问:
Maximizing a combinatorial identity
So here is my problem. If we have a vector $\textbf{x}=(x_1,...,x_n)$ where $x_j \in \mathbb{N}$ for each $j \in \{1,...,n\}$, then is there a way to maximize the value of the following combinatorial identity:

$\sum_{j=1}^{n}\binom{2x_j}{2}$?

The problem here is that we do not know anything about the vector $\textbf{x}$ except for the fact that the entires are all natural numbers and that the constraint is $\sum_{j=1}^{n}x_{j}=k$ where $k\in \mathbb{N}$ is fixed.

答:
Note that whenever $x$ and $y$ are positive integers,
$$
\binom{2(x+y)}2> \binom{2x}2+\binom{2y}2
$$
Therefore, if you have a vector $(x_1,\dots,x_n)$ where there are two nonzero components, then you get a larger value of $\sum_i \binom{2x_i}2$ by combining those two nonzero components into one. This means that the vector which attains the optimum must have only one nonzero component, so it is some permutation of $(k,0,0,\dots,0)$.

Title: Maximizing a combinatorial identity


问:
Quick way to see that $\sqrt{1+x^2+y^2}^{-4}$ is $L^2$-integrable
Is there a quick trick to determine that a function like $$f(x,y):=\sqrt{1+x^2+y^2}^{-4}$$ (which is essentially a negative power of the norm on $\mathbb R^d$ with the singularity at $0$ removed) is in $L^2(\mathbb R^2)$? $f$ is clearly bounded (it takes valeus in $[0,1]$), but that doesn't help here.

答:
You can split the integral of $f^2$ into two pieces,
$$
\int |f(x)|^2dx=\int_{|x|\le 1}|f(x)|^2\ dx+\int_{|x|>1} |f(x)|^2 \,dx
$$
where $x=(x_1,x_2)\in\mathbb{R}^2$.

You have no problem for the first piece $\int_{|x|\le 1}$ since, as you observed, $f$ is bounded. For the second piece, note that
$$
|f(x)|^2=\frac{1}{(1+|x|^2)^4}\le \frac{1}{|x|^8}\tag{1}
$$

It is fairly easy to see by polar coordinates that
$$
\int_{|x|>1} \frac{1}{|x|^8}\,dx<\infty
$$

ContentLicense: CC BY-SA 4.0


问:
Recursion theoretic definition of Kolmogorov complexity
In Kikuchi's paper Kolmogorov complexity and the second incompleteness theorem the Kolmogorov Complexity (KC) of $x$ is defined s

$$ K(x) = \mu e (\varphi_e(0) \simeq x) \, . $$

This seems to give exponentially larger outcomes then the more common (rough) definition of $K(x)$ as "the length of the smallest computer program running on some fixed universal TM that returns $x$". One example where this influences the theory of KC is in the simple and often used lemma "$K(x) \leq \log x + c$ for some fixed constant $c$", which appears false under Kikuchi's definition.

How does Kikuchi's definition match up with the usual KC definition? To what degree does the chosen admissible enumeration of the partial computable functions play a role? I don't see an easy fix to the above problem. Kikuchi's definition seems fundamentally different, already because no two strings can have the same KC under it. Or is it just a typo and did he actually mean $ K(x) = \mu |e| (\varphi_e(0) \simeq x) $? It doesn't seem that way however, as then Lemma 2.1 wouldn't make any sense.

I cross posted a version of this question on MO, since I don't expect an answer here anymore.

答:
Well, I don't think this is an  answer, I have a lot of doubts about it, I would make a comment, but it is too large. I will share some thoughts about this expanding in the direction you are thinking. I don't understand fully the topic, so I would like to know if more people agree with the following.

I will use $K$ for this complexity you mentioned and $C$ for the usual one.

We can turn a partial computable universal function multi-variable function like $\phi(x,y)$ into a single variable function $\phi(c(x,y))$ with a suitable coding like: $1^x0y$, or $1^{|x|}0xy$. We are thinking of strings and number as the same thing here, because of the lexicographical bijection between $\mathbb{N}$ and the strings $\{0,1\}^*$ (note that on this bijection 0 goes to the empty string $\varepsilon$).

So, If we have a universal $\phi$ that works with let say the first of the coding above, we have:

$$\phi_p(c(0)) = \phi(c(p,0) ) = \phi(c(p))$$

The first equality by the enumeration theorem, the second one because in this coding $0 = \epsilon$. We are gonna drop the $c$ assume we can do $\phi= \psi \circ c$, with a $\psi$ that works codding the input. With this we can also define  $C$ as additively optimum ($C(x) \leq C_\phi(x)+x$) in the usual way, as described by Li and Vitányi book.

The $\mu$ operator on partial recursion functions operates like: if the sentence $...x...$ is true for some $x$, then $\mu x(...x...)$ is the smallest $x$ for which $...x... $ is true.

So we have:

$$ K(x) = \mu p( \phi_p(0) =x) = \mu p(\phi(p) = x ) = \min \{ p : \phi(p) =x \}$$

Well the last sentence is by definition the minimal program $x^*$ with respect to this $\phi$. And we have:

$$C(x)  = | x^*|= | K(x) |$$

And we know that by the lexicographic bijection that ($\log$ is the floor log function of base 2 like the one on Li Vitańyi text):

$$C(x) = |K(x)| = \log (K(x))$$

So yes, $K(x)$ behaves exponentially with respect to $C(x)$.

Raatkainen uses this complexity mesure in his article: On Interpreting Chaitin's Incompleteness Theorem, p. 574, and says that $K(x)$ coincides with usual way to define $C(x)$. I think he is saying that with respect only to incompleteness results. Because the  function $\log$ is monotonic, formulas using $\leq$ of the type $(...K(x)... \leq ...)$ could be translated to $(... C(x) ... \leq...)$ changing every number of the kind $y$ by $\log(y)$. In this sense, we could argue that if we proof that: $K(x) > c$  is unprovable on a theory $T$, then $C(x) >  \log(c)$ would also be unprovable. So I think it could be argued that the results with $K(x)$ could be translated to the conventional $C(x)$ with respect to incompleteness, if we translate it properly.

I can't see how the admissible enumeration will play a role here. If we have a enumeration $\phi_0,...,\phi_n,...$ we couldn't argue that there is an admissible system of index $\psi_0,...,\psi_n$ such that the indexes $0,...,n...$ are the size of the programs. For we have more than one program of the same size that gives the same output, any computable $f$ that goes from the system of indices for $\psi$ to the system $\phi$ would not be surjective.


问:
Unitary diagonalization and eigenspace dimensions
I was trying to diagonalize the matrix:

$\left(\begin{array}{ccc}
0 & 0 & i\\
0 & i & 0\\
i & 0 & 0
\end{array}\right)$

I got two eigenvalues, $\lambda_{1}=i$ and $\lambda_{2}=-i$, and found the eigenspaces:

$V_{\lambda_{1}=i}=\mathrm{span}\left\{ \left(\begin{array}{c}
1\\
0\\
1
\end{array}\right)\right\} $

$V_{\lambda_{2}=-i}=\mathrm{span}\left\{ \left(\begin{array}{c}
1\\
0\\
-1
\end{array}\right),\left(\begin{array}{c}
0\\
1\\
0
\end{array}\right)\right\} $

Then I composed a unitary matrix after orthonormalising the eigenvectors:
$U=\left(\begin{array}{ccc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0\\
0 & 0 & 1\\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0
\end{array}\right)\Rightarrow U^{*}=\left(\begin{array}{ccc}
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}}\\
0 & 1 & 0
\end{array}\right)
 $

$U^{*}AU$ does produce a diagonal matrix.

The thing is, I thought the main diagonal should have eigenvalues according to the dimensions $\dim V_{\lambda_{1}=i}=1$ and $\dim V_{\lambda_{2}=-i}=2$:

$\left(\begin{array}{ccc}
i & 0 & 0\\
0 & -i & 0\\
0 & 0 & -i
\end{array}\right)$

But that's not the right diagonal matrix.
Can anyone explain where my error is here?

Thanks.

答:
$\left(\begin{array}{c}
0\\
1\\
0
\end{array}\right)$ is an eigenvector for the eigenvalue $i$, not $-i$.</p>

ContentLicense: CC BY-SA 3.0


问:
Tricky integral limit $\lim_{n \to \infty} \int_0^x \frac{(x-t)^{n}}{(1-t)^{n+1}} dt=0$
I have $x \in (-1,1)$ and I need to show that
$$\lim_{n \to \infty} \int_0^x \frac{(x-t)^{n}}{(1-t)^{n+1}} dt=0$$
How do I even start?

The context to this problem is the remainder of the Taylor series for $\ln{(1-x)}$. If there is a way to show that the Taylor series is identical to the function in $(-1,1)$ without taking this limit: I'm also interested.

答:
Make the substitution $t=xs$ to see that your integral equals
$$x^{n+1}\int_0^1 \frac{(1-s)^{n}}{(1-xs)^{n+1}}\,ds.$$

This is bounded above by

$$\tag 1\frac{x^{n+1}}{1-x}\int_0^1 \frac{(1-s)^{n}}{(1-xs)^{n}}\,ds = \frac{x^{n+1}}{1-x}\int_0^1 \left (\frac{1-s}{1-xs}\right )^n\,ds.$$

The last integrand is $\le 1$ and thus $(1)$ is bounded above by $\dfrac{x^{n+1}}{1-x}\cdot1 \to 0.$


问:
How to prove that $f(x, y)=\frac{2xy}{x^2 + y^2}$ is continuous on $\mathbb{R^2} \smallsetminus \{(0, 0)\}$?
In a small part of a school assignment I have to prove that $f$ is continuous on $\mathbb{R^2} \smallsetminus \{(0, 0)\}$:

$$f(x, y)=\frac{2xy}{x^2 + y^2}$$

I think by using the quotient rule I can prove this, since the quotient of 2 continuous functions is also a continuous function. As long as the denominator is not zero!

As silly it may be sounds, I get stuck proving that these separate functions $2xy$ and $x^2 + y^2$ are continuous functions. It is obvious that they are continuous, but I don't know how to write a good prove of functions with 2 different variables.

答:
Taking any point not equal to $(0,0)$ you have $x^2+y^2$ continuous and not zero in it. Then knowing $2xy$ is continuous everywhere you divide last on former and get continuous ratio on $\mathbb{R}^2 \setminus (0,0)$.

Answering comment about $2xy$ we can consider it as product of continuous functions $f(x,y)=x, g(x,y)=y, h(x,y)=2$. Same trick works for $x^2+y^2$.

Now about continuity of $f(x,y)=x$ as function $\mathbb{R}^2 \to \mathbb{R}$: inequality $|x-x_0|\leqslant \sqrt{(x-x_0)^2+(y-y_0)^2}$ gives continuity in   any $(x_0,y_0)$, as in definition of continuity we can simply consider $\varepsilon=\delta$.


问:
solve this differential equation using substitution
By observing that $y=x$ is a solution of

$$x^2\cos(x)y'' + (2\cos x - x\sin x)(y-xy') = 0$$

find the general solution in the region $x\in(0,{\pi\over 2})$

答:
HINT: Try a solution of the form $y(x)=xf(x)$ after you showed $y=x$ worked. You'll get a new simpler differential equation for $f(x)$.

EDIT: Filling in $y(x)=x f(x)$ you should get

$$x^3\cos x f''(x)+x^3 \sin x f'(x)=0 $$

or after simplifying somewhat and a first integration

$$\ln(f'(x))=\ln(\cos x) + c \; .$$


问:
How to take indices and name of a variable in Sage?
I define the following in Sage.

```<code>rank=8
R=PolynomialRing(QQ, 'a' ,rank+1)
a=R.gens()
</code>```

I would like to define a function f which returns the index of a[3]: f(a[3])=3 and define another function g such that g(a[3])=a. How to do this in Sage? Thank you very much!

答:
This is something generic about lists in python so

