问:
Stochastic process, Gaussian, with zero mean is a Wiener process
Let $(\Omega, \mathcal F , \mathbb P)$ be a probability space and let $\mathcal F = \{\mathcal F_t\}_{t\ge} $ a ﬁltration. Let $W=\{W_t;t ≥ 0\}$ be a stochastic process adapted to $\mathcal F$. Suppose that W is a zero-mean Gaussian process with covariance function $$E(W_s W_t) = \min(s, t), \quad ∀s,t ∈ [0, ∞).$$
Show that W is a standard Wiener process.

In other words, I have to prove that the following conditions are satisfied:

$W_0=0$

$W_t-W_s$ is Gaussian with zero mean and $Var=|t-s|$

increments of W corresponding to non-overlapping time intervals are independent

答:
Hints

Choose $t=s=0$ in $\mathbb{E}(W_s \cdot W_t) = \min\{s,t\}$. (Note that $N(0,0)$ equals -by definition- $\delta_0$.)

Let $s<t$. By assumption, the random vector $X:=(W_s,W_t)$ is Gaussian with mean $0$ and covariance matrix $$C:=\begin{pmatrix} s & s \\  s & t \end{pmatrix}$$ Since $X$ is Gaussian, we know that $\ell^T \cdot X = \ell_1 \cdot W_s+\ell_2 \cdot W_t$ is Gaussian for $\ell \in \mathbb{R}^2$. There are known formulas how to calculate the mean and variance of $\ell^T \cdot X$. Find a suitable $\ell$.

Let $0:=t_0 < t_1 < \ldots < t_n$. Note that $$\Delta := \begin{pmatrix} W_{t_1}-W_{t_0} \\ \vdots \\ W_{t_n}-W_{t_{n-1}} \end{pmatrix} = \underbrace{\begin{pmatrix} -1 & 1 & 0 & 0 &\ldots & 0 \\ 0 &-1 & 1 & 0 & \ldots & 0 \\ \vdots & & \ddots & \ddots & & \\ 0 & 0 & 0 & \ldots & -1 & 1 \end{pmatrix}}_{=:M} \cdot \begin{pmatrix} W_{t_0} \\ \vdots \\ W_{t_n} \end{pmatrix}$$
Since $(W_{t_0},\ldots,W_{t_n})$ is Gaussian, by assumption, we conclude that $\Delta$ is Gaussian. Therefore, it suffices to show that the covariance matrix of $\Delta$ is a diagonal matrix. Again, there are known formulas how to calculate the covariance matrix of a linear transformation of a Gaussian random vector.

Remark In general, one requires that the paths $t \mapsto W(t,w)$ are continuous for almost all $w$. The theorem of Kolmogorov-Chentsov shows that, under the given assumptions, there exist always a modification $(\tilde{W}_t)_t$ of the process $(W_t)_t$ such that the sample paths $t \mapsto \tilde{W}(t,w)$ are continuous almost surely.


问:
$\sum_{n=1}^\infty \frac{(2n+1)!}{2^{2n}(n!)^2}$ converges?
Does the series $$\sum_{n=1}^\infty \frac{(2n+1)!}{2^{2n}(n!)^2}$$ converges? I tried D' Alembert's quotient criterion but the $\displaystyle\lim_n \frac{a_{n+1}}{a_n}$ equals to $1$ and then by Cauchy's n-th root criterion $\displaystyle\lim_n \sqrt[n]{a_n}$ it gives an expession that is $\frac{\infty}{\infty}$ so I use the inequality $\sqrt[n]{n!}>2,\ \forall n\geq 4$ as a result $\displaystyle\lim_n \sqrt[n]{a_n}=+\infty$, and consequently the series diverges. However, I don't know if it is correct (if yes, then somebody may confirm my thoughts). Also, does anybody see another way to decide if the previous series converges or not? Thanks

答:
Note that for $n\ge 1$ we have

$$\begin{align}
\frac{(2n+1)!}{(n!)^2}&=\frac{(2n+1)(2n)(2n-1)(2n-2)\cdots (n+1)n(n-1)(n-2)\cdots4\cdot 3\cdot 2}{(n!)^2}\\\\
&>\frac{(2n)^2 (2n-2)^2 \cdot (2)^2}{(n!)^2}\\\\
&=4^n
\end{align}$$

Hence, the general term of the series, $\displaystyle a_n=\frac{(2n+1)!}{4^n(n!)^2}$, is bounded below by $1$ and fails to approach $0$ as $n\to \infty$.  We conclude that the series diverges.

NOTE:  Using Stirling's Formula shows that the terms of the series are $O\left(\sqrt{n}\right)$ as $n\to \infty$.  So, not only do the general terms fail to approach $0$, they are unbounded.

ContentLicense: CC BY-SA 4.0


问:
Values for x for which distance between vector and subspace is maximal

Given $A=\left ( 1, 1, 1 \right )$ be vector in $\mathbb{R}^{3}$. Let $L$ be a subspace of $\mathbb{R}^{3}$ spanned by the vectors $\left ( 1, 0, -1 \right )$ and $\left ( 3, 5, x \right )$. Find the values of x for which the distance between $A$ and $L$ is maximal.

The approach for the above formulation was calculating the proj$_{L}$$A$ i.e. the orthogonal projection of $A$ on $L$ subspace. Then the distance $d=\left \| A-proj_{L}A \right \|$. The calculations are getting tedious in this process as in the first step for getting the projection matrix, the basis matrix is formed for the given spanned vectors which is then used for further calculations. I am getting x=-8 and the square of distance as 3.

Can you please let me know if the above approach followed for the formulation is correct?

答:
Your approach is correct, but I'll show you another way:

Let's reason about the problem geometrically. You have a fixed vector and a plane that can move, and you want to tilt the plane in such a way that the tip of the vector is as far away from the plane as possible. Maybe you can convince yourself that the optimal configuration is for the plane and the vector to be made orthogonal to each other. (The shadow of the vector on the plane will be as short as possible, making the distance between them as large as possible).

Now, can we make $A$ perpendicular to $L$? It suffice to make $A$ perpendicular to both of the basis vectors $v_1 := (1, 0, -1)$ and $v_2 := (3, 5, x)$. Luckily for us, we can observe that $v_1$ and $A$ are already orthogonal. But what about $A$ and $v_2$? It suffice to impose that their dot product is zero:

$$A \cdot v_2 = (1, 1, 1) \cdot (3, 5, x) = 8 + x = 0,$$
and $x = -8$ appears! Now that you know the optimal position of the plane, your procedure can get the orthogonal projection and desired distance.


问:
find $\theta_{MLE}$ for a function
For
$$
f(x;\theta)=(\theta+1)x^{-\theta-2}
$$
find the maxmimum likelihood estimators (MLEs) for $\theta$ based on a random sample of size $n$.
My work so far:
$$
\begin{align}
\prod_{i=1}^n \log(f(x_i;\theta)) &= \sum \log(\theta+1)-\log(x_i)(\theta+2) \\
&=n\log(\theta+1)-(\theta+2)\sum \log(x_i)
\end{align}
$$
Now take the derivative, and set it equal to 0:
$$
\begin{align}
\frac{d}{dx} \log(f(x;\theta)) &= 0-(\theta+2)\sum \frac{1}{x_i} =0 \\
&=\sum\frac{1}{x_i}=0
\end{align}
$$
I cant make any sense of this. What exactly have I done here? Did something go wrong? Shouldnt I get an equation with $\theta$ in it?</p>

答:
You should take the derivative with respect to $\theta$.

Also, $\prod_{i=1}^n \log(f(x_i;\theta))$ should be $\sum_{i=1}^n \log(f(x_i;\theta))$

ContentLicense: CC BY-SA 3.0

Title: find $\theta_{MLE}$ for a function


问:
Lie algebra of groups' element centralizer

Let $G$ be a Lie group. Let $Z_G(g)$ be a centralizer of $g \in G$. Prove that $Z_G(g)$ is an closed Lie subgroup of $G$ and prove that Lie algebra of $Z_G(g)$ is $\mathfrak  z = \{x \in \mathfrak g \mid \operatorname{Ad}_g(x) = x\}$.

The first part is easy since we can show that $Z_G(g)$ is a stabilizer of $g$ of action of $G$ on itself by conjugations ($h \mapsto hgh^1$).

The second part is trickier. Let $x \in \mathfrak g$ be such that $\operatorname{Ad}_g (x) = x$. If $\phi : G \rightarrow G$ ($h \mapsto ghg^-1$), then we can use functoriality of exponential map

$\require{AMScd}$
$$\begin{CD}
\mathfrak g @>\operatorname{Ad}_g>> \mathfrak g \\
@V\operatorname{exp}VV @VV\operatorname{exp}V \\
G @>>\phi> G
\end{CD}$$

to deduce that $\operatorname{exp}(x) \in Z_G(g)$. But how can I see that $\mathfrak z$ is exactly fixed points of $\operatorname{Ad}_g$?

答: Suppose that $x\in \mathfrak{z}$ for every real $t$, $g\exp(tx)g^{-1} = \exp(tx)$. This implies that $\left.\frac{d}{dt}\right|_{t=0} g\exp(tx)g^{-1} = \left.\frac{d}{dt}\right|_{t=0}\exp(tx)$. This is equivalent to saying that $\operatorname{Ad}_g(x) = x$.


问:
Help with Prove or disprove question
Prove or disprove:

If $f,g$ are differentiables on $\mathbb R$. Assume that there is a $c\in \mathbb R$ such that $f'(x)=g'(x)$ for every $x<c$. Then $f'(c)=g'(c)$.

I tried to find an example to disprove it, but I was told the statement is correct. In that case I'd like to know what I did wrong. $$$$
$f(x)=3$ therefore $f'(x)=0$ for every $x \in \mathbb R$. $g(x)=0$ for every $x<c$ and $g(x)=x$ for every $x\geq c$. Therefore if $c=0$ then $g'(x)=f'(x)$ for every $x<c$ whereas $g'(c)=1\neq f'(c)=0$.

Thanks!

答:
By considering $F(x)=f(x)-g(x)$ we see that $F'(x)=0$ for $x<c$ and we need to prove that $F'(c)=0$.

We know that $F'(c)$ exists and it equals
$$
\lim_{x\to c}\frac{F(x)-F(c)}{x-c}=\lim_{x\to c^-}\frac{F(x)-F(c)}{x-c}
$$
Since l'Hôpital’s theorem applies and
$$
\lim_{x\to c^-}\frac{F'(x)}{1}=0
$$
we can conclude that $F'(c)=0$.

Title: Help with Prove or disprove question


问:
Proving a conditional probability relevant to Bayesian Inference
Is it possible to prove the following?

$p(A|B,D)=\int{p(A|B,C})p(C|D)dC$

I have made a following attempt to expand the RHS for an intuitive feel. While it appears intuitively like some sort of marginalisation hence the LHS equates to RHS, I am unable to mathematically simplify the RHS to equate to LHS.

$p(A|B,D)=p(A|B,C=0)p(C=0|D) + ...  + p(A|B,C=j)p(C=j|D) $ for $C=0,...,j$

The motivation for my question is the equation in the following link regarding the first equation under the section Bayesian Inference.

Thank you in advance!

答:
Technically it should be $\displaystyle p(A|B,D)=\int p(A|B,C,D)P(C|D)dC$ because you are still given $D$.

I read a little bit about the article and it talks about estimating $y^*$ based on a new observed $x^*$ and observed data $D$. The true parameter $\theta$ is not known, and based on the data $D$, there is a certain probability of every single $\theta$, i.e. $\theta$ has a "posterior distribution" depending on the observed $D$. The formula is obtained by marginalizing the distribution of $y^*$ given $\theta$ over the posterior distribution of $\theta$ given $D$.

So: $\underbrace{p(y^*|x^*,D)}_{\text{density of $y^*$ given a new data point $x^*$}}=\displaystyle \int \underbrace{p(y^*|x^*,D,\theta)}_{\text{density of $y^*$ given $\theta$}}\underbrace{p(\theta|D)}_{\text{density of $\theta$ given observed Data}}\underbrace{d\theta}_{\text{over all possible $\theta$}}$.

Note that you can always marginalize out a random variable by summing/integrating all its possible values.

If C is discrete:

$P(A|B,D)=\displaystyle \sum_C P(A|B,C,D)P(C|D)$

If C is continuous:

$f(A|B,D)=\displaystyle \int_C f(A|B,C,D)f(C|D)dC$

This is called the law of total probability for random variables.

$\displaystyle f_X(x)=\sum_yf_{X|Y}(x|y)f_Y(y)$

$\displaystyle f_X(x)=\int_yf_{X|Y}(x|y)f_Y(y)dy$


问:
Form of Ito's formula for Ito diffusions
Suppose we have an Ito diffusion ($Z_{t, z}$) satisfying
$$
Z_{t, z} = z + \int_0^t b(Z_{s,z})ds + \int_0^t \sigma(Z_{s,z})dW_s
$$
where $z\in\mathbb{R}^d, b:\mathbb{R}^d\to\mathbb{R}^d, \sigma:\mathbb{R}^d \to \mathbb{R}^{d\times d}$ and $W_s$ is a $d$-dimensional Brownian motion.
According to this thesis (Eqn. 4.7), Ito's formula looks like this:
$$
f(t, Z_{t, z}) - f(s, Z_{s, z}) = \int_s^t \nabla_x f(r, Z_{r,z})dW_r + \int_s^t \left( \frac{1}{2}\Delta_x f(r, Z_{r,z})
 + \frac{\partial}{\partial t}f(r, Z_{r,z}) \right) dr
$$
where $\Delta_x$ is the Euclidean Laplace operator w.r.t. the second coordinate $x$.
I've never seen it in this form, so I wanted to write out the r.h.s. more explicity. How are $\nabla$ and $\Delta$ interpreted correctly here? I'm thinking
$$\nabla_xf(r, Z_{r, z}) = f'(r, Z_{r, z}) \cdot Z_{r, z}'$$
with $f'(r, Z_{r, z}) \in \mathbb{R}^{1\times d}, Z_{r, z} \in \mathbb{R}^{d\times d}$? But is $Z_{r, z}$ even differentiable? And how does it look like? From the same thesis, it seems that
$$
\nabla_xf(r, Z_{r, z}) = f'(r, Z_{r, z}) \cdot \sigma(Z_{r,x})?
$$
I'm having the same problems understanding $\Delta_x$. Any help is appreciated!

