问:
Does every field extension of a field containing irreducible polynomial have a root?
Suppose $F$ is a field and $K/F$ a field extension of $F$. Assume we have a polynomial $p(x) \in F[x]$ that is irreducible over $F$. My question is:

From these assumptions, does it follow that $p(x)$ has a root $c$ in $K$? If not, what would be a counter-example?

I have previously shown that $F[x] /( p(x) )$, where $( p(x) )$ denotes the principal ideal of $p(x)$, is a field extension of $F$ and contains a root of $p(x) $. Maybe this could be of help.

Thanks

答: A counter-example would be $x^2+1$, which is irreducible in $\Bbb Q[x]$ and which has no root in $\Bbb R$.


问:
Why are groups more important than semigroups?
This is an open-ended question, as is probably obvious from the title. I understand that it may not be appreciated and I will try not to ask too many such questions. But this one has been bothering me for quite some time and I'm not entirely certain that it's completely worthless, which is why I've decided to ask it here.

Why are groups so immensely more important in mathematics and its applications than semigroups are? I know little of group theory, mathematics and its applications, so I cannot understand how much more important they are. But I know they are because I see how many more people are interested in groups than in semigroups.

I wouldn't ask this question if this fact didn't seem a little strange to me. I know that groups are associated with symmetries, or with automorphisms of structures. Cayley's Theorem tells me that every group can be seen as a set (closed with respect to taking compositions and inverses) of automorphisms of a structure with no constants, functions or relations, i.e. a plain set. I know that the automorphisms of a vector space, a module or a group form a group. Obviously, automorphisms are important.

Then we have inverse semigroups. These are less popular but this I can understand. They are associated with partial symmetries, by the Wagner-Preston Theorem. Partial functions do seem much less used than functions.

But then come semigroups. Just like in the two previous cases, there is an "embedding theorem", which says that every semigroup can be embedded in a semigroup of maps from a certain set into itself. And, analogously to the case of groups, the endomorphisms of a vector space, a module or a group form a semigroup.

This seems to say that semigroups are to endomorphisms what groups are to automorphisms. The "conclusion" would be that

$\frac{\mbox{the importance of semigroups}}{\mbox{the importance of endomorphisms}}=\frac{\mbox{the importance of groups}}{\mbox{the importance of automorphisms}}$

Assuming that endomorphisms are about as important as automorphisms, we get that semigroups are about as important as groups. My feeling is that the assumption is correct.

I understand that I must be oversimplifying something at some point. But what am I oversimplifying and where?

I realize that this is possibly a very dumb question, but my confusion is genuine.

Edit: I have realized, after reading your answers and comments, that I made the mistake of using a very vague term without even attempting to define it. The problem seems to be what importance is. Is it popularity or usefulness, or being used a lot, or interesting or being interesting, or something else, or a mixture of many traits? I'm not going to force my understanding of the word now. But if someone still wants to answer this question, perhaps it would be good idea if the answer contained an attempt at clearing this up. (Or maybe not. I'm not sure!)

Edit: A question similar to mine is dealt with here.

答:
There is another structure that is weaker than groups, but is important. Those are quasigroups. Quasigroups are informally "magmas with inverses".

I think it would be unfair to compare two structures, when one has more axioms than the other. That's why we should look more at why quasigroups provide a richer theory than that of semigroups - which they certainly do.

Quasigroups connect to combinatorics, for example, Cayley table of any finite quasigroup is a Latin square, and conversely. They have geometric interpretations. The overall theory I think is weaker than that of groups, it might be a combination of reasons, I think it's because they are looked down upon, are harder to work with, and the results are more technical and more combinatorial. I don't think the theory can rival that of groups, but it's certainly rich enough to be useful, which I don't think semigroup theory is.

The associativity of groups is in my opinion just a missing link, a very convenient simplification.

Concluding, I think the reason why groups give a much better theory than semigroups is the operation of inverse.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
How can I prove this reduction formula for $\int^1_0{x(1-2x^4)^n}dx$
The exercise in my textbook states

You are given that $$I_n=\int^1_0{x(1-2x^4)^n}dx$$

Show that $$I_n=\frac{(-1)^n}{4n +2} + \frac{2n}{2n+1}I_{n-1}$$

I have started out by splitting the power
$$\int^1_0{x(1-2x^4)^n}dx=\int^1_0{x(1-2x^4)(1-2x^4)^{n-1}}dx$$
$$\int^1_0{x(1-2x^4)^{n-1}}dx + \int^1_0{-2x^4(1-2x^4)^{n-1}}dx$$
$$I_n=I_{n-1} + \int^1_0{-2x^5(1-2x^4)^{n-1}}dx$$

Then setting up the second integral for integration by parts

$$ I_n=I_{n-1} + \frac{1}{2}\int^1_0{x^2*-4x^3(1-2x^4)^{n-1}}dx $$

Differentiating $x^2$ and integrating $-4x^3(1-2x^4)^{n-1}$

$$I_n=I_{n-1} + \frac{1}{2}\left(\left[\frac{x^2(1-2x^4)}{n}\right]^1_0 -\frac{2}{n}\int^1_0{x(1-2x^4)^n}dx\right)$$
Evaluating $\left[\frac{x^2(1-2x^4)}{n}\right]^1_0$
$$I_n=I_{n-1} + \frac{1}{2}\left(\frac{(-1)^n}{n} -\frac{2}{n}I_n \right)$$
Collecting like terms
$$I_n=I_{n-1} +\frac{(-1)^n}{2n} -\frac{1}{n}I_n$$

$$I_n=\frac{1}{1+\frac{1}{n}}I_{n-1} + \frac{(-1)^n}{2n(1+\frac{1}{n})}$$
Simplifying the fractions

$$I_n=\frac{n}{1+n}I_{n-1} + \frac{(-1)^n}{2n+2}$$

However this is not equivelant to the expression in the question, have I messed up somewhere or missed something out?

答:
Let $J = \int_{0}^{1}x^{5} (1 - 2x^{4})^{n-1} dx$.

Integration by parts gives: $$I_n=\int^1_0{x(1-2x^4)^n}dx = \frac{(-1)^{n}}{2} + 4nJ$$.

Separating the integral as you have done gives:

$$I_n=\int^1_0{x(1-2x^4)^n}dx =\int^1_0{x(1-2x^4)(1-2x^4)^{n-1}}dx=I_{n-1} -2 J$$

Solving for $J$ in the first and substitute it into the second gives the desired reduction formula for $I_{n}$.


问:
Can Trigonometric Substitution be used to derive the anti-derivative $\int \frac{1}{x^2 -a^2}dx = \frac{1}{2a}\ln|\frac{x-a}{x+a}| + C$?
My textbook (Calculus by James Stewart 8th edition) uses partial fraction decomposition to derive the anti-derivative

$$\int \frac{1}{x^2 -a^2}dx = \frac{1}{2a}\ln|\frac{x-a}{x+a}| + C$$

But I'm wondering if it's possible to use trigonometric substitution as well.

I tried the following trig. substitution by letting
$$x = a\sec(\theta)$$
$$dx = a\sec(\theta)\tan(\theta)d\theta$$

$$\int \frac{1}{x^2 -a^2}dx = \int \frac{1}{a^2\sec^2(\theta)-a^2} a\sec(\theta)\tan(\theta)d\theta$$

$$= \int \frac{1}{a^2\tan^2(\theta)} a\sec(\theta)\tan(\theta)d\theta$$

$$= \frac{1}{a}\int \frac{1}{\tan(\theta)} \sec(\theta)d\theta$$

$$= \frac{1}{a}\int \frac{1}{\frac{\sin(\theta)}{\cos(\theta)}} \frac{1}{\cos(\theta)}d\theta$$

$$= \frac{1}{a}\int \frac{1}{\sin(\theta)}d\theta $$

$$= \frac{1}{a}\int \csc(\theta) $$

$$= \frac{1}{a} \ln |\csc(\theta)-\cot(\theta)|+ C$$

Then using the relation established by the trig. substitution cosecant and cotangent can be written in terms of $x$.

$$\csc(\theta) = \frac{x}{\sqrt{x^2-a^2}}$$

$$\cot(\theta) = \frac{a}{\sqrt{x^2-a^2}}$$

Now plugging these in results in

$$= \frac{1}{a} \ln|\frac{x-a}{\sqrt{x^2-a^2}}| + C$$

I'm not sure how to proceed from here to get the same results as the book, or if I did something wrong along the way. Any help would be appreciated.

Body: Use $\frac{x-a}{x+a}=\frac{1-\cos\theta}{1+\cos\theta}=\frac{(1-\cos\theta)^2}{\sin^2\theta}=(\csc\theta-\cot\theta)^2$ or $\left|\frac{x-a}{\sqrt{x^2-a^2}}\right|=\sqrt{\frac{x-a}{x+a}}$.


问:
How would we proceed to determine the chromatic polynomial and the chromatic number?
Question:

Being provided with the graph below:

Given Graph

How should we go about finding its the chromatic polynomial and chromatic number? How would its minimal proper coloring look like?

I am thinking that we should be making use of p(G, λ )= p(Ge, λ) - p(Ge', λ) here.

If anyone could kindly assist me on how to go about solving it, and providing me with a set of detailed steps required to solve it, it would be more appreciated.

Any help appreciated.

Thanks.

答: You could easily do it without using that equation (usually written like $p(G,\lambda)=p(G-e,\lambda)-p(G/e,\lambda)$), using known properties of chromatic polynomials. Can you regularly color it with 0, 1, 2 colors? Probably not, so it's $p(G,\lambda)=\lambda(\lambda-1)(\lambda-2)(\lambda^2+a\lambda+b)$ (5 nodes meaning degree 5, leading coefficient 1). Moreover, it must start like $\lambda^5-8\lambda^4+\ldots$ (8 edges), what can you conclude concerning the value of $a$? Have you tried to color it with 3 colors? Then, you must have found there are 6 ways to do that, and $p(G,3)=6$ gives you the value of $b$.


问:
Expressing a point in two coordinate systems
Let $(O,e_1,e_2,e_3)$ and $(O',e_1',e_2',e_3')$ be two coordinate systems. Let $\overline{OO'}=2e_1-e_2+3e_3$, $e'_1=e_1-e_2+3e_3$, $e'_2=e_1+e_2+e_3$ and $e'_3=e_1-e_2-e_3$.

a) Find the coordinates of a point A in $(O',e_1',e_2',e_3')$ if the coordinates
of A in $(O,e_1,e_2,e_3)$ are $(2,3,4)$.

b) Find the coordinates of a point A in $(O,e_1,e_2,e_3)$ if the coordinates
of A in $(O',e_1',e_2',e_3')$ are $(2,3,4)$.

Where should I use the information about $\overline{OO'}$? Isn't the answer of a) $(11,9, -5)$?

答:
You have equations expressing $e_i$'s in terms of $e_i$, you need inverse equations expressing $e_i$ in terms of $e_i$'s, so express your equations as a matrix
$$\left(
\begin{array}{c}
e_1'\\
e_2'\\
e_3'
\end{array}\right)=\left(
\begin{array}{ccc}
1 &-1&3\\
1&1&1\\
1&-1&-1
\end{array}
\right)\left(
\begin{array}{c}
e_1\\
e_2\\
e_3
\end{array}
\right)$$
Taking the inverse matrix you get
$$\left(
\begin{array}{c}
e_1\\
e_2\\
e_3
\end{array}\right)=\frac{1}{4}\left(
\begin{array}{ccc}
0 &2&2\\
-1&2&-1\\
1&0&-1
\end{array}
\right)\left(
\begin{array}{c}
e_1'\\
e_2'\\
e_3'
\end{array}
\right)$$
which gives you expressions of $e_i$'s in terms of $e_i'$'s.

Now, part a): you have that the coordinates
of A in $(O,e_1,e_2,e_3)$ are $(2,3,4)$, hence
$$A=O+2e_1+3e_2+4e_3$$

$$=O'+\overline{O'O}+2e_1+3e_2+4e_3$$
$$=O'-\overline{OO'}++2e_1+3e_2+4e_3$$
$$=O'-(2e_1-e_2+e_3)+2e_1+3e_2+4e_3$$
$$=O'+4e_2+3e_3$$
Now replace $e_2$ and $e_3$ with their expressions in terms of $e_2'$ and $e_3'$ giving
$$A=O'-\frac{3}{4}e_1'+2e_2'-\frac{5}{4}e_3'$$
so $A$ has coordinates $(-\frac{3}{4},2,-\frac{5}{4})$ in the coordinate system $(O',e_1',e_2',e_3')$.

Similarly for part b).


Title: How can I prove this reduction formula for $\int^1_0{x(1-2x^4)^n}dx$

Body:
The exercise in my textbook states

You are given that $$I_n=\int^1_0{x(1-2x^4)^n}dx$$

Show that $$I_n=\frac{(-1)^n}{4n +2} + \frac{2n}{2n+1}I_{n-1}$$

I have started out by splitting the power
$$\int^1_0{x(1-2x^4)^n}dx=\int^1_0{x(1-2x^4)(1-2x^4)^{n-1}}dx$$
$$\int^1_0{x(1-2x^4)^{n-1}}dx + \int^1_0{-2x^4(1-2x^4)^{n-1}}dx$$
$$I_n=I_{n-1} + \int^1_0{-2x^5(1-2x^4)^{n-1}}dx$$

Then setting up the second integral for integration by parts

$$ I_n=I_{n-1} + \frac{1}{2}\int^1_0{x^2*-4x^3(1-2x^4)^{n-1}}dx $$

Differentiating $x^2$ and integrating $-4x^3(1-2x^4)^{n-1}$

$$I_n=I_{n-1} + \frac{1}{2}\left(\left[\frac{x^2(1-2x^4)}{n}\right]^1_0 -\frac{2}{n}\int^1_0{x(1-2x^4)^n}dx\right)$$
Evaluating $\left[\frac{x^2(1-2x^4)}{n}\right]^1_0$
$$I_n=I_{n-1} + \frac{1}{2}\left(\frac{(-1)^n}{n} -\frac{2}{n}I_n \right)$$
Collecting like terms
$$I_n=I_{n-1} +\frac{(-1)^n}{2n} -\frac{1}{n}I_n$$

$$I_n=\frac{1}{1+\frac{1}{n}}I_{n-1} + \frac{(-1)^n}{2n(1+\frac{1}{n})}$$
Simplifying the fractions

$$I_n=\frac{n}{1+n}I_{n-1} + \frac{(-1)^n}{2n+2}$$

However this is not equivelant to the expression in the question, have I messed up somewhere or missed something out?

