问:
Find Tangent Line Common to Two Hippopedes Curves
Hello and forgive me for my lack of formality or perhaps poor wording. I am trying to derive the equations for tangent lines common to two Hippopedes ovals with different origins (see Booth's Curve):

$$(x^2+y^2)^2 = a^2x^2 + b^2y^2
$$

and

$$((x-h)^2+(y-k)^2)^2 = a^2(x-h)^2 + b^2(y-k)^2
$$

Here is an example:

I believe the solution steps should be something along the lines of:

Find derivatives of each curve ($y'_1$ and $y'_2$)

Set derivatives equal to each other (Eq. 1)

Write slope of line as $y_1-y_0/x_1-x_0$.

Set slope equal to either $y'_1$ or $y'_2$. (Eq. 2)

then solve Eq. 1 and 2 simultaneously. I'm anticipating the math will be cumbersome, so any help would be greatly appreciated! Thank you.

答:
These ovals have parameterisations

$(x(t),y(t))=(b\frac{ab\cos(t)}{b^2(\cos(t))^2+a^2(\sin(t))^2},a\frac{ab\sin(t)}{b^2(\cos(t))^2+a^2(\sin(t))^2})$ and $(x(t),y(t))=(h+b\frac{ab\cos(t)}{b^2(\cos(t))^2+a^2(\sin(t))^2},k+a\frac{ab\sin(t)}{b^2(\cos(t))^2+a^2(\sin(t))^2}).$

The dual curves are given by $(p(t),q(t))=(\frac{-y'(t)}{x'(t)y(t)-x(t)y'(t)},\frac{x'(t)}{x'(t)y(t)-x(t)y'(t)}).$

Now find the intersection points $P_i$ of the dual curves. Then the common tangents are the lines $x x(P_i)+y y(P_i)=1.$

The picture below shows only a few of the common tangents:

Here are the dual curves. Note that they have nodes corresponding to the bitangents.

To accentuate the node bitangent correspondence, here are both the curve and its dual, the point and the dual line (bitangent).

The red intersection point $P_3: (l,m)$ corresponds to the red common tangent through the correspondence $(l:m:-1)$ to $lx +my-1=0.$

