问:
Convex function of independent vs. correlated random variables redux
Let $f\colon\mathbb R \to \mathbb R$ be convex and $X_1,\ldots,X_n$ be i.i.d. random variables (e.g., uniform). Furthermore, suppose $g_1,\ldots,g_n\colon\mathbb R\to\mathbb R$ are increasing. Is it true that $$\mathbb E f(g_1(X_1) + g_2(X_1) + \cdots + g_n(X_1)) \ge \mathbb E f(g_1(X_1) + g_2(X_2) + \cdots + g_n(X_n))? $$

Like the previous version of this question (Convex function of independent vs. correlated random variables), which is the special case where the $g_i$ are all the identity, it seems like one approach is to show that $g_1(X_1) + \cdots + g_n(X_1)$ is a mean-preserving spread of $g_1(X_1) + \cdots + g_n(X_n)$. As before, I am also curious if there is a name for this sort of result/inequality.

答:
Here is my proof using second-order stochastic dominance:

The inequality follows if we show that $\sum_i g_i(X_1)\succeq \sum_i g_i(X_i)$ in the sense of second-order stochastic dominance (see, e.g., https://www.sciencedirect.com/science/article/pii/0165489687900618). Clearly, $\mathbb E \sum_i g_i(X_1) = \mathbb E \sum_i g_i(X_i)$.
So it remains to check that that the integrated quantile functions satisfy the second-order stochastic dominance property.

For a random variable $X$, let $Q_X$ denote its quantile function. Let $Y = \sum_i g_i(X_1)$ and $Z = \sum_i g_i(X_i)$. Then, since each $g_i$ is increasing,
$$ \int_\tau^1 Q_Y(t) \,dt = \sum_i \int_\tau^1 Q_{g_i(X_1)}(t)\,dt.$$
We have that $\int_\tau^1 Q_X(t) = \sup_{R\in\mathcal{R}_\tau} \mathbb E(RX)$, where $\mathcal{R}_\tau$ is the set of non-negative random variables (defined on the same probability space as $X$) that take values between $0$ and $1$ and have mean at most $\tau$. This lets us deduce the claimed stochastic dominance:
\begin{align*}
 \int_\tau^1 Q_Y(t) &= \sum_i \int_\tau^1 Q_{g_i(X_1)}(t)\,dt \\
  &= \sum_i \sup_{R\in\mathcal{R}_\tau} \mathbb E(R\cdot g_i(X_1)) \\
  &= \sum_i \sup_{R\in\mathcal{R}_\tau} \mathbb E(R\cdot g_i(X_i))
\\ &\ge \sup_{R\in\mathcal{R}_\tau}\mathbb E\Big(R\sum_i g_i(X_i)\Big) \\
&= \int_\tau^1 Q_Z(t)\,dt.
\end{align*}


问:
Exponential simplification help
Why is it true that

$$
\frac{exp(x)}{1+exp(x)} = \frac{1}{exp(-x) + 1}
$$

Why did we negate $x$?

I tried to divide the left-hand side with $exp(x)$ and got
$$
\frac{exp(x)}{exp(x) + (exp(x))^2}
$$
but I cannot get forward.

答:
Or multiply numerator and denominator by $\exp(-x)$, then simplify using the properties of exponentials: $e^ae^b=e^{a+b}$, and $e^0=1$:
$$
\frac{e^x}{1+e^x}=\frac{e^{-x}}{e^{-x}}\cdot\frac{e^x}{1+e^x}=\frac{e^{-x}e^x}{e^{-x}\cdot 1 + e^{-x}e^x}=\frac1{e^{-x}+1}
$$

ContentLicense: CC BY-SA 4.0

Title: Exponential simplification help


问:
Find the Maximum Trigonometric polynomial coefficient $A_{k}$

Let $n,k$ be given positive integers and $n\ge k$. Let $A_i, i=1, 2, \cdots, n$ be given real numbers. If for all real numbers $x$ we have $$A_{1}\cos{x}+A_{2}\cos{(2x)}+\cdots+A_{n}\cos{(nx)}\le 1$$
Find the maximum value of $A_{k}$.

I don't know if this question has been studied

If  $n=2$ it is easy to solve it.

答:
NTstrucker@AoPS's answer:

It is a known result:
$$\max A_k = 2 \cos \frac{\pi}{\lfloor \frac{n}{k} \rfloor + 2}, \ 1\le k \le n.$$
See e.g. Theorem 16.2.4 in [1].

So, $\max A_1 = 2\cos \frac{\pi}{n + 2}$,

$\max A_m = 1$ for $\lfloor \frac{n}{2}\rfloor + 1 \le m \le n$,

$\max A_{\lfloor \frac{n}{2}\rfloor} = \sqrt{2}$.

[1] Qazi Ibadur Rahman and Gerhard Schmeisser, “Analytic Theory of Polynomials”, 2002.


问:
String wrapping around 7 cylinders
If 7 cylinders of R=0.5 cm are tightly wrapped around with a tight  string, what will the string's length be? The figure clearly shows symmetry, any simple and quick way to get the answer?

答: Here is a hint which might help you

Title: String wrapping around 7 cylinders

Body: Here is a hint which might help you


问:
What should I learn next for pure math?
I can do basic single variable calculus which is essentially all you do at A Level in the UK. I also just read "What is mathematics?" by Richard Courant which I found very good. I would like to know where to go after I finish my proof writing and basic discrete maths books I am currently reading. Should I study real analysis? What book? Linear algebra?

I would appreciate advice on where to go next and what book would be good. I am currently 14 and my end goal would to become a pure mathematician.

The reason I ask this question is because I was trying to create a structure and I thought about the Gerard t'Hooft physics one or even the Pure mathematician or statiscian plan but they don't seem particularly specific in what you are meant to do where i.e they call a topic calculus and then a later one vector calculus and then analysis but as a pure mathematician I want to go straight into analysis, of reals obviously. In conclusion, I am asking this because I am going to go by steps now and choose the next topic as it comes.

Thank you in advance.

EDIT:

Would this work:

In order...

Elementary Discrete Maths,
Real Analysis,
Linear Algebra,
ODE's,
Probability,
Fourier Analysis,
Complex Analysis,
PDE's,
Graduate stuff which I will get to when needed.

I'm not sure if probability is necessary but I think it'll be interesting.

答: Start with calculus and after that do linear algebra. For calculus I can recommend you Calculus Early Transcendentals by Stewart, and for Linear Algebra I recommend Linear Algebra: A modern Introduction by Poole. I'd start with calculus though. That'll be challenging enough at this stage.</p>

ContentLicense: CC BY-SA 3.0

Title: What should I learn next for pure math?


问:
Proof verification: sequence $ a_n = (-1)^n $ does not converge
Theorem: Show that the sequence $ a_n = (-1)^n $ for all $ n \in \mathbb{N}, $ does not converge.

Proof: Suppose that there exists a limit $L$ such that $ a_n \rightarrow L  $. Specifically, for $ \epsilon = 1 $ there exists $ n_0 $ s.t. for all $ n > n_0$ then $|(-1)^n-L|<1$, from this we can also infer that $|(-1)^{n+1}-L|<1$ for all $ n > n_0 $.

Let $ n > n_0 $ be arbitrary, then $ 2 = | (-1)^n - (-1)^{n+1} | \leq | (-1)^n - L | + | L - (-1)^{n+1} | < 1 + 1 = 2  $, so since $ n > n_0 $ was arbitrary , hence for all $ n>n_0 $ we have 2<2 which is false, hence we get a contradiction.

My question: Regarding to when I wrote " $ n>n_0 $, we have 2<2 which is false, hence we get a contradiction. "

I'm trying to understand with respect to what there is a contradiction; we get that any statement of the form $ \forall n>n_0 (P(n) )$ is false ( since we got $ \forall n>n_0 . 2<2 ) $ [ and $ P(n) $ is a statement that depends on $ n $ ], and since we already have in the proof the statement " for all $ n > n_0$ then $|(-1)^n-L|<1$ " , then we get a contradiction. Is this correct?

Note: I know there are answers on this sequence's divergence, but they didn't really help me find an answer to my question.

答:
Your statement $P(n):= 2< 2$ does not depend on $n$.

Although if $\lim (-1)^n = L$ then there would be an $n_0$ so that $n > n_0\implies 2< 2$.

But no such $n_0$ can exists because $P(n)$ is false independent of $n$.

.....

Alternatively (although this is a bit abstract for my taste) $n > n_0\implies 2< 2$ and $2< 2$ is false; implies $n > n_0$ is false and we are (in theory) claiming there is an $n_0$ that is is an upper bound for all integers.


答:
Curved lengths sum to the length of perimeter of the circle plus six diameters.

$$ 6 \left(\dfrac{2 \pi R}{6}+2 R\right)$$

Title: String wrapping around 7 cylinders

Body: If 7 cylinders of R=0.5 cm are tightly wrapped around with a tight  string, what will the string's length be? The figure clearly shows symmetry, any simple and quick way to get the answer?


问:
Problem with operations on exponents
Okay, this is a very basic doubt, and I would like to know where exactly the mistake is.

Let's take the equality $1^2=1$

Now I do it this way Step 1: $1^2 = 1^\frac{4}{2}$

Step2: $\implies (1^4)^\frac{1}{2}$

Step3: $\implies 1^\frac{1}{2} = \pm 1$

The answer should be only one but I am also getting minus one.
I know it obviously is a mistake, but I would like to know which exact step and in which particular operation the mistake is....

答:
Cameron's answer is probably the best one to give, but there's another aspect which I think is worth pointing out.

Look at the following argument:
$$x=0\implies x(x-1)=0\implies x=0,1$$

This is completely true! There's nothing actually wrong with the conclusion that "IF $x=0$ THEN $x=0,1$". You're just being less specific. What you can't do is reverse the logic.

Title: Problem with operations on exponents


问:
Which grows faster $n^{\log(\log n)}$ or $(\log n)^{\log n}$
I usually just try playing around with properties of log, but it doesn't help in this case.</p>

答: they are the same...............really...</p>


问:
How do you compute the value of a multivariable limit?
I've been wondering this as I've tried to solve problems by parameterizing variables to f.e x = $\lambda(t)t$ and y = $\mu(t)t$ and reducing by dividing common t factors. I've looked online for methods to compute multivariable limits and I have found no one who shows how to do it and almost all of them just "assume" a limit and use an epsilon-delta proof. Also some use x = $rcos(\theta)$ and y = $rsin(\theta)$, but since $sin^2+cos^2 = 1$ doesn't this not work?

As an example take $\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^6+y^2}$. I then find $\frac{t^4\lambda^3(t)\mu(t)}{t^2(\mu^2(t)+t^4\lambda^6(t))}$ which reduces to $\frac{t^2\lambda^3(t)\mu(t)}{\mu^2(t)+t^4\lambda^6(t)}$ and thus the limit is $\frac{0}{\mu^2(t)+0} = 0$ as t approaches $0$. This limit doesn't exist however as it is path dependent.

Thus my question is: Is there a way to compute multivariable limits without "guessing" them beforehand, and if so how is it done and what is/isn't a correct method?

Thank you in advance!

答:
I don’t believe there is a rule that fits all multi variable limits. Generally, one looks for continuity at the point and then failing that will apply various different techniques until something sticks (such as polar coordinates, trying many different paths then generalising, and Taylor series).

This is a good thread detailing several ways to approach such problems, hopefully it helps a bit!

Is there a step by step checklist to check if a multivariable limit exists and find its value?

If you’re looking to figure out what the limit is to aid proving it, I would try several paths and if all paths tend to the same point then I would assume that to be the result I try to prove. 3D graphing calculators can help with this!!


答:
I don't think there's a particularly generic approach that will always work. There are a good number of tricks. I'll discuss the $(x, y) \to (0, 0)$ case for simplicity.

Fix $m$, substitute $y=mx$, and let $x \to 0$. If the result depends on $m$, the limit does not exist. This is often a quick-and-dirty way to see if a limit "obviously" doesn't exist, but it cannot prove the limit does exist.

Example: $\lim_{(x, y) \to 0} \frac{xy}{x^2 + y^2}$ does not exist since $\lim_{x \to 0} \frac{x(mx)}{x^2 + (mx)^2} = \frac{m}{1+m^2}$ depends on $m$.

Polar: substitute $x=r\cos\theta, y=r\sin\theta$ and let $r \to 0$. If the result is the same no matter what $\theta$ is doing, the limit exists and is that value. Otherwise the limit does not exist. This approach will often give insight even if it doesn't immediately work out.

Example: $\lim_{(x, y) \to 0} \frac{x^2 y}{x^2 + y^2} = 0$ since $\lim_{r \to 0} \frac{r^3\cos^2\theta\sin\theta}{r^2\cos^2\theta+r^2\sin^2\theta} =
lim_{r \to 0} r (\cos^2\theta\sin\theta) = 0$.

Series expansions. L'Hopital's Rule is fundamentally an example of this philosophy, so I won't write out an example. While L'Hopital's Rule is a single-variable phenomenon, you could still try a multivariable Taylor expansion. I haven't personally seen that in practice.

Inspired substitutions. This requires "insight", whatever that means. Often the idea is something like, "well, if $x$ were really really small compared to $y$, but not too small, then this term would dominate, and the quotient would be basically 1, so..."

