问:
I need help to unravel this proposition, I'm stuck because I don't understand what happens when multiplying the adjacent matrix $k$ times.
Let $L$ be the adjacency matrix of a graph. For all pair of vertices $v_i , v_j $, the L input k (i, j) of the kth power of L represents the number of walks of length k between these vertices.

答:
This is a classic graph theory result which procedes by induction on $k$. Let $G$ be our graph, with associated adjacency matrix $A$. Our claim is that $A^k(i,j) = $ the total number of walks from $v_i$ to $v_j$ which use exactly $k$ edges.

Base Case: $k = 1$

The number of walks of length $1$ from $v_i$ to $v_j$ is just $1$ is they are adjacent and $0$ if not, which is precisely how $A = A^1$ is defined.

Inductive Hypothesis: Suppose the result holds for $A^{k-1}$

Now lets think about all possible $k$-length walks from any $v_i$ to $v_j$ (there may be none, but it does not matter). We can break up such walks into a $k-1$ length walk from $v_i$ to $v_h$ for any $v_h$ and a $1$ length walk from $v_h$ to $v_j$. Let $N(v_j)$ denote the neighborhood set of $v_j$. Then we have that the number of length $k$ walks from $v_i$ to $v_j$ is given by
$$
\sum_{v_h \in N(v_j)}\text{the number of $k-1$ length walks from $v_i$ to $v_h$}
$$
By the inductive hypothesis, this is precisely equal to
$$
\sum_{v_h \in N(v_j)} A^{k-1}(i,h)
$$
If $G$ has $n$ vertices, we can rewrite this as
$$
\sum_{h = 1}^nA^{k-1}(i,h)A(h,j)
$$
because $A(h,j)$ will be $1$ if $v_h\in N(v_j)$ and $0$ otherwise by the base case. This exactly how we define matrix multiplication, so we have
$$
\sum_{h = 1}^nA^{k-1}(i,h)A(h,j) = A^k(i,j)
$$
as desired.


问:
Normal Distribution, Probability a point is inside of a circle in $\mathbb{R^2}$
We have $X$~$Normal(0,1)$,$Y$~$Normal(0,3)$. What is the Probability that $M \in [(x,y) \in \mathbb{R^2} : 1 \leq 3x^2 + y^2 \leq 3]$?
\n

What I did, I set $\tilde{Y} := \frac{Y}{\sqrt{3}},\tilde{Y}$~$Normal(0,1)$. Now we have $\frac{1}{3} \leq x^2 + \tilde{y}^2 \leq 1$, a donut. I'm lost here, my professor hinted polar co-ordinates but I don't really see how, because if I set

$$
X:= \cos\theta, Y:= \sin\theta,
$$
$\theta$ would have 2 different distribution function.

答: You need to integrate over the green region:

ContentLicense: CC BY-SA 4.0


问:
Asymptotic behavior of $(1-1/n)^{n/\log_2 n}$
I am reading Garban & Steif's book on Boolean Functions and while doing an exercise about the influence of the tribes example, with the sixe of each tribe being $\log_2n$ I need to find the asymptotical behaviour of
$$\left(1-\frac{1}{n}
 \right)^{n/\log_2n}.$$

I'm having trouble to determine this behaviour because of the $\log_2 n$ factor. If anyone could help me I will be very grateful.

答:
Your expression can be written as
$$\exp\left[\left(\frac{n}{\log_2(n)}\right) \,\log\left(1-\frac{1}{n}\right)\right].$$
Now, using the fact that $\log\left(1-\frac{1}{n}\right)=-\frac{1}{n}+\mathcal{O}\left(\frac{1}{n^2}\right)$, you can simplify it to
$$\exp\left[-\frac{1}{\log_2(n)}+\mathcal{O}\left( \frac{1}{n\log_2(n)}\right) \right]=\exp\left(-\frac{1}{\log_2(n)}\right)+\mathcal{O}\left( \frac{1}{n\log_2(n)}\right).$$


问:
Generalized Euler Constants
According to Wikipedia, the generalized Euler constants are $$\gamma_\alpha=\lim_{n\to \infty} \Big(\sum_{k=1}^n \frac{1}{k^\alpha} - \int_1^n \frac{1}{x^\alpha} \,dx \Big)$$
for $0<\alpha<1$. I am curious why we don't consider $\alpha >1$? For example,
$$\lim_{n\to \infty} \Big(\sum_{k=1}^n \frac{1}{k^2} - \int_1^n \frac{1}{x^2} \,dx \Big) = -1 +\lim_{n\to \infty} \Big(\frac{1}{n}+ \sum_{k=1}^n \frac{1}{k^2} \Big) = \zeta(2)-1=\frac{\pi^2}{6}-1.$$
Or in general $$\lim_{n\to \infty} \Big(\sum_{k=1}^n \frac{1}{k^\alpha} - \int_1^n \frac{1}{x^\alpha} \,dx \Big) = \zeta(\alpha)-\frac{1}{\alpha-1}$$ for natural number $\alpha \geq 2$. Is it because they are easily describable in terms of the zeta function?

答: There certainly isn't any mathematical reason why we cannot consider $\alpha > 1$, since in such a case, both the sum and the integral are finite, thus $\gamma_\alpha$ is well-defined for $\alpha > 1$.  But that's kind of the issue:  it is somewhat trivial or to put it more mildly, not so interesting to consider these cases.  What makes the generalized $\gamma_\alpha$ interesting is precisely the case where $0 < \alpha < 1$, since on this interval, both the sum and integral are divergent, but their difference is finite in a way that is not trivial and may lead to more interesting mathematics.

Title: Generalized Euler Constants


问:
Is Ito Integral of $W_t^{-1}$ well defined? Can Ito's lemma be used?
Let $X_t$ be defined as follow: $$X_t:=\int_{h=0}^{h=t}W_h^{-1}dW_h$$ Is $X_t$ well defined? Can the integral be computed?

We know that for a general function $y=\frac{1}{x}$, we can integrate to $ln(x)$, so I thought about defining a function $F(W_t):=ln(W_t)$ and applying Ito's lemma to $F$, but the problem is that $W_t$ can go negative, so that makes me think that we cannot do that (i.e. log of negative number is undefined).

As another example, if I were to integrate $X_t:=\int_{h=0}^{h=t}W_h^{-2}dW_h$, I'd proceed as follows:

We have:

$$W_t=W_0+\int_0^ta(W_t,t)_{=0}dh+\int_0^tb(W_t,t)_{=1}dW_h$$

So that $W_t$ is an Ito process. Let $F(W_t):=W_t^{-1}$, then, using Ito's Lemma:

$$F(W_t)=\int_{0}^t \left( \frac{\partial F}{\partial t}_{=0}+\frac{\partial F}{\partial W_t}_{=-W_h^{-2}}*a(W_t,t)_{=0} + 0.5\frac{\partial^2 F}{\partial W_t^2}_{=2W_h^{-3}}*b(W_t,t)^2_{=1} \right)+\int_0^t\left(\frac{\partial F}{\partial W_t}_{=-W_h^{-2}}b(W_t,t)_{=1}\right)dW_h=\\=\int_0^tW_h^{-3}dh-\int_0^tW_h^{-2}$$

So that:

$$\int_0^tW_h^{-2}dW_h=\int_0^tW_h^{-3}dt-W_t^{-1}$$

The problem above is again that $W_t^{-1}$ goes to infinity, whenever $W(t)$ approaches zero, so the integrals might not be well defined in any case and I am not sure if they converge: i.e. if I can even use Ito's Lemma as above?

Are integrals of negative powers of Brownian motion well defined?

答:
No, this integral is not well-defined.  Recall that $\int_0^t H_s dW_s$ is only well-defined if there exist a sequence of stopping times $(\tau_n) \nearrow t$ such that $\mathbb{E}[\int_0^{\tau_n} H_s^2 ds] < \infty$, but we can show that $\int_0^t (W_s^{-1})^2ds = \infty$ almost surely for every $t>0$:

Fix $\omega \in \hat \Omega$, by the law of the iterated logarithm there exists $T = T(\omega) > 0$ such that $\frac{|W_s(\omega)|}{\sqrt{2s \log (\log (1/s))}} \le 2$ for all $s \le T$, where $\hat \Omega \subset \Omega$ is a set with probability 1 on which the law of iterated logarithm holds.  Rewriting, this implies $\frac{1}{|W_s(\omega)|^2} \ge \frac 1{8 s \log(\log(1/s))}$ for all $s \le T$ and hence

\begin{align*}
\int_0^t |W_s(\omega)^{-1}|^2ds \ge \frac 18 \int_0^t \frac 1{s \log(\log(1/s))} ds = \infty
\end{align*}

for any $t \le T$.  Since this is the integral of a non-negative function, it is non-decreasing in $t$, so this shows that we actually do have $\int_0^t |W_s(\omega)^{-1}|^2ds = \infty$ for all $t>0$ and $\omega \in \hat \Omega$.  In particular, if $\tau$ is any stopping time that is non-zero with positive probability we have $\mathbb{E}[\int_0^{\tau} (W_s^{-1})^2 ds] = \infty$ and therefore $\int_0^t (W_s)^{-1}dW_s$ is not well-defined.


问:
Show that $ A_G \simeq \mathbb{Z} \otimes_{\mathbb{Z}[G]}A $ for $G$-modules.
I found this ring theory exercise in an algebra textbook.  Let $G$ be a group and then $\mathbb{Z}[G]$ is a module and let $A$ be a $G$-module.

$$ A_G \simeq \mathbb{Z} \otimes_{\mathbb{Z}[G]}A $$
Here the group of

$G$-invariants is: $ A^G = \{ a \in A  : g \cdot A = a \text{ for all } g\in G, a \in A\} $

$G$-coinvariants is $A_G = A/I_G A$ the part of $A$ that is fixed by $G$

The Wikipedia article on tensor product has a lot of discussion but also some warnings, so there is also the article on Stacks Project.

答:
$\Bbb{Z}$ is a $\Bbb{Z}[G]$ module through $g.n=n$

$A\to \mathbb{Z} \otimes_{\mathbb{Z}[G]}A$ is surjective

$1\otimes g.a=g.1\otimes a=1\otimes a$ so

$$\mathbb{Z} \otimes_{\mathbb{Z}[G]}A\cong \mathbb{Z} \otimes_{\mathbb{Z}[G]}(A/I_GA)$$

where $I_G=\{ \sum n_g g\in \Bbb{Z}[G], \sum n_g=0\}$ is the two-sided ideal generated by the $1-g$.

Clearly since $g$ acts trivially on $A/I_GA$ $$\mathbb{Z} \otimes_{\mathbb{Z}[G]}(A/I_GA)\cong A/I_GA$$


问:
Order -Statistics
The random variables $X_1, X_2, . . . , X_n, Y_1, Y_2, . . . , Y_n$ are iid $\mathcal{U}(0, a)$. Determine the distribution of $$Z_n = n \log\bigg(\frac{\max\{X(n), Y(n)\}}{\min\{X(n), Y(n)\}}\bigg)$$
Should I find the joint distribution of $\max$ and $\min$ and then find distribuation of $Z_n$, since we have a two different random variable I do not know how to do that!

答: First observe that the random vector $\Big(X_{(n)},Y_{(n)}\Big)$ is supported on $(0,a)^2$. Suppose $f$ is its joint density. Since $X_{(n)}$ and $Y_{(n)}$ are independent, we have $$f(x,y)=f_{X_{(n)}}(x)f_{Y_{(n)}}(y)=n^2a^{-2n}x^{n-1}y^{n-1}$$ for any $(x,y)\in (0,a)^2$. Also observe how $Z_n$ is supported on $[0,\infty)$, which means for any $z\geq 0$ we have $$F_{Z_n}(z)=P(Z_n\leq z)=P\Big(Z_n\leq z,X_{(n)}<Y_{(n)}\Big)+P\Big(Z_n\leq z,X_{(n)}\geq Y_{(n)}\Big)$$ With a little algebra we have $$F_{Z_n}(z)=P\Big(X_{(n)}<Y_{(n)}\leq e^{z/n}X_{(n)}\Big)+P\Big(Y_{(n)}\leq X_{(n)} \leq e^{z/n}Y_{(n)}\Big)$$ This probability can be written as the double integral $$F_{Z_n}(z)=2\int_0^{a} \int_{\frac{x}{e^{z/n}}}^{x}f(x,y)dydx=1-e^{-z}$$ which shows $Z_n \sim \text{Exp}(1)$.

Title: Order -Statistics


问:
Is there a non-constant function $f: \mathbb{R}_{>0} \to \mathbb{R}$ such that $f(x) = f(x + 1/x)$?
I am looking for a non-constant function $f: \mathbb{R}_{>0} \to \mathbb{R}$ such that $f(x) = f(x + 1/x)$, or a proof that no such function exists.

Replacing $x$ by $1/x$ shows we must have $f(x) = f(1/x)$.

I am most interested in (non-)existence of smooth non-constant $f$.

答:
There should be infinitely many continuous solutions, one for each continuous function $g:[1,2]\to \mathbb{R}$ with $g(1)=g(2)$. After imposing appropriate boundary and differentiability conditions on  $g$, we can make the function smooth.

Let $x_1=1$ and $x_{n+1}=x_n+\frac{1}{x_n}$. Then $1\le x_n\le n$ and by the divergence of the harmonic series, $x_n\to\infty$ as $n\to \infty$. Since $h:t\mapsto t+\frac{1}{t}$ is strictly increasing on $[1,\infty)$, each $x\in[1,\infty)$ belongs to exactly one $[x_{n+1},x_{n+2})$ and $x=h^n(y)$ for exactly one $y\in[1,2)$. Then we define $f(x)=g(y)$. Using the relation $f(x)=f(1/x)$, this extends to $(0,\infty)$. It is continuous since it is continuous on each $[x_n,x_{n+1}]$ and agrees at end points.


问:
Computing determinant $a_{ij = 1}, a_{ii} = 0$
So the question is in the title (the determinat of a matrix $(a_{ij})$ and diagonal elements are zeros) such that all the off-diagonal elements are $1$. I have no idea how to approach this. I am looking for a hints. Solutions are also good but hints would be better.

答:
Hint: Note that $A+I$ is a matrix of all ones, which clearly has rank-$1$. Hence, $0$ is an eigenvalue of $A+I$ with multiplicity $n-1$, and $?$ is the last eigenvalue with multiplicity $1$. It's not hard to work out what this last eigenvalue is if you think about what happens when you multiply $A+I$ by a vector of all ones.