Body:
$$I_n=\int_0^1x(1-2x^4)^ndx$$
$u=x^2\Rightarrow dx=\frac{du}{2x}$ and so:
$$I_n=\frac12\int_0^1(1-2u^2)^ndu$$

$$2I_n=\int_0^1(1+\sqrt{2}u)^n(1-\sqrt{2}u)^ndu$$
If we do IBP:
$$a'=(1+\sqrt{2}u)^n\Rightarrow a=\frac{1}{\sqrt{2}(n+1)}(1+\sqrt{2}u)^{n+1}$$
$$b=(1-\sqrt{2}u)^n\Rightarrow b'=-\sqrt{2}n(1-\sqrt{2}u)^{n-1}$$
so:
$$2I_n=\frac{1}{\sqrt{2}(n+1)}\left[(1-2u^2)^n(1+\sqrt{2}u)\right]_0^1+\frac{n}{n+1}\int_0^1(1-2u^2)^{n-1}(1+\sqrt{2}u)^2du$$
$$2I_n=\frac{(-1)^n(1+\sqrt{2})-1}{\sqrt{2}(n+1)}+\frac{n}{n+1}\left(I_{n-1}+2\sqrt{2}\int_0^1u(1-2n^2)^{n-1}du+2\int_0^1u^2(1-2u^2)^{n-1}du\right)$$


问:
If the line at infinity is a secant line of a conic then the conic is a hyperbola?
I'm learning perspective geometry and I have a question about the line at infinity. Is it true that if the conic has the line at infinity as its secant then the conic represents a hyperbola in Euclidean geometry?

If it is, how can I imagine the shape hyperbola using the line at infinity? I think I can lift up the line to infinity but it seems like a parabola than a hyperbola.

I draw a conic going through $A, B, C, D, E$, and I choose AB as the line at infinity. Then I lift it up and it seems like a parabola

P.S: I'm quite new to this field so I may use the wrong jargon. Thank you

答:
Suppose you have an infinite horizontal table. Some distance above the table you have a point light. You're holding a metal ring and observing it's shadow on the table.

If the ring is fully below the horizontal plane through the light, you get an ellipse. That horizontal plane is the 3d counterpart to the line at infinity, and you're not intersecting that.

If the ring touches the horizontal plane in a single point, you get a parabola which intersects the line at infinity in a single point (of algebraic multiplicity two) so the line at infinity is a tangent.

If the ring is partially below and partially above the horizontal plane is the light, only the part below casts a shadow on the table. You get one component of a hyperbola, with the two points where the ring intersects the plane of the light as points at infinity, defining the asymptotic directions of the hyperbola. The comparison to projective geometry is suffering a bit here because the physical model uses rays where the mathematical setup uses lines. To better model that you'd take a pair of rings with the light as a center of point reflection. That way you get both arms of the hyperbola.

This explanation may not be clearer than the other answers when you just read it, but it should be possible to actually try this out with a real lamp, (finite) table and wire hoop. So perhaps that experiment can help your intuition.


问:
What is the definition of "local equation(s)" for a subscheme?
Hartshorne mentions "local equations" a few times without (so far as I can tell) actually defining them anywhere.  As best as I can guess, the definition would be something like this:

If $Y \subseteq X$ is a closed subscheme, then "local equations for $Y$" consist of an open affine set $U \subseteq X$ and a finite set of generators $f_1, \ldots, f_n \in \mathcal{O}_X(U)$ of the ideal sheaf $\mathscr{I}_Y(U)$ considered as an $\mathcal{O}_X(U)$-module.

(Here I've attempted to adapt the definition of local equations for a subvariety given in Shafarevich.)  Is this the accepted definition?  Or should the assumption that $U$ is affine or that $Y$ is a closed subscheme be weakened?  Or is there something else wrong with it?

What should be done if $X$ is non-noetherian?  Might a closed subscheme simply not have local equations in that case?

Or is "local equation" defined somewhere in Hartshorne?

答:
Let $X$ be an arbitrary scheme and $i : Y \hookrightarrow X$ be a closed immersion corresponding to the quasi-coherent ideal $I \subseteq \mathcal{O}_X$ . If $U \subseteq X$ is an open subset, the "local equations of $i$ on $U$" actually mean the ideal $I|_U \subseteq \mathcal{O}_U$ (in particular they exist). Often it also means any set of its global generators, i.e. a family of elements $(f_s)_{s \in S}$ of $\Gamma(U,I)$ such that $\oplus_{s \in S} \mathcal{O}_U \to I$ is an epimorphism. If $U$ is affine, this means that $\Gamma(U,I)$ is generated by the $f_s$. If $X$ is noetherian, one can choose $S$ to be finite.

When $X$ is projective, we often use another definition. Let's say $X=\mathrm{Proj}(S)$ with a nice graded algebra $S$. Then closed subschemes $Y$ of $X$ come from graded ideals $I$ of $S$ via $Y=\mathrm{Proj}(S/I)$. Then generators of $I$ are also called the (homogeneous) equations for $Y \hookrightarrow X$. We have the obvious notion of local (homogeneous) equation when $S$ is a sheaf of graded algebras on a base scheme.


问:
n digit number made up of 1s and 2s. Divisibility by $2^n$
If an n-digit number is made up of $1$s and $2s$ (for $n\geq2$), show that there is at least one such number that is divisible by $2^n$. Moreover, is there a general way to represent such a number that is divisible by $2^n$?

Although the solution can be solved with induction, I am trying to find if there is a general way to represent the n digit number that is divisible by $2^n$, or if some sort of algorithm could be generated whereby the sequence of $1$s and $2$s in the number is always divisible by $2^n$

答:
You can start trying to find values for $n=2,3,4$. You can find easily $12,112,2112$. Checking OEIS with these values you find the sequence A053312 and the formula for an $n$-digit number ($n \ge 1$) divisible by $2^n$:

$$a(n) = a(n-1)+10^{n-1}\left[2-\left(\frac{a(n-1)}{2^{n-1}} \mod 2\right)\right]$$

with $a(1)=2$.

If $a(n-1)$ is divisible by $2^n$ then the formula becomes:

$$a(n) = a(n-1) + 10^{n-1} \cdot 2 = 2^n\cdot k + 2^n \cdot 5^{n-1}=2^n(k+5^{n-1})$$

If $a(n-1)$ is not divisible by $2^n$ then the formula becomes:

$$a(n) = a(n-1)+10^{n-1} = 2^{n-1}(2k+1)+2^{n-1}5^{n-1}=2^{n-1}(2k+1+5^{n-1})=2^n\cdot m$$


问:
Is this a mistake $\int_{0}^{1}[(1-x^r)^{n/r}-nx^{n-1}(1-x^r)^{1/r}] \mathrm dx=0?$
Can anyone show that,
$$\int_{0}^{1}[(1-x^r)^{n/r}-nx^{n-1}(1-x^r)^{1/r}] \mathrm dx=0\tag1$$
Suppose $\ n\ $ and $\ r\ $ are integers.

I can't understand why it should be zero.

making a basic substitution:

$u=1-x^r$

$\mathrm dx=\frac{\mathrm du}{r(1-u)}$

$$\int_{0}^{1}\frac{u^{n/r}-n\cdot u^{1/r}(1-u)^{(n-1)/r}}{r(1-u)}\mathrm du$$

答:
$$I_1=\int_0^1(1-x^r)^{n/r}$$
$u=x^r\Rightarrow du=rx^{r-1}dx\therefore dx=\frac{du}{rx^{r-1}}=\frac{du}{ru^{(r-1)/r}}$ and so:
$$I_1=\frac1r\int_0^1u^{(1-r)/r}(1-u)^{n/r}du$$

$$I_2=n\int_0^1x^{n-1}(1-x^r)^{1/r}dx$$
$$I_2=\frac nr\int_0^1u^{(n-1)/r}u^{(1-r)/r}(1-u)^{1/r}du$$
$$I_2=\frac nr\int_0^1u^{(n-r)/r}(1-u)^{1/r}du$$
both can be expressed in terms of the beta function (and then the gamma function) and you have $I=I_1-I_2$


问:
Simplifying differentiation with a constant
I have a quick question. Take this:

d(C-x)/dt

Where C is a constant and x is a variable. Is there a way in which I can simplify that expression to get out of the bracket the constant?

答:
I suppose that $x$ is a function of $t$, that is $x=x(t)$.
Since the derivative is linear, you have that
$$\frac{\text{d}}{\text{d}t}(C-x(t))=\frac{\text{d}}{\text{d}t}(C)-\frac{\text{d}}{\text{d}t}x(t)=0-x^\prime(t)=-x^\prime(t)$$
If $x$ is not a function of $t$, then both $C$ and $x$ are constants with respect of $t$ and so the derivative is $0$.


问:
Finite group having a $p$-subgroup has nontrivial center
Let $G$ be a finite group, $H\le G$ a $p$-subgroup. I want to show that $Z(G)$ is non-trivial. I let $G$ act on $H$ by conjugation - or vice versa - but I every time it ended up with $$Z(G)\subseteq Z(H).$$
To add, by the class equation I can only say that $Z(H) \ge p$. How can I show that $Z(G)$ non-trivial? An appropriate action would be enough.

答: This is false. Consider $G = S_3$ and $H = \{(1),(123),(132)\}$. Then, $H \le G$ and $H$ is a 3-subgroup of $G$ but $Z(S_3) = \{(1)\}$.


问:
Asymptotics of a function involving factorials
I want to understand the asymptotics of a function which depends on a parameter $N$ which is a large integer.

The function is of the form (variable $x$ is of order 1)
$$ F_N(x)=\sum_{n=0}^\infty f_{N,n}x^n.$$

I suspect that the coefficient $f_{N,N}$ dominates the asymptotics, so that $F_N(x)\approx f_{N,N}x^N$, in some sense. But I would like to actually prove this, and estimate the order of the error.

The coefficients are given by
$$ f_{N,n}=\sum_{k=0}^{{\rm min}(n,N)-1}\frac{k!(n-1-k)!(N-1-k)!}{(N+n-1-k)!}.$$

What I have done: I can see that if $n$ is small, then $f_{N,n}\sim N^{-n+1}$. On the other hand, if $n=N$ then the contribution to the sum from $k=N-1-m$ is of order $N^{-2m}$, so the sum is dominated by $m=0$ and we get a value of order 1.

Using the Stirling approximation blindly, without considering whether each factorial is large or not, I get an approximation for the summand which is
$$ \frac{k^k(n-k-1)^{n-k-1/2}(1-\frac{k+1}{N})^{N-k-1/2}}{N^n(1+\frac{n-k-1}{N})^{N+n-k-1/2}}$$
which could be useful. But this approximation is not valid in particular for the case $k=n-1=N-1$ which I think is the most important one.

How do I carry out a rigorous asymptotic analysis here?

答:
Just a hint as a possible way to go, through the Beta Function.
$$
\eqalign{
  & f_{\,N,\,n}  = \sum\limits_{k = 0}^{\min (N,n) - 1}
 {{{k!\left( {n - 1 - k} \right)!\left( {N - 1 - k} \right)!} \over {\left( {N + n - 1 - k} \right)!}}}  =   \cr
  &  = n\sum\limits_{k = 0}^{\min (N,n) - 1}
 {{{\left( {k + 1 - 1} \right)!\left( {n - k - 1} \right)!\left( {n - 1} \right)!\left( {N - k - 1} \right)!}
 \over {\left( {n + 1 - 1} \right)!\left( {N + n - k - 1} \right)!}}}  =   \cr
  &  = n\sum\limits_{k = 0}^{\min (N,n) - 1}
 {{\rm B}\left( {k + 1,n - k} \right){\rm B}\left( {n,N - k} \right)}  \cr}
$$
Now, the integral representation of Beta looks to get clumsy.

On the other hand, Beta is rapidly decaying at growing of either of its arguments, so
it might be possible to try by truncating the sum.

Another way is that $F_{\, N}(x)$ actually reads
$$
\eqalign{
  & F_{\,N} (x) = \sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^{\min (N,n) - 1}
 {{{k!\left( {n - 1 - k} \right)!\left( {N - 1 - k} \right)!}
 \over {\left( {N + n - 1 - k} \right)!}}} \;x^{\,n} }  =   \cr
  &  = \sum\limits_n {\sum\limits_k
 {{{\left[ {0 \le k < n} \right]\left[ {0 \le k < N} \right]k!\left( {n - 1 - k} \right)!
\left( {N - 1 - k} \right)!} \over {\left( {N + n - 1 - k} \right)!}}x^{\,n} } }  =   \cr
  &  = \sum\limits_k {\left[ {0 \le k < N} \right]k!\left( {N - 1 - k} \right)!
\sum\limits_n {{{\left[ {k < n} \right]\left( {n - 1 - k} \right)!}
 \over {\left( {N + n - 1 - k} \right)!}}x^{\,n} } }  =   \cr
  &  = \sum\limits_{0 \le k < N} {k!\left( {N - 1 - k} \right)!
\sum\limits_{k < n} {{{1^{\,\overline {\,n - 1 - k\,} } }
 \over {1^{\,\overline {\,N + n - 1 - k\,} } }}x^{\,n} } }  =   \cr
  &  = \sum\limits_{k = 0}^{N - 1} {k!\left( {N - 1 - k} \right)!
\sum\limits_{k + 1 \le n} {{{1^{\,\overline {\,n - 1 - k\,} } }
 \over {1^{\,\overline {\,n - 1 - k\,} } \left( {n - k} \right)^{\,\overline {\,N\,} } }}x^{\,n} } }  =   \cr
  &  = \sum\limits_{k = 0}^{N - 1} {k!\left( {N - 1 - k} \right)!
\sum\limits_{k + 1 \le n} {{{x^{\,n} } \over {\left( {n - k} \right)^{\,\overline {\,N\,} } }}} }  =   \cr
  &  = \left( {N - 1} \right)!\left( {\sum\limits_{k = 0}^{N - 1} {{{x^{\,k} }
 \over {\left( \matrix{  N - 1 \cr   k \cr}  \right)}}} } \right)
\left( {\sum\limits_{1 \le j} {{{x^{\,j} } \over {j^{\,\overline {\,N\,} } }}} } \right) \cr}
$$
where:

$[P]$ denotes the Iverson bracket;

$ x^{\,\overline {\,k\,} } $ represents the  Rising Factorial.

