问:
Probability of Power Rangers, please correct me if I am wrong
There are five Power Rangers.  Of the five,  two are male(Red, Black) and three are female (Pink, Yellow,blue.)I randomly select two of the Power Rangers, without replacement.

(a)  What is the probability that the black Ranger is one of the two selected Power Rangers?

```<code>pretending that you only randomly select one power ranger, the
probability of picking the black would just be 1/5, so if you were
to pick another it would be 1/4 so it would be 1/5 * 1/4 = .05 so 5% chance?
it just seems low
</code>```

(b)  What is the probability at least one of the two selected Power Rangers is female?

```<code>probability that one is female is 3/5, so if you were to pick
another power ranger at random would it be 3/5*3/4
</code>```

(c)  Given that at least one of the two selected Power Rangers is male, what is the conditional probability the Pink Ranger is selected?

```<code>conditional probability, wouldnt it just be 33.3% that the pink is selected given theres only three females?
</code>```

(d)  Are the events “the Pink Ranger is one of the two selected” and “at least one of the two selected Power Rangers is female” independent?  Explain why or why not.

```<code>Yes, event A effects event B.

</code>```

答:
As an alternative to directly using probabilities, Bayes' Theorem, etc., the way I personally think about these questions is first through counting. And then at the end as the last step, perform a division to get the probability. I find that this helps to avoid thinking about "first" make a selection and "second" make a different selection, and all the traps my brain could fall into when thinking that way.

With 5 people to choose 2 from, there are $\binom{5}{2}=\frac{5\cdot4}{2!}=10$ ways to just choose 2 people, regardless of one being pink or not, or whatever. These 10 possible pairs are all equally likely.

How many pairs of 2 people do have the pink ranger? Easy, that's just 4 since you have 4 options for who pink's partner would be.

So out of 10 equally likely possible pairs, 4 qualify. The probability is $\frac{4}{10}$.

For (b), how many pairs have one or more females? Well, how many possible pairs are all male? That's $\binom{3}{2}=\frac{3\cdot2}{2!}=3$. So $7$ of the pairs have at least one female, and the probability is $\frac{7}{10}$.

I don't want to continue laying out answers to (c) and (d). But do you see how to apply the same process to understand an answer for (c)?


问:
sum of the eigenvalues = trace($A$)?
Is it true that for a square matrix $A$, all of whose eigenvalues exist in the base field, sum of the eigenvalues = trace($A$)?

The result holds in all the matrices I've studied.

答: Here is the way I would show it. $A = VDV^{-1}$ which means $tr(A) = tr(VDV^{-1}).$ But trace has a cyclic property. so $tr(VDV^{-1}) = tr(DV^{-1}V) = tr(D)$ so I think your assertion is correct!</p>

ContentLicense: CC BY-SA 3.0

Title: sum of the eigenvalues = trace($A$)?


问:
Set of adherence values of a sequence equal to $\mathbb{N}$
I'm thinking about how to construct, if possible, a sequence $(x_n)$ with the set of adherence values of $(x_n)$ equal to $\mathbb{N}$.

I already constructed a sequence $(y_n)$ with the set of adherence values equal to $[0,1]$, that being an enumeration of the rationals in $[0,1]$, since every real number is a limit of a sequence of rational numbers. So I'm thinking that it is possible to build such sequence with $\mathbb{N}$ being it's set of adherence values, since $\mathrm{card } ([0,1]) > \mathrm{card } (\mathbb{N})$.

Can anyone give me a hint? (Just a hint, not the answer please)

Thanks in advance.

答:
HINT: For each $n\in\Bbb N$ let $\sigma_n$ be a sequence converging to $n$. Now combine the sequences $\sigma_n$ into a single sequence. A more specific hint is hidden below.

 More specifically, consider the set $$\left\{n+\frac1m:n\in\Bbb N\text{ and }m\in\Bbb Z^+\right\}\,.$$

ContentLicense: CC BY-SA 4.0


答:
Hint:

For every integer $n$ construct a sequence $u_n$ whose only limit is $n$.
Let $p_n$ be the $n$-prime number. Consider $v_{p_n^m}=u_m$ its limit is $n$. Write $A_{p_n}=\{p_n^m\}$

Let $A=\cup A_{p_n}$ $A$ is numerable, there exists a bijection $f: \mathbb{N}\rightarrow A $.

Define $w_n=v_{p_n^m}$ where $f(n)=p_n^m$.

Title: Set of adherence values of a sequence equal to $\mathbb{N}$

Body:
I'm thinking about how to construct, if possible, a sequence $(x_n)$ with the set of adherence values of $(x_n)$ equal to $\mathbb{N}$.

I already constructed a sequence $(y_n)$ with the set of adherence values equal to $[0,1]$, that being an enumeration of the rationals in $[0,1]$, since every real number is a limit of a sequence of rational numbers. So I'm thinking that it is possible to build such sequence with $\mathbb{N}$ being it's set of adherence values, since $\mathrm{card } ([0,1]) > \mathrm{card } (\mathbb{N})$.

Can anyone give me a hint? (Just a hint, not the answer please)

Thanks in advance.


问:
Find a one-dimensional sufficient statistic for theta
I am try to find the one-dimensional sufficient statistic for $\theta$. Although, I'm a bit confused about how you would arrive to the sufficient statistic.

Here is the common density function:

$f\left(x_{i} ; \theta\right)=\frac{1}{\theta \sqrt{\pi}} e^{\frac{-x_{i}^{2}}{\theta^{2}}}$, $-\infty<x_{i}<\infty, i=1,2, \ldots, n,$ where $\theta>0$ is unknown parameter.

Please let me know if you need me to clarify anything.

答:
Yes, to use the factorization theorem, which says that if $f(x|\theta)=u(x)v[r(x),\theta]$, then r(x) is a sufficient statistic for theta.

$$f(x)=\frac{1}{\theta\sqrt \pi}e^{-\frac{x^2}{\theta^2}}$$

Then for a sample,

$$\begin{split}f(x)&=\prod\frac{1}{\theta\sqrt \pi}e^{-\frac{x^2}{\theta^2}}\\
&=\underbrace{1}_{u(x)}\cdot \underbrace{\frac{1}{\theta^n\pi^{n/2}}e^{-\frac{1}{\theta^2}\sum x^2}}_{=v[r(x),\theta]}\end{split}$$

Hence $\sum x^2$ is sufficient for theta.


问:
Proof of De la Vallée-Poussin's Test
Here's what I'm trying to prove.

Suppose that $A \subseteq \mathbb{R}^n$ and $J = [c,+\infty)$. Let $f: A \times J \to \mathbb{R}^p$ be continuous and suppose that there exists a function $g: J \to [0,+\infty)$ such that:

$$\forall (x,t) \in A \times J: \|f(x,t)\| \leq g(t)$$

Suppose that $\int_{c}^{\infty} g(t) \ dt$ is convergent. Then, $\int_{c}^{\infty} f(x,t) \, dt$ is uniformly convergent for $x \in A$.

Here's the definition of uniform convergence (I'm pretty sure that this is really just an extension of the definition of improper integrals on the real line).

Again, let $f$ be as given above. Then, define $F(x) = \int_{c}^{\infty} f(x,t) \, dt$. We say that $f$ is uniformly convergent for $x \in A$ iff:

$$\forall \epsilon > 0: \exists D > c: x \in A \land d \geq D \implies \left\| F(x)-\int_{c}^{d} f(x,t) \right\| < \epsilon$$

Proof Attempt:

Observe that for any fixed $x \in A$, we have:

$$\forall i \in \{1,2,\ldots,p\}: |f_i(x,t)| \leq \left\|f(x,t)\right\| \leq g(t)$$

$$\implies \forall d > c: \int_{c}^{d} |f_i(x,t)| \ dt \leq \int_{c}^{d} g(t) \ dt$$

Then, since $\int_{c}^{d} g(t) \ dt$ converges as $d \to \infty$, it follows that $\int_{c}^{d} |f_i(x,t)| \ dt$ converges as $d \to \infty$ by comparison. Then, this implies that $\int_{c}^{\infty} f_i(x,t) \ dt$ converges and since the integral of each component function of $f$ converges, it follows that $\int_{c}^{\infty} f(x,t) \ dt$ converges. $\Box$

The thing that I'm somewhat worried about is if I'm missing something crucial over here. Like, I'm just wondering if I've overlooked something or if I would actually have to do a more formal argument using the definition directly. I'd appreciate it if someone could have a look at my working.

答:
Using

$$\int_c^d |f_i(x,t)| \, dt \leqslant\int_c^d g(t) \, dt,$$

and the fact that the integrand on the LHS is nonnegative, you can only conclude pointwise convergence, that is for each $x \in A$,

$$\lim_{d \to \infty} \int_c^d |f_i(x,t)| \, dt = \limsup_{d \to \infty}\int_c^d |f_i(x,t)| \, dt \leqslant \lim_{d \to \infty}\int_c^d g(t) \, dt$$

To prove uniform convergence, you need to show that

$$\lim_{d \to \infty} \sup_{X \in A}\left|\int_d^\infty f_i(x,t) \, dt \right| = 0$$

We have,

$$\left|\int_d^R f_i(x,t) \, dt \right| \leqslant\int_d^R |f_i(x,t)| \, dt \leqslant\int_d^R g(t) \, dt \leqslant \int_d^\infty g(t) \, dt$$

Taking the limit as $R \to \infty$ we get

$$\left|\int_d^\infty f_i(x,t) \, dt \right|\leqslant  \int_d^\infty g(t) \, dt, \\\sup_{x \in A}\left|\int_d^\infty f_i(x,t) \, dt \right|\leqslant  \int_d^\infty g(t) \, dt$$

Since the improper integral of $g$ converges, we get as desired

$$\lim_{d \to \infty}\sup_{x \in A}\left|\int_d^\infty f_i(x,t) \, dt \right|=\lim_{d \to \infty}  \int_d^\infty g(t) \, dt = 0$$

Title: Proof of De la Vallée-Poussin's Test


问:
Cohomology ring of complex grassmanian as a quotient
This is about the complex version of this question,

I am interested in understanding the integral cohomology ring of $\mathrm{Gr}_k(\mathbb{C}^n)$ as the following quotient of the integral cohomology ring of $\mathrm{Gr}_k(\mathbb{C}^{\infty})$:
$$
H^*(\mathrm{Gr}_k(\mathbb{C}^n))\cong H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))/\langle \bar{c}_{n-k+1},\dots,\bar{c}_n\rangle.
$$
Here the $\bar{c}_i$ are polynomials in the Chern classes $c_i$ of the tautological $k$-plane bundle over $\mathrm{Gr}_k(\mathbb{C}^{\infty})$, and are defined by the relation
$$
c\bar{c}=1.
$$
(Recall that $H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))\cong \mathbb{Z}[c_1,\dots,c_k].$)

The approach suggested for the real case (with $\mathbb{Z}/2$ coefficients) in the linked question is to show that

the inclusion $\mathrm{Gr}_k(\mathbb{C}^{n})\subset \mathrm{Gr}_k(\mathbb{C}^{\infty})$ induces a surjection $H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))\to H^*(\mathrm{Gr}_k(\mathbb{C}^n))$;

the polynomials $\bar{c}_{n-k+1},\dots,\bar{c}_n$ are zero in $H^*(\mathrm{Gr}_k(\mathbb{C}^n))$, hence the above map induces a surjection from the quotient
$H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))/\langle \bar{c}_{n-k+1},\dots,\bar{c}_n\rangle \to H^*(\mathrm{Gr}_k(\mathbb{C}^n))$;

use a dimension count to deduce that the latter map is also injective, hence an isomorphism.

Question: Since we are working with integral cohomology, which is merely a ring (not a $\mathbb{Z}/2$-vector space), my question is how one could carry out or alter step (3) to obtain injectivity.

Thanks for the help. Any references are appreciated.

答: For each $i$, $H^i(\mathrm{Gr}_k(\mathbb{C}^n))$ is a finitely generated free abelian group, as is the $i$th degree of $H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))/\langle \bar{c}_{n-k+1},\dots,\bar{c}_n\rangle$.  So, if you can show they have the same rank, any surjective homorphism between them is automatically injective as well (any surjection $\mathbb{Z}^n\to\mathbb{Z}^n$ is injective; more generally, this is true  for any Noetherian module over a ring, by considering the kernels when you iterate the homomorphism).  If you like, this means you can count their dimensions after tensoring with $\mathbb{Q}$ (or equivalently, use cohomology with coefficients in $\mathbb{Q}$) and check that they are the same.


问:
Show that $S=\{ A\in GL_n(K)\mid AJA^t=J\}$ is a subgroup of $GL_n(K)$, where $J\in K^{n\times n}$.
I have a question about the following problem: show that $$S=\{ A \in GL_n(K) \mid AJA^t = J \}$$ is a subgroup of $GL_n(K)$, where $J\in K^{n\times n}$.

I have shown that the identity element $e$ is part of $S$, but I don't know to prove the second criteria for a subgroup. Thanks in advance!

答:
Assume that $A,B \in S$. This means $A,B \in GL_n(K)$ and $AJA^t = BJB^t = J$. We wish to show $AB \in S$. We have $AB \in GL_n(K)$ and
$$(AB)J(AB)^t = (AB)J(B^tA^t) = A(BJB^t)A^t = AJA^t = J$$
so indeed $AB \in S$.


问:
Proving the solution to linear programming problem is always the boundaries
Show that the solution $\mathbf{y}$ to the following linear problem:

$$\min c_1x_1 + c_2x_2 + \cdot \cdot \cdot + c_nx_n$$
$$\text{subject to} \ \ a_i\leq x_i\leq b_i \ \ \ \ i = 1,2,....n$$

is:

$$
\mathbf{y}_i = \left\{
        \begin{array}{ll}
            a_i & \quad c_i > 0 \\
            b_i & \quad c_i < 0
        \end{array}
    \right.
$$

Any idea or suggestion on how to go about this probelm?

答:
Rewrite the constraints as $x_i \ge a_i$ and $-x_i \ge -b_i$, construct the dual linear program with dual variables $\alpha_i \ge 0$ and $\beta_i \ge 0$, and note that the dual solution $\alpha^*_i = \max\{c_i,0\}$ and $\beta^*_i = \max\{-c_i,0\}$ is dual feasible and has the same objective value as your proposed primal feasible solution.

