问:
Proving compact $m$-manifolds can be embedded in $\mathbb{R}^{nm}$ for some cover $\{(B_i,\varphi_i)\}_{i=1}^n$.
When a smooth $m$-manifold $M$ is compact, choose $\{(B_i,\varphi_i)\}_{i=1}^n$ a covering by coordinate balls. Use bump functions $\chi_i$ so that ${\chi_i}|_{B_i}=1$. Define
\begin{equation}
    F(p)=(\chi_1 \varphi_1(p),\dots, \chi_n \varphi_n(p))\in\mathbb{R}^{nm}.
\end{equation}
We can show this is a smooth injective immersion $M\to\mathbb{R}^{nm}$, and hence an embedding since $X$ is compact. My question is, how does one show this is an immersion?

答: All you have to do is show that the derivative is injective at any $p$. So WLOG suppose $p \in B_1$, then in that neighbourhood, $F(q) = (\varphi_1(q),...,\chi_n \varphi_n(q))$ since $\chi_1 = 1$ there. We see that $dF_p$ has an invertible $m \times m$ sub-matrix corresponding to $\varphi_1$ since it is a diffeomorphism, and since the matrix is $n \times m$, this makes it an immersion (note that you need $n > m$, so you need a sufficiently "fine" partition of unity).


问:
Computing a projection-induced map on homology of the surface of genus $g$
Let $M_g$ be obtained from a sphere $S$ by gluing $g$ tori to it along disks, in such a way that (if one allows for slight abuse of notation):

$M_g/S\cong \bigvee^gM_1$

Observe that $H_2(M_g)\cong \mathbb{Z}$, and $H_2(\bigvee^gM_1)\cong \mathbb{Z}^{g}$.

Let $\pi:M_g\rightarrow\bigvee^gM_1$ be the projection.
Then we have an induced map on homology $\pi_*:\mathbb{Z}\rightarrow \mathbb{Z^g}$.
My geometric intuition tells me this must be the diagonal map $n\mapsto (n,\ldots,n)$, but I am struggling to come up with a rigorous argument for this.

In particular, I've tried to construct generators represented by singular chains which interact nicely with $\pi$, but so far my attempts end up depending on geometric intuition anyway.

答:
It's not really precise to say "gluing $g$ tori to a sphere $S^2$". Instead, it's possible to state something like "attaching $g$ tori to $S=S^2\setminus\bigsqcup_{i=1}^gB_i^2$" (removing $g$ open disks from the the 2-sphere and attach tori to the boundary by a homeomorphism that identifies the boundary of the attaching tori to $S$ counterclockwise)

With this new definition of $S$, $(M_g,S)$ becomes a good pair, meaning that $H_p(M_g, S)\cong \tilde{H}_p(M_g/S)$. Also, note that $M_g/S\approx\bigvee_{i=1}^g T^2$.

The homology exact sequence says
\begin{align}0\to H_2(S)\overset{i_*}{\to} H_2(M_g)\overset{\pi_*}{\to}H_2(M_g,S)\overset{\partial_*}{\to} H_1(S)\overset{j_*}{\to} H_1(M_g)\to H_1(M_g,S)\\
\to H_0(S)\to H_0(M_g)\twoheadrightarrow H_0(M_g,S)\to 0
\end{align}

By definition, $S\simeq\bigvee_{i=1}^{g-1}S^1$, which implies $H_2(S)\cong 0$ and $H_1(S)\cong \bigoplus_{i=1}^{g-1}\Bbb Z$. Hence, $\pi_*$ is injective. Now, we need to look at the boundary map $\partial_*:\Bbb Z^{\oplus g}\to\Bbb Z^{\oplus g-1}$. This map is surjective because $j_*$ is trivial. If $\beta\in H_2(M_g,S)$ is a class of relative cycle, then $\partial_*(\beta)$ is the class $\partial\beta\in H_1(S)$. For the usual basis of $H_2(M_g, S)$, each $\beta_i\mapsto \alpha_i, i=2,...,g$, where $\alpha_i$ is a generator of $H_1(S)$, but $\beta_1\mapsto \alpha_1\simeq\sum_{i=2}^{g}(-\alpha_i)$ (the cycle enclosing the first removed disk is homologous to the combination of other generating cycles in $S$). So, $\partial_*(\sum_{i}\beta_i)=0\implies\ker(\partial_*)=\langle(\beta_1,\ldots, \beta_g)\rangle=\operatorname{im}(\pi_*)$, meaning that $\pi_*$ is the diagonal map.

ContentLicense: CC BY-SA 4.0


问:
Why is the answer to an integral different for |x|<1 despite the original answer working for this condition as well?
I can't seem to understand why the answer for this same integral yields a different answer with the condition of |x|<1 even though this condition works for both answers regardless?

Example given in notes:

Question on the HW with what the answer is supposed to be on the right]:

In general, I don't seem to understand how to change my answer for such conditions other than trying to use algebra on the answer to eliminating the part that leads to undefined.

答:
Both the integrals are different.

In the first integral you have $\color{blue}{\arctan(x)}$ and $x^2$ and both are defined for all $x \in (-\infty,\infty)$. So the integral can be evaluated for $x \in (-\infty, \infty)$.

Whereas in the second integral you have $\color{blue}{\arcsin(x)}$ and $x^2$. Although $x^2$ exists for all $x \in (-\infty,\infty)$, $\color{blue}{\arcsin(x)} $ is defined only for $\color{blue}{-1\le x\le 1}$ or $\color{blue}{|x|\le 1}$, so the integral is limited to $|x| \le 1$


问:
Almost sure convergence of L1 bounded sequence implies L1 convergence
I wanted to ask whether the following statement is true:

Assume random variables $X_n \to X$ almost surely, and that $\left\lvert X_n\right\rvert \leq Y$ with $Y\in L^1$, then $X_n \to X$ in $L^1$

答: You have $|X_n|\to|X|$ almost surely, and $|X_n|\le Y$ with $Y$ integrable. So by dominated convergence, conclude $E|X_n|\to E|X|$. Now you can apply Scheffé's lemma.


问:
Joint distribution of two uniform RV's
Let $X \sim U(0,\theta_1) \text{ and } Y \sim U(0,\theta_2)$ be independent RV's. We're interested in $\mathbb{P}[X-Y\le z]$.

I've done the convolution (or double integrals), denote $Z=X-Y$
\begin{align}
\mathbb{P}[Z\le z] &= \int_{[x-y<z]} dF_X dF_Y \\
&=\int_0^{\theta_2} \int_0^{z+y} \frac{1}{\theta_1} dx \frac{1}{\theta_2} dy\\
&=\frac{z}{\theta_1}+\frac{\theta_2}{2\theta_1}
\end{align}

is this correct?

for $\theta_1=\theta_2 = 1$ we're getting $\mathbb{P}[X-Y\le z] = z+\frac12 $, which intuitively makes sense.

I'm also curious about the joint pdf. We get the pdf by differentiating the cdf w.r.t. $z$. So $f_Z(z) = \frac{1}{\theta_1}$. Had we chosen the convolution the other way around $$\int_0^{\theta_1} \int_0^{z+x} \frac{1}{\theta_2} dy \frac{1}{\theta_2} dx $$ we would have obtained the joint pdf $f_Z(z) = \frac{1}{\theta_2}$. Isnt' that weird?

答:
Assuming that $X$ and $Y$ are independent, the joint distribution is

$f(x,y) = \frac{1}{\theta_1 \theta_2}$ if $0 \le x \le \theta_1, 0 \le y \le \theta_2$ and $0$ otherwise.

The correct way to do this is to integrate $f(x,y)$ over the region $x-y\le z$. The easiest way to compute this integral is to do it geometrically. Compute the area of the intersection of the region $x-y\le z$ and the rectangle $[0, \theta_1] \times [0, \theta_2]$ and divide it by $\theta_1 \theta_2$.

You can also use the convolution to compute the distribution of $Z = X - Y$ and find $P\{Z \le z\}$ by integrating $\int_0^z f_Z(s) ds$.

ContentLicense: CC BY-SA 3.0

Title: Joint distribution of two uniform RV's


问:
Floating point division resulting in a value exceeding 1 but should be equal to 1...maybe!
I am computing the apparent magnitude of comets and minor planets using data from the Minor Planet Center.  The formula I use has a division and that result is passed to the arccosine function (in Python).

Occasionally the result of the division is slightly greater than 1, for example

6.272370848320445 / 6.2723697617591405 = 1.0000001732297912

I assume the result is not actually greater than 1; rather the result is an artefact of floating point arithmetic, particularly given the number of zeros after the decimal point.

If my assumption is correct, I want to catch this sort of situation and treat the result as if the division yielded exactly 1 (and so can safely be passed to arccosine).

If I make an arbitrary choice of say five decimal points and truncate the division result, then if the result is 1 I'm good, otherwise, drop the result.

I'm not sure if what I'm proposing is correct mathematically, so looking for guidance please.  I have assumed this issue arises due to floating point arithmetic, but if that assumption is wrong, then I'm not sure what else I can do.  It is possible the source data from the Minor Planet Center contains spurious data which is causing the actual error (and thus I shouldn't be trying to work around it but just let it fail).

The formula can be found here (using the H,G model).

答:
I looked at the link you mention in your question and I don't
have time to study all of the details. However, the expression

```<code>beta = acos((rp*rp + rho*rho - rsn*rsn)/(2*rp*rho));
</code>```

seems like an instance of
Law of cosines.
This can be rewritten as

```<code>beta = acos(1 -(rsn+rho-rp)(rsn-rho+rp)/(2*rp*rho));
</code>```

The problem in computing this is that rsn+rho-rp or
rsn-rho+rp loses precision if rsn and |rho-rp|
are of comparable value and may become negative. This
happens when the triangle is close to degenerate where
one of the angles becomes very close to zero. In these
cases you can use the haversine
idea and use the formula

```<code>beta = 2*asin(sqrt(max(0,(rsn+rho-rp)(rsn-rho+rp)/(4*rp*rho)));
</code>```

or the approximation when the angle is close to zero

```<code>beta = sqrt(max(0,(rsn+rho-rp)(rsn-rho+rp)/(rp*rho)));
</code>```

where the max(0,.) ensures that we don't attempt to
take the square root of a negative number.

Note that the acos(x) formula is great when $x$ is
close to zero, but the asin(x) formula is better when
$x$ is close to one in absolute value.

Body:
I looked at the link you mention in your question and I don't
have time to study all of the details. However, the expression

beta = acos((rp*rp + rho*rho - rsn*rsn)/(2*rp*rho));

seems like an instance of
Law of cosines.
This can be rewritten as

beta = acos(1 -(rsn+rho-rp)(rsn-rho+rp)/(2*rp*rho));

The problem in computing this is that rsn+rho-rp or
rsn-rho+rp loses precision if rsn and |rho-rp|
are of comparable value and may become negative. This
happens when the triangle is close to degenerate where
one of the angles becomes very close to zero. In these
cases you can use the haversine
idea and use the formula

beta = 2*asin(sqrt(max(0,(rsn+rho-rp)(rsn-rho+rp)/(4*rp*rho)));

or the approximation when the angle is close to zero

beta = sqrt(max(0,(rsn+rho-rp)(rsn-rho+rp)/(rp*rho)));

where the max(0,.) ensures that we don't attempt to
take the square root of a negative number.

Note that the acos(x) formula is great when $x$ is
close to zero, but the asin(x) formula is better when
$x$ is close to one in absolute value.


问:
Is there a strategy for coming up with examples of operators for these types of problems?
"Give an example of an operator $T$ from $\mathbb{C}^{3}$ such that $2$ and $3$ are the only eigenvalues of $T$ and $T^2-5T+6 \neq 0$."

I was able to find an example after some trial, but I was wondering if there is a general strategy for this problem. Is there something I should be looking out for or trying first here?

答:
It's all about testing your understanding of connections between characteristic polynomial, annihilating polynomial and minimal polynomial.

On one hand, since $2$ and $3$ are the only eigenvalues of $T$, the minimal polynomial of $T$ has to be of the form $d(\lambda) = (\lambda - 2)^{m_1}(\lambda - 3)^{m_2}$, where $m_1 \geq 1, m_2 \geq 1$, $m_1 + m_2 \leq 3$ (because the eigenvalues are zeros of $d(\lambda) = 0$). This means the only possibilities of $(m_1, m_2)$ combinations are $(m_1, m_2) = (1, 1), (m_1, m_2) = (2, 1), (m_1, m_2) = (1, 2)$.

On the other hand, $T^2 - 5T + 6 \neq 0$ means $\lambda^2 - 5\lambda + 6 = (\lambda - 2)(\lambda - 3)$ is not an annihilating polynomial of $T$. Thus rules out the possibility $(m_1, m_2) = (1, 1)$ (because $d(\lambda)$ is an annihilating polynomial of $T$).

Therefore, $T$ can be any operator that is similar to

\begin{align*}
J_1 = \begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix}, \quad \text{corresponding to } (m_1, m_2) = (2, 1),
\end{align*}
or any operator that is similar to
\begin{align*}
J_2 = \begin{pmatrix}
2 & 0 & 0 \\
0 & 3 & 1 \\
0 & 0 & 3
\end{pmatrix}, \quad \text{corresponding to } (m_1, m_2) = (1, 2).
\end{align*}


问:
Isomorphism between $H_1(X)$ and $\pi_1(X)^{ab}$
I'd like to understand the proof of the following theorem, born from the conciseness of the proof given. It's also called Hurewicz's theorem in the special case $n=1$. The essential doubts about the proof, which doesn't go in detail about, are :

$\textbf{1}.$ Is $l'$ well defined ? It seems to me that the $u(x)$ is a generic path, which I think is not what we want since doesn't give the property $lh = id$. For example given a loop $\sigma = h([\gamma])$ with basis in $x_0$, we could consider instead of $u(x_0) = c_{x_{0}}$ the costant loop in $x_0$, a random loop so that we have a "$3$-loop" which doesn't return in $[\gamma]$ with $l'$.

