问:
Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $
I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k}  = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series

$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}  = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$

Does it converge?   If so, what is its sum?

答:
Grant Sanderson, aka 3Blue1Brown, has a good explanation of this in one of his Lockdown Math videos. His explanation, summarized:

$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots = f(1)$, where $f(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} -\ldots$

$\frac{df}{dx} = 1 - x + x^2- x^3 + x^4 - \ldots = \frac{1}{1+x}$ (when $-1 ≤ x ≤ 1$)

$f(x) = \ln(1+x)$

$\therefore f(1) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots = \ln(2)$


问:
how is this function a sum of these two functions?
So chui's book on wavelets has this diagram:

I don't understand what he means by the highlight part here; aka I don't really see how the first two walsh basis functions are equal to the sums he's talking about.

答:
Let
$$\phi_H(t) = \begin{cases}1 & \text{, if } t \in [0,1)\\
0 & \text{, otherwise}
\end{cases}$$
and
$$\psi_H(t) = \begin{cases}1 & \text{, if } t \in [0,\frac 12)\\
-1 & \text{, if } t \in [\frac 12,1)\\
0 & \text{, otherwise}
\end{cases}.$$
The top left graph of Fig. 1.4 corresponds to $$\phi_H(t)+\phi_H(t-1) = \begin{cases}1 & \text{, if } t \in [0,2)\\
0 & \text{, otherwise}
\end{cases}$$
periodically extended with a period of $2$.

Similarly the top right graph of Fig. 1.4 corresponds to the $2$-periodic extension of $$\psi_H(t)+\psi_H(t-1) =
\begin{cases}
1 & \text{, if } t \in [0,\frac 12) \cup [1,\frac 32)\\
-1 & \text{, if } t \in [\frac 12,1) \cup [\frac 32,2)\\
0 & \text{, otherwise}
\end{cases}.$$

ContentLicense: CC BY-SA 4.0


问:
Predicate logic for statements about functions?
Given the relation $R$ defined on the cartesian map $A\times A$ where $|A|=n$.

How to use the predicate logic to express the statements about functions?

Examples.

The relation R corresponds to a function from A to A.

The relation R corresponds to an injective function from A to A.

The relation R corresponds to and bijective function from A to A.

where for each part above respectively I think them as

$\forall a\in A$, there exist $c\in A,s.t.(a,b)\in R$.

$\forall a,b\in A$, there exist $c,d\in A,s.t. (a,c),(b,d)\in R\& (a\ne b\to c\ne d)$

i8t is surjective with the statement in ii).

Are the above interpretations right?

答:
No, they aren't:

The first statement you wrote would be satisfied if the set $A$ were ${1,2}$ and $R$ were defined as ${(1,1),(1,2),(2,2)}$. That is, you need to require also that $b$ is unique.

For the second statement, I think you almost have it but I've usually seen it phrased the other way, that is you start by asserting that $R$ is a function, (using what you said for the first statement) and then say that $c=d \to a=b$. But if you fix your first statement, your way works too.

For the third statement, yes, I'd just say the second statement and surjective (which is just $\forall b \in A, \exists a \in A {\rm s.t.} (a,b) \in R$ )

ContentLicense: CC BY-SA 3.0


问:
Quotienting $\mathbb Z[x]$ by a maximal ideal gives a finite field.
Let $R = \mathbb Z[x]$, a polynomial ring over $\mathbb Z$, and $\mathfrak m$ be any maximal ideal of $R$. How do we show that $R / \mathfrak m$ is a finite field? I know that the fact that it is a finite field directly follows, but not sure about the finite part.

答:
Let $A=R/\mathfrak{m}$ and $F$ be its prime subfield. Assume $F$ is finite, then $A$ is generated over $\mathbb{Z}$ by a single element, so the same holds for $A$ over $F$. In other words, $A$ is a quotient of $F[x]$ by a maximal ideal (as $A$ is a field) so is finite.

So we only need to show that $F$ cannot be $\mathbb{Q}$. Indeed, assume for the sake of contradiction that $F=\mathbb{Q}$. Then $\mathfrak{m}\mathbb{Q}[x]$ is a maximal ideal of $\mathbb{Q}[x]$ (because it’s a localization of $\mathbb{Z}[x]$ at a multiplicative subset not meeting $\mathfrak{m}$), and the quotients are equal.

There is an irreducible polynomial $f(x) \in \mathfrak{m}$, and $(f) \subset \mathbb{Q}[x]$ is a maximal ideal, so $\mathfrak{m}\mathbb{Q}[x]=(f)$, so $\mathfrak{m}=f(x)\mathbb{Q}[x] \cap \mathbb{Z}[x]$.

Now, let $g(x)=h(x)/D$ be a rational polynomial with $D > 1$ integer, $h(x) \in \mathbb{Z}[x]$ primitive (ie its coefficients do not have a nontrivial common divisor – note that $f$ is also primitive, since it’s irreducible) and assume $f(x)g(x) \in \mathbb{Z}[x]$. Then, if $p$ prime divides $D$, $f(x)h(x)=0$ mod $p$, so $f(x)=0$ mod $p$ or $h(x)=0$ mod $p$, both impossible. So $\mathfrak{m}=f\mathbb{Q}[x] \cap \mathbb{Z}[x]=f\mathbb{Z}[x]$.

But then, let $n \geq 1$ be an integer, then $n$ is invertible mod $\mathfrak{m}$, so there is a polynomial $p$ such that $f|np-1$ in $\mathbb{Z}[x]$. Take $n=|f(N)|$ where $N$ is a large integer: $f(x)|np(x)-1$ so in particular $f(N)|np(N)-1$, thus $f(N)|1$, hence a contradiction.


问:
Prove $\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$
Proof of the integral
$$\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$$
I try to substitute $u = \ln x$. Then $x = e^u,\>\mathrm{d}x = e^u\mathrm{d}u$ and the limits $(0,\infty)\to (-\infty,\infty)$.
The integral becomes $$\int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1}\mathrm{d}u.$$

答:
Carrying on from where the OP left off, we have

$$\begin{align}
\int_{-\infty}^\infty{ue^u\over e^{3u}-1}du
&=\int_0^\infty{ue^u\over e^{3u}-1}du+\int_{-\infty}^0{ue^u\over e^{3u}-1}du\\
&=\int_0^\infty{ue^{-2u}\over1-e^{-3u}}du+\int_\infty^0{ue^{-u}\over e^{-3u}-1}du\\
&=\int_0^\infty{ue^{-2u}\over1-e^{-3u}}du+\int_0^\infty{ue^{-u}\over1-e^{-3u}}du\\
&=\int_0^\infty u(e^{-2u}+e^{-5u}+e^{-8u}+\cdots+e^{-u}+e^{-4u}+e^{-7u}+\cdots)du\\
&={1\over2^2}+{1\over5^2}+{1\over8^2}+\cdots+1+{1\over4^2}+{1\over7^2}+\cdots\\
&=\left(1+{1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots \right)-\left({1\over3^2}+{1\over6^2}+{1\over9^2}+\cdots \right)\\
&=\left(1+{1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots \right)-{1\over9}\left(1+{1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots \right)\\
&={8\over9}\zeta(2)\\
&={8\over9}{\pi^2\over6}\\
&={4\pi^2\over27}
\end{align}$$

Remark: The final step(s) require knowing that $\zeta(2)=\pi^2/6$, which may not be in the OP's toolbox. All other steps are standard manipulations of (improper) integrals and infinite series, with the geometric series rearing its beautiful head in the middle.


问:
How would I prove that n^2 is O(n!) in terms of witnesses
How would I use c and n0 to prove this relationship where n0 - 1 and c are NOT witnesses, I can't think of a pair that would satisfy this relationship.</p>

答:
Do you mean like this?

For $c=1$ we have
$$
n^2 < n!\qquad\text{for all }n \ge 4
$$
but
$$
3^2 \not< 3!
$$


答:
You can prove by induction that $n^{2} \leq n!$ for $n \geq 4$. Here, $n_{0} = 4$ and $C = 1$.

Note that this only provides that $n^{2} \in O(n!)$. It does not show that $n^{2} \not \in \Theta(n!)$. To obtain that $n^{2} \not \in \Theta(n!)$, you would want to use the limit comparison test. Namely, show that:
\begin{align*}
\lim_{n \to \infty} \frac{n^2}{n!} = 0.
\end{align*}

This implies that $n^{2} \in o(n!)$. So you obtain that $n^{2} \in O(n!)$ and $n^{2} \not \in \Theta(n!)$.

Title: How would I prove that n^2 is O(n!) in terms of witnesses

Body: How would I use c and n0 to prove this relationship where n0 - 1 and c are NOT witnesses, I can't think of a pair that would satisfy this relationship.</p>


问:
Second order hyperbolic PDE with constant coefficients (analytical solution)
In the process of solving a continuous-time Markov chain problem, I derived the following PDE for the probability distribution $P_0(t,T)$, $t\in[0, T], T\in \mathbb{R}$:
\begin{align}
\frac{\partial^2 P_0}{\partial T^2} + \frac{\partial^2 P_0}{\partial t \, \partial T} + (k_{\text{on}} + k_{\text{off}}) \frac{\partial P_0}{\partial T} + k_{\text{on}} \frac{\partial P_0}{\partial t} = 0.
\end{align}
The rates $k_{\text{on}}$ and $k_{\text{off}}$ are positive numbers. This equation is augmented by the condition
\begin{align}
P_0(0,T) = p_0^{\text{ss}} e^{-k_{\text{on}}T}.
\end{align}
I don't have much experience solving PDEs, so I looked up some worked out examples and following  their approach reduced the equation into its canonical form given by
\begin{align}
\frac{\partial^2 P_0}{\partial \eta \partial t} + (k_{\text{on}} + k_{\text{off}}) \frac{\partial P_0}{\partial \eta} + k_{\text{on}} \frac{\partial P_0}{\partial t} = 0,
\end{align}
where $\eta = T-t$. Then, I used the separation of variables approach ($P_0(t, \eta) = Q(t)E(\eta)$), which is what I would normally do to solve, say, the diffusion equation. Unfortunately, the solution I get leads to unphysical results when I impose the 'initial' condition. Skipping some of the steps, I find that
\begin{align}
Q(t)E(\eta) = C \exp\left( -\frac{k_{\text{on}} + k_{\text{off}}}{1+c_e}t + \frac{k_{\text{on}}}{c_e} (T-t) \right),
\end{align}
where $C$ and $c_e$ are constants. Imposing the condition leads to $C=p_0^{\text{ss}}$ and $c_e=-1$. This second number is what blows the solution. It suggests that $1+c_e=0$ and the denominator in the exponent makes the whole expression diverge.

Any suggestions on how to approach this problem? Can it be that the separation of variables does not apply in this case? What other methods can I try? Thanks in advance!

[Edit] There is also a second condition that can be helpful in solving this: $P_0(T,T)=0$.

答:
Simplifying your notation to make the typing a bit easier you have
$$P_{TT} + P_{tT} + (k_1 + k_2)P_T + k_1P_t = 0$$
subject to $P(0, T) = p_0 e^{-k_1 T}$. You're on the right track separating
the variables, $P(t, T) = A(t)B(T)$.

Doing this I get $B' + cB = 0$, where $c$ is some constant, and $ A'' + (k_1 + k_2 - c)A' - ck_1 A = 0$.
Using the initial condition, the former tells us $c = k_1$, so the latter becomes
$A'' + k_2 A' - k_1^2 = 0$.

Solve the characteristic equation associated with this ordinary differential equation. You'll
see it has roots $-\tfrac{1}{2}k_2 \pm \tfrac{1}{2}\sqrt{k_2^2 + 4k_1^2}$. Since your problem
seems to deal with probablility distributions I imagine you'll want to pick the negative root
so that $A \rightarrow 0$ as $t \rightarrow \infty$,
so with $\kappa = \tfrac{1}{2}k_2 + \tfrac{1}{2}\sqrt{k_2^2 + 4k_1^2}$ your solution is
$$
P(t, T) = p_0 e^{-k_1T} e^{-\kappa t}.
$$


答:
Not exactly.

i) $\phi:=$"$\forall a\,\exists b:aRb\land \,\big(\forall c: (aRc\Rightarrow b=c)\big)$"

ii) $\psi:=\phi\land$ "$\forall a,b,c: (aRc\land bRc)\Rightarrow (a=b)$"

iii) $\vartheta:=\psi\land$ "$\forall b\,\exists a: aRb $"

Title: Predicate logic for statements about functions?

Body:
Given the relation $R$ defined on the cartesian map $A\times A$ where $|A|=n$.

How to use the predicate logic to express the statements about functions?

Examples.

The relation R corresponds to a function from A to A.

The relation R corresponds to an injective function from A to A.

The relation R corresponds to and bijective function from A to A.

where for each part above respectively I think them as

$\forall a\in A$, there exist $c\in A,s.t.(a,b)\in R$.

$\forall a,b\in A$, there exist $c,d\in A,s.t. (a,c),(b,d)\in R\& (a\ne b\to c\ne d)$

i8t is surjective with the statement in ii).

Are the above interpretations right?


问:
Change of a scalar field/vector field
In my book the following is written:

The change of a scalar field $du$ in an arbitrary direction, given by an infinitesimal vector with an arbitrary direction $d\vec r$ is calculated:
$$du=u(\vec r + d\vec r) - u(\vec r)= d\vec r\cdot\nabla u.$$
And for the vector field $\vec v$ we have:
$$d\vec v= \vec v(\vec r + d\vec r) - \vec v(\vec r)= (d\vec r \cdot\nabla) \vec v.$$

Can someone explain:

How do we derivate this!

A visualization of both of them! or a link to the asked topic.

