问:
Continuity of $g(t):=f(tx,ty)$

Let $f:\Bbb R^2 \to \Bbb R$ be a function and $M$ be a positive real number, such that for every $(x,y)\in \Bbb R^2$, the function $g(t):=f(tx,ty)$ is differentiable and $|g'(t)|\le M$. Then prove that $f$ is continuous at $(0,0)$.

We have,
\begin{align*}
|g'(0)|&=\left|\lim_{h\to 0}\frac{g(h)-g(0)}{h}\right|\\
&=\left|\lim_{h\to 0}\frac{f(hx,hy)-f(0,0)}{h}\right|
\end{align*}
Since $|g'(x)|\le M$, so $\left|\lim_{h\to 0}\frac{f(hx,hy)-f(0,0)}{h}\right|\le M$. Then how to proceed to show the continuity of $f$ ? Any hint. please?

答: By MVT we get $|f(tx,ty)-f(0,0)|=|g(t)-g(0)| \leq M|t|$. This is true for all $t$ and all $x,y$. Now let $u=\frac x {\|(x,y)\|}$ and $v=\frac y {\|(x,y)\|}$. Then $|f(x,y)-f(0,0)| =|f(tu,tv)-f(0,0)|$ where $t=\|(x,y)\|$ Hence, $|f(x,y)-f(0,0)| \leq t M=M\sqrt {x^{2}+y^{2}}$ from which continuity of $f$ at $(0,0)$ follows.

Title: Continuity of $g(t):=f(tx,ty)$


问:
Uniqueness of simultaneous diagonalization via congruence
My question is about the uniqueness of simultaneous diagonalization via congruence: I have two symmetric matrices A and B of dimension d. One of them (say A) is positive-definite, the other one (B) is positive semi-definite and has rank $k<d$. Say we want to diagonalize A and B simultaneously via congruence, i.e. we want to find a regular matrix $C$ of dimension $d$ such that $$ C^TAC=I_d \quad\text{and}\quad C^TBC=D$$
where $I_d$ is the $d$-dimensional identity matrix and $D$ is some diagonal matrix. There are at least two solutions:

Since A is positive-definite, we can compute the inverse of its square root $A^{-1/2}$. Then $M_1:=A^{-1/2}BA^{-1/2}$ is symmetric and has eigendecomposition $Q_1\Lambda_1 Q_1^T$. Taking $C_1:=A^{-1/2}Q_1$ does the job since: $$C_1^TAC_1=Q_1^TA^{-1/2}AA^{-1/2}Q_1=Q_1^TQ_1=I_d\quad \text{and}\quad C_1^TBC_1=Q_1^TA^{-1/2}BA^{-1/2}Q_1=Q_1^TM_1Q_1=\Lambda_1.$$

Another solution is to use the inverse of the Cholesky factor of $A$ instead of the inverse of its square root: Let $A=R^TR$ the Cholesky decomposition of $A$ where $R$ is upper triangular. Then perform the spectral decomposition of $M_2:=R^{-T}BR^{-1}$ to obtain $Q_2\Lambda_2Q_2^T$ and define $C_2:=R^{-1}Q_2$. The matrix $C_2$ does the job as well since:
$$ C_2^TAC_2=Q_2^TR^{-T}R^TRR^{-1}Q_2=Q_2^TQ_2=I_d\quad\text{and}\quad C_2^TBC_2=Q_2^TR^{-T}BR^{-1}Q_2=Q_2^TM_2Q_2=\Lambda_2.$$

I have implemented both procedures and observe that the diagonal matrices $\Lambda_1$ and $\Lambda_2$ coincide exactly. Furthermore, I observe that the first $k=rank(B)$ columns of the two transformation matrices $C_1$ and $C_2$ coincide (up to orientation, i.e. up to multiplication with -1). My question is whether there are any uniqueness statements in this context. Is the resulting diagonal matrix $D$ unique? Are the first $k=rank(B)$ columns of $C$ unique? Or may there be further matrices $C$ that diagonalize both $A$ and $B$ in the above sense but where the first $k$ columns will not coincide with those of $C_1$ and $C_2$? If the transformation matrix $C$ is unique (or its first $k$ columns), how can one prove that?

答:
Let $Q=A^{1/2}C$. The conditions $C^TAC=I$ and $C^TBC=D$ are then equivalent to $Q^TQ=I$ and $Q^T(A^{-1/2}BA^{-1/2})Q=D$. Therefore, the apparent simultaneous diagonalisation of two matrices by congruence is just the usual orthogonal diagonalisation of a single real symmetric matrix $S=A^{-1/2}BA^{-1/2}$.

Since the diagonal entries of $D$ are the eigenvalues of $S$, they are unique up to permutation. Since the columns of $Q$ are mutually orthonormal eigenvectors of $S$, they are also unique (up to permutation and multiplications by $\pm1$) when $S$ has distinct eigenvalues. In this case, as $C=A^{-1/2}Q$, the columns of $C$ are also unique (up to permutation and multiplications by $\pm1$).

$C$ is not unique in general if $S$ has repeated eigenvalues. An obvious counterexample is the case where $A=I$ and $B=0$. Every orthogonal matrix $C$ will diagonalise $A$ and $B$ by congruence.

ContentLicense: CC BY-SA 4.0


问:
Proofs involving the cardinality of the set of continuous functions
I am trying to prove the following result.

Let $F$ denote the set of all functions $f: \mathbb{R} \to \mathbb{R}$ and $C \subset F$ denote the subset of all continuous functions. Prove that $|\mathbb{R}| = |C| < |F|$. (Hint: use the fact that a continuous function on $\mathbb{R}$ is determined by its values on the rational numbers $\mathbb{Q} \subset \mathbb{R}$.)

The first thing I'm trying to understand is the hint. If I am understanding it correctly, it says this. The rationals are dense in the reals, so I can always construct a sequence of rationals converging to any real. In fact, we can construct the reals as the limits of Cauchy sequences of rationals. So if I have two functions $f,g$ that send the rationals to the same points, it has to be the case that they send all the reals to the same points, so the function is ``uniquely determined'' by its outputs on the rationals, which we can safely enumerate as $q_1, \ldots, $ because they're countable. Is this right?

Formally, if $x \in \mathbb{R}$ and $(q_n) \to x$ where $q_n$ rational, we have by continuity of $f$
\begin{align*}
f(x) = f\left(\lim\limits_{n \to \infty} q_n\right) = \lim\limits_{n \to \infty} f(q_n).
\end{align*}
Is this the right idea? Am I missing any formalism?

As for the actual proof, I only have an idea for how to prove that $|\mathbb{R}| = |C|$. I want to use the Schroeder-Bernstein theorem to construct injections in both directions and then conclude that there exists a bijection. I can write an injection from $\mathbb{R}$ to $C$ rather easily by mapping $c \in \mathbb{R}$ to the function $f(x) = c$ for all $x$, which is constant, hence continuous. In the opposite direction, I can enumerate the rationals $q_1, q_2, \ldots$ and map a continuous function $f$ to a sequence of its values $(f(q_1), f(q_2), \ldots)$, but that gives an injection to sequences of real numbers, $\mathbb{R}^{\mathbb{N}}$. I would need to then inject $\mathbb{R}^{\mathbb{N}}$ to $\mathbb{R}$ and compose injections to get an injection from $C$ to $\mathbb{R}$.

I do not have an idea for how to prove $|C| < |F|$. The notation is non-standard (since the cardinalities aren't finite), but I believe this typically means that there does not exist a surjection from $C$ to $F$, which makes sense, as intuitively there are plenty more total functions that aren't continuous. I considered a diagonal argument, but can't fully work out the details.

Any help, especially on this last part, would be very much appreciated.

答:
What the determined by values on $\Bbb Q$ hint implies is that when two continuous real functions agree on $\Bbb Q$, they agree on $\Bbb R$.

So the map $r: C \to \Bbb R^{\Bbb Q}$ defined by $r(f)= f\restriction_{\Bbb Q}$ is 1-1. The right hand set is the set of all functions from $\Bbb Q$ to $\Bbb R$, as usual.

Let $\mathfrak{c}=|\Bbb R| = 2^{\aleph_0}$ then basic set theory (cardinal arithmetic) tells us

$$|C| \le | \Bbb R^{\Bbb Q} | = \mathfrak{c}^{\aleph_0} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0 \times \aleph_0} = 2^{\aleph_0}= |\Bbb R|$$

As $|F| = \mathfrak{c}^{\mathfrak{c}} =  2^{\mathfrak{c}} > |\Bbb R|$ by Cantor's theorem, we're done: $\mathfrak{c} =|C| < |F|$.


问:
Find the first two terms of the perturbation series solution to the IVP
Here is the equation for the initial value problem, $y'=1+(1+\epsilon)y^2$ with initial condition $y(0)=1$ with $t>0$ where $0<\epsilon\ll1.$. Find exact solution and compare approximation.

My attempt:

Suppose $y=y_0+\epsilon y_1+O(\epsilon^2)$ Then $y_0'+\epsilon y_1'+O(\epsilon^2)=1+(1+\epsilon)(y_0+\epsilon y_1+O(\epsilon^2))^2$, after distributing and combining like terms. $\implies y_0'+\epsilon y_1'+O(\epsilon^2)=1+y_0^2+2\epsilon y_0y_1+\epsilon y_0^2+O(\epsilon^2)$.

Grouping terms $\\\implies (y_0'-y_0^2-1)+\epsilon(y_1'-2y_0y_1-y_0^2)+O(\epsilon^2)=0
\\$.

Finally obtaining two sets of ODE's (This is where I stopped)$ \implies y_0'-y_0^2-1=0, \space y_1'-2y_0y_1-y_0^2=0 \implies \arctan(y_0)=t+C$

Using initial value $y(0)=1$ we get $C=\frac{\pi}{4}$. So now we establish that $y_0=\tan(t+\frac{\pi}{4})$.

So this implies that, $y_1'-2\tan(t+\frac{\pi}{4})y_1-\tan^2(t+\frac{\pi}{4})=0$.

Am I going in the right direction? Solving the second set of ODE's just seems overly complicated so I used Wolfram Alpha.

答:
I see no error in your derivation.

Your last equation can be interpreted as
$$
y_1'+2\frac{u'}{u}y_1=\frac{u'^2}{u^2},~~~u(t)=\cos(t+\pi/4)
$$
which has the surprisingly simple integrating factor $u^2$, so that
$$
\Bigl(\cos^2(t+\pi/4)y_1(t)\Bigr)'=\sin^2(t+\pi/4)=\frac12(1-\cos(2t+\pi/2))
$$
This now should be easy to integrate.


答:
To inject $\Bbb R^{\Bbb N}\to\Bbb R$, I'd identify $\Bbb R$ with $\mathcal P(\Bbb N)$. But then how do we turn a sequence of subsets $A_1,A_2,\ldots $of $\Bbb N$ into a single subset of $\Bbb N$? Simply partition $\Bbb N$ in to infinitely many copies $2^n(2\Bbb N+1)$ of itself! So the sequence $A_1,A_2,\ldots$ maps to the set $A$ where $m\in A$ iff it can be written as $m=2^k(2l+1)$ and $l\in A_k$.

Assumes there is a surjection $f\colon\Bbb R\to F$.
So for $x\in \Bbb R$, we have some function $f_x\in F$, which in particular itself maps $x$ to some real number $f_x(x)$. Define $g\colon \Bbb R\to \Bbb R$ as $g(x):=f_x(x)+1$. Then there is no $\xi\in\Bbb R$ with $g=f_\xi$, for that would imply $g(\xi)=f_\xi(\xi)$ as well as $g(\xi)=f_\xi(\xi)+1$.

Title: Proofs involving the cardinality of the set of continuous functions

Body:
I am trying to prove the following result.

Let $F$ denote the set of all functions $f: \mathbb{R} \to \mathbb{R}$ and $C \subset F$ denote the subset of all continuous functions. Prove that $|\mathbb{R}| = |C| < |F|$. (Hint: use the fact that a continuous function on $\mathbb{R}$ is determined by its values on the rational numbers $\mathbb{Q} \subset \mathbb{R}$.)

The first thing I'm trying to understand is the hint. If I am understanding it correctly, it says this. The rationals are dense in the reals, so I can always construct a sequence of rationals converging to any real. In fact, we can construct the reals as the limits of Cauchy sequences of rationals. So if I have two functions $f,g$ that send the rationals to the same points, it has to be the case that they send all the reals to the same points, so the function is ``uniquely determined'' by its outputs on the rationals, which we can safely enumerate as $q_1, \ldots, $ because they're countable. Is this right?

Formally, if $x \in \mathbb{R}$ and $(q_n) \to x$ where $q_n$ rational, we have by continuity of $f$
\begin{align*}
f(x) = f\left(\lim\limits_{n \to \infty} q_n\right) = \lim\limits_{n \to \infty} f(q_n).
\end{align*}
Is this the right idea? Am I missing any formalism?

As for the actual proof, I only have an idea for how to prove that $|\mathbb{R}| = |C|$. I want to use the Schroeder-Bernstein theorem to construct injections in both directions and then conclude that there exists a bijection. I can write an injection from $\mathbb{R}$ to $C$ rather easily by mapping $c \in \mathbb{R}$ to the function $f(x) = c$ for all $x$, which is constant, hence continuous. In the opposite direction, I can enumerate the rationals $q_1, q_2, \ldots$ and map a continuous function $f$ to a sequence of its values $(f(q_1), f(q_2), \ldots)$, but that gives an injection to sequences of real numbers, $\mathbb{R}^{\mathbb{N}}$. I would need to then inject $\mathbb{R}^{\mathbb{N}}$ to $\mathbb{R}$ and compose injections to get an injection from $C$ to $\mathbb{R}$.

I do not have an idea for how to prove $|C| < |F|$. The notation is non-standard (since the cardinalities aren't finite), but I believe this typically means that there does not exist a surjection from $C$ to $F$, which makes sense, as intuitively there are plenty more total functions that aren't continuous. I considered a diagonal argument, but can't fully work out the details.

Any help, especially on this last part, would be very much appreciated.


问:
Does the square root function behave the same when calculating complex roots?
This is incredibly basic, yet I cannot for the life of me confirm it. Let's say in the topic of complex numbers we are told to calculate the roots of:

$
z^4 = \sqrt 9
$

Are we to assume $ z^4 = 3 $ and calculate 4 roots or should we proceed with " $ z^8 = 9 $ ", and calculate 8 roots ? Since we are not operating in real numbers, the fact that a negative square root can exist in this instance confused me. Do we just take it as +3 since the square root function is specific to non-negative real numbers? I know the professor wanted us to calculate 4 roots, but I would like to make sure.

