问:
Seemingly simple inequality between $L_p$ norms (When is $\|f\|_p \leq \|f\|_2$?)
I've found the following statement on a book:

It is easy to see that given any $p < 2$, if a function $h$ on a probability space has very small support, then its $L_p$ norm is much smaller than its $L_2$ norm.

What I believe "very small support" means is: Given a $h:(\Omega, \Sigma, \mu) \to (\mathbb{R},B(\mathbb{R})$ measurable function, where $\mu(\Omega) = 1$. Define
$$S = supp(h) = \{\omega \in \Omega : h(\omega) \neq 0\}.$$
Then "very small support" would mean that
$$\mu(S) \mbox{ is close to 0}.$$

I'd like to know that, if we have a given $\varepsilon>0$ such that $\mu(S)<\varepsilon$, than that would imply what about $\|h\|_p$ and $\|h\|_2$? It means that there's a $\delta>0$ small such that

$$\left(\int_\Omega |h(\omega)|^pd\mu(\omega)\right)^{\frac{1}{p}}\leq \delta \left(\int_\Omega |h(\omega)|^2d\mu(\omega)\right)^{\frac{1}{2}}?$$
(that is $\|h\|_p \leq \delta \|h\|_2$)

And if so, why is that true? The most I can do is

$$\|h\|_p\leq \left[\sup_{\omega \in S}\{|h(\omega)|^p\}\mu(S)\right]^{\frac{1}{p}}\leq \sup_{\omega \in S}\{|h(\omega)|\}\mu(S)^{\frac{1}{p}}\leq \sup_{\omega \in S}\{|h(\omega)|\}\mu(S)^{\frac{1}{2}}$$

then I can't go back to the $L_2$ norm. It seems it should be simple to prove but I couldn't do it.

This topic appeared in the chapter 5 about "Hypercontractivity" on the book "Noise sensitivity of Boolean functions and percolation [Garban, Steif]".

Thanks in advance.

答:
We want to have an estimate of the form
$$ \def\norm#1{\left\|#1\right\|}
\norm{f}_p \le C\norm{f}_2
$$
(Generalized) Hölder comes to mind. We have
$$ \frac 12 + \frac 1r = \frac 1p \iff
   \frac 1r = \frac 1p - \frac 12 = \frac{2-p}{2p}
  \iff r = \frac{2p}{2-p}
$$
Therefore, as $f = f\chi_S$
$$ \norm{f}_p \le \norm f_2 \norm{\chi_S}_{r}
$$
Now
$$ \norm{\chi_S}_r = \mu(S)^{1/r}.$$
Altogether:
$$ \norm f_p \le \mu(S)^{(2-p)/2p} \norm f_2 $$


问:
How to show in a clean way that $z^4 + (x^2 + y^2 - 1)(2x^2 + 3y^2-1) = 0$ is a torus?
How to show in a clean way that the zero-locus of $$z^4 + (x^2 + y^2 - 1)(2x^2 + 3y^2-1) = 0$$ is a torus? </p>

答:
This took me longer than it should have, given the simplicity of the answer.

First, analyze the equation for when it has solutions: clearly, we require $z^4 \geq 0$, and thus, $(x^2 + y^2 - 1)(2x^2 + 3y^2 - 1) \leq 0$.  Thus, one should be positive and the other negative.  Each factor vanishes on a plane curve that's a conic section, respectively the unit circle at $(0,0)$ and the ellipse centered at $(0,0)$ with its major axis along the $x$-axis of length $1/\sqrt{2}$ and its minor axis along the $y$-axis of length $1/\sqrt{3}$.  The points $(x,y)$ with a corresponding $z$ above them are exactly those in one of those figures and outside the other; since the ellipse is completely contained in the circle, the corresponding set of points is the circle with the ellipse removed, an annular region.

Now, for any $(x,y)$ in the annulus, the corresponding $z$ is a fourth root of some associated nonnegative number; any positive real number has exactly two real fourth roots, one positive and one negative, while zero of course has only one, which is zero.

Putting this together, literally, we see that the implicit equation describes a shape that's obtained from the annulus as follows:

Make two copies, superimposed;

Glue their edges to the $xy$-plane;

With one copy, "inflate upwards", and with the other, "inflate downwards".  The edges remain fixed but the interiors flex away from the $xy$-plane.

The result is, in short, two annuli glued in the natural way along their boundaries, which is a topological torus.

Of course, I didn't use any details from the question other than to verify that the "shadow" of the torus was an annulus.  In fact, that's all that matters: that the equation is of the form

$$z^2 + f(x,y) g(x,y) = 0$$

where the zero sets of $f$ and $g$ are closed plane curves one contained in the interior of the other.  The resulting surface will always be a torus (and you won't always be able to use polar coordinates to figure it out).

(edit: Just noticed I wrote $z^2$ there rather than $z^4$.  Actually, that would work too, as would any even power of $z$.)

ContentLicense: CC BY-SA 3.0


问:
Find integers $x > 1, y > 1$ such that $(y^x+1) \mid (x^y+1)$
A problem from quora.

Find integers $x > 1, y > 1$
such that
$(y^x+1) \mid (x^y+1)$.

I have found some
infinite classes of solutions,
but none with $x$ and $y$
relatively prime.

Here are the ones I have found
so far.
I know how I could construct more,
but the algebra gets annoying.

$$(x, y)
=(2n+1, (2n+1)^{2n+1}),\\
(2(2n+1), (2(2n+1))^2),\\
(3(2m+1), (3(2m+1))^3).
$$

ContentLicense: CC BY-SA 4.0

Body:
We want to find integer solution to
$$\quad \frac{x^y + 1}{y^x + 1}$$
Here are some solutions

$$(2,2)\quad (4,2)\quad (3,3)\quad (2,4)\quad (4,4)\quad (5\le x\le 26,x)\\ (\{3,49,18,27\},27)\qquad
(\{11,14,28\},28)\qquad (\{7,8,9,10,11,29\},29)\\
(\{6,7,8,9,10,11,12,13,29\},30)\qquad
(\{5,6,7,8,9,10,11,12,13,14,15,31\},31)\\
(\{4,5,6,7,8,9,10,11,12,13,14,15,16,32\},32)\quad $$
I cannot see a pattern but they are easy to find in a spreadsheet.


问:
Rank-1-update for a low-rank matrix
Context. Suppose that we have the following eigendecomposition:
$$ A = Q D Q^T,$$
where $D$ is diagonal, non-negative, non-decreasing and the last $r - 1$ out of $p$ eigenvalues are nonzero. Here $Q$ is uniquely defined for the last $r-1$ eigenvectors, and the first $p-r+1$ columns in $Q$ can be anything that's orthonormal and spans the orthogonal complement of the last $r-1$ ones.

Consider a rank-1-update $A \rightarrow A + w w^T = U F U^T$, where $F$ is diagonal with non-negative, non-decreasing eigenvalues on the diagonal, with at most $r$ non-zero eigenvalues. Again, only the last $r$ columns of $U$ matter.

Note that $U F U^T = Q (D + vv^T) Q^T = (QV) F (QV)^T$, so that $U = QV$ and $v=Q^Tw$; here the last $r$ (and the only needed) columns of $V$ are given by the Bunch-Nielsen-Sorensen formula.

Question. How can we compute the last $r$ columns in $U$ using only the last $r-1$ columns in $Q$ and the last $r$ columns in $V$?

Solution complexity. We assume that $r \ll p$, and we're looking for an $O(rp^2)$ solution tops. In particular, taking any orthogonal complement of the last $r-1$ rows in $Q$ is unacceptable, since that will give an $O(p^3)$. Having an extra multiplicative $\log$ term in addition to $rp^2$ would be unpleasant, but manageable.

Clarification. I'm writing $Q$ as a matrix to make things simpler, but perhaps it should be viewed as a union of $r-1$ columns, plus the space that orthogonally complements them. We don't know much about that space, and we don't want to compute an ONB for it, because that is $O(p^3)$-expensive.

Thank you in advance!

答:
Let $q_1,\dots,q_p$ denote the columns of $Q$. Let $\lambda_1,\dots,\lambda_p$ denote the diagonal elements of $D$, in order (which are each eigenvalues of $Q$).

Why is it that the first $p-r+1$ columns in $Q$ make no difference?

Note that we have
$$
A = QDQ^T = \sum_{i=1}^p \lambda_i \,q_iq_i^T.
$$
Note that if $\lambda_i = 0$, then $q_i$ does not affect the resulting sum.

How can we compute the last $r$ columns in $U$ from the last $r-1$ columns in $Q$ and the last $r$ columns in $V$?

Thoughts so far:

Partition the matrix $Q$ into $Q = [Q_1\ \ Q_2]$, where $Q_2$ has $r-1$ columns. Let $D_2$ denote the bottom-right size $r-1$ submatrix of $D$.
We are interested in the eigenvectors $A + ww^T$. Equivalently, it suffices to consider the eigenvalues of $Q^T(A + ww^T)Q$; as you note, this can be written as $D + vv^T$ where $v = Q^Tw$. However, we want to proceed while computing as little of $Q_1$ as we can.

It should be sufficient to compute its last column.

Decompose $w$ into the sum $w = w_1 + w_2$, where $w_2$ is the projection of $w$ onto the column-space of $A$. That is,
$$
w_2 = Q_2Q_2^Tw.
$$
Denote $v_1 = Q_1^Tw_1 = Q_1^Tw, v_2 = Q_2^Tw_2 = Q_2^Tw$. We have
$$
Q^T(A + ww^T)Q = \pmatrix{Q_1 & Q_2}^T(A + ww^T)\pmatrix{Q_1 & Q_2}\\
= \pmatrix{
Q_1^T(A + ww^T)Q_1 & Q_1^T(A + ww^T)Q_2\\
Q_2^T(A + ww^T)Q_1 & Q_2^T(A + ww^T)Q_2}
\\ =
\pmatrix{
v_1v_1^T & v_1v_2^T\\
v_2v_1^T & D_2 + v_2v_2^T}.
$$
Note that the specific value of $v_1$ depends on $Q_1$; in fact, it can be taken to be any vector with length equal to that of $w_1$ whose last $r-1$ entries are zero. A particularly nice choice, then, is to take $v_1$ to be the vector with a its length in the $r$th entry and zeros elsewhere. In order to achieve this, set the final column of $Q_1$ to be equal to the unit vector $\hat w_1 = w_1/\|w_1\|$. Accordingly, repartition $Q$ as
$$
Q = \pmatrix{Q_1' & \hat w_1 & Q_2}.
$$
In the resulting matrix $Q^T(A + ww^T)Q$ will have all of its non-zero entries in the bottom-right size $r$ square submatrix, and this submatrix is a rank-1 update of a diagonal matrix. It should suffice to apply the BNS formula to this submatrix.

Title: Rank-1-update for a low-rank matrix


问:
How to reduce the following second order PDE to its canonical form?
I got this PDE: $u_{xx} + xu_{yy}=0$ on the domains $D_{-}=\{(x,y): x<0\},D_{+}=\{(x,y): x>0\}. $ I see that the PDE is hyperbolic on $D_{-}$ and ellipctic on $D_{+}.$ I'm having trouble finding its canonical form though. Any help, with a bit of detail would be greatly appreciated!

答:
The characteristics are given by the equation

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-B \pm \sqrt{B^2-4AC}}{2A},$$

where $A = 1$, $B = 0$, and $C = x$. Substituting and integrating gives the two complex characteristics $\xi = \frac{2i}{3}x^{3/2}+y$, and $\eta = -\frac{2i}{3} x^{3/2}+y$. By using the chain rule, i.e., recalling that $\partial_x = \xi_x \partial_\xi + \eta_x \partial_\eta$ and $\partial_y = \xi_y \partial_\xi + \eta_y \partial_\eta$ one arrives at the canonical form

$$ 6 (\eta-\xi) u_{\xi\eta} + u_\xi - u_\eta = 0.$$

I hope you find this useful.


问:
Is it true that $ \sum_{n=m+1}^{k-1} A(n) \int_{n}^{n+1} f'(t)dt = \sum_{n=m+1}^{k-1} \int_{n}^{n+1} A(t) f'(t)dt $?
In Introduction to Analytic Number Theory by Apostol, section 4.3 (Relations to $\vartheta(x)$ and $\pi(x)$), there is the following step in the proof for Theorem 4.2 (Abel's identity):

$$
\sum_{n=m+1}^{k-1} A(n) \int_{n}^{n+1} f'(t)dt = \sum_{n=m+1}^{k-1} \int_{n}^{n+1} A(t) f'(t)dt.
$$

Here $k$ and $m$ are integers. They were defined to be $ k = [x] $ and $ m = [y] $ where $ 0 < y < x $ and $ f $ has a continuous derivative on the interval $ [y, x] $.

The function $ A $ is defined as follows: For any arithmetical function $ a(n) $ let $ A(x) = \sum_{n \le x} a(n) $ where $ A(x) = 0 $ if $ x < 1 $.

How does this step hold true?

答:
Hint: Note that $a(n)$ in the definition of
\begin{align*}
A(x)= \sum_{n \le x} a(n)
\end{align*}
is an arithmetic function $a:\mathbb{N}\to\mathbb{C}$. This means, that for $k\in\mathbb{N}$ we have
$$A(x)=A(k)=const.\qquad\qquad k\leq x<k+1$$
It follows
\begin{align*}
A(k)=A(k)\int_0^1dx=\int_0^1 A(x)\,dx
\end{align*}


问:
How do we know that if we keep going with a step of two in the natural numbers we will land on a perfect square, either odd or even?
I really need to use this for another proof that I have. Been struggling with this. Can I say that it is trivial?
It feels very common sense but would I need to specifically explain it if I do some proof?</p>

答: Mh, perfect squares are natural. If you try all even [odd] naturals, you will land on all even [odd] squares. So yes, trivial.</p>


问:
Does convergence in $C(\Bbb R)$ imply convergence in $\mathcal D'(\Bbb R)$
Sequence of continuous functions $(f_k)$ which converge to $f$ in $C(\Bbb R)$. Does it imply that $f_k$ will converge to $f$ in $\mathcal D'(\Bbb R)$ as well? Or we need something more to make this assertion?

