问:
Truth with Universal and Existential Quantifiers
New to both math.stackexchange, please let me know if I am breaking any rules with this question. Also, new to logic, so sorry if this is a stupid question.

I am trying to get my head around Quantifiers. Say you have a statement that can be expressed as:

$$\exists x (Rx \rightarrow \forall y Ry)$$

If the statement is written in such a way that $\forall yRy$ could be both True and False, would this statement be both true and false, and if not, how do you determine the truth of it?

For example, with "there is someone who is eating such that if they are eating then everyone else is eating", $Rx$ will always be true, but will $Ry$ both true and false as we do not know if the other people are actually eating?

答:
OK so you are pretty close on this.

Imagine a society in which, when the queen eats, everybody else then eats. She would be that $x$ there could be other such people (e.g. the king) and still the statement would be true. Are all the people actually eating? Who knows? The statement just says that there is a person such that when they eat, everybody eats. It's not true of our society but it may be true of some (in practice they would have to consist of relatively few people :)).

OK so having nosed around the previous similar question, I now realise this is true in all cases and can be proved by pure logic. It goes like this, if $Ry$ is true for all $y$ then we can choose any $x$ as $P \rightarrow \top$ is true for all propositions $P$. If there is an $x$ for which $Rx$ is false then choose $x$ to be that. In this case, $Ry$ is not true for all $y$, neither is $Rx$ and so we have $\bot \rightarrow \bot$ which is true. In either case, the statement is true!


问:
How many times harder is to think in 3D as compared with 2D?
I'm a chemist, currently going through a course about molecular symmetry and group theory applied to Chemistry. This subject is very demanding in terms of visualization in 3D space. To really grasp the subject, one must be able to see, for example, how the orbitals around a given nucleus transform when subjected to the different symmetry operations in space.

I'm stumped by the sheer difficulty I have at it, specially given I deem myself good at reasoning with geometric concepts in the 2D plane. Add just one dimension and now it feels like to be in a quagmire.

That makes me wonder, can you put hard numbers in this increased overhead when going from 2D to 3D? How much more processing power is necessary to reason about 3D space? A linear scaling can be surely ruled out as, at least for me, 3D doesn't feel like just 50% more difficult. It feels like a several fold increase in complexity. Could be the case it scales like area or volume, as a power of the number of dimensions, say, n³? What about a generalization to higher dimensions? How much more difficult a 4D being would have reasoning about 4D as compared with 3D?

I would not be surprised if grasping something like a 4D were very difficult or even impossible for us, as we don't experience 4D spatial dimensions. But we do live in 3D space, and yet thinking in 3D can be very hard.

答:
This answer is based on my experience.

Usually for me three-dimensional things are easier to imagine than two-dimensional those. This is because to imagine the first I have just to reproduce a slightly modified image from my daily life experience. The latter is mostly three-dimensional. In particular, I feel solid things more real than plain those.

This situation is typical. Nicholas Bourbaki wrote that “the mathematician does not work like a machine, nor as the workingman on a moving belt; we can not over-emphasize the fundamental role played in his research by a special intuition, which is not the popular sense-intuition, but rather a kind of direct divination (ahead of all reasoning) of the normal behavior, which he seems to have the right to expect of mathematical beings, with whom a long acquintance has made him as familiar as with the beings of the real world”.

So I think that for people like me visualization depends not on a processing power (especially on quantifiable one) but on experience. Also it can require some kind of psychical energy. For instance, for me can be hard to provide a detailed visualization when I am tired.

Hard cases can require to detailed study the object first, in order to “saturate” it in mind and then sometimes an image can be forced to flash in mind.

Probably a visualization skill can also be developed by performing visualization exercises, which can be found, for instance, in some parapsychological books.

A real visualization is hard for me. I mean, for instance, to imagine an object not as some related spatial intuition but a concrete image, for instance, a small red ball, which is seen as in real life or dream.

On visualization of high-dimensional things see my recent answer here.

ContentLicense: CC BY-SA 4.0


问:
Confusion in the definition of a generator set
Let $V$ be a vector space and $B\subset V$ a base of $V$. Now let $S \subset V$ be a subspace of $V$, such that $B \cap S = \emptyset$. My question is: can I say that $B$ generates $S$? Or, in order to say that a set $A$ is a generator of a subspace, does $A$ have to be contained in that subspace?

I know this is a bit of a silly question, but I really got confused with the definition... Any help is greatly appreciated!

Thanks in advance!

答:
See this example to grasp:

We know that $e_1=(1,0)'$ and $e_2=(0,1)'$ are generators for $\mathbb R^2$ and if $S$ the subspace generated by $A=\{e_1+e_2\}$, then this shows how the basis are related: $B\cap S=\varnothing$ and $A\subset S$.


问:
Interesting counter examples to "If a sequence tends to zero, must the series converge?"
We know that the sequence

$ 1, 1/2, 1/3, 1/4 \cdots $

converges to zero, yet most of us have seen that the "Harmonic series"

$ \sum_{n=1} \frac{1}{n}  $

is unbounded above and does not converge

I'm looking for more interesting counterexamples other than this one in general - so other sequences that tend to zero, yet their sum are unbounded.

答:
Let me atone for my sin.

Following Jorge's answer, for any $c\in (0,1)$ you have that the series with $a_n=\frac{1}{n^c}$ is divergent but the general term still converges to zero.

You can take $H\subset\Bbb (0,1)$ a $\Bbb Q$-linearly independet set with cardinality $\mathfrak{c}$ (the cardinality of continuum.

Now $V:={\rm span}\,\{(1/n^c)_n:\, c\in H\}$ is a vector space of dimension $\mathfrak{c}$ such that every nonzero member is a sequence tending to zero but such that its series is divergent.

Convergence to zero of any nonzero lineaar combination of that sequebnces is trivial.

The divergence of any linear combination is tricky, but roughly speaking, if you have $$\alpha_1\frac{1}{n^{c_1}}+\ldots+\alpha_p\frac{1}{n^{c_p}}$$ with $\alpha_j \ne 0$ and $0<c_1<\ldots<c_p$, for $n$ big enough the series behaves like $\frac{1}{n^{c_1}}$ wich is divergent.


问:
A question about standard models
As I understand it a standard model is a model in which the relation is the $\in$ on the actual set of sets constituting the model.

(i) Hence theories that aren't in the language of set $L_S$ generally won't have a standard model because the binary relation $\in$ can't model binary functions like $+$ in group theory for example. Is this right?

(ii) What do non-standard models of set theory look like? Would someone show me an example? If possible as simple as possible.

Many thanks for your help.

答:
A standard model is often a model which is minimal in a certain sense, like the constructible universe. The interesting question is in what sense it is minimal, because a countable model of ZF would seem to be "more minimal" than the constructible universe. A seemingly total different characterization of standard model from "The seven virtues of simple type theory" just requires that the higher-order axioms (which have been approximated by first order axioms schemes) are true with respect to the standard semantics of higher-order logic. You can imagine this as a condition which maximizes the "dual-space", such that the "primary-space" gets minimized. Or imagine a Galois-connection where "maximizing" one domain will "minimize" the other domain.

Now you may believe that higher-order axioms are devoid of meaning, but this isn't completely true. If you only have a successor function, the first order induction scheme can't "restrict" the model as much as if you also have an addition operator, a multiplication operator, an exponentiation operator, an ...  Because higher-order logic allows you to define all these operators and all other "definable" functions and relations, the possible models get "restricted" quite strongly before we even get near to the unclear boundaries of the higher-order axioms.

ContentLicense: CC BY-SA 3.0

Title: A question about standard models


问:
Confusion about minimizer(s) of convex functions
I was thinking about convex functions and kind of got myself confused.

Let's say that our function is $f(x) = x^TAx + b^Tx$. And A is a symmetric either positive semidefinite or positive definite matrix (question on this later). And we want to minimize $f(x)$.

$\Delta f(x) = 2Ax + b$

and the Hessian is $= 2A$

Okay, so we know that if the Hessian is positive semidefinite then f is convex. Then for $\tilde x$ if $\Delta f(\tilde x) = 0$ then $\tilde x$ is our global minimizer.

Here comes the confusion part:

If A is not positive definitive but positive semidefinite then there can be some eigenvalues that are 0. This means A is not full rank. Then the nullspace is not empty.

We have an $\tilde x$ such that $\Delta f(\tilde x) = 0$. But for another $x_1$ that is in the nullspace of A, cant we add it to $\tilde x$ and get the same result? Since $Ax_1 = 0$?

Then how can we have a global minimizer if we have infinitely many solutions to the derivative?

Then I thought, maybe when the hessian is positive definitive our global minimizer is a point. But if the hessian is only positive semidefinite the global minimizer is a line, plane, etc. But I couldn't figure out the relation between the Hessian's nullspace and $\tilde x$. For instance, what happens if the Hessian is not full rank and there is no solution to $ \Delta f(x) = 0$. Can this happen? For the example above, I think it can happen if the vector $b$ is not in the column space of A. Then even though our function is convex, do we have no minimizers? How can this happen?

答:
You could see this problem as the sum of two convex function. First one is a $(N-1)$ dimension quadratic function while the other one is a $N$ dimensional affine function as
$$
f(x) = f_1(x) + f_2(x)
$$
where $f_1(x) = x^HAx$ and $f_2(x) =b^Hx $.

Assume in a 2-dimensional case, we would have $ x = [x_1 , x_2]^T $ and $A$ as
$$
A = \left[\begin{matrix}
    \lambda_{11}       & 0  \\
    0       & 0  \\
\end{matrix}\right](1)
$$

On the other hand, for positive semi definite the inequality holds

$$
 x^HAx \geq 0
$$
for every x. If we substitute $A$ from equation (1) into the inequality:
$$
x^HAx = [x_1 , x_2]\left[\begin{matrix}
    \lambda_{11}       & 0  \\
    0       & 0  \\
\end{matrix}\right] \left[ \begin{matrix} x_1 \\ x_2\end{matrix}\right]
$$

$$
= [x_1\lambda_{11} , 0]\left[ \begin{matrix} x_1 \\ x_2\end{matrix}\right]  = x_1^2\lambda_{11}
$$
As $\lambda_{11}$ is positive, the inequality holds for all the values of $x_1$. In fact, for $f_1(x)$ the variable $x_2$ does not have any influence on the output of optimization and the reason why is that there is no need for an additional subspace. One could define an arbitrary number of variables for a convex problem and yet the key parameters (the ones with non-zero $\lambda$) would be unique regardless of the answer of the other ones.
In this example, let's say the global minimizer is $x = [x_1^*,x_2^*]$ so
$$
f_1(x^*) = (x^*)^HA(x^*) = (x_1^*)^2\lambda_{11}
$$

If we assume that there is another point $\widetilde{x} = [x_1^*,\tilde{x_2}] $:

$$
f_1(\widetilde{x}) = (x_1^*)^2\lambda_{11} = f_1(x^*)
$$
For $f_2(x)$ as it is an affine function, there is no minimum on the additional subspace. since it could be written as
$$
 b_1^*x_1^*+b_2^*\tilde{x_2} <  b_1^*x_1^*+b_2^*x_2^*
$$

$$
 \tilde{x_2} <  x_2^*
$$

which means for all the selected values of $x_2$, a lower one could be found.

It could be concluded that the values chosen in the null space of Hessian become crucial only when there exists additional subspace for affine function. (In this example when parameter $b_2$ has non-zero values)


答:
A generator of a vector subspace $S$ is by definition a subset $A\subset S$ such that every vector $w\in S$ can be written as a linear combination of elements of $A$. Therefore, if $B\cap S=\emptyset$, we don't have $B\subset S$, so $B$ can't be a generator of $S$.

A different thing is to say that $B$ can also generate the elements of $S$, which is trivially true, since $B$ is a generator of $V$ (bases are generators), so it can generate any element of $V$, hence the same for $S$, since it is a vector subspace.

It's a matter of terminology, but the common thing is to talk about a generator of a vector subspace, also called a spanning set, in which case we must have it contained in the vector subspace it's generating.

Title: Confusion in the definition of a generator set

Body:
Let $V$ be a vector space and $B\subset V$ a base of $V$. Now let $S \subset V$ be a subspace of $V$, such that $B \cap S = \emptyset$. My question is: can I say that $B$ generates $S$? Or, in order to say that a set $A$ is a generator of a subspace, does $A$ have to be contained in that subspace?

I know this is a bit of a silly question, but I really got confused with the definition... Any help is greatly appreciated!

Thanks in advance!


问:
Show that any discrete or continuous random variable $X$ satisfies $E(X) = \int_{0}^\infty P(X > x)dx - \int_{0}^\infty P(X < -x)dx$
Show:

$$E(X) = \int_{0}^\infty P(X > x)dx - \int_{0}^\infty P(X < -x)dx$$

$E[X]$ is the expectation value of the Random Variable $X$

$P$ is the probability

We know

$$E(X) = \int_{-\infty}^\infty xf(x)dx$$
$$f(x)dx = \frac{\mathrm{d}}{\mathrm{d}x}P(X\le x)$$

So,
$$
E(X) =-\int_0^\infty x\,\frac{\mathrm{d}}{\mathrm{d}x}P(X\le x)\,\mathrm{d}x
$$

$$
E(X) =-\int_0^\infty x\,\mathrm{d}P(X\le x)
$$
Using Integration by parts:
$$
\mathrm{E}(X)=-\lim_{x\to\infty}x\,P(X\le x)+\int_0^\infty P(X\le x)\,\mathrm{d}x
$$

However, I do not know how to proceed further. Help would be greatly appreciated.

答:
Let's take the first integral (by parts)

$$\int_0^{+\infty}\mathbb{P}[X>x]dx=\int_0^{+\infty}[1-F_X(x)]dx=\underbrace{x[(1-F_X(x)\Big]_0^{+\infty}}_{=0}-\int_0^{+\infty}-xf_X(x)dx=\int_0^{+\infty}xf_X(x)dx$$

In fact

$$\lim\limits_{x \to +\infty}x[1-F_X(x)]=\lim\limits_{x \to +\infty}\frac{x}{\frac{1}{1-F_X(x)}}=\frac{+\infty}{+\infty}\xrightarrow{\text{Hôpital}}=\lim\limits_{x \to +\infty}\frac{1}{\frac{f_X(x)}{[1-F_X(x)]^2}}=\lim\limits_{x \to +\infty}\frac{[1-F_X(x)]^2}{f_X(x)}=0$$

With similar reasoning for the other integral you get

$$ \bbox[5px,border:2px solid black]
{
\int_0^{+\infty}\mathbb{P}[X>x]dx-\int_{-\infty}^{0}\mathbb{P}[X\leq x]dx=\int_{-\infty}^{+\infty}xf_X(x)dx=\mathbb{E}[X]
\
}
$$

Graphically, the expectation is the difference of the following areas

Thus you are all set!


