ which involves a combination of some members of $K$ along with operations like $+, -, \times, /, \sqrt[n] {. } $ then the coefficients of $f$ themselves being members of $K$ can not be visually located in the formula. Any member of $K$ can for example easily be written as a combination of any given number of members of $K$ using just the field operations.

You are perhaps trying to think of an example where the coefficients are literals like in case of $x^2+ax+b$ and $K=\mathbb{Q} $, but again this is wrong. In such case the field should be $K=\mathbb{C} (a, b) $.

Let us then assume that we have a literal polynomial $$f(x)= x^n+a_1x^{n-1}+\dots+a_{n-1}x+a_n$$ over field $K=\mathbb{C} (a_1,a_2,\dots,a_n)$. If $f$ is solvable by radicals over $K$ then formula for roots involves arithmetic operations and radicals (nested if needed) applied on members of $K$ and it does include the literal coefficients of $f$ because they are what $K$ is made of. This is easily seen to be the case in case of quadratic or cubic equations which are known to be solvable.

Thus the coefficients always enter the formula for roots if there is a formula available.

Also note the well known fact (established by Abel well before Galois) that the polynomials with literal coefficients are solvable over their field of coefficients ($K=\mathbb{C} (a_1,a_2,\dots,a_n)$) if and only if $n<5$.

To summarize such an example which you are seeking does not exist.

I have tried to discern the meaning of the comment by reuns and it appears related to the treatment of solvable quintic given by Dummit and Foote in his Abstract Algebra.

They describe a criterion to check whether a given quintic $$f(x) =x^5+ax^4+bx^3+cx^2+dx+e\in\mathbb{Q}[x]$$ is solvable over $\mathbb{C} $. The idea is to form a complicated polynomial of degree 6 in $\mathbb{Q} [x] $ with coefficients made using coefficients of $f$ and checking whether it has a rational root or not.

And if the polynomial of degree 6 mentioned above does have a rational root then $f$ is solvable by radicals over $\mathbb{C} $. You perhaps want to check (for this case) if there is a formula for roots based on elements of $K=\mathbb {C} (a, b, c, d, e) $. I think there is such a formula but I am not sure.

Usually when we consider the problem of solvability of a polynomial $f(x) \in K[x] $, the field $K$ is the smallest field containing the coefficients of $f$. In this case if the polynomial is solvable by radicals over $K$ then the roots can be expressed in terms of coefficients of $f$ via arithmetic operations and radicals.

Enlarging the field $K$ to some extension $L$ and checking solvability over $L$ makes the problem simpler (trivial if $L$ is splitting field of $f$).

Also if we consider the scenario where $f(x) \in K[x] $ is solvable by radicals over $K$ and $F\subset K$ is the smallest field containing the coefficients we need to investigate the problem of solvability of $f$ over $F$ separately and one can't deduce anything from its solvability over $K$.

Thus your problem makes sense only in the usual setting where the solvability is checked over the field of coefficients and then (to repeat what I said earlier) the kind of example you seek does not exist.


问:
Help solving the limit of the sequence: $\left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) }$
Given the following sequence,

$$\left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) }$$
I am asked to determine to which $l \in \mathbb{R}$ the sequence converges.

This is what I have tried so far:

$$
\lim_{n} \left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) } \implies
\left( \frac{2n^2 + 1}{2n^2 - 1} \right) ^ { - \left( \frac{2n^3 - n}{n + 3} \right) }  \implies
\left( \frac{2n^2}{2n^2 - 1} + \frac{1}{2n^2 - 1} \right) ^ { - \left( \frac{2n^3 - n}{n + 3} \right) } \implies
\left( 1 + \frac{1}{2n^2 - 1} \right) ^ { - \left( \frac{2n^3 - n}{n + 3} \right) } \implies ? $$

My objective was to algebraically modify the sequence such that I could be able to reduce it to the form: $$\lim _{x\to +\infty }\left(1+{\frac {1}{x}}\right)^{x}=e$$
but seem to be stuck at the ? point.

Any suggestion? Thank you.

答:
Hint

$$=\left(\left(1-\dfrac2{2n^2+1}\right)^{-(2n^2+1)/2}\right)^{-\frac{2(2n^3-n)}{(2n^2+1)(n+3)}}$$

Alternatively

$$=\left(\dfrac{(1-1/2n^2)^{2n^2}}{(1+1/2n^2)^{2n^2}}\right)^{\dfrac{2n^3-n}{2n^2(n+3)}}$$

Now the numerator converges to $e^{-1}$

What about the denominator & the exponent?


问:
Finiteness is like compactness+discreteness?
I was reading this interesting post about compactness. I understood that the outcome of the story was that finiteness is the same as compactness+discreteness. I understand that discreteness+compactness $\implies$ finiteness. Also it is clear that finiteness implies compactness. But why finiteness $\implies$ discreteness? If I consider the finite space $X = \{x_1,x_2,x_3,x_4,x_5\}$ with the coarse topology $T=\{\emptyset,X\}$ it is not discrete no?

答: When I wrote that answer I had in mind a “pre-theoretical” understanding of finiteness, and also as Law Nagi says in the comments the OP’s question was about subspaces of $\mathbb{R}^n$. You can interpret me to have been restricting attention to metric spaces or Hausdorff spaces if you want a precise statement.


问:
Proving that the the sequence $x_n = \frac{n^{100}}{2^n}$ converges using only elementary methods
I realise that there are already questions on the site about this sequence or related sequences, but I'm looking for a completely elementary method that doesn't use:

L'Hopital's Rule

Tests like the comparison test or ratio test

Logarithms

Statements like "exponentials grow faster than polynomials"

I'm trying to show that the sequence $$x_n = \frac{n^{100}}{2^n}$$ converges as $n \to \infty$, where $n$ is a positive integer. I haven't been able to find a way to do this directly using the epsilon definition for convergence. Is there a completely elementary way to show that this converges (not necessarily needing to show what it converges to) without using any of the things mentioned above?

答: As an elementary method, note that for $n>10$, we have $n+1< 1.1n$ hence the numerator of $x_n$ increases at a rate of at most $1.1$ and the denominator increases at a rate of exactly $2$, a sufficient condition for $x_n\to0$.


问:
Lyapunov Function for Nonlinear System
Which Lyapunov function should I choose to show the stability (or instability) of equilibrium points? With $k>0$, $K>0$, $\delta >0$. The system is Hurwitz (asymptotically stable) when $k>K$. Standard quadratic Lyapunov function
\begin{equation}
    V(x_1,x_2,x_3,x_4) = \frac{k(x_1-x_2)^2}{2}+\frac{x_3^2}{2}+\frac{x_4^2}{2}-\frac{K(x_1-x_2)^2}{2}
\end{equation}
does not work.
Linearized system
\begin{equation}
\begin{cases}
\dot{x}_1=x_3,\\
\dot{x}_2=x_4,\\
\dot{x}_3=-k(x_1-x_2)-K x_2,\\
\dot{x}_4=k(x_1-x_2)-\delta x_4,
\end{cases}
\end{equation}
gives
\begin{equation}
    A =
\begin{pmatrix}
0& 0& 1& 0\\
0& 0& 0& 1\\
-k& k-K& 0& 0\\
k& -k& 0& -\delta
\end{pmatrix}.
\end{equation}
By the Lyapunov's converse theorems I know that there exists a unique symmetric positive definite matrix $P$ as a solution to the Lyapunov equation with $\boldsymbol Q=\boldsymbol I$:
$$\boldsymbol{PA}+\boldsymbol{A^TP}=-\boldsymbol{Q}$$
I have tried to solve it symbolically via Matlab, but did cope with it. I have also tried to differentiate Lyapunov function without knowing explicit form for $\boldsymbol {P}$ but the result of $\boldsymbol {\dot{V}} = \boldsymbol x^T \boldsymbol P \boldsymbol x$ gives the system of 16 equations which I could not solve as well.