答:
$z^4=\sqrt9$ which implies: $z^4=3$ and $z^4=-3$

Solving: $z^4=3$

Let $u=z^2$, then $u^{2}=3 \rightarrow u=\pm \sqrt 3$

Now, $z^2=\sqrt3$ and $z^2=-\sqrt3$. Thus, there are four roots $$z=\pm\sqrt[4]{3}i \,\,\,\&\,\,\, z=\pm\sqrt[4]{3}$$

Solving $z^4=-3$

Let $u=z^2$, thus $u^{2}=-3 \rightarrow u=\sqrt3i \,\,\,\&\,\,\, u=-\sqrt3 i$

Hence, $z^2=\sqrt3i$ and $z^2=-\sqrt3 i$

Let $z=x+iy$, for $x,y \in \mathbb{R}$. Then $z^2=\left(x^2-y^2\right)+i(2xy)$.

Case 1: $z^2=\sqrt 3 i$

Now, the real part equals the real part, and the imaginary part equals the imaginary part. Hence, $$x^2-y^2=0\\ x=\pm y\\x=y$$ and $$2xy=\sqrt 3 \\2x^2=\sqrt 3 \\x=\pm\sqrt\frac{\sqrt 3}{2}\\ x=\pm\frac{\sqrt[4]{3}\sqrt{2}}{2}$$ and plugging back in these $x$-values into $2xy=\sqrt3$, we find that for $x=\frac{\sqrt[4]{3}\sqrt{2}}{2}$, $y=-\sqrt[4]{\frac{3}{4}}$ and for $x=-\frac{\sqrt[4]{3}\sqrt{2}}{2}$, $y=\sqrt[4]{\frac{3}{4}}$.
Hence writing them in terms of $z$, we get: $$z=-\frac{\sqrt[4]{3}\sqrt{2}}{2}+\sqrt[4]{\frac{3}{4}}i,\:z=\frac{\sqrt[4]{3}\sqrt{2}}{2}-\sqrt[4]{\frac{3}{4}}i$$

Case 2:

$z^2=-\sqrt 3 i$. Similarly, for $x,y \in \mathbb{R}$ $x^2-y^2=0\\ x=\pm y\\ x=-y$ and $2xy=-\sqrt 3\\-2x^2=-\sqrt 3\\ \therefore x= \pm \frac{\sqrt[4]{3}\sqrt{2}}{2}$ corresponding to $y=\pm \sqrt[4]{\frac{3}{4}}$. Hence the solution in terms of $z$ are: $$z=-\frac{\sqrt[4]{3}\sqrt{2}}{2}+\sqrt[4]{\frac{3}{4}}i,\:z=\frac{\sqrt[4]{3}\sqrt{2}}{2}-\sqrt[4]{\frac{3}{4}}i$$

Overall, there are $8$ roots to this equation: $$z=\pm\sqrt[4]{3}i \,\,\,\&\,\,\, z=\pm\sqrt[4]{3} \,\,\,\&\,\,\, z=-\frac{\sqrt[4]{3}\sqrt{2}}{2}+\sqrt[4]{\frac{3}{4}}i \,\,\,\&\,\,\,\:z=\frac{\sqrt[4]{3}\sqrt{2}}{2}-\sqrt[4]{\frac{3}{4}}i $$


问:
Homomorphisms vs Functors
What really is the difference between a Homomorphism between categories and a Functor? I understand that they are both structure preserving mapping of the objects and arrows between the categories or am i missing something ?</p>

答:
You are correct, functors are the structure preserving entities between categories. They could have been called homomorphisms. I don't know much about the reason for the terminology used, but I think that Mac Lane and Eilenberg, when inventing category theory, borrowed terminology from philosophy where category and functor are known (at least to some).

Polymorphism is common in mathematics. For instance, the structure preserving entities between topological spaces (though some may say these should actually be considered to be between frames) are called continuous rather than homomorphisms. And an invertible continuous function with continuous inverse is called a homeomorphism rather than an isomorphism. A structure preserving entity between vectors spaces is called a linear transformation. A structure preserving mappings between metric spaces is called a short map. And an invertible one is called an isometry rather than an isomorphism. Clearly, historical reasons play a role.

Specifically regarding categories, since categories don't have to be small, one can argue that a functor is not really a function. One can argue that a homomorphism must be a function that preserves structure, and thus one may claim that functors should not be called homomorphisms. In any case, the terminology, while not set in stone, is certainly well-established but that does not mean other possibilities do not make sense. Sometimes they make more sense.

ContentLicense: CC BY-SA 3.0

Title: Homomorphisms vs Functors


问:
When I build a representative matrix, I don't understand when to "transpose" its components
From my understanding when I build a representative matrix for a linear transformation I take the transformations product as a linear combination of the bases vectors and place it transposed as the column of the matrix. I accidentally thought that a check for linear dependancy in which the vectors were taken as rows constituted a representative matrix - which they didn't.
 here are two examples -  I didn't understand the approach to one as opposed to the other.
 https://ibb.co/tz4TSbx
 https://ibb.co/PGXFhCQ

Edit - clarification: I didn't understand why when writing ImT as is done in the comments we take the vectors as columns  while in the solution added below now we use them as rows.
https://ibb.co/HtrR8bN

答:
I don't see the relation between the first and second picture, since the representative matrix of $T$ should be $3\times3$ (it acts on a three-dimensional vector space). But let's see how to derive the representative matrix of $T$ from the first picture you sent:

Let $T$ be a linear transformation. The representative matrix $A$ of $T$ is built as such:

The $i-$th column of $A$ corresponds to the image of the $i-th$ basis vector by $T$.

For example, as in your picture, let $T: a+bx+cx^2 \mapsto (-2a + b -2c) + (-a -2c)x + (-a + b -3c)x^2$

Your vector space is the space of polynomials of degree $\leq 2$, so a basis is $\{1, x, x^2\}$.

We can see that $T(1) = -2 - x - x^2$, which in the above basis can be written $\begin{pmatrix} -2 \\ -1 \\ -1\end{pmatrix}$.

Similarly, $T(x) = 1 + x^2$, which can be written $\begin{pmatrix} 1 \\ 0 \\ 1\end{pmatrix}$. And $T(x^2)$ is $\begin{pmatrix} -2 \\ -2 \\ -3\end{pmatrix}$.

So the representative matrix of $T$ is $\begin{pmatrix} -2 & 1 & -2 \\ -1 & 0 & -2 \\ -1 & 1 & -3\end{pmatrix}$


问:
Is this proof of the radius of convergence correct?
Claim: Suppose that the coefficients of the power series $\sum a_nz^n$ are integers, infinitely many of which are distinct from $0$. Prove that the radius of convergence is at most 1. (question from Rudin)

My Proof: Let $R$ denote the radius of convergence and $\alpha=\limsup \sqrt[n]{|a_n|}$. Assume, for contradiction, that $R>1$. Then, we have $\frac{1}{\alpha}>1 \implies \limsup \sqrt[n]{|a_n|} < 1\implies \limsup|a_n|<1$. Since $a_n$ is an integer, $\limsup a_n=0$. Thus, there exists $N$ such that $n \geq N \implies a_n=0$. However, this contradicts the fact that there are infinitely many distinct integers in the range of $(a_n)$. $\square$

This is different from other proofs I could find so I'm wondering if this is also correct. Thanks!

答:
Your proof is correct.

My proof: there is a subsequence $(a_{n_k})$ such that $|(a_{n_k})| \ge 1$ for all $k$.

This shows that $(a_n)$ does not converge to $0$. Hence $ \sum a_n$ is divergent.

Thus,the power series $\sum a_nz^n$ is divergent at $z=1.$ Consequence: $R \le 1.$


问:
How find the number of $z$,such that$ |a^2-b^2-b+1|\le 10$
let $f:C\longrightarrow C$ and $f(z)=z^2+zi+1$,find the number of $z$,such $Im(z)>0$
and $|Im(f(z))|\le 10,|Re(f(z))|\le 10$,and $ Im(f(z)) ,Re(f(z)) $are integer numbers.
where $i=\sqrt{-1}$,

$A:399,   B:401,  C:413,   D:431,   E:441$

my idea:let $z=a+bi,b>0$, then
$$f(z)=z^2+iz+1=(a+bi)^2+i(a+bi)+1=(a^2-b^2-b+1)+(2ab+a)i$$

so:
$|a^2-b^2-b+1|\le 10,|2ab+a|\le 10$,and
$a^2-b^2-b+1\in Z,2ab+a\in Z,b>0$
then this problem equalitly find the number pairs$(a,b)$? But I can't find .

Thank you everyone,

答:
If this is a competition or homework or exam and you will use anything below, you should clearly indicate it in your answer. It is a matter of honor. Or stop reading.

We are told that the real and imaginary parts of $f(z)$ are integers not greater than 10 in the absolute value and the imaginary part of $z$ is positive. So the imaginary part of $f(z)$ is between $-10$ and $10$ (21 possibilities). The real part is between $-10$ and $+10$ (21 possibilities). When multiplied, we have $21\times 21=441$ possibilities for what $f(z)$ may be. But we're not finished yet.

Now, we want to  count the possibilities for $z$. Here, $z^2+zi+1=f(z)$ for a fixed $f(z)$ is a quadratic equation that typically has two distinct solutions for $z$; the map is mostly two-to-one and about one of the two solutions has $f(z)\gt 0$. There are a few subtleties, however. In some cases, the two solutions may coincide. And we must take care of the special case ${\rm Im}(z)\sim 0$; some of those values of $z$ have to be eliminated (see below).

First, the quadratic equation can't have double (degenerate) roots because the discriminant $b^2-4ac$ only vanishes for $f(z)=5/4$ which hasn't an integral real part so this is not a problem. Also, the pair of solutions for one choice of $f(z)$ is always different from any element of any pair for another value of $f(z)$ because the solutions are given as $1/2(-i\pm \sqrt{4f(z)-5})$. It is self-evident that different values of $f(z)$ produce different values of $z$ symmetrically distributed from $-i/2$.

So the only problem is to look at the values of $f(z)$ for which the imaginary part of $z$ isn't positive exactly for one of the two solutions. But that may only occur if $|{\rm Im}(\sqrt{4f(z)-5})|\leq 1$ (the factors of $1/2$ cancel). That may occur if the square root is close enough to the real axis. It means that $4f(z)-5$ must be close enough to the positive real semiaxis.

One has to go through the few dozens of numbers near the positive real semiaxis. We eliminate all the roots for $f(z)$ equal to $1,2,3,\dots 10$ (ten values of $f(z)$), and $2\pm i,3\pm i,4\pm i,\dots 10\pm i$ (eighteen values of $f(z)$ in total here), and $5\pm 2i,6\pm 2i,\dots 10\pm 2i$ (twelve values), $10\pm 3i$ (two values). Note that all the initial numbers $1;2\pm i; 5\pm 2i; 10\pm 3i$ were eliminated "marginally" and would stay there is one allowed all non-negative ${\rm Im}(z)$ rather than the strictly positive ones.

So for $10+18+12+2=42$ values of $f(z)$, there are no acceptable solutions $z$ because of the condition on the imaginary part of $z$. For the remaining $441-42=399$ of $f(z)$ values, there is a unique solution per point and they're all distinct. So the right answer is A:399.

Your approach was counterproductive. There's no reason to divide the complex number $z$ to its real and imaginary part; on the contrary, it's right to view this number as a unified complex number for as long as possible a part of the calculation. The simple grid-like integrality condition affected $f(z)$ and not $z$ which is why we looked at the grid in the $f(z)$ plane and not the $z$ plane as you proposed.


问:
Does every vertex has at least one outgoing edge implies the graph contains a cycle?
I can proof it for finite vertex because $v_i\rightarrow v_{i+1}$ (outgoing edge) condition. But what about the infinite case? Is there any counter example for infinite case?

答: Suppose you have vertices $v_k$ for each integer $k$. And $v_k \rightarrow v_{k+1}$, and there are no other edges. Then every vertex has an outgoing edge, but there's no cycle.


答:
What you did is correct. Then you have to solve
$$y_1'-2\tan(t+\frac{\pi}{4})y_1-\tan^2(t+\frac{\pi}{4})=0$$.
With initial condition you should obtain :
$$\boxed{y'_1=-\frac12\tan(x+\frac{\pi}{4})+\frac14\left(1+\tan(x+\frac{\pi}{4})\right)(2x+1)}$$
It is easy to check in solving $y'=1+(1+\epsilon)y^2$ which is a separable ODE :
$$t=\int\frac{dy}{1+(1+\epsilon)y^2}\quad\implies\quad t=\frac{1}{\sqrt{1+\epsilon}}\tan^{-1}(\sqrt{1+\epsilon}\:y)+c$$
And with condition $y(0)=1$ :
$$y= \frac{1}{\sqrt{1+\epsilon}}\tan\left(\sqrt{1+\epsilon}\:t+\tan^{-1}(\sqrt{1+\epsilon}) \right)$$
Expending into power series of $\epsilon$ :
$$y=\tan(t+\frac{\pi}{4})+\left(-\frac12\tan(x+\frac{\pi}{4})+\frac14\left(1+\tan(x+\frac{\pi}{4})\right)(2x+1)\right)\epsilon+...$$
This is consistent with the above results.

Title: Find the first two terms of the perturbation series solution to the IVP

Body:
Here is the equation for the initial value problem, $y'=1+(1+\epsilon)y^2$ with initial condition $y(0)=1$ with $t>0$ where $0<\epsilon\ll1.$. Find exact solution and compare approximation.

My attempt:

Suppose $y=y_0+\epsilon y_1+O(\epsilon^2)$ Then $y_0'+\epsilon y_1'+O(\epsilon^2)=1+(1+\epsilon)(y_0+\epsilon y_1+O(\epsilon^2))^2$, after distributing and combining like terms. $\implies y_0'+\epsilon y_1'+O(\epsilon^2)=1+y_0^2+2\epsilon y_0y_1+\epsilon y_0^2+O(\epsilon^2)$.

Grouping terms $\\\implies (y_0'-y_0^2-1)+\epsilon(y_1'-2y_0y_1-y_0^2)+O(\epsilon^2)=0
\\$.

Finally obtaining two sets of ODE's (This is where I stopped)$ \implies y_0'-y_0^2-1=0, \space y_1'-2y_0y_1-y_0^2=0 \implies \arctan(y_0)=t+C$

Using initial value $y(0)=1$ we get $C=\frac{\pi}{4}$. So now we establish that $y_0=\tan(t+\frac{\pi}{4})$.

So this implies that, $y_1'-2\tan(t+\frac{\pi}{4})y_1-\tan^2(t+\frac{\pi}{4})=0$.

Am I going in the right direction? Solving the second set of ODE's just seems overly complicated so I used Wolfram Alpha.


问:
I don't know how to solve $\sin(x) + \sin(2x) + \sin(3x) = -3$
I have no idea how to solve this equation $\sin(x) + \sin(2x) + \sin(3x) = -3$. I've tried to use the formulas for $\sin(2x) = 2\sin(x)\cos(x)$ and for $\sin(3x) =  3\sin(x) - 4\sin^{3}(x)$, but all for nothing.

答:
Note that to attain $-3$, we require

$$\sin(x)=\sin(2x)=\sin(3x)=-1$$

I will leave the rest to you.


