问:
What Are The Necessary Conditions For a Function to be Separable?
A common step in solving many PDEs is to write a multivariable solution function as a product of two more or single variable  functions.  For example, if given,

$$
\alpha\,\frac{\partial U(x,t)}{\partial x} = \frac{\partial U(x,t)}{\partial t},
$$

one may begin be writing,

$$
U(x,t) = X(x) T(t).
$$

My question is, what are the necessary conditions for writing a multi variable function as a product of  several single variable functions?  Furthermore, are there any cases where this cannot be done?

答:
A sufficient and necessary condition is the following.

A function $F:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}:(x,y)\mapsto F(x,y)$ can be written as $F(x,y)=f(x)g(y)$ for some functions $f,g:\mathbb{R}\rightarrow\mathbb{R}$ if and only if for every $x,y,w,z\in\mathbb{R}$ it holds that
$$F(x,y)F(z,w)=F(x,w)F(z,y).$$

The implication "$\implies$" follows directly from the commutativity of multiplication and the other implication is obtained by first noticing that the zero function $F=0$ can obviously be written as a product of functions of a single variable, and if there exists some $(a,b)\in\mathbb{R}^2$ such that $F(a,b)\neq 0$, then we get that $\forall x,y\in\mathbb{R}$ it holds
$$F(x,y)F(a,b)=F(x,b)F(a,y),$$ which implies that

$$F(x,y)=F(x,b)\frac{F(a,y)}{F(a,b)}=f(x)g(y).$$


问:
Linear algebra practice quiz question considering subsets and dimensions of the intersections of subsets
Let $\Bbb R^{(2,2)}$ be the vector space of real matrices with two rows and two columns.

Let us consider the subsets of $\Bbb R^{(2,2)}$:

$$V=\left\{\begin{pmatrix}0&a\\b&0\end{pmatrix}:a,b\in\Bbb R\right\}$$

and

$$W=\left\{\begin{pmatrix}0&k\\h&d\end{pmatrix}:k,h,d\in\Bbb R\right\}\,.$$

Which of the following is correct?

(a) $\dim(V \cup W)=1$.

(b) $\dim(V \cap W)=0$.

(c) $V$ is not a vector subspace of $\Bbb R^{(2,2)}$.

(d) $V \subset W$.

The correct answer is (d).

I apologise first of all for not knowing how to write the problem using mathjax I tried and I couldn't do it!
I found this problem in some of my Linear Algebra practice tests that we have in our university website! I tried for some time and I couldn't figure it out how to solve this! Can you help me?

答:
Hint.

Observe that $V\cup W=W$ and $V\cap W=V$. So in order to see whether (a) and (b) are true, find $\dim(V)$ and $\dim(W)$.

For (c), go over the concept of "subspaces", which tells you that $V$ is a subspace of $\mathbb{R}^{(2,2)}$.

For (d), $V\subset W$ means every matrix $A\in V$ is an element of $W$.

ContentLicense: CC BY-SA 4.0


问:
Cauchy filters in metric spaces
Some terminology: Let $(X,d)$ be a metric space.

A filter $\mathcal F \subseteq \mathcal P (X)$ is Cauchy if $\forall \epsilon >0 \exists x\in X: B_\epsilon(x)\in \mathcal F$.

A filter $\mathcal F \subseteq \mathcal P (X)$ is round if $\forall F\in \mathcal F \exists \epsilon>0 \forall x\in X:B_\epsilon(x)\in \mathcal F \implies B_\epsilon(x)\subseteq F$.

A Cauchy filter $\mathcal F$ is said to be minimal if no filter that is properly contained in it is Cauchy.

I found an article that remarks, as a matter of fact, that round Cauchy filters and minimal Cauchy filters coincide. Showing that a round Cauchy filter must be minimal is straightforward. But I can't immediately see how to prove the other implication. So, how does one show that a minimal Cauchy fiter must be round? Also, any references for the standard constructions of metric theory convergence by using filters (I know it's not strictly required, as sequences suffice, but still, I'm curious) are welcome. Thanks.

答:
I suppose it would be easy to solve this by referring to the sequential completion, but that might be considered cheating.

I will use $N_\epsilon(A)$ to denote the $\epsilon$-neighbourhood of a set $A$.
For a given filter $\mathcal{F}$ the family
$\{ N_\epsilon(A) \mid \epsilon > 0,\, A \in \mathcal{F} \}$ is the base of a
filter $\mathcal{F}^\circ \subset \mathcal{F}$. Clearly if $\mathcal{F}$ is
a Cauchy filter, then so is $\mathcal{F}^\circ$.

I claim that $\mathcal{F}^\circ$ is also round.

Proof:
Take an arbitrary $V \in \mathcal{F}^\circ$. By construction, there are $\epsilon > 0$ and $A \in \mathcal{F}$ such that $N_\epsilon(A) \subset V$.
Put $\delta = \epsilon / 3$. Now if there is some $x$ such that
$B_\delta(x) \in \mathcal{F}^\circ$, then $B_\delta(x)$ meets $N_\delta(A)$, which means that $x \in N_{2\delta}(A)$, and therefore $B_\delta(x) \subset N_\epsilon(A) \subset V$.

To conclude: for every Cauchy filter there is a smaller, round Cauchy filter,
therefore any minimal Cauchy filter is round.

P.S.: I have not really investigated it, but it seems this might apply more generally to uniform spaces.

ContentLicense: CC BY-SA 3.0

Title: Cauchy filters in metric spaces


问:
Computing the Shapley Value for this two-player game
It is well known that the Shapley Value for any player $i\in N$ in any game $v:2^N\to\mathbb{R}$ is defined by the following formula:
\begin{gather}
S_i(\nu)=\sum_{S\subseteq N\backslash\{i\}}\frac{|S|!\times[|N|-|S|-1]!}{|N|!}\times[\nu(S\cup\{i\})-\nu(S)]
\end{gather}

Consider the following two-player game: $\nu(\{i,j\})=1,\nu(\{i\})=0,\nu(\{j\})=-1$. It is straightforward to see that the $S_i(\nu)=1$ and $S_j(\nu)=0$. However, applying the formula above for player $i$ yields:
\begin{align}
S_i(\nu)&=\frac{1!\times0!}{2!}\times[1-(-1)]\\
&=\frac{1}{2}\times2\\
&=1
\end{align}

And for player $j$, applying the formula yields:
\begin{align}
S_j(\nu)&=\frac{1!\times0!}{2!}\times[1-0]\\
&=\frac{1}{2}\times1\\
&=\frac{1}{2}
\end{align}

It is obvious to me that I'm not understanding the formula above and I am thus not applying it correctly. Can anybody please spot and explain my mistake?

Thank you all!

答:
It took me more time than expected, but I finally figured it out myself. Instead of deleting the question, I post here the answer because it may help others in the future.

The error in my reasoning is the omission of the emptyset. It is generally assumed that $\nu(\{\emptyset\})=0$. Therefore, the correct way to apply the formula above is as follows. For player $i$,
\begin{align}
\mathcal{S}_i(\nu)&=\frac{1!\times0!}{2!}\times[1-(-1)]+\frac{1!\times0!}{2!}\times[0-0]\\
&=\frac{1}{2}\times2+\frac{1}{2}\times0\\
&=1
\end{align}

Applying the formula for player $j$,
\begin{align}
\mathcal{S}_j(\nu)&=\frac{1!\times0!}{2!}\times[1-0]+\frac{1!\times0!}{2!}\times[0-1]\\
&=\frac{1}{2}\times1+\frac{1}{2}\times(-1)\\
&=0
\end{align}

Hence, $\mathcal{S}_i(\nu)=1$ and $\mathcal{S}_j(\nu)=0$. And we're done!


问:
Semiprime ideal of non-Noetherian Lie algebra
As in this paper On Prime Ideals of Lie Algebras

A Lie algebra $L$ satisfy the maximal condition for ideals, if for each , ascending chain $H_{1} \subseteq H_{2} \subseteq \ldots $ an index $m$ exists such that $H_{i}=H_{k}$ if $m<i$, $m<k.$ We say in short: $ L\in{\rm Max}-\triangleleft$.

An ideal $Q$ of $L$ is called semi-prime if $H^2 \subseteq Q$ with $H$ an ideal of $L$ implies $H\subseteq Q$.

$r(H)$ denote the intersection of all the prime ideals of $L$ containing $H$.

Corollary 8 from the paper State that:
If $ L\in{\rm Max}-\triangleleft$, then $r(Q)=Q$ if and only if $Q$ is semiprime ideal.

My question:-
If $ L \not \in{\rm Max}-\triangleleft$,
Could someone give an example of a semiprime ideal $Q$ such that $r(Q) \neq Q$, please?

答:
I think that Example 5.1 may be useful in this matter, but I hope that you help me to find semi-prime ideal $Q$ but $r(Q)\not=Q$.
enter link description here


问:
Confusion with flat functors
Let $F: C \to E$ be a functor, where $C$ is small and $E$ is complete and cocomplete. Recall that $F$ is called representably flat if one (hence any) of the following equivalent conditions hold:

Fact: The following are equivalent:

For every $e \in E$, the category $(e \downarrow F)^{op}$ is filtered.

For every $e \in E$, the functor $Hom(e,F): C \to Set$ is flat (in the sense that $Elts(Hom(e,F): C \to Set)^{op}$ is filtered.

For every $e \in E$, the left Kan extension of $Hom(e,F): C \to Set$ along the Yoneda embedding is a functor $Psh(C) \to Set$ which preserves finite limits.

The left Kan extension of $F: C \to E$ along the Yoneda embedding is a functor $Psh(C) \to E$ which preserves finite limits.

I am in the following quandary: I believe the proofs I've read showing that the above conditions are equivalent. And yet it seems there is a counterexample.

"Counterexample?"

Take $C = \{0,1\}$ to be the discrete category with two objects, $E = Psh(\{0,1\})$ to be the corresponding presheaf category, and take $F: \{0,1\} \to Psh(\{0,1\})$ to be the Yoneda embedding. Then

The left Kan extension along Yoneda of $F$ is the identity functor $Psh(\{0,1\}) \to Psh(\{0,1\})$, which preserves finite limits, so that (4) is satisfied.

But

In (1)/(2), take $e = \emptyset$ to be the initial object (i.e the empty presheaf). The functor $Hom(\emptyset,F): \{0,1\} \to Set$ is the constant functor at 1. (The opposite of) its category of elements is equivalent to $\{0,1\}$ again, which is not filtered, so (1)/(2) is not satisfied.

So unfortunately, this is one of those questions where my question is:

Question: Where am I screwing up here?

答:
You have shown that (4) is not equivalent to (1)-(2)-(3). Indeed, in the nLab article you cite, the closest analogue of (4) is instead about the left Kan extension functor $\textbf{Psh} (\mathcal{C}) \to \textbf{Psh} (\mathcal{E})$, and $\mathcal{E}$ is required to be small, not complete.

For the purposes of understanding geometric morphisms $\mathcal{D} \to \mathcal{C}$, where $\mathcal{C}$ and $\mathcal{D}$ are sites, the correct notion of "flat" is "covering-flat", and a functor $\mathcal{C} \to \mathcal{D}$ is covering-flat if and only if the left Kan extension $\textbf{Psh} (\mathcal{C}) \to \textbf{Sh} (\mathcal{D})$ preserves finite limits.
(To get a functor $\textbf{Sh} (\mathcal{C}) \to \textbf{Sh} (\mathcal{D})$, we require $\mathcal{C} \to \mathcal{D}$ to be cover-preserving.)

Title: Confusion with flat functors


问:
Prove that the minimum of a functional doesn't exist
Prove that there is no smooth solution ho the minimization problem:

$$\mathcal{L} (u)= \int_{0}^1 e^{-u'}+u^2 dx$$

Where the admissible space is $X =\{ u \in \mathcal{c}^2 [0,1] | u(0)=0,  u(1)=1
       \} $

UPDATE:

GOAL: I am trying to define a sequence of functions $u_n(x)$ in $X$ s.t their integrals $\mathcal{L}(u_n) \to 0$

What I have done so far:

Define $u_n(x)$ in the following manner:
$$u_n (x)= \begin{cases}
0, \quad 0 \leq x\leq 1-\frac{1}{n} \\[2ex]
ax^2 +bx+ c, \quad 1-\frac{1}{n}<x \leq 1
\end{cases}$$

where $n=1,2,3,..$. Require that $u_n$ satisfy:
\begin{align}
 u_n(1-\frac{1}{n})=0\\
 u_n(1)=1\\
 u_n'(1-\frac{1}{n})=0\\
\end{align}

We then have:
\begin{align}
a+b+c=1\\
a(1-\frac{1}{n})^2+b(1-\frac{1}{n})+c=0\\
2a(1-\frac{1}{n})+b=0
\end{align}

Find $a,b,c$ then:

$$u_n (x)= \begin{cases}
0, \quad 0 \leq x\leq 1-\frac{1}{n} \\[2ex]
n^2x^2 -2n(n-1)x+ (n-1)^2, \quad 1-\frac{1}{n}<x \leq 1
\end{cases}$$

\begin{align}
\int_0^1 e^{-u_n(x)'} + u_n^2(x) dx &= \int_0^{1-\frac{1}{n}} e^{0} + 0 dx +  \int_{1-\frac{1}{n}}^1 e^{-(2n^2 x -2n(n-1))} + (n^2x^2 -2n(n-1)x+ (n-1)^2)^2 dx\\& = 1- \dfrac{1}{n}+ \dfrac{\mathrm{e}^{-2n}\left(\left(2n+5\right)\mathrm{e}^{2n}-5\right)}{10n^2}
\end{align}

$$\dfrac{\mathrm{e}^{-2n}\left(\left(2n+5\right)\mathrm{e}^{2n}-5\right)}{10n^2} \to 0$$ and $$1-\dfrac{1}{n} \to 1$$

答:
Observe that you cannot have $\mathcal{L}(u)=0$, since that would imply $u=0$ a.e., which contradicts the fact that $u(1)=1$ (+continuity). Therefore, $\mathcal{L}(u)>0$ for all $u\in X$.
As someone already mentioned, you can consider the sequence $u_n(x)=x^n$, which satisfies $\mathcal{L}(u_n)\to 0$ (by the dominated convergence theorem).
This implies that $\mathcal{L}$ does not attain its minimum on $X$.