[Here $\mathcal D'(\Bbb R)$ is a linear continuous functional on $\mathcal D(\Bbb R)$ ; collection of all linear maps: from $\mathcal D(\Bbb R)$ to $\Bbb R$)

and we say $f\in$ $\mathcal D'(\Bbb R)$, is continuous in the sense that
for all $\phi_n\to \phi$ in $\mathcal D(\Bbb R)$ => $(f,\phi_n)\to(f,\phi)$ in $\mathcal D'(\Bbb R)$]

答:
It seems that the type of convergence considered is: $f_n\to f$ in $C(\mathbb{R})$ if for every fixed compact $K\subset \mathbb{R}$ we have that $f_n\to f$ uniformly in $K$. By using this definition we have that $f_n\to f$ as a distribution, i.e. if we define $T_n(\phi)=\int_\mathbb{R} f_n\phi$ and $T(\phi)=\int_\mathbb{R}f\phi$, $\phi\in \mathcal{D}(\mathbb{R})$ then $T_n\to T$ in $\mathcal{D}'(\mathbb{R})$

To prove the claim in the last paragraph, let $K=\overline{\operatorname{support}(\phi)}$, where by definition $K$ is a compact set. Hence, $f_n\to f$ uniformly in $K$, which implies that

$T_n(\phi)=\int_\mathbb{R}f_n\phi=\int_K f_n\phi\to \int_Kf\phi=\int_\mathbb{R} f\phi=T(\phi)$ for every $\phi\in \mathcal{D}(\mathbb{R})$


问:
Beta Reduction in Lambda Calculus
I came across the definition of beta reduction in Lambda Calculus which is :
$$(λx.M)N \rightarrow_β Μ[\space x:= N \space]$$ under the constraint that the $FV(N)$ are still free after the substitution.

I also came across three examples and the third one confuses me :

1st example : $$(λx.zx)w \rightarrow_β zw $$

2nd example : $$(λy.zy(λx.xy))w \rightarrow_β zw(λx.xw) $$

3rd example : $$λz.(λf .λx. f z x) (λy.y) \\  \rightarrow_β \\ λz.λx.(λy.y) z x \\ \rightarrow_β \\   λz.λx.z x \\$$

At first I thought that the left most bound variable gets substituted but in the 3rd example this does not seem to be the case.

Then I thought that given the fact that $N = (λy.y)$ is an abstraction maybe the $λf$ notation implies that it's argument will be an abstraction and so the bound variable $z$ will be "skipped".

Is my process here right and if that's the case why does the bound $z$ variable gets "skipped"? $z$ could be a notation for an abstraction as well.

Edit : Changed the word function to abstraction because as far as I understand everything can be represented via a function in $λ$-calculus.

答:
Traditionally $\lambda$-abstraction "associates right", while function evaluation "associates left". That is, if we see $\lambda x. \text{stuff}$, then we assume $x$ is bound for the entirety of "stuff". If we want to explicitly show where the scope of the $\lambda x$ ends, we wrap the term in parentheses.

For example:

$$(\lambda f. \lambda x . f x) w \to_\beta \lambda x . w x$$

While

$$\lambda f . ((\lambda x . f x) w) \to_\beta \lambda f . f w$$

If there aren't any parentheses, then (by convention) we assume we're in the second case. Which agrees with the example in your question. Notice in examples $1$ and $2$, the author explicitly wraps the left $\lambda$-term in parentheses so that we know to apply it.

I hope this helps ^_^

Title: Beta Reduction in Lambda Calculus


问:
analytic expression of eigenvalue of A.
The following square matrix of order $N$ coming out of discretization of pde. Now I want to compute the analytic expression of eigenvalue of $A.$

$$A=\begin{pmatrix}
-2&2&0&0&0&0&0\\
1&-2&1&0&0&0&0\\
0&1&-2&1&0&0&0\\
\ddots&\ddots&\ddots&\ddots&\ddots&\ddots&\ddots\\
0&0&0&1&-2&1&0\\
0&0&0&0&1&-2&1\\
0&0&0&0&0&1&-2
\end{pmatrix}$$

The problem creates $a_{1,2}$, if $a_{1,2}=1$ we get the eigenvalue by Toeplitz matrices structure.

My thought: Let $\lambda$ be the eigenvalue of $A$, then we have $AU=\lambda U.$ From that we can get $u_{i-1}-2u_i +u_{i+1}=\lambda u_i  ,\  \text{for}\, i=2,2,...,N-1$ and $-2u_{1}+u_{2}=\lambda u_1 $
and $u_{N-1}-2u_{N}=\lambda u_N $

so we have the recurrence relation $u_{i+1}=(\lambda +2)u_i -u_{i-1}  ,\  \text{for}\, i=2,2,...,N-1$

Let $u_i=r^{i}$ then we have $r^2-(\lambda+2)r+1=0$ and from that I can get get the eigen vector representation of the form $\alpha r^i+\beta r^{-i}$. But I could not find the value of $\alpha,\beta$ using other two equation and so I havent get the eigenvalue.

Is there a bettere way to get the analytic expression of eigenvalue?

If I use recurrance relation to get the eigenvalue, how I find out $\alpha, \beta$?

Thanks.

答:
I'll take the liberty of writing $A_N$ to denote the $N \times N$ version of the concerned matrix.

We are interested in the eigenvalues of $A_N$, i.e. the roots of the characteristic polynomial $p_N(\lambda) = |A_N - \lambda I_N|$. For simplicity, let's introduce the variable $t = 2 + \lambda$ such that
$$
p_N(t) =
\begin{vmatrix}
-t & 2 & 0 & \dots & \\
1 & -t & 1 & \dots & \\
& & \ddots & & \\
& \dots & 1 & -t & 1 \\
& \dots & 0 & 1 & -t
\end{vmatrix}.
$$
Expanding this determinant about the final row leads us to the observation that $p_N(t) = -t p_{N-1}(t) - p_{N-2}(t)$. By explicitly computing $p_2$ and $p_3$ and working backwards we can assign values to $p_1$ and $p_0$, resulting in the recurrence relation
\begin{align}
p_N &= -t p_{N-1} - p_{N-2}, \\
p_0 &= 2, \\
p_1 &= -t.
\end{align}

The solution takes the form
$$
p_N(t) = c_+ r_+^N + c_- r_-^N,
$$
where $c_\pm$ are constants soon to be determined and $r_\pm = \frac{-t \pm \sqrt{t^2 - 4}}{2}$ are the roots of the polynomial $r^2 + t r + 1$ (i.e. the characteristic polynomial of the recurrence). Considering the cases $N=0$ and $N=1$ and using the known values for $p_0$ and $p_1$ we find that $c_+ = c_- = 1$. Thus, we have
$$
p_N(t) = \left( \frac{-t + \sqrt{t^2 - 4}}{2} \right)^N + \left( \frac{-t - \sqrt{t^2 - 4}}{2} \right)^N.
$$
We now make the substitution $t = -2 \cos{(\theta)}$ so that our expression simplifies to
$$
p_N(\theta) = (\cos{(\theta)} + i \sin{(\theta)})^N + (\cos{(\theta)} - i \sin{(\theta)})^N = e^{i N \theta} + e^{-i N \theta} = 2 \cos{(N \theta)}.
$$
Now, recall that we are looking for $N$ eigenvalues of $A_N$, i.e. for $N$ roots of $p_N$. We find them where $\cos{(N \theta)} = 0$:
\begin{align}
\theta &= \frac{\pi}{N} \left( k + \frac{1}{2} \right) \\
\Rightarrow t &= -2 \cos{ \left( \frac{\pi}{N} \left( k + \frac{1}{2} \right) \right)} \\
\Rightarrow \lambda &= -2 \left( 1 + \cos{ \left( \frac{\pi}{N} \left( k + \frac{1}{2} \right) \right)} \right),
\end{align}
for $k = 0, \dots, N-1$.

Title: analytic expression of eigenvalue of A.


答: You could very well call this trivial depending on the context, but you may also supply a quick proof if the situation demands it. Suppose you start at $n$ ($n\in \mathbb{N}$). Then $n^2 \geq n$. Since squaring preserves parity, you will reach $n^2$ by successively adding $2$ in a finite number of steps.

Title: How do we know that if we keep going with a step of two in the natural numbers we will land on a perfect square, either odd or even?

Body:
I really need to use this for another proof that I have. Been struggling with this. Can I say that it is trivial?
It feels very common sense but would I need to specifically explain it if I do some proof?</p>


问:
If $\Lambda_k = \lambda I + \sum_{i=1}^{k} \phi_i \phi_i^\top $ where $\|\phi_i\|\leq 1$, then $\det(\Lambda_k) \leq (k+\lambda)^d$
Consider a matrix $\Lambda_k = \lambda I + \sum_{i=1}^{k} \phi_i \phi_i^\top$ where $\phi_i \in \mathbb{R}^d$ and $\|\phi_i\|\leq 1$. Prove that $\det(\Lambda_k) \leq (k+\lambda)^d$. Here $\lambda > 0$ and $d \in \mathbb{N}$.

How can I approach this problem?

答:
my sense is you are supposed to follow my hint and then apply $\text{GM}\leq \text{AM}$.  As currently stated, this is a rather loose bound.  I give a somewhat different proof, below.

$\Lambda_k$ is real symmetric and is equal to a PD matrix plus PSD matrices$\implies \Lambda_k\succ \mathbf 0$

1.) $\text{trace}\big(\Lambda_k\big) = \text{trace}\big(\lambda I_d\big) + \sum_{i=1}^{k} \text{trace}\big(\phi_i \phi_i^\top\big)= d\cdot \lambda +\sum_{i=1}^{k} \big \Vert \phi_i \big\Vert_2^2= \eta \leq d\cdot \lambda + k$

note: this is rather misleading notation as a subscript of $k$ typically indicates that the matrix is $k\times k$ -- e.g. $I_d$ does mean it is $d\times d$.  Also $\Lambda$ typically refers to a diagonal matrix with eigenvalues $\lambda_i$ though it doesn't here.

2.) $\Lambda_k$ is orthogonally similar to a matrix with constant diagonal, $C$, i.e.

$c_{i,i}=\frac{1}{d}\cdot \text{trace}\big(\Lambda_k\big)=\frac{\eta}{d}\leq \lambda + \frac{k}{d}$

(this can be done over reals, but it is more convenient to extend $\mathbb R$ to $\mathbb C$, view $\Lambda_k$ as Hermitian PD, diagonalizable by (real) unitary matrix $Q$, then $C$ is similar to $A$ via $\big(FQ\big)$, where $F$ is the unitary matrix known as Discrete Fourier Transform.  If the determinant bound holds over $\mathbb C$ then it must hold over $\mathbb R$ since $\mathbb R\subset \mathbb C$.)

3.) $0\lt \det\big(\Lambda_k\big)=\det\big(C\big)\leq \big(\frac{\eta}{d}\big)^d \leq \big(\lambda + \frac{k}{d}\big)^d \leq \big(\lambda + k\big)^d$

as desired.

The inequalities are (i) Hadamard Determinant Inequality, and (ii) term by term dominance of positive numbers, raised to the $d$th power


问:
Is it possible to know the sum of the digits of a number (in base 10), without knowing the digits?
Let's say that you have a really big power of 2, that's so big that you can't print it out on a computer. Would it still be possible to find the sum of its digits?

There is a similar result that is weaker. If we denote sum of digits by the function $s(n)$, then $s(2^{n+1}) \equiv 2\cdot s(2^n)\pmod 9$. So if we were interested in sum of digits mod 9, we could iteratively apply the identity to find this sum for super large numbers, without knowing the digits themselves.

答:
There is a formula for the sum of digits of $n$:
$$s(n) = n -9\sum_{k=1}^\infty\left\lfloor\frac{n}{10^k}\right\rfloor$$

Note that the sum finite since $\left\lfloor\frac{n}{10^k}\right\rfloor = 0$ for $k$ big enough.

I doubt this formula simplifies enough to calculate the sum of digits of numbers of the form $2^n$.


问:
Calculating a coefficient for a formal power series
My textbook has a whole bunch of exercises on finding some coefficient inside a formal power series. Unfortunately, there aren't any examples on how to do so, especially since many of the series expand to infinite power series and the binomial theorem seems to not apply.

For example, I'm asked to find the coefficient of $x^8$ in the series expansion of $(1-x)^{-7}$, and to find the coefficient of $x^{10}$ in the series expansion of $x^6(1-2x)^{-5}$. How would I proceed? I'd like some general method to do these problems, not just the numerical answer for these two problems.

Sorry if this seems to be a really dumb question.

答:
The following techniques should suffice for problems as simple as the ones you list.

Use binomial theorem to find any coefficient of $(a + bx)^n$, for any $n \in \mathbb{Z}$.
A simpler approach is to factor out the $a^n$ to get $(1 + \frac{b}{a}x)^n$, and then to treat $(\frac{b}{a}x)$ as a single unit.
With this approach, you really only have to know two expansions:

For $n > 0$, $\displaystyle (1 + x)^n = \sum_{k = 0}^n {n \choose k}x^k$.

For $n > 0$, $\displaystyle (1 - x)^{-n} = \sum_{k = 0}^\infty {{n + k - 1} \choose {k}} x^k$.