答:
With the help of MATHEMATICA.

$$
A = \left(
\begin{array}{cccc}
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1 \\
 -k & -k-K & 0 & 0 \\
 k & -k & 0 & -\delta  \\
\end{array}
\right)
$$

and the command

```<code>P = LyapunovSolve[A, Transpose[A], -IdentityMatrix[4]]
</code>```

we obtain

$$
P = \left[
\begin{array}{cccc}
 -\frac{\delta ^2 \left(3 k K+k (3 k+2)+K^2+K\right)+K (k+K) (2 k+K+1)}{2 \delta  k (k+K) (2 k+K)} & -\frac{\delta ^2 k+\delta ^2 K+K}{4 \delta  k^2+2
   \delta  k K} & -\frac{1}{2} & -\frac{k+1}{2 (k+K)} \\
 -\frac{\delta ^2 k+\delta ^2 K+K}{4 \delta  k^2+2 \delta  k K} & \frac{\delta ^2 (k+K)+K (2 k+K+1)}{2 \delta  (k+K) (2 k+K)} & \frac{k+1}{2 (k+K)} &
   -\frac{1}{2} \\
 -\frac{1}{2} & \frac{k+1}{2 (k+K)} & -\frac{\frac{\delta ^2 \left(2 k^2+2 k K+k+K^2\right)}{k+K}+(k+1) K}{2 \delta  k} & \frac{K}{2 \delta } \\
 -\frac{k+1}{2 (k+K)} & -\frac{1}{2} & \frac{K}{2 \delta } & \frac{k K+K}{2 \delta  k+2 \delta  K} \\
\end{array}
\right]
$$

Title: Lyapunov Function for Nonlinear System

Body:
With the help of MATHEMATICA.

$$
A = \left(
\begin{array}{cccc}
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1 \\
 -k & -k-K & 0 & 0 \\
 k & -k & 0 & -\delta  \\
\end{array}
\right)
$$

and the command

P = LyapunovSolve[A, Transpose[A], -IdentityMatrix[4]]

we obtain

$$
P = \left[
\begin{array}{cccc}
 -\frac{\delta ^2 \left(3 k K+k (3 k+2)+K^2+K\right)+K (k+K) (2 k+K+1)}{2 \delta  k (k+K) (2 k+K)} & -\frac{\delta ^2 k+\delta ^2 K+K}{4 \delta  k^2+2
   \delta  k K} & -\frac{1}{2} & -\frac{k+1}{2 (k+K)} \\
 -\frac{\delta ^2 k+\delta ^2 K+K}{4 \delta  k^2+2 \delta  k K} & \frac{\delta ^2 (k+K)+K (2 k+K+1)}{2 \delta  (k+K) (2 k+K)} & \frac{k+1}{2 (k+K)} &
   -\frac{1}{2} \\
 -\frac{1}{2} & \frac{k+1}{2 (k+K)} & -\frac{\frac{\delta ^2 \left(2 k^2+2 k K+k+K^2\right)}{k+K}+(k+1) K}{2 \delta  k} & \frac{K}{2 \delta } \\
 -\frac{k+1}{2 (k+K)} & -\frac{1}{2} & \frac{K}{2 \delta } & \frac{k K+K}{2 \delta  k+2 \delta  K} \\
\end{array}
\right]
$$


答:
Hint:

Linearise the sine: $\;\sin^2\frac{\pi x}{L}=x^2\frac 12\bigl(1-\cos\frac{2\pi x}L\bigr)$, and use that
$$\tfrac 12x^2\Bigl(1-\cos\frac{2\pi x}L\Bigr)=\tfrac 12x^2\operatorname{Im}\Bigl(1-\mathrm e^{\tfrac{2i\pi x}L}\Bigr),$$
to calculate first the antiderivative of $\enspace x^2\mathrm e^{\tfrac{i\pi x}L}$, which has the form
$$(ax^2+bx+c)\mathrm e^{\tfrac{2i\pi x}L}  \quad (a,b,c\in\mathbf C).$$

Title: Solving integrals of combination of $x^2$ and $\sin^2(x)$

Body:
How do I solve the integral:   $\frac{2}{L}\int^{L}_{0}x^2 (\sin(\frac{\pi x}{L}))^2dx$

The answer is given as $\frac{L^2}{3}-\frac{L^2}{2\pi ^2} $

I know one solve using integration by parts, but I was wondering if there formula one can use to simplify this .


答:
I'm assuming your function is defined by cases, otherwise asking if it is continuos in a point where it's not even defined would be meaningless.
A function $f:A\subseteq\mathbb{R}^2\rightarrow R$ is continuos if and only if it's sequentially continuos, that means:
\begin{equation}
\lim_{x\to x_0}f(x)=f(x_0)\iff\forall\{x_n\}\subseteq A: x_n\rightarrow x_0 \implies f(x_n)\rightarrow f(x_0) \tag{1}
\end{equation}
Consider the sequence $\{(\frac{1}{n},\frac{1}{n^2})\}_{n\geqslant1}$ that converges to $(0,0)$. If you compute:
\begin{equation}
\lim_{n\to +\infty}f(x_n)=\lim_{n\to +\infty}\frac{\frac{1}{n}\frac{1}{n}}{\frac{1}{n^4}+\frac{1}{n^4}}=\lim_{n\to +\infty}\frac{\frac{1}{n^2}}{\frac{1}{n^4}}=\lim_{n\to +\infty} n^2=+\infty
\tag{2}
\end{equation}
If your function were continuos at $(0,0)$, you would have that:
\begin{equation}
\lim_{x\to x_0}f(x)=f(x_0)
\end{equation}
that, considering $(1)$ is clearly in contradiction with $(2)$.

Title: How to solve this multivariable limit?

Body:
I want to know if that function is continuous at $(0,0)$. So I tried to evaluate the limit at this point but I don't know how to solve it.
\begin{equation}
\lim_{(x,y)\to(0,0)} \left(\frac{x\sqrt{y}}{x^4+y^2}\right)
\end{equation}
Thanks in advance


Title: Help solving the limit of the sequence: $\left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) }$

Body:
Given the following sequence,

$$\left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) }$$
I am asked to determine to which $l \in \mathbb{R}$ the sequence converges.

This is what I have tried so far:

$$
\lim_{n} \left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) } \implies
\left( \frac{2n^2 + 1}{2n^2 - 1} \right) ^ { - \left( \frac{2n^3 - n}{n + 3} \right) }  \implies
\left( \frac{2n^2}{2n^2 - 1} + \frac{1}{2n^2 - 1} \right) ^ { - \left( \frac{2n^3 - n}{n + 3} \right) } \implies
\left( 1 + \frac{1}{2n^2 - 1} \right) ^ { - \left( \frac{2n^3 - n}{n + 3} \right) } \implies ? $$

My objective was to algebraically modify the sequence such that I could be able to reduce it to the form: $$\lim _{x\to +\infty }\left(1+{\frac {1}{x}}\right)^{x}=e$$
but seem to be stuck at the ? point.

Any suggestion? Thank you.

Body: Use$$\ln l=\lim_{n\to\infty}2n^2\underbrace{\frac{1-1/(2n^2)}{1+3/n}}_{\sim1}\underbrace{\ln\underbrace{\frac{1-1/(2n^2)}{1+1/(2n^2)}}_{1-1/n^2+o(1/n^2)}}_{\sim-1/n^2}=-2\implies l=e^{-2}.$$


问:
Formula for how natural numbers n is x divisible by $2^n$ when $2^n\leq x$
I am currently trying to find a formula that does the same job as:
$$\sum_{n=1}^{\log_2(x)}{f(x, n)}$$ $$f(x, n)=\begin{cases}1 &\text{if } x \text{ divisible by } 2^n \\[4px] 0 &\text{if } x \text{ not divisible by } 2^n \end{cases}$$
The closest I have been able to find online is tricks finding how many numbers are divisible by n which work with methods using binary however I was not able to apply it to this.

答:
If $x$ has a decomposition of $x=2^q\cdot m$ for some non-negative integers $q$ and $m$ where $m$ is odd, then $f(x,n)=1$ only if $n\in\{1,2,\cdots, q\}$. Also $\log_2x=q+\log_2 m$ hence
$$
\sum_{n=1}^{\log_2 x}f(x,n)=q
$$


问:
Find $(A_{t})_{t\geq 0}$ such that $(B_{t}^4+A_{t})_{t\geq 0}$ is a martingale for Brownian motion $(B_t)_{t \geq 0}$
Let $(B_{t})_{t\geq 0}$ be a Brownian motion and $(\mathcal{F}_{t})_{t\geq 0}$ its canonical filtration.  Find a Stochastic Process $(A_{t})_{t\geq 0}$ such that $(B_{t}^{4}+A_{t})_{t\geq 0}$ is a Martingale.

I don't want to settle for a single $(A_{t})_{t\geq 0}$ process, I would like to be able to find a family of processes.

My attempt:
Let $X_{t}:=B_{t}^{4}+A_{t}$ and $t>s$, then we have
\begin{align}
\mathbb{E}[X_{t}-X_{s}|\mathcal{F}_{s}]&= \mathbb{E}[B_{t}^{4}+A_{t}-B_{s}^{4}-A_{s}|\mathcal{F}_{s}]\\
&=3(t-s)^{2}+6(t-s)B_{s}^{2}+\mathbb{E}[A_{t}-A_{s}|\mathcal{F}_{s}]
\end{align}
