问:
Simple Module Over a Product Ring
I am trying to prove the following:

If $A$ and $B$ are commutative, unital rings and $M$ is a simple module over $A \times B$. Then $M$ is a simple module over $A$ or $B$.

Suppose that $M$ is not a simple module over $B$. Then there exists a proper sub-$B$-module $N_0 \subseteq M$. Let $N_1 \subseteq M$ be a non-zero sub-$A$-module. I want to argue that $N_1 =M$, thereby proving $M$ is a simple $A$-module. I was thinking of showing that either $N_0 + N_1$ or $N_0 \cap N_1$ are non-zero sub-$A \times B$-modules, but I don't see how to do this. I don't even know if this is the right approach.

答:
As Eric Wofsey points out, there is no guarantee that we can transport an $A\times B$-module structure on $M$ to an $A$-module or $B$-module structure on $M$. This is related to that fact that, if we assume ring homomorphisms are unital, there are not canonical ring homomorphisms $A\rightarrow A\times B$ or $B\rightarrow A\times B$ in general. Indeed, if we have any ring morphism $f:R\rightarrow S$, there will in general be a natural way to turn an $S$-module into an $R$-module (let $r\cdot m=f(r)\cdot m$), but not the other way around.

Now, this is the statement in the chat that you linked to:

"if we have a simple module $M$ over a product ring $A\times B$, then one can show that either $AM = 0$ or $BM=0$ and in the first case $M$ is a simple $B$-module and in the second case a simple $A$-module"

So, the content of what Lukas is saying is that, if $M$ is a simple $A\times B$-module, then there will be a way of transporting the $A\times B$-module structure on $M$ to either an $A$-module structure or a $B$-module structures on $M$, and that the induced module structure in question will be simple. Let's prove it. We take $M$ to be a simple left $A\times B$-module.

Now, we cannot realize $A$ or $B$ as unital subrings of $A\times B$, but we can realize them as the respective ideals $A\times\{0\}$ and $\{0\}\times B$ of $A\times B$. We will use $\tilde{A}$ to refer to $A\times\{0\}\leqslant A\times B$ and $\tilde{B}$ to refer to $\{0\}\times B\leqslant A\times B$; you may have been confused because Lukas was referring to $\tilde{A}$ as "$A$" and $\tilde{B}$ as "$B$".

In general, if $R$ is a ring, $M$ is a (left) $R$-module, and $I$ is a (left) ideal of $R$, we let $I\cdot M$ be the submodule of $M$ generated by the set $\{i\cdot m:i\in I,m\in M\}$. You should check yourself that $I\cdot M$ is equivalently the abelian subgroup of $M$ generated by this set; this fact follows immediately from the fact that $I$ is closed under left-multiplication by elements of $R$. In any case, this construction behaves well with respect to addition of ideals: if $J$ is another left ideal of $R$, then we have $I\cdot M+J\cdot M=(I+J)\cdot M$. (Verify this yourself!) Thus, because $A\times B=\tilde{A}+\tilde{B}$, and because $M$ is non-zero, we cannot have $\tilde{A}\cdot M=\{0\}=\tilde{B}\cdot M$. Hence without loss of generality we may suppose $\tilde{B}\cdot M\neq 0$. Since $M$ is simple, this therefore means that $\tilde{B}\cdot M=M$.

We claim now that $\tilde{A}\cdot M=\{0\}$; this follows immediately from the fact that $\tilde{A}\tilde{B}=\{0\}$. Indeed, for any $m\in M$, because $M=\tilde{B}\cdot M$, there are some $(0,b_1),\dots,(0,b_k)\in\tilde{B}$ and $m_1,\dots,m_k\in M$ such that $m=\sum_{i=1}^k(0,b_i)\cdot m_i$. Then, for any $a\in A$, we have $$(a,0)\cdot m=(a,0)\cdot\sum_{i=1}^k(0,b_i)m_i=\sum_{i=1}^k(a,0)\cdot(0,b_i)m_i=\sum_{i=1}^k(0,0)\cdot m_i=0,$$ as desired.

Now we are ready to show that the action $b\cdot m:=(0,b)\cdot m$ makes $M$ into a $B$-module. As I mentioned in my comment, the issue you run into without the assumption that $M$ is a simple $A\times B$-module is that there is no guarantee that $(0,1)\cdot m$ equals $m$. In fact, if you look at the list of axioms for a module, it should be easy to verify all but that one, even in the case where $M$ is not simple. Go through all of these axioms and carefully check them yourself!

If $M$ is simple though, that final axiom does hold, and we can show it using the fact that $\tilde{A}\cdot M=\{0\}$. Indeed, let $m\in M$. Because $M$ is an $A\times B$-module, we have $(1,1)\cdot m=m$. On the other hand, we know that $(1,0)\cdot m=0$, so this means that $$(0,1)\cdot m=((1,1)-(1,0))\cdot m=(1,1)\cdot m-(1,0)\cdot m=m.$$ Thus $(0,1)$ acts as the identity on $M$ and so declaring $b\cdot m=(0,b)\cdot m$ indeed gives a well-defined $B$-module structure on $M$.

Finally, we just need to show that, equipped with this structure, $M$ is a simple $B$-module. This again follows immediately from the fact that $\tilde{A}\cdot M=\{0\}$. Indeed, let $m\in M$ be non-zero, and let $n\in M$ be arbitrary. We wish to find an element $b\in B$ such that $n=b\cdot m=(0,b)\cdot m$. Because $M$ is a simple $A\times B$-module, there is an element $(a,b)\in A\times B$ such that $(a,b)\cdot m=n$. On the other hand, we know that $(a,0)\cdot m=0$, and so we have $$(0,b)\cdot m=((a,b)-(a,0))\cdot m=(a,b)\cdot m-(a,0)\cdot m=n,$$ just as desired.

Title: Simple Module Over a Product Ring


问:
What is the value of $e^\frac{-1}{e}$?
As the title suggests, what is the value of $e^\frac{-1}{e}$?? I never understood how things to the power of irrational numbers are calculated. The more broad question would be what is the minimum value for the function $y=x^x$, such that $x \geq 0$? I have tried taking the derivative to get $x^x(\ln(x)+1) = 0$, then simplifying to get $\ln(x)+1=0$, and $\ln(x)=-1$, so $x = \frac{1}{e}$. This simplifies to $e^\frac{-1}{e}$. What do I put for my final answer? Any help would be appreciated.

答:
As the replies indicate, you should put $e^{\frac{-1}{e}}$ since there is no simpler form.

Nevertheless, you seem to seek an intuition into irrational exponents, namely $\frac{-1}{e}$.

The basic idea is that every real number is the limit of a sequence of rational numbers, in this case with the help of the Taylor series of $e^x$
$$\frac{-1}{e}=-e^{-1}=-\sum_{n=0}^\infty \frac{(-1)^n}{n!}
= \sum_{n=0}^\infty \frac{(-1)^{n+1}}{n!}
=-1+\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+\cdots$$

Which follows
$$ \exp\left(\frac{-1}{e}\right)
= \exp\left(\sum\limits_{n=0}^\infty \frac{(-1)^{n+1}}{n!} \right)
= \prod\limits_{n=0}^\infty \exp\left( \frac{(-1)^{n+1}}{n!} \right)
= e^{-1}\cdot e^{\frac{1}{1!}}\cdot e^{-\frac{1}{2!}}\cdot e^{\frac{1}{3!}}\cdot e^{-\frac{1}{4!}}\cdot\dots
$$
Which translates into a product of roots of $e$ and $\frac{1}{e}$.

$$e^{\frac{-1}{e}} = e^{-1}\cdot e\cdot \sqrt{e^{-1}}\cdot \sqrt[3!]{e}\cdot
\sqrt[4!]{e^{-1}}\cdot \ldots$$

While the obtained expression might make sense, it is not mathematically significant.

However, one might use this expression to write a small program to compute $e^\frac{-1}{e}$ in the desired precision.

Check out Real exponents on Wikipedia.

ContentLicense: CC BY-SA 4.0

Title: What is the value of $e^\frac{-1}{e}$?


问:
Find image $~u^2 + v^2 \leq 1, ~~x = au ,~~ y = bv$
I'm trying to find the image of $~u^2 + v^2 \leq 1,~~ x = au ,~~ y = bv$

I'm pretty sure what I did is wrong, but I don't know why or how to know if the answer is correct.

I know $u^2 + v^2 \leq 1$ is a circle, $r = 1$.

$$ u \leq \sqrt{1-v^2}$$

$$x = a\sqrt{1-v^2}$$

$$v = \frac{b}{y}$$

$$\text{img} = x =a\sqrt{1-(\frac{b}{y})^2}$$

答:
If you let $x = au$ and $y = bv$, the proposed relation becomes
\begin{align*}
\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}\leq 1
\end{align*}

which is the closure of an ellipse. Are you acquainted to such locus?