```<code>a.index(a[3])
</code>```

for the first (you can write f = a.index if you like), and for the second if you really want a function rather than using [3] you can do

```<code>def g(n):
    return a[n]
</code>```

Body:
This is something generic about lists in python so

a.index(a[3])

for the first (you can write f = a.index if you like), and for the second if you really want a function rather than using [3] you can do

def g(n):
    return a[n]


问:
Closed subschemes and quasi-coherent sheaves of ideals
Let me quote two results from Qing Liu's Algebraic Geometry and Arithmetic Curves:

Lemma 2.2.23: Let $X$ be a ringed space, $\mathcal{J}$ be a sheaf of ideals on $X$, $V(\mathcal{J}) = \{x\in X\;:\;\mathcal{J}_x \neq \mathcal{O}_{X,x}\}$, and $j:V(\mathcal{J})\to X$ be the inclusion. Then $V(\mathcal{J})$ is a closed subset of $X$, $(V(\mathcal{J}),j^{-1}(\mathcal{O}_X/\mathcal{J}))$ is a ringed space, and $(j,j^\#)$ is a closed immersion where $$j^\#:\mathcal{O}_X\to j_*j^{-1}(\mathcal{O}_X/\mathcal{J}) = \mathcal{O}_X/\mathcal{J}$$ is the canonical projection.

The very next proposition states the converse, that is a closed immersion $Y\to X$ gives rise to a sheaf of ideals (namely the kernel) whose closed subspace is isomorphic to $Y$. Explicitly,

Proposition 2.2.24: Let $f:Y\to X$ be a closed immersion of ringed spaces, $\mathcal{J}:=\operatorname{ker}f^\#$, and $Z = V(\mathcal{J})$. Then $f$ factors into an isomorphism $Y\simeq Z$ followed by the canonical closed immersion $Z\to X$.

Together, these imply that there is a one-to-one correspondence between sheaves of ideals $\mathcal{J}$ of $\mathcal{O}_X$ and closed subschemes $Z$ of $X$ via $\mathcal{J}\mapsto V(\mathcal{J})\mapsto \ker(\mathcal{O}_X\to\mathcal{O}_X/\mathcal{J}) = \mathcal{J}$. Am I interpreting this correctly?

I believe this is all good so far. What confuses me is what happens one introduces quasi-coherence.

Proposition 5.1.15: Let $X$ be a scheme, $Z$ be a closed subscheme, and $j:Z\to X$ denote the canonical closed immersion. Then $\operatorname{ker}j^\#$ is a quasi-coherent sheaf of ideals on $X$, and the map $Z\mapsto\operatorname{ker}j^\#$ establishes a bijection between closed subschemes of $X$ and quasi-coherent sheaves of modules on $X$.

Here are my two questions:

When working over a scheme $(X,\mathcal{O}_X)$, are all sheaves of ideals of $\mathcal{O}_X$ quasi-coherent?

Proof? If $\mathcal{J}$ is any sheaf of ideals of $\mathcal{O}_X$, then the canonical map $(j,j^\#):(V(\mathcal{J}),j^{-1}\mathcal{O}_X/\mathcal{J})\to(X,\mathcal{O}_X)$ is a closed immersion, i.e. $(V(\mathcal{J}),j^{-1}\mathcal{O}_X/\mathcal{J})$ is a closed subscheme of $X$, hence by Proposition 5.1.15, $\ker j^\# = \mathcal{J}$ is a quasi-coherent sheaf of ideals.

Could someone kindly provide an example of a locally ringed space $(X,\mathcal{O}_X)$ and a sheaf of ideals $\mathcal{J}$ that is not quasi-coherent?

Assuming 1. is actually correct, it ought to be the case that $X$ cannot be a scheme, right?

答:
The answer to question 1 is no, and the counterexample we produce is also valid for question 2. Let $X=\Bbb A^1_k$, let $U$ be the complement of the origin, and consider $j_!\mathcal{O}_U$, the sheaf which takes the value $\mathcal{O}_U(W)$ for $W$ an open subset contained in $U$ and $0$ else. This is a sheaf of ideals, but it's not quasi-coherent: it has no nonzero global sections, despite not being the zero sheaf.

The correct statement is that there is a one-to-one correspondence between quasi-coherent sheaves of ideals on $X$ and closed subschemes of $X$.


问:
Find the minimum value of $p$ so that the series diverges
I was doing this particular series question:

Find the minimum value of $p$ for which the series:

$$\sum_{n=1}^{\infty}(1n^{27}+8)^{27}(1+27n^{29}+8n^{30})^p$$

is divergent.

How would I do this? My idea was the use the comparison test and then use that fact for large values of $n$, this behaves like:

$$n^{729}*n^{30p}$$ so we get $30p+729<1$.

Is that the right idea? Or is that not the right approach?

答:
The general term of the series is positive, So we can use limit comparison test.

but

$$u_n\sim 8^p n^{27\times 27}.n^{30p}\sim 8^p.n^{729+30p}$$
So, the series converges if $$-729-30p>1$$

or
$$p<\frac{-73}{3}$$


问:
Confusion about lift in the context of tangent bundle
A lift is defined here: https://mathworld.wolfram.com/Lift.html as a tangent vector field $X$ on a manifold, the same way that a section of the tangent bundle gives us $X$ in the context $\dot g = X(g)$. This answer to a different question: https://mathoverflow.net/a/111198/172470 suggests that a path $g$ can be lifted into the tangent bundle space as $\tilde g \in T_x M$, and is a smooth curve there. I am confused about this, because of two reasons:

Isn't $T_x M$ the set of vectors tangent to a specific point $x$ (the "tangent space" at $x$)? How can we have a curve there, especially a closed loop, like the cited answer suggests?

If we can, how is this lift a vector field?

Thanks for any help you can give, including any clarification of definitions I may be wrong about. Pictures are nice too since english isn't my second language, and I've already gone down the wrong road with some ideas in differential geometry because I had formed the wrong picture. Thank you!

答:
A lift of an injective path $\gamma:I\to X$ is as you suggest a map $\widetilde{\gamma}:I\to TX$ so that $\pi:TX\to X$ satisfies $\pi\circ \widetilde{\gamma}=\gamma$. Actually, such a lifting is equivalent to specifying a tangent vector along each point of the curve $\gamma$, so you can think of this as a section of the bundle $TX$ restricted along $\gamma$ (aka $\gamma^*TX$). Anyway, the point is that such a lift is the smooth choice of a tangent vector at each point to the curve, i.e. a vector field along a curve like one sees in multivariable calculus.

Now, if we have the additional structure of an affine connection $\nabla$ on $TX$, then we get the definition of which vector fields are parallel along $\gamma$, meaning they satisfy a differential equation $\nabla_{\gamma'}X=0$, where $\gamma'$ is the trajectory vector field of $\gamma$. If we start with a loop $\gamma:I\to X$, (such that $\gamma(0)=\gamma(1)$ and let's say without other self-intersection) then we might ask when a lifting $\widetilde{\gamma}:S^1\to TX$ is a loop in the tangent bundle. In particular, this occurs exactly when we can define a vector field on the image of $\gamma$ by $X_{\gamma(t)}=\widetilde{\gamma}(t)$.

On the other hand, if we fix a loop $\gamma:I\to X$ and choose a tangent vector $X_{\gamma(0)}\in T_{\gamma(0)}X$, parallel transport allows us to define $\widetilde{\gamma}(t)$ to be the unique parallel lift of $\gamma$ with initial condition $\widetilde{\gamma}(0)=X_{\gamma(0)}$. Very often, $\widetilde{\gamma}(1)\ne \widetilde{\gamma}(0)$ and as a consequence we do not get a well-defined vector field along the image of $\gamma$. We can however define a linear transformation of $T_{\gamma(0)}X$ by sending $X_{\gamma(0)}\mapsto \widetilde{\gamma}(1)$, where $\widetilde{\gamma}$ is the lifting with initial condition $X_{\gamma(0)}=\widetilde{\gamma}(0)$. In particular, a loop at $x$ defines a map $P_\gamma\in \operatorname{GL}(T_xX)$ by the above procedure for $x=\gamma(0)$. This leads to the notion of the holonomy group of $X$ at $x$:
$$\operatorname{Hol}_x=\{P_\gamma:\gamma\:\text{is a loop based at}\:x\}.$$
If $X$ is a Riemannian manifold with metric $g$ and Levi-Civita connection $\nabla$ then the holonomy interacts with the curvature. Intuitively speaking, nonzero holonomy detects curvature. The idea being that curvature causes parallel transport of curves to deviate from the identity on $T_xX$. The following picture from Wikipedia illustrates exactly the idea:

If we start at the north pole, and move due south to the equator in any direction, then due west along the equator and then due north back to the north pole, the nontrivial curvature in the region bounded by our path of travel causes the parallel-translated vector from the north pole to rotate.

As a last comment, there is an interaction with representation theory when the connection $\nabla$ is flat. In this case, the transformation depends only on the homotopy class of the loop $\gamma$ based at $x$. Hence, we get a group homomorphism $\pi_1(X,x)\to \operatorname{GL}(T_xX)$ called the monodromy representation of $\pi_1(X,x)$, and defined by $[\gamma]\mapsto P_{\gamma}$.


答:
First: proving that $f(x, y) = x^2 + y^2$ is continuous. Let $(a, b)$ and $\epsilon > 0$ be arbitrary. Then $|x^2 + y^2 - a^2 - b^2| = |(x + a)(x - a) + (y - b)(y + b)| \leq |x - a||x - b| + |y - b||y + b|$ by the triangle inequality. Take $(x, y)$ closer than $\delta$ to $(a, b)$. Then $|x - a|, |y - b| < \delta$, which means that also $|x + a| < \delta + 2|a|$, $|y + b| < \delta + 2|b|$ (by the triangle inequality again). So

$$|f(x, y) - f(a, b)| \leq \delta(2\delta + 2|a| + 2|b|)$$

Can you pick a $\delta$ depending only on $a, b, \epsilon$ to make this smaller than $\epsilon$?

For $2xy$, just use the product rule and that projections are continuous.

Actually, now that I think of it the product and sum rule are enough to prove that $x^2 + y^2$ is continuous as well.

Title: How to prove that $f(x, y)=\frac{2xy}{x^2 + y^2}$ is continuous on $\mathbb{R^2} \smallsetminus \{(0, 0)\}$?

Body:
In a small part of a school assignment I have to prove that $f$ is continuous on $\mathbb{R^2} \smallsetminus \{(0, 0)\}$:

$$f(x, y)=\frac{2xy}{x^2 + y^2}$$

I think by using the quotient rule I can prove this, since the quotient of 2 continuous functions is also a continuous function. As long as the denominator is not zero!

As silly it may be sounds, I get stuck proving that these separate functions $2xy$ and $x^2 + y^2$ are continuous functions. It is obvious that they are continuous, but I don't know how to write a good prove of functions with 2 different variables.


答:
Let be $0 < x < 1$. The function
$$t\longmapsto\frac{x - t}{1 - t}$$
is decreasing in $[0,x]$ (why?), so
$$t\in[0,x]\implies 0\le\frac{x - t}{1 - t}\le x
\implies
0 \le \frac{(x - t)^n}{(1 - t)^{n+1}}\le\frac{x^n}{1 - x},$$
And
$$
0\le\int_0^x\frac{(x-t)^n}{(1-t)^{n+1}}\,dt\le
\frac{x^{n+1}}{1 - x}.
$$
When $-1 < x < 0$, the bounding is similar.

Title: Tricky integral limit $\lim_{n \to \infty} \int_0^x \frac{(x-t)^{n}}{(1-t)^{n+1}} dt=0$

Body:
I have $x \in (-1,1)$ and I need to show that
$$\lim_{n \to \infty} \int_0^x \frac{(x-t)^{n}}{(1-t)^{n+1}} dt=0$$
How do I even start?

The context to this problem is the remainder of the Taylor series for $\ln{(1-x)}$. If there is a way to show that the Taylor series is identical to the function in $(-1,1)$ without taking this limit: I'm also interested.


问:
Proof of continuous and differentiable of a given function with two variables
I am trying to figure out why my function is differentiable and therefore continious.

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

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

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

The limit value investigation at a certain point
from the left and from the right only makes sense for a function with only one variable.

How can I proof if my function is differentiable?

答:
Let $(a,b)$ be any point in $\mathbb{R^2}$. Starting from where you left off, we need to show $f$ is differentiable at $(a,b)$. We calculate:
$f(a+h,b+k) - f(a,b)-((3a^2-3b^2)h+(-6ab)k)= (a+h)^3-3(a+h)(b+k)^2-a^3+3ab^2-3a^2h+3b^2h+6abk=3ah^2-3ak^2-6bhk+h^3-3hk^2$. Thus put $A = \left(3a^2-3b^2, -6ab\right)\implies \dfrac{\left|f(a+h,b+k) - f(a,b) - A\cdot \begin{pmatrix} h\\k \end{pmatrix}\right|}{|(h,k)|}=\dfrac{|3a(h^2-k^2)+ h(h^2-6bk-3k^2)|}{\sqrt{h^2+k^2}}\le \dfrac{|3a||(h^2+k^2)}{\sqrt{h^2+k^2}}+ \dfrac{|h|}{\sqrt{h^2+k^2}}\cdot|h^2-6bk-3k^2|\le |3a|\sqrt{h^2+k^2} + |h^2-6bk-3k^2|\to 0$ when $(h,k) \to (0,0) \implies f'(a,b) = A$ as claimed.


问:
If $\operatorname{Im}(x) \perp \operatorname{Im}(y)$, then $\|x+y\| = \max\{\|x\|, \|y\|\}$.
Let $x,y \in B(H)$ be bounded operators on the Hilbert space $H$ such that
$xy=yx = 0$ and $\operatorname{Im}(x) \perp \operatorname{Im}(y)$. Moreover, assume that
$\operatorname{Im}(y)$ has finite-dimension. Then is it true that
$$\|x+y\| = \max\{\|x\|, \|y\|\}?$$

I tried to mimique the proof of Pythagoras theorem but did not manage to prove this result. I'm not even sure it is true, but it looks like a result that is used in a proof I'm reading.

答:
Edit: ruy noted that the original example didn't work.

This is not true, even if $\dim H<\infty$. For instance consider
$$
x=\begin{bmatrix}0&0&0\\ 1&0&0\\0&0&0\end{bmatrix},\qquad y=\begin{bmatrix} 0&0&0\\0&0&0\\1&0&0\end{bmatrix}.
$$
Then $xy=yx=0$, ranges are orthogonal, and
$$
\|x\|=\|y\|=1,\qquad\|x+y\|=\sqrt2.
$$
The equality does hold when $x$ and $y$ are also selfadjoint. Or, more generally, when $x^*y=yx^*=0$. Because in this case one has
$$\|x+y\|^2=\|(x+y)^*(x+y)\|=\|x^*x+y^*y\|=\max\{\|x\|^2,\|y\|^2\}$$ by the selfadjoint case.


答:
Or, use the following formula:
$$y_2(x)=y_1(x)\int\frac{\text{e}^{-\int p(x)dx}}{y_1^2(x)}$$ wherein your OE is $y''+P(x)y'+Q(x)y=0$.</p>

Title: solve this differential equation using substitution

Body:
By observing that $y=x$ is a solution of

$$x^2\cos(x)y'' + (2\cos x - x\sin x)(y-xy') = 0$$

find the general solution in the region $x\in(0,{\pi\over 2})$


问:
Analog of $C^{\infty}$ multiplication for discrete "vector fields"
To make the problem simple, consider an undirected graph $G$ with vertices $V$ and edges $E$.

A discrete "scalar function" (0-cochain) can be defined on the vertices, taking values $f(v)$ on each vertex $v$.