答:
In light of the comment that the original expression was $z^4 = \sqrt{\frac {-9}{i^2}}$ that... this is almost more a case for a lawyer than a mathematician.

The convention in complex analysis is that $\sqrt[p]{\alpha}:= \{z\in \mathbb C| z^p = \alpha\}$ with a practical (yet arbitrary) exception that if  $\alpha \in \mathbb R^+$ then it does not mean the set of all roots be merely $\sqrt[p]{\alpha} = $ the single element $w\in \mathbb R^+$ so that $w^p=\alpha$.

So $z^4 = \sqrt{9}\implies z^4 =3\implies$ 4 complex roots.

!BUT! now the lawyer:  If $z^4 = \sqrt{\frac {-9}{i^2}}$ it is not declared and (for some values of "clear") not immediately clear then $\frac {-9}{i^2} \in \mathbb R^+$ (it is, but it's not immediately apparent).  As the exception for $\alpha \in \mathbb R^+$ was a practical one to mesh with the conventions of "regular" non-complex math, the expression $\frac {-9}{i^2}$ clearly indicates that we are doing with the full range of complex numbers.

So I'd interpret $\sqrt{\frac {-9}{i^2}} = \{z|z^2 =\frac {-9}{i^2}\}=\{3,-3\} \ne 3 = \sqrt 9$. And that $z^4 =\sqrt{\frac {-9}{i^2}}\implies z^4 \in \{3,-3\}\implies $ 8 complex roots.

And we'd have a really weird case where $\alpha = w$ but $\sqrt{\alpha} \ne \sqrt w$ which... well, that's a deliberate manipulation of contextual meaning taken deliberately out of context.  It doesn't count.

Not all books will agree with my assessment.  Ask your professor.

========

I think most text books consider that if $w> 0$ (which implies $w \in \mathbb R$) that $\sqrt[k]{w}$ or $w^{\frac pq}$ refers to the principal positive root, While for any other case, $\sqrt[k]{w}$ or $w^{\frac pq}$ would refer to the set of all roots.

Thus that would mean $\sqrt 9 = 3$. And you are being asked nothing more than to solve $z^4 = 3$>

But... you should ask your professor.

After all, if it is consistent for if $w \not \in \mathbb R^+$ to have $\sqrt[k]{w}$ or $w^{\frac pq}$ be that multi-valued set of roots, a text could argue for consistency it would mean that for $w \in \mathbb R^+$ as well.

If so the $\sqrt{9} = \{3, -3\}$ and you are being asked to solve ($z^4 =3$ OR $z^4 = -3$) which, as you point out would be the same as solving $z^8 = 9$.

.....

So which does your text and professor do?

I'd say in about 80% of the cases, it would mean $\sqrt{9} = 3$ as $9\in \mathbb R^+$; a conventional inconsistency.

.....

But if so... why is this question even being asked?  Why not just say $z^4 = 3$.  Is this part of something more complicated?

If you had $z^4 = \sqrt{\alpha};\alpha \in \mathbb C$ you have $8$ roots.  But then if we did further calculations and found $\alpha =9$.... well the original intent of $\sqrt{alpha}$ was the set of all roots, and suddenly stating $z^4 = \sqrt 9$ is a bait and switch and the meaning has changed.  We can't just change definitions mid calculation just because our number falls into an inconsistant and arbitrary notational convention. so the solution should be $z^4 =\pm{\sqrt 9}$ has eight roots.

.......

On third reading of Levinson/Reheffer "Complex Variables" They state the $z$ so that $z^q = \alpha^p$ are called the $q$th roots all of these is designated by $\sqrt[q]{\alpha^p}$ or by $\alpha^{\frac pq}$. They then state that if $\alpha > 0$ (I don't know why they didn't just say if $\alpha$ is positive real) then $\alpha^{\frac pq}$ is reserved for the one with $\arg$ of $0$ (why the didn't just say the positive real one I don't know) unless stated otherwise.

I see that they did not make the same claim for the notation $\sqrt[q]{\alpha^p}$.  I do not know if that was an oversight or if the specifically mean that the $\sqrt{}$ symbol is not reserved for the positive value.

That'd mean $\sqrt{9} = \pm 3$ and $8$ roots.

I may have been inaccurate in my assessment that 80% of text would believe $\sqrt 9 = 3$.  It could be that they interpret $9^{\frac 12} =3$ but $\sqrt{9} = \{3,-3\}$.

After we kill all the lawyers we should kill all the mathematicians.

Title: Does the square root function behave the same when calculating complex roots?

Body:
This is incredibly basic, yet I cannot for the life of me confirm it. Let's say in the topic of complex numbers we are told to calculate the roots of:

$
z^4 = \sqrt 9
$

Are we to assume $ z^4 = 3 $ and calculate 4 roots or should we proceed with " $ z^8 = 9 $ ", and calculate 8 roots ? Since we are not operating in real numbers, the fact that a negative square root can exist in this instance confused me. Do we just take it as +3 since the square root function is specific to non-negative real numbers? I know the professor wanted us to calculate 4 roots, but I would like to make sure.


问:
If $d$ is a metric, is $d(x,\;\cdot\;)$ continuous from above?
Let $(E,d)$ be a metric space, $x\in E$ and $$d(x,B):=\inf_{y\in B}d(x,y)\;\;\;\text{for }B\subseteq E.$$

Let $(B_n)_{n\in\mathbb N}$ be nonincreasing and $B:=\bigcap_{n\in\mathbb N}$. Are we able to show that $d(x,B_n)\xrightarrow{n\to\infty}d(x,B)$?

Since $(B_n)_{n\in\mathbb N}$ is nonincreasing, we clearly obtain $$d(x,B_n)\xrightarrow{n\to\infty}\sup_{n\in\mathbb N}d(x,B_n)\in[0,\infty]\tag1.$$

Body: Let $B_n=\{0\}\cup (1,1+\frac 1 n)$. Then $B_n$ decreases to $B=\{0\}$. $d(2,B_n) \to 1$ whreas $d(2,B)=2$.


问:
Limiting behavior of a function defined by a Lambert-type series
Fix a positive real constant $\omega$, and let $\left\{ c_{n}\right\} _{n\geq1}$ be a sequence of real numbers so that the series: $$\sum_{n=1}^{\infty}\frac{c_{n}}{n^{\omega}}$$ converges conditionally to a finite value $S$. Next, let: $$L\overset{\textrm{def}}{=}\lim_{x\uparrow1}\left(1-x\right)^{\omega}\sum_{n=1}^{\infty}c_{n}\frac{1-\left(1-x^{n}\right)^{\omega}}{\left(1-x^{n}\right)^{\omega}}$$ Is it true that $L=S$? That is, is the interchange of limit and sum:$$\lim_{x\uparrow1}\left(1-x\right)^{\omega}\sum_{n=1}^{\infty}c_{n}\frac{1-\left(1-x^{n}\right)^{\omega}}{\left(1-x^{n}\right)^{\omega}}\overset{?}{=}\sum_{n=1}^{\infty}c_{n}\lim_{x\uparrow1}\left(1-x\right)^{\omega}\frac{1-\left(1-x^{n}\right)^{\omega}}{\left(1-x^{n}\right)^{\omega}}=\sum_{n=1}^{\infty}\frac{c_{n}}{n^{\omega}}$$ valid? I already know that this is true whenever the sum defining S is absolutely convergent. I've tried using a summation-by-parts argument to deal with the conditionally convergent case, which gives:

$$L-S=\lim_{x\uparrow1}\left(1-x\right)^{\omega}\sum_{n=1}^{\infty}\left(\left(\frac{n+1}{1-x^{n+1}}\right)^{\omega}-\left(\frac{n}{1-x^{n}}\right)^{\omega}+n^{\omega}-\left(n+1\right)^{\omega}\right)\sum_{k=n+1}^{\infty}\frac{c_{k}}{k^{\omega}}$$ and I suspect that Dirichlet's test for series convergence—namely, that: $$\sum_{n=1}^{\infty}a_{n}b_{n}$$ converges whenever the $b_{n}$s are a non-increasing sequence of real numbers that tend to $0$  as $n\rightarrow\infty$, and whenever the $a_{n}$s are complex numbers for which: $$\sup_{N\geq1}\left|\sum_{n=1}^{N}a_{n}\right|<\infty$$ but I can't quite see how to make the argument work.

Anyhow, how might I show that $L=S$ (using Dirichlet's test or some other method), or, is there a counterexample that will dash my hopes to pieces? If there is a counterexample, might there be some minor stipulation to be made for the $c_n$s so as to guarantee $L=S$?

答:
Yes, it holds true. There's a known criterion of "regularity of Abelian summation":

Let $f_n:(a,b)\to\mathbb{R}$ meet $\lim\limits_{x\to b}f_n(x)=1$ for each $n$. Then, to have $$\sum_{n=1}^\infty a_n=\lim_{x\to b}\sum_{n=1}^\infty a_n f_n(x)$$ for any convergent $\sum_{n=1}^\infty a_n$, it is necessary and sufficient to have $$\sup_{x\in(c,b)}\sum_{n=1}^\infty\big|f_n(x)-f_{n+1}(x)\big|<\infty\tag{*}\label{maincond}$$ for some $c\in(a,b)$.

A proof is in section $4.7$ of Divergent Series by G.H.Hardy. We only need the sufficiency (which is fairly easy to prove, basically following your idea of summation by parts), but the necessity implies that we have the only way to go. So we check \eqref{maincond} with $(c,b)=(0,1)$ and $$f_n(x)=\big(n(1-x)\big)^\omega\big((1-x^n)^{-\omega}-1\big).$$ It turns out that $\sup\limits_{x\in(0,1)}=\lim\limits_{x\,\uparrow\,1}$ here, and a computation of this value (call it $S_\omega$) is sketched below.

For a fixed $x\in(0,1)$, the map $n\mapsto f_n(x)$ is monotonic if $\omega\leqslant 1$ and unimodal if $\omega>1$.
This implies  $\color{blue}{S_\omega=2L_\omega-1}$, where $L_\omega=\sup_{x\in(0,1)}\sup_n f_n(x)$, and $L_\omega=1$ if $\omega\leqslant 1$.

To show the above, in rough steps, we set $g_\omega(x)=(-\log x)^\omega\big((1-x)^{-\omega}-1\big)$ so that $$f_\color{blue}{n}(x)=\left(\frac{1-x}{-\log x}\right)^\omega g_\omega(x^{\color{blue}{n}}).$$ Now if $\omega>1$ then $g_\omega'(x)=0\iff 1-(1-x)^\omega=(-x\log x)/(1-x)$ has a unique solution $x=x_\omega\in(0,1)$ [consider the "inverse" function $\omega=\omega(x)$], giving eventually $\color{blue}{L_\omega=g_\omega(x_\omega)}$.

And if $\omega\leqslant 1$ then $g_\omega'(x)>0$ for $x\in(0,1)$ [shown similarly], giving the expected $\color{blue}{L_\omega=1}$.


问:
If $S_1=0$ and $S_2=0$ represent two non intersecting circles, what does $S_1+\lambda S_2=0$ represent?

If $S_1=0$ and $S_2=0$ represent two non intersecting circles, what does $S_1+\lambda S_2=0$ represent?

I know that $S_1+\lambda S_2=0$ represent the family of circles that are coaxial with $S_1=0$ and $S_2=0$, (where $\lambda\ne -1$, because then it would become the radical axis) but I cannot see it geometrically. If the circles are intersecting, then it is clear that $S_1+\lambda S_2=0$ represents the family of circles passing through their common points of intersection and hence they would have the same radical axis, that is, their common chord. How do I prove that for non intersecting circles?

Secondly, I want to ask if we are given a line $L=0$ and a circle $S=0$, what does $S+\lambda L=0$ represent if they are non intersecting? For intersecting case, it is clearly that it would be family of circles passing through their points of intersection. I think it would represent the circles $S'=0$ such that $S=0$ and $S'=0$ have the radical axis $L=0$, but there is no intuition, because intersection points are imaginary.

Maybe it is just that everything is right in front of me, but I am not able to understand it clearly. I would prefer a complete answer rather than comments.

答:
Your two questions are in fact one, because the pencil for circles $C_1+\lambda C_2$ contains the radical axis of $C_1,C_2$ (the one that you rejected), and conversely, the pencil $C+\lambda L$ defined by a circle and a line is such that $L$ is the radical axis of $C and another circle in the pencil.

WLOG, consider

$$\begin{cases}S_1\equiv x^2+y^2=1,\\S_2\equiv(x-d)^2+y^2=r^2.\end{cases}$$ By subtraction, the line in the pencil is

$$L\equiv d(2x-d)=1-r^2$$ or

$$x=\frac d2+\frac{1-r^2}{2d}.$$

The distances of the intersection with the axis $x$ to the centers are

$$\pm\frac d2+\frac{1-r^2}{2d}$$ and we do have by Pythagoras

$$\left(\frac d2+\frac{1-r^2}{2d}\right)^2-1^2=\left(-\frac{d}2+\frac{1-r^2}{2d}\right)^2-r^2,$$

which proves that the lengths of the tangent segments to the circles from this point have equal lengths and the line is the radical axis.

https://en.wikipedia.org/wiki/Radical_axis#/media/File:Radical_axis_orthogonal_circles.svg

These families of circles are well illustrated by the Appolonius' coordinates system:


答: A definition of the square root $x$ of a given $y$ is $x$ such that $x\cdot x=y$. This definition works over any field.


问:
Limit with regularised incomplete Beta function
Let $\mathcal{I}_x(a,b)$ denote the regularised incomplete Beta function:
$$
\mathcal{I}_x(a,b)=\frac{\Gamma(a+b)}{
\Gamma(a)\Gamma(b)
}\int_0^x t^{a-1}(1-t)^{b-1}dt,
$$
where $x \in (0,1)$ and $a,b>0$. Let $q>0$ and, for any $y>0$, denote $\log^qy=(\log y)^q$. What is the asymptotic behaviour of
$$
\mathcal{I}_{\frac{\log^qn}{\sqrt{n}}}(a/\log n, b/ \log n)
$$
as $n\to \infty$? I could not find any useful reference. I know that, as $n \to \infty$,
$$
\frac{\Gamma((a+b)/\log n)}{
\Gamma(a/\log n)\Gamma(b /\log n)
}\sim \frac{ab}{a+b}\frac{1}{\log n};
$$
as for the remaining factor, i.e.
$$
\int_0^{\frac{\log^qn}{\sqrt{n}}} t^{a/\log n-1}(1-t)^{b/\log n-1}dt,
$$
I've only managed to obtain the not so sharp lower bound $\left(\frac{\log^qn}{\sqrt{n}}\right)^{a/\log n} \to e^{-a/2}$ and upper bound
$$
\left(1-\frac{\log^qn}{\sqrt{n}}\right)^{b/\log n-1} \frac{\log n}{a} \left(
\frac{\log^qn}{\sqrt{n}}
\right)^{a/\log n}\sim \frac{\log n}{a e^{a/2}}.
$$
Thus I can conclude the limit of $\mathcal{I}_{\frac{\log^qn}{\sqrt{n}}}(a/\log n, b/ \log n)$ is finite: is it zero?