问:
Natural Deduction Proof: {A v B, ¬A v C} ⊢ B v C
Prove using natural deduction: $(A \lor B), (\lnot A \lor C) ⊢ (B \lor C)$.
My work so far:

```<code>1     A v B
2     A v C
3      A      :AS
4       ¬A      :AS
5       _|_     :¬E 3,4
6    ¬¬A      :¬I 4-5
7  --
8  ¬A     :AS
</code>```

This is the work that I have come up with, for the past 5 hours. Can you please help me?

答:
You are setting up for disjunction elimination, so when you derive contradiction, use explosion to derive the target conclusion.$$\def\fitch#1#2{~~\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{~~1.~A\lor B\hspace{10ex}\textsf{Premise}\\~~2.~\lnot A\lor C\hspace{8.5ex}\textsf{Premise}}{\fitch{~~3.~A\hspace{12.5ex}\textsf{Assumption}}{\fitch{~~4.~\lnot A\hspace{8.5ex}\textsf{Assumption}}{~~5.~\bot\hspace{10ex}\textsf{Negation Elimination}\\~~6.~B\lor C\hspace{5.75ex}\textsf{Explosion (aka }\textit{Ex Falso Quodlibet}\textsf{)}}\\~~\vdots\\~~m.~B\lor C }\\~~\vdots\\~n.~B\lor C}$$

I am sure you can fill out the rest of this "proof by cases".


问:
Number Theory : Solve the system of congruence $28x+17y\equiv 18 \pmod{41}$ and $31x+11y\equiv 35\pmod{41}$
Number Theory : Solve the system of congruence

(1)    $28x+17y\equiv 18 \pmod{41}$

(2)   $31x+11y\equiv 35\pmod{41}$

Attempt :

we know that equation (1) and (2) are in the same$\pmod{41}$. so we can use Modular arithmetic

lets multiply the first equation (1) by $31$ and the second equation(2) by $28$.

(1)    $31\cdot(28x+17y)\equiv 31\cdot18 \pmod{41}$

(1) $868x+527y\equiv 558\pmod{41}$

(2)   $28\cdot(31x+11y)\equiv 28\cdot35\pmod{41}$

(2) $868x+308y\equiv 980\pmod{41}$

So finally we can subtract equation (1) from (2) we get:

$219y\equiv -422\pmod{41}$

lets check the $\gcd(219,41)$ by Euclidian algorithm :

$219 = 41\cdot 5 + 14$

$41= 14\cdot 2 + 13$

$14= 13\cdot 1 + 1$

$\gcd(219,41)=1$

Hence, because the $\gcd$ is equal to $1$ we can find the Inverse and multiply the equation by the Inverse to find $y$.

$219y\equiv 1\pmod{41}$

$219a = 1+41k$ , the $41k$ must end with the digit of $8$ for $1+$digit $8$ will be $9$ so $k$ must be multiply of number with end digit of $8$.

I don't know how to continue from here .

Body:
$$-13x+17y\equiv 18 \pmod{41}\tag1$$
$$-10x+11y\equiv -6 \pmod{41}\tag2$$

$(1)\times 10-(2)\times 13$
$$
27y\equiv 12 \pmod{41} \implies y \equiv \frac{12}{27} \equiv \frac{4}{9} \equiv \frac{20}{45} \equiv \frac{20}{4} \equiv 5 \pmod{41}
$$

$$
x \equiv \frac{11y+6}{10}\equiv \frac{61}{10} \equiv \frac{20}{10} \equiv 2 \pmod{41}.\blacksquare
$$


问:
How would I geometrically prove this situation of a rectangle and a triangle? I.e., how is theta used twice?
Here is a situation of Euclidian geometry, a proof of some sorts.

I am quite unsure how to explain what was done. This is from an AP Physics B (Part 1) prep book, and this was the first step in the answering process, which made too many logical steps. The prompt was this:

The prompt.

What I am asking of is theta; I noted that it was used twice, and I too am aware that some geometric proof is involved in relating the two triangles.

I am sorry for the vagueness of this situation, and I hope the information is sufficient enough. How would I prove that theta of the top triangle is theta on the side? (Perhaps this should be moved to the Physics exchange if too vague in jargon.)

答:
Soo, it sounds like you just need to be convinced that both angles marked as $\theta$ are indeed equal? This is because the two triangles they are parts of are similar:

They clearly both have a right angle.

The unmarked angle in the upper triangle is $180 - (90+\theta) = 90-\theta$, since the triangle's angles sum to $180$. The unmarked angle in the lower triangle makes $180$ degrees together with the original $\theta$ (from the upper triangle) and a right angel; so that is also $90-\theta$.

And so the third angle of the two triangles must be equal as well.


问:
The sum $(1+i)^x + (1-i)^x = 0$
I recently saw that $(1+i)^x + (1-i)^x = 0$  for $x = 2\pmod 4$, but can't figure out how its true. Any help/hints would be greatly appreciated!

Title: The sum $(1+i)^x + (1-i)^x = 0$

Body:
Notice that

$$1+i =\sqrt{2}e^{i\frac{\pi}{4}}$$
$$1-i =\sqrt{2}e^{-i\frac{\pi}{4}}$$

for $x=4n+2$, $n\in\mathbb{N}$, we have

$$
(1+i)^x+(1-i)^x=2^{2n+1}\Big(e^{in\pi+i\frac{\pi}{2}} + e^{-in\pi-i\frac{\pi}{2}} \Big)=2^{2n+1}\Big((-1)^ni-(-1)^ni\Big)$$


Title: The sum $(1+i)^x + (1-i)^x = 0$

Body: I recently saw that $(1+i)^x + (1-i)^x = 0$  for $x = 2\pmod 4$, but can't figure out how its true. Any help/hints would be greatly appreciated!

Body:
If $x \equiv 2\pmod 4$, then let $x = 4n + 2$ for some integer $n$.

$$\begin{align*}
(1+i)^x + (1-i)^x &= (1+i)^{4n+2} + (1-i)^{4n+2}\\
&= \left[(1+i)^2\right]^{2n+1} + \left[(1-i)^2\right]^{2n+1}\\
&= \left(1+2i + i^2\right)^{2n+1} + \left(1-2i + i^2\right)^{2n+1}\\
&= (2i)^{2n+1} + (-2i)^{2n+1}\\
&= (2i)^{2n+1} - (2i)^{2n+1}\\
&= 0
\end{align*}$$


问:
The sum of $(1+i)^x + (1-i)^x$ in a proof
My friend sent me an interesting question:
For all non-negative integers $x$, show that:

$$(1+i)^x + (1-i)^x = 2 \sum_{j=0}^{\lfloor x/2\rfloor}\binom{x}{2j}(-1)^j$$

I've tried induction on $x$, but that didn't seem to do the trick. Some hints as to how to approach this proof would be greatly appreciated.

答:
The explicit steps in processing the above relation -- which @xpaul has left out, as they are quite simple -- are as follows:

$$\begin{align}
&(1+\mathrm{i})^n+(1-\mathrm{i})^n\\
&=\sum_{k=0}^n\binom{n}{k}\left(\mathrm{i}^k+(-\mathrm{i})^k\right)\\
&=\sum_{\substack{0 \leqslant k \leqslant n\\k \in 2\mathbb{N}}}\binom{n}{k}\left(\mathrm{i}^k+(-\mathrm{i})^k\right)+\sum_{\substack{0 \leqslant k \leqslant n\\k \in 2\mathbb{N}+1}}\binom{n}{k}\left(\mathrm{i}^k+(-\mathrm{i})^k\right)\\
&=\sum_{h=0}^{\left[\frac{n}{2}\right]}\binom{n}{2h}\left(\mathrm{i}^{2h}+(-\mathrm{i})^{2h}\right)\\
&=\sum_{h=0}^{\left[\frac{n}{2}\right]}\binom{n}{2h}\left((-1)^h+(-1)^h\right)\\
&=2\sum_{h=0}^{\left[\frac{n}{2}\right]}(-1)^h\binom{n}{2h},
\end{align}$$
where we need to bear in mind two important justifications for the above:

for odd $k \in \mathbb{N}$ we obviously have $\mathrm{i}^k+(-\mathrm{i})^k=\mathrm{i}^k-\mathrm{i}^k=0$

the map:
$$\begin{align}
\left\{h \in \mathbb{N} \mid h \leqslant \left[\frac{n}{2}\right]\right\} &\to \{k \in \mathbb{N} \mid 2 \mid k \wedge k \leqslant n \}\\
h&\mapsto 2h
\end{align}$$
is a bijection on the grounds of which we can perform a bijective change of summation index.

P.S. The relation that you seek to prove is valid in a more general setting, as it relies on the polynomial identity:
$$(X+\mathrm{i}Y)^n+(X-\mathrm{i}Y)^n=2\sum_{k=0}^{\left[\frac{n}{2}\right]}(-1)^k\binom{n}{2k}X^{n-2k}Y^{2k},$$
considered in the polynomial ring $\mathbb{C}[X, Y]$.

However, given your particular choice of complex numbers $1\pm \mathrm{i}$ it would be a pity to pass the occasion of noticing a remarkable property they exhibit and which enables us to give yet another description of the general sum you are studying. For simplicity let us write $\varepsilon=\frac{1+\mathrm{i}}{\sqrt{2}}$ and notice that:
$$\varepsilon^2=\frac{1+\mathrm{i}^2+2\mathrm{i}}{2}=\mathrm{i},$$
which furthermore entails the relations $\varepsilon^4=-1$ respectively $\varepsilon^8=1$. In other words, $\varepsilon$ is a so-called primitive $8$-th root of unity.

With this in mind we infer that:
$$\begin{align*}
(1+\mathrm{i})^n+(1-\mathrm{i})^n &=\left(\sqrt{2}\varepsilon\right)^n+\left(\sqrt{2}\overline{\varepsilon}\right)^n\\
&=\sqrt{2}^n\left(\varepsilon^n+\frac{1}{\varepsilon^n}\right)\\
&=\sqrt{2}^n\frac{\varepsilon^{2n}+1}{\varepsilon^n}\\
&=\sqrt{2}^n\frac{\mathrm{i}^n+1}{\varepsilon^n}\\
&=\sqrt{2}^n\begin{cases}
2, &n \equiv 0 \pmod{8}\\
\sqrt{2}, &n \equiv 1, 7  \pmod{8}\\
0, &n \equiv 2, 6 \pmod{8}\\
-\sqrt{2}, &n \equiv 3, 5 \pmod{8}\\
-2, &n \equiv 4 \pmod{8}
\end{cases}\\
&=\begin{cases}
2^{\frac{n}{2}+1}, &n \equiv 0 \pmod{8}\\
2^{\frac{n+1}{2}}, &n \equiv 1, 7 \pmod{8}\\
0, &n \equiv 2, 6 \pmod{8}\\
-2^{\frac{n+1}{2}}, & n \equiv 3, 5 \pmod{8}\\
-2^{\frac{n}{2}+1}, & n \equiv 4 \pmod{8},
\end{cases}
\end{align*}$$

relations from which you can in particular obtain an answer to the related question you posted separately on this forum.


问:
How many trees of $e$ edges in a labelled clique
Given a labelled clique $G=(E,V)$, I would like to enumerate the number of trees that can be made for a given number of edges $e$, such that the all nodes in the tree are connected.

For instance, let $G$ be the 4-clique graph. I would like to find the number of trees with 3 edges that can be formed as subgraphs.

I have tried to research this, but couldn't really get to grips with the literature. This MSE post looks very similar to my question, but I am struggling with the answer.

Here is my attempt at this problem. Firstly we calculate the total number of ways of selecting 3 edges from the 4-clique. There are 6 edges in total. There are $6\cdot 5 \cdot 4 = 120 $ different ways to make the choices; however, a lot of these are equivalent to one another. In fact, given any three choices of edge, $\{e_1,e_2,e_3\}$, the order that they are picked does not matter. So we must divide 120 by the number of permutations of a 3 letter word, which is 6. Hence, the number of graphs that can be made from the choice of any 3 edges from a 4-clique is $120/6= 20$ which is also the binomial coefficient

$$
\binom{6}{3}
$$

The number of trees must of course be less than this number. My naive approach is to think that a tree of size 3 will always contain 4 nodes. Research has shown that there are 16 such trees in the case of the 4 clique by employing
$$
n^{n-2}=4^2=16
$$
However, this method is only valid in this case. For larger cliques, how can I find the number of trees with $e$ edges, where $e\neq n-1$?

The particular case of $e=3$ is my main objective for arbitrary $n$.

答:
A tree with $e$ edges has $e+1$ nodes. Given a set of $n$ nodes, there are $\binom n{e+1}$ ways to choose the set of nodes for the tree, and then by Cayley's formula there are $(e+1)^{e-1}$ trees on that set of nodes, so the final answer is
$$\binom n{e+1}(e+1)^{e-1}.$$


问:
K-theory proof of index theorem - some minor confusion
I am trying to understand the general approach to the $K$-theory proof of the Atiyah-Singer index theorem, using this https://arxiv.org/pdf/math/0504555.pdf paper. I ran into some confusion on page 29, where the following is stated:

"It only remains to show that the analytic index commutes with the Thom isomorphism $\phi:K(X)\to K(V)$ where $V$ is a complex vector bundle over $X$. [...] This problem is considerably simplified if we consider trivial bundles which can be expressed as the product $V = X \times\mathbb{R}^n$."

On the same page, it goes on to consider a vector bundle $Y$ which seems to be the associated bundle of some principal $G$-bundle, but the author again considers $P\times_{O(n)} \mathbb{R}^n$, that is, a real vector bundle. I don't quite understand how this makes sense, if we want to prove something for complex vector bundles. I get that we can view a complex vector bundle as a real vector bundle by just "forgetting" about the complex structure, but since the Thom isomorphism (at least in the paper) is only defined for complex vector bundles, I think I am missing something more important. I cannot quite put my finger on it, so if someone could explain the construction on page 29, that would be greatly appreciated.

答:
Recall that if $X$ and $Y$ be are compact smooth manifolds and  $i\colon X\hookrightarrow Y$ and is a smooth embedding, we want to define a "shriek map":

$$i_!\colon K_c(TX)\to K_c(TY),$$
where $K_c$ is $K$-theory with compact supports.

The first step (cf. p. 16 of the G. Landweber's article or pp. 497-8 of the original M. Atiyah and I. Singer's The Index of Elliptic Operators: I) is to take take a tubular neighborhood $N\subseteq Y$ of $X$. You can identify it with the normal bundle $N\to X$, which is of course a real vector bundle over $X$. Now observe that $Ti\colon TX\to TY$ is an embedding and that $TN$ is the tubular neighborhood of $TX$. In other words: $TN\to TX$ is a real vector bundle.

But we can say even more. It turns out, that if $\pi\colon TX\to X$ is the projection, then $TN\simeq \pi^*(N\oplus N)$. As $N\oplus N\to X$ can be treated as a complex vector bundle (namely, $N\otimes_\mathbb R \mathbb C)$, we conclude that $TN\to TX$ can be treated as a complex vector bundle as well. In particular it makes sense to consider the Thom homomorphism $K_c(TX)\to K_c(TN)$.

The excision axiom allows us to define the "analytical index" for $N$ as a map $K_c(TN)\to \mathbb Z$. (Note that this "analytical index" is defined via embeddings into compact manifolds, so its meaning is different than in the compact case). We want to show that this analytical index commutes with the Thom homomorphism defined above. To do that we observe that $N$, as a normal bundle over $X$, can be written as $P\times_{O(n)} \mathbb R^n$, where $P$ is a principal $O(n)$-bundle and $X=P/O(n)$. Then one uses the multiplicative axiom of the analytical index. (This is the most advanced part of the proof and in fact motivates the use of equivariant $K$-theory in this case.
However, if $N$ is a trivial bundle, $O(n)$ can be replaced by the trivial group $1$, and the equivariance is not needed. Similarly, for orientable $X$, it suffices to consider the group $SO(n)$, what slightly simplifies the proof).


答:
Just to give a variation on Oliver Diaz' answer. Recall that the modulus of a product is the product of the moduli, and argument of a product is the sum of arguments. You can easily see here that
$$
\arg(1+i) = \frac\pi4, |i+1| = \sqrt2
$$
and
$$
\arg(1-i) = -\frac\pi4, |i+1| = \sqrt2
$$
Therefore
$$
\arg(1+i)^x = \frac\pi 4 x, |(i+1)^x| = (\sqrt2)^x
$$
and
$$
\arg(1-i)^x = -\frac\pi 4 x, |(i-1)^x| = (\sqrt2)^x
$$
Now think about when the sum of these will cancel out.

 They need to have equal modulus, and there argument must differ by an odd multiple of $\pi$.

Title: The sum $(1+i)^x + (1-i)^x = 0$


问:
Definite integration of a trigonometric function

How to integrate $$\int_0^{\pi/2}\!\dfrac{2a \sin^2 x}{a^2
\sin^2 x +b^2 \cos^2 x}\,dx $$

my first step is $$\frac{2}{a} \int_0^{\pi/2}\!\dfrac{a^2 \sin^2 x}{a^2 +(b^2 - a^2) \cos^2 x}\, dx $$

I would kind of want to do some sort of $u=\cos x$ substitution, to get at $\arctan u $ but no idea what to do with the sine in the numerator.

答:
Often, this type of integral is workable using the substitution
$$
z=\tan(x/2)\quad\text{and}\quad\mathrm{d}x=\frac{\mathrm{2\,d}z}{1+z^2}\\
\sin(x)=\frac{2z}{1+z^2}\quad\text{and}\quad\cos(x)=\frac{1-z^2}{1+z^2}
$$
A couple of identities that simplify this integral are
$$
\sin^2(x)=\frac{1-\cos(2x)}{2}\quad\text{and}\quad\cos^2(x)=\frac{1+\cos(2x)}{2}
$$
Then, assuming $a,b\gt0$,
$$
\begin{align}
\int_0^{\pi/2}\frac{2a\sin^2(x)}{a^2\sin^2(x)+b^2\cos^2(x)}\,\mathrm{d}x
&=\int_0^{\pi/2}\frac{2a(1-\cos(2x))}{a^2(1-\cos(2x))+b^2(1+\cos(2x))}\,\mathrm{d}x\\
&=\int_0^\pi\frac{a(1-\cos(x))}{(b^2+a^2)+(b^2-a^2)\cos(x)}\,\mathrm{d}x\\
&=\int_0^\infty\frac{az^2}{b^2+a^2z^2}\frac{2\mathrm{d}z}{1+z^2}\\
&=\int_{-\infty}^\infty\frac{az^2}{b^2+a^2z^2}\frac{\mathrm{d}z}{1+z^2}\\
&=2\pi i\left(\frac1{2i}\frac{a}{a^2-b^2}-\frac1{2i}\frac{b}{a^2-b^2}\right)\\
&=\frac\pi{a+b}
\end{align}
$$
where we use the contour running along the real axis from $-\infty$ to $\infty$, circling back in the upper half-plane. The residue at $z=i$ is $\dfrac1{2i}\dfrac{a}{a^2-b^2}$ and at $z=i\dfrac{b}{a}$ is $-\dfrac1{2i}\dfrac{b}{a^2-b^2}$.

Removing the assumption on the signs of $a$ and $b$, we get
$$
\int_0^{\pi/2}\frac{2a\sin^2(x)}{a^2\sin^2(x)+b^2\cos^2(x)}\,\mathrm{d}x
=\mathrm{sgn}(a)\frac\pi{|a|+|b|}
$$

ContentLicense: CC BY-SA 3.0


问:
How to solve $ \lfloor x+2\rfloor-1=0 $
I need some help to solve the next equation:
$$
\lfloor x+2\rfloor-1=0
$$
Where $\lfloor .\rfloor$ is the floor function.

What I've tried:
$$
\lfloor x+2\rfloor-1=0 \Rightarrow \lfloor x\rfloor=-1 \Rightarrow x \in(0,1)
$$

Is that right??

答:
\begin{align}
[x+2]-1=0 &\iff [x+2]=1 \\
& \iff 1 \leqq x+2 < 2\\
& \iff -1 \leqq x <0.
\end{align}

You can see the graph of $y=[x+2]$ with wolfram alpha.

If you see the point of $y=1$, you will be able to realize the range of $x$.

https://www.wolframalpha.com/input/?i=floor%28x%2B2%29&lang=ja

Title: How to solve $ \lfloor x+2\rfloor-1=0 $


答:
So I'm labelling that 2nd angle as $\beta$. So we want to show that $\beta = \theta$

So looking at the triangle with the angle $\alpha$ and $\beta$, we know the sum of the angles of a triangle are 180

So

$\alpha+\theta+90 = 180$

$\theta = 90-\alpha$

We also know that since angles on a straight line add to 180

$\alpha+90+\beta = 180$

So $\beta = 90-\alpha$

Therefore $\beta = \theta$ so that's why they're labeled the same.

Title: How would I geometrically prove this situation of a rectangle and a triangle? I.e., how is theta used twice?

Body:
Here is a situation of Euclidian geometry, a proof of some sorts.

I am quite unsure how to explain what was done. This is from an AP Physics B (Part 1) prep book, and this was the first step in the answering process, which made too many logical steps. The prompt was this:

The prompt.

What I am asking of is theta; I noted that it was used twice, and I too am aware that some geometric proof is involved in relating the two triangles.

I am sorry for the vagueness of this situation, and I hope the information is sufficient enough. How would I prove that theta of the top triangle is theta on the side? (Perhaps this should be moved to the Physics exchange if too vague in jargon.)


问:
Bound on sum of logarithms
Consider a natural number $m$ and let $m_i, i=1,2,...,\lfloor{\sqrt{m}}\rfloor$ be natural numbers such that:
$$\sum_{i=1}^{\lfloor{\sqrt{m}}\rfloor} m_i = m$$

My question regards the following sum:
$$\sum_{i=1}^{\lfloor{\sqrt{m}}\rfloor} \log(\frac{m_i}{i}) $$

What can we say for the maximum size of this sum in comparison to $m$? Will it be of the order of $\sqrt{m}$?

答:
By the AM-GM inequality,
$$
\prod_i m_i
\le \left(\frac{\sum_i m_i}{\lfloor \sqrt m \rfloor}\right)^{\lfloor \sqrt m \rfloor}
= \left(\frac{m}{\lfloor \sqrt m \rfloor}\right)^{\lfloor \sqrt m \rfloor}
\approx (\sqrt m)^{\sqrt m}
$$
(I'm too tired to care about the floor thingy right now). Therefore
$$
\sum_i \log\frac{m_i}{i}
= \log\prod_i\frac{m_i}{i}
\le \log \frac{(\sqrt m)^{\sqrt m}}{(\sqrt m)!}
= \sqrt m \log (\sqrt m) - \log(\sqrt m)!
$$
From looking a bit on the wiki page for Stirling's approximation, we find
$$
\log n! = n\log n + \Theta(n)
$$
If we plug that into the above we get
$$
\sum_i \log\frac{m_i}{i} \in O(\sqrt m)
$$
Note also that the AM-GM inequality is an equality in case all the $m_i$ are equal. That means that the above order can be attained. So we have in the worst case
$$
\sum_i \log\frac{m_i}{i} \in \Theta(\sqrt m)
$$
i.e. we can't improve on the order in the general case. This was like you guessed! We can make the bound more precise by adding more terms of the Stirling approximation.

Title: Bound on sum of logarithms


答:
Let $k=\lfloor \sqrt{m}\rfloor$. The sum is
$$\sum_{i=1}^k\log\left(\frac{m_i}{i}\right)=\log\left(\frac{m_1m_2\cdots m_k}{k!}\right) $$
By AM-GM
$$m_1m_2\cdots m_k\leq \left(\frac{m_1+m_2+\cdots+m_k}{k}\right)^k=\left(\frac{m}{k}\right)^k $$
For $k!$, we need a lower bound. Among various options:
$$k!\geq \left(\frac{k}{e}\right)^k$$
Hence,
$$\frac{m_1m_2\cdots m_k}{k!}\leq\left(\frac{me}{k^2}\right)^k$$
$$\log\left(\frac{m_1m_2\cdots m_k}{k!}\right)\leq k\log\left(\frac{me}{k^2}\right)\leq\sqrt{m}\log\left(\frac{me}{k^2}\right)$$
Note that $m/k^2 \to 1$ as $m\to\infty$, so $\log\left(me/k^2\right)\to \log(e)=1$ and $\log\left(me/k^2\right)<2$ for sufficiently large $m$. Alternatively, you can try showing that $m/k^2\leq 4$ for all $m$, so $\log\left(me/k^2\right)\leq \log(4e)=\log(4)+1$. The point is to show that $\log\left(me/k^2\right)$ is bounded by a constant.

Title: Bound on sum of logarithms

Body:
Consider a natural number $m$ and let $m_i, i=1,2,...,\lfloor{\sqrt{m}}\rfloor$ be natural numbers such that:
$$\sum_{i=1}^{\lfloor{\sqrt{m}}\rfloor} m_i = m$$

My question regards the following sum:
$$\sum_{i=1}^{\lfloor{\sqrt{m}}\rfloor} \log(\frac{m_i}{i}) $$

What can we say for the maximum size of this sum in comparison to $m$? Will it be of the order of $\sqrt{m}$?


问:
Complex numbers $z$ and $w$ proof
How should I go about proving the statement, $\forall w, z \in \mathbb{C}, |z+iw|=|z-iw|$ if and only if $z\bar w \in \mathbb{R}$?

I'm having some trouble understanding complex numbers. This reminds me of a conjugate, should I use the conjugate to multiply then for the forwards implication?

答:
Note that $$|z+iw|^2-|z-iw|^2=(z+iw)\overline{(z+iw)} - (z-iw)\overline{(z-iw)}\\
=(z+iw)(\bar z - i \bar w)-(z-iw)(\bar z +i \bar w) = 2i(\bar z w-z \bar w) = 2i (\overline{z\bar w}-z\bar w)
$$

Therefore $$|z+iw| = |z-iw| \iff \overline{z\bar w} - z \bar w =0 \iff z \bar w \in \mathbb R. \blacksquare$$

Title: Complex numbers $z$ and $w$ proof


问:
Solve the recurrence $\ T(n) = \sqrt{n}\cdot T(\sqrt{n}) + n\cdot \log(n) $
Solve the following recurrence:

$$\ T(n) = \sqrt{n}\cdot T(\sqrt{n}) + n\cdot \log(n)$$

I tried to just unroll it to see if I can understand, but it is too complicated .

$\ T(n) = n^{\frac{1}{2}} \cdot T(n^{\frac{1}{2}}) +n\cdot \log(n)\\
= n^{\frac{1}{2}} \cdot \left( n^{\frac{1}{4}} \cdot T(n^{\frac{1}{4}} ) + \frac{n}{2} \cdot \log(\frac{n}{2}) \right) + n \cdot \log(n) \\
= n^{\frac{1}{2}} \left( n^{\frac{1}{4}} \cdot \left( n^{\frac{1}{8}} \cdot T(n^{\frac{1}{8}}) +  \frac{n}{4}\cdot \log(\frac{n}{4})  \right) + \frac{n}{2} \cdot \log(\frac{n}{2}) \right) + n\cdot \log(n) $

Not really clear how to proceed here? I also tried recurrence tree, but couldn't see a clear pattern. I've found a similar question here but I could not understand the answer

答:
We have

$$
\frac{T(n)}{n}=\frac{T(\sqrt{n})}{\sqrt{n}}+\ln n
$$

or with $\mathcal{T}(n)=\frac{T(n)}{n}$

$$
\mathcal{T}(n) = \mathcal{T}(\sqrt{n})+\ln n
$$

or

$$
\mathcal{T}(e^{\ln n})=\mathcal{T}(e^{\ln \sqrt{n}})+\ln n
$$

and now calling

$$
\cases
{
\mathtt{T}(\cdot) = \mathcal{T}\left(e^{(\cdot)}\right)\\
z = \ln n
}
$$

we follow with

$$
\mathtt{T}(z)=\mathtt{T}\left(\frac z2\right)+z
$$

This recurrence has the solution

$$
\mathtt{T}(z) = c_1+2(z-1)
$$

and now going backwards

$$
\mathcal{T}(n) = c_1+2(\ln n-1)
$$

and finally

$$
T(n) = n\left(c_2+2\ln n\right)
$$


答:
We have for $(\lnot A \lor C)$:

$1:\text{Premise}; (\lnot A \lor C) : \top$

$2:\text{Assumption}; A : \top$

$3:\text{Negation}(2); \lnot A : \bot$

$4:\text{Insertion}(3\to1); (\bot \lor C) : \top$

$5:\text{Implication}(4); C : \top$

and for $(A \lor B)$

$6:\text{Premise}; (A \lor B) : \top$

$7:\text{Assumption}; A : \bot$

$8:\text{Insertion}(7\to6); (\bot \lor B) : \top$

$9:\text{Implication}(8); B : \top$

and then:

$10:\text{Assumption}; A \lor \lnot A : \top$

$11:\text{Inference}(2,5); (A \supset C) : \top$

$12:\text{Inference}(7,9); (\lnot A \supset B) : \top$

$13:\text{Inference}(10,11,12); (B \lor C) :\top$

Title: Natural Deduction Proof: {A v B, ¬A v C} ⊢ B v C

Body:
Prove using natural deduction: $(A \lor B), (\lnot A \lor C) ⊢ (B \lor C)$.
My work so far:

```<code>1     A v B
2     A v C
3      A      :AS
4       ¬A      :AS
5       _|_     :¬E 3,4
6    ¬¬A      :¬I 4-5
7  --
8  ¬A     :AS
</code>```

This is the work that I have come up with, for the past 5 hours. Can you please help me?


问:
Prove that $(2,1+\sqrt{-5})^2 = (2)\,$ in $\,\Bbb Z[\sqrt{-5}]$

Let $p$ be the ideal $\{2a+(1+\sqrt{-5})b\mid a,b\in\mathbb{Z}[\sqrt{-5}]\}$ in $\mathbb{Z}[\sqrt{-5}]$. Prove that $p^2$ is the principal ideal $(2)$.

I tried multiplying the ideal with itself and tried to simplify to see if I could relate it to the principal ideal $(2)$ but I can't seem to get it. This is what I have so far.

$(2a+(1+\sqrt{-5})b)^2=4a^2+4(1+\sqrt{-5})ab+(1+\sqrt{-5})^2b^2$

$=4a^2+(4+4\sqrt{-5})ab+(2\sqrt{-5}-4)b^2$

$=2(2a^2+(2+2\sqrt{-5})ab+(\sqrt{-5}-2)b^2)$

If $x\in p^2$, then $x$ is a multiple of $2$, so $x\in(2)$. Thus $p^2\subseteq (2)$.