I am sure you can continue by yourself.


问:
Chain Rule in Multiple Variables
Say I have a function
$$z(s,t,x,y)=f(s(x(t,s),y(t,s)),t(x(t,s),y(t,s)))$$

How would I find $$\frac{\partial z}{\partial x}$$

The chain rule here is confusing me.

答:
The notation $\frac{\partial}{\partial x}$ can be ambiguous. Or it means
$$\frac{\partial z}{\partial x}(s,t,x,y):=\lim_{h\to 0}\frac{z(s,t,x+h,y)-z(s,t,x,y)}{h}$$
provided that the limit exists, that is it is the derivative in the $3$-rd variable (and I would prefer the notation $\partial_3$); or it means
$$\frac{\partial z}{\partial x}(s,t,x,y):=\lim_{h\to0}\frac{z(s(x+h),t(x+h),x+h,y(x+h))-z(s(x),t(x),x,y(x))}{h}$$
provided that the limit exists, that is you compute the total derivative of $z$ in the direction $x$ when $s,t$ and $y$ depends on $x$.

In general, $\frac{\partial}{\partial x}$ denotes the partial derivative in the $x$ coordinate (here in the $3$-rd variable), the last situation above being written as $\frac{\mathrm{d} z}{\mathrm{d} x}$ i.e. the coefficient in front the coordinate linear form $\mathrm{d} x$ in the expression of the differential $\mathrm{d} z$.

Title: Chain Rule in Multiple Variables


问:
Does $a \circ b = a$ imply $b$ is the identity morphism?
Let $a$ and $b$ be two morphisms in category $C$ satisfying the following equation:
$a \circ b = a$.

Does it follow that $b$ is the identity morphism?

Obviously, it it true when $a$ is mono. But I doubt it is true in general and can't find a counterexample.

答:
As a simple class of counterexamples, there are plenty of non-identity functions $f:X\to X$ with $f\circ f=f$. Such as
$$
f:\Bbb R\to\Bbb R\\
f(x)=0
$$
But yes, for mono $a$ it does follow, as we are given $a\circ b=a\circ\operatorname{Id}$, and this by definition of mono gives $b=\operatorname{Id}$.


问:
Can anyone help me with this modular manipulation in system of equations?
So i have to solve this system but i got stuck while computing and I need help.

```<code>m = 928377461;
x0 = 21380413;
x1 = 32564732;
x2 = 803330610;
a*x0+c === x1 mod m
a*x1+c === x2 mod m
</code>```

I know the x's and m but I can't figure out reasonable equation to get a and c, because they have to be integers.

I considered:

```<code>a = ((x1 - (x1 * x1) - (x2 * x0)) / (x1 - x0)) mod m
c = (((x1 * x1) - (x2 * x0)) / (x1 - x0)) mod m
</code>```

But they are not giving correct result (It's taken from linear congruential generator).

I would be really thankful if anyone could give me the correct formulas, cause i don't have a clue how am i supposed to solve it and everywhere i searched this part was described as a basic arithmetics, which seems to be too hard for me.

答:
Hint:
From the two congruence equations, you can eliminate $c$ to obtain
$$a(x_1-x_0)\equiv x_2-x_1\pmod m.$$
Next, supposing $x_1-x_0$ is coprime to $m$, it has a modular inverse that you can find with the extended Euclidean algorithm. Multiplying both sides by this modular inverse will give you the congruence class of $a$. Then it is easy to deduce the congruence class of $c$.

Here is how it starts:

$x_1-x_0=11\,184\,319$
\begin{array}{rrrl}
r_i\qquad&u_i&v_i&q_i\\\hline
m=928\,377\,461&0&1\\
x_1-x_0=11\,184\,319&1&0&83\\\hline
78\,984&-83&1&1 \\
47\,575 & 84&-1&1\\
31\,409&-167&2&1\\
16\,166&251&-3&1\\
\vdots\quad&\vdots\enspace&\vdots\,&\vdots\end{array}


问:
Algebra with derivatives operators
Suppose that $y=ax$. This is an often used "calculation", especially in physics:
$$
\frac{d^2}{dx^2}=\frac{d}{dx}\frac{d}{dx}=\left(\frac{dy}{dx}\frac{d}{dy}  \right)\left(\frac{dy}{dx}\frac{d}{dy}  \right)= \left(a\frac{d}{dy} a\frac{d}{dy}  \right)=a^2 \frac{d^2}{dy^2}
$$
But $\frac{d}{dx}$ is only a symbol, not a fraction. So, how we can justify this result and in what conditions is it true ?

答:
Consider a function $f:U\to\mathbb{R}$ for some open subset $U\subseteq\mathbb{R}$. If $f$ is differentiable, we denote its derivative simply as $\frac{df}{dx}$. Now, suppose we had some auxiliary function $y(x)=ax$. If we compose $f$ with $y$, we get a new differentiable function $g:V\to\mathbb{R}$ (where $V$ depends on $U$ and $a$) where $g=f\circ y$. In particular, we write

$$g(x)=f(y(x))$$

To take the derivative of this function, we use the chain rule and get

$$g'(x)=f'(y(x))y'(x)$$

If we think of $f$ as taking some independent argument $y$, the term $f'(y(x))$ is instead lazily written as $\frac{df}{dy}$. Then we have

$$\frac{dg}{dx}=\frac{df}{dy}\frac{dy}{dx}=a\frac{df}{dy}$$

If we further apply a second derivative just to drive the point home, we get

$$\frac{d^2g}{dx^2}=a^2\frac{d^2f}{dy^2}\implies\frac{d^2}{dx^2}(f\circ y)=a^2\frac{d^2}{dy^2}(f)$$

The connection between these symbols as we have presented implies that they cannot be exactly the same, right? Each operation is performed on a different function, either $f$ or $f\circ y$. Can these be used interchangeably?

Basically, these operators are not identical, which is clear based on the fact that they operate on different functions with different domains. In some abstract sense, however, in physics (where the domains are usually all of $\mathbb{R}^n$ or some convenient subset) the transformation of $y$ is often a valid way of changing a view of the space. For example, transformations like these often appear when nondimensionalizing equations, which sort of just stretch or squash the domain in question. This equation basically says "if we change the way we look at the domain, then we can view $f$ either as a function of its original variable or as a function of some new variable I have prescribed."

Here's a silly example. Suppose I have a function $f:\mathbb{R}\to\mathbb{R}$ with $f(y)=y^2$. We then define a function $y(x)=2x$, so $f(y(x))=4x^2$. If I write the symbol $\frac{df}{dx}$, I might be implying that I actually want the derivative of $f\circ y$, since I used the symbol $x$ in the composition, so maybe it's correct to say $\frac{df}{dx}=8x$ in a certain context. However, if I asked you what $\frac{df}{dy}$ is, it should be clear that $\frac{df}{dy}=2y$, but the "symbols" of $\frac{df}{dx}$ and $\frac{df}{dx}$ only mean "the derivative of $f$", so why would they be different? They're different because $\frac{df}{dx}=8x$ is an abuse of notation and conceptualization. It's actually a garbage statement, but it gets the point across as to what I am looking for.

Hopefully this helps! This is sort of an annoying foray into how physicists abuse math however they want, but it often works, so they keep doing it...

Title: Algebra with derivatives operators


Title: Can anyone help me with this modular manipulation in system of equations?

Body:
So i have to solve this system but i got stuck while computing and I need help.

```<code>m = 928377461;
x0 = 21380413;
x1 = 32564732;
x2 = 803330610;
a*x0+c === x1 mod m
a*x1+c === x2 mod m
</code>```

I know the x's and m but I can't figure out reasonable equation to get a and c, because they have to be integers.

I considered:

```<code>a = ((x1 - (x1 * x1) - (x2 * x0)) / (x1 - x0)) mod m
c = (((x1 * x1) - (x2 * x0)) / (x1 - x0)) mod m
</code>```

But they are not giving correct result (It's taken from linear congruential generator).

I would be really thankful if anyone could give me the correct formulas, cause i don't have a clue how am i supposed to solve it and everywhere i searched this part was described as a basic arithmetics, which seems to be too hard for me.

Body:
Ooookay...

$x_2 - x_1 \equiv ax_1 - ax_0 \equiv a(x_1-x_0)\pmod m$.

If $\gcd(x_1-x_0, m) =1$ then $x_1-x_0$ is invertible. So

$a \equiv (x_2 - x_1)(x_1-x_0)^{-1} \pmod m$

And the $c \equiv x_1-ax_0 \pmod m$.

So...

$x_1-x_0 = 11184319$ now use Euclid's Algorithm to find $\gcd(x_1-x_0, m)$ and solve $(x_1-x_0)^{-1}$.

$m = 928377461$

$928377461= 83*11184319 + 78984$

$11184319 = 141*78984+ 47575$ etc....


答: Let $C$ be any category with a terminal object $x$. Let $y$ be an object in $C$, $a\colon y\rightarrow x$ be the unique morphism from $y$ to $x$ and then for any $b\colon y\rightarrow y$ we will have $a\circ b=a$, because both are morphisms $y\rightarrow x$ and there only is one such morphism. But of course $y$ can have many non-identity endomorphisms $b$ (for an explicit example of this occurring, take $C=\mathbf{Set}$, then any singleton $\{\ast\}$ is a terminal object and any set with at least two elements has a non-identity endomorphism, e.g. by switching two elements).

Title: Does $a \circ b = a$ imply $b$ is the identity morphism?

Body:
Let $a$ and $b$ be two morphisms in category $C$ satisfying the following equation:
$a \circ b = a$.

Does it follow that $b$ is the identity morphism?

Obviously, it it true when $a$ is mono. But I doubt it is true in general and can't find a counterexample.


问:
Find the best constant in this complex analysis problem
I stumbled upon a problem that gives me trouble and is quite interesting but I can't do it.
Here it goes.

Let $(z_1, z_2, ... z_n)\in \mathbb{C^n}$, $J \subset$ {$1,2,..n$} for $\forall n \in \mathbb{N}$ and $ S_J := |\sum_{j \in J}z_j$|

Clearly $S_J\leq \sum_{j\in J}|z_j|\leq \sum_{j=1}^{n}|z_j|:=S$

For $n=2$, prove that there exists $J$, such that $S_J\geq aS$ and $a\in \mathbb{R}$. Prove that $a=\frac{1}{2}$ is the best constant.

For $n=3$, prove that there exists $J$, such that $S_J\geq bS$ and $b\in \mathbb{R}$. Prove that $b=\frac{1}{3}$ is the best constant.

What's the best constant if $n\geq 4$ ?

答:
You want to write $\left|\sum_{j \in J} z_j\right|$ as $S_J$, not $S_j$: $j$ is just a "dummy index".

For $n=2$, $S_{\{1\}} + S_{\{2\}} = |s_1| + |s_2| = S$ so $\max(S_{\{1\}}, S_{\{2\}}) \ge S/2$.  Similarly for $n=3$, $\max(S_{\{1\}}, S_{\{2\}}, S_{\{3\}}) \ge S/3$,
and in general $\max(S_{\{1\}}, \ldots, S_{\{n\}}) \ge S/n$.

To see that $a = 1/2$ is the best constant for $n=2$, you can take $z_1 = 1$ and $z_2 = -1$.  To see that $a=1/3$ is the best for $n=3$, you can take $z_1, z_2, z_3$ the three cube roots of $1$.

I don't know the best constants when $n > 3$.

EDIT: see this


问:
Inverse Laplace Transform of the following.
What is the inverse laplace transform of $ F(s) = \frac{1}{2}\ln \left ( \frac{s^{^{2}}+b^{2}}{s^{^{2}}+a^{2}} \right ) $ with $ a,b \ \epsilon \ \mathbb{R} $ ?