Example: $\lim_{(x, y) \to 0} \frac{x^3 y}{x^6 + y^2}$ does not exist since when $y=mx^3$, we have $\lim_{x \to 0} \frac{x^3 (mx^3)}{x^6 + (mx^3)^2} = \lim_{x \to 0} \frac{m}{1+m^2}$ depends on $m$.

Title: How do you compute the value of a multivariable limit?

Body:
I've been wondering this as I've tried to solve problems by parameterizing variables to f.e x = $\lambda(t)t$ and y = $\mu(t)t$ and reducing by dividing common t factors. I've looked online for methods to compute multivariable limits and I have found no one who shows how to do it and almost all of them just "assume" a limit and use an epsilon-delta proof. Also some use x = $rcos(\theta)$ and y = $rsin(\theta)$, but since $sin^2+cos^2 = 1$ doesn't this not work?

As an example take $\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^6+y^2}$. I then find $\frac{t^4\lambda^3(t)\mu(t)}{t^2(\mu^2(t)+t^4\lambda^6(t))}$ which reduces to $\frac{t^2\lambda^3(t)\mu(t)}{\mu^2(t)+t^4\lambda^6(t)}$ and thus the limit is $\frac{0}{\mu^2(t)+0} = 0$ as t approaches $0$. This limit doesn't exist however as it is path dependent.

Thus my question is: Is there a way to compute multivariable limits without "guessing" them beforehand, and if so how is it done and what is/isn't a correct method?

Thank you in advance!


问:
If $f$ is continuous and its left or right derivative is bounded then $f$ is Lipschitz continuous
Problem: Suppose that $$f:\mathbb{R}\rightarrow\mathbb{R}$$ is continuous. Furthermore, suppose that the right derivative exists everywhere and is bounded, i.e., $|f'_{+}|<M$. (or suppose that the left derivative exists everywhere and is bounded). Then $f$ is Lipschitz continuous.

I tried to prove this by all I managed to show is that $f$ is locally Lipschitz continuous.

My Attempt: From the definition of the right derivative we have
\begin{align}
|f'_{+}(x)| =\left|\lim_{h\rightarrow0^+} \frac{f(x+h)-f(x)}{h}\right|\leq M
\end{align}
Fix $\epsilon>0$. Then from the above we know that there exists $h>0$ such that for all $h'<h$ we have
\begin{align}
\left| \frac{f(x+h')-f(x)}{h'}\right|\leq M+\epsilon
\end{align}
It follows that for all $y\in\mathbb{R}$ such that $|y-x|<h$ we have
\begin{align}
\left| \frac{f(y)-f(x)}{y-x}\right|\leq M+\epsilon
\end{align}
So this shows that $f$ is locally Lipschitz continuous. But I cannot see how to extend this approach to the whole domain.

Further context: I am self-studying Royden and a similar result appears in the chapter on absolutely continuous functions (Chapter 5 in 3rd edition) as an exercise and I was not able to crack it either.

答:
Consider $g(x)=f(x)+Mx$; continuous, right differentiable and $g_{+}'(x)>0$. But now for a fixed $x$ there is $h_x>0$ st for $y\in (x, x+h_x)$ we have $g(y)>g(x)$ so $g$ is locally increasing (from the right); a standard argument implies that $g$ is (non necesarily strictly) increasing on the real line

(assume $y>x, g(y) < g(x)$; let $w=\inf_{z\in [x,y]}g(z) < g(x)$; by the above $x \ne w$ and then $g(x)<g(w-\epsilon)$ so $g(x)=g(w)$ but by definition of $w$, there is $z_n \to w,z_n >w, g(z_n)<g(x)=g(w)$ contradicting the existence of $h_w$ above)

So $f(x)+Mx \le f(y)+My, y>x$ so $f(x)-f(y) \le M(y-x)$

Consider $h(x)=f(x)-Mx$; the same reasoning shows that one is (not necessarily strictly) decreasing on the real line, so $f(x)-Mx \ge f(y)-My, y>x$ so $f(y)-f(x) \le M(y-x)$

Together the two inequalities show $|f(x)-f(y)| \le M|x-y|$


问:
The number of planes that hold the edges of a pyramid
The number of planes which hold the edges of the pentagonal pyramid is .....

My attempt:

There is one plane for the base
And there are $5$ planes for the faces and there is a plane for each non-adjacent pair of edges
Which are $5$ pairs so the number of planes is $1 +5 + 5$ = $11$

Is this correct ?

答: Each of the five triangular faces is in a plane.  Those five planes hold all the edges.  You can do better by taking the base and three of the triangles that do not share an edge.  That gets you down to four planes.</p>


答:
Take the log of each and apply $\log(x^y) = y \log x$ to obtain
$$\log \left(n^{\log(\log n)}\right) = \log(\log n) \log n$$
and
$$\log \left((\log n)^{\log n}\right) = \log n \log(\log n).$$
Because $\log$ is monotonic, the original expressions are also equal to each other.

Title: Which grows faster $n^{\log(\log n)}$ or $(\log n)^{\log n}$

Body: I usually just try playing around with properties of log, but it doesn't help in this case.</p>


问:
How prove this $\sum_{k=1}^{m-1}\dfrac{(-1)^{k+1}}{C_{m}^{k}}=\dfrac{(-1)^{m}+1^m}{m+2}$
prove that
$$\sum_{k=1}^{m-1}\dfrac{(-1)^{k+1}}{C_{m}^{k}}=\dfrac{(-1)^{m}+1^m}{m+2}$$

where $C_{m}^{k}=\dfrac{m!}{(m-k)!k!}$

This problem is my frend ask me,I think we can prove by case$:m=2n,m=2n+1$

Thank you everyone can nice methods.

答:
Since
$$\frac{1}{C^m_k}=(m+1)\frac{\Gamma(m-k+1)\Gamma(k+1)}{\Gamma(m+2)}=(m+1)B(m-k+1,k+1),$$
the sum can be rewritten as
\begin{align}
&(m+1)\int_0^1\left(\sum_{k=1}^{m-1}(-1)^{k+1}\left(\frac{x}{1-x}\right)^k\right)(1-x)^mdx=\\
=&(m+1)\int_0^1\left[x(1-x)^m+(-1)^m(1-x)x^m\right]dx=\\=&(m+1)\left(1+(-1)^m\right)B(2,m+1)=\\
=&\frac{1+(-1)^m}{m+2}.
\end{align}
Here $B(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$ denotes Euler beta function, and the summation over $k$ was performed using geometric series.</p>


问:
Prove Liouville theorem without using mean value property
How can I prove the following Liouville theorem without using the mean value property?

If $u$ is harmonic on $\mathbb{R}^n$ and $\int_{\mathbb{R}^n}|D u|^2 dx \leq C$ for some $C > 0$, then $u$ is constant.

The proof that I know indeed uses the mean value property for harmonic functions.

答:
Let me first say that the statement $$ \int_{\mathbb{R}^n} \vert Du \vert^2 d x = -\int_{\mathbb{R}^n} u \Delta u d x$$ which is in the comments is false - integration by parts only works on bounded sets. As a simple counter example to the above claim take $u(x) = x_1$.

One way to prove this for $n>2$ without the Mean Value Formula makes use of a monotonicity formula. This is kind of similar to the Mean Value Formula in spirit (you can think of the MVF as a monotonicity formula where the monotone quantity is in fact constant).

Let $B_r$ be a ball of radius $r$. Then $$\phi :  r\mapsto \frac{1}{r^{n-2}} \int_{B_r} \vert Du \vert^2 d x $$ is nondecreasing.

Observe that \begin{align*}
\frac{d}{dr} \bigg ( \frac{1}{r^{n-2}} \int_{B_r} \vert Du \vert^2 d x\bigg ) &= - \frac{n-2}{r^{n-1}} \int_{B_r} \vert Du \vert^2 d x + \frac 1 {r^{n-2}} \int_{\partial B_r} \vert Du \vert^2 d \mathcal{H}^{n-1}.
\end{align*} It is possible to show that for harmonic functions \begin{equation}(n-2)\int_{B_r} \vert Du \vert^2 d x = r \int_{\partial B_r} \vert Du \vert^2 - 2 \bigg ( \frac{\partial u}{\partial \nu }\bigg )^2d \mathcal{H}^{n-1}. \label{eq:2} \end{equation} (To prove this calculate $ \int_{B_r} \mathrm{div} (x \vert Du \vert^2) dx $ in two different ways: first by using the divergence theorem and secondly by expanding $\mathrm{div} (x \vert Du \vert^2)$ then applying integration by parts to one of the terms. Another way to prove this is using Neother's Theorem, see Chapter 8.6.2  in Partial Differnetial Equations by Lawrence C. Evans.) Hence, it follows $$\frac{d}{dr} \bigg ( \frac{1}{r^{n-2}} \int_{B_r} \vert Du \vert^2 d x\bigg ) = \frac 2 {r^{n-2}} \int_{\partial B_r} \bigg ( \frac{\partial u}{\partial \nu }\bigg )^2d \mathcal{H}^{n-1} \geqslant 0.$$ However, by our assumption $$ 0 \leqslant \phi(r) \leqslant \frac{C}{r^{n-2}} \to 0$$ as $r \to \infty$ provided $n>2$. Hence, $\phi(r) = 0$ for all $r >0$, so $u$ is a constant. The only issue with this is it doesn't apply to the case $n=2$.


问:
Where can I make (2D and) 3D cartesian coordinate system?
I'm in need of a program that makes 3D cartesian coordinate system. I have to make a presentation and I have to make my own referentials. I've tried programs like Paint 3D and desmos and I couldn't do it, because I can never make one right. I needed one that would give me the model, where I can add coordinates or not, label with what I want, where I can rotate them as I want or change places and have several in the same image.

I had images of what I meant but I don't have yet 10 reputation to do it, but I think I made myself clear anyway.

答: I would definitely recommend GeoGebra 6 (https://www.geogebra.org/)</p>


问:
Limits and ordered sequences

I'm having trouble proving this for an arbitrary metric space. It's fairly simple in the R case but I'm unable to work out the trick needed for general cases.
I'd appreciate any help. The concept seems intuitive but tough to prove.

答:
If $t \lt s$, then $A_t \cap B_s = \varnothing.$  Let $d(s, t)=2k$.  Then by the triangle inequality and trichotomy of total orders, $(A_t \cup B(t, k)) \cap (B_s \cup B(s, k))=\varnothing.$

But the left side is an open set containing $t$ so $\exists N_t \in \Bbb N~\forall n \gt N_t~(t_n \in A_t \cup B(t, k))$.  Similarly, the right-hand side is an open set containing $s$, so $\exists N_s \in \Bbb N~\forall n \gt N_s~(s_n \in B_s \cup B(s, k))$.

Choose $N= \max(N_t, N_s)$.  Then $n \gt N \Rightarrow t_n \lt s_n$, contradicting our assumption that $\forall n \in \Bbb N~(s_n \leq t_n)$.  Thus, it can't be the case that $t \lt s$, so it must be the case that $s \leq t$.

You need the assumption that you're working in a metric space (or some other separation axiom) to provide some kind of topological "separation" between $s$ and $t$.  Otherwise it's possible that every open set containing $s$ also contains $t$, or vice versa.

Title: Limits and ordered sequences


问:
How many homomorphisms are there of $\mathbb{Z}/{3\mathbb{Z}} $ into $S_3$?
Studying groups of order $12$, specifically, the semi-direct products that appear in these groups, I come across something that I do not fully understand.

In Dummit (pag. 183 groups order 12) you need to see the amount of homomorphisms that there between $T\simeq \mathbb{Z}/{3\mathbb{Z}}$ and ${\rm Aut}(V)\simeq S_3$. The argument they give is the following:

$\mathbb{Z}/{3\mathbb{Z}}$ is cyclic. Then let $T=\langle y\rangle $and there is a unique subgroup of ${\rm Aut}(V)$ of order $3$, say $\langle \gamma\rangle$ . This, there are three possible homomorphisms from $T$ into ${\rm Aut}(V)$:

$\varphi_i:T\to{\rm Aut}(V)$ defined by $\varphi_i(y)=\gamma^i$, $i=0,1,2.$

Question: Why are there only three possible homomorphisms from $T$ into ${\rm Aut}(V)$?

I know it has to do with the order of the domain of the homomorphism but I don't see why the possible homomorphisms need to reach a subgroup with the same order of their domain (for the homomorphism to be well defined)

答:
Say $G$ is cyclic of order $n$ with generator $g$ and $T \colon G \to K$ a group morphism. Then $T(g)^n = T(g^n) = T(1) = 1$, so the order of $T(g)$ divides $n$. Likewise, if $x \in K$ has order divisible by $n$, you can check that $T(g^k) := x^k$ yields a morphism.

Hence $\hom_{\mathsf{Grp}}(\mathbb{Z}/n\mathbb{Z}, K)$ is in correspondence with the elements of $K$ whose order is divisible by $n$. When $n=3$ there are not many options: these are precisely the elements of order $1$ or $3$.

Now, an element $x$ of order $3$ generates a subgroup $\langle x\rangle = \{1,x,x^2\}$ of order three. If there is only one such group in $K$, then all elements of order three are just $x$ and $x^2$. Together with $1$, they determine all maps $\mathbb{Z}/3\mathbb{Z} \to K$.


