问:
Recurrence Relations with single roots
I have the following recurrence: $a_{n+3}=3a_{n+2}-3a_{n+1}+a_n$
with initial values $a_1 = 1, a_2 = 4, a_3 = 9$

I have found the characteristic equation to be $x^3-3x^2+3x-1$ and the root to be 1.

My text book is not helpful in how I should go about solving this when I have a single root and don't have the $a_0$ value given.

Any tips on how I could move forward to solve this?

答:
Use Wilf's "generatingfunctionology". Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply your recurrence by $z^n$ and sum over $n \ge 0$ to get:
$$
\frac{A(z) - a_0 - a_1 z - a_2 z^2}{z^3}
  = 3 \frac{A(z) - a_0 - a_1 z}{z^2} - 3 \frac{A(z) - a_0}{z} + A(z)
$$
Using the recurrence "backwards" gives $a_0 = 0$. Solving for $A(z)$ and expanding into partial fractions:
$$
A(z) = \frac{1}{1 - z} - \frac{3}{(1 - z)^2} + \frac{2}{(1 - z)^3}
$$
By the expansions:
$$
(1 - u)^{-m} = \sum_{k \ge 0} \binom{-m}{k} (-u)^k
             = \sum_{k \ge 0} \binom{k + m - 1}{m - 1} u^k
$$
(the binomial coeffiecients are just $m - 1$ degree polynomials in $k$) you get:
$$
a_n = 1 - 3 \binom{n + 1}{1} + 2 \binom{n + 2}{2}
    = 1 - 3 \frac{(n + 1)}{1} + 2 \frac{(n + 1) (n + 2)}{2}
    = n^2
$$</p>

Title: Recurrence Relations with single roots


问:
Simplifying $\lnot((\lnot P \land Q) \lor \lnot(R \land \lnot S))$
The statement is
$$\lnot((\lnot P \land Q) \lor \lnot(R \land \lnot S))$$
and the answer is
$$(p \land r) \lor (p \land \lnot s) \lor (\lnot q \land r) \lor (\lnot q \land \lnot s)$$
according to the dCode Boolean Expressions Calculator.

I'm trying to get to the solution myself but I got stuck. I don't know what else to do past what I've done below:

Start with $\lnot((\lnot P \land Q) \lor \lnot(R \lor \lnot S))$

De morgan's law $\lnot(\lnot P \land Q) \land \lnot(\lnot(R \lor \lnot S))$

De morgan's law $\lnot(\lnot P) ∨ \lnot Q \land \lnot(\lnot(R \lor \lnot S))$

Double negation law $P \lor \lnot Q \land R \lor \lnot S$

Distributive laws $(P \lor \lnot Q) \land (P \lor \lnot R) \lor \lnot S$

答:
The result of the second DeMorgan should be $$\color{red}(\neg (\neg P) \lor \neg Q\color{red}) \land \neg (\neg (R \lor \neg S))$$ which by two Double Negations gives you $$(P \lor \neg Q) \land (R \lor \neg S)$$

Now do Distributive Laws (which in this case works just like the FOIL principle, if you're familiar with that), and you're right at the correct answer.

The important point is: you are dropping parentheses, but those parentheses are really important!  For example, it is not clear if a statement like $$P \lor Q \land R$$  means $$P \lor (Q \land R)$$ or  $$(P \lor Q) \land R$$  Those are two different statements, and so you really need to use parentheses to disambiguate!

ContentLicense: CC BY-SA 4.0


问:
Proving that the $\chi^{2}$ Distribution is a PDF
The $\chi^{2}$-distribution is given by $$f(x; k)\equiv \frac{1}{2^{\frac{k}{2}} \Gamma\left(\frac{k}{2}\right)}x^{\frac{k}{2}- 1}e^{-\frac{x}{2}}.$$

If the $\chi^{2}$-distribution is a probability-density function (PDF), then it must hold that $$
\int_{-\infty}^{+\infty} f(x; k) \, dx = 1. $$ However, when trying to prove this, I am stuck:

$$\int_{-\infty}^{+\infty} f(x;k) \, dx = \int_{-\infty}^{+\infty}\frac{1}{2^{\frac{k}{2}}\Gamma\left(\frac{k}{2}\right)}x^{\frac{k}{2}- 1}e^{-\frac{x}{2}} \, dx = \frac{1}{2^{\frac{k}{2}}\Gamma\left(\frac{k}{2}\right)}\int_{-\infty}^{+\infty}x^{\frac{k}{2}-1}e^{-\frac{x}{2}} \, dx$$

Looking only at the last integral, we get: $$\int_{-\infty}^{+\infty}x^{\frac{k}{2}-1}e^{-\frac{x}{2}}dx = \int_{-\infty}^{0} x^{\frac{k}{2}-1}e^{-\frac{x}{2}}dx + \underbrace{\int_{0}^{\infty} x^{\frac{k}{2}-1}e^{-\frac{x}{2}}dx}_{= \Gamma\left(\frac{k}{2}\right)}$$

However, I am not yet sure on what to do with the first term. Any feedback would be appreciated.

答:
The density of the chi-square distribution is
$$
f(x) = \begin{cases} \displaystyle \frac{1}{2^{k/2} \Gamma(k/2)}x^{k/2- 1}e^{-x/2} & \text{for } x>0, \\[6pt]
0 & \text{for } x<0.
\end{cases}
$$
And so
$$
\int_{-\infty}^{+\infty} f(x)\,dx = \int_{-\infty}^0 f(x)\, dx + \int_0^{+\infty} f(x) \, dx = 0 + \int_0^{+\infty} f(x)\, dx.
$$

And then
$$
\int_0^{+\infty} \left( \frac x 2 \right)^{k/2-1} e^{-x/2} \left(\frac{dx} 2 \right) = \int_0^{+\infty} u^{k/2-1} e^{-u} \, du = \Gamma\left( \frac k2 \right).
$$


问:
How to find a formula for horizontal composition?
For natural transformations $F_1 \xrightarrow{\alpha} F_2$ and $G_1 \xrightarrow{\beta} G_2$, I have proven that $(\beta \star \alpha)_X := G_2(\alpha_X) \circ \beta_{F_1X}$ is a natural transformation from $G_1 \circ F_1$ to $G_2 \circ F_2$.

Imagine you are Saunders / MacLane creating Category Theory. You want to work out as much as possible in short time and to find a natural transformation for composed functors. But how do you find this concrete formula for the horizontal composition in a clever way?

Since I am new in Category Theory, my (elaborated) proof of naturality consists of one and a half page and needed some hours. The formula consist of multiple characters. Hence brute-force (generate all combinations, check if the domains / codomains agree, if yes, try to prove naturality) seems not to be the most time efficient way to work.

Searching math.SE, I found questions for applications of horizontal composition and for the idea of horizontal composition itself as well as a proof of natural equivalence to another formula, but not how to develop the concrete definition of $(\beta \star \alpha)_X$.

答:
One approach is to draw diagrams and see how to make sense of them.

In this picture you have to think first of what $\beta \circ F_1$ might mean, and what $G_2 \circ \alpha$ might mean as natural transformations. Draw more diagrams and there is really only one way to go at that point.
Of course there is another diagram you can draw to start with. This gives another answer and you can then think if the two answers are equal. Having a plausible picture as a prompt is usually going to be much more productive than generating combinations of symbols in a brute force fashion.


问:
Equivalence of full and sequential minimization
I would like to show that two problems are equivalent, so that by solving one of them I get the optimal solution to the second. Formally, I'm trying to prove that:
$$
\min_{\mathbf{x,y}\in\mathbf{X\times Y}}\; f(\mathbf{x,y})\; =\; \min_{\mathbf{x}\in\mathbf{X}}\; \min_{\mathbf{y}\in\mathbf{Y}} f(\mathbf{x,y})
$$
where $f$ is a closed, proper, convex function and $\mathbf{X,Y}$ are closed convex sets. Is this generally possible? So far I have the "easy" direction:
$$
 \min_{\mathbf{x}\in\mathbf{X}}\; \min_{\mathbf{y}\in\mathbf{Y}} f(\mathbf{x,y})=\min_{\mathbf{x}\in\mathbf{X}}\; \min_{\mathbf{x,y}\in\mathbb{R}^n\times\mathbf{Y}} f(\mathbf{x,y})\leq  \min_{\mathbf{x}\in\mathbf{X}}\; \min_{\mathbf{x,y}\in\mathbf{X\times Y}} f(\mathbf{x,y}) = \min_{\mathbf{x,y}\in\mathbf{X\times Y}} f(\mathbf{x,y})
$$
where the first inequality is due to the fact that $\big(\mathbf{X\times Y}\big)\subseteq\mathbb{R}^n\times\mathbf{Y}$ and the second equality is since we already minimized w.r.t. $\mathbf{x}$, hence the inner minimization produced a solution which is already optimal in $\mathbf{x}$.

However I'm not sure how to get the opposite direction. Does anyone have any ideas? Maybe a different technique should be used?

答:
To show the other direction, we consider the optimal vector $(\mathbf{x}^*,\mathbf{y}^*)\in\text{argmin}_{\mathbf{x,y}}f(\mathbf{x,y})$. Then we have the relation:
$$
\min_{\mathbf{x,y}}f(\mathbf{x},\mathbf{y})\leq f(\mathbf{x}^*,\mathbf{y}^*)=\min_{\mathbf{x}}\min_{\mathbf{y}}f(\mathbf{x}^*,\mathbf{y}^*)\leq\min_{\mathbf{x}}\min_{\mathbf{y}}f(\mathbf{x},\mathbf{y})
$$
Coupled with the direction in the question, this proves that $\min_{\mathbf{x,y}}f(\mathbf{x},\mathbf{y})=\min_{\mathbf{x}}\min_{\mathbf{y}}f(\mathbf{x},\mathbf{y})$.


问:
Calculating tangent spaces of Lie groups such as $SO(n)$ or $O(n)$
I was asking myself what was, given either $O(n)$ or $SO(n)$ as manifolds immersed in $\mathbb{R}^k$ for some natural $k$, the tangent space at a generic point $p$ of such a manifold.
I know that I can, given a smooth map between manifolds $f: X \rightarrow Y$ and $q$ a regular value for $f$, express the tangent space at a point $p$ of $f^{-1}(q)$ as $Ker(d_{p}f)$ where $d_{p}f$ is the differential of $f$ at $p$; however, even if this lemma has proven using in a bunch of cases (like calculating such spaces when $p=I$), it could not lead me to a satisfying answer in the general one. I was wondering then if there exists some general way of describing such tangent space as a subset of $M(n)$ for a generic point $p$.</p>

答:
The method you mention is actually very useful, it is just that too often, the devil is in the details.

Let $f:\mathbb{R}^{n\times n} \to S_n$ be defined by $f(A)=AA^\intercal$, where $S_n$ is the space of symmetric matrices. Then $f$ is a smooth function for which the identity matrix $I$ is a regular value, and so
$$O(n)=f^{-1}(I)$$ is a manifold. To find the tangent space at a generic point $P$ (a matrix), you could compute the directional derivative at $P$ as follows:

\begin{align*}\ker(df_P)=&\left \{A\in \mathbb{R}^{n\times n}\;|\; \lim_{t \to 0}\frac{f(P+tA)-f(P)}{t}=0\right \}\\
=&\left \{A\in \mathbb{R}^{n\times n}\;|\; \lim_{t \to 0}\frac{(P+tA)(P+tA)^{\intercal}-PP^\intercal}{t}=0\right \}\\
=&\left \{A\in \mathbb{R}^{n\times n}\;|\; PA^\intercal+AP^\intercal=0\right \}\\
\end{align*}
Thus, the tangent space of $O(n)$ at $P$ is collection of matrices $A$ such that $PA^\intercal$ is skew-symmetric.

To find it for $SO(n)$, it may be easier to find it for $SL(n)$ first. Using a similar method, define
$g:\mathbb{R}^{n\times n} \to \mathbb{R}$ by $g(A)=\det(A)$. This is another $C^\infty-$ smooth function having $1$ as a regular value, and $$SL(n)=g^{-1}(1).$$
To find the tangent space, we do as before:
\begin{align*}
\ker(dg_P)=&\left \{A\in \mathbb{R}^{n\times n}\;|\; \lim_{t \to 0}\frac{\det(P+tA)-\det(P)}{t}=0\right \}\\
=&\left \{A\in \mathbb{R}^{n\times n}\;|\; \lim_{t \to 0}\frac{\det(P^{-1})\det(P+tA)-\det(P)}{t}=0\right \}\\
=&\left \{A\in \mathbb{R}^{n\times n}\;|\; \lim_{t \to 0}\frac{\det(I+tP^{-1}A)-1}{t}=0\right \}\\
=&\left \{A\in \mathbb{R}^{n\times n}\;|\; \lim_{t \to 0}\frac{1+\text{tr}(P^{-1}A)t+O(t^2)-1}{t}=0\right \}\\
=&\left \{A\in \mathbb{R}^{n\times n}\;|\; \text{tr}(P^{-1}A)=0\right \},
\end{align*}
where we used that $\det(P^{-1})=1$ and that $\det(I+tM)=1+\text{tr}(M)t+O(t^2)$ as $t\to \infty$ (see here: https://en.wikipedia.org/wiki/Determinant).

Finally, to find the tangent space of $SO(n)=O(n)\cap SL(n)$, note that

$$T_P\big(O(n)\cap SL(n)\big)=T_P(O(n))\cap T_P(SL(n)),$$ so

$$T_p(SO(n))=\{A\in \mathbb{R}^{n\times n}\;|\; PA^\intercal+AP^\intercal=0 \; \land \; \text{tr}(P^{-1}A)=0\},$$
and we are done.

ContentLicense: CC BY-SA 3.0


问:
Force between two wires
Eight points, A,B,C,D and A',B',C',D' are arranged in space as the vertices of a cube of side a, with the faces ABCD and A'B'C'D' opposite, and A nearest to A' etc. Currents I and I' flow in current carrying wires around the sides of these two faces. What is the force between the two wires.

I know that I need to consider all the adjacent pairings of wires and the result will be zero unless the edges are parallel. This leaves 8 pairs, 4 with edges adjacent like AB and A'B' and 4 with edges opposite like AB and D'C'. I have the equation for a force between two wires, but I'm not sure if I can reduce my workload more and then use it, but I'm also not sure what I actually need to integrate to find the force...

