答:
By substitution $u=\tan(\frac{x}{2})$ then $du=\frac{1}{2}\sec^2(\frac{x}{2})$.
Using the trigonometric formula, $\sin (x)=\frac{2u}{1+u^2}, \quad \cos(x)=\frac{1-u^2}{1+u^2}$ and $dx=\frac{2 du}{1+u^2}$.

When  you  do the required substitutions, you find a nice integral with rational expression, which can be solved by decomposition into simple elements.

Title: Evaluate $\int_0^{\pi/2} \frac{1+2\cos x}{(2+\cos x)^2}$

Body:
I don’t know how to begin solving. Can I get a hint?

My failed attempt

Let $t =\tan (x/2)$

Then $$I =2 \int_0^1 \frac{(3-t^2)}{(3+t^2)^2 }$$
which I am not able to solve


问:
Ex. 1.2.56 - Topology of metric spaces by Kumaresan
Let be $X := \mathcal{C}[0,1]$ withe the $\sup$ norm metric $|| \ \ ||_{\infty}$. Let $E$ be the set of all functions in $X$ that do not vanish (that is, they do not take the value $0$) at $t = 0$. Is $E$ open in $X$?

I think that $E$ is open in $X$. Next, my attempt:

Define $A_f := \{ |f(t)| \in \mathbb{R} \ ; \ f(t) \neq 0 \}$ and $R_f := \inf A_f$ (clearly, $R_f$ > 0). If $f \in E$, then exists $R_f > 0$ such that $B(f,R_f) \subset E$. Indeed, given $g \in B(f,R_f)$, i.e., $\sup_{t \in [0,1]} | g(t) - f(t) | < R_f$, then $|g(t) - f(t)| < \inf A_f$ for all $t \in [0,1]$, i.e., $f(t) - \inf A_f < g(t) < f(t) + \inf A_f$ for all $t \in [0,1]$, in particular, $f(0) - \inf A_f < g(0) < f(0) + \inf A_f$. Now, we analyze two cases:

Case 1 - f(0) > 0:

In this case, we have $f(0) + \inf A_f > 0$ and $f(0) - \inf A_f \geq 0$, then $0 \leq f(0) - \inf A_f < g(0) < f(0) + \inf A_f$ and we have $g \in E$ because $g(0) > 0$

Case 2 - f (0) < 0:

In this case, we have $f(0) - \inf A_f < 0$ and $f(0) + \inf A_f$ $\textbf{???}$

but now I getting stuck from $\textbf{???}$. Is it true that $E$ is open in $X$? I would like to receive a hint how to prove that $E$ is open in case 2 or a counter example that show that $E$ is not open in $X$.

Thanks in advance!

EDIT:

I think it's not true that $E$ is open because $f(t) - \inf A_f < g(t) < f(t) + \inf A_f$ for all $t \in [0,1]$ is an particular case of this question:

Question

then is not true that $E$ is open in $X$ in general.

答:
Let $\phi : X \to \mathbb{R}$ defined by $\phi(f) = f(0)$, clearly the mapping is well defined.

Claim: $f_n \to f$ in $C[0,1] \implies f_n \to f$ uniformly on $[0,1]$

Now for any $f_n \to f $ in $C[0,1]$ we have for any $\epsilon>0$ there exists $n_0 \in \mathbb{N}$ such that $\|f_n - f\|_{\infty}<\epsilon$ for $n \geq n_0$

i.e $\sup_{x \in [0,1]}|f_n(x) - f(x)|< \epsilon$ for $n\geq n_0$ and since $|f_n(x) - f(x)| \leq \sup_{x \in [0,1]}|f_n(x) - f(x)|$ for all $x\in [0,1]$. We see that $|f_n(x) - f(x)| < \epsilon$ for all $n\geq n_0$ and for all $x\in [0,1]$.

So $f_n  \to f$ converges uniformly on $[0,1]$, this implies $f_n \to f$ converges pointwise on $[0,1]$, in particular $f_n(0) \to f(0)$.

Now for any $f_n \to f$ in $C[0,1] \implies f_n(0) \to f(0)$ so $\phi(f_n) = f_n(0) \to f(0) = \phi(f)$ i.e $\phi$ is continuous on $C[0,1]$.So its inverse image of an open set is open. Therefore $E = \phi^{-1}\{(-\infty,0) \cup (0,\infty)\}$ is open.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Existence of Differential Equations
We are given two functions $x^3$
and $x^2|x|$
Clearly they are LI though they have Wronskian zero.

So Does there exist continuous functions p(x) and q(x)
Such that the $x^3$ and $x^2|x|$
Satisfy the ODE $y" + py'+qy =0$

If this Question is already answered please share the link I will delete this
And please don't vote down

答: No, neither function is solution of a second order DE as at $x=0$ both have initial conditions zero, so at the same time have to be identical to the zero function.

Title: Existence of Differential Equations


问:
How to prove $\operatorname{Pic}(\mathbb{P}_X^n)\cong\operatorname{Pic}(X)\times\mathbb{Z}$?
Let $X$ be a Noetherian regular scheme. Then how can one prove $\operatorname{Pic}(\mathbb{P}_X^n)\cong\operatorname{Pic}(X)\times\mathbb{Z}$? I want to use this specific case for the more general case where one replaces $\mathbb{P}_X^n$ with $\mathbb{P}(\mathcal{E})$ and $\mathcal{E}$ a locally free sheaf of rank $n+1$, where the result is still true. So the proof shouldn't use this more general result. Is there a simpler proof for $\mathbb{P}_X^n$ than for $\mathbb{P}(\mathcal{E})$?

答:
Let me first note that you should require $X\newcommand{\Pic}{\operatorname{Pic}} \newcommand{\Spec}{\operatorname{Spec}} \newcommand{\PP}{\mathbb{P}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\Cl}{\operatorname{Cl}}$ connected - if not, you can get something silly like $X=\Spec k \sqcup \Spec k$ with trivial Picard group and $\PP^1_X=\PP^1_k \sqcup \PP^1_k$, which has Picard group $\ZZ^2$.

There's an obvious choice for the map $\Pic(X)\times\ZZ\to \Pic(\PP^n_X)$: send $$(\mathcal{L},d)\mapsto (p^*\mathcal{L})\otimes(q^*\mathcal{O}(d))$$ where $p:\PP^n_X\to X$ and $q:\PP^n_X\to \PP^n$ are the projections, so we show this map is injective and surjective.

Injectivity: Suppose $p^*\mathcal{L}\otimes q^*\mathcal{O}(d)\cong \mathcal{O}_{\PP^n_X}$. By a relative version of the fact that $\mathcal{O}_{\Bbb P^n_A}(d)$ is the $d$-graded piece of $A[x_0,\cdots,x_n]$, we see that $p_*(p^*\mathcal{L}\otimes q^*\mathcal{O}(d))\cong \mathcal{O}_X$, so by the projection formula we have that $\mathcal{L}\otimes p_*q^*\mathcal{O}(d)\cong\mathcal{O}_X$, so $d=0$ and $\mathcal{L}\cong\mathcal{O}_X$.

Surjectivity: Pick an affine open cover $\{U_i=\Spec A_i\}$ of $X$ with each $A_i$ an integral domain. Then each $U_i$ and hence each $U_i\times\PP^n$ is regular, noetherian, integral, and separated, so $$\Pic (U_i\times\PP^n) \cong \Cl (U_i\times\PP^n) \cong (\Cl U_i) \times\ZZ \cong (\Pic U_i)\times\ZZ$$ by the results of Hartshorne chapter II section 6, where the map is the same as we've picked above. It remains to check that these isomorphisms patch together correctly.

So let $U_i,U_j$ be two of the opens from above. From the previous isomorphism of Picard groups, we see that the restriction of our line bundle $\mathcal{L}$ on $X\times\PP^n$ is of the form $p_i^*\mathcal{L}_i\otimes q_j^*\mathcal{O}(n_i)$ on $U_i\times\PP^n$ and $p_j^*\mathcal{L}_j\otimes q_j^*\mathcal{O}(n_j)$ on $U_j\times\PP^n$. Since the restriction of both of these to the intersection $U_i\times\PP^n\cap U_j\times\PP^n$ is the same, we see again from the first result in the proof of injectivity that $n_i=n_j$. Untwisting by $n_i=n_j$ along $U_i\times\PP^n$ and $U_j\times\PP^n$, we see that we get a line bundle trivial in the $\PP^n$ direction on $U_i\times\PP^n$ and $U_j\times\PP^n$. The claim is that this comes from a line bundle on $X$: after taking the gluing data for $\mathcal{L}$, applying the above untwisting and the projection formula gives us that it is exactly the gluing data for a line bundle on $X$ which pulls back correctly. So we've shown surjectivity.

Note that this same proof works for $\PP(\mathcal{E})$ by requiring $\mathcal{E}$ to be trivial along the $U_i$, which is always possible.


问:
Choosing point uniformly on surface of sphere
I saw the following statement posted by Fermat's Library on Twitter:

The way to correctly generate a random point on the surface of a unit sphere is not to pick uniform distributions $\theta$ in $[0,2\pi)$ and $\phi$ in $[0,\pi)$.

Instead, choose $u$ and $v$ from uniform distributions on $[0,1)$. Then,

$\phi = \cos^{-1}(2v-1)$

$\theta = 2\pi u$

I can see why uniformly choosing $\theta$ in $[0,2\pi)$ and $\phi$ in $[0,\pi)$ would lead to points near the poles being more likely to be chosen than points near the equator.

But I can't think of an intuitive explanation for why the given solution correctly gives a uniform distribution. Is there a good intuitive explanation for this?

答:
The length of the ring at latitude $\phi$ is proportional to $\cos\phi$; this is also proportional to the pdf of the random variable that is the latitude of a point uniformly distributed on a sphere.

The integral of the pdf is the cdf; since the pdf expression contains $\cos\phi$ its integral should also be a sine wave. If we use the angular difference from a pole as variable, this wave should go from $0$ at $0$ to $1$ at $\pi$.

The inverse of the cdf is the quantile function, which is what we really want. Since the cdf was a sine wave, the quantile function should be the $\cos^{-1}$ or $\sin^{-1}$ of something; looking at the required values ($0$ at $0$, $\pi$ at $1$) we infer that the quantile function is $\cos^{-1}(1-2x)$. To generate the latitude for a uniform random point on the sphere, $x$ should be a uniform $U(0,1)$ variate; it is easy to see that both $1-2x$ and $2x-1$ are uniformly distributed over $[-1,1]$, so we have $\phi=\cos^{-1}(2x-1)$.


问:
Evaluate $\int_0^{\pi/2} \frac{1+2\cos x}{(2+\cos x)^2}$
I don’t know how to begin solving. Can I get a hint?

My failed attempt

Let $t =\tan (x/2)$

Then $$I =2 \int_0^1 \frac{(3-t^2)}{(3+t^2)^2 }$$
which I am not able to solve

Body:
$$\int_0^\tfrac{\pi}{2} \frac{1 + 2\cos x}{(2 + \cos x)^2}{\rm d}x \\
= \int_0^\tfrac{\pi}{2} \frac{1 + 2\cos x}{(2 + \cos x)^2}\frac{/\sin^2 x}{/\sin^2 x} {\rm d}x \\
= \int_0^\tfrac{\pi}{2} \frac{\csc^2 x + 2\cot x \csc x}{(2\csc x + \cot x)^2}{\rm d}x \\
= \int_0^\tfrac{\pi}{2} \frac{-1}{(2\csc x + \cot x)^2}{\rm d}(2\csc x + \cot x) \\
= \left[\frac{1}{(2\csc x + \cot x)}\right]_0^\tfrac{\pi}{2}\\
= \left[\frac{\sin x}{(2 + \cos x)}\right]_0^\tfrac{\pi}{2} \\
= \frac{1}{2}$$
 Hope it helps.


问:
How to denote all possible sequences?
Take a sequence A. Imagine a sequence C=(An)n∈J, where J ⊆ U. U is an index set. J ⊆ U means J is a proper subsequence of U. Therefore, various possible sequences C are denoted by restricting the original sequence A to a smaller index set.

When C is defined as above, will simply showing the variable C inherently signify all possible sequences or just one possible sequence? How can I make sure to formally denote all possible sequences C can be? Is there a symbol commonly used for this?

答:
I'm assuming you have defined $\mathbb A = (A_n)_{n\in \mathbf U}$. Then any subsequence of $\mathbb A$ can be defined as
$$\mathbb C=(A_n)_{n\in \mathbf J},\mathbf J\subseteq \mathbf U$$
The collection of all possible subsequences of $\mathbb A$ is then given as $$\mathcal C=\{(A_n)_{n\in \mathbf J}: \mathbf J\subseteq \mathbf U \}$$

Title: How to denote all possible sequences?


问:
How to write in $2^x=5$ in logarithmic form?
How do I write:
$$2^x = 5$$

In a logarithmic form?

I've looked for a solution for some time now, so I decided to try here.

答:
Any equation of the form

$$
\text{base}~^{\textbf{power}} = \text{answer},
$$

can be rewritten as a logarithm as follows:

$$
\log_{~\text{base}}\text{answer} = \textbf{power}.
$$

So, $2^x = 5$ quickly becomes $\log_2 5 = x$

Who wants to do all that messing around with logs of both sides and change of base rules?


问:
Solve this recurrence relation $a_n=-a_{n-2}$
So far I have this relation of recurrence:
$$a_n=-a_{n-2}$$
With $a_1=1$ and $a_2=2$.

I know I shoud replace $C \times r^n $ for $a_n$ and $C \times r^{n-2}$ for $a_{n-2}$

Then the relation gets here:
$$Cr^{n}+Cr^{n-2}=0$$

If I divide the left side by $Cr^{n-2}$
The equation will be like this $r^2+1=0$, thus $r=-i$ or $r=i$

After this a can't solve it because I forgot complex numbers
Can you guys please help me?

答:
Your sequence is $$1,2,-1,-2,1,2,-1,-2,....$$

So $$a_n =\cases{(-1)^{n-1\over 2}& if n  odd \\ -2(-1)^{n\over 2}&if n even }$$

Or, as you put, $$a_n = ai^n +b(-i)^n$$ for some complex $a$ and $b$. From boundary values we have \begin{align}ai-bi &= 1\\ -a-b&=2\end{align}

So $a= {1-2i\over 2}$ and $b={1+2i\over 2}$.

