答:
$p: X \to Y$ is a quotient map iff

$C \subseteq Y$ is closed iff $p^{-1}[C]$ is closed in $X$ $(a)$.

(a reformulation in terms of closed sets; follows from the standard definition as $X\setminus p^{-1}[O] = p^{-1}[Y\setminus O]$ for any $O \subseteq Y$ and any map $p$)

So if $p$ is continuous and onto, we can show $\Rightarrow$ in $(a)$ by noting that $p$ is continuous and $\Leftarrow$ by noting that $p[p^{-1}[C]] = C$ as $p$ is onto and closed as $p$ is closed. So $(a)$ holds for $p$ and so $p$ is quotient.

Title: A surjective, continuous and closed map is a quotient map

Body:
I want to prove that a surjective, continuous and closed map is a quotient map. I know that a surjective map $p: X \longrightarrow Y$ is a quotient map if every set $V \subset Y$ is open $\Longleftrightarrow$ $p^{-1}(V)\subset X$ is open. Please note that $p^{-1}$ is the pre-image

Since we have continuity, one only has to prove the implication $\Longleftarrow$

Let $V\subset Y$ so that $p^{-1}(V)\subset X$ is open. $X\backslash p^{-1}(V)$ is closed and since $p$ is a closed map, $p(X\backslash p^{-1}(V))$ is closed. Here is where I get confused. If I knew that $p$ was injective, then I could keep going like this:
$$p(X\backslash p^{-1}(V))=p(X)\backslash p(p^{-1}(V))$$
and since $p$ is surjective:
$$p(X)\backslash p(p^{-1}(V))=Y\backslash V$$
and then we'd be done, because $Y\backslash V$ is closed and therefore $V$ is open. However, as I don't know if $p$ is injective I can't follow this path. Could someone help me?


问:
Gaussian like sum with binomial coefficients
Is there any way to calculate expressions of the form

$$
f(q)=\sum_{n=0}^N\binom{n}{k}\exp({2\pi i\tfrac{n}{N}\cdot  q})
$$

where $q\in\mathbb{Q}$?

It reminds me quite alot of Gauss sums, but I don't know much about those.

Addendum:
Please find my more general question here: Geometric sum with binomial coefficient

答:
I presume $k$ is a positive integer, $N \ge k$, and the terms for $n < k$ are taken as $0$.

Maple says
$$ \sum_{n=k}^N {n \choose k} r^n = \frac{r^k}{(1-r)^{k+1}} - {N+1 \choose k} r^{N+1} {}_2F_1(1,N+2; N-k+2; r) $$

ContentLicense: CC BY-SA 4.0


问:
Why Conditional Probability is used
Question:

Parcels from sender 'S' to receiver 'R' pass sequentially through 2 post-offices.
Each post-office has probability 1/5 of losing an incoming parcel, independently of all other parcels.
Given that a parcel is lost,  what is the probability that it was lost by the second post-office.

Why do we have to use conditional probability in this problem?

Shouldn't the answer just be:

P(lost at 2nd) = P(not lost at 1st) * P(lost at 2nd)

P(lost at 2nd) = 4/5 * 1/5

P(lost at 2nd) = 4/25

But the answer is 4/9

答:
P(lost somewhere) $=1-(4/5)^2=9/25$

P(lost at second | lost somewhere) $=\frac{4/25}{9/25}=4/9$

Title: Why Conditional Probability is used


问:
Removing the Worst Swimmer Using Linear programming
Given a table of personal best records of 5 people in 4 different swimming styles (freestyle stroke, backstroke, breaststroke, butterfly stroke) how can I choose 4 out of 5 said swimmers for an upcoming competition (Using Linear Programming).

There are many ways I can approach this question but none of them seem to be related to linear algebra in any way, I feel like this question is more related to probability.

First I tried to choose based on their total average, then I tried to choose the guy most likely to get the most losses, But as I said before none of them seem to be related to linear algebra.

any help will be appreciated.

答:
Assume you have to chose one swimmer for each style and no swimmer can perform in more than one style. This is an unbalanced assignment problem. You can frame the linear programming problem as follows:

$$\begin{array}{|c|c|c|c|c|}
\hline
\text{Style\Swimmer}&1&2&3&4&5\\\hline
1&x_{11}&x_{12}&x_{13}&x_{14}&x_{15}\\\hline
2&x_{21}&x_{22}&x_{23}&x_{24}&x_{25}\\\hline
3&x_{31}&x_{32}&x_{33}&x_{34}&x_{35}\\\hline
4&x_{41}&x_{42}&x_{43}&x_{44}&x_{45}\\\hline
\end{array}
$$

In the above table I have arranged the swimmers column-wise and strokes row-wise. $x_{ij}=1$ iff swimmer $j$ is selected to compete in stroke $i$ and $0$ otherwise. Let $t_{ij}$ denote the personal best of swimmer $j$ in stroke $i$.

Since one swimmer can be selected for at-most one stroke, we require all column-sums to be $\le1$. Thus one set of constraints is $\forall j\left(\sum_{i=1}^4x_{ij}\le1\right)$.

For each stroke we must select exactly one swimmer, so the row-sum must be $1$. Thus another set of constraints is $\forall i\left(\sum_{j=1}^5x_{ij}=1\right)$.

The final set of constraints is that $0\le x_{ij}\le1$ and integral (i.e. $x_{ij}=0,1$). The objective, in my opinion, should be to minimize the total time $T=\sum_i\sum_jx_{ij}t_{ij}$. You can solve this LPP using integer programming or the Hungarian method to solve assignment problems.


问:
Expectation of Itô stochastic integral
$$E\left[\int_0^\infty \left|\int_0^t(W(s))^2 \, dW(s)\right|^{\,2} e^{-t} \, dt\right]$$

Not sure if, by Itô Isometry, it is possible to pull the expectation operation inside in the integral because of the $$e^{-t}$$ term.

Any insight is appreciated.

答:
Let $X_t = \int_0^t W_s^2 dW_s$. You want to compute $C = \mathbb{E}[\int_0^\infty X_t^2 e^{-t} dt]$. Since $X_t^2 e^{-t} \geq 0$, by Fubini's Theorem we have
\begin{align*}
C =& \int_0^\infty \mathbb{E}[X_t^2 e^{-t}] dt \\
=& \int_0^\infty e^{-t} \mathbb{E}\left[\int_0^t W_s^4 ds\right] dt \\
=& \int_0^\infty e^{-t} \int_0^t \mathbb{E}[W_s^4] ds dt \\
=& \int_0^\infty \int_0^t 3e^{-t}s^3 ds dt \\
=& \int_0^\infty t^3 e^{-t} dt
\end{align*}
where the second equality follows from the Ito Isometry and the third by another application of Fubini.

Title: Expectation of Itô stochastic integral


问:
Show $\overline{\oint_{\gamma}f(z)dz} =- \oint_{\gamma}\overline{f(z)}\cdot z^{-2}dz$
Let $U \subset \Bbb C$ and $f:U \to \Bbb C$ continuous. Furthermore let $\gamma: [0,2 \pi] \to U$ be a curve given by $\gamma(t)=exp(it)$. To show is:
$$\overline{\oint_{\gamma}f(z)dz} =- \oint_{\gamma}\overline{f(z)}\cdot z^{-2}dz$$
I could show:
$\overline{\int_\gamma \! f(z) \,dz}=\overline{\int_0^{2\pi} \! f(e^{it})\cdot ie^{it} \,dt}=\int_0^{2\pi} \! \overline{f(e^{it})\cdot ie^{it}} \,dt$ but now I can't end the proof. Can someone helps me?

Body: Note that $$-\int_\gamma\overline{f(z)}z^{-2}dz=-\int_0^{2\pi}\overline{f(e^{it})}e^{-2it}ie^{it}dt=\int_0^{2\pi}\overline{f(e^{it})}(-i)e^{-it}dt=\int_0^{2\pi}\overline{f(e^{it})ie^{it}}dt=\cdots$$


问:
Is every direct product of (finite) cyclic groups abelian?
Long story short I'm halfway through a proof and have hit a step where I show that a group order $p^2$ is abelian. I did this by splitting into the $C_{p^2}$ case and the $C_p\times C_p$ case, the first being trivial as all cyclic groups are abelian. The second case I proved by using the definition of the direct product and showing it directly from the definition of the two $C_p$ groups.

However this feels odd. Something feels off - especially added to the fact that my proof seemingly holds for any two cyclic groups, implying the direct product of any two cyclic groups is abelian, and yet I can't find anything online talking about this. Is it just so trivial nobody mentions it? The only thing I consistently see brought up is that "every cyclic group is an abelian group" (trivial), and "every finitely generated abelian group is a direct product of cyclic groups", which is the converse of what I'm trying to show.

答:
Yes.

If $G\cong \langle a\mid a^m\rangle, H\cong\langle b\mid b^n\rangle$, then

$$G\times H\cong\langle a,b\mid a^m, b^n, ab=ba\rangle.$$

See here for details.


问:
A matrix polynomial equation when power goes to infinity.
$$A=\pmatrix{\frac2{10}&\frac3{10}\\\frac3{10}&\frac5{10}},\\B=\pmatrix{2&3\\4&5},\\XA+XA^2+XA^3+...=B$$ what is $X$?
I think it is kind of like the transition matrix, but the sum of each row is not 1. So how should I get started?

答:
Your matrix $A$ has eigenvalues of absolute value $< 1$, so the series $\sum_{n=1}^\infty A^n$ converges to $A(I-A)^{-1}$.

But please don't call it a "polynomial of infinite degree".


问:
$\arctan x$ vs $1/\tan x$
Why is $\arctan x = (\tan x)^{-1} = \tan^{-1}x$ but is not equal to $1/\tan x$, even though $\tan^{2}x = (\tan x) \cdot (\tan x)$?

Is there not another way to write $1/\tan x$?

答:
Strangely, nobody seems to have answered the second part of your question:

Is there not another way to write  $1/\tan x$?

The answer is yes:  the multiplicative inverse of the tangent function is called the cotangent function, and is denoted $\cot x$.  You can also write this as $(\tan x)^{-1}$, if you prefer.  Notice that the parentheses are playing an important role, here!

You may notice that in the sentence above I used the phrase "multiplicative inverse" to describe the relationship between $\tan$ and $\cot$.  This phrase means something different from "inverse function", which describes the relationship between $\tan$ and $\tan^{-1}$, and I think this distinction is at the heart of your question, so let me take a minute to explain the underlying issue.

The set of all functions (or, if you prefer, all functions defined on a specific interval) is not just a set of individual things, like buttons on a calculator; there are relationships among them, and operations that can be performed on them.  For example, you can add two functions $f$ and $g$ together, to get a new function $f + g$.  You can also multiply two functions together to get a new function $fg$.  And there is a third operation:  you can compose two functions to get $f\circ g$, defined by the rule $(f\circ g) (x) = f(g(x))$.

With respect to each of these operations, you can ask some important questions:

Is there an identity function with respect to this operation?

If there is an identity function for a particular operations, do functions have inverses with respect to that operation?

Let's consider how we answer these questions for the three operations I've mentioned: addition, multiplication, and composition.

1.  Function addition:

An "additive identity" would be a function $k$ with the property that for any $f$, we would have $f + k = f$. Because of how function addition is defined, this is equivalent to $f(x) + k(x) = f(x)$, which in turn is equivalent to $k(x) = 0$.  So the constant function $k(x) = 0$ is the additive identity function.

Now what about additive inverses?  The question we ask is, given any function $f$, is there a "counterpart" function $g$ with the property that $f + g = k$, where $k$ is the additive identity function we just found?  The answer is yes:  we define $g$ by $g(x) = -f(x)$.  Then $g$ is the additive inverse of $f$.

2.  Function multiplication:

A "multiplicative identity" would be a function $i$ with the property that for any $f$, we would have $f \cdot i = f$. Because of how function multiplication is defined, this is equivalent to $f(x) \cdot i(x) = f(x)$, which in turn is equivalent to $i(x) = 1$.  So the constant function $i(x) = 1$ is the multiplicative identity function.

Now what about multiplicative inverses?  The question we ask is, given any function $f$, is there a "counterpart" function $g$ with the property that $f \cdot g = i$, where $i$ is the multiplicative identity function we just found?  The answer is yes, provided that $f(x)$ is always nonzero:  we define $g$ by $g(x) = \frac{1}{f(x)}$.  Then $g$ is the multiplicative inverse of $f$.  (If $f(x) = 0$ for one or more values of $x$ then $g$ would only be defined on a restricted domain; in that case we can say that $$g(x) = \frac{1}{f(x)}$ is a partial multiplicative inverse.)

3.  Function composition:

A "compositional identity" would be a function $e$ with the property that for any $f$, we would have $f \circ e= f$ and $e \circ f = f$.  (We need both requirements here because composition, unlike multiplication and addition, is not commutative). This is equivalent to requiring $f(e(x)) = f(x)$ and $e(f(x)) = f(x)$.  It is not too difficult to see that the function $e(x) = x$ has both of these properties, and in fact is the only function with both of these properties.  So the constant function $e(x) = x$ is the compositional identity function.   (People usually just call it the "identity function", and drop the word "compositional", but I'm going to keep it here because the distinction between function multiplication and function composition is at the core of your question.)

Now what about compositional inverses?  We want to know if, for any function $f$, there exists some $g$ with the property that $f \circ g = e$ and $g \circ f = e$; you might be better able to recognize these requirements in the form $f(g(x)) = x$ and $g(f(x)) = x$.  Such a function might not exist -- in particular, the second requirement requires that $f$ be a one-to-one function (you may recognize this as the "horizontal line test").  If $f$ is not invertible, we may be able to restrict the domain of $f$ to one in which it is one-to-one, in which case we can define a compositional inverse function, which we denote $f^{-1}(x)$.

Notice what has happened:  for each operation, there is an identity function, and a different kind of inverse function.

For addition, the identity is $k(x) = 0$, and the additive inverse of $f(x)$ is $-f(x)$.

For multiplication, the identity is $i(x) = 1$, and the multiplicative inverse of $f(x)$ is $\frac{1}{f(x)}$ (where defined).