Then, you can get the eigenvalues of $A$ by simply subtracting $1$ from each eigenvalue of $A+I$. Finally, the determinant of $A$ is the product of the eigenvalues of $A$.

EDIT: This is basically what Will Jagy suggested in the comments a minute earlier.


问:
Question about basis changing (linear algebra)
I have a question about basis changing and the matrix associated with it. A book I am following has the following example:
Find the matrix associated with the change of basis from $E = \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}$ (the usual base) to $S = \{(1, 0, 1), (2, 1, 2), (1, 2, 2)\} $.

The author goes on by explaining that the answer is the matrix
$$M =
\begin{bmatrix}
1 & 2 & 1 \\
0 & 1 & 2 \\
1 & 2 & 2 \\
\end{bmatrix}$$

I disagree with that assertion and I have the impression that this matrix actually corresponds to the converse change (ie. from $S$ to $E$).

For example, say that we have $[x]_S = (1, 0, 0)^T$. That means that we are having a unit of the $i$ that corresponds to $S$ and therefore the vector in $E$ would be $[x]_E = (1, 0, 1)^T$. We can verify this by verifying that by multiplying $[x]_E = M \circ [x]_S$. This kinda justifies that $M$ actually translates vectors from $S$ to $E$, in contrast to the author who states that's the converse.

What am I missing? Thanks!

Optional Note

A simpler example to further illustrate the way I am approaching the concept is the change of basis from $A = \{(1, 0), (0, 1)\}$ to $B = \{(2, 0), (0, 2)\}$, which basically stretches both the standard vectors by $2$.

Say, on the coordinates of $B$, we have that $i_B = 3$ and $j_B = 2$. That is, we have $[x]_B = (3, 2)$. This means that, on the standard coordinates, both $i$ and $j$ are half the coordinates of $B$, therefore, in order to find the coords of our example on our standard language, we should multiply both the coordinates by a factor of $2$. So, we have $[x]_A = (6, 4)$.

This can be done exactly by the matrix:
$$M^B_A =
\begin{bmatrix}
2 & 0 \\
0 & 2 \\
\end{bmatrix}$$

and the converse is:
$$(M^B_A)^{-1} = M^A_B =
\begin{bmatrix}
1/2 & 0 \\
0 & 1/2 \\
\end{bmatrix}$$, which follows my reasoning as on the above example

答:
You are perfectly right, and indeed the coordinate transformation $[x]_E\mapsto [x]_S$ is given by $M^{-1}$.

However, here it seems basis transformation means (the linear extension of) the map $e_i\mapsto s_i$, and that is simply $M$, coordinated in the basis $E$.


问:
Prob question - If the hats are both red, what is the probability that it was Factory A?
Factory A makes: 40% red hats, 60% black hats
Factory B makes: 60% red hats, 40% black hats
Two hats are sampled at random from one of the factories, also selected at random.

If the hats are both red, what is the probability that it was Factory A?

I am struggling with this..
For Factory A, prob would be 2/5
For Factory B, prob would be 3/5
Am I right?

So then we ll have to add the prob of picking a factory randomly, so 1/2
Is it correct?

Then we have to add the prob that both hats will be red?
Any help, I am super lost :)

答:
Suppose each factory provides $100$ random samples of $2$ hats. Both provide the same number, so there is the same chance of selecting from each factory. From factory A there is $0.4$ chance a hat is red, so there is $0.16$ chance both hats are red, and so $16$ of the samples have both hats red. Similarly from factory B $36$ samples have both hats red.

We know that a sample of two red hats was chosen, so there were $16+36 = 52$ samples which could have been chosen. Conditional probability of 2 red from A given that there were 2 red is

Pr(factory A) = $\frac{16}{52} = \frac{4}{13}$.


答:
Credit to Shivering Soldier for pointing out the cleaner argument given below.

Lemma: For any ring $R$, ideal $I\leqslant R$, and left $R$-module $M$, we have $M\big/IM\cong R/I\otimes_R M$.

Proof: Define a map $\phi:M\rightarrow R/I\otimes_RM$ by $m\mapsto (1+I)\otimes m$. That $\phi$ is a morphism of abelian groups is immediate from the definition of the tensor product; indeed, for $m,n\in M$, we have $$\phi(m-n)=(1+I)\otimes(m-n)=(1+I)\otimes m-(1+I)\otimes n,$$ as needed. Furthermore, for any $i\in I$ and $m\in M$, we have $$\phi(i\cdot m)=(1+I)\otimes i\cdot m=(1+I)\cdot i\otimes m=(i+I)\otimes m=(0+I)\otimes m=0,$$ so $IM\subseteq\ker{\phi}$ and hence $\phi$ descends to a map $\overline\phi:M\big/IM\rightarrow R/I\otimes_R M$.

We now define $\overline\psi:R/I\otimes_R M\rightarrow M\big/IM$, an inverse to $\overline\phi$, via the universal property of tensor products. First define $\psi:R/I\times M\rightarrow M\big/IM$ by $(r+I,m)\mapsto rm+IM$. To see that this is well-defined, note that $\psi(i+I,m)=im+IM=0+IM$ for any $i\in I$, as needed. $\psi$ is clearly additive in each variable, and we claim that it is $R$-balanced. Indeed, for $x\in R$, $r+I\in R\big/I$, and $m\in M$, we have $$\psi((r+I)\cdot x,m)=\psi(rx+I,m)=rxm+IM=\psi(r+I,x\cdot m),$$ as desired.

Thus $\psi$ descends to a group homomorphism $\overline\psi:R/I\otimes_R M\rightarrow M\big/IM$. We claim now that $\overline\psi$ and $\overline\phi$ are mutual inverses. To see that $\overline\psi\circ\overline\phi=\operatorname{id}_{M/IM}$, let $m\in M$. Then $$\overline\psi(\overline\phi(m+IM))=\overline\psi((1+I)\otimes m)=1m+IM=m+IM,$$ as desired. To see that $\overline\phi\circ\overline\psi=\operatorname{id}_{R/I\otimes M}$, it suffices to show that it acts as the identity on simple tensors. Thus let $r+I\in R\big/I$ and $m\in M$. We have $\overline\psi((r+I)\otimes m)=rm+IM$, and $$\overline\phi(rm+IM)=(1+I)\otimes r\cdot m=(1+I)\cdot r\otimes m=(r+I)\otimes m,$$ again as desired. Hence $\overline\phi$ and $\overline\psi$ are mutual inverses and so we have the desired isomorphism.

The result you desire is now an immediate corollary of the above lemma. Just to make all notation clear: $\mathbb{Z}\otimes_{\mathbb{Z}[G]} A$ is an abelian group, where $\mathbb{Z}$ is considered as a (right) $\mathbb{Z}[G]$-module on which every $g\in G$ acts as the identity. We can define $\iota:\mathbb{Z}[G]\rightarrow \mathbb{Z}$, the "augmentation map", by extending $g\mapsto 1$ linearly for each $g\in G$. Then $I_G$, the "augmentation ideal", is defined to be $\ker{\iota}$, and $A_G:=A\big/I_GA$.

Note that $\iota$ is in fact surjective (for instance, $\iota(n1_G)=n$ for any $n\in\mathbb{N}$, where $1_G\in G$ is the identity element). Thus, by the first isomorphism theorem, we have $\mathbb{Z}\cong\mathbb{Z}[G]\big/I_G$. This induces the chain of isomorphisms $$\mathbb{Z}\otimes_{\mathbb{Z}[G]} A\cong\mathbb{Z}[G]/I_G\otimes_{\mathbb{Z}[G]}A\cong A\big/I_GA,$$ where the second isomorphism follows from the lemma above, and this gives the desired result.

Title: Show that $ A_G \simeq \mathbb{Z} \otimes_{\mathbb{Z}[G]}A $ for $G$-modules.

Body:
I found this ring theory exercise in an algebra textbook.  Let $G$ be a group and then $\mathbb{Z}[G]$ is a module and let $A$ be a $G$-module.

$$ A_G \simeq \mathbb{Z} \otimes_{\mathbb{Z}[G]}A $$
Here the group of

$G$-invariants is: $ A^G = \{ a \in A  : g \cdot A = a \text{ for all } g\in G, a \in A\} $

$G$-coinvariants is $A_G = A/I_G A$ the part of $A$ that is fixed by $G$

The Wikipedia article on tensor product has a lot of discussion but also some warnings, so there is also the article on Stacks Project.


问:
Moment generating function of a stochastic integral
Let $(B_t)_{t\geq 0}$ be a Brownian motion and $f(t)$ a square integrable deterministic function. Then:
$$
\mathbb{E}\left[e^{\int_0^tf(s) \, dB_s}\right] = \mathbb{E}\left[e^{\frac{1}{2}\int_0^t f^2(s) \, ds}\right]
$$
Now assume $(X_t)_{t\geq 0}$ is such that $\left(\int_0^tX_sdB_s\right)_{t\geq 0}$ is well defined. Does
$$
\mathbb{E}\left[e^{\int_0^tX_s \, dB_s}\right] = \mathbb{E}\left[e^{\frac{1}{2}\int_0^tX_s^2 \, ds}\right]
$$
still hold?</p>

答:
No, in general this is not correct.

Applying Itô's formula, one can actually show that $$(t,w) \mapsto M(t,w) := \exp \left( \int_0^t X_s \, dB_s - \frac{1}{2} \int_0^t X_s^2 \, ds \right)$$ is a local martingale. There are several sufficient conditions (on the process $(X_t)_t$) implying that $(M_t)_{t \geq 0}$ is indeed a martingale. In this case, one obtains

$$1=\mathbb{E}M_0 = \mathbb{E}M_t = \mathbb{E}  \exp \left( \int_0^t X_s \, dB_s - \frac{1}{2} \int_0^t X_s^2 \, ds \right)$$

Note that this is not equivalent to $$\mathbb{E} \exp \left( \frac{1}{2} \int_0^t X_s^2 \, ds \right) = \mathbb{E}\exp \left( \int_0^t X_s \, dB_s  \right)$$

(It works fine for deterministic functions, since in this case the left-hand side of the last equation does not depend on $\omega$.)

ContentLicense: CC BY-SA 3.0


问:
filter on a topological space- how to define filter
So $(X,\tau$) is a topological space. Let $\mathcal{C}$ be the collection of closed sets in $X$. Then a $\textit{filter}$, on $\mathcal{C}$, call it $\mathcal{F}$, is a sub collection of these closed sets such that

$(1)$ $\emptyset \notin \mathcal{F}$

$(2)$ closed under intersection, that is, if $C_i,C_j \in \mathcal{F}$, then $C_i \cap C_j \in \mathcal{F}$.

($3)$ closed under supersets?, that is if $C_i \subset C_j$ where $C_i \in \mathcal{F}$ and $C_j \in \mathcal{C}$, then $C_j \in \mathcal{F}$.

My problem says if every filter on $\mathcal{C}$ has non-empty intersection, prove that $(X,\tau)$ is compact.

I'd like how to rigorously define what it means for a filter to have nonempty intersection, so what intersection is nonempty? what is the subscript? where are the elements being pulled from?

so $\bigcap \mathcal{F} \neq \emptyset$ but I am so lost when it comes to the subscripts here, that is, the subscript of my $\bigcap$. Basically want to know how to formally define a filter on a top space. I have never heard of filters til now. Are they a way of sort of sifting out smaller sets and only including "bigger" sets? And can one put a filter on $\tau$ as well as the closed subsets? What is the purpose of a filter? Feel like once I get a hang on what a filter is I could possibly prove this statement, so I would like to know about this filter business not a solution to my problem I want to still attempt it myself I do not like spoon feeding.

答:
You don’t need any subscripts: by definition

$$\bigcap\mathscr{F}=\{x\in X:\forall F\in\mathscr{F}(x\in F)\}\,.$$

It is simply the intersection of the family $\mathscr{F}$ of sets, the set of points of $X$ that belong to every member of the family $\mathscr{F}$.

If you choose to index $\mathscr{F}$ by some index set $I$, so that $\mathscr{F}=\{F_i:i\in I\}$, then

$$\bigcap\mathscr{F}=\bigcap_{i\in I}F_i\,,$$

but there is generally no need to do so.


问:
How do you solve $y^2+2xyy' = e^x$?
How would one solve this ODE?

$$y^{2}+2xyy'=e^{x}$$

I tried dividing everything by $x$:

$$\frac{y^{2}}{x} + 2yy' = \frac{e^{x}}{x}$$

and then I did a substitution $v=y^2 1/2ydv=dy$

and then i end up with

$$\frac{v}{x} + v' = \frac{e^{x}}{x}$$

and then I'm not sure what to do from there.

答:
Hint: the left hand side is $\frac{d(xy^2)}{dx}$.

Added: your last equation is essentially $\frac{d(xv)}{dx} = e^x$ (with $v = y^2$), hence $xy^2(x) = e^x + C$.

Title: How do you solve $y^2+2xyy' = e^x$?


问:
prove a statement with implies and limits using epsilon.
If a function is that $\lim_{x \to 4} f(x)=7, \implies\lim_{x \to 4}[(f(x))^{2}+f(x)+1]=57$. how do I use the $\epsilon$ and $\delta$ definition to prove this. I know if $\lim_{x \to 4}f(x)=7,$ then $|f(x)-7|<\epsilon$. The implies, statement is a quadratic equation. If $f(x)=a$, then $\lim_{x \to 4}[(f(x))^{2}+f(x)+1]=57$ is the same to $a^{2}+a+1$. And,  then I can make it $a^{2}+a-56$. But what to do now? Can I have some assistance?

答:
Since $\lim_{x \to 4} f(x)=7$, $\forall \epsilon_0 > 0$ there exists $\delta_0$ such that when $|x-4| < \delta_0, |f(x)-7| < \epsilon_0$.

Now $\forall \epsilon >0$, we pick $\epsilon_0 = \min (1,\frac{\epsilon_0}{16})$, therefore $(f(x))^2+f(x)+1 > f(x) > 0$ when $|x-4|<\delta_0$. Then we set $\delta= \delta_0$, and if $|x-4| < \delta=\delta_0, 6 < f(x) < 8$, we have
$$\left||(f(x))^2+f(x)+1| - 57\right| = |(f(x))^2+f(x)+1 - 57|\\ = (f(x)+8)|f(x)-7| < 16 |f(x)-7| \le \epsilon.$$


问:
Evan PDE exercise 7 chapter 7
Exponential decay estimate for $(x,t) \in U_T$

I have question from the above mentioned problem:

I take $v=e^{\gamma t}u(x,t)$, then

$v_t= u_t e^{\gamma t}+ \gamma u e^{\gamma t}$

$\Delta v= e^{\gamma t} \Delta u+ 2 \gamma u_t+\gamma^2e^{\gamma t} u$

So, $v_t -\Delta v-(\gamma -c)v=-\gamma^2u e^{\gamma t}- \gamma u_te^{\gamma t}$.

