问:
Is the Converse of the Cayley-Hamilton Theorem true or not?
I was thinking about taking the Cayley-Hamilton Theorem in the opposite direction.
Does this hold : If for $F\in$ End$(V)$ $\exists P\in K[t]$ such that $P(F)=0$ then
$P$ is the characteristic polynomial of $F$ ?

If not could you give a counterexample?

答:
No. If $p_F$ is the characteristic polynomial of $F$ and $q$ is any other polynomial, then
$$P(F) = p_F(F) q(F) = 0 \cdot q(F) = 0.$$
So $P$ is a polynomial with $P(F) = 0$ that is not the characteristic polynomial of $F$ (just take $q$ to be nontrivial).

We can also have something like the following happen: Let
$$F = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$
Then
$$p_F(t) = (1 - t)^2.$$
But of course if
$$q(t) = 1 - t,$$
we have
$$q(F) = 0.$$


问:
Preservation Hamiltonian vector field
Suppose I have the Hamiltonian vector field $X_H$ on the symplectic manifold $(M, \omega)$. Consider the symplectic transformation $P: M \rightarrow M$. Will the linear terms of $X_H$ be preserved (up to symplectic linear transformation) under any symplectic transformation $P$?

I am not sure, but I would think so.

答: It is not true. There is a matrix associated to the linear terms, and you should compute the eigenvalues of the matrix, all the eigenvalues of the matrix are independent of coordinates transformations, which are, by means of your symplectic transformations.</p>

ContentLicense: CC BY-SA 3.0

Title: Preservation Hamiltonian vector field


问:
Modular form with integral Fourier coefficients
Suppose I am given a sequence $a_0, a_1,a_2 \cdots a_{k-1} \in \mathbb{Z}$. Does there exist a modular form $f$ of weight $k$ such that

$$
f(q) = \sum_{n=0}^{\infty}{a_n q^n}
$$

is its Fourier series ?

Where $\Gamma = \left\{\begin{pmatrix} a & b \\ c & d \end{pmatrix} : \; a,b,c,d \in \mathbb{Z}, \; ad-bc =1 \right\}$

My idea was to find a suitable basis of $M_k(\Gamma)$ but I didn't find any which would help here.

Would appreciate any hints (!) as to how to approach this problem.

答:
Let $a_0,a_1,a_2,\dots a_{k-1}\in\mathbb Z$ be $k\geq 1$ integers. The  modular forms on $\Gamma=SL(2,\mathbb Z)$ form a ring. For fixed $l\geq 4$ even, the dimension of the vector space $M_l(\Gamma)$ of modular forms of weight $l$ on $\Gamma$ has complex dimension $\lfloor\tfrac{l}{12}\rfloor$ if  $l\equiv 2\mod 12$ and $\lfloor\tfrac{l}{12}\rfloor+1$ otherwise. Set now $l$ such that $\dim_{\mathbb C}M_l(\Gamma)\geq k$. As the ring of modular forms on $\Gamma$ is generated by the Eisenstein series $E_4$ and $E_6$, there is a polynomial
$$P_l=\sum_{m,n\geq0: 4m+6n=l}E_4^mE_6^n$$
in $E_4$ and $E_6$ with weight $l$. By linear algebra, one can easily find the coefficients of $P_l$ such that
$$P_l(\tau)=\sum_{n=0}^\infty a_nq^n,$$
with the first $k$ coefficients $a_n$ as given above.

ContentLicense: CC BY-SA 4.0


问:
If $|G/Z(G)|= p$ for a prime $p$, Can we have $|Z(G)|=p^2$?
I came across this theorem in many of algebra books.

If $G/Z(G)$ is cyclic, then $G$ is abelian.

I tried to apply it when $G=p^3$ and $Z(G)=p^2$. What I got is $|G/Z(G)|=p$ so is cyclic. Therefore, $G$ is abelian according to the theorem which yields $G=Z(G)$ contradiction!
So I thought either theorem doesn't apply here or we cannot have the centre of degree $p^2$. Am I right? If so, Which one is correct?

Thanks.

答: Every group of order $p$ is cyclic. Therefore, by the theorem you cited, $G/Z(G)$ cannot be of order $p$. Hence, $|G|=p^{3}$ and $|Z(G)|=p^{2}$ is impossible.


问:
How to determine Two-tailed or One-tailed test?
For example while testing difference of two means $\mu_1-\mu_2$ as the numerator in the test statistic, is this left-tailed or right tailed? What if I take $\mu_2-\mu_1$ in the numerator? Will the left right be swapped? Another question is suppose if the $z$ value is 6.58 for a left tailed test at 1% significance level, this should be accepted right? Since the value is to the right of critical value.

答:
the understanding if the text is one or two tails depends mostly on the context:

Example 1:

An electronic device has a mean lifetime of 10 hours. After some improvements, the factory delcares that the new device has a longer lifetime. To prove this hypothesis, 4 items are tested with the following results:

$\{10.1;9.8;9.4;11.5\}$

The sample mean is $10.2>10$

This is a One tail test, as you must understand if actually $10.2>10$ or the difference is due only to the randomness

Example 2:

To verify that a coin is fair, the coin is tossed 10 times giving the following results

$\{T,T,H,T,T,H,T,H,T,T\}$

this is a Two tails Test.


问:
Tough Integral in a separable ODE
Denote $y(x) = y$ for simplicity. I am trying to calculate a solution to:

$$\frac{\sqrt{1-k^2(1+y)^2}}{k(1+y)}  = y'$$

Separating I get:
$$\int  \frac{k(1+y)}{\sqrt{1-k^2(1+y)^2}} dy = \int dx$$, but from here I cannot calculate the LHS integral.

答:
$$\int  \frac{k(1+y)}{\sqrt{1-k^2(1+y)^2}} dy = \int dx$$
Substitute $u=k(1+y)$:
$$\int  \frac{u \ du}{\sqrt{1-u^2}}  = k\int dx$$
$$\int  \frac{du^2}{\sqrt{1-u^2}}  = 2k\int dx$$
Substitute $v=u^2$:
$$\int  \frac{dv}{\sqrt{1-v}}  = 2k\int dx$$
$$\int  \frac{d(1-v)}{\sqrt{1-v}}  = -2k\int dx$$
Where $v=k^2(1+y)^2$.
This is easy to calculate now.

Title: Tough Integral in a separable ODE


问:
Can an Integral Domain be considered a Vector Space?
If $D$ is an integral domain that contains a field $F$. Can $D$ be considered a Vector Space over $F$?

答: Assuming that $D$ contains $F$ as a subring then yes, it is absolutely an $F$ vector space. It is an abelian group under addition and is closed under multiplication by $F$. All the axioms of vector space like distributivity come from $F \subseteq D$ being a subring. More specifically, we say that $D$ is an $F$ algebra, meaning that it has the structure of an $F$ vector space and a ring in a compatible way.


问:
Suppose $AB=AC$, where $B$ and $C$ are matrices and $A$ is invertible. Show that $B=C$. Is this true, in general, when $A$ is not invertible?
Suppose $AB=AC$, where $B$ and $C$ are matrices, and $A$ is invertible. Show that $B$=enter image description here$C$. Is this true, in general, when $A$ is not invertible?
What can be deduced from the assumptions that will help to show $B=C$?

答:
If the matrix $A$ is invertible, there exists an inverse matrix $A^{-1}$ with $AA^{-1}=I=A^{-1}A$, where $I$ is the identity matrix. The inverse matrix is uniquely determined.

If $AB=AC$ and $A$ is invertible, then multiplying with $A^{-1}$ from the left, using associativity and that $A^{-1}A=I$ gives $B=C$.


问:
Finding the double integral using polar coordinates.

Above is the graph I have obtained from the equation.

Hi there so I've just been introduced to using polar co-ordinates to find double integrals and Im having a hard time working out how to tackle questions step by step.

Im currently looking at the question:

$I_{6}=\int\int_{Q}x/\pi (x^{2}+y^{2}) dxdy$ with $Q=\{(x,y)|(x-3/2)^2 +y^2 \leq9/4, y\geq0\}$

And Im struggling to make a start to this as I dont know how to obtain the limits of the double integral from $Q$ .

All I know is that I should eventually obtain the answer 3/4.

Any help would be greately appreciated.

Update:

So I now have the double integral

$\int_{0}^{\pi/2}\int_{0}^{3cos\theta}(rcos\theta/\pi)*((rcos\theta)^{2}+(rsin\theta)^{2}) rdrd\theta$

答:
On your specific question about how the equation in polar form becomes $r = 3 \cos \theta$,

In polar form, $x = r \cos \theta, y = r \sin \theta$

Now your curve is $(x-3/2)^2 +y^2 = 9/4 \implies x^2 + y^2 = 3x$

$r^2 cos^2 \theta + r^2 \sin^2 \theta = 3r \cos \theta$

$r^2 = 3r \cos \theta \implies r = 3 \cos \theta$

Here is a diagram showing how $r$ changes with $\theta$ with $x$-axis.


答:
If $A$ is invertible, then:

$$AB = AC \Rightarrow A^{-1}AB = A^{-1}AC \Rightarrow B = C.$$

If $A$ is not invertible, $AB = AC \Rightarrow B =C $ does not work.

Consider for instance:

$$A = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}, B = \begin{bmatrix}1 & 1 \\ 2 & 3\end{bmatrix}~\text{and}~ C = \begin{bmatrix}1 & 1 \\ -5 & 1\end{bmatrix}.$$

Notice that $A$ is not invertible. Moreover:

$$AB = AC = \begin{bmatrix}1 & 1 \\ 0 & 0\end{bmatrix},$$

but $B \neq C.$

Title: Suppose $AB=AC$, where $B$ and $C$ are matrices and $A$ is invertible. Show that $B=C$. Is this true, in general, when $A$ is not invertible?

Body:
Suppose $AB=AC$, where $B$ and $C$ are matrices, and $A$ is invertible. Show that $B$=enter image description here$C$. Is this true, in general, when $A$ is not invertible?
What can be deduced from the assumptions that will help to show $B=C$?


问:
Find logarithmic equation from a graph but line only clearly passes through one point
I'm working through an online textbook, currently 'graphs of logarithmic functions' on this page, try it #11 right underneath example 11.

For the example before the question, the page shows an example of solving by taking two points on the graph:

This works because there are two clear points on the graph. Then the book provides an opportunity to reinforce the learning with an exercise, #11:

This function only passes through one integer point, (-2,-1). Given that, how can I fins the function using the method I was just shown in example 11?

答:
This question asks for a natural logarithm, whereas the first question asks for a common logarithm.  I think that the intent here is that the general form of the logarithm would be $y=\ln(x+b)+c$ and not $y=a\ln(x+b)+c$.  The reason is that if the coefficient in front of the $\ln$ is not $1$, then the function could be rewritten as a logarithm with respect to a different base.  In particular, the second form could be written as
$$
y=\log_{e^a}(x+b)+c.
$$
This assumption is not stated in the book (through a simple search), but it makes the problem possible (and provides some ideas for the alternate phrasing).  Therefore, we'll consider the form
$$
y=\ln(x+b)+c.
$$
In this case, it seems that there is enough data in the problem (using the asymptote and the point on the curve).

My solution:

 It appears that $x=-3$ is a vertical asymptote, so $b=3$ in our general form and we're interested in $y=\ln(x+3)+c$.

 The point $(-2,-1)$ is on the logarithm, so $-1=\ln(-2+3)+c$ or that $-1=\ln(1)+c$, which simplifies to $-1=c$, since $\ln(1)=-1$.


问:
Proving spheres are orthogonal

Given two spheres in $\mathbb{R}^3$:

$x^2+y^2+z^2=2ax; \ \  \ x^2+y^2+z^2 = 2by$

and $a,b>0$, and $\gamma$ the intersection of the spheres, show that for any $p_0 \in \gamma$, the spheres are orthogonal at $p_0$.

I am not sure I fullly understand the problem but here is what I tried:

First of all assuming $p_0$ is some point s.t each sphere can be represented as a function $z_1, z_2 : \mathbb{R}^2 \to \mathbb{R}$ in a neighborhood of $p_0$ (if it isn't we can use another variable), what I think we want to show is that the tangant spaces to $z_1, z_2$ are orthogonal at this point, i.e $\langle\nabla z_1(p_0), \nabla z_2(p_0)\rangle=0.$ But when I calculate this I get that it does not equal 0:

$z_1 = \sqrt{2ax-x^2-y^2}, z_2 = \sqrt{2by-x^2-y^2}$

$\nabla z_1 = (\frac{2a-2x}{2 \sqrt{2ax-x^2-y^2}}, \frac{-2y}{2 \sqrt{2ax-x^2-y^2}})$

$\nabla z_2 = (\frac{-2x}{2 \sqrt{2by-x^2-y^2}}, \frac{2b-2y}{2 \sqrt{2by-x^2-y^2}})$

$\langle\nabla z_1(p_0), \nabla z_2(p_0)\rangle \neq 0.$

Can someone explain where is my mistake?

答:
The gradients are not applied properly.

Let $f(x,y,z)=x^2+y^2+z^2-2ax$ and $g(x,y,z)=x^2+y^2+z^2-2by$.

Then the spheres are given by $f(x,y,z)=0$ respectively $g(x,y,z)=0$.

The gradients of $f$ and $g$ are orthogonal to their tangent spaces. So to verify if the tangent spaces are orthogonal, we need that for every $(x,y,z)$ on both spheres that $\langle \nabla f(x,y,z),\nabla g(x,y,z)\rangle=0$.