答:
For $x\to 0+$, we can expand the factor $(1-t)^{b-1}$ about $t=0$ to find
$$
I_x (\alpha ,\beta ) = \frac{{\Gamma (\alpha  + \beta )}}{{\Gamma (\alpha )\Gamma (\beta )}}\frac{{x^\alpha  }}{\alpha }\left( {1 +  \mathcal{O}(x)} \right).
$$
With your specific parameters, this gives
$$
\frac{b}{{a + b}}\left( {\frac{{\log ^q n}}{{\sqrt n }}} \right)^{a/\log n} \left( {1 + \mathcal{O}\!\left( {\frac{{\log ^q n}}{{\sqrt n }}} \right)} \right) = \frac{{be^{ - a/2} }}{{a + b}}\left( {1 + \mathcal{O}\!\left( {\frac{{\log \log n}}{{\log n}}} \right)} \right),
$$
as $n\to +\infty$.


答: The second example was to check linear dependancy and not to make a rep matrix, so in that case we do take it as rows and not columns, while when we do work on rep. matrices we always take them as columns. I was confused - hope I didn't mislead anybody or caused to much headaches</p>

Title: When I build a representative matrix, I don't understand when to "transpose" its components

Body:
From my understanding when I build a representative matrix for a linear transformation I take the transformations product as a linear combination of the bases vectors and place it transposed as the column of the matrix. I accidentally thought that a check for linear dependancy in which the vectors were taken as rows constituted a representative matrix - which they didn't.
 here are two examples -  I didn't understand the approach to one as opposed to the other.
 https://ibb.co/tz4TSbx
 https://ibb.co/PGXFhCQ

Edit - clarification: I didn't understand why when writing ImT as is done in the comments we take the vectors as columns  while in the solution added below now we use them as rows.
https://ibb.co/HtrR8bN


问:
Proof: Characterization of Extreme Points
I'm having difficulty following a proof in the book Nonlinear Programming Theory and Algorithms, Third Edition by Bazaraa, Sherali, and Shetty on page 68 section 2.6.4. The proof is reproduced below, up to the part of the proof I'm confused about.

My question concerns the author's statement:

But this implies that $x_{11} = x_{21} = B^{-1}b$.

I don't see how the author makes that connection.

Theorem (Existence of Extreme Points)

Let $S = \{ x : Ax = b, x \geq 0\}$, where $A$ is an m x n matrix and $b$ is an m-vector. A point $x$ is an extreme point of $S$ if and only if $A$ can be decomposed into $[B, N]$ such that

$x = \begin{bmatrix}
x_B \\ x_N
\end{bmatrix} =
\begin{bmatrix}
B^{-1}b \\ 0
\end{bmatrix}$

Proof:

Suppose that $A$ can be decomposed into $[B, N]$ with $x = \begin{bmatrix} B^{-1}b \\ 0 \end{bmatrix}$ and $B^{-1}b \geq 0$. It is obvious that $x \in S$. Now suppose that $x = \lambda x_1 + (1 - \lambda)x_2$ with $x_1$, $x_2 \in S$ for some $\lambda \in (0,1)$. In particular, let $x^{t}_{1} = (x^{t}_{11}, x^{t}_{12})$ and $x^{t}_{2} = (x^{t}_{21}, x^{t}_{22})$. Then

$\begin{bmatrix} B^{-1}b \\ 0 \end{bmatrix} = \lambda \begin{bmatrix} x_{11} \\ x_{12} \end{bmatrix} + (1 - \lambda) \begin{bmatrix} x_{21} \\ x_{22} \end{bmatrix}$

Since $x_{12}, x_{22} \geq 0$ and $\lambda \in (0,1)$, it follows that $x_{12} = x_{22} = 0$. But this implies that $x_{11} = x_{21} = B^{-1}b$ and hence $x = x_1 = x_2$. This shows that $x$ is an extreme point of $S$.

答:
Note that $x_1,x_2 \in S$ and hence $Ax_1 = b$ and $Ax_2 = b$. Hence, you get $x_{11} = x_{21} = B^{-1}b$ and hence $x_1 = x_2$.

To elaborate a bit, since $x_{1},x_{2} \geq 0$ and $\lambda \in (0,1)$, we have $x_{12} = 0 = x_{22}$.

Now, $Ax_1 = b \implies Bx_{11} + Nx_{12} = b \implies B x_{11} = b \implies x_{11} = B^{-1}b$.

Similarly, $Ax_2 = b \implies Bx_{21} + Nx_{22} = b \implies B x_{21} = b \implies x_{21} = B^{-1}b$.

Hence, $x_1 = x_2$ and both equals $x$.


问:
finding the umvue of variance of exponential distribution
I'm suppose to find the UMVUE for the variance $\theta^2$ of the exponential distribution $Exp(\theta) \sim f(x,\theta)=\frac{1}{\theta}\exp(-x/\theta)$

The hint is to use $\bar{X}^2$.

In the brief solution, it expressed $E(\bar{X}^2)=\frac{n+1}{n}\theta^2$

Then it stated that after bias correction, by lehmann-scheffe: $\frac{n(\bar{X}^2)}{n+1}=\frac{T^2}{n(n+1)} = $ UMVUE of variance($\theta^2$)

My problem is, I don't follow what bias correction step it applied before Lehmann-Scheffe? What did it use as the initial unbiased estimator to apply L-S Theorem? How did it turn $E(\bar{X}^2)$ to $\bar{X}^2$? How L-S Theorem would be needed to have $\bar{X}^2 = T^2/n^2$

答:
The hint is correct but you can use also this one that, in my humble opinion, looks more suitable (the more basic is the estimator the better it is):

$$T=\left(\Sigma_i X_i\right)^2$$

This because $Y=\Sigma_i X_i$ is complete and sufficient (CSS), thus the goal now is to construct an unbiased estimator based on the CSS.

$$\mathbb{E}[T]=\int_0^{\infty} y^2\frac{1}{\theta^n \Gamma(n)}y^{n-1}e^{y/\theta}dy=$$

$$=\frac{\Gamma(n+2)\theta^2}{\Gamma(n)}\underbrace{\int_0^{\infty} y^2\frac{1}{\theta^{n+2} \Gamma(n+2)}y^{(n+2)-1}e^{y/\theta}dy}_{=1}=(n+1)n\theta^2$$

thus the UMVUE is

$$T_{\text{UMVUE}}=\frac{\left(\Sigma_i X_i\right)^2}{(n+1)n}$$

It is self evident that this estimator matches with the one stated in you textbook

$$\frac{n(\overline{X})^2}{n+1}$$


问:
Is intersection of a subset with a net a subnet?
Say $\{x_i\}_{i \in I}$ is a net in a set $X$ and $Y\subset X$ then, is $\{x_i\}_{i \in I}\cap Y$ a subnet? intuitively this seems true but I don't know how to approach the proof.

答:
Be careful in your notations: the net is a fucntion $x: I \to X$ (often denoted for short by $\{x_i\}_{i \in I}$ and I can interpret your abuse of notation as

$x^{-1}[Y] = \{i \in I: x(i) \in Y\}$, and then you probably mean $x\restriction_{x^{-1}[Y]}$ as the subnet. It is indeed a map into $Y$, but for a net we need the domain to be directed. And for a subnet we need $x^{-1}[Y]$ to be cofinal in $I$, and then the directedness will follow from that of $I$:

So assume $x^{-1}[Y]$ is cofinal i.e.

$$\forall i \in I \exists j \in I: j \ge i \land x_j=x(j) \in Y$$

If this condition is satisfied then $x^{-1}[Y]$ is directed: if $i_1, i_2 \in x^{-1}[Y]$ we can find $i_3 \in I$ with $i_3 \ge i_1$ and $i_3 \ge i_2$, and then we have $i_4 \in x^{-1}[Y]$ so that $i_4 \ge i_3$ so by transitivity we have $i_4 \ge i_1,i_2$ as required. The inclusion map from $x^{-1}[Y]$ into $I$ shows then that $x\restriction_{x^{-1}[Y]}$ is indeed a subnet of $x$ (in both the Kelly and Willard sense, so in most texts, where one of these two is used as the definition of subnet..)

So yes iff $x^{-1}[Y]$ is cofinal in $I$.


答:
No. In order for something to be a subnet it has to:

Be a net. How exactly $\{x_i\}_{i\in I}\cap Y$ is a net? It is a set, not a map from a directed set. A natural choice would be to define $J=\{i\in I\ |\ x_i\in Y\}$ and $j\mapsto x_j$. But then we need that $J$ to be a directed set. I.e. it has to satisfy: if $x_i\in Y$ and $x_j\in Y$ then there exists $k\in I$ such that $k\geq i$, $k\geq j$ and $x_k\in Y$. With that your $\{x_i\}_{i\in I}\cap Y$ is indeed a net.

There is a final monotonic function $f:J\to I$, such that our net composed with $f$ is the subnet. In other words subnets arise only by modifying the directed domain via final monotonic functions.

With that we have a simple counterexample: assume that $Y$ is such that $\{x_i\}_{i\in I}\cap Y$ is a singleton, while $I$ is a directed poset with no maximal element (e.g. $I=\mathbb{N}$ and our net is actually a sequence). Then there is no final monotonic function $J\to I$ at all, and so the induced net $(x_j)_{j\in J}$ is not a subnet of $(x_i)_{i\in I}$

Title: Is intersection of a subset with a net a subnet?

Body: Say $\{x_i\}_{i \in I}$ is a net in a set $X$ and $Y\subset X$ then, is $\{x_i\}_{i \in I}\cap Y$ a subnet? intuitively this seems true but I don't know how to approach the proof.


问:
Help with Cartesian product subsets
I want to prove that if  $A \subseteq C\,$ and $\,B \subseteq D,\,$ then $\,A \times B \subseteq C \times D.$

I know that $A \subseteq C \iff a \in A \rightarrow a \in C$ and that $B\subseteq D\iff b \in B \rightarrow  b \in D$  I also know that $A \times B = \{(a, b)\mid a\in A, b\in B\}$ and that $C\times D = \{(c, d) \mid c \in C, d \in D\}$.

So keep that in mind, how do I connect the dots?

答:
You have all of the pieces.

You want to show that every element of $A\times B$ is an element of $C\times D$, so start with an arbitrary element of $A\times B$. What does such an element look like? It’s $\langle a,b\rangle$ for some $a\in A$ and $b\in B$. Since $a\in A$, you know that $a\in C$, and since $b\in B$, you know that $b\in D$. Now what can you conclude about $\langle a,b\rangle$?

Title: Help with Cartesian product subsets


问:
Let $X=\{1,2,3,4,5\}$. Find the number of ordered pairs $(Y,Z)$ such that $Y, Z\subset X$ and $Y\cap Z=\emptyset$
Basically the question is asking the number of ways in which 2 elements can be selected such that the order pair formed by them doesn’t have common elements

So if one element is picked from the 5, only 4 elements remain, and the pairs themselves can be arranged in two ways
$$\binom 51 \binom 41 2!$$ $$=40$$

The correct answer is $243$. What am I missing here?

答:
Label every point of $\{1,2,3,4,5\}$ by $1$ (goes into set $A$), $2$ (goes into set $B$) or $3$ (goes into neither).

Every labelling yields a good pair $(A,B)$ and vice versa.

So we have $3^5 = 243$ such pairs.


问:
Probability of $2$ people leaving their rooms
$7$ people enter a hotel and go into $4$ rooms numbered $1$ to $4$. Each person chooses
at random a room to go into, independently of other people.

Exactly two people leave their rooms in the first day. What is the probability
that the two people who left the hotel were not from the same room?

How do I start doing this kind of problems? Please help! :(

Update: We are not assuming that each room has to have 1 person in it. We can have empty rooms.

答:
Hint: Let the first person leave the room $X$. What is the probability that the second person was in the same room?

 The probability is $\frac14$ so that the answer to your question is $1-\frac14=\frac34$. Important is only the fact that people choose the room independently and at random. Also important is the fact that there are at least two people in the hotel. Everything else (e.g. exact number of the people) is irrelevant.


问:
Find all positive solutions for equation:
$nx^{(n+1)}-(n+1)x^n+1=0$

There's nothing told about $n$, I guess $ n \in N $. I would like any kind explanations, thanks! I appreciate your time.

答:
Let $f(x)=nx^{(n+1)}-(n+1)x^n+1$ $n\in N,x>0$

$f'(x)=n(n+1)x^n-n(n+1)x^{n-1}=n(n+1)x^{n-1}(x-1)$

So we have $f'(x)>0 \forall x>1$

and we have $f'(x)<0\forall 0<x<0$

This implies that the function is increasing for all $x>1$ and decreasing for $0<x<1$

Now $f(1)=0$ as $f(x)$ is increasing for $x>1$ so we have $f(x)>f(1)=0,\forall x>1$

And as it is decresing for $0<x<1$ so we have $f(x)<f(1)$ for $0<x<1$.

So the only (positive)solution of this equation is $x=1$


问:
Find Cross-Tangents to Offset Ellipses
Question:

I am stuying projective geometry and would like to find an analytic solution to my problem. This is a follow-up to my older question.

Given two ellipses with the same major and minor axes, but different origins (written in homogeneous coordinates):

$$
\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1 = 0
$$

and

$$
\frac{(x_2-h)^2}{a^2} + \frac{(y_2-k)^2}{b^2} - 1 = 0
$$

what are the equations of the cross-tangents to the two ellipses? Assume $h, k, a,$ and $b$ are all knowns.

My Attempt:

Following the method from my older question, it makes sense to find the intersection points of the dual curves to each ellipse. Let $f_1$ be the implicit homogeneous function defining our first ellipse (i.e. $f_1(x_1, y_1, z_1) = 0$) and $f_2$ that of our second, respectively. The dual curve of an ellipse is an ellipse (at least, AFAIK, for an ellipse at the origin), which Wikipedia shows nicely.

Hence, our first dual is

$$
a^2 X_1^2 + b^2 Y_1^2 = 1.
$$

I am having trouble finding the second dual (namely, that of the ellipse centered at $(h,k)$). Could someone please help me in deriving this? Is this the simplest way to solve this problem?