问:
Compute $\lim_{n \rightarrow \infty} \left(1 + \frac{i}{n^2 + in}\right)^n$
I have a feeling that

$$\lim_{n \rightarrow \infty} \left( 1 + \frac{i}{n^2 + in}\right)^n = 1,$$

but I don't know how to justify it. If $i$ was a real number $r > 0$, I know how to compute this: take logs and then solve
$$\lim_{n \rightarrow \infty} n \ln \left( 1 + \frac{r}{n^2 + rn}\right)$$
which after applying L'Hopital's rule gives $0$, and hence an original limit of $e^0 = 1$.
But since complex numbers are involved I'm don't even know if I'm allowed to apply the same logic.

What theorems / results can be applied to efficiently solve the limit, now that complex numbers are involved?

答: It works almost like the case when $i$ is replaced by a real number. You can check that the real part of $1+\frac i {n^{2}+in}$ is positive. Hence, its principal logarithm $Log (1+\frac i {n^{2}+in})$ is well defined. Also, $Log (1+z)\sim z$ as $z \to 0$. Hence, $nLog (1+\frac i {n^{2}+in}) \to 0$ and the given limit is $e^{0}=1$.


问:
Non-linear first order differential equation
I've found this particular equation rather tough, can you give me some hints on how to solve
$$\dot{y}+t\cos\frac{\pi y}{2}+1-t=y$$
Thanks a lot.</p>

答:
$\dot{y}+t\cos\dfrac{\pi y}{2}+1-t=y$

$\dfrac{dy}{dt}=\left(1-\cos\dfrac{\pi y}{2}\right)t+y-1$

Let $u=y-1$ ,

Then $y=u+1$

$\dfrac{dy}{dt}=\dfrac{du}{dt}$

$\therefore\dfrac{du}{dt}=\biggl(1-\cos\dfrac{\pi(u+1)}{2}\biggr)t+u$

$\left(\left(1-\cos\left(\dfrac{\pi u}{2}+\dfrac{\pi}{2}\right)\right)t+u\right)\dfrac{dt}{du}=1$

$\left(\left(1+\sin\dfrac{\pi u}{2}\right)t+u\right)\dfrac{dt}{du}=1$

Let $v=t+\dfrac{u}{1+\sin\dfrac{\pi u}{2}}$ ,

Then $t=v-\dfrac{u}{1+\sin\dfrac{\pi u}{2}}$

$\dfrac{dt}{du}=\dfrac{dv}{du}+\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{\left(1+\sin\dfrac{\pi u}{2}\right)^2}-\dfrac{1}{1+\sin\dfrac{\pi u}{2}}$

$\therefore\left(1+\sin\dfrac{\pi u}{2}\right)v\left(\dfrac{dv}{du}+\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{\left(1+\sin\dfrac{\pi u}{2}\right)^2}-\dfrac{1}{1+\sin\dfrac{\pi u}{2}}\right)=1$

$\left(1+\sin\dfrac{\pi u}{2}\right)v\dfrac{dv}{du}+\left(\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{1+\sin\dfrac{\pi u}{2}}-1\right)v=1$

$\left(1+\sin\dfrac{\pi u}{2}\right)v\dfrac{dv}{du}=\left(1-\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{1+\sin\dfrac{\pi u}{2}}\right)v+1$

$v\dfrac{dv}{du}=\left(\dfrac{1}{1+\sin\dfrac{\pi u}{2}}-\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{\left(1+\sin\dfrac{\pi u}{2}\right)^2}\right)v+\dfrac{1}{1+\sin\dfrac{\pi u}{2}}$

This belongs to an Abel equation of the second kind.

In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $v=\dfrac{1}{w}$ ,

Then $\dfrac{dv}{du}=-\dfrac{1}{w^2}\dfrac{dw}{du}$

$\therefore-\dfrac{1}{w^3}\dfrac{dw}{du}=\left(\dfrac{1}{1+\sin\dfrac{\pi u}{2}}-\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{\left(1+\sin\dfrac{\pi u}{2}\right)^2}\right)\dfrac{1}{w}+\dfrac{1}{1+\sin\dfrac{\pi u}{2}}$

$\dfrac{dw}{du}=-\dfrac{w^3}{1+\sin\dfrac{\pi u}{2}}+\left(\dfrac{\dfrac{\pi u}{2}\cos\dfrac{\pi u}{2}}{\left(1+\sin\dfrac{\pi u}{2}\right)^2}-\dfrac{1}{1+\sin\dfrac{\pi u}{2}}\right)w^2$

Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2


答:
More generally, if $\lim_{n\to \infty} ng(n)=0$ then $$\lim_{n\to\infty} (1+g(n))^n = 1.$$

This follows from a simple application of the binomial theorem.

Proof: When $|ng(n)|<1$ you have:

$$\begin{align}\left|(1+g(n))^n - 1\right| &\leq \sum_{k=1}^n \binom{n}{k}\left|g(n)\right|^k\\
&\leq \sum_{k=1}^n  n^k|g(n)|^k \\&= \sum_{k=1}^\infty (n|g(n)|)^k\\&=\frac{n|g(n)|}{1-|ng(n)|}
\end{align}$$

So $(1+g(n))^n\to 1$ since $\frac{ng(n)}{1-ng(n)}\to 0$.

See this answer for more about applying this result.

In your case, $g(n)=\frac{i}{n^2+in}$ then $ng(n)=\frac{i}{n+i}\to 0.$

Title: Compute $\lim_{n \rightarrow \infty} \left(1 + \frac{i}{n^2 + in}\right)^n$

Body:
I have a feeling that

$$\lim_{n \rightarrow \infty} \left( 1 + \frac{i}{n^2 + in}\right)^n = 1,$$

but I don't know how to justify it. If $i$ was a real number $r > 0$, I know how to compute this: take logs and then solve
$$\lim_{n \rightarrow \infty} n \ln \left( 1 + \frac{r}{n^2 + rn}\right)$$
which after applying L'Hopital's rule gives $0$, and hence an original limit of $e^0 = 1$.
But since complex numbers are involved I'm don't even know if I'm allowed to apply the same logic.

What theorems / results can be applied to efficiently solve the limit, now that complex numbers are involved?


问:
Nested sums - reciprocal of a sum : Find exact value
$$\text{Find:} ~~~~~~ \sum_{k=1}^{\infty} \frac{1}{ \left (
 \sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}}  \right )^2}$$

(Beware of the bounds of $j$, it does not always start from $1$, but starts at $k$ and goes to $k^2$)

At first I thought it has something to do with Riemann's zeta function $\zeta$ and thus the solution is:

$$ \zeta \left( \sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}} \right ) $$

However this totally seems incorrect because we cannot find this sum as we have an unknown, $k$.

Which also is being summed to $\infty$.

I thought (however was unsure) of the fact that:

$$ \left ( \sum f \right )^2 \ge  \sum f^2  $$

Thus (maybe) it is trivial the sum is converging to some value, which according to a little program I wrote is about: $$ \sum_{k=1}^{\infty} \frac{1}{ \left (
 \sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}}  \right )^2} \approx 1.596$$

But this is not a rigorous proof, is there a way to solve this so we get a solution by hand - calculating the exact value of this sum if it is finite (if not, why? )

Any mathematical tools (not programs) are acceptable, because this is not a question from a specific course ( I don't have context for this).

答:
(This is not a solution.)

The question of convergence is raised in the comments, so I show a proof here.

From $k > 0$ and
$$  k \leq j \leq k^2 $$
we have
$$ \frac{1}{\sqrt{k}} \geq \frac{1}{\sqrt{j}} \geq \frac{1}{k}  \text{,}  $$
using the monotonicity of the square root and standard results about reciprocals and inequalities.  So $1/\sqrt{j}$ is lower bounded by $1/k$.  Then
\begin{align*}
\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}  &\geq  \sum_{j=k}^{k^2} \frac{1}{k} = (k^2 - k + 1)\frac{1}{k}  \text{,}  \\
    \left(\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}\right)^2  &\geq \frac{(k^2 - k + 1)^2}{k^2}  \text{,}  \\
    \frac{1}{\left(\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}\right)^2}  &\leq \frac{k^2}{(k^2 - k + 1)^2}  \text{, and}  \\
    \sum_{k=1}^\infty \frac{1}{\left(\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}\right)^2}  &\leq \sum_{k=1}^\infty\frac{k^2}{(k^2 - k + 1)^2}  \text{.}
\end{align*}
(This last sum can be expressed in terms of polygamma functions, but let's not.)

Factoring out "the big part", we have
$$  \frac{k^2}{(k^2 - k + 1)^2} = \frac{k^2}{k^4} \cdot \frac{1}{1 - 2/k + 3/k^2 - 2/k^3 + 1/k^4}  \text{.}  $$
This is $\frac{1}{k^2} \cdot \text{[some bounded expression]}$.  Let $f(k_) = \frac{1}{1 - 2/k + 3/k^2 - 2/k^3 + 1/k^4}$.  For $k \geq 1$, $f'(k) = 0$ at $k = 2$ and $f''(2) > 0$.  So $f'(k) < 0$ on $[1,2)$ and $f'(k) > 0$ on $(2,\infty)$.  This means the value of $f(k)$ on $[1,\infty)$ is bounded by
$$
\max \{f(1), \lim_{k \rightarrow \infty} f(k)\} = \max \{1,1\} = 1  \text{.}  $$
So \begin{align*}
&\sum_{k=1}^\infty \frac{1}{\left(\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}\right)^2}  \\
&\leq \sum_{k=1}^\infty \frac{1}{k^2} \cdot \frac{1}{1 - 2/k + 3/k^2 - 2/k^3 + 1/k^4}  \\
&\leq \sum_{k=1}^\infty \frac{1}{k^2} \cdot 1  \\
&= \frac{\pi^2}{6}  \text{.}
\end{align*}
(This last is the Basel problem.  Alternatively, one can also use the comparison test with $\int^\infty \frac{\mathrm{d}x}{x^2}$ to show this last sum converges.)

So the sum converges.  There's a bit of a gap between this upper bound and the partial sum you cite, but our first approximation (replacing all the $1/\sqrt{j}$ terms with the extremal term in their sum) is a loose step and pretending the bounded constant near the end is $1$ is also loose.


问:
A simpler way to prove that $\frac 32-x-\frac{1}{x+1}$ is negative for $x\geq1$, instead of computing derivative?
I have this function
$$f(x)=\frac 32-x-\frac{1}{x+1}$$
I want to prove that $f(x)$ is negative for $x\geq1$. We can easily prove this by calculating the first derivative. Is there a simpler way to prove that? Instead of taking derivative?

答:
Just note that, for $x \ge 1$,
$$-x - \frac{1}{x + 1} \le -\frac{3}{2} \quad \Leftrightarrow \quad 0 \le \frac{2x^2-x-1}{2(x + 1)}\quad \Leftrightarrow \quad 0 \le 2x^2 - x -1 = (x - 1)(2x+1)$$
and the last inequality clearly holds.


答:
\begin{align} \left(-x-\frac{1}{x+1}\leq -\frac{3}{2}\right)\quad \text{ and }\quad  x\geq1\\
\\
\iff\quad \left(x+\frac{1}{x+1}\geq \frac{3}{2}\right)\quad \text{ and }\quad  x\geq1  \\
\\
\iff \quad \left(x(x+1) + 1 \geq \frac{3}{2}(x+1)\right)\quad \text{ and }\quad  x\geq1\\
\\
\iff \quad \left(x^2 - \frac{1}{2}x - \frac{1}{2} \geq 0\right)\quad \text{ and }\quad  x\geq1\\
\\
\iff\quad \left(\left(x-\frac{1}{4}\right)^2 \geq \frac{9}{16}\right)\quad \text{ and }\quad  x\geq1\\
\\
\iff\quad \left(x - \frac{1}{4} \geq \frac{3}{4}\quad \text{ or }\quad x - \frac{1}{4} \leq -\frac{3}{4}\right)\quad \text{ and }\quad  x\geq1\\
\\
\iff\quad \left(x\geq1\quad \text{ or }\quad x\leq -\frac{1}{2}\right)\quad \text{ and }\quad x\geq1\\
\\
\iff\quad x\geq1\\
\end{align}

Title: A simpler way to prove that $\frac 32-x-\frac{1}{x+1}$ is negative for $x\geq1$, instead of computing derivative?

Body:
I have this function
$$f(x)=\frac 32-x-\frac{1}{x+1}$$
I want to prove that $f(x)$ is negative for $x\geq1$. We can easily prove this by calculating the first derivative. Is there a simpler way to prove that? Instead of taking derivative?


问:
Linear fuzzy system
I need help if any one can help me
How to define a crisp number or crisp system?
Whats the difference between crisp number and fuzzy number?
And if there is any references. </p>

答:
Crisp logic is something like Boolean logic or what we know as classical logic. Either a statement is true or it is not: truth-value as dichotomous.

Fuzzy logic(s) recognize that truth-value may not be dichotomous: it aims to capture the degree to which something is true (i.e., it sees truth-value probabilistically). It may be the case that a fuzzy number is a number assigned to measure the degree to which a statement is true.

For more on Fuzzy Logic/fuzzy sets/ etc, see Wikipedia for a summary and additional links.

In parallel, we are seea distinction between crisp sets (set theory in the tradition of Cantor) and fuzzy sets see, e.g., Zadeh, 1965. See Wikipedia for more on Fuzzy Sets.. It seems that fuzzy numbers or fuzzy number systems are values /number system defined within fuzzy set theory. See more at Wikipedia: Fuzzy Number.

You might appreciate this youtube video: "From Crisp Logic to Fuzzy Logic by George Klir": it captures a lecture given at a Workshop discussing what some take to be "The Paradigm Shift in Mathematics."

Title: Linear fuzzy system


答:
As suggested in my comment, to prove convergence (this approach can only however help you approximate the value of the sum, not get it exactly — I doubt there is a nice closed form), you can write
$$
\int_{j}^{j+1} \frac{dx}{\sqrt{x}} \leq \int_{j}^{j+1} \frac{dx}{\sqrt{j}}  = \frac{1}{\sqrt{j}} = \int_{j-1}^{j} \frac{dx}{\sqrt{j}} \leq \int_{j-1}^{j} \frac{dx}{\sqrt{x}}
$$
and so, for $k\geq 2$,
$$
2k-2\sqrt{k}\leq \int_{k}^{k^2+1} \frac{dx}{\sqrt{x}}  = \sum_{j=k}^{k^2} \int_{j}^{j+1} \frac{dx}{\sqrt{x}} \leq \sum_{j=k}^{k^2} \frac{1}{\sqrt{j}} \leq \sum_{j=k}^{k^2} \int_{j-1}^{j} \frac{dx}{\sqrt{x}}  = \int_{k-1}^{k^2} \frac{dx}{\sqrt{x}}  \leq 2k \tag{1}
$$
using that $\int_{a}^{b} \frac{dx}{\sqrt{x}} = 2(\sqrt{b}-\sqrt{a})$. Since $k-\sqrt{k}\geq k/2$ for $k\geq 4$, we have
$$
\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}} \geq k
$$
for $k\geq 4$, and therefore
$$
\sum_{k=4}^\infty \frac{1}{\left(\sum_{j=k}^{k^2} \frac{1}{\sqrt{j}}\right)^2} \leq \sum_{k=4}^\infty \frac{1}{k^2} < \infty \tag{2}
$$
which shows convergence.

