问:
Sufficient Conditions for Borel-Cantelli
The first Borel-Cantelli Lemma states

If $\sum \mathbf{P}(A_n) < \infty$ then $\mathbf{P}(A_n i.o.) = 0$

Question :

If $\mathbf{P}(A_n) \rightarrow 0$ and $\sum \mathbf{P}(A_{n+1} / A_{n}) < \infty$ then $\mathbf{P}(A_n i.o.) = 0$

Now of course $\sum \mathbf{P}(A_{n+1} / A_{n}) < \infty$ does not imply $\sum \mathbf{P}(A_n) < \infty$. For example we can take $\mathbf{P}(A_n) = 1/n$.

So how can we takle such a problem.

My approach:
We can find indices $n_k$ s.t. $\sum_{n_k}^{n_{k+1}}\mathbf{P}(A_n) < 1/k^2$ $k = 1,2,3..$ Define an event $B_k = \cup_{n_k}^{n_{k+1}}\mathbf{P}(A_n) $ Then By Borel Cantelli $\mathbf{P}(B_n i.o.) = 0$ Thus as only a finite number of $B_n$s happen we conclude on a finite number of $A_n$s happen.

Any help is appreciated

答:
Set $B_n := A_{n+1} \setminus A_n$. Then by Borel-Cantelli we obtain $$\mathbb{P}(B_n \quad i.o.) = 0 $$ since $\sum_{n \in \mathbb{N}} \mathbb{P}(B_n) < \infty$. This means that for almost all $\omega \in \Omega$ there are only finitely many $n \in \mathbb{N}$ such that $\omega$ is in $A_{n+1}$ but not in $A_n$. Thus we can choose $N$ sufficiently large such that either for all $n > N$ it is $\omega \in A_n$ or for all $n > N$ it is $\omega \not \in A_n$. Intuitively $\omega$ cannot change infinitely often between being inside of $A_n$ or being outside of $A_n$.

It remains to show that the first is impossible, i.e. that for almost all $\omega$ it is not $\omega \in A_n$ for all $n > N$. For that we use the assumption $\lim_{n \rightarrow \infty} \mathbb{P}(A_n) = 0$, which yields $$ \mathbb{P}(\bigcap_{n>N} A_n) \leq \mathbb{P}(A_n) \rightarrow 0 \quad (n \rightarrow \infty).$$ Thus for almost all $\omega \in \Omega$ it is eventually $\omega \not \in A_n$.


问:
Resultant of two polynomials f and g
I know how to calculate the resultant of two polynomials.. but I am  little confused with an example I have come across in a past paper...

The type of example I am used to are f= $5X^3-185X-420$ and g=$-2X^3+218X-840$

I know the initial matrix would be

$\begin{bmatrix} 5 & 0 & 0 & -2 & 0 & 0 \\0 & 5 & 0 & 0 &-2 & 0 \\-185 & 0 &5&218&0&-2\\-420&-185&0&-840&218&0\\0&-420&-185&0&-840&218\\0&0&-420&0&0&-840 \end{bmatrix}$

But then I have an example where the two have different largest powers

f=$2X^3-8X^2-40X+96$ and g=$3X^2-17X-6$

Would the first matrix be...

$\begin{bmatrix} 2&0&0&0&0&0 \\ -8&2&0&3&0&0\\-40&-8&2&-17&3&0\\96&-40&-8&-6&-17&3\\0&96&-40&0&-6&-17\\0&0&96&0&0&-6 \end{bmatrix}$

Im very confused by this and the way in which the matrix should look for g

Any help wouldbe greatly appreciated as I have the exam tomorrow

答: No; the size of the Sylvester matrix is the sum of the degrees. In your second case, you should have $$\begin{pmatrix} 2 & 0 &  3 & 0 & 0 \\ -8 & 2 & -17 & 3 & 0 \\ -40 & -8 & -6 & -17 & 3 \\ 96 & -40 & 0 & -6 & -17 \\ 0 & 96 & 0 & 0 & -6 \end{pmatrix}$$ that is, write the degree $m$ polynomial $n$ times, the degree $n$ polynomial $m$ times.</p>

ContentLicense: CC BY-SA 3.0

Title: Resultant of two polynomials f and g


问:
Visualizing lower bound for the Happy Ending Problem (small cases).
I'm studying the Happy Ending Problem now, which states that

For any positive integer $s$, any sufficiently large finite set of points in the plane in general position has a subset of $s$ points that form the vertices of a convex polygon.

Let $g(s)$ denote the minimum number of points in general position whose set must contain a convex $s-$gon. It's known that $g(s) = 2^{s-2}+1$ for $s\le 6$, but the exact value for $s>6$ is unknown. However, it has been shown that $g(s) > 2^{s-2}$ for all $s$ through an example of a set of $2^{s-2}$ points without any convex $s-$gon. I've seen the construction of this example, but I have a hard time visualizing it because it involves organizing sets of points very distant from each other recursively.

I'm looking for a geometric visualization of sets like that for small cases of $s$. It is easy to find the setting for $s = 4$ and $s = 5$ (which are as follows) in the online articles on the subject, but I was unable to find any illustration for $s\ge 6$. An example for the case $s = 7$ would satisfy me a lot.

Above, examples for $s=4$ (violet) with $4$ points for and $s=5$ (green) with $8$ points.

Edit: Going with my intuition, I have found the following set of $16$ points which seems to not contain any convex hexagon. There are too many polygons to check, though, despite of all the symmetries, so I'm not sure about it.

In the figure I'm using a polar grid with an angle of $\dfrac{\pi}{40}$, in case someone wants to extract the exact coordinates.

I think those examples for $4\le s\le 6$ (if the last one is correct) consisting of concentric $s$-gons are pretty neat, but my hopes for this pattern to continue are not very high.

ContentLicense: CC BY-SA 4.0

Body:
If you can track down a copy of the paper "On some extremum problems in elementary geometry" by Erdős and Szekeres (Ann. Univ. Sci.
Budapest. Eotvos Sect. Math. 3-4 (1961) 53-63.), they describe such a construction of $2^{n-2}$ set that is void of convex $n$-gons. It is also reprinted in "Paul Erdős: the art of counting. Selected writings". Alas, there is no visualization there in the paper but an algorithm to construct such sets. I don't have copies of these on hand, but you may try to find them. Best of luck!

Edit.

If I didn't implement it incorrectly, we have for $n=6$ the following $16$ points in cartesian coordinates without a convex hexagon:

   1      1
   2301   -459
   5305  -1210
   5306  -1208
   5307  -1201
   5308  -1170
   6592  -1639
   6593  -1630
   6594  -1628
   6595  -1567
   6596  -1565
   6597  -1558
   6938  -1812
   6939  -1764
   6940  -1755
   6941  -1753

Which is kind of hard to see from its plot:

And for $n=7$ the following $32$ points:

        1        1
    16135    -2688
    63595   -12180
    63596   -12178
    63597   -12171
    63598   -12140
    63599   -11975
    96151   -20319
    96152   -20310
    96153   -20308
    96154   -20247
    96155   -20245
    96156   -20238
    96157   -19499
    96158   -19497
    96159   -19490
    96160   -19459
   109624   -24810
   109625   -24762
   109626   -24753
   109627   -24751
   109628   -23990
   109629   -23981
   109630   -23979
   109631   -23918
   109632   -23916
   109633   -23909
   111766   -25881
   111767   -25581
   111768   -25533
   111769   -25524
   111770   -25522


问:
Get $\frac{d^2y}{dx^2}$ from $2x(\frac{dy}{dx}) = \frac{dy}{dt}$
I stumbled upon this on a "worked solutions manual"
But I dont quite understand the part I selected in red.
[![This is the working out they show][1]][1]

How can $ \frac{dy}{dt}$ be converted to $\frac{d^2y}{dx^2}$  when differentiating with respect to $x$?

Im probably missing something...
[1]: https://i.stack.imgur.com/YZ3FC.jpg

Body:
You have $$\frac{dy}{dx}=\frac{1}{2x}\frac{dy}{dt},~\frac{dx}{dt}=2x$$
Then $$\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})=\frac{d}{dt}(\frac{dy}{dx})(\frac{dt}{dx})=\frac{d}{dt} \left(\frac{1}{2x} \frac{dy}{dt}\right)\frac{1}{2x}$$
$$=\left(\frac{-1}{2x^2}\frac{dx}{dt}\frac{dy}{dt}+\frac{1}{2x}\frac{d^2y}{dt^2}\right)\frac{1}{2x}=\frac{1}{4x^2}\left(\frac{d^2y}{dt^2}-2\frac{dy}{dt}\right)$$


问:
Using Cauchy's integral formula to evaluate the integral
I have got the following question in my homework at which I am stuck.

Use Cauchy's Integral Formula to evaluate the following integral:

$$\int_{|z|=1}\frac{z^2-z+1}{z-1}dz$$

Now, here $z=1$ is a point on the boundary of the given curve $|z|=1$, at which the integrand has a singularity. And Cauchy's integral formula is applicable for points in the interior of the curve. But the question specifies to use Cauchy's integral formula.

I am really confused as to how solve this and whether Cauchy's Theorem can be used here which states that:

If f(z) is holomorphic in a simply connected open region R and $\gamma$ is a rectifiable closed curve contained in R then, $\int_{\gamma} f(z)dz$ = 0

答:
The integral on a path that passes through a pole will likely not exist in the normal sense.  However, one convention is instead to calculate its principal value.  This amounts to calculating the integral on the curve excluding a small symmetrical segment on either side of the pole and then letting the size of the excluded portion approach zero.  If the limit exists then that is the principal value.

It can be evaluated using the "Indentation Lemma" which says that for $f$ holomorphic except for a simple pole at $z_0$, integration along an incomplete circular shaped arc around $z_0$, given by $\gamma= z_0+\varepsilon e^{i\theta}$, $\theta_1 \leqslant \theta \leqslant\theta_2$ has limiting value $i \cdot(\theta_2-\theta_1)\cdot\text{res}(f, z_0)$ as $\varepsilon \to 0$.

Then, to proceed in your case, evaluate the integral around $|z|=1$ except take a small (nearly) semi-circular detour around the outside of $z=1$.  By Cauchy's residue theorem this integral is equal to $2\pi $ times the sum of the residues inside the curve, which includes that at $z=1$ by virtue of the detour.  You can use the Indentation lemma to calculate how much the detour contributes in the limit, an by subtraction you obtain the limiting value of the integral along the curve $|z| = 1$ excluding a small segment.

The result will be $2\pi i \cdot \sum \text{res}(f, |z|<1)$, zero in this case, plus, $\pi i\cdot \text{res}(f,z=1)$


问:
"Big O" notation
To work out the order of a function $f(x)$, I used to just look at the leading term of $f(x)$ (we call $g(x)$) and we would have $\forall x \quad \exists C$ such that $f(x) < Cg(x)$.

I was just given the definition:

Given two functions $F(t)$ and $G(t)$, we say that $F(t) = O(G(t))$ if $\exists C, \epsilon$ between $(0, \infty)$ such that $|t|<\epsilon \implies |F(t)| \leq C|G(t)|$.

Using this definition, $t^2$ is $O(t)$ and $t$ is not $O(t^2)$ - is this right? I would've said the other way around.

答:
With your definition:

$F(t) = O(G(t)) \iff $ there is a neighborhood $U$ of $0$ such that $\frac{F(t)}{G(t)}$ is bounded on $U.$

$ \frac{t^2}{t}=t$ is bounded in a neighborhood of $0$, hence $t^2 $ is $O(t)$.

$ \frac{t}{t^2}= \frac{1}{t}$ is not bounded in a neighborhood of $0$, hence $t$ is not $O(t^2)$.

Title: "Big O" notation


问:
List of all $\mathbb{C}[X]$ modules?
Let $M$ be a $\mathbb{C}[X]$-Module.

The task ist to list all $\mathbb{C}[X|$-modules $M$ with $\dim_\mathbb{C}M=1$, $\dim_\mathbb{C}M=2$, $\dim_\mathbb{C}M=3$ up to isomorphism.

I'm not sure I understand this question. In general defining a $K[X]$-Module $M$ is equivalent to defining a $K$ linear map $M\longrightarrow M$, so I would just answer "any matrix with 1 component, any $2x2$ matrix, any $3x3$ matrix even with $0$ determinant". This whole thing doesn't make much sense to me, so I'd be glad if anyone could clarify the question.

答:
I will do the case $\dim_{\Bbb C}M=2$ and leave the rest to you.

First note that such a module $M$ is necessarily finitely generated over $\Bbb C[X]$ and hence is of the form $$M=R^n\oplus\bigoplus_{i=1}^kR/(f_i)$$
with $R=\Bbb C[X],0\ne f_i\in R$ monic. Since $\dim_{\Bbb C}R=\infty$ we necessarily have $n=0$. Now we have $$\dim_{\Bbb C}M=\sum_{i=1}^k\deg f_i$$
So by eliminating those $f_i$ that are units, i.e. for which $R/(f_i)=0$, we see that $M$ is of the form $$M=R/(f_1)\oplus R/(f_2)$$ with $\deg f_1=\deg f_2=1$ or $$M=R/(f)$$ with $\deg f=2$. Let us look at the first case: $M$ is of the form $$M=\Bbb C[X]/(X-\alpha)\oplus\Bbb C[X]/(X-\beta)$$
If $\alpha,\beta\in\Bbb C$ are distinct then we can write this as $M=\Bbb C[X]/((X-\alpha)(X-\beta))$ and we are reduced to the second case. If $\alpha=\beta$ we get $$M=(\Bbb C[X]/(X-\alpha))^2$$

So in total we get two types of modules: The first is of the form $(\Bbb C[X]/(X-\alpha))^2$ and the second type is of the form $\Bbb C[X]/((X-\alpha)(X-\beta))$ for $\alpha,\beta\in\Bbb C$. By looking at the annihilators we see that for different choices of $\alpha,\beta$ these modules will not be isomorphic (as $\Bbb C[X]$-modules)

If you prefer to think about these modules as $\Bbb C$-vectorspaces with an endomorphism $f$, the first type correspond to those endomorphisms whose minimal polynomial has degree one (or those that are diagonalizable with exactly one eigenvalue) and the second type correspond to those whose minimal polynomial has degree two (in this case the homomorphism is diagonalizable iff $\alpha\ne\beta$)

Title: List of all $\mathbb{C}[X]$ modules?