Following Wikipedia, about the closest I get to is the following and I don't know how to eliminate $\lambda$:

$$
\frac{a^2}{2\lambda} \left( X_2 + \frac{\lambda h}{a^2} \right)^2 + \frac{b^2}{2\lambda} \left( Y_2 + \frac{\lambda k}{b^2} \right)^2 - \left( \frac{\lambda}{2} \right) \left( \left( \frac{h}{a} \right)^2 + \left( \frac{k}{b} \right)^2 \right)
 + 1 = 0$$

Title: Find Cross-Tangents to Offset Ellipses

Body:
The dual conics are for $ax^2+bxy+cy^2+dx+ey+f=0$ given by $$ (-\frac14e^2+cf)x^2+(\frac12de-bf)xy+(-\frac14d^2+af)y^2+(-cd+\frac12be)x+(\frac12bd-ae)y-\frac14b^2+ac=0$$ or $\begin{pmatrix}x&y&1\end{pmatrix}\begin{pmatrix}a&\frac{b}{2}&\frac{d}{2}\\\frac{b}{2}&c&\frac{e}{2}\\\frac{d}{2}&\frac{e}{2}&f\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}$ to $$\begin{pmatrix}x&y&1\end{pmatrix}\begin{pmatrix}-\frac14e^2+cf&\frac{\frac12de-bf}{2}&\frac{-cd+\frac12be}{2}\\\frac{\frac12de-bf}{2}&-\frac14d^2+af&\frac{\frac12bd-ae}{2}\\\frac{-cd+\frac12be}{2}&\frac{\frac12bd-ae}{2}&-\frac14b^2+ac\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}$$

from which you can see it's to do with the matrix of a conic and it's adjoint matrix which is the inverse when divided by the determinant. In your special case:

k^2*y^2-b^2*y^2+2*h*k*x*y+2*k*y+h^2*x^2-a^2*x^2+2*h*x+1=0
(-b^2*y^2)-a^2*x^2+1=0

Now solve for intersection points

solve([((b^2*h^2+a^2*k^2)*y^2-h^2)*((b^2*h^2+a^2*k^2)*y^2+4*a^2*k*y+4*a^2-h^2),(2*a^2*b^2*h^3+2*a^4*h*k^2-2*a^2*h^3*k^2)*x+(b^4*h^4*k+2*a^2*b^2*h^2*k^3+a^4*k^5)*y^3+(-b^4*h^4+2*a^2*b^2*h^2*k^2+3*a^4*k^4)*y^2+(2*a^2*b^2*h^2*k-b^2*h^4*k+2*a^4*k^3-3*a^2*h^2*k^3)*y+b^2*h^4-3*a^2*h^2*k^2],[x,y]);

[[x = k/sqrt(a^2*k^2+b^2*h^2),y = -h/sqrt(a^2*k^2+b^2*h^2)],
     [x = -k/sqrt(a^2*k^2+b^2*h^2),y = h/sqrt(a^2*k^2+b^2*h^2)],
     [x = (k*sqrt(a^2*k^2+b^2*h^2-4*a^2*b^2)-2*b^2*h)/(a^2*k^2+b^2*h^2),
      y = -(h*sqrt(a^2*k^2+b^2*h^2-4*a^2*b^2)+2*a^2*k)/(a^2*k^2+b^2*h^2)],
     [x = -(k*sqrt(a^2*k^2+b^2*h^2-4*a^2*b^2)+2*b^2*h)/(a^2*k^2+b^2*h^2),
      y = (h*sqrt(a^2*k^2+b^2*h^2-4*a^2*b^2)-2*a^2*k)/(a^2*k^2+b^2*h^2)]]

So the cross tangents are $$x  \frac{k\sqrt{a^2k^2+b^2h^2-4a^2b^2}-2b^2h}{a^2k^2+b^2h^2}+y \frac{ -(h\sqrt{a^2k^2+b^2h^2-4a^2b^2}+2a^2k)}{a^2k^2+b^2h^2}+1=0\\
 x \frac{-(k\sqrt{a^2k^2+b^2h^2-4a^2b^2}+2b^2h)}{a^2k^2+b^2h^2}+
  y \frac{h\sqrt{a^2k^2+b^2h^2-4a^2b^2}-2a^2k}{a^2k^2+b^2h^2}+1=0.$$


问:
Recurrent solution to $v'_{n+1}+\alpha v_{n+1} = -\beta/(n+1) v'_n$
Looking for a series solution to a second order PDE (namely, of the form $f(x,t) = \sum_n v_n(x) t^n$), I came across the following 'difference-differential' equation (of a kind I didn't encounter before):
\begin{align}
v'_{n+1} + \alpha v_{n+1} = - \frac{\beta}{n+1}v'_{n}.
\end{align}

I'm wondering if there is a general recurrent relation that will relate the $n=0$ terms (function and its derivatives) to the $N^{\rm th}$ terms (function and/or its derivatives). Somehow I'm unable to express this relation in the form of a matrix equation, no matter what vector form I choose. The last one I tried was $\vec{v}_n = [v'_{n+1}, v_{n+1}, v'_n, v_n]$ with a matrix equation $\mathbf{B}\vec{v}_{n+1} = \mathbf{A}_n \vec{v}_n$, specifically,
\begin{align}
\begin{pmatrix}
1 & \alpha  & 0 & 0\\
0 & 0  & 1 & \alpha \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\begin{bmatrix}
v'_{n+1} \\
v_{n+1} \\
v'_{n} \\
v_{n}
\end{bmatrix}
=
\begin{pmatrix}
- \frac{\beta}{n+1}  & 0 & 0 & 0 \\
0 & 0 & -\frac{\beta}{n} & 0 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0
\end{pmatrix}
\begin{bmatrix}
v'_{n} \\
v_{n} \\
v'_{n-1} \\
v_{n-1}
\end{bmatrix}
\end{align}
The problem is that $\mathbf{B}$ is not invertible, so I cannot really do recursion on this. Any suggestion on how to approach the problem is appreciated.

答:
Hint.

Applying the Laplace transform we have

$$
(s+\alpha) V_n(s) = -\frac{\beta}{n+1}\left(sV_{n-1}(s)-v_{n-1}(0)\right)+v_n(0)
$$

then giving an $v_0(t)$ we can build all the sequence. For instance, with null initial conditions

$$
V_n(s) = \frac{(-\beta)^{n-1}}{(3)_{n-1}}\left(\frac{  s}{\alpha +s}\right)^{n-1}V_0(s)
$$

where $(3)_{n-1}$ is the Pochhammer function.


问:
Usage of (VIII) in mendelson for proving (IX).
This is from "Introduction to Mathmatical Logic-forth edition" by Elliot Mendelson.

In page 61-62 , the book has propositon (VIII) and (IX). In the solution manual (I mean the "solutions to selected exercises" section of the book) , they used proposition (VIII) to prove proposition (IX).But I am kind of skeptical of the proof.I my point of view , (VIII) can only tell what will will happen if the free variables of a wf $\mathscr B$  occur in a list $x_{i_1},...,x_{i_k}$ .It can't tell anything about what will happen if  the free variables of a wf $\mathscr B$  don't occur in a list $x_{i_1},...,x_{i_k}$.

In the case of the closed wf $\mathscr B$ in (IX) where there is no free variables , the free variables don't occur in a list $x_{i_1},...,x_{i_k}$ (or in any arbitrary list).So , we can't use (VIII) to prove that this closed wf $\mathscr B$ that it is either true or false under a given interpretation (this is proposition (IX) ).But the solution manual uses (VIII) for proving (IX).

$1$.Is my reasoning wrong ?

$2$.If my reasoning isn't wrong , can someone show me an alternate proof of (IX) ?

答:
The Lemma "sounds" like (in the meta-theory):

for every formula $\mathscr B$, for every sequence $s,s'$, for every variable $x_i$: if $x_i$ occures in the "list" and $s$ and $s'$ "agree", then ($s \vDash \mathscr B \text { iff } s' \vDash \mathscr B)$.

If $\mathscr B$ is closed, there are no variables occurring in it. Thus, we can use the "empty list" for the Lemma.

Whatever variables $x_i$ we consider, we have that $x_i$ does not occur in the (empty) list, and thus the conditional of the Lemma is vacuously satisfied.

If you are uncomfortable with "vacuous truth", you can reverse the approach: if two sequence $s$ and $s'$ disagree on some free variable, they can assign different truth values to the same formula.

Consider my preferred example: $(x_1=0)$ and the two sequences: $s(x_1)=0$ and $s'(x_1)=1$.

The two sequences do not satisfy the antecedent of the Lemma, and thus the result is different.

Now, what happens with a closed formula? There is no free variable such that two sequences can disagree on. Thus, the result will be the same for every sequence.


答:
Let $(x,y)$ be an arbitrary point in $A \times B$.you need to show that $(x,y)$ is in $C \times D$.

now $(x,y)$ be an arbitrary point in $A \times B \implies x \in A \subseteq C ,y \in B \subseteq D $

Title: Help with Cartesian product subsets

Body:
I want to prove that if  $A \subseteq C\,$ and $\,B \subseteq D,\,$ then $\,A \times B \subseteq C \times D.$

I know that $A \subseteq C \iff a \in A \rightarrow a \in C$ and that $B\subseteq D\iff b \in B \rightarrow  b \in D$  I also know that $A \times B = \{(a, b)\mid a\in A, b\in B\}$ and that $C\times D = \{(c, d) \mid c \in C, d \in D\}$.

So keep that in mind, how do I connect the dots?


问:
If there is a 1% chance with 100 repetitions, and a 5% chance with 20 repetitions, why aren't the probabilities of an occurence the same?
So if we have a 1% chance and 100 instances there's a ~63% chance of an occurence. If we have a 5% chance and 20 instances, the probability of an occurence is ~64%. And if we have a 33.3% chance with 3 instances the probablity is ~70%. What is causing this difference? It seems counterintuitive, though I understand that it has to grow too because 100% chance with one instance is a 100% chance of occurence after all. What am I missing?</p>

答:
If there are $n$ independent attempts each with probability $p$ of success then the probability of at least one success is $1$ minus the probability of no successes, so $$1-(1-p)^n$$

If $p=\frac1n$ then this is $$1-\left(1-\frac1n\right)^n$$ which tends towards $1-e^{-1}\approx 0.63212$ as $n$ increases.  We can even give a series: $$1-\left(1-\frac1n\right)^n\approx 1- \frac1e + \frac1{2e n} + \frac5{24e n^2}+ \frac5{48e n^3}+\cdots$$ so you can see it is slightly higher for small $n$


问:
If $gcd(a,b)=d$ and $a^2+b^2=c^2$, show that $gcd(a,c)=gcd(b,c)=d$.
If $gcd(a,b)=d$ and $a^2+b^2=c^2$, show that $gcd(a,c)=gcd(b,c)=d$.

This is what I have so far. We know that $d|a$ and $d|b$ which implies that $d|c$, but that is not enough to show that $gcd(a,c)=gcd(b,c)=d$, only that they are at least $d$. Also $a^2=c^2-b^2$ and $b^2=c^2-a^2$. Any hints or solutions are greatly appreciated.

Body: $\color{#c00}{(a,c)^2} \overset{\rm \color{#0a0}{FD}}= (a^2,c^2) = (a^2,a^2\!+\!b^2) = (a^2,b^2) \overset{\rm \color{#0a0}{FD}}= \color{#c00}{(a,b)^2}\,$ so $\,\color{#c00}{(a,c) = (a,b)}.\,$ $\overbrace{\text{Same for $(b,c)$}}^{\text{ by $\,(a\ b)\,$ symmetry}}$.


问:
Evaluating $\lim_{x \to a} x^{x^x}$
I know that $\lim_{x \to 0}x^x = 1$

But how do I apply the above to $\lim_{x \to 0}x^{x^x}$

$\lim_{x \to 0}x^{x^x} = e^{\lim_{x \to 0}x^xlnx}$

(Given that we can only apply limit laws when both $\lim_{x \to a}f(x)$ and $\lim_{x \to a}g(x)$ exists such that

$\lim_{x \to a}f(x)g(x) = \lim_{x \to a}f(x) . \lim_{x \to a}g(x)$)

答:
If you are working on set of real nos. $\mathbb R$, then you can apply limit rules.But if you consider set of extended reals, that is the set $\mathbb R \cup \{-\infty, \infty\}$ (here $\infty $ is assumed to be a no. for which it is true that if $y\in \mathbb R$, then $y\lt \infty $. Similarly, $-\infty $ is defined as a no. which is smaller than any given real number), then limit rules don't always work. One example is $\lim_{x\to 0}\frac 1{|x|}+\lim_{x\to 0}(1-\frac 1{|x|})=\infty -\infty\ne \lim_{x\to 0}(\frac 1{|x|}+1-\frac 1{|x|})=1.$

Now, $\lim_{x\to \infty}x^x\ln x$ does not exist in $\mathbb R$, in other words the limit does not exist finitely.

But the limit exists in $\mathbb R \cup \{-\infty, \infty\}$ and infact $\lim_{x\to 0}x^x\ln x=-\infty$.

What existence of this limit means is that given any $t\lt 0$,there exists a $\delta\gt 0$ such that $0\lt x\lt \delta\implies x^x \ln x\lt t\implies e^{x^x\ln x}\lt e^t$. Now note that for any $\epsilon\gt 0$, there exists a $t_\epsilon\lt 0$ such that $e^{t_\epsilon}\lt \epsilon$

Therefore, for any $\epsilon\gt 0$, there is a $\delta_\epsilon\gt 0$ such that $0\lt x\lt \delta_\epsilon\implies e^{x^x\ln x}\lt \epsilon\implies \lim_{x\to 0}e^{x^x\ln x}=0.$

In other words, $e^{x^x\ln x}$ tends to $0$ as $x\to 0$.

Title: Evaluating $\lim_{x \to a} x^{x^x}$


问:
A complete graph with $n$ vertices is $n$-chromatic
It is true that a complete graph with $n$ vertices is $n$-chromatic.But is it true that if a graph with $n$ vertices has chromatic number $n$, then it is complete?

答:
We can prove that by showing that a (simple, undirected, finite) graph of order $n$ with chromatic number $n$ has $\frac{n(n-1)}{2}$ edges (the maximum number of edges possible in such a graph).

Assume $G$ with $n$ vertices is $n$-chromatic, then there are $n$ disjoint independent sets, each representing one of the $n$ colour classes of $G$. Since $G$ is $n$-chromatic, the $n$ colour classes must be pairwise adjacent, or otherwise we could reuse colours. (trivially these independent sets consists of a single vertex ...)

By simple application of the Handshaking lemma, the number of edges follows and therefore $G$ is complete.

Actually, the general result for $k$-chromatic graphs is discussed in this network here.


问:
Arranging a partial product
Why is it that for real numbers $0<x_i<1$ $(1-x_1)(1-x_2)\cdots(1-x_n)(1-x_{n+1})=1-x_1-x_2-\cdots-x_n-x_{n+1}+x_{n+1}(x_1+\cdots+x_n)$?

The most I was able to do was factor out the $x$ terms individually, \begin{align*} &= (1-x_1)(1-x_2)\cdots (1-x_n)(1-x_{n+1}) \\ &= x_1(\frac{1}{x_1}-1)x_2(\frac{1}{x_2}-1)\cdots x_n(\frac{1}{x_n}-1)x_{n+1}(\frac{1}{x_{n+1}}-1) \\ &= (x_1\cdots x_{n+1})(\frac{1}{x_1}-1)\cdots (\frac{1}{x_{n+1}}-1)\end{align*}

答:
The identity you give is true for $n=1$:

$$(1-x_1)(1-x_2)=1-x_1-x_2+x_2x_1$$

but for $n>1$, it is erroneous in general.