So $$\boxed{a_n = {1\over 2}\Big(i^n+(-i)^n\Big) -i\Big(i^n-(-i)^n\Big)}$$


问:
I've a small problem with this matrix (to find the determinant)

Find the following determinant. $$\det\begin{bmatrix} 1 & 0 & 2 & -1\\ 0 & 1 & 4 & -2 \\ 2 & -1 & 3 & 1 \\ 2 & -1 & -1 & 2\end{bmatrix}$$

Theoretically I know how to find the determinant of a $m \times n$ matrix, as follow:

Gaussian elimination,

Laplace method,

Gaussian elimination + Laplace method,

Sarrus's rule (this method is valid only for $3 \times 3$ matrix),

$2 \times 2$ matrix (easiest way).

I used the 3rd method (because if I had used only the second one, it would have  be a long and tedious process).

here's my attempt to solve it:

I reduced the matrix using Gaussian elimination and the result is as follows:

$$\begin{bmatrix} 1 & 0 & 2 & -1\\ 0 & 1 & 4 & -2 \\ 0 & -1 & -1 & 2 \\ 0 & -1 & -5 & 4\end{bmatrix}$$
This is enough in order to compute the determinant of this matrix, using the first column.

I used Laplace method and here's what I got:
$$(-1)^2 \det\begin{bmatrix} 1 & 4 & -2\\ -1 & -1 & 2\\ -1 & -5 & 4\end{bmatrix}$$

My result (using Sarrus's rule to compute the determinant of the minor) is 6, but the exercise has these possible solutions:

(1) -7

(2) 0

(3) 7

(4) 5

答: The $(3,4)$ entry of your matrix after step $1$ should be a $3$.


问:
Incandescent light bulb lifetime - exponential distribution?
It's been said that incandescent light bulbs lifetime has exponential distribution. As I understand, this means a 10,000 hours time-to-failure has the same probability no matter how long the light bulb was used until now (i.e. the "memorylessness" property).

On the other hand, Wikipedia says that "(a light bulb's) Lifetime is approximately proportional to $V^{−16}$". Doesn't this mean that if the lifetime is 10,000 hours and 5,000 hours have passed, failure in 10,000 hours is less likely than in 5,000 hours?

答: I would not expect light bulbs to have an exponential distribution, as there is a wear out mechanism involved-evaporation of the filament material.  The Wikipedia statement that the lifetime is proportional to $V^{-16}$ is for steady state conditions.  Higher voltage leads to higher operating temperature, which leads to increased evaporation.  "Long-life" bulbs have cooler filaments, as you can see because they generate less (and redder) light for the same power consumption.  All this ignores the stresses generated in on/off cycles, and the fact that the power when the bulb is turned on is much higher than steady state due to the lower resistance at lower temperature.</p>


问:
Limit of a recursive sequence with $u_n$
It is given that $u_{n+1} =1+\dfrac{1}{u_n}$ and $u_1 =1$. Find the limit of $u_n$ as $n\to\infty$.

The limit is $\frac{\sqrt{5}+1}{2}$ from a calculator. Is there an algebraic way to determine this? You can also determine that sequence is bounded but not monotonic.

答:
If you want to be rigorous you should proove that the limit exists.

You can do that showing that the odd terms form a decreasing sequence  while the even terms form an increasing sequence (induction). Then show that those sequences are bounded below/above.


问:
Asymptotic formula/closed form for $\sum_{n=1}^{x}\frac{1}{\log n}$
I was trying to get an asymptotic formula for the sum
$$\sum_{n=2}^{x}\frac{1}{\log n}$$
this looks a bit tricky, but I still tried it. We have

\begin{align}
\sum_{n=2}^{x}\frac{1}{\log n}&=-\sum_{n=2}^{x}\frac{1}{\log(nt)}\Bigg{|}_{t=1}^{t=\infty}\\
&=\sum_{n=2}^{x}\int_{1}^{\infty}\frac1{t\log^2(nt)}dt
\end{align}
Now I don't know what to do further. Any help would be appreciated. Any asymptotic formula or closed form would help.

Note: I will probably update this question to post more of my work.

Update: I proved that this sum is $O(x/\log x)$, and this was also mentioned in the comments. But I want to improve this. How can one improve the error term?

答:
Note that
$$
\sum\limits_{n = 2}^x {\frac{1}{{\log n}}}  \le \frac{1}{{\log 2}} + \sum\limits_{n = 3}^x {\int_{n - 1}^n {\frac{{dt}}{{\log t}}} }  = \frac{1}{{\log 2}} + \int_2^x {\frac{{dt}}{{\log t}}}
$$
and
$$
\sum\limits_{n = 2}^x {\frac{1}{{\log n}}}  \ge \sum\limits_{n = 2}^x {\int_n^{n + 1} {\frac{{dt}}{{\log t}}} }  \ge \sum\limits_{n = 2}^{x - 1} {\int_n^{n + 1} {\frac{{dt}}{{\log t}}} }  = \int_2^x {\frac{{dt}}{{\log t}}} .
$$
Thus, by the known asymptotic expansion of the logarithmic integral,
$$
\sum\limits_{n = 2}^x {\frac{1}{{\log n}}}  = \int_2^x {\frac{{dt}}{{\log t}}}  + \mathcal{O}(1) \sim \frac{x}{{\log x}}\left( {1 + \frac{{1!}}{{\log x}} + \frac{{2!}}{{\log ^2 x}} +  \ldots } \right)
$$
as $x\to +\infty$.


问:
How is $i^3$ = $\sqrt{-i}$.
I new to complex numbers.So please don’t mind.

We know $i^2$ = -1.

Then $i^3$ is written as $= -i$

Then does it’s steps of writing follow  this:

$-1^2$= $i^3$= 1

So in terms i = It says in my book

= -  $\sqrt{-i}$.

How did we get this ?

I have done calculation mistake while writing this question.I have now answered this question.

答:
$i = \sqrt{-1~}$ so $i^2 = -1$ and $i^3 = i^2 \cdot i = -1 \cdot \sqrt{-1~} = -\sqrt{-1~}$
Hope This helps!

Title: How is $i^3$ = $\sqrt{-i}$.


Body:
Continuing from your method is messy but possible:

Let $\frac{(3-t^2)}{(3+t^2)^2 } = \frac{A}{(3+t^2)} + \frac{B}{(3+t^2)^2}$. Then $3-t^2 = (3+t^2)A + B$. When $t = 0, 3 = 3A + B$, and when $t = 1, 2=4A+B$, so $A = -1, B= 6$. Thus:

$$I = -2 \int_0^1 \frac{1}{3+t^2} \ dt + 12 \int_0^1 \frac{1}{(3+t^2)^2} \ dt $$

For the second integral, substitute $t = \sqrt{3} \tan u, dt = \sqrt{3} \sec^2 u$. Then $(3+t^2)^2$ $ =(3(1 + \tan^2 u))^2 = 9 \sec^4 u$, and so:

$$I_2 = \frac{\sqrt{3}}{9} \int_0^1 \cos^2u \ du = \frac{\sqrt3}{18} \int_0^1 1+\cos2u \ du= \left[\frac{\sqrt3}{18} (u+\frac{1}{2} \sin 2u)\right]_0^1 = \left[\frac{\sqrt3}{18} (\tan^{-1} \frac{t}{\sqrt3}+\frac{1}{2} \sin(2(\tan^{-1}\frac{t}{\sqrt3}))\right]_0^1$$

and therefore:

$$I = -2 \left[\frac{1}{\sqrt 3} \tan^{-1} \frac{t}{\sqrt3}\right]_0^1+12\left[\frac{\sqrt3}{18} (\tan^{-1} \frac{t}{\sqrt3}+\frac{1}{2} \sin(2(\tan^{-1}\frac{t}{\sqrt3}))\right]_0^1$$
$$=-2(\frac{1}{6\sqrt3} \pi - 0) + 12 \cdot \frac{\sqrt{3}}{18}(\frac{\pi}{6}+\frac{\sqrt3}{4}-0) = \frac{1}{2}.$$


答:
Hint

Like How to integrate $\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$,

find $$\dfrac{dy}{dx}$$ where $$y=\dfrac{a\sin x+b\cos x+c}{2+\cos x}$$

Compare the coefficients of $\sin x,\cos x,\sin x\cos x$ and the constant to find the arbitrary constants $a,b,c$


答:
So you have $a_r=Ai^r+B(-i)^r$  as a general solution and you want $a_1=1, a_2=2$ so $$a_1=Ai-Bi=1$$ and if we multiply this by $i$ we get $B-A=i$

And also $$a_2=Ai^2+B(-i)^2=2=-A-B$$so that $A+B=-2$

We add the two equations to get $2B=-2+i$ and subtract first from second to get $2A=-2-i$

I've just sketched this out to show that there is nothing to be afraid of - it is just simple arithmetic.

Title: Solve this recurrence relation $a_n=-a_{n-2}$

Body:
So far I have this relation of recurrence:
$$a_n=-a_{n-2}$$
With $a_1=1$ and $a_2=2$.

I know I shoud replace $C \times r^n $ for $a_n$ and $C \times r^{n-2}$ for $a_{n-2}$

Then the relation gets here:
$$Cr^{n}+Cr^{n-2}=0$$

If I divide the left side by $Cr^{n-2}$
The equation will be like this $r^2+1=0$, thus $r=-i$ or $r=i$

After this a can't solve it because I forgot complex numbers
Can you guys please help me?


问:
D'alembert's approach for boundary value problems
I have the following problem

$\left\{\begin{array}{l}
u_{t t}=u_{x x}+2 u, \quad t>0, \quad 0<x<\pi \\
\left.u\right|_{t=0}=0,\left.\quad u_{t}\right|_{t=0}=0 \\
\left.u_{x}\right|_{x=0}=1,\left.u_{x}\right|_{x=\pi}=1
\end{array}\right.$

It is clear that it needs to use D'Alembert's approach:

$u(x, t)=\frac{\varphi(x+a t)+\varphi(x-a t)}{2}+\frac{1}{2 a} \int_{x-a t}^{x+a t} \psi(\alpha) d \alpha$

where

$u(x, 0)=\varphi(x), \quad u_{t}(x, 0)=\psi(x)$

In my case I have $\varphi(x) = \psi(x) = 0$. What id does mean and how to solve the problem in this case? And also, how to treat this term $2u$ in the equation?

答:
D'Alembert's solution is hardly adapted to solve the wave equation with an additional term $2u$. Instead, let us use separation of variables. To do so, we need to substract the stationary solution
$$
u^*(x) = \frac{\sin(x\sqrt{2}) - \tan\left(\frac{\pi}{\sqrt{2}}\right) \cos(x\sqrt{2})}{\sqrt{2}}
$$
by introducing $v = u^* + u$. Thus, $v$ satisfies the PDE problem $v_{tt} = v_{xx} + 2 v$ with homogeneous Neumann boundary conditions, and with the initial conditions $v = u^*$ and $v_t = 0$ at $t=0$.
Separation of variables $v(x,t) = X(x)T(t)$ gives
$$
T'' + \lambda T = 0, \qquad  X'' + (\lambda+2)X = 0
$$
where $\lambda$ is the separation constant. For $\lambda >0$, solutions are of the form
\begin{aligned}
T(t) &= A \cos(t\sqrt{\lambda}) + B \sin(t\sqrt{\lambda})\\
X(x) &= C \cos(x\sqrt{\lambda+2}) + D \sin(x\sqrt{\lambda+2})
\end{aligned}
Hope you can take it from here.


问:
Given that the family has at least one girl, determine the probability that the family has at least one boy.
Suppose that a family has exactly n children (n ≥ 2). Assume that the probability that any child will be a girl is 1/2 and that all births are independent. Given that the family has at least one girl, determine the probability that the family has at least one boy.

This is how I went about it:

Since we know that the family has at least one girl, we now have to determine the probability that the family has at least one boy from a pool of n-1 children.

Thus, probability is equal to:

1-Probability of having no boys

=1-(1/2)^(n-1)

But the answer is (1-(1/2)^(n-1))   /    (1-(1/2)^(n))

The structure of the answer suggests that Bayes theorem is being used in the correct answer, but even the Bayes theorem should lead to my answer. I do not know where I am going wrong?

Body:
Define $Y$ to be the numbe of boys, $X$ the number of girls. You need:
$$
P(Y \geq 1 |X \geq 1)=1-P(Y=0|X \geq 1) = 1- \frac{P(Y=0|X \geq 1)P(X \geq 1)}{P(X \geq 1)}\\
=1 -\frac{P(Y=0 \cap X \geq 1)}{1-P(X=0)}=1-\frac{\frac{1}{2^n}}{1-\frac{1}{2^n}}=\frac{1-\frac{1}{2^{n-1}}}{1-\frac{1}{2^n}}
$$


问:
Write down the set in set build form $\{-1, 2, -3, 4, -5, 6...\}$
$$\{-1, 2, -3, 4, -5, 6...\}$$

I have been wondering how to rewrite the following set using the set build method because I have never seen such from before, the odd index is negative number while the even index is positive number

答: That set is equal to$$\{n\in\Bbb Z\mid(\exists k\in\Bbb N):n=(-1)^kk\}.$$


问:
Calculate the $F^{−1}$ of a function
Let $V:=\{x \in \mathbb{R}:-5 \leq x \leq 5\} .$ Consider the function $F: V \rightarrow V$ defined by
$$
F(x)=\left\{\begin{array}{ll}
-1, & \text { if }-5 \leq x \leq 1 \\
2, & \text { if } 1<x \leq 3 \\
0, & \text { if } 3<x<4 \\
x, & \text { if } 4 \leq x \leq 5
\end{array}\right.
$$

$I:=\{x\in\mathbb{R}\ :\ -5\leq x\leq1\}$. Choose the set that $F^{−1}(I)$ is equal to.

答:
Of all the  possible values of $F$ the only value in $(0,4)$ is $2$. Hence, $F^{-1}(I)=\{x: F(x) \in (0,4)\}=(1,3]$. The correct answer is b).

For  the edited version of the question $F^{-1}(I)=[-5,1] \cup (3,4)$.

Title: Calculate the $F^{−1}$ of a function


问:
Connected graph with minimum degree $k \geqslant 3$ and path between two vertices
In a connected graph with every vertex has a degree of at least $k \geqslant 3$, there exist a path $P$ between any two vertex $x,y$ such that $|V(P)| \leqslant \frac{3 |V(G)|}{k−2}$.

My thought is the path can have only one $N(x)$, and one $N(N(x))$, so on and so forth, but I don't know what to do beyond this point.

答:
I will show that, in a connected graph where every vertex has degree at least $k\ge3$, any two vertices $x$ and $y$ are connected by a path $P$ with $|V(P)|\le\frac{3|V(G)|-2}{k+1}\lt\frac{3|V(G)|}{k-2}$.

