问:
Linearization of arbitrary equation
If we define linearization of equation as here:

http://academic.macewan.ca/physlabs/Linearization.pdf,

is it possible to linearize any arbitrary equation?

If so, is there an algorithm, I could code let's say in Mathematica, that could always take some equation and output, equation for $x$ axis, and equation for $y$ axis?

答:
It is always possible to linearize equations (given some technical requirements), but not always useful to do so. To understand the how and why, note that any equation in $x$ can be put into the form $F(x)=0$. If the equation has more than one variable in it, then $F$ would depend on these other variables as well. The simplest way to see that all equations can be put into this form is to write the equation as $(something)=(something else)$ and move the right hand side to the left: $(something)-(something else)=0$ so $F=(something)-(something else)$.

Given this, and assuming $F$ is an analytic function (meaning its Taylor series exists), then we may Taylor expand about any point $x_0$:
$$
0=F(x)=F(x_0)+F^\prime(x_0)(x-x_0)+\frac{1}{2}F^{\prime\prime}(x_0)(x-x_0)^2+\cdots.
$$
Truncating this expansion at first order (it is interesting to think about when this is reasonable to do) we would find
$$
F(x_0)+F^\prime(x_0)(x-x_0)=0
$$

but since $x_0$ is some fixed number, $F(x_0)$ and $F^\prime(x_0)$ are constants with respect to $x$, and so this is a linear equation. We would say that this is the linearization of the original equation.

I will note that what I have described here shows linearization to, in general, be an approximation to the true equation, and hence any solutions thereof will only be approximate.

The link in the question only considers examples where $F(x)$ is an invertible function. In this special case, there exists a function $F^{-1}(x)$ such that $F^{-1}(F(x))=x$. For example if $F(x)=\ln x$, then $F^{-1}(x)=e^x$. With this we are able to apply $F^{-1}$ to both sides of $F(x)=0$ and obtain
$$
F^{-1}(0)=F^{-1}(F(x))=x,
$$
which is not only linear, but is in fact the solution.

OwnerDisplayName: GalZoidberg

Title: Linearization of arbitrary equation

OwnerDisplayName: Richard Myers


问:
Bound on the shortest non-zero vector in any full rank n-dimensional lattice $\Lambda \subseteq \mathbb{R}^n$ with respect to the $1$-norm.
How can i prove
$$\lambda_1 \; \leq \; (n! \; det(\Lambda))^{\frac{1}{n}} \approx \frac{n \; det{(\Lambda)}^\frac{1}{n}}{e}.$$

Here $\Lambda_1$ is shortest non-zero vector. My initial thought was using Minkowski theorem (choosing $S =$ n-Ball of radius $\sqrt n \frac{\lambda_1}{n}$) and proof by contradiction (assuming $\lambda_1 \; > \; (n! \; det(\Lambda))^{\frac{1}{n}}$ and contracting with minimality of $\lambda_1$).

[Minkowski’s convex body theorem] : Let $\Lambda$ be a full dimensional lattice.  If $S \subset \mathbb{R}^n$ is a symmetric convex body of volume $vol(S) > 2^n det(\Lambda)$ ,  then $S$ contains a  non-zero lattice point.

$1-$norm for a vector $x$ : $\sum{}{}{|x_i|}$.

答:
I would like to add my answer too, somebody may find it useful. We show that every (full rank, n-dimensional) lattice $\Lambda$ always contains a non-zero vector $x$ such that
$||x||_1 \leq \; (n! \; det(\Lambda))^{\frac{1}{n}}$.
Let
$\lambda_1 = min\{||x||_1 : x \in \Lambda \setminus \{0\}\}$ and assume for contradiction $l > (n! \; det(\Lambda))^{\frac{1}{n}}$. As @reuns  mentioned take
$S = \{ x : ||x||_1 < l \}$. Notice that $S$ is convex, symmetric and has volume
$vol(S) = 2^n \frac{l^n}{n!} > 2^n det(\Lambda)$. So, by Minkowski's theorem, $S$ contains a non-zero lattice vector $x$. By definition of $S$ ,we have$||x||_1 < l$, a contradiction to the minimality of $l$.

ContentLicense: CC BY-SA 4.0


问:
Might the derivative be a measure of a shrinking curve instead of a point?
I see a lot of posts here questioning the validity of ‘instantaneous rate of change’ and most responses point to the epsilon–delta definition of a limit (or at least some informal version of it). My understanding is that with this definition we calculate the average change of a function across some non-zero interval, and then continue to decrease the interval to no end, at least in principle.

Since this process never ends, the object of study is always a curve, not a point. Albeit, it is an ever shrinking curve but it is a curve nonetheless. In one breath we stress that the interval never actually becomes zero, but then in the next breath we claim that the derivative corresponds to the slope of the tangent line at a point. To me this seems to be a contradictory view.

I have a view on this issue for which I would love to hear your criticisms. I argue that the issue stems from us constructing continua from points when we should be constructing points from continua. To explain my view, consider how we draw graphs. We start with a blank piece of paper (a continuum of sorts) and draw lines on it (i.e. axis lines, grid lines, functions, etc.). Only once these lines intersect can we identify points. With this construction, points are emergent, not fundamental. Since we can continue adding lines to no end, we’ll never completely fill the paper with ink. There will always be continua in between the lines. Might the derivative describe not the points, but the shrinking continua adjacent to the points?

My apologies if I was unclear in the last paragraph, I provided more details on this view in a YouTube video if you are interested.

I believe that continuum-based constructions don’t face the problem of completeness and paradoxes of actual infinity present with point-based constructions. As such, it’s a view that I find hard to let go of but I want to let go of it if it’s wrong. So I would appreciate it if you can point out problems with this view.

Thanks for your time.

答:
the interval never actually becomes zero, but…slope of the tangent line at a point. To me this seems to be a contradictory view.

If $f$ is differentiable at $a$, then "the tangent line at $a$" or "the tangent line through $a$" refers to a particular line through the point $(a,f(a))$ (specifically, the one with slope $f'(a)$). There's at most one tangent line for each point $(b,f(b))$ on the graph, so we can speak of the tangent line at a point, if it exists. I don't see any problem with this terminology.

I'm not sure, but it seems like you might be objecting to the fact that defining the derivative (the slope of the tangent line) involves secant lines which always pass through two relevant points on the graph. My personal gut reaction is "That's why we don't call it the 'secant' line at the point. The different word 'tangent' is used for that reason."

the issue stems from us constructing continua from points when we should be constructing points from continua.

I'm not certain I understand what you're getting at here, but it seems like your idea might not work well for an example like $f(x)=\begin{cases}x^{2} & \text{if }x\in\mathbb{Q}\\0 & \text{else}\end{cases}$. There are holes in the parabola part of that graph, and it's not anything like a simple contiguous curve near $(0,0)$, but it's still differentiable at $0$ and so most would say it still has a tangent line there.

Since we can continue adding lines to no end, we’ll never completely fill the paper with ink.

I'm having trouble understanding what this part of the discussion has to do with the derivatives part, but this depends sensitively on how (or "how fast") we add lines. For example, if I want to fill the unit square with vertical line segments, let's just say I fill $x=0$ through $x=t$ by time $t$, so that by time $0.1$ I've already added a continuum's worth of lines. Then I'll completely fill the square by time $1$.


问:
Generalizing the example 1.23 of Hatcher's Algebraic Topology to $\mathbb R^n\setminus\mathbb S^1$
In Hatcher's Algebraic Topology, you can see the example 1.23 in page 46, where they calculate the fundamental group of the complement in $\mathbb R^3$ of the circle. To sum up the computations, they prove that $\mathbb R^3\setminus\mathbb S^1$ deformation retracts into $\mathbb S^1\vee\mathbb S^2$, and then $\Pi_1(\mathbb R^3\setminus\mathbb S^1)\cong\Pi_1(\mathbb S^1\vee\mathbb S^2)\cong\mathbb Z$.

I was wondering how can I generalize that result to calculate the fundamental group of $\mathbb R^n\setminus\mathbb S^1$, for $n>3$, using Seifert-Van Kampen's theorem as well. Someone told me that I could use this result, but I don't know how exactly. Can anyone please help me out?

答:
I suppose you could use the result here to see that $\mathbb{R}^n \setminus S^1$ is simply connected for $n > 3$.

Or you could visualize the same deformation retraction in Hatcher in higher dimensions to see that $\mathbb{R}^n \setminus S^1 \simeq S^{n-1} \cup_{S^{n-3}} D^{n-2} \simeq S^{n-1} \vee S^{n-2}$, and $\pi_1(S^{n-1} \vee S^{n-2}) = 0$.


问:
If $a+b$ is even, then $a-b$ is even
Prove that for all $a,b \in\mathbb Z$, if $a+b$ is even then $a-b$ is even.

I started off by assuming that a and $b$ are both odd integers, $2k+1$ and $2m+1$ such that their sum is $2(k+m+1)$ which is an even integer. I then concluded that $a-b$ is $2(k-m)$ which is also even, hence the statement is true.
My question is, is this enough to prove that the statement is true or do I also have to proof it when both $a$ and $b$ are even, and when one of $a$ and $b$ is even and the other is odd. Thanks.

答:
$a+b$ is even.

Assume $a-b$ is odd.

Add: $(a+b)+(a-b)=2a;$

Adding an odd number $(a-b)$ to an even number $(a+b) $ gives an even number $2a$. A contradiction.

Title: If $a+b$ is even, then $a-b$ is even


问:
How to represent number that is the smallest possible value larger than another number?
How would I represent the number, i, that is one increment larger than $x$, where an increment is infinitely small.

For example, if $s \in (1,2)$, $i$ would be the smallest number $s$ can take on if $x=1$. I thought about making it $i= x+ (\lim_{n\to\infty} \frac{1}{n})$ but this evaluates to $x$. How would I represent this number $i$ for a given $x$?

I could represent it as $i=\text{min}(s)$ for $s \in (x,\infty)$ but this seems messy. I'm looking for a cleaner way to write this if possible

答:
i+1/∞

Why?

Short: Infinity is a deferred loop stop.

Infinity (∞) is a loop in an information processor (brain, computer),
that does never stop, like while True: ....
Since the looping never stops it never gets into existence as such.
This kind of ∞ therefore does not exist.

Consequently ∞ is an infinite loop, where the stop is deferred to later.
The number of actual loops is thus defined by the need of the context.

The real numbers have a double infinite loop:
dense (arbitrarily precise) and infinite (arbitrarily large).

√2,... and also 1, 2,... ∈ℝ
do not exist other than
"depending on the situation we want this or that precision".
In the same sense ∞ means
"depending on the situation we want a number large enough".

So i+1/∞ means stop when close enough to i from above,
where the reader can decide what enough means in a situation.


问:
Prove that any continuous integer-valued function of a real variable is constant.
I'm stuck on a question which asks to prove that any continuous integer-valued function of a real variable is constant. In my lecture notes we are told that a map is continuous if the preimage of any open set is open.

How do I use this to answer the question?
A full answer would be greatly appreciated as I can't seem to figure out how to solve it in my
notes, in analysis books or anywhere online.

答:
Hint

$$f^{-1}\big(\{f(0)\}\big)$$ is clopen in $\mathbb R$.

ContentLicense: CC BY-SA 3.0


问:
on the commutator subgroup of a special group
Let $G'$ be the commutator subgroup of a group $G$ and $G^*=\langle g^{-1}\alpha(g)\mid g\in G, \alpha\in Aut(G)\rangle$.

We know that always $G'\leq G^*$.

It is clear that if $Inn(G)=Aut(G)$, then $G'=G^*$.

Also if $G$ is a non abelian simple group or perfect group, then  $G'=G^*=G$.

Now do there exist a group such that  $Inn(G)\neq Aut(G)$ and $G'=G^*\neq G$?

Thank you

答:
Serkan's answer is part of a more general family and a more general idea.

The more general idea is to include non-inner automorphisms that create no new subgroup fusion. The easiest way to do this is with power automorphisms, and the simplest examples of those are automorphisms that raise a single generator to a power.

The more general family is parameterized by pairs $(n,d)$ where $n$ is a positive integer and $1 \neq d \neq n-1$, but $d$ divides $n-1$. Let $A=\operatorname{AGL}(1,\newcommand{\znz}{\mathbb{Z}/n\mathbb{Z}}\znz)$ consist of the affine transformations of the line over $\znz$. The derived subgroup $D=A'$ consists of the translations. Choose some element $\zeta$ of multiplicative order $d$ mod $n$, and let $G$ be the subgroup generated by multiplication by $\zeta$ and by the translations. Explicitly:
$$\begin{array}{rcl}
A &=& \left\{ \begin{bmatrix} \alpha & \beta \\ 0 & 1 \end{bmatrix} : \alpha \in \znz^\times, \beta \in \znz \right\} \\
D &=& \left\{ \begin{bmatrix} 1 & \beta \\ 0 & 1 \end{bmatrix} : \beta \in \znz \right\} \\
G &=& \left\{ \begin{bmatrix} \zeta^i & \beta \\ 0 & 1 \end{bmatrix} : 0 \leq i < d, \beta \in \znz \right\} \\
\end{array}$$

For example, take $n=5$ and $d=2$, to get $D$ is a cyclic group of order 5, $G$ is dihedral of order 10, and $A$ is isomorphic to the normalizer of a Sylow 5-subgroup ($D$) in $S_5$.

Notice that $Z(G)=1$ since $d\neq 1$, so $G \cong \operatorname{Inn}(G)$. We'll show $A \cong \operatorname{Aut}(G)$ so that $D=G'=G^*$ as requested. Since $d \neq n-1$, $A \neq G$.