If you multiply both sides of $x_i \ge a_i$ by $\alpha^*_i$ and both sides of $-x_i \ge -b_i$ by $\beta^*_i$ and then add these up, you obtain
$$\sum_i (\alpha^*_i - \beta^*_i) x_i \ge \sum_i (\alpha^*_i a_i - \beta^*_i b_i). \tag1$$
By the choice of $\alpha^*_i$ and $\beta^*_i$, we have $\alpha^*_i - \beta^*_i = c_i$, and $(1)$ becomes
$$\sum_i c_i x_i \ge \sum_{i:c_i > 0} c_i a_i + \sum_{i:c_i < 0} c_i b_i,$$
and this lower bound on the objective function certifies the optimality of your proposed solution $\mathbf{y}$.


问:
Show that if $z$ is the midpoint of the line segment $[x,y]$ between $x,y\in\mathbb R^n$, then $T(z)$ is the midpoint of $[T(x),T(y)]$.
What does [x,y] represent? Is it a one-dimensional line segment? How is a line segment bounded by x and y if x and y are n-tuples (not numbers on the real number line, for instance)? And what is the "midpoint"?

Moreover, I'm looking for a hint/nudge in the right direction.

Let $T:\mathbb R^n\rightarrow\mathbb R^m$ be a linear map. Show that if $z$ is the midpoint of the line segment $[x,y]$ between $x,y\in\mathbb R^n$, then $T(z)$ is the midpoint of $[T(x),T(y)]$.

答: Since $z$ is midpoint of $[x,y]$ we have $z = {1\over 2}(x+y)$. Since $T$ is linear we have $$T(z) = {1\over 2}T(x+y) = {1\over 2}(T(x)+T(y))$$ so $T(z)$ is midpoint of $[T(x),T(y)]$.


问:
Index of the vector field $X(z) = p(z) \cdot e^{-\| z \|^2}$ projected on $\mathbb{S}^2$
Define the vector field $X:\mathbb{C} \to \mathbb{C}$ as $X(z) = p(z) \cdot e^{-\| z \| ^2}$ where $p$ is a polynomial. I'm trying to compute the index in $\infty$ of the vector field obtained projecting $X$ on $\mathbb{S}^2$.

This was my attemp: if I consider a sufficiently small circumference along a parallel, this is projected by stereographic projection on a circumference in $\mathbb{C}$ containing all the zeros of the vector field $X$ except $\infty$. Then is well known that the degree of the map $\frac{X(z)}{\lVert X(z) \rVert} = \frac{p(z)}{\lVert p(z) \rVert}$ is equal to the number of zeros (with
multiplicity) of the polynomial that are inside the circumference. Thus my claim is $ind_{X}(\infty) = deg(p)$.

Is my claim true? I'm not sure about the fact that the degree of the projection of $X$ normalized and restricted to a circumference is the same as the degree of $X$ normalized and restricted to the projection of the circumference.

答:
Let us define $\psi : \mathbb{C} \to \mathbb{S}^2$ the stereographic projection using the North Pole and $\varphi : \mathbb{C} \to \mathbb{S}^2$ the other one. Let $x_0$ be the North Pole and $x_1$ the South Pole.

The vector field induced in $\mathbb{S}^2$ and the one of which you want the index is the following:

$$ Z(y)=
\begin{cases}
0 \mbox{ if }z \mbox{ is the North Pole } \\
d \psi \circ X \circ \psi^{-1}(z) \mbox{ otherwise}
\end{cases}$$

Because you want to compute $ind_{Z}(x_0)$, let us find a parameterization of a neighborhood of $x_0$. We observe that we can take $\varphi$.

Thus by definition $ind_Z(x_0)=ind_{d \varphi ^{-1} \circ Z \circ \varphi}(0)$. The key point is that $\psi ^{-1} \circ \varphi = \frac{z}{\|z\|^2}=\frac{1}{\overline{z}}$. Thus we have reduced the problem to the following:

Consider the map $Y: \mathbb{C} \to \mathbb{C}$ defined in this way: $Y(z)=X(\frac {1}{\overline{z}})$ where $X(z) = p(z) \cdot e^{-\| z \| ^2}$ and $W:\mathbb{C} \to \mathbb{C}$ defined as $W(z)=\frac{1}{\overline{Y(z)}}$. Suppose $p(z)=z^n+a_{n-1}z^{n-1}+...+a_0$.

We want to compute $ind_W(0)=-ind_Y(0)$.

Now take $r >0$ sufficiently small such that $r^{-n}>|a_{n-1}|r^{-(n-1)}+...+|a_0|$.

Then define for $z \in \partial B(0,r)$.

$$H(t,z)=\frac{[t\overline{z}^{-n}+(1-t)p(\frac {1}{\overline{z}})]e^{-\frac{1}{r^2}}}{\|[t\overline{z}^{-n}+(1-t)p(\frac {1}{\overline{z}})]e^{-\frac{1}{r^2}}\|}=\frac{t\overline{z}^{-n}+(1-t)p(\frac {1}{\overline{z}})}{\|t\overline{z}^{-n}+(1-t)p(\frac {1}{\overline{z}})\|}$$

which is an homotopy between $\frac{p(\frac {1}{\overline{z}})}{\|p(\frac {1}{\overline{z}})\|} : \partial B(0,r) \to \mathbb{S}^1$ and $r^{-n}\overline{z}^{-n} : \partial B(0,r) \to \mathbb{S}^1$ which has degree $n$.

We conclude that $ind_X(\infty)=-n$.


问:
Integral over fundamental solution in Green's formula
Consider Green's representation formula
\begin{equation}
u(x) = \int_{\Omega} \Gamma(x-y)\left(-\Delta u(y)\right)\mathrm{d}y+\oint_{\partial\Omega}\Big(\Gamma(x-y)\nabla u(y)-u(y)\nabla_y(\Gamma(x-y)) \Big)\cdot\nu\,\mathrm{d}S,
\end{equation}
where $\Gamma: \mathbb{R}^n\setminus\{0\}\rightarrow \mathbb{R}$ is the fundamental solution of the Laplace equation, i.e.
\begin{equation}
\Gamma(x) = -\frac{1}{2}\log|x|
\end{equation}
for $n=2$ and
\begin{equation}
\Gamma(x)=-\frac{1}{n(n-2)\alpha(n)|x|^{n-2}}
\end{equation}
for $n\geq 3$, $\Omega$ is a bounded Lipschitz domain, $u\in C^2(\bar{\Omega})$ and $x\in\Omega$.

My question is why the first integral in this formula exists at all, as we integrate over the singularity of the fundamental solution $\Gamma$ at $y=x$. Is this an integral in the classical sense? In the proof, one considers Green's identity on $\Omega\setminus B_\epsilon(x)$ for $\epsilon>0$. At one step in the proof, we also perform the first integral in the representation formula over $B_\epsilon(x)$ in order to show that it vanishes for $\epsilon\to 0$. When integrating over $B_\epsilon(x)$, it is also possible that $y=x$. At the end of the proof, we obtain
\begin{equation}
\int_{\Omega} \Gamma(x-y)\left(-\Delta u(y)\right)\mathrm{d}y = \lim_{\epsilon\to 0}\int_{\Omega\setminus B_\epsilon(x)} \Gamma(x-y)\left(-\Delta u(y)\right)\mathrm{d}y.
\end{equation}
Is this how one defines the integral on the left-hand side? Or am I just confused and the singularity of $\Gamma$ does not matter for the integration as it is only a null set?

答:
In $\Bbb{R^n}$ we have that

$$\int_{B(0,1)} \frac{dx}{|x|^p} = C_n \int_0^1 \frac{r^{n-1}dr}{r^p}$$

where $C_n$ is the "surface area" of the unit $n$-ball by switching over to $n$ dimensional spherical coordinates. The integral converges if $p-n+1<1$ by $p$ test.


问:
Torsion filtration of coherent sheaves: Looking for a simple example.
Let X be a scheme and let E be a coherent sheaf on X.

A torsion filtration (according to Huybrecht's "Geometry of Moduli of Sheaves") for $E$ is the unique filtration $0 \subset T_0(E) \subset T_1(E) \subset \dots \subset T_d(E)=E$, where $d=\dim(E)$ and $T_i(E)$ are the maximal subsheaves of $E$ of dimension $ \le i$.

Unfortunately, no simple example of a torsion filtration is given and I am failing to come up with one on my own.

Are there any good, instructive examples of a torsion filtration?

答:
In the affine case (so $E$ is just a module over some ring $A$), $T_i(E)$ is just the submodule of $E$ consisting of elements $x\in E$ such that $A/\operatorname{Ann}(x)$ has dimension $\leq i$.  So for a really simple example, you could take $A=k[x,y]$ and $E=A/(x,y)\oplus A/(x)\oplus A$.  Then $T_0(E)$ would be just $A/(x,y)$, $T_1(E)$ would be $A/(x,y)\oplus A/(x)$, and $T_2(E)$ would be all of $E$.

For another example, if $A$ is a domain of dimension $n$, then $T_{n-1}(E)$ is just the submodule of elements of $E$ whose annihilator is nontrivial, i.e. it is the entire torsion submodule of $E$.  For smaller values of $i$, $T_i(E)$ consists of elements that are "more strongly" torsion, such that their annihilator is large enough that when you mod it out you reduce the dimension down to $i$ (or lower).


问:
Prove this inequality of integrals:
Let $(X, M, \mu\ )$ be a measure space. Let $f$ be a positive measurable function with $\int_X f d \mu\ < \infty\ .$  Then for every $\epsilon\ > 0$, there is a set $E \in M$ such that $\mu(E)<\infty$ and
\begin{equation}
\int_X f d \mu\ \leq \int_E f d \mu\ + \epsilon
\end{equation}

This problem appears on Bartle's Elements of Integration and Lebesgue Measure. I couldn't prove it for the life of me. And obviously, $E$ has to be a proper subset of $X$, otherwise, this is just trivial. Any idea?

答:
Since the result is trivial if $\int_X {fd\mu }  = 0$ or $\mu(X) < \infty$, assume that  $\int_X {fd\mu }  > 0$ and $\mu(X) = \infty$.

Given $\varepsilon > 0$, it follows from the definition
$$
\int_X {fd\mu }  = \sup \bigg\{ \int_X {sd\mu } :0 \le s \le f,s \; {\rm simple}\bigg \}
$$
that there exists a simple function $0 \leq s \leq f$ such that
$$
\int_X {fd\mu }  - \varepsilon  \le \int_X {sd\mu }.
$$
We can suppose that, for some $n \in \mathbb{N}$, $s$ takes positive values $\alpha_i$, $i=1,\ldots,n$, on some measurable subsets $E_i$ of $X$ with (necessarily) $\mu(E_i) < \infty$ (since $\int_X {sd\mu }  < \infty $), and $s$ is zero on the complement of $E_1  \cup  \cdots  \cup E_n $. Noting that $s \leq f$, we then have
$$
\int_X {sd\mu } = \int_{E_1  \cup  \cdots  \cup E_n } {sd\mu }  \le \int_{E_1  \cup  \cdots  \cup E_n } {fd\mu } ,
$$
from which the desired result follows with $E = E_1  \cup  \cdots  \cup E_n$.

Title: Prove this inequality of integrals:


问:
Why does the z-ultrafilter Stone-Čech compactification construction have the universal property?
For the notions of $z$-filter, prime $z$-filters and $z$-ultrafilters see A Prime $\mathcal P$-filter is contained in a unique $\mathcal P$-ultrafilter?

Let $X$ be a Tychonoff space. Let $BX$ be the set of $z$-ultrafilters of $X$. For each zero-set $Z$, let $Z^*=\{p\in BX: Z\in p \}$, then the sets $Z^*$ form a base of closed sets for a topology on $BX$. Now consider $K$ a compact Hausdorff space and $f:X\rightarrow K$ a continuous function.

If $p\in BX$, it is easy to see that $\mathcal{F}_p=\{Z\subseteq K: Z$ is a zero-set and $f^{-1}[Z]\in p\}$ is a prime $z$-ultrafilter, then using that $K$ is $T_3$ one can show that $F_p$ has a unique cluster point $q$, as prime $z$-filters are contained in unique $z$-ultrafilters in $T_3$ spaces, then define $\bar{f}(p)=q$. I want to show $\bar{f}$ is continuous. To do this, it is enough to prove that for any zero-set $B$ of $K$, $\bar{f}^{-1}[B]$ is closed in $BX$, since $K$ is $T_3$.

I have shown that if $p\notin \bar{f}^{-1}[B]$, then $p\notin (f^{-1}[B])^*$, so I only need to show  ${(f^{-1}[B])^*}^c\cap \bar{f}^{-1}[B]=\emptyset$, that is, $\bar{f}^{-1}[B]\subseteq (f^{-1}[B])^*$, since then ${(f^{-1}[B])^*}^c$ would be an open set such that $p\in {(f^{-1}[B])^*}^c$ disjoint with $\bar{f}^{-1}[B]$. The problem is that I do not know how to prove that $\bar{f}^{-1}[B]\subseteq (f^{-1}[B])^*$.

So, is this true?, and, is there an easier way to prove that $BX$ has the universal property with respect to compact Hausdorff spaces?

Thanks

答:
Take a zero-set neighbourhood $Z_1$ of $q = \bar{f}(p)$ (as $Y$ is Tychonoff, these form a base), we need to find a neighbourhood of $p$ that maps (under $\bar{f}$) into $Z_1$). Let $C$ be a co-zeroset (complement of a zero-set) neighbourhood of $p$ contained in $Z_1$ and such that $C \subset Z_1$. Again, this uses Tychonoff. Now set $Z_2 = Y \setminus C$, and now $Y = Z_1 \cup Z_2$, a union of two zero-sets, only one of which ($Z_1$) is a neighbourhood of $q$.

Now, $f^{-1}[Z_1]$ and $f^{-1}[Z_2]$ are also zero-sets in $X$, whose union is $X$ so the union of $(f^{-1}[Z_1])^{\ast}$ and $(f^{-1}[Z_2])^{\ast}$ equals $BX$.