$\textbf{2}.$ I don't how $\triangle^2$ being contractible implies that $\tau_2 \star \tau_1 \sim \tau_1$. I think I can prove the statement using that the simplex is convex but I don't see how the contractbile property of the space implies such homotopy

$\textbf{3}.$ I think the question three is linked to question one, I'd like to understand why the last equality holds : $[\sigma] = [u(\sigma_0)]+[\sigma] - [u(\sigma_1)] =  [(u(\sigma_0)\star \sigma)\star u(\sigma_1)^-]$

I'm currently reading the proof on Algebraic Topology of Tammo Tom Dieck, I add below a picture of the proof of the theorem.

答:
For question 1, $l'$ is well-defined because $C_1(X)$ is a free abelian group generated by the 1-simplices of $X$, so you get a well-defined map from $C_1(X)$ to any other abelian group by specifying where the generators go. If you have a loop based at $x_0$, then the key is that you have to use the same path $u$ from $x_0$ to $x_0$ at both ends in the definition of $l'$, traveling along $u$ once in each direction, so $l h$ will indeed be the identity map.

For question 2, the statement is accurate, but maybe to make it easier, the author should have said that $\Delta^2$ strongly deformation retracts to any of its vertices (for example because of convexity). But see Path homotopy in contractible space if you just want to use contractibility.

For question 3, it is not the case that $[\sigma] = \dots$. Rather, the extra terms $\sum a_{\sigma}([u(\sigma_0)] - [u(\sigma_1)])$ taken together give precisely the boundary of $\sum a_{\sigma} \sigma$ and therefore add up to zero, since the original sum is assumed to be a cycle. For the last equality, $u(\sigma_0) + \sigma - u(\sigma_1)$ should be homologous to $(u(\sigma_0)*\sigma)*u(\sigma_1)^{-}$, so the sums are equal when working in $H_1$.


问:
Path-connectedness of $\mathbb{C}\setminus \lbrace z_1, ..., z_n \rbrace$
Today I was thinking about the fact that $\mathbb{C} \setminus \lbrace z_1, ..., z_n \rbrace$ is path-connected. I know how to prove that, just constructing a path joining two points, but I was wondering : is there a way to prove this using the fact that the image of a path-connected space by a continuous function is path-connected ?

More precisely, it is possible prove that $\mathbb{C} \setminus \lbrace z_1, ..., z_n \rbrace$ is path-connected by defining explicitely a surjective continuous function $f : X \rightarrow  \mathbb{C} \setminus \lbrace z_1, ..., z_n \rbrace$, where $X$ is a path-connected topological space ?

For the case $n=1$, using $\exp : \mathbb{C} \rightarrow \mathbb{C}^*$ works, but I was not able to generalize that in the case you remove more than one point.

Thanks for your help !

答:
First of all, the explanation "just constructing a path joining two points" proves absolutely nothing. I suggest, you write a careful proof of path-connectedness of $Z^c$ which is the complement to a given finite subset $Z\subset {\mathbb C}$ and compare complexity of the proof to the complexity of your proof that ${\mathbb C}^*$ is path-connected (it hinges upon proving continuity and surjectivity of the complex exponential function $\exp: {\mathbb C}\to  {\mathbb C}^*$: Proving this from scratch is a nontrivial challenge).

No, let's prove path-connectivity of $Z^c= {\mathbb C}\setminus Z$ along the lines of your request.

First, consider a finite subset $W\subset  {\mathbb R} \subset {\mathbb C}$. Then $W^c$ is the union of two subsets $U^+, U^-$, where
$$
U^+=\{z: Im(z)\ge 0\}\setminus W$$
and
$$
U^-=\{z: Im(z)\le 0\}\setminus W.$$
In fact, both sets are homeomorphic via the complex conjugation map. I claim that $U^+$ is path-connected. Indeed, it is star-like with respect to the point $i$: For each $z\in U^+$ the line segment $[i, z]$ between $i$ and $z$ is contained in $U^+$. I will leave it for you to verify that each star-like subset of  ${\mathbb C}$ is path-connected. (Concatenate two segments $[p, i], [i,q]$ to connect points $p, q\in U^+$.) It follows that $U^-$ is path-connected as well. Observe that $U^+\cap U^-\ne \emptyset$, which implies path-connectivity of  the union $U^+\cup U^-=W^c$.

Second, I will use the fact that if $W, Z$ are two subsets of the same cardinality in ${\mathbb C}$ then there exists a homeomorphism $h: {\mathbb C}\to {\mathbb C}$ sending $W$ bijectively to $Z$, see for instance here for an explicit construction.

Thus, given aby finite subset $Z\subset {\mathbb C}$, I find a finite subset $W\subset
 {\mathbb R} \subset {\mathbb C}$ of the same cardinality and, hence, a homeomorphism
$$
f: X:= W^c\to Z^c
$$
obtained by restricting the homeomorphism $h$. This is the continuous surjective function from a connected space $X$ to $Z^c$ you asked for.

Needless to say, this is much more complex than the standard direct proof of path-connectivity of $Z^c$.


问:
Let X, Y, Z and W be sets defined on the universal set $U = N $ as follows: How can I solve this Venn Diagram?
So I have the following assertions and I have to illustrate this on a Venn Diagram.