A discrete "1-form" (1-cochain) can be defined as an anti-symmetric function on pairs of vertices $u(v_1,v_2) = -u(v_2,v_1)$ whose value is 0 if $v_1$ and $v_2$ are not connected by an edge. We can naturally think of the non-zero values of $u$ as being encoded on the edges of the graph supplemented with an orientation of each bond, i.e. $u$ is essentially a weighted, directed version of $G$. For example, by choosing an orientation of each bond one could define $u$ to be the finite difference gradient of $f$.

For concreteness, I am considering $f$ and $u$ to be real or complex valued.

A lot of intuitive ideas from differential geometry can be brought over (and generalized further if we consider for example simplicial complexes instead of graphs), like boundary/coboundary etc. My question is, is there a natural analog of $C^\infty$ multiplication in the discrete case? I don't see an obvious way to achieve this because the scalar functions are defined on vertices while the discrete 1-cochains are defined on the edges, so we can't just multiply them pointwise like we can on a manifold.

答:
This somewhat depends on what features of the module structure $\Omega^0M\times\Omega^1M\to\Omega^1M$ you're interested in replicating. One common choice is the cup product, which replicates the properties of the wedge product of differential forms in simplicial cohomology. It is defined in various ways in various contexts, and oftentimes one is only interested in the induced product on cohomology. Using your definition of simplicial cochains, we could, for instance, define
$$
(a^p\smile b^q)(v_0,\cdots,v_{p+q}) \\
=\frac{1}{(1+p+q)!}\sum_{\pi\in\mathcal{P}(0,\cdots,p+q)}\operatorname{sgn}(\pi)a^p(v_{\pi(0)},\cdots,v_{\pi(p)})b^q(v_{\pi(p)},\cdots,v_{\pi(p+q)})
$$
where $a^p$ is a $p$-cochain, $b^q$ is a $q$-cochain, $v_0,\cdots,v_{p+q}$ are vertices of of a $p+q$-simplex, $\mathcal{P}$ denotes the set of permutations of a sequence, and $\operatorname{sgn}$ denotes the sign of a permutation. This product has a lot of the properties of the wedge product, for instance being graded-commutative, and making the coboundary operator an antiderivation.

For $0$ and $1$-cochains, for instance, this would give
$$
(a^0\smile b^1)(v_0,v_1)=\frac{1}{2}(a^0(v_0)+a^0(v_1))b^1(v_0,v_1)
$$


问:
$D_1 \Sigma D_1 \succeq D_2 \Sigma D_2$ with $\Sigma$ being positive semi-definite and $D_1 \succeq D_2$ diagonal matrices
Let $\Sigma \in \mathbb{R}^{n \times n}$ be a positive semi-definite matrix, and $D_1, D_2  \in \mathbb{R}^{n \times n}$ be two diagonal matrices with $(D_1)_{ii} \geq (D_2)_{ii} \geq 0$ for any diagonal entry $i \leq n$.

I wonder if it is true that $D_1 \Sigma D_1 \succeq D_2 \Sigma D_2$, i.e.,
$D_1 \Sigma D_1 - D_2 \Sigma D_2$ is a positive semi-definite matrix?

I feel it is true because $D_1 y$ just scales a vector $y$ more in some directions.

答:
The answer is no. As a counterexample, consider
$$
\Sigma = \pmatrix{1&1\\1&1}, \quad D_1 = \pmatrix{1&0\\0&1}, \quad D_2 = \pmatrix{1&0\\0&0}.
$$


问:
Prove there is a full measure set s.t Birkhoff averages of any continuous function converge on that set
Let $X$ be a compact metric space equipped with a Borel probability measure $\mu$. If $T$ is a continuous ergodic transformation, prove there is a full measure set s.t Birkhoff averages of any continuous function converge on that set.

I am only looking for hints.

My work: First of all, $C(X)$ is separable, so there is a dense sequence $f_k$. It is clear that there is a set of full measure s.t for each $k$ the Birkhoff averages converge to $\int f_k$. At this point I tried to do some limit argument to no avail. I get a double limit and I just cannot seem to justify why you can exchange the two limits. Thus I think I am looking at this problem the wrong way. Especially since I do not see why continuity of $T$ is needed here.

I would much appreciate advice/hints rather than flat out solutions.

答:
Your first intuition is right. As $C(X)$ is separable you can find a dense sequence $f_k$. For every $f_k$ there is a full measure set $X_k$ for which the Birkhoff sums converge to the integrable.

Then a countable intersection of full measure set is a full measure set. So on the set $Y = \cap X_k$ every $f_k$ has its Birkhoff sums converging to its average.

No take a $g \in C(X)$. There is a subsequence $\{g_k\} \subset \{f_k\}$ such that for every $\epsilon$, there is a $K$ such that for every $k>K$, $|f-g_k|_\infty \leq \epsilon$.

So for every $x \in Y$, $|\frac{1}{N}\sum_{i=1}^{N}g(T^i(x))-\int g|=|\frac{1}{N}\sum_{i=1}^{N}g(T^i(x))-g_k(T^i(x))+\frac{1}{N}\sum_{i=1}^{N}g_k(T^i(x))-\int g_k+\int (g_k-g)|$

But we have that $$
|\int (g_k-g)| \leq \epsilon Vol(X)
$$

$$
|\frac{1}{N}\sum_{i=1}^{N}g(T^i(x))-g_k(T^i(x))|\leq \epsilon
$$
And $$
\lim \frac{1}{N}\sum_{i=1}^{N}g_k(T^i(x))-\int g_k = 0
$$

So applying the triangular inequality to detach this three pieces we get that $$
\limsup |\frac{1}{N}\sum_{i=1}^{N}g(T^i(x))-\int g| \leq \epsilon (1+ Vol(X))
$$
as this is true for every $\epsilon$ we have that $$
\limsup |\frac{1}{N}\sum_{i=1}^{N}g(T^i(x))-\int g| =0
$$
And then $$
\lim |\frac{1}{N}\sum_{i=1}^{N}g(T^i(x))-\int g| =0
$$


问:
Locally finite vs. Borel measures on $\sigma$-compact Polish spaces
Let $E$ be a Polish space, and let $\mu$ be a measure on $E$. Define the following properties:

$E$ is $\sigma$-compact if $E$ is the countable union of compact sets.

$E$ is locally compact if every $x \in E$ has an open neighborhood $U$ whose closure is compact.

$\mu$ is locally finite if for every $x \in E$, there is an open set $U \subset E$ containing $x$ with $\mu(U) < \infty$.

$\mu$ is a Borel measure if for every compact $K \subset E$, we have $\mu(K) < \infty$.

Clearly a locally finite measure is Borel. And if $E$ is locally compact (not even necessarily separable or complete), Borel measures are necessarily locally finite. But are there more general conditions for which Borel measures are locally finite?

Question: If $E$ is a $\sigma$-compact Polish space and $\mu$ is a Borel measure, is $\mu$ locally finite?

I can’t think of a counter example to this, but I’m having trouble proving it. My original strategy was to prove that a $\sigma$-compact Polish space is locally compact. However, as the comments demonstrate, $\sigma$-compact Polish spaces are not necessarily locally compact, so that strategy doesn’t work. A counter example is the subset $X \subset \ell^2$ given by the union of the lines $X_k = \{\lambda e_k : \lambda \in \mathbb R\}$, where $\{e_k\}_{k \geq 1}$ is the standard orthonormal basis of $\ell^2$. Then $X$ is $\sigma$-compact, but not locally compact (at the origin specifically).

But I’m not sure of a measure $\mu$ on this space that is Borel, but not locally finite; the problem is there are compact subsets of $X$ containing $0$ and intersecting infinitely many of the $X_k$. Can anyone think of a counter example?

答:
I think that indeed the idea from the common thread works:

Let $e_n, n \in \Bbb N$ be the standard orthonormal base of Hilbert space $\ell^2$. Let $X = \{\lambda e_n \mid n \in \Bbb N, \lambda \in \Bbb R\}$. Then $X$ is $\sigma$-compact Polish but not locally compact (at $0$).

Let $L_k = \{\lambda e_k: \lambda \neq 0\}$ and $\delta_A$ be the Dirac measure with carrier $A$: $\delta_A(B) = 1$ iff $A \cap B \neq \emptyset$ and $0$ otherwise, we can define $\mu = \displaystyle_{k=1}^\infty \delta_{L_k}$, which is a Borel measure (not $\sigma$-finite, and not locally finite at $0$), but (I think) finite at compacta.

I haven't checked all the nitty gritty details. Others might feel inclined to add to this.


问:
Examples of continuous functions that are not homotopic to each other.
Let $f,g:X \to Y$ be continuous functions. Then, consider $F:X \times I \to Y$ defined as follows:$$F(x,t) = f(1-t)+gt$$

How does one check whether $F$ is continuous(w.r.t the product topology)?
Is there any other way to define a homotopy between two functions?

As an example, take $f,g :  \mathbb{R} \to \mathbb{R}$ to be $f(x) = x, g(x) = sin(x)$. Then how does one PROVE that the homotopy map as defined above is continuous?

Moreover, what are some examples of maps that are not homotopic?

答:
The homotopy you describe for maps to $\mathbb{R}$ (or maybe $\mathbb{R}^n$) is continuous because addition and multiplication on $\mathbb{R}$ are continuous operations, along with the functions $f$ and $g$ being continuous.

In general, maps $f$ and $g$ are not homotopic if some $f(x)$ and $g(x)$ are in different path components (essentially using that if $f$ and $g$ were homotopic then so would be the compositions $\{x\} \hookrightarrow X \to Y$). This tells you for example that the identity of $\mathbb{R} - 0$ is not homotopic to the map $x \mapsto -x$.

Looking at such compositions with maps $Z \to X$ from other 'test' spaces $Z$ can often detect if $f$ and $g$ are not homotopic. For example looking at all maps $S^n \to X$ composed with $f$ and $g$ is the same as looking at the maps $\pi_n(X) \to \pi_n(Y)$ induced by $f$ and $g$.


问:
Direct Proof: For all n(n is a natural number) 5 divides n^5 - n
In order for $n^5 - n$ divisible by $5$, $n^5 - n = 5 x + 0$ (for some $x$, $x$ is a natural number)
I simplified the $(n^5 - n) = n(n^2+1)(n+1)(n-1)$ and I do not know what to do next.

And I tried something else like $(n^5 - n) = n (n^4 - 1)$, but now I need to show $(n^4 - 1)$ is divisible by $5$, how do I do that? Thanks in advance.

答:
Well, $n\equiv 0,\pm 1,\pm 2\pmod 5$.

If $n \equiv 0 \pmod 5$ then $5|n$ and $5|n(n^2 + 1)(n+1)(n-1)$.

If $n \equiv \pm 1 \pmod 5$ then $5|n\mp 1$ and $5|n(n^2+1)(n+1)(n-1)$.

And if $n \equiv \pm 2$ then $n^2 \equiv 4 \equiv -1$ and $5|n^2+1$ and $5|n(n^2+1)(n+1)(n-1)$.

..... alternative let $n = 5k + i; i = 0,1,2,3,4$

$(5k+i)((5k+i)^2 + 1)(5k + i + 1)(5k + i - 1)\equiv $

$i(i^2 + 1)(i+1)(i-1)\equiv $

$i(i^2 + 1)(i^2-1)\equiv $

$i(i^4 - 1) \equiv $

$i^5 - i\pmod 5$.

It's now a matter of showing $i^5 \equiv i \pmod 5$ for $i = 0,1,2,3,4$.

If you know Fermat's Little Theorem you are done.  FLT says precisely that.

If you don't,  there are only $5$ cases to test:

$0^5\equiv 0; 1^5 \equiv 1; 2^5\equiv 32 \equiv 2; 3^5 \equiv 3^4\cdot 3\equiv 81\cdot 3\equiv 3;  4^5 \equiv 4^2\cdot 4^2\cdot 4 \equiv 16\cdot 16\cdot 4\equiv 1\cdot 1\cdot 4 \equiv 4$.

That's it.


问:
Stirling Numbers of the first kind diagonal closed form, understanding indexing.
So for the signed stirling numbers of the first kind, I wanted a formula for the diagonals

On Wikipedia they describe the formula as the below. I don't quite understand how the indexing works for the sum (why is there a sum inside the index and what is the small k's)

Here the link to it. Any help is much appreciated!

答:
The notation

$$\sum_{0\le k_1,\ldots,k_p:\sum_1^pmk_m=p}f(k_1,\ldots,k_p)$$

means that we’re summing the general term $f(k_1,\ldots,k_p)$ over all $p$-tuples $\langle k_1,\ldots,k_p\rangle$ of non-negative integers such that $\sum_{m=1}^pmk_m=p$. Each of these $p$-tuples describes a partition of $p$: specifically, each $k_m$ for $m=1,\ldots,p$ is the number of parts of size $m$ in the partition. Thus, if $p=11$, the $11$-tuple $\langle 3,2,0,1,0,0,0,0,0,0,0\rangle$ describes the partition $11=4+2+2+1+1+1$.

The summation could also be written

$$\sum\left\{f(k_1,\ldots,k_p):0\le k_1,\ldots,k_p\text{ and }\sum_{m=1}^pmk_m=p\right\}\,.$$


问:
Sequence/series divergence proof
Suppose $\{x_n\}$ is a sequence (for $n \geq 1$) that diverges and let $s_n = |x_{n+1} - x_n|$. Prove that $\sum_{n=1}^{\infty} s_n$ diverges.

My attempt:

Let us use a contrapositive argument:

Assume that $\sum_{n=1}^{\infty} s_n$ converges. This shows that $\lim_{n\to\infty} s_n = 0$. By the definition of limits, for all $\epsilon > 0$, there is some $N \in \mathbb{N}$ so that for $n \geq N$, we have $|s_n| < \epsilon \implies |x_{n+1} - x_n| < \epsilon$.

We need to show that $\lim_{n\to\infty} x_n = x$ for some $x$. So for all $\epsilon_2 > 0$ we need to find some $N_2$ such that for all $n \geq N_2$, we have $|x_n - x| < \epsilon_2$.

(What I thought to do at this point: Because $\lim_{n\to\infty} x_n = \lim_{n\to\infty} (x_n + 1)$ by change of index we can let $\epsilon =$ something?)

Notice that

$||x_{n+1}| - |x_n|| \leq |x_{n+1} - x_n| < \epsilon$ - the reverse triangle inqlty

Not too sure what to do from here and I'm hoping that somebody can point me in the right direction. Any assistance is appreciated.

答:
Recall this theorem: every absolutely convergent series is convergent. We are going to use it in the proof.

Suppose that $\sum_n s_n$ converges, then the series
$$\sum_{n=1}^{\infty} (x_{n+1} - x_n)$$
is absolutely convergent. In particular, it is convergent to some real number $A$. This is a telescoping series, and for all $N$ you have
$$\sum_{n=1}^N (x_{n+1} - x_n) = x_{N+1}-x_1$$
Indeed
$$x_{N+1} -x_1= (x_{N+1} - x_N ) + (x_{N} - x_{N-1} ) + (x_{N-1} - x_{N-2} ) + \cdots +(x_2-x_1)$$
If you take the limit of the RHS you get a real number
$$A = \lim_{N \to \infty} \sum_{n=1}^N (x_{n+1} - x_n) = \sum_{n=1}^\infty (x_{n+1} - x_n)$$
Hence $\{ x_N\}$ converges to $A+x_1$: in particular it is convergent, and this completes the proof.

Title: Sequence/series divergence proof


问:
Find all values for a, b, c, and d, such that the following matrix equation is consistent.
I need to find the values for the given variables to make the below equation consistent. I've tried setting the corresponding entries equal to each other and solving the system for a, b, c, d, but I'm not sure how to do that if there are many other x variables. How do I go about answering this question?