For composition, the identity is $e(x) = x$, and the composition inverse of $f(x)$ is $f^{-1}(x)$ (if defined).

Let's take a specific example, related to the one you asked about.  With $f(x) = \sin x$,

the additive inverse is $-\sin x$

the multiplicative inverse is $\frac{1}{\sin x}$, which is called the cosecant function, denoted $\csc x$

the compositional inverse is $\sin^{-1}(x)$, which is also denoted $\arcsin x$, and is called the arcsine function.

Confusion between the two operations of "function multiplication" and "function composition" is commonplace and, to some extent, natural.  We use almost the same notation for both (compare $f \cdot g$ and $f \circ g$), and the notations $f^{-1}(x)$ and $(f(x))^{-1}$ are almost identical. I think at a deeper level the confusion between multiplication and composition also underlies common errors, like writing
$$\sqrt{x + y} = \sqrt{x} + \sqrt{y}$$
and other versions of what is called the "Freshman's Dream".  At a formal level, the error here is thinking that
$f \circ (g + h) = f \circ g + f \circ h$, i.e. believing that composition, like multiplication, distributes across addition.

Title: $\arctan x$ vs $1/\tan x$


问:
Hodge decomposition seems to say harmonic and exact is zero (and possibly (harmonic and co-exact) and (exact and co-exact))

What I understand. Please correct if wrong:

Part of Hodge Decomposition Theorem says that for a compact oriented Riemannian (smooth) $m$−manifold $(M,g)$ (I think M need not be connected, but you may assume connected if need be or you want) and for a smooth $k$-form $\omega$, i.e. $\omega \in \Omega^k(M)$

A. $\omega$ decomposes into exact, co-exact and harmonic: $\omega =$ (respectively) $\omega_d \oplus \omega_\delta \oplus \omega_\Delta $$\in \Omega^k(M) = B^kM \bigoplus \mathscr H^k(M,g) \bigoplus image(\delta_{k+1})$

B. $\omega$ is zero if $\omega$ is at least 2 of the ff: harmonic, exact, co-exact. Specifically:

B.1. 'harmonic and exact is zero'

B.2. 'harmonic and co-exact is zero'

B.3. 'exact and co-exact is zero'

C. I think (B.1) is equivalent to the injectivity of the map $\phi: \mathscr H^k(M,g) := \ker(\Delta_k) \to$$ H^k_{dR}M := \frac{Z^kM}{B^kM}$, which maps a harmonic form to its de rham cohomology class, $\phi(\omega)=[\omega]:=\omega + B^kM$ (or '$\omega\bmod B^kM$').

Proof: $\phi$ is the restriction of $\Phi:Z^kM \to H^k_{dR}M$ (from closed $k$-forms) to harmonic $k$-forms. Then $\ker(\phi) = \ker(\Phi) \cap Domain(\phi) = B^kM \cap \mathscr H^k(M,g) = $$image(d_{k-1}) \cap \ker(\Delta_k)$. QED

C.1. Note on notation: I personally would like to have $\phi$ and $\Phi$ to have subscripts $k$ since their domains and ranges depend on $k$, but I'm omitting the subscript $k$ because I'm following notation in the following powerpoint.

Questions about this powerpoint by Ryan Vaughn:

What exactly is going on in slide 46? This is part of proving that $\phi$ is injective.

1.1. What I think: Here, Ryan Vaughn seems to conclude, for the decomposition of harmonic $\omega = \omega_d \oplus \omega_\delta \oplus \omega_\Delta$, that $\omega_{\Delta}$ is zero because $\omega_{\Delta}$ is exact and because of (B.1). However, I really don't think decomposition is necessary because the other components of $\omega$ will obviously be zero.

1.2. I assume that the decomposition itself (I mean without the 'orthonormal' part) doesn't rely on (B.1) (or (B.2)-(B.3)), otherwise I think this would be circular. I'm guessing that (B.1) (or (B.2)-(B.3)) is used in the 'ortho(normal)' part of the decomposition theorem. Thus, it makes sense to talk about $\omega = \omega_d \oplus \omega_\delta \oplus \omega_\Delta$ even if we don't know (B.1) (or (B.2)-(B.3)).

How does Ryan Vaughn prove (B.1), if Ryan Vaughn did? Otherwise, how does one prove (B.1) (of course without assuming the full Hodge Decomposition Theorem; soooo of course proving (B.1) is part of proving Hodge Decomposition Theorem)?

What I tried so far:

2.1 - I think the following is somehow relevant, if true:

a smooth $k$-form is harmonic if and only if closed and co-closed, i.e. $\ker (\Delta_k) = $$ \ker(d_k) \cap \ker(\delta_k) = Z^kM \cap \ker(\delta_k)$

2.2 - If true: It seems 'harmonic and exact' is equivalent to 'co-closed and exact'. Please explain how to show 'co-closed and exact is zero'.

答:
On slide $43$ he writes $[\omega] = [\omega_d] + [\omega_{\delta}] + [\omega_{\Delta}]$. Then on slide $44$, he replaces $\omega_{\delta}$ by $0$ because a form which is both $d$-closed and $d^*$-exact on a closed manifold is zero. To see this, note that if $\alpha = d^*\beta$ and $d\alpha = 0$ then

$$\|\alpha\|^2 = \langle\alpha,\alpha\rangle = \langle\alpha, d^*\beta\rangle = \langle d\alpha, \beta\rangle = \langle 0, \beta\rangle = 0$$

so $\alpha = 0$. On slide $45$, he replaces $\omega_d$ by $0$ because a form which is both $d^*$-closed and $d$-exact on a closed manifold is zero (which follows from a similar argument). Therefore $\phi(\omega) = [\omega_{\Delta}]$.


问:
rewriting equation in terms of $y$
From Stewart, Precalculus, 5th ed, P98, Q.45

$$x^2+xy+y^2=4$$

how can I re-write this equation in terms of $y$? I want to put this equation into graphing software but don't know to put $y$ on one side and $x$ on the other.

Any suggestion would be great.

答:
rewrite in quadratic form in x

$$x^2+xy+(y^2-4)=0$$

any quadratic equation $$ax^2+bx+c=0$$ has a two roots

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

 so that

$$x=\frac{-y\pm\sqrt {16-3y^2}}{2}$$

ContentLicense: CC BY-SA 3.0

Title: rewriting equation in terms of $y$


问:
Why are distributions only defined on maps with compact support?
Why do we restrict ourselves to maps of compact support, what’s wrong with merely smooth functions or some more general space.</p>

答:
Re: "Why do we restrict ourselves to maps of compact support", we don't. The space of smooth functions $\mathscr{E}(\Omega)$ is perfectly fine as a space of test functions. It has a natural Frechet space topology which is the locally convex topology define by the collection of seminorms
$$
f\longmapsto\ \sup_{x\in K} \max_{|\alpha|\le n} |\partial^{\alpha}f(x)|
$$
where $K$ ranges over compact subsets of $\Omega$ and $n$ ranges over nonnegative integers.
The topological dual $\mathscr{E}'(\Omega)$ is the space of compactly supported distributions which can be seen as a subset of $\mathscr{D}'(\Omega)$.

Elementary introductions to the theory of distributions don't mention these spaces much but you can find more information in more substantial treatments like the book by Schwartz or the one by Horváth.


问:
Find the mean and the variance of an F random variable with $r_1$ and $r_2$ degrees of freedom.

First find $E(U), E(\frac{1}{V}), E(U^2),E(\frac{1}{V^2})$.

When I consider finding $E(U)$ I feel as though integrating over the pdf of the F distribution multiplied by $u$ will leave me with a spare $u$. Is there a better strategy to consider? I know that the F distribution is made up of two independent Chi-Square distributions, perhaps I should calculate the expected values separately?

答:
In answer to your doubt:

When I consider finding E(U) I feel as though integrating over the pdf of the F distribution multiplied by u will leave me with a spare u.

Well here's the 'direct' way which I tried:

If $U\sim F\left(r_1,r_2\right)$, then,
$$\begin{align}
E(U)=&\int_0^\infty \frac{u\left(\frac{r_1}{r_2}\right)^{r_1/2}u^{\left(r_1/2\right)-1}}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)\left(1+\frac{r_1}{r_2}u\right)^{\left(r_1+r_2\right)/2}}du\\
=&\frac{\left(\frac{r_1}{r_2}\right)^{r_1/2}}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)}\int_0^\infty \frac{u^{\left(r_1/2\right)}}{\left(1+\frac{r_1}{r_2}u\right)^{\left(r_1+r_2\right)/2}}du\\
\end{align}$$

Now, substitute $\frac{r_1}{r_2}u=t$. Then,  we're getting:

$$\begin{align}
E(U)=&\frac{\left(\frac{r_1}{r_2}\right)^{r_1/2}}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)}\int_0^\infty \frac{\left(\frac{r_2}{r_1}t\right)^{r_1/2}}{\left(1+t\right)^{\left(r_1+r_2\right)/2}}\left(\frac{r_2}{r_1}\right)dt\\
=&\frac{\left(\frac{r_2}{r_1}\right)}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)}\int_0^\infty \frac{t^{r_1/2}}{\left(1+t\right)^{\left(r_1+r_2\right)/2}}dt\\
=&\frac{\left(\frac{r_2}{r_1}\right)B\left(\frac{r_1}{2}+1,\frac{r_2}{2}-1\right)}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)}\int_0^\infty \underbrace{\frac{t^{\left(r_1/2\right)+1-1}}{B\left(\frac{r_1}{2}+1,\frac{r_2}{2}-1\right)\left(1+t\right)^{\overline{\left(r_1/2\right)+1}+\overline{\left(r_2/2\right)-1}}}}_{\text{pdf of a Beta distribution}}dt\\
=&\frac{\left(\frac{r_2}{r_1}\right)B\left(\frac{r_1}{2}+1,\frac{r_2}{2}-1\right)}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)}\\\end{align}$$

Now, if I use the relationship of the Beta function with the Gamma function, we'll end up getting something like this:

$$\begin{align}
E(U)=&\left(\frac{r_2}{r_1}\right)\frac{\Gamma\left(\frac{r_1}{2}+1\right)\Gamma\left(\frac{r_2}{2}-1\right)}{\Gamma\left(\frac{r_1+r_2}{2}\right)}\frac{\Gamma\left(\frac{r_1+r_2}{2}\right)}{\Gamma\left(\frac{r_1}{2}\right)\Gamma\left(\frac{r_2}{2}\right)}\\
=&\left(\frac{r_2}{r_1}\right)\frac{\Gamma\left(\frac{r_1}{2}+1\right)\Gamma\left(\frac{r_2}{2}-1\right)}{\Gamma\left(\frac{r_1}{2}\right)\Gamma\left(\frac{r_2}{2}\right)}\\
=&\left(\frac{r_2}{r_1}\right)\left(\frac{r_1}{2}\right)\frac{1}{\frac{r_2}{2}-1}\\
=&\frac{r_2}{r_2-2}\\
\end{align}$$

This gives us the expectation of $U$ if $U\sim F\left(r_1,r_2\right)$.

Similarly, you can obtain $E\left(U^2\right)$ and ultimately find the $Var(U)=E\left(U^2\right)-(E(U))^2$.

I would still recommend Michael Hardy's method as that is far more efficient than this one.

P.S. Pardon me for any formatting errors that might've crept while writing the answer.


答:
You need conditional probability because of the way the question was phrased.

You also need to actually understand the question. It's not about the probability that a package gets lost, nor the that a randomly chosen package gets lost by the second post office. It's about only those parcels that actually are lost.

$$
\begin{array}{|ccccccc|}
\hline
& & \text{lost by 1st p.o.} & & & & \frac 1 5 = \frac 5 {25} \\ & \nearrow \\ \text{start} & & & & & \text{lost by 2nd p.o.} & \frac 4 5 \times \frac 1 5 = \frac 4 {25} \\
& \searrow & & & \nearrow \\
& & \text{not lost by 1st p.o.} \\
& & & & \searrow \\
& & & & & \text{not lost by 2nd p.o.} & \frac 4 5 \times \frac 4 5 = \frac{16}{25} \\ \hline
\end{array}
$$
The probability of being lost is $\dfrac{5+4}{25}.$

Among those $9$ cases out of $25$ that are lost, $5$ are lost by the first post office and $4$ by the second.

Or you can express it like this: Suppose $25$ parcels are mailed. The $5$ are lost by the first post office, so $20$ go on to the second. Of those $20$ the number lost by the second post office is $4$. Thus $9$ parcels have been lost: $5$ by the first post office and $4$ by the second.

Title: Why Conditional Probability is used

Body:
Question:

Parcels from sender 'S' to receiver 'R' pass sequentially through 2 post-offices.
Each post-office has probability 1/5 of losing an incoming parcel, independently of all other parcels.
Given that a parcel is lost,  what is the probability that it was lost by the second post-office.

Why do we have to use conditional probability in this problem?

Shouldn't the answer just be:

P(lost at 2nd) = P(not lost at 1st) * P(lost at 2nd)

P(lost at 2nd) = 4/5 * 1/5

P(lost at 2nd) = 4/25

But the answer is 4/9


问:
Recreation math : Fill the box type problem with equal sum of row and column

Today I solve an interesting recreational math problem which took some time.

First I will briefly explain the situation of the problem.  Look above figure.

(1) a,b,c,d,e,f,g,h,i,j are one of 1,2,3,4,5,6,7,8,9,10

(2) the sum of each row and columns are the same. (i.e., a+b+c+d = a+e+g, ...)

Find a,b,c,d,...

I tried to solve this by making equations but for me, that direction was too complicated.

Instead, I noticed that e and f should be big, and putting $x$ be the sum of each row or column
and make 4x = 55 + a + b + c + d, hence $x=\frac{55}{4} + \frac{a+b+c+d}{4}$, and from the assumption I have $14\leq x \leq 22$. letting $e=10, f=9$ and solve $x = a+10+g = d+ 9 + j$ with many trials and errors I obtain $x=18$ and $a=2, g=6, d=8, j=1, b=5, c=3, h=4, i=7$.

Is there any simple way to solve this problem? Please let me know the simple strategy.