I used $a=0.5, b=2.3, h=2.95, k=0.85$ with the following to get the plot

```<code>curve(((a^2*b*sin(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))  /(a^2*sin(t)^2+b^2*cos(t)^2)^2  -(a^2*b*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2))  /((a^2*b*sin(t)        *((-(a*b^2*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2))         -(a*b^2*cos(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))          /(a^2*sin(t)^2+b^2*cos(t)^2)^2))   /(a^2*sin(t)^2+b^2*cos(t)^2)   -(a*b^2*cos(t)      *((a^2*b*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2)       -(a^2*b*sin(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))        /(a^2*sin(t)^2+b^2*cos(t)^2)^2))    /(a^2*sin(t)^2+b^2*cos(t)^2)),((-(a*b^2*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2))  -(a*b^2*cos(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))   /(a^2*sin(t)^2+b^2*cos(t)^2)^2)  /((a^2*b*sin(t)        *((-(a*b^2*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2))         -(a*b^2*cos(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))          /(a^2*sin(t)^2+b^2*cos(t)^2)^2))   /(a^2*sin(t)^2+b^2*cos(t)^2)   -(a*b^2*cos(t)      *((a^2*b*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2)       -(a^2*b*sin(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))        /(a^2*sin(t)^2+b^2*cos(t)^2)^2))    /(a^2*sin(t)^2+b^2*cos(t)^2)),t,0,2pi)
curve(((a^2*b*sin(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))  /(a^2*sin(t)^2+b^2*cos(t)^2)^2  -(a^2*b*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2))  /(((a^2*b*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2)+k)   *((-(a*b^2*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2))    -(a*b^2*cos(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))     /(a^2*sin(t)^2+b^2*cos(t)^2)^2)   -((a^2*b*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2)    -(a^2*b*sin(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))     /(a^2*sin(t)^2+b^2*cos(t)^2)^2)    *((a*b^2*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2)+h)), ((-(a*b^2*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2))  -(a*b^2*cos(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))   /(a^2*sin(t)^2+b^2*cos(t)^2)^2)  /(((a^2*b*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2)+k)   *((-(a*b^2*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2))    -(a*b^2*cos(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))     /(a^2*sin(t)^2+b^2*cos(t)^2)^2)   -((a^2*b*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2)    -(a^2*b*sin(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))     /(a^2*sin(t)^2+b^2*cos(t)^2)^2)    *((a*b^2*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2)+h)),t,0,2pi)
</code>```

There are ways to get the implicit equations defining the dual curves (I'd need to read up on them), and you could further use grobner bases.

The first has equation $a^4b^{10}x^6+3a^6b^8x^4y^2+3a^8b^6x^2y^4+a^{10}b^4y^6+(8a^6b^6-8a^4b^8-a^2b^{10})x^4+(-20a^8b^4+38a^6b^6-20a^4b^8)x^2y^2+(-a^{10}b^2-8a^8b^4+8a^6b^6)y^4+(16a^8b^2-32a^6b^4+8a^4b^6+8a^2b^8)x^2+(8a^8b^2+8a^6b^4-32a^4b^6+16a^2b^8)y^2-16a^6b^2+32a^4b^4-16a^2b^6=0$

Body:
These ovals have parameterisations

$(x(t),y(t))=(b\frac{ab\cos(t)}{b^2(\cos(t))^2+a^2(\sin(t))^2},a\frac{ab\sin(t)}{b^2(\cos(t))^2+a^2(\sin(t))^2})$ and $(x(t),y(t))=(h+b\frac{ab\cos(t)}{b^2(\cos(t))^2+a^2(\sin(t))^2},k+a\frac{ab\sin(t)}{b^2(\cos(t))^2+a^2(\sin(t))^2}).$

The dual curves are given by $(p(t),q(t))=(\frac{-y'(t)}{x'(t)y(t)-x(t)y'(t)},\frac{x'(t)}{x'(t)y(t)-x(t)y'(t)}).$

Now find the intersection points $P_i$ of the dual curves. Then the common tangents are the lines $x x(P_i)+y y(P_i)=1.$

The picture below shows only a few of the common tangents:

Here are the dual curves. Note that they have nodes corresponding to the bitangents.

To accentuate the node bitangent correspondence, here are both the curve and its dual, the point and the dual line (bitangent).

The red intersection point $P_3: (l,m)$ corresponds to the red common tangent through the correspondence $(l:m:-1)$ to $lx +my-1=0.$

I used $a=0.5, b=2.3, h=2.95, k=0.85$ with the following to get the plot

curve(((a^2*b*sin(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))  /(a^2*sin(t)^2+b^2*cos(t)^2)^2  -(a^2*b*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2))  /((a^2*b*sin(t)        *((-(a*b^2*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2))         -(a*b^2*cos(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))          /(a^2*sin(t)^2+b^2*cos(t)^2)^2))   /(a^2*sin(t)^2+b^2*cos(t)^2)   -(a*b^2*cos(t)      *((a^2*b*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2)       -(a^2*b*sin(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))        /(a^2*sin(t)^2+b^2*cos(t)^2)^2))    /(a^2*sin(t)^2+b^2*cos(t)^2)),((-(a*b^2*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2))  -(a*b^2*cos(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))   /(a^2*sin(t)^2+b^2*cos(t)^2)^2)  /((a^2*b*sin(t)        *((-(a*b^2*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2))         -(a*b^2*cos(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))          /(a^2*sin(t)^2+b^2*cos(t)^2)^2))   /(a^2*sin(t)^2+b^2*cos(t)^2)   -(a*b^2*cos(t)      *((a^2*b*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2)       -(a^2*b*sin(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))        /(a^2*sin(t)^2+b^2*cos(t)^2)^2))    /(a^2*sin(t)^2+b^2*cos(t)^2)),t,0,2pi)
curve(((a^2*b*sin(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))  /(a^2*sin(t)^2+b^2*cos(t)^2)^2  -(a^2*b*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2))  /(((a^2*b*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2)+k)   *((-(a*b^2*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2))    -(a*b^2*cos(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))     /(a^2*sin(t)^2+b^2*cos(t)^2)^2)   -((a^2*b*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2)    -(a^2*b*sin(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))     /(a^2*sin(t)^2+b^2*cos(t)^2)^2)    *((a*b^2*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2)+h)), ((-(a*b^2*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2))  -(a*b^2*cos(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))   /(a^2*sin(t)^2+b^2*cos(t)^2)^2)  /(((a^2*b*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2)+k)   *((-(a*b^2*sin(t))/(a^2*sin(t)^2+b^2*cos(t)^2))    -(a*b^2*cos(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))     /(a^2*sin(t)^2+b^2*cos(t)^2)^2)   -((a^2*b*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2)    -(a^2*b*sin(t)*(2*a^2*cos(t)*sin(t)-2*b^2*cos(t)*sin(t)))     /(a^2*sin(t)^2+b^2*cos(t)^2)^2)    *((a*b^2*cos(t))/(a^2*sin(t)^2+b^2*cos(t)^2)+h)),t,0,2pi)

There are ways to get the implicit equations defining the dual curves (I'd need to read up on them), and you could further use grobner bases.

The first has equation $a^4b^{10}x^6+3a^6b^8x^4y^2+3a^8b^6x^2y^4+a^{10}b^4y^6+(8a^6b^6-8a^4b^8-a^2b^{10})x^4+(-20a^8b^4+38a^6b^6-20a^4b^8)x^2y^2+(-a^{10}b^2-8a^8b^4+8a^6b^6)y^4+(16a^8b^2-32a^6b^4+8a^4b^6+8a^2b^8)x^2+(8a^8b^2+8a^6b^4-32a^4b^6+16a^2b^8)y^2-16a^6b^2+32a^4b^4-16a^2b^6=0$


问:
What is the Name of this Point?
I am trying to find the name of the point at which the derivative of a function at that point is zero but it is not a local max or min within any interval. For instance, take the function $y=(x+1)(x-1)^3$. The roots of the first derivative are -0.5, one, and one but the function at $x=1$ is neither a local max or a local min. What is the name of this point? Is it just a Point of Inflection or is there another name?
<img src="https://i.stack.imgur.com/mxanE.png" alt="$y=(x+1)(x-1)^3$" />

答:
A colleague of mine from the late 1990s used the term terrace point for a point $x=a$ where the first derivative is zero and the first derivative doesn't change sign as you pass through $x=a,$ and I liked the idea so much that I subsequently used it in all my calculus classes. Of course, this was only used in simple introductory calculus examples where the zeros of the derivative are isolated from each other.

A few years later (in July 2002; see 6. TERRACE POINTS IN THE FIRST DERIVATIVE TEST here) I was told that the term "terrace point" is in Ostebee/Zorn's Calculus book, although I've never looked at a copy to be sure, and I don't know whether the term was in both the 1994 1st and 2002 2nd editions or only in the 2002 2nd edition.

Anyway, when I last wrote about this term (as far as I can recall), there wasn't nearly as much on the internet as there is now, and google-books searching wasn't available. A google search shows that the term "terrace point" is now fairly widely used. Also, a google-books search shows that it is definitely used in the 2002 2nd edition of Ostebee/Zorn's book, as well as in several other books. Interestingly, the term also appears on p. 40 (line −10) of William Richard Ransom's 1915 Early Calculus. As far as I can determine, the only use of "terrace point" in one of the math oriented Stack Exchanges is this answer from 24 October 2019.

One of the reasons I liked having a name for this this notion is that it allows you to label all four of the possibilities that can show up on a first derivative sign chart where the derivative is zero (and isolated from the other zeros):

$$ ++++0++++ \;\;\;\;\; \text {ter} $$

$$ ++++0----  \;\;\;\;\; \text {max} $$

$$ ----0++++  \;\;\;\;\; \text {min} $$

$$ ----0----  \;\;\;\;\; \text {ter} $$

ContentLicense: CC BY-SA 4.0

Title: What is the Name of this Point?


问:
Closed, unbounded subset of a cardinal.
I missed two lectures in my set theory course, and now I don't understand the homework problems.

One is this: let $\kappa$ be a regular uncountable cardinal. Show that the following sets are closed and unbounded in $\kappa.$

$\{\alpha<\kappa\,:\,\alpha \text{ is a limit ordinal}\},$

$\{\alpha<\kappa\,:\,\mathrm{cf}(\alpha)=\omega\}$ for $\kappa=\omega_1,$

$\{\lambda<\kappa\,:\,\lambda\text{ is a cardinal}\}$ for $\kappa$ inaccessible.

I don't understand what closed or unbounded subsets of $\kappa$ is. Could you explain this to me and give some pointers on solving the problem?

答:
An unbounded subset of $\kappa$ is some $A\subseteq\kappa$ such that for all $\beta<\kappa$, there is some $\alpha\in A$ such that $\beta<\alpha$.

A closed subset of $\kappa$ is some $A\subseteq\kappa$ such that for all $0<\alpha<\kappa$, if $\sup(A\cap\alpha)=\alpha,$ then $\alpha\in A$.

For the first one, do you know the result that all limit ordinals are of the form $\omega\cdot\alpha$ for some non-zero $\alpha$? In particular, you'll want to show that the limit ordinals less than $\kappa$ are those of the form $\omega\cdot\alpha$ for $0<\alpha<\kappa$. Use continuity of ordinal multiplication to show that this set is closed, and the fact that $\kappa$ is a cardinal to show that it's unbounded.

The second one is just a special case of the first one. (Why?)

The third one won't be too tricky. Do you know the recursive definition of the alephs? (Also, don't forget that the natural numbers are cardinals, too.)

ContentLicense: CC BY-SA 3.0

Title: Closed, unbounded subset of a cardinal.


问:
Limit of $(3n+1)^{1/n}$
I am working through a question that requires me to prove that a certain limit tends to 1. I have reached an endpoint where I have:

$$\lim\limits_{n\to \infty} (3n+1)^{1/n}$$

I know that and can use the fact that $\lim\limits_{n\to \infty}( n^{1/n}) = 1$

However, I am not sure if I can just use this fact to say that it applies for $(3n+1)^{1/n} $ and I also am stuck on proving it.

答:
$$L=\lim_{n\to \infty} (3n+1)^{1/n}$$
Apply D'Alembert-Cauchy criterion:
$$L=\lim_{n\to \infty} \dfrac {3n+4}{3n+1}$$
$$\implies L=1$$

Title: Limit of $(3n+1)^{1/n}$


问:
Checking an "equivalence of products" in the notes on deformation quantization by Allen C. Hirshfeld and Peter Henselder
These notes are a great introduction to deformation quantization but I failed to check the validity of the statement p.9, right before (5.18).

Context: let $(\mathcal{A},+,\mu)$ be an algebra. $\mu:\mathcal{A}\times \mathcal{A} \to \mathcal{A}$ standing for multiplication. Deformation consists in considering a family (paramatrized by $\nu$ in a yet to be chosen space) of product on $\mathcal{A}[[\nu]]$ (formal power series with coefficients in $\mathcal{A}$) generically given by
$$ \forall\ f,g\in \mathcal{A},\quad  \mu_{\nu}(f,g) := \mu(f,g) + \sum_{k=1}^{+\infty} \nu^k \mu_k (f,g) \label{1}\tag{1}$$
i.e. by a family of bilinear maps $\mu_k:\mathcal{A}\times \mathcal{A} \to \mathcal{A}$ satisfying some conditions and extended to elements $F, G\in \mathcal{A}[[\nu]]$ of the form $F=\sum_{k=1}^{+\infty} \nu^k f_{k},\ f_k \in \mathcal{A}$ by $\mathbb{K}[[\nu]]$-bilinearity. (the idea behind formal power series, as far as I understand, is to ignore convergence issues but still have a structure where one can compare terms of the same degree in $\nu$").

Two of these star-product are equivalent if there exists an invertible algebra isomorphism (transition map) $T:(\mathcal{A}[[\nu]],+,\mu_{\nu}) \longrightarrow (\mathcal{A}[[\nu]],+,\rho_{\nu})$, i.e. a map such that
$$ \forall\ F,G \in \mathcal{A}[[\nu]], \quad T\big(\mu_{\nu}(F,G)\big)= \rho_{\nu}(T(F),T(G))$$

Question: Let $\mathcal{A}=\mathcal{C}^{\infty}(\mathbb{R}^2)$ and denote $(a, \overline{a})$ or $(b, \overline{b}$) the variables of the functions. I want to check that the normal product ( (5.4) p.8; with the more usual notation for products)
$$f \ast_N g := \sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \frac{\partial^k f}{\partial a^k} \frac{\partial^k g}{\partial \overline{a}^k} = f\, e^{\hbar \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}}}\, g \label{2}\tag{2}$$
is equivalent to the Moyal product ((5.15) p.9, one can consider $\hbar$ as the deformation parameter... although there is usually the factor as in (\ref{5}))
$$ f \ast_M g := \sum_{k=0}^{+\infty} \left(\frac{\hbar}{2} \right)^k \frac{1}{k!}  \left. \left( \frac{\partial }{\partial a} \frac{\partial }{\partial \overline{b}} - \frac{\partial }{\partial \overline{a}} \frac{\partial }{\partial b}\right)^k f\big(a, \overline{a}\big) g\big(b, \overline{b}\big)  \right|_{\genfrac{}{}{0pt}{1}{a=b}{\overline{a}=\overline{b}}} = f\, e^{\frac{\hbar}{2}\big( \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}} - \overleftarrow{\partial}_{\overline{a}} \overrightarrow{\partial}_{a} \big)}\, g \label{3}\tag{3}$$
i.e.
$$ T\big(f \ast_N g \big) = T(f) \ast_M T(g)\quad \text{with}\quad T = \genfrac{}{}{0pt}{0}{"}{}\!\!\exp\left(-\frac{\hbar}{2} \frac{\partial^2}{\partial a\, \partial \overline{a}} \right)\genfrac{}{}{0pt}{0}{"}{} \label{4}\tag{4}$$

Remarks:

In fact I already checked (\ref{4}) up to second order in $\hbar$ but it did not work at order 3 (although I'm not sure as the calculations were quite tedious...). It was not a priori clear that (\ref{4}) hold, one could have the other way round $ T\big(f \ast_M g \big) = T(f) \ast_N T(g)$ instead but this seems to fail at order 1. I only want to check the first few orders, but I would gladly take a proof for all order. I will soon write what I have done, but as I mentionned, it's tedious.

The Moyal product is first defined in the text (3.5-3.6) p.5 by
$$ f \ast_M g := \sum_{k=0}^{+\infty} \frac{\nu^k}{k!} \underbrace{\left(\frac{\partial }{\partial q_1} \frac{\partial }{\partial p_{2}} - \frac{\partial }{\partial p_{1}} \frac{\partial }{\partial q_2} \right)^k f(q_1,p_1)\, g(q_2,p_2)}_{\mu_k(f,g)}\left.\vphantom{\frac{T}{T}}\right|_{\genfrac{}{}{0pt}{1}{q_1=q_2}{p_{1}=p_{2}}}\quad \text{with}\quad \nu = \frac{i\hbar}{2} \label{5}\tag{5}$$
and it does coincide with (\ref{3}) via (these are the correct $\sqrt{2}$ factors...)
$$ \left\lbrace \begin{aligned}
 a & := \frac{1}{\sqrt{2}} \left(q  + i\hspace{.5pt} p \right) \\
 \overline{a} & := \frac{1}{\sqrt{2}} \left( q - i\hspace{.5pt} p \right)
 \end{aligned} \right. \enspace \Longrightarrow\quad
 \left\lbrace \begin{aligned}
 \frac{\partial}{\partial\hspace{.7pt} q} & = \frac{\partial\hspace{.7pt} a}{\partial\hspace{.7pt} q} \frac{\partial}{\partial\hspace{.7pt} a} + \frac{\partial\hspace{.7pt} \overline{a}}{\partial \hspace{.7pt} q} \frac{\partial}{\partial\hspace{.7pt} \overline{a}} =\frac{1}{\sqrt{2}} \left( \frac{\partial}{\partial\hspace{.7pt} a} +  \frac{\partial}{\partial\hspace{.7pt} \overline{a}} \right) \\
 \frac{\partial}{\partial\hspace{.7pt} p} & = \frac{\partial\hspace{.7pt} a}{\partial\hspace{.7pt} p} \frac{\partial}{\partial\hspace{.7pt} a} + \frac{\partial\hspace{.7pt} \overline{a}}{\partial \hspace{.7pt} p} \frac{\partial}{\partial\hspace{.7pt} \overline{a}} = \frac{i}{\sqrt{2}} \left( \frac{\partial}{\partial\hspace{.7pt} a} - i\, \frac{\partial}{\partial\hspace{.7pt} \overline{a}}\right)
 \end{aligned} \right. $$

To make (\ref{3}) (same for (\ref{5})) more explicit, let me write the $k=2$ term: (notation $\displaystyle \partial_a=\frac{\partial}{\partial a},\ \partial_{ab}= \frac{\partial^2}{\partial a \partial b}$ etc.)
$$\begin{split}
 \mu_2(f,g) &= \Big(\partial_{aa\overline{b}\overline{b}} - 2 \partial_{a\overline{a}b\overline{b}} + \partial_{\overline{a}\overline{a}bb} \Big) f(a,\overline{a})g(b,\overline{b})\left.\vphantom{\frac{T}{T}}\right|_{\genfrac{}{}{0pt}{1}{a=b}{\overline{a}=\overline{b}}}\\
 &= (\partial_{aa}f)(\partial_{\overline{a}\overline{a}}g) - 2 (\partial_{a\overline{a}}f)(\partial_{b\overline{b}}g) + (\partial_{\overline{a}\overline{a}}f) (\partial_{aa}g)
\end{split} \label{6}\tag{6}$$
One can also use the $\overleftarrow{\partial}$ or $\overrightarrow{\partial}$ notations or a tensorial notation.

答:
Your reference has this explained in its ref [26], namely Zachos (2000) J Math Phys 41, 5129–5134, hep-th/9912238.

In any case, it is straightforward to prove your (4) through elementary Fourier analysis. That is, use test/sample functions
$$
f=\exp (ma+n\bar a), ~~~~ g=\exp (ka+s\bar a),
$$
so  (4) presents as
$$
\exp (\hbar ms -\hbar(m+k)(n+s)/2) ~ \overset{?}{=}  ~ \exp (-\hbar mn/2-\hbar ks/2 +\hbar(ms-nk)/2),
$$
indeed, an identity.

You might, or might not, appreciate the geometrical features associated with it.

For the mainstream review of all such moves, see this booklet.


问:
Left multiplication commuting functions (proof verification)
Let $G$ be a group and $g \in G$. Define $\lambda_g: G \rightarrow G$ by $\lambda_g(x) = gx$ for all $x \in G$. Similarly, define $\tau_g: G \rightarrow G$ by $\tau_g(x) = xg$ for all $x \in G$. If $\theta: G \rightarrow G$ is a bijection such that $\lambda_g \theta = \theta \lambda_g$ for all $g \in G$, show that $\theta = \tau_h$ for some $h \in G$.

My proof is as follows:

Since $\lambda_g \theta = \theta \lambda_g$ for all $g \in G$, we have
$$ g \theta(x) = \lambda_g(\theta(x)) = \theta (\lambda_g(x)) = \theta(gx) $$
for all $g, x \in G$. In particular, we have $$x^{-1} \theta(x) = \theta(x^{-1}x) = \theta(1)$$ or
$\theta(x) = x\theta(1)$ for all $x \in G$. Hence $\theta = \tau_{\theta(1)}$ as required.

It looks pretty simple, but I don't seem to use the condition that "$\theta$ is a bijection".

Is my proof correct? Thank you.

答:
The proof looks correct to me. As to the condition of $\theta$ being a bijection, rather than being an assumption this property is implied by the condition that $\lambda_g\theta=\theta\lambda_g$ for all $g\in G$.

Indeed, let $x\in G$. Setting $g=x(\theta(x))^{-1}$, we have
$$x = x(\theta(x))^{-1}\theta(x) = g\theta(x) = \lambda_g\theta(x) = \theta(\lambda_g(x)),$$
hence $x\in\theta(G)$. Thus, $\theta$ is surjective.

And from what you do, if $\theta(x)=\theta(y)$, then
$$\begin{align*}
\theta(1) &= \theta(x^{-1}x) = \theta\lambda_{x^{-1}}(x) = \lambda_{x^{-1}}\theta(x) = x^{-1}\theta(x)\\
\theta(1) &= \theta(y^{-1}y) = \theta\lambda_{y^{-1}}(y) = \lambda_{y^{-1}}\theta(y) = y^{-1}\theta(y) = y^{-1}\theta(x).
\end{align*}$$
Thus, $y^{-1}\theta(x)=x^{-1}\theta(x)$, so $y=x$. Thus, $\theta$ is also one-to-one.


问:
Subsequence of $G$?
Say we had a sequence or function $$G_x=\frac{1}{x+1}$$ Would $$Y_x=\frac{1}{x+1}$$ be considered a sub sequence. Also, would $$Z_x=\frac{1}{x+2}$$ be considered a sub sequence? I feel that the 2nd one is a sequence for sure, but I am not too sure about $Y_x$

答:
Yes, both $Y_x$ and $Z_x$ can be sub sequences of $G_x$. This is of course, dependent on how you define a sub sequence. Let's say we have a sequence $A=1,2,3$. Would the sequence $B=1,2,3$ be a sub sequence to $A$?

About $Z_x$, it is simply $G_x$ with a "shift". Simply,
$$ Z_x = G_{x+1} $$
Which is definitely a sub sequence of $G_{x}$

Title: Subsequence of $G$?


问:
Three dimensional Fourier Transform
I have a function
$$
\operatorname{g}\left(r\right) =
\left\{\begin{array}{lcl}
{\displaystyle 0} & \mbox{if} &
{\displaystyle r < \rho}
\\
{\displaystyle 1 + \mathrm{e}^{-kr}} &&
\mbox{otherwise}
\end{array}\right.
$$
And I have a question that asks: What is the value of $S(\vec{k}) = \int d^3 r [g(\vec{r})-1] e^{i \vec{k} \cdot \vec{r}}\ ?$.

The question said it was a tri-dimensional fourier transform, so I checked on internet what is that and a found that for a tri-dimensional transform

$f(\vec{x}) = \int d^3   k \space \space \mathscr{F}(k) e^{i \vec{k} \cdot \vec{r}} /(2 \pi)^3$

So I assumed that I only need to inverse transform $[g(r)-1]$ and multiply by $1/(2\pi)^3$

Is that right? Cause I got a non-infinity value for $S$, but I had to assume that $r>0$, and the value of $S$ is one-dimensional and only depends of the absolute value of the vector.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\operatorname{g}\left(r\right) =}
\left\{\begin{array}{lcl}
{\displaystyle 0} & \mbox{if} &
{\displaystyle r < \rho}
\\
{\displaystyle 1 + \mathrm{e}^{-kr}} &&
\mbox{otherwise}
\end{array}\right.$

\begin{align}
\on{S}\pars{\vec{k}} & \equiv
\bbox[5px,#ffd]{\iiint_{\mathbb{R}^{3}}
\bracks{\on{g}\pars{r} - 1}\expo{\ic\vec{k}\,{\large\cdot}\,\vec{r}}\dd^{3}\vec{r}}
\\[5mm] & =
\int_{0}^{\infty}\bracks{\on{g}\pars{r} - 1}\
\underbrace{\pars{\int_{\Omega_{\,\vec{\!r}}}
\expo{\ic\vec{k}\,{\large\cdot}\,\vec{r}}\,{\dd\Omega_{\,\vec{r}} \over 4\pi}}}
_{\ds{\sin\pars{kr} \over kr}}
4\pi r^{2}\,\dd r
\\[5mm] & =
{4\pi \over k^{3}}\left[%
\int_{0}^{k\rho}\pars{-1}
\sin\pars{\theta}\,\theta\,\dd\theta\right.
\\[2mm] & \phantom{{4\pi \over k^{3}}\left[\,\,\,\,\,\,\,\right.}
\left. + \int_{k\rho}^{\infty}\expo{-\ic\theta}
\sin\pars{\theta}\,\theta\,\dd\theta\right]
\end{align}

The $\ds{\underline{\mbox{first}}}$ integral is equal to $\ds{\quad k\rho\cos\pars{k\rho} - \sin\pars{k\rho}\quad}$ while the
$\ds{\underline{\mbox{second}}}$ one
$\ds{\color{red}{\tt diverges}}$.

Title: Three dimensional Fourier Transform


问:
How could you evaluate this limit without L'Hopitals Rule?
How could someone in high-school solve this limit without using L'Hopitals Rule?
$$\lim _{x \rightarrow 0} \frac{\sin x \cos x-\sin x}{x^{2}} = 0$$
I assume that it involves the fact that
$$
\lim _{x \rightarrow 0} \frac{\sin (x)}{x}=1
$$
but I'm not seeing how.

答:
Hint:

$$\sin x\cos x-\sin x=-\sin x(1-\cos x)=-\dfrac{\sin^3x}{1+\cos x}$$

as $(1+\cos x)(1-\cos x)=\sin^2x$


答:
Notice that $$\lim_{x\to 0} \frac{\sin x \cos x - \sin x}{x^2} = \lim_{x\to 0} \frac{\sin x}{x} \cdot \frac{\cos x - 1}{x} = \lim_{x\to 0} \frac{\sin x}{x} \cdot \lim_{x\to 0} \frac{\cos x - 1}{x}$$

I'm assuming you already know the derivatives of sine and cosine.

Title: How could you evaluate this limit without L'Hopitals Rule?

Body:
How could someone in high-school solve this limit without using L'Hopitals Rule?
$$\lim _{x \rightarrow 0} \frac{\sin x \cos x-\sin x}{x^{2}} = 0$$
I assume that it involves the fact that
$$
\lim _{x \rightarrow 0} \frac{\sin (x)}{x}=1
$$
but I'm not seeing how.


Body: Hint:we have  $$\lim_{x \to 0} \frac{\sin x}{x}.\frac{(-2\sin^2 (x/2))}{x^2/4}.4x=?$$


问:
Numerical errors in specifying covariance of two (or more) random signals
I am looking at conditioning randomly generated signals that produce very precise first and second order statistics.

Lets consider two randomly generated, 1-D signals $[X,Y]$ of arbitrary, but finite length. And we condition these signals to have zero mean and unit variance.

Lets begin with 100 elements, generated with some python code:

```<code>import numpy as np
np.random.seed(10101)

slen = 100 #signal length

#each column of x is a signal, X = x[:,0], Y = x[:,1]
x = np.random.uniform(low=-1,high=1,size=(slen,2))

#Zero mean
x = x-np.mean(x, axis=0)
#Unit variance
x = x/np.sqrt(np.mean(x**2,axis=0))

</code>```

With zero mean and unit variance, we can create a covariance matrix $R_{ij}$ that specifies our desired variance and covariance between our two signals. Following a standard Cholesky decomposition to pre-multiply our signals by in order to give them the desired second order statistics. Just use some arbitrary values here.