\begin{bmatrix}x_{1}-3x_{2}+2x_{4}&-2x_{4}+5x_{5}+b\\3x_{1}-x_{3}+7x_{5}&9x_{2}+6x_{4}-d\end{bmatrix} = \begin{bmatrix}-2x_{5}+a&-2x_{1}+6x_{2}+x_{3}\\9x_{2}+c&3x_{1}-2x_{3}+8x_{5}\end{bmatrix}

Body:
Here what to do for $a$

You have $x_{1}-3x_{2}+2x_{4}=-2x_{5}+a$

so $a=x_{1}-3x_{2}+2x_{4}+2x_{5}$.

You can try $b,c,d$


问:
I can't find the piecewise linear function given the following set of dependent and independent variables.
So, let's say we're in a parking lot and it costs 5 cents per minute to park at the airport with the rate dropping to 3 cents per minute after 9:00 p.m. Assume that t is the number of minutes after 8 p.m.

Find and graph the cost function c(t) for values of t satisfying $0 <= t <= 120 $

What I think the answer is :

So for the first part of this piecewise function is to take 0.05t for $0 <= t <= 60
It is now 9:00 p.m.

For the second part of this piecewise function is where I fall short.

The typical procedure of finding the function involving two points is to take the slope of the two points. $\frac{dependent_2 - dependent_1}{independent_2 - independent_1}$

Our dependent variable is 0.03 and 0. Our independent variables are 120 and 60.

So m should logically be $\frac{0.03}{60} = 0.0005$

So our function should be d - 0.03 = 0.0005(t - 120)

Problem is the ending fucntion in the back of my book says $p(t) = 0.03t + 1.2$.

I'm not sure how it came to this and I'm coming to you, my friends at math stackexchange for help.

答:
The first part of your answer (for the hour between 8 and 9) is correct.

For the time after 9, don't try to use the
typical procedure for the line through two points. Think directly about the slope. It will be
$$
0.03 \frac{\text{\$}}{\text{minute past } 9}
$$
so the formula for the time after 9 will be the full cost for the first hour plus the cost for the added time. That is
$$
3.00 + 0.03(t-60).
$$

That simplifies algebraically to the answer in the book.


问:
Probability that points are on a straight line
I am looking at a formula to calculate the probability that $n$ points are on a straight line between point $1$ and point $n$ in 2d Euclidean space.

If the points are exactly on the line, the probability should be $1$, and I guess if point #$1$ is $(0,0)$ and point $n$ is $(m, m)$ then if the points are on the line $(0,0; 0,m)$ and $(0,m; m,m)$ then the probability would be $0$.

I can calculate the distance to the line between $(0,0)$ and $(m,m)$ but how can I use that as a measure? This would be very dependent on the length of the line and the amount of points.

If somebody could point me in the right direction that would be great :)

Thanks

答:
thanks to fgp I found my answer; when googling collinearity measure I found on wikipedia correlation, which makes perfect sense :)

If the points are on a straight line, the abs(correlation) will be 1 and if there is a large curve between points, it will be close to 0 which is exactly what I was looking for.

Thanks for you help and describing in good way what I was looking for :)


问:
Evaluate an absolute monster integral $\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$
I want to figure out a way to evaluate $$\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$$
I tried to substitute $x = u^2$ and cancel the square root in the denominator, getting $$2\int\limits_{0}^{1} \frac{\log(1-u^2+u^4)}{1+u^2}\mathrm{d}u.$$But now I am stuck again.

Body:
Note

$$I=\int\limits_{0}^{1} \frac{\ln(1-x+x^2)}{\sqrt{x}(1+x)}{d}x
\overset{t=\sqrt x} = 2\int\limits_{0}^{1} \frac{\ln(1-t^2+t^4)}{1+t^2}dt
$$

Let $J(a) = \int_0^1 \frac{\ln(\frac14(1+t^2)^2 \sec^2a -t^2)}{1+t^2}dt
$.
Then, $J(0) = \int_0^1 \frac{2\ln\frac{1-t^2}2}{1+t^2}dt=
-2G$

$$J’(a)= \int_0^1 \left(\frac{\tan a}{t^2+2t\cos a+1} + \frac{\tan a}{t^2-2t\cos a+1} \right)dt=\frac\pi2 \sec a
$$

$$J(\frac\pi6) = \int\limits_{0}^{1} \frac{\ln(1-t^2+t^4)-\ln3}{1+t^2}dt=\frac12I-\frac\pi4\ln3
$$

Thus
\begin{align}
I= \frac\pi2\ln3+2J(\frac\pi6)
= \frac\pi2\ln3+2J(0) +2\int_0^{\pi/6} J’(a)da
=\pi \ln3-4G
\end{align}


问:
Where does $\frac{3}{2}$ in $1<s\cdot\zeta(s+1) < \frac{3}{2}$ come from?
In his short note on Euler and the prime harmonic series, Professor of Mathematics Paul Pollack, makes the following intermediate statement about the Riemann zeta function $\zeta(s)$ for $s>1$, without explanation:

$$
1<s\cdot\zeta(s+1) < \frac{3}{2}
$$

Question: Where does that $\frac{3}{2}$ comes from?

My thinking ...

He says it follows from a lemma which in his note is the well known inequality:

$$
\frac{1}{s-1} <\zeta(s)< \frac{s}{s-1}
$$

.. which he expresses as

$$1<(s-1)\cdot \zeta(s)<s$$

Substituting $(s+1)$ for $s$ would lead to the following inequality:

$$1 < s \cdot \zeta(s+1) < s+1$$

答: In the document linked, Lemma 3 provides the general result that $$1<(s-1)\zeta(s)<s$$ for every $s$, but, the Lemma 4 statement which contains the $\frac 32$ restricts $s$ to $(0,\frac12)$. Indeed, in this case, we have that for each $s\in(0,\frac12)$: $$1<s\zeta(s+1)<\frac12\zeta \left(\frac32\right)<\frac32$$


问:
What is a technical argument (in contrast to a non-technical argument)?
I often see arguments referred to as "technical" but I'm not sure what that really means. Does it mean "unintuitive"? Or does it mean "tedious"? Or something else entirely?

I have sometimes seen it used to contrast with an intuitive argument, but I find that even these "technical" arguments can be very intuitive once understood.

Is there an agreed (or mostly agreed) upon definition of "technical"?

I would also really appreciate some lucid examples of technical vs non-technical arguments.

答:
I would think "technical" is a rather subjective term because what one person deems technical another might regard as fundamental. But in most cases it refers to a situation where a rather intricate argument is required to establish a certain statement, and usually one uses this term in the situation where this extra effort doesn't necessarily lead to any greater clarity of the big overall idea.

One example I can think of is the Fubini-Tonelli theorem in measure theory. I think intuitively this theorem makes a lot of sense "if you add non-negative things up in two different ways you should get the same result", but of course if you've seen the proof of this theorem it is no trivial task. For example the proof I know (the one in Folland's book) requires the use of the Monotone class lemma; I would count this as a rather technical lemma. It also requires $\sigma$-finite measure spaces, and so just by reading the proof one immediately sees that there's a lot of facts to be verified, and while all of them are plausible sounding, none of them is immediately obvious to prove.

Also as is common in measure theory, many results are usually easily proven in the case where the measure is finite, and then one extends the argument to $\sigma$-finite spaces. And in certain situations, one can then extend the argument to yet more general situations. For example, consider the duality theorem:

For any measure space $(X,\mathfrak{M},\mu)$ and any $1<p<\infty$ with $q$ the conjugate Holder exponent, the mapping $L^q(\mu)\to (L^p(\mu))^*$ defined by $g\mapsto \left(f\mapsto \int_X fg\, d\mu\right)$ is an isometric isomorphism of Banach spaces.

The real crux of this theorem lies in the case when $\mu$ is a finite measure. From here, extending to the case of $\sigma$-finiteness is rather simple (we're just adding up countably many results and patching everything together nicely). Then finally, one can extend the theorem to any arbitrary measure space by a rather intricate argument which essentially reduces is back down to the $\sigma$-finite case. Also, in trying to establish the more general cases, there are no "new ideas" involved in the proof.

So, one could say this theorem requires a rather technical argument to establish in full generality (and also, since most people work with $\sigma$-finite spaces anyways, this last bit of extra generality, while nice to know, will certainly be deemed as additional technicalities).

An example of a technical definition, rather than a technical argument is in the definition of integration on manifolds. For this, one commonly uses a partition of unity, and as you can see from the definition of a partition of unity itself, proving the existence (in the smooth case) is not a trivial task at all, and requires a rather intricate argument. Moreover, the definition of integration on manifolds using partitions of unity is hardly useful in actual computations (though very useful for theoretical proofs), and as such it is regarded as a very technical definition.

Also, take a look at the usual proofs of the multivariable change of variables theorem as presented in Spivak or Munkres... god that's an awful theorem to prove (sure, one can try to finesse the proof using some measure theory, using Radon-Nikodym derivatives etc, but this is, I think, one of the most basic yet technical theorems in integral calculus).

In fact, depending on how far along one is in their mathematical journey, even the proof that every continuous function on a compact interval $[a,b]$ is Riemann-integrable (a result which most people take for granted) is somewhat technical because it uses the notion of uniform continuity (well I'm sure you can avoid it as Spivak does in his calculus text, but still, this proof is not at all transparent). (I realize that so far all my examples relate to integration; this is unintentional... that's all that immediately springs to mind immediately).

Another example I can think of is the inverse function theorem from differential calculus:

If $f:\Bbb{R}^n\to\Bbb{R}^n$ is $C^r$ and $a\in\Bbb{R}^n$ is a point such that $Df_a$ is an isomorphism then there exist open neighborhoods $A$ of $a$ and $B$ of $f(a)$, such that the resitrction $f:A\to B$ has a $C^r$ inverse $f^{-1}:B\to A$.

Conceptually, this is simple enough to understand: a derivative by definition is a local linear approximation to the function. So, if the derivative at a point is invertible then it would be nice to expect that the function itself inherits this same nice property. In the linear case, what this is saying is that if $M$ is an invertible matrix then the equation $y=Mx$ can be solved for $x$ as $x=M^{-1}y$.

Now, in the course of proving the inverse function theorem, what one usually does is start by a sequence of approximations; then one has to invoke certain technical lemmas such as (Banach's I think) fixed point theorem, and then one has to show that the limit of the sequence of approximations actually converges to the right thing etc. These are all I think somewhat technical details.  I think more important is realizing that this is one of many instances of "behavior of derivative implies something about behavior of function", and it is also important to recognize and know how to prove the equivalence between the inverse and implicit function theorems.


问:
How does a Gaussian Process define a probability distribution in the functions space?
I am studying Gaussian Process Regression. I will post a text from the book Gaussian Process for Machine Learning, by C. E. Rasmussen & C. K. I. Williams:

We first consider a simple 1-d regression problem, mapping from an input x to an output f (x). In Figure 1.1 (a) we show a number of sample functions drawn at random from the prior distribution over functions specified by a particular Gaussian process which favors smooth functions. This prior is taken to represent our prior beliefs over the kinds of functions we expect to observe, before seeing any data.

The definition of a Gaussian process that I know of is given by:

Definition (Gaussian process) A random process $\left(X_{t}\right)_{t \in T}$ is called a Gaussian process if, for any finite subset $T_{0} \subset T,$ the random vector
$$
\left(X_{t}\right)_{t \in T_{0}}
$$
has normal distribution. Equivalently, $\left(X_{t}\right)_{t \in T}$ is Gaussian if every finite linear combination
$$
\sum_{t \in T_{0}} a_{t} X_{t}
$$
is a normal random variable.

I would like to better understand the connection of this definition with the possibility of defining a distribution in the functions space, because to draw functions according to a distribution, I would have to define sigma algebras in the function space and a measure of probability in this sigma algebra. Or perhaps there is another meaning when we say "sample functions drawn at random from the prior distribution over functions specified by a particular Gaussian process".

Does anyone help to build this bridge?

答:
The fact that a stochastic process defines a function-valued random variable is simple enough. Say you have $X : \Omega \times T \to \mathbb{R}$. Then you can consider the mapping $\omega \mapsto (t \mapsto X(\omega,t))$ to be a function-valued random variable: for each fixed $\omega$ you have a function that maps $T$ into $\mathbb{R}$. In shorthand you would call this function $X(\omega,\cdot)$.

It's the other matters, about what the $\sigma$-algebra on the function space is and what the probability measure is, that are more technical. Rather than going into all these details here, I will simply say that:

the usual $\sigma$-algebra on function spaces is the one generated by the so-called cylindric sets, which take the form $\{ \omega : X(\omega,t_0) \in I \}$ where $t_0$ is a fixed element of $T$ and $I$ is an interval.

the probability measure on the cylindric sets can be shown to exist by using the so-called Kolmogorov extension theorem, once you have the finite dimensional distributions of the process.


问:
Is it possible to use the matrix element-wise max norm when checking if a vector-valued map is a contraction?