答:
The gradient and Laplacian only act on $f$.  The process $Z$ is composed with these functions.  In other words, $\nabla_{x} f(r,Z_{r,z})$ might be clearer if we instead wrote $(\nabla_{x} f)(r,Z_{r,z})$.  Similarly, $\Delta_{x} f(r,Z_{r,z})$ is $(\Delta_{x} f)(r,Z_{r,z})$.  Thus, there is no chain rule involved and no need for any derivatives of $Z$.

(It's like if I wrote $f'(t^{2})$ in single variable calculus.  $t^{2}$ is the argument of $f'$.  The notation is not meant to mean $\frac{d}{dt}\{f(t^{2})\}$.)

$\nabla_{x}f$ is the gradient of $f$ with respect to the $x$ variable.  Its values are vectors in $\mathbb{R}^{1 \times d}$.  $\Delta_{x} f$ is the Laplacian of $f$.  It is a real-valued function.

In (4.7) in the text, it looks like the author is restricting attention to a Brownian motion.  (Notice in the last two integrals on the right-hand side there is a $B_{r}$ in $f$ instead of $X_{r}$.  I don't exactly know what $X_{r}$ is supposed to be, but I suspect it should be $B_{r}$ throughout (4.7).)  In general, if $Z_{r,z}$ is the solution of the SDE
\begin{equation*}
dZ_{t,z} = b(Z_{t,z}) dt + \sigma(Z_{t,z}) dB_{t}, \quad Z_{0,z} = z,
\end{equation*}
then Ito's formula for the composition $t \mapsto f(t,Z_{t,z})$ gives, in this case,
\begin{align*}
df(t,Z_{t,z}) &= \nabla_{x}f(t,Z_{t,z}) d Z_{t,z} + \frac{1}{2} D^{2} f(t,Z_{t,z}) \cdot \langle Z_{\cdot,z} \rangle_{t} + \frac{\partial f}{\partial t}(t,Z_{t,z}) dt \\
&= b(Z_{t,z}) \cdot \nabla_{x} f(t,Z_{t,z}) dt + \sigma(Z_{t,z})^{*} \nabla_{x}f(t,Z_{t,z}) \cdot dB_{t} + \left( \frac{1}{2} \text{tr} (\sigma(Z_{t,z}) \sigma(Z_{t,z})^{*} D^{2}f(t,Z_{t,z})) + \frac{\partial f}{\partial t}(t,Z_{t,z}) \right) dt
\end{align*}
In integral form, this reads:
\begin{equation*}
f(t,Z_{t,z}) = f(0,z) + \int_{0}^{t} \left( b(t,Z_{t,z}) \cdot \nabla_{x}f(t,Z_{t,z}) + \frac{1}{2} \text{tr} (\sigma(t,Z_{t,z}) \sigma(t,Z_{t,z})^{*} D^{2}_{x}f(t,Z_{t,z})) + \frac{\partial f}{\partial t}(t,Z_{t,z}) \right) \, dt + \int_{0}^{t} \sigma(t,Z_{t,z})^{*} \nabla_{x}f(t,Z_{t,z}) \, dB_{t}.
\end{equation*}


问:
Calculate $\lim_{x \to 0^+} \frac{3x + \sqrt{x}}{\sqrt{1- e^{-2x}}}$
I need to calculate:
$$\displaystyle \lim_{x \to 0^+} \frac{3x + \sqrt{x}}{\sqrt{1- e^{-2x}}}$$

I looks like I need to use common limit:
$$\displaystyle \lim_{x \to 0} \frac{e^x-1}{x} = 1$$

So I take following steps:

$$\displaystyle \lim_{x \to 0^+} \frac{3x + \sqrt{x}}{\sqrt{1- e^{-2x}}} = \displaystyle \lim_{x \to 0^+} \frac{-3x - \sqrt{x}}{\sqrt{e^{-2x} - 1}}$$

And I need to delete root in the denominator and make nominator equal to $-2x$. But I don't know how.

答: Note that\begin{align}\lim_{x\to0^+}\frac{\left(3x+\sqrt x\right)^2}{1-e^{-2x}}&=\lim_{x\to0^+}\frac{9x^2+6x\sqrt x+x}{1-e^{-2x}}\\&=-\lim_{x\to0^+}\frac x{e^{-2x}-1}\left(9x+6\sqrt x+1\right)\\&=-\frac1{\lim_{x\to0^+}\frac{e^{-2x}-1}x}\times\lim_{x\to0^+}\left(9x+6\sqrt x+1\right)\\&=-\frac1{-2}\times1\\&=\frac12.\end{align}Therefore, the limit that you're after is $\sqrt{1/2}$.


问:
Given a characteristic polynomial of a Matrix, prove a certain Matrix is equal to 0
I've been trying to solve this issue:

Let $A$ be a diagnosable $n\times n$ matrix. It's characteristic polynomial is: $p(t) = a_n t^n +a_{n-1}t^{n-1}+...+a_1t+a_0$.

Given the square $n\times n$ matrix $p(A)=a_nA^n+a_{n-1}A^{n-1}+...+a_1A+a_0$, prove that $p(A)=0$

We can deduce that $a_n = 1$, and that $a_{n-1}=-tr(A)$, or the minus the sum of all eigenvalues.
But I'm not sure what more can we find out about the given characteristic polynomial.

Any help would be appreciated.

EDIT: Thanks to the help of Sam Freedman, a solution was found.

Matrix A is diagnosable, which means it is similar to a diagonal Matrix D:

$A = P^-1DP$

That means we can define $p(A)$ with matrix D:

$p(P^-1DP) = P^-1(D^n +a_{n-1}D^{n-1}+...+a_1D+a_0)P$

We can conclude that matrix $p(A)$ is similar to a diagonal matrix, described above.

Matrix $D$ can be broken down to: $diag(d_1,...,d_n)$. And the diagonal matrix $p(A)$ is similar to can will have the value: $d_i^n+a_{n-1}d_i^{n-1}+...+a_1d_i +a_0$, which means those are the eigenvalues of $p(A)$.

Now, thanks to knowing the characteristic polynomial of $A$, we know that the group $d_1,...,d_n$ are the roots of said polynomial, meaning they send it to 0.

And the form of each eigenvalue of $p(A)$ is equal to the form of said characteristic polynomial. Which means each eigenvalue of $p(A)$ is equal to 0.

The diagonal matrix $p(A)$ is similar to must be the zero matrix. And the zero matrix is the only matrix which is similar to itself, and therefore we can conclude and claim:

$p(A) = 0$

Many thanks, Sam Freedman!

答:
This is the famous Cayley-Hamilton theorem, but we can also prove this special diagonalizable case directly.

Writing $A = P^{-1} D P$, we use a standard trick to compute out
\begin{align*}
p(A) = p( P^{-1} D P) &= ( P^{-1} D P)^n + a_{n - 1}( P^{-1} D P)^{n - 1} + \dots + a_1 ( P^{-1} D P) + a_0I \\
&=  P^{-1} D^n P + a_{n - 1}P^{-1} D^{n - 1} P + \dots + a_1 ( P^{-1} D P) + a_0I\\
&= P^{-1}( D^n + a_{n - 1} D^{n - 1} + \dots + a_1 D + a_0I ) P.
\end{align*}

Now we've reduced the problem to the case of a diagonal matrix. Given what you know about the characteristic polynomial of a diagonal matrix, can you go from here?


问:
$H_{n}:Comp(\mathcal{C}) \to \mathcal{C}$ is an additive functor implies $H_{n}:K(\mathcal{C}) \to \mathcal{C}$ is an additive functor.
We know that if $\mathcal{A}$ is an abelian category, then the homology functor $H_{n}:Comp(\mathcal{C}) \to \mathcal{C}$ is an additive functor for every $n \in \mathbb{Z}$. One can find a prove of this statement in Rotman´s Homological Algebra Proposition 6.8. But sometimes why the same holds for $H_{n}:K(\mathcal{C}) \to \mathcal{C}$,  where $K(\mathcal{C})$ is the homotopy category ? Maybe my question is simple and silly but for example if I have $[f],[g] \in K(\mathcal{C})$ I need to prove among other things that $H_{n}([gf])=H_{n}([g])H_{n}([f])$ or that as  $K(\mathcal{C})$ is also an additive category then $H_{n}([f+g])=H_{n}([f])+H_{n}([f])$. Is there a way to easily prove this if its already proved in the category of complexes. Also how I know $H_{n}([f])$ is well defines if I already know that the assignation $H_{n}(f)$ is well defined. Thanks.

答:
I'm a little more comfortable working with cohomology, so I'm going to answer this question in that language; of course the concepts are dual, so simply dualizing gives you the corresponding result for homology. The easiest way (in my opinion, anyway) to see that the cohomology functors
$$
H^n:K(\mathscr{A}) \to \mathscr{A}
$$
are additive for any Abelian category $\mathscr{A}$ is to use the universal property of the homotopy category. By definition, the category
$$
K(\mathscr{A}) = \mathbf{Ch}(\mathscr{A})_{/\text{homotopy}}
$$
is a strict localization, i.e., it is given by inverting a congruence relation. It also follows from the bilinearity of chain homotopies (i.e. if $f \simeq g$ and $h \simeq k$ then $f + g \simeq f + k \simeq g + h \simeq h + k$) that the quotient functor
$$
q:\mathbf{Ch}(\mathscr{A}) \to K(\mathscr{A})
$$
is also additive. Now recall that applying the universal property of a localization of categories says that in our case if you have a functor $F:\mathbf{Ch}(\mathscr{A}) \to \mathscr{C}$ which kills homotopy (so if $g \simeq h$ then $Fg = Fh$) then there exists a unique functor $\overline{F}:K(\mathscr{A}) \to \mathscr{C}$ making the diagram
$$
\begin{array}{ccc}
 \mathbf{Ch}(\mathscr{A}) & \xrightarrow{F} & \mathscr{C} \\
 q\downarrow & \nearrow & \\
 K(\mathscr{A})
\end{array}
$$
where the diagonal arrow is $\overline{F}$. In the case we actually care about, we use the functor $H^n:\mathbf{Ch}(\mathscr{A}) \to \mathscr{A}$ and produce the commuting diagram of categories
$$
\begin{array}{ccc}
\\
 \mathbf{Ch}(\mathscr{A}) & \xrightarrow{H^n} & \mathscr{A} \\
 q\downarrow & \nearrow & \\
 K(\mathscr{A})
\end{array}
$$
where in this case the diagonal arrow is the cohomology functor on the homotopy category (which I will denote $\overline{H}^n$ for the moment. To see that this is additive, we just need to prove that $\overline{H}^n$ preserves finite products. For this we first consider the zero object $0_K \in K(\mathscr{A})_0$ and note that $0_K \cong q(0_C)$ for $0_C$ the zero object in $\mathbf{Ch}(\mathscr{A})$. Then
$$
\overline{H}^n(0_K) \cong \overline{H}^n(q(0_C)) = H^n(0_C) \cong 0_{\mathscr{A}}.
$$
Now fix two arbitrary complexes $A^{\bullet}, B^{\bullet} \in \mathbf{Ch}(\mathscr{A})_0$ and consider the product $A^{\bullet} \times B^{\bullet}$. We note that because $q$ is additive and acts as the identity on objects, we have that
$$
A^{\bullet} \times B^{\bullet} = q(A^{\bullet}) \times q(B^{\bullet}) \cong q(A^{\bullet} \times B^{\bullet}).
$$
Now using the additivity of $H^n$ and the fact that $H^n = \overline{H}^n \circ q$ gives
$$
\overline{H}^n(A^{\bullet} \times B^{\bullet}) \cong \overline{H}^n(q(A^{\bullet} \times B^{\bullet}) = H^n(A^{\bullet} \times B^{\bullet}) \cong H^n(A^{\bullet}) \times H^n(A^{\bullet}) \cong \overline{H^n}(q(A^{\bullet})) \times \overline{H^n}(q(B^{\bullet})) = \overline{H}^n(A^{\bullet}) \times \overline{H}^n(B^{\bullet}),
$$
which proves that $\overline{H}^n$ preserves finite products. We can thus conclude that $\overline{H}^n$ is an additive functor, as was desired.


问:
how to solve this multivariate quadratic equation?
Any hope to acquire an analytic solution to such equations:

Solve:
$$\sum_{j=1}^n a_{ij} x_i x_j = b_i$$

for $i=1,\ldots,n$, where $a_{ij}$'s and $b_i$'s are known constants and $x_i$'s are unknowns to be solved.

Thanks a lot!

P.S. Thanks for alex.jordan's interesting comment! Let's consider this problem in a positive setting, i.e. let's require all coefficients($a_{ij}$'s and $b_i$'s) to be positive so that the solution seems to exist. Also $a$ is symmetric, i.e. $a_{ij}=a_{ji}$. If there could be any fast numerical solution it's also useful.

P.S. Just for the information of all readers: the original form of this problem is formulated as:
$$ x_i\cdot\left(\sum_{j=1}^n a_{ij}x_j\right) = \sum_{j=1}^n a_{ij}q_{ij} $$ where $q_{ij}\in(0, 1)$. I simplified the RHS into $b_{ij}$ but now it seems better not to do so(sorry for that!) to at least give the existence of real solutions a better chance.

答:
As "everything in sight is positive" write your system in the form
$$\sum_{j=1}^n a_{ij} x_j={b_i\over x_i}\qquad(1\leq i\leq n)$$
and use Newton's method: Given an approximate solution $x=(x_1,\ldots, x_n)$ find a better approximation $x+\Delta$ by solving
$$\sum_{j=1}^n a_{ij} (x_j+\Delta_j)={b_i\over x_i+\Delta_i}\doteq{b_i\over x_i}\left(1-{\Delta_i\over x_i}\right)\qquad(1\leq i\leq n) $$
for $\Delta$. This means solving the linear system
$$\sum_{j=1}^n \left(a_{ij}+\delta_{ij}{b_i\over x_i^2}\right)\>\Delta_j={b_i\over x_i} -\sum_{j=1}^n a_{ij}x_j\qquad(1\leq i\leq n)$$
for the $\Delta_j$.</p>


Title: Calculate $\lim_{x \to 0^+} \frac{3x + \sqrt{x}}{\sqrt{1- e^{-2x}}}$

Body:
I need to calculate:
$$\displaystyle \lim_{x \to 0^+} \frac{3x + \sqrt{x}}{\sqrt{1- e^{-2x}}}$$

I looks like I need to use common limit:
$$\displaystyle \lim_{x \to 0} \frac{e^x-1}{x} = 1$$

So I take following steps:

$$\displaystyle \lim_{x \to 0^+} \frac{3x + \sqrt{x}}{\sqrt{1- e^{-2x}}} = \displaystyle \lim_{x \to 0^+} \frac{-3x - \sqrt{x}}{\sqrt{e^{-2x} - 1}}$$

And I need to delete root in the denominator and make nominator equal to $-2x$. But I don't know how.

