问:
Orbit of a point under the action of the stabiliser of another point
I'm wondering, for a group $G$ acting on a set $X$, what can be said about the orbit of a point under the action of a stabiliser subgroup of $G$. Let $x\in X$ and let $H = {\rm Stab}_G(x)$. Obviously $|H\cdot x|$ is going to be $1$, but what can we say about $|H\cdot y|$ for some other $y\in X$, perhaps in relation to $|G\cdot y|$?  If that not much can be said in general, feel free to put interesting restrictions on $y$, e.g., $y$ in $G\cdot x$.

Edit. I guess the original question was far too vague. I am still interested in the general question, but in the specific examples I am thinking of, $X$ is the set of vertices of a finite graph and $G$ acts by graph automorphisms. Does this extra information tell us more about $H\cdot y$?

答:
There is a lot one can say about $\text{Aut}(X)$ acting on $X$. Much of Algebriac Graph Theory is devoted to this. The Wikipedia page (https://en.wikipedia.org/wiki/Graph_automorphism#Graph_families_defined_by_their_automorphisms) is a good starting point for the highlights.

There are a couple natural families with which to start, including:

Vertex-Transitive Graphs (i.e., Graphs $X$ where $\text{Aut}(X)$ acts transitively on $V(X)$.)

Cayley graphs, which are defined as follows. Let $G$ be a group, and let $S \subset G$ such that $1 \not \in S$. The Cayley graph $\text{Cay}(G, S)$ has vertex set $G$. There is an edge $(g, h)$ iff there exists $s \in S$ such that $gs = h$. If $S = S^{-1}$, then we consider $\text{Cay}(G, S)$ to be undirected (rather than having directed edges going both ways). Otherwise, $\text{Cay}(G, S)$ is directed. Cayley graphs are vertex transitive.

For a comprehensive treatment, I'd start with the Algebraic Graph Theory text by Godsil and Royle.


问:
Why is $d(x^2)= 2xdx$?
This seems a bit obvious, but I'm not clear on one part. My teacher had told us $d(x^2) =2xdx$.
However on asking why, we were told $d(x^2)= (x+dx)^2-x^2= 2xdx+(dx)^2$. We were told that we can ignore the $(dx)^2$ because its small.
For obvious reason I feel this step is wrong, why can we ignore $(dx)^2$, is their any other way of getting the result?

答:
If you think about it in terms of differentials, let $y = x^2$:

$$ dy = \frac{dy}{dx} dx $$

$$ d\left(x^2\right) = \left(\frac{d}{dx}x^2\right) dx $$

$$ d(x^2) = 2xdx $$

ContentLicense: CC BY-SA 4.0

Title: Why is $d(x^2)= 2xdx$?


问:
Limit problem. Do not understand how do I evaluate it.
What do I do with it?

I cannot use L'Hopital's rule to equaliate limits right now, thats what the teacher said.

答: There's no need to that. If you use the Superpolynomials (Go to part 1.2 in page 2 of the following pdf: https://gofile.io/d/neAPjM ) The degree of the Superpolynomial in the denominator is equal to the degree of the Superpolynomial in the numerator. Hence this limit is equal to $\frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}}=\frac{-2}{-1}=2$ this is much simpler than evaluating the limit with L'hopital rule.


问:
The set $\{x \mid x \in \mathbb{Z}, x^2 \leq 25\}$ is equal to $\{0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5\}.$ Why is that so?
I encountered this statement as a multiple choice question in a test. My answer was $\{x \mid x \in \mathbb Z, x \leq 5\text{ or }x \geq -5\}.$ But as seen in the title, the correct answer is $\{0, \pm 1,  \pm 2, \pm 3, \pm 4, \pm 5 \}?$

Can someone please explain why?

答:
I think the issue is the use of "or". Note that $x=7$ satisfies the condition that "$x\geq -5\textbf{ or } x\leq 5$", as $7\geq 5$. When we say "or" we generally mean that either one or both of the conditions is satisfied.

If you had instead written $S=\{ x| x\geq -5 \text{ and } x\leq 5\}$ then any element $z\in S$ would satisfy both $z\geq -5$ and $z\leq 5$ simultaneously.


问:
Show $T$ is compact operator if $\langle Te_n,e_n \rangle$ tend to zero.
Suppose $\mathcal{H}$ is a Hilbert space, and $T\in B(\mathcal{H})$. If for each orthonormal (norm 1) basis $\{e_n\}\subseteq \mathcal{H}$, we have $\langle Te_n, e_n \rangle \rightarrow 0$. Can we deduce $T$ is compact?

I guess it may use spectra decomposition. Take adjoin, use the fact that $\langle Ae_n,e_n\rangle \rightarrow 0$ iff $\langle (A+A^*)e_n,e_n \rangle \rightarrow 0$ and $\langle (A-A^*)e_n,e_n\rangle \rightarrow 0$ , we can assume $T$ is self-adjoint. So $T$ can be viewed as a multiple operator(multiple by a $L^{\infty}(X,d\mu)$ function) on $L^2(X,d\mu)$, via unitary equivalence. Here $d\mu$ is an abstract $\sigma$-finite Borel measure on X. But I don’t know how to use the condition.. And, another way, if we use the spectra decomposition via $T=\int z dE$, E is the corresponding spectra measure. To show $T$ is compact, it is sufficient to prove the projection $E(-\infty,-\epsilon)$ and $E(\epsilon, +\infty)$ are all finite rank, for every $\epsilon>0$. And I feel the condition may can be use together with something like dominate converge theorem? But I fail.
Any help or hint? Thanks.

答:
No. Take the real space $H=l^2$ and define $Tx$ for $x=(x_1,x_2,\dots)$ by
$$
Tx=(x_2,-x_1,x_4,-x_3, \dots).
$$
Then $\langle Tx,x\rangle=0$ for all $x$, but $T$ is bijective and cannot be compact.


问:
Convergence of total variation given uniform convergence of variation in open sets
Let $\Theta$ be a metric space, and let $\{\mu_\theta\}_{\theta \in \Theta}$ be a collection of probability measures over the Borel $\sigma$-algebra $\mathcal B(X)$ of some metric space $(X, \rho)$. Further, suppose that for every $\theta_0 \in \Theta$,
$$\lim_{\theta \to \theta_0}\sup_{A \in \tau} |\mu_{\theta_0}(A) - \mu_\theta(A)| = 0$$
where $\tau$ denotes the collection of open sets in $X$. Does it follow that the total variation
$$\lim_{\theta \to \theta_0}\sup_{A \in \mathcal B(X)} |\mu_{\theta_0}(A) - \mu_\theta(A)| = 0$$
also converges to zero?

Some information that I think may be helpful:

I believe this question can be seen as a generalization of If two Borel measures coincide on all open sets, are they equal?

Rather than coinciding, for any $\theta$ in some sufficiently small neighborhood of $\theta_0$, we have that the measures $\mu_\theta$ almost coincide with $\mu_{\theta_0}$ on all open sets. We need to show that we can find a similar neighborhood where the measures almost coincide on all measurable subsets.

In the above question, it was resolved that if $X$ is a union of an increasing sequence of open sets on which the two measures are finite (which is the case for us), then the answer is "yes" due to the monotone class theorem. I'm not familiar with this theorem, but perhaps a generalization of that argument will work in this case.

答:
An even stronger result holds.

Let $A \in \mathcal B(X)$ and $\theta \in \Theta$.
Because any finite Borel measure on a metrizable topological space is outer regular, for each $n \in \mathbb N$ there is an open $U_n \supset A$  such that $\mu_{\theta_0}(A \triangle U_n)<1/n$ and an open $V_n \supset A$ such that $\mu_\theta(A \triangle V_n)<1/n$. Let $W_n = U_n \cap V_n \in \tau$. Then,

\begin{align}
|\mu_{\theta_0}(A) - \mu_\theta(A)| &\leq \mu_{\theta_0}(A \triangle W_n) + \sup_{W \in \tau}|\mu_{\theta_0}(W) - \mu_\theta(W)| + \mu_\theta(A \triangle W_n)\\
&\to_{n \to \infty} \sup_{W \in \tau}|\mu_{\theta_0}(W) - \mu_\theta(W)|.
\end{align}

It follows from this that
$$\sup_{A \in \mathcal B(X)}|\mu_{\theta_0}(A) - \mu_\theta(A)| \leq \sup_{A \in \tau}|\mu_{\theta_0}(A) - \mu_\theta(A)|,$$
and the reverse inequality is trivial, so
$$\sup_{A \in \mathcal B(X)}|\mu_{\theta_0}(A) - \mu_\theta(A)| = \sup_{A \in \tau}|\mu_{\theta_0}(A) - \mu_\theta(A)|.$$


问:
Solution of an SDE
Can someone guide me on how to solve the SDE

\begin{equation}
dX_{t} = \gamma(a-\beta X_{t})dt + \delta X_{t}dW_{t}
\end{equation}
where $a,\beta,\gamma,\delta$ are all positive constants? I tried to approach like an Ornstein-Uhlenbeck SDE but I don't think it is the right way.

答:
Another Idea can be:
Take $g(x,t)=e^{\gamma\beta t}\times X$,apply It$\hat o$ dif. rule.
$$dg=\frac{\partial g}{\partial t}dt+\frac{\partial g}{\partial x}dX+\frac 12\frac{\partial^2 g}{\partial X^2}(dX)^2\\
dg=\gamma\beta e^{\gamma\beta t}Xdt+e^{\gamma\beta t}dX+\frac12(0)\\$$now put $dX_{t} = \gamma(a-\beta X_{t})dt + \delta X_{t}dW_{t} $so
$$dg=\gamma\beta e^{\gamma\beta t}Xdt+e^{\gamma\beta t}dX\\
dg=\gamma\beta e^{\gamma\beta t}Xdt+e^{\gamma\beta t}\big(\gamma(a-\beta X_{t})dt + \delta X_{t}dW_{t})\big)\\dg=e^{\gamma\beta t}\gamma adt+e^{\gamma\beta t}\delta X_{t}dW_{t}$$left hand side is complete differential ,so apply integration to both sides
$$\int dg=\int e^{\gamma\beta t}\gamma adt +\int e^{\gamma\beta t}\delta X_{t}dW_{t}\\g(x,t)-g(x_0,0)=\gamma a\int_0^t e^{\gamma\beta s}ds+\delta \int_0^t e^{\gamma\beta s} X_{s}dW_{s}\\
e^{\gamma\beta t}\times X-X_0=\gamma a\int_0^t e^{\gamma\beta s}ds+\delta \int_0^t e^{\gamma\beta s} X_{s}dW_{s}\\
e^{\gamma\beta t}\times X-X_0=\gamma a(\frac{1}{\gamma\beta})(e^{\gamma\beta t}-1)+\delta \int_0^t e^{\gamma\beta s} X_{s}dW_{s}\\
e^{\gamma\beta t}\times X=X_0+(\frac{a}{\beta})(e^{\gamma\beta t}-1)+\delta \int_0^t e^{\gamma\beta s} X_{s}dW_{s}\\\times e^{-\gamma\beta t} \\
X=X_0e^{-\gamma\beta t}+\frac{a}{\beta}(1-e^{-\gamma\beta t})+\delta e^{-\gamma\beta t}\int_0^t e^{\gamma\beta s} X_{s}dW_{s}$$

ContentLicense: CC BY-SA 3.0

Title: Solution of an SDE


问:
Compute $v,W,k$ such that the following is true
$$
\left\{ x \in \mathbb{Z}^4 |
\begin{pmatrix}
5 & 3 & 7 & 0 \\
2 & -4 & 6 & 5
\end{pmatrix}
x =
\begin{pmatrix}
5 \\0
\end{pmatrix}
\right\} = \left\{ v + Wy \ | \ y \in \mathbb{Z}^k \right\}
$$

I denoted $x=(x_1,x_2,x_3,x_4)^T$ and tried expressing $x_1$ and $x_2$ in terms of $x_3$ and $x_4$ but i could not proceed. I am studying for linear programming (optimization) exam and i found this exercise. Any suggestions?

答:
Hints (if I guessed correctly your notation and part of your problem):

$$\begin{pmatrix}5&3&7&0\\
2&\!\!-4&6&5\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}=\binom 50\iff\begin{cases}5x_1+3x_2+7x_3=5\\{}\\2x_1-4x_2+6x_3+5x_4=0\end{cases}$$

Two particular solutions to this system are

$$\begin{pmatrix}\;15\\\;\;0\\\!\!-10\\\;\;6\end{pmatrix}\;,\;\;\begin{pmatrix}\;\;0\\\!\!-10\\\;\;5\\\!\!-14\end{pmatrix}$$

Now I'd need to know what  $\,v,y,W\,$ are expected to be and whether you insist all is done over a ring that is not a field (this makes things very hard!) . Also, is it $\,k=4\,$ in the rightmost set?


问:
Trignometric functions and their polynomial forms
The trigonometric values also has a infinite series which I had learnt from the topics of Limits

Sin x = x - x³/ 3! + x⁵/ 5!.......

There are formulas like these for other trigonometric functions too. But here's where I don't understand. If we substitute 'π' instead of 'x' , then by using this formula , we get sin π not zero when it is zero. Similarly for all sin values of nπ ,where n ∈ Z get into the same trap.

sinπ = π - π³/3! + π⁵/5!...

0    = π - π³/3! + π⁵/5! ...

L.H.S ≠ R.H.S

This shouldn't happen, right ?

Where did I go wrong, then ?

Body:
Compute the partial sums
$$S_p=\sum_{n=0}^p (-1)^n \frac{\pi^{2n+1}}{(2n+1)!}$$ For $p=4$, $S_4=0.00692527$, for $p=5$, $S_5=-0.00044516$, for $p=6$, $S_6=0.0000211426$.


问:
Scaling a Poisson distribution
Is it possible to scale a Poisson distribution and receive the same result. Lets say that I have bridge A. On average 10 cars drive over bridge A per hour, thus if I want to calculate the probability that at most 4 cars drove over bridge A after a given hour I would take the Poisson CDF with lambda 10, in R ppois(4,10), which is roughly 3%. However, lets say instead I would like to see the probability that 2 cars drove over the bridge in 30 minutes instead, same methodology as above i.e. ppois(2,5) which gives roughly 12.5%.

I would initially think that you would be able to scale, but thinking about it, does this happen as with fewer instances there is a larger chance that relatively speaking more events deviate?

答:
The answer to your question is twofold.

Clarifying cdf: The cumulative distribution function (cdf) is a probability of a series of events. In R, ppois(k1,rt) is the dpois(k,rt) summed over all k<=k1. Here, r is the rate of the event per unit time t. Therefore, cdf is a sum of probabilities.