问:
What are some applications for inverting matrices?
I would like to know some of the real-world (preferably not in the mathematics field) applications for inverting matrices, specifically doing it fast. It would be great if you could provide online resources to help me answer this question.

Thanks.

答:
Matrix inversion is useful whenever you are working with matrices, and the actual applications of this are actually pretty great these days (e.g. fields like machine learning and statistics readily use the language of linear algebra, matrices, etc.). Just like we want to be able to solve equations like $xy = 10x$ when $x$ is a nonzero real number (clearly $y = 10$), inverting provides the theory to solve things like $AB = 10A$ when $A$ and $B$ are matrices.

For an example of an application, least squares regression uses matrix inversion. If you have some data arranged in a matrix $X$ that predicts response values $\mathbf{y}$, the "best linear estimate" (in a least squares sense) in the form $\widehat{\mathbf{y}} = X \widehat{\beta}$ has parameters calculated by $\widehat{\beta} = (X^T X)^{-1} X^T \mathbf{y}$. (The first part $(X^T X)^{-1} X^T \mathbf{y}$ is known as the "hat" or "projection matrix".)

So I think this is a very "real-world" application of matrix inversion. Doing it fast is of practical importance, since very large datasets correspond to very large matrices, and computing matrix inverses quickly is a challenging problem. (For large matrices and when speed is an issue, it seems that there are different approaches that try and avoid the inversion step as much as possible when speed and efficiency is of importance -- things like QR/SVD/Cholesky factorisation, etc. -- that you might search up if you are interested.)


问:
Algorithm for inversion in truncated polynomial ring
I have a paper that describes an algorithm for finding the inverse of a polynomial in $(\mathbb{Z}/3\mathbb{Z})[X]/(X^N-1)$:

I want to implement it in code, but I'm having difficulties to understand it. I hope someone could answer me the following questions to clarify mu dubts:

The paper specifies that all computations are done modulo 3, and thus the coefficients are choosen from the set $\{-1,0,1\}$. Is it because we are talking about $\mathbb{Z}/3\mathbb{Z}$? Some implementations in code seem to use coefficients in the set $\{0,1,2\}$. Is there any difference or are these sets equivalent?

Step 4 involves a division and multiplication by $X$. I think that dividing by $X$ means multiplying by the inverse of $X$, but I don't know if that inverse should be calculated modulo 3 or if it is the inverse of $X$ in the stated ring. Same question applies to the star multiplication by $X$.

Thanks in advance for any answer.

答:
$X$ is not a variable ranging over $\Bbb{Z}_3$. It is an indeterminate. So division by $X$ should be polynomial division (division modulo three is meaningless). Think about the situation in Step 3. You have a polynomial $f(X)$. If $f_0=0$ it means that the constant term of $f(X)$ is zero.
So at this step you are required to, for example, divide
$$f(X)=X-X^2+X^5-X^6$$
by $X$, and come up with $f(X)/X=1-X+X^4-X^5$. Or, taking into account the do-while loop, divide something like
$$f(X)=X^4+X^5+X^6-X^{11}$$ by $X$ four times to end up with $1+X+X^2-X^7$. The point is that when exiting this loop $f(X)$ must have a non-zero constant term.

If you store a polynomial as an array of its coefficients, division by $X$ amounts to stripping the leading zeros from that array (and padding them to the high degree end).

In the step where you are to multiply $c(X)$ by $X$ you must take care not to let the array of coefficients of $c(X)$ overflow from the high end. The algebra is done modulo $X^N-1$, meaning that if $c(X)*X$ get a term of degree $N$, you reduce it modulo $X^N-1$, whence $X^N=1$. So if you represent
$$c(X)=c_0+c_1X+c_2X^2+\cdots+c_{N-1}X^{N-1}$$
with the array $(c_0,c_1,\ldots,c_{N-1})$, then $c(X)*X$ becomes
the array $(c_{N-1},c_0,c_1,\ldots,c_{N-2})$. In other words, multiplication by $X$ amounts to a cyclic shift to the right for the list of coefficients.


问:
Validity of Jordan's lemma
I'm trying to calculate the following integral

$$ \int_0^\infty dx \, e^{ix}$$

with contour integration. By choosing a quarter circle contour $\gamma$, by the residue theorem we get

\begin{align}
0 =\oint_\gamma dz \, e^{iz} =& \int_0^\infty dx\, e^{ix} + \int_{\mathrm{Arc}} dz \, e^{iz} -i\int_0^\infty dy \, e^{-y} \\
=& \int_0^\infty dx\, e^{ix} + \int_{\mathrm{Arc}} dz \, e^{iz} -i
\end{align}

therefore, if the integral over the arc is zero, the original integral becomes $i$.

Can Jordan's lemma, of a variation thereof, be applied in this case?

Edit: the parametrization would be $z=R e^{i\theta}$ with $\theta \in [0, \frac{\pi}{2}]$.

Aside: it seems to me that the original integral can be interpreted as a distribution as the Fourier transform of the theta function at $k=1$, which would give $i$ as well.

答: That integral is not well-defined because for any horizontal line $z=x+iy,e^{iz}$ follows the circle of radius $e^{-y}$ in the complex plane. Therefore, symmetry shows us that the integral $\int_0^{2n\pi}e^{ix}dx=0$ but $\int_0^{(2n+1)\pi} e^{ix}dx=\left. \frac{1}{i}e^{ix}\right|_0^\pi=-i(-1-1)=2i$. Jordan's lemma will not allow you to solve the improper integral, because no solution exists. The reason for this is that the proof of Jordan's lemma relies on the symmetry $\sin\theta=\sin(\pi-\theta)$, so only the half-circular arc in the upper half-plane can apply it. However, if you want to apply a similar result, you could instead consider the maximum modulus of the function over that interval (in this case, that approach fails because $|e^{ix}|=1$ for all real $x$).

Title: Validity of Jordan's lemma


答:
I am not sure if I understand your definition.

The correct formal definition is:
$$f \in O(g) \iff \exists C > 0 \; \exists x_0 > 0 \; \forall x > x_0: |f(x)| \leq C \cdot |g(x)|$$

If I now use your example we get the following:

$f(t) = t$

$g(t) = t^2$

Clearly $f(t) \in O(g(t)) \iff t \in O(t^2)$ because:

$C = 2 > 0 $

$x_0 = 1 > 0$

$\forall x > x_0 = 1: |t| \leq 2 \cdot |t^2|$, since every number $\forall t \geq 1: t \leq t^2$.

For showing that $t^2 \in O(t)$, we cannot find a $C > 0$ and a $x_0 > 0$, such that $\forall x > x_0: |t^2| \leq C \cdot|t|$, since the quadratic growth is just to big.
So $t^2 \not \in O(t)$.

Title: "Big O" notation

Body:
To work out the order of a function $f(x)$, I used to just look at the leading term of $f(x)$ (we call $g(x)$) and we would have $\forall x \quad \exists C$ such that $f(x) < Cg(x)$.

I was just given the definition:

Given two functions $F(t)$ and $G(t)$, we say that $F(t) = O(G(t))$ if $\exists C, \epsilon$ between $(0, \infty)$ such that $|t|<\epsilon \implies |F(t)| \leq C|G(t)|$.

Using this definition, $t^2$ is $O(t)$ and $t$ is not $O(t^2)$ - is this right? I would've said the other way around.


问:
Finding convex conjugate of a bounded function
The convex conjugate of a function $f:\mathcal{X}\mapsto \mathbb{R}$ is formally defined as $$f^\star\left(y\right)=\sup_{x\in\mathcal{X}}\ \left\langle x,y\right\rangle-f\left(x\right).$$

In cases where $f\left(\cdot\right)$ is bounded, the supremum can become unbounded (e.g., $\mathcal{X}=\mathbb{R}^n$). To avoid this problem, $x$ is further constrained to a bounded set as I've read in textbooks and papers.

My question is that if there's a principle on how to bound $x$, because different constraints result in different convex conjugate functions. For instance, to find convex conjugate of $f\left(x\right)=\left\Vert x\right\Vert_0=\sum_{i=1}^n 1(x_i\neq 0)$ with $\mathcal{X}=\mathbb{R}^n$, it does make a difference to choose $\left\Vert x\right\Vert_{+\infty}\leq 1$ or $\left\Vert x\right\Vert_2 \leq 1$ as the bounding constraint.

答:
Now that I understand the context better I am going to promote my own comments to an answer.

There is no general principle for artificially restricting the domain of a function when computing its convex conjugate. In fact, you should not do so unless you have a compelling and mathematically sound reason in your specific application. The domain of the conjugate function---that is, the set of $y$ for which the supremum is finite---is an important part of the conjugate itself, and should not be discarded. For example, when computing the dual of a convex optimization problem with an objective $f(x)$, the domain information imposes implicit constraints on the dual variables. If you remove that information, the dual problem is incorrect; it no longer provides bounds for the primal problem.

So what's going on in Chapter 5 of Ms. Fazel's thesis, or in the case of the so-called $\ell_0$ norm? Note that the interest is not in the conjugate of these functions, but rather in their convex envelopes. It just so happens that the convex envelope of a function can computed using the conjugate of the conjugate. Unfortunately, the convex envelopes of $f(x)=\mathop{\textbf{card}}(x)$ or $g(X)=\mathop{\textbf{rank}}(X)$ are not very interesting---in fact, I think they are identically zero. On other hand, the modified, extended-valued functions
$$\bar{f}(x) = \begin{cases} \mathop{\textbf{card}}(x) & \|x\|_\infty \leq 1 \\ +\infty &\|x\|_\infty > 1 \end{cases}, \qquad
\bar{g}(X) = \begin{cases} \mathop{\textbf{rank}}(x) & \|X\|_2 \leq 1 \\ +\infty &\|X\|_2 > 1 \end{cases}$$
have non-trivial convex envelopes.

As for why one would want these convex envelopes, it is because they provide some theoretical justification for why $\bar{f}^{**}(x)=\|x\|_1$ and $\bar{g}^{**}(X)=\|X\|_*$ are effective convex proxies for their non-convex counterparts. As you know, Ms. Fazel's thesis is entitled Matrix Rank Minimization with Applications, and it makes heavy use of trace minimization and nuclear norm minimization to find low-rank matrices that satisfy the modeling conditions. Section 5.1 is devoted to providing justifications for the convex heuristics that she uses throughout the work.

Added to clarify: you have expressed an interest in a comment above in finding the "tightest" convex envelope of a function. It's very important to note that if you truly want the tightest convex envelope for the entire function, you cannot restrict the domain in any way before taking the double conjugate. When you impose a domain restriction, you are computing the convex envelope of a different function. It will no longer serve as a lower bound for the original function. It will, however, be a tighter envelope over the domain you have selected.

You can see this for yourself: look at $f(x)=\mathop{\textbf{card}}(x)$ and $g(x)=\|x\|_1$. Clearly, there are values of $x$ for which $f(x)<g(x)$. So $g$ cannot be the convex envelope for $f$. It is, however, the convex envelope if you restrict the domain of $f$ to $\{x\,|\,\|x\|_\infty \leq 1\}$ (as we did with $\bar{f}$ above).


问:
If $\lim_{x \to a} f\left(x\right) = L$ then does $\lim_{x \to L} f^{-1}\left(x\right) = a$
It seems intuitively true to be based upon logarithm and exponential, as well as tangent and inverse tangent, but is this always true?</p>

答:
Assuming that $f$ is only known to be bijective (so that $f^{-1}$ exists) the answer is no. Consider for example $f:[0,1]\cup(2,3]\to[0,2]$ with $f(x)=x$ if $x\in [0,1]$ and $f(x)=x-1$ if $x\in (2,3]$. Then it is clear that $f$ is bijective and satisfies $$\lim_{x\to 1}f(x)=1$$
However the inverse function $f^{-1}:[0,2]\to[0,1]\cup(2,3]$ is given by $f^{-1}(y)=y$ if $y\in [0,1]$ and $f(y)=y+1$ if $y\in (1,2]$. So the limit $$\lim_{y\to 1}f^{-1}(y)$$
doesn't exist because $\lim_{y\to1^+}f^{-1}(y)=2$ and $\lim_{y\to1^-}f^{-1}(y)=1$.

(Basically $f$ glues together the two intervals to one ($[0,2]$) while $f^{-1}$ tears that interval apart into two pieces and is therefore not continuous)

Edit: A positive result: If $f:I\to \Bbb R$ is defined on an interval $I$ (may be unbounded) and is continuous and injective with image $f(I)$, then the inverse function $f(I)\to I$ is also continuous.


问:
Mathematical rigor for engineers
An engineer with a master's degree, I would like to improve my level of mathematical rigor. For example, I would like to come to grasp

what measure theory has to do with the Fourier series and

what group theory has to do with the eigenvalue,

among others.  I would like to understand why Lebesgue integration is important (the idea seems sort of trivial to me, which naturally means that I don't yet really get it), why the formality of Weierstrassian tests is preferable to the mere evasion of the problem by, say, the extraction of Hadamard's finite part, and so on.

To this end, I have been reading Georgi E. Shilov's Elementary Real and Complex Analysis, which I like; but at the rate Shilov is going it would take like 10,000 pages to advance as far as the Fourier series.  Therefore, I have looked instead on Amazon and Google Books for the book Mathematical Rigor for Engineers.

Unfortunately, no one seems to have written that book.

What are my options?

(To clarify: I do not mean to ask the opinion-based question, "What should I do?" Rather, I mean to ask the experience-based question, "What are my options?" Maybe the only answer is that what I seek resembles Financial Analysis for Kindergarteners or Transoceanic Logistics for Retail Clerks, but since I happen to possess an applicationist's acquaintance with mathematical productions like the Fourier Series and the eigenvalue, not to mention Hadamard's finite part, I thought that I would ask.)

答:
See Roman's 2-volume book [1] below. Although it's possibly the best reference I know for what you're asking (which has been asked fairly often over the years in various online math forums), his 2-volume book is almost never mentioned. The books [2] and [3] are rather well known and possibly others will mention them, the book [4] is fairly advanced but sufficiently reader-friendly to be worth looking at from time-to-time, and [5] is a bit less known (and a bit idiosyncratic).