Body:
$$\lim_{x \to 0^+} \frac{3x + \sqrt{x}}{\sqrt{1- e^{-2x}}} =\lim_{x \to 0^+}\frac{\sqrt{x}}{\sqrt{1- e^{-2x}}} \cdot(3\sqrt{x} + 1) = \lim_{x \to 0^+} \sqrt{\frac{x}{1- e^{-2x}}} \cdot\lim_{x \to 0^+} (3\sqrt{x} + 1)$$

$$\lim_{x \to 0^+} \sqrt{\frac{x}{1- e^{-2x}}} = \lim_{x \to 0^+} \sqrt{\frac{-2x}{(2)( e^{-2x} -1)}}= \frac{1}{\sqrt{2}}$$


问:
$f \in L^p(]0,1[)$ almost everywhere.
Why are $L^p$ functions defined almost everywhere? As if we are defining an $L^p$ function on a set $\Omega=[0,1]$ , so we say $f \in L^p(]0,1[)$ almost everywhere. And not on $[0,1]$ !

答:
That's because of the way we construct the $L^p$ spaces.
First you look at the vector space (let $\Omega \subseteq \mathbb{R}$)
$$
\mathcal{L}^p = \left\{ f\colon \Omega \to \mathbb{R} : f \text{ measurable and} \int_\Omega |f(x)|^p dx < \infty\right\}
$$
and the mapping
$$
q\colon \mathcal{L^p} \to [0,\infty[,\ f\mapsto \Bigg(\int_\Omega |f(x)|^p dx\Bigg)^{\frac{1}{p}}
$$
and notice that it defines a seminorm (use Minkowski's inequality).
Next you look want to "convert" this into a norm.
When you have a seminorm you can look at the set
$$
N = \left\{f \in \mathcal{L}^p | q(f)=0\right\} = \left\{f \in \mathcal{L}^p : \int_\Omega |f(x)| dx = 0\right\}.
$$
Now you define $L^p := \mathcal{L}^p / N$ as the quotient vector space.

What this essentially does is the following:
We have
$$
f=g \text{ a.e.} \Leftrightarrow \int_\Omega |(f-g)(x)| dx= 0\\
\Leftrightarrow q(f-g)= \Bigg(\int_\Omega |(f-g)(x)|^p dx\Bigg)^{\frac{1}{p}} = 0.
$$

so every time $f=g$ almost everywhere, they land in the same equivalence class. This means, while they may be different functions (they can differ on a null set), they are "the same thing" in $L^p$. When we integrate, we can't tell the difference.

For example you could look at $L^p([0,1])$ and the functions
$$
f\colon [0,1]\to \mathbb{R},\ f(x) = \begin{cases}0,\ \text{if } x\in \mathbb{Q},\\ 1,\ \text{otherwise.}\end{cases}\\
g\colon [0,1]\to \mathbb{R},\ g(x) = 1.
$$
Now $g$ and $f$ only differ on a set of measure null (they are equal almost everywhere). But this means they lie in the same equivalence class. Therefore
they are the same element in $L^p$.
So although we might have different functions $f$ and $g$, if we can't tell them apart by our integral ($q(f)=q(g)$), we say they are the same element in $L^p$.
That's why it's okay to define them almost everywhere. We can always set them to $0$ or basically anything we want on the null set and get a function that represents the same thing and is fully defined.

In your case $\{0,1\}$ is null set, therefore it doesn't matter if we talk about $[0,1]$ or $[0,1]\setminus \{0,1\} = ]0,1[$ and define the function only on the smaller set.

[I'm working with the Lebesgue measure in this post]

edit regarding your comment:
Well, we know that for $x\in [0,\infty[$ and $r\in ]0,\infty[$
$$
x^r = 0 \Leftrightarrow x=0,
$$
therefore
$$
q(f) = 0 \Leftrightarrow \int_\Omega |f|^p dx = 0.
$$
By the same argument we know $|f(x)|=0 \Leftrightarrow |f(x)|^p = 0\ (\forall x\in\Omega)$
and thus
$$
\int_\Omega |f|^p dx = 0 \Leftrightarrow \int_\Omega |f| dx = 0.
$$

Title: $f \in L^p(]0,1[)$ almost everywhere.


Body:
$$\sqrt{1-e^{-2 x}}\sim \sqrt{2} \sqrt{x};\;\text{as }x\to 0$$

$$\lim_{x \to 0^+} \frac{3x + \sqrt{x}}{\sqrt{2} \sqrt{x}}=\lim_{x \to 0^+}\left(\frac{3}{\sqrt 2}\sqrt x +\frac{1}{\sqrt 2} \right)=\frac{1}{\sqrt 2} $$


问:
Equivalence classes of the relation that the largest digit of integer a = largest digit of integer b.
Define the relation $\mathcal{R}$ on the set of all positive integers by: for all positive integers $a$ and $b$, $a\,\mathcal{R}\,b$ if and only if the largest digit of $a$ is equal to the largest digit of $b$. for example, $271\,\mathcal{R}\,770$ because the largest digit of $271$ is $7$ which is also the largest digit of $770$.

a) Find the number of equivalence classes of $\mathcal{R}$.

b) Find and simplify the number of positive integers between $100$ and $1000$ which are in the equivalence class $[271]$.

I have an idea of where to begin but I think I am wrong. For (a) would there be $9$ equivalence classes?
And for (b) I got my answer to be $56$.

答:
You're right on target for (a).

You can find the answer to (b) as follows:

Let's start by taking care of the numbers in $[271]$ whose first digit is $7$. The last two digits of such a number will need to be at most $7$, so we have $8$ options for each of the last two digits, and so there are $8\cdot 8=64$ ways we can pick the last two digits of such a number. Thus, there are $64$ numbers in $[271]$ between $100$ and $1000$ whose first digit is $7$.

Now, if the second digit is $7$ and the first digit is not $7$, then how many options are there for the first digit? How many options for the last? Multiply these together to get the number of elements of $[271]$ between $100$ and $1000$ with second digit $7$ and first digit not $7$.

If the last digit is $7$ and neither the first two digits is $7$, then how many options are there for the first digit? How many for the second? Multiply these two numbers together to get the number of elements of $[271]$ between $100$ and $1000$ with last digit $7$ and first two digits not equal to $7$.

You should be able to show that every element of $[271]$ between $100$ and $1000$ is of exactly one of the three types we discussed above, so we just add up the numbers of each type of element to get the answer.


问:
Convolution of a compactly supported $f \in L^1(\mathbb{R}^n)$ with a function $g \in L^1_{\text{loc}}(\mathbb{R}^n)$
Suppose $f$ and $g$ are two $L^1_{\text{loc}}(\mathbb{R}^n)$-functions. Their convolution is well-defined under many conditions, three of which are:

When $f$ is compactly supported and continuous, and $g$ is just $L^1_{\text{loc}}(\mathbb{R}^n)$.

When $f$ and $g$ are in $L^1(\mathbb{R}^n).$

When $f$ and $g$ are in $L^2(\mathbb{R}^n).$

There are many more. However, I'm wondering whether the following is true:

If $f,g \in L^1_{\text{loc}}(\mathbb{R}^n)$ are such that $f$ is compactly supported (thus also in $L^1(\mathbb{R}^n)$), then $f * g$ is well-defined almost everywhere. Also, do I then have $f * g \in L^1_{\text{loc}}(\mathbb{R}^n)$?

The proof of $1.$ doesn't go through: it's based on the fact that $f$ is bounded, and I don't have that if $f$ is not continuous. The proof of $2.$ that I've found goes along the following lines: let $H(x,y) := f(x-y)g(y).$ Then $$\int_{\mathbb{R}^{2n}} |H(x,y)|dxdy = \int_{\mathbb{R}^n} \left( \int_{\mathbb{R}^n} |f(x-y)|dx \right) |g(y)| dy = ||f||_1 \cdot ||g||_1, $$so by Fubini, the integral $\int_{\mathbb{R}^n} f(x-y)g(y)dy$ exists and is integrable for a.e. $y \in \mathbb{R}^n.$ However, in my case, $H$ is not necessarily integrable.

It's hard to think of a counterexample, because we'd need $f$ to be so discontinuous as to ruin the local integrability of $g$ on a set of positive measure! I can easily do it at one point, (for example $f(y) = g(y) = \frac{1}{\sqrt{y}}$ at $x = 0$), or even at a countable set of points (think of infinitely many "volcanoes" of the above type converging to zero), but doing it for a large enough set of points seems very hard to me.

答:
(Hoping I didn't make a mistake here..) If $f$ has compact support and is locally integrable, then $f$ is actually integrable (take compact $K$ such that $\operatorname{supp}(f)\subseteq K$) which will make $f*g$ locally integrable. Let $K$ be compact, then

$$\int_K |(f*g)(x)|\,dx = \int_K \bigg|\int_{\operatorname{supp}(f)}f(y)g(x-y)\,dy\bigg|\,dx \le \int_K \int_{\operatorname{supp}(f)}|f(y)||g(x-y)|\,dy\,dx.$$

By Fubini-Tonelli, we can interchange these even if it diverges to infinity, so

$$\int_K |(f*g)(x)|\,dx \le \int_{\operatorname{supp}(f)}|f(y)|\int_K |g(x-y)|\,dx\,dy.$$

The $x$ integral is finite since $g$ is locally integrable and $K+y$ is also a compact set. The $y$ integral is finite since $f$ is actually just straight up integrable.


问:
Truncated Taylor series of the exponential
Let $N \in \mathbb N^*$, $\delta > 0$, $t > 0$ and consider
\begin{equation}
f(t,\delta, N)= e^{-t/\delta} \sum_{k=0}^{N-1}\frac{(t/\delta)^k}{k!}. \qquad \qquad \qquad (1)
\end{equation}
Let now $N = \lceil\delta^{-\gamma}\rceil$, with $0 < \gamma < 1$. Can we conclude that
\begin{equation}
\lim_{\delta \to 0} f(t,\delta,\lceil \delta^{-\gamma}\rceil)=0,
\end{equation}
for all $t > 0$?

The reasoning is that if $N \to \infty$ (for $\delta$ fixed), then the sum tends to $e^{t/\delta}$ and therefore $f(t,\delta,N)\to 1$. If on the other hand $\delta \to 0$, for $N$ fixed, we have that $f(t,\delta,N)\to 0$ since the exponential goes faster to zero than the polynomial to infinity. In this case, we let $\delta \to 0$ and simultaneously $N \to \infty$, but with a "slower pace". Can we conclude that the limit is zero? If yes, how?

Edit:

After some reasoning I came up with
$$
f(t,\delta,N)= \frac{\Gamma(N, t/\delta)}{(N-1)!},
$$
with $\Gamma(\cdot, \cdot)$ the upper incomplete Gamma function. I implemented this in Matlab, and the interesting behaviour is that if $N=\delta^{-1}$, then the limit (1) is $1/2$ if $0 <t \leq 1$ and it is 0 if $t > 1$. If $N=\delta^{-\gamma}$, with $\gamma \in (0, 1)$, the limit is 0, and if $N = \delta^{-\gamma}$, with $\gamma > 1$, then the limit is 1, independently of $t$.

答:
For simplicity let $x=t/\delta$ for the moment.

The main point is that $\frac{x^k}{k!}$ is within a polynomial factor of $\left ( \frac{xe}{k} \right )^k$, by Stirling. (It's within a bounded factor of $\frac{(xe)^k}{k^{k+1/2}}$ but this is overkill for this application.)

So for $0<c<1$, the sum of all the terms with $k>x(1+c)$ behaves basically like $\left ( \frac{e}{1+c} \right )^{x(1+c)} \frac{c+1}{c}$. This is the dominant term times $\frac{1}{1-r}$ where $r=\frac{1}{1+c}$ is the common ratio of the bounding geometric series.

Similarly for the first $x(1-c)$ terms you have a bound like $\left ( \frac{e}{1-c} \right )^{x(1-c)} \frac{1}{c}$.

Returning to your setting, if $\gamma>1$ then you exclude only the terms past $x(1+c)$ where $c \to \infty$ as $\delta \to 0$, and so the error goes to zero.

If $\gamma<1$ then you exclude all the terms past $x(1-c)$ for $c \to 1$ as $\delta \to 0$ and so the terms you are including become negligible relative to $e^x$, i.e. the error goes to $1$.

If instead you look at $N \sim C \delta^{-1}$ then the above bounds properly come into play.


问:
Recurrence relations for strings of $0$'s and $1$'s of length $n$ where no $k$ consecutive $1$'s appear.
This question is related to this one, but the answers there do not explain my confusion.

This is exercise 20 in page 28 in the book generatingfunctionology:

Let $f(n,m,k)$ be the number of strings of $n$ $0$'s and 1's that contain exactly $m$ $1$'s, no $k$ of which are consecutive. Then

(a) Find a recurrence formula for $f$. It should have $f(n,m,k)$ on the left side, and exactly three terms on the right.

(b) Find, in simple closed form, the generating functions

$$ F_k(x,y)=\sum_{n,m\geq 0}f(n,m,k)x^ny^m\quad(k=1,2,...)$$

Here's what I have:

I didn't get exactly three terms in the recurrence, but I did come up with one plausible. Well, we want to use $n-m$ $0$'s to separate $m$ $1$'s. That's the same as putting $m$ balls into $n-m+1$ ordered boxes, and each box can at most contain $k-1$ balls. We divide the cases by how many balls the last box has: it could be $0,1,...,k-1$. Hence we have the recurrence relation:

$$f(n,m,k)=\sum_{i=1}^{k}f(n-i,m-i+1,k)$$

This recurrence should hold for arbitrary $n,m,k$ as long as we specify the appropriate initial values, or at least it should hold for $k$ large enough. (This might be where I ran into problems but I can not see it immediately).

However, when I carry this to the next part I am confused, because

\begin{aligned}
 F_k(x,y)& = \sum_{n,m\geq 0}f(n,m,k)x^ny^m \\
& = \sum_{n,m\geq 0}\sum_{i=1}^{k}f(n-i,m-i+1,k)x^ny^m \\
& = \sum_{i=1}^k\sum_{n,m\geq 0}f(n-i,m-i+1,k)x^ny^m \\
& = \sum_{i=1}^kx^iy^{i-1}\sum_{n,m\geq 0}f(n-i,m-i+1,k)x^{n-i}y^{m-i+1}\\
& = \sum_{i=1}^kx^iy^{i-1}F_k(x,y)
\end{aligned}
The second to bottom equality holds in my mind because $f(m,n,k)=0$ if either of $m$ or $n$ is $0$. However, it confuses me since at the end the equation implies $F_k(x,y)=0$?