I stuck here, how do I claim $v_t -\Delta v-(\gamma -c)v \geq 0$. I want to show this to apply maximum principle.

Can anyone suggest some hint to handle the last step?

答:
The Laplacian is only in terms of the spatial variables. In particular $\Delta v= e^{\gamma t}\Delta u$. This, together with your formula for $v_t$, gives
$$
v_t-\Delta v= -cue^{\gamma t} + \gamma e^{\gamma t} u= (\gamma-c)v.
$$
Therefore $v_t-\Delta v + (c-\gamma)v =0$. Since $c-\gamma\geq 0$, we can apply the weak maximum and minimum principles.

Title: Evan PDE exercise 7 chapter 7


问:
Cohomology vs. Reduced Cohomology in homotopy type theory
I just read Mike Shulman's blog post the other night, discussing the definition of cohomology in Homotopy Type Theory. I really enjoyed the new (to me) perspective, but I had a few questions on technicalities.

Given types $X$ and $Y$, Mike defines

$$H^n(X; Y) :\equiv \| X \to \Omega^{-n} Y \|_0.$$

This agrees with the definition given in the nLab article on cohomology. Here is where my first question comes up:

Although it's not stated in the article, I assume that for $n \neq 0$ we need $Y$ to be a pointed type, since (if I understand correctly) looping/delooping requires a choice of basepoint? Or does it not matter?

(Mike does explain in the article that for $n>0$ we need to specify choices of deloopings, since deloopings might not exist and might not be unique, and that we usually just take $Y$ to be a spectrum so that this works.)

However, in section 1.1 of this paper on Eilenberg-MacLane spaces in HoTT by Daniel Licata and Eric Finster they seem to be saying that the definition above actually produces reduced cohomology, since they write:

On the other hand, in his blog post Mike writes defines reduced cohomology for pointed types $X$ and $Y$ by
$$\tilde{H}^n(X; Y) :\equiv \| \operatorname{Map}_*(X, \Omega^{-n} Y) \|_0.$$
(He technically defines it only in the case that $Y$ is a spectrum, but I'm extrapolating here.) So this leads me to my second question:

What is the correct definition for reduced cohomology in HoTT?

答:
In order for the notation $\Omega^{-n} Y$ to make sense for $n \ge 1$, $Y$ must be a grouplike $E_n$-space, which in particular means it has a basepoint (the identity of the $E_n$ multiplication). For $n \le -1$ we also need a basepoint to define the based loop space. However note that $X$ does not need to have a basepoint.

As far as I can tell, the excerpt from Licata-Finster is incorrect; it gives ordinary cohomology, not reduced cohomology. Mike Shulman's definition is the correct one.


问:
Describing Domain of Integration (Triple Integral)
I'm really struggling to go about starting the following problem:

This question concerns the integral,

$\int_{0}^{2}\int_0^{\sqrt{4-y^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{8-x^2-y^2}}\!z\ \mathrm{d}z\ \mathrm{d}x\ \mathrm{d}y$.

Sketch or describe in words the domain of integration.

I've tried sketching the projections of this integral's domain onto the $xy$-plane, the $yz$-plane and the $xz$-plane. I established that the projection onto the $xy$-plane is the portion of the circle $x^2+y^2=4$ on $0\le x\le2,\ y\ge0$; that the projection onto the $yz$-plane is the portion of the circle $y^2+z^2=8$ above the line $z=y$, on $0\le y\le2$; and that the projection onto the $xz$-plane is the portion of the circle $x^2+z^2=8$ above the line $z=x$, on $0\le x \le2$.

I'm not completely confident that these findings are useful in any way, and if they are, I am struggling to see the link which will allow me to solve the problem.

Do you care to get me started?

Thank you very much.

答: +1 for the @Jared, cause he pointed out what you need to do the problem. I am posting a graph maybe you find out better that points. The circle projected on $xy$ plane is the same as the intersection of two equations (See another answer).


问:
How can I prove $f(x) = \frac32 x − \frac23$ is surjective?
So I'm trying to prove the function $f(x) = \frac32 x − \frac23$ is bijective (injective and surjective)...

For injection I can easily write $\frac32x - \frac23 = \frac32y - \frac23$; that gives me $x=y$ so it's injective.

So the one more step to prove is to show function is surjective. I can say $f(x)=y$ which gives me $\frac32 x - \frac23 = y$ or $\frac{9x - 4}{6} = y$,
which finally gives me $x = \frac{6y+4}{9}$.

But there is a problem when I plug $x$ in function $\frac32 x - \frac23$ I can't get that $y$ value... Basically I get $\frac32 (6y + 4)\frac19 - \frac23 \implies 6y + \frac23 - \frac23$ which is $6y$ but I need $y$? I saw on Wolfram that this function must be surjective but I don't know how to prove it. Can anyone help me with that. And almost forgot to mention $f :\mathbb R \to\mathbb R$.

Thanks in advance :)

答:
The function is surjective, you just made a small computational error.

\begin{align*}
\frac{3}{2} \frac{6y+4}{9} - \frac{2}{3} &= \frac{6y+4}{6} - \frac{2}{3}\\
&= y + \frac{2}{3}-\frac{2}{3}\\
&= y.
\end{align*}


问:
Values of the zeta function of a modular form at negative integers
Suppose $f$ is an eigenform of level $N$ and weight $k$. Then, we can define a zeta function $L(s,f)$ attached to $f$. There are lots of studies about the values of $L(s,f)$ at integers between 1 and $k-1$. However, I have not been able to find out some references about the values of $L(s,f)$ outside this range. I am particularly interested in the values of $L(s,f)$ at negative integers.

There is only one case that I am quite familiar with. That is when $f$ corresponds to a CM elliptic curve. In this case, $L(-n,f)=0$ so it is not very interesting.

So my question is: what do we know about $L(-n,f)$?

答:
Those things are in Diamond and Shurman.

If $f\in S_k(\Gamma_1(N))$,

$L(s,f)$ is the Dirichlet series $\sum_n a_n(f)n^{-s}$, from the easy bound $|a_n(f)|=O(n^{k/2})$ it converges at least for $\Re(s)> k/2+1$ and $$(2\pi)^{-s} \Gamma(s) L(s,f)=\mathcal{M}[f(iy)](s)$$ Due to the exponential decay at the cusps $\mathcal{M}[f(iy)](s)$ is entire so that the analytic continuation of $L(s,f)$ is entire and has a zero at each pole of $\Gamma(s)$.

Also (Fricke involution)  $$\pmatrix{0&1\\-N&0}^{-1}\pmatrix{1+Na&b\\Nc&1+Nd}\pmatrix{0&1\\-N&0}=\pmatrix{1+Nd&b\\Nc&1+Na}$$
gives that $$f(z)=(Nz)^{-k} f^*(-1/(Nz))$$ where $f^*$ is again in $S_k(\Gamma_1(N))$, which gives the functional equation $$(2\pi)^{-s} \Gamma(s) L(s,f)=\mathcal{M}[f(iy)](s)$$ $$=\mathcal{M}[(iyN)^{-k} f^*(i/(Ny))](s)=(iN)^{-k}N^{k-s}\mathcal{M}[f^*(iy)](k-s)$$


问:
Combinatorial proof of $ \sum\limits_{k=0}^n(-1)^k\frac{2^{n-k}\binom{n}{k}}{(m+k+1)\binom{m+k}{m}}=\sum\limits_{k=0}^n\frac{\binom{n}{k}}{m+k+1}$
While answering the following question : Solution to a definite integral as $\int_0^{1}(x^{m}\left(1+x)^{n}dx\right)$, I came accross a strange identity involving  binomials.

Using the binomial theorem, we can find our first expression to that integral : $ \int_{0}^{1}{x^{m}\left(1+x\right)^{n}\,\mathrm{d}x}=\sum\limits_{k=0}^{n}{\frac{\binom{n}{k}}{m+k+1}} $, while solving it the way I did leads to another expression : $ \int_{0}^{1}{x^{m}\left(1+x\right)^{n}\,\mathrm{d}x}=\sum\limits_{k=0}^{n}{\left(-1\right)^{k}\frac{2^{n-k}\binom{n}{k}}{\left(m+k+1\right)\binom{m+k}{m}}} $.

I was wondering if we could prove, using combinatorials, that for any $ n,m\in\mathbb{N} $ : $$ \sum\limits_{k=0}^{n}{\left(-1\right)^{k}\frac{2^{n-k}\binom{n}{k}}{\left(m+k+1\right)\binom{m+k}{m}}}=\sum\limits_{k=0}^{n}{\frac{\binom{n}{k}}{m+k+1}}$$

答:
Here is an algebraic proof by way of enrichment. We seek to verify that

$$\sum_{k=0}^n (-1)^k
\frac{2^{n-k} {n\choose k}}{(m+k+1) {m+k\choose k}}
= \sum_{k=0}^n \frac{{n\choose k}}{m+k+1}.$$

We can re-write this as

$$\frac{m! n!}{(n+m+1)!} 2^n
\sum_{k=0}^n (-1)^k 2^{-k} {n+m+1\choose n-k}
= \sum_{k=0}^n \frac{{n\choose k}}{m+k+1}$$

or

$$2^n \sum_{k=0}^n (-1)^k 2^{-k} {n+m+1\choose n-k}
= (m+1) {n+m+1\choose n}
\sum_{k=0}^n \frac{{n\choose k}}{m+k+1}.$$

We get for the LHS

$$2^n \sum_{k=0}^n (-1)^k 2^{-k} {n+m+1\choose m+k+1}
= 2^n \sum_{k=0}^n (-1)^k 2^{-k} [z^{n-k}]
\frac{1}{(1-z)^{m+k+2}}
\\ = 2^n [z^n] \frac{1}{(1-z)^{m+2}}
\sum_{k=0}^n (-1)^k 2^{-k} z^k \frac{1}{(1-z)^k}.$$

Here the coefficient extractor enforces the range
and we find

$$2^n [z^n] \frac{1}{(1-z)^{m+2}}
\sum_{k\ge 0} (-1)^k 2^{-k} z^k \frac{1}{(1-z)^k}
= 2^n [z^n] \frac{1}{(1-z)^{m+2}}
\frac{1}{1+z/(1-z)/2}
\\ = [z^n] \frac{1}{(1-2z)^{m+2}}
\frac{1}{1+z/(1-2z)}
= [z^n] \frac{1}{(1-2z)^{m+1}}\frac{1}{1-z}.$$

On the other hand we have

$${n+m+1\choose n} {n\choose k}
= \frac{(n+m+1)!}{(m+1)! \times k! \times (n-k)!}
= {n+m+1\choose n-k} {m+k+1\choose m+1}$$

which gives for  the RHS

$$\sum_{k=0}^n {n+m+1\choose n-k}
\frac{m+1}{m+k+1} {m+k+1\choose m+1}
= \sum_{k=0}^n {n+m+1\choose m+k+1} {m+k\choose m}
\\ = \sum_{k=0}^n {m+k\choose m}
[z^{n-k}] \frac{1}{(1-z)^{m+k+2}}
= [z^n] \sum_{k=0}^n {m+k\choose m}
\frac{1}{(1-z)^{m+k+2}} z^k.$$

We once more have the coefficient extractor enforcing the range
and we get

$$[z^n] \frac{1}{(1-z)^{m+2}} \sum_{k\ge 0} {m+k\choose m}
\frac{1}{(1-z)^k} z^k
\\ = [z^n] \frac{1}{(1-z)^{m+2}} \frac{1}{(1-z/(1-z))^{m+1}}
= [z^n] \frac{1}{1-z} \frac{1}{(1-2z)^{m+1}}.$$

The LHS is the same as the RHS which concludes the argument. The
coeffcient extractor evaluates to

$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=0}^n {k+m\choose m} 2^k.}$$


问:
Find the sum using properties of binomial coefficients
I need to find the following sum using properties of binomial coefficients:$$\sum\limits_{k=0}^{n} k^3 {n\choose k}^2.$$ I transformed this sum first in:$n^2\sum\limits_{k=1}^{n} k{n-1\choose k-1}^2$.
Then after substitution $j=k-1$ I get $n^2\sum\limits_{j=0}^{n-1}(j+1){n-1\choose j}^2$.
Then, after separating this in two sums for first one I get: $n^2\sum\limits_{j=0}^{n-1}j {n-1\choose j}^2$. And for the second $\sum\limits_{j=0}^{n-1}{n-1 \choose j}^2$. I am not sure how to solve these two. Probably is has something to do with Vandermonde convolution, but honestly, I don't know to use it properly.

答:
Once you reduce it to

$$n^2\sum_{k=0}^{n-1}(k+1)\binom{n-1}k^2\,,\tag{1}$$

you can finish the calculation combinatorially.

Suppose that you have a pool of $n$ women and $n$ men, and you want to form a committee of $n+1$ people. The committee is to have a man and a woman as co-chairs, and it is to have a treasurer, who must be a women and who can be the female co-chair. There are $n^2$ ways to choose the co-chairs. Once that has been done, there are $$\binom{n-1}k\binom{n-1}{n-1-k}=\binom{n-1}k^2$$ ways to choose the remaining $n-1$ members of the committee in such a way that there are altogether $k+1$ women on the committee, and there are $k+1$ ways to choose which woman is to be the treasurer, so the number of different outcomes is given by $(1)$.