Let $G$ be a connected (finite, simple) graph with minimum degree $\delta(G)\ge k\ge3$. Consider any vertices $x,y\in V(G)$. The case $x=y$ is trivial, so we assume $x\ne y$. Let $P$ be a minimum length path from $x$ to $y$. Let $|V(G)|=n$ and $|V(P)|=p$. Let $e$ be the number of edges between $P$ and $V(G)\setminus V(P)$.

Since each endpoint of $P$ has at least $k-1$ neighbors in $V(G)\setminus V(P)$, while each internal vertex of $P$ has at least $k-2$ neighbors in $V(G)\setminus V(P)$, we have
$$e\ge2(k-1)+(p-2)(k-2).\tag1$$
On the other hand, since each vertex in $V(G)\setminus V(P)$ has at most $3$ neighbors in $V(P)$ (or else we could use it to get a shorter path between $x$ and $y$), we have
$$e\le3(n-p).\tag2$$
From $(1)$ and $(2)$ we get
$$2(k-1)+(p-2)(k-2)\le3(n-p)$$
which simplifies to
$$(k+1)p\le3n-2$$
and finally
$$p\le\frac{3n-2}{k+1}.$$

P.S. In the same way you can show that, if $G$ is a finite simple graph with minimum degree $\delta(G)\ge2$, then $G$ contains a cycle of length $\le\max\left\{4,\left\lfloor\frac{|V(G)|}{\delta(G)-1}\right\rfloor\right\}$.


问:
Find a minimal polynomial of a primitive 5-th root of unity over $\mathbb{Q}_3(\sqrt{-7})$
Let $K=\mathbb{Q}_3(\sqrt{-7})$ be the base field and $L=K(\zeta_5)$ where $\zeta_5$ is a primitive $5$-th root of unity. I know that the extension $L/K$ is unramified of degree $2$.

Now I would like to find a minimal polynomial of $\zeta_5$ over $K$. I know that such a minimal polynomial must divide $x^4 + x^3 + x^2 + x + 1 = \frac{x^5-1}{x-1}$, i.e. it will split into two quadratic polynomials (which both are potential minimal polynomials of $\zeta_5$). But aside from that, I don't know how to compute the actual factors.

Could you please help me with this problem?

答:
$$(1+2\zeta_5+2\zeta_5^9)^2=5$$
(expand the LHS then use $\sum_{n=0}^4 \zeta_5^n=0$)

Thus the minimal polynomial is $$(x-\zeta_5)(x-\zeta_5^9)=x^2-\frac{1+\sqrt5}{2} x+1$$
Where
$$\sqrt5 = \sqrt{-7} \sqrt{-5/7}=\sqrt{-7} \sum_{k=0}^\infty {1/2\choose k}(-1-5/7)^k$$ $$=\sqrt{-7} \sum_{k=0}^\infty {1/2\choose k}(-4/7)^k 3^k \in \Bbb{Z}_3[\sqrt{-7}]$$

Note: in fact it is $x^2-\frac{1 \pm \sqrt5}{2} x+1$ and $\sqrt5= \pm \sqrt{-7} \sqrt{-5/7}$ and which pair of 5-th root of unity I chose is implied by the sign I chose


问:
analysis of one example in shortest path tree
Consider connected and undirected Graph $G$.

shortest path tree for each vertexes of $G$ was given.

if we makes union of all these shortest path trees then  order (number)  of edges in this union at worst case is $O(n^2)$ !!

how this achieved? is there any idea?

My Idea:

I think a complete graph with $n$ vertex, each vertex has shortest path tree at most of $n-1$ edges in worst case and then $O(n^2)$.

答:
Welcome to MSE!

Your idea is exactly right. More formally:

We have a shortest path tree $T_v$ for each vertex $v \in G$.

Each $T_v$ is a tree, and has $n$ vertices (since $G$ is connected).

This means each $T_v$ has $n-1$ edges.

So we have $n$ choices of $v$, and $n-1$ edges in each $T_v$

So the total number of edges in all the trees is $n \cdot (n-1)$.

but this is $n^2 -n$, which is $O(n^2)$.

I hope this helps ^_^


问:
If a curve $y=a\sqrt x +bx$ passes through $(1,2)$ and the area bounded by the curve, line $x=4$ and the $x$ axis
The way solve this is by integrating the function from 0 to 4 ie
$$\int_0^4 f(x) =8$$ since the given curve clearly intersects $(0,0)$

But if try to find a point where $y=0$ by setting $y=0$

Then
$$0= a\sqrt x +bx$$
$$\sqrt x =\frac{-a}{b}$$ which doesn’t give $x=0$ since according to the answer, $a\not =0$

What is the reason behind this anomaly?

答: Assuming that $x\geqslant0$ and that $b>0$, then\begin{align}a\sqrt x+bx=0&\iff\sqrt x\left(a+b\sqrt x\right)=0\\&\iff\sqrt x=0\vee a+b\sqrt x=0\\&\iff x=0\vee\sqrt x=-\frac ab.\end{align}


答:
$i^3$ =   $\sqrt{-1}$. *  $\sqrt{-1}$.*  $\sqrt{-1}$.

Therefore it equals = -1 *  $\sqrt{-1}$.

Or we can write it also using the property he listed in his question.

$i^2$ * i = -i.

Remember:When a and b both are negative and are multiplied.

Therefore , it equals - $\sqrt{ab}$.

For the OP.

Title: How is $i^3$ = $\sqrt{-i}$.

Body:
I new to complex numbers.So please don’t mind.

We know $i^2$ = -1.

Then $i^3$ is written as $= -i$

Then does it’s steps of writing follow  this:

$-1^2$= $i^3$= 1

So in terms i = It says in my book

= -  $\sqrt{-i}$.

How did we get this ?

I have done calculation mistake while writing this question.I have now answered this question.


问:
Fundamental theorem of calculus over a point of discontinuity
Let $\mathbb{P}$ be a probability over the Borel sets of $\mathbb{R}$ and let $X$ be a real random variable with this distribution. Consider that $m_{0}$ is a point of discontinuity of the function $\mathbb{P}(X\leq x)$. I want to compute the following:

$$\cfrac{\partial}{\partial c}\left(\int_{-\infty}^{c}\mathbb{P}(X\leq t)dt\right) (m_{0})$$.

If we take $b < m_{0}$, we know that $\mathbb{P}(X\leq x)$ is continuous and bounded in $(-\infty, b]$, then, applying the fundamental theorem of calculus we have that
$$\cfrac{\partial}{\partial c}\left(\int_{-\infty}^{c}\mathbb{P}(X\leq t)dt\right) (b) = \mathbb{P}(X\leq b)$$

My question is whether it is correct the following computing:

$$\cfrac{\partial}{\partial c}\left(\int_{-\infty}^{c}\mathbb{P}(X\leq t)dt\right) (m_{0}) = \lim_{x\rightarrow m_{0}^{-}}\cfrac{\partial}{\partial c}\left(\int_{-\infty}^{c}\mathbb{P}(X\leq t)dt\right) (x) = \mathbb{P}(X < m_{0})$$

As the integrand is continuous in $(-\infty,m_{0})$ and bounded, it may help the dominated convergence theorem or something like that.

Thanks in advance.

答:
Take a simple example: $\mathbb P(X=0)=1$

Then $\mathbb P(X\le t)=0$ when $t<0$, and $\mathbb P(X\le t)=1$ when $t \ge 0$, and the discontinuity is at $t=0$

So $\int\limits_{-\infty}^{c}\mathbb{P}(X\leq t)\,dt = 0$ when $c \le 0$, and $\int\limits_{-\infty}^{c}\mathbb{P}(X\leq t)\,dt = c$ when $c \ge 0$, so this is continuous

Thus $\cfrac{\partial}{\partial c}\left(\int\limits_{-\infty}^{c}\mathbb{P}(X\leq t)\,dt\right) (b) =0$ when $b <0$, and $\cfrac{\partial}{\partial c}\left(\int\limits_{-\infty}^{c}\mathbb{P}(X\leq t)\,dt\right) (b) =1$ when $b >0$, and it is not defined when $b=0$, i.e. at the original point of discontinuity


问:
Using Bayes' Theorem with three events
As a short seatwork in probability, we were given this problem but I am not exactly sure if we're supposed to use Bayes' Theorem for this one:

If you get to have a nap for at least 3 hours, then you can stay awake at midnight 80% of the time. If you get 1-3 hours of nap, then you can stay awake at midnight 60% of the time. Lastly, if you get 1 hour or less, then you can only stay awake at midnight 20% of the time.

Let there be 10% chance that you get to nap for at least 3 hours, 70% chance to nap 1-3 hours, and 20% chance to nap for at most 1 hour.

If you fell asleep through the whole night, what is the probability that you got to nap for at least 3 hours?

How do we solve this kind of problem?

答:
Some hints:

First formalize by defining events:

$A$ - stays awake $\Rightarrow A'$ sleeps through

$E_1$ - gets at most 1-hour nap

$E_2$ - gets 1-3 hours nap

$E_3$ - gets at least 3-hours nap

Now, the data given is:

$P(E_1)=0.2$, $P(E_2)=0.7$, $P(E_3)=0.1$

$P(A|E_1) = 0.2$, $P(A|E_2) = 0.6$, $P(A|E_3) = 0.8$

The question asks for

\begin{eqnarray*}P(E_3|A')
& = & \frac{P(E_3\cap  A')}{P(A')}\\
& = & \frac{P(E_3) - P(E_3\cap  A)}{1-P(A)} \\
& \stackrel{P(E_i \cap A)=P(A|E_i)P(E_i)}{=} & \frac{P(E_3) - P(A|E_3)P(E_3)}{1-\sum_{i=1}^3 P(A|E_i)P(E_i)}
\end{eqnarray*}

Can you take it from here?

Title: Using Bayes' Theorem with three events


问:
Integrating: $\int_0^\infty \frac{\sin (ax)}{e^x + 1}dx$
I am trying to evaluate the following integral using the method of contour which I am not being able to. Can anyone point out what mistake I am making?
$$\int_0^\infty \frac{\sin ax}{e^x + 1}dx$$

I am considering the following contour. And function $\displaystyle f(z):= \frac{e^{iaz}}{e^z + 1}$

The pole of order $1$ occours at odd multiple of $i\pi$. By considering above contour there is no singularity. The integral can be broken down into six parts.

$$\int_0^R \frac{e^{iax}}{e^x + 1} dx + i \int_0^{2\pi} \frac{e^{ia(R + iy)}}{e^{R + iy} + 1} dy + \int_{R}^{0}\frac{e^{ia(x+2\pi i)}}{e^{x + 2 \pi i } + 1} dx +  \\ i \int_{2 \pi }^{\pi + \epsilon} \frac{e^{ai( iy)}}{e^{ iy } + 1}dy + \int_\gamma \frac{e^{iaz}}{e^z + 1} dz + i \int_{ \pi -\epsilon}^{0} \frac{e^{ia( iy)}}{e^{ iy } + 1}dy$$

First and third gives $\displaystyle (1 - e^{-2 a\pi})\int_0^R\frac{e^{iax}}{e^x + 1} dx$. Second goes to $0$ as $R \to \infty$

For fifth integral, $$\int_\gamma \frac{e^{iaz}}{e^z + 1} dz = \int_{-\pi/2}^{\pi/2} \frac{e^{ia\pi + a\epsilon e^{i\theta}}}{e^{i\pi + \epsilon e^{i\theta}+1}}i \epsilon i e^{i\theta }d\theta \to 0 \text{ as } \epsilon \to 0$$

The real part of fourth and sixth integral does not converge. But since my original integral is imaginary, it suffices to take imaginary part. As $\epsilon \to 0$, I get
$$i\int_{2\pi }^0 \Re \left [\frac{e^{-ay}}{e^{iy} + 1} \right] dy = i \int_{2\pi}^0 \frac{e^{-ay}}{2}dy = i \frac{e^{-2\pi a} - 1}{2a}$$

Finally using residue theorem, I am geting which is incorrect.
$$(1 - e^{-2 a\pi})\int_0^\infty \Im \left [\frac{e^{iax}}{e^x + 1} \right ] dx +\frac{e^{-2\pi a} - 1}{2a} = 0$$

Can anyone point out my mistake or give worked out solution?? Thanks in advance!!

ADDED::
I evaluated fifth integral incorrectly
$$\int_\gamma \frac{e^{iaz}}{e^z + 1} dz = \int_{-\pi/2}^{\pi/2} \frac{e^{ia(i\pi + \epsilon e^{i\theta})}}{e^{i\pi + \epsilon e^{i\theta}}+1}i \epsilon  e^{i\theta }d\theta = ie^{-a\pi}\int_{\pi/2}^{-\pi/2}\frac{e^{ia\epsilon e^{i\theta}}}{-e^{\epsilon e^{i\theta}} + 1} \epsilon e^{i\theta}d\theta = i \pi e^{-a\pi}$$
So the total sum should be
$$(1 - e^{-2 a\pi})\int_0^\infty \Im \left [\frac{e^{iax}}{e^x + 1} \right ] dx +\frac{e^{-2\pi a} - 1}{2a} +\pi e^{-a\pi}= 0 $$
After slight manipulation we find that
$$\int_0^\infty \frac{\sin ax}{e^x + 1}dx = -\frac{\pi}{2\sinh (\pi a)} +\frac{1}{2a}$$

答:
Combine the 4th and 6th integrals to get the Cauchy principal value:

$$PV \int_{\pi}^0 dy \frac{e^{-a y}}{1+e^{i y}}  = PV \int_{\pi}^0dy \frac{e^{-a y}}{2 \cos{(y/2)}}  e^{-iy/2}  $$

You still need to evaluate the imaginary part of the integral.


问:
Dimension Notation for Topological Spaces
Lots of families of topological spaces get superscripts denoting dimension: $\mathbb{R}^n$, $B^n$, $D^n$, $\Delta^n$, $S^n$, $\mathbb{R}P^n$, $T^n$. There's a niceness to this: $S^n$ is the boundary of $D^{n+1}$, which is a bit awkward at first, especially if you think about how you would define their dimensions as subsets of $\mathbb{R}^n$ as a vector space, but under any applicable topological definition of dimension, their superscripts represent their dimensions. For once, there is a nice continuity of notation. However, there are multiple topological notions of dimension: topological dimension, both inductive dimensions, dimension of CW-complexes, dimension of manifolds... So are there any explicit rules for what the superscript $n$ is actually telling us about a space, or is it just a guide that should wind up being equivalent under different definitions for sufficiently nice spaces?

答:
I normally use the dimension-number to describe the number of orthogonal lines supported at an instant of space.  In other words, if one imagines 2-cloth being shaped into any thing, it's still 2D.

Not with standing, there are a number of issues here.

