问:
Prove that the number of common tangents to two circles exterior to one another is 4
Given two circles outside each other, the maximum number of common tangents to the two circles is 4 according to Wikipedia. How can this be proven?

That's a picture above showing what I'm talking about. How can we prove that no other common tangents can be drawn aside those four?

答:
Any common tangent line must fall in one of the the following two cases:
the centers of the circles, $A$ and $B,$ are on the same side of the tangent line, or they are on opposite sides of the tangent line.

When the centers are on the same side,
first consider the case where the line connecting the centers of the circles intersects the tangent line at $C$.
The point $C$ cannot be between $A$ and $B$ (because then the centers would be on opposite sides of the tangent line);
without loss of generality, suppose the points $A,$ $B,$ and $C$ occur in that order along their common line.
Since the tangent is perpendicular to the radius at the tangent point,
we have two similar right triangles, one with hypotenuse $AC$ and one with hypotenuse $BC,$ where the ratio $AC:BC$ is the ratio of the radii of the circles.

Using this ratio and the order of the points on their common line, we can construct the location of the point $C$ uniquely.
The point of tangency of the circle $B$ with the tangent through $C$ is the intersection of circle $B$ with the circle constructed on the diameter $BC.$
There are exactly two such intersections, so we can construct exactly two tangent lines in this case.
We observe that the radii are unequal in this case.

In the case where the centers are on the same side of the tangent and the tangent does not intersect the line between the centers,
the tangent is parallel to the line between the centers.
The radius from $B$ to the tangent point is perpendicular to the line between the centers; so we construct a perpendicular to $AB$ through $B$, which intersects the circle $B$ at two points, which produce the two possible tangent lines in this case.
We observe that the radii are equal in this case,
so the two cases where the centers are on the same side of the tangent are mutually exclusive; we get only two tangent lines from that general case.

In the case where the centers are on opposite sides of the tangent line,
the line connecting the centers must intersect the tangent line
between the centers.
Let $D$ be that intersection point.
Again we have two similar right triangles, one with vertices at $A$, $D$, and the point of tangency on circle $A$, the other with vertices at $B$, $D$, and the point of tangency on circle $B$,
so $AD:BD$ is the ratio of the two radii.
This lets us uniquely construct the position of $D.$
The point of tangency with circle $B$ is then the intersection of that circle with the circle constructed on the diameter $BD,$ which gives two possible points of tangency and two possible tangent lines.

So we have at most two tangent lines that have both centers on the same side and at most two tangent lines that have the centers on opposite sides.
The arguments above do not show that either set of tangent lines exist;
to do that, we would have to consider whether each circle is inside, outside, or intersecting the other.
But if you accept the figure in the question as demonstrating the existence of four tangent lines in the case where the circles are mutually outside each other,
we have shown there are no other tangents.


问:
How to integrate $ \int x^n e^x dx$?
How can I solve this indefinite integral for an arbitrary integer $n>0$?

$$ \int{x^n e^x dx}$$

I could partially integrate it for small $n$, but that's not really a solution.

Edit: (TB) This question is closely related to: Is there a closed form solution for $\int x^n e^{cx}$?, but it is more elementary, because $n$ is an integer here.

答:
$  \int{x^ne^xdx}=e^x\sum^n_{k=0}{{\left(-1\right)}^{n-k}\frac{n!}{k!}x^k}+C$

Solution:

$  \int{xe^xdx}=\ ?$

$  u=x\to \ \frac{du}{dx}=1\to du=dx$

$  dv=e^xdx\to \frac{dv}{dx}=e^x\to \ \int{\frac{dv}{dx}dx}=\int{e^xdx}+C{{\stackrel{C=0}{\longrightarrow}}}v=e^x$

$  \frac{duv}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}\to \ \int{\frac{duv}{dx}dx}=\int{v\frac{du}{dx}dx}+\int{u\frac{dv}{dx}dx}+C_a\to \int{udv}=uv-\int{vdu}+C_b$

$  \int{xe^xdx}=xe^x-\int{e^xdx}+C_1=xe^x-e^x+C=e^x\sum^1_{k=0}{{\left(-1\right)}^{1-k}\frac{1!}{k!}x^k}+C$

$  \int{x^2e^xdx}=?$

$  u=x^2\to du=2xdx$

$  dv=e^xdx{{\stackrel{C=0}{\longrightarrow}}}v=e^x$

$  \int{x^2e^xdx}=x^2e^x-2\int{{xe}^xdx}+C=x^2e^x-2\left(xe^x-e^x+C\right)=x^2e^x-2xe^x+2e^x+C_1=e^x\sum^2_{k=0}{{\left(-1\right)}^{2-k}\frac{2!}{k!}x^k}+C_1$

Now we suppose:

$  \int{x^ke^xdx}=e^x\sum^k_{i=0}{{\left(-1\right)}^{k-i}\frac{k!}{i!}x^k+C}$

$  \int{x^{k+1}e^xdx=\ ?}$

$  u=x^{k+1}\to du=\left(k+1\right)x^kdx$

$  dv=e^xdx{{\stackrel{C=0}{\longrightarrow}v=e^x}}$

$  \int{x^{k+1}e^xdx=\ x^{k+1}e^x-\int{\left(k+1\right)x^ke^xdx}}+C=x^{k+1}e^x-\left(k+1\right)\int{x^ke^xdx}+C=x^{k+1}e^x-\left(k+1\right)e^x\sum^k_{i=0}{{\left(-1\right)}^{k-i}\frac{k!}{i!}x^k+C}=e^x\sum^{k+1}_{i=0}{{\left(-1\right)}^{k-i+1}\frac{(k+1)!}{i!}x^{k+1}+C}$

$  k+1=n$

$  \int{x^ne^xdx}=e^x\sum^n_{i=0}{{\left(-1\right)}^{n-i}\frac{n!}{i!}x^n+C}$

ContentLicense: CC BY-SA 3.0

Title: How to integrate $ \int x^n e^x dx$?

ContentLicense: CC BY-SA 4.0


问:
Why it is important to write a function as sum of even and odd functions?
For the function $f(x)$ we can write it as sum of even and odd functions:

$$f(x)=\underbrace{\frac{f(x)+f(-x)}{2}}_{\text{Even}}+\underbrace{\frac{f(x)-f(-x)}{2}}_{\text{Odd}}$$

My question is why it is important for us to write a function as sum of these two even and odd functions? Is there any application of that?

答:
A famous example of a decomposition in odd and even functions is given by Euler's formula
\begin{align*}
\color{blue}{e^{iz}}&\color{blue}{=}\color{blue}{\cos z+i\sin z}\\
&=\frac{e^{iz}+e^{-iz}}{2}+\frac{e^{iz}-e^{-iz}}{2}\qquad\qquad z\in\mathbb{C}\\
\end{align*}
which is used in many applications.


问:
Can an operator not be self-adjoint even though its representation is?
Let $V = P_2(\mathbb R)$ be the space of all polynomials of degree not exceeding 2, endowed with the inner product $\langle f(t),g(t) \rangle$ = $\int_0^1 f(t)g(t)dt$. Suppose that $T : V → V$ is the linear map defined by $T(a_0 + a_1 t + a_2 t^2 ) = a_1 t$.

(a) Demonstrate that $T$ is not self-adjoint by considering $\langle T(1),t \rangle$.

(b) Let $C = \{1,t,t^2\} $ and $B = [T]_C$ . Show that $B^∗ = B$.

(c) Although $B^∗ = B$, $T$ is not self-adjoint as observed in part (a). Why is this not a contradiction? Justify.

I got the first and second parts in this. The third part however, I'm not sure if I got the right answer. My reasoning was that if the operator $T$ is self-adjoint, then its representations are hermitian, and the converse need not be true. Any suggestions to make this reasoning more rigorous would be really helpful.

答:
The equivalence

$T$ is self-adjoint $\Longleftrightarrow$ $[T]_C$ is a Hermitian matrix

is only true when $C$ is an orthonormal basis. If $C$ is any basis, neither $\Longrightarrow$ nor $\Longleftarrow$ need to hold.


问:
Definition 7.8 of embedding in "General Topology" by Willard
In the text "General Topology" by Willard, def. 7.8 says

If X and Y are topological spaces, a function $f$ from X to Y is a homeomorphism iff $f$ is one-one, onto and continuous and $f^{-1}$ is also continuous. In this case, we say X and Y are homeomorphic. If $f$ is everything but onto, we call it an embedding of X into Y, and say that X is embedded in Y by $f$. Thus, X is embedded in Y by $f$ iff $f$ is a homeomorphism between X and some subspace of Y.

My question is regarding the phrase in the middle: "If $f$ is everything but onto". In such case, $f^{-1}$ is no longer a function, and then how would the earlier requirement "$f^{-1}$ is also continuous" apply?

Note that my question is not about the meaning of embedding, which has already been explained again in the last sentence in the paragraph and I find no confusion there.

My question is about how to interpret the phrase "If $f$ is everything but onto". One way I can think of is rephrasing it as something like (this is also like an elaboration of the last sentence in the original text):

If $f$ is everything but onto, and the bijection $\hat{f}:X\rightarrow f[X]$ that has the same graph as $f$, where $f[X]$ is the image of $X$ under $f$ equipped with the subspace topology induced by $Y$, is a homeomorphism, we call it ...

but this seems adding too much additional stuff. I wonder if there is any more straightforward and natural way of fixing/interpreting the original text to justify the way it's written, or if the original text is indeed a little sloppy here.

答: Your interpretation is correct. It is confirmed by the author who states that an embedding is an homeomorphism from $X$ onto $f[X]$.


答:
That's fine, although it's not necessary to go into graphs and all that, you can simply word it in terms of the rather common concept of "range restriction", like this:

If $f$ is one-to-one and continuous, and if the range restriction $f : X \to f[X]$ has continuous inverse $f^{-1} : f[X] \to X$ with respect to the subspace topology on $f[X] \subset Y$, then we call $f$ an embedding.

Title: Definition 7.8 of embedding in "General Topology" by Willard

Body:
In the text "General Topology" by Willard, def. 7.8 says

If X and Y are topological spaces, a function $f$ from X to Y is a homeomorphism iff $f$ is one-one, onto and continuous and $f^{-1}$ is also continuous. In this case, we say X and Y are homeomorphic. If $f$ is everything but onto, we call it an embedding of X into Y, and say that X is embedded in Y by $f$. Thus, X is embedded in Y by $f$ iff $f$ is a homeomorphism between X and some subspace of Y.

My question is regarding the phrase in the middle: "If $f$ is everything but onto". In such case, $f^{-1}$ is no longer a function, and then how would the earlier requirement "$f^{-1}$ is also continuous" apply?

Note that my question is not about the meaning of embedding, which has already been explained again in the last sentence in the paragraph and I find no confusion there.

My question is about how to interpret the phrase "If $f$ is everything but onto". One way I can think of is rephrasing it as something like (this is also like an elaboration of the last sentence in the original text):

If $f$ is everything but onto, and the bijection $\hat{f}:X\rightarrow f[X]$ that has the same graph as $f$, where $f[X]$ is the image of $X$ under $f$ equipped with the subspace topology induced by $Y$, is a homeomorphism, we call it ...

but this seems adding too much additional stuff. I wonder if there is any more straightforward and natural way of fixing/interpreting the original text to justify the way it's written, or if the original text is indeed a little sloppy here.


问:
Show that $f$ is a modular form for the congruence subgroup $\Gamma_0(N)$

Let $f\in M_k(\text{SL2}(\mathbb{Z}))$ and
$$
f_N(\tau) :=\sum_{r=0}^{N-1}{f(\frac{\tau +r}{N})}
$$
then $f\in M_k(\Gamma_o(N))$

To check that $f\in M_k(\Gamma_o(N))$ I have to check that

$$
f_{\vert k \gamma} = f \; \text{for all }\gamma \in \Gamma_o(N)
$$

and $$f_{\vert k \alpha^{-1}} = \sum_{n=0}^{\infty}{a_n \exp{(\frac{2\pi i n}{N}})} \text{ for all }\gamma \in \text{SL2}(\mathbb{Z})\; .$$

By definition of $f_N$ it's N-periodic and therefore has a fourier series of the form

$$
\sum_{n=0}^{\infty}{c_n\exp(\frac{2\pi i n \tau}{N})}
$$

I had a hard time checking the conditions above as "simply" plugging in $\tau = \gamma \tau$ and using the transformation for modular forms of weight $k$ didn't help. Whats the best way to check both conditions ?

Would appreciate any help/hint

答:
$$U_N f=\frac1N \sum_{r=0}^{N-1}{f(\frac{z+r}{N})} = \sum_n a_{nN}(f) e^{2i\pi nz}, \qquad U_N = \prod_{p^k \| N} U_p^k$$

For $f\in M_k(\Gamma_0(m))$ $$ U_p f= T_p f-1_{p\ \nmid\ m} p^{k-1}f(pz)$$
where $T_p$ is the Hecke operator $M_k(\Gamma_0(m))\to M_k(\Gamma_0(m))$,

$$T_p f\in M_k(\Gamma_0(m)),\qquad f(pz) \in M_k(\Gamma_0(pm))$$ so that $U_p f\in  M_k(\Gamma_0(pm))$ and $$U_N f\in M_k(\Gamma_0(Nm))$$
In fact $U_N f\in M_k(\Gamma_0(m\prod_{p| N} p))$


问:
Does $M \oplus N \cong M$ imply that $N = 0$ for modules $M,N$?
Let $R$ be a commutative ring and $M,N$ be $R$-modules. Does $M \oplus N \cong M$ imply that $N = 0$?

I would think that  $M \oplus N \cong M$ implies that $(M \oplus N)/M \cong M/M = 0$, but is that true? And if yes, how would the first isomorphism induce the second one? Sorry if this question has been answered before, I could not find an answer.

Edit: And what if $M$ is assumed to be finitely generated?

答:
Nope, that is in general not true. Not even if both of them are free and over a field. Consider $M=\mathbb{R}^\mathbb{N}=N$.

Added:
In case you also assume that $M$ is finitely generated, then it becomes true if you have no torsion and $R$ is an integral ring. We can define $n(M)=:L$ to be the maximal cardinality of a linearly indepedent (over $R$) family in $M$ ($L$ is finite as $M$ is finitely generated and $R$ an integral ring). One readily checks that this number is preserved under isomorphism. However, if $N$ was different from $\{0\}$, then we can take $((x_1,0), \dots, (x_L,0), (0,s))$ with $s\neq 0$. As we have no torsion, we get that $n(M)=n(M\oplus N)\geq n(M)+1$, which gives you a contradiction.