Let $f:\mathbb R^2 \to \mathbb R^2$. Using the MVT for vector-valued function
$$
f(x)-f(y)= \nabla f(c) \cdot (x-y),
$$
for some $c$ on a curve connecting $x$ and $y$.

If
$$
\| \nabla f(c) \|_{max} = \max_{i,j} \left| (\nabla f(c))_{i,j} \right| \le \lambda,
$$
for any $c$, can I infer that $f$ is Lipschitz with Lipschitz constant $\lambda$?
That is, to conclude
$$
\|f(x) - f(y) \| \le \lambda \|x-y\|?
$$
I asked because some reference infer that $f$ is Lipschitz using a matrix norm on the Jacobian,
while wikipedia mentioned that the Jacobian should only be bounded.

For example, for
$$
f(x,y)=\left( \frac{2}{1/x+1/y},\sqrt{xy}\right), \text{ with } x,y \in [1,2],
$$
then, element-wise,
$$
\nabla f \le \begin{pmatrix} \frac{8}{9} & \frac{8}{9} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{pmatrix},
$$
so
\begin{align}
\|\nabla f \|_{max} &= \frac{\sqrt{2}}{2} < 1, \\
\|\nabla f \|_1 & =\frac{8}{9}+\frac{\sqrt{2}}{2} > 1 \\
\|\nabla f \|_\infty & =\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} > 1
\end{align}
So that only by using $\|\cdot\|_{max}$, I can conclude that $f$ is a contractive map.

That is I would like to conclude
$$
\|f(x)-f(y)\|_1 \le \frac{\sqrt{2}}{2} \|x-y\|_1.
$$

答:
some reference infer that f is Lipschitz using a matrix norm on the Jacobian, while wikipedia mentioned that the Jacobian should only be bounded.

To show that a differentiable map $f:\mathbb R^2\mathbb R^2$ is Lipschitz, it is sufficient to prove a uniform bound for some norm of its Jacobian matrix. It does not matter which norm you choose: $\max_{ij} |a_{ij}|$, $\sum_{ij} |a_{ij}|$,  $\left(\sum_{ij} |a_{ij}|^2\right)^{1/2}$, the operator norm $\max_{\|x\|=1} \|Ax\|$, or something else. Any two norms on a finite dimensional vector space are comparable.

But if want to determine the Lipschitz constant of $f$, the right norm to use is the operator norm $\max_{\|x\|=1} \|Ax\|$. Indeed, the operator norm of the Jacobian matrix at point $a$ is equal to $\limsup_{x\to a}\dfrac{\|f(x)-f(a)\|}{\|x-a\|}$. From here (and the mean value theorem) one obtains that the supremum of the operator norm is equal to the  Lipschitz constant of $f$.


问:
Proving vector space property $(-c)*v=c*(-v)$
Let $V$ be a vector space with a field $F.$

Prove that $\forall v \in V$ and $c \in F$ $(-c)*v=c*(-v)$ using properties of vector spaces and fields.

$(-c)*v = (-1*c)*v = (c*-1)*v = c*(-1*v) = ...$

I'm not how to show $-1*v=-v$

答:
To prove $-1\cdot v=-v$, you can just notice that $v+(-1\cdot v)=0$, doing the following:
$$v+(-1\cdot v)=(1-1)\cdot v=0\cdot v=v,$$
so you have proven that $v+(-1\cdot v)=0$, so this means that $(-1\cdot v)$ is the inverse by addition of the element $v$ (commonly called opposite), denoted $-v$, so we can conclude that
$$-1\cdot v=-v$$


问:
Characterization of $H^{1/2}(\partial \Omega)$ norm?
I can't understand why $\|g\|_{H_{\partial}^{1/2}} = \|u\|_{H^1} $ where $u$ solves the problem
\begin{equation*}
- \Delta u + u = 0, \quad \Omega\\
u|_{\partial \Omega} = g
\end{equation*}

The weak form should be to find $w \in H^1_0$:

\begin{equation*}
(w,v)_{H^1} = -(G, v)_{H^1} , \quad \forall v \in H^1_0
\end{equation*}

where $ u = w+G$ and $G$ is the lifting of $g$.

$(.,.)_{H^1}$ is the $H^1$ inner-product.

Now I can only get this estimate from the Lax-Milgram lemma:

$$
\|u\| -\|G\| \leq \|w\| \leq \|G\|
$$

And so:

$$
\|g\| \leq \|u\| \leq 2\|g\|
$$

As the $H^{1/2}$ norm is the $\inf$

But how to get equality?

答:
Recall that $H^{1/2}(\Gamma):=\gamma_0(H^1(\Omega))$, where $\gamma_0(u):=u|_{\partial \Omega } $. Then we have that $\ker \gamma = H_0^1(\Omega)$. By the orthogonal descomposition $H^1(\Omega) = H^1_0(\Omega) \oplus H_0^1(\Omega)^{\perp}$.  The restricted trace operator $\tilde {\gamma}_0:H_0^1(\Omega)^{\perp} \to H^{1/2}(\Gamma)$ is a linear isomorphism.

Now for any $g \in H^{1/2}(\Gamma)$ there is a $u = u_0+u^{\perp} \in H^1_0(\Omega) \oplus H_0^1(\Omega)^{\perp}$  with $g = \gamma_0(u)$. In particular $g=\gamma_0(u^{\perp})$, immediately we obtain $\tilde{\gamma}_0^{-1}(g)=u^{\perp}$, then

$$\|g\|_{1/2} = \inf_{u \in H^1(\Omega):\gamma_0(u)=g} \|u\|_{H^1}= \inf_{u \in H^1(\Omega):\gamma_0(u)=g} \|u_0+\tilde{\gamma}^{-1}(g)\|_{H^1}.$$
This infimum reachs when $u_0=0$, then
$$\|g\|_{1/2} = \|\tilde{\gamma}^{-1}(g)\|_{H^1}=\|u^{\perp}\|_{H^1}$$

Finally, is not too complicate to show that $u^{\perp}$ is the weak solution, because
\begin{align}
H_0^1(\Omega)^{\perp} & = \{u^{\perp}:\int_{\Omega}(u^{\perp}w+\nabla u^{\perp}\cdot \nabla w ) = 0 \quad \forall w \in H^1_0(\Omega) \} \\
&= \{u^{\perp} \in \mathcal{D}'(\Omega):-\Delta u^{\perp}+u=0\}.
\end{align}


问:
Why $\frac{1 + \sqrt{d}}{2}\notin \{a+b\sqrt{d}\in\mathbb{C}:a,b \in\mathbb{Z}\} $?
Let $d\in \mathbb{Z}-\{0\}$ be an Square-free integer ${\displaystyle d\equiv 1{\pmod {4}}}$. Why $\frac{1 + \sqrt{d}}{2}\notin \{a+b\sqrt{d}\in\mathbb{C}:a,b \in\mathbb{Z}\} $ ?

答:
If $\frac{1 + \sqrt{d}}{2} = a + b \sqrt{d}$ then
$$\sqrt{d} = \frac{2a-1}{1-2b}
$$
If $a,b \in \mathbb Z$ this proves $\sqrt{d}$ is rational (using that $b \ne \frac{1}{2}$). But if $d$ is square free then $\sqrt{d}$ is not rational.


问:
Specific question about example of Family of sets & Set builder notation
I encountered the following example whilst studying Equivalence Classes from How to prove it, Velleman. :

Let $ B= \{(p, q) ∈ P × P | \text{the person p has the same birthday as the person q } \} $

Let D be the set of all possible birthdays. In other words,$ D = \{Jan. 1, Jan. 2, Jan. 3, . . . ,Dec. 30, Dec. 31\}. $

Now for each $d ∈ D$, let $P_d = \{p ∈ P | \text{the person p was born on the day d } \} $ .

Then the family $ F = \{P_d | d ∈ D\} $  is an indexed family of
subsets of P.

We can then redefine B as:

$B = \{(p, q) ∈ P × P | ∃d ∈ D(p ∈ P_d \land q ∈ P_d )\}$

$= \{(p, q) ∈ P × P | ∃d ∈ D((p, q) ∈ P_d × P_d )\}$

$ = ∪_{d∈D}(P_d × P_d ). $

My question:
Could I write $ \{(p, q) ∈ P × P | ∃d ∈ D((p, q) ∈ P_d × P_d )\}$ as $ \{(p, q) ∈ P_d × P_d | d ∈ D\}$?

I think yes ( but I want to be sure ). My reasoning is since inside the set builder notation we have $ (p,q) ∈ P × P \land (p,q) ∈ P_d × P_d $ then $ (P × P) \land (P_d × P_d) = (P_d × P_d)$ Since $   P_d × P_d \subseteq P × P  $.

答:
This is not standard set builder notation and I would advise against using it. The general scheme is:
$$
B:=\{x\in A\mid\varphi(x)\}
$$
Here the collection $A$ serves as an upper bound on where the $x$'s come from, so that you immediately know that $B\subseteq A$. This notation is a direct implementation of the axiom of restricted comprehension, which tells us that if $A$ is a set, then $B$ as defined above is a set as well.

Crucially, the set $A$ is fixed before defining $B$! In your notation
$$
\{(p, q) ∈ P_d × P_d | d ∈ D\}
$$
however, the role of $A$ is taken by $P_d \times P_d$ which depends on the parameter $d$ which in turn varies over $D$. While it is clear what you want to define, your relaxed notation makes it very easy to write down set builder terms which denote proper classes, for example:
$$
\{x\in A\mid A\text{ is a set with at most $2$ elements}\}
$$


问:
Integral involving exponentials of cosh functions
I recently encountered the following integral

$$I_p\equiv \int\limits_0^{+\infty}\,\int\limits_{-\infty}^{+\infty}\,e^{-p\,[c+(c^2+1)\cosh x]}\,\mathrm{d}x\,\mathrm{d}c\,,$$

with $p>0$. Can $I_p$ be calculated in terms of elementary functions?

Update:

Using the following integral representation for the Bessel function

$$
\int\limits_0^{+\infty}e^{-a\cosh x}\;\mathrm{d}x=K_0(a)\,,
$$

one can bring $I_p$ to the following form:

$$
I_p=2 \int\limits_0^{+\infty}e^{-c\,p}\;K_0\left[\left(c^2+1\right) p\right]\;\mathrm{d}c\,.
$$

Which shows that $I_p$ can be seen as a Laplace transform. However, I haven't managed to do the last integral either.

答:
Given the mess that results in doing just one of the integrals, I think it would be shocking if your integral had a closed form in elementary functions. We will find the large $p$ behavior. Let

$$
I(p)=\int\limits_0^\infty dx \ e^{-px} K_0\left[p(x^2+1)\right]
$$

Using the asymptotic expansion for $K_0$

$$
K_0(z)\sim\sqrt{\frac{\pi}{2z}}e^{-z}\sum\limits_{n=0}^\infty c_n z^{-n} \ \ , \ z \to \infty
$$

We will not need the form of the $c_n$ here. When we substitute $z=p(x^2+1)$, we will get a Gaussian $e^{-p(x^2+1)}$; we will use (essentially) Laplace's method to do the integral for large $p$. When we do, all the factors of $(x^2+1)$ outside the exponent will be evaluated at the maxima $x=0$. We will be left with

$$
\sqrt{\frac{\pi}{2p}}e^{-p(x^2+1)} \sum\limits_{n=0}^\infty c_np^{-n}
$$

But this simplifies because the sum is known

$$
K_0(p)\sim\sqrt{\frac{\pi}{2p}}e^{-p}\sum\limits_{n=0}^\infty c_np^{-n} \ \ , \ \ p \to  \infty
$$

So we get

$$
I(p)\sim \int\limits_0^\infty dx \ e^{-px} e^{p-p(x^2+1)} K_0(p) \ \ , \ \ p \to \infty
$$

The integral is an error function

$$
I(p) \sim \frac{1}{2}\sqrt{\frac{\pi}{p}}e^{p/4}\operatorname{erfc}\left(\frac{\sqrt{p}}{2} \right) K_0(p) \ \ , \ \ p \to \infty
$$

We can simplify further (thanks @Gary). Expand $K_0$ and $\operatorname{erfc}$ for large arguments, yielding

$$
I(p) \sim \sqrt{\frac{\pi}{2}} p^{-3/2}e^{-p} \quad , \quad p \to \infty
$$

Here is a plot of the approximations versus numeric result for 'large' values of $p$ less than $1$:


问:
Solving a wave equation with boundary conditions on a circle?
How might one solve a wave equation which had boundary conditions on a circle?

i.e. given $$(\partial^2_x+\partial^2_y)\phi(x,y)=0$$

And known values, $f$ on a circle:

$$\phi(\cos(\theta),\sin(\theta)) = f(\theta)$$

How would one find a solution for $\phi$ in terms of $f$?

答:
This is a suplimentary answer.
Taking Michael's answer above and feeding back in the Fourier components of $f(\theta)$ using e.g.

$$C_n = \frac{1}{\pi}\int f(\theta) \sin(n\theta) d\theta$$
$$D_n = \frac{1}{\pi}\int f(\theta) \cos(n\theta) d\theta$$

Then doing the summation, (which the main part is a simple power series in:

$$2\Re[ \sum_{n=0}^{\infty} (r \exp(i(\theta-\beta))^n]-1$$

I find that I can write the solution as:

$$\phi(r,\theta) =  \frac{1}{2\pi}\int\limits_0^{2\pi}\left(\frac{1 -  r^2 }{1-2r \cos(\theta-\beta)+ r^2}\right)f(\beta)d \beta$$

which works everywhere except at the boundary where you must take the limit as $r\rightarrow 1$.


问:
Divergence of $\frac{x}{||x||^n}$ identical zero?
Currently I am stuck at an exercise about the Divergence Theorem
in which I am asked to prove:

$$
\int_{\partial A} \frac{\cos(\alpha(x, \nu))}{||x||^{n-1}}dS(x) = \operatorname{Vol}_{n-1}(S^{n-1}),
$$

where $A$ is a compact subset of $\mathbb{R}^n$ with "the origin in its center",
$\alpha(x,\nu)$ is the angle between $x$ and $\nu$, the normal unit vector on $\partial A$.

We should prove it by applying the Divergence Theorem on $F(x) = \frac{x}{||x||^n}$ on $A_\epsilon = \{x \in A|\,\,||x|| > \epsilon\}$.

I see how applying the Divergence Theorem on $F$ yields the statement
to prove, but when I calculate the Divergence of $F$, I always get
identical zero.
I am pretty sure I am missing something obvious, but I don't have any idea.
I hope you can help me!

答:
This is not a strict math proof - rather an illustration.

We have to evaluate $\int_{\partial A} \frac{\cos(\alpha(x, \nu))}{||x||^{n-1}}dS(x)=\oint_{\partial A} \frac{(\vec{r},d\vec{S})}{r^{n}}$, where $\vec{r}$ is radius vector from the origin, $r=|\vec{r}|$, and $d\vec{S}=\vec{n}dS$, where $\vec{n}$ is a normal unit vector on $\partial{A}$.

We can divide the subset $A$ into arbitrary layers covering the origin ($A=A_1\cup{A}_2\cup{A_3}...$) and, therefore we also get internal boundaries of these layers. We can notice that $\oint_{\partial A_k} \frac{(\vec{r},d\vec{S})}{r^{n}}=0$, as soon as $A_k$ does not include the origin.

Indeed, for all external layers (which covers but do not contain the origin) $\oint_{\partial A_k} \frac{(\vec{r},d\vec{S})}{r^{n}}=\int_{A_k} (\vec\nabla\frac{\vec{r}}{r^{n}})dV$

and in Cartesian coordinate system (summation over repeated indices is implied)

$$\vec\nabla\frac{\vec{r}}{r^{n}}=\frac{\partial}{\partial{x}_i}\frac{x_i}{r^n}=\frac{r^2\delta_{ii}-nx_ix_i}{r^{n+1}}=\frac{r^2n-n{r}^2}{r^{n+1}}=0$$

($\delta_{ik}$ - Kronecker symbol; $\delta{ii}=\sum_{i=1}^n1=n$).

On the other hand

$\oint_{\partial A_k} \frac{(\vec{r},d\vec{S})}{r^{n}}=\int_{\partial A_k (external)} \frac{(\vec{r},d\vec{S})}{r^{n}}+\int_{\partial A_k (internal)} \frac{(\vec{r},d\vec{S})}{r^{n}}$ - difference of total flows through external and internal surfaces of the layer $A_k$.

But $\int_{\partial A_k (internal)} \frac{(\vec{r},d\vec{S})}{r^{n}}=-\int_{\partial A_{k-1} (external)} \frac{(\vec{r},d\vec{S})}{r^{n}}$ - for the external boundary of second layer which is closer to the origin and shares the boundary with $A_k$ (we get a different sign because the normal for each layer points outward).

Due to arbitrary choice of $A_k$ it means that the integral flow through any external boundary of any layer is the same - as soon as the layer covers the origin. We can move as close to the origin as we want, and choose an arbitrary surface covering the origin - for example, it can be a shere $S_{n-1}$ of some radius $\epsilon\to0$

$$\oint_{\partial A} \frac{(\vec{r},d\vec{S})}{r^{n}}=\int_{\partial{R^n_\epsilon}}\frac{1}{\epsilon^{n-1}}dS_{\epsilon}=\frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\frac{\epsilon^{n-1}}{\epsilon^{n-1}}=\frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})} =\operatorname{Vol}_{n-1}(S^{n-1})$$

(surface area of a unit ball $R^n$ of dimension $n$)


问:
Notation $\overline{K}(C)^*$ in Silverman's "The Arithmetic of Elliptic Curves"
