答:
You try to conclude that $\inf S=4$ by a wrong argument.

Using $m$ to denote $\inf S$ is confusing, so I'll use $\inf S$. You claim that if $r$ is any rational number such that $4<r<\inf S$, then $r\in S$.

This is to be proved, not simply stated (and I think it's even false, because no rational number with even numerator and odd denominator can be written in the form $\frac{m^2+4n^2}{mn}$).

The infimum is actually a minimum, because choosing $m=2$ and $n=1$ you have
$$
\frac{m}{n}+\frac{4n}{m}=\frac{2}{1}+\frac{4}{2}=4
$$

The part about the supremum is correct, but I'd avoid limits: if you fix any $K$, choose a natural number $m$ such that $m>K$. Then
$$
K<\frac{m}{1}+\frac{4\cdot1}{m}\in S
$$
and so $S$ is not upper bounded.

Title: Find the supremum and infimum (if exist) of the set $\{\frac{m}{n} + \frac{4n}{m} \mid m,n \in \Bbb N\}$ and $\{x:x^2+x+1>0\}$.

Body:
Find the supremum and infimum (if exist) of the set
$S = \{\frac{m}{n} + \frac{4n}{m} \mid m,n \in \Bbb N\}$ and $T = \{x:x^2+x+1>0\}$.

Attempt:
First, notice that $S = \{x \in \Bbb R \mid x \gt 4\}$.

For the set $S$, by $AM-GM$ inequality, we have
\begin{align*}
\frac{m}{n} + \frac{4n}{m} &\ge 2\sqrt{\frac{m}{n} \cdot \frac{4n}{m}} = 2\cdot \sqrt{4} = 2\cdot 2 = 4.
\end{align*}
Thus, $4$ is a lower bound of $S$. We claim that $\inf(S)=4$. Let $m$ be an any lower bound of $S$.
We'll show that $m \le 4$. For the sake of contradiction, suppose $m \gt 4$. Let $r$ be an arbitrary rational number such that $4 < r < m$. Then, we have $r \in S$ and $m \gt r$. This is impossible since $m$ is a lower bound of $S$. Hence, $m \le 4$ and therefore, $\inf(S) = 4$.

Next, set $n=1$ and $m \to \infty$. Then, $\frac{m}{n} + \frac{4n}{m} \to \infty$.
Hence, $\sup(s)$ is doesn't exist.

Am I true?

${}$

For the set $T$, notice that $x^2+x+1$ has two distinct complex roots. How to process next?
What if $x$ there is in real set? What if $x$ there is in complex set?

Thanks in advanced.


问:
Stuck on last step showing that $\mathbb{R}/\mathbb{Z}$ is homeomorphic to $S^1$, and stating what the homomorphism is.
Let $q:\mathbb{R}\to\mathbb{R}/\mathbb{Z}$ be the quotient map, and let $\phi:\mathbb{R}\to S^1$ be the map given by $\phi(\theta)=(\cos(2\pi\theta),\sin(2\pi\theta)).$

I proved in a previous problem that this will induce a unique continuous bijection, $\hat{\phi}:\mathbb{R}/\mathbb{Z}\to S^1$ that satisfies $\phi = \hat{\phi}\circ q$. I've already shown that the mapping $$[\theta]\mapsto (\cos(2\pi\theta),\sin(2\pi\theta))$$ satisfies this property, and thus this mapping must be equal to $\hat{\phi}$.

So what I've currently shown is that $\hat{\phi}([\theta]) = (\cos(2\pi\theta),\sin(2\pi\theta))$ is a well defined continuous bijection between the quotient and $S^1$. But to show that this is a homeomorphism, I need to show that its inverse is continuous, and that's where I'm stuck
I wanted to say something like;

Since we need to prove the inverse is continuous, we pick some open $U$ in $\mathbb{R}/\mathbb{Z}$, then we need to show that $\hat{\phi}^{-1}(U)$ is open in $S^1$, but I can't see why this should be true. I'd like to use the openness of $U$ in $\mathbb{R}/\mathbb{Z}$ along with $q$ to move this into an open set in $\mathbb{R}$, then apply $\phi$ and use its continuity to say something about how $U$ maps back into $S^1$ in under $\hat{\phi}$. But I can't quite get that to work; unfortunately the image of an open set isn't open under a continuous function, so I'm stuck on this last part.

Any thoughts on how to prove that $\hat{\phi}^{-1}$ is continuous would be greatly appreciated.

Thanks in advance.

答:
I think conceptually the best approach is to show that $\phi : \mathbb R \to S^1, \phi(t) = e^{2 \pi i t}$, is a quotient map. Then the following applies:

Let $f : X \to Y$ and $g : X \to Z$ be quotient maps and $h : Y \to Z$ be a bijection such that $h \circ f = g$. Then $h$ is a homeomorphism.

Let us show that $\phi$ is an open map (which implies that it is a quotient map). Since the open intervals $(a,b)$ form a basis for the topology on $\mathbb R$, it suffices to show that all $\phi((a,b))$ are open in $S^1$.

This is trivial for $b - a > 1$ because then $\phi((a,b)) = S^1$.

For $b - a \le 1$ the closed interval $[a,a+1]$ contains $(a,b)$. The set $K = [a,a+1] \setminus (a,b)$ is compact, hence $\phi(K) \subset S^1$ is compact, thus closed in $S^1$. Therefore $S^1 \setminus \phi(K)$ is open in $S^1$.  We have $S^1 = \phi([a,a+1]) = \phi(K \cup (a,b)) = \phi(K) \cup \phi((a,b))$. But $K$ and $(a,b)$ are disjoint, thus $s \in K$ and $t \in (a,b)$ cannot have the same image under $\phi$ (note that the only two distinct points in $[a,a+1]$ having the same image under $\phi$ are $a$ and $a+1$ which are both contained in $K$). We conclude that $\phi(K)$ and $\phi((a,b))$ are disjoint, hence $\phi((a,b)) = S^1 \setminus \phi(K)$ which is open in $S^1$.

See also the following related questions:

Is it an open function?

Open sets on the unit circle $S^1$

ContentLicense: CC BY-SA 4.0


问:
Is every projective variety the zero set of a system equations which are homogeneous polynomials of the same degree?
Is every projective variety the zero set of a system equations which are homogeneous polynomials of the same degree?

答:
Yes, you can assume that all defining equations have the same degree! Here is the trick:

Suppose that the variety $V\subset \mathbb P^n_k$ is defined by $s$ homogeneous polynomials $f_i(x_0,\cdots,x_n)\in k[x_0,\cdots,x_n]$
of  degrees $\deg f_i=d_i$, namely $$V=Z(f_i\vert i=1,\cdots,s)$$ Now let $d=\max d_i$. The trick is then to realize that each equation $f_i=0$ is equivalent to the system of $n+1$ equations $$f_i\cdot x_j^{d-d_i}=0 \;(j=0,\cdots,n)$$ We thus see that $V$ is defined by the system of $s(n+1)$ equations $f_i\cdot x_j^{d-d_i}=0$ all of degree d:
$$V=Z(f_i\cdot x_j^{d-d_i}\vert i=1,\cdots17\;;j=0,\cdots,n)$$
(And note that this trick  works whether "irreducible" is part of your definition of variety or not.)


问:
Spectral radius greater than 1
The eigenvalues of the matrix $$A=\begin{pmatrix}0 & 1/2 & -1 \\ -1/2 & 0 & 1 \\ -1 & -1 & 0\end{pmatrix}$$ are relatively complicated (roots of the polynomial $x\mapsto x^3+x/4+1$). Is there an easy way to see that the spectral radius of this matrix satisfies $\rho(A)>1$, without explicitly calculating the eigenvalues?

答: Suppose $\rho(A)\le1$ so $|\lambda_i|\le1$ for all $i$. But the product of eigenvalues is $-1$ i.e. $|\lambda_1||\lambda_2||\lambda_3|=1$. This means that all eigenvalues have unit modulus, and in particular the one real eigenvalue is either $1$ or $-1$. Can you check that $-1,1$ is not an eigenvalue?

Title: Spectral radius greater than 1


问:
Why $L(f,P_n) \leq \frac{b^2}2 \leq U(f, P_n)$ for $f(x)=x$?
I'm reading the Integrals chapter in Spivak's Calculus, he's taking integrals using lower and upper sums for equal intervals, often saying that it is clear that $L(f,P_n) \leq a \leq U(f, P_n)$ where $a$ is the limit to which the upper and lower sums are approaching as $n$ becomes very large, $n$ is the number of equal intervals.

For example, for $f(x)=x$, he's finding explicit formulas for the sums using the Gauss sum formula:

$$L(f,P_n)=\frac{n-1}n\cdot \frac{b^2}{2}$$

$$L(f,P_n)=\frac{n+1}n\cdot \frac{b^2}{2}$$

both are approaching $\frac{b^2}{2}$ for large $n$.

Then he says:

It is clear, first of all, that

$L(f,P_n) \leq \frac{b^2}2 \leq U(f, P_n)$

How is it clear?

I've checked these particular functions for both sums, lower is increasing and upper is decreasing, hence the inequality is indeed true. Is that what he meant by clear or I'm missing something?

答:
If we take the uniform subdivision $P_n$ of the interval $[0,b]$, i.e.
$$0<\frac{b}{n}<\frac{2b}{n}<\dots<\frac{(n-1)b}{n}<b$$ then
$$L(f,P_n)=\sum_{k=0}^{n-1}\frac{kb}{n}\cdot \frac{b}{n}=
\frac{b^2}{n^2}\sum_{k=0}^{n-1}k=\frac{b^2}{n^2}\cdot\frac{n(n-1)}{2}=\frac{n-1}{n}\cdot\frac{b^2}{2}.$$

Then it is clear that the inequality $n-1<n$ implies
$$L(f,P_n)< \frac{b^2}{2}.$$
Similarly, the inequality $n+1>n$ implies
$$U(f,P_n)=\sum_{k=1}^{n}\frac{kb}{n}\cdot \frac{b}{n}=\frac{b^2}{n^2}\cdot\frac{n(n+1)}{2}=\frac{n+1}{n}\cdot\frac{b^2}{2}>\frac{b^2}{2}.$$


问:
How do I show that the binary representation on the real line is uncountable?
How do I show that the binary representation on the real line is uncountable? Thanks.</p>

答: It sounds like you're asking to show that the set of all infinitely long binary strings is uncountable. This can be seen intuitively by considering a string $x=x_1x_2\ldots$ where $x_i \in \{0,1\}$. The letters in this string have an obvious bijection to $\mathbb{N}$, taking $1 \to x_1$, $2 \to x_2$ and so on (so there are countably many characters in this string). Then, we have $2$ options for each position in the string, meaning there are $2^\mathbb{N}$ possible infinite binary strings which is uncountable by Cantor diagonalization. </p>

ContentLicense: CC BY-SA 3.0


问:
$x^3 + y^3 +3x^2 y^2 =x^3y^3$ Find all possible values to $\frac{x+y}{xy}$
$x,y \in \mathbb{R}\setminus\{0\}$

Indeed the original question said :Find all possible values to: $$\frac{1}{x} + \frac{1}{y}$$
But it’s the same thing.

My Attempt:
$$x^3 + y^3 +3x^2 y^2 =x^3y^3 \iff (x+y)(x^2-xy+y^2)=x^3y^3 -3x^2y^2$$
$$\frac{x+y}{xy}=\frac{x^2y^2-3xy}{x^2-xy+y^2}$$

But i don’t know what to do now.
Please if you know the key to this type of problems ‘Find all possible values’ post it.

Body:
Hint:

Let $\dfrac{x+y}{xy}=k$

$$(xy)^3=3(xy)^2+(x+y)^3-3xy(x+y)$$

$$\iff(xy)^3(1-k^3)-3(xy)^2+3xy(kxy)=0$$

$$\iff(xy)^3(1-k^3)+3(xy)^2(k-1)=0$$

As $xy\ne0$ $$(xy)(1-k^3)=3(1-k)$$

Clearly, $k=1$ is a solution.

Otherwise,  $$xy=\dfrac3{1+k+k^2}$$


问:
Embedding of separable $C^*$-algebra
Let $A$ be a separable $C^*$-algebra. Can $A$ embedded into a separable, simple $C^*$-algebra?

答:
In B. Blackadar, Weak expectations and nuclear C -algebras. Indiana Univ.
Math. J. 27 (1978), 1021-1026. Blackadar shows that every separable subalgebra of a simple $C^*$-algebra is contained is a simple separable $C^*$-subalgebra of $A$. In particular, every separable $C^*$-algebra can be embedded into a simple separable $C^*$-algebra (since it can be embedded into the Calking algebra).

Title: Embedding of separable $C^*$-algebra


问:
A subgroup containing a kernel of a group homomorphism into an abelian group is a normal subgroup.
Let $ f \colon G \rightarrow H $ be a group homomorphism, where $ H $ is an abelian group.

If $\ker f \subset N $ for some $ N \subset G $, then $ N $ is a normal subgroup of $ G $.

I don't know how to start to prove the statement. Any hint?

答:
$f:G\rightarrow~H$,  by the first isomorphism theorem,

$$G/\ker f \cong f(G)$$

So you need to create a correspondance $G/N \rightarrow f(G)$ ,

now $f^{-1}(f(h'h))=f^{-1}(f(hh'))$ $f^{-1}(f(h')f(h))=f^{-1}(f(h)f(h'))$then $gN=Ng$ if ker(f) $\subset$ N, you can map element $e^H$ to ker(f) inside N for all elements.


问:
Intuition behind picking group actions and Sylow
A common strategy in group theory for proving results/solving problems is to find a clever group action. You take the group you are interested in (or perhaps a subgroup), and find some special set that your group can act on, usually by left multiplication or conjugation. Somehow studying this action makes the result you want clearer to see.

The most obvious example of this is with proving the Sylow theorems.

To prove the First Sylow theorem (that a $p$-subgroup exists) you can let $G$ act by left multiplication on all subsets of $G$ that have size $p^n$.

To prove the Second Sylow theorem (that Sylow $p$-subgroups are conjugate) you let $Q$ be any $p$-subgroup, $P$ a Sylow $p$-subgroup, and let $Q$ act on $G/P$ by left multiplication.

To prove the Third Sylow theorem (that the number of Sylow $p$-subgroups is $1\pmod p$ and divides $|G|$) you can consider both $G$ and some Sylow $p$-subgroup $P$ acting on $\operatorname{Syl}_p(G)$ by conjugation.