答:
It's a direct application of the chain rule for multi-variable calculus. Perhaps using summatory notation clarifies further the point:
$$
dv^{i}=v^{i}\left(\boldsymbol{r}+d\boldsymbol{r}\right)-v^{i}\left(\boldsymbol{r}\right)\simeq\sum_{k}dr^{k}\frac{\partial v^{i}}{\partial r^{k}}
$$
As to an intuitive picture of what it is, it is how much the i-th component of $\boldsymbol{v}$ varies in the direction of the small displacement vector $d\boldsymbol{r}$, so it's natural to project gradient of $v^i$ along such small vector.

OwnerDisplayName: imbAF

Title: Change of a scalar field/vector field

OwnerDisplayName: joigus


问:
Determine the following sum
Let $S(n)$ be the sum of the digits of n in its binary representation. For example the binary notation for 19 is $10011$ and $S(19)=3$. Determine $\sum_{n=1}^{\infty} \frac{S(n)}{n(n+1)}$?

I've made several tables trying to pinpoint any useful pattern, but I've had minimal luck. The best I'd came to was determining that the number of digits is $\left \lceil{\frac{ln(n)}{ln(2)}}\right \rceil$ which gives a bound for $S(n)$. What I'd picked up from those tables is $S(2^k-1)=k$ and $S(2^k)=1$. However, breaking the sum up into those n's that are powers of 2 proved to be a bit of a disaster.

Title: Determine the following sum

Body:
The key idea is to expand over digits and find a recurrence.  Let $$\require{cancel}F(N)=\sum_{n=1}^{2^N}{\frac{S(n)}{n(n+1)}}=\sum_{n=1}^{2^N}{S(n)\left(\frac{1}{n}-\frac{1}{n+1}\right)}$$ and \begin{align*}
G(N)&=\sum_{n=0}^{2^{N-1}-1}{\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)}-\left(\frac{1}{2^N+1}-\frac{1}{2^N+2}\right) \\
&=\sum_{n=1}^{2^N}{\frac{(-1)^{n+1}}{n}}-\frac{1}{(2^N+1)(2^N+2)}\text{;}
\end{align*} we know that $\lim_{N\to\infty}{G(N)}=\ln{(2)}$ and want to find $\lim_{N\to\infty}{F(N)}$.

We can compute \begin{align*}
F(N)&=\sum_{n\in[2^N],2\mid n}{S(n)\left(\frac{1}{n}-\frac{1}{n+1}\right)}+\sum_{n\in[2^N],2\nmid n}{S(n)\left(\frac{1}{n}-\frac{1}{n+1}\right)} \\
&=\sum_{n=1}^{2^{N-1}}{S(2n)\left(\frac{1}{2n}-\frac{1}{2n+1}\right)}+\sum_{n=0}^{2^{N-1}-1}{S(2n+1)\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)} \\
&=\sum_{n=1}^{2^{N-1}}{S(n)\left(\frac{1}{2n}-\frac{1}{2n+1}\right)}+\sum_{n=0}^{2^{N-1}-1}{(1+S(n))\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)} \\
&=\sum_{n=1}^{2^{N-1}}{S(n)\left(\frac{1}{2n}-\cancel{\frac{1}{2n+1}}+\cancel{\frac{1}{2n+1}}-\frac{1}{2n+2}\right)}-\left(\frac{1}{2^N+1}-\frac{1}{2^N+2}\right)+ \\
&\phantom{{}={}}\quad\sum_{n=0}^{2^{N-1}-1}{\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)} \\
&=\frac{F(N-1)}{2}+G(N) \\
&=\frac{F(0)}{2^N}+\frac{G(1)}{2^{N-1}}+\dots+\frac{G(N-1)}{2}+G(N) \\
&=\frac{1}{2^{N+1}}+\sum_{k=0}^{N-1}{\frac{G(N-k)}{2^k}}
\end{align*}  Thus $$|F(N)-\ln{(2)}|\leq\frac{1}{2^{N+1}}+\sum_{k=0}^{N-1}{\frac{|G(N-k)-\ln{(2)}|}{2^k}}+\sum_{k=N}^{\infty}{\frac{\ln{(2)}}{2^k}}$$  The outer two terms are clearly tending to $0$ as $N\to\infty$, so it suffices to show the same for the inner.

Fix $\epsilon$.  There exists $K$ such that, for $k\geq K$, we have $|G(k)-\ln{(2)}|\leq\epsilon$; let $M=\sup_{k<K}{G(k)}$.  Then \begin{align*}
\sum_{k=0}^{N-1}{\frac{|G(N-k)-\ln{(2)}|}{2^k}}&=\sum_{k=1}^N{\frac{|G(k)-\ln{(2)}|}{2^{N-k}}} \\
&\leq\sum_{k=0}^{K-1}{\frac{M}{2^{N-k}}}+\sum_{k=K}^N{\frac{\epsilon}{2^{N-k}}} \\
&\leq\frac{(2^K-1)M}{2^N}+\frac{\epsilon}{2^{K-1}}
\end{align*}  Thus $\limsup_{N\to\infty}{|F(N)-\ln{(2)}|}\leq\epsilon$; now take $\epsilon\to0^+$ to get $$\lim_{N\to\infty}{F(N)}=\ln{(2)}$$

Of course, we selected a subsequence of partial sums to get $F(N)$, so theoretically the unselected partial sums could exhibit $\Theta(1)$ variation from $F(N)$.  But each term in your original sum is nonnegative, so in fact that can't happen.


问:
Is $\int_{-1}^9 (2x-1)^4\mathrm{d}x=\frac{1}{2}\int_{-1}^9 (2x-1)^4\mathrm{d}(2x-1)$ correct?
When solving $\displaystyle \int_{-1}^9 (2x-1)^4\mathrm{d}x$, I often find someone doing $\displaystyle \frac{1}{2}\int_{-1}^9 (2x-1)^4\mathrm{d}(2x-1)$. Are the boundary values still correct? Or should we change them as $\displaystyle \frac{1}{2}\int_{-3}^{17} (2x-1)^4\mathrm{d}(2x-1)$?

答:
For changing variable, in appropriate conditions, we have general formula
$$\int\limits_{\phi(\alpha)}^{\phi(\beta)}f(x)dx=\int\limits_{\alpha}^{\beta}f(\phi(t))\phi'(t)dt$$
in your case taking $x=\phi(t)=\frac{t+1}{2}$ we have mapping $\phi \colon [-3, 17]\to [-1,9]$ with $\phi(-3)=-1$ and $\phi(17)=9$. Obviously $t=\phi^{-1}(x)=2x-1$.


问:
Is the graph of $f$, $G_{r}(f)$, is connected?
Consider the Cantor set $C$ in $[0,1]$. And a function $f: [0,1] \rightarrow [-1,1]$ defined by

$$f(x) = \frac{2(x-a)}{b-a}-1 \text{ if } x \in [a,b] \text{ where }(a,b) \text{ is a contiguous interval of } C, \quad f(x) = 0 \text{ otherwise }$$

Is the $G_{r}(f)$ is connected ?, where $G_{r}(f)=\{(x,f(x)): x \in [0,1]\}$. We know that $f$ is not continuous.

I supposed it doesn't. if we define $U= \{(x,0): x \in C \}$ and $V=\{ (x, f(x)): x \in [a,b], (a,b) \text{ contiguous interval }\}$

we can check $\overline{U} \cap V = U \cap \overline{V}= \emptyset$ and $G_{r}(f)=U \cup V$ then $G_{r}(f)$ is disconnected.

Could you check if I am right? Thanks for your help.

答:
If $x$ is an endpoint of one of the deleted intervals, $\langle x,0\rangle$ is in your set $U$ but is not in $G_r(f)$. I think that you meant $U$ to be the set of $\langle x,0\rangle$ such that $x\in C$ and $f(x)=0$; if $E$ is the set of endpoints of the deleted intervals, and $C_0=C\setminus E$, this set would be $U=\{\langle x,0\rangle:x\in C_0\}=G_r(f)\setminus V$.

With that fixed, let’s take a look at your proposed separation.

Let $\langle x,y\rangle\in V$. Then there is a deleted interval $(a,b)$ such that $a\le x\le b$. If $a<x<b$, let $W=(a,b)\times\Bbb R$; if $x=a$, so that $y=-1$, let $W=\Bbb R\times(-2,0)$, and if $x=b$, so that $y=1$, let $W=\Bbb R\times(0,2)$. In each case $W\cap G_r(f)$ is a relatively open nbhd of $\langle x,y\rangle$ in $G_r(f)$, and $W\cap U=\varnothing$, so $\langle x,y\rangle\notin\operatorname{cl}U$, and therefore $V\cap\operatorname{cl}U=\varnothing$. (All closures are taken in $G_r(f)$.) So far so good.

Now let $\langle x,0\rangle\in U$, and suppose that $\langle x,0\rangle\notin\operatorname{cl}V$; then there is an $\epsilon\in(0,1)$ such that if $W=(x-\epsilon,x+\epsilon)\times(-\epsilon,\epsilon)$, then $W\cap V=\varnothing$. But this is impossible: there is a deleted interval $(a,b)$ such that $x-\epsilon<a<b<x+\epsilon$, and clearly

$$\left\langle\frac{a+b}2,0\right\rangle\in W\cap V\,.$$

Thus, not only is $U$ not disjoint from $\operatorname{cl}V$, but in fact $U\subseteq\operatorname{cl}V$, and hence $\operatorname{cl}V=G_r(f)$. And that means that if we could show that $V$ is connected, it would follow that $G_r(f)$ is also connected, so perhaps we should try that instead.

Unfortunately, $V$ is not connected: if $x\in C_0$, $W_0=(x-1,x)\times\Bbb R$, and $W_1=(x,x+1)\times\Bbb R$, then $W_0\cap V$ and $W_1\cap V$ are a separation of $V$. However, each of the line segments making up $V$ is connected and must therefore lie entirely in one piece of any separation of $V$, so we might reasonably suspect that this is the only kind of separation of $V$. Let’s see if we can use this idea to show that $G_r(f)$ is connected anyway.

Suppose that $G_r(f)=A\cup B$, where $A$ and $B$ are disjoint, relatively clopen subsets of $G_r(f)$. We may assume that $V\cap A\ne\varnothing$. If $V\subseteq A$, then

$$U\subseteq\operatorname{cl}V\subseteq\operatorname{cl}A=A$$

as well, so $A=G_r(f)$, and $A$ and $B$ do not form a separation of $G_r(f)$. Suppose, then, that $V\cap B\ne\varnothing$ as well. Let

$$T_A=\{x\in C:\langle x,-1\rangle,\langle x,1\rangle\in A\}$$

and

$$T_B=\{x\in C:\langle x,-1\rangle,\langle x,1\rangle\in B\}\,;$$

both of these sets are non-empty. (Why?) Without loss of generality there are $x_A\in T_A$ and $x_B\in T_B$ such that $x_A<x_B$. Let $x_0=\sup\big(T_A\cap[x_A,x_B]\big)$. There are three possibilities: $x_0\in T_A$, $x_0\in T_B$, or $x_0\in C_0$.

Show that if $x_0\in T_A$, then $\langle x_0,1\rangle\in A\cap\operatorname{cl}B$, which is impossible.

Show that if $x_0\in T_B$, then $\langle x_0,-1\rangle\in B\cap\operatorname{cl}A$, which is impossible.

Show that if $x_0\in C_0$, then $\langle x_0,0\rangle\in(\operatorname{cl}A)\cap\operatorname{cl}B$, which is impossible.

Conclude from this that $G_r(f)$ is connected.

Obviously this is not a neatly packaged argument such as you might find in a textbook. For the most part I’ve simply followed my own thought processes on encountering the problem in your question, in hopes that seeing how someone with more experience attacked the problem, including a false start, might be more illuminating than a sanitized version.


问:
How do you respond to "I was always bad at math"?
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.

答:
Since a vast majority of people are under the impression that mathematics is like the curricula they had in primary and secondary school, this is not too surprising.

Really they don't have any qualification to say they are bad at math. (I know I didn't see any real math until college.) They might have been bad at the curriculum, but that's understandable given how boring and stupid it is. That is what happens when you teach mathematical tasks as behaviors to be trained in.

Unfortunately, teaching mathematics in an effective way would be more time consuming and less testable than the current curriculum. Moreover, very few people in the position of teaching those age groups are probably not suitable for teaching real mathematics. In short, the way people expect young people to be educated is at odds with actual effective instruction.

For anyone who hasn't already read it, Lockhart's lament is relevant to this and is worth reading once.


问:
Probability of $X > \max\{Y,Z\}$ exponentials
Let $X,$ $Y$ and $Z$ be exponentially distributed with $E[X] = 1/\lambda_X$, $E[Y] = 1/\lambda_Y$, $E[Z] = 1/\lambda_Z$ respectively.

How do I calculate $P(X > \max\{Y,Z\})$?