In Joe Silverman's "The Arithmetic of Ecliptic Curves" he talks a lot about the integral domain $\overline{K}(C)$. On page $27$ he suddenly decides to chose an element $f$ from some set $\overline{K}(C)^*$. What does this notation mean? After page $27$ he begins using $\overline{K}(C)^*$ a lot, and I haven't been able to pick up what it is from context clues.

答:
$\bar{K}(C)$ is the function field of $C$ over the algebraically closure of $K$ (as recalled in the beginning of chapter).

$\bar{K}(C)^*$ is the multiplicative group of units of $\bar{K}(C)$.

The notation $R^*$ for the multiplicative group of units of a ring $R$ is standard.  If $R$ is a field then $R^*=R-\{0\}$.


答:
More generally, show:

Lemma: If $d\in\mathbb Z$ is not a perfect square, and $a_1,a_2,b_1,b_2\in \mathbb Q$ and $$a_1+b_1\sqrt d =a_2+b_2\sqrt d$$ then $a_1=a_2$ and $b_1=b_2.$

Hint: Use that $\sqrt d$ is not rational. It is either an irrational real number if $d>0$ or imaginary if $d<0.$

In particular, if $a_1=b_1=\frac12,$ then there are no integers $a_2,b_2,$ since $\frac 12\notin \mathbb Z.$

So, you don’t need $d\equiv 1\pmod 4,$ nor do you need $d$ square-free (it works for $d=12,$ or $d=-4,$) but you do need to exclude $d=1.$

Indeed, if $d=1$ then:

$$\frac12+\frac12\sqrt 1=1+0\sqrt 1.$$

Title: Why $\frac{1 + \sqrt{d}}{2}\notin \{a+b\sqrt{d}\in\mathbb{C}:a,b \in\mathbb{Z}\} $?

Body: Let $d\in \mathbb{Z}-\{0\}$ be an Square-free integer ${\displaystyle d\equiv 1{\pmod {4}}}$. Why $\frac{1 + \sqrt{d}}{2}\notin \{a+b\sqrt{d}\in\mathbb{C}:a,b \in\mathbb{Z}\} $ ?


问:
How do I expand this to a power series?
I am given the function

$$f(x) = \dfrac{6x^2}{(16+2x^3)^2}$$

I am stuck trying to expand $f(x)$ to a power series with center of $0$. I cannot seem to figure how to get $f(x)$ to a clean geometric series form or manipulate it and am confused how to go about it. Thanks!

Image 1

答:
A Maclaurin series is given by

$$f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}x^k$$

In our case we have:
$$f\left(x\right) \approx P\left(x\right) = \sum\limits_{k=0}^{5}\frac{f^{(k)}\left(a\right)}{k!}x^k$$

I stop me for $n=5$. You can calculate the derivates and you will have

$$f\left(x\right)\approx\frac{0}{0!}x^{0}+\frac{0}{1!}x^{1}+\frac{\frac{3}{64}}{2!}x^{2}+\frac{0}{3!}x^{3}+\frac{0}{4!}x^{4}+\frac{- \frac{45}{64}}{5!}x^{5}\iff P\left(x\right) = \frac{3}{128}x^{2}- \frac{3}{512}x^{5}$$

At the end:

$$\frac{6 x^{2}}{\left(2 x^{3} + 16\right)^{2}}\approx P\left(x\right)=\frac{3}{128}x^{2}- \frac{3}{512}x^{5}$$

Title: How do I expand this to a power series?


问:
Solving a cubic equation on an interval
$d(t)=-3t^3+45t^2+600t$ where $0\leq t \leq 10$

What is the value of $t$ if $d(t) = 3750$?

Practice question from my textbook I'm having a little trouble with for some reason. Would like to have the steps clarified In case this kind of question shows up in a future test. Thanks.

答:
You have to solve the cubic equation $-3t^3+45t^2+600t = 3750$ which is equivalent to
$$t^3 -15t^2 -200t +1250 = 0. \tag{1}$$

In my opinion the best strategy is to verify first whether it has a rational root. Here all coefficients are integers and it is well-known that a rational root (if it exists) must be an integer which divides $1250$. We have $1250 = 2 \cdot 5^4$ and a simple trial shows that $t = 5$ is a root. For the other roots you need to solve
$$t^2 - 10t -250 = 0$$
which gives $t = 5 \pm 5\sqrt{11}$. Among these three roots $t = 5$ is the only one between $0$ and $10$.

Cubic equations can also be solved via the Cardano formula. However, we know that $(1)$ has three real roots, thus we are in the so-called casus irreducibilis. This means that the roots of $(1)$ are represented as sums of two complex numbers - and this is not very nice.

Title: Solving a cubic equation on an interval


问:
Bounded operator such that image of ball is $(1-\varepsilon)$-dense is surjective.
This is not a homework problem, I found it in some notes and I am curious on how to prove it. The statement is as follows:

Let $T:X\to Y$ be a bounded linear operator between Banach spaces. If there is $\varepsilon>0$ and $R<\infty$ such that $T(B_X(R))$ is $(1-\varepsilon)$-dense in $B_Y(1)$ then $T$ is surjective.

Here $(1-\varepsilon)$-dense means that for every point in $B_Y(1)$ the ball of radius $(1-\varepsilon)$ intersects $T(B_X(R))$.

What I've thought so far:

It is enough to show that $T(B_X(R))$ contains some ball around $0\in Y$.

Assume that this is not the case so that for every $n$ the ball $B_Y(\frac{1}{n})$ is not contained in $T(B_X(R))$. By the $(1-\varepsilon)$-density we have that for any $y\in B_Y({\frac{1}{n}})\setminus T(B_X(R))$ there is some $z\in T(B_X(R))$ such that $d(y,z)<(1-\varepsilon)$. Furthermore, we have $d(0,z)\le d(0,y)+d(y,z)<\frac{1}{n}+(1-\varepsilon)$. If we let $n$ be large enough so that $(1-\varepsilon)+\frac{1}{n}<1$ then $z\in B_Y(1)$.

I don't know if this works or makes any sense and I don't know how to continue from here. Any help, hints or solutions would be appreciated.

答:
For simplicity adapt the statement to closed unit balls and $≤$ instead of $<$. Let $y\in Y$ lie in the unit ball and find some $z_0=T(x_0)\in T(B(R)_X)$ so that $\|y-z_0\|≤1-\epsilon$. Define $k_0:=\|y-z_0\|$. Then you've got a $z_1=T(x_1)\in T(B(R)_X)$ with
$$\|\frac{y-z_0}{k_0}-z_1\|≤1-\epsilon$$
which implies
$$\|y-z_0 - z_1\cdot k_0\|≤(1-\epsilon)k_0≤(1-\epsilon)^2$$
continue like this to get a sequences $z_n =T( x_n)$ and $k_n$ with:
$$\|y-z_0-k_0z_1-k_0k_1z_2-... -k_0\cdot...\cdot k_{n-1}z_n\|≤(1-\epsilon)^n$$
whence the sum
$$\sum_{n=0}^\infty z_n\prod_{i=0}^{n-1}k_i$$
converges to $y$. But each of the $k_i$ are less than $1-\epsilon$ and the $z_n$ are contained in $T(B(R)_X)$, so the norms of the summands are bounded by
$$\|T\|\,R\,(1-\epsilon)^{n-1}$$
and the sum converges absolutely. Indeed the sum
$$\sum_n x_n\prod_i k_i$$
converges in $X$ by the same argument. If you denote the limit with $x$ you then get with continuity of $T$:
$$T(x) = T(\sum x_n\prod k_i) = \sum T(x_n)\prod k_i = \sum z_n\prod k_i = y$$
and any $y$ in the unit ball of $Y$ admits a pre-image under $T$.


问:
Use Newton's Binomial Theorem to prove series
Verify that for all integer $n\ge 2$,
\begin{eqnarray*}
\sum_{k=0}^n\binom{\frac{1}{2}}{k}\binom{\frac{1}{2}}{n-k}=0.
\end{eqnarray*}
(Hint: Consider Newton's Binomial Theorem.)

=================================

EDIT:

From Newton's Binomial Theorem:
\begin{eqnarray*}
(1+x)^p &=& \sum_{n=0}^\infty \begin{pmatrix}
p \\n
\end{pmatrix} x^n.
\end{eqnarray*}
So,
\begin{eqnarray*}
A: (1+x)^\frac{1}{2} &=& \sum_{k=0}^\infty \begin{pmatrix}
\frac{1}{2} \\k
\end{pmatrix} x^k\\
B: (1+x)^\frac{1}{2} &=& \sum_{k=n}^\infty \begin{pmatrix}
\frac{1}{2} \\{n-k}
\end{pmatrix} x^{n-k}
\end{eqnarray*}

I assume finding $A-B$ will lead to the answer $0$ but I'm not sure how to continue from here

答:
Hint:

We know from the binomial theorem that,
$$
(1+x)^{1/2}=\sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}x^k
$$
(for suitable $x$).

This means, in particular, that
$$
1+x=(1+x)^{1/2}\cdot(1+x)^{1/2}=\left[\sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}x^k\right]^2.
$$
Note that for any $n\geq 2$, the coefficient by $x^n$ on the left side is $0$.  Therefore the coefficient by $x^n$ (where $n\geq2$) must also be $0$ on the right side.

The coefficient by $x^n$ in the right side can be computed using the Cauchy product.


问:
Prove $(A_1 \smallsetminus B) \cup (A_2 \smallsetminus B) = (A_1 \cup A_2) \smallsetminus B$

Prove $(A_1 \smallsetminus B) \cup (A_2 \smallsetminus B) = (A_1 \cup A_2) \smallsetminus B$

If I used the set identity to prove it and it seems it goes on and on, did I make a mistake or ...

$(A_1 \cap \lnot B) \cup (A_2 \cap\lnot B)$ and then I use the distributive law and ...

$((A_1 \cap \lnot B) \cup A_2) \cap (A_1 \cap \lnot B) \cup \lnot B)$

答:
Isn't the distributive law just $(M\cap N) \cup (S\cap N)= (M\cup S)\cap N$?

So $(A_1\setminus B)\cup (A_2\setminus B) = $

$(A_1\cap B^c)\cup (A_2\cap B^c) =$

$(A_1\cup A_2)\cap B^c =$

$(A_1\cup A_2)\setminus B$.

Doesn't seem to go "on and on".


答:
You're solving a cubic here. While there is a general solution, it's no picnic. For most "elementary" (school-style) problems, you won't be expected to apply the full cubic solution. Instead you'll be expected to apply some combination of Rational Root Theorem, Factor Theorem and Polynomial Long Division (or a shortcut like Synthetic Division) to reduce the degree to a quadratic you can solve with methods you already know. But this requires the cubic to have at least one "nice" rational root, which often (but not always) turns out to be an integer.

Let's proceed on the assumption that this is a problem specifically designed to be solvable at an elementary level.

Rearrange the cubic and make it monic (lead coefficient one):

$t^3 - 15t^2 - 200t + 1250 = 0$

By Rational Root Theorem, any rational root (if it exists) will need to divide $1250$. You'll have to consider both positive and negative possibilities for each number.

$1250 = (2^1)(5^4)$

This means it has a total of $(1+1)(4+1)= 10$ factors, considering only the positive ones. Double this for the negatives, and you're looking at $20$.

At this point, you're going to have to "appeal" to the good sense of the question setter and hope the root you're looking for is within that narrow range they stipulated. Which means you'll start by testing $1, 2, 5, 10$. You don't even need to get to the end of that list because you'll hit paydirt at $5$.

Once you know $t=5$ is a root, you can immediately deduce that $(t-5)$ is a factor of the cubic polynomial (this is Factor Theorem). You can then divide the cubic by $(t-5)$ and get a quadratic you can easily solve by any method you've learned (factorisation by inspection, completing the square or quadratic formula).

The division can be simplified by Ruffini's method of synthetic division. The quadratic you get as the result will be $t^2 - 10t - 250$, for which the roots (by formula) are $5(1 \pm \sqrt{11})$. Neither of these values is in the required range, so you can reject them after finding them. But finding them is a must, for rigour.

Now, what would've happened if, in the original application of Rational Root Theorem, you couldn't find a root among the candidates $1,2,5,10$? Technically you have a few choices. You can go on to test the remaining $20-4 = 16$ possibilities. It is possible you'll get a nice rational root outside the stipulated range that, after polynomial division, gets you to a quadratic which does have one or two root(s) in the stipulated range. But that would have to be a pretty sadistic question setter. Not impossible, just unpleasant.

But it's still a more pleasant undertaking than actually solving the cubic using the general solution, which I won't be going into. It's far too tedious and involved.

Anyway, the only solution for your question in the required range is $t=5$.

Title: Solving a cubic equation on an interval

Body:
$d(t)=-3t^3+45t^2+600t$ where $0\leq t \leq 10$

What is the value of $t$ if $d(t) = 3750$?

Practice question from my textbook I'm having a little trouble with for some reason. Would like to have the steps clarified In case this kind of question shows up in a future test. Thanks.


问:
Example of a group which is abelian and has finite (except the $e$) and infinite order elements.

Exercise 7: Show that the elements of finite order in an abelian
  group $G$ form a subgroup of $G$

I just solved this exercise but I can't find example of a group which is abelian and has finite (except the $e$) and infinite order elements. Are there any well known such groups?

For not abelian group with other properties group of matrixes I know.

答:
Any infinite finitely generated abelian group that is not a free group will satisfy this property. This follows from the classification theorem, as any such group will have the form

$\mathbb{Z}_{m_{1}}\times\cdots\mathbb{Z}_{m_{r}}\times\mathbb{Z}^{n}$

for some $m_{1},\ldots,m_{r}\geq 2$ and some $n\geq 1$, and $m_{1}\vert m_{2}\vert\cdots\vert m_{r}$.

The example $\mathbb{Z}_{2}\times \mathbb{Z}$ given by pritam above is the the easiest such example and is a great one to show why the given property is satisfied.


答:
Distribution is the next step, but you have to distribute out the common factor.  You may be familiar with distributing like this. $$(A_1\cup A_2)\cap\lnot B_1 ~=~ (A_1\cap\lnot B)\cup(A_2\cap\lnot B)$$  However, note that distribution is an equivalence -- it works in both directions.
$$(A_1\cap\lnot B)\cup(A_2\cap\lnot B) ~= ~(A_1\cup A_2)\cap\lnot B_1$$

Body:
Prove $(A_1 \smallsetminus B) \cup (A_2 \smallsetminus B) = (A_1 \cup A_2) \smallsetminus B$

If I used the set identity to prove it and it seems it goes on and on, did I make a mistake or ...

$(A_1 \cap \lnot B) \cup (A_2 \cap\lnot B)$ and then I use the distributive law and ...

$((A_1 \cap \lnot B) \cup A_2) \cap (A_1 \cap \lnot B) \cup \lnot B)$


问:
Two sets have the same cardinal proof
Suppose that $S, T$ are sets such that $S$ is infinite and $T = \{t_i: i = 1, ..., n\}$ with $T$ having $n \geq 1$ distinct elements. If $S \cap T = \{\}$, show that $card(S) = card(S \cup T)$.

Hint: Construct a bijective function $\phi: S \rightarrow S \cup T$.

I tried defining various functions but cannot seem to construct one that satisfies the requirements above. I would appreciate if someone could provide such a bijective function.

答:
If $S$ is infinite then there exists $C\subset S$ such that $C$ is countable infinite.