Multiplying by a factor of $x^m$ is easy -- this just shifts all the terms over.  For instance, if you want to find the coefficient of $x^8$ in $x^3 \cdot (\text{other stuff})$, just find the coefficient of $x^5$ in $(\text{other stuff})$.  The $x^5$ term will then be multiplied by $x^3$, resulting in the desired $x^8$ term.


问:
Why is $\lim_{n\rightarrow \infty}\sqrt[n]{n!} = \infty$?
Why is $$\lim_{n\rightarrow \infty}\sqrt[n]{n!} = \infty$$

if :
$$\lim_{n\rightarrow \infty}\frac{n!}{n^n} = 0 \implies n^n >n! \quad\text{for "big" n}$$

I don't understand it.

答:
Let $a_n = n!$.

The limit $\lim \sqrt[n]{a_n}$ is related to the root test for series convergence.
In this context, it is well known that if $\lim \dfrac{a_{n+1}}{a_n}$ exists, then so does $\lim \sqrt[n]{a_n}$ and they are equal. Let's compute the ratios then:
$$
\lim \frac{a_{n+1}}{a_n} = \lim (n+1) = \infty
$$
Therefore, $\lim \sqrt[n]{a_n}= \infty$.


问:
What is the rational homology of the unordered configuration space of points in the plane?
What is the rational homology of the unordered configuration space of $n$ points in the Euclidean plane?

In the unordered configuration space, I know there is torsion in homology, but I just want to know the Betti numbers, i.e. what is
$$\beta_j = \dim H_j (\mbox{Conf}_n(\mathbb{R}^2),\mathbb{Q} ),$$
as a function of $n$ and $j$?

For the ordered configuration space, there is no torsion in homology and we have the Poincaré polynomial
$$\beta_0 + \beta_1 t + \beta_2 t + \dots = (1+t)(1+2t) \dots (1+(n-1)t).$$

答:
This is classical and in some form is due to Arnol'd (The cohomology ring of the colored braid group, which unfortunately is in Russian) but I am having trouble locating a precise statement and reference for the unordered as opposed to the ordered configuration spaces. Since $\text{Conf}_n(\mathbb{R}^2)$ is the classifying space of the braid groups $B_n$ you are equivalently asking about their (rational) group homology.

$\text{Conf}_n(\mathbb{R}^2)$ has a Galois cover by the ordered configuration space, which I will denote $\text{OConf}_n(\mathbb{R}^2)$, with Galois group $S_n$. (These spaces are classifying spaces of the pure braid groups $PB_n$.) It follows that the rational homology of $\text{Conf}_n(\mathbb{R}^2)$ satisfies

$$H_{\bullet}(\text{Conf}_n(\mathbb{R}^2), \mathbb{Q}) \cong H_{\bullet}(\text{OConf}_n(\mathbb{R}^2), \mathbb{Q})^{S_n}$$

so we can do this computation by understanding not only the rational homology of the ordered configuration spaces but also the action of $S_n$ on them.

The ordered configuration spaces are homotopy equivalent to the spaces of the little $2$-disks operad $E_2$, and accordingly their rational homology is the rational homology $H_{\bullet}(E_2, \mathbb{Q})$. This rational homology forms another operad (in graded vector spaces) and it's known exactly what this operad is:

Theorem: The rational homology of the little $d$-disks operad is the $d$-Poisson operad $\text{Pois}_d$ of (graded) commutative algebras equipped with a Poisson bracket of homological degree $d-1$ (often stated as cohomological degree $1-d$; I hope I am negating this correctly).

You can check out Dev Sinha's The homology of the little disks operad for more details here. When $d = 2$ we get the Gerstenhaber operad. The Poisson bracket comes from the generator of $H_{d-1}(\text{OConf}_2(\mathbb{R}^d), \mathbb{Q})$; here $\text{OConf}_2(\mathbb{R}^d)$ is homotopy equivalent to $S^{d-1}$. Besides being a compact way of describing all of the homology at once this description also tells us the action of $S_n$: it's given by permuting the operations in the Poisson operad. For example we have the graded skew-symmetry $\{ x, y \} \mapsto \{ y, x \} = -(-1)^{d-1} \{ x, y \}$ of the Poisson bracket itself coming from the action of $S_2$ on $\text{OConf}_2(\mathbb{R}^d)$; the sign here corresponds to the action of the antipode of $S^{d-1}$ on $H_{d-1}(S^{d-1})$.

Now it suffices to determine for $d = 2$ which of the $n$-ary operations in the Poisson operad are invariant under the action of $S_n$. In degree $0$ the only operation (up to scale)  is the $n$-ary commutative product $x_1 \dots x_n$ which is always invariant; this corresponds to $H_0 \cong \mathbb{Q}$. In degree $1$ we have the operation

$$\sum_{i < j} \{ x_i, x_j \} \prod_{k \neq i, j} x_k$$

which is also invariant (the Poisson bracket commutes and $\{ x_i, x_j \}$ has degree $1$ but the other $x_k$ still have degree $0$ so the action of $S_n$ on monomials is just by permutation with no signs) and which is also the unique invariant (up to scale).

In higher degrees the operations are spanned by "monomials" in the commutative product and the Poisson bracket which involve either products $\{ x_i, x_j \} \{ x_k, x_{\ell} \}$ of two or more Poisson brackets or nested brackets $\{ x_i, \{ x_j, x_k \} \}$. (By repeated application of the Leibniz rule we can at least assume WLOG that commutative products are never nested inside Poisson brackets.) Here I believe what happens is that "signs cancel": that is, summing any monomial over its $S_n$-orbit gives $0$, which implies that there are no $S_n$-invariant operations. It is basically clear in my head how this works but I am having trouble writing down an explicit rigorous proof; in any case, assuming this (I will try to fix this gap later), we have:

Corollary: The rational homology of $\text{Conf}_n(\mathbb{R}^2)$, and hence of the braid groups, is given for $n \ge 2$ by

$$H_i(\text{Conf}_n(\mathbb{R}^2), \mathbb{Q}) \cong \begin{cases} \mathbb{Q} & \text{ if } i = 0, 1 \\ 0 & \text{ otherwise}. \end{cases}.$$

(The cases $n = 0, 1$ can be handled separately; in both cases we get $\mathbb{Q}$ in degree $0$.) In other words, maybe somewhat surprisingly, the unordered configuration spaces have the rational homology of a circle.

Fun aside: this computation turns out to be related, via the Grothendieck trace formula, to the fact that the probability that a random monic polynomial over $\mathbb{F}_q$ is squarefree is $1 - \frac{1}{q}$, which implies that the computation can actually be redone (using the Grothendieck trace formula, together with a comparison theorem between $\ell$-adic and singular cohomology) by counting that there are $q^n - q^{n-1}$ monic squarefree polynomials of degree $n$; I learned this from Jordan Ellenberg's blog post The braid group, analytic number theory, and Weil's three columns.

Edit: Here is the promised calculation. I still believe the "signs cancel" approach above will work, but since I'm not sure how to organize the calculation nicely, let's do the following instead. Since the action of a finite group on a rational vector space is completely reducible, invariants are canonically isomorphic to coinvariants. And we have a trick available for calculating coinvariants: namely, if $O(n)$ is any operad in graded vector spaces, the free $O$-algebra on a graded vector space is $V$ is

$$\bigoplus_{n \ge 0} V^{\otimes n} \otimes_{S_n} O(n).$$

When $V$ is the $1$-dimensional vector space in degree $0$, $V^{\otimes n}$ is the trivial $1$-dimensional representation of $S_n$ in degree $0$, so this calculates exactly the $S_n$-coinvariants $O(n)_{S_n}$. To keep track of the answers for different values of $n$ we will introduce a second grading for $V$; let's call it the "vector grading," and we'll call the other grading the homological grading.

The upshot is that we've reduced to calculating the free Gerstenhaber algebra on a generator $x$ in homological degree $0$ (and vector degree $1$). This is actually quite straightforward. First, we get the Poisson (really I should say "Gerstenhaber" here) bracket $\{ x, x \}$, which lives in homological degree $1$ and vector degree $2$. The graded Jacobi identity gives $\{ x, \{ x, x \} \} = 0$, so no nested brackets appear. And since $\{ x, x \}$ is odd, it anticommutes with itself with respect to ordinary multiplication, so $\{ x, x \} \{ x, x \} = 0$. So all we can do is take the product of at most $1$ copy of $\{ x, x \}$ with any number of copies of $x$, and that's all.

So the free Gerstenhaber algebra on $x$ has a basis given by $x^i$ (which lives in homological degree $0$ and vector degree $i$) and $x^i \{ x, x \}$ (which lives in homological degree $1$ and vector degree $i + 2$), for $i \in \mathbb{Z}_{\ge 0}$. The correspondence with the rational homology of the configuration spaces is that the subspace in homological degree $i$ and vector degree $n$ is $H_i(\text{Conf}_n(\mathbb{R}^2), \mathbb{Q})$. This exactly reproduces the previous calculation (and as a bonus corrects an error I had previously made, which is to forget that the cases $n = 0, 1$ behave differently).

Fun aside #2: this free Gerstenhaber algebra on a generator in degree $0$ occurs "in nature" as the rational homology $H_{\bullet}(\Omega^2 S^2, \mathbb{Q})$ of the double loop space of the $2$-sphere, which is in turn the "free grouplike $E_2$ space" on a point. The generator $x$ itself corresponds to the generator of $\pi_2(S^2) \cong \pi_0(\Omega^2 S^2)$, embedded into $H_0(\Omega^2 S^2, \mathbb{Q})$, while the Poisson bracket $\{ x, x \}$ corresponds to the Whitehead bracket of this generator with itself, which I believe is something like twice the Hopf fibration as an element of (the rationalization of) $\pi_3(S^2) \cong \pi_1(\Omega^2 S^2)$, embedded into $H_1(\Omega^2 S^2, \mathbb{Q})$. The connection between double loop spaces and braid groups is also classical (I think) but I don't know what a good reference is. The nLab page on iterated loop spaces has some references.


问:
The average weight of $46$ weightlifters is $106$ kg; how many can weigh $125$ kg if none may drop below $82$ kg?
The average weight of $46$ weightlifters is $106$ kg. If none of the lifters weighs less than $82$ kg, at most how many of them can weigh more than $125$ kg?

We are given that $$\overline{X}=\dfrac{x_1+x_2+x_3+\ldots+x_{46}}{46}=106.$$ From here we can conclude $$x_1+x_2+x_3+\ldots+x_{46}=106\times46=4876 \text{ kg }$$ So the total weight of the weightlifters is $4876$ kg. We also know that $x_n\ge82$ kg for every $n$. I don't know what to do next. Thank you in advance!

答:
$
\newcommand{\ve}{\varepsilon}
$
We can pair them off, one at a time. Say $x_1 = 82$ (the minimum weight), and then $x_2 = 125 + \varepsilon$, where $\ve$ is some small but positive number. The pair in total weigh $207 + \ve$, and there are $23$ such pairs. Thus, we have a total weight, at this point, of $4761 + 23 \ve$, but we also know to this to be equal to $4876$. Thus,

$$4761 + 23 \ve = 4876 \implies \ve = 5$$

So among them, there are $5$ kg extra per pair, and $23 \times 5 = 115$ in total spare. We can redistribute this to some pairs, with the lower-weight person receiving the weight. They each need $43 + \mu$  for some small positive $\mu$, and thus we can only give two people enough ($115/43 \approx 2.67$).

Thus, in total, $25$ people may exceed $125$ kg in weight (and $21$ do not).

To see that $26$ is not viable, suppose the remaining $20$ are at minimum weight. Then $26 \times 125 + 20 \times 82 = 4890$, just in excess of $4876$, with an average of $106.3 \ne 106$.


问:
Existence and the value of $\lim_{x \to 0} \int^{3x}_x \frac{\sin t}{t^2}dt$
I need the show the existence of the following limit and then calculate the limit

$$\lim_{x \to 0} \int^{3x}_x \frac{\sin t}{t^2}dt$$

Since the antiderivative of $\frac{\sin t}{t^2}$ was not nice, I tried to use the approximation $\sin x \approx x$ for $x$ close to $0$. Then, I can integrate and find the limit as $\ln3$.

Is this a valid solution and can I solve this without using such approximation?

答:
Hint: You can apply an integration for parts taking $u=\sin t$ and $v'=1/t^2$ and you will found

$$=\left[-\frac{\sin t}{t}-\int -\frac{\cos t}{t}dt\right]^{3x}_x$$

Remember that exist the cosine integral named $\operatorname{Ci}(x)$ https://en.wikipedia.org/wiki/Trigonometric_integral:

$$\operatorname{Ci}(x) = -\int_x^\infty \frac{\cos t}{t}dt = \gamma + \ln x - \int_0^x \frac{1 - \cos t}{t}dt$$


问:
Does every category have a functor?
Is there any one (or more) categories that doesn't have a functor? Functors go between categories, so is there any category that only has an identity functor but no other functor that maps it to another category?</p>

答:
Every category has at least 2 outgoing functors: the identity functor plus at least another one.
Actually every category has infinitely many outgoing functors.
You can easily see this by considering the cases of:

the empty category which goes to every other one (it is the initial category, after all)

the non empty categories which go trivially or non trivially to every other non empty and/or non terminal category

However the empty category has only one ingoing functor. The identity functor. Every other category has more than one ingoing functor

Title: Does every category have a functor?


答:
Since $\sin(t)=t+O(t^3)$,

$\int^{3x}_x \frac{\sin t}{t^2}dt
=\int^{3x}_x \frac{t+O(t^3)}{t^2}dt\\
=\int^{3x}_x (\frac1{t}+O(t))dt\\
=\ln(3)+O(x^2)\\
\to \ln(3)\\
$

You can use
$t-t^3/6 < \sin(t) < t$
to get explicit bounds for the integral.