答:
Some restrictions could be derived:
if $$x = a+b+c+d = g+h+i+j,$$
then
$$
2x+e+f = 55,
$$
so
$$
18\le x \le 26.
$$

But:

 $x=26$ doesn't work, since then $e+f=3$, so $e=1$ (if $e<f$), and $a+g=25$.

 $x=25$ doesn't work, since then $e+f=5$, so $e\le 2$ (if $e<f$), and $a+g\ge 23$.

 $x=24$ doesn't work, since then $e+f=7$, so $e\le 3$ (if $e<f$), and $a+g\ge 21$.

 $x=23$ doesn't work, since then $e+f=9$, so $e\le 4$ (if $e<f$), and $a+g\ge
19$, which is possible for $e=4$, but isn't possible for corresponding $f=5$ and $d+j=18$.

Actually there are ten solutions (with sums $18$, $19$, $20$, $22$) (with $a<g$, $e<f$, $b<c$, $h<i$).

Therefore there is no unique short algorithm to determine "proper" solution.

There should be some "trial-and-error" part here, I guess.

Just few examples :

```<code> x=18:
 1 4 7 6
 9     10
 8 3 5 2

 x=18:
 4 1 7 6
 9     10
 5 3 8 2

 x=19:
 1 5 6 7
 8     9
10 2 4 3

 x=20:
 4 1 5 10
 7     8
 9 3 6 2

 x=22:
 8 2 3 9
 4     7
10 1 5 6
</code>```

Body:
Some restrictions could be derived:
if $$x = a+b+c+d = g+h+i+j,$$
then
$$
2x+e+f = 55,
$$
so
$$
18\le x \le 26.
$$

But:

 $x=26$ doesn't work, since then $e+f=3$, so $e=1$ (if $e<f$), and $a+g=25$.

 $x=25$ doesn't work, since then $e+f=5$, so $e\le 2$ (if $e<f$), and $a+g\ge 23$.

 $x=24$ doesn't work, since then $e+f=7$, so $e\le 3$ (if $e<f$), and $a+g\ge 21$.

 $x=23$ doesn't work, since then $e+f=9$, so $e\le 4$ (if $e<f$), and $a+g\ge
19$, which is possible for $e=4$, but isn't possible for corresponding $f=5$ and $d+j=18$.

Actually there are ten solutions (with sums $18$, $19$, $20$, $22$) (with $a<g$, $e<f$, $b<c$, $h<i$).

Therefore there is no unique short algorithm to determine "proper" solution.

There should be some "trial-and-error" part here, I guess.

Just few examples :

 x=18:
 1 4 7 6
 9     10
 8 3 5 2

 x=18:
 4 1 7 6
 9     10
 5 3 8 2

 x=19:
 1 5 6 7
 8     9
10 2 4 3

 x=20:
 4 1 5 10
 7     8
 9 3 6 2

 x=22:
 8 2 3 9
 4     7
10 1 5 6


问:
question about conditional probability formula
I know that $P(A|B)=0.2, P(A|\bar{B})=0.3, P(B)=0.8$ and I need to find $P(A)$ and $P(B|A)$.

I started with formula: $P(A|B) = \frac{P(A\cap B)}{P(B)} \implies P(A\cap B) = 0.8\cdot 0.2$

Now can I write that $P(A\cap B) = P(A)\cdot P(B)$ and from here to find $P(A)$ ?

答:
Law of Total Probability:

$$\Pr(A)=\Pr(A\cap B)+\Pr(A\cap \overline{B}) = \Pr(A\mid B)P(B) + \Pr(A\mid \overline{B})\Pr(\overline{B})$$

Note: $\Pr(A\cap B)$ is NOT equal to $\Pr(A)\cdot \Pr(B)$ in general.  This is true if and only if $A$ and $B$ are independent events, so your proposed continuation was incorrect.


答:
Given your question, see if this helps.

Say there are $100$ letters sent.

You expect first post office to lose $20$ letters $( = \frac{1}{5} \times 100)$.

You expect second post office to lose another $16$ letters $( = \frac{1}{5} \times (100-20)$).

So we expect $64$ letters to reach and $36$ letters to be lost.

Now we are told that our sample space is no more $100$ letters. It is those $36$ letters that are lost.

Probability that it is Post office $2$ that loses the letter if the letter is lost

$\displaystyle  = \frac{16}{36} = \frac{4}{9}$

Title: Why Conditional Probability is used


答:
As you did you can also easy calculate $P(A \cap \overline{B})$

Then note that

$$P(A)= P(A \cap B)+   P(A \cap \overline{B})$$

Title: question about conditional probability formula

Body:
I know that $P(A|B)=0.2, P(A|\bar{B})=0.3, P(B)=0.8$ and I need to find $P(A)$ and $P(B|A)$.

I started with formula: $P(A|B) = \frac{P(A\cap B)}{P(B)} \implies P(A\cap B) = 0.8\cdot 0.2$

Now can I write that $P(A\cap B) = P(A)\cdot P(B)$ and from here to find $P(A)$ ?


问:
The basis of $\mathbb{Z}_{2}/\left \langle x^{2}+x+1 \right \rangle$ is $\left \{ [1], [x] \right \}$

Show that $A=\left \{ [1], [x] \right \}$  is a basis of $\mathbb{Z}_{2}/\left \langle x^{2}+x+1 \right \rangle$ over $\mathbb{Z}_{2}$

The first thing I did was see that the elements of the ring were $[0], [1] ,[x]$ and $[x+1]$. Then to show that it's a basis, it must be proved that the set A span the ring and that the elements of A are linearly independent.

To see that it span the ring, I must see that it generates the elements $[0]$ and $[x + 1]$, so $[x + 1] = 1\cdot [x] + 1 \cdot [1]$, but $[0] = 0 + p(x)$ , ($p (x) \in \mathbb{Z}_{2}[x]$)

I don't know if I can say that $[0]=0\cdot [x]+p(x)\cdot [1]$, because $[1] = (1 + q (x))$ and thus $[0] = p (x) + p (x) q (x)$ which becomes a polynomial again

I appreciate any comments

答:
Something a bit more general is true. Take $k$ a field (such as $\mathbb{Z}_2$) and $$f = a_nX^n + a_{n-1}X^{n-1} \ldots + a_1 X + a_0 \in k[X]$$ with $a_n \neq 0$. Then $B = \{[1], [X], \ldots, [X]^{n-1}\}$ is a $k$-basis for $k[X]/(f)$ as a $k$-vector space.

To see this: pick $[g] \in K[X]/(f)$. Since $k$ is a field, we know that $k[X]$ has a division algorithm. Namely, there exist unique $q,r \in k[X]$ with $r = 0$ or $\deg r < n$ such that $g = qf + r$; and so $[g] = [qf]+[r] = [r]$. But since either $r = 0$ or $r$ has degree less than $n$, we have

$$
r = b_0 + b_1 X + \ldots + b_{n-1}X^{n-1}
$$

with $b_1, \ldots, b_{n-1} \in k$ (some or even all of these may be zero). Consequently,

$$
[g] = [r] = b_0[1] + b_1[X] + \cdots +b_{n-1}[X^{n-1}]
$$

and thus $B$ is a set of generators. Finally, to see that $B$ is linearly independent, suppose that

$$
0 = b_0[1] + b_1[X] + \cdots +b_{n-1}[X^{n-1}] = [b_0 + b_1 X + \ldots + b_{n-1}X^{n-1}]
$$

for some $b_i \in k$. This means that there exists $g$ such that

$$
b_0 + b_1 X + \ldots + b_{n-1}X^{n-1} = gf.
$$

But that's impossible unless $b_0 = \cdots = b_{n-1} = 0$ and $g = 0$, as otherwise the right hand side has degree $\deg gf \geq \deg f = n$.


问:
How To Learn Mathematics Basics
I wanna learn mathematics from zero to advanced which books should I read and which websites should I visit?</p>

答:
Courant-Robbins, "what is Mathematics"

https://www.ams.org/notices/200111/rev-blank.pdf
https://www.google.com/books/edition/What_is_Mathematics/_kYBqLc5QoQC?hl=en
https://projecteuclid.org/download/pdf_1/euclid.bams/1183504823

Title: How To Learn Mathematics Basics


问:
Converting cartesian equation to polar issue
I have tried converting the equation $ y = x^3 $ to a polar equation using the following steps but the graph of the resulting polar equation also appears to include $ y=-x^3 $.

$ r^2 = x^2 + y^2 $

$ r^2 = x^2 + (x^3)^2$

$ r^2 = r^2\times cos(\theta)^2 + r^6cos(\theta)^6 $

$ 1 = cos(\theta)^2+r^4\times cos(\theta)^6$

$ \frac{sin(\theta)^2}{cos(\theta)^6}=r^4$

$ r = (\frac{sin(\theta)^2}{cos(\theta)^6})^{1/4}$

However, the graph of this is not $ y = x^3 $.

I am wondering why converting a cartesian equation to a polar equation does not appear to work this way. I understand that $ y=x^3$ can be easily converted to polar by substituting $ y=rsin(\theta)$ and $x=rcos(\theta)$ into $y=x^3$ but I don't understand why I do not get the same answer when substituting $y = x^3$ for $x$ in $r^2 = x^2+y^2$.

Thanks.

答:
$y=x^3$

$r\sin\theta=r^3\cos^3\theta$

cancel $r$ and solve

$r^2=\frac{\sin\theta}{\cos^3\theta}$

to solve for $r$ we must have $\frac{\sin\theta}{\cos\theta}\ge 0\to \tan \theta \ge 0$ that is $0\le \theta<\pi/2\lor \pi\le \theta<3\pi/2$

$r=\sqrt{\frac{\sin\theta}{\cos^3\theta}};\;\theta\in [0,\pi/2)\cup[\pi,3\pi/2)$


问:
On the Hurwitz Zeta Function
In my mathematics course in Uni. (I'm a physics student) my prof. gave us the following exercise: to express the Hurwitz Zeta function $\zeta(2k+1,\frac{1}{4})$ with $k=1,2,3,\dots$ in terms of the Riemann zeta function. He says there is a closed form for this, something like
$\zeta(2k+1,\frac{1}{4})=C(k)\zeta(2k+1)$. With $C(k)$ some elementary function of $k$.

After some basic calculations I found the following

$\zeta(2k+1,\frac{1}{4})=2^{2k+1}(2^{2k+1}-1)\zeta(2k+1)-\zeta(2k+1,\frac{3}{4})$

but I don't know what to do about $\zeta(2k+1,\frac{3}{4})$. Trying the same method on this function, I stumble upon $\zeta(2k+1,\frac{1}{8})$. And I'm going in a loop getting nothing like my prof. said. Also I've looked through books and articles on the Hurwitz function and found nothing.

Title: On the Hurwitz Zeta Function

Body:
\begin{equation}
\begin{array}{c}
\left.
\begin{array}{c}
\zeta (2n+1,\frac{1}{4}) \\
\zeta (2n+1,\frac{3}{4})%
\end{array}%
\right\} =2^{2n}(2^{2n+1}-1){\zeta }(2n+1) \\
\pm \frac{1}{2\pi }\left( 2n+2+4^{2n+2}\right) {\zeta }(2n+2)-2\sum
\limits_{l=0}^{n-1}4^{2n-2l}{{{\zeta }(2n-2l)\zeta }}(2l+2)%
\end{array}
\tag*{(1)}
\end{equation}

\begin{equation}
\begin{array}{c}
\left.
\begin{array}{c}
\zeta (2n+1,\frac{1}{3}) \\
\zeta (2n+1,\frac{2}{3})%
\end{array}%
\right\} =\frac{{3^{2n+1}-1}}{2}{\zeta (2n+1)} \\
\pm \frac{\sqrt{3}}{2\pi }\left( \left( 2n+2+3^{2n+2}\right) {\zeta }%
(2n+2)-2\sum\limits_{l=0}^{n-1}3^{2n-2l}{{{\zeta }(2n-2l)\zeta }}%
(2l+2)\right)%
\end{array}
\tag*{(2)}
\end{equation}

\begin{equation}
\begin{array}{c}
\left.
\begin{array}{c}
\zeta (2n+1,\frac{1}{6}) \\
\zeta (2n+1,\frac{5}{6})%
\end{array}%
\right\} =\frac{{6^{2n+1}-{3^{2n+1}}-{{2^{2n+1}}+1}}}{2}{\zeta (2n+1)} \\
\pm \frac{1}{2\sqrt{3}\pi }\left( 6^{2n+2}-3^{2n+2}\right) {\zeta }%
(2n+2)-2\sum\limits_{l=0}^{n-1}\left( 6^{2n-2l}-3^{2n-2l}\right) {{{\zeta }%
(2n-2l)\zeta }}(2l+2)%
\end{array}
\tag*{(3)}
\end{equation}


问:
Given a Lebesgue measurable set I ⊂ R so that m(I) = π, there is a subset A ⊂ I so that m(A) = e (euler number).
Problem: I have to solve this problem about the Lebesgue measure, where given a Lebesgue measurable set I ⊂ R so that m(I) = π, I need to prove that there exists a subset A ⊂ I so that m(A) = e (euler number).

My idea: I have been thinking about it, and I thought that maybe those numbers are just there to distract you from the real problem. I guessed that maybe, given any set I ⊂ R so that m(I) = a, for every 0≤b≤a, there should be a Lebesgue measurable subset A ⊂ I so that m(A) = b.
I think it makes sense, but it's just an idea, and I don't know how to prove it or even if I'm right.
Any help will be much appreciated, thank you.

答:
Suppose $0<a<b$ and $m(I)=b.$

Let $F(x) = m(I\cap(-\infty, x]). $

See if you can prove that $F$ is continuous and $F(x)\to0$ as $x\to-\infty$ and $F(x)\to a$ as $x\to+\infty.$

Then apply the intermediate value theorem.


答: Use continuity of the function $f(x)=m(I \cap (-\infty,x])$ and the intermediate value theorem

Title: Given a Lebesgue measurable set I ⊂ R so that m(I) = π, there is a subset A ⊂ I so that m(A) = e (euler number).

Body:
Problem: I have to solve this problem about the Lebesgue measure, where given a Lebesgue measurable set I ⊂ R so that m(I) = π, I need to prove that there exists a subset A ⊂ I so that m(A) = e (euler number).

My idea: I have been thinking about it, and I thought that maybe those numbers are just there to distract you from the real problem. I guessed that maybe, given any set I ⊂ R so that m(I) = a, for every 0≤b≤a, there should be a Lebesgue measurable subset A ⊂ I so that m(A) = b.
I think it makes sense, but it's just an idea, and I don't know how to prove it or even if I'm right.
Any help will be much appreciated, thank you.


问:
$f:(0,\infty) \rightarrow \mathbb{R}$ satisfies $f(x)-f(y)=f\left(\dfrac{x}{y}\right)$ for all $x,y \in (0,\infty)$ and $f(1)=0.$
Suppose that $f:(0,\infty) \rightarrow \mathbb{R}$ satisfies $f(x)-f(y)=f\left(\dfrac{x}{y}\right)$ for all $x,y \in (0,\infty)$ and $f(1)=0.$

$(a)$. Prove that $f$ is continuous on $(0,\infty)$ if and only if $f$ is continuous at $1$

$(b)$. Prove that $f$ is differentiable  on $(0,\infty)$ if and only if $f$ is differentiable at $1$

$(c).$ Prove that $f$ is differentiable at $1$, then $f^{\prime}(x)=\dfrac{f^{\prime}(1)}{x}$ for all $x\in (0,\infty)$.

$\underline {Attempt}$

$(a).$  If $f$ is continuous on $(0,\infty)$  then $f$ is also continuous at $1$.

$\space$ $ \space $$\space$ $\space$ If $f$ is continuous at $1,$ We have  $$ \forall \space \varepsilon \space \exists \space \delta \space\text{such that} \space|x-1|<\delta \space \text{whenever} \space|f(x)-f(1)|=|f(x)|< \varepsilon $$