问:
A function on the unit circle that is not homotopic to the identity must be a reflection at some point
I have a problem in the theory of homotopy which I have just started learning. The question states:

If we have a function
$$
f:\mathbb{S}^1\rightarrow \mathbb{S}^1 \;\;\; \text{s.t} \;\;\;\text{$f$ is not homotopic to the identity map}
$$
then at some point $x \in \mathbb{S}^1$, $f(x)=-x$.

To try and prove this I tried to use the contrapositive and tried to show that if $f(x)\neq -x \;\;\forall x \in \mathbb{S}^1 $ then $f$ must be homotopic to the identity. However I constructed the homotopy
$$
H(x,t) = x+tf(x)-tx
$$
which seems to be continuos and satisfy the necessary conditions, but I cannot see where to use the assumption, or how this homotopy breaks down in the case $f(x)=-x$.

Any hints or answers appreciated, thanks!

答:
Okay, so using the hint provided by @Zarrax, I think I have a solution. The problem with my homotopy was that its image was not actually a subset of $\mathbb{S}^1$. In other words, for H to be a homotopy, its norm should always be 1 so that for every $(x,t)$ we are still in the unit circle. I then altered by homotopy by normalising it:

$$
H(x,t) = \frac{x+tf(x)-tx}{||x+tf(x)-tx||}
$$
Now the vector has norm 1 at all times but we still have:

$$
H(x,0) = \frac{x}{||x||} = x\;\;\;\;\;H(x,1) = \frac{f(x)}{||f(x)||}=f(x)
$$

However to show that this is a homotopy, we need to show it in continuous and thus that the denominator is never 0:

Suppose that it is, then $x-tf(x)-tx=0$. This can be rearranged to give
$$
\frac{t-1}{t} = \frac{f(x)}{x}
$$
By taking norms we can see,
$$
\frac{|t-1|}{|t|}=\frac{|f(x)|}{|x|}=1
$$

We then have to consider two cases:

$\frac{t-1}{t}=1$ which is of course impossible

$\frac{t-1}{t} =-1$ which then implies $\frac{f(x)}{x}$=-1 and thus $f(x)=-x$ which we assumed to be false.

As a result, we have found a homotopy and proved the contrapositive.


问:
Simplify $\log_23\ \log_34\ \log_4 5\ \log_5 6\ \log_6 7\ \log_7 8$
How do I evaluate the product:

$$\log_23\ \log_34\ \log_4 5\ \log_5 6\ \log_6 7\ \log_7 8$$

I know that $$\log_ba=\frac{\log\ a}{\log\ b}$$

How can I apply it?

Thanks!

答:
$log_23\ log_34\ log_4 5\ log_5 6\ log_6 7\ log_7 8$
= $\frac{log3}{log2} \times \frac{log4}{log3} \times \frac{log5}{log4} \times \frac{log6}{log5} \times \frac{log7}{log6} \times \frac{log8}{log7}$

log 4 = 2log 2
log 8 = 3log 2
By putting these values and simplifying you get : 3 as answer.


答: None. For example, let $M = \oplus_{n\in\mathbb N}\mathbb Z$ and $N = \mathbb Z$.

Title: Does $M \oplus N \cong M$ imply that $N = 0$ for modules $M,N$?

Body:
Let $R$ be a commutative ring and $M,N$ be $R$-modules. Does $M \oplus N \cong M$ imply that $N = 0$?

I would think that  $M \oplus N \cong M$ implies that $(M \oplus N)/M \cong M/M = 0$, but is that true? And if yes, how would the first isomorphism induce the second one? Sorry if this question has been answered before, I could not find an answer.

Edit: And what if $M$ is assumed to be finitely generated?


问:
How do I prove that $A^{T} A$ is not invertible when n<m?
$A$ is an $ \times $ matrix.
Prove that if $n < m$ then $A^{T}A$ is not invertible.

I know that when $Ax=0$ does not have trivial solution the matrix is not invertible.

So by my understanding I need to show that $x$ has a non trivial solution.

I have trouble proving this with out using dimensions and ranks since I have not yet learned those  in detail.

答:
Use the fact that $C(AB) \subset C(A)$ or $R(AB) \subset R(A)$. (C is column space and R is row space)

So $R(A^TA) \subset R(A)$. This means that rank of $A^TA$ is at most rank of $A$. But rank of $A$ is at most $min\{m,n\}=n$.

However, $A^TA$ is $m \times m$ matrix and rank is at most $n$ which is strictly less than $m$. Thus $A^TA$ cannot be invertible.


问:
Gagliardo–Nirenberg–Sobolev inequality notations
I was looking up the Gagliardo–Nirenberg–Sobolev inequality. In its formulation

\begin{equation}
\|u\|_{L^{p^*}(\mathbf{R}^n)}\leq C \|Du\|_{L^{p}(\mathbf{R}^n)}.
\end{equation}

I didn't really understand what $Du$ was. Is it the gradient of $u$? Just a partial derivative? Something else?
And what is the $L^p$ norm $\lVert Du\rVert$ for this vector valued function?

答:
$Du$ denotes the transposed gradient. The norm is usually given by
$$
\lVert Du \rVert_{L^p(\mathbb{R}^n)} = \left(\sum_{j = 1}^n \int_{\mathbb{R}^n}  \lvert \partial_{x_j} u(x) \rvert^p ~\mathrm{d}x \right)^\frac{1}{p}
$$
or
$$
\lVert Du \rVert_{L^p(\mathbb{R}^n)} = \sum_{j = 1}^n \left(\int_{\mathbb{R}^n} \lvert \partial_{x_j} u(x) \rvert^p ~\mathrm{d}x \right)^\frac{1}{p}.
$$
Both are equivalent norms. They are known as sobolev norms.

This notation does, admittedly, take some time getting used to but it is very common in functional analysis and PDE theory.


问:
Conceptual doubt in taking elements for integration
In physics, we were taught to find the center of mass using calculus for different shapes. I tried to find the volume of a sphere using similar concepts.

I took a cylindrical element to integrate whose radius is $R\sin \theta$ where $\theta$ is the angle made with the vertical to the edge of the element . And took the height to be $Rd\theta$ when I integrated all the cylindrical elements varying theta from o,$\pi$ and did not get the right answer.

I realize the error is taking $Rd\theta$ as height but why is it that we can take the "extension" to be $rd\theta$ when finding the surface area of the sphere.

To state my question more clearly:

Why cant the height of the element be $Rd\theta$ when finding volume since it works when finding surface area.

What I did to find volume:

$$\int_0^{\pi} \pi (R\sin \theta)^2 (R d\theta)$$

What I did to find surface area:

$$\int_0^{\pi}2\pi R\sin \theta(Rd\theta)$$

答:
The difference in the method of integration is that in the case of area,
you need to measure a slant distance along the surface of the sphere
from one circular edge of the area element to the other,
whereas in the case of volume you need to measure the perpendicular distance from one circular face of the cylinder to the other.

The differential $R \,\mathrm d\theta$ accurately represents the slant distance along the surface of the sphere, which is perfect for measuring the surface area, but the surface of the sphere is not perpendicular to the circular faces of the cylindrical element
(except when $\theta = \frac\pi2$),
and therefore $R \,\mathrm d\theta$ is not generally an accurate representation of the height of the cylindrical element.

The plane that cuts the sphere at an angle $\theta$ from the axis is at a distance $R \cos(\theta)$ from the center.
If you increase the angle to $\theta + \Delta\theta$ then the distance of the plane from the center is $R \cos(\theta + \Delta\theta)$.
The cylindrical element for the change in angle $\Delta\theta$ is between these two planes, so the height of the cylindrical element is
$\lvert R \cos(\theta + \Delta\theta) - R \cos(\theta)\rvert$.

To use this in an integral, take the limit
$$
\lim_{\theta\to 0}
\frac{\lvert R \cos(\theta + \Delta\theta) - R \cos(\theta)\rvert}
{\Delta \theta} = R \sin\theta,
$$
which tells you that the perpendicular distance between circular cross-sections is represented by $R \sin\theta\, \mathrm d\theta.$

So the correct integral for the volume is
$$\int_0^{\pi} \pi (R\sin \theta)^2 (R \sin\theta\, \mathrm d\theta).$$

If you still do not see how to correctly translate the angle $\theta$ into the height of a cylindrical cross-section, try integrating by a variable that translates directly into the height of each cylinder.
Put the sphere at the origin of a Cartesian system of coordinates and integrate along the $z$ axis. The radius of the cylindrical element for any given value of $z$ is $\sqrt{R^2 - z^2}$ and the height of the element is $\mathrm dz,$ so the integral is

$$ \int_{-R}^{R} \sqrt{R^2 - z^2} \mathrm dz. $$

Make the trig substitution $z = -R \cos\theta$ to get an integral in terms of $\theta.$
Note that in this substitution, you will replace $\mathrm dz$
with $R \sin\theta\, \mathrm d\theta.$

Also see the explanation of how the measurement of volume differs from the measurement of area in this answer.


问:
Examples of properties that hold almost everywhere, but such that explicit examples are not known
In measure theory one makes rigorous the concept of something holding "almost everywhere" or "almost surely", meaning the set on which the property fails has measure zero.

I think it is very interesting that there are some properties which hold almost surely for which it is either very difficult to construct an example, an example requires the axiom of choice, or for which no known examples exist.

I was looking to put together a list of some of these properties.
More precisely, I am looking for (interesting) examples of measure spaces $(X,\mathscr{A},\mu)$ and $S\subseteq X$ such that there exists $N \in \mathscr{A}$ with $S^c \subseteq N$ and $\mu(N)=0$, but that it is hard to find/construct an explicit example of an element of $S$.

A good place to start is of course $(\mathbb{R}^n,\mathcal{B}^n,\lambda)$.

Here's one to get us started:

Theorem. Take $X=(0,1]$ with Lebesgue measure. Writing $\frac{p_n}{q_n}$ to be the $n$th continued fraction approximate of $x\in(0,1]$ in reduced terms, we have
  $$
\lim_{n \to \infty} \frac{\log q_n}{n} = \frac{\pi^2}{12 \log 2}
$$
  almost surely.
  Note: The result fails for all rationals.

答:
The Wikipedia article on transcendental numbers lists $15$ different classes.  An example of a 'class' of transcendental numbers is all numbers of the form $\ln(a)$, where $a$ is an algebraic number different from $0$ and $1$.  The moral is that transcendental numbers are very hard to write down.

Then again, pick a random real number, and it will be transcendental.


问:
Solving $x^{\log(x)}=\frac{x^3}{100}$
How do I find the solution to:

$$x^{\log(x)}=\frac{x^3}{100}$$

So I multiplied 100 both sides getting:

$$100x^{\log(x)}=x^3$$

Now what should I do?

答: Hint: Apply log on both side and try to solve</p>

Title: Solving $x^{\log(x)}=\frac{x^3}{100}$


问:
Prove or disprove $\lim_{n \to +\infty}\mathbb{P}(X_n \in B) = \mathbb{P}(X\in B)$
Let $\{X_n : n\geq 1\}$ and X be random variables om some probability space with probability density $f_n$ and $f$ respectively.

If $f_n \to f$ almost everywhere then for every $B \in \mathscr{B}(\mathbb{R})$ ( $\mathscr{B}$ for Borel set), prove or disprove $\lim_{n \to +\infty}\mathbb{P}(X_n \in B) = \mathbb{P}(X \in B)$.

I appreciate it if you can tell me if the statement holds so that I can work on a proof or a counterexample.

Thanks

答: This statement is true. If you want an approach, I leave one in a comment.</p>


问:
$Tx=(\lambda_k x_k)_{k=1}^\infty.$ Well defined, bounded and compact.

Fix a sequence of real numbers $(\lambda_k )_{k=1}^\infty$, and define the linear operator $T: \ell_2 \to \ell_2$ by $$Tx=(\lambda_k x_k)_{k=1}^\infty.$$ For what multiplier sequences $(\lambda_k)$ is the operator $T$

well defined?

bounded?

compact?

In general if the sequence $(\lambda_k)$ is not bounded, then our linear operator $T$ is not bounded. Assume $(\lambda_k)$ is bounded. Then
$$\left\| Tx \right\|^2 =
\left\|\sum_{k=1}^\infty \lambda_k x_k \right\|^2 \leq
\sum_{k=1}^\infty \left\|\lambda_k \right\|^2 \left\|x_k\right\|^2 \leq
\left\|\lambda \right\|^2 \left\|x\right\|^2$$ and so the linear operator is bounded with $\left\|T\right\|=\left\|\lambda \right\|_{l_2}$.

For compactness,
$$\left\| Tx-T_n x \right\|^2= \sum_{k=n+1}^\infty \left\|\lambda_k \right\|^2 \left\|x\right\|^2$$ and so
$$\left\| T-T_n  \right\| \leq \sqrt{\left \|{\lambda_k} \right\|^2} \to 0$$ iff $\lambda_k \to 0$.

Are my arguments clear enough for boundedness and compactness? In any case, are my solutions getting close to what the task asks?

答:
The operator is well-defined and bounded if and only if the sequence $(\lambda_k)$ is bounded. Clearly if it is bounded with bound $M$ then for a sequence $x$ we have $||Tx|| \leq M ||x ||$ so it is well-defined and bounded.

If $(\lambda_k)$ is not bounded, then for all $n \in \mathbb{N}$ there is a $k_n$ such that $\lambda_{k_n} > n$. Note that we can choose the $k_n$ to be increasing. Let $x$ be the sequence that is $1/n$ at $k_n$ and $0$ elsewhere. Clearly, $x \in l_2$. However, $Tx \notin l_2$.

I don't know about compact though.


答:
Hint: Take the log of both sides. You will get a quadratic equation in $\log x$.
 The equation is even "nice."

Title: Solving $x^{\log(x)}=\frac{x^3}{100}$

Body:
How do I find the solution to:

$$x^{\log(x)}=\frac{x^3}{100}$$

So I multiplied 100 both sides getting:

$$100x^{\log(x)}=x^3$$

Now what should I do?


问:
Multiplication of two polynomials
Let us say we are given two polynomials $p=\sum_{k=0}^{m}p_kA^k$ and $q=\sum_{j}^{l}q_jA^j$ where we say that for $m<l$, we set $p_{m+1}=...=p_{m+l}=0$. The same goes for $l>m$.