Now check that $BX \setminus (f^{-1}[Z_2])^{\ast}$ is a neighbourhood of $p$ that maps into $Z_1$ under $\bar{f}$.


问:
How do you set up the space $L^2(S^1)$ and what is integration on this space.
The Set up:

Let $S^1=\{e^{2\pi ix}\|x\in[0,1) \}$. Now we endow this space with the Lebesgue measure, which is defined as follows. $\mu(A)=m(f^{-1}(A))$ where $A \subseteq S^1$ $f(x)=e^{2\pi i x}$ and $m$ is the lebesgue measure on $[0,1)$. Now let $g:S^1 \to \mathbb{C}$. Then we can write $g(z)=h(z)+ik(z)$ then $\int_{S^1}g(z)d\mu=\int_{S^1}h(z)d\mu+i\int_{S^1}k(z)d\mu$ Now $h,z$ are real valued so we can the standard way to assign values to this integrals (using simple functions)

The problem:

I thinks all of this works in theory. But I do not know how to apply this in practice. Let us take the function $z^n$ what is $\int_{S^1}z^n$ ? All of this confusion was sparked by trying to understand why $\{z^n,n\in \mathbb{Z}\}$ is an orthonormal basis of $L^2(S^1)$ which made me realize I do not fully understand how to compute the integral $\int_{S^1}|z|^n\overline z=\int_{S^1}\overline z$ (since $|z|=1$). I would appreciate any clarification.

答:
Per my comment above I assumed an answer would be soon forthcoming, but it wasn't, so here goes. A few introductory comments.

Comment 1.  The Wikipedia page https://en.wikipedia.org/wiki/Pushforward_measure gives an OK explanation of aspects of the concept with reference to further resources and proofs.  Note that it even gives your example as an example (although I don't know how you could have found it if you didn't know the terminology).  If you have a textbook on measure theory (or a textbook treating integration in some degree of generality, perhaps just on $\mathbb{R}^n$ or $\mathbb{C}$), you might want to compare the Wikipedia treatment to your own resource and assure yourself that you understand the differences or similarities, particularly if this is for purposes that involve presentation to others who have not visited other resources and may not want to be referred to resources they have not seen.  I would caution anybody against using Wikipedia or results in random textbooks as "black box" authority where you can just plug in inputs and receive results; if the point of the problem is to engage with the technical particulars of push-forward measures (or what in this context might sometimes be called a "change of variables") directly, the above page and its references may nevertheless guide you further and help you deal with that.

Comment 1.5. In the context of nice mappings from simple spaces to one another, this whole "push forward" measure thing is the good old calculus change of variables formula. It might be worth thinking more about this later.  Sometimes measure theory is taught as though it's completely separate from anything that may have preceded it in a course sequence.  It's really not.

Comment 2.  I note that while your question refers to a set as being an "orthonormal" basis of $L^2(S^1)$, and hence implicitly references an inner product and notion of orthogonality, your question does not explicitly specify the inner product at issue.  This is fine, because the inner product on the $L^2$ space of a measure space is a very standard thing.  But should you have a textbook or course notes, it is possible that more information on this topic and citable results other than Wikipedia will be found around the definition of the inner product, whether in $L^2$ spaces in general or in $L^2(S^1)$ in particular.

OK, no more comments.  From its definition, the measure $\mu$ as defined in your problem is seen to be the "pushforward" (as defined on Wikipedia) of Lebesgue measure via the map $f$.  So for any $g$ on $\mathbb{T}$ we have
$$
\int_{S^1} g \, d\mu = \int_0^1 g \circ f \, d\lambda
$$
Where $\lambda$ is the usual Lebesgue measure.  And of course integrating a continuous function with respect to that (whether real or complex valued) is just boring old calculus integration, which as you note in the complex-valued case can be broken into real and imaginary parts of even more boring old real-valued calculus integration.

If $z$ is the usual identity function on $\mathbb{T}$ then for any integers $n, m$ you have
$$
\int_{S^1} z^n \overline{z^m} \, d\mu = \int_0^1 e^{2\pi int} e^{-2 \pi imt} \, dt
$$
Because $\overline{z^m} = 1/z^m$ on $\mathbb{T}$ and when you plug $f(t)$ into $z^n$ and $1/z^m$ and use properties of the exponential function you get $(e^{2\pi i t})^n = e^{2 \pi i nt}$ and $1/(e^{2\pi i t})^m = e^{-2\pi imt}$ respectively.  By more properties of the exponential function the integrand "simplifies" to $e^{2\pi i (n-m) t}$.

Now we're in the territory of previously answered questions such as Showing that complex exponentials of the Fourier Series are an orthonormal basis, in which the need to distinguish the case $n \neq m$ arises from the fact that one needs $k$ nonzero to use $\frac{1}{k} e^{kt}$ for an antiderivative for $e^{kt}$.  The calculation and comments there explain why when $n \neq m$ the result is $0$.  When $n=m$, the calculation is simpler: we're integrating $e^0 = 1$ from $0$ to $1$ $dt$ and that's $1$.  That's the desired orthogonality relation.


问:
Proof of $\sum_{n=1}^\infty a_nb_n^2c_n^3<\infty$.
Let $a_n \geq 0, b_n \geq 0, c_n \geq 0 \; \forall n \in \mathbb{N}$ and suppose that $\sum_{n=1}^\infty a_n^3b_n^3, \;\sum_{n=1}^\infty b_n^3c_n^3, \; \sum_{n=1}^\infty c_n^6$ are convergent. Prove that the series $\sum_{n=1}^\infty a_nb_n^2c_n^3$ is convergent.

I saw that $(a_n^3b_n^3 \cdot b_n^3c_n^3 \cdot c_n^6)^\frac{1}{3}=a_n b_n^2 c_n^3$ and here I stopped.
I was thinking about using Cauchy-Schwarz inequality, but I don't really know how to apply it (especially having $x_n^3$, not $x_n^2$).

I would be grateful for some hints! :)

答: Apply the Arithmetic Mean - Geometric Mean inequality: $$\frac{a_i^3b_i^3 +b_i^3c_i^3+c_i^6}3\geqslant \sqrt[3]{a_i^3b_i^6c_i^9}=a_ib_i^2c_i^3$$ So your sequence is bounded.


问:
$2yy^{\prime} = y^2 +x -1 , y(0) = 1$
$$2yy^{\prime} = y^2 +x -1 ,\ \ \ \ \ y(0) = 1$$

I have tried the substitution $u = y^2 + x -1 $ but i dont know how i get $u^{\prime}$

答:
$$u = y^2 + x -1$$
Just differentiate:
$$u'=2yy'+1$$
And substitute in the differential equation  $2yy'=u'-1$. Note that an easier substitution could be $u=y^2+x$ and $u'=2yy'+1$.
$$2yy' = y^2 +x -1$$
$$2yy'+1=y^2+x$$
$$(y^2+x)'=y^2+x$$
$$u'=u$$
This is a separable differential equation.

Title: $2yy^{\prime} = y^2 +x -1 , y(0) = 1$


答:
You could also recognize $2yy'$ as $(y^2)'$ so that your differential equation is

$$(y^2)' = y^2 + x - 1.$$

Letting $u = y^2$, you get

$$ u' = u + x - 1 $$

which is easily solved and then you can solve for $y$ from there.

Title: $2yy^{\prime} = y^2 +x -1 , y(0) = 1$

Body:
$$2yy^{\prime} = y^2 +x -1 ,\ \ \ \ \ y(0) = 1$$

I have tried the substitution $u = y^2 + x -1 $ but i dont know how i get $u^{\prime}$


问:
Why does $i^3$ equal $-i$ if you multiply the numbers inside a radical?
$i^3$ equals $-i$. Since $i$ is $\sqrt{-1}$, and you can multiply the number inside radicals that are being multiplied together, wouldn't $i^3$ equal $\sqrt{-1×-1×-1}$, which is $\sqrt{-1}$, which is $i$?

答:
In complex numbers it is wrong to write $i$ as $\sqrt{-1}$. For any complex number $z \neq 0$ there are two distinct values for $\sqrt{z}$ and, for $z \notin [0, \infty)$ there is no choice which is compatible with the algebraic properties you learned about roots.

If you want to use roots, square root is a multivalued function over the complex numbers and
$$
\sqrt{-1} = \pm i
$$
which makes your formula correct.


问:
How to calculate the length of the negative parts of $ \cos \alpha x-\cos 2 \alpha x $?
I have this periodic function
$$  f(x) = \cos (\alpha  x)-\cos (2 \alpha  x)     $$
defined for $x>0$ and for $\alpha>0$.

How can I find the length of the graph of $f$ along one of its period when $f(x) < 0$?

Here I have plotted the function:

$\hspace{3.5cm}$

答:
Too long for a comment:

The function is periodic in $x$ with period $T = -\frac{2\pi}{a}$.

Find the $x$ for which $f(x) < 0$ that are for example $x_1:=\frac{2\pi}{3a} \le x \le \frac{4\pi}{3a}=:x_2$. Then using the usual formula for evaluating the lenght of a curve you get:

$$L = \int_{x_1}^{x_2} \sqrt{1+f'(x) ^ 2}dx = \dots$$

[I used Wolfram for the calculations in an hurry]


答:
In complex analysis, as the other response indicates, the square root function is ambiguous.

In complex analysis, $i = e^{i\pi/2} = \cos(\pi/2) + i\sin(\pi/2).$

Using DeMoivre's theorem, $i^3 = e^{3i\pi/2}.$

This equates to $-i = e^{-i\pi/2}$ precisely because

$3\pi/2 \equiv -\pi/2 \pmod{2\pi}.$

Note that $(-i)$ is the complex conjugate of $(i)$, which explains why arg$(-i)$ = -arg$(i)$, within a modulus of $2\pi.$

Title: Why does $i^3$ equal $-i$ if you multiply the numbers inside a radical?

Body: $i^3$ equals $-i$. Since $i$ is $\sqrt{-1}$, and you can multiply the number inside radicals that are being multiplied together, wouldn't $i^3$ equal $\sqrt{-1×-1×-1}$, which is $\sqrt{-1}$, which is $i$?


答: If $a=\sqrt{b}\sqrt{c}$ with $a,\,b,\,c\in\Bbb C\supseteq\Bbb R$ then $a^2=\sqrt{b}\sqrt{c}\sqrt{b}\sqrt{c}=\sqrt{b}\sqrt{b}\sqrt{c}\sqrt{c}=bc$ so $a=\pm\sqrt{bc}$. If $b,\,c\ge0$, we define the square roots of $b,\,c,\,bc$ to be non-negative, so the $\pm$ sign can be dropped. In complex numbers, we can't do this. Famously $i^2=-1\ne\sqrt{(-1)^2}$; you've just encountered a more complicated counterexample.


答:
$2yy^{\prime} = y^2 +x -1 $ , rewrite it as $$y^{\prime}-\frac{1}{2}y =\frac{1}{2}(x -1)y^{-1}$$

This is a Bernoulli differential equation with $n=-1$ let $z=y^{1-n}.$ Can you continue?

Title: $2yy^{\prime} = y^2 +x -1 , y(0) = 1$


问:
Laplace transform of solution of an integral equation for non-recursive logistic map computations
Consider the logistic map,

$x_n = r \: x_{n-1} \left( 1 - x_{n-1} \right)$

If we generalize this to a complex function $f : \mathbb{C} \mapsto \mathbb{C}$, we get,

$f \left( z \right) = r \: f \left( z-1 \right) \left( 1- f \left( z-1 \right)\right)$

$$f \left( z \right) = r \left( f \left( z-1 \right)- \left( f \left( z-1 \right) \right)^2 \right) \: \dots \: \left( 1 \right) $$

Assume that there exists a function $u: \mathbb{C} \times \mathbb{R} \mapsto \mathbb{C} \times \mathbb{R} $, such that,

$\displaystyle{f \left( z \right) = \mathcal{L} \{ u \left( z, t \right) \} \left( s \right) = \int_0^\infty u \left( z, t\right) e^{-st} dt }$

As per integration by parts,

$$\mathcal{L} \{ u \left( z, t \right) \} \left( s \right) = \frac{1}{s} \left( u \left( z, 0 \right) + \mathcal{L} \{ \dot{u} \left( z, t \right) \} \left( s \right)\right) \: \dots \: \left( 2 \right)$$

Where $\dot{u} \left( z, t\right) = \frac{\partial u \left( z,t \right)}{\partial t}$

Notice that equations $\left( 1 \right)$ and $\left( 2 \right)$ have the same L.H.S. They are identical in form if we take,

$s = \frac{1}{r}, \\ u \left( z, 0 \right) = f \left( z-1 \right), \\ \mathcal{L} \{ \dot{u} \left( z, t \right) \} \left( s \right) = - \left( f \left( z-1 \right) \right) ^2 \text{, or,}$

$\mathcal{L} \{ \dot{u} \left( z, t \right) \} \left( s \right) = - \left( u \left( z, 0 \right) \right)^2$

Replacing $s$ with $\frac{1}{r}$,

$$ \mathcal{L} \{ \dot{u} \left( z, t \right) \} \left( \frac{1}{r} \right) = - \left( u \left( z, 0 \right) \right)^2 \dots \left( 3 \right) $$

The same equation may be rewritten in integral form,

$\displaystyle{ \left( u \left( z,0 \right) \right)^2 + \int_0^\infty \frac{\partial u \left( z,t \right)}{\partial t} e^{-\frac{t}{r}} dt = 0 }$

Applying the Leibniz integral rule,

$\displaystyle{ \left( u \left( z,0 \right) \right)^2 + \frac{d}{dt} \int_0^\infty u \left( z,t \right) e^{-\frac{t}{r}} dt = 0 }$, or,

$\displaystyle{ \left( u \left( z,0 \right) \right)^2 + \frac{d}{dt} f \left( z \right) = 0 }$

But $\frac{d}{dt} f \left( z \right) = 0$, so,

$ \left( u \left( z,0 \right) \right)^2 = 0 \implies u \left( z,0 \right) = 0 $

$$ u \left( z,0 \right) =0 \: \dots \: \left( 4 \right) $$

But something is wrong in the above derivations because according to the above result, $f \left( z-1 \right) = u \left( z,0 \right) = 0$ which yields a constant function.