My question: is there any intuitive or natural way to come up with these group actions? To me, all three proofs seem magical -- if someone had told me which group action to consider, I could probably have completed the proofs myself, but I would never have come up with the appropriate action myself.

More generally, are there any ways to "see" which group action might help for solving a specific problem?

答:
Well, there's only a few group actions to consider. The ones that worked made it into the proofs. That said, there are a priori reasons to check these actions used in the proofs first.

In investigating these actions, having a fundamental understanding of orbit-stabilizer is critical.

For Sylow I, we don't know from the get-go there is any subgroup of order $p^n$. But of course there are subsets of size $p^n$, and the subgroups are distinguished among these in that they are closed under multiplication. This closure property can be rephrased in terms of stabilization and fixed points. Indeed, if a subset has stabilizer $S$ (under left multiplication), then it is a union of right cosets of $S$, so in particular stabilizers must be $p$-subgroups. By orbit-stabilizer, a subset has full Sylow stabilizer iff its orbit size is not divisible by $p$. Which suggests a possible way to prove Sylows exist by contradiction by considering the sum of orbit sizes mod $p$!

For Sylow II, recall some basic facts about group actions a la (again) orbit-stabilizer. Every $G$-set is a disjoint union of orbits, all orbits are eqiuvalent to $G/H$ for stabilizers $H$. But these coset spaces are only inequivalent up to conjugacy class of subgroup, because $\mathrm{Stab}(g\omega)=g\mathrm{Stab}(\omega)g^{-1}$. This can be rephrased as follows: $H,K\le G$ are conjugate iff $K$ has a fixed point acting on $G/H$ (by left-multiplication).

For Sylow III, after we already know Sylow subgroups are conjugate we automatically know they carry a conjugation action from the group. Orbit-stabilizer gets you $G/N_G(P)$. A trick we used for Sylow I to get a numerical result is to mod by $p$, so we can do that here to investigate the size of this index, which means we want to consider the action of a $p$-group for modding out to yield results. Might as well a Sylow subgroup, and further might as well $P$ so we have a guaranteed fixed point.


问:
Does any probability distribution have an entropy defined?
There are lot of probability distribution having infinite moments, for example Cauchy distribution has even the first moment infinite. Therefore, often we cannot calculate second moments used in measures of dispersion or uncertainty (variation and standard deviation). In some application, for example finance, we could replace such measures with entropy. However, firstly I would like to know whether any probability distribution has defined entropy (in particular Shannon entropy).

In case of discrete distributions with finite number of possible outcome the entropy is
$$
H = -\sum_{i=1}^{n} p_i \log p_i,
$$
where $p_i$ is probability of ith outcome. Since $p_i > 0$ and number of terms in the sum is finite, the sum is defined and it is finite.

But I am getting stuck with case $n \rightarrow +\infty$. In this case I need to prove that the sum  under condition that $\sum_{i=1}^{+\infty}p_i = 1$ converges.

Similarly in case of continuous distribution, I would need to prove that integral
$$
H = -\int_{\mathbb{R}} f(x) \log f(x) \mathrm{d}x
$$
for any real function satisfying $f(x) > 0\,\, \forall \in \mathbb{R}$ and $\int_\mathbb{R}f(x)\mathrm{d}x=1$ exists and it is finite.

Is it possible to prove these statements?

答:
It is possible to have infinite Shannon entropy when dealing with countably many outcomes.  For example,
$$
p_n = \frac{c}{n (\log n)^{1+\epsilon}}\quad n\geq 3
$$
where the constant $c=1/\sum_{n\geq 3}\frac1{n(\log n)^{1+\epsilon}}$ is chosen to satisfy $\sum_{n=3}^\infty p_n=1$.  Since $\sum\frac1{n(\log n)^{1+\epsilon}}$ converges only for $\epsilon>0$ and
$$
-p_n\log p_n=\frac{c}{n(\log n)^{1+\epsilon}}(-\log c+\log n+(1+\epsilon)\log\log n)
\sim \frac{c}{n(\log n)^\epsilon}
$$
we see that for $0<\epsilon\leq 1$ the sum $H=-\sum_n p_n\log p_n$ diverges.


问:
Show that $\mathcal H_0 \cap \mathcal H_1 \neq (0).$

Let $\mathcal H_0$ and $\mathcal H_1$ be closed subspaces of $\mathcal H$ with $\dim \mathcal H_0 + \dim \mathcal H_1 \gt \dim \mathcal H.$ Then show that $\mathcal H_0 \cap \mathcal H_1 \neq (0).$

For finite dimensional case it is fine. Because for finite dimensional case Hilbert space dimension coincides with the vector space dimension and if the result fails to hold then $\mathcal H_0 \oplus \mathcal H_1 \subseteq \mathcal H,$ which contradicts the hypothesis.   How to tackle it for infinite dimensional case? If $\mathcal H$ is infinite dimensional then by the hypothesis it follows that $2 \dim \mathcal H \gt \dim \mathcal H.$ This strict inequality can only hold if $\dim \mathcal H \lt \infty.$ Hence according to me it turns out that if the given hypothesis holds then $\mathcal H$ has to be a finite dimensional Hilbert space. Am I missing something?

答:
I think you're more or less there, though you should explain where the $2 \dim \cal H$ inequality comes from.

It should be emphasized that it is possible to have strict inequalities between transfinite numbers (there are strictly more real numbers than integers), but not between such a number and twice itself.


问:
Error Propagation in Successive Least Square Adjustment
I have a certain problem in surveying in which I'm trying to do some error analysis. I'll layout the problem first.

My current goal is to evaluate the variance-covariance matrix for the
position of a point T on the ground.

Now to first calculate the position of T, I observe this point from 4
other points (S1, S2, S3, S4).

The observation is done in terms of (Ri,θi,φi) and is converted to
(xi,yi,zi) position of T.

Now using simple least square adjustment I calculate the adjusted
position of (X,Y,Z).

I need to find out the net variance-covariance matrix of (X,Y,Z)
provided I know the following information

Available information

Var-Covar matrix of the positions of S1, S2, S3, S4.

Var-Covar matrix of the observations (Ri,θi,φi) that is converted into the matrix for (xi,yi,zi)

Var-Covar matrix of (X,Y,Z) obtained through the co-factor matrix in least square adjustment procedure - this matrix is only a by-product of the adjustment - it assumes no uncertainty in S1, S2, S3, S4.

Question

I need to know how can I propagate the errors in the positions of S1, S2, S3, S4 into the errors of point T.

答:
Here are the general ideas:

1) Treat errors coming from different sources as independent with mean 0.

Moral: When two errors add together, the covariance matrices merely add up.
The least square procedure in your case merely returns the average of 4 observations, so the final matrix $A=\frac 1{16}\sum_i A_i$ and each $A_i$ is the sum of the covariance matrix of $S_i$ plus the covariance matrix of the vector going from $S_i$ to $T$.

2) When having non-linear dependences and small errors, linearize near the observed values.

You have the conversion from polar to Cartesian here. I'm not 100% sure what notation you use for polar, but let's say the conversion is
$$
x=R\sin\theta\cos\phi, y=R\sin\theta\sin\phi, z=R\cos\theta
$$

Suppose you want to compute some covariance for the coordinates of the Cartesian direction vector, say $E[dx dz]$ where $dx$ is the error in $x$, etc.
Linearizing yields
$$
dx=(\sin\theta\cos\phi)dR+(R\cos\theta\cos\phi)d\theta-(R\cos\theta\sin\phi)d\phi,
\\
dz=(\cos\theta)dR-(R\sin\theta)d\theta
$$
Now multiply out and open parentheses. You will get various expressions of the kind
$$
F(R,\theta,\phi) dR d\phi
$$
(or (dR)^2, or...). Read the expectations $E[dR d\phi]$ from the polar covariance matrix you have, plug them in, and do the arithmetic.

That's all it takes.

P.S. This is done under the assumption that S_i were independently measured. If they themselves were obtained by measurements from the same point, their errors are not completely independent any more and the picture gets more involved. Let me know if this effect is of importance in your setting.


问:
Prove that for a continuous function $q(t)\leq-1$ the boundary problem has only the trivial solution

Prove that for a continuous function $q(t)\leq-1$ the boundary problem has only the trivial solution

$x''(t)+q(t)x(t)=0,x(a)=0,\sqrt{3}x'(b)=x(b),a<b$

I Know that $q(t)\leq-1$ so I tried to solve $x''(t)-x(t)=0$ which I get $x(t)=c_1e^{-t}+c_2e^t$ and I got from the $x(a)=0,\sqrt{3}x'(b)=x(b)$ that $c_1,c_2=0$ but I don't think I've done this right

答:
You can shift the whole situation so that $a=0$. As the equation and boundary conditions are homogeneous, scalar multiples of solutions are also solutions. Thus fix $x'(0)=1$. If $x$ had another root, it would also have a maximum $c$ where $x''(c)\le0<-q(c)x(c)$ which is impossible.

The obvious function to compare the solution $x(t)$ against is $\sinh(t)$. In the Liouville theory this is often done via a Wronski-like expression. Set
$$
w(t)=\cosh(t)x(t)-\sinh(t)x'(t).
$$
Then
$$
w'(t)=\sinh(t)(x(t)-x''(t))=\sinh(t)(1+q(t))x(t)\le 0
$$
which implies $w(t)\le w(0)=0$ and thus
$$
0<x(t)\le\tanh(t)x'(t)< x'(t)<\sqrt3 x'(t)
$$
Equality is thus never possible.


答:
The result that you want looks very unlikely, and in fact one can prove that it is false by the following argument: let ${\varphi} \left(x\right) =-x \ln  \left(x\right)$. For ${\alpha} \in  \left(0 , 1\right)$, let us take $n$ numbers ${u}_{1} , \ldots  , {u}_{n}$, all equal to $\frac{{\alpha}}{n}$. We have

\begin{equation}
\sum _{i = 1}^{n} {u}_{i} = {\alpha} \qquad  \sum _{i = 1}^{n} {\varphi} \left({u}_{i}\right) =-{\alpha} \ln  \left(\frac{{\alpha}}{n}\right) = {\varphi} \left({\alpha}\right)+{\alpha} \ln  \left(n\right)
\end{equation}

By choosing $n  \geqslant  {e}^{1/{\alpha}}$, we can ensure that
$\sum _{i = 1}^{n} {\varphi} \left({u}_{i}\right)  \geqslant  1$.

Let us now consider the sequence ${{\alpha}}_{k} = {2}^{{-k}}$ for $k  \geqslant  1$.
let ${n}_{k} = \left\lceil {e}^{1/{{\alpha}}_{k}}\right\rceil $ and let us choose the sequence ${p}_{i}$ by the following algorithm: the ${n}_{1}$ first terms are equal to ${{\alpha}}_{1}/{n}_{1}$, then the ${n}_{2}$ next terms are equal to ${{\alpha}}_{2}/{n}_{2}$, etc.

It follows from this construction that

\begin{equation}
\sum _{i = 1}^{\infty } {p}_{i} = 1 \qquad  \sum _{i = 1}^{\infty } {\varphi} \left({p}_{i}\right) = \infty
\end{equation}

Title: Does any probability distribution have an entropy defined?

Body:
There are lot of probability distribution having infinite moments, for example Cauchy distribution has even the first moment infinite. Therefore, often we cannot calculate second moments used in measures of dispersion or uncertainty (variation and standard deviation). In some application, for example finance, we could replace such measures with entropy. However, firstly I would like to know whether any probability distribution has defined entropy (in particular Shannon entropy).

In case of discrete distributions with finite number of possible outcome the entropy is
$$
H = -\sum_{i=1}^{n} p_i \log p_i,
$$
where $p_i$ is probability of ith outcome. Since $p_i > 0$ and number of terms in the sum is finite, the sum is defined and it is finite.

But I am getting stuck with case $n \rightarrow +\infty$. In this case I need to prove that the sum  under condition that $\sum_{i=1}^{+\infty}p_i = 1$ converges.

Similarly in case of continuous distribution, I would need to prove that integral
$$
H = -\int_{\mathbb{R}} f(x) \log f(x) \mathrm{d}x
$$
for any real function satisfying $f(x) > 0\,\, \forall \in \mathbb{R}$ and $\int_\mathbb{R}f(x)\mathrm{d}x=1$ exists and it is finite.

Is it possible to prove these statements?


问:
Understanding Taylor's Theorem for multivariate functions
Let $f:\mathbb{R}^n \to \mathbb{R}$. Consider Taylor's theorem up to 3rd term.
\begin{align*}
    f(\vec{x}) &= f(\vec{a}) + \sum_{i=1}^n \frac{\partial f(\vec{a})}{\partial x_i} (\vec{x}_i - \vec{a}_i) + \frac{1}{2} \sum_{i_1, i_2 = 1}^n \frac{\partial^2 f(\vec{a})}{\partial x_{i_1} \partial x_{i_2}} (\vec{x}_{i_1} - \vec{a}_{i_1}) (\vec{x}_{i_2} - \vec{a}_{i_2}) + \sum_{|\beta|=  3} R_3(\vec{x})(\vec{x} - \vec{a})^3
\end{align*}
where the last term is written using multi-index notation. Now in my understanding
\begin{align*}
    \sum_{|\beta|=  3} R_3(\vec{x})(\vec{x} - \vec{a})^3 = \frac{1}{3!} \sum_{i_1, i_2, i_3 = 1}^n \frac{\partial^3 f(\vec{y})}{\partial x_{i_1} \partial x_{i_2} \partial x_{i_3}}  (\vec{x}_{i_1} - \vec{a}_{i_1}) (\vec{x}_{i_2} - \vec{a}_{i_2}) (\vec{x}_{i_3} - \vec{a}_{i_3}) \tag1
\end{align*}
where $\vec{y} = \vec{a} + t(\vec{x} - \vec{a})$ for some $t \in [0, 1]$.

But from wikipedia here, it seems that
\begin{align}
    \sum_{|\beta|=  3} R_3(\vec{x})(\vec{x} - \vec{a})^3 &=  \sum_{i_1, i_2, i_3=1}^n  \left [ \frac{3}{3!}\int_0^1 (1-t)^{2} \frac{\partial^3 f(\vec{a} + t(\vec{x} - \vec{a}))}{\partial x_{i_1} \partial x_{i_2} \partial x_{i_3}} dt\right ] (\vec{x}_{i_1} - \vec{a}_{i_1}) (\vec{x}_{i_2} - \vec{a}_{i_2}) (\vec{x}_{i_3} - \vec{a}_{i_3}) \tag2
\end{align}

How do I reconcile $(1)$ and $(2)$? Am I misreading the formula on wikipedia?