Spaces of more positive curvature are usually represented in euclidean spaces of n+1 or so dimensions.  The most common of these is to represent S2 in E3 (ie the surface of the sphere against its volume).  The reality is that any space will appear as a ball in a space of higher dimension and lesser curvature.  Euclidean space E3 appears in hyperbolic space H4 as a surface bounding a horoglome (or horochoron).

The point in the PG is to distinguish between the fabric against what it is made into.  For example a 'hedrix' is a 2d fabric, and one describes E2 as a horohedrix (horizon-centred 2d fabric), the sphere as a glomohedrix (round 2d space), and the hyperbolic space H2 as a bollohedrix (negative-curvature 2d fabric).

If one talks of 'polyhedron' or 'apeirohedron', then this refers to something sewn together from patches, a polyhedron is (many [closed] 2d patched), an apeirohedron is [without + space-boundary (fence) + 2d + patches] or a tiling in any 2d space.  A {3,6} is a horohedron or (2d patches centred on the horizon).

The products of draught are rooted at -1 dimensions.  This means that the exponents on the algebraic products defined by them, is equal to the number of vertices a simplex has, so 3P is the same as 2D.  The products of repetition are rooted in 0 dimensions, so the usual dimension is the same as the repetitions.

Most mathematicians have not caught up with the drawn product, so it's less of a matter to them.

The explicit rule is that you should refer to the dimension of 'all-space', which is to say, that the surface of a sphere treated as spherical geometry is S2, and leaving such S2 is to go into 'hyperspace'.  If one is looking at a sphere as a solid, then it is an object in E3 or S3 or H3.  The space is normally referred to in terms of the repetitions of a line, rather than any drawn product.  So a triangle is 2-dimensional, although it occurs as a³ in products of draught.

PG = polygloss at http://www.os2fan2.com/gloss/index.html


问:
Continuous Markov Chain
Let $\left(\xi_n\right)_{n\geqslant1}$ be a sequence of independent identically distributed random variables with distribution $\mathcal{N}(0,1)$. Let $X_0$ be a random variable in $\mathbb{R}$ with distribution $\mu$, independent from the sequence $\left(\xi_n\right)_{n\geqslant1}$. Let us define the sequence $\left(X_n\right)_{n\geqslant1}$ by $$X_n=\alpha X_{n-1}+\xi_n$$
where $\alpha\in(0,1)$.

Show that $\left(X_n\right)_{n\geqslant0}$ is a Markov chain and compute its transition kernel. Find $\sigma^2>0$ such that $\mathcal{N}\left(0,\sigma^2\right)$ is a stationnary distribution for this chain.

My work

I have computed $X_n=\alpha^nX_0+\sum_{j=1}^n \alpha^{n-j}\xi_j$, however when writing
$$\mathbb{P}\left(X_0\in C_0,\dots,X_n\in C_n\right)$$
for $C_0,\dots,C_n$ borel sets of $\mathbb{R}$, I do not know how to use independence to make it look like
$$\int_{\substack{x_0\in C_0\\\dots\\x_n\in C_n}}\mu(dx_0)P(x_0,dx_1)\dots P(x_{n-1},dx_n)$$

答: Observe that $X_0,X_1,..,X_{n-1}$ depend only on $\xi_1,\xi_2,...,\xi_{n-1}$.  Hence, $P(X_n\leq x|X_0,X_1,..,X_{n-1})=P(\xi_n\leq x-\alpha X_{n-1}|X_{n-1})$. By independence of $\xi_n$ and $X_{n-1}$ you can compute this as  $\Phi (x-\alpha X_{n-1})$.

Title: Continuous Markov Chain


答:
I would say there are no explicit rules. For me the point is that you have a sequence of “similar” spaces naturally indexed by $n$. For the examples you mentioned it feels natural to align the indexes with the dimension (where for these spaces various notions of dimension coincide). If one wants to directly discuss some dimension of a space, I'd suggest being explicit: $\operatorname{somedim}(X) = n$ while $\operatorname{otherdim}(X) = n'$.

There are more examples of the general pattern not directly related to dimension: $ℤ_n$ for cyclic group od order $n$, $S_n$ and $A_n$ for the symmetric and alternating group; $\ell^p$ and $L^p$ (also denoted by $\ell_p$ and $L_p$) for the Lebesgue spaces.

Note that in $ℝ^n$ the superscript is also a genuine operation – the Cartesian power. (Maybe the similarity with $ℝ^n$ is the reason why $S^n$ instead of $S_n$ is used for spheres, etc, but that is just a speculation.)

Title: Dimension Notation for Topological Spaces

Body: Lots of families of topological spaces get superscripts denoting dimension: $\mathbb{R}^n$, $B^n$, $D^n$, $\Delta^n$, $S^n$, $\mathbb{R}P^n$, $T^n$. There's a niceness to this: $S^n$ is the boundary of $D^{n+1}$, which is a bit awkward at first, especially if you think about how you would define their dimensions as subsets of $\mathbb{R}^n$ as a vector space, but under any applicable topological definition of dimension, their superscripts represent their dimensions. For once, there is a nice continuity of notation. However, there are multiple topological notions of dimension: topological dimension, both inductive dimensions, dimension of CW-complexes, dimension of manifolds... So are there any explicit rules for what the superscript $n$ is actually telling us about a space, or is it just a guide that should wind up being equivalent under different definitions for sufficiently nice spaces?


问:
$a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$
I have not solved inequalities in a while, so I am a little rusty. Could you help me with this inequality I have found?
$$a^2 + b^2 + c^2 + 6 \ge 3(a + b + c),$$ where $a, b, c > 0$ and $abc = 1$

My initial idea was $a ^ 2 + 2 \ge 2\sqrt 2a$ and the inequalities with $b $ and $c$, then adding these 3, we get, $a ^ 2 + b ^ 2 + c ^ 2 + 6 \ge 2\sqrt 2(a + b+ c)$, but then we get to $2\sqrt2 > 3$, which is false.

Edited: I found some variants of the original problem.

Problem 1: Let $a, b, c > 0$. Prove that $a^2 + b^2 + c^2 + 6 + (abc - 1) \ge 3(a+b+c)$.

Problem 2: Let $a, b, c$ be reals with $abc \le 1$. Prove that $a^2 + b^2 + c^2 + 6 \ge 3(a+b+c)$.

答:
My second solution:

WLOG, assume $c = \min(a,b,c)$.

If $c \le \frac{1}{2}$, we have
$$a^2+b^2+c^2 + 6 - 3(a+b+c) = (a-\tfrac{3}{2})^2 + (b - \tfrac{3}{2})^2 + c^2 - 3c + \tfrac{3}{2}
\ge 0.$$

If $c > \frac{1}{2}$, noting that $x^2 + 2 - 3x + \ln x \ge 0$ for all $x > \frac{1}{2}$
(see the remark at the end),
we have
$$a^2+b^2+c^2 + 6 - 3(a+b+c) = \sum_{\mathrm{cyc}} (a^2 + 2 - 3a + \ln a) \ge 0.$$

We are done.

Remark: Let $f(x) = x^2 + 2 - 3x + \ln x$. We have $f'(x) = \frac{(2x-1)(x-1)}{x}$.
Thus, $f(x)$ is strictly decreasing on $(\frac{1}{2}, 1)$, and strictly increasing on $(1, \infty)$.
Also $f(1) = 0$. Thus, we have $f(x) \ge 0$ for all $x > \frac{1}{2}$.


问:
Graduate Commutative algebra book or lecture notes in German
It may look strange but I am looking for a Graduate level book / lecture notes on Commutative algebra in German.
I wish the content to be as close as it is possible to Eisenbud's "Commutative Algebra with a view towards Algebraic Geometry".

Any suggestions are welcome.

答: Kunz's Einführung in die kommutative Algebra und algebraische Geometrie is fairly standard, and (as far as I can tell) corresponds quite neatly to the book by Eisenbud. If you're looking for something freely available online, I believe this Skript (lecture notes) is quite comprehensive.


答:
What about a simple one?
$\{1,2,3,4,5,\cdots\}$ is simply $\{n:\,n\in\mathbb{N}\}$.

Now, see: this one. $\{-1,1,-1,1,\cdots\}$ should be $\{x:\,x=(-1)^n\,,n\in\mathbb{N}\}$, right?

Combine both and the quoted set is thus
$$\{x:\,x=n(-1)^n\,, n\in\mathbb{N}\}$$

Hope this helps :)

Title: Write down the set in set build form $\{-1, 2, -3, 4, -5, 6...\}$

Body:
$$\{-1, 2, -3, 4, -5, 6...\}$$

I have been wondering how to rewrite the following set using the set build method because I have never seen such from before, the odd index is negative number while the even index is positive number


问:
Question related to probability density function
A continuous random variable X has the probability density function

$f(x)=
\begin{cases}
 px+q & 1 \leq x \leq 4 \\
 0 & \text{otherwise}
\end{cases}$

If $P(X<2)=\dfrac{1}{9}$ , find the values of $p$ and $q$. $\left[ans: p= \dfrac{1}{9},\ q=\dfrac{1}{18}  \right] $

This is my attempt,

\begin{aligned}
P(X < 2) &= \dfrac{1}{9}\\
P(X \leq 2) &= \dfrac{1}{9} \\
\int_1^2 (px+q) \,dx &= \dfrac{1}{9} \\
\left[\dfrac{px^2}{2}+qx \right]_1^2 &= \dfrac{1}{9} \\
2p+2q- \left( \dfrac{p}{2}+q\right) &= \dfrac{1}{9} \\
\dfrac{3}{2} +q &= \dfrac{1}{9} \qquad(1) \\
\end{aligned}

\begin{aligned}
\int_{-\infty}^{\infty} f(x) \,dx &= 1 \\
\int_{1}^{4} (px+q) \,dx &= 1 \\
\left[\dfrac{px^2}{2}+qx \right]_1^4 &= 1 \\
8p+4q- \left( \dfrac{p}{2}+q\right) &= 1 \\
\dfrac{15}{2} +3q &= 1  \qquad(2)  \\
\end{aligned}

By solving both $(1)$ and $(2)$, I get $p=\dfrac{2}{9} \text{ and } q=- \dfrac{2}{9}$

Can someone check my work? Thanks.

Body:
Hint:please check (1) and (2)

$$\frac{3p}{2}+q=\frac{1}{9}(1)$$

$$\frac{15p}{2}+3q=1(2)$$

$$p=\frac{2}{9} , q= \frac {-2}{9}$$

same answer


问:
Number of permutations of the word ABCDEFGHI which contains the string CDE; the strings AC and CF; the strings ABC and ECG
How many permutations of the letters ABCDEFGHI contain

a) the string CDE?

b) the strings AC and CF?

c) the strings ABC and ECG?

Here is my solution:

a) We treat the mentioned string as one single letter. Thus we have the possible letter A,B,CDE,F,G,H,I.