Note that using (1), you could approximate the original sum arbitrarily by using the inequalities from (1) starting at any given $k=K$ of your choosing, and computing exactly the value of the terms for $k\leq K$. That does not seem very practical, however.

Title: Nested sums - reciprocal of a sum : Find exact value

Body:
$$\text{Find:} ~~~~~~ \sum_{k=1}^{\infty} \frac{1}{ \left (
 \sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}}  \right )^2}$$

(Beware of the bounds of $j$, it does not always start from $1$, but starts at $k$ and goes to $k^2$)

At first I thought it has something to do with Riemann's zeta function $\zeta$ and thus the solution is:

$$ \zeta \left( \sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}} \right ) $$

However this totally seems incorrect because we cannot find this sum as we have an unknown, $k$.

Which also is being summed to $\infty$.

I thought (however was unsure) of the fact that:

$$ \left ( \sum f \right )^2 \ge  \sum f^2  $$

Thus (maybe) it is trivial the sum is converging to some value, which according to a little program I wrote is about: $$ \sum_{k=1}^{\infty} \frac{1}{ \left (
 \sum_{j=k}^{k^2} \frac{1}{\sqrt{~j~}}  \right )^2} \approx 1.596$$

But this is not a rigorous proof, is there a way to solve this so we get a solution by hand - calculating the exact value of this sum if it is finite (if not, why? )

Any mathematical tools (not programs) are acceptable, because this is not a question from a specific course ( I don't have context for this).


问:
Proof Concerning Indexed family of Sets and Intervals
I have been stuck on the following problem for some time.

I know to prove two things are equal, you use the proof where you show they are both subsets. I think I was able to show that the set on the left, (which I called A) is a subset of the interval through simple logic, but to show the converse I am having some issues.

I attached my working thus far for anyone to look at it, but please note that after the line " $ \Leftrightarrow \alpha \geq -1 = \alpha + 1 \geq 0 and  \alpha \leq -2 ....$" the rest is tata (lol- but an attempt to work towards what I know would be the end of the proof, showing that a belongs also to A)

I have tried using the hint, but then I end up with something along the lines of '$n > \alpha + 1$ so  $n + 1 > \alpha$', which does not bring me closer to either -x, or x+1 of the interval on the left.

Any help or suggestions would be greatly appreciated. Thanks in advance.

答:
I tried the case with y>2 and got
y >2 $\iff$ y-1>1. letting b= y-1 $\implies$ (-(y-1), y] = (-b, b+1] which would imply that y$\in$ (-(y-1), y] = (-b,b+1]. Thus, it is possible for y$\notin$ [-1,2] but y$\in$ U$_{x>1}$ for all x. This would disprove the statment. Am I going wrong?


问:
Finding exponent of the number 10
Finding exponent of the number 10 in 100! .

Now , some explains in my book show this.

So , it says that expel the of 5 and 10 are same but in my other book where there is a Q to Finding exponent of the number 10 in 100!.

Answer for 5 is 24 and for 10 is 11. Why is that ?

Title: Finding exponent of the number 10

Body:
$100! =$

9332621544394415268169923885626670049071596826438162146859296389521759999322991
5608941463976156518286253697920827223758251185210916864000000000000000000000000

which has $24$ zeros at the end.  There are more $2$s than $5$s and so the number of $5$s determines the number of $10$s.

Arguing there are just eleven $10$s would be like saying the $5!$ is not a multiple of $10$. But since it is $120$ there is clearly one $10$ which come from the $5$ and an earlier $2$.


问:
Trigonometric system with different coefficients
I have a trigonometric system as below and I want to solve it for $x$.

\begin{gather}
A\cos(x)+B\cos(2x)+C\sin(2x)  = D \\
-C\cos(2x)-A\sin(x)+B\sin(2x) = F
\end{gather}

Can anyone suggest a solution?
Regards.

答:
Note $(1)$ the first equation and $(2)$ the second. Multiplying $(1)$ by $C$ and $(2)$ by $B$ the sum of two resulting equations give $$A\cos(\theta-x)+\cos(2x)=\frac{BD-CF}{B^2+C^2}=M$$ Similarly you get $$A\sin(\theta-x)+\sin(2x)=\frac{CD-BF}{B^2+C^2}=N$$ where $\sin(\theta)=\dfrac CB$. Now taking the square and adding you get
$$A^2(1)=(M-\cos(2x))^2+(N-\sin(2x))^2$$ so you have
$$M\cos(2x)+N\sin(2x)=\frac{M^2+N^2+1-A^2}{2}$$ which is a standard equation you can solve. In fact you do have
$$\sin(\phi+x)=\frac{M^2+N^2+1-A^2}{2(M^2+N^2)}$$ where $\sin(\phi)=\dfrac{M}{M^2+N^2}$.


问:
Minimal $k$-colouring of the line graph of $K_n$, $L(K_n)$
I'm trying to find $\chi(L(K_{10}))$ which is the minimal coloring of the vertices of $L(K_n)$.
By Vizing, we know that $\Delta(K_n)\leq \chi(L(K_n))\leq \Delta(K_n)+1$ and this yields the inequality $$ 9\leq \chi(L(K_{10})) \leq 10$$
How do I proceed from here on? Test a 9-colouring and see if it is proper?

答:
All considered graphs are finite, simple and undirected.

Consider such a graph $G$. We first observe that the chromatic index $\chi'(G)$ is equal to the chromatic number of its line graph $L(G)$.

The following is due to results of Behzad et. al in "The colour numbers of complete graphs. Journal of the London Mathematical Society 1.1 (1967): 226-228."

Let $K_n$ be a complete graph on $n$ vertices, then

$$
 \chi'(K_n)= \left\{
\begin{array}{c}
n \text{ for odd } n \geq 3\\
n-1 \text{ for } n \text{ even}
\end{array}
\right.
$$

The proof can be found in the above paper.

Hence, $\chi(L(K_{10}))=\chi'(K_{10}) =9$.

Using SageMath to confirm:

```<code>G = graphs.CompleteGraph(10)
h = G.line_graph()
print(h.chromatic_number())
</code>```

Of course the above yields to $9$.

The above theorem is discussed in this network here.

Body:
All considered graphs are finite, simple and undirected.

Consider such a graph $G$. We first observe that the chromatic index $\chi'(G)$ is equal to the chromatic number of its line graph $L(G)$.

The following is due to results of Behzad et. al in "The colour numbers of complete graphs. Journal of the London Mathematical Society 1.1 (1967): 226-228."

Let $K_n$ be a complete graph on $n$ vertices, then

$$
 \chi'(K_n)= \left\{
\begin{array}{c}
n \text{ for odd } n \geq 3\\
n-1 \text{ for } n \text{ even}
\end{array}
\right.
$$

The proof can be found in the above paper.

Hence, $\chi(L(K_{10}))=\chi'(K_{10}) =9$.

Using SageMath to confirm:

G = graphs.CompleteGraph(10)
h = G.line_graph()
print(h.chromatic_number())

Of course the above yields to $9$.

The above theorem is discussed in this network here.


问:
Differentiation Rules
Can these rules be extended to $\Bbb R^n$ with $n>1$ using norms? Or does that not work? From Calculus 1, we know these rules are applicable because we are working with 1 variable functions but how about when we have higher dimensions?

答:
The classic generalization of derivatives most intuitively similar to higher dimensions is partial derivatives.

The notion of a derivative is inherently (up through real analysis) tied to the idea of a limit. Recall
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
defines the derivative of $f$ at $x$.
If we extend to $\mathbb R^n$ ($n>1$) and set $x, h \in \mathbb R^n$ for $f: \mathbb R^n \to \mathbb R^n$, This limit is not defined as we don't have a rigorous notion of dividing two length-$n$ vectors.

Your question about using norms to define derivatives doesn't specify a specific construct, so let's look at a few.

In general, we want the derivative operator to return real numbers (both positive and negative) and ensure that the 1-dimensional case can "fall out" under simplification.

$$ f'(x) = \lim_{h \to 0} \frac{||f(x+h) - f(x)||}{||h||}$$
We run into a first immediate problem with signs - all the 'derivatives' here must be positive (as both norms are always positive), so we could never have a negative derivative.

If we want Calculus 1 rules to fall out, this type of construct simply won't work. You can try this out with your favorite polynomial using $|| \cdot || = |\cdot|$ as a norm on $\mathbb R$.

The closest generalization of a derivative that uses norms is the total derivative of $f: \mathbb R^n \to \mathbb R^m$, defined implicitly as $f'(x)$ such that
$$ \lim_{\mathbf h\to \mathbf 0} \frac{||f(x+h) - (f(x) + f'(x) \mathbf{h}) ||}{|| \mathbf{h} ||} = 0$$

where $h \in \mathbb R^n$ is a vector. As $f(x)$ is a vector, $f'(x)$ is a $m \times n$ matrix. If $n=m=1$, then we have a derivative in in the classical sense using $| \cdot |$ as a norm. How do we compute $f'(x)$ then? It's actually a jacobian matrix, so we can build this concept using partial derivatives.

Title: Differentiation Rules


问:
Finding function with discontinuity at $x = 0$
I am trying to find a function $f: \mathbb{R} \to \mathbb{R}$ that is continuous everywhere except $x = 0$. The only example that comes to mind is $f(x) = \frac{1}{x}$, but this isn't even a candidate, I think, because it is undefined at $x = 0$, so I can't really talk about "continuity" at $x = 0$. Another option is a function with a jump discontinuity at $x = 0$: something like $f(x) = x$ for $x \neq 0$ and $f(x) = 1$ when $x = 0$. Then I can take a sequence of points, either from above or below, converging to $0$, but not to $f(0)$.

Is this incorrect? Is there a more elegant solution than this?

答:
What about

$$f(x)=\begin{cases}\frac{\sin x}x,&x\neq0\\{}\\-4,&x=0\end{cases}\;\;\;?$$


问:
Prove that $QBYZ$ is cyclic
Circles $\omega_1$ and $\omega_2$intersect each other at points $A$ and $B$. Point $C$ lies on
the tangent line from $A$ to $\omega_1$ such that $\angle ABC = 90^{°}$. Arbitrary line
passes through $C$ and cuts $\omega_2$ at points $P$ and $Q$. Lines $AP$ and $AQ$ cut
$\omega_1$ for the second time at points $X$ and $Z$ respectively. Let $Y$ be the foot
of altitude from $A$ to $PQ$.

Prove that $QBYZ$ is cyclic.

I proved that $AYBC$ and $PYBX$ are cyclic but I have no idea how to prove that $QBYZ$ is cyclic.

答:
Below when I use arcs I am referring to the circle $\omega_1$. By $\overset{\huge\frown}{AB}$ I mean the arc on $\omega_1$ that passes from $A$ through $Z$ to $B$.
$$\widehat{BZQ} = \frac{\overset{\huge\frown}{AZ}+\overset{\huge\frown}{ZB}}{2} = \frac{\overset{\huge\frown}{AB}}{2} \qquad(1)$$
$$\widehat{BYQ} = 180^o - \widehat{BYP}$$
And as you proved, $PYBX$ is cyclic. Therefore:
$$\widehat{BXP} = 180^o - \widehat{BYP}$$
$$\therefore \widehat{BYQ} = \widehat{BXP} = \frac{\overset{\huge\frown}{AB}}{2} \qquad(2)$$
From (1) and (2) we have $\widehat{BZQ} = \widehat{BYQ}$ which means $QBYZ$ is cyclic.

Title: Prove that $QBYZ$ is cyclic


问:
How did they get from (5/3)^(log3(n)) to n^log3(5/3)
Here's the note I got from class:

So I was wondering how my prof got from
$$\biggr(\frac{5}{3}\biggl)^{\log_3n}$$
to
$$n^{\log_3(5/3)}.$$

答:
Is not hard to prove that $a^{\log_m(b)}=b^{\log_m(a)}$. Note that:$$\log_m(b)\cdot \log_m(a) =\log_m(a) \cdot \log_m(b) $$
$$ \log_m(a^{\log_m(b)})=\log_m(b^{\log_m(a)})$$
$$ a^{\log_m(b)}=b^{\log_m(a)}$$

This last step is done applying exponentials in both sides. Also you con prove it using the formula for change of base:
$$ \log _{a}b={\frac {\log _{c}b}{\log _{c}a}} $$
In your example this is $a=5/3,b=n$ and $m=3$.


问:
proving that $\alpha_1+\alpha_2<-3$ which are the negative roots of $f(x)$
I am stuck on the following problem:

Given
\begin{align*}
f(x) &= x^4 + (-m - 2n + 6)x^3 + (mn - 5m - 8n + 10)x^2\\
&\quad + (3mn - 7m - 8n)x + mn-m-n-7,
\end{align*}
if $\alpha_1,\alpha_2$ are two negative roots of $f(x)=0$, show that $\alpha_1+\alpha_2< -3$ for $m,n\ge 4$.

I took some values of $m,n$ and found the following:

$m=n=4$, $\alpha_1=-2.476, \alpha_2=-0.621$;

$m=5,n=4$, $\alpha_1=-2.49, \alpha_2=-0.59$

I tried some other values of $m,n$,
In all the above cases, I am getting $\alpha_1+\alpha_2<-3$.

But I am not able to prove it. Can someone please help me out?

NOTE:

I found the following. I obtained $-1<\alpha_1<-k$ and $-3<\alpha_2<-3+k$
where $k$ is some function of $m$ and $n$.

I took $m=4,n=5$ and I got $\alpha_1=-2.492<-3+0.55, \alpha_2=-0.592<-0.55$.
I took some other values of $m,n$ and obtained similar stuff.
So I think my guess is correct.

Is there any way to find $k$ such that $k$ is a function of $m,n$?

答:
Let $x_1 < x_2 < 0$ be two negative solutions and I will obtain a bound:
$$x_2 <-\dfrac{3}{10}(5+\sqrt{10})\approx-2.4487.$$

First, rewrite the polynomial as:
\begin{align}
f(x,m,n) = (x^2+3x+1)mn
\\-(x^3+5x^2+7x+1)m
\\-(2x^3+8x^2+8x+1)n
\\+(x^4+6x^3+10x^2-7)
\end{align}
As my previous update and @RiverLi's comment said, we can obtain the simple bound:
\begin{equation}
\lambda_1 < x_1 < x_2 < \lambda_2 < 0,
\end{equation}
where $\lambda_{1,2} = \dfrac{-3+\sqrt{5}}{2}$ are the roots of the equation $x^2+3x+1 = 0,$ by simply considering $f(\lambda_i,m,n).$ But this is not strong enough because we want an upper bound on $x_1.$ Therefore, for a small $\varepsilon > 0$ to be chosen later, consider the equation:
$$x^2+3x+1 = -\varepsilon\,\,(1)$$
and call its roots $x_{1,2}^{\varepsilon} = \dfrac{-3\pm\sqrt{5-4\varepsilon}}{2},\,x_1^{\varepsilon}<x_2^{\varepsilon} .$ The nice thing about this is that if $x < 0$ is root of $(1),$ then we have a nice form for:
\begin{align}
f(x,m,n) = -\varepsilon mn
\\+(x\varepsilon+2\varepsilon +1)m
\\+(2x\varepsilon+2\varepsilon+1)n
\\+(\varepsilon^2+\varepsilon-7-3x)
\end{align}
which we will call $(2).$
We want to choose $\varepsilon$ so that $f(x_1^{\varepsilon})$ is negative, which will give $x_1 < x_1^{\varepsilon}.$

One can immediately see that:
$$2x_1^{\varepsilon}\varepsilon+2\varepsilon+1 = 1-\varepsilon -\varepsilon\sqrt{5-4\varepsilon}<0\iff 0.344446\approx\varepsilon_0 < \varepsilon < 1.25$$
where $\varepsilon_0$ is the unique positive solution of $1-\varepsilon -\varepsilon\sqrt{5-4\varepsilon}=0.$ Choosing this $\varepsilon = \varepsilon_0$ will give a slightly better bound, but when I was doing it by hand I found $0.35$ works, which gives the quadratic equation:
$$20x^2+60x+27 = 0\implies x_1^{\varepsilon} = -\dfrac{3}{10}(5+\sqrt{10})\approx-2.4487.$$

For this chosen value, the only thing left to show is that $f(x_1^\varepsilon) < 0.$ We automatically know that the coefficient of $n$ is negative by our choice and the free coefficient is trivially negative as well. For the remaining part:
$$-\varepsilon mn+(x\varepsilon+2\varepsilon +1)m \leq m\left(1+x\varepsilon-2\varepsilon\right)=m\cdot\dfrac{2-7\varepsilon - \varepsilon\sqrt{5-4\varepsilon}}{2}<0,$$
which can be verified by hand for $\varepsilon = 0.35.$

To conclude, we then have:
$$x_2+x_1 < -\dfrac{3}{10}(5+\sqrt{10}) +\dfrac{-3+\sqrt{5}}{2}\approx -2.83.$$
But all of this was done to obtain an explicit upper bound on $x_1$, which we previously do not have. So now by considering $x^2+3x+1 = \varepsilon$ for small, suitable $\varepsilon >0,$ I hope to better the not-so-tight upper bound $x_2:$
$$x_2<\lambda_2 = \dfrac{-3+\sqrt{5}}{2}\approx -0.381,$$
next time I got some time on my hand.


问:
Non-linear system of differential equations
I am trying to solve this differential equation which popped up in an engineering problem.

\begin{align}
&a\dot{V}(t) + b P(t) &= x(t)\\
&V(t)P(t) &= y(t)
\end{align}

The values $a$ and $b$ are known constants and $x(t)$ and $y(t)$ are known functions.  We are asked to find $V(t)$ and $P(t)$.
The equation was more complicated in the beginning and this is as much as I managed to simplify it, however  I am stuck at this point, and I haven't solved differential equations in a while.

It's not hard to reduce this to
$$
\frac{x(t)}{b}V(t) - \frac{a}{b}V(t)\dot{V}(t) = y(t)
$$
But this is non-linear and I don't know how to solve it.

答:
Your ODE is an Abel equation of the second kind. Here's a resource for finding your solution!
http://eqworld.ipmnet.ru/en/solutions/ode/ode0125.pdf</p>


问:
Number of ways to rearrange in the word UNSUCCESSFULLY with the given requirements?
Question: Given the word UNSUCCESSFULLY, calculate the number of ways to rearrange word with the given requirements below:

Letter 'Y' appears after all the vowels in the word?

The first 'S' appears before the first 'U'?

Both U's and L's appear consecutively at the same time?

C's don't appear consecutively?

Note: Each question is separate from the others.

My attempted solution:

For question 1, I placed the 4 vowels and the letter 'Y' first and then rearranged the word and got:
$${\binom{14}{5}} (\frac{4!}{3!})(\frac{9!}{2!2!3!3!})$$

For question 2, I did the exact same and got
$${\binom{14}{6}} (\frac{4!}{2!2!})(\frac{8!}{2!2!})$$
I'm honstly a bit confused of what the difference is between question 1 and 2. I might not be right but I used the same approach in the two answers I got above.

For question 3, I bascially combined the 2 L's and also the 3 U's and treaten them as one single letter. This resulted in the word having 11 letters. And then I got the answer:
$${\binom{11}{3}}{\binom{8}{2}}{\binom{6}{1}}{\binom{5}{1}}{\binom{4}{1}}{\binom{3}{1}}{\binom{2}{1}}{\binom{1}{1}}$$

For question 4, I caclculated the max possible ways to rearrange the letter and subtracted the ways at which three C's are together from that. I got:
$${\binom{14}{3}}{\binom{11}{1}}{\binom{10}{3}}{\binom{7}{2}}{\binom{5}{1}}{\binom{4}{1}}{\binom{3}{2}}{\binom{1}{1}} - {\binom{13}{3}}{\binom{10}{1}}{\binom{9}{3}}{\binom{6}{1}}{\binom{5}{1}}{\binom{4}{1}}{\binom{3}{2}}{\binom{1}{1}}$$

I all honesty, I only followed the steps to which I think it might be correct and most of these are likely to be incorrect. I'd appreciate any help!

答:
Total number of combinations of UNSUCCESSFULLY:

```<code>> factorial(14)/1/1/1/2/2/6/1/6
[1] 605404800
</code>```

Answers:

First see how many combinations there are of UEUUY where Y appears last. There are 4. Choose 5 slots for it in the 14 slots. Then, rearrange the remaining 9 letters anyhow you want.

$$4*{14\choose 5}*\begin{pmatrix}9\\1,1,2,2,3\end{pmatrix}$$

```<code>> 4*choose(14,5)*factorial(9)/1/1/2/2/6
[1] 121080960
This is 20%.
</code>```

See how many combinations of SSSUUU with the first S appearing before the first U. The possibilities are SU_ _ _ _ , SSU _ _ _ and SSSUUU. Choose 6 slots for the S's and U's. Fill in the remaining 8 slots any way you want.

$$\left(\begin{pmatrix}4\\2\end{pmatrix}+3+1\right)*\begin{pmatrix}14\\6\end{pmatrix}*\begin{pmatrix}8\\1,1,1,1,2,2\end{pmatrix}$$

```<code>> (factorial(4)/2/+3+1)*choose(14,6)*factorial(8)/2/2
[1] 151351200
This is 25%.
</code>```

I assume you meant all 3 U's appear consecutively and all 2 S's. Treating the 3 U's as one block and similarly the 2 S's, the problem reduces to.

$$\begin{pmatrix}11\\1,1,1,1,1,1,2,3\end{pmatrix}$$

```<code>> factorial(11)/2/6
[1] 3326400
This is roughly .549%.
</code>```

This is easiest done by subtracting the number of times C's do appear consecutively from the total number of arrangements.

$$\begin{pmatrix}14\\1,1,1,1,2,2,3,3\end{pmatrix}-\begin{pmatrix}12\\1,1,1,1,2,3,3,\end{pmatrix}*13$$

```<code>> factorial(14)/2/2/6/6-factorial(12)/2/6/6*13
[1] 518918400
This is roughly 85.7%.
</code>```

Body:
Total number of combinations of UNSUCCESSFULLY:

> factorial(14)/1/1/1/2/2/6/1/6
[1] 605404800

Answers:

First see how many combinations there are of UEUUY where Y appears last. There are 4. Choose 5 slots for it in the 14 slots. Then, rearrange the remaining 9 letters anyhow you want.

$$4*{14\choose 5}*\begin{pmatrix}9\\1,1,2,2,3\end{pmatrix}$$

> 4*choose(14,5)*factorial(9)/1/1/2/2/6
[1] 121080960
This is 20%.

See how many combinations of SSSUUU with the first S appearing before the first U. The possibilities are SU_ _ _ _ , SSU _ _ _ and SSSUUU. Choose 6 slots for the S's and U's. Fill in the remaining 8 slots any way you want.

$$\left(\begin{pmatrix}4\\2\end{pmatrix}+3+1\right)*\begin{pmatrix}14\\6\end{pmatrix}*\begin{pmatrix}8\\1,1,1,1,2,2\end{pmatrix}$$

> (factorial(4)/2/+3+1)*choose(14,6)*factorial(8)/2/2
[1] 151351200
This is 25%.

I assume you meant all 3 U's appear consecutively and all 2 S's. Treating the 3 U's as one block and similarly the 2 S's, the problem reduces to.

$$\begin{pmatrix}11\\1,1,1,1,1,1,2,3\end{pmatrix}$$

> factorial(11)/2/6
[1] 3326400
This is roughly .549%.

This is easiest done by subtracting the number of times C's do appear consecutively from the total number of arrangements.

$$\begin{pmatrix}14\\1,1,1,1,2,2,3,3\end{pmatrix}-\begin{pmatrix}12\\1,1,1,1,2,3,3,\end{pmatrix}*13$$

> factorial(14)/2/2/6/6-factorial(12)/2/6/6*13
[1] 518918400
This is roughly 85.7%.


Title: How did they get from (5/3)^(log3(n)) to n^log3(5/3)

Body:
Here's the note I got from class:

So I was wondering how my prof got from
$$\biggr(\frac{5}{3}\biggl)^{\log_3n}$$
to
$$n^{\log_3(5/3)}.$$

Body:
\begin{align*}
\left( \frac{5}{3} \right)^{\log_3 n}
    &= \left( 3^{\log_3 5/3} \right)^{\log_3 n}  \\
    &= 3^{(\log_3 5/3)(\log_3 n)}  \\
    &= 3^{(\log_3 n)(\log_3 5/3)}  \\
    &= \left( 3^{\log_3 n}\right)^{\log_3 5/3}  \\
    &= \left( n\right)^{\log_3 5/3}  \text{.}
\end{align*}


问:
Does any theory of infinite quantities provide info for $-1$ raised to infinite power, its absolute value, finite part and a series or integral form?
I've heard about surreal numbers, hyperreal numbers, Hardy fields, nonstandard analysis, cardinal ariththmetic, ordinal arithmetic, games, etc.

My impression is that neither of them can exactly show the result of the following operation: $(-1)^w$, where $w$ is some infinite number from that theory.

The only theory that exactly gives a result is extended real line ${\overline {\mathbb {R} }}$, where, of course, $(-1)^\infty=0$ (because of Cesàro mean).

Am I correct in my skepticism about all these theories?

答:
In the context of hyperreal numbers (defined using ultrapowers, for instance), $(-1)^w$ is defined for any infinite hypernatural number $w$.  Specifically, $(-1)^w=1$ if $w$ is even and $(-1)^w=-1$ if $w$ is odd.  More generally, $(-1)^w$ can be defined as a hypercomplex number for any hyperreal $w$ in the same way as it is defined for reals, as $\exp(i\pi w)$.  (Of course, just as for real numbers, this is only one possible "branch" of the exponential; you could similarly say $(-1)^w$ is $\exp((i\pi+2ki\pi)w)$ for any hyperinteger $k$.)

I would also mention that $(-1)^\infty$ is usually considered to be undefined in the context of the extended real numbers.  There may be contexts where it is convenient to define it to be $0$, but this is certainly not a widespread convention.


问:
Application of Hahn-Banach Theorem in $\Bbb{R}^2$
I'm trying to solve

Let consider the space $\Bbb{R}^2$ with the norm $\|(x,y)\|_p=(|x|^p+|y|^p)^{1/p}$ and let $Z$ be the subspace $$Z=\{(t,mt);t\in\Bbb{R}\},$$ for some $m\in\Bbb{R}$. Let $f\colon Z\to\Bbb{R}$ be defined by $f(t,mt)=t$.

(a) For $1<p<\infty$ find the unique linear extension $\tilde{f}$ of $f$ defined over $\Bbb{R}^2$ such that $\|\tilde{f}\|=\|f\|$.

(b) For $p=1$, find two distincts extensions of $f$. Do the same for $p=\infty$.

I'm stucked in (a), and neither try (b) yet. I was thinking that the extension which works in (a) is $\tilde{f}(x,y)=y/m$. But I'm with big difficult to see what to do with the norms.

For any $(t,mt)\in Z$, we have that $$\|(t,mt)\|_p=|t|(1+|m|^p)^{1/p},$$ and $$|f(t,mt)|=|t|\leqslant|t|(1+|m|^p)^{1/p}=\|(t,mt)\|_p,$$ hence $\|f\|\leqslant1$. So the candidate to be the norm of $f$ is $1$. But I can't find the vector in $Z$ that does the work, neither usying some sequence in $Z$ that makes the norm be $1$.

So, I'm thankful for any help.

答:
First of all, as it has correctly been observed, $\|f\|_{Z^*}=\frac{1}{(1+|m|^p)^{1/p}}$.

Note that $Z$ is a line that passes through the origin. Take the vertical line to $Z$ which is the subspace $W=\{(-mt,t):t\in\mathbb{R}\}$. In order to define an extension of $f$ on $\mathbb{R}^2$ it suffices to define a value for the extension at the point $(-m,1)$, because $(-m,1)$ and $(1,m)$ are linearly idependent and span $\mathbb{R}^2$. The extensions are in one to one correspondence with the possible values of $(-m,1)$, so let's take an extension that assigns the value $c\in\mathbb{R}$ to the point $(-m,1)$. Let's denote this (unique) extension by $F_c$, so $F_c(-m,1)=c$ and $F_c(1,m)=f(1,m)=1$.