The equation goes as follows:
$$\sum_{k=0}^{m}p_kA^k\sum_{j}^{l}q_jA^j=\sum_{k=0}^{m}\sum_{j=0}^{l}p_kq_jA^{k+j}\overset{(1)}{=}\sum_{k=0}^{m+l}\sum_{j+l=k}p_kq_{j}A^k\overset{(2)}{=}\sum_{k=0}^{m+l}\sum_{j=0}^{k}p_jq_{k-j}A^k$$

I really do not get the $(1)$ and $(2)$ equality, or rather I do not know how to get there and whenever I tried something I do not even get to the $(1)$ .
Any help is appreciated!

答:
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^{m}p_kA^k\sum_{j=0}^{l}q_jA^j}
&=\sum_{k=0}^{m}\sum_{j=0}^{l}p_kq_jA^{k+j}\\
&=\sum_{{0\leq k\leq m}\atop{0\leq j\leq l}}p_kq_jA^{k+j}\tag{1}\\
&=\sum_{n=0}^{m+l}\left(\sum_{{{0\leq k\leq m}\atop {0\leq j \leq  l}}\atop {k+j=n}}p_kq_j\right)A^{n}\tag{2}\\
&\,\,\color{blue}{=\sum_{n=0}^{m+l}\left(\sum_{k=0}^{n}p_kq_{n-k}\right)A^{n}}\tag{3}\\
\end{align*}

Comment:

In (1) we use another convenient notation to write the double sum.

In (2) we rearrange the summands by increasing powers of $A^n$. Note the exponent $n$ of $A^n$ is coupled with the indices $k,j$ by explicitly writing $k+j=n$.

In (3) we eliminate the index $j$ by using $k+j=n$. Since $p_k=0$ if $n>m$ and $q_j=0$ if $n>l$ the representation of the inner sum in (3) is fine.

Title: Multiplication of two polynomials


问:
Let $z\bar z -(3-4i)z-(3+4i)\bar z + 21=0$. If M and m are max and min values of $\frac{z-\bar z}{i(z+\bar z)}$, then find $\frac 1M +\frac 1m$
Let $z=x+iy$

Solving the above equation we get

$$x^2 +y^2-6x-8y+21=0$$
Which indicates a circle centered at $(6,8)$ of radius $2$

Also the require value of $|\frac yx|$

How do I find the max and min of this?

答:
Note that $x^2 +y^2-6x-8y+21=0$ represents a circle of center $(3,4)$ and radius $2$. Parametrize the circle with $x-3=2\cos t$ and $y-4=2\sin t$. Then
$$\frac{z-\bar z}{i(z+\bar z)}=\frac yx = \frac{4+2\sin t}{3+2\cos t}=f(t)\tag1$$

Set $f’(t) =0 $ to get $4\sin t+3\cos t+2=0$, leading to $\tan\frac t2 = 4\pm \sqrt{21}$. Substitute into $f(t)$ in (1) expressed in the form

$$f(t) = \frac{4(1+\tan\frac t2 +\tan^2\frac t2)}{5+ \tan^2\frac t2}
$$
and obtain
$$f_{max} =M= \frac25 (6+\sqrt{21}),\>\>\>\>\>
f_{min}=m = \frac25 (6-\sqrt{21})$$
Thus,

$$\frac1M+\frac1m= 2
$$


问:
Related rates problem, swimming pool
I am having issues getting the correct answer for this question. I'll provide my attempted solution, as well as the book's answer.

Question

A swimming pool is 25 ft wide, 40 ft long, 3 ft deep at one end and 9 ft deep at the other, the bottom being an inclined plane. If water is pumped into the pool at the rate of 10 ft$^3$/min, how fast is the water level rising when it is 4 ft deep at the deep end?

Solution (Attempt)

Around the 4 ft mark the volume is given by a triangular prism, i.e.
$$
V = \frac{1}{2}(25xy)
$$
where $y$ is the depth of the water, and $x$ is the length as in the picture. Thus
$$
\frac{dV}{dt} = \frac{25}{2}\left(y\frac{dx}{dt} + x\frac{dy}{dt}\right)
$$
Furthermore, by similar triangles we get that
$$
\frac{y}{x} = \frac{9}{40}
$$
that is
$$
x = \frac{40}{9}y
$$
so that
$$
\frac{dx}{dt} = \frac{40}{9}\frac{dy}{dt}
$$
Also by similar triangles we get that when $y=4$ then $x=\frac{160}{9}$.

Inserting the values for $x$ and $y$, as well as making use of the equation for $\frac{dx}{dt}$ into the equation for the rate of change for the volume, and using the fact that the change in volume should be 10, we get
$$
10 = \frac{25}{2}\left(4\frac{40}{9}\frac{dy}{dt} + \frac{160}{9}\frac{dy}{dt}\right)
$$
i.e.
$$
\frac{dy}{dt} = \frac{9}{400}
$$

However, the answer in the back of the book reads $\frac{3}{200}$.

答: I incorrectly thought the left side was 9 ft, and not 6 ft. Changing this solves the problem.</p>

Title: Related rates problem, swimming pool


问:
Sampling inspection - Joint distribution
I am self-learning probability theory from William Feller's Introduction to Probability theory and its applications. I would like to ask for some help in deriving the correct solution to the below very interesting problem.

Problem IX.12

Suppose that items with a probability $p$ of being acceptable are subject to inspection in such a way, that the probability of an item being inspected is $p'$. We have four classes, namely, "acceptable and inspected", "acceptable but not inspected" and so forth with probabilities $pp'$, $pq'$, $p'q$ and $qq'$. We are concerned with double Bernoulli trials. Let $N$ be the number of items passing the inspection desk (both inspected and uninspected) before the first defective is found, and let $K$ be the (undiscovered) number of defectives among them.  Find the joint distributions of $N$ and $K$, and the marginal distributions.

Solution (My Attempt).

We have,

\begin{array}{c|cc}
& \text{Acceptable} & \text{Defective}\\
\hline
\text{Inspected} & pp' & qp'\\
\text{Undiscovered} & pq' & qq'\\
\end{array}

The first defective item is found at trial number $(n+1)$, if the preceding $n$ items were either acceptable or uninspected.

$P\{\text{Acceptable} \cup \text{Uninspected}\} = P\{\text{Defective},\text{Inspected}\}^C = 1 - qp'$

$N$ is the waiting time to the first defective. $N$ follows a geometric distribution.

$P\{N = n\} = (1 - qp')^n qp' \tag{1}$

$K$ is the number of undiscovered defectives among these $n$ trials. So, given that we waited for time $n$ to find the first defective, the probability that the number of defectives equals $k$ is given by,

$P\{K = k \vert N = n\} = {n \choose k} (qq')^k p^{n-k}$

Thus, the joint distribution

\begin{align*}
P\{K = k, N = N\} &= P \{K = k \vert N =n \} \cdot P \{N = n\}\\
&= {n \choose k} (qq')^k p^{n-k} (1 - qp')^n qp'
\end{align*}

However, the textbook states the expression for the joint distribution as,

$$
P\{K = k, N = n\} ={n \choose k}(qq')^k p^{n-k} qp'
$$

Also, how to sum over all $n$ to derive an expression for the marginal distribution of $N$?

答:
For the joint distribution note that

$$P(n,k)=P(\text{k defects over n trials})P(\text{k not detected})P(\text{defect detected at n+1})$$

$$={n \choose k}p^{n-k}q^k \times (q')^k\times q p' $$

$$={n \choose k}p^{n-k}\times (qq')^k\times q p' $$


答:
There are no solutions in the real numbers.

Edit: If, as RecklessReckoner suggests, the question was misstated and the intent was to use $\log_{10},$ then the solution can be found easily by taking logarithms.

Title: Solving $x^{\log(x)}=\frac{x^3}{100}$


问:
Is this a valid proof using Pigeon Hole Principle?
Pigeon Hole Question from last semester

S = { 1, 2, 3, . . . 100 }

examples of sets that add to 3: {1 , 2}

examples of sets that add to 4: {1 , 3}

examples of sets that add to 5: {3 , 2}, {4 , 1}

examples of sets that add to 6: {4 , 2}, {5 , 1}

examples of sets that add to 7: {4 , 3}, {5 , 2}, {6 , 1}

examples of sets that add to 8: {2 , 6}, {5 , 3}, {7 , 1}

examples of sets that add to 9: {1 , 8}, {2 , 7}, {5 , 4}, {3 , 6}

There are 10 pigeons and 9 holes. Therefore by PHP there are 2 elements in a subset.

From the above eamples we see that there exists two or more subsets that can be made

which add to the same number and as the sum increases so does the possible subsets.

Body:
No need for PHP. Just note that
$$1+3+4+5+6+7+8+9+10+12 = 2+3+4+5+6+7+8+9+10+11$$


问:
Reverse statement of Kato-Rellich Theorem
The Kato-Rellich Theorem states: If $T$ is a self-adjoint operator in the hilbert space $\mathcal{H}$  and if $A$ is symmetric and $T-\text{bounded}$ with $T$-bound $< 1$. Then $T+A$ is self-adjoint. I´m wondering if the following is true:

$$\text{If}~~T~~\text{is s.a. in}~~\mathcal{H}~~\text{and if}~~A~~\text{is symmetric and if it holds that}~~\overline{T+A} = \left(T+A\right)^{*},~~\text{then}~~A~~\text{is}~~T-\text{bounded}.$$

答: A counterexample is the quartic quantum anharmonic oscillator $H+V$ with $V=x^4$. On the Hermite function basis $\{e_n\}$, $He_n = ne_n$ while $\vert Ve_n \vert \gt$ const.$\times n^2$. The latter is obvious from the creation/annihilation representation. There are various proofs of e.s.a. for $H+V$.


答: Hint:  $log_23=\frac{log3}{log2}, log_34=\frac{log4}{log3}$ and so on, find anything fishy?</p>

Title: Simplify $\log_23\ \log_34\ \log_4 5\ \log_5 6\ \log_6 7\ \log_7 8$

Body:
How do I evaluate the product:

$$\log_23\ \log_34\ \log_4 5\ \log_5 6\ \log_6 7\ \log_7 8$$

I know that $$\log_ba=\frac{\log\ a}{\log\ b}$$

How can I apply it?

Thanks!


问:
Existence of epimorphism from a Ring to a Module
So the problem is the following, given a conmutative ring $R$ and a free $R$-module let's say $M$, such that $\dim(M)=2$.

Prove there is no epimorphism from $R$ to $M$

I've been struggling for a while to solve this problem. I believe the solution has to with what elements in the ring are sent to the basis of $M$ but I literally have no clue of where to start.

Any help would be appreciated

答:
Suppose that $f:R \to R \oplus R$ is a surjective $R$-module homomorphism.

Then, let $(a_1,a_2)$ be $f(1)$. By surjectivity, that there must exist elements $b_1$ and $b_2$ of $R$ for which $f(b_1)=(b_1a_1,b_1a_2)=(1,0)$ and $f(b_2)=(b_2a_1,b_2a_2)=(0,1)$.

Using commutativity of $R$, one then gets that $b_2a_1b_1=0b_1=0$, $b_2a_1b_1=b_2b_1a_1=b_21=b_2$, $b_1a_2b_2=0b_2=0$, and $b_1a_2b_2=b_1b_2a_2=b_11=b_1$, so $b_1=b_2=0$. But then, that would mean that $(1,0)=(0,1)=(0,0)$, so $R$ is the zero ring.


问:
Closed form of the sum $\sum_{n=1}^{\infty}\frac{H_n}{n^x}$
Some days ago I derived the identity
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2}=2\zeta(3)$$
where $H_n$ is the $n$th Harmonic number. Other related identities include
$$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac{\pi^4}{72}$$
$$\sum_{n=1}^{\infty}\frac{H_n}{n^4}=\frac{-1}{6}\pi^2\zeta(3)+3\zeta(5)$$
$$\sum_{n=1}^{\infty}\frac{H_n}{n^5}=\frac{1}{540}(\pi^6-270\zeta(3)^2)$$
(I proved all except the last one) Now I wondered if there is a closed form for the generalized form
$$\sum_{n=1}^{\infty}\frac{H_n}{n^x}$$
For some real number $x$. Here is my try: The well known identity $\psi(n+1)=H_n-\gamma$ gives
$$\sum_{n=1}^{\infty}\frac{H_n}{n^x}=\sum_{n=1}^{\infty}\frac{\psi(n+1)}{n^x}-\gamma\zeta(x)$$
But I don't know what to do further. Any help would be appreciated.

答:
Long Comment:

The sum can be written in terms of unsigned Stirling numbers of the first kind,  $s_{n}^{(k)}=\left[_{k}^{n} \right]$

$$\sum_{n=1}^{\infty}\frac{H_n \, x^n}{n^{\alpha}}=\sum _{n=1}^{\infty } \frac{ s_{n+1}^{(2)} x^n}{n!\; n^{\alpha}}$$

Since $H_n=\left[_{\,\,\,2}^{n+1} \right]\frac{1}{n!}$

The well known recurrence relation for unsigned Stirling numbers of the first kind is

$$\left[_{\,\,\,k}^{n+1} \right]=n\left[_{k}^{n} \right] +\left[_{k-1}^{\,\,n} \right]$$

which immediately leads to

$$\sum_{n=1}^{\infty}\frac{H_n \, x^n}{n^{\alpha}}=\sum _{n=1}^{\infty } \frac{ s_{n}^{(2)} x^n}{n!\; n^{{\alpha}-1}}+\sum _{n=1}^{\infty } \frac{  x^n}{ n^{{\alpha}+1}}\tag{1}$$

which can (according to Mathematica) be rewritten simply as (with $x=1$)

$$\sum_{n=1}^{\infty}\frac{H_n}{n^{\alpha}}=S_{{\alpha}-1,2}(1)+\text{Li}_{{\alpha}+1}(1)=S_{{\alpha}-1,2}(1)+\zeta({\alpha}+1)$$
where $S_{n,p}(z)=\frac{(-1)^{n+p-1}}{(n-1)! p!}\int _0^1\frac{ t \log ^{n-1} \left(\log ^p (1-t z)\right)}{t} \,dt$ is the Nielsen generalized polylogarithm function and  $\text{Li}_n$ is the polylogarithm function.

[Update: The full infinite series form for the Nielsen generalized polylogarithm according to Mathematica is

$$S_{n,p}(z)=\sum _{k=1}^{\infty } \frac{  s_k^{(p)}}{k! \,k^n}z^k$$

However I cannot find this result online.]

The Nielsen generalized polylogarithm function is utilised in the study of quantum electrodynamics according to information available online, so its properties have been quite well studied apparently. I didn't have time to look further, but hopefully this gives one potential starting point in your quest for an answer.