答:
As we know:
$$\int\limits_{0}^{1}(1-t)^2dt=\frac{1}{3}$$
So it's enough to use mean value theorem for definite integrals
$$\int\limits_{a}^{b}f(x)g(x)dx=g(c)\int\limits_{a}^{b}f(x)dx$$
where $\exists c \in (a,b)$


问:
Essentially self adjoint operators: a verification procedure
$\langle \ .,. \rangle : \scr H \! \times \! \scr H \rightarrow \mathbb C$ is the inner product.

Let $T : {\cal D}(T) \rightarrow \scr H $ be a linear symmetric
operator, so:

${\cal D} (T)$ is dense in $\scr H$

$\langle \phi, T\psi \rangle = \langle T\phi, \psi \rangle \quad \forall \phi, \psi \in {\cal D}(T)$

The following properties are equivalent:

$T$ is essentially self-adjoint: $\ T^* = \overline T $

${\rm Ker} (T^* \pm \lambda  I ) \equiv \{ 0\} $

If I have understood correctly, point 2. of the theorem is telling us that $(T^* \pm \lambda I)\psi = 0 \Longleftrightarrow \psi =0 $, or, to put it in other terms, the only eigenvector satisfying $$\ T^* \psi = \mp \lambda \psi $$ is the null vector.

My professor: "It's therefore evident that if we want to check if any symmetric operator $T$ admits a unique self-adjoint extension - its closure - we just have to make sure $\pm i$ are not eigenvalues of $T^*$"

Is this really enough? And what in the world has it to do with the kernel triviality?

答:
Suppose $T : \mathcal{D}(T)\subset \mathcal{H}\rightarrow\mathcal{H}$ is closed, densely-defined and symmetric. (It is reasonable to assume that $T$ is closed because, if $T$ is symmetric, $T$ has a closure that is also symmetric.) Then the graph $\mathcal{G}(T)$ of $T$ is a closed subspace of $\mathcal{G}(T^*)$. So there is a closed subspace $M$ of $\mathcal{H}\times\mathcal{H}$ such that one has the following orthogonal decomposition in $\mathcal{H}\times\mathcal{H}$:
$$
                 \mathcal{G}(T)\oplus M=\mathcal{G}(T^*).
$$
If $(m,n)\in M$, then $(m,n)\in\mathcal{G}(T^*)$, which gives $n=T^*m$. And $(m,T^*m)\perp\mathcal{G}(T)$ iff
$$
                  \langle (x,Tx),(m,T^*m)\rangle_{\mathcal{H}\times\mathcal{H}}=0,\;\;\; x\in\mathcal{D}(T).
$$
That is,
$$
             \langle x,m\rangle+\langle Tx,T^*m\rangle=0,\;\; x\in\mathcal{D}(T).
$$
This is equivalent to
$$
               \langle Tx,T^*m\rangle = \langle x,-m\rangle,\;\;\;  x\in\mathcal{D}(T).
$$
Therefore, $T^*m\in\mathcal{D}(T^*)$ and $(T^*)^2m+m=0$. Any such $m$ may be written as $m_{+}+m_{-}$ where $T^*m_{+}=im_{+}$ and $T^*m_{-}=-im_{-}$. The decomposition is
$$
               m=\frac{1}{i}(iI+T^*)m+\frac{1}{i}(iI-T^*)m, \\
                m_{+}=\frac{1}{i}(iI+T^*)m,\;\;\; m_{-}=\frac{1}{i}(iI-T^*)m.
$$
This gives the following orthogonal decomposition decomposition in $\mathcal{H}\times\mathcal{H}$:
$$
           \mathcal{G}(T)\oplus\mathcal{N}(T^*+iI)\oplus\mathcal{N}(T^*-iI)=\mathcal{G}(T^*).
$$
From this it becomes evident that $T$ is self-adjoint iff $\mathcal{N}(T^*\pm iI)=\{0\}$.
You can substitute $T$ with $\frac{1}{\Im\lambda}(T-\Re\lambda I)$ in order to obtain the result with a more general $\lambda\notin\mathcal{R}$, not just with $\lambda=i$.


问:
What's the probability of a number being chosen if it can be chosen twice?
Given a set of {1,2,3,...,n} numbers and a number can be chosen randomly and there is also the possibility of assigning the same number chosen by two different participants.

My thought process is, since 2 people can choose the same number then there is $2^n$ ways each number could be chosen.

Suppose I want to choose the number k, then the probability someone is assigned that number k is $\dfrac{2^{n-1}}{2^n}$?

Please let me know if I am incorrect. Thank you!

答:
If there are $m$ independent and equally probable selections from $n$ numbers, then the probability a particular number is not selected is $\left(1-\frac1n\right)^m$

so the probability a number is selected is $1-\left(1-\frac1n\right)^m$

For $m=2$ this is $\frac2n - \frac1{n^2}$


问:
Prove that $\mathbb{Z} ( p^\infty) \cap \mathbb{Z} ( q^\infty) = (\overline{0})$

Let $ G=\mathbb{Q}/\mathbb{Z}$. Let $p$ be a prime integer and $\bar{<\frac{1}{p^n}>} $ is the subgroup generated by $\bar{\frac{1}{p^n}}$. Let $\mathbb{Z} ( p^\infty)= \bigcup_{p>0} \bar{<\frac{1}{p^n}>}$. Prove the following:

If $p$ and $q$ are distinct primes, then $\mathbb{Z} ( p^\infty) \cap \mathbb{Z} ( q^\infty) = (\overline{0})$.

Attempt:

Clearly, $\overline{0}\in \mathbb{Z} ( p^\infty) \cap \mathbb{Z} ( q^\infty)$ because $\overline{0}=\overline{\frac{p}{p}} =\overline{\frac{q}{q}}$.

So $(\overline{0}) \subset\mathbb{Z} ( p^\infty) \cap \mathbb{Z} ( q^\infty) $.

Now, suppose that $x\in \mathbb{Z} ( p^\infty) \cap \mathbb{Z} ( q^\infty)$. We want to show that $x \in \overline{(0)}$.

$x=\overline{\frac{a}{p^n}}= \overline{\frac{b}{q^m}}$ for some $n$ and $m$.

$\overline{\frac{a}{p^n}}- \overline{\frac{b}{q^m}}=\overline{0}$

$\overline{\frac{aq^m-bp^n}{p^n q^m}} =\overline{0}$

How can I complete this, please. Any help would be appreciated. Thanks

答:
Hint: without loss of generality, you can assume $p \not \mid a$ and $q \not \mid b$. Your last equality implies, however, that

$$
aq^m-bp^n = p^nq^m c \tag{$\star$}
$$

for some $c \in \mathbb Z$. What happens if both $n$ and $m$ are positive?

 First, let us prove the "without loss of generality" part. Write $a = p^k a'$ with $\mathbf{gcd}(p,a') = 1$. If $k > n$, then $a/p^n$ is an integer, its class is zero, and there is nothing to prove. Otherwise, we can write $a/p^n = a'/p^{n-k}$. Now, if $n,m \geq 1$ then $(\star)$ implies that $p \mid aq^m$, so $p\mid q^m$. Hence $p \mid q$, a contradiction. We see in this way that either $n = 0$ or $m = 0$. But then either $a/p^n \in \mathbb Z$ or $b/p^m \in \mathbb Z$, in either case their classes (which coincide) are the class of $0$.


问:
Truncation and rounding error bound derivation for finite difference approximation to the first derivative
Finite difference  approximation to the first derivative:
$$
f'(x)= \frac{f(x+h) - f(x)}{h}
$$

Heath's book on Scientific computing Section 1.2.4, Example 1.3 says truncation error for the finite difference approximation is bounded by $\frac{|f''(t)|h}{2}$ for $t$ near $x$, and the rounding error in computing the finite difference is bounded by $\frac{2\epsilon}{h} $

How are these bounds calculated?

答:
By the linear Taylor expansion with quadratic remainder term, and

by assuming that the evaluation of $f$ has a seemingly random error bounded in size by $ϵ$.

See Computing differentiation rule with error bound for a plot of the error against the exact derivative value in different approximation schemes for the first derivative.


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

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

答:
Each set $\displaystyle{A_n=\left\{x:|f(x)|>\frac{1}{n}\right\}}$ has finite measure.  Each function $f\cdot\chi_{A_n}$ is dominated by $f$, and $f\cdot\chi_{A_n}\to f$ pointwise ($\chi_{A_n}$ denotes the characteristic function of $A_n$).  By Lebesgue's dominated convergence theorem, $\int_{A_n}fd\mu\to \int_X fd\mu$ as $n\to\infty$.  Therefore $E$ can be taken to be $A_n$ for sufficiently large $n$.

Here you could also just use monotone convergence, because $(f\cdot \chi_{A_n})$ increases to $f$.

Title: Prove this inequality of integrals:


问:
What does the discriminant of an algebraic number field mean intuitively?
If $E/F$ is a finite extension of fields and $\alpha_1,\ldots, \alpha_n$ is a basis
of $E/F$, the discriminant of $\{\alpha_1,\ldots, \alpha_n\}$ is $$\det(\operatorname{Tr}_{E/F}(\alpha_i\alpha_j))$$

I want to know if there is any other interpetration of this quantity? What information is actually conveyed by this quantity?

答:
Let $K$ be a number field and $\mathcal{O}_K$ its ring of integers. Let $\sigma_1, ... \sigma_n$ be the complex embeddings of $K$, where $n = r + 2s$, $r$ is the number of real embeddings, and $s$ is the number of conjugate pairs of complex embeddings. The $\sigma_i$ furnish an embedding

$$K \to \mathbb{R}^r \times \mathbb{C}^s$$

and the image of $\mathcal{O}_K$ in $\mathbb{R}^r \times \mathbb{C}^s$ is a lattice. The square root of the discriminant of $K$ measures the volume of a fundamental domain of this lattice (with respect to a suitable volume form).

In this setting the prime factorization of the discriminant also provides important information about ramification. In particular, the primes ramifying in $\mathcal{O}_K$ are precisely the primes dividing the discriminant. See, for example, Wikipedia.


问:
Solving SDE with sign function in drift term?
Consider the following SDE with $X_0 = 1$,
$$
dX_t = X_t\operatorname{sign}(X_t) \, dt + X_t \, dW_t,
$$
where $\operatorname{sign}(x) = \mathbb{1}_{\{x \ge 0\}} -\mathbb{1}_{\{x < 0\}}$. How am I supposed to solve this SDE?

答:
We assume that $X_t = f(t,W_t)$ is an Ito process, so that we can assume the Ito formula to find a candidate for the solution of the equation, by matching the $dt$ and $dW_t$ drifts. Write $f \equiv f(t,x)$. We proceed to do some rough work in the next part.

Note that by the Ito formula, we have :
$$
dX_t = \left(\frac{\partial f}{\partial t} + \frac{1}{2}\frac{\partial f}{\partial x^2}\right) dt + \frac{\partial f}{\partial x} dW_t
$$

Therefore, we must find $f$, such that $\frac{\partial f}{\partial x} = f$ and $\frac{\partial f}{\partial t} + \frac{1}{2}\frac{\partial f}{\partial x^2} = f \mbox{sign}\{f\}$, along with $f(1,0) = 1$. Indeed, from the first condition we obtain that $f(t,x) = g(t)e^{x}$ for some $t$, then by the second condition we obtain that $g(t) = e^{\frac t2}$ fits, therefore we get $X_t = e^{\frac t2 + W_t}$ to fit in the boundary condition.

Note that $\mbox{sign}(X_t) = 1$ a.s. since $X_t =e^{R_t}$ for a real-valued random variable. Hence that particular term never contributes to the equation.

Of course, this is rough work since we did not rigorously solve the ODEs above, for example. But then we have a guess : one can easily verify now that the $X_t$ obtained from the rough work is indeed a strong solution using the Ito formula.

This solution is unique, since the coefficients satisfy the Lipschitz and linear growth conditions.


Title: $x^3 + y^3 +3x^2 y^2 =x^3y^3$ Find all possible values to $\frac{x+y}{xy}$

Body:
$x,y \in \mathbb{R}\setminus\{0\}$

Indeed the original question said :Find all possible values to: $$\frac{1}{x} + \frac{1}{y}$$
But it’s the same thing.

My Attempt:
$$x^3 + y^3 +3x^2 y^2 =x^3y^3 \iff (x+y)(x^2-xy+y^2)=x^3y^3 -3x^2y^2$$
$$\frac{x+y}{xy}=\frac{x^2y^2-3xy}{x^2-xy+y^2}$$

But i don’t know what to do now.
Please if you know the key to this type of problems ‘Find all possible values’ post it.

Body:
Because $x,y \ne 0$, from $x^3 + y^3 +3x^2 y^2 =x^3y^3$, we have
$$\frac{1}{x^3}+\frac{1}{y^3} +\frac{3}{xy} =1 \iff\frac{1}{x^3}+\frac{1}{y^3}  =1-\frac{3}{xy} \tag{1}$$
But we have also

$$(\frac{1}{x}+\frac{1}{y})^3 = (\frac{1}{x^3}+\frac{1}{y^3}) + \frac{3}{xy}(\frac{1}{x}+\frac{1}{y}) \tag{2}$$
From (1) and (2), we deduce
\begin{align}
&(\frac{1}{x}+\frac{1}{y})^3 = (1-\frac{3}{xy}) + \frac{3}{xy}(\frac{1}{x}+\frac{1}{y}) \\
&\iff (\frac{1}{x}+\frac{1}{y})^3- 1=\frac{3}{xy}\left( \frac{1}{x}+\frac{1}{y} - 1  \right) \\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( (\frac{1}{x}+\frac{1}{y})^2 +(\frac{1}{x}+\frac{1}{y})+1   \right)=\frac{3}{xy}\left( \frac{1}{x}+\frac{1}{y} - 1  \right) \\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( (\frac{1}{x}+\frac{1}{y})^2 +(\frac{1}{x}+\frac{1}{y})+1  -\frac{3}{xy} \right)=0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( (\frac{1}{x}+\frac{1}{y})^2 -\frac{4}{xy} +(\frac{1}{x}+\frac{1}{y})+1 +\frac{1}{xy}   \right)=0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( (\frac{1}{x}-\frac{1}{y})^2  +\left( 1+\frac{1}{x}   \right)\left( 1+\frac{1}{y} \right)  \right)  =0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( ((1+\frac{1}{x})-(1+\frac{1}{y}))^2  +\left( 1+\frac{1}{x}   \right)\left( 1+\frac{1}{y} \right)  \right)  =0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( \left(1+\frac{1}{x} \right)^2+\left(1+\frac{1}{y} \right)^2  -\left( 1+\frac{1}{x}   \right)\left( 1+\frac{1}{y} \right)  \right)  =0\\
\end{align}

The last equation happens if and only if $\frac{1}{x}+\frac{1}{y} =1$  or $\left( 1+\frac{1}{x}, 1+\frac{1}{y} \right) = (0,0) $. Hence, the two possible values of $\frac{1}{x}+\frac{1}{y}$ is $1$ and $-2$.