答:
$$F(s)=\mathcal{L}[f(t)](s)=\int_0^\infty f(t)e^{-st}dt$$
$$F'(s)=-\int_0^\infty tf(t)e^{-st}dt$$

You have:
$$F(s)=\frac12\ln\left(\frac{s^2+b^2}{s^2+a^2}\right)$$
$$F'(s)=\frac{s}{s^2+b^2}-\frac{s}{s^2+a^2}$$
now we know that:
$$\mathcal{L}^{-1}\left(\frac{s}{s^2+b^2}-\frac{s}{s^2+a^2}\right)=\cos(bt)-\cos(at)$$
and from what I showed above we can say that:
$$-tf(t)=\cos(bt)-\cos(at)$$
$$f(t)=\frac{\cos(at)-\cos(bt)}{t}$$


问:
Question Regarding the Standard Picture of $K_{0}(A)$ for a C$^{*}$-Algebra $A$
I am working through Rordam's book and I am stuck on exercise 4.4. Suppose we have a (not necessarily unital) C$^{*}$-algebra $A$. We are asked to prove that every element of $K_{0}(A)$ can be written as
$$
[p]_{0}-\left[\begin{pmatrix}1_{n} & 0_{n}\\ 0_{n} & 0_{n}\end{pmatrix}\right]_{0},
$$
for some projection $p\in M_{2n}(\widetilde{A})$ satisfying
$$
p-\begin{pmatrix}1_{n} & 0_{n}\\ 0_{n} & 0_{n}\end{pmatrix}\in M_{2n}(A).
$$

Following one of the proofs in the book (a quarter-way down page 64), I took an arbitrary $g\in K_{0}(A)$ and expressed it as $[e]_{0}-[f]_{0}$ for some projections $e,f\in M_{n}(\widetilde{A})$. Then, I put
$$
p=\begin{pmatrix}1_{n}-f & 0\\ 0 & e\end{pmatrix}\qquad\text{and}\qquad q=\begin{pmatrix}1_{n} & 0_{n}\\ 0_{n} & 0_{n}\end{pmatrix}.
$$
As in the proof in the book, it follows that $g=[p]_{0}-[q]_{0}$ and that
$$
[s(p)]_{0}=[q]_{0},
$$
where $s\colon\widetilde{A}\to\widetilde{A}$ is the map given by $s(a+\lambda\cdot1_{\widetilde{A}})=\lambda\cdot 1_{\widetilde{A}}$. So, we have $g=[p]_{0}-[s(p)]_{0}$, but to guarantee that  $p-\begin{pmatrix}1_{n} & 0_{n}\\ 0_{n} & 0_{n}\end{pmatrix}\in M_{2n}(A)$, I need $s(p)=q$, which obviously does not hold in general.

I've been thinking for hours how to modify this construction to produce the result, but I haven't been able to come up with anything else. Any help is really appreciated.

Thank you.

答:
I would like to give another answer $\underline{\text{which is essentially the same with the one above}}$, but phrased in a different manner and more fit to the approach of Rordam's book.

So let $x\in K_0(A)$ and write $x=[q]_0-[s(q)]_0$ for some projection $q\in P_n(\tilde{A})$. Mimicking the proof of the standard picture of $K_0$, we set $p'=q\oplus(1_n-s(q))$, which is a projection in $P_{2n}(\tilde{A})$ and note that $s(p')=s(q)\oplus(1_n-s(q))\sim_01_n\sim_01_n\oplus0_n$. It is very easy to verify that $x=[p']_0-[s(p')]_0$. Observe that $s(p')$ and $1_n\oplus0_n$ are projections of the C*-algebra $M_{2n}(\mathbb{C}\cdot1_{\tilde{A}})$ and they are Murray-von Neumann equivalent.

Now for any $k\geq1$ we have an obvious *-isomorphism $M_k(\mathbb{C})\cong M_k(\mathbb{C}1_{\tilde{A}})$. It is true that when we have two projections $r_1,r_2\in M_k(\mathbb{C})$ of the same rank (i.e. of the same trace) then they are similar, i.e. we have a unitary $u\in M_k(\mathbb{C})$ so that $ur_1u^*=r_2$. Passing this result through our *-isomorphism, since $s(p')$ and $1_n\oplus0_n$ are Murray-von Neumann equivalent (thus the corresponding projections through the isomorphism have the same trace, i.e. they are of the same rank), there exists a unitary $u\in M_{2n}(\mathbb{C}\cdot1_{\tilde{A}})$ so that $u\cdot s(p')\cdot u^*=1_n\oplus0_n$.

Note that $u$ is a scalar element, i.e. $u=s(u)$. We set $p=u\cdot p'\cdot u^*$. Now $p$ is a projection in $M_{2n}(\tilde{A})$, we have $s(p)=s(up'u^*)=s(u)s(p')s(u^*)=us(p')u^*=1_n\oplus0_n$ and it is obvious that $p\sim_0 p'$ and $s(p)\sim_0 s(p')$, so $[p']_0-[s(p')]_0=[p]_0-[s(p)]_0$, as we wanted.


问:
example of a flat but not faithfully flat ring extension
I am learning commutative algebra and there is a definition about faithfully flat modules or ring extensions. I can't think of an example of a flat but not faithfully flat ring extension or module. Can someone help me with it? thanks</p>

答: $0$ is always flat, but not faithfully flat (unless the base is $0$).</p>


问:
on a limit of a composition of functions
If $\lim_{x\rightarrow o^{-}} f(x)=a$, $\lim_{x\rightarrow o^{+}} f(x)=b$, and $f(0)=c$ with $a\neq b, a\neq c, b\neq c$, then, what can we say about $\lim_{x\rightarrow o^{-}} f(x^2-x)$ ? Does this limit necessarily exist ? If so, how can we show that by the $\epsilon-\delta$ argument ? Any answer will be appreciated.

Body:
hint

$$x^2-x=x(x-1)$$
So, when $ x\to 0^- $,

$x<0 $ and. $ x-1<0 $, thus
$$x^2-x \to 0^+$$
and
$$f(x^2-x)\to b$$
******** * $\epsilon-\delta \; $ proof *****

Given $ \epsilon>0 $,
$$\lim_{x\to 0^+}f(x)=b\implies$$
$$\exists \alpha>0\;:\; (0<x<\alpha \implies $$
$$|f(x)-b|<\epsilon)$$

and
$$\lim_{x\to 0^-}(x^2-x)=0^+\implies$$
$$\exists \delta>0 \;:\; (-\delta<x<0\implies $$
$$0<x^2-x<\alpha)$$

thus
$$\exists \delta>0 \;:\; (-\delta<x<0\implies$$
$$|f(x^2-x)-b|<\epsilon)$$

Which proves that $\lim_{x\to0^-}f(x^2-x)=b$.


问:
Proofing that det$\begin{pmatrix}A&B\\ C&D\end{pmatrix}$ = $\det(DA -BC)$ When B Invert And $BD=DB$
I have the matrices $A,B,C,D \in\mathcal{M}_{n\times n}(\mathbb{R})$
lets $X = \begin{pmatrix}A&B\\ C&D\end{pmatrix}$ and I know that $BD=DB$ and B is invertible.

How can I prove that $\det(X)=\det(DA -BC)$?

答:
Notice first that

$$X=\begin{bmatrix}A & B \\ C & D \end{bmatrix}=
\begin{bmatrix}0 & B \\ I & D \end{bmatrix}\begin{bmatrix}C - D B^{-1} A & 0\\
B^{-1} A& I \end{bmatrix}$$

Also

$$\det \begin{bmatrix}0 & B \\ I & D \end{bmatrix}=(-1)^n \det \begin{bmatrix}B& 0  \\  D & I \end{bmatrix} = (-1)^n \det(B) = \det(-B)$$

Then $$\det(X) = \det(-B) \det(C-DB^{-1}A)=\det(B (DB^{-1}A-C))=\det(DA-BC)$$

because $BD=DB$.


问:
Spivaks proof on the intermediate value theorem
I am a student trying to learn calculus and I've been working my way through spivaks calculus book and I am quite confused on his proof of the intermediate value theorem.

I understand how to use the intermediate value theorem but in his proof he states that if $f$ is continuous on $[a,b]$ and $f(a)<c<f(b)$ then there is some $x$ in $[a,b]$ such that $f(x)=c$. His proof says that let $g=f-c$, then $g$ is continuous and $g(a)<0<g(b)$, then by theorem 1 (which is a theorem that I understand) there is an $x$ in $[a,b]$ such that $g(x)=0$, which means $f(x)=c$.

I looked online everywhere for a similar version of this proof to no avail. My question is how can we say that $g$ is continuous? It does not seem like a rigorous proof?

答:
You can think of $ c $ as a constant function, so it’s still a function and also a “very continuous” function. $ f $ is continuous by hypothesis. The linear combination, so sum, difference and multiplication by a number, of continuous functions is still a continuous function. If you want to understand why this is true, you can directly prove it using limits and their properties: remember that a function $ f $ is continuous at $x_0$ if and only if $ \lim _{ x \to x_0} = f(x_0) $.
It comes almost automatically by the properties of limits.


问:
Center Manifold Theorem
Let $f:M \rightarrow M$ be a partially hyperbolic diffeomorphism of $M$ with the usual definition that at each $p$ tangent space splits to $Df$ invariant subspaces: $T_pM = E^s_p \oplus E^c_p \oplus E^u_p$ with the relevant constants on the norm of $Df$ restricted to these spaces. I just need one which is $\mu^n < |Df^n_{E^u}|$.

I have some confusions about how people define the center-stable manifolds etc. In the case of just partially hyperbolic fixed point $p$ in $R^n$, Robinson in his book defines the center stable manifold of the fixed point as $$ W^{cs}(p) = \{x | \lim_{n\to \infty}\frac{1}{\mu^n}d(f^n(x),p)=0\} $$ (after extending the map to whole of $\mathbb{R}^n$, if its not already defined everywhere) and says that although if not unique, such center manifolds exists and are tangent to $E^c$. In the case of partially hyperbolic diffeomorphism of manifolds, I have never seen this characterization. They basically define $W^{cs}(p)$ as a submanifold tangent to $E^{cs}$.

My question is for a partially hyperbolic set. Is it not necessary that a submanifold $N(p)$ passing through $p$ which satisfies
$$N(p) = \{x | \lim_{n\rightarrow \infty}\frac{1}{\mu^n}d(f^n(x),p)=0\} $$
will be tangent to $E^c$? I was able to show that the unstable manifold has to be transversal to this manifold but transversality is not enough to show that tangent space of this manifold has to be $E^c \oplus E^s$

Why is it not preferred to define central stable behaviour in this manner for this setting? Is there also a center stable manifold theorem for this setting in the manner which says, for all $p$ there is always a submanifold passing through $p$, which satisfies the convergence criteria I wrote above and is tangent to $E^s \oplus E^c$?

答:
The reason is simply that the "right" rates for a partially hyperbolic set may depend on $p$. Of course, since you tell us in your second paragraph that you only rates of a certain type, then it may be fine to use your second "definition".

A second aspect, more important, but also more technical, concerns the regularity of the manifolds and that depends on somewhat delicate relations between your rate $\mu$ and other measuring the top and bottom rates on each fiber.

Up to these points, the answer to your last question is yes.

Title: Center Manifold Theorem


答:
$f$ is a continuous on $[a,b]$.  $c$ is a constant.

Lemma: for any continuous function $f$  on $[a,b]$ and constant $c$, the function $g(x) =f(x) -c$ is continuous on $[a,b]$.

Pf:  Let $d \in [a,b]$.  $f$ is continuous at $x=d$ so $\lim_{x\to d} f(x) = f(d)$.  That means that for any $\epsilon > 0$ there is a $\delta>0$ so that $|x-d| < \delta$ will imply $|f(x) - f(d)| < \epsilon$.

But that means $|g(x) - g(d)| = |(f(x) -c) -(f(d) -c)| = |f(x) -f(d)| < \epsilon$.

That means using the same $\delta$ and $\epsilon$ that $\lim_{x\to d} g(x) = g(d)$.

Which means $g$ is continuous at $x = d$ for all $d \in [a,b]$.

Title: Spivaks proof on the intermediate value theorem

Body:
I am a student trying to learn calculus and I've been working my way through spivaks calculus book and I am quite confused on his proof of the intermediate value theorem.

I understand how to use the intermediate value theorem but in his proof he states that if $f$ is continuous on $[a,b]$ and $f(a)<c<f(b)$ then there is some $x$ in $[a,b]$ such that $f(x)=c$. His proof says that let $g=f-c$, then $g$ is continuous and $g(a)<0<g(b)$, then by theorem 1 (which is a theorem that I understand) there is an $x$ in $[a,b]$ such that $g(x)=0$, which means $f(x)=c$.

I looked online everywhere for a similar version of this proof to no avail. My question is how can we say that $g$ is continuous? It does not seem like a rigorous proof?


问:
Double integral $\iint_A x dx dy$ bounds with algebra
Given $A=\{(x,y)\in\mathbb{R}^2\ |\ x^2+(y-1)^2 \leq 1, \ x+y \geq 2\}$, I must evaluate
$$\iint_A xdxdy$$
I used a change of coordinates, letting $x=r \cos t$ and $y=r \sin t+1$, so the set of integration became $B=\{(r,t)\in\mathbb{R}^2\ | \ 0 \leq r \leq 1, \ r(\cos t + \sin t) \geq 1\}$ and then I have to evaluate
$$\iint_B r^2 \cos t drdt$$
I have two main doubts:

sketching $A$ it is obvious that $0 \leq x \leq 1$ and $1 \leq y \leq 2$ and so $t \in [0,\pi/2]$, but how can I find out with algebra that $0 \leq x \leq 1$ and $1 \leq y \leq 2$?

My attempt is this: Since $x^2+(y-1)^2 \leq 1 \implies x^2 \leq 1 \ \land \ (y-1)^2 \leq 1 \iff -1 \leq x \leq 1 \ \land \ 0 \leq y \leq 2$ it follows that $y \geq 0$ and that $-2 \leq -y \leq 0$, hence from $x+y \geq 2 \iff x \geq 2-y \geq 2-2=0 \implies x \geq 0$.

So combining $-1 \leq x \leq 1 \ \land x \geq 0$ it follows that $0 \leq x \leq 1$.

Hence it is $-1 \leq -x \leq 0$ and then $x+y \geq 2 \iff y \geq 2-x \geq 2-1=1 \implies y \geq 1$; so combining $y \geq 1 \ \land \ y \leq 2$ it is $1 \leq y \leq 2$. Can someone check if it is correct?

I would like to deduce the angle and the fact that we are in the first quadrant from polar coordinates too: so in $B$ it is both $0 \leq r \leq 1$ and $r(\cos t + \sin t) \geq 1$, now I would like to divide for $\cos t + \sin t$ but I don't know its sign and so I don't know how to proceed; can someone help me figure this out?