Write a point $(x,y)$ of $\mathbb{R}^2$ as a linear combination of $(1,m)$ and $(-m,1)$, say $(x,y)=a\cdot(1,m)+b\cdot(-m,1)$. We have to solve the system (solve for $a,b$)
$$a-mb=x$$
$$ma+b=y$$
and by Cramer's rule we get that $a=\frac{x+my}{1+m^2}$ and $b=\frac{y-mx}{1+m^2}$. Therefore we have that
$$F_c(x,y)=\frac{x+my}{1+m^2}F_c(1,m)+\frac{y-mx}{1+m^2}F_c(-m,1)=\frac{x+my+c(y-mx)}{1+m^2}$$
Now $$\|F_c\|=\sup_{(x,y)\neq0}\frac{|F(x,y)|}{\|(x,y)\|}=\frac{1}{1+m^2}\cdot\sup_{x,y\neq0}\frac{|(1-mc)x+(c+m)y|}{(|x|^p+|y|^p)^{1/p}}\;\;\;\;(\star).$$

Question 1

Let $q$ be the conjugate exponent of $p$, i.e. $1/p+1/q=1$. Then Using the H"older inequality we have that $|(1-mc)x+(c+m)y|\leq\|(1-mc,c+m)\|_q\cdot\|(x,y)\|_p$, so $$\|F_c\|\leq\frac{\|(1-mc,c+m)\|_q}{1+m^2}$$. On the other hand, take the vector $(x,y)=((1-mc)\cdot|1-mc|^{q-2},(c+m)\cdot|c+m|^{q-2})$. Note that $\|(x,y)\|_p=\|(1-mc,c+m)\|_q^{q/p}$ so
$$\|F_c\|\geq\frac{|F_c(x,y)|}{\|(x,y)\|_p}=\frac{1}{1+m^2}\cdot\frac{\|(1-mc,c+m)\|_q^q}{\|(1-mc,c+m)\|_q^{q/p}}=\frac{\|(1-mc,c+m)\|_q}{1+m^2} $$
So we conclude that $$\|F_c\|=\frac{\|(1-mc,c+m)\|_q}{1+m^2}$$
Now in order to have an isometric extension we must choose $c$ so that $\|F_c\|=\|f\|_{Z^*}$, i.e.
$$\frac{\|(1-mc,c+m)\|_q}{1+m^2}=\frac{1}{(1+|m|^p)^{1/p}},$$
or equivalently
$$|1-mc|^q+|c+m|^q=\bigg(\frac{1+m^2}{(1+|m|^p)^{1/p}}\bigg)^{q}  $$
So one has to verify that this equation has a unique solution for $c\in\mathbb{R}$.

Question 2

Suppose that $p=1$. Then the obvious inequality $|(1-mc)x+(c+m)y|\leq\max\{|1-mc|,|c+m|\}\cdot(|x|+|y|)$ together with $(\star)$ for $p=1$ yield
$$\|F_c\|\leq\frac{\max\{|1-mc|,|c+m|\}}{1+m^2}.$$
On the other hand, applying $F_c$ at $(1,0)$ and at $(0,1)$ yield $\|F_c\|\geq\frac{|1-mc|}{1+m^2}$ and $\|F_c\|\geq\frac{|c+m|}{1+m^2}$, so we have that $$\|F_c\|=\frac{\max\{|1-mc|,|c+m|\}}{1+m^2}.$$

we want the extension to be isometric, so we want $\|F_c\|=\|f\|_{Z^*}=\frac{1}{1+|m|}$, so we want
$$\frac{\max\{|1-mc|,|c+m|\}}{1+m^2}=\frac{1}{1+|m|}$$

so one has to verify that this equation has at least two solutions for $c\in\mathbb{R}$.

Suppose that $p=\infty$. In that case $\|f\|_{Z^*}=\frac{1}{\max\{1,|m|\}}$. Also, $$|F_c(x,y)|=\frac{|(1-mc)x+(c+m)y|}{1+m^2}\leq\frac{(|1-mc|+|c+m|)\max\{|x|,|y|\}}{1+m^2}$$
so $\|F_c\|\leq\frac{|1-mc|+|c+m|}{1+m^2}$. On the other hand, $$\|F_c\|\geq\frac{|x(1-mc)+y(c+m)|}{1+m^2}$$
for all $x,y\in\{-1,1\}$. Depending on the order of $1,m,c$ we can choose proper values of $x,y\in\{-1,1\}$ to establish the inequality
$$\|F_c\|\geq\frac{|1-mc|+|c+m|}{1+m^2}$$
so we conclude that $$\|F_c\|=\frac{|1-mc|+|c+m|}{1+m^2}.$$
We want an isometric extension, so we are looking for the values of $c$ so that
$$\frac{|1-mc|+|c+m|}{1+m^2}\equiv\|F_c\|=\|f\|_{Z^*}\equiv\frac{1}{\max\{1,|m|\}}$$

and one has to verify that this equation has at least two solutions for $c\in\mathbb{R}$.

A final comment: This seems too complicated to me. If I were you I would only care for $m=0$, since any other line can be obtained from the canonical axis system with a mere rotation. Anyway, I am unsure how one would solve the equations that arise in each case, all those absolute values make things somehow complicated. I believe the idea and the norm computations are correct though. If I have any mistakes, please point them out.


问:
Find the chromatic polynomials to the wheel graph W6
I am starting with graph theory and have this exercise of wheel graphs, but I do not really understand how to do it.

Could anyone explain me how to find the chromatic polynomial of this graph? The book suggests using deletion/addition-contraction but I do not understand.

This is the wheel graph:

W6

答:
The method of deletion/contraction

The idea of this method of counting colourings is as follows.

Delete an edge, joining vertices $u$ and $v$ say, of a simple graph. When you count the colourings $D$ of this simpler graph you will overcount because you will include colourings where $u$ and $v$ have been given the same colour.

These unwanted colourings are precisely the number of colourings $C$ where  $u$ and $v$ are forced to have the same colouring by contracting edge $uv$ so that  $u$ and $v$ become the same vertex.

So there are $D-C$ colourings of the original graph.

Example: For the cycle graphs $C_n$

If you delete an edge of $C_n$ then, using $k$ colours, the simpler graph can be coloured in $k(k-1)^{n-1}$ ways since you can colour an end vertex in $k$ ways and then colour each successive vertex in $k-1$ ways.

We must now subtract the number of colourings when the edge is contracted. In which case the graph is $C_{n-1}$.

Now the chromatic polynomial for $C_3$ is clearly $k(k-1)(k-2)$.

So the chromatic polynomial for $C_4$ is $k(k-1)^3-k(k-1)(k-2)=k(k-1)(k^2-3k+3)$.

The chromatic polynomial for $C_5$ is $k(k-1)^4-k(k-1)(k^2-3k+3)=k(k-1)(k^3-4k^2+6k-4)$.

You can probably see the pattern now but you only require $C_5$ for your problem.

Your problem: W_6

Use the method suggested by @bof together with the result for $C_5$.


问:
Show that the mapping $w (z) = z^2 $ takes the upper-half complex plane $\text{Im} (z) \ge 0$ into the whole plane.
I am supposed to Show that the mapping $w (z) = z^2 $ takes the upper-half complex plane $\text{Im} (z) \ge  0$ into the whole plane.
$w = u+iv$ and $z=x+iy$

I tried doing it as follows:

$$w = u+iv = (x+iy)(x+iy) = x^2-y^2 + i2yx  $$

This implies that  $$u = x^2-y^2  \space\space\space\space\space  \text{and} \space\space\space\space\space v=2yx$$

The condition $\text{Im}(z) \geq 0$ would mean that $v \geq0$ and that $x^2 \geq u$.
Since x can take any value from the condition we see that u can be any positive value.

This would mean that we have transferred the initial complex number into the upper right quadrant but not the entire quadrant. I probably did something wrong but am very confused.

Any help would be appreciated.

答: The best way to see that is to use polar coordinates, $z=re^{iu}$, where $\in[0,2\pi]$ take $z'=\sqrt re^{iu/2}$, remark that ${z'}^2=z$ and since $u/2\in [0,\pi], z'=\sqrt r(\cos(u/2)+i\sin(u/2))$ and $\sin(u/2)\geq 0$.


问:
The spectrum norm of difference of two matrix converging to 0
Let $\Theta$ be a positive-definite matrix, and $\hat\Theta$ be a matrix with the same dimension as $\Theta$. If the spectrum norm of $\Theta - \hat\Theta$ converges to 0 with large probability, why can we say that $\hat\Theta$ is also positive-definite with large probability?

This fact is used in many papers, e.g. this one says Theorem 1 implies that are $\hat\Theta$ and $\hat\Omega$ are positive-definite with large probability. Why does this hold? What is the intuition behind it? Will the same result hold if we replace the spectrum norm with any other norms?

答:
Matrix perturbation theory might explain it. The intuition is that two close matrix also have close eigenvalue.

And since all matrix norms are equivalent up to a constant multiplier, this conclusion should also hold under other norms.


答:
The "jump discontinuity" approach is very common.
$$ f(x) = \frac{1}{x}$$
is indeed not defined at $0$, so it is not an example of the desired function.

The canonical example is called a $\delta$ function, which there are two (Dirac and Kronecker). In a set-theoretic application, the Kronkecker delta function is also called an indicator function.

Take
$$
\delta_a(x) =  \begin{cases}1 & x = a \\ 0 &  x \neq a\end{cases}
$$

In topology, the "topologist's sine curve" can be extended onto all of $\mathbb R$ by

$$
T(x) = \begin{cases} \sin\left( \frac \pi x\right) & x \neq 0 \\ 0 & x = 0
\end{cases}
$$
$T$ has some interesting topological properties when restricted to $(0,1]$.

Title: Finding function with discontinuity at $x = 0$

Body:
I am trying to find a function $f: \mathbb{R} \to \mathbb{R}$ that is continuous everywhere except $x = 0$. The only example that comes to mind is $f(x) = \frac{1}{x}$, but this isn't even a candidate, I think, because it is undefined at $x = 0$, so I can't really talk about "continuity" at $x = 0$. Another option is a function with a jump discontinuity at $x = 0$: something like $f(x) = x$ for $x \neq 0$ and $f(x) = 1$ when $x = 0$. Then I can take a sequence of points, either from above or below, converging to $0$, but not to $f(0)$.

Is this incorrect? Is there a more elegant solution than this?


答: It is well-known that each complex number $w = u +iv$ has a root. This can be explicitly computed. If $w \ne 0$, then there are exactly two distinct roots $z_1, z_2$ which differ by a factor $-1$. Thus at least one of the roots has $\text{Im}(z_i) \ge 0$. This shows that the set all all $z^2$ with $\text{Im}(z_i) \ge 0$ gives all $w \in \mathbb C$.

Body:
I am supposed to Show that the mapping $w (z) = z^2 $ takes the upper-half complex plane $\text{Im} (z) \ge  0$ into the whole plane.
$w = u+iv$ and $z=x+iy$

I tried doing it as follows:

$$w = u+iv = (x+iy)(x+iy) = x^2-y^2 + i2yx  $$

This implies that  $$u = x^2-y^2  \space\space\space\space\space  \text{and} \space\space\space\space\space v=2yx$$

The condition $\text{Im}(z) \geq 0$ would mean that $v \geq0$ and that $x^2 \geq u$.
Since x can take any value from the condition we see that u can be any positive value.

This would mean that we have transferred the initial complex number into the upper right quadrant but not the entire quadrant. I probably did something wrong but am very confused.

Any help would be appreciated.


问:
Prove that $SO_2$ is a normal subgroup of $O_2$
I have been working on this for a few days. I can show the conditions that $SO_2$ is a group, but I cannot show the normal subgroup condition.

答: Hint: The kernel of the determinant restricted to $O(2)$ is $SO(2)$


问:
The formulas of prostapheresis: memorization technique
This question is related purely for my students of an high school and indirectly for me. The formulas below are the formulas of prostapheresis,

\begin{cases}
\sin\alpha+\sin\beta=2\,\sin \dfrac {\alpha+\beta}{2}\, \cos \dfrac {\alpha-\beta}{2} \\
\sin\alpha-\sin\beta=2\sin \dfrac {\alpha-\beta}{2} \,\cos \dfrac {\alpha+\beta}{2}\\
\cos\alpha+\cos\beta=2\cos \dfrac {\alpha+\beta}{2}\,\cos \dfrac {\alpha-\beta}{2}\\
\cos\alpha-\cos\beta=-2 \,\sin \dfrac {\alpha+\beta}{2} \,\sin \dfrac {\alpha-\beta}{2}
\end{cases}

and while I am able to find them, I am not able to find a technique to memorize them.

Is there a technique to be able to memorize them?