问:
Thales theorem to show half-circle encloses largest area over a line given a circumference
I'm working on a proof for the isoperimetric inequality, and want to show that for all curves over a line, the half-circle encloses the maximum area with that line given a fixed circumference.

I was given a hint to use Thales's theorem.

The hint confuses me more than it helps me, so I was hoping that someone might be able to shed some light on it. What I don't get is why the right triangles from Thales's theorem are useful if they are obviously not disjoint, and I don't think I can use them to approximate the area.

Does someone see the relationship?

答:
You should first review this page:

http://www.cut-the-knot.org/do_you_know/isoperimetric.shtml

Specifically, the section beginning with "Proof of Statement 1" contains Lemma 2, which is the part you are working on.

For the sake of completeness, I have reproduced it below.

Lemma 2.  Consider all the arcs with a given length and the endpoints $S$ and $T$ on a fixed straight line. The curve that encloses the maximum area between the curve and the straight line is a semicircle.

Proof.  Suffice it to show that every angle inscribed into the arc is right. If there is a point on the arc, say $P$, for which $\angle SPT$ is not right, slide either point $S$ or point $T$ along the straight line $ST$ until the angle becomes right. Let pieces of the arc move along. As an exercise prove that among all triangles with two given sides the one whose sides enclose a right angle has the largest area. Since the area of the two red regions did not change but the area of the triangle grew, the whole area between the new curve $SPT$ and the line $ST$ has increased. This shows that unless the curve is a semicircle we can always increase the area in question by moving points $S$ and $T$. This proves Lemma 2 and with it Statement 1.

However, note the remark that follows this proof, in which it is pointed out that there is a flaw in the line of reasoning for Statement 1.  This does not invalidate the proof of Lemma 2 which you are specifically interested in, but it does mean that the argument in which Lemma 2 plays a role is incomplete.

The missing piece in Lemma 2 is the "exercise" which is actually quite simple to show, which is why I have omitted the proof.


问:
Reducing modulo equations
How do I reduce the following and remove the $11$ from LHS,
$$
11x \equiv 4 \ \ \text{mod }50
$$

Title: Reducing modulo equations

Body:
$11x = 50k +4$

Multiplying the equation by 9,

$99x = 450k + 36$

$-x = 50(9k - 2x) + 36$

$x = 50(2x - 9k -1) + 14$

$x \equiv 14 \pmod {50}$


问:
Leibniz test and $\sum\frac{(-1)^k k^2}{\sqrt{k!}}$
I would like to check if the following series $\sum\frac{(-1)^k k^2}{\sqrt{k!}}$ is convergent.

My original idea was to employ Leibniz's criterion, the one that says: An alternating series $(-1)^k a_k$ such that $a_k$ is a non-increasing sequence that satisfies $\lim a_k=0$ then the series converges, however I'm having difficults when showing that my $a_k$ is non-increasing through induction.

I would appreciate some help.

答:
$$a_n=\frac{k^2}{\sqrt{k!}}\implies \log(a_n)=2\log(k)-\frac 12\log(k!)$$ Using Stirling approximation
$$\log(a_n)=\frac 12(1-\log(k))k+\cdots$$
$$a_n=e^{\log(a_n)}=e^{\frac{1}{2} k (1-\log (k))}~~~\to ~~~0$$


问:
Showing a infinite product is compact if each factor is homeomorphic to Sierpinski space
"Using (i) and Proposition 9.2.5, show that if $(X_i , T_i)$, for each $i \in N$, is homeomorphic to the Sierpinski Space, then $\prod_i(X_i , T_i)$ is compact."

i) There is a continuous image from $[0,1]$ to the Sierpinski space.
Proposition 9.2.5) If for each $i,$ $f_i$ is a continuous mapping from $X_i$ to $Y_i$, then $f$ defined by $f=(f_1,f_2,f_2...)$ is continuous.

We have not proven Tychonoff's theorem otherwise this would be trivial. We can use the fact that the Cantor space is compact and the Cantor Space is homeomorphic to infinite products of itself and $[0,1]$ is a continuous image of the Cantor space, so I just need to make sure I am connecting the dots correctly.

So if each factor $X_i$ is homeomorphic to the Sierpinski space, there is a continuous map from $[0,1]$ to each factor. Now from proposition 9.2.5, $f$ is continuous each $f_1$ is continuous and each $[0,1]$ is homeomorphic to the cantor space which so the product topology of $[0,1]$ is homeomorphic to the product topology of the cantor space $[0,1]$ and the cantor space is homeomorphic to any number of cartesian product of itself. The Cantor space is compact, and $f$ is a continuous mapping so compactness is preserved thus $\prod_i(X_i , T_i)$ is compact.

答:
The Cantor set is not needed. Every Sierpiński space is the continuous image of $[0,1]$, so combining these into a product map, we get that the product of Sierpiński spaces is the continuous image of the Hilbert cube, of which you know it is compact. So your product is too (continuous image of compact is compact). That’s all.

If you only know the Cantor cube is compact then use the trivial continuous maps from the discrete two-point space to Sierpiński  space instead.


问:
How to find coordinates of reflected point?
How can I find the coordinates of a point reflected over a line that may not necessarily be any of the axis?

Example Question:

If P is a reflection (image) of point (3, -3) in the line $2y = x+1$, find the coordinates of Point P.

I know the answer is $(-1,5)$ by drawing a graph but other than that, I cannot provide any prior workings because I don't know how to start...

答:
Similar answer to @Vrisk, but a bit faster

Consider the line $L=Ax+By+C=0$ and  find image of point $(u,v)$  assuming $Au + Bv + C \neq 0 $

If we extend the space which the equation exists into $R^3$ , we will find that the equation $Ax+By+C=0$ denotes a plane with a unit normal vector as : $$ \hat{n} = \frac{}{\sqrt{A^2 +B^2} } $$

Now, consider the unsigned perpendicular distance of this point from the line/plane :

$$ d =  \frac{Au +Bv +C}{\sqrt{A^2 +B^2}}$$

Now, depending on sign of the above quantity we can see which part of the plane/ line which the point is (see here).

If, from our point we were translate in twice the direction of unit normal scaled up by the distance from line to point, we would reach the image point. Hence, the image coordinates$(I)$ are given by the vector:

$$I =- \frac{2(Au+Bv+c)}{A^2 +B^2} $$

Or more concisely:

$$I=  -2d \hat{n}$$

Here is a useful diagram:

Body:
Similar answer to @Vrisk, but a bit faster

Consider the line $L=Ax+By+C=0$ and  find image of point $(u,v)$  assuming $Au + Bv + C \neq 0 $

If we extend the space which the equation exists into $R^3$ , we will find that the equation $Ax+By+C=0$ denotes a plane with a unit normal vector as : $$ \hat{n} = \frac{<A,B>}{\sqrt{A^2 +B^2} } $$

Now, consider the unsigned perpendicular distance of this point from the line/plane :

$$ d =  \frac{Au +Bv +C}{\sqrt{A^2 +B^2}}$$

Now, depending on sign of the above quantity we can see which part of the plane/ line which the point is (see here).

If, from our point we were translate in twice the direction of unit normal scaled up by the distance from line to point, we would reach the image point. Hence, the image coordinates$(I)$ are given by the vector:

$$I =<u,v>- \frac{2(Au+Bv+c)}{A^2 +B^2} <A,B>$$

Or more concisely:

$$I= <u,v> -2d \hat{n}$$

Here is a useful diagram:


问:
$n$ such that $ n^2 \mid \varphi(n)^{\varphi(n)} - 1 $. Can $n$ be a prime number $n>2$ or a Carmichael number?
For all $n$ such that $\large n^2 \mid \varphi(n)^{\varphi(n)} - 1 $ with $\varphi()$ denotes Euler's totient function.

It can be shown that $n$ must be a cyclic number. That means $n$ is coprime to $\varphi(n)$.

Implying further that $n > 2$ is odd and squarefree.

Besides the two trivial solutions $n=1$ and $n=2$ OEIS states five nontrivial solutions below $10^{12}$:

$19043= 137 \cdot 139$,

$289627=13 \cdot 22279$,

$6674419= 571 \cdot 11689$,

$49865347= 7 \cdot 173 \cdot 41177$,

and

$185014655= 5 \cdot 7 \cdot 17  \cdot 310949 $.

As Carmichael numbers are also cyclic numbers, I tested the first $1,401,644$ of them below $10^{18}$ from a database. None of them gives a solution.

If $n$ is a prime number then  $n^2 \mid (n-1)^{n-1} - 1$.

Hence $n$ must be a Wieferich prime to base $n-1$.

Can $n$ be a prime number $n>2$ or a Carmichael number?

答:
Partial answer :

$n$ cannot be an odd prime.

Proof :

Let $p$ be an odd prime and denote $m:=p-1$. Then we have

$$m^m-1=\prod_{d\mid m} \Phi_d(m)$$

We have $\Phi_1(m)=m-1=p-2$ , which is not divisible by $p$

We have $\Phi_2(m)=m+1=p$

It remains to show that for $d>2$, $\Phi_d(m)$ cannot be divisible by $p$. First
note $$\Phi_d(m)\equiv \Phi_{d}(-1)\mod p$$

Let $r$ be the radical of $d$ (product of the distinct prime factors of $d$). Using the well-known formula $\Phi_d(b)=\Phi_r(b^{d/r})$ , applied for $-1$ , we get $$\Phi_d(-1)=\Phi_r(\pm 1)$$ depending on the parity of $\frac{d}{r}$.

In the case $\Phi_d(-1)=\Phi_r(1)$ we have $\Phi_d(-1)=1$ or a prime factor of $r$ and hence of $d$.

So, let us assume
$\Phi_d(-1)=\Phi_r(-1)$. We have $3$ cases :

$r=2$ , then $\frac{d}{r}$ is even, hence $\Phi_d(-1)=\Phi_r(1)=2$

$r$ even and greater than $2$. Then, $\Phi_r(-1)=\Phi_{r/2}(1)$ and $\frac{r}{2}>1$, so $\Phi_r(1)$ is either $1$ or a  prime factor of $r$ and hence of $d$

$r$ odd. Then $\Phi_r(-1)=\Phi_{2r}(1)=1$

In none of the cases $p$ divides $\Phi_d(-1)$, which completes the proof.


问:
Prove the minimal polynomial $a$ over $\mathbb{Q}$ is equal to the minimal polynomial $\overline{a}$ over $\mathbb{Q}$
Let $a \in \mathbb{C}$ be such that $a \notin \mathbb{R}$ and $a$ is algebraic over $\mathbb{Q}$. Let $\overline{a}$ be the complex conjugate of $a$.

Let $f(x)$ be the minimal polynomial $a$ over $\mathbb{Q}$ and $g(x)$ be the minimal polynomial $\overline{a}$ over $\mathbb{Q}$.

(1) Prove that $f(x) = g(x)$.

(2) Prove or disprove that $[\mathbb{Q}(a) : \mathbb{Q}]$ is an even integer.

$\textbf{My Attempt:}$

Question (1):

Since, $f(x)$ is the minimal polynomial $a$ over $\mathbb{Q}$, if $f(x)$ is monic polynomial and irreducible over $\mathbb{Q}$ and $f(a) = 0$.

Similarly $g(x)$ is the minimal polynomial $\overline{a}$ over $\mathbb{Q}$, if $g(x)$ is monic polynomial and irreducible over $\mathbb{Q}$ and $g(\overline{a}) = 0$.

Then, $f(a) = 0$ and $g(\overline{a}) = 0$.

So, $f(a) = g(\overline{a})$ which means $f(x)$ and $g(x)$ has exactly same roots.

Therefore, $f(x) = g(x)$.

Question (2):

Want to prove that $[\mathbb{Q}(a) : \mathbb{Q}]$ is an even integer.

Since, notice that $a \in \mathbb{C}$ be such that $a \notin \mathbb{R}$ and $a$ is algebraic over $\mathbb{Q}$.

Which means $a$ must equal to some terms with adding, substracting, multiplying or dividing $i = \sqrt{-1}$ somewhere.

So, in order to get $f(x)$, we must square both side to get rid of $i = \sqrt{-1}$.

So, by square both side, $f(x)$ will have degree of a even number.

Since, notice that $[\mathbb{Q}(a) : \mathbb{Q}] = $ the degree of $f(x)$.

So, $[\mathbb{Q}(a) : \mathbb{Q}]$ is even.

$\textbf{My Question:}$

For question (1): I don't think the prove is correct, but I have no other idea on proving $f(x) = g(x)$.

For question (2): Is this correct ? (if so, are there ways to make this prove more theoretical ?).

答:
Your logic going from $f(a)=g(\bar{a})$ and concluding that they must have the same roots is flawed. How do you know that? The way to prove the first part is to take the minimal polynomial of $a$, $f(x)$, and showing that
$$
0=f(a)=\overline{f(a)}=f(\overline{a})\ ,
$$
and concluding that $f$ is also the minimal polynomial of $\overline{a}$. You can go figure out the details, in particular, why we can say $\overline{f(a)}=f(\overline{a})$ and why $f(\overline{a})$ is not only a polynomial with $\overline{a}$ as a root, but must also be its minimal polynomial.

EDIT: My original answer to the second part was completely wrong. Here is a counterexample which shows that the answer to part $2$ of the question is 'No'.

Consider the polynomial $f(X)=X^3-2$ over $\mathbb{Q}$. This is the go-to example for a non-normal extension of $\mathbb{Q}$. Let $\alpha=\sqrt[3]{2}$ and $\omega=e^{2\pi i/3}=\frac{1}{2}+\frac{\sqrt{3}}{2}i$, then the roots of $f$ are $\alpha, \omega\alpha$ and $\omega^2\alpha$. It can be shown that $\mathbb{Q}(\alpha)$, $\mathbb{Q}(\omega\alpha)$ and $\mathbb{Q}(\omega^2\alpha)$ are all isomorphic and that all their degrees over $\mathbb{Q}$ are $3$. Taking $a=\omega\alpha$ gives the desired result.


问:
How to prove that $\mathrm{T^1}\mathbb{H} \cong\mathrm{PSL}_2(\mathbb{R})$
Ergodic Theory with a view towards Number Theory, Chapter 9, Page 281.