Hence, probability that two cars drove over the bridge in half an hour represented by the probability mass function dpois(2,5) (not the cdf ppois()).

Sum of Poisson distributed random variables: First consider the probability that exactly 4 cars pass through a bridge where on average 10 cars pass. This is represented by dpois(4,10).
When split into half hour intervals, the average rate drops to 5 per half hour. However, the two half hour intervals can combine in several combinations to create the 4 cars in one hour: 0,4 1,3 2,2 3,1 4,0 respectively in fist half and the second half hour. Therefore, the probabilities for all these half-hour events combine to equate the former one hour event. dpois(4,10)=dpois(0,5)*dpois(4,5)+dpois(1,5)*dpois(3,5)+dpois(2,5)*dpois(2,5)+dpois(3,5)*dpois(1,5)+dpois(4,5)*dpois(0,5).

Title: Scaling a Poisson distribution


问:
Length of nested tuples
Assume we have the following tuple of tuples: S = ((1,2), (2,3,4,5), (4,5,6,7,8)).

There are three nested tuples. To write the length of S (the number of nested tuples), is it correct notation to write $\lvert S \rvert = 3$?

How do one then write the notation for the length of a nested tuple? $\lvert S_i \rvert, i=1,2,..,\lvert S \rvert$? Examples:

$\lvert S_1 \rvert = 2$

$\lvert S_2 \rvert = 4$

$\lvert S_3 \rvert = 5$

答: This notation is not frequently used to denote the length of a tuple. Now it depends where you want to define "length of a tuple". In linear algebra, analysis,$\cdots$ we usually define a natural number $n\in\mathbb{N}$ to denote dimension of Euclidean space $\mathbb{R}^{n}$ or for a vector space $V$ we sometimes use $n=\dim(V)$

Title: Length of nested tuples


答:
Question: solve

\begin{align}\ln x&=\tfrac12(x-1)\tag{1}\label{1}\end{align}

$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$

Equation \eqref{1} has one obvious root $x=1$.
For the other roots we can divide \eqref{1} by $(1-x)$ to get

\begin{align}
\frac{\ln x}{(1-x)}&=-\tfrac12
\tag{2}\label{2}
,
\end{align}
a recognizable form of
parametric representation of the real branches $\Wp,\Wm$ of the Lambert $\W$ function,
which has only one solution $x>1$:

\begin{align}
x&=\frac{\Wm\left(-\tfrac12\exp(-\tfrac12)\right)}
{\Wp\left(-\tfrac12\exp(-\tfrac12)\right)}
=\frac{\Wm\left(-\tfrac12\exp(-\tfrac12)\right)}
{-\tfrac12}
=-2{\Wm\left(-\tfrac12\exp(-\tfrac12)\right)}
\approx 3.5128624
\tag{3}\label{3}
.
\end{align}

$\endgroup$

Title: Find the two solutions of $\log x=\frac{1}{2}(x-1)$

Body:
Question: solve $\log x=\frac{1}{2}(x-1)$

With the following I only get one solution (apparently with $W_0$), but I can't find the solution with $W_{-1}$ this way. What is my mistake?

$$\log x=\frac{1}{2}(x-1)$$
$$\log x-\frac{1}{2}x=-\frac{1}{2}$$
$$e^{\log(x)-\frac{1}{2}x}=e^{-\frac{1}{2}}$$
$$xe^{-\frac{1}{2}x}=e^{-\frac{1}{2}}$$
$$-\frac{1}{2}xe^{-\frac{1}{2}x}=-\frac{1}{2}e^{-\frac{1}{2}}$$
$$-\frac{1}{2}x=W(-\frac{1}{2}e^{-\frac{1}{2}})$$
$$x=-2W(-\frac{1}{2}e^{-\frac{1}{2}})$$


问:
Can a quadratic equation not equal to zero
Well I wanted to know whether or not $y = x^2 + x + 7$ is a quadratic equation since the general form is $ax^2 + bx + c = 0$ here the equation
$y=x^2+x+7$. Isn't equal to zero so I'm a bit confused

答:
It could be taken as both a relationship between $y$ and $x$ or as a quadratic equation, depending on the context of the problem.

Most of the time the form $y=ax^2+bx+c$ is used to indicate the relationship between $y$ and $x$. This is called is a quadratic function. But if it is given that $y$ is a constant, then you could simply rewrite it as $ax^2+bx+(c-y)=0$ and this is a quadratic equation.

The difference between these two is that $y$ in a quadratic function could have any value depending on the value of $x$. Whereas $y$ in a quadratic equation is independent of $x$.

Hope that clears it up.


答: When treating the differential operator like a fraction, you might see this:  $$\frac d {\color{blue}{dx}}(x^2)=2x\implies d(x^2)=2x\,\color{blue}{dx}.$$

Title: Why is $d(x^2)= 2xdx$?

Body:
This seems a bit obvious, but I'm not clear on one part. My teacher had told us $d(x^2) =2xdx$.
However on asking why, we were told $d(x^2)= (x+dx)^2-x^2= 2xdx+(dx)^2$. We were told that we can ignore the $(dx)^2$ because its small.
For obvious reason I feel this step is wrong, why can we ignore $(dx)^2$, is their any other way of getting the result?


问:
Solve equation $\sqrt{s+13} - \sqrt{7-s} = 2$
Solve the equation

$$\sqrt{s+13}-\sqrt{7-s} = 2$$

I moved the $-\sqrt{7-s}$ to the right side

Thus, I had
$$\sqrt{s+ 13} = 2 +\sqrt{7-s}$$

I then squared both sides
$$\sqrt{s+ 13}^2 = \left(2 +\sqrt{7-s}\right)^2$$

Using the product formula $(x + y)^2 = x^2 + 2xy + y^2$

I got  $$s + 13 = 4 + 4\sqrt{7-s}+  7 – s$$

I then combined like terms
$$2s + 2= 4 \sqrt{7-s}$$

I’m stuck at this point. Does anyone have an idea how to solve this equation?

答:
From where you left off, squaring both sides again yields

$(2s + 2)^2 = (4 √(7-s))^2$ Which becomes

$4s^2 +8s +4 = 16(7-s)$

And you have a quadratic to solve, take all terms to one side and factorise or use the quadratic formula to find solutions to $s$, there will be two solutions of course. And then plug values of $s$ found back into your original equation to make sure you dont have a negative value under the square root.


答: We have $\Delta(x^2) = 2x \Delta x + (\Delta x)^2$. Note that $x^2 = y$ so, we have $\Delta(x^2) = \Delta y$. Thus, we have $\Delta y = 2x\Delta x +(\Delta x)^2$. Divide both sides by $\Delta x$ and we get $\frac{\Delta y}{\Delta x} = 2x + \Delta x$. Now, we take the limit as $\Delta x$ approaches $0$, giving us $dy/dx = 2x$. Note that $dy/dx$ is the limit as $\Delta x$ goes to $0$ of $\Delta y/\Delta x$ by definition.

Title: Why is $d(x^2)= 2xdx$?


问:
How do you find the inverse of $\ -log(\frac{x}{y})$?
There's this equation in chemistry called the hasselbalch equation. I looked at how it was derived, and it isn't making sense to me. $\ pKa -log(\frac{x}{y})$ equals to $\ pKa + log(\frac{y}{x})$, which confuses me because I thought the inverse would be $\ pKa -\frac{1}{log(\frac{x}{y})}$. Where did I go wrong, and could you please show me how to do it correctly.

This is my first time using MathJax (:

答: You're confusing additive and multiplicative inverses, and in any case you didn't change the overall sign, so all you've done is replace $a-b$ with $a-b^{-1}$ instead of $a+b^{-1}$ (which would still be wrong, but would at least result from confusing one "inverse" of $b$ with another). Note$$-\log\tfrac{x}{y}=-(\log x-\log y)=\log y-\log x=\log\tfrac{y}{x}.$$


问:
Mathematics language/syntax/grammar Cheat sheet
As the hack fraud mathemagician that I am, I use mathematics typically only in engineering related uses...It is almost always practically applied...and once it gets large (or tedious) enough I am almost always using software to solve different problems. ODE's, PDE's, linear algebra and etc such as mathematica or rarely matlab.

However as of recently, (particularly in writing for others), but also when having to learn a new subject, or sub-sub-subject of a particular field I am often confronted with a wall of new things to read and attempt to understand.

While I surely know many puzzle like solving methods to apply basic mathematical structures/rules to solve a particular equation or apply some method, I find myself always forgetting basic mathematical syntax. Often times the problem I have is simply not knowing how to actually read and correctly understand a statement written in the mathematics grammar/syntax/symbology more so than the actual use or application of such systems.

Is there a particularly nice 'cheat sheet' of sorts for just mathematics symbols and their syntax/use?

I mean things simply as the correct use of $\setminus$ or $\in$ or the the actual meaning of {}, (), []
or (] used in context, Relationship symbols, or others such as $\Rightarrow$ or more, less frequent symbols such as $\forall$ (for me less frequent that is...as I understand this symbol isn't just shorthand for "for all" and must also be used in a specific way) or $\exists!$

I'm not looking for a cheatsheet on how to write said symbols in LaTeX, nor something like this websites fairly nice set of mathematics cheatsheets , but one that says or explains, in english what they mean with context of example, this answer Is sort of in the direction I'd prefer, though it's more far more in depth, than a simple sort of cheat sheet I'd be prefer to find.

$$a \in R  \setminus\{0\} \text{ Represents a is element of R excluding 0}$$

With a continued list of other meanings and etc.

I'd appreciate if anyone is aware of such a thing, or maybe a tiny syntax handbook (an actual hand book, not the incredibly useful but unwieldy 2kg Bronstein mammoth). I would love to become more proficient in mathematics and less hacky in its use, however I certainly need to get up to speed in actually reading it, before that can happen.

Thanks for the help.

答:
I believe that this source may help you
https://en.wikipedia.org/wiki/Glossary_of_mathematical_symbols
Also, this widget allows you to draw a symbol. Once you have identified what the symbol is called, you can search it up to learn more about it. http://detexify.kirelabs.org/classify.html</p>


答:
You're almost at the end result:

After achieving $ 2s + 2 = 4 \sqrt{7-s} $, square both sides again to generate a quadratic equation.

$$ 4 ( s + 1 )^2 = 16 ( 7 - s ) = 4 (s^2 + 2s+ 1)$$
Solving which is quite easy.

$$\begin{align}
s^2 + 2s + 1 &= 28 - 4s \\
s^2 + 6s - 27 &= 0 \\
(s + 9) (s - 3) &= 0
\end{align}$$

NOTE

You have to neglect $ s = -9 $ from the solutions as it'll give you $ \sqrt{-9 + 13} - \sqrt{ 7 - (-9)} = \sqrt{16} = +2 - (+4) \neq 2 $

Title: Solve equation $\sqrt{s+13} - \sqrt{7-s} = 2$

Body:
Solve the equation

$$\sqrt{s+13}-\sqrt{7-s} = 2$$

I moved the $-\sqrt{7-s}$ to the right side

Thus, I had
$$\sqrt{s+ 13} = 2 +\sqrt{7-s}$$

I then squared both sides
$$\sqrt{s+ 13}^2 = \left(2 +\sqrt{7-s}\right)^2$$

Using the product formula $(x + y)^2 = x^2 + 2xy + y^2$

I got  $$s + 13 = 4 + 4\sqrt{7-s}+  7 – s$$

I then combined like terms
$$2s + 2= 4 \sqrt{7-s}$$

I’m stuck at this point. Does anyone have an idea how to solve this equation?


问:
Solve a Second Order Differential Equation
how would I solve the following:

$$
v_{yy} + \frac{1}{2}(y-2t)v_y + \frac{1}{2}v = 0
$$

where we treat $t$ as a constant and $v(y)$

There are no initial values.

The solution says to "integrate once" and then the equation becomes a Riccati equation.
I don't understand how the integration is done.

Body:
$$v_{yy} + \frac{1}{2}(y-2t)v_y + \frac{1}{2}v = 0$$
$$2v_{yy}  -2tv_y +yv_y+v = 0$$
$$2v_{yy}  -2tv_y +(yv)' = 0$$
Then integrate since $t$ is a constant.
$$2v_{y}  -2tv +yv = C$$


问:
What would $\min(f(x))$ do?
Would $\min(f(x))$ give one the minimum extremum of the function, or, would it give the $y$-value of the function with the smallest $x$-value possible? Or would it do neither? Is this notation even allowed?

答: Altough not so common, $\min(f(x))$ would mean the minimum value of $f(x)$, i.e smallest $y$-value as you called it, while $\underset{x}{\operatorname{argmin}} (f(x))$ is the x value that will give you that smallest value. You can think of it as a key-value relation. Though keep in mind, that when in doubt, be specific and write exactly what you mean.

Title: What would $\min(f(x))$ do?


问:
Help with Linear Algebra Question, Don't Understand Different Notations
I have a Homework Problem I'm working on and I really don't recall my Linear Algebra very well. I can't get started on it because I don't understand a lot of the notation.

What does the bold capital M mean?

Is there any significance to saying that the coordinate system is rotated?

What does the prime or apostrophe signify?

What does the -1 subscript mean?

Any tips for getting started on this?

答:
Bold letters are often used to denote matrices. In this case, $\bf{M}$ is the same matrix that appears earlier on that line: $\bf{M}=\begin{pmatrix}a&b\\c&d\end{pmatrix}$. Similarly, $\bf{R}$ in the following line is the $2 \times 2$ matrix with the trig functions. Looks like they forgot to make it bold, but don't let that throw you off. There's no difference here between $R$ and $\mathbf{R}$, it's just an aesthetic typesetting choice.

This is a broad question; you might want to ask it separately.

Primes typically indicate that a variable is somehow related to the non-prime version; here, $x'$ is like writing $x_\textrm{new}$. The notation is more or less arbitrary; they could have used $\hat{x}$, for example.

It denotes the inverse of the matrix.

Practise multiplying some matrices; go back to the fundamentals of linear algebra; learn how inverses work. It might be informative to try some simple values to see what's actually happening: if $\bf{M}=\begin{pmatrix}2&0\\0&1\end{pmatrix}$, for example, then plot a few points $\begin{pmatrix}x\\y\end{pmatrix}$ as well as their image $\bf{M}\begin{pmatrix}x\\y\end{pmatrix}$ and see what $M$ is actually doing. What about $M=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$? And so on.

Good luck!


答:
The bold capital $M$ represents the matrix.

I am confused with this question. What do you mean significant? In reference to what?

The prime is way of denoting the image of a transformation (or the new values as a consequence of the transformation).