问:
Starting to study the Model Theory
I would like to start studying Model Theory and I would like some suggestions of books. And I would like to know: Is it better to make a mathematical logic review after start model theory?</p>

答:
I would suggest:

Model theory by Chang / keisler

Model theory: an introduction by Marker

Title: Starting to study the Model Theory


问:
Prove that supremum of one function of $x$ is less than or equal to another function of $x$
Let $X \subset \mathbb{R}$ and $f,g:X\to\mathbb{R}$ with g bounded above and $f(x)<g(x)$ for all $x \in X$. Prove that $sup(f) \leq sup(g)$.

Also is it necessarily true that $sup(f) < sup(g)$?

答:
Both $f,g$ are bounded above so $\sup f,\sup g$ exist.

Since $f<g\le\sup g,\sup g$ is an upper bound for $f(X)$. Thus $\sup f\le\sup g$.

It is not necessary that $\sup f<\sup g$. Take, for example, $g(x)=k,f(x)=k-1/x,X=\Bbb R^+$. $g$ is bounded above by $k,f<g$ over $\Bbb R^+$ but $\sup g=\sup f=k$.


问:
improper integral question - $\int_{1}^{\infty}\!e^{-x}\ln x\,dx$
I ran into this integral question:

does this integral converge:

$$\int_{1}^{\infty}\!e^{-x}\ln x\,dx$$
?

Thank you very much in advance,

Yaron

答:
Note that for large enough $x$, we have $\ln x\lt e^{x/2}$. So for large enough $x$, our integrand is $\lt e^{-x/2}$. And it is well known that $\int_1^\infty e^{-x/2}\,dx$ converges: indeed we can easily find an explicit expression for it. So by Comparison, our integral also converges.

Remark: The point is that $\ln x$ grows very feebly in comparison to $e^{x}$, so the $e^{-x}$ term crushes poor $\ln x$.

If we want to be formal about the relationship between $e^{x/2}$ and $\ln x$, consider $\dfrac{\ln x}{e^{x/2}}$. By L'Hospital's Rule, or in other ways, we can show that this $\to 0$ as $x\to \infty$.


问:
A question on the construction of the Lebesgue integral in Rudin
In Rudin's Real and Complex Analysis (page 19, definition 1.23) we take a measurable function $f:X\rightarrow [0,\infty]$ in which $(X,\mathcal M,\mu)$ is a measure space with $\sigma-$algebra $\mathcal M$ and measure $\mu$. We define the integral of this function over $E\in \mathcal M$ to be: $$\int_E f\text{ }d\mu=\sup\int_E s\text{ }d\mu \tag{1},$$ in which $s$ is a simple function, $s:X\rightarrow [0,\infty)$, defined as: $$s(x)=\sum_{i=1}^n \alpha_i\chi_{A_i}, \tag{2}$$ in which $\chi_{A_i}(x)$ is the characteristic function on $X$. Rudin states that the supremum in $(1)$ is being taken over all simple measurable functions such that $0 \leq s\leq f$, I do not know what this means. Is this a pointwise comparison of the functions between $0$ and $f$?

Standard mandatory disclaimer: I am a physicist not a mathematician.

答:
Break it into pieces:

$1). S$ is the collection of simple functions such that $s(x)\le f(x)$ for all $x\in X$.

$2).\ I:= \{\int_X s:s\in S\}$ is a set of numbers, with each $\int_X s$ defined as in your question.

$3).\ $ Since $I$ is a set of numbers, we can take $\sup I=\sup\{\int_Xs:s\in S\}.$ The result is a number in the extended reals, and this is the integral of $f$ by definition.


答:
I would suggest:

A Course in Model Theory, by Tent and Ziegler

Title: Starting to study the Model Theory

Body: I would like to start studying Model Theory and I would like some suggestions of books. And I would like to know: Is it better to make a mathematical logic review after start model theory?</p>


问:
Finding a polynomial of degree $n$ when value of $f(k)$ is equal to some value
Problem :

If $f(x)$ is a polynomial of degree $n$ and if $f(k) = \frac{k}{k+1}$ where $k =0,1,2,\ldots,n$, find $f(x)$.

Can we go like this : Let the polynomial be $a(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)......(x-\alpha_n)$ where $\alpha_1, \alpha_2....\alpha_n$ are the roots of the $f(x)$ and $a$ is the leading coefficient.

Please guide how to proceed in such question.

答:
Hint: Consider the polynomial $g(x) = (x+1) f(x) - x$. What is the degree of $g(x)$? What are the zeros of $g(x)$? What is the value of $g(-1)$?

More details added.

The degree of $g(x)$ is equal to the degree of $f(x)$ plus 1 (since we multiplied by $x+1$), which is equal to $n+1$.

The zeros of $g(x)$ are (at least) $0, 1, 2, \ldots n$, so by the Remainder-Factor Theorem,

$$g(x) = A(x) \prod_{i=0}^n (x-i)$$

Since $g$ has degree $n+1$, it follows that $A(x)$ is a constant polynomial, which we will denote by $A$.

The value of $g(-1)$ is $g(-1) = (-1 + 1) f(-1) - (-1) = 1 $. Hence

$$1 = g(-1) = A \prod_{i=0}^n (-1-i)$$

This allows us to determine that $A = \frac{1}{(n+1)!}$.

Hint: $\frac{ g(x) + x) } { x+1}$ is indeed a polynomial, and is our $f(x)$.


问:
Write the solution as a Fourier Series. $u_{tt}=u_{xx},~ u(t,0)=u(t,\pi)=0,~u(0,x)=1, u_t(0,x)=0$
I'm running into a roadblock on this IBV problem

$$u_{tt}=u_{xx},~ u(t,0)=u(t,\pi)=0,~u(0,x)=1, u_t(0,x)=0$$

Assuming separability:

$$u(t,x)=w(t)v(x) \implies  w^{\prime\prime}(t)=\lambda w(t)~~\mbox{and}~~v^{\prime\prime}(x)=\lambda v(x)$$

If $\lambda=-\omega^2<0$ then
\begin{align}
w(t) &= \sin(\omega t) \ \mbox{or} \ w(t) = \cos(\omega t) \\ v(x) &= \sin(\omega x) \ \mbox{or} \ v(x) = \cos(\omega x)
\end{align}

If $\lambda=0$ then
\begin{align}
w(t) &= 1 \ \mbox{or} \ w(t) = t \\
v(x) &= 1 \ \mbox{or} \ v(x) = x
\end{align}

If $\lambda=\omega^2$ then
\begin{align}
w(t) &= e^{-\omega t} \ \mbox{or} \ w(t) = e^{\omega t} \\
v(x) &= e^{-\omega x} \ \mbox{or} \ v(x) = e^{\omega x}
\end{align}

By initial boundary condition $u(t,0) = u(t,\pi) = 0$, we have $v(0) = v(\pi) = 0 \implies v(x) = \sin(\omega x)$.

Here's where I encounter my roadblock. By the IC $u(0,x)=1$, we have $w(0)=\frac{1}{v(x)}$ (which is not a possible solution for $w(t)$. Am I to conclude in this scenario that the function $u(t,x)$ is not separable and I must seek another strategy for finding the solution?

Further Analysis:
Based on $v(x)=\sin(\omega x)$ in conjunction with $u(0,x)=1\implies \sin(\omega x)=1$ at $w(0)=1$.  Thus, $\omega=n\pi$ and $w_n(t)=\cos(n\pi t)$ and therefore $$u_n(t,x)=\cos(n\pi t)\sin(n\pi x)~~~n=1,2,3,...$$

I think I may have solved my own problem.  If I'm making an error here, please correct me.

This brings me to a new stumbling block.  Translating this solution into a Fourier Series solution:

Logically (in my mind at least) $u(t,x)=\sum_{n=1}^{\infty}u_n(t,x)=\sum_{n=1}^{\infty}b_n \cos{(nt)}\sin(nx)$

Solving for $b_n$ is a little tricky for me here.  I'm using v(x) to find the $b_n$ values in the following integral.  I don't see how to put the result back into the Fourier Sum.

$$b_n=\frac{2}{\pi}\int_0^\pi{\sin(\omega x)\sin(kx)}=2~~~\forall ~\omega=k$$

End Further Analysis

This problem is from Chapter 4.2 of "Introduction to Partial Differential Equations" by Peter Olver.  The section is on Separation of Variables, hence the approach I am trying to use.

答:
I think there is a bit of confusion in your working, so I will outline a general approach to solve the problem. You have assumed separability and determined ODEs correctly. Next, you need to solve the ODEs (as you have done, but with some caveats). The general solution to

$$v'' = -\omega^{2} v$$

is

$$v = A \cos(\omega x) + B \sin(\omega x)$$

You can't say apriori it is $\cos(\omega x)$ or $\sin(\omega x)$; it could also be a sum of both (which is what you get when you have periodic boundary conditions). The boundary conditions determine the particular eigenfunction(s). Applying the boundary condition $v(0) = 0 \implies A = 0$. Applying the other boundary condition $v(\pi) = B \sin(\omega \pi) = 0$, we see that if $B = 0$ we get the trivial solution. Hence, we assume $B \ne 0$ and so $\sin(\omega \pi) = 0 \implies \omega = n, \ n \ge 1$ and so the spatial eigenfunction and eigenvalue are determined to be $v = B_{n} \sin(n x)$ and $\lambda = -n^{2}$. Going to the ODE in $t$, you need to solve

$$w'' = -n^{2}w$$

which has solution

$$w = C_{n} \cos(nt) + D_{n} \sin(nt)$$

The general solution is then a superposition

$$u = \sum_{n \ge 1} [C_{n} \cos(nt) + D_{n} \sin(nt)] \sin(nx)$$

which you can check satisfies the original PDE. It is at this point that you apply the ICs. You have

$$u(0,x) = 1 = \sum_{n \ge 1} C_{n} \sin(nx)$$

To determine the coefficients $C_{n}$, multiply both sides by $\sin(mx)$ and integrating over the domain $x \in [0, \pi]$

$$\implies \int_{0}^{\pi} 1 \cdot \sin(mx) dx = \sum_{n \ge 1} C_{n} \int_{0}^{\pi} \sin(nx) \sin(mx) dx$$

which yields the orthogonality relations. Once you have determined $C_{n}$ explicitly, you can use the same approach to determine $D_{n}$.


问:
Can this definite integral be solved in terms of Bessel functions as a function of the constant $a$?
This Integral is used to derive isotropic antenna Directivity of a uniform array.

$$\int_{\phi =0}^{2 \cdot \pi} \int_{\theta =0}^{\pi}\cos(a \cdot \sin(\theta)\cdot \cos(\phi))\cdot \sin(\theta)\cdot d\theta \cdot d\phi $$
Where $a$ is a constant.

答:
Integrate over $\phi$ first.  I get
$$ \int_0^{2\pi} \cos(a \sin(\theta) \cos(\phi)) \; d\phi = 2\pi J_0(a \sin(\theta)) $$
So now your integral is

$$\eqalign{ 2 \pi \int_0^\pi J_0(a \sin(\theta)) \sin(\theta)\; d\theta &=
2\pi \sum_{k=0}^\infty \frac{(-1/4)^k a^{2k}}{k!^2} \int_0^\pi \sin^{2k+1}(\theta)\; d\theta\cr
&= 4 \pi \sum_{k=0}^\infty \frac{(-1)^k a^{2k}}{(2k+1)!}\cr
&= 4 \pi \frac{\sin(a)}{a} }$$


问:
little-oh notation
Assume I have a function $ f:\mathbb{R}^{m}\to\mathbb{R}^n $, and let $ h\in \mathbb{R}^m $.

Does the notation $ f\left(x_0+h\right)=o\left(h\right) $ (for fixed $x_0 $)
mean that
$$ \lim_{h\to0}\frac{||f\left(x_{0}+h\right)||_{\mathbb{R}^{n}}}{||h||_{\mathbb{R}^{m}}}=0 ?$$

Because I've seen in a few places that lecturers wrote just $ \lim_{h\to0}\frac{f\left(x_{0}+h\right)}{||h||}=0 $, which doesn't make any sense to me. I'll be glad for a clarification. What's the acceptable definition worldwide? Thanks in advance

答:
Writing

$$\lim_{h\to0}\frac{f\left(x_{0}+h\right)}{||h||_{\mathbb R^m}}=0$$  does make sense. For $h \neq 0$, the map $g(h) = \frac{f(x_0+h)}{\Vert h \Vert_{\mathbb R^m}}$ is well defined. Stating that its limit is equal to $0$ means that

$$\lim\limits_{h \to 0} \left\Vert \frac{f(x_0+h)}{\Vert h \Vert_{\mathbb R^m}} - 0\right\Vert_{\mathbb R^n} = \lim\limits_{h \to 0} \left\Vert \frac{f(x_0+h)}{\Vert h \Vert_{\mathbb R^m}} \right\Vert_{\mathbb R^n} = \lim\limits_{h \to 0}  \frac{\Vert f(x_0+h) \Vert_{\mathbb R^n}}{\Vert h \Vert_{\mathbb R^M}}=0.$$

So the two definitions are indeed equivalent.

Title: little-oh notation


答:
I would never say $f(x_0 + h) = o(h)$ when $f$ has values in $\mathbb R^n$ rather than $\mathbb R$.

But if I were going to say it, it would mean one of the two limits you wrote. Those limits are equivalent: $\frac{f(x_0 + h)}{\|h\|} \to 0$ (that is, $0 \in \mathbb R^n$) exactly when $\frac{\|f(x_0 + h)\|}{\|h\|} \to 0$ (that is, $0 \in \mathbb R$).

That's because the definition of a limit in $\mathbb R^n$ means that the first of these limits holds if and only if $\left\|\frac{f(x_0 + h)}{\|h\|} - 0\right\| \to 0$, and this simplifies to the second limit.

Title: little-oh notation

Body:
Assume I have a function $ f:\mathbb{R}^{m}\to\mathbb{R}^n $, and let $ h\in \mathbb{R}^m $.

Does the notation $ f\left(x_0+h\right)=o\left(h\right) $ (for fixed $x_0 $)
mean that
$$ \lim_{h\to0}\frac{||f\left(x_{0}+h\right)||_{\mathbb{R}^{n}}}{||h||_{\mathbb{R}^{m}}}=0 ?$$

Because I've seen in a few places that lecturers wrote just $ \lim_{h\to0}\frac{f\left(x_{0}+h\right)}{||h||}=0 $, which doesn't make any sense to me. I'll be glad for a clarification. What's the acceptable definition worldwide? Thanks in advance


问:
Permutation representation of Quaternion group $Q_8$, What determines cycles elements are mapped to?