Then we have 7 letters of which we want to select 7 letters.

```<code>N=7

R=7

Permutation: P(7,7)=7!/(7-7)!

                   =7!=5040
</code>```

b)   A string cannot contain both AC and CF, since the string

can only contain C once and C can not be followed by A and F.

Thus there are 0 string that contain both AC and CF.

c)  A string cannot contain both ABC and ECG, since the string

can only contain C once and C can not be followed by A and E.

Thus there are 0 string that contain both ABC and ECG.

Is my solution right? or need any modification?

答:
As saulspatz indicated in the comments, your answers to parts (a) and (c) are correct.

In part (b), I think the source of your confusion is that you are assuming that AC and CF must be separate strings.  Since you only have one copy of each letter available, for the strings AC and CF to both appear, they must overlap, meaning the permutation must contain the string ACF.  Thus, the number of permutations of the letters ABCDEFGHI is the number of permutations which contain the string ACF.  Thus, we have seven objects to permute: ACF, B, D, E, G, H, I.  The seven objects can be permuted in $7! = 5040$ ways.


问:
Let $A$ be a $5\times 3$ matrix and let $rank(A)=3$ then which from these two statements is right:
a)for every matrix $B$ thats $3\times 3$, and $rank(B)=3$, for the equation $(AB)x=0$ theres a unique solution.

b)for every matrix $B$ thats $5\times 5$, and $rank(B)=5$, for the equation $(BA)x=0$ theres infinitely number of solutions.

What I know is that the rank of $(AB)$ or $(BA)$ can be at most 3. which makes me think that they both can be correct. but the answer is a and I can't understand why.

 any assistance or explanation is much appreciated.

答: Observe that from the associativity $(BA)x = B(Ax)$ . Since rank of $B$ is $5$,the only solution for the equation $By=0$ is $y=0$, so it is enough to check the equation $Ax = 0$. Similarly, since rank of $A$ is $3$ ,the only solution of  $Ax = 0$ is $x=0$. Thus, b) is not correct. Using the same idea, you can show that a) is correct as well.


问:
Set of homomorphisms of abelian group is noncommutative ring
This is question I faced

Question : Prove that the set of group homomorphism for abelian group $A$ is a noncommutative ring with 1.

So, I showed every conditions for ring with 1. But, is it true that for every abelian group $A$, there exists two homomorphisms $f, g$ such that $fg !=gh$?

It is easy if $f, g$ are just functions, not homomorphisms. How can I show it?

答: The result is not always true. $\mathrm{Hom}(\mathbb Z / 2 \mathbb Z)$ has two elements, the trivial homomorphism and the identity that commute.


问:
Proving that an injective and surjective linear map is invertible
I'm trying to understand how a surjective and injective linear map is invertible. I'm using Axler's book but I found the proof there hard to follow (in one of the directions only; I can see why an invertible linear map is surjective and injective).

Say $T: V \to W$ is a linear map from vector space $V$ to vector space $W$. It is injective and surjective. I want to prove that there exists a linear map $S:W \to V$ such that $TS = I, ST = I$ where each $I$ is the identity map in $W$ and $V$ respectively.

Axler's proof introduces $S$ as:

For each $w \in W$, define $Sw$ to be the unique element of $V$ such
that $TSw$ = $w$ (the existence and uniqueness of such an element
follow from the surjectivity and injectivity of $T$).

First of all, what is $S$ here? Is it a function or some sort of operator? Or is it a linear map $S:W \to V$?

But the proof ends with a final section:

To complete the proof, we need to show that $S$ is linear

So $S$ was not a linear map, then what was it?

If $S$ can be taken as a linear map

If I assume $S:W \to V$, I can use the surjective $T$ to say that $\exists v \in V$ such that $Tv = w \forall w \in W$.

Additionally with injective $T$, $\forall w_1, w_2 \in W$ such that $w_1 \neq w_2$ we know $v_1, v_2 \in V$ exists such that $Tv_1 = w_1, Tv_2 = w_2$. If $v_1 = v_2$ then $Tv_1 = Tv_2 \implies w_1 = w_2$ which contradicts our assumption that $w_1 \neq w_2$. So $v_1 \neq v_2$. So $w_1 \neq w_2 \implies v_1 \neq v_2$.

Define $S$ as a map from each $w \in W$ to a unique (proved to exist in previous point) $v \in V$ such that $Sw = v$ where $Tv = w$. So $TSw = Tv= w$.

Is this a valid way to interpret Axler's bracketed text?

答:
What is happening is the following.

Let $T: V \to W$ be a linear map that is both injective and surjective. We want to construct an inverse $S: W \to V$. (

Given $w \in W$, how would we define $Sw \in V?$ Well, since $T$ is injective and surjective, there exists a unique element $v \in V$ with $Tv = w$. This uniqueness allows us to define $Sw:= v$.

Axler used suggestive notation and immediately wrote $v = Sw.$

Note that you only need to check that $S$ is an inverse of $T$ and not that it is linear. The inverse of a linear map is always automatically linear (exercise!).


答:
Here, $S$ is a map from $W$ to $V$. Asserting that $T\colon V\longrightarrow W$ is invertible means that there is a linear map from $W$ to $V$ such that:

$(\forall v\in V):S(T(v))=v$;

$(\forall w\in W):T(S(w))=w$.

Title: Proving that an injective and surjective linear map is invertible

Body:
I'm trying to understand how a surjective and injective linear map is invertible. I'm using Axler's book but I found the proof there hard to follow (in one of the directions only; I can see why an invertible linear map is surjective and injective).

Say $T: V \to W$ is a linear map from vector space $V$ to vector space $W$. It is injective and surjective. I want to prove that there exists a linear map $S:W \to V$ such that $TS = I, ST = I$ where each $I$ is the identity map in $W$ and $V$ respectively.

Axler's proof introduces $S$ as:

For each $w \in W$, define $Sw$ to be the unique element of $V$ such
that $TSw$ = $w$ (the existence and uniqueness of such an element
follow from the surjectivity and injectivity of $T$).

First of all, what is $S$ here? Is it a function or some sort of operator? Or is it a linear map $S:W \to V$?

But the proof ends with a final section:

To complete the proof, we need to show that $S$ is linear

So $S$ was not a linear map, then what was it?

If $S$ can be taken as a linear map

If I assume $S:W \to V$, I can use the surjective $T$ to say that $\exists v \in V$ such that $Tv = w \forall w \in W$.

Additionally with injective $T$, $\forall w_1, w_2 \in W$ such that $w_1 \neq w_2$ we know $v_1, v_2 \in V$ exists such that $Tv_1 = w_1, Tv_2 = w_2$. If $v_1 = v_2$ then $Tv_1 = Tv_2 \implies w_1 = w_2$ which contradicts our assumption that $w_1 \neq w_2$. So $v_1 \neq v_2$. So $w_1 \neq w_2 \implies v_1 \neq v_2$.

Define $S$ as a map from each $w \in W$ to a unique (proved to exist in previous point) $v \in V$ such that $Sw = v$ where $Tv = w$. So $TSw = Tv= w$.

Is this a valid way to interpret Axler's bracketed text?


问:
Minina and maxima (2-variable function)
I need to find a minima and maxima of a function $z = x^2 - 12x + y^2 - 2y$ that is limited by points $A(-7,-5)$, $B(5;-5)$ and $C(5,10)$ but i do not clearly understand the algorithm $><$

What should i do? Especially with equation of $CA$ line ($y = 1.25x + 3.75$)

答: One way would be setting the Jacobian Matrix equal to $0$ and solve the systems of equation. In this way you will find all points that could be potentially minima or maxima. Now you have to check wich of these points meet the condition of beeing limited by $A,B,C$. Anyway are those points $(x,y),(y,z)$ or $(x,z)$ ?</p>

Title: Minina and maxima (2-variable function)


答:
If $T:V\to W$ is a linear map then also it is a function.

It is well known that a function is bijective if and only if it has an inverse.

Also it is well known that this inverse is unique.

Denoting this inverse as $S:W\to V$ we have $T\circ S=\mathsf{id}_W$ and $S\circ T=\mathsf{id}_V$.

What remains now is proving that this $S$ is a linear map.


问:
Evaluating $\int \sqrt{16-x^2} \,dx$
Are the steps below correct? Can you tell me another way to solve it, just hint?

$   \int \sqrt{16-x^2}   \,dx $

$ x = 4 \sin u \rightarrow  u=\arcsin(\frac{x}{4}) \rightarrow  dx = 4\cos u \,dx$

$   \int  \sqrt{16-x^2}   \,dx   = 16 \int \cos^2 u \,du = 16 \int \frac{\cos 2u +1}{2} \,du  $

$ v = 2u  \Rightarrow  dv = 2 du \Rightarrow du = \frac{1}{2}dv  $

$  16 \int \frac{\cos 2u +1}{2} \,du = 16 \int \frac{\cos v}{2} +\frac{1}{2} \,dv = 16(\frac{\sin v}{2} +\frac{v}{2}) = 8\sin v +8v = 8\sin 2u +16u  $

$ \Downarrow $

$8\sin(2\arcsin(\frac{x}{4})) + 4\arcsin\frac{x}{4} $

答:
The procedure is fine. Just be careful with the substitution in line 4, it is not necessary, but it is not wrong to do it. Nevertheless, you make a mistake since you forget to divide by 2.

Another way to solve it may be to substitute at the beginning for $x=4cos(u)$, the procedure is the same but formally is another method.

Title: Evaluating $\int \sqrt{16-x^2} \,dx$


问:
Solve a linear equation system, while enforcing a unit vector solution
I have an equation system of the form Aix + Biy + Ciz = Di, where (x,y,z) is a unit vector, and (Ai, Bi, Ci, Di) are sets of measurements from a noisy system (with typically 3-5 independant readings).

My first intuition to solve this problem was to pose this as an overdetermined linear equation system AX = B where X = (x,y,z), and to solve for X. However, with that approach, I have no way to enforce that the solution for vector X is a unit vector.

Is there an elegant (or standard) solution to that problem, or should I simply dive into non-linear equation solving solutions?

答:
This problem can be solved using singular value decomposition. The original problem can be stated as
$$min\|Ax-b\|_2, \|x\|_2=1$$
By substituting $$A'=[A;-b], x'=w x; w$$ and choosing $w=\frac{\|x\|_2}{\|x;1\|_2}$ we get $\|x'\|_2=w\|x;1\|_2=\|x\|_2$, so the problem becomes
$$min\|A'x'\|_2, \|x'\|_2=1$$
which is same as in this question: Minimizing $\|Ax\|_2$ subject to $\|x\|_2 = 1$
It can be solved by decomposing $A'$ into $U \Sigma V^T$ and then taking the last column of $V$, which corresponds to the smallest singular value of $A'$.

Finally $x$ is calculated by dividing $x'$ by the last element, taking elements of $x'$ except the last one and then normalizing it.

ContentLicense: CC BY-SA 2.5


问:
Largest interval of a differential equation
I was able to find the general solution of the given differential equation:
$ \frac{dy}{dx} + \frac{cos(x)}{sin(x)}y = \frac{1}{cos^2(x)sin(x)}$.

The general solution is y = sec(x) + csc(x)C.

Now I don't get it how to find the largest interval(interval of validity). All I know are sec(x)= 1/cos(x) is undefined if cos(x) is equal to 0, that is, the value $ \frac{\pi}{2} + \pi n $ for all integers n, and csc(x)=1/sin(x) is undefined if sin(x) is equal to 0, that is, the value $\pi n$ for all integers n.

答:
If you multiply with $\sin x$, you get the integrable equation
$$
(\sin(x)y(x))'=\frac1{\cos^2x}\implies \sin(x)y(x)=\tan(x)+C
$$
which gives directly the cited solution.

You have correctly identified the singularities of that solution formula. An ODE solution has its domain restricted such that the function is continuously differentiable over this domain. This means that the available domains are $(k\frac\pi2,(k+1)\frac\pi2)$, $k\in\Bbb Z$. The exact domain depends on where the initial point is given.

It is in general true that solutions for
$$
y^{(n)}(x)+a_{n-1}(x)y^{(n-1)}(x)+...+a_1(x)y'(x)+a_0(x)y(x)=r(x)
$$
have as maximal domain the maximal open interval that contains the initial time or position and such that all coefficients $a_k(x)$ and the right side $r(x)$ are continuous over this interval.


答: The endomorphism ring of a cyclic group is always commutative.</p>

Title: Set of homomorphisms of abelian group is noncommutative ring

Body:
This is question I faced

Question : Prove that the set of group homomorphism for abelian group $A$ is a noncommutative ring with 1.

So, I showed every conditions for ring with 1. But, is it true that for every abelian group $A$, there exists two homomorphisms $f, g$ such that $fg !=gh$?

It is easy if $f, g$ are just functions, not homomorphisms. How can I show it?


问:
Infinite Variance Probability Distribution
I have data $\vec{X}=(x_{1}, ..., x_{n})$ with size $N$, sampled from a physical phenomenon. The data takes discrete values in the range $[-N,N]$. As $N$ is increased, the sample variance and fourth order moments explode. I obtained a theoretical result about the problem that implies as $N\rightarrow +\infty$, then $Variance(\vec{X})\rightarrow \infty$. The mean remains finite. If I am not mistaken, this means that as the distribution converges to be continuous, then the variance tends to $+\infty$. I know that a discrete distribution cannot have infinite variance, of course.

The following figure (1) shows one example sample obtained with $N=100$. The y-axis shows the relative frequencies and the x-axis the value of $\vec{X}$.
This figure shows another sample with a kernel estimate 2.

I am looking for a distribution that could model this type of data, whether in its continuous limit or discrete. Distributions with infinite variance I am aware of are continuous and exhibit some type of power law, which my estimated samples PMF do not appear show.
My question is: Do you know any distributions that could model this type of data?
Please tell me if I am missing any important information. Any reference or hint would be extremely helpful. Thank you!

答: Given the information you have provided it is hard to tell which limiting distribution best describes your data. However, e.g. the student's t-distribution with $1<v\leq 2$ degrees of freedom has a finite mean yet inifite variance. Its support is also the whole real line.


问:
Prove if $f(x) \rightarrow \infty$ as $ x \rightarrow \infty$, then $\frac{\sin(x^2+x+1)}{f(x)} \rightarrow 0$ as $ x \rightarrow \infty$
Prove if $f(x) \rightarrow \infty$ as $ x  \rightarrow \infty$, then $\frac{\sin(x^2+x+1)}{f(x)} \rightarrow 0$ as $ x  \rightarrow \infty$

attempt

we know $\forall x \in \mathbb{R}, -1\leq \sin(x^2+x+1)\leq1$ so

$$-\frac{1}{f(x)}\leq \frac{\sin(x^2+x+1)}{f(x)} \leq \frac{1}{f(x)} $$

since $f(x) \rightarrow \infty$ as $ x  \rightarrow \infty$, $\frac{1}{f(x)} \rightarrow 0$ as $ x  \rightarrow \infty$
therefore according to squeeze theorem $\frac{\sin(x^2+x+1)}{f(x)} \rightarrow 0$ as $ x  \rightarrow \infty$

Is my attempt correct? if not give me some hints.

答: Your attempt is right. In general when encountering functions like $\dfrac{g(x)}{f(x)}$ and $g(x)$ is bounded from both above and below and $f(x)$ is not bounded, we usually use the squeeze theorem to get the desired results.


问:
According to what rigorous definition of "polynomial" is $\sum_{k=0}^n a_k D^k$ a polynomial, where $D$ is the derivative operator?
Let $Df$ be the derivative of $f\in C^{\infty}$.
\begin{align}
D^k\colon C^{\infty}&\to C^{\infty}\\
f&\mapsto f^{(k)}=\begin{cases}f&\text{ if }k=0\\
Df^{(k-1)}&\text{ if }1\leq k
\end{cases}
\end{align}
I was told that if $(a_0,\ldots,a_n)\in\mathbf{C}^{n+1}$, $\sum_{k=0}^n a_k D^k$ is a polynomial and can thus be factorized. That is, there is a $(b_1,\ldots,b_n)\in\mathbf{C}^n$ such that
\begin{equation}
\sum_{k=0}^n a_k D^k = a_n \prod_{i=1}^{n}(D^1 - b_iD^0).
\end{equation}

According to what rigorous definition of "polynomial" is that true and where can I find more information?

Unfortunately, I didn't find the Wikipedia article helpful and I'm not sure if the definition as sequences in some ring is what I'm looking for, as I have no background in abstract algebra.

答:
Let $P = \sum_k a_k X^k \in \mathbb{C}[X]$. The sum $P(D) := \sum_k a_k D^k$ is an endomorphism of the vector space $C^{\infty}$ (see the theory of endomorphism polynomials).

One has, when $E$ is a vector space over the field $\mathbb{K}$, $R, Q \in \mathbb{K}[X]$ are polynomials and $u$ an endomorphism, that: $R(u) \circ Q(u) = (RQ)(u)$.

Over $\mathbb{C}$, the fundamental theorem of algebra gievs us the existence of coefficients $b_1, \dots, b_n$ such that $P = a_n \prod (X - b_k)$.

Thus: $P(D) = a_n \prod \left( (X - b_k)(D) \right) = a_n \prod \left( D - b_k Id_{C^{\infty}} \right)$.


问:
Proper constants for $\alpha, \beta$
Here is the problem:

For what values of $\alpha$ and $\beta$, the function

$$\mu(x,y)=x^{\alpha}y^{\beta}$$

is an integrating factor for the OE $$ydx+x(1-3x^2y^2)dy=0.$$ I am working on it just knowing the definition. :(

答:
Just a good point:

If your OE is of form $$yf(xy)dx+xg(xy)dy=0$$ then you can assume that the integrating factor is $$\mu(x,y)=\frac{1}{xy(f-g)}$$ It means you have it here $$\mu=\frac{1}{xy(1-1+3x^2y^2)}$$

Title: Proper constants for $\alpha, \beta$


问:
Problem about uniform continuity on $[0,\infty)$
Can you help me to prove this statement? I found it in this book http://www.amazon.com/Selected-Problems-Translations-Mathematical-Monographs/dp/0821809539/ref=sr_1_1?ie=UTF8&qid=1391546817&sr=8-1&keywords=selected+problems+in+real+analysis.

$\textbf{Problem.}$ Prove that if $f\in C[0,\infty)$ and $f(x+h)-f(x)\to 0$ as $x\to \infty$, for any $h\in\mathbb{R}$, then $f(x+h)-f(x)\rightarrow 0$ ($\Rightarrow$ denote uniform continuity) as $x\to \infty$ on each finite interval and, hence, function $f$ is uniformly continuous on $[0,\infty)$.

$\textbf{Hint.}$ Prove that the assertion
$$
f(x+h) - f(x) \rightarrow 0 \quad\text{as}\quad x\to \infty
$$
on any finite interval is equivalent to the assertion that for all $\varepsilon>0$ there exists $a_{\varepsilon}$ and $b_{\varepsilon}$ such that $a_{\varepsilon}<b_{\varepsilon}$ and
$$
\overline{\lim\limits_{x\to\infty}} \sup\limits_{a_{\varepsilon}\leqslant h\leqslant b_{\epsilon}} |f(x+h)-f(x)|\leqslant\varepsilon.
$$
Prove the last assertion by arguing by contradiction and constructing a sequence $\{\triangle_n\}$ of nested closed intervals and a numerical sequence $x_n$ such that $x_n\to \infty$ and $|f(x_n+h)-f(x_n)|\geqslant\varepsilon$ for $h\in\triangle_n$.

答:
It's not a full answer, but may be useful.

Lemma. For all $[a,b]$ and all $\epsilon > 0$ $\exists [a_{\epsilon}, b_{\epsilon}] \subset [a,b]$: $$\overline{lim}_{x \to \infty} \sup_{h \in [a_{\epsilon}, b_{\epsilon}]} |f(x+h) - f(x)| \le \epsilon.$$

Proof. Let us prove it by contradiction. Suppose $\exists$  $[a_0,b_0]$ and $\epsilon_0 > 0$ such that $\forall [c,d] \subset [a_0,b_0] $
$$\overline{lim}_{x \to \infty} \sup_{h \in [c,d]} |f(x+h) - f(x)| > \epsilon_0.$$

Consider $[c,d] = [a_0, b_0]$. We have $\overline{lim}_{x \to \infty} \sup_{h \in [a_0, b_0]} |f(x+h) - f(x)| > \epsilon_0$ and hence $\exists x_1$ such that $\sup_{h \in [a_0, b_0]} |f(x_1+h) - f(x_1)| > \epsilon_0$. Condider a function $g_1(h) =  |f(x_1+h) - f(x_1)|$ for $h \in [a_0, b_0]$. It's a continious function and it's maximal value is bigger than $\epsilon_0$. Thus $\exists$ a closed interval $\Delta_1 \subset [a_0, b_0]$ such that $|f(x_1+h) - f(x_1)| \ge \epsilon_0$ for all $h \in \Delta_1$.

Further consider $[c,d] = \Delta_1$. We have $\overline{lim}_{x \to \infty} \sup_{h \in  \Delta_1} |f(x+h) - f(x)| > \epsilon_0$ and hence $\exists x_2 > x_1+1$ such that $\sup_{h \in \Delta_1} |f(x_2+h) - f(x_2)| > \epsilon_0$. Condider a function $g_2(h) =  |f(x_2+h) - f(x_2)|$ for $h \in \Delta_1$. It's a continious function and it's maximal value is bigger than $\epsilon_0$. Thus $\exists$ a closed interval $\Delta_2 \subset \Delta_1$ such that $|f(x_2+h) - f(x_2)| \ge \epsilon_0$ for all $h \in \Delta_2$.

Continuing the construction further, we get $x_n$ and $\Delta_n$ such that $x_{n+1} > x_n + 1$, $\Delta_{n+1} \subset \Delta_n$ and $|f(x_n+h) - f(x_n)| \ge \epsilon_0$ for all $h \in \Delta_n$.

The set $\cap_n \Delta_n$ is not empty and thus $\exists h_0 \in \cap_n \Delta_n$. We know $|f(x_n+h_0) - f(x_n)| \ge \epsilon_0$ and $x_n \to \infty$. Hence, $f(x+h) - f(x)$ doesn't converge to $0$ as $x \to \infty$. We got a contradiction. Lemma is proved.

It follows form lemma that for all $[a,b]$ and all $\epsilon > 0$ $\exists [a_{\epsilon}, b_{\epsilon}] \subset [a,b]$: $\overline{lim}_{x \to \infty} \sup_{h \in [a_{\epsilon}, b_{\epsilon}]} |f(x+h) - f(x)| < 2\epsilon$ and hence
$$ |f(x+h) - f(x)| < 2\epsilon$$
for all $h \in [a_{\epsilon}, b_{\epsilon}]$ and $x \ge x_0(\epsilon, a, b)$.

Let us show that $f$ is uniformly continuous on $[0, \infty)$.

Fix $\epsilon > 0$. Let $y, z$ be such real numbers that $x_0(\epsilon, 1, 2) + a_{\epsilon}  \le y \le z \le y + b_{\epsilon} - a_{\epsilon}$. Put $x =y -a_{\epsilon} $, $h_1= y-x$, $h_2 = z-x$. Thus $h_1 = a_{\epsilon} \in [a_{\epsilon}, b_{\epsilon}]$ and $ a_{\epsilon} = h_1 \le h_2 = z-x \le y + b_{\epsilon} - a_{\epsilon} -x = b_{\epsilon}$, $h_2 \in [a_{\epsilon}, b_{\epsilon}]$. We have $x \ge x_0(\epsilon, 1, 2)$,

$$|f(y) - f(x)| = |f(x + (y-x)) - f(x)| = |f(x+h_1) - f(x)| \le 2\epsilon,$$
$$|f(z) - f(x)| = |f(x + (z-x)) - f(x)| = |f(x+h_2) - f(x)| \le 2\epsilon,$$
$$|f(z) - f(y)| = |(f(z) - f(x)) - (f(y)-f(x))| \le |f(z) - f(x)| + |f(y) - f(x)| \le  4\epsilon.$$
But $[a_{\epsilon}, b_{\epsilon}] \subset [a,b] = [1,2]$ and thus $a_{\epsilon} \le 2$. Put $\delta_1(\epsilon) = b_{\epsilon} - a_{\epsilon}$. We proved that
if $y, z$ are such that $x_0(\epsilon, 1, 2) + 2  \le y,z$ and $|y-z| \le \delta_1$ then $|f(z) - f(y)| \le 4\epsilon.$

Moreover, we know that  $f$ is uniformly continuous on $[0, x_0(\epsilon, 1, 2) + 3]$. Hence $\exists \delta_2(\epsilon) > 0$ such that if $y, z \in [0, x_0(\epsilon, 1, 2)+3]$ and $|y-z| \le \delta_2$ then $|f(z) - f(y)| \le \epsilon$.

Put $\delta = \min(\delta_1, \delta_2, 1)$. If $|y-z| \le \delta$ then either $y,z \in [0, x_0(\epsilon, 1, 2) + 3]$ or $y, z \ge  x_0(\epsilon, 1, 2) + 2$. Thus,  $|f(z) - f(y)| \le 4\epsilon$ and $f$ is uniformly continuous on $[0, \infty)$.


问:
Write down the set in set build form $\{-1, 2, -3, 4, -5, 6...\}$
$$\{-1, 2, -3, 4, -5, 6...\}$$

I have been wondering how to rewrite the following set using the set build method because I have never seen such from before, the odd index is negative number while the even index is positive number

答: For another way to write it down, it's $$\{(-1)^nn | n \in \mathbb{N}\}$$


答:
Let $I = \int\sqrt{16-x^2} \ dx$. Then applying integration by parts with $u = \sqrt{16-x^2}, du = -\frac{x}{\sqrt{16-x^2}}$ and $dv = 1, v = x$:

$$I = x \sqrt{16-x^2} + \int \frac{x}{\sqrt{16-x^2}} \ dx$$

and now let $w = 16-x^2$.

Title: Evaluating $\int \sqrt{16-x^2} \,dx$

Body:
Are the steps below correct? Can you tell me another way to solve it, just hint?

$   \int \sqrt{16-x^2}   \,dx $

$ x = 4 \sin u \rightarrow  u=\arcsin(\frac{x}{4}) \rightarrow  dx = 4\cos u \,dx$

$   \int  \sqrt{16-x^2}   \,dx   = 16 \int \cos^2 u \,du = 16 \int \frac{\cos 2u +1}{2} \,du  $

$ v = 2u  \Rightarrow  dv = 2 du \Rightarrow du = \frac{1}{2}dv  $

$  16 \int \frac{\cos 2u +1}{2} \,du = 16 \int \frac{\cos v}{2} +\frac{1}{2} \,dv = 16(\frac{\sin v}{2} +\frac{v}{2}) = 8\sin v +8v = 8\sin 2u +16u  $

$ \Downarrow $

$8\sin(2\arcsin(\frac{x}{4})) + 4\arcsin\frac{x}{4} $


问:
Heun's method for a ball and beam system
I have a problem with a homework request by the professor at the university but I really can't solve the equation. I have just to resolve the equation using Heun method and choosing appropriately
the discretization step (the sampling time interval), adequately motivating the choices made.
It's a ball and beam system that after a bit of steps give me these equations:

$\dot x=\begin{bmatrix} \dot x_1\\  \dot x_2\\ \dot x_3 \\ \dot x_4\end{bmatrix}=\begin{bmatrix} x_2\\  x_1x_4^2-g\sin x_3\\ x_4 \\ \frac{-2mx_1x_2x_4-mgx_1\cos x_3+\tau}{mx_1^2+J}\end{bmatrix}= f(x,\tau)$

initial state condition given by the vector:

$ x_0=\begin{bmatrix} x_1 \\ 0 \\ 0 \\ 0\end{bmatrix}$

The only thing I understood is that I have to reduce that to a first grade equation imposing $ \frac{\partial x}{\partial t} = y$ but I don't know how to resolve then. Don't know how to proceed solving that problem with Heun method. Because I never used it and even if I see a lot of video online about it I can't resolve that problem up here. Only started using Matlab but don't know how to use it even if the prof said to us to use it if we want to resolve that. I never used finite difference methods unfortunately...

答:
You write the right side of your equation in Matlab format