$ (X - Y) \cap  Z = ${1,2,3,4}

$ Y =  $ {5,6}

$Z \cap Y = \emptyset $

$ W \cap (X - Z) =$ {7,8}

$ X \cap W \cap Z = $ {2,4}

So I did the following code on Python,

```<code>import matplotlib.pyplot as plt
from matplotlib_venn import venn3

v = venn3(subsets = (1,2,3,4,5,6,0,7,8,2,4), set_labels = ('X', 'Y', 'Z', 'W'))

plt.show()

</code>```

but it's not working because I need Four-Ellipse Venn Diagram to solve it. And I have to specify when two sets intersects and when I have the diference between them.

I'm very stuck and I'm not able to find the logic in this exercise anymore.
Any tips on how to solve it?
I also tried to draw Venn Diagram on paper but I also failed.

答:
here is what i did.
first I get a big paper and draw three regions, like shown at this link:
https://commons.wikimedia.org/wiki/File:Intersection_of_3_circles_0.svg

these regions I label X, Z, W. I think it is better to wait to draw region Y, because it is not clear where to put it initially. Then I look at the bullets (the requirements) individually.

bullet 5 is easy to satisfy. Now look at bullet 3; introduce the final circle for region Y so that bullet 3 is happy.

Next is bullet 4.  You have to put 7,8 somewhere. There are two possible places to put 7,8. Which one is correct?


问:
Solve for element of a matrix
Say we have a system of equations

$$\mathbf{xA}=\mathbf{y}$$

where $\mathbf{A}$ is an $n\times n$ matrix and $\mathbf{x}$ and $\mathbf{y}$ are both $1\times n$. Say we wanted to write an expression for element $a_{ij}$ of this matrix. Is this possible?

答: This problem is very much underdetermined.  Look at some simple examples, say $\begin{bmatrix} x_1 & x_2 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} y_1 & y_2 \end{bmatrix}$: The only information you have on $a$ is that $x_1 a + x_2 b = y_1$: If $x_1$ and $x_2$ are nonzero then this has infinitely many solutions for $a$, as you can choose $a$ freely and adjust by choice of $b$.

Title: Solve for element of a matrix


问:
Find this limit without using L'Hopital's rule or Taylor Series
I want to solve $$\lim_{x\to 2}{{\log_2(\sqrt{4x-x^2})-\sqrt{\cos{\pi x}}}\over(\frac{x}{2})^{(\sin{\pi x})}-1}$$
without using differentiation in any context. Basically without using L'Hopital's rule and Taylor's theorem.

I have solved the given limit by using Taylor's theorem in order to find equivalent functions and I have gotten the correct answer. However I am required to do so, without actually differentiating. I am guessing that already established equivalent function formulae and substitutions are the only two tools that I am allowed to use.

Any hints or help will be very much appreciated.

答: As suggested in comments let's put $x/2=1+t$ so that $$4x-x^2=8(1+t)-4(1+t)^2=4(1-t^2)$$ and thus the expression under limit becomes $$\frac{2+\log_2(1-t^2)-2\sqrt{\cos 2\pi t}}{2((1+t)^{\sin 2\pi t}-1)}$$ Since both terms of denominator tend to $1$ we can safely replace terms by their logarithm (justify this!!) to get $$\frac{2+\log_2(1-t^2)-2\sqrt{\cos 2\pi t}} {2\sin 2\pi t\log(1+t)}\tag{1}$$ This can be split into two terms the first of which is $$\frac{1-\sqrt{\cos 2\pi t}} {\sin 2\pi t\log(1+t)}=\frac{1-\sqrt {\cos 2\pi t}} {1-\cos 2\pi t} \cdot \frac{1-\cos 2\pi t} {(2\pi t) ^2}\cdot\frac{2\pi t} {\sin 2\pi t} \cdot\frac{2\pi t} {\log(1+t)}$$ and this tends to $$A=(1/2)(1/2)(1)(2\pi)=\frac{\pi}{2}$$ The other term in expression $(1)$ is $$\frac{\log_2(1-t^2)}{2\sin 2\pi t\log(1+t)}=\left(-\frac{1}{4\pi\log 2}\right)\cdot\frac{\log(1-t^2)}{(-t^2)}\cdot\frac{2\pi t} {\sin 2\pi t} \cdot\frac{t}{\log(1+t)}$$ which tends to $$B=-\frac{1}{4\pi\log 2}$$ The desired limit is thus $$A+B=\frac{\pi} {2}-\frac{1}{4\pi\log 2}$$


问:
My son's Sum of Some is beautiful! But what is the proof or explanation?
My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into the following equation:
$$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}.$$

Why is this so? What is the proof or explanation? Is it true for any $n$?

His own presentation of his finding:

Every one of these numbers is two times the number before it.
  $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.

  I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that.

  If I then skip three ones and read the fourth, that one equals my sum!

Body: Factor out the $2^n$ and you get: $2^n (100+20+8) = 2^n 128 = 2^{n+7}$ since $2^7 = 128$


问:
Calculate left and right Haar measures and modular character of affine group
I would like to calculate the left and the right Haar measures of the group
\begin{equation*}
    G = \left \{ \begin{pmatrix} a & b \\ & 1 \end{pmatrix} \:| \: a \in \mathbb{C}^\times, b \in \mathbb{C} \right \}
\end{equation*}
and its modular character.

I have seen this calculation for the affine group in one dimension given by
\begin{equation*}
    H = \left \{ \begin{pmatrix} a & b \\ & 1 \end{pmatrix} \:| \: a > 0, b \in \mathbb{R} \right \},
\end{equation*}
however I am stuck in this complex setting.
I was hoping that you could explain how to carry out this calculation.

Thank you very much in advance.

答:
Suppose there is some density on $G=\mathrm{Aff}_1\mathbb{C}$ which integrates to the measure:

$$ \mu(X)=\int_X \rho(x)\,\mathrm{d}x . $$

(Here, $\mathrm{d}x$ is from the usual Lebesgue measure on $\mathbb{C}^{\times}\times\mathbb{C}$ interpreted as a subspace of $\mathbb{R}^4$.)

Then left-invariance $\mu(X)=\mu(gX)$ may be expressed as

$$ \begin{array}{ll}
\displaystyle \int_X \rho(x)\,\mathrm{d}x & \displaystyle =\int_{gX} \rho(x)\,\mathrm{d}x \\[5pt] & \displaystyle =\int_X \rho(gx)\,\mathrm{d}(gx) \\[5pt] & \displaystyle = \int_X\rho(gx)J_{g,x}\,\mathrm{d}x \end{array} $$

where we use a change of variables and $J_{g,x}$ is the Jacobian determinant of left-multiplication-by-$g$. Since this is true for all $X$, the integrands must match, so set $x=e$ and $\rho(e)=1$ so $\rho(g)=J_{g,x}^{-1}$.

Set $g=(a,b)=(a_1+a_2i,b_1+b_2i)$ and $x=(x_1+x_2i,y_1+y_2i)$ so that

$$ gx=\big((a_1x_1-a_2x_2)+(a_2x_1+a_1x_2)i,(a_1y_1-a_2y_2+b_1)+(a_2y_1+a_1y_2+b_2)i\big) $$

with corresponding determinant

$$ J=\det\begin{pmatrix} a_1 & -a_2 & 0 & 0 \\ a_2 & a_1 & 0 & 0 \\ 0 & 0 & a_1 & -a_2 \\ 0 & 0 & a_2 & a_1 \end{pmatrix}=(a_1^2+a_2^2)^2. $$

So we conclude $\rho(a,b)=1/|a|^4$.

I'll let you figure out the right-invariant Haar measure and modular function.


问:
Create N points that are spaced as far as possible within a D dimensional sphere
I'm not a big geometry buff, but I was wondering if you're give a dimension $D$ and a number of points $N$ and a radius $R$ of a sphere in that dimension. How would you generate $N$ points on the edge or in the sphere such that their magnitude and cosine similarity between each point is as far as possible? The dimensional space could be something as large as 512.

The ultimate goal is to achieve something like: Assume we're placing planets in a galaxy that is spherical, but we have no idea how big each planet is, so we want to place all the origins of each planet as far as possible from each other in order to avoid overlapping even if the planets are very large.

答:
In one dimension, you can place two points at $x = 1$ and $x = -1$. If you need more points, the best you can do is to evenly space them between the endpoints because the entire boundary is occupied.

In two dimensions, you can use the corners of a regular n-gon inscribed in the unit circle, to get an arbitrarily large number of equally spaced points.  (I'm not completely clear what your criterion is, but if you wanted interior points as well, I'm not sure yet how you would optimize for closest to equal distance between all pairs of points; it may be that the n-gon is still optimal even in that case.

In three dimensions, I have always found it interesting from VSPER molecular geometry in chemistry, that you can equally space two, or three, or four, or six atoms around a central atom, but for five the closest you can get is the trigonal bipyramidal form, with two on the "North Pole" and "South Pole" and three equally spaced around the "equator".  Likewise the compound $IF_7$ demonstrates that the lowest energy state for seven has the poles and five equally spaced around the equator.  In the chemistry case, what is being optimized is the lowest energy state of the mutual repulsion of the electron domains around the central atom, which sounds similar to the condition you want.  Using point groups based on the Platonic solids (tetrahedron, cube, octahedron, dodecahedron, icosahedron) you can get a few more with total symmetry.  Buckminsterfullerene $(C_{60})$ has a higher symmetry, probably the maximum that can be achieved in 3 dimensions.

In higher dimensions, with $N \leq D$, you can use the lower dimensional arrangements, which remain optimal in the higher dimensional setting.  For $N = D + 1$ the generalization of a tetrahedron has total symmetry. I've never worked out that formula myself but I am sure it has been done.  It reminds me of the formula for the content (generalization of length, area, volume) of a hypersphere in n dimensions.

If you can state the exact condition you are trying to optimize, it might help others to home in more precisely on what you need.


问:
Does $\int_{0}^{\infty} \frac{\sin (\tan x)}{x} \ dx $ converge?
Does $ \displaystyle \int_{0}^{\infty} \ \frac{\sin (\tan x)}{x} dx $ converge?

$ \displaystyle \int_{0}^{\infty} \frac{\sin (\tan x)}{x} \ dx = \int_{0}^{\frac{\pi}{2}} \frac{\sin (\tan x)}{x} \ dx + \sum_{n=1}^{\infty} \int_{\pi(n-\frac{1}{2})}^{\pi(n+\frac{1}{2})} \frac{\sin (\tan x)}{x} \ dx $

The first integral converges since $\displaystyle \frac{\sin (\tan x)}{x}$ has a removable singularity at $x=0$ and is bounded near $ \displaystyle \frac{\pi}{2}$.

And $ \displaystyle \int_{\pi(n-\frac{1}{2})}^{\pi(n+\frac{1}{2})} \frac{\sin (\tan x)}{x} \ dx$ converges since $\displaystyle \frac{\sin (\tan x)}{x}$ is bounded near $\pi(n-\frac{1}{2})$ and $\pi(n+\frac{1}{2})$.

But does $ \displaystyle \sum_{n=1}^{\infty} \int_{\pi(n-\frac{1}{2})}^{\pi(n+\frac{1}{2})} \frac{\sin (\tan x)}{x} \ dx$ converge?

答:
Yes.

Note that
$$\begin{align} I_n:=\int_{\pi(n-\frac12)}^{\pi(n+\frac12)}\frac{\sin(\tan(x))}{x}\,\mathrm dx&=\int_0^{\frac\pi2}\left(\frac1{n\pi+x}-\frac1{n\pi-x}\right)\sin(\tan(x))\,\mathrm dx\\&=\int_0^{\frac\pi2}\frac{-2x}{n^2\pi^2-x^2}\sin(\tan(x))\,\mathrm dx\\
\end{align}$$
With $A:=\int_0^{\frac\pi2}\max\{-2x\sin(\tan (x)),0\}\,\mathrm dx$, $B:=\int_0^{\frac\pi2}\min\{-2x\sin(\tan (x)),0\}\,\mathrm dx$ (which both converge), we can thus estimate
$$  \frac{1}{n^2\pi^2-\pi^2/4}B+\frac{1}{n^2\pi^2-0}A\le I_n\le \frac{1}{n^2\pi^2-\pi^2/4}A+\frac{1}{n^2\pi^2-0}B,$$
The difference between these bounds and $\frac1{n^2\pi^2}(A+B)$ is governed by
$$\frac{1}{n^2\pi^2-\frac{\pi^2}{4}}-\frac1{n^2\pi^2}=\frac1{4\pi^2 n^4-\pi^2 n^2},$$
hence
$$ I_n=\frac1{n^2\pi^2}(A+B)+O(n^{-4}).$$
We conclude that $\sum I_n$ converges at least as good as $\sum\frac1{n^2}$.


问:
How to show two compatible charts are of the same dimension?
I encounter a statement in Lie Algebras and Lie Groups by Serre that if two charts $(U,\phi,n)$ and $(U', \phi', n')$ on $X$ are compatible and $V = U\cap U' \ne \emptyset$, then $n=n'$.

A chart $c=(U,\phi,n)$ on $X$ satisfies

$U\subseteq X$ is open

$n\in \mathbb{Z}$ and $n \ge 0$

$\phi:U\to \phi(U)\subseteq k^{n}$ is open and $\phi$ is a homeomorphism, where $k$ is a field.

Two charts $(U,\phi,n)$ and $(U',\phi',n')$ are said to be compatible if, setting $V = U\cap U'$, the maps $\phi'\circ \phi^{-1}\mid_{\phi(V)}$ and $\phi\circ\phi'^{-1}\mid_{\phi'(V)}$ are analytic.

Most of the materials simply include $n=n'$ implicitly in their definitions, so I have difficulty probing the statement above.

EDIT: The complete definition of analytic for arbitrary $k$ is as follows.

For $\alpha=(\alpha_{1},\cdots, \alpha_{n}), x= (x_{1},\cdots, x_{n})$, denote $x^{\alpha} = (x_{1}^{\alpha_{1}}, \cdots, x_{n}^{\alpha_{n}})$. For any open subset $k^{n}$, we say that a function $f = (f_{1},\cdots,f_{m}):U\to k^{m}$ is analytic at $x\in U$ if there exists $r=(r_{1},\cdots,r_{n})\in \mathbb{R}_{+}^{n}$ s.t. $f_{j}(y) =\sum\limits_{\alpha \in \mathbb{N}^{n}}c_{\alpha}^{(j)} (y-x)^{\alpha}$ for any $y$ in $B(x,r)=\{y\in k^{n}:|y_{i}-x_{i}|<r_{i} \}$. And the following series converges
\begin{equation}
\sum\limits_{\alpha\in\mathbb{N}^{n}}|c_{\alpha}^{(j)}|r^{\alpha} < \infty
\end{equation}
We say $f$ is analytic if it is analytic everywhere on $U$.

答:
Thanks to @guidoar, I figure out a proof.

Let $F=\phi'\circ \phi\mid_{\phi(V)}$, then $F^{-1}=\phi\circ(\phi')^{-1}\mid_{\phi'(V)}$ and both $F$ and $F^{-1}$ are analytic. By $id_{\phi(V)}=F^{-1}\circ F$ and the chain rule, we have $I_{n} = J_{F^{-1}\circ F}(p)=J_{F^{-1}}(F(p))\cdot J_{F}(p)$ for all $p\in \phi(V)$. Similarly, by $id_{\phi'(V)}$, we get $I_{n'} = J_{F}(p)\cdot J_{F^{-1}}(F(p))$ for all $p\in \phi(V)$.

According to the property
\begin{equation}
tr(AB)=tr(BA)
\end{equation}
we have $n=tr(I_{n}) = tr(J_{F^{-1}}(F(p))\cdot J_{F}(p)) = tr(J_{F}(p)\cdot J_{F^{-1}}(F(p))) = tr(I_{n'}) =n'$.


问:
Determine the equations of the tangents to the curve $y=2x^2+3$ that pass through point $(2,3)$
I am unsure of how to answer this question. All I have found is the slope of the tangents: $y'= 4x$.

答:
Hint:

The equation of tangent at $P(h,2h^2+3)$

will be $$\dfrac{y-(2h^2+3)}{x-h}=4h$$

Now this line must pass through $(2,3)$


问:
How does the tangent line according to calculus correlates to the classic intuition of a line that only passes through one point of the curve?