In this case, if I assume $\cos t +\sin t >0$ I can divide and get $\frac{1}{\cos t +\sin t} \leq r \leq 1$, but sometimes there are two lower bounds for $r$ and I don't know how to proceed. So the questions is: if we imagine (for the sake of a better comprehension for me) that we don't know if the sign of $\frac{1}{\cos t +\sin t}$ is always positive, I can't deduce $0 \leq \frac{1}{\cos t +\sin t} \leq r \leq 1$ just because $r \geq 0$ and it is $\frac{1}{\cos t+ \sin t} \leq r$ (it would be different if it was $\frac{1}{\cos t+ \sin t} \geq r$, because $\frac{1}{\cos t+ \sin t} \geq r \geq 0 \implies \frac{1}{\cos t+ \sin t} \geq 0$), so I have to consider the values of $t$ that makes $\frac{1}{\cos t +\sin t}<0$ too? Or ignore them? Thanks.

答:
$A=\{(x,y)\in\mathbb{R}^2\ |\ x^2+(y-1)^2 \leq 1, \ x+y \geq 2\}$

$x + y \geq 2$ indicates your region is part of the circle above the line $x + y = 2$.

For your question $1$ on not going by the sketch, you can find the intersection points of the circle and the line.

$x^2+(y-1)^2 = 1$ and $x + y = 2 \implies y = 2 - x$

$x^2 + (1-x)^2 = 1 \implies 2x^2 - 2x = 0 \implies x = 0, 1$

Plugging in $y = 2 - x$, we get two intersection points $(1, 1), (0, 2)$.

Now on your question $2$, given the points you know $\theta = 0, \frac{\pi}{2}$ but if you want to plug into the polar equation and check -

$x = \cos \theta = 1, y = 1 + \sin \theta = 1 \implies \theta = 0^0 \,$ ($r = 1$ on the circle)

Similarly check for point $(0, 2)$.


问:
Prove that a continuous function on a compact subset of ${\mathbb{R} }^n$ is almost like a Lipschitz function.
Suppose I have $K \subset { \mathbb{R} }^n $ a compact subset,and a continuous function $f : K \to \mathbb{R} $.
Given some $\epsilon >0$ prove there exist some $L$ such that for all $x,y \in X$
I have $$|f(x)-f(y) | \leq L |x-y | + \epsilon $$

Now I was able to find some proofs for a similar problem where $K$ is convex but couldn't find a proof that doesn't use convexity.

One thing I considered is to approximate the derivative of $f$ like this $$Df \approx \frac{f(x)-f(y)}{x-y} - \epsilon $$ and somehow prove that the approximation is bounded but couldn't come up with a proof, maybe the mean value theorem may help.

How can I show this?

答:
Now given $\epsilon$, there is $\delta$ so that $|x-y|<\delta$ implies
$|f(x)-f(y)|<\epsilon$: such $\delta$ exists since $K$ is compact so $f$
is uniformly continuous; also note $f$ is bounded, so $|f|\leq M$ for
some $M$.
So we can let $L=\frac{2M}{\delta}$: if $|x-y|<\delta$, then $|f(x)-f(y)|<\epsilon\leq L|x-y|+\epsilon$; if $|x-y|\geq\delta$, then
$$
|f(x)-f(y)|\leq |f(x)|+|f(y)|\leq 2M=\frac{2M}{\delta}\delta=L\delta
\leq L|x-y|<L|x-y|+\epsilon.
$$

This estimate looks quite weak.


问:
Conversion: dec to bin
We have:
$$\begin{align}
A&=\frac{19}{32} \\[0.3em]
B&=\frac{21}{32}\\[0.3em]
C&=\frac{19}{64}\\[0.3em]
D&=\frac{21}{64}.\\[0.3em]
\end{align}$$
In binary it is:
$$\begin{align}
A&=0,10011\strut \\
B&=0,10101\strut\\
C&=0,010011\strut\\
D&=0,010101\strut\\
\end{align}$$
Is it correct?
Does it mean that the denominator says how many positions I have after comma?</p>