Title: Existence and the value of $\lim_{x \to 0} \int^{3x}_x \frac{\sin t}{t^2}dt$

Body:
I need the show the existence of the following limit and then calculate the limit

$$\lim_{x \to 0} \int^{3x}_x \frac{\sin t}{t^2}dt$$

Since the antiderivative of $\frac{\sin t}{t^2}$ was not nice, I tried to use the approximation $\sin x \approx x$ for $x$ close to $0$. Then, I can integrate and find the limit as $\ln3$.

Is this a valid solution and can I solve this without using such approximation?


答:
Suppose you are given the natural number $r$, then $r^2-r=r(r-1)$ is even and $$r^2=r+2\cdot\frac {r(r-1)}2$$ gives an explicit construction of a square which you will hit in due course.

It is trivial, as others have noted, that you will hit a square, but if you need a specific example of a square, this is an easy way ...


问:
Steps for solving $T= (\frac{xr}{1000} n) + \frac{xc}{100} $ for $x$
I have a problem, a correct solution and not nearly enough tools to get from one to the other.

I had this formula ...

$$T= (\frac{xr}{1000} n) + \frac{xc}{100} $$

... and I needed to get here ...

$$x = \frac{ 1000 * T }{10 c + n r }$$

... so eventually I cheated by means of Wolfram Alpha using ...

$$T=(((x*r)/1000)*n) + ((x*c)/100)$$ solve for $x$.

How "should" I have got there? I will continue to try and get there on my own - but a "right" way would be really useful for me to be able to see and study. I'm not after a full algebra lesson or anything - just a clever human's steps maybe.

Sorry if this type of question isn't preferred here (?) - but if anyone wants to show how they would have done it I would greatly appreciate your time :)

Update: I can't believe that after just a few minutes I got not only a sensible correction to the question title but so much help.
I'm choosing the answer that helped my brain to engage the quickest and gave me the starting point I needed. Thank you all - moved by the help. Off to study.

UPDATE 2: The Internet. Wolfram Alpha. StackExchange. What a time to be alive.

Body:
Can you get from

$$ 7 = (\frac{11x}{1000} \cdot 13) + \frac{17x}{100} $$

to $ x = \frac{7000}{170 + 11 \cdot 13} $ ?


问:
Prove that the sum of the digits of $9^n$ is not equal to $9$ for $n \ge 3$
It may seem really easy to solve but almost all of us are stuck on this one.

Seems really weird:

Prove that, for all $3\leqslant n$, $S(9^n)\not=9$.

$S(n)$ is the sum of digits of $n$ base $10$.

َAlso, there's no need to check for odd $n$ because in that case the solution is as follows:

First of all it's clear that $S(n)$ is divisible by $9$ so if it's not $9$ we're done.

Furthermore, if $n$ is odd, the rightmost digit of $9^n$ is $9$, and there's at least one other non-zero digit, so sum of digits of $n$ is more than $9$, and that's it.

But for even $n$ I'm seriously stuck.

答:
Adapting from @ACheca suggested paper "The Decimal Expansions of Powers of 9" by Sapir, Lossers & Montgomery (independently) 1999:

First as stated note that for $\underline{\text{odd }n>1}$, since $9^n\equiv 9 \bmod 10$ we know that the last digit is $9$ and since $9^n> 9$ there are other non-zero digits, so $S(9^n)>9$

Thus for $\underline{\text{even }n}$ we are looking to show that $S(81^k)>9,$ where $\underline{k=n/2>1}$

Let $M(a)$ be the value $(a \bmod 99999)$ with $0\leq M(a)\leq 99999$. Then we can show that $S(M(a))\leq S(a)$ by the following process:

Find $M(a)$ from $a$ by taking block of 5 digits and summing the blocks. This process is invariant $\bmod 99999$ and repeated as necessary will produce $M(a)$. However this summing process never increases the digit sum - any carrying effects only reduce the digit sum. Thus $S(M(a))\leq S(a)$.

Now the values of $M(81^k)$ cycle through the following $30$ values:
$C =(81,6561,31446,$ $47151,19269,60804,$ $25173,39033,61704,$ $98073,43992,63387,$ $34398,86265,87534,$ $90324,16317,21690,$ $56907,9513,70560,$ $15417,48789,51948,$ $7830,34236,73143,$ $24642,96021,77778)$. By inspection the digit sum of all of these except the first are greater than $9$.

Thus:

For $k \not\equiv 1 \bmod 30$, we have $S(81^k)\geq S(M(81^k))>9$

For $k\equiv 1 \bmod 30$ , we have $81^k\equiv 81 \bmod 100$ and recalling that $k>1$ we have $81^k > 81$ and there are other non-zero digits, so for this case also $S(81^k)>9$

... proving the case $n$ even also.


问:
Tensor product of regular modules is regular
Let $G$ and $H$ two algebraic linear groups, $V$ a regular $G$- module and $W$ a regular $H$ - module. I want to prove that $V \otimes W$ is a regular $G \times H$-module. If $V$ and $W$ are regular modules, it means that there exist $\phi : G \times V \to GL(V)$, with $\phi(g, v)= \phi(g)v$ and $\psi : H \times W \to GL(W)$, with $\psi(h, w) = \phi(h)w$ and with $\psi$ and $\phi$ are both regular. That means that they are both polynomials in certain variables (the entries of a matrix $x_{ii}$ and $\frac{1}{det(x)}$). So we can define $\omega : (G \times H) \times (V \otimes W) \to GL(V \otimes W)$ such that $\omega((g, h), (v, w)) = \phi(g)\psi(h)(v \otimes w)$, which is again regular because the product of polynomials is a polynomial. Is this correct?

答:
$\require{AMScd}$As far as I understand and remember, you are in the following situation: there are polynomial maps $P,Q$
$$
\begin{array}{ccc}
\mathbb{A}^m & \overset P\to & \mathbb{A}^n\\
\cup && \cup \\
G &\underset{\phi}\to & \text{GL}(V)
\end{array}
$$
and
$$
\begin{array}{ccc}
\mathbb{A}^r & \overset Q\to & \mathbb{A}^s\\
\cup && \cup \\
H &\underset{\phi}\to& \text{GL}(W)
\end{array}
$$ where the integers $m,n,r,s\ge 1$ are determined by the dimensions of $G,H,V,W$; now, the tensor representation $\phi\otimes\psi : G\times H \to \text{GL}(V\otimes W)$ is defined as $(g,h)\mapsto \phi(g)\otimes\psi(h)$, which is the unique automorphism $V\otimes W \to V\otimes W$ induced by the bilinear map $V\times W \to V\otimes W : (v,w)\mapsto \phi g(v)\otimes \psi h(w)$ using the universal property of $V\otimes W$:
$$
\begin{CD}
V\times W @>(\phi g,\psi h)>> V\otimes W \\
@VVV @| \\
V\otimes W @>>\phi g\otimes\psi h> V\otimes W
\end{CD}
$$
The endomorphism $\phi g\otimes \psi h$ now can be concretely expressed as the Kronecker product of the two matrices $\phi g, \psi h$.

Regularity (say, for $\phi$) should amount to the request that each $\phi(g)$ can be written as a $d\times d$ matrix ($d = \dim V$) whose entries are polynomials in $k[T_1,\dots,T_m,\frac{1}{\det}]$ ($k$ the field where you take the varieties).

Similarly, $\psi(h)$ is a polinomial in $k[Y_1,\dots,Y_r,\frac{1}{\det}]$; from this, it follows that $\phi(g)\otimes\psi(h)$, i.e. the matrix obtained as Kronecker product of two matrices whose entries are polynomials, is again a polynomial; furtim, you are exploiting the natural isomorphism $k[E]\otimes k[F]\cong k[E\sqcup F]$, valid for sets $E,F$.

This can be proved directly, or deduced from the fact that $k[\_]$ is the free $k$-algebra functor ${\sf Set} \to {\sf Alg}_k$, and as such it sends coproducts (of sets, i.e. disjoint unions) into coproducts of $k$-algebras ($k$-bilinear tensor products).

It is this last remark that allows you to say that an abstract tensor $\phi g\otimes \psi h$, i.e. an element of $k[T_1,\dots,T_m,\frac{1}{\det}]\otimes k[Y_1,\dots,Y_r,\frac{1}{\det'}]$ "is" a polynomial: it's canonically identified to a polynomial in $k[T_1,\dots,T_m,\frac{1}{\det}, Y_1,\dots,Y_r,\frac{1}{\det'}]$.


问:
Would this method to eliminate extraneous solutions be correct?
In algebra I, they taught us that when solving equations with radicals, you just had to plug the solutions you received back into the original equation to check if they were legit solutions or extraneous solutions. However, this can be time consuming. This is my method.

Take an equation like $\sqrt{10-x}$ = $3x$. If you square both sides, you get $10-x$ = $9x^2$. Then, you get the polynomial $9x^2+x-10$. Solving this polynomial, you get solutions $x=1$ & $x= -\frac{10}9$. However, if you plug the negative solution, the answers do not match up. Here's my way to solve this equation without checking for extraneous solutions.

First, note that $\sqrt{10-x}$ $\geq$ $0$ for all real x because square root is positive. $3x$ $\leq$ $0$ for all negative x. Since $\sqrt{10-x}$ $\geq$ $0$ for all real x, but $3x$ $\lt$ $0$ for all negative x, that means our solution can't be negative or the sides won't be equal. Thus, we can eliminate all negative solutions easily without plugging them in to check.

This method will save a lot of time for higher degree polynomials where you have to check more solutions. So, is this method mathematically valid? If so, why don't we learn it in school?

答: Yes, it is valid. However, it is a lot harder to use this in a lot of cases than to simply plug things in, because the equation may be way more complicated, so you can't tell if the functions are positive or negative. Also, $x^2$ stays positive everywhere.


答:
I think you might be making this more complicated than it needs to be.
$$T= \left(\frac{xr}{1000} n\right) + \frac{xc}{100} = x\left(\frac{nr}{1000} + \frac{c}{100}\right),$$
which means that $$x = \frac{T}{\frac{nr}{1000} + \frac{c}{100}}.$$  Now, you can make this look nicer by finding a common denominator for the two fractions in the denominator:
$$x = \frac{T}{\frac{nr + 10c}{1000}} = \frac{1000T}{nr+10c}.$$

Title: Steps for solving $T= (\frac{xr}{1000} n) + \frac{xc}{100} $ for $x$

Body:
I have a problem, a correct solution and not nearly enough tools to get from one to the other.

I had this formula ...

$$T= (\frac{xr}{1000} n) + \frac{xc}{100} $$

... and I needed to get here ...

$$x = \frac{ 1000 * T }{10 c + n r }$$

... so eventually I cheated by means of Wolfram Alpha using ...

$$T=(((x*r)/1000)*n) + ((x*c)/100)$$ solve for $x$.

How "should" I have got there? I will continue to try and get there on my own - but a "right" way would be really useful for me to be able to see and study. I'm not after a full algebra lesson or anything - just a clever human's steps maybe.

Sorry if this type of question isn't preferred here (?) - but if anyone wants to show how they would have done it I would greatly appreciate your time :)

Update: I can't believe that after just a few minutes I got not only a sensible correction to the question title but so much help.
I'm choosing the answer that helped my brain to engage the quickest and gave me the starting point I needed. Thank you all - moved by the help. Off to study.

UPDATE 2: The Internet. Wolfram Alpha. StackExchange. What a time to be alive.


答:
On problems like this, it's usually easiest if you can get rid of the fractions first. The least common multiple of 1000 and 100 is 1000, so we start by multiplying both sides by that:

$$ T = \frac{xr}{1000}n+\frac{xc}{100} $$
$$ 1000*T = 1000*\left[\frac{xr}{1000}n+\frac{xc}{100}\right] $$

Then, we can distribute 1000 on the right-hand side:
$$ 1000T = 1000\frac{xr}{1000}n+1000\frac{xc}{100} $$

The 1000's in the first term cancel, and $\frac{1000}{100}$ in the second simplifies to 10:
$$ 1000T = xrn+10xc $$

Then, we need to isolate $x$, but there's a factor of $x$ in both terms on the right, so we need to factor it out first:
$$ 1000T = x(rn+10c) $$

To isolate $x$ now, we just need to divide both sides by $(rn+10c)$ (that factor will cancel on the right):
$$ \frac{1000T}{(rn+10c)} = \frac{x(rn+10c)}{(rn+10c)} $$

$$ \frac{1000T}{(rn+10c)} = x $$


问:
Prove that the product topology in $\Bbb C^n$ is equal to the usual one
So it is well know that the function $\tau_n:\Bbb C^n\times\Bbb C^n\rightarrow\Bbb C$ defined by the condition

$\tau_n(x,y):=\sum_{i=1}^n x_i\overline y_i$

for any $x,y\in\Bbb C^n$ is an inner product. So I ask to prove that the product topology on $\Bbb C^n$ induced by the inner product $\tau_1$ is equal to the topology $\tau _n$ as above defined. I point out that I need of this result to show that the linear functions between two topological vector spaces are continuous and so to show that all topologies in a finite dimensional topological vector space are equivalent and thus I  courteously request to not give what just said as answer. So could someone help me, please?

答:
The product topology is generated by the norm

$$N_\infty(x)=\max(\vert x_1\vert, \dots \vert x_n\vert)$$ where $\vert x \vert = \sqrt{\tau_1(x,x)}$. Denoting

$$N_2(x) = \sqrt{\tau_n(x,x)}$$ we have

$$1/\sqrt{n}N_\infty(x) \le N_2(x) \le \sqrt{n} N_\infty(x)$$ which enables to conclude to the desired result.