Alternatively, after choosing the co-chairs we could choose the treasurer and then the rest of the committee. There are $n^2$ ways to choose the co-chairs. If we decide that the female co-chair is also the treasurer, we can fill out the committee with any $n-1$ of the $2n-2$ remaining people, so there are $n^2\binom{2n-2}{n-1}$ outcomes in which the treasurer is also a co-chair. Otherwise, there are $n-1$ ways to choose one of the remaining women to be the treasurer, after which we can fill out the committee with any $n-2$ of the remaining $2n-3$ people, so there are $n^2(n-1)\binom{2n-3}{n-2}$ outcomes in which the three offices are held by three different people. It follows that

$$n^2\sum_{k=0}^{n-1}(k+1)\binom{n-1}k^2=n^2\binom{2n-2}{n-1}+n^2(n-1)\binom{2n-3}{n-2}\,.$$


问:
prove $\partial_x^2f + \partial_y^2 f = \partial_r^2 g+\frac{1}{r}\cdot \partial_r g + \frac{1}{r^2}\cdot \partial_\Theta^2 g$
Given $g(r, \Theta) = f(x(r,\Theta), y(r,\Theta))$ and $x = r \cdot \cos (\Theta)$ and $y = r \cdot \sin(\Theta)$ (so f is undetermined)

First thing I gotta clear. Is my assumption righ:  $\partial_r^2 g = 2\,\sin\Theta\,\cos\Theta\,\partial_x\partial_y\,f+\cos^2\Theta\,\partial_x^2\,f+\sin^2\Theta \partial_y^2 \,f$

Second thing: Is it accordingly correct: $\partial_\Theta^2\, g = -2\,r^2\,\cos\Theta\,\sin\Theta\,\partial_x\partial_y\,f+r^2\,\sin^2\Theta\,\partial_x^2\,f+ r^2\,\cos^2\Theta \partial_y^2 \,f$

Seems almost correct 'cos $r^2$ and $\frac{1}{r^2}$ cancel out and factoring out seems probable.

But $+ \frac{1}{r}\,\partial_r\,g = \frac{1}{r} \cdot (\cos\Theta\,\partial_x\,g+\sin\Theta\,\partial_y\,g)$ kinda destroys the entire symmetry ...

答:
Note that this is actually the Laplacian in two dimensions which is expressed in polar and Cartesian coordinates. You can easily find the proof, see this and this for example. In three dimensions we have

$$\Delta f = f_{xx} + f_{yy} + f_{zz} $$
And

$$\Delta f =  \frac{1}{r} (rf_r)_{r} + \frac{1}{r^2} f_{\theta \theta} + f_{zz}$$

Which reduces to two dimensions case when $f$ doesn't depend on $z$.


问:
Operation to Find the Ratio Between Two Numbers $x,y$
Background: I'd like to find the fraction between two positive numbers $x,y$. By definition the fraction must be lower than or equal to one. So if $x>y$, I find $R(x,y)=\frac{y}{x}$, ad if $y>x$, $R(x,y)=\frac{x}{y}$.

Motivation: It can be seen as the absolute value for ratios (in contrast to the absolute value for addition). It can also be seen as a metric or a distance defined by ratio.

Question: What operation is this $R$? Does it covered in any literature?

Guess: It is topologically equivalent to logarithm difference: $|\ln(x)-\ln(y)|$. This is a metric while $R$ is only a semimetric.

Does it have a name in literature so I can look it up?

答:
If your numbers are rational, $(x + y) / 2$ is rational and in between.

If they are irrational, you can try the following (here $\lceil z \rceil$ is the integer just larger than $z$, $\lfloor z \rfloor$ is the integer just smaller than $z$):

$\begin{align*}
  a
    &= \frac{1}{\lceil y - x \rceil} + 1 \\
  m
    &= \frac{\lceil a x \rceil + \lfloor a y \rfloor}{2 a}
\end{align*}$


问:
Finding a primitive root of a prime number
How would you find a primitive root of a prime number such as 761? How do you pick the primitive roots to test? Randomly?

Thanks

答:
Finding a primitive root of unity for $\text{modulo-}761$.

For computing purposes, create/calculate the $\text{powers of } 2 \text{ table}$ with 95 rows and 4 columns shown in the second section, corresponding to the subgroup generated by $[2]$; the table contains $378$ elements, missing only $[1]$ and $[760]=[-1]$ from the subgroup.

From the last row of table , where duplicates (two, $[722]$ and $[39]$) are first detected by the computing algorithm,

$\tag 1 \Large2^{-95} \equiv -2^{95} \pmod{761}$

and therefore the order of $[2]$ is $380$ and $[2]$ is not a primitive root of unity.

The number $[3]$ is not in the table but (from the $65^{th}$ row),

$\tag 2 \Large2^{65} \equiv 3^{2} \pmod{761}$

The order of $[2^{65}]$ is $76$ and the order of $[3]$ is $152$ (see this) and

$[3]$ is not a primitive root of unity.

The number $[4]$ is in the table (on the $2^{nd}$ row) and therefore can't be a primitive root of unity.

The number $[5]$ is in the table (on the $50^{th}$ row, $\,\large 2^{-50} = 5$) and therefore

can't be a primitive root of unity.

The number $[6]$ is not in the table but (from the $67^{th}$ row),

$\tag 3 \Large2^{67} \equiv 6^2 \pmod{761}$

The order of $[2^{67}]$ is $380$ and the order of $[6]$ is $760$ and $[6]$ is a primitive root of unity.