Where did I go wrong? Any kind of help would be appreciated.

答:
Your recurrence relation needs a base case: as written, it does not hold when $n=0$.
$$
f(0,0,k) = 1\\\qquad\;\;\;\;\;\;\;\;\;\;
f(0,m,k) = 0\qquad (m>0)
$$
Incorporating this into your generating function calculations should allow you to solve for $F$.


问:
simplifying math equation with factorials
I am trying to understand how simplified that equation:
$$\frac{(x-2)!}{(x-4)!}-46=\frac{x!}{(x-1)!}$$

In my book it was solved like this:

but I really don't understand how it happened? Where did $(x-3)(x-4)(x-5)$ come from  in first fraction? It looks like I missed something fundamental about factorials...

Body:
$$\frac{(x-2)!}{(x-4)!}-46=\frac{x!}{(x-1)!}\iff \frac{(x-2)!}{(x-4)!}-\frac{x!}{(x-1)!}-46=0$$
$$\frac{(x-2)!}{(x-4)!}-\frac{x!}{(x-1)!}-46= \frac{(x-1)(x-2)(x-3)}{(x-1)(x-2)(x-3)}\cdot \frac{(x-2)!}{(x-4)!}-\frac{x!}{(x-1)!}-46=
$$
$$=\frac{[(x-2)(x-3)]\cdot(x-1)!}{(x-1)!}-\frac{x\cdot(x-1)!}{(x-1)!}-\frac{46(x-1)!}{(x-1)!}=(x-2)(x-3)-x-46
$$


问:
almost sure convergence and convergence of expectation implies convergence in expectation
IF $X_n \xrightarrow{\text{a.s.}}X$ and $\mathbf{E}[X_n] \rightarrow \mathbf{E}[X]  $ does this imply $\mathbf{E}[|X_n - X|] \rightarrow 0$?

答:
No.

For instance let $U$ be uniformly distributed on $(0,1)$ and for all $n\in\mathbb N^*$ let $X_n$ be defined by
$$
X_n=-n²1_{\{U<\frac1n\}}+n^21_{\{U>1-\frac1n\}}.
$$

Then $X_n\underset{n\to+\infty}{\overset{\textrm{a.s.}}{\longrightarrow}}0$ and $\mathbb E[X_n]=0\underset{n\to+\infty}{\longrightarrow}\mathbb E[0]$ but $\mathbb E[\vert X_n-0\vert]=2n\underset{n\to+\infty}{\longrightarrow}+\infty$.


问:
Show that Function Compositions Are Associative
My intent is to show that a composition of bijections is also a bijection by showing the existence of an inverse.  But my approach requires the associativity of function composition.

Let $f:  X \rightarrow Y, g:  Y \rightarrow Z, h:  Z \rightarrow W$ be functions.

$((f \circ g) \circ h)(x) = h((f \circ g)(x)) = h(g(f(x)))$, and $(f \circ (g \circ h))(x) = (g \circ h)(f(x)) = h(g(f(x)))$.

However, I am having problems in justifying that the two compositions, $(f \circ g) \circ h$ and $f \circ (g \circ h)$, have the same domain and range.  When I consulted ProofWiki, whose link is at the bottom, I got even more confused.  Specifically, for $(f \circ g) \circ h = f \circ (g \circ h)$ to be defined, ProofWiki requires that dom$g =$ codom$f$ and dom$h =$ codom$g$.

First of all, I think that it should be dom$g =$ range$f$ ....  Moreover, as you can see in the example below, you actually have to adjust domains and ranges of $f, g, h$ for the requirement to hold true.

Let $f: \mathbb R \rightarrow \mathbb R$ be $f(x) = 2x$, $g: \mathbb R^+ \rightarrow \mathbb R$ be $g(y) = ln(y)$, $h: \mathbb R \rightarrow \mathbb R$ be $h(z) = z - 10$.
  Then $((f \circ g) \circ h)(x) = ln(2x) - 10 = (f \circ (g \circ h))(x)$, with dom$((f \circ g) \circ h) = \mathbb R^+$ = dom$(f \circ (g \circ h))$.  As a result, we need to set dom$f = \mathbb R^+$, range$f = \mathbb R^+$; dom$g$, range$g$, dom$h$, and range$h$ remain the same.  Am I allowed to do that?

This adjustment implies that when we say dom$f = X$, $f$ must be defined for all elements in $X$, but $X$ may not be the entire set of elements for which $f$ is defined.

http://www.proofwiki.org/wiki/Composition_of_Mappings_is_Associative

答:
Perhaps it's EVEN easier (clearer?) to reason about a more general construction (with heavy inspiration both from the definition of a category, the excellent answer of mcg256, and the helpful comment of Milan)?

We can view a relation R(A,B) as a collection of paths so that each path has a unique initial point in the set A and a unique terminal point in the set B.

In the special case of a relation a path is determined by its endpoints, but this is NOT a necessary assumption.

Thus, to define composition more generally, we can view the starting data as two collections of arrows M(A,B) and M(B,C), every arrow in M(X,Y) has source in X and target in Y, and we can concoct a third collection of arrows M(A,B)*M(B,C)=M(A,B,C) via cancatenation,  we can combine two arrows precisely if the target of the first is the source of the second.

To check associativity, given the starting data M(A,B) M(B,C) and M(C,D), we must show that M(A,B,C)*M(C,D) ``='' M(A,B)*M(B,C,D). We are at risk of having a canonical isomorphism rather than equality, and hence the quotes around the symbol =.

Precisely the same input is needed to manufacture a typical arrow in each side of the previous equation.

The pictures one might draw to illustrate a typical arrow in the respective calculations are

(A->B-->C)--->D and A->(B-->C--->D).

Thus, if we define M(A,B,C,D) as the collection of all diagrams A->B-->C--->D, (so that the target of -> is the source of --> etc..) then the 3 collections at hand are all canonically isomorphic.

Have we shown that composition is associative?

Perhaps we have at least gotten to the heart of the matter, and established a precise sense in which 3 formally different things are the same.

It is tempting to say that in the special case of functions and relations, that the 3 formaly different things are precisely the same, but even here perhaps we should be careful, since we are perhaps relying on canonical isomorphisms among products of finite sets:

(AxB)xC is canonically isomorphic to Ax(BxC) which in turn is canonically isomorphic to the collection of all ordered triples (a,b,c) from the respective factors.

It is also tempting to say that in practice, no confusion arises, if our biggest crime is conflating canonically isomorphic things.

However the vector (a,b,c) reminds me of a purported proof of a VERY famous conjecture in number theory, which according to my extremely pedestrian grasp of the opinion of experts, is perhaps problematic, due in part the conflation of: thing versus thing up to isomorphism or perhaps isomorphic copy of thing.

Attempted summary of the post at hand: The candidate isomorphism from the types (xy)z and x(yz) is in fact an isomorphism, rendering a sense in which xyz is unambiguous.

But, much like the composer of this post, https://explainingmaths.wordpress.com/2012/04/01/is-composition-of-functions-associative/,

I am unsatisfied with the post at hand. It is obvious that function composition is/(must be) associative, but attempts at modest generalization, leave me questioning where, if anywhere, there might be a bit of sleight of hand.

It would be fun to have a long thoughtful discussion leading to the agreement that indeed all the matters at are trivially obvious, obviously trivial, and that nowhere has there been any sleight of hand after all.


问:
Does an elementary solution exist to $x^2+1=y^3$?
Prove that there are no positive integer solutions to $$x^2+1=y^3$$
This problem is easy if you apply Catalans conjecture and still doable talking about Gaussian integers and UFD's. However, can this problem be solved using pre-university mathematics?

I am talking about elementary number theoretical solutions. Do they exist?

答:
An alternative proof and a generalisation

As stated in the accepted answer, it is sufficient to prove that $3a^4+3a^2+1=b^2$ has  no positive integer solution. With a change of notation we shall prove the following generalisation by using the same method used for Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution?

Theorem

No equation of either of the forms
$$ Ax^4-6x^2y^2+Cy^4=z^2,AC=-3\tag{1}$$ $$ ax^4+3x^2y^2+cy^4=z^2,ac=3\tag{2}$$
has a positive integer solution.

Proof

First note that, for either equation, we can suppose that $x,y,z$ are pairwise coprime  since a common factor of any pair of $x,y,z$ would be a factor of all and cancellation can occur.

Also note that precisely one of $A$ and $C$ is divisible by $3$. Without loss of generality we can suppose that $3$ is a factor of $C$ and that $A\in \{\pm 1\}.$ Then $z^2\equiv A\pmod 3$ and so $A=1$. Similarly, we can suppose $a=1$.

An equation of form (1)

$x^4-6x^2y^2-3y^4=z^2$ can be rewritten, using completing the square, as $$\left (\frac{x^2-3y^2-z}{2}\right )\left (\frac{x^2-3y^2+z}{2}\right)=3y^4.$$
Since the two bracketed factors, $L$ and $M$ say, differ by the integer $z$ and have integer product, they are both integers. Furthermore, if $q$ is a prime common factor of $L$ and $M$, then $q$ would be a factor of both $z$ and $y$, a contradiction.

Therefore $\{L,M\}=\{au^4,cv^4\}$, where $ac=3$ and $y=uv$, with $u$ and $v$ coprime. Then $$au^4+cv^4=x^2-3y^2=x^2-3u^2v^2.$$ Therefore $au^4+3u^2v^2+cv^4=x^2$, $ac=3$, an equation of form (2).

It is important to note that the mapping $(x,y,z)\rightarrow (u,v,w)$ is invertible. Only one solution set can map to $(u,v,w)$ by this process.

An equation of  form (2)

Let $u,v,x$ be a pairwise coprime solution of  $ u^4+3u^2v^2+3v^4=x^2$ and let $t$ be the greatest common divisor of $v$ and $2$. Then $(U,V,W)=(\frac{2u}{t},\frac{v}{t},\frac{4x}{t^2})$ is a pairwise coprime solution of $U^4+12U^2V^2+48V^4=W^2$.

This can be rewritten, using completing the square, as $$\left (\frac{U^2+6V^2-W}{2}\right )\left (\frac{U^2+6V^2+W}{2}\right)=-3V^4.$$
The bracketed factors, $L$ and $M$,  are again coprime integers.Therefore $\{L,M\}=\{aX^4,cY^4\}$, where $ac=-3$ and $V=XY$, with $X$ and $Y$ coprime. Then $$aX^4-6X^2Y^2+cY^4=U^2,ac=-3,$$ an equation of form (1). Again, this mapping of solutions is invertible.

Conclusion

We have seen that any positive integer solution $(x,y,z)$ of an equation of form (1) leads to another positive integer solution $(X,Y,Z)$, where $$Y=\frac{y}{tuX}.$$
Any monotonically strictly decreasing series of positive integers must terminate and so the above process must lead to solutions with $tuX=1$.

It is now straightforward to plug $u=X=1$ and thus $U=2$ back into the equations to obtain   the contradiction
$$1-6Y^2-3Y^4=4.$$


问:
Needs help with numerical analysis programming
I'm trying to implement, in Matlab, the following methods for root finding:

Fixed point iteration with g(x) = x-αf(x)
Newton's method,
The secant method,
Steffensen's method.

For f(x)=x^2+x-1, with x0 = 1

However, since I haven't used Matlab before, my loops for each algorithm are full of bugs. Can someone please show me a correct way to do them?