Lemma 9.2. The action of $\mathrm{PSL}_2(\mathbb{R})$ on $\mathrm{T}^1\mathbb{H}$ is simply transitive.

Notice that this allows us to describe the unit tangent bundle to $\mathbb{H}$ as
$$\mathrm{T}^1\mathbb{H}\cong\mathrm{PSL}_2(\mathbb{R}).\tag{9.4}$$

In order to do this, we have to choose an arbitrary reference vector $(z_0,v_0)$ in $\mathrm{T}^1\mathbb{H}$ which corresponds to $I_2\in\mathrm{PSL}_2(\mathbb{R})$; the identification is then given by $g\mapsto\mathrm{D}g(z_0,v_0)$. We will make the convenient choice $z_0=\mathrm{i}$ and $v_0=\mathrm{i}$. That is, the reference vector is the upward unit vector based at the imaginary unit $\mathrm{i}\in\mathbb{H}$. Under the resulting identification the action of $\mathrm{PSL}_2(\mathbb{R})$ on $\mathbb{H}$ is conjugated to the action of $\mathrm{PSL}_2(\mathbb{R})$ by left multiplication on $\mathrm{PSL}_2(\mathbb{R})$.

How do we have that $$\mathrm{T^1}\mathbb{H} \cong\mathrm{PSL}_2(\mathbb{R})$$
And the last three lines in the attached image above (Under the resulting......$\mathrm{PSL}_2(\mathbb{R})$).

答:
With the Iwasawa decomposition $SL_2\mathbb{R}=KAN$, where $K=SO(2)$, $A$ consists of diagonal elements, and $N$ consists of unitriangular elements, we can show $SL_2\mathbb{R}$ acts transitively on $\mathbb{H}$. Indeed, the subgroup $AN$ itself (upper triangular elements) acts transitively: you can use $a\in A$ to make the imaginary part of a point whatever you need it to be, then use $n\in N$ to make the real part whatever you want it to be.

Indeed, given any $(z,v)$ (with $v$ the unit tangent vector), you can use $g\in G$ to turn this into $(i,w)$ for some tangent vector $w$. Then, you can pick a $k\in K=SO(2)=\mathrm{Stab}(i)$ to apply to get $(i,i)$. Thus you can get $(i,i)$ from any $(z,v)$ (and thus also conversely).

The orbit-stabilizer theorem says if $G\curvearrowright\Omega$ is transitive then $\Omega\cong G/\mathrm{Stab}(\omega)$. In the case of a sharply transitive action (i.e. a regular action), this means $\Omega\cong G$ are equivalent $G$-sets. Picking the correspondence between $G$ and $\Omega$ amounts to picking a $\omega\in\Omega$ to correspond to $e\in G$. (In the case of $G$ a vector space and $\Omega$ an affine space, this means picking an "origin.") Then $g\mapsto g\omega$ is the equivalence $G\cong\Omega$. We have $G$ acting on itself by left-multiplication, of course.

Your source is choosing $\omega=(z_0,v_0)$ as a "reference vector" - specifically $(i,i)$ is nice.


答:
It is enough to show that the terms $a_k$ are non-increasing for all $k$ larger than a fixed finite number (Can you see why?). For $k \ge 3$,
$$
\frac{{\frac{{(k + 1)^2 }}{{\sqrt {(k + 1)!} }}}}{{\frac{{k^2 }}{{\sqrt {k!} }}}} = \left( {1 + \frac{1}{k}} \right)^2 \frac{1}{{\sqrt {k + 1} }} < \left( {1 + \frac{1}{3}} \right)^2 \frac{1}{{\sqrt {3 + 1} }} = \frac{8}{9} < 1
$$
Thus
$$
\frac{{(k + 1)^2 }}{{\sqrt {(k + 1)!} }} < \frac{{k^2 }}{{\sqrt {k!} }}
$$
for $k\geq 3$, i.e., the terms are decresing starting from the third term.

Title: Leibniz test and $\sum\frac{(-1)^k k^2}{\sqrt{k!}}$

Body:
I would like to check if the following series $\sum\frac{(-1)^k k^2}{\sqrt{k!}}$ is convergent.

My original idea was to employ Leibniz's criterion, the one that says: An alternating series $(-1)^k a_k$ such that $a_k$ is a non-increasing sequence that satisfies $\lim a_k=0$ then the series converges, however I'm having difficults when showing that my $a_k$ is non-increasing through induction.

I would appreciate some help.


问:
Computing the integral $I=\int_{0}^{1}\sqrt{1-x^{\pi}}dx$
Let $I=\int_{0}^{1}\sqrt{1-x^{\pi}}dx$ If we use $u=x^{\pi}$ then we get $dx=\frac{du}{\pi^{\frac{\pi-1}{\pi}}}$ are there any better identites or substitution that can make integral easier to compute?

答:
[Note] : You can use the beta function :

Beta Function : $\displaystyle \mathfrak{B}(x,y):=\int_{0}^{1}(1-\mu)^{x-1}\mu^{y-1}\;\text{d}\mu=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$

Let $\mu:=x^{\pi}$. Then :
\begin{align*}
x^{\pi}=\mu&\implies\text{d}\mu=\pi x^{\pi -1}\\ \\
&\iff\text{d}x=\frac{\text{d}\mu}{\pi\cdot\mu^{\displaystyle\frac{\pi-1}{\pi}}} =\frac{1}{\pi}\mu^{\displaystyle\frac{1-\pi}{\pi}}\text{d}\mu
\end{align*}
Therefore :
\begin{align*}
I&=\frac{1}{\pi}\int_{0}^{1}\sqrt{1-\mu}\cdot\mu^{\displaystyle\frac{1-\pi}{\pi}}\text{d}\mu\\ \\
&=\frac{1}{\pi}\int_{0}^{1}(1-\mu)^{\frac{3}{2}-1}\mu^{\frac{1}{\pi}-1}\;\text{d}\mu\\ \\
&=\frac{1}{\pi}\mathfrak{B}\left(\frac{3}{2},\frac{1}{\pi}\right)\\ \\
&=\frac{1}{\pi}\frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{\pi}\right)}{\Gamma\left(\frac{3}{2}+\frac{1}{\pi}\right)}\\ \\
&\approx 0.846941\cdots
\end{align*}


问:
$n$ balls into $n$ urns until there is no empty urn.
We have $n$ balls and $n$ urns.

Step 1. Throw $n$ balls into $n$ urns randomly.

Check each urn and if there is more than one ball in an urn, choose only one and keep in the urn and remove all the other balls.

Step 2. Suppose there are $k$ balls ''not placed'' in an urn in Step 1. Throw these $k$ balls into the urns randomly.

And move on in this fashion until there is no empty urn.

What is the expected number of urns that are empty in each step?

Answer. Step 1 is easy. For any urn $u_{j}$, define $X_{j}$ as $X_{j}=1$ if no ball hits and $X_{j}=0$, otherwise. Then, expected number of empty urns is just $E(X)$ where $X=X_{1}+X_{2}+...+X_{n}$. Note that $E(X_{j})=Pr(X_{j}=1)=(\frac{n-1}{n})^n$. Hence, $E(X)=n(\frac{n-1}{n})^n$.

After Step 1, things get complicated. A heuristic approach for Step 2: The expected number of empty urns (and hence unplaced balls) is just $n(\frac{n-1}{n})^n$. Thus, for large $n$, the expected number of empty urns in Step 2 is $n\left( \frac{n-1}{n}\right) ^{n}\left( \frac{n-1}{n}\right) ^{n\left(\frac{n-1}{n}\right) ^{n}}$ (within some bound?)

Is this approach correct? Can this be made rigorous?

答:
Let label the states with the number of the filled urns. Our aim is to find the probability of transition from the state $m$ to the state $l$ ($0\le m\le l\le n$). This probability is:
$$\begin{align}
P_{ml}(n)&=\frac1{n^{n-m}}\binom{n-m}{l-m}
\sum_{i=l-m}^{n-m}\binom{n-m}i(l-m)!{i \brace l-m}m^{n-m-i}\\
&=\frac1{n^{n-m}}\frac{(n-m)!}{(n-l)!}
\sum_{i=l-m}^{n-m}\binom{n-m}i{i \brace l-m}m^{n-m-i},\tag1
\end{align}$$
where the symbols count the following numbers of way:

$n^{n-m}$ - to place $n-m$ balls into $n$ urns;

$\binom{n-m}{l-m}$ - to choose $l-m$ "active" urns to be filled out of $n-m$ empty ones;

$\binom{n-m}i$ - to choose $i$ balls out of $n-m$ ones;

$(l-m)!{i \brace l-m}$ - to place $i$ balls into $l-m$ "active" urns in such a way that there is at least one ball in each of these urns. Here  ${i \brace l-m}$ is the Stirling number of second kind;

$m^{n-m-i}$ - to place the rest balls into $m$ already filled urns.

After the transition matrix $P(n)$ is constructed the expected value of the empty urns after $k$ steps can be computed as:
$$
\mathsf E_k(n)=n-\sum_{i=1}^n i[P^k(n)]_{0i}.\tag2
$$

Example.

For $n=5$ one obtains:

$$P(5)=\left(\begin{array}{cccccc} 0 & \frac{1}{625} & \frac{12}{125} & \frac{12}{25} & \frac{48}{125} & \frac{24}{625} \\ 0 & \frac{1}{625} & \frac{12}{125} & \frac{12}{25} & \frac{48}{125} & \frac{24}{625} \\ 0 & 0 & \frac{8}{125} & \frac{57}{125} & \frac{54}{125} & \frac{6}{125} \\ 0 & 0 & 0 & \frac{9}{25} & \frac{14}{25} & \frac{2}{25} \\ 0 & 0 & 0 & 0 & \frac{4}{5} & \frac{1}{5} \\ 0 & 0 & 0 & 0 & 0 & 1 \\\end{array}\right).$$

The corresponding values $\mathsf E_k (5)$ for $k=1,\dots,10$ are respectively:
$$
\begin{array}{l} 1.6384 \\ 1.07168 \\ 0.782347 \\ 0.599572 \\ 0.470231 \\ 0.372795 \\ 0.297015 \\ 0.237173 \\ 0.18958 \\ 0.151607 \\\end{array}
$$

One can easily demonstrate that the 0th and the 1st rows of the transition matrix are identical, and the 0th column is $0$. Therefore one can for the sake of simplicity drop the 0th row and column from the matrix, so that $1\le m,l\le n$ in (1). The only required change after the redefinition of $P(n)$ is to replace the index $0i $ in the formula (2) with $1i $.


问:
A quite difficult limit: $\lim _{x\to 0}\frac{\exp(x^2)-\cos x-x\sin x}{\sinh^2x}$
I tried this limit with Taylor series. However, proceeding in this manner I obtained a series over another series. Furthermore, I do not know how to continue as long as we have $x$ tends to $0$. Were it to be $x$ tends to infinite, this limit would be simpler.

This is what I got

$$\begin{align}&
\lim _{x\to 0}\frac{\exp(x^2)-\cos x-x\sin x}{\sinh^2x} \\[6pt]
=\;&\frac{\frac12x^2+\frac{5}{8}x^4+\frac{23}{144}x^6+\frac{241}{5760}x^8+\frac{3359}{403200}x^{10}+\cdots}{{x^2+\frac{1}{3}x^4+\frac{2}{45}x^6+\frac{1}{315}x^8+\frac{2}{14175}x^{10}+\cdots}}
\end{align}$$
I can't continue.

Body:
Hint

$$e^{x^2}-\cos(x)-x\sin(x)=\frac{x^2}{2}+o(x^2)$$
and
$$\sinh(x)^2=x^2+o(x^2).$$


问:
How does family of curves formula exactly work?
I am familiar with the following equations

$$S_1 + \lambda S_2=0$$ for two circles

$$L_1 + \lambda L_2=0$$ for two lines
$$L_1 + \lambda S_1=0$$ for circle and line

But I never really understood how they worked. Now this is a sample questions

A circles touches the parabola $y^2=2x$ at P $(\frac 12, 1)$ and cuts parabola at vertex V. If centre of circle is a Q, find the radius of circle

The formula used here was

$$(x-\frac 12)^2 + (y-1)^2 +\lambda (2x-2y+1)=0$$

Now it’s easy to see that equation is basically hinting at a curve passing through $(1/2, 1)$ and tangent to $(2x-2y+1)$, but how exactly was the form mat determined? How can we tell if this will us gives a circle? Why was the distance formula used in the first part of the equation? Basically I want to know the process of writing such equations .

答:
$C_1:\left(x-\frac 12\right)^2 + (y-1)^2 =0$

is the equation of the circle having centre in $P\left(\frac12,1\right)$ and radius $r=0$ and $C_2:2x-2y+1=0$ is the equation of the line tangent to the parabola $\mathcal{C}:y^2=2x$ at $P$

A linear combination of the two $C_1+\lambda C_2=0$ represents all circles tangent to the parabola $\mathcal{C}$ at $P$ (see image (2) below)
$$\left(x-\frac 12\right)^2 + (y-1)^2+\lambda(2x-2y+1)=0\tag{1}$$
If the circle passes through the origin, the vertex of $\mathcal{C}$, then substitute $(0,0)$ in $(1)$ to get $\lambda=-5/4$.

The circle we are looking for has equation
$$\left(x-\frac{1}{2}\right)^2+(y-1)^2-\frac{5}{4} (2 x-2 y+1)=0$$
simplify
$$2 x^2+2 y^2-7 x+y=0$$
center is $\left(\frac74,-\frac14\right)$ and radius $r=\frac{5\sqrt 2}{4}$

$$...$$


问:
Why we equalize the logarithms those with same bases?
This is not an exam's question , I already have the answer.
I want to know why when we have the logarithm of something with the same base, we can equalize the things together.
Like here:
$\log_{10} x=\log_{10}y^2\quad$
So  $\quad x=y^2$
Why is this?

答:
This is because the logarithm function with a specific base is injective. For an injective function $f$, it holds that if $f(x) = f(y)$, then $x=y$.

You can prove this in many ways, for example that it is the inverse function of the exponential function.


答: The implication you mention is because $x=a^{\log_ax}$, therefore $\log_ay=\log_az$ implies $y=z$.

Title: Why we equalize the logarithms those with same bases?

Body:
This is not an exam's question , I already have the answer.
I want to know why when we have the logarithm of something with the same base, we can equalize the things together.
Like here:
$\log_{10} x=\log_{10}y^2\quad$
So  $\quad x=y^2$
Why is this?