$R_{ij}=
\begin{bmatrix}
1 & 0.25 \\
0.25 & 1
\end{bmatrix}$

$L_{ij}=
\begin{bmatrix}
1 & 0 \\
0.25 & 0.96824584
\end{bmatrix}$

```<code>Rij = np.array([[1,0.25],[0.25,1]])
Lij = np.linalg.cholesky(Rij)
</code>```

We can now pre-multiply our signals by $L_{ij}$ and should return two signals with the given stats in $R_{ij}$.

$\begin{bmatrix}
x'\\
y'
\end{bmatrix} = L_{ij}
\begin{bmatrix}
x^{T}\\
y^{T}
\end{bmatrix}$

```<code>xp = np.matmul(Lij, x.T).T
</code>```

If we check the stats of the actual signal, we (not surprisingly) get something close-ish.

```<code>print('Signal stats')
print('R_xx = ', np.mean(xp[:,0]**2))
print('R_xy = ', np.mean(xp[:,0]*xp[:,1]))
print('R_yy = ', np.mean(xp[:,1**2]))
</code>```

Resulting in

```<code>Signal stats
R_xx =  0.9999999999999996
R_xy =  0.19102763783107032
R_yy =  0.9705138189155358
</code>```

Obviously $R_{xx}$ is dead on, and $R_{xy}$ and $R_{yy}$ are clearly in the ballpark, but not as accurate as most would like. (It turns out that it is ALWAYS $R_{yy}$ that is less accurate than $R_{xx}$, more on that later).

I am fairly sure this stems from the fact that our 'randomly' generated signal of just 100 elements is not statistically converged. I.e. the conditioned $X$ and $Y$ vectors do not have zero covariance as required by this Cholesky decomp technique to work. Indeed we can check this:

```<code>print('Random vector stats')
print('R_xy = ', np.mean(x[:,0]*x[:,1]))
</code>```

yields

```<code>Random vector stats
R_xy =  -0.06090639375114042
</code>```

We can improve these results by increasing the length of the originally random signals, so say, 10,000 elements, and the results do improve:

```<code>Signal stats
R_xx =  1.000000000000001
R_xy =  0.2632783322133913
R_yy =  1.0066391661066976
</code>```

And the error will scale with the non-zero covariance error.

This leads to my first question:

 Can I condition the original signals X and Y to truly have zero covariance and give a better level of accuracy in this procedure?

My second question comes from practical observation involving the exact same procedure with variable $R_{ij}$ values.

Swapping in a smaller $R_{xy}$ value to something like

$R_{ij}=
\begin{bmatrix}
1 & 0.001 \\
0.001 & 1
\end{bmatrix}$

We get the following results

```<code>Signal stats
R_xx =  0.9999999999999996
R_xy =  -0.059906363297935916
R_yy =  0.9998781872734045
</code>```

So the accuracy of the procedure also depends on the level of covariance we attemp to set.

Playing around with values of $R_{xy}$ yields:

```<code>Smaller $R_{xy}$ yields more accurate $R_{yy}$
</code>```

So my second question is:

```<code>What is the source of this pattern and can it be improved? (Preconditioning?)
</code>```

Thanks for taking a look!

答:
Here is the best link to "de-correlating" or "whitening" the signals I have found

https://courses.media.mit.edu/2010fall/mas622j/whiten.pdf


问:
Is the following claim about distance inequalities true?
I'm working on a real analysis homework problem and my approach depends on the following claim being true:

Let $(S,d)$ be a metric space, $(z_n)$ a sequence in $S$ converging to $x$ for which $d(z_n,x)\geq d(z_{n+1},x)$ for all $n$ (i.e. $z_n$ is always getting closer to $x$), and let $r_n=\sup\{d(z_m,z_n): m\geq n\}$. Then
$$\overline{B}(z_n, r_n) \supset \overline{B}(z_{n+1}, r_{n+1}),$$
i.e.
$$d(y,z_{n+1})\leq r_{n+1} \Rightarrow d(y,z_n)\leq r_n$$
for all $y,n$. Intuitively (i.e. by sketching examples in $\mathbb{R}$), this seems true, but I'm having trouble proving it. Any hints would be much appreciated!

答:
Claim: If $(x_n)_{n=0}^\infty$ a Cauchy sequence, then for any $N > 0$ we have $(x_n)_{n=N}^\infty$ is Cauchy as well.

Proof: Fix $\epsilon > 0$. Since $(x_n)_{n=0}^\infty$ Cauchy, there exists $M$ so that for $n,m \geq M$ we have $d(x_n, x_m) < \epsilon$. If $M \geq N$, then choose the same $M$ for $(x_n)_{n=N}^\infty$. If $N > M$, then notice that for all $n,m \geq N > M$ we have that $d(x_n, x_m) < \epsilon$, so there still exists such an $N$. So if we let $N' = \max\{N, M\}$, then we have that for all $n,m \geq N'$ $d(x_n, x_m) < \epsilon$, proving the subsequence $(x_n)_{n=N}^\infty$ is Cauchy.

Claim: Suppose that for every sequence of nested closed balls $\overline{B}(x_{n+1}, r_{n+1}) \subseteq \overline{B}(x_n, r_n)$ with $r_n \rightarrow 0$, we have $\bigcap_{n=1}^\infty \overline{B}(x_n, r_n) \neq \varnothing$. If $(x_n)$ is Cauchy, then there is an $x$ so that $x_n \rightarrow x$.

Proof:
Take $(x_n)$ a Cauchy sequence. Notice there exists an $N_1$ so that for all $n,m \geq N_1$, we have $d(x_n, x_m) < 1/2$. Let $z_1 = x_{N_1}$. Notice that $(x_n)_{n=N_1}^\infty \subseteq \overline{B}(z_1, 1)$ is a Cauchy sequence by the above claim. Thus there exists an $N_2 > N_1$ so that for all $n,m \geq N_2$, we have $d(x_n, x_m) < 1/4$. Let $z_2 = x_{N_2}$. We claim that $\overline{B}(z_2, 1/2) \subseteq \overline{B}(z_1, 1)$. Taking $y \in \overline{B}(z_2, 1/2)$, we see

$$d(y,z_1) \leq d(y,z_2) + d(z_2, z_1) \leq 1/2 + 1/2 = 1,$$
so $y \in \overline{B}(z_1, 1)$. Continue inductively like this. Suppose we have $z_r$ and $N_r$ so that $(x_n)_{n=N_r}^\infty \subseteq \overline{B}(z_r, 1/2^r)$ and for all $n,m \geq N_r$ we have $d(x_n, x_m) < 1/2^{r+1}$. This sequence is Cauchy, and so we find $N_{r+1} > N_r$ so that for $n,m \geq N_{r+1}$ we have $d(x_n, x_m) < 1/2^{r+2}$. Letting $z_{r+1} = x_{N_{r+1}},$ we take $\overline{B}(z_{r+1}, 1/2^{r+1})$. We claim that $\overline{B}(z_{r+1}, 1/2^{r+1}) \subseteq \overline{B}(z_r, 1/2^r)$. Take $y \in \overline{B}(z_{r+1}, 1/2^{r+1})$, and note

$$d(z_r, y) \leq d(z_r, z_{r+1}) + d(z_{r+1}, y) \leq \frac{1}{2^{r+1}} + \frac{1}{2^{r+1}} = \frac{1}{2^r},$$
so $y \in \overline{B}(z_r, 1/2^r)$.

Using this, we have constructed a decreasing sequence of closed balls, so using the property we get that $\bigcap_{r=0}^\infty \overline{B}(z_r, 1/2^r) \neq \varnothing$. Take $x$ in the intersection. We claim $x_n \rightarrow x$. This will follow if we show that the subsequence $z_r \rightarrow x$. To see this, fix $\epsilon > 0$. There exists $R$ sufficiently large so that $1/2^R < \epsilon$, so for $r \geq R$ we have that $d(z_r, x) \leq 1/2^r < \epsilon$ (since $x \in \overline{B}(z_r, 1/2^r)$). Thus $z_r \rightarrow x$, and since we have a convergent subsequence of a Cauchy sequence this implies that $x_n \rightarrow x$.

For this last fact, see Cauchy sequence is convergent iff it has a convergent subsequence.


问:
If a characteristic function is infinitely divisible, it is nonzero everywhere
Let $E$ be a $\mathbb R$-Banach space and $\mathcal M_1(E)$ denote the set of probability measures on $(E,\mathcal B(E))$. Remember that if $\mu\in\mathcal M_1(E)$, then $$\varphi_\mu:E'\to\mathbb C\;,\;\;\;x'\mapsto\int\mu({\rm d}x)e^{{\rm i}\langle x',\:x\rangle}$$ is the characteristic function of $\mu$. Let $$C_1(E):=\left\{\phi_\mu:\mu\in\mathcal M_1(E)\right\}.$$ Remember that $\varphi\in C_1(E)$ is called infinitely divisible if $$\forall k\in\mathbb N:\exists\psi\in C_1(E):\varphi=\psi^k\tag1.$$

Now let $\varphi\in C_1(E)$ be infinitely divisible. How can we show that $\varphi(x')\ne0$ for all $x'\in E'$?

The idea should be to take $(\psi_n)_{n\in\mathbb N}$ with $$\varphi=\psi_n^n\;\;\;\text{for all }n\in\mathbb N\tag2$$ and observe that $$\left|\psi_n(x')\right|=\left|\varphi(x')\right|^{\frac1n}\xrightarrow{n\to\infty}\begin{cases}1&\text{, if }\varphi(x')\ne0\\0&\text{, otherwise}\end{cases}\tag3$$ for all $x'\in E'$, but how do we need to argue further?

答:
Let $(E,\| \cdot \|)$ be a Banach space.

If $\varphi$ is infinitelly divisible, then for any $n \in \mathbb N$ you have $\varphi_n$ such that $\varphi_n^n \equiv \varphi$.

Now, since $\varphi$ is a characteristic function, it's continuous and $\varphi(0)=1$. In particular, $ \| \varphi \|^2 \ge \frac{1}{2} $ on some small ball $B(0,\delta)$ in $E^*$.

Hence $$ \|\varphi_n(x)\|^2 = \varphi_n \cdot \bar{\varphi_n}(x) \to 1 $$ for $x \in B(0,\delta)$. But now, given a characteristic function $\psi$ of probability measure $\mu$, then function $\psi \cdot \bar{\psi}$ is a characteristic function (and it corresponds to $X-Y$ where $X,Y$ are independent with $\mu$ distribution)

Hence we showed that $(\varphi_n \cdot \bar{\varphi_n})_n$ is a sequence of characteristic functions that converge to continuous function on $B(0,\delta)$ and by Levy Continuity theorem we get that $\varphi_n \cdot \bar{\varphi_n}$ converges to some $\psi$ which is a characteristic function such that $\psi(x)= 1$ in a neighbourhood of $0$ (for $x \in B(0,\delta)$).

Now, the point is, the only characteristic function being equal to $1$ on some neighbourhood of $0$ is a characteristic function of $\delta_0$ measure. Indeed, by taking any $z \in E^* \setminus \{0\}$ we can introduce function $\eta_z:\mathbb R \to \mathbb C$ by formula $\eta_z(t) = \psi(tz)$. It is easy to see that if $\psi$ corresponds to random variable $X$ on $(E,\mathcal B(E))$, then $\eta_z$ corresponds to random variable $z(X)$ on $(\mathbb R,\mathcal B(\mathbb R))$ (due to $\eta_z(t) = \psi(tz) = \mathbb E[\exp(i\cdot (tz)(X))] = \mathbb E[\exp(it\cdot z(X))] = \varphi_{z(X)}(t)$). But then $\eta_z$ is a characteristic function corresponding to real random variable such that $\eta_z(t) = 1$ for some $t \in (-\delta,\delta)$ (if $\psi(x) = 1$ for $\|x\| \le r$ then $\delta = \frac{r}{\|z\|}$ for example). Hence it is enough to prove that real characteristic function has such property, which is rather known (if you're not familiar with it, I'll provide a proof)

But if $\varphi(x) = 0$ for some $x \in E^*$, then $\varphi_n(x) \to 0$ and that would contradict the fact that norms of $\varphi_n(x)$ need to converge to $1$.


问:
What does $\langle x,3\rangle$ mean if I'm talking about $\mathbb Z[x]$
I know this sounds like a basic question but I'm really confused. What does the notation $\langle x,3\rangle$ refer to for $\mathbb Z[x]$? Can someone write out what this is?</p>

答:
The notation means the ideal of $\mathbb{Z}[x]$ generated by the elements $3$ and $x$ (see here)
$$\langle x,3\rangle=\{xf+3g\mid f,g\in\mathbb{Z}[x]\}=\{a_nx^n+\cdots+a_1x+a_0\mid a_i\in\mathbb{Z}\text{ and }3\text{ divides }a_0\}.$$</p>


问:
What is the coefficient of $\sqrt{2}$ in $f(\sqrt{2})$ where $f(x) = 3 - x + 2x^2 - 5x^3 +4x^4?$

What is the coefficient of $\sqrt2$ in $f(\sqrt2)$ where $f(x) = 3 - x + 2x^2 - 5x^3 +4x^4?$

Hi
I'm studying polynomial rings.
I don't get what the question is asking me to do.
Should I just plug in $\sqrt2 $ in $f(x)$?
Isn't the definition of coefficient the number in front of a variable? How can there be a coefficient of a number, i.e. $\sqrt2$?