答:
I quickly threw together a code that uses Newton's method, I hope it's correct. In any case you should be able to see how the syntax of Matlab works.</p>
```<code>f = @(x) x^2 + x -1; % function
df = @(x) 2*x + 1;   % derivative needed for newtons method

e = 10^-7;           % error margin
x = 0;               % initial value

n = 0;               % number of iterations

while abs(f(x)) > e  % apply newton's method until error small enough
    x = x - (f(x)/df(x));
    n = n+1;
end

fprintf('root: %i\n', x)
fprintf('number of iterations: %i\n', n)
</code>```

Body:
I quickly threw together a code that uses Newton's method, I hope it's correct. In any case you should be able to see how the syntax of Matlab works.

f = @(x) x^2 + x -1; % function
df = @(x) 2*x + 1;   % derivative needed for newtons method

e = 10^-7;           % error margin
x = 0;               % initial value

n = 0;               % number of iterations

while abs(f(x)) > e  % apply newton's method until error small enough
    x = x - (f(x)/df(x));
    n = n+1;
end

fprintf('root: %i\n', x)
fprintf('number of iterations: %i\n', n)


问:
Can the second term of the Schur complement of a symmetric matrix be undefined?
Given the next symmetric matrix conformably partitioned

$$\begin{bmatrix}
A &B  \\ B^T  &C
\end{bmatrix}$$

I know that $A$ and $C$ are positive definite matrices.

The Schur complement is $S=C-B^TA^{-1}B$

What can I say about $B^TA^{-1}B$? Is this undefined in general? or it is positive/negative (semi)definite.

Probably it is an easy question, but I do not see why.
Thanks in advance.

答: If $A \succ 0$, $S$ is positive semidefinite iff the block matrix is positive semidefinite.</p>


问:
How to prove that the quadratic variation of $X_t = W_t^4 - 6tW_t^2 + 3t^2$ is integrable?
My task is to prove that $X_t = W_t^4 - 6tW_t^2 + 3t^2$ is a martingale using Itô's lemma.

So far I've proven that it is a local martingale by applying the lemma to the function $f(t,x) = x^4 - 6tx^2 + 3t^2$ and finding that $dX_t = (4W_t^3 - 12tW_t)dW_t = \sigma_t dW_t$, showing that there is no drift term. Is this correct?

If I understand correctly, to prove that this local martingale is a martingale, I need to prove that it's quadratic variation is integrable. This is what I have so far:

$$
\langle X\rangle _t = \int_0^t \sigma_s^2 ds = 16 \int_0^t W_s^6ds - 96\int_0^tW_t^4ds + 144 \int_0^t sW_s ds
$$
Looking at the first integral now. We use integration by parts:
$$
d(W_t^6t) = (W_t^6 + 15W_t^4)dt + 6W_t^5dW_t \\
W_t^6dt = d(W_t^6t) - 15W_t^4dt - 6W_t^5dW_t \\
\int_0^t W_s^6ds = W_t^6t - 15\int_0^tW_s^4ds + 6\int_0^t W_s^5 dW_s
$$

For fixed $t$, the first term is constant. I need to prove that the two integrals exist. My feeling here is that if I can prove that the third integral exists, I can apply the same logic recursively to the second integral. To my understanding:
$$
\int_0^tW_s^5dW_s = \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{n-1}W_{t_{i+1}}^5(W_{t_{i+1}} - W_{t_i})
$$

How do I go about proving that this limit exists? Is what I'm doing so far correct at all?

答:
You have $dX_t=\sigma_t\,dW_t$, and you need to prove that
$$
\mathbb E[\langle X\rangle_t]=\int_0^t\mathbb E[\sigma_s^2]\,ds<+\infty.
$$

Since $\sigma_s^2=16W_s^6-96sW_s^4+144s^2W_s^2$ and $W_s$ has the same distribution as $\sqrt t G$, for $G\sim\mathcal N(0,1)$, we have that
$$
\mathbb E[\sigma_s^²]=\left(16\mathbb E[G^6]-96 \mathbb E[G^4]+144\mathbb E[G^2]\right)s^3,
$$
whose integral on $[0,t]$ is clearly finite.


问:
Continuous functions and preservation of topological structure.
In some book I read "continuous fuctions between topological spaces preserve their topological structure" and they say that this is similar to the case when homomorphims perserve algebraic structures. But thinking about it, it seems to me that this structure preservation is made backwards.

For example, given $f:X \Rightarrow Y$, both $X$ and $Y$ topological spaces, it is true that the inverse image of finite intersection of open sets in $Y$ will be open in $X$. But this not happen with direct images. I might conclude that topological structure of $Y$ is preserved when the inverse images of $f$ send it to $X$.

It that true? Thanks in advance.

答:
It’s complicated. Continuous functions do preserve many topological characteristics, among them compactness, connectedness, separability, and limits of convergent sequences. However, there are many topological characteristics that they do not necessarily preserve, even things as basic and important as being Hausdorff. This really is somewhat like the behavior of homomorphisms in algebra, which preserve some structure of the domain but can also lose a great of that structure. And just as isomorphisms are especially nice homomorphisms, homeomorphisms (which are clearly the topological analogue of algebraic isomorphisms) are especially nice continuous functions. But the analogy has some real limitations. For instance, a bijective homomorphism is an isomorphism, but a continuous bijection need not be a homeomorphism. I’d say that the analogy is qualitatively reasonable but cannot be pushed very far at a more detailed level.

It is true that continuous functions inversely preserve some topological characteristics, but note that if $f:X\to Y$ is a continuous function, the inverse map in question is not from $Y$ to $X$: it’s the function $\wp(Y)\to\wp(X):A\mapsto f^{-1}[A]$. This map does take open sets to open sets and closed sets to closed sets, for instance. But overall it transfers few topological properties from $Y$ to $X$, even when $X$ and $Y$ have the same underlying set: no matter what the topology on $Y$, $f$ will be continuous if $X$ has the discrete topology. You might note that something similar happens in the algebraic case: if $h:G\to H$ is a homomorphism, and $K$ is a subgroup of $H$, then $h^{-1}[K]$ is a subgroup of $G$. That is, homomorphisms, like continuous functions, inversely preserve some ‘nice’, structurally important subsets — but it’s still the maps themselves that we consider structure-preserving.


问:
An equation of unit vectors
Find u, v and w such that all are unit vectors and $\textbf{u} + 2\textbf{v} + \textbf{w} = 0$.

I can guess a solution, but u and w are the same in my solution, of course the problem has not made it explicit that u and w must be different, but I'm not sure if my solution is a good one: u and w = $(cost, sint, 0)$, and v= $(-cost, -sint, 0)$. What are other possible answers? Also another issue is that I've guessed these values for u, w and v, is there a way to solve such an equation?

答:
I think it is not possible to find another solution (where $u\ne w$).

Assume all vectors are (pairwise) different. Since $u+2v+w=0$, they form a triangle with sides $|u|=1$, $|2v|=2$ and $|w|=1$. This is not possible unless $u=w=-v$ (in a nondegenerate triangle, the sum of any two sides must be strixctly greater than the other).

Title: An equation of unit vectors


答: HINT: If $\mathbf u+\mathbf w$ is twice a unit vector, you should be able to deduce that $\mathbf u\cdot\mathbf w = 1$. Now what does that tell you?

Title: An equation of unit vectors

Body:
Find u, v and w such that all are unit vectors and $\textbf{u} + 2\textbf{v} + \textbf{w} = 0$.

I can guess a solution, but u and w are the same in my solution, of course the problem has not made it explicit that u and w must be different, but I'm not sure if my solution is a good one: u and w = $(cost, sint, 0)$, and v= $(-cost, -sint, 0)$. What are other possible answers? Also another issue is that I've guessed these values for u, w and v, is there a way to solve such an equation?


问:
Is OEIS A248049 an integer sequence?
The OEIS sequence A248049 is defined by

$$ a_n \!=\! \frac{(a_{n-1}\!+\!a_{n-2})(a_{n-2}\!+\!a_{n-3})}{a_{n-4}} \;\text{with }\; a_0\!=\!2, a_1\!=\!a_2\!=\!a_3\!=\!1.$$

is apparently an integer sequence but I have no proofs. I have numerical evidence using PARI/GP and Mathematica only. It is a real
problem because its companion
OEIS sequence A248048 has the same
recursion with $\,a_0=-1, a_1=a_2=a_3=1\,$ but now $\,a_{144}\,$ has a denominator of $2$. There is a
resemblance to the Somos-4 sequence but that probably won't help with an integrality proof.

I have some interesting unproven observations about its
factorization algebraically and $p$-adically for a few small values of $p$, but nothing that would prove integrality. For example, if
$\,x_0,x_1,x_2,x_3\,$ are indeterminates, and
we use initial values of
$$ a_0=x_0,\; a_1=x_1,\; a_2=x_2,\; a_3=x_3 \;\text{ and }\;
x_4 := x_1+x_2,$$ with the same recursion, then $\,a_n\,$ has denominator a monomial in
$\,x_0,x_1,x_2,x_3,x_4\,$ with
exponents from OEIS sequence A023434.
Since $\,x_0=x_4=2\,$ with the original sequence I can't prove that the numerator
has enough powers of $2$ to compensate. Another
example is that $\,a_{12n+k}\,$ is odd for $\,k=1,2,3\,$ and
even for the other residue classes modulo $12$. I also have
some further observations about its $2$-adic valuation
behavior which I can't prove.

By the way, the sequence grows very fast. My best
estimate is $\,\log(a_n) \approx 1.25255\, c^n\,$
where $\,c\,$ is the plastic constant OEIS sequence A060006. Note
that $$x^4-x^3-x^2+1 = (x-1)(x^3-x-1) $$ and
$\,c\,$ is the real root of the cubic factor.

Can anyone give a proof of integrality of A248049?

答:
A proof of integrality can be based on elementary algebra and some fortunate observations. The Darij Grinberg comments reminded me of some of my work I did in 2013 and which I did
not follow up on adequately.

Factorization of the $\,a_n\,$ sequence suggests the Ansatz

$$ a_n = p_n p_{n+1} p_{n+2} p_{n+3} $$

where $\,p_n\,$ is some sequence yet to be determined.
The sequence $\,a_n\,$ is supposed to satisfy a recurrence. For example, we must have

$$ a_4a_0 = (a_1+a_2)(a_2+a_3). $$

Rewriting this equation in terms of $\,p\,$ and solving for
$\,p_7\,$ gives the rational solution

$$ p_7 = \frac{(p_1 + p_5)(p_2 +p_6) p_3 p_4}{p_0 p_1 p_6}. $$

Rewrite this as a polynomial equation to get

$$ p_6p_0p_7p_1 = (p_1 + p_5)p_3(p_2 + p_6)p_4. $$

Now suppose that $\,p_n\,$ satisfies the recurrence

$$ p_n = p_{n-3}\frac{p_{n-1} + p_{n-5}}{p_{n-6}}. $$

Check that this recurrence satisfies the polynomial
equation for $\,p_7.\,$

From the $\,p_n\,$ recurrences for $\,n=9\,$ and
$\,n=6\,$ we have

$$ p_9p_3 = (p_4 + p_8)p_6 \quad \text{ and }
\quad p_6p_0 = (p_1 + p_5)p_3. $$

Combine the two equations to simply get

$$ (p_0 + p_4 + p_8)p_6 = (p_1 + p_5 + p_9)p_3. $$

This implies that the number

$$ c := \frac{ p_0 + p_4 + p_8 }{p_3 p_4 p_5} =
  \frac{ p_1 + p_5 + p_9 }{p_4 p_5 p_6}$$

is constant and thus, the sequence $\,p_n\,$ satisfies the
equation

$$ p_{n}+p_{n-4}+p_{n-8} = c\,p_{n-3}p_{n-4}p_{n-5}. $$

By the way, defining another constant

$$ s := \sqrt{(a_2+a_1)a_2a_0/(a_3a_1)} $$

implies the equation

$$ c = s\frac{(a_0+a_1+a_2)(a_1+a_2+a_3)}{a_0a_2(a_1+a_2)}, $$

or more symmetrically, this can be written as

$$ c = \frac{(a_0+a_1+a_2)(a_1+a_2+a_3)}
   {\sqrt{a_0a_1a_2a_3(a_1+a_2)}}. $$

Given values of $\,p_0\,$ and $\,p_1\,$ then
$\,p_2 = s/p_0\,$ and $\,p_3 = a_0/(p_1s)\,$
while the two sequences are related by
$\,p_n = p_{n-4}a_{n-3}/a_{n-4}.\,$

If the sequence terms $\,p_0, p_1,\dots, p_7\,$ are integers
and the constant $\,c\,$ is an integer, then this implies that
$\,p_n\,$ is an integer sequence, and also $\,a_n\,$ using the
Ansatz. In our case, $\,c=6\,$ and the sequence $\,p_n\,$
begins $\,1,1,1,1,1,1,2,3,4,10,33,140,\dots.\,$
This sequence was known to me in 2013 but I
do not think I connected it to A248049 at that time.

A simpler example of a sequence similar to $\,p\,$
is OEIS A064098 with
$$ a_na_{n-3} = a_{n-1}^2 + a_{n-2}^2 $$
and now with a constant
$$ c := \frac{a_n^2+a_{n+1}^2+a_{n+2}^2}
{a_na_{n+1}a_{n+2}} $$
such that the sequence $\,a_n\,$ also satisfies
$$ a_n + a_{n-3} = c\,a_{n-1}a_{n-2}. $$

Title: Is OEIS A248049 an integer sequence?


问:
how do I write the function obtained from these two graphs?

I would like to design a function that has the maximums of the first image as maximums and the minimums of the second.

in the first image f(x) is $$ \cos ^ {- 1} (\cos (x)) + x / 3 $$ while the second is $$ \cos ^ {- 1} (\cos (x)) + x / 6 $$ the lines are $$ x / 3 + \pi $$ (the blue one) and $$ x / 6 $$ the other.

答:
Let call $f(x)=\arccos(\cos(x))$.

Since the minimum of $f(x)$ is always $0$ then we have to go $f(x)+\dfrac x6$ to keep the minima at the right height.

Now the maxima are occuring at $x_M=(2n+1)\pi\iff n=\dfrac 12\left(\dfrac {x_M}\pi-1\right)$

And their value is $\quad\overbrace{f(x_M)}^{\pi}+\dfrac{x_M}3=\dfrac{(2n+4)\pi}3$

So we are going to multiply $f(x)$ by a coefficient such that the maxima coincide.

$C\times f(x_M)=\dfrac{(2n+4)\pi}3-\dfrac{(2n+1)\pi}6=\dfrac{(2n+7)\pi}6$

We can now rewrite $Cf(x)+\frac x6$ in term of $x$ by eliminating $n$

$$g(x)=\left(\frac x{6\pi}+1\right)\arccos(\cos(x))+\frac x6$$

Edit:

This is maybe a bit hidden, but the technique is very similar to what we do for scaling interval $[0,1]$ to interval $[min,max]$ with the formula below:

$$g(t)=min+(max-min)\,t\quad\text{for }0\le t\le 1$$

Here we have $\begin{cases} min=\frac x6\\ max=\frac x3+\pi\\ t=\frac 1\pi f(x)\end{cases}$

Giving $g(t)=\frac x6+(\frac x3+\pi-\frac x6)\frac 1\pi f(x)=(\frac x{6\pi}+1)f(x)+\frac x6$