The -1 refers to the inverse matrice.

If you have trouble with these symbols, you may want to go back and review linear algebra as a whole. To get started on the notation, you can visit this link: http://linear.ups.edu/html/notation.html

Good luck!

Title: Help with Linear Algebra Question, Don't Understand Different Notations

Body:
I have a Homework Problem I'm working on and I really don't recall my Linear Algebra very well. I can't get started on it because I don't understand a lot of the notation.

What does the bold capital M mean?

Is there any significance to saying that the coordinate system is rotated?

What does the prime or apostrophe signify?

What does the -1 subscript mean?

Any tips for getting started on this?


答:
$$2s+2=4\sqrt{7-s}$$
Divide by $2$ both sides
$$s+1=2\sqrt{7-s}$$
Square both sides
$$s^2+1+2s=28-4s$$
Put everything on the left
$$s^2+6s-27=0$$
Now solve by radicals
$$s_{1,2}=\frac{-6\pm\sqrt{6^2-4(-27)}}{2}$$
giving you the two solutions
$$s_1=3\qquad;\qquad s_2=-9$$

Finally discard $s_2$ since it is not a solution of your title equation


问:
Showing that $\sum_{r=1}^{N} \exp\left(\frac{2\pi irs}{N}\right)=0$ when $0<s<N$
I am trying to show that

$$\sum_{r=1}^{N} \exp\left(\frac{2\pi irs}{N}\right)=0$$ when $0<s<N$. This result is stated in a paper by A. Turing (see page $39$). I can see how it is a geometric progression (as suggested by Turing), but do not understand how the result follows.

答:
$\begin{align*}
\sum_{r=1}^{N}\exp(\frac{2i\pi rs}{N}) &= \sum_{r=1}^{N}{\exp(\frac{2i\pi s}N)}^{r}\\
&=\underbrace{\exp(\frac{2i\pi s}N)}_{= \alpha}\sum_{r=0}^{N-1}{\exp(\frac{2i\pi s}N)}^r\\
&= \alpha \frac{1-\alpha^N}{1-\alpha} \text{ because } \alpha \neq 1 \text{ as } 0 < s < N
\end{align*}$

Now, we see that $\alpha^N = {\exp(\frac{2i\pi s}N)}^N = \exp(2i\pi s) = 1$ because $s$ is an integer, so the sum is indeed equal to $0$.

Of course, other interpretations given in the comments, like geometric considerations of how this amounts to compute some kind of barycenter are completely valid, but usually require a bit more "hand-waving" and might be hard to understand if you're not too familiar with how complex numbers/the complex plane work. The "bruteforce" computation of the geometric sum, on the other hand, requires little visualisation or intuition, and is (in my opinion) easier to generalise to similar problems.


问:
show that : $\mathcal{S}\cap\mathcal{B}(\mathbb{R})$ is the smallest $\sigma$-algebra on $\mathbb{R}$ making the function $x \mapsto x^2$ measurable.
Let $\mathcal{S}$ be the set of symmetrical subset of $\mathbb{R}$ $\big($ i.e. $\mathcal{S}:= \{A\subset\mathbb{R}:\forall x \in\mathbb{R}~~,~~ x \in A \Leftrightarrow -x \in A\}$$\big)$.

Problem:

I want to show that : $\mathcal{S}\cap\mathcal{B}(\mathbb{R})$ is the smallest $\sigma$-algebra $\mathcal{A}$ on $\mathbb{R}$ making the function $f:x \mapsto x^2$ from $\mathbb{R}$ to $\mathbb{R}$, $(\mathcal{A}, \mathcal{B} (\mathbb{R}))$ - measurable.

My effort

We have $$\sigma \big(f^{-1}(\mathcal{B} (\mathbb{R})\big)=\sigma\bigg(f^{-1}\bigg(\sigma\bigg(\big\{]-\infty,q]:q\in\mathbb{Q}\big\}\bigg)\bigg)\bigg)=\sigma\bigg(\bigg\{f^{-1}\big(]-\infty,q]\big):q\in\mathbb{Q}\big\}\bigg)\bigg)\bigg)=\sigma(\{[-q,q]:q\in\mathbb{Q}\})$$

Then $\mathcal{A}=\sigma(\{[-q,q]:q\in\mathbb{Q}\})$ is the smallest $\sigma$-algebra on $\mathbb{R}$ making the function $f:x \mapsto x^2$ from $\mathbb{R}$ to $\mathbb{R}$, $(\mathcal{A}, \mathcal{B} (\mathbb{R}))$ - measurable.

Is what I wrote true? If so, you can help me show that $\sigma(\{[-q,q]:q\in\mathbb{Q}\})=\mathcal{S}\cap\mathcal{B}(\mathbb{R})$?

An idea please.

答:
There is a much simpler approach:

We want to show that if $\mathcal{M}$ is a $\sigma$-algebra which makes $x^2$ measurable, then $\mathcal{S} \cap \mathcal{B} \subseteq \mathcal{M}$.

So let $E \in \mathcal{S} \cap \mathcal{B}$. We want to show $E \in \mathcal{M}$ too.

Well, there's only one fact we know about $\mathcal{M}$ (it renders $x^2$ measurable) and one fact we know about $E$ ($x \in E \iff -x \in E)$, so we have to use those facts somehow.

Now $x^2$ being $\mathcal{M}$-measurable means that $(x^2)^{-1}(B) \in \mathcal{M}$ for any borel set $B$. This is useful, since we want to show $E \in \mathcal{M}$. If we can we show that $E = (x^2)^{-1}(B)$ for some borel set $B$ then we're done!

The dream now is to write

$$
E =
(x^2)^{-1}(B) =
\{ x ~|~ x^2 \in B \} =
\{ \pm \sqrt{x} ~|~ x \in B \}.
$$
But here is where we use our assumption on $E$! There is an obvious reason that as set wouldn't be of this form: this set is symmetric. But thankfully we know that $E$ is symmetric too, so this is not an issue!

Reversing the definition above shows $B = \{ x ~|~ \sqrt{|x|} \in E \}$ works. Do you see why $E = (x^2)^{-1}(B)$? More importantly, do you see why this set $B$ is borel? (Here it is important that $E$ is borel too, so we use our last remaining assumption.)

So at the end of the day, if $E \in \mathcal{S} \cap \mathcal{B}$, we've seen that $E = (x^2)^{-1}(B)$ for some borel $B$ as defined above. But then $E \in \mathcal{M}$ too since $x^2$ is $\mathcal{M}$-measurable.

I hope this helps ^_^


问:
Distance between highest and lowest point on rotated rectangle
I have a rectangle and I want to know the vertical distance between the highest point and the lowest point. The rectangle is 3×8 cm, and is rotated by $\theta$.

So for example in this picture I would need to know the length of the blue line (H, F):

Does anyone know a formula to calculate this?

答:
In the figure, the height is equal to $AG + BH$.

We have $AG = AB \cos \theta$ and $BH = BC \sin \theta$.


问:
Atomless?: product measure of an atomless measure and a measure which has atoms
I wonder whether the product measure of an atomless measure and a measure which has some atoms is atomless.

Let us recall that given a measure space $(X, \mathscr A, \mu)$, the set $A \in \mathscr A$ is an atom of $\mu$ if $\mu(A)>0$ and every measurable subset $B \subset A$ has measure either $0$ or $\mu(A)$, and a measure that has no atoms is called atomless.

答:
No.  Trivial example: suppose $\mu$ is Lebesgue measure and $\nu(S)=\begin{cases}0&|S|=\omega\\\infty&|S|\geq\omega_1\end{cases}$.  Then $\mu\times\nu=\nu$ (morally — the product is defined by the same formula), which is atomic.

However, if one requires $\mu$ and $\nu$ to be ($\sigma$-)finite measures, then the product is indeed atomless.

Let $\mu$ be atomless and $\nu$ arbitrary, and write $\lambda=\mu\times\nu$.  We will show $\lambda$ lacks atoms, by transfinite induction on the rank over the generators.1  For this induction, it is useful to adopt a slightly stronger-appearing definition of atomlessness:

$\lambda$ is atomless if, for any $0<\delta<\epsilon<1$ and measurable $A$, there exists $B\subseteq A$ such that $\delta\lambda(A)\leq\lambda(B)\leq\epsilon\lambda(A)$.

Note that either definition of non-atomicity is preserved under finite unions/intersections/complements, so the only challenge is countable ones.  Proving that the definitions are equivalent is a bit tricky, but luckily Wikipedia's page on atomic measures includes a sketch, so I'll defer to them and just jump straight to the induction.

For a base case, suppose $A=B\times C$ for some $B$, $C$.  If $A$ is null, the claim is trivial, so suppose $0<\lambda(A)=\mu(B)\nu(C)$, i.e. $\mu(B)>0$.  Since $\mu$ is atomless, there exists $D\subseteq B$ with relative measure at most between $\delta$ and $\epsilon$; taking Cartesian products preserves relative measure, so $D\times C$ witnesses that $A$ is not atomic.  Thus no rank-0 sets are atomic.

If $\alpha$ is a limit ordinal, then any set of rank at most $\alpha$ in fact has lower rank, so it remains to show the successor ordinal case.

Suppose all rank-$\alpha$ sets are not atoms and $A$ is rank-$(\alpha+1)$ and has positive measure.  Either $A=\bigcup_{j\in\omega}{A_j}$ or $A=\bigcap_{j\in\omega}{A_j}$, where each $A_j$ has rank at most $\alpha$.  Since our definition of atomic is symmetric under taking complements, it suffices to assume the former.  Moreover, we can disjointify the $\{A_j\}_j$: let $\tilde{A}_0=A_0$ and $\tilde{A}_J=\left(\bigcup_{j=0}^J{A_j}\right)\setminus\tilde{A}_{J-1}$ for $J>1$.  This doesn't preserve rank, but it does preserve atomicity (see above).  By hypothesis, each $A_j$ is not atomic; we now take them disjoint, so that $\lambda(A)=\sum_{j\in\omega}{\lambda(A_j)}$.

Fix $\delta$ and $\epsilon$: for each $j$, there exists $B_j\subseteq A_j$ such that $\delta\lambda(A_j)\leq\lambda(B_j)\leq\epsilon\lambda(A_j)$.  By disjointness, taking a union in $j$ adds measures, so that $\delta\lambda(A)\leq\lambda\left(\bigcup_{j\in\omega}{B_j}\right)\leq\epsilon\lambda(A)$, as desired.

1 A quick reminder of rank for $\sigma$-algebras, where I adopt the slightly nonstandard convention that taking complements does not increase rank: let $\mathcal{F}$ be a $\sigma$-algebra generated by $\mathcal{G}$.  Define the (transfinite) sequence \begin{align*}
\mathcal{F}_0&=\mathcal{G}\cup\{G^{\mathsf{c}}:G\in\mathcal{G}\} \\
\mathcal{F}_{\alpha+1}&=\mathcal{F}_{\alpha}\cup\left\{\bigcup_{j\in\omega}{F_j}:\forall j\,F_j\in\mathcal{F}_{\alpha}\right\}\cup\left\{\bigcap_{j\in\omega}{F_j}:\forall j\,F_j\in\mathcal{F}_{\alpha}\right\} \\
\mathcal{F}_{\alpha}&=\bigcup_{\beta<\alpha}{F_{\beta}}\quad\quad\alpha\textrm{ limit ordinal}
\end{align*}  It is a fact that $\mathcal{F}=\mathcal{F}_{\omega_1}$ (exercise!).

If $F\in\mathcal{F}_{\alpha}$, then we say $F$ has rank (over $\mathcal{G}$) at most $\alpha$; the rank of $F$ is the minimal such $\alpha$.


问:
show that the hypothesis $\mu(X) < \infty$ can not be omitted
Let $(X,\mathfrak{M},\mu)$ be a measure space and $\mu(X)<\infty$ and $\{f_n\}$ be a sequence of bounded complex measurable functions on $X$, and $f_n\rightarrow f$ uniformly on $X$  then  $\lim_{n\to\infty}\int_X f_n d\mu =\int_X f d\mu$.

show that  the  hypothesis $\mu(X)  < \infty$  can not be omitted

My attempt : I was  thinking about constant function  $f_n= c $

Is there any other example ?

答:
Notation: $M$ is $\mathfrak M.$

Even if $\int_X|f|d\mu\le 1$ for all $n$: Take Lebesgue measure on $\Bbb R.$ For $n\in \Bbb N$ let $f_n(x)=0$ if $x\not \in [n,2n]$ and let $f_n(x)=1/n$ if $x\in [n,2n].$

In general suppose $\mu$ is atomless, i.e. if $A\in M$ and $\mu(A)>0$ then there exists $B\in M$ with $B\subset A$ and $0<\mu(B)<\mu(A).$ Now if $\mu(X)=\infty$ then there is a set $\{B_n:n\in\Bbb N\}\subset M$  with $n\le\mu(B_n)<\infty,$ so let $f_n=\frac {\chi_{B_n}}{\mu(B_n)}.$


问:
Curve fitting: $a f_1(x)+b f_2(x)+c f_3(x)+d f_4(x)$
I want to fit my data $y$ with a method such as
$$a f_1(x)+b f_2(x)+c f_3(x)+d f_4(x)$$
However I don't know how to obtain coefficients $(a,b,c,d)$ of the known functions $(f_1(x),f_2(x),f_3(x),f_4(x))$. I don't know least squares very well. Therefore I want a matrix formula that I can get the coefficients for any data easily, without getting into learn least squares since I don't have time. Could you please help me?</p>

答:
Let's do least squares, as it gives you an equation as wished. $\def\R{\mathbb R}\def\Mat{\mathop{\mathrm{Mat}}\nolimits}$Let $A \in \Mat_{n,4}(\R)$ and $y \in \R^n$ be given by
$$ A = \begin{pmatrix} f_1(x_1) & f_2(x_1) & f_3(x_1) & f_4(x_1) \\ & &\vdots & \\ f_1(x_n) & f_2(x_n) & f_3(x_n) & f_4(x_n)\end{pmatrix}, y =\begin{pmatrix} y_1 \\ \vdots \\ y_n\end{pmatrix} $$
where $(x_i, y_i)$, $1 \le i \le n$ are your data. We are asked to minimize
$$ A \def\x{\begin{pmatrix} a \\b\\c\\d\end{pmatrix}}\x - y $$
The this is - supposing that $A$ has full rank - exactly the case this residual in orthogonal to the image of $A$, that is, iff
$$ A^tA \x = A^t y, $$
so
$$ \x = (A^tA)^{-1}A^ty. $$</p>


问:
Maximum of a convex function in an interval
Let $f:[a,b]\rightarrow \mathbb{R}$ a convex function. Does $f$ have a maximum in $[a,b]$?.