问:
Evaluate $\int_0^\infty \dfrac{e^{-2t}\sin(-t)+e^{-3t}\sin(-t)}{-t}\; dt$ by Laplace transform
$\require{cancel}$

Note: All logarithmic functions in the post include the absolute value function (this doesn't change the convergence of the variables).

For an entrance exam i got the integral

$$\int_0^\infty \dfrac{e^{-2t}\sin(-t)+e^{-3t}\sin(-t)}{-t}\; dt$$

The exercise is to calculate it by the Laplace transform.

By the frequency domain integration theorem (only in unilateral), it holds that:

$$\mathcal{L}\left[\frac{f(t)}{t} \right]= \int_s^\infty F(\omega)\; d\omega = \int_s^\infty \mathcal{L}(f(t))\; d\omega $$
$\omega$ being our dummy variable here

Let our $f(t)$ time-domain function be $e^{-\alpha t}\sin(-t)$

$$\mathcal{L}\left[\frac{f(t)}{t}\right]= \mathcal{L}\left[\frac{e^{-\alpha t}\sin(-t)}{t} \right] =\int_s^\infty \mathcal{L}\left[{e^{-\alpha t}\sin(-t)}\right]\; d\omega $$

$$\mathcal{L}\left[e^{-\alpha t}\sin(\beta t)\right]= \dfrac{\beta}{(\omega+\alpha)^2+\beta^2} \hspace{10mm}\alpha,\beta \in \mathbb{R} $$

If $\beta=-1$
$$\mathcal{L}\left[e^{-\alpha t}\sin(-t)\right]= \dfrac{-1}{(\omega+\alpha)^2+(-1)^2} = \dfrac{-1}{(\omega+\alpha)^2+1} $$

$$\int_s^\infty \dfrac{-1}{(\omega+\alpha)^2+1}\; d\omega = -\int_s^\infty \frac{1}{\omega^2 + 2\alpha\omega + \alpha^2 +1}\; d\omega =
- \int_s^\infty \frac{\frac{-i}{2}}{\omega-\alpha-i}\;d\omega \; - \int_s^\infty \frac{\frac{i}{2}}{\omega-\alpha+i}\;d\omega=$$

$$=-\frac{i}{2} \left\{ \int_s^\infty -\frac{1}{\omega-\alpha-i}\;d\omega + \int_s^\infty \frac{1}{\omega-\alpha+i}\;d\omega  \right\}= -\frac{i}{2} \bigg( \left[ -\ln(\omega-\alpha-i) \right]_s^\infty + \left[\ln(\omega-\alpha+i)\right]_s^\infty       \bigg)= $$

$$\stackrel{\color{red}{\Delta}}{=}-\frac{i}{2} \bigg(  \left[\cancelto{-\infty}{-\ln(\infty)}+\ln(s-\alpha-i)\right] + \left[\cancelto{\infty}{\ln(\infty)}-\ln(s-\alpha+i)\right] \bigg)= \boxed{-\frac{i}{2}\ln\left(\frac{s-\alpha-i}{s-\alpha+i} \right)} $$

Thus the integral

$$\int_0^\infty \frac{e^{-2t}\sin(-t)+e^{-3t}\sin(-t)}{-t}\; dt = -\int_0^{\infty}\frac{e^{-2t}\sin(-t)+e^{-3t}\sin(-t)}{t}\; dt$$

is

$$-\left[-\frac{i}{2}\ln\left(\frac{s-2-i}{s-2+i}\right) + -\frac{i}{2}\ln\left(\frac{s-3-i}{s-3+i}\right)\right]= \frac{i}{2}\ln \left(\frac{s-2-i}{s-2+i}\cdot \frac{s-3-i}{s-3+i} \right)=$$

$$= \boxed{\frac{i}{2}\ln\left(\frac{\left(s^2-5s+5\right)+i\left(5-2s\right)}{\left(s^2-5s+5\right)-i\left(5-2s\right)} \right)} $$

The problem here is that i cannot really take the Inverse Laplace Transform of a rational function which is inside a logarithm.

How could i proceed?

If someone is confused by the $\color{red}{\Delta}$ step:
$$\color{red}{\Delta}:=-\frac{i}{2} \bigg( \left[ -\ln(\omega-\alpha-i) \right]_s^\infty + \left[\ln(\omega-\alpha+i)\right]_s^\infty  \bigg)=-\frac{i}{2} \left\{ \int_s^\infty -\frac{1}{\omega-\alpha-i}\;d\omega + \int_s^\infty \frac{1}{\omega-\alpha+i}\;d\omega  \right\}=$$

$$=-\frac{i}{2} \bigg( \left[ -\ln(\omega-\alpha-i) \right]_s^\infty + \left[\ln(\omega-\alpha+i)\right]_s^\infty\bigg) = \frac{i}{2}\left(\Bigg[-\ln(\infty-\alpha-i)\underbrace{--}_{+}\ln(s-\alpha-i)\Bigg]+\Bigg[\ln(\infty-\alpha+i)-\ln(s-\alpha+i)\Bigg]\right)=$$

$$=-\frac{i}{2}\left(\color{blue}{\lim_{\sigma\to\infty}\left(-\ln(\sigma-\alpha-i)+\ln(\sigma-\alpha+i)\right)}+\ln(s-\alpha-i)-\ln(s-\alpha+i)\right)=$$

$$=-\frac{i}{2}\left(\color{blue}{\lim_{\sigma\to\infty}\ln\left(\dfrac{\sigma-\alpha-i}{\sigma-\alpha+i}\right)}+\ln\left(\dfrac{s-\alpha-i}{s-\alpha+i}\right)\right)$$

$$=-\frac{i}{2}\left(\color{blue}{\underbrace{\ln\underbrace{\lim_{\sigma\to\infty}\left(\dfrac{\sigma-\alpha-i}{\sigma-\alpha+i}\right)}_{1}}_{0}}+\ln\left(\dfrac{s-\alpha-i}{s-\alpha+i}\right)\right)$$

答:
This is easily done by doing the following.

Note that $$\mathcal{L} \left( \frac{\sin t}{t} \right) = \tan^{-1} \left( \frac{1}{s} \right) $$

(use the formula $\mathcal{L} \left( \frac{f(t)}{t} \right) = \int_s ^\infty \mathcal{L} (f)(w) \ dw$ and an arctan identity), so we can do the following:

\begin{align*}
\int_0^\infty \dfrac{e^{-2t}\sin(-t)+e^{-3t}\sin(-t)}{-t}\; dt &= \int_0^\infty \dfrac{-e^{-2t}\sin(t)-e^{-3t}\sin(t)}{-t}\; dt \\
&= \int_0^\infty \dfrac{e^{-2t}\sin(t)+e^{-3t}\sin(t)}{t}\; dt \\
&= \int_0^\infty \frac{e^{-2t} \sin t}{t} \ dt + \int_0 ^\infty \frac{e^{-3t} \sin t}{t} \ dt \\
&= \mathcal{L} \left( \frac{\sin t}{t} \right) \Big\vert_{s = 2} + \mathcal{L} \left( \frac{\sin t}{t} \right) \Big\vert_{s = 3} \\
&= \tan^{-1} \left( \frac {1}{2} \right) + \tan^{-1} \left( \frac {1}{3} \right) \\
&= \frac{\pi}{2}.
\end{align*}

For the last line, use the tangent sum identity - let $\tan \alpha = 1/2$ and $\tan \beta = 1/3$. Then

\begin{align*}
\tan(\alpha + \beta) &= \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \\
&= \frac{1/2 + 1/3}{1 - (1/2)(1/3) } \\
&= \frac{5/6}{5/6} \\
&= 1
\end{align*}

so $\alpha + \beta = \frac{\pi}{2}.$

TL;DR, You weren't asked to take the Laplace transform of the integrand, rather, you were to recognize this as a Laplace transform evaluated at a point.


问:
imaginary numbers - how can I understand them - especially as they occur in 'roots' of polynomials?
In another question here, about roots of equations being imaginary, the accepted answer
said something interesting about "imaginary" (as a technical word in math) not meaning "not real".  I understand that imaginary numbers are "manipulable and usable to do all kinds of things".   And thus, "they exist" at least, in the realm of mathematics.

So my question is this. I was playing around in wxMaxima, and solving polynomials, to get their roots.  In maxima, $\sqrt{-1}$ is represented as %i.

Editor's note: all "%i"s have been replaced with $i$.

Here's an example of a polynomial, say $3x^6 + 4x^4$:

solve($3x^6 + 4x^4$,$x$);

Maxima's results:
[$x=\displaystyle -\frac{2i}{\sqrt{3}}$,$x=\displaystyle \frac{2i}{\sqrt{3}}$,$x=0$]

My exact question:

Is it common to have the same imaginary result twice with only the sign different (both q and -q are imaginary roots)?  (At first I thought it had reported the same root twice as I overlooked the negative sign.)

One can easily plot a real root and see how it intercepts the x axis.  How does one "visualize" or think of an imaginary root -- does it correspond to a local minimum that does not occur at the x axis?

What uses do imaginary roots of polynomials have in math?

答:
For 1:  It's not only common, it's always the case: if you have a polynomial with real coefficients, then the "complex roots" come in what are called "conjugate pairs: if $a+bi$ is a root, with $a,b$ real numbers and $b\neq 0$, then $a-bi$ is also a root.

The explanation has to do with something called "complex conjugation". Complex conjugation is an operation on complex numbers that sends the complex number $a+bi$ to the complex number $a-bi$. This is denoted with a line over the number, that is, $\overline{z}$ denotes the complex conjugate of $z$:
$$\overline{a+bi} = a-bi,\qquad a\text{ and }b\text{ real numbers.}$$
Conjugation respects sums and products:
$$\overline{z+w} = \overline{z}+\overline{w}\text{ and }\overline{z\times w}=\overline{z}\times\overline{w}\text{ for all complex numbers }z\text{ and }w.$$

Also, a number $a+bi$ is real (that is, has $b=0$) if and only if $\overline{a+bi} = a+bi$.

Now suppose you have a polynomial with real coefficients,
$$p(x) = \alpha_nx^n + \cdots + \alpha_1x + \alpha_0$$
and that there are real numbers $a$ and $b$ such that $a+bi$ is a root of this polynomial. Plugging it in, we get $0$:
$$p(a+bi) = \alpha_n(a+bi)^n + \cdots + \alpha_1(a+bi) + \alpha_0 = 0.$$
Taking complex conjugates on both sides, and applying the properties mentioned above (complex conjugate of the sum is the sum of the complex conjugates; complex conjugate of the product is the product of the complex conjugate; complex conjugate of a real number like $\alpha_i$ is itself) we have:
$$\begin{align*}
0 &= \overline{0}\\
&= \overline{\alpha_n(a+bi)^n + \cdots + \alpha_1(a+bi) + \alpha_0}\\
&= \overline{\alpha_n}\overline{(a+bi)}^n + \cdots + \overline{\alpha_1}\overline{(a+bi)} + \overline{\alpha_0}\\
&= \alpha_n(a-bi)^n + \cdots + \alpha_1(a-bi) + \alpha_0\\
&= p(a-bi).
\end{align*}$$
So if $p(a+bi)=0$, then $p(a-bi)=0$ as well.

Complex roots don't correspond in any reasonable way to "maxima and minima" of the polynomials. They correspond to irreducible quadratic factors. It's hard to visualize, from a graph of the polynomial in the real axis, where (or even whether) it has complex roots.

The polynomial you tried, $3x^6+4x^4$, can be written as
$$3x^6 + 4x^4 = x^4(3x^2 + 4).$$
The product is equal to $0$ if and only if one of the factors is $0$, so either $x=0$ or $3x^2+4 = 0$; for the latter, you would need $x^2 = -\frac{4}{3}$, which is impossible with real numbers; that is, $3x^2 + 4$ is an irreducible quadratic, which is where the complex roots (in this case, purely imaginary) come from. Since you want a (complex) number whose square is $-\frac{4}{3}$, one possibility is to take a real number whose square is $\frac{4}{3}$, namely $\frac{2}{\sqrt{3}}$, and then multiply it by a complex number whose square is $-1$. There are two such numbers, $i$ and $-i$, so the two complex roots are $\frac{2}{\sqrt{3}}i$ and $\frac{2}{\sqrt{3}}(-i) = -\frac{2}{\sqrt{3}}i$.

Each distinct irreducible quadratic factor will give you a pair of conjugate complex roots.

As a consequence of the Fundamental Theorem of Algebra, every polynomial with real coefficients can be written as a product of polynomials that are either degree $1$, or irreducible quadratics.

Might as well mention one application here: one reason people were interested in knowing that the Fundamental Theorem of Algebra was true (that every real polynomial could be written as a product of linear and irreducible quadratic polynomials) was to ensure that the method of partial fractions would always be available to solve an integral of a rational function (a function given as the quotient of a polynomial by another polynomial).

Other uses are too many to mention, but as Yuval mentions, this question and its answers may get you started.


问:
birth and death processes
Suppose we have a system of N balls, each of which can be in one of two
boxes. A ball in box I stays there for a random amount of time with exponential(lambda)
distribution and then moves instantaneously to box II. A ball in box II stays there
for a random amount of time with exponential(lambda) distribution and then moves
instantaneously to box I. All balls act independently of each other. Let Xt be the
number of balls in box I at time t.

a) I'm trying to show that X is a birth and death process and specify
the birth and death rates.

b) How can we find the stationary distribution of the process.