Am I missing something..

答: The square root doesn't survive even powers.  As $-x-5x^3=-x(1+5x^2)$ the coefficient is $-(1+5\cdot2)$.


问:
Understanding infinitely many primes proof.
Why does it follow that $p_i$ must divide $P-p_1p_1...p_n$?

答: Because $P$ is greater than the putative greatest prime, it must be composite, and is thus divisible by some $p_i$. Because $p_i$ is included in the product $p_1 \dots p_n$, it divides $p_1 \dots p_n$. Then, $p_i$ divides both $P$ and $p_1 \dots p_n$, so $p_i$ divides their difference.


问:
Continuity of primitive function
Suppose $f : [a,b] \times [c,d]\to\Bbb{R}$ is continuous. Let $F:[c,d]\to \Bbb{R}$ and $$F(y)=\int_a^bf(x,y) \space\space dx.$$

Show $F(y)$ is continuous.

I try the following : Let $y,t \in [c,d]$.  Suppose $F(y)$ is not continuous $\implies$ $F(y)$ is not uniformly continuous. We have that $\forall \delta>0\exists \epsilon >0$, s.t $$\vert y-t\vert<\delta\implies\vert F(y)-F(t)\vert=\Biggr\vert\int_a^bf(x,y)-\int_a^bf(x,t)\Biggr\vert=0\geq\epsilon.$$

A contradiction. I feel like this is wrong, since I didnt invoke the continuity of $f$ at any point. What should I take a look at to understand this problem?

答:
In your attempt, the statement:

$$
|F(y)-F(t)| = \left|\int_a^b(f(x,y) - f(x,t))dx\right|=0
$$
doesn't follow from what you assumed, that is, neither the form of $F$ nor that it is not uniformly continuous. Thus you can't get the contradiction $0\ge \epsilon$.

By definition of continuity of $F$, let $\epsilon >0$. We need to show there is some $\delta>0$ s.t for $y, t \in [c, d]$,
$$
|y-t|<\delta \implies |F(y)-F(t)|<\epsilon
$$

Since $f$ is continuous on $[a,b]\times[c,d]$, it is uniformly continuous there. Hence, in particular, given $\epsilon >0$, there is $\delta >0$ s.t. for all $x\in[a,b]$,
$$
|(x,y) - (x,t)|=|y-t| <\delta \implies |f(x,y)-f(x,t)|<\epsilon/(b-a)
$$

Thus:
$$
|F(y)-F(t)| = \left|\int_a^b(f(x,y) - f(x,t))dx\right|\\
           \leq \int_a^b \left|f(x,y)- f(x,t)\right| dx\\
           \leq \epsilon(b-a)/(b-a) = \epsilon
$$

Title: Continuity of primitive function


答: Because by the fundamental theorem of arithmetic (which is usually proved before the infinitude of primes) every integer greater than $1$ either is a prime number itself or can be represented as the product of prime numbers.

Title: Understanding infinitely many primes proof.

Body: Why does it follow that $p_i$ must divide $P-p_1p_1...p_n$?


问:
If $\int|f|^p\ d\mu=K$, then $\int \lvert f \rvert^p\ \chi_E\ d\mu\le K\mu(E)$
Let $f\in L_p$, so

$$\lVert f \rVert_p <+\infty \Longrightarrow \int\vert f\rvert^p\ d\mu=K<\infty$$

For a demonstration of an exercise I assumed that,

$$\int \lvert f \rvert^p\ \chi_E\ d\mu\le\int K\chi_E\ d\mu=K\mu(E)$$

But a colleague questioned me about that statement, and I ended up not so sure about it anymore, because it’s an old proof and I don’t remember exactly what I thought when I wrote this. I was unable to demonstrate this statement or find a counterexample.

So, my question is whether the above implication is true or not.

答:
No it's not. Consider for example
$$f(x)=\sum_{n=1}^{\infty} n \chi_{[\frac{1}{n^3+1}, \frac{1}{n^3})}$$
for $x\in (0,1)$ and $f\equiv 0$ elsewhere.

Then $f$ is in $L^1$, and lets say it has norm $\infty>K>0$. Choose $n$ large enough so that $n>K$. Then for $E=[\frac{1}{n^3+1}, \frac{1}{n^3})$ you have that
$$\int_E f > \int_E K$$


问:
Calculating the variance of a random variable that has an infinite range
A man has $7$ distinct balls. The experiment is to drop $1$ ball at a time down a chute to his friend (each drop is an independent trial), who chose which ball he wanted to receive before the experiment started. If he receives his desired ball, the experiment ends. If he doesn't receive his desired ball, he sends the ball back up the chute and the man at the top adds the ball back into his pile to choose from (so this man picks balls with replacement). However, the man picking balls is drunk and therefore immediately forgets whatever ball he had thrown down the chute last (so the probability of each ball leading to success is always $\frac 17$).

Let $X$ be the number of trials the man needs until he succeeds.

So far, I have $\operatorname{Range}(X) = \{1,2,3,\ldots\}$ and the expected value $E(X) = 7$ (because the distribution is geometric and therefore $E(X) =\frac 1p = \frac 1{1/7} = 7$).

How do I solve for the variance of $X$?

Also, how would I solve for the variance of $X$ if the experiment were exactly the same except without replacement?

答:
Variance is defined as $\mathbb{E}\left[\left(X-\mu\right)^2\right] = \mathbb{E}\left[X^2\right]-\mathbb{E}\left[X\right]^2$. The PMF is given by the probability of failing the previous times and succeeding on the $n$ roll, so it is $$P(X=n) = (1-p)^{n-1}p$$

Then $\mathbb{E}\left[X\right]$ is $\frac{1}{p}$, as you got already, and $$\mathbb{E}\left[X^2\right] = \sum_{n=1}^{\infty} P(X=n)n^2 = \sum_{n=1}^{\infty} (1-p)^{n-1}pn^2$$

Equivalently, this sum is $$p\sum_{n=0}^{\infty}(1-p)^nn^2 + 2p\sum_{n=0}^{\infty}(1-p)^nn + p\sum_{n=0}^{\infty}(1-p)^n$$

Using that $\sum_{n=0}^{\infty}x^n = \frac{1}{1-x}$ and differentiating, this is equal to $$\frac{2-p}{p^2}$$

So the variance would be $$\frac{2-p}{p^2} - \frac{1}{p^2} = \frac{1-p}{p^2}$$
which is what is expected from a geometric distribution.

Edit: For $p = \frac{1}{7}$ as in the question, the variance is $42$.


问:
Induction proof with binomial coefficients and inequalities
I am attempting to prove that for some non-negative integers
p ≥ q ≥ r, ${p \choose r} \geqslant {q \choose r}$.

My base cases were p=q=r=0 and p=q=r=1, and my induction hypothesis is to assume that the statement holds for some arbitrary non-negative values of p, q, and r. I get stuck, of course, at the inductive step. Right now I have:

Consider p + 1, q + 1, e.g.
${p+1 \choose r} ≥ {q+1 \choose r}$
I'm attempting to use Pascal's Identity and the fact that ${(r - 1)! = \frac{r!}{r}}$, but I end up with a very circular argument, where I'm saying that

From ${p+1 \choose r} \geqslant {(q+1) \choose r}$
we get
${p \choose r} + {p \choose r-1} ≥ {q \choose r} + {q \choose r-1}$

then
$$\frac{p!}{r!(p-r)!} + \frac{p!}{(r-1)!(p-r-1)!} ≥ \frac{q!}{r!(q-r)!}+\frac{q!}{(r-1)!(q-r-1)!}$$

And here's where I get stuck, because if this equation reduces to $${p+1 \choose r} \geqslant {q+1 \choose r}$$
then I've just shown something that was a given, and I need to change my inductive step to be more robust? Or would this be sufficient, and I'm overthinking it?

Thank you in advance!

答:
Hint:

$$
\eqalign{
  & 0 \le r \le q \le p\quad  \Rightarrow \quad {{q^{\,\underline {\,r\,} } } \over {r!}}
 \le {{p^{\,\underline {\,r\,} } } \over {r!}}\quad  \Rightarrow \quad q^{\,\underline {\,r\,} }
  \le p^{\,\underline {\,r\,} } \quad  \Rightarrow   \cr
  &  \Rightarrow \quad \left( {q + 1} \right)q^{\,\underline {\,r\,} }
  = \left( {q + 1} \right)^{\,\underline {\,r + 1\,} }  \le \left( {p + 1} \right)^{\,\underline {\,r + 1\,} }
  = \left( {p + 1} \right)p^{\,\underline {\,r\,} } \quad  \Rightarrow   \cr
  &  \Rightarrow \quad \left( {q + n} \right)^{\,\underline {\,r + n\,} }  \le \left( {p + n} \right)^{\,\underline {\,r + n\,} }  \cr}
$$

so we are moving diagonally up ...


问:
Gradient iff conditions on vector-valued Lipschitz function on $R^n$
Suppose that $f\in C^1(\mathbb{R})$.

$f\colon \mathbb{R}\to \mathbb{R}$ is Lipschitz if and only if $|f'|$ is bounded on $\mathbb{R}$.

Is it still true in multidimensional and vector-valued situation?

Say, $f\colon \mathbb{R}^n \to \mathbb{R}$. Is it true that $||\nabla f||_2$ bounded $\Longleftrightarrow f$ is Lipschitz?

Say, $f\colon \mathbb{R}^n \to \mathbb{R}^m$. Is it true that $||Jf||_2$ bounded $\Longleftrightarrow f$ is Lipschitz, where $Jf$ is the Jacobian of $f$ and $\|\,\|_2$ is matrix induced norm in L2 sense?

If not, what are some good iff/if conditions by using gradients?

Example, let $f(\mathbf{x}) = x_1x_2$, where $\mathbf{x} = [x_1\,x_2]^T\in \mathbb{R}^2$. Is this $f(\mathbf{x})$ Lipschitz? It intuitively looks like Lipschitz, but its gradient is unbounded.

答:
In your example $f(xy)=xy,$ $f$ is not Lipschitz. Proof: $|f(t,t)-f(0,0)|=t^2,$ and

$$\frac{t^2}{|(t,t)-(0,0)|}\to \infty$$

as $t\to \infty.$

But yes, it is true that if $f:\mathbb R^n\to \mathbb R^m,$ then $f$ is Lipschitz iff $\|Jf(x)\|_2$ is bounded. Let's take $m=1$ for simplicity. To prove $f$ Lipschitz implies $\|Jf(x)\|_2$ is bounded, note the latter is bounded iff each $D_kf(x)$ is bounded. If $D_kf(x)$ is not bounded, then there is a sequence $x_j$ such that $|D_kf(x_j)|>j.$  That implies

$$\left |\frac{f(x_j+he_k)-f(x_j)}{h}\right| > j\text {  for small positive } h.$$

That violates the Lipschitz condition.

I'll leave the other direction and the case $m>1$ to you for now.

OwnerDisplayName: anon


问:
All involutive isometry has a fix point
Let $T:\mathbb R^n\to \mathbb R^n$ an isometry. Then $T(x)=Mx+v$ for some orthogonal matrix $M\in \mathbb R^{n\times n}$ and some vector $v\in\mathbb R^n$. We suppose $T$ involutive, that mean that $T^2=id$. I have to prove that $T$ has a fix point.

Attempts

So, if $c=-(M-I)^{-1}v$, then $$T(c)=c.$$
Indeed,

$$T(c)=-M(M-I)^{-1}v+v=-(M-I)(M-I)^{-1}v+v-(M-I)^{-1}v=-(M-I)^{-1}v=c.$$

But this only hold whenever $(M-I)^{-1}$ invertible. Since $M$ is orthogonal, it may be not the case that $M-I$ is invertible. So, how can I do ? Since I didn't use the fact that $T$ is involutive whenever $M-I$ is invertible, I suppose that we need this assumption here. But I don't know how I can use it. The only thing I see it's that since $T$ is involutive, $$M^2x+Mv+v=x.$$
But this doesn't really help. Can someone help here ?

答:
If $M-I$ not invertible, there is $u\neq 0$ s.t. $(M-I)u=0$. Since $T$ is involutive (and as you remarked),
\begin{align*}
M^2x+Mv+v=x&\iff (M^2-I)x+Mv+v=0\\
&\iff (M+I)(M-I)x+Mv+v=0.\tag{E}
\end{align*}

In particular, evaluating $(E)$ for $x=u$ yields $Mv=-v$. Therefore, $c=\frac{1}{2}v$ is a fixed point.

Title: All involutive isometry has a fix point


问:
Application of Triple Integrals
Any easy way to solve this integration?