If $(X_{t})_{t\geq 0}$ is a martingale, then we should have
$$0=3(t-s)^{2}+6(t-s)B_{s}^{2}+\mathbb{E}[A_{t}-A_{s}|\mathcal{F}_{s}]$$
and and we should also have that $(A_{t})_{t\geq 0}$ is adapted to filtration $(\mathcal{F}_{t})_{t\geq 0}$. Therefore, we have
$$A_{s}=3(t-s)^{2}+6(t-s)B_{s}^{2}+\mathbb{E}[A_{t}|\mathcal{F}_{s}] \tag{$\bigstar$}$$

Therefore, it all comes down to solving equation ($\bigstar$).

答:
(I take it that you are looking for a non-trivial solution - otherwise you could simply take $A_t := -B_t^4$, as was pointed out in a comment.)

Using that $(B_t)_{t \geq 0}$ has independent and stationary increments and the fact that $B_t \sim N(0,t)$, it follows that

$$M_t^{(\lambda)} := \exp \left(\lambda B_t - \frac{1}{2} \lambda^2 t \right)$$

is a martingale for any $\lambda \in \mathbb{R}$. Equivalently,

$$\int_F M_s^{(\lambda)} \, d\mathbb{P} = \int_F M_t^{(\lambda)} \, d\mathbb{P} \tag{1}$$

holds for all $s \leq t$ and $F \in \mathcal{F}_s$. Now the idea is to differentiate $(1)$ with respect to $\lambda$. For the first derivative, we get

$$\int_F (B_s-\lambda s) M_s^{(\lambda)} \, d\mathbb{P} = \int_F (B_t-\lambda t) M_t^{(\lambda)} \, d\mathbb{P}. \tag{2}$$

Since this holds in particular for $\lambda=0$, we see that $\int_F B_t \, d\mathbb{P} = \int_F B_s \, d\mathbb{P}$ for all $s \leq t$ and $F \in \mathcal{F}_s$, which means that $(B_t)_{t \geq 0}$ is a martingale (no surprise here; we know this already). By calculating higher derivatives, we can increase the "power" of the martingale. E.g. by differentiating $(2)$ once more again with respect to $\lambda$, we find that

$$\int_F ((B_s-\lambda s)^2-s) M_s^{(\lambda)} \, d\mathbb{P} = \int_F ((B_t-\lambda t)^2-t) M_t^{(\lambda)} \, d\mathbb{P}.$$

Putting $\lambda=0$, we see that $(B_t^2-t)_{t \geq 0}$ is a martingale. To get a martingale involving $B_t^4$ we need two differentiate two more times. Doing so yields

$$\int_F (B_s-\lambda s)((B_s-\lambda s)^2-3s) M_s^{(\lambda)} \, d\mathbb{P} = \int_F (B_t-\lambda t)((B_t-\lambda t)^2-3t) M_t^{(\lambda)} \, d\mathbb{P}$$

(which shows, for $\lambda=0$, that $(B_t^3-tB_t)_{t \geq 0}$ is a martingale) and

\begin{align*} &\int_F \left[((B_s-\lambda s)^2-s)((B_s-\lambda s)^2-3s)-2s(B_s-\lambda s)^2\right] M_s^{(\lambda)} \, d\mathbb{P} \\
&= \int_F\left[((B_t-\lambda t)^2-t)((B_t-\lambda t)^2-3t)-2t(B_t-\lambda t)^2 \right] M_t^{(\lambda)} \, d\mathbb{P},\end{align*}

which shows, for $\lambda=0$, that

$$(B_t^2-t)(B_t^2-3t)-2t B_t^2=B_t^4-6tB_t+3t^2$$

is a martingale.

Remark: There is a close connection to Hermite polynomials, see e.g. this question.


问:
Is it possible for an increasing function to tend to zero?
I have to plot a graph of a real function $f(x)$ such that
$$\lim_{x\to\infty}f(x)=0$$
and for $x\geq3$, $f$ increases ($x=3$ is supposed to be the minimum point of $f$).

So, is it even possible? For me, since $f$ increases, then
$$\lim_{x\to\infty}f(x)=\infty\neq0$$

答:
Other people have given explicit examples, but I wanted to explain what is wrong with the reasoning that

Since $f$ increases, then $$ \lim_{x \to \infty} f(x) = \infty \, .$$

Consider the function $f$ defined by
$$
f(x)=-\frac{1}{x} \, ,x>0\, .
$$
This doesn't satisfy your requirement of having a minimum at $x=3$, but notice that the function is strictly increasing. As $x$ gets larger, $f(x)$ gets closer and closer to $0$, but does not ever reach it.


问:
What is the center and radius of convergence of this complex series?
$$\sum_{n=0}^{\infty} (\frac{2-i}{1+5i}) z^n$$

So my book says the center = 0 and the radius $= \sqrt{\frac{26}{5}}$

I don't know how we go there.

So I can see how the above can be written: $$\sum_{n=0}^{\infty} (\frac{2-i}{1+5i}) (z - 0)^n$$

and that $a = \frac{2-i}{1+5i}$ and that $a_{n+1} = \frac{2-i}{1+5i}$ too so by Cauchy Hadamard doesn't that mean: $\lim_{n \to \infty} \left | \frac{2-i}{1 + 5i} \cdot \frac{1 + 5i}{2-i}\right | = 1 = R$ so where does the $\sqrt{\frac{26}{5}}$ come from?

答: You shouldn't define $x_n={2-i\over 1+5i}$ and apply the ratio test since you have obviously ignored $z^n$. Try to apply the ratio test over the full summand of the $\sum$ operator, i.e. $x_n=({2-i\over 1+5i})z^n$ to obtain what you wish for.


问:
Prove that $f(1)-f(1/e)\le \int_0^1 \sqrt{x} f'(x) dx$
Let $f:[0,1]\rightarrow \mathbb{R}$ be a differentiable function such that
$$f(x^2)+f(y^2)\le2 f(\sqrt{x y}), \space x,y\ge0 $$

Prove that

$$f(1)-f(1/e)\le \int_0^1 \sqrt{x} f'(x) dx$$

Where should I start from?

答:
This is more of a comment. You could probably tweak this to get the answer.

Note that
$$\int_0^1 \sqrt{x} f'(x)dx = \left.\sqrt{x} f(x) \right\vert_0^1 - \int_0^1\dfrac{f(x)}{2\sqrt{x}}dx = f(1) - \int_0^1 \dfrac{f(x)}{2\sqrt{x}}dx$$
Hence, it is enough to prove that
$$f(1/e) \geq \int_0^1 \dfrac{f(x)dx}{2\sqrt{x}}$$
From the given inequality, we have $$f(x^2) \leq f(x) \leq f(0)$$
Hence,
$$\int_0^1 \dfrac{f(x)}{2\sqrt{x}}dx \leq f(0) \int_0^1 \dfrac{dx}{2\sqrt{x}} \implies f(0) \geq \int_0^1 \dfrac{f(x)}{2\sqrt{x}}dx$$

You can stretch this a bit more. I have weakened it by using the fact that $f(x) \leq f(0)$, instead of $f(x^2) \leq f(x)$. You might try instead to somehow exploit just the fact that $f(x^2) \leq f(x)$. This seems feasible but now I have to run for a meeting.


问:
how would you write taylor series for higher order dimension in this case?
Let $x$ and $y$ be vectors where $x=(x_1,\dots,x_d)$ and $y=(y_1,\dots,y_d)$ and let $g(λ)=f_i(x+λ(y-x))$ for $λ\in\mathbb R$ and $1\le i\le d$. How can I write a Taylor expansion for $g(λ)$ giving the remainder term at 2nd order? Any help would be appreciated.

答:
Hint

Use may use $$
{d\over dt} f(u+tv)=\nabla f(u+tv)^T\cdot v
$$
where $u$ and $v$ are vectors and the equality actually comes from a simple partial differentiation by hand.


问:
For which values of $α \in \mathbb R$ is the matrix $A$ diagonally dominant. Justify.
Let the matrix $A \in R
^{n×n}$. Then:

• $A$ is strictly diagonally dominant if $|A(i, i)| > \sum_{j=1,j \not= i}^n |A(i, j)|$ for all $i = 1, 2,..., n.
$

• $A $is diagonally dominant if  $|A(i, i)| \geq \sum_{j=1,j \not= i}^n |A(i, j)|$ for all $i = 1, 2,..., n$.

• If $A$ is symmetric strictly diagonally dominant matrix with positive diagonal entries $(A(i, i) > 0, i = 1 : n)$,
then $A$ is positive definite.

Let $A = C+α||C||_ ∞I$ where $C$ is any real $n×n$ matrix. For which values of $α \in \mathbb R$ is the matrix $A$ diagonally
dominant. Justify. For which values of $α \in \mathbb R$ is the matrix $A$ strictly diagonally dominant. Justify.

My attempt:

$|A(i, i)| = |C(i, i) + \alpha||C(i, i)||_ ∞I(i, i) | = |C(i, i) + \alpha||C(i, i)||_ ∞|$ since $I(i, i) = 1$

And $|A(i, j)| = |C(i, j) + \alpha||C(i, j)||_ ∞I(i, j) | = |C(i, j)|$ since $I(i, j) = 0$ for $i \not= j$.

Then $A$ is diagonally dominant if $|C(i, i) + \alpha||C(i, i)||_ ∞| \geq \sum_{j=1,j \not= i}^n|C(i, j)|$

And $A$ is strictly diagonally dominant if $|C(i, i) + \alpha||C(i, i)||_ ∞| > \sum_{j=1,j \not= i}^n|C(i, j)|$

But I didn't know how to proceed from there to get $\alpha$. Any help please?

答:
Note that $\|C\|_\infty = \max_{1\le i \le n} \sum_{i=1}^n |C(i,j)|$. For $|\alpha| \ge 1$ we have

$$|A(i,i)| = |C(i,i)+\alpha \|C\|_\infty| \ge |\alpha|\|C\|_\infty - |C(i,i)| \ge \|C\|_\infty - |C(i,i)| \ge\\
\ge \sum_{i=1}^n |C(i,j)| - |C(i,i)| = \sum_{i=1,i\ne j}^n |C(i,j)| =  \sum_{i=1,i\ne j}^n |A(i,j)|$$

so $A$ is diagonally dominant. Similarly, for $\alpha > 1$, assuming $C \ne 0$, we see that $A$ is strictly diagonally dominant.

If $|\alpha| < 1$ then $A$ in general doesn't have to be diagonally dominant, e.g. with $$C = \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}.$$


问:
Contour integration around pole and essential singularity
I've got the following integral:

$$
\frac{1}{2\pi i}\oint_\limits{|z|\,=\,R} \frac{z^5}{z-1} e^\frac{2}{z} dz,
$$

where $R$ - sufficiently large number.

I've tried to evaluate it with residues. The integrand has a simple pole at $z = 1$ and an essential singularity at $z=0$.

Evaluation of the residue about the pole is trivial: it equals to $e^2$. I tried to deal with the singularity:

$$
\frac{z^5}{z-1} e^\frac{2}{z} = z^4 \frac{1}{1-\frac{1}{z}} e^\frac{2}{z} = \left[z^4 + z^3 + z^2 + z + 1 + \frac{1}{z} + ... \right]\left[1 + \frac{2}{z} + \frac{2}{z^2} + \frac{4}{3z^3} + \frac{2}{3z^4} + \frac{4}{15z^5} + ... \right].
$$

After term-by-term multiplication and collecting terms with $\frac{1}{z}$ I got the residue $\frac{109}{15}$. Therefore, the integral equals to $e^2 + \frac{109}{15}$.

But I am told it's wrong, and I can't find a problem for a long time. Can someone tell, is that solution wrong? Very grateful in advance for your help and time. Thanks!

答: There are two errors. A minor one is that you forgot to divide the residue by $2\pi i$. And a major one is that, near $0$, you have$$\frac{z^5}{z-1}=-z^5 - z^6 - z^7 -\cdots$$instead of$$\frac{z^5}{z-1}=z^4+z^3+z^2+\cdots$$


问:
Recommended textbook on function approximation on computer
There is a cephes math library on the Internet to provide accuracy computation of math function, e.g. sin,cos,tan,erf, gamma, lgamma, etc.

It implements the function in an intelligent way. For example, it uses different asymptotic series on the different domain when computing gamma function and switches to Stirling's approximation when the argument is too large.

Basically, the library uses all means of methods, e.g. Asymptotic series, Numerical Integration, Newton's method, Pade approximation, continued fraction.

If I want to fully understand what it is going on and study how to deduce such formula and implement such numerical algorithms, what are the recommended textbook?

For example, I want to compute an efficient and accurate specialized lgamma_exp(x) = lgamma(exp(x)) with argument in exp(x) rather than in x. There is no way except computing exp(x) first, which might overflow on the computer.

答:
In addition to the books by J.-M. Muller and co-authors already mentioned in other answers, I provided some additional pointers to relevant literature in an answer to a related question.

In the specific case of lgamma_exp, extensive literature research does not seem necessary. For the basic mathematical formulas we can rely on NIST's online resource Digital Library of Mathematical Functions (DLMF). Required knowledge about IEEE-754 floating-point arithmetic can be gleaned from the following seminal paper:

David Goldberg, "What Every Computer Scientist Should Know About Floating-Point Arithmetic", ACM Computing Surveys, Vol. 23, No. 1, March 1993, pp. 5-48 (online)

In addition, we need to know that polynomial minimax approximations are an adequate way to approximate most simple special functions, and that tools like Mathematics, Maple, and the free Sollya tool can generate these.

From DLMF §5.7 we learn of the series $\frac{1}{\Gamma(z)} = z + \gamma z^{2} + \cdots$, which tells us that near the origin, $\Gamma(x) \approx \frac{1}{x}$. Since $\exp(38) \approx 3.2\cdot10^{16}$, for IEEE double precision, $\frac{1}{x}$ is accurate to full precision for $x < -38$, therefore lgamma_exp(x) = -x for $x < -38$. From the Stirling approximation in DLMF §5.11.1 we likewise see that for sufficiently large $x$ we have $\Gamma(x) \approx (x-1)\exp(x)$, which is accurate to full double precision for $x >38$.

The only other interval requiring attention is the vicinity of the zeros of $\log\Gamma(x))$ in the positive half plane at $x=1$ and $x=2$. In finite precision floating-point precision, for all $x < \epsilon$, $\exp(\epsilon) = 1$, so we cannot use the naive computation lgamma (exp (x)) if we desire accurate results. Here we can use polynomial minimax approximations $\mathrm{P}_{1}(\exp(x)-1)$ and $\mathrm{P}_{2}(\exp(x)-2)$. Since I am lazy, I establish the bounds of suitable intervals for these approximations numerically, using an arbitrary-precision library as the reference to determine where the error of the naive computation exceeds three ulps. I determined as suitable intervals: $[-0.188, 0.405465)$ for $\mathrm{P}_{1}$ and  $[0.405465, 1.1]$ for $\mathrm{P}_{2}$.

The computations of $\exp(x)-1$ and $\exp(x)-2$ suffer from subtractive cancellation. For $\exp(x)-1$ this is easily addressed by use of the expm1 standard math function available in most computing environments. For the other interval, we can compute $\exp(x) - 2$ as 2 * expm1 (x - log(2)), where log(2) is represented to quadruple precision by a pair of double-precision constants, so the computation is accurate when $x \approx \log(2)$.

I generated the polynomial minimax approximations used in the ISO-C code below  specifically for this answer using a proprietary implementation of the Remez algorithm. The polynomials are evaluated using a second-order Horner scheme for higher instruction-level parallelism which improves performance on many platforms. In preliminary testing against an arbitrary-precision library (R.P. Brent's MP library: functional but outdated; not recommended for new development), the maximum error found so far is about 3.5 ulp when the code is compiled with the Intel C/C++ compiler on an x86_64 system with strict IEEE-754 conformance (/fp:strict).

