问:
Probability Axioms: why must the Probability, $Pr$, of and event $E_i$, be between $0$ and $1$; that is, $0\leq Pr(E_i) \leq 1$?
I think the question (stated in the title) is self explanatory, but here is the complete question:

Consider the triple: $\langle S, \mathscr S, Pr \rangle$

Axiom 1: $Pr(S)=2$, $Pr(\emptyset)=0$

Axiom 2: $Pr(s)+Pr(s^c)=2$ for each $s\in \mathscr S$.

Basically, this is all I am asking: if we push the bound up (for the first axiom), and likewise tweak the second axiom will we have the same Probability Theory we have right now (perhaps with slight numeric differences, but consistent nonetheless)?

Further Explanation and Context:

Given a coin with even density, we know that the probability of landing $\{H\}$ or $\{T\}$ is $\frac{1}{2}$. Now, if we modify axiom 1, as suggested, $Pr\{H\}$ would be $1$, likewise for $\{T\}$; however, modifying the second axiom will, fortunately, help us reformulate conditional Probability $Pr\{H\}$ or $Pr\{T\}$ given we land either $\{H\}$ or $\{T\}$, and we get $Pr\{H\}=\frac{1}{2}$ and $Pr\{T\}=\frac12$ given we land either $\{H\}$ or $\{T\}$.

Gae. S. rightfully pointed out that for countable and uncountable sets the measure might stop being $\sigma-$finite.

This is the essence of my question: Most "physical" spaces are finite; will such a measure, axiomatized by what was stated, above be sufficient? My guess is that it will work out just fine, but I am not quite sure if I can prove this.

Note:
I found these two questions to be similar to what I am asking, but I don't think they were helpful to me.

Probability Axioms

Why must the probability of an event be between 0 and 1?

答:
Every statement in the usual probability theory of the space $(S, \mathscr{S}, \mathbb{P})$ can be changed into a statement in your alternative theory of the space $(S,\mathscr{S},Pr)$ by changing $\mathbb{P} \mapsto \frac{Pr}{2}$.  Similarly every statement in your alternative theory becomes a statement of the usual theory by $Pr \mapsto 2\mathbb{P}$.  In other words, they are equivalent and one is consistent if and only if the other is.

That said, changing $1$ to $2$, in my mind, just adds needless complication.  When dealing with products, for instance, the usual theory holds that $$\mathbb{P}_{S_1\times S_2}(A_1\times A_2) = (\mathbb{P}_{S_1}\otimes \mathbb{P}_{S_2})(A_1\times A_2) = \mathbb{P}_{S_1}(A_1)\mathbb{P}_{S_2}(A_2)$$ which extends how we compute area as length times width.  In your theory, the equivalent statement would be
$$Pr_{S_1\times S_2}(A_1\times A_2) = \frac{1}{2} Pr_{S_1}(A_1)Pr_{S_2}(A_2)$$ and it becomes worse with higher powers
$$Pr_{\prod_{i=1}^n S_i}\left(\prod_{i=1}^nA_i\right) = \frac{1}{2^{n-1}}\prod_{i=1}^nPr_{S_i}(A_i),$$ which directly complicates every statement involving independence, most likely with mixed powers of $2$.

No doubt there's a philosophical razor we could use here to reject your alternative theory as being more complicated with no worthwhile payoff.

To create an alternative theory that may result in an actual benefit, you would need to do something more drastic like taking as an axiom $\widetilde{Pr}(S) = x$ is a fixed variable and work in the ring $\mathbb{R}[x,x^{-1}]$ or some extension thereof.  This might force the theory to keep track of another form of "dimension", but I'm skeptical that the payoff, if it ever materializes, would be worth the massive headache caused.


问:
How to recover a closed convex set from its support function?
Let $A\subset\mathbb R^n$ closed and convex.
Its support function is defined as $h_A(u)=\sup_{x\in A}\langle x,u\rangle$, for every $u\in \mathbb R^n$.

Being $A$ closed, there must be, for each $u$, some $x_u\in A$ such that $h_A(u)=\langle x_u,u\rangle$.

For example, the support function of a unit circle is $h(n_\theta)=1$, with $h(n_\theta) = \langle x_\theta,n_\theta\rangle$ with $x_\theta=n_\theta$.

In $\mathbb R^2$, we can parametrise $u$ via its polar angle (sticking to vectors with unit norm). Denote such unit vectors with $n_\theta$.
Then we write $h_A(n_\theta) = \langle x_\theta,n_\theta\rangle$ for some $x_\theta\in A$ (there might be more than one choice of $x_\theta$, so let's say we use some criterion to fix one). We want to recover the points $x_\theta$ corresponding to each $\theta$, and thus $\partial A$.
Assuming $\theta\mapsto h_A(n_\theta)$ is differentiable around $\theta$, we have
$$\partial_\theta h(n_\theta) = \langle \partial_\theta x_\theta, n_\theta\rangle + \langle x_\theta,\hat\theta\rangle,$$
where $\partial_\theta n_\theta=\hat\theta$ is the unit versor orthogonal to $n_\theta$.

If $\langle \partial_\theta  x_\theta,n_\theta\rangle=0$, one can easily recover $x_\theta$, whose polar coordinates are $h(n_\theta)$ and $\partial_\theta h(n_\theta)$. This is the case whenever $\partial_\theta x_\theta=0$ (as happens for example for polygons), or when $\partial_\theta x_\theta$ is orthogonal to $n_\theta$ (this is the case e.g. for the case of the unit circle, in which $x_\theta=n_\theta$).

Does $\partial_\theta h_A(n_\theta)$ return the polar coordinate of $x_\theta$ also in the general case? In other words, does $\langle \partial_\theta x_\theta,n_\theta\rangle=0$ always hold?
And what about the case of higher dimensions? Does this technique generalise straightforwardly to $\mathbb R^n$?

I asked a closely related question here, and used the same technique discussed here in an answer. This question differs from that one in that it uses a different language/formalism, and focuses on the validity of a specific relation (whether $\partial_\theta x_\theta$ is orthogonal to $n_\theta$) to show that the derivatives of the support function return the coordinates of the corresponding point in its support hyperplane.

Another way to derive the convex region corresponding to a given support function is mentioned in this other answer, but the focus is there on using the convex conjugate (which as far as I understand is not as directly applicable in practice as the method discussed here).

答:
Assuming some smoothness, for example, that the convex body  has a strictly convex boundary and that every boundary point has a unique normal, then it follows that the support function is differentiable, and its gradient evaluated at a unit vector $u$ is equal to the unique boundary point of the convex body it supports, where $u$ is a normal vector. Therefore the gradient map provides an effective parametrization of the boundary via the unit Euclidean sphere , and you can use it to actually draw the boundary in two and three dimensions, if you have an explicit expression for the support function.

By a convex body I mean a convex, compact with non-empty interior subset.

ContentLicense: CC BY-SA 4.0


问:
If $m,n\in N$ Prove that there is such a positive integer k, such that $(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}$
If $m,n\in N$ Prove that there is such a positive integer k, such that $(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}$

I attempted to solve this question using binomial coefficients, saying:

$(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}=\sum\limits_{a=0}^n {n\choose a}*\sqrt{m}^{n-a}*\sqrt{m+1}^a$

and from here I was thinking that I had to do something with $\sqrt{m}^{n-a}$ with $\sqrt{k}$ and do something else with $\sqrt{k+1}$ and $\sqrt{m+1}^a$. Unfortunately I couldn't think of what to do with this circumstance.

I then thought of maybe taking $(\sqrt{k}+\sqrt{k+1})^2=(\sum\limits_{a=0}^n {n\choose a}*\sqrt{m}^{n-a}*\sqrt{m+1}^a)^2$, but this immediately over-complicated the question.

Could you please explain to me how to solve this question and how to solve similar questions in the future?

答:
Let $ x = \sqrt{m+1} + \sqrt{m}$ and $ y = \sqrt{m+1} - \sqrt{m}$.

Observe that (IE Fill in the gaps to prove these statements. If you're stuck, explain what you've tried.)

$x \geq 1$

$xy = 1$

Since $ f(a) = \sqrt{a+1} + \sqrt{a}$ is strictly increasing with image $[1, \infty)$, so there is a unique real (not necessarily integer) $k_n$ such that $x^n = \sqrt{ k_n+1} + \sqrt{k_n}$. It remains to show that $k_n$ is indeed an integer.

$\sqrt{ k_n+1} - \sqrt{k_n} = \frac{1}{ \sqrt{k_n+1} + \sqrt{k_n}}  = \frac{1}{x^n} = y^n .$

$\sqrt{k_n } = \frac{1}{2} (x^n - y^n) $

$k_n = \frac{1}{4} (x^{2n} + y^{2n} - 2 ) =  \frac{1}{2} \left[  -1 + \sum_{i=0}^n {2n \choose 2i}m^i (m+1)^{n-i}           \right] $.

Now, show that this is an integer, because the numerator is even.

 Consider seperate cases of $m$, $m+1 $ even.

 Notice that all but one of the terms in the summation are even.

Notes

Having read the official solution, I'm amazed this (natural to me) approach worked out so quickly but they didn't use it.

Defining $y$ is pretty natural given $x$, and it's a useful trick in such situations. (EG It's also in the official solution.)

Bonus: $ y^n = \sqrt{ k_n + 1 } -\sqrt{k_n}$

$k_1 = m, k_2 = 4m^2 + 4m, k_3 = 16m^3 + 24m^2 + 9m$

I'm slightly amazed that there is a closed form expression for $k_n$. I looked at initial $k_n$ and couldn't spot a generalizable pattern.


问:
Is there a possible geometric method to find length of this equilateral triangle?

Problem Given that $AD \parallel BC$, $|AB| = |AD|$, $\angle A=120^{\circ}$, $E$ is the midpoint of $AD$, point $F$ lies on $BD$, $\triangle EFC$ is a equilateral triangle and $|AB|=4$, find the length $|EF|$.

Attempt At first glance, I thought it could be solved using a geometric method. I considered the law of sines/cosines, similar triangles, Pythagorean theorem, even Menelaus' theorem, however, got properties which contributed nothing to calculate $|EF|$.

What I've got after draw a line perpendicular to $BC$ through $E$

$\triangle ABH$ and $\triangle AHD$ are both equilateral triangles of length 4.

$\triangle EFD \sim \triangle GEH$

$|EH|=2\sqrt{3}$

Algebraic method Eventually, I've changed my mind to embrace algebra. I found it is easy to coordinate $E,A,B,D$ and $C$ is related to $F$ (rotation) and $B$ (same horizontal line). Make $E$ as the origin, $AD$ points to $x$-axis, $HE$ points to $y$-axis, we got

$E = (0,0)$

$A = (-2,0)$

$B = (-4,-2\sqrt{3})$

$D = (2,0)$

Point $(x, y)$ in line $BD$ has $y=\frac{1}{\sqrt{3}}(x-2)$. Assume $F=(x_0,y_0)$, $C=(x_1,y_1)$, we can obtain $C$ by rotating $F$ around pivot $E$ $60^{\circ}$ counter-clockwise

$$
\begin{bmatrix}
x_1 \\ y_1
\end{bmatrix}
=
\begin{bmatrix}
\cos{\theta} & -\sin{\theta} \\
\sin{\theta} & \cos{\theta}
\end{bmatrix}
\begin{bmatrix}
x_0 \\ y_0
\end{bmatrix}
$$
, also we know that $BC$ is parallel to $x$-axis, then
$$
\begin{align*}
y_1
 & = \sin{60^{\circ}} x_0 + \cos{60^{\circ}} y_0 \\
 & = \sin{60^{\circ}} x_0 + \cos{60^{\circ}} \frac{1}{\sqrt{3}}(x_0-2) \\
 & = -2\sqrt{3}
\end{align*}
$$
, thus $F=(-\frac{5}{2}, -\frac{3\sqrt{3}}{2})$, and finally $|EF|=\sqrt{13}$

Thoughts afterwords I noticed that $F$ (through its coordinate) is actually the midpoint of $BK$. It may be a key point in geometric method, but I cannot prove it either.

Graph I made it in GeoGebra and it is shared. Please go and edit it to save your time if you have any idea.
Link: https://www.geogebra.org/graphing/yqhbzdem

答:
This can be solved in your imagination.  It takes a lot of words to describe it, but you don't need these words when you imagine it.

Imagine moving $F$ back and forth along $BD$, while holding $E$ fixed, so $C$ (defined as the third point of the equilateral triangle) moves around.  $C$ is always a 60° counter-clockwise rotation of $F$ (rotating around $E$), so the set of points visited by $C$ is a 60° counter-clockwise rotation of $BD$ (around $E$).  So $C$ moves vertically.  When $F$ is at $D$, then the equilateral triangle is small and $C$ is above the midpoint of $DE$.  So we see that $C$ is always on the perpendicular bisector of $DE$.

Now we return to the diagram as shown.  The distance between $AD$ and $BC$ is $\sqrt{4^2-2^2=12}$, and since half of $ED$ is 1, we have EF$\;=\;$EC$\;=\sqrt{12+1^2=13}$.


问:
Reference request for ordered groups
I've been reading Pete Clark's notes on commutative algebra, and I especially liked section 17 on valuation rings, and ordered groups in particular.

I'm looking for more introductory material regarding the ordering of groups, monoids and vectorspaces. Could anyone suggest something?

I'm interested in printed works as well as online notes and courses. I read some of this book online and it seems good ,but I want to see if anyone has a more informed opinion.