Notice that $D=G'$ (not just $D=A'$) so $D$ is characteristic in $G$, so automorphisms of $G$ restrict to automorphisms of the cyclic group $D$. Let $f$ be an automorphism of $G$, and choose $\beta, \bar\alpha, \bar\beta \in \znz$ so that
$$\begin{array}{rcl}
f\left( \begin{bmatrix} 1 & 1 \\0 & 1 \end{bmatrix} \right)
&=& \begin{bmatrix} 1 & \beta \\ 0 & 1 \end{bmatrix} \\
f\left( \begin{bmatrix} \zeta & 0 \\0 & 1 \end{bmatrix} \right)
&=& \begin{bmatrix} \bar\alpha & \bar\beta \\ 0 & 1 \end{bmatrix}
\end{array}$$
Notice that $f$ is determined by these numbers since $G$ is generated by those two matrices, and that $\beta, \bar\alpha \in \znz^\times$. Now
$$
\begin{bmatrix} \zeta & 0 \\0 & 1 \end{bmatrix} \cdot
\begin{bmatrix} 1 & 1 \\0 & 1 \end{bmatrix} \cdot
{\begin{bmatrix} \zeta & 0 \\0 & 1 \end{bmatrix}}^{-1}
= \begin{bmatrix} 1 & \zeta \\0 & 1 \end{bmatrix}
$$
so applying $f$ we get
$$
\begin{bmatrix} 1 & \bar\alpha\cdot\beta \\0 & 1 \end{bmatrix}
=
\begin{bmatrix} \bar\alpha & \bar\beta \\0 & 1 \end{bmatrix} \cdot
\begin{bmatrix} 1 & \beta \\0 & 1 \end{bmatrix} \cdot
{\begin{bmatrix} \bar\alpha & \bar\beta \\0 & 1 \end{bmatrix}}^{-1}
= \begin{bmatrix} 1 & \zeta\cdot\beta \\0 & 1 \end{bmatrix}
$$
hence $\bar\alpha \beta = \zeta\beta$ and since $\beta \in \znz^\times$ (lest $D \cap \ker(f) \neq 1$), we get $\bar\alpha=\zeta$. Hence every automorphism of $G$ is conjugation by an element of $A$, namely $$\bar f = \begin{bmatrix} 1/\beta & \bar\beta/(\beta(\zeta-1)) \\ 0 & 1 \end{bmatrix} \in A. $$


问:
Equivalent definition of orthogonal support projections

The above sreentshot is from Takesaki's book.

I have three questions

(1).how to get $\|\varphi_{+}\|=\langle s(\varphi_{+}), \varphi\rangle$?

(2) How to conclude that $\|\varphi_{+}\|=\langle s(\varphi_{+}), \varphi_1\rangle$?

(3)How to verify that $s(\varphi_1)\leq s(\varphi_{+})$?

$s()$ denote by the support projection of a normal positive linear functional

答:
The functional $\varphi_+$ is positive, so (since $s(\varphi_+)\leq e$ and so $f\,s(\varphi_+)=0$),  $$\|\varphi_+\|=\varphi_+(1)=\varphi_+(s(\varphi_+))=\varphi(e\,s(\varphi_+))=\varphi(e\,s(\varphi_+)-f\,s(\varphi_+))=\varphi(s(\varphi_+)).$$

You have that $\|\varphi_+\|\leq\|\varphi_1\|$, $\|\varphi_-\|\leq\|\varphi_2\|$, and $\|\varphi_+\|+\|\varphi_-\|=\|\varphi_1\|+\|\varphi_2\|$.

From$\varphi_1(s(\varphi_+))=\|\varphi_1\|$. As $s(\varphi_1)$ is the smallest projection $p$ with $\|\varphi_1\|=\|\varphi(p)\|$, you get that $s(\varphi_1)\leq s(\varphi_+)$.


问:
Integration by (further) substitution?
Question

The function $f$ is defined on $[0, 1]$ by $$f(x) = e^{2\sqrt x}\ .$$

Use the substitution $x = u^2$ to find the exact area of the region enclosed by the graph of $f$, the axes and the line $x = 1$, then use your answer to deduce the exact value of $$\int_{1}^{e^2} (\ln y)^2\ \mathrm{d}y\ .$$

My working

Let the required area be $A$.

\begin{align}
A & =
\int_{0}^{1} e^{2\sqrt x}\ \mathrm{d}x
\\[5 mm] & =
\int_{0}^{1} 2ue^{2u}\ \mathrm{d}u
\end{align}

Let $s = 2u$ and $t' = e^{2u}\ .$

$\implies s' = 2$ and $t = \frac 1 2 e^{2u}\ .$

\begin{align}
A & =
[ue^{2u} - \frac 1 2 e^{2u}]^{u = 1}_{u = 0}
\\[5 mm] & =
\frac 1 2 (e^2 + 1)
\end{align}

I am able to complete the first part (finding the area), but I do not know how to proceed with the second part (evaluating the integral). I have tried to use another substitution, where $x = (\ln y)^2$, but I am not getting anywhere. Any help/intuition will be greatly appreciated :)

答:
The required integral is connected with the inverse function
$$y=e^{2 \sqrt{x}}\to 2\sqrt{x}=\ln y\to x=\frac{\ln^2 y}{4}$$
The area of the inverse function (red in the picture) is the area of the rectangle less the blue area calculated by OP.
$$\frac{1}{4}\int_1^{e^2}\ln^2 y\,dy=e^2-\frac{1+e^2}{2}=\frac{1}{2} \left(e^2-1\right)$$
Therefore requested integral is
$$\int_1^{e^2}\ln^2 y\,dy=2\left(e^2-1\right)$$

Title: Integration by (further) substitution?


问:
Prove that $L$ is closed if and only if for any converging sequence $(x_n)$ with $x_n\in L$, the limit $x=\lim x_n$ is also an element of $L$
Could somebody please point me in the right direction to prove $\Leftarrow$?

Here is my proof for $\Rightarrow$:

Let $L\subseteq R$ be closed and $(x_n)$ be a converging sequence with $x_n\in L$.

Show that (1) $L$ is closed gives (2) $\lim x_n=x\in L$: suppose by contradiction that (2) does not hold given (1), i.e. $\lim x_n=x\notin L$; for any arbitrary $\epsilon>0$, there exists $N^\prime\in N$ such that for $n\geq N^\prime,|x_n-x|<\epsilon\Leftrightarrow -\epsilon<x_n-x<\epsilon\Leftrightarrow x-\epsilon<x_n<x+\epsilon$; $x_n\in L$ but $(x-\epsilon,x+\epsilon)\subseteq L^c\Rightarrow$ contradiction; thus, (1) gives (2).

答:
Proposition

Let $(X,d)$ be a metric space, let $E$ be a subset of $X$, and let $x_{0}$ be a point in $X$. Then the following statements are logically equivalent:

(a) $x_{0}$ is an adherent point of $E$.

(b) There exists a sequence $(x_{n})_{n=1}^{\infty}$ in $E$ which converges to $x_{0}$ with respect to $d$.

Proof

Let us prove the implication $(\Rightarrow)$ first.

If $x_{0}$ is an adherent point of $E$, then for every $\varepsilon > 0$ there corresponds $x_{\varepsilon}\in E$ such that $d(x_{\varepsilon},x_{0}) \leq \varepsilon$.

In particular, if we choose $\varepsilon = 1/n$, there corresponds a $x_{n}\in E$ such that $d(x_{n},x_{0})\leq 1/n$.

Taking the limit from both sides we conclude that $x_{n}\to x_{0}$, and we are done.

Let us prove now the implication $(\Leftarrow)$.

Suppose that $x_{n}\to x_{0}\in E$.

Then for every $\varepsilon > 0$, there corresponds a $n_{\varepsilon}\in\mathbb{N}$ such that $d(x_{n},x_{0})\leq \varepsilon$ whenever $n\geq n_{\varepsilon}$.

In other words, for every $\varepsilon > 0$, there corresponds a $x_{n_{\varepsilon}}\in B(x_{0},\varepsilon)\cap E$.

Consequently, $x_{0}$ is an adherent point of $E$.

Solution

At your case, $L$ is considered to be closed. This means that $L$ contains all its adherent points.

Based on the previous result, this means that every $x\in L$ is the limit of a sequence $x_{n}\in L$.

Conversely, if every convergent sequence $x_{n}\in L$ converges within $L$,we conclude that $L\supseteq\overline{L}$.

Since $L\subseteq\overline{L}$, it results that $L$ is closed.

Hopefully this helps!


问:
Why is $a\sin^2(t) = \frac{a}{2} - \cos(2t)$?
I know that $\sin^2(t)=\frac{1−cos2x}{2}$, but I fail to understand what happens with $a$.

Body:
Since:

$$\sin^2(t) = \frac{1−\cos(2t)}{2},$$

then:

$$a\sin^2(t) = \frac{a}{2}\left(1 - \cos(2t)\right).$$


问:
Is there a way to normalize a vector such that it lives on a Poincare Disk?
If I have a vector, by normalizing its length I can make sure such vectors lives on a hypersphere.

Is there a way to do this for Poincare Disk or other hyperbolic geometry?

答: I guess the open disk is homeomorphic to the Euclidean plane. I found my answer on Wikipedia (https://en.wikipedia.org/wiki/Unit_disk).</p>


答:
The deduction is wrong: set $v_2 = -v_1$ your last inequality

$$
(v_1 - v_2)^2 < (v_1^2 - v_1 v_2 + v_2^2) < (v_1 + v_2)^2
$$

becomes

$$
4 v_1^2 < 3 v_1^2 < 0
$$

which is obviously wrong even if you replace the "$<$" signs with "$\leq$" signs.

You can instead rewrite your expression:

$$
v_1^2 - v_1 v_2 + v_2^2 = \begin{pmatrix} v_1 & v_2 \end{pmatrix} \begin{pmatrix} 1 & -0.5 \\ -0.5 & 1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}
$$

The matrix is positive definite so $v_1^2 - v_1 v_2 + v_2^2 > 0$ if $v_1 \neq 0, v_2 \neq 0$.

Title: Is this algebraic deduction true?

Body:
Since $v_{1}^2 - 2v_1v_2 + v_{2}^2 > 0$ because $(v_{1} - v_{2})^2 > 0$, and
since $v_{1}^2 + 2v_1v_2 + v_{2}^2 > 0$ because $(v_{1} + v_{2})^2 > 0$.

Can we deduce that $v_{1}^2 - v_1v_2 + v_{2}^2 > 0$ because $(v_{1} - v_{2})^2$ $<$ $(v_{1}^2 - v_1v_2 + v_{2}^2)$ $<$ $(v_{1} + v_{2})^2$, for every real number $v_{1}$ and $v_{2}$.


问:
Can fractional/decimal radicals/roots exist?
For questions like "What is the 1/2th root of x would the answer be $x^2$?

My logic is that since $$
\sqrt[\cfrac{1}{2}]{x}=x^{1/{(\cfrac{1}{2}})}
$$
Which simplifies to $x^2$.

So as a general rule it could be $$
\sqrt[\cfrac{1}{a}]{x}=x^{1/{(\cfrac{1}{a}})}
=x^a
$$

And with a different denominator $$\sqrt[\cfrac{b}{a}]{x}=x^{1/{(\cfrac{b}{a}})}
=x^{\cfrac{a}{b}}$$

This corresponds to how decimal/fractional exponents denote radicals (their inverse) while fractional radicals are easier shown with exponents.

Example : (2/3rd root of 4)

$$\sqrt[\cfrac{2}{3}]{4}=4^{1/{(\cfrac{2}{3}})}
=4^{\cfrac{3}{2}}= 8$$

Example (22/7th root of π) :

$$\sqrt[\cfrac{22}{7}]{π}=π^{1/{(\cfrac{22}{7}})}
=π^{\cfrac{7}{22}}
\approx 1.439$$

Example (1/2th root of 1/4) :

$$\sqrt[\cfrac{1}{2}]{\cfrac{1}{4}}=\cfrac{1}{4}^{1/(\cfrac{2}{1})}
=\cfrac{1}{4}^{(\cfrac{2}{1})}
=\cfrac{1}{4}^{2}
=\cfrac{1}{16}
$$

答:
The definition is:

$$\sqrt[a]{x} = x^{\frac{1}{a}}.$$

Hence, if $a = \frac{1}{b}$, then:

$$\sqrt[\frac{1}{b}]{x} = x^{b}.$$

As a consequence:

$$\sqrt[\frac{1}{2}]{x} = x^2.$$

Notice that $a = \frac{1}{b}$ implies that, if for example $b = \frac{22}{91}$, then $a = \frac{91}{22}$. If for example $b = \sqrt{3}$, then $a = \frac{1}{\sqrt{3}}$. And so on.


问:
Function that contiguously enumerates all positive integer sequences of length $x$ that sum to $y$
I am looking for a function that contiguously enumerates all positive integer sequences of length $x$ that sum to a positive integer value $y$. By "contiguously enumerate" I mean that the function needs to map all possible sequences to a unique positive integer value, such that there is no output integer value in the range of $[1,n]$ that does not have a corresponding input sequence. $n$ is the number of possible sequences.

More formally, I'm looking for a bijective function $F(x,y,S) \to \mathbb{Z}^{+}$ where:

$S$ is valid sequence. $S$ is valid if:

$|S|=x$

$\forall s \in S, (s \in \mathbb{Z}) \land (s \geq 0)$

$\sum_{i=1}^{x}S_i = y$

$\forall$ valid $S$, $F(x,y,S) \in [1,n]$ where $n=|\{$ all valid sequences $\}|$

Here's an example of the behavior I'm looking for:

```<code>F(3, 3, [0,0,3]) = 1
F(3, 3, [0,1,2]) = 2
F(3, 3, [0,2,1]) = 3
F(3, 3, [0,3,0]) = 4
F(3, 3, [1,0,2]) = 5
F(3, 3, [1,1,1]) = 6
F(3, 3, [1,2,0]) = 7
F(3, 3, [2,0,1]) = 8
F(3, 3, [2,1,0]) = 9
F(3, 3, [3,0,0]) = 10
</code>```

Just to be extra clear, these are sequences, so the order of their elements is significant. [3,0,0] is not treated the same as [0,0,3].

EDIT: Here's an iterative version of @Phicar's solution in Python:

```<code>def F(x, y, S):
    result = 0
    i = 0
    _x = x
    for i in range(_x - 1):
        cc = S[i]
        if cc > 0:
            result += binomial(y + x - 1, x - 1) - binomial(y - cc + x - 1, x - 1)
        x -= 1
        y -= cc
    return result
</code>```

(This doesn't account for the case where $x=1$ or $y=0$)

答:
Recursively it looks like this:

 If your sequence has length $1,$ treat this as a base case and do accordingly, if not, select the first element in $S$ i.e., $S_1.$

 If $S_1=0,$ go to the next index(nothing has gone before this so far). If not, consider $\binom{y-i+x-2}{x-2}$ elements that have to go before this one for a fixed $i$ (as noticed in the comments by Hagen von Eitzen) for $i=0$ all the way to $S_1-1,$ hence you want to push $$\binom{y+x-1}{x-1}-\binom{y-S_1+x-1}{x-1}.$$
This last formula is just a consequence of the hockey-stick identity.