答: For a pair of parallel wires, $dF=\frac{\mu_0II'}{2\pi r}\mathrm{d}l$, where $r$ is the distance between the wires and $\mu_0$ is the permittivity of free space. The direction of the force is away from the other wire if the currents are parallel. Let $I$ and $I'$ be both currents flowing clockwise, or replace $I$, say, with $-I$ if the current goes anticlockwise. For each side  of $ABCD$, the component of that force in the direction perpendicularly away from $A'B'C'D'$ is $F=\frac{\mu_0II'}{2\pi a}a+\frac{\mu_0II'}{2\pi  \sqrt{2}a}a$, giving a total force of $(1+\frac{1}{\sqrt{2}})\frac{\mu_0II'}{2\pi}$</p>

Title: Force between two wires


问:
Geometric distribution from exponential estimation
Problem:

$X \sim \exp(\lambda)$

Now $X$ is descretized to $Y$ using floor function

$Y_1, Y_2, \dots,Y_n$ is available from $Y$ distribution

I need to find method of moment estimator for $\lambda$ using $Y$ sample value

My solution:

$Y \sim$ geometric distribution with parameter $p$

Here $p=1-\exp(-\lambda)$

Now, moment estimator for $p$ is

$$ \hat{p} = \frac{1}{\overline{Y}}. $$

If we keep this $p$ value in above equation to get $\lambda$.
$$\lambda= -\ln\left(1-(1/\overline{Y})\right)$$

But my answer is

$$\lambda= \ln\left(1+(1/\overline{Y})\right)$$

Where am I getting it wrong.

答:
The distribution of $\lfloor X \rfloor$ when $X$ has pdf $\lambda e^{-\lambda x}$ (which I assume is the convention you are using for the exponential distribution) will have PMF $P(Y=y)=\int_y^{y+1} \lambda e^{-\lambda x} \, dx = -e^{-\lambda(y+1)}+e^{-\lambda y}=e^{-\lambda y} \left ( 1-e^{-\lambda} \right )$. Thus it is geometric with success probability parameter $p=e^{-\lambda}$, under the failure-counting convention (so $0$ is a possible value).

The mean of it is $\frac{1}{p}-1=e^\lambda-1$, so $\overline{Y}$ is a consistent and unbiased estimator for $e^\lambda-1$. Therefore $\log(1+\overline{Y})$ is a consistent but probably biased estimator for $\lambda$.

If you use $\hat{p}$ you will arrive at an algebraically equivalent expression, by using $\frac{1}{\hat{p}}-1=\overline{Y}$ and $\hat{p}=e^{-\hat{\lambda}}$ and solving for $\hat{\lambda}$.


问:
Injective Modules Motivation & Intuition
A module $M$  over a commutative ring $R$ is called a
'injective module' if it satisfies certain universal property explaned here.

Question: Is there any intuition how to think concretely
about injective modules? Do them naturally arise as an
attempt go generalize a special class of modules?
I'm asking this because I try to find an analogy to the
dual concept of projective modules.

Although these are formally defined by a similar (but dual) UP
these have a more accessible interpretation: These arise as
a natural generalization of free modules and form
literally finer building blocks of free modules since there is
a fact that a module is projective iff it is a direct
summand of a free module.

Does these exist a similar interpretation for injective modules?
Which class of modules do these naturally generalize and
do they arise also as 'building blocks' of something?

答:
I've always thought of injective modules this way (beyond simply a dual definition to projective modules):

Injective modules are a summand of any module containing them

Every module embeds in an injective module, which is a kind of "completion"

Now, I have not worked with cofree modules, but you may want to check out this entry on cofree modules. On one hand, I believed that such a dual description exists, on the other hand, I don't have personal experience with it, and I don't find wolfram mathworld to be very reliable. So please take it with a grain of salt.

It's also worth noting that the second bullet above does not dualize: the dual would usually be considered to be "every module has a projective cover" but it is not true. However, a famous result is that every module has a flat cover.  Rings for which every right module has a projective cover are known as right perfect rings.


问:
Given that $\frac{a}{1-x}+\frac{b}{1+2x}\approx-3+12x$ for small values of $x$, find the value of $a$ and $b$

Given that $\frac{a}{1-x}+\frac{b}{1+2x}\approx-3+12x$ for small values of $x$, find the value of $a$ and $b$.

This question is very hard i am on the chapter of Binomial expansion of $(1+x)^n$ for values of n that are not positive integers i know how to solve these questions but have no idea how to solve this type of question when involves 2 unknown variables.

答:
Using $\displaystyle\frac{1}{1+y}\approx 1 - y$ for small $y$, we wish to find $a$ and $b$ such that:

$$\frac{a}{1-x} + \frac{b}{1 + 2x} \approx 12x - 3$$

$$a(1 + x) + b(1 - 2x)= 12x - 3$$

$$(a - 2b)x + (a + b)= 12x - 3$$

Now, to make all of the terms match we have the system of linear equations:

$$\begin{cases}a - 2b = 12\\a + b = -3\end{cases}$$

Solving, we get $\boxed{(a, b) = (2, -5).}$


问:
Evaluating $\int_{0}^{1} \frac{\ln^{n} x}{(1-x)^{m}} \, \mathrm dx$
On another site, someone asked about proving that

$$ \int_{0}^{1} \frac{\ln^{n}x}{(1-x)^{m}} \, dx = (-1)^{n+m-1} \frac{n!}{(m-1)!} \sum_{j=1}^{m-1} (-1)^{j} s (m-1,j) \zeta(n+1-j), \tag{1} $$

where $n, m \in \mathbb{N}$, $n \ge m$, $m \ge 2$, and $s(m-1,j)$ are the Stirling numbers of the first kind.

My attempt at proving $(1)$:

$$ \begin{align}\int_{0}^{1} \frac{\ln^{n}x}{(1-x)^{m}} \, dx  &= \frac{1}{(m-1)!} \int_{0}^{1} \ln^{n} x \sum_{k=m-1}^{\infty} k(k-1)  \cdots (k-m+2) \  x^{k-m+1} \ dx \\ &= \frac{1}{(m-1)!} \sum_{k=m-1}^{\infty} k(k-1)  \cdots (k-m+2)  \int_{0}^{1} x^{k-m+1} \ln^{n} x \, dx \\ &= \frac{1}{(m-1)!} \sum_{k=m-1}^{\infty} k(k-1) \cdots (k-m+2) \frac{(-1)^{n} n!}{(k-m+2)^{n+1}}\\ &= \frac{1}{(m-1)!} \sum_{k=m-1}^{\infty} \sum_{j=0}^{m-1} s(m-1,j) \  k^{j}  \ \frac{(-1)^{n} n!}{(k-m+2)^{n+1}} \\ &= (-1)^{n}  \frac{n!}{(m-1)!} \sum_{j=0}^{m-1} s(m-1,j)  \sum_{k=m-1}^{\infty}   \frac{k^{j}}{(k-m+2)^{n+1}} \end{align} $$

But I don't quite see how the last line is equivalent to the right side of $(1)$. (Wolfram Alpha does say they are equivalent for at least a few particular values of $m$ and $n$.)

答:
Here is another approach based on the beta function

$$ F = \int_{0}^{1}x^{\alpha}(1-x)^{-m} dx = \beta( \alpha+1,1-m ),\quad \alpha+1>0,\, 1-m>0. $$

Now, our integral $I$ can be evaluated as

$$ I = \lim_{\alpha \to 0} \frac{d^n}{d \alpha^n} \beta( \alpha+1,1-m ),\quad n>m-1,\,m\geq 2. $$

Here is a useful identity

$$
{\partial \over \partial x} \mathrm{B}(x, y) = \mathrm{B}(x, y) \left( {\Gamma'(x) \over \Gamma(x)} - {\Gamma'(x + y) \over \Gamma(x + y)} \right) = \mathrm{B}(x, y) (\psi(x) - \psi(x + y)),  $$

where $\psi(x)$ is the digamma function.


问:
Best books for self-studying differential geometry
Next semester (fall 2021) I am planning on taking a grad-student level differential topology course but I have never studied differential geometry which is a pre-requisite for the course. My plan is to self-study in the summer and this semester so that I do not have to waste a semester taking a differential geometry course which will ruin my schedule.

I am looking for a book that covers all of the following topics (ideally) but at least most of them:

any suggestions would be welcome.

答: I had a good experience with self study as an undergrad using “Differential Forms and Connections” by Darling. It covers much of what you mention, and does a good job building from curves and surfaces to smooth manifolds, vector bundles etc. I agree with the comments that some of the topics listed in your quote are more in the direction of geometry than topology, and wouldn’t be necessary in a course following, say, Guillemin/Pollack, Tu’s “Introduction to Manifolds” or Hirsch’s “Differential Topology”.</p>


答:
Maybe this is better understood as "homotopies".

A little analogy before going into details. Suppose you have $f_1, f_2 : X \to Y$ honotopic one to another through $\alpha : X \times I \to Y$ and $g_1, g_2 : Y \to Z $ homotopic one to another through $\beta: Y \times I \to Z$.

How would you produce an homotopy between $g_1f_1$ and $g_2 f_2$ ? Well we surely have a map

$$ \gamma: X\times I^2 \to Y \times I \to Z$$

Where the first map applies $\alpha$ to the first two components and second map is just $\beta$. Notice also that $\gamma(x,0,0) = g_1 f_1, \gamma(x,1,1) = g_2 f_2 $. In order for $\gamma$ to be an homotopy it should really start from $X \times I$. Since we want our homotopy to start at $g_1 f_1 $ and finish at $g_2 f_2$, it is sensible to precompose with the diagonal $I \to I^2$.

So now we define

$$( \beta \star \alpha ) (x, t) = \beta( \alpha(x, t), t) $$

And we are done. But how the hell this is relevant to categories?? In the modern viewpoint, categories are thought as 1 dimensional spaces in a sense. Here is the plot twist: we have an "interval object" in categories given by $I = \{ 0 \to 1\}$. It is the category with two objects and one arrow from 0 to 1 (which I denote by $01$). I let you show that it plays the same role as in spaces (with the exception that here homotopies have a direction) : a natural transformation $\alpha : F_1 \to F_2$ between two functors $F_1, F_2 : C \to D $ is the same as a functor $C \times I \to D$.

What does it happen if you repeat the argument of spaces for functors..? Well, we will end up with a formula and guess what..! It is the right one. I think it is very instructive for you to fill in the details. Feel free to comment if you have any doubts!

Title: How to find a formula for horizontal composition?

Body:
For natural transformations $F_1 \xrightarrow{\alpha} F_2$ and $G_1 \xrightarrow{\beta} G_2$, I have proven that $(\beta \star \alpha)_X := G_2(\alpha_X) \circ \beta_{F_1X}$ is a natural transformation from $G_1 \circ F_1$ to $G_2 \circ F_2$.

Imagine you are Saunders / MacLane creating Category Theory. You want to work out as much as possible in short time and to find a natural transformation for composed functors. But how do you find this concrete formula for the horizontal composition in a clever way?

Since I am new in Category Theory, my (elaborated) proof of naturality consists of one and a half page and needed some hours. The formula consist of multiple characters. Hence brute-force (generate all combinations, check if the domains / codomains agree, if yes, try to prove naturality) seems not to be the most time efficient way to work.

Searching math.SE, I found questions for applications of horizontal composition and for the idea of horizontal composition itself as well as a proof of natural equivalence to another formula, but not how to develop the concrete definition of $(\beta \star \alpha)_X$.


问:
Question on Rudin Theorem 3.31 Proof
On page 65, we have:

$e-s_n = \frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}+\dots < \frac{1}{(n+1)!}(1 + \frac{1}{n+1}+\frac{1}{(n+1)^2}+\dots)=\frac{1}{n!n}$

Given $s_n = \sum_{k=0}^{n}{\frac{1}{k!}}$.

What I don't understand is, where does $\frac{1}{n!n}$ come from?

答: It comes from computing the sum of a geometric series:\begin{align}1+\frac1{n+1}+\frac1{(n+1)^2}+\cdots&=\frac1{1-1/(n+1)}\\&=\frac{n+1}n.\end{align}So,$$\frac1{(n+1)!}\left(1+\frac1{n+1}+\frac1{(n+1)^2}+\cdots\right)=\frac{n+1}{(n+1)!n}=\frac1{n!n}.$$

Title: Question on Rudin Theorem 3.31 Proof


答:
Note that since for $ |a| < 1$ (Geometric series formula)
$$
\sum_{i=0}^{\infty}a^i = \frac{1}{1-a},
$$
choosing $a := \frac{1}{n+1}$, on the right hand side it follows that
$$
\frac{1}{(n+1)!}\left(\sum_{i=0}^{\infty}\left(\frac{1}{n+1}\right)^i\right) = \frac{1}{(n+1)!}\left(\frac{1}{1 - \frac{1}{n+1}}\right) = \frac{1}{n!n}
$$

Title: Question on Rudin Theorem 3.31 Proof

Body:
On page 65, we have:

$e-s_n = \frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}+\dots < \frac{1}{(n+1)!}(1 + \frac{1}{n+1}+\frac{1}{(n+1)^2}+\dots)=\frac{1}{n!n}$

Given $s_n = \sum_{k=0}^{n}{\frac{1}{k!}}$.

What I don't understand is, where does $\frac{1}{n!n}$ come from?


问:
Decomposing a sum of symmetric tensors components
Let $\mathbf{T}$ be a fully symmetric tensor of order $3$ and size $N$. Its components can be represented as $T_{ijk}$ for all $1\leq i,j,k\leq N$. By symmetric I mean that if I permute any indices the value stays the same:
\begin{equation}
T_{ijk}=T_{ikj}=\dots=T_{jki}
\end{equation}
Now, I would like to rewrite the following sum, making use of the fact that $\mathbf{T}$ is symmetric:
\begin{equation}
\boxed{\sum_{i}^N\sum_{j}^N\sum_k^{N}T_{ijk}}
\end{equation}

That is, I would like to decompose the sum into:

\begin{equation}
\sum_{ijk}^NT_{ijk}=\alpha\sum_{i\leq j\leq k} T_{ijk}+\beta\sum_{i=j\leq k} T_{iik}+\gamma\sum_{i=j=k} T_{iii}+\text{ something else? }
\end{equation}

I would like to find the values of $\alpha,\beta\gamma$ but I am unsure of the values.

In the case of \textbf{symmetric} matrices I could have written:

\begin{align*}
\sum_{i}\sum_j M_{ij}&=\sum_{i<j}M_{ij}+\sum_i M_{ii}+\sum_{i>j}M_{ij}\\
&=2\sum_{i<j}M_{ij}+\sum_i M_{ii}
\end{align*}

But with tensors I have trouble decomposing this sum.

The bigger scope is trying to compute the following integral:

\begin{equation}
I=\int\left(\prod_{i\leq j\leq k}\mathrm{d}T_{ijk}\right)\exp\left\{-\frac{1}{2}\sum_{ijk}T_{ijk}^2\right\}
\end{equation}

Note the sum over all $ijk$ in the exponential, which is why I tried to decompose the sum into different parts.

答:
$$\sum_{i,j,k}T_{ijk}=6\sum_{i<j<k}T_{ijk}+3\sum_{i<j}T_{iij}+3\sum_{i<j}T_{ijj}+\sum_iT_{iii}$$
The $6$ accounts for all ways to permute $(i,j,k)$. For $(i,i,j)$, there are only $3$ ways, since switching the first two $i$'s has no effect, so you divide by $2$.