答:
Mathscinet gives a favourable review of the book you mention, namely Partially Ordered Groups by A. M. W. Glass. The review says that it `will surely be an instant classic', and it has apparently been cited 78 times. It concludes saying that the book

will get the reader to the forefront of research in the field and would be suitable texts for students in modern algebra, group theory, or ordered structures.

As for other resources, Dale Rolfsen has notes on Ordered Groups and Topology: see http://www.math.ubc.ca/~rolfsen/papers/luminynotes/lum.pdf.

ContentLicense: CC BY-SA 3.0

Title: Reference request for ordered groups


问:
Show that G is not simple where $|G| = 2^2 \cdot 5^3 \cdot 7^9$
Show that if a group $G$ has order $|G|=2^2\cdot5^3\cdot7^9$, then $G$ is not simple.

$\textbf{Def:}$ A group $G$ is $\textbf{simple}$ if and only if $|G|>1$ and $\lbrace 1_G\rbrace$ as well as $G$ are its only normal subgroups.

I tried to use contradiction. First assuming that $G$ is simple. I was under the impression that we have to do this proof by cases with Sylow subgroups $n_2$, $n_5$, and $n_7$. I am stuck here since I thought we have to think about case 1 where $n_2 \equiv 1(\bmod2)$ and $n_2|5^3\cdot7^9 $. The trouble I'm having is wouldn't there be many factors where this case is true? Similarly for the other 2 Sylow subgroups.

答:
Probably the intended solution is the argument in the comments using $n_7 = 50$, but it's worth knowing that we can completely avoid thinking about $n_7$. In fact we have the following.

Proposition: The order of a finite simple group $G$ is divisible by either $8$ or $12$ (assuming Feit-Thompson), or the smallest prime $p$ dividing the order is odd and divides it with multiplicity at least $3$ (not assuming Feit-Thompson, but Feit-Thompson implies this never happens).

Proof. By the Burnside normal $p$-complement theorem, if a Sylow $p$-subgroup $P$ is normal in its centralizer $N(P)$, then it has a normal complement (a subgroup splitting the short exact sequence $1 \to P \to G \to G/P \to 1$), and in particular $G$ is not simple. Let $p$ be the smallest prime dividing $|G|$ and write $\nu_p(G)$ for the exponent of $p$ dividing $|G|$. If $\nu_p(G) \le 2$ then the Sylow $p$-subgroup $P$ is abelian (it must be either $C_p, C_p \times C_p$, or $C_{p^2}$) and the order of its automorphism group is only divisible by primes $\le p$, unless $p = 2$. We now split into cases:

$p$ odd (by Feit-Thompson this never happens): since $p$ is the smallest prime dividing $|G|$, the primes dividing $|N(P)|$ are all greater than $p$. It follows that $P$ is normal in $N(P)$ (since $N(P)/P$ can't act nontrivially on $P$), so $P$ has a normal complement. If $G$ is simple then this can't occur, so taking contrapositives, $\nu_p(G) \ge 3$.

$p = 2$: $|\text{Aut}(P)|$ is only divisible by a prime $\ge 3$ when $P = C_2 \times C_2$, which has automorphism group $S_3$. If $G$ is simple then $P$ can't have a complement, so either $\nu_2(G) \ge 3$ or $P = C_2 \times C_2$ and the conjugation map $N(P)/P \to \text{Aut}(P)$ is surjective, which implies that $12 \mid |N(P)|$ and hence that $12 \mid |G|$. $\Box$


问:
example 2 chapter 32 munkres
In the proof of $S_{\Omega} \times \bar S_{\Omega}$ is not normal in example 2 chapter 32 munkres' topology it claims that because the sequence ${\{x_1, \dots,} \} $ is countable then any of its upper bounds lies in $S_{\Omega}$. Why the least upper bound $b$ cannot be $\Omega $ itself? If I can prove that $b$ can never be $\Omega $ then the whole picture is complete!

PS Sorry the example is two pages not easy to retype, I thought because it is a common book available so maybe no need to include the whole example in here.

答:
Theorem 10.3 in Munkres states explicitly:

If $A$ is a countable subset of $S_\Omega$ then $A$ has an upper bound in $S_\Omega$.

So the lub of $A$, which is the smallest upper bound of $A$, must lie in $S_\Omega$ too (which is a section of $\Omega$ as stated before so contains with any ordinal also all smaller ordinals..

$\Omega$ itself is certainly an upperbound for $A$ in $\overline{S_\Omega}$, but certainly never the smallest one by the application of 10.3

Title: example 2 chapter 32 munkres


问:
Probability of disease transmission to future generations
Assume disease D is transmitted through a father (male) to his children. Let us assume that every family in a society have $C\geq1$ children. Let the father of a family have the disease D. What is the probability that disease D is transmitted to at least one person in the $N$-th generation after him $(N\geq1)$? $N=1$ represents the children of the father, $N=2$ represents the grandchildren, etc.

One can calculate the probability for a given $N$ and $C$, but I was wondering if it possible to find a general closed-form solution.

P.S.: The problem is equivalent to the problem of passing the family-name. In a society where the family-name is passed from father to children, what is the probability of having the family-name of a father in the $N$-th generation after him.

答:
As it has already been pointed out by @darthline and @b-maréchal, there is no closed-form solution to this problem and the answer should be found recursively.

For this specific example, one can find $q_N$ as the probability that the disease is not transferred to any person of the $N$-th generation. The probability of having the disease is then $p_N = 1-q_N$.

To find $q_N$ we have:

$q_1 = 0$

$q_2 = (\frac{1}{2})^C$

$q_N = \sum_{i=0}^{C} {C \choose i} (\frac{1}{2})^C  (q_{N-1})^i, \quad N\geq3$

($i$ represents the number of males among the children of a father)


问:
To show that an integral is 0 a.e. if it is 0 over every subset of measure 2/3.
I'm trying to solve the following past qual problem:

Let $f\in L^1[0,1]$ such that for every $E\subset[0,1]$ with $m(E)=2/3$, $\int_E f=0$. Show that $f=0$ a.e. ($m$ is the Lebesgue measure)

I found a similar problem but the same argument doesn't work: If integral is 0 on any set of measure 1/pi, then the function is 0 a.e.

Any hints are appreciated. Thanks!

答:
If the Lebesgue differentiation theorem is applicable, here is a quick solution.

Let $a, b \in [0, 1]$ be Lebesgue points of $f$, and let $I_1, I_2 \subseteq [0, 1]$ be intervals satisfying $a \in I_1$, $b \in I_2$, and $m(I_1) = m(I_2) < \frac{1}{6}$. Since $m(I_1\cup I_2) < \frac{1}{3}$, we can find $E \subseteq [0,1]\setminus(I_1\cup I_2)$ satisfying $m(E\cup I_1) = m(E\cup I_2) = 2/3$, and so,

$$ \int_{E\cup I_1} f = 0 = \int_{E \cup I_2} f. $$

From this,

$$ \frac{1}{m(I_1)}\int_{I_1} f = \frac{1}{m(I_2)} \int_{I_2} f. $$

So by letting $m(I_1) = m(I_2) \to 0$, it follows that $f(a) = f(b)$.

Since the set of all Lebesgue points of $f$ has full measure in $[0, 1]$, this implies that there exists a constant $c$ such that $f(x) = c$ for a.e. $x \in [0, 1]$. The value of $c$ is easily determined to be zero, and therefore $f$ is zero a.e.


问:
Finding the SVD of particular rank $1$, $4 \times 3$ matrix
I am preparing for a Numerical Analysis exam, and one of the practice questions is the following:

Find both a reduced and full SVD of
$$A = \begin{bmatrix}
1 & 0 & 4\\
1 & 0 & 4\\
1 & 0 & 4\\
1 & 0 & 4\\
\end{bmatrix}$$

I would only have about 10-15 minutes to work on this problem. I have a partial solution below, but I am wondering if there is any faster way to do it, since my method seems long and unreliable (after finding a reduced SVD, I found the full SVD by trial and error).

My Solution: $A$ has rank $1$, so a reduced SVD is of the form
$$\begin{bmatrix}
1 & 0 & 4\\
1 & 0 & 4\\
1 & 0 & 4\\
1 & 0 & 4\\
\end{bmatrix} = \begin{bmatrix}u_1 \\u_2\\u_3\\u_4 \end{bmatrix} [\sigma] \begin{bmatrix}\overline{v_1} & \overline{v_2} & \overline{v_3}  \end{bmatrix} = u \sigma v^*.$$
We know that $\{ u \}$ is a basis for $Col(A) = \langle \begin{bmatrix}1 & 1 & 1 & 1 \end{bmatrix}^T\rangle$, so $u$ is of the form $u = \begin{bmatrix}x & x & x & x \end{bmatrix}^T$. Since $||u||_2 = 1$, we must have $u = \pm \begin{bmatrix}\frac 12 & \frac 12 & \frac 12 & \frac 12 \end{bmatrix}^T$. We will chose the positive sign.

Similarly, $\{v \}$ is a basis for $Col(A^*) = \langle \begin{bmatrix} 1 & 0 & 4 \end{bmatrix}^T \rangle$ , and using $||v||_2 = 1$, we must have $v = \pm \begin{bmatrix} \frac{1}{\sqrt{17}} & 0 & \frac{4}{\sqrt{17}} \end{bmatrix}^T$. We will chose the positive sign. Finally, to find $\sigma$, look at the big SVD factorization of $A$ above. Equating entries $(1, 1)$ on both sides gives $ \frac 12 \sigma \frac {1}{\sqrt{17}}= 1$, which gives $\sigma = \frac {2}{\sqrt{17}}.$ Therefore a reduced SVD of $A$ is
$$A = \begin{bmatrix}  1/2 \\ 1/2\\\ 1/2\\ 1/2 \end{bmatrix} [2\sqrt{17}] \begin{bmatrix}\frac {1}{\sqrt{17}} & 0 & \frac {4}{\sqrt{17}} \end{bmatrix} $$

To find a full SVD, we just have to extend $u$ and $v$ to orthonormal matrices. I did this by basically guessing. An orthonormal extension of $u$ is

$$\frac 12 \begin{bmatrix} 1 & -1 & -1 & 1\\
1 & -1 & 1 & -1\\
1 & 1 & -1 & -1\\
1 & 1 & 1 & 1
\end{bmatrix}$$

and an orthonormal extension of $v$ is
$$\begin{bmatrix}
1/\sqrt{17} & 0 & 5/\sqrt{17}\\
0 & 1 & 0\\
5/\sqrt{17} & 0 & -1/\sqrt{17}
\end{bmatrix}$$

Therefore, a full SVD of $A$ is
$$A = \begin{bmatrix} 1 & -1 & -1 & 1\\
\frac 12 & -\frac 12  & \frac 12  & -\frac 12 \\
\frac 12  & \frac 12  & -\frac 12  & -\frac 12 \\
\frac 12  & \frac 12  & \frac 12  & \frac 12
\end{bmatrix}\begin{bmatrix}2 \sqrt {17} & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0 \end{bmatrix}
\begin{bmatrix}
1/\sqrt{17} & 0 & 5/\sqrt{17}\\
0 & 1 & 0\\
5/\sqrt{17} & 0 & -1/\sqrt{17}
\end{bmatrix}
$$

答:
Your method seems fairly quick. The only thing that I would consider optimizing is the extension to an orthonormal basis.

First of all, an approach to be considered is writing something like

To find the full SVD, it suffices to extend $u$ and $v$ to orthonormal bases $(u,u_2,u_3,u_4)$ and $(v,v_2,v_3)$. With that, we can see that $A$ has full SVD $A = U \Sigma V^T$, with
$$
U = \pmatrix{u & u_2 & u_3 & u_4}, \quad \Sigma = \pmatrix{2\sqrt{17}&0&0\\ 0&0&0\\ 0&0 &0\\ 0&0&0}, \quad V = \pmatrix{v_1&v_2&v_3}
$$

and simply giving this as the "full answer". I would suspect that this would get you most of the partial credit.

That said, if you must compute an orthonormal basis, then the standard approach is the Gram Schmidt process. For instance, we can get the extension to $U$ by applying the procedure to the list $(u,e_1,e_2,e_3)$, where $e_1,\dots,e_4$ denote the standard basis vectors, i.e. the columns of the size-$4$ identity matrix.

However, we can indeed take advantage of the fact that $u$ is the vector of equal elements. In particular, we can always extend such a matrix into an orthonormal basis by using the columns of the DFT matrix. Alternatively, since the dimension of the space is a power of $2$, we could do the same by considering the columns of a scaled Hadamard matrix.

In general, we can also extend $v$ to such a basis in a nice fashion using the fact that $u \in \Bbb R^3$. In particular, for $v_2$ we can select any vector orthogonal to $v$, then take the cross-product $v \times v_2$ as a third vector. However, I prefer your method which simply takes advantage of the zero-entry in $v$.


问:
Finding the length of a chord with an arc length
Given a circle with diameter $d$, is there a way to find the length of the chord f that cuts off an arc of length $l$?

I am trying to understand the relationships between chords and arcs etc, and this concept came up - is there a general formula for it?

(And before I get shut down for this being a homework question, it is not. The concept came up in independent study.)