Permutation representation of Quaternion group $Q_8$, What determines cycles elements are mapped to?

So I made a permutation representation of $Q_8$ acting on itself where I labeled the elements $1=1,2=-1,3=i,4=-i,5=j,6=-j,7=k,8=-k$

And then figured out the permutation representation as elements of $S_8$ and I noticed that the cycle types of elements came out different. I expected that $i,j,k$ should all be products of two $4-$cycles after I did $i$, $\sigma_i=(1324)(5768)$

but instead $j$ gets mapped to a $4-$cycle and two $2$ cycles.

$\sigma_j=(1526)(48)(37)$

My question is just why does this happen? Is there something which determines what kind of cycle you get from an element in the image of a permutation representation?

答: You didn't calculate $\sigma_j$ correctly.  Since $j$ has order 4, $\sigma_j$ must be the product of 4-cycles.


答:
This is not a complete answer.

Reorder the double integral so that the inner integral is something very close to the Bessel Integral (1)

$$J_n(x)=\frac{1}{\pi} \int_0^{\pi} \cos(n \tau - x \sin \tau) \,d \tau \tag{1}$$

Use the substitution $\tau=\phi+\frac{\pi}{2}$ in (1) with $n=0$

thus
$$J_0(x)=\frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \cos(-x \cos \phi) \,d \tau$$

Utilising angular symmetry and $\cos(x)=\cos(-x)$ we have
$$J_0(x)=\frac{1}{2\pi} \int_{-\pi}^{\pi} \cos(x \cos \phi) \,d \tau$$

which leads to your outer final integral according to both Mathematica and Robert Israel being

$$2 \pi\int_0^{\pi }   \sin (\theta ) J_0(a \sin (\theta )) \, d\theta=\frac{4 \pi  \sin ( a)}{a}=4 \pi\, j_0(a)$$

where $j_0(x)$ (the sinc function) is the first spherical Bessel function.

Title: Can this definite integral be solved in terms of Bessel functions as a function of the constant $a$?

Body:
This Integral is used to derive isotropic antenna Directivity of a uniform array.

$$\int_{\phi =0}^{2 \cdot \pi} \int_{\theta =0}^{\pi}\cos(a \cdot \sin(\theta)\cdot \cos(\phi))\cdot \sin(\theta)\cdot d\theta \cdot d\phi $$
Where $a$ is a constant.


问:
Taylor expansion of $\lvert\sin(x)\rvert$
Is it possible to generate a Taylor expansion of $\lvert\sin(x)\rvert$?

I understand that this is not possible around the point $0$, since $\frac{d}{dx}\lvert\sin(x)\rvert$ is undefined at $x = n\pi$. But can we achieve this by generating the series around the points $\frac{n\pi}{2}$ and obtain an expression that is valid for all $0<x<\infty$, except the points $x=n\pi$?

答:
For some neighborhood of any $c \ne k\pi, k \in \Bbb Z$, either $|\sin x| = \sin x$ everywhere in the neighborhood, or $|\sin x| = -\sin x$ in the neighborhood.

All the higher derivatives of $|\sin x|$ at $c$ will therefore be the same as either $\sin x$ or its opposite, and thus the Taylor series at $c$ will either be the Taylor series for $\sin x$ or for $-\sin x$. This Taylor series will indeed converge for all $x$. But it will not converge to $|\sin x|$ everywhere.


问:
Reasoning behind derivative of log function $\log_{a}x$
I was told to convert the following function

$$\log_{a}x$$
into

$$\ln x / \ln a$$
which makes sense. However, I am unsure as to why this would become

$$\frac{1}{x} * \frac{1}{\ln a}$$
instead of the result of the quotient rule:

$$\left[\frac{1}{x}(\ln a) - \frac{1}{a}(\ln x)\right]/(\ln a)^2$$

Is there a logical reason for this exception to the quotient rule? Or is this one of those functions you just have to memorize?

答:
First, $\ln a$ is a constant, so using the quotient rule wastes time.

However, math works, so let's apply the quotient rule:
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} \frac{\ln x}{\ln a} &= \frac{(\ln a)(1/x) - (\ln x)(0)}{(\ln a)^2}  \\
    &= \frac{1}{x \ln a}  \text{.}
\end{align*}

Here, since $a$ is a constant, the rate of change of $a$ (and of $\ln a$) with respect to a change in $x$ is zero.


问:
If $a > b$, is $a^2 > b^2$?
Given $a > b$, where $a,b ∈ ℝ$, is it always true that $a^2 > b^2$?</p>

答:
This answer is the same as Bill Dubuque's, but goes into a bit more detail. The question

Does $a>b$ imply that $a^2>b^2$?

is equivalent to

Does $a-b>0$ imply that $a^2-b^2>0$?

In other words, the answer to the bottom question is the same as the answer to the top question.

By factorising $a^2-b^2$, we see that no, $a-b>0$ does not imply that $a^2-b^2>0$:
$$
a^2-b^2=(a+b)(a-b) > 0 \text{ if, and only if, $a+b>0$} \, .
$$
We know that $a-b>0$, but that is not enough. We also require that $a+b>0$ so that $(a+b)(a-b)>0$. If we instead ask

Does $a>b$ and $a>-b$ imply that $a^2>b^2?$

then the answer is in the affirmative.

Title: If $a > b$, is $a^2 > b^2$?


问:
finding dual basis of vector space of polynomial degree less than or equal to 3
Assume that $V = P(3)$, Describe a basis for $V^*$ and express the linear functional $f : V \to \Bbb R$ given by $f\left(a_3x^3 + a_2x^2 + a_1x + a_0\right) = 2a_3 + a_2 - 5a_0$ as a linear combination of the basis you found for $V^*$.

I am struggling to find the basis for $V^*$, I know that the standard basis for $P(3)$ is $\left\{1, x, x^2, x^3\right\}$ and then the basis for $V^*$ is $f\left\{1, x, x^2, x^3\right\}$. Any help would be greatly appreciated, thank you.

答: You need to find a basis for the dual space $V^*$, $\{f_1,f_2,f_3,f_4\}$  such that $$f_i(v_j)=\begin{cases}1 \ \ \ \mbox{if} \ \ i=j \\0 \ \ \  \mbox{if} \ \ i\ne j \end{cases}. $$ As a small hint, try using a combination of derivatives and evaluation maps.</p>


问:
Let $G$ be a group of order $99$ and $X$ a set with $17$ elements. Show that an action of $G$ on $X$ must have at least $6$ fixed points.
Let $G$ be a group of order $99$ and $X$ a set with $17$ elements. Show that an action of $G$ on $X$ must have at least $6$ fixed points.

I have used the orbit-stabilizer theorem to show that the only possible orbit sizes are $1,3,9$, or $11.$ I have tried applying Burnside's Lemma to get information on the number of orbits but I haven't found anything useful.

答:
OK, I will turn my comment into an answer.

Suppose that $G = C_{11} \times C_3 \times
C_3 = \langle x,y,z \rangle$, where $x,y,z$ have orders $11$, $3$, $3$, respectively.

Then we can define an action of $G$ with no fixed points on a set $X = \{1,2,3, \ldots,17\}$, by letting $x,y,z$ act as the cycles $(1,2,3,\ldots,11)$, $(12,13,14)$, and $(15,16,17)$, respectively (where they all fix all points not in that cycle).


问:
Uniformly Continuous Function and Limits

Question: Given that $f$ is uniformly continuous on $[0,\infty)$, show that $\lim_{x \to \infty} \frac{f(x)}{1+x^2} = 0$.

What I know:

Uniform continuity: For every $\epsilon > 0$, $\exists\,\delta > 0$ such that for all $x,y \in [0,\infty)$, if $|x-y| < \delta$, then $|f(x)-f(y)| < \epsilon$.

Want to show:

$|\frac{f(x)}{1+x^2}| < \epsilon$ as $x \to \infty$.

Attempt #1: I attempted to show that $f$ is Lipschitz but doesn't seem to be the case.

Attempt #2: Tried to re-write $|x-y| < \delta$ as $|x| - |y| < \delta$ and $|f(x)-f(y)| < \epsilon$ as $|f(x)| - |f(y)| < \epsilon$ to manipulate the expression but to no avail.

Also thought of trying to show that $f(x)$ "moves slower" than $1+x^2$ or $x^2$ in the limit but not sure how to put it exactly.

Any help will be appreciated. Thanks!

答:
Since $f$ is uniformly continuous, then for $\varepsilon=1$, there exists a $\delta>0$, such that
$$
|x-y|<\delta \quad\Longrightarrow\quad |f(x)-f(y)|<1.
$$
Pick $h$, with $0<h<\delta$.
So, for every $x>0$, set $n=\big[\frac{x}{h}\big]$. Then
$$
|f(x)-f(0)|\le |f(0)-f(h)|+|f(h)-f(2h)|+\cdots+\\|f\big((n-1)h\big)-f(nh)|+|f(nh)-f(x)| < n+1<\frac{x}{h}+1,
$$
and hence
$$
\frac{|f(x)|}{1+x^2}< \frac{|f(0)|+\frac{x}{h}+1}{x^2+1}\to 0
$$
as $x\to\infty$.


问:
For odd $n$, if $x^2 = a \pmod{n}$ has a solution then at least one solution has a specific representation.
Update:

̶F̶o̶r̶ ̶o̶d̶d̶ ̶$n$,̶ ̶i̶f̶ ̶ $x^2 = a \pmod{n}$ ̶h̶a̶s̶ ̶a̶ ̶s̶o̶l̶u̶t̶i̶o̶n̶ ̶t̶h̶e̶n̶ ̶a̶t̶ ̶l̶e̶a̶s̶t̶ ̶o̶n̶e̶ ̶s̶o̶l̶u̶t̶i̶o̶n̶ ̶h̶a̶s̶ ̶a̶ ̶s̶p̶e̶c̶i̶f̶i̶c̶ ̶r̶e̶p̶r̶e̶s̶e̶n̶t̶a̶t̶i̶o̶n̶.̶

See the counterexample provided by lonza leggiera.

Also, see my conjecture (stated on November 10, 2020 and still open on November 14),

$\quad$ A new method for finding a solution to $x^2 = a \pmod p$?

arising from this initial work.

Here is the proposed representation,

For odd $n$, if $x^2 = a \pmod{n}$ has a solution then there is at least one solution $b$ such that there exists a $k \ge 0$ such that

$\tag 1 \text{The residue, } \Large 0 \lt r \lt n, \normalsize \text{ of } \, \Large 4^k  a \, \normalsize \text{ is a square}, \Large r = t^2$

$\tag 2 \LARGE b \equiv 2^{-k}  t \pmod{n}$

Is the above proposed representation valid?

My work

What got me going on this can be found in my answer to

$\quad$ How to solve $x^2 \equiv 12 \pmod {13}$?

I then looked at this

$\quad$ How to solve $x^2\equiv 4\pmod{143}$?

and found that the representation works for all solutions.

Example 1:

We have $141^2 = 4 \pmod{143}$. Also,

$\quad 4^{38} \cdot 4 \equiv 25 \pmod{143}$

and

$\quad 141 \equiv 2^{-38} \cdot 5 \pmod{143}$

Example 2:

There are two solutions, $3$ and $4$ for $x^2 \equiv 2 \pmod{7}$; the representation works for $4$ but not for $3$.

Example 3:

The representation, if sound, could be used to show that $x^2 \equiv 3 \pmod{7}$ has no solutions.

Our 'lookup table of squares', $S$, is defined by

$\quad S = \{1,4\}$

$\; 4^0 \cdot 3 \equiv 3 \pmod{7} \text{ and } 3 \notin S$

$\; 4^1 \cdot 3 \equiv 5 \pmod{7} \text{ and } 5 \notin S$

$\; 4^2 \cdot 3 \equiv 6 \pmod{7} \text{ and } 6 \notin S$

$\; 4^3 \cdot 3 \equiv 3 \pmod{7} \text{ and } 3 \notin S \text{ and the cycle repeats} $

and we could argue that $x^2 \equiv 3 \pmod{7}$ has no solutions.

Example 4:

Find (if it exists) a solution to $x^2 \equiv 7 \pmod{9}$.

Our 'lookup table of squares', $S$, is defined by

$\quad S = \{1,4\}$

$\; 4^0 \cdot 7 \equiv 7 \pmod{9} \text{ and } 7 \notin S$

$\; 4^1 \cdot 7 \equiv 1 \pmod{9} \text{ and } 1 \in S$

There is a solution, $\large x \equiv 2^{-1} \times 1 \equiv 5 \pmod{9}$.

Example 5 (motivated by this question on quadratic reciprocity):

Find (if it exists) a solution to $x^2 \equiv 3\pmod {10007}$.