问:
Intuition on non-unique ideal factorization if not integrally closed
I'm aware of the result that integral domains admit unique factorization of ideals iff they are Dedekind domains.

It's clear that $\mathbb{Z}[\sqrt{-3}]$ is not a Dedekind domain, as it is not integrally closed.  I'm having difficulty demonstrating directly that $\mathbb{Z}[\sqrt{-3}]$ does not admit unique factorization of ideals.

Obviously we only need one example of an ideal that doesn't have a unique factorization into prime ideals, but a good answer would provide either a process or some intuition in finding such an example.

答:
Let me take an example with no fractions, Z[2i]=Z[x]/(x^2+4).  Now (2+2i)*(2-2i) = 8, but 2+2i is irreducible.  So that is a little crazy.

What about the ideal (2)?  In the quotient ring we have Z[x]/(2,x^2+4) = Z[x]/(2,x^2) is not a domain, so (2) is not prime.  How does it factor into prime ideals?

I claim it doesn't.  It ought to factor into (1+i)*(1-i), but there are no such elements in Z[2i].  How to prove it?  Well if J = P*Q, then at least J ≤ P, so let's look for primes above (2).

Z[x]/(2,x^2) is a nice local ring with a unique maximal ideal (2,x)/(2,x^2), which pulls back to (2,2i).  Since there is only one prime above J=(2), we must have J is some power of P=(2,2i), or else J has no such prime factorization.

Well P*P = (4,4i,-4) = (4,4i) does not contain 2, so that is done.  J has no prime factorization.

In case you want to stick with the evil Dr. √−3 and his house of lies, Z[√−3] = Z[x]/(x^2+3), here is an example there:

(1+√−3)(1-√−3) = 4, but 2, 1+√−3, and 1-√−3 are all irreducible.  Evil.

What about the ideal (2)?  In the quotient ring we have Z[x]/(2,x^2+3) = Z[x]/(2,(x+1)^2) is again not a domain, so (2) is not prime.  How does it factor into prime ideals?

I claim it doesn't.  (I am little unclear on what it should factor into, since (1+√−3)/2 is a unit, rather than a prime).  At any rate, we look for prime ideals above it.

Since Z[x]/(2,(x+1)^2) is a nice local ring with unique maximal ideal (2,x+1)/(2,(x+1)^2) that pulls back to (2,1+√−3), we see that if J=(2) has a factorization into prime ideals, it must be a power of P=(2,1+√−3).  However,
P*P = (4,2+2√−3,−2+2√−3) = (4,2+2√−3) does not contain 2.  Hence J has no such factorization into prime ideals.


问:
How do we identify a probability problem as a conditional probability problem?
NOT a conditional probability problem

Sixty percent of the students at a certain school
wear neither a ring nor a necklace. Twenty percent
wear a ring and 30 percent wear a necklace.
If one of the students is chosen randomly, what is
the probability that this student is wearing
(a) a ring or a necklace?
(b) a ring and a necklace?

A conditional probability problem

All bags entering a research facility are screened. Ninety-seven percent of the bags
that contain forbidden material trigger an alarm. Fifteen percent of the bags that
do not contain forbidden material also trigger the alarm. If 1 out of every 1,000
bags entering the building contains forbidden material, what is the probability
that a bag that triggers the alarm will actually contain forbidden material?

How do we distinguish the two?

For the second problem(before I knew it was a conditional probability problem) I interpreted it as

$A$= triggers the alarm

$F$= contains forbidden material

$P(A \bigcap F)$ =.97

$P(A \bigcap F^C)$ =.15

$P(F)$ = $\frac{1}{1000}$

答:
NOT a conditional probability problem:

Sixty percent of the students at a certain school wear neither a ring nor a necklace. Twenty percent wear a ring and 30 percent wear a necklace. If one of the students is chosen randomly, what is the probability that this student is wearing (a) a ring or a necklace? (b) a ring and a necklace?

In this question, you are not talking about conditional probability because all the specific sub-groups referred to are "symmetric."  In other words you the subgroups of wearing a ring or necklace have equal status.  This can be demonstrated in a venn box diagram:

\begin{array}
{|c|c|c|c} \hline & R & R' &  \\ \hline N& & &30\\ \hline N'& & 60&\\ \hline & 20 & & 100\\
\hline . \end{array}

You know the venn box diagram above is a magic table, so you can complete it as such:

\begin{array}
{|c|c|c|c} \hline & R & R' &  \\ \hline N& 10& 20&30\\ \hline N'&10 & 60 & 70\\ \hline & 20 & 80 & 100\\
\hline . \end{array}

So part (a) is asking:  $Pr(R \cup N)$.  By the Inclusion Exclusion Principle you know:

$Pr(R \cup N)=Pr(R) + Pr(N) - Pr(R\cap N)=20+30-10=40.$

part (b):  $Pr(R \cap N) = 10$ from the table above.

You see in both questions, they are just asking probabilities about specific groups.

A conditional probability problem

All bags entering a research facility are screened. Ninety-seven percent of the bags that contain forbidden material trigger an alarm. Fifteen percent of the bags that do not contain forbidden material also trigger the alarm. If 1 out of every 1,000 bags entering the building contains forbidden material, what is the probability that a bag that triggers the alarm will actually contain forbidden material?

In this case,  you can view the subgroups as a tree:

Observe that whether the Alarm gets setoff is a "subset" of F or F'.  In other words, there is a narrowing of specific subgroups within a group.  This is what makes this problem a conditional probability.


问:
Why is the dimension of U one less than $P_4(F)$?
Problem from Linear Algebra Done Right by Axler, Exercise 2.C.4a

Problem:

Let $U =$ { $p \in P_4(F) : p(6) = 0$}. Find a Basis of U.

The solution to 4a is found here: https://linearalgebras.com/2c.html

In particular, the solution shown for 4a I can get by taking a known basis of $P_4(F)$, $b_1=(1,x,x^{2},x^{3},x^{4})$.
Because of the condition of U, $p(x)$ must be divisble by $x-6$ and scalar multiplying it by $x-6 \in F$ produces $b_2=(x-6,x^{2}-6x,x^{3}-6x^{2},x^{4}-6x^{3},x^{5}-6x^{4})$. The fifth vector in $b_2$ is dropped because the degree of that polynomial is 5, and the fact that $p \in P_4(F)$ means the basis can only contain a polynomial of degree 4 or less.

Hence we now have a vector $b_3=(x-6,x^{2}-6x,x^{3}-6x^{2},x^{4}-6x^{3})$. Now if we prove linear independence and $span(b_3) = U$, we are done with 4a. I'm comfortable that this is a linearly independent set of vectors because they are all polynomials of different degrees, but what is not clear is how to prove $span(b_3) = U$.

Any help on intuition of why U is one degree less than $P_4(F)$ (as well as any problems with my work so far) is greatly appreciated.

答:
Assume $p\in U$. The key is what you mentioned, that this is equivalent to saying $(x-6)|p$ or, equivalently, $p=(x-6)q$ for some $q\in P_3(F)$.

Take the coefficients of $q$ to express $p$ in basis $b_3$.


问:
multivariate function multiplied by its partial derivatives
$$g\left[\frac{\partial g}{\partial x}\mathbf{i} + \frac{\partial g}{\partial y}\mathbf{j} + \frac{\partial g}{\partial z}\mathbf{k}\right] = \frac{\partial}{\partial x}\left(\frac{1}{2}g^2\right)\mathbf{i} + \frac{\partial}{\partial y}\left(\frac{1}{2}g^2\right)\mathbf{j} + \frac{\partial}{\partial z}\left(\frac{1}{2}g^2\right)\mathbf{k} $$

Could somebody explain this step? I believe it is chain rule but I am slightly lost in the jump.

答:
You are correct in thinking this is an application of the chain rule.  It's a bit easier to see going right to left.  Looking just at the first term:
$$\frac{\partial}{\partial x}\left(\frac{1}{2}g^2\right)\mathbf{i} = \frac{1}{2}\frac{\partial}{\partial x}g^2\mathbf{i} = \frac{1}{2}\cdot 2\cdot g\cdot \frac{\partial g}{\partial x}\mathbf{i} = g\frac{\partial g}{\partial x}\mathbf{i}.$$  Doing the same thing for each of the terms gives us
\begin{align}
\frac{\partial}{\partial x}\left(\frac{1}{2}g^2\right)\mathbf{i} + \frac{\partial}{\partial y}\left(\frac{1}{2}g^2\right)\mathbf{j} + \frac{\partial}{\partial z}\left(\frac{1}{2}g^2\right)\mathbf{k} &=g\frac{\partial g}{\partial x}\mathbf{i} + g\frac{\partial g}{\partial y}\mathbf{j} + g\frac{\partial g}{\partial z}\mathbf{k}\\
&=g\left[\frac{\partial g}{\partial x}\mathbf{i} + \frac{\partial g}{\partial y}\mathbf{j} + \frac{\partial g}{\partial z}\mathbf{k}\right].
\end{align}


答:
By using the substitution $u = \frac{t}{x}$, you get
$$\int^{3x}_x\frac{\sin(t)}{t^2}dt = \int^3_1\frac{\sin(ux)}{u^2x^2}\cdot x \cdot du = \int^3_1\frac{\sin(ux)}{u^2x}du$$
Now, since
$$\left|\frac{\sin(ux)}{u^2x}\right| \leq \left|\frac{1}{u^2x}\right|$$
which is integrable and bounded on $u \in [1, 3]$, you can use dominated convergence and swap the limit and the integral. For the inner limit, note that by using l'Hospital's rule it follows that
$$\lim_{x\rightarrow 0} \frac{\sin(ux)}{u^2x} = \lim_{x\rightarrow 0} \frac{u\cdot \cos(ux)}{u^2} = \frac{1}{u}.$$
Plugging this back into the integral yields:
$$\lim_{x\rightarrow 0} \int^{3x}_x \frac{\sin(t)}{t^2}dt = \int^3_1 \frac{1}{u}du = \ln(3) - \ln(1) = \ln(3)$$


问:
Let $G=\langle \mathbb R \times \mathbb R, + \rangle$ and $H=\{(x,y): x^2+y^2>0 \}$. Show that $H$ is a subgroup of $G$.
Let $G=\langle \mathbb R \times \mathbb R, + \rangle$ and $H=\{(x,y): x^2+y^2>0 \}$. Show that $H$ is a subgroup of $G$.

It appears that the identity element would be the ordered pair $(0,0)$. However that would imply that $0^2+0^2=0+0=0\not> 0$. So does this mean that $H$ is not a subgroup of $G$. I feel like this makes sense, but I also feel like I'm missing something and I don't understand the problem. Any solutions, hints, and advice is greatly appreciated.

答: As noted by David Scholz, you are correct that this implies that $H$ is not a subgroup of $G$. I'm not sure exactly what further insight you're looking for, but one thing to note is that given $G = (\mathbb{R}\times\mathbb{R},+)$ has identity element $(0,0)$, any subgroup of $G$ must have that same identity element. I bring this up only to clarify that in order to prove that a subset $H$ of a group $G$ is not a subgroup (under the same operation), it is sufficient to show that $H$ does not contain the identity element of $G$.


问:
Example of a covering map that lift some loop to loop but not all.
I want to find a covering map $p: X\rightarrow Y$ and two loops $f,g$ such that $f$ is lifted to a loop in $X$ but $g$ is not lifted to a loop in $X.$ So it seems like two loops $f$ and $g$ must have some different trait to make a different consequence. I was thinking about maybe $f$ can be a loop that can be deformed to a point but $g$ cannot? But I am not sure. Any help is appreciated.

答:
Yes, you're entirely right.  You could consider contractible and non-contractible loops lifting under a universal covering.  The "different trait" you reference is the following result:

${\bf Fact:}$ For a based covering $p \colon (\hat{X},\hat{b}) \to (X,b)$, a loop $l$ in $X$ lifts to a loop if and only if $[l] \in p^\ast\pi_1(\hat{X},\hat{b}) \leqslant \pi_1(X,b)$.


问:
Evaluating the line integral $\int_c = 4x^2ds$ from $(-2,-1)$ to $(1,2)$
Here is what I've done.

We know that to solve this we must evaluate $\int_c = f(x(t),y(t)\sqrt{(\frac{dx}{dt})^2}+(\frac{dy}{dt})^2dt$

So we parametrize our x and y values.

$r_0 = (-2,-1)$ and $r_1 = (1,2)$

$r(t) = ((1-t) \times (-2,-1)) + (t(-1,2)$

$= x = -2+2t-t$ and $y = -1+t+2t$

So $\int_c = f(x(t),y(t)\sqrt{(\frac{dx}{dt})^2}+(\frac{dx}{dt})^2 = $$\int_{0}{1} = f(4(-2+2t-t)\sqrt{(12x^2)^2}dt$

$=[4(-2+2t-t) \times (\sqrt{12x^2})^2]_01$

$=4[(-2+2(1)-1) \times (\sqrt{12(1)^2})^2 ] - 4[(-2+2(0)-0) \times (\sqrt{12(0)^2})^2 ]$

I'm not sure over which interval is the integral evaluated? Would it just simply be between 0 and 1? As this is what I've done, but not sure whether thats right.

Is my parametrization right? I know the formula is $(1-t) \times$ the starting point $+t \times$ the end point. But I get numbers as well and I'm not sure whether I've classified them right for x and y.