Another valid way to write this is
$$\sum_{i,j,k}T_{ijk}=6\sum_{i<j<k}T_{ijk}+3\sum_{i\neq j}T_{iij}+\sum_iT_{iii}$$


答:
As a (fill the gap) re the other responses:

$(1-x)(1 + x + x^2 + \cdots + x^n) = 1-x^{(n+1)} \approx 1$, for small values of $x$.  In fact, when $|x| < 1$, the infinite series $(1 + x + x^2 + \cdots)$ converges to $\frac{1}{1-x}.$

Further, for small values of $x$, it is reasonable to discard all but the first two terms of $(1 + x + x^2 + \cdots).$

This implies that for small values of $x$, $\frac{1}{1-x}$ can be approximated by $(1 + x).$

Similarly, $(1 + 2y)\times [1 - 2y + 4y^2 - 8y^3 + \cdots - \cdots + (-1)^n (2y)^n] = 1 + [-2y]^{(n+1)}.$

Therefore, for $|y| < \frac{1}{2}$, the infinite series

$[1 - 2y + 4y^2 - 8y^3 + \cdots - \cdots ]$ converges to $\frac{1}{1 + 2y}$.

Thus, similar to the analysis in the first part of this answer, for small values of $y,~ \frac{1}{1 + 2y}$ may be approximated by $(1 - 2y).$

The above analysis begs the question: how do you know what approximations to take.

Answer

Look at the original problem.

Since the RHS is a first degree polynomial, a blind guess is that you want to find first degree polynomials to approximate the LHS, for small values of $x$.

Unfortunately, this blind guess requires the abstract vision that such a LHS approximation will result in two (solvable) linear equations in two unknowns for the variables $a,b.$

It is reasonable to surmise that the premise that $x$ is a small value is authorizing polynomial approximations to the fractions.  It is then reasonable to question what degree polynomials should be used for the LHS approximations.

You are then supposed to question as follows: how could the LHS polynomial approximations be of degree higher than $1$, and still be equivalent to the RHS polynomial?

Body:
Given that $\frac{a}{1-x}+\frac{b}{1+2x}\approx-3+12x$ for small values of $x$, find the value of $a$ and $b$.

This question is very hard i am on the chapter of Binomial expansion of $(1+x)^n$ for values of n that are not positive integers i know how to solve these questions but have no idea how to solve this type of question when involves 2 unknown variables.


问:
Relationship Between Power Series ODE Solution Techniques?
When solving an ODE via a power series at an ordinary (nonsingular) point, the initial guess is $y = \sum_{n = 0}^\infty a_n x^n$.

When solving an Euler ODE, the second order equation $x^2 y'' + pxy' + qy = 0$ where $p$ and $q$ are constant coefficients for example, the initial guess is $y = x^s$.

When solving an ODE via The Method Of Frobenius at a regular singular point, the initial guess is said to be the product of the two prior guesses, $y = x^s \sum_{n = 0}^\infty a_n x^n$.

My question is whether The Method Of Frobenius is a generalization of each of the two prior solution techniques, and if not, what its relationship is to those techniques.  To break this down:

Does applying The Method Of Frobenius to an ODE at an ordinary point always produce $s = 0$ with algebraic multiplicity equal to the order of the ODE?

Does applying The Method Of Frobenius to an ODE and finding that $s = 0$ with algebraic multiplicity equal to the order of the ODE entail that the point approximated around was ordinary?

Does applying The Method Of Frobenius to an Euler ODE always produce a power series factor $\sum_{n = 0}^\infty a_n x^n$ in the guess identically equal to $1$?

Does applying The Method Of Frobenius to an ODE and finding that the power series factor $\sum_{n = 0}^\infty a_n x^n$ in the guess is identically equal to $1$ entail that the ODE was an Euler ODE?

If the answers are not all "yes," then The Method Of Frobenius is not a generalization of each of the other solution techniques.  In this case, what is the correct, high-level way to think about how these solution techniques relate, aside from symbolic similarity?

答:
I'm going to first remark that the respective assumptions $x^s\sum a_nx^n$ and $\sum a_nx^n$ are equivalent, and that $x^s$ is a relevant special case of either.
I'll also remark that if we accept the notion of a generalization to have some structural relevance as opposed to simply looking like it is more general, then the series assumption and the Frob. assumption are the same. For example, I could say $1+x^s$ is more general than $1+x$, and in a naïve way I would be correct by seeing that $s=1$ produces $1+x$--but the keen eye will note that by changing $x\to x^s$, we get the same form.

To 3.: Clearly for the Euler ODE, if we assume $y$ is some polynomial of degree $n$, then the first term $x^2y''$ remains degree $n$, and the term $pxy'$ remains degree $n$, and the term $qy$ remains degree $n$. Since our $x$ is not a root and is independent, one must solve for the coefficients such that each new coefficient at the $n$th term is $0$. In other words, assuming $y=a_0+a_1x+\cdots + a_nx^n$ produces $x^2y'' +pxy'+qy = A_nx^n+A_{n-1}x^{n-1}+\cdots + A_0=0$ and so each $A_k=0$. More exactly, $$x^2y'' + pxy' +qy = \sum [n^2+(p-1)n+q]a_nx^n$$ and so $$n^2+(p-1)n+q=0$$ Therefore $n$ only takes on two values for the sum to be valid. This is the same if I had assumed $x^s$ in the original equation to obtain $$s^2+(p-1)s+q=0$$ which obviously produces $$y=C_1x^{s_1}+C_2x^{s_2}.$$

And clearly, it is not true that in the guess $y=x^s\sum a_nx^n$ will have the term $\sum a_nx^n=1$, but rather some combination of indices telling you essentially the same thing.

To 4.: Your statement in (4) is clearly false unless you call all ODEs of the form $$x^ny^{(n)}+a_{n-1}x^{n-1}y^{(n-1)}+\cdots =0$$ Euler ODEs. However, it's rather arbitrarily false since what is called the general Euler-Cauchy ODE is the second-degree case $n=2$. Your statement (4) is, respectfully, not a requirement for the guess $y=x^s\sum a_nx^n$ to be a more general kind of guess. It's actually true that the guess $y=\sum a_nx^n$ suffices for an Euler ODE, because you'll find the conditions necessary for the index.

Lastly, the assumption $y=uv$ or knowing that $u'v+v'u = (uv)'$ is probably one of the most important in differential equations, alongside using those assumptions to bend the equation to your will, into the suitable forms you choose or which make the problem more obvious for you with respect to your solution goals. And this is why the Frob. method works its magic in certain relevant cases.

I hope this may have answered your questions.


问:
Derivative and inverse
Is there a function whos derivative of its inverse equals the inverse of its derivative? This may also hold on a specific interval only.
$$\frac{d}{dx}f^{\langle-1\rangle}(x)=\left(\frac{d}{dx}f(x)\right)^{\langle-1\rangle}$$
Considering that $f(x)=x$ is its own inverse and $e^x$ is its own derivative, neither of them satisfies the equation, but both are the only functions with the respective properties (over the reels). Is this already a proof that there is no solution to the above equation?

Also, I dont know much about complex analysis but if a complex-valued function meets the criteria I'd be happy to know too :)

答:
Disclaimer. In view of comments issued, and by growing insight, this answer must be considered as a partial one.

The easiest way is to start with considering the derivatives. Let $g(x)=f'(x)$.
Then $g(x)$ must have an inverse $g^{-1}(x)$. This means that $y=g(x)$ is monotonically increasing or decreasing.
$$
\mbox{Increasing:} \quad \begin{cases}
x_1 \lt x_2 \quad \Longleftrightarrow \quad y_1 \lt y_2 \\
x_1 \gt x_2 \quad \Longleftrightarrow \quad y_1 \gt y_2
\end{cases} \\
\mbox{Decreasing:} \quad \begin{cases}
x_1 \lt x_2 \quad \Longleftrightarrow \quad y_1 \gt y_2 \\
x_1 \gt x_2 \quad \Longleftrightarrow \quad y_1 \lt y_2
\end{cases}
$$
The inverse $g^{-1}(x)$ is obtained (graphically) by mirroring $g(x)$ in the line $y=x$, thus by exchanging $x$ and $y$.
From this it clear that $g(x)$ and $g^{-1}(x)$ must be both monotonically increasing or both be monotonically decreasing.

The same considerations are valid for $f(x)$ and $f^{-1}(x)$ as well, because these two are each others inverse too.

But $g(x)$ is also a derivative. If this derivative would become zero somewhere, then $f(x)$ would no longer be monotonic at that place.
So $g(x)$ must be nonzero everywhere in its domain.
For the inverse $g^{-1}(x)$ this means that there cannot be an intersection with the $y$-axis.
Without loss of generality, we may assume that both $g(x)$ and $g^{-1}(x)$ are defined in the first quadrant, for $x \gt 0$ and $y \gt 0$.
Other cases are covered by re-defining $g$ using symmetry: $g(x) := g(-x)$ , $g(x) := -g(x)$ , $g(x) := -g(-x)$.

Before proceeding, a few examples will be presented. First example:
$$
f'(x) = g(x) = 1/x \quad ; \quad [f'(x)]^{-1} = g^{-1}(x) = 1/x \\
f(x) = \int_1^x\frac{dt}{t} = \ln(x) = y \quad ; \quad x = \ln(y) \; \Rightarrow \; f^{-1}(x) = e^x \quad ; \quad [f^{-1}]'(x) = e^x
$$
It is noticed that $[f'(x)]^{-1}$ is monotonically decreasing while $[f^{-1}]'(x)$ is monotonically increasing.
Second example:
$$
f'(x) = g(x) = \sqrt{x} \quad ; \quad [f'(x)]^{-1} = x^2 \\
f(x) = \int_0^x\sqrt{t}\,dt = \frac{2}{3}x^{3/2} \quad ; \quad x = \frac{2}{3}y^{3/2}
\; \Rightarrow \; f^{-1}(x) = \left(\frac{3x}{2}\right)^{2/3} \quad ; \quad [f^{-1}]'(x) = \left(\frac{3x}{2}\right)^{-1/3}
$$
It is noticed that $[f'(x)]^{-1}$ is monotonically increasing while $[f^{-1}]'(x)$ is monotonically decreasing.

A pattern is emerging. Let's consider now the general case for $g(x)=f'(x)$ monotonically decreasing. It will be assumed
that the reader is able to construct the analogue for monotonically increasing $g(x)$.
Then $[f'(x)]^{-1}$ is monotonically decreasing as well. Define the function $f(x)$ as a definite integral
in statu nascendi (a Riemann sum). For example as:
$$
\sum_i g\left(x_i\right)\left(x_{i+1}-x_i\right) \approx \int_a^x g(t)\,dt = f(x)
$$
And select the subdivisions in such a way that
$$
g\left(x_i\right)\left(x_{i+1}-x_i\right) = \mbox{constant} = dA
$$
Then we can construct the inverse function $f(x)$ by numerical approximation.
It is clear thar if $g(x_i)$ is decreasing, then the area $dA$ can only kept constant by increasing the intervals $\left(x_{i+1}-x_i\right)$.
This means that the intervals of $f^{-1}(x_i)$ must be increasing, because of $\left(x_{i+1}-x_i\right)\to\left(y_{i+1}-y_i\right)$.
But then the derivative of $f^{-1}(x_i)$ is equal to $\left(y_{i+1}-y_i\right)/dA$, which means that $[f^{-1}]'(x)$ is increasing numerically,
therefore analytically when taking limits. So here comes our end-result,
apart from technicalities to filled in by the questioner eventually.

Theorem.

$[f'(x)]^{-1}$ is monotonically decreasing if and only if $[f^{-1}]'(x)$ is monotonically increasing.

$[f'(x)]^{-1}$ is monotonically increasing if and only if $[f^{-1}]'(x)$ is monotonically decreasing.

Therefore it is impossible to find (real valued) functions $f$ with $[f'(x)]^{-1}=[f^{-1}]'(x)$.

EDIT.

Below is a visualization of the technique employed, for the special example
$f'(x)=[f'(x)]^{-1}=1/x$ , $\color{red}{f(x)=\ln(x)}$ , $\color{green}{f^{-1}(x)=[f^{-1}]'(x)=e^x}$ .

Further specifications: $-0.1 \le x \le 9$ ; $-0.1 \le y \le 9$ ; $dA = 1/10$ ;

thick colored lines numerically, thin colored lines analytically.

BUG FIX.

The domains of interest are much trickier than I thought.

Counter example. Consider $f(x)=1/x$, defined for $x\gt 0,y\gt 0$. Then $f'(x)=-1/x^2$ is defined for $x\gt 0,y\lt 0$;
$[f'(x)]^{-1}=\sqrt{-1/x}$ is defined for $x\lt 0,y\gt 0$ and increasing; $f^{-1}(x)=1/x$ is defined for $x\gt 0,y\gt 0$;
$[f^{-1}]'(x)=-1/x^2$ is defined for $x\gt 0,y\lt 0$ and increasing.

So we have a problem with the first two statements of our Theorem.

In general. Consider $f(x)$, defined for $x\gt 0,y\gt 0$ and monotonically decreasing.
Then $f'(x)$ is defined for $x\gt 0,y\lt 0$ (i.e. negative) and monotonically increasing.
$[f'(x)]^{-1}$, because of the mirroring in $y=x$, is defined for $x\lt 0,y\gt 0$ and monotonically increasing as well.
$f^{-1}(x)$, because of the mirroring in $y=x$, is defined for $x\gt 0,y\gt 0$ and monotonically decreasing.
So $[f^{-1}]'(x)$ is defined for $x\gt 0,y\lt 0$ and monotonically increasing.

However, because of the two different domains $[f^{-1}]'(x)$ cannot be identical to $[f'(x)]^{-1}$.

Conclusion: the last statement of our Theorem remains valid, but the first two statements, in general, are wrong.

Note. A better numerical approximation of the integral sometimes can be obtained with the trapezium rule:
$$
\frac{g\left(x_i\right)+g\left(x_{i+1}\right)}{2}\left(x_{i+1}-x_i\right) = \mbox{constant} = dA = \frac{1}{n}
$$
But it does work only if $g\left(x_{i+1}\right)$ can be predicted, given the value of $dA$. In case $g(x)$ is an orthogonal hyperbola, prediction is successful:
$$
x_{i+1}-x_i=\frac{1}{n}\frac{2}{1/x_{i+1}+1/x_i}=\frac{1}{n}\frac{2x_{i+1}x_i}{x_{i+1}+x_i} \quad \Longleftrightarrow \\
n\left(x_{i+1}-x_i\right)\left(x_{i+1}+x_i\right)-2x_{i+1}x_i=nx_{i+1}^2-2x_ix_{i+1}-nx_i^2=0 \quad \Longleftrightarrow \\
x_{i+1} = \frac{x_i}{n}\pm\sqrt{\left(\frac{x_i}{n}\right)^2+x_i^2}\quad \Longrightarrow\quad
x_{i+1}=x_i\left(\frac{1}{n}+\sqrt{1+\frac{1}{n^2}}\right)
$$

Title: Derivative and inverse


问:
Write $AS$ as a multiple of $AD$ and $CS$ as a multiple of $CE$.
In a triangle $ABC$ the point $D$ is on $BC$ such that $\overline{BD}=\frac{2}{3}\cdot \overline{BC}$ and the point  $E$ is on $AB$ such that $\overline{AE}=\frac{3}{5}\cdot \overline{AB}$.