This means that our assumptions in the beginning were wrong. However, is there a way to proceed with the problem (of expressing logistic maps non-recursively) along similar lines of thinking?

答:
Taking a look at the assumption $ f ( z ) = \mathcal L _ t \{ u ( z , t ) \} ( s ) $, one can see that $ s $ does not appear on the left-hand side of the equation, which means that the Laplace transform of $ u $ is assumed to be constant (when talking about a fixed $ z $). It is well-known that no such function exists, and you need to generalize the notion of functions to find candidates for $ u $. Doing that, it is also well-known that $ u $ must be a multiple of Dirac $ \delta $ function. More precisely, you must have $ u ( z , t ) = f ( z ) \delta ( t ) $.

The point is that generalized functions do not exactly act like functions. Dirac $ \delta $ function, for example, on one hand acts like a function that that take the value zero at any point other than the origin, but on the other hand, has positive integral; for any $ \epsilon > 0 $ you have $ \int _ { - \infty } ^ { - \epsilon } \delta ( t ) \ \mathrm d t = \int _ \epsilon ^ { + \infty } \delta ( t ) \ \mathrm d t = 0 $, while $ \int _ { - \infty } ^ { + \infty } \delta ( t ) \ \mathrm d t = 1 $. This cant happen for any genuine function, even if you allow the function to take the value $ + \infty $; considering a function $ f : ( - \infty , + \infty ) \to [ - \infty , + \infty ] $ such that $ f ( x ) = 0 $ for any $ x \in ( - \infty , 0 ) \cup ( 0 , + \infty ) $ and $ f ( 0 ) = + \infty $, you still have $ \int _ { - \infty } ^ { + \infty } f ( t ) \ \mathrm d t = 0 $, in the sense of Lebesgue integration.

That said, it should be obvious that you have to avoid treating $ u $ like a function that takes definite values at given points $ z $ and $ t $. As long as you're keeping a distance from $ t = 0 $ though, it acts like a constant zero function, which may explain why you ended up with $ u ( z , 0 ) = 0 $: because $ u ( z , \epsilon ) $ behaves like the constant zero function for $ \epsilon \ne 0 $, even for arbitrarily small values of $ \epsilon $. But note that this explanation is rather loose, as $ u ( z , 0 ) $ has no meaning, because Dirac $ \delta $ function has no definite value at the origin.


问:
prove closure $\overline{f(V)}$ is a neiborhood of origin in Topological vector space
Let X,Y be topological vector space,Y is Baire space,let $f$ be a continuous linear map $f:X\to Y$,assume $f(X)$ has non empty interior.

Prove given any neiborhood $V$ of origin for $X$, $\overline{f(V)}$ is neiborhood of origin for $Y$

(I think my proof is almost done,but not very clear)
My attempt,first since $f(X)$ has non empty interior the map is onto,now the question is how to use the fact that $Y$ is Baire space?I know we need to construct something like $\bigcup U_n$ which each of them is closed and no interior.But I don't know how to construct this sequence $U_n$ (It may be an ascending sequence?)

each of them seems be the set $\overline{f(V)}$ which is closed,and assume no interior contains in them,then if countable union to covers $Y$.it must has empty interior,hence contradiction?

To construct the countable union,note that any $V$ exist balanced absorbing neiborhood $A$ contains in it if we can show for this $A$ we have $\overline{f(A)}$ is neiborhood of origin,then it will also holds for $\overline{f(V)}$ Since absorbing we know for this $A$ ,$\bigcup_n nA$ covers $X$ hence $f(\bigcup_n nA) = Y $

To complete the proof,we need to show to fact:

if $\overline{f(A)}$ contains no interior then $n\overline{f(A)}$ also contains no interior,since they are homeomorphic.

if zero is not interior for $\overline{f(A)}$ then all point can not be interior for it.

答:
You are right that since $f(X)$ (which is a linear subspace of $Y$) has no empty interior then $f(X)=Y$: indeed, let $y_0\in f(X)$ so that there exists an open neighborhood $V\subset f(X)$. If $y\in Y$ then we can take the points $y_0+\lambda y$, $\lambda\in\mathbb{C}$. As $\lambda\to0$ we have that $y_0+\lambda y\to y_0$, so there exists some non-zero $\lambda\in\mathbb{C}$ so that $y_0+\lambda y\in V$. But then $y_0+\lambda y\in f(X)$, so $y\in f(X)$.

We give a definition: If $Z$ is a vector space and $E\subset Z$ we say that $E$ is absorbent when $Z=\bigcup_{r>0}rE$. We say that $E$ is balanced when for any $x\in E$ and $\alpha\in\overline{\mathbb{D}}$ we have that $\alpha x\in E$.

Note that if $Z$ is a TVS and $E$ is balanced then $\text{int}(E)$ is also balanced: indeed, if $x\in\text{int}(E)$ then there exists an open $V\subset E$ so that $x\in V$. If $\alpha\in\mathbb{C}$ with $|\alpha|\leq1$, we have that $\alpha\cdot x\in\alpha\cdot V\subset E$ and $\alpha\cdot V$ is open, thus $\alpha\cdot x\in\text{int}(E)$.

Lemma: Let $Z$ be a Baire TVS. If $E\subset Z$ is absorbent, then $E$ is a Baire space with the subspace topology.

Proof of the lemma: Let $(F_n)$ be a sequence of closed sets such that $E\subset\bigcup_nF_n$. For $k\in\mathbb{N}$ we have that $kE\subset\bigcup_nkF_n$, therefore since $Z=\bigcup_kkE$ (because $E$ is absorbent) we have that $Z=\bigcup_{n,k}kF_n$. Since $Z$ is a Baire space, there exists a pair of integers $k,n$ so that $\text{int}(kF_n)\neq\emptyset$. But $kF_n$ is homeomorphic to $F_n$, so $\text{int}(F_n)\neq\emptyset$. This shows that $E$ is a Baire space.

Claim: Let $X$ be a TVS and let $V$ be a neighborhood of $0$ in $X$. Then $V$ contains a balanced neighborhood of $0$.

Proof of the claim: Indeed, since scalar multiplication is a continuous operation $m:\mathbb{C}\times X\to X$ we have that $m^{-1}(V)$ is open in $\mathbb{C}\times X$ and contains $(0,0)$ so there exists $r>0$ and a neighborhood $W\subset X$ of $0$ such that $D(0,r)\times W\subset m^{-1}(V)$, so $m(D(0,r)\times W)\subset V$. But $m(D(0,r)\times W)=\{\alpha\cdot x: |\alpha|<r, x\in W\}$ is open, contains $0$ and if $\beta\in\mathbb{C}$ satisfies $|\beta|\leq1$ then for any $\alpha\in D(0,r)$ and $x\in W$ we have that $\beta\cdot(\alpha\cdot x\in m(D(0,r)\times W)$, since $|\alpha\beta|\leq r$.

So let $V$ be a neighborhood of $0$ in $X$ and take $U$ a balanced neighborhood of $0$ so that $U\subset V$. Now let $y\in Y$. As we said, $f$ is onto, so $y=f(x)$ for some $x\in X$. Since $\lambda x\to0$ as $|\lambda|\to0$, we have that there exists some $\lambda\in\mathbb{C}$ so that $\lambda x\in U$. So $y=\frac{1}{\lambda}f(\lambda x)\in\frac{1}{\lambda}f(U)\subset\frac{1}{\lambda}\overline{f(U)}$ and therefore $\overline{f(U)}$ is absorbent in $Y$ and it is also balanced, since $U$ is balanced. By the lemma we have that $\overline{f(U)}$ is a Baire space, so $\text{int}(\overline{f(U)})\neq\emptyset$. But since $\overline{f(U)}$ is balanced, its interior is also balanced, thus $0\in\text{int}(\overline{f(U)})\subset\text{int}(\overline{f(V)})$ which is what we wanted to show.


问:
2-colorable belongs to $\mathsf P$
To show that 2-colorable belongs to $\mathsf P$, I have a straightforward mental description in mind that I don't think will be considered as a formal proof. Hence I am interested to know how this must be said as an answer to this question. Here's what I think: Say we have 2 colors and n vertices. 2-coloring will be applied such that no two adjacent vertices have the same color. If a vertex is in color A, the other is B and so on. Is my reasoning correct?

Appendix: A graph G is said to be k-colourable if and only if a k-colouring of G exists.

A k-colouring is an assignment of k colours to the vertices of a graph G such that no edge joins two vertices of the same colour.

答:
You have the idea, but "the other" what?
One should present a more explicit algorithm and then investigate if it is polynomial. So, to elaborate your idea

Prepare an empty queue $Q$ and assign colour $0$ to each vertex.

If $Q$ is empty go to step 7; otherwise pop $(v,c)$ from $Q$

If the current colour of $v$ is $c$, go to step 2; if it is $-c$, the algorithm terminates: No 2-colouring exists

Assign colour $c$ to vertex $v$

Enumerate all neighbours of $v$ (that is all edges $vw$ with one end $v$). For each neighbour $w$ of $v$, push $(w,-c)$ to $Q$

Go to step 2.

Select any vertex $x$ that has not been coloured yet. If no such $x$ exists the algorithm terminates and the graph is 2-coloured.

push $(x,+1)$ to the queue $Q$ and go to step 2.

Considering memory, a little refinement of the above can get along with one bit plus one pointer per vertex to realize the queue. Can you determine the time complexity of the algorithm?

Title: 2-colorable belongs to $\mathsf P$


答:
I will use the one-step subgroup test.

Clearly $\varnothing\neq S\subseteq GL_n(K)$.

Let $A,B\in S.$ Then $AJA^t=J=BJB^t$. Observe that
$$\begin{align}
B^{-1}J(B^t)^{-1}&=B^{-1}BJB^t(B^t)^{-1}\\
&=J.
\end{align}$$

We aim to show that $AB^{-1}\in S$. Indeed,

$$\begin{align}
(AB^{-1})J(AB^{-1})^t&=(AB^{-1})J((B^{-1})^tA^t)\\
&=A(B^{-1}J(B^t)^{-1})A^t\tag{1}\\
&=AJA^t\\
&=J.
\end{align}$$

Hence $S\le GL_n(K)$.

$(1)$: See here.

Title: Show that $S=\{ A\in GL_n(K)\mid AJA^t=J\}$ is a subgroup of $GL_n(K)$, where $J\in K^{n\times n}$.

Body:
I have a question about the following problem: show that $$S=\{ A \in GL_n(K) \mid AJA^t = J \}$$ is a subgroup of $GL_n(K)$, where $J\in K^{n\times n}$.

I have shown that the identity element $e$ is part of $S$, but I don't know to prove the second criteria for a subgroup. Thanks in advance!


问:
What is the difference between a Partial Order Set and a Group
According to my what I read everywhere...

A partially ordered set (also poset) formalizes and generalizes the intuitive concept of an ordering, sequencing, or arrangement of the elements of a set. A poset consists of a set together with a binary relation indicating that, for certain pairs of elements in the set, one of the elements precedes the other in the ordering.

And,

A group is a set equipped with a binary operation that combines any two elements to form a third element in such a way that three conditions called group axioms are satisfied, namely associativity, identity and invertibility.

So in both cases there is a set with respect to a binary operation. Yet in posets we explore topics such as lattice and boolean algebra while in group theory there are semi-groups, monoids etc.

I just dont get the subtle difference between the two.

答:
No, these are very different structures with very different requirements.

An operation is different than a relation. An operation is a function $A \times A \to A$ while a relation is simply a subset of $A \times A$. We can think of any relation also as a function $A \times A \to \{0, 1\}$. This way it becomes obvious that an operation is not the same as a relation.

To be more concrete for us to have a group noone requires from us to have ordering (even partial) between the elements of the group. While for a partially ordered set this ordering requirement is essential.


问:
sum of the eigenvalues = trace($A$)?
Is it true that for a square matrix $A$, all of whose eigenvalues exist in the base field, sum of the eigenvalues = trace($A$)?

The result holds in all the matrices I've studied.

答:
Let $A \in M_n(k)$ and let $\overline{k}$ be an algebraic closure of $k$. Because the charateristic polynomial $p_A$ of $A$ splits into linear factors in $\overline{k}[X]$, $A$ is trigonalizable in $M_n(\overline{k})$: there exists $P \in GL_n(\overline{k})$ such that $A=PTP^{-1}$ for some triangular matrix $T \in M_n(\overline{k})$. Now $\text{tr}(A)=\text{tr}(T)$ is the sum of the eigenvalues of $T$, and a fortiori of $A$.

Notice that $\text{tr}(A)$ is a symmetric polynomial with respect to the eigenvalues of $A$ so you can express $\text{tr}(A)$ with respect to the elementary symmetric polynomials and finally with respect to the coefficients of $p_A$. Because $p_A \in k[X]$, $\text{tr}(A) \in k$.

Title: sum of the eigenvalues = trace($A$)?


问:
Meaning of integration with respect to a multivariable function
I'm learning the joint distribution of 2 random variables and I see the notation below:   $$\int_{[0,1]^2} xy\,dF(x,y)$$

I would like to ask 2 questions:

What does it mean to integrate with respect to $dF(x,y)$ ? Is the above notation a double integration ?

Could you please suggest me some books or some keyword so that I could read about the meaning and the underlying theory of the above notation (especially the part with respect to dF(x,y))?

If I set:
$f(x,y) = \frac{\partial F(x,y)} {\partial x .\partial y}$

Can I replace $dF(x,y)$ by $f(x,y) \, dxdy$ to get the equation:

$$\int_{[0,1]^2} xy\,dF(x,y) = \int_0^1 \int_0^1 x\,y \, f(x,y) \, dxdy$$
If yes/no, could you please explain why can (can't) we do this?

答:
The total derivative looks like this:
$$dF(x,y)=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy$$
so maybe it meant this?


问:
Is my symbolic representation of "nobody loves everybody" correct?
Suppose we want to represent "nobody loves everybody" and the universe of discourse is all people. Would a correct representation be:

$$\require{enclose}\enclose{horizontalstrike}{(\forall x)(\sim \!x\Rightarrow(\forall y)(y\Rightarrow (\sim\!x) \; \text{loves} \; y )) \; \; \; \text{(Universe is all People)}}$$

Edit: I made a second attempt

$$\sim\!\!(\forall x)(x\Rightarrow(\forall y)(y\Rightarrow x \; \text{loves} \; y )) \; \; \; \text{(Universe is all People)}$$

According to the answer key, the answer is

$$(\forall x)\!(\sim\!\!(\forall y)(x \; \text{loves} \; y))$$

Am I correct? If not then why?