$\int_{0}^{3} \int_{0}^{\frac{6-2x}{3}} \int_{0}^{\frac{6-3y-2x}{2}} xyz \ dzdydx$

My work

$\int \int \int_D xyz \ dV = \dfrac{1}{2}\int_{0}^{3} \int_{0}^{\frac{6-2x}{3}}  xy(z^2 )\biggr|_{0}^{\frac{6-3y-2x}{2}} \ dy \ dx$

$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} \int_{0}^{\dfrac{6-2x}{3}}  xy(6-3y-2x)^2 \ dy \ dx$

$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} x \left [y\dfrac{(6-3y-2x)^3}{3(-3)} -\dfrac{(6-3y-2x)^4}{4(-3)3(-3)} \right ]\biggr|_{0}^{\dfrac{6-2x}{3}} \ dx$

$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} x \left [y\dfrac{(6-3y-2x)^3}{3(-3)} -\dfrac{(6-3y-2x)^4}{4(-3)3(-3)} \right ]\biggr|_{0}^{\dfrac{6-2x}{3}} \ dx$

The terms are expanding, any way to find a solution in easy case. please help

答:
The domain of integration is $D$, the set of $(x,y,z)$ with $x,y,z\ge 0$ and $2x+3y+2z\le 6$. Well, to have a simpler, symmetric formula, we can change variables. For instance $X=2x/6$, $Y=3y/6$, $Z=2z/6$. Then the domain $D$ is transformed into the simplex (cone) $E$ of all $(X,Y,Z)$ with $X,Y,Z\ge 0$ and $X+Y+Z\le 1$. Formally, $x=3X$, $y=2Y$, $z=3Z$, so $dx\; dy\; dz=3\cdot 2\cdot 3\; dX\; dY\; dZ$. We can compute in a few easy lines, explicitly, the integral $J$ (notation) from the OP:
$$
\begin{aligned}
J &=\iiint_D xyz\; dx\; dy\; dz\\
&=\iiint_E (3X)(2Y)(3Z)\cdot (3\cdot 2\cdot 3)\; dX\; dY\; dZ\\
&=324\iiint_{\substack{0\le X,Y,Z\\X+Y+Z\le 1}}XYZ\; dX\; dY\; dZ\\
&=324\int_0^1X\;dX\int_0^{1-X}Y\; dY\int_0^{1-X-Y}Z\; dZ\\
&=324\int_0^1X\;dX\int_0^{1-X}Y\; dY\cdot\left[\frac 12Z^2\right]_0^{1-X-Y}\\
&=324\cdot \frac 12\int_0^1X\;dX\int_0^{1-X}Y(1-X-Y)^2\; dY\\
&=324\cdot \frac 12\int_0^1X\;dX\int_0^{1-X}\Big(\ (1-X)(1-X-Y)^2-(1-X-Y)^3\ \Big)\; dY\\
&=324\cdot \frac 12\int_0^1X\;dX\left[
-\frac 13(1-X)(1-X-Y)^3+\frac 14(1-X-Y)^4\right]_0^{1-X}
\\
&=324\cdot \frac 12\int_0^1X\;dX\left(
\frac 13(1-X)^4-\frac 14(1-X)^4\right)
\\
&=324\cdot \frac 12\int_0^1X\cdot\frac 13\cdot \frac 14(1-X)^4\;dX
\\
&=324\cdot \frac 12\cdot\frac 13\cdot \frac 14\int_0^1(1-(1-X))\cdot(1-X)^4\;dX
\\
&=324\cdot \frac 12\cdot\frac 13\cdot \frac 14\int_0^1\Big((1-X)^4-(1-X)^5\Big)\;dX
\\
&=324\cdot \frac 12\cdot\frac 13\cdot \frac 14\left(\frac 15-\frac 16\right)\;dX
\\
&=324\cdot \frac 12\cdot\frac 13\cdot \frac 14\cdot \frac 15\cdot \frac 16\\
&=\frac{18^2}{6!}=\frac 9{20}\ .
\end{aligned}
$$
$\square$

Note: I saw no quick way to proceed using Stokes...

Homework: Try to find a generalization. What is for instance the integral of $XYZW$ w.r.t. volume $dX\; dY\; dZ\;dW$ on the simplex $0\le X,Y,Z,W\le X+Y+Z+W\le 1$?

Title: Application of Triple Integrals


问:
Finding a comparison series for $\frac{\sqrt[3]{k^2+1}}{\sqrt{k^3+2}}$
I've tried a number of other convergence tests to no avail and I have a hunch that either an ordinary/limit comparison test is probably where I need to be looking. I'm not looking for a direct answer here, rather some guidance in how to choose comparison series in situations like this.

$$\frac{\sqrt[3]{k^2+1}}{\sqrt{k^3+2}}$$

My initial thought was to look at $$\frac{k^\frac{2}{3}}{k^\frac{3}{2}}$$ as it follows the leading terms of $a_n$ but it doesn't match the conditions for the ordinary comparison test. I'm at a bit of a loss for the process of finding a $b_n$ as we didn't do much of this sort of work in class.

Thanks!

答:
Your initial thought was correct, but you have to use some elementary asymptotic analysis: as a polynomial is asymptotically equivalent to its leading term, we  have $k^2+1\sim_\infty k^2$, $\:k^3+2\sim_\infty k^3$, so
$$ \sqrt[3]{k^2+1}\sim_\infty k^{\tfrac23},\enspace\sqrt{k^3+2}\sim_\infty k^{\tfrac32}, \quad\text{whence}\quad \frac{\sqrt[3]{k^2+1}}{\sqrt{k^3+2}}\sim_\infty \frac{ k^{\tfrac23}}{k^{\tfrac32}}=\frac 1{k^{\tfrac56}},$$
which is a divergent $p$-series.


问:
The area of a triangle, if the vertices are complex numbers
I need to check for this task, if my logic is correct ?

The task is:

Complex numbers are given, which represent the points of a triangle.

$z_1=2+i$

$z_2=i$

$z_3=\lambda-2i$

$z_1$ and $z_2$ represent fixed points, while $z_3$ depends on $\lambda$.

Calculate the area of the triangle

I tried this:

I took that $\lambda$ for example $1$.

$P_{\triangle z_1z_2 z_3}=\frac{1}{2}\left|\begin{vmatrix} 2 & 1 & 1\\ 0 & 1 & 1\\ 1 & -2 & 1 \end{vmatrix}\right|=\frac{1}{2}\cdot (2+1+0-(1-4+0))=\frac{1}{2}\cdot 6=3$

By taking any other number for the $\lambda$, the same result is obtained.

Can it be done in any other way?

Thanks in advance !

答:
The area of $\triangle ABC$ in terms of the coordinates of its vertices
$(A_x,A_y)$, $(B_x,B_y)$, $(C_x,C_y)$
is  found as
\begin{align}
S&=
\tfrac12\,|(B_x-A_x)(C_y-A_y)-(B_y-A_y)(C_x-A_x)|
,
\end{align}

so, for $A=(2,1)$, $B=(0,1)$, $C=(\lambda,-2)$

\begin{align}
S&=\tfrac12\,|(0-2)(-2-1)-(1-1)(\lambda-2)|
=3
.
\end{align}

In this case $\lambda$ vanishes.


问:
Close Form Expression of The Cartesian Product of Linearly Ordered Sets
I was reading a combinatorics textbook,and I came across this problem.

Let the $product$ of partially ordered sets $(S_1, \preceq_1),..., (S_n, \preceq_n)$ be the partial order $\preceq$ on the cartesian product $S_1 \times S_2 \times ... \times S_n$ given by $(s_1,...,s_n) \preceq (t_1,...,t_n)$ iff $s_i \preceq_i t_i \; \forall i \in [n]$.
Also, let $(S, \leq)$ be a linearly ordered set with $m$ elements and $(S^n, \leq^n)$ be its $n^{th}$ power in the above sense. What is the closed form expression for $|\leq ^n|$?.

My intuition on this is that this $n-$ary cartesian power of $(S, \leq)$ is isomorphic to the space of all functions from $[m]$ to $[n]$, which is $n^m$. In other words, to me it seems like for every element $(s_1, s_2,...,s_m)$, any $s_i$ for $1\leq i \leq m$ can be chosen $n$ times regardless of other elements. Since $(S, \leq)$ is a linearly ordered set with $m$ elements, then $|\leq ^n| = n ^m$.

Is there any error in my reasoning? I would appreciate any help on this!

答:
We can take $S$ to be $[m]$ with the usual order, in which case $|S^n|=m^n$, but that’s not what you’re asked to find: what’s wanted here is the cardinality of the order relation itself, i.e., the number of pairs

$$\big\langle\langle k_1,\ldots,k_n\rangle,\langle \ell_1,\ldots,\ell_n\rangle\big\rangle\in[m]^n\times[m]^n$$

such that $k_i\le\ell_i$ for $i=1,\ldots,n$.

When $n=1$ this is $\sum_{k=1}^mk=\frac{m+1}2$, since $|\{\ell\in[m]:\ell\le k\}|=k$ for each $k\in[m]$.

When $n=2$ it is

$$\begin{align*}
\sum_{k=1}^m\sum_{\ell=1}^mk\ell&=\sum_{k=1}^mk\sum_{\ell=1}^m\ell\\
&=\sum_{k=1}^mk\binom{m+1}2\\
&=\binom{m+1}2\sum_{k=1}^mk\\
&=\binom{m+1}2^2\,,
\end{align*}$$

since there are $k\ell$ pairs $\langle i,j\rangle\in[m]^2$ such that $i\le k$ and $j\le\ell$.

Can you take it from there?


问:
Limit of $(3n+1)^{1/n}$
I am working through a question that requires me to prove that a certain limit tends to 1. I have reached an endpoint where I have:

$$\lim\limits_{n\to \infty} (3n+1)^{1/n}$$

I know that and can use the fact that $\lim\limits_{n\to \infty}( n^{1/n}) = 1$

However, I am not sure if I can just use this fact to say that it applies for $(3n+1)^{1/n} $ and I also am stuck on proving it.

答:
\begin{align*}
\lim_{n \rightarrow \infty} (3n+1)^{1/n}
    &= \lim_{n \rightarrow \infty} \exp \left( \ln \left( (3n+1)^{1/n} \right) \right)  \\
    &= \lim_{n \rightarrow \infty} \exp \left( \frac{1}{n} \ln \left( 3n+1 \right) \right)  \\
    &= \exp \left( \lim_{n \rightarrow \infty}  \frac{\ln \left( 3n+1 \right)}{n} \right)  \\
    &= \exp \left( \lim_{n \rightarrow \infty}  \frac{\ln \left( 3n \left( 1 + \frac{1}{3n} \right) \right)}{n} \right)  \\
    &= \exp \left( \lim_{n \rightarrow \infty}  \frac{\ln (3) + \ln (n) + \ln \left( 1 + \frac{1}{3n}  \right)}{n} \right)  \\
    &\overset{?_1}= \exp \left( \lim_{n \rightarrow \infty}  \frac{\ln (3)}{n} + \lim_{n \rightarrow \infty} \frac{\ln (n)}{n} + \lim_{n \rightarrow \infty} \frac{\ln \left( 1 + \frac{1}{3n}  \right)}{n} \right)  \\
    &= \exp \left( 0 + \lim_{n \rightarrow \infty} \ln n^{1/n} + \lim_{n \rightarrow \infty} \frac{\ln \left( 1 + \frac{1}{3n}  \right)}{n} \right)  \\
    &\overset{?_2}= \exp \left( \ln \lim_{n \rightarrow \infty} n^{1/n} + \lim_{n \rightarrow \infty} \frac{\ln \left( 1 + \frac{1}{3n}  \right)}{n} \right)  \\
    &= \exp \left( \ln 1 + \lim_{n \rightarrow \infty} \frac{\ln \left( 1 + \frac{1}{3n}  \right)}{n} \right)  \\
    &\overset{?_3}= \exp \left( 0 + \frac{\lim_{n \rightarrow \infty}  \ln \left( 1 + \frac{1}{3n}  \right)}{n} \right)  \\
    &\overset{?_4}= \exp \left( \frac{ \ln \lim_{n \rightarrow \infty} \left( 1 + \frac{1}{3n}  \right)}{n} \right)  \\
    &\overset{?_5}= \exp \left( \frac{ \ln \left( \lim_{n \rightarrow \infty} 1 + \lim_{n \rightarrow \infty} \frac{1}{3n} \right) }{n} \right)  \\
    &= \exp \left( \frac{ \ln \left( 1 + 0 \right) }{n} \right)  \\
    &= \exp \left( \frac{ \ln 1 }{n} \right)  \\
    &= \exp \left( \frac{ 0 }{n} \right)  \\
    &= \exp \left(0 \right)  \\
    &= 1  \text{.}
\end{align*}

We use the continuity of the exponential on its domain without further comment.

$?_1$ : The limit of a sum is a sum of the limits if those limits exist.  We subsequently show that they do, so this equality is valid.

$?_2$ : The logarithm is continuous on its domain, so this equality holds if the new limit is in $(0,\infty)$.  It is $1$, so this equality is valid.

$?_3$ : Again w exchange the limit and the logarithm.  Again, the new limit is in $(0,\infty)$, so this equality is valid.

$?_4$ : Division is continuous in its numerator without any restriction on domain, so this equality is valid.  (Notice that the domain of the denominator is restricted, so there can be something to check here.)

$?_5$ : The limit of a sum is the sum of the limits, if those limits exist.  As we subsequently show, those limits do exist, so this equality is valid.

Title: Limit of $(3n+1)^{1/n}$


问:
Are projective modules extended under surjective map
Let A and B be commutative rings with unity. Given a surjective ring homomorphism
$$\pi : A \rightarrow B$$
Is every projective B module extended from a projective A module? That is given a B-projective module P, does there exist a projective A-module Q such that
$$ Q \otimes_A B = P$$

My attempt:
Use the relation for an A-module M
$$A/I \otimes_A M \simeq M/IM.$$
I can consider B isomorphic to A/I. Then any B-module P is annihilated by I. Does this mean P is isomorphic to Q/I for a A module Q?

答: Let $\mathbb{Z}$ be the ring of integers. Then the natural  ring homomorphism  $\pi:\mathbb{Z}\to\mathbb{Z}/\langle 5\rangle$ is a surjective ring homomorphism. Since $\mathbb{Z}/\langle 5\rangle$ is a field, every $\mathbb{Z}/\langle 5\rangle$-module is free and so is projective. Thus, $\prod_{I} \mathbb{Z}/\langle 5\rangle$ is a projective $\mathbb{Z}/\langle 5\rangle$-module, where $I$ is an infinite set, and since $\mathbb{Z}$ is a PID, every projective $\mathbb{Z}$-module is free, that is, every projective  $\mathbb{Z}$-module is of the form $\bigoplus_{J}\mathbb{Z}$ for some set $J$.  Also, $(\bigoplus_{J}\mathbb{Z})\bigotimes_\mathbb{Z}\mathbb{Z}/\langle 5\rangle\cong\bigoplus_{J}\mathbb{Z}/\langle 5\rangle$. But if $I$ is an infinite set we have $\prod_{I} \mathbb{Z}/\langle 5\rangle\not\cong\bigoplus_{J} \mathbb{Z}/\langle 5\rangle$ for each set $J$.


问:
Why is right-continuity assumed in the definition of stochastic processes?
Let $X_t$, $t \geq 0$ be a continuous-time stochastic process taking values on an enumerable set $I$ and assume that $X_t(\omega)$ is right-continuous function of $t$ for any $\omega$ in the probability space.

I read in a book that right continuity is important since it allows to use the following type of identities, for $i \in I$,
$$
\mathbb{P}(X_t = i \mbox{ for some } t \in [0, \infty)   ) \\ = 1 - \lim\limits_{n \rightarrow \infty} \sum\limits_{j_1, \ldots, j_n \neq i   } \mathbb{P} \big (  X_{q_1} = j_1, \ldots, X_{q_n} = j_n \big )
$$
where $q_1$, $\ldots$, $q_n$ is an enumeration of the rationals.

Where is right-continuity used in the above expression and why do we need to consider only the rationals?