Title: Proving spheres are orthogonal


答: Let$$f(x,y,z)=(x-a)^2+y^2+z^2\quad\text{and}\quad g(x,y,z)=x^2+(y-b)^2+z^2.$$Then your spheres are the surfaces$$\{(x,y,z)\in\Bbb R^3\mid f(x,y,z)=a^2\}\quad\text{and}\quad\{(x,y,z)\in\Bbb R^3\mid g(x,y,z)=b^2\}.$$If $p=(x_0,y_0,z_0)\in\gamma$ then the tangent plane to the first sphere at $p$ is orthogonal to$$\nabla f(x_0,y_0,z_0)=\bigl(2(x_0-a),2y_0,2z_0\bigr)$$and the tangent plane to the second sphere at $p$ is orthogonal to$$\nabla g(x_0,y_0,z_0)=\bigl(2x_0,2(y_0-b),2z_0\bigr).$$And we have\begin{align}\bigl\langle\nabla f(x_0,y_0,z_0),\nabla g(x_0,y_0,z_0)\bigr\rangle&=\bigl\langle\bigl(2(x_0-a),2y_0,2z_0\bigr),\bigl(2x_0,2(y_0-b),2z_0\bigr)\bigr\rangle\\&=4x_0^{\,2}+4y_0^{\,2}+4z_0^{\,2}-2ax_0-2by_0\\&=2(x_0^{\,2}+y_0^{\,2}+z_0^{\,2}-ax_0)+2(x_0^{\,2}+y_0^{\,2}+z_0^{\,2}-by_0)\\&=0+0\text{ (since $p\in\gamma$)}\\&=0.\end{align}And, since the vectors $\nabla f(x_0,y_0,z_0)$ and $\nabla g(x_0,y_0,z_0)$ are orthogonal, then so are tangent planes at $p$, since one of them is orthogonal to the first vector, whereas the other one is orthogonal to the second vector.

Title: Proving spheres are orthogonal

Body:
Given two spheres in $\mathbb{R}^3$:

$x^2+y^2+z^2=2ax; \ \  \ x^2+y^2+z^2 = 2by$

and $a,b>0$, and $\gamma$ the intersection of the spheres, show that for any $p_0 \in \gamma$, the spheres are orthogonal at $p_0$.

I am not sure I fullly understand the problem but here is what I tried:

First of all assuming $p_0$ is some point s.t each sphere can be represented as a function $z_1, z_2 : \mathbb{R}^2 \to \mathbb{R}$ in a neighborhood of $p_0$ (if it isn't we can use another variable), what I think we want to show is that the tangant spaces to $z_1, z_2$ are orthogonal at this point, i.e $\langle\nabla z_1(p_0), \nabla z_2(p_0)\rangle=0.$ But when I calculate this I get that it does not equal 0:

$z_1 = \sqrt{2ax-x^2-y^2}, z_2 = \sqrt{2by-x^2-y^2}$

$\nabla z_1 = (\frac{2a-2x}{2 \sqrt{2ax-x^2-y^2}}, \frac{-2y}{2 \sqrt{2ax-x^2-y^2}})$

$\nabla z_2 = (\frac{-2x}{2 \sqrt{2by-x^2-y^2}}, \frac{2b-2y}{2 \sqrt{2by-x^2-y^2}})$

$\langle\nabla z_1(p_0), \nabla z_2(p_0)\rangle \neq 0.$

Can someone explain where is my mistake?


问:
Guides/tutorials to learn abstract algebra?
I recently read up a bit on symmetry groups and was interested by how they apply to even the Rubik's cube. I'm also intrigued by how group theory helps prove that "polynomials of degree $\gt4$ are not generally solvable".
I love set theory and stuff, but I'd like to learn something else of a similar type. Learning about groups, rings, fields and what-have-you seems like an obvious choice.

Could anyone recommend any informal guides to abstract algebra that are written in (at least moderately) comprehensible language? (PDFs etc. would also be nice)

答:
"Abstract Algebra: Theory and Applications" could interest you. It is a free textbook and while it is not written informally, it is easier to understand than usual Abstract Algebra textbooks. I think it is a good way to get into Abstract Algebra.

http://abstract.ups.edu/


问:
Is the set of complex numbers a ring?
Is the set of complex numbers of the form $\frac{x+y\sqrt{D}}{2}$ ring, where $D$ - fixed, square-free integer, $x,y$ - integers equal parity?

The answer is: This set is ring when $D \equiv 1 \ (mod4)$, but why?

答:
It's straightforward to check our set is closed under addition.

We need to make sure the set is closed under multiplication.

$$(\frac{x+y\sqrt{D}}{2})(\frac{a+b\sqrt{D}}{2})=$$

$$(\frac{xa+Dyb}{4}) + (\frac{xb+ya}{4})\sqrt{D}$$

Since x,y have the same parities and a,b have the same parities... xb and ya have the same parities. So $xb+ya$ is even.

So we know $\frac{xb+ya}{4} = \frac{n}{2}$ for some integer n

Now we need to make sure $xa+Dyb$ is similarly divisible by 2

We know $xa$ and $yb$ have the same parity using similar reasoning as before. So if D is odd this condition is satisfied. On the other hand if D is even, then if x,y,a,b are all odd, we break this condition. So D must be odd.

Finally we need to make sure that

$\frac{xa+Dyb}{2}$ and $\frac{xb+ya}{2}$ have the same parity

We just need to ensure the difference is divisible by 2:

$\frac{x(a-b)+y(Db-a)}{2}$

In other words we require.

$x(a-b)+y(Db-a) \equiv 0 (\mod 4)$

We already know D has to be odd... so D can only be 1 or 3 mod 4.

Assume $D\equiv 3 \mod 4$

We need to check $x(a-b)+y(3b-a) (\mod 4)$

substitute in a choice of $x=y=a=b=1$

this comes out to $2 \mod 4$. So $D \equiv 3\mod 4$ doesn't satisfy our condition.

On the other hand suppose $D \equiv 1 \mod 4$

$x(a-b)+y(1b-a) \equiv (x-y)(a-b) (\mod 4)$

Since (x-y) and (a-b) are both even, this comes out to $0 \mod 4$ for all valid choices of x,y,a,b

So we've proved that our set is closed under multiplication only for $D \equiv (1 \mod 4)$

So our set is a ring only for $D \equiv (1 \mod 4)$

Title: Is the set of complex numbers a ring?


问:
How to show that $P(\mathbb R^n$) admits a positive sectional curvature?
I saw from wikipedia (https://en.wikipedia.org/wiki/Real_projective_space) that $P(\mathbb{R}^n)$ admits a positive scalar curvature metric. Then it also describes that "For the standard round metric, this has sectional curvature identically $1$."

My question is to show that "$P(\mathbb{R}^n)$ admits a positive sectional curvature which are constant for two planes $\sigma$ at any $Tp(P(\mathbb{R}^n))$. I am not sure what metric I should consider. Is it the "standard round metric" ? (I am not sure either if this is the metric induced from $S^{n+1}$ ?) Also, sectional curvature is quite hard to calculate for any possible $2-$plane. How can one show the statement like "all sectional curvature is positive, and moreover, they are all equal !"

答: The standard projection $\pi : S^n \to P(\mathbb{R}^{n+1})$ is the universal covering. Given the round metric on the sphere, there exists a unique metric on the right that makes this projection a riemannian covering, that is, a local isometry (this is because the round metric is unvariant under $x\mapsto -x$.) This is what we actually mean by "the round metric of the projective space". It has constant sectional curvature $1$, because so has the sphere and because the projection is a local isometry, hence it has constant (positive) scalar curvature.


问:
How to solve this third order recurrence?
I have the following recurrence
$$
\left(4-3 n^3\right) \mu _{n+1}+n ((n-3) n+3) \mu _n+(3 n (n (n+3)+3)+4) \mu _{n+2}+6\\
=(n+2)^3 \mu _{n+3}
$$
with initial values
$$
\left\{0,1,\frac{9}{8},\frac{281}{216},\frac{9955}{6912},\frac{1340387}{864000},\frac{8519417}{5184000},\frac{3061810421}{1778112000}\right\}$$
starting from $n=1$.

I have tried to use Sigam to solve it. But it can only find one inhomogenious solution
$$
\frac{6}{11}\left(\underset{\iota _1=2}{\overset{n}{\sum}}\frac{1}{\iota _1-1}\right)
$$
Is it possible to find an homogenious solution?

If not possible, how can we get an asymptotic expansion of $\mu_n$?

I am able to estimate numerically that
$$
\mu_n \approx \frac{6}{11}\left(\underset{\iota _1=2}{\overset{n}{\sum}}\frac{1}{\iota _1-1}\right) + 0.308492+0.0565532 \left(n^{-3+i \sqrt{2}}+n^{-3-i \sqrt{2}}\right)
$$
using the Asymptotics package. I wonder if it is possible to find a closed form for constant term.

答:
The constants you sought may be expressed in terms of (generalized) hypergeometric functions. As the results are quite involved, I shall only sketch the procedure.

Put
$$\mu_n=\nu_n+\frac6{11}\sum_{\iota=1}^{n-1}\frac1\iota,$$
$\nu_n$ satisfies the homogeneous recurrence. Let
$$f(z)=\sum_{n=1}^\infty \nu_nz^n$$
be the generating function. The recurrence translates to the inhomogeneous DE
$$z^3 (1-z)^3 f'''(z)+z (1-z)^3 f'(z)-(1-2 z+7 z^2) f(z)=z^2 (u+4vz),$$
where
$$\begin{aligned}
&u=\nu_2-\nu_1,&&v=2\nu_3-\nu_2-\nu_1
\end{aligned}$$
depend on initial values. The singularities of this DE lie at $z=0$, $1$, $\infty$. The asymptotic behavior of $\nu_n$ is reflected on the singularity of $f(z)$ at $z=1$. At this point the DE has a holomorphic particular integral $P(z)$, and a set of fundamental solutions of the homogeneous DE are
$$\begin{aligned}
&\frac{z}{1-z},&&F_1(z)=(1-z)^{2+i\sqrt{2}}(1+c_1(1-z)+\cdots),&&F_2(z)=(1-z)^{2-i\sqrt{2}}(1+c_2(1-z)+\cdots).
\end{aligned}$$
If we have
$$f(z)=P(z)+C\frac{z}{1-z}+C_1F_1(z)+C_2F_2(z),$$
then the standard procedure (Darboux's method) yields
$$\nu_n\sim C+C_1\frac{\sinh (\sqrt{2} \pi ) \Gamma (3+i \sqrt{2})}{\pi  i}n^{-3-i\sqrt{2}}-C_2\frac{\sinh (\sqrt{2} \pi ) \Gamma (3-i \sqrt{2})}{\pi  i}n^{-3+i\sqrt{2}}.$$
In general the coefficients $C$'s can only be determined numerically, as they depend on the special values of these solutions.

In the present case, however, a reduction to the hypergeometric DE is possible. Put
$$f(z)=\frac{z}{1-z}\int g(z)dz,$$
it leads to
$$z^2 (1-z)^2g''(z)+3 z (1-z) g'(z)+(1+4 z+z^2)g(z)=u+4vz.$$
If we further substitute
$$g(z)=z^{-1}(1-z)^{2+i\sqrt{2}}h(z),$$
we find
$$z (1-z) h''(z)+(1-2(1+i \sqrt{2}) z) h'(z)+(1-i \sqrt{2}) h(z)=(1-z)^{-3-i\sqrt{2}}(u+4vz).$$
The left hand side is hypergeometric DE, so we can find a set of fundamental solutions of the homogeneous DE of $g(z)$
$$G_1(z)=\frac{(1-z)^{2+i \sqrt{2}}}{z}\, _2F_1\left(\frac{1}{2}+i \sqrt{2}-\frac{i \sqrt{3}}{2},\frac{1}{2}+i \sqrt{2}+\frac{i \sqrt{3}}{2};1+2 i \sqrt{2};1-z\right),$$
$$G_2(z)=\frac{(1-z)^{2-i \sqrt{2}}}{z}\, _2F_1\left(\frac{1}{2}-i \sqrt{2}-\frac{i \sqrt{3}}{2},\frac{1}{2}-i \sqrt{2}+\frac{i \sqrt{3}}{2};1-2 i \sqrt{2};1-z\right).$$
By variation of constants, we find
$$\begin{aligned}
g(z)&=G_1(z)\int_0^z\frac{iz(u+4vz)}{2\sqrt{2}(1-z)^5}G_2(z)dz-G_2(z)\int_0^z\frac{iz(u+4vz)}{2\sqrt{2}(1-z)^5}G_1(z)dz\\
&=Q(z)+c_1G_1(z)+c_2G_2(z),
\end{aligned}$$
where
$$Q(z)=-G_1(z)\int_z^1\frac{iz(u+4vz)}{2\sqrt{2}(1-z)^5}G_2(z)dz+G_2(z)\int_z^1\frac{iz(u+4vz)}{2\sqrt{2}(1-z)^5}G_1(z)dz$$
is holomorphic at $z=1$, and
$$\begin{aligned}
&c_1=\int_0^1\frac{iz(u+4vz)}{2\sqrt{2}(1-z)^5}G_2(z)dz,&&c_2=-\int_0^1\frac{iz(u+4vz)}{2\sqrt{2}(1-z)^5}G_1(z)dz.
\end{aligned}$$
Note that these integrals actually diverges, but this difficulty can be circumvented by analytic continuation or by other means.

Finally,
$$f(z)=\frac{z}{1-z}\int_0^zg(z)dz=\frac{z}{1-z}\int_0^1g(z)dz-\frac{z}{1-z}\int_z^1(Q(z)+c_1G_1(z)+c_2G_2(z))dz.$$
It follows that
$$\begin{aligned}
&C=\int_0^1g(z)dz,&&P(z)=-\frac{z}{1-z}\int_z^1Q(z)dz,&&C_iF_i(z)=-c_i\frac{z}{1-z}\int_z^1G_i(z)dz.
\end{aligned}$$
In particular,
$$\begin{aligned}
&C_1=\frac{-3+i \sqrt{2}}{11}c_1,&&C_2=\frac{-3-i \sqrt{2}}{11}c_2.
\end{aligned}$$

In summary, the second constant 0.0565532 can be expressed by integrals involving $\, _2F_1$, or by values of $\, _3F_2$ at $1$. The first constant 0.308492 is expressed by double integrals involving $\, _2F_1$, which I do not know whether it can be reduced further or not.

I retain the dependence on initial values for there is a special case that renders a more exact treatment. If $u=4v=3k$ or $\nu_2=\nu_1+3k$, $\nu_3=\nu_1+15k/8$, the DE of $g(z)$ has a particular solution $Q(z)=k/z$, and the constants in the asymptotic expansion can be expressed by Gamma functions.


问:
For which values of t is the following subset of $\mathbb{R}^3$ linearly independent?
The 3 vectors are $v_1 = (2, -3, 1)~~ v_2 = (-4, 6, -2)~~ v_3 = (t, 1, 2)$

So I put them all into a 3x3 matrix. Calculated the determinant, which came to $0$. So surely they're dependent? Or am I doing something wrong. Thanks

答:
You are not doing something wrong.

$v_2=-2v_1$, so $v_1, v_2, $ (and $v_3$) are linearly dependent no matter what $t$ is.