Sorry if my English is wrong.

In calculus, given a function $f$, derivable at $x_0$, the tangent line to the curve at $x_0$ is

$$t(x) = f(x_0) + f'(x_0)(x-x_0)$$

How can I convince myself that this line corresponds to the intuition of "a line is tangent to a curve if it intersects the curve one and only one time". I think you can get a segment of line that verifies this for all non-linear functions.

The reason I say segment of line, and not line is because, for example if $f(x)=x^3$, the tangent line to the graph of $f$ at $x_0=-0.5$, $t(x)=\frac{\left(3x\ +\ 1\right)}{4}$intersects the curve at $x=-0.5$ and $x=1$.

How to prove that $t$ is such line, and no other? Thank you.

答:
That's not actually the definition of the tangent line, because for example the line $x = 0$ touches $y = x^3$ at only one point, but is not a tangent line.

A better version might be "the slope of the line passing through two points of the curve, in the limit as the two points approach each other and become one."  It's a much more specific criterion.


答:
Let the point of intersection of the tangent and curve be $(x_{1},y_{1})$

$y'=4x_{1}$

Also $y'=\frac{y_{1}-3}{x_{1}-2}\ \ \ $ ....as $(2,3)$ doesn't lie on the curve

$y_{1}=2x_{1}^2 + 3$

$\implies 4x_{1}= \frac{2x_{1}^2 +3-3}{x_{1}-2}$

$4x_{1}\cdot(x_{1}-2)=2x_{1}^2$

Now you could solve for $x_{1}$ and find $y_{1}$ and subsequently, you could determine the  equations too!

Title: Determine the equations of the tangents to the curve $y=2x^2+3$ that pass through point $(2,3)$

Body: I am unsure of how to answer this question. All I have found is the slope of the tangents: $y'= 4x$.


答: It's a classic quadratic equation problem in disguise. Let the tangent line be $y = mx+b$ as usual. This line passes through the point $(2,3)$ implies $3 = 2m+b \implies b = 3 - 2m\implies y = mx + 3 - 2m$. We now determine $m$ and hence the line is known. Since the line is a tangent line, this means it intersects the parabola given by $y = 2x^2+3$ at only $1$ point. It means that the equation: $2x^2+3 = mx + 3-2m$ or $2x^2 -mx + 2m = 0$ has exactly one real solution. Thus $\triangle = m^2 - 16m = 0\implies m = 0,16$. Thus there are $2$ tangent lines to the parabola that goes through the point $(2,3)$, namely: $y = 3$, and $y = 16x-29$.


问:
Solve system of first order differential equations
I have to solve differential systems like this:
$$
\left\{
\begin{array}{c}
x' = 3x - y + z \\
y' = x + 5y - z \\
z' = x - y + 3z
\end{array}
\right.
$$

Until now I computed the eigenvalues $k = \{2,4,5\}$ by solving the equation resulted from this determinant of this matrix:
$$
\begin{pmatrix}
3 && -1 && 1 \\
1 && 5 && -1 \\
1 && -1 && 3
\end{pmatrix} - kI_3 =
\begin{pmatrix}
3-k && -1 && 1 \\
1 && 5-k && -1 \\
1 && -1 && 3-k
\end{pmatrix}
$$
I don't know what to do next.

NOTE: This is an example but from it I want to learn the method to solve any system of this kind. I learn better from particular examples than directly from generalization.

答:
The idea here is just like a single ODE with constant coefficients: Assume the solutions are given by $x = \exp(\alpha t)$, $y = \exp(\beta t)$, $z = \exp(\gamma t)$. Your system is linear, so you can mix and match solutions just like a single ODE. Substitute, and you'll get a set of equations for the $\alpha, \beta, \gamma$, and take it from there.

Or you could differentiate the first, and substitute the derivatives from the others, and keep going until you get a single third order ODE. Solve that, and work backwards. The result will be exactly the same as above, the cubic to solve will be the same too. Just different paths. (Yes, I did prove this by sheer curiosity when I was at school.)


问:
Sets shattered by $\mathcal{F}=\{\emptyset, \{2\},\{3\},\{1,2\},\{1,2,3\}\}$
I have worked through this problem and determined that the $\mathcal{F}$ listed in the title is shattering-extremal and shatters the following sets: $\emptyset, \{2\},\{3\},\{2,3\},\{1,3\}$.

Now I was fairly confident that I had correctly found the collection of shattered sets, however it misaligns with a supposedly true necessary and sufficient condition. I'm hoping that someone can verify that I have correctly identified all the sets shattered by $\mathcal{F}$. I know there can't be more than 5 by the Sauer inequality, however maybe I mixed one up. A second pair of eyes would be very useful here!

答: The Sauer inequality says that $\mathscr{F}$ cannot shatter fewer than $5$ sets. If I’m not mistaken, it shatters six and so is not extremal: in addition to the five that you named, it shatters $\{1\}$: $\varnothing=\{1\}\cap\varnothing$, and $\{1\}=\{1\}\cap\{1,2\}$, for instance.


问:
If $A - B \succ 0$, is $AB^{-1} -I \succeq 0$?
Let matrices $A, B \in \Bbb C^{n \times n}$ be Hermitian, and $A - B \succeq 0$. Is $A B^{-1} - I \succeq 0$? If so, why?

What about the positive definite case? ($A - B \succ 0$)

答:
That's not true. I'm assuming $A -B \succeq 0$ means $A-B$ is Hermitian positive semi-definite (and definite for $A-B \succ 0)$. Take $A=I_2$,
$B=\begin{pmatrix}
0.5 & 0 \\\\
0  & -2
\end{pmatrix}$.


问:
Proof that the sum of exterior angles of a complex polygon or a non-planar polygon is greater than 2 Pi
In the context of writing a ray-tracing program, I must be able to determine if a polygon in three-dimensional space is a planar simple convex polygon. My conjecture is that if I take the sum of the exterior angles (absolute values, always in the range $[0, \pi]$) then this sum is equal to $2\pi$ if and only if it is a planar simple convex polygon.

I firmly believe this to be true, but I seek somewhat rigorous proof. I already managed to show that the sum equals $2\pi$ for convex polygons and that it exceeds $2\pi$ for concave ones. However, I do not have any ideas on how to continue to prove that complex and non-planar polygons have a sum of exterior angles greater than $2\pi$.

For complex planar polygons, I managed to show (w.l.o.g.) that the exterior angles which "turn to the right" minus those that "turn to the left" must be a $2\pi n$ for some natural $n$. From there it's trivial to show that if $n>1$ then the total must be greater than $2\pi$, but for $n=0$ and $n=1$ I do not know how to proceed.

For non-planar polygons, I thought about what would happen if we project it to 2-dimensions then somehow show that the projection has an exterior angle sum strictly between $2\pi$ and that of the non-planar version, but I once again cannot fathom how to show that. (Edit: this idea does not work since there are counterexamples of non-planar polygons whose projections onto two-dimensional space have a larger exterior angle sum.)

答:
After many days of thought, I have come up with a proof for complex polygons. I will not accept my answer until I either edit it with a proof for non-planar polygons too or someone else is able to answer that part of my question.

The first thing to do is to disqualify any polygon with overlapping colinear (not necessarily consecutive) edges. If a walk through the edges would demonstrate that these edges are walked through in the same direction twice that means that a total turn of at least $2\pi$ has already been made between the two edges, yet the traversal has not yet concluded. Otherwise, if the edges are walked over in opposite directions, that would require a total turn of at least $\pi$ to turn around at each end of these edges. If that is all the turns that there are, then this sequence of points has only two significant points and does not form a polygon. The number of significant (non-colinear) vertices will have to be checked separately.

Now that we've gotten that out of the way, we may proceed with the assumption that no edges overlap. Note, this also means that no interior angle is equal to 0, $\pi$ or $2\pi$ radians, which means that no exterior angle is $\pm\pi$ or 0 radians, which in turn means that no turn (the absolute value of the exterior angle) is 0 or $\pi$ radians.

For any complex polygon $P$, we are able to find a sequence of consecutive vertices of $P$ which when appended to an intersection point of $P$ forms a simple polygon in the following way.

We choose any arbitrary point on the polygon's bounding path. Then from that point we trace the bounding path, marking any intersection points we encounter. Once we arrive at an intersection point which we've previously encountered, we stop. Let this intersection point be called $I$.

The closed path which we traced from the first encounter with $I$ to our second encounter with $I$ is a simple polygon (see figure).

Figure: Extracting a simple polygon from a complex one. This complex polygon is a pentagram with vertices $v_0$ through $v_4$. $S$ is the arbitrarily chosen starting point. The orange arrowhead marks the bounding edges of a simple polygon $E = (v_0, v_4, v_3, I)$ embedded in the complex pentagram. $I$ is the first (and necessarily the only) intersection point of the pentagram which we encountered twice.

The embedded polygon $E$ found with this procedure must be simple since we stop our trace at the first intersection point we encounter twice. Thus all of $P$'s intersection points are touched only once by $E$, except for point $I$ which is the start and end point, so is not an intersection point of $E$ either.

Let $\alpha \in (0, 2\pi) - \{\pi\}$ be the interior angle of $E$ at vertex $I$, and let $\beta = \pi - \alpha \in (-\pi, \pi) - \{0\}$ be the exterior angle of the same.

Suppose now we were to 'walk' a cycle through the edges of $P$ starting at the midpoint of the line segment formed by $I$ and one of the two vertices adjacent to $I$ which is common to both $P$ and $E$ (either $v_0$ or $v_3$ in the figure). Suppose we start walking in the direction away from $I$ and are only allowed to walk forward and turn.

The first time we reach $I$, we will have already turned all the turns of $E$ but one --- $T_E(I)$ (the turn of polygon $E$ at vertex $I$), which is equal to $|\beta|$. We immediately pass $I$ and enter a section of the path which is no longer part of $E$. Let us analyze the current position.

Since $I$ is an intersection point of $P$, we will in future be forced to encounter it again before arriving finally at the start. However, since we've just passed $I$, it is positioned a small distance directly behind us. This means that in order to reach it in future, our path must force us to make some turns which sum to at least $\pi$ radians.

This now allows us to formulate our first inequality. ($\sum T_P$ means the total turn of the polygon $P$, that is the sum of the turn at each of its vertices)
\begin{equation}
 \sum T_P \geq \sum T_E - |\beta| + \pi
\end{equation}

Since we know that $\sum T_E \geq 2\pi$ due to the simplicity of $E$, and that $\beta \neq \pm\pi$ from the first paragraph of the proof, we can from the previous inequality come to the following conclusion.
\begin{equation}
 \sum T_P \geq 3\pi - |\beta| > 2\pi
\end{equation}


问:
prove this ODE solution is genuine?
Prove that the solution to:
$y'-2y\tan x=2x\sec x$ is $$y=\frac{2\left(x\sin\left(x\right)+\cos \left(x\right)\right)+C}{\cos^2\left(x\right)}$$

I have found the derivative: $$\frac{2x\cos^3\left(x\right)+\sin \left(2x\right)\left(2\left(x\sin \left(x\right)+\cos \left(x\right)\right)+C\right)}{\cos ^4\left(x\right)}$$

But cannot substitute derivative into original equation to prove

答:
Treating it as a linear first order ODE, then proceed to find the integrating factor: $$\mu = e^{\int -2\tan xdx}= e^{2\ln (\cos x)}= \cos^2 x $$

$$\implies (y\cos^2x)' = 2x\sec x\cdot \cos^2x=2x\cos x$$

$$\implies y\cos^2x= \displaystyle \int 2x\cos xdx$$

$$ =  \int 2xd(\sin x)$$
$$= 2x\sin x- \int 2\sin xdx$$
$$= 2x\sin x+2\cos x+C$$
$$\implies y = \dfrac{2x\sin x+2\cos x +C}{\cos^2x}$$ as claimed above.

Title: prove this ODE solution is genuine?


问:
How to evaluate integrals of type $\int\frac{\cos^2x}{\cos^2x+ 4\sin^2x}\,dx$
I think it is by substitution but I don't  have any clue on how to go further. Should I substitute $$\sin^2x=t~?$$

$$\int\frac{\cos^2x}{\cos^2x+ 4\sin^2x}\,dx$$

答: Simplify the denominator by using $\sin^2x+\cos^2x=1$ and then try to express the resultant wholly in terms of $\tan x$. Then substitute $u= \tan x$. Its pretty much just applying standard identities from there onwards.


问:
Defining loops: why is divisibility and identitiy implying invertibility?
Wikipedia contains the following figure (to be found, e.g. here) in order to visualize the relations between several algebraic structures. I highlighted a part that I find especially interesting.

It seems to suggest that a loop can be defined either

as a magma with identity and invertibility, or

as a magma with identity and divisibility.

It is easy to see that from 1. follows 2. But I have trouble proving the converse. Here is a proof for the case that the loop is finite:

By divisibility $xa=xb\implies a=b$ for all $x,a,b\in L$. Thus, the map $a\mapsto xa$ is injective. If the loop is finite, the map is also surjective. Thus, there is a $x'\in L$ with $xx'=e$. This $x'$ is therefore the inverse of $x$.

But what if $L$ is not finite?

答:
It should be pointed out that the accepted answer's definition of "has division" is not how the term is used in quasigroup theory or universal algebra. A magma $(M,\cdot)$ (or binar or groupoid in the old noncategorical sense) is said to be a division magma if for every $a,b\in M$, there exist $x,y\in M$ such that $a\cdot x = b$ and $y\cdot a = b$. Uniqueness is not assumed. A quasigroup is a cancellative, division magma.