Title: A quite difficult limit: $\lim _{x\to 0}\frac{\exp(x^2)-\cos x-x\sin x}{\sinh^2x}$

Body:
I tried this limit with Taylor series. However, proceeding in this manner I obtained a series over another series. Furthermore, I do not know how to continue as long as we have $x$ tends to $0$. Were it to be $x$ tends to infinite, this limit would be simpler.

This is what I got

$$\begin{align}&
\lim _{x\to 0}\frac{\exp(x^2)-\cos x-x\sin x}{\sinh^2x} \\[6pt]
=\;&\frac{\frac12x^2+\frac{5}{8}x^4+\frac{23}{144}x^6+\frac{241}{5760}x^8+\frac{3359}{403200}x^{10}+\cdots}{{x^2+\frac{1}{3}x^4+\frac{2}{45}x^6+\frac{1}{315}x^8+\frac{2}{14175}x^{10}+\cdots}}
\end{align}$$
I can't continue.

Body:
$$\frac{0.5x^2+\frac{5}{8}x^4+\frac{23}{144}x^6+\frac{241}{5760}x^8+\frac{3359}{403200}x^{10}+\ldots \:}{{x^2+\frac{1}{3}x^4+\frac{2}{45}x^6+\frac{1}{315}x^8+\frac{2}{14175}x^{10}+\ldots \:}}=\frac{0.5+\frac{5}{8}x^2+\frac{23}{144}x^4+\frac{241}{5760}x^6+\frac{3359}{403200}x^{8}+\ldots \:}{{1+\frac{1}{3}x^2+\frac{2}{45}x^4+\frac{1}{315}x^6+\frac{2}{14175}x^{8}+\ldots \:}} \to 1/2$$

as $ x \to 0.$


问:
Definition : Weak Topology , nuance between "continuous" terms.
Definition of weak topology i could find : Let X be a Banach space and X' being it's topological dual. The weak topology over X is the smallest topology such as continuous elements (to the sense of the norm) of X' remains continuous.

Does the last "continuous" refers to the topological sense of continuity ? f is continuous to the topological sense in X' if for all neighborhood V of f(x) with x $\in X$ , $f^{-1}(V)$ is a neighborhood of x.

答:
Yes, of course this refers to the topological definition of continuity. This continuity depends on the topology of $X$ and that of the scalars (so $\Bbb R$ here).

But it turns out that if $f$ is a functional ( so a linear map to the scalars) then it’s equivalent to be norm continuous or weak continuous.


问:
Number comparison: $5^{152}<2^{353}$ and $2^{1413}<3\cdot 5^{608}$
Is it possible to prove that $5^{152}<2^{353}$ and $2^{1413}<3\cdot 5^{608}$ without using a calculator or logarithms (middle school math only recommended)?

My idea for the first one was to use the obvious $5^3<2^7$ and then raise to the power of $50$ to get $5^{150}<2^{350}$. But since $5^2>2^3$, I couldn't use this approach to get the desired result.

Can you please help me find a relatively short proof for these inequalities? Thank you.

答:
$$\begin{array} {}
 5^{152} &\overset?\lt & 2^{353} \\
 {5^2 \over 2^3} &\overset?\lt& {2^{350} \over 5^{150}} \\
 {25 \over 8} &\overset?\lt& \left({2^7  \over 5^3}\right)^{50} \\
 3+{1\over 8} &\overset?\lt& \left(1+{3 \over 125}\right)^{50} & \qquad \text{      //binomial} & \text{expansion} \\
 3+{1\over 8} &\overset?\lt& 1+ 50 \cdot {3 \over 125} & + \binom{50}{2}\cdot {9 \over 125^2} &+ \cdots\\
 2+{1\over 8} &\overset?\lt&  0+ {6 \over 5} & + \binom{50}{2}\cdot {9 \over 125^2} &+ \cdots\\
 1-{3\over 40} &\overset?\lt&   & + {50 \cdot 49 \over 2}\cdot {9 \over 125^2} &+ \cdots\\
 1-{3\over 40} &\overset?\lt&   &  25 \cdot (50-1) \cdot {9 \over 25^3} &+ \cdots\\
 {37\over 40} &\overset?\lt&   &   (50-1) \cdot {9 \over 25^2} &+ \cdots\\
 {37\over 8} &\overset?\lt&   &   {18 \over 5} -  {9 \over 5^3} &+ 5\cdot (\cdots)\\
 4+{5\over 8} &\overset?\lt&   &   3+{3 \over 5} -  {9 \over 5^3} &+ 5\cdot (\cdots)\\
 1+{5\over 8} &\overset?\lt&   &   {3 \over 5} -  {9 \over 5^3} &+ 5\cdot (\cdots)\\
 {65\over 40} &\overset?\lt&   &   {24 \over 40} -  {9 \over 125} &+  5\cdot ( \cdots) \\ {41\over 40} + {9 \over 125} &\overset?\lt&   &    &+ 5 \cdot  \binom{50}{3}\cdot {27 \over 125^3} & + 5 \cdot(\cdots)\\
 \vdots \end{array}$$
After this, my patience ran out, but only with that 4'th term of the binomial expansion the result would have been achieved, so this problem is surely solvable by paper&pen ...

Binomial expansion to 4 terms:

25/8 < 1 + 6/5 + 441/625 + 21168/78125 + (...) = 248168/78125 + (...)


答:
Do you know what the logarithm is?

$$
\log_{10}(x) = a \Leftrightarrow 10^x=a
$$

In other words, $\log_{10}(x)$ asks the question: What do I need to put into the exponent to get $10^{\log_{10}(x)}=x$

Thus, if you now look at
$$\log_{10}(x)=\log_{10}(y^2)$$

you need to ask these questions again. But, because $10^x$ or any other exponential function with a positive base is injective (that means, every y-value is uniquely in correspondence with a fitting x-value), these two questions get answered by the same number and thus
$$x=y^2$$ holds true.


问:
How do you prove that a particle with the position $(\cos t, \sin t)$ has unit speed?
Consider a particle that's motion is given by the vector function
$$
S(t)=(\cos t,\sin t) \, ,
$$
where $t$ is measured in radians. How do you prove that the particle has unit speed, i.e. $|S'(t)|=1$? At first, I thought this would be easy, since $\cos t$ and $\sin t$ are simply the $x$- and $y$-coordinates of a particle that has traced an arc of $t$ units anticlockwise around the unit circle. This means that between two points $S(t+h)$ and $S(t)$, the particle must have travelled a distance of $h$ units. Hence,
$$
|S'(t)|=\lim_{h \to 0}\frac{h}{h}=1 \, .
$$
However, I'm unsure if this solution is actually correct. If I understand correctly, $|S'(t)|$ is the limit of the displacement from the initial position, i.e.
$$
|S'(t)| := \lim_{h \to 0}\frac{\sqrt{(\cos(t+h)-\cos t)^2+(\sin(t+h)-\sin t)^2}}{h} \, .
$$
In other words, although the particle will have travelled a distance of $h$ units in going from $S(t)$ from $S(t+h)$, its displacement would be something else. However, according to Wolfram Alpha, the above limit does not even exist. So it seems that I have some misunderstanding of what $|S'(t)|$ actually means, but I'm unsure what it is.

答:
Your mistake is in dropping the absolute value.

$$
|S'(t)| := \lim_{h \to 0}\frac{\sqrt{(\cos(t+h)-\cos t)^2+(\sin(t+h)-\sin t)^2}}{|h|} =1.
$$


问:
Show for the centroid $M$ of a triangle $ABC$ that $\vec{OM}=\frac13(\vec{OA}+\vec{OB}+\vec{OC})$
Show for the centroid $M$ of a triangle $ABC$ that $$\vec{OM}=\dfrac13\left(\vec{OA}+\vec{OB}+\vec{OC}\right)$$
where $O$ is an arbitrary point.

I haven't studied position vectors and am very new to vectors so I would like a simple solution. Any help would be appreciated. I don't even know how to start. Maybe we can use the fact that the centroid divides each median in a ratio of $2:1$? Thank you!

答:
We take the triple $(A,B,C)$ as barycentric coordinate system for the plane. That means: any point $P$ of the plane is written

$$ P = \alpha A + \beta B + \gamma \,C \qquad \text{ with }\;\alpha+\beta+\gamma=1, \tag{1} $$

which is equivalent to

$$ P-O = \alpha(A-O) + \beta(B-O) + \gamma\,(C-O) \qquad \big[\alpha+\beta+\gamma=1\big]  \tag{2} $$

for any point $\,O\,$ of the plane (see Note 1 below).

The numbers $\;\alpha,\beta,\gamma\;$ in $(1)$ are the barycentric coordinates of $\;P\;$ with respect to the given coordinate system.

We have $\;A_1-C=B-A_1\;$ and hence $A_1=\frac12(B+C)$. Similarly for others, so, in barycentric coordinates:

$$ A_1=\frac12(B+C)\qquad B_1=\frac12(A+C)\qquad C_1=\frac12(A+B). \tag{3} $$

The median $AA_1$ is therefore described by equation $\;tA+(1-t)A_1\quad[t\in\Bbb R],\;$ i.e. $\;tA+(1-t)\frac12(B+C)$, and similarly for others medians:

$$ AA_1:\qquad tA+\frac12(1-t)(B+C) \qquad \big[t\in\Bbb R\big] \tag{4}$$
$$ BB_1:\qquad sB+\frac12(1-s)(A+C) \qquad \big[s\in\Bbb R\big] \tag{5}$$
$$ CC_1:\qquad rC+\frac12(1-r)(A+B) \qquad \big[r\in\Bbb R\big]. \tag{6}$$

The lines $(4), (5)$ intersect in a point $M$ iff there exist $t,s\in\Bbb R\;$ such that

$$ M=tA+\frac12(1-t)(B+C) = sB+\frac12(1-s)(A+C) \tag{7}$$

i.e. (see Note 2 below) iff:

$$ t=\frac12(1-s) \qquad\quad \frac12(1-t)=s \qquad\quad \frac12(1-t)=\frac12(1-s) $$

which gives $\;t=s=\frac13$. A similar work on the other medians give $\;t=s=r=\frac13,$ therefore the medians are concurrent in $\;M$, and from $(7)$ we get

$$ M=\frac13A+\frac13B+\frac13C, \tag{8} $$

i.e. the barycentric coordinates of the centroid $M$ are $(\frac13,\frac13,\frac13).$

The equation $(8)$ immediately give our thesis

$$ M-O = \frac13\Big((A-O) + (B-O) + (C-O)\Big). $$

$\quad$

Note 1.$\quad$ We prove that $(2)$ is independent of $\,O$. If $\;O'\;$ is any other point, we have

\begin{align}
P-O' &= (P-O) +(O-O') =\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\[1ex]
&= \alpha(A-O) + \beta(B-O) + \gamma\,(C-O) + (O-O') =\\[1ex]
&= \alpha\Big((A-O')+(O'-O)\Big) + \beta\Big((B-O')+(O'-O)\Big) +\\[1ex] &\qquad\qquad +\gamma\Big((C-O')+(O'-O)\Big) + (O-O')=\\[1ex]
&= \alpha(A-O')+\beta(B-O')+\gamma\,(C-O')+(\alpha+\beta+\gamma)(O'-O)+(O-O') =\\[1ex]
&= \alpha(A-O')+\beta(B-O')+\gamma\,(C-O'),\\
\end{align}
being $\;\alpha+\beta+\gamma=1$.

The point $P$ in $(2)$, independent of $O$, is indicated as in $(1)$.

$\quad$

Note 2.$\quad$ If $\;\alpha+\beta+\gamma = \alpha'+\beta'+\gamma' = 1$, we have
$$ \alpha A+\beta B+\gamma\,C = \alpha'A+\beta'B+\gamma'C \quad\implies\quad \alpha=\alpha', \;\;\beta=\beta', \;\;\gamma=\gamma' $$

In fact, by choosing $\;O=A\;$, the preceding equality give

$$ A+\alpha(A-A)+\beta(B-A)+\gamma\,(C-A) = A+\alpha'(A-A)+\beta'(B-A)+\gamma'(C-A) $$

hence

$$ \beta(B-A)+\gamma\,(C-A) = \beta'(B-A)+\gamma'(C-A). $$

As $\;B-A,\;C-A\;$ are linearly independent, we deduce $\beta=\beta'\;$ and $\;\gamma=\gamma'$; and finally $\;\alpha=1-\beta-\gamma=1-\beta'-\gamma'=\alpha'.$


答:
You may use some standard limits as follows:

\begin{eqnarray*}\frac{e^{x^2}-\cos x - x\sin x}{\sinh^2 x}
& = & \left(\frac{e^{x^2}-1}{x^2}+ \frac{1-\cos x}{x^2}-\frac{\sin x}{x}\right)\cdot \left(\frac{x}{\sinh x}\right)^2\\
& \stackrel{x\to 0}{\longrightarrow} & \left(1+\frac 12 - 1\right)\cdot 1^2 = \frac 12
\end{eqnarray*}


问:
the set of generators of the ideal $I = (2, 1 + \sqrt{-5})$ of $R = \mathbb Z[\sqrt{-5}]$
In this question here Finding the set of generators of the ideal $I = (2, 1 + \sqrt{-5})$ of $R = \mathbb Z[\sqrt{-5}]$ @Noah has told me that yes $\{ 2, 1 + \sqrt{-5} \}$ are generators of my ideal $I$ of $R$ and @Lubin has written those 2 comments to me:

1 - We can see that, if $\{ 2, 1 + \sqrt{-5} \}$ is a set of generators, then so is $\{ 2, n + \sqrt{-5} \}$ for $n$ any odd integer .

2-our two generators $\{ 2, 1 + \sqrt{-5} \}$ definitely are not $R$-linearly independent.

But I am unable to prove these 2 comments, could someone show me the details of the proofs? why specifically an odd $n$ in $ 1 + \sqrt{-5} $ instead of the $1$ will make it also a generator, why the other generator $2$ should be left as it is? Is that related to the norm function of $R$?
Also,how to prove that our two generators $\{ 2, 1 + \sqrt{-5} \}$ definitely are not $R$-linearly independent?

答:
Generally, the ideal generated by two elements $a,b$ is the same as the ideal generated by $a$ and $a+b$, since it must also contain $b=(a+b)-a$.

Applying it repeatedly (or proving right away in the above manner), we get
$$(a,b)=(a,b+na)$$
for every $n\in\Bbb Z$.

We have
$$2\cdot(1+\sqrt{-5})-(1+\sqrt{-5})\cdot 2=0$$
with both coefficients $2,\ (1+\sqrt{-5})$ being in $R$.