问:
Proving $3\times3$ nilpotent matrix satisfies $A^3=0$
Let $A$ be a $3$ $\times$ $3$ matrix. Prove that if there exists $n$ $>$ $3$ such that $A^n$= $0$, then $A^3$ = $0$. Can someone help explain how to prove this? I know that since A is nilpotent, its only eigenvalue ($\lambda$) is $0$. So, I was thinking that I could argue the following: $A^3$$\dot{\vec{v}}$=$\lambda^3$$\dot{\vec{v}}$=$\dot{\vec{0}}$, which implies that $A^3$=$0$. I don't think this proof is fully correct, however, because it doesn't use the fact that $n>3$. I could have made the same argument to suggest that $A^2=0$, which is not necessarily true when a $3$ $\times$ $3$ matrix A is nilpotent.

答: Because $p(A)=0$ where $p(\lambda)=\lambda^3$, then the minimal polynomial $m(\lambda)$ must divide $p$, forcing $m(\lambda)=\lambda^n$ for either $n=0,1,2$ or $3$. The conclusion is that $A$ is nilpotent or $0$.


问:
Ito process that is a submartingale
Could anyone give me an example of an Ito process that is submartingale? Thanks a lot</p>

答:
How about $B_t^2$, with $B$ a Brownian motion.

More generally, for an Ito process like $X_t=X_0+\int_0^t H_s\,ds+\int_0^t K_s\,dB_s$, if $H_s\ge 0$ a.s. for all $s$, then $X$ is a submartingale.

Title: Ito process that is a submartingale


答:
I suppose $\log$ means $\log_{10}$? I'm not familiar with this sort of notation.
Take logarithm on both sides, and you will get $2+\log^2x=3\log x$. Substitute $\log x$ with t. And you get $t^2-3t+2=0$, therefore $(t-1)(t-2)=0$. That should do it.</p>

Title: Solving $x^{\log(x)}=\frac{x^3}{100}$


答:
The subsets of size $10$ have a sum that ranges from $55$ to $955$, and presumably all sums are possible, hence $901$ distinct values.

But there are $\displaystyle\binom{100}{10}$ ways to form the sums, which is a much larger number.

Title: Is this a valid proof using Pigeon Hole Principle?

Body:
Pigeon Hole Question from last semester

S = { 1, 2, 3, . . . 100 }

examples of sets that add to 3: {1 , 2}

examples of sets that add to 4: {1 , 3}

examples of sets that add to 5: {3 , 2}, {4 , 1}

examples of sets that add to 6: {4 , 2}, {5 , 1}

examples of sets that add to 7: {4 , 3}, {5 , 2}, {6 , 1}

examples of sets that add to 8: {2 , 6}, {5 , 3}, {7 , 1}

examples of sets that add to 9: {1 , 8}, {2 , 7}, {5 , 4}, {3 , 6}

There are 10 pigeons and 9 holes. Therefore by PHP there are 2 elements in a subset.

From the above eamples we see that there exists two or more subsets that can be made

which add to the same number and as the sum increases so does the possible subsets.


问:
Writing complex numbers in algebraic form
Could someone give me one of these tasks as a "recipe"? I'm lost here.....

Body:
COMMENT.-Note that $i^2=-1,i^3=-i$ and $i^4=1$
$$z_1=\frac{(3+2i)(1+3i)}{13}+\frac{2i(3-i)}{10}\\
z_2=1-5(2i)+10(2i)^2-10(2i)^4+5(2i)^4-(2i)^5\\$$
For $z_3$ you have$$\frac{3-i^{99}}{i}=\frac{3i-i^{100}}{-1}\text{ and }\frac{2}{3+i^9}=\frac{2}{3+i}$$


问:
$\sigma$-weak topology versus $\sigma$-strong topology on the bounded operators $B(H)$.
Let $H$ be a Hilbert space and let $T(H)$ be the traces class operators on $H$. Denote the bounded operators on $H$ by $B(H)$. We define the normal functionals on $B(H)$ by
$$B(H)_*:= \{\omega \in B(H)^*: \exists x \in T(H): \forall y \in B(H): \omega( y) = \text{Tr}(xy)\}$$
Let $B(H)_*^+$ denote the positive normal functionals.

We can introduce the $\sigma$-weak and the $\sigma$-strong topology on the space $B(H)$. The former is the locally convex topology generated by the family of seminorms
$$ B(H)\ni x \mapsto |\omega(x)|, \  \omega\in B(H)_*$$
The latter is the locally convex topology generated by the family of seminorms
$$B(H)\ni x \mapsto  \omega(x^*x)^{1/2}, \ \omega \in B(H)_*^+$$

I want to show that the $\sigma$-weak topology is weaker than the $\sigma$-strong topology (as the name suggests!). More precisely, suppose that $x_\alpha \to x$ in the $\sigma$-strong topology. Then I want to show also that $x_\alpha \to x$ in the $\sigma$-weak topology. How can I show this?

We need to show
$$\forall \omega \in B(H)_*: |\omega(x_\alpha)-\omega(x)| \to 0$$

using something about positive normal functionals but I'm unsure how to approach this.

答:
We for $\omega\in B(H)_*^+$ we have, $$|\omega(x-x_\alpha)|=|\omega(1^*(x-x_\alpha)|\leq \omega(1^* 1)^{1/2} \omega((x-x_\alpha)^*(x-x_\alpha))^{1/2}$$.

by applying Cauchy-Schwarz to the sesquilinear form $(x, y) \mapsto \omega(y^*x)$.

For a general $\omega\in B(H)_*$ we pick $x\in B(H)$ such that $\omega=\text{Tr}(x\cdot)$ and write $x$ as a linear combination of positive elements $x=\sum \alpha_i x_i$. Then $\omega=\sum \alpha_i\text{Tr}(x_i \cdot)$ and it easy to see that the functionals $\text{Tr}(x_i\cdot)$ are positive: write $x_i=z^*z$, then we have $\text{Tr}(x_i(y^*y))=\text{Tr}(z^*zy^*y)=\text{Tr}(yz^*zy^*)=\text{Tr}((zy^*)^*zy^*)\geq 0$ for all $y\in B(H)$. We conclude that $\omega$ is a linear combination of elements in $B(H)_*^+$.


问:
Constructing elements of $\mathbb{Z}_p[\Gamma_n]$ from $\mathbb{Z}_p[\operatorname{Gal}(\mathbb{Q}(\zeta_{p^{n+1}})/\mathbb{Q})]$
Fix a prime $p$ and let $G_n=\operatorname{Gal}(\mathbb{Q}(\zeta_{p^{n+1}})/\mathbb{Q})$. We can identify $G_n$ with $(\mathbb{Z}/p^{n+1}\mathbb{Z})^\times$ by mapping $a$ mod $p^{n+1}$ to $\sigma_a:\zeta_{p^{n+1}} \mapsto\zeta_{p^{n+1}}^a$. There is a (unique, if $p$ is odd) decomposition $G_n=\Delta\times \Gamma_n$, where $\Delta\cong (\mathbb{Z}/p\mathbb{Z})^\times$ and $\Gamma_n\cong\mathbb{Z}/p^n\mathbb{Z}$. One can now define an 'obvious' map $\mathbb{Z}_p[G_{n}]\rightarrow \mathbb{Z}_p[\Gamma_n]$ as follows. Let $\omega :(\mathbb{Z}/p\mathbb{Z})^\times\rightarrow \mathbb{Z}_p^\times$ be the mod $p$ cyclotomic character and $\langle a \rangle$ the projection of $a \in (\mathbb{Z}/p^{n+1}\mathbb{Z})^\times$ onto $\mathbb{Z}/p^n\mathbb{Z}$, so that $\sigma_{\langle a \rangle}\in \Gamma_n$. We then have a map
$$
\omega^i:\mathbb{Z}_p[G_n]\rightarrow \mathbb{Z}_p[\Gamma_n]
$$
given by $\sum_{a \in (\mathbb{Z}/p^{n+1}\mathbb{Z})^\times}c_a\sigma_a\mapsto \sum_{a \in (\mathbb{Z}/p^{n+1}\mathbb{Z})^\times}\omega^i(a)c_a\sigma_{\langle a \rangle}$, where $c_a\in \mathbb{Z}_p$ and we are extending $\omega$ to $(\mathbb{Z}/p^{n+1}\mathbb{Z})^{\times}$ by precomposing with reduction mod $p$.

There is also another way to define an element of $\mathbb{Z}_p[\Gamma_n]$ from $\mathbb{Z}_p[G_n]$. We can view $\mathbb{Z}_p[G_n]$ as a $\mathbb{Z}_p[\Delta]$-module and take the decomposition coming from orthogonal idempotents. More specifically, if
$$
\epsilon_i=\frac{1}{p-1}\sum_{\sigma\in \Delta}\omega^i(\tau)\tau^{-1}
$$
and if $\vartheta\in \mathbb{Z}_p[G_n]$ we get that $\epsilon_i\vartheta$ lies in the $\omega^i$-eigenspace of $\mathbb{Z}_p[G_n]$, i.e, a copy of $\mathbb{Z}_p[\Gamma_n]$ where $\tau\in \Delta$ acts by multiplication by $\omega^i(\tau)$.

My intuition tells me that these two methods of defining an element of $\mathbb{Z}_p[\Gamma_n]$ by 'projection' from  $\mathbb{Z}_p[G_n]$ should be the same, i.e., that
\begin{equation}\tag{1}
\epsilon_i\vartheta=\omega^i(\vartheta),
\end{equation}
but for some reason I'm having a hard time showing this. Is this equation (1) true?

I think this is either easy and I'm getting myself confused or possibly it's just not true... In any case, here are some thoughts: In general, one has $\tau\epsilon_{\chi}=\chi(\tau)\epsilon_{\chi}$ for a character $\chi\in \hat \Delta$. Thus, since we can decompose $\sigma_a=\tau_a\sigma_{\langle a\rangle}$ for some $\tau_a\in \Delta$, we have
$$
\epsilon_i \vartheta= \epsilon_i \sum_{a\in (\mathbb{Z}/p^{n+1}\mathbb{Z})^\times}\omega^i(\tau_a)c_a \sigma_{\langle a \rangle}=\frac{1}{p-1}\sum_{\tau\in \Delta}\sum_{a\in (\mathbb{Z}/p^{n+1}\mathbb{Z})^\times}\omega^i(\tau\tau_a)c_a \sigma_{\langle a \rangle}\tau^{-1}.
$$
After staring at this for far too long and fiddling with reindexing, I am still not sure why this should be equal to $\omega^i(\vartheta)=\sum_{a \in (\mathbb{Z}/p^{n+1}\mathbb{Z})^\times}\omega^i(a)c_a\sigma_{\langle a \rangle}$...

答:
For $p$ odd prime with $g$ of order $p-1$ and $1+p$ of order $p^{n-1}$ the generators of $G_n=\Bbb{Z/p^nZ}^\times$. Let $r(p-1)=1\bmod p^{n-1}$ so that $\sigma\to \sigma^{r(p-1)}$ is the natural projection $G_n\to G_n^{p-1}=\Gamma_n$.

Let
$$e_j = \frac1{p-1}\sum_{k=1}^{p-1} g^k \zeta_{p-1}^{kj}\in \Bbb{Z}_p[G_n], \qquad e_j^2 = e_j, \qquad \omega(g^m (1+p)^r) = \zeta_{p-1}^m\in \Bbb{Z}_p$$

Then for $\sigma\in G_n$ $$e_j \sigma  =e_j\sigma^{r(p-1)}\omega(\sigma)^{-j} \in \Bbb{Z}_p[G_n]$$

The map (non-unital ring endomorphism of $\Bbb{Z}_p[G_n]$) $$\sigma \to e_j \sigma^{r(p-1)}\omega(\sigma)^{-j} $$
factors through the ring homomorphism $$\sigma\to \sigma^{r(p-1)}\omega(\sigma)^{-j} ,\qquad \Bbb{Z}_p[G_n]\to \Bbb{Z}_p[G_n^{p-1}]$$


问:
When does Busy Beaver surpass TREE(3)?
I saw a question which asked, "When does Busy Beaver surpass TREE(n)?" I am asking a somewhat different question, about a specific value of TREE. I know that TREE(3) is an unimaginably vast number, far more than even Graham's number. I would be satisfied with even a good upper bound when Busy Beaver surpasses TREE(3).</p>

答:
As far as I know, there is no known upper bound for $TREE(3)$. To have at least a slight idea how huge this gap is , you should study the "fast growing hierarchy".

Graham's number has level $\omega+1$.

Conway chains far surpass Graham's number, already $4\to4\to4\to4$ is vastly larger. The level of Conway chains of length $n$ is about $\omega^2$

Bowers arrays far surpass Conway chains. $[3,3,3,3,3]$ is already inexpressible with Conway chains.

Even Bowers planes only reach $\omega^{\omega^2}$-level and with $3$ dimensions $\omega^{\omega^3}$.

This is still far away from the $\epsilon_0$ level and this is FAR surpassed by the $\Gamma_0$ level.

And even this level is FAR too low for $TREE(3)$.

I think you begin to understand that $TREE(3)$ is not just another league , the difference in magnitude is barely comprehensible even in googology-standards.

The busy beaver function however should surpass the TREE - function relatively early, but I have no idea where. I think, it does it below $n=100$.

Title: When does Busy Beaver surpass TREE(3)?


问:
Looking for the name of a Rising/Falling Curve
I'm looking for a particular curve algorithm that is similar to to a bell curve/distribution, but instead of approaching zero at its ends, it stops at its length/limit. You specify the length of the curve of the curve and its maximum peak, and the plot will approach its peak at the midpoint of length(the middle) and then it curves downward to its end. As a math noob, I may not be making any sense. Here's an image of the curve I'm looking for:

答: The curve which you are looking for is a parabola. When I plugged in the equation $$f(x) = -(x-3.9)^{2} + 4$$ I got this figure, which some what resembles what you are looking for.


问:
4 equations with specific set of numbers
I must make 1 addition (x+y=z), 1 subtraction (x-y=z), 1 multiplication (x*y=z), and 1 division (x/y=z) equation with the following numbers. All the numbers must be used to fill x, y, and z of each equation. x, y, and z can consist be 1, 2, or 3-digit numbers.

0-0 available

1-13 available

2-3 available

3-1 available

4-2 available

5-1 available

6-1 avilalble

7-1 available

8-2 available

9-2 available

Thanks!

Body:
Finally got it:

1111*1=1111

89/89=1

12+12=24

76-35=41


问:
Show that $\int_0^\infty {1\over{x^4+1}}\,dx=\int_0^\infty {x^2\over{x^4+1}}\,dx$
Could someone give me a hint how to show $$\int_0^\infty {1\over{x^4+1}}\,dx=\int_0^\infty {x^2\over{x^4+1}}\,dx?$$

I know how to do both integrals separately, but this question leads into another way of evaluating them and requires this to be shown first. As such I want to show the equivalence by manipulating the integral as the question intends rather than evaluating both separately.