问:
Why is the first eigenvalue on the principal component always superior to the second eigenvalue?
I'm learning PCA and I wanted to know why the first eigenvalue on the principal component is always superior at second eigenvalue. Thanks for helping me

Ps:  I tried to unterstand before asking this question but to be honest, I had difficult to unterstand what I was reading on book.
I would like just an easy answer who can allow me to unterstand it better

答:
This is the way it is conventionally presented, and for good measure, since the higher eigenvalues explain a higher percentage of the variance in the dataset. This is sometimes reflected in PCA analysis in what is called a scree plot, depicting the percentage of the variance accounted for by each of the eigenvalues ordered from higher to lower, and with the intent of summarizing the data selecting only the first (highest) eigenvalues, which explain the majority of the variance with reduced dimensionality.

Here is an intuitive graphical representation of what we try to achieve by selecting the eigenvectors of the largest eigenvalues of the covariance matrix:


问:
Why $-(\Delta)^{-1}$ is compact operator?
Assume
$$
A= -(\Delta)^{-1}: L^2(\Omega)\rightarrow L^2(\Omega),
$$
I know $A$ is self-adjiont. Then why the space $H^1_0(\Omega)$ can be compactly imbedded in $L^2(\Omega)$ by means the fact that  $A$ is compact operator ?

From the point of view of notation, by  $H^1_0(\Omega)$ I mean the Sobolev space $\mathring W^{1,2}(\Omega)$.

答:
I suppose that $\Omega$ is a bounded open subset of $\mathbb R^n$ and $\Delta$ is defined as the Friedrichs extension of Laplace operator $\Delta: \mathscr D(\Omega) \rightarrow L_2(\Omega)$. In this case $-\Delta$ is positive densely defined operator with $D_\Delta \subset H^1_0( \Omega)$. Operator $A = \Delta^{-1}$ is defined and is continuous in the following sense: $A: L_2(\Omega) \rightarrow H^1_0(\Omega)$. Now we discuss the compactness of $A:L_2(\Omega) \rightarrow L_2(\Omega)$.

If you assume Rellich theorem ($j:H^1_0(\Omega) \rightarrow L_2(\Omega)$ is a compact embedding), then $A:L_2(\Omega) \rightarrow L_2(\Omega)$ is equal to a composition of continuous operators $L_2(\Omega) \xrightarrow{A} H^1_0(\Omega) \xrightarrow{j} L_2(\Omega)$, where the second operator is compact. Thus $A$ is compact.

If you assume that $A:L_2(\Omega) \rightarrow L_2(\Omega)$ is compact, then $\sqrt{A}:L_2(\Omega) \rightarrow L_2(\Omega)$ is also compact. It is known that $\sqrt{A}$ is an isomorphism betweeen $L_2(\Omega)$ and $H^1_0(\Omega)$. Therefore $(\sqrt{A})^{-1}:H^1_0(\Omega) \rightarrow L_2(\Omega)$ is a continuous operator. Now you can represent the embedding $H^1_0(\Omega) \rightarrow L_2(\Omega)$ as the composition $H^1_0(\Omega) \xrightarrow{(\sqrt{A})^{-1}} L_2(\Omega) \xrightarrow{\sqrt{A}} L_2(\Omega)$, where the second operator is compact. Thus, the embedding is compact.


问:
$ \underset{r \rightarrow 0+}{\lim} \int ^1 _r x^{2013}(\ln x)^{1001} dx$
This integral is in my book, but I don't know how to solve it using simply methods. It's possible to do it clever.

$$ \underset{r \rightarrow 0+}{\lim} \int ^1 _r x^{2013}(\ln x)^{1001} dx$$

Could You help me, how?

答:
Let $x = e^{-t}$. This gives us $dx = -e^{-t} dt$. We then get
$$\lim_{r \to 0^+}\int_r^1 x^{2013} \ln^{1001}(x) dx = -\int_0^{\infty} t^{1001}e^{-2014t}dt$$
Now make the substitution $2014t = y$ and see how it relates to the $\Gamma$-function.</p>


问:
How to start learning knot theory?
Knot theory really sounds cool and I'm very interested in it. But I'm wondering what basic knowledge it is required and how I should start learning about it. Thanks</p>

答:
If you're willing to work on an intuitive level, you can get pretty far with very little background. Colin Adams' The Knot Book is an excellent and elementary introduction at this level.

If you want more rigor and depth, you'll want to learn some point-set topology and maybe a little group theory and algebraic topology. If you don't have a solid undergraduate background in mathematics this will take awhile and I don't recommend it if you have just a passing interest in knot theory. Knot theory gets very deep. The modern theory has unexpected connections to various fields, and to appreciate these you'll want to learn some abstract algebra leading into some representation theory, among other things... but just read The Knot Book first.

Title: How to start learning knot theory?


问:
Given the graphic sequence of a simple graph, how to construct the adjacency matrix?
Suppose I have the graphic sequence (7,6,6,6,5,4,4,2), how do I get the adjacency matrix out of it?</p>