答:
The values given are indeed correct.  64 is 1,000000

```<code>        1/64                        1/32
19 * 0.000 001 = 0. 010 011       0. 10011
21 * 0.000 001 = 0. 010 101       0. 10101
</code>```

Dividing by 1000000 in binary is the same as in decimal:  shift the comma six places to the left (with zero-filling as needed).  So to make 1 000 00 into one, you shift the comma five places to the right, and to divide 10101 by the same amount, shift the comma five to the left, ie 0,10101.

If you need further places, zero-fill to the left.  Divide 10011 by 1000000, you need an extra place before the comma, so eg 000000 10011, becomes 00000,0 10011, and delete unnecassary zeros -> 0,010011.

Title: Conversion: dec to bin

Body:
The values given are indeed correct.  64 is 1,000000

        1/64                        1/32
19 * 0.000 001 = 0. 010 011       0. 10011
21 * 0.000 001 = 0. 010 101       0. 10101

Dividing by 1000000 in binary is the same as in decimal:  shift the comma six places to the left (with zero-filling as needed).  So to make 1 000 00 into one, you shift the comma five places to the right, and to divide 10101 by the same amount, shift the comma five to the left, ie 0,10101.

If you need further places, zero-fill to the left.  Divide 10011 by 1000000, you need an extra place before the comma, so eg 000000 10011, becomes 00000,0 10011, and delete unnecassary zeros -> 0,010011.


问:
Analytic continuation of square root with branch cut along the negative imaginary axis by using a suitable logarithmic branch.
We are working with real integrals and the complex residue theorem. In order to solve the following integral:

$$ \int_0 ^\infty \frac{\sqrt{x}}{1+x^2},$$

I will have to think about how a square root continues "along the complex plane". I am a bit confused about the hint I have gotten, it confuses me more than that it actually helps me. The problem is as follows:

Prove that the function $x \mapsto \sqrt{x}$, defined on $(0,\infty)\to \mathbb R$ has a unique continuation to the analytic function $w: \mathbb C \setminus (-\infty, 0)i \to \mathbb C$

The hint is as follows: use a suitable branch of the logarithmic function. Observe that the notation means we leave out the negative imaginary axis.

We have covered some identities in class with analytic continuation, often involving identities with sines and cosines or exponential laws. The approach was then to show that some function was zero on $\mathbb C$ (using Liouville) and the identity emerged for $\mathbb C$. All the material involving this topic that I have found online talks about branch cuts and about the angles that are then permitted. I am not quite sure how to put it together formally. Essentially we want to define for $z=re^{\phi i}$ and $\phi \in (-\pi/2, 3\pi/2)$:
$$ w(z)=\sqrt{r} e^{\frac{\phi i}2}$$
Where do logarithms come in?

答:
In my lecture notes I found the following definition of the power of a complex number. First we choose a suitable branch of the logarithm, in our case we look at the interval $(\pi/2 - \pi, \pi/2+\pi)$,then we define:
$$ z^{\frac{1}{2}}:=e^{\frac{1}{2} \log_{\pi/2}(z)}.$$
Where $\log_{\pi/2}$ has a branch cut along $\{ R e^{\pi/2} | R\in \mathbb R_{\geq 0}\}$.

After this it is indeed shown that such a function is complex differentiable on its domain (which is open, hence this function is holomorphic, hence analytic by Goursat)

The comment by sasquires thus answered my question and I wish to post this so I can close my question as solved.


问:
Eigenvectors of Normal matrix (orthogonality)
Let's suppose A is normal matrix. I know that the eigenvectors of normal matrix and its conjugate are the same, and that for the fixed eigenvector the eigenvalues are conjugate. Let's fix some $e_1$ normal eigenvector $(||e_1||=1)$, so $Ae_1= λ_1 e_1$ and $A^*e_1=\bar{λ_1}e_1 $.
$$
S_1 = span\{e_1\}
$$
$$
T_1 = \{x : (x,e_1)=0\}
$$
How can I show that there is an eigenvector $e_2$ which belongs to $T_1$?

P.S. I know that if $x∈T_1$ then $Ax∈T_1,A^*x∈T_1$, because
$$
(Ax,e_1)=(x,A^*e_1)=(x,\bar{λ_1}e_1)={λ_1}(x,e_1)=0
$$

答: So $L_1$ subspace is $A$-invariant, due to the last thing that you prove. And it is a famous theorem about linear operators, that every linear operator has at least one eigenvalue, so there exists some $e_2$ eigenvector which belongs to $T_1$. Here you can find it page 5.


问:
Closed form of the sum $\sum_{r\ge2}\frac{\zeta(r)}{r^2}$
Note:This is the same question, but it doesn't answer my question, the answer doesn't give a closed form. In fact, the answer is not accepted. Moreover, I  don't think that a 9 month old inactive question would get answers.

I have been working on various sums involving the zeta function (which come up frequently in my research), and it turned out that many of them had nice closed forms. Today, I was trying to evaluate
$$\sum_{r\ge2}\frac{\zeta(r)}{r^2}$$
which turned out to be a bit harder than the others. Wolfram Alpha and Mathematica both cannot give the answer, they only give an approximate value of $0.835998$. After half an hour of work, I turned this sum to
$$-\int_{0}^{\infty}\frac{t}{e^t}\psi(1-e^{-t})dt-\gamma$$
where $\gamma$ is the Euler–Mascherni constant and $\psi$ is the Digamma function. Now I don't know what to do further. Does this integral have any closed form? Any help would be appreciated. By the way, some of my ideas to evaluate the integral would be to use some integral or sum representation of the digamma function, and in this case we can interchange the sums and integrals safely.

Note: My work is too long to be stated here, so I cannot write it.

Update: This integral can be further turned to
$$\lim_{n\to\infty}H_n-\gamma-\sum_{k=1}^{n}\mathrm{Li}_2\left(\frac1k\right)$$
where $\mathrm{Li}$ is the polylogarithm. I turned this to
$$\lim_{n\to\infty}H_n-\gamma-\sum_{r=1}^{\infty}\frac{H_{n,r}}{r^2}$$
where $H_{n,r}$ are generalized harmonic numbers.

答:
I've 3 different idea's, will work them out later but maybe it might be helpful.

The easiest is by using similair known generating functions e.g.:
$$\sum_{m=2}^{\infty} (1+(-1)^m) \frac{\zeta(m)}{m^2}=-\int_{x=0}^{\pi} \frac{\ln\bigg(\frac{sin(x)}{x}\bigg)}{x}$$

Another one is generating functions but on the negative side of the zeta function and somehow try to extend that/find a relation.
$$\int_{n=1}^{\infty} \frac{ln(n!)}{n}-\ln(n)+1 -\frac{ln(2\pi)}{2n}-\frac{\ln(n)}{2n}=-\sum_{m=1}^{\infty} \frac{\zeta(-m)}{m^2}$$

and the third one, is writing it out like you did and see if you can manipulate it there. But there are too much roads to go to know what works, just an example.

The part with 'p' goes to 0 as p goes to infinity.
$$ \frac{c^n}{n^2} = \sum_{j=0}^p c^n\frac{(j)!(n-1)!}{(n+j+1)!}+\frac{(n-1)!p!}{(n+p+1)! n}$$
$$\sum_{n=1}^{\infty} \sum_{j=0}^p (j)! c^n\frac{(n-1)!}{(n+j+1)!}=$$
$$=-\ln{(c)}\ln(1-c)+\sum_{s=1}^p c^{1-s} (-1)^{s+1}\sum_{m=s}^{p}\frac{(m-1)!}{(m-s+1)!}\sum_{n=s}^m \frac{1}{n}=$$

And then adjust for the altercation ofcourse and summed over c. But there are most likely a lot of different approaches here.

And if you are lucky the sum might just show up somewhere without expecting it :)


问:
Examples of functions with a removable discontinuity
I recently came across the following example of a function which is discontinuous at a point:

The cost of a sushi buffet is calculated by weighing your plate, where you are then charged £10 per pound. As part of a promotion, if you manage to fill your plate so that the weight is precisely 1lbs, you get your meal for free.

This is a nice way to motivate a function like
$$f(x) = \begin{cases}10\,x\hfil&\text{$x\neq 1$}\\\hfil0\hfil&\text{$x=1$}\end{cases}$$
which are nice for illustrating the idea of a limit to students.

Are there any other real-world examples of functions like this, for the sake of explaining limits? Ideally one where you would actually want the limiting value, and not the assigned value, to motivate the concept further. (Apart from the obvious difference quotient.)

答:
The most common example I can think of are characteristic functions. Given sets $A\subset B$, we define $\chi_A:B\to\{0,1\}$ by $$\chi_A(x)=\begin{cases}1&x\in A\\0&x\notin A\end{cases}$$

With an appropriate choice of $A$ as a discrete set in $B$, the characteristic function has a removable discontinuity at every point in $A$.

An actual real-world example is sprinting. In sprinting, we care a lot about initial acceleration, i.e., acceleration off the blocks. However, if we plot acceleration vs time in the sprint, the value of acceleration at time $0$ (the starting gun shot) is $0$, but it immediately jumps to a large value and then maintains a high value for the rest of the race. We don't really care about the fact that the acceleration at the moment the gun is shot is $0$, rather, we care about the initial push.

As a generalization of this idea, you will be able to find other examples in things like acceleration, jerk, etc, since they are not inherently continuous.


问:
Integration problems
Can anyone help me with these:-

(a)Prove by induction: $\displaystyle\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$

(b) By explicitly calculating upper and lower Riemann sums on a uniform partition and
passing to the limits, calculate $\displaystyle\int_0^1 x^2 \mathrm{d}x$.

答:
$P(k)=\sum k^2=\dfrac{n(n+1)(2n+1)}{6}$

Check for the base, $P(1)=\dfrac{1 \cdot 2 \cdot 3}{6}=1$

Assume $P(k)$ is true, $P(k)=\dfrac{n(n+1)(2n+1)}{6}$

$P(k+1)=\sum (k+1)^2=\sum k^2+2k+1=\dfrac{n(n+1)(2n+1)}{6}+2 \dfrac{n(n+1}{2}+n$.

You will see that its true. Finally you will get $\dfrac{(n+1)(n+2)(2n+3)}{6}$.

My $\sum$ means $\sum_1^n$

Title: Integration problems


问:
If $\gcd(a,b)=1$ then $\gcd(a+b, a^2-ab+b^2)$ is equal to $1$ or $3$.
I need help to prove this:
If $\gcd(a,b)=1$ then $\gcd(a+b, a^2-ab+b^2)$ is equal to $1$ or $3$.
I have done this:

Let $d$ be the g.c.d. of $(a+b, a^2-ab+b^2)$, then $d$ divides $a+b$ and $d$ divides $a^2-ab+b^2$. That implies $d$ divides $(a+b)n + (a^2-ab+b^2)m$, for some $n,m$ integers. Let $n$ and $m$ be $a$ and $1$ respectively, then $d$ divides $2a^2+ b$. With the same argument but with $n=b$ we get $d$ divides $a^2+2b$. Then $d$ divides $3a^2+3b^2$. That implies $3a^2+3b^2 \geq d$, and we get that $3 \geq d$, because $\gcd(a^2+b^2)=1$. So $d$ must be $3$ or $1$ because if $d =2$, $d$ has to divide $a^2+2b$ and $2a^2+ b$, but we see that no. So $d=3$ or $1$.
I don't know if I did it well.

答:
Well try to factor $a^2 -ab +b^2$ into terms of $a+b$ or $a,b$.

$a^2 -ab + b^2 = a^2 + 2ab + b^2 - 3ab= (a+b)^2 -3ab$ so

$\gcd(a+b,a^2 -ab + b^2) =$

$\gcd(a+b, (a+b)(a+b) - 3ab) = $

$\gcd(a+b, -3ab)=\gcd(a+b,3ab)$.

Now... $\gcd(a,b) =1$ so any factor of $a$ will not be a factor of $b$ so it will not be a factor of $a+b$.  And no factor of $b$ will be a factof so it will not be a factor of $a+b$.  And the only prime factors of $3ab$ are either prime factors of $a$, or prime factors of $b$ or $3$.  No prime factor of $a$ or of $b$ are factors of $a+b$ so the only possible prime factors of $3ab$ and $a+b$ are (possibly) $3$.  Or there are no common prime factors of $3ab$ and $a+b$.

If there are no common prime factors then $\gcd(a+b,3ab) = 1$.

If the only common prime factor is $3$ then $3$ is a factor of $a+b$ but not of either $a$ nor of $b$.  So $\gcd(a+b, 3ab) =3$.


问:
If A, B are n by n matrices, and x is a vector, does (Ax)(Bx) = (AB)x?
It says it all in the title. Just a simple question about multiplication of matrices/vectors. I saw this in an old youtube video and in one of the steps, they multiplied the following two equations together:

Ax=cx

Bx=dx

Where c and d are scalars, to obtain

(Ax)(Bx) = (cx)(dx)

And then they simplified it to (AB)x=(cd)x

I didn't think you could even multiply the equations together since the dimensions don't match (Ax and Bx are both n by 1 matrices), and on top of that shouldn't you get (x)^2? Or am I missing something here and it's just a property of matrix/vector distribution?

答:
You are right that the expression
$$
A(x)B(x)
$$
is nonsense.

But this is true:
$$
(AB)(x) = A(B(x)) = A(dx) = cdx.
$$

It's probably what was intended.


答:
You started right. Let $d =\gcd(a+b, a^2-ab+b^2)$, then $d$ divides $a+b$ and $a^2-ab+b^2$. But from here it is pretty messy.

We can say $$d\mid (a+b)^2- (a^2-ab+b^2) = 3ab$$

So if prime $p\mid d$ we have $p\mid 3$ or $p\mid a$ or $p\mid b$. Say $p\nmid 3$ and say $p\mid a$. Since $p\mid a+b$ we have then $p\mid b$, a contradiction. So $p\mid 3$ and thus $d=3^n$.

Say $n\geq 2$, then $9\mid 3ab$ so $3\mid ab$ so $3\mid a$ or $3\mid b$. In both cases we have that $3$ divides then also the other number since $3\mid a+b$, a contradiciton. So $n\leq 1$ and we are done.

Body:
I need help to prove this:
If $\gcd(a,b)=1$ then $\gcd(a+b, a^2-ab+b^2)$ is equal to $1$ or $3$.
I have done this:

Let $d$ be the g.c.d. of $(a+b, a^2-ab+b^2)$, then $d$ divides $a+b$ and $d$ divides $a^2-ab+b^2$. That implies $d$ divides $(a+b)n + (a^2-ab+b^2)m$, for some $n,m$ integers. Let $n$ and $m$ be $a$ and $1$ respectively, then $d$ divides $2a^2+ b$. With the same argument but with $n=b$ we get $d$ divides $a^2+2b$. Then $d$ divides $3a^2+3b^2$. That implies $3a^2+3b^2 \geq d$, and we get that $3 \geq d$, because $\gcd(a^2+b^2)=1$. So $d$ must be $3$ or $1$ because if $d =2$, $d$ has to divide $a^2+2b$ and $2a^2+ b$, but we see that no. So $d=3$ or $1$.
I don't know if I did it well.


问:
Proving that the MLE remains invariant under measurable functions
I'm working through my course notes for Mathematical Statistics and I've come across an example which uses this without proving it (in fact without even explicitly specifying it). If $\hat{\theta}$ is the MLE for $\theta$ and $g$ a measurable function, why is $\hat{g(\theta)}$ MLE of $g({\theta})$.

答: Write $\eta:=g(\theta)$. Suppose the likelihood on some dataset is $L(\theta)=L^\ast(\eta)$, where our notation hides dependence on any other parameters. Let $\hat{\theta},\,\hat{\eta}$ be respective MLEs of $\theta,\,\eta$ so$$L^\ast(\hat{\eta})=\sup_\eta\sup_{\{\theta|g(\theta)=\eta\}}L(\theta)=\sup_\theta L(\theta)=L(\hat{\theta})=\sup_{\{\theta|g(\theta)=g(\hat{\theta})\}}L(\theta)=L^\ast(g(\hat{\theta})).$$This is theorem 7.2.10 in Casella and Berger's Statistical Inference.


问:
Writing this concatenating decimal as a series/fraction
I am trying to work out a way to write the number $0.123456789101112...$ as a series, I thought at first doing something like:
$$\sum_{n=1}^N\frac{n}{10^n}$$
but when we get into the double digits this doesn't work, so how should I do this? Also is there a way to evaluate sums of the form above? Thanks

答:
This number is known as the Champernowne constant, and is known to have a few summation identities, neither of which are pleasant:

$$\sum_{n=1}^\infty\frac n{10^{~\left(\sum\limits_{k=1}^n\left\lceil\log_{10}(k+1)\right\rceil\right)}} \qquad \sum_{n=1}^\infty\frac n{10^{\left(n+\sum\limits_{k=1}^{n-1}\left\lfloor\log_{10}(k+1)\right\rfloor\right)}}$$

Also is there a way to evaluate sums of the form above?

An infinite sum of the form you describe ($\sum n/10^n$) can be evaluated by taking derivatives of the geometric series. Recall that

$$\sum_{k=0}^\infty x^k = \frac{1}{1-x}$$

whenever $|x|<1$. Take the derivative with respect to $r$ of both sides -- you can do it on the left side term by term due to absolute convergence -- and then multiply by $x$. You get

$$\sum_{k=1}^\infty kx^k = \frac{x}{(1-x)^2}$$

In your case, $x=1/10$.


问:
How to identify a summed number given the total and the position (or index) of the summation?
I'm not a mathematician, I'm only a programmer with not so good math skills trying to implement something that need this math.
Imagining that somehow a sequence of numbers where given to you, randomly, and while been generated they are summed (through a function, anyway) in a manner that after all you have only the total and you can enter an index (the order or position) and is returned what number was summed in that position.