[1] Paul Roman, Some Modern Mathematics for Physicists and Other Outsiders, Volume 1 (1975): An Introduction to Algebra, Topology, and Functional Analysis (Volume 1 contents) AND Volume 2 (1975): Functional Analysis with Applications (Volume 2 contents -- click on amazon.com's "look inside" for Volume 1; Volume 2 contents are on pp. x-xi)

Review in Physics Today
Volume 30 #5 (May 1977), pp. 72 & 74; review by Andrew Lenard (1927-2020)

Review in Computers and Mathematics with Applications
Volume 3 #1 (1977), pp. 83-84; review by  Wilhelm Ornstein (1905-2002)

[2] George F. Simmons, Introduction to Topology and Modern Analysis (1963)

[3] Thomas A. Garrity and Lori Pedersen, All the Mathematics You Missed: But Need to Know for Graduate School (2001)

[4] Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis. A Hitchhiker's Guide (2006)

[5] Robert Hermann, Lectures in Mathematical Physics Volume 1 (1970) and Volume 2 (1970)

Title: Mathematical rigor for engineers


问:
Does there exist a transitive action on a set with ... elements?
I'm interested in finding out on which sets $X$ it is possible to define an action of $G=\mathbb{F}_{p} ^\times$ or $S_n$.

For any $x \in X$ follows by the orbit-stabilizer theorem
$$|G/Stab_G(x)|=|Gx|=|X|$$
Hence $|X|$ divides $|G|=p-1$ and for example it is not possible to define such an action on a set with an odd number of elements.

However I don't know if/why in general it is possible to define a transitive action on a set which doesn't contradict the relation above. Could anyone enlighten me?

答:
Using the relation you wrote yourself, you can clearly see that $G$ acts transitively on a set with $m$ elements if and only if $G$ has a subgroup of index $m$.

Since $\mathbb{F}_p^\times\simeq \mathbb{Z}/(p-1)\mathbb{Z}$ is cyclic, this is equivalent to $m$ dividing $p-1$.

For $G=S_n$, I have no idea if we know the possible indices of subgroups, but I would not be surprised if this were an open problem (there is a lot we do not know about the subgroups of $S_n$, like their number for instance).

Something that we can say, though, is that if $S_n$ acts transitively on a set with $m$ elements with $m\leqslant n$, then $m$ must be $1$ (trivial action), $2$ (action though the sign of the permutation) or $n$ (standard tautological action), except if $n=4$, in which case there is an exceptional additional $m=3$ allowed (given by the fact that $S_4$ has the $V_4$ subgroup of double transpositions, such that $S_4/V_4\simeq S_3$). There is also an exceptional phenomenon when $n=6$: there is a transitive action of $S_6$ on a set of $6$ elements which is not equivalent to the standard action (this corresponds to the exterior automorphism of $S_6$); this does not add new values of $m$ though.

When $m>n$ I have no idea what can be said.


问:
Average of number of dice rolls untill getting specific number
If we roll a dice till getting the number 5 (or any other number) - how can I calculate the average number of dice rolls?

Thanks

答:
Let $a$ be the average number of rolls.

When you roll the die the first time you have probability $1/6$ of obtaining a 6 and probability $5/6$ of needing to roll again a further (on average) $a$ times.

So we have a simple equation:
$$a=\frac{1}{6}.1+\frac{5}{6}.(a+1).$$

This gives $a=6$.


问:
How to find the limit $\lim_{n\to \infty}\left(n-\frac{n}{e}\left(1+\frac{1}{n}\right)^n\right)$
To find the limit

$$\displaystyle\lim_{n\to \infty}\left(n-\frac{n}{e}\left(1+\frac{1}{n}\right)^n\right)$$

答:
To get even more than the limit itself, compose Taylor series one piece at the time $$y=n-\frac{n}{e}\left(1+\frac{1}{n}\right)^n$$
$$a=\left(1+\frac{1}{n}\right)^n \implies \log(a)=n \log\left(1+\frac{1}{n}\right)$$
$$\log(a)=n\left(\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3 n^3}+O\left(\frac{1}{n^4}\right)\right)$$
$$a=e^{\log(a)}=e-\frac{e}{2 n}+\frac{11 e}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ making
$$y=\frac{1}{2}-\frac{11}{24 n}+O\left(\frac{1}{n^2}\right)$$


问:
Find the biggest area of an isosceles triangle in an unit circle
An isosceles triangle is inscribed in the unit circle. Determine the largest area of the triangle can assume.

My solution attempt:

An isosceles triangle has two sides of equal size, and the sides will be at some point on the circumference of the unit circle during suggestion the $x$-axis. The height will be $1 + h \le 2$ and width $b + 1 \le 2$

Another way would be to make the vectors of the three points that will be on the periphery (assuming that it can provide the largest area on the periphery)

I really do not know how to solve it.

答:
First note that the two angles in the figure below must be equal (the two triangles in the lower part of the picture are congruent by side-angle-side). Denote their common value by $\theta$.

The area of the triangle is then $A(\theta) = \frac{\sin2\theta}{2}+\sin\theta$, where  $0<\theta<\pi$.

Now find the maximum of $A$ on its domain.


答: I dont think this is necessarily true. In general if this were true we would call such a map homeomorphic. That is both function and its inverse is continuous. A sufficient condition for this to happen is the domain needs to be compact. You can construct non compact domains on which the inverse is not continuous.</p>

Title: If $\lim_{x \to a} f\left(x\right) = L$ then does $\lim_{x \to L} f^{-1}\left(x\right) = a$

Body: It seems intuitively true to be based upon logarithm and exponential, as well as tangent and inverse tangent, but is this always true?</p>


问:
Way to tackle probability odds problems
I have checked out other questions, but still haven't got satisfactory answer as conditions in those questions were slightly different.
Disclaimer: Only for educational purposes.
We have a fair dice with 10 faces (1-10) and $100. Each roll is completely random. If the number you chose doesn't show up you make 10% profit on your bet, but if the number shows up you lose your entire bet.
How to tackle this problem? What are the ways to optimize profits and limit losses?</p>

答:
If I am interpreting the problem correctly, you are rolling a single die, which has 10 possible outcomes.  Then, the math inherent in the comment of Pixel is pertinent.  That is, each bet that you make has a negative expectation.

Based on this, it is impossible in the long run to show a profit.  Further, if you never bet, then (in effect), you break even, which is the absolute best that you can do.  Therefore, the way to minimize losses is to bet as little as possible.  The closer that you can come to betting nothing, the more that your expectation will approach break-even, from below.

Title: Way to tackle probability odds problems


问:
Prove that linear transformation is isomorphic, given the characteristic polynomial
I've been trying to solve this question:

Let $T :\mathbb R^2 \rightarrow \mathbb R^2 $ be a linear transformation with the following characteristic polynomial:
$p(x) = x^2 +2x -3$

Prove that the linear transformation $2T+I$ is an isomorphism

What is the characteristic polynomial of $T^3$?

I've found that the eigenvalues of the transformation $T$ are $\lambda_1 =1, \lambda_2 =-3$.

Therefore, T is an isomorphism. Can I claim that because T can have a representing matrix, the multiplication by scalar and addition of the identity matrix keep it an isomorphism?
$2·[T]_B+I$

Thank you!

Body: $T^2+2T-3I=0$ implies that $T(T+2I)=3I$ and ${1\over 3}T(T+2I)=(2T+I){1\over 3}T=I$


问:
Qualitative study of the solution of a system
I am having serious troubles understanding what I need exactly to solve this exercise. Here it is.

Let x($\cdot$),y($\cdot$): (${t_-},{t_+})\mapsto\mathbb{R}^2$ be the maximal solution of the system:
\begin{aligned}
 & x'(t)= -x(t)+y^2(t)\\
 & y'(t)= x(t)-y^2(t)
\end{aligned}
with initial condition $({x_0},{y_0})\in$ ${R_+}\times{R_+}$

I proved that x(t)>0 and y(t)>0 for every t in [0,${t_+}$) and now it says to prove, knowing that $V(x,y)=x+y$  is a first integral of the system ( i.e. that $\frac{dV}{dt}$=0) that ${t_+}=\infty$ and that x($\cdot$),y($\cdot$) are monotone functions.

Here is my doubt. What should I do with a first integral? What informations it gives to me? I actually do not understand the intuition behind that. I read it suggests us something about the orbits, but I do not understand why and I to use the information about the orbits to answer this question. Can someone suggest me an idea? Or telling me where I can learn more about that, I find nothing at all on internet.

答:
You can use $x+y=c$ to make the second equation a scalar one,
$$
y'=c-y-y^2.
$$
As $c>0$ the right side has a positive and a negative root. By general principles for scalar equations any solutions that starts above the negative root extends to $+\infty$. Then insert into the first equation which is linear in $x$ with continuous coefficients and thus extends as far as $y$ exists.


问:
Theorem about mean value of iterations of isometries
Let $\mathcal{H}$ be a Hilbert space and let $T\in \mathcal{L}(\mathcal{H})$ be an isometry, and for every integer n let
$$A_n=\frac{1}{n} \sum_{j=0}^{n-1}T^j.$$
Then $A_nv$ converges to $P_Sv$ as n diverges, for every $v\in \mathcal{H}$, where $P_S$ is the orthogonal projection on the space $S:=$Ker$(\mathbb{I} - T)$, i.e. of vectors which are $T$-invariant.

I am not able to make any real progress on proving this statement, any hints?

My only observations:

I saw that if $v\in S$ then, by definition, $Tv=\mathbb{I}v=v$, so that
$$A_nv=\frac{1}{n} \sum_{j=0}^{n-1}T^jv = \frac{1}{n} \sum_{j=0}^{n-1}v=\frac{n}{n}v=v$$
for all $n\in \mathbb{N}$, so that $A_nv \to v=P_Sv$.

The struggle starts when I consider $v\not \in S$.

答:
Consider the subspace $\mathrm{Im}(I - T)$. We can check that $\mathrm{Im}(I - T)^\perp = \mathrm{Ker}(I - T) = \mathrm{Im}(P_S) = S$. Therefore $\mathcal{H} = \overline{\mathrm{Im}(I - T)} \oplus S$ (where overline is the closure). You checked that it holds for $v \in S$.

Let $v \in \mathrm{Im}(I - T)$. Then $v = w - Tv$ for some $w \in \mathcal{H}$. We can check that $P_S v = 0$. Then
$$
\frac{1}{n} \sum^{n-1}_{j=0} T^j v = \frac{1}{n} \sum^{n-1}_{j=0} (T^j w - T^{j+1} w) = \frac{1}{n} (w - T^nw) .
$$
(We have a telescoping series.) Since $T$ is an isometry, we can apply the triangle inequality to show that
$$
\frac{1}{n} \sum^{n-1}_{j=0} T^j v \to 0 , \quad n \to \infty .
$$
Since $P_S v = 0$, the claim is also true for $v \in \mathrm{Im}(I - T)$.

To finish the argument, we have to check that this holds in the closure as well. To do this, for $v \in \overline{\mathrm{Im}(I - T)}$, we can take a sequence $v_n \in \mathrm{Im}(I - T)$ such that $v_n \to v$ as $n \to \infty$, and use what we have just shown (again, using the fact that $T$ is an isometry).


问:
A linear functional $f$ is continous if and only if...
Let be $E$ a topological vector space and a linear functional $f$. How to show that $f$ is continous if and only if exist an oper set $U$ and a real value $t$ such that $t\notin f(U)$.

答:
If $f$ is continuous: Then $f$ is bounded so let $U$ be any non-empty bounded open set. Then $f(U)$ is bounded so let $t=1+\sup \{|f(u)|:u\in U\}.$

If $f$ is discontinuous: Then $f$ is unbounded so there exists $\{e_n: n\in \Bbb N\}\subset E$ with $\{1\}=\{f(e_n):n\in \Bbb N\}$ and $\lim_{n\to \infty}\|x_n\|=0.$

Let $p\in U\subset E$ where $U$ is open in $E.$ Let $B$ be an open ball in $E$ centered at $p,$ with radius $r>0,$  such that $B\subset U.$

For $t\in \Bbb R$ take $n\in\Bbb N$ large enough that $\|x_n\|\cdot |t-f(p)|<r.$  Then $p+x_n(t-f(p))\in B\subset U$ and $f( p+x_n(t-f(p)))=t.$


问:
Likelihood function of two i.i.d. normal samples
Let $X_1,...,X_{n_1}$ be an i.i.d. sample from $N_p(\mu_1,\Sigma)$ and let $Y_1,...,Y_{n_2}$ be an independent sample from $N_p(\mu_2,\Sigma)$, for some $\mu_1,\mu_2 \in \mathbb{R}^p$ and some invertible, $p\times p$ positive definite matrix $\Sigma$.

I'd like to find the likelihood function $L(\mu_1,\mu_2,\Sigma)$ of the commbined sample:

In my book, the likelihood function of $X_1,...,X_n \sim N_p(\mu,\Sigma)$ is given by

$$\frac{1}{(2\pi)^{np/2}\text{det}(\Sigma)^{n/2}}\exp\biggl(-1/2\bigl(\sum^n_{i=1}(x_i-\mu)^T\Sigma^{-1}(x_i-\mu)\bigr)\biggr)$$

So, $$L(\mu_1,\mu_2,\Sigma)=$$
$$\frac{1}{(2\pi)^{p(n_1+n_2)/2}\text{det}(\Sigma)^{\frac{n_1+n_2}{2}}}\exp\biggl(-1/2\sum^{n_1}_{i=1}(x_i-\mu_1)^T\Sigma^{-1}(x_i-\mu_1)-1/2\sum^{n_2}_{i=1}(y_i-\mu_2)^T\Sigma^{-1}(y_i-\mu_2)\biggr)$$

Would this be correct?