$\space$ $ \space $$\space$ $\space$ Now let $a \in (0,\infty)$ and if $|x-a|<\delta_1 $,

$$|f(x)-f(a)|=\left|f\left(\dfrac{x}{a}\right)\right| <\varepsilon$$

$\therefore$  $f$ is continuous on $(0,\infty)$ if and only if $f$ is continuous at $1$

$(b).$ If $f$ is differentiable $(0,\infty)$ then $f$ is also differentiable at $1$.

$\space$ $ \space $$\space$ $\space$ If $f$ is differentiable  at $1$, We have
$$f^{\prime}(1)=\lim_{h\to 0 } \frac{f(1+h)-f(1)}{h}$$

$\space$ $ \space $$\space$ $\space$ set $h=\dfrac{t}{x}$ implies when $h \to 0$, $t \to 0$ so $$f^{\prime}(1)=\lim_{h\to 0 } \frac{f(1+h)}{h}=\lim_{t\to 0 } \frac{f\left(1+\dfrac{t}{x}\right)}{\dfrac{t}{x}}=x\lim_{t\to 0} \frac{f(x+t)-f(x)}{t} $$

$\therefore$  $f$ is differentiable  on $(0,\infty)$ if and only if $f$ is differentiable at $1$

$(c).$ If $f$ is differentiable at $1$,We have
$$f^{\prime}(1)=\lim_{h\to 0 } \frac{f(1+h)-f(1)}{h}$$
$\space$ $ \space $$\space$ $\space$ this implies,
$$f^{\prime}(1)=\lim_{h\to 0 } \frac{f(1+h)}{h}=x\lim_{t\to 0} \frac{f(x+t)-f(x)}{t}=xf^{\prime}(x) $$

I referred Functional equation $f(xy)=f(x)+f(y)$ and differentiability and Show a function for which $f(x + y) = f(x) + f(y) $ is continuous at zero if and only if it is continuous on $\mathbb R$ but I don't know exactly my attempt is correct or not if not give some advises.

Thank you!

答:
The OP is clearly about whether the solution provided is ok, and not about a solution, I address the former.

Answer to OP: no!!

Because you neither specify what $\delta_1$ is, nor do you prove, by other means, that such a $\delta_1$ exists. It was fixed by @Paramanand Sigh in the comments. Another possible addition to OP will do the job. Add: For a given $a>0$ there is a $\delta_1>0$ such that
$$
|x-a|<\delta_1 \implies |\frac{x}{a}-1|<\delta \, .
$$

To emphasize: the fact that a $\delta$ exists does not automatically prove a $\delta_1$ exists. Notice that $\delta_1$ must depend on the point where you consider continuity.

Advice: When writing a proof, assume that a computer will read it. So, the first thing you do is to make sure every parameter you introduce is clearly defined either beforehand, or immediately after a comma.


问:
Why convolution regularize functions?
There is a tool in mathematics that I have used a lot of times and I'm still not confortable with. In fact I can't figure out (by this I mean that I cannot understand it geometrically) why does convolution regularize things. It is know for example that if $u\in L_{loc}^1(\mathbb{R}^N)$ and $f\in C_0^k(\mathbb{R}^N)$, then $f\ast u\in C^k(\mathbb{R}^N)$ and $$D^\alpha (f\ast u)=(D^\alpha f)\ast g$$

Well, I have a lot of questions about convolution that might help me understand it better and I would appreciated one who points me out references to read and some insight about it.

I - For what reason was convolution implemented in mathematics? Was it for regularizing functions? Who used it the first time?

II - Is there any intuitive way to find out why convolution make functions more smooth?

I can understand the analytical proof of II, but in the end it is just algebraic manipulations and I dont get the essence of the thing. Thank you for your help.

答:
Perhaps this might soothe some of your discomfort.

Smoothing Action

There are many ways that convolution is useful in mathematics. First of all, as you have noted,
$$
\mathrm{D}^\alpha\left(f\ast g\right)=\left(\mathrm{D}^\alpha f\right)\ast g\tag{1}
$$
This is simply repeated changes of the order of integration and differentiation:
$$
\frac{\mathrm{d}}{\mathrm{d}x}\int f(x-t)\,g(t)\,\mathrm{d}t=\int f'(x-t)\,g(t)\,\mathrm{d}t\tag{2}
$$
This step can be justified in different ways depending on the context. For instance, if the limit which defines the derivative of $f$,
$$
f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\tag{3}
$$
converges uniformly, then $(2)$ is valid for all $g\in L^1$.

Convolution combines the smoothness of two functions. That is, if both $f$ and $g$, and their first derivatives are in $L^1$, then the second derivative of their convolution is in $L^1$. This is because $f\ast g = g\ast f$, and so we can use $(2)$ twice to get
$$
\frac{\mathrm{d}^2}{\mathrm{d}x^2}(f\ast g)=f'\ast g'\tag{4}
$$

Fourier Analysis

Convolution plays an important role in Fourier Analysis. The key formulas demonstrate the duality between convolution and multiplication:
$$
\mathscr{F}(f\ast g)=\mathscr{F}(f)\mathscr{F}(g)\quad\text{and}\quad\mathscr{F}(fg)=\mathscr{F}(f)\ast\mathscr{F}(g)\tag{5}
$$
There also exists a duality between decay at $\infty$ and smoothness. Essentially, one derivative of smoothness of $f$ corresponds to one factor of $1/x$ in the decay of $\mathscr{F}(f)$, and vice versa.

The product of decaying functions decays even faster; e.g. $x^{-n}x^{-m}=x^{-(n+m)}$. The duality demonstrated in $(5)$ then says that the convolution of smooth functions is even smoother.

The Riemann-Lebesgue Lemma says that for $f\in L^1$,
$$
\lim_{|x|\to\infty}\mathscr{F}(f)(x)=0\tag{6}
$$
However, this is simply decay with no quantification. About all that can be said about $f,g\in L^1$ is that $f\ast g\in L^1$. However, if $f,g\in L^2$, then $f\ast g$ is continuous.

Summing Dice

Perhaps one of the earliest uses of convolution was in probability. If $f_n(k)$ is the number of ways to roll a $k$ on $n$ six-sided dice, then
$$
f_n(k)=\sum_jf_{n-1}(k-j)f_1(j)\tag{7}
$$
That is, for each way to achieve $k$ on $n$ dice, we must have $k-j$ on $n-1$ dice and $j$ on the remaining die. Equation $(7)$ represents discrete convolution.

The distribution function for the roll of a single six-sided die is evenly distributed among $6$ possibilities. This has discontinuities at $1$ and $6$ ($n=1$). The distribution function for the sum of two six-sided dice is the convolution of two of the one die distributions. This is continuous, but not smooth ($n=2$). The distribution function for the sum of three six-sided dice is the convolution of the one die and two dice distributions. This is smooth ($n=3$). For each die we add, we convolve one more of the one die distributions and the function gets smoother.

$\hspace{8mm}$

As $n\to\infty$, the distribution approaches a scaled version of the normal distribution: $\frac1{\sqrt{2\pi}}e^{-x^2/2}$.

Title: Why convolution regularize functions?


问:
Example of functions satisfying certain little-oh condition
Let $f:\mathbb{Z}_{>0} \rightarrow \mathbb{C}$ be a function. Now require $f$ to satisfy the condition
$$
|f(n)|\leq n^{o(1)}
$$
where $o(1)$ is any function $g(n)$ satisfy $\frac{g(n)}{1}\rightarrow 0$ as $n\rightarrow\infty$.

Can someone give some example of such $f(n)$ besides constant functions?

答:
$f(n)=\log n, n \ge 2$ and $g(n)=(\log n)^{-1/2}=o(1), n \ge 2$; $n^{g(n)}=e^{(\log n)^{1/2}}$ and we have $(\log n)^{1/2} \ge \log \log n$ so $n^{g(n)} \ge e^{\log \log n}=\log n$

(if needed of course we can use a small constant and take $f(n)=C\log n$ to take care of small cases but I think the inequality $(\log n)^{1/2} \ge \log \log n$ holds for all $n \ge 2$ not only for large $n$ where it is trivial)


问:
If $\mathcal O(G)$ is semi-simple then $G$ is finite.
I want to show that if $\mathcal O(G)$ is semi-simple, for $G$ an affine algebraic group and $\mathcal O(G)$ its coordinate ring, then $G$ is finite but I don't really see how to do that. I tought maybe use Artin-Wedderburn's theorem to deduce that $\mathcal O(G)$ is a finite dimensionnal $k$-vector space (where $k$ is an algebrically closed fields) and then use the correspondence between coordinate rings and affine algebraic sets $G \cong \text{Hom}_{\text{Alg}_k} (\mathcal O(G), k) \subset \mathcal O(G)^* =  \text{Hom}_{k} (\mathcal O(G), k)$ but I don't know how to conclude that $G$ has to be finite from here. Is it the right way ? Does anyone know how to do that ?

答:
Using Artin-Wedderburn as you suggest will work. If $\mathcal{O}(G)$ is a semisimple commutative algebra over an algebraically closed field $k$ then Artin-Wedderburn implies that it's a finite direct product of field extensions of $k$, which are either equal to $k$ or transcendental. Since $G$ is an affine algebraic group $\mathcal{O}(G)$ must be finitely generated as a $k$-algebra which implies (by the Nullstellensatz) that the transcendental extensions can't occur, so $\mathcal{O}(G)$ is a finite direct product of copies of $k$.

I don't know exactly what you mean by "finite" here. You might just mean that the set of $k$-points $G(k)$ is finite, which is clear from here since the number of $k$-points of a finite direct product $\prod_{i=1}^n k$ is exactly $n$ (exercise).  The standard definition of "finite" in the context of affine group schemes, as far as I know, is that $\mathcal{O}(G)$ is finitely generated as a $k$-module, which is also clear here.


问:
How can I find some $r\in\mathbb{R}$ such that the following matrix has a full rank?
I have to task to determine whether the following matrix $A_r\in\mathbb{R}^{39\times 39}$ defined as

$$
A_r := \begin{pmatrix}
2+4!+r^2 & 2r & -1 & 0 & \ldots & \ldots & 0 \\
2^5 & 2+5!+r^2 & -2r & 1 &\ddots &\ddots & \vdots\\
0 & 2^6 & 2+6!+r^2 & 2r & -1 &\ddots & \vdots \\
\vdots &\ddots & \ddots & \ddots & \ddots &\ddots & 0\\
\vdots & \ddots & \ddots &2^{40} & 2+40!+r^2 & 2r & -1 \\
\vdots &\ddots &\ddots & \ddots &2^{41} & 2+41!+r^2 & -2r\\
0 &\ldots &\ldots &\ldots & 0 &2^{42} & 2+42!+r^2
\end{pmatrix}
$$
Does there exist some $r\in\mathbb{R}$ for that $A_r$ has a full rank? If yes, define these $r\in\mathbb{R}$.

My ansatz was to find two vectors which are linearly dependent, but it's difficult to choose two vectors and check whether they are linearly independent or not. To determine $\det(A_r)$ is not a good choice or to find $(A_r)^{-1}$. Is it possible to find some $r\in\mathbb{R}$ for that the matrix $A_r$ has a full rank? What should I do to answer the above questions?

答:
The matrix has full rank since it is strictly diagonally dominant.
This result is known as the Levy–Desplanques theorem.
https://en.wikipedia.org/wiki/Diagonally_dominant_matrix</p>


问:
Open Cover / Real Analysis
I have the next question: Let $K \subset $ $R^1$ consist of $0$ and the numbers 1/$n$, for $n=1,2,3,\ldots$ Prove that $K$ is compact directly from the definition (without using Heine-Borel).

I'm trying to understand compact sets so I would be grateful if someone could give me some examples of open covers and subcovers. Thank you!

答:
As you're trying to understand this, I'll try not to give too much away so you can work it out for yourself.

Let $\mathcal{U}$ be a cover of the set $K$. As $\mathcal{U}$ is a cover of $K$, there must exist some open $U_0\in\mathcal{U}$ such that $0\in U_0$.

From this, can you construct a finite subcover of $\mathcal{U}$ for $K$? I would suggest putting the set $U_0$ in your finite subcover.

I'll try to help you establish why the set $U_0$ is so important in this proof.

Suppose that $K'=K\setminus\{0\}$. If we tried to do something similar to the above, we would notice that we can not be entirely certain that there exists some open set in $\mathcal{U}$ which contains $0$. This really is crucial. With this information, we can quite easily construct an infinite cover of $K'$ which has no finite subcover.

Let $\mathcal{U}=\{U_n\}$ where $U_n=\left(\frac{1}{n}-\frac{1}{2}(\frac{1}{n(n+1)}),\frac{1}{n}+\frac{1}{2}(\frac{1}{n(n+1)})\right)$.

Now, the $U_n$ are defined so that $\frac{1}{n}\in U_n$ and no two $U_n$ and $U_{n'}$ intersect in a non-empty subset for $n\neq n'$. So there are no proper subcovers of $\mathcal{U}$ for $K'$, let alone a finite subcover. It follows that $K'$ is not compact.

Because $0$ is no longer in our set, the possible open covers of our set can now include covers which include no open sets which cover all but finitely many elements, and this was the key to the proof for the original $K$ which included $0$.