I have thought that if $f$ is not continous at $x=a$ or $x=b$ then I can find a counterexample for which it does not exist a global maximum. The same thing for minimum. Is it right?

答:
If $f$ is convex, then it is continuous (and therefore upper semicontinuous) on $(a,b)$.

I claim that $f$ is upper semicontinuous at $a$ and $b$. If $x \to a$ from the right, we can write $x = \lambda a + (1-\lambda) b$, then by convexity
$$f(x) \leq \lambda f(a) + (1-\lambda) f(b),$$
and therefore
$$\limsup_{x \to a} f(x) \leq \limsup_{\lambda \to 1}(\lambda f(a) + (1-\lambda) f(b)) = f(a).$$
A similar argument works for $x \to b$ from the left.

Since $f$ is upper semicontinuous on the compact set $[a,b]$, it attains a maximum. (Intuitively, the idea here is that if $f$ is discontinuous at an end point, the discontinuity must be an "upward jump" in order to preserve convexity.)

But $f$ need not attain a minimum. For example, consider
$$f(x) = \begin{cases} (x-b)^2, \ \ x \in [a,b)\\ 1, \ \ x=b.  \end{cases}$$


问:
Example of a non-nilpotent linear transformation that for every $v$ gives $A^n(v)=0$ for some $n$
The question is as follows:
Give an example of a non-nilpotent linear transformation for some vector space with the following property, for every $v \in V$ there is an $n$ for which $A^n(v)=0$.

I really don't know how such a matrix would exist, since you can find a basis for this vector space, let's say $v_1,v_2,...$. Then we know for every $v_i$ there is an $n_i$ for which $A^{n_i}(v_i)=0$.

Then write $v$ as $\sum_{i}^{k} \alpha_i v_i$ and consider $A^{\sum_{i}^{k} n_i} (v)$, which gives $\sum_{i}^{k} \alpha_i \cdot A^{\sum_{i}^{k} n_i} (v_i)$ which should give zero for every $v_i$.

Maybe they meant an infinite vector space, however I doubt it since we never touched the infinite vector spaces in our algebra course.

Then you can consider the basis $B=v_1,v_2, ...$ an infinite basis over some field and let $A(v_1)=0$ and $A(v_i)=v_{i-1}$ then since there exist online finite amount of non-zero coordinates then there must be an $n$ for which $A^n(v)=0$, however the transformation is not nilpotent.

So my question is whether it's really true that such a transformation doesn't exist for finite-dimensional vector spaces.

答: The vector space is neceessarily infinite dimensional. Consider the space of all real sequences $(x_n)$ such that $x_n=0$ for $n$ sufficiently large. Define $A(x_n)=(x_2,x_3,..)$. If $x_n=0$ for $n>N$ then $A^{N}(x_n)=0$. But, for any $N$,  $Ae_{N+1} \neq 0$ so $A$ is not nilpotent. [$e_i$ is the sequence which has $1$ in the $i-$th coordinate and $0$ elsewhere].


问:
Area of shaded triangular region from absolute value functions
So, I got this question from my teacher. I've tried solving it but to no avail. I cannot work out a method either, which can help me solve this.

So, it gives us two absolute value functions
$g(x) = 4|x-3|+3$
$f(x) = -6|x-3|+9$

There are two points, A & B as shown in the figure given below.
I don't understand, when I find the points of intersection by solving
$g(x) = f(x)$
then what are those $x$ values. Are they the coordinates of point A and B?
And what would be the next steps leading to the complete solution of this question.

I'd appreciate if anyone points out the steps or just solves it for my ease.

Question FIGURE

答:
From your diagram, we can see that we must use $A,B$, and the vertex of $f(x)$ to calculate the area of our triangle. $A, B$ are the intersection points of $f(x)$ and $g(x)$.

Thus, we set $g(x) = f(x)$ to solve for our intersection points, and we get $4|x-3| +3 = -6|x-3|+9$. Simplifying, we get $10|x-3| = 6$, so $|x-3| = 0.6$. The two values that have an absolute value of $0.6$ are $0.6 $ and $-0.6$. Thus, we have $x-3 = 0.6$ or $x-3 = -0.6$. Solving for $x$, in each case, we get $x=2.4, 3.6$.

We plug these values into either $f(x)$ or $g(x)$, it doesn't matter because both functions are equal at our intersection points. So, we have $g(2.4) = 4|2.4-3| +3 = 5.4$ and $g(3.6) = 4|3.6-3|+3=5.4$. Thus, our intersection points are $(2.4,5.4)$ and $(3.6,5.4)$.

Now, we have to find the vertex of $f(x) = -6|x−3|+9$. Looking at our diagram, we see that the vertex of $f(x)$ is when it is the biggest. Since we are subtracting $-6|x−3|$ in $f(x)$, we try to make it as small as possible to make $f(x)$ as big as possible. Since absolute value is never negative, the smallest we can make it is $0$. We achieve $-6|x−3|=0$ when $x=3$. So, the vertex of $f(x)$ has $x$-coordinate $3$, and the $y$-coordinate is just $f(3) = 9$ because we made $-6|x−3|=0$. Thus, the vertex of $f(x)$ is $(3,9)$.

So, we have all the points of our triangle $(2.4,5.4), (3.6,5.4), (3,9)$. The base of our triangle is between the points $(2.4,5.4)$ and $(3.6,5.4)$ and the distance between them is $3.6-2.4 = 1.2$. So $b=1.2$. The height of our triangle is the distance from $(3,9)$ to our base. The $y$-coordinate of $(3,9)$ is $9$, and the $y$-coordinate of our base is $5.4$. The difference between the $y$ is $3.6$, so height $h = 3.6$. Use area of triangle formula $A =bh/2$ to get $A = 1.2*3.6/2 =2.16$. Thus, the area of the shaded traingle is $2.16$.


问:
Topological conjugacy between linear maps in dimension $1$
Let $A_{\alpha}: \mathbb{R} \to \mathbb{R}$ denote the linear map

\begin{align}
A_{\alpha}(x)=\alpha x.
\end{align}

By definition two homeomorphisms $f:X \to X$ and $g:X \to X $ are topologically conjugate to each other if there is a homeomorphism $h:X \to X $ such that $hf=gh$.

I wish to prove that if $ 0 < \alpha < 1$ and $ 0 < \beta < 1 $ then $A_{\alpha}$ and $A_{\beta}$ are topologically conjugate. Could anyone help me find this homeomorphism?

答:
You want to find $h$ such that $h(\alpha x)=\beta h(x)$. By symmetry, we may as well work on $[0,\infty)$.

Can you find $\gamma$ for which $h(x)=x^\gamma$ works? (Which means $h(x)=\mathrm{sgn}(x)|x|^\gamma$ on all of $\mathbb{R}$.)


问:
finding the mean in normal distribution
I am struggling on this question where I need to find the mean given the standard deviation.
The question is the following:

```<code>A certain brand of flood lamps has a length of life that is normally distributed,
with a standard deviation of 392 hours.
If 8 % of the lamps last more than 5600 hours, what is the mean length of life?
</code>```

This is what I've tried so far

let $X$ = length of light

$X\sim\mathcal{N}(\text{mean = unknown }(u), SD = 392)$

$P(X > 5600) = 0.8$

$1 - P(X < 5600) = 0.8$ (stuck after this part)

$1 - P(5600-u/392 < 5600) = 0.8$ ( incorrect)

I am having troubles working backwards to the find the mean.
Could someone point me in the right direction.

答:
$$\mathbb{P}[X>5600]=8\%=0.08$$

$$\mathbb{P}[X\leq 5600]=92\%=0.92$$

that is

$$\Phi\Bigg[\frac{5600-\mu}{392}\Bigg]=0.92$$

Now using Z-table you get

$$\frac{5600-\mu}{392}=1.4051$$

that is

$$\mu=5049$$

How to read a Z-table

I used a calculator. With Paper -table you get that the quantile corresponding to the probability of $0.92$ is between $1.40$ and $1.41$, more or less exactly in the middle...

Title: finding the mean in normal distribution


答:
Define a grid of $m$ values $x_i$ corresponding to your data $y_i$. Then you get an overdetermined system of linear equations:

$$\mathbf{Ax}=\mathbf{y}$$

where $\mathbf{A}$ is a $m\times 4$ matrix with elements $a_{ij}=f_j(x_i)$, $\mathbf{x}=[a,b,c,d]^T$ is your vector of unknowns, and the vector $\mathbf{y}$ contains your data. You can solve this system in a least squares sense. Depending on the software you use, you can either directly solve it like
$$\tt{x=A \backslash b}$$
in Matlab or Octave. Otherwise you need to compute the pseudo-inverse and obtain the least squares solution like this:

$$\mathbf{x}=(\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T\mathbf{y}$$

Title: Curve fitting: $a f_1(x)+b f_2(x)+c f_3(x)+d f_4(x)$

Body:
I want to fit my data $y$ with a method such as
$$a f_1(x)+b f_2(x)+c f_3(x)+d f_4(x)$$
However I don't know how to obtain coefficients $(a,b,c,d)$ of the known functions $(f_1(x),f_2(x),f_3(x),f_4(x))$. I don't know least squares very well. Therefore I want a matrix formula that I can get the coefficients for any data easily, without getting into learn least squares since I don't have time. Could you please help me?</p>


问:
Complex conjugation map is not properly homotopic to the identity map of complex plane
Definition: A map $f:X\to Y$ between two topological spaces $X$ and $Y$ is said to be proper if $f^{-1}(K)\subseteq_\text{compact} X$ for every $K\subseteq_\text{compact} Y$. A homotopy $H:X\times [0,1]\to Y$ is said to be proper homotopy if $H$ is a proper map.

Problem: Consider the plane $\Bbb R^2$ and two maps $\text{Id},\overline{\text{Id}}:\Bbb R^2\to \Bbb R^2$ defined as
follows $\text{Id}(x,y)=(x,y)$ and $\overline{\text{Id}}(x,y)=(x,-y)$
for all $(x,y)\in \Bbb R^2$. Does there exist a proper homotopy
$H:\Bbb R^2\times [0,1]\to \Bbb R^2$ such that $H(-,0)=\text{Id}$ and
$H(-,1)=\overline{\text{Id}}$?

Note that any two maps of $\Bbb R^2=\Bbb C$ are homotopic via a convex combination. For example $G:\Bbb R^2\times [0,1]\to \Bbb R^2$ defined by $G\big((x,y),t\big)=\big(x,(1-2t)y\big)$ is a homotopy between $\text{Id}$ and $\overline{\text{Id}}$, but $G$ is not a proper homotopy.

Note also that, $\overline{\text{Id}}\circ \overline{\text{Id}}=\text{Id}$. Now, if we use compactly-supported cohomology then $H^2_\text{c}(\Bbb R^2;\Bbb Z)=\Bbb Z$ and we can talk about degree of a proper map: $\deg_\text{c}(\text{Id})=1$ and $\deg_\text{c}(\overline{\text{Id}})\cdot\deg_\text{c}(\overline{\text{Id}})=1$ i.e. $\deg_\text{c}(\overline{\text{Id}})=\pm 1$. Note that $\deg_\text{c}$ is invariant under proper homotopy. My guess is that $\deg_\text{c}(\overline{\text{Id}})=-1$, but I have no proof. Is there any other approach?

Thank you.

答:
Suppose we have a proper homotopy $f_t:\mathbb C\times I\rightarrow \mathbb C$ with $f_0(z)=z$ and $f_1(z)=\bar z$.

We can extend this to a homotopy of the Riemann sphere $\mathbb C\mathbb P^1$ by defining $f_t(\infty)=\infty$ for all $t\in I$. Then continuity of $f_t:\mathbb C\mathbb P^1\times I\rightarrow \mathbb C\mathbb P^1$ follows from the properness of our original homotopy. But now $f_0$ is the identity and $f_1$ is a reflection of the Riemann sphere, so they have degrees $1$ and $-1$, respectively. This is a contradiction because degree is a homotopy invariant.


答:
\begin{align}
X &\sim \mathcal{N}(\mu, 392^2) \\
P(X \le 5600) &= 0.92 \\
P \left( Z \le \frac{5600 - \mu}{392} \right) &= 0.92 \tag{normalize} \\
\mu \approx 5600 - 392 (1.41) &= 5047.28 \tag{check standard normal table}
\end{align}

P.S. My answer used $\approx$ instead of $=$ as the value is only an approximation.

Title: finding the mean in normal distribution

Body:
I am struggling on this question where I need to find the mean given the standard deviation.
The question is the following:

```<code>A certain brand of flood lamps has a length of life that is normally distributed,
with a standard deviation of 392 hours.
If 8 % of the lamps last more than 5600 hours, what is the mean length of life?
</code>```

This is what I've tried so far

let $X$ = length of light

$X\sim\mathcal{N}(\text{mean = unknown }(u), SD = 392)$

$P(X > 5600) = 0.8$

$1 - P(X < 5600) = 0.8$ (stuck after this part)

$1 - P(5600-u/392 < 5600) = 0.8$ ( incorrect)

I am having troubles working backwards to the find the mean.
Could someone point me in the right direction.


问:
Bounded Sequence help
A series $\{S_n\}$ is defined as follows:
$$S_n = \sum_{n=1}^\infty \frac{1}{3^n+n}.$$

Evidently this sequence is increasing as for every $a_n = (3^n+n)^{-1}$ is less than $a_{n+1}$.

However, how do you show that the sequence is bounded? Bounded below is simple as since it is an increasing function, it is bounded below by $S_1$. But what about bounded above? The calculator shows about $0.39$ as the upper bound.

答:
Welcome to MSE!

Hint:

Since $3^n \leq 3^n + n$, we have the following inequality (working term by term):

$$\sum \frac{1}{3^n + n} \leq \sum \frac{1}{3^n}$$

Do you know how to show this new sum is bounded? In fact, you can compute the new sum exactly and see it's finite.

I hope this helps ^_^

Title: Bounded Sequence help


问:
How to prove that $\lim_{n\to\infty}\frac{n^k}{3^{\sqrt{n}}}=0$
I know we can apply the L'Hospital's rule repeatedly $k$ times but $3^{\sqrt{n}}$ becomes unwieldy while taking derivative $k$ times. Is there another way to solve this?

答:
Lemma: a single application of l'Hospital's Rule shows that when $b>1$, then $\lim\limits_{x\to\infty}\frac{x}{b^x}=0$. (This can be shown easily without l'Hospital as well.)