I have A, B, C and D for 3 cash boxes in a queueing system. Consider D the one waiting. I want to find the probability D is the last one to leave. The service times of the C.Boxes are indep. Exp with mean 5 (same mean for c.box 1, 2 and 3) That's how I'm trying to solve this: P(D last one) = P(T1 > max{T2,T3})*P(T1 < min{T2,T3}) + P(T2 > max{T1,T3})*P(T2 < min{T1,T3}) + P(T3 > max{T1,T2})*P(T3 < min{T1,T2})

And then:
$$(\frac{1/5}{1/5+1/5})^4+(\frac{1/5}{1/5+1/5})^4+(\frac{1/5}{1/5+1/5})^4 = = 3(\frac{1}{2})^4 = 3\frac{1}{16} = \frac{3}{16}$$

答:
Comment:. When you get your general answer, you can compare
it with specific results, approximated by
simulation, for the case: $\lambda_x = 1/4, \lambda_y = 1/3, \lambda_z = 1/2.$

```<code>set.seed(213)
x = rexp(10^6, 1/4)
y = rexp(10^6, 1/3)
z = rexp(10^6, 1/2)
w = pmax(y,z)
mean(x > w)
[1] 0.469116
mean((x > y) & (x > z))
[1] 0.469116
</code>```

Body:
Comment:. When you get your general answer, you can compare
it with specific results, approximated by
simulation, for the case: $\lambda_x = 1/4, \lambda_y = 1/3, \lambda_z = 1/2.$

set.seed(213)
x = rexp(10^6, 1/4)
y = rexp(10^6, 1/3)
z = rexp(10^6, 1/2)
w = pmax(y,z)
mean(x > w)
[1] 0.469116
mean((x > y) & (x > z))
[1] 0.469116


问:
Difference between separable and linear? Differentials

My understanding was that a separable equation was one in which the x values and y values of the right side equation could be split up algebraically.  I tried this once before and got the wrong answer. Can someone help me?

答:
Consider the equation $\frac{dy}{dt} = ry$. This equation is a separable differential equations since we can rewrite this in the form of $\frac{dy}{y} = rdt$.

Consider the fact that this is also a linear equation since $\frac{dy}{dt} - ry = 0$ all the derivatives are attached to purely functions of t, and 0 is also a function of t. Also, y is raised to the power of 1.

On the other hand, take a look at the first equation in your picture. That equation is nonlinear since y is raised to the power of 2, but it is separable.


问:
If $x+y+z+a+b+c=1$, $xyz+abc=1/36$, what is the max value of $abz+bcx+cay$?
Question : If $x+y+z+a+b+c=1,xyz+abc=1/36,$ What is the max value of $abz+bcx+cay$?  ($x, y, z, a, b, c ∈ R$ (non-negative))

I tried $abz+bcx+cay≥3abc$ (because of AM-GM) to make it similar to $xyz+abc=1/36$, but I'm stuck with expanding.

答:
For the first equation, we can test the waters with a naive solution
$x+y+z+a+b+c=1\qquad=6\left(\dfrac16\right)
=\dfrac16+\dfrac16+\dfrac16+\dfrac16+\dfrac16+\dfrac16 \tag{1}$
and it appears that at least $\space 2\space $ of $\space a,b,c,x,y,z$ must be other than $\space \frac16\space $ for the other equation to work. Again, we suggest a non-working naive solution.

$$\text{So we let}\quad xyz+abc=\dfrac{1}{36}
=\frac{1}{2\cdot6\cdot6}+\frac{1}{2\cdot6\cdot6}???\\
\implies 36(xyz+abc)=1\\
\implies x+y+z+a+b+c=36(xyz+abc)\tag{2}$$
$$\implies
\big( a + b + c -36 a b c\big) +\big(x + y + z - 36 x y z\big)=0\\
\tag{3}$$
Assuming that both addends in $(3)$ equal zero.
$$( a + b + c - 36 a b c)  = 0\\
\implies c = \frac{a + b}{36 a b - 1}\\
\implies 36 a b - 1>0 \implies b>\frac{1}{36a}\tag{4}
$$

I found $\space \big(a=1/12\quad b=1/3\quad c=1/2  \quad x=y=z=1/36\big)\space $ for equation $(1)$ with a,b interchangeable but a,b violate the condition of $\space b>\dfrac{1}{36a}$ and the x,y,z values of $1/36$ do not work with the the right part of equation $(3)$.  This means that the two parts of equation $(3)$ are not separately zero an combinations must make them equal but with opposite signs.

Forgive me but it is past my bedtime and I cannot finish this tonight.

I can suggest is to trying the $2,3,4,6,9,12,18,36$- factors of $36$ as denominators to see which add up to $\space 1\space $ and then seeing which satisfy the second equation. From there,  you can select the combinations that yield the highest value for the final equation in your question.

Perhaps an even faster approach would be to use equation $(4)$ equal to a non-zero factor of $36$ and  test when the swapping of $a,b$ yields an opposite sign. That is the approach I would take if I did not have to get to bed. Good luck.


答:
Let $=$ be the equality predicate. Then the relation $R$ is a function if
$$
\forall x \forall y \forall z: ((x, y) \in R \land (x, z) \in R) \to (y = z).
$$
It is a total function if it is a function and
$$
\forall x \exists y: x \in A \to (x, y) \in R.
$$
Usually we omit "total" and assume that all "functions" are total. (Those that are not total are given the adjective "partial".) So I believe you need both statements in your problem.

In order for $R$ to be an injective function, it must be a (total) function and satisfy
$$
\forall x \forall y \forall z: ((x, z) \in R \land (y, z) \in R) \to (x = y).
$$

Finally, in order for $R$ to be a bijective function, it has to be injective and satisfy
$$
\forall y \exists x: y \in A \to (x, y) \in R.
$$

If your quantifiers are always over $A$, predicates for surjectivity and totality can be shortened a little bit. Here's the summary:

Well-defined: $\forall x \forall y \forall z: ((x, y) \in R \land (x, z) \in R) \to (y = z)$

Total: $\forall x \exists y: (x, y) \in R$

Injective: $\forall x \forall y \forall z: ((x, z) \in R \land (y, z) \in R) \to (x = y)$

Surjective: $\forall y \exists x: (x, y) \in R$

Being a (total) function: Total and well-defined

Being an injective (total) function: Total, well-defined and injective

Being a bijection: Total, well-defined, injective and surjective


问:
If $x+2y=8$, find the minumum of $x+y+\frac{3}{x}+\frac{9}{2y}$ (x, y ∈ R+)
Question : If $x+2y=8$, find the minumum of $x+y+\dfrac{3}{x}+\dfrac{9}{2y}$ $(x, y \in \mathbb{R}^+)$

I tried to use $x=8-2y$, and I thought I can plug in. But I think I'm not doing it well. I thought about AM-GM, and how can I expand this  using  $x+2y=8$?

答:
The AM-GM inequality, as in the post tag, gives us a lower bound.

Indeed,
$$x+\frac{3}{x} + y +\frac{9}{2y}=2\frac{x+\frac{3}{x}}{2} + 2\frac{y +\frac{9}{2y}}{2}.$$

Since $x,y>0$ we can apply the AM-GM inequality to obtain

$$2\frac{x+\frac{3}{x}}{2} + 2\frac{y +\frac{9}{2y}}{2}\geq 2\sqrt{3}+2\sqrt{\frac{9}{2}}=2(\sqrt 3 + \sqrt\frac{9}{2})\sim 7.71.$$

However, this is just a lower bound.

To find the minimum, we have $$x+y+\frac{3}{x}+\frac{9}{2y}=x+2y-y + \frac{3}{x+2y-2y}+\frac{9}{2y}$$
which is equal to $$8-y +\frac{3}{8-2y}+\frac{9}{2y}.$$

Note that since $x=8-2y>0$, we want $0<y<4$.

A local minimum occurs at $y=3$.

One can see this by solving for when the derivative of $8-y +\frac{3}{8-2y}+\frac{9}{2y}$ is $0$.

At $y=3$, we have $x=2$ so that $2+3+\frac{3}{2}+\frac{9}{6}=8$ is the minimum desired.


问:
Why does adding an extra column to a matrix not change the RREF of the first columns?
This resource (2.2.7) points out that deleting a column from a matrix would not change the RREF of the initial undeleted columns.

For example, the RREF form of this $3\times3$ matrix is the $3\times3$ identity matrix.
\begin{bmatrix}1&1&1\\1&0&1\\1&2&3\end{bmatrix}

The RREF form of this matrix (the above with third column deleted) would be the $3\times 3$ identity matrix with the third column deleted, i.e. the RREF form of the first two colums are preserved.
\begin{bmatrix}1&1\\1&0\\1&2\end{bmatrix}

Is there a way to geometrically, or numerically indicate why the RREF form of individual columns are not changed by adding/deleting other columns? Intuitively, I believe this has something to do with the fact that adding another column to a matrix doesn't change the linear independence of the other columns with respect to each other. Furthermore, how does this connect to deleting or adding rows? Would messing with rows change the RREF form of columns?

答:
Welcome to MSE!

When you bring a matrix into RREF the pivots correspond to the linearly independent columns of your original matrix. See here for how to determine bases of the column space, it's not super crazy. You can translate this question into a basis problem about the column space of the matrix.

Here are two examples that might be of interest to you, you might not get the exact RREF by deleting a column, I chose to work with the $4\times 4$ matrix given from wikipedia:

$$
    \begin{pmatrix}
    1 & 3 & 1 & 4\\
    2 & 7 & 3 & 9\\
    1 & 5 & 3 & 1\\
    1 & 2 & 0 & 8\\
    \end{pmatrix} \rightsquigarrow_{RREF}
\begin{pmatrix}
    1 & 0 & -2 & 0\\
    0 & 1 & 1 & 0\\
    0 & 0 & 0 & 1\\
    0 & 0 & 0 & 0\\
    \end{pmatrix}
$$

If you delete the linear dependent column
$\begin{pmatrix}
    1 & 3 & 3 & 0 \\
    \end{pmatrix}^{T} =
(-2)\cdot\begin{pmatrix}
    1 & 2 & 1 & 1 \\
    \end{pmatrix}^{T}+
\begin{pmatrix}
    3 & 7 & 5 & 2 \\
    \end{pmatrix}^{T} $ then
$$
    \begin{pmatrix}
    1 & 3  & 4\\
    2 & 7  & 9\\
    1 & 5  & 1\\
    1 & 2  & 8\\
    \end{pmatrix} \rightsquigarrow_{RREF}
\begin{pmatrix}
    1 & 0  & 0\\
    0 & 1  & 0\\
    0 & 0  & 1\\
    0 & 0  & 0\\
    \end{pmatrix}
$$
has the same RREF as if you had just deleted the third column. But say I deleted the second column instead, then

$$
    \begin{pmatrix}
    1 & 1  & 4\\
    2 & 3  & 9\\
    1 & 3  & 1\\
    1 & 0  & 8\\
    \end{pmatrix} \rightsquigarrow_{RREF}
\begin{pmatrix}
    1 & 0  & 0\\
    0 & 1  & 0\\
    0 & 0  & 1\\
    0 & 0  & 0\\
    \end{pmatrix} \text{not}
\begin{pmatrix}
    1 & -2  & 0\\
    0 & 1  & 0\\
    0 & 0  & 1\\
    0 & 0  & 0\\
    \end{pmatrix}
$$
Deleting linear dependent columns doesn't affect the linear dependence/independence of the remaining columns.

In your example all of the columns were linearly independent (given that you get the RREF is the identity $I_{3\times 3}$), from a basis point of view, removing a vector from a linear independent set does not affect the linear independence of the remaining vectors. Also, the last column was deleted. Having the same RREF with a column deleted 'should' work if all the column vectors are linearly independent.

What I mean by 'should' is, if you deleted the second column instead of the third, your RREF would be

$$
\begin{pmatrix}
    1   & 0\\
    0   & 1\\
    0   & 0\\
    \end{pmatrix} \text{not}
\begin{pmatrix}
    1   & 0\\
    0   & 0\\
    0   & 1\\
    \end{pmatrix}
$$

It wouldn't be exactly the same as just deleting the second column, however we still get the information conveyed that the two column vectors are linearly independent.

If you add a column and performed the same row operations you did to get the RREF of the matrix minus the column, then you would get the same RREF with the column added.

For rows, you can always choose to work with the transpose of the matrix, and then analyze column-wise.

Hope this helps!


问:
Masters Theorem $T(n) = 2T(\frac{n}{4}) + (\log(n))^2$
I have a problem that I am trying to understand using the master theorem since it applys.  The problem at hand is, $T(n) = 2T(\frac{n}{4}) + (\log(n))^2$.   I am not sure I am correct in my final solution.  The $\log(n)^2$ is what is causing me some doubt.  I believe this recurrence relation satisfies case 1, see below.  Any guidance or correction would be greatly appreciated!  Here is my work:

$T(n) = 2T(n/4) + (\log(n))^2$

a = 2, b = 4, k = 0, p = 2

following, $\log_b a$ or $\log_4 2 = .5$

Since .5 > k = 0, we satisfy case 1

Thus it is: $\Theta(n^{\log_4 2})$

Is this correct, and if not, where did I go wrong?

答:
Step 4 could use some more work. Namely, you need to show that $(\log(n))^{2} \in O(n^{c})$ for some $c < 1/2$. This shouldn't be too hard. For example, take $c = 1/3$. We have that:
\begin{align*}
\lim_{n \to \infty} \frac{(\log(n))^{2}}{n^{1/3}} = 0.
\end{align*}

I'll leave it to you to fill in the details for the limit computation.

So $(\log(n))^{2} \in O(n^{1/3})$ (actually, $(\log(n))^{2} \in o(n^{1/3}))$.

Your reasoning is otherwise fine.


答:
I know this might be a bit stupid, but, I plotted it with:

$$x = (x_1, x_2)$$

$$x \leftarrow x - \vert \nabla^{2} f(x) \vert^{-1} \nabla f(x)$$

And got:

$$y = 3\ \ \ \text{and} \ \ x = 2$$

Title: If $x+2y=8$, find the minumum of $x+y+\frac{3}{x}+\frac{9}{2y}$ (x, y ∈ R+)

Body:
Question : If $x+2y=8$, find the minumum of $x+y+\dfrac{3}{x}+\dfrac{9}{2y}$ $(x, y \in \mathbb{R}^+)$

I tried to use $x=8-2y$, and I thought I can plug in. But I think I'm not doing it well. I thought about AM-GM, and how can I expand this  using  $x+2y=8$?