The lines $AD$ and $CE$ intersect in $S$.

Write $AS$ as a multiple of $AD$ and $CS$ as a multiple of $CE$.

We have a triangle as follows, right?

But how can we get an information for $AS$ and $CS$ ? Could you give me a hint? Do we maybe use vector operations?

$$$$

EDIT :

Using vectors we have $$\vec{AS}+\vec{SD}=\vec{AD} \Rightarrow \vec{AS}=\vec{AD}-\vec{SD} \Rightarrow \vec{AS}=\vec{AD}+\vec{DS}$$ Does this help us ? How can we continue?

答:
Here is a method using vectors.

Let position vectors of the vertices be $A,B,C$. Then
$$D=\frac{B+2C}{3} \quad ; \quad  E=\frac{2A+3B}{5}$$

Let $AS=\lambda AD$ and $CS=\mu CE$, where $\lambda, \mu$ are to be found.

Then $S$ divides $AD$ in ratio $\lambda : (1-\lambda)$ and $CE$ in ratio $\mu  : (1-\mu)$, we can write position vector of $S$ in two ways as

$$S=\lambda D+(1-\lambda)A=(1-\mu)C+\mu E$$
$$\Rightarrow (1-\lambda)A+\frac{\lambda}{3}B+\frac{2\lambda}{3}C=\frac{2\mu}{5}A+\frac{3\mu}{5}B+(1-\mu)C$$

This can be solved for $\lambda, \mu$ by equating the coefficients.


问:
Asymptotic of incomplete Gamma function, $\Gamma(n,n/a)$
Let $a>1$. I need to approximate (for n large)
$$
\Gamma(n,n/a) \approx f(n)
$$

Approximations I found are for the case which the second argument is fixed. Is there any simple formula for this asymptotic?

答:
If $0<\alpha<1$ and $\alpha$ is bounded away from $1$ (e.g., $1 - n^{ - 1/2} \gg \alpha$), then
$$
\Gamma (n,\alpha n) \sim \Gamma (n) + \alpha ^n \frac{{n^{n - 1} e^{ - \alpha n} }}{{1-\alpha}}\left( {1 - \frac{\alpha }{{n(1-\alpha)^2 }} + \frac{{\alpha (2\alpha  + 1)}}{{n^2 (1-\alpha  )^4 }} -  \cdots } \right)
$$
as $n\to +\infty$ (cf. http://dlmf.nist.gov/8.11.E6). If $\alpha$ is close to $1$, one may use
\begin{align*}
\frac{{\Gamma (n,n - \beta \sqrt n )}}{{\Gamma (n)}} \sim 1 & - \frac{1}{2}\operatorname{erfc}\left( {\frac{\beta }{{\sqrt 2 }}} \right) \\ &+ \frac{1}{{\sqrt {2\pi n} }}\exp \left( { - \frac{{\beta ^2 }}{2}} \right)\left( {\frac{{\beta ^2  - 1}}{3} - \frac{{\beta (2\beta ^4  - 11\beta ^2  + 3)}}{{36\sqrt n }} +  \cdots } \right)
\end{align*}
as $n\to +\infty$ with $0 < \beta \ll n^{1/6}$. Here $\operatorname{erfc}$ denotes the complementary error function (cf. https://doi.org/10.1007/s00365-018-9445-3).


问:
How to multiply Roman numerals?
How to multiply Roman numerals? I need an algorithm of multiplication of numbers written in Roman numbers. Help me please. </p>

答:
Take the first number you want to multiply and break it down into parts, any way you choose.
E.G. 43 = XLIII = XL + III, or X + X + X + X + III, etc.
Do the same with the second number in the multiplication.
E.G. 15 = XV = X + V, or V + V + V, etc.

Take any pair of those combinations and set them within a table, with the higher row showing the first number's parts, and the lower row showing the second number's parts.
Multiply starting from the lower right to the upper row, then the lower left  to the upper row. Place the result to each equation underneath, vertically.

Total them.

Title: How to multiply Roman numerals?


问:
Cosine Fourier series
Given the following function
$$f(x)=\left\{\begin{array}{ll} 0, & 0\leq x\leq1 \\ 1, & 1<x\leq2 \end{array}\right.$$
I have to find its cosine Fourier series. I expanded it in a even way such that
$$2L=4\Rightarrow L=2$$
So,
$$a_{0}=\frac{2}{2}\int_{0}^{2}f(x)dx=1$$
and
$$a_{n}=\frac{2}{2}\int_{0}^{2}f(x)\cos\left(\frac{n\pi x}{2}\right)dx=-\frac{2}{n\pi}\sin\left(\frac{n\pi}{2}\right)$$

I know $\displaystyle\sin\left(\frac{n\pi}{2}\right)=0$ if $n$ is even, but if $n$ is odd, it will be equal to $-1$ or $1$. My problem is: how can I rewrite this as $(-1)$ power to something? I mean, I tried
$$\sin\left(\frac{n\pi}{2}\right)=(-1)^{n}$$
or even
$$\sin\left(\frac{n\pi}{2}\right)=(-1)^{2n+1}$$
But I always get $1$.

答:
The correct way to write what you want is
$$\sin\left(\frac{(2m+1)\pi}{2}\right)=(-1)^m\qquad \text{and}\qquad \sin\left(\frac{2m\pi}{2}\right)=0$$
for any $m\in\mathbb Z$.

Title: Cosine Fourier series


问:
What do you take as the Particular Integral when is cancels out
I'm trying to solve the second order ODE $\ddot{x} + \omega^2x = A\cos(\omega t)$ and when I try to solve the PI it cancels out to be just $0 = A\cos(\omega t)$.  What do I take for the PI?

答:
When working with hyperbolic/trigonometric functions we never just take one of them.
Always use them in pairs, that way you won't lose data.

$$f(t)=A\cos(\omega t)+B\sin(\omega t)$$

where $A$ and $B$ are constants.

Particular Solutions for the Inhomogeneous term

If the homogenous solution has a similar term to the inhomogeneus one (resonance), you should multiply your 'testing' function by a $t$.


问:
Prove that AM-GM-HM recursive function converge to the same value.
I am trying to do this question Let $0<a_1 < b_1 < c_1$ $$ a_n = HM(a_{n-1}, b_{n-1}, c_{n-1})$$$$  b_n = GM(a_{n-1}, b_{n-1}, c_{n-1})$$$$  c_n = AM(a_{n-1}, b_{n-1}, c_{n-1})$$
Prove that ${a_n}, {b_n}, {c_n}$ converges and converge to the same value.

From HM-GM-AM inequality, I managed to show that $a_n< b_n< c_n$ and that $a_{n}< a_{n+1} < b_{n+1} <c_{n+1} < c_n$. This implies that $a_n$ and $c_n$ converges due to being bounded and monotonic. However, I am unable to show that they converge to the same limit or that $b_n$ converges. Any help would be appreciated.

答:
Hint

$$c_n=\frac{a_{n-1}+b_{n-1}+c_{n-1}}{3}\to b_{n-1}=3c_n-a_{n-1}-c_{n-1}.$$

If $a_n$ converges to $A$ and $c_n$ converges to C, do $n\to \infty$ in the above equation and find that $b_n$ also converges. Now call its limit $B$ and uses the three relations (for $a_n,b_n,c_n$) to find the solution of a system in $A,B,C$ and conclude that $A=B=C$.


问:
Show $\frac{x_1}{x_n} + \frac{x_2}{x_{n-1}} + \frac{x_3}{x_{n-2}} + \dots + \frac{x_n}{x_1} \geq n$
I was recently asked this question which stumped me.

How can you show $\dfrac{x_1}{x_n} + \dfrac{x_2}{x_{n-1}} + \dfrac{x_3}{x_{n-2}} + \dots + \dfrac{x_n}{x_1} \geq n$ for any positive reals $x_1, x_2, \dots, x_n$?

Body: $$\dfrac{x_1}{x_n} + \dfrac{x_2}{x_{n-1}} + \dfrac{x_3}{x_{n-2}} + \dots + \dfrac{x_n}{x_1}- n =\left(\dfrac{\sqrt{x_1}}{\sqrt{x_n}}-\dfrac{\sqrt{x_n}}{\sqrt{x_1}} \right)^2+\left(\dfrac{\sqrt{x_2}}{\sqrt{x_{n-1}}}-\dfrac{\sqrt{x_{n-1}}}{\sqrt{x_2}} \right)^2+...$$


问:
Does the polynomials span the subspace of $p_2$?
There is a problem in my Linear algebra book, that I have some trouble with. I have looked around on the forum, but cant seem to get the hang of it, because of the difference in polynomials, the degree of the polynomials etc.

Hope there is a kind stranger that can give me a tip!

Let $p_3(R)$ be a real vector space of polynomials: $p=ax+bx^2+c$ s.t.  $a,b,c\in\mathbb R$ of degree $2$ or less.
Let $S=\{p \in p_3(R):p(1)=0\}$.

Sow that $p_1=1-X^2$, $p_2=X-X^2$ is a spanning set of S.

答: Hint: If $p(x)\in S$, then $p(1)=0$, and therefore, since $\deg p(x)\leqslant2$, $p(x)$ can be written as $(x-1)(ax+b)$, for some real numbers $a$ and $b$. But then\begin{align}p(x)&=a(x-1)x+b(x-1)\\&=a(x^2-x)+b(x-1)\\&=a(x^2-x)+a(x-1)+(b-a)(x-1)\\&=a(x^2-x)+b(x-1).\end{align}


答:
From $a_n<a_{n+1}<c_{n+1}<c_n$ and $b_n<c_n$ for all $n$ we can deduce
$$\begin{align*}c_{n+1}-a_{n+1} &< c_{n+1}-a_n\\
&= \frac{1}{3}a_n+\frac{1}{3}b_n+\frac{1}{3}c_n-a_n \\
&< \frac{1}{3}a_n+\frac{1}{3}c_n+\frac{1}{3}c_n-a_n \\
&= \frac{2}{3}c_n-\frac{2}{3}a_n \\
&= \frac{2}{3}(c_n-a_n)
\end{align*}$$
hence $c_n - a_n \to 0$ and the limits of $(a_n)$, $(c_n)$ are the same, which must then also be the limit of $(b_n)$.

Title: Prove that AM-GM-HM recursive function converge to the same value.

Body:
I am trying to do this question Let $0<a_1 < b_1 < c_1$ $$ a_n = HM(a_{n-1}, b_{n-1}, c_{n-1})$$$$  b_n = GM(a_{n-1}, b_{n-1}, c_{n-1})$$$$  c_n = AM(a_{n-1}, b_{n-1}, c_{n-1})$$
Prove that ${a_n}, {b_n}, {c_n}$ converges and converge to the same value.

From HM-GM-AM inequality, I managed to show that $a_n< b_n< c_n$ and that $a_{n}< a_{n+1} < b_{n+1} <c_{n+1} < c_n$. This implies that $a_n$ and $c_n$ converges due to being bounded and monotonic. However, I am unable to show that they converge to the same limit or that $b_n$ converges. Any help would be appreciated.


问:
Probability of stepping on n-th field in game where you move using dice

We play a game where in each round we roll a six-sided dice and move the appropriate number of squares. How likely are we to step on the $n$-th field during the game, if we started with the number $0$?

The answer is sufficient in the form of a recurrent formula, which refers to a constant number of values for the lower one. Can you use only $2$ values for lower?

I am kinda lost here. I did bit know how even approach this.