```<code>double my_lgamma_exp (double x)
{
    const double log2_hi = 6.9314718055994529e-1;
    const double log2_lo = 2.3190468138462996e-17;
    double r, s, t;
    if (fabs (x) > 38) {
        if (x < 0) {
            r = 0 - x;
        } else {
            r = (x - 1) * exp (x);
        }
    } else if ((x >= -0.188) && (x <= 1.1)) {
        if (x < 0.405465) {
            /* minimax polynomial around exp(x)=1; 2nd-degree Horner scheme */
            x = expm1 (x); // exp(x) - 1
            s = x * x;
            r =       - 4.7091965873693646e-3;
            t =         1.9912834549094705e-2;
            r = r * s - 4.1005903342038519e-2;
            t = t * s + 5.7612615780147633e-2;
            r = r * s - 6.6582808312903363e-2;
            t = t * s + 7.1835727184955733e-2;
            r = r * s - 7.7041643077651434e-2;
            t = t * s + 8.3353766668420856e-2;
            r = r * s - 9.0949364918343004e-2;
            t = t * s + 1.0009890150433524e-1;
            r = r * s - 1.1133433459692958e-1;
            t = t * s + 1.2550968527263201e-1;
            r = r * s - 1.4404989656400469e-1;
            t = t * s + 1.6955717682057603e-1;
            r = r * s - 2.0738555102576703e-1;
            t = t * s + 2.7058080842860699e-1;
            r = r * x + t;
            r = r * x - 4.0068563438654731e-1;
            r = r * x + 8.2246703342411209e-1;
            r = r * x - 5.7721566490153287e-1;
            r = r * x;
        } else {
            /* minimax polynomial around exp(x)=2; 2nd-degree Horner scheme */
            x = 2 * expm1 ((x - log2_hi) - log2_lo); // exp(x) - 2
            s = x * x;
            r =         9.7009080515552103e-9;
            t =       - 6.6363290060543801e-8;
            r = r * s + 2.1575134587124466e-7;
            t = t * s - 4.8610795617393086e-7;
            r = r * s + 9.7870572069649216e-7;
            t = t * s - 2.0308648833487155e-6;
            r = r * s + 4.3609845401391367e-6;
            t = t * s - 9.4380453602222448e-6;
            r = r * s + 2.0510699022734775e-5;
            t = t * s - 4.4927200874090269e-5;
            r = r * s + 9.9457052285033636e-5;
            t = t * s - 2.2315458092948354e-4;
            r = r * s + 5.0966955797798817e-4;
            t = t * s - 1.1927539271577999e-3;
            r = r * s + 2.8905103294554332e-3;
            t = t * s - 7.3855510280415159e-3;
            r = r * s + 2.0580808427809637e-2;
            t = t * s - 6.7352301053207803e-2;
            r = r * x + t;
            r = r * x + 3.2246703342411304e-1;
            r = r * x + 4.2278433509846719e-1;
            r = r * x;
        }
    } else {
        r = lgamma (exp (x));
    }
    return r;
}
</code>```