答:
The angle of the corresponding circular sector is $\frac{2l}{d}$. If you trace a line from the center through the midpoint of the chord and two more lines from the center to the chord's endpoints, you will have two right triangles. It follows from basic trigonometry that:
$$ \frac{f}{2} = \frac{d}{2} \sin\left(\frac{l}{d}\right)  $$
and therefore
$$ f = d \sin\left(\frac{l}{d}\right)$$


问:
Normal Distribution and Dice game
Suppose I use the following rules to generate random numbers. I roll a fair $D_6$ twice. The first roll I call $\mu$ and the second $\sigma^2$. I then Generate a normal random variable with mean $\mu$ and variance $\sigma^2$. I call this $X$

In more concise notation: $X \sim N(U_1,U_2)$ where $U_1,U_2$ are identical independent discrete uniform R.V's on $\{1,2,3,4,5,6\}$

Find an $x \in \mathbb{R} $ such that $\mathbb{P}[X\leq x] = \frac{1}{2}$

I think $x=3.5$.

Reasoning: Symmetry. More precisely: $\mathbb{P}[X \leq x] = \sum\limits_{i=1}^6\mathbb{P}[X\leq x|U_1=i]\cdot\mathbb{P}[U_1=i]=\sum\limits_{i=1}^6\mathbb{P}[X\leq x|U_1=i]\cdot\frac{1}{6}$

Now examine each term in the sum, $\mathbb{P}[X\leq x|U_1=i]\cdot\mathbb{P}[U_1=i]$. Let us take $i=1 $ and $i=6$ as $3.5$ is in the middle of these two values we have that $\mathbb{P}[N(1,U_2)\leq3.5]=\mathbb{P}[N(6,U_2)\geq3.5]$ from symmetry of normal dist. And then finally $\mathbb{P}[N(6,U_2)\leq3.5] = 1- \mathbb{P}[N(6,U_2)\geq3.5] = 1 - \mathbb{P}[N(1,U_2)\leq3.5]$

Hence: $\mathbb{P}[N(1,U_2)\leq3.5]+\mathbb{P}[N(6,U_2)\leq3.5] = \mathbb{P}[N(1,U_2)\leq3.5] + 1- \mathbb{P}[N(1,U_2)\leq3.5]=1 $

So if we split the sum into the $3$ pairs of integers that have $3.5$ in the middle, i.e sum to $7$ i.e $\{(1,6) , (2,5) , (3,4) \}$ we end up with:

$\mathbb{P}[X \leq x] = \sum\limits_{i=1}^6\mathbb{P}[X\leq x|U_1=i]\cdot\frac{1}{6} = \frac{1}{6}(1+1+1)=\frac{1}{2}$

Is this correct? :)

The second part of the question is to now have $X \sim N(U_1^2,U_2)$ where $U_1,U_2$ are identical independent discrete uniform R.V's on $\{1,2,3,4,5,6\}$ Find an $x \in \mathbb{R} $ such that $\mathbb{P}[X\leq x] = \frac{1}{2}$

And then finally to have  $X \sim N(U_1^n,U_2)$ for some $n \in \mathbb{N}$
and where $U_1,U_2$ are identical independent discrete uniform R.V's on $\{1,2,3,4,5,6\}$ Find an $x \in \mathbb{R} $ such that $\mathbb{P}[X\leq x] = \frac{1}{2}$

答:
You describe a mixture distribution of six normal distributions
$\mathsf{Norm}(\mu=i, \sigma = \sqrt{i}),$ for $i = 1,2,3,4,5,6.$ Perhaps see Wikipedia on mixture distributions, under 'Moments'.

Simulation in R of a million iterations of your procedure, from which we may expect about 2-place accuracy:

```<code>set.seed(103)
m = 10^6
u1 = sample(1:6, m, rep=T)
u2 = sample(1:6, m, rep=T)
x = rnorm(m, u1, sqrt(u2))
median(x)
[1] 3.497238  $ aprx 3.5

hist(x, prob=T, col="skyblue2", main="Simulated Mixture Distribution")
 curve(dnorm(x, mean(y), sd(y)), add=T, col="orange", lwd=2)
</code>```

Note: In this case the mixture distribution is approximately normal.
For example, human heights are often modeled as normal, but may actually be mixture distributions of heights of men and women in diverse ethnic groups.

However, in general, mixtures of normal distributions are not (necessarily) anywhere near normal. The 50:50 mixture of
$\mathsf{Norm}(100, 10)$ and $\mathsf{Norm}(130, 10)$ is bimodal because the difference in means is more than two standard deviations.

```<code>set.seed(2021)
m = 10^6
x1 = rnorm(m, 100, 10);  x2 = rnorm(m, 130, 10)
p = rbinom(m, 1, .5);  y = p*x1 + (1-p)*x2
hist(y, prob=T, col="skyblue2")
</code>```

Title: Normal Distribution and Dice game

Body:
You describe a mixture distribution of six normal distributions
$\mathsf{Norm}(\mu=i, \sigma = \sqrt{i}),$ for $i = 1,2,3,4,5,6.$ Perhaps see Wikipedia on mixture distributions, under 'Moments'.

Simulation in R of a million iterations of your procedure, from which we may expect about 2-place accuracy:

set.seed(103)
m = 10^6
u1 = sample(1:6, m, rep=T)
u2 = sample(1:6, m, rep=T)
x = rnorm(m, u1, sqrt(u2))
median(x)
[1] 3.497238  $ aprx 3.5

hist(x, prob=T, col="skyblue2", main="Simulated Mixture Distribution")
 curve(dnorm(x, mean(y), sd(y)), add=T, col="orange", lwd=2)

Note: In this case the mixture distribution is approximately normal.
For example, human heights are often modeled as normal, but may actually be mixture distributions of heights of men and women in diverse ethnic groups.

However, in general, mixtures of normal distributions are not (necessarily) anywhere near normal. The 50:50 mixture of
$\mathsf{Norm}(100, 10)$ and $\mathsf{Norm}(130, 10)$ is bimodal because the difference in means is more than two standard deviations.

set.seed(2021)
m = 10^6
x1 = rnorm(m, 100, 10);  x2 = rnorm(m, 130, 10)
p = rbinom(m, 1, .5);  y = p*x1 + (1-p)*x2
hist(y, prob=T, col="skyblue2")


问:
Grid Combinations with Positive and Negative Numbers
If there is an $n$ by $n$ matrix where each element is either 1 or -1, how many unique matrices are there such that each row and each column multiplies to 1?

I solved for the trivial case of $n = 2$, which is 2. However, for larger $n$, I'm not sure how to find a systematic way to count. I have tried using the fact that you can transform an existing correct matrix into another correct matrix by selecting 4 points that form a rectangle to all become negative, but I am not sure how I could count these in an organized way.

答: Each row can have $1$'s & $-1$'s in $2^{n-1}$ ways. Fill in the first $n-1$ rows & then complete the last row to give each column a positive parity. So there are $\color{red}{2^{(n-1)^2}}$ ways to construct these matricies.


答:
The length of an arc of a circle is $l=r\theta$, where $r=d/2$ is the radius and $\theta$ is the central angle (in radians) subtended by the arc. So from the known arc length and radius (or diameter), you can find the angle.

Then drawing the radii to the endpoints of the chord (and the arc) gives us an isosceles triangle with the angle of $\theta$ at the vertex at the center of the circle. Drop the height/median from this vertex at the center to cut this triangle into two right triangles, and use trigonometry to find the desired length of the base. You should see that $\displaystyle \sin(\theta/2)=\frac{f/2}{r}$, from which you can find $f$.

Title: Finding the length of a chord with an arc length

Body:
Given a circle with diameter $d$, is there a way to find the length of the chord f that cuts off an arc of length $l$?

I am trying to understand the relationships between chords and arcs etc, and this concept came up - is there a general formula for it?

(And before I get shut down for this being a homework question, it is not. The concept came up in independent study.)


问:
Would a circle overlap a parabola's bottom by more than just its vertex?
I mean, out of the condition that a circle actually crosses the parabola.
My question is when a circle is "inside" a parabola, would it touch part of the parabola other than just the parabola's vertex at the bottom?

So just the curiosity brought me crazy...but would this be possible?? (There was an image to show but my reputation is not enough since I'm a freshmeat :P)

Thanks xx

答:
Consider the equations for the circle and parabola. For simplicity, we'll have the circle centred at $(0,0)$, and the parabola tangent to the circle at $(0,-a)$, where $a$ is the radius of the circle.

From this, we have the equation
$$
x^2+y^2=a^2
$$
for the circle, and
$$
y=bx^2-a
$$
for the parabola. Now, if they coincide at any point, then the $x$ and $y$ values must be the same. So, substituting the second equation into the first one, we get
$$
x^2+(bx^2-a)^2=a^2
$$
or
$$
x^2+b^2x^4-2abx^2=0
$$
Now, at $x=0$, we have the "vertex" point, so we can assume that $x\neq 0$ for other solutions. This gives
$$
b^2x^2 = 2ab-1
$$
or
$$
x = \pm\frac{\sqrt{2ab-1}}b
$$
As such, if $2ab-1>0$, then there are more intersection points than just the vertex, given by
$$
\left(\pm\frac{\sqrt{2ab-1}}b,\frac{ab-1}b\right)
$$
If $2ab-1=0$, then the parabola follows the circle's trajectory in the vicinity of the vertex, but does not coincide. If $2ab-1<0$, then the "intersection points" occur at imaginary values, and thus do not exist in the reals.


问:
Show that $|\phi(a)| \mid|G|$
Let $G,H$ be groups and $\phi\colon G\to H$ be a one-to-one group homomorphism. Let $a\in G$. Show that $|\phi(a)| \mid |G|$. Not sure how to attack this proof.

Do we use the First Isomorphism Theorem and Lagrange's Theorem? I am unsure if the canonical homomorphism is helpful for this proof.

答:
By Lagrange, the order of $a$ divides the order of $G$.  By the homomorphism property, the order of $\phi(a)$ divides the order of $a$.  Now by transitivity of divisibility, the result follows.

You don't need injectivity.

However, injectivity could be considered to simplify the proof.  Because $\phi(a)$ will be an element of the image.  Then apply the first isomorphism theorem and Lagrange.

Title: Show that $|\phi(a)| \mid|G|$


问:
Proving the formula for $\prod_{k=1}^n \cos(kx)$
I'm trying to find a closed-form for the following product:

$$ \prod_{k=1}^n \cos(kx)$$

Here is my attempt at it:

$$ \prod_{k=1}^n \cos(2kx) = \prod_{k=1}^n \frac{ e^{i2kx} + e^{-i2kx} }{2} = \frac{(e^{2ix} - e^{-2ix}) \prod_{k=1}^n  e^{i2kx} + e^{-i2kx}  }{2^n (e^{i2x}- e^{-i2x})}= \frac{1}{2^n (e^{i2x}- e^{-i2x})} ( e^{4inx} - e^{-4inx} )= \frac{1}{2^n} \frac{\sin(4nx)}{\sin(2x)} $$

The above was simplified by using the algebraic identity $(a+b)(a-b)=a^2 -b^2$ repeatedly in the product. Now sub:$  x \to \frac{x}{2}$

$$ \prod_{k=1}^n \cos( kx)= \frac{1}{2^n} \frac{\sin(2nx)}{\sin x}$$

I'm not sure if I'm doing something wrong, but when I plug $ x= \frac{\pi}{n}$, I'm not getting the result as shown in this MSE post

Edit: Whoops! The algebraic manipulation doesn't hold after k=3 as correctly pointe od out by @leon. However, still am looking for a closed form.