For part (a), if we can show it satisfies a Yule process, this is essentially what we're trying to do. And for part (b), I also want some clarification on the detailed balanced equations for this problem.

答:
As the comments mention, this is indeed a continuous time version of the Ehrenfest model and there are papers written considering this problem like

Karlin, Samuel, and James McGregor. Ehrenfest urn models. Journal of Applied Probability 2.2 (1965): 352-376.

The model is a birth-death process, births when there are $i$ balls in box I are at rate $(N-i)\lambda$ and deaths at rate $i\lambda$. These formulas cover the special cases too. When box I is empty then there are no deaths, and births happen at rate $N\lambda$, while when box I is full there are no births and deaths occur at rate $N\lambda$.

The balance equations (writing $\pi_i$ for the probability of being in state $i$) are
$$ \begin{align}
(N\lambda)\pi_0 &= (N\lambda)\pi_1\\
(N\lambda)\pi_1 &= \lambda \pi_0 + (N-1)\lambda \pi_2 \\
(N\lambda)\pi_2 &= 2\lambda \pi_1 + (N-2)\lambda \pi_3 \\
& \vdots\\
(N\lambda)\pi_{N-1} &= (N-1)\lambda \pi_{N-2} + \lambda \pi_N \\
(N\lambda)\pi_N &= N\lambda \pi_{N-1} \\
\end{align}$$

which you then need to solve along with the constraint that $\sum_{i=0}^N \pi_i = 1$ to find the stationary probability distribution.

Title: birth and death processes


答:
Note that maps $\mathbb{Z} \to G$ correspond exactly to elements of $G$ - $f$ corresponds to $f(1)$. Note that maps $\mathbb{Z} / 3\mathbb{Z} \to G$ correspond exactly to maps $f : \mathbb{Z} \to G$ such that $3 \mathbb{Z} \subseteq \ker(f)$. And note that $3 \mathbb{Z} \subseteq \ker(f)$ iff $f(3) = e$. So maps $\mathbb{Z} / 3 \mathbb{Z} \to G$ correspond to elements $x \in G$ such that $x^3 = e$.

Thus, the real question is: what automorphisms $f \in {\rm Aut}(V)$ are there such that $f^3 = e$? Or equivalently, we must find $f$ such that the order of $f$ is a factor of 3 - that is, such that the order is either 1 or 3. There are exactly 3 such maps.

Title: How many homomorphisms are there of $\mathbb{Z}/{3\mathbb{Z}} $ into $S_3$?

Body:
Studying groups of order $12$, specifically, the semi-direct products that appear in these groups, I come across something that I do not fully understand.

In Dummit (pag. 183 groups order 12) you need to see the amount of homomorphisms that there between $T\simeq \mathbb{Z}/{3\mathbb{Z}}$ and ${\rm Aut}(V)\simeq S_3$. The argument they give is the following:

$\mathbb{Z}/{3\mathbb{Z}}$ is cyclic. Then let $T=\langle y\rangle $and there is a unique subgroup of ${\rm Aut}(V)$ of order $3$, say $\langle \gamma\rangle$ . This, there are three possible homomorphisms from $T$ into ${\rm Aut}(V)$:

$\varphi_i:T\to{\rm Aut}(V)$ defined by $\varphi_i(y)=\gamma^i$, $i=0,1,2.$

Question: Why are there only three possible homomorphisms from $T$ into ${\rm Aut}(V)$?

I know it has to do with the order of the domain of the homomorphism but I don't see why the possible homomorphisms need to reach a subgroup with the same order of their domain (for the homomorphism to be well defined)


问:
Number of paths on $\mathbb Z^d$
Given two points $x,y \in \mathbb Z^d$ I am curious whether there is a formula for the number of paths $P(x,y,n)$ of length $n$(=number of steps) between $x$ and $y.$

Although an explicit expression would be nice, I am particularly curious to know if there exist sharp upper bounds on the scaling of $P(x,y,n)$ as a function of $n$. In particular, is it polynomial, exponential, factorial etc.?

So to be precise, the paths are allowed to be self-intersecting (and you can go back and forth,yes) and can move horizontically and vertically.

Please let me know if you have any questions.

答:
Here is an algorithm to compute $P(x,y,n)$ in $O(n^{\lceil d/2\rceil})$ time. It suffices to compute $P(0,a,n)$, which I abbreviate as $P_d(a,n)$. with a subscript to emphasize the dimension. We can show that
$$
P_d(a,n)=\sum_{k=0}^n\binom{n}k P_2((a_1,a_2),k)\cdot P_{d-2}((a_3,a_4,\dots,a_n),n-k)
$$
The summation variable $k$ represents the number of steps which are in either the $x_1$ or the $x_2$ direction. We choose which of the $n$ steps are in those two directions in $k$ ways, then spend those $k$ steps changing the first two coordinates to $(a_1,a_2)$, and finally use the remaining $n-k$ steps to change the remaining coordinates to $(a_3,\dots,a_n)$.

Now, we can compute $P_2((a_1,a_2),k)$ in $O(1)$ time, since
$$
P_2((a_1,a_2),k)=\binom{k}{\frac12(k+a_1+a_2)}\cdot \binom{k}{\frac12(k+a_1-a_2)}
$$
with the convention that this is zero when the lower index is fractional. (For a proof, see Number of N step Random walks on 2D infinite grid or lattice). This will require precomputing the first $n$ rows of Pascal's triangle so these binomial coefficients can be looked up in constant time. (For $d\le 2$, the precomputation is not necessary).

All that remains is to compute $P_{d-2}((a_3,a_4,\dots,a_n),n-k)$ for each $k$. There are $O(n)$ terms to compute, and by induction each can be computed in $O(n^{\lceil d/2\rceil-1})$ time, so the entire sum is doable in $O(n^{\lceil d/2\rceil})$ time.

Title: Number of paths on $\mathbb Z^d$


问:
Sequence of critical points converging to $0$
I would like an example of a function with the following properties,

A differentiable function $f:\mathbb{R}\to\mathbb{R}$ such that $0$ is the limit of a sequence of critical points of $f$, and $f'(0)>0$.

The examples I was able to think of, always fail to be differentiable at $0$, would anyone have any idea of an example?

I believe that the function must oscillate infinitely times close to $0$ so that it is possible to create the sequence of critical points, so I am finding it difficult to think of a function that is differentiable at zero, because in its neighborhood, the function is strange.

$c$ is a critical point if $f'(c)=0$

答:
$$
f(x) = \begin{cases}
x+x^2\sin \frac{1}{x},\qquad & x \ne 0,
\\
0,\qquad & x = 0
\end{cases}
$$

addd

The derivative exists at zero.
$$
f'(0) = \lim_{x \to 0}\frac{f(x)}{x} = \lim_{x \to 0} 1+ x \sin\frac{1}{x} = 1 .
$$
But the derivative is not continuous at zero.


答: I'd suggest Calcplot3D and Math3D, both of which allow you to add coordinates, label with what you want, where you can rotate it, etc.

Title: Where can I make (2D and) 3D cartesian coordinate system?

Body:
I'm in need of a program that makes 3D cartesian coordinate system. I have to make a presentation and I have to make my own referentials. I've tried programs like Paint 3D and desmos and I couldn't do it, because I can never make one right. I needed one that would give me the model, where I can add coordinates or not, label with what I want, where I can rotate them as I want or change places and have several in the same image.

I had images of what I meant but I don't have yet 10 reputation to do it, but I think I made myself clear anyway.


问:
Exponential simplification help
Why is it true that

$$
\frac{exp(x)}{1+exp(x)} = \frac{1}{exp(-x) + 1}
$$

Why did we negate $x$?

I tried to divide the left-hand side with $exp(x)$ and got
$$
\frac{exp(x)}{exp(x) + (exp(x))^2}
$$
but I cannot get forward.

答:
Perhaps you should tell us what $\exp(x)$ is defined as.

The simplest explanation is that you know there is a number $e$ and you know what raising a base to a power is and $\exp(x)$ is simply defined as $\exp(x)= e^x$.  And therefore $\exp(-x) = e^{-x}= \frac 1{e^x} = \frac 1{exp(x)}$ (and we know that $e^{-x}= \frac 1{e^x}$ because $e^{-x}e^{x} = e^{-x+x} = e^0 = 1$).

But that assume you know what raising numbers to a power means.  We usually use the concept of $\exp(x)$ to define what raising a number to a power is by $b^x = \exp(x\cdot \ln b)$ where $\ln b$ is defined as $\int_1^b \frac 1t dt$ and $\exp(x) = \ln^{inverse}(x)$ (i.e. $\exp(x)$ is the value $k$ so that $\ln (k) = x$.

Then its a matter of proving that $\exp(x+y) = \exp x\cdot \exp y$.  If we know that then we have to know that $\exp(x)\exp(-x) = \exp(x-x)=\exp(0) = 1$ and therefore $\exp(-x) =\frac 1{\exp(x)}$.

I may provide a proof of that in a few hours.

Title: Exponential simplification help


问:
Uniform convergence of $f_n(x)=\frac{x^n}{n!}$
Let $f_n: [0,1]\rightarrow \mathbb{R}$ be a sequence of functions defined by $$f_n(x)=\frac{x^n}{n!}.$$
Prove that $f_n(x)$ converges uniformly to $0$.

We know that $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$ is this enought to prove that $f_n \rightarrow 0$?

答:
The answer by lorenzo was what I had in mind with my comment. But if you really want to beat this problem with a hammer, then here goes:
\begin{align}
\sum_{n=0}^{\infty}\sup_{x\in[0,1]}\left|\frac{x^n}{n!}\right| &=\sum_{n=0}^{\infty}\frac{1}{n!} = e < \infty.
\end{align}
Hence, by the vanishing criterion for convergent series, it follows that $\lim\limits_{n\to\infty}\sup\limits_{x\in[0,1]}\left|\frac{x^n}{n!}\right| = 0$, which is precisely the definition (or atleast it's very easy to see this is equivalent) of uniform convergence to $0$ over the the interval $[0,1]$.


问:
What should I learn next for pure math?
I can do basic single variable calculus which is essentially all you do at A Level in the UK. I also just read "What is mathematics?" by Richard Courant which I found very good. I would like to know where to go after I finish my proof writing and basic discrete maths books I am currently reading. Should I study real analysis? What book? Linear algebra?

I would appreciate advice on where to go next and what book would be good. I am currently 14 and my end goal would to become a pure mathematician.

The reason I ask this question is because I was trying to create a structure and I thought about the Gerard t'Hooft physics one or even the Pure mathematician or statiscian plan but they don't seem particularly specific in what you are meant to do where i.e they call a topic calculus and then a later one vector calculus and then analysis but as a pure mathematician I want to go straight into analysis, of reals obviously. In conclusion, I am asking this because I am going to go by steps now and choose the next topic as it comes.

Thank you in advance.

EDIT:

Would this work:

In order...

Elementary Discrete Maths,
Real Analysis,
Linear Algebra,
ODE's,
Probability,
Fourier Analysis,
Complex Analysis,
PDE's,
Graduate stuff which I will get to when needed.

I'm not sure if probability is necessary but I think it'll be interesting.

答:
You should study both, so you I would advise you to start studying the one that interests you the most.

I prefer analysis so I'll recommend you a couple of books about the subject, I took a course in real analysis last year and one recommended book was the one by Bartle and Sherbert (http://www.amazon.co.uk/Introduction-Real-Analysis-Robert-Bartle/dp/0471433314/ref=sr_1_1?ie=UTF8&qid=1369842686&sr=8-1&keywords=bartle+sherbert), it seemed to me a good book, it introduces some concepts of topology which are really interesting and help you to work with abstract spaces and develops some important result in convergence of series of functions (ie, exchanging the order of operation if you have a good enough kind of convergence) that have been quite useful in all the courses I took in many branches of mathematics and physics (ie, ordinary differential equations), you might find some lacks but I think it's a good book to start. I'd also recommend you Calculus by Spivak (which is useful especially for practising with more "practical" exercises, not just theorethical ones).

I can't recommend you good books of algebra because I'm spanish and I used books which were written in spanish (and I don't think there are english translations). EDIT: I just saw in one of the comments the book of algebra by Hoffmann and Kunze and it reminded me that I tried to use it while studying the subject, it's a hard book but at the same time it can give you a complete understanding of what are your goals while developing all the important results of linear algebra.

But my most important advice should be that you don't have to have any hurry, you're quite young yet and you don't have to feel dissappointed or discouraged if you can't get as much from those books as you had planned originally, you have to work hard, try again or look in a different book if you didn't understand something. There isn't a perfect book, to achieve a good understanding of the subject is helpful looking for information in a bunch of different places but I recommend you the book I stated previously as a good place to start into the subject of analysis of a real variable and to help you get deeper later.

I hope this has been helpful.

Title: What should I learn next for pure math?


问:
Classification problem
I have this question in the assignment and I just don't know how to proceed.

答:
Here, you are asked to find the most likely class of $x$, or in other words, to perform Maximum Likelihood Estimation.