Our 'lookup table of squares', $S$, is defined by

```<code>1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2704, 2809, 2916, 3025, 3136, 3249, 3364, 3481, 3600, 3721, 3844, 3969, 4096, 4225, 4356, 4489, 4624, 4761, 4900, 5041, 5184, 5329, 5476, 5625, 5776, 5929, 6084, 6241, 6400, 6561, 6724, 6889, 7056, 7225, 7396, 7569, 7744, 7921, 8100, 8281, 8464, 8649, 8836, 9025, 9216, 9409, 9604, 9801, 10000
</code>```

Calculating,

$\; 4^0 \cdot 3 \equiv 3 \pmod{10007} \text{ and } 3 \notin S$

$\; 4^1 \cdot 3 \equiv 12 \pmod{10007} \text{ and } 12 \notin S$

$\; 4^2 \cdot 3 \equiv 48 \pmod{10007} \text{ and } 48 \notin S$

$\; 4^3 \cdot 3 \equiv 192 \pmod{10007} \text{ and } 192 \notin S$

$\; 4^4 \cdot 3 \equiv 768 \pmod{10007} \text{ and } 768 \notin S$

$\; 4^5 \cdot 3 \equiv 3072 \pmod{10007} \text{ and } 3072 \notin S$

$\; 4^6 \cdot 3 \equiv 2281 \pmod{10007} \text{ and } 2281 \notin S$

$\; 4^7 \cdot 3 \equiv 9124 \pmod{10007} \text{ and } 9124 \notin S$

$\; 4^8 \cdot 3 \equiv 6475 \pmod{10007} \text{ and } 6475 \notin S$

$\dots\quad$(and writing several lines of python code)

$\; 4^{482} \cdot 3 \equiv 4664 \pmod{10007} \text{ and } 4664 \notin S$

$\; 4^{483} \cdot 3 \equiv 8649 \pmod{10007} \text{ and } 8649 \in S$

and $8649 = 93 \times 93$.

There is a solution, $\large x \equiv 2^{-483} \times 93 \equiv 1477 \pmod{10007}$.

答: The representation is not always possible. If $\ n=63\ $ and $\ a=58\ $, then the equation $\ x^2=a\pmod{n}\ $ has the solutions $\ x=11\ $ and $\ x=52\ $.  If $\ r\ $ is a residue of $\ 4^ka\pmod{n}\ $, then its only possible values are $\ 58\,(k\equiv0\pmod{3})\ $, $\ 43\, (k\equiv1\pmod{3})\ $, and $\ 46\,(k\equiv2\pmod{3})\ $, none of which is a square.


问:
Proof of $\frac{1}{e^{\pi}+1}+\frac{3}{e^{3\pi}+1}+\frac{5}{e^{5\pi}+1}+\ldots=\frac{1}{24}$
I would like to prove that $\displaystyle\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{n}{e^{n\pi}+1}=\frac1{24}$.

I found a solution by myself 10 hours after I posted it, here it is:

$$f(x)=\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{nx^n}{1+x^n},\quad\quad g(x)=\displaystyle\sum_{n=1}^{\infty}\frac{nx^n}{1-x^n},$$

then I must prove that $f(e^{-\pi})=\frac1{24}$. It was not hard to find the relation between $f(x)$ and $g(x)$, namely $f(x)=g(x)-4g(x^2)+4g(x^4)$.

Note that $g(x)$ is a Lambert series, so by expanding the Taylor series for the denominators and reversing the two sums, I get

$$g(x)=\sum_{n=1}^{\infty}\sigma(n)x^n$$

where $\sigma$ is the divisor function $\sigma(n)=\sum_{d\mid n}d$.

I then define for complex $\tau$ the function
$$G_2(\tau)=\frac{\pi^2}3\Bigl(1-24\sum_{n=1}^{\infty}\sigma(n)e^{2\pi in\tau}\Bigr)$$ so that
$$f(e^{-\pi})=g(e^{-\pi})-4g(e^{-2\pi})+4g(e^{-4\pi})=\frac1{24}+\frac{-G_2(\frac i2)+4G_2(i)-4G_2(2i)}{8\pi^2}.$$

But it is proven in Apostol "Modular forms and Dirichlet Series", page 69-71 that $G_2\bigl(-\frac1{\tau}\bigr)=\tau^2G_2(\tau)-2\pi i\tau$, which gives $\begin{cases}G_2(i)=-G_2(i)+2\pi\\ G_2(\frac i2)=-4G_2(2i)+4\pi\end{cases}\quad$. This is exactly was needed to get the desired result.

Hitoshigoto oshimai !

I find that sum fascinating. $e,\pi$ all together to finally get a rational. This is why mathematics is beautiful!

Thanks to everyone who contributed.

答:
The calculation of the Mellin transform of $f(x)$ is not present in the above answer, so I will show it here.

$$\mathfrak{M}\left(\frac{1}{e^{\pi x}+1};s \right) =
\int_0^\infty \frac{1}{e^{\pi x}+1} x^{s-1} dx =
\int_0^\infty \frac{1}{e^{\pi x}} \frac{1}{1+e^{-\pi x}} x^{s-1} dx \\=
\int_0^\infty \frac{1}{e^{\pi x}} \sum_{q\ge 0} (-1)^q e^{-\pi q x} x^{s-1} dx =
\int_0^\infty \sum_{q\ge 0} (-1)^q e^{-\pi (q+1) x} x^{s-1} dx \\
= \Gamma(s) \sum_{q\ge 0} (-1)^q \frac{1}{\pi^s (q+1)^s}
= \frac{1}{\pi^s} \Gamma(s) (\zeta(s) - 2 \times 2^{-s} \times \zeta(s))
= \frac{1}{\pi^s} \Gamma(s) (1 - 2\times 2^{-s}) \zeta(s).$$
It now follows from the definition of the Mellin transform that
$$\mathfrak{M}\left(\frac{x}{e^{\pi x}+1};s \right) =
\frac{1}{\pi^{s+1}} \Gamma(s+1) (1 - 2^{-s}) \zeta(s+1).$$


问:
Min-Max recurrence relations
Now here I am not getting any idea whether it's linear homogeneous or linear non-homogeneous equation. How can this equation be solved correctly?</p>

答:
Hint:

Play with the first few $T(n)$'s and you'll discover that everything depends on a simple relationship between $T(1)$ and $T(2)$ in a consistent way.

!
If $2x+2>y$, the first few $T(n)$'s are: $$x, y, x+y+2, 2y+2, x+2y+4, 3y+4, $$
If $2x+2\leqslant y$, the first few $T(n)$'s are: $$ x, y, x+y+2, 2x+y+4, 3x+y+6,4x+y+8,$$
Do you see the pattern?

Title: Min-Max recurrence relations


问:
By using the properties of definite integrals, evaluate $\int_0^{\pi}\frac{x}{1+\sin x}dx$

By using the properties of definite integrals, evaluate $\int_0^{\pi}\frac{x}{1+\sin x}dx$.

My attempt:

(Using the property $\int_0^{2a}f(x)dx=\int_0^a(f(x)+f(2a-x))dx$)

$$\int_0^{2\frac{\pi}{2}}\frac{x}{1+\sin x}dx=\int_0^{\frac{\pi}{2}}(\frac{x}{1+\sin x}+\frac{\pi-x}{1+\sin x})dx$$$$=\pi\int_0^{\frac{\pi}{2}}\frac1{1+\sin x}dx=\pi\int_0^{\frac{\pi}{2}}\frac{1-\sin x}{\cos^2x}dx$$$$=\pi\int_0^{\frac{\pi}{2}}(\sec^2x-\sec x\tan x)dx=\pi[\tan x-\sec x]_0^{\frac{\pi}{2}}$$

Now I am stuck. I understand there might be other ways of solving it, but what's wrong in my method? Why am I not getting the answer?

答: $1-\sin x=0$ for $x=\pi/2$ and so you're dividing by zero when multiply and divide by that expression.


问:
Proving that a composition is a linear transformation

Let $T: U \rightarrow V$ be an isomorphism.

Given that $S: V \rightarrow U$ satisfies $T \circ S = Id_V$ (where
$Id_V (v) = v$, for all $v \in V$), show that $S$ is a linear
transformation.

I honestly have no clue in solving this exercise. I would really appreciate a hint. How can I even start?

答:
Observe that $S$ is in fact the inverse transformation of $T$. If this does not appear completely obvious, we can show it like this: Observe that since $T$ is an isomorphism, it is invertible. We apply it to the equation from the question and get

$$T^{-1} \circ (T \circ S) = T^{-1} \circ \text{id}_V$$

$$(T^{-1} \circ T) \circ S = T^{-1}$$

$$ \text{id}_V \circ S = S = T^{-1}$$

Now, since $T$ is an isomorphism, its inverse is so as well, and in particular it is linear. You can also show this from the definition, if you want to.


问:
About the eigenvalues and singular values of this matrix
Let be
$T_n=\begin{bmatrix}0&...&0&1&...&1\\ \vdots&&\vdots&\vdots&&\vdots\\0&...&0&1&...&1\\1&...&1&0&...&0\\\vdots&&\vdots&\vdots&&\vdots\\1&...&1&0&...&0 \end{bmatrix}$, proof that the eigenvalues of $T_n$ are $0,-n,n$.

My proof

$p(\lambda)=det(\lambda I-T_n)=det\begin{bmatrix}
\lambda^2-n&-n&...&-n\\-n&\lambda^2-n&...&-n&\\ \vdots&&\ddots&\vdots\\-n&&...&-n
\end{bmatrix}$
Let $D=diag(\lambda^2,...,\lambda^2)$ and $v$ the column vector with $n$ ones, note that
$\begin{bmatrix}
\lambda^2-n&-n&...&-n\\-n&\lambda^2-n&...&-n&\\ \vdots&&\ddots&\vdots\\-n&&...&-n
\end{bmatrix}=D-nvv^T$, we have that $det(D-nvv^T)=det(D)-nv^TAdj(D)v=\displaystyle\prod_{j=1}^{n}\lambda^2-n\displaystyle\sum_{i=1}^{n}\displaystyle\prod_{k\neq i}^{n}\lambda^2=\\=\lambda^{2n}-n(n\lambda^{2(n-1)})=\lambda^{2n}-n^2\lambda^{2(n-1)}=\lambda^{2(n-1)}(\lambda^2-n^2)$

Question

the afirmation $det(A+uv^T)=det(A)+v^TAdj(A)u$ I found in Wikipedia Matrix determinant lemma, but I don't have able to proof it then I try to proof that if $\lambda$ is nonzero then $\begin{bmatrix}
\lambda^2-n&-n&...&-n\\-n&\lambda^2-n&...&-n&\\ \vdots&&\ddots&\vdots\\-n&&...&-n
\end{bmatrix}$ is nonsingular and so I can use that $det(A+uv^T)=det(A)(1+v^TA^{-1}u)$ that is easier to proof, I think the problem is calculate $det\begin{bmatrix}
\lambda^2-n&-n&...&-n\\-n&\lambda^2-n&...&-n&\\ \vdots&&\ddots&\vdots\\-n&&...&-n
\end{bmatrix}$, there exist a alternative way to calculate it?

答:
We can find $2n$ eigenvectors and the corresponding eigenvalues.

$$
v_1=(1,1,\ldots,1) \quad\text{corresponds to the eigenvalue}\quad \lambda_1=n
$$
$$
v_2=(1,1,\ldots,1,-1,-1,\ldots,-1) \quad\text{corresponds to the eigenvalue}\quad \lambda_1=-n
$$
If
$$
A=\left(\begin{array}{ccccccccccccc}
1&1&1&\cdots &1&1 \\
1&1&1&\cdots &1&1 \\
\vdots & \vdots & \vdots &&\vdots&\vdots  \\
1&1&1&\cdots &1&1
\end{array}\right)
$$
and $Aw=0$, then
$$
T_n\binom{w}{w}=T_n\binom{w}{-w}=0.
$$
There are $n-1$ linearly independent such $w$'s:
$$
(1,-1,0,0,\ldots,0),
(1,0,-1,0,\ldots,0),
(1,0,0,-1,,\ldots,0),\ldots,
(1,0,0,,\ldots,-1,0),
(1,0,0,,\ldots,0,-1)
$$
and hence $2n-2$ linearly independent vectors corresponding to the eigenvalue 1.


问:
Prove that the mapping is order preserving
$ P=Q=\{ \mathcal{P}(S); \subseteq \} $ and $\phi$ from $P \rightarrow Q$ is defined by
$\\  \phi (U)= \left\{ \begin{array}{ll}
\{1\} & \text{if} \; 1 \in U \\
\{2\} & \text{if} \; 2\in U, \; 1 \notin U\\
\emptyset & \text{otherwise} \\
\end{array} \right. $

答: It's not order preserving, since $\{2\}\subseteq\{1,2\}$, but $\phi(\{2\})=\{2\}\nsubseteq\{1\}=\phi(\{1,2\}).$


问:
Convergent series $\iff$ absolutely convergent
Consider the measure space $(\mathbb{N}, 2^{\mathbb{N}},\#)$, where $\#$ is the counting measure.

It is easy to show that for $f: \mathbb{N} \rightarrow \mathbb{R}_{\geq 0}$ we have $\int_{\mathbb{N}}f d\#=\sum_nf(n)$.

Now, suppose $f: \mathbb{N} \rightarrow \mathbb{R}$. Then $f \in L^1$ if and only if $\int_{\mathbb{N}}f^+ d\#$ and $\int_{\mathbb{N}}f^- d\#$ are finite. So, if $f\in L^1$ we can write $\int_{\mathbb{N}}f d\#=\sum_nf^+(n)-\sum_nf^-(n)=\sum_nf(n)$, for the properties of convergent series.

This, combined with $f \in L^1 \iff |f| \in L^1$, yields that convergence is equivalent to absolute convergence, for series.

Can you help me spot the mistake?

答:
The statement "This, combined with $f \in L^1 \iff |f| \in L^1$, yields that convergence is equivalent to absolute convergence, for series." is false.

If $\sum_n f(n)$ converges and $\sum_n |f(n)|=+\infty$, then you have proved nothing for $f$, since it does not satisfy the hypoteses of the properties you have proved (in fact in this case $f$ is not in $L^1(\mathbb{N}, 2^{\mathbb{N}},\#)$).

According to measure therory, $\int_{\mathbb{N}} f d\#$ is defined for nonnegative $f$ first and it equals $\sum f(n)$. Then, the definition is extended for some functions $f$ assuming also negative values: in this case $\int_{\mathbb{N}} f d\#$ is defined as $\sum_nf^+(n)-\sum_nf^-(n)$ if at least one of the terms $\sum_nf^+(n), \sum_nf^-(n)$ is finite. If $\sum f(n)$ is convergent but not absolutely convergent, then $\int_{\mathbb{N}} f d\#$ is undefined. Have a look at Adam's answer.


问:
Describing sample space of given experiments

The question is to describe the sample space for the experiment of
randomly picking an arc from a circle and this defining of $\Omega$
should be done in two ways.

I really don't understand the experiment. I can choose an arc in infinitely many ways. The length of an arch can be given by

$$\ell=\frac{\theta 2\pi r}{2\pi} = r\theta.$$

Thus the sample space becomes

$$\Omega=\{r,\theta : \theta\in(0,2\pi], \ r=\text{constant in } \mathbb{R}^+\}.$$

Is this what they are looking for? Also, what is the other way?

答:
The central angle of an arc uniquely determines its length, but not its position. Minor arcs between $0,\pi/4$ and $\pi/4,\pi/2$ have the same length but they are not the same arc.

Choosing two bounding angles of the arc does not indicate whether we are talking about the minor or major arc bounded between the angles.

Both the above methods don't differentiate between an arc that contains its end points and one which doesn't.

One method to create the sample space is to use the bounding angles with an interval-esque notation along with explicitly stating that we are measuring the arcs anti-clockwise.

So $(\pi/4,\pi)$ denotes the arc from $\pi/4\to\pi$ where the end points are not included, like an open interval. But $[\pi/4,\pi]$ is the same arc plus its end points. $[\theta,\theta]$ denotes the minor arc that contains only one point, $[\theta,\theta),(\theta,\theta]$ denote the entire circle, $(\theta,\theta)$ denotes the major arc excluding one point. The issue here is that the entire circle is denoted by multiple sample space points.