A better way to think about it is in terms of the translation maps $L_a : M\to M; x\mapsto a\cdot x$ and $R_a :M\to M; x\mapsto x\cdot a$ for all $a\in M$. In a division magma, these are all surjective. In a cancellative magma, they are injective. In a quasigroup, they are bijective.


问:
The asymptotics of the following expression
I wonder the asymptotics of the expression $$\sum_{k=1}^{n} \frac{n!}{kn^k(n-k)!}$$ when $n$ tends to be infinity and don't know how to do it. I have tried Stirling approximation but I think it doens't work. I can figure out that it's in $ O(\log n)$ scale but can't figure out its relation with $ \Theta(\log n / \log\log n)$. Can anybody give a guide on how to get its asymptotics? I conjecture that it doesn't exceed $ \Theta(\log n / \log\log n)$. Thanks a lot for any help!

答:
Your $k$-th term is equal to

$$\frac1k\prod_{i=0}^{k-1} (1-\frac{i}{n}).$$

The product is strictly smaller than one, so the sum is dominated by $\sum_{k=1}^n \frac1k = H_n \sim \log n,$ bearing out your $O(\log n)$ statement.

If $k > n/\log n,$ then the product is of order $o(1/k),$ so is negligeable, and for smaller $k$ it can be estimated by taking logs and approximating by the integral, so you can get the precise constant (if you use the Euler-Mclaurin formula, you will see that the integral approximation is very good).


问:
What does $f \lor g$ mean where $f,g$ are functions?
I've encountered a problem asking me to prove that $f \lor g$ is lower-semi-comtinuous if $f,g$ are so. But I don't know what $f\lor g$ means... Any answer would be appreciated!

答: The notation you are searching for is: $ f \lor g = \dfrac{f+g+|f-g|}{2}$. Can you manage to define the $f \land g$ ?


答:
This notation comes from lattice theory, where $\wedge$ means meet (i.e., greatest lower bound) and $\vee$ means join (i.e., least upper bound). In this context, $f \vee g$ is the function given by the pointwise maximum of $f$ and $g$, while $f \wedge g$ is their pointwise minimum.

(Personally, I don't know why some people prefer this notation over the clearer $\min(f,g)$ and $\max(f,g)$; the symbols look like they're pointing the wrong way to me. $f \vee g$ points down, but it means maximum.)

Title: What does $f \lor g$ mean where $f,g$ are functions?

Body: I've encountered a problem asking me to prove that $f \lor g$ is lower-semi-comtinuous if $f,g$ are so. But I don't know what $f\lor g$ means... Any answer would be appreciated!


答: As pointed out in the comments, this means the maximum $\max(f,g)$ between $f$ and $g$. The notation is common when dealing with posets, where $x\vee y$ is called the "join" of $x$ and $y$.


答:
After dividing by $$cos^2x$$ I got $$\int1/(1+4tan^2x)\,dx$$

And solving it, my answer is $$(arctan(2tanx))/2$$. Is this correct? The answer given in textbook is different. It is
$$(-1/3)arctan(tanx)-(2/3)arctan(2tanx)$$

I graphed my solution and the one in textbook. They are different. Then how to know, which substitution will give me a correct answer?

Title: How to evaluate integrals of type $\int\frac{\cos^2x}{\cos^2x+ 4\sin^2x}\,dx$

Body:
I think it is by substitution but I don't  have any clue on how to go further. Should I substitute $$\sin^2x=t~?$$

$$\int\frac{\cos^2x}{\cos^2x+ 4\sin^2x}\,dx$$


问:
Does any sense that a is many times greater than b, with negative numbers?
Let $a,b \in R^+$ and $b\le a$. I want to calculate how many times  greater or less these numbers are, by taking $b$ as reference and using: $  x = \frac{a}{b}$, and I have no doubt in this case. But when $a,b \in R $. Does the $x$ value make sense with

Both $(a,b)$ negative numbers

One negative $(-b)$ and other positive $(a)$

Example:

$ x = \frac{10}{2} = 5$, is 5 five times greater, this is OK

$ x = \frac{10}{-2} = -5$, this is no necessary $|-5|$ times greater because de difference is $ 10-(-2) = 12$, only if I take absolutes values for that

$ x = \frac{-5}{-8}=0.6$, I'm not sure if this has any meaning

So I'm not sure if $x=\frac{a}{b}$ only if $ a,b \in R^+ $. Nevertheless what does the people when work with temperatures and wants to know how many times greater is a temperature given a reference, for instance: reference $-15^{\circ}$ and the other temperature $-4^{\circ}$, maybe, in this case I can apply reciprocal of $\frac{a}{b}$ and use $x = \frac{b}{a}$.

Sorry if I've  written something wrong, I'm newbie in math  but I want to try use the correct notation. I started to read an arithmetic book where this question came to me. I read related these questions before
this,
this
and this

答:
I find your example of temperature interesting. There is a unit of temperature called Kelvin that avoids negative values entirely. Kelvin temperature is also called "absolute temperature". It seems absolute temperature is proportional to energy, and it seems reasonable to say things like "three times the energy".

Conversely, there is nothing wrong with saying "LeBron James is three times the height of my cousin." But I could introduce a ridiculous new unit of length in which Lebron is -102 units while my cousin is -168.


问:
If polynomials agree on roots, then they have the same irreducible factors
Suppose $K$ is a field and $f_1,f_2 \in K[x_1,\dots,x_n]$ are polynomials such that $f_1(a)=0 \iff f_2(a)=0$ for any element $a=(a_1,\dots,a_n) \in \tilde{K}^n$, the algebraic closure of $K^n$. I would like to show that $f_1$ and $f_2$ have the same irreducible factors in $K[x_1,\dots,x_n]$. I am not really sure where to begin. I know that a variety is irreducible if its ideal is prime and any variety has
a unique decomposition as a finite union of irreducible subvarieties. Any reference or hint is appreciated.

答: The Nullstellensatz precisely says that $f_1$ and $f_2$ have the same zero locus if and only if the ideals $(f_1)$ and $(f_2)$ have the same radical. You can easily see however (using unique factorization in $K[x_1,\cdots,x_n]$) that the radical of $(f)$ is the ideal generated by the product of the distinct irreducible factors of $(f)$. Hence $f_1$ and $f_2$ must have the same irreducible factors.


问:
How to get the general solutions?
Three solutions of a certain second order non-homogeneous linear differential equation are $$y_1(x)=1+xe^{x^2} , y_2(x)= (1+x)e^{x^2}-1 , y_3(x)=1+e^{x^2}$$
Which of the following are general solutions of the differential equation ?

1.$(C_1+1)y_1+(C_2-C_1)y_2-C_2y_3. $

2.$ C_1(y_1-y_2)+C_2(y_2-y_3).$

3.$ C_1(y_1-y_2)+C_2(y_2-y_3)+C_3(y_3-y_1).$

4.$C_1(y_1-y_3)+C_2(y_3-y_2)+y_1 $ ,  Where $C_1 , C_2$ and $C_3$ are arbitrary constant.

My efforts is that  ,  option third will be wrong (since the  second order differential equation is given, So in the general solution , the number of constant should be two only.)

答:
Note: Any linear combination of solutions of a non-homogeneous second order linear differential equation is again a solution if the sum of the coefficients is one.

For option $\bf {(1)}$ : If the solution is
$$(C_1+1)y_1+(C_2-C_1)y_2-C_2y_3$$Sum of coefficients is $$(C_1+1)+(C_2-C_1)-C_2=1$$Hence option $(1)$ is correct.

For option $\bf {(2)}$ : If the solution is
$$C_1(y_1-y_2)+C_2(y_2-y_3)=C_1y_1+(C_2-C_1)y_2-C_2y_3$$ Sum of coefficients is $$C_1+(C_2-C_1)-C_2=0$$Hence option $(2)$ is incorrect.

For option $\bf {(3)}$ : If the solution is
$$C_1(y_1-y_2)+C_2(y_2-y_3)+C_3(y_3-y_1)=(C_1-C_3)y_1+(C_2-C_1)y_2+(C_3-C_2)y_3$$Sum of coefficients is $$C_1-C_3+C_2-C_1+C_3-C_2=0$$Hence option $(3)$ is again incorrect.

For option $\bf {(4)}$ : If the solution is
$$C_1(y_1-y_3)+C_2(y_3-y_2)+y_1=(C_1+1)y_1+(-C_2)y_2+(C_2-C_1)y_3$$Sum of coefficients is $$C_1+1-C_2+C_2-C_1=1$$Hence option $(4)$ is correct.

Title: How to get the general solutions?


答:
The notation $a \lor b$ means the maximum of $a$ and $b$. $a \land b$ means the minimum of $a$ and $b$.

This explanation is mnemonic; from a historical perspective it is not accurate.

There are many ways to remember this and explain the choice of symbol, but I like thinking of the minimum and maximum reading of $\land$ and $\lor$ as generalizations of their meaning in logic.

If we think of classical logic and use $0$ to mark false and $1$ to mark true, then $a \;\text{or}\; b$ the maximum of $a$ and $b$. This means we can think of the maximum in general as a generalization of $\lor$. The case for $\land$ is analogous.


问:
What's the preferred term researchers like to use in the theory of magmas/groupoids?
As we know, mathematicians like to avoid the term "groupoid" to refer to a set with binary operation. This term, as we know, originates from the works of Brandt, so called Brandt groupoid. A Brandt groupoid is a groupoid in the sense of category theory, however for some (unknown to me) reason in history the term began to also be the name for a set with binary operation.

For the historical reasons, and the fact that the term is somewhat ambiguous, I saw a lot of people prefer the term "magma".

However, there is another side of the argument, and that is, what do people that actually work in the field call their objects? Is it magma, groupoid, or maybe something else entirely? In some fields, like quasigroup or semigroup theory, I saw a lot of people refer to those objects as groupoids. Is this possible that researchers actually prefer the term groupoid in their work?

References:

"The Algebraic Theory of Semigroups" A. H. Clifford, G. B. Preston

"Elements of Quasigroup Theory and Applications" V. Shcherbacov

"Universal Algebra" S. Burris, H. P. Sankappanavar

答:
Speaking as a quasigroup theorist, I can say that it is definitely true that even today more people in the area use "groupoid" in the Hausmann-Ore sense (set with binary operation) than in the categorical sense. A lot of the preference is both geographically and generationally based. For instance, that usage is prevalent among older quasigroup theorists from former Eastern Bloc countries.

In semigroup theory (my other area of interest), "groupoid" is usually meant in the categorical sense, because the notion is heavily used in the theory of inverse semigroups.  However, many semigroup theory papers treat groupoids as algebraic structures with a partially defined operation rather than as a special kind of category.


问:
annuity- find the present value of varying payments
I am trying to find the present value of an annuity, where we have a $20$-year annuity immediate where the first payment is  $1000$, and each subsequent payment is increased by either $50$ for
the first $10$ payments or by $100$ for the last $10$ payments.

The 1st payment= $1000$ (at time 1)

Y2 TO Y10 payment = $1000 + 50$ = $1050$

Y11 TO Y20 payment = $1000 + 50 + 100$ = $1150$

I did the present value of each individual payment using the principle
$PV = Pν + (P + Q)ν^{2} + (P + 2Q)ν^{3} + ... + (P + (n − 1)Q)ν^{n}$

$PV = 1000v +1050v^{2} + 1100v^{3} + ... + 1450v^{10} + 1600v^{11} + 1750v^{12} + 1900v^{13} +...+ 4300v^{20}$

then

$$\require{enclose} PV = 1000 [ 50a_{\enclose{actuarial}{10}i} + 150v^{10}  a_{\enclose{actuarial}{10}i}]  \tag{1}$$

where $i = 10$%

I wasn't sure if I derived the formula correctly. I appreciate your help.

答:
The way you describe the cash flow does not match the equation of value you wrote.

You say that the payments are increased by $50$ for each year for the first ten years, and then by $100$ for each year for the last ten years.  But the way you wrote it, you're increasing the payments by $150$ in the last ten years.  Which is it?

My interpretation of the question would have the payments be:
$$1000, 1050, 1100, 1150, 1200, 1250, 1300, 1350, 1400, 1450, \\
1550, 1650, 1750, 1850, 1950, 2050, 2150, 2250, 2350, 2450.$$  Note that the total increase from the first payment to the last is $50(9) + 100(10) = 1450 = 2450 - 1000$.

In such a case, there are multiple ways to write the equation of value.  One way is to take the level payments of $1000$ out, and then what is left are two increasing annuities, one deferred by one year, and the other deferred by $10$ years.  Refer to the following table:

$$\begin{array}{c|c|c|c|c}
\text{Year} & \text{Level Payment} & \text{Increasing 1} & \text{Increasing 2} & \text{Total} \\
\hline
1 & 1000 & 0 & 0 & 1000 \\
2 & 1000 & 50 & 0 & 1050 \\
3 & 1000 & 100 & 0 & 1100 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
10 & 1000 & 450 & 0 & 1450 \\
11 & 1000 & 500 & 50 & 1550 \\
12 & 1000 & 550 & 100 & 1650 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
20 & 1000 & 950 & 500 & 2450 \\
\end{array}$$
This gives us the equation of value
$$\require{enclose}
\begin{align}
PV
&= 1000(v + v^2 + \cdots + v^{20}) + (50v^2 + 100v^3 + \cdots + 950v^{20}) \\ &\qquad + (50v^{11} + 100v^{12} + \cdots + 500v^{20}) \\
&= 1000 a_{\enclose{actuarial}{20} i} + 50 v (Ia)_{\enclose{actuarial}{19}i} + 50 v^{10}(Ia)_{\enclose{actuarial}{10}i}.
\end{align}$$

Alternatively, we can structure the payments as follows:
$$\begin{array}{c|c|c|c|c}
\text{Year} & \text{Level Payment} & \text{Increasing 1} & \text{Increasing 2} & \text{Total} \\
\hline
1 & 950 & 50 & 0 & 1000 \\
2 & 950 & 100 & 0 & 1050 \\
3 & 950 & 150 & 0 & 1100 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
10 & 950 & 500 & 0 & 1450 \\
11 & 1450 & 0 & 100 & 1550 \\
12 & 1450 & 0 & 200 & 1650 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
20 & 1450 & 0 & 1000 & 2450 \\
\end{array}$$
This gives the equation of value
$$PV = 950 a_{\enclose{actuarial}{10}i} + 1450 v^{10} a_{\enclose{actuarial}{10}i} + 50 (Ia)_{\enclose{actuarial}{10}i} + 100v^{10} (Ia)_{\enclose{actuarial}{10}i}.$$