Thanks to anyone who can help me guide to right direction

I made this:

```<code>| 1 | 1                                             |
| 2 | 1+1,2                                         |
| 3 | 1+1+1, 2+1, 3                                 |
| 4 | 1+1+1+1, 1+1+2, 1+3, 2+2, 4                   |
| 5 | 1+1+1+1+1, 1+1+1+2, 1+2+2, 1+1+3, 1+4, 2+3, 5 |
</code>```

But I can not see any pattern or how I can calculate probability from it. Does order of numbers matter ?

答:
For $n=1\dots 6$ we have:
$$
p_n=\frac16+\frac16\sum_{k=1}^{n-1}p_{n-k}\implies p_n=\frac16\left(\frac76\right)^{n-1}\tag1
$$

Indeed (1) obviously holds for $p_1$ and by induction:
$$
p_n=\frac16+\frac1{6}\sum_{k=1}^{n-1}\frac1{6}\left(\frac76\right)^{k-1}
=\frac16+\frac1{36}\frac{\left(\frac76\right)^{n-1}-1}{\frac76-1}=
\frac16\left(\frac76\right)^{n-1}.
$$

For $n>6$ the probabilities can be computed by the following formula:
$$
p_n=\frac16\sum_{k=1}^{6}p_{n-k}=\frac{7p_{n-1}-p_{n-7}}6.\tag2
$$
The rightmost expression in $(2)$ is valid for $n>7$. It can be used for $n=7$ with the convention $p_0=1$.


问:
Do we have $(f^{-1}(\mathcal O_{\operatorname{Spec}A}))(\operatorname{Spec}B)=A$?
Given a morphism of schemes $f:\operatorname{Spec}B\to\operatorname{Spec}A$, do we have $(f^{-1}(\mathcal O_{\operatorname{Spec}A}))(\operatorname{Spec}B)=A$?

答:
I believe the following holds: If $f: X \rightarrow S$ is a map of schemes and if $p \neq q\in X$ are points with $f(p)=f(q):=s \in S$ you get a canonical isomorphism

$$f^{-1}(\mathcal{O}_S)_p \cong \mathcal{O}_{S,s}\cong f^{-1}(\mathcal{O}_S)_q.$$

Since $p \neq q$ we do not have an isomorphism

$$ \mathcal{O}_{X,p}\neq \mathcal{O}_{X,q}$$

in general. Hence $f^{-1}(\mathcal{O}_S) \neq \mathcal{O}_X$ in general.

Question: "Do we have $f^{-1}(Spec(A))(Spec(B))=B$?"

Answer: We have a canonical map

$$f(Spec(B)): f^{-1}(\mathcal{O}_{Spec(A)})(Spec(B)\rightarrow \mathcal{O}_{Spec(B)}(Spec(B))\cong B,$$

but since the sheaves $f^{-1}(\mathcal{O}_{Spec(A)})$ and $\mathcal{O}_{Spec(B)}$ differ the map $f(Spec(B))$ is not an isomorphism in general.

Note: If there is a spelling mistake in the question, and we replace $A$ with $B$
the answer is positive when the map is an open immersion:

There seems to be some confusion on an exercise on adjoint functors.
If you look in Hartshorne Chapter II.1 you will find for any map of topological space $f: X\rightarrow Y$ (exercise 1.18) an isomorphism

$$M1.\text{   }Hom_X(f^{-1}G, F)\cong Hom_Y(G, f_*F).$$

Hence if $F:=(\tilde{f}, f^{\#}): X \rightarrow S$ is a map of schemes there is canonically a map of sheaves of rings on $X$:

$$f: \tilde{f}^{-1}(\mathcal{O}_S)\rightarrow \mathcal{O}_X.$$

The map $f$ corresponds to the map

$$f^{\#}: \mathcal{O}_S \rightarrow \tilde{f}_*\mathcal{O}_X$$

via the isomorphism $M1$. If $X:=Spec(B), S:=Spec(A)$ it follows the map $f$ is a map of sheaves of rings on $X$, not on $S$. Hence it does not make any sense to evaulate $f$ at $S$, since $S$ is not an open set in $X$. It only makes sense to evaluate $f$ at $X:=Spec(B)$. Moreover $A$ is not a subring of $B$ hence you never recover $A$ from $\mathcal{O}_X(U)$ or $\mathcal{O}_X(X)$.

Example 1. Let $S:=Spec(A), X:=D(f)$ for $f\in A$. We get an inclusion of the basic open set $D(f)$ into $S$:

$$\tilde{i}: X:=D(f) \rightarrow S$$

and a canonical map

$$i: \tilde{i}^{-1}(\mathcal{O}_S) \rightarrow \mathcal{O}_{D(f)},$$

and the map $i$ is an isomorphism: You may check that $\tilde{i}^{-1}(\mathcal{O}_S)\cong \mathcal{O}_{D(f)}$. In this special case you get an isomorphism

$$ i(D(f)): \tilde{i}^{-1}(\mathcal{O}_S)(D(f)) \cong A_f \cong\mathcal{O}_{D(f)}(D(f)),$$

but you do not get an isomorphism in general.

The natural question to ask is: for a general map of rings $\phi: A \rightarrow B,$

Question: Is the canonical map

$$f(X):\tilde{f}^{-1}(\mathcal{O}_S)(X) \rightarrow \mathcal{O}_X(X) \cong B$$

an isomorphism?"

Answer: The answer is "no", by the argument given in the previous post.

Note: if you take the "pull back"

$$F^*(\mathcal{O}_S)$$

you get

$$F^*(\mathcal{O}_S):= \mathcal{O}_X \otimes_{\tilde{f}^{-1}(\mathcal{O}_S)} \tilde{f}^{-1}(\mathcal{O}_S) \cong \mathcal{O}_X$$

hence

$$F^*(\mathcal{O}_X)(X)\cong \mathcal{O}_X(X) \cong B.$$

To people reacting negatively to this, denouncing it as "vandalism": Do exercise II.1.18 in HH and consult the definition of the notion "morphism of schemes" in Chapter II in HH.


问:
First hitting time of a symmetric random walk
Definitions:

Let $\xi_n$ be a symmetric random walk, i.e.,
$$
\xi_n=\eta_1+\eta_2+\dots+\eta_n,
$$
where $\{\eta_n\}$ is a sequence of i.i.d. random variables such that
$$
P\{\eta_n=1\}=P\{\eta_n=-1\}=\frac{1}{2}.
$$
Furthermore, we define the first hitting time to be $$\tau=\min\left\{n:|\xi_n|=K\right\},$$
where $K$ is a positive integer.

I was reading a book on stochastic processes and here we want to show that $\tau<\infty$ a.s. The book proves this as follows

We want to show that $P\{\tau=\infty\}=0.$ To this end we shall estimate $P\{\tau>2Kn\}.$ Notice that $$P\{\tau>2Kn\}\le \left(1-\frac{1}{2^{2K}}\right)^n\longrightarrow 0$$ as $n\to\infty.$ Thus, we have \begin{align}
P\{\tau=\infty\}&=\bigcap_{n=1}^\infty P\{\tau>2Kn\} \\ &=\lim_{n\to\infty} P\{\tau>2Kn\}=0.
\end{align}

After spending some time, I could not figure out how to get the inequality $$P\{\tau>2Kn\}\le \left(1-\frac{1}{2^{2K}}\right)^n$$ in the first line of the proof. Can someone help me understand why this inequality holds?

Many thanks in advance.

答:
$\frac{1}{2^{2K}}$ is the probability of $2K$ successive $+1$s in a row in an attempt of $2K$ steps. If this happens you must either hit the upper boundary or must have started below the lower boundary, since they are $2K$ apart.

$1-\frac{1}{2^{2K}}$ is the probability of not having $2K$ successive $+1$s in a row  in an attempt of $2K$ steps.

$\left(1-\frac{1}{2^{2K}}\right)^n$ is the probability of not having $2K$ successive $+1$s in a row in any of $n$ independent attempts of $2K$ steps ($2Kn$ steps in total).

That last probability is therefore less than or equal to the probability of never hitting the boundary in $2Kn$ steps, which is $P\{\tau>2Kn\}$.

Tighter bounds are possible but unnecessary.


问:
Prove that certain elements are not in some ideal
I have the following question:

Is there a simple way to prove that

$x+1 \notin \langle2, x^2+1\rangle_{\mathbb{Z}[x]}$ and

$x-1 \notin \langle2, x^2+1\rangle_{\mathbb{Z}[x]}$

without using the fact that $\mathbb{Z}[x]/\langle2, x^2+1\rangle$ is an integral domain?

Thanks for the help.

答:
Perhaps the simplest way to try to address these questions is tro try to characterize the elements in the ideal. In this case

$$f(x)\in\langle 2,x^2+1\rangle\iff f(x)=2g(x)+(x^2+1)h(x)\;,\;\;g(x),h(x)\in\Bbb Z[x]$$

and this means that any non-zero element in the ideal has either all its coefficients even and/or it is a multiple of $\,x^2+1\,$ , and this means either $\,f\,$ has all its coefficients even or its degree is at least two

Well, since neither $\,x-1\,,\,x+1\,$ fulfills the above neither belongs to the ideal.


问:
Semi-continuity for extended value function
Given a topological vector space $(X,\tau)$ a function $f:(X,\tau)\to\overline{\mathbb{R}}:=\mathbb{R}\cup\{\pm\infty\}$ is said to be lower-semicontinuous at $x_0\in X$ if

$$
f(x_0)\le\liminf_{x\to x_0}f(x):=\sup_{V\in\mathscr{V}(x_0)}\inf_{x\in V}f(x),
$$
where $\mathscr{V}(x_0)$ stands for the system of neighborhoods of $x_0$.

Similarly, the upper-semicontinuity at $x_0$ is defined with "$\ge\limsup$" instead of "$\le\liminf$".

My question are: If a function is identically equal to $+\infty$ is it lower and upper semicontinuous everywhere?

Is such a function continuous everywhere? (Continuity requires $f(x_0)\in\mathbb{R}$)

My questions are more about formalism and asks about a discussion on basic definitions. As stated somewhere in a book,  a function is continuous at $x_0$ if and only if it is both upper and lower semi-continuous there.

答:
If $f \equiv +\infty$, and $x_0 \in X$ then whatever $V \in \mathscr{V}(x_0)$ is, $\displaystyle \inf_{x \in V} f(x)=+\infty$ and so $f$ is trivially lsc at $x_0$. That $f$ is also usc is checked the same way.

BTW it's perfectly fine to talk about continuity at $x_0$ if $f(x_) = +\infty$. Maybe your text book thinks differently but $\overline{\Bbb R}$ has an (even metrisable) order topology, so continuity can be discussed...


问:
Show that the curve is contained in a plane.

If all osculating planes of a regular curve pass through one point, show that the curve is contained in a plane.

The osculating plane equation is as follows:
$$\begin{vmatrix}
X-x & Y-y & Z-z \\
x' & y' & z' \\
x'' & y'' & z''
\end{vmatrix}=0$$
or $(R-r, r', r'') = 0$ where $R = (X,Y,Z)$ and $r = r(t)$ is the curve equation.

The first idea came to my mind was showing that the torsion is zero everywhere, but I do not see how to observe this fact.

Let $M(A,B,C)$ be the point that all those planes pass through. Then, for any value of $t$, we have
$$\begin{vmatrix}
A-x & B-y & C-z \\
x' & y' & z' \\
x'' & y'' & z''
\end{vmatrix}=0$$
Expanding the determinant does not seem to give anything. I got stuck here.

答:
Just apply Serret-Frenet.

WLOG the common point is the origin and the curve is parametrised by arclength $s$.  The given condition says $\mathbf{r}\cdot(\mathbf{r}'\times\mathbf{r}'')=0$.

Since $\mathbf{r}'\times\mathbf{r}''=\mathbf{T}\times\mathbf{T}'=\kappa\mathbf{T}\times\mathbf{N}=\kappa\mathbf{B}$, either we have $\kappa=0$ on an interval (so we are done), or $\mathbf{r}\cdot\mathbf{B}=0$ on an interval.

Differentiating $\mathbf{r}\cdot\mathbf{B}=0$ gives $\mathbf{r}\cdot\tau\mathbf{N}=0$ so again either zero torsion on an interval (so we are done) or $\mathbf{r}\cdot\mathbf{N}=0$ on an interval.  Since $\mathbf{r}\cdot\mathbf{N}=\mathbf{r}\cdot\mathbf{B}=0$ we have $\mathbf{r}=(\mathbf{r}\cdot\mathbf{T})\mathbf{T}$ so $\kappa=0$.


问:
Can a partially ordered set $D$ have a function that maps every chain to a strict upper bound?
My Claim:

Let (D,$\le$) be a poset and f a function from the set of linearly ordered subsets of D to D. Assume that for every linearly ordered subset U of D we have $\forall$x $\in$ U, x $\lt$ f(U). Then there is no maximal element in D.

My proof:

Suppose for a contradiction x is a maximal element in D. Clearly, {x} $\subset$ U is linearly ordered. Then x $\lt$ f({x}) $\in$ U. This contradicts with the assumption x being a maximal element in D.

On the other Zorn's Lemma implies that D has a maximal element because every linearly ordered subset U of D is upper bounded by f(U) $\in$ D.

Clearly, there is something wrong here. There are three possibilities that might explain these contradicting results.

My proof of the claim is incorrect.

The use of Zorn's Lemma is incorrect.

There exist no such function f and thus there is no contradiction.

What do you think about this?

答:
Working in $\sf ZFC$, every partial order has a maximal chain. Your requirement about $f$ is that $f$ is a strict upper bound, but a maximal chain cannot have a strict upper bound, since adding a strict upper bound to a chain will result in a strictly larger chain.

So assuming $\sf ZFC$, there can be no such $f$. But the same can be said in $\sf ZF$, actually. What $f$ is doing is to choose an upper for every chain, and it is easy to see that it is exactly the choice function needed for the proof of Zorn's lemma, so by assuming it exists we circumvent to appeal to the Axiom of Choice. Consequently, it means that by applying $f$ recursively, we can create a maximal well-ordered chain, which again by maximality cannot have a strict upper bound.

So in fact such $f$ cannot exist, even without assuming the Axiom of Choice. So in either case, your proof is completely vacuous.


问:
Six ducks in a pond
This puzzle has got me stumped.</p>

<p>$6$ ducks are swimming on a pond radius $5$. Show that at any moment there are two ducks a distance at most $5$ apart. </p>

答: Let's see if this answer makes sense.  Let's say the ducks spread out as far as possible.  It would make sense that to do this, they spread evenly around the circumference.  Draw a circle of radius 5 and place 6 points along the perimeter to represent the ducks.  Line segments connecting each point to its closest 2 neighbors creates a regular hexagon.  Connecting these points to the center of the circle creates 6 equilateral triangles with a side length equal to the radius of 5. </p>

Title: Six ducks in a pond


问:
What are the gradient, divergence and curl of the three-dimensional delta function?
The three-dimensional delta function is defined as follows:

$$\delta(\mathbf{r}-\mathbf{r'})= 0 \;\; \mathrm{for} \;\;\mathbf{r}\neq\mathbf{r'} $$
$$\delta(\mathbf{r}-\mathbf{r'})= \infty \;\; \mathrm{for} \;\;\mathbf{r}=\mathbf{r'} $$

$$\int_V\delta(\mathbf{r}-\mathbf{r'})\;dV= 1 .$$

By definition also:
$$\int_V f(\mathbf{r})\delta(\mathbf{r}-\mathbf{r'})\;dV= f(\mathbf{r'}) $$.