答:
Wolfram Mathematica for Product[Cos[k x],{k,n}] returns

```<code>2^(-1 - n)*E^((I/2)*n*(1 + n)*x)*QPochhammer[-1, E^((-2*I)*x), 1 + n]
</code>```

which in usual mathematical notation using
the Q-Pochhammer symbol is
$\; 2^{-1-n}e^{i n(n+1)x/2} (-1;e^{-2ix})_{n+1}. $

Once the closed form of the product is known, it is not hard to
prove it using the definition of the Q-Pochhammer symbol
$$ (a;q)_n := \prod_{k=0}^{n-1}(1-aq^k). $$

Body:
Wolfram Mathematica for Product[Cos[k x],{k,n}] returns

2^(-1 - n)*E^((I/2)*n*(1 + n)*x)*QPochhammer[-1, E^((-2*I)*x), 1 + n]

which in usual mathematical notation using
the Q-Pochhammer symbol is
$\; 2^{-1-n}e^{i n(n+1)x/2} (-1;e^{-2ix})_{n+1}. $

Once the closed form of the product is known, it is not hard to
prove it using the definition of the Q-Pochhammer symbol
$$ (a;q)_n := \prod_{k=0}^{n-1}(1-aq^k). $$


问:
The Lebesgue Dominated Convergence Theorem: Example.
The following question is from Real Analysis by Royden (4th Edition). More precisely, it is problem 36 of chapter 4. Further, the problem follows right after the section over The Lebesgue Dominated Convergence Theorem, hence, that is what should be used. Below is the problem followed by my attempted solution which I have a question about.

Problem:

Let $f$ be a real-valued function of two variables $\left(x,y\right)$ that is defined on the following unit square
$Q:=\{\left(x,y\right): 0\leq x\leq 1, 0\leq y\leq 1\}$ and is a measurable function of $x$ for each fixed value of
$y$. For each $\left(x,y\right)\in Q$ let the partial derivative a $\partial f/\partial y$ exist. Suppose there is a function $g$ that is integrable over $[0,1]$ and such that,

$$\left\vert\frac{\partial f}{\partial y}\left(x,y\right)\right\vert \leq g(x) \text{ for all } \left(x,y\right)\in Q.\tag{*}$$

Prove that

$$\frac{d}{dy}\left[\int_0^1 f(x,y)dx\right] = \int_0^1\frac{\partial f}{\partial y}(x,y)dx, \text{ for all } y\in[0,1].$$

(My) Proof:

Fix $y\in[0,1]$ and consider the sequence of functions $f_n\colon[0,1]\to\mathbb{R}$ defined by,

$$f_n(x):=\frac{f(x,y+1/n) - f(x,y)}{1/n}.$$

Then, $\{f_n\}$ is a sequence of measurable functions ($f$ is a measurable function of $x$ for each fixed value of $y$) dominated by $g(x)$ that converge pointwise almost everywhere to $\partial f/\partial y$ on $[0,1]$ (for each $(x,y)\in Q$ the partial derivative $\partial f/\partial y$ exists). Therefore, by The Lebesgue Dominated Convergence Theorem,

$$\lim_{n\to\infty}\int_0^1f_n=\int_0^1f$$
$$\lim_{n\to\infty}\int_0^1\frac{f(x,y+1/n) - f(x,y)}{1/n}dx =
\int_0^1\frac{\partial f}{\partial y}(x,y)dx.\tag{1}$$

Notice,

$$\lim_{n\to\infty}\int_0^1\frac{f(x,y+1/n) - f(x,y)}{1/n}dx =
\lim_{h\to 0}\frac{\int_0^1f(x,y+h)dx-\int_0^1f(x,y)dx}{h}=
\frac{d}{dy}\left[\int_0^1f(x,y)dx\right].\tag{2}$$

Thus, combining (1) and (2),

$$\frac{d}{dy}\left[\int_0^1f(x,y)dx\right] = \int_0^1\frac{\partial f}{\partial y}(x,y)dx,$$

which was to be shown.

My main concern with my proof is that, I am not sure how the sequence $\{f_n\}$ are dominated by $g(x)$. I know by (*) that the limit of the sequence $\{f_n\}$ is bounded by $g(x)$. Also, if there are any mistakes in the proof, please let me know!

Edit:

Does the boundedness of the $\{f_n\}$ come from the fact that every convergent sequence is a bounded sequence?

答:
Consider $y \in [0,1]$ and any sequence $y_n \in [0,1]$ such that $y_n \to y$. By the mean value theorem, there exists $\xi_{x,n}$ between $y$ and $y_n$ such that

$$|f_n(x)| = \left|\frac{f(x,y_n) - f(x,y)}{y_n - y} \right| = \left|\frac{\partial f}{\partial y}(x,\xi_{x,n})\right| \leqslant g(x)$$

We have

$$\lim_{n \to \infty}\frac{f(x,y_n) - f(x,y)}{y_n - y} = \frac{\partial f}{\partial y}(x,y),$$

and by the dominated convergence theorem

$$\lim_{n \to \infty}\int_0^1\frac{f(x,y_n) - f(x,y)}{y_n - y}\, dx = \int_0^1\lim_{n \to \infty}\frac{f(x,y_n) - f(x,y)}{y_n - y}\, dx = \int_0^1 \frac{\partial f}{\partial y}(x,y) \, dx$$


问:
Computing the specific relative error
Suppose that $x$ is the exact value of some quantity and $\hat{x}$ is the approximate value of this quantity. The numbers $$\Delta(\hat x):=|x-\hat x|,$$ $$\delta(\hat x):=\frac{\Delta(\hat x)}{|\hat x|}$$ are called absolute and relative errors, respectively.

For example, the Gravitational constant is $G=(6,67259\pm0,00085)\times 10^{-11} $.

My book says that in this example relative errors does not exceed $13\times 10^{-5}$
.

Can anyone clarify it please? The question seems really stupid but anyway I'd be thankful for clarification!

答:
The approximated value of $G$ is $6.67259\times 10^{-11}$, but the real value could be anything in the interval $[6.67174\times 10^{-11},6.67344\times 10^{-11}]$. You do no really know it, but the maximum absolute error would be achieved when it is one of the endpoints. Therefore, the maximum relative error is:
$$\delta(\hat x)=\frac{\Delta(\hat x)}{|\hat x|}=\frac{0.00085\times 10^{-11}}{6.67259\times 10^{-11}}\approx13\times 10^{-5}$$

Title: Computing the specific relative error


答:
Let $\phi(G) = \{ \phi(g) : g \in G\}$ be the image of $G$ under $\phi$.  It is a subgroup of $H$ (check this), and in particular, $\phi(G)$ is a group.  Because $\phi(G)$ is a group containing the element $\phi(g)$, Lagrange's theorem tells us that the order of $\phi(g)$ divides the order of $\phi(G)$.

Now because $\phi$ is assumed to be one-to-one, what does that tell you about the relationship between the order of $G$ and the order of $\phi(G)$?

Title: Show that $|\phi(a)| \mid|G|$

Body:
Let $G,H$ be groups and $\phi\colon G\to H$ be a one-to-one group homomorphism. Let $a\in G$. Show that $|\phi(a)| \mid |G|$. Not sure how to attack this proof.

Do we use the First Isomorphism Theorem and Lagrange's Theorem? I am unsure if the canonical homomorphism is helpful for this proof.


问:
The negative square root of $-1$ as the value of $i$
I have a small point to be clarified.We all know $ i^2 = -1 $ when we define complex numbers, and we usually take the positive square root of $-1$ as the value of "$i$" , i.e, $i =  (-1)^{1/2} $.

I guess it's just a convention that has been accepted in maths and the value $i =  -[(-1)^{1/2}] $ is neglected as I have never seen this value of "$i$" being used. What I wanted to know is, if we will use $i =  -[(-1)^{1/2}] $ instead of $ (-1)^{1/2} $, would we be doing anything wrong? My guess is there is nothing wrong in it as far as the fundamentals of maths goes. Just wanted to clarify it with you guys. Thanks.

答:
... and we usually take the positive square root of −1 as the value of "$i$".

There's no such thing as a positive or negative complex number. By treating $\mathbb{R}$ as a subset of $\mathbb{C}$ you can call some complex numbers positive or negative, but only those which actually are real numbers. There's no way to define a total order on $\mathbb{C}$ which would behave like the usual order on $\mathbb{R}$ does.

What happens is that one simply picks any root of $-1$, i.e. any solution of $x^2 + 1 = 0$, and calls it $i$. There's then always a second solution, and that solution is $-i$. You can't even say which solution you picked for $i$ - the two solutions of $x^2 + 1=0$ are algebraically indistinguishable, i.e. you cannot tell them apart with algebraic means.

Imagine a friend hands you a bag containing two balls, of equal size and material. You can pick one arbitrarily and call it "ball 1", and take a pen and mark it with a big "1". The other is then "balls 2", and gets marked with a big "2". Now, imagine you had picked the other ball. Would you end up in a different situation? No! You'd still have two balls, one marked "1" and one marked "2", and indistinguishable otherwise. Now, after you've marked the balls, they are of course different. You can now for example put both back into the bag, let your friend pick one, and ask "Which ball have you picked?". But that only works after you marked them!


问:
$x_n \rightharpoonup x$ then $\sigma_n= \frac{x_1+x_2+...+x_n}{n} \rightharpoonup x$
This is the exercise 3.2 of Brezis book. Let $E$ a Banach space and let $(x_n)$ be a sequence such that $x_n\rightharpoonup x$ in the weak topology $\sigma(E,E^*)$. Set $\sigma_n=\frac{x_1+x_2+...+x_n}{n}$. Prove that $\sigma_n\rightharpoonup x$ in the weak topology $\sigma(E,E^*)$.

I want to show that for
every $f\in E^* \quad |<f,\sigma_n>-<f,x>| \rightarrow 0$.

$|<f,\sigma_n>-<f,x>|=\displaystyle{\left| \frac{f(x_1-x)}{n}+\frac{f(x_2-x)}{n}+...+\frac{f(x_n-x)}{n}\right|}$ and I don't know how to continue, I was thinking about $f(x_n-x)$ is bounded but how can I use this? and the most important fact that $|<f,x_n>-<f,x>| \rightarrow 0$. Thanks for helping.

答:
Suppose $|\langle f, x_n \rangle - \langle f, x \rangle| < \varepsilon$ for all $n \ge  N$.

Then for $m \ge N$, split $\sigma_m$ into two parts:

$$\sigma_m = \frac{x_1 + \dots + x_N}{m} + \frac{x_{N + 1} + \dots + x_m}{m}.$$

As $m \to \infty$, the first half the expression goes to $0$ because the numerator is fixed. For the other half, you can use the bound $|\langle f, x_n \rangle - \langle f, x \rangle| < \varepsilon$ from the beginning.


问:
Solving heat equation using radial solution
Find a solution to the heat equation with the initial values

$\begin{cases} u_t - 4u_{xx} = 0 &   -\infty < x < \infty, \; t>0 \\ u(x,0)= 2e^{-x^2} &     -\infty < x < \infty \end{cases}$

I know how to do this in separable method. But I want the solution in terms of radial method.

I would start by assuming $u(x,t)= v(r)$.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
There is not any cylindrical or spherical symmetry which justify your initial claim !!!.

Hereafter, you'll see a straightforward approach:

Multiply both member of the Heat Equation by
$\ds{\expo{-st}}$ and integrate over $\ds{t > 0}$. Namely,
\begin{align}
0 & =
\int_{0}^{\infty}\expo{-st}\partiald{\on{u}\pars{x,t}}{t}\,\dd t -
4\,\partiald[2]{}{x}\
\overbrace{\int_{0}^{\infty}\on{u}\pars{x,t}\expo{-st}\,\dd t}
^{\ds{\hat{\on{u}}\pars{x,s}}}
\\[5mm] & \stackrel{\substack{{\rm IBP}
\\[0.5mm] {\rm first\ integral}}}{=}\,\,\,\,\,\,
-\on{u}\pars{x,0} + s\,\hat{\on{u}}\pars{x,s}
-4\,\partiald[2]{\,\hat{\on{u}}\pars{x,s}}{x}
\\[5mm] \implies &
\pars{\partiald[2]{}{x} - {1 \over 4}\,s}\hat{\on{u}}\pars{x,s} =
-\,{1 \over 4}\,\on{u}\pars{x,0} = -\,{1 \over 2}\expo{-x^{2}}
\\[5mm] & \mbox{which has the solution}
\\[2mm] &
\bbx{\hat{\on{u}}\pars{x,s} = \hat{\on{u}}_{p}\pars{x,s}
-\,{1 \over 2}\int_{-\infty}^{\infty}\on{G}\pars{x,x'}
\expo{-x\,'^{2}}\,\,\dd x\,'} \\ &
\end{align}
$\ds{\hat{\on{u}}_{p}\pars{x,s}}$ is a particular solution -a homogeneous equation solution- which is usually chosen to satisfies the boundary conditions. In this way, $\ds{\on{G}\pars{x,x'}}$ satisfies homogeneous boundary conditions. Namely,
\begin{align}
& \left\{\begin{array}{rcl}
\ds{\pars{\partiald[2]{}{x} - {1 \over 4}\,s}\on{G}\pars{x,x'}} & \ds{=} & \ds{0\,,\quad x \not= x\,'}
\\[1mm]
\ds{\on{G}\pars{0,x'}} & \ds{=} & \ds{0}
\\[1mm]
\ds{\left.\partiald{\on{G}\pars{x,x'}}{x}
\,\right\vert_{\,x\ = x\ '^{\,-}}^{\,x\ = x\ '^{\,+}}} & \ds{=} & \ds{1}
\\[3mm]
\substack{\ds{\on{G}\pars{0,x'}}\ \mbox{continuous}
\\ \mbox{at}\ \ds{x = x'}} &&
\end{array}\right.
\end{align}
Then,
\begin{align}
\on{G}\pars{x,x\ '} & =
\left\{\begin{array}{lcl}
\ds{0} & \mbox{if} & \ds{x < x\ '}
\\
\ds{A\sinh\pars{{1 \over 2}\root{s}\pars{x - x\ '}}} & \mbox{if} & \ds{x > x\ '}
\end{array}\right.
\\[5mm]
\left.\partiald{\on{G}\pars{x,x'}}{x}
\,\right\vert_{\,x\ = x\ '^{\,-}}^{\,x\ = x\ '^{\,+}} & = 1\
\implies A = {2 \over \root{s}}
\end{align}
Then,
\begin{align}
\hat{\on{u}}\pars{x,s} & = \hat{\on{u}}_{p}\pars{x,s}
-\int_{-\infty}^{x}{\sinh\pars{\root{s}\bracks{x - x'}/2} \over \root{s}}\expo{-x\,'^{2}}\,\,\dd x\,'
\\[5mm] & = \hat{\on{u}}_{p}\pars{x,s} -
{\expo{\root{s}x/2} \over 2\root{s}}
\int_{-\infty}^{x}\exp\pars{-x\ '^{2} - \root{s}x\ '/2}\,\dd x\ '
\\[2mm] &
+ {\expo{-\root{s}x/2} \over 2\root{s}}
\int_{-\infty}^{x}\exp\pars{-x\ '^{2} + \root{s}x\ '/2}\,\dd x\ '
\end{align}
I'll leave the OP the evaluation of the remaining integrals.


问:
Show that the mapping $I : C([a,b]) \to \Bbb{R}$, $I(f) = \int_{a}^{b} f(s) \ ds$ is Lipschitz continuous.

Let $C([a,b])$ be the norm space of continuous functions $f:[a,b] \to \Bbb{R}$ with the metric $\|f\|_{\infty}= \max_{s\in[a,b]} |f(s)|$. Show that the mapping $I : C([a,b]) \to \Bbb{R}$ $$I(f) = \int_{a}^{b} f(s) \ ds$$ is Lipschitz continuous.

Letting $f,g \in C([a,b])$ I have that $$|I(f)-I(g)| = |\int_{a}^{b} f(s) \ ds-\int_{a}^{b} g(s) \ ds| = \int_{a}^{b}|f(s)-g(s)| \ ds \leqslant \int_{a}^{b} \max_{s\in [a,b]} |f(s)-g(s)| \ ds$$