For example, in the case $n=2$, if $x_1=x_2=x_3=1/2$:

$$\underbrace{(1-x_1)(1-x_2)(1-x_3)}_{\tfrac{1}{8}}\ne \underbrace{1-x_1-x_2-x_3+x_3(x_1+x_2)}_0$$

Title: Arranging a partial product


问:
A linear operator $T$ is closed if and only if whenever $x_n\rightarrow x$ and $Tx_n\rightarrow y$, $y=Tx$.
Let $X,Y$ be Banach spaces, and $T:X\longrightarrow Y$ be a linear operator. I want to understand the following fact.

The graph $\Gamma(T) = \{(x,Tx)\;|\;x\in X\}$ is closed in $X\oplus Y$ if and only if for every sequence $x_n\longrightarrow x$ such that $Tx_n$ converges, $Tx_n\longrightarrow Tx$.

I take the first part of this statement as the definition of a closed operator, i.e. $T$ is closed if $\Gamma(T)$ is closed. All the sources I have tried so far (Yosida's book, Lax's book, Wikipedia...) simply mention this as a self-evident fact, so I assumed it would be easy to see, but I just was't able to warp my head around it.

Does anyone have a proof for this? Please refrain from invoking the closed graph theorem, as it is a further development. Thanks!

答:
Hint

In a metric space, a set is closed if and only if it's sequentially closed.


答:
If the condition holds and $(x_n,Tx_n)$ is a sequence in the graph converging to some point $(u,v)$ then $x_n \to u$ and $Tx_n \to v$. By the given condition this implies $Tx_n \to Tu$. But $Tx_n \to v$ also.  This  gives $v=Tu$ so $(u,v)=(u,Tu) $ belongs to the graph. Hence,, the graph is closed.

Conversely. if the graph is closed and $x_n \to x, Tx_n \to z$ then $(x_n,Tx_n) \to (x,z)$ so $(x,z) $ belongs to the graph of $T$. Hence, $z=Tx$.

Title: A linear operator $T$ is closed if and only if whenever $x_n\rightarrow x$ and $Tx_n\rightarrow y$, $y=Tx$.

Body:
Let $X,Y$ be Banach spaces, and $T:X\longrightarrow Y$ be a linear operator. I want to understand the following fact.

The graph $\Gamma(T) = \{(x,Tx)\;|\;x\in X\}$ is closed in $X\oplus Y$ if and only if for every sequence $x_n\longrightarrow x$ such that $Tx_n$ converges, $Tx_n\longrightarrow Tx$.

I take the first part of this statement as the definition of a closed operator, i.e. $T$ is closed if $\Gamma(T)$ is closed. All the sources I have tried so far (Yosida's book, Lax's book, Wikipedia...) simply mention this as a self-evident fact, so I assumed it would be easy to see, but I just was't able to warp my head around it.

Does anyone have a proof for this? Please refrain from invoking the closed graph theorem, as it is a further development. Thanks!


问:
Question on applying L'Hopital's rule--Calculus I
In the below picture, how does the equality sign work out?

It seems that using L'Hopital's rule, the x on the bottom turns to 1, and the derivative of the expression on the numerator side is equal to the expression on the right-handed side, but I don't see how..

This is from the article: https://www.mdpi.com/2073-8994/10/9/380

Thank you.

Body:
$$\begin{align}\lim_{x\to0^+}\frac{\ln\sum_{k=1}^n\lambda_kc_k^x}{x}
&=\lim_{x\to0^+}\frac{\left(\ln\sum_{k=1}^n\lambda_kc_k^x\right)'_x}{(x)'_x}\\
&=\lim_{x\to0^+}\frac{\sum_{k=1}^n\lambda_kc_k^x\ln c_k}{\sum_{k=1}^n\lambda_kc_k^x}\\
&=\frac{\sum_{k=1}^n\lambda_k\ln c_k}{\underbrace{\sum_{k=1}^n\lambda_k}_{=1}}.\\
\end{align}$$


问:
Is the product of countable metrizable spaces with box-topology still metrizable?
I am currently working on product and box topology.

Textbook has given an example:'$R^{\omega}$ is not metrizable with box-topology.

So I have a suppose:

If X is metrizable, will $X^{\omega}$ be metrizable?

Even more strict, if $X_{\alpha}$ is metrizable for each $\alpha$, will the countable product of $X_{\alpha}$ metrizable with box-topology?

答:
No, if $X$ is any non-discrete metric space, $X^\omega$ in the box topology will not be first countable, so in particular not metrisable.

Sketch of proof: let $p \in X$ be a non-isolated point and pick a local base $U_n$ which obeys $\overline{U_{n+1}} \subsetneq U_n$ for all $n$ and then prove (via a diagonal argument) that $(p,p,p,\ldots)$ does not have a countable local base in $X^\omega$. For more info on box products, see the chapter about them in the Handbook of Set-theoretic Topology. You can also see there that $X^\omega$ will rarely be compact, connected and many other nice properties. It's a rarely used topology for that reason, except to construct examples.


问:
intuition for the closed form of the fibonacci sequence
I'm trying to picture this closed form from Wikipedia visually:

The idea is, if you take $\phi^n / \sqrt{5}$ and round it to the nearest integer, you'll get the $n$th Fibonacci number. I see how it works out on paper, but is there an intuitive way to understand this? Specifically I'm trying to wrap my brain around how $\phi^n$ would lead to the correct answer... if I want the 8th Fibonacci number, for example, how does it help to multiply $\phi$ by itself 8 times?

答:
The defining relation is $F_n=F_{n-1}+F_{n-2}$ with $F_1=F_2=1$, and this clearly defines the whole sequence.

To see why powers arise, look first at a much simpler problem, with $$A_n=cA_{n-1}$$ and $A_0=1$, so that $$A_1=c, A_2=c^2 \dots A_n=c^n \dots$$ - this is easy to prove by induction. And if $A_0$ were instead to be $d$, say, then we'd have $A_n=dc^n$. So in this simple recurrence the power of $c$ arises simply and naturally.

Now let's tackle $$B_n=cB_{n-1}+tb^n$$ with $b\neq c$. This is a linear recurrence, and if we have a particular solution $B_n$ then it is easy to see that $B_n+kA_n$ is another solution to the recurrence. An initial value will determine the unknown constant $k$.

To find a solution, we try $B_n=rb^n$, whence $$rb^n=cb^{n-1}+tb^n$$ which gives $$rb=c+tb$$ so that $r=t+\frac cb$ and the solution $B_n=kc^n+rb^n$ involves two different powers.

We now have the tools to tackle a recurrence of the form $$C_n=(\alpha+\beta)C_{n-1}-\alpha\beta C_{n-2} $$ or $$ C_n-\alpha C_{n-1}=\beta (C_{n-1}-\alpha C_{n-2})$$

We set $D_n=C_n-\alpha C_{n-1}$ which gives us $$D_n=\beta D_{n-1}$$ We know that this has solution $$D_n=t\beta^n$$ and this can be written in terms of $C_n$ as $$C_n-\alpha C_{n-1}=t\beta^n$$ which is the form we had before for $B_n$ and which is solved by $C_n=k\alpha^n+r\beta^n$ where $k$ and $r$ are fully determined by two initial values.

Now we note that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2-(\alpha+\beta)x+\alpha\beta=0$

So to put $F_n=F_{n-1}+F_{n-2}$ into the form we can solve we need $\alpha+\beta=1$ and $\alpha\beta=-1$.

So we need to find the roots of $x^2-x-1$, which are $\phi, -\frac 1{\phi}$ and we expect $F_n$ to be the sum of powers of these roots ($k\phi^n+r(-\phi)^{-n}$, where $k$ and $r$ can be determined from the initial values).

Finally we note that $\phi \gt 1, \phi^{-1} \lt 1$ so that the term in $\phi^{-n}$ gets much smaller as $n$ increases.

This is a somewhat extensive explanation of how powers arise.


问:
Number of distinct right triangles formed by connecting vertices of a unit cube
Suppose we have a unit cube in ${\mathbb{R}^3}.$

We want to count the total number of distinct right triangles formed by connecting vertices of the unit cube. I can see that the total number of right triangles simply on a face of the cube will be $4$ and since there are $6$ faces we have $4 \times 6 = 24$ right triangles on the faces alone. Of course there are others across the cube's diagonals. I think the total will be $48$, due a symmetry across the diagonals, but am not entirely sure of my reasoning here, so I may be wrong.

If someone could elucidate or provide a better/more rigorous answer it would be appreciated.

答:
You are correct in determining that there would be $48$ right triangles sharing at least one side with the cube. These can be counted as
$$\underbrace{4\times 6}_{\text{sharing 2 edges}} + \underbrace{2 \times 12}_{\text{sharing 1 edge}}=48$$

Observe that these are all the triangles any edge makes with six remaining vertices.

Now here is an argument that these are the only right triangles with vertices on the cube.

The distance between any two vertices of the cube belongs to the set $\{1,\sqrt{2}, \sqrt{3}\}$. This allows for right triangles only with sides $(1,1, \sqrt{2})$ and $(1,\sqrt{2}, \sqrt{3})$. That is, one of the sides of the right triangle must be $1$, an edge of the cube. And we have counted all of these. $ \qquad \square$


问:
Tangent bundle is an oriented manifold
Recall that the tangent bundle $TM$ of a manifold $M$ consists of all pairs $(x, \overrightarrow{v})$ where $x \in M$ and $\overrightarrow{v}$ is the tangent space $T_xM$ of $M$ at $x$. Show that $TM$ is an oriented manifold.

An orientable manifold is one such that there exists a continuous vector field. If $(\phi, U)$ is a chart for $M$ we can obtain a natural chart $(\Phi, TU)$ for the tangent bundle by defining $\Phi: TU\rightarrow \mathbb{R}^n \times \mathbb{R}^n$ by $\Phi(x, y) = \phi(x), (D\phi_a^{-1})^{-1}y)$ where $D$ is the derivative map. We just need to check the jacobian of the transition maps have positive determinant. If $(U, \Phi)$ and $(V, \Psi)$ where $\Psi(x, y) = (\psi(x), (D\psi_a^{-1})^{-1}(y))$ are two overlapping charts of $TM$ then the transition maps is for $(x, y) \in \mathbb{R}^{2n}$.
\begin{equation}
     \Phi \circ \Psi^{-1}(x, y) = (\phi\circ\psi^{-1}(x), (D\phi_a^{-1})^{-1} \circ D\psi_a^{-1}(y))
\end{equation}

Thus, the Jacobian of the transition map $\mathcal{J}(\Phi \circ \Psi^{-1}(x, y))$
\begin{equation}
    \left\lvert\begin{matrix}
        \mathcal{J}(\phi\circ\psi^{-1}(x)) & 0\\
        0 & \mathcal{J}((D\phi_a^{-1})^{-1} \circ D\psi_a^{-1})(y))
    \end{matrix}\right\rvert
\end{equation}

From here I'm not sure how to prove that this determinant is positive. Some hints would be greatly appreciated.

答:
It seems to me that you got the computation of the derivative wrong. As you have written (with a small simplification as observed in the comment by @guidoar and using the chain rule), you get
$$
\Phi\circ\Psi^{-1}(x,v)=((\phi\circ\psi^{-1})(x), D(\phi\circ\psi^{-1})(x)(v)).
$$
(I have changed the notion to emphasize that $v$ is a tangent vector, while $x$ is a point in $M$.) But when differentiating this, you have to distinguish wheter you differentiate in $x$-directions or in $v$-directions: Differentiating the second component in $x$ directions, you get a second derivative, whereas the fact that the derivative is linear in $v$ implies that you just get back the first derivative when differentiating in $v$-directions. More formally this means that
$$
D(\Phi\circ\Psi^{-1})(x,v)(w_1,w_2)=(D(\phi\circ\psi^{-1})(x)(w_1),D^2(\phi\circ\psi^{-1})(x)(w_1,v)+D (\phi\circ\psi^{-1})(x)(w_2)).
$$
In matrix notation, this leads to a block form as
$$
\begin{pmatrix} D(\phi\circ\psi^{-1})(x) & 0\\
 D^2(\phi\circ\psi^{-1})(x)(\_,v) & D(\phi\circ\psi^{-1})(x) \end{pmatrix}
$$
Taking the determinant, the off-diagonal block does not play a role, so you conclude that $\mathcal J(\Phi\circ\Psi^{-1})=\mathcal J(\phi\circ\psi^{-1})^2>0$, which implies the claim.

Title: Tangent bundle is an oriented manifold


答: The sense you're getting that they all ought to be the same is reasonable, because all of those distributions have the exact same Expected Value for their number of successes, that being 1.  The difference is that the more repetitions, the higher the chance of getting more than one success, which counts multiple times for the expected value, but only once for the binary "was there at least one success" question.</p>

Title: If there is a 1% chance with 100 repetitions, and a 5% chance with 20 repetitions, why aren't the probabilities of an occurence the same?

Body: So if we have a 1% chance and 100 instances there's a ~63% chance of an occurence. If we have a 5% chance and 20 instances, the probability of an occurence is ~64%. And if we have a 33.3% chance with 3 instances the probablity is ~70%. What is causing this difference? It seems counterintuitive, though I understand that it has to grow too because 100% chance with one instance is a 100% chance of occurence after all. What am I missing?</p>


问:
$p_{i + k} - p_i \neq \text{const}$ for any $k \geq 1$ where $p_i = i$th prime number.
Let $p_i$ be the $i$th prime number.  This should be simple to prove:

$$
\forall k \geq 1, c \in \Bbb{Z}, \\
p_{i + k} - p_i \neq c, \\
$$
for some $i \geq 2$.  But for example:

$$
11 - 5 = 6 \\
13 - 7 = 6 \\
17 - 11 = 6 \\
19 - 13 = 6 \\
23 - 17 = 6 \\
$$

So long strips do exist, I'm saying the whole diagonal strip (a diagonal line in the table for $p_i - p_j$) cannot be constant.

I specified $i \geq 2$ because I'm only interested in odd primes, but lowering to $i \geq 1$ should also work.

It should be easy to prove, but I'm stuck on it.

If $p_{i + k} - p_i = c$ for all $i \geq 2$ then in particular $3 + cz$ is prime for all $z \geq 0$, which would include $z \in 3 \Bbb{Z}$?  Does that work as a proof?

My next question is how can we generalize the statement? I got a little too specific with the indices.

I think I've got it now.