You might ask, why would we write it this way instead of the previous?  Well, notice that even though there is one more annuity symbol, there's actually only two different ones, rather than three in the first equation of value.  In other words, things factor more nicely in the second equation:
$$PV = 50\left( (19 + 29v^{10})a_{\enclose{actuarial}{10}i} + (1 + 2v^{10})(Ia)_{\enclose{actuarial}{10}i}\right).$$
In both cases, the answer is the same, approximately $11843.639297$.  But no matter how you write the equation of value, you must use an increasing annuity symbol $(Ia)_{\enclose{actuarial}{n}i}$.


问:
How do I know if this limit even converges?
So, there is this question which I found quite new according to my experience, and it is as follows.

Assuming that interchange of limit and integration is permissible, evaluate the following: $$\lim_{n \to \infty} {\int_0^1 \frac{nx^{n -1}}{1 + x}dx}, \quad 0 < x < 1$$

Now, I first tried integration by parts, and it doesn't give me more than an expression which is also nearly impossible to evaluate (maybe I'm doing it wrong, though) -

$$\lim_{n \to \infty} \left|\frac{1}{1+x}\int nx^{n - 1}dx  +  \int\frac{x^n}{(1 + x)^2}dx\right|_0^1$$

I'm not sure how to proceed from there, so I tried this instead:

$$\lim_{n \to \infty} {\int_0^1 {nx^{n - 1}(1 + x)^{-1}}dx}$$

Expanding $(1 + x)^{-1}$ using Taylor Series:

$$\lim_{n \to \infty} {\int_0^1 {nx^{n - 1}(1 - x + x^2 - x^3 + x^4 - x^5 \ldots)}dx}$$
$$ = \lim_{n \to \infty} {\int_0^1 {(nx^{n - 1} - nx^n + nx^{n + 1} - nx^{n + 2} + nx^{n + 3} - n^{x + 4} \ldots)}dx}$$
$$\lim_{n \to \infty} \left(1 - \frac{n}{n+1} + \frac{n}{n+2} - \frac{n}{n +3} + \frac{n}{n +4} - \frac{n}{n + 5}\ldots\right)$$

I think that diverges. Then, I tried putting the expression in Desmos and increased the $n$ value slowly upto $500$, $1000$ and similar big numbers - which all returned values around $0.5$.

The question was an multiple-choice based and there were both of them listed as options.

I'm not sure who is correct, the calculator, or me.

It would be a huge help if someone looks into this. Thank you in advance.

答: OK. Use integration by parts. Rewrite $I_n = \displaystyle \int_{0}^1 \dfrac{d(x^n)}{1+x}= \dfrac{x^n}{1+x}|_{x=0}^1+ \displaystyle \int_{0}^1\dfrac{x^n}{(1+x)^2}dx= \dfrac{1}{2}+J_n$, and $0 \le J_n \le \displaystyle \int_{0}^1 x^ndx= \dfrac{1}{n+1}$. Hence $J_n \to 0$, and $I_n \to \dfrac{1}{2}$. So if you got $0.5$ then  you are right.


问:
Integration change of variables
I have the following integral to solve
$$\int_a^b e^{\sigma z}e^{-\frac{1}{2}z^2}dz $$
where $\sigma,a,b$ are some constants.
I rewrite this as $$e^{\sigma^2/2}\int_a^b e^{-\frac{1}{2}(z-\sigma)^2}dz $$
By substituting $u=z-\sigma$, we have $du=dz$ and the limits become $a+\sigma$ to $b+\sigma$. Then,
$$e^{\sigma^2/2}\int_{a+\sigma}^{b+\sigma}e^{-\frac{1}{2}u^2}dz = e^{\sigma^2/2}\big(N(b+\sigma)-N(a+\sigma) \big)  $$
where $N(\cdot)$ is the normal density CDF. Am I using the correct limits (or is it $a-\sigma, b-\sigma$ for the lower and upper limits respectively)?

答:
In your last integral, you forgot to change $dz$ to $du$. Maybe that is why you are getting confused. When $z$ is $a$, $u$ is $a-\sigma$.

When $z$ is $b$, $u$ is $b-\sigma$.

Title: Integration change of variables


问:
how do I justify this $\int_{0}^{\infty}\frac {e^{-2x}-e^{-x}}{x}dx=\lim_ {s\to1^{-}} \int_{0}^{\infty}\frac{e^{-2x}-e^{-x}}{x^s}dx$
I was trying to compute
$\displaystyle\int_{0}^{\infty}
\frac{{\rm e}^{-2x} - {\rm e}^{-x}}{x}\,{\rm d}x$ when I had an idea, the gamma function is $\Gamma(1+z)=\int_{0}^{\infty}x^ze^{-x}dx$ and for $s<1$ the integral $\int_{0}^{\infty}\frac{e^{-2x}-e^{-x}}{x^s}dx$
converges, so call
$$I(s)=\int_{0}^{\infty}\frac{e^{-2x}-e^{-x}}{x^s}dx$$
and hope that it is continuous from the left at $s=1$.

Assuming that it is, for $s<1$ we have
$$I(s)=(2^{s-1}-1)\Gamma(1-s)$$
and so:
$$I(1)=\int_{0}^{\infty}\frac {e^{-2x}-e^{-x}}{x}dx = \lim_{s\to1^-}I(s)= \lim_{s\to1^-}(2^{s-1}-1)\Gamma(1-s)$$

Now, write $$2^{s-1}-1=1+\ln(2)(s-1)+o((s-1)^2)-1=\ln(2)(s-1)+o((s-1)^2)$$
and so:
$$\lim_{s\to1^-}(2^{s-1}-1)\Gamma(1-s) = -\ln(2)\lim_{s\to 1^-}(1-s)\Gamma(1-s)+\lim_{s\to 1^-}o((1-s)^2)\Gamma(1-s)$$

Here comes the cool part, use the reflection formula ($\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$) and obtain:
$$(1-s)\Gamma(1-s) = \frac {\pi (1-s)}{\sin(\pi s)\Gamma(s)}$$
and:
$$o((1-s)^2)\Gamma(1-s) = o(1-s)\frac {\pi (1-s)}{\sin(\pi s)\Gamma(s)}$$

Now, $\Gamma$ is continuous at 1 and $\Gamma(1)=1$ so:
$$\lim_{s\to 1^-}\frac {\pi (1-s)}{\sin(\pi s)\Gamma(s)} = \lim_{s\to 1^-}\frac {\pi (1-s)}{\sin(\pi s)}$$

this limit is finite and is equal to 1 (with l'hopital) and so
$$\lim_{s\to 1^-}o((1-s)^2)\Gamma(1-s) = 0$$

And we can finally finish up and say $I(1)=-\ln(2)$.
Now, I don't know how to justify the fact that $I(s)$ is continuous at $s=1$ from the left and so I don't know if my solution is correct.

BUT! the answer is correct which gives me hope that this solution is correct with some additional justifications about the continuity of $I$ at $s=1^-$ which is why I am looking for help.

答:
I thought it might be of interest to present a way to show directly, using elementary inequalities only, that the integral of interest, $I(s)=\int_0^\infty \frac{e^{-2x}-e^{-x}}{x^s}\,dx$, is right continuous at $s=1$.  To that end we proceed.

Note that for $0<s\le1$, we have the estiamtes

$$\begin{align}
|I(s)-I(1)|&=\left|\int_0^\infty \frac{e^{-2x}-e^{-x}}{x^s}\,dx-\int_0^\infty \frac{e^{-2x}-e^{-x}}{x}\,dx\right|\\\\
&=\int_0^\infty (e^{-x}-e^{-2x})\left|\frac1{x^s}-\frac1x\right|\,dx\\\\
&= \int_0^\infty \left(\frac{e^{-x}-e^{-2x}}{x}\right)|x^{1-s}-1|\,dx\\\\
&=\int_0^\infty \left(\frac{e^{-x}-e^{-2x}}{x}\right)\left|e^{(1-s)\log(x)}-1\right|\,dx\\\\
&= \int_0^1  \left(\frac{e^{-x}-e^{-2x}}{x}\right)\left(1-e^{(1-s)\log(x)}\right)\,dx\\\\
&+\int_1^\infty \left(\frac{e^{-x}-e^{-2x}}{x}\right)\left(e^{(1-s)\log(x)}-1\right)\,dx\tag1\\\\
&\le (1-s)\int_0^1  \left(\frac{e^{-x}-e^{-2x}}{x}\right)\left|\log(x)\right|\,dx\\\\
&+\int_1^\infty \left(\frac{e^{-x}-e^{-2x}}{x}\right)\left(e^{(1-s)x}-1\right)\,dx\tag2\\\\
&=(1-s)\int_0^1  \left(\frac{e^{-x}-e^{-2x}}{x}\right)\left|\log(x)\right|\,dx\\\\
&+\int_1^\infty \left(\frac{e^{-sx}-e^{-(1+s)x}}{x}\right)\left(1-e^{-(1-s)x}\right)\,dx\tag3\\\\
&\le (1-s)\int_0^1  \left(\frac{e^{-x}-e^{-2x}}{x}\right)\left|\log(x)\right|\,dx\\\\
&+(1-s)\int_1^\infty \left(e^{-sx}-e^{-(1+s)x}\right)\,dx\tag4\\\\
\end{align}$$

Inasmuch as the first integral on the right-hand side of $(4)$ exists (in fact, it is easy to show that it is less than $\int_0^1 |\log(x)|\,dx=1$) and the second integral exists and is a bounded function of $s\in (\delta,1]$ for $1>\delta>0$ (in fact, the integral is less that $1/(s(1+s))$), we conclude that the integral of interest, $I(s)$ is right continuous at $s=1$.

And we are done!

NOTES:

In going from $(1)$ to $(2)$ we used the inequality $e^x\ge 1+x$ and the inequality $\log(x)\le x$.

In going from $(3)$ to $(4)$, we used the inequality $1-e^{-x}\le x$.


问:
$x'' + x''\cdot x^2 + x'^2 \cdot x + x - k = 0$
$x'' + x'' x^2 + x'^2 x + x - k = 0$ where $k \in \mathbb R$.

Usually, when I come across differential equations like this, I try to solve it through mathematica or some other math tool or find similar equations online. However, these don't help me much in this case (I will admit that I am not very adept with solving second order nonlinear differential equations.) I am wondering if there is really any analytical solution for this differential equation. Note that I am not familiar with how to approach this problem numerically.

答:
Maple finds an implicit general solution involving an integral:

$$\pm {{\int}}_{}^{x \left(t \right)}\frac{s^{2}+1}{\sqrt{\left(s^{2}+1\right) \left(2 k s -s^{2}+C_1 \right)}}{d}s -t -C_2
 = 0
$$

The integral can be expressed in terms of elliptic integral functions, but it is not pretty.


问:
Locus of a point of intersecting lines
The perpendicular from the centre of the ellipse $$b^2x^2+a^2y^2=a^2b^2$$ to the tangent at any point $P$ meets the line joining  point $P$ to the focus $(ae,0)$ at $Q$. Prove that the locus of $Q$ is $$x^2+y^2-2aex=b^2$$

My attempt

Tangent to an ellipse at $P(\theta)$

$$bx\cos\theta +by\sin\theta=ab\tag{1}$$

slope $m_1$ of a normal line to $(1)$

$$m_1=\frac{a\sin\theta}{b\cos\theta}$$

Equation of the normal passing through origin

\begin{align*}
y=m_1x
&=\frac{a\sin\theta}{b\cos\theta}x\tag{2}
\end{align*}

Equation of a line through $P$ and $S$

$$\begin{align*}
y=m_2(x-ae)&=\frac{b\sin\theta}{\left(a\cos\theta-ae\right)}(x-ae)
\tag{3}
\end{align*}$$

Point $Q(h,k)$ satisfies $(2)$ and $(3)$.

How to eliminate $\sin\theta$ and $\cos\theta$ to obtain the required locus?

答:
You are almost there. Now equating both,

$\displaystyle \frac{a}{b\cos\theta} x = \frac{b}{\left(a\cos\theta-ae\right)} (x-ae)$

$(x-ae+ae) (a^2\cos\theta-a^2e) = b^2\cos\theta(x-ae)$

$(x-ae) (a^2 \cos \theta - b^2 \cos\theta - a^2e)= a^3e^2 - a^3e \cos\theta$

We will use the fact that $a^2 - b^2 = a^2e^2$ now and again later.

$x-ae = \displaystyle \frac{a\cos\theta-ae}{1 - e \cos\theta}$

$x = \displaystyle \frac{b^2\cos\theta}{a(1-e\cos\theta)}$

$y = \displaystyle \frac{b \sin\theta}{1-e\cos\theta}$

$x^2 + y^2 - 2aex = (x-ae)^2 + y^2 - a^2e^2$

$ = \displaystyle \frac{a^2 \cos^2\theta+a^2e^2-2a^2e\cos\theta + (a^2-a^2e^2)\sin^2\theta}{(1-e\cos\theta)^2} - a^2e^2$

$ = a^2 - a^2e^2 = b^2$

Title: Locus of a point of intersecting lines


问:
Find which two bottles are contaminated

King Bubba III has 1024 potions, 1022 of which are just water while two of them are magical gives him superpowers (the special potions are colorless, odorless, etc.). Bubba, who is an intellectual, has a machine that tells him whether a particular sample of liquid is magical or not. However, he has calculated that he can mix the potions and still use the machine - if he takes a set of potions and inputs it into his machine, the machine will return positive if there is magic in any of the bottles and negative if none of the bottles are magic. There is, however, two limitations on the machine - firstly, because mixing too many potions dilutes magic, he can mix at most 32 potions an receive an accurate result from the machine, and secondly, the machine will return all the results at once, in 7 days (which means Bubba cannot base future tests off of the result of past tests). What is the minimal amount, up to a factor of 2, of times he needs to use the machine and identify both magical potions?

I think the answer actually should just be $1024$, because I've tested many other strategies you can use but still we must consider the least optimal case, which makes reducing this number significantly rather difficult. (there is a slight optimization you can use to reduce the number of uses to $1023$ but I really dont care

I believe the a similar, well-known variant of this problem has been posted here and here, but I'm not sure how the ideas can be cross applied - there are extra restrictions in this problem.

答:
The significance of the numbers $1024$ and $32$ is important here.  Imagine arranging all of the bottles in a $32x32$ square.  Make a mix for every row, and make a mix for every column. That's 64 tests.  That's almost enough information right there--if there were only one magic bottle, it would be enough.  If you get rows $a$ and $b$ to test positive, and columns $c$ and $d$ to test positive, then there are only two possible arrangements: $(a,c)$ and $(b,d)$, or $(a,d)$ and $(b,c)$.  If I understand the problem wording right, then 64 is the answer.

If not, the question now is how to distinguish them without knowing in advance which ones to test.  Imagine making a mix of every down diagonal, which is 63 more tests.  That would enable you to distinguish which of the two options applies.  There is probably an even more efficient way to do it, but that's what I've got so far.

Title: Find which two bottles are contaminated


答:
$$x'' + \color{red}{ x''\cdot x^2 + x'^2 \cdot x} + x - k = 0$$
$$x'' + \color{red}{x(x'x)'} + x - k = 0$$
Multiply by $x'$:
$$x'x'' + x'x(x'x)' + xx' - kx' = 0$$
Integrate:
$$x'^2 + (x'x)^2 + x^2 - 2kx = C$$
$$x'^2 (1+ x^2) + x^2 - 2kx = C$$
$$x'^2 =\dfrac  {C -x^2 + 2kx }{1+ x^2}$$
This is separable:
$$\int \sqrt { \dfrac  {1+ x^2}{C -x^2 + 2kx }}dx=\pm t+C_2$$
But as Robert Israel pointed out in his answer the integral is not nice.

Title: $x'' + x''\cdot x^2 + x'^2 \cdot x + x - k = 0$

Body:
$x'' + x'' x^2 + x'^2 x + x - k = 0$ where $k \in \mathbb R$.

Usually, when I come across differential equations like this, I try to solve it through mathematica or some other math tool or find similar equations online. However, these don't help me much in this case (I will admit that I am not very adept with solving second order nonlinear differential equations.) I am wondering if there is really any analytical solution for this differential equation. Note that I am not familiar with how to approach this problem numerically.


答:
\begin{gather*}
\int \frac{\cos^{2} x}{\cos^{2} x+4\sin^{2} x} dx\\
I=\int \frac{1}{1+4\tan^{2} x} dx\\
Let\ \tan x\ =t\\
\sec^{2} x\ dx=dt\\
dx=\frac{dt}{1+t^{2}}\\
I=\ \int \frac{dt}{\left( 1+t^{2}\right)\left( 1+4t^{2}\right)} =\frac{1}{3}\int \left(\frac{dt}{t^{2} +\frac{1}{4}} -\frac{dt}{t^{2} +1}\right)\\
\end{gather*}
The last expression is attained by using the method of partial fractions, and is solvable by standard integration techniques.
Hope this helps!


问:
Reproducing kernel Hilbert spaces and the isomorphism theorem