问:
A specific type of orthogonal bases
In $\mathbb{R}^n$, can we find an orthogonal basis $v_1,v_2,\ldots,v_n$ with $\|v_i\|=1$ such that
$$\|\bar{v}_1\|=\|\bar{v}_2\|= \cdots =\|\bar{v}_n\|$$
where $\bar{v}_i\in\mathbb{R}^m$ is the vector formed by the first $m$ coordinates of $v_i$ with respect to the standard basis and $m$ is a given integer $m<n$.</p>

答:
The answer is yes:  Assume that $m=n-1$. Same construction could be done for other $m$ as well.
Denote by $W$ the subspace of $V=\mathbb{R}^n$ spanned by $e_1,..,e_m$, the first $m$ vectors of the standard basis (hence $W \cong \mathbb{R}^m$).

Denote by $p:V\to W$ the projection (i.e. $p(a_1,...,a_n)=(a_1,...,a_m,0,...,0)$ w.r.t the standard basis). Let the unit sphere in $V$ be denoted by $S_1=\{{v\in V : ||v||=1\}}$.

Denote $U={\rm Span}(e_2-e_1,...,e_n-e_1)$. And finally, $A=e_1+U$, which is an affine subspace, passing through the endpoints of $e_1,..,e_n$.

Find a linear operator (rotation) $T:V\to V$ such that $\forall v\in  V,||T(v)||=||v||$ and $T(U)=W$, $T(e_1)\in W^\perp$. Hence, denoting $v=T(e_1)$, $v+W=T(A)$ (This makes $A$ "paralel" to $W$). Such an operator exists since $U$ is of co-dimension $1$, and hence can be rotated to coincide with any other sub-space of co-dimension 1.

Now consider $T(e_1),...,T(e_n)$. Since $T$ is a rotation, they form an orthonormal basis of $V$.

Now, if $w_1,w_2\in S_1\cap T(A)$ then $||w_1||=||w_2||=1$ and $w_1=T(e_1)+T(u_1)$, $w_2=T(e_1)+T(u_2)$ where $u_1,u_2\in U$. From the choice of $v\in W^\perp$, we have $p(w_1)=w_1-v$, $p(w_2)=w_2-v$. Hence $$||p(w_1)||^2=||w_1-v||^2=||w_1||^2+||v||^2=||w_2||^2+||v||^2=||p(w_2)||^2$$
(Pythagoras theorem).

Since $T(e_1),...,T(e_n)\in S_1\cap T(A)$, we have $||p(T(e_1))||=...=||p(T(e_n))||$, as required.

Title: A specific type of orthogonal bases


问:
Simplifying a Product of Summations
I have, for a fixed and positive even integer $n$, the following product of summations:

$\left ( \sum_{i = n-1}^{n-1}i \right )\cdot \left (  \sum_{i = n-3}^{n-1} i \right )\cdot \left ( \sum_{i = n-5}^{n-1}i \right )\cdot ... \cdot \left (\sum_{i=5}^{n-1}i \right )\cdot \left (\sum_{i=3}^{n-1}i \right )\cdot \left (\sum_{i=1}^{n-1}i \right )$

Where there are $\frac{n}{2}$ groups of summations multiplied together.

For example, consider the case where $n=4$ :

$\left ( \sum_{i = 3}^{3}i \right )\cdot \left (  \sum_{i = 1}^{3} i \right ) = \left ( 3 \right )\left ( 1+2+3 \right ) = 18$

I have tried in vain to simplify the product. Perhaps there are identities I could make use of.

Edit : I can expand the product to clarify:

$$\left ( n-1 \right )\cdot \left [ (n-3)+(n-2)+(n-1) \right ]\cdot \left [ (n-5)+...+(n-1) \right ]\cdot ... \cdot\left [3+4+...+(n-1)\right ]\cdot \left [1+2+...+(n-1) \right ]$$

From where I can see a $(n-1)^{\frac{n}{2}}$ term, but the others are quite jumbled.

Title: Simplifying a Product of Summations

Body:
Let $n=2m$. Your expression is then

$$\begin{align*}
&(2m-1)\cdot\frac32(2n-4)\cdot\frac52(2n-6)\cdot\ldots\cdot\frac{n-1}2n\\\\
&\qquad=1\cdot(2m-1)\cdot3(2m-2)\cdot5(2m-3)\cdot\ldots\cdot(2m-1)m\\\\
&\qquad=\prod_{k=1}^m(2k-1)(2m-k)\\\\
&\qquad=(2m-1)!!\frac{(2m-1)!}{(m-1)!}\\\\
&\qquad=\frac{(2m)!}{2^mm!}\cdot\frac{(2m-1)!}{(m-1)!}\\\\
&\qquad=\frac{(2m)!}{2^m}\binom{2m-1}m\\\\
&\qquad=\frac{n!}{2^{n/2}}\binom{n-1}{n/2}\;.
\end{align*}$$


问:
Ex 4.1.10 from topology without tears

Let (Y $\tau_y$) be a subspace of (X $\tau$)

Show $\tau_Y \subseteq \tau$ iff Y $\in \tau$

(Hint :Y$\in\tau_Y$)

Rewrite of proof with help from others

Let $\tau_Y$={$O\cap Y$: $O\in\tau$}

(=>) Assume $\tau_Y \subseteq \tau$

Let $Y\in \tau_Y$ $\exists$ Y open in X s.t Y $\cap X$=Y
as X $\in\tau $ So Y$\in \tau_Y\subseteq\tau$ Thus Y$\in \tau$

(<=) Let Y $\in \tau.$ Let U=$\tau_Y$ be arbitrary open set of Y

As Y,O $\in\tau, \exists O\in Y$ s.t O$\cap Y$=U So U$\in\tau$.

Since U is arbitrary $\tau_Y\subseteq\tau$

答:
Your argument for $(\Rightarrow)$ simply doesn’t make sense. You write:

$O\cap Y$ implies that $(O\cap Y)\cap Y\subset(O\cap X)\cap Y$

Thus $Y\in\tau$

First, $O\cap Y$ is not a statement, so it cannot imply anything. Secondly, the fact that $Y\subseteq X$ alone is enough to imply that $(O\cap Y)\cap Y$ and $(O\cap X)\cap Y$ are both equal to $O\cap Y$, whether or not $Y\in\tau$, so the conclusion that $Y\in\tau$ is a non sequitur.

The hint actually tells you how to prove this implication: $Y=X\cap Y$, and $X\in\tau$, so by definition $Y\in\tau_Y$. For this implication you’re assuming that $\tau_Y\subseteq\tau$, and you know that $Y\in\tau_Y$, so $Y\in\tau$. That’s all there is to it.

Your argument for $(\Leftarrow)$ is stated clumsily enough that I can’t tell for sure whether you have the right idea; here’s a cleaned up version.

Suppose that $Y\in\tau$. Let $U\in\tau_Y$ be arbitrary; then there is an $O\in\tau$ such that $U=O\cap Y$. And $O$ and $Y$ are both in $\tau$, so $U=O\cap Y\in\tau$. Since $U$ was an arbitrary member of $\tau_Y$, $\tau_Y\subseteq\tau$.

Title: Ex 4.1.10 from topology without tears


答:
First, the forward direction, $\implies$, is much simpler than you are making it to be (and the argument you give is nonsensical): since $\tau_Y$ is a topology on $Y$, then $Y\in\tau_Y$. If $\tau_Y\subseteq \tau$, then $Y\in\tau_Y\subseteq \tau$, so $Y\in\tau$. Done and done.

Second: what you are "supposing" in the backward direction, $\Longleftarrow$, is not correct. You do not "suppose"  that the equivalence of "$Y\in \tau$"  and "$Y$ is open in $X$". That's not a supposition. That's a definition. What you need to assume is that $Y\in\tau$. Just that.

Then you are trying to prove that $\tau_Y\subseteq\tau$, and you start with something in $\tau$... well, no. You need to prove that if $A\in\tau_Y$, then $A\in\tau$. You never do that. So, you are correct: something is missing, namely, the proof of the "if" part. Instead, you seem to be arguing that if $O\in\tau$, then $O\cap Y\in \tau\cap\tau_Y$; that does not prove what you want to prove. All you are arguing (since we already know that $O\in\tau_Y$ by definition) is that if you assume that $Y$ is open is open in $X$, and $O$ is open in $X$, then $O\cap Y$ is open in $X$. But we know that just from the properties of topological spaces; you aren't proving anything about the subspace topology.

How do the proof of the $\Longleftarrow$ part?  You want to prove $\tau_Y\subseteq \tau$, so you should start by taking something in $\tau_Y$ and proving it is in $\tau$.

To that end, let $A\in\tau_Y$. That means that $A=\mathcal{O}\cap Y$ for some $\mathcal{O}\in\tau$. Now you need to show that $A$ is in $\tau$.

Title: Ex 4.1.10 from topology without tears

Body:
Let (Y $\tau_y$) be a subspace of (X $\tau$)

Show $\tau_Y \subseteq \tau$ iff Y $\in \tau$

(Hint :Y$\in\tau_Y$)

Rewrite of proof with help from others

Let $\tau_Y$={$O\cap Y$: $O\in\tau$}

(=>) Assume $\tau_Y \subseteq \tau$

Let $Y\in \tau_Y$ $\exists$ Y open in X s.t Y $\cap X$=Y
as X $\in\tau $ So Y$\in \tau_Y\subseteq\tau$ Thus Y$\in \tau$

(<=) Let Y $\in \tau.$ Let U=$\tau_Y$ be arbitrary open set of Y

As Y,O $\in\tau, \exists O\in Y$ s.t O$\cap Y$=U So U$\in\tau$.

Since U is arbitrary $\tau_Y\subseteq\tau$


问:
Matrices and linear maps End($V$) of infinite dimension
I am struggling to understand the following concept required to solve a task that I found in my Algebra book.

The task: Let $V$ be (a possibly infinite-dimensional) vector space and $A$ an endomorphism. Furthermore let $\sigma(A)$ be the set of Eigenvalues of $A$ (analogously for the endomorphism $B$). Show that $\sigma(AB)\setminus\{0\}=\sigma(BA)\setminus\{0\}$.

My initial instinct was to represent $A$ and $B$ as $n\times n$ matrices and do some brute-force calculation and show that the characteristic polynomials of $AB$ and $BA$ are similar. However, I am having second doubts regarding the representation of $A$ and $B$ as $n\times n$ matrices as I cannot represent the basis of $V$ in only $n$ variables (it might be infinite-dimensional).

This concern left me in doubt whether I should solve it with matrices at all or somehow express the eigenvalues through $(A\circ B)(v)=\lambda v$. Any clarification would be much appreciated!

答:
The idea or motivation from the finite dimensional case is that $AB$ and $BA$ are "almost similar." Like, we have $B(AB)B^{-1} = BA$ except $B$ might not be invertible. So you have to work around the non-invertibility by assuming you're not in $\ker B$ or something like that.

So suppose $v$ is an eigenvector of $AB$ corresponding to a non-zero eigenvalue $\lambda$. Then what we would like to do is define a vector $w$ such that $B^{-1}w = v \iff w = Bv$ and then we have

$$BAw = B(AB)B^{-1}w = B(AB)v = B(\lambda v) = \lambda Bv = \lambda w.$$

Now this works fine if $B$ is invertible. If $B$ is not invertible then all we need to do to fix this is to cut out the second expression:

$$BAw = B(AB)v = B(\lambda v) = \lambda Bv = \lambda w.$$

Therefore $\lambda$ is an eigenvalue of $BA$ corresponding to the non-zero (!!) eigenvector $w = Bv$.


答:
I'm not really sure what is happening in your attempt to be honest. But I feel like it is much simpler than you might think, especially if you take the hint into account.

$"\Rightarrow"$: Suppose that $\tau_Y \subseteq \tau$. We have that $Y \in \tau_Y$, either by using the hint or by writing $Y = Y \cap X$, as $X \in \tau$. Then we have that $Y \in \tau_Y \subseteq \tau$, therefore $Y \in \tau$.

$"\Leftarrow"$: Suppose now that $Y \in \tau$. Let $A \in \tau_Y$ be an arbitrary open set of $Y$. So we can write $A = O \cap Y$ for some $O \in \tau$. Now we have that $A$ must also be in $\tau$ since it is the finite intersection of elements in $\tau$. Since $A$ was arbitrary, this shows that $\tau_Y \subseteq \tau$.

Title: Ex 4.1.10 from topology without tears


问:
Help me with Goldschmidt Division Algorithm please
I am trying to implement a program that performs division by using Goldschmidt Algorithm. But first, I need to understand the mathematical explanation of this algorithm and I am stuck in a little detail of this algorithm.

https://lauri.xn--vsandi-pxa.com/hdl/arithmetic/goldschmidt-division-algorithm.html

In this link
\begin{equation*}
D_{-1} = 7_{10} = 00000111.000000000000_{2}
\end{equation*}
and "Initial reciprocal is the inverse of divisor which is calculated by shifting bits around the fraction point:"
\begin{equation*}
F_{-1} = \frac{1}{D_{-1}} \approx 0.0546875_{10} = 00000000.000011100000_{2}
\end{equation*}

I have one doubt about how to find $F_{-1}$ here.

If $F_{-1} = \frac{1}{D_{-1}}$, then $F_{-1} = \frac{1}{7} = 0.142$

Please help me to understand how to find $F_{-1}$ for any given number of bits of integer and fraction digits. Thanks.

答:
Look at the code for the "Reciprocal Approximator", about halfway down this related page:
https://lauri.xn--vsandi-pxa.com/hdl/arithmetic/goldschmidt-divisor-implementation.html

It looks like a simple "count how many leading zero bits in the mantissa, put 16 - that many zeroes, then tack on '11'", at least in that implementation. So it's sort of like reflecting the binary representation around the "binary" point, with a little shifting.


问:
Is $\int_{0}^{1}\frac{\|f\|_{L^p(0,x)}}{x}dx $ controlled by $\|f\|_{L^p(0,1)}$
Let $f\in L^{p}(0,1)$ for some $1<p<\infty$.

Is $$\int_{0}^{1}\frac{\|f\|_{L^p(0,x)}}{x}dx $$ controlled by $\|f\|_{L^p(0,1)}$  ?

Equivalently:

Does $\frac{\|f\|_{L^p(0,x)}}{x}$ behave like
$\frac{\|f\|_{L^p(0,1)}}{x^{\alpha}}$ for some $\alpha<1$
as $x\rightarrow 0^+$ ?

I think it is stupid to try to prove the pointwise inequality
$\|f\|_{L^p(0,x)}\leq C x^{1-\alpha}\|f\|_{L^p(0,1)}$. Simply because for each fixed "very small" $x$ we can find a function supported in $[0,x]$ so that $\|f\|_{L^p(0,x)}=\|f\|_{L^p(0,1)}$ and the inequality fails.

答:
I think I figured this out.  A counterexample:

Let $f(x):=\frac{\chi_{[0,1]}(x)}{x^{\frac{1-\epsilon}{p}}}$. Then
$\int_{0}^{1}\frac{\|f\|_{L^p(0,x)}}{x}dx=\frac{C}{\epsilon^{1+\frac{1}{p}}}$.
While $\|f\|_{L^p(0,1)}=\frac{C}{\epsilon^{\frac{1}{p}}}$.


Title: Simplifying a Product of Summations

Body:
I have, for a fixed and positive even integer $n$, the following product of summations:

$\left ( \sum_{i = n-1}^{n-1}i \right )\cdot \left (  \sum_{i = n-3}^{n-1} i \right )\cdot \left ( \sum_{i = n-5}^{n-1}i \right )\cdot ... \cdot \left (\sum_{i=5}^{n-1}i \right )\cdot \left (\sum_{i=3}^{n-1}i \right )\cdot \left (\sum_{i=1}^{n-1}i \right )$

Where there are $\frac{n}{2}$ groups of summations multiplied together.

For example, consider the case where $n=4$ :

$\left ( \sum_{i = 3}^{3}i \right )\cdot \left (  \sum_{i = 1}^{3} i \right ) = \left ( 3 \right )\left ( 1+2+3 \right ) = 18$

I have tried in vain to simplify the product. Perhaps there are identities I could make use of.

Edit : I can expand the product to clarify:

$$\left ( n-1 \right )\cdot \left [ (n-3)+(n-2)+(n-1) \right ]\cdot \left [ (n-5)+...+(n-1) \right ]\cdot ... \cdot\left [3+4+...+(n-1)\right ]\cdot \left [1+2+...+(n-1) \right ]$$

From where I can see a $(n-1)^{\frac{n}{2}}$ term, but the others are quite jumbled.

Body: Note that $ \displaystyle \sum_{i=n - 2k +1}^{n-1} i = (2k - 1)(n-k) $. Let $ n = 2m $ $$ \begin{align*}\prod_{k = 1}^{\frac{n}{2}} (2k - 1)(n- k) &= \prod_{k = 1}^{m} (2k - 1) \prod_{k=1}^m (2m - k)  \\ &=(2m-1)!! \frac{(2m-1)!}{(m-1)!} \end{align*}$$