I am wondering what the vector calculus operators are for the delta function. For example:

Curl:

$\nabla \times \delta(\mathbf{r}-\mathbf{r'})$

Divergence:

$\nabla \cdot \delta(\mathbf{r}-\mathbf{r'})$

Gradient:

$\nabla \delta(\mathbf{r}-\mathbf{r'})$

Any insight or references are appreciated. Please note that I am a geophysicist by training and it has been a few years since I have taken any vector calculus classes!

答:
The answer to your question becomes quite easy if you are able to build the correct mathematical framework. Note that I try to build an answer adapted to the OP background, whence it will not be strictly rigorous.

First of all, let me try to explain the definition of the delta "function". In mathematics, the Dirac delta is an example of what we call distributions (or generalized functions), roughly speaking mappings (functionals) that assign to each smooth function a real number; in other words $T$ is a distribution if

$$ T : \{ \text{smooth functions vanishing at infinity} \} \rightarrow \mathbb{R}, $$

it is linear and has a continuity property I won't write explicitly. The vanishing at infinity condition should also be understood in a suitable sense, but let me go on. For a  fixed $\mathbf{r}_0 \in \mathbb{R}^3$, the Dirac delta $\delta_{\mathbf{r}_0}$ acts on smooth functions $f : \mathbb R^3 \rightarrow \mathbb R$ as

$$\delta_{\mathbf{r}_0}(f) = \langle \delta_{\mathbf{r}_0}, f \rangle = f(\mathbf{r}_0) \in \mathbb{R}. \qquad (*) $$

Note that the smoothness of $f$ ensures that the pointwise evaluation makes sense. This reminds the last identity you wrote in the question, with the bracket notation (which is common in the theory of distributions) instead of the integral since the Delta is not a function in the common sense, thus we can't integrate it in a classical sense.

Other examples are distributions defined by means of an integration, for instance

$$ f \mapsto \int_{\mathbb{R}^3} f g \quad \text{or } \ \ \ f \mapsto \int_{\mathbb{R}^3}  (\partial_{xy}f)g, $$

where say $g$ is another fixed integrable function (note that here the condition at infinity can be crucial).

So what's interesting about this concept? The fact that we can do some calculus with distributions, in particular there is a notion of derivative for distributions which yields other distributions! The idea is rather simple: since distributions act on smooth functions that can be differentiated as many times as we like, we define the derivative of a given distribution dumping the derivation on the smooth function. If $T$ is any distribution, $\partial_x T$ (same for other derivatives) is the ditribution defined via:

$$(\partial_x T) (f) := - T (\partial_x f); \qquad (**) $$

notice that the evaluation on the right hand side has a meaning since $T$ is a distribution and $\partial_x f$ is smooth because $f$ is.

We already have the tools to compute the derivative of the Dirac delta, indeed by $(*)$ and $(**)$:

$$(\partial_x \delta_{\mathbf{r}_0})(f) = - \delta_{\mathbf{r}_0}(\partial_x f) = -(\partial_x f)(\mathbf{r}_0).$$

The notion of gradient of a distribution (and hence of a Dirac delta) is quite similar to $(**)$, although not completely intuitive since such gradient is not a "vector", it rather acts on vectors (i.e., colletions of smooth functions!):

$$\nabla T : \{ \text{smooth functions vanishing at infinity} \}^3 \rightarrow \mathbb{R}, \qquad \nabla T (f_1, f_2, f_3) := - ( T(\partial_x f_1) + T(\partial_y f_2) +T(\partial_z f_3)  ).$$

Now you can try to put the pieces together and compute the (distributional) gradient of a Dirac delta. The notions of curl and divergence can be defined in a similar way, and for the Delta will result in distributions that act as combinations of pointwise evaluations of the derivatives of the "test function/vector" $f$.


问:
When does the isomorphism $G\simeq \ker(\phi)\times \operatorname{im}(\phi)$? hold?
Suppose you have a group isomorphism given by the first isomorphism theorem:

$$G/\ker(\phi) \simeq \operatorname{im}(\phi)$$

What can we say about the group $\ker(\phi)\times \operatorname{im}(\phi)$? In particular, when does the following hold:

$$G\simeq \ker(\phi)\times \operatorname{im}(\phi)?$$

I ask this question because i want to prove that $GL_n^+(\mathbb{R}) \simeq SL_n(\mathbb{R}) \times \mathbb{R}^*_{>0}$, with $GL_n^+(\mathbb{R})$ the group of matrices with positive determinant. I proved that $SL_n(\mathbb{R})$ is a normal subgroup and that $GL_n^+(\mathbb{R})/  SL_n(\mathbb{R}) \simeq \mathbb{R}^*_{>0}$, using the surjective homomorphism $\det(M)$. I tried something with semidirect products but I got stuck.

答:
In your special case you actually have a morphism $GL_n ^+ \to SL_n \times \mathbb{R}^+ $ given by

$$ M \mapsto (M/(\det M)^{1/n}, \det M) $$

the inverse being given by $(N, t) \mapsto t^{1/n} N$.


问:
Why consider specific pmf when calculating expected value from cumulative distribution function?
$$ F_i(x)= \begin{cases}
       0 &\quad x <-1\\
       \frac{1}{2i} &\quad x \in [-1,1)\\
       \frac{1}{2} +\frac{1}{2}(x-1) &\quad x \in [1,2]\\
       1 &\quad x>2 \\
     \end{cases}$$

Hello, I want to calculate the expected value of the corresponding random variable (X) of F2. The solution that was presented to me is:  $$ EX= -1 * P(X=-1)+ 1* P(X=1) + \int_{1}^{2} \frac{1}{2}x \,dx$$

Now, my question is why do we need to add P(X=1) and do we still need to consider it when  calculating the expected value of the random variable corresponding to F1 where there's no jump at 1?

答:
When considering $F_1$

$$\mathbb{P}[X=1]=0$$

so nothing changes if you consider it or not.

Observe that, as $F_1$ is concerned, the drawing is the following

The mean is the areas with (+) minus the area with (-). Being the areas of the two rectangle identical, they cancel thus the mean is exactly the area of the triangle marked with (+).

Thus $E(X)=\frac{1}{4}$ without doing any calculation

Generalizing the formula for any random variable you get

$$\mathbb{E}[X_i]=-\frac{1}{2i}+\left(1-\frac{1}{2i}\right)+\frac{1}{4}$$

$i=1,2,3,...$

this is awesome too...


问:
Understanding steps of a proof of a positive matrix being symmetric
In the solution it's previously proven through weak formulation theorem that if we can show
$<Px,x>=
$, then a positive matrix is symmetric.

Here is the written proof. We assume that $A$ is a positive matrix.

$<Ax,x>=(Ax)^*x$

$=((Ax)^*x)^*$

$=x^*(Ax)$

$=<x,Ax>$

$=$

I have two basic questions about the steps:

I don't understand the first step $<Ax,x>=(Ax)^*x$. How do we end up there?

My second question is is on the second line where it's stated that $(Ax)^*x=((Ax)^*x)^*$. Is this a property of symmetric matrices and why aren't the $x$:es affected by this?

答:
$<Ax,x>=(Ax)^*x$ is the definition of the inner product for two vectors $<x,y> = x^* y$

$(Ax)^* x$ is a number so that is why is equal to it transpose $(Ax)^*x=((Ax)^*x)^*$


问:
Hilbert space on Graph
Assume we have a finite graph with edges $E$ and nodes $V$. How can one in a sensible way define an inner product on this graph for functions $f:V\rightarrow \mathbb{R}$ in the nodes  ?  I have seen some different settings: one where we take simply the euclidean inner product - however this does take into account the graph structure, and others which are defined by for example $f' L f$ where $L$ is the graph laplacian. In the latter case we have that the constant functions on fully connected subgraphs are in the null space of $L$ -> so this might not be an inner product will it ?
Do you have any resources I could look at for Hilbert space theory on graphs?

答:
The most common inner product to use is just the ordinary Euclidean product; it is used even when it's not the best choice. It's in this setting that you do linear algebra with things like the graph Laplacian, just writing expressions like $\langle f, Lf\rangle$ or possibly normalizing to get $\frac{\langle f, Lf\rangle}{\langle f, f\rangle}$.

This inner product works well when the graph is regular. When the graph is not regular, you see people still using this inner product, and throwing around matrices like the symmetric normalized Laplacian $L^{\text{sym}} = D^{-1/2}L D^{-1/2}$. Instead of doing this,  we could keep our ordinary Laplacian, but change our inner product to something nicer!

An inner product that does this could be written as a weighted average
$$
   \frac1{2|E(G)|} \sum_{v \in V} \deg(v) f(v) g(v)
$$
but that doesn't explain why it's nice. To do that, consider a random walk on $G$. The stationary distribution $\pi$ of the random walk visits vertex $v$ with probability $\pi(v) = \frac{\deg(v)}{2|E(G)|}$. Then, define the inner product $\langle f,g\rangle_{\pi}$ as the expected value $$\langle f,g\rangle_\pi = \mathbb E_{\boldsymbol v \sim \pi} [f(\boldsymbol v) g(\boldsymbol v)]
$$
where by $\boldsymbol v \sim \pi$ I mean that $\boldsymbol v$ is a random vertex sampled from the stationary distribution $\pi$.

Title: Hilbert space on Graph


问:
what are the steps to get this answer?
If $\frac ab+\frac ba=1$, and if "a" and "b" are not equal to zero then what would be the value of $a^3-b^3$? It would be helpful if the answer is given in the form of steps leading to the value.</p>

答:
Expanding on Lab's comments,

Making the assumption that $\frac{a}{b} + \frac{a}{b} = -1 $,

Hint: Clear denominators for the equation. It states that

$$a^2 + b^2 = -ab$$

Hint: $$a^3 -b^3 = (a-b)(a^2 + b^2 + ab) = ??$$

If we want $a^3 + b^3$ instead, we can proceed in a similar manner.

Title: what are the steps to get this answer?


问:
Difference of the Stirling cycle numbers and the Stirling set numbers
Denote by $\left\langle\!\! \left\langle k\atop  j\right\rangle\!\! \right\rangle$ the second-order Eulerian numbers A340556. Define
$$ \left| n\atop  k\right| = \sum_{j=0}^k \left( \binom{n + j - 1}{2k} - \binom{n + k - j}{2k} \right) \left\langle\!\! \left\langle k\atop  j\right\rangle\!\! \right\rangle .$$
Observe that the difference between the Stirling cycle numbers and the Stirling set numbers can be expressed as:
$$ \left[ n\atop k\right] -  \left\{ n\atop  k\right\} = \left| n\atop n- k\right| \quad (0 \le k \le n)$$

It is surprising that the difference between the two Stirling numbers had no entry in the OEIS -- until today! Now they are A341102. But what we still need is an independent combinatorial interpretation of these numbers. Who knows one?

Who is willing to delight the combinatorists among us with a proof of this identity?

答:
We seek to show that with $0\le k\le n$ the following identity holds:

$${n\brack n-k} - {n\brace n-k} =
\sum_{j=0}^k \left({n+j-1\choose 2k} - {n+k-j\choose 2k}\right)
\left\langle\!\! \left\langle k\atop  j
\right\rangle\!\! \right\rangle.$$

We will use the following  two representations of the Eulerian numbers
of the second  order in terms of associated  Stirling numbers (consult
MSE                                                              link
4037172):

$$ \sum_{j=0}^{k} (-1)^{k-j} {n-j \choose k-j}
\left\{ \!\! \left\{ n+j\atop j\right\} \!\! \right\} =
\left\langle\!\! \left\langle n\atop  k\right\rangle\!\! \right\rangle
=\sum_{j=0}^{n-k+1} (-1)^{n-k-j+1} {n-j\choose k-1}
\left[\! \left[ n+j\atop j\right] \! \right] $$

We get for the first piece

$$\sum_{j=1}^k {n+j-1\choose 2k}
\sum_{p=0}^{k-j+1} (-1)^{k-j-p+1} {k-p\choose j-1}
\left[\! \left[ k+p\atop p\right] \! \right]
\\ = \sum_{p=0}^k
\left[\! \left[ k+p\atop p\right] \! \right]
(-1)^{k-p+1}
\sum_{j=1}^{k+1-p} (-1)^j {n+j-1\choose 2k}
{k-p\choose j-1}
\\ = \sum_{p=0}^k
\left[\! \left[ k+p\atop p\right] \! \right]
(-1)^{k-p} [z^{2k}] (1+z)^n
\sum_{j=1}^{k+1-p} (-1)^{j-1} (1+z)^{j-1}
{k-p\choose j-1}
\\ = \sum_{p=0}^k
\left[\! \left[ k+p\atop p\right] \! \right]
(-1)^{k-p} [z^{2k}] (1+z)^n
(1-(1+z))^{k-p}
\\ = \sum_{p=0}^k
\left[\! \left[ k+p\atop p\right] \! \right]
[z^{k+p}] (1+z)^n
= \sum_{p=0}^k
\left[\! \left[ k+p\atop p\right] \! \right]
{n\choose k+p}.$$

Now this last piece evaluates  combinatorially to ${n\brack n-k}$ when
written  as $\left[\!  \left[ k+p\atop  p\right] \!  \right] {n\choose
n-k-p}$ namely we choose $n-k-p$  fixed points and split the remaining
$k+p$ elements  into $p$ cycles  of size at least  two for a  total of
$n-k$ cycles.  Here we  must have  $k+p\ge 2p$ or  $p\le k.$  (We have
classified by the number of fixed points).

We get for the second piece

$$\sum_{j=1}^k {n+k-j\choose 2k}
\sum_{p=0}^{j} (-1)^{j-p} {k-p\choose j-p}
\left\{\! \left\{ k+p\atop p\right\} \! \right\}
\\ = \sum_{p=0}^k
\left\{\! \left\{ k+p\atop p\right\} \! \right\}
(-1)^p \sum_{j=p}^k (-1)^j
{n+k-j\choose 2k} {k-p\choose j-p}
\\ = \sum_{p=0}^k
\left\{\! \left\{ k+p\atop p\right\} \! \right\}
\sum_{j=0}^{k-p} (-1)^j
{n+k-j-p\choose 2k} {k-p\choose j}
\\ = \sum_{p=0}^k
\left\{\! \left\{ k+p\atop p\right\} \! \right\}
[z^{2k}] (1+z)^{n+k-p} \sum_{j=0}^{k-p} (-1)^j
(1+z)^{-j} {k-p\choose j}
\\ = \sum_{p=0}^k
\left\{\! \left\{ k+p\atop p\right\} \! \right\}
[z^{2k}] (1+z)^{n+k-p}
\left(1-\frac{1}{1+z}\right)^{k-p}
\\ = \sum_{p=0}^k
\left\{\! \left\{ k+p\atop p\right\} \! \right\}