Therefore $C=\{c_1,c_2,...\}$. Now, consider $\phi(z)=z$ for $z\notin C$, $\phi(c_k)=t_k$ for $k\in\{1,2,...,n\}$ and finally $\phi(c_{n+k})=c_{k}$ for $k\geq 1$.

This function is surjective because $\phi(S\setminus C)=S\setminus C$, $\phi(\{c_1,c_2,...,c_n\})=\{t_1,t_2,...,t_n\}$ and finally $\phi(\{c_{n+1},c_{n+2},...\})=\{c_1,c_2,...\}=C$.

Let $a,b\in S$. If both are in $C$ then we have $\phi(a)=\phi(b)\Rightarrow a=b$. If $a\in C$ and $b\notin C$ then their image can not be equal because $\phi(a)\in C\cup T$ and $\phi(b)\in S\setminus C$ and $(C\cup T) \cap (S\setminus C)=\emptyset$.

We are left to prove that if $a,b\in C$ then $\phi(a)=\phi(b)\Rightarrow a=b$. Let $a=c_r$ and $b=c_s$ and consider $r<s$ without loss of generality.

If $s\leq n$ then $\phi(a)=\phi(b)\iff\phi(c_r)=\phi(c_s) \iff t_r=t_s$and so they are indeed different.

If $r\geq n+1$ then $\phi(a)=\phi(b)\iff\phi(c_r)=\phi(c_s) \iff c_{r-n}=c_{s-n}$ and so they are indeed different.

If $r\leq n< s$ then $\phi(a)=t_r$ and $\phi(b)=c_{s-n}$ which are different.

I hope this can help you.

Title: Two sets have the same cardinal proof


答:
Along the lines of @Mark Viola 's hint, we can rewrite as
$$f(x)=\frac{6x^2}{16^2}\cdot \frac{1}{\left(1-\frac{-x^3}{8}\right)^2}=\frac{3x^2}{128}\cdot g\left(-\left(\frac{x}{2}\right)^3\right)$$
Where of course
$$g(t):=\frac{1}{(1-t)^2}$$
Notice that
$$g(t)=\frac{\mathrm{d}}{\mathrm{d}t}\left\{\frac{1}{1-t}\right\}$$
Can you continue?

Title: How do I expand this to a power series?

Body:
I am given the function

$$f(x) = \dfrac{6x^2}{(16+2x^3)^2}$$

I am stuck trying to expand $f(x)$ to a power series with center of $0$. I cannot seem to figure how to get $f(x)$ to a clean geometric series form or manipulate it and am confused how to go about it. Thanks!

Image 1


问:
Functions and natural ordering
Imagine that we have the following function:
$$
Y=f(X)
$$
where you can imagine $Y$ to be anything- say quantity of food consumed,
and $X$ is income. Now, when we think about counting objects, an
important property is that of order irrelevance, i.e. it does not
matter in which order you count, the amount you reach to is the same.
In the case of functions, you can imagine, say, my saying that if
$X=5,$ I consumed 10 units of food. Here, I don't see a natural ordering
of food or income. The 10 dollars I have could be counted in any which
way, as long as they add up to 10. On the other hand, when looking
at the derivative, $\frac{dY}{dX}$, a natural ordering is in fact
formed. This is because now, fixing $X$, as we increase $X$ by $\epsilon$,
$Y$ changes. I can now think of $X$ being,say the \emph{first five
}units of food, and now consuming the next little bit. My question
is this: is there an implied natural ordering in functions? In other
words, is there a 1:1 mapping between $X$ and the order of the objects?

My answer: You can imagine, say, my saying that real numbers have a natural order. Here, I don't see any functions. On the other hand, when you look at a function, the order lets you do calculus. Now when we think about using real numbers to represent anything- say quantity of food consumed, then you have implicitly fixed an ordering.  Is this correct?

答: Functions do not require an ordering on the argument.  I can define a function from houses to the country they are in.  This does not impose any order at all.  When the domain of a function has a natural order, like the reals, the function values get a natural order based on the order of the variable values.  The reals have much more than a natural order, they are a metric space, so it makes sense to talk of the distance between two reals.  You need that if you are to take a derivative.</p>

Title: Functions and natural ordering


问:
Why is $\int_0^t g(s)dB_s \sim \mathcal{N}\left(0, \int_0^t g^2(s) ds\right)$
I've found in different references that $\int_0^t g(s)dB_s$ is a Wiener integral, where $g(s)$ is a bounded, continuous function and $dB_s$ is the standard Brownian Motion. Also, that $\int_0^t g(s)dB_s \sim \mathcal{N}(0, \int_0^t g^2(s) ds)$. Why is this?

My attempt so far, is to approximate it as a series of normal random variables. Since the increments $B_{t_{i+1}}-B_{t_i}$ have a distribution $\mathcal{N}(0, t_{i+1}-t_i)$ and they are pairwise independent. Then, thinking it as a Riemann integral $\sum_{i=0}^{\infty} g_i(s) (B_{t_{i+1}}-B_{t_i}) \sim \mathcal{N}(0, \sum_{i=0}^{\infty} g_i^2(s)(t_{i+1}-t_i))$ for some $s\in [t_i, t_{i+1}]$. My question is, is it posible to take the limit $\lim_{t_{i+1} \to t_i} \sum_{i=0}^{\infty} g_i^2(s) (t_{i+1}-t_i) = \int_0^t g^2(s)ds$ in this case? Why?

答:
We consider the general case that $g: [0,t] \longrightarrow \mathbb{R}$ is a square integrable function.

Choose a sequence of simple/elementary functions $(g_n)_{n\in\mathbb{N}}$
that converge in $L^2([0,t])$ to $g$. Then, by Ito's Isometry
one has
$$
\mathbb{E}\left[ \left(\int_0^t g(s) d B_s - \int_0^t g_n(s) d B_s  \right)^2 \right] =
\mathbb{E}\left[ \left(\int_0^t (g(s)-g_n(s)) d B_s\right)^2\right] =
\|g_n-g\|_{L^2}^2 \to 0,
$$
as $n\to\infty$. This implies that $\left(\int_0^t g_n(s) d B_s\right)_{n\in\mathbb{N}}$ converges in $\int_0^t g(s) d B_s$ in
distribution.

Next, calculate the characteristic function of $\int_0^t g_n(s) d B_s$.
As you have seen, $\int_0^t g_n(s) d B_s$ is simply a sum of independent Normal-distributed variables and you should see that this is the characteristic function of
a $\mathcal{N}(0,\int_0^t g_n^2(s) d s)$-distributed random variable.
Use $L^2$-convergence of $(g_n)_{n\in\mathbb{N}}$ to $g$ to show that the characteristic function of $\int_0^t g_n(s) d B_s$ converges point-wise to
the characteristic function of a $\mathcal{N}(0,\int_0^t g^2(s)ds)$-distributed random variable.  This
implies
$$
  \operatorname{law}(\int_0^t g_i(s) d B_s) \Longrightarrow \mathcal{N}(0,\int_0^t g^2(s) d s)
$$
as $n\to\infty$. Since the limit in distribution is unique one has
$$
   \operatorname{law}(\int_0^t g(s) d B_s) = \mathcal{N}(0,\int_0^t g^2(s) d s).
$$


答:
$S$ is of infinite cardinal, i.e.
$$ |S| \geq \aleph_0 $$
We define $\varphi : S \to S \sqcup T$ as it follows:

for $1\leq i \leq n$, there exits $s_i^0\in S$ s.t. $\varphi(s_i^0) = t_i$

recursively we define, for $1\leq i \leq n$, $s_i^{k+1} \in S\setminus \cup_{l=0}^k \{s_1^l, \cdots, s_n^l\}$, s.t. $\varphi(s_i^{k+1}) = s_i^{k}$

We put $S' = \cup_{k\geq 0}\{s_1^k,\cdots, s_n^k\}$, and for $s\notin S'$,
we define $\varphi(s) = s$

Title: Two sets have the same cardinal proof

Body:
Suppose that $S, T$ are sets such that $S$ is infinite and $T = \{t_i: i = 1, ..., n\}$ with $T$ having $n \geq 1$ distinct elements. If $S \cap T = \{\}$, show that $card(S) = card(S \cup T)$.

Hint: Construct a bijective function $\phi: S \rightarrow S \cup T$.

I tried defining various functions but cannot seem to construct one that satisfies the requirements above. I would appreciate if someone could provide such a bijective function.


问:
What's the difference between Spatial velocity and Body velocity?
Let us suppose a rigid body $\{B\}$ is undergoing rotational and translational motion,and the spatial frame is $\{S\}$. In this case, we say that its configuration space is the Special Euclidean group, i.e.,$T:= SE(3)$

Spatial velocity is defined by
$$\dot{T}T^{-1} =\left[ \begin{array} &\dot{R} &\dot{p}  \\0 &0 \end{array} \right] \left[\begin{array}{cc} {R}^T & -R^T{p}  \\0 &1 \end{array}\right] = \left[\begin{array}{cc} \dot{R}{R}^T & \dot{p}-\dot{R}R^T{p}  \\0 &0 \end{array}\right]= \left[\begin{array}{cc} [\omega_s] & v_s  \\0 &0 \end{array}\right]$$

The corresponding body velocity is
$$T^{-1}\dot{T} =\left[\begin{array}{cc} {R}^T & -R^T{p}  \\0 &1 \end{array}\right]\left[ \begin{array} &\dot{R} &\dot{p}  \\0 &0 \end{array} \right]  = \left[\begin{array}{cc} {R}^T\dot{R} & R^T\dot{p}  \\0 &0 \end{array}\right]= \left[\begin{array}{cc} [\omega_b] & v_b  \\0 &0 \end{array}\right]$$

From first formula, $v_s=\dot{p}-\dot{R}R^T{p},$ it is not the linear velocity of the rigid body frame expressed in the spatial frame(indeed,quantity would be $\dot{p}$).

I cann't understand the physical interpretation of $v_s$.

Reference:Murray, Richard M. A mathematical introduction to robotic manipulation. CRC press, 2017.

and related question What's the difference between Spatial angular velocity and Body angular velocity?

答:
"Spatial velocity" v.s. "body velocity":
It's rather confusing to think of "body velocity" as some mathematical magic of swapping $\dot{T}$ and $T^{-1}$ in the definition. Its correct definition is "spatial velocity as observed in a frame that coincides with the current body frame".

Consider a different fixed spatial frame $S'$. Then $T_{S'S}$ is constant and $T_{S'B}(t) = T_{S'S}T_{SB}(t) \ \forall t$. The spatial velocity observed in $S'$ is
$$\dot{T}_{S'B}(t)T_{S'B}(t)^{-1} = T_{S'S}(\dot{T}_{SB}(t) T_{SB}(t)^{-1}) T_{S'S}^{-1}$$
Now imagine that $S'$ coindicdes with $B$ at time $t$, i.e. $T_{S'S} = T_{BS}(t) = T_{SB}(t)^{-1}$. Then the above becomes
$$T_{SB}(t)^{-1}\dot{T}_{SB}(t), \mbox{ or simply } T^{-1}\dot{T}$$

So "spatial velocity" and "body velocity" are derived from the same concept.

Spatial velocity is indepedent of the choice of the body frame. Imagine another frame $B'$ that moves together with the body, so $T_{BB'}$ is constant. The spatial velocity for $B'$ is
$$\dot{T}_{SB'}T_{SB'}^{-1} = \dot{T}_{SB}(T_{BB'}T_{BB'}^{-1})T_{SB}^{-1} = \dot{T}_{SB}T_{SB}^{-1}$$
This is great, because the spatial velocity should describe the motion of the entire body, without relying on which body frame is chosen.

Physical meaning of $v_s$: You probably wonder why $v_s \not= \dot{p}$. It's because if the body rotates, $\dot{p}$ depends on which body frame you choose. Therefore $v_s$ is defined from only the frame $S$ itself.

In fact, imagine a point $c$ that (1) moves together with the body, (2) coincides with the origin of $S$ at time $t$. Then its linear velocity observed in $S$ at time $t$ is $\dot{p} - w_s \times p = \dot{p} - \dot{R}R^Tp = v_s$.


问:
Finding combinations of 81 discrete values that add up to a known value
First, I'm a mechanical engineer, and not a programmer. So for practical purposes, the answer may be as simple as "Excel can't do that without VB", or even "Excel can't do that". This is just a thought experiment for me, but I've run out of brainpower. Anyone that wants to solve this for me will get a heartfelt "NEAT!" which, I think we can agree, is pretty much the pinnacle of engineering accolades...

I watched a video that showed a set of gauge blocks of various sizes, from .0500 to 4.0000" thick, 81 unique blocks in total. The author made reference to the large number of combinations that could be made with such a set. That got me wondering if I could use Excel to figure out all the combinations (and sums) of the set of 81 blocks.

To bound the problem, the sum of all the blocks added together is 26.5295", so that's the maximum, and the minimum increment is 0.0001", so that's 265,295 possible options for sums. I quickly calculated the number of combinations of 81 blocks, C(81,r) where r is 0 thru 81. There are 81 combos of 1 block, 1 combo of 81 blocks, and in between, there are 2.1E23 combos of 41 blocks. All told, there are 2.42E24 combinations of blocks, and thus 2.4 septillion sums (admittedly there will be a lot of repeated sums). I think this rules out Excel...

Alternatively, if you start with each of the 265295 possible sums, someone with the programming knack might be able to work the other way. I can picture this workflow in my mind, but I don't have the chops to program it. For some value, say 3.9762 (copied from video), you start by subtracting the correct 1/10000 block, in this case 0.1002. That leaves 3.8760. Subtract the correct 1/1000 block, here there could be multiple solutions, I would use the smallest appropriate block 0.1060. That leaves 3.7700. Next I'd use 0.1200 to get to 3.6500. Then 0.6500 to get 3.0000. Finally, 3.0000 to get to 0.0000 remaining.

The 0.000X space is either 0, or it requires 0.1001 to 0.1009 blocks.
The X.0000 space for X<10 can be dealt with by 1, 2, 3, 4 blocks. Beyond X=10, some wacky combo of smaller 0.0500" increment blocks is required.

If anyone is interested in the specific values of the blocks, they are:

4 blocks 1.000 to 4.000 in 1.0000 increments

19 blocks 0.0500 to 0.9500 in 0.0500 increments

49 blocks from 0.1010 to 0.1490 to 0.0010 increments

9 blocks from 0.1001 to 0.1009 in 0.0001 increments

Sauce, for those interested:
Video: Origins of Precision, Relevant discussion around 26:34

答:
You have 81 blocks and you either include or exclude each one. (This results in the possibility of a sum of $0$ but it seems you do not want to count this one, so there are $2^{81}-1$ possibilities in total).

So you can just use a recursion, where if $b_n$ is the $n$th and the counts of possibilities adding up to $x$ after $n$ blocks is $f_n(x)$, then $$f_n(x)=f_{n-1}(x)+f_{n-1}(x-b_n)$$ starting with nothing, i.e. $f_0(0)=1$ and $f_0(x)=0$ for $x \not=0$.

This is easier enough to program and runs quickly.  The following is in R, including a plot of frequencies. You could use Excel. Note:

R starts counting at $1$ not $0$

There is an offset to enable adding $f_{n-1}(x-b_n)$ for negative $x-b_n$ but that offset and the value corresponding to $0$ are dropped at the end from the results

It is is easier and sensible to work in integers, and avoids some potential rounding errors, so all your lengths are multiplied by $10000$

The plot appears to show two lines: the higher line is for lengths which are a multiple of $5$ thanks to a very small bias introduced by the interaction of your last nine blocks.

.