I am unsure of how to show that $(2)\subseteq p^2$.

答:
Let's call this ideal $\mathfrak{p}$. Note that $2 \in \mathfrak{p}^2$, since
$$
2 = (1+\sqrt{-5})(1 - \sqrt{-5}) \  - \ 2\cdot2.
$$
On the other hand, if $m \in \mathfrak{p}^2$, then $m$ is a sum of terms of the form
\begin{align}
(2a + (1 + \sqrt{-5})b)&(2a'+(1+\sqrt{-5})b') = \\
             & 4aa' + 2ab' + 2a'b + bb' - 5bb' + \bigg(2ab' + 2a'b + bb'+bb'\bigg)\sqrt{-5}.
\end{align}
and everything is divisible by $2$ when we group like terms.

This is a bit of a miracle proof; you can redo it conceptually by observing that $\mathfrak{p}$ is the ideal of all elements of the form $c + d\sqrt{-5}$ where $c$ and $d$ have the same parity.


问:
Calculate gas-station probabilities
I would like to calculate probabilities for the next exercise:

Knowing the average amount of cars that drive per minute into a gas-station is 3.

** How can I calculate the probability of arriving at least 12 cars into the station in a period of 5 minutes?

** And the probability of a car arriving before two minutes go by?

Thank you all very much in advance.

答:
We use a Poisson model.  So if $X$ is the number of cars in randomly chosen $1$ minute interval, then $X$ is assumed to have Poisson distribution with parameter $3$.

Thus by general theory, the number $Y$ of cars in a $5$ minute interval has Poisson distribution with parameter $\lambda=15$. thus the probability of fewer than $12$ cars in a $5$ minute interval   is
$$\sum_{k=0}^{11}e^{-\lambda}\frac{\lambda^k}{k!}.\tag{$1$}$$
This is a somewhat unpleasant calculation. Maybe compute from $k=11$ down. After a while, the terms you are adding become negligible. The probability of at least $12$ cars is $1$ minus the probability computed in $(1)$.

The number $Z$ of cars in a $2$ minute interval has Poisson distribution with parameter $6$. We want the probability of at least one car, which is $1$ minus the probability of no cars. The probability of no cars is $e^{-6}$.  Alternately, for the no cars problem one can use the relationship between the Poisson and the exponential.

Title: Calculate gas-station probabilities


问:
Sequence in $c_0$ (related with $\ell_1$).
I need some help with this cuestion:

Let b a sequence $a=\{a(n)\}$ in $\ell_1$. We define, for each $x \in c_0$, the sequence $$T_a: c_0 \longrightarrow c_0, \hspace{0.15cm} T_a(x)(n)=\sum_{k=n}^{\infty}a(k)x(k), n \in \mathbb{N}.$$ Give the properties of the sequences $a \in \ell_1$ for which $T_a$ reach its norm.

I think that I have to give an if-and-only-if property, but I don't know exactly how to do it. I'm trying to prove that maybe $\exists k \in \mathbb{N}$ such that $|a(k)|=||a||_{\infty}$, or something about $||a||_1$, but I don't arrive to any conclusion.

答:
First, let us calculate the norm of $T$. To do this, I suggest extending $T$ to $\ell^\infty$.

Let $S$ be the natural extension of $T$ to $\ell^\infty$, i.e. the same definition, just with sequences from $\ell^\infty$ instead. Let $k \in \ell^\infty$ be the sequence
$$k_n = \begin{cases} \dfrac{|a_n|}{a_n} & \text{if } a_n \neq 0 \\ 1 & \text{if } a_n = 0. \end{cases}$$
Note that $|k_n| = 1$ for all $n$, hence $\|k\|_\infty = 1$. We also have, for any $x \in \ell^\infty$ with $\|x\|_\infty \le 1$,

$$\|S(x)\| = \sup_{n \in \Bbb{N}} \left|\sum_{k=n}^\infty a_k x_k \right| \le \sup_{n \in \Bbb{N}} \sum_{k=n}^\infty |a_k| |x_k| \le \sup_{n \in \Bbb{N}} \sum_{k=n}^\infty |a_k| = \sum_{k=1}^\infty |a_k| = \|S(k)\|.$$

Therefore $S$ achieves its norm on $\ell^\infty$ with this sequence $k$.

Turning our attention back to $T$ and $c_0$, define a sequence $(x^m)_{m \in \Bbb{N}}$ of points in $c_0$ defined by
$$x^m_n = \begin{cases} k_n & \text{if } n \le m \\ 0 & \text{if } n > m. \end{cases}$$
Note that $\|T\| \le \|S\| = \|a\|_1$ and $\|x^m\|_\infty = 1$. Then,
$$\|T(x^m)\| = \max_{1 \le n \le m} \sum_{k = n}^m |a_k| = \sum_{k=1}^m |a_k| \to \sum_{k=1}^\infty |a_k| = \|S(k)\|,$$
thus proving $\|T\| = \|S\| = \|a\|_1$.

Now, we just need to ask, when is there a sequence $x \in c_0$ such that $\|x\|_\infty \le 1$ and $\|T(x)\| = \|a\|_1$? Well, one sufficient condition would be for $a_n$ to eventually become $0$, so that $\|T(x^m)\| = \|a\|_1$ for sufficiently large $m$. Is this necessary too?

Well, yes, it must be. For every $n$ such that $a_n \neq 0$, in order for a point $x \in c_0$ to achieve the norm, we need $|x_n| = 1$. If this happens infinitely often, then this contradicts $x \in c_0$. So, our necessary and sufficient condition is $a \in c_{00}$, the space of sequences which are eventually $0$.


问:
Infinitude of primes which are inert in one extension and completely split in other for extensions of prime degree.
I want to prove the following:

Let $L_1, \cdots , L_n$ be extensions of number field $K$ of prime degree $p$ such that each $L_i$ is disjoint from the compositum of $L_j$s for $ j\neq i $. Then there are infinitely many primes of $K$ that are inert in $L_1$ and split completely in $L_i$ for $i \geq 2 $.

I need to prove this to prove the second inequality of Global Class Field Theory (so I do not want to use the Chebotarev density theorem because that can give me the second inequality directly instead).

I have a couple of facts, consequences of the first inequality, at my disposal, which I think might be useful here.

In a non-trivial finite abelian extension of $K$, infinitely many primes of $K$ do not split completely.

In a cyclic prime power degree extension of $K$, infinitely many primes remain inert.

I first try to prove the theorem in the case where we just have $L_1$ and $L_2$, hoping that the proof will generalize easily.

In $L_1L_2$ there are infinitely many primes which do not split completely. There are infinitely many of these for which the prime lying below in $L_1$ is inert, let $S$ be set of such primes. We should show that infinitely many of primes $S$ when contracted to $L_2$ are completely split. I am stuck here.

答:
Right... I’m not entirely sure how to generalize this (and I’m not 100 percent sure it works), but in the case $n=2,L_i/K$ Galois, so that $L_1L_2$ is the tensor product and is Galois of group $\mathbb{F}_p^2$ (it’s of order $p^2$ and consider its subgroups).

Let $\mathfrak{q}$ be a prime of $L_2$ (above $\mathfrak{p}$ of $K$) which is inert in $L_1L_2$. There are infinitely many such $\mathfrak{p}$ that are unramified in $L_1L_2$.

Assume $\mathfrak{p}$ inert in $L_2$, then $f_{L_1L_2/K,\mathfrak{p}}=p^2$ by multiplicativity. It implies in particular that $\mathfrak{p}$ is inert in $L_1$. But then $L_1 \otimes K_{\mathfrak{p}}=L_2\otimes K_{\mathfrak{p}}$ (they’re unramified of same degree over $K_{\mathfrak{p}}$ – because Galois and prime) so $L_1L_2$ splits, a contradiction.

So such a $\mathfrak{p}$ splits in $L_2$. Assume that it splits in $L_1$, too. Then $(L_1L_2) \otimes K_{\mathfrak{p}}=L_1 \otimes (L_2 \otimes K_{\mathfrak{p}} \cong L_1 \otimes K_{\mathfrak{p}}^p=(L_1 \otimes K_{\mathfrak{p}})^p=K_{\mathfrak{p}}^{p^2}$ thus $\mathfrak{p}$ splits totally in $L_1L_2$, a contradiction.

So the primes $\mathfrak{p}$ of $K$ unramified in $L_1L_2$ such that there are primes $\mathfrak{q}$ of $L_2$ above them that are inert in $L_1L_2$ are infinitely many solutions to your question.

I think a similar argument (with $L_2\ldots L_n$ instead of $L_2$) works in general if each $L_i/K$ is Galois (with some minor adaptations) but I’m not sure.


问:
How do you solve this calculus maxima minima problem graphically, without calculus?
A rectangular paddock is to be fenced using a river as one boundary. The paddock is to have an area of $8000$ m. Let one side of the paddock be $$ then use a suitable graph to determine the minimum amount of fencing required.

Do I introduce another variable? I am confused because it only tells me to introduce one ($x$).So I have $xy = 8000$ and $x + 2y = P$. I sketched $y = \frac{8000}{x}$ and I assume for $P$ to be minimized, it must touch at a tangent. Please help me find these coordinates as this is where I get stuck - there are three variables now ?! And there isn't enough information to label my graphs

答:
If you put an imaginary wall perpendicular to the river through the center of a paddock, you will find two areas of 4000 m² with two identical fence walls with lengthes x and y, so $xy=4000$. The half-perimeter of the whole fence is $P=x+y$.

On a coordinate plane, $x+y=P$ gives you a line at $45°$, whereas $xy=4000$ gives you a hyperbola. When $P$ is small, line doesn't cross hyperbola, when you increase $P$ the line will move parallelly and at some point will touch it. From the symmetry of both curves, one can see that it will happen at point $x=y$, thus $x=y=\sqrt{4000}$. And perimeter is $2(x+y)$


问:
Defining "tuples" of $n \geq 2$ as SETS of ordered pairs (indexing strategy) versus single ordered pairs (Kuratowski strategy)
After reading some posts on the forum (Definition of the ordered triple (a, b, c) according to Kuratowski's Set Theory. and Associativity of cartesian product and nested ordered n-tuples to name a few), I learned that there are two common ways of defining elements of Cartesian Products of the form $X \times X \times X \times ...$

The first way involves purely applying the Kuratowski definition (recursively if $n \gt 2$), which will produce a single ordered pair, and the second way involves an indexed set ( $\{0,1,...,n\}$) mapping, which ends up producing a collection (set) of ordered pairs.

Let $X \times X$ and $X \times X \times X$ be my two working examples.

Consider an element $\langle a,b \rangle \in X \times X$

"Purely" Kuratowski Definition - $\langle a, b \rangle := \{\{a\},\{a,b\}\}$

Indexed Set Definition (followed by Kuratowski) - $\langle a,b \rangle := \{\langle 0,a\rangle,\langle1,b\rangle\} =  \Big \{\big \{\{0\},\{0,a\}\big\}, \big \{\{1\},\{1,b\}\big\}\Big\}$

Consider an element $\langle a,b,c \rangle \in X \times X \times X$

"Purely" Kuratowski Definition - $\langle a, b, c \rangle := \langle \langle a, b \rangle, c \rangle = \Big \{ \big \{ \big \{ \{a\},\{a,b\}\big \} \big \}, \big \{ \big \{ \{a\}, \{a,b \} \big \}, c \big \} \Big \}$

Indexed Set Definition (followed by ...) - $\langle a, b, c \rangle :=\{\langle 0,a\rangle,\langle1,b\rangle\,\langle 2,c\rangle\} =  \Big \{\big \{\{0\},\{0,a\}\big\}, \big \{\{1\},\{1,b\}\big\},\big \{\{2\},\{2,c\}\big\}\Big\}$

Now, my question is as follows:

For a given Cartesian Product "dimension" ($2$ in the case of $X \times X$ and $3$ in the case of $X \times X \times X$), are these definitions "bijective" (maybe isomorphic is the appropriate term here)?

I am guessing the answer is yes, but I am not really seeing what that might look like. I've always thought of bijective functions of finite sets as requiring each set to have the same number of elements. In the $2$D example, this is certainly true...but in the $3$D example, this is not true. So I am uncertain of what to think.

Any insight is appreciated! Thank you.

答:
In each case one can uniquely recover the components and their order from the $n$-tuple. Suppose, for example, that $t\in A\times B\times C$ is a Kuratowski ordered triple. Then there are a unique $p\in A\times B$ and $c\in C$ such that $t=\langle p,c\rangle$, and there are unique $a\in A$ and $b\in B$ such that $p=\langle a,b\rangle$. Thus, we can read off the first, second, and third components of $t$ as $a,b$, and $c$, respectively: $t=\langle a,b,c\rangle$.

Mind you, actually describing how to recover $a$ and $b$ from $p$, for instance, is a bit of a pain, but it can be done. First, $a$ is the unique $x\in\bigcup p$ such that $\forall y\in p\,(x\in y)$. If $a$ is the only element of $\bigcup p$, then $b=a$; otherwise, $b$ is the unique element of $\left(\bigcup p\right)\setminus\{a\}$.