答:
The way right continuity is used here is that if $X_t(\omega) = i$ then there exists some interval $[t, t+\varepsilon]$ such that $X_s(\omega) = i$ for all $s \in [t, t+\varepsilon]$ (this also uses the fact that $X$ takes values in $I$, but it's a similar idea for more general processes).  Since every interval has to contain a rational number, this implies that we can just check rational times because if $X_t(\omega) = i$ for some $t \in [0, \infty)$ then we must also have that $X_t(\omega) = i$ for some rational time $t \in [0,\infty)$.  Hence \begin{align*} \{\omega : X_t(\omega) = i \text{ for some } t \in [0,\infty)\} &= \{\omega : X_t(\omega) = i \text{ for some rational } t \in [0,\infty)\} \\
 &= \bigcup_{q_n \in \mathbb{Q}} \{\omega : X_{q_n}(\omega) = i \}.\end{align*}


问:
Demonstrate that every positive integer can be expressed as the sum of distinct non-negative integer powers of 2
Demonstrate that every positive integer can be expressed as the sum of distinct non-negative integer powers of 2. In other words, prove that for every positive integer can be re-written as $2^{b_0}+2^{b_1}+...2^{b_r}$ such that $b_0\lt b_1\lt...b_r$

I'm not sure if I'm doing this right but so far this is what I have:

Base Case: n=1, which can be written as $2^0=1$. Assume that for some integer $k\ge 1$, $k$ can be written as $2^{b_0}+2^{b_1}+...2^{b_r}$. We will demonstrate that $k+1$ can be written as the sum of distinct powers of $2$.

Case 1: $k+1$ is odd

$k+1=2^0+k$; since k is already a sum of distinct powers of 2, for any odd number $k+1$ the claim is true.

Case 2: $k+1$ is even

Since $k+1$ is even, then $\frac{k+1}{2}$ is an integer which is also strictly less than k. By applying the inductive hypothesis, $\frac{k+1}{2}\le 2^{b_0}+2^{b_1}+...2^{b_r}$. Multiplying 2 on both sides, we get $k+1=2^{b_0+1}+...2^{b_r+1}$

Since for both conditions the claim is held true, it is true that for any integer n, it can be written as a sum of distinct powers of 2

Uhhh I'm quite certain that I've made some sort of mistake because I'm not jiggy with this proof

I'm not really satisfied with my explanation for case 1 - I'm more confident about case 2.

I'm also just assuming that for all numbers less than k the claim is true which I feel is some sort of logical flaw?

答:
This is (essentially) a correct argument. You are using what's known as strong induction when you assume the inductive hypothesis for all previous cases, not just the immediately previous case.

See https://undergroundmathematics.org/glossary/strong-induction

Note: you are proving that every positive integer has a  base $2$ representation.


答:
The domain of integration is a tetrahedron with vertices at $(0,0,0),(3,0,0),(0,2,0),(0,0,3)$. The integrand is symmetric with respect to permuting variables. Thus we are allowed to rotate the domain of integration so that slices perpendicular to the $z$-axis are right isosceles triangles and the integral is equal to
$$\int_0^2\int_0^{(6-3z)/2}\int_{(6-3z-2y)/2}xyz\,dx\,dy\,dz$$
We move the $z$ as far out as it will go:
$$\int_0^2z\int_0^{(6-3z)/2}\int_{(6-3z-2y)/2}xy\,dx\,dy\,dz$$
Then rewrite to make the slices of right isosceles triangles clearer. $T(k)$ is the triangle with vertices at $(0,0),(k,0),(0,k)$:
$$\int_0^2z\iint_{T((6-3z)/2)}xy\,dA\,dz$$
Now let's work out what the inner integral is before we substitute $(6-3z)/2$. In other words, what is
$$\iint_{T(k)}xy\,dA=\int_0^k\int_0^{x-k}xy\,dy\,dx$$
This evaluates to
$$\int_0^k x[y^2/2]_0^{x-k}\,dx=\frac12\int_0^k x(x-k)^2\,dx=\frac12[x^4/4-(2/3)kx^3+k^2x^2/2]_0^k=k^4/24$$
So our original integral becomes
$$\frac1{24}\int_0^2z\left(\frac{6-3z}2\right)^4\,dz=\frac{81}{24×16}\int_0^2z(2-z)^4\,dz$$
$$=\frac{27}{128}[8z^2-(32/3)z^3+6z^4-(8/5)z^5+z^6/6]_0^2$$
$$=\frac{27}{128}×\frac{32}{15}=\frac9{20}$$

Title: Application of Triple Integrals

Body:
Any easy way to solve this integration?

$\int_{0}^{3} \int_{0}^{\frac{6-2x}{3}} \int_{0}^{\frac{6-3y-2x}{2}} xyz \ dzdydx$

My work

$\int \int \int_D xyz \ dV = \dfrac{1}{2}\int_{0}^{3} \int_{0}^{\frac{6-2x}{3}}  xy(z^2 )\biggr|_{0}^{\frac{6-3y-2x}{2}} \ dy \ dx$

$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} \int_{0}^{\dfrac{6-2x}{3}}  xy(6-3y-2x)^2 \ dy \ dx$

$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} x \left [y\dfrac{(6-3y-2x)^3}{3(-3)} -\dfrac{(6-3y-2x)^4}{4(-3)3(-3)} \right ]\biggr|_{0}^{\dfrac{6-2x}{3}} \ dx$

$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} x \left [y\dfrac{(6-3y-2x)^3}{3(-3)} -\dfrac{(6-3y-2x)^4}{4(-3)3(-3)} \right ]\biggr|_{0}^{\dfrac{6-2x}{3}} \ dx$

The terms are expanding, any way to find a solution in easy case. please help


问:
$x^4-6x^3-73x^2+kx+m=0$, $\alpha\beta=\gamma\delta=4$, $\alpha^2-3(1+\sqrt{10})\alpha+4=0$, equations satisfied by $\beta$, $\gamma$ and $\delta$

The equation $x^4-6x^3-73x^2+kx+m=0$ has two positive roots, $\alpha$,
$\beta$ and two negative roots $\gamma$, $\delta$. It is given that
$\alpha\beta=\gamma\delta=4$. Show that
$\alpha^2-3(1+\sqrt{10})\alpha+4=0$, and find similar quadratic
equations satisfied by $\beta$, $\gamma$ and $\delta$.

Unsure how to approach this question. So far I have:

Found $k=-24$ and $m=16$

Shown $(\alpha+\beta)(\gamma+\delta)=-81$

Found quadratic equation with roots $\alpha+\beta$ and $\gamma+\delta$ to be $x^2-6x-81=0$

Found $\alpha+\beta=3+3\sqrt{10}$ and $\gamma+\delta=3-3\sqrt{10}$

Thanks for any help!

答:
Almost there. You showed that $$\alpha+\beta=3+3\sqrt{10}$$ and you are given $$\alpha\beta=4$$
So a quadratic equation with roots $\alpha$ and $\beta$ is $$x^2-(\alpha+\beta)x+\alpha\beta=0$$
or $$x^2-3(1+\sqrt{10})x+4=0$$
Since you know $\alpha$ is root:
$$\alpha^2-3(1+\sqrt{10})\alpha+4=0$$
You have the same equation for $\beta$. Apply similar reasoning for the other parameters.

Note you can multiply the above equation by $\beta$ and you have a linear equation for $\alpha$, in case you want to solve for $\alpha$ and you find quadratics difficult :)


问:
Is a majorable subset in lp is relatively compact?
Let A be a nonempty set in the sequence space lp with 1 ≤ p < +∞.
I want to show that if A is majorable then it is relatively compact.
A is majorable by definition if there exists y ∈ lp such that
|x(k)| ≤ |y(k)|, for all x ∈ A, k ∈ N.
I know that A is relatively compact if and only if it is totally bounded and if and only if every sequence in A has a subsequence that converges in lp.
I thought that i could use this theorem to prove my thesis but i couldn't come up with a solution.</p>

答:
Let $\{x_n\}_n$ be any sequence in $A$.  For every  $k$ one has that the sequence of scalars
$$
  \{x_n(k)\}_{n\in {\mathbb N}}
  $$
is  bounded by $|y(k)|$, so it admits a converging sub-sequence.  By doing this for each $k$, and using Cantor's diagonal
argument, we may find  a sequence $n_1<n_2<\cdots \quad$ of natural numbers, such that
$$
  z(k) := \lim_{i\to \infty } x_{n_i}(k)
  $$
exists for every $k$.  Since $|z(k)|\leq |y(k)|$, it is obvious that $z := (z(k))_k$ lies in $\ell^p$.  Moreover
$$
  \lim_{i\to \infty }\|x_{n_i} - z\|^p =
  \lim_{i\to \infty }\sum_{k=1}^\infty |x_{n_i}(k) - z(k)|^p = 0,
  $$
by Lebesgue's Dominated Convergence Theorem.  This proves that every sequence in $A$ has a converging sub-sequence (to a
point in $\ell^p$), so $A$ is relatively compact.


答:
For large $k$, $1$ is tiny compared to $k^2$ and $2$ is tiny compared to $k^3$. We can write

$$\frac{\sqrt[3]{k^2+1}}{\sqrt{k^3+2}}=k^{-5/6}\frac{\sqrt[3]{1+\dfrac1{k^2}}}{\sqrt{1+\dfrac1{k^3}}}.$$

The factor $k^{-5/6}$ carries the main contribution of the expression, and the ratio tends to $1$.

It is not very hard to find two limits such that

$$a\le\frac{\sqrt[3]{1+\dfrac1{k^2}}}{\sqrt{1+\dfrac1{k^3}}}\le b$$ as of some $k$. For example, $\dfrac12$ and $2$.

Title: Finding a comparison series for $\frac{\sqrt[3]{k^2+1}}{\sqrt{k^3+2}}$

Body:
I've tried a number of other convergence tests to no avail and I have a hunch that either an ordinary/limit comparison test is probably where I need to be looking. I'm not looking for a direct answer here, rather some guidance in how to choose comparison series in situations like this.

$$\frac{\sqrt[3]{k^2+1}}{\sqrt{k^3+2}}$$

My initial thought was to look at $$\frac{k^\frac{2}{3}}{k^\frac{3}{2}}$$ as it follows the leading terms of $a_n$ but it doesn't match the conditions for the ordinary comparison test. I'm at a bit of a loss for the process of finding a $b_n$ as we didn't do much of this sort of work in class.

Thanks!


问:
How to make this bound true if p=2
I'm trying to complete the proof of this theorem, pag. 234. At the end of the proof, it says that for $p=2$, the proof is only slightly different. The problem with $p=2$ is that the argument made using the Legendre formula in $(5)$:

$$
v_p(j!)=\sum_{i=1}^\infty\left\lfloor\frac{n}{p^i}\right\rfloor\leq\frac{j}{p}+\frac{j}{p^2}+\frac{j}{p^3}+\cdots=\frac{j}{p-1}\leq j-1
$$

would not hold. So without that inequality, you cannot assure that $$\frac{a^n-1}{a-1}$$ is divisible by $p^\alpha$ ($2^\alpha$ in this case). The Legendre formula is a key part of the proof because he uses to count the number of times $p$ can appear in $j$.

I know I have left out a lot of details because they are explained in the proof. If you need more information, please let me know.

答:
If you consider the Legendre formula in its alternate form (see here) you have:

$$ \nu_2(j!) = j - s_2(j) \le j - 1$$

and also:

$$ \nu_2(j) \le \log_2{j} \le j-1 $$

for $j \gt 1$.

Title: How to make this bound true if p=2


问:
Application of the Extreme Value Theorum
I understand what the Theorem means and how to apply it effectively. For example, the theorem holds true at intervals of $f(x)= \sin(x)$ at $[-2,2]$. But one thing is confusing me is if the maxima AND minimum values exist on the END points of the interval. For example, if I said that at $$f(x): \sin(x) for [\frac{-\pi}{2},\frac{\pi}{2}]$$Would the theorem still apply? The way I have been taught that it shouldn't because it is at the boundaries, but what is the point of the $[\ ]$, shouldn't it say $( \ )$ or was I taught wrong?

答:
The [ ] means the boundaries are included in looking for the maximum and minimum values. Don't get confused. The way the theorem is formulated is:
$$ f(c) \geq f(x) \geq f(d), \forall x \in [a,b] $$
Note that a can be equal to c or d, and b can be equal to c or d.


问:
Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways.

Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways.($a^2+b^2$, is same $b^2+a^2$), $n \in \mathbb{N}.$

I have proved the above question which appeared in one of the Math-Olympiad. And I do know the solution. Sharing the question only because the question has a cute solution.

答:
Let $n=t^2$

$n(n+1)=(t^2)^2+t^2$

Also
$t^4+t^2=(t^2-1)^2+3t^2-1$ and $3t^2-1=k^2$ for infinitely many k . (It is a Pell like equation and 3 is odd.)


问:
Finding all integer solutions of $\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = \frac12$
If we have integers $x,y,z$ such that $x,y,z \ge 3,$ find all solutions to $$\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = \frac12.$$

I was thinking of first expanding out this, and then simplifying from there, but the equation got very messy. Is there a better method?

答:
WLOG let $x\leq y$. Then, we have that $x = 3$ because we are given that $x\geq 3$, and if $x\geq 4$, $\frac{1}{x}+\frac{1}{y}\leq \frac{1}{2}$.

Then, we must solve $\frac{1}{y}-\frac{1}{z} = \frac{1}{6}$. Note that $y = 3,4,5$ because they are the only values such that $y\geq 3$ and $\frac{1}{y}>\frac{1}{6}$. Then, we can go into casework:

For $y = 3$, we get $\frac{1}{z} = \frac{1}{6}$ and thus $z = 6$.

For $y = 4$, we get $\frac{1}{z} = \frac{1}{12}$ and thus $z = 12$.

Finally, for $y = 5$ we get $\frac{1}{z} =\frac{1}{30}$ and thus $z = 30$.

We must permute $x$ and $y$ when they are different to get all of the solutions because of the earlier WLOG.

Thus, the only solutions are $\boxed{(3,3,6),\  (3,4,12),\ (3,5,30),\ (4,3,12),\text{ and }(5,3,30).}$


问:
Square root is operator monotone
This is a fact I've used a lot, but how would one actually prove this statement?

Paraphrased: given two positive operators $X, Y \geq 0$, how can you show that $X^2 \leq Y^2 \Rightarrow X \leq Y$ (or that $X \leq Y \Rightarrow \sqrt X \leq \sqrt Y$, but I feel like the first version would be easier to work with)?

Note: $X$ and $Y$ don't have to commute, so $X^2 - Y^2$ is not necessarily $(X+Y)(X-Y)$.

答:
Here is a proof which works more generally for $x,y\ge 0$ in a $C^*$-algebra, where the spectral radius satisfies $r(z)\le \|z\|=\sqrt{\|z^*z\|}$ for every element $z$, and $r(t)=\|t\|$ for every normal element $t$. The main difference with @user1551's argument is that we will use the invertibility of $y$. Other than that, the idea is essentially the same.

If $y$ is not invertible, replace $y$ by $y+\epsilon1$ in the following argument, and then let $\epsilon$ tend to $0$ at the very end.