Now using $a_n=\frac{n^k}{3^{\sqrt{n}}}$, consider

$$
\lim_{n\to\infty}\sqrt[2k]{a_n}=\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt[2k]{3}^{\sqrt{n}}}
=\lim_{x\to\infty}\frac{x}{\left(\sqrt[2k]{3}\right)^{x}}=0
$$

Since $\lim\limits_{n\to\infty}\sqrt[2k]{a_n}=0$, it follows that $\lim\limits_{n\to\infty}a_n=0$.


问:
Is the set $A = \{1,1/2,1/3,\ldots,1/n,\ldots\}$ open or closed
I don't think it is closed because $0$ is a limit point of the set (by the Archimedean principle) but $0$ is not in $A$ but $A$ is clearly not open, does than mean it's not open nor closed. Is that even possible?

答:
Yes, that is deffinitely possible. A set can be both closed, open, open and closed and neither.

The only relation between the two is that the complement of a set is closed if and only if that set is open, and the complement of a set is open if and only if that set is closed.

This can be checked as a theorem in Abbott's book "A set O is open if and only if $O^c$ is closed. Likewise, a set $F$ is closed if and only if $F^c$ is open."


问:
Proving $x^n$ is continuous for all $n$
I'm trying to prove that $f(x) = x^n$ is continuous for all $n$. One option is the Binomial theorem, but it seems slightly messy, so I've opted for induction. Here is my attempt.

Let $a \in \mathbb{R}$ be given and induct on $n$. If $n=1$, given $\epsilon = 0$, set $\delta = \epsilon$. Then $|x-a| < \delta$ implies $|f(x) - f(a)| = |x - a| < \delta = \epsilon$, so $f(x)$ is continuous at $a$ and thus on $\mathbb{R}$.

If $n = 2$, we need to show that $f(x) = x^2$ is continuous at $a$. This is where I am having trouble, because the scratchwork falls apart. If $|x-a| < \delta$, then
$$|f(x) - f(a)| = |x^2 - a^2| = |(x+a)(x-a)| = |x+a||x-a| < \delta |x+a|.$$
$\delta$ can depend on $a$, but not on $x$, so I don't know what kind of additional restrictions I can place on $\delta$.

The induction is straightforward from here, but only if I don't use the definition, but rather the fact that $x^k$ is continuous by assumption, as is $x$, and the product of two continuous functions is continuous. I'd prefer to do this from the definition, however, as this seems to trivialize the result. I could have proved that $x^2$ is continuous just by using the $n=1$ case.

Help would be appreciated.

答:
Hint

Continuity at $0$ is simple as for $\vert x \vert \le 1$ your have $\vert x^n \vert \le \vert x \vert$.

And following your idea of induction to prove continuity at $a \neq 0$

$$\vert x^{n+1} - a^{n+1} \vert = \vert (x-a)(x^n + x^{n-1}a + \dots ++ a^{n}) \vert \le \vert x - a \vert (n+1)2^n \vert a \vert^n$$ for $ \vert x \vert \le 2 \vert a \vert$.

You can apply that to the case $n \ge 2$. Based on the inequality, proof of continuity by induction is straightforward.

Title: Proving $x^n$ is continuous for all $n$


答: Let $V$ be the space of all polynomial functions from $\Bbb R$ into $\Bbb R$ and let $A\bigl(p(x)\bigr)=p'(x)$. If $\deg p(x)=n$, then $A^{n+1}\bigl(p(x)\bigr)=0$, but $A$ is not nilpotent: for any $n\in\Bbb N$,$$A\left(x^{n+1}\right)=(n+1)!\ne0.$$

Title: Example of a non-nilpotent linear transformation that for every $v$ gives $A^n(v)=0$ for some $n$

Body:
The question is as follows:
Give an example of a non-nilpotent linear transformation for some vector space with the following property, for every $v \in V$ there is an $n$ for which $A^n(v)=0$.

I really don't know how such a matrix would exist, since you can find a basis for this vector space, let's say $v_1,v_2,...$. Then we know for every $v_i$ there is an $n_i$ for which $A^{n_i}(v_i)=0$.

Then write $v$ as $\sum_{i}^{k} \alpha_i v_i$ and consider $A^{\sum_{i}^{k} n_i} (v)$, which gives $\sum_{i}^{k} \alpha_i \cdot A^{\sum_{i}^{k} n_i} (v_i)$ which should give zero for every $v_i$.

Maybe they meant an infinite vector space, however I doubt it since we never touched the infinite vector spaces in our algebra course.

Then you can consider the basis $B=v_1,v_2, ...$ an infinite basis over some field and let $A(v_1)=0$ and $A(v_i)=v_{i-1}$ then since there exist online finite amount of non-zero coordinates then there must be an $n$ for which $A^n(v)=0$, however the transformation is not nilpotent.

So my question is whether it's really true that such a transformation doesn't exist for finite-dimensional vector spaces.


问:
Suppose $CA=I_n$ (the $n \times n$ identity matrix. Show that the equation $A\textbf{x}=\textbf{0}$ has only the trivial solution
Suppose $CA=I_n$ (the $n \times n$ identity matrix. Show that the equation $A\textbf{x}=\textbf{0}$ has only the trivial solution.

The textbook solution is as follows:

If x satisfies $A\textbf{x} = \textbf{0}$, then $CA\textbf{x} = C\textbf{0} = \textbf{0}$ and so $I_n\textbf{x} = \textbf{0}$ and $\textbf{x} = \textbf{0}$. This shows that the equation $A\textbf{x} = \textbf{0}$ has no free variables. So every variable is a basic variable and every column of A is a pivot column.

Now I get that $I_n\textbf{x} = \textbf{0}$ would mean $\textbf{x}$ must be the trivial solution since there are no free variables and you have a pivot position at every column. However, I don't get how that implies that  $A\textbf{x}=\textbf{0}$ has the trivial solution since $A \neq I_n$. I don't quite get how $C$ comes into play for solving this problem.

答: The second sentence in the texbook solution is incomplete. What they have proved is $Ax=0$ implies $x=0$ so no vector other than $0$ is  a solution.


答:
Let $f$ and $g$ be two functions on some open domain $D$, continuous at $x_0\in D$, and let $h=fg$. Let's prove $h$ is continuous at $x_0$.

For any $\varepsilon>0$, let $\varepsilon'=\dfrac{\min(\sqrt{\varepsilon},\varepsilon)}{3\max(1,|f(x_0)|,|g(x_0)|)}$ (to simplify later).

Then since $f$ is continuous at $x_0$ there is a $\delta_1>0$ such that $|x-x_0|\le\delta_1\implies |f(x)-f(x_0)|\le\varepsilon'$.

Likewise, since $g$ is continuous at $x_0$ there is a $\delta_2>0$ such that $|x-x_0|\le\delta_2\implies |g(x)-g(x_0)|\le\varepsilon'$.

Now let $\delta=\min(\delta_1,\delta_2)$, we have, for $|x-x_0|\le\delta$,

$$|f(x)g(x)-f(x_0)g(x_0)|=|(f(x)-f(x_0))(g(x)-g(x_0))\\+f(x_0)(g(x)-g(x_0))+g(x_0)(f(x)-f(x_0))|\le \varepsilon'^2+|f(x_0)|\varepsilon'+|g(x_0)|\varepsilon'\le\varepsilon$$

Hence $fg$ is continuous at $x_0$.

Now, since $x\to1$ and $x\to x$ are trivially continuous everywhere, you can prove by induction that $x\to x^n$ is continuous everywhere, for all $n\in\Bbb N$.

Title: Proving $x^n$ is continuous for all $n$

Body:
I'm trying to prove that $f(x) = x^n$ is continuous for all $n$. One option is the Binomial theorem, but it seems slightly messy, so I've opted for induction. Here is my attempt.

Let $a \in \mathbb{R}$ be given and induct on $n$. If $n=1$, given $\epsilon = 0$, set $\delta = \epsilon$. Then $|x-a| < \delta$ implies $|f(x) - f(a)| = |x - a| < \delta = \epsilon$, so $f(x)$ is continuous at $a$ and thus on $\mathbb{R}$.

If $n = 2$, we need to show that $f(x) = x^2$ is continuous at $a$. This is where I am having trouble, because the scratchwork falls apart. If $|x-a| < \delta$, then
$$|f(x) - f(a)| = |x^2 - a^2| = |(x+a)(x-a)| = |x+a||x-a| < \delta |x+a|.$$
$\delta$ can depend on $a$, but not on $x$, so I don't know what kind of additional restrictions I can place on $\delta$.

The induction is straightforward from here, but only if I don't use the definition, but rather the fact that $x^k$ is continuous by assumption, as is $x$, and the product of two continuous functions is continuous. I'd prefer to do this from the definition, however, as this seems to trivialize the result. I could have proved that $x^2$ is continuous just by using the $n=1$ case.

Help would be appreciated.


问:
Poisson Approximation to calculate the probability that at most 2 persons in 500 will have a birthday on Christmas
This is from Hoel's Probability book. Use the Poisson Approximation to calculate the probability that at most 2 persons in  500 will have a birthday on Christmas. Assume 365 days in the year.
On average $\frac{500}{365}=1.36$ of people have a birtday every day.
This is what I've got at the moment:
$$ P\left(X\leq2\right)=\sum_{x=0}^{2}\frac{e^{-\mu}\mu^{x}}{x!}=f\left(0\right)+f\left(1\right)+f\left(2\right)$$

$$P\left(X\leq2\right)=0.2566+0.3490+0.1186=0.7242$$

答:
Following @DanielAdams' Comment:

Binomial probability: $X \sim \mathsf{Binom}(n=500, p=1/365),$  This is exact assuming independence and
365 equally likely birthdays.

Find $P(X \le 2) = 0.8317.$ [Computation in R.]

```<code>pbinom(2, 500, 1/355)
[1] 0.8316545
</code>```

Poisson probability: $ Y \stackrel{aprx}{\sim}\mathsf{Pois}(\lambda = 500/365).$ This is an approximation.

Find $P(Y \le 2) = 0.8407.$