From the other definition, of course, we recover $a,b$, and $c$ as $t(0),t(1)$, and $t(2)$, respectively. The bijection between the two versions of $A\times B\times C$ is then defined in terms of the components: if $\langle a,b,c\rangle$ is a Kuratowski triple, the indexed set triple corresponding to it under the bijection is the unique $t$ in the indexed set product such that $t(0)=a$, $t(1)=b$, and $t(2)=c$.

All of this generalizes readily to $n$-fold products for arbitrary $n\in\Bbb Z^+$.


问:
Inequality about holomorphic functions on the disk
I'm trying to show that $ |g(z) +g(-z)| \leq 2|z|^{2}, z\in \mathbb{D}, g:\mathbb{D} \rightarrow\mathbb{D}, g(0)=0$

I applied Schwarz's lemma and got $|g(z)| \leq |z|$ and $g(-z)\leq |-z|=|z|$.

$$|g(z)+g(-z) | \leq |g(z)| +|g(-z)| \leq 2|z|$$

the first inequality follows from the triangle inequality. But, I'm now stuck as I cannot just say that $2|z|\leq 2|z|^2$ since $z\in \mathbb{D}$ where it it is not true. Where do I go from here?

答:
Let $h(z)=\frac{g(z)+g(-z)}{2}$; we notice that $h(z)=h(-z)$ which means (as the Taylor series of $h$ has only even powers) that $h(z)=u(z^2)$ for some holomorphic function $u: \mathbb D \to \mathbb D , u(0)=0$ as clearly $h$ satisfies same conditions too by the properties of $g$.

Applying Schwarz lemma to $u$ we get $|u(z)| \le |z|$ or $|h(z)| =|u(z^2)| \le |z|^2$ and that is precisely what we need to prove


问:
Find the cosets in $\mathbb{Z}[\sqrt{-5}]/Q_1$
Find the cosets in $\mathbb{Z}[\sqrt{-5}]/Q_1$ where $Q_1=\{3a+(1+\sqrt{-5})b|a,b\in\mathbb{Z}[\sqrt{-5}]\}$. I am studying this and am having trouble understanding how to find cosets of quotient rings. I understand that it means the congruence classes modulo $Q_1$ and I know in this case there should be 3 distinct cosets. But how do I find the cosets? Please explain how you found them. Thank you.

答: Consider the ring isomorphism $\frac{\mathbb{Z}[x]}{(x^2+5)}\cong \mathbb{Z}[\sqrt{-5}]$ which comes from the evaluation homomorphism $f:\mathbb{Z}[x]\rightarrow \mathbb{C}$, $p(x)\mapsto p(\sqrt{-5})$. Now, consider the preimage of the ideal $Q_1=(3,1+\sqrt{-5})$ under the isomorphism $\bar{f}:\frac{\mathbb{Z}[x]}{(x^2+5)}\rightarrow \mathbb{Z}[\sqrt{-5}]$ induced by $f$ above. This, in turn, induces an isomorphism $$\frac{\frac{\mathbb{Z}[x]}{(x^2+5)}}{f^{-1}(Q_1)}\cong \frac{\mathbb{Z}[\sqrt{-5}]}{(3,1+\sqrt{-5})}.$$ Now, you can check that $f^{-1}(Q_1)=\frac{(3,1+x,x^2+5)}{(x^2+5)}$. Hence, by one of the isomorphism theorems you have $$\frac{\frac{\mathbb{Z}[x]}{(x^2+5)}}{\frac{(3,\,1+x,\,x^2+5)}{(x^2+5)}}\cong \frac{\mathbb{Z}[x]}{(3,1+x,x^2+5)}\cong \frac{\frac{\mathbb{Z}[x]}{(3)}}{\frac{(3,1+x,x^2+5)}{(3)}}\cong \frac{\frac{\mathbb{Z}}{3\mathbb{Z}}[x]}{(1+x)},$$ where in the last isomorphism, it is used that $x^2+5=x^2-1=(x+1)(x-1)$ in $\frac{\mathbb{Z}}{3\mathbb{Z}}[x]$. Clearly, the last ring is the field with 3 elements. Therefore, if you keep track of the cosets under the isomorphisms above, you can see an explicit description of your cosets in the original ring.


问:
Neighbourhood of an element in a product space $\mathbb{R}\times\mathbb{R}$
Let $A$ be the topological subspace $\left([0,\infty)\right)\times \mathbb{R}) \cup (\mathbb{R}\times \{0\})$ of $\mathbb{R}\times\mathbb{R}$

If $U$ is an open subset in $A$ and $(x,0)\in U$, why does there exist $\varepsilon>0$ such that $(x-\varepsilon,x+\varepsilon)\times\{0\}\subseteq U$

I know that $(x,0)$ is contained in a neighbourhood $N\subseteq U$ because of the open property of $U$ but I don't know how to go whittle that down to the conclusion as $N$ is a neighbourhood in product space.

答:
Let $p=\langle x,0\rangle$. You know that there is an open $V$ in $\Bbb R^2$ such that $U=V\cap A$: that’s just the definition of the subspace topology on $A$. You also know that there is an $\epsilon>0$ such that $B(p,\epsilon)\subseteq V$, where $B(p,\epsilon)$ is the open ball of radius $\epsilon$ centred at $p$. Thus,

$$B(p,\epsilon)\cap A\subseteq V\cap A=U\,.$$

Next, it’s clear that

$$(x-\epsilon,x+\epsilon)\times\{0\}\subseteq B(p,\epsilon)\,.$$

Finally,

$$(x-\epsilon,x+\epsilon)\times\{0\}\subseteq\Bbb R\times\{0\}\subseteq A\,,$$

so

$$(x-\epsilon,x+\epsilon)\times\{0\}\subseteq B(p,\epsilon)\cap A\subseteq U\,.$$


答:
You can use arithmetic-geometric mean inequality.

You know that

$P = x+2y$

$P = \frac{2x+4y}{2}$

So P is the arithmetic mean of $2x$ and $4y$.

The geometric mean of $2x$ and $4y$ is

$\sqrt{(2x)(4y)} = \sqrt{8xy} = \sqrt{64000} = 80\sqrt{10}$

We know the arithmetic mean is greater than or equal to the geometric mean. So the minimum possible fencing is $80\sqrt{10}$.

The minimum occurs when $2x=4y$ ie: $x=2y$.

Title: How do you solve this calculus maxima minima problem graphically, without calculus?

Body:
A rectangular paddock is to be fenced using a river as one boundary. The paddock is to have an area of $8000$ m. Let one side of the paddock be $$ then use a suitable graph to determine the minimum amount of fencing required.

Do I introduce another variable? I am confused because it only tells me to introduce one ($x$).So I have $xy = 8000$ and $x + 2y = P$. I sketched $y = \frac{8000}{x}$ and I assume for $P$ to be minimized, it must touch at a tangent. Please help me find these coordinates as this is where I get stuck - there are three variables now ?! And there isn't enough information to label my graphs


问:
Prove that the function is Riemann-Integrable.
This is an incorrect attempt! Look below for @Laithy 's advice and my second attempt.

I tried this problem but I'm not sure whether I am getting it right. I am having trouble with the very last step where I am supposed to show that $I^-(f) = I^+(f)$. Here is my attempt. Please point out any errors and help me justify why $I^-(f)$ and $I^+(f)$ are equal. Thank you!

$
f(x) =\begin{cases}
      1 & 0\leq x < \frac{1}{2} \\
      0 & x = \frac{1}{2} \\
      1 & \frac{1}{2} < x \leq 1
   \end{cases}
$

Let $P$ be the following partition $P = \{[0, \frac{1}{2}], [\frac{1}{2}, 1]\}$ of $[0, 1]$.

Consider $I_1 = [0, \frac{1}{2}]$. It follows that $m_1 = 0$ and $M_1 = 1$.

This implies $m_1(x_1 - x_0) = 0$ and $M_1(x_1 - x_0) = 1(\frac{1}{2} - 0) = \frac{1}{2}$.

Now consider $I_2 = [\frac{1}{2}, 1]$. It follows that $m_2 = 0$ and $M_2 = 1$.

This implies $m_2(x_2 - x_1) = 0$ and $M_2(x_2 - x_1) = 1(1 - \frac{1}{2}) = \frac{1}{2}$.

All this gives the following:

$S^- = 0$ and $S^+ = 1$.

Therefore...

$I^-(f) = \sup\{S^-, \sigma\in G[0, 1]\} = \sup\{0, \sigma\in G[0, 1]\}$

and

$I^+(f) = \inf\{s^+, \sigma\in G[0, 1]\} = \inf\{1, \sigma\in G[0, 1]\}$

How do I decide whether $I^-(f)$ equals $I^+(f)$?

答:
I don't quite understand your definition of $I_-$ and $I^+$. This should be the supremum of the lower sum over all partitions and the infimum of the upper sum over all partitions respectively. The partition you defined is not a good choice. You want to choose a partition that makes the upper sum and lower sum arbitrarily close.

Let $\varepsilon >0$. Consider the partition $P = \left\{ [0,\frac{1}{2} - \delta], [\frac{1}{2} - \delta, \frac{1}{2} + \delta], [\frac{1}{2} + \delta, 1] \right\} $
for some $\delta>0$.

Choose an appropriate $\delta >0$ so that $U(f,P) - L(f,P) < \varepsilon$

And so $0 \leq I^+ - I_- < \varepsilon$. Note that $I^+$ and $I_-$ do not depend on $\varepsilon$ and that this inequality holds for all $\varepsilon>0$. This implies that they are equal.


问:
Showing a function is Lipschitz continuous and finding the Lipschitz constant.
Consider the function $H: \mathbb{R}^2 \to \mathbb{R}$ defined by

$$H(x,y) = \frac{-1}{2} \left[ a - x^2 + \sqrt{a^2 + x^4 -2ax^2 + 4y^2}\right]^{\frac{1}{2}}$$

where $a > 0$ is fixed.

This function is clearly well defined and $C^{\infty}$. One can easily see that $H$ is lipschitz on any ball centered at $(0,0)$ with radius $R$ since $|\nabla H|$ is bounded there. I am having trouble computing the Lipschitz constant $C_R$. I want to find how it depends on $R$. Does it go to infinity as $R$ goes to infinity? If so, at what rate?

I computed the derivative $|\nabla H|$ and tried to find a bound for it, but it gets very messy. I get an upper bound for $C_R$ but I think it's very weak, and there probably is a way to get a sharper upper bound on $C_R$. Any help is appreciated.

答:
I do not know if it is helpful, but the proposed formula resembles the $\sinh^{-1}$ function.

Precisely speaking, if $z > 0$, one has that
\begin{align*}
\sinh^{-1}\left(\frac{w}{z}\right) = \ln\left(\frac{w + \sqrt{w^{2} + z^{2}}}{z}\right)
\end{align*}

At your case, one has that
\begin{align*}
a - x^{2} + \sqrt{a^{2} + x^{4} - 2ax^{2} + 4y^{2}} & = a - x^{2} + \sqrt{(a-x^{2})^{2} + (2y)^{2}}\\\\
& = \left(\frac{a - x^{2} + \sqrt{(a-x^{2})^{2} + (2y)^{2}}}{2y}\right)\times 2y\\\\
& = \exp\left(\sinh^{-1}\left(\frac{a-x^{2}}{2y}\right)\right)\times 2y
\end{align*}

Hopefully this helps!


问:
Integral of $(x+c)^n e^{-x^2}$
As part of some convolution, I am desperately trying to solve the following integral:

$$
\int_a^b (x+c)^ne^{-x^2} dx \,\,\,\, \text{with} \,\,\,\, n \in \mathbb{N}.
$$

I am looking forward to any input! Many thanks!

答:
Using the binomial expansion
$$(x+c)^n=\sum_{k=0}^n \binom{n}{k} c^{n-k} x^k$$
$$I_n=\int (x+c)^ne^{-x^2}\,dx =\sum_{k=0}^n \binom{n}{k} c^{n-k}\int x^k e^{-x^2}\,dx$$
$$I_n=-\frac 12\sum_{k=0}^n \binom{n}{k} c^{n-k} \,\Gamma \left(\frac{k+1}{2},x^2\right)$$
$$J_n=\int_a^b (x+c)^ne^{-x^2}\,dx$$ $$J_n=\frac 12\sum_{k=0}^n \binom{n}{k} c^{n-k} \left(\Gamma \left(\frac{k+1}{2},a^2\right)-\Gamma
   \left(\frac{k+1}{2},b^2\right)\right)$$

Title: Integral of $(x+c)^n e^{-x^2}$


问:
Subsequence such that integrals converge over any Borel set in $[0,1]$
I was reading this question: Existence of subsequence such that integration converge

The idea is this. I have a sequence of uniformly bounded measurable functions $\{f_{n}\}$ on $[0,1]$ and I want to find a subsequence $f_{n_{j}}$ such that $\lim_{n \to \infty} \int_{A} f_{n_{j}}$ exists for all Borel sets $A$. I can show the following:

(1) If $\{S_{i}\}_{i}$ is a countable collection of Borel sets, then we can find a subsequence so that $\int_{S_{i}} f_{n_{j}}$ has a limit for all $S_{i}$.