Ex.:
The computer generated randomly the numbers [6, 45,97,12, 26] and summed through this magical function and returned 186 (or other result). Now, with this result I can enter 4 and the algorithm  will return the value 12, because the fourth number added was 12.

Are there some way I can calculate this?

Please, be nice with that symbols.

Edit.

What I'm trying to says is that, when you add numbers, due to the distributive aspects you can not retrieve the exact same numbers you added. If you have 4 + 8, the result, 12, can be expressed as 8+4, 6+6, 4+4+4, 2+10, etc. But if you have only the result you will never know what the original numbers generated the result. To me, the final result doesn't matter that much since I have some way to make a math operation with may sequence of numbers and having the result and the formula I can search for the original elements of the computation with the original order. Imagine you need to send a message, a sequence of numbers, and your receiver not only need to have the original numbers, but he needs to be able to look the number in the position it was added, without the need of reveal all the number at once, but see the given number in the exact order it was added or the operation he needs to do will fail, the message will not be correct.
So you have thousands of numbers, randomly generated, and you need to transmit, but your receiver will not get thousands of numbers, he needs to get one or two number and a key (the formula) with wich he is able to search, say, he needs to look the numbers added (or processed any way) in the position 637 to the 856 in a moment, so he will get the sum, inform the position he needs and the result will be the number added in that position.
I saying sum, but is a formula, some way to take number in a specific order, compute, get a result and with the result be able to look what was placed in a specific position.

答:
There are many ways to do this.  One of the most elegant is to encode your sequence as follows:
$$
(a_1, a_2, \ldots, a_k)\rightarrow F(a_1,a_2,\ldots, a_k)=2^{a_1} 3^{a_2} \cdots p_k^{a_k},
$$
where $p_k$ is the $k$-th prime number.  To retrieve the $i$-th element, just factor the output and read off the power to which $p_i$ is raised.

Of course, you can encode any string as a number directly: an alternate approach is to treat "6,45,97,12,6" as a string of digits in base $11$ (where the comma is the $11$th digit), and encode your sequence that way.  Then retrieving the $i$-th element is as simple as writing the output in base $11$, splitting on the commas, and reading off the digits in the $i$-th block.


问:
Analytic subbasis for a locally convex topology
Let $E$ be a $\mathbb R$-vector space, $P$ be a nonempty family of seminorms on $E$, $$U_p:=\{x\in E:p(x)<1\}\;\;\;\text{for }p\in P$$ and $\tau$ denote the topology on $E$ generated by $P$.

I was able to show that \begin{equation}\begin{split}\mathcal B_P&:=\left\{\varepsilon\bigcap_{p\in F}U_p:F\subseteq P\text{ is finite and }\varepsilon>0\right\}\\&=\left\{\bigcap_{p\in F}U_{\frac1\varepsilon p}:F\subseteq P\text{ is finite and }\varepsilon>0\right\}\end{split}\end{equation} is an analytic basis$^1$ for $\tau$.

Now I was wondering whether $$\mathcal S_P:=\left\{\varepsilon U_p:p\in P\text{ and }\varepsilon>0\right\}=\left\{U_{\frac1\varepsilon p}:p\in P\text{ and }\varepsilon>0\right\}$$ is an analytic subbasis$^2$ for $\tau$.

By definition, this would imply that $$\mathcal B'_P:=\left\{\bigcap\mathcal F:\mathcal F\subseteq\mathcal S_P\text{ is finite}\right\}\tag1$$ is an analytic basis for $\tau$.

Does this in turn imply that $\mathcal B_P=\mathcal B'_P$?

The inclusion $\mathcal B_P\subseteq\mathcal B'_P$ is clearly trivial. Let's consider the other inclusion: Let $B\in\mathcal B_P'$. Then there is a finite $\mathcal F\subseteq\mathcal S_P$ with $$B=\bigcap\mathcal F.$$ Assume $n:=|\mathcal F|\in\mathbb N$ (i.e. $\mathcal F\ne\emptyset$). Since $\mathcal F\subseteq\mathcal S_P$, $$\mathcal F=\left\{\varepsilon_1U_{p_1},\ldots,\varepsilon_nU_{p_n}\right\}$$ for some $p_1,\ldots,p_n\in P$ and $\varepsilon_1,\ldots,\varepsilon_n>0$. If $$p:=\max_{1\le i\le n}\frac1{\varepsilon_i}p_i,$$ then clearly $$B=\bigcap_{i=1}^n\varepsilon_iU_{p_i}=U_p\tag2.$$

However, that doesn't immediately yield $B\in\mathcal B_P$, unless $p\in P$. So, maybe the desired claim doesn't hold and $\mathcal B_P$ and $\mathcal B'_p$ are distinct analytic bases for $\tau$. If that's the case, is my initial guess that $\mathcal S_P$ is in fact an analytic subbasis for $\tau$ correct at all?

$^1$ $\mathcal B$ is called analytic basis for $\tau$ if $\mathcal B\subseteq2^E$ and $$\tau=\left\{\bigcup\mathcal A:\mathcal A\subseteq\mathcal B\right\}.$$

$^2$ $\mathcal S$ is called analytic subbasis for $\tau$ if $\mathcal S\subseteq2^E$ and $$\left\{\bigcap\mathcal F:\mathcal F\subseteq\mathcal S\text{ is finite}\right\}$$ is an analytic basis for $\tau$.

答:
As you noticed, it doesn't have to be $\mathcal{B}_P = \mathcal{B}_P'$ in general. However, $\mathcal{B}_P'$ is an analytic basis for $\tau$. Indeed, clearly we have
$$\mathcal{B}_P \subseteq \mathcal{B}_P' \subseteq \tau$$
because $\varepsilon U_p$ are in $\tau$ and $\tau$ is closed under finite intersections. In particular

for any $\mathcal{A} \subseteq \mathcal{B}_P'$ we have $\bigcup \mathcal{A} \in \tau$ since $\mathcal{B}_P' \subseteq \tau$ and $\tau$ is closed under unions.

for any $U \in \tau$ since $\mathcal{B}_P$ is an analytic basis there exists $\mathcal{A} \subseteq \mathcal{B}_P \subseteq \mathcal{B}_P'$ such that $U =\bigcup\mathcal{A}$.


问:
Surjective map inducing homeomorphism between spectra

Let $\phi: A \to B$ be a surjective map of rings such that $\phi^*: \operatorname{Spec}{B} \to \operatorname{Spec}{A}$ is a homeomorphism, then $\operatorname{ker}{\phi} \subseteq \operatorname{nil}{(A)}$, where $\operatorname{nil}{(A)}$ is the nilradical of $A$.

I tried assuming that this is false, but then I got confused and didn't know where to go. I'm sure there is a nice, geometric proof of this, but I couldn't see it. Can anyone help me?

答: Let $I = \ker(\phi)$. Since $\phi \colon A \to B$ is a surjective map of rings, it follows that the canonical induced morphism $A/I \to B$ is an isomorphism, and $\phi^{\ast}$ is a closed immersion of $\operatorname{Spec}(B)$ onto the (closed) subset $V(I)$ of $\operatorname{Spec}(A)$. By assumption, $\phi^{\ast}$ is a homeomorphism, and so it follows that $V(I) = \operatorname{Spec}(A)$. In particular, this means that $I$ is contained in every prime ideal of $A$, whence $I \subset \operatorname{Nil}(A)$.


答:
Are you asking “given the cumulative sums, how to recover the summands”? In which case, to find the k-th summand, you should do (k-th cumulative sum) minus ((k-1)-th cumulative sum). The convention is that the 0-th cumulative sum is zero.

Example: original sequence is 3 1 2 4. The cumulative sums are (0) 3 4 6 10. So, to find the 2nd number (1), you should do 4 minus 3.

Title: How to identify a summed number given the total and the position (or index) of the summation?

Body:
I'm not a mathematician, I'm only a programmer with not so good math skills trying to implement something that need this math.
Imagining that somehow a sequence of numbers where given to you, randomly, and while been generated they are summed (through a function, anyway) in a manner that after all you have only the total and you can enter an index (the order or position) and is returned what number was summed in that position.

Ex.:
The computer generated randomly the numbers [6, 45,97,12, 26] and summed through this magical function and returned 186 (or other result). Now, with this result I can enter 4 and the algorithm  will return the value 12, because the fourth number added was 12.

Are there some way I can calculate this?

Please, be nice with that symbols.

Edit.

What I'm trying to says is that, when you add numbers, due to the distributive aspects you can not retrieve the exact same numbers you added. If you have 4 + 8, the result, 12, can be expressed as 8+4, 6+6, 4+4+4, 2+10, etc. But if you have only the result you will never know what the original numbers generated the result. To me, the final result doesn't matter that much since I have some way to make a math operation with may sequence of numbers and having the result and the formula I can search for the original elements of the computation with the original order. Imagine you need to send a message, a sequence of numbers, and your receiver not only need to have the original numbers, but he needs to be able to look the number in the position it was added, without the need of reveal all the number at once, but see the given number in the exact order it was added or the operation he needs to do will fail, the message will not be correct.
So you have thousands of numbers, randomly generated, and you need to transmit, but your receiver will not get thousands of numbers, he needs to get one or two number and a key (the formula) with wich he is able to search, say, he needs to look the numbers added (or processed any way) in the position 637 to the 856 in a moment, so he will get the sum, inform the position he needs and the result will be the number added in that position.
I saying sum, but is a formula, some way to take number in a specific order, compute, get a result and with the result be able to look what was placed in a specific position.


问:
Induction proof for $V - E + F = 2$ (Connected planar graphs)
This might be a bit of a silly question, but I'm struggling to understand this proof for $V - E + F = 2$, where $V$ = number of vertices, $E =$ number of edges and $F = $ number of faces.

I understand the first part, but I'm confused as to why we can assume true for a number of faces but then subtract from $F$, if we've proved it for $F = 1$.

An explanation would be much appreciated. Thanks.

答:
For an easy to picture example, consider the complete graph on 4 vertices. Initially $F=3, V=4, E=6.$ Notice that $V-E+F=2$. Now imagine removing the edge joining the bottom-left vertex to the center vertex (this edge is appropriate because the proof requires us to remove an edge that belongs to a cycle.) How has the picture changed? We haven't removed any vertices, so $V=4$ still. We have removed one edge, so our new $E=5$. We have also combined two faces into 1 new face, so now there are two remaining, so our new  $F=2$. Notice that it is still the case that $V-E+F=2$.

The idea is that this is what will always happen when you remove an edge that belongs to a cycle in a planar graph - doing so will reduce the number of edges and the number of faces both by 1, so the value of $V-E+F$ will not change. In this way, the author is reducing the problem for a graph with $n$ faces to a graph with $n-1$ faces. They are saying that because the value of $V-E+F$ is the same in each case, and we have assumed in the $n-1$ case that $V-E+F=2$, we have shown that the same is true in the case with $n$ faces.


问:
Existence of topological space whose open sets are anti-closed under infinite intersection
Is there an infinite topology $T$, such that whenever $I$ is an infinite set of open sets of $T$ which does not contain the empty set, then $\bigcap I$ is not an open set?

答:
The cofinite topology suggested by Yuval is almost right, but not quite, because you can have an infinite intersection be empty even if none of the sets being intersected is itself empty. But a small twist to it will fix it.

Let $Y$ be an infinite set, let $\infty$ be something not in $Y$, and let $X=Y\cup\{\infty\}$. We define a topology $\tau$ on $X$ as follows: for $A\subseteq X$, $A\in\tau$ if and only if $A=\varnothing$, or else
$$\infty\in A\text{ and }|X\setminus A|\lt\infty.$$

This is a topology: it includes $\varnothing$ by definition;  it includes $X$ itself. If $A$ and $B$ are in $\tau$ then $A\cap B\in\tau$, since the complement of $A\cap B$ is the union of $X\setminus A$ and $X\setminus B$, two finite sets. And if $\{A_i\}_{i\in I}$ is a family of open sets, at least one of which, say $A_j$, is nonempty, then $\cup A_i$ contains $\infty$ and has complement contained in $X\setminus A_j$, hence finite.

Now assume that $\{B_i\}_{i\in I}$ is an infinite family of nonempty open sets. I claim that if $\cap B_i$ is open, then there are actually only finitely many distinct $B_i$ (and as a consequence, if the family contains infinitely many pairwise distinct $B_i$, then the intersection is not open).

Indeed, $\cap B_i$ is not empty since $\infty\in \cap B_i$. If $\cap B_i$ is open, then it has finite complement, so there exists a finite set $F$ such that $X\setminus (\cap B_i) = F$. Since
$$F = X\setminus (\cap B_i) = \cup(X\setminus B_i),$$
it follows that there are only finitely many possibilities for $X\setminus B_i$, which in turn means there are only finitely many possibilities for the $B_i$. Thus, our infinite family consists of only finitely many distinct sets.

By contrapositive, if the family includes infinitely many pairwise distinct sets, then $\cap B_i$ must have infinite complement, and since $\cap B_i\neq\varnothing$, it follows that it is not open.

Thus, this topology satisfies the required conditions.


问:
Verify that ~(P ↔ Q) ≡ P ↔ ( ~Q) using logical equivalencies (not truth tables)
I have been able to verify this via truth tables but not with logical equivalencies yet. I understand some of the basic principles of logical equivalencies but I cannot seem to get to the end of this problem where one side equals the other.</p>

答:
Notice that $A\vee(B\wedge C)\equiv (A\vee B)\wedge(A\vee C)$.

On the left hand side, we have

$\neg(P\leftrightarrow Q)\equiv \neg[(P\to Q)\wedge(Q\to P)]\equiv(P\wedge\neg Q)\vee(Q\wedge\neg P)\equiv[(P\wedge\neg Q)\vee Q]\wedge[(P\wedge\neg Q)\vee\neg P]\\\equiv[(P\vee Q)\wedge(\neg Q\vee Q)]\wedge [(P\vee\neg P)\wedge(\neg Q\vee\neg P)]\equiv(P\vee Q)\wedge(\neg Q\vee\neg P)$

Consider the negation of the right hand side and we have

$\neg[(P\to\neg Q)\wedge(\neg Q\to P)]\equiv(P\wedge Q)\vee(\neg Q\wedge\neg P)$

Now we negate it back

$\neg[(P\wedge Q)\vee(\neg Q\wedge\neg P)]\equiv[\neg(P\wedge Q)]\wedge[\neg(\neg Q\wedge\neg P)]\equiv(\neg P\vee\neg Q)\wedge(Q\vee P)$

Hence it gives the result that we are asked to verify.


问:
Why truth table is not used in logic?
One day, I bought Principia Mathematica and saw a lot of proofs of logical equations, such as $\vdash p \implies p$ or $\vdash \lnot (p \wedge \lnot p)$. (Of course there's bunch of proofs about rel&set in later)

After reading these proofs, I suddenly thought that "why they don't use the truth table?". I know this question is quite silly, but I don't know why it's silly either (just my feeling says that).

My (discrete math) teacher says that "It's hard question, and you may not understand until you'll become university student," which I didn't expected (I thought the reason would be something easy).

Why people don't use truth table to prove logical equations? (Except for study logic (ex: question like "prove this logic equation using truth table"), of course.)

PS. My teacher is a kind of people who thinks something makes sense iff something makes sense mathematically.

答:
In Principia, the authors wanted to produce an explicit list of purely logical ideas, including an explicit finite list of axioms and rules of inference, from which all of mathematics could be derived.  The method of truth tables is not such a finite list, and in any case would only deal with propositional logic.

The early derivations of Principia are quite tedious, and could have been eliminated by adopting a more generous list of initial axioms.  But for reasons of tradition, the authors wanted their list to be as small as possible.