The algorithm, in sage, then looks like :

```<code>def tal(a,b,c):
    if len(c)==1:
        if a==1:
            if b==c[0]:
                return 1
            return 0
        return 0
    d = [0]*(len(c)-1)
    for i in range(1,len(c)):
        d[i-1]=c[i]
    cc = c[0]
    push = 0
    if cc>0:
        push = binomial(b+a-1,a-1)-binomial(b-cc+a-1,a-1)
    return push+tal(a-1,b-cc,d)
print(tal(3,3,[2,0,1]))
</code>```

where $a$ is the original $x,$ $b$ is $y$ and $c$ is the sequence.

Body:
Recursively it looks like this:

 If your sequence has length $1,$ treat this as a base case and do accordingly, if not, select the first element in $S$ i.e., $S_1.$

 If $S_1=0,$ go to the next index(nothing has gone before this so far). If not, consider $\binom{y-i+x-2}{x-2}$ elements that have to go before this one for a fixed $i$ (as noticed in the comments by Hagen von Eitzen) for $i=0$ all the way to $S_1-1,$ hence you want to push $$\binom{y+x-1}{x-1}-\binom{y-S_1+x-1}{x-1}.$$
This last formula is just a consequence of the hockey-stick identity.

The algorithm, in sage, then looks like :

def tal(a,b,c):
    if len(c)==1:
        if a==1:
            if b==c[0]:
                return 1
            return 0
        return 0
    d = [0]*(len(c)-1)
    for i in range(1,len(c)):
        d[i-1]=c[i]
    cc = c[0]
    push = 0
    if cc>0:
        push = binomial(b+a-1,a-1)-binomial(b-cc+a-1,a-1)
    return push+tal(a-1,b-cc,d)
print(tal(3,3,[2,0,1]))

where $a$ is the original $x,$ $b$ is $y$ and $c$ is the sequence.


问:
Upper function/bound for $y'(x)=x^3+y^3(x)$
The function is $y'(x)=x^3+y^3(x)$ for $x>0$ and $y(0)=1$

I was asked to find an upper & lower function so that $y_1'(x)<y'(x)<y_2'(x)$

I've found a lower bound for $y'(x)$ by seeing that $y_1'(x)=y^3(x)<x^3+y^3(x)=y'(x)$ with $y_1(0)=1$

With that, I got $y_1(x)=\frac{1}{\sqrt{1-2x}}$ as solution and now I can deduce that $0<x<\frac{1}{2}$

But not I'm having problems constructing the upper function for that problem.

My initial guess was $y_2(x)=1+y^3(x)$ but that proofed hard to solve. Can someone give me a hint for a better estimate?

答:
If one adds more complications to the (first, for $x<\frac14$, now improved) strategy of the answer of Martin R, one could observe that
$$
y'(x)\ge x^3+1
\implies
y(x)\ge 1+x+\frac{x^4}4\ge 1+x+\frac{x^4}8+\frac{x^4}8
\ge 1+3\sqrt[3]{x·\frac{x^4}8·\frac{x^4}8}\ge\frac34 x^3
$$
(exact at $x=2$, factor can be improved to slightly above $0.8$) so that one obtains an upper bound via the differential inequality
$$
y'(x)\le \frac43y(x)+y(x)^3.
$$
This can be solved via the Bernoulli approach to get
$$
y(x)^{-2}+\frac34\ge(1+\frac34)e^{-\frac83x}
$$
which gives a singularity-free region for $x<-\frac38\ln\frac37=0.317736697...$

Reducing the initial estimate to the interval $[0,\frac12]$ one has there
$$
y(x)\ge x^3(x^{-3}+x^{-2}+\tfrac14x)>12x^3.
$$
This gives the upper bound a differential inequality of the Bernoulli type
$$
y'(x)\le \frac1{12}y(x)+y(x)^3
\implies
y(x)^{-2}+12\ge(1+12)\exp(-\tfrac16x).
$$
This bound prevents divergence to infinity for $x<6\ln(1+\frac1{12})=0.4802562460...$


问:
Bound for $\Vert\sum_iA_i^\dagger A_i\Vert_\infty$ given that $\Vert\sum_i A_i\Vert_\infty$ is small
Let $A_i$ be matrices such that

$$\left\Vert \sum_i A_i \right\Vert_\infty \leq \varepsilon,$$

where $\Vert\cdot\Vert_\infty$ is the operator norm and is equal to the largest singular value of its argument and $\dagger$ denotes the conjugate transpose. Can one say anything about an upper bound for

$$\left\Vert \sum_i A^\dagger_iA_i \right\Vert_\infty$$

in terms of $\varepsilon$? Moreover, is there a converse statement too i.e. given

$$\left\Vert \sum_i A^\dagger_iA_i \right\Vert_\infty\leq \delta$$

is there an upper bound for $\left\Vert \sum_i A_i \right\Vert_\infty$ in terms of $\delta$?

答:
Suppose that we have finitely many matrix $A_1,\dots,A_n$ to consider. Let $\tilde A$ denote the matrix
$$
\tilde A = \pmatrix{A_1\\ \vdots\\ A_n}.
$$
We see that $\sum_{i=1}^n A_i^\dagger A_i = \tilde A ^\dagger \tilde A$, so that
$$
\left\| \sum_{i=1}^n  A_iA_i^\dagger\right\|_{\infty}
= \left\|\tilde A^\dagger \tilde A \right\|_\infty
= \left\| \tilde A\right\|_\infty^2 \leq \delta.
$$
On the other hand, we can write $\sum_{i=1}^n A_i$ as the product $M \tilde A$, where
$$
M = \pmatrix{I & \cdots & I}.
$$
We find that $\|M\|_\infty = \sqrt{n}$, so that
$$
\left\|\sum_{i=1}^n A_i \right\|_\infty  = \| M \tilde A\|_\infty \leq \|M\|_\infty \cdot \|\tilde A\|_\infty  = \sqrt{n} \cdot \|\tilde A\|_\infty
\leq \sqrt{n\delta}.
$$


问:
Does power rule holds for time functions of unit quaternions?
I'm trying to understand quaternion calculus. Let's suppose I have a time function of unit quaternion: $q(t) = (\cos \theta(t), (\sin \theta(t)) \mathbf{a}(t)), \mathbf{a}(t) \in \mathbb{R}^3$, $||\mathbf{a}(t)|| = 1$ $\forall$ $t$ to the power $f(t) \in \mathbb{R}$. Would someone clarify if $\dfrac{d\left(q(t)^{f(t)}\right)}{dt}$ $$= \left(−\sin(fθ) \dfrac{d(f\theta)}{dt}, \mathbf{a}\cos(fθ)\dfrac{d(f\theta)}{dt} + \dfrac{d(\mathbf{a})}{dt}\sin(fθ)\right)$$ or $$
= q(t)^{f(t)} \times \dfrac{d}{dt} \left[f(t) \ln q(t)\right] = q(t)^{f(t)} \times \dfrac{d}{dt} \left[f(t) \cdot (0, \theta(t) \mathbf{a}(t)) \right] \\
= (\cos (f\theta), (\sin (f\theta))\mathbf{a}) \times (0, f'\theta\mathbf{a} + f\theta'\mathbf{a} + f\theta\mathbf{a}') $$ or both are correct? (BTW, do we have a general formula for $\mathbf{a} \times \mathbf{a}'$?) Thanks a lot.

答:
Technically the power $q^f$ is ill-defined, but I guess it's ok shorthand for $\exp(f\theta\mathbf{a})$.

Your first application of the chain rule is correct.

Your second application seems to be using the idea that $(e^\mathbf{v})'=e^\mathbf{v} \mathbf{v}'$. This is wrong. First of all, there's no reason to expect $e^\mathbf{v}\mathbf{v}'$ any more than $\mathbf{v}'e^\mathbf{v}$, but if these were the same then $e^\mathbf{v}\mathbf{v}'e^{-\mathbf{v}}=\mathbf{v}'$ which means rotating $\mathbf{v}'$ around $\mathbf{v}$ by an angle of $\|\mathbf{v}\|$ fixes $\mathbf{v}'$, which is easy to cook up a counterexample to.

Indeed, writing $\mathbf{v}=\theta\mathbf{u}$ with $\mathbf{u}$ a unit vector (so $f=1$ here), we have

$$ (e^\mathbf{v})'=(e^{\theta\mathbf{u}})'
=(\cos\theta+\sin\theta\,\mathbf{u})'
=\color{Green}{(-\sin\theta+\cos\theta\,\mathbf{u})\theta'}+\color{Blue}{\sin\theta\,\mathbf{u}'} $$

whereas

$$ e^\mathbf{v}\mathbf{v}'
=(\cos\theta+\sin\theta\,\mathbf{u})(\theta'\mathbf{u}+\theta\mathbf{u}')
=\color{Green}{(-\sin\theta+\cos\theta\,\mathbf{u})\theta'} +\color{Purple}{\theta\cos\theta\,\mathbf{u}'}+\color{Red}{\theta\sin\theta\mathbf{u}\mathbf{u}'}. $$

Notice $\color{Blue}{\sin\theta\,\mathbf{u}'}$ doesn't match $\color{Purple}{\theta\cos\theta\,\mathbf{u}'}$ and $\color{Red}{\theta\sin\theta\mathbf{u}\mathbf{u}'}$ doesn't match anything.

Keep in mind $\mathbf{u},\mathbf{u}',\mathbf{u}\mathbf{u}'$ are three orthogonal vectors. (Differentiating $\mathbf{u}\cdot\mathbf{u}=1$ yields $\mathbf{u}\cdot\mathbf{u}'=0$, so we know $\mathbf{u}$ and $\mathbf{u}'$ are orthogonal, which means $\mathbf{u}\mathbf{u}'=\mathbf{u}\times\mathbf{u}'$ is orthogonal by properties of the cross product; also note the quaternion product of vectors is $\mathbf{ab}=-\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\times\mathbf{b}$.)


问:
Why is $\sum_{i=0}^{n - 1}2^i = 2 ^ n - 1$?
Probably another simple question and possibly not asked too well. I'd like to know why this is true (I'm looking for a simple, possibly combinatorial explanation with a real-world example):

$$\sum_{i=0}^{n - 1}2^i = 2 ^ n - 1$$

To add more detail, I don't get why $2^{16}$ is the same as $2^{15}+2^{14}+2^{13}...+2^{0} + 1$.

答:
This comes from the mid-school formula:
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\dots+ ab^{n-2}+b^{n-1}),$$
with $a=x$ and $b=1$.


问:
Is the absolute value function an equivalence relation?
Let $f \colon \mathbb{R} \to \mathbb{R}$ be the well-known absolute value function: $f(x) = \lvert x \rvert$. Show that f, considered as a
relation, is neither reflexive nor symmetric. Show also that it is transitive.

I'm not sure how to proceed on this one. Is the absolute value function not an equivalence relation?

答:
This relation is defined by

$$(\forall x,y\in\Bbb R)\; (x \mathcal R y\iff y=|x|)$$

So, it is not reflexive since
$-1$ is not in relation with $ -1 $ because $|-1|\ne -1$.

It is not symetric since
$-1 $ is in relation with $ 1$ but $ 1 $ is not.

It is transitive :

$$x\mathcal Ry \quad \text{ and } \quad y\mathcal R z\implies$$
$$|x|=y \quad \text{ and } \quad|y|=z \implies |x|=|y|=z$$

$$\implies x \mathcal R z$$


问:
Using Ito calculus to prove that $\int_0^t W_s^2dW_s = \frac{1}{3} W_t^3 - \int_0^t W_s d_s$
I am busy trying to teach myself some stochastic calculus and have come across a statement that I am trying to prove.

How can I prove that
\begin{align}
\int_0^t W_s^2dW_s = \frac{1}{3} W_t^3 - \int_0^t W_s d_s
\end{align}
where $W \in \mathbb{R} $ is Brownian motion, using Ito's rule?

答:
Hint: apply Ito's formula to $f(x)=x^3$.

 $f'(x)=3x^2$, $f''(x)=6x$ so $dW^3=3W^2dW+3Wdt$


问:
How much will the height of the water level rise when Satoshi immerses a saké bottle with diameter 7.5 cm?
I've been struggling with this problem for an hour and I have no idea how to start it:

Satoshi likes warm saké. He places 1 litre of water in a sauce pan
with diameter 17 cm. How much will the height of the water level rise when
Satoshi immerses a saké bottle with diameter 7.5 cm?

Hint: You’ll need the volume conversion ratio 1 litre = 1000 cm³

So far this section of the book had simple geometry problems. My intuition tells me this is more of a physics problem, and the Archimedes principle comes to my mind. I know there must be a way though.

(1) Could someone point me in the right direction without actually solving the problem?
(2) Could someone solve the problem (and let me know he/she is going to solve it) in case (1) is not enough?

Thank you

答:
The volume of water initially is equal to the area of the pan, multiplied by the initial height.
$$V=A_p\cdot h_0=\pi R^2h_0$$
When you put the bottle in, I assume that it goes all the way to the bottom. The water in the pan will now raise to $h_1$. The volume of the water is $$V=(A_p-A_b)\cdot h_1=\pi(R^2-r^2)h_1$$
Here $R=17/2\ cm$, $r=7.5/2\ cm$, and $V=1000\ cm^3$. You can find $h_0$ and $h_1$.


答: If you assume the bottle is cylindrical and tall enough to continue above the surface of the water, even after immersion, all you need to know is that the base has changed from a circle of diameter 17 to an annulus (flat washer shape), outer diameter 17 and inner diameter 7.5. How many square centimeters are there in this smaller base?</p>

Body:
I've been struggling with this problem for an hour and I have no idea how to start it:

Satoshi likes warm saké. He places 1 litre of water in a sauce pan
with diameter 17 cm. How much will the height of the water level rise when
Satoshi immerses a saké bottle with diameter 7.5 cm?

Hint: You’ll need the volume conversion ratio 1 litre = 1000 cm³

So far this section of the book had simple geometry problems. My intuition tells me this is more of a physics problem, and the Archimedes principle comes to my mind. I know there must be a way though.

(1) Could someone point me in the right direction without actually solving the problem?
(2) Could someone solve the problem (and let me know he/she is going to solve it) in case (1) is not enough?

Thank you


问:
Finding the volume of the cap of a sphere

Let's say I have a sphere of radius 2 and we drill a hole with the
shape of a sphere of radius 1. How shall we find the volume of the cap
of the sphere?

What I did was finding the spherical coordinates for the surface:

$\Phi(\theta, \phi)=2(\cos\theta \sin \phi, \sin \theta \sin \phi, \cos \phi)$ with $\theta \in [0, 2\pi]$

To find the boundaries for $\phi$, I used the Pythagorean theorem:

To find $h$: $(\frac{h}{2})^2=2^2+1^2$, so $h=\sqrt{20}$

So I end up with this triangle:

Then $\sin \phi = \frac{1}{2}$ so, $\phi = \frac{\pi}{6}$.

Setting up the integral:

$\int _{0} ^{\pi/6} \int _{0} ^{2\pi} \int _{0} ^{2} \rho^2 \sin \phi d\rho d\theta d\phi=-\frac{8}{3}(\sqrt{3}-2)\pi$

Which is wrong... I've done a lot of steps, so I can't really tell where I made a mistake

答:
Applying Pythagoras, $h^2 = 2^2 - 1^2$. You instead did addition. (I am using $h$ instead of $h/2$ to keep things simpler).

So you know $h = \pm \sqrt3$.

You calculated the angle correctly $(0 \leq \phi \leq \frac{\pi}{6})$ for the spherical cap above $z$ axis. You can multiply by two if you need to find the volume for both spherical caps.

Now coming to your integral, the volume that you found is of the spherical cone and not of the spherical cap. As you need to find the volume of the spherical cap, you need to make sure that $\rho$ is only between $\frac{\sqrt{3}}{\cos \phi}$ and $2$. So your integral becomes -

$V = \displaystyle \int_{0}^{2\pi} \int_{0}^{\pi/6} \int_{\sqrt{3}/cos{(\phi)}}^{2} r^2 \sin \phi \, dr \, d\phi \, d\theta \approx 0.431$. This is for one spherical cap.

EDIT: see a 2D diagram.


问:
Can't find error in $\int{xe^{-(x-\theta)}}dx$
I haven't done integrals in some time and I'm kinda rusty. This is a very easy one, but for some reason I keep getting a slightly different result than the correct one. There has to be a sign error somewhere which I really can't find.

$$\int{xe^{-(x-\theta)}}dx $$
Let $t = x-\theta$. Then the above is:
$$\int{(t+\theta)e^{-t}dt = \int{te^{-t}dt}+\theta\int{e^{-t}dt} =-e^{-t}(t+1) -\theta e^{-t} = -e^{-t}(t-\theta+1) + c = -e^{\theta-x}(x-2\theta+1) + c}$$

However the correct results is $-e^{\theta-x}(x+1) +c$. The two thetas should cancel out but they don't. Where's my mistake?

Any help is appreciated.

答:
Note that:

$$\int e^{-t} dt=-e^{-t}+C$$

Body:
Note that:

$$\int e^{-t} dt=-e^{-t}+C$$


问:
Use an infinite series to approximate to four decimal places: integral from 0 to 1 sin(7x)/x (finding number of terms without calculator)
So I managed to solve this problem by simply taking the Maclaurin series of sin(7x), which was

$$\sum_{n=0}^∞  \frac{(-1)^n(7x)^{2n+1}}{(2n+1)!}$$

and divided the summation by x, which gave me

$$\sum_{n=0}^∞  \frac{(-1)^n(7x)^{2n}}{(2n+1)!}$$

I integrated this and got $$\sum_{n=0}^∞  \frac{(-1)^n(7x)^{2n+1}}{(2n+1)(2n+1)!}$$ and set the bounds from 0 to 1. I plugged in 1 to get

$$\sum_{n=0}^∞  \frac{(-1)^n(7)^{2n+1}}{(2n+1)(2n+1)!}$$

Now, this is where my problem came in. I managed to use a calculator to figure out what the series converges to. However, I want to know how to solve this by finding the number of terms needed to find the sum. How can one find the number of terms necessary to get this sum?

答:
In an alternating series the error is less than the absolute value of the first term neglected. Thus to compute the value with four decimal places, we must make an error less than $10^{-4}$
$$s=\sum_{n=0}^{\infty} \frac{(-1)^n\,7^{2n+1}}{(2n+1)(2n+1)!}$$
This means  finding  an index $n$ such that
$$\frac{7^{2n+1}}{(2n+1)(2n+1)!}\le 10^{-4}\to n=11$$


问:
$x+y-1$ divide $n$.
Find all $n>1$  odd number s.t. for any $x, y$ divisors of $n$ with $gcd(x,y)=1$ we have $x+y-1$ divide $n$.

It's intuitive that $n$ is of the form $p^k$ with $p$ prime.

My idea:

I consider $n=p_1^{a_1}\cdot p_2^{a_2}\cdot ...\cdot p_k^{a_k}$ with $p_1<p_2<...<p_k$.

$k\geq 2$

I let $x=p_1$ and $y=p_2^{a_2}\cdot...\cdot p_k^{a_k}$.

But $p_2,...,p_k$ don't divide $x+y-1$ because of minimality of $p_1$. So $x+y-1$ should be a power of $p_1$. But this is not necessary false and I am stuck.

答:
Only prime powers can satisfy the desired hypothesis.
As you rightly observed, if $p$ is the smallest prime factor of $n$, with $p^m\mid\mid n$, and $n$ is not a prime power, then with $x=p$ and $y=np^{-m}$ such that $(x+y-1)\mid n$, it follows that $x+y-1= p^l$ for some natural number $l\le m$; equivalently,
$$n=(1-p+p^l)p^m\,.$$ Clearly, we must have that $l\ne 1$, thus in particular $m\ge 2$. We claim that we can find a different pair $(x,y)$ failing the desired hypothesis. This time, let $$x=p^2\,,~\,~\,~\,~\text{and}\,~\,~\,~\,~y= 1-p+p^l\,.$$
In particular, $n=xyp^{m-2}$. It is now not difficult to check that $(x+y-1)\nmid n$; indeed, a common prime factor of both $n$ and $x+y-1=-p+p^2+p^l$ must either divide $y-1$ and $x$ (which is simply $p$) or must divide $y$ and $x-1=p^2-1=(p+1)(p-1)$, which, because $p$ is odd, cannot exist because it would necessarily be smaller than $p$, contrary to the minimality of $p$. We therefore conclude that if $(x+y-1)|n$, then $p$ is the only prime factor of $x+y-1=p(-1+p+p^{l-1})$, which is blatantly absurd!

Remarks

The argument shows that if one allows for $n$ to be even, the last part would require that $p=2$ such that $(2^{l-1}+1)|(2^l-1)$. This is only possible if $l=2$, leading to the only potential even non-prime power cases satisfying your hypothesis as $n\in\{2^{m}\cdot 3:m\ge 2\}$. Of these, letting $x=2^m$ and $y=3$ shows that necessarily $m=2$, and thus $n=12$ is the only admissible solution.

Title: $x+y-1$ divide $n$.


问:
Does $(\sim {[p \Rightarrow (\sim q)] \vee (r \Leftrightarrow s)}) \wedge (r \vee \sim q)$ translate to $NKKNCpNqErsKrqRq$ in Polish Notation?
In translating $$(\sim {[p \Rightarrow (\sim q)] \vee (r \Leftrightarrow s)}) \wedge (r \vee \sim q)$$

to Polish notation, I got $$NKKNCpNqErsKrq$$

If anyone could give me clarification if this answer is correct or not, I would appreciate it.

答:
Translate $NKKNCpNqErsKrq$ back to the usual notation.

$Krq$ is $r\land q$.

$Ers$ is $r\leftrightarrow s$.

$Nq$ is $\neg q$.

$CpNq$ is $p\to\neg q$.

$NCpNq$ is $\neg(p\to\neg q)$.

$KNCpNqErs$ is $\neg(p\to\neg q)\land(r\leftrightarrow s)$.

$KKNCpNqErsKrq$ is $\big(\neg(p\to\neg q)\land(r\leftrightarrow s)\big)\land(r\land q)$.

$NKKNCpNqErsKrq$ is $$\neg\left[\big(\neg(p\to\neg q)\land(r\leftrightarrow s)\big)\land(r\land q)\right]\,.$$

As you can check by making a truth table, this is not equivalent to the original expression

$$\left(\neg\left\{\big[p\to(\neg q)\big]\lor(r\leftrightarrow s)\right\}\right)\land(r\lor\neg q)\,.\tag{1}$$

If you have $A$ for disjunction, you can convert $(1)$ as follows:

$r\lor\neg q$ is $ArNq$.

The expression inside the curly braces is $ACpNqErs$.

Negating it yields $NACpNqErs$.

And the final result is $KNACpNqErsKrq$.

If you don’t have $A$ available, you’ll have to use DeMorgan’s laws to convert $x\lor y$ to $\neg(\neg x\land\neg y)$ and then do the conversion to Polish notation.


问:
Differentiation of $x^TAx$ (Multivariate Calculus)
Suppose $A$ is an $n\times n$ symmetric matrix. Consider a function $f : \mathbb{R}^n \to \mathbb{R}$ such that $$f(x) = \langle Ax , x \rangle = x^TAx\quad\quad\text{for all}~~~ x \in \mathbb{R}^n$$ I want to find its derivative, i.e. $Df$ .

Most probably, it can be done by expanding $x^TAx$ by usual matrix multiplication. But I'm looking for a procedure which doesn't involve this sort of expansion. However, any elementary proof will help. Thanks in advance. Sorry if someone had asked this problem before.

答:
There are a number of notations and methods used by different authors in discussing vector calculus.

One simple approach to computing derivatives like this is to expand out the function $f(x+\Delta x)$ and truncate the result to $f(x)$ plus a term that is linear in $\Delta x$.  The coefficient of $\Delta x$ is then the derivative. Higher-order terms involving $\Delta x$ are ignored.

$(x+\Delta x)^{T}A(x+\Delta x)=x^{T}Ax+(\Delta x)^{T}Ax+x^{T}A\Delta x+\Delta x^{T}A\Delta x.$

Note that $\Delta x^{T}Ax=x^{T}A^{T}\Delta x$.  Thus

$(x+\Delta x)^{T}A(x+\Delta x)=x^{T}Ax+x^{T}(A+A^{T})\Delta x+\Delta x^{T}A\Delta x$

The $\Delta x^{T}A \Delta x$ term is higher order and can be ignored in computing the gradient.  The derivative is the coefficient of $\Delta x$ in the linear term, $x^{T}(A+A^{T})$.

However, the gradient is always presented as a column vector, so

$\nabla (x^{T}Ax)=(A+A^{T})x.$

If $A$ is symmetric, this is commonly written as $2Ax$.


问:
What is 'Operator valued function is holomorphic' ? Does Neumann series imply holomorphic?
Let $H_1$ and $H_2$  be Hilbert spaces and $K(w):H_1 \rightarrow H_2$ be an operator for given $w\in\Omega\subset\mathbb{C}$.(like a resolvent. It would not be resolvent.)

In complex analysis, a function $f:\Omega \rightarrow \mathbb{C}$ is holomorphic if
$$f'(w_0):=\lim_{w\rightarrow w_0} \frac{f(w)-f(w_0)}{w-w_0}$$
exists for every $w_0\in\Omega$. And its limit takes modulus.
However, since an operator valued function $w \mapsto K(w) $ is not complex value, $\left|\frac{K(w)-K(w_0)}{w-w_0}\right|$ cannot be defined.
So, I can GUESS its limit is defined in operator norm, i.e. if $K'(w_0)$ is defined such that
$$\left\|\frac{K(w)-K(w_0)}{w-w_0} - K'(w_0)\right\|_{H_1 \rightarrow H_2} \rightarrow 0 \quad as \quad w \rightarrow w_0,$$ $K(w_0)$ is holomorphic for $w_0 \in \Omega$. In my book, there is no mention about this detail.

By the way, consider the Neumann series,
$$(I-K(w))^{-1} = \sum_k^\infty (K(w))^k.$$
In order to show $(I-K(w))^{-1}$ is holomorphic, I can think about 'analytic' first (since analytic and holomorphic is equivalent in $\mathbb{C}$) and Neumann series say '$K(w)$ is analytic for each $w$'. But, again,  it is not complex number.

Questions

What is definition of 'Operator valued function is holomorphic' ? Is my guess right ?

Does it suffice to show existence of its Neumann series ? If then, why ?

答:
I think it is wise to be uneasy about whether or not Cauchy (-Goursat) complex function theory works for operator-valued functions the same as it works for complex-valued functions. Happily (some details below) it does, and in a fashion which is basically idiot-proof! As often happens, the proofs that everything works amazingly well are more sophisticated than the assertions themselves.

(To carefully distinguish, I'd prefer to use complex_differentiable for existence of a complex derivative, and complex_analytic for existence of a (convergent) power series expression.)

First, yes, we can talk about the operator norm topology on operators, making the space a Banach space. But, also, the same discussion applies to the strong operator topology (given by seminorms $\mu_x(T)=\sup_{|x|\le 1}|Tx|$), and other topologies on operators. These other topologies are fancier than Banach spaces, but still do have necessary completeness properties to make vector-valued Cauchy-Goursat theory succeed. Maybe this is not the main point for you, but it does illustrate the robustness of the truth of the analogues.

As in Abel's and others' proofs about convergent power series giving complex differentiable functions... the same argument (with obvious adjustments) proves the same for Banach-space valued power series. (The fact that they're Banach spaces of operators doesn't enter.)

The converse, that complex differentiability of Banach-space valued functions implies a Cauchy theorem, Cauchy formula, etc., is less trivial, but holds. A very general case of vector-valued holomorphic functions was treated by Grothendieck in the early 1950s. Again, some of the robustness of the situation is illustrated by the theorem that a weakly holomorphic $V$-valued function $f$ is holomorphic: that is, if for all continuous linear functionals $\lambda$ on $V$ the composite $\lambda\circ f$ is a holomorphic scaler-valued function, then $f$ is holomorphic $V$-valued. This holds for quasi-complete, locally convex spaces $V$, so applies to Frechet spaces, spaces of test functions, weak duals of these, strong duals, and nearly anything that arises naturally with any reasonable completeness.

Many functional analysis books at least treat the Banach-space-valued case.


问:
Calculate a function with the use of a development point and geometric series
Assuming I have this function :
$$ f(x) = \frac{1 + x^3}{2-x}  $$

The question was to calculate this with the use of the geometric series and  the development point $x_{0} = 0$.

Well setting first the function to a geometric series is quite clear here .