So the sample space is$$S=\{[\theta_1,\theta_2],(\theta_1,\theta_2):\theta_{1,2}\in[0,2\pi)\}\cup\{[\theta_1,\theta_2),(\theta_1,\theta_2]:\theta_{1,2}\in[0,2\pi),\theta_1\ne\theta_2\}\cup\{[0,0)\}$$

I am not sure how relevant differentiating between an arc that does and does not contain its end points is to your example.


答:
Your final integral is an improper integral, so you do not evaluate the antiderivative you obtain at the endpoints, you take limits approaching the endpoints.

\begin{align*}
\lim_{x \rightarrow 0^+} (\tan x - \sec x) &= -1  \\
\lim_{x \rightarrow (\pi/2)^-} (\tan x - \sec x) &= 0  \text{.}
\end{align*}

So you obtain $\pi(0--1) = \pi$, which is the correct answer.  (The first limit is not actually necessary, since the antiderivative is continuous at $x = 0$.)

Body:
By using the properties of definite integrals, evaluate $\int_0^{\pi}\frac{x}{1+\sin x}dx$.

My attempt:

(Using the property $\int_0^{2a}f(x)dx=\int_0^a(f(x)+f(2a-x))dx$)

$$\int_0^{2\frac{\pi}{2}}\frac{x}{1+\sin x}dx=\int_0^{\frac{\pi}{2}}(\frac{x}{1+\sin x}+\frac{\pi-x}{1+\sin x})dx$$$$=\pi\int_0^{\frac{\pi}{2}}\frac1{1+\sin x}dx=\pi\int_0^{\frac{\pi}{2}}\frac{1-\sin x}{\cos^2x}dx$$$$=\pi\int_0^{\frac{\pi}{2}}(\sec^2x-\sec x\tan x)dx=\pi[\tan x-\sec x]_0^{\frac{\pi}{2}}$$

Now I am stuck. I understand there might be other ways of solving it, but what's wrong in my method? Why am I not getting the answer?


问:
Reversing the order of integration for this double integral
$\int_{0}^{2}\int_{2x}^{4x-x^2}1\,dy\,dx $

For this problem, I've graphed the region and determined the bounds for $y$ as $ 0 \leq y \leq 4$. I'm trying to determine the functions of $y$ that bound the $x$ values. Well, I know the upper bound is $x = y/2$.

答: When $x$ goes from $0$ to $2$, both $2x$ and $4x-x^2$ goes from $0$ to $4$, and $4x-x^2\geqslant2x$. So, when $y$ goes from $0$ to $4$, $x$ goes from $2-\sqrt{4-y}$ to $\frac y2$. Therefore, the answer is$$\int_0^4\int_{2-\sqrt{4-y}}^{y/2}1\,\mathrm dx\,\mathrm dy.$$


问:
Proof that the trace of a matrix is the sum of its eigenvalues
I have looked extensively for a proof on the internet but all of them were too obscure. I would appreciate if someone could lay out a simple proof for this important result. Thank you.</p>

答: By the Schur decomposition, any matrix $A$ is unitarily similar to an upper triangular matrix $T$. Being similar, $A$ and $T$ have the same trace and the same eigenvalues. Moreover, the diagonal entries of $T$ are equal to its eigenvalues (since $T$ is triangular). The stated result follows by calculating the trace of $T$. See https://www.statlect.com/matrix-algebra/properties-of-eigenvalues-and-eigenvectors.


问:
Prove that the matrices $H_\phi^{(n)}$ and $H_\psi^{(n)}$ are similar.
For any map $f: \mathbb{N}\to \mathbb{R}$ and any strictly positive integer $n$, We denote by $H_f^{(n)}$ the matrix of $\mathcal{M}_n$ given by:
$$
H_f^{(n)}=(f(i + j -2))_{1\leq i,j\leq n} =
$$
Let $n\geq 3$ and $\phi$ and $\psi$ are  the functions defined on $\mathbb{N}$ by:
$$
\forall k\in  \mathbb{N},\qquad \phi (k) = (- 1)^ k (k + 1) \qquad,\qquad \psi (k) =k + 1
$$

Problem

Prove that the matrices $H_\phi^{(n)}$ and $H_\psi^{(n)}$ are similar.

An idea please

答: Each entry of $H_\phi^{(n)}$ has the same modulus as its counterpart in $H_\psi^{(n)}$. The only difference between the two matrices is a checkerboard sign pattern. Therefore $H_\phi^{(n)}=DH_\psi^{(n)}D^{-1}$ where $D=D^{-1}=\operatorname{diag}(1,-1,1,-1,\ldots,(-1)^{n-1})$, and the two matrices are similar.


答:
Let $v_1,v_2\in V$. Then, for any scalar $a$,
\begin{align}
T(S(av_1+v_2)) &= av_1+v_2 \\
&= aT(S(v_1)) + T(S(v_2)) \\
&= T(aS(v_1)+S(v_2)) \qquad \textrm{since $T$ is linear}
\end{align}
and then, injectivity implies that $S(av_1+v_2) = aS(v_1)+S(v_2)$, that is, $S$ is linear.

Title: Proving that a composition is a linear transformation

Body:
Let $T: U \rightarrow V$ be an isomorphism.

Given that $S: V \rightarrow U$ satisfies $T \circ S = Id_V$ (where
$Id_V (v) = v$, for all $v \in V$), show that $S$ is a linear
transformation.

I honestly have no clue in solving this exercise. I would really appreciate a hint. How can I even start?


问:
Doob's martingale inequality. Why does $(2)$ hold true?

Doob's Martingale Inequality Let $M=\left(M_n\right)_{n\ge0}$ be a martingale or a positive submartingale. Set $M^*_n=\sup_{j\le n}|M_j|$. Then
$$\mathbb{P}\left(M_n^*\ge \alpha\right)\le\frac{\mathbb{E}\left\{|M_n|\right\}}{\alpha}\tag{1}$$

Does $(1)$ imply that for all $p\ge1$:
$$\mathbb{P}\left(M_n^*\ge \alpha\right)\le\frac{\mathbb{E}\left\{|M_n|^{\color{red}{p}}\right\}}{\alpha^{\color{red}{p}}}\tag{2}$$?

 If so, does that simply follow from the fact that:
$$\mathbb{P}\left(M_n^*\ge \alpha\right)=\mathbb{P}\left((M_n^*)^p\ge (\alpha)^p\right)\le\frac{\mathbb{E}\left\{|M_n|^\color{red}{p}\right\}}{\alpha^\color{red}{p}}\tag{3}$$?

答: $\phi(x)=|x|^p$ is a positive convex function therefore $( |M_n|^p,n \ge 0)$ is also a submartingale (as long as it is in $L^p$). Hence, your implication is true.


问:
Every image of an immersion is an image of an injective immersion?
Let $M,N$ be smooth connected* manifolds, with or without boundary.
Let $f:M \to N$ be a smooth immersion. Can we realize $f(M)$ as an image of some injective immersion into $N$?

That is, does there exist a manifold $\tilde M$ and an injective smooth immersion $j:\tilde M \to N$, such that $j(\tilde M)=f(M)$?

I am particularly interested in the case where $\dim M=\dim N$, and $\partial M \neq \emptyset$. (If $\partial M = \emptyset$, the image is open, hence an embedded submanifold of $N$).

*As commented below, if we assume $M$ is not connected, there probably are examples where the image cannot be realized as an image of an injective immersion.

答:
This is a counterexample for manifolds without boundary: consider the map $\varphi : M = \mathbb{D}_1^2 \sqcup \mathbb{D}_2^2 \to \mathbb{R}^3$ given by $(u,v) \mapsto (u, v, 0)$ in the open unit disk $\mathbb{D}_1^2$ and $(u, v) \mapsto (u + 1, 0, v)$ in another open unit disk $\mathbb{D}_2^2$. Clearly $\varphi$ is an immersion and I claim that $\varphi(M)$ is not the image of an injective immersion $\psi : L \to \mathbb{R}^3$. If so, for every point $p \in \varphi(M)$ there’s a unique $q \in L$ with $\psi(q) = p$, so around $q$, $\psi$ defines a small embedded disk contained in $\varphi(M)$, which therefore can be either parallel to the $z=0$ plane or to the $y=0$ plane. This dichotomy allows us to partition the segment $I = \{(x, 0, 0) \mid -\frac{1}{2} \leq x \leq \frac{3}{2}\}$ into two sets $V_1$ and $V_2$. Of course these two sets are nonempty, so if we prove they are open, this will contradict the fact that $I$ is connected. If for $p = \psi(q) \in V_i$ we have that $p \in \psi(U) \subseteq \varphi(M)$ is a small embedded disk around $p$, then for any $p' \in \psi(U) \cap I$, $\psi(U)$ also serves as an embedded disk around $p'$, so $\psi(U) \cap I \subseteq V_i$, and we are done because $\psi(U)$ is open in $\varphi(\mathbb{D}_i^2)$ (thanks to the inverse function theorem) and so $\psi(U) \cap I$ is open in $I$.

If you insist that $M$ is connected, you can just add a strip joining the two disks away from the part where they intersect.


问:
Integration with a Fourier transform
Consider a function $f$ that has smooth Fourier transform $\widehat{f}$ with compact support. In  a particular problem I am considering, it would be useful to be able to compute the integral

$$\int_{\mathbb{R}} \widehat{f}(u) \ \frac{(e^{h-2\pi i u}-1)}{h-2\pi i u} \ du$$

where $h >0$. Is anything known about this integral? In particular can we compute this integral for general $f$ as above?

Any help/ideas/comments will be appreciated, cheers.

答:
$$
      \int_{-\infty}^{\infty}\widehat{f}(u)\left.\frac{e^{t(h-2\pi iu)}}{h-2\pi iu}\right|_{t=0}^{1}du \\
      = \int_{-\infty}^{\infty}\widehat{f}(u)\int_0^1e^{t(h-2\pi iu)}dt du \\
      = \int_{0}^{1}e^{th}\int_{-\infty}^{\infty}\widehat{f}(u)e^{-2\pi itu} du dt \\
      = \int_{0}^{1}e^{th}f(-t)dt
$$

Title: Integration with a Fourier transform


问:
Uniform continuity over subsets of closed intervals
Could someone please verify my solution to the second part of Carothers, Ch. 8 Exercise 50?

If $f$ is uniformly continuous over $(0, 2)$ and on $(1,3)$ is $f$ uniformly continuous on $(0,3)$? If $f$ is uniformly continuous on $[n, n+1]$ for every $n\in \mathbb{Z}$ is $f$ necessarily uniformly continuous on $\mathbb{R}$?

I believe the first part is already answered here, but the closest answer I could find for the second part was here, which is actually its converse.

Answer: If $f$ is uniformly continuous on $[n, n+1]$ for every on $n\in \mathbb{Z}$, the function is not necessarily uniformly continuous on $\mathbb{R}$. For instance, the function taking the values $f(x) = x^2$ is not uniformly continuous over $\mathbb{R}$ but over the interval $[n, n+1]$, it is (which I try to show below).

Because $f'$ increases monotonically, $|f'(x)| = 2|x|$ attains a maximum at $n$ or $n+1$ and so it is bound by $2\max(|n|, |n+1|)$. The maximum change that can occur in a given function over the chosen interval is $|f(x) - f(x+\delta)| \leq|\delta|(2\max(|n|, |n+1|)$ and so $\frac{|f(x)-f(x+\delta)|}{2\max(|n|, |n+1|)} \leq \delta$ and so it should be sufficient to choose
$$|\delta| < \min\Big(\frac{\epsilon}{2\max(|n|, |n+1|}, x-n, n+1-x \Big)$$
where $|\delta| < |x-n|$ and $|n+1-x|$ to ensure that $|x|$ and $|x-\delta|$ remain in the interval $[n, n+1]$. Because $\delta$ does not depend on $x$ (besides restricting it to $[n, n+1]$), the function $f$ is uniformly continuous over $[n, n+1]$.

Is this solution ok? I tried to plug $\delta$ into $|x^2 - (x-\delta)^2|$ but started getting confused with all the inequalities.

Thank you!

答: For the second, part consider $f(x)=\exp(x)$.


问:
Can a disconnected set disconnect a connected set 2?
See Can a disconnected set disconnect a connected set?. In the given counter examples $F$ is unbounded. I change a little the same question. Suppose $V$ is a connected open set in $\mathbb{R}^n$ with $n>1$ and $F$ a closed set in $\mathbb{R}^n$. If $V\setminus F$ is disconnected, can $F$ have a "bounded component" meaning  $\partial V\cap F=\emptyset$? ($\partial V$ is the boundary of $V$)

答: Take $V$ to be the ball of radius $2$ in $\mathbb{R}^2$ and $F = \mathbb{S}^1 = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 = 1\} \subset V$. Then $V \setminus F$ has two components and $\partial V \cap F = \varnothing$.


问:
Iterative Methods for Solving a PDE?
Given the following PDE: $-\Delta u + u = 1$ in the domain $[0,1]^2$ and $u = 0$ as the boundary condition, I want to solve for the solution via an iterative method (Gauss-Seidel / Jacobi, for example).

If I let $h = 1/(n+1)$ the unknowns in the following equation are $x_{i,j}$ for $\space 1, ..., n$ that satisfy:
$$(4+h^2)x_{i,j} - x_{i-1,j} - x_{i,j-1} - x_{i+1,j} - x_{i,j+1} = h^2$$ for $i,j = 1, ... , n$

and
$$x_{0,j} = x_{n+1,j} = x_{i,0} = x_{i,n+1} = 0, \quad \text{for} \quad i,j = 1, ..., n$$

Question:

I believe the unknowns $x$ are a $n \times n$ matrix, which would make $A$ in $Ax = b$ a 4th order tensor, but does $A$ need to be constructed in order to implement an iterative method? If not why? If so, how would $A$ look? Thanks!

答:
This is a partial answer to address what $A$ can look like.
I don't know iterative methods.
Suppose we arrange the $x_{ij}$ into a column vector with
$x_{11},\ldots,x_{1n}$ as the first $n$ entries, followed by
$x_{21},\ldots,x_{2n}$ and so forth down to $x_{nn}$.
Then most rows of $A$
have $d = 4+h^2$  on the diagonal with a $-1$ to each side, and two more $-1$'s located $n$ places to left and right, as below. But this has to be modified for
points $(i,j)$ adjacent to the boundary. The $4\times 4$ case is
representative:
$$
\begin{smallmatrix}
 d &-1 & 0 & 0 &-1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \cr
-1 & d &-1 & 0 & 0 &-1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \cr
 0 &-1 & d &-1 & 0 & 0 &-1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \cr