A reproducing kernel Hilbert space is a Hilbert space in which the evaluation functional

$L_x : f \rightarrow f(x)$ is continuous. By continuity, the Riesz representation theorem says that this functional can be represented as an inner product.

I have a feeling there's something fundamental I've misunderstood here. If any two real Hilbert spaces of the same cardinality are isomorphic, then why is it that $l_2$ is a RKHS, but $L^2[0,1]$ is not?

答:
To wrap up what Theo and Jonas already said: Two (complex or real) Hilbert spaces are isomorphic if they have orthonormal bases with the same cardinality, as Hilbert spaces. So, every statement that makes use of the Hilbert space structure only, and is true or false for one space, will be true or false for the other.

But a concrete Hilbert space may have more structure than just the Hilbert space structure. When you look at the statement "A reproducing kernel Hilbert space is a Hilbert space in which the evaluation functional..." then the "evaluation functional"-part presupposes that the Hilbert space under consideration has (real or complex valued) functions as elements. This is an additional property that some Hilbert spaces have and some have not.

The space $L^2[0, 1]$ for example consists of equivalence classes instead of functions, and the "evaluation functional" cannot be well defined because it depends on the representative of an equivalence class $[f]$. In fact, for any $x \in \mathbb{R}$ and for every real number $y$ including $\infty$ and $-\infty$, every equivalence class has an element $f$ such that $f(x) = y$.  And I can also define an abstract complex Hilbert space by saying that it is the space spanned by an orthonormal basis $(e_n)_{n \in \mathbb{N}}$. Now one cannot make sense of the term "evaluation functional", because the elements of this Hilbert space are not functions.

On the other hand, the Hardy space of the unit disk (Wikipedia) consists of holomorphic functions, therefore the evaluation functional is well defined. It is possible to prove that it is also continuous, but the proof makes use of the fact that the elements of this Hilbert space are holomorphic functions on the unit disk, which, as I said before, is additional structure that happens to exist for the Hardy space.

All of these examples are isomorphic as separable complex Hilbert spaces, but this isomorphism does not say anything about any structure that may exist beyond the Hilbert space structure.


问:
Showing $f$ is continuous
Let $x_1, x_2, ... , x_n$ be linearly independent vectors of a normed space $X$, so for any $ x \in X$ set $x = \alpha_1 x_1 + \alpha_2 x_2 + ... + \alpha_n x_n$ for real numbers $\alpha ... \alpha_n$. Prove that the function $f: \mathbb{R}^n \rightarrow \mathbb{R}^{\geq 0}$ defined by
$$f(\alpha_1, \alpha_2, ... \alpha_n) = || \alpha_1 x_1 + \alpha_2 x_2 + ... + \alpha_n x_n ||$$ is continuous.

Do I have to show that this is a linear operator and then show that is is bounded to imply continuity? The confusing part is this is a general normed space-- I don't have an explicit norm definition. How do I proceed?

答:
You just need to look at the axioms of a norm (for instance here).
With these axioms, you can show that if the input of the function gets close to $a$, for any norm of $\mathbb{R}^n$, like euclidean or maximum norm, then the output gets close to $f(a)$.

More formally, for any $a\in\mathbb{R}^n$ and $\varepsilon>0$, there is $h>0$ such that for all $b\in\mathbb{R}^n$, if $||a-b||\leq h$ then $|f(a)-f(b)|\leq \varepsilon$.

Title: Showing $f$ is continuous


问:
Boundaries on double integral help
New to double integrals I'm not understanding how to apply the domain on this so I can solve.

$\iint_D \frac{x}{\sqrt{x^2+y^2}} \ln(x^2+y^2)\ dA$

$D=\{(x,y)\ |\ x>0, y>0, \text{ and } 1<x^2+y^2<2\}$

答:
\begin{gather*}
\iint \frac{x}{\sqrt{x^{2} +y^{2}}}\ln\left( x^{2} +y^{2}\right) dxdy\\
Convert\ ot\ polar\ coordinates,\ \\
x=r\cos \theta \\
y=r\sin \theta \\
The\ given\ intgeral\ then\ becomes\\
I=\iint \frac{r\cos \theta }{r^{2}} \cdotp \ln\left( r^{2}\right) \cdotp rdrd\theta \\
where\ 1< r< \sqrt{2}\\
0< \theta < \frac{\pi }{2}\\
I=\left(\int ^{\sqrt{2}}_{1} 2\ln rdr\right)\left(\int ^{\frac{\pi }{2}}_{0}\cos \theta d\theta \right)
\end{gather*}

Can you solve these two integrals and complete the answer?

Title: Boundaries on double integral help


问:
Proving that $G$ is a hermitian matrix

Consider the set $P_n$ of all polynomials of degree $\le n$ with complex coefficients. Any such polynomial $p$ can be represented as a coefficient vector $[p]\in \Bbb C^{n+1}$ via $$p(x) = \sum_{i=0}^n a_ix^i \Longleftrightarrow [p] = \left[\begin{matrix}a_0 \\a_1\\ \vdots \\a_n \end{matrix}\right]$$
Suppose we define an inner product on polynomials as $(p,q) = \int_{-1}^1 \overline{p(x)} q(x)dx$. Thus, two polynomials $p,q$ are orthogonal if $\int_{-1}^1 \overline{p(x)} q(x)dx = 0$.

(a) Show that there exists a Hermitian matrix $G$ such that $(p,q) = [p]^*G[q]$ for all polynomials $p,q\in P_n$.

I know that that this is the polynomial analogue of QR factorization, but I cannot see how to get started. I thought of proving the general definition by somehow but I don't understand how to deal with the LHS that is $(p,q)$.

答: You can write down the inner product explicitly. What is $(x^n, x^m)?$ It will be the $n,m$ entry of your mysterious $G.$

Title: Proving that $G$ is a hermitian matrix


问:
Why is this result true for complex numbers?
I read that for complex number if $\,0,z_1,z_2\,$ represent vertices of an equilateral triangle then

$$ z_1^2 +z_2^2 = z_1 z_2 $$

On plunging values $z_1=x+iy$  what value of $z_2$ should I use to proof this result. Is there a  more appropriate way to prove this

答:
I'm assuming your third vertex in the triangle is the origin.

Since the third vertex is the origin, $|z_1-0|=|z_2-0|$. Let this be $\def\E{\mathrm{e}}R\in\mathbb R$, and write $z_1=R\E^{i\theta} (\theta\in\mathbb R)$. Switching $z_1$ and $z_2$ if necessary, we see that $z_2=R\E^{i(\theta+\frac\pi3)}$. Now you should be able to simplify the equation to $\omega^2+1=\omega$ where $\omega=\E^{i\frac\pi3}$. This is true because $0=\omega^3+1=(\omega+1)(\omega^2-\omega+1)$ and $\omega\neq-1$.


问:
What's the consequence of telling isomorphic graphs apart for the complexity class of a graph property?
A graph  property is  a  class  of  graphs  with  the property  that  any  two  isomorphic  graphs  either  both  belong  to  the  class,  or both do not belong to it.

A graph invariant is any function $S:G \to {R}^+_{0}$ such that for isomorphic graphs $G,G'$, i.e. $G \sim G'$, holds $S(G)=S(G')$.

If there is now an algorithm being able to tell isomorphic graphs apart, $S(G) \neq S(G') \Leftrightarrow G \not\sim G$,
that is between $P$ and $NP$, how can I then interpret that a graph property is in $NP$?

I am somehow lacking here a piece that telling graphs apart is between $P$ and $NP$, and hence any graph property should be bound by such a complexity, and still having properties in $NP$.

答:
Even being instantly able to tell whether two graphs are isomorphic doesn't give you an obvious test for some graph properties.

Consider the property "$G$ has a Hamiltonian cycle": testing for this is one of the classic NP-complete problems. You could try to test for it by testing if $G$ is isomorphic to any graph on a list of all $n$-vertex graphs with a Hamiltonian cycle. However, there are exponentially many graphs on any such list. So you will not get an efficient algorithm in this way.


问:
Let $\mu$ be a regular outer measure on $X$, $\mu(X) < \infty$. If $A \subseteq X $ is set such that $\mu(A)+\mu(A^c)=\mu(X)$ then $A$ is measurable
A set $A \subset X$ is measurable if $\forall T \subset X$
$$
\mu(T) = \mu(A \cap T)+\mu(A^c \cap T)
$$
By a regular measure I mean a measure $\mu$ such that for every $A \subset X$ there exists $B$ measurable such that $\mu(A) = \mu(B)$.

Edit: Forgot to add that $\mu(X) < \infty$

答:
I think you meant:

A set $B \subseteq X$ is measurable if $\forall T \subseteq X$
$$
\mu(T) = \mu(B \cap T)+\mu(B^c \cap T)
$$
By a regular outer measure it is meant an outer measure $\mu$ such that for every $A \subseteq X$ there exists $B$ measurable such that $A \subseteq B$ and $\mu(A) = \mu(B)$.

The result to prove is

Let $\mu$ be a regular outer measure on $X$, $\mu(X) < \infty$. If $A \subseteq X $ is set such that $\mu(A)+\mu(A^c)=\mu(X)$ then $A$ is measurable.

Proof:
Let $A \subseteq X $ such that $\mu(A)+\mu(A^c)=\mu(X)$.

For any measurable subset $B$ of $X$, we have  $\mu(A)=\mu(A \cap B) + \mu(A \cap B^c)$

and $\mu(A^c)=\mu(A^c \cap B) + \mu(A^c \cap B^c)$. And we also have, by sub-additivity:
$\mu(B) \leqslant \mu(A \cap B)+ \mu(A^c \cap B) $ and $\mu(B^c) \leqslant \mu(A \cap B^c)+ \mu(A^c \cap B^c) $.

Since $B$ is measurable, we also have $\mu(X)= \mu(B) + \mu(B^c)$. Since $\mu(X)< \infty$, we have $\mu(B^c) < \infty$ and so we have $\mu(B)= \mu(X) - \mu(B^c)$.

Now, using those equalities, we have:

\begin{align*}
\mu(B) & \leqslant \mu(A \cap B)+ \mu(A^c \cap B) \leqslant \\
& \leqslant \mu(A \cap B)+ \mu(A^c \cap B) + ( \mu(A \cap B^c)+ \mu(A^c \cap B^c) -\mu(B^c) ) =\\
& = (\mu(A \cap B) + \mu(A \cap B^c)) + ( \mu(A^c \cap B) + \mu(A^c \cap B^c)) - \mu(B^c) = \\
& = \mu(A) + \mu(A^c) - \mu(B^c) = \\
& = \mu(X) - \mu(B^c) = \\
& = \mu(B)
\end{align*}
So, any measurable subset $B$ of $X$, we have
$$ \mu(B) = \mu(A \cap B)+ \mu(A^c \cap B)$$

Now, given any $T \subseteq X$, since $\mu$ is a regular outer measure, there is  $B$ measurable such that $T \subseteq B$ and $\mu(T) = \mu(B)$. So, we have
\begin{align*}
\mu(T) & \leqslant  \mu(A \cap T)+ \mu(A^c \cap T )\leqslant
  \mu(A \cap B)+ \mu(A^c \cap B) =
 \mu(B) = \mu(T)
\end{align*}
So, given any $T \subseteq X$, we have
$$ \mu(T) = \mu(A \cap T)+ \mu(A^c \cap T ) $$
It means, $A$ is measurable.


问:
Let $()$ be continuous on $[0,∞)$ and twice differentiable on $(0,∞)$
Let $()$ be continuous on $[0,∞)$ and twice differentiable on $(0,∞)$. If $′()> 0$ and $″()<0$ on $[1,∞)$ prove that for all $ ≠  ≥ 1$, $|()−()|<′(1)|−|$.

Hello, can anyone help me with this question? Thank you!

答: Simple. Use the MVT which says: $f(x) - f(y) = f'(c)(x-y)\implies |f(x)-f(y)| = |f'(c)||x-y|=f'(c)|x-y|< f'(1)|x-y| $ since $f'(c) > 0$, and $f'(c) < f'(1)$ since $c > x \ge 1$ as $f'(x)$ is a decreasing function.


问:
From convex set to convex function
The definition of a convex set is geometrically intuitive. But the definition of convex function doesn't seem so intuitive: $S \subset \mathbb{R}^n$ is convex if given $x,y\in S$ the line segment joining $x,y$ is in $S$.

Let $f$ be a real valued function from an open interval $I$. Consider the graph of $f$ in plane: set theoretically it is
$$\{(x,f(x))\,|\, x\in I\}.$$
For any $a,b\in I$, consider the line joining $(a,f(a))$ and $(b,f(b))$. Then one of the following happens:

The line lies above the graph of $f$.

The line lies below the graph of $f$.

None of these hold.

Q. Suppose you know the definition of convex set and let $f:I\rightarrow \mathbb{R}$ be a function which does not satisfy (3). This means $f$ satisfies (1) or (2). What is intuition way to define function to be convex or concave?

For example, in Wikipedia, it  says that a function is convex if graph above $f$ is convex. But if we are trying to give intuitive definition of convex function based on convex set, according to convexity of region above $f$ or below $f$, what is intuitive way to decide one of them?

Since $f$ is satisfying (1) or (2), so the words convex function and concave function are reserved for it; we can assign any word to any case without intuition. But, considering definition of convex set can we get intuition to define convex function? Note that almost everyone knows geometric explanation of the standard definition of convex function.

答:
I'm refer Boris T. Polyak in his book 'INTRODUCTION TO OPTIMIZATION', page 8.

Definition of convex function

A scalar function $f(x)$ on $\mathbb R^n$ is said to be convex if

$$f(\lambda x+(1-\lambda)y)\leq \lambda f(x) + (1-\lambda)f(y)$$

for any $x,y\in \mathbb R^n$ and $0\leq \lambda \leq 1$.

Theorem

Let $f:\mathbb R^n \to \mathbb R$. $f$ is a convex function if and only if $H_{\gamma}=\{(x,\gamma)|\gamma \geq f(x)\}$ is a convex set.

Now, look this theorem who is so powerful.

Theorem

A function $f$ is convex if and only if $\nabla^2f(x)\geq 0$.

$\nabla^2f(x)$ are Hessian matrix and $\nabla^2f(x)\geq 0$ refer nonnegative definite.

Example

Let $g(x,y) = x^2-xy+y^2$.

$$\implies \nabla g(x,y) = (2x-y,-x+2y)$$
Then
\begin{equation}
\nabla^2g(x,y) =
\left(\begin{smallmatrix}
2 & -1\\
-1 & 2
\end{smallmatrix}\right)
\end{equation}

$$\implies \nabla^2g(x,y)\geq 0,\forall x,y\in \mathbb R$$

$$\therefore g(x,y) \text{ is a convex function}$$

Note that if $\nabla^2f(x)$ are not a constant matrix but is nonnegative definite you can not say $f(x)$ are a convex function because this needs to be true on all variables of the function by definition of convex function.

Title: From convex set to convex function


问:
Space spanned by matrices
I have a set of $5 \times 5$ matrices, $M_1, M_2,\dots, M_{19}, M_{20}$. I want to try to find a basis from this set and also to find relationships between these matrices.

This is how I think I should approach the problem. First I want to look at the dimension of the space spanned by these $20$ matrices. This information will allow me to determine how many matrices I need to use as a basis. Once I have said basis I can write a computer program to solve the analogous question of $AX=b$.