For any $i\in\{1,2,3\}$, by applying Bayes' theorem, we have that
$$\begin{align}\mathbb P(Y=i\mid X=x) &=\frac{f_{X\mid Y=i}(x) \times \mathbb P(Y=i)}{f_X(x)} \\
&= \frac{f_{X\mid Y=i}(x) }{3\times f_X(x)}\\
&\propto f_{X\mid Y=i}(x)\end{align}$$
We do not need to explicitly compute the terms whose value are independent of the value of $Y$.

Now, we know that conditional on $Y = i$, $X$ follows a multivariate normal distribution with mean $\mu_i$ and covariance matrix $\Sigma_i$. The conditional density of $X\mid Y=i$ is thus
$$f_{X\mid Y=i}(x) = \frac{\exp\left(-\frac{1}{2}(x-\mu_i)^T\Sigma_i^{-1}(x-\mu)\right)}{\sqrt{(2\pi)^2|\Sigma_i|}} $$
To classify $x$, you have to pick the class $i$ that maximizes $\mathbb P(Y=i\mid X=x) $.

We see that it is equivalent to choosing the class $i$ that maximizes $f_{X\mid Y=i}(x) $, so now you have all the elements you need to classify $x$.

Title: Classification problem


问:
Prove that there exist four polynomials $p_1,p_2,p_3,p_4$ in $x,y,z$ so that $(x^2+y^2+z^2)^3-8(z^3x^3+x^3y^3+y^3z^3)=p_1^2+p_2^2+p_3^2+p_4^2$

Prove that there exist four polynomials $p_{1}, p_{2}, p_{3}, p_{4}$ in $x, y, z$ so that
$$\left ( x^{2}+ y^{2}+ z^{2} \right )^{3}- 8\left ( z^{3}x^{3}+ x^{3}y^{3}+ y^{3}z^{3} \right )= p_{1}^{2}+ p_{2}^{2}+ p_{3}^{2}+ p_{4}^{2}$$
Source: AoPS/@Ji_Chen_ on.AoPS

I like Ji Chen's sum of squares, which is very hard to make decomposition like they said. Here is the way of thinking of mine about this problem_

if $f= ab+ c= -ad+ e$ with $b, c, d, e\geq 0\Rightarrow f\geq 0,$ we have a new $f$ is an SOS$,\quad f:=\dfrac{cd+ e}{b+ d}$

I think finding SOS like this is very funny. One day, you try substitutions and you succeed, however, another day, that trick is not useful anymore. I found this formula_ on.StackMath 4 years ago. Very amazing result, I have thought of it till now, that's a travelled task.

Body:
The four polynomials $p_i(x,y,z)$ should be a combination of the following terms:
\begin{align}
 \left\{xyz,yx^2,y^2x,zx^2,z^2x,zy^2,z^2y,x^3,y^3,z^3\right\}
\end{align}
other terms would create the different left hand side. I have tried to use brutal force with the help of the deterministic global optimizer BARON starting with 40 unknowns coefficients by minimizing the objective of the form
\begin{align}
((a_{003}^2+b_{003}^2+d_{003}^2+c_{003}^2)-1)^2+
((c_{030}^2+b_{030}^2+d_{030}^2+a_{030}^2)-1)^2+\dots \nonumber
\end{align}

(where $a_{ijk},b_{ijk},c_{ijk},d_{ijk}$ are coefficients to polynomials $p_1,p_2,p_3,p_4$). The calculation recovered the following form of the polynomials $p_1,p_2,p_3,p_4$:
\begin{align}
p_1(x,y,z)&=a_1(yz(y+z)-y^3-z^3) + a_2x^2(y+z)-a_3x(y^2+z^2) \\
p_2(x,y,z)&=b_1yz(z-y)+b_2x^2(y-z)+b_3x(z^2-y^2) \\
p_3(x,y,z)&=c_1yz(y-z)+c_2x^2(y-z)+c_3x(z^2-y^2)+c_4(y^3-z^3) \\
p_4(x,y,z)&=d_1xyz+d_2x(y^2+z^2)-x^3+d_3(y^3+z^3-yz(y+z))
\end{align}
where
\begin{align}
a_1&=0.44778749472139178777752 \\
a_2&=1.6556891275393716966846 \\
a_3&=1.2079562282512847914973 \\
b_1&=-1.9014217828335169269138 \\
b_2&=-0.55113503404381225525555 \\
b_3&=-0.5511281347996555002311 \\
c_1&=1.1890222040768996247806 \\
c_2&=0.75843958935644073537929 \\
c_3&=1.3269533603627028384153 \\
c_4&=0.56886478882077529117822 \\
d_1&=1.8623260525794982367387 \\
d_2&=0.31014344808027582978127 \\
d_3&=0.68984000724248439873065
\end{align}
with error of the objective function $0.43$E-$014$. Hope it helps further. However, analytical expressions of these coefficients should be recovered to check if this solution is really correct.


问:
Arzela - Ascoli application
Let $a>0$ and the sequence $f_n:[-a,a] \rightarrow \mathbb{R}$. Suppose that the following conditions are true:

$\bullet$ There is $k \in \mathbb{N}\cup\{0\}$ such that $f_n$   is $(k+1)$-differentiable in $[-a,a]$;

$\bullet$ $f_n^{(i)}(0)=0 \ \forall \ i \in \{0,1,\dots,k\}$

$\bullet$ $|f_n^{k+1}(x)|\leq C.$

Prove that there is a subsequence in $(f_n)_{n \in \mathbb{N}\cup \{0\}}$ that is uniformly convergent in $[-a,a]$.

I know that this must be like this question Ascoli's Theorem, but i'm having some hard time one this.

答:
Hint: if $f$ is differentiable and $|f'| \leq C$, then $|f(x)-f(y)| \leq C|x-y|$. Moreover, $|f(x)| = |f(x)-0| \leq C|x-0| \leq Ca$.

Apply this to $f_n^k$ for each $n$. The bound on their derivatives is the same for all of them, so the Lipschitz constant you get is the same for all. The proceed likewise unitl you reach each $f_n$, keeping track of how the bound changes. It will be uniform; this will allow you to conclude that $\{f_n\}$ is uniformly bounded and uniformly equicontinuous. Then apply Arzelà-Ascoli.

Title: Arzela - Ascoli application


问:
Dimension of $L^p$ space
Let $X = \{a,b,c\}$, $\mu$ is a measure on $2^X$ with $\mu(\{a\})=0, \mu(\{b\})=1, \mu(\{c\})=\infty$.

What is the dimension of $L^1(\mu)$, $L^2(\mu)$ and $L^\infty(\mu)$?

Consider a function $f$ on $2^X$, then we have

$||f||_{L^1} = |f(a)|\mu(\{a\})+|f(b)|\mu(\{b\})+|f(c)|\mu(\{c\}) = |f(b)|+|f(c)|\cdot\infty$.

Similarly,

$||f||_{L^2} = (|f(b)|^2+|f(c)|\cdot\infty)^{\frac{1}{2}}$, and $||f||_{L^\infty}=\text{esssup}\{|f(a)|,|f(b)|,|f(c)|\}=\sup\{|f(b)|,|f(c)|\}$.

For $f\in L^1(\mu)$ or $f\in L^2(\mu)$, $|f(b)|<\infty$, $f(c)=0$, $f(a)$ can be finite or infinite (two choices), so $L^1(\mu)=L^2(\mu)=3$.

For $f\in L^\infty(\mu)$, $|f(b)|<\infty$, $|f(c)|<\infty$, so $L^\infty(\mu)=4$.

Is this correct?

答:
Note that any function space on $X$ is a subset (up to equivalence classes) of $\mathbb R^X$. The collection $\{\mathbb 1_{\{a\}},\mathbb 1_{\{b\}},\mathbb 1_{\{c\}}\}$ forms a basis for this space, so your dimension of any $L^p$ space is at most $3$. Also your functions shouldn't be allowed to have infinite values technically if you want to talk about the standard dimension of a vector space, because the extended real line is not a field.

Now let $f\in L^1(\mu)$. Then as you say $|f(a)|\mu(\{a\})+|f(b)|\mu(\{b\})+|f(c)|\mu(\{c\})<\infty$. This is only possible if $f(c)=0$. Thus we can write $f=f(a)\mathbb 1_{\{a\}}+f(b)\mathbb 1_{\{b\}}$. However remember that $L^1(\mu)$ is actually the quotient space $\mathcal L(\mu)/\ker(\|\cdot\|_1)=\mathcal L(\mu)/\operatorname{span}(\mathbb 1_{\{a\}})$, so the dimension of $L^1(\mu)$ must be $1$. I leave the rest to you.

Title: Dimension of $L^p$ space


问:
Ordinary differential equation in the plane
Suppose $V(x,y) = \left(\frac{x}{\sqrt{x^2+y^2+1}}, \frac{-y}{\sqrt{x^2+y^2+1}}\right)$ is a vector field in the plane.

Can its integral curve (t) through an arbitrary initial condition (x0, y0) be expressed explicitly in terms of familiar functions? I.e., '(t) = V((t)) for all t, with (0) = (x0, y0).

答: Suppose the initial position of our particle $(x_0,y_0)$ belongs in the first quadrant, $\mathbb{Q}_1$. If $(x(t),y(t))$ is the position of the particle at time $t$, it's not hard to see that $x(t)y(t)=x_0y_0$. In other words, our trajectory $(x(t),y(t))$ is bounded to the portion of the hyperbola $xy=x_0y_0$ that's contained in $\mathbb{Q}_1$. This means $x(t)$ must satisfy the differential equation $$\frac{dx}{dt}=\frac{x^2}{\sqrt{x^4+x^2+x_0^2y_0^2}}$$ Unless you're able to solve this DE for $x(t)$ explicitly, I don't see how you'll be able to express the trajectory for $x(t)$ in terms of familiar functions. Wolfram produces a nasty expression for $\int \frac{\sqrt{x^4+x^2+x_0^2y_0^2}}{x^2}dx$ when we consider $(x_0,y_0)=(3,1)$.


答:
I'm not exactly too sure on what you mean by "basic single variable calculus", but for those who are not from the UK, I figure it's worth briefly outlining the calculus involved in A-Level maths (for those who aren't in the know, A-levels are (usually) spread out over two years, so I'll list them as under years 1 and 2):

Year $1$ - Basic differentiation and integration of polynomials and functions of the form $x^a$ for $ a \neq -1$ (as in most pupils will not even learn the definition of differentiation, just the rule), calculation of stationary points for cubic points and sufficient conditions for the turning point to be a local maxima/minima/stationary point

Year $2$ - Differentiation and integration of functions of the form $ x^{-1}$, trigonometric functions, the exponential function, and the logarithmic function. In this, the chain, product and quotient rules are covered, along with the methods of integration by substitution and integration by parts (without proof).

While the above list is not completely exhaustive, I feel like it covers the majority of what I covered at A-level (which was only a year ago, but a year at university studying maths is a long time!)

Anyways, as to answering the question, I feel like a few good places to start are:

Calculus by Spivak - This is a book which acts as a good introduction to real analysis, building the whole theory of single valued calculus from first principles (this including a treatment of the real number systems)

The Pleasures of Counting by T. W. Körner - While this isn't a textbook as such, I feel like its worth a read by any aspiring mathematician (the reader will also find that the author has a very pleasant writing style, which makes the read very enjoyable)

If you haven't done Further Maths, then looking at the textbooks for the Further Pure courses should also be of interest - if you aren't home schooled then you should be able to get copies from your school, else I would personally recommend to get the MEI FP1  through to FP3 books, along with possibly the differential equations book.

As for linear algebra, I can't think of a really good textbook to do with it, but maybe someone else will have a good idea.

Title: What should I learn next for pure math?

Body:
I can do basic single variable calculus which is essentially all you do at A Level in the UK. I also just read "What is mathematics?" by Richard Courant which I found very good. I would like to know where to go after I finish my proof writing and basic discrete maths books I am currently reading. Should I study real analysis? What book? Linear algebra?

I would appreciate advice on where to go next and what book would be good. I am currently 14 and my end goal would to become a pure mathematician.

The reason I ask this question is because I was trying to create a structure and I thought about the Gerard t'Hooft physics one or even the Pure mathematician or statiscian plan but they don't seem particularly specific in what you are meant to do where i.e they call a topic calculus and then a later one vector calculus and then analysis but as a pure mathematician I want to go straight into analysis, of reals obviously. In conclusion, I am asking this because I am going to go by steps now and choose the next topic as it comes.

Thank you in advance.

EDIT:

Would this work:

In order...

Elementary Discrete Maths,
Real Analysis,
Linear Algebra,
ODE's,
Probability,
Fourier Analysis,
Complex Analysis,
PDE's,
Graduate stuff which I will get to when needed.

I'm not sure if probability is necessary but I think it'll be interesting.


问:
Sketching complex curves in the w- plane for using Cauchy's theorem
I'm trying to understand what is being discussed in the page-386 of Tristan Needham's Visual Complex Analysis Book:

Figures 7:

Figure 8:

I'm having a hard time understanding what he is talking about the turning angle idea. He says that if the angle of $w$ i.e: point on the output curve changes by $t_i$ , then turning angle also increases by $t_i$.. but what does it mean?

What even is $\tilde{\phi}$? Can someone explain the whole idea of what Tristan Needham is trying to convey in simple words? I'm really confused as to how to approximate the turning of the curve in output plane from looking at input plane.