问:
Proving that $(M_{n \times n}(\mathbb K),\tau_M)$ is a topological space
I'm trying to conclude whether the following statement is true or not:

Let $(\mathbb K, \tau)$ be a topological field and let $M_{n \times n}(\mathbb K)$ be the set of all $n \times n$ matrices over $\mathbb K$.

Let $\tau_M \subseteq \mathcal{P}(M_{n \times n}(\mathbb K))$ with $\emptyset \in \tau_M$.

Let $A \subseteq M_{n \times n}(\mathbb K)$. Then $A \in \tau_M$ if an only if: $$\forall i,j \in \{1,...,n\}, \bigcup_{M \in A} \{M_{ij}\} \in \tau$$

Then $(M_{n \times n}(\mathbb K),\tau_M)$ is a topological space.

I was able to prove the following:

$\emptyset,M_{n \times n}(\mathbb K) \in \tau_M$

If $A_i \in \tau_M$ with $i \in I$, then $\bigcup_{i \in I} A_i \in \tau_M$

But I'm having some trouble proving that if $A, B \in \tau_M$ then $A \cap B \in \tau_M$. Yesterday I posted this post asking help for this specific part. With the approach that I used, I was not able to conclude anything and I was wondering if this statement is indeed true and there's another way of proving this, or if it's false.

答:
Consider $M_{2\times 2}(\Bbb R)$, with the usual topology $\tau$ on $\Bbb R$. Let

$$U=\left\{r\begin{pmatrix}1&1\\1&1\end{pmatrix}:r\in\Bbb R\right\}$$

and

$$V=\left\{r\begin{pmatrix}1&-1\\-1&1\end{pmatrix}:r\in\Bbb R\right\}\,;$$

then $U,V\in\tau_M$, but

$$U\cap V=\left\{\begin{pmatrix}0&0\\0&0\end{pmatrix}\right\}\,,$$

which is not in $\tau_M$.


问:
almost sure convergence and convergence of expectation implies convergence in expectation
IF $X_n \xrightarrow{\text{a.s.}}X$ and $\mathbf{E}[X_n] \rightarrow \mathbf{E}[X]  $ does this imply $\mathbf{E}[|X_n - X|] \rightarrow 0$?

答: As Will's answer above shows, the answer is no in general, but if you also assume that all of the $X_n$ are non-negative, then the answer is yes.  Proof: apply Fatou's lemma to $X_n + X - |X_n - X|$.


问:
If the set of specific sequences of 0's and 1's is countable then it can be used to count the set of all infinite sequences of 0's and 1's. Right?
I am seeking some help to understand why an infinite sequence of $\{0, 1\}$ is uncountable. While there are similar questions here with detailed answers, I wasn't able to resolve the contradiction below:

Given an alphabet $A$ composed of only $0$ and $1$, the set of all possible sequences of alphabet characters of the form $$a_1, a_2, a_3, ..., a_k,$$ where $a_i \in A$, is countably infinite. How is it possible that at the same time the set of infinite sequences of $0$'s and $1$'s is uncountably infinite?

First, let's prove the following:

Given an alphabet $A$ composed of only $0$ and $1$, the set of all possible sequences of alphabet characters of the form $$a_1, a_2, a_3, ..., a_k,$$ where $a_i \in A$, is countably infinite.

Proof 1

The set that is composed of described sequences, including zero-length sequence (an empty sequence), can be denoted as $A^*$. Let's correspond a zero-length sequence with the number $1$. Then all single-character sequences will take numbers from $2$ to $n + 1$, where $n$ is the number of characters in $A$ (which are unique by alphabet's definition). All two-character sequences will take numbers from $n+2$ to $n^2 + n + 1$. All three character sequences are getting numbers from $n^2 + n + 2$ to $n^3 + n^2 + n + 1$. This pattern can be continued infinitely. Therefore, it is possible to enumerate every element of $A^*$. Thus, I conclude that the set $A^*$ is countably infinite since there is a correspondence between $A^*$ and the set of all natural numbers $\mathbb{N}$.

Contradiction

At the same time, the set of infinite sequences of $0$'s and $1$'s (let's call it $T$) is proved to be uncountably infinite (here is a link to Wikipedia with the proof using Cantor's diagonal argument). If proof 1 is correct, then I can make an infinite sequence of zeros and substitute the first $k$ elements of it with any sequence from $A^*$ of the length $k$ which is enumerated. Since $|k| = |\mathbb{N}|$, with this arrangement, it should be correct (but I guess it is not) to state that now every element of $T$ is numbered, hence $T$ is countably infinite.

Reflection

OBVIOUSLY, there is a problem somewhere in my line of thought, but it is hidden from me. My goal is to properly understand why $A^*$ cannot be used to count all the elements of $T$. The answer to this question will also help me to deepen my understanding of Cantor's diagonal argument and explain why it cannot be applied to $\mathbb{N}$.

I sort of understand that Cantor's diagonal method will not work with $A^*$ because it is impossible to construct an argument that is not counted. But on the other hand, if a set of all infinite sequences of $0$'s and $1$'s is said to be infinite, doesn't it mean that any argument one can make must already be in the set?

Any comments are very much appreciated. Thanks!

答:
In Cantor's argument, there are no finite-length strings. You do have them; in fact, if I read it correctly, that's all you use. That's how you get your contradiction.

There is a translation of Cantor actual paper at http://www.logicmuseum.com/cantor/diagarg.htm. He first assumes that he has a list of infinite length strings. He doesn't assume this list is complete, or even that each listed element is different. Just that each element of the list has infinite length.

He then constructs, from that list, an infinite length string a that cannot be in the list. This is actually a lemma that he uses in the second part of the proof.

That's when he assumes that it is possible to make a list of all such strings. The lemma proves that there is a string a that is not in this new list. The assumption of "all" says that a is in the list. This contradiction proves that the assumption is false.


问:
Prove that some group is infinite based on its presentation.
Suppose $G$ is a group with presentation $$G = \langle a, b, c\mid a^2=b^2=c^2=(ab)^3=(bc)^3=(ca)^3=1\rangle.$$

I want to show that this group is of infinite order.

I think since the element $abc$ is of infinite order, we can conclude that $G$ is infinite. I feel that for any $n \in\mathbb{N},$ we have that $(abc)^n = (abc)(abc) \cdots (abc)$ (with $n$ factors) cannot be reduced further, as we do not have any relations involving $a, b,$ and $c$ consecutively. But I also think that this reasoning is not that convincing, and hence, I am asking for help.

答:
You want to give a homomorphism from $G$ to a suitable group, such that the image of $abc$ has infinite order. For $G$ I cannot think of one without having to go into nasty matrix coefficients, but the following works:

Let $U=\langle b,abc\rangle$. This subgroup has index $[G:U]=3$. Then (by Reidemeister rewriting) $U$ has a presentation
$$
\langle b,d=abc\mid b^2,d^2bd^{-2}b\rangle
$$
in these two generators, which means that for the infinite cyclic group $C_\infty=\langle z\rangle$ the map $U\to C_\infty$, $b\mapsto 1$, $d\mapsto z$ is a homomorphism that proves that the order of $abc$ must be infinite.

Addendum: Actually, by inducing this representation up to $G$, we get a homomorphism into a matrix group, in which the image of $abc$ has infinite order:

Take the homomorphism $G\to\mbox{GL}_3(\mathbb{Q})$ given by
$$
a\mapsto \left(\begin{array}{rrr}%
0&1&0\\%
1&0&0\\%
0&0&1\\%
\end{array}\right)%
,\qquad
b\mapsto \left(\begin{array}{rrr}%
1&0&0\\%
0&0&2\\%
0&1/2&0\\%
\end{array}\right)%
,\qquad
c\mapsto \left(\begin{array}{rrr}%
0&0&1\\%
0&1&0\\%
1&0&0\\%
\end{array}\right).
$$

It is easy to verify that the images satisfy the relations, so it is a homomorphism. But it maps $abc$ to
$$
\left(\begin{array}{rrr}%
2&0&0\\%
0&0&1\\%
0&1/2&0\\%
\end{array}\right)%
$$
which clearly has infinite order.


问:
Automorphic forms and periodic functions
A lot of resources I looked at, including Wikipedia, thinks of automorphic forms as some sort of generalized periodic functions with respect to the action of discrete groups satisfying some extra conditions (growth conditions, eigenfunction properties, etc.). What comes after as a "prime example" is always something related to modular forms, Poincare half disk, Eisenstein series, etc; the group considered is typically something like $SL(2,\mathbb{R})$.

Is there a reason why the "trivial" example of automorphic forms is not, for instance, simply about eigenfunctions of the Laplacian operator on $\mathbb{R}\times S^1$? Loosely speaking, the cylinder is obtained by discrete action as a coset space $\mathbb{R^2}/\mathbb{Z}$, so the eigenfunctions of the Laplacian operator in two-dimensional Euclidean plane becomes periodic functions in the cylinder.

The only thing I can think of is that either (1) this example is so trivial that it is not worth anyone's time, or (2) the usual "periodic boundary conditions" on the wave equation (equivalent to the action of discrete group $\mathbb{Z}$ on the plane) does not satisfy the requirements of automorphic form framework.

I am new to automorphic forms, so if (1) is true, I would appreciate an explicit description of how the "trivial" example fits into the language of automorphic forms (e.g. how should I think of the generators, growth conditions, etc.), and how (possibly) this fits well into covering space theory. If (2), I would like to know what failed in the construction. Essentially I want to know why, if (2) is true, we should think of automorphic forms as generalized periodic functions at all.

答:
In my opinion, yes, any sort of periodic function is a kind of "automorphic form/function", because they are functions (or sections of vector bundles, or ...) on quotients $\Gamma\backslash G$ or $\Gamma\backslash G/K$ for topological groups $G$ (Lie groups, adele groups, ...) and discrete subgroups $\Gamma$ (and maybe compact subgroups $K$).

The most-classical ideas about periodic functions, say on $\mathbb R$, of period $1$, or of period $2\pi$, or... already have substantial analytical challenges, but the group theory and the associated differential operators, etc., are not the difficulties. Most often, people "doing" automorphic forms are not focusing on analytical difficulties, but the group-theoretic ones, so the very simple classical cases (Fourier series, Fourier transforms) don't shed light.

Also, although I do not claim to be expert on this, I have the impression that the representation-theoretic and harmonic-analytic behavior of unipotent/nilpotent Lie groups (for example) is significantly different from that of reductive groups, such as $SL_2$ or $GL_2$.

Then $SL_2$ is really the smallest semi-simple or reductive group to consider, if we pass over abelian groups and nilpotent/solvable/etc groups.

And the Harish-Chandra isomorphism for the center of the universal enveloping algebra only really applies to semi-simple (or adapted to reductive). Similarly, results about Hecke algebras of various sorts. So there simply aren't good or illuminating analogues in those other cases.

But, yes, I would claim that many basic (and significant) analytical issues are indeed already non-trivial in the classical, abelian cases. Still, their description in most elementary terms will look different from the reductive case.


问:
If $v$ is an incompressible, irrotational fluid satisfying Euler's equation, show $\overline{v}$ is holomorphic
This feels like it'll be simple once I see it, but I've been stuck for a while now. The setup is from Sijue Wu's 1997 paper Well-posedness in Sobolev spaces of the full water wave problem in 2D. $v=(v_1,v_2)$ is the fluid velocity, $p$ is the pressure, and $\Omega(t)$ is essentially the water region. Then
$$
\begin{cases}
v_t + v\cdot \nabla v & = -(0,1)-\nabla p\\
\operatorname{div}(v) & = 0\\
\operatorname{curl}(v) & = 0
\end{cases}
$$Viewing $z=(x,y)$ as a complex number, if we write $z$ in Lagrangian coordinates with parameter $\alpha$
$$
z_t(\alpha,t) = v(z(\alpha,t),t),
$$the claim is that the last two equations imply $\overline{v}(\alpha,t) = v_1(\alpha,t)-i v_2 (\alpha,t)$ is holomorphic. I had thought to use the Cauchy-Riemann equations to try to show
$$
\frac{\partial v_1}{\partial \alpha} = -\frac{\partial v_2}{\partial t};\qquad \frac{\partial v_1}{\partial t} = -\frac{\partial v_2}{\partial \alpha}
$$but I'm embarassed to say I got mixed up doing the Chain Rule. Am I on the right track? I feel like I'm  not fully using the information about being incompressible and irrotational, particular the curl equation as it should relate the partial derivatives.

答:
You have a 2D flow that is incompressible and irrotational.  This implies that the velocity components everywhere satisfy

$$\tag{*}\nabla \cdot \mathbf{v} = 0 \implies\frac{\partial v_x}{\partial x}= - \frac{\partial v_y}{\partial y}, \quad \\ \nabla \times \mathbf{v} = 0 \implies \frac{\partial v_x}{\partial y}=  \frac{\partial v_y}{\partial x}$$

Also $\mathbb{v} = \nabla \phi$, the gradient of a potential, and there exists a streamfunction $\psi$ such that the complex potential $f(z) = \phi + i \psi$ is holomorphic.  By the Cauchy-Riemann equations, we have

$$v_x = \frac{\partial \phi}{\partial x} = \frac{\partial \psi}{\partial y}, \quad v_y = \frac{\partial \phi}{\partial y} = -\frac{\partial \psi}{\partial x} $$

The complex derivative of $f$ is the complex velocity given by

$$\frac{df}{dz}= \frac{\partial \phi}{\partial x}+ i \frac{\partial \psi}{\partial x}= v_x - i v_y$$