$\quad \small 2^n \;\,\, 2^{-n} \quad\; -2^n \, -2^{-n}$

```<code>002 381     759 380
004 571     757 190
008 666     753 095
016 333     745 428
032 547     729 214
064 654     697 107
128 327     633 434
256 544     505 217
512 272     249 489
263 136     498 625
526 068     235 693
291 034     470 727
582 017     179 744
403 389     358 372
045 575     716 186
090 668     671 093
180 334     581 427
360 167     401 594
720 464     041 297
679 232     082 529
597 116     164 645
433 058     328 703
105 029     656 732
210 395     551 366
420 578     341 183
079 289     682 472
158 525     603 236
316 643     445 118
632 702     129 059
503 351     258 410
245 556     516 205
490 278     271 483
219 139     542 622
438 450     323 311
115 225     646 536
230 493     531 268
460 627     301 134
159 694     602 067
318 347     443 414
636 554     125 207
511 277     250 484
261 519     500 242
522 640     239 121
283 320     478 441
566 160     195 601
371 080     390 681
742 040     019 721
723 020     038 741
685 010     076 751
609 005     152 756
457 383     304 378
153 572     608 189
306 286     455 475
612 143     149 618
463 452     298 309
165 226     596 535
330 113     431 648
660 437     101 324
559 599     202 162
357 680     404 081
714 340     047 421
667 170     094 591
573 085     188 676
385 423     376 338
009 592     752 169
018 296     743 465
036 148     725 613
072 074     689 687
144 037     617 724
288 399     473 362
576 580     185 181
391 290     370 471
021 145     740 616
042 453     719 308
084 607     677 154
168 684     593 077
336 342     425 419
672 171     089 590
583 466     178 295
405 233     356 528
049 497     712 264
098 629     663 132
196 695     565 066
392 728     369 033
023 364     738 397
046 182     715 579
092 091     669 670
184 426     577 335
368 213     393 548
736 487     025 274
711 624     050 137
661 312     100 449
561 156     200 605
361 078     400 683
722 039     039 722
</code>```

Body:
Finding a primitive root of unity for $\text{modulo-}761$.

For computing purposes, create/calculate the $\text{powers of } 2 \text{ table}$ with 95 rows and 4 columns shown in the second section, corresponding to the subgroup generated by $[2]$; the table contains $378$ elements, missing only $[1]$ and $[760]=[-1]$ from the subgroup.

From the last row of table , where duplicates (two, $[722]$ and $[39]$) are first detected by the computing algorithm,

$\tag 1 \Large2^{-95} \equiv -2^{95} \pmod{761}$

and therefore the order of $[2]$ is $380$ and $[2]$ is not a primitive root of unity.

The number $[3]$ is not in the table but (from the $65^{th}$ row),

$\tag 2 \Large2^{65} \equiv 3^{2} \pmod{761}$

The order of $[2^{65}]$ is $76$ and the order of $[3]$ is $152$ (see this) and

$[3]$ is not a primitive root of unity.

The number $[4]$ is in the table (on the $2^{nd}$ row) and therefore can't be a primitive root of unity.

The number $[5]$ is in the table (on the $50^{th}$ row, $\,\large 2^{-50} = 5$) and therefore

can't be a primitive root of unity.

The number $[6]$ is not in the table but (from the $67^{th}$ row),

$\tag 3 \Large2^{67} \equiv 6^2 \pmod{761}$

The order of $[2^{67}]$ is $380$ and the order of $[6]$ is $760$ and $[6]$ is a primitive root of unity.

$\quad \small 2^n \;\,\, 2^{-n} \quad\; -2^n \, -2^{-n}$

002 381     759 380
004 571     757 190
008 666     753 095
016 333     745 428
032 547     729 214
064 654     697 107
128 327     633 434
256 544     505 217
512 272     249 489
263 136     498 625
526 068     235 693
291 034     470 727
582 017     179 744
403 389     358 372
045 575     716 186
090 668     671 093
180 334     581 427
360 167     401 594
720 464     041 297
679 232     082 529
597 116     164 645
433 058     328 703
105 029     656 732
210 395     551 366
420 578     341 183
079 289     682 472
158 525     603 236
316 643     445 118
632 702     129 059
503 351     258 410
245 556     516 205
490 278     271 483
219 139     542 622
438 450     323 311
115 225     646 536
230 493     531 268
460 627     301 134
159 694     602 067
318 347     443 414
636 554     125 207
511 277     250 484
261 519     500 242
522 640     239 121
283 320     478 441
566 160     195 601
371 080     390 681
742 040     019 721
723 020     038 741
685 010     076 751
609 005     152 756
457 383     304 378
153 572     608 189
306 286     455 475
612 143     149 618
463 452     298 309
165 226     596 535
330 113     431 648
660 437     101 324
559 599     202 162
357 680     404 081
714 340     047 421
667 170     094 591
573 085     188 676
385 423     376 338
009 592     752 169
018 296     743 465
036 148     725 613
072 074     689 687
144 037     617 724
288 399     473 362
576 580     185 181
391 290     370 471
021 145     740 616
042 453     719 308
084 607     677 154
168 684     593 077
336 342     425 419
672 171     089 590
583 466     178 295
405 233     356 528
049 497     712 264
098 629     663 132
196 695     565 066
392 728     369 033
023 364     738 397
046 182     715 579
092 091     669 670
184 426     577 335
368 213     393 548
736 487     025 274
711 624     050 137
661 312     100 449
561 156     200 605
361 078     400 683
722 039     039 722


问:
Point and a normal vector question
I'm trying to calculate a normal vector given three points in the plane.  My results do not agree with what I expect, and I would appreciate guidance.  I'm using the following Wiki definition:

$n = (p_2 - p_1) \times (p_3 - p_1)$

Points:
$p_1 = (1, 1, 1), p_2 = (1, 0, 0), p_3 = (0, 1, 0)$

$p_2 - p_1 = (0, -1, -1)$

$p_3 - p_1 = (-1, 0, -1)$

It's my understanding that $p_2 - p_1$ and $p_3 - p_1$ represent two perpendicular vectors specifying the plane through the origin.  We then take the cross product to get the norm and the determinant can be used to calculate the cross product.

$\begin{vmatrix}
i & j & k \\
0 & -1 & -1 \\
-1 & 0 & -1 \\
\end{vmatrix}$

$i \begin{vmatrix}
   -1 & -1 \\
   0 & -1 \\
   \end{vmatrix}
 + j \begin{vmatrix}
   0 & -1 \\
   -1 & -1 \\
   \end{vmatrix}
 - k \begin{vmatrix}
   0 & -1 \\
   -1 & 0 \\
   \end{vmatrix}
$

$i(-1*-1 - 0) + j(-1*-1 - 0) - k(-1*-1 - 0)$

$i + j - k$

This vector does not appear normal to the plane when I look at a graph.  Does anyone know if I made an error?

答:
Does anyone know if I made an error?

It should be $i(-1\times-1 - 0) - j(0 - (-1\times-1)) + k(0 - (-1\times-1))= i \color{red}+ j - k$.

Title: Point and a normal vector question


问:
what is the relation of smooth compact supported funtions and real analytic function?
What is the major difference between real analytic and test function (smooth compact supported functions). Can we find a real analytic function $f$ on $\mathbb R^n$ which is also smooth compact supported? If not then need a proof.

答:
Assume $f$ is a real-analytic test function which is not identically zero; this will lead to a contradiction. Let $K$ denote the support of $f$. The idea (also present in OP's attempt) is to expand $f$ into a power series centered at some boundary point $x_0$ of the support; find that the coefficients are all zero, obtain a contradiction. Key steps:

A nonempty bounded set must have nonempty boundary (because the only sets with empty boundary are $\varnothing$ and $\mathbb R^n$).

Let $x_0$ be a boundary point of $K$. Note that $x_0\in K$ and $x_0$ is also   a limit point of the complement of $K$.

All derivatives of $f$ are identically zero on the complement of $K$

They are also continuous on all of $\mathbb R^n$. By continuity they are equal to zero at $x_0$.

The power series of $f$ centered at $x_0$ is $0+0(x-x_0)+\dots$

Therefore, $f$ is equal to zero  in some neighborhood of $x_0$. But this contradicts the definition of support of $K$.


问:
finding the mean and standard deviation of randomly answering questions correctly
2 students each write the same multiple choice test. The test has 5 questions, and each question has 5 answers (exactly one of which is right). The students answer the questions randomly.

What is the probability that they both get the same number of answers correct? (ANSWER: 0.3198)

What is the probability that their papers are identical, assuming that they have answered independently of one another? (ANSWER: 0.00032)

For the first part I tried taking the joint probability of X=Y

For the second part I took the total number of ways to answer 5 questions (3125) and put that under 1 (1/3125)

Does my reasoning make sense?

答:
Your reasoning for the second part is correct.  The idea is to observe that, no matter what the first student answers, the second student has to match their answers exactly, and this is only possible if for each question, they choose the same answer as the first student.  Since for a single question that happens with probability $1/5$, for all five, it is $1/5^5 = 1/3125$.

The first part only requires that they get the same number of questions correct.  This is given by the sum $$\sum_{k=0}^5 \left( \binom{5}{k} (1/5)^k (1 - 1/5)^{5-k} \right)^2 = \frac{3122577}{9765625} \approx 0.319752.$$


问:
Digital Logic 3 Light Switch In A Room Question
There are three doors in a room with a light switch next to each one. You can turn on the light from any door when you enter, and you can turn-off the light from any door when you leave.

a) Write the truth table for the function

b) Use the Karnaugh map to see if it can be simplified

c) Draw logic circuit for the function

d) Check if it is possible to realize function with a single logic gate

Especially d is the hardest one for me to solve. Can you help me?
If you wanna see the picture of question, click here.

答:
Labelling the switches by each door as $A$, $B$, and $C$, and using digital electronics notation,

Logical OR is represented with a $+$,

AND is represented with multiplication. $A$ AND $B$ is $AB$,

Negation is represented with an overline. Not $A$ is $\overline{A}$,

"Exclusive or" (XOR) is represented as $\oplus$,

a Karnaugh map would look something like:

\begin{array}{c|cccc}
&\overline{A}\overline{B}&\overline{A}B&AB&A\overline{B}\\ \hline
\overline{C}&0&1&0&1\\
C&1&0&1&0
\end{array}

Unfortunately this doesn't simplify on the Karnaugh map in an obvious way. It can be simplified using the laws of Boolean algebra though.

\begin{align}
\overline{A}B\overline{C}+ABC+\overline{A}\overline{B}C+A\overline{B}\overline{C}\\
A(BC+\overline{B}\overline{C})+\overline{A}(B\overline{C}+\overline{B}C)\\
A(\overline{B\oplus C})+\overline{A}(B\oplus C)\\
A\oplus(B\oplus C)\\
\end{align}

Although it may look like you can't implement a single logic gate to realise function, since it is an application question, you could just recommend the 3-input XOR gate bellow:

XOR logic gates with more than 2 inputs are sometimes called "odd functions” or “modulo-2 sum functions", since they return TRUE whenever an odd number of their inputs are true, precisely the requirements for a light to turn on in your question.


问:
Mean and Variance Verification
Let X be a random variable with the pdf

$f(x) = \begin{cases}
-2e^{-2x}+2e^{-x} &: 0 <x\\
0 &: \text{else}
\end{cases}$

and suppose we want to find the mean and variance.

First to find the mean,

$\mu=\int _0^{\infty }\:x\cdot f\left(x\right)dx=\int _0^{\infty }\:x\cdot \left(-2e^{-2x}+2e^{-x}\right)dx=\int _0^{\infty }\:-2xe^{-2x}+2xe^{-x}dx=\frac{5}{2}$

and then to find the variance,

$\sigma^2(X)=E(X)^2-(E(X))^2$

$\quad \qquad=\int _0^{\infty }\:x^2\cdot \left(-2e^{-2x}+2e^{-x}\right)dx-\left(\frac{5}{2}\right)^2$

$\quad \qquad = \int _0^{\infty }\:-2x^2e^{-2x}+2x^2e^{-x}dx-\left(\frac{5}{2}\right)^2$

$\quad \qquad =\frac{7}{2}-\left(\frac{5}{2}\right)^2$

$\quad \qquad = -\frac{11}{4} =-2.75$

However this is obviously incorrect, as the variance shouldn't be negative. Can anyone tell me what I did wrong here? Thanks.

答:
Your calculation of the mean is incorrect, therefore making the variance calculation wrong.

Instead, the mean is $3/2$.  You can see this by observing that $2e^{-2x}$ is the density for an exponential distribution with mean $1/2$, and $e^{-x}$ is the density for an exponential distribution with mean $1$.  Therefore, the linear combination of these has mean $2(1) - \frac{1}{2} = \frac{3}{2}$.  Your second moment calculation is correct, so you would get a variance of $\frac{7}{2} - (\frac{3}{2})^2 = \frac{5}{4}$.

Title: Mean and Variance Verification


问:
derivative of following function defined in proper domain w.r.t x
I want to find derivative of following function defined in proper domain w.r.t x

$$
e^{3x} \log{2x}
$$

答: Use the product rule which says that if $f$ and $g$ are differentiable functions then $(f(x)g(x))' = f(x) \cdot g'(x) + g(x) \cdot f'(x)$ and note that the derivative of $e^{ax}= ae^{ax}$ for some $a \neq 0$ and derivative of $\log{x}$ is $\frac{1}{x}$</p>


问:
Moment generating function of a stochastic integral
Let $(B_t)_{t\geq 0}$ be a Brownian motion and $f(t)$ a square integrable deterministic function. Then:
$$
\mathbb{E}\left[e^{\int_0^tf(s) \, dB_s}\right] = \mathbb{E}\left[e^{\frac{1}{2}\int_0^t f^2(s) \, ds}\right]
$$
Now assume $(X_t)_{t\geq 0}$ is such that $\left(\int_0^tX_sdB_s\right)_{t\geq 0}$ is well defined. Does
$$
\mathbb{E}\left[e^{\int_0^tX_s \, dB_s}\right] = \mathbb{E}\left[e^{\frac{1}{2}\int_0^tX_s^2 \, ds}\right]
$$
still hold?</p>

答:
If $X$ and $B$ are independent, yes (use the first result to compute the expectation conditional on $X$, then take the expectation).

Otherwise, no. For a counterexample, consider $X=B$ and use Itô's formula $\mathrm d (B^2)=2B\mathrm dB+\mathrm dt$.


问:
Short exact sequence of tensors
Consider the following sequence

Where $\gamma: n \mapsto 2n $ and $\varphi: n \mapsto \overline{n}$. When using the properties of tensors we get the sequence of $\mathbb{Z}/2\mathbb{Z}$, which is clearly not exact. On the other hand, $\text{im}(\gamma\otimes1) = \{2k \otimes l \mid k \in \mathbb{Z}, l \in \mathbb{Z}/2\mathbb{Z} \} = \{0\otimes 0 \}$ and $\ker(\varphi\otimes 1) = \{k\otimes l \mid \varphi(k) = \overline{0} \text{ or } l = \overline{0} \} = \{0\otimes 0\}$. Therefore, $\text{im}(\gamma\otimes1) = \ker(\varphi\otimes 1)$, which gives us that the sequence is exact.  Is my image or kernel wrong?

答: The sequence is indeed exact in the middle (though your method of checking this really requires more justification--in general, not every element of a tensor product $A\otimes B$ has the form $a\otimes b$, and moreover $a\otimes b$ can be $0$ even if neither $a$ nor $b$ is $0$).  The place it fails to be exact is on the left: $$0\to\mathbb{Z}\otimes\mathbb{Z}/2\mathbb{Z}\stackrel{\gamma\otimes1}{\to}\mathbb{Z}\otimes\mathbb{Z}/2\mathbb{Z}$$ is not exact because the kernel of $\gamma\otimes 1$ is nontrivial.

Title: Short exact sequence of tensors


答:
You have indeed defined the equivalent for multiplication of absolute value for addition.

I know of no particular name for this function.

It defines the same topology as the absolute value of the difference of the logarithms. That might be a sensible way to measure "distance defined by ratio". Whether it's useful for you depends on the application.

Title: Operation to Find the Ratio Between Two Numbers $x,y$

Body:
Background: I'd like to find the fraction between two positive numbers $x,y$. By definition the fraction must be lower than or equal to one. So if $x>y$, I find $R(x,y)=\frac{y}{x}$, ad if $y>x$, $R(x,y)=\frac{x}{y}$.

Motivation: It can be seen as the absolute value for ratios (in contrast to the absolute value for addition). It can also be seen as a metric or a distance defined by ratio.

Question: What operation is this $R$? Does it covered in any literature?

Guess: It is topologically equivalent to logarithm difference: $|\ln(x)-\ln(y)|$. This is a metric while $R$ is only a semimetric.

Does it have a name in literature so I can look it up?


问:
Prove that $\log X < X$ for all $X > 0$
I'm working through Data Structures and Algorithm Analysis in C++, 2nd Ed, and problem 1.7 asks us to prove that $\log X < X$ for all $X > 0$.

However, unless I'm missing something, this can't actually be proven. The spirit of the problem only holds true if you define several extra qualifiers, because it's relatively easy to provide counter examples.

First, it says that $\log_{a} X < X$ for all $X > 0$, in essence.

But if $a = -1$, then $(-1)^{2} = 1$. Therefore $\log_{-1} 1 = 2$. Thus, we must assume
$a$ is positive.

if $a$ is $< 1$, then $a^2 < 1$.  Therefore we must assume that $a \geq 1$.

Now, the book says that unless stated otherwise, it's generally speaking about base 2 for logarithms, which are vital in computer science.

However, even then - if $a$ is two and $X$ is $\frac{1}{16}$, then $\log_{a} X$ is $-4$. (Similarly for base 10, try taking the log of $\frac{1}{10}$ on your calculator: It's $-1$.) Thus we must assume that $X \geq 1$.

...Unless I'm horribly missing something here. The problem seems quite different if we have to prove it for $X \geq 1$.

But even then, I need some help solving the problem. I've tried manipulating the equation as many ways as I could think of but I'm not cracking it.

答:
If the proposition is true then $\frac{\ln x}{x} < 1$ for $x > 0$.

Let $x = e^z$. The condition $x > 0$ translates to $z \in \mathbb R$: no restriction on $z$.

Then $\frac{\ln e^z}{e^z} = \frac{z\ln e}{e^z} = \frac{z}{e^z} < 1$, for all $z$.

For the negative $z$, this is true by inspection. $e^z$ is positive, and so $\frac{z}{e^z}$ is negative and thus less than $1$.

The proposition is true at $z = 0$, where $\frac{z}{e^z}$. Furthermore, $\lim_{z\to\infty}\frac{z}{e^z} = 0$ because $e^z$ grows much faster than $z$.

In between these extremes, the function reaches a maximum, which we can find using calculus.

$$\frac{d}{dz}\frac{z}{e^z} = 0$$

$$\rightarrow\frac{e^z - ze^z}{e^{2z}} = 0$$

$$\rightarrow e^z - ze^z = 0$$

$$\rightarrow e^z(1 - z) = 0$$

$$\rightarrow z = 1$$

At this maximum, the value is $1/e$ which is less than 1. So in fact not only $\frac{z}{e^z} < 1$ but in fact $\frac{z}{e^z} \le \frac{1}{e}$.

From this we can work backwards to $\frac{\ln x}{x} < 1$, which shows that $\ln x$ < $x$.

Can we generalize this to other bases, to discover whether this is true for some smaller bases $b$, $1 < b < e$. Base 2 is in this range.

Let's repeat the argument for a generalized base $b$, starting with the proposition $\frac{z}{b^z} < 1$.

Aagin, this proposition is true at $z = 0$ in the same way, and has the same limit $0$ out to infinity, and some maximum value where the derivative is zero:

$$\frac{d}{dz}\frac{z}{b^z} = 0$$

$$\rightarrow\frac{b^z - zb^z\ln b}{b^{2z}} = 0$$

$$\rightarrow b^z - zb^z\ln b = 0$$

$$\rightarrow b^z(1 - z\ln b) = 0$$

$$\rightarrow z = \frac{1}{\ln b}$$

At $z = \frac{1}{\ln b}$, the function's value is $\frac{1/{\ln b}}{e^{1/\ln b}}$. As long as this is less than 1, then the proposition we are trying to prove is true for base $b$:

$$\frac{1/{\ln b}}{e^{1/\ln b}} < 1$$

Raise both sides to the power of $\ln b$:

$$\left(\frac{1/{\ln b}}{e^{1/\ln b}}\right)^{\ln b} < 1$$

$$\frac{1/\left({\ln b}\right)^{\ln b}}{e} < 1$$

$$\frac{1}{e\left({\ln b}\right)^{\ln b}} < 1$$

$$\left({\ln b}\right)^{\ln b} > \frac{1}{e}$$

At this point we could treat ${\ln b}$ as a base, and take the log in that base of both sides but that doesn't seem to lead to any obvious separation of variables.

Never mind, we can switch to numeric techniques for probing different values of $b$.

It is true for $b = 2$, and hence the proposition holds for base two logarithms. It also holds for $b = 1 + \epsilon$ for various small $\epsilon$. The limit appears to be 1. $\left({\ln b}\right)^{\ln b}$ appears to approach 1 as $b$ approaches $1$ from above, and it is $1$ at $b = e$. Between $1$ and $e$ it hits a minimum, but not lower than $1/e$. After $e$ it increases.

So it look like $\log_b x < x$ holds for all bases $b > 1$.

Title: Prove that $\log X < X$ for all $X > 0$


答:
You've gotten a lot of feedback on how to evaluate the sum without Vandermonde's convolution
$$\sum_{k=0}^r \binom{m}{k} \binom{n}{r-k} = \binom{m+n}{r}$$
but if you want to use it, this is how you would proceed:
$$\begin{align}
\sum_{j=0}^n \binom{n}{j}^2 &= \sum_{j=0}^n \binom{n}{j} \binom{n}{n-j} = \binom{2n}{n}.
\end{align}$$
We simply choose $m = r = n$, but before doing so, we use the binomial coefficient reflection identity $$\binom{n}{j} = \frac{n!}{j!(n-j)!} = \binom{n}{n-j}.$$

Title: Find the sum using properties of binomial coefficients

Body:
I need to find the following sum using properties of binomial coefficients:$$\sum\limits_{k=0}^{n} k^3 {n\choose k}^2.$$ I transformed this sum first in:$n^2\sum\limits_{k=1}^{n} k{n-1\choose k-1}^2$.
Then after substitution $j=k-1$ I get $n^2\sum\limits_{j=0}^{n-1}(j+1){n-1\choose j}^2$.
Then, after separating this in two sums for first one I get: $n^2\sum\limits_{j=0}^{n-1}j {n-1\choose j}^2$. And for the second $\sum\limits_{j=0}^{n-1}{n-1 \choose j}^2$. I am not sure how to solve these two. Probably is has something to do with Vandermonde convolution, but honestly, I don't know to use it properly.


问:
Function continuous at $ (0,0) $ but not differentiable at $ (0,0) $
In a book I ran into the following problem:

Let $ f: \mathbb {R}^2 \rightarrow \mathbb{R} $ defined as $ \displaystyle \frac{| x | y}{\sqrt {x ^ 2 + y ^ 2}} $ if $ (x , y) \neq (0,0) $ and $ 0 $ if $ (x, y) = (0,0) $.

They ask me to prove that $ f $ is continuous at $ (0,0) $, that it has all its directional derivatives in $ (0,0) $ but that $ f $ is not differentiable at $ (0,0) $, I It seems strange that it is continuous at that point but at the same time is not derivable. Is this correct? If so, how is it resolved?

答:
HINT:

The function is continuous at $(0,0)$ if $\displaystyle\lim_{(x,y)\to(0,0)}f(x,y)=f(0,0)$. So for this, you should look at the following limit: $$\lim_{(x,y)\to(0,0)}\frac{|x|y}{\sqrt{x^2+y^2}}$$

The function is differentiable at $(0,0)$ if the tangent plane at $(0,0)$ provides a "good" approximation for $f(x,y)$ near $(0,0)$. You should calculate $f_x(0,0)$ and $f_y(0,0)$ to get the tangent plane.


问:
probability of passing a multiple choice test
The test has 5 questions each with 5 answers. what is the probability of getting more than 3 questions right (4/5, 5/5)?

What I know: Each question has 5 selections, and the probability of getting one wrong is 4/5. I thought it would be as simple as the multiplicative rule, but there's more to it apparently. I think it has to do with a formula as you go along in the questions, but it states that each probability is equal, as a 1/5 chance.

Any help?

答:
Hints:

What's the probability of getting all five questions right? That's a straightforward application of the multiplicative rule.

What's the probability of getting exactly four questions right? This is slightly different, as there are five ways to do so: ****-, ***-*, **-**, *-***, -****. The probability for each of these is the same, so calculate the probability $p$ for ****- and multiply it by $5$.


问:
What real number is exactly one less than its cube?
And does it have any of the special properties that the golden ratio (one less than its square) has?</p>

答: Actually, this is fairly well studied, although not as extensively as the Golden Ratio. As such, it does indeed have a name, the Plastic number. It's also called the <b>silver ratio</b> but as the linked page says, this is somewhat improper. </p>

Body: Actually, this is fairly well studied, although not as extensively as the Golden Ratio. As such, it does indeed have a name, the Plastic number. It's also called the silver ratio but as the linked page says, this is somewhat improper.


问:
Evaluate $ \lim_{n \to \infty} \int_{-\infty}^\infty \frac{1}{1+x^2e^{nx^2}}\,dx$
I am trying to evaluate the following limit:

$$ \lim_{n \to \infty} \int_{-\infty}^\infty \frac{1}{1+x^2e^{nx^2}}\,dx$$

Now seems we can use the fact that $$ \int_\mathbb{R}\frac{1}{1+x^2}dx = \pi$$ since this is the definite integral of $\arctan(x)$.

I tried finding a clever way to re-write the denominator but I am not getting anywhere.

答:
$\lim\limits_{n\rightarrow +\infty}\frac{1}{1+x^2e^{nx^2}}=0$ if $x\neq 0$ and $1$ if $x=0$, the dominated convergence tells you that $$\lim\limits_{n\rightarrow +\infty}\int_{-\infty}^{+\infty}\frac{dx}{1+x^2e^{nx^2}}=0$$
(using the domination $\frac{1}{1+x^2e^{nx^2}}\leqslant\frac{1}{1+x^2}$)


问:
The famous prime race and generalizations
So I was messing around with the famous prime race that comes down to this:

We make a list of primes. The list has two rows; the top row is for primes $1\mod 4$ and the bottom row for primes $3\mod 4$. Our list, up to the 10th prime:

```<code>5  13  17  29
3   7  11  19  23  31
</code>```

The idea is that the two rows are "racing" to be longest, and the bottom row always seems to be winning.

An intuitive visual representation of this would be, we start at $0$ on the number line, and start going through the primes. Whenever we encounter a prime $1\mod 4$ we do a step right, and whenever the prime is $3\mod 4$, we do a step left. This is not quite such an interesting visualization however; it's just a line. However, this becomes more interesting when we generalize it.

The generalization.

We generalize this by picking a number, and do the prime race $\mod n$. Apart from a couple (certainly finite number of) exceptions, all primes fit in one of the $\varphi(n)$ categories. With this, I mean, let $U(n)=\{u_1,u_2,\cdots,u_{\varphi(n)}\}$ with $0\leq u_i<n$, and $i<j\iff u_i<u_j$, and $\gcd(u_i,n)=1$, then each $u_i$ represents a category; all primes $p$ (except for the ones that are divisors of $n$) must have $p\equiv u_i\mod n$ for exactly one $u_i$. Now we start going through the primes, and whenever we encounter a prime $p\equiv u_i\mod n$, we take a step length $1$ in the direction $\tfrac{2i\pi}{\varphi(n)}$.

An example, $n=8$.
This is a relatively easy example to try on grid paper, since $\phi(8)=4$, so we have $4$ directions to go in. For the first $10^3$, $10^4$ and $10^5$ primes, this looks like this (the red dot is the starting point, the blue dot is the end. Open the image in a new tab to see it at full resolution)

Interestingly enough, the red dot, the start, is in all images the right-most point in the entire thing, even though it is seemingly random. So this is the case for $n=4$, and for $n=8$. Maybe it's the powers of $2$, let's try $n=10$ (which happens to have $\phi(10)=4$ as well). We get (again, for the first $10^3$, $10^4$ and $10^5$ primes):

We see this time it's tending more to the bottom-left rather than just the left; but again, one side (the top-right) remains untouched. Let's try this again, but this time with an $n$ without $\phi(n)=4$. We try $n=11$ through $n=15$:

$n=11$

$n=12$

$n=13$

$n=14$

and $n=15$

We see $n=12$ is very convincingly going to the left and not touching the right, and $n=15$ is leaning towards the top-left, and leaves the bottom-left almost untouched. $n=11$, $n=13$ and $n=14$ however there's no specific way the blob seems to lean.

Question.

Why do the blobs with $\varphi(n)=4$ lean so convincingly to one side, while not even touching the other? Do there exist similar patterns for other $n$ with different $\varphi$? Are they related to the fact that the $\varphi$ is a power of $2$ (since $n=15$ has similar behaviour)?

I expected them to stay around the middle a lot more since the density of the primes $\mod n$ is evenly spaced, that is, picking any prime, it has chance $\frac{1}{\varphi(n)}$ to be $i\mod n$ (where $\gcd(i,n)=1$).

答:
The phenomenon that most of the time, there are more primes $p=3\pmod 4$ than $1\pmod 4$ is known as Chebyshev's Bias. Let $P_{a,b}(N)$ denote the number of primes congruent to $a \pmod b$ $ ≤ N$. Hardy and Littlewood showed that $P_{1,4}(N) - P_{3,4}(N) = 0$ for infinitely many $N$. In other words, $P_{1,4}(N)$ and $P_{3,4}(N)$ switch leads in the prime race infinitely often. However, one residue does not have any siginifant lead over another residue, and as $N → ∞$, $P_{3,4}(N) - P_{1,4}(N) = 0$. This also holds for any integers $a$ and $c$ relatively prime to $b$: $P_{a,b}(N) - P_{c,b}(N) = 0$ as $N → ∞$.

A generalization of Chebyshev's Bias states that most of the time, $P_{a,b}(N) > P_{c,b}(N)$ if $a$ is a quadratic non-residue $\pmod b$ and $c$ is a quadratic residue $\pmod b$. This usually works best if $b$ is prime, but if $b$ is composite, there is more than one way to group the elements $\pmod b$ into quadratic residue and non-residue groups.

For example, let $b=8$.

The elements $1,3,5,7$ are all coprime to $8$, and each odd prime will fall into one of these $4$ categories. Following from a generalization of Chebyshev's Bias, we can assume either of $3$ scenarios:

$P_{1,8}+P_{5,8} < P_{3,8}+P_{7,8}$ (this is the same as the original Chebyshev's Bias, categorized by $-1$ being a quadratic residue $\pmod p$)

$P_{1,8}+P_{3,8} < P_{5,8}+P_{7,8}$ (categorized by $-2$ being a quadratic residue $\pmod p$)

$P_{1,8}+P_{7,8} < P_{3,8}+P_{5,8}$ (categorized by $2$ being a quadratic residue $\pmod p$)

In either case, $1\pmod 8$ appears on the left hand side $3$ of $3$ times, so it would make sense that it might be behind in the race when individually compared to the rest of the residues. As for comparing $3,5,7 \pmod 8$, all of them fall into exactly one residue class, having $-2,-1,2$ as quadratic residues, respectively. Therefore, there appears to be no bias for these three residues and can evidently be seen here.

This phenomenon can be observed $\pmod {12}$ and $\pmod {24}$: $1$ trails far behind the rest of the residues. The more highly composite the modulus $b$ is, the more bias there will be.

This also appears to be the case when comparing primes solely congruent to $3\pmod 4$ with other primes congruent to $3\pmod 4$, with a different modulus $b$. For example, there should be no bias between $P_{7,12}$ and $P_{11,12}$. The argument is simple.

If we look at $-3$ being a quadratic non-residue $\pmod p$, then we can assume that most of the time, $P_{7,12} > P_{11,12}$. However, if we look at $3$ being a quadratic non-residue $\pmod p$, then most of the time, $P_{7,12} > P_{11,12}$, a contradiction. Therefore, there should be no bias between $P_{7,12}$ and $P_{11,12}$.

For general $b$, if $p=3\pmod 4$, then either $b$ is a quadratic residue, or $-b$ is a quadratic residue $\pmod p$, so no residue (already assuming all primes in the race are $3 \pmod 4$) should have any lead over the other.

Edit: I have investigated the prime race $\mod 8$ more extensively. It appears that up to a bound $10^{10}$, there is no integer $1 < N < 10^{10}$ such that .$P_{1,8} > [P_{3,8}$ and $P_{5,8}$ and $P_{7,8}]$ (In other words, $1 \pmod 8$ does not lead the prime race below $10^{11}$). I did compare $P_{1,8}$ with $P_{[[{3,5,7}]],8}$ individually up to $10^{11}$. The only time(s) when $P_{1,8}$ has a lead over any of the other residues is at $N=588067889$, when $P_{1,8} > P_{5,8}$.

I have also attempted to study more generalizations. From numeric evidence, it appears that the more residues $\pmod b$ there are, the less likely that $P_{1,b}$  will be the most behind in the race, but it will still have a tendency to fall behind in the race when compared to the rest of the residues:

For example, $\mod 120$, the only $a$ with no integer $N < 10^{10}$ such that $P_{1,120} > P_{a,120}$ is $a=59$. It is also observed that most of the time, $P_{1,120}$ is not the least behind in the race, it is more likely that it is $2$nd, $3$rd, $4$th least.

List the coprime elements $\pmod b$ from left to right based on the value of $P_{a,b}$ from least to greatest, notice that $1$ will have a tendency to gradually shift toward the right as the modulus $b$ increases,
explaining the more darker grids for $n=11,12,13,14,15$ as for your original question.


答:
This problem is a straightforward application of the dominated convergence theorem. For all $n \ge 1$, we have $\dfrac{1}{1+x^2e^{nx^2}} \le \dfrac{1}{1+x^2}$, and $\displaystyle\int_{-\infty}^{\infty}\dfrac{1}{1+x^2}\,dx = \pi < \infty$.

Hence, we can use the dominated convergence theorem to say that $$\lim_{n \to \infty}\left[\int_{-\infty}^{\infty}\dfrac{1}{1+x^2e^{nx^2}}\,dx\right] = \int_{-\infty}^{\infty}\left[\lim_{n \to \infty}\dfrac{1}{1+x^2e^{nx^2}}\right]\,dx$$ provided that $\displaystyle\lim_{n \to \infty}\dfrac{1}{1+x^2e^{nx^2}}$ exists for all $x \in \mathbb{R}$. Can you evaluate this limit for all $x \in \mathbb{R}$? Then all you have to do is integrate that limit over $\mathbb{R}$.

Title: Evaluate $ \lim_{n \to \infty} \int_{-\infty}^\infty \frac{1}{1+x^2e^{nx^2}}\,dx$

Body:
I am trying to evaluate the following limit:

$$ \lim_{n \to \infty} \int_{-\infty}^\infty \frac{1}{1+x^2e^{nx^2}}\,dx$$

Now seems we can use the fact that $$ \int_\mathbb{R}\frac{1}{1+x^2}dx = \pi$$ since this is the definite integral of $\arctan(x)$.

I tried finding a clever way to re-write the denominator but I am not getting anywhere.


问:
How do I calculate the greatest guarantee period through normal distribution?
The full question says:

"If the life of a car engine, calculated in miles, is normally distributed, with a mean of 240,000 miles and a standard deviation of 18,500 miles, what is the greatest the guarantee period should be if the company wants less than 2% of the engines to fail while under warranty?"

I've been learning standard distribution for the past couple weeks but have really struggled when I actually get questions. I understand the z score equation to solve it:

Z score equation

I also know how to use the table but I need to have a better understanding of questions like this. Any help would be very much appreciated :)

答:
Assuming that engine life is normally distributed (which is questionable in practice), you have $X \sim \mathsf{Norm}(\mu = 240, \sigma =18.5),$ in thousands of miles. You want $k$ such
that $P(X <k) = 0.02.$