```<code>ppois(2, 500/365)
[1] 0.8407328
</code>```

The three Poisson probabilities summed are as follows:

```<code>dpois(0:2, 500/365)
[1] 0.2541418 0.3481394 0.2384517
</code>```

Note: In R pbinom is a binomial CDF, ppois is a Poisson CDF, and dpois is a Poisson PDF.

Body:
Following @DanielAdams' Comment:

Binomial probability: $X \sim \mathsf{Binom}(n=500, p=1/365),$  This is exact assuming independence and
365 equally likely birthdays.

Find $P(X \le 2) = 0.8317.$ [Computation in R.]

pbinom(2, 500, 1/355)
[1] 0.8316545

Poisson probability: $ Y \stackrel{aprx}{\sim}\mathsf{Pois}(\lambda = 500/365).$ This is an approximation.

Find $P(Y \le 2) = 0.8407.$

ppois(2, 500/365)
[1] 0.8407328

The three Poisson probabilities summed are as follows:

dpois(0:2, 500/365)
[1] 0.2541418 0.3481394 0.2384517

Note: In R pbinom is a binomial CDF, ppois is a Poisson CDF, and dpois is a Poisson PDF.


问:
Looking for an intuitive explanation for a basic combinatorial identity
I'm looking for an intuitive explanation for this identity:
$${n \choose k} = \frac{n}{k}{n-1 \choose k-1}$$
for $0 < k \leq n$.

The math adds up, but I can't see why it's true. I'd expect that choosing $k$ elements from an $n$-set would be like choosing $k-1$ elements from an $(n-1)$-set, then add an $n$th element to the set, and choose another element from the $n - k + 1$ not-yet-chosen elements. But I guess I'm missing something.

Can anyone provide some intuition?

答: It is equivalent to $${n\choose r}{r\choose 1}={n\choose 1}{n-1\choose r-1}$$


答:
$$ A\mathbf{x} = \mathbf{0} $$

$$ \Rightarrow CA\mathbf{x} = C\mathbf{0} $$

$$ \Rightarrow I_n\mathbf{x} = C\mathbf{0} $$

$$ \Rightarrow \mathbf{x} = C\mathbf{0} $$

$$ \Rightarrow \mathbf{x} = \mathbf{0} $$

Body:
Suppose $CA=I_n$ (the $n \times n$ identity matrix. Show that the equation $A\textbf{x}=\textbf{0}$ has only the trivial solution.

The textbook solution is as follows:

If x satisfies $A\textbf{x} = \textbf{0}$, then $CA\textbf{x} = C\textbf{0} = \textbf{0}$ and so $I_n\textbf{x} = \textbf{0}$ and $\textbf{x} = \textbf{0}$. This shows that the equation $A\textbf{x} = \textbf{0}$ has no free variables. So every variable is a basic variable and every column of A is a pivot column.

Now I get that $I_n\textbf{x} = \textbf{0}$ would mean $\textbf{x}$ must be the trivial solution since there are no free variables and you have a pivot position at every column. However, I don't get how that implies that  $A\textbf{x}=\textbf{0}$ has the trivial solution since $A \neq I_n$. I don't quite get how $C$ comes into play for solving this problem.


答:
As $3^n$ quickly exceeds $n$, you can get tighter and tighter bounds by evaluating the first terms exactly and replacing the tail of the summation with larger terms $\dfrac1{3^n}$.

$$\frac1{3}\frac32=\frac12=0.5$$
$$\frac14+\frac1{3^2}\frac32=\frac5{12}=0.416666\cdots$$
$$\frac14+\frac1{11}+\frac1{3^3}\frac32=\frac{157}{396}=0.3964646\cdots$$
$$\frac14+\frac1{11}+\frac1{30}+\frac1{3^4}\frac32=\frac{2333}{5940}=0.3927609\cdots$$

These are guaranteed upper bounds.

Title: Bounded Sequence help

Body:
A series $\{S_n\}$ is defined as follows:
$$S_n = \sum_{n=1}^\infty \frac{1}{3^n+n}.$$

Evidently this sequence is increasing as for every $a_n = (3^n+n)^{-1}$ is less than $a_{n+1}$.

However, how do you show that the sequence is bounded? Bounded below is simple as since it is an increasing function, it is bounded below by $S_1$. But what about bounded above? The calculator shows about $0.39$ as the upper bound.


问:
How do i find the quotient and remainder of this problem?
$x^3 - a^3$ divided by $x - a$

Would really appreciate a step by step explanation of how to find the quotient and remainder because I am still learning this concept.

Thanks so much!

答: There is a famous factorization called the difference of cubes factorization. It is $c^3-d^3 = (c-d)(c^2+cd+d^2)$. Your numerator is a difference of cubes with $c=x$ and $d=a$. Plugging these values into the factorization, we get $x^3-a^3 = (x-a)(x^2+xa+a^2)$. So, your fraction now becomes $\frac{(x-a)(x^2+xa+a^2)}{x-a}$. The quotient is $(x^2+xa+a^2)$ with no remainder.


问:
Finding $\int -2\arcsin(4x)dx$
Can someone point out my mistake? I can't seem to spot what I did wrong.

Here's my work, using integration by parts with these variables: $u = \arcsin(4x)$, $du = \frac{1}{\sqrt{1-x^2}}dx$, $dv = dx$, and $v = x$.

$$\int -2\arcsin(4x)dx = -2(\arcsin(4x) - \int[\frac{x}{\sqrt{1-x^2}}dx])$$
Using substitution to find the second term, I arrive at:
$$=-2(x\arcsin(4x)+\sqrt{1-x^2})+C$$

Title: Finding $\int -2\arcsin(4x)dx$

Body:
$$I=-2\int \sin^{-1} (4x) dx$$ Integrate it by parts treating $\sin^{-1} x$ 1 as tyhe first function.
$$I=-2\left(x\sin^{-1} (4x) -\int \frac{4x}{\sqrt{1-16x^2}}dx\right)$$
$$I=-2\left(x\sin^{-1} (4x) +\frac{1}{8}\int \frac{dt}{\sqrt{t}}\right), t=1-16x^2.$$
$$I=-2x\sin^{-1} (4x) -\frac{1}{2} \sqrt{1-16x^2}+C$$


问:
Do contraction mappings contract everything by the same amount?
The definition of a contraction mapping is:

"A contraction mapping $T$ on $(X,d)$ is a continuous function $T:X\to X$ such that $\forall x,y\in X$ we have that $d(T(x),T(y))≤qd(x,y)$ where $q\in(0,1)$"

So if I'm reading this correctly, there are some points $x,y$ such that the distance shrinks by a mulpiple of $q$ after the mapping, but there may be another pair of points that shrinks by even more, so not all points shrink by the same ratio, is that correct?

答:
so not all points shrink by the same ratio, is that correct?

Yes, that's correct. Here's an example.

For a given function $f:X\to X$, assuming $x\neq y$ the ratio would be

$$r(x,y)=\frac{d(T(x),T(y))}{d(x,y)}$$

So your question can be restated as: $r(x,y)$ need not be constant, right? Yes.

Consider $X=[1,\infty)$ and $f(x)=\sqrt{x}$. The ratio is then

$$r(x,y)=\frac{|\sqrt{x}-\sqrt{y}|}{|x-y|}=$$

$$=\frac{|\sqrt{x}-\sqrt{y}|\cdot |\sqrt{x}+\sqrt{y}|}{|x-y|\cdot|\sqrt{x}+\sqrt{y}|}=\frac{|x-y|}{|x-y|\cdot|\sqrt{x}+\sqrt{y}|}=\frac{1}{|\sqrt{x}+\sqrt{y}|}$$

Since $x,y\in[1,\infty)$ the ratio above cannot exceed $\frac{1}{2}$. Meaning $\sqrt{x}$ is a contraction on $[1,\infty)$ with $q=\frac{1}{2}$. But the ratio $r(x,y)$ clearly is not constant, it depends on $x$ and $y$, e.g. $r(4,9)=\frac{1}{5}$, $r(9,16)=\frac{1}{7}$ and so on. In fact in this case $r(x,y)$ can be any number in $(0,\frac{1}{2})$.


问:
Help solving $\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}$
Reading an article online I came across this integral I've been trying to solve since yesterday morning and I literally tried every integration method I know and I still can't solve it. I am a beginner to integrals so help will be appreciated!

I tried substitution, integration by parts and I can't seem to find the correct way!

$$\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}$$

Does anyone know how to proceed from here? What to do? Because I am clueless. Thanks!

答:
Try this.

\begin{align}
\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}
&={\displaystyle\int}\left(\dfrac{\sqrt{9x^2+9x+2}}{8x^2+9x+2}-\dfrac{x}{8x^2+9x+2}\right)\mathrm{d}x \\
{\displaystyle\int}\dfrac{\sqrt{9x^2+9x+2}}{8x^2+9x+2}\,\mathrm{d}x
&={\displaystyle\int}\dfrac{\sqrt{\left(3x+\frac{3}{2}\right)^2-\frac{1}{4}}}{8x^2+9x+2}\,\mathrm{d}x\tag{solve for 1st integral} \\
&=\frac12 {\displaystyle\int}\dfrac{\sqrt{9\left(2x+1\right)^2-1}}{8x^2+9x+2}\,\mathrm{d}x \\
&=\frac12{\displaystyle\int}\dfrac{\sqrt{9u^2-1}}{4u^2+u-1}\,\mathrm{d}u \tag{$u = 2x+1$, $\mathrm{d}x=\dfrac{1}{2}\,\mathrm{d}u$} \\
&=\frac12 {\displaystyle\int}\dfrac{\sqrt{\sec^2\left(v\right)-1}\sec\left(v\right)\tan\left(v\right)}{3\left(\frac{4\sec^2\left(v\right)}{9}+\frac{\sec\left(v\right)}{3}-1\right)}\,\mathrm{d}v \tag{$u=\dfrac{\sec\left(v\right)}{3}$, $v=\operatorname{arcsec}\left(3u\right), \mathrm{d}u=\dfrac{\sec\left(v\right)\tan\left(v\right)}{3}\,\mathrm{d}v$} \\
&=\frac32 {\displaystyle\int}\dfrac{\sec\left(v\right)\tan^2\left(v\right)}{4\sec^2\left(v\right)+3\sec\left(v\right)-9}\,\mathrm{d}v \\
&=\frac32 {\displaystyle\int}\dfrac{\frac{\left(\tan^2\left(\frac{v}{2}\right)+1\right) \,\cdot\, 4\tan^2\left(\frac{v}{2}\right)}{\left(1-\tan^2\left(\frac{v}{2}\right)\right)^3}}{\left(\frac{4\left(\tan^2\left(\frac{v}{2}\right)+1\right)^2}{\left(1-\tan^2\left(\frac{v}{2}\right)\right)^2}+\frac{3\left(\tan^2\left(\frac{v}{2}\right)+1\right)}{1-\tan^2\left(\frac{v}{2}\right)}-9\right)}\,\mathrm{d}v \\
&
\end{align}


问:
Examples of failure of excision for homotopy groups ($\pi_k(X, A)$ is not $\pi_k(X/A, *)$)
Let $A$ be a subcomplex of CW-complex $X$. The excision axiom for homology implies that $H_i(X, A)\cong H_i(X/A, *)$, and it is widely known that homotopy groups don't have this property. However, they satisfy a significantly weaker Blakers-Massey theorem. One of its consequences is that when $A$ is $n$-connected and the inclusion $A\to X$ is an isomorphism on the first $s$ homotopy groups, $\pi_k(X, A) \cong \pi_k(X/A, *)$ for $k < s + n - 1$.

Thus to devise an example where $\pi_k(X, A) \not\cong \pi_k(X/A, *)$ it's necessary to take large enough $k$, and the calculation of both sides becomes hard. The original paper of Blakers and Massey claims there are simple examples, but I wasn't able to make them up myself. What are some simple examples of the pairs $(X, A)$ and $(X/A, *)$ with different homotopy groups? Preferably with both $A$ and $X$ simply connected.

答: Probably, the easiest example is pair $(D^2,S^1)$ ($S^1$ is a boundary circle): by homotopy exact sequence we have $\ldots\to\pi_n(S^1)\to\pi_n(D^2)\to \pi_n(D^2,S^1)\to\pi_{n-1}(S^1)\to\ldots $ where easy to see that if $n\geqslant 3$ all the groups $\pi_n(D^2,S^1)$ are zero, while groups $\pi_n(D^2/S^1,*)\cong \pi_n(S^2,*)$ are not trivial (in particular, $\pi_3(S^2)\cong \mathbb{Z}$). Doesn't even need any "high theory".


问:
How to compare two rotations represented by axis-angle rotation vectors?
I currently have two rotation vectors r1 and r2 using axis-angle representations, i.e. they are 3d vectors, their norms are the rotation angles and the normalized unit vectors are rotation axes respectively.

I am confused how to compare r1 and r2. In an optimization process of a regression system, if r1 is the ground truth, and r2 is the current result produced by the system, how do I define a proper loss to encourage the system to converge to r1? Can I just use L1/L2 loss between r1 and r2?

Thanks!

答:
There are multiple ways one could define the error between two rotation, see for example: Huynh, Du Q. "Metrics for 3D rotations: Comparison and analysis." Journal of Mathematical Imaging and Vision 35.2 (2009): 155-164. That source mainly defines errors between rotation matrices and quaternions, but your representation can be converted to those relatively easily. In my opinion the most geometrically meaningful error measure would be the smallest angle the second rotation representation would have to be rotated by in order to match the first rotation. This angle can be calculated from the rotation separating the two, which can be obtained by first representing $r_1$ and $r_2$ as rotation matrices or quaternions and divide one by the other.

Your axis-angle representation can be converted into a rotation matrix using

$$
R(r) = I + \frac{\sin(\|r\|)}{\|r\|}S(r) + \frac{1-\cos(\|r\|)}{\|r\|^2}S(r)^2, \tag{1}
$$

with $I$ the three by three identity matrix, $r = \begin{bmatrix}r_1 & r_2 & r_3\end{bmatrix}^\top$ the vector of the axis-angle representation and

$$
S(r) =
\begin{bmatrix}
0 & -r_3 & r_2 \\
r_3 & 0 & -r_1 \\
-r_2 & r_1 & 0
\end{bmatrix}. \tag{2}
$$

Rotation matrices can be "divided" by each other by multiplying one by the inverse of the other and the inverse of a rotation matrix is the same as the transpose of that rotation matrix. Taking the transpose of a matrix is equivalent to negating its skew-symmetric part. From $(1)$ it can be noted that   $S(r)$ is always skew-symmetric, but $I$ and $S(r)^2$ are symmetric. Using this and that the scalar parts of $(1)$ are even or odd functions of $\|r\|$ yields that when taking the transpose of $(1)$ is equivalent to using $R(-r)$. Therefore, the relative rotation in your case could be obtained using

$$
R_e = R(r_1)\,R(-r_2), \tag{3}
$$

whose associated angle in the interval $[0,\pi]$ can be calculated with

$$
\theta = \cos^{-1}\left(\frac{\text{tr}(R_e) - 1}{2}\right). \tag{4}
$$

It is also possible to calculate the same using unit quaternions. The axis-angle representation can be converted into a quaternion using the scalar vector representation with

$$
q(r) = \left(\cos\left(\frac{\|r\|}{2}\right),\sin\left(\frac{\|r\|}{2}\right)\frac{r}{\|r\|}\right). \tag{5}
$$

Quaternion can be divided by each other by multiplying one by the inverse of the other. Here the inverse of unit quaternions can be obtained by negating the vector part. However, when multiplying the quaternions together one has to use the Hamilton product. When one only want to recover the angle in the interval $[0,\pi]$ from the resulting unit quaternion it is sufficient to only calculate the scalar part. After substituting in $(5)$ it can be shown that the resulting angle can be calculated with