My first question: This problem is straightforward with vectors, but now that matrices are involved I am lost. Can I create a vector out of each matrix?

To find relationships among the $20$ matrices, I will look at the space spanned by the kernel of these $20$ matrices. Again, I am not sure how to do this with matrices.

答:
Yes, it's enough to look at the matrices as vectors (by concatenating the columns, for example): as $K$-vector spaces, $K^{n \times n}$ (matrices) and $K^{n^2}$ (vectors) are isomorphic.

If you already know an algorithm for the problem with vectors, there's no need to modify it here.

Title: Space spanned by matrices


问:
Uniformly bounding the difference of two logarithms
Consider two logarithmic functions defined as $g(\epsilon) = \log \epsilon $ and $ f(\epsilon) = \log (\epsilon + \epsilon^2 )$ for $0<\epsilon <1$. Can I find a $\delta > 0$ such that the difference $g(\epsilon) - f(\epsilon)$ is uniformly bounded on the interval $(0,\delta)$: $$ \sup_{\epsilon\in (0,\delta)} |g(\epsilon)-f(\epsilon)| = M < \infty ? $$

答: Obser that $|g(\epsilon) - f(\epsilon)| = \log(1+\epsilon) < \epsilon < 1 =\delta $ or use $\log(1+\epsilon) < \log 2 = M$. Either one works.


问:
Can anyone explain me the solutions of this $|x-5| + |x+6| = 11$?
I have an intuition of this using number line but I want to know what mathematics goes into solving these type of problems.</p>

答:
Try different cases, for example,

when x>5 and x<6;

when x<5;

when x>6.

Substitute the expression for the absolute value in each case, solve the resultant linear equation (if it exists) and then verify whether these are in the bounds of your assumptions.


问:
Rate of change of position vector of point on a circle with a twist
I am wondering how I would calculate the x & y components of a position vector of a point on a circle of radius 1, where:

the origin of the position vector is at the bottom of the circle (or the bottom of the circle is at point (0,0) on the x-y axis)

the position vector is rotating counterclockwise at a given constant angular velocity or rate of change of angle, from 0 to 180 degrees

Picture a circle of radius 1 on the x-y plane, such that its center is coincident with the y-axis and the bottom of the circle is tangent to the x-axis. The origin of the position vector is at origin of x-y plane.

答:
If the center of the unit circle is (0,1), then the position vector   with given angular velocity $\omega$ :

$\vec{OP}\\$ = $\vec{OA}+\vec{AP}$=$<0,1>+<\cos \omega t,\sin \omega t$>= <$\cos \omega t,1+\sin \omega t>$


问:
Questions on proof of Gödel's Completeness Theorem in A Friendly Introduction to Mahematical Logic by Leary and Kristiansen
In Proposition 3.2.6, the authors are trying to prove that the structure $\mathfrak{U}$, whose universe is the equivalence classes of the variable free terms, models the set of sentences $\Sigma'$ using induction.

They use the usual trick of completing the proof by doing induction on complexity of the formula like they do here in Theorem 1.4.2. But when it comes to proving the base cases which is showing that statement $\sigma \in \Sigma^\prime \: \text{if and only if} \: \mathfrak{A} \models \sigma$ holds for the relations, they only consider the variable free terms, not all of the terms. How do I justify this? Shouldn't we prove for all terms instead of just the variable free terms?

Also, when proving $\sigma \in \Sigma^\prime \: \text{if and only if} \: \mathfrak{A} \models \sigma$ for statements of the form $\forall x \varphi$, the authors say that $\varphi^x_{t}$ is less complex than $\forall x \varphi$ so the induction hypothesis holds for $\varphi^x_{t}$. How do I make all of these rigorous?

I'm looking for a proper justification of all these arguments. What should I be doing induction on?

答:
The ability to avoid free variables comes from the fact that all the elements of the structure we're looking at are named by closed terms: to check that $\mathfrak{A}\models\forall x\varphi(x)$ we just need to check that $\mathfrak{A}\models\varphi(t)$ for each closed term $t$. This is not true for all structures in general - it's a very special feature of the structure $\mathfrak{A}$ built in the proof.

(On the other hand, every structure $\mathfrak{B}$ has an expansion $\hat{\mathfrak{B}}$ gotten by adding new constant symbols naming all the elements. We then can just work with sentences in the $\hat{\mathfrak{B}}$-context rather than with formulas in the $\mathfrak{B}$-context. This language-expansion trick is really just a repackaging of variable assignments, but it can frequently be convenient; I personally prefer it.)

As to the measure of complexity, this is the same as (in my experience anyways) all the usual arguments by induction on complexity: we count logical operators, that is, Booleans and quantifiers. From "$\forall x\varphi(x)$" to "$\varphi(t)$" is clearly a step down, regardless of how complicated the term $t$ is, since terms cannot use logical operators.


答: The root of $x-5=0$ and $x+6=0$ are $5,-6$ respectively. And these two roots partition the reals into $3$ regions: $(-\infty,-6]\cup(-6,5]\cup (5,\infty)$, and you can consider the equation in each region at a time and combine all solutions you obtain. Note that $-6,5$ are the solutions already.

Title: Can anyone explain me the solutions of this $|x-5| + |x+6| = 11$?

Body: I have an intuition of this using number line but I want to know what mathematics goes into solving these type of problems.</p>


答: Yes, you can treat each $n \times n$ matrix as vector of length $n^2$. Many ideas in linear algebra are related to this equivalency: for example, the inner product on $\mathbf{S}^n_+$ is defined as $\langle X, Y \rangle = \mbox{Tr} (XY)$, which is equivalent to the inner product on $\mathbb{R}^n$, $\langle x,y \rangle = x^T y$ if the matrices were written as vectors e.g. $x = \mbox{vec}(X)$.</p>

Title: Space spanned by matrices

Body:
I have a set of $5 \times 5$ matrices, $M_1, M_2,\dots, M_{19}, M_{20}$. I want to try to find a basis from this set and also to find relationships between these matrices.

This is how I think I should approach the problem. First I want to look at the dimension of the space spanned by these $20$ matrices. This information will allow me to determine how many matrices I need to use as a basis. Once I have said basis I can write a computer program to solve the analogous question of $AX=b$.

My first question: This problem is straightforward with vectors, but now that matrices are involved I am lost. Can I create a vector out of each matrix?

To find relationships among the $20$ matrices, I will look at the space spanned by the kernel of these $20$ matrices. Again, I am not sure how to do this with matrices.


问:
How can you calculate $\int_{0}^{1} \exp(\sin(x))dx$ exactly?
I'm taking a computational statistics class and I have to calculate $\int_{0}^{1} \exp(\sin(x))dx$ numerically (Monte Carlo integration) and compare the numerical result to the exact result. The problem is, how can one canculate the exact value of the integral? The antiderivative doesn't seem to exist in terms of elementary functions. Mathematica gives me this:

Not only that, Mathematica also cannot calculate the definite integral:

And since Mathematica cannot do it, I don't think other softwares can, though I'm not sure. Of course, Mathematica can do the numerical integration

But that's not what I need! Thanks in advance for any help.

Body:
As @Bumblebee commented, for the definite integral, compose Taylor series for the integrand
$$e^{\sin(x)}=1+x+\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^5}{15}-\frac{x^6}{240}+\frac{x^7}{90}+\frac{31
   x^8}{5760}+\frac{x^9}{5670}-\frac{2951 x^{10}}{3628800}-\frac{x^{11}}{3150}+\frac{181
   x^{12}}{14515200}+\frac{2417 x^{13}}{48648600}+\frac{58913
   x^{14}}{4151347200}-\frac{5699 x^{15}}{2554051500}+O\left(x^{16}\right)$$

Integrate termwise and use the bounds to get
$$\int_0^1 e^{\sin(x)}\,dx\sim \frac{60970121917}{37362124800}=1.63186977\cdots$$ while numerical integration gives
$1.63186961$

Edit

The next term in the expansion is $-\frac{52635599 }{20922789888000}x^{16}$ that is to say that its contribution to the integral is $-\frac{52635599}{355687428096000} \sim -1.48\times 10^{-7}$ which is not much.

Update (almost eight years later)

Using
$$e^{\sin(x)}=\sum_{n=0}^\infty \frac{\sin^n(x)}{n!}$$
$$I_n=\int \sin^n(x)\,dx=-\cos (x) \, _2F_1\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};\cos ^2(x)\right)$$ where appears the gaussian hypergeomtric function.
$$J_n=\int_0^t \sin^n(x)\,dx=\frac{\sqrt{\pi }\,\, \Gamma \left(\frac{n+1}{2}\right)}{2 \Gamma
   \left(\frac{n+2}{2}\right)}-\cos (t) \,
   _2F_1\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};\cos ^2(t)\right)$$

For $t=1$, the partial sums converge very fast
$$\left(
\begin{array}{cc}
p & \sum_{n=0}^p \frac {J_n}{n!} \\
 0 & 1.0000000 \\
 1 & 1.4596977 \\
 2 & 1.5960355 \\
 3 & 1.6258589 \\
 4 & 1.6310267 \\
 5 & 1.6317681 \\
 6 & 1.6318589 \\
 7 & 1.6318686 \\
 8 & 1.6318695 \\
 9 & 1.6318696
\end{array}
\right)$$


问:
Validity of simple intuition for $\sum 1/p \sim \log \log n$ work?
I want to show the following using the simplest arguments for students without university training in mathematics.

$$\sum_{p<n} 1/p \sim \log \log n$$

I'd like to check the validity of my rationale.

Step 1:

We start with the Euler product formula which we derive elsewhere [ref].

$$\sum_{n}\frac{1}{n^{s}}=\prod_{p}(1-\frac{1}{p^{s}})^{-1}$$

We take the logarithms of both sides.

Step 2:

On the RHS we have:

$$\begin{align}
\ln\left(\prod_{p}\frac{1}{(1-\frac{1}{p^{s}})}\right) &= \sum_{p}\ln\frac{1}{(1-\frac{1}{p^{s}})}\\
&=-\sum_{p}\ln(1-\frac{1}{p^{s}})\end{align}$$

We can use $\ln(1-x)=-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\ldots$ to expand $\ln(1-\frac{1}{p^s})$.

$$\begin{align}
-\sum_{p}\ln(1-\frac{1}{p^{s}})&=\sum_{p}(\frac{1}{p^{s}}+\frac{1}{2p^{2s}}+\frac{1}{3p^{3s}}+\frac{1}{4p^{4s}}\ldots)\\&=\sum_{p}\frac{1}{p^{s}}+C
\end{align}$$

We know $C$ converges because we know $\sum1/n^s$ converges for $s>1$, and the sum over primes $\sum 1/p^s$ is a subset of the sum over integers $\sum 1/n^s$.

Step 3:

We start with the logarithm of the partial harmonic series.

$$
\ln(\ln(n+1)) < \ln\left(\sum_{1}^{n}\frac{1}{x}\right) < \ln(\ln(n)+1)
$$

The inequality is from the integral comparison tests for finite harmonic partial sums (ref). The lower and upper bounds are like $\ln(\ln(n))$

The sum $\ln\left(\sum_{1}^{n}\frac{1}{x}\right)$ is the LHS of the Euler product in the limit as $n\rightarrow \infty$, and $s\rightarrow1^{+}$ for the Euler product.

Step 4: This is the step I am not sure about.

In this step we try to combine the results from step 2 and 3.

We consider the limit $s\rightarrow1^{+}$ for the expression from step 2:

$$\sum_{p}\frac{1}{p}+C$$

Similarly we consider the limit $n\rightarrow \infty$ for the expression in step 3 and conclude that the logarithm of the harmonic series diverges but it does so as $\ln(\ln(n))$.

Therefore we can say the prime harmonic series diverges as

$$\sum_{p<n} 1/p \sim \ln(\ln(n))$$

The key idea I think I'm using is to enable comparison of two expressions by taking the limits $n\rightarrow \infty$  and $s\rightarrow1^{+}$ to make them equivalent.

Is there a flaw in this argument? I would appreciate explanations that didn't assume advanced knowledge.

答:
In Step 2, you should write $C$ as $C_s$: it depends on $s$.  You gloss over this point in Step 4 where you write $C$ and let $s \to 1$.  You in fact should check (i) $C_s$ converges for $s > 1/2$ and (ii) $C_s$ is continuous in $s$, so as $s \to 1^+$, $C_s$ tends to $C_1$.  They're not all the "same" $C$.

The rest of the reasoning in Step 4 is even less justified (or more flawed, as you suggest may be the case). Analysis is full of limits taken in different ways, and proving the two methods can be related may be a very hard problem or even an unsolved problem. One source of heuristics in number theory is to take an iterated limit and just switch the order without justification to see what happens, and if the result makes sense, claim that should be the result of carrying out the limits in the initial order even if there is no proof of this yet.

In the setting you are writing about, there are two kinds of series with a limit that you are interested in:
$$
\lim_{s \to 1^+} \sum_p \frac{1}{p^s} \text{ and } \lim_{n \to \infty} \sum_{p < n} \frac{1}{p}.
$$
The first series runs over all prime numbers for each $s > 1$, while the second runs over finitely many primes for each $n$, so these are fundamentally very different kinds of series over the primes. Using the pole of the zeta-function at $s = 1$, it is "not hard" to show $\sum_{p} 1/p^s \sim \log(1/(s-1))$. Converting knowledge of the $s$-limit of a series over all prime numbers to knowledge of the $n$-limit of a series of finitely many prime numbers (for each $n$) is not something I thought you would be able to do at the level you are trying to pitch your presentation. But I was mistaken. See Paul Pollack's paper here, which uses estimates on $\zeta(s)$ and $\log \zeta(s)$ for $s$ near $1$ from the right to show $\sum_{p < n} 1/p$ differs from $\log(\log n)$ by at most 6 for large $n$, and having $\sum_{p < n} 1/p - \log(\log n)$ be bounded is stronger than $\sum_{p < n} 1/p \sim \log(\log n)$.  You'll have to judge how much of Pollack's argument can be made accessible to your students.