Body:
Segment from $(-2,-1)$ to $(1,2)$ has parametric equations

$C:(x=3 t-2,y=3 t-1), \;0\le t\le 1$

$ds=\sqrt{x'^2+y'^2}=\sqrt{9+9}\,dt=3\sqrt 2\,dt$

$$\int_C 4x^2\,ds=\int_0^1 4(3t-2)^2 \cdot 3\sqrt 2\,dt=12 \sqrt{2}$$


问:
Compactness of the moduli space of representations of fundamental group
Let $M$ be a compact manifold and $G$ a compact Lie group. I am trying to deduce that the moduli space of flat connections on $M$ is compact, and for that I have shown that there is a correspondence between flat $G$-connections modulo gauge equivalence and equivalence classes of representations of the fundamental group of $M$, but now I can't see why this space of representations is compact.

I do not have much background in representation theory or algebraic topology, so I am mostly looking for a reference for this result. If this is something that should be clear from more basic results (if I had the background), a sketch of the proof would also be appreciated.

答:
(I'm assuming $M$ is connected; the disconnected case is not much more complicated since there are only finitely many connected components. Also, presumably $M$ is a smooth manifold here so you can talk about flat connections.)

The moduli space of flat connections is $\text{Hom}(\pi_1(M), G)/G$ (quotienting by the conjugation action of $G$). $\text{Hom}(\pi_1(M), G)$ is a closed subspace of $G^{\pi_1(M)}$ (cut out by the axioms of a homomorphism) which is compact Hausdorff by Tychonoff's theorem, so $\text{Hom}(\pi_1(M), G)$ is also compact Hausdorff. Hence its quotient by the conjugation action of $G$ is also compact (but not necessarily Hausdorff).

As far as I can tell, this argument does not require that $M$ is compact. If $M$ is compact we learn more, though: in that case $\pi_1(M)$ is finitely presented so $\text{Hom}(\pi_1(M), G)$ can be cut out of a finite product of copies of $G$ by a finite number of relations, and we learn that the moduli space of flat connections is "finite-dimensional" and even "algebraic" in a suitable sense.

This follows, for example, from the fact that compact manifolds have the homotopy type of finite CW complexes (which in particular have finitely many $0$-cells, $1$-cells, and $2$-cells); see, for example, this MO question. For compact smooth manifolds we can instead use Morse theory but I'm not sure what the precise statements are.


问:
Determine whether the following definition defines a group action.
Let $G$ be a group and $\Omega$ be a nonempty set.

Determine whether the following definition of $\omega \cdot g$ (for $\omega \in \Omega$ and $g \in G$) defines a group action.

$\Omega = \{H \mid H \leq G\}$, and $\omega \cdot g = C_G(\omega)$

I have the following solution and am curious if anyone could explain it in further detail. Are we choosing an $\omega = H$ since $\Omega$ is the set of $H$ elements? Why does this break one of the conditions of $G$ acting on $\Omega$?

$\textbf{Definitions:}$ $G$ $\textbf{act}$ on $\Omega$ means $\Omega \neq \emptyset$. $G$ $\textbf{acts}$ on $\Omega$ means the following conditions hold:
(i) $\omega \cdot 1 = \omega$ $\forall \omega \in \Omega$
(ii) $(\omega \cdot g) \cdot h = \omega \cdot (gh)$ $\forall g,h \in G$ and $\omega \in \Omega$
where $\cdot$ is the binary operation on $\Omega$ and $G$ uses right multiplication.

If $G$ is non abelian, then for some $g \in G$, the set $C_G(g)=\{x \in G \mid xg=gx\}$ of all elements that commute with g. We call $C_G(g)$ the $\textbf{centralizer}$ of $g$ in $G$. Note: $C_G(g) \leq G$.

The $\textbf{centralizer of a set}$ $X \leq G$, $C_G(X) =\{y \in G \mid xy=yx \,\, \forall x \in X\}$. Thus, $C_G(X)=\bigcap\limits_{x \in X} C_G(x)$. So $X=G \implies C_G(G)$ is the $\textbf{center}$ of $G$ and denoted $Z(G)$.

$\textbf{Solution:}$ $H \cdot g = C_G(\omega)$. So, $H=H \cdot 1 = C_G(H)$.

$\therefore$ This definition does not define a group action.

答: You are missing neutrality of $1$, since $H$ may be not the same as its centralizer in $G$.


问:
If $Z\subset Y\subset X$ and $Z$ is closed in $Y$, then $Z=Y\cap\overline{Z}$
Suppose $X,Y,Z$ are topological spaces such that $Z\subset Y\subset X$ and $Z$ is closed in $Y$. Is it then true that $Z=Y\cap\overline{Z}$? (Here $\overline{Z}$ is the closure of $Z$ in $X$).

If $Z$ is closed in $Y$, then $Z=C\cap Y$ for some set $C$ closed in $X$. I'm not sure where to go from here though.

答:
Since $\def\Bar{\overline} Z=C\cap Y$ then we have $\Bar Z\subseteq \Bar C\cap \Bar Y=C\cap\Bar Y$, so
$$\Bar Z\cap Y\ \subseteq\ C\cap\Bar Y\,\cap\, Y\ =\ C\cap Y\ =\ Z\,,$$
and clearly $Z\subseteq \Bar Z\cap Y$.


问:
Is the area of a line = 1?
I tried to teach my son multiplication using a rectangle.
(e.g. 3cm * 4cm = 12cm^2).  Now I have 12 little squares.  But how do I explain where the "little square" came from?

My best guess: If I cut a little square in two, I can get a rectangle 2cm long * 1/2cm deep.  Then I cut that rectangle in two, I get 4cm long * 1/4 cm deep.  I keep going and say I get a rectangle 100,000cm long by 1/100,000 cm deep.

This is very long and very thin.  And I deduce, that the limit would be a line.  And so I conclude that the area of a line = 1.  (units become irrelevant, infinity takes care of that).

I also like this to explain why the integral of y=0x+1 (for x from 0 to 1) is 1.
Note: this should be a 1 unit square, and integrals are an area but we say the answer is 1 not 1 unit^2.

My question: Do you agree? (if not, why not?)

答:
Sometimes it is possible to formalize an intuition like this and sometimes it isn't.  In this case it isn't, because if you were to start with a $2 \times 2$ square instead of a $1 \times 1$ square and then perform the same process, the "limit" would be the exact same line.  Therefore your argument, if it were valid, would show that the same line has an area of 1 and also an area of 4.

The problem is that the function assigning area to figures is not continuous in the appropriate sense, so the area of the limit is not the limit of the areas.

Title: Is the area of a line = 1?


问:
Normal Bundle Orientable and inclusion null homotopic then M is orientable
I am trying to do the following exercise :

Let $M$ be a submanifold of $N$, both without boundary. Show that if the normal bundle of $M$ in $N$ is orientable and $i: M\rightarrow N$ is null-homotopic then $M$ is orientable.

I am bit lost and have no clue where to start. Since the normal bundle is orientable I guess proving the orientability of $M$ would be equivalent to that of $N$ but I am not sure this helps. Also I have no idea where to use the fact that $i: M\rightarrow N$ is null-homotopic, thought about using degree theory but this is only defined when we are working with orientable manifolds.

So we have a map $H:M\times [0,1] \rightarrow N$ such that $H(0,m)=i(m)$ and $H(1,m)=p$ for some $p\in N$. I have no idea what to do with this affirmation , I have thought about tubular neighborhoods , or doing the pull-back by the homotopy but none of this seems useful.

Any hint to get me started is appreciated , just trying to understand how to tackle this.

Thanks in advance.

答:
If $M$ is a submanifold of $N$ with inclusion $i$, then you have the isomorphism $TN|_M \cong TM \oplus \nu$ where $\nu$ is the normal bundle of $M$ in $N$ and $TN|_M$ is the restriction of $TN$ to $M$. Another way to write this last part is that $TN|_M = i^*(TN)$, the pullback of the tangent bundle of $N$ with respect to the inclusion map.

Now, using the fact that homotopic maps yield isomorphic pullback bundles, since $i \simeq c$ for $c$ a constant map, you have $i^*(TN) \cong c^*(TN)$ and hence  $c^*(TN) \cong TM \oplus \nu$. From here, you should try to show that $c^*(TN)$ is orientable (what's the pullback of a bundle by a constant map?) and that this together with the fact that $\nu$ is orientable will force $TM$ to be an orientable bundle, which is equivalent to orientability of $M$.


答:
If you have four linearly independent vectors in a four-dimensional space, they automatically span the space. The space is four-dimensional because it can't be five-dimensional (as then it would be all of $P_4(F)$).

The intuition is that you have a subspace of $5$-space cut out by a single linear equation, so the resulting space has dimension $5-1 = 4$ (it's a so-called hyperplane). This is analogous to the situation in three dimensions, where a subspace defined by a single equation $ax + by + cz = 0$ yields a plane, i.e., you lose one dimension. The same holds generically: if you have $k$ independent equations in $n$-space, then the subspace consisting of all solutions has dimension $n-k$.

Title: Why is the dimension of U one less than $P_4(F)$?

Body:
Problem from Linear Algebra Done Right by Axler, Exercise 2.C.4a

Problem:

Let $U =$ { $p \in P_4(F) : p(6) = 0$}. Find a Basis of U.

The solution to 4a is found here: https://linearalgebras.com/2c.html

In particular, the solution shown for 4a I can get by taking a known basis of $P_4(F)$, $b_1=(1,x,x^{2},x^{3},x^{4})$.
Because of the condition of U, $p(x)$ must be divisble by $x-6$ and scalar multiplying it by $x-6 \in F$ produces $b_2=(x-6,x^{2}-6x,x^{3}-6x^{2},x^{4}-6x^{3},x^{5}-6x^{4})$. The fifth vector in $b_2$ is dropped because the degree of that polynomial is 5, and the fact that $p \in P_4(F)$ means the basis can only contain a polynomial of degree 4 or less.

Hence we now have a vector $b_3=(x-6,x^{2}-6x,x^{3}-6x^{2},x^{4}-6x^{3})$. Now if we prove linear independence and $span(b_3) = U$, we are done with 4a. I'm comfortable that this is a linearly independent set of vectors because they are all polynomials of different degrees, but what is not clear is how to prove $span(b_3) = U$.

Any help on intuition of why U is one degree less than $P_4(F)$ (as well as any problems with my work so far) is greatly appreciated.


问:
Complete Cayley table for a field

Let $M=\{0,1,a,b,c\}.$  Can I complete the Cayley tables so that $M$ is a field?

My thought was that $(M,+)$ must be an abelian group. This abelian group must be isomorphic to $(\mathbb Z_5,+).$ Correct? I managed to find that if $a=3, b=4$ and $c=2$ it works for addition.

But this does not work for the multiplication table. Does that mean that I can´t complete the Cayley table so that $M$ is a field?

答:
I think you're right; this can't be a field of five elements.

If $a+c=0$ and $a\cdot a=c$, then $a=-c$, so $a\cdot a=-a$, so $a\cdot a+a=a(a+1)=0$,

so $a=0$, which doesn't make sense,

or $a=-1$, which also doesn't make sense because then $c=1$.

Title: Complete Cayley table for a field


问:
How can I derive the Legendre function of first kind in terms of the hypergeometric function?
I was reading in Wikipedia about Legendre's differential equation. I was particularly interested in the simple case of the equation given by
$$
\left(1-x^2\right)y'' -2xy' + \lambda(\lambda+1)y = 0 \tag{1}
$$
In the article, it is stated that one of the two linearly independent solutions to $(1)$ is given by
$$
P_{\lambda}(x) =  \,_2F_1 \left(-\lambda, \lambda+1; 1; \frac{1-x}{2}\right) \tag{2}
$$
where $\,_2F_1$ represents the hypergeometric function. Since by definition, the hypergeometric function ${}_2F_1(a,b;c;t)$ is a solution to the hypergeometric differential equation given by
$$
t(1-t) \ \ddot{y} + \left[c-(a+b+1)t \right] \dot{y} - ab\,y = 0 \tag{3}
$$
if $(2)$ is a solution to $(1)$ then I should be able to re-write $(1)$ to look like $(3)$ with the corresponding parameters. With this in mind, taking
\begin{align*}
a &= - \lambda\\
b &= \lambda +1\\
c &= 1\\
t & =  \frac{1-x}{2}
\end{align*}
From where we also get that
$$
\dot{y} = \frac{dy}{dt} =  \frac{dy}{dx} \cdot  \underbrace{\frac{dx}{dt}}_{\color{blue}{-2}} = -2y'
$$
which also tells us that
$$
\ddot{y} = -2 y''
$$
Substituting this into $(3)$ gives
\begin{align}
\frac{1-x}{2}\left(1-\frac{1-x}{2}\right)\ddot{y} + &\left[1-\left(-\lambda+(\lambda +1)+1\right)\frac{1-x}{2} \right] \dot{y} - (-\lambda)(\lambda +1)y = 0\\
\implies  \frac{1}{4}\left(1-x^2 \right)\left(-2 y''\right) +& \left[1-2\frac{1-x}{2} \right] \left(-2 y'\right) +\lambda(\lambda +1)y = 0\\
\implies \left(1-x^2 \right)y'' &+ 4x y' -2 \lambda(\lambda +1)y = 0 \tag{4}
\end{align}
and although $(4)$ looks really similar to $(1)$, they're not the same equation.

Could anyone tell me where did I make a mistake? Or alternatively, does anyone know another simple way to show that the Legendre function can be written in terms of the hypergeometric function? Thank you!

答: From Abramowitz and Stegun, 'Legendre functions are those special cases of the hypergeometric function for which a quadratic transformation exists.'  There is then a list of some twenty formulas, eq. 15.4.7-15.4.26.</p>


答:
Note that for a group $G$ and a set $A$, a group action is a function $\cdot: G \times A \to A$ (or $A \times G \to A$  such that it satisfies

1) $e \cdot a = a, \forall a \in A$

2) $g_1 \cdot (g_2 \cdot a) = (g_1g_2)\cdot a, \forall a \in A, g_1,g_2 \in G$

Now, in your example, we have a group $G$ and set $\Omega$, which is the set of all subgroups of $G$. And notice that 1) must hold for all elements in the set, on which the group acts. So, can we find a group where centralizer of every subgroup $H \le G$ is $H$, that is, $C_G(H) = H, \forall H \le G$?