答:
I know a technique that my high school teacher taught me, and that could be particularly useful for Italian students, quite famous where I come from. It is the following: if in the second formula you apply commutativity and swipe $\cos$ with $\sin$ in the second member of the equation, obtaining $2\cos\big(\frac{\alpha+\beta}{2}\big) \sin\big(\frac{\alpha-\beta}{2}\big)$, then in the second member for all equations in the first argument $\alpha+\beta$ appears, while in the second argument $\alpha-\beta$ appears (reading from left to right). Now, there is a sentence that says "Cento rose sono meno belle" that presents vocals in the order by which you find $\cos$ and $\sin$ reading from left to right and top-down second members of formulas, reminding that in Italian $\sin$ can be used also as sen (meno (= less) indicates the sign of the second member of the last line). So using "Cento rose sono meno belle" the difficulty is reduced a lot, because it only needs to remember factors $2$ at the beginning and factor $\frac{1}{2}$ in the argument. For the first members of the equations I think that notice that they are pretty ordered (four $\sin$ then four $\cos$ with ordered $\alpha$  $\beta$ and alternated signs $+-+-$) should make it easy to remember them.

```<code>                         "Cento rose sono meno belle"
</code>```

Body:
I know a technique that my high school teacher taught me, and that could be particularly useful for Italian students, quite famous where I come from. It is the following: if in the second formula you apply commutativity and swipe $\cos$ with $\sin$ in the second member of the equation, obtaining $2\cos\big(\frac{\alpha+\beta}{2}\big) \sin\big(\frac{\alpha-\beta}{2}\big)$, then in the second member for all equations in the first argument $\alpha+\beta$ appears, while in the second argument $\alpha-\beta$ appears (reading from left to right). Now, there is a sentence that says "Cento rose sono meno belle" that presents vocals in the order by which you find $\cos$ and $\sin$ reading from left to right and top-down second members of formulas, reminding that in Italian $\sin$ can be used also as sen (meno (= less) indicates the sign of the second member of the last line). So using "Cento rose sono meno belle" the difficulty is reduced a lot, because it only needs to remember factors $2$ at the beginning and factor $\frac{1}{2}$ in the argument. For the first members of the equations I think that notice that they are pretty ordered (four $\sin$ then four $\cos$ with ordered $\alpha$  $\beta$ and alternated signs $+-+-$) should make it easy to remember them.

                         "Cento rose sono meno belle"


问:
Accumulation points with respect to Zariski Topology on $\mathbb{R}$.
Consider the topology on $\mathbb{R}$  : the closed sets are given by the finite sets and $\mathbb{R}$ (basically the Zariski Topology on $\mathbb{R}$).

Let $(x_n)_{n \in \mathbb{N} }$ be a sequence in $\mathbb{R}$.

How do we show : if $l$ is a limit of the sequence $(x_n)_{n \in \mathbb{N} }$  if and only if for all $x \neq l$, the number of indices $n$ such that $x_n = x$ is finite.

Determine the accumulation points of the sequence $(x_n)_{n \in \mathbb{N} }$ w.r.t. Zariski topology

答:
Suppose that $\lim x_n=l$ and $x\neq l$. $U_x=\mathbb{R}-\{x\}$ is an open subset which contains $l$, there exists $N$ such that $n>N$ implies that $x_n\in U_x$. This implies that $x_n\neq x$ for $n>N$.

Conversely suppose that $x\neq l$ implies that the set of $x_n$ such that $x_n\neq x$ is finite. Let $U$ be an open subset containing $l$, $U=\mathbb{R}-\{u_1,...,u_p\}$. Let $A\subset \mathbb{N}$ such that $a\in A$ implies that $x_a=x_i$. $A$ is finite. Let $N=\max A$. For every $n>N$, $x_n\neq u_i$ implies that $x_n\in U$.


问:
Properties of Logarithms (evaluate)
use $\log 4= 0.602$ and $\log 12=1.079$ to evaluate the logarithm

$\log 3$

I'm very confused on how I will evaluate this one I've tried other things but I'm not sure

Title: Properties of Logarithms (evaluate)

Body: $\log(12) - \log(4) = \log\left(\frac{12}{4}\right) = \log(3)$


答:
You made some errors, the reason for which I can hardly explain. So the correct answers:

Q.1
$${\binom{14}{5}} \frac{4!}{1!3!}\frac{9!}{1!1!2!2!\color{red}{3!}}$$

Q.2
$${\binom{14}{6}} \frac{\color{red}{5!}}{2!\color{red}{3!}}\frac{8!}{1!1!1!1!2!2!}$$

Q3 and Q4 are correct but it is possibly better to write:
$$
\frac{11!}{1!1!1!1!1!1!2!3!}.
$$

Note that I write all factors in the denominator. So one can easier check, that the sum of the numbers is equal to the number in numerator.

Title: Number of ways to rearrange in the word UNSUCCESSFULLY with the given requirements?

Body:
Question: Given the word UNSUCCESSFULLY, calculate the number of ways to rearrange word with the given requirements below:

Letter 'Y' appears after all the vowels in the word?

The first 'S' appears before the first 'U'?

Both U's and L's appear consecutively at the same time?

C's don't appear consecutively?

Note: Each question is separate from the others.

My attempted solution:

For question 1, I placed the 4 vowels and the letter 'Y' first and then rearranged the word and got:
$${\binom{14}{5}} (\frac{4!}{3!})(\frac{9!}{2!2!3!3!})$$

For question 2, I did the exact same and got
$${\binom{14}{6}} (\frac{4!}{2!2!})(\frac{8!}{2!2!})$$
I'm honstly a bit confused of what the difference is between question 1 and 2. I might not be right but I used the same approach in the two answers I got above.

For question 3, I bascially combined the 2 L's and also the 3 U's and treaten them as one single letter. This resulted in the word having 11 letters. And then I got the answer:
$${\binom{11}{3}}{\binom{8}{2}}{\binom{6}{1}}{\binom{5}{1}}{\binom{4}{1}}{\binom{3}{1}}{\binom{2}{1}}{\binom{1}{1}}$$

For question 4, I caclculated the max possible ways to rearrange the letter and subtracted the ways at which three C's are together from that. I got:
$${\binom{14}{3}}{\binom{11}{1}}{\binom{10}{3}}{\binom{7}{2}}{\binom{5}{1}}{\binom{4}{1}}{\binom{3}{2}}{\binom{1}{1}} - {\binom{13}{3}}{\binom{10}{1}}{\binom{9}{3}}{\binom{6}{1}}{\binom{5}{1}}{\binom{4}{1}}{\binom{3}{2}}{\binom{1}{1}}$$

I all honesty, I only followed the steps to which I think it might be correct and most of these are likely to be incorrect. I'd appreciate any help!


Title: Properties of Logarithms (evaluate)

Body:
use $\log 4= 0.602$ and $\log 12=1.079$ to evaluate the logarithm

$\log 3$

I'm very confused on how I will evaluate this one I've tried other things but I'm not sure

Body:
$$\log(b/a) = \log(b) - \log(a)$$
$3=\frac {12}{4}$ implies $\log(3) = \log(12)-\log(4)$


问:
How to show frechet characterization differentiability 2
Let $U\subset R^{n} $ and $f:U\to R$. Show that: $f$ is differetiable in $x_0\in U$ iff exist $A\in L(R^{n},R)$ and exist $r>0$:

$\lim_{t\rightarrow 0^{+}}\frac{f(x_0 +tv)-f(x_0)}{t}=Av$ uniformly in $v\in r\mathcal{S}^{n-1}$ with $\mathcal{S}^{n-1}$ is unitary sphere.

Attempt

$\Rightarrow$ Suppose that $f$ is differentiable in $x_0\in U$, then exist $A_{x_0}:R^{n}\to R$ linear transformation such that $f(x_0 +y)=f(x_0)+A_{x_0}(y)+||y||E_{x_0}(y)$ where $E_{x_0}(y)\rightarrow 0$ when $y\rightarrow 0$, equivalently

$\frac{f(x_{0} + y)-f(x_{0})}{||y||}-\frac{A_{x_0}(y)}{||y||}=E_{x_0}(y)$, now if i take $z=r\frac{y}{||y||}$ then $\frac{y}{||y||}\in \mathcal{S}^{n-1}$ and $y=\frac{||y||}{r}z$, if we will call $t=\frac{||y||}{r}$ imply

$\frac{f(x_{0} + tz)-f(x_{0})}{tr}-A_{x_0}(\frac{z}{r})=E_{x_0}(tz)$ and when i take limit when $t\rightarrow 0$ tehn

$\lim_{t\rightarrow 0}\left( \frac{f(x_{0} + tz)-f(x_{0})}{tr}-A_{x_0}(\frac{z}{r}) \right)=\frac{1}{r}\left[\lim_{t\rightarrow 0}\frac{f(x_{0} + tz)-f(x_{0})}{t}-A_{x_0}(z)\right]=\lim_{t\rightarrow 0}E_{x_0}(tz)=0$ hence

$\lim_{t\rightarrow 0}\frac{f(x_{0} + tz)-f(x_{0})}{t}=A_{x_0}(z)$ with $z\in r\mathcal{S}^{n-1}$.

$\Leftarrow$ Now by hypothesis $\lim_{t\rightarrow 0}\frac{f(x_{0} + tv)-f(x_{0})}{t}=A_{x_0}(v)$ with $v\in r\mathcal{S}^{n-1}$ uniformly, then $v=rw$ with $r>0$ and $w\in \mathcal{S}^{n-1}$ so

$\lim_{t\rightarrow 0}\frac{f(x_{0} + trw)-f(x_{0})}{t}=A_{x_0}(rw)=rA_{x_0}(w)$ or the same way

$\lim_{t\rightarrow 0}\frac{f(x_{0} + trw)-f(x_{0})}{rt}=A_{x_0}(w)$, but if we choose $h=tr$ when $t\rightarrow 0$ hence $h\rightarrow 0$ imply

$\lim_{h\rightarrow 0}\frac{f(x_{0} + hw)-f(x_{0})}{h}=A_{x_0}(w)$
and this is a definition of differentiablility in $x_0$.

My profesor say that i m not using the condition of uniformly no anywhere and my proof is confuse, therefore i have to proof that exist $r$, i just take one but i dont know how to show that exist,  can somebody to help me please, thank you

答:
Let $U\subset \Bbb R^{n} $ and $f:U\to \Bbb R$. Show that: $f$ is differentiable at $x_0\in U$ iff exist $A\in L(\Bbb R^{n},\Bbb R)$ and exist $r>0$:

$\lim_{t\rightarrow 0^{+}}\frac{f(x_0 +tv)-f(x_0)}{t}=Av$ uniformly in $v\in r\mathcal{S}^{n-1}$ with $\mathcal{S}^{n-1}$ is the unitary sphere.

Proof: $(\Rightarrow)$ Since $f$ is differentiable at $x_0\in U$, exist $A\in L(R^{n},R)$, such that

$$\lim_{h\to 0} \frac{\|f(x_0+h)-f(x_0)-Ah\|}{\| h\|}=0$$

So, for all $\varepsilon >0 $, there is $\delta_\varepsilon >0$ such that, for all $h\in \Bbb R^n$, if $\|h\|<\delta_\varepsilon$ then
$$\frac{\|f(x_0+h)-f(x_0)-Ah\|}{\| h\|} < \varepsilon$$

It follows, immediately that, for any $r>0$,  for all $\varepsilon >0 $, there is $\delta'= \frac{\delta_{\varepsilon/r}}{r} >0$ such that, for all $t>0$, if $t<\delta'$, then, for all $v\in r\mathcal{S}^{n-1}$, $\|tv\|=t\cdot r < \delta_{\varepsilon/r}$  and so

$$\frac{\|f(x_0+tv)-f(x_0)-tAv\|}{t \cdot r} < \varepsilon/r$$
So,
$$\left \| \frac{f(x_0+tv)-f(x_0)}{t}  - Av  \right \| =\frac{\|f(x_0+tv)-f(x_0)-tAv\|}{t} < \varepsilon$$

Note that $\delta'$ independs on $v$. So, we have proved that $\lim_{t\rightarrow 0^{+}}\frac{f(x_0 +tv)-f(x_0)}{t}=Av$ uniformly in $v\in r\mathcal{S}^{n-1}$.

$(\Leftarrow)$ Suppose exists $A\in L(\Bbb R^{n},\Bbb R)$ and exists $r>0$ such that
$\lim_{t\rightarrow 0^{+}}\frac{f(x_0 +tv)-f(x_0)}{t}=Av$ uniformly in $v\in r\mathcal{S}^{n-1}$.

Then, for all   for all $\varepsilon >0 $, there is $\delta_\varepsilon >0$ such that, for all $t>0$, if $t<\delta_\varepsilon $, then, for all $v\in r\mathcal{S}^{n-1}$,

$$\left \| \frac{f(x_0+tv)-f(x_0)}{t}  - Av  \right \| < \varepsilon$$

So, for all $\varepsilon >0$, there is $\delta'=r \delta_{r\varepsilon}>0$, such that for all $h\in \Bbb R^n$, if $\|h\|<\delta'=r \delta_{r\varepsilon}$, then $t= \frac{\|h\|}{r} <\delta_{r\varepsilon}$ and $v=r \frac{h}{\|h\|} \in r\mathcal{S}^{n-1}$ and $h=tv$.  So

\begin{align*}
\frac{\|f(x_0+h)-f(x_0)-Ah\|}{\| h\|} &= \left \| \frac{f(x_0+tv)-f(x_0)- tAv }{t r}   \right \| =\\
&= \frac{1}{r}\left \| \frac{f(x_0+tv)-f(x_0)}{t}  - Av  \right \|
< \frac{1}{r} r\varepsilon = \varepsilon
\end{align*}
So, we have proved that
$$\lim_{h\to 0} \frac{\|f(x_0+h)-f(x_0)-Ah\|}{\| h\|}=0$$
It means, $f$ is differentiable in $x_0\in U$.

Remark: In fact, we proved a stronger result. We proved that

Let $U\subset \Bbb R^{n} $ and $f:U\to \Bbb R$. Then the following three statements are equivalent:

There is $A\in L(\Bbb R^{n},\Bbb R)$ and there is $r>0$ such that
$\lim_{t\rightarrow 0^{+}}\frac{f(x_0 +tv)-f(x_0)}{t}=Av$ uniformly in $v\in r\mathcal{S}^{n-1}$ (where $\mathcal{S}^{n-1}$ is the unitary sphere).

$f$ is differentiable at $x_0\in U$

There is $A\in L(\Bbb R^{n},\Bbb R)$ and, for all $r>0$,
$\lim_{t\rightarrow 0^{+}}\frac{f(x_0 +tv)-f(x_0)}{t}=Av$ uniformly in $v\in r\mathcal{S}^{n-1}$ (where $\mathcal{S}^{n-1}$ is the unitary sphere).

In the first part of our proof, we proved that $(2 \Rightarrow 3)$. In the second part of our proof, we proved that $(1 \Rightarrow 2)$. The fact that $(3 \Rightarrow 1)$ is trivial.


问:
How do I average out to a particular number if I have to add another number to get that average?
So, lets say, that you have the following numbers (4, 15, 3, 22) and you have to add the fifth number (which I would call X) and they must average out to say 20.

What do you have to do?

I figured the formula is something like (4+15+3+22+X)/5=20.

But for some reason I can't solve it. It just doesnt click in. Can anyone tell me if at least I have the formula right?

Body: For such $X$  you have $$4+15+3+22+X = 5\cdot 20,$$ that is $$X = 100 -4 - 15-3-22=... $$


问:
How do you respond to "I was always bad at math"?
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.

答:
I respond with "what was your favorite subject?"   If I was bad at it then I would say so.  If I liked it even though I was bad at it, I would move the conversation to that topic.  If I didn't like it, I would say so and ask "why was it your favorite subject?"

The "what do you do?" question is an attempt at finding a commonality and a mutually interesting topic. Keep moving in that direction.  If the disclosure came in some other context, well, I'd just say "most people don't like math" and go back to the previous topic.


问:
Continuous and residual spectrum of a multiplication operator on $\ell^p$ in the non-Hilbert case $p\ne2$
Let $a\in\ell^\infty$, $p\in[1,\infty)$ and $$T:\ell^p\to\ell^p\;,\;\;\;x\mapsto ax.$$ It's easy to show that $\sigma_p(T)=\{a_n:n\in\mathbb N\}$.

How can we determine $\sigma_c(T)$ and $\sigma_r(T)$? And is there a nice characteriation of $\sigma_p(T')$?

It really easy to show that $\mathbb C\setminus\overline{\sigma_p(T)}\subseteq\rho(T)$.

But beyond that, I'm only able to answer the question in the case $p=2$. In that case $T$ is self-adjoint and hence $\sigma_r(T)=\emptyset$ from which we can conclude that $\sigma_c(T)=\overline{\sigma_p(T)}\setminus\sigma_p(T)$.

答:
The crucial fact that makes the Hilbert space case easy to deal with  is  that
$$
  \text{Ker(T)}^\perp =   \overline{\text{Ran}(T^*)}.
  $$
So, when $T$ is self-adjoint and $\lambda $ is real, it immediately follows that $T-\lambda $ is injective iff $\text{Ran}(T-\lambda )$ is dense.

Even though there is no notion of self-adjointness on a Banach space such as $\ell ^p$, when $p\neq 2$,  the same principle can be
easily adapted, as long as we are dealing with  diagonal operators,  that is,  operators of the form $T(x)=ax$,  for
$a\in \ell ^\infty $,  as in the original post.

In fact, for such an operator, it is very easy to prove that the following are equivalent:

$T$ is injective,

$a_n\neq 0$,  for every $n$,

$T$ has dense range.

Noticing that   $T-\lambda $ is a diagonal operator, whenever $T$ itself is, this shows that the residual spectrum
of a diagonal operator
is always  empty, so the conclusion follows just as easily as in the case of self-adjoint operators on Hilbert spaces.

EDIT.  Let me break down my answer in order to try to  make it a bit clearer.  Before I start,  let me say that the
arguments below are completely elementary and, in particular, make no use of  transposed operators.

Lemma.  Let  $1\leq p<\infty $,  and let
$T$ be a diagonal operator on $\ell ^p$,  namely an operator of the form
$$
  T(x_1, x_2, x_3, \ldots ) = (a_1x_1, a_2x_2, a_3x_3, \ldots ),
  $$
where $a = (a_1, a_2, a_3, \ldots ) \in  \ell ^\infty $.  Then the following are equivalent:

$T$ is injective,

$a_n\neq 0$,  for every $n$,

$T$ has dense range.

Proof.  (1) $\Rightarrow$ (2). For each $n$, let $e_n$ the $n^{th}$ canonical basis vector of
$\ell ^p$.  Then, since $T$ is injective, we have that
$$
  0\neq T(e_n) = a_ne_n,
  $$
so clearly $a_n\neq 0$.

(2) $\Rightarrow$ (1). Suppose that $T(x)=0$.  Then $a_nx_n=0$, for all $n$, and,  since $a_n\neq 0$, we deduce that $x_n=0$,
whence $x=0$, as well.

(2) $\Rightarrow$ (3). Observing that
$$
  T\left(\frac{e_n}{a_n}\right) = e_n,
  $$
we see that $e_n$ lies in the range of $T$.  Consequently also
$$
  \text{span}\{e_n:n\geq 1\}\subseteq  \text{Ran}(T),
  $$
so it follows that $\text{Ran}(T)$ is dense.

(3) $\Rightarrow$ (2).  For each $n$,  let $E_n$ be the subspace of $\ell ^p$ given by
$$
  E_n=\{x\in  \ell ^p: x_n=0\}.
  $$
Notice that $E_n$ is the kernel of the continuous linear functional
$$
  x\in  \ell ^p\mapsto x_n\in  \mathbb R,
  $$
so    $E_n$ is a proper closed subspace.

If $a_n=0$, then clearly
$$
  \text{Ran}(T)\subseteq E_n,
  $$
so $\text{Ran}(T)$  cannot be dense,    contradicting (3).  QED

Back to the question,  observe that if $T$ is a diagonal operator,  as in the above Lemma,  then so is $T-\lambda I$,  for
every scalar $\lambda $.  Therefore $T-\lambda I$ is injective iff it has a dense range.  We then conclude,  as in the case of
self-adjoint operators on Hilbert's space,  that the residual spectrum of $T$ is empty.


问:
Solving a Diophantine equation involving fourth powers and squares
I am trying to show that $x^4 - 3y^4=z^2$ has no solutions over the positive integers. I tried working on $\mathbb{Z}[\sqrt{3}]$ but couldn't go far. Factorizing didn't do much either.

答:
Consider a positive integer solution, with $y$ minimal, of

$$ x^4-3y^4=z^2\tag{1}$$

We can suppose that $x,y,z$ are pairwise coprime  since a common factor of any pair of $x,y,z$ would be a factor of all and cancellation could occur. Then, considered modulo $8$ we see that $y$ is even. We rewrite the equation as

$$\left (\frac{x^2-z}{2}\right )\left (\frac{x^2+z}{2}\right)=12t^4.$$
Since the two bracketed factors, $L$ and $M$ say, differ by the odd integer $z$ and have integer product, they are both integers. Furthermore, if $q$ is a prime common factor of $L$ and $M$, then $q$ would be a factor of both $x$ and $z$, a contradiction.

Therefore $\{L,M\}=\{au^4,cv^4\}$, where $ac=12$ and $t=uv$, with $u$ and $v$ coprime, and $$au^4+cv^4=x^2.$$ Since $x$ is odd the only possibility is
$u^4+12v^4=x^2.$ (Note that $4u^4+3v^4=x^2$ is not possible modulo $4$). Then
$$\left (\frac{u^2-x}{2}\right )\left (\frac{u^2+x}{2}\right)=-3v^4.$$
Since the two bracketed factors, $L$ and $M$ say, differ by the  integer $x$ and have integer product, they are both integers. Furthermore, if $q$ is a prime common factor of $L$ and $M$, then $q$ would be a factor of both $x$ and $u$, a contradiction.
Therefore $\{L,M\}=\{aU^4,cV^4\}$, where $ac=-3$ and $v=UV$, with $U$ and $V$ coprime, and $$aU^4+cV^4=u^2.$$ Modulo $3$ the only possibility is $\{a,c\}=\{1,-3\}.$ From a solution $(x,y,z)$ of  equation (1) we have therefore deduced a second solution $(X,Y,Z)$, where $$Y=\frac{y}{2UX}<y,$$ a contradiction.

NOTE

This is the method used on

Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution?


问:
Finding Points of Inflection from First Derivative Graph
If I am finding the inflection points of a function using the first derivative graph, I recognize that it exists where the first derivative changes from increasing to decreasing or vice versa. In the example below, however, the graph is stationary from 3 to 4, so what is my inflection point (imagine that it is a first derivative graph, despite the label)? Is it 3, since that is where the graph of the derivative is no longer decreasing? Or is it 4, since that is where the graph of the derivative begins to increase?

Example of my graph.

Thanks for your help.

答:
An inflection point is a point where the curve changes concavity, from up to down or from down to up.  It is also a point where the tangent line crosses the curve.  The tangent to a straight line doesn't cross the curve (it's concurrent with it.)  So none of the values between $x=3$ to $x=4$ are inflection points because the curve is a straight line.

Is $x=3$ and inflection point?  No.  The concavity doesn't change from up to down or down to up.  It changes from being there to not being there.  The 2nd-derivative doesn't exist at that point, so it can't be an inflection point.  It's a theorem that an inflection point is a point where the 2nd-derivative exists AND the curve changes concavity.  Ditto for $x=4$.

The only inflection point for your function is at $x=6.$


问:
Calculate the area of a $N \times N$ grid using a $2D$ random walk.
Is there a formula to determine the approx. steps needed to calculate the area of a $N \times N$ grid?

Suppose we have a discrete random walk on $\mathbb{Z}^2$ starting at $(0,0)$ and each "step" is determined by adding one of the vectors $(1,0), (−1,0), (0,1), (0,−1)$. The walk continues in the same manner for many steps and cannot step outside its boundary edges or corners.

Anytime the walk completes a unit square perimeter, it's added to the total area so far. This walk continues until every unit square has been completed. As the comment section noted, this is equivalent to walking every segment.

Example

A random walk on a $150\times150$ grid in progress...

EDIT

Finally wrote a C program to traverse $N \times N$ grids.

For example, the $10 \times 10$ grid is iterated $100$ times and the average steps taken.