答:
Needham is attempting to construct a geometric interpretation of the Riemann Sum in the complex plane. For each piece of the Riemann Sum in $\mathbb{R^2}$, he proposes an analogous geometric description in $\mathbb{C}$. $$A=\sum_{i}^{n}f(x_{i})\Delta_{i}$$ Now what happens instead when we have $$A=\sum_{i}^{n}f(z_{i})\Delta_{i}$$
In the previous pages, he discussed the midpoint Riemann Sum tending to a limiting value, in this case, $$\int_{K}f(z)dz$$ and a justification for it. $$f(z)=w$$ Therefore, the curve in the w-plane is a mapping of the curve in the z-plane. Therefore if we take the Riemann sum along the path in the z-plane, we have $$A=\sum_{i}^{n}w_{i}\Delta_{i}$$ where $\Delta_{i}$ are partitions along the path and $w_{i}$ are the midpoints on the path. There should be one more diagram after this that gives a representation of what the ϕ~  are. The key point is that when you multiply two complex numbers, you add their arguments one's magnitude is scaled by the other. Therefore the Riemann Sum in the complex plane is the sum of the vectors/complex numbers $w_{i}\Delta_{i}$ added tail to head. Therefore when the argument of $w_{i}$ rotates by a factor, the argument of $w_{i}\Delta_{i}$ rotates by that same factor. Also, note that in the original curve not all the $\phi_{i}$ are positive.
As an illustration of this, in the next diagram in the book, note that the longest arrow is first, corresponding to the fact that all the $\Delta_{i}$ are the same and that $f(a)$ is the farthest away from the origin and that they're also adding the arguments in the manner I described earlier.


问:
In what sense do isomorphisms "preserve all logical properties"?
My background in mathematical logic, model theory, etc. is patchy, so I'm looking for a clearer way to think about this. (Edit: I'm not asking what an isomorphism is, I'm asking how to formalize the idea of "preserving all logical properties" in order to state the principle described below in its full power.)

Motivating example: we know that $\Bbb N$ and $\Bbb Z$ are not order-isomorphic, because $\Bbb N$ has a least element whereas $\Bbb Z$ does not. The principle underlying this argument seems to be that an order isomorphism preserves the truth values of (first-order) sentences: if $(A,<_A)\cong(B,<_B)$ then $\operatorname{Th}(A,<_A)=\operatorname{Th}(B,<_B)$. Since the property "there is a least element" corresponds to the sentence $\exists x\forall y. x\le y$, the principle applies. The principle also seems to apply to higher-order properties. For example, we can conclude that $(\Bbb R\setminus\{0\},<)\not\cong(\Bbb R,<)$ by checking the least-upper-bound property, which is second-order.

More generally (if we think of set theory as modeling higher-order logic in first-order logic), I'm tempted to claim that the whole "set theory of <" (I'm not sure what this is exactly - a two-sorted first-order logic with relation symbols $<,\in$, maybe?) is preserved by isomorphisms. I think I can convince myself of this fact by observing that when $f:A\to B$ is an isomorphism, certain substitution rules involving $f$ in various syntactic contexts are valid, allowing us to prove equivalences between propositions about $<_A$ and the corresponding propositions about $<_B$.

I'm calling this a "principle" instead of a "theorem" because it seems easier to apply in specific cases (by reasoning directly about the elements and the isomorphism, without formalizing the notion of sentences) than to express in its full generality. My claim that "the set theory of $<$ is preserved" seems closely related to the fact that set theory is my metatheory (if that's the right word), so some kind of circularity or self-reference is appearing when I attempt to state the principle, and I wonder whether any formal version of the principle can fully eliminate the need to apply the principle directly.

So I guess my questions are:

In what sense (if any) is there a most general version of this principle, applying to all properties preserved by order isomorphisms?

Can it be expressed as a theorem in the deductive system where we want to apply it, or is there a theoretical obstacle to this?

Am I getting anything wrong? Or missing a clarifying perspective?

Thanks!

答:
Here is an outline of a theorem which, I think, addresses the situation pretty well (incidentally, this MO question is related):

Most logic books will include a proof that first-order logic is isomorphism-invariant. But what about non-first-order properties? For example, the isomorphism-invariance of (say) Archimedeanness in the context of ordered fields is not a mere case of the isomorphism-invariance of first-order logic, even though "obviously" it holds for exactly the same reason.

There is in fact a framework for extending isomorphism-invariance from first-order logic to all "reasonable" logics, in a precise sense. The logics beyond first-order logic that we actually consider are really just "first-order logic in a superstructure." E.g. saying that a relation $R$ is definable over a structure $\mathcal{A}$ in monadic second-order logic is the same as saying that it is definable over $\mathcal{A}\sqcup\mathcal{P}(\mathcal{A})$ (being a bit vague about what that is, in the interest of brevity) in first-order logic.

This motivates the following idea. Working in $\mathsf{ZFC}$, let $V$ as usual denote the whole set-theoretic universe. There is a standard machinery for developing, first-order-definably inside $V$, a "version of $V$ with urelements" where the urelements correspond to elements of a given structure; the details are a bit tedious, but see e.g. Barwise's book. Call the resulting object "$V_\mathcal{A}$." Being a bit sloppy about set/class issues, we can prove in $\mathsf{ZFC}$ that $\mathcal{A}\cong\mathcal{B}$ implies $V_\mathcal{A}\cong V_\mathcal{B}$. This, together with the fact that isomorphism preserves first-order truth, gives the following result:

Isomorphisms preserve truth with respect to every "set-theoretically-definable" logic.

Here the idea is that each sentence $\varphi$ in a set-theoretically-definable logic should correspond to a first-order sentence $\hat{\varphi}$ such that $$\mathcal{A}\models\varphi\quad\iff\quad V_\mathcal{A}\models\hat{\varphi}$$ (or at worst, a first-order sentence $\hat{\varphi}$ with a parameter from $V$ - note that $V$ sits inside $V_\mathcal{A}$ in a canonical way, so a single $V$-parameter makes sense across all $V_\mathcal{A}$s independently of $\mathcal{A}$; we need to allow parameters to handle logics with "too many sentences," such as infinitary logic).

So what we have here is a way to "bootstrap" the isomorphism-invariance of first-order truth to all logics which are set-theoretically definable in a precise sense.

I'm skipping a lot of detail in the above, since quite frankly it's rather tedious. If you're interested, let me know and I'll expand on this. But I think it is worth mentioning, even if only very briefly, the existence of a precise theorem of the above type.


问:
For non-square non-negative matrix $A$, under what conditions does a non-negative matrix $B$ exist such that $AB = I$?
For non-negative matrix $A$ with dimensionality $m \times n$ where $m < n$, I am trying to find all possible non-negative matrices $B$ that are its pseudoinverses:

$AB = I$

We can find $B = A^T(AA^T)^{-1}$ that sometimes satisfies this condition. But is there only one unique $B$ that satisfies $AB = I$ for a given $A$? And is that $B$ non-negative? How do I go about rigorously approaching this question? How do I enforce the non-negativity constraint on $B$?

答:
You want the columns $b_1, \ldots, b_n$ of $B$ to satisfy

$$
AB = (Ab_1 \mid \cdots \mid Ab_n) = I.
$$

That's the same as looking for solutions to

$$
Ab_i = e_i. \tag{1}
$$

Assuming such a solution $b_i$ for $(1)$ exists, all solutions are given by $b_i+x$ with $Ax = 0$. Hence all possible matrices would be of the form

$$
B = (b_1+x_1\mid \cdots \mid b_n+x_n)
$$

where $Ab_i = e_i$ and $Ax_i = 0$ for all $i$.

By size considerations, you matrix $A$ is not injective (meaning there are non-zero $x$ for which $Ax = 0$). So, if you have such a matrix $B$, there will be infinitely many. And the existence can be phrased as: are $e_1,\ldots,e_n$ in the image of $A$? In turn, this means for $A$ to have full rank.

The coordinate free approach would be as follows: you matrix a defines a lineal map $f \colon \Bbbk^m \to \Bbbk^n$, you want $g$ such that $fg = \mathrm {id}$. This is the same as having $fg(e_i) = e_i$ for all $i$. Also, we can define $g$ on a basis, so the problem reduces to picking $y_i \in f^{-1}(e_i)$ for each $i$ and defining $g(e_i) := y_i$.

Of course, such an element need not be unique and the non-uniqueness is measured by $\ker f$, which by dimension considerations is a non trivial subspace.

On the other hand, existence is equivalent to having $\mathrm{im} f = \langle e_1, \ldots, e_n\rangle = \Bbbk^n$, i.e. to $f$ being surjective.


问:
Compute close-form of $\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}$
I came across the improper trigonometric integral recently shared in a Chinese web forum
\begin{align}
\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}\\
\end{align}
which amuses me because of its appearance. What is more amusing is the claim, without providing the proof, that is has the close-form result
\begin{align}
\frac{\sqrt{8\sqrt2+2\sqrt7}}{7^{3/4} }\tanh^{-1}  \frac{\sqrt{2\sqrt7(4\sqrt2-5)}}{4\sqrt2-5 +\sqrt7}
+\frac{\sqrt{8\sqrt2-2\sqrt7}}{7^{3/4} }\tan^{-1}\frac{\sqrt{2\sqrt7(4\sqrt2-5)}}{4\sqrt2-5 - \sqrt7}
\end{align}
I was able to verify it numerically; yet rather curious in how it could ever be derived. Based on my knowledge and experience, I do not assume it would be easy.

I was able to find an unsolved post here from a long time ago, which only reexpresses the corresponding  indefinite integral via the tangent half-angle substitution as
$$\int\frac{2t(1-t)}
{2 t^4-3 t^3+3 t^2+t+1}dt$$
but gives up due to the complexity in partial fractionalization. I am doubtful that this approach would lead to the explicit expression claimed above. I am interested in any suitable methods producing the close-form.

答:
Evaluate
$$\int_0^{\pi / 2} \frac{dx}{\sin x + \cos x + \tan x + \sec x + \csc x + \cot x} .$$

As in the linked question, applying the classical Weierstrass substitution, $$x = 2 \arctan t, \qquad dx = \frac{2 \,dt}{t^2 + 1},$$ transforms the integral to
$$
\int_0^1 \frac{2 t (1 - t)}{2 t^4 - 3 t^3 + 3 t^2 + t + 1}.
$$

One can show that the Galois group of the polynomial $$f(t) := 2 t^4 - 3 t^2 + 3 t^2 + t + 1$$ in the denominator of the integrand
is $D_8$, suggesting applying a transformation that takes advantage of that symmetry.*

The following shows one way of doing so, in particular transforming the integrand to one to which the Method of Partial Fractions can be applied somewhat more easily than the one in $t$. I don't know whether the particular substitution used here is in any sense the nicest choice.

Let $\alpha := \cot \frac{3 \pi}{8} = \sqrt{2} - 1$. Applying to the original integral the substitution
$$x = \frac{\pi}{4} - 2 \arctan (\alpha u), \qquad dx = -\frac{2 \alpha \,du}{\alpha^2 u^2 + 1} ,$$
simplifying considerably, and taking advantage of the evenness of the integration in $u$ gives that the integral is equal to
$$\frac{2 (8 + 5 \sqrt{2})}{7} \int_0^1 \frac{1 - u^2}{u^4 + \beta^4} du , \qquad \textrm{where } \beta := \frac{\sqrt{5 + 4 \sqrt{2}}}{\sqrt[4]{7}} .$$ Remark 1 The substitution $x \rightsquigarrow u$ is related to the Weierstrass substitution, $x \rightsquigarrow t$ by the linear fractional transformation $$t = \frac{\alpha (1 - u)}{\alpha^2 u + 1} .$$

Remark 2 Perhaps despite appearances the substitution $x \rightsquigarrow u$ is not particularly clever: It is the composition of a translation that centers the domain of integration on the origin, the classical Weierstrass substitution, and a dilation to make the coefficients and limits in the resulting integrand nicer.

Over $\Bbb Q(\beta) = \Bbb Q(\beta, \sqrt{2})$ (or just $\Bbb R$), the denominator of the integrand in $u$ factors into irreducible polynomials as
$$u^4 + \beta^4 = (u^2 + \sqrt{2} \beta u + \beta^2) (u^2 - \sqrt{2} \beta u + \beta^2),$$ so the rest of the computation can be handled with standard techniques: Applying the Method of Partial Fractions gives the decomposition
$$\frac{1 - u^2}{u^4 + \beta^4} = \frac{A u + B}{u^2 + \sqrt{2} \beta u + \beta^2} + \frac{C u + D}{u^2 - \sqrt{2} \beta u + \beta^2} .$$
Evenness of the integrand implies that $C = -A, D = B$, reducing the equation in the unknown coefficients to a $2 \times 2$ system in $A, B$, and some elementary algebra gives
$$A = -\frac{\beta^2 + 1}{2 \sqrt{2} \beta^3}, \qquad B = -\frac{1}{\beta^2} .$$

Separating each summand into the sum of a scalar multiple of $$\frac{u}{u^2 \pm \sqrt{2} \beta u + \beta^2} \qquad \textrm{and one of} \qquad \frac{1}{u^2 \pm \sqrt{2} \beta u + \beta^2} ,$$ and integrating respectively gives antiderivatives
$$\frac{1}{2} \log \left[\sqrt{2} \beta u \pm (\beta^2 + u^2)\right] \qquad \textrm{and} \qquad \arctan \left(1 \pm \frac{\sqrt{2} u}{\beta}\right) .$$ We can combine the two $\log$ terms with the identity $\operatorname{artanh} x = \frac{1}{2} \log \left(\frac{1 + x}{1 - x}\right)$, yielding
$$\operatorname{artanh} \left(\frac{u^2 + \beta^2}{\sqrt{2} \beta u}\right),$$
and we can combine the two $\arctan$ terms with the arctan sum identity, $\arctan x + \arctan y = \arctan \left(\frac{x + y}{1 - xy}\right)$, yielding
$$
\arctan \left(\frac{\beta^2}{u^2}\right) .
$$

Putting this all together gives essentially the given exact form, and a numerical evaluation suggests that the expressions indeed agree, with value $0.1760244214\ldots$.