Body:
In addition to the books by J.-M. Muller and co-authors already mentioned in other answers, I provided some additional pointers to relevant literature in an answer to a related question.

In the specific case of lgamma_exp, extensive literature research does not seem necessary. For the basic mathematical formulas we can rely on NIST's online resource Digital Library of Mathematical Functions (DLMF). Required knowledge about IEEE-754 floating-point arithmetic can be gleaned from the following seminal paper:

David Goldberg, "What Every Computer Scientist Should Know About Floating-Point Arithmetic", ACM Computing Surveys, Vol. 23, No. 1, March 1993, pp. 5-48 (online)

In addition, we need to know that polynomial minimax approximations are an adequate way to approximate most simple special functions, and that tools like Mathematics, Maple, and the free Sollya tool can generate these.

From DLMF §5.7 we learn of the series $\frac{1}{\Gamma(z)} = z + \gamma z^{2} + \cdots$, which tells us that near the origin, $\Gamma(x) \approx \frac{1}{x}$. Since $\exp(38) \approx 3.2\cdot10^{16}$, for IEEE double precision, $\frac{1}{x}$ is accurate to full precision for $x < -38$, therefore lgamma_exp(x) = -x for $x < -38$. From the Stirling approximation in DLMF §5.11.1 we likewise see that for sufficiently large $x$ we have $\Gamma(x) \approx (x-1)\exp(x)$, which is accurate to full double precision for $x >38$.

The only other interval requiring attention is the vicinity of the zeros of $\log\Gamma(x))$ in the positive half plane at $x=1$ and $x=2$. In finite precision floating-point precision, for all $x < \epsilon$, $\exp(\epsilon) = 1$, so we cannot use the naive computation lgamma (exp (x)) if we desire accurate results. Here we can use polynomial minimax approximations $\mathrm{P}_{1}(\exp(x)-1)$ and $\mathrm{P}_{2}(\exp(x)-2)$. Since I am lazy, I establish the bounds of suitable intervals for these approximations numerically, using an arbitrary-precision library as the reference to determine where the error of the naive computation exceeds three ulps. I determined as suitable intervals: $[-0.188, 0.405465)$ for $\mathrm{P}_{1}$ and  $[0.405465, 1.1]$ for $\mathrm{P}_{2}$.

The computations of $\exp(x)-1$ and $\exp(x)-2$ suffer from subtractive cancellation. For $\exp(x)-1$ this is easily addressed by use of the expm1 standard math function available in most computing environments. For the other interval, we can compute $\exp(x) - 2$ as 2 * expm1 (x - log(2)), where log(2) is represented to quadruple precision by a pair of double-precision constants, so the computation is accurate when $x \approx \log(2)$.

I generated the polynomial minimax approximations used in the ISO-C code below  specifically for this answer using a proprietary implementation of the Remez algorithm. The polynomials are evaluated using a second-order Horner scheme for higher instruction-level parallelism which improves performance on many platforms. In preliminary testing against an arbitrary-precision library (R.P. Brent's MP library: functional but outdated; not recommended for new development), the maximum error found so far is about 3.5 ulp when the code is compiled with the Intel C/C++ compiler on an x86_64 system with strict IEEE-754 conformance (/fp:strict).

double my_lgamma_exp (double x)
{
    const double log2_hi = 6.9314718055994529e-1;
    const double log2_lo = 2.3190468138462996e-17;
    double r, s, t;
    if (fabs (x) > 38) {
        if (x < 0) {
            r = 0 - x;
        } else {
            r = (x - 1) * exp (x);
        }
    } else if ((x >= -0.188) && (x <= 1.1)) {
        if (x < 0.405465) {
            /* minimax polynomial around exp(x)=1; 2nd-degree Horner scheme */
            x = expm1 (x); // exp(x) - 1
            s = x * x;
            r =       - 4.7091965873693646e-3;
            t =         1.9912834549094705e-2;
            r = r * s - 4.1005903342038519e-2;
            t = t * s + 5.7612615780147633e-2;
            r = r * s - 6.6582808312903363e-2;
            t = t * s + 7.1835727184955733e-2;
            r = r * s - 7.7041643077651434e-2;
            t = t * s + 8.3353766668420856e-2;
            r = r * s - 9.0949364918343004e-2;
            t = t * s + 1.0009890150433524e-1;
            r = r * s - 1.1133433459692958e-1;
            t = t * s + 1.2550968527263201e-1;
            r = r * s - 1.4404989656400469e-1;
            t = t * s + 1.6955717682057603e-1;
            r = r * s - 2.0738555102576703e-1;
            t = t * s + 2.7058080842860699e-1;
            r = r * x + t;
            r = r * x - 4.0068563438654731e-1;
            r = r * x + 8.2246703342411209e-1;
            r = r * x - 5.7721566490153287e-1;
            r = r * x;
        } else {
            /* minimax polynomial around exp(x)=2; 2nd-degree Horner scheme */
            x = 2 * expm1 ((x - log2_hi) - log2_lo); // exp(x) - 2
            s = x * x;
            r =         9.7009080515552103e-9;
            t =       - 6.6363290060543801e-8;
            r = r * s + 2.1575134587124466e-7;
            t = t * s - 4.8610795617393086e-7;
            r = r * s + 9.7870572069649216e-7;
            t = t * s - 2.0308648833487155e-6;
            r = r * s + 4.3609845401391367e-6;
            t = t * s - 9.4380453602222448e-6;
            r = r * s + 2.0510699022734775e-5;
            t = t * s - 4.4927200874090269e-5;
            r = r * s + 9.9457052285033636e-5;
            t = t * s - 2.2315458092948354e-4;
            r = r * s + 5.0966955797798817e-4;
            t = t * s - 1.1927539271577999e-3;
            r = r * s + 2.8905103294554332e-3;
            t = t * s - 7.3855510280415159e-3;
            r = r * s + 2.0580808427809637e-2;
            t = t * s - 6.7352301053207803e-2;
            r = r * x + t;
            r = r * x + 3.2246703342411304e-1;
            r = r * x + 4.2278433509846719e-1;
            r = r * x;
        }
    } else {
        r = lgamma (exp (x));
    }
    return r;
}


问:
Do lay persons really believe that all of mathematics is known?
I once found a professor of medicine to be under the impression that everything in mathematics is already known.

It's probably commonplace to hear that the masses labor under that misconception.

In some contexts, one is expected to cite authoritative sources to support factual statements.  If one asserts that the belief that everything in mathematics is already known is widespread among the general population, are there authoritative sources one could cite to support that?

It seems to me that if one were taking a poll, it might be difficult to phrase a suitable question in a way that doesn't give away hints.

答:
It is a common misconception in regard to math, but this kind of misconception is common everywhere.

It's caused by the recursive properties of knowledge. The more you learn, the more you realize how little you know.; i.e:

Since it is known that less informed individuals see fields as more finite than informed individuals; While laymen may relate to the field of Carpentry as "If someone is skilled in carpentry; Then that someone builds houses," while a carpenter may relate to the field as such: "Carpentry is a complex occupational field which includes, but is not limited to:  Formwork carpentry, framing, coopering, and cabinetmakers."

I think the lack of authoritative sources on the misconception that everything in mathematics is already known, simply is a result of the fact that this is true for any topic, profession, occupation, ect.

I haven't come across any sources which provides specifically what you've requested, but there are studies supporting what I am describing. (I'll append a link to a paper if I can find it again.)


问:
Can we still prove $\mathbb{E}[Z] = \mathbb{E}[N]\mathbb{E}[X_1]$ for a sequence of iid variables $X1,X2,\ldots$ when N is not independent
Let $X_1, X_2,\ldots$ be a sequence of independent and identically distributed non-negative integer valued random variables, and let $N$ be a non-negative integer valued random variable which is independent of the sequence $X_1,X_2,\ldots$. Let $Z = X_1+X_2+\ldots+X_N$ (where we take $Z=0$ if $N=0$).

(a) Show that $\mathbb{E}[Z] = \mathbb{E}[N] \mathbb{E}[X_1]$.

(b) Suppose we remove the condition that $N$ is independent of the sequence ($X_i$). Is it still necessarily the case that $\mathbb{E}[Z] = \mathbb{E}[N] \mathbb{E}[X_1]$? Find a proof or a counterexample.