Body:
Assuming $x, y>0;$

$\log_{10} x=\log_{10}y^2;$

$\log_{10} x-\log_{10} y^2=0;$

$\log_{10} (x/y^2)=0;$

$\Rightarrow x/y^2=1.$


答:
Here's an elementary solution,(I mean elementary as I have expressed the limit in question, to various standard limits that a beginner in calculus faces.)

The limit in question is taken to be $L$

$L=\displaystyle\lim_{x\rightarrow 0} \frac{(e^{x^2}-1)+(1-\cos x)-x\sin x}{\sinh^2x}$

$L=\displaystyle\lim_{x\rightarrow 0} \frac{(e^{x^2}-1)}{\sinh^2x}+\frac{(1-\cos x)}{\sinh^2x}-\frac{x\sin x}{\sinh^2x}$

$$L=\displaystyle\lim_{x\rightarrow 0} \frac{(e^{x^2}-1)}{x^2}\frac{x^2}{\sinh^2x}+\frac{2\sin^2\left(\frac{x^2}{2}\right)}{\sinh^2x}-\frac{x\sin x}{\sinh^2x}=1+\frac{1}{2}-1=\frac{1}{2}$$


问:
$x^3 + y^3 +3x^2 y^2 =x^3y^3$ Find all possible values to $\frac{x+y}{xy}$
$x,y \in \mathbb{R}\setminus\{0\}$

Indeed the original question said :Find all possible values to: $$\frac{1}{x} + \frac{1}{y}$$
But it’s the same thing.

My Attempt:
$$x^3 + y^3 +3x^2 y^2 =x^3y^3 \iff (x+y)(x^2-xy+y^2)=x^3y^3 -3x^2y^2$$
$$\frac{x+y}{xy}=\frac{x^2y^2-3xy}{x^2-xy+y^2}$$

But i don’t know what to do now.
Please if you know the key to this type of problems ‘Find all possible values’ post it.

Body:
$$x^3 + y^3 + 3 x^2 y^2 = x^3 y^3\tag{1}$$
$$\frac{x+y}{xy}\tag{2}$$
$$x^3 + y^3 + 3 x^2 y^2 - x^3 y^3=0$$
can be factorized$^1$ as
$$(x+y-xy) \left(x^2 y^2+x^2 y+x^2+x y^2-x y+y^2\right)=0$$
$x^2 y^2+x^2 y+x^2+x y^2-x y+y^2=0$ has solution$^2$ given by $x=y=-1$ we have $$\frac{x+y}{xy}=-2$$, the solutions of $(1)$ are only those of $x+y-xy=0$ that is $x+y=xy$ and finally
$$\frac{x+y}{xy}=1$$

$(^1)$

$$x^3 + y^3 + 3 x^2 y^2 - x^3 y^3=(x+y)^3-3x^2y-3xy^2+ 3 x^2 y^2 - x^3 y^3=$$
$$=\left[(x+y)^3- x^3 y^3\right]+\left[-3x^2y-3xy^2+ 3 x^2 y^2 \right]=$$
$$=(x+y-xy)\left[(x+y)^2+xy(x+y)+x^2y^2)\right]-3 x y (x+y-x y)=$$
$$=(x+y-xy)\left[(x+y)^2+xy(x+y)+x^2y^2)-3xy\right]=(x+y-xy)\left(x^2 y^2+x^2 y+x^2+x y^2-x y+y^2\right)$$

$(^2)$

$$x^2 y^2+x^2 y+x^2+x y^2-x y+y^2=0$$
set $y=kx$. We get
$$k^2 x^4+k^2 x^3+k^2 x^2+k x^3-k x^2+x^2=0\to x^2 \left(k^2 x^2+k^2 x+k x+k^2-k+1\right)=0$$
As $x\ne 0$ we have
$$k^2 x^2+k^2 x+k x+k^2-k+1=0$$
the discriminant of this equation is $D=-k^4+2 k^3-k^2=-k^2(k-1)^2$

$D\ge 0\to k=1 $ (solution $k=0$ is discarded)

if $k=1$ we have $x^2+2x+1=0\to x=-1$ thus the solutions are
$$x=-1;\;y=-1$$


问:
Video lectures for Commutative Algebra
Are there any good video lectures for learning commutative algebra at level of Atiyah-Macdonald? </p>

答: Yes, Borcherds has a series of lectures on commutative algebra forming an online course, roughly following Eisenbud's book.</p>

Title: Video lectures for Commutative Algebra


问:
Reducing modulo equations
How do I reduce the following and remove the $11$ from LHS,
$$
11x \equiv 4 \ \ \text{mod }50
$$

答:
Variant using the particularity of multiplication by $11$.

Generally $11\times \overline{ab}=\overline{asb}=100a+10s+b\, $ where $\, s=a+b$ (provided $s<10$).

Notice that when $s=5$ we have $\begin{cases}100a\equiv 0\pmod {50}\\10s\equiv 0\pmod{50}\\11\times \overline{ab}\equiv b\pmod{50}\end{cases}$

Since we want $b=4$ then $a=1$ is forced and $$11\times 14\equiv 4\pmod{50}$$

Note that it works too for finding the inverse of $11$ (this time $b=1$ and $a=4$), thus $11^{-1}\equiv 41\pmod{50}$.

Title: Reducing modulo equations


Title: Nasty integrals involving polylogarithms

Body:
Is there a general method how to solve integrals including polylogarithms? I have, for example, a following peculiar integral:

$$\int_0^1 \frac{\log (1-q)}{(1-q)^{19}} \big[19122427017242392409179196696957 -2335312823676226739258108857200 \pi ^2-290355604888434431371406862640901 q+32737624286330384009900385643800 \pi ^2
   q+2089178147864338015378359942349146 q^2-217908181258375090989657061033200 \pi ^2 q^2-9513735107311961105716228372823832 q^3+923058652929127659169477549327800 \pi ^2
   q^3+30860609840601979343569796243664360 q^4-2809234642150007839007261169271200 \pi ^2 q^4-75991613070307679674084684299763488 q^5+6560138082212689647470512926380400 \pi ^2
   q^5+147604422934589574241111368769031514 q^6-12225064353406525792601847884728800 \pi ^2 q^6-231661749521735051694587401454488374 q^7+18614430483751971866343908315754000 \pi
   ^2 q^7+298158588677298781555571530676473698 q^8-23467372646806215931232745831715200 \pi ^2 q^8-317258462849638359532191702700757050 q^9+24645135112158586773581379231285600
   \pi ^2 q^9+279929745859987197112521991075600990 q^{10}-21578983058687084626530891022466400 \pi ^2 q^{10}-204525346750399538308156387121441190
   q^{11}+15702582554471282285512462526560800 \pi ^2 q^{11}+123049277403587430345156505609140720 q^{12}-9430354354307541344225967691260000 \pi ^2
   q^{12}-60342329966544036497302398449157240 q^{13}+4622406501952432638430248014413200 \pi ^2 q^{13}+23738252991464115594141538592264430
   q^{14}-1818869679546039116051640669453600 \pi ^2 q^{14}-7312550577333389628853566957018834 q^{15}+560636064795229527948416643193200 \pi ^2
   q^{15}+1699354671638585927650255051925577 q^{16}-130382484017845412425019267350800 \pi ^2 q^{16}-280173803150789489806907495704617 q^{17}+21513060264068052978453696664200 \pi
   ^2 q^{17}+29212691355365195449956724538624 q^{18}-2244803977761215443782290650800 \pi ^2 q^{18}-1448394095369360889798167890490 q^{19}+111380159353759803311707564200 \pi ^2
   q^{19}-864839351159251344060440256000 \pi ^2 \log (q)+9614284037777341680385275648000 \pi ^2 q \log (q)-48596947298066064189458694528000 \pi ^2 q^2 \log
   (q)+149016194597643828944407014096000 \pi ^2 q^3 \log (q)-311515153604740784777334800544000 \pi ^2 q^4 \log (q)+472522854921446523354235979904000 \pi ^2 q^5 \log
   (q)-540131604846004142202578675328000 \pi ^2 q^6 \log (q)+476835369490723415021630300520000 \pi ^2 q^7 \log (q)-330431763957397218290451028752000 \pi ^2 q^8 \log
   (q)+181570501304738210584142319096000 \pi ^2 q^9 \log (q)-79469952891751512800087794416000 \pi ^2 q^{10} \log (q)+27618312749376667202083351152000 \pi ^2 q^{11} \log
   (q)-7469792604527284717674468480000 \pi ^2 q^{12} \log (q)+1441811254291252370952714288000 \pi ^2 q^{13} \log (q)-47548846798971176802583440000 \pi ^2 q^{14} \log
   (q)-242992965510346357364274216000 \pi ^2 q^{15} \log (q)+315652971889245500921659248000 \pi ^2 q^{16} \log (q)-286658885174869914970791528000 \pi ^2 q^{17} \log
   (q)+190547414679020492013416496000 \pi ^2 q^{18} \log (q)-91723410310948362363008592000 \pi ^2 q^{19} \log (q)+31251841079492830435486848000 \pi ^2 q^{20} \log
   (q)-7178196236973842421012672000 \pi ^2 q^{21} \log (q)+1000683546793492378825008000 \pi ^2 q^{22} \log (q)-64152027504348590478672000 \pi ^2 q^{23} \log (q)+116396280 \log
   (1-q) \left((-1+q)^2 \left(-257586022317616038442244+3541119036337161972270259 q-22985344007557567401249245 q^2+94080232161007231115696965 q^3-272947169011867587813604265
   q^4+596855442237497586986976649 q^5-1019309093239565006695750169 q^6+1388155171326881304329317345 q^7-1524633136813377308169137885 q^8+1356305480530438789865344115
   q^9-975804898738457785419837691 q^{10}+563741882769447393607104851 q^{11}-257984481217331023678963375 q^{12}+91454555996476460965835495 q^{13}-24229819048292743306257775
   q^{14}+4517255786475215854979111 q^{15}-528777243429852450045991 q^{16}+29245925849336575939850 q^{17}\right)+17463600 \left(2552782515875592-28378884658059936
   q+143445643657566096 q^2-439857339564018222 q^3+919512319237712508 q^4-1394765491161108528 q^5+1594329068484771696 q^6-1407494920941509015 q^7+975349269884284614
   q^8-535949249427887597 q^9+234574786644003962 q^{10}-81522129873123914 q^{11}+22048899523933360 q^{12}-4255854634983666 q^{13}+140351921539830 q^{14}+717252508230687
   q^{15}-931726090681386 q^{16}+846143031204471 q^{17}-562446780417522 q^{18}+270743829850494 q^{19}-92247367548336 q^{20}+21188182319304 q^{21}-2953759514706
   q^{22}+189360225054 q^{23}\right) \log ^2(q)\right)+2032698075408000 \left(2552782515875592-28378884658059936 q+143445643657566096 q^2-439857339564018222
   q^3+919512319237712508 q^4-1394765491161108528 q^5+1594329068484771696 q^6-1407494920941509015 q^7+975349269884284614 q^8-535949249427887597 q^9+234574786644003962
   q^{10}-81522129873123914 q^{11}+22048899523933360 q^{12}-4255854634983666 q^{13}+140351921539830 q^{14}+717252508230687 q^{15}-931726090681386 q^{16}+846143031204471
   q^{17}-562446780417522 q^{18}+270743829850494 q^{19}-92247367548336 q^{20}+21188182319304 q^{21}-2953759514706 q^{22}+189360225054 q^{23}\right) \log (q) \text{Li}_2(1-q)\big] \mathrm dq$$

Body:
$$7000862380461139672573968144870 \zeta
   (3)-\frac{25106433798599975547209462384350566086427971}{3127445521200}-\frac{15277122989236784320739564512904737 \pi
   ^2}{58344}+36405909335514889429116587730 \pi ^4$$

should be the answer


问:
Solve a trigonometric equation
I'm trying to solve this trigonometric equation: $3\tan x+5=2$.
This is how I'm trying to solve it:
$$
3\tan x+5=2
\Leftrightarrow \tan x+5=\frac{2}{3}
\Leftrightarrow x+5=0,588+k.\pi
\Leftrightarrow x=-4,412+k.\pi
$$
When I check on photomath it tells me I'm correct however in my math book they say the answer is $-\frac{\pi}{4}+k.\pi$

答: Can you re-arrange this to show that you have $\tan x = -1$ ? And then take it from there?