答: Let $L(x,\,y)$ mean "$x$ loves $y$". Working with quantifiers over people, the statement is $\neg\exists x\forall y(L(x,\,y))$ or something equivalent. The cleanest notation may be $\forall x\exists y(\neg L(x,\,y))$.


问:
Counterexample on two identical random variables with Poisson sum
Raikov's theorem states that if $X_{1}$ and $X_{2}$ are two independent random variables, such that $X_{1}+X_{2}$ has a Poisson distribution, then these two variables also have Poisson distribution.

Is the same true when $X_{1}$ and $X_{2}$ are not necessarily independent, but identically distributed and have a symmetric joint probability distribution, i.e., $\mathbb{P}(X_{1} = i, X_{2} = j) = \mathbb{P}(X_{1} = j, X_{2} = i)$ for all $i,j$?

答:
The fact that $X_1$ and $X_2$ have a symmetric joint distribution implies that they are identically distributed because
$$
P(X_1 = i) = \sum_{j \ge 0} P(X_1 = i, X_2 = j) = \sum_{j \ge 0} P(X_2 = i, X_1 = j) = P(X_2 = i).
$$
If you see the joint distribution of $X_1$ and $X_2$ as a non-negative function on $\mathbb N^2$, then the distribution of $X_1$ is determined by adding up entire rows and the distribution of $X_2$ by adding up entire columns. The symmetry implies equality of those two distributions.

So if you see it that way, you basically have a positive function on $\mathbb N^2$ whose sum along diagonals $i+j = k$ for $k$ fixed is determined (by $e^{-\lambda} \lambda^k / k!$), but on this diagonal the only restriction is symmetry. One way to achieve this without having $X_1$ or $X_2$ follow Poisson distributions is to consider the following joint distribution:
$$
\mathbb P(X_1 = i, X_2 = j) = \begin{cases}
0 & \text{ if } 0 < i,j  0 \\
\frac{e^{-\lambda} \lambda^j}{2j!} & \text{ if } j = 0, i > 0 \\
e^{-\lambda} & \text{ if } i=j=0.
\end{cases}
$$
This gives the following distribution for $X_1+ X_2$:
$$
\forall k > 0, \quad \mathbb P(X_1+X_2 = k) = \underset{i+j=k}{\sum_{i,j \ge 0}} \mathbb P(X_1=i,X_2=j) \\
= \mathbb P(X_1=0, X_2=k) + \mathbb P(X_1=k, X_2=0) \\
= 2 \left( \frac{e^{-\lambda} \lambda^k}{2k!} \right) \\
= \frac{e^{-\lambda} \lambda^k}{k!}
$$
and for $k = 0$, clearly $\mathbb P(X_1+X_2 = 0) = \mathbb P(X_1=0,X_2=0) = e^{-\lambda}$. This means $X_1+X_2 \sim \mathrm{Poi}(\lambda)$, and clearly the joint distribution of $X_1$ and $X_2$ is symmetric by construction, so in particular they are identically distributed by the above argument. But
$$
\forall i > 0, \quad \mathbb P(X_1 = i) = \sum_{j \ge 0} \mathbb P(X_1=i, X_2 = j) = \mathbb P(X_1=i, X_2=0) = \frac{e^{-\lambda} \lambda^i}{2i!}
$$
and for $i = 0$,
$$
\mathbb P(X_1 = 0) = \sum_{j \ge 0} \mathbb P(X_1=0, X_2 = j) \\
= e^{-\lambda} + \frac 12 \left( \sum_{j \ge 1} \frac{e^{-\lambda} \lambda^j}{j!}\right) \\
= e^{-\lambda} + \frac 12 (1 - e^{-\lambda} ) \\
= \frac{1+e^{-\lambda}}2.
$$

Clearly $X_1$ is not Poisson. You can also clearly see from this argument that the space of distributions $X_1$ satisfying $X_1 + X_2 = \mathrm{Poi}(\lambda)$ where $X_2$ has the same distribution as $X_1$ is an infinite-dimensional manifold with boundary since you can do whatever you want with the values $p_{ij}$ for $i,j \ge 0$, $i+j=k$ and $i \le \lfloor k/2 \rfloor$, as long as these values give you the corresponding required sum of $e^{-\lambda} \lambda^k/k!$.

Hope that helps,

Body:
The fact that $X_1$ and $X_2$ have a symmetric joint distribution implies that they are identically distributed because
$$
P(X_1 = i) = \sum_{j \ge 0} P(X_1 = i, X_2 = j) = \sum_{j \ge 0} P(X_2 = i, X_1 = j) = P(X_2 = i).
$$
If you see the joint distribution of $X_1$ and $X_2$ as a non-negative function on $\mathbb N^2$, then the distribution of $X_1$ is determined by adding up entire rows and the distribution of $X_2$ by adding up entire columns. The symmetry implies equality of those two distributions.

So if you see it that way, you basically have a positive function on $\mathbb N^2$ whose sum along diagonals $i+j = k$ for $k$ fixed is determined (by $e^{-\lambda} \lambda^k / k!$), but on this diagonal the only restriction is symmetry. One way to achieve this without having $X_1$ or $X_2$ follow Poisson distributions is to consider the following joint distribution:
$$
\mathbb P(X_1 = i, X_2 = j) = \begin{cases}
0 & \text{ if } 0 < i,j < i+j \\
\frac{e^{-\lambda} \lambda^i}{2i!} & \text{ if } i = 0, j > 0 \\
\frac{e^{-\lambda} \lambda^j}{2j!} & \text{ if } j = 0, i > 0 \\
e^{-\lambda} & \text{ if } i=j=0.
\end{cases}
$$
This gives the following distribution for $X_1+ X_2$:
$$
\forall k > 0, \quad \mathbb P(X_1+X_2 = k) = \underset{i+j=k}{\sum_{i,j \ge 0}} \mathbb P(X_1=i,X_2=j) \\
= \mathbb P(X_1=0, X_2=k) + \mathbb P(X_1=k, X_2=0) \\
= 2 \left( \frac{e^{-\lambda} \lambda^k}{2k!} \right) \\
= \frac{e^{-\lambda} \lambda^k}{k!}
$$
and for $k = 0$, clearly $\mathbb P(X_1+X_2 = 0) = \mathbb P(X_1=0,X_2=0) = e^{-\lambda}$. This means $X_1+X_2 \sim \mathrm{Poi}(\lambda)$, and clearly the joint distribution of $X_1$ and $X_2$ is symmetric by construction, so in particular they are identically distributed by the above argument. But
$$
\forall i > 0, \quad \mathbb P(X_1 = i) = \sum_{j \ge 0} \mathbb P(X_1=i, X_2 = j) = \mathbb P(X_1=i, X_2=0) = \frac{e^{-\lambda} \lambda^i}{2i!}
$$
and for $i = 0$,
$$
\mathbb P(X_1 = 0) = \sum_{j \ge 0} \mathbb P(X_1=0, X_2 = j) \\
= e^{-\lambda} + \frac 12 \left( \sum_{j \ge 1} \frac{e^{-\lambda} \lambda^j}{j!}\right) \\
= e^{-\lambda} + \frac 12 (1 - e^{-\lambda} ) \\
= \frac{1+e^{-\lambda}}2.
$$

Clearly $X_1$ is not Poisson. You can also clearly see from this argument that the space of distributions $X_1$ satisfying $X_1 + X_2 = \mathrm{Poi}(\lambda)$ where $X_2$ has the same distribution as $X_1$ is an infinite-dimensional manifold with boundary since you can do whatever you want with the values $p_{ij}$ for $i,j \ge 0$, $i+j=k$ and $i \le \lfloor k/2 \rfloor$, as long as these values give you the corresponding required sum of $e^{-\lambda} \lambda^k/k!$.

Hope that helps,


问:
Identifying the contrapositive of this statement $x>0$, then $∃_{y>0}$ such that $x>y$
Statement: If $x>0$, then $∃_{y>0}$ such that $x>y$. (x and y are real numbers)

Contrapositive: If $∀_{y>0}$ $x<y$, then $x≤0$.

Is the contrapositive correct?

答:
Your first statement should be

Let $ x\in \Bbb R$.

$$x>0\implies (\exists y>0 : x>y)$$

The contrapositive is

$$(\forall y>0\;\; x\le y) \implies x\le 0$$


问:
Find Value of Y given one Stationary Point and Equation with Unknown Variable
I am doing a question from my Differentiation work.

Given that y = $x^3-5x^2+kx$ has a stationary point where x = 2. Find value of y coordinate at stationary point.

my knowledge so far for this question:
$dy/dx$ = 0 (gives stationary points)

Any help will be appreciated also any recommendations of good resources to learn math from? I am in high school

答: To find the value of y when x=2 (written in function of k) it has to substitute the value x=2 to the expression of $y=x^3-5x^2+kx=8-20+2k=-12+2k$. To eliminate the k parameter we need another equation: we can use the information that the first derivative has to be $0$ when x=2. So $\frac{dy}{dx}=3x^2-10x+k=0$ when $x=2$ so: $-8+k=0$ $\to$ $k=8$. Now, we substitute $k$ to the previous equation for $y$: $y=-12+2\cdot 8=4$.


问:
Finding the determinant of $2A+A^{-1}-I$ given the eigenvalues of $A$

Let $A$ be a $2\times 2$ matrix whose eigenvalues are $1$ and $-1$. Find the determinant of $S=2A+A^{-1}-I$.

Here I don't know how to find $A$ if eigenvectors are not given. If eigenvectors are given, then I can find using $A=PDP^{-1}$. Please solve for $A$; the rest I can do.

Body:
Hint:

If $A=PDP^{-1}$ then  $S=P(2D+D^{-1}-I)P^{-1}$.


问:
$ u(x) = g(x) + \int_0^x f(x,t, u(t)) dt $
$f: [0,k] \times [0,k] \times R \rightarrow R\ \ \ $ is continous in $ \{ (x,t,z) \in [0,k] \times [0,k] \times R , 0 \leq t \leq x \leq k \} $

$ g: [0,k] \rightarrow R \ \ $ is continous

And $\ \ \ | f(x,t,z) - f(x,t,y)| \leq C |z-y| \ \ \ \ $ for a $ 0 \leq C $

Now i have to show that the equation:
$$ u(x) = g(x) + \int_0^x f(x,t, u(t)) dt $$
has exactly one continous solution in $[0,k]$.

I have tried using the Banach fixed-point theorem but i dont know how.
is it the right approach at all?

答:
Let $X =  C[0, k]$. Let us define $\Phi:X \rightarrow X$ by
\begin{align}
\Phi(w)(x) = g(x) = \int^x_0 f(x, t, w(t))\ dt.
\end{align}
Indeed, $\Phi$ maps $X$ to $X$, since for any $0\le x \le y\le k$
\begin{align}
|\Phi(w)(x)-\Phi(w)(y)| \le&\ |g(x)-g(y)| + \left|\int^x_0 f(x, t, w(t))\ dt-\int^y_0 f(y, t, w(t))\ dt \right|\\
\le&\  |g(x)-g(y)| + \int^y_0|f(x, t, w(t))-f(y, t, w(t))|\ dt + \int^x_y|f(x, t, w(t))|\ dt \\
\le&\  |g(x)-g(y)| + \int^y_0|f(x, t, w(t))-f(y, t, w(t))|\ dt + \|f\|_\infty|x-y|.
\end{align}
Hence, by continuity of $f$, we see that $\Phi(w)(x)$ is also continuous which means $\Phi$ maps $X$ to $X$.

Next, let us show that $\Phi: X\rightarrow X$ is a contraction on $X$. Notice that

\begin{align}
\|\Phi(u)-\Phi(w)\|_\infty =&\ \sup_{x \in [0, k]}|\Phi(u)(x)-\Phi(w)(x)|\\
\le&\ \sup_{x \in [0, k]} \int^x_0 |f(x, t, u(t)) - f(x, t,w(t))| \ dt\\
\le&\ C\sup_{x, \in [0, k]} \int^x_0 |u(t) -w(t)|\ dt\\
\le&\ C k \|u -w\|_\infty.
\end{align}

However, we can't conclude that $\Phi$ is a contraction since $Ck$ doesn't necessarily need to be less than $1$. So, instead of proving $\Phi$ is a contraction on $X$, we can prove that $\Phi$ is a contraction on $[0, k_1]$ for some $k_1<k$ and $Ck_1<1$. Then, by Banach fixed point theorem, we know that there exists a unique $u$ defined on $[0, k_1]$ such that
\begin{align}
u(x) = g(x) + \int^x_0 f(x, t, u(t))\ dt
\end{align}
for $x \in [0, k_1]$. Repeat the argument on the interval $[k_1, 2k_1]$. Since $[0, k]$ is compact, then we only need to repeat the argument a finite number of times. This shows that there is a unique $u \in C[0, k]$ that solves the above integral equation.


问:
this is equivalent to the condition that all nodes in some cut set of the collection of all paths leading
I can't understand what a'cut set' is, can you please explain in simple form?

A node is implicitly granted in X mode if all of its parents are (implicitly or explicitly) granted to the transaction in X mode. By induction, this is equivalent to the condition that all nodes in some cut set of the collection of all paths leading from the node to the roots of the graph are explicitly granted to the transaction in X mode and all ancestors of nodes in the cut set are explicitly granted in IX or SIX mode.
Jim Gray; Raymond A. Lorie; G. R. Putzolu; Iriving L. Traiger (1976). "Granularity of locks and degrees of consistency in a shared data base".

答:
The basic notion is that of a graph cut, but "cut set" seems to be used in a slightly different way in your article, where it refers to a vertex set instead of an edge set.

In any case, the idea is that the recursively-defined condition, that $n$ is IGX if each of its parents is either EGX or IGX, is equivalent to the (non-recursively-defined) condition that every path from $n$ to a root goes through an EGX node (abbreviating "explicitly/implicitly granted in X mode" as EGX/IGX). The cut set is the set of these EGX nodes.