答:
The graph has $8$ vertices, and $\tfrac{1}{2}(7+6+6+6+5+4+4+2)=20$ edges by the Handshaking Lemma.  We can exhaustively generate the non-isomorphic $8$-vertex $20$-edge graphs with vertex degrees between $2$ and $7$ using geng which comes with nauty using:

```<code>geng 8 20:20 -d2 -D7
</code>```

I wrote a script that filters out the ones that don't have degree sequence (7,6,6,6,5,4,4,2) leaving two remaining.  The two graphs are as follows:

I've marked the vertices with their degrees, and their corresponding adjacency matrices are given below.

$\begin{array}{|cccccccc|} \hline 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 &
1 & 1 \\ 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 &
1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \\ \hline \end{array}
 \quad\quad\quad
\begin{array}{|cccccccc|} \hline 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 &
1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 &
1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \\ \hline \end{array}$

Body:
The graph has $8$ vertices, and $\tfrac{1}{2}(7+6+6+6+5+4+4+2)=20$ edges by the Handshaking Lemma.  We can exhaustively generate the non-isomorphic $8$-vertex $20$-edge graphs with vertex degrees between $2$ and $7$ using geng which comes with nauty using:

geng 8 20:20 -d2 -D7

I wrote a script that filters out the ones that don't have degree sequence (7,6,6,6,5,4,4,2) leaving two remaining.  The two graphs are as follows:

I've marked the vertices with their degrees, and their corresponding adjacency matrices are given below.

$\begin{array}{|cccccccc|} \hline 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 &
1 & 1 \\ 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 &
1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \\ \hline \end{array}
 \quad\quad\quad
\begin{array}{|cccccccc|} \hline 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 &
1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 &
1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \\ \hline \end{array}$


问:
limiting value of the angle between two vectors
Let $θ$ be the angle between the vectors $A =
(1,1,...,1)$ and$ B = (1,2,...,n)$ in $\mathbb{R}^n$. Find the limiting value of θ as n → ∞.

For this question, I want to apply the equation for the angle between two vectors: $θ=arccos\frac{A·B}{\|A\|\|B\|}$. For A·B, I use $A·B=\sum_{k=1}^{n}a_kb_k$. And for the norm of A and B, I simply use the definition to get $\|A\|=\sqrt{n}$ and $\|B\|=\sqrt{\sum_{k=1}^{n}k^2}$. But I'm stuck here. I don't know how to calculate the  limit of θ as n→ ∞ by these equations.

答:
Knowing that

$$\begin{cases}
\sum_{k=1}^n k &= \frac{n(n+1)}{2}\\
\sum_{k=1}^n k^2 &= \frac{n(n+1)(2n+1)}{6}
\end{cases}$$

You get

$$\frac{A_n \cdot B_n}{\Vert A_n \Vert \Vert B_n \Vert}  = \frac{n(n+1)/2}{\sqrt n \sqrt{n(n+1)(2n+1)/6}} \to \frac{\sqrt 3}{2}$$ as $n \to \infty$ and therefore

$$\theta = \frac{\pi}{6}$$ as $\arccos$ is continuous.


问:
When we say a limit exists, does $\infty$ counts as existence?
This is such a basic question but I'd like to get some clarify. When we say the "limit exists" for $\lim_{x \rightarrow c} f(x)$, does it count when $\lim_{x \rightarrow c} f(x) = \infty$?

答:
Usually, no, when we say a limit exists, that includes a requirement that it is finite. Such limits can, for instance, sensibly be used in further calculations, so it is a very natural distinction to make.

Compared to a limit of $\infty$, which is still, in some sense, a definite answer to what the limit is, but it's not really a sensible result as an intermediate calculation.

Ultimately, though, it is a matter of taste. And some do include infinite limits in the limits that they say exist. Any author ought to make it clear to their readers which way they mean it.


问:
preimage of a closed set under continuous function
We have the following statement:

Let be $f$ a continuous function. The preimage of a closed set is again a closed set.

Let be $f:(0,\infty)\times [0,\infty)\to\mathbb{R}$ where $f(x,y)=x+y$.

(Where "$[$" means that $0$ is included and "$($" means that $0$ is excluded).

If I consider the preimage $f^{-1}(1)$ or in other words $D:=\{(x,y)\in (0,\infty)\times [0,\infty)\mid f(x,y)=1\}$ then this would mean that it is closed. However, it is easy to construct a counter-example to see that not all limit points of $D$ are an element of $D$ (e.g. the limit point $(0,1)$). So $D$ can't be a closed set.

How is this possible? Where is my mistake?

答: The set $D$, which is equal to$$\{(x,y)\in(0,\infty)\times[0,\infty)\mid x+y=1\}=\{(x,1-x)\mid x\in(0,1]\},$$is a closed subset of $(0,\infty)\times[0,\infty)$. In fact, it is not a closed subset of $\Bbb R^2$, but that's not relevant here.


问:
Is $\mathbb{Z}+\mathbb{Z}\alpha$ dense in $\mathbb{R}$ if $\alpha\in\mathbb{R}\setminus\mathbb{Q}$?
Is $\mathbb{Z}+\mathbb{Z}\alpha$ dense in $\mathbb{R}$ if $\alpha\in\mathbb{R}\setminus\mathbb{Q}$? I know this should be easy, but I am at a loss right now.

答:
$G=\mathbb{Z}+\mathbb{Z}\alpha$ is an additive subgroup of $\mathbb R$. And the additive subgroups of the real numbers are either discrete or dense.

If $G$ was discrete, it would exist $0 \neq \beta \in \mathbb R$ and $a,b \in \mathbb Z$ such that

$$\begin{cases}
1 &= b \beta\\
\alpha &= a \beta
\end{cases}$$

Therefore, $\alpha = a/b$ would be rational. A contradiction.

Hence $G$ is dense.


问:
Evaluating the integral: $\lim_{R \to \infty} \int_0^R \frac{dx}{x^2+x+2}$
Please help me in this integral:

$$\lim_{R \to \infty} \int_0^R \frac{dx}{x^2+x+2}$$
  I've tried as usually, but it seems tricky. Do You have an idea?

Thanks in advance!

答:
We can rewrite the integrand as

$$\frac{1}{\frac{7}{4}+(x+\frac{1}{2})^2}$$

so that the integral becomes

$$\int_{0}^{\infty}\frac{1}{\frac{7}{4}+(x+\frac{1}{2})^2}=\frac{2}{\sqrt7}\tan^{-1}\left(\frac{2x+1}{\sqrt{7}}\right)\Big|_0^{\infty}=\frac{1}{\sqrt{7}}\left(\pi-2\tan^{-1}\left(\frac{1}{\sqrt{7}}\right)\right)$$


答: Once you have some basic exposure to point-set topology and group theory, Crowell and Fox's "Introduction to Knot Theory" is surprisingly nontechnical and intuitive. It develops most of the needed algebraic topology in it only as far as necessary to understand some basic knot invariants. It is only 146 pages and will allow you to compute basic knot polynomials.</p>

Title: How to start learning knot theory?

Body: Knot theory really sounds cool and I'm very interested in it. But I'm wondering what basic knowledge it is required and how I should start learning about it. Thanks</p>


问:
Getting the mean without having the total population
I have the following question:

$10$% of applications for a job possess the right skills. A company has
3 positions to fill and they interview applicants one at a time until
they fill all $3$ positions. The company takes $3$ hours to interview an
unqualified applicant and $5$ hours to interview a qualified applicant.
Calculate the mean of the time to conduct all interviews.

I got a bit lost considering that they wanted to mean of all interviews however I just did $0.10*3+0.90*5=3.2$ which seemed too simplistic, any help in correction/verification would be appreciated.

答:
I think you were assuming that only one interview was done.

What happens if the company conducts interviews until 3 qualified applicants are found ?


问:
How to calculate basis knowing only coordinates?
I am working on this exercise our teacher gave us:

So basically I am given nine coordinates in one base and nine in another and I should then figure out a matrix used to convert $e$ coordinates into $\tilde{e}$ coordinates? In my book there is this guide to solving problems like this I think.

Trying this, I got the change-of-basis matrix to be:

\begin{bmatrix}
13\quad & -7\quad & 37\\
-2\quad & 1\quad & -6\\
6\quad & -3\quad & 17
\end{bmatrix}

Am I thinking straight? There was another exercise similar to this that I used this exact method for, but the teacher said I was way off and I asked him why and he gave me an answer that I honestly did not understand nor didn't help me solve the problem and because of covid it would probably take at least a week, probably two weeks to get an answer to a question :( and I do not have that time. How do I solve these kind of problems and what is going on here?

答:
This is straightforward.

What's going on is that an invertible matrix can always be interpreted as the change of basis between the standard basis and the basis consisting of the columns.

So, you should get $\begin{pmatrix}1&1&-1\\1&3&2\\-2&-1&3\end{pmatrix}\begin{pmatrix}-1&1&1\\0&-1&0\\-3&2&2\end{pmatrix}^{-1}=\begin{pmatrix}-1&-2&0\\8&-1&-3\\5&4&-1\end{pmatrix}$.

Your basis is the set of columns of the last matrix.

The method you referenced will work, but you need to apply it twice, with the standard basis as a point of reference, and multiply the two transition matrices as I have done.

I just remember the fact I mentioned about the columns, which saves time.  This may seem like a lot of hot air, but the method you referred to doesn't give the product in the right order, when you try to do it all at once.


问:
Find the magnitude of the required vector using the given information.

If $\vec a, \vec b, \vec c$ are non coplanar vectors such that $\vec b \times \vec c=\vec a$, $\vec c \times \vec a=\vec b$ and $\vec a \times \vec b=\vec c$, then find $|\vec a+\vec b+\vec c|$

I dont know how to begin the solution because no data regarding the magnitude or direction of vectors is given, and the answer is numeric. Can I get a hint to start the solution?

答:
Hint:

prove that $|a|=|b|=|c|$ (this might be a bit difficult)

prove that $\vec{a},\vec{b},\vec{c}$ are mutually perpendicular

from here deduce $|a|=|b|=|c|=1$

now use $${|a+b+c|}^2=|a|^2+|b|^2+|c|^2+2a\cdot b+2b\cdot c+2a\cdot c$$

For the first step you may use the fact that

 $$[a\space b\space c]=[b\space c\space a]=[c\space a\space b]$$ where $[a b c]$ dentes the scalar triple product of $\vec{a},\vec{b},\vec{c}$


问:
Proving spheres are orthogonal

Given two spheres in $\mathbb{R}^3$:

$x^2+y^2+z^2=2ax; \ \  \ x^2+y^2+z^2 = 2by$

and $a,b>0$, and $\gamma$ the intersection of the spheres, show that for any $p_0 \in \gamma$, the spheres are orthogonal at $p_0$.

I am not sure I fullly understand the problem but here is what I tried:

First of all assuming $p_0$ is some point s.t each sphere can be represented as a function $z_1, z_2 : \mathbb{R}^2 \to \mathbb{R}$ in a neighborhood of $p_0$ (if it isn't we can use another variable), what I think we want to show is that the tangant spaces to $z_1, z_2$ are orthogonal at this point, i.e $\langle\nabla z_1(p_0), \nabla z_2(p_0)\rangle=0.$ But when I calculate this I get that it does not equal 0:

$z_1 = \sqrt{2ax-x^2-y^2}, z_2 = \sqrt{2by-x^2-y^2}$

$\nabla z_1 = (\frac{2a-2x}{2 \sqrt{2ax-x^2-y^2}}, \frac{-2y}{2 \sqrt{2ax-x^2-y^2}})$

$\nabla z_2 = (\frac{-2x}{2 \sqrt{2by-x^2-y^2}}, \frac{2b-2y}{2 \sqrt{2by-x^2-y^2}})$

$\langle\nabla z_1(p_0), \nabla z_2(p_0)\rangle \neq 0.$

Can someone explain where is my mistake?

答:
First solution:

Here is a quick analytic solution first.
Let $p_0=(x_0,y_0,z_0)$ be a point on both spheres.

Then the tangent plane in $p_0$ to the one or the other sphere has the equation:
$$
\begin{aligned}
2x_0(x-x_0) + 2y_0(y-y_0) + 2z_0(z-z_0) - 2a(x-x_0) &=0\ ,\\
2x_0(x-x_0) + 2y_0(y-y_0) + 2z_0(z-z_0) - 2b(y-y_0) &=0\ ,
\end{aligned}
$$
so that two (unnormalized) normal vectors to the two planes are $(x_0-a,y_0,z_0)$ and $(x_0, y_0-b,z_0)$. Their scalar product is:
$$
x_0(x_0-a)+y_0(y_0-b)+z_0^2=\frac 12\Big( \ (x_0^2+y_0^2+z_0^2-2ax_0)+(x_0^2+y_0^2+z_0^2-2by_0)\ \Big)=0\ .
$$

Second solution:

Trying to go as the OP goes is also possible, even if complicated. The locally defined function $z_1$ is only a part of the parametrization
$$
(x,y)\to(x,y,z_1(x,y))\ .
$$
Letting $x$ variate around $x_0$, and keeping $y=y_0$ we obtain a tangent vector $v_x$ to the sphere, similarly keeping $x=x_0$ and letting $y$ variate around $y_0$ we get an other tangent vector $v_y$. These two vectors, and the normal vector $n_1=v_x\times v_y$,  are:
$$
\begin{aligned}
v_x &= (1,0,\nabla_x z_1(x_0,y_0)) =\left(1, 0, \frac{a-x_0}{\sqrt{2ax_0-x_0^2-y_0^2}}\right)\ ,\\
v_y &= (0,1,\nabla_y z_1(x_0,y_0)) =\left(0, 1, \frac{0-x_0}{\sqrt{2ax_0-x_0^2-y_0^2}}\right)\ ,\\
n_1 &= v_x\times v_y
=
\left(
\ -\frac{a-x_0}{\sqrt{2ax_0-x_0^2-y_0^2}}\ ,
\ -\frac{0-y_0}{\sqrt{2ax_0-x_0^2-y_0^2}}\ ,
\ 1\
\right)
\\
&\sim \left(\ x_0-a\ ,\ y_0\ ,\ \sqrt{2ax_0-x_0^2-y_0^2}\ \right)
\\
&=(\ x_0-a\ ,\ y_0\ ,\ z_0\ )\ .
\end{aligned}
$$
The computation of the corresponding normal vector using the second parametrization delivers analogously the value
$$
(\ x_0\ ,\ y_0-b\ ,\ z_0\ )\ .
$$
As in the first solution, these vectors are orthogonal.

Third solution:

A geometrical solution is as follows.

The first sphere is a sphere $(S_a)$ centered in $(a,0,0)$ with radius $a$.

The second sphere is a sphere $S_b$ centered in $(0,b,0)$ with radius $b$.

The inversion with center $O=(0,0,0)$ and power $2ab$ transforms $S_a$ in the plane with equation $x=b$, and $S_b$ in the plane with equation $y=a$. Obviously, these planes intersect orthogonally, the inversion is an orthogonal transformation, we apply it again, so $S_a$ and $S_b$ intersect orthogonally.

Title: Proving spheres are orthogonal


问:
How to find roots of $X^5 - 1$?
How to find roots of $X^5 - 1$? (Or any polynomial of that form where $X$ has an odd power.)</p>

答:
Edit 2: Since the equation $x^n-1=0$ is equivalent to $x=\sqrt[n]{1}$, it  has the following $n$ solutions:

$$e^{i\dfrac{2\pi (1+k)}{n}}=\cos\left( \dfrac{2\pi (1+k)}{n}\right)+i\sin\left( \dfrac{2\pi (1+k)}{n}\right),\qquad k=0,1,\ldots ,n-1,$$

where $n$ is a positive integer, odd or even.

See comment by Gerry Myerson indicating how the Galois group for an irreducible quartic determines whether the method exposed below works.

Although late I present another algebraic solution, less elegant but more "automatic", in the sense that no clever trick is necessary. [Edit in response to Gerry Myerson's comment] We are going to factor it into two factors, one being linear and the other quadratic. This method works for quintic polynomials, as long as we are able to find one solution and solve the quadratic polynomial factor. If it fails, in principle we can solve the quadratic polynomial by means of a resolvent cubic equation (also in this post in Portuguese) [end of edit].   For the general polynomial of 7th degree or higher it is not applicable, because it depends on finding one root by inspection (or numerically) and the remaining ones algebraically.

By inspection we see that $x=1$ is a zero of $x^{5}-1$. By long division or Ruffini's rule, we find

$$x^{5}-1=\left( x-1\right) \left( x^{4}+x^{3}+x^{2}+x+1\right) .$$

Now we factor $x^{4}+x^{3}+x^{2}+x+1$ into two quadratic polynomials

$$x^{4}+x^{3}+x^{2}+x+1=\left( x^{2}+bx+c\right) \left( x^{2}+Bx+C\right) ,$$

whose (real) coefficients have to be found by comparing LHS with the expanded RHS. Since

$$\left( x^{2}+bx+c\right) \left( x^{2}+Bx+C\right)$$

$$ =x^{4}+\left( B+b\right) x^{3}+\left( C+bB+c\right) x^{2}+\left(bC+cB\right) x+cC,$$

the coefficients must satisfy

$$\left\{
\begin{array}{c}
B+b=1 \\
C+bB+c=1 \\
bC+cB=1 \\
cC=1%
\end{array}%
\right. \Leftrightarrow \left\{
\begin{array}{c}
B=1-b \\
1/c+b\left( 1-b\right) +c=1 \\
b/c+c\left( 1-b\right) =1 \\
C=1/c%
\end{array}%
\right. $$

One of the solutions of $b/c+c\left( 1-b\right) =1$ is $c=1$. By
substitution we find the remaining coefficients:

$$\left\{
\begin{array}{c}
B=\frac{1}{2}\mp \frac{1}{2}\sqrt{5} \\
b=\frac{1}{2}\pm \frac{1}{2}\sqrt{5} \\