答:
There is a simple trick that helps to deal with this likelihood. Set
$$
\bar{x}:=\frac{1}{n_1}\sum_{i=1}^{n_1}x_{i} \quad\text{and}\quad S_x:=\sum_{i=1}^{n_1}(x_{i}-\bar{x})(x_{i}-\bar{x})^{\top},
$$
and, similarly, $\bar{y}$ and $S_y$. Then
\begin{align}
\ln\mathcal{L}(\mu_1,\mu_2,\Sigma)&=-\frac{(n_1+n_2)p}{2}\ln(2\pi)+\frac{(n_1+n_2)}{2}\ln|\Sigma^{-1}| \\
&\quad-\frac{1}{2}\operatorname{tr}(\Sigma^{-1}(S_x+S_y)) \\
&\quad-\frac{n_1}{2}(\bar{x}-\mu_1)^{\top}\Sigma^{-1}(\bar{x}-\mu_1)-\frac{n_2}{2}(\bar{y}-\mu_2)^{\top}\Sigma^{-1}(\bar{y}-\mu_2)
\end{align}
because
\begin{align}
&\sum_{i=1}^{n_1}(x_{i}-\mu)^{\top}\Sigma^{-1}(x_{i}-\mu_1)=\sum_{i=1}^{n_1}\operatorname{tr}\left((x_{i}-\mu_1)^{\top}\Sigma^{-1}(x_{i}-\mu_1)\right) \\
&\qquad=\sum_{i=1}^{n_1}\operatorname{tr}\left(\Sigma^{-1}(x_{i}-\mu_1)(x_{i}-\mu_1)^{\top}\right)=\operatorname{tr}\left(\Sigma^{-1}\sum_{i=1}^{n_1}(x_{i}-\mu_1)(x_{i}-\mu_1)^{\top}\right)\\
&\qquad=\operatorname{tr}\left(\Sigma^{-1}(S_x+n_1(\bar{x}-\mu_1)(\bar{x}-\mu_1)^{\top})\right) \\
&\qquad=\operatorname{tr}(\Sigma^{-1}S_x)+n_1(\bar{x}-\mu_1)^{\top}\Sigma^{-1}(\bar{x}-\mu_1).
\end{align}
Now it is easy to see that the likelihood is maximized when $\mu_1=\bar{x}$ and $\mu_2=\bar{y}$ ($\because \Sigma^{-1}$ is positive definite).


问:
Evaluate a sum involving n choose r
Evaluate: $\sum_{k=0}^6 (-1)^k \binom{6}{k}$ where $\binom{n}{r}= \frac{n!}{r!(n-r)!}$.

I'm unsure how to compute the part with $\binom{6}{k}$, it should be something along the lines of $\frac{6!}{k!(6-k)!}$ can we simplify this down?

I think there was an expansion method for factorials I could used in the denominator that was somewhere along the lines of $(6-k)(5-k)(4-k)(3-k)(2-k)(1-k)$ ?

答:
The most transparent way to look at this is to use the binomial theorem, as both earlier answers suggest,
$$
(p+q)^n=\sum_{k=0}^n\binom{n}{k}p^{n-k}q^k
$$
and then set $n=6$, $p=1$, $q=-1$.</p>

Title: Evaluate a sum involving n choose r


答:
Q1

Let $S=2T+I$, then $4T^2+8T-12I=0$ and so $(S-I)^2+4(S-I)-12I=0$ i.e. $S^2+2S-15I=0$. Therefore $S$ has characteristic polynomial $x^2+2x-15$ and is an isomorphism since $15\ne 0$.

Q2

Since $T$ is similar to $\begin{pmatrix}1&0\\0&-3\\\end{pmatrix}$,  $T^3$ is similar to $\begin{pmatrix}1&0\\0&-27\\\end{pmatrix}$ and therefore its characteristic polynomial is $(x-1)(x+27).$

N.B. Regarding your idea of adding $I$

This is not valid. As a simple example, imagine that your original matrix was $-I$. Adding $I$ would then reduce it to $0$.

Title: Prove that linear transformation is isomorphic, given the characteristic polynomial

Body:
I've been trying to solve this question:

Let $T :\mathbb R^2 \rightarrow \mathbb R^2 $ be a linear transformation with the following characteristic polynomial:
$p(x) = x^2 +2x -3$

Prove that the linear transformation $2T+I$ is an isomorphism

What is the characteristic polynomial of $T^3$?

I've found that the eigenvalues of the transformation $T$ are $\lambda_1 =1, \lambda_2 =-3$.

Therefore, T is an isomorphism. Can I claim that because T can have a representing matrix, the multiplication by scalar and addition of the identity matrix keep it an isomorphism?
$2·[T]_B+I$

Thank you!


问:
How to find $\mu(X\setminus A)?$
This question is  highly motivated from this

I was  searching the example .After that this example given below came in my mind

Given that $f_n :X \to [0,\infty]$ is measurable for each positive integer  $n$

let $f_n = \chi_{[n,n+1]}$ where  $A=[n,n+1]$

now $\int_{X}\chi_A d\mu=\mu(A)=n+1-n=1$.Then $\int_X{\chi_{X\setminus A}}=\mu(X\setminus A)=?$

My question is that

How to find $\mu(X\setminus A)?$

答:
$X$ supposed to be $\mathbb{R}$

since  $A \subseteq X$, So $X$ is the disjoint union of $A$ and $X \setminus A$. Hence,
$\mu(X) = \mu(A) + \mu(X\setminus A)$.

$\mu(X\setminus A)=\mu(X) - \mu(A)=\infty-1=\infty$

Title: How to find $\mu(X\setminus A)?$


问:
How to calculate this trigonometric sum
In this exercise i have to calculate this sum :

$$S=1+\frac{\cos(x)}{\cos(x)}+\frac{\cos(2x)}{\cos^2(x)}+\ldots+\frac{\cos(nx)}{\cos^n(x)}$$

There is no hint in the exercise.
I tried to use trigonometric identities but i didn't find the solution

答:
Let
$$C = 1 + \frac{\cos(x)}{\cos(x)} + \frac{\cos(2x)}{\cos^2(x)} + \cdots + \frac{\cos(nx)}{\cos^n(x)}$$
$$S = \frac{\sin(x)}{\cos(x)} + \frac{\sin(2x)}{\cos^2(x)} + \cdots + \frac{\sin(nx)}{\cos^n(x)}$$
So
$$\begin{align}C+iS&=1+\frac{\cos(x)+i\sin(x)}{\cos(x)}+\frac{\cos(2x)+i\sin(2x)}{\cos^2(x)}+...+\frac{\cos(nx)+i\sin(nx)}{\cos^n(x)}
\\&=1+\frac{e^{ix}}{\cos(x)}+\frac{e^{2ix}}{\cos^2(x)}+\cdots+\frac{e^{nix}}{\cos^n(x)}\\
&=1+\left(\frac{e^{ix}}{\cos(x)}\right)+\left(\frac{e^{ix}}{\cos(x)}\right)^2+\cdots+\left(\frac{e^{ix}}{\cos(x)}\right)^n\\
&=\frac{\left(\frac{e^{ix}}{\cos(x)}\right)^{n+1}-1}{\left(\frac{e^{ix}}{\cos(x)}\right)-1}\\
&=\frac{e^{(n+1)ix}-\cos^{n+1}(x)}{\cos^{n+1}(x)}\times\frac{\cos(x)}{e^{ix}-\cos(x)}\\
&=\frac{1}{\cos^n(x)}\left(\frac{e^{(n+1)ix}-\cos^{n+1}(x)}{e^{ix}-\cos(x)}\right)\end{align}$$
Now try simplifying this and equating real and imaginary parts. If you need any more help please don't hesitate to ask.

Title: How to calculate this trigonometric sum


问:
If $f$ is even and $y'=f(y)$ then $y$ is odd

Let $f\in C^1(\mathbb{R}, \mathbb{R})$ be an even function.

Consider the maximal solution $y\colon\left]\alpha ,\beta\right[\to \mathbb{R}$ of the IVP $$y'=f(y),\ y(0)=0$$

Prove that $y$ is an odd function and $\beta =-\alpha$.

To be able to prove that $y$ is odd, I first need its domain to be symmetric ($x\in \left]\alpha ,\beta\right[\implies -x\in \left]\alpha ,\beta\right[$), from here I can conclude that $\alpha=-\beta$. But how to prove the domain is symmetric?

And how to prove that $y(-x)=-y(x)$ for all $x\in \left]\alpha,\beta\right[$? I suspect it has something to do with the fact that the derivative of an even function is odd and $y'=f(y)$, but I can't see how to get the desired result.

答:
Define $h(x)=-y(-x)$. Then
$$
h'(x)=y'(-x)=f(y(-x))=f(-h(x))=f(h(x)),
$$
where we have used the eveness of $f$. Now, notice that $h(0)=-y(0)=0=y(0)$. As the solution should be unique you obtain that $y$ is odd.</p>


问:
Is disjunction elimination a more general form of modus ponens?
It seems to me that disjunction introduction and disjunction elimination implies modus ponens. To be clear, I'm defining disjunction introduction as:
$$p\vdash (p\lor q)$$

disjunction elimination as:
$${\displaystyle \{p\lor q,p\to r,q\to r\}\vdash r}$$

and modus ponens as:

$$\{p,p\to q\}\vdash q$$

The proof of modus ponens ($\{A, A\to B\} \vdash B$) would be as follows:

$A$ (Premise)

$A \lor A$ (From (1) by disjunction introduction)

$A \to B$ (Premise)

$B$ (From (2) and (3) by disjunction elimination)

Am I missing something? Isn't modus ponens just a special case of disjunction elimination where $p = q$?

答:
Your reasoning is fine, but you are using a nonstandard form of disjunction elimination. The usual natural deduction formulation is:

From $\vdash p\lor q$ and $p\vdash r$ and $q\vdash r$ infer $\vdash r$.

Compared to yours, this formulation uses the derivability relation $\vdash$ instead of the logical implication $\rightarrow$. It is an easy exercise to see that your version of disjunction elimination derives the standard one using the rule of $\rightarrow$-introduction.

Conversely, here is a derivation of your disjunction elimination from the standard one:

$\vdash p\lor q$ (premise)

$\vdash p\rightarrow r$ (premise)

$\vdash q\rightarrow r$ (premise)

$p,p\rightarrow r\vdash r$ (MP)

$p\vdash r$ (from (2) and (4) by cut)

$q,q\rightarrow r\vdash r$ (MP)

$q\vdash r$ (from (3) and (6) by cut)

$\vdash r$ (from (1), (5) and (7) using standard disjunction elimination)

Crucially, this derivation makes use of the modus ponens rule. So one could say that your version of disjunction elimination has some built-in modus ponens, and therefore your initial observation is not surprising.

On the other hand it is not possible to derive modus ponens from the standard disjunction elimination rule. Indeed, if your only rule involving $\rightarrow$ is the $\rightarrow$-introduction rule

From $p\vdash q$ infer $\vdash p\rightarrow q$

then the meaning of $\rightarrow$ is underspecified: you cannot rule out that the interpretation of $p\rightarrow q$ is something silly like a fixed tautology $\top$.


问:
Proving Theorem 19.2 in <Topology> by Munkres

Theorem 19.2 in Munkres Suppose the topology on each space $X_{\alpha}$ is given by a basis $\mathcal{B}_{\alpha}$. The collection of all sets of the form $$\prod_{\alpha \in J} B_{\alpha}$$ where $B_{\alpha} \in \mathcal{B}_{\alpha}$ for each ${\alpha}$, will serve as a basis for the box topology on $\prod_{\alpha \in J} X_{\alpha}$

I am trying to prove this theorem but I am wondering if my proof is solid. (For reminder, the definition, lemma, or any theorem used in the proof will be the ones mentioned in the Munkres.)

The way I tried to prove it is as follows:

proof. Use Lemma 13.2 in Munkres. Let $\mathcal{C}$ be the collection of all sets of the form $\prod_{\alpha \in J} B_{\alpha}$ (mentioned in the Theorem 19.2.). Then $\mathcal{C}$ is collection of open sets of the box topology since each $B_{\alpha}$ is open in $X_{\alpha}$. ($\because$ Box topology is the topology having as basis all sets of the form $\prod U_{\alpha}$, where $U_{\alpha}$ is open in $X_{\alpha}$.)

Let $U=\prod_{\alpha \in J} U_{\alpha}$ be open set in $\prod X_{\alpha}$ and $x\in U$. For $\alpha \:th$ coordinate of $x$ (denoted as $x_{\alpha}$), there is $B_{\alpha}$ s.t. $x_{\alpha} \in B_{\alpha}\subset U_{\alpha}$. ($\because \: U_{\alpha}$ is open in $X_{\alpha}$ and $B_{\alpha}$ is a basis element of $X_{\alpha}$. Thus by property of basis, the inclusion holds.) Since this can be done for all coordinates of $x$, we can say that $x$ is contained in some set $B$ which is an element of the $\prod_{\alpha \in J} B_{\alpha}$ mentioned in the Theorem 19.2 and $B$ is contained in the open set $U$ ($x\in B \subset U)$.

Hence by Lemma 13.2 in Munkres, the collection of all sets of the form $\prod_{\alpha \in J} B_{\alpha}$ is basis for the box topology on $\prod_{\alpha \in J} X_{\alpha}$. $$\tag*{$\blacksquare$}$$

The part that I am especially concerned about is the part where is start with "Let $U=\prod_{\alpha \in J} U_{\alpha}$ be open set in $\prod X_{\alpha}$" Even though I think this seems okay (since box topology is defined as the topology having as basis all sets of the form $\prod U_a$ where $U_a$ is open in $X_a$ for each $a$. I thought then arbitrary open set in the box topology will be of the form $\prod_{a \in J} U_{a}$ for open set is union of basis of the topology), I am still wondering if it was a hasty statement.

Thank you for your time.

答:
Simpler: Let $O$ be open in the box topology and $x \in O$. Then by the definition of the standard base of the box topology, there are open sets $O_\alpha \subseteq X_\alpha$, $\alpha \in J$ such that $$x \in \prod_{\alpha \in J} O_\alpha \subseteq  O$$

and so we have for each coordinate $\alpha$ that $x_\alpha \in O_\alpha$. We apply the fact that each $\mathcal{B}_\alpha$ is a base for $X_\alpha$ to give us $B_\alpha \in \mathcal{B}_\alpha$ such that $$x_\alpha \in B_\alpha \subseteq O_\alpha$$

for each $\alpha \in J$. Now clearly

$$x \in \prod_{\alpha \in J} B_\alpha \subseteq \prod_{\alpha \in J} O_\alpha \subseteq O$$

and the first set is in the "product base" collection defined by Munkres, and as $O$ and $x$ were arbitary, that collection has been shown to be a base for the box topology (by 13.2).