*The discriminant of the irreducible quartic $f$ is $\Delta := 2^5 7^3$; now, (a) $f$ remains irreducible over $\Bbb Q(\sqrt{\Delta}) = \Bbb Q(\sqrt{14})$, and (b) the cubic resolvent of $f$ factors over $\Bbb Q$ as a product of an irreducible quartic and a linear polynomial (up to a scalar multiple, $(2 x − 1) (4 x^2 − 4 x − 13)$), and these two facts together imply that $\operatorname{Gal}(f) \cong D_8$.


问:
Weighting a cubic hermite spline
I am trying to figure out a function behind the software's curve drawing algorithm. Originally, each node comes with 3 parameters : time, value, and tangent. I have found that it fits cubic Hermite spline, and confirmed that using the equation from the Wikipedia gives me equal result to the software's evaluation.

The following curve is when both tangents are 0.

The following curve is when both tangents are 1 (45 degree).

But this software has "weight" value on each node, which ranges from 0 to 1. I am trying to figure out how the weight could affect the spline empirically. Here's my observation :

The weight ranges from 0 to 1. In this gif below, dragging the handle to the right increase the weight. Dragging stretch up do not affect the weight (however you are changing the tangent)

When both tangents are 0, with weight applied and both weight are 0.3333333, it results in the same shape as if no weights are applied. (The smooth S shape, the first image) I think this is the biggest hint, 0.3333333 may has to do something with the "cubic" function. Below is an image of this.

When both tangents are 0, with weight applied and both weight are at maximum 1, the curve skewed further in X axis to meet at the center between 2 points. Indicating that, weight 1 doesn't mean unweighted like a weight function would have behave but rather really maximum possible weight. Below is an image of this.

When both tangents are 0, with weight applied and both weight are 0, results in a linear graph as if their tangents are 1. Below is an image of this.

When both tangents are 1, weight do not affect their shape at all no matter the value. It stays linear. Suggesting that weight do something to the components which became tangent, but nullified when both components are equal. I guess it did something to the cos component? Since with maximum weight the graph skewed further in X axis.

However if the other tangent is not 1, changing the weight of the side that has tangent 1 do affect the shape. Indicating that the weight is not simply "weighting that side's tangent".

If someone could figure out where the weight should be applied (or any example spline you know with weight parameter), it is very much appreciated. Thank you.

答:
Here's a solution in C# which I believe duplicates the function of the Unity AnimationCurve according to the answer by @Mario Carneiro.

I've used doubles instead of floats for reasons particular to my own need to solve this problem.  The value of Eps would need to be changed to the corresponding Machine Epsilon for floats.  It might be useful to also have a check for excessive iterations and throw to avoid infinite loops.

I have a check which short circuits the rootfinding problem to solve the cubic hermite problem, which depends specifically on the weights matching the floating point value 1/3.0 precisely, which integrates with the rest of my API, one might consider checking for equality to machine precision there.  Also if anyone uses the floating point code directly, be careful of converting from 1/3.0f to 1/3.0d if converting from unity AnimationCurves in order to have the short circuit work correctly.

Using Halley's method over Newton-Raphson gave a significant speedup, using the third order Householder method gave another ~10% boost.  Cardano's method of solving the cubic equation is probably slower due to the need to do several square/cube roots which would all be iterative methods.

For some reason it performs faster with doubles than floats for me(!) and is comparable to the Unity AnimationCurve performance (presumably done in C++).