Title: Open Cover / Real Analysis


问:
$\mathbb{Q}[\sqrt{2}]$ is not a complete space
I am trying to prove that $\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in\mathbb{Q}\}$ is not a complete space (with the standard metric). For that purpose, I am looking for a Cauchy sequence in $\mathbb{Q}[\sqrt{2}]$ which is not convergent in $\mathbb{Q}[\sqrt{2}]$. Can somebody help me? Thanks!

答: Consider $1, 2, \dfrac32, \dfrac53, \dfrac85, \dfrac{13}8, ...$


问:
Why is this sequence of events increasing?
Asymmetric random walk $S_n$ with $p=P(X_i=1)=1-P(X_i=-1) \in (\frac{1}{2},1)$ and, given a < 0 < b, define the stopping time $T_{a,b}=\inf\{n\geq 0: S_n = a \text { or } b\}$. I don't see why the event $\{S_{T_{-1,n}}=-1\}$ with $n \geq b$ would be included in every event $\{S_{T_{-1,n+k}}=-1\}$ for any $k\geq 1$.

答:
Suppose $\omega \in \{S_{T_{-1,n}} = -1 \}$, then $S_{T_{-1,n}(\omega)}(\omega) = -1$, that is for each $t < T_{-1,n}(\omega)$ we have $-1 < S_t(\omega) < n$. For $k \ge 1$ now $-1 < S_t(\omega) < n+k$, all $t < T_{-1,n}(\omega)$. Therefore, as $S_{T_{-1,n}(\omega)}(\omega) = -1$, we have $T_{-1,n+k}(\omega) = T_{-1,n}(\omega)$. This gives $S_{T_{-1,n+k}(\omega)}(\omega) = -1$ or $\omega \in \{S_{T_{-1,n+k}} = -1 \}$.

By the same argument, you can show that the sequence of events is increasing, that is
$$
\{S_{T_{-1,n}} = -1 \} \subseteq \{S_{T_{-1,n+1}} = -1 \} \subseteq \{S_{T_{-1,n+2}} = -1 \} \subseteq \cdots
$$


答: Note that $\mathbb{Q}$ is dense in $\mathbb{R}$. Hence, there exists a sequence in $\mathbb{Q}$ which converges to $\sqrt{3}$. This sequence is also contained in $\mathbb{Q}[\sqrt{2}]$ (and it's Cauchy, because it converges) but clearly $\sqrt{3}$ is not.

Title: $\mathbb{Q}[\sqrt{2}]$ is not a complete space

Body: I am trying to prove that $\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in\mathbb{Q}\}$ is not a complete space (with the standard metric). For that purpose, I am looking for a Cauchy sequence in $\mathbb{Q}[\sqrt{2}]$ which is not convergent in $\mathbb{Q}[\sqrt{2}]$. Can somebody help me? Thanks!


问:
Show that a Star graph is balanced
A star graph $S_{k}$ is the complete bipartite graph $K_{1,k}$. One bipartition contains 1 vertex and the other bipartition contains $k$ vertices. Wikipedia Article

A graph G is balanced if the average degree of every subgraph H is less than or equal to the average degree of G. In other words $\bar{d}(H) \leq \bar{d}(G)$.

Show that a star graph is balanced.

I have been able to prove this, however in an extremely ugly and long way using many different cases. I was wondering if there is any easy way to prove this. Any ideas?

答:
By direct computation,

$$
\widetilde{d}(S_k) = \frac{k + \overbrace{1 + \cdots  + 1}^{k}}{k+1} = 2k/(k+1).
$$

Let's enumerate the vertices of $S_k$ as $v_0, \ldots, v_k$ with $d(v_0) = k$ and $d(v_i) = 1$ otherwise.

Pick a subgraph $H \leq S_k$. Suppose that there are $v,w \in H$ which are not connected in $H$ but they in fact are connected in $S_k$. Then adding such an edge to $H$ can only increase its average degree. Thus, for our objective, this means that we can wlog assume our subgraph $H$ is full.

There are only two possibilities: either $H$ contains $v_0$, and is isomorphic to $S_j$ for $j \leq k$, or $H$ has average degree zero.

Hence, we are left with showing that $2j/(j+1)$ is an increasing sequence: indeed,

$$
2j/(j+1) \leq 2k/(k+1) \iff 2jk+2j \leq 2kj+2k \iff j \leq k.
$$

Title: Show that a Star graph is balanced


答: Every open cover has at least one element that covers $0$.  Only finitely many members of $K$ are not in that one member of the cover.</p>

Title: Open Cover / Real Analysis

Body:
I have the next question: Let $K \subset $ $R^1$ consist of $0$ and the numbers 1/$n$, for $n=1,2,3,\ldots$ Prove that $K$ is compact directly from the definition (without using Heine-Borel).

I'm trying to understand compact sets so I would be grateful if someone could give me some examples of open covers and subcovers. Thank you!


问:
Interchanging roles of $A$ and $B$ in geometric Hahn-Banach
Consider the following version of the geometric Hahn-Banach theorem: let $X$ be a locally convex topological vector space over the real or the complex numbers, $A$ and $B$ two non-empty, disjoint and convex subsets of $X$ with $A$ compact and $B$ closed, then there is a continuous linear functional $f$ on $X$ such that $$ \sup_{x\in A} \Re f (x) < \inf_{x \in B} \Re f (x).$$

What if I want the reverse the inequality? Is it true that there is a continuous linear functional $g$ on $X$ such that
$$ \sup_{x\in B} \Re g (x) < \inf_{x \in A} \Re g (x)?$$

My naive idea is to take $g=-f$, but does this work and why?

答:
Just let $g = -f$. As $f$ is linear continuous, $g$ is also. As we have
\begin{align*}
  \Re g(x) &= \Re (-f)(x) \\
   &= - \Re f(x)\\
  \sup_{x\in B} \Re g(x) &= \sup_{x\in B} -\Re f(x)\\
      &= -\inf_{x\in B} \Re f(x) \\
      &< -\sup_{x\in A} \Re f(x) & \text{by the inequality for $f$}\\
      &= \inf_{x\in A} -\Re f(x)\\
      &= \inf_{x \in A} \Re g(x)
\end{align*}


问:
Representation of algebra v.s. modules over algebra
I got confused when dealing with these two situations. If we consider the same algebra, and also require the module itself is a real/complex vector space,

Are they equivalent definitions?

How much are the two theory equivalent? e.g. Is simple module the same as irreducible representation?

答:
Yes, a representation and a module are the same thing (e.g. of a group algebra), the theory is equivalent, and irreducible representations are the same thing as simple modules.

I don't know exactly what is responsible for the difference in terminology but it likely comes from different communities of researchers working in representation theory and (non)commutative algebra or similar, respectively. I don't think it was understood initially that representations of e.g. finite groups could be subsumed into module theory (which did not yet exist), and not all representation theories can, e.g. representations of algebraic groups are comodules rather than modules.


答:
It is clear that
$$\bar{d}(S_k) = \frac{k+k\cdot1}{k+1} = \frac{2k}{k+1} = 2-\frac{2}{k+1}$$

Now, in $H$, we can consider only two cases:

Case 1: Center vertex $v \notin H$ (vertex with degree $k$ in star graph). Then, $H$ has no edges so $\bar{d}(H) = 0$,

Case 2: Center vertex $v \in H$. If $H$ has all the vertices but missing some of the edges, then clearly $\bar{d}(H) < \bar{d}(S_k)$ since we are decreasing the total degree while keeping the number of vertices same. So, suppose $H$ has $n$ vertices of degree $1$ with $n \le k$ (here, note that $H$ still may be missing some edges but it is enough to check the maximal case, in which we have a star graph $S_n$). Then, we have
$$\bar{d}(H) = \frac{2n}{n+1} = 2 - \frac{2}{n+1}$$

So, all that's left is to compare $2-\dfrac{2}{k+1}$ and $2-\dfrac{2}{n+1}$ where $n \le k$. But, it is easy to see that

$$2-\dfrac{2}{k+1} \ge 2-\dfrac{2}{n+1}$$

So, we are done.

Title: Show that a Star graph is balanced

Body:
A star graph $S_{k}$ is the complete bipartite graph $K_{1,k}$. One bipartition contains 1 vertex and the other bipartition contains $k$ vertices. Wikipedia Article

A graph G is balanced if the average degree of every subgraph H is less than or equal to the average degree of G. In other words $\bar{d}(H) \leq \bar{d}(G)$.

Show that a star graph is balanced.

I have been able to prove this, however in an extremely ugly and long way using many different cases. I was wondering if there is any easy way to prove this. Any ideas?


问:
Is $F=(yx,x+z,yz)$ conservative?
I need a check on the following exercise:

Let $S$ be the surface $$S= \{(x,y,z):y=x^2+z^2, y \in [0,1] \}$$ and $F(x,y,z)=(yx,x+z,yz)$ a vector field. Compute $$\oint_{\partial S} F \cdot dr$$ Is $F$ conservative?

The boundary $\partial S$ is parametrized by $\gamma(t)=(\cos(t),1,\sin(t))$, so $$\int_0^{2 \pi} F(\gamma(t)) \cdot \gamma'(t)dt =\int_0^{2 \pi} (\cos(t),\cos(t)+\sin(t),\sin(t)) \cdot  (-\sin(t),0,\cos(t)) dt = 0$$

Unfortunately, this is not enough to say that it's conservative, because we need $\oint_{\gamma} F =0$ for every closed curve. I just noticed that since $\mathbb{R}^2$ is simply connected the fact that $\nabla \times F=\vec{0}$ is equivalent to the fact that $F$ is conservative. However,here $\nabla \times F = [z-1,0,1+x]$ is not zero, except for $z=1$ and $x=-1$. Can I say it's not conservative?

答:
$curl F = \nabla \times F = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ yx & x+z & yz\end {vmatrix} = \, (z-1,0,1-x) \ne 0$

So yes it is not a conservative vector field.

Your working is correct and the way the vector field is set up, $x$ and $z$ components will cancel out over any circle parallel to $XZ$ plane. But that may not be the case if the circle was not parallel to $XZ$ plane or if it was another closed curve, say ellipse.

Also be careful in orientation of the curve to make sure it matches the orientation of the surface. In this case it does not matter as the integral is zero and the question does not state orientation for the surface explicitly.

Title: Is $F=(yx,x+z,yz)$ conservative?


问:
Adding copy of row to "almost normalized" matrix does not change $X^\top X$ product
Good evening, I have a hard time with the following problem, I found how to start, but don't see how to continue:
Imagine we have a $D$x$N$ matrix $X$ such that all rows are normalized (zero mean and variance one), except for the last row, that has variance 2. Suppose that we also have a second matrix, say $\tilde{X}$, which is defined as a $(D+1)$x$N$ matrix with the same rows as X, except row $D$ is now normalized, and row $D + 1$ is a copy of row $D$.

So to summarize, $X$ is almost normalized, only last row has variance 2 and $\tilde{X}$ is perfectly normalized, but has two equal rows. Now I have to show that $X^\top X = \tilde{X}^\top\tilde{X}$.

I started by writting $\tilde{X}$ as $KX$ where K is a $(D+1)$x$D$ matrix with the first $D$ rows forming an identity matrix, and the last row being $\frac{1}{\sqrt{2}} * \textbf{e}_n$ where $\textbf{e}_n$ is the unit vector with a 1 in position n and zeros everywhere else, so that we have a normalized row in $\tilde{X}$. After that, I still cannot show the result...

Any help would be greatly appreciated

答:
Suppose that $X$ is a matrix with rows $r_1,r_2,\dots,r_D$ (each of size $1 \times N$). We have
$$
X^\top X = \sum_{i=1}^D r_i^\top r_i = r_D^\top r_D + \sum_{i=1}^{D-1}r_i^\top r_i.
$$
Now, the modified matrix $\tilde X$ has the same rows except that the final row $r_D$ is replaced with the normalized $\tilde r_D = r_D/\sqrt{2}$ and a row $\tilde r_{D+1} = \tilde r_D$ is added. We then find that
$$
\begin{align}
{\tilde X}^\top \tilde X &= \tilde r_{D+1}^\top \tilde r_{D+1} + \tilde r_D^\top \tilde r_D + \sum_{i=1}^{D-1}r_i^\top r_i
\\ & =
[r_D/\sqrt{2}]^\top [r_D/\sqrt{2}] + [r_D/\sqrt{2}]^\top [r_D/\sqrt{2}] +
\sum_{i=1}^{D-1}r_i^\top r_i
\\ & =
\frac 12 r_D^\top r_D + \frac 12 r_D^\top r_D + \sum_{i=1}^{D-1}r_ir_i^\top
\\ & = r_D^\top r_D + \sum_{i=1}^{D-1} r_i^\top r_i = X^\top X.
\end{align}
$$

Alternatively, note that we can write $\tilde X = RX$, where
$$
R = \pmatrix{
I_{(D-1) \times (D - 1)} & 0_{(D - 1) \times 1}\\
0_{1 \times (D-1)} & 1/\sqrt{2}\\
0_{1 \times (D-1)} & 1/\sqrt{2}}.
$$
Verify that $R^\top R = I$, so that
$$
[\tilde X]^\top [\tilde X] = [RX]^\top [RX] =
X^\top R^\top R X = X^\top I X = X^\top X.
$$


问:
Laurent series of complex rational function.
I want to calculate the Laurent series of function :
$$
f(z) = \frac{z}{(z-1)^3(z+1)}$$
in $z =1$.

I know that point $z=-1$ is the pole of this function. So I have to calculate Laurent series for $|z-1|<2$ and $|z-1|>2$. I tried this way:

$f(z)=\frac{z}{(z-1)^3(z+1)}= \frac{1}{8(z+1)}-\frac{1}{8(z-1)}+\frac{1}{4(z-1)^2}+\frac{1}{2(z-1)^3}$.

$f_1(z)=-\frac{1}{8(z-1)}=\frac{1}{16}\frac{1}{1-\frac{z+1}{2}}=\frac{1}{16}\sum(\frac{z+1}{2})^n$.

By analogy,