问:
little inequality conjecture
proof or disproof for $n\geq2$ even and $x>0$

$$\sum\limits_{i=0}^{n}x^i\geq \frac{(1+2\sum\limits_{i=1}^{ \frac{n}{2} }x^i)^2}{ \frac{n}{2} (x+1)+1}$$

I came up with this little inequality while playing with some standard polynomial expressions.
I checked my proof; I'm pretty sure  that it's right but it would be nice from you to find an alternative proof or apporach.

(I posted it earlier with some errors , so I apologize for that)

答:
Write $$1+2\sum_{i=1}^{n/2}x^i$$ as the dot product of the two vectors $$[x^{1/2},x^{3/2},\cdots,x^{(n-1)/2},1,x,x^2,\cdots,x^{n/2}]$$ and
$$[x^{1/2},x^{1/2},\cdots,x^{1/2},1,1,1,\cdots,1]$$ and apply Cauchy-Schwarz inequality.

Good problem.

Title: little inequality conjecture


问:
Express each of the following in modulus-argument form, where theta is between zero and pi over 2.

(a) Express each of the following in modulus-argument form, where $0<\theta<\frac{\pi}{2}$:
$$\text{i)}\ 1+i\tan\theta,\quad\text{ii)}\ 1+i\cot\theta,\quad\text{iii)}\ \frac{1}{\sin\theta}+\frac{1}{\cos\theta}i.$$
(b) Hence simplify each of the following:
$$\text{i)}\ (1+i\tan\theta)^2,\quad\text{ii)}\ (1+i\cot\theta)^{-3},\quad\text{iii)}\ \frac{1}{\sin\theta}-\frac{1}{\cos\theta}i.$$

I can answer all the questions in my math textbook except for this question. And there are no specific examples of this question in my textbook. Any help is appreciated.
THANKS

答:
Most of your examples are of the form
$$a(\cos\theta+i\sin\theta)$$
For example,
$$1+i\tan\theta=\frac{1}{\cos\theta}(\cos\theta+i\sin\theta)$$
Can you do the rest? Please tell me if you need any more help.

Note that as $0<\theta<\frac{\pi}{2}$ the values of both $\sin\theta$ and $\cos\theta$ will always be positive.


问:
Union of an open set with a boundary point
Let S be the union of  E = {$x^2 + y^2 < 1$}  and the point P = {(0, 1)}.
Is the set S open or closed or neither of them?

I know it should be trivial but I'm not sure about the answer.
I would say it is closed, since the point P is on the boundary but belongs to the set, hence the definition "set D is open if every point of D is an interior point of D" does not apply.
However I don't think it makes much sense.

Thank you for helping me.

答:
It's not open as $P$, a boundary point is not in the interior.

It's not closed as $(0,-1)$, another boundary point, is not in the set.

So "neither" is correct. If we would have added all boundary points of $E$ result would have been closed. Adding just one doesn't do it.


问:
Is this calculation of a change of variable correct?
I have a metric with the form

$$\left( \begin{array}{ccc}
\dfrac{\alpha\beta}{\gamma} \cos^2\theta+\dfrac{\beta\gamma}{\alpha}\sin^2\theta & \dfrac{\beta}{\gamma\alpha} (\gamma^2-\alpha^2)\cos\theta\sin\theta & 0\\
\dfrac{\beta}{\gamma\alpha} (\gamma^2-\alpha^2)\cos\theta\sin\theta & \dfrac{\alpha\beta}{\gamma} \sin^2\theta+\dfrac{\beta\gamma}{\alpha}\cos^2\theta&0\\
0 & 0 & \dfrac{\alpha\gamma}{\beta}\sin^2\theta
\end{array} \right)$$

which can be brought to a diagonal form with the change of variables:
$$d\theta \cos\theta+dr\sin\theta=dX$$
$$d\theta \sin\theta - dr \sin\theta= dY$$

Is it correct to say that the variable $X$ and $Y$ are related to $\theta$ and $r$ by:
$$X=r\sin\theta$$
$$Y=-r\cos\theta \, ?$$

I think that one can go from the second expression to the differential one because the correct differential for $\theta$ is $\dfrac{1}{r}d\theta$. Is this correct?

答:
There is no change of variable $X=X(\theta,r)$ and $Y=Y(\theta,r)$ that gives you that differential.

Recall that $$dX=\frac{\partial}{\partial \theta}X(\theta,r)d\theta+\frac{\partial}{\partial r}X(\theta,r)dr$$ and $$dY=\frac{\partial}{\partial \theta}Y(\theta,r)d\theta+\frac{\partial}{\partial r}Y(\theta,r)dr$$
You can check that these equations have no solutions in your case.
Another way to see it is recalling that if you differentiate twice you must get zero $d^2X=0$. So in general, in order to have $df=\Sigma_ig_idx_i$ for some functions $g_i(x_1,\dots,x_n)$, $i=1,\dots,n$, these needs to satisfy the conditions $$\frac{\partial}{\partial x_i}g_j=\frac{\partial}{\partial x_j}g_i$$
In this case this amounts to ask $$\frac{\partial}{\partial \theta}\sin\theta=\frac{\partial}{\partial r}\cos\theta$$ and $$-\frac{\partial}{\partial \theta}\sin\theta=\frac{\partial}{\partial r}\sin\theta$$
In other words we can say that the forms $$d\theta\cos\theta+dr\sin\theta$$ and $$d\theta\sin\theta-dr\sin\theta$$ are not exact.


问:
Striking applications of linearity of expectation
Linearity of expectation is a very simple and "obvious" statement, but has many non-trivial applications, e.g., to analyze randomized algorithms (for instance, the coupon collector's problem), or in some proofs where dealing with non-independent random variables would otherwise make any calculation daunting.

What are the cleanest, most elegant, or striking applications of the linearity of expectation you've encountered?

答:
Here is an application from theoretical computer science: recall that a 3-SAT formula is a CNF (conjunctive normal form) formula where each clause has three literals: that is, $\phi = C_1\land C_2\land \dots \land C_k$, where each $C_k$ (a clause) is an $\lor$ of 3 (distinct) literals.

Theorem. For every 3-SAT formula formula over $n$ variables, there exists an assignment $x\in\{0,1\}^n$ satisfying at least $7/8$ of the clauses.

Proof. Consider a 3-SAT formula $\phi = C_1\land C_2\land \dots \land C_k$ (we can assume wlog that no clause is trivially true). Letting $I_j$ be the indicator r.v. of whether the $j$-th clause $C_j$ is satisfied, we have that the number $N$ of clauses satisfied by a uniformly random assignment is
$$
\mathbb{E}[N] = \mathbb{E}\left[\sum_{j=1}^k I_j\right]
= \sum_{j=1}^k \mathbb{E}[I_j]
= \sum_{j=1}^k \mathbb{P}\{C_j \text{ satisfied}\}
= \sum_{j=1}^k \frac{7}{8} = \frac{7}{8}k
$$
since each clause has 3 (distinct) literals, and therefore is satisfied with probability $1-1/2^3=7/8$.

Now, since $N$ has expectation $\frac{7}{8}k$, we have that $\mathbb{P}\{N\geq \frac{7}{8}k\}>0$; that is, there exists some realization of $x$ satisfying at least a $7/8$ fraction of the clauses.


问:
Probabilistic modelling
This question is about how to make sense of a probabilistic model I'm reading about:

You have three random variable $A$, $B$ and $C$ that are all real-valued. You want to model an input-output relationship with these random variables, by assuming the following relationships between them:

It is assumed that $A$ is the constant random variable that only takes the value $a\in \mathbb
{R}$, where $a$ is thought of as input.

Furthermore, $P(B=b | A=a)=\mathcal{N}(b;3\cdot e^a,1)$ and  $P(C=c | B=b)=\mathcal{N}(c;2\cdot b^2,1)$, where $\mathcal{N}(x;\mu,1)$ denotes the probability density of a normal distribution with mean $\mu$ and variance $1$.

My questions are:

Does an equation like $P(B=b | A=a)=\mathcal{N}(b;3\cdot e^a,1)$ even make sense?

If we takes samples, on the left we have a number between $0$ and $1$ but on the right we have a number that can be larger, as the pdf does not need to stay within $[0,1]$, right?

What is the output?

In the (not online available) text I'm reading, the above is all that is mentioned. I would assume that to arrive at the output, one needs to proceed in the following way:

Given $a$, draw a number $b$ from a normally distribution with mean $3\cdot e^a$ and unit variance and then similarly draw $c$, the output, from a normally distribution with mean with mean $2\cdot b^2$ and unit variance. But I'm not sure if that's correct; also, it is not clear to me, in case this interpretation is indeed correct, why conditional probabilities in the model formulation are needed.

答:
I think the question is almost correct except for a few notational issues

The first:

Note $B|A \sim \mathcal{N}(3e^a,1)$. Therefore $\mathbf{P}(B=b|A) = 0$ This follows from a basic definition of continous random variables. What would be true is the density function that is $f_{B|A}(b) = 1/\sqrt{2\pi}e^{\frac{-(b - 3e^a)^2}{2}}$

The second the output is ofcourse C
The output is a probabilistic output that is it's a random variable whose density function conditioned on the input is
$f_{C|A}(c) = \int_\mathbf{R}f_{C|B(x)}f_{B(x)|A}dx$
$ = \int_\mathbf{R}1/\sqrt{2\pi}e^{\frac{-(c - 2x^2)^2}{2}}1/\sqrt{2\pi}e^{\frac{-(x - 3e^a)^2}{2}}dx$
$= 1/2\pi \int_\mathbf{R}e^{\frac{-(c - 2x^2)^2 -(x - 3e^a)^2}{2}}dx$

EDIT Since you mentioned about the credibility, I really dont know what to say. Do we have to show our academic transcript or something?

Title: Probabilistic modelling


问:
Black Jack Game Probability
I am reviewing for my upcoming test but I have difficulty in solving one problem because the way the define the variable X causing my confusion.I really need your help to build up a good solution for this problem.

Jack plays a game which assigns a number that is uniformly distributed
between 0 and 5. If the game assigns a number less than or equal to
k then he loses 1 dollar. On the other hand, if the game assigns a number larger than k, then he will gain 1 dollar.

(a) Find the expected profit of the game.

(b) What is the k value, which minimizes the variance of the
profit obtained in this game?

(c) If you were to play this game 10 times, what is the probability
that you gain 2 dollars?