$$
\theta = 2\cos^{-1}\left(\left|\cos\left(\frac{\|r_1\|}{2}\right) \cos\left(\frac{\|r_2\|}{2}\right) + \sin\left(\frac{\|r_1\|}{2}\right)\sin\left(\frac{\|r_2\|}{2}\right)\frac{r_1^\top\,r_2}{\|r_1\|\,\|r_2\|}\right|\right). \tag{6}
$$

Both $(4)$ and $(6)$ should give the same result (up to machine accuracy), but $(6)$ is probably faster to calculate.


答:
Multiply by the conjugate first to make
$$\int\frac{dx}{x+\sqrt{9x^2-9x+2}}=\int\frac{\sqrt{9 x^2-9 x+2}-x}{8 x^2-9 x+2} \,dx$$
$$I_2=\int\frac{x}{8 x^2-9 x+2} \,dx=\frac 1{16}\int\frac{16x-9+9}{8 x^2-9 x+2} \,dx$$
$$I_2=\frac 1{16}\int\frac{16x-9}{8 x^2-9 x+2} \,dx+\frac 9{16}\int\frac{dx}{8 x^2-9 x+2} $$ The first one is obvious and the second one is simple (if required, use partial fractions).

Now, we are left with
$$I_1=\int\frac{\sqrt{9 x^2-9 x+2}}{8 x^2-9 x+2} \,dx$$ which looks to be much more difficult. However,  partial fraction decomposition, we have
$$\frac 1{8 x^2-9 x+2}=\frac{1}{8 (a-b) (x-a)}-\frac{1}{8 (a-b) (x-b)}$$
and you can find the required integral in the  "Table of Integrals, Series, and Products"
(seventh edition) by I.S. Gradshteyn and I.M. Ryzhik.

The global result will not look very nice.

Title: Help solving $\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}$

Body:
Reading an article online I came across this integral I've been trying to solve since yesterday morning and I literally tried every integration method I know and I still can't solve it. I am a beginner to integrals so help will be appreciated!

I tried substitution, integration by parts and I can't seem to find the correct way!

$$\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}$$

Does anyone know how to proceed from here? What to do? Because I am clueless. Thanks!


问:
Is the matrix $A^4+A^3-3A^2-3A$ invertible?

Let $A$ be a real $5 \times 5$ matrix satisfying $A^3-4A^2+5A-2I=O$. Is the matrix $A^4+A^3-3A^2-3A$ invertible?

Consider the polynomial $t^3-4t^2+5t-2=(t-1)^2(t-2)$. The minimal polynomial of $A$ divides $(t-1)^2(t-2)$. Let $p(t)$ be the minimal polynomial of $A$.

If $p(x)=t-1$ or $t-2$, then $A$ is a scalar multiple of the identity and $A^4+A^3-3A^2-3A=-4I \text{ or }6I$, so $A^4+A^3-3A^2-3A$ is invertible.

If $p(t)=(t-1)^2(t-2)$ or $(t-1)(t-2)$, then $A$ is similar to a upper triangular matrix $J$ with diagonal entries $1,2$. Then $A^4+A^3-3A^2-3A=P(J^4+J^3-3J^2-3J)P^{-1}$ and $\det(A^4+A^3-3A^2-3A)=\det(J^4+J^3-3J^2-3J)$. Since $J^4+J^3-3J^2-3J$ is upper triangular and the diagonal entries do not vanish, $\det(J^4+J^3-3J^2-3J) \ne 0$.

If $p(t)=(t-1)^2$, then by the same reasoning, $A^4+A^3-3A^2-3A$ is invertible.

Is there more efficient way to solve this kind of problems or I have to discuss any possible situations every time?

答:
Since $A^3-4A^2+5A-2=0$, $A$ must be invertible ($0$ is not an eigenvalue, else we get $-2v=0$.) Since $$A^4+A^3-3A^2-3A=A(A^3+A^2-3A-3),$$ it is enough to show that $A^3+A^2-3A-3$ is invertible. This follows from \begin{align}
A^3+A^2-3A-3&=(A+1)(A+\sqrt3)(A-\sqrt3)\end{align} Since $-1$, $\pm\sqrt3$ are not eigenvalues of $A$ (do not satisfy the polynomial in $A$), this polynomial in $A$ is invertible.


答:
If $P$ is a polynomial, the matrix $P(A)$ is invertible if and only if $P$ is prime to the minimal polynomial $\mu_A$.

Proof. If $P$ is prime to $\mu_A$, then $UP+V\mu_A=1$ for some polynomials $U,V$. Now evaluate in $A$.
Conversely, if $D$ is a non constant common divisor, then $P=QD, \mu_A=RD$, so $\mu_A(A)=R(A)D(A)=0$. Now, since $D$ is not constant, the degree of $R$ is strictly less than the degree of $\mu_A$, and $R(A)$ is not $0$.Now $P(A)=Q(A)D(A)$, so $P(a)R(A)=0$ and $P(A)$ cannot be invertible  (otherwise $R(A)=0$)

In your case, performing Euclid algorithm for example, you get that your polynomials are coprime, so you have invertibility.

Body:
Let $A$ be a real $5 \times 5$ matrix satisfying $A^3-4A^2+5A-2I=O$. Is the matrix $A^4+A^3-3A^2-3A$ invertible?

Consider the polynomial $t^3-4t^2+5t-2=(t-1)^2(t-2)$. The minimal polynomial of $A$ divides $(t-1)^2(t-2)$. Let $p(t)$ be the minimal polynomial of $A$.

If $p(x)=t-1$ or $t-2$, then $A$ is a scalar multiple of the identity and $A^4+A^3-3A^2-3A=-4I \text{ or }6I$, so $A^4+A^3-3A^2-3A$ is invertible.

If $p(t)=(t-1)^2(t-2)$ or $(t-1)(t-2)$, then $A$ is similar to a upper triangular matrix $J$ with diagonal entries $1,2$. Then $A^4+A^3-3A^2-3A=P(J^4+J^3-3J^2-3J)P^{-1}$ and $\det(A^4+A^3-3A^2-3A)=\det(J^4+J^3-3J^2-3J)$. Since $J^4+J^3-3J^2-3J$ is upper triangular and the diagonal entries do not vanish, $\det(J^4+J^3-3J^2-3J) \ne 0$.

If $p(t)=(t-1)^2$, then by the same reasoning, $A^4+A^3-3A^2-3A$ is invertible.

Is there more efficient way to solve this kind of problems or I have to discuss any possible situations every time?


问:
Can anyone show me, how to solve these system of Equations:
$$\begin{align*}
x+y+z &= 2 \\
(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y) &= 1 \\
x^2(y+z)+y^2(z+x)+z^2(x+y) &= -6
\end{align*}$$

Can anyone explain me the solution. I asked it in mathoverflow but god knows why it was closed. Please help and detail the process in step by step. It is a RMO question (The Pre- Indian team selection exam for IMO)

答:
Generally proceed as:

Is calculated $s=x+y+z, q=xy+yz+zx, p=xyz;$

Solve the equation $t^3-st^2+qt-p=0.$


答:
In general, the eigenvalues (with algebraic multiplicity) of $P(A)$ for any polynomial $P$ are the values $P(\lambda)$ for every eigenvalue $\lambda$ of $A$. In particular, the first condition gives you that all eigenvalues of $A^3-4A^2+5A-2I$ are $0$, so
$$(\lambda-1)^2(\lambda-2)=\lambda^3-4\lambda^2+5\lambda-2=0$$
for every eigenvalue $\lambda$ of $A$, so $A$ can only have the eigenvalues $1$ and $2$. As a result, every eigenvalue of $A^4+A^3-3A^2-3A$ will be
$$\lambda^4+\lambda^3-3\lambda^2-3\lambda$$
for $\lambda\in\{1,2\}$, and you can manually check each $\lambda$.

Title: Is the matrix $A^4+A^3-3A^2-3A$ invertible?


问:
Finding the value of a variable given a equation
The equation $$\frac{24x^2+25x-47}{ax-2} = -8x-3-\frac{53}{ax-2} $$ is true for all values of $$ x \neq \frac{2}{a} $$ where a is a constant.

What is the value of a?

Please someone help me solve this question? This is a question from the RD Sharma Book which I was not able to solve. Actually I solved but I want a more better way to solve it. I putted x as 0 and then got the value of a as -3.

Please note this has been edited so the equation in the question accords with the accepted answer.

答:
It appears the initial equation has been edited several times, I believe this might be what you intended originally:

$$ \frac{24x^2+25x-47}{ax-2} = -8x-3-\frac{53}{ax-2} , \ x \neq \frac{2}{a} . $$

If that's the case, then we have

$$ \frac{24x^2+25x-47}{ax-2} = -8x-3-\frac{53}{ax-2} , \ x \neq \frac{2}{a} $$

$$ \Rightarrow \frac{24x^2+25x-47+53}{ax-2} = -8x-3 $$

$$ \Rightarrow \frac{24x^2+25x+6}{ax-2} = -8x-3 $$

$$ \Rightarrow \frac{(8x+3)(3x+2)}{ax-2} = -8x-3 $$

$$ \Rightarrow (8x+3)(3x+2) = -(8x+3)(ax-2) $$

$$ \Rightarrow 3x+2 = -(ax-2) $$

$$ \Rightarrow a = -3 $$


问:
Compute integral using residue theorem: $\int_{0}^{\infty}\frac{x}{1+x^6}dx$.
I am trying to compute this integral $\int_{0}^{\infty}\frac{x}{1+x^6}dx$ using the residue theorem. To do so, I am integrating $f(z)=\frac{z}{1+z^6}$ in the frontier of this sector of a circle: $\{z:|z|<R, 0<arg(z)<\pi/3\}$.

I kwo how to deal with the integrals over the horizontal segment of the sector and over the arc of the circle. My problem is the "diagonal segment". When I parametrize it, I do not get something easy to integrate. How could I approach this?

(The path of integration was suggested in my book, so I do not think that is the problem).

Thanks in advance.

答: The paramertic form is $z=r(\frac {1 +\sqrt 3} 2), 0<r<R$. The integral on this line becomes $(\frac {1 +\sqrt 3} 2) \int_0^{R} \frac r {1+r^{6}}dr$ since $(\frac {1+\sqrt 3 } 2)^{6}=1$. Now you get $[1-\frac {1 +\sqrt 3} 2)]\int_0^{\infty} \frac r {1+r^{6}}dr=$ $2\pi i$ times the sum of the residues.


问:
Am I right or is Wolfram right?
Let ${a_n}$ be a sequence whose corresponding power series $A(x)=\sum_{i\geq 0}a_ix^i$ satisfies

$$A(x)=\frac{6-x+5x^2}{1-3x^2-2x^3}$$

Determine a recurrence relation that ${a_n}$ satisfies.

I multiplied it out to

$$(1-3x^2-2x^3)A(x)=6-x+5x^2$$
$$a_0+(a_1-3a_0)x+(a_2-3a_1-2a_0)x^2+\dots = 6-x+5x^2$$
thus
$$a_0=6$$
$$a_1=-1+3a_0$$
$$a_2=5+3a_1+2a_0$$
$$a_{n\geq 3}=3a_{n-1}+2a_{n-2}$$
but Wolfram Alpha instead gives the power series expansion as
$$6-x+23 x^2+9 x^3+67 x^4+73 x^5+\cdots$$which does not satisfy my relation. Which is correct?

答: Hint: When multiplying $A(x)$ by the denominator, it seems you took $1-3x-2x^2$ instead of $1-3x^2-2x^3$.</p>

Title: Am I right or is Wolfram right?


问:
Is it possible to get Sign of an integer nonnegative number using 4 basic operations?
I need to get Sign of an integer nonnegative number using 4 basic math operations (summation, subtraction, multiplication, division). Abs or other functions are not allowed. How can I do this?</p>

答: Any such function would be, by definition, the restriction to the natural numbers of some rational function $f:\Bbb R\setminus\{\text{a few points}\}\to \Bbb R$, and it ought to be $1$ on all natural numbers except $0$, and be $0$ in $0$. This cannot be the case, because the constant $1$ is a rational function and two distinct rational functions can only coincide on finitely many points.


问:
Interesting Identity: $\lim_{m\to\infty}\lim_{n\to\infty}\prod_{k=1}^{\infty}\prod_{j=2}^{m}(1+\phi_{n})^{nk^{-j}\Phi_{j}}=e^{\gamma}$
Define the following identities:
$$
\phi_{n}:=\frac{4}{\pi}\int_{0}^{\infty}\frac{\coth(nx^{-1})-xn^{-1}}{n(1+x^{2})^{2}}\;dx \qquad
\text{and}\qquad
\Phi_{n}:=\frac{\cos(n\pi)}{n}
$$
$\forall n\in\mathbb{N}$. I want to show that :
$$
\lim_{m\to\infty}\lim_{n\to\infty}\prod_{k=1}^{\infty}\prod_{j=2}^{m}(1+\phi_{n})^{ nk^{-j}\Phi_{j}}=e^{\gamma}
$$
Where $\gamma$ is the Euler - Mascheroni constant.

What I did is that I used the linearity of integrals for $\phi_{n}$
$$
\frac{\pi}{4}\phi_{n}=\int_{0}^{\infty}\frac{\coth\left(\frac{n}{x}\right)}{n(1+x^{2})^{2}}\;dx-\int_{0}^{\infty}\frac{x}{n^{2}(1+x^{2})^{2}}\;dx
$$
The first one I define it as $I_{1}$ and the second one I define it as $I_{2}$. Afterwards I defined:
$$
\lambda_{n}(x):=\frac{\coth\left(\frac{n}{x}\right)}{n(1+x^{2})^{2}}\leq\left|\frac{1}{n(1+x^{2})^{2}}\right|\leq\left|\frac{1}{2n(1+x^{2})}\right|=:V_{n}(x)
$$
so that:
$$
\lim_{n\to\infty}\int_{0}^{\infty}V_{n}(x)\;dx=\lim_{n\to\infty}\frac{\pi}{4n}=0
$$
Applying the Lebesgue Dominated Convergence Theorem I get:
$$
\lim_{n\to\infty}\int_{0}^{\infty}\lambda_{n}(x)\;dx=0
\tag 1$$
Now note that:
$$
I_{2}=\int_{0}^{\infty}\frac{x}{n^{2}(1+x^{2})}\;dx=\frac{1}{2n^{2}}\quad\implies\quad\lim_{n\to\infty}I_{2}=0
\tag 2$$
$(1)$ and $(2)$ verify that $\lim_{n\to\infty}(1+\phi_{n})^{n}\leq\lim_{n\to\infty}(1+I_{1}-\displaystyle I_{2})^{n}=\lim_{n\to\infty}\left(1+\frac{1}{n}-\frac{2}{n\pi^{2}}\right)^{n}=e$

Now all I have to evaluate $\displaystyle\exp\left(\lim_{m\to\infty}\sum_{k=1}^{\infty}\sum_{j=2}^{\infty}\frac{\cos(j\pi)}{k^{j}j}\right)$ but this seems a bit confusing and misleading when I try it which is what I am stuck on.

答:
The expression under the limit(s) is equal to ${a_n}^{b_m}$, where $$a_n=(1+\phi_n)^n,\qquad b_m=\sum_{k=1}^\infty\sum_{j=2}^m\frac{(-1)^j}{jk^j}.$$

Let $f(t)=\coth(1/t)-t$, then $n\phi_n=(4/\pi)\int_0^\infty(1+x^2)^{-2}f(x/n)\,dx$. Observe that $f$ is decreasing (to check, take the derivative and use $\sinh z>z$ for $z>0$). Also, $f(0)=1$ if understood as the limit.

Thus, $n\mapsto f(x/n)$ is increasing for each fixed $x$, so DCT (or even MCT) is applicable, and $\lim\limits_{n\to\infty}n\phi_n=(4/\pi)\int_0^\infty(1+x^2)^{-2}f(0)\,dx=1$, hence $\lim\limits_{n\to\infty}a_n=e$ as you know.

As for $b_m$, the corresponding $\sum_{k=\color{red}{2}}^\infty\sum_{j=2}^{\color{red}{\infty}}$ is absolutely convergent, so that we're free to interchange the summations, hence to take $m\to\infty$ directly under the outer sum: $$\lim_{m\to\infty}b_m=\sum_{k=1}^\infty\sum_{j=2}^\infty\frac{(-1/k)^j}{j}=\sum_{k=1}^\infty\left[\frac1k-\log\left(1+\frac1k\right)\right]\\=\lim_{n\to\infty}\sum_{k=1}^n[\ldots]=\lim_{n\to\infty}\left[\left(\sum_{k=1}^n\frac1k\right)-\log(n+1)\right].$$ As $\lim\limits_{n\to\infty}\big(\log(n+1)-\log n\big)=0$, we then have $\lim\limits_{m\to\infty}b_m=\gamma$ "almost by definition".

Finally, the given limit is $e^\gamma$ because $(a,b)\mapsto a^b$ is continuous at $(e,\gamma)$.


问:
Orbit of a point under the action of the stabiliser of another point
I'm wondering, for a group $G$ acting on a set $X$, what can be said about the orbit of a point under the action of a stabiliser subgroup of $G$. Let $x\in X$ and let $H = {\rm Stab}_G(x)$. Obviously $|H\cdot x|$ is going to be $1$, but what can we say about $|H\cdot y|$ for some other $y\in X$, perhaps in relation to $|G\cdot y|$?  If that not much can be said in general, feel free to put interesting restrictions on $y$, e.g., $y$ in $G\cdot x$.