I have tried working with both sides and I feel like I'm missing a trick. Using integration by parts increases the power of the denominator and no nice cancellation happens(except an unrelated reduction formula). Cannot see a great substitution either.

答:
Note that by enforcing the substitution $x\mapsto 1/x$, we find

$$\begin{align}
\int_0^\infty \frac{1}{1+x^4}\,dx&\overbrace{=}^{x\mapsto 1/x}\int_\infty^0 \frac1{1+1/x^4}\,\left(-\frac1{x^2}\right)\,dx\\\\&=\int_0^\infty \frac{x^2}{1+x^4}\,dx
\end{align}$$

And we are done!


答:
Basically you want to prove that

$$\int_0 ^\infty \frac{1-x^2}{1+x^4} dx = 0$$

Consider the integral in the $(1,\infty)$ interval. Applying the change of variables $y = 1/x$ we get

$$\int_0 ^1 \frac{1-x^2}{1+x^4}dx - \int_0^1 \frac{1-y^2}{1+y^4} dy = 0$$

which is clearly true.

Title: Show that $\int_0^\infty {1\over{x^4+1}}\,dx=\int_0^\infty {x^2\over{x^4+1}}\,dx$

Body:
Could someone give me a hint how to show $$\int_0^\infty {1\over{x^4+1}}\,dx=\int_0^\infty {x^2\over{x^4+1}}\,dx?$$

I know how to do both integrals separately, but this question leads into another way of evaluating them and requires this to be shown first. As such I want to show the equivalence by manipulating the integral as the question intends rather than evaluating both separately.

I have tried working with both sides and I feel like I'm missing a trick. Using integration by parts increases the power of the denominator and no nice cancellation happens(except an unrelated reduction formula). Cannot see a great substitution either.


问:
Bijection between open and partially closed interval
i need some help finding an explicit bijection between $(0,1)$ and $[0,2]\cup[3,5]$. I know it exists because of Cantor's theorem but im struggling to find it.

I've tried with dividing the problem to use an already known bijection from $[0,2]$ to $(0,\frac{1}{2})$ ,then finding a bijection from $(3,5]$ to $(\frac{1}{2},1)$ while sending 3 into $\frac{1}{2}$.

My last try was the following but it ended up not being onto:

\begin{equation}
S={x|x=\frac{1}{n+2}+\frac{1}{2},n\in N^{+}}\\
f(x)=
\begin{cases}
       5(\frac{1}{n+1}+\frac{1}{2}), & x\in S \\
      4x+1, & x\notin S
\end{cases}
\end{equation}
I'd appreciate some help on this please.

答:
I'll describe the map $[0,2]\cup [3,5]\to (0,1)$ because it may look simpler:
$$ f(x)=\begin{cases}1-\frac1{n+1}=\frac1{2-x}&x=1-\frac1n\\
1+\frac1{n+1}=2-\frac{1}{x}&x=1+\frac1n\\x&\text{otherwise}\end{cases}$$
would biject $[0,2]\to (0,2)$.
$$ f(x)=\begin{cases}4+\frac1{n+1}=5-\frac1{x-3}&x=4+\frac1n
\\x&\text{otherwise}\end{cases}$$
would biject $[3,5]\to [3,5)$. Combine and scale a bit to arrive at
$$
 f(x)=\begin{cases}\frac1{2-x}&x<1, \frac 1{1-x}\in \Bbb N\\
2-\frac{1}{x}&1<x\le 2, \frac 1{x-1}\in\Bbb N\\
5-\frac1{x-3}&x>4,\frac 1{x-4}\in\Bbb N\\
\frac{x-1}4&x\ge 3,(x=4\text{ or }\frac 1{x-4}\notin\Bbb N \\
\frac x4&\text{otherwise}\end{cases}$$


问:
What is $\lim_{n \to \infty} e^{-n} f(n)$?
Let $(a_n)$ be a sequence that converges to $a$. And $f(x) = \sum_{k=0}^{\infty} \frac{a_k x^k}{k!}$

From previous parts, I know that $e^{-x}f(x) -a= e^{-x} \sum_{k=0}^{\infty} \frac{(a_k-a) x^k}{k!}$

Prove that $\lim_{n \to \infty} e^{-n} f(n)$ exists and find its value.

I'm having a lot of trouble with this. For one I'm not sure what the limit is. I thought mabey the limit is 0 but I got stuck:

$|e^{-n}\sum_{k=0}^{\infty} \frac{a_k n^k}{k!} - 0|=|e^{-n}| *|\sum_{k=0}^{\infty} \frac{a_k n^k}{k!}| \leq |e^{-n}| * |A|\sum_{k=0}^{\infty} |n^k/k!|$ where A is bound for $a_n.$ Then  $\leq |e^{-n}|*|A|e^{-n} = |A|$. But that wouldn't be $< \epsilon.$

Then I though maybe it converges to a because from previous part we have $e^{-x}f(x) =a+ e^{-x} \sum_{k=0}^{\infty} \frac{(a_k-a) x^k}{k!}$. Then

$|a+ e^{-n} \sum_{k=0}^{\infty} \frac{(a_k-a) n^k}{k!} -a| = |e^{-n} \sum_{k=0}^{\infty} \frac{(a_k-a) n^k}{k!}| \leq |e^{-n}|*\sum_{k=0}^{\infty} |\frac{(a_k-a) n^k}{k!}| \leq |e^{-n}| * |2A| \sum_{k=0}^{\infty} |\frac{n^k}{k!}| \leq |2A|$. But again not not sure how to make it $< \epsilon.$

答:
For a constant sequence $a_n = a$ we have $f(n)e^{-n} = a$. So one might suspect that
$$
\lim_{n \to \infty} f(n)e^{-n} = a = \lim_{k \to \infty} a_k\, .
$$
holds also in the general case.

That is indeed true. The “trick” is to split the series into two parts, one finite part and an infinite part whose terms are known to be small.

For given $\epsilon > 0$ there is an index $K$ such that $|a_k - a| < \epsilon$ for all $k > K$. Then
$$
 |f(n)e^{-n} - a| \le e^{-n}\left | \sum_{k=0}^K \frac{(a_k-a) n^k}{k!}\right| + \epsilon e^{-n} \sum_{k=K+1}^\infty \frac{n^k}{k!} \le \sum_{k=0}^K \frac{|a_k - a| n^k}{k!e^n} + \epsilon
$$
It follows that
$$
 \limsup_{n \to \infty} |f(n)e^{-n} - a| \le \epsilon
$$
because $e^n$ grows faster than any polynomial. This holds for all $\epsilon > 0$, so that $\limsup_{n \to \infty} |f(n)e^{-n} - a| = 0$, i.e.
$$
\lim_{n \to \infty} f(n)e^{-n} = a \, .
$$


问:
Condition number of a diagonal matrix
Let $\|\cdot\|$ be any norm on $\mathbb{C}^n$. Let $A\in \mathbb{C}^{n\times n}$ We define the matrix norm by $||A||=\max_{||x||=1}||Ax||$. If $A=diag(\lambda_1,...,\lambda_n)$ and it is invertible, then do we always have $$||A||\cdot||A^{-1}||=\frac{\max_i|\lambda_i|}{\min_i|\lambda_i|}?$$</p>

答:
This is not true in general!

Consider the norm $\|x\|_M := \|Mx\|_\infty$ where M is some invertible matrix.

Then $\|A\|_M = \| M A M^{-1}\|_\infty$ so $\kappa(A)_M = \| M A M^{-1}\|_\infty\| M A^{-1} M^{-1}\|_\infty$.

Take

$
D = \begin{pmatrix}1&0\\0&2\end{pmatrix}.
$

Then $\kappa(D)_\infty = 2$ but if

$
M = \begin{pmatrix}1&2\\3&4\end{pmatrix}
$

$\kappa(D)_M = \|\begin{pmatrix}4&-1\\6&-1\end{pmatrix}\|_\infty\|\begin{pmatrix}-0.5&0.5\\-3&2\end{pmatrix}\|_\infty = 35$.

It is true for any p-norm however.

Title: Condition number of a diagonal matrix


问:
Truncated expresion of $e^x$ is an irreducible polynomial
I need to prove that the truncated expression of the Taylor development of the exponential function is irreducible.

$ 1 + \frac{1}{1!}x + \frac{1}{2!}x^2+... + \frac{1}{p!}x^p \in \mathbb{Q}[x]$

where p is prime. Searching for answers I have found some proofs that use Galois theory, but that is very complex for me to understand. I know about Eisenstein's and reduction criterion.

Eisenstein's criterion

Suppose we have the following polynomial with integer coefficients.
$
{\displaystyle Q(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}}
$

$
{\displaystyle Q(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}}
$

If there exists a prime number p such that the following three conditions all apply:

$p$ divides each $a_i$ for $0 ≤ i < n$,

$p$ does not divide $a_n$, and

$p^2$ does not divide $a_0$,

Reduction criterion
(converse of reduction criterion?)

Any help would be appreciated. Thank you :).

答:
Let us denote $$P(x)=1 + \frac{1}{1!}x + \frac{1}{2!}x^2+ \dots + \frac{1}{p!}x^p.$$ Then we have that

$$p!P(x)= p!+ p!x + \frac{p!}{2!}x^2+ \dots + p(p-1)x^{p-2}+px^{p-1}+x^p$$ satisfies the Eisenstein's criterion, from where it follows that $p!P(x)$ is irreducible. And thus $P(x)$ is irreducible.


问:
Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$
Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?

$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$

Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell without touching our calculators?

答:
IMO, the most obvious first thing to do when comparing $\sqrt{999} + \sqrt{1001}$ with $2 \sqrt{1000}$ is to split the right hand side into a sum of two copies of $\sqrt{1000}$ and compare $\sqrt{1000}$ with $\sqrt{999}$ and $\sqrt{1000}$ with $\sqrt{1001}$.

By looking at the shape of the graph of $\sqrt{x}$, we can see that it grows faster with smaller numbers. The difference between $\sqrt{999}$ and $\sqrt{1000}$ should be slightly larger than the difference between $\sqrt{1000}$ and $\sqrt{1001}$. Thus, we get a greater loss from replacing one copy of $\sqrt{1000}$ with $\sqrt{999}$ than the gain we get by replacing the other copy of $\sqrt{1000}$ with $\sqrt{1001}$, and so should expect

$$ \sqrt{999} + \sqrt{1001} < 2\sqrt{1000}$$

Okay, but how can we be sure that our reasoning is actually correct? We may have overlooked something, or our intuition might be flat out wrong! How can we turn this argument into a rigorous one?

Normally, this would be a wonderful job for calculus; the main idea of the proof is very closely related to the fact that the second derivative of $f(x) = \sqrt{x}$ is negative. But since you tagged this precalculus, let's find another way....

There are a number of algebraic tricks for doing this sort of thing without the help of calculus, e.g. by manipulating the problem to eliminate square roots. Ad-hoc methods often work well for simple problems, as in the other answers.

But what if we don't find something ad hoc? Here is something more direct.

In order to understand $\sqrt{1000} - \sqrt{999}$, we can multiply by it its conjugate to simplify (and divide it out so as not to change the overall value)

$$ \begin{align} \sqrt{1000} - \sqrt{999} &= \frac{(\sqrt{1000} - \sqrt{999})(\sqrt{1000} + \sqrt{999})}{\sqrt{1000} + \sqrt{999}}
\\&= \frac{1000 - 999}{\sqrt{1000} + \sqrt{999}}
\\&= \frac{1}{\sqrt{1000} + \sqrt{999}}
\end{align}$$

Multiplying by conjugates is very commonly useful. While we still might not be able to fully understand $\sqrt{1000} + \sqrt{999}$, it is much easier to understand than $\sqrt{1000} - \sqrt{999}$, because its most significant parts combine constructively rather than cancel each other out. This tells us the difference is very nearly

$$ \frac{1}{2 \sqrt{1000}}$$

We'll get a similar result by applying the same method to $\sqrt{1001} - \sqrt{1000}$ (in fact, we can use the exact forms of the result directly, but let's assume we didn't notice that). So we need to go a step further and figure out how much the true difference differs from this estimate:

$$\begin{align}
\left(\sqrt{1000}- \sqrt{999} \right) - \frac{1}{2 \sqrt{1000}}
&= \frac{1}{\sqrt{1000} + \sqrt{999}} - \frac{1}{2 \sqrt{1000}}
\\&= \frac{2 \sqrt{1000} - \sqrt{1000} - \sqrt{999}}{2 \sqrt{1000} (\sqrt{1000} + \sqrt{999})}
\\&= \frac{\sqrt{1000} - \sqrt{999}}{2 \sqrt{1000} (\sqrt{1000} + \sqrt{999})}
\\&= \frac{1}{2 \sqrt{1000} (\sqrt{1000} + \sqrt{999})^2}
\end{align}$$

Normally I would have had to use the "multiply by conjugate" trick in the numerator again, but we've already seen how to deal with $\sqrt{1000} - \sqrt{999}$ so I could just plug in what we already knew.

Similarly

$$
\left(\sqrt{1001}- \sqrt{1000} \right) - \frac{1}{2 \sqrt{1000}}
= -\frac{1}{2 \sqrt{1000} (\sqrt{1001} + \sqrt{1000})^2}
$$

And now the two numbers are easily distinguishable: one is positive and one is negative!

And so

$$ \left(\sqrt{1000}- \sqrt{999} \right) - \frac{1}{2 \sqrt{1000}}
> \left(\sqrt{1001}- \sqrt{1000} \right) - \frac{1}{2 \sqrt{1000}} $$

which can be simplified to give the desired answer.


问:
Calculate the nth Fourier transform of the Fejer kernel, i.e., $N\int\Big(\frac{\sin \pi xN}{\pi xN}\Big)^2 e^{-2\pi ixn}dx$
I want to calculate the Fourier transform of the Fejer kernel $\mathcal{F}_{N}$ on the real line, which is given by
$$
\mathcal{F}_N\left(x\right) =
\begin{cases}
N\left[\,{\frac{\sin\left(\,{\pi xN}\,\right)}{\pi xN}}\,\,\right]^{2},  & \text{if}\ x\neq 0 \\[2ex]
N, & \text{if}\ x = 0
\end{cases}
$$

I want to find
$$
\hat{\mathcal{F}}_{N}\left(n\right) =
N\int\left[\frac{\sin\left(\pi xN\right)\,\,}
{\pi xN}\right]^{2}
\mathrm{e}^{-2\pi\mathrm{i}xn}\,\,\,\mathrm{d}x\,,
$$
but so far I can't think of any transformations to compute this integral. I would greatly appreciate any help.