but I cannot find any $M$ for the Lipschitz condition, I guess it comes from the integral, but I'm not sure how?

答:
It might help you to use a different dummy variable for that $\max$, to avoid a clash of variables under the integral sign:
$$\int_a^b \max_{t \in [a,b]} |f(t)-g(t)| \, ds
$$
And the value of that $\max$ is, of course, a constant independent of $s \in [a,b]$, equal to $\|f\|_\infty$. Perhaps you can finish now?


问:
Seeking upper bound on largest prime factor of colossally abundant numbers
I recently came across an older post where the OP referred to papers giving upper bounds on the size of the largest prime factor in superabundant and colossally abundant numbers. I asked for links to those papers in the comments. Am still hoping for a response, however maybe it's a good stand-alone question.

Does anyone know of results on the upper bound on the size of the largest prime factor of colossally abundant numbers? It seems that the answer will roughly be slightly less than log(n) but I'd be grateful if there's some known explicit results out there. Thanks.

Edit
The poster of the original question has answered this question. Thanks Ahmad! Here is the original post.

答:
The main paper i rely on in my question was "A Class of Equivalent Problems Related to the Riemann Hypothesis" by Sadegh Nazardonyavi in which in proposition 3.33 he proves that $ \sqrt{p} < \ln n -\theta(p) < \sqrt{3p}$ for large enough super abundant (SA) number $n$ and $p$ being the largest prime divisor of $n=2^k \cdots p$.

Other important papers are for Erdos, Nicolas on highly composite numbers.

Also one can show from PNT that any counter-example for RH must have these properties :

$n =2^k \cdots p$ a multiplication of primes in ascending order and the powers in descending order.

And that $\ln n \approx \theta(p) \approx p$.

I did it 2 years ago and never looked back to it, so i don't have any papers on my pc nor my own proof for the statements above but you should find them in Sadegh Nazardonyavi,Erdos works.

I tried to make them sharp as a way to attack RH, but i couldn't improve on the bounds you should get from PNT.


问:
Universal meaning of 'normal'
The word 'normal' is used in many different areas of math, with about 20 different uses identified by Wikipedia. Is there any overarching meaning explaining why the same word is used for, say,  'perpendicular', 'normal subgroup' and 'normal distribution'?</p>

答:
In parts, perhaps this question could also belong on History of Science and Mathematics Stack Exchange and/or English Language and Usage, but I think it can fit here as well.

General Meanings/Origin

"Normal" comes to English ultimately from Latin "nōrma", meaning "carpenter's square" which also had a meaning like "rule, pattern" even in Latin. See Wikitionary or the Online Etymology Dictionary, for instance.

This combination basically explains all related meanings you come across in English mathematics.

Perpendicular

Carlo Beenakker
's answer to How did "normal" come to mean "perpendicular"? on Math Overflow quotes a fifteenth century commentary:

"Angulus normalis est idem qui angulus rectus" = "a normal angle is the same as a right angle"

And Brian M. Scott's answer to Etymology of the word "normal" (perpendicular) quotes "OED Third Edition, an entry updated December 2003" (Here OED refers to the Oxford English Dictionary, not the aforementioned Online Etymology Dictionary.):

classical Latin normālis right-angled

Normal Subgroup

Martin Brandenburg's answer to Why are normal subgroups called "normal"? quotes Jeff Miller's Earliest Known Uses of Some of the Words of Mathematics page:

According to The Genesis of the Abstract Group Concept (1984) by Hans Wussing, "The German Normalteiler (normal subgroup) goes back to Weber [H., Lehrbuch der Algebra, vol. 1, Braunschweig, 1895. p.511] and is possibly linked to Dedekind's term Teiler (divisor), which was employed in ideal theory" [Dirk Schlimm].

If true, then "normal" for "normal subgroup" may have come through German, but Wikitionary says that the German word "normal" has essentially the same source and meanings as the English.

user810157 conjectured in their answer:

"normal" means "inducing some regularity/order" and hence "some structure": think of the group structure induced in the quotient when the subgroup is (indeed) "normal". Of course this is just my guess.

Normal Distribution

Wikipedia cites Probability Theory: The Logic of Science, Ch 7 by Edwin J. Jaynes, which I quote here:

The literature gives conflicting evidence about the origin of the term "Normal distribution"...However, the term had long been associated with the general topic: given a linear model $y=X\beta+e$ where the vector $y$ and the matrix $X$ are known, the vector of parameters $\beta$ and the noise vector $e$ unknown, Gauss (1823) called the system of equations $X'X\hat\beta=X'y$ which give the least squares parameter estimates $\hat\beta$, the "normal equations" and the ellipsoid of constant probability density was called the "normal surface." It appears that somehow the name got transferred from the equations to the sampling distribution that leads to those equations.

Presumably, Gauss meant "normal" in its mathematical sense of "perpendicular" expressing the geometric meaning of those equations. The minimum distance from a point (the estimate) to a plane (the constraint) is the length of the perpendicular.

If this is the ultimate reason why people started calling it the "normal distribution", then it is a coincidence that the central limit theorem makes this distribution occupy a central role which could seem related to the meanings "standard"/"usual" that the word "normal" can have. In fact, Jaynes says:

This leads many to think — consciously or subconsciously — that
all other distributions are in some way abnormal.

Actually, it is quite the other way; it is the so-called 'normal' distribution that is abnormal in the sense that it has many unique properties not possessed by any other. Almost all of our experience in inference has been with this abnormal distribution, and much of the folklore that we must counter here was acquired as a result. For decades, workers in statistical inference have been misled, by that abnormal experience, into thinking that methods such as coence intervals, that happen to work satisfactorily with this distribution, should work as well with others.

Title: Universal meaning of 'normal'


问:
The representation of closed set
We already know that closed set is the complement of open set.

But I want to know the representation of closed set without the fact above in the usual topology.

For instance, the integers $\mathbb{Z}$ is closed due to the fact that $\mathbb{Z}$ can be represented by the union of one point sets $\{n\}$
(I don't want to use the fact that "the union of open intervals $(n, n+1)$ is open").

答:
Here is one representation: A closed set is the set with all its accumulations.

Note that $Z$ is closed not because that it is the union of one point set. If a set is a finite union of one point set in the $T_1$ space, then it must be closed. However, if it is a countble union of one point set, it need not be closed.

Example: $\mathbb R$ with usual topology, the subset $\mathbb Q$ is the union of one point set, however it is not closed.

Title: The representation of closed set


问:
Are two graphs isomorphic?
Are the two graphs isomorphic?

$$G_1=\begin{bmatrix}
a & b & c & d & e & f \\
b & a & a & c & d & a \\
c & c & b & e & f & d \\
f &   & d & f &   & e
\end{bmatrix}\
\quad\quad
G_2=\begin{bmatrix}
u & v & w & x & y & z \\
v & u & v & u & x & u \\
x & w & x & w & z & w \\
z &   & z & y &   & y
\end{bmatrix}$$

$a,b,c$ creates a triangle in $G_1$, but no triangle is created in $G_2$.

Is that enough to claim that they are not isomorphic?

答:
This is a good example of two graphs with the same degree sequence $(3,3,3,3,2,2)$ which are not isomorphic. As the question states, the existence of the two triangles in $G_1$ and their absence in $G_2$ is proof that the two graphs are not isomorphic.

Here they are in graphical form:

Title: Are two graphs isomorphic?


问:
$\prod_{i=1}^n[(N-x_i)!]=\prod_{i=1}^n[(N-y_i)!]$ for all $N\in\mathbb{Z}^+$ implies $x_{(i)}=y_{(i)}$, for all $i\in\{1,\cdots,n\}$
Suppose $$\prod_{i=1}^n[(N-x_i)!]=\prod_{i=1}^n[(N-y_i)!]$$ for all $N\in\mathbb{Z}^+$.

Why does it imply $x_{(i)}=y_{(i)}$, for all $i\in\{1,\cdots,n\}$? Here $X_{(i)}$ is the order statistic.

I am trying to find a minimal sufficient statistic for $\{X_i\}_{i=1}^n\sim\text{Binomial}(N,\frac{1}{2})$. The last step is showing $\prod_{i=1}^n[(N-x_i)!]=\prod_{i=1}^n[(N-y_i)!] \implies x_{(i)}=y_{(i)}$.

Body:
\begin{equation} \prod_{i=1}^n[(N-x_i)!]=\prod_{i=1}^n[(N-y_i)!]\iff\prod_{i=1}^n[(N-x_{(i)})!]=\prod_{i=1}^n[(N-y_{(i)})!] \end{equation}   $(N-x_{(i)})!$ is a polynomial of $N$ to the degree of $N-x_{(i)}$.

$\prod_{i=1}^n[(N-x_{(i)})!]$ is a polynomial of $N$ to the degree of $nN-\sum_{i=1}^nx_{(i)}$. Its powers include $nN-\sum_{i=1}^{n} x_{(i)}$, $(n-1)N-\sum_{i=1}^{n-1} x_{(i)}$, $(n-2)N-\sum_{i=1}^{n-2} x_{(i)},\cdots$.

This means $$\sum_{i=1}^m x_{(i)}=\sum_{i=1}^m y_{(i)}, \forall m\in\{1,\cdots,n\}\iff x_{(i)}=y_{(i)},\forall i\in\{1,\cdots,n\}$$

--- Update:

$(N-x_{(i)})!$ is like $N^{N-x_{(i)}}+c_{i,1}N^{N-x_{(i)}-1}+c_{i,2}N^{N-x_{(i)}-2}+\cdots$

$\prod_{i=1}^n[(N-x_{(i)})!]$ is like $(N^{N-x_{(n)}}+c_{n,1}N^{N-x_{(n)}-1}+c_{n,2}N^{N-x_{(n)}-2}+\cdots)\times(N^{N-x_{(n-1)}}+c_{n-1,1}N^{N-x_{(n-1)}-1}+c_{n-1,2}N^{N-x_{(n-1)}-2}+\cdots)\times(N^{N-x_{(n-2)}}+c_{n-2,1}N^{N-x_{(n-2)}-1}+c_{n-2,2}N^{N-x_{(n-2)}-2}+\cdots)\underset{\text{n terms product}}{\cdots}$

Thus the largest degree term of $\prod_{i=1}^n[(N-x_{(i)})!]$ is $N^{nN+\sum_{i=1}^nx_{(i)}}$. Also it has terms like  $N^{(n-1)N+\sum_{i=1}^{n-1}x_{(i)}}$, $N^{(n-2)N+\sum_{i=1}^{n-2}x_{(i)}}$, $\cdots$

If $\prod_{i=1}^n[(N-x_{(i)})!]=\prod_{i=1}^n[(N-y_{(i)})!],\forall N\in\{1,2,\dots\}$, their terms must match. Thus this implies $$\sum_{i=1}^mx_{(i)}=\sum_{i=1}^my_{(i)},\forall m\in\{1,2,\dots\,n\}$$


答:
The largest possible factor of a number will be $\lfloor\sqrt{x}\space\rfloor$ but these numbers have many factors by definition so the largest factor will be much less and one might think it would be the $x^{th}$ root of the number where x is the number of digits but this proves to be far too aggressive with larger values of n.

Your conjecture appears correct. For instance, the largest prime factor of $224403121196654400$ is $37$ and
log $224403121196654400\approx 40.$ A discussion of this and other 'abundant numbers is here.

Title: Seeking upper bound on largest prime factor of colossally abundant numbers

Body:
I recently came across an older post where the OP referred to papers giving upper bounds on the size of the largest prime factor in superabundant and colossally abundant numbers. I asked for links to those papers in the comments. Am still hoping for a response, however maybe it's a good stand-alone question.

Does anyone know of results on the upper bound on the size of the largest prime factor of colossally abundant numbers? It seems that the answer will roughly be slightly less than log(n) but I'd be grateful if there's some known explicit results out there. Thanks.

Edit
The poster of the original question has answered this question. Thanks Ahmad! Here is the original post.


问:
What does commutativity of partial derivatives imply about geometry about the surface of a function?
Suppose we have a function $f(x,y)$ and it has the property that:

$$  \partial_x \partial_y f(x,y) = \partial_y \partial_x f(x,y)$$

What does this imply about the geometry of the surface described the function?

答:
Comment/Remark:

For a continuous surface $z(x,y)$ cross derivative is related to torsional /shear curvature at  a point. The rate of  change of slope taken along parameter x  changing with respect to $y$  and it does not change for surface continuity. Torsional curvature of a line is uniquely defined by either order in exchange for commutativity to hold good.

The commutativity can be looked at from a more general /  functionally comprehensive complex variable viewpoint.

Taking a complex holomorphic differentiable function

$$ \phi(x,y) =  f(x,y)+i g(x,y)$$

On  the first of Cauchy- Riemann relations partial differentiate w.r.t $v$

$$ f_{u}= g_{v };\; f_{u,v}= g_{v,v}\tag1$$

and on the second C-R relation differentiate w.r.t $u$
$$ f_{v}=- g_{u }\; f_{v,u}= -g_{u,u}\tag2$$

Subtracting
$$f_{u,v}-f_{v,u}=0\tag{3.1}$$

$$ g_{u,u}+g_{v,v}=0 \tag{3.2}$$