I thought about a short circuit for when the problem fits a quadratic exactly, but that is probably unlikely to be hit as an actual use case.

```<code>    public const double Eps = 2.22e-16;

    public static double AnimationCurveInterpolant(double x1, double y1, double yp1, double wt1, double x2, double y2, double yp2, double wt2, double x)
    {
        double dx = x2 - x1;
        x = (x - x1) / dx;
        double dy = y2 - y1;
        yp1 = yp1 * dx / dy;
        yp2 = yp2 * dx / dy;
        double wt2s = 1 - wt2;

        double t = 0.5;
        double t2;

        if (wt1 == 1 / 3.0 && wt2 == 1 / 3.0)
        {
            t  = x;
            t2 = 1 - t;
        }
        else
        {
            while (true)
            {
                t2 = (1 - t);
                double fg = 3 * t2 * t2 * t * wt1 + 3 * t2 * t * t * wt2s + t * t * t - x;
                if (Math.Abs(fg) < 2*Eps)
                    break;

                // third order householder method
                double fpg = 3 * t2 * t2 * wt1 + 6 * t2 * t * (wt2s - wt1) + 3 * t * t * (1 - wt2s);
                double fppg = 6 * t2 * (wt2s - 2 * wt1) + 6 * t * (1 - 2 * wt2s + wt1);
                double fpppg = 18 * wt1 - 18 * wt2s + 6;

                t -= (6 * fg * fpg * fpg - 3 * fg * fg * fppg) / (6 * fpg * fpg * fpg - 6 * fg * fpg * fppg + fg * fg * fpppg);
            }
        }

        double y = 3 * t2 * t2 * t * wt1 * yp1 + 3 * t2 * t * t * (1 - wt2 * yp2) + t * t * t;

        return y * dy + y1;
    }
</code>```

Title: Weighting a cubic hermite spline

Body:
Here's a solution in C# which I believe duplicates the function of the Unity AnimationCurve according to the answer by @Mario Carneiro.

I've used doubles instead of floats for reasons particular to my own need to solve this problem.  The value of Eps would need to be changed to the corresponding Machine Epsilon for floats.  It might be useful to also have a check for excessive iterations and throw to avoid infinite loops.

I have a check which short circuits the rootfinding problem to solve the cubic hermite problem, which depends specifically on the weights matching the floating point value 1/3.0 precisely, which integrates with the rest of my API, one might consider checking for equality to machine precision there.  Also if anyone uses the floating point code directly, be careful of converting from 1/3.0f to 1/3.0d if converting from unity AnimationCurves in order to have the short circuit work correctly.

Using Halley's method over Newton-Raphson gave a significant speedup, using the third order Householder method gave another ~10% boost.  Cardano's method of solving the cubic equation is probably slower due to the need to do several square/cube roots which would all be iterative methods.

For some reason it performs faster with doubles than floats for me(!) and is comparable to the Unity AnimationCurve performance (presumably done in C++).

I thought about a short circuit for when the problem fits a quadratic exactly, but that is probably unlikely to be hit as an actual use case.

    public const double Eps = 2.22e-16;

    public static double AnimationCurveInterpolant(double x1, double y1, double yp1, double wt1, double x2, double y2, double yp2, double wt2, double x)
    {
        double dx = x2 - x1;
        x = (x - x1) / dx;
        double dy = y2 - y1;
        yp1 = yp1 * dx / dy;
        yp2 = yp2 * dx / dy;
        double wt2s = 1 - wt2;

        double t = 0.5;
        double t2;

        if (wt1 == 1 / 3.0 && wt2 == 1 / 3.0)
        {
            t  = x;
            t2 = 1 - t;
        }
        else
        {
            while (true)
            {
                t2 = (1 - t);
                double fg = 3 * t2 * t2 * t * wt1 + 3 * t2 * t * t * wt2s + t * t * t - x;
                if (Math.Abs(fg) < 2*Eps)
                    break;

                // third order householder method
                double fpg = 3 * t2 * t2 * wt1 + 6 * t2 * t * (wt2s - wt1) + 3 * t * t * (1 - wt2s);
                double fppg = 6 * t2 * (wt2s - 2 * wt1) + 6 * t * (1 - 2 * wt2s + wt1);
                double fpppg = 18 * wt1 - 18 * wt2s + 6;

                t -= (6 * fg * fpg * fpg - 3 * fg * fg * fppg) / (6 * fpg * fpg * fpg - 6 * fg * fpg * fppg + fg * fg * fpppg);
            }
        }

        double y = 3 * t2 * t2 * t * wt1 * yp1 + 3 * t2 * t * t * (1 - wt2 * yp2) + t * t * t;

        return y * dy + y1;
    }


问:
What is the minimum weighting trials to be made with a two pan scale to find the lighter marble?
The puzzle is as follows:

Over a table there are three bags with marbles, on the first one there are 100 blue marbles, on the second one there are 100 red marbles and on the third one there are 100 green marbles. All the marbles inside the bags are in equal weight, except two of them whose weight is slightly less from the others and identical, which are on different bags. Considering all of this, How many weighing trials at minimum must be made by using a two pan scale to find with certainty those two mysterious lighter marbles?

The choices given are as follows:

$\begin{array}{ll}
1.&\textrm{11 trials}\\
2.&\textrm{8 trials}\\
3.&\textrm{9 trials}\\
4.&\textrm{10 trials}\\
\end{array}$

I'm not sure exactly how to solve this problem, my best effort goes as this:

Assuming that those bags fit in the two pan scale (hence marbles are sufficiently small) then you put in the left pan one bag and in the other the other bag.

The best thing that it can happen is that right off the bat one bag is heavier than the other because it contains the heavier marbles. The worst thing is that they are equal in weight because they both contain those mysterious lighter marbles. But either one way or another. Assuming we know beforehand there are two marbles whose weight is less than the others and those two are in two separate bags then this means that if those two bags weight the same, then those two bags contains the marbles we're looking for. While the other bag has the heavier marbles.

All of this account for one weighing trial.

Now that we discriminated those bags, we know our two bags which contains the lighter marbles we split them in groups of 50. This accounts for four groups.

Assuming that we do not mix those, we put those two groups in one pan and the other two in the other pan.

The worst that can happen is that they are even, and we still cannot tell on which side there are the ligther msrbles.

Thus a new weighing trial is needed.

This accounts for a second trial.

In the next trial, we switch the groups by fixing two groups. After doing this, the scale will reveal which side is the lighter one. Thus the heavier marbles can now be returned back to the group of the marbles which do weight more.

This accounts for our third trial.

We repeat the cycle again. But now we have 100 marbles remaining, as the other 100 went back with their set of heavier marbles. Thus, we split those in 4 groups of 25.

Assuming that it happens the same as before, we can't tell the different first, so a new weighing trial is needed.

This accounts for our fourth trial.

We repeat the weighing trial, switching the groups, and we know for sure 50 marbles are heavier and the other 50 must contain the lighter marbles.

This accounts for the fifth trial.

We now have 50 marbles whose weight must contain those two marbles. We again repeat the cycle. In this case we cannot divide 25 by 4. Remember that all the time we are keeping track on which side belongs to which bag, so we have took care not to mix the groups. So we use one of these groups of 25 (and save the other for later) from which we know it contains a ligther marble. But because 25 cannot be split in two. We take out from that group a marble, and we have two groups of 12.

We perform a new weighing trial. The best thing it can happen is both sides are even and we got our lighter marble. The worst thing is that we need a new weighing trial.

This accounts for a sixth trial.

We collect the heavier marbles as this was revealed by the scale plus the marble what we took out first and we end up with 12 ones from which we don't know where is the mysterios marble and perform a new weighing trial. This time splitting in six and six.

Again we assume that the worst thing that can happen is that we are still unable to know where is the mysterious marble.

From those we know that six are heavier as the scale revealed this to us by tilting it to one side.

This account for a seventh trial.

We now have six from which we don't know where is the ligther marble. We split them in groups of 3 and 3 and perform a new weighing trial, from it we do know the heavier ones.

This accounts for the eight trial.

Now it comes finally our last 3 marbles. From these we must take out one. In any case it will be revealed which is ligther. If the scale is even, then the one which we took out is the lighter. If one side is heavier the other resting in the other pan is the ligther one, which belongs to one bag.

Remember that we saved one group from before performing the sixth trial. Thus as we required 3 steps to discover which marble was the ligther one, this adds to the already weighing trials that we accounted.

This means $8+3=11$

Thus I conclude that at minimum we do need $11$ weighing trials for that, or choice 1. But of course, I could be overlooking something.

Can someone please check my logic?. I am not sure if it can done in less steps or in a different way, but to my understanding this is the most logical way to do it.

Thus can someone help me here?.

答:
It certainly can't be done in fewer than $10$ weighings.  There are $3\cdot10^2\cdot10^2$ possible answers.  Each weighing gives $3$ possible answers: the pans balance, the left-hand pan is heavier, or the right-hand pan is heavier.  $9$ weighings have only $3^9<3\cdot10^4$ possible outcomes, so they can't reliably distinguish between all possible answers.

I believe that it can't be done in $10$ weighings, either, but I can't say my argument is truly a rigorous proof.

To begin with, it seems like weighing one bag against another must be the way to start.  This eliminates $100$ marbles in one shot, and we can't expect to do any better.  Let's say that the blue marbles are all standard, and there's one light red and one light green.  Then it seems to me that we can just weigh the red and green marbles separately.  If after the first weighing, and oracle tells you which is the light green marble, that doesn't help you find the light red marble.  Now $3^4<100$, so we'd need $5$ weighings at least to find the light red marble.  In the absence of an oracle, we also need $5$ weighings to find the light green marble, so $1+5+5=11$ weighing in all.

There are a couple of points where the reasoning seems a bit iffy to me, but I'm going to post it anyway.  Maybe someone can clean it up.

EDIT

As antkam's answer shows, the above reasoning is false, and $10$ weighings are possible.  I'm not going to delete or edit the post, however, because antkam's answer refers to it.


问:
Derivation of Dual Curve for Parametric Equations
I am studying projective geometry and am stuck on understanding the following:

For a parametric curve

$$
x = x(t), y = y(t)
$$

the dual curve is given by

$$
X=\frac{-y′}{xy′-yx'},
Y=\frac{x′}{xy′−yx′}
$$

This is explained on Wikipedia, in Chapter 1, Section 2 of Discriminants, Resultants, and Multidimensional Determinants by Gelfand, Kapranov, and Zelevinsky (referred by @Jan-Magnus-Økland; see this question), and in this document.

I feel like something so simple and fundamental is escaping me...how was this derived? Is Cramer's Rule involved?

答:
Let

$$
Xx + Yy + 1 = 0 \tag{1}
$$

be the equation of a line in line coordinates. A point $(x)$ and line $[X]$  (in the notation of Coxeter) are said to be incident in projective geometry if their inner product is zero:

$$
\{xX\} = 0.
$$

The tangent to our parametric equation at point $t_0$ is given by $<x'(t_0),y'(t_0)>$. Hence, the point $(x')$ is incident on our line if

$$
\{x'X\} = 0.
$$

That is,

$$
Xx' + Yy' = 0 \tag{2}
$$

Thus, we have two equations in two unknowns ($(1)$ and $(2)$), which via Cramer's rule yield

$$
X=\frac{-y'}{xy'-yx'},
Y=\frac{x'}{xy'-yx'}.
$$

Q.E.D.


问:
Relation between determinants and co planar vectors
I have the following question here:

Let $$\vec{u}=(u_1,u_2,u_3),\vec{v}=(v_1,v_2,v_3),\vec{w}=(w_1,w_2,w_3)$$
be $3$ vectors in $\mathbb{R}^3$. Define a matrix whose rows are these three vectors:
$$\begin{bmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3\\ w_1 & w_2 & w_3\\
\end{bmatrix}$$
a) Show that $$\det(A)=\vec{u} \cdot (\vec{v} \times \vec{w})$$

This proof was trivial. Simply compute $\vec{v} \times \vec{w} = \begin{vmatrix} i & j & k \\ v_1 & v_2 & v_3\\ w_1 & w_2 & w_3\\
\end{vmatrix}$ and dot the result with $\vec{u}=(u_1,u_2,u_3)$. Compute the determinant (I used cofactor expansion) and you'll get both sides to match.

b) Suppose that $\vec{v},\vec{w}$ are not colinear. Show that $\vec{v} \times \vec{w}$ is normal to the plane given by $\text{Span}\{\vec{v},\vec{w}\}$.

Since $\vec{v}$ and $\vec{w}$ are not collinear, this implies any vecotor can be written as the span of the two vectors. Let $\vec{u}=\alpha\vec{v}+\beta\vec{w}$ where $\alpha, \beta$ are real numbers. If $\vec{u}$ is normal to $\vec{v} \times \vec{w}$ then $\vec{u}\cdot(\vec{v} \times \vec{w})$ implies:

$\vec{u}\cdot(\vec{v} \times \vec{w})$

$(\alpha\vec{v}+\beta\vec{w})\cdot(\vec{v} \times \vec{w})$

$(\alpha\vec{v})\cdot(\vec{v} \times \vec{w})+(\beta\vec{w})\cdot(\vec{v} \times \vec{w})=0$

It's $0$ since two of the rows are the same so the determinant (Since the is the same as part a) evaluates to $0$ as well.

c) Use parts (a) and (b) to argue that $\vec{u},\vec{v},\vec{w}$ all lie on the same plane if and only if $\det(A)=0$.

I can do this geometrically quite easily since this is just the volume of the parallelipiped spanned by the three vectors $\vec{u}$,$\vec{v}$ and $\vec{w}$.

Is there a way way to argue this algebraically though? I keep reading various links about how this relates to the normal vector but I can't seem to put it together.

The most common answer I see is:

"If $\mathbf{u}, \mathbf{v}, \mathbf{w}$ are coplanar, their triple product $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$ is $0$ since $\mathbf{u}$ must be perpendicular to $\mathbf{v} \times \mathbf{w}$."

Why though? I feel silly asking this but why does $\vec{u}$ have the be orthogonal to $\vec{v}$ and $\vec{w}$? I feel like I'm missing a very simple and obvious step here.

Can someone draw out some kind of visual for this?

答:
This could also follow by the point-normal definition of a plane: $\mathbf{\vec{n}}\cdot \vec{x} = 0$.

Given two points $x_1,x_2 \in \mathcal{P}$ then the vector $\vec{x} = \mathbf{x_2} - \mathbf{x_1 }\in \mathcal{P}$, and the normal has the property that is perpendicular to all vectors in $\mathcal{P}$:

If $\vec{x} \in \mathcal{P}$ (the plane), then $\mathbf{\vec{n}}\cdot \vec{x} = 0$.

The normal of the plane $\mathcal{P}_{\vec{v},\vec{w}}$ made by noncolinear vectors $\vec{v},\vec{w}$ is $\mathbf{\vec{n}} = \vec{v} \times \vec{w}\not= \vec{0}$, and if $\vec{u} \in \mathcal{P}_{\vec{v},\vec{w}}$ then $\det(A)=(\vec{v} \times \vec{w})\cdot \vec{u}= \mathbf{\vec{n}}\cdot \vec{u} =0$.

If $\vec{v},\vec{w}$ are colinear then $\mathbf{\vec{n}}=\vec{0}$ and certainly $\vec{0}\cdot\vec{u} =0$.

There's an image of the normal vector to a plane in this question.


问:
Need help with showing divergence
I'd be glad if someone could help me once again :)

Given is the series:

$$\sum_{n=1}^\infty \frac{n+1}{n^2+n+1}$$ I need to show the divergence. Which way is the right? (Am I correct that the series $$\sum_{n=1}^\infty (-1)^n \cdot \frac{n+1}{n^2+n+1}$$ converges? Tried it using Leibnitz)

and:

$$\sum_{n=1}^\infty \frac{2^n}{2^n+3^n}$$ (again: correct that the series $$\sum_{n=1}^\infty \frac{(-2)^{n}}{2^n+3^n}$$ converges? Tried it using Leibnitz)

Thank you very much!

EDIT: The second series was the wrong one, someone already helped me :)

答:
Let $a_n=\frac{n+1}{n^2+n+1}$. To show the convergence of $\sum_{n=1}^\infty (-1)^n a_n$, we verify that this is an alternating series. The alternation of signs is clear, as is the fact that $\lim_{n\to\infty} a_n=0$. We also need to show that the sequence $(a_n)$ is non-increasing.

One can do this in various ways. For example, one can show that the derivative of $\frac{x+1}{x^2+x+1}$ is negative when $x\gt 0$. Or else we can note that $a_n=\frac{1}{n+\frac{1}{n+1}}$. If we increment $n$ by $1$, the term $n$ in the denominator is incremented by $1$, and the term $\frac{1}{n+1}$ is decremented by less than $1$, so the denominator increases.

The series $\sum_1^\infty \frac{2^n}{2^n+3^n}$ converges. For $\frac{2^n}{2^n+3^n}\lt \left(\frac{2}{3}\right)^n$, and the geometric series  $\sum_1^\infty\left(\frac{2}{3}\right)^n$ converges.

Because absolute convergence implies convergence, it follows that $\sum_1^\infty (-1)^n\frac{2^n}{2^n+3^n}$ converges.

Title: Need help with showing divergence


问:
How to expand this tensorial Taylor expansion to the $n$th term?
Wikipedia describes the use of the Hessian matrix in a Taylor series expension. I've noted that the first term is written in terms of the original function, the second term uses the gradient of the function, and the third term uses the Hessian of the function. While each term is a scalar, I see a pattern of using higher rank tensor operations as the order of approximation increases.

$$y = f(\vec{x} + \Delta \vec{x}) \approx f(\vec{x}) + \nabla f(\vec{x}) \Delta \vec{x} + \frac{1}{2} \Delta \vec{x}^T H(\vec{x})\Delta \vec{x}$$

How can we expand this approximation to an arbitrary $n$th term? I am expecting, or hoping, there is a way to do this in terms of tensors.

答: This answer is not 'tensorial', but there is a series expansion given on wikipedia for several variables. Leaving aside notions of vector spaces and their dual spaces when considering contravariant or covariant indices, you can see each term as an array of derivatives.</p>


问:
A new type of Cauchy sequence

I have very little idea about how to approach this question. It doesn't make any sense to me how this can be Cauchy. Any insight into it would be greatly appreciated.

答: HINT: Suppose that $\langle a_n:n\in\Bbb N\rangle$ is Daudhy, and let $\epsilon>0$. There is a $T\in\Bbb N$ such that $d(a_j,a_k)<\frac{\epsilon}2$ whenever $T<j<2T<k$. Use the triangle inequality to show that $d(a_j,a_k)<\epsilon$ whenever $j,k>2T$ and conclude that the sequence is Cauchy.

Title: A new type of Cauchy sequence


问:
Measurabilty question
Consider following problem:

Let $f:S^1 \rightarrow S^1$ be a rotation.
Show that a subset $M \subset S^1$ is measurable if and only $f(M)$ is measurable.

答:
Let us recapitulate here

Consider the unit circle $S^1 \subset \mathbb C$ as a submanifold of $\mathbb R ^2.$
Choose an atlas for $S^1$. Choose a partition of unity $\{p_i\}_{i \in I}$ that is subordinate to the domains of the charts of that atlas (i.e. $\operatorname{supp}(p_i) \subset V_i$ for a domain $V_i$).
Choose an embedding $\gamma: \Omega \rightarrow \mathbb R^2$ with an open $\Omega \subset \mathbb R$.
We call a subset $M \subset S^1$ measurable if for all $i \in I$ the function $(\chi_M \cdot p_i )\circ \gamma \cdot \sqrt{g^\gamma}: \mathbb R \rightarrow \mathbb R$ is measurable with respect to the Lebesgue measure on $\mathbb R$. Here $\chi_M$ denotes the characteristic function of $M$ and $g^\gamma$ is the Gram determinant of $d \gamma$. (One can show that this definition is independent of the choices made.)

We want to prove that

Let $f:S^1 \rightarrow S^1$ be a rotation.
Show that a subset $M \subset S^1$ is measurable if and only $f(M)$ is measurable.

Proof: We can write $(\chi_M \cdot p_i )\circ \gamma \cdot \sqrt{g^\gamma} = \chi_{\gamma^{-1}(M)}\cdot (p_i \circ \gamma)\cdot \sqrt{g^\gamma}$. The Gram determinant is non-zero because $\gamma$ is an embedding. Because of that and because the product of measurable functions is measurable, $(\chi_M \cdot p_i )\circ \gamma \cdot \sqrt{g^\gamma}$ is measurable if and only if $\chi_{\gamma^{-1}(M)}\cdot (p_i \circ \gamma)$ is measurable.

So we have:

$M$ is measurable if and only if, for all $i \in I$,  $\chi_{\gamma^{-1}(M)}\cdot (p_i \circ \gamma)$ is measurable (which independs of the athas chosen, the $\{p_i\}_{i \in I}$ partition of unity  that is subordinate to the domains of the charts of the atlas and the embedding $\gamma: \Omega \rightarrow \mathbb R^2$).

In the same way,

$f(M)$ is measurable if and only if, for all $i \in I$,  $\chi_{\gamma^{-1}(f(M))}\cdot (p_i \circ \gamma)$ is measurable, (which independs of the athas chosen, the $\{p_i\}_{i \in I}$ partition of unity  that is subordinate to the domains of the charts of the atlas and the embedding $\gamma: \Omega \rightarrow \mathbb R^2$).

It is easy to see that $f^{-1}\circ \gamma$ is an embedding. So by item 1, we have that  $M$ is measurable if and only if, for all $i \in I$,  $\chi_{\gamma^{-1}(f(M))}\cdot (p_i \circ f^{-1}\circ \gamma)$ is measurable. But $\{p_i \circ f^{-1}\}_{i \in I}$ is a partition of unity  that is subordinate to the domains of the charts of the atlas rotated by $f$. Let, for all $i\in I$, $p_i'= p_i \circ f^{-1}$. Then, using item 2,  we have that  $M$ is measurable if and only if, for all $i \in I$,  $\chi_{\gamma^{-1}(f(M))}\cdot (p_i'\circ \gamma)$ is measurable  if and only if $f(M)$ is measurable.
So, $M$ is measurable if and only if $f(M)$ is measurable.

Title: Measurabilty question