问:
Conditional expectation of the number of tails when coins are removed after $m$ tails
In a coin tossing game, I have access to infinite number of unfair coins. The game follows as described below.

We take a coin and toss it till we get $m$ tails. Once we get $m$ tails, we remove that coin, sample a new coin and continue the game. The game continues till total number of trials hit $n$. I am trying to find the expected number of tails conditioned on the number of coins used.

The random variable $Y_i=1$ indicates $i^{th}$ coin is the last coin with $m$ tails. The random variable $X_i = 1$ indicates the probability of selecting the $i^{th}$ coin.
Let $C$ be the random variable denoting the number of coins used and for $C=c$, the expected number of tails is

$$
\mathbb{E}(T|C=c) = m\big(P(X_1=1) + P(X_2=1,Y_1=1) +  \cdots + P(X_{c-1} = 1,Y_{c-2}=1)\big) + (n-m(c-1))P(X_c = 1,Y_{c-1}=1)
$$

Now,
$$
P(Y_1=1) = P(X_1=1) = \frac{c}{c+1} \\
P(Y_2=1) = P(X_2 = 1,Y_1=1) = P(X_2=1|Y_1=1)P(Y_1=1) = \frac{c-1}{c} \frac{c}{c+1} = \frac{c-1}{c+1}\\
P(Y_3=1) = P(X_3 = 1, Y_2=1) = P(X_3=1|Y_2=1)P(Y_2=1) = \frac{c-2}{c-1}\frac{c-1}{c+1} = \frac{c-2}{c+1}$$

Following this way, I get

$$
\mathbb{E}(F|C=c) = m\Big(\frac{c}{c+1} + \frac{c-1}{c+1} \dots \frac{2}{c+1}\Big) + n-(c-1)m\frac{1}{c+1}
$$

Is this calculation correct ? Thanks in advance

答:
For convenience I will use $c$ for the number of removed coins. Assuming the coins being fair the probability of $k$ tails is proportional to the number of ways to permute them with heads. Besides we know that $mc\le k<m(c+1)$. Therefore the expected number of tails is:
$$
mc+\frac{\sum_{i=0}^{m-1}i\binom n{mc+i}}{\sum_{i=0}^{m-1}\binom n{mc+i}}.
$$


问:
Sparsity of $L^1$ Norm Regularization vector
Consider a generic optimal control problem
$$
\min\limits_{u(t)}\int\limits_0^T f(x(t), u(t))\, dt + \mathcal{R}(u)
$$
where the evolution of $x(t)$ is given by a generic ODE system $\dot x(t) = g(x(t),u(t))$. In my case functions $f$ and $g$ do not depend on time. I used the following expressions
$$
\mathcal{R}_2(u) = \frac{1}{2}\eta\:\|u\|_2^2
\qquad
\text{and}
\qquad
\mathcal{R}_1(u) = \frac{1}{2}\eta\:\|u\|_1
$$
as regularization terms. Numerically speaking I used Forward-Backward sweep method. Consider the problem with $\mathcal{R}_2(u)$. At each FB iteration I compute gradients and use gradient descent method. Instead with the other expression $\mathcal{R}_1(u)$ to minimize the functional at each iteration I used MATLAB function $\texttt{fminunc}$.

Then I would like to compare the results to know if using $\ell^1$ norm I can reach some sparsity of the control vector. So let $u$ be the solution of the optimal control problem using $\mathcal{R}_2(u)$ and let $v$ be the solution using $\mathcal{R}_1(u)$. A first idea to check the sparsity of the vectors was to compare the two $\ell^1$ norms. In other words compare $\|u\|_1$ vs $\|v\|_1$. Using this method the norms are really near for certain parameters of the problem, instead for other parameters the $\ell^1$ norm seems to be lower. Another idea was to check entries inside $v$ if some of them are equal to $0$ or below a certain threshold.

Is there a particular method to study the sparsity or to compare the terms of the two control vectors $u$ and $v$?

答:
I'll assume the problem has been discretized so that the optimization variable is a vector $u \in \mathbb R^n$. Your optimization problem has the form
$$
\tag{1} \text{minimize} \quad f(u) + g(u)
$$
where $f$ is differentiable and $g$ is a "simple" regularization term. I'll assume that $g(u) = \eta \| u \|_1$.

It is common to solve problems of the form (1) with the forward-backward method, also called the proximal gradient method. The iteration for the proximal gradient method with step size $t > 0$ is
$$
u^{k+1} = \text{prox}_{tg}(u^k - t \nabla f(u^k)).
$$
In words, we take a step in the direction of steepest descent for $f$, then we apply the proximal operator of $g$, which is defined by
$$
\text{prox}_{tg}(\hat u) = \arg \min_u g(u) + \frac{1}{t} \|u - \hat u \|_2^2.
$$
If $g(u) = \eta \|u\|_1$, then it can be shown that the proximal operator of $g$ simply shrinks each component of $\hat u$ towards the origin by a distance $t \eta$. (If you hit the origin, you stop.)

I believe if you use this optimization algorithm you'll obtain a truly sparse solution to your optimization problem.


问:
Is there any way to know the numbers which are NOT a term in a particular arithmetic progression??
For example say: 5n+2, 5=d, n = x-1 (for any x), a = 2. So, we can know that 3,5,8, etc. are definitely not terms in this progression, by hit and trial. But, is there any direct way by which can represent the terms which are NOT any term in the Arithmetic Progression?? The particular progression I'm looking for is:

3n+1 + (x-1)(2n+1). I want to know the numbers which are not a part of this progression. Thank you.

This is my first time posting here, please bare with me and spare me for my naivety.

答:
OK, given your response to my question, I think what you're asking is better rephrased like this:

Consider the arithmetic progression $a_x = a_1 + (x-1)d$, for $x \in \mathbb{Z}^+$. Define the progression using $a_1 = 3n + 1$ and $d = 2n + 1$. We can write an infinite number of progressions for all $n \in \mathbb{Z}^+$.

Define $A_n$ as the set of numbers in the progression for each $n$. Then $A$ is the union of all of those sets--the integers whose values are part of at least of of the progressions. Now we ask: what is left if we remove $A$ from $\mathbb{Z}^+$ ?

(I specified positive integers because if we allow $n = 0$, then the progression is the trivial $1 + (x-1) \cdot 1$, which is simply all of the non-negative integers.)

I like to just look at the first few possible values to get a feel for things. For $ n = 1, 2, 3, 4, 5$ we have

$$ A_1 = \{4,7,10,13, 16...\} \\
A_2 = \{7, 12, 17, 22, 27...\} \\
A_3 = \{10, 17, 24, 31, 38...\} \\
A_4 = \{13, 22, 31, 40, 49...\} \\
A_5 = \{16, 27, 38, 49, 60...\}$$

Huh. There's a quite pleasing symmetry in there! In all cases, $a_x$ in $A_n$ is also $a_n$ in $A_x$! Maybe we can use that to our advantage later.

What else can we see in here? Well, all of these progressions increase by odd numbers, which makes sense, because since $2n + 1$ is always odd. But that means every one of these numbers has to be the sum of $4$ and some subset of the odd numbers $2n + 1 \geq 3$. Hey hey hey, now we're getting somewhere. In fact, we could probably stop here if we wanted, and say "All numbers not expressible as the sum of 4 and some list of odd numbers (not necessarily different) cannot be part of any of these progressions." Maybe. But is that right?

Well, I don't see $9$ anywhere up there, so we can't add just $4+5$. There must be some boundaries to what we can do. Let's collect some terms, though, to make some things easier. If we have added $m$ odd numbers to $4$, we can say that all the values in $A$ must be:

$$a = 4 + (2k_1 +1) + (2k_2 + 1) + \cdots + (2k_m +1) = 4+ 2\sum k_i+m = 4 + 2K + m$$

Where $K$ is just the sum of the sequence $k_n$, and the sequence has length $m$. Let's make that initial evaluation of the sets into a diagram, taking advantage of symmetry to use just a triangle:

$\require{AMScd}$
\begin{CD}
4 @>1>>7@>1>>10@>1>>13@>1>>16@>1>>19\\
@.@VV2V@VV3V@VV4V@VV5V@VV6V\\
@.12@>>2>17@>>2>22@>>2>27@>>2>32\\
@.@.@VV3V@VV4V@VV5V@VV6V\\
@.@.24@>>3>31@>>3>38@>>3>45\\
@.@.@.@VV4V@VV5V@VV6V\\
@.@.@.40@>>4>49@>>4>58\\
@.@.@.@.@VV5V@VV6V\\
@.@.@.@.60@>>5>71\\
@.@.@.@.@.@VV6V\\
@.@.@.@.@.80\\
\end{CD}

Whoa-hoa-hoakay, that looks complex. But it's not, really. The main numbers are what we had before, though I extended it on the right. The arrows are the "moves" we can make to get to another number, and the number on each arrow is a $k$. Each "move" adds $2k_i +1$.
Note that every row and column of arrows has the same $k_i$ as all the others in that row or column.

Now we can decide on $m$, and see what possible sequences of moves are available.

For $m = 0$, well that's just zero moves, not interesting.
For $m = 1$, we only have one choice: $k_1 = 1$.
For $m = 2$, we start having choices. Our sequence can be $(1,1)$ or $(1,2)$. Let's look at the next few values of $m$:

$$\begin{array}{cccccccc}
m=3& (1,1,1) & (1,2,2) & (1,1,3) \\
m = 4 & (1,1,1,1) & (1,1,1,4) & (1,2,2,2) & (1,2,2,3) & (1,1,3,2) & (1,1,3,3) \\
\end{array}$$

Well that'll get big really fast. But we can get something from it. For $m = 1, K = 1$. For $m = 2, K \in \{2,3\}$. For $m = 3, K \in \{3,5\}$, and for $m = 4, K \in \{4,7,8\}$.

And now... I'm going to leave you hanging just a bit. Figure out for yourself what values of $K$ are allowed for the next few values of $m$, up to maybe $m=7$ or $8$ (because typing them here will take a long time). You'll have to extend the diagram to the right once or twice. Then look for the pattern. And there is a pattern!

(Hint: you'll need another variable, I called mine $i$, and you'll end up writing K in terms of $i$ and $m$. The possible values of $i$ are limited by $m$ in some way.)

I'll come back in a couple of days in hopes you'll find the pattern!

Alas, you didn't come back to comment... but here is the remainder of the solution:

If you look at the numbers that pop out for various values of $m$, two values are always there: $m$ itself and $2m-1$. Then we see other initially random-appearing values, like $2m$ for $m=4$. But wait... $2m = 3m - 4$ when $m=4$. Could there be something here dealing with squares? $m=4$ has $4,7,8$ available. $7 = 4m - 9$... and $4 = 5m - 16$, even!

If we consider each of these cases separately, what we find is that for $1 \leq i \leq m+1$, we can write $K = im - (i-1)^2$. That means we can take our old friend $a = 4 +2K + m$ and rewrite:

$$a = 4 + 2\left(im - (i-1)^2\right) -m \\
a = 2+m+4i+2mi-2i^2$$

For $m \geq 0$ and $m+1 \geq i \geq 1$. The $m+1$ in the last constraint allows us to use $m = 0, i = 1$ to get the original $4$.

So: all numbers not expressible in this form are not in the progression.


答: The negation of $x\gt y$ is $x\le y$, not $x\lt y$, so the correct contrapositive is "If $\forall_{y\gt0}x\le y$ then $x\le 0$," not "If $\forall_{y\gt0}x\lt y$ then $x\le 0$." Your concern with choosing $x=y$ is illogical; $y$ here is a bound variable (bound by the $\forall$ quantifier), so it makes no sense to "choose" to set it equal to the (free) variable $x$.

Title: Identifying the contrapositive of this statement $x>0$, then $∃_{y>0}$ such that $x>y$

Body:
Statement: If $x>0$, then $∃_{y>0}$ such that $x>y$. (x and y are real numbers)

Contrapositive: If $∀_{y>0}$ $x<y$, then $x≤0$.

Is the contrapositive correct?


答:
Edit: I made a second attempt

$$\sim\!\!(\forall x)(x\Rightarrow(\forall y)(y\Rightarrow x \operatorname{loves} y )) \; \; \; \text{(Universe is all People)}$$

Firstly, $x$ and $y$ as propositions do not make sense.   It would if you we using the proposition $x\in\mathcal U$, where $\mathcal U$ is some explicit domain or other restriction.  However, as you have an implicit domain of discourse, you do not need the implication.   So we shall drop them:

$$\sim\!\!(\forall x)(\forall y)(x \operatorname{loves} y ) \; \; \; \text{(Universe is all People)}$$

Secondly, this says "not everyone loves everyone", which is still not what you require, that is "not someone loves everyone."

Remember this: No-one is short for "Not Someone"; it is the claim there does not exist an example of this.

So we shall change that to an existential. We can also use the duality of quantifies to 'move' the negation inwards, once or twice.

$${\sim\!\!(\exists x)(\forall y)(x \operatorname{loves} y ) \; \; \; \text{(Universe is all People)}\\(\forall x)({\sim}\forall y)(x\operatorname{loves}y)\\(\forall x)(\exists y)\;{\sim}(x\operatorname{loves}y)}$$

Which says, "Everyone has someone they do not love."   Which is equivalent to "No-one loves everyone".

Title: Is my symbolic representation of "nobody loves everybody" correct?

Body:
Suppose we want to represent "nobody loves everybody" and the universe of discourse is all people. Would a correct representation be:

$$\require{enclose}\enclose{horizontalstrike}{(\forall x)(\sim \!x\Rightarrow(\forall y)(y\Rightarrow (\sim\!x) \; \text{loves} \; y )) \; \; \; \text{(Universe is all People)}}$$

Edit: I made a second attempt

$$\sim\!\!(\forall x)(x\Rightarrow(\forall y)(y\Rightarrow x \; \text{loves} \; y )) \; \; \; \text{(Universe is all People)}$$

According to the answer key, the answer is

$$(\forall x)\!(\sim\!\!(\forall y)(x \; \text{loves} \; y))$$

Am I correct? If not then why?


问:
Integration of a radial function over a bounded domain
Let $\Omega$ be a bounded domain in $\mathbb{R}^N$. Let $f(x)=|x|^\alpha$. Then $f\in L^1(\Omega)$ if $\alpha>-N$.