问:
Calculating Likelihood of Regression Outputs
I am given the equation $y = mx + c + z$ where $z \sim Normal(0,\sigma^2)$ is our error term of the linear regression.

I am given a sample with just one point $(x_1, y_1)$, and I've been asked to find the likelihood of this sample.

Additionally, I am also asked to find the likelihood of another sample where there are $n$ of $(x_i, y_i)$ points. I've learned in the past in econometrics about how to find the maximum likelihood estimators of $\beta_1$ and $\beta_0$ in $y = \beta_1x + \beta_0$, but I am struggling to see how I should approach this one. If anyone could guide me on how to approach the first question, I would greatly appreciate it.

Thanks.

答:
You're given that y=mx+c+z, with $z\sim N(0, \sigma^2)$. You can view this as a constant plus a normal random variable. Also each observation of y should be considered as independent from one another. We expect y to follow a normal distribution with mean $E(y)=E(mx+c+z)=mx+c+E(z)=mx+c$ and variance $var(y)=var(mx+c+z)=var(z)=\sigma^2$. Now there are y's to keep in mind, the y that you observed and the $\hat y$, predicted y from your x value.

The likelihood for a single observation is given by $$p(y|x)=\frac 1 {\sqrt{2\pi\sigma}}\exp-\frac 1 {2\sigma^2}\left(y-(mx+c)\right)^2$$

The likelihood for a sample is given by $$\begin{split}p(\textbf y|\textbf x)&=\prod\frac 1 {\sqrt{2\pi\sigma}}\exp-\frac 1 {2\sigma^2}\left(y-(mx+c)\right)^2\\
&=\left(\frac 1 {\sqrt{2\pi\sigma}}\right)^n\exp-\frac 1 {2\sigma^2}\sum_{i=1}^n\left(y_i-(mx_i+c)\right)^2\end{split}$$


问:
Stokes' Thm. $\vec F = <2y,xz,x+y>$ through intersection of $x^2+y^2=1$ and $z=y+2$
I am trying to evaluate $\int_C \vec F \cdot d \vec r$ where

$\vec F = <2y,xz,x+y>$ and $C$ is the intersection of $x^2+y^2=1$ and $z=y+2$.

I believe that $$curl \vec F = <1-x, -1, z-2> $$

The surface is an ellipse, on  the plane $z=y+2$ so the unit normal vector should be

$$\vec n = \frac{1}{\sqrt{2}}<0,-1,1>$$

because the dot product is $\frac{1}{\sqrt 2} \left(z-1\right)$ I am thinking that we are to evaluate

$$\int \int_S \frac{1}{\sqrt 2} \left(z-1\right) dS$$

I used $x= \cos\theta, \space y= \sin\theta, z=z$ to evaluate this and got the expression

$$\frac{1}{\sqrt2} \int_0^{2\pi} \int_1^3 z-1 \space dz d\theta$$

where I got $4\sqrt{2}\pi$.

Supposedly the answer should be $\pi$.

Can any one help with what my error is?

答:
Your curl is correct, as is your normal vector and your dot product.
The problem I believe is in the integral; I'm not sure where you got that integral or those limits.

The area below the ellipse is the unit circle. With the factor of $\sqrt{2}$ converting $dS$ to $dA$, I would have expected $$ \int_{0}^{2\pi} \int_{0}^{1}(1 + r sin {\theta})r dr d\theta$$ which gives
$$ \int_{0}^{2\pi}  \frac{1}{2} + \frac{1}{6}sin {\theta} d\theta = \pi$$


Body:
$$x+2y=8  \implies x = 8 - 2y\tag{1}$$

Let $$f(x, y) = x+y+\frac{3}{x}+\frac{9}{2y} \tag{2}$$

Using $(1)$ in $(2)$,

$$f(x, y) = 8 - 2y +y+\frac{3}{8 - 2y}+\frac{9}{2y} = 8 - y +\frac{3}{8 - 2y}+\frac{9}{2y} \tag{3}$$

Differentiate $(3)$ wrt $y$, set it equal to $0$ and solve for $y$

$$\frac{-9}{2y^2} + \frac{3}{2(4-y)^2} - 1 = 0$$

The real root is $y = 3$ and the second derivative $$\left[\frac{9}{y^3} + \frac{3}{(4-y)^3} \right]_{y = 3} = \frac{10}{3} > 0$$

From $(1)$, $x = 2$ and the minimum value is $f(2, 3) = 8$


问:
A basic proof on Morse Function
The questions is to show if $f_t$ is a homotopic family of functions on $R^k$, show that if $f_0$ is Morse in some neighborhood of a compact set $K$, then so is every $f_t$ for sufficiently small t.

I get the previous question leads to the proof, but couldn't reach the question:

det$(H)^2 + \sum_{i=1}^k (\partial f/ \partial x_i)^2 > 0$

And also, on a compact manifold:

det$(H)^2 + \sum_{i=1}^k (\partial f/ \partial x_i)^2 >\epsilon$

Thanks for your help!

答: G&P have you prove that the first inequality is equivalent to $f$ Morse. If the quantity is $>\epsilon$ for $f_0$, then by smoothness, for small enough $t$ it'll be $>\epsilon/2$ for $f_t$. </p>

Title: A basic proof on Morse Function


问:
Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $
I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k}  = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series

$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}  = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$

Does it converge?   If so, what is its sum?

答:
Proving that this series converges can be done using the alternating series test: any series that alternates forever between positive and negative terms, where each term is smaller than the preceding term, and the terms approach a limit of 0 converges.

Why this test works:

The odd partial sums decrease forever. Every odd partial sum (except the first) is less than the one before it: $S_{2n+1} = S_{2n-1} + \left(-\frac{1}{2n} + \frac{1}{2n+1} \right)$ (where $S_n$ is the $n$th partial sum) and $-\frac{1}{2n} + \frac{1}{2n+1} < 0$ for all $n$.

Similarly, the even partial sums increase forever.

Every odd partial sum is greater than the (even) partial sum before it, since all the terms at odd positions are positive.

Similarly, every even partial sum is less than the (odd) partial sum before it.

Since the odd partial sums decrease forever but never go below the first even partial sum $\left( 1 - \frac{1}{2} \right)$, they must approach a limit.

Similarly, since the even partial sums increase forever but never go above the first term $\left( 1 \right)$, they must approach a limit.

The odd partial sums and the even partial sums must approach the same limit, since the difference between each odd partial sum and the even partial sum before it approaches 0: $\lim\limits_{n \to \infty}\frac{1}{n} = 0$. Similarly, the difference between each even partial sum and the odd partial sum before it approaches 0.

∴ the series converges.


问:
What is comparability in relations?
So I've just started trying to teach myself some topology and I've been reading through topology by munkres.
In the first chapter he defines order relations being a binary relation satisfying transitivity, nonreflexivity and comparability this has caused me a lot of confusion and I am struggling to grasp the concepts.

My main confusion is to do with that of comparability what exactly is it on Wikipedia the definition of a total  order (which is what munkres is talking about in his book) is a binary relation satisfying transitivity,antisymmetry and connexity.

my question is what is connexity is it the same as comparability and how are these 2 definitions equivalent?.

Thanks in advance.

答:
The difference in definitions is that Wikipedia defines non-strict total orders (things like $x\le y$ where elements relate to themselves) and Munkres defines strict total orders (things like $x<y$ where elements do not relate to themselves). Wikipedia's connexity condition states that for any $x,y$ either $x\le y$ or $y\le x$. This is not true for strict total orders in the case where $x=y$, so Munkres' definition instead states for any distinct $x,y$ either $x<y$ or $y<x$. Annoyingly "total order" as used in math is ambiguous, we precede it by "strict" or "non-strict" if we want to distinguish which we are talking about.

For what it's worth "connex" is quite rare, I don't think I've ran into it outside Wikipedia. According to this talk page it was introduced because the previous language (calling this a "total" relation) was ambiguous.

Title: What is comparability in relations?


答:
Stokes Theorem says $\displaystyle \int_C \vec F \cdot dr = \iint_S curl \vec{F} \cdot dS$

where $S$ is a surface with the boundary and $C$ is the boundary curve of the surface.

You have been asked to find the line integral over the elliptic curve which is intersection of cylinder $x^2+y^2= 1$ and plane $z = y + 2$.

You are applying Stokes' Theorem and instead finding the right hand side of the equation. The easiest way to do so is to consider a disk which has the same boundary as in the question and find flux of the curl of the vector field over the disk.

$\vec{n} = (0,-1,1)$

If you normalize the vector then you have to note that $dS = \sqrt2 dA$  where $dA$ is the projection of $dS$ in XY plane.

Parametrization of the disk surface:

$x = r \cos \theta, y = r \sin\theta, z = y + 2 = 2 + r \sin \theta$.

So the integral becomes,

$\displaystyle \int_0^{2\pi} \int_0^1 (r^2 \sin \theta + r) \ dr \ d\theta = \pi$

Title: Stokes' Thm. $\vec F = <2y,xz,x+y>$ through intersection of $x^2+y^2=1$ and $z=y+2$

Body:
I am trying to evaluate $\int_C \vec F \cdot d \vec r$ where

$\vec F = <2y,xz,x+y>$ and $C$ is the intersection of $x^2+y^2=1$ and $z=y+2$.

I believe that $$curl \vec F = <1-x, -1, z-2> $$

The surface is an ellipse, on  the plane $z=y+2$ so the unit normal vector should be

$$\vec n = \frac{1}{\sqrt{2}}<0,-1,1>$$

because the dot product is $\frac{1}{\sqrt 2} \left(z-1\right)$ I am thinking that we are to evaluate

$$\int \int_S \frac{1}{\sqrt 2} \left(z-1\right) dS$$

I used $x= \cos\theta, \space y= \sin\theta, z=z$ to evaluate this and got the expression

$$\frac{1}{\sqrt2} \int_0^{2\pi} \int_1^3 z-1 \space dz d\theta$$

where I got $4\sqrt{2}\pi$.

Supposedly the answer should be $\pi$.

Can any one help with what my error is?


问:
Does every Cesaro summable sequence have bounded partial sums, is the sums are all positive.
If $a_n \in \mathbb{R}$, $A_N = \sum_{n=1}^N a_n$,
all partial sum so $A_N$ are positive, but their sequence oscillates and doesn't not converge,
It probably down to matter but all $a_n$ are not necessarily positive,
and $\lim_{K \to \infty} \frac{1}{K+1} \sum_{k=1}^K A_k \to L < \infty$ exists,
Then are partial sums ${A_N}$ bounded?

Ignoring $\{a_n\}$ sequence which is a distraction, we may consider only the sequence $\{A_k\}$. This sequence has positive real numbers with a finite average. The average must grow at a rate less than $k^{\epsilon}$. So it could grow at a rate $\log(k)$.

答:
No, $A_{N}$ can be unbounded.
Define $a_{n}$ like this:

if $n=2^{l}$ then $a_{n} = l$

if $n = 2^{l} + k, k = 1,2, ...,l$  then $a_{n} = -1$

if $n= 2^{l} + k, k = l+1, l+2, ... 2^{l+1} -1$ then $a_{n} = 0$
Note that it follows from the definition that

$A_{n} = l$ if $n=2^{l}$

$A_{n} = l-k$ if $n=2^{l} + k, k =1,2,3, ... l$

$A_{n} = 0$ if $n=2^{l} + k, k = l+1, l+2, ... 2^{l+1} - 1$

Clearly $A_{n} >=0$ and $A_{n}$ is unbounded, however Cesaro means of $A_{n}$ tend to $0$.

To see this, for given $K$ take $l$ such that $2^{l} < K <= 2^{l+1}$, we get
$$\frac{1}{K+1} \sum_{n=1}^K A_n <= \frac{1}{K+1} \sum_{n=1}^{2^{l+1}} A_n =
\frac{1}{K+1}\sum_{n=0}^{l}\sum_{k=2^{n}}^{2^{n+1}-1} A_{k} + \frac{1}{K+1} A_{2^{n+1}} $$
The last term tends to $0$ of course.

But the sum in each internal block, is just sum of numbers from $n$ down to $1$ so it has order of $n^{2}$. Then we sum these numbers from $n=1$ to $n=l$ so the sum will have order of $l^{3}$.

But $K$ is bigger than $2^{l}$ so the ration tends to $0$.


答:
$$
x + y + \frac{3}{x} + \frac{9}{2y} = \frac{1}{4}(x+2y) + \left (\frac{3}{4}x + \frac{1}{2}y \right) + \left( \frac{3}{x} + \frac{9}{2y} \right)
= 2 + \left (\frac{3}{4}x + \frac{1}{2}y \right) + \left( \frac{3}{x} + \frac{9}{2y} \right)
$$
And

$$
\left (\frac{3}{4}x + \frac{1}{2}y \right) + \left( \frac{3}{x} + \frac{9}{2y} \right) \geq 2 \sqrt{\left (\frac{3}{4}x + \frac{1}{2}y \right) \left( \frac{3}{x} + \frac{9}{2y} \right)}
$$ (AM-GM, equality: $\left (\frac{3}{4}x + \frac{1}{2}y \right) = \left( \frac{3}{x} + \frac{9}{2y} \right)$)

Further,
$$
\left (\frac{3}{4}x + \frac{1}{2}y \right) \left( \frac{3}{x} + \frac{9}{2y} \right) \geq \left( \frac{3}{2} + \frac{3}{2} \right)^2 = 9
$$ (Cauchy– Schwarz inequality, equality: $\frac{4}{x^2} = \frac{9}{y^2}$)

Observe that the equalities hold when $x=2$ and $y=3$, and the minimum is $2+2*(\sqrt{9}) = 8$


问:
expected value of minimum of sum random variable uniformly distributed variables
Let $U_1$, $U_2$, $U_3$...  random variables uniformly distributed over the interval [0,1].
Define $n(x)$ as $n(x)$= $\min$ $\{$$n$| $\sum_1^n$ $U_i>x$$\}$. I need to find $\Bbb E[n(x)]$
I tried to find $\Bbb E[n(x)]$ conditioning on the value of  $U_1$ than i use the fact