```<code>  function dotx = ballbeam(t,x)
      dx2 = x(1)*x(4)^2 - g*sin(x(3));
      dx4 = (-2*m*x(1)*x(2)*x(4)-m*g*x(1)*cos(x(3))+tau)/(m*x(1)^2+J);
      dotx = [ x(2), dx2, x(4), dx4 ];
  end
</code>```

and then feed this into a generic Heun loop

```<code>  function x = odeheun(f,t,x0)
      x = [x0];
      for i=1:(length(t)-1)
          h = t(i+1)-t(i);
          k1 = h*f(t(i),x(i,:));
          k2 = h*f(t(i+1),x(i,:)+k1);
          x(i+1,:) = x(i,:)+0.5*(k1+k2);
      end%for
  end%function
</code>```

and then call it like (tested in octave)

```<code>  x0 = [x1_ini, 0, 0, 0 ];
  t = t0:h:tf;
  x = odeheun(@(t,x)ballbeam(t,x),t,x0); % possibly only @ballbeam in matlab
  plot(t,x(:,1));
</code>```

Body:
You write the right side of your equation in Matlab format

  function dotx = ballbeam(t,x)
      dx2 = x(1)*x(4)^2 - g*sin(x(3));
      dx4 = (-2*m*x(1)*x(2)*x(4)-m*g*x(1)*cos(x(3))+tau)/(m*x(1)^2+J);
      dotx = [ x(2), dx2, x(4), dx4 ];
  end

and then feed this into a generic Heun loop

  function x = odeheun(f,t,x0)
      x = [x0];
      for i=1:(length(t)-1)
          h = t(i+1)-t(i);
          k1 = h*f(t(i),x(i,:));
          k2 = h*f(t(i+1),x(i,:)+k1);
          x(i+1,:) = x(i,:)+0.5*(k1+k2);
      end%for
  end%function

and then call it like (tested in octave)

  x0 = [x1_ini, 0, 0, 0 ];
  t = t0:h:tf;
  x = odeheun(@(t,x)ballbeam(t,x),t,x0); % possibly only @ballbeam in matlab
  plot(t,x(:,1));


问:
Find the $1717$th term of the sequence of positive integers whose binary representation reads the same left to right as right to left.
The question is to find the number of $1$'s in the $1717$th term.

I solved it but my method is not elegant IMO and it goes like this.

Let the number of digits in one binary representation in this pattern be $x$. Let $f(x)$ be the amount of representations having $x$ digits.

Then $f(1)=f(2)=1$, $f(3)=f(4)=2$, $f(5)=f(6)=4=2^2, f(7)=f(8)=8=2^3 \cdots \cdots$

So, $2^n$ gives the amount of the binary representations with $2n+2$ and $2n+1$ digits.

And $1+1+2+2+\cdots 2^n+2^n=2(2^{n+1}-1)$ gives the total number of the binary representations from $1$-digit representation to $2n$-digit representations.

Hence, $2(2^{n+1}-1)=1717$, so $512\le 2^{n+1}=859.5\le 1024$. That means $8\le n\le 9$.

Until $n=8$, there are altogether $2(2^9-1)=1022$ numbers, and until $n=9$ there are altogether $2046$ numbers.

So, for the $1717$th term, it corresponds to $n=9$. Since there are $1024$ terms in the case where $n=9$, which is corresponding to $512$ representations with $19$ and another $512$ representations with $20$ digits, the $1717$th term should have $20$ digits, and it is in fact the $1717-512-1022=183$th term starting from the first $20$-digit representation.

Then since every representation is symmetric, we only need to consider what is going on in the first 10 digits.

Since it must start with $1$, and for each digit, it can be either $0$ or $1$, the $k$ in $2^k<183, k\in \mathbb Z$ implies where the "highest" $1$ is other than the starting $1$.

Since $k=8$, it is 1 _ 1 _ _ _ _ _ _ _.

Then since $2^{k_1}<183-128$ that means $k_1=5$, the representation is 1 _ 1 _ _ 1 _ _ _ _.

Similarly, I finally got 1 _ 1 _ _ 1 1 _ 1 1. That means the $1717$th term will have $12$ $1$'s.

Ok now could anyone offer a more elegant method?

答:
You need to find a way of calculating exactly the first half of the binary representation knowing the place, in this case $1717$. If you create the sequence of numbers that give you only the first half you find this one: A162751. There are some good properties about it, the most important one is that for $i \geq 2$ we have $a(2^i-1) = 2^{i-1}$. You could prove this by induction, having in mind that each "cycle" that we see in the sequence is just writing numbers with odd $0$ and $1$s in its binary representation first (which we know there is a particular power of $2$ of them) and after that the same repeated sequence but for ones with even number of $0$ and $1$. So you end up ending each cycle after a power of $2$, so it's natural to have a formula like the one before.