Let $b,c,k \in \Bbb{N}$ be positive numbers.  Then:

$$
p_b + c = p_{b + k} \\
p_{b + k} + c = p_{b + 2k} \\
p_{b + 2k} + c = p_{b + 3k} \\
\vdots
$$

would imply
$$
p_b + c = p_{b + k} \\
p_b + 2c = p_{b + 2k} \\
p_b + 3c = p_{b + 3k} \\
\vdots
$$
or in general $p_b + cz = p_{b + zk}$ for all $z \geq 0$.  Thus choose any $z \in p_{b}\Bbb{N}$.  For more tightness we'd only have to choose the minimal composite in $w \in p_b + c \Bbb{N}$, and then take $z = \dfrac{w - p_b}{c}$

Thus to generalize, $b + zk$ was in the place of $i + k$ so:

$$
b + zk = i + k = b + \dfrac{w - p_b}{c} k \implies \\
\exists i, b \leq i \leq b + k(\dfrac{w - p_b}{c} - 1) : \\
p_{i + k} - p_i \neq c \\
$$

答:
It's a standard argument that any non-constant polynomial $g(n) \in \mathbb Z[n]$ takes (arbitrarily large) composite values.  Your problem is just the case of a polynomial of degree $1$.

Pick any value of $g(n_0)$ that you like (pick $n_0$ large enough so that $|g(n_0)|>1$).  If it's composite, then great.  If it's a prime $p$, then note that $g(n_0 + kp) \equiv g(n_0) \equiv 0 \pmod p$ for any $k\ge 1$.  Now it's possible that $g(n_0 + kp)$ is exactly equal to $\pm p$ and hence still prime, but this can only happen for finitely many values of $k$.  Thus $g(n_0 + kp)$ will eventually be always composite for large enough $k$.


问:
Can the value of sin theta be greater than 1?
For example, is it possible for sin theta to be 1.06 or 1.2 under any circumstances?
Or is it possible for it to be exactly 1?</p>

答:
Recall the definition of the sine function:

As you can see, $c > a$, so $\sin(\alpha) \leq 1$.


问:
$\lim\limits_nx_n=0,\forall n \in \mathbb{N},\phi_X(x_n)=e^{-\sigma^2x^2_n} \implies \forall x \in \mathbb{R}, \phi_X(x)=e^{-x^2\sigma^2} $
$X$ is a random variable such that there exist a sequence $(x_n)_n$ in $\mathbb{R}^*,\lim_nx_n=0,$ and for all $n \in \mathbb{N},\phi_X(x_n)=e^{-\sigma^2x^2_n}, \sigma \in \mathbb{R}.$

Prove that $X$ is normally distributed (possible degenerate).

The case where $\sigma=0$ (degenerate distribution) was asked here.

Any suggestions on how to deal with $\sigma \neq 0$?

答:
Let $Y$ be a random variable with characteristic function $\phi_Y:t\mapsto e^{-\sigma^2t^2}$, i.e. $Y$ is Gaussian (possibly degenerate).

The hypothesis in the problem states that $\phi_X$ and $\phi_Y$ coincide along a sequence $(x_n)$ of non-zero reals such that $x_n\to 0$. If $\phi_X$ were analytic in a neighborhood of $0$, so would be $\phi_X - \phi_Y$ and since the zeroes of an analytic function are isolated, this would imply $\phi_X - \phi_Y = 0$, as wanted.

Instead of showing directly that $\phi_X$ is analytic, we show that $X$ has moments of all orders. This will imply that $\phi_X$ is infinitely differentiable. We show next that $\forall n \geq 0$, $\phi_X^{(n)}(0)=\phi_Y^{(n)}(0)$. Since $Y$ is uniquely determined by its moments (see here or here), this will yield $\phi_X=\phi_Y$.

For $n\geq 1$, let $\displaystyle f_n:t\mapsto \frac{2-e^{ix_n t}-e^{-ix_n t}}{x_n^2}$ and note that it is real-valued, non-negative and by a Taylor expansion $\forall t\in \mathbb R, f_n(t) \to t^2.$ By Fatou's lemma,
$$\begin{align}
E(X^2)
&\leq \liminf_n E(f_n(X)) \\
&=\liminf_n \frac{2-\phi_X(x_n)-\phi_X(-x_n)}{x_n^2} \\
&=\liminf_n \frac{2-\phi_X(x_n)-\overline{\phi_X(x_n)}}{x_n^2} \\
&=\liminf_n \frac{2-\phi_Y(x_n)-\phi_Y(-x_n)}{x_n^2} \\
&= -\phi_Y''(0)
\end{align}
$$
where the last equality follows from a Taylor expansion of $\phi_Y$. Hence $X$ has a second moment, and by a classical result $\phi_X$ is twice-differentiable over $\mathbb R$.

Since $\phi_Y$ is even we may assume WLOG that all the $x_n$ are positive, and also that $(x_n)$ is strictly decreasing (by taking a subsequence if needed). By Rolle's theorem applied to $\phi_X-\phi_Y$, there is some sequence $(y_n)$, strictly decreasing, with $y_n\to 0$ and $\forall n\geq 1, \phi_X'(y_n) = \phi_Y'(y_n)$. Another application of Rolle's provides some sequence $(z_n)$, strictly decreasing, with $z_n\to 0$ and $\forall n\geq 1, \phi_X''(z_n) = \phi_Y''(z_n)$.

Defining a new $f_n$ and performing a similar reasoning, we prove successively that $E(X^4)<\infty$, $\phi_X^{(3)}$ and $\phi_Y^{(3)}$ coincide along a sequence that goes to $0$ and similarly for $\phi_X^{(4)}$ and $\phi_Y^{(4)}$. This continues indefinitely. $\phi_X$ is therefore $C^\infty$, each $\phi_X^{(m)}$ is continuous and taking limits along the aforementioned sequences we find that $\forall m\geq 1$, $\phi_X^{(m)}(0)=\phi_Y^{(m)}(0)$.

Note that the proof follows through if $Y$ is replaced by any random variable that is uniquely determined by its moments. For example, it is sufficient that $\limsup_n \frac{E(|Y|^n)^{1/n}}{n} <\infty$.


答:
No and yes.

No:
If you let $\theta$ be an angle in a right angled triangle, we know that $\sin(\theta)$ is equal to $\frac{\text{Opposite}}{\text{Hypotenuse}}$. We know that the Hypotenuse is never shorter than the line Opposite the angle $\theta$, so this fraction can never exceed $1$.

Yes:
You can use complex numbers. So if $\theta$ is complex, then it can exceed $1$. For example, $\sin(1.57080 - 0.344701i) = 1.06$ (correct to 5dp at least).

Presumably you're at a stage where you aren't considering complex numbers, so the 'Yes' response here is a bit of a cheat. And I don't know of many cases where people go through explicit computations of $\sin(\theta)$ for complex $\theta$..

Title: Can the value of sin theta be greater than 1?

Body:
For example, is it possible for sin theta to be 1.06 or 1.2 under any circumstances?
Or is it possible for it to be exactly 1?</p>


答:
A key insight is that the only instance in which three vertices are selected that do not form a right triangle, is when those vertices form an equilateral triangle--i.e., no two vertices share an edge.

Since the cube circumscribes two distinct regular tetrahedra (cf. stella octangula), there are a total of $8$ such triangles.  Since there are a total of $$\binom{8}{3} = 56$$ ways to choose any three distinct vertices, and each choice corresponds to a unique triangle, the number of right triangles is $56 - 8 = 48$.

Title: Number of distinct right triangles formed by connecting vertices of a unit cube

Body:
Suppose we have a unit cube in ${\mathbb{R}^3}.$

We want to count the total number of distinct right triangles formed by connecting vertices of the unit cube. I can see that the total number of right triangles simply on a face of the cube will be $4$ and since there are $6$ faces we have $4 \times 6 = 24$ right triangles on the faces alone. Of course there are others across the cube's diagonals. I think the total will be $48$, due a symmetry across the diagonals, but am not entirely sure of my reasoning here, so I may be wrong.

If someone could elucidate or provide a better/more rigorous answer it would be appreciated.


答: There are $\binom{8}{3}=56$ triangles in total, and the only non-right triangles are the equilateral triangles formed by the vertices exactly distance $1$ from a given vertex, and there are $8$ of these, so the answer is $56-8=48$.


答:
Well, when you defined the map $\overline{\phi}: TU \to \phi(U) \times \mathbb{R}^{n}$, you didn't write it the right way, because it is given by $\overline{\phi}(p,v) = (\phi(p), D_p\phi(v))$ (you need to take into account the derivative of the second component with respect to the point on the manifold, and not just the vector in the tangent space), so when you take the jacobian of a change of coordinates $\overline{\phi}_{\alpha} \circ \overline{\phi}_{\beta} ^{-1} : V_{\beta} \times \mathbb{R}^n \to V_{\alpha}\times \mathbb{R}^n$, if you call $(x_1,...,x_n)$ the coordinates in $V_{\beta}$, and $(x_{n+1},...,x_{2n})$ the coordinates in the $\mathbb{R}^n$ of the domain and do the same with $y$'s for the co domain, to calculate the jacobian, we need to write $\frac{\partial y_j}{\partial x_i}$ for any $i,j = 1,...,2n$. Now, obviously, it is zero when $1\leq j \leq n$ and $n+1 \leq i \leq 2n$ so the upper right quadrant in the matrix representation is zero, but the rest of them won't in general, so it will actually be a lower triangular block matrix which will have determinant equal to the product of the determinants of $[\frac{\partial y_j}{\partial x_i}]_{i,j=1,...,n}$ and $[\frac{\partial y_j}{\partial x_i}]_{i,j=n+1,...,2n}$ as you can check here Determinant of a block lower triangular matrix.

Now, if you notice, the second matrix I pointed out in the product of the determinants, will be exactly $det (J(D_p \phi_{\beta}^{-1} \circ D_p \phi_{\alpha}))$, where $p \in U$ is now fixed because we're calculating partial derivatives with respect to the variables indexed by $n+1,...,2n$. Now, $D_p \phi_{\beta}^{-1} \circ D_p \phi_{\alpha}$ is a linear map from $\mathbb{R}^n$ to itself, and as we all know, in that case, the derivative of a linear map is itself, so the jacobian is the matrix representation of $D_p \phi_{\beta}^{-1} \circ D_p \phi_{\alpha}$ which in turn is exactly $J(\phi_{\beta}^{-1}\circ \phi_{\alpha})$.
We have now concluded that the determinant of the jacobian of $\overline{\phi}_{\alpha} \circ \overline{\phi}_{\beta} ^{-1}$ is exactly $det(J(\phi_{\beta}^{-1}\circ \phi_{\alpha}))^2 >0$.

Title: Tangent bundle is an oriented manifold

Body:
Recall that the tangent bundle $TM$ of a manifold $M$ consists of all pairs $(x, \overrightarrow{v})$ where $x \in M$ and $\overrightarrow{v}$ is the tangent space $T_xM$ of $M$ at $x$. Show that $TM$ is an oriented manifold.

An orientable manifold is one such that there exists a continuous vector field. If $(\phi, U)$ is a chart for $M$ we can obtain a natural chart $(\Phi, TU)$ for the tangent bundle by defining $\Phi: TU\rightarrow \mathbb{R}^n \times \mathbb{R}^n$ by $\Phi(x, y) = \phi(x), (D\phi_a^{-1})^{-1}y)$ where $D$ is the derivative map. We just need to check the jacobian of the transition maps have positive determinant. If $(U, \Phi)$ and $(V, \Psi)$ where $\Psi(x, y) = (\psi(x), (D\psi_a^{-1})^{-1}(y))$ are two overlapping charts of $TM$ then the transition maps is for $(x, y) \in \mathbb{R}^{2n}$.
\begin{equation}
     \Phi \circ \Psi^{-1}(x, y) = (\phi\circ\psi^{-1}(x), (D\phi_a^{-1})^{-1} \circ D\psi_a^{-1}(y))
\end{equation}

Thus, the Jacobian of the transition map $\mathcal{J}(\Phi \circ \Psi^{-1}(x, y))$
\begin{equation}
    \left\lvert\begin{matrix}
        \mathcal{J}(\phi\circ\psi^{-1}(x)) & 0\\
        0 & \mathcal{J}((D\phi_a^{-1})^{-1} \circ D\psi_a^{-1})(y))
    \end{matrix}\right\rvert
\end{equation}

From here I'm not sure how to prove that this determinant is positive. Some hints would be greatly appreciated.


问:
cardinality of units in a commutative ring of cardinality odd
How to prove the cardinality of units is even, if the ring has odd cardinality.?

Since |R| is odd then $|M_{i}|$ is odd for all maximal ideals $M_{i}$ in R.So $U(R)=R\backslash \bigcup_{i=1}^{n}M_{i}=|\bigcap_{i=1}^{n}R \backslash M_{i}|$.

how to pove after this

答:
If the ring has odd cardinality, its characteristic is different from $2$.

If $u$ is a unit in $R$, then also $-u$ is a unit.


问:
Reference Request for Fourier Analysis Dedicated to Signal Processing
Some background about myself, I am an electrical and computer engineering in my junior year. I am interested in mathematical analysis in various fields $($real, complex, functional$)$ and have self studied them. I am currently being introduced to Fourier analysis in continuous-time and discrete-time domain in my signal processing course.

Therefore, I was wondering if someone can recommend me some excellent books/notes that maintain a rigorous introduction on Fourier Analysis with theorems and proofs and its applications dedicated to only signals and systems.

The purpose for this request is that I have seen many books (even famous ones like Stein's book) but they kind of diverge their focus and interest on PDEs mainly and other physical systems.

Edit : I have accepted Manlio's answer since it satisfies the requirements for my question. However, Any other recommendations are still welcomed and I would be also grateful. I also hope this question is beneficial to anyone who shares my question.

答:
A possibility:

Gasquet, Claude; Witomski, Patrick, Fourier analysis and applications. Filtration, numerical calculus, wavelets, Enseignement des Mathématiques. Paris: Masson. xi, 355 p. (1995). ZBL0914.94001.

It explores several aspects of the theory, but has a lot of focus on applications and algorithms.


问:
Derivative of $E\left(\int_t^T S_udu|\mathcal{F}_t\right)$
Calculate the derivarive of $E\left(\int_t^T S_udu|\mathcal{F}_t\right)$ in function of $t$
$$\frac{d}{dt}E\left(\int_t^T S_udu|\mathcal{F}_t\right)$$
where $S_t$ is a stochastic process( I try to avoid specifying its form).

My attempt:
We have
$$E\left(\int_t^T S_udu|\mathcal{F}_t\right) = \int_t^T E\left(S_u|\mathcal{F}_t\right)du$$
then
$$\frac{d}{dt}E\left(\int_t^T S_udu|\mathcal{F}_t\right) = \frac{d}{dt}\left(\int_t^T E\left(S_u|\mathcal{F}_t\right)du\right) $$
But after that, I don't know how to calculate
$\frac{d}{dt}\left(\int_t^T E\left(S_u|\mathcal{F}_{\color{red}{t}}\right)du\right) $ because of the subscript in red color.

答:
Why do you think it's derivable ? Take for example $(S_t)$ being a Brownian motion. Then $$\mathbb E\left[\int_t^T S_u\,\mathrm d u\mid \mathcal F_t\right]=S_t(T-t),$$

which is not derivable.