Since $f$ is everywhere holomorphic, so is $\frac{df}{dz}$.  This also can be seen from condition (*) which shows that the real and imaginary parts of the complex velocity also satisfy the Cauchy-Riemann equations.


问:
Equality condition from functions between the disk and the sphere

Let $f,g:D\rightarrow \mathbb{S}^2$ where $D$ is the disk $\{(x,y)|x^2+y^2\le 1\}$, such that for $(x,y) \in \mathbb{S}^1$ these functions are given by: $f(x,y)=(x,y,0)$ and $g(x,y) = (-y,x,0)$ show that there is a point $(x,y) \in D$ such that $f(x,y)=\pm g(x,y)$.

For context, this is a question on the consequences of $\pi_1(\mathbb{S}^1)=\mathbb{Z}$, my attempts were into finding a function $h:D \rightarrow D$(by projecting some composition of $f$ and $g$) such that it having a fixed point by Brower would imply the result, but I really can't see how the other conditions come into play in that case.

Another shot I've tried was to morph the disk to the upper hemisphere and finding $F,G:\mathbb{S}^2 \rightarrow\mathbb{S}^2$ extending $f,g$ through oddness but I can't see an easy way from which this would imply the result either.
The main problem arises from the co-domain being the sphere, most of the results I've seen use that as the domain and prove the results arriving at a contradiction with the fundamental group of the circle.

答:
Suppose for sake of contradiction that $f(x,y) \neq \pm g(x,y)$ for all $(x,y)$, i.e., $f(x,y)$ and $g(x,y)$ are linearly independent.  Then $(f,g)$ defines a map $F$ from $D^2$ to the Stiefel manifold $V_2(\mathbb{R}^3)$ of $2$-frames in $\mathbb{R}^3$.

Based on the boundary conditions, we see $F|_{\partial D^2}: S^1 \cong \partial D^2 \to V_2(\mathbb{R}^3)$ represents a nontrivial generator of $\pi_1(V_2(\mathbb{R}^3))$ as it is modelled by the standard inclusion $SO(2) \hookrightarrow SO(3)$.  However, by construction $F|_{\partial D^2}$ extends to the disk $D^2$ providing a nullhomotopy. This gives the desired contradiction.

Answering a question in the comments:

As I alluded to above, the map $F|_{\partial D^2}: \partial D^2 \to V_2(\mathbb{R}^3) \simeq SO(3)$ is given in coordinates by $$(x,y) \mapsto \begin{pmatrix} x & - y & 0 \\ y & x & 0 \\ 0 & 0 & 1 \end{pmatrix},$$ i.e., it is the inclusion of the stabilizer of the $z$-axis in $SO(3)$.  Since $SO(3)$ acts transitively on $S^2$, this inclusion fits into a fiber sequence $$S^1 \cong SO(2) \to SO(3) \to S^2.$$  Using the long exact sequence in homotopy, we see that the map $\pi_1(S^1) \to \pi_1(SO(3))$ is onto, and hence $F|_{\partial D^2}$ represents a generator of $\pi_1(SO(3))$.

I don't see how a torus is relevant.  There may be a clever trick to visualize the loop inside $SO(3)$ for this problem, but I'll have to think more about it.


问:
Calculate $\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$
Calculate:
$$\displaystyle  \lim_{x \to 3}  \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$$

The problem with that case is that the roots are in different powers so multiplication in nominator and denominator by conjugate is not an option (at least I think it's not).

答:
The conjugation/rationalization approach works, but is somewhat tedious. The key is to introduce the right terms in order to force the numerator and denominator into a differences of integer powers:

$$\begin{align}
\frac{\sqrt{19-x}-2\sqrt[4]{13+x}}{\sqrt[3]{11-x}-x+1}&=\frac{\left((19-x)^2\right)^{\frac14}-\left(2^4(13+x)\right)^{\frac14}}{(11-x)^{\frac13}-\left((x-1)^3\right)^{\frac13}}\\[1ex]
&=\frac{a^{\frac14}-b^{\frac14}}{c^{\frac13}-d^{\frac13}}\\[1ex]
&=\frac{a^{\frac14}-b^{\frac14}}{c^{\frac13}-d^{\frac13}}\times\frac{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}}{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}}\times\frac{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}\\[1ex]
&=\frac{\left(a^{\frac14}\right)^4-\left(b^\frac14\right)^4}{\left(c^{\frac13}\right)^3-\left(d^\frac13\right)^3}\times\frac{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}}\\[1ex]
&=\frac{a-b}{c-d}\times\frac{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}}
\end{align}$$

(where I hope the replacements of $a,b,c,d$ are obvious)

Upon simplification (with $x\neq3)$, we have

$$\frac{a-b}{c-d}=\frac{153-54x+x^2}{12-4x+3x^2-x^3}=\frac{(51-x)(3-x)}{(3-x)(4+x^2)}=\frac{51-x}{4+x^2}$$

and the remaining fraction of rational powers is continuous at $x=3$. Then the limit is

$$\lim_{x\to3}\frac{\sqrt{19-x}-2\sqrt[4]{13+x}}{\sqrt[3]{11-x}-x+1}=\frac{48}{13}\times\frac{12}{256}=\boxed{\frac9{52}}$$


问:
Does ``Converge to'' and ``Strict Equality'' Always Mean the Same Thing? If Not, is This a Paradox?
Consider the geometric series with a = 1 and r = 1/10.

Then, we have

$$
\sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^{n} = \sum_{n=0}^{\infty}9\left(\frac{1}{10}\right)^{n} - 9 = 9\left[ \lim_{k\to\infty}\left(\frac{ar^{k}}{r - 1}\right) + \frac{a}{1-r}  - 1 \right] = 9\left[ \lim_{k\to\infty}\left(\frac{1\cdot(1/10)^{k}}{1/10 - 1}\right) + \frac{1}{1-1/10}  - 1 \right] = 9\left(0 + \frac{10}{9} - 1 \right) = 9 \left( \frac{1}{9} \right) = 1.
$$

Hence, we say that the geometric series with a = 1 and r = 1/10 converges to 1.

Now, (correct me if I am wrong) the given series will never achieve the sum of 2 (rather, approach arbitrarily close) to 2. I conclude such because the limit of the quotient after the second equals sign never achieves its limit of 0.

Furthermore, there are many generally accepted ways to show that $0.\overline{9} = 1$; that is identically equal to 1.

One conventional way, of course, is to let $x=0.\overline{9}$ and subtract it from $10x$ giving $9x = 9$, and thus $x = 1$. And I've heard nobody complain about that---even though we must subtract an infinite number of terms in the exact same manner as we do a finite number of terms.

But, that means

$$
\underbrace{1 = 0.\overline{9} = \sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^{n}}_{\text{exactly equal to}} = \overbrace{\sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^{n} = 1}^{\text{arbitrarily close to}}.
$$

Doesn't it?

It seems to me that the only way the above equalities can hold is if converges to always means exactly equal to.

Otherwise, do we have a paradox here?

Please explain.

Thank you.

答:
I'll try to explain some concepts super-explicitly, and you can say whether or not this clarifies your questions:

A series (or sequence) isn't a number. It's actually an infinite set of numbers (indexed by the positive integers). So a series is never equal to a number. A series has a limit (sometimes), and that limit is a number.

The phrase "the series converges to $a$" is just a different way of saying the limit of the series exists and is equal to the number $a$.

The phrase "the series gets arbitrarily close to $a$" is yet another way of saying that the limit exists and is equal to $a$, one that gestures towards the $\epsilon$ - $N$ definition of limit.

A fact about infinite sequences is that they can have a limit that is not equal to any of their terms. So "the given series will never achieve the sum of ..." isn't a problem.

The notation "$0.\overline{9}$" should be strictly thought of as a limit of the infinite sequence it describes. Notationally it works just like $\lim_{n \rightarrow \infty} ...$ or $\sum_{i=1}^\infty$.

Responding to comment:

1 - When you say "the limit of that sequence of partial sums merely approaches (but does not hit) its limit" you're going wrong.

The sequence of partial sums approaches a number.

The limit of the sequence of partial sums is a number.

2 - You're smart to be wary of proofs that manipulate infinite sequences (or any limits) just like finite arithmetical expressions. One can get into a lot of trouble that way. Many of those proofs that $0.\overline{9} = 1$ aren't rigorous and should be left behind -- let's cal them middle school proofs.

Firstly, theorems about manipulating infinite series usually start with "If certain limits exist, then ...." and the middle school proofs don't prove existence before doing the manipulations.

Secondly, remember that $0.\overline{9}$ (or, indeed, any non-terminating decimal) is notation for writing the limit of an infinite series. Limits of infinite series are too complicated for middle school, so we gloss over that distinction. But if you want to be really sure of what you're doing you have to deal with that.


问:
If $R$ is a Householder reflection matrix, show that $(R+I)$ is always singular
If $R$ is a Householder reflection matrix, show that $(R+I)$ is always singular, and that $(R + \lambda I)$ is invertible for any $\lambda \not\in \{1, −1\}$. The Householder matrix is defined as $R = I - 2vv^T$

答:
Hint: If $R = I - 2vv^T$ (for a unit vector $v$), show that $(R + I)v = 0$.

Regarding the second part of the question: one approach is to consider eigenvalues. The eigenvalues of $R$ are the values of $\lambda$ for which $R + \lambda I$ is invertible. To see that $\pm 1$ are the only eigenvalues, it suffices to note that $\operatorname{span}\{v\}$ and $\operatorname{span}\{v\}^\perp$ form a complete decomposition of $\Bbb R^n$ into eigenspaces.

Another approach: I claim that for $\lambda \notin \{\pm 1\}$, $R + \lambda I$ has an inverse of the form $a I + b vv^T$. Indeed, note that $R + \lambda I = (1 + \lambda) I - 2vv^T$. We have
$$
(a I + bvv^T)((1 + \lambda)I - 2vv^T) = 1\cdot I + 0 \cdot vv^T \iff\\
[(1 + \lambda)a] I + [(1 + \lambda)b - 2a - 2b]vv^T \iff\\
\begin{cases}
(\lambda + 1) a = 1\\
(\lambda - 1)b - 2a = 0
\end{cases}
\iff\\
\begin{cases}
a = \frac 1{\lambda + 1}\\
b =  \frac{2a}{\lambda - 1} = \frac{2}{\lambda^2 - 1}.
\end{cases}
$$
So indeed: if $\lambda \notin \{\pm 1\}$, then $R + \lambda I$ is invertible.


问:
How to solve this integral on the unit interval
I'm trying to solve the following integral
$$
I(x)= \int_0^1  \frac{[1-(1-w)^2][1-w^2]}{[1-w(1-w)]^2}\exp\left( -
\frac{1}{xw(1-w)}
\right)dw.
$$
for $x>0$. The aim is to obtain some closed form solution, or a solution in terms of some special function depending on $x$.

I've attempted a solution through the change of variable $w=\frac{1}{2}(1+\tanh(t))$ and the properties of hyperbolic functions. This has led me to
\begin{equation}
\begin{split}
I(x)&= 2\int_{-\infty}^{\infty}\frac{5+2(e^{2t}+e^{-2t})}{(e^{2t}+e^{-2t}+1)(e^{2t}+e^{-2t}+2)^2} \exp\left(
-\frac{1}{x} (e^{2t}+e^{-2t})
\right)dt\\
&=\int_0^\infty \frac{1}{s}\frac{5+2(s+s^{-1})}{(s+s^{-1}+1)(s+s^{-1}+2)^2}\exp\left(
-\frac{1}{x}(s+s^{-1})
\right)ds\\
&=:\int_0^\infty f(s) ds
\end{split}
\end{equation}
Afterwards, I've tried to split the integral as $\int_0^1f(s)ds$ and $\int_1^\infty f(s)ds$ and perform the change of variable $y=s+1/s$, but this hasn't produced anything useful (apparently). Any suggestion would be much appreciated.

答:
With $w=\frac{1+t}2$, the integral simplifies to

$$I(x) =\int_0^1 \left( 1- \left(\frac{4t}{t^2+3}\right)^2\right)e^{-\frac4{x(1-t^2)}}dt
$$

which is unlikely to yield a close-form result.


问:
I don't know where to begin with this functions question (one-to-one, onto)
a) Suppose that $f:\Bbb Z\to  \Bbb Z$ is a one-to-one function. Define a function $g:\Bbb Z\to  \Bbb Z$ by: for all integers $x$, $g(x)= -f(x)$. Prove that $g$ is also one-to-one.

b) Suppose $f:\Bbb Z\to  \Bbb Z$ is an onto function. Define a function $g:\Bbb Z\to  \Bbb Z$ by: for all integers $x$, $g(x) = f(x) + 4$. Prove that $g$ is also onto.

c) Suppose that $f$ and $g$ are one-to-one functions from $\Bbb Z$ to $\Bbb Z$. Define the function $h: \Bbb Z\to \Bbb Z$ by $h(x)=f(x)+g(x)$ for all integers $x$. Must $h$ be one-to-one? Give a proof or counterexample.

答:
For $(a)$, note that $m=n\iff -m=-n$.

For $(b)$, you know $f$ is onto, so for each $m$ there exists $n$ such that $f(n)=m$. But then for each $m+4$ there exists $n$ such that $g(n)=f(n)+4=m+4$, and $h(m)=m+4$ is a bijection in the integers.

For $(c)$, try $f(n)=-n$ and $g(n)= n$.


问:
Rank of a matrix and its properties
Given A is a 4 x 5 matrix with rank 2, are any of the below statements not always true or are all 4 always true?

Any row echelon form of A has 3 non-pivot columns.

Any row echelon form of A has 2 zero rows.

A is not full rank.

The first two rows of A are non-zero.

Based on what I've been taught in class, 4 is not always true. For 1, rank is a number of pivot-columns of the REF of A, so if rank is 2, it the REF should have 3 non-pivot columns. For 3, full rank means rank(A) less than or equal to the smaller of its size. Since rank is 2 and not 4, A is not full rank.

答:
Everything that you have said is correct. 1 and 3 are always true, but 4 is not always true. You didn't say anything about 2; statement 2 is also always true.

So, 4 is the only statement that is not always true.

Title: Rank of a matrix and its properties


Title: Calculate $\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$

Body:
Calculate:
$$\displaystyle  \lim_{x \to 3}  \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$$

The problem with that case is that the roots are in different powers so multiplication in nominator and denominator by conjugate is not an option (at least I think it's not).

Body:
Let $y=x-3$.

Then it's $\lim\limits_{y\to0}\dfrac{\sqrt{16-y}-2\sqrt[4]{16+y}}{\sqrt[3]{8-y}-y-2}.$

Using binomial series, that's $\dfrac{(4-\frac18y\dots)-2(2+\frac1{32}y\dots)}{(2-\frac1{12}y...)-y-2}=\dfrac{-3/16}{-13/12}=\dfrac9{52.}$