```<code>10x10 Grid, Total Steps: 2455
10x10 Grid, Total Steps: 2206
10x10 Grid, Total Steps: 1872
10x10 Grid, Total Steps: 2390
10x10 Grid, Total Steps: 2044
10x10 Grid, Total Steps: 2411
10x10 Grid, Total Steps: 2976
10x10 Grid, Total Steps: 2718
10x10 Grid, Total Steps: 2669
10x10 Grid, Total Steps: 3641
10x10 Grid, Total Steps: 1879
10x10 Grid, Total Steps: 2288
10x10 Grid, Total Steps: 1906
10x10 Grid, Total Steps: 1688
10x10 Grid, Total Steps: 1857
10x10 Grid, Total Steps: 2571
10x10 Grid, Total Steps: 3085
10x10 Grid, Total Steps: 2274
10x10 Grid, Total Steps: 3288
10x10 Grid, Total Steps: 2096
10x10 Grid, Total Steps: 2415
10x10 Grid, Total Steps: 1850
10x10 Grid, Total Steps: 2113
10x10 Grid, Total Steps: 1926
10x10 Grid, Total Steps: 3149
10x10 Grid, Total Steps: 2189
10x10 Grid, Total Steps: 1873
10x10 Grid, Total Steps: 1662
10x10 Grid, Total Steps: 3776
10x10 Grid, Total Steps: 3109
10x10 Grid, Total Steps: 1723
10x10 Grid, Total Steps: 2088
10x10 Grid, Total Steps: 2726
10x10 Grid, Total Steps: 1897
10x10 Grid, Total Steps: 1935
10x10 Grid, Total Steps: 1719
10x10 Grid, Total Steps: 2358
10x10 Grid, Total Steps: 2501
10x10 Grid, Total Steps: 3118
10x10 Grid, Total Steps: 2870
10x10 Grid, Total Steps: 2459
10x10 Grid, Total Steps: 3090
10x10 Grid, Total Steps: 1862
10x10 Grid, Total Steps: 2370
10x10 Grid, Total Steps: 1905
10x10 Grid, Total Steps: 2063
10x10 Grid, Total Steps: 2255
10x10 Grid, Total Steps: 2484
10x10 Grid, Total Steps: 2861
10x10 Grid, Total Steps: 1535
10x10 Grid, Total Steps: 2026
10x10 Grid, Total Steps: 3210
10x10 Grid, Total Steps: 3116
10x10 Grid, Total Steps: 2013
10x10 Grid, Total Steps: 2204
10x10 Grid, Total Steps: 1668
10x10 Grid, Total Steps: 1614
10x10 Grid, Total Steps: 2347
10x10 Grid, Total Steps: 2694
10x10 Grid, Total Steps: 4518
10x10 Grid, Total Steps: 2213
10x10 Grid, Total Steps: 3105
10x10 Grid, Total Steps: 2483
10x10 Grid, Total Steps: 1900
10x10 Grid, Total Steps: 2732
10x10 Grid, Total Steps: 3267
10x10 Grid, Total Steps: 2392
10x10 Grid, Total Steps: 2540
10x10 Grid, Total Steps: 2805
10x10 Grid, Total Steps: 2872
10x10 Grid, Total Steps: 5039
10x10 Grid, Total Steps: 3851
10x10 Grid, Total Steps: 2909
10x10 Grid, Total Steps: 2081
10x10 Grid, Total Steps: 2621
10x10 Grid, Total Steps: 2561
10x10 Grid, Total Steps: 2296
10x10 Grid, Total Steps: 1655
10x10 Grid, Total Steps: 3922
10x10 Grid, Total Steps: 3123
10x10 Grid, Total Steps: 4437
10x10 Grid, Total Steps: 2631
10x10 Grid, Total Steps: 2175
10x10 Grid, Total Steps: 2606
10x10 Grid, Total Steps: 2214
10x10 Grid, Total Steps: 2831
10x10 Grid, Total Steps: 2565
10x10 Grid, Total Steps: 3157
10x10 Grid, Total Steps: 2268
10x10 Grid, Total Steps: 1704
10x10 Grid, Total Steps: 2817
10x10 Grid, Total Steps: 2695
10x10 Grid, Total Steps: 2102
10x10 Grid, Total Steps: 4982
10x10 Grid, Total Steps: 2618
10x10 Grid, Total Steps: 2219
10x10 Grid, Total Steps: 4198
10x10 Grid, Total Steps: 3987
10x10 Grid, Total Steps: 1868
10x10 Grid, Total Steps: 2375
10x10 Grid, Average Steps: 2564
</code>```

Similar grid and graph results follow:

```<code>10x10   Grid, Average Steps: 2564
20x20   Grid, Average Steps: 13829
30x30   Grid, Average Steps: 34849
40x40   Grid, Average Steps: 69704
50x50   Grid, Average Steps: 112535
60x60   Grid, Average Steps: 178182
70x70   Grid, Average Steps: 249028
80x80   Grid, Average Steps: 354800
90x90   Grid, Average Steps: 456223
100x100 Grid, Average Steps: 575570
</code>```

QUESTION

Given an $N \times N$ grid, is there a way to devise a formula to find the approx. number of random steps needed to compute its area based upon these program results?

Maybe a Lagrange Interpolating Polynomial like this:

$f(x) = \\ \begin{array}{c}  \frac{31381 x^9}{8640000000000}-\frac{1427173 x^8}{806400000000}+\frac{1549913 x^7}{4200000000}-\frac{123941687 x^6}{2880000000}+\frac{886369883 x^5}{288000000}-\frac{1602458819 x^4}{11520000}+\frac{17031670657 x^3}{4320000}-\frac{67368570503 x^2}{1008000}+\frac{1020468443 x}{1680}-2208005 \end{array}$

Thanks.

```<code>// Calculate the area of a N×N grid using a 2D random walk.