Now you can use this to obtain $a(1023) = 512$. We know the sequence of $512$ numbers after this one repeats, and we notice $1717$ is in the second half. We need to know where exactly it ends. You can see that the cycle repeats exactly in $2^i - 1 + 2^{i-1}$, in this case $i=10$ and it repeats in $1535 = 1023+512$, so $a(1535) = 512$. Until $1717$ the sequence just adds one, so $a(1717) = a(1535) + 1717-1535 = 512+182 = 694$. If you count the number of this number in binary ($1010110110$) you're counting the number of $1$ in the first half, so take that number, in this case six $1$, and double it to obtain $12$ ones.


问:
Does every ordinal have cardinality no greater than $\aleph_\mathbb{0}$?
My notes say that the ordinals $\omega + 1, \omega + 2, ... , 2 \omega, ... , 3 \omega, ... \omega^2, ... $ are all countable, and hence have cardinality equal to $\omega = \aleph_\mathbb{0}$. So I was wondering if it's fair to say that every ordinal has cardinality no greater than $\aleph_\mathbb{0}$?

Alternatively, I guess it's possible that that the sequence of infinite ordinals listed above, does not include some of the ordinals with a cardinality greater than the naturals.. but I wasn't sure.

答:
The ordinal $\omega_1$ has cardinality $\aleph_1$ which is greater than $\aleph_0$. Similarly, $\omega_2$, $\omega_3$,.... has cardinalities which are all greater than $\aleph_0$.

You can say that every countable ordinal has cardinality which is not greater than $\aleph_0$, in fact the cardinality is equal to $\aleph_0$.

Note that the ordinals $\omega + 1, \omega + 2, ... , 2 \omega, ... , 3 \omega, ... \omega^2, ... $ are all countable ordinal, however $\omega_1$ is the first uncountable ordinal.


问:
Write $x$ in terms of $y$ in inequality ${x}^{2}+x\le y$
How can I write the inequality ${x}^{2}+x\le y$ such that $?\le x\le ?$ (in terms of $y$) where $-2\le y\le 1$ ?

I tried $x(x+1)\le y\Rightarrow x\le \frac{y}{x+1}$ then became stuck.

Any help will be appreciated.

答: Rewrite the inequality $$x^2 + x \le y $$ as follows: $$ x^2 + x - y \le 0 $$ Then, by discriminant formula we get $$x_{1,2} = \frac{-1 \pm \sqrt{1+4y}}{2} $$ and therefore $$\left(x - \frac{-1 + \sqrt{1+4y}}{2}\right)\left(x - \frac{-1 - \sqrt{1+4y}}{2}\right) \le 0$$ Thus, $$\frac{-1 - \sqrt{1+4y}}{2} \le x \le \frac{-1 + \sqrt{1+4y}}{2}$$


答: $$x^2+x+\frac 14 \le y +\frac 14\\ (x+\frac 12)^2 \le y+\frac 14 \\ -\sqrt{y+\frac 14}\le x+\frac 12 \le \sqrt{y+\frac 14} \\-\frac 12 -\sqrt{y+\frac 14}\le x \le \sqrt{y+\frac 14}-\frac 12 $$

Title: Write $x$ in terms of $y$ in inequality ${x}^{2}+x\le y$

Body:
How can I write the inequality ${x}^{2}+x\le y$ such that $?\le x\le ?$ (in terms of $y$) where $-2\le y\le 1$ ?

I tried $x(x+1)\le y\Rightarrow x\le \frac{y}{x+1}$ then became stuck.

Any help will be appreciated.


问:
Let f be a Lebesgue integrable function on (0,1). Show that $g(x)=\int_{x}^{1} (\frac{f(t)}{t})dt $ is Lebesgue measurable.
The actual question asks to show that g is a Lebesgue integrable function on (0,1), but am I correct in assuming that I have to show that g is, first of all, a measurable function?
Or is it trivial in a way i'm not seeing?

I'm trying to show that the preimage of (a,+inf) through g is lebesgue-measurable for every real a, but i'm just going nowhere with it.
Thanks in advance for your help.

答:
You need to be careful in writing the domain of the integral and in explaining all the variables.

If the question is to show a function is integrable over a domain, yes, you should always check it is measurable. Usually, this follows from the fact the function is continuous, but this is not necessary.


问:
$A_n$ Can be Generated by 3-Cycles with Consecutive Elements
Let $ A_n $ be the alternating group on $n$ elements. I have already proven that $A_n$ can be generated by the set of 3-cycles with distinct elements. However, I am not sure how to further prove that $A_n$ can be generated by the set of 3-cycles with consecutive elements i.e. $ A_n = \langle (k,k+1,k+2) : 1 \leq k \leq n-2 \rangle$.

The natural thing to do is to show that an arbitrary 3-cycle with distinct elements can be generated by 3-cycles with consecutive elements, but I don't know how to do this.

答:
Induction on $n$ plus transitivity of $A_n$, $n\ge3$, settle this.

Base Case.

The claim is clear when $n=3$.

Inductive Step.

I will use the natural identification of elements of $A_n$ with those elements of $A_{n+1}$ that map the number $n+1$ to itself.

Assume that $n\ge3$ and that the claim holds for $A_n$. We show that the claim holds for $A_{n+1}$ as well.

Let $\sigma\in A_{n+1}$ be arbitrary. Let $d=\sigma(n+1)$. If $d=n+1$, then $\sigma\in A_n$. And by the induction hypothesis, it is a product of 3-cycles of consecutive elements involving $\{1,2,\ldots,n\}$ only.

Otherwise $d<n+1$. Because $A_n$ is transitive, there is an element $\alpha\in A_n$ such that $\alpha(d)=n$. It follows that
$$
\beta:=(n-1;n;n+1)\alpha\sigma\in A_n
$$
because it maps the number $n+1$ to itseld.

Solving for $\sigma$ from this gives
$$
\sigma=\alpha^{-1}(n-1;n;n+1)^2\beta.
$$
The induction hypothesis says that as elements of $A_n$ both $\beta$ and $\alpha^{-1}$ can be written in terms of the prescribed generators. Hence so can $\sigma$. As $\sigma$ was an arbitrary element of $A_{n+1}$, the inductive step is valid.

An alternative, possibly closer to what the OP had in mind, could be the following:

Brute force it for $A_4$.

In the inductive step we can then assume that $n\ge4$. Thus we know that $A_n$ is doubly transitive.

In the inductive step it suffice to show that all 3-cycles of the form $(a;b;n+1)$ can be generated. But, by double transitivity there exists an element $\alpha\in A_n$ such that $\alpha(n-1)=a$ and $\alpha(n)=b$. The usual conjugation rule of cycles says that
$$
\alpha (n-1;n;n+1) \alpha^{-1}=(a;b;n+1).
$$
Here both $\alpha$ and its inverse can be written in terms of the prescribed generators, so $(a;b;n+1)$ is included, and we are done.


问:
An equivalent condition for Normal spaces
Let $X$ be a normal space. By definition it is a $T_1$ space and any pair of disjoint closed sets can be separated by open sets in the sense that they have disjoint neighborhoods.

What I want to prove is

$X$ is normal iff each nbd of a closed set contains the closure of some nbd of the set

To prove the first implication I took a closed set $A$ and nbd $U$ of $A$. Since $U'$ is closed and disjoint from $A$ there exist nbd $H_A$ of $A$ and nbd $H_{U'}$ of $U'$ such that they are disjoint. I feel that closure of $H_A$ is subset of $U$ but not able to prove it.

For converse part, let $A$ and $B$ be closed disjoint sets in $X$. Then for any nbd $U_A$ and $U_B$ of $A$ and $B$ respectively there exist nbd $V_A$ and $V_B$ of $A$ and $B$ such that
$$
A \subseteq V_A \subseteq \bar{V_A} \subseteq U_A
$$
and

$$
B \subseteq V_B \subseteq \bar{V_B} \subseteq U_B
$$
If $U_A$ and $U_A$ are disjoint then we are done. If not how can I proceed further?

Please suggest any alternative approach also if possible.

答:
In this answer a neighborhood of a set $A$ is defined to be an open set that contains $A$ as subset.

You could also define a neighborhood of a set $A$ more broadly as a subset $N$ that satisfies $A\subseteq O\subseteq N$ for some open set $O$.

But it does not matter which of the definitions is used by defining normality.

Also what is proved below can easily be modified in such a way that we arrive at a proof applying to the broader definition.

$H_A\cap H_{U^{\complement}}=\varnothing$ tells us that $H_A\subseteq(H_{U^{\complement}})^{\complement}$ and $(H_{U^{\complement}})^{\complement}$ is closed.

So we conclude that: $$\mathsf{cl}(H_A)\subseteq(H_{U^{\complement}})^{\complement}$$

Here $U^{\complement}\subseteq H_{U^{\complement}}$ or equivalently $(H_{U^{\complement}})^{\complement}\subseteq U$ so we found that:$$\mathsf{cl}(H_A)\subseteq U$$

Let $A$ be a closed set and let $U_A$ be a neighborhood of $A$.

The sets $A$ and $U_A^{\complement}$ are closed and disjoint, so disjoint open sets $U, V$ exist with $A\subseteq U$ and $U_A^{\complement}\subseteq V$.

Then: $$A\subseteq U\subseteq V^{\complement}\subseteq U_A$$

Here $U$ is a neighborhood of $A$ and this with $\mathsf{cl}(U)\subseteq V^{\complement}$ so that also:$$A\subseteq U\subseteq \mathsf{cl}(U)\subseteq U_A$$


问:
Levy measure - $dx$ symbol
Hey I have problem with understanding notation of levy measure. For example for Variance Gamma process most authors writes $$\Pi_{VG}\left(dx\right)=C\frac{e^{-Mx}}{x}1_{\left\{ x>0\right\} }dx+C\frac{e^{Gx}}{\left\vert x\right\vert }1_{\left\{ x<0\right\} }dx$$ but what it is $dx$? Measure works on sets. Can anyone explain how to go from Levy measure to Levy density?

答: $$\Pi_{VG}\left(dx\right)=C\frac{e^{-Mx}}{x}1_{\left\{ x>0\right\} }dx+C\frac{e^{Gx}}{\left\vert x\right\vert }1_{\left\{ x<0\right\} }dx$$  means $$\Pi_{VG}\left(E\right)=C\int_E \frac{e^{-Mx}}{x}1_{\left\{ x>0\right\} }dx+C\int_E\frac{e^{Gx}}{\left\vert x\right\vert }1_{\left\{ x<0\right\} }dx$$ for all $E$ where $\int f(x)dx$ is integration w.r.t. Lebesgue measure.

Title: Levy measure - $dx$ symbol


答:
In probability theory, the following notation is commonly used:

Let $f \geq 0$ be a function. Then if one writes $\nu = f dx$ where $\nu$ is a measurue, it means that  $$\nu(A) = \int_A f dx$$
In your situation,
$$\Pi_{VG}(A) = \int_A C \frac{e^{-Mx}}{x} 1_{\{x > 0\}} dx + \int_A C \frac{e^{Gx}}{|x|} 1_{\{x < 0\}} dx$$

Title: Levy measure - $dx$ symbol

Body: Hey I have problem with understanding notation of levy measure. For example for Variance Gamma process most authors writes $$\Pi_{VG}\left(dx\right)=C\frac{e^{-Mx}}{x}1_{\left\{ x>0\right\} }dx+C\frac{e^{Gx}}{\left\vert x\right\vert }1_{\left\{ x<0\right\} }dx$$ but what it is $dx$? Measure works on sets. Can anyone explain how to go from Levy measure to Levy density?


问:
Why this ODE shows large values of errors in numerical solution??
As you can see in the figure, the numerical solution (RK4) vs analytical

for this function:

with exact solution :

,although the step size is increased, the error is so high when compared to exact solution.

答:
Your exact solution does not fit the differential equation. Divide by $(t+1)^3$ to get
$$
\frac{d}{dt}\frac{y(t)}{(t+1)^3}=\frac1{(t+1)^2}
$$
then integrate to
$$
\frac{y(t)}{(t+1)^3}=-\frac1{t+1}+1\implies y(t)=t(t+1)^2.
$$
As you can see, there is no exponential involved.

The given function is more a solution to
$$
\frac{d}{dt}\frac{y(t)}{t^2}=e^t\iff y'(t)=\frac{2y(t)}{t}+t^2e^t,~~y(1)=0
$$
(This equation is singular at $t=0$, so initial points have to be at $t\ne 0$, for instance where $(e^t-e)=0$.)


问:
Construction of a sequence of theorems with increasing and unbounded "difficulty"?
Let's define the "difficulty" of a theorem as the logarithm of the size of its shortest proof divided by the logarithm of the size of the theorem itself.  For example, if a theorem has difficulty less than 2, then it has a proof smaller than than the square of its size.  (Is there a standard terminology for this or a similar concept?)

Assume the theory is consistent, effective, and essentially incomplete (e.g. PA, ZFC).  Its difficulty is unbounded, because otherwise we would have an algorithm to determine provability (in NP even!), contradicting Church's theorem.

How to construct a specific sequence of theorems with increasing and unbounded difficulty?  My best idea is to iteratively apply Rosser's provability predicate, but I can't figure out how to prove that this works.  Any references for this or related problems?

答:
Perhaps you can use a Gödel statement stating "I am not provable with a proof of length N", where N is some huge number with a short definition.

By the way, the corresponding question for quantifier-free fragments (i.e. propositional calculus) is open - it's equivalent to NP=coNP.


问:
(Follow Up) Checking the solutions of a quadratic polynomial
I'm following up from this question:

Solve a polynomial involving geometric progression?

I have had trouble with this question:

"Solve the equation $8x^3−38x^2+57x−27=0$" if the roots are in geometric progression.

Any help would be appreciated.

I eventually solved the equation (please check @lab bhattacharjee's answer), but I got two answers for $r$, which were 3/2 and 2/3. How exactly do I know which is correct? I've been stuck for ages, so I'd prefer a full explanation rather than hints.

Any help would be appreciated.
Thanks.

答:
Solution :

Since it is a cubic polynomial you can find one of the root in the constant term of the equation : which is here 27 , so factors are 1,3 only. So by putting 1 in  equation it gives x -1 as one factor of the equation.