 0 & 0 &-1 & d & 0 & 0 & 0 &-1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \cr
-1 & 0 & 0 & 0 & d &-1 & 0 & 0 &-1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \cr
 0 &-1 & 0 & 0 &-1 & d &-1 & 0 & 0 &-1 & 0 & 0 & 0 & 0 & 0 & 0 \cr
 0 & 0 &-1 & 0 & 0 &-1 & d &-1 & 0 & 0 &-1 & 0 & 0 & 0 & 0 & 0 \cr
 0 & 0 & 0 &-1 & 0 & 0 &-1 & d & 0 & 0 & 0 &-1 & 0 & 0 & 0 & 0 \cr
 0 & 0 & 0 & 0 &-1 & 0 & 0 & 0 & d &-1 & 0 & 0 &-1 & 0 & 0 & 0 \cr
 0 & 0 & 0 & 0 & 0 &-1 & 0 & 0 &-1 & d &-1 & 0 & 0 &-1 & 0 & 0 \cr
 0 & 0 & 0 & 0 & 0 & 0 &-1 & 0 & 0 &-1 & d &-1 & 0 & 0 &-1 & 0 \cr
 0 & 0 & 0 & 0 & 0 & 0 & 0 &-1 & 0 & 0 &-1 & d & 0 & 0 & 0 &-1 \cr
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &-1 & 0 & 0 & 0 & d &-1 & 0 & 0 \cr
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &-1 & 0 & 0 &-1 & d &-1 & 0 \cr
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &-1 & 0 & 0 &-1 & d &-1 \cr
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &-1 & 0 & 0 &-1 & d
\end{smallmatrix}
$$
Here in rows 4,5,8,9,12,13, or more generally in matrix rows $kn,(k+1)n$,
you omit one $-1$ because you are at the end of a row of nodes
adjacent to a zero boundary value.
Also in the first $n$ and last $n$  matrix rows you omit one or two
$-1$'s because you are on the first and last row of nodes that have
zero boundary values adjacent.

That $4\times 4$ can be expressed as
$$
A = I\otimes\begin{bmatrix}d & -1 & 0 & 0 \cr
                              -1 & d & -1 & 0 \cr
                               0 & -1 & d & -1 \cr
                               0 & 0 & -1 & d\end{bmatrix}
 +\begin{bmatrix}0 & 1 & 0 & 0 \cr
                 1 & 0 & 1 & 0 \cr
                 0 & 1 & 0 & 1 \cr
                 0 & 0 & 1 & 0 \end{bmatrix}\otimes I.
$$

Title: Iterative Methods for Solving a PDE?


问:
On Riesz mean ergodic theorem(questions)
I am reading Riesz mean ergodic theorem proof and have some questions.

I don't understand where $||p_x-z||<\epsilon$ comes from.I know that if $A\subset R$ and a=infA then for every $\epsilon$>0 i can find another element a' $\in A$ such that $a'<a+\epsilon$ but i don't think this is what he used.Using this i can derive the upper bound of $||z||$ but the author used the fact that $||p_x-z||<\epsilon$ and $||z||=||z-p_x+p_x||$,then apply triangle inequality and get $||z||\leq \mu +\epsilon$.My question is how does one conclude that $||p_x-z||<\epsilon$?

My second question is about the lower of bound of ||z||:Is he using the boundedness of T ?
For example $||T^2x||=||T(Tx)||\leq||Tx||\leq ||x||$ and so on for all $T^nx$ .From that we get using the triangle inequality :
$||\frac{z+Tz+...+T^nz}{n+1}||\leq ||\frac{n+1||z||}{n+1}||=||z||$ ,is this what he did?

My last question is why the highlighted limit vanishes?

I think that if $T^nx$ converges then it must converge to the fixed point.If $T^nx$ converges to some y then since T is bounded,T is also continuous and so :

$Ty=T(\lim_{n}T^nx)=\lim_{n}T(T^nx)=\lim_{n}T^{n+1}x=y$.Is this what the author implies?

答:
Not so few questions. I will try to answer each of them.

Regarding the last term, this one vanishes because there is the term $\frac{1}{n+1}$ in the limit, which converges to $0$.

Regarding $\Vert p_x -z \Vert< \epsilon$, we can choose such point and such representation of $z$ because $C$ is the closure of a convex set formed by finite linear (positive) combination of the points $\left(T^n x , n \ge 0 \right)$
Concerning the term bounded above by $\Vert z \Vert $, yes, you're right, that's it.

About your last question concerning the idea of author, my answer is "No, he's not meant that". The convergence of $T^n$ needs not to happen everytime, what we can be sure of is the convergence the mean $T^n$. And if we denote $P$ for the mean operator. The author was trying to show that: $TP=P$. I.e $Range(P) $ is a set containing only fixed points of $T$( in this case the vice versa is also true).


问:
Proof that any tangent to an hyperbola bisects the angle of the point with the foci
I'm trying to solve this problem in my text book.

The approach I want to try is as follows:

The tangent vector at P is $\vec{v_h} = (1, y')$. Where y' is the derivative of the hyperbola function.

Given $\vec{u_1} = \frac{\vec{F'P}} {||\vec{F'P}||}$ and $\vec{u_2} = \frac{\vec{FP}} {||\vec{FP}||}$, these dot products should be equal $\vec{v_h} \cdot \vec{u_1} = \vec{v_h} \cdot \vec{u_2} $ because they are supposed to form the same angle with the tangent line.

My attempt:

Start with the equation of the hyperbola:
$$
b^2x^2 - a^2y^2=a^2b^2
$$

In this formula, c is half the distance between the foci.

Then isolate y:

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

Then we compute the derivative of y which tells us the slope of the function.

$$
y' = \frac{b^2}{a^2} \frac{x}{y}
$$

With the derivative we can get a direction vector for the slope.
$$
\vec{v_h} = (1, y')
$$

And we compute the unit vectors that go from the foci to P.

$$
\vec{u_1} = \frac{1}{\sqrt{(c+x)^2 + y^2}} (c+x, y)
$$
$$
\vec{u_2} = \frac{1}{\sqrt{(c-x)^2 + y^2}} (c-x, y)
$$

If the angles between he tangent and the foci are the same, the cosine of the angle is too, and in this case, the dot product is the cosine of the angle scaled by the length of $\vec{v_h}$.

$$
\vec{v_h} \cdot \vec{u_1} = \vec{v_h} \cdot \vec{u_2}
$$

$$
\frac{c+x + y y'} {\sqrt{(c+x)^2 + y^2}} = \frac{c-x + y y'}{\sqrt{(c-x)^2 + y^2}}
$$

From this point, I have tried many things but I'm unable to prove that both sides are equal. Can you please help me?

答:
It took me a while but managed to solve it.

This equation was wrong.

$$
\frac{c+x + y y'} {\sqrt{(c+x)^2 + y^2}} = \frac{c-x + y y'}{\sqrt{(c-x)^2 + y^2}}
$$

It should be

$$
\frac{c+x + y y'} {\sqrt{(c+x)^2 + y^2}} = \frac{x-c + y y'}{\sqrt{(x-c)^2 + y^2}}
$$

We know that

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

$$
y' = \frac{b^2}{a^2} \frac{x}{y}
$$

We are going to substitute the $b$ using this equation

$$
b^2 = c^2 - a^2
$$

Then

$$
y = \pm \sqrt{\frac{c^2 - a^2}{a^2}x^2 - (c^2 - a^2)}
$$

$$
y' = \frac{c^2-a^2}{a^2} \frac{x}{y}
$$

If we substitute $y'$ in the first equation, $y$ cancels out nicely.

$$
\frac{c+x + \frac{c^2-a^2}{a^2}x} {\sqrt{(c+x)^2 + y^2}} = \frac{x-c + \frac{c^2-a^2}{a^2}x}{\sqrt{(x-c)^2 + y^2}}
$$

Rearange fractions
$$
\frac{\sqrt{(x-c)^2 + y^2}}{\sqrt{(x+c)^2 + y^2}}=
\frac{\frac{c^2}{a^2}x-c}{\frac{c^2}{a^2}x+c}
$$

Square both sides

$$
\frac{(x-c)^2 + y^2}{(x+c)^2 + y^2}=
\left[\frac{\frac{c^2}{a^2}x-c}{\frac{c^2}{a^2}x+c} \right] ^2
$$

Expand the squares

$$
\frac{x^2 + c^2 - 2cx + y^2}{x^2 + c^2 + 2cx + y^2} = \frac{\frac{c^4}{a^4}x^2 + c^2 -2\frac{c^3}{a^2}x}{\frac{c^4}{a^4} x^2 + c^2 +2\frac{c^3}{a^2}x}
$$

On the right hand side, divide numerator and denominator by $c^2$

$$
\frac{x^2 + c^2 - 2cx + y^2}{x^2 + c^2 + 2cx + y^2} = \frac{\frac{c^2}{a^4}x^2 + 1 -\frac{2c}{a^2}x}{\frac{c^2}{a^4} x^2 + 1 +\frac{2c}{a^2}x}
$$

Move denominators to the other side
$$
(x^2 + c^2 - 2cx + y^2) (\frac{c^2}{a^4} x^2 + 1 +\frac{2c}{a^2}x) =
\\
(x^2 + c^2 + 2cx + y^2) (\frac{c^2}{a^4}x^2 + 1 -\frac{2c}{a^2}x)
$$

Now we do the multiplications. Most terms are equal on both sides so they cancel out.

$$
\frac{2c}{a^2} x^3 + \frac{2c^3}{a^2} x - \frac{2c^3}{a^4} x^3 - 2cx + \frac{2c}{a^2}xy^2 =
\\
-\frac{2c}{a^2} x^3 - \frac{2c^3}{a^2} x + \frac{2c^3}{a^4} x^3 + 2cx - \frac{2c}{a^2}xy^2
$$

Move everything to the same side and divide by $4cx$

$$
\frac{x^2}{a^2} + \frac{c^2}{a^2} - \frac{c^2}{a^4}x^2 - 1 + \frac{y^2}{a^2} = 0
$$

Gather fractions
$$
\frac{x^2 + c^2 + y^2 - a^2}{a^2} - \frac{c^2}{a^4}x^2 = 0
$$

Multiply by $a^2$, then substitute $y^2$

$$
x^2 + c^2 +\frac{c^2 - a^2}{a^2}x^2 - c^2 +a^2 -a^2 -\frac{c^2}{a^2}x^2 = 0
$$

Now we see a few terms cancel out.

$$
x^2 +\frac{c^2 - a^2}{a^2}x^2 -\frac{c^2}{a^2}x^2 = 0
$$

Take $x^2$ as common factor

$$
\frac{c^2 - a^2}{a^2}x^2 -(\frac{c^2}{a^2}-1)x^2 = 0
$$

Substract the 1 to the fraction

$$
\frac{c^2 - a^2}{a^2}x^2 -\frac{c^2 - a^2}{a^2}x^2 = 0
$$

Finally

$0 = 0$

So the angles are the same.


问:
Bayesian Posterior Density Derivation
Assume that $Y_{i} \sim N\left(\theta x_{i}^{2}, \sigma^{2}\right)$ for $i = 1,\ldots,n$ independently. Consider a Bayesian model with a known $\sigma = 1$ and with a prior distribution $\theta \sim N\left(0, \sigma_{0}^{2}\right)$ for some fixed value $\sigma_0 > 0$.

Show that the posterior density of $\theta$ is $$\theta \mid \mathbf{y} \sim N\left(\tilde{\theta},\left(\sum_{i=1}^{n} x_{i}^{4}+\sigma_{0}^{-2}\right)^{-1}\right)$$

and find $\tilde{\theta}$

By using the fact
$p(\theta\mid \textbf{y}) \propto {f (\textbf{y}|\theta)p(\theta)}$, how do i go about deriving the desired posterior density?

Thank you

答:
Ok Let's start!

$$p(\theta|\mathbf{y}) \propto Exp\Bigg\{-\frac{\theta^2}{2\sigma_0^2}\Bigg\}Exp\Bigg\{-\frac{1}{2}\Sigma_i(y_i-\theta x_i^2)^2\Bigg\}$$

Now let's expand the exponent throwing away any element not depending on $\theta$ (it will be part of the normalization constant)

$$\propto Exp\Bigg\{-\frac{\theta^2}{2\sigma_0^2}+\frac{2\theta\Sigma_i y_ix_i^2}{2}-\frac{\theta^2\Sigma_i x_i^4}{2}\Bigg\}$$

$$\propto Exp\Bigg\{-\frac{1}{2}\Big[\theta^2\Big(\sigma_0^{-2}+\Sigma_i x_i^4\Big)-2\theta\cdot \Sigma_i y_ix_i^2\Big]\Bigg\}$$

$$\propto Exp\Bigg\{-\frac{1}{2\Big(\sigma_0^{-2}+\Sigma_i x_i^4\Big)^{-1}}\Bigg[\theta^2-2\theta\cdot\underbrace{\frac{\Sigma_iy_i x_i^2}{\sigma_0^{-2}+\Sigma_i x_i^4}}_{=\tilde{\theta}}+?\Big]\Bigg\}$$

As you can see in the exponent there is a missing square. To complete the gaussian is enough to sum and subtract the missing element, $\pm(\tilde{\theta})^2$, and multiply all by the normalization constant (but it is not necessary to answer to your question).

The mean you are requesting to shoe, in fact, is exactly the expression of $\tilde{\theta}$ I underbraced.

The variance they ask to show is evidently written...

I hope this help.

Title: Bayesian Posterior Density Derivation


问:
Order preserving bijection between $(a,b)$ and $(c,d)$ to $(e,f)$ on $\Bbb Q$
I need to prove $((0,1)\cap\mathbb{Q})\cup((2,3)\cap\mathbb{Q}) $ is isomorphic to $(0,10)\cap\mathbb{Q}$. I have tried by looking for a bijective relation between sets and Yi think my best try has been a piecewise function:

\begin{equation}
f(x)=
\begin{cases}
       x, & 0\lt x\lt 1 \\
      9x-17, & 2\lt x\lt 3
\end{cases}
\end{equation}
The problem is $1$ belongs to $(0,10)$ but it is not on the image of $f$ so it doesn't work as a bijection. I'd really appreciate some help on this problem. Thanks
Edit: I need an order-preserving bijection.

答:
Let $a_n=1-\frac1n$ and partition $(0,1)$ as
$$(0,1)=\bigcup_{n=1}^\infty (a_n,a_{n+1}].$$
Likewise, let $b_n=2+\frac1n$ and partition
$$(2,3)=\bigcup_{n=1}^\infty [b_{n+1},b_n).$$

Next, fix a real number $\alpha$ between $0$ and $10$.
Let $\{c_n\}_n$ be a strictly increasing sequence of rational numbers that converge to $\alpha$ and starts with $c_1=0$.
Similarly, let $\{d_n\}_n$ be a strictly decreasing sequence of rational numbers that converge to $\alpha$ and starts with $d_1=10$.

Note that
$$ (0,10)=\bigcup_{n=1}^\infty (c_n,c_{n+1}]\;\cup\{\alpha\}\cup \;\bigcup_{n=1}^\infty [d_{n+1},d_n)$$

Now we can readily biject $(a_n,a_{n+1}]\to (c_n,c_{n+1}]$. This preserves rationality as all $a_n$ and all $c_n$ are rational. We can do the same with $[b_{n+1},b_n)\to [d_{n+1},d_n)$. Ultimately, this gives us an order-preseving bijection
$$\bigl((0,1)\cap \Bbb Q\bigr) \cup \bigl((2,3)\cap \Bbb Q\bigr)\to \bigl((0,10)\cap \Bbb Q\bigr)\setminus\{\alpha\}.$$
But we can ignore that spurious "$\setminus\{\alpha\}$" if we pick $\alpha$ to be irrational!


答:
First, find an irrational "hole" in $(0,10)$, let's say $Z=\pi$.

Can you now construct an order-preserving bijection of $(0,1)\cap\mathbb Q$ to $(0,Z)\cap\mathbb Q$? For example:

Make a (strictly) monotonically increasing sequence of rational numbers $a_i\in(0,1)$ such that $\lim_{n\to\infty}a_i=1$, e.g. $0.9, 0.99, 0.999,\ldots$.

Also, make another (strictly) monotonically increasing sequence of rational numbers $b_i\in(0,Z)$ such that $\lim_{n\to\infty}b_n=Z$, e.g. $3, 3.1, 3.14, 3.141\ldots$.