c=1 \\
C=1%
\end{array}%
\right. $$

Thus

$$x^{4}+x^{3}+x^{2}+x+1=\left( x^{2}+\left( \frac{1}{2}+\frac{1}{2}\sqrt{5}%
\right) x+1\right) \left( x^{2}+\left( \frac{1}{2}-\frac{1}{2}\sqrt{5}%
\right) x+1\right) $$

and finally

$$x^{5}-1=\left( x-1\right) \left( x^{2}+\left( \frac{1}{2}+\frac{1}{2}\sqrt{5%
}\right) x+1\right) \left( x^{2}+\left( \frac{1}{2}-\frac{1}{2}\sqrt{5}%
\right) x+1\right) .$$

Hence the roots of $x^{5}-1$  are the roots of these three factors. They are, respectively, $x_1=1$ and

$$x_{2,3}=-\frac{1}{4}-\frac{1}{4}\sqrt{5}\pm \frac{1}{4}\sqrt{-10+2\sqrt{5}},\quad x_{4,5}=-\frac{1}{4}+\frac{1}{4}\sqrt{5}\pm \frac{1}{4}\sqrt{-10-2\sqrt{5}}.$$

Title: How to find roots of $X^5 - 1$?


问:
Is the Converse of the Cayley-Hamilton Theorem true or not?
I was thinking about taking the Cayley-Hamilton Theorem in the opposite direction.
Does this hold : If for $F\in$ End$(V)$ $\exists P\in K[t]$ such that $P(F)=0$ then
$P$ is the characteristic polynomial of $F$ ?

If not could you give a counterexample?

答:
For an extreme example in which this fails, consider the zero matrix $\left(0\right) \in M_n(k)$.  Here, the characteristic polynomial $\chi_0(t)=t^k$, corresponding to the fact that $0$ occurs as an eigenvalue of $(0)$ with multiplicity $n$.  Obviously, though, $(0)$ is annihilated by many other endomorphisms, e.g. the identity map.

Your proposition happens to be true if and only if the characteristic polynomial of $A$ equals the minimal polynomial of $A$ (the polynomial that truly satisfies the statement of your proposition), and if you replace "is the characteristic polynomial" with "is a multiple of the characteristic polynomial".

Edit: In the comments below, it's pointed out that $\chi_A$ need not be irreducible for this to work.  Of course, if $\chi_A$ is irreducible, then this is true (and what I meant, originally.)


问:
elementary matrices and eigenvalues
consider the elementary matrix (which is $n*n$) $E=I-\alpha xy^T$ show that this matrix has at least n-1 eigenvalues that equal 1.

答: Calculate the matrix $xy^T$ for non zero vectors $x,y$. Observe its rank. From the rank deduce that it has $n-1$ eigenvalues as zero and the last one being $x^Ty$.

Title: elementary matrices and eigenvalues


问:
Circle related problem-Circle and family of lines
The equation of the circle (Radius >0) whose every tangent is perpendicular to exactly one of the member of the family of the $x+y-2+\lambda (7x-3y-4)=0$ at the point of contact and the circle touches only one member of the family $(2x-3y)+\mu (x-y-1)=0$, is given by $x^2+y^2+ax+by+c=0$, then $|a+b+c|$=________.

My approach is as follow.
The point of intersection of $x+y-2+\lambda=0$ and $7x-3y-4=0$ is (1,1)

The point of intersection of $2x-3y=0$ and $x-y-1=0$ is (3,2).

We assume that (1,1) is the point where the tangent intersect at $90^o$.

and (3,2) refer to the point where the line touches the circle viz. tangent

Let $(h,k)$ be the centre of the circle anf $r$ be the radius
$(h-3)^2+(k-2)^2=r^2$

The locus of the point of intersection of perpendicular tangents refer to the director circle.

If $x^2+y^2=r^2$ refer to the circle then its director circle is $x^2+y^2=2r^2$

Hence $r^2=(3-1)^2+(2-1)^2=5$.

Locus of the centre of the circle is $(x-3)^2+(y-2)^2=5^2$

How do we find the circle.

答:
HINT.

The circle is centred at $(1,1)$ and passes through $(3,2)$.


问:
Show that the eigenvalues of the matrix are symmetric around origin

Let $A \in M_n (\mathbb{R})$ be a matrix whose all entries are $0$ except those on the diagonals above and below the principal diagonal, which are all $1$. That is, $a_{ij} = 1$ if $|i-j|=1$ and $0$ otherwise.
Show that the eigenvalues of the matrix are symmetric around origin

I have to show the eigenvalues of this matrix are symmetric with respect to origin.
If I understand correctly, I'm required to show if $\lambda$ is an eigenvalue of $A$ then so is $-\lambda$. My idea is to somehow show $ \text{Ker}(A+\lambda I)$ is non trivial when I know $ \text{Ker} (A-\lambda I)$ is non trivial, but I'm not being able to deduce this.

答:
Let $\lambda$ be an eigenvalue of $A$ corresponding to the eigenvector $v$ of $A$, then $Av=\lambda v$ where
$$v=\begin{bmatrix} x_1\\ x_2\\x_3\\ x_4 \\\dots \\ x_{n-1}\\
x_n\end{bmatrix}.$$

Now take $w=\begin{bmatrix} x_1\\ -x_2\\x_3\\ -x_4 \\\dots \\ x_{n-1}\\
-x_n\end{bmatrix}$.
Then $Aw=-\lambda w$ and thus the proof follows.


问:
Any functions $f(x, y)$ so that two limits exist
I need to find for any functions $f(x, y)$ so that
$$\lim_{\{x, y\}\rightarrow\{0, 0\}}\frac{f(x, y)- \frac{y}{x}}{1+ \frac{f(x, y)y}{x}}, \lim_{\{x, y\}\rightarrow\{0, 0\}}f(x, y){\,\,\it{exist}}$$
I used Wolfram|Alpha to find, so many choices, I spent my times, but maybe there's no $f(x, y)$ at all

I use that for the following calculating
$$\lim_{\{x, y\}\rightarrow\{0, 0\}}x\tan^{-1}\frac{y}{x}$$

答:
Let $f(0,0):=\lim f(x,y)$.

If we look at the first limit along the path $y=0$, it evaluates to $f(0,0)$.
If we look at the first limit along the path $y=x$, it evaluates to $\frac{f(0,0)-1}{1+f(0,0)}$.
If we look at the first limit along the path $y=-x$, it evaluates to $\frac{f(0,0)+1}{1-f(0,0)}$.
If all these shall be equal, we must have
$f(0,0)=\pm i$.


问:
Proof by contradiction : about the assumption that there are finitely many primes.
To prove by contradiction that there are infinitely many primes, we first assume there are finitely many primes.

It sounds to me that we can consider all primes in $\mathbb{N}$.

Is this assumption possible beacause of a ZFC axiom, that is, we take the set of all primes $\{x\in\mathbb{N}:x\text{ is a prime}\}$ and assume the set is finite?

(I did not know Euclid's proof is a proof by cases, a direct proof method.)

答:
Yes, the axioms of ZFC guarantee the existence of the set of all primes in $\mathbb{N}$.

Indeed, in ZFC the axiom scheme of specification (also known as axiom scheme of separation, or axiom scheme of comprehension) says that:

$$\tag{1}
    \forall w_{1},\ldots ,w_{n}\,\forall A\,\exists B\,\forall x\,(x\in B\Leftrightarrow [x\in A\land \varphi (x,w_{1},\ldots ,w_{n},A)])
$$

where $\varphi$ is any formula in the language of set theory with free variables among $x, w_1, ..., w_n, A$ ($B$ does not occur free in $φ$).

Now, instantiate $(1)$ with $n=0$ and $A = \mathbb{N}$ (it exists in ZFC because of the axiom of infinity), and with the formula $\varphi(x)$ that says that $x$ is prime (it can easily be expressed in the language of set theory).
It yields
$$
\exists B  \, \forall x (x \in B \Leftrightarrow (x \in \mathbb{N} \land x \text{ is prime}))
$$
So, in ZFC there exists a set $B$ that is nothing but $\{x \in \mathbb{N} : x \text{ is a prime}\}$.

In ZFC, the axiom scheme of specification is not assumed as axiom because it can be derived from another axiom scheme in ZFC, the axiom scheme of replacement, as explained here.


问:
A boundary version of Cauchy's theorem
I am looking for a reference for the following theorem (or something like it) that is not Kodaira's book.

Let $D$ be a domain and $\overline{D}$ be it's closure. Suppose that $f:\overline{D} \rightarrow \mathbb{C}$ is holomorphic in $D$ and continuous on $\overline{D}$. If the boundary of $D$, denoted $\partial D$ is composed of piecewise $C^1$ curves then
    $$f(w) = \frac{1}{2\pi i}\oint_{\partial D}\frac{f(z)}{z-w}dz$$
    for all $w\in D$.

This is stronger than the classical Cauchy integral theorem, because here we can allow the path we integrate over to be the boundary. This appears in Kodaira's book, but the book seems sloppily written and I found an egregious error in the first few pages, so I'd like another reference.

This does not appear in Ahlfors or any other standard text I looked at. I am looking for any proof that extends to usual Cauchy's theorem to allow paths that lie on the boundary of a region, instead of requiring them to lie entirely in the open region like the usual version does.

I managed to find a paper that sketches a high-powered proof when the boundary is a rectifiable Jordan arc (or some multiply connected region where each boundary component is) and gives a reference to a more elementary proof of the same result. I think a simpler proof should be possible in this case, where the boundary is assumed to be $C^1$.

This is closely related to an old question that did not get much attention.

答:
For domains with (piecewise) $C^1$ boundary, there is a simple proof using Stokes' theorem, at least if you assume that $f$ extends to $C^1$ on $\bar D$. A little computation will show that for $f \in C^1(\bar D)$,

$$2\pi i f(w) = \int_{\partial D} \frac{f(z)}{z-w}\,dz + \iint_D \frac{\frac{\partial f}{\partial \bar z}}{z-w}\,dz\wedge d\bar z,$$
from which your version of Cauchy's theorem follows.

I'll try to dig up a reference for the $C(\bar D)$ case.

Title: A boundary version of Cauchy's theorem


答:
First, we can note that, no matter how many applicants are interviewed, $3$ qualified applicants will be interviewed, taking a total of $3*5=15$ hours.

Next, we need to find how many unqualified applicants are interviewed. Since each applicant can be assumed to be independently equally likely to be qualified, and you stop interviewing after a fixed number of qualified applicants, the number of unqualified applicants interviewed is given by a Negative Binomial random variable, with parameters $p=0.9$ (the probability of an unqualified candidate) and $r=3$ (the number of qualified candidates). The mean of this is given by $\dfrac{pr}{1-p}=\dfrac{0.9*3}{1-0.9}=27$.

Therefore, the mean time taken for all interviews is $3$ hours for each unqualified candidate plus $5$ hours for each qualified candidate, giving a total of
$$27*3 + 3*5 = 96$$

Title: Getting the mean without having the total population

Body:
I have the following question:

$10$% of applications for a job possess the right skills. A company has
3 positions to fill and they interview applicants one at a time until
they fill all $3$ positions. The company takes $3$ hours to interview an
unqualified applicant and $5$ hours to interview a qualified applicant.
Calculate the mean of the time to conduct all interviews.

I got a bit lost considering that they wanted to mean of all interviews however I just did $0.10*3+0.90*5=3.2$ which seemed too simplistic, any help in correction/verification would be appreciated.


答: Let $x$ be an eigenvector of $E$ with eigenvalue $1$. So $Ex=x$. You have to find that there are at least $n-1$ linearly independent vectors satisfying $Ex=x$. This is equivalent to $(E-I)x=0 \rightarrow -\alpha x y^T = 0$. So the vectors satisfying $Ex=x$ are in $y^\perp$. Since $y^\perp$ is $n-1$ dimensional, we can find linearly independent $n-1$ eigenvectors. This is true for $\alpha$ is nonzero. The case $\alpha$ is zero is obvious.

Title: elementary matrices and eigenvalues

Body: consider the elementary matrix (which is $n*n$) $E=I-\alpha xy^T$ show that this matrix has at least n-1 eigenvalues that equal 1.


问:
Taylor vs. Laurent series.
So I'm trying to figure out methods to get a Laurent series instead of a Taylor series. Here is the initial problem:

$$\frac{1}{(z-1)}$$

So if I'm trying to find the Laurent series and I know they need negative exponents, what's the strategy? I know the answer is to factor out a $\frac{1}{z}$ but how would I get there?

$$\frac{1}{(z-1)} = \frac{1}{z} \cdot \frac{1}{1 - \frac{1}{z}} = \frac{1}{z}(1 + \frac{1}{z} + \frac{1}{z^2} + ... )$$

The above Laurent series (it has negative exponents) converges when $|z| > 1$ right?

And what's the corresponding Taylor series representation? Is the takeaway here that you can represent the same function with two different series?

答: Assuming that you want to have your series centered at $0$, then it turns out that the Laurent series of $\frac1{z-1}$ in $D(0,1)$ is equal to its Taylor series there, which is$$-1-z-z^2-z^3-\cdots$$On the other hand, its Laurent series on $\Bbb C\setminus\overline{D(0,1)}$ is indeed what you got:$$z^{-1}+z^{-2}+z^{-3}+\cdots$$

Title: Taylor vs. Laurent series.


问:
Prime counting function as ratio of polynomials.
I came across this interesting question in an advanced Analytic number theory book

Prove that there are no polynomials $M(x)$ and $N(x)$ such that $\pi(x) = \frac{M(x)}{N(x)}$ for all $x = 1,2,3,\dots$

Suppose there are polynomials $M$ and $N$ such that $\pi(x) = M(x)/N(x)$ , since prime number theorem says that

$$\lim_{x \to \infty} \frac{\pi(x)\log(x)}{x} = 1.$$

Now using this and our previous assumed expression and by using elementary method to calculate limits, we can conclude following things about $M$ and $N$.

$M$ and $N$ have same degree.

$M$ and $N$ have same leading coefficient and obviously $N$ should divide $M$ for each $x$.

This would force me to conclude prime counting function is a constant polynomial since $M$ is multiple of $N$ and they have same degree which is a contradiction to infinitude of prime numbers.

Is this rigorous enough to solve the problem?

I observe here that we used prime number theorem which is very sophisticated tool and certainly non intuitive. Is there any method that doesn't use PNT to solve this problem?

答:
Yes. There is an easy way to do this.

Let $m$ and $n$ be the degree of $M$ and $N$. Then clearly $m \geq n$. Now we know that $\pi(x) < x $. So $\frac{M(x)}{N(x)} < x $ implies that $m \leq n + 1$.

So we must have $m = n$ or $m = n + 1$. If $m =n$ then as you pointed out we will have a contradiction. If $\pi(x)$ is linear polynomial then it is certainly not possible since $\pi(0) = 0$ implies that the constant term is $0$ and $\pi(2) = 1$ implies that the coefficient of $x$ is half which is certainly not possible.

More generally if $\pi(x)$ is some $p^{th}$ degree polynomial, since there are infinite number of primes certainly there will be more than $p+1$  number of linear equations in terms of coefficients of the polynomial and there will be no solution to this system of linear equations. So $\pi(x)$ cannot be represented by any polynomial.


问:
Existence of surjective but not injective morphism between rings imply being non isomorphic?
Let R and R' be two rings and we already know that they are isomorphic if there is a one to one and onto map between them (of course this map is a ring homomorphism). I have a question at this point.

If we find a surjective but not injective ring morphism between R and R', can we directly say they are not isomorphic?

For example, let $k$ be a field and $k[x,y]$ be polynomial ring with two variables. Let $I$ be the ideal generated by $xy-1$. I want to show $\frac{k[x,y]}{I}$ is not isomorphic to $k[x]$.

Let $\phi:\frac{k[x,y]}{I} \rightarrow k[x]$ map $f(x)+I$ to $f(x)$. Easily one can show $\phi$ is surjective. However, for example, $x+y+I$ has image $h(x)$, so $\phi(x+y+I)=\phi(h(x)+I)$, by showing $x+y-h(x) \notin I$, we can show $\phi$ is not injective. Can we say $\frac{k[x,y]}{I}$ is not isomorphic to $k[x]$.

答: In general, the existence of a surjective but non-injective ring homomorphism does not imply that two rings are not isomorphic. Consider e.g. the ring $\mathbb{Z}(x_1,x_2,\dots)$ of integer polynomials on countably many variables and the map that sends $x_1$ to itself and any other variable $x_j$ to $x_{j-1}$. This results in a surjective but non-injective ring homomorphism, yet any ring is isomorphic with itself.


问:
What is the usual definition of a proof?

I saw this on a book. It says, $X$ is provable if every branch(connected by $\vee$) of $\sim X$ implies contradiction.

However, I thought that a proof is a sequence of formulas $p_n$, where each $p_n$ either is an axiom or 'follows' from $p_{m_1}=q$ and $p_{m_2}=q\to p_n$ for some $m_1,m_2<n$.