Actually, I have thought that I will put X as the number of numbers which is less than or equal to k. Due to uniform distribution, I think I can build up a Probability distribution Function (PDF) can be expressed

```<code>
        1/6, if X in [0, k]

F(X) =  5/6, if X in (k, 5]

          0, otherwise.
</code>```

For this case, I can put Y according to X which shows the profit of Jack playing this game. However, I think the way I solve is wrong due to the expression of X.

答:
Denote profit by $\pi$. Then

(a) $E[\pi] = (-1)\cdot P(\pi=-1)+(+1)\cdot P(\pi=1) = (-1)\cdot \frac{k}{5}+(+1)\cdot \frac{5-k}{5} = \frac{5-2k}{5}$

(b)$Var[\pi] = E[\pi^2] - \left(\frac{5-2k}{5}\right)^2 = \frac{k}{5} + \frac{5-k}{5} - \frac{(5-2k)^2}{25} = 1 - \frac{(5-2k)^2}{25} \to \min$

Solution is $k^* = 0$ or $k^* = 5$ (both yield $Var[\pi] = 0$)

(c) It is possible only if we have exactly 6 successes. Hence $$P(6 \text{ successes } ) = \binom{10}{6}\left(1-\frac{k}{5}\right)^6\left(\frac{k}{5}\right)^4$$

Title: Black Jack Game Probability


问:
How to find an unknown angle in a triangle with many sections?
I’m not doing so well with trigonometry! I once again would really appreciate some help with this question. I have been given a number of different angles in this triangle but I can’t figure out how to solve for the specific angle.

答: $\alpha = \arctan{\frac{\overline{DF}}{\overline{FE}}} = \arctan{\frac{\overline{AF}\tan{10°}}{\overline{FB}\tan{40°}}} = \arctan{\frac{\tan{30°}\tan{10°}}{\tan{40°}}} = 6.92°$ (since $\angle{AFB}=90°$)


问:
Is there any way to change $\int\limits_0^c {f\left( x \right){e^{ - ax}}dx} \to \int\limits_0^\infty {f\left( u \right){e^{ - au}}du}$?
Is there any way to make a change of variable such that ?

$\int\limits_0^c {f\left( x \right){e^{ - ax}}dx}  \to \int\limits_0^\infty  {f\left( u \right){e^{ - au}}du}$

Thank you very much !

答:
$$\int_0^{\infty} f(u) e^{-au} du = \int_0^{c} f(u) e^{-au} du \ + \int_c^{\infty} f(u)e^{-au} du$$

What you want is only possible, hence, when $$\int_c^{\infty} f(u) e^{-au} du =0$$


问:
Conjugation action on a semi-direct product $N\rtimes H$

Let $G$ be a  finite group, with $H\le G$ and $N\trianglelefteq G$ such that $G=N\rtimes H$.

Suppose that the conjugation action of $H$ on $N$ induces $2$ orbits in $N$. Prove the following requests:

a) $\exists p$ prime  $\forall 1\ne n\in N$ such that $o(n)=p.$

b) SHow that $N$ is abelian.

I said that if $G=N\rtimes H$, hence we have the conditions $(*)\begin{cases}N \cap H=\{1_G\}\\NH=G \end{cases}$. For a generic element, $n \in N$ the stabilizer of $n$ in the action $\varphi:H\to\operatorname{Aut}(N)$ is such that $h^{-1}nh=n\iff h\in C_H(n)$, whose order divides $H$.

Furthermore, if the number of orbits is two, the quotient $H/Ker(\varphi)\cong \varphi(H)\le\operatorname{Aut}(N)$ has order two, with $Ker(\varphi):=\underset{n\in N}\bigcap C_H(n)$. I'm confusing things because the map induced by the semi-direct product has the same form of the map induced by the conjugation action and I'm not sure about how using the Hypothesis $(*)$.

If $N$ is abelian it means that its centralizer is the entire $G$ (with $C_G(H)\le Z(G)$), and in particular, considering tha action restricted to $H$, $C_H(N)$ should corespond to $H$.

Thank for the help and for your patience.

答:
Under conjugation one orbit is just the identity. So all other elements of $N$ are conjugate in $G$ and therefore have the same order.

Let this order be a multiple of the prime $p$, then the $p$th power of an element of $N$ is still in $N$ but has a smaller order i.e. it has to be the identity.

Since $N$ is a $p$-group, $Z(N)$ is non-trivial and is a characteristic subgroup of $N$. Therefore $Z(N)$ is fixed by $H$ and so all elements of $N$ are in $Z(N)$.


问:
Some questions on invariant factor decomposition of modules
Let $M$ be the $\mathbb{Z}$-module $F/N$ where $F=\mathbb{Z}^2$ and $N=\langle(6,4),(4,8),(4,0)\rangle\le\mathbb{Z}^2$. Consider the homomorphism $\varphi:\mathbb{Z}^3\to\mathbb{Z}^2$ such that the matrix of the homomorphism w.r.t. standard bases is $A=\begin{bmatrix}6 & 4 & 4\\4 & 8 & 0\end{bmatrix}$. Then there is an equivalent matrix $D$ such that $XAY=D$ where $X=\begin{bmatrix}1 & 0\\2 & 1\end{bmatrix}$, $D=\begin{bmatrix}2 & 0 & 0\\0 & 8 & 0 \end{bmatrix}$, $Y=\begin{bmatrix}1 & -2 & 2\\-1 & 2 & -1\\0 & 1 & -2\end{bmatrix}$. Since $D$ represents $\varphi$ and $N=\operatorname{im}(\varphi)$, we conclude that $M=\mathbb{Z}^2/\operatorname{im}(\varphi)\cong\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$.

I am trying to understand why $N=\operatorname{im}(\varphi)$ and $\mathbb{Z}^2/\operatorname{im}(\varphi)\cong\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$.

As stated in the first line of question, how can we conclude immediately that $N=\operatorname{im}(\varphi)$?

In the final step, how can we conclude from "$D$ represents $\varphi$ and $N=\operatorname{im}(\varphi)$" to "$M=\mathbb{Z}^2/\operatorname{im}(\varphi)\cong\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$" ? How exactly do we get 2 and 8? What I know is that 2 and 8 are invariant factors of $M$, but not entirely sure how do we derive that.

Thanks!

答:
$\operatorname{im}(\varphi)=\{\varphi(x_1,x_2,x_3):(x_1,x_2,x_3)\in\mathbb Z^3\}=\{x_1\varphi(1,0,0)+x_2\varphi(0,1,0)+x_3\varphi(0,0,1):$ $(x_1,x_2,x_3)\in\mathbb Z^3\}=\{x_1(6,4)+x_2(4,8)+x_3(4,0):(x_1,x_2,x_3)\in\mathbb Z^3\}$ $=$ $\langle(6,4),(4,8),(4,0)\rangle=N$.

The form of the canonical diagonal matrix $D$ shows that there exists a $\mathbb Z$-basis $\{f_1,f_2\}$ of $\mathbb Z^2$ such that $\{2f_1,8f_2\}$ is a $\mathbb Z$-basis of $N$. Then $$\mathbb Z^2/N=f_1\mathbb Z\oplus f_2\mathbb Z/2f_1 \mathbb Z \oplus 8f_2\mathbb Z\simeq f_1\mathbb Z/2f_1\mathbb Z\oplus f_2\mathbb Z/8f_2\mathbb Z\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}.$$


问:
Improper even integral
I have this improper integral:

$$I = \int_{-\infty}^{\infty} x^4 e^{\frac{-x^2}{2t}} \ dx$$

I used a substitution to get ($u=\frac{x^2}{2t}$):

$$I = \int_{-\infty}^{\infty} u^{\frac{3}{2}} e^{-u} \ du$$

(note I have taken the t's and the 2's to the front and left them out for now as they are constants)

However I am having trouble solving it. Ideally I was thinking of using the gamma integral however the limits are not correct. I need them to be between $0$ and $\infty$, I was thinking of doing this by trying to make the function even so I can change the limits. I think this should be right as this would yield $2*\Gamma(\frac{5}{2})$ which is the answer I wanted. However the function above is not even to my knowledge so I am stuck!

Could anyone help me out? I don't need an answer just leave a comment please!

答:
The function is even, hence you can write it as

$$2\int_0^{+\infty} x^4 e^{-x^2/2t}\ \text{d}t$$

Using your substitution (please be careful of what you get), you can deal it via Gamma function, obtaining the result
$$3 \sqrt{2 \pi } t^{5/2}$$

Details on the substitution

$$z = \dfrac{x^2}{2t} ~~~~~~~ x = \sqrt{2tz} ~~~~~~~ \text{d}x = \dfrac{t^{1/2}}{\sqrt{2z}}\ \text{d}z$$

Hence the integral becomes

$$2\int_0^{+\infty} 4 t^2z^2 e^{-z}\left(\dfrac{t^{1/2}}{\sqrt{2z}}\right)\ \text{d}z$$

$$\dfrac{8}{\sqrt{2}}t^{5/2}\int_0^{+\infty} z^{3/2}e^{-z}\ \text{d}z$$

Whence the above result.

Title: Improper even integral


问:
why does integration allow multiple solutions?
When integrating $\frac 1x$ or $\frac{n}{nx}$ where n is a constant both produce $\ln(x)$ and $\ln(nx)$.
Does this mean $\ln(x)$ is equal to $\ln(nx)$?
Am I making a mistake somewhere?

答:
$$\ln nx = \ln x +\ln n $$

The solutions only differ by a constant, which is allowed. ($\ln n$ is a constant as $n$ is a constant)


答:
$$
I = \int_{-\infty}^\infty u^{3/2}e^{-u}du
\\
\implies I = \int_0^\infty u^{3/2}e^{-u}du + \int_{-\infty}^0 u^{3/2}e^{-u}du
$$
Changing the sign of the limits on the second term and swapping them has no effect, as both introduce one minus sign, so we can write
$$
I = 2\int_0^\infty u^{3/2}e^{-u}du = 2\Gamma(5/2).
$$

Title: Improper even integral

Body:
I have this improper integral:

$$I = \int_{-\infty}^{\infty} x^4 e^{\frac{-x^2}{2t}} \ dx$$

I used a substitution to get ($u=\frac{x^2}{2t}$):

$$I = \int_{-\infty}^{\infty} u^{\frac{3}{2}} e^{-u} \ du$$

(note I have taken the t's and the 2's to the front and left them out for now as they are constants)

However I am having trouble solving it. Ideally I was thinking of using the gamma integral however the limits are not correct. I need them to be between $0$ and $\infty$, I was thinking of doing this by trying to make the function even so I can change the limits. I think this should be right as this would yield $2*\Gamma(\frac{5}{2})$ which is the answer I wanted. However the function above is not even to my knowledge so I am stuck!

Could anyone help me out? I don't need an answer just leave a comment please!


问:
Proof of greatest element of a non-empty set
I am working from Elliott Mendelson's "Number Systems and the Foundations of Analysis," which I am thoroughly enjoying thus far.

The proof I am struggling with comes at the end of a section on order relations in the chapter on natural numbers:

By a greatest element of a set $B \subseteq P$ we mean an object $z$ such that $z\in B$ and $(\forall u)(u\in B \implies u \leq z)$.

If $\emptyset \neq A \subseteq P$ and $A$ is bounded above (that is, $(\exists w)(\forall u)(u \in A \implies u \leq w)$), then $A$ has a greatest element. (Hint: Let $B = \{w: (\forall u)(u\in A \implies u \leq w)\}$. By hypothesis, $B \neq \emptyset$. Apply the least number principle.)

The hint basically prescribes how to satisfy the requirement of a greatest element possessing the property that there exists some $z$ such that $(\forall u)(u\in A \implies u \leq z)$. In particular, define $B$ as given in the hint; since $A$ is bounded above, $B$ is non-empty. Since $B$ is, in turn, non-empty, it possesses a least element by the least number principle. This least element of $B$ is clearly the "$z$" referred to above, i.e., the greatest element of $A$. I have been struggling for quite a while with demonstrating that this "$z$" is, indeed, contained in $A$. I have attempted the method of proof by contradiction considering the alternative case that $z \notin A$, but can't seem to figure a means to arrive at a contradiction. I've also considered induction proofs as a means of reaching a contradiction, but likewise can't reach any contradictions. Can anyone provide an extra tip or approach to explore, please?

答:
Here is a write-up which I think only uses concepts up to that point in the book:

Let $z$ be the smallest element of $B$.
Assume $\forall a \in A \; (a \not = z)$.

This means $a<z$.

Now notice that $z \not = 1$ since $A$ is not empty.

This means $z = S(w)$ for some $w$. (Theorem 1.1 in that chapter of the book).

But this says

$$a < S(w) = z$$

meaning

$$a \leq w < z$$

so $w \in B$ and $z$ is not minimal.


问:
Generalized Lagrange interpolation
Find a polynomial of degree smaller than five such that:

$P(1)=2$, $P(2)=-1$, $P'(2)=-1$,$P'(1)=3$,$P''(2)=1$

I know I need to use a combination of Lagrange Interpolation and Taylor series but the notes i have a very unclear on this and any help would be appreciated.

答:
If you don't know how to mix these two methods =, you can always solve it the hard way, by defining $P(X)=aX^4+bX^3+cX^2+dX+e $, and write the above conditions as a system and solve in terms of $(a,b,c,d,e)$, here is what it looks like in terms of matrices :

$$\underbrace{\begin{pmatrix}
1 & 1 &1&1&1\\16&8&4&2&1\\4&3&2&1&0\\32&12&4&1&0\\48&12&2&0&0
\end{pmatrix}}_{A} \cdot\begin{pmatrix}a\\b\\c\\d\\e \end{pmatrix}=\underbrace{\begin{pmatrix}2\\-1\\3\\-1\\1 \end{pmatrix}}_{B}$$

Then you can compute $A^{-1}$ and then compute $A^{-1}\times B $ :

$$A^{-1}=\begin{pmatrix}-3&3&-1&-2&0.5\\20&-20&7&13&-3\\-48&48&-18&-30&6.5\\48&-48&20&29&-6\\-16&17&-8&-10&2 \end{pmatrix} $$

Finally, you get :

$$\boxed{\begin{pmatrix}a\\b\\c\\d\\e \end{pmatrix}=\begin{pmatrix}-9.5\\65\\-161.5\\169\\-61 \end{pmatrix} }$$

Title: Generalized Lagrange interpolation


问:
Commuting idempotents repel each other
From a Banach algebra $A$ there are elements $e,f$ satisfying $e^2=e,f^2=f,ef=fe,e\ne f$, prove that $\|e-f\|\ge1$;

I can prove$\|e+f|\ge 1$by$\|e-f\|=\|e^2-f^2\|=\|(e-f)(e+f\|\le\|e-f\|\cdot\|e+f\|$, how to substitute$-
$for$+$?

答: Hint: Suppose $\|e-f\| <1$. Then $\sum_n (e-f)^{2n+1} $ is convergent. Compute the power using the fact that $e^{k}=e$ and $f^{k}=f$ for all $k \geq 1$. Can you finish?

Title: Commuting idempotents repel each other


答: Your substitution is invalid as it stands because $x^2/2t$ is not bijective on the integration domain. The integral is, however, twice the integral from $0$ to $\infty$, whereby the substitution and gamma definition can be applied and the result is $(2t)^{5/2}\cdot\frac{3\sqrt\pi}4$.

Title: Improper even integral


问:
I cant solve this problem... Im 17 and I dont know calculus. The topic that this problem was in is Polynomial zeros.
Write the polynomial $p$. $p$ is a 3rd degree polynomial. $p(x)$ is divisible by $(x-0,5)$. $p(x)$ is divisible by $(x+0,5)$. $p(x)$ is divisible by $(x+2)$. When we divide $p$ by $(x-2)$ we get a remainder $-3$.