If $G$ is trivial group, then there won't be any problem and we will indeed have a group action. But for the non-trivial case, one can easily show that the centralizer of trivial subgroup $\{e\}$ will be $G$ itself so we have $\{e\} \cdot e = C_G(\{e\}) = G \ne \{e\}$ since $G$ is non-trivial.

Title: Determine whether the following definition defines a group action.

Body:
Let $G$ be a group and $\Omega$ be a nonempty set.

Determine whether the following definition of $\omega \cdot g$ (for $\omega \in \Omega$ and $g \in G$) defines a group action.

$\Omega = \{H \mid H \leq G\}$, and $\omega \cdot g = C_G(\omega)$

I have the following solution and am curious if anyone could explain it in further detail. Are we choosing an $\omega = H$ since $\Omega$ is the set of $H$ elements? Why does this break one of the conditions of $G$ acting on $\Omega$?

$\textbf{Definitions:}$ $G$ $\textbf{act}$ on $\Omega$ means $\Omega \neq \emptyset$. $G$ $\textbf{acts}$ on $\Omega$ means the following conditions hold:
(i) $\omega \cdot 1 = \omega$ $\forall \omega \in \Omega$
(ii) $(\omega \cdot g) \cdot h = \omega \cdot (gh)$ $\forall g,h \in G$ and $\omega \in \Omega$
where $\cdot$ is the binary operation on $\Omega$ and $G$ uses right multiplication.

If $G$ is non abelian, then for some $g \in G$, the set $C_G(g)=\{x \in G \mid xg=gx\}$ of all elements that commute with g. We call $C_G(g)$ the $\textbf{centralizer}$ of $g$ in $G$. Note: $C_G(g) \leq G$.

The $\textbf{centralizer of a set}$ $X \leq G$, $C_G(X) =\{y \in G \mid xy=yx \,\, \forall x \in X\}$. Thus, $C_G(X)=\bigcap\limits_{x \in X} C_G(x)$. So $X=G \implies C_G(G)$ is the $\textbf{center}$ of $G$ and denoted $Z(G)$.

$\textbf{Solution:}$ $H \cdot g = C_G(\omega)$. So, $H=H \cdot 1 = C_G(H)$.

$\therefore$ This definition does not define a group action.


问:
How to show this fact?
For any event $A$, how to prove that bound:
$$\mathbb{P}(A) \leq \mathbb{P}(A|B) + \mathbb{P}(\overline{B}) ?$$

Thanks

答:
$\mathbb{P}(A)=\mathbb{P}(A|B)\mathbb{P}(B)+\mathbb{P}(A|B^c)\mathbb{P}(B^c)\leq\mathbb{P}(A|B)+\mathbb{P}(B^c)$

The "$\leq$" holds because $\mathbb{P}(B)\leq1$ and $\mathbb{P}(A|B^c)\leq 1$. Probability is always between $[0,1]$.

Title: How to show this fact?


问:
How to show in a clean way that $z^4 + (x^2 + y^2 - 1)(2x^2 + 3y^2-1) = 0$ is a torus?
How to show in a clean way that the zero-locus of $$z^4 + (x^2 + y^2 - 1)(2x^2 + 3y^2-1) = 0$$ is a torus? </p>

答: Not an answer, but here's a picture.  Cheers.


问:
When giving an angle measure in radians, should we explicitly state that it's in radians?
Let's say you are solving the equation $\arcsin(1) = x$ in radians. Would you write the answer as $\frac {\pi}2$ radians or just $\frac {\pi}2$? I see both methods being used, and I am aware that radians are a dimensionless quantity. However, it seems odd to just say, "The angle is $2$" rather than "The angle is $2$ radians."

答: People write $\arcsin1=\frac{\pi}{2}$, if only because any alternative approach would be unwieldy in more complicated expressions, such as if you evaluate $\int_0^x\arcsin tdt$ with integration by parts. But if you need to specify an angle in radians without it seeming like it's no angle at all, you could say $\frac{\pi}{2}\operatorname{rad}$. (You'll find other symbols here.)


答:
Yes, if you are describing some property of a geometrical object, like angle, it is more common to say the units in what angle is actually measured.

But when you solve an equation like
$$\arcsin(1)=x$$
you can just write $x=\frac{\pi}{2}$ since, firstly, everybody understand that it is not $\frac{\pi}{2}$ degrees(which is weird). And secondly, nobody uses degrees solving equations.

Title: When giving an angle measure in radians, should we explicitly state that it's in radians?

Body: Let's say you are solving the equation $\arcsin(1) = x$ in radians. Would you write the answer as $\frac {\pi}2$ radians or just $\frac {\pi}2$? I see both methods being used, and I am aware that radians are a dimensionless quantity. However, it seems odd to just say, "The angle is $2$" rather than "The angle is $2$ radians."


问:
Automorphism of tensor product of Hopf algebras
Let $A$ be a commutative Hopf algebra over the commutative ring $R$, that is we have comultiplication $\Delta:A\to A\otimes_{R}A$, counit $\epsilon: A\to R$ and coinverse (or antipode) $S: A\to A$. I want to prove that the map $a\otimes b\mapsto (a\otimes 1)\cdot \Delta(b)$ is an automorphism of $R$ algebras of $A\otimes_{R}A$.
I found a similar exercise on Milne's notes on affine group schemes, though my context is just that of commutative algebra.
My first idea was to prove directly injectivity and surjectivity, but I'm completely stuck. Can anyone give me some hints on which is the correct way to follow?

答:
You want to write down an inverse explicitly using the other operations.

Since $A$ is commutative $G = \text{Spec } A$ is an affine group scheme so we can think about this map in terms of the dual map $G \times G \to G \times G$ of affine group schemes, which is easier to think about. Its action on a pair of $S$-points $g, h : A \to S$ of $G(S)$, for any $R$-algebra $S$, sends $(g, h) \in G(S) \times G(S)$ to some other pair of points in $G(S) \times G(S)$, which we can calculate by restricting the given map to the first and then the second copy of $A$. Restricting to the first copy gives

$$a \mapsto (a \otimes 1) \Delta(1) = a \otimes 1$$

which reveals that the first component of our map is $(g, h) \mapsto g$. Restricting to the second copy gives

$$b \mapsto (1 \otimes 1) \Delta(b) = \Delta(b)$$

which reveals that the second component of our map is $(g, h) \mapsto gh$. Altogether, our map is

$$G(S) \times G(S) \ni (g, h) \mapsto (g, gh) \in G(S) \times G(S).$$

This is much easier to think about: its inverse is just $(g, h) \mapsto (g, g^{-1} h)$.


问:
Eating cake probability problem
Question: We want to divide a cake between Mike and Jack. First, Mike takes a portion of the cake. Then, Jack takes a piece of the remaining cake. Both of the pieces are random variables uniformly distributed over what is currently available. What is the probability that Jack gets more than half of the cake?

Here is my approach:

$X_M \sim Uni(0,1)$

$X_J \sim Uni(0,1)$

$P[ \ X_M(1- X_J) > 0.5 \ | \ X_J] = 1 - P[ \ X_M(1- X_J) \leq 0.5 \ | \ X_J]$

$ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 - F_{X_m}(\frac{0.5}{1- X_J} |
 X_J)$

$ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $

At this point I don't know how to proceed and I am not sure if I have modelled the problem correctly. Any help would be appreicated.

答:
If $X_M$ and $X_J$ are i.i.d. $U(0,1)$, then
$P(X_M(1-X_J) > 0.5)$ seems to be what you need; I am not sure why you are conditioning on $X_J$.

Since the joint distribution of $(X_M, X_J)$ is uniform on the square $[0,1] \times [0,1]$, you can compute the probability by computing the area of the region $\{(x,y) \in \mathbb{R}^2  : x(1-y) > 0.5\}$ that intersects the unit square $[0,1] \times [0,1]$. The region looks like this and the area can be computed with an integral.

Title: Eating cake probability problem


问:
Prove that the ring is Artinian and find a composition series of ideals
I came across the following problem

Let $k$ be a field and $R=k[t^2,t^3]/(t^4)$. The ring $k[t^2,t^3]$ is defined as the set $\{f(t^2,t^3)|f \in k[x,y]\}$ with the obvious ring structure.

prove that the ring R is Artinian

find ideals $A_0, \dots , A_n \unlhd R$ such that $$R=A_0 \supseteq \cdots \supseteq A_n$$ with each $A_i/A_{i+1}$ be a simple $R$ module.

for the first part, I know that in a short exact sequence of R-modules, the middle R-module is Artinian iff the two others are.

I thing that $R \simeq k[x,y,z]/(x-z^2, y-z^3, z^4)$ but I don't now how is this helpful.

for the second part I don't have a single clew how to proceed.

Any hints, useful remarks or theorems that may help are appreciated.

答:
If we denote by $t^n$ the class of $t^n$ in $R$, the elements of $R$ can be written as $a_0+a_2t^2+a_3t^3+a_5t^5$, with $a_i\in{k}$, and a composition series is $0\subset(t^5)\subset(t^2)\subset(t^2,t^3)$.

Observe that $R$ is not a principal ideal ring; indeed the ideal $(t^2,t^3)$ is not principal.

Otherwise, the ring $k[x,y,z]/(x-z^2,y-z^3,z^4)\cong{k(z)/(z^4)}$ is a principal ideal ring; so it is not isomorphic to $R$.


问:
$\lim f'(x) = l$
Let $f : \left]0, +\infty\right[ \to \mathbb R$ a differentiable and bounded function such that
$$\lim_{x \to +\infty} f'(x) = l$$
Show that $l = 0$.

My attempt is the following :
Suppose that $l>0$, then $\forall \epsilon > 0, \exists x_0 > 0, \text{ such that } \forall x>x_0, \lvert f'(x) - l \rvert \leq \epsilon $.
Taking $\epsilon = \frac{l}{2}$, we have that $\frac{l}{2} \leq f'(x)$ $\forall x > x_0$.

Then by the mean value theorem, we can say that $\forall x > x_0, f(x) \geq \frac{l}{2}(x-x_0) + f(x_0)$.

Hence, $\lim_{x \to +\infty}f(x)= +\infty$.
I guess we can do the same for $l < 0$ and deduce that $\lim_{x \to +\infty}f(x)= -\infty$. So if $f$ is bounded then $l=0$.

Is it correct ?
Is there any direct proof ?

答:
That is correct. If $\lim_{x \to +\infty} f'(x)$ exists and is not zero then $f$ diverges to $+\infty$ or $-\infty$ for $x \to \infty$ and is not bounded.

As an alternative (direct) proof you can apply the mean-value theorem to intervals $[n, 2n]$ with positive integers $n$:
$$
 \frac{f(2n)-f(n)}{n} = f'(c_n)
$$
with some $c_n \in (n, 2n)$. Then note that for $n\to \infty$, the left-hand side converges to zero (because $f$ is bounded) and the right-hand side converges to $\lim_{x \to +\infty} f'(x) = l$. It follows that $l=0$.

Title: $\lim f'(x) = l$


问:
Initial value problem with $t$ and $x'$
Really stuck with this,

\begin{cases}
 2tx'(t)-1=x^2t \\
 x(t=1)=1 \\
 \end{cases}

I don't understand the variables. What does it mean that $x$ differentiated have $t$ as a function value, like this: $x'(t)$? Don't get it :(

答:
$x'(t)$ simply means the derivative of the function $x$ with respect to $t$. It means exactly the same thing as
$$\frac{dx}{dt}$$

Title: Initial value problem with $t$ and $x'$


问:
Existence and the value of $\lim_{x \to 0} \int^{3x}_x \frac{\sin t}{t^2}dt$
I need the show the existence of the following limit and then calculate the limit

$$\lim_{x \to 0} \int^{3x}_x \frac{\sin t}{t^2}dt$$

Since the antiderivative of $\frac{\sin t}{t^2}$ was not nice, I tried to use the approximation $\sin x \approx x$ for $x$ close to $0$. Then, I can integrate and find the limit as $\ln3$.

Is this a valid solution and can I solve this without using such approximation?

答:
MVT for integrals

$((\sin s)/s)\displaystyle{\int_{x}^{3x}}(1/t)dt=$

$((\sin s)/s)[\log 3+\log x -\log x] ;$

where $s \in [x,3x].$

Note $\lim x \rightarrow 0$ implies $\lim s \rightarrow 0$.

Take the limit.


问:
multivariate function multiplied by its partial derivatives
$$g\left[\frac{\partial g}{\partial x}\mathbf{i} + \frac{\partial g}{\partial y}\mathbf{j} + \frac{\partial g}{\partial z}\mathbf{k}\right] = \frac{\partial}{\partial x}\left(\frac{1}{2}g^2\right)\mathbf{i} + \frac{\partial}{\partial y}\left(\frac{1}{2}g^2\right)\mathbf{j} + \frac{\partial}{\partial z}\left(\frac{1}{2}g^2\right)\mathbf{k} $$

Could somebody explain this step? I believe it is chain rule but I am slightly lost in the jump.

答:
The $$\frac{\partial g}{\partial x}\cdot \vec{i} + \frac{\partial g}{\partial y} \cdot \vec{j} + \frac{\partial g}{\partial z} \cdot \vec{k}$$
is called the gradient of a function. It is a vector contains partial derivatives as components.