$\tag{2}\Bbb E \left[ n(x) | U_1=y\right] = \cases{1,& $y>x$\cr 1+\Bbb E(x-y),& $y\le x $}.$ but I'm struggling to build the integral, I don't know how to proceed any further than that. I appreciate any help. Thank you in advance.

答:
For the sake of notation, let's denote $U_0 =0$.
The probability of the event $\{n(x)=k \}$ for $k\in \mathbb{N^*}$ is
\begin{align}
P(n(x)=k) &= P(\{\sum_{i=0}^{k-1}U_k<x\}\cap  \{\sum_{i=0}^{k}U_k \ge x \}) \\
&= P(\{S_k-U_k<x\}\cap  \{S_k \ge x \}) \\
\end{align}
with $S_k$ defined as $S_k=\sum_{i=0}^{k}U_k$.

We have $S_k$ follows the Irwin-Hall distribution with the density function $f(s;k)$. Hence
\begin{align}
P(n(x)=k) &= P(\{S_k-U_k<x\}\cap  \{S_k \ge x \}) \\
&= P(x\le S_k \le x +U_k ) \\
&= \int_0^1 \left( \int_{x}^{\min\{x+u,k\}} f(s;k)ds\right)du \\
\end{align}

And the expectation $E(n(x))$ is then calculated as

$$E(n(x)) = \sum_{k=1}^{+\infty}\left(\int_0^1  \int_{x}^{\min\{x+u,k\}} f(s;k)dsdu\right)$$

PS: I think the solution of leonbloy is better than mine.


问:
help with Integral manipulation for equality
$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{x^2-m+m}{\sigma} e^{-\frac{x^2}{2\sigma^2}} \, dx$$

trying to get this into the form $\int_{-\infty}^\infty e^{-\frac{x^2}{2}} \, dx$

so I can make an equality. can't seem to get to that. I think I'm close but I can't get the $\sigma^2$ out of the e.

EDIT: I'm getting closer I think. I managed to get to this $\frac{1}{\sqrt{2\pi}}\cdot3\int_{-\infty}^{\infty}x^{2}x^{2}e^{-\frac{\left(x\right)^{2}}{2}}dx$ need to deal with the $x^2x^2$ now.

答:
In the first place,
\begin{align}
& \int_{-\infty}^{+\infty} \frac{x^2}\sigma e^{-x^2/(2\sigma^2)} \, dx \\[8pt] = {} & \sigma^2 \int_{-\infty}^{+\infty} \frac{x^2}{\sigma^2} e^{-x^2/(2\sigma^2)} \, \frac{dx} \sigma \\[8pt]
= {} & \sigma^2 \int_{-\infty}^{+\infty} u^2 e^{-u^2/2} \, du.
\end{align}
That is scaling.
\begin{align}
& \int_{-\infty}^{+\infty} u \Big( e^{-u^2/2} u\,du \Big) = \int u \,dv \\[8pt]
= {} & uv-\int v\, du \quad\text{This is integration by parts.} \\[8pt]
= {} & \Big[ ue^{-u^2/2} \Big]_{-\infty}^{+\infty} + \int_{-\infty}^{+\infty} e^{-u^2/2} \, du \\[8pt]
= {} & 0 + \sqrt{2\pi}.
\end{align}


问:
Describe the complex function $f(z) = z + 1/z$.
A question from my homework asks me to find the region where $f(z) = z+ \frac{1}{z}$ is real, and the region where $Im f(z) > 0$, the region where $Im f(z) < 0$.

But I don't know where to start, how do you tackle this kind of problems in general?
I am always very bad at this kind of visualizing problems on complex functions, any help would be appreciated.

What I would usually do is solve for
\begin{equation}
0 = Im ( (a + bi) + \frac{a-bi}{a^2 + b^2})
\end{equation}
But the algebra is usually  hideous. Is there any slick approach?

答:
Im$\left(a+b\color{blue}i+\dfrac{a-b\color{blue}i}{a^2+b^2}\right)=b-\dfrac b{a^2+b^2}=b\left(1-\dfrac1{a^2+b^2}\right)=0$

$\implies b=0$ or $a^2+b^2=1$, so $z+\dfrac1z$ is real when $z\in\mathbb R$ or $z=e^{i\theta}$ for $\theta\in\mathbb R$.


答:
You seem on the right track.

Letting $g(x) = E[n(x)]$, applying the total law of expectation and using the fact that $f_U(u)=1 [0\le u \le 1]$ I get:

$$ g(x) =\begin{cases}
1 + \int_0^x g(u) du &0\le x\le1\\
1 + \int_{x-1}^x g(u) du & x >1 \\
\end{cases}
$$

This gives $g(x) = \exp(x)$ for $0\le x\le1$

In general, $g(x)$ can be expressed as a sum of piecewise differentiable functions in each interval $[k,k+1]$, $g(x) = \sum_{k=0}^\infty g_k(x) $ where

$$g_k(x)= \exp(x-k) h_k(x-n) \quad [ k\le x < k+1] $$
and $h_k(\cdot)$ is a polynomial of degree $k$

The first values are $h_0(x)=1$, $h_1(x)=e-x$ , $h_2(x)=\frac{{{x}^{2}}}{2}-e x+{{e}^{2}}- e$

For large $x$, $g(x) \approx 2x + \frac23$ (empirically)

Added: The functions $h_k(x)$ are produced by this recursion:

$$ h_k(x) = e \,h_{k-1}(1) - \int_0^x h_{k-1}(u) du $$

Title: expected value of minimum of sum random variable uniformly distributed variables

Body:
Let $U_1$, $U_2$, $U_3$...  random variables uniformly distributed over the interval [0,1].
Define $n(x)$ as $n(x)$= $\min$ $\{$$n$| $\sum_1^n$ $U_i>x$$\}$. I need to find $\Bbb E[n(x)]$
I tried to find $\Bbb E[n(x)]$ conditioning on the value of  $U_1$ than i use the fact
$\tag{2}\Bbb E \left[ n(x) | U_1=y\right] = \cases{1,& $y>x$\cr 1+\Bbb E(x-y),& $y\le x $}.$ but I'm struggling to build the integral, I don't know how to proceed any further than that. I appreciate any help. Thank you in advance.


问:
Is inversion continuous on normed (not necessarily Banach) algebras?
There is a nice proof here that the map $x\mapsto x^{-1}$ is continuous on the group of invertible elements in a Banach algebra.

I am wondering if the same result holds for normed algebras only? In particular, if $X$ is a normed algebra and $G(X)$ denotes the group of invertible elements in $X$, then for any sequence $(x_n)$ where $G(X)\supset(x_n)_{n\in\mathbb N}\to x\in X$, is it true (1) that $x$ is invertible, and (2) that $({x_n}^{-1})\to x^{-1}?$ Also, are my conditions (1) and (2) equivalent to continuity of $x\mapsto x^{-1}$ on $G(X)$? I think yes, but I'm a bit confused because we are talking about a subspace topology on $G(X)$ induced by a not-necessarily complete norm.

The proof in the linked answer explicitly uses completion in the assumption that a series converges, so it is not sufficient to guarantee continuity in general normed algebras.

Edit: After more thought, I think my condition (1) is equivalent to $G(X)$ being a topologically closed subset of $X$, and (2) is equivalent to continuity of the inverse map, if we only consider cases where $(x_n)$ converges in $G(X)$. At any rate, I would still like to know whether both (1) and (2) are satisfied.

答:
As requested by the OP, here is a proof that does not use completeness.

Let $A$ be a (not necessarily complete)  unital normed algebra.

Lemma.  Given  $r$ in $A$, such that $1-r$ is invertible, and $\|r\|<1$,   then:

The series
$
  \displaystyle  \sum_{k=0}^\infty  r^k
  $
converges to   $(1-r)^{-1}$,

$\displaystyle \|(1-r)^{-1}-1\| \leq  \frac{\|r\|}{1-\|r\|}$.

Proof.  The  series in (1)  is absolutely convergent but, unfortunately, in a normed space this is not enough to guarantee
convergence.  On the bright side,   there are other methods for proving the convergence of a series which do not rely on completeness!

Observe that, for every $n$, one has that
$$
  (1-r)\sum_{k=0}^n r^k = 1-r^{n+1},
  $$
so
$$
  \lim_{n\to \infty }\sum_{k=0}^n r^k =
  \lim_{n\to \infty }(1-r)^{-1}(1-r^{n+1}) =
  (1-r)^{-1},
  $$
where we have used that $\|r^{n+1}\| \leq  \|r\|^{n+1}$.

To prove (2) we have
$$
  \|(1-r)^{-1}-1\| =   \Big\|\sum_{k=1}^\infty  r^k\Big\| \leq  \sum_{k=1}^\infty  \|r\|^k = \frac{\|r\|}{1-\|r\|}.
  \tag*{\square}
  $$

Remark.  In a Banach algebra the absolute convergence of $\sum_{k=0}^\infty  r^k$ implies convergence, and hence one may prove that $1-r$ is invertible.  On the other hand, the hypothesis that $1-r$ is invertible was the key factor making the above proof work, as we are unable to deduce convergence from absolute convergence.

From (2) it is easy to see that inversion is continuous at 1, and hence at every invertible element  $a$ in $A$ because
if $\{a_n\}_n$ is a sequence of invertible elements converging to $a$, then
$$
  \|a^{-1}-a_n^{-1}\| \leq
  \|a^{-1}\|\|1-aa_n^{-1}\| =
  \|a^{-1}\|\|1-c_n^{-1}\|,
  $$
where
$c_n:=a_na^{-1}$, and clearly $c_n\to 1$.


问:
SVD and frobenius norm
If I have an SVD, $A = X \Sigma Y$, where $\Sigma \in \mathbb{R}^{m \times n}$ has the singular values on the diagonal and $X \in \mathbb{R}^{m \times m}$ and $Y \in \mathbb{R}^{n \times n}$ are orthogonal matrices, then does the squared frobenius norm $||A||^2$ give the summation of the squared singular values of $A$?

答:
Yes.  This is a consequence of the invariance of the Frobenius norm under orthogonal transformations.

If $Q$ is an orthogonal matrix, then $\| QA \|_{F}= \| A \|_{F}$.  Similarly, $\| AQ \|_{F}=\| A \|_{F}$.

Since $U$ and $V$ in the SVD are orthogonal, $\| A \|_{F}=\| \Sigma \|_{F}$.

Title: SVD and frobenius norm


问:
Binomial sum of $n$ terms in closed form
Can we calculate the given combinatorial sum in closed form?

$$ \frac{\binom{2}{0}}{1}+\frac{\binom{4}{1}}{2}+\frac{\binom{8}{2}}{3}+\frac{\binom{16}{3}}{4}+\cdots+\frac{\binom{2^n}{n-1}}{n}$$

答:
It is doubtful that a closed form exists for this sum.
However, the terms grow very quickly.
For $n$ large the sum is well approximated by its last term,
$$\sum_{k=1}^n \frac{1}{k}{2^k\choose k-1} \sim \frac{1}{n}{2^n\choose n-1}.$$
For $n=13$ the error due to this approximation is one part in a million.
For $n=20$ the error is one part in ten billion. </p>


问:
Why doesn't $f(x) = x^2$ generate a counter-example to $f(-x) = -f'(x)$?
According to the chain rule, the derivative of $f(-x) = -f'(x)$, since

$$
\frac{d}{dx}(f(-x)) = \frac{df(u)}{du} \frac{du}{dx} = f'(-x)*-1 = -f'(-x)
$$

where $u = -x$.  But doesn't the case of $f(x) = (x)^2$ generate a counter-example to this?

$$
f'(-x) = \frac{d}{dx}(-x)^2 = \frac{d}{dx}x^2 = 2x \ne -(2x) = -f'(-x)
$$

答:
Chain rule is
$$ \left(\frac d{dx} (f\circ g)\right) (x) = \left(\frac{df}{dx}\circ g\right)(x)\frac{dg}{dx}(x),$$
or using the $f'$ notation:
$$ (f\circ g)'(x) = (f'\circ g)(x) g'(x).$$
When we apply this with $g(x) = -x$, we see that the correct rule is
$$\frac{d}{dx} (f(-x)) = -f'(-x).$$
(This is what your derivation says, even if you stated it wrong in the title and the line before your derivation.)

We can verify this for $f(x)=x^2$ by calculating each side. On the one hand, since $f$ is even, $f\circ g= f$ so $\text{LHS}=(f\circ g)'(x)=f'(x)=2x$. On the other hand, $\text{RHS}=2(-x)\times -1=2x$, which is what I got for $\text{LHS}$.

It seems that the important error$^1$  is in saying that $-f'(-x)$ is $-2x$. In fact,
$$ f'(x) = 2x \implies f'(-x) =2(-x) = -2x \implies -f'(-x) = 2x.$$

Here are the two paths you could take
$$ f(x) \begin{cases} \xrightarrow{\text{evaluate at $-x$}} f(-x) \xrightarrow{d/dx} \frac{d}{dx} [f(-x)]\overset{\substack{Chain\\ Rule}}= -\frac{df}{dx}(-x) \\  \xrightarrow{d/dx}   \frac{df}{dx}(x) \xrightarrow{\text{evaluate at $-x$}} \frac{df}{dx}(-x) \end{cases} $$

Some of the confusion could be stemming from confusing notation. I'll list here the notations that I know: for the first I can only think of $ \frac{d}{dx}(f(-x)) = (f(-x))' =2x$
which would be the derivative of $g(y)=f(-y)=y^2$ at the point $y=x$. This is the situation in which Chain rule applies.

The other case can be written in a number of ways, $f'(-x)=\frac{df}{dx}(-x) = \left(\frac{d}{dx} f\right)(-x)  = \left.\frac{d}{dy} (y^2)\right|_{y=-x} = -2x$, these mean the function $f'(y)=2y$ evaluated at $y=-x$.  The difference here is that you differentiate $f$ first before you evaluate.

In the comments to this answer the notation $\frac{d}{d(-x)}f(-x)$ is used, but I don't recognize it. If I had to guess, I would guess that this also means $\left.\frac{d}{dy} f(y)\right|_{y=-x}=-2x$.

$^1$  as mentioned by David K and zkutch.


问:
Recovering a polynomial from its product with its reciprocal
Let $p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in\mathbb{Z}[x]$ be an integer polynomial. We can form the reciprocal polynomial $p^\ast(x)=a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$. The product $p(x)p^\ast(x)$ is a palindromic polynomial.

Can we recover $p(x)$ from $p(x)p^\ast(x)$?

The answer is no, since replacing $p(x)$ with any of $\{p(x),-p(x),p^\ast(x),-p^\ast(x)\}$ does not change the value of $p(x)p^\ast(x)$.

Can we recover $p(x)$ from $p(x)p^\ast(x)$, up to negation and reciprocal?

This can be phrased in fancier language. There is a function $s\colon\mathbb{Z}[x]\to\mathbb{Z}[x]$ given by $p(x)\mapsto p(x)p^\ast(x)$. The Klein four-group $\langle-,\ast\rangle$ acts on $\mathbb{Z}[x]$, and looking at the orbits gives an equivalent relation $\sim$ on $\mathbb{Z}[x]$. Then $s$ is constant on equivalent classes of $\sim$, so we obtain a function $\overline{s}\colon(\mathbb{Z}[x]/\sim)\to\mathbb{Z}[x]$. Is $\overline{s}$ injective?

It is conceivable that the answer is yes, since if you just look at polynomials of degree at most $n$, then this boils down to a system of $n+1$ equations in $n+1$ unknowns.
For example, if you just look at quadratic polynomials then you have to solve the system of equations