I know that if polynomial p is divisible polynomial q we can write $p(x) = (x)\cdot q(x) + r(x)$ where $k(x)$ is a quotient and $r(x)$ is the remainder.
I don't know how to solve the problem however.
Any help would be appreciated.
Thank you.

答:
If the polynomial is divisible by $(x-\frac 12)$, $(x+\frac 12)$, $(x-2)$ and has degree 3, you have that
$$
p(x) = c(x-\frac 12)(x+\frac 12)(x+2)
$$

The constant $c$ can be computed from the last condition...


答:
According to your information, we know that :
$$P(X)=a(X-0.5)(X+0.5)(X+2) $$

Then you can use your last information to get $a$ :

$$P(X)=Q(X)(X-2)-3 $$

So you know that $P(2)=-3 $

$$a\times 1.5\times 2.5\times 4 =-3$$

So you get :
$$\boxed{a=-0.2} $$

Title: I cant solve this problem... Im 17 and I dont know calculus. The topic that this problem was in is Polynomial zeros.

Body:
Write the polynomial $p$. $p$ is a 3rd degree polynomial. $p(x)$ is divisible by $(x-0,5)$. $p(x)$ is divisible by $(x+0,5)$. $p(x)$ is divisible by $(x+2)$. When we divide $p$ by $(x-2)$ we get a remainder $-3$.

I know that if polynomial p is divisible polynomial q we can write $p(x) = (x)\cdot q(x) + r(x)$ where $k(x)$ is a quotient and $r(x)$ is the remainder.
I don't know how to solve the problem however.
Any help would be appreciated.
Thank you.


问:
A peculiar fact about studying the rate of change of physical quantities
This question arose from me trying to understand the link between variations of physical quantities with respect to different, but related, variables. However, the question is more about understanding the meaning of the math behind it.

Say we have a function $$R(\rho)=k\rho$$ (physically $R$ could be the electrical resistance of a material, $\rho$ its electrical resistivity, and $k$ a proportionality constant.) Then, we have $\frac{dR}{d\rho} = k$; the rate-of-change of $R$ (wrt $\rho$) is constant and independent of $\rho$.

Now, say I write $R$ as a function of the variable $\sigma=\frac{1}{\rho}$ (physically this would be the electrical conductivity of the material), i.e.,
$$R(\sigma) = \frac{k}{\sigma}$$
Then we have $\frac{dR}{d\sigma} = \frac{-k}{\sigma^2}$; the rate-of-change of $R$ (wrt $\sigma$) is not only not constant but also dependents (nonlinearly) on $\sigma$.

This Boggles my mind!
Even though the math is quite simple, clear, understandable, and all you want... I fail to see how can this shift in perspective in what variable we study the function under (this seemingly naive change of variable; from resistivity to conductivity, which from a physics point-of-view say the same thing about what's happening in the material...), leads to this much difference between the description of variation of the function itself.

Any insights on that are welcome.

答:
It's not mind boggling.

The resistance $R$ is linearly dependent on the resistivity $\rho$, whilst it is inversely proportional to the conductivity.

$\rho$ and $\sigma$ are two very different physical quantities with a different physical meaning. It's rather obvious that the rate of change of $R$ with respect to $\rho$ behaves in a very different way than the rate of change of $R$ wrt $\sigma$.

It's like to have

$$f(x) = kx$$ and

$$f(x) = \dfrac{k}{x}$$

The first function is a straight line; the second one is an equilateral hyperbola. The rate of change of $f(x)$ wrt x clearly differs.


问:
Convergence of semi-alternating harmonic series
I'm having trouble showing whether or not the following series converges:
$$\frac{1}{1}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}...$$
I've tried using several methods, such as creating two leibniz series from the terms, or using comparison tests, but failed to prove anything.  Does this series converge and why? Thank you.

答:
Hint

Ignore all but the 1st, 4th, 7th, 10th, ... terms, with the idea that the remainder of the terms (i.e. the ones being ignored) have a positive sum.  Then, compare the terms not being ignored to (1/3) of the harmonic series, which is known to be divergent.


问:
How to calculate large exponents by hand?
How to calculate large exponents by hand like they did in ancient times?
Is it something to do with Prosthaphaeresis? for example calculate $2^{15}$.

答:
Use logarithms perhaps?

$$\log_{10} 2^{15} = 15\log_{10} 2 \sim 15\cdot0.3 =4.5$$
So that:
$$2^{15} \sim 10^{4.5} = 10^4\sqrt{10} \sim3\cdot 10^4$$
Which is just 7% percent away, and can be done in your head as long as you remember that $\log_{10} 2 \sim 0.3$, which most engineers probably do.


答:
I don’t see a contradiction. All $$\frac{dR}{d\sigma} =-\frac{k}{\sigma^2} $$ is saying is that as $\frac{1}{\rho} $ changes by a small amount, the amount which $R$ changes by is dependent on the value of $\rho$.

Saying $$\frac{dR}{d\rho} = k$$ is entirely different, as in this case we look at small changes to $\rho$ instead, and this tells us that $R$ will change by the same amount, no matter the value of $\rho$.

Title: A peculiar fact about studying the rate of change of physical quantities

Body:
This question arose from me trying to understand the link between variations of physical quantities with respect to different, but related, variables. However, the question is more about understanding the meaning of the math behind it.

Say we have a function $$R(\rho)=k\rho$$ (physically $R$ could be the electrical resistance of a material, $\rho$ its electrical resistivity, and $k$ a proportionality constant.) Then, we have $\frac{dR}{d\rho} = k$; the rate-of-change of $R$ (wrt $\rho$) is constant and independent of $\rho$.

Now, say I write $R$ as a function of the variable $\sigma=\frac{1}{\rho}$ (physically this would be the electrical conductivity of the material), i.e.,
$$R(\sigma) = \frac{k}{\sigma}$$
Then we have $\frac{dR}{d\sigma} = \frac{-k}{\sigma^2}$; the rate-of-change of $R$ (wrt $\sigma$) is not only not constant but also dependents (nonlinearly) on $\sigma$.

This Boggles my mind!
Even though the math is quite simple, clear, understandable, and all you want... I fail to see how can this shift in perspective in what variable we study the function under (this seemingly naive change of variable; from resistivity to conductivity, which from a physics point-of-view say the same thing about what's happening in the material...), leads to this much difference between the description of variation of the function itself.

Any insights on that are welcome.


问:
Seeing a distribution is integrable
I am trying to do the following exercise :

Let $\alpha$ and $\beta$ be $1-$forms on a compact manifold $M^n$ such that $(\alpha\wedge \beta)_p \neq 0, \forall p\in M$ and such that $\alpha\wedge \beta$ is closed. Then the distribution generated by $\alpha$ and $\beta$ is integrable.

Well I think the idea would be use to use the frobenius theorem and try and show that we have a differential ideal. We know that $d(\alpha \wedge \beta)=0=d\alpha\wedge \beta-\alpha\wedge d\beta$ and so $d\alpha\wedge \beta=\alpha\wedge d\beta$. I have no clue where we can use the fact that $(\alpha\wedge \beta)_p \neq 0$ is relevant to check that $d\alpha =\alpha \wedge \eta$ or $\beta\wedge \eta$ for some $\eta$ and analogously for $d\beta$.

Any step towards the right direction is appreciated. Thanks in advance.

答:
The requirement that $\alpha\wedge\beta$ be nonvanishing is there to ensure that the kernel $\mathcal{D}=\{v\in TM:\alpha(v)=\beta(v)=0\}$ is indeed a distribution; otherwise it may have nonconstant rank.

Let $I$ be the ideal generated by $\alpha$ and $\beta$ (or equivalently the set of forms which vanish on $\mathcal{D}$). Another characterization more useful to this problem is that $\omega\in I$ iff $\alpha\wedge\beta\wedge\omega=0$. Can you show this?

(Hint: Consider a local coframe containing $\alpha$ and $\beta$.)

Title: Seeing a distribution is integrable


答:
$$I=\int x^4 e^{-x^2/2t}\ \text{d}t=\\Integrate -by- parts{\int}\mathtt{f}\mathtt{g}' = \mathtt{f}\mathtt{g} - {\int}\mathtt{f}'\mathtt{g}\\=-tx^3\mathrm{e}^{-\frac{x^2}{2t}}-{\displaystyle\int}-3tx^2\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x$$Now solving ${\displaystyle\int}-3tx^2\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x==-\class{steps-node}{\cssId{steps-node-1}{3t}}{\cdot}{\displaystyle\int}x^2\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x$
$$=-tx\mathrm{e}^{-\frac{x^2}{2t}}-{\displaystyle\int}-t\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x$$
Now solving ${\displaystyle\int}-t\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x$

Substitute
$u=\dfrac{x}{\sqrt{2}\sqrt{t}}$

$${\displaystyle\int}-t\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x=-\class{steps-node}{\cssId{steps-node-2}{\dfrac{\sqrt{{\pi}}t^\frac{3}{2}}{\sqrt{2}}}}{\displaystyle\int}\dfrac{2\mathrm{e}^{-u^2}}{\sqrt{{\pi}}}\,\mathrm{d}u$$ note that ${\displaystyle\int}\dfrac{2\mathrm{e}^{-u^2}}{\sqrt{{\pi}}}\,\mathrm{d}u$ This is a special integral ,Gauss error function which called $=erf(u)$so $$-\class{steps-node}{\cssId{steps-node-3}{\dfrac{\sqrt{{\pi}}t^\frac{3}{2}}{\sqrt{2}}}}{\displaystyle\int}\dfrac{2\mathrm{e}^{-u^2}}{\sqrt{{\pi}}}\,\mathrm{d}u==-\dfrac{\sqrt{{\pi}}t^\frac{3}{2}\operatorname{erf}\left(\frac{x}{\sqrt{2}\sqrt{t}}\right)}{\sqrt{2}}$$so plug in the second step
$$-tx^3\mathrm{e}^{-\frac{x^2}{2t}}-{\displaystyle\int}-3tx^2\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x=\\=\dfrac{3\sqrt{{\pi}}t^\frac{5}{2}\operatorname{erf}\left(\frac{x}{\sqrt{2}\sqrt{t}}\right)}{\sqrt{2}}-tx^3\mathrm{e}^{-\frac{x^2}{2t}}-3t^2x\mathrm{e}^{-\frac{x^2}{2t}}$$ an ,finally
$${\displaystyle\int}x^4\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x=\\=\dfrac{3\sqrt{{\pi}}t^\frac{5}{2}\operatorname{erf}\left(\frac{x}{\sqrt{2}\sqrt{t}}\right)}{\sqrt{2}}-tx^3\mathrm{e}^{-\frac{x^2}{2t}}-3t^2x\mathrm{e}^{-\frac{x^2}{2t}}+C$$ now apply limit bounds to find numerical value
$$I=I = \int_{-\infty}^{\infty} x^4 e^{\frac{-x^2}{2t}} \ dx=\dfrac{3\sqrt{{\pi}}t^\frac{5}{2}}{\sqrt{2}}$$

Title: Improper even integral


答:
The isolating of the first, fourth etc terms is the first thing I thought of, as in another proposed answer.

Another way is to group the terms other than the first in threes and study $$\frac 1{3n-1}-\frac 1{3n}+\frac 1{3n+1}$$ which (from the curvature of the reciprocal function I expect to be greater than $\dfrac 1{3n}$ which will do for comparison - indeed even if close to $3n$ a comparison is likely to be possible).

Title: Convergence of semi-alternating harmonic series

Body:
I'm having trouble showing whether or not the following series converges:
$$\frac{1}{1}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}...$$
I've tried using several methods, such as creating two leibniz series from the terms, or using comparison tests, but failed to prove anything.  Does this series converge and why? Thank you.


问:
My LHS does not match the LHS in Apostol's Introduction to Analytic Number Theory while applying Abel's identity
The book "An Introduction to Number Theory" by Tom A. Apostol states in its Theorem 4.2, the Abel's identity:

For any arithmetical function $ a(n) $ let $ A(x) = \sum_{n \le x} a(n) $, where $ A(x) = 0 $ if $ x < 1 $. Assume $ f $ has a continuous derivative on the internal $[y, x]$, where $0 < y < x$. Then we have
$$
\sum_{y < n \le x} a(n) f(n) = A(x) f(x) - A(y) f(y) - \int_y^x A(t) f'(t)\,dt.
$$

But then in Theorem 4.12, it applies Abel's identity in the following step like this:

Let $$ A(x) = \sum_{p \le x} \frac{\log p}{p} $$
and let $$ a(n) = \begin{cases} 1 & \text{ if } n \text{ is prime},\\ 0 & \text{ otherwise.}\end{cases}$$
Then
$$
\sum_{p \le x} \frac{1}{p} = \sum_{n \le x} \frac{a(n)}{n} \; \text{ and } \;
A(x) = \sum_{n \le x} \frac{a(n)}{n} \log n.
$$
Therefore if we take $ f(t) = 1/\log t$ in Theorem 4.2 we find, since $ A(t) = 0 $ for $ t < 2 $,

$$ \sum_{p \le x} \frac{1}{p} = \frac{A(x)}{\log x} + \int_2^x \frac{A(t)}{t \log^2 t}\, dt $$

The whole theorem 4.12 is not important for my question. Only the above step is important for my question. How did the LHS become $ \sum_{p \le x} \frac{1}{p} $?

When I try to apply Abel's identity to $ a(n) $ defined above and $ f(t) = 1/\log t $, I get this as the LHS:

$$
LHS = \sum_{y < n \le x} a(n) f(n) = \sum_{0 < n \le x} \frac{a(n)}{\log n} = \sum_{p \le x} \frac{1}{\log p} \ne \sum_{p \le x} \frac{1}{p}.
$$

The RHS matches fine when I apply Abel's identity:

\begin{align*}
RHS & = A(x) f(x) - A(y) f(y) - \int_y^x A(t) f'(t)\,dt \\
& = \frac{A(x)}{\log x} - 0 - \int_0^x A(t) \frac{-1}{\log^2 t} \left( \frac{1}{t} \right)\,dt \\
& = \frac{A(x)}{\log x} + \int_2^x \frac{A(t)}{t \log^2 t}\,dt.
\end{align*}

What mistake did I make that my LHS does not match the book's LHS?

答:
I was finally able to straighten it out. While the book says, "take $ f(t) = 1/\log t$" while applying Abel's identity, it says nothing about what $ a(n) $ should be and that was the missing clue.

Due to the missing clue, I erroneously assumed $$ a(n) = \begin{cases} 1 & \text{ if } n \text{ is prime},\\ 0 & \text{ otherwise.}\end{cases} $$ but with this definition we are never going to get $ A(x) = \sum_{n \le x} a(n) = \sum_{n \le x} \frac{a(n)}{n} \log n $.