$f_2(z)=\frac{1}{4}\frac{1}{(z-1)^2}=-\frac{1}{8}\sum(\frac{z+1}{2})^{2n}$ and $f_3(z)=\frac{1}{2}\frac{1}{(z-1)^3}=-\frac{1}{4}\sum(\frac{z+1}{2})^{3n}$. And finally, $$f(z)=\frac{1}{8(z+1)}+f_1(z)+f_2(z)+f_3(z)$$
Why I have to consider two cases $|z-1|<2$ and $|z-1|>2$ ? How is that different?

Body:
With partial fraction decompose:
$$\frac{z}{(z-1)^3 (z+1)}=\frac{1}{2 (z-1)^3}+\frac{1}{4 (z-1)^2}-\frac{1}{8 (z-1)}+\frac{1}{8 (z+1)}$$

$$\frac{1}{z+1}=\sum _{n=0}^{\infty } \frac{(-1)^n (z-1)^n}{2^{n+1}}$$

Hope this helps


问:
rewriting equation in terms of $y$
From Stewart, Precalculus, 5th ed, P98, Q.45

$$x^2+xy+y^2=4$$

how can I re-write this equation in terms of $y$? I want to put this equation into graphing software but don't know to put $y$ on one side and $x$ on the other.

Any suggestion would be great.

答:
You can think of $x$ (or $y$) as a constant, then you get something of the form:
$$ay^2 + by + c = 0.$$
The solution to this $2$. degree equation will give you $y$ as a function of $x$. You will get two solutions (plus/minus square root) which can both be plotted.

More specifically you will get:
$$y^2 + xy + x^2 - 4 = 0$$

Solution is then:
$$y = -\frac{x}{2} \pm \frac{\sqrt{x^2 - 4x^2 + 16}}{2}$$

Title: rewriting equation in terms of $y$


问:
Number of possible sums of the form $a+b$ where a and b are non negative integers less than a fixed n
Let $A = \{x_1,x_2,...,x_n\}$ where $x_i$ are non-negative integers. Let $A+A := \{x_i + x_j | 1\leq i,j \leq n\}. $ Then what is $|A+A|?$
Even if an exact expression cannot be determined, are there any tight bounds I can achieve here?
I have observed that if $A = \{0,1,...,n\}$ then $|A+A| = 2n+1.$ However when $A$ is not so specifically chosen then $|A+A|$ need not be equal to $2n+1.$

Thanks!

答:
The example where $A = \{0,1,2,\dots,n\}$, or more generally, when $A$ is an arithmetic progression, gives the least possible size of $A+A$: it gives $|A+A|=2|A|-1$. That's the only case where this lower bound is achieved.

This lower bound is easy to prove: if $x_1 < x_2 < \dots < x_n$, then $A+A$ has at least the $2n-1$ distinct elements $$x_1 + x_1 < x_1+x_2 <\dots < x_1 + x_n < x_2 + x_n < \dots < x_n + x_n.$$

On the other hand, a Sidon set such as $\{2^1, 2^2, 2^3, \dots, 2^n\}$ does not have any repeated sums. If $A$ is a Sidon set, we achieve the upper limit of $|A+A| = \binom{|A|+1}{2}$.

Freiman's theorem is one possible result that tells us that $A$ must be very rigidly structured for $|A+A|$ to be small (as a function of $|A|$).


问:
How to solve the recurrence $()=5(/2)+^3()$ using iteration method
How do we solve the recurrence $()=5(/2)+^3()$ using the Iteration method?

I solved the recurrence using a master method - master method

Now using the iteration method

$()=5(/2)+3() = 5(5T(/4)+(n/2)^3)*log(n/2))+n^3*logn$ $=$ ... $= 5^i(n/2^i)+n^3*log(n) * ∑(n^3*(5^k) *log(n/2^k) ) / 8^k) $

How is it equal to $Θ(n^3log(n))$ ?

答:
Let $u_k:=T(2^k)$. Then substituting $n=2^k$ in the given equation, $$u_k=5u_{k-1}+8^kk$$
This can be solved in the usual way, $$u_k=\tfrac{1}{3} 8^{k+1} (k-\tfrac{5}{3}) + C5^k$$ Substituting $k=\log_2n$, $5^k=5^{\log_2n}=n^{\log_25}$, $$T(n)=\tfrac{8}{3}n^3(\log_2 n-\tfrac{5}{3})+Cn^{\log_2 5}=O(n^3\log_2 n)$$


问:
Showing that the Klein-four group is associative without checking 64 options

Showing that the Klein-four group is associative without checking 64 options

I don't know how to go about this. I could use the fact it's abelian to eliminate a few duplicates, but many options remain. Is there an easier way of proving that the Klein-four is associative?

答:
It's a Monday afternoon, so let's have some fun:

Let's start with the assumption that there is a set $S = \{e, a, b, c\}$ with binary operation $*$ having Cayley table:
$$\begin{array} {|r|r|}\hline \cdot & e & a & b & c \\ \hline e & e & a & b & c \\ \hline a & a & e & c & b \\ \hline b & b & c & e & a \\ \hline c & c & b & a & e \\ \hline  \end{array}$$

We're going to use Light's associativity test. Because $S$ is generated by $a$ and $b$ (i.e. $e = a * a$ and $c = a * b$), it suffices to show that for all $x, y \in S$ we have both $x * (a * y) = (x * a) * y$, and $x * (b * y) = (x * b) * y$.

Let's check the first of these claims: start by defining binary operations $x \star y = x * (a * y)$ and $x \circ y = (x * a) * y$. We'll be down when we show that the Cayley tables for $\star$ and $\circ$ agree. There are a few optimizations we can take based on the circumstances:

As the Cayley table is symmetric about the diagonal, we know that $(S, *)$ is abelian. It follows that $x \star y = y \circ x$ (write it out!). Therefore $x \star x = x \circ x$. Moreover, if $x \star y = x \circ y$ then also $y \star x = y \circ x$ (again, convince yourself!).

We don't need to include the identity element in our table, because if either $x = e$ or $y = e$, then $x \star y = x \circ y$.

So now the question is how much of the two Cayley tables do we have to check? Using the above remarks, you get down to something like the below, where "No" means "No, you don't need to check it":
$$\begin{array} {|r|r|}\hline \star & e & a & b & c \\ \hline e & No & No & No & No \\ \hline a & No & No & ? & ? \\ \hline b & No & No & No & ? \\ \hline c & No & No & No & No \\ \hline  \end{array}$$

When the smoke clears, you see that it suffices to check:
$$a \star b = a \circ b,\ a \star c = a \circ c,\ \text{and } b \star c = b \circ c.$$

To check each equation, you need to do 4 multiplications (i.e. Cayley table look-ups). For instance,
\begin{align*}
a \star b &= a * (a * b) = a * c = b,\\
a \circ b &= (a * a) * b = e * b = b.
\end{align*}
To conclude: to check whether the Klein four-group is associative, for each of its 2 generators, it suffices to check 3 "associativity equations" in $(S, *)$.

I'm curious whether there are further simplifications one can make by using algebraic properties specific to the Klein four-group table, but this is all the mileage I can get out of Light's test alone.


问:
Difficulties solving this integral: $ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $ by differentiation under the integral sign
So in the book Advanced Calculus Explored, by Hamza E. Asamraee. The next integral appears as an exercise to solve by differentiating under the integral sign:

$$ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $$

I have solved this integral before by substitution and change in the limits of integration, but in this chapter the book asks to solve it by differentiation under the integral sign. I have tried several ways of solving this, but the only one that i thought it was leading me somewhere was:

$$f(a) = \int_0^1 \frac{\ln(x+a)} {x^2 + 1} \, \mathrm{d}x $$

So that:

$$f'(a) = \int_0^1 \frac{1} {(x+a)(x^2 + 1)} \, \mathrm{d}x $$

Then i tried to separate this last integral by partial fractions, my result on this was:

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

And the integral reduces to:

$$f'(a) = \int_0^1 \frac{1} {a^2 + 1} \left(\frac {1} {x+a} - \frac{x-a} {x^2+1} \right) \, \mathrm{d}x $$

Then this last expression evaluates to:

$$f'(a) = \frac{1} {a^2 + 1} (\ln(a+1) - \ln(a) - \ln(4)+ \frac{π}{4} a)$$

Then integrating from 0 to 1 with respect to $a$ we will get:

$$f(1) - f(0) = \int_0^1 \frac{\ln(a+1)} {a^2 + 1} \, \mathrm{d}a - \int_0^1 \frac{\ln(a)} {a^2 + 1} \, \mathrm{d}a $$

(The last two terms of $f'(a)$ cancel each other after the integration so i didn't wrote them)

But then the two integrals on the right hand side are equal to $f(1) - f(0)$ so the differentiation under the integral led nowhere.

Do i need some other approach? Or did i made any mistake?
Any help is appreciated.

Body:
This would work:
$$I(a) = \int_0^1 \frac{\ln(ax+1)}{x^2+1} dx \\
I’(a) =\int_0^1 \frac{x}{(x^2+1)(ax+1)} dx \\ \overset{\text{partial fractions}}= \\ \frac{-2\ln |ax+1| +\ln(x^2+1)+2a\tan^{-1} x}{2(a^2+1)} \bigg |_0^1 \\ =-\frac{\ln(a+1)}{a^2+1}+\frac{\ln 2}{2a^2+2}+\frac{\pi}{4} \frac{a}{a^2+1}
$$
Integrating from $0$ to $1$, $$I(1)-I(0) = -I(1) +\int_0^1 \left(\frac{\ln 2}{2a^2+2}+\frac{\pi}{4} \frac{a}{a^2+1}\right) da$$

Hopefully you can finish.


问:
maximal ideal of PIR
PIR means principal ideal ring.This is extended concept of PID, we do not assume integral domain.

Is it true that every nonzero prime ideal is maximal, even in PIR ?
Or, do we need to restrict the condition ?（For example, non nilpotent prime ideal）
Thank you for your help.

答:
I think the best one can say is that the class of commutative rings for which the nonzero prime ideals are maximal can be classified in two pieces:

domains of Krull dimension $1$ (which come in a large variety and can't really be described more simply)

rings of Krull dimension $0$ (also called $\pi$-regular rings) which can be characterized as

Rings for which chains of the form $(x)\supseteq (x^2)\supseteq (x^3)\supseteq\ldots$ stabilize for all $x$

$R/N$ is von Neumann regular, where $N$ is the nilradical of $R$.

Principal ideal domains fall under the first item, but since taking finite ring products increases the Krull dimension but preserves the property of being a PIR, you will quickly leave the class (as mentioned in the comments.)

Any Artinian PIR is zero dimensional, though.

Title: maximal ideal of PIR


Title: Difficulties solving this integral: $ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $ by differentiation under the integral sign

Body:
So in the book Advanced Calculus Explored, by Hamza E. Asamraee. The next integral appears as an exercise to solve by differentiating under the integral sign:

$$ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $$

I have solved this integral before by substitution and change in the limits of integration, but in this chapter the book asks to solve it by differentiation under the integral sign. I have tried several ways of solving this, but the only one that i thought it was leading me somewhere was:

$$f(a) = \int_0^1 \frac{\ln(x+a)} {x^2 + 1} \, \mathrm{d}x $$

So that:

$$f'(a) = \int_0^1 \frac{1} {(x+a)(x^2 + 1)} \, \mathrm{d}x $$

Then i tried to separate this last integral by partial fractions, my result on this was:

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

And the integral reduces to:

$$f'(a) = \int_0^1 \frac{1} {a^2 + 1} \left(\frac {1} {x+a} - \frac{x-a} {x^2+1} \right) \, \mathrm{d}x $$

Then this last expression evaluates to:

$$f'(a) = \frac{1} {a^2 + 1} (\ln(a+1) - \ln(a) - \ln(4)+ \frac{π}{4} a)$$

Then integrating from 0 to 1 with respect to $a$ we will get:

$$f(1) - f(0) = \int_0^1 \frac{\ln(a+1)} {a^2 + 1} \, \mathrm{d}a - \int_0^1 \frac{\ln(a)} {a^2 + 1} \, \mathrm{d}a $$

(The last two terms of $f'(a)$ cancel each other after the integration so i didn't wrote them)

But then the two integrals on the right hand side are equal to $f(1) - f(0)$ so the differentiation under the integral led nowhere.

Do i need some other approach? Or did i made any mistake?
Any help is appreciated.

Body:
$$f(x)=\frac{1}{(x+a)(x^2+1)}=$$
$$\frac{A}{x+a}+\frac{Bx+C}{x^2+1}$$

we find that $$A=\frac{1}{a^2+1}$$

and

$$Bi+C=\frac{1}{a+i}=\frac{a-i}{a^2+1}$$

So,

$$B=-\frac{1}{a^2+1}\;,\;C=\frac{a}{a^2+1}$$


问:
Proving invariance of a functional under transformation
Suppose we have a functional of the form
$$
J(y)= \int_{x_{0}}^{x_{1}} f(x,y,y')dx
$$
and a smooth transformation $X=\theta (x,y: \epsilon)$ , $Y=\psi (x,y: \epsilon)$ where $\epsilon$ is a parameter. I have the following definition of invariance under a transformation:

Definition. The functional $J$ is invariant under the above transformation if, for all $\epsilon$ sufficiently small, in any subinterval $[a,b] \subset [x_{0},x_{1}]$ we have that
$$
\int_{a}^{b} f(x,y,y')dx= \displaystyle \int_{a_{\epsilon}}^{b_{\epsilon}} f(X,Y,Y')dX
$$
for all smooth functions $y$ defined on $[a,b]$. Here, $a_{\epsilon}=\theta (a,y(a): \epsilon)$, $b_{\epsilon}=\theta (b,y(b): \epsilon)$

Once said that, I'm interested in proving the invariance of the functional $$
J(y)=\int_{x_{0}}^{x_{1}} xy'^2dx
$$