The above fact seems to hold for the following reason.

Since $\Omega$ is bounded, there exists $R>0$ such that $\Omega\subset B(0,R)$. Then, we have
\begin{align*}
I&=\int_{\Omega}f(x)dx\\
&\leq\int_{B(0,R)}|x|^\alpha dx\\
&=\frac{r^{\alpha+N}}{\alpha+N}\Bigg|_{0}^{R}\\
&=\frac{R^{\alpha+N}}{\alpha+N},
\end{align*}
where $\alpha+N>0$.

Then we get $f\in L^1(\Omega)$.

Kindly inform me, if the above argument seems fine with you.

Thanks.

答:
The idea behind the proof is correct, but I believe there is a computation error (off by a scalar multiple). Note that the $N$-dimensional ball is a union of concentric $(N-1)$-dimensional spherical shells, hence we have that
$$\frac{dV_N(r)}{dr}=S_{N-1}(r)=r^{N-1} S_{N-1}(1) \tag{$\ast$}$$
where $V_n(r)$ and $S_{n}(r)$ are the volume and surface area of the $n$-ball and $n$-sphere of radius $r$ respectively. We use hyperspherical coordinates $(r,\varphi_1,\varphi_2,\dots,\varphi_{N-1})$ to evaluate the integral. Note that one can simply plug in $dV=r^{N-1} S_{N-1}(1)~dr$ (see the earlier computation in $(\ast)$, and note the abuse in notation) due to radial symmetry of the integrand. Hence, for $R>0$ such that $\Omega\subset B(0,R)$, we have that
$$I=\int_{\Omega} f(x)~dV\leq \int_{B(0,R)} |x|^{\alpha}~dV=\int_{0}^R r^{\alpha}\cdot r^{N-1} S_{N-1}(1)~dr=S_{N-1}(1)\cdot \frac{R^{\alpha+N}}{\alpha+N},$$
which is finite for $\alpha>-N$, hence $f\in L^1(\Omega)$.

In case you are interested, an explicit formula for $S_n(1)$ for $n\in \mathbb{N}$ is given by
$$S_n(1)=\frac{2\pi^{(n+1)/2}}{\Gamma((n+1)/2)},$$
where $\Gamma$ is the gamma function.

Note: If radial symmetry does not hold, the volume element is given by (see the first link)
$$dV= r^{N-1}\sin^{N-2}(\varphi_1)\sin^{N-3}(\varphi_2)\cdots \sin(\varphi_{N-2})\, dr\,d\varphi_1 \, d\varphi_2\cdots d\varphi_{N-1}.$$

Note 2: Your $f$ is undefined for $\alpha<0$ at $x=0$. To resolve this problem you can set $f$ to be
$$f(x)=\begin{cases} |x|^{\alpha},&\mathrm{if}~x\neq 0, \\ 0 ,&\mathrm{if}~x=0, \end{cases}$$
and use the same proof.


问:
Prove $\{z^n|n \in \mathbb{Z}\}$ is an orthonormal basis of $L^2(S^1)$ with Haar measure
Prove $\{z^n|n \in \mathbb{Z}\}$ is an orthonormal basis.

We show that this is an orthonormal set.

All we need to show is that $\int_{S^1} z^n d\mu=0$ since $\mu$ is a Haar measure $\int_{S^1} z^n d\mu=\int_{S^1} (az)^n d\mu=a^n\int_{S^1} z^n d\mu$, but if $a$ is not a root of unity, this implies the integral is $0$. Thus the set is orthonormal.

We show the  the span is dense in $C(S^1)$ which in turn is dense in $L^2(S^1)$ which proves it is an orthonormal basis.

The Span is clearly an algebra so all we have to show is that it separates points. But that is fairly obvious. $x\not = y$ then take the identity map.

Is this correct?

答: You missed another condition. You need the fact that the span contains the complex conjugate of each of its elements. For this note that the conjugate of $z^{n}$ is $\frac 1 {z^{n}}=z^{-n}$ when $|z|=1$.


问:
Arbitrary constant C, when to resolve it to C?

in the picture on top, we resolve the arbitrary constant from $-2C_1$ to C, by why don't we do so when it is $\frac{10}{C}$ (below)?
In essence, why isn't $y=\frac {10}{e^x}+C$ correct?

答:
They are very different functions. Try to sketch them. As $x\rightarrow\infty$, one tends to $C$ and the other tends to $10/C$.

This basically comes down to the fact that you can’t write

$$\frac{b}{a+c}=\frac{b}{a}+\frac{b}{c}.$$

Sure, if, on the other hand, you had

$$\frac{e^x+C}{10}=\frac{e^x}{10}+\underbrace{\frac{C}{10}}_{C_1}$$

you could have relabelled the constants.


答:
Look at the results. The solution space of a differential equation is not always of the form "some function plus a constant", especially if a $y$ term appears on the right-hand side. In other words,
$$
\frac{\partial^2 (y(x) + C)}{\partial x^2}  = \frac{\partial^2 y(x)}{\partial x^2}
$$
but
$$
-\frac{e^x (y(x)+C)^2}{10} \neq -\frac{e^x y(x)^2}{10}
$$
(Maybe it's true for some values of $C$ such as $C=0$, but I'm writing that this is not an equality for all $C$.)

So the reasoning is different. So far, this has probably not been very helpful to you though. The thing that you need to remember when you're computing is that you can only replace $kC$ by $C$ if you're looking at the final solution, i.e. $y(x) = f(x) + kC$; in this case it is fine to replace $kC$ by $C$ since $C$ is arbitrary and the family $\{f(x) + kC \, | \, C \in \mathbb R \}$ is the same family as the family $\{f(x) + C \, | \, C \in \mathbb R\}$ because those are just two parametrizations of the same family.

Similarly, if $y(x) = e^C f(x)$ for some constant $C$ (as a solution of some other differential equation), you could re-write this as $y(x) = C f(x)$ for $C > 0$ since the families $\{e^Cf(x) \, | \, C \in \mathbb R \}$ and $\{Cf(x) \, | \, C > 0 \}$ parametrize the same solutions. So this statement about $C$ is really about re-parametrizing the solution space to simplify the description of the family, so when you do it, you need to check if you are describing the same space of solutions.

Hope that helps,

Title: Arbitrary constant C, when to resolve it to C?

Body:
in the picture on top, we resolve the arbitrary constant from $-2C_1$ to C, by why don't we do so when it is $\frac{10}{C}$ (below)?
In essence, why isn't $y=\frac {10}{e^x}+C$ correct?


答: Yes, that $C$ is also an arbitrary constant. And therefore you can write it as, say, $10K$, in which case $\frac{10}y=e^z+C$ becomes $\frac{10}y=e^z+10K$, and therefore$$y=\frac1{e^z/10+K}.$$But you will never get $\frac{10}{e^z}+C$ by this process.


答:
Take the Jordan form and all it's easy:

$$A=\begin{pmatrix}1&0\\0&\!\!-1\end{pmatrix}$$

Title: Finding the determinant of $2A+A^{-1}-I$ given the eigenvalues of $A$

Body:
Let $A$ be a $2\times 2$ matrix whose eigenvalues are $1$ and $-1$. Find the determinant of $S=2A+A^{-1}-I$.

Here I don't know how to find $A$ if eigenvectors are not given. If eigenvectors are given, then I can find using $A=PDP^{-1}$. Please solve for $A$; the rest I can do.


问:
What is the value of $c$ so that $f$ becomes continuous over the whole domain?
Let $f$ be defined as follows
\begin{align*}
f(x) =
\begin{cases}
\dfrac{4x^{2} - 39x + 27}{x - 9} &\text{if}\quad x\neq 9,\\
c &\text{if}\quad x = 9
\end{cases}
\end{align*}

I can't really grasp how to choose the constant $c$ so that the $f(x)$ becomes continuous.

Could someone help me understand this?

答:
Hint:

You have to show the fraction has a limit when $x\to 9$, which means $x-9$  is a factor of $4x^2-39x+27$. Hence perform the Euclidean division  by $x-9$.


问:
how to solve the following integral (maybe by integration by parts)?
Could you please help me solve this integral?
I already have the answer but I don't understand it. I know that I can use integration by parts. But I can't make sense of it.
$$
g = \int_{1}^{\infty} (t-1)^j e^{-t^2} dt
$$

答:
I will assume that $j$ is a positive integer.

Hints: Call the integral $I(j)$. Then $$I(j)=\int_1^{\infty} (t-1) (t-1)^{j-1} e^{-t^{2}} dt$$ $$=\int_1^{\infty} t(t-1)^{j-1} e^{-t^{2}} dt-\int_1^{\infty}  (t-1)^{j-1} e^{-t^{2}} dt$$ The second term is $I(j-1)$. In the first term write $te^{-t^{2}}$ as the derivative of $-\frac 1  2 e^{-t^{2}}$  and integrate by parts. This term becomes $-\frac 1  2 e^{-t^{2}}(t-1)^{j-1}|_1^{\infty} +(j-1)\frac 1  2 \int_1^{\infty}(t-1)^{j-2}e^{-t^{2}}dt$. Now you get a recurrence relation and you have to solve this to find $I(j)$.


问:
Can we compute the fugitive integers k(n)
Is this function $k:\mathbb{N}\to\mathbb{N}$ computable, which is defined as:

$$
k(n)=\begin{cases}\text{the position of the first $n$ consecutive 9's in $\pi$}\\0\text{ if there is no such position}\end{cases}
$$

The term "fugitive integer" is taken from

Intuitionism: An Inspiration?

Wim Veldman - 2021

https://arxiv.org/abs/2102.01561

答:
It is conjectured that $\pi$ is a normal number, which would imply that your function $k(n)$ is computable; since there would always exist a sequence of $n$ consecutive $9$'s in the digits of $\pi$; in particular, it would never take the value $0$ and be non-decreasing. Since very little progress has been made on this conjecture, I believe we don't have an answer to your question as of now.

The reason why this is conjectured is because we know that the set of real numbers which are not normal has measure zero, and so if you pick a real number in the interval $[a,b]$ at random, you pick a normal number with probability $1$, implying that the function $k(n)$ for this number would be computable. So we know it's almost always the case for a random number, but when it comes to specific numbers, we know very little.

Hope that helps,


问:
The global section of direct sum of twisted sheaves
I am working with the Exercise 5.14 in Hartshorne's GTM52. And its problem (a) is stated roughly as follows.

Let $X$ be a closed subscheme of $\mathbb{P}^r_k$, a projective $r$-space over an algebraically closed field $k$, let $S$ be the homogeneous coordinate ring of $X$, i.e. $S=k[x_0,\dots,x_r]/I$, where $I=\Gamma_*(\mathscr{I}_X)$, and let $S'=\bigoplus_{n\ge0}\Gamma(X,\mathscr{O}_X(n))$.

Show that $S$ is a domain and that $S'$ is its integral closure.

Here my questions don't involve the $S$. I just want to ask something about the $S'$.

The hint of the problem (a) tells that one can regard the ring $S'$ as the global sections of such a sheaf of rings $\mathscr{S}=\bigoplus_{n\ge0}\mathscr{O}_X(n)$ on $X$, a direct sum of infinitely many sheaves. That is, there is an equation about them:
$$\Gamma(X,\mathscr{S})=\Gamma(X,\bigoplus_{n\ge0}\mathscr{O}_X(n))=\bigoplus_{n\ge0}\Gamma(X,\mathscr{O}_X(n)).$$
Does it true? I know that the arbitrary direct sum of sheaves $\{\mathscr{F}_i\}$ is defined to be the associated sheaf of the presheaf $U\mapsto\bigoplus_{i}\mathscr{F}_i(U)$. In the finite case, such a presheaf is just a sheaf. And else, it may be not, because we can give a counter-example (see here: Direct and inverse limits of sheaves). So I want to query if the global sections of $\mathscr{S}$ is just $\bigoplus_{n\ge0}\Gamma(X,\mathscr{O}_X(n))$, and if the process of sheafification doesn't change the global sections (up to an isomorphism).

Thanks in advance.

答:
The presheaf direct sum of sheaves should be a sheaf as long as $X$ is Noetherian. Clearly it is a separated presheaf, so all we need to show is the gluability axiom. If I have an infinite collection $s_i \in \bigoplus_{j} \mathcal{F}_j(U_i)$ that agree on overlaps, then they have a natural gluing $s \in \prod_j \mathcal{F}_j(\bigcup_i U_i)$. By Noetherianness, the union of the $U_i$ is quasicompact, so in fact this section $s$ can be viewed as gluing together FINITELY many $s_i$, hence is nonzero in only finitely many entries and lives in the subgroup $\bigoplus_j \mathcal{F}_j(\bigcup_i U_i)$.

In particular, in your case above this holds and so the direct sum commutes with global sections.


问:
Volume of solid of revolution about a line other than the axis - using Cylindrical Shells method
I am struggling with a concept of finding the radius in the shell method of finding the volume of the solid when revolved around a line other than the axis.  Maybe the issue here is that I am looking for a "rule" instead of actually visualizing the problem.  It was my understanding that when revolving around a line other than the axis a formula could be used as follows: (this example uses revolution about a vertical line other than the $y$ axis)
the radius would be $|verticalvalue|$ $+/-$ $x$.  Use $+$ if the region of revolution is to the right of the line and use $-$ if the region is to the left.

I thought I had worked many problems using this concept and gotten the correct answer.  However, doesn't work in this problem:

Use the method of cylindrical shells to find the volume generated by rotating $y=x^2$, $y=0$, $x=1$, $x=2$ about the line $x=1$.

My integral, using my above "concept" was to form the following integral:

$V=2\pi\int(1+x)(x^2) dx$ evaluated from lower limit of 1 to upper limit of 2.  I used the absolute value of $1$ PLUS (since the region of revolution is to the right of the vertical line of revolution.

This does not produce the book answer and the book used $x-1$ for the radius instead of $x+1$, obviously producing very different results.

My question:  is there a rule for writing the radius?  Why didn't my rule work?