(2) That this holds for all half-open half-closed intervals $(a_{i}, b_{i}]$ with rational endpoints.

e know that the collection of half-open half-closed intervals with rational endpoints is countable and generates the Borel $\sigma$-algebra, so the idea is now to approximate every Borel set using sets in this algebra and show that the result holds for them. In particular if $A \subset [0,1]$ is a Borel subset then we can find a sequence $I_{i}$ of half-open half-closed intervals with rational endpoints such that $I_{i} \downarrow A$, but I'm not able to proceed further. Is it true that if $\int f_{n_{j}}$ has a limit on each $I_{i}$, and $I_{i}$ is a decreasing sequence of sets, then $\int f_{n_{j}}$ has a limit on $\bigcap_{i} I_{i}$?

答:
So what you have is a (countable) family $(A_m)$ of subsets of $[0,1]$ and a subsequence $g_n$ of the $(f_n)$ such that
$$(\int_{A_m} g_n)_n$$
is convergent for every $m$.

( you got that $g_n$ with the diagonal process).

Now, you can take $(A_m)$ such that for every $A$ measurable and $\epsilon > 0$ there exists $A_m$ such that $\mu(A\Delta A_m) < \epsilon/3$. For instance, take $A_m$'s to be the finite union of intervals with rational ends.

Since $\int_{A_m} g_n$ is convergent there exists $N$ such that
$$|\int_{A_m} g_n - \int_{A_m} g_{n'}| < \epsilon/3 $$
for all $n,n'> N$.

Note that since $|g_n|\le 1$ for all $n$ we have
$$\int_{A_m} g_n - \int_{A} g_n| < \epsilon/3$$

Now use the triangle inequality to conclude
$$|\int_{A} g_n - \int_{A} g_{n'}| < \epsilon$$
for all $n,n'> N$.

We conclude $\int_A g_n$ is convergent.

Note that this implies $g_n$ converges in measure, so there exists a subsequence of it that converges a.e.to a function.


问:
Calculate max/min of $x_1 x_2+y_1 y_2+z_1 z_2+w_1 w_2$
What is a good way to calculate max/min of
$$x_1 x_2+y_1 y_2+z_1 z_2+w_1 w_2$$
where $x_1+y_1+z_1+w_1=a$ and $x_2+y_2+z_2+w_2=b$ and $x, y, z, w, a, b \in \mathbb{N} \cup \{0 \}$, and please explain your answer (how your result comes out).</p>

答:
We can start with the inequality

\begin{equation}
0 \leq x_1x_2+y_1y_2+z_1z_2+w_1w_2 \leq (x_1+y_1+z_1+w_1)(x_2+y_2+z_2+w_2) = ab
\end{equation}

Since you want a solution in $\mathbb{N}$ including $0$, you get equality only when $x_1 =a$ and $x_2=b$. The other variables will be zero. You get similar solutions by shifting the variables around.

To minimize the expression, you will have $x_1 = a$ and either $y_2$, $z_2$ or $w_2$ set to $b$ and all other variables set to zero.


问:
Are these 2 graphs isomorphic?

They meet the requirements of both having an $=$ number of vertices ($7$).

They both have the same number of edges ($9$).

They both have $3$ vertices of degree $2$ and $4$ of degree $3$.

However, graph two has $2$ simple circuits of length $3$ whereas graph one has only $1$ of length $3$.

Is this not a valid method for checking isomorphism?

My guide says that these two figures are isomorphic.

答:
Once you know, as pointed out in this answer, that
$$f(A)=7,\: f(B)=4,\: f(C)=3,\: f(D)=6,\: f(E)=5,\: f(F)=2,\: f(G)=1$$ isomorphism, you can create an animation illustrating how to morph one graph into the other.  Let's say that ${vc}_1$ is a list of vertex coordinates for one and ${vc}_2$ is the corresponding list of vertex coordinates for the other.  (It's important that the order of the vertex coordinates be dictated by the isomorphism.)  We can then morph from one graph to the other using a function like

$$p(t) = t \, {vc}_2 + (1-t) {vc}_1.$$

Here's the result:

Since several folks asked for code generating the animation in the comments, you can find it here:

In Mathematica,

In Javascript, and

In LaTeX.

Title: Are these 2 graphs isomorphic?


问:
Trigonometric Integral Limit
My problem is to evaluate the following limit:

$$\lim_{n \to \infty}{\int^{n}_{0}{|\sin(x)|^n dx}}$$

I have no idea where to begin, but as you can tell the area under $\sin(x)^n$ decreases as $n$ approaches $\infty$. WolframAlpha can't evaluate this for $n>134$, where it equals about $9.29$. If I had to guess I would say it's probably $\infty$.

One way to tackle this could be to find a representation of $$\int^{2\pi}_{0}{|\sin(x)|^n dx}$$ and multiply by $\frac{n}{2\pi}$. This should approximate the integral for large $n$.

答:
Your idea works with some modifications. The integral turns out to be nice if $n$ is even but annoying if $n$ is odd, and we can reduce to the even case since $\int_0^{\pi} |\sin x|^n \, dx$ is monotonically decreasing as a function of $n$. (As Greg Martin observes in the comments, $|\sin x|$ is $\pi$-periodic, so we only have to integrate over an interval of length $\pi$.)

To begin, we'll rewrite the integral as

$$I(n) = \int_0^{\pi} |\sin x|^n \, dx = \int_0^{\pi} \sin^n x \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^n x \, dx$$

which is a bit easier to reason about; writing $\cos x = \frac{e^{ix} + e^{-ix}}{2}$ then gives

$$I(n) = \frac{1}{2^n} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \sum_{k=0}^n {n \choose k} e^{i(n-2k)x} \right) \, dx = \frac{1}{2^n} \sum_{k=0}^n {n \choose k} \left( \frac{e^{i(n-2k)\frac{\pi}{2}} - e^{-i(n-2k)\frac{\pi}{2}}}{i(n-2k)} \right)$$

which simplifies further to

$$I(n) = \frac{1}{2^{n-1}} \sum_{k=0}^n {n \choose k} \frac{\sin (n-2k)\frac{\pi}{2}}{n-2k}.$$

(We need a special case if $n$ is even and $n = 2k$ since then we're dividing by zero: in that case $e^{i(n-2k)x} = 1$ and the corresponding term above should be replaced with its limit as $k \to \frac{n}{2}$, which is $\frac{\pi}{2}$.)

If $n$ is even then every term disappears except the term $n = 2k$, which gives

$$I(2k) = \frac{1}{2^{2k}} {2k \choose k} \pi.$$

If $n$ is odd then the terms alternate in sign which is annoying but by the monotonicity observation from before we have $I(2k+1) \ge I(2k+2)$ which is good enough. We have

$$\int_0^n |\sin x|^n \ge \int_0^{\pi \lfloor \frac{n}{\pi} \rfloor} |\sin x|^n \, dx = \left\lfloor \frac{n}{\pi} \right\rfloor I(n)$$

so to prove that the limit is $\infty$ it suffices to prove that $I(n)$ grows faster than $\frac{1}{n}$. In fact it grows like $\frac{1}{\sqrt{n}}$: for example, Stirling's formula can be used to prove that the central binomial coefficients grow asymptotically like

$${2k \choose k} \sim \frac{2^{2k}}{\sqrt{\pi k}}$$

which gives

$$\boxed{ I(2k) \sim \sqrt{\frac{\pi}{k}} }$$

(and the same asymptotic for $I(2k+1)$ by monotonicity), and this is enough to conclude. As a sanity check, for $n = 134$ this asymptotic gives

$$\int_0^{134} |\sin x|^{134} \, dx \approx \frac{134}{\pi} \sqrt{ \frac{\pi}{67} } = 9.2 \dots$$

which roughly agrees with your calculation.

An alternative and much more general strategy for estimating integrals of this form (integrating a large power of some function) is to apply some form of Laplace's method; the idea here is that for large $n$ nearly all the contribution to the integral of $|\sin x|^n$ comes from the regions where $\sin x \approx 1$, since anything else falls off exponentially sharply as $n$ increases. So it suffices to estimate these contributions, which can generally be done using a Gaussian integral; this is where the square-root behavior comes from.

This actually gives an alternative proof of the asymptotic for the central binomial coefficients which avoids Stirling's formula, and Stirling's formula itself can be proven using a variation of this method applied to the integral $n! = \int_0^{\infty} x^n e^{-x} \, dx$, as explained e.g. in this blog post by Terence Tao.

Edit: A simple version of this idea, which I guess is an elaboration on Greg Martin's comment, already works. We'll switch to analyzing the integral $I(n) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^n x \, dx$ as before. As above, most of the contribution to this integral comes from the part where $\cos x \approx 1$ or equivalently $x \approx 0$. For $x < \sqrt{2}$ we have

$$\cos^n x \ge \left( 1 - \frac{x^2}{2} \right)^n \ge e^{- n \frac{x^2}{2}}$$

by Bernoulli's inequality, which gives

$$I(n) \ge \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} e^{-n \frac{x^2}{2}} \, dx.$$

(It doesn't matter much but it's convenient here that $\sqrt{2} < \frac{\pi}{2}$.) This integral can be lower bounded by $\delta$ where $\delta \le \sqrt{2}$ has the property that $e^{-n \frac{x^2}{2}} \ge \frac{1}{2}$ for all $x \in [-\delta, \delta]$. This gives $\delta = \sqrt{ \frac{2 \log 2}{n} }$, which is always less than $\sqrt{2}$, so we conclude that

$$I(n) \ge \sqrt{ \frac{2 \log 2}{n} } $$

which is within a multiplicative constant of the true asymptotic and good enough. We have $2 \log 2 = 1.386 \dots$ compared to the true asymptotic  $2 \pi = 6.283 \dots$ (here we're comparing the constants under the square root).

Just for fun, we can attempt to slightly improve the asymptotic above by optimizing the constant a bit. If we ask for $e^{-n \frac{x^2}{2}} \ge r$ where $r \in [0, 1]$ we get $\delta = \sqrt{ \frac{-2 \log r}{n} }$ and

$$I(n) \ge 2r \sqrt{ \frac{-2 \log r}{n} }.$$

Now we can try to optimize the constant as a function of $r$. It suffices to minimize $r^2 \log r$, which has derivative $2r \log r + r = r(2 \log r + 1)$; setting this equal to zero gives $r = e^{- \frac{1}{2} } = 0.606 \dots$ so $\delta = \frac{1}{\sqrt{n}}$ (as suggested by Greg!) and we've improved the constant under the square root from $2 \log 2 = 1.386 \dots$ to $4r^2 = \frac{4}{e} = 1.471 \dots$, giving

$$I(n) \ge \sqrt{ \frac{4}{en} }.$$

Title: Trigonometric Integral Limit


问:
For what value of $\alpha$ does the series $\sum(\frac{1}{n^\alpha}\times\sum_{k=1}^{n}\ln(k))$ converge?
I don't see how i can find an equivalent to the sequence, I have tried using Bertrand series.</p>

答:
Do you know about Riemann sums? If so, you know that, as $n\rightarrow+\infty$,  $$\frac 1 n \sum_{k=1}^n \ln\left ( \frac k n\right) \rightarrow \int_0^1\ln x dx=-1$$

Since $\ln \frac k n = \ln k - \ln n$, you can rewrite that as $$\left (\frac 1 {n}\sum_{k=1}^n \ln k\right) - \ln n \rightarrow -1$$
And thus
$$\left (\frac 1 {n}\sum_{k=1}^n \ln k\right)\sim\ln n$$
This implies that
$$\frac 1 {n^\alpha}\sum_{k=1}^n \ln k \sim \frac {\ln n}{n^{\alpha - 1}}$$
So your series converge iff $\alpha > 2$.


问:
Has this phenomenon been discovered and named?
If $$x-\frac{x}{2}=\frac{x}{2},$$ and $$\frac{x}{\sqrt{x}}=\sqrt{x},$$
and $$x-\uparrow(x-\uparrow^22)=x-\uparrow^22$$
when $(x\uparrow^n-[A])\uparrow^nA=x$, where $A$ is some constant, and one uses standard Knuth up-arrow notation[where ($\uparrow^{-1}x$)($-\uparrow^{-1}x$), ($\uparrow^0x$)($-\uparrow^{0}x$), ($\uparrow{x}$)($-\uparrow{x}$), ($\uparrow^2x$), and ($-\uparrow^2x$) all mean addition(and subtraction), multiplication(and division), exponentiation(and roots), and tetration(and super-roots) in that order],will
$$x-\uparrow^n(x-\uparrow^{n+1}2)$$
always yield
$$x-\uparrow^{n+1}2$$

for every degree $n$? If so, could someone please let me know where I can find an explanation of this identity?

Another, possibly more mathematically concise definition could be given by $$x-\uparrow^n\sum_{k=1}^{z-1}{x-\uparrow^{n+1}z}=x-\uparrow^{n+1}z$$

where the sigma could signify the sum, product, etc. depending on the degree of the up-arrow. $x\uparrow^az$ could be the $z$-th hyperpower($a$-power) of $x$ ($x$ $(n+2)$-ated to the $z$-th), or inversely, $x-\uparrow^az$ is the $z$-th hyperroot of x. Or in a different notation (MphLee's),

$$Hrt_n(x,\sum_{k=1}^{z-1}{Hrt_{n+1}(x,z))}=Hrt_{n+1}(x,z)$$

Where the $\sum$ still doesn't nessecarily represent the sum, but $z-1$ iterations of hyperoperations of degree $n$, and rank $z$.