under the transformation
$$
\begin{cases}
X=x+2\epsilon x \ln(x)\\
Y=(1+\epsilon)y
\end{cases}
$$

My attempt:

This would be easier if we could solve explicitly for $x$ in the first equation of the transformation, but, since we can't, suppose $x=\Theta (X; \epsilon)$ such inverse function exist(locally and for $\epsilon$ sufficiently small because of the inverse function theorem). On the other hand, we easily get $y=1/(1+\epsilon) Y$. Therefore,
$$
y'(x)=\displaystyle \frac{dy}{dx}=\displaystyle \frac{Y'(X)}{(1+ \epsilon) \Theta'(X)}
$$
which implies that
$$
xy'dx= \frac{\Theta(X)}{\Theta '(X)} \frac{Y'(X)^2}{(1+\epsilon)^2}dX
$$
So, as you can see, if I could prove
$$
\frac{\Theta(X)}{(1+\epsilon)^2 \Theta '(X)}=X
$$
I'd be done, but I could not prove it, and actually, it doesn't seem true.

Any help? Thanks in advance.

Update:

Of course we can also try to express $XY'(X)^2 dX$ in terms of $x$ and $y'(x)$. In this case, we get
$$
XY'(X)^2 dX= \displaystyle \frac{x+2 \epsilon x ln(x)}{1+ 2 \epsilon ln(x)+2 \epsilon} (1+ \epsilon)^2 y'(x)^2 dx
$$
In any case, it is not clear to me how to reconcile this expressions with the definition :( .

答:
Actually, this transformation is NOT a variational symmetry. From your computations above we have that,
$$
\displaystyle \int_{a_{\epsilon}}^{b_{\epsilon}} XY'(X)dX= \displaystyle \int_a^b xy'^2 dx + \displaystyle \int_a^b \displaystyle \frac{y'^2(x \epsilon ^2+2 \epsilon ^3 x ln(x))}{1+2 \epsilon ln(x)+2 \epsilon} dx
$$
And this definitely does not agree with the definition. On the other hand, even if we ignore this, if this transformation were a variational symmetry, because of the definition we should have that
$$
\displaystyle \frac{d}{d \epsilon} \left( \displaystyle \int_{a_{\epsilon}}^{b_{\epsilon}} f(X,Y,Y') dX  \right)=0
$$
But, in this case, this is obviously false.

Remark:

Althought this is not a variational symmetry, it's certainly true that the functional is infinitesimally invariant under the transformation, but it's no the same thing, check the definition in the book Advanced Engineering Analysis by Leonid P. Lebedev.


问:
To sum $1+2+3+\cdots$ to $-\frac1{12}$
$$\sum_{n=1}^\infty\frac1{n^s}$$
only converges to $\zeta(s)$ if $\text{Re}(s)>1$.

Why should analytically continuing to $\zeta(-1)$ give the right answer?

答:
Here is a variant on Lubos Motl's answer:

Let $S = \sum_{n=1}^{\infty} n$.  Then $S - 4 S = \sum_{n = 1}^{\infty} (-1)^{n-1} n.$   We will evaluate this latter expression with a regularization similar to Lubos Motl's.

Namely, consider $$\sum_{n=1}^{\infty} (-1)^{n-1} n t^n
 = -t \dfrac{d}{dt} \sum_{n=1}^{\infty} (-t)^n = -t \dfrac{d}{dt} \dfrac{1}{1+t}
= \dfrac{t}{(1+t)^2}.$$

Letting $t \to 1,$ we find that $-3 S = \dfrac{1}{4}$, and hence that
$S = \dfrac{-1}{12}.$

To see the relationship between this approach and Lubos Motl's, note that if
we write $t = e^{\epsilon},$ then $t\dfrac{d}{dt} = \dfrac{d}{d\epsilon},$
so in fact the arguments are essentially the same, except that Lubos doesn't
perform the initial step of replacing $S$ by $S - 4S$, which means that
he has the pole $\dfrac{1}{\epsilon^2}$ which he then subtracts away.

As far as I know, this trick of replacing $\zeta(s)$ by $(1-2^{-s+1})^{-1}\zeta(s)$ is due to Euler, and it is a now standard method for replacing
$\zeta(s)$ by a function which carries the same information, but
does not have a pole at $s = 1$.   The evaluation of $\zeta(s)$ at negative integers by passing to $(1-2^{-s+1})\zeta(s)$ and then performing Abelian regularization as above is also due to Euler, I believe.  It is easy to see
the Bernoulli numbers appearing in this way, for example.

Of course, taken literally, the series $\sum_{n=1}^{\infty} n$ diverges to $+\infty$, so any attempt to assign it a finite value will involve some form of regularization.  Analytic continuation of the $\zeta$-function is one form of regularization, and the Abelian regularization that Lubos Motl and I are making is another.  I can't quote a precise theorem to this effect (although maybe others can), but with such a simple expression as $\sum_{n = 1}^{\infty} n,$
I'm reasonably confident that any sensible regularization will necessarily yield the same value of $\dfrac{-1}{12}$.  (Lubos Motl makes the same assertion in his answer.)

OwnerDisplayName: perplexed

Title: To sum $1+2+3+\cdots$ to $-\frac1{12}$


问:
Open Cover / Real Analysis
I have the next question: Let $K \subset $ $R^1$ consist of $0$ and the numbers 1/$n$, for $n=1,2,3,\ldots$ Prove that $K$ is compact directly from the definition (without using Heine-Borel).

I'm trying to understand compact sets so I would be grateful if someone could give me some examples of open covers and subcovers. Thank you!

答: Let $\{U_\alpha\}$ be a open covering of $K$. Since it is a covering, there exists a $\beta$ such that $0 \in U_\beta$. Since $U_\beta$ is open, there exists an $N$ such that $B_\frac{1}{N}(0) \subseteq U_\beta$. Hence for all $n > N$, $\frac{1}{n} \in B_\frac{1}{N}(0) \subseteq U_\beta$. Since $\{U_\alpha\}$ covers, there exists $\gamma_1, \ldots, \gamma_N$ such that $\frac{1}{i} \in U_{\gamma_i}$. Hence $U_\beta, U_{\gamma_1}, \ldots, U_{\gamma_N}$ is a finite subcover of $K$. </p>

Title: Open Cover / Real Analysis


问:
Problem similar to Collatz conjecture
Let $f:\mathbb{Z} \rightarrow \mathbb{Z}$ be given by $f(x) = \frac{n}{2}$ if $n$ is even and $f(n) = n^2 -1$ if $n$ is odd. Consider the sequence $(a_n)$ given by $a_n = f(a_{n-1})$ with $a_0$ a positive integer. Will the sequence eventually reach $1$? ( i.e does there exist an $N$, such that for any $m>N$, $a_m = 1$ no matter what the value of $a_0$ is ?)

At first glance, it does seem to be the case, but taking $a_0 = 13$, it is not clear whether the sequence reaches $1$. Coding it in python, after $1000$ iterations the sequence got exponentially large, to the point that squaring woulnd not be accurate enough in python. I tried looking for patterns in $mod(8)$ but couldn't really make much progress with this approach.

答:
An idea rather than an answer

This can be analysed as a function on odd numbers, $$ F(2k+1)={Od}(k(k+1)),$$ where
$Od$ is the odd part function.

Unless $k$ or $k+1$ is a power of $2$, ${Od}(k(k+1))\ge 3k$ and so $F$ is an increasing function unless and until a number of the form $2^n\pm 1$ is reached. From such numbers we immediately have a decreasing chain of numbers of the same form until 1 is reached.

The problem thus becomes one of analysing the rare cases when $ F(2k+1)$ is of the form $2^n\pm 1$.

Example  Of the first few odd numbers, $11$ and $13$ are the first not to be of the form  $2^n\pm 1$ and $11$ immediately maps to $15=2^4-1.$

Title: Problem similar to Collatz conjecture


答:
All trees are balanced, so in particular stars are balanced.

An $n$-vertex tree has average degree $2-\frac 2n$. Any subgraph of average degree $2$ can be reduced to a subgraph of minimum degree $2$ by removing leaves, but a subgraph of minimum degree $2$ would contain a cycle. Trees don't have cycles, so any subgraph of a tree has average degree less than $2$. However, for a $k$-vertex subgraph, the largest possible average degree less than $2$ is $2 - \frac 2k \le 2 - \frac 2n$.

(In fact, all trees are strictly balanced, since the only way to get $2 - \frac2k = 2 - \frac2n$ is to have $k=n$, taking the entire tree as a subgraph.)

Title: Show that a Star graph is balanced


问:
Details of the proof of the Exit Time Theorem
Theorem 1.29. Let $V_{A}=\inf \left\{n \geq 0: X_{n} \in A\right\} .$ Suppose $C=S-A$ is finite, and that $P_{x}\left(V_{A}<\infty\right)>0$ for any $x \in C .$ If $g(a)=0$ for all $a \in A,$ and for $x \in C$ we have
$$
g(x)=1+\sum_{y} p(x, y) g(y)
$$
Then $g(x)=E_{x}\left(V_{A}\right)$

Proof. It follows from Lemma 1.3 that $E_{x} V_{A}<\infty$ for all $x \in C .(1.27)$ implies that $g(x)=1+E_{x} g\left(X_{1}\right)$ when $x \notin A .$ The Markov property implies

$$
g(x)=E_{x}\left(V_{A} \wedge n\right)+E_{x} g\left(X_{V_{A} \wedge n}\right)
$$
We have to stop at time $T$ because the equation is not valid for $x \in A .$ It follows from the definition of the expected value that $E_{x}\left(V_{A} \wedge n\right) \uparrow E_{x} V_{A} .$ Since $g$ is bounded and $g(a)=0$ for $a \in A,$ we have $E_{x} g\left(X_{V_{A}  \wedge n}\right) \rightarrow 0$

This theorem 1.29 is from Richard Durrett's 《Essentials of Stochastic the Processes Third Edition》(Page 62-63).In this theorem, g(x) is expected exit time with initial state x.The hint is simply to use the Markov property,but I finally can't prove it.Could someone show me the details of the proof that  $g(x)=1+E_{x} g\left(X_{1}\right)$ is equivalent to $
g(x)=E_{x}\left(V_{A} \wedge n\right)+E_{x} g\left(X_{V_{A} \wedge n}\right)
$?Thank you very much！

答:
One way to proceed is by induction. The case $n=0$ is clear. Suppose the stated identity has been shown for $n$, and let's check it for $n+1$. First,
$$
E_xg(V_A\wedge (n+1)) = E_x[g(X_{V_A}); V_A\le n]+E_x[g(X_{n+1}); V_A>n].
$$
The first term on the right vanishes because $X_{V_A}\in A$ on $\{V_A<\infty\}$. In dealing with the second, notice that $\{V_A>n\}$ is $\mathcal F_n$-measurable. Here is where to use the Markov property to see that
$$
\eqalign{
E_x[g(X_{n+1}); V_A>n]
&=E_x\left[ E_x\left[g(X_{n+1})\mid \mathcal F_n\right]; V_A>n\right]\cr
&=E_x\left[\sum_y p(X_n,y) g(y); V_A>n\right]\cr
&=E_x\left[g(X_n)-1; V_A>n\right]\cr
&=E_x[g(X_{V_A\wedge n}]-E_x[g(X_{V_A}; V_A\le n]-P[V_A>n]\cr
&=E_x[g(X_{V_A\wedge n}]-P[V_A>n]\cr
}
$$
Turning to the left side, we can break the expectaion in two as before:
$$
\eqalign{
E_x[V_A\wedge (n+1)]
&=E_x[V_A; V_A\le n]+E_x[n+1: V_A>n]\cr
&=E_x[V_A\wedge n]+P_x[V_A>n]\cr
}
$$
Adding these two and using the induction hypothesis you see that the identity holds for step $n+1$.


问:
Combinatorics problem IMO style, existence type problem
I've tried and failed solving this problem for a week now; it doesn't seem too difficult and I feel like I'm missing something obvious.  What is the solution?

EDIT:
I tried constructing some sequences that would work and would divide 2001, for example 23 divides 2001 and if you can make a sequence of coins with length 21 you could just use the sequence again and again, but I couldn't find such sequence that wouldn't interfere itself (by for example having a 3 on the outside); also I just noticed that my calculations were wrong and the length of the sequence doesn't perfectly divide 2001. I tried induction and strong induction but that didn't work; finally I tried using the hostile neighbors principle (which I read about in a book and I'm pretty sure the author made the name up)

答:
Hints:

Notice that any $4$ consecutive coins can have at most one $3$-kopeyka.

Show that if you choose any $4$ consecutive coins, say, $a_k, a_{k+1}, a_{k+2}, a_{k+3}$, then exactly one of them is $3$-kopeyka.


答: Let $\{U_\alpha\}_{\alpha \in A}$ be a cover of $K$.  Then there exists some $\alpha_0 \in A$ such that $U_{\alpha_0}$ contains $0$.  Since $U_{\alpha_0}$ is open, for some $\epsilon > 0$, we have an open ball $B(0,\epsilon)$ centered at $0$ such that $B(0,\epsilon) \subset U_{\alpha_0}$.  Since $\epsilon >  0$, we can find $n \in \mathbb{N}$ such that $\epsilon > 1/n$.  Thus, $B(0,\epsilon)$ contains $\{0,1/n, 1/(n+1), \ldots\}$.  Now take elements of the open cover containing $1/2, \dots, 1/(n-1)$.  Since only finitely many elements are not in $U_{\alpha_0}$, we only need finitely many open sets in our cover.</p>

Title: Open Cover / Real Analysis


问:
In each cell of a $7\times 7$ grid is a number. The sum of all the numbers in each $2\times 2$ square and each $3\times 3$ square is zero.
In each cell of a $7\times 7$ grid is a number. The sum of all the numbers in each $2\times 2$ square and each $3\times 3$ square is zero. Prove that the sum of the numbers in the 24 bordering cells is $0$ too.

答:
Consider a $3 \times 3$ square such as this one.

You can see that we can put a $2\times 2$ square in it (in white here).