答: " Maybe the issue here is that I am looking for a "rule" instead of actually visualizing the problem". I am not sure if that is a good idea. However, what you CAN do is, perform a horizontal translation (a shift of $1$ units to the left) on all functions in question. So "Use the method of cylindrical shells to find the volume generated by rotating $y=x^2$,$ y=0$, $x=1$, $x=2$ about the line $x=1$, now becomes "rotating the region formed by$y=(x+1)^2$, $y=0$, $x=0$, $x=1$ about $x=0$. This way you can now apply the "standard" formulas for cylindrical shells since the axis of rotation is now a coordinate axis. I would still make a graph of the situation, try to understand why this works


问:
What significance is there if a theorem can be proved in the same system in more than one way?
Doesn't that imply some sort of redundancy or point to some inner connection between the axioms? Or is it the case that when all theorems used in the proofs are broken down to their axioms that the two proofs must in fact be equivalent?</p>

答: Complementary to Robert's answer, you might like to be aware of the proofs-as-programs or propositions-as-types paradigm, also known as the Curry-Howard correspondence. Under this correspondence, a logical statement (such as $\alpha \to (\beta \to \alpha)$) corresponds to a program (or function) that represents a proof of that statement (such as $x \mapsto y \mapsto x$ or $\lambda x \cdot \lambda y\cdot x$, if you prefer $\lambda$-notation). Under the Curry-Howard correspondence, the existence of multiple proofs of the same statement has very interesting consequences. For example, the proofs of the statement $(\alpha \to \alpha) \to (\alpha \to \alpha)$ provide a model for the natural numbers (the Church numerals).


问:
Pugh exercise 1.18
I am trying to solve the below exercise.

Prove that real numbers correspond bijectively to decimal expansions to decimal expansions not terminating in an infinite string of nines, as follows. The decimal expansion of $x \in \mathbb{R}$ is $N.x_1 x_2 \ldots$, where $N$ is the largest integer $\leq x$, $x_1$ is the largest integer $\leq 10(x-N)$, $x_2$ is the largest integer $\leq 100(x-(N+ x_1/10)$, and so on.

(a) Show that each $x_k$ is a digit between $0$ and $9$.

(b) Show that for each $k$ is there an $\ell \geq k$ such that $x_{\ell} \neq 9$.

(c) Conversely, show that for each expansion $N.x_1 x_2 \ldots$ not terminating in an infinite string of nines, the set
$$\{N, N + \frac{x_1}{10}, N + \frac{x_1}{10} + \frac{x_2}{100}, \ldots \}$$
is bounded and its least upper bound is a real number $x$ with decimal expansion $N.x_1 x_2 \ldots$.

First, I'm having trouble understanding the construction. So $x = N.x_1 x_2 \ldots$. It makes sense to define $N = \lfloor{x} \rfloor$. I don't fully understand the digits $x_k$, though I know what they, intuitively, should represent.

I think part (a) can be done by induction on $k$, but the statement sounds as though it follows vacuously from the definition of the floor function. I tried to solve part (b) by contradiction, supposing that there exists a $k$ such that for all $\ell \geq k$, $x_{\ell} = 9$. Since $N \leq x$, $N - x \leq 0$, and
$$x = N.x_1 x_2 \ldots = N + \frac{x_1}{10} + \frac{x_2}{100} + \ldots,$$
we have
$$N - (N + \frac{x_1}{10} + \frac{x_2}{100} + \ldots) \leq 0.$$
By the assumption:
$$N - (N + \frac{x_1}{10} + \frac{x_2}{100} + \ldots + \frac{x_k}{10^k} + \frac{9}{10^{k+1}} + \frac{9}{10^{k+2}} + \ldots$$
I should be able to find a contradiction, but I can't figure out where to go from here.

For part (c): since each $x_k \leq 9$ by (a), we have
\begin{align*}
N + \frac{x_1}{10} + \frac{x_2}{10} + \ldots & \leq N + \frac{9}{10} + \frac{9}{100} + \ldots \\
& = N + 9 \sum\limits_{k=1}^{\infty} \left(\frac{1}{10}\right)^k \\
& = N + 9 \cdot \frac{1}{9} \\
& = N + 1
\end{align*}
Non-emptyness seems vacuous: it's defined to be nonempty. So it has a supremum. I can't figure out how to proceed from here.

Any help on this problem would be appreciated. I will update this first post as I work on this problem more.

答:
This will not be a complete solution, but it should help make clear what’s going on, give you a push in the right direction for finishing off all three parts.

To see how the digits $x_k$ are being calculated, imagine that we already knew that the decimal expansion of $x$ was $N.x_1x_2x_3\ldots\;$. Clearly $N=\lfloor x\rfloor$, and $x-N$ is the fractional part: $N-x=0.x_1x_2x_3\ldots\;$. Multiplying this by $10$ just shifts the decimal point one place to the right to yield $x_1.x_2x_3\ldots\,,$ and taking the floor picks off the digit $x_1$: $x_1=\lfloor 10(x-N)\rfloor$.

To get $x_2$, we need to apply the same idea to $0.0x_2x_3\ldots$: multiply it by $10^2$ to shift the decimal point immediately to the right of $x_2$, getting $x_2.x_3x_4\ldots\,,$ and take the floor. To do this, we have to have $0.0x_2x_3\ldots\,,$ so what is it? It’s $x-\left(N+\frac{x_1}{10}\right)$, so

$$x_2=\left\lfloor10^2\left(x-\left(N+\frac{x_1}{10}\right)\right)\right\rfloor\,.\tag{1}$$

In general, if we’ve determined $x_1,\ldots,x_n$, we have

$$N.x_1x_2\ldots x_n=N+\sum_{k=1}^n\frac{x_k}{10^k}\,,$$

and

$$x-\left(N+\sum_{k=1}^n\frac{x_k}{10^k}\right)=0.\underbrace{00\ldots00}_nx_{n+1}x_{n+2}\ldots\,,$$

so that we expect that

$$\begin{align*}
x_{n+1}&=\lfloor x_{n+1}.x_{n+2}x_{n+3}\ldots\rfloor\\
&=\left\lfloor 10^{n+1}\left(x-\left(N+\sum_{k=1}^n\frac{x_k}{10^k}\right)\right)\right\rfloor\,.
\end{align*}$$

Part (a) does require an argument: it is not vacuously true. Remember, we don’t actually have a decimal expansion: we’re showing that one exists. From the definition of the floor function we know that $0\le x-N<1$, so $0\le 10(x-N)<10$, and it follows immediately that $x_1$ is an integer, and $0\le x_1\le 9$.

We can now argue that $x_1\le 10(x-N)<x_1+1$, so $\frac{x_1}{10}\le x-N<\frac{x_1+1}{10}$, so

$$0\le x-\left(N+\frac{x_1}{10}\right)<\frac1{10}\,,$$

and therefore

$$0\le 10^2\left(x-\left(N+\frac{x_1}{10}\right)\right)<10\,.$$

It follows immediately that $x_2$, as defined by $(1)$, is an integer, and $0\le x_2\le 9$.

Now see if you can use the step from $x_1$ to $x_2$ as a model for the general induction step from $x_n$ to $x_{n+1}$.

Your idea for (b) is basically sound. You will need to show by induction on $n$ that

$$x\ge N+\sum_{k=1}^n\frac{x_k}{10^k}$$

for all $n\in\Bbb Z^+$. You can then conclude that if $x_\ell=9$ for each $\ell\ge k$, then

$$\begin{align*}
x&\ge N+\color{red}{\sum_{\ell=1}^{k-1}\frac{x_\ell}{10^\ell}}+\sum_{\ell\ge k}\frac9{10^\ell}\\
&=N+\color{red}{\sum_{\ell=1}^{k-1}\frac{x_\ell}{10^\ell}}+\frac1{10^{k-1}}\,,
\end{align*}$$

where the red sum is $0$ if $k=1$. If $k=1$, this says that $x\ge N+1$, which we know is false, since $N=\lfloor x\rfloor$. If $k>1$, we can rewrite it as

$$x\ge N+\sum_{\ell=1}^{k-2}\frac{x_\ell}{10^\ell}+\frac{x_{k-1}+1}{10^{k-1}}$$

and show that this contradicts the definition of $x_{k-1}$ as

$$\left\lfloor 10^{k-1}\left(x-\left(N+\sum_{\ell=1}^{k-2}\frac{x_\ell}{10^\ell}\right)\right)\right\rfloor\,.$$

For (c) you should first be showing that $N$ and each of the numbers $N+\sum_{k=1}^n\frac{x_k}{10^k}$ for $n\in\Bbb Z^+$ is bounded above by $N+1$; you should not yet be dealing with the infinite decimal. The calculation is almost identical to the one that you actually made: the only difference is the lefthand side. This shows that the set has a least upper bound. Let $y$ be that least upper bound; you’ll be done if you can show that $y$ has the decimal expansion $N.x_1x_2x_3\ldots\;.$

One approach is to show first that

$$N+\sum_{k=1}^{n-1}\frac{x_k}{10^k}+\frac{x_n+1}{10^n}$$

is an upper bound for the set for each $n\in\Bbb Z^+$, so that for each $n\in\Bbb Z^+$ we have

$$N+\sum_{k=1}^n\frac{x_k}{10^k}\le y<N+\sum_{k=1}^{n-1}\frac{x_k}{10^k}+\frac{x_n+1}{10^n}\,;$$

this follows from the definition of the digits $x_k$, and you can use it to show that the $x_k$ are the digits of the decimal expansion of $y$.

Title: Pugh exercise 1.18


问:
Equality involving cosine and binomial coefficients - check if true for every $n$
While doing some computations with complex numbers I think I proved that

$$(\cos{ \frac{n \cdot \pi}{4} )\cdot 2^{n/2}} = \binom{n}{0} - \binom{n}{2} + \binom{n}{4} - \binom{n}{6} + ... $$

How? Well, these are two different forms of

$$\frac{1}{2} \cdot ((1+i)^n + (1-i)^n)$$

But because of the late hour, I am not sure if I didn't make some mistake.

Is this equality really true for every natural n?

Also, I wonder how can one prove it without using complex numbers. I mean is these some sort of elementary (high-school) proof?

答:
Can be proven via DeMoivre's theorem.

$[\cos(\pi/4) + i\sin(\pi/4)]^n = [\cos(n\pi/4) + i\sin(n\pi/4)]$.

The alternative computation, which must yield identical results, is via binomial expansion:

$[\cos(\pi/4) + i\sin(\pi/4)]^n
= \sum_{k=0}^n \binom{n}{k}\cos(\pi/4)^k ~~i^{(n-k)}\sin(\pi/4)^{(n-k)}.$

The (true) conjecture, which is close to the conjecture that you are focusing on yields to:

If two complex analysis expressions are equivalent, the real portion of each expression must be equivalent.

In the binomial expansion, the real terms will be precisely those terms where $(n-k)$ are even.

Since $\cos(\pi/4) = (1/\sqrt{2}) = \sin(\pi/4)$, you have that for all $k \in \{0,1,2,\cdots n\}$,
$~\cos(\pi/4)^k \sin(\pi/4)^{(n-k)} = (1/\sqrt{2})^n.$

So the actual (true) conjecture, which will hold for any positive integer $n$ is that

$$[\cos(n\pi/4)] \times 2^{(n/2)}
~=~ \sum_{k=0}^n \binom{n}{k}i^{(n-k)}$$

with the understanding that only those terms in the RHS summation above where $(n-k)$ is even will be counted.

The intended RHS summation can be more tidily expressed by noting that $\binom{n}{k} = \binom{n}{n-k}$ and letting $c$ denote the largest integer such that $c \leq \frac{n}{2}.$

Then the re-expression of the RHS summation becomes

$$\sum_{k=0}^c \binom{n}{2k}(-1)^k.$$


问:
Show that a nonempty neighborhood exists around x for a given function that is 0 at x with nonzero derivative at the same point.
Let $f : [0,1] \rightarrow \mathbb{R}$ and $x$ a fixed point in $(0,1)$. Assume $f(x)=0$ and $f$ is differentiable at $x$ with $f'(x)\neq 0$.

Prove: $\exists \delta \gt 0$ such that $f(t)\ne 0 \ \ \forall t$ with $0 \lt |{t-x}| \lt \delta$.

I'm pretty sure that I need to invoke continuity of $f$ in some way, but I don't see the immediate path forward. Any help is appreciated!

答:
Suppose otherwise. Then for any $\;n\in\Bbb N\;$ there exists $\;t_n\in A_n:=\{t\in\Bbb R\;|\;|t-x|<\frac1n\}\;$ s.t. $\;f(t_n)=0\;$ , but then by definition

$$f'(x)=\lim_{t\to x}\frac{f(t)-f(x)}{t-x}\stackrel{\text{since the limit exists!}}=\lim_{n\to \infty\\t_n\in A_n}\frac{f(t_n)-f(x)}{t_n-x}=0$$

and we get a contradiction.


问:
Explain why the following statements are logically equivalent (2 examples)
I have the following 2 separate statements.

(pvq)-> q = ~(p^~q)

r->(~pvq)=(pvr)->q

I have proved they are both equivalent with a truth table, but I am asked to briefly explain (in words not with a theorem chart) why they are equivalent, and I have no idea why they are. I have looked at the Laws of Algebra of Propositions and they don't really address more 'complex' statements. Could I have some help and maybe someone knows of a more advanced chart (or I will learn to extrapolate better)

Thanks

答:
The first equivalence can be proven as follows:
\begin{align*}
(p\vee q)\rightarrow q & \Longleftrightarrow (\neg(p\vee q))\vee q\\\\
& \Longleftrightarrow (\neg p\wedge\neg q)\vee q\\\\
& \Longleftrightarrow (\neg p\vee q)\wedge(\neg q\vee q)\\\\
& \Longleftrightarrow \neg p\vee q\\\\
& \Longleftrightarrow \neg(p\wedge\neg q)
\end{align*}

The second equivalence can be proven as follows:
\begin{align*}
r\to(\neg p\vee q) & \Longleftrightarrow \neg r\vee(\neg p\vee q)\\\\
& \Longleftrightarrow (\neg p\vee\neg r)\vee q\\\\
& \Longleftrightarrow (\neg(p\wedge r))\vee q\\\\
& \Longleftrightarrow (p\wedge r)\to q
\end{align*}

and we are done. Please let me know if my interpretation of the second equivalence is correct, since the obtained result is different.