\begin{align*}
a_0a_2&=b_0,\\
a_1(a_0+a_2)&=b_1,\\
a_0^2+a_1^2+a_2^2&=b_2.
\end{align*}
Here $b_0,b_1,b_2$ are known (the coefficients of $p(x)p^\ast(x)$), and $a_0,a_1,a_2$ are unknown (the coefficients of $p(x)$).

答:
Not quite.

$$p^*(x)=x^np(1/x)$$
So the roots of $p(x)p^*(x)$ come in pairs $r,1/r$.  Pick one of them for $p$, then the other one goes to $p*$.

If there are no repeated roots, you have $2^n$ choices for $p's$ roots, and I think the sign gives another factor of $2$.


问:
Uniform Convergence of series expansion of $\frac{\log(1+x)}{x}$
We know that the series expansion of $\frac{\log(1+x)}{x}$ is
$$\frac{\log(1+x)}{x}=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}....$$
It is also known that the value of dilogarithm at 1 is
$$\int_{-1}^{1} \frac{\log(1+x)}{x}=2\sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}=\frac{3}{2}\zeta(2)=\frac{3}{2}Li_2(1)$$
For this to be true we must show that the RHS of the first equation that is $1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}....$ is termwise integrable. To prove that this series is termwise integral we must show that it's uniformly convergent in $(-1,1)$. Although it is simple to prove that the series converges by the ratio test.

How to show that $1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}....$ is uniformly convergent in $(-1,1)$?

答:
For $x \in [-1+\delta,1-\delta]$ where $0 < \delta < 1$, we have

$$\left|(-1)^k \frac{x^k}{k+1}\right|\leqslant \frac{(1-\delta)^k}{k+1} \leqslant (1-\delta)^k$$

As the geometric series $\sum_{k \geqslant 0} (1-\delta)^k$ converges, it follows by the Weierstrass M-test that we have uniform convergence on $[-1+\delta,1+\delta]$ of the series

$$\frac{\log (1+x)}{x} = \sum_{k=0}^\infty(-1)^k \frac{x^k}{k+1}$$

Now the series may be integrated termwise over $[-1+\delta,1-\delta]$ to obtain

$$\begin{align} \int_{-1+\delta}^{1-\delta}\frac{\log(1+x)}{x} \, dx &= \sum_{k=0}^\infty (-1)^k \frac{(1-\delta)^{k+1} - (-1+\delta)^{k+1}}{(k+1)^2}\\ &= \sum_{k=0}^\infty (-1)^k \frac{(1-\delta)^{k+1} - (-1)^{k+1}(1-\delta)^{k+1}}{(k+1)^2} \\ &= \sum_{k=0}^\infty \frac{(1-\delta)^{k+1} + (-1)^{k}(1-\delta)^{k+1}}{(k+1)^2}\\&= 2\sum_{j=0}^\infty \frac{(1-\delta)^{2j+1}
}{(2j+1)^2}\end{align}$$

The series on the RHS is uniformly convergent for $\delta \in [0,1]$, and, consequently we may interchange the limit as $\delta \to 0+$ with the sum to obtain

$$ \int_{-1}^{1}\frac{\log(1+x)}{x} \, dx =  \lim_{\delta \to 0+}\int_{-1+\delta}^{1-\delta}\frac{\log(1+x)}{x} \, dx  = 2\sum_{k=0}^\infty \frac{1
}{(2j+1)^2}$$


问:
Volume of solid of rotation - can you always get same answer using shell or disk-washer?
The question is to find the volume of revolution of the region bonded by $x=0$ and $y=10$ and $y=8$ and $xy=8$ when rotated about the $x$ axis.

Using the Shell method is almost trivial:

$V=2 \pi \int y (8/y)dy $ with bounds from $8$ to $10$.  The y variables cancel and the answer is $32 \pi$.

I'm having difficulty doing this with the disk-washer method.

Is it always possible to achieve the correct answer using both methods?

答:
For disk method, you are considering radius along $y$ axis and that is not defined by the same curves throughout $8 \leq y \leq 10$.

$xy = 8 \implies x = 1 $ when $y = 8$.

$xy = 8 \implies x = \frac{8}{10} $ when $y = 10$.

For $0 \leq x \leq \frac{4}{5}$, the radius of revolution is defined by lines $y = 8$ and $y = 10$.

For $\frac{4}{5} \leq x \leq 1$, the radius of revolution has lower bound of $y = 8$ and upper bound of $y = \frac{8}{x}$ which is hyperbola curve.

So volume of revolution using disk method,

$\displaystyle \int_0^{4/5} \int_8^{10} 2\pi r \ dr \ dx + \int_{4/5}^{1} \int_8^{8/x} 2\pi r \ dr \ dx$


答:
Without loss of generality (WLOG), $z \neq 0.$

Further, WLOG Im$(z) \neq 0$, else the entire problem (both questions) becomes trivial.

$$z + \frac{1}{z} = z + \frac{\overline{z}}{z\overline{z}}.\tag1$$

Further, $z + \overline{z} \in \mathbb{R}.$

Therefore, it is immediate that the RHS of equation (1) above will be real if and only if the denominator in the 2nd term of the RHS of equation (1) happens to equal $1$.

For the other question (WLOG) $|z| \neq 1$ :

Case 1

Im$(z) > 0.$

Im$(\overline{z})$ which happens to equal $(-1) \times ~$Im$(z)$, is being scaled by $\frac{1}{z\overline{z}} = \frac{1}{|z|^2}.$

Further $|z|^2 > 1 \iff |z| > 1.$

Therefore Im$\left(z + \frac{1}{z}\right) > 0 \iff |z| > 1.$

Case 2

Here, with Im$(z) < 0$, the situation is symmetrically opposite to Case 1.

Therefore Im$\left(z + \frac{1}{z}\right) < 0 \iff |z| > 1.$

Title: Describe the complex function $f(z) = z + 1/z$.

Body:
A question from my homework asks me to find the region where $f(z) = z+ \frac{1}{z}$ is real, and the region where $Im f(z) > 0$, the region where $Im f(z) < 0$.

But I don't know where to start, how do you tackle this kind of problems in general?
I am always very bad at this kind of visualizing problems on complex functions, any help would be appreciated.

What I would usually do is solve for
\begin{equation}
0 = Im ( (a + bi) + \frac{a-bi}{a^2 + b^2})
\end{equation}
But the algebra is usually  hideous. Is there any slick approach?


问:
What happens to the infinite monkey theorem when there are an infinite number of keys on the typewriter?
What happens to the infinite monkey theorem when there are an infinite number of keys on the typewriter? So what is the probability of a finite string of keys like the works of Shakespeare being typed up. Thanks for any insights.</p>

答: You need to pick a probability distribution over the keys. If the number of keys is at most countable and each key occurs with positive probability, then the infinite monkey theorem continues to hold and the proof is the same (any finite string only uses finitely many keys so you can ignore the rest). If one of the keys occurs with probability zero and it also occurs in your string, then the string occurs with probability zero.</p>


问:
If $K\subset \mathbb{R}^n$ compact and $U\subset \mathbb{R}^n$ open, $K \subset U$.Prove that $U \cap \left(\mathbb{R}^n -K \right) \neq \emptyset$

Let $K\subset \mathbb{R}^n$ be compact subspace and let $\ U\subset \mathbb{R}^n\ $ be open, non-empty and containing $\ K.\ $ Prove that $\ U \cap \left(\mathbb{R}^n -K \right) \neq \emptyset$

Attempt

The first idea that comes to my mind was start to compare $U$ with $\left(\mathbb{R}^n -K \right)$ and see if is possible that the sets have a intersecton.

Let $U$ open notice that $\mathbb{R}^n-U \subset \mathbb{R}^n-K$ since $U$ is open
then $\mathbb{R}^n-U$ is closed. Now since $K \subset U$ then $U-K \neq \emptyset $ and
in fact $U-K \subset \mathbb{R}^n-K$.From here I affirm that $U\cap  \mathbb{R}^n-K$. The problem is that   I never use that $K$ is compact.
Someone can give me a hint or advice at the moment of prove future propositions that involve Complements.

答:
Method 1. $\Bbb R^n$ is a connected space. It is also a non-compact Hausdorff space, so a compact  $K$ is   closed, and  $\Bbb R^n\setminus K$ is non-empty & open.

For any $v\in \Bbb R^n$ we have $v\not\in U\implies v\not \in K\implies v\in \Bbb R^n\setminus K.$ So $$\Bbb R^n=U\cup ( \Bbb R^n\setminus K).$$ But now if  $U\cap (\Bbb R^n\setminus K)$ is empty then $\Bbb R^n$ is the union of the disjoint pair $\{U,\, \Bbb R^n\setminus K\}$ of non-empty open sets, contrary to the connectedness of  $\Bbb R^n.$

This will hold verbatim under the weaker condition on $K$ that $\overline K=K\ne \Bbb R^n.$

Appendix. The connectedness of $\Bbb R$ implies that a path-connected space is a connected space. Now if $u,v\in \Bbb R^n$ then $f(r)=(1-r)u+rv$ is continuous from $[0,1]$ into $\Bbb R^n$ with $f(0)=u$ and $f(1)=v.$ So  $\Bbb R^n$ is path-connected.

Method 2. (i). If $U=\Bbb R^n$ then $U\cap (\Bbb R^n\setminus K)=\Bbb R^n\setminus K\ne \emptyset.$

(ii). If $\emptyset \ne U\ne\Bbb R^n$ then $\overline U\ne U$ (...because $U$ is open and $\Bbb R^n$ is connected...) so let $p\in \overline U\setminus U.$ Then $p\not \in \overline K$ (...because $p\not\in U$ and $\overline K=K\subset U$...) so there exists an open $V$ with $p\in V$ and $K\cap V=\emptyset.$ But $p\in \overline U$ so there must exist $q\in V\cap U.$ So $q\in U\cap (\Bbb R^n\setminus K).$


问:
Choose $4$ students out $15$ but Andy can only be selected if Jordan is chosen as treasurer. What is the probability of this event happening?
A committee of four people will be selected among a group of 15 students . The committee will have the following positions: president, secretary, treasurer and marshal. Andy, one of the students , will accept a position in a committee as long as Jordan is selected as the treasurer. What is the probability that Andy serves in a committee? Your answer must be rounded to the nearest hundredth.

In my answer I first found the total ways to select 4 out of 15 students. Next since Jordan takes one position there are three positions left and if one of the three is given to Andy, only two are left. Thus, this equals to:
(12C2)(3!)/(15C4)

答: First we calculate the number of possible committees. We have $15$ choices for pres, $14$ for secret, $13 $ for treasure, and $12$ for marshal. Thus, the possible committees are $15*14*13*12$. Now, we need to find the number of VALID committees. We know that we need both Andy and Jordan on our committee. We should note that Jordan MUST BE treasurer. So, how do we arrange Andy and two other members amongst president, marshall, and secretary? Well, we have to choose $2$ other members besides out of the remaining $13$. We may do this $13\choose2$ ways. Since there are $3$ positions, we may arrange these members and Andy $3!$ times. Thus, the number of valid committees is $3! {13\choose2}$. The total number of committees is $15*14*13*12$. Thus the probability is $3! {13\choose2} \over 15*14*13*12$. After some cancelations and multiplications, we get $\frac{1}{70}$.


问:
if $ a_{n+3}=a_{n+2}+a_{n+1}+a_{n}$,show that $a_{n+168}=a_{n}\pmod {13}$
as before my question:problems: there's one step I don't understand
$$a_{0}=0,a_{1}=a_{2}=1,a_{n+3}=a_{n+2}+a_{n+1}+a_{n}$$
then $$a_{n+168}=a_{n}\pmod {13}$$
it is said can use calculating polynomial discriminant .But I can't Thanks

答:
Here's an idea from algebraic number theory. We solve the recursion in $\overline{\mathbb F_{13}}.$ Note that our characteristic polynomial in $\mathbb F_{13}[x]$ factors as
$$x^3-x^2-x-1=(x+6)\left(x^2+6x+2\right).$$
Now fix $\alpha=-6$ with $\beta$ and $\gamma$ the roots of the quadratic. Because $\alpha,\beta,\gamma\in\mathbb F_{13^2}$ (they are, at worst, quadratic irrationals) we remark that it suffices to work in $\mathbb F_{13^2}$ from here on. In particular, we can solve the recursion as
$$a_n=A\alpha^n+B\beta^n+C\gamma^n$$
for some constants $A,B,C\in\mathbb F_{13^2}.$ To finish, we note that $\alpha,\beta,\gamma\in\mathbb F_{13^2}^\times,$ so they have multiplicative order dividing into $\#\mathbb F_{13^2}^\times=168.$ Thus,
$$a_{n+168}=A\alpha^{n+168}+B\beta^{n+168}+C\gamma^{n+168}=A\alpha^n+B\beta^n+C\gamma^n=a_n$$
in $\mathbb F_{13^2}.$ This finishes the proof.


问:
Is an integrated Wiener process recurrent or transient?
Like the title says, if I take an integrated Wiener process / Brownian motion $\int ^t _0 W_s ds$, will it be recurrent or transient? Or, under what conditions will it be one or the other?

I know that, for any $t$, the integral is a normal variable ~$N(0,\frac{t^3}{3})$. So it'll diverge to $\infty$ as $t\rightarrow \infty$. And it's not a martingale. But would it still be recurrent? And how would I show that, or show that it's not?

(If there are any theorems or discussions of this out there, even just a link to that would be great!)

答:
When reading about Wiener processes on Wikipedia, I encountered the statement: " Like the random walk, the Wiener process is recurrent in one or two dimensions (meaning that it returns almost surely to any fixed neighborhood of the origin infinitely often) whereas it is not recurrent in dimensions three and higher," which cited Pólya's Random Walk Constants. This page leads to a whole sort of rabbit whole of many things that are pretty unknown to a current student of stochastic processes.