Since the second equation (3.2) is the Laplace equation that includes differentiability  and smoothness,  (3.1 ) must be also valid along with it, so the order of partial differentiation of $f$ with respect to $ {u,v} $ should not matter.

Similarly

$$ g_{u,v}= g_{v,u}; \; f_{u,u}+f_{v,v} =0 \tag4$$

Here also order of partial differentiation of $g$ also with respect to $ u,v $does not matter.


问:
Is an argument with all false premises and a false conclusion valid?
I have this truth table:

And I'm trying to determine the validity of the argument, which is:

(B v A), (A ^ ¬A) therefore C

I've got one row at the bottom, where (B v A) and (A ^ ¬A) are both false, but C is true.
And the one below that has all false premises and a false conclusion.

Maybe this is a dumb question, but can an argument still be valid even if either (1) all the premises are false but still has a true conclusion or (2) all premises and conclusion are false?

Furthermore, does an argument lacking any row in which all premises are true + conclusion is true automatically mean that the argument is invalid?

I would like to make it clear that I do understand that if a truth table contains a row in which all premises are true but the conclusion is false the argument is invalid.

Thanks.

(Note: the table has (A v ¬A), but it is meant to read as (A ^ ¬A). My bad!

答:
I can't quite make sense of the title and your first question -- true or false where, i.e. under which interpretation? What we need to check is that truth is preserved from the premises to the conclusion under all possible interpretations:

An argument is valid iff

there is no row where all premises are true and the conclusion is false.

We can not determine from a single row with false premises whether the argument is valid or not. Rows where at least one of the premises is false count positive towards the validity, the truth value of the conclusion does not matter in these cases. The only thing that must not happen is for there to be a row where all premises are true but the conclusion is false: We have "if and only if a truth table contains a row in which all premises are true but the conclusion is false the argument is invalid".

This means in particular that if there is no row that makes all premises true to begin with, because the premises are contradictory, then there can be no counterexample. In this case, the argument is (vacuously) valid.

So your argument is valid because there is no counter example where all premises are true but the conclusion is false.


问:
Prove the ratio of convergence implies $\frac{U_{h}(f)-U_{h/b^2}(f)}{U_{h/b^2}(f)-U_{h/b^4}(f)} \approx b^k$
I need to prove the following statement:

Given a $U_h(f)$ which is an approximation with a composite quadrature of an integral $I(f)$, with $h$ being the length of the subintervals and a ratio of convergence $O(h^k)$. Then

$$\frac{U_{h}(f)-U_{h/b^2}(f)}{U_{h/b^2}(f)-U_{h/b^4}(f)} \approx b^k$$

is true.

I'm pretty lost here. I don't know how what kind of information does $O(h^k)$ gives to help prove the statement. Or if I would decompose the left hand side to get to the right hand side. Thanks for the help.

答:
In reality, you do not have enough information to make any progress towards your goal. You need a stronger assumption. Specifically, there must exists an asymptotic error expansion of the form
$$I(f) - U_h(f) = \alpha h^k + O(h^q), \quad h \rightarrow 0, \quad h > 0$$
where $\alpha \not = 0$ is a constant independent of $h$ and $0 < k < q$. Now let $\rho>0$. By replacing $h$ with $\rho h$ we have $$I - U_{\rho h} = \alpha \rho^k h^k + O(h^q),\quad h \rightarrow 0, \quad h > 0.$$
It follows that
$$U_h - U_{\rho h} = \alpha(\rho^k - 1)h^k + O(h^q), \quad h\rightarrow 0, \quad h > 0.$$
By once again replacing $h$ with $\rho h$ we have
$$ U_{\rho h} - U_{\rho^2h} = \alpha (\rho^k-1) \rho^k h^k + O(h^q), \quad h\rightarrow 0, \quad h > 0.$$
It follows that
$$ \frac{U_h - U_{\rho h}}{U_{\rho h} - U_{\rho^2h}} = \frac{\alpha(\rho^k - 1)h^k + O(h^q)}{\alpha (\rho^k-1) \rho^k h^k + O(h^q)} = \frac{1 + O(h^{q-k})}{\rho^k + O(h^{q-k})} \approx \rho^{-k}$$
A common choice of $\rho$ is $\rho = \frac{1}{2}$ which corresponds to halving the length of each subinterval. This allows us to recycle the function values that have already been computed. In this case
$$ \frac{U_h - U_{h/2}}{U_{h/2} - U_{h/4}} \approx 2^k.$$
Your left-hand side corresponds to the choice of $\rho = b^{-2}$ and your right-hand side should have been $b^{2k}$ rather than $b^k$ but this is not a serious issue.

In the case of the trapezoidal rule and a sufficiently smooth integrand the existence of a suitable asymptotic error expansion follows from the Euler-Maclaurian summation formula and $$(k,q) = (2,4).$$ If the integrand is not smooth, then you may still have an asymptotic error expansion, but the exponents $k$ and $q$ are not necessarily integers. Be mindful of the fact that you will never see the numbers $U_h$, only the computed value $\hat{U}_h$ which is affected by rounding errors and subtractive cancellation is an issue when computing $U_h - U_{\rho h}$ for small values of $h$. It is possible to say substantially more about the behavior of your fraction and the impact of rounding errors, but that is is outside the scope of your question.


问:
Partial fractions when numerator polynomial has greater degree than denominator
Consider the 'The Big example' as shown in this site, in it we are tasked to split:

$$ \frac{x^2 +15}{(x^2+3)(x+3)^2} $$

They split it as:

$$ \frac{A_1}{x+3} + \frac{A_2}{(x+3)^2} + \frac{Bx+C}{(x^2+3)}$$

Suppose, the degree of the numerator polynomial was greater than degree of expanded polynomial in numeartor. For example say it was $x^7+15$, it seems clear to me that the above split fails. So, does there exist a split which is applicable even in such cases?

答: If the degree of the numerator is greater than or equal to the degree of the denominator, perform polynomial long division to obtain $$P(x)+\frac{A_1}{x+3}+\frac{A_2}{(x+3)^2}+\frac{Bx+C}{x^2+3},$$ where $P(x)$ is a polynomial.


问:
Are $x^y$ and $x^z$ "Powers" of $x$?
Or would "powers of $x$" in those examples refer to $y$ and $z$, respectively?

答:
In the expression "$x^y$", $y$ is the "exponent"/"power" and $x$ is the "base".

But if $x$ is an arbitrary real number, I would not generally call $x^y$ "a power of $x$", though I might call it "$x$ raised to a power".

"Powers of $n$" where $n$ is a particular positive integer (or rarely, perhaps a positive rational) refers to the values $n^m$ where $m$ is an integer. In an integer-only context, the "powers of $n$" might require $m$ to be nonnegative or even positive. But there are also contexts where we allow $m$ to be a negative integer (the famous film(s) Powers of Ten is one example). Curiously, a phrase like "negative powers of $n$" is common, because context makes it clear that it must be the exponent $m$ which is negative.

It's difficult to provide a definitive source for the meaning of this phrase because it is not often carefully defined in textbooks. The English Wikipedia page for "Power of two" has some references, though.

Title: Are $x^y$ and $x^z$ "Powers" of $x$?


问:
Combinations or Permutations of bits
I am a computer science major and was explaining to someone how a computer uses bits to represent numbers.  If you have $1$ bit, you can have $0$ or $1$.  With $2$ bits, you can have $00, 01, 10, 11$ or $0, 1, 2, 3$ in decimal.  And so on, up to $32$ bits, where you can have about $4.2$ billion.... what?  Combinations or Permutations?

答:
neither of the two, but states

Here's why? say we have 4 bits then the sum of configurations (states) from possible combinations for it would be $$\sum_{n=0}^4 4C_n $$ from $$ \sum_{n=0}^N NC_n  = 2^N $$
$$ where\,N = no.\,of\,bits  \\where\,n = no.\,of\,set\,bits\\and\,C = combination\,operator $$
$$i.e. 4C_0 + 4C_1 + 4C_2 + 4C_3 + 4C_4 = sum\,of\, the\,5\,possible\,combinations =  2^4  states $$
 as per the addition of $$4C_0[1]  = \begin{cases}
0000
\end{cases} \\ \\ 4C_1[4]  = \begin{cases}
0001 \\
0010 \\
0100 \\
1000 \\
\end{cases} \\ 4C_2 [6] = \begin{cases}
0011 \\
0101 \\
0110 \\
1001 \\
1010 \\
1100 \\
\end{cases} \\ 4C_3 [4] = \begin{cases}
0111 \\
1011 \\
1101 \\
1110 \\
\end{cases}\\ 4C_4 [1] = \begin{cases}
1111
\end{cases} $$

to give 16 states in total

Title: Combinations or Permutations of bits


问:
Product measure of two different measures
I have a probability measure space:
$(\mathbb{R},\mathcal{B},Q)$ and random variable $X: \mathbb{R} \to \mathbb{R}$ with $X(\omega) =\omega$ and consequently $P_X=Q$

Now a new probability measure  $((\mathbb{R}^{2},\mathcal{B}^{2})$ with property :
$$ P(A \times B)=Q(A)Q(B)$$ for $A,B \in \mathcal{B}$ i.e the product measure $(P = Q \otimes Q)$ with random variable $Y:\mathbb{R}^{2} \to \mathbb{R}:Y(\omega_{1},\omega_{2})=\omega_{1}$

Showing that $P_X = P_Y=Q$ the only thing that someone can do is to take
the intersection of $P(A) \cap P(B) =P_X \cap P_Y= (\omega_{1},\omega_{1},\omega_{2},\omega_{3},\dots)=\omega_{1}=Q$

or I am wrong since this property

$Y(\omega_{1},\omega_{2})=\omega_{1}$ holds ?

答:
Here is how you can prove that $P_X=P_Y=Q$.

Since $X: \mathbb{R} \to \mathbb{R}$, with $X(\omega) =\omega$, we have $P_X$ is a probability defined on $\{A \subseteq \mathbb{R} : X^{-1}(A) \in \mathcal{B} \} = \mathcal{B}$  and, for every $A \in \mathcal{B}$,
$$P_X(A) = Q(X^{-1}(A)) = Q(A)$$

On the other hand,  $Y:\mathbb{R}^{2} \to \mathbb{R}$, with $Y(\omega_{1},\omega_{2})=\omega_{1}$, we have $P_Y$ is a probability defined on $$\{A \subseteq \mathbb{R} : Y^{-1}(A) \in \mathcal{B}^2 \}= \{A \subseteq \mathbb{R} : A \times \mathbb R \in \mathcal{B}^2 \} = \mathcal{B}$$  and, for every $A \in \mathcal{B}$, $$P_Y(A) = (Q\otimes Q)(Y^{-1}(A)) = (Q\otimes Q)(A \times \mathbb R)= Q(A)Q(\mathbb R)=Q(A)$$
where we used that $Q(\mathbb R)=1$, because $Q$ is a probability.


答:
Maybe this is a dumb question, but can an argument still be valid even if either (1) all the premises are false but still has a true conclusion or (2) all premises and conclusion are false?

Yes. Consider:

All animals are blue

All blue things are mortal

Therefore, all animals are mortal

Yes. Consider:

All animals are blue

All blue things have 56374 legs

Therefore, all animals have 56374 legs

Title: Is an argument with all false premises and a false conclusion valid?

Body:
I have this truth table:

And I'm trying to determine the validity of the argument, which is:

(B v A), (A ^ ¬A) therefore C

I've got one row at the bottom, where (B v A) and (A ^ ¬A) are both false, but C is true.
And the one below that has all false premises and a false conclusion.

Maybe this is a dumb question, but can an argument still be valid even if either (1) all the premises are false but still has a true conclusion or (2) all premises and conclusion are false?

Furthermore, does an argument lacking any row in which all premises are true + conclusion is true automatically mean that the argument is invalid?

I would like to make it clear that I do understand that if a truth table contains a row in which all premises are true but the conclusion is false the argument is invalid.

Thanks.

(Note: the table has (A v ¬A), but it is meant to read as (A ^ ¬A). My bad!


答:
This comes out of Klaus Janich's Topology:

Alternative Definition for Topological Spaces:

A topological space is a pair$(X,\mathscr{A})$ consisting of a set $X$ and a set $\mathscr{A}$ of subsets of $X$ called closed sets such that the following hold:

A1. Any Intersection of closed sets is closed

A2. The union of any two closed sets is closed

A3.X and $\emptyset$ are closed.

Title: The representation of closed set

Body:
We already know that closed set is the complement of open set.

But I want to know the representation of closed set without the fact above in the usual topology.

For instance, the integers $\mathbb{Z}$ is closed due to the fact that $\mathbb{Z}$ can be represented by the union of one point sets $\{n\}$
(I don't want to use the fact that "the union of open intervals $(n, n+1)$ is open").


问:
'Unusual' examples of graded rings?
Almost all examples I have seen of graded rings are some variants of polynomial rings; either localisations or quotients, etc. I am looking for more 'unusual' examples, such as the cohomology example given here: https://en.wikipedia.org/wiki/Graded_ring#Basic_examples . What kind of graded rings do people encounter in surprising places? Are there 'organic' examples where the index monoid is something other than $(\mathbb{N},+)$? I am interested in any examples people might have.