I got $ f(x)=$ $\sum_{n=1}^\infty \frac{x^n}{2^{n+1}}(1+x^3) $ as a result . What's the meaning of calculating an infinite series here ? By calculating does this mean that I can choose to calculate the first terms of the series (For example from n=0 to n=3 and then   $  '+\ldots '$ ) ?

Plus why do I need to use the development point $x_{0}$ . I can already calculate the function with the geometric series only . Won't that be enough or is there another idea to implement this?

答:
We are looking for a representation
\begin{align*}
f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n=\sum_{n=0}^\infty a_nx^n\tag{1}
\end{align*}
where the series expansion is at the point $x_0=0$.

We obtain
\begin{align*}
\color{blue}{f(x)}&\color{blue}{=\frac{1+x^3}{2-x}}
=\frac{1}{2}\cdot\frac{1+x^3}{1-\frac{1}{2}x}\\
&=\frac{1}{2}\sum_{n=0}^\infty\frac{1}{2^{n}}x^n(1+x^3)\\
&=\sum_{n=0}^\infty\frac{1}{2^{n+1}}x^n+\sum_{n=0}^\infty\frac{1}{2^{n+1}}x^{n+3}\\
&=\sum_{n=0}^\infty\frac{1}{2^{n+1}}x^n+\sum_{n=3}^\infty\frac{1}{2^{n-2}}x^{n}\\
&\,\,\color{blue}{=\frac{1}{2}+\frac{1}{4}x+\frac{1}{8}x^2+\sum_{n=3}^\infty\frac{9}{2^{n+1}}x^n}
\end{align*}
with the last line being a representation according to (1).


问:
Show the the image of correspondence is bounded.
Show that if $\Gamma : X \to Y$ is compact-valued and upper hemi continuous, then for any compact set $K\subseteq X$, the set $\Gamma(K) \subseteq Y$ is also compact. [ Hint: To show that $\Gamma(K)$ is bounded, suppose the contrary. Let $\{y_n\}$ be a divergent sequence in $\Gamma(K)$, and choose $\{x_n\}$ such that $y_n \in \Gamma(x_n)$, all $n$.] ($X, Y \subseteq \mathbb{R}^n$)

In my book, the definition of u.h.c. is that for every $(x_n) \to x$ and every $(y_n)$ such that $y_n \in \Gamma(x_n)$, there exists a subsequence of $(y_n)$ converging to the limit in $\Gamma(x)$.

I can prove the closedness. For boundedness, following the hint, I first construct a divergent sequence $(y_n)$ such that $y_n \in \Gamma(x_n)$. Since $K$ is compact, there exists a subsequence $(x_n^*)$ of $(x_n)$ converging to $x \in K$. We als have a corresponding sequence $(y_n^*)$. Since $\Gamma$ is u.h.c, there exists a subsequence $(y_n^{**})$ of $(y_n^*)$ converging to $y \in \Gamma(x)$. I am stuck here because I don't know how to reach a contradiction from here. I would appreciate if you give some help.

答: You're almost there. When the hint says to let $\{y_n\}$ be divergent, they want $\|y_n\| \to \infty$, not just $y_n$ not to converge converge. So, the convergent subsequence $y^{**}_n$ converging to some $y$ is indeed a contradiction, as it would suggest $\|y^{**}_n\|$ would have some finite limit.


问:
Neat way to prove that $x^2\gt (1+x)(\ln(1+x))^{2}, ~\forall x\gt 0$
I am supposed to prove the following inequality. One standard method that can be put to use is considering a function $f(x)=x^2-(1+x)(\ln(1+x))^2$ and take its derivative to comment on the behaviour of its increase and arrive at concluding the inequality using the fact that $f(0)=0$.

$$x^2\gt (1+x)(\ln(1+x))^2,~~ \forall x\gt 0$$

I was wondering if there was a neater way to come up with the proof, especially as this is a contest-practice problem. Any hints are appreciated. Thanks.

答:
First, take the square root on both sides, turning the inequality into $x\gt\sqrt{x+1}\ln(x+1)$. Next, let $u^2=x+1$ and rewrite the inequality to prove as

$${u^2-1\over u}\gt\ln(u^2)\quad\text{for }u\gt1$$

Finally, let $f(u)=u-{1\over u}-2\ln u$, noting that $f(1)=0$. We see that

$$f'(u)=1+{1\over u^2}-{2\over u}={u^2-2u+1\over u^2}=\left(u-1\over u \right)^2\gt0$$

for $u\gt1$ and thus $f(u)\gt0$ for $u\gt1$, which proves the desired inequality.


问:
Prove a formula to evaluate Legendre Symbol $(\frac{-3}{p})$ using quadratic Gauss Sums (i.e. without using quadratic reciprocity)
I've been trying to find a question similar to this, but have failed to find any. The full statement is let $\zeta = e^{\frac{2\pi i}{3}}$, and use the fact that $(2 \zeta + 1)^2 = -3$ and quadratic Gauss sums to prove a formula to evaluate $(\frac{-3}{p})$. I understand how to do this with reciprocity very easily, and I have a vague understanding of the proof of reciprocity using roots of unity and quadratic Gauss sums. Using reciprocity, this is what I came up with:

$$ (\frac{-3}{p}) = (\frac{-1}{p})(\frac{3}{p}) $$

CASE 1: let $p \equiv 1 \text{ mod } 4$

$$  (\frac{-1}{p}) = 1 \\
(\frac{3}{p}) = (\frac{p}{3}) \\
\therefore (\frac{-1}{p})(\frac{3}{p}) = (\frac{p}{3})$$

CASE 2: let $p \equiv 3 \text{ mod } 4$

$$  (\frac{-1}{p}) = -1 \\
(\frac{3}{p}) = -(\frac{p}{3}) \\
\therefore (\frac{-1}{p})(\frac{3}{p}) = -1 \cdot -(\frac{p}{3}) = (\frac{p}{3})$$

Thus, it would be enough to perform quadratic reciprocity on $(\frac{3}{p})$ and evaluate the Legendre Symbol where $p \equiv 1 \text{ mod } 3$ and $p \equiv 2 \text{ mod } 3$. Knowing this, I try to formulate a claim using Gauss sums. Let $g(a,p)$ be the quadratic Gauss sum $\Sigma_{n=0}^{p-1} (\frac{n}{p}) \zeta_{p}^{an} $, and $\zeta_p = e^{\frac{2 \pi i}{p}}$, then

$$ g(-3,p) = (\frac{-3}{p})g(1,p) \\
\implies \frac{g(-3,p)}{g(1,p)} = (\frac{-3}{p})$$

Is this the correct way to go about proving a formula? And I am unsure when to use the statement $\zeta = e^{\frac{2\pi i}{3}}$. I know I somehow want to relate this to the fact that reciprocity shows I can evaluate this Legendre Symbol knowing if $p \equiv 1 \text{ or } 2 \text{ mod } 3$. Thanks for any help!

答:
(for $p\ge 5$) $-3$ is a square iff $\zeta\in \Bbb{F}_p$ iff $3| p-1$, where $\zeta$ is a primitive 3rd root of unity, ie. a root of $x^2+x+1\in \Bbb{F}_p[x]$, either in $\Bbb{F}_p$ (if $x^2+x+1$ is reducible) or in $\Bbb{F}_{p^2}=\Bbb{F}_p[t]/(t^2+t+1)$.

In fact together with $(\frac{-1}{p})= (-1)^{(p-1)/2}$ this is how we find $(\frac{3}{p})$.


问:
Is this the correct payoff function for mixed strategy matrix games?
Consider a mixed strategy matrix game where the payoff matrix is (for simplicity)

\begin{bmatrix}
 & a & b\\
a & u^1_{aa},u^2_{aa} & u^1_{ab},u^2_{ab}\\
b & u^1_{ba},u^2_{ba} & u^1_{bb},u^2_{bb}
\end{bmatrix}

Then we create the matrices $A,B$

where,

$ A =
\begin{bmatrix}
u^1_{aa} & u^1_{ab}\\
u^1_{ba} & u^1_{bb}
\end{bmatrix}$ $\quad B =
\begin{bmatrix}
u^2_{aa} & u^2_{ab}\\
u^2_{ba} & u^2_{bb}
\end{bmatrix}$

Let $x, y$ be the mixed strategy for player 1 and 2 respectively. Is the following the correct payoff function for player 1 and player 2?

Player 1 payoff: $x^T Ay$

Player 2 payoff: $y^T B^T x$

If so, I have some questions:

why is $x^TAy$ for player 1? Why is it not $y^TAx$?

why is there a transpose on $B$ for player 2's payoff?

答:
Both are correct. Recall that the payoff is a number so

$$ y^TB^Tx=(y^TB^Tx)^T=x^T(B^T)^T(y^T)^T=x^TBy$$

Personally, I would keep the same notation ($x$ always on the left with $^T$ sign) for both payoffs.


问:
Is it right that $\cup_{g\in G}gHg^{-1}\subsetneq G$, for every $H$, proper subgroup of infinite group $G$?
Let $G$ be a group and $H$, a proper subgroup of $G$ with $[G:H]<\infty$. Is it  right that $\bigcup\limits_{g\in G}gHg^{-1}$ is a proper subset of $G $? If $G$ is a finite group we can prove this claim, but is this right for infinite groups?

答:
It is false for some infinite groups. (Edit: The "false" here refers to the question asked in the title, not in the body where there is the additional condition of $H$ having finite index in $G$.) The simplest counterexample is $G = {\rm GL}_2(\mathbf C)$ and $H$ is the subgroup of upper-triangular matrices $\begin{pmatrix}a&b\\0&c\end{pmatrix}$ where $a, c \in \mathbf C^\times$.

Every $A \in {\rm GL}_2(\mathbf C)$ has an eigenvector in $\mathbf C^2$, say $v$ with eigenvalue $\lambda$: $Av = \lambda v$ and $v \not= \binom{0}{0}$. Let $w$ be a vector in $\mathbf C^2$ that is outside the line $\mathbf C v$. We can write $Aw = zv + z'w$ for $z$ and $z'$ in $\mathbf C$.  (The matrix $A$ might not have an eigenvector linearly independent of $v$, i.e., not all $2 \times 2$ complex matrices are diagonalizable, so we need not be able to pick $w$ as an eigenvector of $A$.)  The matrix representation of $A$ with respect to the basis $\{v,w\}$ of $\mathbf C^2$ is $\begin{pmatrix}\lambda &z\\0&z'\end{pmatrix}$, so $A$ is conjugate by an invertible matrix in $G$ to a $2 \times 2$ matrix in $H$. That proves $G = \bigcup_{g \in G} gHg^{-1}$.


问:
Solve cubic equation $x^3-9 x^2-15x-6 =0$ without going Cardano

Solve the cubic equation for $x\in\mathbb{R}$ $$x^3-9 x^2-15x-6 =0$$

Note that the only real solution is $x=3+2\sqrt[3]{7}+\sqrt[3]{7^2}$. Given the regularity of this solution, can we solve for it constructively, without going full Cardano?.

Also, can we prove that there is only one real solution without using the discriminant?

答:
Substitute $x=\frac1{t-1}$ to translate $x^3-9 x^2-15x-6 =0$ into
$$6t^3-3t^2-3t-1=0$$
and then rewrite it as $7t^3=(t+1)^3$,
which yields $t=\frac1{\sqrt[3]7-1}$ and in turn the solution
$$x= \frac1{t-1} =\frac{\sqrt[3]7-1}{2-\sqrt[3]7}=\sqrt[3]{49}+\sqrt[3]7+3
$$


问:
Wrong Reasoning about the problem of breaking a stick in $2$ points and build a triangle with the $3$ parts.
For homework I was asked to solve this classical problem "If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4" and ok, it must result $1/4$. But I can't figure out why. This is my reasoning (obviously wrong somewhere!). Let $X,Y$ be the points (picked uniformy at random on the stick)
There are $3$ case:

1) $0<X<Y<1$ with probability $\frac{1}{2}$ to happen

2) $0<Y<X<1$ with probability $\frac{1}{2}$ to happen

3) $0<Y=X<1$ which has probability $0$ to happen.

Consider the first case: ($X$ indicate now the length of the segment $[0,X]$)
the triangle inequalities leads to:

$X<(Y-X)+(1-Y) = 1-X \longrightarrow X<\frac{1}{2}$

$Y-X < (X) + (1 - Y) \longrightarrow Y-X < \frac{1}{2}$

$1-Y < Y \longrightarrow Y > \frac{1}{2}$

which leads to a probability of "success" of $\frac{1}{8}$

and the "global" probability of this event is $\mathbb{P}(\text{case} \ 1)*\mathbb{P}(\text{ I can build a triangle in this case})=$ $\frac{1}{16}$.
for the second case is the same.
the third case has probability $0$, so it doesn't give any contribution to the probability. According to my reasoning the global probability is $\frac{1}{8} \neq \frac{1}{4}$.

I can't find why my reasoning is wrong. I don't want a solution of the problem, but WHERE and WHY I've made an error. I know there are a lot of questions about this problem, but I need a correction in this reasoning not a complete solution, and I didn't found an answer in the other questions :) Thank you in advance.

答:
There's nothing wrong with your (corrected) calculation.  The triangular region specified by the three inequalities $x<\frac 12$, $y<x+\frac 12$, and $y>\frac 12$, does indeed have area $\frac{1}{8}$.  However, the total area under consideration is the portion of the $[0,1]$ square subject to $x<y$, which has area $\frac 12$.  Hence the conditional probability is their ratio, or $\frac 14$.

You've made things more complicated by splitting into cases this way.  Looking at the $[0,1]$ square, there are two triangular regions where the stick-breaking leads to a triangle; their total area is $\frac 14$.


问:
$y(t)=-(V+2)e^{-3t}+(V+3)e^{-2t}$ nonnegative for which $V$?
From $(V+3)e^{-3t}-(V+2)e^{-2t}\geq 0$, we get $(V+3)e^{-t} \geq V+2$. The answer in the textbook says $V\geq -3$. How does one come to this?

Context: this is problem 2.2.9 of Braun's Differential Equations and Their Applications.

Let $y(t)$ be the solution of the initial-value problem $$y''+5y'+6y=0; \ y(0)=1,\ y'(0)=V.$$ For what values of V does $y(t)$ remain nonnegative for all $t\geq0$.

Edit.
After @NeatMath pointed out in the comment, I saw that I misread my own notes. :)

We have $y(t)=c_1e^{-3t}+c_2e^{-2t}$.

Inserting the initial conditions, we have $y(0)=c_1+c_2=1$ and $-3c_1-2c_2=V$. Substituting $c_1=1-c_2$, into the first equation, we have
\begin{align}
-3(1-c_2) -2c_2 &= V \\
-3+c_2 &= V \\
c_2 &= V+3 \\
c_1&= 1-c_2 = 1-(V+3) = -(V+2).
\end{align}
Hence $y(t)=-(V+2)e^{-3t} + (V+3)e^{-2t}$.

答:
Using $y(t) = -(V + 2)e^{-3t} + (V + 3)e^{-2t}$, as pointed out in the comments:

$$y(t)\geq 0$$

$$(V + 3)e^{-2t} - (V+2)e^{-3t} \geq 0$$

$$(V + 3)e^{-2t} \geq (V+2)e^{-3t}$$