I had no problem proving part (a), but am struggling with part (b). Any advice on how to approach part (b) would be greatly appreciated.

答:
Here is a counterexample:

Suppose $X_1$ is a symmetric Bernoulli random variable and $N = 1 - X_1$.

Then $E[N] = E[X_1] = \frac{1}{2}$, but $Z$ is always $0$.


问:
Is $0 \neq 1$ a necessary ring axiom?
In the listing of the ring axioms, is $0 \neq 1$ a necessary axiom? I see it in some versions, but not others. Also, is the existence of a multiplicative identity necessary in listing the axioms? Thanks.

答:
As far as I am concerned, the only ring in which $0=1$ is the null ring $A={0}$.
As this ring is not of interest, many books include $0 \neq 1$ as an axiom.
For your second question, the existence of a multiplicative identity is a characteristic of unitary rings. Not all the rings have an identity element.

Title: Is $0 \neq 1$ a necessary ring axiom?


答:
This is a bad axiom and you should not have it. The zero ring is a perfectly fine ring, and without it the category of rings fails to have some very natural constructions, such as tensor products. In algebraic geometry this would mean that affine schemes would fail to be closed under fiber products which would be very silly. You want to be able to talk about empty intersections.

I am also firmly of the opinion that rings should have multiplicative identities by default. This convention is, as far as I know, universal in fields such as algebraic geometry and algebraic topology. It is less universal in functional analysis / operator algebras.

Title: Is $0 \neq 1$ a necessary ring axiom?

Body: In the listing of the ring axioms, is $0 \neq 1$ a necessary axiom? I see it in some versions, but not others. Also, is the existence of a multiplicative identity necessary in listing the axioms? Thanks.


问:
Generalisations of the Gronwall's lemma
Suppose we have the following differential inequality

$F''(w)\le \frac{p-1}{p}\frac{(F'(w))^2}{F(w)}$ on $w\in(w_0,w_0+\varepsilon)$, $w_0>0$, $\varepsilon>0$, $p>1$. In addition, $F(w)>0$, $F'(w)>0$ for $w\in[w_0,w_0+\varepsilon)$.

Let's consider the solution of the Cauchy problem

$\begin{cases}G''(w)= \frac{p-1}{p}\frac{(G'(w))^2}{G(w)},& w\in(w_0,w_0+\varepsilon),\\G'(w_0)=F'(w_0),&\\G(w_0)=F(w_0),&\end{cases},$ which can be written explicitly:

$G(w) = F(w_0)\left(1+\frac{F'(w_0)}{F(w_0)}\frac{w-w_0}{p}\right)^p$.

Is it possible to compare the functions $F$ and $G$ in any meaningful way? This case fits neither the canonical version of the Gronwall's lemma nor its generalisation by Bihari, because it involves second derivatives.

I'd be glad to hear all suggestions!

答:
For the sake of having an answer, I repost here a solution based on the idea proposed by Hicham at mathoverflow.

In fact, the first differential inequality leads to $\frac{d^2}{dw^2}((F(w))^{1/p})\le 0$, or $(F(w))^{1/p}$ being concave. In the same time, $(G(w))^{1/p}$ is a tangent line to $(F(w))^{1/p}$ in $w_0$, therefore $(F(w))^{1/p}\le (G(w))^{1/p}$ and $ F(w) \le  G(w) $.

Title: Generalisations of the Gronwall's lemma


问:
How to solve an integrand that diverge?
In the following Eq. 1, the integral diverges. However, it is part of my equation for the Expectation of rate and it should not diverge. I will really appreciate any help as I am stuck on this for over 4 days now.

$$\begin{align}
E = \int_r^\infty\dfrac{\mathrm{e}^{-x}}{1+ \mathrm{e}^{b-cx}}dx \tag{1} \label{1}
\end{align} $$

In the following, I found I can solve it using the My Take below, but even that diverges.

My Take:

--> following the properly $(1+z)^{-1} =  \sum_{j=1}^{\infty} (-1)^{j-1}(z)^{j-1}$

$E = \int_r^\infty\mathrm{e}^{-x}\sum_{j=1}^{\infty} (-1)^{j-1}(e^{b-cx})^{j-1}$

$E$ diverge even now.

How can I solve this?

答:
Following your "My Take" and continue, you will get

$E = \int_r^\infty e^{-x}[\sum_{k=0}^{\infty} (-1)^{k} e^{(b-cx)k}]\space dx$

$=\sum_{k=0}^\infty(-1)^k e^{bk}\int_r^\infty e^{-(kc+1)x} dx$

$=\sum_{k=0}^\infty \frac{(-1)^k e^{bk}}{kc+1} e^{-(kc+1)r}$.

It converges.

OwnerDisplayName: Kashan

Title: How to solve an integrand that diverge?


答:
It is not a necessary axiom, but, as far as I know, there is no benefit in allowing such a thing. The additional structure that you would get is quite boring. Assuming that $0=1$ in $R$, we get that, for every element $a \in R$,
$$a=1 \cdot a = 0 \cdot a=0,$$
so $R$ is a ring of only one element.

Title: Is $0 \neq 1$ a necessary ring axiom?


答:
In a ring with $1$ the assumption $1\ne0$ is not necessary and, in fact, the standard modern definition allows the zero ring to exist. The way I personally make sense of this is that the zero ring is the terminal object of the category of rings with $1$, and the reason rings with $1$ should have a terminal object is that they naturally have products of every non-empty family. Therefore it's just natural that the product of the empty family should be there as well.

For the record, allowing the zero ring to exist makes the hypothesis $f(1)=1$ for homomorphisms come to its full fruition: in fact, there is no homomorphism $\{0\}\to R$ for any $R\ne \{0\}$, because it would send $1_0=0$ to $0_R\ne 1_R$.

To my knowledge, the current standard definition is that $1\ne 0$ in fields and, commonly, in integral domains.

For the second question, the nomenclature for a "ring without $1$" varies; some names in use are: ring, non-unital ring, pseudo-ring, rng. Likewise, possible terms for a ring with $1$ are: ring with 1, unital ring, ring.

Title: Is $0 \neq 1$ a necessary ring axiom?


问:
Definition of the flow of an ODE and its inverse
At lesson, the teacher considers a flow $\Phi$ given by the solutions of the ode system for $t\in[0, T]$ and $x\in\mathbb R^d$,
$$
\begin{cases}
y'(s)=b(y(s), s),&s\leq T\\
y(t)=x
\end{cases},\qquad(\star)
$$
that is $\Phi(x, t, s)=y(s)$ solving $(\star)$. He said that we will be mostly concerned with $\Phi(\cdot, 0, \cdot)$. The field $b$ is assumed to be Lipschitz continuous in both variables and bounded.

Then, he intoduces the inverse $\Psi$ of the above flow as follows: $\Psi(x, 0, s)=y(s)$ satisfying
$$
\begin{cases}
y'(s)=-b(y(s), t-s),&s<t\leq T\\
y(0)=x
\end{cases},\qquad(\triangle)
$$
and he said that $\Psi$ is such that
$$
\Phi(\Psi(x, 0, s), 0, s)=x,\quad \Psi(\Phi(x, 0, s), 0, s)=x.\qquad (\star\star)
$$
I do not understand $(\star\star)$. Can someone help me? Maybe is the definition of the inverse wrong?

Thank you

答:
After @lutz-lehmann pointed out that my arguments would apply only to an autonomous system, I will try to make an analogous argument for the related autonomous system.