The sum of the numbers in the "boomerang" shape colored in red is the sum of the numbers in the $3 \times 3$ square minus the sum of the numbers in the $2 \times 2$ square which is zero.

As such the sum of the numbers in every boomerang is zero.

Let's put a boomerang in every corner of our $7 \times 7$ grid below (colored in blue). It just remains to prove that the numbers in the 4 central cells of the grid's sides sum to zero.

Consider the $4 \times 4$ square in the lower left corner.
The sum of its numbers is zero because it can be separated into four $2 \times 2$ squares.

However, we can see it can also be separated into a boomerang (in blue), a $3\times 3$ square (in red) and two of the central cells (in green).
As such the sum of these two central cells is zero.

The sum of the other two central cells is similarly zero.

As such the sum of the numbers in the border is zero.


问:
Evaluate $\int\frac{x^4}{(x^2-1)^3}dx$
I tried:$$I=\int\frac{x^4}{(x^2-1)^3}dx=\int\frac{(x^2-1)(x^2+1)+1}{(x^2-1)^3}dx=\int\frac{x^2+1}{(x^2-1)^2}+\frac{1}{(x^2-1)^3}dx$$
For first fraction we can write it as $\frac{1}{x^2-1}+\frac{2}{(x^2-1)^2}$. therefor we have:

$$I=\frac12\ln\left|\frac{x-1}{x+1}\right|+\int\frac{2}{(x^2-1)^2}dx+\int\frac{1}{(x^2-1)^3}dx$$

I don't know how to evaluate remaining integrals.

答: Another approach: with $x=-\csc t$ the original integral becomes $\int\sec^5tdt$, then use this together with $\sec t\tan t=\frac{x}{1-x^2},\,(\sec t+\tan t)^2=\frac{2x-1-x^2}{1-x^2}$.

Title: Evaluate $\int\frac{x^4}{(x^2-1)^3}dx$


答:
Using the $K_4$ properties:

: elements are self-inverse

: operation is commutative

Consider $x,y,z \in K_4$, not necessarily distinct. We ask if $x\cdot (y \cdot z) = (x\cdot y) \cdot z$.

If any of $x,y,z$ is the identity, the equality is immediately true.

If $x,y,z$ are distinct, the result is the identity (since inverse is unique):

$x\cdot (y \cdot z) = x\cdot x = e$

$(x\cdot y) \cdot z = z\cdot z = e$

For $x=z$, commutativity gives:

$(x\cdot y)\cdot x =(y\cdot x)\cdot x =x\cdot (y\cdot x )$

If $y=x\neq z$,

$(x\cdot y) \cdot z = e\cdot z = z$ and

$x\cdot (y \cdot z) = y\cdot w = z$ where $w$ is the other element, $w\neq (x=y),z$

and similarly if $y=z\neq x$.

Body:
Showing that the Klein-four group is associative without checking 64 options

I don't know how to go about this. I could use the fact it's abelian to eliminate a few duplicates, but many options remain. Is there an easier way of proving that the Klein-four is associative?


问:
Draw the derivative of a function
I am trying to draw the graph of the derivative of this function:

Since the function has a maximum at $x = 0$, the derivative at this point must be zero. In addition, the function is increasing up to $x = 0$ and decreasing from this value, so the derivative has $y> 0$ for $x <0$ and $y <0$ for $x> 0$. When $x$ tends to $+\infty$, the graph tends to zero, so $f'(x) = 0$. My problem is to understand the shape of the concavity, or if there is a concavity in the graph. The graph I made was as follows:

Could you tell me if the concavities and the shape of the graph are right and why do you have the concavities that way?

答:
Your graph appears to look good. However, note that the derivative for values for $x \to -\infty$, the derivative does not reach an asymptote, as you assume, instead it goes to infinity. The slope is very big if $x \to -\infty$, as indicated by the graph.

So yes, it should grow "like $x^2$ up.", because the slope is very big. Additionally, the left side of the graph seems close to $y = ax^3 + 1$, whose derivative is $3ax^2$, a quadratic function.

The "trick" says things about the concavity of $f^\prime$. To find the concavity, you have to check the sign of second derivative of the function. In this case, the function is the first derivative of your original function. $(f^\prime)^{\prime \prime}$ = $f^{\prime \prime \prime}$.

Title: Draw the derivative of a function


问:
$f_n(x) = \log(1 + n^3 x^2)/n^2$ (uniform converge)
$$
\mbox{Let}\quad
\operatorname{f}_{n}\left(x\right) =
\frac{\log\left(1 + n^{3}x^{2}\right)}{n^{2}}
$$
Show that $\left(\operatorname{f}_{n}\right)_{n}$ converges uniformly to the zero function on $\left[0,1\right]$.

I know that uniform convergence mean

 $$
\left\vert\operatorname{f}_{n}\left(x\right)−\operatorname{f}\left(x\right)\right\vert <
\varepsilon, \qquad \forall n \geq N,\ \forall x\in \left[0,1\right].$$

Can someone help me prove it.

答:
For $ n>0$,

$$x\mapsto 1+n^3x^2 $$
is increasing from $ [0,1] $ to $[1,+\infty)$

$$x \mapsto \ln(x) $$
is increasing at $ [1,+\infty)$,

So, $ f_n $ is increasing at $ [0,1]$ and

$$M_n=\sup_{x\in [0,1]}|f_n(x)|=f_n(1)$$
$$=\frac{\ln(1+n^3)}{n^2}$$
with

$$0\le M_n\le \frac{\ln(n^3+n^3)}{n}=\frac{\ln(2)+3\ln(n)}{n}$$
and by squeezing,

$$\lim_{n\to +\infty}M_n=0$$

the convergence to zero is uniform at $ [0,1]$.


答: Hint: $\sup\limits_{x\in [0,1]}|f_n(x)-0|\rightarrow 0$ as $n\rightarrow\infty.$

Title: $f_n(x) = \log(1 + n^3 x^2)/n^2$ (uniform converge)

Body:
$$
\mbox{Let}\quad
\operatorname{f}_{n}\left(x\right) =
\frac{\log\left(1 + n^{3}x^{2}\right)}{n^{2}}
$$
Show that $\left(\operatorname{f}_{n}\right)_{n}$ converges uniformly to the zero function on $\left[0,1\right]$.

I know that uniform convergence mean

 $$
\left\vert\operatorname{f}_{n}\left(x\right)−\operatorname{f}\left(x\right)\right\vert <
\varepsilon, \qquad \forall n \geq N,\ \forall x\in \left[0,1\right].$$

Can someone help me prove it.


问:
Existence of a similar nilpotent matrix in a general sense.
Let there be a finite dimensional vector space $V$ over $\mathbb{C}$ of dimension $n$ with $L : V \rightarrow V$ a nilpotent operator with index $k$. If $L$ is diagonalizable then I know that it is similar to the $0$ matrix. I was curious in an extension of this, that is does there exists a finite dimensional (say m) $\mathbb{C}$ vector space $W$ and  $L' : W \rightarrow W$ another nilpotent operator such that $$L \oplus L'\;is\;similar\;to\;0 \oplus L'. $$
Now if $L$ was diagonalizable then $L \oplus L'$ is similar to $0 \oplus L'$ for any $L'$. Because I can take the invertible matrix as the block diagonal matrix
$\begin{pmatrix}
[P_1] & [0] \\
[0] & [I_m]
\end{pmatrix}$ where $P_1 L P_1^{-1} = [0]$. So what I am trying to find is does there exists a pair ($W,L'$) of an $m$ dimensional vector space $W$ and a nilpotent endomorphism $L'$, and an $(n+m \times n+m)$ matrix $Q$ such that $$Q
\begin{pmatrix}
[L] & [0] \\
[0] & [L']
\end{pmatrix} Q^{-1} = \begin{pmatrix}
[0] & [0] \\
[0] & [L']
\end{pmatrix} $$ given a pair $(V,L)$ of an $n$ dimensional vector space and a nilpotent endomorphism $L$. I was trying with the pair $(\mathbb{C}^2,\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} )$ but having no luck. So if anyone could help me I will be highly obliged.

答:
This is false for every nonzero nilpotent operator $L$ (on a finite-dimensional vector space). Rank is additive with respect to direct sum and invariant under similarity, so $\text{rank}(L \oplus L') = \text{rank}(L) + \text{rank}(L')$ has strictly larger rank than $\text{rank}(0 \oplus L') = \text{rank}(L')$ and so the two can't be similar.

(Also, you don't need to say "similar to the zero matrix": the zero matrix is the unique matrix similar to itself so a matrix similar to the zero matrix is just identical to the zero matrix.)


问:
If $X$ and $Y$ are not independent, is $E\left(X^2Y\right) = E\left(X^2E\left(Y|X\right)\right)$?
If $X$ and $Y$ are not independent, does the following hold by law of iterated expectation?

$$E\left(X^2Y\right) = E\left(E\left(X^2Y|X\right)\right) = E\left(X^2E\left(Y|X\right)\right)$$

I'm not too sure about the second equality. The inner expectation is wrt $Y$, holding $X$ constant. However, if $X$ depends on $Y$, does the second equality still hold?

答:
Assuming the expectations exist,

$E[X^2 Y \mid X=x] = E[x^2 Y \mid X=x]= x^2 E[Y \mid X=x]$

so $E[X^2 Y \mid X] = X^2 E[Y \mid X]$

and $E\left[E[X^2 Y \mid X]\right] = E\left[X^2 E[Y \mid X]\right]$


问:
Check whether a given graph is bipartite except for one edge
Given any graph $G=(V,E)$, how to decide if deleting just one of its edges can make it bipartite?

Trivial solution: check for each edge whether deleting it makes the graph bipartite. This is in $O(|E|\cdot(|E|+|V|))$.

Are there more efficient ways?

答:
We seem to lack inspiration here, so let me detail an $O(|V|\cdot(|E|+|V|))$ solution.

Assume there is an odd cycle in $G$; if there is none then $G$ is bipartite. Then, pick any vertex $v$ and consider a BFS tree from it. If an edge exists between two vertices at the same distance from $v$, then we have an odd cycle. Conversely, notice that the existence of an odd cycle implies that such an edge exists too. Therefore, we obtain an odd-length cycle in $O(|E|)$.

Any cycle has at most $|V|$ edges, and if $G$ is bipartite except for one edge then the additional edge is in all odd-length cycles. We therefore try to remove each edge of our odd-length cycle in turn and check if the obtained graph is bipartite, as proposed in the initial question for all edges in $E$. This makes an $O(|V|\cdot(|E|+|V|))$ solution.

But there must be better ways!


问:
Find $a,b,c$ of a progression

The numbers $a,b,c$ are in an arithmetic progression with $a+b+c=124$

If $a,b,c$ are the third, thirteenth and fifteenth terms of an arithmetic progression, find $a,b,c$

There are no hints in the book regarding this question. A friend of mine suggests that the problem is worded wrongly and the first sentence should be:

The numbers $a,b,c$ are in a geometric progression with $a+b+c=124$

If I consider $a,b,c$ as being in a geometric progression as the initial information I can solve the problem. I get the result $r = 1/5$, $a=100$, $b=20$, $c=4$.

I can't solve the other way around (if $a,b,c$ are in an arithmetic progression as the first sentence). Something doesn't really click in my brain so that's why I am here for help.

To sum it all up, I am interested in finding out:

a. if the exercise is wrong and it should be geometric instead of arithmetic in the first sentence

or b. a hint to solve the original exercise if it turns out to be correct

答:
The exercise is wrong: both ways of interpreting the question lead to unexpected answers.

Suppose "In arithmetic progression" means that $b-a = c-b$. Using the general term of an arithmetic progression:

$$T(n) = x + (n-1)d$$

where $x$ is the first term and $d$ the common difference, we obtain

$$a=x+2d, b=x+12d, c=x+14d \leadsto b-a = 10d=c-b=2d$$

giving $d=0, a=b=c=\frac{124}3$, which is weird to say the least.

Now suppose "In arithmetic progression" gives no extra information beyond the "3rd, 13th, 15th term" condition. Then we have:

$$x+2d+x+12d+x+14d = 3x+28d = 124$$

which is a single equation with two unknowns, which has an infinite number of solutions, and a unique solution cannot be obtained without additional information (e.g. they are actually in geometric progression).

With the extra information, we can write

$$\frac {10d}{x+2d} = \frac{2d}{x+12d} \leadsto5(x+12d) = x+2d \leadsto 4x+58d=0$$

which combined with $3x+28d=124$ gives the solution you found ($x=116, d= -8$).

Title: Find $a,b,c$ of a progression


答:
Integrate by parts to reduce the remaining integrals

$$\int \frac{dx}{(x^2-1)^3}=\int \frac1{4x^3}d\left( \frac{-x^4}{(x^2-1)^2}\right)= -\frac14 \frac{x}{(x^2-1)^2}-\frac34 \int \frac{dx}{(x^2-1)^2}
$$

$$\int \frac{dx}{(x^2-1)^2}=\int \frac1{2x}d\left( \frac{-x^2}{x^2-1}\right)= -\frac12 \frac{x}{x^2-1}-\frac12\int \frac{dx}{x^2-1}
$$

Title: Evaluate $\int\frac{x^4}{(x^2-1)^3}dx$

Body:
I tried:$$I=\int\frac{x^4}{(x^2-1)^3}dx=\int\frac{(x^2-1)(x^2+1)+1}{(x^2-1)^3}dx=\int\frac{x^2+1}{(x^2-1)^2}+\frac{1}{(x^2-1)^3}dx$$
For first fraction we can write it as $\frac{1}{x^2-1}+\frac{2}{(x^2-1)^2}$. therefor we have:

$$I=\frac12\ln\left|\frac{x-1}{x+1}\right|+\int\frac{2}{(x^2-1)^2}dx+\int\frac{1}{(x^2-1)^3}dx$$

I don't know how to evaluate remaining integrals.