Dividing by the positive value $e^{-2t}$:

$$V + 3\geq (V + 2)e^{-t}$$

Here, we will split into cases:

Case 1: $V < -2$

If this is the case, then $V+ 2$ is negative. Dividing by this value:

$$\frac{V+ 3}{V + 2}\leq e^{-t}$$

$$1 + \frac{1}{V + 2}\leq e^{-t}$$

$e^{-t}$ may become arbitrarily close to $0$, and $V$ is independent of $t$, so we must have:

$$1 + \frac{1}{V + 2}\leq 0$$

$$-\frac{1}{V + 2}\geq 1$$

$$V + 2\geq -1$$

$$V\geq -3$$

Thus, $-3\leq V<-2$ works.

Case 2: $V\geq -2$

$V + 2$ is now positive. Dividing by this value:

$$\frac{V + 3}{V + 2}\geq e^{-t}$$

$$1 + \frac{1}{V + 2}\geq e^{-t}$$

Since $t\geq 1$, $e^{-t}\leq 1$. Combining this with $\frac{1}{V+ 2}>0$, we see that this inequality is always true, and thus $V\geq -2$ works.

Combining both cases, we see that $\boxed{V\geq -3}$


答:
Your mistake as at the end of your solution:
$-e^{-t}(t+1)-\theta e^{-t} = -e^{-t}(t + 1 + \theta)$, not $-e^{-t}(t + 1- \theta)$, as you write, so now $-e^{-t}((x-\theta)+1+\theta)=-e^{-t}(x+1)$.

Title: Can't find error in $\int{xe^{-(x-\theta)}}dx$

Body:
I haven't done integrals in some time and I'm kinda rusty. This is a very easy one, but for some reason I keep getting a slightly different result than the correct one. There has to be a sign error somewhere which I really can't find.

$$\int{xe^{-(x-\theta)}}dx $$
Let $t = x-\theta$. Then the above is:
$$\int{(t+\theta)e^{-t}dt = \int{te^{-t}dt}+\theta\int{e^{-t}dt} =-e^{-t}(t+1) -\theta e^{-t} = -e^{-t}(t-\theta+1) + c = -e^{\theta-x}(x-2\theta+1) + c}$$

However the correct results is $-e^{\theta-x}(x+1) +c$. The two thetas should cancel out but they don't. Where's my mistake?

Any help is appreciated.


问:
How do we prove the continuity of the inner product?
Let $\lim x_{k}=a$, $\lim y_{k}=b$ in $\mathbb{R^n}$ then it must be proved that $\lim \langle x_{k}, y_{k}\rangle = \langle a, b\rangle$, where $\langle a, b \rangle$ is the inner product in $\mathbb{R^n}$

The idea I have is that $ \langle x_{k}, y_{k} \rangle$ $=$ $\sum_{i=1}^{n} x_{i}^k y_{i}^k$ then $\lim \langle x_{k},y_{k}\rangle = \lim \sum_{i=1}^{n} x_{i}^k y_{i}^k$ $=$ $\lim x_{1}^ky_{1}^k+...+\lim x_{n}^ky_{n}^k$

That would be fine and being like that how I continue?

答:
Show that $f(x,y)=\|x+y\|^2$ and $g(x,y)=\|x-y\|^2$ are continuous from $(\Bbb R^n)^2$ to $\Bbb R.$

The inner product $<x,y>=\frac {f(x,y)-g(x,y)}{4}.$


问:
Greatest Integer Function/Floor Function Definition? (Discrete Mathematics)
I came across this set builder definition for the greatest integer function (which is also equal to the floor function) in my Discrete Mathematics course indicated below:

${[[x]]} = {\lfloor{x}\rfloor}$

${\lfloor{x}\rfloor} = {\max{\left\{{{{m}\in{\mathbb{Z}}}\mid{{m}\leq{x}}
}\right\}}}$

My question is - is this equivalent to the following?:

${\max{\left\{{{{m}\in{\mathbb{Z}}}\mid{{m}\leq{x}}}\right\}}} \overset{?}= {\max{\left({{{m}\in{\mathbb{Z}}}\mid{{m}\leq{x}}}\right)}}$

More directly, is this equivalent to the following?:

${\lfloor{x}\rfloor} \overset{?}= {\max{\left({{{m}\in{\mathbb{Z}}}\mid{{m}\leq{x}}}\right)}}$

答:
Technically, the $\max$ is taken over a set, which is why braces are used in the definition given.

In many situations, where it's clear, the set description is omitted. For instance,

$$\max_{x\in[0,1]} f(x)$$

is used to mean $\max\{f(x)\,|\,x\in[0,1]\}$. Parentheses are not commonly used.


问:
Evaluating real integral using complex analysis.
I'm trying to compute the following integral:
$$\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx$$

I'll not write down everything I've done, but choosing the branch cut on the positive real axes we have that:

$$\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx=\pi i \sum_{z_i}Res(f,z_i) \qquad z_i\in\{\pm \sqrt{i},\pm\sqrt{-i}\}$$

So we have to compute four residues.
My thought was changing the branch cut by putting it on the negative imaginary axes. We can do it by choosing $arg(z) \in (-\frac{\pi}{2},\frac{3\pi}{2}]$. So we have that:

$$(1+i)\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx=2\pi i \sum_{z_i}Res(f,z_i) \qquad z_i\in\{e^{i\frac{\pi}{4}},e^{i\frac{3\pi}{4}}\}$$

By doing this, we now need to compute only two residues. But I'm really finding difficulties in computing those residues: in fact I can't obtain the result I'm expecting. Can you please show me the computation and tell me if my argument was clear and correct?

Thanks in advance.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\int_{0}^{\infty}{\root{x} \over 1 + x^{4}}\,\dd x = {1 \over 4}\pi\sec\pars{\pi \over 8}} = {1 \over 2}\pi\root{1 - {\root{2} \over 2}} \approx 0.8501: {\Large ?}}$. Hereafter, I'll perform an evaluation of $\ds{\oint_{\cal C}{\root{z} \over 1 + z^{4}}\,\dd z}$ where $\ds{\cal C}$ is defined in each particular case for the chosen $\ds{\root{z}}$-branch cut.

$\ds{\Large\left.\mbox{a}\right)}$
The $\ds{\root{z}}$-branch cut is given by
$$
\root{z} = \root{\verts{z}}\expo{\ic\arg\pars{z}/2}\,,\quad
0 < \arg\pars{z} < 2\pi\,\quad z \not= 0
$$
which is the OP choice. Poles are given by
$\ds{p_{n} =
\expo{n\pi\ic/4}\ \mbox{with}\  n = 1,3,5,7}$.
Then,
\begin{align}
&\bbox[5px,#ffd]{\oint_{}{\root{z} \over
1 + z^{4}}\,\dd z} =
2\pi\ic\sum_{\braces{p_{n}}}{\root{p_{n}} \over 4p_{n}^{3}}
\\[5mm] = &\
-\,{1 \over 2}
\,\pi\ic\sum_{\braces{p_{n}}}p_{n}\root{p_{n}}
=
\pi\root{1 - {\root{2} \over 2}}\label{1}\tag{1}
\end{align}
Also,
\begin{align}
&\bbox[5px,#ffd]{\oint_{}{\root{z} \over
1 + z^{4}}\,\dd z} =
\int_{0}^{\infty}{\root{x} \over
1 + x^{4}}\,\dd x
\\[2mm] + & \require{cancel}
\cancel{\mbox{integration over arc with}\ \pars{\mbox{radius}\ \to \infty}}
\\[2mm] & + \int_{\infty}^{0}{\root{x}\expo{\ic\pi} \over
1 + x^{4}}\,\dd x =
2\int_{0}^{\infty}{\root{x} \over
1 + x^{4}}\,\dd x\label{2}\tag{2}
\end{align}
With (\ref{1}) and (\ref{2}):
\begin{align}
\bbox[5px,#ffd]{\int_{0}^{\infty}{\root{x} \over
1 + x^{4}}\,\dd x} & =
\bbx{{1 \over 2}\pi\root{1 - {\root{2} \over 2}}}
\approx 0.8501 \\ &
\end{align}

$\ds{\Large\left.\mbox{b}\right)}$ I guess the following evaluation is the simplest one because it involves just ONE pole: The integration is performed along a quarter circle in the complex plane first quadrant. The $\ds{\root{z}}$-branch cut is given by
$$
\root{z} = \root{\verts{z}}\expo{\ic\arg\pars{z}/2}\,,\quad
-\pi < \arg\pars{z} < \pi\,\quad z \not= 0
$$
which is the principal one. The contour encloses the pole
$\ds{p = \expo{\pi\ic/4}}$.
Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
{\root{x} \over 1 + x^{4}}\dd x}
\\[5mm] = &\
2\pi\ic\,{p^{1/2} \over 4p^{3}} -
\int_{\infty}^{0}
{\root{y}\expo{\pi\ic/4} \over 1 + y^{4}}\,
\ic\,\dd y
\\[5mm] = &\
-\,{1 \over 2}\,\pi\ic\,
\expo{3\pi\ic/8} +
\ic\expo{\pi\ic/4}\int_{0}^{\infty}
{\root{y} \over 1 + y^{4}}\,\dd y
\\[5mm] \implies &\
\int_{0}^{\infty}
{\root{x} \over 1 + x^{4}}\dd x =
{\pars{-\pi\ic/2}
\expo{3\pi\ic/8} \over 1 - \ic\expo{\pi\ic/4}}
\\[5mm] = &\
\bbx{{1 \over 2}\pi\root{1 - {\root{2} \over 2}}}
\approx 0.8501 \\ &
\end{align}

$\ds{\Large\left.\mbox{c}\right)}$

Ramanujan's Master Theorem:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
{\root{x} \over 1 + x^{4}}\dd x}
\,\,\,\stackrel{x^{4}\ \mapsto\ x}{=}\,\,\,
{1 \over 4}\int_{0}^{\infty}
{x^{\color{red}{3/8} - 1} \over 1 + x}\dd x
\end{align}
Note that
$\ds{{1 \over 1 + x} =
\sum_{k = 0}^{\infty}\pars{-x}^{k} =
\sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{1 + k}}{\pars{-x}^{k} \over k!}}$.

Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
{\root{x} \over 1 + x^{4}}\dd x} =
{1 \over 4}\bracks{\Gamma\pars{3 \over 8}
\Gamma\pars{1 - {3 \over 8}}}
\\[5mm] = &
{1 \over 4}\,{\pi \over \sin\pars{3\pi/8}} =
{1 \over 4}\,\pi\sec\pars{\pi \over 8}
\\[5mm] = &\
\bbx{{1 \over 2}\pi\root{1 - {\root{2} \over 2}}}
\approx 0.8501 \\ &
\end{align}


问:
Relation between set and inverse transformation applied on set
I have problem with the following exercise. Let $f:M \rightarrow N$ a map and $M' \subseteq M$ and $N' \subseteq image(f) \subseteq N$. Show $M' \subseteq f^{-1}(f(M'))$. Intuitively, I can imagine the to be proven statement is true but I can't show how this is possible. So I try:

$f^{-1}: N \rightarrow M$

Let $x \in M' \rightarrow x \in M$.

$f(M') \subseteq N \rightarrow f^{-1}(f(M')) \subseteq M$

I thought if i can somehow prove $M \subseteq f^{-1}(f(M'))$, then I can also prove the statement. However,  $M \subseteq f^{-1}(f(M'))$ doesn't make sense to me ... Any tips or suggestions would be great.

答: We have to show $M' \subset  f^{-1}(f(M'))$. Let $x$ be an element of $M'$. Let's show that $x \in f^{-1}(f(M'))$. To show this, we must show that $f(x) \in f(M')$. This is obvious as $x$ belongs to $M'$. That's the proof!


问:
Is my $\epsilon$-$\delta$ calculation correct?
I have to show that $\lim_{x \to 1} x^4-1 =0$. Here is how i have done it:
$\mid x^4-1 \mid = \mid x-1 \mid\mid x+1 \mid\mid x^2+1 \mid < \epsilon \qquad$ and since we are close to 1, we can assume that the $\delta$-neighborhood of $c=1$ must be havea radius of max $\delta =1$ which implies that :   $\mid x+1 \mid \le 2 \quad and \mid x^2+1 \mid \le 2 \quad \forall x \in V_{\delta}(c) \quad$

We now choose $\delta=min \left \{1,\frac{\epsilon}{4}\right\} \quad$ and we can conclude that if $\mid x-1 \mid < \delta$, it follows that $\mid x^4-1 \mid = \mid x-1 \mid\mid x+1 \mid\mid x^2+1 \mid < 4\frac{\epsilon}{4} =\epsilon$. Is this calculation correct? Do I miss something? Or some details?

答:
You want
$$
|x^2+1||x+1||x-1|<\varepsilon
$$
near $x=1$.

First step is to control the quantity $|x^2+1||x+1|$ near $x=1$. So first
restrict $x$ so that $|x-1|<1=\delta_1$. This means
$0<x<2$, and hence $|x+1|<3$ while $x^2+1<5$. Hence
$$
|x-1|<1=\delta_1 \quad\Longrightarrow\quad |x^2+1||x+1|<15,
$$
and thus
$$
|x-1|<1=\delta_1 \quad\Longrightarrow\quad |x^4-1|=|x^2+1||x+1||x-1|<15|x-1|,
$$
Now, we have
$$
|x-1|<\delta=\min\Big\{1,\frac{\varepsilon}{15}\Big\}\quad\Longrightarrow\quad
|x^4-1|<15|x-1|<15\cdot\frac{\varepsilon}{15}=\varepsilon.
$$


问:
Problem solving a system of equations
How can I solve this equation system for x and y?

$\frac{xy}{x+1} = a$

$\frac{xy^2}{x+2} = b$

$a>0$  $b>0$

I've tried to isolate $y$ of both equations and then equal the result to solve $x$ but then I can't isolate $x$. I've also tried to combine the two equations but I get nothing.

答:
Note that
$$
y=\frac{a(x+1)}{x}\ \text{and}\
y^2=\frac{b(x+2)}{x}
$$
Therefore,
$$
\frac{a^2(x+1)^2}{x^2}=\frac{b(x+2)}{x}
$$
that is
$$a^2(x+1)^2=bx(x+2)$$
Hence
$$
(a^2-b)x^2+2(a^2-b)x+a^2=0
$$
Now solve this quadratic equation.

Title: Problem solving a system of equations


问:
If $\dim (V/ W) = 1$, what can we say about $\dim(\overline{V}/\overline{W})$
Let $X$ be a Banach space and $W\subset V \subset X$ are two subspaces of $X$. Suppose now we know $\dim(V/ W) = 1$, then what can we say about the dimension $\dim(\overline{V}/\overline{W})$? Here $\overline{V}$ and $\overline{W}$ are the closures of $V$ and $W$ in $X$, respectively. Is it true that $\dim(\overline{V}/\overline{W})\leq1$?

Any hint will be appreciated. Thanks!

答:
Indeed it follows that $\dim (\overline{V}/\overline{W}) \leqslant 1$.

Let $x \in V \setminus W$. Then, since $\dim (V/W) = 1$, we have $V = W + \mathbb{K}\cdot x$ (where $\mathbb{K}$ is the scalar field, be that $\mathbb{R}$ or $\mathbb{C}$). Note that this is just the vector space sum, it need not be a topological sum.

Now, the sum of a closed subspace and a finite-dimensional subspace is closed, hence $U = \overline{W} + \mathbb{K}\cdot x$ is a closed subspace of $X$. And $V \subseteq U$, thus $\overline{V}\subseteq U$. On the other hand we have $\overline{W} \subseteq \overline{V}$ and $\mathbb{K}\cdot x \subseteq V \subseteq \overline{V}$, hence $U \subseteq \overline{V}$. Together it follows that $\overline{V} = \overline{W} + \mathbb{K}\cdot x$, whence $\overline{V}/\overline{W}$ is generated by $x + \overline{W}$, hence its dimension  at most $1$.

More precisely, we see that $\dim (\overline{V}/\overline{W}) = 1$ if and only if $W$ is relatively closed in $V$, i.e. $W = \overline{W} \cap V$, and $\overline{W} = \overline{V}$ otherwise.