答:
Let's make a little list of interesting graded rings.

The ring of modular forms (https://en.wikipedia.org/wiki/Ring_of_modular_forms). In some cases this is (isomorphic to) a polynomial ring, but in general we only know details about generators of modular forms for congruence groups. And modular forms are nice objects anyways.

Let $k$ be a field and $R = k[x_1, ..., x_n]$ a polynomial ring. Write $x^a = x_1^{a_1}\cdot ... \cdot x_n^{a_n}$, and let $\deg(x^a) = a = (a_1,...,a_n)$. This is the standard multigrading on $R$, and is among other things the setting for an algebraic combinatorics (or combinatorial algebra). Multigraded Hilbert series and Betti numbers are used to assess properties of combinatorial objects such as abstract simplicial complexes.

The associated graded ring (https://en.wikipedia.org/wiki/Associated_graded_ring). Let's look at an example which has applications to solving partial differential equations, and which is mentioned as "multiplicative filtrations" in the wikipedia article (details are taken from Jan-Erik Björk's book Rings of Differential Operators). Let $k$ be a field of characteristic zero, and $R = A_n(k)$ the Weyl algebra of dimension $n$ over $k$. Let $F_i$ be the $k$-subspace of $R$ spanned by elements $x^a\partial^b$ (in the multivariate notation as above) such that $|a| + |b| \leq i$. The filtration $F = (F_i)$ is called the Bernstein filtration on $R$, and one checks that $\cup_i F_i = R$ and that $F_iF_j \subset F_{i+j}$. The vector space $\operatorname{gr}_F(R) = \oplus_i F_i/F_{i+1}$ is in fact a ring, the graded ring associated to $F$. Disappointingly, this graded ring is isomorphic to $k[y_1, ..., y_{2n}]$, but in return one can use its algebraic properties to study solutions to partial differential equations, which is a beautiful marriage of algebra and analysis. Note that $A_n(k)$, as well as many other important rings in this theory, are non-commutative.

For something with negative degree, let $R$ be a graded commutative unitary ring, $S$ a multiplicative subset of homogeneous elements, and $S^{-1}R$ the localisation. Then one can get negative degrees by declaring $\deg(f/g) = \deg(f) - \deg(g)$ on the localisation.

Title: 'Unusual' examples of graded rings?


问:
Formula for a square of any number ending with $5$
In an exercise I'm asked the following:

Note that: $\underline {1}5^2= \underline{2}25,\ \underline {3}5^2= \underline{12}25,\ \underline {8}5^2= \underline{72}25,\ \underline {10}5^2= \underline{110}25$. Find a rule to evaluate the square of any natural number that ends with $5$ and prove it.

So, looking at the parted I noticed the following: We take the number before the final $5$ (let's call it $n$), and then the square will be $n^2 + n$ followed by $25$. For example: $\underline{65}5^2 \to 65^2 + 65 = 4290$ so $655^2 = \underline{4290}25$. Although I found a general formula I'm not being able to prove this. I thought of the following:

Let $n = 5 + \sum_{k=1}^n a_k10^k$. Let's define $\alpha:= \sum_{k=1}^n a_k10^{k-1}$. Then: $n^2 = 100({\alpha}^2+ \alpha) + 25$. I'm not sure if this is the right way to put this pattern, but how can I prove this?

答:
As Arthur pointed out in the comments, all you need is to write $n = 10 \alpha + 5$. Then,

$n^2=(10 \alpha + 5)^2 = 100 \alpha^2 + 100 \alpha + 25 = 100( \alpha^2 + \alpha) + 25.$


答:
The rule for the square of any number ending in $5$ is to take all the digits preceding the $5$, multiply it by itself incremented by $1$, and append to the result the digits $25$.  For example, $$65^2=\underline{6}5^2=\underline{42}25\qquad(6\times7=42).$$

Proof.  A number $n$ ending in $5$ satisfies $$n=5+10k,\quad k\in\Bbb Z.$$
As such, we have
\begin{align}
n&=5+10k\\
n^2&=(5+10k)^2\\
&=25+100k+100k^2\\
&=25+100k(1+k)
\end{align}
as desired.

Title: Formula for a square of any number ending with $5$

Body:
In an exercise I'm asked the following:

Note that: $\underline {1}5^2= \underline{2}25,\ \underline {3}5^2= \underline{12}25,\ \underline {8}5^2= \underline{72}25,\ \underline {10}5^2= \underline{110}25$. Find a rule to evaluate the square of any natural number that ends with $5$ and prove it.

So, looking at the parted I noticed the following: We take the number before the final $5$ (let's call it $n$), and then the square will be $n^2 + n$ followed by $25$. For example: $\underline{65}5^2 \to 65^2 + 65 = 4290$ so $655^2 = \underline{4290}25$. Although I found a general formula I'm not being able to prove this. I thought of the following:

Let $n = 5 + \sum_{k=1}^n a_k10^k$. Let's define $\alpha:= \sum_{k=1}^n a_k10^{k-1}$. Then: $n^2 = 100({\alpha}^2+ \alpha) + 25$. I'm not sure if this is the right way to put this pattern, but how can I prove this?


问:
Partial diffeq from generating function
I have a two-variable recurrence relation which was given to me, and I've been playing around with it. I am familiar with the method of generating functions, but not so much with differential equations. This was one representation that I derived, and while I believe I have something else in one variable that may be more promising, I was wondering if this was a known form:

$$Q(x,y) = x^2\left[\frac{\partial}{\partial x} Q(x,y)\right] +xy \left[\frac{\partial}{\partial y}Q(x,y)\right] + x^2y\left[\frac{\partial}{\partial x} Q(x,y)\right]$$

I have no experience in PDE's, so I have no grounds for anticipating this to be intractable v. well-known.

Thanks!

答:
You can use the Method of Characteristics - Lagrange-Charpit Equations.
You need to solve these differential equations:

$$\dfrac {dx}{x^2(1+y)}=\dfrac  {dy}{xy} =\dfrac {dQ}{1}$$
The first differential equation is separable:
$$\dfrac {dx}{x^2(1+y)}=\dfrac  {dy}{xy} $$
$$\int \dfrac {dx}{x}=\int \dfrac  {1+y}{y} dy $$
$$\ln x= \ln y +y +k$$
$$x=C_1ye^y \implies C_1=\dfrac {x}{ye^y}$$
Now solve the second DE:
$$\dfrac  {dy}{xy} =\dfrac {dQ}{1}$$
$$\int \dfrac  {dy}{y^2e^y} =C_1Q$$
This is not easy to integrate.

Title: Partial diffeq from generating function


问:
Web based app for differential geometry visualization
Is there by any chance a web based app for viewing parametric curves and surfaces that arise in differential geometry to help visualization?
Thanks.</p>

答:
I assume you are talking about curves/surfaces that are immersed in $\mathbb R^2$ or $\mathbb R^3$. More abstract or higher dimensional curves/surfaces would be a bit harder to work with.

Static Plots

As HallaSurvivor mentioned in a comment, Wolfram|Alpha's plotting and graphics capabilities include static plots of parametric surfaces and curves in 3d, or curves in 2d.

2D Animations/Interactives

If you want something more interactive, Desmos's graphing calculator works well for 2D. The Desmos learn page on "parametric equations" links to an example page including a family of parametric curves with an extra parameter $a$ that can be modified with a slider to display different curves. And the Desmos learn page on "points" links to an example page including a parametric curve shown one point at a time where the parameter can be controlled with a slider. Pressing the play button ▶ will have the parameter increase at a fixed speed so the nature of the parametrization can be understood in motion.

3D Animations/Interactives

Finally, the Math3D app has many features similar to Desmos, plus some additional features built in (e.g. vectors), and can handle interactive/animated parametrized curves and surfaces in 3D, as shown in the Math3D example for a ruled hyperboloid (of one sheet).

Edit: Thanks to Semiclassical's comment for reminding me about the GeoGebra 3d calculator which has similar functionality (and can do a lot more with some complicated inputs). Geogebra has curve and surface commands for parametric constructions. You can see both with some sliders in a demonstration by matheagle. You can even code a 2d slider to control a pair of parameters for a surface, as exemplified in Lenore Horner's demonstration.

Programming

I didn't mention this originally because there is a much steeper barrier to entry, but CoCalc has a free plan and could be used for graphing on the web if you're already comfortable with a relevant programming language. See one collection of examples (including a graph of a parametrized surface) using SageMath and one collection (including a parametrized triangulated Möbius strip) using matplotlib in Python.


问:
$f$ is continuous on $\mathbb{R}$ and $f(x) \cdot f(−x)\leq 0$ for all $x$. Is $f$ a onto function and does $f$ have a zero?
Question is:
Given that $f$ is continuous on $\mathbb{R}$ and $f(x) \cdot f(−x)\leq 0$ for all $x$.

Is $f$ always an onto function ?

Does $f$ has zero in all examples?

My thoughts:

Firstly I think $f$ is an odd function so its onto. Can you give an anti example for this? I couldn't find.

Frankly, I don't fully understand this part. What is mean that $f$ has zero? Could you explain?

答:
$f(x)=\sin x$ satisfies the hypothesis, but it is not onto.

Putting $x =0$ we get $0 \leq f(0)^{2} \leq 0$ so $f(0)$ must be $0$.


问:
having trouble intuiting analyticity
My textbook seems to suggest that the analytic functions are precisely the functions that can be written in terms of $z$ alone (no $x$ or $y$ or conjugate-$z$).

Am I inferring correctly?

Does this mean that $\sin (z+x)$ is not analytic?   [where $z=x+iy$]

答: More precisely, an analytic function can be expressed locally in terms of its power series.  Here one must distinguish between two related notions: real analyticity and complex analyticity.  Thus, the function you mentioned is real analytic (of two variables), but is not complex-analytic as a function of $z$.</p>

Title: having trouble intuiting analyticity


问:
Odd perfect numbers and egyptian fraction conjecture
I have reached the conclusion that the non-existence of odd perfect numbers is related to the following

Conjecture Let it be $R=\{d_1,d_2,...,d_n\}$ the set of distinct proper divisors less than $\sqrt{O}$ of some odd positive integer $O$, such that $d_1<d_2<...<d_n$, and such that for some divisor $d_k\in R$ holds that $d_k\leq d_n<3d_k$. Then, it exist no solution to the egyptian fraction $$\frac{2}{d_k}+\frac{1}{d_1}+\frac{1}{d_2}+...=1$$

The process to reach the conjecture is the following (for you to check it is correct):

Let $P$ be some odd perfect number.

Let $R=\left\{ d_{1},d_{2},...,d_{n}\right\}$  be the set of proper divisors of $P$ less than $\sqrt{P}$ excluding $1$, and $T=\left\{ \frac{P}{d_{1}},\frac{P}{d_{2}},...\frac{P}{d_{n}}\right\}$  be the set of proper divisors of $P$ greater than $\sqrt{P}$ excluding $P$.

As $P$ is a perfect number,

$$1+d_{1}+d_{2}+...+d_{n}+\frac{P}{d_{n}}+...+\frac{P}{d_{2}}+\frac{P}{d_{1}}=P (1)$$

Operating,

$$1+d_{1}+d_{2}+...+d_{n}=P-\frac{P}{d_{1}}-\frac{P}{d_{2}}-...-\frac{P}{d_{n}}$$

$$1+d_{1}+d_{2}+...+d_{n}=P\left(1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}\right)$$

$$\frac{1+d_{1}+d_{2}+...+d_{n}}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}=P$$

As $1+d_{1}+d_{2}+...+d_{n}$ is an integer, it follows that $\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}$ must be integer. Therefore, $P$ is a perfect number only if both $1+d_{1}+d_{2}+...+d_{n}$ and $\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}$ are proper divisors of $P$.

It is trivial to show that one of the two expression is less than $\sqrt{P}$, and therefore belongs to $R$, and the other is greater than $\sqrt{P}$ and belongs to $T$.

It is clear that $1+d_{1}+d_{2}+...+d_{n}$ is greater than the greatest element of $R$, so subsequently we can state that

$$\left(\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}\right)=d_{k}\in R (2)$$

$$1+d_{1}+d_{2}+...+d_{n}=\frac{P}{d_{k}}\in T (3)$$

The proof showing that $d_k\leq d_n<3d_k$ is a bit long, so I skip it for the shake of briefness. And operating with $(2)$, the conjecture follows. I have found (thanks to some help provided here in MathStackExchange) examples of odd positive integers for which $(3)$ alone holds, but not any example of an odd positive integer for which $(2)$ holds with the restrictions exposed.

Any help on how to attack this conjecture would be welcomed. Otherwise, if someone is able to show any counterexample to the conjecture (that is, provide some odd positive integer for which $(1)$ holds), it would be more than welcomed to save efforts proving it. I have not seen this conjecture elsewhere, so it is possible that it is mistaken.

Thanks in advance!

Edit

I post the proof showing that $d_k\leq d_n<3d_k$, as it is fundamental for the conjecture and for the solution I have provided for verification.