So, your question was, why this equality holds $$g\cdot \frac{\partial g}{\partial x}=\frac{\partial (\frac{1}{2}g^2)}{\partial x}?$$

Let's find the derivative of $\frac{1}{2}g^2$:
$$
\frac{\partial (\frac{1}{2}g^2)}{\partial x}=\frac{\partial (\frac{1}{2}g^2)}{\partial g} \cdot \frac{\partial g}{\partial x}=g \cdot \frac{\partial g}{\partial x}
$$
(as you have mentioned, chain rule is useful here). And we are done!


问:
$f(f(f(x)))=x$, does it necessarily imply $f(x)=x$?

[TIFR GS-2017] Let $f$ be a continuous function on $\mathbb R$ satisfying the relation
$$f(f(f(x))) = x\ \forall x \in \mathbb R$$ Prove or disprove that $f$ is the identity function.

There is a solution to this question in here, but I was trying in this way...

Suppose $f$ is continuous and not identity function. Then we can find some point $c \in \mathbb R$ such that there is a neighbourhood $N_\delta(c)$, where $f(x) \neq x\ \forall x \in \mathbb N_\delta(c)$.

Then two cases arise.

$\underline{Case (i)}:$ $f(f(x))=f(x)$, but that means $f(f(x))=f(x)\neq x$, hence $f(f(f(x))) = f(f(x)) =f(x) \neq x$.

Hence we have a contradiction.

$\underline{Case (ii)}:$ $f(f(x)) \neq f(x)$.

Suppose $p$ be such a point where it holds. Then $f(f(p)) \neq f(p)$. Suppose $m \in \mathbb R$ s.t. it is in between $f(f(p))$ and $f(p)$. Then by IMVP, there is some $q$ between $p$ and $f(p)$ s.t. $f(q) = m$.

After this scene, I am out of ideas to proceed further. We can't say that $f(p) \in N_\delta(c)$. Any help would be appreciated.

答:
Continuity is important

As the others pointed out, the requirement that the function f is continuous is very important here.

If we formally answer not the question in your text, "Let f be a continuous function ...", but the question in the title, where continuity is not mentioned, then many solutions are possible.

Here is an example. Unlike other examples, where function is undefined in some points, following function is defined for every value of x:
$$f(x) = floor(x) + frac(frac(x) + 1/3)$$

For any x following holds:
$$f(f(f(x))) = x$$

But this function is not continuous and thus does not prove anything.

Proof of not existence

You wanted to have a proof based on contradiction. It is possible, but not exactly the way you started. First I'd suggest to show that if f(x) exists, then it must be monotone. Then you can consider separate cases how f(x) can be located relative to y(x) = x. There are only a few types possible. If you look at each case closer, you will come to contradiction.


问:
Solve Diophantine equation: $2^x=5^y+3$ for non-negative integers $x,y$.
Today my friend asked a question for help:

Find all solutions of $2^x=5^y+3$ for non-negative integers $x,y$.

It is obvious that the solutions are $(x,y)=(2,0),(3,1),(7,3)$, and I think there is no more solution. However, we can't prove that these are the only solutions. We have already tried to mod many numbers and still get "may" possible solutions other than the $3$ solutions I have written. I have surfed the Net and still can't find solutions. I hope you guys can help my friend solve. Thank you very much!

答:
From WolframAlpha here we get

$$x = \frac{\log{5^y + 3}}{\log{2}}\quad\text{for}\quad
 5^y\in\mathbb{Z}\quad\land\quad5^y >=-2$$
The restrictions are overkill but simply indicate $y\in\mathbb{Z}$

Applying this in a spreadsheet for $-15000\le y \le 15000$ we get only $3$ pairs.

$$(2,0)\quad
(3,1)\quad
(7,3)\quad$$


问:
If $T$ is a linear operator. Prove that this 3 conditions are equivalent
I have to prove this problem. I would like you to check if my proof is correct (or not) please.

Let $V$ be a vector space over $\mathbb{R}$ and $T$ a linear operator
on $V$. Prove that the following conditions are equivalent:

$T^{2}= \text{id}_V$,

$V$ is the direct sum of the null space (kernel) of $T- \text{id}_V$ and the null space of $T+ \text{id}_V$,

there exist two subspaces, $W$ and $X$, of $V$ such than $V=W \oplus X$ and $T(w+x)=w-x$ for all $w \in W$ and all $x \in X$.

Here, my attempt:

3 $\Longrightarrow$ 1

$\forall \alpha \in V$, we have that $\alpha=w+x $ with $ w\in W$, and $x\in X$

\begin{align}
\text{If} \ \ \ \  T(\alpha)=T(w+x)&=w-x\\ \Rightarrow \ \ \ T^{2}(\alpha)=T^{2}(w+x)&=T(T(w+x))\\&=T(w-x)\\&=T(w+(-x))\\&=w-(-x)\\&=w+x\\&=\alpha \\ \therefore T^{2}=id_{V}
\end{align}

1 $\Longrightarrow$ 2

Firt, we have to see this
\begin{align}
T^{2}=id_{V} &\iff (T^{2}-id_{V})=0\\&\iff(T-id_{V})(T+id_{V})=0\\&\iff (T-id_{V})=0 \ \ \ \ \text{or} \ \ \ \  (T+id_{V})=0\\&\iff \alpha \in N(T-id_{V}) \ \ \ \ \text{or} \ \ \ \ \alpha \in N(T+id_{V})
\end{align}
Now, we suppose that any $\alpha \in V$ is of the form $\alpha=w+x$ with $w\in N(T-id_{V})$, and $x \in N(T+id_{V})$.
\begin{align}
T^{2}(\alpha)=T^{2}(w+x)=T^{2}(w)+T^{2}(x)=0 & \iff \alpha =0 \\& \iff w=x=0\\
\end{align}

\begin{align}
\therefore V=N(T-id_{V}) \oplus N(T+id_{V})
\end{align}

2 $\Longrightarrow$ 3

First, we have that $V=N(T-id_{V}) \oplus N(T+id_{V})$

$\Longrightarrow$ There exists two subspaces, $W,X$ such that $V=W \oplus X$

Now, if $w \in N(T-id_{V})$

\begin{align}
T(w)-id_{V}(w)= 0 \iff T(w)=w
\end{align}

And, if $x \in N(T+id_{V})$

\begin{align}
T(x)+id_{V}(x)=0 \iff T(x)=-x
\end{align}

By last, any vector $\alpha \in V$ is of the form $\alpha=w+x$ (because of the direct sum)

\begin{align}
\therefore T(\alpha)=T(w+x)=w-x
\end{align}

I hope you can help me to check this, and I really really appreciate your help.

答:
$1\implies2$ doesn't look correct. $(T-I)(T+I)=0\not\Rightarrow T-I=0\vee T+I=0$. Remember that the product of two non-zero matrices may be zero matrix.

First you need to show that $\ker(T-I)\cap\ker(T+I)=\{0\}$. Suppose $v\in\ker(T-I)\cap\ker(T+I)$, then $(T-I)v=0=(T+I)v$. This gives $Tv=v=-v\iff v=0$.

Next you need to show that every $\alpha\in V$ can be written as the sum of some $v^-\in\ker(T-I)$ and $v^+\in\ker(T+I)$. From $T^2\alpha=I\alpha$ we get $(T-I)(T+I)\alpha=0$. Thus $(T+I)\alpha=u_1\in\ker(T-I)$ which gives$$\alpha=u_1-T\alpha$$Similar we also have $(T+I)(T-I)\alpha=0$. Thus $(T-I)\alpha=u_2\in\ker(T+I)$ which gives$$\alpha=T\alpha-u_2$$Adding the two equations we get$$\alpha=\frac{u_1-u_2}2$$where $u_1/2=v^-\in\ker(T-I)$ and $-u_2/2=v^+\in\ker(T+I)$.


问:
Write the integral $\int_0^\infty \left(\frac{1}{1 + x^2} \right)^\alpha x^\beta dx$ in terms of the Euler Beta Function?
How can I write the integral $$\int_0^\infty \left(\frac{1}{1 + x^2} \right)^\alpha x^\beta dx$$
with $\alpha, \beta >0$, in terms of the Euler Beta Function? For which $\alpha, \beta$ does the integral actually converge? Maybe some change of variable can do the trick?

答: With $x=\tan t$ the integral becomes$$\int_0^{\pi/2}\sin^\beta t\cos^{-2\alpha-\beta-2}t\mathrm dt=\tfrac12\operatorname{B}\left(\tfrac{\beta+1}{2},\,\alpha-\tfrac{\beta+1}{2}\right)$$(if this evaluation isn't obvious, use $u=\sin^2t$). I'll leave you to deduce convergence conditions.


问:
Finding a suitable comparison to find the limit of $(5n!+5)/(n^{100}+n!)$ as $n \to \infty $
I'm struggling to find a suitable comparison for this limit. Could you help me out with your thought process?</p>

答:
I would just rearrange a little,
\begin{align*}
\lim_{n\rightarrow \infty} \frac{5!+5}{n^{100}+n!} &= \lim_{n\rightarrow \infty} \left( \frac{5n!}{n^{100}+n!} + \frac{5}{n^{100}+n!}\right) \\
&= \lim_{n\rightarrow \infty} \frac{5n!}{n^{100}+n!} + \lim_{n\rightarrow \infty}\frac{5}{n^{100}+n!}\\
&= \lim_{n\rightarrow \infty} \frac{5}{n^{100}/n! + 1} + 0 \qquad\qquad \text{(divide by n! on top and bottom)}\\
&=  \frac{5}{0+1} = 5
\end{align*}
so that the limit is $5$. Nottice that I'm assuming your notation means $5(n!)$, and not $(5n)!$, and I'm also using that $\lim_{n\rightarrow \infty} n^{100}/n! = 0$, which you could use some easier test to prove.


问:
showing the direct sum of an endomorphism $f$ where $f$ is a projection $f^2=f$
Endomorphism $f$ of vector space $V$ is a projection such that $f^2=f$. Show that $f=f_0 \oplus f_1$ where $f_0$ is the zero-map on $E_0$ and $f_1$ is the identity map on $E_1$.

Attempt thus far:

I know that first I have to show that $V$ is the direct sum of $E_0$ and $E_1$.

$E_1=ker(f-id_v)=im(f)$ and $E_0=ker(f)$

To prove $V=im(f)\oplus ker(f)$ I have to show that:

$im(f)+ker(f)=V$

$im(f) \cap ker(f)=\{0\}$

I know how to do all this but where I get stuck is understand how $f=f_0 \oplus f_1$

If anyone can explain this in a clear way, I'd much appreciate this.

答:
You need to do the following (in order of increasing difficulty):

Show that $f|_{E_0} = f_0$

Show that $f|_{E_1} = f_1$

Show that $E_0 \cap E_1 = 0$. In other words, if $(f - \operatorname{id}_V)x = 0$ and $fx = 0$, then it must hold that $x = 0$.

Show that $V = E_0 + E_1$. That is, show that for every $x \in V$, there are vectors $x_0 \in E_0$ and $x_1 \in E_1$ for which $x = x_0 + x_1$. Hint: take $x_1 = f(x)$.

Facts 3 and 4 imply that $V = E_0 \oplus E_1$. From there, the fact that $f|_{E_0} = f_0$ and $f|_{E_1} = f_1$ allow you to conclude that (by definition of the direct sum of two linear maps) $f = f_0 \oplus f_1$.


问:
UMVUE of $\lambda$ in $\operatorname{Exp}(\lambda)$ distribution
Suppose $(X_1,...,X_n)$ is a sample from a $Exp(\lambda)$ population. Try to find an UMVUE of $\lambda$. Remember $S = \sum_{i=1}^n X_i$ is both complete and sufficient for the Exponential Family.

I'm trying to use the Lehmann-Scheffé theorem, but I can't manage to get it done easily. I also tried to use Laplace-inverse transforms but all I get is a Dirac Delta function from WolframMathematica.

答:
Well, if you find a statistic $T=t(S)$ that is a function  of $S=\sum_{i=1}^{n} X_i$ and which is an unbiased estimator of $\lambda$, you've got your UMVUE by Lehmann-Scheffe.

For $T$ to be an unbiased estimator we want: $E[(t(S))]=\lambda$, of course we know that for a $X\tilde{} exp(\lambda)$ we already have $E[X]=\lambda$ and in general $E[\bar X]=E[X]$.

So, we see that $\bar X$ is an unbiased estimator of $\lambda$ and because $\bar X$ is an one-to-one function of $S$ (by $\bar X=S/n$), we're done.


答:
To make the algebra a little easier, let $X$ be the fraction that Mike doesn't eat (which is still $\mathcal U[0,1]$) and $Y$ the fraction that Jack eats. We want $\mathbb P(XY>1/2)$. Now do the integral:
$$\mathbb P(XY>1/2)=\int_0^1\mathbb P(Y>1/2x)\cdot f_X(x)\mathop{}\!\mathrm{d}x=\int_{1/2}^1\left(1-\frac{1}{2x}\right)\mathop{}\!\mathrm{d}x=\frac{1-\log2}{2}.$$

Title: Eating cake probability problem

Body:
Question: We want to divide a cake between Mike and Jack. First, Mike takes a portion of the cake. Then, Jack takes a piece of the remaining cake. Both of the pieces are random variables uniformly distributed over what is currently available. What is the probability that Jack gets more than half of the cake?

Here is my approach:

$X_M \sim Uni(0,1)$

$X_J \sim Uni(0,1)$

$P[ \ X_M(1- X_J) > 0.5 \ | \ X_J] = 1 - P[ \ X_M(1- X_J) \leq 0.5 \ | \ X_J]$

$ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 - F_{X_m}(\frac{0.5}{1- X_J} |
 X_J)$

$ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $

At this point I don't know how to proceed and I am not sure if I have modelled the problem correctly. Any help would be appreicated.