Therefore by dividing the complete polynomial with x -1 you get $8x^2-30x+27$ which after solving gives you two more factors

x = $\frac{3}{2}$ and  x =$\frac{9}{4}$ Threfore, 1, $\frac{3}{2} ; \frac{9}{4}$ are the roots of equation.

we can see roots are in G.P with common difference $\frac{3}{2}$


答:
The function $$h(t,x)=\frac{f(t)}{t}\boldsymbol 1_{\{(x,1)\}},$$

is Lebesgue meausrable on $(0,1)\times (0,1)$. So, if $h$ is integrable on $(0,1)\times (0,1)$, then, by Fubini theorem, $g$ will be Lebesgue measurable. Therefore, in this context, you don't have to worry about the measurability of $g$ and really see about integrability.

Title: Let f be a Lebesgue integrable function on (0,1). Show that $g(x)=\int_{x}^{1} (\frac{f(t)}{t})dt $ is Lebesgue measurable.

Body:
The actual question asks to show that g is a Lebesgue integrable function on (0,1), but am I correct in assuming that I have to show that g is, first of all, a measurable function?
Or is it trivial in a way i'm not seeing?

I'm trying to show that the preimage of (a,+inf) through g is lebesgue-measurable for every real a, but i'm just going nowhere with it.
Thanks in advance for your help.


问:
Finding the probability to get red balls in a basket of red and blue balls given that blue balls are returned when picked
In a box of 35 blue balls and 15 red balls, we draw a ball one by one and return the picked ball when it is not red.

What is the probability that it will take exactly 5 draws to pick a red ball?

What is the expected number of draws that it will take to finally get a red ball?

We had this seatwork and I'm not sure how to go around the problem now that it is essentially "returning" an item. Does the probability remain the same?

Lastly, what probability distribution can I use to model this problem?

答: Since the blue balls are returned, and you are looking for the first red ball to be extracted, then the situation is like if all the balls were returned, i.e. a sequence of Bernoulli trials.</p>


答:
For the first question: This follows a geometric distribution since this is a process involving Bernoulli trials  where picking a red ball is defined as success and picking any other ball is defined as failure, there is no replacement, and we need to find the probability that the first success is on the $k^{th}$ trial

Applying the geometric distribution formula we get the result as:

$(\frac{35}{50})^4 \cdot \frac{15}{50}$

Since we have $35+15=50$ balls in total, $35$ are blue and $15$ are red

The expected value of a geometric distribution is $\frac{1}{p}$ where $p$ is the probability of success which is $ \frac{15}{50}$ in our case so the expected value is $ \frac{50}{15}=\frac{10}{3}$  .

Title: Finding the probability to get red balls in a basket of red and blue balls given that blue balls are returned when picked

Body:
In a box of 35 blue balls and 15 red balls, we draw a ball one by one and return the picked ball when it is not red.

What is the probability that it will take exactly 5 draws to pick a red ball?

What is the expected number of draws that it will take to finally get a red ball?

We had this seatwork and I'm not sure how to go around the problem now that it is essentially "returning" an item. Does the probability remain the same?

Lastly, what probability distribution can I use to model this problem?


问:
Order of elements in a group.

Corollary 4.6.8. There is a group $G$ of order $n^3$ given by $G = \{b^ic^ja^k | 0 ≤ i, j, k < n\}$, where $a$, $b$, and $c$ all have order $n$, and $b$ commutes with $c$, $a$ commutes with $c$, and $aba^{−1} = bc$. Thus,

$$(b^ic^ja^k) (b^{i'}c^{j'}a^{k'}) = b^{i+i'}c^{j+j'+ki'}a^{k+k'}.$$

a) Consider the groups constructed in Corollary 4.6.8 for $n$ a prime number $p$. When $p>2$, show that every element in this group has exponent $p$. (Why is this different from the case $n=2$?) Recall that in Problem 3 of Exercises 2.1.16 it was shown that any group of exponent $2$ must be abelian. As we see here this property is not shared by primes $p>2$.

b) Now consider the groups of Corollary 4.6.8 for general values of $n$. For which values of $n$ does the group in question have exponent $n$? If the exponent is not equal to $n$, what is it?

This is what I did:

I first tried to see if there was any pattern that I could detect by multiplying the elements $b^ic^ja^k = b^ia^kc^j$ over an over again. This is what I got:

$(bac)(bac) = b^2a^ac^{2+1}$

$(b^2a^ac^{2+1})(bac) = b^3a^3c^{4+2}$

$(b^3a^3c^{4+2})(bac) = b^4a^4c^{7+3}$

$(b^4a^4c^{7+3})(bac) = b^5a^5c^{11+4}$

But I wasn't sure if this would help me in finding the order of bca...however, the corollary tells us that $aba^{-1}=bc$, so $c=b^{-1}aba^{-1}$, right?

So I tried using that, and we have:

$(bca) = bb^{-1}aba^{-1}a = ab$. However, this doesn't really help much either.

So I was wondering if anybody could help me with this...

Thank you in advance.

答: I make it $(a^ic^jb^k)^m = a^{mi}b^{mk}c^{mj+m(m-1)ki/2}$. So the exponent is $n$ when $n$ is odd and $2n$ when $n$ is even.</p>

Title: Order of elements in a group.


问:
Find the minimum value of the ratio $\frac{S_1}{S_2}$ using given data

3 points $O(0, 0) , P(a, a2 ) , Q(b, b2 )$ are on the parabola $y=x^2$. Let S1 be the area bounded by the line PQ and the parabola and let S2 be the area of the triangle OPQ, then find min of $\frac{S1}{S2}$

$$S_1 = \frac 12 (a+b)(a^2+b^2) -\int_{-b}^a x^2.dx$$
$$=\frac 16 (a^3+b^3) +\frac 12 (ab^2+a^2b)$$

And $$S_2=\frac 12 (a^2b+ab^2)$$
How do I solve from here?

Body:
$$
\dfrac{S_1}{S_2}=\frac{\frac{1}{2} \left(a^2 b+a b^2\right)+\frac{1}{6} \left(a^3+b^3\right)}{\frac{1}{2} \left(a^2 b+a b^2\right)}=\frac{(a+b)^2}{3 a b}\geq \dfrac{4ab}{3ab}=\dfrac{4}{3}
$$


问:
Show that the midpoints form a parallelogramm
Let $\triangle ABC$ be a triangle with centroid $S$.

Show that :

(a) The midpoints of $\overline{AS}, \overline{BS}, \overline{BC}, \overline{AC}$ form a parallelogramm.

(b) If $\triangle ABC$ has two medians of the same length, then it is isosceles.

$$$$

For (a) :

Let $M_{AS}, M_{BS}, M_{BC}, M_{AC}$ the midpoints.

Then we have to show that $M_{AS}M_{AC}=M_{BS}M_{BC}$ and that $M_{AS}M_{AC} \parallel M_{BS}M_{BC}$, right?

For that do we have to consider similar triangles?

答: All you have to use that red segments are parallel and that bigger is twice as smaller (because of similarity of triangles $ABC$ and $ADE$).


答: In general, if you have a fraction involving $a,b$ where each term in the numerator and denominator has the same degree $d$, you can divide through by say $b^d$ to get an expression in $\frac{a}{b}$. Here, this would look like $$\frac{S_1}{S_2} =\frac{ \frac 16 (a^3+b^3) +\frac 12 (ab^2+a^2b)}{\frac 12 (a^2b+ab^2)}  =\frac{ \frac 16 ((a/b)^3 +1) +\frac 12 (a/b+a^2/b^2)}{\frac 12 (a^2/b^2+a/b)} $$ Let $a/b =t$, then $$\frac{S_1}{S_2} = \frac{ t^3/6 +1/6 + t/2 +t^2/2}{t^2/2 +t/2} $$ And then you can follow the usual optimization procedure.

Title: Find the minimum value of the ratio $\frac{S_1}{S_2}$ using given data

Body:
3 points $O(0, 0) , P(a, a2 ) , Q(b, b2 )$ are on the parabola $y=x^2$. Let S1 be the area bounded by the line PQ and the parabola and let S2 be the area of the triangle OPQ, then find min of $\frac{S1}{S2}$

$$S_1 = \frac 12 (a+b)(a^2+b^2) -\int_{-b}^a x^2.dx$$
$$=\frac 16 (a^3+b^3) +\frac 12 (ab^2+a^2b)$$

And $$S_2=\frac 12 (a^2b+ab^2)$$
How do I solve from here?


问:
Application of derivative - Want to know the relation of graph with this problem
Problem :

For the curve $y = 3 \sin \theta \cos \theta, x = e^\theta \sin \theta, 0 \leq  \theta \leq \pi $ ; the tangent is parallel to x-axis when $\theta$  is

a) $0$

b) $\dfrac{\pi}{2}$

c)  $\dfrac{\pi}{4}$

d) $\dfrac{\pi}{6}$

Solution : We will find $\dfrac{dy}{d\theta}$ and $\dfrac{dx}{d\theta}$ with the help of this we will find $\dfrac{dy}{dx}$ which should be equal to 0 so that slope will be parallel to x axis.

$\dfrac{dy}{d\theta} = -3(\cos^2\theta -\sin^2 \theta)$ ;

$\dfrac{dx}{d\theta} = e^{\theta} \cos\theta +\sin\theta e^{\theta}$

$\Rightarrow \dfrac{dy}{dx} = \dfrac{-3(\cos^2\theta-\sin^2\theta}{e^{\theta} \cos\theta +\sin\theta e^{\theta}}$

If we put $\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -3(\cos\theta-\sin\theta) = 0  \Rightarrow \cos\theta = \sin\theta $

$\therefore \theta = \dfrac{\pi}{4}$

Now my question is how can we show $\dfrac{dy}{d\theta} $ and $ \dfrac{dx}{d\theta}$ graphically.   Thanks...

答:
Consider the parameter $\theta$ of the curve as time, and the curve itself at a given moment $\theta$ as the location of an ant, say. Then the vector $(\frac{dx}{d\theta},\,\frac{dy}{d\theta})$ is just the velocity vector of the moving of the ant.

A path can be walked through in many-many ways, choosing other magnitudes for velocities, however, it is always true that at the point $(x(\theta),\,y(\theta))$, the velocity vector is tangent to the path.


问:
Identity between characteristic polynomial of matrix and its derivative (using the determinant lemma)
Let $A = \text{diag}(\lambda_1,\dots,\lambda_n)$ be a diagonal $n \times n$ matrix and $J:=\textbf{1}\textbf{1}^T$ be the all-ones square matrix. Using the determinant lemma, we can write

$$\det(xI-A-J)=\det(xI-A)\cdot(1-\textbf{1}^T(xI-A)^{-1}\textbf{1})$$

Let $p(x):=\det(xI-A)$ be the characteristic polynomial of $A$. What can I do to build a relation (equality) between $p$ and $p'$ (the first derivative of $p$). Concretely, I am looking  to show

$$\det(xI-A-J)=p(x)-p'(x)$$

which thanks to the lemma is equivalent to showing

$$p'(x)=p(x)\cdot\sum\limits_{i=1}^n\frac{1}{x-\lambda_i}$$

but I don't know how to tackle this. I cannot compare coefficients because a priori the RHS is not a polynomial. Any ideas/directions I can go?

答:
Write polynomial as $$ p(x) = a(x-x_1)^{n_1}(x-x_2)^{n_2}\cdots (x-x_k)^{n_k}$$ Now take it natural logarithm:

$$ \ln p(x) = \ln a +n_1\ln (x-x_1) + n_2\ln (x-x_2)+\cdots + n _k\ln (x-x_k) $$

Now, calculate the derivative of it...


问:
Can I say if $A$ is a projection then $\ker A= (\text{Im } A)^{\perp}$?

Suppose $A\in M_{n \times n}$ and $A^2=A$. Prove that $\ker A= (\text{Im } A)^{\perp}$.

I want to show this statement is true or not. I know that $A$ is a orthogonal projection $\iff$ $A=A^2=A^*$.

Body:
Consider
$$
\pmatrix{1 & 1 \\ 0 & 0}.
$$


问:
$a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$
I have not solved inequalities in a while, so I am a little rusty. Could you help me with this inequality I have found?
$$a^2 + b^2 + c^2 + 6 \ge 3(a + b + c),$$ where $a, b, c > 0$ and $abc = 1$

My initial idea was $a ^ 2 + 2 \ge 2\sqrt 2a$ and the inequalities with $b $ and $c$, then adding these 3, we get, $a ^ 2 + b ^ 2 + c ^ 2 + 6 \ge 2\sqrt 2(a + b+ c)$, but then we get to $2\sqrt2 > 3$, which is false.

Edited: I found some variants of the original problem.

Problem 1: Let $a, b, c > 0$. Prove that $a^2 + b^2 + c^2 + 6 + (abc - 1) \ge 3(a+b+c)$.

Problem 2: Let $a, b, c$ be reals with $abc \le 1$. Prove that $a^2 + b^2 + c^2 + 6 \ge 3(a+b+c)$.

答:
My third solution:

Actually $a^2 + b^2 + c^2 + 6 - 3(a+b+c) + (abc - 1) \ge 0$ for all $a, b, c \ge 0$.

(pqr method)

Let $p = a + b + c$, $q = ab + bc + ca$ and $r = abc$.

By Schur's inequality $a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b) \ge 0$
which is written as $p^3 - 4pq + 9r \ge 0$, we have $\frac{p^3 + 9r}{4p}\ge q$.

We need to prove that $p^2 - 2q + 6 - 3p + r - 1 \ge 0$.

It suffices to prove that
$p^2 - 2 \cdot \frac{p^3 + 9r}{4p} + 6 - 3p + r - 1 \ge 0$ or
$$\frac{(2p-9)r}{2p} + \frac{1}{2}p^2 - 3p + 5 \ge 0. \tag{1}$$

If $2p- 9 \ge 0$, clearly (1) is true.

If $2p - 9 < 0$, since $\frac{p^3}{27} \ge r$, it suffices to prove that
$$\frac{(2p-9)\frac{p^3}{27}}{2p} + \frac{1}{2}p^2 - 3p + 5 \ge 0$$
that is
$$\frac{1}{27}(p+15)(p-3)^2 \ge 0.$$

We are done.


问:
Jensen Inequality Euclidean Norm
I'm trying to prove $$\mathbb{E}\|X\|\geq \|\mathbb EX\| $$ for any random matrix $X$ and the spectral norm $\|\cdot \|$. To finish my proof all I need is the same statement for the euclidean norm $|\cdot|$ of a random vector $v$:
$$\mathbb{E}|v|\geq |\mathbb Ev|$$
But I am stuck trying to prove this. I am familiar with the one-dimensional Jensen inequality. Am I simply missing the right application of it here?

答: For any unit vector $u$ we have $\mathbb E |v| \geq \mathbb E  \langle u,v \rangle = \langle u,\mathbb Ev \rangle $. The supremum of the right side over all unit vectors $u$ is equal to $|\mathbb E v|$.

Title: Jensen Inequality Euclidean Norm