[z^{2k}] (1+z)^{n} z^{k-p}
= \sum_{p=0}^k
\left\{\! \left\{ k+p\atop p\right\} \! \right\}
{n\choose k+p}.$$

With this  piece we get exactly  the same reasoning as  with the first
one, namely it evaluates to ${n\brace n-k}$. We write it as $\left\{\!
\left\{ k+p\atop  p\right\} \! \right\} {n\choose  n-k-p}$ in choosing
the number  of singletons, of  which there are $n-k-p.$  The remaining
$k+p$ elements are distributed into $p$  disjoint sets of at least two
elements for  a total of $n-k$  sets. We once more  have the condition
that $k+p\ge  2p$ or $p\le  k.$ (We have  classified by the  number of
singleton sets.)

This concludes the argument.


问:
Problem with Mean Value Theorem and Limit of Derivative
Consider the function $f(y)=\sin(y)$.
Then on the interval $[\sqrt x,x]$ by MVT $$\exists t\in(\sqrt x,x) \text{ s.t. } f'(t)=\frac {f(x)-f(\sqrt x) } {x-\sqrt x}. $$
Then by taking $x\to \infty \implies t\to \infty$
$$ \lim_{t \to \infty} f'(t) = \lim_{x \to \infty} \frac {f(x)-f(\sqrt x) } {x-\sqrt x} $$
But then this implies that $$ \lim_{t \to \infty} \cos(t) = \lim_{x \to \infty} \frac {\sin(x)-\sin(\sqrt x) } {x-\sqrt x} = 0 $$
Where have I gone wrong?

答:
Because is $t_{x}$ actually. We have that for each $x$ there exist some $t_{x} \in (\sqrt{x},x)$ such that $cos(t_{x}) = \frac{\sin(x)-\sin(\sqrt{x})}{x-\sqrt{x}}$

May be $t_{x} = \frac{\pi}{2} + 2m_{x}\pi$ for all $x$, with $m_{x}\in \mathbb{Z}$ such that $m_{x} \to \infty$ when $x\to \infty$.

The point is, $cos(t_{x}) \to 0$ when $x\to \infty$ not implies that $cos(t) \to 0$ when $t \to \infty$


问:
Limit of $\sqrt n (\sqrt{n+1}-\sqrt n)$
How should I go about calculating $$\lim_{n \to \infty} \sqrt n (\sqrt{n+1}-\sqrt n)$$

I found empirically it approaches $\frac 1 2$ (and fast), but it's been decades since I last took a formal calculus class, and I don't know how to approach this.

答:
Hint.

$$\sqrt n \left(\sqrt{n+1}-\sqrt n\right) = \frac{\sqrt n}{\sqrt{n+1} + \sqrt n} = \frac{1}{\sqrt{1 + \frac1n} + 1}$$

Title: Limit of $\sqrt n (\sqrt{n+1}-\sqrt n)$


问:
Sigma and Series involving factorials
Can someone please try this problem? It is kinda complex as it has factorial in the sigma and I could not see any telescopic addition possible here.
Find the Sum given below:

$\sum_{n=0}^{\infty}  \frac{16n^2 +20n +7}{(4n+2)!} $

Title: Sigma and Series involving factorials

Body:
Please note that $$16n^2+20n+7=(4n+2)(4n+1)+2(4n+2)+1$$

So,
$$\frac{16n^2 +20n +7}{(4n+2)!}=\frac{1}{(4n)!}+\frac{2}{(4n+1)!}+\frac{1}{(4n+2)!}$$$$=\frac{1}{(4n)!}+\frac{1}{(4n+1)!}+\frac{1}{(4n+2)!}+\frac{1}{(4n+3)!}+\left(\frac{1}{(4n+1)!}-\frac{1}{(4n+3)!}\right)$$

Therefore,
$$\sum_{n=0}^{\infty}  \frac{16n^2 +20n +7}{(4n+2)!}=\left(\sum_{n=0}^{\infty}  \frac{1}{n!}\right)+\left(\frac{1}{1!}-\frac{1}{3!}+\frac{1}{5!}-+\ldots \right)$$
$$=e+\sin1$$


问:
Proof that the largest eigenvalue of a stochastic matrix is $1$
The largest eigenvalue of a stochastic matrix (i.e. a matrix whose entries are positive and whose rows add up to $1$) is $1$.

Wikipedia marks this as a special case of the Perron-Frobenius theorem, but I wonder if there is a simpler (more direct) way to demonstrate this result.

答: Here's a really elementary proof (which is a slight modification of Fanfan's answer to a question of mine). As Calle shows, it is easy to see that the eigenvalue $1$ is obtained.  Now, suppose $Ax = \lambda x$ for some $\lambda > 1$.  Since the rows of $A$ are nonnegative and sum to $1$, each element of vector $Ax$ is a convex combination of the components of $x$, which can be no greater than $x_{max}$, the largest component of $x$.  On the other hand, at least one element of $\lambda x$ is greater than $x_{max}$, which proves that $\lambda > 1$ is impossible.</p>


问:
A question related to Riemann zeta function
Does anyone know why the following statement is correct?

Let $f(x)$ be the function whose value on the interval $m\pi<x<(m+1)\pi, m=0,1,2,\cdots$, is $(-1)^m\frac{\pi}{4}$.  Let $0<s<1$. Then $$\int_0^\infty x^{s-1}f(x)\,dx$$ represents an analytic function for $\Re s<1$ and is equal to $$2(1-2^{s+1})\zeta(1-s)$$ for $\Re s<0$.
Here $\zeta(\cdot)$ is the Riemann zeta function.

Edit. I think now I partially understand the first part. Let $\alpha(x)$ be the function defined by $$\int_0^x u(t)\,dt$$ where $u(t)=(-1)^m$ on $[m\pi,(m+1)\pi], m=0,1,2,\cdots.$ Then the integral $\int_0^\infty x^{s-1}f(x)\,dx$ is equal to the Riemann-Stieltjes integral $\frac{\pi}{4}\int_0^\infty x^{s-1}\,d\alpha$. For $0<s<1$, this can be written as $$\frac{\pi}{4}\left(\int_0^\pi x^{s-1}\,dx-\pi^s-(s-1)\int_{\pi}^\infty\frac{\alpha(x)}{x^{2-s}}\,dx\right),$$ which is analytic in $s$ for $0<\Re s<1$.

Now I need to know how to do it for $\Re s\le 0$ and how to make it equal to $2(1-2^{s+1})\zeta(1-s)$ for $\Re s<0$.

答:
Note: Pb. 21 in Andrews-Askey-Roy contains a few misprints and is formulated inaccurately.

The sum over $n$ should start from $n=0$ and not from $1$.

As you might have noticed, the functional relation for $\zeta$ in the 1st line of the problem (correct) does not coincide with the relation (incorrect) implicitly stated in question (b). Namely, instead of $\zeta(1-s)$ one should have $\zeta(-s)$ - this is precisely what is obtained below.

as for the prefactor $\frac{\pi^{1+s}}{2s}$ in my formula, it should also be there (check this by computing the lhs in AAR problem) if one want correct functional relation for $\zeta$.

By definition of $f$,
\begin{align}\int_0^{\infty}x^{s-1}f(x)dx&=\frac{\pi}{4}\sum_{m=0}^{\infty}(-1)^m\int_{m\pi}^{(m+1)\pi}x^{s-1}dx.
\end{align}
Introduce instead of this
$$F(a)=\frac{\pi}{4}\sum_{m=0}^{\infty}(-1)^m\int_{ma}^{(m+1)a}x^{s-1}dx.$$
Differentiating w.r.t. parameter $a$, we have
\begin{align}\frac{\partial F}{\partial a}&=\frac{\pi a^{s-1}}{4}\left(1+\sum_{m=1}^{\infty}(-1)^m\left[(m+1)^{s}-m^{s}\right]\right)=-\frac{\pi a^{s-1}}{2}\sum_{m=1}^{\infty}(-1)^m m^s.\tag{1}
\end{align}
Recall that the "odd" part of the Riemann $\zeta$-function can be determined from
$$\zeta(s)=\underbrace{\sum_{m=0}^{\infty}(2m+1)^{-s}}_{\zeta_{\mathrm{odd}}(s)}+\underbrace{2^{-s}\zeta(s)}_{\zeta_{\mathrm{even}}(s)}.$$
But our sum in (1) is then just $$\zeta_{\mathrm{even}}(-s)-\zeta_{\mathrm{odd}}(-s)=\left(2^{1+s}-1\right)\zeta(-s).$$
Using this, integrating (1) back with respect to $a$, taking into account that $F(0)=0$, and setting $a=\pi$, we get
$$F(\pi)=\frac{\pi^{1+s}}{2s}\left(1-2^{1+s}\right)\zeta(-s).$$


问:
Finding the Expected Value and Variance
The distribution function of a random variable $X$ is given by

$P (X \leq x)$ =
\begin{cases}
0 & \text{if $x<0$}, \\
x^{k} & \text{if $0 \leq x \leq 1, \,\,\,\,\,\, k\geq1$}, \\
1     & \text{if $x > 1$}.
\end{cases}

Determine the mean and variance of $X$.

Workings:

I am aware that I must start by calculating $\sum_{-\infty}^{\infty} x \,\cdot p_X(x)$, but because this is a continuous distribution, I will instead have to calculate $\int_{-\infty}^{\infty} x \, \cdot p_X(x) \,dx$. However, once I integrate it, I am confused as to what further steps as to which I am meant to take given that I get infinity as the expected value. Could someone please tell me if it is possible for the expected value to be infinite and if I have made a mistake in my workings? Thank you.

\begin{align}
\text{E(X)} & = \int_{-\infty}^{\infty} x \, \cdot p_X(x) \,dx \\
            & = \int_{-\infty}^{0} x \, \cdot 0 \,dx \: + \int_{0}^{1} x \, \cdot x^{k} \,dx \: + \int_{1}^{\infty} x \,dx \\
& = C + \bigg{[}\frac{x^{k+2}}{k+2}\bigg{]}_{0}^{1} + \bigg{[}\frac{x^{2}}{2}\bigg{]}_{1}^{\infty}
\end{align}

答: You are using the CDF in your integral, and you should instead use the PDF, which is the derivative $$p_X(x) = \begin{cases}0 & x \le 0 \\ k x^{k-1} & 0 \le x \le 1 \\ 0 & x > 1\end{cases}$$

Title: Finding the Expected Value and Variance


问:
$(a_n)$ diverges and $(b_n)$ converges, then $(a_n b_n)$ diverges
If $(a_n)$ diverges and $(b_n)$ converges, then $(a_n b_n)$ diverges. How can I prove that? I think this is true, but I can't prove it.

答: Simply False. Take $a_n=(-1)^n$ and $b_n=\frac{1}{n}$. Then $a_nb_n \to 0$ !!


答:
Take $a_n=n$ and $b_n=\frac{1}{n}$.
The product converges

Body: If $(a_n)$ diverges and $(b_n)$ converges, then $(a_n b_n)$ diverges. How can I prove that? I think this is true, but I can't prove it.


问:
Confused with a proof about paradoxical groups.
I am reading about the Banach-Tarski paradox, however, there are a couple of things about theorem 6 that confuse me. The image of theorem 6 and the proof can be found below. The first thing that confuses me are the braces marked by red in the image: $A_i^* = \{\bigcup_{g \in A_i} g(R)\}$. Are they really needed? If the braces are included then $A_i^*$ just becomes a set inside a set, which seems unnecessary. The second thing is the last sentence marked by red in the image. How does $X = \bigcup_{i = 1}^n g_i(A_i(R))$ follow directly from the fact that $G = \bigcup_{i = 1}^n g_i(A_i)$? Any help would be highly appreciated since I literally have no one to ask questions to.

答:
Ladies and gentlemen, I am happy to announce that we've had a breakthrough! I have finally understood this proof and I am here to share my thoughts. Let's first define $A_i(R)$ to be
$$A_i(R) := \bigcup_{g \in A_i}gR  \qquad  (1)$$
so $\bigcup_{i = 1}^n g_i(A_i(R))$ simplifies to $\bigcup_{i = 1}^n g_i(A_i^*)$, as mentioned in the picture in my post/question. We also know that the cosets of $R$ partition $X$, i.e $X = \bigcup_{g \in G}gR$. We also know that $G$ is a paradoxical group, so we can "choose" $g \in G$ in two different ways! To use this fact we will need to use the notation above (like we defined $A_i(R)$ in equation (1)). So $X = G(R)$. Now, since $G$ is paradoxical, we can substitute two different $G$'s in this equation and then simplify to see that $X$ indeed is $G$-paradoxical!

I have also concluded that the braces in red in the picture are a typo since (1) would not be equal to $A_i^*$.


问:
KVPY Question : Series and Progression
There are $20$ urns such that the first urn contains $5$ balls, the second contains $10$ balls and in general the $k^{th}$ urn contains $2k+1$ balls more than that in the $(k-1)^{th}$ urn. Then the total number of balls in all the urns is:

I know that this question has been asked before (What kind of sequence/progression is this? What will be the answer to the question?) , but I have quite not understood the solution and how the $n^{th}$ term has been found.

My approach goes forward like this, but I'm getting stuck at a point:

Given : In general the $k^{th}$ urn contains $2k+1$ balls more than that in the $(k-1)^{th}$ urn.

Using this info:

$3^{rd}$ urn contains : $10 + (2 \times 3 + 1)$ balls, $i.e.$ $17$ balls.

Let $k_i$ represent the natural number $i$.

Using this, the balls in $20^{th}$ urn are:

$5 + (2k_2+1) + (2k_3+1) + (2k_4+1) + (2k_5+1) + (2k_6+1) + ... +  (2k_{18}+1) + (2k_{19}+1)+ (2k_{20}+1)  $

Therefore total balls are:

$5\times 20 + 19(2k_2+1) + 18(2k_3+1) + 17(2k_4+1) + 16(2k_5+1) + 15(2k_6+1) + ... +  3(2k_{18}+1) + 2(2k_{19}+1)+ (2k_{20}+1)$

Which can further be written as:

$100 + \sum_{n=1}^{19} r +$

Not getting further. Please help.

答:
HINT

"Let $k_i$ represent the natural number $i$" is a step in the wrong direction.  Just call it $i$.

First thing to do is to get a formula for the number of balls in urn $k$.  Urn $1$ has $5$ balls, urn $2$ has $5+5$, urn $3$ has $5+5+7$, urn $4$ has $5+5+7+9$, and so on.  We can see that urn $1$ doesn't quite fit the pattern, but if $k>1$ then urn $k$ contains $$5+\sum_{i=2}^k(2i+1)$$ balls.  Can you compute this?

Title: KVPY Question : Series and Progression


问:
prove $x+1 = 1+x$ using Peano axioms
I wanna prove that $x+1 = 1+x$ (without considering "$x+0=x$",and Im using the old definition of Peano axioms)

This is my try:

Using this basis:

$(1):1+x = x^+$

$(2):x^+ +y=(x+y)^+$

Actually my idea is if numbers succesor are equal then actual numbers are equal too. (based on Peano axioms)
$$(1+x)^+ = 1^+ + x = (1+1)+x$$

and
$$(x+1)^+ = x^+ +1 = (1+x)+1$$

But I stucked in how to show that this two are equal.

Somebody help :)

答:
We will prove by induction that $x+1=x^+$.

Basis: $1+1=1^+$, from (1).

Induction step: assume $x+1=x^+$ and prove that $x^++1=(x^+)^+$.

By (2): $x^++1 = (x+1)^+ = (x^+)^+$ using induction hypotheses and substitution for equality.

Having proved that $x+1=x^+$, for every $x$, we use (1) and transitivity and symmetry of equality to conclude that:

$x+1=1+x$.

Title: prove $x+1 = 1+x$ using Peano axioms


答:
You are integrating the distribution function and not the probability density

https://en.wikipedia.org/wiki/Probability_density_function

The two are related because the distribution function gives you the probability that $X$ is in $]-\infty,x]$ and the integral of the probability density between $a$ and $b$ gives you the probability of finding $X$ in $[a,b]$, so the distribution function $P(X\le x)$ is related to the probability density $p_X(x)$ by the relation