Ex.
$$Hrt_{1}(x,\sum_{k=1}^{2}{Hrt_2(x,3)})$$

Here, $Hrt_1$ represents a difference between $x$ and $\sum_{k=1}^{2}{Hrt_2(x,3)}$. Sigma ($\sum$) is operating in the first degree; addition, or $H_1$ (note that although we are using $Hrt$ and not $H$, only the subscript or 'degree' matters when defining Sigma ($\sum$)), and is the sum of 2 ($\sum_{}^{2}$) identical terms: $$Hrt_2(x,3)$$ that is $\frac{x}{3}$. Therefore,$$Hrt_{1}(x,\sum_{k=1}^{2}{Hrt_2(x,3)})=x-((\frac{x}{3})_1+(\frac{x}{3})_2)=x-2(\frac{x}{3})=\frac{x}{3}$$.

I am fairly certain that this equation is 'clumsy' to some extent, but if anyone could suggest a better form of notation, I would greatly appreciate it.

Here is a better rendering:
$$Hrt_n(x,H_{n+1}(Hrt_{n+1}(x,z),[z-1]))=Hrt_{n+1}(x,z)$$
and if $Hrt_{n+1}(x,z)=T$, then$$Hrt_n(x,H_{n+1}(T,[z-1]))=T$$

答:
Sorry but I'm really getting confused with your up-arrow's  notation of the inverses (root-type inverse), I'll try to write this problem with this notation:

$H_n$ is the $n$-th hyperoperation and $Hrt_n$ is its hyper-root

$H_1(x,a)=x+a=y$ and its root is $$Hrt_1(y,a):=y-a=x$$

$H_2(x,a)=x\times a=y$ and its root is $$Hrt_2(y,a):=y/a=x$$

$H_3(x,a)=x^a=y$ and its root is $$Hrt_3(y,a):=\sqrt[a]{y}=x$$

$H_4(x,a)=x\uparrow \uparrow a=y$ (that is tetration) and its root is
  $$Hrt_4(y,a):=srrt_a(y)=x$$ ($a$-th super-root)

and if we have $H_n(x,a)=y$ than
$$Hrt_n(y,a):=x$$

$\mathcal Y$our statement:

$$x-\uparrow^n(x-\uparrow^{n+1}2)=x-\uparrow^{n+1}2$$ or
  $$x-\uparrow^nr=r$$ and $r=x-\uparrow^{n+1}2$

It can now be written as

$$Hrt_n(x,Hrt_{n+1}(x,2))=Hrt_{n+1}(x,2)$$ or $$Hrt_n(x,r)=r$$ and
  $r=Hrt_{n+1}(x,2)$

Now, we know that for the definiton of hyper-root must hold $H_n(Hrt_n(x,a),a)=x$, where $H_n$ is the $n$-th hyperoperation.

so if we apply this to your formula we get

$Hrt_n(x,r)=r$

$H_n(Hrt_n(x,r),r)=H_n(r,r)$

$x=H_n(r,r)$

Since for all Hyperoperations ($n\ge 1$) we have $H_n(r,r)=H_{n+1}(r,2)$ and $r=Hrt_{n+1}(x,2)$

$x=H_n(r,r)=H_{n+1}(r,2)=H_{n+1}(Hrt_{n+1}(x,2),2)=x$

$\mathcal O$bservation:

Here in the proof of your statement I'm using some theorems that I have not proved here, and I have instead assumed them to be true:

I assumed that $H_n(r,r)=H_{n+1}(r,2)$ for $n\ge 1$ that is same as $H_n(r,H_{n+1}(r,0))=H_{n+1}(r,1)=r$ and we can see that holds:

$r+(r\times 0)=r$

$r\times(r^0)=r$

$r^{(r\uparrow \uparrow 0)}=r$

and that $H_n(Hrt_n(x,a),a)=Hrt_n(H_n(x,a),a)=x$

the last affermation we know that does not hold always, for example it does't hold for the sqare-root or the $a$-th root ($\sqrt[a]{y}$) when we are working with negative (or complex) numbers because the function $f(x)=H_n(x,a)=y$ is not invertible than $f^{-1}(y)=Hrt_n(y,a)$ is not a fucntion, but it works for the naturals.


问:
How can I simplify the following sum : $\sum_{k=0}^{N}{\binom{mk}{j}\binom{N}{k}x^{mk}y^k}$
$j,m,N\in\mathbb{N}^\ast$, and $x,y\in\mathbb{R}^\ast_+$.

Any help or tips will be very much appreciated.

Body:
I am not sure that e general formula does exist. In any manner, this looks like hypergeometric functions.

There are quite clear patterns for
$$f_m=\sum_{k=0}^{N}{\binom{mk}{j}\binom{N}{k}x^{mk}y^k}$$
$$f_1=\binom{0}{j} \, _2F_1(1,-N;\frac{1-j}1;-x y)$$
$$f_2=\binom{0}{j} \,
   _3F_2\left(\frac{1}{2},\frac{2}{2},-N;\frac{1-j}{2},\frac{2-j}{2};-x^2 y\right)$$
$$f_3=\binom{0}{j} \,
   _4F_3\left(\frac{1}{3},\frac{2}{3},\frac{3}{3},-N;\frac{1-j}{3},\frac{2-j}{3},\frac{3-j}{3};-x^3 y\right)$$
$$f_4=\binom{0}{j} \,
   _5F_4\left(\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{4}{4},-N;\frac{1-j}{4},
   \frac{2-j}{4},\frac{3-j}{4},\frac{4-j}{4};-x^4 y\right)$$


答:
Okay, I tried this again and I hope @Laithy can help me figure out whether I got it correct this time. Feel free to chime in give your opinions, thanks!

$
f(x) =\begin{cases}
      1 & 0\leq x < \frac{1}{2} \\
      0 & x = \frac{1}{2} \\
      1 & \frac{1}{2} < x \leq 1
   \end{cases}
$

Let $\epsilon\in\mathbb{R}^+$, choose $\delta = \frac{\epsilon}{4}$ and let $P$ be the following partition of $[0, 1]$... $P = \{[0, \frac{1}{2}-\delta], [\frac{1}{2} - \delta, \frac{1}{2} + \delta], [\frac{1}{2} + \delta, 1]\}$.

Consider $[0, \frac{1}{2} - \delta]$. It follows that $m_1 = 1$ and $M_1 = 1$.

This implies: $m_1(x_1 - x_0) = \frac{1}{2} - \delta$ and $M_1(x_1 - x_0) = \frac{1}{2} - \delta$.

Now consider $[\frac{1}{2} - \delta, \frac{1}{2} + \delta]$. It follows that $m_2 = 0$ and $M_2 = 1$.

This implies: $m_2(x_2 - x_1) = 0$ and $M_2(x_2 - x_1) = 2\delta$.

Consider the final sub-division $[\frac{1}{2} + \delta, 1]$. It follows that $m_3 = 1$ and $M_3 = 1$.

This implies: $m_3(x_3 - x_2) = \frac{1}{2} - \delta$ and $M_3(x_3 - x_2) = \frac{1}{2} - \delta$.

All this gives the following results:

$U(f, P) = \frac{1}{2} - \delta + 2\delta + \frac{1}{2} - \delta = 1$

and

$L(f, P) = \frac{1}{2} - \delta + 0 + \frac{1}{2} - \delta = 1 - 2\delta$

Therefore...

$U(f, P) - L(f, P) = 1 - (1 - 2\delta) = 2\delta = 2\big(\frac{\epsilon}{4}\big) = \frac{\epsilon}{2} < \epsilon$

$f$ is Riemann-Integrable.

Title: Prove that the function is Riemann-Integrable.

Body:
This is an incorrect attempt! Look below for @Laithy 's advice and my second attempt.

I tried this problem but I'm not sure whether I am getting it right. I am having trouble with the very last step where I am supposed to show that $I^-(f) = I^+(f)$. Here is my attempt. Please point out any errors and help me justify why $I^-(f)$ and $I^+(f)$ are equal. Thank you!

$
f(x) =\begin{cases}
      1 & 0\leq x < \frac{1}{2} \\
      0 & x = \frac{1}{2} \\
      1 & \frac{1}{2} < x \leq 1
   \end{cases}
$

Let $P$ be the following partition $P = \{[0, \frac{1}{2}], [\frac{1}{2}, 1]\}$ of $[0, 1]$.

Consider $I_1 = [0, \frac{1}{2}]$. It follows that $m_1 = 0$ and $M_1 = 1$.

This implies $m_1(x_1 - x_0) = 0$ and $M_1(x_1 - x_0) = 1(\frac{1}{2} - 0) = \frac{1}{2}$.

Now consider $I_2 = [\frac{1}{2}, 1]$. It follows that $m_2 = 0$ and $M_2 = 1$.

This implies $m_2(x_2 - x_1) = 0$ and $M_2(x_2 - x_1) = 1(1 - \frac{1}{2}) = \frac{1}{2}$.

All this gives the following:

$S^- = 0$ and $S^+ = 1$.

Therefore...

$I^-(f) = \sup\{S^-, \sigma\in G[0, 1]\} = \sup\{0, \sigma\in G[0, 1]\}$

and

$I^+(f) = \inf\{s^+, \sigma\in G[0, 1]\} = \inf\{1, \sigma\in G[0, 1]\}$

How do I decide whether $I^-(f)$ equals $I^+(f)$?


问:
Does left-multiplication by compact operators turn strong-convergence into norm-convergence?
If $\{T_i\}_{i\in I}$ is a bounded net of operators on a Hilbert space $\mathscr H$,  converging strongly to some operator $T$, and
if $K$ is  a compact operator on $\mathscr H$, then the net
$\{T_iK\}_i$ is known to converge in norm to $TK$.

Question. Is it also true that   $\{KT_i\}_i$  converges in norm to $KT$?

PS:

The present question arouse in the comments following this answer and, while I can't remember ever questioning it, neither do I remember this being discussed anywhere.  After a while I
finally figured out the answer and I thought it would be nice to record it here.

An affirmative answer to my question is implicitly assumed in the statement of this question.

答:
This may come as a surprise but the answer is no.  In fact it is no in a very strong way!!

There exists a sequence $\{T_n\}_n$ of bounded operators, strongly converging to zero, such that $\{KT_n\}_n$ does not converge to zero in norm for every nonzero bounded operator $K$, regardless of whether $K$ is compact or even finite rank!

Let $S:\ell ^2\to \ell ^2$ be the so called unilateral shift given by $$ S(x_0, x_1, x_2, x_3, \ldots ) = (0, x_0, x_1, x_2, \ldots ).  $$ The adjoint of $S$ turns out to be given by $$ S^*(x_0, x_1, x_2, \ldots ) = (x_1, x_2, x_3, \ldots ), $$ from where one has that $S^*S$ is the identity operator.

Now let $T_n={S^*}^n$, so that $$ T_n(x_0, x_1, x_2, \ldots ) = (x_n, x_{n+1}, x_{n+2}, \ldots ), $$ and hence it is clear that $T_n\to 0$ strongly.

If $K$ is any bounded operator (compact or not) we have that $$ \|K\| = \|K{S^n}^*S^n\| \leq \|K{S^n}^*\|\|S^n\| = \|K{S^n}^*\| = \|KT_n\|, $$ so we can't have $\|KT_n\|\to 0$, unless $K=0$.

As mentioned in the original post, if $K$ is compact, it is well known that "$T_n\to 0$ strongly" implies "$T_nK\to 0$ in norm".

If $T_n^*\to 0$ strongly (e.g. if $T_n\to 0$ in the $^*$-strong topology, or if $T_n$ is self-adjoint and converges to zero strongly) then $T_n^*K^*\to 0$ in norm by (1) and hence $KT_n\to 0$ in norm as well.


问:
Parametric equation for Taubin heart surface
We know the implicit equation for building a Taubin's heart surface: $$\\\left(x^2+\frac{9y^2}{4}+z^2-1\right)^3-x^2 z^3-\frac{9y^2 z^3}{80}=0$$
Can we convert to parametric equation equivalent?

What tools can we use to solve this?

If the answer is positive, i am interested about parametric equation for this surface

答:
I am not sure, if it's too late for helping, but you can try below an alternative equation to plot a parametric heart surface.