I am also comfused by the quantificational rules:

$A_\alpha^x$ means the formula constructed by replacing all 'free' $x$ with $a$. 'with proviso' means that $a$ does not occur in $A$, I guess.

It seems that I can not use De Morgan's laws here, since then there will be $\frac{A_\alpha^x}{(\exists x )A}$  and $\frac{(\forall x )A}{(\exists x )A}$ which are definitely not correct, right?

答:
There are different proof systems, which give rise to different definitions of a proof. What the definition of a proof is depends on which proof system you are talking about.

The definition of proof as a sequence of formulas etc. is what you get for axiomatic proof systems, where you have a bunch of axioms and few inference rules and combine them to a linear shaped proof with one inference step per line.

The tableau calculus is a different proof system. There are no axioms in the tableau calculus. Also a tableau proof is not a linear sequence, but a tree. And the proof strategy is not a direct proof of the conclusion, but proof by refutation, i.e. starting with the assumption that the conclusion does not hold and deriving a contradiction from that assumption.

Since proofs look different in the tableau calculus vs. an axiomatic proof system, the definition of a proof in that system is a different one. Neither of the two definitions of a proof is wrong; they just talk about different proof systems.

And yes, you can not use De Morgan's laws in a tableau proof, because there is no De Morgan rule in the tableau calculus. Everything is derived from the positive and negative rules for each connective.

But the tableau calculus is complete, meaning that every valid inference can be derived in it, so you will be able to find a tableau proof of the De Morgan laws using the A, B, C and D rules. E.g. to prove $\neg \forall x A(x) \vdash \exists x \neg A(x)$ in the tableau system, you start by assuming $\neg \forall x A(x)$ and $\neg \exists x \neg A(x)$, infer $\neg A(a)$ from the first assumption with rule D and $\neg \neg A(a)$ from the second one with rule C, which gets you the contradiction $\neg A(a), \neg \neg A(a)$, thereby completing the single-branch proof.

I don't know how you arrive at the deductions in your last paragraph, but yes, they are not proofs in the tableau system, because they don't match the rules of the tableau calculus.

The motivation behind the C rules is that if you know that $A$ holds of all $x$, it holds of whatever $a$ you insert for $x$. The idea of the D rules is that if $A$ holds of some $x$, you can give that object which you know exists a name and insert that name for $x$, provided you don't make any additional assumptions about that object by identifying it with some other previously introduced name of which other sentences are already established to hold. The proviso is that  $a$ is new in the sense that is does not yet occur in any formulas on the branch (not just not in $A$). An analogous reasoning applies to the negated quantifiers.


问:
Is an irrational number odd or even?
My sister just asked this question to me: "Is an irrational number odd or even?" I told her that decimals are not odd or even and that does imply that not recurring and non repeating decimals will also not be odd or even. But I want a rigorous math proof for this. Can someone help me with this proof? Also, can I do this proof by contradiction?</p>

答:
All odd numbers on the pascal triangle are in a Sierpinski triangle, which is  a fractal.

The binomial coefficient $\binom{n}{k}$ is defined for integers, but it can be extended to any real number.

$$\binom{\alpha }{k}= \frac{\alpha ^{\underline{k}}}{k!} $$

Another way is by using the Gamma function:
$$\binom{\alpha }{k}=\frac{\Gamma \left ( \alpha +1 \right )}{\Gamma \left ( k +1 \right )*\Gamma \left ( \alpha-k+1 \right )}$$

The thing with the Sierpinski triangle, is that since it is a fractal, it has infinite detail, if you zoom in.

So the concept of being odd can be extended to all numbers  $\binom{n}{k}$ whose coordinates (n,k) fall in the Sierpinski triangle, in the sense that they share a property of odd numbers.

I don't know if the white triangles (made of even numbers on the pascal triangle) also contain odd numbers for non integer coordinates, but if they do, and also conform a fractal, the same concept apply.

Title: Is an irrational number odd or even?


问:
Show that a certain complex power series satisfies $f'(w)=\frac{f(w)}{2w}$
Suppose $$a_{n+1}=\left(\frac 1 2 - n\right)a_n$$ with $a_0=1$ defines a real sequence of numbers.

I am then given that the following complex power series is differentiable on $D(1,1)$:
$$ f(w)= \sum_{n=0}^\infty \frac{a_n}{n!}(w-1)^n.$$
I know how to differentiate complex power series on the disk with radius $1$. Its derivative is given by:
$$f'(w)=\sum_{n=1}^\infty n \frac{a_n}{n!}(w-1)^{n-1}.$$

$$f'(w)=\sum_{n=1}^\infty \frac{a_n}{(n-1)!}(w-1)^{n-1}.$$
An index shift gives us:
$$f'(w)=\sum_{k=0}^\infty  \frac{a_{k+1}}{k!}(w-1)^{k}.$$
Which almost is the original series again, apart from an incremented coefficient.

Now I am asked to show the following relation:

$f'(w)=\frac{f(w)}{2w}$

But I do not see how I can possibly get there from what I have. I tried binomial expansion and using the recursion, but by using the recursion I get two terms and I'm struggling to make it one single series again. Any insights/hints?

答:
Let $g(z)=f(1+z)$. Then the identity to be shown is
$$2 g'(z) = \frac{g(z)}{1+z}\text{.}$$

To show this identity, consider the the Chu-Vandermonde identity:

$$\binom{\alpha+\beta}{n}=\sum_{k=0}^n \binom{\alpha}{k}\binom{\beta}{n-k}\text{.}$$

Let $\alpha=\tfrac{1}{2}$, $\beta=-1$. Since

$$\binom{-1}{k}=(-1)^k$$
it follows that

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

Then
$$\sum_{n=0}^{\infty}\binom{-\tfrac{1}{2}}{n}z^n=\sum_{n=0}^{\infty}\left(\sum_{k=0}^n\binom{\tfrac{1}{2}}{k}(-1)^{n-k}\right)z^n\text{,}$$
an equation which is equivalent to the required identity.


问:
Finding Blowup of an ODE with maple
I've been interested in finite blowups of odes lately. I'm looking for a good way to calculate blowup times in maple. Right now all I do is to use dsolve, plot the ode, and then look where the solution appears to blowup. Unfortunately, this is rather cumbersome for my computer to handle for the odes I'm interested in. Just wondering if someone knew of another way to calculate the blowup times. I can use other math programs as well, I'm just most comfortable with maple. Thanks in advance for your help.</p>

答:
Have you tried matlab? I calculate the follows: dsolve('D4y+D2y+y^3=0','y(0)=0','Dy(0)=1','D2y(0)=0','D3y(0)=0')

I also have tried sevral ICs, but it said there is no solutions. Also ODE23 ODE45 can be used. Good luck.

Title: Finding Blowup of an ODE with maple


问:
If $\;\lim\limits_{x\rightarrow 10}g(x)=a\;$ and if$\;\lim\limits_{x \rightarrow10}\dfrac{f(x)}{g(x)}=b$, find $\lim\limits_{x\rightarrow10}f(x).$
If $\;\lim\limits_{x \rightarrow10}g(x)=a\;$ and if $\;\lim\limits_{x\rightarrow10}\dfrac{f(x)}{g(x)}=b\;,\;$ where $\;a,b\neq 0\;,\;$ what's $\;\lim\limits_{x\rightarrow10}f(x)\;?$

The answer is insufficient information to determine. But I thought there was a "limit multiplication rule" I could use to determine it's $ab$? Can someone give an example when this doesn't hold?

[EDIT] based on below discussion, I guess the question boils down to, can we conclude $\lim\limits_{x\rightarrow10}f(x)$ exists based on above conditions?

I am also curious if the answer changes if $b=0$ or $a=0$?

答:
If $\;\lim\limits_{x \rightarrow10}g(x)=a\;$ and if $\;\lim\limits_{x\rightarrow10}\dfrac{f(x)}{g(x)}=b\;,\;$ where $\;a,b\neq 0\;,\;$ what's $\;\lim\limits_{x\rightarrow10}f(x)\;?$

Since $\;a\ne0\;,\;$ there exists $\;\delta>0\;$ such that $\;g(x)\ne0\;$ for any $\;x\in\left]10-\delta,10+\delta\right[\;.$

Consequently,

$f(x)=g(x)\cdot\dfrac{f(x)}{g(x)}\quad\forall\;x\in\left]10-\delta,10+\delta\right[\;,$

and, by applying the limit multiplication rule, we get that

$\begin{align}
\lim\limits_{x\rightarrow10}f(x)&=\lim\limits_{x\to10}\left(g(x)\cdot\dfrac{f(x)}{g(x)}\right)=\\
&=\lim\limits_{x\to10}g(x)\cdot\lim\limits_{x\to10}\dfrac{f(x)}{g(x)}=a\cdot b\;.
\end{align}$

Now we are going to prove the following property

If there exists $\;\delta>0\;$ such that $\;g(x)\ne0\;$ for any $\;x\in\left]10-\delta,10+\delta\right[\;,\;$ if $\;\lim\limits_{x \rightarrow10}g(x)=a\;$ and if $\;\lim\limits_{x\rightarrow10}\dfrac{f(x)}{g(x)}=b\;,\;$ then $\;\lim\limits_{x\rightarrow10}f(x)=a\cdot b\;.$

Since $\;f(x)=g(x)\cdot\dfrac{f(x)}{g(x)}\quad\forall\;x\in\left]10-\delta,10+\delta\right[\;,$

by applying the limit multiplication rule, we get that

$\begin{align}
\lim\limits_{x\rightarrow10}f(x)&=\lim\limits_{x\to10}\left(g(x)\cdot\dfrac{f(x)}{g(x)}\right)=\\
&=\lim\limits_{x\to10}g(x)\cdot\lim\limits_{x\to10}\dfrac{f(x)}{g(x)}=a\cdot b\;.
\end{align}$


问:
How to modify a formula to suit this application
I have an problem that I am unsure how to solve.

Hopefully someone can edit this question to get the fancy equations up as I don't know how.

I have a formula to find a variable, say $$a = \frac{(b * c)} {d}$$

Now if d = 1000 but due to my application in software (its a long story), I cannot divide by 1000 but I can divide by a number in the range of 2^n, say 2,4,8,16...1024, basically so I can shift right to divide. So,

How can I modify the above equation so that d = 1024 instead of 1000 and still get the correct answer?

Hope I made this clear, if not I will edit my question.

Thanks for any help.

Body: $\frac{b c}{1000}=\frac{b c x}{1024}$ if $x=1.024$. You can multiply $b$ or $c$ by $1.024$


问:
$\int _{-\infty }^{\infty }\dfrac{dx}{x^{2}+1}$ with using Residue theorem
I solve this question with using Residue theorem.
$$\int _{-\infty }^{\infty }\dfrac{dx}{x^{2}+1}$$

I think

\begin{align}P\cdot V\int _{-\infty }^{\infty }\dfrac{dx}{x^{2}+1}&=2\pi i\left(\operatorname{Res}_{z=i}\left\{ \dfrac{1}{z^{2}+1}\right\}+\operatorname{Res}_{z=-i}\left\{ \dfrac{1}{z^{2}-1}\right\}\right)\\&=2\pi i\left( \lim _{z\rightarrow i}\left\{ \dfrac{1}{z+i}\right\} +\lim _{z\rightarrow -i}\left\{ \dfrac{1}{z-i}\right\} \right)\\&=2\pi i\left( \dfrac{1}{2i}+\left( -\dfrac{1}{2i}\right) \right) \\ &=0\end{align}
But answer is $\pi$. This is answer.
I don't know where I wrong. Please tell me.

答: You should only take into account those singularities with imaginary part greater than $0$. So, you will get $2\pi i\times\frac1{2i}$, which is indeed equal to $\pi$.


问:
Solving non-linear equations of the form $\mathbf x = A f(\mathbf x)$
Let $A$ be a real, invertible $n\times n$ matrix. I am interested in finding the vectors $\mathbf{x}\in\mathbb R^n$ that solve the following equation:

$$\mathbf x = A \tanh(\mathbf x)$$

where the $\tanh$ is applied element-wise. More generally, we can consider other kinds of non-linearities instead of the $\tanh$ (but always applied element-wise).

Is there a generic approach to studying the solutions of these type of equations? Probably exploiting the eigen decomposition of $A$?

I added the tag "reference-request" in case someone can suggest relevant references to the literature.

答:
In the 2D case, the equation takes the form $$\begin{cases}x=a f(x)+bf(y),\\y=cf(x)+df(y)\end{cases}$$

and after elimintation of $y$, we get a univariate nonlinear equation
$$\frac{x-af(x)}b=f\left(cf(x)+\frac db(x-af(x)\right).$$ We don't see any particular simplification nor connection with the Eigenvalues.

I have seen numerical cases with four distinct positive solutions.


问:
Sum of Bernoulli random variables with different success probabilities
Let $X_{i} \in \{0,1\}$ be Bernouli random variable with probability of success $p_{i}$, i.e., $P(X_{i}=1) = p_{i}$ and $P(X_{i}=0) = 1-p_{i}$ and let $Y=\sum_{i=1}^{n}X_{i}$ for $n>0$. Is it relatively simple to estimate the pmf $P(Y=y)$ when $p_{i} \neq p_{j}$ for $i \neq j$?</p>

答:
What you are describing is called a Poisson Binomial Distribution (PBD). In what sense do you want to estimate it?

If you are looking to some approximate learning of the distribution, based on samples drawn from it, you can have a look at this paper. It gives a (randomized) algorithm which, given access to i.i.d. draws from a PBD $p$ and a parameter $\varepsilon>0$, uses $\tilde{O}(\varepsilon^{-3})$ samples (no dependence on $n$), runs in polynomial time (in $\log n$ and $\varepsilon^{-1}$) and with high probability, outputs a representation of a probability distribution which is close (in terms of statistical distance — that is, $L_1$ distance, up to a factor 2) to the distribution $p$.

The distribution outputted by the algorithm is not a PBD, though; if you insist on getting a PBD, this is possible with the same number of samples, but a blowup in the running time of the algorithm (now superpolynomial in $\varepsilon^{-1}$).

Remarks: $\tilde{O}(\varepsilon^{-3})$ means $O(\varepsilon^{-3}\mathrm{poly}(\log(\varepsilon^{-1})))$ (the tilde "hides" the logarithmic factors).


问:
Recursively Defined Functions for string
Let be a recursively defined language over the alphabet Σ = {0,1} as follows. Basis case: ε ∈ L, where ε denotes the empty string. Recursive step: if s ∈ L
Recursively Defined Functions

can any one help me please ?
it's really complicated for me

答:
a)
$$
\epsilon
\\
00
\\
0110
\\
010010
\\
01000010
$$
Note in each step we replace $ \epsilon $ by some rule such as $ 0s0 \ or \ 0 \epsilon 0 $


问:
Does the master theorem apply for ≤ inequalities?
Does the master theorem specifically apply when the recurrence contains ≤, ie. $T(n) ≤ 4T(n/2) + n$ ?

I cannot seem to find a definition containing inequalities.

答: Well, the Master Theorem lets you find the Big-O upper bound of a function $T(n)$ as $n \to \infty$. And it's easy to see that if $T(n) = aT(n/b) + f(n)$ implies $T(n) = \Theta(g(n))$, then $T(n) \leq aT(n/b) + f(n)$ would imply $T(n) = O(g(n))$ by induction on $n$.


答: If I may expand on @JoséCarlosSantos's point, the idea is to use an "infinite semicircular contour", either counterclockwise on $\Im z\ge0$ or clockwise (thereby inducing a $-$ sign) on $\Im z\le0$. We thereby enclose the pole at $i$ or $-i$ respectively; either can be used, so "only one pole counts" isn't an arbitrary or special-pleasing statement. The pole at $\pm i$ can be used to evaluate the integral as$$\pm2\pi i\lim_{z\to\pm i}\frac{z\mp i}{z^2+1}=\pm2\pi i\lim_{z\to\pm i}\frac{1}{z\pm i}=\frac{\pm2\pi i}{\pm2i}=\pi.$$In fact, your original calculation can be salvaged: if we induce the sign change required of the contour enclosing the pole at $-i$, we're counting the integral twice (once with each contour), so the correct reuslt is $\tfrac122\pi i\left(\frac{1}{2i}\color{blue}{+}\frac{1}{2i}\right)=\pi$.

Title: $\int _{-\infty }^{\infty }\dfrac{dx}{x^{2}+1}$ with using Residue theorem

Body:
I solve this question with using Residue theorem.
$$\int _{-\infty }^{\infty }\dfrac{dx}{x^{2}+1}$$

I think

\begin{align}P\cdot V\int _{-\infty }^{\infty }\dfrac{dx}{x^{2}+1}&=2\pi i\left(\operatorname{Res}_{z=i}\left\{ \dfrac{1}{z^{2}+1}\right\}+\operatorname{Res}_{z=-i}\left\{ \dfrac{1}{z^{2}-1}\right\}\right)\\&=2\pi i\left( \lim _{z\rightarrow i}\left\{ \dfrac{1}{z+i}\right\} +\lim _{z\rightarrow -i}\left\{ \dfrac{1}{z-i}\right\} \right)\\&=2\pi i\left( \dfrac{1}{2i}+\left( -\dfrac{1}{2i}\right) \right) \\ &=0\end{align}
But answer is $\pi$. This is answer.
I don't know where I wrong. Please tell me.


问:
Matrix for rotation
I am trying to code an AES encryption program. I have a 4x4 matrix of values and I need to rotate each of the rows individually such that:

```<code>[[A B C D]        [[A B C D]       :  ROTATE 0
 [E F G H]   =>    [H E F G] i.e.  :  ROTATE 1
 [I J K L]         [K L I J]       :  ROTATE 2
 [M N O P]]        [N O P M]]      :  ROTATE 3
</code>```

Is there a simple matrix that I could use to multiply my original to achieve this result?

答:
No, there is no such matrix. For if
$$
\pmatrix{
A & B & C & D\\
E & F & G & H \\
I & J & K & L \\
M & N & O & P } S =
\pmatrix{
A & B & C & D\\
H & E & F & G  \\
K & L & I & J \\
 N & O & P & M }
$$
for any $A, B, \ldots P$, then it must be the case that
$$
\pmatrix{
1 & 0 & 0 & 0\\
1 & 0 & 0 & 0\\
1 & 0 & 0 & 0\\
1 & 0 & 0 & 0 } S =
\pmatrix{
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1}
$$
The matrix on the right is the identity, and has rank 4; the first matrix on the left has rank 1. The rank of a product is no greater than the max of the rank of the factors, so we have a contradiction. (The same argument, slightly adjusted, works for when $S$ is multiplied on the left rather than the right.)

NB: I've use the product-of-ranks theorem here, but there's a simpler argument. In
$$
PS = Q
$$
for $4 \times 4$ matrices $P, S,$ and $Q$, we can say that the first row of $P$, multiplied by $S$, gives the first row of $Q$. So in the equation above, we'd have
$$\pmatrix{1&0&0&0} S = \pmatrix{1&0&0&0}.$$

But the same thing works for the second row, so we'd have
$$\pmatrix{1&0&0&0} S = \pmatrix{0&1&0&0}.$$

Now $\pmatrix{1&0&0&0} S$ can't possibly have two different values, so there's no such matrix $S$.

Title: Matrix for rotation


问:
Prove that the Galois group $G = Gal(x^5-2; \mathbb{Q})$ is a group metacyclic.
Prove that the Galois group $G = Gal(x^5-2; \mathbb{Q})$ is a group metacyclic, that is, $G$ contains a normal cyclic subgroup $N$ whose quotient $G/N$ is also cyclic.

MY ATTEMPTY: I was trying to follow this idea I found, which is:

You have a group $G$ that has a cyclic normal subgroup $K$ such that its quotient $G/K$ is also cyclic. Now take a quotient group $G/N$ of $G$ by a normal subgroup $N$. The image $KN/N$ of $K$ in $G/N$ is clearly also cyclic, as a homomorphic image of the cyclic group $K$. And for the quotient we have
\begin{equation}
(G / N) / (K N / N) \cong G / K N \cong (G /K) / (KN / K),
\end{equation}
which is then also cyclic, as a homomorphic image of the cyclic group $G/K$.

However, first I need to prove that the mentioned Galois Group is metacyclic.

Would you help me?

答: Hint. Show that $N=\operatorname{Aut}(L/\mathbf{Q}(\zeta_5))$ satisfies these properties using the Galois correspondence and exhibiting the explicit automorphisms in $G=\operatorname{Gal}(\mathbf{Q}(\zeta_5,\sqrt[5]{2})/\mathbf{Q})$.


问:
Australian Maths Competition: An Expected-Value Problem
A friend and I play a game. We each start with two coins. We take it in turns to toss
a coin; if it comes down heads, we keep it, if tails, we give it to the other. I always
go first, and the game ends when one of us wins by having all four coins.
If we play this game 840 times, what is the expected number of games that I would
win?

I found the probability of me winning in a single game to be 1/4. I multiplied 840 by 1/4 to have an answer of 210. However, it turns out to be wrong since 360 was written in the key. How was that happened?

答:
Say that my probability to win is $p$, and I want to find the value of $p$.

If I get head in the first flip (0.5 chance), my friend can win with $p$ probability or lose with $(1-p)$ probability. $\frac{1-p}{2}$

If i get tail in the first flip (0.5 chance), then my friend and I keep getting heads until eventually he get tail, the game reset to the start and I can win with $p$ probability. $\sum_{i=1}^{\infty}{\frac{p}{4^{i}}}$

$$
\begin{align}
p&=\frac{1-p}{2}+\sum_{i=1}^{\infty}{\frac{p}{4^{i}}}\\
&=\frac{1-p}{2}+\frac{p}{3}
\end{align}
$$

Solve for $p$ and you will get $p=\frac{6}{14}$


问:
Is this stopping time finite a.s ? bounded?
Let $B$ and $W$ be two independent Brownian motions on $(\mathcal{F}_{t})$. Is $\tau$ defined as:
$$\tau = \inf \{ t \ge 0: B_{t} \ge W_{t} + e^{-t} \}$$ almost surely finite ? Bounded ?

I would say that it can not be finite due to Brownian motion and therefore can't be bounded. But I don't know how to prove it.
What do you think guys ?

答:
Edit 02/12/2020: upon reflection, I believe my original proof was incorrect, albeit it led to the correct result. I present an amended proof below, and leave the original proof in the appendix for reference.

========================

Supposing that Doob's Optional Stopping Theorem holds, we should have that:

$$\mathbb{E}[B_{\tau}]=\mathbb{E}[B_{0}]$$

Now, because of continuity of Brownian motion, and because $e^{0}=1$, we must have that:

$$\tau = \inf \{ t \ge 0: B_{t} \ge W_{t} + e^{-t} \}=\inf \{ t \ge 0: B_{t} = W_{t} + e^{-t} \}$$

Therefore:

$$B_{\tau}=W_{\tau}+e^{-\tau}$$

So:

$$\mathbb{E}[B_{\tau}]=\mathbb{E}[W_{\tau}+e^{-\tau}]=\mathbb{E}[W_\tau] + e^{-\tau} $$

But:

$$\mathbb{E}[B_{0}]=0$$

So for Doob's theorem to hold, we must have:

$$0=\mathbb{E}[W_{\tau}]+e^{-\tau}$$

Or:

$$\mathbb{E}[W_{\tau}]=-e^{-\tau}$$

Given that $W_0=0$, we have $\mathbb{E}[W_t|W_0]=0$ $\forall t \in \mathbb{R}$, so that $\mathbb{E}[W_{\tau}|W_0]_{=0}=-e^{-\tau}$ is not true for any $\tau\in \mathbb{R}$. Therefore we have a contradiction and Doob's optional stopping theorem doesn't hold, meaning that $\mathbb{E}[\tau]$ is not finite.

For completeness: Doob's optional stopping theorem states three conditions, any of which is sufficient for some stochastic process $X_{\tau}$ to be well defined and for the following to be true: $\mathbb{E}[X_{\tau}]=\mathbb{E}[X_0]$.

One of these three conditions is that $\mathbb{E}[\tau]<\infty$.

Conversely, if one can show that $\mathbb{E}[X_{\tau}]\neq\mathbb{E}[X_0]$, then none of the three conditions of the stopping theorem are true, including the one that stipulates $\mathbb{E}[\tau]<\infty$.

PS: most problems do not require proving the stopping theorem from scratch. In most problems, one just needs to apply the theorem (so the trick is to memorize the conditions of the theorem and the statement $\mathbb{E}[X_{\tau}]=\mathbb{E}[X_0]$, and then just using the statement to prove the conditions or using the conditions to prove the statement).

If you also need to prove the Stopping theorem, it's a bit more difficult, but the proof can be found here, for example.

==============================

Appendix with original proof:
Supposing that Doob's Optional Stopping Theorem holds, we should have that:

$$\mathbb{E}[B_{\tau}]=\mathbb{E}[B_{0}]$$

Now, because of continuity of Brownian motion, and because $e^{0}=1$, we must have that:

$$\tau = \inf \{ t \ge 0: B_{t} \ge W_{t} + e^{-t} \}=\inf \{ t \ge 0: B_{t} = W_{t} + e^{-t} \}$$

Therefore:

$$B_{\tau}=W_{\tau}+e^{-\tau}$$

So:

$$\mathbb{E}[B_{\tau}]=\mathbb{E}[W_{\tau}+e^{-\tau}]=\mathbb{E}[W_\tau] + e^{-\tau} $$

But:

$$\mathbb{E}[B_{0}]=0\neq \mathbb{E}[W_0]+e^{0}=1$$

(the line above is wrong, because the statement doesn't need to be true for Doob's theorem to hold: counterexample would be: $\tau:=inf \left\{t>0: B_t = 1-t\right\}$. With $B_t$ being a standard Brownian motion, clearly $\mathbb{E}[\tau]$ is finite (and in fact equals to 1), but we don't have $\mathbb{E}[B_0]=1$)

So that:

$$\mathbb{E}[B_{\tau}]\neq \mathbb{E}[B_0]$$

And we have a contradiction (again, this conclusion is wrong, because it cannot be reached just based on the statements above)


答:
If you want to know why, in applying the residue theorem to calculate the principal value, you only need take the residue at the pole with positive imaginary part, You need to appl the residue theorem with care.  You first choose a closed contour that traces the real axis, say from $-M$ to $+M$, and then circle back from $+M$ to $-M$ following a large semicircle or rectangle either above the real axis or below the real axis.  In doing so it will enclose one or other of the two poles at $z = \pm i$ but not both.  If I call this curve $\gamma$ you then have, from the residue theorem,

$$\int_{-M}^M \frac{1}{1+z^2} dz \pm \int_\gamma \frac{1}{1+z^2} dz = \pm 2\pi i \times
R \tag{1} \label{E1}$$
where $R$ is the residue at whichever pole is enclosed, either at $z=i$ or $z=-i$.   The  sign ambiguity reflects the fact that in one choice, where  $\gamma$ is taken above the real axis, the curve is traced counter clockwise so the plus sign is appropriate, and in the other the curve is traced clockwise so the minus sign becomes appropriate.

You then need to show that as $M \to \infty$ the integral along $\gamma$ has limit zero.  This allows you to conclude that the integral along the axis, in the limit, is $\pm 2\pi i R$.

The trick is that the $\gamma$ integral often only converges to zero in one case and not the other.  However, in this case you are lucky and the $z^2$ means you get a zero limit along the curve in either case and hence we should end up with either sign in \eqref{E1}, take the plus sign and $R$ to be the residue at $z = i$ or the minus sign and the residue at $z=-i$.  This then gives the answer you wanted.


问:
Recurrence relation $a_{n+2}=3a_{n+1}-2a_n$
For the recurrence relation, $a_{n+2}=3a_{n+1}-2a_n$ with $a_0=2$ and $a_1=3$, compute the first six terms of the sequence and derive a closed form formula for this sequence.

So I'm totally lost with this question because every recurrence relation I have tried was sum of two previous terms. Any clear explanation would be greatly appreciated so that I can do this type of question in the future.

答:
Use Wilf's "generatingfunctionology" methods. Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply your recurrence by $z^n$ and sum over $n  \ge 0$ to get:
$$
\frac{A(z) - a_0 - a_1 z}{z^2} = 3 \frac{A(z) - a_0}{z} - 2 A(z)
$$
Solve for $A(z)$ and split into partial fractions:
$$
A(z) = \frac{1}{1 - 2 z} + \frac{1}{1 - z}
$$
This is just two geometric series:
$$
a_n = 2^n - 1
$$</p>


问:
Statistics of a random process derived from multinomial distribution
Suppose we have a random process defined on set with $S=\{1, 2, \ldots, N\}$, where at each time step $1 \leq t \leq M$, you select a random variable $ X_t \in S$ with equal probability. I am interested in distribution of the number of unique elements $u$ that have been selected at least once after $M$ trials, let's call this $f(u; N, M)$.

For example, $f(u=6; N=10, M=15)$ would give the probability that 6 out of $N=10$ distinct variables have been selected, after $M=15$ trials. We see that the domain of $u$ is restricted to $1 \leq u \leq \min(N, M)$.

What I have tried so far:
Let $(x_1, \ldots, x_N)$ be the number of times each of the variables are selected, subject to the constraint $\sum_{i=1}^{N} x_i = M$. The statistics of $f(u)$ seems to be obtainable from the multinomial distribution which with equal probabilities for all outcomes,
\begin{equation}
f_M(x_1, \ldots, x_{N}) = \frac{1}{N^M} \binom{M}{x_1, \ldots, x_{N}}.
\end{equation}
Here, the multinomial coefficient $\binom{M}{x_1, \ldots, x_{N}}$ gives the number of realizations that leads to the outcome $(x_1, \ldots, x_N)$ and $N^M$ is the total number of realizations.

Now we want to restrict ourselves to $(x_1, \ldots, x_{N})$ where only $u$ of the $x_i$ are larger than zero and the rest is zero. If we order these non-zero values such that $(x_1 >0, \ldots, x_u >0, x_{u+1}=0, \ldots, x_{N} =0)$, then the probability in this ordering equals
\begin{equation}
\sum_{x_1 + \ldots + x_u = M} f_M(x_1, \ldots, x_u, x_{u+1}=0, \ldots, x_{u}=0) = \sum_{x_1 + \ldots + x_u = M} \frac{1}{N^M} \binom{M}{x_1, \ldots, x_{N}} = \left(\frac{u}{N}\right)^M.
\end{equation}
Now in general the $u$ non-zero outcomes do not have be the first $u$ outcomes, but can be any of the $\binom{N}{u}$ ways of selecting these variables. Hence my final expression is
\begin{equation}
f(u; N, M) = \binom{N}{u} \left(\frac{u}{N}\right)^M.
\end{equation}
However, it is easily verified that this is incorrect for small $N$ and $M$ because the $\sum_{u=1}^{ \min(N,M) } f(u) \neq 1$. What am I missing here?