$$P(X\le x)=\int_{-\infty}^xp_X(t)dt$$

Title: Finding the Expected Value and Variance

Body:
The distribution function of a random variable $X$ is given by

$P (X \leq x)$ =
\begin{cases}
0 & \text{if $x<0$}, \\
x^{k} & \text{if $0 \leq x \leq 1, \,\,\,\,\,\, k\geq1$}, \\
1     & \text{if $x > 1$}.
\end{cases}

Determine the mean and variance of $X$.

Workings:

I am aware that I must start by calculating $\sum_{-\infty}^{\infty} x \,\cdot p_X(x)$, but because this is a continuous distribution, I will instead have to calculate $\int_{-\infty}^{\infty} x \, \cdot p_X(x) \,dx$. However, once I integrate it, I am confused as to what further steps as to which I am meant to take given that I get infinity as the expected value. Could someone please tell me if it is possible for the expected value to be infinite and if I have made a mistake in my workings? Thank you.

\begin{align}
\text{E(X)} & = \int_{-\infty}^{\infty} x \, \cdot p_X(x) \,dx \\
            & = \int_{-\infty}^{0} x \, \cdot 0 \,dx \: + \int_{0}^{1} x \, \cdot x^{k} \,dx \: + \int_{1}^{\infty} x \,dx \\
& = C + \bigg{[}\frac{x^{k+2}}{k+2}\bigg{]}_{0}^{1} + \bigg{[}\frac{x^{2}}{2}\bigg{]}_{1}^{\infty}
\end{align}


问:
Estimating $\hat{p}$

let $X\sim Bin(n,p)$  and $\hat{p} =\frac{X}{n}$

a) Find a constant c such that $E[c\hat{p}(1-\hat{p})]=p(1-p)$

My work:
$$
\begin{align}
cE[\hat{p}(1-\hat{p})] &=E[\frac{X}{n}]-E[\frac{X^2}{n^2}]\\
&= \frac{1}{n}E[X]-\frac{1}{n^2}E[X^2] \\
\end{align}
$$
And I continue with $E[X] = p$ and $E[X^2] = Var(X)+E[X]^2$ but this gives me
$$
= \frac{p}{n} - \frac{1}{n^2}np(1-p)-\frac{p^2}{n^4}
$$
Which doesnt seem righ. Am I doing something wrong?

答:
Since $X$ is a binomial random variable, $E[X]=np$ and $Var(X)=np(1-p)$.

Hence,
$$\frac{1}{n}E[X]-\frac{1}{n^2}E[X^{2}]=\frac{np}{n}-\frac{np(1-p)+n^{2}p^{2}}{n^2}$$
Simplifying,
$$=p-p^{2}-\frac{1}{n}p(1-p)=\left(1-\frac{1}{n}\right)p(1-p)$$

So we simply let $c=\frac{1}{1-\frac{1}{n}}=\frac{n}{n-1}$

Title: Estimating $\hat{p}$


问:
Finding the sum of the three roots of the polynomial
Without solving the corresponding equation, find the sum of the three roots of the polynomial

$$2x^3 − 8x^2 + 20x − 19$$

I know that sum of the roots is $-b/a$ which is $-(-8)/2 = 4$ however I'm not confident in my answer because it seems to simple, could anyone please help explain.

note: still new to this not too sure how to format polynomials correctly if anyone could help with that as well

答:
Let $a,b,c$ the complex roots. Then the factorization of $p$ is

$p=2(x-a)(x-b)(x-c)=2x^3-2(a+b+c)x^2+2(ab+bc+ac)x-2abc$.


问:
Show that $\alpha(t)$ lies on a sphere
I'm asked to prove that the trace of the curve
$\alpha(t)=(r\sin^2(t), r \sin(t),r \cos(t))$
lies on a sphere.

What I tried so far:

Letting $\alpha(t)=(x(t),y(t),z(t))$, we have that $x^2(t)+y^2(t)+z^2(t)=r^2(\sin^4(t)+1)$ which does not satisfy the equation of a sphere. Also, I plotted the curve and I'm not sure whether the trace does actually lie on a sphere.

Any hints? Thanks!

答:
Suppose that $C=(a,b,c)$ is the center of the sphere on which $\alpha$ is supposed to lie. The square of the distance $d(C,\alpha)$ has to be constant. Therefore by differentiation

$$\langle \alpha^\prime(t), \alpha(t) - C \rangle$$ has to be equal to zero for all $t \in \mathbb R$. I.e.

$$\begin{aligned}
2\sin t \cos t(a - r\sin^2 t) + \cos t(b -r \sin t) - \sin t(c - r \cos t)&=\\
2\sin t \cos t(a - r\sin^2 t) + b \cos t - c\sin t=0
\end{aligned}$$

Using $t = 0$ we get $b=0$ and $c =0$ follows from plugging in $t= \pi/2$. We're left with the equation $$\sin t \cos t(a - r\sin^2 t) = 0$$ which should be valid for all $t \in \mathbb R$. And that can't be as it would imply $a - r\sin^2 t=0$ for all $t\in (0,\pi/2)$.

Therefore, the curve doesn't lie on a sphere.

Title: Show that $\alpha(t)$ lies on a sphere


问:
$x^p-c$ has no root in a field $F$ if and only if $x^p-c$ is irreducible?
Hungerford's book of algebra has exercise $6$ chapter $3$ section $6$ [Probably impossible with the tools at hand.]:

Let $p \in \mathbb{Z}$ be a prime; let $F$ be a field  and let $c \in
 F$. Then $x^p - c$ is irreducible in $F[x]$ if and only if $x^p - c$
  has no root  in $F$. [Hint: consider two cases: char$(F) = p$ and
  char$(F)$ different of $p$.]

I have attempted this a lot. Anyone has an answer?

答:
Perhaps the simplest tool I can think of is the following:

Let $F$ be a field and $f(x)$ an irreducible polynomial over $F$, then there is a field $K\geq F$ where $f(x)$ has a root; as $f(x)$ is irreducible in $F[x]$, a principal ideal domain, then $\langle f(x)\rangle$ is a maximal ideal of $F[x]$, hence $K=F[x]/\langle f(x)\rangle$ is a field, $\bar{x}$ is a root of $f(x)$, and it is easy to see how to embed $F$ into $K$. Now given a polynomial $f(x)\in F[x]$ it is clear how to construct a field $K\geq F$ such that $f(x)$ factors into linear polynomials in $K[x]$.

Now your question can be answered as follows:

Let $K\geq F$ be a field where $x^p-c$ factors into linear polynomials, say $x^p-c=(x-z_1)\cdots(x-z_p)$. Suppose $x^p-c$ is not irreducible in $F[x]$, then there are polynomials $f(x),g(x)\in F[x]$ of degree $\geq 1$ such that $x^p-c=f(x)g(x)$, then we may assume $f(x)=(x-z_1)\cdots(x-z_n)$, where $\deg f(x)=n<p$.

Put $z=z_1\cdots z_n$, then $z$ is $\pm$ the constant term of $f(x)$, so $z\in F$, and  $z^p=(z_1\cdots z_n)^p=z_1^p\cdots z_n^p=c^n$. As $p$ is prime there are integers $a,b$ such that $1=ap+bn,$ then $$(c^az^b)^p=c^{ap}z^{bp}=c^{ap}c^{bn}=c,$$
but $c^az^b\in F$, so $x^p-c$ has a root in $F$.


答:
Make use of
$$\sqrt n (\sqrt {n+1}-\sqrt n)=\sqrt n (\sqrt {n+1}-\sqrt n)\frac{ \sqrt {n+1}+\sqrt n}{\sqrt {n+1}+\sqrt n}$$

This simplifies to
$$\sqrt n\frac{ (\sqrt n+1)^{2}-\sqrt n^{2}}{\sqrt n+1)+\sqrt n}=\sqrt n\frac{n+1-n}{\sqrt{n+1}+\sqrt n}=\frac{\sqrt n}{\sqrt {n+1}+\sqrt n}=\frac{1}{\sqrt \frac{n+1}{n}+1}$$

This can be taken to the limit since $\lim_{n\rightarrow \infty}\frac{n+1}{n}=1$. There are several alternatives for this direct path. $\lim_{n\rightarrow \infty}\frac{n+1}{n}=\lim_{n\rightarrow \infty}1+ \frac{1}{n}=1$. This is calulating with limits. The denominator can not be developed for example into a Taylor series since  one is not small.

So a substitution is well working: $\frac{n+1}{n}=\left(\frac{m+1}{m}\right)^{2}$. This takes some time to write down. Then the square root is gone and the limit again can be shown to be $\frac{1}{2}$.
$$\lim_{m\rightarrow \infty}\frac{1}{2+\frac{1}{m}}=\frac{1}{2}$$

Title: Limit of $\sqrt n (\sqrt{n+1}-\sqrt n)$

Body:
How should I go about calculating $$\lim_{n \to \infty} \sqrt n (\sqrt{n+1}-\sqrt n)$$

I found empirically it approaches $\frac 1 2$ (and fast), but it's been decades since I last took a formal calculus class, and I don't know how to approach this.


问:
Apollonius circle
I'm given two points, $A$ and $B$, and two lengths, $b$ and $c$.  I need to find the locus of point $C$ such that $BC:AC=b:c$.

This link describes Apollonius circle of first type, but I can't seem to find anywhere how to construct the center of the Apollonius circle.

答:
Assuming that length $ AB< b+c$ draw arcs of lengths $(b,c)$  with centers at A,B to find point C. Let the internal /external angle green bisectors intersect line AB at P,Q.

The required locus is a circle on diameter PQ. Find center point of PQ and complete the Apollonian Circle... in the classical way.

Title: Apollonius circle


问:
If $G_1, G_2, G_3$ are abelian groups and $0 \to G_1 \to G_2 \to G_3 \to 0$ is exact, then $G_2 \simeq G_1 \oplus G_3$

If $G_1, G_2, G_3$ are abelian groups and $0 \to G_1 \xrightarrow{\varphi_1} G_2 \xrightarrow{\varphi_2} G_3 \to 0$ is exact, then $G_2 \simeq \ker(\varphi_2) \oplus \text{im}(\varphi_2) \simeq G_1 \oplus G_3$

The above is the conclusion of Example 12.2 in Nonlinear Analysis and Semilinear Elliptic Problems, by Ambrosetti and Malchiodi.

I worked out as follows:

By exactness, ${0} = \ker(\varphi_1)$, so $G_1 \simeq \text{im}(\varphi_1) = \ker(\varphi_2) \lhd G_2$, so it makes sense to consider $G_2/G_1$. On the other hand, again by exactness, $\text{im}(\varphi_2) = G_3$.  By the First Isomorphism Theorem,
$$
G_2/ \ker(\varphi_2) \simeq G_2/G_1 \simeq G_3.
$$

What one would like to do now is to "multiply both sides by $G_1$ and cancel out in the left-hand side". My question is, how to do it in a rigorous way?

I tried to write $G_2/G_1$ explicitly, but it was a dead end. I also tried to follow the hint by Najib Idrissi in this question, but failed.

Thanks in advace.

答:
This is unfortunately false. Consider the standard counterexample

$$0 \to \mathbb{Z} \xrightarrow{\varphi} \mathbb{Z} \xrightarrow{\pi} \mathbb{Z}/2\mathbb{Z} \to 0$$

where $\varphi(x)=2x$ and $\pi(x)=x+2\mathbb{Z}$ is the quotient map. The sequence is exact but clearly $\mathbb{Z}$ is not a direct sum $\mathbb{Z}\oplus(\mathbb{Z}/2\mathbb{Z})$.

For that to hold we need the $0 \to G_1 \xrightarrow{\varphi_1} G_2 \xrightarrow{\varphi_2} G_3 \to 0$ sequence to be a split exact sequence (see also: splitting lemma). This is for example true whenever $G_3$ is free abelian, i.e. $G_3\simeq \bigoplus\mathbb{Z}$.


问:
Why is the function differentiable in the point $(0,0)$?
I am trying to figure out why my function is differentiable and therefore continuous in the point $(0,0)$ which is also a critical point and a saddle point.

Considering my function: $f(x, y) = x^3 - 3xy^2$

The partial derivative of $x$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial x}(x,y) = 3x^2-3y^2 $$
The partial derivative of $y$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial y}(x,y) = -6xy $$

Therefore I get: $$\nabla f(x,y) = \begin{pmatrix} \frac{\partial f }{\partial x}\\ \frac{\partial f }{\partial y} \end{pmatrix} = \begin{pmatrix} 3x^2-3y^2 \\ -6xy  \end{pmatrix} $$

How can I show that my function is point-wise differentiable in $(0,0)$?

答: This is just a polynomial in $x$ and $y$, so it's a sum of products of infinitely differentiable functions starting with the projections $p(x,y) = x$ and $q(x,y) = y$. That makes it infinitely differentiable.


问:
Same winding number implies homotopic
In the lecture we defined the winding number $\omega$ to be the unique integer $d$ such that for a continuous map $f:S^{n}\to S^{n}$ it holds that $f_{\ast}:H_{n}(S^{n})\to H_{n}(S^{n}), [a]\mapsto d[a]$, where $H_{n}(S^{n})$ denotes the singular homology.

Is there an easy way to prove that if $\omega(f)=\omega(g)$ for some continuous satisfying $f,g:S^{n}\to S^{n}$ such that $f(p_{0})=g(p_{0})=q_{0}$ where $p_{0},q_{0}\in S^{n}$ then $f$ and $g$ are homotopic relative to $\{p_{0}\}$?

The inverse statement is clear, by the homotopy invariance of $\omega$, which directly follows by the homotopy invarianve of induced maps. But the statement above seems to be not so trivial.

(This is not a homework)

答:
This is not such an easy theorem to prove.

The proof in Hatcher's Topology appears in Corollary 4.25, as an application of the Freudenthal Suspension Theorem stated in Corollary 4.24, which itself depends on a restricted version of the excision theorem that holds in the setting of homotopy groups (the full version of the excision theorem for homology groups does not generalize to homotopy groups).

It is also a consequence of the Hurewicz theorem, although if you're not careful that can lead you into a circular argument, given that some proofs of the Hurewicz theorem actually assume the statement you are asking to prove.

I do believe there exists a "low technology" proof, where one first homotopes an arbitrary continuous function $S^n \mapsto S^n$ to one which is a close approximation possessing better local properties (i.e. smooth; or cellular with respect to a CW decomposition). Next one uses that approximation to compute the degree $d$ of the map. Finally one uses the local structure as a guideline for constructing a homotopy from the given continuous function to some "standard" example of a continuous function of degree $d$.

Title: Same winding number implies homotopic


问:
Does $CD$ bisect $AE$?
Construct triangle $ABC$.choose a $D$ point on the side $AB$ such that $\frac{BD}{AD}=\frac{2}{1}$.Now draw median $AE$. $AE$ intersects $CD$ at $O$ .İf area of triangle $AOD$ is 20, what is the area of triangle $ABC$? İ have tried using formula $S=\frac{1}{2}AB\cos \alpha$ for triangles $AOD$ and $AEB$ and comparing them but the side $AO$ is different than $AE$ so im so confused

答:
Since this question is about ratio of areas, and such ratios are preserved by linear transformations, we can choose convenient coordinates and angles for the vertices.

Choose $A=(0,0)$, $B=(1,0)$, $C=(0,1)$, $D=(\tfrac{1}{3},0)$, $E=(1,1)/2$.

Then point $O$ is the intersection of the line $y=x$ and $y=1-3x$, that is, $x=1/4=y$.

Thus the area of $AOD$ is $\tfrac{1}{2}\times\tfrac{1}{3}\times\tfrac{1}{4}=\tfrac{1}{24}$, while that of $ABC$ is $\tfrac{1}{2}$. Thus the ratio $\frac{AOD}{ABC}=\frac{1}{12}$. Since the actual area of $AOD$ is 20, that of $ABC$ is 240.

Title: Does $CD$ bisect $AE$?