Now assume that $y$ is invertible and note that
$$x^2\le y^2\quad\Rightarrow\quad y^{-1}x^2y^{-1}\le y^{-1}y^2y^{-1}=1\quad\Rightarrow\quad\left\|y^{-1}x^2y^{-1}\right\|\le 1.$$

Since $t=y^{-1/2}xy^{-1/2}$ and $z=xy^{-1}$ are similar, they have the same spectral radius and therefore
$$\left\|t\right\|=r(t)=r\left(z\right)\le\|z\|=\sqrt{\|z^*z\|}=\sqrt{\left\|y^{-1}x^2y^{-1}\right\|}\le 1.$$
It follows that $y^{-1/2}xy^{-1/2}\le 1$, whence $x\le y$ as desired.

Title: Square root is operator monotone


问:
Distribution of size of connected components in Erdos-Renyi random graphs in the subcritical case
In Alon and Spencer's The Probabilistic Method, in the subcritical case, we have theorems 11.4.2 and 11.6.1 stating that in $G(n,p=c/n)$ where $c<1$, for a constant $k$
$$\lim_{n\to\infty}\Pr[|C(v)|=k]=\Pr[T_c^{po}=k]=\frac{e^{-ck}\cdot (ck)^{k-1}}{k!}$$

I know that in the subcritical case the largest connected component is w.h.p. of size $O(\log n)$, so can we say something about $\Pr[|C(v)|=k]$ when $k=k(n)$ is super constant, e.g. $k=a\cdot \log n$ for some constant $a>0$?

答:
Here is just what we can extract from the Poisson limit.

As $k \to \infty$, $\Pr[T_c^{po} = k] \to 0$. So, for any $\epsilon>0$, there is $K = K(\epsilon)$ such that $\Pr[T_c^{po} > K] < \epsilon$ and therefore $$\lim_{n \to \infty} \Pr[|C(v)| > K] \le \epsilon.$$

Let $\mathbf X = |\{v : |C(v)| > K\}|$: the number of vertices in components larger than $K$. Then $\mathbb E[\mathbf X] \le \epsilon n$ by the above and by linearity of expectation, so
$$
   \Pr[\mathbf X > \sqrt{\epsilon} n] \le \frac{\epsilon n}{\sqrt{\epsilon} n} = \sqrt{\epsilon}
$$
which can be made arbitrarily small. This can be summarized as "with high probability, all but $o(n)$ of the vertices are in components of size $O(1)$" but that's a lot of implied limits in a very short sentence.

This analysis by itself leaves open a range of possibilities: there could be a single component of size $\Theta(\log n)$, or many, as long as there are $o(n)$ vertices in such components.

This is not really the most interesting statement about $|C(v)|$. Theorems 11.6.2 and 11.6.3 in the same section are more helpful, because they do not require $k$ to be constant.

In particular, the probability that a vertex is in a component of size $\ge K \log n$ is sandwiched between two branching process probabilities:
$$
   \Pr[T^{bin}_{n-K\log n,p} \ge K \log n] \le \Pr[|C(v)| \ge K \log n] \le \Pr[T^{bin}_{n-1,p} \ge K \log n].
$$
Alon and Spencer leave out any argument that $T^{bin}_{n,p}$ and $T^{po}_{np}$ (as they mention at the end of section 11.6) so I will also just assume this because I am even more lazy than they are. Then $(n-K\log n)p$ and $(n-1)p$ are $c + o(1)$, so
$$
    \Pr[|C(v)| \ge K \log n] = \Pr[T^{po}_{c+o(1)} \ge K \log n] = \sum_{k=K \log n}^\infty \frac{e^{-(c+o(1))k}((c+o(1))k)^{k-1}}{k!}.
$$
This gives you what you need to be very very careful, but now I'm going to be very sloppy and say that for large $k$, this is approximately a geometric series, which yields
$$
   \Pr[|C(v)| \ge K \log n] = n^{K(1 - c + \log c + o(1))}.
$$


问:
Conditions under which the Lattice Generated by a Subset of Lattice Vectors is Equivalent to the Original Lattice
In general, the sublattice $L_2$ generated by subset of vectors of $L_1$ need not have the same rank as $L_1$.  Even if it does, it may be a proper sublattice of $L_1$.  However, if the rank and determinant of $L_2$ matches those of $L_1$, then is it possible to conclude that $L_2 = L_1$?  If not, what are the minimal necessary conditions on the subset?  Clearly, if the subset includes a basis of $L_1$, it's sufficient, but I'm looking for conditions which are easier to test.

答: sigh $\;$ Answered my own question again.  If the rank and determinants are equal, then the index of $L_2$ in $L_1$ is 1, i.e. they're the same lattice, perhaps with a different choice of basis.  This page has more details.


问:
Smart method for computing the determinant of the matrix defined by $A_{ii}=a+b$ and $A_{ij}=a$?
We are given the following $n\times n$ matrix, whose terms are:

$a+b$ for the terms in the diagonal ($i=j$).

$a$ for the rest of the elements.

I know we can simply compute this and try to find a pattern, but is there a faster way? Maybe using some properties of the determinant of a matrix?

答:
The given matrix may be written as $bI_n+\mathbf a\mathbf1^T$ where $\mathbf x$ denotes a vector consisting of $n$ copies of $\mathbf x$. The matrix determinant lemma gives the required determinant as
$$(1+\mathbf 1^T(I_n/b)\mathbf a)\det bI_n=(1+na/b)b^n=b^n+nab^{n-1}$$


答:
You can use the matrix determinant lemma.

You have

$$C = b I + a {\bf u}  {\bf u}^t$$

where ${ \bf u}$ a all-ones column vector.

Then $$\det(C) = (1 + a {\bf u}^t({b}I)^{-1}  {\bf u}) \det(b I) $$

Can you go on from here?

Title: Smart method for computing the determinant of the matrix defined by $A_{ii}=a+b$ and $A_{ij}=a$?

Body:
We are given the following $n\times n$ matrix, whose terms are:

$a+b$ for the terms in the diagonal ($i=j$).

$a$ for the rest of the elements.

I know we can simply compute this and try to find a pattern, but is there a faster way? Maybe using some properties of the determinant of a matrix?


问:
Probability of event, How should I continue from here?
Let $R>1$ be real number. Let $\mathbb{P}$ be probability measure  on probability space $$\Omega=\mathbb{N}\smallsetminus\{0\}=\{1,2,3,...\}$$  such that $$\mathbb{P}[ \{ n \} ] = C\cdot n^{-R}$$ for every $$n\in\Omega$$ and $C>0$ is constant. Define the event $A_m$ as a given number that is a multiple of $m$ $$A_m = \{ k\cdot m \  : \ k\in\mathbb{N}\smallsetminus\{0\} \} = \{m,2m,3m,...\}$$

calculate $P[A_m]$

My attempt:

I argue that $A_m$ is union of disjoint sets thus $$P[A_m]= \Sigma^{k=\infty}_{k=1}C\cdot(km)^{-R}$$ $$C\cdot m^{-R}\Sigma^{k=\infty}_{k=1}\frac{1}{k^{R}}$$

I dont know how should I procedd from here.

答: Hint: You are very close. Note that $C \sum_{k=1}^\infty \frac{1}{k^R} = \sum_{k=1}^\infty P(\{k\})$


问:
The area of a triangle, if the vertices are complex numbers
I need to check for this task, if my logic is correct ?

The task is:

Complex numbers are given, which represent the points of a triangle.

$z_1=2+i$

$z_2=i$

$z_3=\lambda-2i$

$z_1$ and $z_2$ represent fixed points, while $z_3$ depends on $\lambda$.

Calculate the area of the triangle

I tried this:

I took that $\lambda$ for example $1$.

$P_{\triangle z_1z_2 z_3}=\frac{1}{2}\left|\begin{vmatrix} 2 & 1 & 1\\ 0 & 1 & 1\\ 1 & -2 & 1 \end{vmatrix}\right|=\frac{1}{2}\cdot (2+1+0-(1-4+0))=\frac{1}{2}\cdot 6=3$

By taking any other number for the $\lambda$, the same result is obtained.

Can it be done in any other way?

Thanks in advance !

Body:
kludginess says hello via https://en.wikipedia.org/wiki/Heron%27s_formula.

$$A = \left(\frac{1}{4}\right)
\sqrt{4a^2b^2 - (a^2 + b^2 - c^2)^2}
.$$

$z_1 = (2+i), ~z_2 = (i), ~z_3 = (\lambda - 2i).$

$a = |z_1 - z_2| = 2.$

$b = |z_1 - z_3| = |(2 - \lambda) - 3i|
= \sqrt{(\lambda - 2)^2 + 9}.$

$c = |z_2 - z_3| = |\lambda - 3i| = \sqrt{\lambda^2 + 9}.$

$$A = \left(\frac{1}{4}\right)
\sqrt{4(2)^2(\lambda^2 - 4\lambda + 13)
- \left[(4 + (\lambda^2 - 4\lambda + 13) - (\lambda^2 + 9)\right]^2}
$$

$$= \left(\frac{1}{4}\right)
\sqrt{4(2)^2(\lambda^2 - 4\lambda + 13)
- \left[(8 - 4\lambda)\right]^2}
$$

$$= \sqrt{(\lambda^2 - 4\lambda + 13)
- \left[(2 - \lambda)\right]^2}
$$

$$= \sqrt{(\lambda^2 - 4\lambda + 13)
- \left[4 - 4\lambda + \lambda^2\right]}
$$

$$= \sqrt{(13)
- [4]} = 3.
$$


问:
Showing that $(\langle T(b_{i}),b_{j}\rangle )_{1\leq i,j\leq n}=G(b_{1},...,b_{n}) [T^{*}]_{B}^{B}$
Let $V$ be a vector space with a finite Dimension above $\mathbb{C}$ or $\mathbb{R}$.

Let $B=(b_{1},...,b_{n})$ be a basis of V.

I need your help proving this:

$(\langle T(b_{i}),b_{j}\rangle )_{1\leq i,j\leq n}=G(b_{1},...,b_{n}) [T^{*}]_{B}^{B}$

when $G(b_{1},...,b_{n})$ is a Gram matrix, and  $T^{*}$ is the adjoint transformation of $T$.

Thank you

答:
Calculate $G(b1,..,bn)[T^*]_B^B\cdot e_i$ for any $1\leq i \leq n$, where $e_i=(0,...,0,1,0,...,0)$ ($i$'th place).

You get $[T^*(b_i)]_B=(\alpha_1,...,\alpha_n)$ (i.e. $T^*(b_i)=\alpha_1b_1+...+\alpha_nb_n$ by definition of $[T^*]_B^B$). Multiplying $G(b1,..,bn)(\alpha_1,...,\alpha_n)$ (column), in the $j$'th row we get $$\alpha_1\langle b_1 , b_j \rangle+...+\alpha_n\langle b_n , b_j\rangle= \langle \alpha_1b_1+...+\alpha_nb_n,b_j\rangle=\langle T^*(b_i),b_j\rangle$$
On the other hand, multiply $(\langle T(b_i),b_j \rangle)_{1\leq i,j\leq n} e_i$ and look at the $j$'th row. They should be equal, thus proving the equality.

(Either I messed with the multiplication or you should have $(\langle b_i,T(b_j) \rangle)_{1\leq i,j\leq n}$)


问:
Norm inequality for positive element of C* algebra
Let $\mathcal A$ be a unital C* algebra and $A\in \mathcal A$ a positive element $A\geq \mathbb 0$, i.e. A is self-adjoint and the spectrum satisfies $\sigma(A)\subset [0,\infty)$.

Let $\alpha \in [0,1]$ Is it true that

$$\vert\vert\, \alpha A - \vert\vert A \vert\vert \,\vert\vert \leq \vert\vert A \vert\vert $$

I just know the triangle inequality which is not sharp enough. What other means are there to look for such an inequality?

答:
Let $B:=\alpha A-\|A\| 1$. Then $A$ and $B$ are both self-adjoint.

The spectrum of $B$ is
$$\sigma(B)=\alpha\sigma(A)-\|A\|=\{\alpha \lambda-\|A\|\;;\; \lambda\in \sigma(A)\}$$ as a special case of the spectral mapping theorem: $\sigma(f(A))=f(\sigma(A))$, which is trivial in the case of a degree $1$ polynomial $f$ like here.

Since $\sigma(A)\subseteq[0,\|A\|]$ we get
$$
\sigma(B)\subseteq [-\|A\|,(\alpha-1)\|A\|]\subseteq [-\|A\|,0].
$$
Therefore the spectral radius is $\rho(B)\leq \|A\|$.

Now for a normal element, whence for a self-adjoint element, $\|B\|=\rho(B)$.

It follows that $\|B\|\leq \|A\|$.

Alternative: it is convenient to use the partial order on self-adjoint elements $C\leq D$ if $D-C$ is positive. Then
$$
0\leq \alpha A\leq A\leq \|A\|1\quad\Rightarrow \quad-\|A\|1\leq \alpha A-\|A\|1\leq 0.
$$
Finally, for $B$ self-adjoint, we have $\|B\|\leq m$ if and only if $-m\cdot 1\leq B\leq m\cdot 1$.


问:
Solving $\sin^2x=1-\cos x$

How would I solve the following trig equation?

$$\sin^2x=1-\cos x$$

I have to write the solution in radians.

答:
We rewrite

$\sin^2(x)=1-\cos(x)$

as

$1-\cos^2(x)=1-\cos(x),$

so

$-\cos^2(x)=-\cos(x)$

$\cos^2(x)=\cos(x)$

$\cos^2(x)-\cos(x)=0$

$\cos(x)(\cos(x)-1)=0.$

Now:

${\cos(x)=0}\quad\Rightarrow\quad{x=\dfrac{\pi}{2}+k\pi}\ ,\ {k\in\mathbb Z},$

${\cos(x)-1=0}\quad\Rightarrow\quad{\cos(x)=1}\quad\Rightarrow\quad{x=2k\pi}\ ,\ {k\in\mathbb Z}.$

Title: Solving $\sin^2x=1-\cos x$


问:
What is a typical basis for the finite complement topology $\cal{T}_f$? (James R. Munkres "Topology 2nd Edition")
I am reading "Topology 2nd Edition" by James R. Munkres.

On p.77, Example 3:

Let $X$ be a set; let $\cal{T}_f$ be the collection of all subsets $U$ of $X$ such that $X-U$ either is finite or is all of $X$. Then $\cal{T}_f$ is a topology on $X$, called the finite complement topology.

I wonder what a typical basis for $\cal{T}_f$ is.

I only know $\cal{T}_f$ is a basis for $\cal{T}_f$.

Please tell me a typical basis for $\cal{T}_f$.