答:
Square the first equation
\begin{eqnarray*}
\frac{x^2y^2}{(x+1)^2}=a^2.
\end{eqnarray*}
Now rewrite the second equation
\begin{eqnarray*}
\frac{x^2y^2}{(x+1)^2}  \times\frac{(x+1)^2}{x(x+2)} =b.
\end{eqnarray*}
So we have the quadratic
\begin{eqnarray*}
a^2(x+1)^2=bx(x+2)
\end{eqnarray*}
which is easily solved.

Title: Problem solving a system of equations

Body:
How can I solve this equation system for x and y?

$\frac{xy}{x+1} = a$

$\frac{xy^2}{x+2} = b$

$a>0$  $b>0$

I've tried to isolate $y$ of both equations and then equal the result to solve $x$ but then I can't isolate $x$. I've also tried to combine the two equations but I get nothing.


问:
Continuous of $\sqrt{z}$ in $\mathbb C$
Consider the function $f(z)=\sqrt{z}$, $f:\mathbb C\to \mathbb C$, where by $\sqrt{z}$ we mean the unique $w\in\mathbb C$ such that $z=w^2$ and $0<Re(w)$ if $w\notin(-\infty,0]$ or $w=i\sqrt{-z}$ for $w\in(-\infty,0]$.

I know that $f$ is analytic in every $z$ such that $0<Re(z)$. Is $f$ is continues in every point in $\mathbb C$?

答:
You have chosen the branch cut that gives you the following data:

$z = r \mathrm{e}^{i \varphi}, \quad r \ge 0, \quad -\pi < \varphi \le \pi,\quad \sqrt{z} = \sqrt{r} \mathrm{e}^{i \varphi/2}\ $

So, $(r_n,\varphi_n)=(1,\pi-1/n)\to (1,\pi)$ and $\sqrt z_n\to (1,\pi/2).$

And $(r_n,\varphi_n)=(1,-\pi-1/n)\to (1,-\pi)$ and $\sqrt z_n\to (1,-\pi/2).$

But passing to rectangular coordinates, this means that as $z_n\to (-1,0)$ along the arc of the unit circle in the second quadrant of $\mathbb C,$ $\sqrt z_n\to (1,i)$, whereas as $z_n\to (-1,0)$ along the arc of the unit circle in the third quadrant of $\mathbb C,\ \sqrt z_n\to (1,-1).$

This argument works for any $r>0,$ so the root function, with the branch cut you chose, is not continuous on the negative $x-$ axis.

Title: Continuous of $\sqrt{z}$ in $\mathbb C$


问:
For a complex measure m, why is m(B = E - A) = m(E) - m(A)?
So, this should be a little obvious, but not to me at the moment.

Why is $m(B = E - A) = m(E) - m(A)$?

Of course, all given sets are measurable.

Thanks.

答: In general $m(B-A) \neq m(B)-m(A)$. If $B$ and $A$ are disjoint and $m(B)=m(A)>0$, then $m(B)=m(B-A)>0$ yet $m(B)-m(A)=0$.</p>


问:
Every manifold admits a vector field with only finitely many zeros
Let $M$ be a smooth manifold. I am trying to prove that $M$ admits a vector field with only finitely many zeros.

This will follow if we can find a function $f : M\rightarrow \mathbb R$ such that $df$ has only finitely many zeros, but I cannot find such a function with this property either. My initial idea was to try to embed $M$ in $\mathbb R^N$ for some $N$ and look at $x\mapsto u \cdot x$ for fixed $u\in \mathbb R^N$, but I could not find a way to prove that there must be a $u$ such that the differential of this map has only finitely many zeros.

Does anyone have an elementary construction of such a vector field (or function)?

答: A classic method - I think of Steenrod -  is to triangulate the manifold then form the vector field whose singularities are the barycenters of the triangulation.For instance on a triangle the field flows away from the barycenter of the triangle towards the vertices and the centers of the edges. along the edges the field flows away from the centers towards the vertices. Draw a picture. It is easy to see.</p>


问:
How to get the integral of $\log(\det(A + Bt))$ w.r.t variable t?
Suppose we have two positive definite matrices $A$ and $B$, now I want to get the integral of:

\begin{align}
\int_{a}^{b} \log(\det(A + Bt)) dt ~~~~~~~~~~~~\text{for } a, b > 0
\end{align}
Analytic solution is better, but good approximation is also acceptable here. I tried to approximate this integral via interpolation, but as $A$, $B$ are of large size (i.e, 100 × 100), that maybe too slow to calculate $\det(A + Bt)$ every time. When I seek the analytics solution, I met the problem of finding: $\frac{d(\det(A + Bt))}{dt}$. If you have any ideas or inference on this question, welcome to your answer and suggestions, thank so much!!

答:
In general, $\det X$ is the product of the eigenvalues of $X$ (counted with multiplicity). Note the eigenvalues of $I+X$ are $1+\lambda$ where $\lambda$s are the eigenvalues of $X$. Therefore

$$ \begin{array}{ll} \ln\det(A+Bt) & =\ln\big[\det A\,\det(I+A^{-1}Bt)\big] \\
& =\ln\det A+\ln\prod(1+\lambda t) \\ & = \ln\det A+\sum\ln(1+\lambda t). \end{array}$$

where $\lambda$s run over the eigenvalues of $X=A^{-1}B$.

A constant $\ln\det A$ is easy to integrate and $\ln(1+\lambda t)$ can be integrated elementarily, yielding

$$ t\ln\det A+\sum\left[\frac{1+\lambda t}{\lambda}\ln(1+t\lambda) - t\right]+C. $$

If $X$ is diagonalizable, this can be written as

$$ t\big(\ln\det A-n\big)+\mathrm{tr}\left[(B^{-1}A+t)\ln(I+tA^{-1}B)\right]+C, $$

presumably a density argument implies this also holds if $X$ is not diagonalizable (i.e. the expression above is continuous, and diagonalizable matrices are dense, so...)


答:
Hint

When you took $$\delta=\min(\color{red}{1},...)$$

this means that you assumed that
$$|x-1|<\color{red}{1}$$

which gives
$$-1<x-1<1 \iff 1<x+1<3$$

So, your inequality $|x+1|<2$ should be replaced by $ |x+1|<3$

and

$|x^2+1|<2 $ by $ |x^2+1|<5$.

Title: Is my $\epsilon$-$\delta$ calculation correct?

Body:
I have to show that $\lim_{x \to 1} x^4-1 =0$. Here is how i have done it:
$\mid x^4-1 \mid = \mid x-1 \mid\mid x+1 \mid\mid x^2+1 \mid < \epsilon \qquad$ and since we are close to 1, we can assume that the $\delta$-neighborhood of $c=1$ must be havea radius of max $\delta =1$ which implies that :   $\mid x+1 \mid \le 2 \quad and \mid x^2+1 \mid \le 2 \quad \forall x \in V_{\delta}(c) \quad$

We now choose $\delta=min \left \{1,\frac{\epsilon}{4}\right\} \quad$ and we can conclude that if $\mid x-1 \mid < \delta$, it follows that $\mid x^4-1 \mid = \mid x-1 \mid\mid x+1 \mid\mid x^2+1 \mid < 4\frac{\epsilon}{4} =\epsilon$. Is this calculation correct? Do I miss something? Or some details?


问:
converges to infinity in complex plain
Let $(z_1,z_2,z_3,\ldots)$ be a sequence of complex numbers. Is it reasonable to define $\lim_{n\to\infty} z_n=\infty$ whenever $\lim_{n\to\infty}|z_n|=\infty$?

Under this definition some results which are true in $\mathbb R$ are no longer true in $\mathbb C$. For instance, it is no longer true that $\infty+\infty=\infty$ since $z_n=(-2)^n+2^n$ have no limit, while $2^n$ and $(-2)^n$ converges to $\infty$.

答: Yes, it absolutely makes sense. It is called the Alexandroff extension or one-point compactification of the space. In case of the complex plane you obtain the Riemann sphere, which is well-known and an important tool in complex analysis. There are also other compactifications like the radial compactification and the Stone-Čech compactification, where you add more than just one point at infinity. But you are right that either way you can't reasonably extend your usual operations to the extension.</p>

Title: converges to infinity in complex plain


问:
Graph theory - The clients can still communicate with all the remaining servers
I have the following graph problem that I have no idea how to solve. Could you help me please? I need a proof.

In a computer network there are two types of nodes: client and server. Each client is directly
connected (using network cables) to at least one server and there are no direct connections between
clients. Suppose that each server can route messages to other servers with which it is directly connected
and the entire network is connected. In order to make the network more reliable, a scenario is considered:

Every client is directly connected to at least 2 servers and every pair of servers having the
distance between them 2 are connected using a direct cable.

Prove that, in this new network, if
a server becomes unavailable the clients can still communicate with all the remaining servers.

答:
Let $S$ be the server the became unavaible.

If a client was not connected to $S$, it's still connected to at least $2$ other servers, and each of those is connected to all other servers.

If a client was connected to $S$, it's still connected to at least one other server, and each of those is connected to all other servers.

Regardless of the situation, the client can communicate to all other servers.


问:
Calculate expected value of Y=max{X, 1/2} in a uniform distribution
Random variable $X$ has uniform distribution over $[0, 1]$. Random variable $Y=\max(X, 1/2)$. Calculate the expected value of $Y$.

I have PDF, so:

$f(x) = 1$ for the $x \in [0, 1]$

$f(x) = 0$ for the $x \notin [0, 1]$

Also, I calculated the expected value of $X$:

$E(X) = \int_{0}^{1} f(x) x\,dx = 1/2* x |_{0}^{1} = 1/2 $

So I'm not sure how to handle $\max()$. I'm thankful for any tips.

At the side of the questions, while searching for tips I encountered that $\Pr(\max(A, B) \leq x)=\Pr(A \leq x)\cdot \Pr(B \leq x) = x^2$ over uniform distribution, but I can't see why it's $x^2$.

答:
There are several approaches to this problem. I think the simplest would be to condition on the event that $X$ is smaller than $0.5$ (so $Y=0.5$). Denote this event by $A$.
Clearly, $\Pr(A)=0.5$ and $E(Y\vert A)=0.5$.

Now, given $\overline{A}$, we know that $X>0.5$ and by standard conditioning*, $X$ is uniform $(0.5,1)$ with expectation $0.75$.

$$E(Y)=E(Y\vert A)\Pr(A)+E(Y\vert \overline{A})\Pr(\overline{A})=0.5\cdot 0.5+0.75\cdot 0.5=\tfrac{5}{8}$$

*The conditioning. For $z\in(0.5,1)$, $\Pr(X<z\vert \overline{A})=\tfrac{\Pr(X<z \cap \overline{A})}{\Pr(\overline{A})}=\tfrac{\Pr(X<z \cap x>0.5)}{0.5}=\tfrac{z-0.5}{0.5}$ which is the CDF of a uniform random variable over $(0.5,1)$.


问:
Where does the error lie here?
Initially assume that the area $a$ and circumference $c$ of a radius $R>0$ circle are given axiomatically as:

$a=πR^2$ and $c=2πR$

Now in order to calculate the surface area of the top hemisphere (without its base) of a radius $R$ sphere, we approximate it with the following sum:

$$A=\sum_{i=1}^n2\pi r_iR\Delta\theta _i$$

Where $r_i$ is the radius of the disk on the plane $z=Rcos{\theta}_i$ and ${\theta}_i$ is the complementary to the polar angle of a point that is both on that plane and the hemisphere surface.

$A'$ as it can easily be proven, when our partition length $\Delta\theta_i$ tends to zero for every $i$, converges to the integral:

$$\int_{0}^{\frac{π}{2}}2πR^2cos\theta d\theta=2πR^2$$

The surface area of the entire sphere is naturally twice that, hence the familiar formula:

$$A=4πR^2$$

Now it would be tempting to approximate the volume of the hemisphere by partitioning it as indicated by the following sum:

$$V'=\sum_{i=1}^nπ{r_i}^2R\Delta{\theta}_i$$

We get the area of the disk at a given point and we multiply that by the arc length that is directly in-between that disk and the one directly after it, with respect to our partitioning. Basically, a direct continuation of the process we followed for the derivation of the surface area of the sphere, instead there we had the circumference of the disk, whereas now we substitute that for its area.

But if we get the Riemann integral that corresponds to the sum $V'$, we get an obviously false result for the volume of the hemisphere:

$$V=πR^3\frac{π^2}{4}$$

When we know that actually: $$V=πR^3\frac{2}{3}$$

(Again, the last two values are associated with the hemisphere, ie half the sphere)

So, it is evident that there is something wrong with the assumption that we can generalize the first process, to derive the volume of the hemisphere. My question in a few words is:

Where, exactly, does that assumption fail?

答: When you calculated the area, $R\Delta\theta_i$ was a segment along the surface, "perpendicular" to the circle of radius $r_i$. If you follow the same procedure, you would need to multiply the area of the disk with radius $r_i$ with a segment perpendicular to the disk. $R\Delta\theta_i$ is tilted at an angle $\pi/2-\theta$. To get the perpendicular component, you need to multiply by $\sin\theta$. Then $$V=\int_0^{\pi/2}\pi R^3\sin^2\theta\sin\theta d\theta=\pi R^3\frac 23$$

Title: Where does the error lie here?


问:
Verbal Subgroups of $\mathbb{Z}^2$
So I want to try to show that all the verbal subgroups of an abelian group $G$ are of the form $G(X^n), n \geq 1$. I want to start with $\mathbb{Z}$ and $\mathbb{Z}^n$. So I worked with $\mathbb{Z}$, I know that all the subgroups of $\mathbb{Z}$ are $m\mathbb{Z} = $ $\langle {m} \rangle$ $ = \mathbb{Z}(X^m)$, so then all the subgroups of $Z$ are verbal.

Now if I look at $\mathbb{Z}^2$, the subgroups of this are:

$(0,m) \mathbb{Z} = \langle (0,m) \rangle$

$(m,0) \mathbb{Z} = \langle (m,0) \rangle$

$(a,b) \mathbb{Z} = \langle (a,b) \rangle$ $ a,b \in \mathbb{Z}-0$

$(a,b) \mathbb{Z} + (c,d) \mathbb{Z} = \langle (a,b) \rangle + \langle (c,d) \rangle$

Now I know that a verbal subgroup must be fully characteristic, so i know the first 3 subgroups are not verbal under the endomorphism $f(x,y) = (y,x)$. In the third case if $a = b$ then it fails under the endomorphism $g(x,y) = (x+1,y+2)$.

The last case, I'm not really sure about, I could show it fails in general with $f$ as above but perhaps in certain cases it would work? But then how do I show it is verbal , I'm not even sure how to begin to write any of these subgroups as $\mathbb{Z}(X^n)$, it doesn't seem like you can, at least not as easily as we did for $\mathbb{Z}$.

Am I missing a subgroup of $\mathbb{Z}^2$? How could I extend this to a general $\mathbb{Z^n}$?

Edit: Definition of verbal subgroup. Let $W$ be a set of words, then the verbal subgroup $G(W)$ of a group G, is the subgroup of G generated by the words $w \in W$, written with elements of G.

i.e. $G(X^2) = \langle g^2 | g \in G\rangle$ or $G(X^2, XYX) = \langle g^2, ghg | g,h \in G\rangle$

答:
I wouldn't bother with $\mathbb{Z}^n$, and instead start at full generality.

First off, any word in $x_1,\cdots,x_m$ can be simplified to the form $x_1^{a_1}\cdots x_m^{a_m}$ because $G$ is abelian. Second, we can set all but one of the $x_i$s to be $e$ to say the verbal subgroup contains any of the powers $x^{a_i}$, and conversely if it contains any of those powers then of course it contains the word $x_1^{a_1}\cdots x_m^{a_m}$. Therefore, the verbal subgroup is WLOG generated by a bunch of powers $x^{a_1},\cdots,x^{a_m}$. Now, we can take these themselves to powers and multiply them together to get any $x^{a_1b_1+\cdots+a_mb_m}$ (for given choice of $x$). This in turn means the verbal subgroup contains all powers $x^g$ where $g=\gcd(a_1,\cdots,a_m)$, but conversely all $a_i$s are multiples of $g$ so...

Title: Verbal Subgroups of $\mathbb{Z}^2$


问:
People sitting versus one another
Find number of ways to place 8 people around circular table so two specific persons sit versus one another.
Correct answer is $6!$. I tried placing two of them versus one another, and then I have exactly $3$ persons between them. Those $3$ we can choose on $6\choose3$ ways, but I am not able to get $6!$.

答:
Suppose the people are Alice and Bob. After sitting them you can count six places (clockwise) from Alice to Bob (3 to Bob, 3 from Bob to Alice). Sitting the 6 people in these 6 seats is exactly $6!$ options.

For your method: you need to choose the 3 that will sit between then (from Alice to Bob CW) then sit them, then sit the rest: ${6 \choose 3}\cdot 3! \cdot 3!$

Title: People sitting versus one another


答: The number of circular permutations of six people is $\frac{6!}{6}=5!.$ Then according to your defintion of "versus" meaning opposing one another, there are $6$ such ways to place the remaining two people in between the other six people. So the answer is $6\cdot 5!=6!.$

Title: People sitting versus one another

Body:
Find number of ways to place 8 people around circular table so two specific persons sit versus one another.
Correct answer is $6!$. I tried placing two of them versus one another, and then I have exactly $3$ persons between them. Those $3$ we can choose on $6\choose3$ ways, but I am not able to get $6!$.


问:
$y(t)=-(V+2)e^{-3t}+(V+3)e^{-2t}$ nonnegative for which $V$?
From $(V+3)e^{-3t}-(V+2)e^{-2t}\geq 0$, we get $(V+3)e^{-t} \geq V+2$. The answer in the textbook says $V\geq -3$. How does one come to this?

Context: this is problem 2.2.9 of Braun's Differential Equations and Their Applications.

Let $y(t)$ be the solution of the initial-value problem $$y''+5y'+6y=0; \ y(0)=1,\ y'(0)=V.$$ For what values of V does $y(t)$ remain nonnegative for all $t\geq0$.

Edit.
After @NeatMath pointed out in the comment, I saw that I misread my own notes. :)

We have $y(t)=c_1e^{-3t}+c_2e^{-2t}$.

Inserting the initial conditions, we have $y(0)=c_1+c_2=1$ and $-3c_1-2c_2=V$. Substituting $c_1=1-c_2$, into the first equation, we have
\begin{align}
-3(1-c_2) -2c_2 &= V \\
-3+c_2 &= V \\
c_2 &= V+3 \\
c_1&= 1-c_2 = 1-(V+3) = -(V+2).
\end{align}
Hence $y(t)=-(V+2)e^{-3t} + (V+3)e^{-2t}$.

答:
$e^t(V+3)\geqslant V+2 \Rightarrow V \geqslant \frac{2-3e^t}{e^t-1}, \forall t>0 \Rightarrow V \geqslant \lim_{t\to \infty} \frac{2-3e^t}{e^t-1} = -3$.

On the other hand if $V\geqslant -3$, then $e^t(V+3)\geqslant V+2.$


问:
finding the probability density function of $ dY_t = - Y_t X_t dW_t$
Could someone point me to where I can learn how to derive the stationary distribution for the martingale $Y_t$ which itself has stochastic volatility drive by $X_t$:

\begin{align}
dY_t &= - Y_t\ X_t\ dW^Y_t  \\
dX_t &= - \kappa X_t dt + \sigma dW^X_t
\end{align}

where the brownians $W_t^Y$ and $W_t^X$ are uncorrelated. Many thanks in advance !

答:
I am not sure I can give a reference for this particular question, but it can be done easily enough.

If the two Brownian motions are independent you can find the probability distribution for $Y_t$ conditioned on a path of $X_t$. You can write
$$ \log Y_t = \log Y_0 - \int_0^t X_t dW^Y_t - \frac12 \int_0^t X_t^2\,dt. $$
The point is that $\int_0^t X_t^2\,dt = Z_t$ is just a random variable and, conditioned on a particular path of $X_t$, the first integral is an Ito integral of a deterministic function, so
$$ \log Y_t = \log Y_0 - N(0, Z_t) - \frac12 Z_t.$$
So the distribution of $\log Y_t$ conditioned on the value of $Z_t=\int_0^t X_t^2\,dt$ is a normal distribution, as above, and the distributions of $X_t$ and $Z_t$ you can calculate independently of this question.

The particular $X_t$ that you wrote down is an Ornstein-Uhlenbeck process. See Wikipedia, for example, which also gives the pdfs for $X_t$, $X_t^2$, etc., from which you can find the distribution of $Z_t$.


问:
Is it right that $\cup_{g\in G}gHg^{-1}\subsetneq G$, for every $H$, proper subgroup of infinite group $G$?
Let $G$ be a group and $H$, a proper subgroup of $G$ with $[G:H]<\infty$. Is it  right that $\bigcup\limits_{g\in G}gHg^{-1}$ is a proper subset of $G $? If $G$ is a finite group we can prove this claim, but is this right for infinite groups?

答:
Yes it is correct. Consider the action of $G$ by left multiplication on the finite set $X$ of left cosets of $H$ in $G$. This action is transitive. The action defines a homomorphism $\phi:G \to {\rm Sym}(X)$.

Then image $P$ of $\phi$ is a transitive group group on the finite set $X$, and there must exist an element $a \in P$  without fixed points. (This follows from the result that the average number of fixed points in a transitive action on a finite set is $1$, and the identity fixes more than one point - since $H$ is a proper subgroup we have $|X| > 1$).

Now, if $b \in gHg^{-1}$ for some $g \in G$, then $b(gH) = gH$, so $\phi(b)$ fixes $gH$. Hence an element $b \in G$ with $\phi(b) = a$ does not lie in $\cup_{g \in G} gHg^{-1}$.


答:
You can also express $x^4-1$ in powers of $x-1$ as in
$$|x^4-1| = |((x-1)+1)^4-1| = |(x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)|.$$
If we assume $|x-1| < \delta$, the triangle inequality gives
$$|x^4-1| \le |x-1|^4+4|x-1|^3+6|x-1|^2+4|x-1| < \delta^4+4\delta^3+6\delta^2+4\delta.$$
Now we want $\delta^4+4\delta^3+6\delta^2+4\delta \le \varepsilon$. Since $$\delta^4+4\delta^3+6\delta^2+4\delta =\delta(\delta^3+4\delta^2+6\delta+4)$$
we only have to bound the second factor since the first is merely $\delta$. If we set $\delta = \min\left\{1, \frac\varepsilon{15}\right\}$, we have
$$|x^4-1| \le \delta(\delta^3+4\delta^2+6\delta+4) \le \delta \cdot (1+4+6+4) = 15\delta \le 15\cdot \frac\varepsilon{15} = \varepsilon.$$


问:
Applying reduction formula to integrals
Question

For $n \geq 0$, let $$I_n = \int_{0}^{8} x^n(8 - x)^\frac 1 3\ \mathrm{d}x\ .$$

Show that, for $n \geq 1$, $$I_n = \frac {24n} {3n + 4} I_{n - 1}\ .$$

My working

Let $u = x^n$ and $v' = (8 - x)^{\frac 1 3}$

$\implies u' = nx^{n - 1}$ and $v = -\frac 3 4 (8 - x)^{\frac 4 3}$

$\implies I_n = [-\frac 3 4 x^n(8 - x)^{\frac 4 3}]^{x = 8}_{x = 0} + \frac 3 4 n \int_{0}^{8} x^{n - 1}(8 - x)^\frac 4 3\ \mathrm{d}x$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac 3 4 n \int_{0}^{8} x^{n - 1}(8 - x)^\frac 1 3(8 - x)\ \mathrm{d}x$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac 3 4 n \int_{0}^{8} [8x^{n - 1}(8 - x)^\frac 1 3 - x^n(8 - x)^{\frac 1 3}]\ \mathrm{d}x$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = 6nI_{n - 1} - \frac 3 4 n I_n$

$\implies (1 + \frac 3 4)I_n = 6nI_{n - 1}$

$\therefore I_n = \frac {24n} {3n + 4} I_{n - 1}$

I try to use integration by parts, but with my choices of $u$ and $v'$, I do not seem to be getting anywhere. I am currently taking an introductory module to calculus in college. In class, we have only gone through the reduction formula for $\sin$ and $\cos$ and I have been stuck at this question for quite some time. Any hints/intuitions as to how to proceed will be greatly appreciated :)

Edit

Following some help from Phicar, I have finally got it! :) (I have edited my post to include my updated working)

答:
$$I_n=\int_0^8x^n(8-x)^{1/3}dx$$
$x=8u, dx=8du$
$$I_n=2\int_0^1\left(\frac u8\right)^n(1-u)^{1/3}du=2^{1-3n}\int_0^1u^n(1-u)^{1/3}du=2^{1-3n}B\left(n+1,\frac43\right)$$
you can use this as a quick sanity check for any values :)

Title: Applying reduction formula to integrals


问:
How to calculate the unconditional probability of a R.V. given the the probability of the R.V. conditioned on two events?
Consider the following problem:

Let $X \sim \mathcal{N}(0,1)$. $Y$ is conditioned on a flip of a fair
coin (independent from  $X$). In the case of Heads, we let  $Y=X$. In
the case of Tails, we let $Y=−X$. Determine $f_Y$

Here's my attempt:

\begin{eqnarray}
f_Y(y) &=& f_{Y|Heads}\mathbb{P}(Heads) + f_{Y|Tails}\mathbb{P}(Tails)\\
\end{eqnarray}

Considering the fact that $f_{Y|Tails} = - f_{Y|Heads}$

\begin{eqnarray}
f_Y(y) &=& f_{Y|Heads} \left[ \mathbb{P}(Heads) - \mathbb{P}(Tails)\right]\\
f_Y(y) &=& f_X(x) \left[ \mathbb{P}(Heads) - \mathbb{P}(Tails)\right]\\
\end{eqnarray}

Since it is a fair coin, $\mathbb{P}(Heads) = \mathbb{P}(Tails)$, then

\begin{eqnarray}
f_Y(y) &=& 0\\
\end{eqnarray}

However, I couldn't fully make sense of this solution. Have I made any mistakes throughout the resolution? If so, how could I have approached this problem differently?

答:
$F_Y(t)=\Pr(Y\leq t)=\Pr(X\leq t)\Pr(H) + \Pr(-X\leq t)\Pr(T)=0.5[\Pr(X\leq t) + \Pr(X\geq -t)]$

From symmetry, $\Pr(X\geq -t)=\Pr(X\leq t)$ so $F_Y(t)=\Phi(t)$ which means that $Y\sim N(0,1)$.


问:
Finding the limits to minus and plus infinity
I need to calculate the $\lim_{x\to-\infty}$ and $\lim_{x\to+\infty}$ of the function:

$$f(x)=\begin{cases} \dfrac{x^2-3x^3e^{2x}}{x^4+3},x<1 \\ \dfrac{3x^2+\cos(3x+1)}{x^2+2},x≥1 \end{cases}$$

I've attempted to use my knowledge that $-1\le \cos(x)\le1$ to conclude that:

$$\dfrac{3x^2-1}{x^2+2}\le\dfrac{3x^2+\cos(3x+1)}{x^2+2}\le\dfrac{3x^2+1}{x^2+2}$$

And since the second case should only be interested in positive values, then I've also conclude that:

$$\lim_{x\to+\infty}\dfrac{3x^2+1}{x^2+2} = 3 \Rightarrow \lim_{x\to+\infty} f(x) = 3$$

Not sure if I'm correct on this second case, so if you find my thought process incorrect I'm all ears.

As for the first case, I'm completely lost and would really appreciate some hints on the best way to approach this.

答:
How about applying the same type of approach as you did for the second part. Since $e^{2x}$ is an increasing function, you have $$0<e^{2x}<e^2$$
Can you take it from here?