Here is what I did to straighten this out.

In Abel's identity replace $ a(n) $ with $ \frac{a(n) \log n}{n} $. Then take $ f(t) = 1/\log t $ like the books suggest. With these two things, everything begins to match.

Now indeed $ A(x) = \sum_{n \le x} \frac{a(n) \log n}n$.

LHS of Abel's identity now becomes:

$$
LHS = \sum_{n \le x} \frac{a(n) \log n}{n} \cdot \frac{1}{\log n} = \sum_{n \le x} \frac{a(n)}{n} = \sum_{p \le x}\frac{1}{p}.
$$


问:
If $x$ is an ordinal, then so is $x\,\bigcup\,\{x\}$
I wish to show that $x$ is an ordinal, then so is $x\,\bigcup\,\{x\}$ (with $y<x$ for all $y\in x$).

I have seen a proof in Cameron's Sets, Logic and Categories (p.41), but I find it unsatisfactory. He writes:

The set $a = x\,\bigcup\,\{x\}$ (with order as specified in the the statement of the theorem) has as sections all the sections of $x$ and one additional one, namely $a_x$. But since all the elements of $x$ are smaller than $x$, we have $a_x = x$.

I don't see why it necessarily follows from this that $a_x = x$. By definition, $a_x = \{y\in a : y\subset x\}$, so $a_x\subseteq x$, but this is different than $a_x = x$.

答:
In my original post, I wrote $a_x=\{y\in a:y\subset x\}$, but the definition Cameron actually gives is $a_x=\{y\in a:y < x\}$. Does this make a difference?

Either way, it seems that:

$$\begin{align}
a_x &= \big\{y\in a:y<x\big\} \\
&= \big\{y\in x\cup\{x\}:y<x\big\} \\
&= \big\{y\in x:y<x\big\}\cup\big\{x\in\{x\}:y<x\big\} \\
&= x\,\cup\,\varnothing \\
&= x
\end{align}$$

where $\big\{x\in\{x\}:y<x\big\} = \varnothing$ because $x\nless x.$


问:
Proving $\tan(90^\circ- x) = \cot x$
How do you prove $\tan(90^\circ-x)=\cot x$ using the compound angle formulae?

答:
$$\tan(90^\circ-x)=\frac{\tan 90^\circ-\tan x}{1+\tan 90^\circ \tan x}=\lim_{z\to \infty}\frac{z-\tan x}{1+z \tan x}=\lim_{z \to \infty} \frac{1-\tan x/z}{1/z+\tan x}=\cot x.$$
Note that $\tan 90^\circ=\infty.$

Alternate way $$\tan(90^\circ-x)=\frac{\sin(90^\circ-x)}{\cos (90^\circ-x)}=\frac{\cos x}{\sin x}=\cot x$$

Title: Proving $\tan(90^\circ- x) = \cot x$


问:
Find the $3 \times 3$ matrix
Question is
$$
\text{adj} B =
\left[
\begin{array}{ccc}
         -45& 33& -28 \\
         32& -10& -16\\
        -4& -24& 2\\
\end{array}
\right]
$$

If $\det B = −202$, then find the $3\times 3$ matrix $B$, whose adjoint is given as above.

First, I wrote the $3x3$ matrix as $$\begin{bmatrix}a & b & c\\
                                 d & e & f\\
                                 g & h & i\end{bmatrix}$$. Then I found the cofactor matrix and typed its transpose. As a result,
$$
\begin{cases}
-fh&+ei&=-45 \\
 -id&+fg&=32 \\
 dh&-eg&=-4 \\
 -ib&+ch&=33 \\
 ia&-cg&=-10 \\
 -ah&+bg&=-24 \\
 bf&-ec&=-28 \\
 -af&+cd&=-16 \\
 ea&-bd&=2
\end{cases}
$$

came from the transactions I made. Then I applied the determinant formulas and replaced what I had just found. $$\begin{cases}-45a+32b-4c=-202 \\
                 33d-10e-24f=-202\\
                -28g-16h+2i=-202\end{cases}$$   But I got stuck here. How can I find $a, b, c, d, e, f, g, h, i$ here? I will be grateful if you could help me.

答:
You don't have to solve it backward. See this solution, please.
Given

$ \mathbf {detB} = −202$, and

$
\mathbf{adj} B =
\left[
\begin{array}{ccc}
         -45& 33& -28 \\
         32& -10& -16\\
        -4& -24& 2\\
\end{array}
\right]
$,

then, $ \mathbf{B^{-1}} = \frac{1}{detB} . adjB$.

This implies that,
$\mathbf {B^{-1}} = -\frac{1}{202}
\left[
\begin{array}{ccc}
         -45& 33& -28 \\
         32& -10& -16\\
        -4& -24& 2\\
\end{array}
\right]
$.

Now take $\mathbf{B} =
\left[
\begin{array}{ccc}
         a& b& c\\
         d& e& f\\
         g& h& i\\
\end{array}
\right]
$.

Using invertibility relation, $\mathbf{BB^{-1}} = \mathbf{B^{-1}B} = \mathbf{I}$, i.e.,
$$
-\frac{1}{202}
\left[
\begin{array}{ccc}
         -45& 33& -28 \\
         32& -10& -16\\
        -4& -24& 2\\
\end{array}
\right]
\left[
\begin{array}{ccc}
         a& b& c\\
         d& e& f\\
         g& h& i\\
\end{array}
\right]
=
\left[
\begin{array}{ccc}
         1& 0& 0\\
         0& 1& 0\\
         0& 0& 1\\
\end{array}
\right]
$$.

Using simple matrix multiplication, you'll get the following system of equations by using $B^{-1}$ against the first, second, and third columns of the matrix $B$ in that order.

$$
 (\mathbf{1st}) \rightarrow
-\frac{1}{202}
\left[
\begin{array}{ccc}
         -45& 33& -28 \\
         32& -10& -16\\
        -4& -24& 2\\
\end{array}
\right]
\left[
\begin{array}{ccc}
         a\\
         d\\
         g\\
\end{array}
\right]
=
\left[
\begin{array}{c}
         1\\
         0\\
         0\\
\end{array}
\right]
$$

$$
 (\mathbf{2nd}) \rightarrow
-\frac{1}{202}
\left[
\begin{array}{ccc}
         -45& 33& -28 \\
         32& -10& -16\\
        -4& -24& 2\\
\end{array}
\right]
\left[
\begin{array}{ccc}
         b\\
         e\\
         h\\
\end{array}
\right]
=
\left[
\begin{array}{c}
         0\\
         1\\
         0\\
\end{array}
\right]
$$

$$
 (\mathbf{3rd}) \rightarrow
-\frac{1}{202}
\left[
\begin{array}{ccc}
         -45& 33& -28 \\
         32& -10& -16\\
        -4& -24& 2\\
\end{array}
\right]
\left[
\begin{array}{ccc}
         c\\
         f\\
         i\\
\end{array}
\right]
=
\left[
\begin{array}{c}
         0\\
         0\\
         1\\
\end{array}
\right]
$$

Now, you can multiply both sides, in each of the equations, by $-202$ to eliminate the $-\frac{1}{202}$. Then, solve the 3 equations to get the first, second, and third columns of the matrix $B$. If you solve it correctly, you'll get the matrix $B$ below
$$
\mathbf (1st) \rightarrow
\left[
\begin{array}{ccc}
         a\\
         d\\
         g\\
\end{array}
\right]
=
\left[
\begin{array}{c}
         2\\
         0\\
         4\\
\end{array}
\right]
$$
$$
\mathbf (2nd) \rightarrow
\left[
\begin{array}{ccc}
         b\\
         e\\
         h\\
\end{array}
\right]
=
\left[
\begin{array}{c}
         -3\\
         1\\
         6\\
\end{array}
\right]
$$
$$
\mathbf (3rd) \rightarrow
\left[
\begin{array}{ccc}
         c\\
         f\\
         i\\
\end{array}
\right]
=
\left[
\begin{array}{c}
         4\\
         8\\
         3\\
\end{array}
\right]
$$

Hence,
$$
\mathbf B =
\left[
\begin{array}{ccc}
         2& -3& 4 \\
         0& 1& 8\\
        4& 6& 3\\
\end{array}
\right]
$$

Title: Find the $3 \times 3$ matrix


问:
Showing that for any composite binary number $b > 1$, $bb$ is also composite.
Say $b$ is the binary form of a composite number $d > 1$, I want to prove that $bb$ (concatenation of $b$ with itself) is also a composite number.

My approach:

Given a binary number $b$, $bb$ is obtained by a series of transformations picked from $2b, 2b + 1$ (since appending a 0 is equivalent to $2b$ and appending a 1 is equivalent to $2b + 1$)

We can rule out all series of transformations whose last transformation is appending a $0$ because these are even numbers

Are there any prime numbers of the form $2n + 1, n > 1$? Turns out there is (for $n=3$ and probably more). So that seems like a dead-end.

答: If $n$ is the bit-length of $b$, $\overline{bb}=b(2^n+1)$. Since $d>1$, $n\ge2$ and $b>1$. Thus $\overline{bb}$ is composite.


问:
Idempotents in rings without unity

Suppose there are non-trivial idempotents in the ring without unity. Is it right that all of them are zero divisors?

If we're given unitary ring with unity $e$ and $a$ is non-trivial idempotent then $e-a \neq 0$. But $a(e-a) = 0$ so $a$ is zero divisor.

But i'm not sure about the case when ring doesn't have unity.

EDIT
In the non-commutative case i wonder whether each non-zero idempotent must be SOME kind of zero divisors, i.e. left or right, or perhaps both.

答:
A partial answer:

If the ring is commutative and finite, then the answer is yes. Take $e\in R$ an idempotent and define $f:R\to R$ by $f(a)=ea$. If $e$ is not a zerodivisor then $f$ is injective and therefore bijective. Now prove that $e$ is the unity of $R$: if $b\in R$ then there exists $a\in R$ such that $b=ea$ and we have $eb=e(ea)=ea=b$.

Title: Idempotents in rings without unity


问:
Can we identify the pole of the Gamma function with the limit of the harmonic numbers?
The expansion of the gamma function around $x=0$ is
$$\Gamma(x)=\frac{1}{x}-\gamma+O(x).$$
The Euler constant $\gamma$ is defined by $$\gamma=\lim_{n\to\infty}(H_n-\log n)$$ where $H_n$ is the $n$th harmonic number. On the other hand, $\gamma$ can also be defined as $$\gamma=\lim_{x\to 0}\left(\frac{1}{x}-\Gamma(x)\right).$$

However, $\lim_{n\to\infty} \log n$ and $\lim_{x\to 0} 1/x$ can be sort of identified in the following way:
$$\lim_{x\to 0^+} \frac{1}{x} = \lim_{x\to 0^+}\lim_{\delta\to 0}\int_\delta^1 t^{-1+x} dx$$
Now I will be sketchy and interchange the limits.
$$\lim_{x\to 0^+} \frac{1}{x} = \lim_{\delta\to 0}\lim_{x\to 0^+}\int_\delta^1 t^{-1+x} dt = \lim_{\delta\to 0}\int_\delta^1 t^{-1} dt
=-\lim_{\delta\to 0}\log\delta=\lim_{n\to\infty} \log n .$$
Then comparing to the two definitions of $\gamma$ above, it seems like one could again "identify" the pole of $\Gamma(x)$ with the limit of the harmonic numbers.

We can do a similar procedure to try to establish that.
$$\lim_{x\to 0^+} \Gamma(x)=\lim_{x\to 0^+} \lim_{\delta\to 0} \int_\delta^\infty t^{x-1} e^{-t} dt$$
Again being sketchy with the limits:
$$\lim_{x\to 0^+} \Gamma(x)=\lim_{\delta\to 0}  \int_\delta^\infty t^{-1} e^{-t} dt = \lim_{\delta\to 0}\Gamma(0,\delta)$$
Here $\Gamma(s,z)$ is the upper incomplete gamma function, which for $s=0$ has the following expansion around $z=0$.
$$\Gamma(0,z)=-\gamma-\log z+O(z)$$
Comparing this to the definition of $\gamma$ we obtain
$$\lim_{x\to 0^+} \Gamma(x) = -\lim_{n\to\infty}H_n.$$

My questions are: How legitimate is this? Is this a well-known statement? (I haven't seen it anywhere.) If it is legitimate, what is its significance?

答:
What is true is that the two equations
$$ H_n = \log(n) + \gamma + \frac1{2n} + \cdots \tag{1}$$
and
$$ \Gamma(1/n) = n - \gamma + \frac1n + \cdots \tag{2} $$
together imply that
$$ \lim_{n\to\infty} \Gamma(1/n) + H_n - n - \log(n) = 0.
\tag{3} $$


问:
Calculate rotation matrix that flips the frame up
Let's start with the standard basis frame: [1, 0, 0], [0, 1, 0], and [0, 0, 1].

Imagine this frame goes through some arbitrary rotation R, with one constraint, that we know that after the R is applied to the frame, the z-axis becomes [0, 0, -1].

I want to calculate a rotation that "flips" the [0, 0, -1] back to the [0, 0, 1] vector with the smallest angle possible. Note: Inverting the R is NOT correct, because we want the smallest angle, and inverting R would always completely erase the rotation to give us the identity matrix.

The top answer here: Calculate Rotation Matrix to align Vector A to Vector B in 3d? does not apply, because I am describing the edge case that Jur van den Berg says is not supported.

Geometrically, if we start with the orientation corresponding with the z-axis at [0, 0, -1], I think to "flip" the frame we need to swap the corresponding x and y axises, and simply multiply [0, 0, -1] by -1 to get [0, 0, 1]. You can understand what I'm saying if you use your hand to form a frame by right hand rule, then 'flip' your thumb to negate the z-axis, and then 'flip' it back. How do I calculate this 'flip' relative rotation for any arbitrary orientation?

答: To do the rotation that you need, choose any axis perpendicular to the z-axis, and rotate by $\pi$.  All of the rotations will get you there and cannot be distinguished by the "size" of the rotation as the angles are all the same.


问:
Two basic logic questions
I'm pretty sure that these two logical moves are okay, but I just wanted to be sure.  If there are any subtleties (e.g. they are okay in first-order logic but not in more complex systems), please let me know.

1) Can you universally instantiate a negated proposition?  I.e. can you move from:

$ \left ( \forall x \right )\left ( x \rightarrow Y \right ) $

to

$ \neg b \rightarrow Y $?

2) Can you use rules of replacement within the scope of an operator?  I.e. can you move from:

$ \left ( \forall x \right )\left ( x \vee Y \right ) $

to

$ \left ( \forall x \right ) \neg \left (\neg x \& \neg Y \right ) $?

I think these two moves are okay but just want to be sure.  Thanks!

答:
In the first case, it would be better to instantiate a variable or a constant. You could define: $a =\lnot b$, in which case you could use universal instantiation $a$ as a witness, to get $a \rightarrow Y$, and then use rule of replacement by identity to get $\lnot b \rightarrow Y$.

Yes you can, in the second case; anything within the scope of the quantifier should be legitimate. However, it would probably be best to universally instantiate, using a witness, say $a$, to first get $a \lor Y$, and then $\lnot(\lnot a \land \lnot Y)$, assuming your formal system includes rules of replacement, based on logical identities, e.g., by DeMorgan's. That allows you to use the original universally quantified statement more directly (instantiating, for example $b \lor Y$, if you should choose to do so for another witness.

Title: Two basic logic questions