Edit. I guess the original question was far too vague. I am still interested in the general question, but in the specific examples I am thinking of, $X$ is the set of vertices of a finite graph and $G$ acts by graph automorphisms. Does this extra information tell us more about $H\cdot y$?

答:
One interesting result connected to this question is the Sims Conjecture, dating from 1967, which was eventually proved using the classification of finite simple groups.

If $G$ acts faithfully and primitively on $X$ and $H = {\rm Stab}_G(x)$ with $x \in X$, then $|H|$ is bounded above as a function of $|H \cdot y|$, for any $y \in X \setminus \{x\}$.


问:
Discrete approximation to continuous normal
Im reading Gelman's "Regression and other stories" book and I'm finding it difficult to wrap my head around one idea: in chapter 3 (p. 46) he approximates a discrete density using a normal density function. The question he's asking is what is the probability of tied election if we assume that the election outcome is forested to be N(0.49, 0.04). He than goes on and, assuming a population of voters of n=200000, calculated that P(tied election) = 1/n * f, where f is the PDF of this forecast. Where does the 1/n comes from in this formula? He uses the same rationale in his " Estimating the Probability of Events That Have Never Occurred: When Is Your Vote Decisive?" Gelamn et al. (1998) paper (J of the American Statistical Association):

Pr(your vote is decisive in your state)
= Pr(np = .5n) = Pr(p = .5), which is approx.  f(.5)/n

答:
Conclusion

The division by n takes into account the fact that there are other voters in the state. If you were the only voter in your state, the probability of your vote being decisive would be $f_v(.5)$, the density/probability of the state being at .5 democratic share. If there are 100 voters, certainly the probability of your vote being decisive is not the same as in a state with just one person, because any voter can be the “decisive vote” hence the division by n. The authors distract the reader by including misleading details like “n is the number of voters excluding yourself”.

tl;dr

It’s talking about the probability that a randomly selected persons vote will be decisive in a state.

Shoddy derivation of the formula (page 3)

From the paper there are 3 places where formula is discussed. They depend on $n_i=\text{number of voters in your state excluding yourself}$, and $v_i=\text{democratic share of the two party vote in the state excluding you}$. The first is $P(\text{your vote is decisive in your state})=P(nv =.5n)=P(v=.5)\approx f(.5)/n$. A more correct formula would be $P(\text{your vote is decisive in your state})=P(v=.5)/n=f_v(.5)/n$.

Using model parameters (page 5)

The second hint that a Normal approximation is used for binomial vote counts is that for each simulated parameter vector, using data from previous elections, this expression is modeled to be $\frac 1 n  N(.5, X\beta+\gamma+\delta, n^{-\theta}\sigma^2)$

Using some made-up values (page 6)

The third is a discussion of the actual values. Suppose n people in your state vote and the forecast has mean $\mu$ and std $\tau$. Then the probability a single vote will be decisive is $(\sqrt{2\pi}\tau n)^{-1}exp (-(\mu-.5)^2/2\tau^2)$. The probability of a tie is maximized at $\mu=.5$ (i.e. the model predicts a tie). A minimum bound on $\tau$ is about 2%, since it’s hard to forecast an election to more accuracy than that. This gives 20/n as an upper bound that a randomly selected persons vote will be decisive in a relatively close election. A typical value of n is 200,000 according to Congress, giving an upper bound of 1 in 10,000 that a randomly selected vote will be decisive. The authors also says another way to look at it is it’s not possible to forecast an election to closer than 10,000 votes.

etc.

Your vote has much higher chance of being decisive if you come from a small state, say n=2, than if n=10,000. The f is the probability that the state will be decisive, namely there are an equal number of democratic and republican ballots cast. And the probability that any particular vote is the decisive one is that divided by the number of votes.


问:
For $S$ a simple object, $\text{End}_\mathcal C(S)$ is a division ring.
Let $\mathcal C$ a semi-simple abelian category and $S \in \mathcal C$ a simple object. I'm trying to show that $\text{End}_\mathcal C(S) = \text{Hom}_\mathcal C(S, S)$ is a division ring, i.e. each $f \in \text{End}_\mathcal C(S)$ is either $0$ or an isomorphism. This statement looks like Schur's Lemma so I though maybe use the same kind of argument to prove this but I am not able to do it. We define a simple object in $\mathcal C$ as an object $S$ such that for each monomorphism $f : X \to S$ then $f = 0$ or $f$ is an isomorphism, but how can I show that each $f \in \text{End}_\mathcal C(S)$ is at least a monomorphism ?

答: Pick a nonzero endomorphism $f \colon S  \to S$. We then have a mono $\ker f \to S$ which is not an isomorphism, thus it must be the zero map. This shows that $f$ is mono. Using simplicity again we conclude that $f$ must be an iso.


问:
What does it mean to evaluate an integral over a volume?
I have a homework problem where I need to evaluate an integral and my calculus skills are really rusty.

The problem:

Evaluate the following integral
$$\int r^3z^2\mathrm{d}V$$
over a volume defined by a cylinder of legth $L$ and radius $R$, oriented along the $z$-axis and centered at the origin.

Hint: first integrate over a circle and then integrate along the $z$-axis.

Honestly, I'm confused by the language of the problem. What does it mean to evaluate the integral over a volume, and what does it represent? I know the equation of the volume of a cylinder, but what do I do with that formula and the given equation $r^3 z^2$. How do I relate the two to start to solve the problem?

image of the problem

答:
When you are asked to integrate a function over the volume of the region, please use the same bounds for your triple integral that you would, to find the volume of the region. In this case it is a cylinder of length $L$ and radius $R$ oriented along $z-$axis and with center at the origin.

So, $0 \leq r \leq R, -\frac{L}{2} \leq z \leq \frac{L}{2}$.

In cylindrical coordinates, $dV = r \ dr \ dz \ d\theta$

So your integral is $\displaystyle \int_0^{2\pi}\int_{-L/2}^{L/2}\int_0^R  r^4 z^2 \ r \ dr \ dz \ d\theta $.


问:
Solve:$(1-y^2 +\frac{y^4}{x^2})p^2- (2\frac{y}{x})p + \frac{y^2}{x^2}=0$
I have the differential equation:  $(1-y^2 +\frac{y^4}{x^2})p^2- (2\frac{y}{x})p + \frac{y^2}{x^2}=0$ where $p=\frac{dy}{dx}$

I have solved this up until $\frac{dy}{dx}$, but I am not able to reduce it to an exact differential equation.

Steps until here:

$$\frac{1}{x^2}\Bigl((x^2-x^2 y^2 +y^4)p^2- (2xy)p + y^2\Bigr)=0$$
As $\frac{1}{x^2}$ cannot equal zero,
$$(x^2-x^2 y^2 +y^4)p^2- (2xy)p + y^2=0$$
Applying the quadratic formula:
$$p=\frac{2xy \pm \sqrt{4x^2y^2-4x^2y^2+4x^2y^4-4y^6}}{2(x^2-x^2 y^2 +y^4)}$$
$$p=\frac{xy \pm y^2\sqrt{x^2-y^2}}{x^2-x^2 y^2 +y^4}$$
$$p=\frac{y\Bigl(x \pm y\sqrt{x^2-y^2}\Bigr)}{x^2-y^2(x^2-y^2)}$$
$$p=\frac{y\Bigl(x \pm y\sqrt{x^2-y^2}\Bigr)}{\Bigl(x-y\sqrt{x^2-y^2}\Bigr)\Bigl(x+y\sqrt{x^2-y^2}\Bigr)}$$
Thus,
$$p=\frac{dy}{dx}=\frac{y}{\Bigl(x-y\sqrt{x^2-y^2}\Bigr)}$$
Or,
$$p=\frac{dy}{dx}=\frac{y}{\Bigl(x+y\sqrt{x^2-y^2}\Bigr)}$$

From here, we can write the two equations as:
$$xdy-ydx=y\sqrt{x^2 - y^2}dy$$
$$xdy-ydx=-y\sqrt{x^2 - y^2}dy$$

I tried reducing this equation to an exact differential equation, but failed to do so. Kindly help me proceed from here. Thank you!

答:
$$xdy-ydx=-y\sqrt{x^2 - y^2}dy$$
$$yx'-x=y\sqrt{x^2 - y^2}$$
$$\left(\dfrac xy \right)'=\sqrt{\dfrac {x^2}{y^2} -1}$$
$$w'=\sqrt{w^2 -1}$$
Where $w=\dfrac xy$.Then integrate.


问:
Area of the square is $16$ square units. Midpoint of the sides of the square are $a,b,c,d$. Then what is the area of the shaded part?
Area of the square is $16$ square units. Midpoint of the sides of the square are $a,b,c,d$. Then what is the area of the shaded part?

答:
Hint

(assuming the four arcs are symmetric - the diagram does not seem precise and the midpoints are not in the middle)


问:
How to calculate the limit of zeta function
Suppose $f(x)>0$,$f''(x)\leqslant0$,and $\lim\limits_{x\to+\infty}f(x)=+\infty$ on$[0,+\infty)$.prove that
$$\lim\limits_{s\to0^+}\sum\limits_{n=0}^{\infty}\dfrac{(-1)^n}{f^s(n)}=\frac{1}{2}.$$
I tried to do it ,first of all we have
$$\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)}=\sum\limits_{n=0}^{\infty} \left( \frac{1}{f^s(2n)}-\frac{1}{f^s(2n+1)} \right).$$
Use MVT we have
$$ \frac{1}{f^s(2n)}-\frac{1}{f^s(2n+1)}=-\frac{sf'(\xi_n)}{f^{s+1}(\xi_n)}(\xi_n\in(2n,2n+1)).$$
But next, I don't know how to deal with it. I want to ask that this problem can be solved by the property that the function is concave.

答:
We can generalize @zhw's solution from Prove that $\lim\limits_{x\to 0^+}\sum\limits_{n=1}^\infty\frac{(-1)^n}{n^x}=-\frac12$.

Starting with your application of the mean-value theorem (note that you have a sign error),
$$
\frac{1}{f^s(2n)}-\frac{1}{f^s(2n+1)}=\frac{sf'(\xi_n)}{f^{s+1}(\xi_n)}
\le \frac{sf'(2n)}{f^{s+1}(2n)}
$$
since $f$ is increasing and $f'$ is decreasing. It follows that
$$
\sum_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)} \le \sum_{n=0}^{\infty} \frac{sf'(2n)}{f^{s+1}(2n)}  \le \frac{sf'(0)}{f^{s+1}(0)} + s\int_0^\infty \frac{f'(2t)}{f^{s+1}(2t)} \, dt = \frac{sf'(0)}{f^{s+1}(0)} + \frac{1}{2f^s(0)} \, .
$$
Similarly,
$$
\sum_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)} \ge \sum_{n=0}^{\infty} \frac{sf'(2n+1)}{f^{s+1}(2n+1)} \ge  s\int_1^\infty \frac{f'(2t)}{f^{s+1}(2t)} \, dt = \frac{1}{2f^s(1)} \, .
$$
The result now follows by squeezing these estimates for $s \to 0$.


答:
The polynomials

$$ p(x) = x^3 -4 x^2 + 5x -2 = (x-1)^2(x-2) $$
$$ q(x) = x^4 + x^3 - 3 x^2 -3x = x(x+1)(x^2 - 3)$$

are coprime.

From Bézout's identitity, it follows that there exists two polynomials $u(x)$ and $v(x)$ such that

$$ u(x) p(x) + v(x) q(x) = 1$$

As $p(A) = 0$, it follows that $q(A)$ is invertible.

A pair of Bézout coefficients $u(x)$ and $v(x)$ can be computed by the extended Euclidean algorithm.


答:
The long division of $x^4+x^3-3x^2-3x$ by $x^3-4x^2+5x-2$ leaves a remainder $12x^2-26x+10$, so your matrix to be checked for invertibility is
$$
2(6A^2-13A+5I)
$$
Suppose $v\ne0$ is in the null space of $6A^2-13A+5I$, so $6A^2v-13Av+5v=0$.

We also know that $A^3v-4A^2v+5Av-2v=0$. Multiplying the former by $A$, the latter by $6$ and subtracting, we obtain $11A^2v-25Av+12v=0$ and an elimination from the two “degree two” relations yields
$$
7Av-17v=0
$$
which means that $A$ should have $17/7$ as an eigenvalue. However, this is not possible, because an eigenvalue $\lambda$ of $A$ must satisfy $\lambda^3-4\lambda^2+5\lambda-2=0$.

The above doesn't use the Jordan canonical form. If you know it, you can prove that the eigenvalues of $p(A)$ (for $p$ a polynomial) have the form $p(\lambda)$ where $\lambda$ is an eigenvalue of $A$. Since the eigenvalues of $A$ must satisfy $(\lambda-1)^2(\lambda-2)=0$, they can only be $1$ or $2$; however
$$
x^4+x^3-3x^2-3x=x(x^2-3)(x+1)
$$
and neither $1$ nor $2$ are roots of this polynomial.


问:
calculating $\lim\limits _{x\to0}\left(\frac{1}{x^{2}}-\frac{e^{x}}{\left(e^{x}-1\right)^{2}}\right)$ using taylor polynomials
I am trying to calculate the limit $$\lim\limits _{x\to0}\left(\frac{1}{x^{2}}-\frac{e^{x}}{\left(e^{x}-1\right)^{2}}\right)$$ using taylor polynomials. I tried using L'Hôpital's rule, but it was tedious. the final answer should be $\frac{1}{12}$.

I tried taylor many times but couldn't make it work. here is my best attempt:

$$\lim\limits _{x\to0}\left(\frac{1}{x^{2}}-\frac{e^{x}}{\left(e^{x}-1\right)^{2}}\right)=\lim\limits _{x\to0}\left(\frac{\left(e^{x}-1\right)^{2}-x^{2}e^{x}}{x^{2}}\right)=\lim\limits _{x\to0}\left(\frac{\left(1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+R_{3}\left(x\right)-1\right)^{2}-x^{2}\left(1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+R_{3}\left(x\right)\right)}{x^{2}}\right)=\lim\limits _{x\to0}\left(\frac{\frac{1}{3}R_{3}\left(x\right)x^{4}+R_{3}^{2}\left(x\right)x^{2}+2x^{2}R_{3}\left(x\right)+\frac{x^{6}}{36}+\frac{x^{4}}{12}}{x^{2}}\right)=\lim\limits _{x\to0}\left(\frac{R_{3}\left(x\right)\left(\frac{1}{3}x^{4}+R_{3}\left(x\right)x^{2}+2x^{2}\right)}{x^{2}}\right)=\lim\limits _{x\to0}\left(R_{3}\left(x\right)\left(\frac{1}{3}x^{2}+R_{3}\left(x\right)+2\right)\right)$$

the number $\frac{1}{12}$ appears here but then goes away since $x^4$ goes to zero. There must be a simpler way then opening the brackets here, and making all those tedious calculations.

答: You have $(e^x-1)^2-x^2e^x=e^{2x}-(x^2+2)e^x+1$ and the first $5$ terms of the Taylor series of this function are simply $\frac{x^4}{12}$. And the first $5$ terms of the Taylor series of $x^2(e^x-1)^2$ are simply $x^4$. Therefore your limit is indeed $\frac1{12}$.