From (3), and operating,

$$1+\sum_{j=1}^{n}d_{j}=\frac{P}{d_{k}}$$

$$P=d_{k}\left(1+\sum_{j=1}^{n}d_{j}\right)$$

As by definition we have that $d_{n}<\sqrt{P}$, just substituting we can set that

$$d_{n}<\sqrt{d_{k}\left(1+\sum_{j=1}^{n}d_{j}\right)}$$

$$d_{n}^{2}<d_{k}\left(1+\sum_{j=1}^{n}d_{j}\right)$$

$$\frac{d_{n}}{d_{k}}<\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}} (4)$$

Other hand, taking (1) and dividing by $\sqrt{P}$, we get that

$$\frac{1}{\sqrt{P}}+\frac{d_{1}}{\sqrt{P}}+\frac{d_{2}}{\sqrt{P}}+...+\frac{d_{n}}{\sqrt{P}}+\frac{P}{d_{n}\sqrt{P}}+...+\frac{P}{d_{2}\sqrt{P}}+\frac{P}{d_{1}\sqrt{P}}=\sqrt{P}$$

$$\frac{1}{\sqrt{P}}+\frac{d_{1}}{\sqrt{P}}+\frac{d_{2}}{\sqrt{P}}+...+\frac{d_{n}}{\sqrt{P}}+\frac{\sqrt{P}}{d_{n}}+...+\frac{\sqrt{P}}{d_{2}}+\frac{\sqrt{P}}{d_{1}}=\sqrt{P}$$

As $\forall d_{j}\,\frac{\sqrt{P}}{d_{j}}>1$, and $1+\sum_{j=1}^{n}d_{j}>\sqrt{P}$, then we get inmediately that $n<\sqrt{P}-1$; otherwise, the sum of all the terms would be greater than $\sqrt{P}$ and $P$ would not be a perfect number. Thus, we can affirm that

$$n\leq\sqrt{P}-2 (5)$$

As the minimum gap between consecutive elements of $R$ is $2$, the maximum sum of elements of $R$ with the minimum gap between them is

$$d_{n}+\left(d_{n}-2\right)+\left(d_{n}-4\right)+...+\left(d_{n}-2\left(n-1\right)\right)$$

Therefore, we can establish that

$$1+\sum_{j=1}^{n}d_{j}\leq d_{n}+\left(d_{n}-2\right)+\left(d_{n}-4\right)+...+\left(d_{n}-2\left(n-1\right)\right)$$

$$1+\sum_{j=1}^{n}d_{j}\leq nd_{n}-\left(\left(n-1\right)^{2}+\left(n-1\right)\right)$$

Subsequently, we get that

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq\frac{nd_{n}-\left(\left(n-1\right)^{2}+\left(n-1\right)\right)}{d_{n}}$$

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq n-\frac{\left(\left(n-1\right)^{2}+\left(n-1\right)\right)}{d_{n}}$$

Substituting $n$ by the inequality obtained in (5), we get that

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq\sqrt{P}-2-\frac{\left(\left(\sqrt{P}-2-1\right)^{2}+\left(\sqrt{P}-2-1\right)\right)}{d_{n}}$$

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq\sqrt{P}-2-\frac{\left(\left(\sqrt{P}-3\right)^{2}+\left(\sqrt{P}-3\right)\right)}{d_{n}} (6)$$

Operating with the numerator of the third term of the right handside of (6), we get that

$$\left(\sqrt{P}-3\right)^{2}+\left(\sqrt{P}-3\right)=$$

$$P+9-6\sqrt{P}+\sqrt{P}-3=$$

$$P-5\sqrt{P}+6$$

As by definition $d_{n}<\sqrt{P}$, we can affirm that

$$\frac{P-5\sqrt{P}+6}{d_{n}}>\sqrt{P}-5+\frac{6}{\sqrt{P}}$$

Therefore, substituting at (6), we get that

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}<\sqrt{P}-2-\left(\sqrt{P}-5+\frac{6}{\sqrt{P}}\right)$$

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}<3-\frac{6}{\sqrt{P}}$$

Subsequently,

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}<3$$

$$1+\sum_{j=1}^{n}d_{j}<3d_{n}$$

Finally, substituting in (4), we get the desired result

$$\frac{d_{n}}{d_{k}}<3$$

答:
I have been able to find a potential proof of the conjecture, I post it for you to check and tell me if you think it is right:

Let us define set $S=\left\{ 1,2,...,n\right\}$ . Operating with the egyptian fraction, we get that

$$\frac{\sum_{\substack{j=1\\j\neq k}}^{n}\left(\prod_{\substack{s\in S\\s\neq j}}d_{s}\right)+2\prod_{\substack{s\in S\\s\neq k}}d_{s}}{\prod_{s\in S}d_{s}}=\frac{\prod_{s\in S}d_{s}}{\prod_{s\in S}d_{s}} (1)$$

$$\sum_{\substack{j=1\\j\neq k}}^{n}\left(\prod_{\substack{s\in S\\s\neq j}}d_{s}\right)+2\prod_{\substack{s\in S\\s\neq k}}d_{s}=\prod_{s\in S}d_{s}$$

It is easy to see that this implies the following:

For each $d_{j\in S}$, we have that

$$d_{j}\mid\prod_{\substack{s\in S\\s\neq j}}d_{s}$$

This property, considered jointly with the condition $d_k\leq d_n<3d_k$ , has the following direct implication:

$d_{k}$ is some composite number

Proof. If $d_{k}$ was some prime number, as $d_{k}\mid\prod_{\substack{s\in S\\s\neq k}}d_{s}$ and all the proper divisors of $O$ are distinct, then some $d_{s\neq k}$ must be some odd composite number multiple of $d_{k}$. As the minimum possible multiple of $d_{k}$ distinct of $d_{k}$ is $3d_{k}$, then we have that some $d_{s\neq k}\geq3d_{k}$. However, as we have that $d_{n}<3d_{k}$, there can not exist any $d_{s\neq k}\geq3d_{k}$. Subsequently, $d_{k}$ must be composite.

Thus, $d_{k}=d_{i}d_{j}$, where $d_{i}$ and $d_{j}$ can be either prime numbers or composite numbers, but in any case such that $3d_{i}\leq d_{k}$ and $3d_{j}\leq d_{k}$.

Now, we are in a position to prove that

if $d_{k}\leq d_{n}<3d_{k}$, it can not exist any solution to the egyptian fraction $\frac{2}{d_{k}}+\frac{1}{d_{1}}+\frac{1}{d_{2}}+...=1$.

Proof. For each $d_{j\in S}$, dividing each of the terms of (1) by all $d_{s\neq k,j}$, we get that

$$d_{j}d_{k}=2d_{j}+d_{k}+d_{j}d_{k}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right) (2)$$

Operating with (2), we get that

$$d_{j}\left(d_{k}-2\right)=d_{k}\left(1+d_{j}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)\right)$$

$$d_{j}=\frac{d_{k}\left(1+d_{j}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)\right)}{d_{k}-2}$$

As $d_{k}$ and $d_{k}-2$ are odd integers, it follows that $\gcd\left(d_{k},d_{k}-2\right)=1$. As each $d_{j}$ is some odd positive integer, then necessarily for each $d_{j}$ we have one of the following two options: either $d_{k}=3$, or $d_{k}-2$ divides $1+d_{j}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)$. For each $d_{j}$ we can discard the first option, as $3$ is a prime number and $d_{k}$ is some composite number. Looking at the second option left, it can be noticed that $\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}<1$, because precisely the original egyptian fraction states that $\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)+\frac{1}{d_{j}}+\frac{2}{d_{k}}=1$. But this implies that, if $d_{k}-2$ divides $1+d_{j}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)$, then necessarily

$$d_{k}-2<1+d_{j}$$

$$d_{k}-3<d_{j}$$

As $d_{k}-3$ is an even number, then

$$d_{k}-2\leq d_{j} (3)$$

As this must be true for each $d_{j}$, we reach a contradiction between the bound set in (3) and the fact that $d_k$ is composite; according to the bound, $d_{k}$ cannot be some composite number of two other divisors $d_{i}$ and $d_{j}$ and must be some prime number. Subsequently, it follows that the egyptian fraction with the condition $d_k\leq d_n<3d_k$ can not exist.


问:
Determine the probability that no guest will receive the proper hat.
Suppose that 4 guests check their hats when they arrive at a restaurant, and that these hats are returned to them in a random order when they leave. Determine the probability that no guest will receive the proper hat.

My attempt:
So I know that there are 4! ways of "arranging" the hats.
The probability of each receiving their hat is 1/4 and the probability that they will not receive the right hat is 3/4.

I think the answer should be something like:

$\frac{}{4!}$

where the 4! is from the total ways of arranging the hats, and the numerator is the number of "correct" ways to arrange the hats (so no one gets their own hat)But I can't seem to think of how to express that mathematically.

How should I think about this problem instead?

So the book gives the answer 3/8 and myclassmate said the answer is

$\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}=\frac{3}{8}$

But I don't understand why this is?

答:
Sorry for the very late reply, lmao.
Anyhow, the answer is to use derangements: an arrangement such that no elements appear in their original position.
Basically, if you map a hat and its owner, you can find an arrangement such that no hat is mapped to its original owner again, which is just derangement.
It is also useful to know when the question asked it "If at least one person leave with their original hats", that would just be the complement of the derangement and be
(1 - Pr(x)), where x is the probability that no one leaves with their hats.

Using derangement, you can see that 9 arrangements fit the category and since there's 4! available permutation, the result is 9/24 = 3/8


问:
Finding the nth non square free number
We all know what squarefree numbers are. I was more interested in checking an efficient way to find nth non square free number.</p>

答: The $n$-th non square integer is given by the formula $n + [ \sqrt{n} + 0.5],$ where $[•]$ denotes floor function. E.g. $n=1$ yields $2$, $n=2$ yields $3$, $n=3$ yields $5$,...and so on. Is this what you are looking for?

Title: Finding the nth non square free number


答:
This expansion results in $$A_1(x^2+3)(x+3)+A_2(x^2+3)+(Bx+C)(x+3)^2=x^2+15$$ $$\implies(A_1+B)x^3+(3A_1+A_2+6B+C)x^2+(3A_1+9B+6C)x+(9A_1+3A_2+9C)=x^2+15$$

Forcing through $x=-3\implies A_2=2$, we create a system: $$\begin{array}{lcr} A_1 & B & C & |& P \\\hline 1 & 1 & 0 &|& 0 \\ 3 & 6 & 1 & |&-1 \\ 3 & 9 & 6 &|& 0 \\ 9 & 0 & 9 & |&9 \end{array}$$
which Gaussian elimination handles. Notice that my left-hand polynomial has an $x^3$ and $x$ term, but the right-hand does not, so I just consider them as $0$ and solve normally.

Title: Partial fractions when numerator polynomial has greater degree than denominator

Body:
Consider the 'The Big example' as shown in this site, in it we are tasked to split:

$$ \frac{x^2 +15}{(x^2+3)(x+3)^2} $$

They split it as:

$$ \frac{A_1}{x+3} + \frac{A_2}{(x+3)^2} + \frac{Bx+C}{(x^2+3)}$$

Suppose, the degree of the numerator polynomial was greater than degree of expanded polynomial in numeartor. For example say it was $x^7+15$, it seems clear to me that the above split fails. So, does there exist a split which is applicable even in such cases?


问:
How do you show that the language L= { $ a^nb^nc^na^kb^lc^m : k,l,m,n >0 $ } is not a cfg?
I understand how for the language $ a^nb^nc^n $ it is a straightforward application of the pumping lemma, but the part after that seems to exclude the possibility of using the pumping lemma. I would like to know whether there is some way of using any closure property to do the same. Also is there some stronger version of the pumping lemma where it could be applied to some subset of the string rather than the entire string? I mean if we could just apply the pumping lemma on the $ a^nb^nc^n $ out of the $ a^nb^nc^na^kb^lc^m $ ?

答: HINT: If $p$ is the pumping length, apply the pumping lemma to the word $a^pb^pc^pabc$ and pump down.


问:
What does "the partition of $\mathcal{P}(A)$ induced by $R$" mean?
I am working on the following question:

Let $A$ be a non-empty set and fix a subset $B$ of $A$. Define a relation $R$ on the set $\mathcal{P}(A)$ of subsets of $A$ as follows:
$$R = \{(X, Y) | X \cap B = Y \cap B \},$$
where $X$ and $Y$ are subsets of $A$.

a) Show that $R$ is an equivalence relation.

b) If $A = \{ 1, 2, 3 \}$ and $B = \{ 1, 2 \}$ find the partition of $\mathcal{P}(A)$ induced by $R$.

c) If $A = \{ 1, 2, 3, 4, 5 \}$ and $B = \{ 1, 2, 3 \}$ find the equivalence class $[X]$ if $X = \{ 1, 3, 5 \}$.

I couldn't understand the meaning of the part (b). What does it mean?

答: If $A = \{1,2,3\}$, then $\mathcal P(A) = \{\varnothing,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},A\}$. Now, for every $X \in \mathcal P(A)$ (the eight possible choices) write explicitly the elements of the set $$[X]_R := \{Y \in \mathcal P(A) : (X,Y) \in R\} = \{Y \in \mathcal P(A) : X \cap \{1,2\} = Y \cap \{1,2\}\}.$$ Now, the partition of $\mathcal P(A)$ induced by $R$ is the set given by $\{[X]_R : X \in R\}$.


问:
How to show $\lvert e^{i2t\sin\theta}\rvert= e^{-2\Im(t\sin(\theta))}$
First, I say sorry if this might looks obvious for you. I just confused since I can't found something related about this.

Here's the problem :

All I know about modulus is
$$\lvert x+iy\rvert = \sqrt{x^2+y^2}$$. And even if I apply this on
$\lvert e^{i2t\sin\theta}\rvert$, this gives me $1$ as the result.

So, where does $\Im(t\sin(\theta))$ come from? Please explain the step with detail. Thanks. Your little help means a lot to me. Thanks in advance.

And btw, $t>0, t\in \Bbb R$ and $0\leq \theta \leq 2\pi$

答:
If $t$ is any real number then indeed $|e^{it}|=1$, you are right. It follows that if $z=x+iy$ is a complex number then:

$|e^{iz}|=|e^{i(x+iy)}|=|e^{ix}e^{-y}|=e^{-y}=e^{-Im(z)}$

Now, when you checked the result in Wolfram you obviously didn't write that $2t\sin(\theta)$ is a real number, so you got an answer in the general case, where it might be any complex number. After all, if $2t\sin(\theta)$ is real then its imaginary part equals $0$, so the answer still remains the same.