问:
Prove if $A\subseteq C$ or $B\subseteq C$ then $A\cap B\subseteq C$
Is it true please? I do not know how to prove it and I need to. There is nothing saying any of sets are not empty, so they probably can be. Can you help, please?

Prove or show counterexample

If $A\subseteq C$ or $B\subseteq C$ then $A\cap B\subseteq C$

答: Let $x \in A \cap B$, so $x \in A$ and $x \in B$. Now one of $A \subseteq C$ or $B \subseteq C$ must hold. Say, $A \subseteq C$. Then from $x \in A$ we get $x \in C$ and we're done. Likewise if $B \subseteq C$ we get $x \in C$ too but now from $x \in B$. So we always have $x \in C$, whatever alternative holds. So $A \cap B \subseteq C$ is proven.


问:
Let $b_n = \sqrt{n ^ 3 + 6n-1} - \sqrt{n ^ 3-10n-11}$ then $\lim_{n → ∞}b_n =?$
I have a problem with the following task:

Let $b_n = \sqrt{n ^ 3 + 6n-1} - \sqrt{n ^ 3-10n-11}$

Then $\lim_{n → ∞}b_n =?$

Attempt:

$(\sqrt{n ^ 3 + 6n-1} - \sqrt{n ^ 3-10n-11})\cdot\frac{\sqrt{n ^ 3 + 6n-1} + \sqrt{n ^ 3-10n-11}}{\sqrt{n ^ 3 + 6n-1} + \sqrt{n ^ 3-10n-11}}$
$= \frac {16n+10}{\sqrt{n ^ 3 + 6n-1} + \sqrt{n ^ 3-10n-11}}$
$= \frac {n(16+\frac{10}{n})}{\sqrt{n ^ 3 + 6n-1} + \sqrt{n ^ 3-10n-11}}$

How exactly do I proceed with the $n^3$ in the denominator, or which $n$ makes sense to exclude there and why ?

I am very happy about your help.^^

Body:
Hint:

As $n>0,$

$$\sqrt{n^3+6n-1}+\sqrt{n^3-10n-11}=n^{3/2}\left(\sqrt{1+\dfrac6{n^2}-\dfrac1{n^3}}+\sqrt{1-\dfrac{10}{n^2}-\dfrac{11}{n^3}}\right)$$


问:
Can someone help me finish the solution to this differential equation?
Find the general solution of $(x-4y-3)dx-(x-6y-5)dy=0.$

The rest of my solution:

$x-4y=3$

$x-6y=5$

The point of intersection is $(-1, -1)$.

Let

$x=u-1$; $dx=du$

$y=v-1$; $dy=dv$

$(u-4v)du-(u-6v)dv=0$

Let $u=vz; du=vdz+zdv$

$(vz-4v)(vdz+zdv)-(vz-6v)dv=0$

$(z-4)(vdz+zdv)-(z-6)dv=0$

$(z^2-5z+6)dv-v(z-4)dz=0$

$\frac{dv}{v}-\frac{z-4}{(z^2-5z+6)}dz=0$

$\int\frac{dv}{v}-\int \frac{z-4}{(z^2-5z+6)}dz=0$

$\int\frac{dv}{v}-2\int \frac{dz}{(z-2)}+\int\frac{dz}{z-3}=0$

$\ln \lvert v\rvert -2\ln \lvert z-2\rvert +\ln \lvert z-3 \rvert + \ln \lvert c \rvert=0$

$\ln\lvert cv(z-3) \rvert =2\ln \lvert z-2 \rvert$

$\ln \lvert (z-2)^2 \rvert = \ln\lvert cv(z-3) \rvert $

I do not know how the final answer became $(x-2y-1)^2=c(x-3y-2).$

Body:
You find :
$$\ln((z-2)^2)=\ln(cv(z-3))$$
$$(z-2)^2=cv(z-3)$$
$z=\frac{u}{v}$
$$(\frac{u}{v}-2)^2=cv(\frac{u}{v}-3)$$
$$(u-2v)^2=cv^2(u-3v)$$
$u=x+1$ and $v=y+1$
$$((x+1)-2(y+1))^2=c(y+1)^2((x+1)-3(y+1))$$
$$(x-2y-1)^2=c(y+1)^2(x-3y-2)$$


问:
Homomorphisms vs Functors
What really is the difference between a Homomorphism between categories and a Functor? I understand that they are both structure preserving mapping of the objects and arrows between the categories or am i missing something ?</p>

答:
If I understand your question and definitions correctly: no, a homomorphism of categories and a functor are not the same thing.

Your definition of "homomorphism of categories" seems to be: given categories $M$ and $N$, a homomorphism $h:M\to N$ is

a mapping from $\mathsf{ob}(M)$ to $\mathsf{ob}(N)$, together with

for every $x,y\in\mathsf{ob}(M)$, a mapping from $\mathrm{Hom}_M(x,y)$ to $\mathrm{Hom}_N(h(x),h(y))$ such that $$h(a\circ b)=h(a)\circ h(b)$$ for all $a,b\in \mathrm{Hom}_M(x,y)$.

A functor is then a homomorphism of categories that satisfies the additional requirement that

$h(\mathrm{id}_x)=\mathrm{id}_{h(x)}$ for all $x\in\mathsf{ob}(M)$.

Thus, being a functor is a strictly stronger property than being a homomorphism of categories. For example, let $M$ be the category with

$\mathsf{ob}(M)=\{x,y\}$

$\mathrm{Hom}_M(x,x)=\{\mathrm{id}_x\},\quad \mathrm{Hom}_M(x,y)=\{a\}, \quad \mathrm{Hom}_M(y,x)=\varnothing,\quad \mathrm{Hom}_M(y,y)=\{\mathrm{id}_y\}$

and let $N$ be the category with

$\mathsf{ob}(N)=\{z\}$

$\mathrm{Hom}_N(z,z)=\{\mathrm{id}_z,c\}$ where $c\circ c=c$.

Then the homomorphism of categories $h:M\to N$ defined by

$h(x)=z,\quad h(y)=z$

$h(\mathrm{id}_x)=c,\quad h(a)=c,\quad h(\mathrm{id}_y)=c$

is not a functor.

Title: Homomorphisms vs Functors


问:
Evaluating $\lim_{x \to a} x^{x^x}$
I know that $\lim_{x \to 0}x^x = 1$

But how do I apply the above to $\lim_{x \to 0}x^{x^x}$

$\lim_{x \to 0}x^{x^x} = e^{\lim_{x \to 0}x^xlnx}$

(Given that we can only apply limit laws when both $\lim_{x \to a}f(x)$ and $\lim_{x \to a}g(x)$ exists such that

$\lim_{x \to a}f(x)g(x) = \lim_{x \to a}f(x) . \lim_{x \to a}g(x)$)

答:
Just for your curiosity.

If we compose Taylor series, we have
$$f(x)=x^{x^x}=x+\sum_{n=2}^\infty \frac {P_{n-2}}{(n-1)!} \big[x \log(x)\big]^n$$ where the $P_n$'s are polynomials of degree $n$ in $t=\log(x)$. The first ones are
$$\left(
\begin{array}{cc}
 n & P_n(t) \\
 2 & 1 \\
 3 & t+1 \\
 4 & t^2+3 t+1 \\
 5 & t^3+6 t^2+7 t+1 \\
 6 & t^4+10 t^3+25 t^2+15 t+1 \\
 7 & t^5+15 t^4+65 t^3+90 t^2+31 t+1 \\
 8 & t^6+21 t^5+140 t^4+350 t^3+301 t^2+63 t+1 \\
 9 & t^7+28 t^6+266 t^5+1050 t^4+1701 t^3+966 t^2+127 t+1
\end{array}
\right)$$ in which we can find some interesting patterns.

Even with few terms, these expansions match quite well the original function for the range $0\leq x \leq 1$. The same for the integral. For illustartion purposes, consider
$$g_p(x)=x+\sum_{n=2}^p \frac {P_{n-2}}{(n-1)!} \big[x \log(x)\big]^n$$ and the norm
$$\Phi_p=\int_0^1 \Big[f(x)-g_p(x)\big]^2\,dx$$ Here are the results
$$\left(
\begin{array}{cc}
 p & \color{red}{10^8} \Phi_p\\
 2 & 2790.43 \\
 3 & 224.866 \\
 4 & 2.50098 \\
 5 & 0.07587 \\
 6 & 0.00219 \\
 7 & 0.00002
\end{array}
\right)$$

Title: Evaluating $\lim_{x \to a} x^{x^x}$


问:
Does this norm inequality hold?
Does this inequality hold?

Let $n \in \mathbb{N}$. Is it true that
$$ \left\lvert \frac{\lVert x \rVert}{\sqrt n \sqrt 3}-\frac{1}{3} \right\rvert \leq  \displaystyle\left\lvert  \frac{\lVert x \rVert ^2}{n}-\frac{1}{3} \right\rvert .$$

Here $\lVert \cdot \rVert$ denotes 2-norm in $\mathbb R ^n$ and $\lvert \cdot \rvert$ the absolute value in $\mathbb R$.

答:
Note that
$$\left| \frac{||x||^2}{n} - \frac{1}{3}\right| = \left| \frac{||x||}{\sqrt n} - \frac{1}{\sqrt 3}\right| \cdot \left| \frac{||x||}{\sqrt n} + \frac{1}{\sqrt 3}\right|$$
while
$$\left| \frac{||x||}{\sqrt 3 \sqrt n} - \frac{1}{3}\right| = \frac{1}{\sqrt 3}\left| \frac{||x||}{\sqrt n} - \frac{1}{\sqrt 3}\right| $$
Thus your inequality is equivalent to
$$\frac{1}{\sqrt 3} \le \left| \frac{||x||}{\sqrt n} + \frac{1}{\sqrt 3}\right| \ \ \mbox{ or } \ \  \left| \frac{||x||}{\sqrt n} - \frac{1}{\sqrt 3}\right|=0$$
which can be rewritten as
$$0 \le ||x||  \ \ \mbox{ or } \ \  ||x|| = \sqrt{\frac{n}{3}}$$
which is clearly true.

Title: Does this norm inequality hold?


问:
Find a function that satisfies the following property
I am finding it hard to find such a function $v:$ { 1,2,...,n } $\rightarrow \mathbb{R}$ which satisfies the following property:

$$v(k) - v(q) > 2^{n-1}(v(k) - v({q+1}))$$

where $$v(1) < v(2) < ... < v(q) < v({q+1}) < ... < v(k) \text{ and } n\geq  k \neq {q+1}.$$

In words, the difference between 2 points (first maximal and second some other point $q$) is strictly greater than the $2^{n-1}$ times the difference of the maximal point and point $q+1$.

答:
It's easier to start from $n$ and work your way backwards. Start by assigning the value 0 to $n$. There is no restriction on $v(n-1)$, so let it be $-1$.
The inequality inequalities for $k=n$ and $q<n-1$ become
$$
\begin{align}
v(n) - v(q) &> 2^{n-1}(v(n)-v(q+1))\\
v(q) &< 2^{n-1}v(q+1).
\end{align}
$$
Inspired by this fact you can take any value $c>2^{n-1}$ and let $v(n-i)=-c^{i-1}$.
Note that this satisfies the condition that $v$ is always increasing.
To prove this works for all other inequalities take $k<n$:
$$
\begin{align}
v(k)-v(q) &= -c^{n-k-1} + c^{n-q-1}\\
&=c(c^{n-q-2} - c^{n-k-2})\\
&>2^{n-1}(c^{n-(q+1)-2}-c^{n-(k+1)-1})\\
&=2^{n-1}(-v(q+1)+v(k+1))\\
&>2^{n-1}(v(k)-v(q+1))
\end{align}
$$

Of course not every sequence $v$ that satisfies these equations is of this form, but these sequences all satisfy the equations.

For $n=3$ you can choose sequence $(-5, -1, 0)$.

For $n=4$ you can choose sequence $(-81, -9, -1, 0)$.

For $n=5$ you can choose sequence $(-4913, -289, -17, 0)$, etc.


问:
Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$
Prove that  $\forall x,y\in \mathbb{R}$ the inequality  $x^2+y^2+1 \geq xy+x+y$ holds.

Attempt

First attempt: I was trying see the geometric meaning, but I´m fall.

Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare $\frac{x}{y}+\frac{y}{x} \geq 2 $ and the equality $(1+\frac{1}{x}) (1+\frac{1}{y})\leq 2$ unfortunelly not is true the last inequality and hence I can´t conclude our first inequality.

Third attempt:comparing  $x^2+y^2$ and $(\sqrt{x}+\sqrt{y})^2$ but unfortunelly I don´t get bound the term $2\sqrt{xy}$ with $xy$.

Any hint or advice of how I should think the problem was very useful.

答:
Let $z=x^2+y^2-xy-x-y$

$$\iff x^2-x(1+y)+y^2-y-z=0$$

As $x$ is real, the discriminant must be $\ge0$ of the above quadratic equation

i.e., $$(1+y)^2\ge4(y^2-y-z)$$

$$\iff4z\ge3y^2-6y-1=3(y-1)^2-4\ge-4$$

$$\implies z\ge-1$$ the equality occurs if $y=1$ and $x=\dfrac{1+y}2=?$


问:
$\mathbb E(X_{T \wedge n}) = \mathbb E(X_0)$, where $ T \wedge n := \min(T, n)$, $\forall n$

If $X$ is martingale and $T$ is a stopping time, then $X_{T \wedge n}$
is a martingale, so that in particular, $\mathbb E(X_{T \wedge n}) = \mathbb E(X_0)$,
where $ T \wedge n := \min(T, n)$, $\forall n$.

Why is $X_{T \wedge n}$ a martingale, and why does $\mathbb E(X_{T \wedge n}) = \mathbb E(X_0)$, $\forall n$?

答:
You need to check integrability and adaptation, but for the martingale property, considering the filtration $(\mathcal{F}_n)_{n \ge 0}$ to which $(X_n)_{n \ge 0}$ is adapted, and since $T$ is a stopping time so that $\{T \le n\}$ and its complementary $\{T > n\}$ belong to $\mathcal{F}_n$, you have
$$ \mathbb{E}[X_{T \wedge (n+1)} |\mathcal{F}_n] =  \mathbb{E}[X_{n+1} \mathbf{1}_{T > n} |\mathcal{F}_n] + \mathbb{E}[X_T \mathbf{1}_{T \le n} |\mathcal{F}_n], $$
which leads to, using that we can take out the indicator functions and that $X_T \mathbf{1}_{T \le n}$ is $\mathcal{F}_n$ measurable,
$$ \mathbb{E}[X_{T \wedge (n+1)} |\mathcal{F}_n] = \mathbf{1}_{T > n}  \mathbb{E}[X_{n+1} |\mathcal{F}_n] + \mathbf{1}_{T \le n}  X_T = \mathbf{1}_{T > n} X_n + \mathbf{1}_{T \le n}  X_T = X_{T \wedge n}. $$