答:
Let us use the identity
$$\frac{\sin ax}{a x} = \frac{1}{2a}\int_{-a}^{a} e^{i x\xi}\,\mathrm{d}\xi
$$
with $a=\pi N$. Subsituting in the definition
\begin{align}
\hat{\mathcal{F}}_N(n)
&= N\int_{-\infty}^{\infty} \left(\frac{\sin \pi x N}{\pi x N}\right)^2 e^{-2\pi i x n} \mathrm{d}x\\
&= N\int_{-\infty}^{\infty}
\left(\frac{1}{2a}\int_{-a}^{a} e^{i x\xi_1}\,\mathrm{d}\xi_1\right)
\left(\frac{1}{2a}\int_{-a}^{a} e^{i x\xi_2}\,\mathrm{d}\xi_2\right)
e^{-2\pi i x n} \mathrm{d}x\\
&= \frac{N}{4a^2}\int_{-\infty}^{\infty}  \mathrm{d}x \int_{-a}^{a}d\xi_1 \int_{-a}^{a}d\xi_2
e^{i x(\xi_1+\xi_2 - 2\pi n)}\,.
\end{align}
Rearranging the order of the integrals we may recognize the $\delta$ function, which may be generally written as
$$
\delta(z-\alpha)=\frac{1}{2\pi} \int_{-\infty}^\infty e^{ik(z-\alpha)}\ dk
=\frac{1}{2\pi} \int_{-\infty}^\infty e^{ik(z-\alpha)}\ dk\,,
$$
with the identification $k\equiv x$ and $z-\alpha \equiv \xi_1+\xi_2-2\pi n$. Since $a=\pi N$, this leads to
$$\hat{\mathcal{F}}_N(n)
= \frac{1}{2\pi N}
\int_{-\pi N}^{\pi N}d\xi_1 \int_{-\pi N}^{\pi N}d\xi_2
\,\delta(\xi_1+\xi_2 - 2\pi n)\,.
$$
The integral is proportional to the length of the diagonal where $\xi_2=2\pi n - \xi_1$ within the square with edges at $\pm\pi N$. The result is zero if $|n|>N$ and nonzero otherwise. To make the calculation simpler we may shift the coordinates by changing the integration variable as $\xi_{1,2}=\xi'_{1,2}-N\pi$,
$$\hat{\mathcal{F}}_N(n)
= \frac{1}{2\pi N}
\int_{0}^{2\pi N}d\xi'_1 \int_{0}^{2\pi N}d\xi'_2
\,\delta\left(\xi'_1+\xi'_2 - 2\pi \left(n+N\right)\right)\,.
$$
This shows that the length of the side is $2\pi(n+N)$ if $0\leq 2\pi(n+N)\leq 2\pi N$ and decreases otherwise if $2\pi N\leq 2\pi(n+N)\leq 4\pi N$
\begin{equation}
\hat{\mathcal{F}}_N(n) =
\left\{
\begin{array}{cl}
0 & {\rm if~} |n|>N\\
\frac{n-N}{N} & {\rm if~} -N\leq n \leq 0\\
1-\frac{n-N}{N} & {\rm if~} 0\leq n \leq N
\end{array}
\right.
\end{equation}
This may be simplified as
\begin{equation}
\hat{\mathcal{F}}_N(n) =
\left\{
\begin{array}{cc}
0 & {\rm if~}  |n| \geq N \\
1-\frac{|n|}{N} & {\rm if~} |n| \leq N
\end{array}
\right.\,.
\end{equation}

We may verify the result for $n=0$ and $n=\pm N$ by direct calculation
\begin{align}
\hat{\mathcal{F}}_N(0)
&= N\int_{-\infty}^{\infty} \left(\frac{\sin \pi x N}{\pi x N}\right)^2  \mathrm{d}x = 1
\end{align}
and
\begin{align}
\hat{\mathcal{F}}_N(\pm N)
&= N\int_{-\infty}^{\infty} \left(\frac{\sin \pi x N}{\pi x N}\right)^2 \cos(\pm 2\pi x N) \mathrm{d}x \\
&= \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{\sin^2 \theta}{\theta^2} \cos(\pm 2\theta) \mathrm{d}\theta \\
&= \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{\sin^2\theta}{\theta^2} (1-2\sin^2\theta) \mathrm{d}\theta\\
&= \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{\sin^2\theta}{\theta^2}  \mathrm{d}\theta
-\frac{2}{\pi}\int_{-\infty}^{\infty} \frac{\sin^4\theta}{\theta^2}  \mathrm{d}\theta = 1 - 1 = 0\,,
\end{align}
as required.


问:
How to show that a group is not simple using a normalizer
Let $G$ be a group of order $7^3p^n$ for some $n\ge3$ and $p>7$. Suppose that $P,Q$ are distinct Sylow $p$-subgroups of $G$. I showed that $$|P \cap Q| = p^{n-1} \text{ and } |N_G(P\cap Q)| \ge p^{n+1}.$$ Now, I want to show that    $G$ is not simple but I am stuck at this point. I thought that  $N_G(P \cap Q) $ may be  normal in $G$ but it is not always the case. How can I show that $G$ is not simple?

Edit. Condition on $p$ is fixed

答:
If $G$ is simple then the number of Sylow $p$-subgroups divides $7^3$ and is congruent to $1$ modulo $p$.

We have $7-1 = 2\times 3$, $7^2-1 = 2^4\times 3$, and $7^3-1 = 2 \times 3^2 \times 19$.

So if $p>7$, then the only possibility is $p=19$ with $343$ Sylow $p$-subgroups.

Now, letting $P$ and $Q$ be distinct Sylow $p$-subgroups as in your argument, the number of Sylow $p$-subgroups of $N_G(P \cap Q)$ must also divide $7^3$, so it must be $343$, and hence $N_G(P \cap Q) = G$, and $G$ is not simple.


问:
Folland Question 6.4 trouble with estimating norm
The question is as follows:

If $1\leq p<r\leq \infty$, prove that $L^p+L^r$ is a Banach space with norm $\lVert f\rVert= \inf\{\lVert g\rVert_p+\lVert h\rVert_r\,|\, f=g+h\in L^p+L^r\}$, and prove that for $p<q<r$, the inclusion map $L^q\to L^p+L^r$ is continuous.

So, I managed to prove $\lVert \cdot\rVert$ really is a norm, and that the result is a Banach space with the given norm, but I'm having trouble with the continuity part. Given $f\in L^q$, I considered the set $A=\{x\in X\, | \, |f(x)|>1\}$, and the functions $f\cdot 1_A\in L^p$ and $f\cdot 1_{A^c}\in L^r$ (I've already shown these inclusions). What I'm having trouble with is estimating the norms appropriately. I found that (assuming $r<\infty$)
\begin{align}
\lVert f\rVert \leq \lVert f\cdot 1_A\rVert_p + \lVert f\cdot 1_{A^c}\rVert_r \leq \lVert f\cdot 1_A \rVert_q^{q/p} + \lVert f\cdot 1_{A^c} \rVert_q^{q/r}
\end{align}
From here, I'm not sure how to get an upper bound of the form $C\lVert f\rVert_q$, for some constant $C$. Note that I've seen this answer, but I'm not sure how the last few estimates arise (particularly, why $|f\cdot 1_A|^p\leq |f\cdot 1_A|^q$ implies $\lVert f\cdot 1_A\rVert_p\leq \lVert f\cdot 1_A\rVert_q$, and likewise for the $r$ term). Any help is appreciated.

答:
You are already quite far!

Recall that ~o show continuity of a linear map, you only need to
show that it is continuous at $0$

If you estimate
$\lVert f\cdot 1_A \rVert_q^{q/p} + \lVert f\cdot 1_{A^c} \rVert_q^{q/r}$
by
$\lVert f \rVert_q^{q/p} + \lVert f\rVert_q^{q/r}$,
then we have
$$
\| f \|_{L^p+L^r} \to 0
\quad\text{for}\; \|f\|\to0.
$$
Thus, the inclusion is continuous at $0$ and therefore continuous.


问:
Find all lines on cubic surface
Problem: Find the singular point of the cubic surface

$$ZX^2 - Y^3 + TZ^2 = 0.$$

And also determine all the lines on the surface.

My attempt: I calculated the singular point by setting the partial derivatives to $0$ and I got $P = (0 : a : 0 : b)$ where $a, b$ are not both zero. Did I get this right?

About calculating the lines, I have no idea where to start. Is there perhaps some sort of technique/method/proposition for doing this? Can someone perhaps show me as an answer how to do this for this problem since I have two more of this type to do and then I can practice on those other two.

As a side question: Is there perhaps some criterion of determining the amount of total lines on the surface?

Note: We are in $\mathbb{P}^{3}$ over an algebraically closed field which does not have characteristic $2$.

答:
Let me give a computational answer. The idea is to form the general line

$$x = p_1 + l q_1, \quad y = p_2 + l q_2, \quad z = p_3 + l q_3, \quad
t = p_4 + l q_4$$

Substitute that into the cubic $f(x,y,z,t) = z x^2 - y^3 + t z^2$, and collect coefficients $a_i$ of $l$. Form an Ideal $I$ from these coefficients $a_i$.

It is an ideal in $\mathbb{Q}[p_i,q_j]$.

Now consider the 6 Plücker coordinates

$$\xi_i : w_i - \psi_i(p_k,q_l)$$

where the $\psi_i$ are the six $2$-minors of

$$\begin{pmatrix} p_1 & p_2 & p_3 & p_4 \\ q_1 & q_2 & q_3 & q_4 \end{pmatrix}$$

Form a ring $T=\mathbb{Q}[p_i, q_j, w_l]$ and take the ideal $J = I + (\xi_i)_i$ of $T$. Eliminate from this ideal the variables $p_i,q_j$, it remains an Ideal $L$ of $T_1=\mathbb{Q}[w_1,\ldots,w_6]$. The zeros of this ideal $L$ are the points in Plücker space corresponding to the lines on $V(f)$. Now to see that only one line exist, do a generic linear projection from

$$\mathbf{P}^5 = \mathrm{proj}( T_1) \to \mathbf{P}^1 = \mathrm{proj} (\mathbb{Q}[u_1,u_2])$$

Compute the ideal of the image by elimination and find out, that it is

$$(a u_1 + b u_2)^{17}$$

with $a,b \in \mathbb{Z}$ depending on the projection choosen. Why $17$ and not $27$? (I don't know).

On Macaulay2 homepage

https://faculty.math.illinois.edu/Macaulay2/

you find a web interface of Macaulay 2 on which you can run the code below (computation time is short):