Remark: Principia is nowadays only of historical interest, since the subject has developed in quite different directions from those initiated by Russell and Whitehead. The idea of basing mathematics (including the development of the usual integers, reals, function spaces) purely on "logic" has largely been abandoned in favour of set-theory based formulations.  And Principia does not have a clear separation between syntax and semantics.  Such a separation is essential to the development of Model Theory in the past $80$ years.

Title: Why truth table is not used in logic?


问:
Develop the next function:$f(x)=\frac{4x+53}{x^2-x-30}$ into power series, Find the radius on convergence and find $f^{(20)}(0)$
Develop the next function:$\displaystyle f(x)=\frac{4x+53}{x^2-x-30}$ into power series, Find the radius on convergence and find $f^{(20)}(0).$

For the first part:

$\displaystyle\frac {4x+53}{(x-6)(x-5)}\to\frac {4x+53}{(x-6)(x+5)}=\frac{A}{(x-6)}+\frac{B}{(x+5)}=\frac{7}{(x-6)}+\frac{-3}{(x+5)}=\frac{7}{-6(1-\frac x6)}+\frac{-3}{5\left(1-(-\frac x5)\right)}$

$\displaystyle=-\frac76\sum_{n=0}^\infty{\left(\frac x6\right)^n}-\frac35\sum_{n=0}^\infty(-1)^n\left(\frac x5\right)^n$

Now i'm having trouble with finding the radius of convergence and $f^{(20)}(0)$.

What does $f^{(20)}(0)$ means? 20 derivatives and $x=0$? That means i should find a some-kind of pattern and then place $x=0$?

答:
Hints:

1) For the radius of convergence: do you know what is the radius of convergence of $\sum_{n=0}^\infty t^n$? if so, find the radius of convergence for each series separately. Then, the radius of convergence for the sum will be the minimal (do you see why?)

2) You just found the taylor series around $x=0$. The general form of Taylor series is $\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$. So, what is $f^{(20)}(0)$?


问:
Prove that a continuous function on a compact subset of ${\mathbb{R} }^n$ is almost like a Lipschitz function.
Suppose I have $K \subset { \mathbb{R} }^n $ a compact subset,and a continuous function $f : K \to \mathbb{R} $.
Given some $\epsilon >0$ prove there exist some $L$ such that for all $x,y \in X$
I have $$|f(x)-f(y) | \leq L |x-y | + \epsilon $$

Now I was able to find some proofs for a similar problem where $K$ is convex but couldn't find a proof that doesn't use convexity.

One thing I considered is to approximate the derivative of $f$ like this $$Df \approx \frac{f(x)-f(y)}{x-y} - \epsilon $$ and somehow prove that the approximation is bounded but couldn't come up with a proof, maybe the mean value theorem may help.

How can I show this?

答:
Fix $\epsilon>0.$ If the claim is false, then there are sequences $(x_n), (y_n)\subseteq K$ such that

$\tag1 |f(x_n)-f(y_n) | > n |x_n-y_n | + \epsilon.$

Since $K$ is compact, there is a convergent subsequence $(x_{n_k})\to x\in K$, of $(x_n)$. For the same reason, there is a convergent subsequence  $(y_{n_{k_l}})\to y\in K.$ Thus, $x_{n_{k_l}}-y_{n_{k_l}}\to x-y.$ Continuity of $f$ implies that $f(x_{n_{k_l}})\to f(x)$ and $f(y_{n_{k_l}})\to f(y)$ and continuity of the absolute value function now implies that

$\tag2 |f(x_{n_{k_l}})-f(y_{n_{k_l}})|\to |f(x)-f(y)|.$

If $x=y$ then $(1)$ and $(2)$ combine to give a contradiction: $0>\epsilon.$

But if $x\neq y$ we also get a contradiction since in this case we have $|f(x_{n_{k_l}})-f(y_{n_{k_l}})|\to |f(x)-f(y)|<\infty$ whereas $n |x_n-y_n |\to \infty.$


问:
Graph Theory, graph and complement question
Which is the easier method to prove such a statement, would it be contradiction or induction?

If in a graph, each vertex has an even degree, how should we go about to prove that in the complement of G, the same parity applies to all vertex degrees?

Any guidance would be really appreciated.

答:
If $d$ is degree for $v$ in $G$ then the degree $d'$ of $v$ in $G'$ is $n-1-d$.

So, if $$d_1 \equiv d_2\equiv \cdots \equiv d_n \equiv 0\pmod 2$$ then $$n-1-d_1\equiv n-1-d_2\equiv \cdots \equiv n-1-d_n \equiv n-1\pmod 2$$


问:
Soft question, limit of $n\mathbb{Z}$
I was trying to think of countable subsets of $\mathbb{R}$ that were, in a sense, small. By that I mean how 'spread out' the terms were. I know this isn't at all concise, but for example, $3\mathbb{Z}$ is 'smaller' (in the manner described above) than say $2\mathbb{Z}$. So this got me thinking, looking at specifically $n\mathbb{Z}$, is there anything meaningful to talk about $\lim_{n\to\infty}{n\mathbb{Z}}$ ?

Firstly I thought, does this even make sense? Is it just a meaningless thing to consider?

What made intuitive sense is, if we consider the extended reals, that the limit is $\{-\infty,+\infty\}$, but that just seemed somehow not quite right.

Is there any resources available, or anything to talk about limits, or limits alike?

答:
The $\liminf$ and $\limsup$ of a sequence of sets are always defined. When they are equal, one can speak of the limit of the sequence.

The definitions are as follows, for a sequence $(A_n)$ of sets.

$$\liminf_{n \to \infty} A_n = \bigcup_{n=1}^{\infty} \bigcap_{m=n}^{\infty} A_m$$
$$\limsup_{n \to \infty} A_n = \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty} A_m$$

In words, $\liminf A_n$ consists of those points that are contained in all $A_n$'s for sufficiently large $n$. And $\limsup A_n$ consists of those points that are contained in infinitely many of the $A_n$'s.

In your case ($A_n = n\mathbb Z$), it's clear that both $\liminf n \mathbb Z$ and $\limsup n \mathbb Z$ contain exactly one point, namely $0$. Therefore,
$$\liminf_{n \to \infty} n \mathbb Z = \limsup_{n \to \infty}n \mathbb Z = \{0\}$$
Since the $\liminf$ and $\limsup$ are equal, it's valid to say that $\lim_{n \to \infty}n \mathbb Z$ exists and equals $\{0\}$.

See set-theoretic limit for more details.

Title: Soft question, limit of $n\mathbb{Z}$


问:
References for Names of Various Parametric Curves
I've been using quite a few 3D parametric curves recently, where the X, Y, and Z components are defined by some function of t. For example:

$t\in\{-\pi,\pi\}$

$x=sin(2t)$

$y=sin(t)$

$z=sin(t)$

To create a simple curve like this:

I'm guessing a curve like this has a name, but I don't know what references I could use to find it. Are there any references that list names for 3d parametric curves like the one above?

答: I believe these are called "Lissajous Curves". A reference you may consider for looking up different curves: A Catalog of Special Plane Curves by J. Dennis Lawrence.</p>


Title: Develop the next function:$f(x)=\frac{4x+53}{x^2-x-30}$ into power series, Find the radius on convergence and find $f^{(20)}(0)$

Body:
Develop the next function:$\displaystyle f(x)=\frac{4x+53}{x^2-x-30}$ into power series, Find the radius on convergence and find $f^{(20)}(0).$

For the first part:

$\displaystyle\frac {4x+53}{(x-6)(x-5)}\to\frac {4x+53}{(x-6)(x+5)}=\frac{A}{(x-6)}+\frac{B}{(x+5)}=\frac{7}{(x-6)}+\frac{-3}{(x+5)}=\frac{7}{-6(1-\frac x6)}+\frac{-3}{5\left(1-(-\frac x5)\right)}$

$\displaystyle=-\frac76\sum_{n=0}^\infty{\left(\frac x6\right)^n}-\frac35\sum_{n=0}^\infty(-1)^n\left(\frac x5\right)^n$

Now i'm having trouble with finding the radius of convergence and $f^{(20)}(0)$.

What does $f^{(20)}(0)$ means? 20 derivatives and $x=0$? That means i should find a some-kind of pattern and then place $x=0$?

Body:
When you have

$$f(x) = \sum_{n=0}^{\infty} a_n x^n$$

then

$$f^{(k)}(0) = k! \, a_k$$

You know $a_{20}$:

$$a_{20} = -\frac{7}{6} \frac{1}{6^{20}} - \frac{3}{5} \frac{1}{5^{20}}$$


问:
Prove for all $m \in \mathbb{N}$ exists $n \in \mathbb{N}$ such that $\varphi(n)-\varphi(n+1)>m$ and $\varphi(n)-\varphi(n-1)>m$
Number Theory :

Prove for all $m \in \mathbb{N}$ exists $n \in \mathbb{N}$ such that :

$\varphi(n)-\varphi(n+1)>m$ and $\varphi(n)-\varphi(n-1)>m$

Attempt:

For $m\in \mathbb{N}$ let $q_m$ be prime number form of $4k+3$ for some $k\in \mathbb{N}$ such that $2m+3<q_m$

答:
Hint: Suppose $n$ is an odd prime. Then $\varphi(n) = n-1$. Also, $n-1$ and $n+1$ are both even, and thus, $\varphi(n-1) \le \dfrac{n-1}{2}$ and $\varphi(n+1) \le \dfrac{n+1}{2}$. Do you see why this is true?

Using these results, how large does $n$ need to be to guarantee that $\varphi(n)-\varphi(n-1) > m$ and $\varphi(n)-\varphi(n+1) > m$?


问:
Formalizing the use of Dirac delta in a PDF
Suppose that you have a CDF $F:\mathbb{R}_+\to\mathbb{R}_+$, with only a single indifferentiability point at $x=1$, where $\lim_{z\to 1^{-} }F(z)=\frac{1}{3}$ but $F(1)=\frac{2}{3}$.  I'd like to define the PDF $f$ associated with $F$ such that $f(x)=F'(x)$ at $x\neq 1$, and $f(1)=\frac{1}{3}{\mathbf\delta}$ where ${\mathbf\delta}$ is the Dirac delta function. However, I was told that this is informal: I cannot formally define the range of $f$ to contain both $\mathbb{R}_+$ and $\{a {\mathbf \delta}: a\in \mathbb{R}_+\}$. Is my definition formal? If no, how can I make it formal? Thank you!

答:
The natural way to make this rigorous is to use measure theory.

A CDF is a way of describing the statistics of a (real-valued) random variable $X$, the idea being that $F(x) = \mathbb{P}\{X \leq x\}$.

Another way, which turns out to be equivalent, but is arguably more natural, is a (Borel) probability measure on $\mathbb{R}$.  Precisely, a (finite, positive) Borel measure $\nu$ assigns to any given Borel set $A \subseteq \mathbb{R}$ a number $\nu(A) \in [0,\infty)$.  (Think of a Borel set as a "nice" subset of $\mathbb{R}$, or one that can be built out of intervals.  Intervals are Borel sets.)  It is a probability measure if $\nu(\mathbb{R}) = 1$.  For us, the idea is $\nu(A) = \mathbb{P}\{X \in A\}$, that is, $\nu$ is a more vivid description of $X$.

It is a theorem of Lebesgue-Stieltjes integration that there is a one-to-one correspondence between Borel probability measures and CDFs through the map $\nu \mapsto F^{\nu}$ given by $F^{\nu}(x) = \nu((-\infty,x])$.  Hence anytime you work with a CDF, there is also an associated probability measure $\nu$.

You can think of $\nu$ as being the "derivative" of $F^{\nu}$ (or $F^{\nu}$ being the "anti-derivative" of $\nu$).  The relevance to your question: this makes sense whether or not $F^{\nu}$ is differentiable in the usual sense.

For example, if $F^{\nu}$ is differentiable at every point in $\mathbb{R}$ and $(F^{\nu})' = f$, then $\nu$ is given (by Lebesgue integration) through the formula:
\begin{equation*}
\nu(A) = \int_{A} f(x) \, dx.
\end{equation*}
A shorthand way to write this is $\nu = f(x) \, dx$ (hence, almost literally,
$\nu$ is the derivative of $F^{\nu}$).

As in your example, there is no reason $F^{\nu}$ should be differentiable.  Basically, your question seems to be: what replaces the decomposition $\nu = f(x) \, dx$ in general?

The Lebesgue Decomposition Theorem says that, given a probability measure $\nu$, there is a function $f$ and a singular measure $\nu^{s}$ such that
\begin{equation*}
\nu = f(x) \, dx + \nu^{s}.
\end{equation*}

The sum is interpreted "set-wise" (that is, pointwise) as $\nu(A) = \int_{A} f(x) \, dx + \nu^{s}(A)$.  The measure $\nu^{s}$ is singular: you can think of that as meaning it does not match our intuition about densities.

In your example, $F = F^{\nu}$ for the probability measure $\nu$ given by
\begin{equation*}
\nu(A) = \int_{A} F'(x) \, dx + \frac{1}{3} \delta_{1}(A)
\end{equation*}
where $\delta_{1}$ is the measure defined by
\begin{equation*}
\delta_{1}(A) = \left\{ \begin{array}{r l}
1, & \text{if} \, \, 1 \in A \\
0, & \text{otherwise}
\end{array} \right.
\end{equation*}
We call $\delta_{1}$ the Dirac mass at $1$; for an arbitrary $x \in \mathbb{R}$, $\delta_{x}$ is defined analogously.

More often than not, $\nu^{s}$ looks something like the previous example (or, as you say, it is a sum of finitely many Dirac deltas).  In most applications, there is are sequences $\{x_{n}\}_{n \in \mathbb{N}} \subseteq \mathbb{R}$ and $\{p_{n}\}_{n \in \mathbb{N}} \subseteq [0,1]$ with $\sum_{n = 1}^{\infty} p_{n} < 1$ such that
\begin{equation*}
\nu^{s}(A) = \sum_{n = 1}^{\infty} p_{n} \delta_{x_{n}}(A) = \sum_{x_{n} \in A} p_{n}
\end{equation*}
(It is worth noting that if $\nu^{s}$ is a probability measure, or $f = 0$ above, then $\sum_{n = 1}^{\infty} p_{n} = 1$ and $\nu^{s}$ is precisely the probability measure describing a discrete random variable with PMF $\{p_{n}\}_{n \in \mathbb{N}}$.  Hence what we are doing unifies "continuous" and "discrete" RV and PDFs and PMFs.)

However, the singular measure $\nu^{s}$ can also have a "continuous" part: for example, it is possible that $\nu^{s}(\{x\}) = 0$ for all $x \in \mathbb{R}$ (so it has no Dirac parts), but still it cannot be represented using a density.  Usually, you won't see such measures unless you are asking about long-time behavior of a stochastic process or an iterative procedure.  For example, if $X_{n} \in [0,1]$ is a RV whose ternary expansion up to order $n$ is equally likely to be any string consisting of $0$'s and $2$'s, but no $1$'s, then $X_{n} \overset{d}{\to} \tilde{X}$, where $\tilde{X}$ is described by such a singular probability measure that cannot be described using Dirac masses.  $\tilde{X}$ is the RV in $[0,1]$ whose ternary expansion is equally likely to have a $0$ or a $2$ at any given digit, but never has a $1$; its CDF is the "Devil's Staircase" or Cantor function.  (Interestingly, the Cantor function has derivative zero at "most" points, yet it remains a CDF.)


问:
Let $A, B, C$ be subgroups of $D$, where $A \le B$, $A \cap C = B \cap C$, and $AC=BC$. Show $A=B$.
Let $A,B, C \le D.$ (i.e.,  $A, B, C$ are subgroups of $D$), where $A \le B$, $A \cap C = B \cap C$, and $AC=BC$. Show $A=B$.

I've entertained commutative diagrams as a way to show this, but I'm having difficulty showing this more formally. Intuitively, it makes sense, but I'm not sure how to incorporate $AC = BC$.

答:
Without the assumption that $AC=BC$, the conclusion that $A=B$ does not follow. For example, in the dihedral group $D_8$ of order $8$, if you take $A=\langle r^2\rangle$, $B=\langle  r\rangle$, and $C=\langle s\rangle$, then $A\leq B$, $A\cap C=B\cap C=\{e\}$, but $A\neq B$. However, that example does not satisfy $AC=BC$.

Since you already know that $A\subseteq B$, you just need to show that $B\subseteq A$. Let $b\in B$. We want to show that $b\in A$. We know that $b\in B\subseteq BC=AC$, so there exists $a\in A$ and $c\in C$ such that $b=ac$. Then $c=a^{-1}b$. Since $a\in A\leq B$, then $a^{-1}b\in B$.

Can you finish it from there?


问:
Closed form of the sum $\sum_{r\ge2}\frac{\zeta(r)}{r^2}$
Note:This is the same question, but it doesn't answer my question, the answer doesn't give a closed form. In fact, the answer is not accepted. Moreover, I  don't think that a 9 month old inactive question would get answers.

I have been working on various sums involving the zeta function (which come up frequently in my research), and it turned out that many of them had nice closed forms. Today, I was trying to evaluate
$$\sum_{r\ge2}\frac{\zeta(r)}{r^2}$$
which turned out to be a bit harder than the others. Wolfram Alpha and Mathematica both cannot give the answer, they only give an approximate value of $0.835998$. After half an hour of work, I turned this sum to
$$-\int_{0}^{\infty}\frac{t}{e^t}\psi(1-e^{-t})dt-\gamma$$
where $\gamma$ is the Euler–Mascherni constant and $\psi$ is the Digamma function. Now I don't know what to do further. Does this integral have any closed form? Any help would be appreciated. By the way, some of my ideas to evaluate the integral would be to use some integral or sum representation of the digamma function, and in this case we can interchange the sums and integrals safely.

Note: My work is too long to be stated here, so I cannot write it.

Update: This integral can be further turned to
$$\lim_{n\to\infty}H_n-\gamma-\sum_{k=1}^{n}\mathrm{Li}_2\left(\frac1k\right)$$
where $\mathrm{Li}$ is the polylogarithm. I turned this to
$$\lim_{n\to\infty}H_n-\gamma-\sum_{r=1}^{\infty}\frac{H_{n,r}}{r^2}$$
where $H_{n,r}$ are generalized harmonic numbers.

答: The simplest form i got is $$\sum_{n\ge 2}\frac{\zeta(n)}{n^2} = \int_0^1\frac{\log\Gamma(t)}{1-t}\,dt - \gamma$$ using the ordinary generating function of $\zeta(n)$.