Using software: In R, where qnorm is a normal quantile function (inverse CDF):

```<code>qnorm(.02, 240, 18.5)
[1] 202.0056
pnorm(202, 240, 18.5)
[1] 0.01998523
pnorm(200, 240, 18.5)
[1] 0.01530284
</code>```

So 202,000 miles is the largest limit. Maybe easier to
say it's a 200,000 mile guarantee.

Using printed normal tables: Using standardization and printed normal CDF tables you would have the following.

$$P(X < k) = P\left(\frac{X-\mu}{\sigma} < \frac{k - 240}{18.5}\right) = 0.02.$$

Then from a printed table find the z-score that cuts probability $0.02$ from the lower tail of the standard normal distribution. Look in the body of the table to find a value (nearly) equivalent to probability  $0.0200$ (of being smaller), and then read the z-score from the margins.
Maybe $z = -2.05.$ Then solve $\frac{k-240}{18.5} = -2.05$
to get a reasonable approximation of $k.$

Body:
Assuming that engine life is normally distributed (which is questionable in practice), you have $X \sim \mathsf{Norm}(\mu = 240, \sigma =18.5),$ in thousands of miles. You want $k$ such
that $P(X <k) = 0.02.$

Using software: In R, where qnorm is a normal quantile function (inverse CDF):

qnorm(.02, 240, 18.5)
[1] 202.0056
pnorm(202, 240, 18.5)
[1] 0.01998523
pnorm(200, 240, 18.5)
[1] 0.01530284

So 202,000 miles is the largest limit. Maybe easier to
say it's a 200,000 mile guarantee.

Using printed normal tables: Using standardization and printed normal CDF tables you would have the following.

$$P(X < k) = P\left(\frac{X-\mu}{\sigma} < \frac{k - 240}{18.5}\right) = 0.02.$$

Then from a printed table find the z-score that cuts probability $0.02$ from the lower tail of the standard normal distribution. Look in the body of the table to find a value (nearly) equivalent to probability  $0.0200$ (of being smaller), and then read the z-score from the margins.
Maybe $z = -2.05.$ Then solve $\frac{k-240}{18.5} = -2.05$
to get a reasonable approximation of $k.$


问:
Find an expression for the $nth$ composition of $T$ with itself $T ^ n$ $(x, y)$. $T (x, y) = (3x - y, 2y)$
Let $$T: \mathbb{R}^2 → \mathbb{R}^2$$ be the linear transformation defined by $$T (x, y) = (3x - y, 2y)$$

Find an expression for the $nth$ composition of $T$ with itself $T ^ n$ $(x, y)$.

I'm stuglying find $T^n$, my teacher told to consider the Base =
$$\{(1, 1), (0, 1)\}$$
and apply some change of base.

Can someone help me?

Hint: consider the base {(1, 1), (1, 0)} and apply some change of base.

答:
Well, let's see.  The matrix rel the standard basis would be $A=\begin{pmatrix}3&-1\\0&2\end{pmatrix}$.  The characteristic polynomial is $C(x)=(x-3)(x-2)$.  The eigenvalues are thus $3$ and $2$.  Note that $T(1,1)=2\cdot(1,1)$, and $T(1,0)=3\cdot(1,0)$.  Thus your hint asks you to consider a basis of eigenvectors.

This basis can be used to diagonalize the matrix.  That is, if we set $P=\begin{pmatrix}1&1\\1&0\end{pmatrix}$, then $P^{-1}AP=\begin{pmatrix}2&0\\0&3\end{pmatrix}$.

Now figuring out the $n$-th power of a diagonal matrix is easy.  In this case it's just $\begin{pmatrix}2^n&0\\0&3^n\end{pmatrix}$.

And you should be able to convince yourself that $A^n=P\begin{pmatrix}2^n&0\\0&3^n\end{pmatrix}P^{-1}$.


问:
Solve for $a, b$ if $|x^2 + ax + b| \le \frac 12, \forall 0 \le x \le 2$.
This is from a recently closed question. I do not think the accepted answer is rigorous because it did not address the cases when the discriminant of the two quadratics is negative, even though the final answer was correct.

My solution relies heavily on plotting several inequalities. I wonder if there is an alternative yet clean way of doing it without the plots. Another idea is whether it is helpful by replacing $x$ with $y+1$ so that $-1 \le y \le 1$ may simplify things a little bit.

My solution:

Denote $f(x)=x^2+ax+b$. It achieves its minimum at $x=-\frac a2$ and $f(-\frac a2)=b-\frac{a^2}{4}$.

First we must have $|f(0)| \le \frac 12 \text{ and } |f(2)| \le \frac 12$, or equivalently
$$|b| \le \frac 12 \tag1$$
$$|4+2a+b|\le \frac 12 \tag2$$

From the plots on the $a,b$-plane, we see that $(a,b)$ must reside in the parallelogram $BCDE$ and that $-\frac a2 \in [\frac{1.5}{2}, \frac{2.5}{2}] \subset [0,2]$, therefore we must also have

$$|f(-\frac a2)|=|b-\frac{a^2}{4}|\le \frac 12 \tag 3$$

After we add $(3)$ to the plots we found the only intersection of $(1), (2), (3)$ is at the point $C$ where $a=-2, b=\frac 12$. And indeed, when $0 \le x \le 2$, we have $|f(x)|=\left|(x-1)^2-\frac 12\right| \le \frac 12$.

答: The polynomial  $x^2 - 2 x + \frac{1}{2}$ takes values $\frac{1}{2}$, $-\frac{1}{2}$, $\frac{1}{2}$ at $0$, $1$, $2$. We conclude that the linear function $(x^2-2x+ \frac{1}{2}) - (x^2+ a x + b)$ takes values $\ge 0$, $\le 0$, $\ge 0$ at $0$, $1$, $2$, so it must be the $0$ function. We get $a=-2$, $b=\frac{1}{2}$.


问:
Show that $-Z$ is also a standard normal random variable.
Show that $-Z$ is also a standard normal random variable; that is, show that $P[-Z < x] = P[Z < x] \,\forall x.$</p>

答:
Since the standard normal is a symmetric distribution about $0$ we have $P[Z<0+x]=P[Z>0-x]\forall x\\ \implies P[Z<x]=P[Z>-x]\forall x$

Again note that $$P[-Z<x] =P[Z>-x]\forall x$$Hence it is proved.


问:
polynomials span p2
p1 = x^2,
p2 = 1+x,
p3 = 1+x^2
Show that these span P2.

I created a matrix A = \begin{bmatrix}0&1&1\\0&1&0\\1&0&1 \end{bmatrix}
I showed that the determinant$\neq$0. that means AX = 0 is consistent and p1,p2 and p3 are linearly independent.I think this is enough to show that the polynomial span P2 as the dimension of P2 = 3 and we have 3 polynomials.

But my professor put a note that I should argue {p1,p2,p3} is a basis. Then I can say that the polynomials span p2.

To be a basis: the polynomials should be linearly independent and span p2. How can I argue the polynomials are a basis without verifying the span? I am confused over here.

答:
I'll answer your question in an actual answer format since that's really where it should go. Lets say we have a vector space $V$ and a set of vectors $\alpha$. By definition, $\alpha$ is a basis if and only if there are linearly independent and span($\alpha) = V$. If you want to show that a given set is a basis, showing that it is linearly independent and spans is a valid method of proving that the given set is in fact a basis. If you already know the dimension of $V$ and $|\alpha| = \dim(v)$, then one only needs to show that $\alpha$ spans $V$ or is linearly independent.

So it is not necessary to prove directly that a given set spans to show that it is a basis, since it can be shown indirectly.

Title: polynomials span p2


问:
Study the uniform convergence of the integral $\int_0^\infty\frac{\sin \alpha x}{\alpha^2+x^2}\mathrm{d}x, \alpha\in(0,+\infty)$

Study the uniform convergence of the integral
$$\int_0^\infty\frac{\sin \alpha x}{\alpha^2+x^2}\mathrm{d}x,\quad \alpha\in(0,+\infty)$$

I have tried the Weierstrass criterion, but $\int_0^\infty\frac{1}{\alpha^2+x^2}\mathrm{d}x$ does not converge. Then I tried the Dirichlet criterion, but $\int_0^A \sin \alpha x$ are not bounded uniformly. So I turned to prove it does not converge uniformly. My attempt are as follows

$\forall A>1$, choose $\alpha=\frac{\pi}{4A}$ then $\forall x\in[A, 2A]$
$$
\left|\int_A^{2A}\frac{\sin \alpha x}{\alpha^2+x^2}\mathrm{d}x\right|>\frac{\sqrt{2}}{2}\frac{1}{\alpha}(\arctan\frac{2A}{\alpha}-\arctan\frac{A}{\alpha})
$$
but I cannot find a lower bound of RHS, so the method seems do not work.

Is my way wrong? You can also suggest your way. Appreciate any help!

答:
You can use Dirichlet's Criterion:
Let $f\colon (a, b) \to \mathbb{R}$ a monotone function on $(a, b)$ and $g\colon (a, b)\to \mathbb{C}$ an integrable function in $(a, t]$, for each $t\in (a, b)$. If $\displaystyle \lim_{x\to b} f(x) = 0$, and $\int_{a}^{t} g(x) dx$ is bounded for each $t \in (a, b)$, then $\int_{a}^{\to b} fg $ converges.

Taking $a = 0$, $b = \infty$, $f = \frac{1}{\alpha^2 + x^2}$ and $g = \sin \alpha x$ we have $$\int_{0}^{\infty} \frac{\sin \alpha x}{\alpha^2 + x^2} \quad \text{converges}$$


问:
The number of solutions to $Ax=b$, is either zero, or if a solution exists, then equal to the size of the null space of $A$
The number of solutions to $Ax=b$, is either zero, or if a solution exists, then equal to the size of the null space of $A$

How can I prove this?

If the system was $Ax=0$ then it is clear to me as it directly follows from the definition. What about $b\neq 0$

Thanks in advance.

答:
This is a basic fact, that follows from the result that $\{x|Ax=b\}= \{x_0+y| y\in\rm{ker} A\}$, where $x_0$ is any particular solution to $Ax=b$.

The proof is straight forward.


问:
How to prove that $\int_0^{\infty} \log^2(x) e^{-kx}dx = \dfrac{\pi^2}{6k} + \dfrac{(\gamma+ \ln(k))^2}{k}$?
I was answering this question: $\int_0^\infty(\log x)^2(\mathrm{sech}\,x)^2\mathrm dx$ and in my answer, I encountered the integral

$$\int_0^{\infty} \log^2(x) e^{-kx}dx$$
which according to WolframAlpha for $k=1,2,3$ and thereby generalizing gives us

$$\int_0^{\infty} \log^2(x) e^{-kx}dx = \dfrac{\pi^2}{6k} + \dfrac{(\gamma+ \ln(k))^2}{k} \tag{$\star$}$$

Also, in general, is there a nice form for

$$\int_0^{\infty} \log^m(x) e^{-kx} dx \tag{$\perp$}$$

EDIT

For $(\star)$, if we let $kx = t$, we then get
$$I = \int_0^{\infty} \log^2(t/k) e^{-t} \dfrac{dt}k = \dfrac1k\int_0^{\infty}\log^2(t) e^{-t}dt-\dfrac{2\log(k)}{k}\int_0^{\infty}\log(t) e^{-t} dt + \dfrac{\log^2(k)}k$$

Considering the above, $(\star)$ and $(\perp)$ boil down to evaluating the integral

$$\color{red}{I(m) = \int_0^{\infty} \log^m(x) e^{-x} dx} \tag{$\spadesuit$}$$

Some values of $I(m)$.
\begin{align}
I(1) & = -\gamma\\
I(2) & = \zeta(2) + \gamma^2\\
I(3) & = -2\zeta(3) - 3 \gamma \zeta(2) - \gamma^3\\
I(4) & = \dfrac{27}2 \zeta(4) +8 \gamma \zeta(3) + 6 \gamma^2 \zeta(2) + \gamma^4\\
I(5) & = -24 \zeta(5) - \dfrac{135}2 \gamma \zeta(4) - 20 \gamma^2 \zeta(3) - 20\zeta(2) \zeta(3) - 10 \gamma^3 \zeta(2) -\gamma^5
\end{align}

答:
As mentioned in this answer
$$
\int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t=\Gamma'(1)=-\gamma
$$
Using the definition
$$
\Gamma(x)=\int_0^\infty t^{x-1}\,e^{-t}\,\mathrm{d}t
$$
and taking the derivative $n$ times, we get
$$
\Gamma^{(n)}(1)=\int_0^\infty\log(t)^n\,e^{-t}\,\mathrm{d}t
$$
In the aforementioned answer, we also have
$$
\Gamma'(x+1)=\Gamma(x+1)\left(-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)\right)\tag{$\ast$}
$$
Taking the derivative of $(\ast)$ at $x=0$ yields
$$
\begin{align}
\Gamma''(1)
&=\Gamma'(1)\left(-\gamma\right)+\Gamma(1)\zeta(2)\\
&=\gamma^2+\zeta(2)
\end{align}
$$
Taking the second derivative of $(\ast)$ at $x=0$ yields
$$
\begin{align}
\Gamma'''(1)
&=\Gamma''(1)(-\gamma)+2\Gamma'(1)\zeta(2)+\Gamma(1)(-2\zeta(3))\\
&=-\gamma^3-3\gamma\zeta(2)-2\zeta(3)
\end{align}
$$
We can use Leibniz rule and
$$
\frac{\mathrm{d}^n}{\mathrm{d}x^n}\left(-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)\right)=(-1)^{n+1}n!\,\zeta(n+1)\quad\text{ for }n\ge1
$$
applied to $(\ast)$, to get the recursion
$$
\Gamma^{(n)}(1)=-\gamma\,\Gamma^{(n-1)}(1)+(n-1)!\sum_{k=2}^n(-1)^k\zeta(k)\frac{\Gamma^{(n-k)}(1)}{(n-k)!}
$$</p>


答:
To begin with, let us take the limit as $(x,y)$ approaches $(0,0)$. In order to do so, notice that
\begin{align*}
y \leq |y| = \sqrt{y^{2}} \leq \sqrt{x^{2}+y^{2}} \Rightarrow \frac{y}{\sqrt{x^{2}+y^{2}}}\leq 1 \Rightarrow \frac{|x|y}{\sqrt{x^{2}+y^{2}}} \leq |x|
\end{align*}
Similarly,

\begin{align*}
y\geq -|y| = - \sqrt{y^{2}} \geq -\sqrt{x^{2}+y^{2}} \Rightarrow \frac{y}{\sqrt{x^{2}+y^{2}}}\geq -1 \Rightarrow \frac{|x|y}{\sqrt{x^{2}+y^{2}}}\geq -|x|
\end{align*}

Due to the sandwich theorem, one concludes that $f$ is continuous. Indeed, one has that
\begin{align*}
\lim_{(x,y)\to(0,0)}f(x,y) = \lim_{(x,y)\to(0,0)}\frac{|x|y}{\sqrt{x^{2}+y^{2}}} = 0 = f(0,0)
\end{align*}

As to the directional derivatives at $(0,0)$ along $v = (a,b)$, we have the following definition
\begin{align*}
\lim_{t\to 0}\frac{f((0,0) + t(a,b)) - f(0,0)}{t} & = \lim_{t\to 0}\frac{f(ta,tb)}{t}\\\\
& = \lim_{t\to 0}\frac{|ta|tb}{t\sqrt{t^{2}a^{2}+t^{2}b^{2}}}\\\\
& = \lim_{t\to 0}\frac{|a|b}{\sqrt{a^{2}+b^{2}}} = \frac{|a|b}{\sqrt{a^{2}+b^{2}}}
\end{align*}

Moreover, according to the definition of the partial derivatives at $(0,0)$, we have that
\begin{align*}
\begin{cases}
\displaystyle f_{x}(0,0) = \lim_{t\to 0}\frac{f(t,0) - f(0,0)}{t} = 0\\\\
\displaystyle f_{y}(0,0) = \lim_{t\to 0}\frac{f(0,t) - f(0,0)}{t} = 0
\end{cases}
\end{align*}

Now suppose that $f$ is differentiable at $(0,0)$. Then we could express
\begin{align*}
Df((0,0)) = (f_{x}(0,0),f_{y}(0,0)) = (0,0)
\end{align*}

Consequently, if we let $v = (a,b)$, the directional derivative along $v = (a,b)$ would be equal to
\begin{align*}
D_{v}f((0,0)) = \langle Df((0,0)),(a,b)\rangle = \langle (0,0),(a,b)\rangle = 0 \neq \frac{|a|b}{\sqrt{a^{2}+b^{2}}}
\end{align*}
which is a contradiction. Thus $f$ is not differentiable at $(0,0)$.

Hopefully this helps!

Title: Function continuous at $ (0,0) $ but not differentiable at $ (0,0) $

Body:
In a book I ran into the following problem:

Let $ f: \mathbb {R}^2 \rightarrow \mathbb{R} $ defined as $ \displaystyle \frac{| x | y}{\sqrt {x ^ 2 + y ^ 2}} $ if $ (x , y) \neq (0,0) $ and $ 0 $ if $ (x, y) = (0,0) $.

They ask me to prove that $ f $ is continuous at $ (0,0) $, that it has all its directional derivatives in $ (0,0) $ but that $ f $ is not differentiable at $ (0,0) $, I It seems strange that it is continuous at that point but at the same time is not derivable. Is this correct? If so, how is it resolved?


问:
linear transformations with parameter

Let
\begin{align*}
S&=\langle(1,0,k,k),(3,−2,−1,k)\rangle \\
H&=\{x \in \Bbb{R}^4 \mid 3x_1 + x_2 − 2x_3 − kx_4=0\}.
\end{align*}
Find all $k \in \Bbb{R}$ such that $S \subset H$.

Okay... I'm a bit confused with linear transformations, I feel like they don't teach them very well. There I have two Subspaces from what I see.
The first thing I did was try to find the basis of $H$..

I have $H\{\langle(1,-3,0,0),(0,2,1,0),(0,k,0,1)\rangle\}$
and now ? I don't know where I'm in the subspace.
I made a drawing so that you understand what I ask.

and sorry if I draw like a child, but I need to understand well why.

答:
Finding a basis for $H$ is not what you want to be doing. Deciding if $\langle w_1, \ldots, w_m \rangle \subset \langle v_1, \ldots, v_n\rangle$ is not nearly as simple as testing to see if $\{w_1, \ldots, w_m \} \subset \{ v_1, \ldots, v_n\}$. Non-trivial finite-dimensional vector spaces have many bases (infinitely many when they're real or complex), and two bases for the same space might have no vectors in common.

What you should do is see when the two spanning vectors of $S$ belong to $H$. Let's call these vectors $v$ and $w$. If we can see that $v, w \in H$, then because $H$ is a subspace, we know that $a v + b w \in H$. That is, every element of $\langle v, w \rangle = S$ belongs to $H$, i.e. $S \subset H$. Conversely, of course, if either of the vectors don't belong to $H$, then neither does all of $S$.

How do we see when these vectors belong to $H$? We just have to check when they each satisfy the definition of $H$. In particular,
$$(1, 0, k, k) \in H \iff 3(1) + (0) - 2(k) - k(k) = 0 \iff 3 - 2k - k^2 = 0.$$
Similarly,
$$(3, -2, -1, k) \in H \iff 3(3) + (-2) - 2(-1) - k(k) = 0 \iff 9 - k^2 = 0.$$
So, $S \subset H$ if and only if $3 - 2k - k^2$ and $9 - k^2 = 0$ at the same time! If you solve both quadratics, you'll find this happens if and only if $k = -3$.

Title: linear transformations with parameter


问:
How to sketch $xyz=0$ near zero
Given set $M=\{(x,y,z) \ | \ xyz=0\}$
how do I go about sketching $M$ near zero. I can't wrap my head around what it should look like. Any help would be great.

答: The condition $xyz = 0$ is equivalent to $x = 0$ OR $y = 0$ OR $z = 0$. Do you know how to sketch $\{(x,y,z) \ | \ x = 0\}$, $\{(x,y,z) \ | \ y = 0\}$, and $\{(x,y,z) \ | \ z = 0\}$ separately? $M$ is just the union of those sets.

Title: How to sketch $xyz=0$ near zero


问:
probability of divisibility
Let S be the sum of k randomly selected integers between 1 and n.

What is the probability of S being divisible q?

Can this be expressed in a closed form?

This is the generalization of one of the problems from this year's hungarian high school final exam - advanced mathematics.

答:
One can use discrete Fourier transform, starting with $z=\mathrm e^{2\mathrm i\pi/q}$ a $q$th root of unity and using the identity
$$
\sum_{u=1}^qz^{un}=q\,\mathbf 1_{q\ \text{divides}\ n}.
$$
Thus,
$$
P[q\ \text{divides}\ S]=\frac1q\sum_{u=1}^qE[z^{uS}].
$$
Now, for every $x$, $E[x^S]=E[x^X]^k$ where $X$ is uniform on $\{1,2,\ldots,n\}$, and, for every $x\ne1$,
$$
E[x^X]=\frac1n\sum_{u=1}^nx^u=\frac{x}n\frac{1-x^n}{1-x},
$$
and whether the formula for $P[q\ \text{divides}\ S]$ all this, when put together, yields, is a "closed form" or not is... well, debatable.</p>

Title: probability of divisibility


问:
Computing the rank of a specific group
The rank of a group is the minimum number of elements needed to generate the group.  In general, the rank of a group is not algorithmically computable for finitely presented groups. I have a specific group $G$ (given by a presentation) that I believe has rank 4 and not smaller rank.

Ideally, I could just plug my group into some preexisting software to approach this question - does anyone know of such software?

Here is how I (not knowing any better) would approach this. I would like to find a surjection $G \to H$ where $H$ is say a finite group of rank 4.  I would like to generate a list of some (hopefully manageably sized) such finite groups $H$ and start going through the list and testing each such group to see if there is a surjection $G \to H$.  How can I easily get my hands on a list of rank 4 finite groups?

Edit:

Here is the group:
$$
\langle u,v,w,x,y,z : uxy^{−1}v^{−1}w, uyz^{−1}w^{−1}x, uzv^{−1}x^{−1}y,uvw^{−1}y^{−1}z,uwx^{−1}z^{−1}v, vxzwy \rangle
$$

(Edit by Derek Holt) And here it is again in a form suitable for copying and pasting:

```<code>< u,v,w,x,y,z |
u*x*y^-1*v^-1*w, u*y*z^-1*w^-1*x,
u*z*v^-1*x^-1*y, u*v*w^-1*y^-1*z,
u*w*x^-1*z^-1*v, v*x*z*w*y >
</code>```

答:
There are many, many finite groups of rank 4, and just very few will be quotients of your group. I think it will be much better to compute finite quotients of your group G (using so-called quotient algorithms, and then try to show that these quotients must have rank 4.

Here are some first steps in GAP. Define the group, and remove redundant generators (that will speed up things):

```<code>gap> f:=FreeGroup("u","v","w","x","y","z");;
gap> rels:=ParseRelators(f,"uxYVw,uyZWx,uzVXy,uvWYz,uwXZv,vxzwy");
[ u*x*y^-1*v^-1*w, u*y*z^-1*w^-1*x, u*z*v^-1*x^-1*y, u*v*w^-1*y^-1*z,
  u*w*x^-1*z^-1*v, v*x*z*w*y ]
gap> g:=f/rels;
<fp group on the generators [ u, v, w, x, y, z ]>
gap> sim:=IsomorphismSimplifiedFpGroup(g);
[ u, v, w, x, y, z ] -> [ u, v, v*y*x^-1*u^-1, x, y, u*x*y^-1*v^-1*x*u*y ]
gap> h:=Image(sim);
<fp group on the generators [ u, v, x, y ]>
</code>```

Now we find the largest abelian quotient:

```<code>gap> AbelianInvariants(h);
[ 5, 5, 5 ]
</code>```

This gives us a minimum of 3 generators. Not good enough. Try for another quotient. We intersect all subgroups of index <=6, and take the quotient by the core of their intersection:

```<code>gap> l:=LowIndexSubgroups(h,6);;
gap> k:=Intersection(l);;
gap> Index(h,k);
2863692069181011181015005016041480695253565440000000000
gap> q:=DefiningQuotientHomomorphism(k);;
gap> p:=Image(q);
<permutation group of size 366552584855169431169920642053309528992456376320000000000000000000000000000000000000 with 4 generators>
</code>```

This is a gargantuan quotient, lets look at its structure. After some playing around, we find that it is a direct product

```<code>gap> res:=PerfectResiduum(p);;
gap> Index(p,res);
125
gap> z:=Centre(p);
<permutation group of size 125 with 3 generators>
gap> Size(Intersection(z,res));
1
gap> p=ClosureGroup(res,z);
true
</code>```

So it is a direct product of $5^3$ with a perfect group. It turns out to also be a direct product and here an ad-hoc decomposition works better than the generic method:

```<code>gap> dom:=MovedPoints(res);;
gap> o:=Orbits(res,dom);;
gap> Length(o);
34
gap> Collected(List(o,Length));
[ [ 5, 7 ], [ 6, 27 ] ]
gap> os:=List(o,x->Stabilizer(res,Difference(dom,x),OnTuples));;
#G  FULL  504889/  55606kb live  2877911/ 333718kb dead    50312/ 131072kb free
#G  FULL  520120/  61305kb live   210758/  64281kb dead    44376/ 131072kb free
gap> Collected(List(os,Size));
[ [ 60, 7 ], [ 360, 27 ] ]
gap> ForAll(os,x->IsNormal(p,x));
true
gap> ForAll(os,IsSimpleGroup);
true
gap> n:=TrivialSubgroup(res);;for i in os do n:=ClosureGroup(n,i);od;
gap> n=res;
true
</code>```

So we have a quotient $5^3\times A_5^7\times A_6^{27}$. Now (see
https://mathoverflow.net/questions/265260/how-many-generators-does-a-direct-product-of-alternating-groups-need   ) the minimal generator number of this moduct is the maximum of the minimal generator number for each of the direct powers. By a famous theorem of Hall, for a finite group $G$, the maximal $m$ such that $G^m$ is $k$-generated is $\phi_k(G)/|Aut(G)|$, where $\phi_k(G)$ is the number of $k$-generator ordered tuples. We can calculate this in GAP as number of quotients of a free group of rank $k$:

```<code>gap> Length(GQuotients(FreeGroup(2),AlternatingGroup(5)));
19
gap> Length(GQuotients(FreeGroup(2),AlternatingGroup(6)));
53
</code>```

Thus the three "coprime" direct factors are all $3$-generated, and thus the whole group, and we would need a different quotient.

Investigating further, however did not find more fruitful quotients. Thus I'm sceptical about thelikeliness of this approach being successful.

Title: Computing the rank of a specific group

Body:
There are many, many finite groups of rank 4, and just very few will be quotients of your group. I think it will be much better to compute finite quotients of your group G (using so-called quotient algorithms, and then try to show that these quotients must have rank 4.

Here are some first steps in GAP. Define the group, and remove redundant generators (that will speed up things):

gap> f:=FreeGroup("u","v","w","x","y","z");;
gap> rels:=ParseRelators(f,"uxYVw,uyZWx,uzVXy,uvWYz,uwXZv,vxzwy");
[ u*x*y^-1*v^-1*w, u*y*z^-1*w^-1*x, u*z*v^-1*x^-1*y, u*v*w^-1*y^-1*z,
  u*w*x^-1*z^-1*v, v*x*z*w*y ]
gap> g:=f/rels;
<fp group on the generators [ u, v, w, x, y, z ]>
gap> sim:=IsomorphismSimplifiedFpGroup(g);
[ u, v, w, x, y, z ] -> [ u, v, v*y*x^-1*u^-1, x, y, u*x*y^-1*v^-1*x*u*y ]
gap> h:=Image(sim);
<fp group on the generators [ u, v, x, y ]>

Now we find the largest abelian quotient:

gap> AbelianInvariants(h);
[ 5, 5, 5 ]

This gives us a minimum of 3 generators. Not good enough. Try for another quotient. We intersect all subgroups of index <=6, and take the quotient by the core of their intersection:

gap> l:=LowIndexSubgroups(h,6);;
gap> k:=Intersection(l);;
gap> Index(h,k);
2863692069181011181015005016041480695253565440000000000
gap> q:=DefiningQuotientHomomorphism(k);;
gap> p:=Image(q);
<permutation group of size 366552584855169431169920642053309528992456376320000000000000000000000000000000000000 with 4 generators>

This is a gargantuan quotient, lets look at its structure. After some playing around, we find that it is a direct product

gap> res:=PerfectResiduum(p);;
gap> Index(p,res);
125
gap> z:=Centre(p);
<permutation group of size 125 with 3 generators>
gap> Size(Intersection(z,res));
1
gap> p=ClosureGroup(res,z);
true

So it is a direct product of $5^3$ with a perfect group. It turns out to also be a direct product and here an ad-hoc decomposition works better than the generic method:

gap> dom:=MovedPoints(res);;
gap> o:=Orbits(res,dom);;
gap> Length(o);
34
gap> Collected(List(o,Length));
[ [ 5, 7 ], [ 6, 27 ] ]
gap> os:=List(o,x->Stabilizer(res,Difference(dom,x),OnTuples));;
#G  FULL  504889/  55606kb live  2877911/ 333718kb dead    50312/ 131072kb free
#G  FULL  520120/  61305kb live   210758/  64281kb dead    44376/ 131072kb free
gap> Collected(List(os,Size));
[ [ 60, 7 ], [ 360, 27 ] ]
gap> ForAll(os,x->IsNormal(p,x));
true
gap> ForAll(os,IsSimpleGroup);
true
gap> n:=TrivialSubgroup(res);;for i in os do n:=ClosureGroup(n,i);od;
gap> n=res;
true

So we have a quotient $5^3\times A_5^7\times A_6^{27}$. Now (see
https://mathoverflow.net/questions/265260/how-many-generators-does-a-direct-product-of-alternating-groups-need   ) the minimal generator number of this moduct is the maximum of the minimal generator number for each of the direct powers. By a famous theorem of Hall, for a finite group $G$, the maximal $m$ such that $G^m$ is $k$-generated is $\phi_k(G)/|Aut(G)|$, where $\phi_k(G)$ is the number of $k$-generator ordered tuples. We can calculate this in GAP as number of quotients of a free group of rank $k$:

gap> Length(GQuotients(FreeGroup(2),AlternatingGroup(5)));
19
gap> Length(GQuotients(FreeGroup(2),AlternatingGroup(6)));
53

Thus the three "coprime" direct factors are all $3$-generated, and thus the whole group, and we would need a different quotient.

Investigating further, however did not find more fruitful quotients. Thus I'm sceptical about thelikeliness of this approach being successful.