Now make the bijection piecewise-linear on: $(0,a_1]\to(0,b_1], [a_1,a_2]\to[b_1,b_2], [a_2,a_3]\to[b_2,b_3]$ etc.

Altogether, these linear mappings will provide an order-preserving bijection of $(0,1)\cap\mathbb Q$ to $(0,Z)\cap\mathbb Q$.

Similarly, construct an order-preserving bijection of $(2,3)\cap\mathbb Q$ to $(Z,10)\cap\mathbb Q$. (You will need strictly monotonically decreasing sequences of rational numbers converging to $2$ and $Z$ "from above".)

Finally, glue those two bijections together.

Title: Order preserving bijection between $(a,b)$ and $(c,d)$ to $(e,f)$ on $\Bbb Q$

Body:
I need to prove $((0,1)\cap\mathbb{Q})\cup((2,3)\cap\mathbb{Q}) $ is isomorphic to $(0,10)\cap\mathbb{Q}$. I have tried by looking for a bijective relation between sets and Yi think my best try has been a piecewise function:

\begin{equation}
f(x)=
\begin{cases}
       x, & 0\lt x\lt 1 \\
      9x-17, & 2\lt x\lt 3
\end{cases}
\end{equation}
The problem is $1$ belongs to $(0,10)$ but it is not on the image of $f$ so it doesn't work as a bijection. I'd really appreciate some help on this problem. Thanks
Edit: I need an order-preserving bijection.


问:
Spectral decomposition of $-\Delta$ the Laplacian
I am currently trying to self learn about an interesting idea that caught my eye in spectral geometry, which is the whole idea of hearing the shape of the drum by solving the wave equation $\Delta \psi = k^2 \psi_{tt}$ on a compact Riemannian manifold $(M,g)$ with dirichlet boundary constraint $\psi|_{\partial M}=0$. The thing is, I don't think I have that of a solid background in functional analysis and PDEs, and I am looking for some help or a rather self contained source which proves that such decomposition exists
$$ -\Delta(\cdot)= \sum_{l=0}^\infty \lambda_l \left \langle \cdot,\phi_l \right \rangle_{L^2(M)}\phi_l$$
What I do know: I wasn't able to find many sources which dive deep into this problem, but the first thing I was able to realize is this decomposition probably isn't on $L^2(M)$ but rather some subspace of the form $V=\left \{ f\ \text{nice enough}|\ f|_{\partial M}=0\right \}$, the reason for this is that we probably need the laplacian to be self adjoint, and looking at one of Green's identities
$$\int_M u\Delta v - v\Delta u   =\int_{\partial M}u\frac{\partial v}{\partial n}-v \frac{\partial u}{\partial n}$$
We need to zero the RHS with the constraint. What is $V$? I saw alot of references to a "Sobolev space $H_0^1(M)$" but couldn't find a coherent definition of it, in addition to the fact that it is equipped with a different inner product than of $L^2(M)$ which makes things more confusing for me.

Anyhow, this tells us automatically why eigenfunctions must be orthogonal, since if we have
$$ -\Delta \phi_1 = \lambda_1 \phi_1,\ -\Delta \phi_2 = \lambda_2 \phi_2$$
for $\lambda_1,\lambda_2 \neq 0$ different then
$$\left \langle \lambda_1\phi_1,\phi_2 \right \rangle_{L^2(M)}=-\left \langle \Delta\phi_1,\phi_2 \right \rangle_{L^2(M)}=-\left \langle \phi_1,\Delta\phi_2 \right \rangle_{L^2(M)}=\left \langle \phi_1,\lambda_2\phi_2 \right \rangle_{L^2(M)}$$
and we get that $\left \langle \phi_1,\phi_2 \right \rangle_{L^2(M)}=0$

I also know that $-\Delta$ is positive definite since $$\int_M u (-\Delta) u=\int_M \left \| \nabla u \right \|_g^2\geq 0$$

What I'm looking for: This topic is pretty weird for me right now and I would like to learn about it thoroughly.

Are my proofs correct or relevant at all?

As I mentioned what is $V$? it's hard for me to see why $C^2(M) \cap C(\bar{M})$ needs to be restricted further.

How do we know that $-\Delta$ is compact if at all? This means that if $f_k \rightarrow f$ in $V$, then $-\Delta f_k$ has a convergent subsequence of functions. I know according to functional analysis this should imply the spectral decomposition but I don't know whether this is true. Edit: appearantly not, $-\Delta^{-1}$ is. Then I am looking at the moment for a proof of that

Assuming 3 is true, we also get for free that the eigenvalues are discrete $0\leq \lambda_0 \leq \lambda_1 \leq \cdots$. However, I saw claims that $\lambda_k \rightarrow \infty$ (when according to functional analysis it should approach to $0$). This doesn't make sense to me, how sums like the spectral decomposition $\Delta f=\sum_{l=0}^\infty  \lambda_l\left \langle f,\phi_l \right \rangle_{L^2(M)}\phi_l  $ converge?

How does one show $\left \{ \phi_l\right \}_{l=0}^\infty$ is complete? meaning we can express any $f\in V$ as a fourier series $f=\sum_{l=0}^\infty \left \langle f,\phi_l \right \rangle_{L^2(M)}\phi_l  $

Sorry for the mess of a question, In the probable case I said something wrong I would love to hear. If anyone knows of a free-access source which adresses this problem please refer me to it!

EDIT: I have read several notes and articles which address this issue. Right now, it all boils down for me to find proofs of the Rellich-Kondrachov theorem and Poincaré inequality for Riemannian manifolds, and I can take it from there using the theory of weak solutions.

答:
After a lot of digging online I have found the following resources very useful

The Calderón problem on Riemannian manifolds by Mikko Salo

PDEs Basic Theory by Michael Taylor

There they treat with no lack of rigour this exact problem, in the spirit of @Chris' answer. Leaving it here since it might help anyone who sees this in the future.


问:
eigen vectors and linear independence
I know that it is the case that if we have eigenvectors $v_1....v_r$ that correspond to distinct eigen values $λ_1....λ_r$ of an $n$ x $n$ matrix, then {$v_1....v_r$} is linearly independent.

But is is it possible to have $a_1$ linearly independent eigen vectors corresponding to $λ_1$, $a_2$ linearly independent eigen vectors corresponding to $λ_2$... $a_p$ linearly independent eigen vectors corresponding to $λ_n$, and the set of all $r$ eigenvectors where $r=a_1+a_2+...+a_p$ be linearly dependent?

For example: consider a $3x3$ matrix. Imagine we have one vector(call it $v_1$) forming a basis for the eigenspace of a one eigen value, but two vectors(call them $v_2$ and $v_3$) forming a basis for different eigen value. are we guaranteed that {$v_1,v_2,v_3$} are linearly independent?

答:
We claim that the set of all $r$ eigenvectors is linearly independent. The main proof ideas are already provided in the comments. For each $i=1,\dots, p$ let $v_{i1},\dots, v_{ia_i}$ be linearly independent system of eigenvectors corresponding to the eigenvalue $\lambda_i$ and  $\mu_{i1},\dots, \mu_{ia_i}$ be scalars such that $$\sum_{i=1}^p \sum_{j=1}^{a_i} \mu_{ij} v_{ij}=0.$$

For each $i$ put $v_i=\sum_{j=1}^{a_i} \mu_{ij} v_{ij}$. Then $v_i$ is an eigenvectors corresponding to $\lambda_i$ and

$$\sum_{i=1}^p v_i=\sum_{i=1}^p \sum_{j=1}^{a_i} \mu_{ij} v_{ij}=0.$$

The linear independence property formulated in the first paragraph of the question implies that $0=v_i=\sum_{j=1}^{a_i} \mu_{ij} v_{ij} $ for each $i$. Then the linear independence property formulated in the second  paragraph of the question implies that $\mu_{ij}=0$ for each $j$.

Title: eigen vectors and linear independence


问:
Uniqueness of adjoint functors up to isomorphism
Suppose we are given functors $F:\mathcal{C}\rightarrow\mathcal{D}$ and $G,G':\mathcal{D}\rightarrow\mathcal{C}$ such that $G$ and $G'$ are both right adjoint to $F$. To show that $G$ and $G'$ are isomorphic, we wish to come up with a natural isomorphism between the two functors. This is how Awodey does it:

For any $D\in\mathcal{D}$ and any $C\in\mathcal{C}$, we have that $Hom(C,GD)\cong Hom(FC,D)\cong Hom(C,G'D)$ simply using the hom-set definition of adjunctions. By the Yoneda principle, this means that $GD\cong G'D$ for all $D\in\mathcal{D}$.

Awodey then states that

But this isomorphism is natural in $D$,  again by adjointness.

I dont' quite understand how the naturality follows from adjointness. Any help is appreciated.

答:
We actually have a composite of natural isomorphisms $\tau:\hom(\bullet,G\bullet)\to \hom(F\bullet,\bullet)\to \hom(\bullet,G'\bullet)$.

In more details, $\tau$ has components $\tau_{C,D}:\hom(C,GD)\to\hom(C,G'D)$ satisfying the naturality condition: for arrows $\gamma:C_1\to C$, $\ \delta:D\to D_1\,$  and $\, f:C\to GD$ we have
$$\tau(G\delta\circ f\circ\gamma)=G'\delta\circ\tau(f)\circ\gamma\,. $$
Now applying the Yoneda lemma basically means to consider the case when $C=GD$ and apply $\tau$ to $1_{GD}$. It leads to a $GD\to G'D$ arrow, verify (using the above equation) that it's natural in $D$, and find its inverse (e.g. by applying the same to $C=G'D$ and $\tau^{-1}$.


问:
Find a regular function $f$ satisfying Re $(f) =u$
$$u = 2x^3 - 6xy^2$$

How does one do this?

I can find a function $f(0) = i$

Not sure what this is asking me to do...

答:
If
$f=u+iv$, then $u_x=v_y$. So
$$
u=2x^3-6xy^2\quad\Longrightarrow\quad u_x=6x^2-6y^2=v_y
\quad\Longrightarrow\quad v=6x^2y-2y^3+h(x)
$$
But $v_x=-u_y$ and hence $h'(x)=0$, and thus $h(x)=c$.
So
$$
f(x+iy)=2x^3-6xy^2+i(6x^2y-2y^3+c)
$$
Since $f(0)=i$, then c=1, and finally
$$
f(x+iy)=2x^3-6xy^2+i(6x^2y-2y^3+1)
$$


问:
Counting One-To-One Functions
How many one-to-one functions are there from the set with $m$ elements to one with $n$ elements?

Solution:

Knowing that $m \le n$ for being one-to-one. Let the domain be $a_1, \cdots, a_m$ and they map to one element in the codomain. When we choose what $a_1$ maps to we cannot use it again. Then for choosing what $a_2$ maps to we have $(n-1)$ ways to choose.

My question is how do we get an expression by the product rule from counting? I am stuck at the part where I left off.

答:
You have $n$ choices for the image of $a_1$, then $n-1$ choices for the image of $a_2$, $\dots$, and finally $n-m+1$ choices for the image of $a_m$, yielding
$$\prod_{k=1}^m (n-k+1)=\frac{n!}{(n-m)!}$$

Title: Counting One-To-One Functions


问:
ODE without solutions
Is it possible to find functions $a,b,c$ defined on $\mathbb{R}$, such that
\begin{equation*}
ay'+by=c
\end{equation*}
has no global solution and $a(x)$ and $b(x)$ are never both zero ?

If $c=0$ there is always $y=0$ but I don't know about the general case. I think there is always a solution but I don't know how to prove it and it doesn't fall into the Cauchy-Lipschitz realm.

My question comes from the theory of pasting solutions: when $a(x)$ is zero the affine subset of solutions can be of dimension 0,1,2 according as to how solutions of homogenous equations paste around the singularity. I wonder if it can be empty.

The condition on $a$ and $b$ is to avoid trickery counter-examples like $x^3y'+3x^2y=2x$ whose solutions $y=\frac{1}{x}+\frac{\lambda}{x^3}$ are not global.

答:
Let $a=x,b=1$ and $c=2x$ for positive $x$, $c=0$ for negative $x$.
Then $y=\frac{\lambda}{x}$ on $\mathbb{R}^{*}_{-}$ and $y=x+\frac{\mu}{x}$ on $\mathbb{R}^{*}_{+}$, so the only hope for pasting is $\lambda=\mu=0$ but then $y$ wouldn't be differentiable in $0$.

Another example can be obtained with $c=\text{sgn}(x)\sqrt{|x|}$ where again $y=\frac{2}{3}\sqrt{x}$ is not differentiable in $0$.

Differentiability fails in both examples, but continuity doesn't. If I am not mistaken you can prove (by using the explicit formulae given by AVK) that there is always a solution that extends by continuity to the zero of $A$ (but there may be pasting problems for several zeroes).

Title: ODE without solutions


问:
Can anyone explain why expressions of the form $\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}$ can be rational?
$$\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}=1$$

Can anyone explain to me how this works? I don't understand why two irrational numbers cube rooted and added together return the number $1$.

I'm trying to find all the positive integer values for $x$ and $y$ in this formula
$$\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}=1$$

Sorry if this is a stupid question.

答:
Writing $c = \sqrt[3]{2-\sqrt{5}}, d = \sqrt[3]{2+\sqrt{5}}, a=c+d$, cubing the expression gives:

\begin{align}(c+d)^3 &= c^3+3c^2d+3cd^2+d^3\\
&=(2-\sqrt 5) + 3cd(c+d) + (2+ \sqrt 5)\\
&=4+3a\sqrt[3]{(2-\sqrt5)(2+\sqrt 5)}\\
&=4 -3a=a^3
\end{align}
Hence $a^3+3a-4 = 0 = (a-1)(a^2+a+4)$.