In googling for an answer to this, I came across this question; so I suppose an indirect answer is one condition for Wiener process to be recurrent is that it must be one- or two-dimensional. Suffice to say I don't really understand why, though.

(I'm unsure if this would be better as a comment, but I don't have enough reputation for that anyway.)


问:
If a sequence in a metric space does not have any convergent subsequence, then its range is closed
"If a sequence $(x_n: n\in\mathbb N)$ of a metric space does not contain any convergent subsequence, then the set $\{x_n: n\in\mathbb N\}$ is closed in the metric space"

I am trying to figure out which definitions make this proof the easiest. Since we are talking about sequences, closed should be best defined as "every sequence in the closed set converges to a point in the set." Now since the sequence $(x_n)$ does not contain any convergent subsequence, it must be unbounded by the Bolzano theorem.

Now can we say that the subset $\{x_n\mid n\in\Bbb N\}$ is closed because you cannot find a sequence in it which converges to a point outside it? (since no sequence even converges to anything at all)

答:
You cannot use Heine-Borel for any arbitrary metric space $X$.

We wish to show that the set $A = \{x_n \mid n \in \Bbb N\}$ is closed in $X$. Note that if $y$ is a limit point of $A$, then there exists a sequence of distinct points in $A$ converging to $y$. However, given a sequence of distinct points in $A$, you can find a corresponding subsequence of $(x_n)_n$.1 However, we know that $(x_n)_n$ has no convergent subsequences. Thus, $A$ has no limit points and is closed.

Note that I've also shown that no point of $A$ is a limit point.

To elaborate more:

Suppose $(z_n)_n$ is a sequence of distinct points in $A$. Then, each $z_n$ is equal to some $x_{m(n)}$. Points being distinct implies that $n \mapsto m(n)$ is one-one.

Now, there are only finitely many naturals which are smaller than $m(1)$. Thus, we can find $n_2 > 1$ large enough so that $m(n_2) > m(1).$ A similar argument shows that we can find $n_3 > n_2$ large enough so that $m(n_3) > m(n_2)$ and so on.

Putting $n_1 = 1$, we see that $\left(z_{n_i}\right)_i$ is indeed a subsequence of $(x_n)_n$. Now, being a subsequence, $\left(z_{n_i}\right)_i$ has the same limit as $(z_n)_n$ and the argument goes through.


答:
Is it always possible to achieve the correct answer using both methods?

It depends on what you mean by "achieve."

Both methods should always give the same answer--when the integral can be evaluated explicitly by both methods.

However, it may be that one antiderivative may not have an elementary expression, or that the limits of integration cannot be expressed by elementary antiderivatives, and so the volume requires the use of special functions to calculate (or approximate) if the "wrong" method is selected.

One simple example would be the area produced by revolving the region $0 \leq x \leq 1, x^5 + x \leq y \leq 2$ about the $y$-axis:

Even writing down the integrand to evaluate this volume via disk integration requires Bring radicals, which are not elementary functions: $$\text{Volume } = \pi \int_0^2 BR(y)^2 dy.$$ It's debatable at best whether computing the volume of this solid via the disk method is "achievable."

Yet if we set this up using shell integration instead, any second-semester high-school calculus student can evaluate the resulting volume integral easily: $$\text{Volume } = 2\pi \int_0^1 x (2 - x^5 - x) dx = 2\pi [x^2 - \frac{x^7}{7} - \frac{x^3}{3}]_0^1 = \frac{22 \pi}{21}.$$ Perhaps there's some clever substitution which could be used to evaluate the disk integral above, in which case we would get the same answer of $\frac{22 \pi}{21}$.

But I'm pretty sure that any such substitution would in fact end up being a thinly (or not-so-thinly) disguised version of the shell integral, regardless.

Title: Volume of solid of rotation - can you always get same answer using shell or disk-washer?

Body:
The question is to find the volume of revolution of the region bonded by $x=0$ and $y=10$ and $y=8$ and $xy=8$ when rotated about the $x$ axis.

Using the Shell method is almost trivial:

$V=2 \pi \int y (8/y)dy $ with bounds from $8$ to $10$.  The y variables cancel and the answer is $32 \pi$.

I'm having difficulty doing this with the disk-washer method.

Is it always possible to achieve the correct answer using both methods?


问:
What is the smallest solution to $x = 2 \operatorname{mod} (6y+1)$, $x = 3 \operatorname{mod} (12y+1)$, $x= 3 \operatorname{mod} (18y+1)$?
I'd like to solve a system of modular equations, but it's a somewhat unusual system. Specifically, what is the smallest natural number $x$ for which $x = 2\pmod{6y+1}$, and $x=3\pmod{12y+1}$, and $1 \pmod{18y+1}$ for some natural number $y$?

If $y=1$, then WolframAlpha tells me that $x=1654$.  But is it possible that a bigger choice of $y$ could yield a smaller value of $x$?  This came up in the context of Godel coding, by the way.

答:
The bases, call them $p,q$ and $r$, are coprime, so try to solve
$$aqr=2\pmod p\\
bpr=3\pmod q\\
cpq=1\pmod r$$
The answer is $aqr+bpr+cpq\pmod{pqr}$


答:
Let's have a sharper statement:

If a sequence $(x_n: n\in\mathbb N)$ of a metric space does not contain any subsequence convergent to any point different from any of $\ x_n,\ $ then the set $\{x_n: n\in\mathbb N\}$ is closed in that metric space.

Proof   For every point $\ y\ $ different from all $\ x_n\ $ there exists a radius $\ r_y>0\ $ such that open ball

$$ B_y\ :=\ \{z: d(y\ z)<r_y\} $$

does not contain any of $\ x_n.\ $ Then the union of such balls,

$$ \bigcup_y\, B_y $$

is an open set, and $\ \{x_n: n\in\mathbb N\} $ is closed because it is the complement of the said union.

Title: If a sequence in a metric space does not have any convergent subsequence, then its range is closed

Body:
"If a sequence $(x_n: n\in\mathbb N)$ of a metric space does not contain any convergent subsequence, then the set $\{x_n: n\in\mathbb N\}$ is closed in the metric space"

I am trying to figure out which definitions make this proof the easiest. Since we are talking about sequences, closed should be best defined as "every sequence in the closed set converges to a point in the set." Now since the sequence $(x_n)$ does not contain any convergent subsequence, it must be unbounded by the Bolzano theorem.

Now can we say that the subset $\{x_n\mid n\in\Bbb N\}$ is closed because you cannot find a sequence in it which converges to a point outside it? (since no sequence even converges to anything at all)


问:
If $x+2y=8$, find the minumum of $x+y+\frac{3}{x}+\frac{9}{2y}$ (x, y ∈ R+)
Question : If $x+2y=8$, find the minumum of $x+y+\dfrac{3}{x}+\dfrac{9}{2y}$ $(x, y \in \mathbb{R}^+)$

I tried to use $x=8-2y$, and I thought I can plug in. But I think I'm not doing it well. I thought about AM-GM, and how can I expand this  using  $x+2y=8$?

答:
Knowing the answer from taking the derivative, I'd cheat by "observing"

$$x+y+\frac 3x + \frac{9}{2y}=\left(\frac x2+y\right) + \left(\frac x2 + \frac 2x\right)+\left(\frac 1x+\frac {9}{2y}\right)$$
$$\ge 4+2\sqrt{\frac x2 \cdot \frac 2x}+\frac{(x+2y)(\frac 1x + \frac{9}{2y})}{x+2y} \tag{AM-GM}$$
$$\ge 4+2+\frac{(1+3)^2}{8}=8 \tag{C-S}$$

Without knowing the answer and barred from using calculus, I'd do

$$x+y+\frac 3x + \frac{9}{2y}=\left(\frac x2+y\right) + \left(\frac x2 + \frac \alpha x\right)+\left(\frac {3-\alpha}{ x}+\frac {9}{2y}\right)\\
$$
and solve for $\alpha$ and $x_0$ such that
$$\frac {x_0} {2}=\frac{\alpha}{x_0} \tag1$$
$$\frac{x_0^2}{3-\alpha} = \frac{(8-x_0)^2}{9}\tag2$$


问:
What are the elements in this set $(-\infty, 4] \cap \mathbb{N}$
I believe the answer would be $\{1,2,3,4\}$ since $\mathbb{N}$ means natural numbers and there can't be any negative integers?

答:
The intersection denoted, $\cap$, refers to the elements in common between both sets.

In this problem, you are asked, what is the intersection of $(-\infty, 4]$ and $\mathbb{N}$ (natural numbers $0,1,2,3,4 \ldots)$

The numbers in common between the two sets are  $\{0,1,2,3,4 \}$


问:
Yet Another Proof of Inequality by Induction
I need to show by induction that for $ n \geq 1$, the following inequality holds:

$$\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}} \geq\frac{1}{2(n+1) \sqrt{n}}$$

My initial approach would be to use $P(n-1)$ :

$$\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}} \geq\frac{1}{2(n) \sqrt{n-1}}$$

and then try to show that $P(n-1) \rightarrow P(n)$. Clearly the base case $P(1)$ holds.

I understand that this should be solved by transforming the terms, e.g. by multiplying  the expression $P(n-1)$ by something like $\sqrt(n-1)$... but after a few "guesses" I'm not making any progress. Any hints on how to transform the expression in order to be able to use the inductive step ? Generally, what is a good routine/rule of thumb for tackling these problems ? Do you just start trying out things to modify the expression ?

答:
The base case $n = 1$ reads: $\dfrac{1}{1} - \dfrac{1}{2} \ge \dfrac{1}{4}$ which is clearly true. Assume the inequality is true for $n = k$, we need to show it is also true for $n = k+1$. This means to prove: $\dfrac{1}{\sqrt{k+1}} - \dfrac{1}{\sqrt{k+1}+1}\ge \dfrac{1}{2(k+2)\sqrt{k+1}}$. We have: $LHS = \dfrac{1}{\sqrt{k+1} \left(\sqrt{k+1}+1\right)}\ge RHS $ because $2(k+2) \ge 1+\sqrt{k+1}$ is quite clear as $2(k+2) = 1+2k+3 \ge 1+k+1 \ge 1 +\sqrt{k+1}$. Thus you are done.

Note: This note should come before the two comments below. But it seems that the induction step was not helpful or not necessary in showing the step for $n = k+1$ and this is why I did not invest time in finding ways to link this inductive step to the final step when $n = k+1$.


问:
Understanding the idea of the proof of "For two ideals $I$ and $J$ prove that prove that $I(R/J)=(I+J)/J$"
I was reading this question, but I could not understand the proof of why for two ideals $I$ and $J$ we have $I(R/J)=(I+J)/J$.

Here is the proof given in this link:

Well, if you use the definition, it's obvious: an element of $I\cdot R/J$ is  the ideal of $R/J$ generated by $I$, so it is the set of finite sums of the form:
$$\Bigl\{\sum_k i_kr_k+J\bigm\vert i_k\in I, r_k\in R\Bigr\}.$$
Now, as $I$ is an ideal in $R$, the set of these finite sums u=is nothing but $I$ itself, whence the formula
$$I\cdot R/J=(I+J)/J.$$
Note that, by the second isomorphism theorem,  the latter quotient is isomorphic to $I/I\cap J$.

Here is what I know:

I know how to show that $IM$ is a submodule of $M$ and hence it is an ideal of $M.$

Here is what I do not understand in the solution above:

1- Why $I \cdot R/J$ must be generated by $I$?

2- Why "as $I$ is an ideal in $R$, the set of these finite sums u=is nothing but $I$ itself" and why that leads to that $I(R/J)=(I+J)/J$ are not we moding out by $J$ in the RHS of this formula?

Could anyone explain these points to me please?

答: Given commutative unital rings $R$ and $S$ such that there exists a ring homomorphism $\varphi : R \to S,$ we define the extension of an ideal $I$ of $R$ to be the ideal $$I^e = \varphi(I) S = {\left \{\sum_{k = 1}^n s_k \varphi(i_k) \,|\, n \geq 0 \text{ is an integer, } i_k \in I, \text{ and } s_k \in S \text{ for all } k \right \}}$$ of $S.$ For an ideal $J$ of $R,$ there is a ring homomorphism $\pi : R \to R/J$ defined by $\pi(r) = r + J.$ Putting this together with the above shows that $I^e = \pi(I)(R/J)$ consists of the elements $$\sum_{k = 1}^n (r_k + J)(i_k + J) = \sum_{k = 1}^n (r_k i_k + J) = {\left(\sum_{k = 1}^n r_k i_k \right)} + J.$$ Often, the map $\pi$ is omitted, and this ideal is simply written $I(R/J).$ From here, it is straightforward to verify that $I(R/J) = (I + J)/J$ by showing the containments $\subseteq$ and $\supseteq.$


答:
Let $\ U\ne\emptyset\ $ be open. If $\ U\ $ were also compact then it would be closed (and open at the same time). Then $\ U\ $ must be empty (but it is not!) or $\ U=\mathbb R^n\ $ because $\ \mathbb R^n\ $ is connected. But $\ \mathbb R^n\ $ is not compact -- a contradiction.

It follows that non-empty open $\ U\ $ is not compact. If $\ K\subseteq U\ $ for a compact K then (since $\ K\ne U)$ we obtain $\ U\setminus K\ne\emptyset.$

Title: If $K\subset \mathbb{R}^n$ compact and $U\subset \mathbb{R}^n$ open, $K \subset U$.Prove that $U \cap \left(\mathbb{R}^n -K \right) \neq \emptyset$

Body:
Let $K\subset \mathbb{R}^n$ be compact subspace and let $\ U\subset \mathbb{R}^n\ $ be open, non-empty and containing $\ K.\ $ Prove that $\ U \cap \left(\mathbb{R}^n -K \right) \neq \emptyset$

Attempt

The first idea that comes to my mind was start to compare $U$ with $\left(\mathbb{R}^n -K \right)$ and see if is possible that the sets have a intersecton.

Let $U$ open notice that $\mathbb{R}^n-U \subset \mathbb{R}^n-K$ since $U$ is open
then $\mathbb{R}^n-U$ is closed. Now since $K \subset U$ then $U-K \neq \emptyset $ and
in fact $U-K \subset \mathbb{R}^n-K$.From here I affirm that $U\cap  \mathbb{R}^n-K$. The problem is that   I never use that $K$ is compact.
Someone can give me a hint or advice at the moment of prove future propositions that involve Complements.


问:
Moment of inertia using calculus
In physics books,the moment of inertia of a uniform rod with negligible thickness is determined using calculus.They take an element $dx$ $x$ distance from the center of the rod and calculuate moment of inertia as $dI=\frac{M}{l}x^2dx$,where $M$ is the mass of whole rod and $l$ is the total length.Then they integrate to find the total moment of inertia.My question is how do we express the integral as the limit of riemann sum or an approximation using $\Delta x$ instead of $dx$?

答: Integrating your given expression, we get $I=\int_a^b\frac{M}{l}x^2dx$. We note that $M$ and $l$ are constants. Now, we examine the Riemann sum definition of the integral which states that $\int_a^b f(x)dx=\lim \limits_{n \to \infty} \sum_{i=1}^n f(x_i)*\frac{b-a}{n} $. We have that $f(x) = \frac{M}{l}x^2$, so our Reimann sum will be $\lim \limits_{n \to \infty} \sum_{i=1}^n \frac{M}{l}{x_i}^2*\frac{b-a}{n}$.

Title: Moment of inertia using calculus


问:
If a sequence in a metric space has no convergent subsequence, then there exist a open ball around each point
"If a sequence in a metric space has no convergent subsequence, then there exist a open ball around each point that does not intersect any other point of the sequence "

I think the important idea for this proof is that if we take any point of the sequence and consider all points less than radius 1, we can take the minimum or half of the minimum and the ball around this point would not intersect any other point of the sequence. Now the reason why the neighborhood around any point must only contain a finite number of other points is because the sequence has no convergent subsequence. Otherwise we can create a convergent subsequence.

Is this correct?

答: If every open ball around point $p$ intersects a point of the sequence other than $p$, then you can construct a subsequence that converges to $p$.


问:
What set theory axioms do I need to believe in uncountable ordinals?
Math people:

The title is the question. I am convinced that uncountable sets exist, thanks to Cantor's diagonal proof.  It is not intuitively clear to me that uncountable ordinals or cardinals should exist.  What axioms of set theory are needed to accept their existence?  I think I have read here that the Axiom of Choice is not necessary.  I am reluctant to accept as "true" any theorems that rely on the Axiom of Choice.

EDIT: This question may be a duplicate: I just found No uncountable ordinals without the axiom of choice? .  I apologize if this is so.

答:
As [1], [2], [3], and [4] may tell you, the axiom of choice is not needed to define $\omega_1$ (in particular [1] and [4]).

Two key axioms are the power set axiom and the replacement axiom schema. In [2] and [3] you can see why the axiom of power set is needed. It is consistent with $\sf ZF$ without the power set axiom that there are only countable ordinals. In particular the set of hereditarily countable sets satisfies that. In fact, it shows that without the power set axiom we cannot prove the existence of any uncountable sets.

But the replacement schema is also essential. We use it in order to map a certain subset of $\mathcal P(\omega\times\omega)$ onto ordinals, and we need the replacement schema to show that the result is a set. Indeed if we consider $\sf ZF$ without the replacement schema, then $V_{\omega+\omega}$ is a model of these axioms, and there are only countable ordinals in that model. It should be remarked that there may still be well-ordered of length $\omega_1$ in $V_{\omega+\omega}$ but the von Neumann ordinal, a transitive set ordered by $\in$ does not exist there. That is to say, it is consistent that the axiom of choice holds, and every set can be well-ordered, but the von Neumann ordinals don't exist beyond $\omega+\omega$. In such model we separate between ordinals as we think about today (von Neumann's definition), and as equivalence classes of order types (which are proper classes, of course).

Of course one uses the axiom of union all the time, as well extensionality. It should be remarked that regularity is not necessarily because we can always limit ourselves to the part of the well-founded sets, where it holds, at least if we have replacement.

The Links:

How do we know an $ \aleph_1 $ exists at all?

Uncountable ordinals without power set axiom

Does the definition of countable ordinals require the power set axiom?

No uncountable ordinals without the axiom of choice?


问:
Linear algebra square root operator example
Hello I'm currently reading Axlers Linear Algebra done right 3rd edition and I'm a little confused by example 7.34 and was wondering if someone could explain it to me. It defines square root as: An operator R is called a square root of an operator T if $R^2=T$.

Example: If $ T \in \mathcal{L}(\mathbf{F^3})$ is defined by $T(z_{1},z_{2},z_{3}) = (z_{3},0,0)$, then the operator $R \in \mathcal{L}(\mathbf{F^3})$ defined by $R(z_{1},z_{2},z_{3}) = (z_{2},z_{3},0)$ is a square root of T.

答:
Since $R$ is an operator, $R^2$ means applying $R$ to the answer you get from applying $R$ once. Also clearly, $R$ just sifts the last two vector entries of any 3-d vector to  the left and makes the last entry zero. So

$R^2(z_1,z_2,z_3)=R(R(z_1,z_2,z_3))=R(z_2,z_3,0)=(z_3,0,0)=T(z_1,z_2,z_3).$

This shows $R$ is indeed the square root of $T$.


问:
How to get vector $\vec{v}$ and on matrix $A$?
I have a surjective matrix such that

$$A * V = A * [1,1,1]$$

$$A = \begin{bmatrix}1&2&3\\2&1&1\end{bmatrix}$$

I need to find V that is of shape (3,).

This is what I did so far

$$\begin{bmatrix}1&2&3\\2&1&1\\0&0&1\end{bmatrix}.\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}=\begin{bmatrix}6\\4\\?\end{bmatrix}$$

I used a value of 2 for the ?

So

$$\begin{bmatrix}1&2&3\\2&1&1\\0&0&1\end{bmatrix}.\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}=\begin{bmatrix}6\\4\\2\end{bmatrix}$$

Solving this got me

$$v_1 + 2v_2 + 3v_3 = 6$$
$$2v_1 + v_2 + v_3 = 4$$
$$v_3 = 2$$

which implies

$$v_1 = -2, v_2 = 6, v_3 = 2$$

But apparently, this is wrong. I am not sure what I am doing wrong. Any help is appreciated. Thanks in advance

Body:
[1 2 3][v1] = [1 2 3][1] = [6]
[2 2 1][v2]   [2 1 1][1]   [4]
       [v3]          [1]

v1+2v2+3v3=6 (1)
2v1+v2+v3=4  (2)

From (2): v3=4-v2-2v1

Plug into (1): v1+2v2+12-3v2-6v1=6
v2=-5v1+6

Plug back in (2): v3=4+5v1-6-2v1
v3=-2+3v1

One possible answer is:
Set v1=2
v2=-4
v3=4

All possible answers of the form:
[0 ] + [1 ]
[6 ]   [-5]x
[-2]   [3 ]


问:
Not every open set in $\mathbb{R}^n$, for $n\geq2$, is a disjoint union of countable open balls.
I know that every open set in $\mathbb{R}$ can be decomposed as the union of countably many disjoint open intervals and I'm trying to show that this property no longer holds in higher dimensions. For this I'm trying to show that there is no countable, pairwise disjoint open balls whose union is the
open square $(0, 1)^2 ⊂ \mathbb{R}^2$. I know that it is because of the corners, but I don't know how to write this in mathematical terms. Thanks for any help.

答: If any connected open set is written as a countbale disjoint union of open bals then all but one of them  must be empty (because of conncetedness) so the original open set must itself be an open ball. So an open rectangle would be a counter-example.</p>


问:
Struggling with the intuitive picture behind the Radon-Nikodym derivative
I understand the math and the formal statements behind this theorem (well perhaps not), but I cannot seem to grasp an intuitive picture behind it like how one can intuitively understand an ordinary derivative as a simple rate of change. I have read through a few answers on here but most seem to focus on probability applications. Can anyone provide an intuitive picture of what is going on behind this theorem?</p>

答:
[Following Masacroso's notation]

There is the common notation $\frac{d\nu}{d\mu}$ for the function $f$ that follows the intuition that this function represents a "rate of change of the density $\nu$ with respect to $\mu$." This notation is quite suggestive, as the equation in Masacroso's comment can be written as $\int_A \, d\nu = \int_A \frac{d\nu}{d\mu} \, d\mu$. Although this isn't exactly a derivative in the usual sense (since $\nu$ and $\mu$ are measures, not functions), the Radon-Nikodym derivative does have many similar properties.

In some instances the Radon-Nikodym derivative is literally a derivative of some function.
For example, if $\nu$ is some measure on the real line with distribution function $F(x) := \nu((-\infty, x])$ satisfying some condition (absolute continuity as a function), then the Radon-Nikodym derivative with respect to the Lebesgue measure is precisely the derivative of $F$, as $\nu((a,b]) = F(b)-F(a) = \int_a^b F'(x) \, dx$ by the fundamental theorem of calculus.


答:
You know that the moment of inertia of a point mass $m$ at distance $x$ from the axis of rotation is $mx^2$.

The rod is rotated about the axis perpendicular to the rod passing through its center. So, we can consider only the momentum of inertia of half of the rod and take the double at the end.

This half rod can now be approximated by $n$ equally distributed point masses with mass $\frac{M}{l}\Delta x^{(n)}$ at distance $x_i=i\cdot \Delta x^{(n)}$ where $\Delta x^{(n)} = \frac l2\cdot \frac 1n $ and with a total momentum of inertia of

$$\frac Ml\sum_{i=1}^nx_i^2\Delta x^{(n)}\stackrel{n\to\infty}{\longrightarrow}\frac Ml\int_0^{\frac l2}x^2\,dx$$

So, for the whole rod you then get

$$I = 2\frac Ml\int_0^{\frac l2}x^2\,dx = \frac{Ml^2}{12}$$

Title: Moment of inertia using calculus

Body: In physics books,the moment of inertia of a uniform rod with negligible thickness is determined using calculus.They take an element $dx$ $x$ distance from the center of the rod and calculuate moment of inertia as $dI=\frac{M}{l}x^2dx$,where $M$ is the mass of whole rod and $l$ is the total length.Then they integrate to find the total moment of inertia.My question is how do we express the integral as the limit of riemann sum or an approximation using $\Delta x$ instead of $dx$?


问:
Comparing coefficients of polynomials
If we have two polynomials, say for example, $a_0+a_1x+a_2x^2 = 1+2x+3x^2$ for all $x$, why must it follow that $a_0 = 1,a_2 = 2,$ and $a_3 = 3$? Can we generalize this for any polynomials and infinite sum polynomials as well?</p>

答:
If equal, subtract them from each other and you know the answer is zero, perfect! So that means
$0 = (a_0 - 1) + (a_1 - 2)x + (a_2 - 3) x^2$, but you already picked up on that. But, and here's where it all happens, there's a LOT of flexibility with what $x$ is. What is it telling you when $x = 0$, for example? And if that's the case, about $a_1$? I think you see the pattern.

Title: Comparing coefficients of polynomials


问:
Unique largest normal pi-subgroup
Let $\pi$ be a set of prime numbers. A finite group is said to be a $\pi$-group if every prime that divides its order lies in $\pi$. If $G$ is finite, show that $G$ has a unique largest normal $\pi$-subgroup (which may be trivial and may be all of $G$).

What I did: suppose $|G|=p_1^{k_1}p_2^{k_2}\ldots p_n^{k_n}$. If $\pi$ contains a prime outside of the $p_i$'s, then the only $\pi$-subgroup is the trivial group. Then consider when $\pi$ contains a subset of the $p_i$'s, say $p_1,p_2,\ldots,p_m$. I don't know what I can say about a subgroup whose order is divisible by $p_1,p_2,\ldots,p_m$.

答: There exists a maximal normal $\pi$-subgroup of $G$ since $G$ is finite. If $H$ and $K$ are two such subgroups what can you say about $HK$?</p>

Title: Unique largest normal pi-subgroup