#include <stdio.h>
#include <stdbool.h>
#include <stdlib.h>
#include <time.h>
#include <random>

unsigned int GRID_SIZE;
std::minstd_rand simple_rand;

typedef struct
{
    int xF;
    int yF;
    int xT;
    int yT;
} FromTo;

typedef struct
{
    FromTo FT;
    bool walked;
} segment;

int xFrom, yFrom, xTo, yTo;

segment* xPaths[1000];
segment* yPaths[1000];

void PathsCheck(segment** Paths)
{
    int i=0, j=0;
    segment* Segments;

    for (i = 0; i < GRID_SIZE; i++)
    {
        Segments = Paths[i];

        for (j = 0; j < GRID_SIZE; j++)
        {
            // Check for reversible walks {(0,0),(1,0)} or {(1,0),(0,0)}
            if ((((Segments[j].FT.xF == xFrom) && (Segments[j].FT.yF == yFrom)) &&
                ((Segments[j].FT.xT == xTo) && (Segments[j].FT.yT == yTo))) ||
                (((Segments[j].FT.xT == xFrom) && (Segments[j].FT.yT == yFrom)) &&
                ((Segments[j].FT.xF == xTo) && (Segments[j].FT.yF == yTo))))
            {
                Segments[j].walked = true;
                return;
            }
        }
    }
}

int PathsStats(segment** Paths)
{
    int i = 0, j = 0, segmentsTraversed = 0;
    segment* Segments;

    for (i = 0; i < GRID_SIZE; i++)
    {
        Segments = Paths[i];

        for (j = 0; j < (GRID_SIZE - 1); j++)
        {
            if (Segments[j].walked)
            {
                segmentsTraversed++;
            }
        }
    }

    return segmentsTraversed;
}

void GenerateRandomWalk(void)
{
    // (0,3) (1,3) (2,3) (3,3)
    // (0,2) (1,2) (2,2) (3,2)
    // (0,1) (1,1) (2,1) (3,1)
    // (0,0) (1,0) (2,0) (3,0)

    xFrom = xTo;
    yFrom = yTo;

    do
    {
        switch (simple_rand() % 4)
        {
        case 0: if (yTo != (GRID_SIZE - 1)) yTo++; break; // up
        case 1: if (yTo != 0) yTo--; break;               // down
        case 2: if (xTo != 0) xTo--; break;               // left
        case 3: if (xTo != (GRID_SIZE - 1)) xTo++; break; // right
        }
    } while ((xFrom == xTo) && (yFrom == yTo));
}

void CreateXsegments(void)
{
    int x, y, i=0, j;
    segment* xSegments;

    // xSegments[0] = {(0,3),(1,3)} {(1,3),(2,3)} {(2,3),(3,3)}
    // xSegments[1] = {(0,2),(1,2)} {(1,2),(2,2)} {(2,2),(3,2)}
    // xSegments[2] = {(0,1),(1,1)} {(1,1),(2,1)} {(2,1),(3,1)}
    // xSegments[3] = {(0,0),(1,0)} {(1,0),(2,0)} {(2,0),(3,0)}

    for (y = (GRID_SIZE - 1); y >= 0; y--)
    {
        xSegments = (segment*)malloc((GRID_SIZE - 1) * sizeof(segment));
        if (!xSegments)
        {
            exit(-1);
        }

        xPaths[i++] = xSegments;

        j = 0;

        for (x = 0; x < (GRID_SIZE - 1); x++)
        {
            xSegments[j++] = { {x,y,x + 1,y},false };
        }
    }
}

void CreateYsegments(void)
{
    int x, y, i = 0, j;
    segment* ySegments;

    // ySegments[0] = {(0,3),(0,2)} {(0,2),(0,1)} {(0,1),(0,0)}
    // ySegments[1] = {(1,3),(1,2)} {(1,2),(1,1)} {(1,1),(1,0)}
    // ySegments[2] = {(2,3),(2,2)} {(2,2),(2,1)} {(2,1),(2,0)}
    // ySegments[3] = {(3,3),(3,2)} {(3,2),(3,1)} {(3,1),(3,0)}

    for (x = 0; x < GRID_SIZE; x++)
    {
        ySegments = (segment*)malloc((GRID_SIZE - 1) * sizeof(segment));

        if (!ySegments)
        {
            exit(-1);
        }

        yPaths[i++] = ySegments;

        j = 0;

        for (y = (GRID_SIZE - 1); y > 0; y--)
        {
            ySegments[j++] = {{x,y,x,y - 1},false};
        }
    }
}

int main(int argc, char *argv[])
{
    int WalksDone;
    int xWalks, yWalks;
    int TotalSteps, AvgSteps;
    int ctr,ctrmax=102,avg,avgmax=100;

    simple_rand.seed((unsigned int)time(0));

    // Compute Grids 10x10, 20x20, ... ,100x100

    for (ctr = 11; ctr < ctrmax; ctr += 10)
    {
        AvgSteps = 0;

        for (avg = 0; avg < avgmax; avg++)
        {
            GRID_SIZE = ctr;
            WalksDone = (GRID_SIZE * (GRID_SIZE - 1));

            TotalSteps = 0;

            xWalks = yWalks = 0;
            xFrom = yFrom = xTo = yTo = 0;

            CreateXsegments();
            CreateYsegments();

            do
            {
                TotalSteps++;

                GenerateRandomWalk();

                if (xWalks != WalksDone)
                {
                    PathsCheck(xPaths);
                    xWalks = PathsStats(xPaths);
                }
                if (yWalks != WalksDone)
                {
                    PathsCheck(yPaths);
                    yWalks = PathsStats(yPaths);
                }

            } while ((xWalks != WalksDone) || (yWalks != WalksDone));

            printf("%ux%u Grid, Total Steps: %d\n", GRID_SIZE - 1, GRID_SIZE - 1, TotalSteps);

            AvgSteps += TotalSteps;

            // Release memory....
            for (int i = 0; i < GRID_SIZE; i++)
            {
                free(xPaths[i]); xPaths[i] = NULL;
                free(yPaths[i]); yPaths[i] = NULL;
            }
        }

        printf("%ux%u Grid, Average Steps: %d\n", GRID_SIZE - 1, GRID_SIZE - 1, AvgSteps/avgmax);
    }

    return 0;
}
</code>```

答:
The number of steps needed for random walk to visit all vertices of an $n$ by $n$ square is asymptotic to $4n^2 \log^2 n/\pi$, see [1]. The upper bound was known earlier, and the result also applies to crossing all edges (with or without periodic boundary conditions.)

[1] Dembo, Amir, Yuval Peres, Jay Rosen, and Ofer Zeitouni. "Cover times for Brownian motion and random walks in two dimensions." Annals of mathematics (2004): 433-464.


问:
Why does continuity on $\mathbb{T}$ imply that $f(-\pi) = f(\pi)$?
In Vretblad's Fourier Analysis and its Applications, it says:

Suppose that $f \in C^1(\mathbb{T})$ , which means that both $f$ and its derivative $f'$ are continuous >on $\mathbb{T}$. We compute the Fourier coefficients of the derivative:

$$ \begin{aligned} c_n(f') &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} f'(t) \ e^{-int} \ dt \\
&= \frac{1}{2 \pi} \Big[ f(t) e^{-int} \Big]_{-\pi}^{\pi} - \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) (-in) e^{-int} dt  \\
&= \frac{1}{2 \pi} \Big( f(\pi)(−1)^n − f(−\pi)(−1)^n \Big) + in c_n(f) \\
&= in c_n(f) \end{aligned}$$
The fact that $f$ is continuous on $\mathbb{T}$ implies that $f(−\pi) = f(\pi)$.

So my questions is: how do we know that $f$ being continuous on $\mathbb{T}$ implies that $f(−\pi) = f(\pi)$?

答:
The book identifies a function that is $2\pi$-periodic as a function on the circle $\mathbf{T}$ (see Section 4.1 of the book):

So when $f$ is a function on the circle, it must be true that
$$
f(-\pi)=f(-\pi+2\pi)=f(\pi).
$$

"Continuity" is actually not quite relevant here, although the author mentioned it.


问:
Limits and substitutions
I have seen the following performed during analysis of the limiting behaviour of real functions. consider
$$
f(x) = \frac{1-x^a}{1-x^b}
$$
with integers $a,b$. At the point $x\rightarrow 1$ the function is undefined. However, the limit can be found from L''Hospital's rule as
$$
\lim_{x\rightarrow 1} f(x) = \frac{ax^{a-1}}{bx^{b-1}}=\frac{a}{b}\bigg|_{x=1}
$$
Now, at this point, a substitution is made such that $x$ is replaced by $\xi =-1/\log x$ to arrive at
$$
\frac{1-e^{-a/\xi}}{1-e^{-b/\xi}}
$$
I am aware that $\log(1) = 0$; however, why can we substitute this expression in, and can any relevant function with the desired behaviour at $x=1$ be used?

答: One of the reasons we like continuous functions is that they preserve limits of functions, so if $f$ is continuous, then $\lim_{x\to a} f(g(x)) = f(\lim_{x\to a}g(x))$ (assuming the limit of $g$ exists). That means in particular, that if $-1/\log{x}$ applied to our function has some nice properties, then you can use it just as you noticed.

Title: Limits and substitutions


答:
One general theorem is following: suppose exists, finite or not, limits $\lim\limits_{x \to a}f(x)=b$ and $\lim\limits_{y \to b}F(x)$. If in some punctured(deleted) neighbourhood of $a$ holds $f(x) \ne b$, then in point $a$ exists limit of composition and holds
$$\lim\limits_{x \to a}F(f(x))=\lim\limits_{y \to b}F(x)$$

In your case, assuming $a,b \gt 1, a\ne b$, after applying L'Hospital you can directly put $x=1$ without any substitution.

If we want to use substitution of variables, without L'Hospital, then, for example, we can write
$$f(x) = \frac{1-x^a}{1-x^b} = \frac{1-(1+t)^a}{1-(1+t)^b}=\frac{\frac{1-(1+t)^a}{t}}{\frac{1-(1+t)^b}{t}}$$
and use limit $\lim\limits_{t\to 0}\frac{(1+t)^a-1}{t}=a$ for substitution $x=1+t$

Title: Limits and substitutions

Body:
I have seen the following performed during analysis of the limiting behaviour of real functions. consider
$$
f(x) = \frac{1-x^a}{1-x^b}
$$
with integers $a,b$. At the point $x\rightarrow 1$ the function is undefined. However, the limit can be found from L''Hospital's rule as
$$
\lim_{x\rightarrow 1} f(x) = \frac{ax^{a-1}}{bx^{b-1}}=\frac{a}{b}\bigg|_{x=1}
$$
Now, at this point, a substitution is made such that $x$ is replaced by $\xi =-1/\log x$ to arrive at
$$
\frac{1-e^{-a/\xi}}{1-e^{-b/\xi}}
$$
I am aware that $\log(1) = 0$; however, why can we substitute this expression in, and can any relevant function with the desired behaviour at $x=1$ be used?


答:
I tend to nod agreeably and say that math isn't instinctive and takes a lot of time to learn, but that I was fortunate to have positive early experiences with math.  Maybe I'll say a few words about what I think were key events (or key things that didn't happen--like never had a teacher who turned it into confusing drudgery), suggest that most people don't have as good of experiences, and suggest that teachers and early reinforcement is really important.

That leaves the conversation in a nice flexible position: I haven't tried to make the person feel stupid, I haven't claimed that math is hard for me (it was nonintuitive to learn, but it was fun and pretty easy, so it was no problem playing with it until I developed some intuition), and I've given easy paths to talk about either the positive aspects of math, the problems with math teaching, or any positive early experience the other person might have had.

The main problem is saying too much: the person probably wasn't looking for a thesis on math ability.  I haven't figured out how to do this briefly enough to flow quickly in casual conversation.  (But I haven't had much practice either--most people I talk to are not bad at math.)

Title: How do you respond to "I was always bad at math"?

Body:
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.


问:
Proof of inequality about $\cot \pi z$.
I am trying to show that for $z=x+iy$ where $|y|\geq 1$,
$$|\cot\pi z|\leq\frac{1+\exp(-2\pi|y|)}{1-\exp(-2\pi|y|)}.$$

Here is my work so far:
\begin{align*}
|\cot\pi z|&=\left\vert\frac{\exp(i\pi z)-\exp(-i\pi z)}{\exp(i\pi z)-\exp(-i\pi z)}\right\vert\\
&=\left\vert\frac{\exp(i\pi x)\exp(-\pi y)+\exp(-i\pi x)\exp(\pi y)}{\exp(i\pi x)\exp(-\pi y)-\exp(-i\pi x)\exp(\pi y)}\right\vert\\
&=\left\vert\frac{\exp(-2\pi y)+\exp(-2i\pi x)}{\exp(-2\pi y)-\exp(-2i\pi x)}\right\vert\\
&\leq\frac{1+\exp(-2\pi y)}{1-\exp(-2\pi y)},
\end{align*}
where the last step is from the triangle inequality in the numerator and the reverse triangle inequality in the denominator. It appears that I have lost the absolute value sign of $y$; did I do anything wrong?

答:
The last step in your calculation is wrong for negative $y$. You have
$$
 |\exp(-2\pi y) + \exp(-2i\pi x)| \le 1 + \exp(-2\pi y) \\
 |\exp(-2\pi y) - \exp(-2i\pi x)| \ge 1 - \exp(-2\pi y)
$$
but the right-hand side of the second inequality is negative if $y < 0$, so that you cannot divide the inequalities.

But your calculation is correct for $y > 0$. For negative $y$ you can write
\begin{align*}
|\cot\pi z|&=\left\vert\frac{\exp(i\pi z)-\exp(-i\pi z)}{\exp(i\pi z)-\exp(-i\pi z)}\right\vert\\
&=\left\vert\frac{\exp(i\pi x)\exp(-\pi y)+\exp(-i\pi x)\exp(\pi y)}{\exp(i\pi x)\exp(-\pi y)-\exp(-i\pi x)\exp(\pi y)}\right\vert\\
&=\left\vert\frac{\exp(2i\pi x)+\exp(2\pi y)}{\exp(2i\pi x)-\exp(2\pi y)}\right\vert\\
&\leq\frac{1+\exp(2\pi y)}{1-\exp(2\pi y)} = \frac{1+\exp(-2\pi |y|)}{1-\exp(-2\pi |y|)}
\end{align*}
behause then $1-\exp(2\pi y) > 0$.

Alternatively, use that $|\cot |$ is an even function, so that for $y < 0$
$$
|\cot \pi (x+iy)| = |\cot \pi (-x-iy)| \leq\frac{1+\exp(-2\pi (-y))}{1-\exp(-2\pi (-y))}
= \frac{1+\exp(-2\pi |y|)}{1-\exp(-2\pi |y|)} \, .
$$

Note that this proves
$$
|\cot\pi z|\leq\frac{1+\exp(-2\pi|y|)}{1-\exp(-2\pi|y|)}.
$$
for all $z=x+iy$ with $y \ne 0$, not only for $|y| \ge 1$.

Title: Proof of inequality about $\cot \pi z$.