I will stick to the $t = 0$ case.

Let $s\mapsto\mathbf{u}(s) = [y(s),~s]^{\mathsf{T}}$ be the solution of
\begin{eqnarray}
\frac{d}{ds}\left[\begin{array}{c}y(s)\\s\end{array}\right] &=& \left[\begin{array}{c}b(y(s),s)\\1\end{array}\right] ~=~ \left[\begin{array}{c}b(\mathbf{u}(s))\\1\end{array}\right],\\
\mathbf{u}(0) &=& \left[\begin{array}{c}x\\0\end{array}\right].
\end{eqnarray}

Since $b$'s argument is $\mathbf{u}(s)$, this system is autonomous.

Let $U$ be the flow for this autonomous system. Then $U(\cdot,\cdot,s)$ will map this example's initial conditions as follows:
\begin{equation}
U(\cdot,\cdot,s):\left[\begin{array}{c}x\\0\end{array}\right]\mapsto\left[\begin{array}{c}y(s)\\s\end{array}\right].
\end{equation}
In other words,
\begin{equation}
U(x,0,s) = \left[\begin{array}{c}\Phi(x,0,s)\\s\end{array}\right].
\end{equation}

Let $s\mapsto\mathbf{v}(s) = [w(s),-s]^{\mathsf{T}}$ be the solution of
\begin{eqnarray}
\frac{d}{ds}\left[\begin{array}{c}w(s)\\-s\end{array}\right] &=& \left[\begin{array}{c}-b(w(s),-s)\\-1\end{array}\right] ~=~ \left[\begin{array}{c}-b(\mathbf{v}(s))\\-1\end{array}\right],\\
\mathbf{v}(0) &=& \left[\begin{array}{c}x\\0\end{array}\right].
\end{eqnarray}
This system is also autonomous.

Let $V$ be the flow of this system. Then $V(\cdot,\cdot,s)$ maps this example's initial conditions as follows:
\begin{equation}
V(\cdot,\cdot,s):\left[\begin{array}{c}x\\0\end{array}\right]\mapsto\left[\begin{array}{c}w(s)\\-s\end{array}\right].
\end{equation}
In other words,
\begin{equation}
V(x,0,s) = \left[\begin{array}{c}\Psi(x,0,s)\\-s\end{array}\right].
\end{equation}

Since $U$ is the flow of an autonomous system, $U(\cdot,\cdot,s)$ pushes a solution $\mathbf{u}$ from its value at any time (due to the unchaning vector field) to its value $s$ time units later. The complication here is that $w(s)$ is not a fixed point.

Note, however, that at each point $(x,t)$, the vector-valued right-hand sides of the two autonomous systems differ only in sign: one is $[b(x,t),~1]^{\mathsf{T}}$ and the other is $[-b(x,t),-1]^{\mathsf{T}}$. That means a solution of one is a solution of the other run backward in time.

This implies that $V(\cdot,\cdot,s) = U(\cdot,\cdot,-s)$. This, in turn, implies that, since $[y(s),~s]^{\mathsf{T}}$ and $[w(s),-s]^{\mathsf{T}}$ started at the same point ($[x,~0]^{\mathsf{T}}$) at time $0$ and evolved according to opposite-pointing vector fields, $w(s) = y(-s)$.

\begin{eqnarray}
\Phi(\Psi(x,0,s),0,s) &=& \Phi(w(s),0,s)\\
&=& \Phi(y(-s),0,s)\\
&=& \Phi(\Phi(x,0,-s),0,s)\\
&=& x
\end{eqnarray}

I'm sure this can be argued with much more sophistication, but I think the rough shape of the argument is clear. I apologize for the completely incorrect previous answer.


问:
show that $\max(d(x,y),d(y,x))$ verifiy triangular inequality
Let $(E,\,d')$  be a metric space such that $d'( x,y) =\frac{d( x,y) +d( y,x)}{2}$ with $d$ a function defined in $\displaystyle \times E$ to $\mathbb{R}^{+}$ verifying the following :

$\begin{array}{{l}}
\forall x\in E,\ d( x,x) =0\\
\forall x,y\in E,\ d( x,y) =0\Leftrightarrow x=y\\
\forall x,y,z\in E;\ d( x,y) \leqslant d( x,z) +d( z,y)
\end{array}$

show that $\begin{array}[t]{l}
d'':\ E\times E\rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \mathbb{R}\\
\ \ \ \ \ \ \ ( x,y)\rightarrow \ \max( d( x,y) ,d( y,x))
\end{array}$ is a distance in $\displaystyle E$.

To prove the triangle inequality axiom for $d'$ i thought since we have $d'$ I can use the definition of $max(a,b)$ that is
$max(a,b) =\frac{a+b-|b-a|}{2}$ since i can write $max( d( x,y) ,d( y,x)$ in function of $d'( x,y)$.

答:
Start from triangle inequality verified by $d$: $\begin{cases}d(x,y)\le d(x,z)+d(z,y)\\d(y,x)\le d(y,z)+d(z,x)\end{cases}$

$d'$ is a linear combination, so it is straightforward:

$\begin{align}2\,d'(x,y)
&=d(x,y)+d(y,x)\\
&\le\big(d(x,z)+d(z,y)\big)+\big(d(y,z)+d(z,x)\big)\\
&=\big(d(x,z)+d(z,x)\big)+\big(d(y,z)+d(z,y)\big)\\
&=2d'(x,z)+2d'(y,z)
\end{align}$

For $d''$ this is not much complicated:

Simply use $a\le\max(a,b)$ for any $b$

$\begin{cases}
d(x,y)\le d(x,z)+d(z,y)\le\max(d(x,z),d(z,x))+\max(d(z,y),d(y,z))\\
d(y,x)\le d(y,z)+d(z,x)\le\max(d(y,z),d(z,y))+\max(d(z,x),d(x,z))\end{cases}$

And notice both RHS are in fact the same quantity $d''(x,z)+d''(z,y)$.

Therefore $d''(x,y)=\max\big(d(x,y),d(y,x)\big)\le d''(x,z)+d''(z,y)$


问:
Voronoi vertex of points along a parabola at it's focus
So I am trying to learn and understand voronoi diagrams. And one thing I was messing around with was computing the voronoi diagram of points that lie along a parabola. And it made me wonder does the voronoi vertex of the points lie at the parabola's focus? I don't know how to even begin proving or disproving this. Any help would be much appreciated!</p>

答:
If you place sites symmetrically on a parabola and form the Voronoi diagram, the vertices of the Voronoi diagram lie on the axis of symmetry of the parabola above the focus. The focus closer to the parabola than any Voronoi vertex will be.

Perhaps the most useful edge case of the Voronoi diagram involving the parabola involves placing one site at the focus of the parabola and any number of sites along the directrix of the parabola. Then all the Voronoi vertices will lie on the parabola and the parabola focus will be adjacent to every point on the directrix.

In both images above, the blue points are the input sites to the Voronoi diagram and the yellow points are the vertices of the Voronoi diagram shown with solid finite lines and dotted infinite lines.

Here is an image of the Voronoi diagram for points only on the positive branch of a parabola. An explicit expression for the locations for the vertices of this Voronoi diagram appears to be pretty complex. But as the input points on the parabola grow arbitrarily close together, the Voronoi vertices approach the curve,
$$
y = \frac{3}{2^{4/3}}x^{2/3} + \frac{1}{2}.
$$