https://www.unimelb-macaulay2.cloud.edu.au/#home

(You can copy the code below with Control-C and insert it with Control-V on line i1 in the web interface.)

```<code>-- We need all these variables for our computation

R = QQ[p_1..p_4,q_1..q_4,l,x,y,z,t]

-- our singular cubic

f = z * x^2 - y^3 + t * z^2

-- as a test a nonsingular cubic
-- f = x^3 + y^3 + z^3 + t^3

-- the variables of the projective Space P^3

vs = {x,y,z,t}

-- we make a substitute list x => p_1 + l * q_1, y => p_2 + l * q_2, and so on

sl = toList apply(1..4, i-> vs_(i-1) => p_i + l * q_i)

-- we apply these substitute list on f getting a polynomial in p_1..q_4 and l
-- we want that it vanishes exactly in l, so we take coefficients with respect to l

cfs = coefficients(sub(f, sl), Variables=>{l})

-- we form the ideal of these coefficients, it is an ideal of QQ[p_1..p_4,q_1..q_4]

iJ = ideal cfs_1

-- we provide explicitly the ring S with only p_i, q_i as variables

S = QQ[p_1..q_4]

-- now we prepare the canonical map from R to S which is identity on p_i, q_j and 0 otherwise

psi=map(S,R)

-- now we transfer our ideal in this smaller ring

iJ1 = psi(iJ)

-- we switch the default input to ring S

use S

-- we prepare to map the solutions of iJ to Plücker-coordinates

mM = matrix {{p_1,p_2,p_3,p_4},{q_1,q_2,q_3,q_4}}

-- the 6 polynomials in iK are the expressions of the Plücker variables

iK = minors(2, mM)

-- w_1..w_6 are our Plücker-coordinates, T has p_i, q_j, w_s as coordinates

T = S[w_1..w_6]

gamma = map(T,S)

-- we form the ideal w_a = psi_a(p_i, q_j) where psi_a are the Plücker expressions

iK1 = ideal toList(apply(1..6, i->w_i - gamma(iK_(i-1))))

-- our goal is to eliminate the p_i, q_j from the ideal made of iJ and iK1
-- we call this ideal iK2

iK2 = iK1 + gamma(iJ1)

-- following line can be ignored
elv = apply({p_1,p_2,p_3,p_4,q_1,q_2,q_3,q_4}, z->gamma(z))

-- we do the elimination by kernel computation

-- T1 is our final coordinate space

T1 = QQ[w_1..w_6]

-- the following two lines do elimination of p_i, q_j by a kernel computation.

rho=map(T/iK2, T1)

-- iL is our solution ideal in Plücker coordinates. It is of dim = 1 (projective dim = 0)
-- and degree 27

iL = ker rho

-- just to get a bit better informed, we calculate the primaryDecomposition of iL

pd1 = primaryDecomposition iL

iL1 = pd1_0

-- we want to project V(iL) to a line with a random linear projection

uw1 = sum toList(apply(1..6, i->random(-20,20) * w_i))

uw2 = sum toList(apply(1..6, i->random(-20,20) * w_i))

-- our ring that corresponds to a projective line

T2 = QQ[u_1,u_2]

-- here we do the projection by elimination

rho1 = map(T1/iL, T2, {uw1,uw2})

-- iL2 contains only one element g

iL2 = ker rho1

g = iL2_0

-- see for yourself, that g1 = l^17 with l linear in u_1,u_2
-- why 17 and not 27: I do not know, perhaps a question for Georges Elencwajg

g1 = factor g
</code>```

Title: Find all lines on cubic surface

Body:
Let me give a computational answer. The idea is to form the general line

$$x = p_1 + l q_1, \quad y = p_2 + l q_2, \quad z = p_3 + l q_3, \quad
t = p_4 + l q_4$$

Substitute that into the cubic $f(x,y,z,t) = z x^2 - y^3 + t z^2$, and collect coefficients $a_i$ of $l$. Form an Ideal $I$ from these coefficients $a_i$.

It is an ideal in $\mathbb{Q}[p_i,q_j]$.

Now consider the 6 Plücker coordinates

$$\xi_i : w_i - \psi_i(p_k,q_l)$$

where the $\psi_i$ are the six $2$-minors of

$$\begin{pmatrix} p_1 & p_2 & p_3 & p_4 \\ q_1 & q_2 & q_3 & q_4 \end{pmatrix}$$

Form a ring $T=\mathbb{Q}[p_i, q_j, w_l]$ and take the ideal $J = I + (\xi_i)_i$ of $T$. Eliminate from this ideal the variables $p_i,q_j$, it remains an Ideal $L$ of $T_1=\mathbb{Q}[w_1,\ldots,w_6]$. The zeros of this ideal $L$ are the points in Plücker space corresponding to the lines on $V(f)$. Now to see that only one line exist, do a generic linear projection from

$$\mathbf{P}^5 = \mathrm{proj}( T_1) \to \mathbf{P}^1 = \mathrm{proj} (\mathbb{Q}[u_1,u_2])$$

Compute the ideal of the image by elimination and find out, that it is

$$(a u_1 + b u_2)^{17}$$

with $a,b \in \mathbb{Z}$ depending on the projection choosen. Why $17$ and not $27$? (I don't know).

On Macaulay2 homepage

https://faculty.math.illinois.edu/Macaulay2/

you find a web interface of Macaulay 2 on which you can run the code below (computation time is short):

https://www.unimelb-macaulay2.cloud.edu.au/#home

(You can copy the code below with Control-C and insert it with Control-V on line i1 in the web interface.)

-- We need all these variables for our computation

R = QQ[p_1..p_4,q_1..q_4,l,x,y,z,t]

-- our singular cubic

f = z * x^2 - y^3 + t * z^2

-- as a test a nonsingular cubic
-- f = x^3 + y^3 + z^3 + t^3

-- the variables of the projective Space P^3

vs = {x,y,z,t}

-- we make a substitute list x => p_1 + l * q_1, y => p_2 + l * q_2, and so on

sl = toList apply(1..4, i-> vs_(i-1) => p_i + l * q_i)

-- we apply these substitute list on f getting a polynomial in p_1..q_4 and l
-- we want that it vanishes exactly in l, so we take coefficients with respect to l

cfs = coefficients(sub(f, sl), Variables=>{l})

-- we form the ideal of these coefficients, it is an ideal of QQ[p_1..p_4,q_1..q_4]

iJ = ideal cfs_1

-- we provide explicitly the ring S with only p_i, q_i as variables

S = QQ[p_1..q_4]

-- now we prepare the canonical map from R to S which is identity on p_i, q_j and 0 otherwise

psi=map(S,R)

-- now we transfer our ideal in this smaller ring

iJ1 = psi(iJ)

-- we switch the default input to ring S

use S

-- we prepare to map the solutions of iJ to Plücker-coordinates

mM = matrix {{p_1,p_2,p_3,p_4},{q_1,q_2,q_3,q_4}}

-- the 6 polynomials in iK are the expressions of the Plücker variables

iK = minors(2, mM)

-- w_1..w_6 are our Plücker-coordinates, T has p_i, q_j, w_s as coordinates

T = S[w_1..w_6]

gamma = map(T,S)

-- we form the ideal w_a = psi_a(p_i, q_j) where psi_a are the Plücker expressions

iK1 = ideal toList(apply(1..6, i->w_i - gamma(iK_(i-1))))

-- our goal is to eliminate the p_i, q_j from the ideal made of iJ and iK1
-- we call this ideal iK2

iK2 = iK1 + gamma(iJ1)

-- following line can be ignored
elv = apply({p_1,p_2,p_3,p_4,q_1,q_2,q_3,q_4}, z->gamma(z))

-- we do the elimination by kernel computation

-- T1 is our final coordinate space

T1 = QQ[w_1..w_6]

-- the following two lines do elimination of p_i, q_j by a kernel computation.

rho=map(T/iK2, T1)

-- iL is our solution ideal in Plücker coordinates. It is of dim = 1 (projective dim = 0)
-- and degree 27

iL = ker rho

-- just to get a bit better informed, we calculate the primaryDecomposition of iL

pd1 = primaryDecomposition iL

iL1 = pd1_0

-- we want to project V(iL) to a line with a random linear projection

uw1 = sum toList(apply(1..6, i->random(-20,20) * w_i))

uw2 = sum toList(apply(1..6, i->random(-20,20) * w_i))

-- our ring that corresponds to a projective line

T2 = QQ[u_1,u_2]

-- here we do the projection by elimination

rho1 = map(T1/iL, T2, {uw1,uw2})

-- iL2 contains only one element g

iL2 = ker rho1

g = iL2_0

-- see for yourself, that g1 = l^17 with l linear in u_1,u_2
-- why 17 and not 27: I do not know, perhaps a question for Georges Elencwajg

g1 = factor g


问:
Term by term differentiation sum and criterion for differentiability for $f(x) =\sum_{1}^{\infty} 2^{-n} |x-1/n|$
Let $f(x) =\sum_{1}^{\infty} 2^{-n} |x-1/n|$

Compute $f'(x)$ for $x \in (1,\infty), (1/2,1)$

Prove that $f$ is not differentiable at $x=1$

For part 1 I have that since $f_n$ is differentiable at each $x\in \{1,1/2,1/3,...\}^c$ and $f$ converges for all $\mathbb{R}$ and $\sum f'_n(x)$ converges uniformly then by term by term differentiablity theorem $f'(x) = \sum f'_n(x).$

To find the sum. I'm a little unsure about this part since i got one of them to sum to 0:

If $x \in (1, \infty)$ then $x> 1/n, \forall n \in \mathbb{N}.$ Thus $f'_n(x) = 2^{-n}$ for all $n,$ which implies $f'(x) = \sum f'_n(x) = \sum_{n=1}^{\infty} 1/2^n$ which is geometric thus sum equals $\frac{1/2}{1-1/2} =1$

If $x \in (1/2, 1)$ then $x<1/n$ only for $n=1$. Thus $f'_1(x) = -2^{-1}$ and $f'_n(x) = 2^{-n}$ for $n>1.$Thus $$f'(x) = \sum_{n=1}^{\infty} f'_n(x) = -2^{-1} + \sum_{n=2}^{\infty}1/2^n = -1/2 +1/2 =0.$$Is this right?

For part 2 I'm sort of stuck. How do i show $f$ is not differentiable at $x=1$? I know that due to $|x-1/n|, f_n$ is not differentiable at $1/n, \forall n \in \mathbb{N}$. Namely $f_1$ is not differentiable at 1. So the term by term differentiability theorem fails here. But I don't think that this failing here implies $f$ is not differentiable here?

How would you show $f$ is not differentiable at $x=1$?

答:
First compute $f(1) = \sum_{n=1}^{\infty}\frac1{2^n}(1-\frac1{n}) = \sum_{n=1}^{\infty}\frac1{2^n}-\sum_{n=1}^{\infty}\frac1{n2^n} = 1-\ln(2)$

because $\sum_{n=1}^{\infty}\frac{x^n}{n} = -\ln(1-x)$ for $|x| < 1$ (in particular the sum follows with $x=\frac1{2}$).

Then notice that since you already know that $f'(x) = 0$ in $(1/2, 1)$ it follows that $f$ must be constant, because $f$ is continuous in $[1/2, 1]$, as the series is uniformly convergent (in this interval) and each term is continuous. Thus $f(x) = 1-\ln(2)$ for all $x$ in $[1/2, 1]$.
It follows that the left-derivative at $x=1$ is $0$. It remains to see what the right-derivative is. For $x > 1$ the sum is $x-\ln(2)$ (the computation is just as above for $f(1)$). Thus: $$f'_+(1) = \lim_{h\to0+}\frac{f(1+h)-f(1)}{h} = \lim_{h\to0+}\frac{(1+h-\ln(2))-(1-\ln(2))}{h} = \lim_{h\to0+}1 = 1$$
This shows that $f$ is not differentiable at $x=1$, since $f'_+(1) = 1 \ne f'_-(1) = 0$.

You can also show that $f$ is constant in $[1/2, 1]$ by directly computing the sum: you will notice that the $x$'s cancel out.


问:
Let $z\bar z -(3-4i)z-(3+4i)\bar z + 21=0$. If M and m are max and min values of $\frac{z-\bar z}{i(z+\bar z)}$, then find $\frac 1M +\frac 1m$
Let $z=x+iy$

Solving the above equation we get

$$x^2 +y^2-6x-8y+21=0$$
Which indicates a circle centered at $(6,8)$ of radius $2$

Also the require value of $|\frac yx|$

How do I find the max and min of this?

答:
The slope of the line $y=kx$ reaches its minimum or maximum when it's tangent to the circle $x^2 +y^2-6x-8y+21=0$. This happens when $x^2+k^2x^2-6x-8kx+21=0$ has duplicate roots, i.e, the discriminant is zero:

$$(1+k^2)x^2-(8k+6)x+21=0 \implies (8k+6)^2 = 4(1+k^2)\cdot 21 \\\implies 5k^2-24k+12=0$$

Therefore $\frac 1m + \frac 1M=\frac{m+M}{mM} = \frac{24/5}{12/5}=2.\blacksquare$


问:
Why the ring of regular functions on a variety at a point is a local ring
In the Hartshorne's book on Algebraic Geometry one reads the following, '' Let Y be a variety. Denote by $O(Y) $ the ring of all regular functions on $Y. $ For $p\in Y, $ we define the local ring of $P$ on $Y, O_P, $ to be the ring of germs of regular functions on $Y$ near $P$''. He then gives an argument, why $O_P $ is a local ring: '' its maximal ideal $m$ is the set of germs of regular functions which vanish at $P. $ For if $f(P) \neq 0,$ then $1/f$ is regular in some neighborhood of $P$ ''. I unfortunately don't understand the argument, why $O_P$ contains a maximal ideal. Can somebody give some more explanations on that? Many thanks for your comment.

答: The ideal $\mathfrak{M}_P = \{f \in \mathcal{O}_X(U): f(p) = 0\}$ is maximal as it is the kernel of the evaluation map at $P$, which gives an isomorphism of $\mathcal{O}_{X, P}/\mathfrak{M}_P \cong k$.


问:
Solve equation $\sqrt{4t + 1} = 3 - 3t$
Solve equation $\sqrt{4t + 1} = 3-3t$

→ I squared both sides and got ► $4t + 1 = 9 - 18t- 3t²$

→ I then moved the 3t² to the left side and combined like pairs and got ► $3t² + 12t - 8 = 0$

I'm stuck at that point. Can someone tell me what I am doing wrong?

答:
You made a mistake in squaring the right hand side:
$$
(3-3t)^2=9-18t+9t^2
$$
However you have also to impose the condition $3-3t\ge0$, because the left hand side is non negative by definition. So you get

$$
9t^2-22t+8=0,\qquad t\le1
$$

The quadratic equation has roots $2$ and $4/9$, as it's easy to see. Hence the unique solution to your equation is $t=4/9$.

If you "back substitute", you can understand what happens: for $t=2$ the left hand side is $\sqrt{4\cdot2+1}=3$, while the right hand side is $3-3\cdot2=-3$.

For $t=4/9$, the left hand side is $\sqrt{16/9+1}=5/3$, and the right hand side is $3-3\cdot(4/9)=3-4/3=5/3$.

The problem is that upon squaring you also add the solutions to $\sqrt{4t+1}=-(3-3t)$ that is another equation.

Title: Solve equation $\sqrt{4t + 1} = 3 - 3t$


问:
Interpretation of a certain general theorem used by Gauss in his work on theta functions.
I'm trying to understand the meaning of a general proposition stated by Gauss in a posthomous paper (this paper is in pp. 470-481 of volume 3 of Gauss's werke) on theta functions, a proposition which seems to serve as a guiding and organizing principle of the vast amount of relations among theta functions that he found.

Gauss's notation and definitions

Denote by $P(x,y),Q(x,y),R(x,y)$ the following functions:

$$P(x,y)=1+x(y+\frac{1}{y})+x^4(y^2+\frac{1}{y^2})+x^9(y^3+\frac{1}{y^3})+...$$
$$Q(x,y)= 1-x(y+\frac{1}{y})+x^4(y^2+\frac{1}{y^2})-x^9(y^3+\frac{1}{y^3})+...$$
$$R(x,y)=x^{\frac{1}{4}}(y^{\frac{1}{2}}+y^{-\frac{1}{2}})+x^{\frac{9}{4}}(y^{\frac{3}{2}}+y^{-\frac{3}{2}})+x^{\frac{25}{4}}(y^{\frac{5}{2}}+y^{-\frac{5}{2}})+...$$

These functions include Jacobi theta functions in their usual meaning as special cases; if $y$ is a complex number whose absolute value is $1$, and $z$ is defined to be a real number such that $y = e^{2iz}$, then we have:

$$P(x,y)=1+2cos(2z)x+2cos(4z)x^4+2cos(6z)x^9+...=\vartheta_3(z,x)$$

which follows from the identity $cos(2nz)= \frac{e^{2inz}+e^{-2inz}}{2}$. In paticular, we have:

$$P(x,1)=1+2x+2x^4+2x^9+...=\vartheta_3(0,x)$$,
So one can understand $P(x,y),Q(x,y),R(x,y)$ as a generalization of Jacobi theta function $\vartheta(z,x)$ from purely real $z$ to a complex $z$ (non-zero imaginary part of z), so that $|y| \ne 1$.

Remark: I'm not very familiar with Jacobi's publications, so it's quite possible that Jacobi's original definition of his theta functions includes also the case when $z$ is complex, so Gauss's functions $P(x,y),Q(x,y),R(x,y)$ are nothing else than simply Jacobi's theta functions with different notation.

Gauss's theorem

On August 6, 1827, Gauss stated the following "general theorem":

$$P(x,ty)\cdot P(x,\frac{y}{t}) = P(x^2,t^2)P(x^2,y^2) + R(x^2,t^2)R(x^2,y^2) $$

and then goes on to derive a multitude of relations from it.

For more comprehensive background on this question, please look at the answer to HSM stackexchange post https://hsm.stackexchange.com/questions/6256/did-gauss-know-jacobis-four-squares-theorem.

Therefore, i'd like to know how to interpret the general theorem stated by Gauss.

答:
Let's use the variables $q, z$ with $q=x, y=e^{2iz}$ so that $$P(x, y) =\vartheta_3(z,q),Q(x,y)=\vartheta_4(z,q),R(x,y)=\vartheta_2(z,q)$$ and we can now transcribe the general theorem of Gauss as $$\vartheta_3(z+w,q)\vartheta_3(z-w,q)=\vartheta_3(2z,q^2)\vartheta_3(2w,q^2)+\vartheta_2(2z,q^2)\vartheta_2(2w,q^2)$$ (with $t=e^{2iw}$) as an identity between Jacobi theta functions.

This is one of the most fundamental identities between theta functions and almost all algebraic relations between theta functions can be derived using this. You may have a look at this paper at arXiv for some identities derived via this general theorem of Gauss

The proof of the same can be given by considering the ratio of left and right sides and showing that it is a doubly periodic functions with no poles. And thus is a constant. It requires some effort to show that the constant is $1$ but can be shown with some algebraic manipulation on the series corresponding to these functions with $z=0,w=0$.

At the moment I don't have a direct algebraic proof of the above identity and will need to check Jacobi Fundamenta Nova to see if Jacobi provided any such proof. Also as you have remarked in your question, Jacobi Theta functions are defined for all complex $z, q$ with $|q|<1$.


答:
Since $a_n\to a$ as $n\to\infty$, for $\forall \epsilon>0$, there is $N_1\in\mathrm N$ and $M>0$ such that
$$ |a_n-a|<\epsilon/2, \forall n> N_1 $$
and
$$ |a_n-a|\le M, \forall n>0. $$
So
\begin{eqnarray}
|e^{-n}f(n)-a|&\le& e^{-n} \sum_{k=0}^{\infty} \frac{|a_k-a| n^k}{k!}\\
&=&e^{-n} \sum_{k=0}^{N_1} \frac{|a_k-a| n^k}{k!}+e^{-n} \sum_{k=N_1+1}^{\infty} \frac{|a_k-a| n^k}{k!}\\
&\le& Me^{-n} \sum_{k=0}^{N_1} \frac{n^k}{k!}+\frac12\epsilon e^{-n} \sum_{k=N_1+1}^{\infty} \frac{n^k}{k!}\\
&\le& Me^{-n} \sum_{k=0}^{N_1} \frac{n^k}{k!}+\frac12\epsilon e^{-n} \sum_{k=0}^{\infty} \frac{n^k}{k!}\\
&\le& Me^{-n}\sum_{k=0}^{N_1} \frac{n^k}{k!}+\frac12\epsilon. \end{eqnarray}
Since $\lim_{n\to\infty}e^{-n}\sum_{k=0}^{N_1} \frac{n^k}{k!}=0$ for fixed $N_1$, there is $N_2>0$ such that
$$ e^{-n}\sum_{k=0}^{N_1} \frac{n^k}{k!}<\frac{\epsilon}{2M}, \forall n>N_2. $$
Let $N=\max\{N_1,N_2\}$. Then for $n>N$,
$$ |e^{-n}f(n)-a|<\frac12\epsilon+\frac12\epsilon=\epsilon, \forall n>N. $$
Namely
$$ \lim_{n\to\infty}e^{-n}f(n)=a. $$

Title: What is $\lim_{n \to \infty} e^{-n} f(n)$?

Body:
Let $(a_n)$ be a sequence that converges to $a$. And $f(x) = \sum_{k=0}^{\infty} \frac{a_k x^k}{k!}$

From previous parts, I know that $e^{-x}f(x) -a= e^{-x} \sum_{k=0}^{\infty} \frac{(a_k-a) x^k}{k!}$

Prove that $\lim_{n \to \infty} e^{-n} f(n)$ exists and find its value.

I'm having a lot of trouble with this. For one I'm not sure what the limit is. I thought mabey the limit is 0 but I got stuck:

$|e^{-n}\sum_{k=0}^{\infty} \frac{a_k n^k}{k!} - 0|=|e^{-n}| *|\sum_{k=0}^{\infty} \frac{a_k n^k}{k!}| \leq |e^{-n}| * |A|\sum_{k=0}^{\infty} |n^k/k!|$ where A is bound for $a_n.$ Then  $\leq |e^{-n}|*|A|e^{-n} = |A|$. But that wouldn't be $< \epsilon.$

Then I though maybe it converges to a because from previous part we have $e^{-x}f(x) =a+ e^{-x} \sum_{k=0}^{\infty} \frac{(a_k-a) x^k}{k!}$. Then

$|a+ e^{-n} \sum_{k=0}^{\infty} \frac{(a_k-a) n^k}{k!} -a| = |e^{-n} \sum_{k=0}^{\infty} \frac{(a_k-a) n^k}{k!}| \leq |e^{-n}|*\sum_{k=0}^{\infty} |\frac{(a_k-a) n^k}{k!}| \leq |e^{-n}| * |2A| \sum_{k=0}^{\infty} |\frac{n^k}{k!}| \leq |2A|$. But again not not sure how to make it $< \epsilon.$


问:
Does $M \oplus N \cong M$ imply that $N = 0$ for modules $M,N$?
Let $R$ be a commutative ring and $M,N$ be $R$-modules. Does $M \oplus N \cong M$ imply that $N = 0$?

I would think that  $M \oplus N \cong M$ implies that $(M \oplus N)/M \cong M/M = 0$, but is that true? And if yes, how would the first isomorphism induce the second one? Sorry if this question has been answered before, I could not find an answer.

Edit: And what if $M$ is assumed to be finitely generated?

答:
I would think that  $M \oplus N \cong M$ implies that $(M \oplus N)/M \cong M/M = 0$, but is that true?

As you've already seen from the other answers, this argument doesn't work. Don't let the notation fool you too much; if you just have some arbitrary isomorphism $\varphi : M \oplus N \cong M$, then when you write "$M/M$" what you really mean is $\varphi^{-1}(M)/M$, or $M/\varphi(M)$. There are two different copies of $M$ here which the notation is fooling you into identifying! Your argument requires the much stronger condition that the isomorphism $M \oplus N \cong M$ is the natural projection $M \oplus N \to M$, so that the two copies of $M$ involved are the same. This is a good example of the difference between a natural isomorphism and an "unnatural" isomorphism.


问:
Mathematical properties of Rank-$N$ tensors where $N$>2
WARNING:

This question might not have all the necessary tags.

I asked about the uses of rank-$N$ tensors in physics on physics stackexchange, but for some reason it was closed saying that my question was not specific enough. Maybe that will also be the case here, but I am asking the question anyway.

For that reason, I have come here to ask about the mathematical meaning and properties of rank-$N$ tensors where $N$>2.

SO.....scalars(Rank-$0$ tensor), vectors(Rank-$1$ tensor), matrices(Rank-$2$ tensor) are all tensors of different ranks. But these are the only tensors I have seen in use. No one talks about tensors of rank-$3$ or higher.

I don't even know how to multiply high rank tensors and also what is the geometric meaning of tensors of rank-$3$ or higher.

So finally the question is that:
What are the mathematical properties, uses and significance of tensors with rank-$3$ or higher?

答:
Tensors are a source of great conceptual confusion among many. The reason is that they are often introduced in a very different way to vectors and matrices. When we first learn about vectors and matrices, they are presented as nothing but a list or array of numbers. In some applications like computer science and programming, this is indeed all they are and all they need to be. But, as we know, there are much more enlightening and useful ways to think about these objects. For instance, we can actually think of a (column) vector $\boldsymbol{v}$ as a linear map that takes in one row (co) vector (hence order 1) and spits out a real number (or whatever other field you want to use). For instance,
$$\begin{bmatrix}
\omega_{1} & \omega_{2}
\end{bmatrix}\begin{bmatrix}
v^{1}\\
v^{2}
\end{bmatrix} =v^{1} \omega_{1} +v^{2} \omega_{2}$$
And we can view covectors as linear maps that take in one vector (hence order 1) and spit out real numbers, if you simply reverse the way you see the multiplication above. We could write
$$\boldsymbol{v}(\boldsymbol{\omega})=\boldsymbol{\omega}\boldsymbol{v}=\boldsymbol{\omega}(\boldsymbol{v})$$
And importantly, these maps are linear, eg
$$\boldsymbol{v}(a\boldsymbol{\alpha}+b\beta)=a\boldsymbol{v}(\boldsymbol{\alpha})+b\boldsymbol{v}(\boldsymbol{\beta})$$
With $a,b$ scalars, $\boldsymbol{\alpha},\boldsymbol{\beta}$ covectors. The same is true for $\boldsymbol{\omega}$.

We can extend this idea to matrices. We can think of a matrix as a bilinear map that takes in one covector and one vector (two arguments total, hence order 2) and outputs a real number. I'll illustrate below:
$$\begin{bmatrix}
\omega_{1} & \omega_{2}
\end{bmatrix}\begin{bmatrix}
M^{1}{}_{1} & M^{1}{}_{2}\\
M^{2}{}_{1} & M^{2}{}_{2}
\end{bmatrix}\begin{bmatrix}
v^{1}\\
v^{2}
\end{bmatrix} =\omega_{1} M^{1}{}_{1} v^{1} +\omega_{1} M^{1}{}_{2} v^{2} +\omega_{2} M^{2}{}_{1} v^{1} +\omega_{2} M^{2}{}_{2} v^{2}$$
Multiplication order is right to left. We could write this as
$$\mathbf M(\boldsymbol{\omega},\boldsymbol{v})=\boldsymbol{\omega}~\mathbf M~ \boldsymbol v$$
REMARK: One can trivially think about a scalar in this way as well - a scalar is a zero order tensor that takes in zero arguments and spits out a real number.

We don't actually need to have numbers arranged in arrays to create similar constructions of the above. We can think of the column vectors as simply members of an arbitrary real vector space $\mathcal{V}$ and covectors as members of its dual space $\mathcal{V}^*$. Then instead of using the matrix multiplication illustrated above, we simply use the more abstract linear maps. In this setting, don't think of a matrix as a box of numbers - simply think of it as a bilinear map that takes in a covector and a vector and outputs a real number. Normally, when we think about the indices of a vector or whatever we are referring to the entries in its corresponding array. However, when we don't have the notion of an array in the general setting we instead have to use the basis of either the vector space or the dual space -
$$v^i=\boldsymbol{v}(\boldsymbol{e}^i)~;~\omega_i=\boldsymbol{\omega}(\boldsymbol{e}_i)$$
Here, $\boldsymbol{e}_1,...,\boldsymbol{e}_n$ is the basis of the $n$ dimensional vector space and $\boldsymbol{e}^1,...,\boldsymbol{e}^n$ is the basis of the $n$ dimensional dual space. We can of course verify this in the usual array setting:
$$v^1=\boldsymbol{v}(\boldsymbol{e}^1)=\begin{bmatrix}
1 & 0
\end{bmatrix}\begin{bmatrix}
v^{1}\\
v^{2}
\end{bmatrix} =v^{1}$$

This leads us to the most often used definition of a $(r,s)$ tensor. Given a real vector space $\mathcal{V}$ and its dual space $\mathcal{V}^*$ a $(r,s)$ tensor is a multilinear map:
$$\mathbf{T}:(\mathcal{V}^*)^r\times \mathcal{V}^s\to\mathbb{R}$$
Where the power denotes Cartesian set product $\times$. This tensor has order $r+s$. If we have some notion of array multiplication like the above, we could write
$$\mathbf{T}(\boldsymbol{\omega}^1,...,\boldsymbol{\omega}^r,\boldsymbol{v}_1,...,\boldsymbol{v}_s)=\boldsymbol{\omega}^1,...,\boldsymbol{\omega}^r~\mathbf{T}~\boldsymbol{v}_1,...,\boldsymbol{v}_s=t$$
Where $t$ is some scalar. The indices of a tensor are defined much in a similar way as to how we went about retrieving indices of vectors:
$$T^{i_1\dots i_r}{}_{j_1\dots j_s}=\mathbf{T}(\boldsymbol{e}^{i_1},...,\boldsymbol{e}^{i_r},\boldsymbol{e}_{j_1},...,\boldsymbol{e}_{j_s})$$

Concerning the geometry, this is kind of hard to pin down. I'm sure there are some nice examples of visualizing the Ricci tensor, etc but its hard to talk intuitively about the geometry of these complex objects. But hopefully my answer does a good job connecting the notions of vectors, matrices, and tensors and you get some benefit out of it. As a final comment, the above only covers "square" tensors - that is the range of all of its indices is the same. We can actually define non-square tensors as maps over arbitrary vector and dual spaces:
$$\mathbf{T}:\mathcal{W}_1^*\times ...\times \mathcal{W}_r^*\times \mathcal{V}_1\times...\times\mathcal{V}_s\to \mathbb{R}$$
But they don't have many applications so they are not talked about much.

BONUS - A nice notation for sets real tensors
When talking about real tensors that we can view as arrays, I like to use the notation
$${}^r_s\mathbb{R}^n$$
As the set of $n$ dimensional $(r,s)$ real square tensors. However if the tensor is non square we might write
$${}^r_s\mathbb{R}^{m_1,...,m_r}_{m_1,...,m_s}$$
I think this is rather nice. More generally I use the notation ${}^r_s\mathcal{V}$ to denote the set of $(r,s)$ tensors over a vector space $\mathcal{V}$.

Sorry about the rambling answer. Please comment any questions.


问:
Closed form of the sum $\sum_{n=1}^{\infty}\frac{H_n}{n^x}$
Some days ago I derived the identity
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2}=2\zeta(3)$$
where $H_n$ is the $n$th Harmonic number. Other related identities include
$$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac{\pi^4}{72}$$
$$\sum_{n=1}^{\infty}\frac{H_n}{n^4}=\frac{-1}{6}\pi^2\zeta(3)+3\zeta(5)$$
$$\sum_{n=1}^{\infty}\frac{H_n}{n^5}=\frac{1}{540}(\pi^6-270\zeta(3)^2)$$
(I proved all except the last one) Now I wondered if there is a closed form for the generalized form
$$\sum_{n=1}^{\infty}\frac{H_n}{n^x}$$
For some real number $x$. Here is my try: The well known identity $\psi(n+1)=H_n-\gamma$ gives
$$\sum_{n=1}^{\infty}\frac{H_n}{n^x}=\sum_{n=1}^{\infty}\frac{\psi(n+1)}{n^x}-\gamma\zeta(x)$$
But I don't know what to do further. Any help would be appreciated.

答:
Euler proved the following result:

Theorem For integer $q\geq 2$, we have $$\sum_{n=1}^\infty \frac{H_n}{n^q}=\left( 1+\frac{q}{2}\right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1) \zeta(q-k)$$

A proof using the residue theorem can be found in the paper: "Euler sums and contour integral representations" by Philippe Flajolet and Bruno Salvy.