Title: Solve a trigonometric equation


问:
The chance of an event happening is the square of the chance of a second event .
Question:
The chance of an event happening is the square of the chance of a second event but the odds against the first are the cubes of the odds against the second.The chances of the events are

The answer is given as
p1=p2^{2}

```<code>\frac{1-p1}{p1}=(\frac{1-p2}{p2})^{3}
</code>```

Substituting p1=p2^2 and
solving we get
p1=1/9
p2=1/3

But p1'=8/9 is not equal to cube of p2'.

8/9 is not equal to 8/27
Where is the mistake?
Thank you.....

答:
Your computations are correct, if a bit difficult to read, but your checking is flawed because you must remember that $p_1$ and $p_2$ are probabilities, not odds.  So when we say "the odds against the first event is equal to the cube of the odds against the second event," that literally means $$\frac{1 - p_1}{p_1} = \left( \frac{1 - p_2}{p_2} \right)^3,$$ and the left-hand side of this equation represents the odds against the first event, which from your solution $p_1 = 1/9$ is $$\frac{1 - 1/9}{1/9} = 8,$$ not $8/9$.  Similarly, the odds against the second event is $$\frac{1 - 1/3}{1/3} = 2.$$  And now $8 = 2^3$ and the conditions are verified.

Note that "odds against" will be a number greater than $1$ if the probability in favor is less than $1/2$; and the smaller the probability in favor, the larger the odds against.  This is consistent with our computations above.


问:
Proving from a sequence of real numbers
I am solving the following problem.

Let $a_n$ be a sequence of nonzero real numbers.

Assume that lim sup $|a_{n+1}/a_n|$ = L

a.  Let L’ be any number bigger than L. Prove that there exists N>0 such that $|a_{n+1}/a_n| < L’$ for any n > N.

b.  Prove that for any n>N we have |$a_n|<(L’)^{n-N}|a_N|$

For a. I know I want to use the definition of convergence which says for every epsilon greater than zero there exists a N>0 such that  n>N $\implies$ $|a_n-a|<\epsilon$ but my sequence is a little different and for that reason I am having a hard time with this problem. Thanks

答:
Because $\left|\frac{a_{n+1}}{a_n}\right| \to L$ and $L' > L$, we can find some $N \in \Bbb{N}$ such that
\begin{align*}
n > N &\implies \left|\left|\frac{a_{n+1}}{a_n}\right| - L\right| < L' - L \\
&\implies \left|\frac{a_{n+1}}{a_n}\right| - L < L' - L \\
&\implies \left|\frac{a_{n+1}}{a_n}\right| < L',
\end{align*}
proving (a). As a bit of intuition, I chose $\varepsilon = L' - L > 0$ because $L' - L$ is the distance between the limit $L$ and $L'$. So, when the sequence gets closer to $L$ than this distance, it has to lie below $L'$.

For part (b), we prove by induction on $n - N$. When $n - N = 1$, then
$$N + 1 > N \implies \implies \left|\frac{a_{N+1}}{a_N}\right| < L' \implies |a_{N+1}| < (L')^1 |a_N|.$$
So, let us suppose the given statement holds for $n - N = k$ for some $k \ge 1$. That is, for this $k$,
$$|a_{N+k}| < (L')^k |a_N|.$$
Then,
\begin{align*}
N + k + 1 > N &\implies \left|\frac{a_{N+k+1}}{a_{N+k}}\right| < L' \\
&\implies |a_{N+k+1}| < L'|a_{N+k}| \\
&\implies |a_{N+k+1}| < L'((L')^k |a_N|) \\
&\implies |a_{N+k+1}| < (L')^{k+1} |a_N|,
\end{align*}
completing the proof by induction.

Title: Proving from a sequence of real numbers


问:
How many solutions $x(t)$ of this ODE satisfy $\lim_{t \to \pm \infty}x(t) =0$?
I don't know how to solve the following problem. How many solutions of $$x^{\prime}(t) = x(t) -e^{-t^2}$$ satisfy the conditions
$\lim_{t \to \pm \infty}x(t) =0$?
Should I look for some explicit formula? Can someone give me a hint? Thanks a lot!

答:
The limit at $-\infty$ is trivial. The linear nature gives an affine linear dependence for large $t$ of $x(t)$ on $x(0)$. Hand-waving one can say that between the solutions diverging to $-\infty$ and $+\infty$ for $t\to+\infty$, there has to be one converging to $0$.

However, that is not really certain, you can only make that precise with the solution formula.


答:
The confusion probably stems from the fact that $3\tan (x) +5 \neq 3\tan (x+5)$.

The intended answer in your math book corresponds to the equation $3\tan (x) +5 = 2$, while you interpreted the question differently and solved the equation $3\tan (x+5) = 2$ instead.

Parentheses matter!

Title: Solve a trigonometric equation

Body:
I'm trying to solve this trigonometric equation: $3\tan x+5=2$.
This is how I'm trying to solve it:
$$
3\tan x+5=2
\Leftrightarrow \tan x+5=\frac{2}{3}
\Leftrightarrow x+5=0,588+k.\pi
\Leftrightarrow x=-4,412+k.\pi
$$
When I check on photomath it tells me I'm correct however in my math book they say the answer is $-\frac{\pi}{4}+k.\pi$


问:
Oscillating Euler DE and its solution
Consider the DE $$y''+\frac{k}{x^2}y=0 \quad \Bbb{for} \quad x>0$$
Despite the fact that $$\int_1^{\infty}dx\;x^{-2}$$ exists, the solutions of this DE may oscillate.

a) Show that they do if k>1/4 and dont if k≤1/4.

b) Show that the solutions of $$y'' + \frac{1}{1+x^2}y = 0 $$ do oscillate by referring to the result of a).

By my understanding, the question is asking me to solve the Euler DE given and obtain the k value as 1/4. But how would I know that the last equation oscillates just from the results of a)? Would it be just because the overall function is the same?

答:
a) Find, as usual for an Euler-Cauchy DE, basis solutions in the form $y(x)=x^r$ with characteristic equation $$0=r(r-1)+k=(r-\tfrac12)^2+k-\tfrac14.$$ In the case of non-real $r$, combine the conjugate solutions into a real basis $\sqrt{x}\cos(\omega\ln x),\sqrt{x}\sin(\omega\ln x)$.

b) Apply the Sturm-Picone comparison theorem, you could even use it twice to get upper and lower bounds from $$\frac1{(x+1)^2}\le \frac1{1+x^2}\le\frac1{x^2}.$$

This means that between any two roots of $\cos(ω\ln(x+1))$ you find at least one root of $y(x)$, and between any two roots of $y(x)$ you find a root of $\cos(ω\ln(x))$. Or more adapted, if $y(a)=0$, then $y$ as another root $b$ before the next root of $\sin(ω(\ln(x+1)-\ln(a+1)))$ and after the root of $\sin(ω(\ln x-\ln a))$, $$ae^{\pi/ω}\le b\le (a+1)e^{\pi/ω}-1.$$ As this gives finite distances between roots, there has to be an infinite number of roots.

Title: Oscillating Euler DE and its solution


问:
Geometric-Variational idea of Sine
I like to see sin(theta) as a property of a line making clockwise angle of theta with a horizontal axis.That property would dictate how much %  the vertical component of each point in the line represents in relation the hiponetuse component of the point. Sin(80degrees) = 0,98 . That is, choosing any point on a line making 80degrees with a horizontal axis, we get that the vertical component is only 98% of the hipotenusa component.

I'm trying to go from there to the graph of a sin function.

Can anyone help me see intuitively why this % ( the values of the sine ) goes smoothly decelerating ( forming that semi-circle in the sin graph ) as we increase the angle towards 90?       Why doesn't it go linearly decelerating for example ?

I don't wanna bring circles in since sin is just one component of the circle.

Thanks a lot.

答:
What you've identified is an important reason mathematicians don't use the classical geometric definition of sine (opposite-over-hypotenuse, SOH-CAH-TOA) as the formal definition - it's not clear what happens after 90 degrees since you can't have an obtuse right triangle, much less a triangle with more than 180 degrees!

You said you didn't want to bring circles in but let me show you a different way of looking at the unit circle that might be more enlightening: Consider a point on a unit circle centered at the origin. Draw the radius from the origin to that point - this will be your hypotenuse. Draw a segment from the point straight down to the x-axis - the length of this leg represents the sine of the angle $\theta$ between the radius and the x-axis. Draw another segment along the x-axis to where you dropped the perpendicular down already.

Now imagine moving from $\theta=0^o$ anticlockwise at a constant rate, say 1 degree per second, around the circle. You should see why the graph is the shape it is. Still confused? Take a look here for what I mean.

Title: Geometric-Variational idea of Sine


Body:
Find the inverse elements for the permutations $(15643)$ and $(143)(256)$.

If $\sigma = (15643)$, then isn't $\sigma^{-1}=(34651)$? Since $(15643)\cdot(34651)=(15643)$?

Also for $(143)(256)$ the inverse would be $(652)(341)$ since $(143)(256) \cdot(652)(341)=(143)(256)?$

Body:
The inverse $\sigma^{-1}$ of $\sigma\in S_n$ is an element of $S_n$ such that $$\sigma^{-1}\sigma=\sigma\sigma^{-1}={\rm id}_{S_n}.$$

The inverse of a cycle $(s_1s_2\dots s_{m-1}s_m)$ is $$(s_ms_{m-1}\dots s_2s_1).$$ (Why?)

Hence $(15643)^{-1}=(34651)$ and $((143)(256))^{-1}=(256)^{-1}(143)^{-1}=(652)(341)$.


问:
integral of a discontinous periodic function?

Background

I am a biology student and I have been struggling with finding the integral of the function:

$$ a(t) = a_0\times e^{-k\times(t- \lfloor t/\tau \rfloor * \tau)}  \space\space\space \text{with}\space (k \neq 0, \tau \neq0)$$

Which (for some arbitrary values for $a_0, k$, $\tau $) looks like this. A fancy version of exponential decay.

My understanding of this function is

periodic (for a $\tau > 0$), therefore I only need to determine the integral over one period.

non-continuous /discontinuous.

What I did:
I found the integral of the "non-periodic" version of $a(t)$, lets call that one $\alpha(t)= a_0*e^{-kt}$. So,$$\int \alpha(t) dt = -\frac{a_0*e^{-kt}}k  $$
I then simply did $$\int_0^{\tau} \alpha(t) dt = -\frac{a_0*e^{-k \times \tau}}k +\frac{a_0}k  $$ but this way I am clearly missing that originally $a(\tau)= a_0$ (depicted by the black points in the image above).

My Question

Can I use  $\alpha(t)$ instead of $a(t)$ as to integrate over the interval fomr 0 to $\tau$? I am asking because obviously $a(tau)\neq\alpha(tau)$, but I really can't think of another way.

Bigger picture: $a(t)$ is a simplified model of the concentration of a "pollutant" in a waterbody. The concentration of this "pollutant" influences the growth of microorganisms. By finding $ \int_0^{\tau} a(t)  $ I ultimately want to determine the respective mean growth of microorganisms over $\tau$. But one step at a time. I want to do this analytically, not numerically.

答:
For $0 \leqslant t \leqslant \tau$, the term $\lfloor t/\tau \rfloor = 0$.  Then your integral is correct save only you have dropped the factor $a_0$.  But if you want to integrate over period greater than $\tau$, you need to consider the integral over the initial period:  you will end up with a sum for a whole number of $\tau$ length periods and a residual.

For instance when $T \geqslant \tau$, let $\lfloor T/\tau \rfloor = n$,
\begin{align}
\int_0^T \alpha(t)~ dt
&= a_0\int_0^{\tau} e^{-kt} dt + a_0\int_{\tau}^{2\tau} e^{-k(t-\tau)}~dt + \cdots \\
&\quad\qquad + a_0\int_{(n-1)\tau}^{n\tau} e^{-k(t-(n-1)\tau)}~dt+ a_0\int_{n\tau}^T e^{-k(t-n\tau)}~dt \\
&=a_0\int_0^{\tau} e^{-kt} dt + a_0\int_{0}^{\tau} e^{-kt}~dt + \cdots \\
&\quad\qquad + a_0\int_{0}^{\tau} e^{-kt}~dt+ a_0\int_{0}^{T-n\tau} e^{-kt}~dt \\
&=a_0\frac{\lfloor T/\tau\rfloor}{k}(1-e^{-k\tau})+\frac{a_0}{k}(1-e^{-k(T-\lfloor T/\tau\rfloor \tau)}).
\end{align}
I hope this is helpful.


答:
A solution of homogeneous equation is
$$
x(t) = Ce^{t}.
$$
Now let $C = C(t)$. Then we obtain an equation on $C(t)$:
$$
C'(t)e^{t} = -e^{-t^2}
$$
or
$$
C'(t) = -e^{-t^2-t} = -e^{\frac{1}{4}}e^{-\left(t+\frac{1}{2}\right)^2}.
$$
So
$$
C(t) = c - \sqrt[4]{e}\int_0^te^{-\left(u+\frac{1}{2}\right)^2}du
$$
or
$$
C(t) = c - \sqrt[4]{e}\int_0^{t+\frac{1}{2}}e^{- u^2}du.
$$
Finally we get
$$
x(t) = ce^t - e^{t+\frac14}\int_0^{t+\frac{1}{2}}e^{- u^2}du.
$$

When $t \to \pm\infty$ the integral
$$
\int_0^{t+\frac{1}{2}}e^{- u^2}du
$$
is bounded. So the asymptotic behaviour of $x(t)$ is the one of $Me^t$. When $t \to -\infty$ $x(t) \to 0$ anyway. When $t \to +\infty$
$$
x(t) \to
\left\{\begin{align*}
+\infty, &\quad M > 0\\
0, &\quad M = 0\\
-\infty, &\quad M < 0
\end{align*}\right. .
$$

The expression for $M$ is
$$
M = c - e^{\frac14}\int_0^{\frac{1}{2}}e^{- u^2}du.
$$

Title: How many solutions $x(t)$ of this ODE satisfy $\lim_{t \to \pm \infty}x(t) =0$?

Body:
I don't know how to solve the following problem. How many solutions of $$x^{\prime}(t) = x(t) -e^{-t^2}$$ satisfy the conditions
$\lim_{t \to \pm \infty}x(t) =0$?
Should I look for some explicit formula? Can someone give me a hint? Thanks a lot!


问:
Obtaining oscillatory solution and corresponding normal form of DE
Consider the constant coefficients DE $$y''+by'+cy=0, \quad  x \in \mathbb{R}$$
a) Use the usual approach to obtain the condition so that its solutions are oscillatory.

b) Find the corresponding normal form DE and derive the same condition from it.

I am having trouble understanding the question because I don't know what they mean by an oscillatory solution. And for b) what do they mean by "normal" form of DE? Do they just mean the equation given?

答:
For a), as mentioned in the comments, you need to find criteria for $b$ and $c$ such that the solution has $\sin$ and/or $\cos$ in it. We know that we get a solution on the form $$y = e^{\mu x}(A\cos\lambda x + B\sin\lambda x)$$ if the roots of the characteristic polynomial are complex $r = \mu \pm \lambda i$. In your case, the characteristic polynomial is $$r^2 + br + c = 0$$ see if you can find the criteria for $b, c$ that ensures that the roots are complex.

For b), normal form just means solving for the highest derivative, that is write your differential equation on the form
$$\frac{d^n f}{dx^n} = F(f, f' , f'', \ldots, f^{(n-1)}).$$


答:
The general solution is $$
x(t) = Ce^t  - \frac{{\sqrt \pi  }}{2}e^{t + 1/4} \operatorname{erf}\left( {t + \tfrac{1}{2}} \right).
$$ Since $\mathop {\lim }\nolimits_{t \to  \pm \infty } \operatorname{erf}\left( {t + \frac{1}{2}} \right) =  \pm 1$ with a rate of about $e^{-t^2}$, the only particular solution that satisfies the requirement is the one with $C=\frac{{\sqrt \pi  }}{2}e^{1/4}$: $$
x(t) = \frac{{\sqrt \pi  }}{2}e^{t + 1/4} \operatorname{erfc}\left( {t + \tfrac{1}{2}} \right).
$$


问:
Checking if a recurrence relation converges or not
Consider the recurrence relation $x_{n+1} = \ln{|x_n|}$ where $x_0$ is a real number that is not equal to any multiple of $e$

Will $x_n$ converge to a finite value as $n$ approaches infinity?

答:
If the recurrence converges, it converges to the root of

$$x=\log(|x|)$$ which is about $x_\infty=-0.5671432904098\cdots$.

But $(\log(|x_\infty|))'=\dfrac1{|x_\infty|}>1$, making convergence impossible.