答:
Notice that when you do $u^M$ you are assuming that you are taking any possibility of assignment in those $u$ outcomes. What if in all of those you choose the same element? The number of chosen ones would be $1$ instead of $u.$ So you need is the number of surjective functions from $M$ to $u,$ this is given by $u!{M\brace u},$ where ${a \brace b}$ denotes the Stirling numbers of second kind (number of ways to partition $[a]$ into $b$ non empty blocks).
so $$f(u,N,M)=\binom{N}{u}\frac{{M\brace u}u!}{N^M}.$$


答:
Let $f(x) = x^5 - 2 \in \mathbb{Q}[x]$. Let $\alpha = 2^{1/5}$ and $\omega = e^{2\pi i/5}$. Then the splitting field of $f$ over $\mathbb{Q}$ is $L = \mathbb{Q}(\alpha, \omega)$. Since $f$ is irreducible and $f(\alpha) = 0$, we have $[\mathbb{Q}(\alpha) : \mathbb{Q}] = \deg f = 5$. Also, $5$ is prime so the $5^\text{th}$ cyclotomic polynomial is $x^4 + x^3 + x^2 + x + 1$, which is irreducible, so $[\mathbb{Q}(\omega) : \mathbb{Q}] = 4$. By the tower law, $4$ and $5$ both divide $[L : \mathbb{Q}]$, so $20$ divides $[L : \mathbb{Q}]$. But by the Tower Law, we have
$$
[L:\mathbb{Q}]=[\mathbb{Q}(\alpha, \omega) : \mathbb{Q}] = [\mathbb{Q}(\alpha, \omega) : \mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}].
$$
Since $[\mathbb{Q}(\alpha, \omega) : \mathbb{Q}(\alpha)]$ is at most $4$, and $[\mathbb{Q}(\alpha):\mathbb{Q}] = 5$, we have $[L:\mathbb{Q}]\leq 20$. But since $20$ divides $[L:\mathbb{Q}]$, we have $[L:\mathbb{Q}] = 20$. The extension is a splitting field, so it is normal, and $\operatorname{char}\mathbb{Q} = 0$, so the extension is separable. Therefore $L/\mathbb{Q}$ is a Galois extension and we have $\lvert G \rvert = 20$.

Let $M = \mathbb{Q}(\omega)$. Then $M$ is the splitting field of $x^5 - 1$ over $\mathbb{Q}$, so $M/\mathbb{Q}$ is Galois. By the Fundamental Theorem of Galois Theory $H = \operatorname{Gal}(L/M)$ is a normal subgroup of $G$. Furthermore, we have
$$
G / H \cong \operatorname{Gal}(M/\mathbb{Q}).
$$
Note that $[L : M] = 5$, so $H$ has order $5$, which is prime, so it is cyclic. It remains to show that $\operatorname{Gal}(M/\mathbb{Q})$ is cyclic.

Note that $M$ is the splitting field of the irreducible polynomial $g(x) = x^4 + x^3 + x^2 + x + 1$ over $\mathbb{Q}$, so there is an element $\sigma$ of $H$ with $\sigma(\omega) = \omega^2$ (since these are both roots of $g$, and the Galois group of an irreducible polynomial acts transitively on the roots). Clearly $\sigma^2(\omega) = \omega^4 \neq \omega$, so since $\lvert H \rvert = 4$, we have that $H = \langle \sigma \rangle$ is cyclic.

Title: Prove that the Galois group $G = Gal(x^5-2; \mathbb{Q})$ is a group metacyclic.

Body:
Prove that the Galois group $G = Gal(x^5-2; \mathbb{Q})$ is a group metacyclic, that is, $G$ contains a normal cyclic subgroup $N$ whose quotient $G/N$ is also cyclic.

MY ATTEMPTY: I was trying to follow this idea I found, which is:

You have a group $G$ that has a cyclic normal subgroup $K$ such that its quotient $G/K$ is also cyclic. Now take a quotient group $G/N$ of $G$ by a normal subgroup $N$. The image $KN/N$ of $K$ in $G/N$ is clearly also cyclic, as a homomorphic image of the cyclic group $K$. And for the quotient we have
\begin{equation}
(G / N) / (K N / N) \cong G / K N \cong (G /K) / (KN / K),
\end{equation}
which is then also cyclic, as a homomorphic image of the cyclic group $G/K$.

However, first I need to prove that the mentioned Galois Group is metacyclic.

Would you help me?


问:
how to take the integral that contains exponential element
Can someone help me to find a solution for the following integral?

$ \int_{0}^{\infty} \frac{(ax+c)}{(ax+b) ^2} e^{-\frac{x}{b}} dx$

Thank you.

答:
Making the problem more general, consider
$$I=\int \frac{ (\alpha  x+\beta)}{(\gamma  x+\delta )^2}\,e^{- \epsilon x}\,dx$$

Concerning the antiderivative first, let
$\gamma  x+\delta=t$ to make
$$I=\frac{e^{\frac{\delta  \epsilon }{\gamma }} }{\gamma^2} \int\frac{\alpha  t +(\beta  \gamma-\alpha  \delta)}{t^2}  {e^{-\frac{t \epsilon }{\gamma }} }\,dt$$

Now, $t=\frac{\gamma }{\epsilon }y$ to make
$$I=\frac{e^{\frac{\delta  \epsilon }{\gamma }} }{\gamma^3} \int \frac {\alpha  \gamma  y +\epsilon(\beta  \gamma  -\alpha  \delta  )  } {y^2 }\,e^{-y}\,dy$$

Now, two integrals
$$I_1=\int \frac {e^{-y}} y \,dy=\text{Ei}(-y)$$
$$I_2=\int \frac {e^{-y}} {y^2} \,dy=-\text{Ei}(-y)-\frac{e^{-y}}{y}$$

Back to $x$
$$I=\frac{ (\alpha  \delta -\beta  \gamma )}{\gamma ^2 (
   \gamma  x+\delta)}e^{- \epsilon x }+\frac{\alpha  \gamma +\epsilon(\alpha  \delta   -\beta  \gamma  ) }{\gamma ^3 }e^{\frac{\delta  \epsilon }{\gamma }} \text{Ei}\left(-\frac{\epsilon  }{\gamma }(\gamma  x+\delta )\right)$$  Now, for the definite integral, there will be conditions to be fulfilled.

For you specific case, the result would be

$$\int_{0}^{\infty} \frac{(ax+c)}{(ax+b) ^2} e^{-\frac{x}{b}}\, dx=\frac{a (c-b)-e^{\frac{1}{a}} \text{Ei}\left(-\frac{1}{a}\right) (a b+b-c)}{a^2   b}$$ if
$$(\Re(a) \Im(b)\neq \Im(a) \Re(b)\lor \Re(a)\geq 0)\land
   \left(\Re\left(\frac{b}{a}\right)\geq 0\lor \frac{b}{a}\notin
   \mathbb{R}\right)\land \Re(b)>0$$


问:
Smallest axis-aligned bounding box of hyper-ellipsoid
Let $E$ be the $n$-dimensional ellipsoid defined by
$$E:=\{x \in \mathbb{R}^n: (x-c)^T A (x-c) \le 1\},$$
where $c \in \mathbb{R}^n$ is the center of the ellipsoid, and $A \in \mathbb{R}^{n \times n}$ is a symmetric positive definite matrix.

Question:
How can one efficiently compute the coordinate-axis-aligned bounding box that just barely contains the ellipsoid?

For a 2D example, see the following picture:

Note:
I ask this question, and answer it myself, because this question (in general form) is surprisingly absent from math.stackexchange even after 10+ years. Good answers to this question are hard to find on the internet more generally. After googling around, I eventually had to figure this out myself, and am posting here to spare future people the same trouble. Many websites discuss the question in the special case of $2D$ and $3D$, but the the format of the ellipse is given in terms of axes and angles rather than SPD matrices, and the formulas do not generalize to n-dimensions. The good answer is given by achilles hui in the comments to the following closed question:
Bounding box of Ellipsoid
but no proof is provided there, and the question is closed so I cannot provide the answer with proof there. Even if that question were reopened, it is focused on the 3D case with axes and angles rather than the n-dimensional case with SPD matrices.

答:
Given vector $\rm{c} \in \Bbb R^n$ and matrix $\rm{Q} \succ \rm{O}_n$, let

$$\mathcal E := \left\{ \rm{x} \in \Bbb R^n \mid \left( \rm{x} - \rm{c} \right)^\top \rm{Q}^{-1} \left( \rm{x} - \rm{c} \right) \leq 1 \right\}$$

Let $g (\rm{x}) := \left( \rm{x} - \rm{c} \right)^\top \rm{Q}^{-1} \left( \rm{x} - \rm{c} \right)$. The vector field orthogonal to the boundary of ellipsoid $\mathcal E$ is

$$\nabla g (\rm{x}) = 2 \, \rm{Q}^{-1} \left( \rm{x} - \rm{c} \right)$$

Let us pick $i \in [n]$ and focus on the $i$-th axis. Let $\rm{P}_i := \rm{e}_i \rm{e}_i^\top$ be the projection matrix that projects onto the $i$-th axis. At the two points where ellipsoid $\mathcal E$ touches the (smallest) bounding box, we have $\rm{P}_i \nabla g (\rm{x}) = \nabla g (\rm{x})$, i.e.,

$$\left( \rm{I}_n - \rm{P}_i \right) \underbrace{ {\rm Q}^{-1} \left( \rm{x} - \rm{c} \right)}_{=: {\rm y}} = 0_n$$

Hence, $y_i$ is free and all other entries of $\rm y$ are zero, i.e., ${\rm y} = t \, {\rm e}_i$, or, ${\rm x} = {\rm c} + t \, {\rm Q} \, {\rm e}_i$. Intersecting this line with the boundary of ellipsoid $\mathcal E$, we obtain

$$t^2 = \left( {\rm e}_i^\top {\rm Q} \, {\rm e}_i \right)^{-1} = q_{ii}^{-1}$$
or, $t = \pm \frac{1}{\sqrt{q_{ii}}}$. Thus, ellipsoid $\mathcal E$ touches the (smallest) bounding box at points

$${\rm x} = {\rm c} + t \, {\rm Q} \, {\rm e}_i = {\rm c} \pm \frac{1}{\sqrt{q_{ii}}} \, {\rm Q} \, {\rm e}_i$$

and, projecting onto the $i$-th axis,

$$x_i = c_i \pm \frac{1}{\sqrt{q_{ii}}} \, {\rm e}_i^\top {\rm Q} \, {\rm e}_i = c_i \pm \frac{q_{ii}}{\sqrt{q_{ii}}} = c_i \pm \sqrt{q_{ii}}$$

Hence, the bounding box is

$$\color{blue}{\left[ c_1 - \sqrt{q_{11}}, c_1 + \sqrt{q_{11}} \right] \times \left[ c_2 - \sqrt{q_{22}}, c_2 + \sqrt{q_{22}} \right] \times \cdots \times \left[ c_n - \sqrt{q_{nn}}, c_n + \sqrt{q_{nn}} \right]}$$


问:
Congruence with powers and primitive roots
Determine all the solutions of the congruence

$x^{85} ≡ 25 \pmod{31}$

using index function in base $3$ module $31$.

It is clear to me that $3$ is primitive root module $31$, but how do I use this information in the solution?

答:
$3^22^2\equiv5 $ and $2^2\equiv(-3)^3$, so $3^5\equiv -5$, and $3^{10}\equiv25\pmod{31}$.

Let $x=3^y$, so you're asking for $(3^y)^{85}\equiv3^{10}\pmod{31},$ which means $85y\equiv10\pmod{30}$

$\iff 17y\equiv2\bmod6\iff y\equiv4\bmod6\iff y\equiv 4, 10, 16, 22, $ or $28\pmod{30}$.

Now do you see how $x^{85}\equiv25\pmod{31}$ can be solved using indices with base $3$?


问:
product of two diffeomorphic maps are diffeomorphic?
Suppose I have a diffeomorphism  $f_i: U_i \subseteq \mathbb{R}^{n_i} \to f_i(U_i) \subseteq \mathbb{R}^{m_i}$ where $U_i$ is an open subset. Does it then follow that
$$
f_1 \times f_2: U_1 \times U_2 \to f_1(U_1) \times f_2(U_2)
$$

is still diffeomorphic? I can see that the map is still bijective, but I was struggling to show that the map and its inverse are continuous. Any explanation would be appreciated!

答:
To see that the Cartesian product of two continuous maps is continuous:

Suppose $f: X \to Y$ and $g: Z \to W$ are continuous, where $X, Y, Z, W$ are (for convenience) metric spaces.  A sequence $(x_n, z_n)$ converges to $(x,z)$ in $X \times Z$ iff $x_n \to x$ in $X$ and $z_n \to z$ in $Y$.  Then $f(x_n) \to f(x)$ in $Y$ and
$g(z_n) \to g(z)$ in $W$, so $(f \times g)(x_n, z_n) = (f(x_n), g(z_n)) \to (f(x),g(z))$ in $Y \times W$.  Thus $f \times g$ is continuous.

As for the inverse, it's easy to see that $(f \times g)^{-1} = f^{-1} \times g^{-1}$.


问:
Prove $x_n=\frac{n}{2^n}$ converges to $0$
Prove $x_n=\frac{n}{2^n}$ converges to $0$.

Since $2^n>\frac{n^2}{2}$ for all $n \in \mathbb{N}$ we have

$$-\frac{2}{n}<\frac{n}{2^n}<\frac{2}{n}.$$

Since $\frac{1}{n} \rightarrow0$ as $n \rightarrow \infty$, by Squeeze theoreom $\frac{n}{2^n}$ converges to $0$.

Is there anything wrong this proof?

答:
Your proof seems legit. However,  your problem can be easy solved using L'Hospital Rule.

$\lim\limits_{n\to\infty}\frac{n}{2^n}=\lim\limits_{n\to\infty}\frac{1}{2^n\log2}=0$


问:
Find all the singularities in the finite plane and the corresponding residues. Show the details.
I am a bit stuck on this problem:

Find all the singularities in the finite plane and the corresponding residues. Show the details.

$$\frac{8}{1+z^2}$$

So I know the residual is going to be the coefficient associated with the first negative exponent in the laurent series.

So the singularities are at $z = i $ because $i^2 = -i$ and that's when the denominator will equal 0.

But how do I find the residuals?

So I remember that:
$$\frac{1}{1+z} = 1 - z + z^2 - ...$$

So I now multiply each element by 8:

$$\frac{8}{1+z} = 8(1 - z + z^2 - ...)t$$

But none of these terms have an exponent of -1 so I cannot find the coefficient?

Sigh, what am. I doing wrong?

Next problem, same as above but different function: $$\frac{1}{1-e^z}$$

Body:
$$\frac{8}{Z^2+1} = \frac{8}{(Z+i)(Z-i)} = \frac{A}{Z+i}+\frac{B}{Z-i}$$

$$=\frac{A(Z-i) + B(Z+i)}{Z^2+1}$$

So $$(A+B)Z + (B-A)i = 8$$

So
$$A+B=0$$
$$(B-A)i=8$$

So
$$B=-4i$$
$$A=4i$$

So, $$\frac{8}{Z^2+1}=\frac{4i}{Z+i} + \frac{-4i}{Z-i}$$