答:
The only natural bases for the cofinite topology $\tau_f$ on an infinite set $X$ are $\tau_f$ and $\tau_f\setminus\{\varnothing\}$, the latter being more common. There are other bases, however. For instance, for any $n\in\Bbb Z^+$ the family

$$\mathscr{B}_n=\{X\setminus F:F\subseteq X\text{ and }n\le|F|<\omega\}$$

of complements of finite sets of cardinality at least $n$ is a base for $\tau_f$. To see this, let $U=X\setminus F\in\tau_f$ with $|F|<n$. Let $D_0$ and $D_1$ be disjoint subsets of $X\setminus F$ such that $|D_0|=|D_1|=n-|F|$, and let $F_i=F\cup D_i$ for $i=0,1$. Then

$$U=X\setminus F=X\setminus(F_0\cap F_1)=(X\setminus F_0)\cup(X\setminus F_1)\,,$$

where $X\setminus F_i\in\mathscr{B}_n$ for $i=0,1$.


问:
consider a uniformly continuous $f:(a,b) to R$, prove $\lim$ at b exists
I choose $x_n$ to be a cauchy sequence in (a,b). Then, for $n, m \geq N$, $|x_n - x_m| < \epsilon$. Since f is uniformly continuous, then $|x_n - x_m| < \epsilon$ gives f is cauchy. But I don't know how to go further.

答:
Choose $x_n$ to be a sequence converging to $b$.

$(x_n)$ is therefore a Cauchy sequence

Your work shows that $(f(x_n))$ is also a Cauchy sequence.

Therefore...


问:
Cardinality of set containing points on a circle such that  = i * 1 radian?
I'm on a conversion MSc and haven't done math in a while, so apologies if this is an easy Q but I'm stumped. I've literally only just learned what radians and sets are- I think I've done a decent job of getting the basics of each, but I'm struggling with applying them to this problem.

Set $A, B$ are fine, but I'm pretty stumped by set $C$. I suppose the relationship $1 / 2\pi$ isn't rational, but I'm lost as to how that helps me figure out the cardinality of the set. I'm guessing that as the relationship between how many points can be drawn on the circumference isn't rational that giving the cardinality isn't as simple as giving a number?

I've got as far as sort of testing a few ideas:

there aren't an integer number of radians in the full circle, so how can I have a finite series of points to give the cardinality?

I don't really want the answer as I'm trying to figure it out for myself, but any hints constructive to my learning would be appreciated!

答:
Putting my comment as a hint in an answer.

There are $2\pi$ radians in a full rotation. Suppose that you have two points $i_1w,i_2w$ with $i_1\ne i_2$. These two points are coincident iff $i_1w−i_2w$ is a multiple of $2\pi$ (angle traversed in one rotation). Is it possible?


答:
If $n$ is a square then $n(n+1)$ is a sum of two squares, $n(n+1)=n^2+n$. If $n$ is a square of the form $n=(k+1)^2$ with $2k+2$ a square then $n(n+1)$ can be written as a sum of two squares in another way: $$n(n+1)=(k+1)^2(k^2+2k+2)=(k+1)^2k^2+(k+1)^2(2k+2).$$
Of course there are infinitely many $k\in\mathbb N$ such that $2k+2$ is a square.</p>

Body:
Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways.($a^2+b^2$, is same $b^2+a^2$), $n \in \mathbb{N}.$

I have proved the above question which appeared in one of the Math-Olympiad. And I do know the solution. Sharing the question only because the question has a cute solution.


问:
Finding a comparison series for $\frac{\sqrt[3]{k^2+1}}{\sqrt{k^3+2}}$
I've tried a number of other convergence tests to no avail and I have a hunch that either an ordinary/limit comparison test is probably where I need to be looking. I'm not looking for a direct answer here, rather some guidance in how to choose comparison series in situations like this.

$$\frac{\sqrt[3]{k^2+1}}{\sqrt{k^3+2}}$$

My initial thought was to look at $$\frac{k^\frac{2}{3}}{k^\frac{3}{2}}$$ as it follows the leading terms of $a_n$ but it doesn't match the conditions for the ordinary comparison test. I'm at a bit of a loss for the process of finding a $b_n$ as we didn't do much of this sort of work in class.

Thanks!

答:
If you don't know equivalents or are afraid to use them, remember that you can often use rough inequalities instead.

$\sqrt[3]{k^2+1}\ge\sqrt[3]{k^2}=k^{2/3}$

$\sqrt{k^3+2}\le\sqrt{k^3+3k^3}=2k^{3/2}$ since obviously $2<3k^3$ for $k\ge 1$.

You get $a_k\ge \dfrac{k^{2/3}}{2k^{3/2}}\ge \dfrac 1{2k^{5/6}}$ which is a term of a divergent series.


问:
If the Symmedian/Lemoine point of triangle $ABC$ lies on the Altitude from vertex $C$, show that either $BC = AC$ or angle $C = 90 ^\circ$.
If the Symmedian/Lemoine point of triangle $ABC$ lies on the Altitude from vertex $C$, show that either $BC = AC$ or angle $C = 90^\circ$.

I have proved it going the other way. When given that $C$ is $90^\circ$, I can show that the Lemoine point lies on the altitude, but I cannot figure out how to go backward or incorporate the $BC = AC$ part.

I have tried using the median from vertex $C$, which I know is an isogonal conjugate of the symmedian from $C$. But then from there I don't know where to go.

Thanks for your help.

答:
Let $CM$, $CK$ and $CH$   be a median, a bisector and an altitude of $\Delta ABC$ respectively.

Since $CH$ is a symmedian, we obtain: $$\measuredangle KCM=\measuredangle KCH,$$ which gives:

$$\measuredangle ACM=\frac{1}{2}\measuredangle ACB-\measuredangle KCM=\frac{1}{2}\measuredangle ACB-\measuredangle KCH=\measuredangle BCH=90^{\circ}-\measuredangle CBA,$$ which says $$\measuredangle ACM+\measuredangle CBA=90^{\circ}.$$

Now, let $\Phi$ be a circumcircle of $\Delta ABC$ and $CM\cap\Phi=\{C,D\}$.

Thus, $$\measuredangle CBD=\measuredangle CBA+\measuredangle ABD=\measuredangle CBA+\measuredangle ACD=90^{\circ},$$ which says that $CD$ is a diameter of the circle.

Now, if also $AB$ is a diameter of $\Phi$, so $\measuredangle ACB=90^{\circ},$ otherwise, $CD\perp AB$, which says $CB=CA$ and we are done!


问:
How many games will you play and how many of those games will you lose?
Each time you play a game, you win with probability p. (The games are independent.) You plan
to play 5 games, but if you win the fifth game, then you will keep on playing until you lose.
(a) Find the expected number of games that you play.
(b) Find the expected number of games that you lose.

I'm not sure how to account for games that you play past the 5 games. I know that if you have a p% chance of winning the game you would only have to check on the 5th game and onward whether you win or not. But how would I go about doing so?

答:
(1) If you win the fifth game, you play on average $\frac1{1-p}$ extra games (the geometric distribution here has "success" probability $1-p$ because in this phase a failure, where we keep going, is a win). Therefore the expected number of games played is $5+\frac p{1-p}$.

(2) In the first four games, you are expected to lose $4(1-p)$ games on average. If you lose the fifth game, with probability $1-p$, that's it; if you win the fifth game there is exactly one loss ahead of you. Thus the number of games lost after game $4$ is constant $1$, and the expected number of games lost is $4(1-p)+1$.


问:
The Funtion Complex - Product Adjunction of simplicial sets and diagram chasing
I run into the following claim in the book Simplicial Homotopy Theory (in the proof of Proposition 5.2.). Given $i:K \rightarrow L$ inclusion of simplicial sets and $p:X\rightarrow Y$ fibration.

Someone explained to me that I can use the exponential law to get the identification, but I don't get his idea, for it involves careful interchanging lim/colim etc. Is there any written down proof for this fact? Moreover, I wish to know that if there is general way to understand the interplay between adjunction and commutative diagram so that I might guess from the first diagram what kind of equivalent diagram should I look for.

答:
I'm just going to write down a proof here, since there seem to be very few written proofs of this, since most sources seem to leave it as an exercise. Fosco of course has linked a paper on the arXiv with a proof, but I think it's best that we have an answer on MSE as well.

Let $\newcommand\calC{\mathcal{C}}\calC$ be a category, let's suppose we have a closed symmetric monoidal structure on $\calC$, with $\otimes$ the tensor, and the internal hom denoted by $[-,-]$.

Now we make an observation. In any category, commuting squares between morphisms $f:A\to B$ and $g:C\to D$ are given by the following pullback:
$$
\require{AMScd}
\begin{CD}
\calC(A,C)\underset{\calC(A,D)}{\times} \calC(B,D) @>>>\calC(B,D)
\\
@VVV @VVf^*V
\\
\calC(A,C) @>g_*>> \calC(A,D).
\end{CD}
$$
It's worth carefully thinking about what this is saying first, it's saying that a commuting square is a morphism $a:A\to C$ and a morphism $b:B\to D$ such that $bf = ga$.

Now we can apply the same idea to your situation. We have three maps, $i:A\to B$, $j:K\to L$, and $p:X\to Y$ ($i:A\to B$ here is taking the place of your horn inclusion, and I've renamed your $i$ to $j$).
I claim both sets of squares in your question can be identified with the set of triples of maps
$$
\begin{align*}
\{
(a,k,y) \in \calC(A\otimes L, X)\times &\calC(B\otimes K, X) \times \calC(B\otimes L, Y) \\
:\quad
& pa = y(i\otimes 1_K), \\
& a(1_A\otimes j) = k(i\otimes 1_K),\\
& pk = y(1_A\otimes j)
\}.
\end{align*}
$$
Note that we can equivalently take the maps to be $a':A\to [L,X]$, $k':B\to [K,X]$, and $y':B\to [L,Y]$ by the adjunction isomorphisms, and the equalities become
$[1_L,p]a' = y'i$, $[j,1_X]a' = k'i$, and $[1_K,p]k' = [j,1_Y]y'$.

To see that both sets of squares are of this form, we could expand out the data of a square like the first one you have:
$$
\begin{CD}
A @>>> [L,X]
\\
@ViVV @VV\widehat{[j,p]}V
\\
B @>>> [K,X]\underset{[K,Y]}{\times}[L,Y].
\end{CD}
$$

The bottom map is a pair of maps $(k' : B\to [K,X],y':B\to [L,Y])$ such that
$[1_K,p]k'=[j,1_Y]y'$. The top map is of course $a' : A\to [L,X]$, and the requirement that the square commute gives the equalities
$[j,1_X]a' = k'i$ and $[1_L,p]a'=y'i$.

Conversely, triples of maps satisfying the equalities give such a square.

We can do exactly the same thing for the bottom square.

Notes

This generalizes to two variable adjunctions more generally.

Moreover it's worth noting that the collection of such squares in both cases is actually given by the cubical analogue of a pullback, the limit over $(0\to 1)^3$ minus its initial vertex (I'll call diagrams of this shape punctured cubes.).  Then the adjunction isomorphisms give isomorphisms between the punctured cubes, which induces isomorphisms on the limits, which are the sets of commuting squares. Finally, a morphism in the arrow category say $i'\to i$, $j'\to j$, or $p\to p'$ induces translations of the punctured cubes which make the commuting diagrams contravariantly functorial in $i$ and $j$ and covariantly functorial in $p$.

Moreover, since the adjunction isomorphisms are natural, this means that we'll end up with a 2-variable adjunction on the arrow category.


问:
Finding all integer solutions of $\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = \frac12$
If we have integers $x,y,z$ such that $x,y,z \ge 3,$ find all solutions to $$\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = \frac12.$$

I was thinking of first expanding out this, and then simplifying from there, but the equation got very messy. Is there a better method?

答:
As an alternative to Joshua Wang's excellent answer:

By symmetry we may assume, without loss of generality, that $x\leq y$. Clearing denominators yields
$$2xz+2yz-2xy=xyz,\tag{1}$$
and a bit of rearranging then shows that
$$xy(z+2)=2(x+y)z.$$
Of course $z+2>z$ and so $xy<2(x+y)$, or equivalently
$$(x-2)(y-2)<4.$$
Because $x\leq y$ we see that $x<4$, so $x=3$ and then $y<6$. Plugging into $(1)$ and rearranging shows that
$$yz+6y-6z=0,$$
or equivalently $z=\frac{6y}{6-y}$, yielding the following three solutions for $(x,y,z)$:
$$(3,3,6),\qquad(3,4,12),\qquad(3,5,30).$$


问:
Show $x \leq y$ iff $y^c \leq x^c$ in a lattice which is a boolean algebra

Show $x \leq y$ iff $y^c \leq x^c$ where $x,y$ are elements of a lattice $(B,\leq,\lor, \land)$. $(B,\leq)$ is a partially ordered set, and $\lor$ and $\land$ refer to the l.u.b and g.l.b operations respectively.

I started off with the definition of the complement $^c$, which is:

$x \land x^c = 0$

$x \lor x^c = 1$

Then, assuming $x \leq y$, I am trying to show $y^c \leq x^c$ - or equivalently, $x^c \leq y^c$ will result in a contradiction. However, I'm unable to proceed. Could someone please help?

答:
What I meant in my comment (before you added that the lattice is a Boolean algebra) is that, for example, with the lattice in the following picture below (where $x'$ is what you denote as $x^c$)

we have $a < b$ but $b^c = d \nleq c = a^c$.

Likewise, $c^c < d^c$ but $d \nleq c$.

So neither of the implications in the equivalence apply.

Of course this is not a Boolean algebra (neither it is a distributive or even modular lattice), and also the de Morgan laws don't apply, since for example,
$$(a \wedge b)^c = a^c = c \neq 1 = c \vee d = a^c \vee b^c.$$


答:
You play exactly $n\ge5$ games iff the $n^\text{th}$ game is a loss and games $5$ through $n-1$ are wins. Games $1$ through $4$ may be wins or losses. The probability associated with playing $n$ games is thus $p^{n-5}(1-p),n\ge5$. You should be able to solve for the expectation.

Suppose you lose exactly $m\ge1$ times in $n$ games. Then games $5$ through $n-1$ are surely wins and the last game is surely a loss. Then games $1$ through $4$ must contain $m-1$ losses i.e. $m-1\le4\implies m\le5$. The probability associated with $m$ losses is $\binom 4{m-1}(1-p)^{m-1}p^{5-m}$. Can you find the expectation?

Title: How many games will you play and how many of those games will you lose?

Body:
Each time you play a game, you win with probability p. (The games are independent.) You plan
to play 5 games, but if you win the fifth game, then you will keep on playing until you lose.
(a) Find the expected number of games that you play.
(b) Find the expected number of games that you lose.

I'm not sure how to account for games that you play past the 5 games. I know that if you have a p% chance of winning the game you would only have to check on the 5th game and onward whether you win or not. But how would I go about doing so?