The quadratic equation $a^2+a+4=0$ has no solutions in the reals. Hence $a=1$.

Hopefully this answer can give some insight on how you can approach the general problem for $\sqrt[3]{x\pm\sqrt y}$.

Update: Since the other answer has proceeded to partially solve your second question, I may as well prove the sufficient condition.

Writing $a = \sqrt[3]{x-\sqrt y}, b =\sqrt[3]{x+\sqrt y}, c=a+b$:

\begin{align}(a+b)^3 &= a^3+3a^2b+3ab^2+b^3\\
&=(x-\sqrt y) + 3ab(a+b) + (x+ \sqrt y)\\
&=2x+3c\sqrt[3]{(x-\sqrt y)(x+\sqrt y)}\\
&=4 - 3c\sqrt[3]{x^2-y}=c^3
\end{align}

From $c^3 + 3c\sqrt[3]{x^2-y} - 2x = 0$ and $c=1$ being a solution we must have $$y= x^2 + \left(\dfrac {2x-1}3\right)^3$$ where $x \equiv 2 \pmod 3$ for $y$ to be an integer.

For positive integer values of $y$ we have $3\sqrt[3]{x^2-y} = 1-2x$. Hence the cubic equation reduces to

$$c^3+(1-2x)c-2x= 0=(c-1)(c^2+c+2x)$$

For $x > \dfrac18$, $c^2+c+2x$ has no solution in the reals. Hence $c=1$, showing that this condition is indeed necessary and sufficient.

One could, of course, obtain the explicit form of $\sqrt[3]{x\pm\sqrt y}$ with our condition, by observing:

$$8 (x \pm \sqrt y) = 1^3 \pm 3\sqrt {\frac {8x-1}3} +3 \sqrt {\frac {8x-1}3}^2 \pm \sqrt {\frac {8x-1}3}^3 = \left(1 \pm \sqrt{\frac {8x-1}3}\right)^3$$

Which gives:

$$\sqrt[3]{x\pm\sqrt y} = \frac{1 \pm \sqrt{\frac {8x-1}3}}2$$

and indeed sums up to $1$, as desired.


问:
Stuck at order preservation of mapping $\phi : P \rightarrow Q$
I had to show that this is order preserving:

$ P=\{ \mathcal{P}(S); \subseteq \}; |S|>1; \; Q=2 $ and $\phi$ from $P \rightarrow Q$ is defined by
$\\  \phi (U)= \left\{ \begin{array}{ll}
1 & U=S \\
0 & U \neq S\\
\end{array} \right. $

My solution:

$ U,V \subseteq P $ such that $\phi (U) = \epsilon; \; \phi(V)=\delta \\$
$\\ \Leftrightarrow \forall x\in S; \epsilon = 1 \Rightarrow \delta=1
\\ \Leftrightarrow \phi(U) \leq \phi(V)$

I do not understand where to use $0$ in this? According to my solution, this is order preserving.

答:
I’m afraid that I can’t make any sense of your solution. You need to show that if $U,V\subseteq S$ and $U\subseteq V$, then $\varphi(U)\le\varphi(V)$. There are basically just two cases:

If $V=S$, then $\varphi(V)=1$; $\varphi(U)$ is either $0$ or $1$, so $\varphi(U)\le 1=\varphi(V)$ no matter what $U$ is.

If $V\ne S$, then $\varphi(V)=0$. Since $U\subseteq V$, we know that $U\ne S$, and therefore $\varphi(U)=0$, too. Thus, once again we have $\varphi(U)\le\varphi(V)$.


问:
Why over $\mathbb{Z}/n\mathbb{Z}$ projectivity, injectivity and flatness coincide for cyclic modules?
Assume $R=\mathbb{Z}/n\mathbb{Z}$ ($n\neq0$) and let $M$ be a cyclic $R$-module. Could you tell me how to prove that $M$ is projective if and only if it is injective if and only if it is flat? And also why $M$ is free if and only if it is cofree?

By cardinality reasons I think I proved that if $M$ is free then it is equal to $R$, but I don't get the rest.

答:
I guess you really mean $n>1$, and that is rather a lot of questions. I'll try to hit them all.

Well the nicest thing about $\Bbb Z/n\Bbb Z$ is that it's a quasi-Frobnius ring meaning that it's Artinian and self-injective. This is in fact equivalent to the projective modules being precisely the injective modules. Actually, all of these are equivalent:

$R$ is quasi-Frobenius

The projective and injective right $R$ modules coincide

The projective right $R$ modules are also injective

The injective right $R$ modules are projective

All the left versions of the above

This is true without commutativity, without the assumption that modules are cyclic.

Since it's Artinian, it is also a perfect ring, and over perfect rings the projective and flat modules coincide.


问:
Consider $T : \mathbb{R^7} \to \mathbb{R^7} $ linear transformation such that $\operatorname{Ker}(T) \cap T(\mathbb{R^7}) = ${0}$ $, show that ...
Consider $T : \mathbb{R^7} \to \mathbb{R^7} $ linear transformation such that  $\operatorname{Ker}(T) \cap T(\mathbb{R^7}) = ${0}$ $, show that $\mathbb{R^7}=\operatorname{Ker}(T) \bigoplus T(\mathbb{R^7} )$

I managed to show that if $\mathbb{R^7}=\operatorname{Ker}(T) + T(\mathbb{R^7} )$ then the sum is straightforward.

For the cases $y\in T(\mathbb{R^7})$  and $y\in\operatorname{Ker}(T)$ It is obvious.

but when $y$ doesn't meet the conditions above i couldn't work anymore, i tried to use the fact that the $ T ^ 2 $ function is bijection , but i couldn't complete anything

答: Note that $\dim\ker T+\dim T(\Bbb R^7)=7$, by the rank-nullity theorem. So, since $\ker T\cap T(\Bbb R^7)=\{0\}$,$$\dim\bigl(\ker T+T(\Bbb R^7)\bigr)=\dim\ker T+\dim T(\Bbb R^7)=7,$$which means that $\ker T+T(\Bbb R^7)=\Bbb R^7$.


答:
For the order to not be preserved, we need to have two sets where $U\subseteq V$ but $\phi(U)\not\le\phi(V)$.

The latter is only possible when $\phi(U)=1$ and $\phi(V)=0$. But the former means that $U=S$, which, with $U\subseteq V$ implies $V=S$ and so $\phi(V)=1$ too - a contradiction!

Title: Stuck at order preservation of mapping $\phi : P \rightarrow Q$

Body:
I had to show that this is order preserving:

$ P=\{ \mathcal{P}(S); \subseteq \}; |S|>1; \; Q=2 $ and $\phi$ from $P \rightarrow Q$ is defined by
$\\  \phi (U)= \left\{ \begin{array}{ll}
1 & U=S \\
0 & U \neq S\\
\end{array} \right. $

My solution:

$ U,V \subseteq P $ such that $\phi (U) = \epsilon; \; \phi(V)=\delta \\$
$\\ \Leftrightarrow \forall x\in S; \epsilon = 1 \Rightarrow \delta=1
\\ \Leftrightarrow \phi(U) \leq \phi(V)$

I do not understand where to use $0$ in this? According to my solution, this is order preserving.


问:
Tensor Product of operator algebras
I want to clearly know the notion of 'algebraic tensor product' for Hilbert spaces or $C^*$-algebras or von Neumann algebras. Take any two objects $A$ and $B.$ In many places people have told to define the algebraic tensor product and then complete it suitably. For example for Hilbert spaces one define the inner product $:=$ and then complete it. But how is it well-defined? I am assuming that by 'algebraic tensor product' one means the usual vector space tensor product. What about the $C^*$ algebra? I have same confusion for min tensor product for $C^*$-algebra.

答:
To see that this inner product is well-defined, see theorem 6.3.1 in Murphy's book "$C^*$-algebras and operator theory". This has a very accessible exposition of the tensor product of Hilbert spaces. All you need for prerequisites  is understanding how the algebraic tensor product of vector spaces is made.

The $C^*$-algebraic case is much more complicated. Let $A$ and $B$ be $C^*$-algebras and write $A \odot B$ to denote their algebraic tensor product. This becomes a $*$-algebra for the obvious multiplication and involution maps. We want to put a norm on $A \odot B$ such that it becomes a $C^*$-algebra, or at least a $C^*$-algebra after taking a completion. In general, many suitable norms are possible, and different choices of norms may give different completions so it must always be specified with which $C^*$-norm one is working.

In practice, one always works with either the minimal or the maximal $C^*$-norms for the reason that these are much-better behaving than the intermediate norms.

Title: Tensor Product of operator algebras


答:
Let's say $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$, where $u,v$ are real functions. You need to make sure the Cauchy-Riemann equations are satisfied:

$\begin{cases}\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}\\\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}\end{cases}$

Since $u=2x^3-6xy^2\Rightarrow\dfrac{\partial u}{\partial x}=6x^2-6y^2=\dfrac{\partial v}{\partial y}\Rightarrow v=\displaystyle\int(6x^2-6y^2)\,\text{d}y\;+\phi(x)=6x^2y-2y^3+\phi(x)$. We are adding a function $\phi(x)$ because when doing $\dfrac{\partial v}{\partial y}$ we would get rid of $\phi(x)$, since it doesn't depend on $y$, so just in case there was a part of $v$ not depending on $y$, we should add it.

Now you have to make sure $\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}$: Try it for yourself first!

$\dfrac{\partial u}{\partial y}=-12xy$ and $-\dfrac{\partial v}{\partial x}=-12xy-\phi'(x)$, so we get $\phi'(x)=0\Rightarrow\phi(x)=C$, with $C$  a real constant. So $v=6x^2y-2y^3+C$.

Therefore

$f(z)=f(x+iy)=2x^3-6xy^2+i(6x^2y-2y^3+C)=2(x^3-3xy^2+3ix^2y-iy^3)+iC=2(x+iy)^3+iC=2z^3+iC$

If we also impose $f(0)=i$ then

 $f(0)=20^3+iC=i\Rightarrow C=1$

So the final answer is

 $f(z)=2z^3+i$

Title: Find a regular function $f$ satisfying Re $(f) =u$

Body:
$$u = 2x^3 - 6xy^2$$

How does one do this?

I can find a function $f(0) = i$

Not sure what this is asking me to do...


问:
Truncating $p$-series
Let $p \in \mathbb{R}$. Series of the form $$S_p = \sum_{k \in \mathbb{N^*}} \frac{1}{k^p}$$ converge if and only if $p > 1$. Let us define $$S_{p, n} \triangleq \sum_{k = 1}^{n} \frac{1}{k^p},$$ the truncation of the $p$-series at its $n$-th term. (It is evident that $S_{p, n} \space \xrightarrow{n\to\infty} \space S_p$.) Is there a closed expression for $S_{p, n}$, at least for some values of $p$ (integers must be easier)?</p>

答:
To the best of my knowledge, there is no closed form expression for $S_{p,n}$. </p>

<p>EDIT: Let me amend my answer to read, to the best of my knowledge, there is no closed form expression in terms of functions familiar to, say, students of 1st-year Calculus. </p>

Title: Truncating $p$-series

Body:
To the best of my knowledge, there is no closed form expression for $S_{p,n}$.

EDIT: Let me amend my answer to read, to the best of my knowledge, there is no closed form expression in terms of functions familiar to, say, students of 1st-year Calculus.


问:
concave function is positive on convex set
I have seen the following sentence somewhere, but I can not prove it.

Let $f$ be a nonnegative concave function  on a convex domain $D$. Assume that $f$ achieves a positive value at some point $x\in D$. Then $f(y)>0$ for all $y\in D^o$.

答:
I don't think this question is that complicated, maybe you have overthought a little bit.

So let $x_0 \in D $ be that positive point, that is $f(x_0)>0$

For any point $y$ in the interior of $D$, we can always choose and point $z \in D^{o}$ ( altogether with a $\lambda \in (0,1)$) such that:

$$(1-\lambda)x_0 +\lambda z = y$$
( Note that $\dfrac{y- (1-\lambda)x_0}{\lambda} \longrightarrow y$ when $\lambda \longrightarrow 1^-$)

Hence :
$$ (1-\lambda)f(x_0) +\lambda f(z) \le f(y)$$
Which is a contradiction if $f(y)=0$ because the $LHS$ is positive( due to the positivity of $f(x_0)$


问:
If T maps R-n ONTO R-m, is the nullity of $T = n-m$?
I'm trying to find the dimension of the kernel of T when T: R-n onto R-m, which is the same as its nullity. I know the nullity of a matrix is equal to the number of free variable, and the standard matrix A of a transformation has columns that span R-m if T is onto. So if A has m pivot positions, then would it be correct to say A has $n-m$ free variables, which is the same as it's nullity and the dimension of the kernel of T?

答:
By the rank nullity theorem, the dimension of the domain of a linear map is equal to the sum of the dimension of its image plus its nullity.

As $T$ is supposed to be onto, the dimension of the image of $T$ is equal to $m$ and indeed the nullity is equal to $n-m$.


问:
Why if a polynomial is divided by x-1 and gives the remainder 3 then p (1) = 3?
Why if a polynomial is divided by $x-1$ and gives the $remainder=3$ then $p(1) = 3$? Does it have to do with ruffini? Could you describe the steps that lead me to affirm this?

答:
Call $P$ your polynomial. By definition of division, $P=Q*(x-1)+3.$ for $Q$ a polynomial.

 Replace at $x=1.$


答:
Hint: Let $x$ be a point in $D$ for which $f(x) > 0$, and suppose (for the purpose of contradiction) that $y \in D^o$ is such that $f(y) = 0$. Note that for $t \geq 1$, we have
$$
f((1 - t)x + ty) \leq (1-t)f(x) + tf(y).
$$

Title: concave function is positive on convex set

Body:
I have seen the following sentence somewhere, but I can not prove it.

Let $f$ be a nonnegative concave function  on a convex domain $D$. Assume that $f$ achieves a positive value at some point $x\in D$. Then $f(y)>0$ for all $y\in D^o$.