```<code>blocks <- c(seq(from=10000, to=40000, by=10000),
            seq(from=  500, to= 9500, by=  500),
            seq(from= 1010, to= 1490, by=   10),
            seq(from= 1001, to= 1009, by=    1))
sumblocks <- sum(blocks)
zerooffset <- max(blocks) + 1
counts <- numeric(sumblocks+zerooffset)
counts[zerooffset] <- 1
for (n in 1:length(blocks)){
    counts[(blocks[n]+1):(sumblocks+zerooffset)] <-
      counts[(blocks[n]+1):(sumblocks+zerooffset)] +
      counts[1:(sumblocks+zerooffset-blocks[n])]
    }
results <- counts[(zerooffset+1):(sumblocks+zerooffset)]
sum(results)
# 2.417852e+24
plot(results, pch = ".")
</code>```

Body:
You have 81 blocks and you either include or exclude each one. (This results in the possibility of a sum of $0$ but it seems you do not want to count this one, so there are $2^{81}-1$ possibilities in total).

So you can just use a recursion, where if $b_n$ is the $n$th and the counts of possibilities adding up to $x$ after $n$ blocks is $f_n(x)$, then $$f_n(x)=f_{n-1}(x)+f_{n-1}(x-b_n)$$ starting with nothing, i.e. $f_0(0)=1$ and $f_0(x)=0$ for $x \not=0$.

This is easier enough to program and runs quickly.  The following is in R, including a plot of frequencies. You could use Excel. Note:

R starts counting at $1$ not $0$

There is an offset to enable adding $f_{n-1}(x-b_n)$ for negative $x-b_n$ but that offset and the value corresponding to $0$ are dropped at the end from the results

It is is easier and sensible to work in integers, and avoids some potential rounding errors, so all your lengths are multiplied by $10000$

The plot appears to show two lines: the higher line is for lengths which are a multiple of $5$ thanks to a very small bias introduced by the interaction of your last nine blocks.

.

blocks <- c(seq(from=10000, to=40000, by=10000),
            seq(from=  500, to= 9500, by=  500),
            seq(from= 1010, to= 1490, by=   10),
            seq(from= 1001, to= 1009, by=    1))
sumblocks <- sum(blocks)
zerooffset <- max(blocks) + 1
counts <- numeric(sumblocks+zerooffset)
counts[zerooffset] <- 1
for (n in 1:length(blocks)){
    counts[(blocks[n]+1):(sumblocks+zerooffset)] <-
      counts[(blocks[n]+1):(sumblocks+zerooffset)] +
      counts[1:(sumblocks+zerooffset-blocks[n])]
    }
results <- counts[(zerooffset+1):(sumblocks+zerooffset)]
sum(results)
# 2.417852e+24
plot(results, pch = ".")


问:
Proving that $\deg(f) = - \infty$ iff $f=0.$
In a $\mathbb Z$-graded integral domain $B = \bigoplus_{i \in \mathbb Z} B_i$ my definition for the degree function is as follows:

Given $f \in B,$ define $\deg : B \rightarrow \mathbb Z \cup \{-\infty \}$ by $$\deg(f) = \mathrm{maxSupp}{(f)} \text{ if } f \neq 0 \text{ and } \deg(0) = -\infty $$

my definition for the  $\mathrm{Supp}$ of $f$ is as follows $\{i \in \mathbb Z| f_i \neq 0\}.$

I am trying to prove that  $\deg(f) = - \infty$ iff $f=0.$ The backward direction is clear by the definition of the $\deg$ function given to me. The forward direction, I am trying to prove it by contra positive i.e. Assume that $f \neq 0$ and I want to prove that  $\deg(f) \neq  - \infty$, does that follows from the definition of the support I have because my $i \in \mathbb Z$ or from what?  Does the  $\mathrm{maxSupp}{(f)}$ always $\geq 0$ if $f \neq 0$? if so, why?

Any help will be appreciated!

答:
Since $B = \bigoplus_{i \in \mathbb Z}B_i$, an element $f \in B$ can be written uniquely as a sum

$$
f = \sum_{i \in \mathbb Z}f_i, \quad f_i\in B_i
$$

with finitely many $f_i$ non zero. In paritcular, the support of $f$ is always a finite set. If $f \neq 0$, then not all $f_i$ are zero, hence $\mathbf{supp} f \neq \emptyset$ and thus $\max \mathbf{supp} f \in \mathbb Z$.

Note that this does not mean that $\deg$ is always non-negative. For example, consider the ring $B = \mathbb{Z}[t,t^{-1}]$ of Laurent polynomials with the grading defined as $B_n = \mathbb{Z}t^n$ for each integer $n$. In this ring, we have $\deg t^{-1} = -1$.


问:
Is there a way to represent the relation $<$ on the real numbers without using multiplication?
I know for $x,y\in\mathbb R$ we have $x<y$ iff there exists a $z\in\mathbb R$, so that $y=x+z^2$. Is there a similar way to „prove“ $x<y$ using only $+$ and $=$? And if not, is there a more or less simple proof for that?

答:
You can't do that.

Your question, in other words, is whether or not $<$ is definable in $(\Bbb R,+)$, and the answer is no. The simplest way to do that is to show that there's an automorphism of $(\Bbb R,+)$ which is not order preserving.

For example, $x\mapsto -x$.


问:
Is there a quick and intuitive way to derive $\sum_{i=2}^{n-1}\frac{1}{i(i+1)}=\frac{1}{2}-\frac{1}{n}$?
Is there a quick and intuitive way to derive the following finite sum for integer $n\geq 3$?

$$\sum_{i=2}^{n-1}\frac{1}{i(i+1)}=\frac{1}{2}-\frac{1}{n}$$

Please, do not suggest proofs by induction, since these require prior knowledge of the result, which makes the derivation circular. I'm curious to see a way to quickly determine what the result is in the first place.

How I come up with the equality:

Define

$$s_n=\sum_{i=2}^{n-1}\frac{1}{i(i+1)}$$

evaluate the first few terms by hand

$$s_3=\frac{1}{6}~,~s_4=\frac{1}{4}~,~s_5=\frac{3}{10}~,~s_6=\frac{1}{3}~,~s_7=\frac{5}{14}~,~s_8=\frac{3}{8}~,~s_9=\frac{7}{18}$$

take an ansatz for what the result possibly might look like

$$S_n=\frac{a_{-3}}{n^3}+\frac{a_{-2}}{n^2}+\frac{a_{-1}}{n}+a_0+a_1 n+a_2 n^2+ a_3 n^3$$

evaluate this ansatz for the same range of explicit $n$ values as above and solve the resulting system of linear equations for the $a_i$. This directly gives:

$$a_{-3}=a_{-2}=a_1=a_2=a_3=0~,~a_{-1}=-1~,~a_{0}=\frac{1}{2}$$

Finally, with the result known, do usual proof by induction.

答:
HINT
$$
\frac{1}{i(i+1)} = \frac1i-\frac1{i+1}
$$
and telescoping series