Julia's parametric heart surface equation:

x=Sin[v](15 Sin[u]-4 Sin[3u])

y=8 Cos[v]

z=Sin[v](15 Cos[u]-5 Cos[2u]-2 Cos[3u]-Cos[2u])

In Wolfram Mathematica Code:

```<code>ParametricPlot3D[{Sin[v](15Sin[u]-4 Sin[3 u]),8Cos[v],Sin[v](15Cos[u]-5 Cos[2u]-2 Cos[3 u]-Cos[2 u])},{u,0,2Pi},{v,0,Pi},Axes->True,AxesLabel-> {"x","y","z"},PlotStyle->Red,PlotLabel->Style["Julia's Parametric Heart Surface",14,Bold],ViewPoint->Front]
</code>```

Julia's parametric heart surface

Body:
I am not sure, if it's too late for helping, but you can try below an alternative equation to plot a parametric heart surface.

Julia's parametric heart surface equation:

x=Sin[v](15 Sin[u]-4 Sin[3u])

y=8 Cos[v]

z=Sin[v](15 Cos[u]-5 Cos[2u]-2 Cos[3u]-Cos[2u])

In Wolfram Mathematica Code:

ParametricPlot3D[{Sin[v](15Sin[u]-4 Sin[3 u]),8Cos[v],Sin[v](15Cos[u]-5 Cos[2u]-2 Cos[3 u]-Cos[2 u])},{u,0,2Pi},{v,0,Pi},Axes->True,AxesLabel-> {"x","y","z"},PlotStyle->Red,PlotLabel->Style["Julia's Parametric Heart Surface",14,Bold],ViewPoint->Front]

Julia's parametric heart surface


问:
The time gap between the two instants, one before and one after 12:00 noon, when the angle between the hour hand and the minute hand is 66°

Question: The time gap between the two instants, one before and one after $12:00$ noon, when the angle between the hour hand and the minute hand is $66^°$ , is

$1.~~~~ 12 ~\text{min}.$
$2.~~~~ 16~\text{min}.$

$3.~~~~ 18 ~\text{min}.$
 $4.~~~~ 24 ~\text{min}.$

My thought: We know that at $12:00$ noon, both the hour and the minute hand are at position $0^°$.
Also after $x$ hours of time, the hour hand travels $x / 12$ rotations around the clock. So after $x$ minutes, it travels $x / (60 \cdot 12) = x / 720$.

After $x$ minutes of time, the minute hand travels $x / 60$ rotations around the clock.

Now how to proceed the further ? Please help.

答: The hour hand travels $360^\circ$ in $12$ hours, or $720$ minutes.  That is $\frac 12^\circ$ per minute.  The minute hand travels $360^\circ$ in $60$ minutes or $6^\circ$ per minute, so the minute hand gains $\frac {11}2$ degrees/minute.  To gain $66^\circ$ therefore takes $12$ minutes.  Similarly, going backwards, the minutes hand loses $66^\circ$ in $12$ minutes, so the times are $24$ minutes apart.


问:
Is there any common method to evaluate $\lim_{x\to +\infty} \left( \frac{\ln^4 x}{4}-\int_0^x \frac{\ln^3 t}{1+t} \, dt \right)$?
So is there any common method to solve this kind of problem?

Calculate that $$\lim_{x\to +\infty} \left( \frac{\ln^4 x}{4}-\int_0^x \frac{\ln^3 t}{1+t} \, dt \right)$$

Before this I have seen a similar problem which was solved by using digamma function or beta function. So is that mean we can always use special functions to solve them?

答:
By using Puiseux series I get that $\int_{0}^{x}\frac{\ln^3 t}{1+t} dt=(\frac{\ln^4 x}{4}-\frac{7\pi^4}{60})+O(\frac{1}{x})$

So it may be $\frac{7\pi^4}{60}$ at last.


问:
Better understanding of the meaning of implication in equations
While solving this equation
$$\sqrt{x^2+1}+x=1$$
I started to have some doubts about my real logical understanding of what I'm doing: I will show the steps to try to communicate better what I mean. We know that adding/subtracting functions that belong to the domain of the original equation gives an equivalent equation (meaning it has the same solutions of the first one), so if I want to add $-x$ both sides the correct logical operator is
$$\sqrt{x^2+1}+x=1\iff\sqrt{x^2+1}=1-x $$
Going on, I know that is true that $a=b\implies a^2=b^2$, so the next step is not an equivalence but an implication; meaning that the correct logic operator is
$$\sqrt{x^2+1}=1-x \implies x^2+1=(1-x)^2$$
Now since $(1-x)^2=x^2-2x+1$, I suppose that an equality is an equivalence (correct me if I'm wrong) so I get
$$x^2+1=(1-x)^2 \iff x^2+1=x^2-2x+1 \iff -2x=0 \iff x=0$$
So, since they are all equivalence and there is an implication, connecting the first and the last I get
$$\sqrt{x^2+1}=1-x \implies x=0$$
So the questions are:

a) since an equation is the problem to determine which unknowns, substituted in the equality, makes the equality true, it seems to me, of course wrongly, that we've obtained the converse, that is "if equation is zero then the unknowns are $*$" while, if I've not understood wrong, we are searching for what unknowns make the equation verified, so we're searching for the converse implication "if unknowns are $*$ these then the equation is verified". What am I missing here?

b) In the equation $\sqrt{x^2+1}=1-x$ I have on the left hand side a non negative quantity while on the right hand side a quantity that can be negative for some values of $x$; so, when arriving to $(2)$, I should add a condition $1-x \geq 0$ to have a real equivalence?

Thanks.

答:
Since an equation is the problem to determine which unknowns,
substituted in the equality, makes the equality true, it seems to me,
of course wrongly, that we've obtained the converse.

Nope. Nothing wrong with that. It just has to be presented properly.

Theorem: $\sqrt{x^2+1}=1-x \iff x=0$

Proof: You have already proven that $\sqrt{x^2+1}=1-x \implies x=0$. Now, it is trivial to prove the converse: $$x=0 \implies (\sqrt{x^2+1}= 1 ~~\land~~ 1-x=1)\implies \sqrt{x^2+1}=1-x ~~\square$$


问:
Maximum flow in a connected digraph
Let $G=(X,U)$ be a connected digraph and $(G,c)$ a network. Let $s$ and $t$ be different vertices such that $d^+(s)>0$ and $d^-(t)> 0 $. Show that if there is no path from $s$ to $t$ in $G$ then the null flow is a maximum flow from $s$ to $t$ in the $(G,c)$ network.

I was thinking of doing it by contradiction, supposing that the maximal flow is not null for this network, so we would have to have a path from $s$ to $t$ which contradicts the hypothesis. I'm not sure, it's just an idea. Does someone have any hint on how can I start the proof?

答: Yes, you are essentially proving the contrapositive, which is a reasonable approach.</p>

Title: Maximum flow in a connected digraph


问:
How to find the point of symmetry for $f(x) = \frac{6}{1+3\cdot(0.4)^x}$ algebraically?
In analyzing the graph of the function $$f(x) = \frac{6}{1+3\cdot(0.4)^x}$$ the textbook lists that it has symmetry around the point (1.2, 3). Is there a way to find the coordinates of this point algebraically, without tracing the graph or using calculus to find the inflection point?

Thank you!

答:
It is so simple with calculus
$$f(x)=\frac{a}{1+b c^x}$$ $$f'(x)=-\frac{a b c^x \log (c)}{\left(b c^x+1\right)^2}\qquad \text{and} \qquad f''(x)=\frac{a b c^x \log ^2(c) \left(b c^x-1\right)}{\left(1+b c^x\right)^3}$$

So, the inflection point is at
$$x_*=-\frac{\log (b)}{\log (c)}\implies f(x_*)=\frac a 2$$

This was the long way.

The short way is to look for $x$ such that
$$\frac{a}{1+b c^x}=\frac a 2\implies 1+b c^x=2\implies b c^x=1\implies x=-\frac{\log (b)}{\log (c)}$$


问:
Build a cheaper highway
We consider a non-directed graph, where each vertex is a city and each edge is a possible road connecting two cities.

I need to build a cheap highway (a path on a graph) connecting all the cities starting and ending with the main city. I know the cost to build each edge that connects two  verteces (connects two cities). This is the only restriction: to visit all vertices of the graph. The goal is to minimize the cost to build this highway.

This problem is similar to the Traveling salesman problem, but here I can visit each vertex of the graph more than once, and also only the first time that I use an edge I pay to build it, but from the second it has no cost.

I'm not sure if I can connect the Traveling salesman problem with my problem, may be there are other way to solve it. I appreciate a lot your help.

答: This is the minimum spanning tree problem, and a greedy algorithm is optimal.</p>

Title: Build a cheaper highway


问:
Spivak's Calculus 12-10(b) Solution seems incorrect
Hi I just want to confirm that this solution given for Chapter 12 Question 10(b) in the Answers Book is incorrect.  The question asks for an expression for the Schwarzian derivative of inverse $f$.  The solution says:

But instead of $1$ we should have:

$0 = \mathcal{D}(\operatorname{id}) = \mathcal{D}(f\circ f^{-1}) = \bigl((\mathcal{D}f)\circ f^{-1}\bigr)\cdot \bigl((f^{-1})'\bigr)^2 + \mathcal{D}(f^{-1}).$  As suggested in this duplicate question: Schwarzian derivative of inverse function..

For reference this is problem 10-17(a):

答:
I just ran into this exercise myself and noticed the same thing. It's nice when MSE confirms a suspected error. For the record, I'm using 3rd Edition.

The Answer Book solution actually seems to contain two other substantial errors. Not Spivak's best outing!

Though the question is old, I'm offering an answer, less for the original poster and more for the next person that comes across this exercise:

Error 1: We need to first show that $f^{-1}$ is differentiable. The Answer Book fails to do this. Without this, neither 10(a) or 10(b) can be completed.

We are told that $\mathcal{D}f(x)$ exists for all $x$. From the definition of the Schwarzian derivative, this means that $f^\prime(x)$ exists and is $\neq 0$.

Thus $f$ is continuous.

Furthermore, suppose $f(a) = f(b)$ for some $a < b$. By Rolle's there exists some $x$ in the interval $(a,b)$ such that $f^\prime(x)= 0$.

Since we know $f^\prime(x)\neq 0$, this cannot be. Therefore, $f(a)\neq f(b)$ for any $a \neq b$, i.e. $f$ is one-one (and either increasing or decreasing).

To summarize, $f$ is a continuous one-one function with derivative $f^\prime(x) \neq 0$ for all $x$. Therefore, by the theorems in the chapter, $f^{-1}$ is continuous, one-one (increasing or decreasing) and differentiable for all $x$ in its domain.

Furthermore, $f^{-1\prime} \neq 0$, because if it were, $f$ would not be differentiable everywhere.

Now that we've demonstrated $f^{-1}$ is differentiable with $(f^{-1})^\prime \neq 0$, we can go about showing it's three times differentiable, as the book solution does. This will show that $\mathcal{D}f^{-1}(x)$ exists for all $x$.

Error 2:
The Book answer for 10(a) lists the third order derivative of $f^{-1}$ as

$$(f^{-1})^{\prime\prime\prime}(x) = \frac{-[f^\prime(f^{-1}(x))]^3 f^{\prime\prime\prime}(f^{-1}(x)) + 3f^{\prime\prime}(f^{-1}(x))[f^\prime(f^{-1}(x))]^2}{[f^\prime(f^{-1}(x))]^7}$$

or, to simplify the expression a bit:

$$= \frac{-[f^\prime]^3 f^{\prime\prime\prime} + 3f^{\prime\prime}[f^\prime]^2}{[f^\prime]^7} \circ f^{-1}(x)$$

This is wrong. There seems to have been a calculation error, with the book missing a power of $f^{\prime\prime}$ in the numerator.

The correct expression, after simplifying is (I think):

$$(f^{-1})^{\prime\prime\prime}(x) = \frac{-f^{\prime\prime\prime}f^\prime + 3(f^{\prime\prime})^2} {[f^\prime]^5} \circ f^{-1}(x)$$

You can use this expression along with those for the first and second order derivatives to calculate part (b).

Error 3: As @helios321 points out, the answer for part (b) incorrectly states that $\mathcal{D}(f \circ f^{-1}) = 1$.

Actually,

$$\mathcal{D}(f \circ f^{-1}) = \mathcal{D}x = 0$$

To complete part (b) we can use this along with the identity from Problem 10-17(a) as suggested by the Answer Book, or simply use the derivatives of $f^{-1}$ and the definition of $\mathcal{D}$ to calculate $\mathcal{D}f^{-1}$.

Either way, (I believe) we should end up with
$$\mathcal{D}f^{-1}(x)= \frac{3f^{\prime\prime2} - 2f^{\prime\prime\prime}f^\prime}{2f^{\prime 4}} \circ f^{-1}(x)$$


问:
Diagonal triangle is autopolar
I was searching things about the diagonal triangle of a quadrilateral. I found in MathWorld's "Diagonal Triangle" entry the following question:

If the quadrangle is a cyclic quadrilateral, then the circle is the polar circle of the diagonal triangle.

How can I prove that? I tried using the fact that is cyclic, but it doesn't leave me to an answer.

答:
Here is a possible solution. First, let us insert a picture with labelled vertices:

We use it for several partial results, some of them involve only a few vertices, please draw the simplified picture on paper.

Recall the properties of pole and polar:
en.wikipedia.org/wiki/Pole_and_polar

Recall the cross ratio and harmonic conjugations:
en.wikipedia.org/wiki/Cross-ratio

Proposition:

Let $ABCD$ be a cyclic quadrilateral. Let $O$ be the center of the circle
$(ABCD)$. Let $Q$ be the intersection $AD\cap BC$. Assume it is an exterior point to the circle $(O)=(ABCD)$. We denote the inversion centered in $O$ that
invariates pointwise the circle $(O)$ by a star. So $Q^*$ is the inverse of $Q$, it can be constructed by drawing the tangents from $Q$ to $(O)$, let $Q_1,Q_2$ be the tangent points, then $Q^*=OQ\cap Q_1Q_2$.

Let $P,R$ be the intersections $P=AC\cap BD$ and $R= AB\cap CD$.

Also draw $X=RP\cap QAD$, $Y=RP\cap QBC$.

Then we have:

(a) The line $Q_1Q_2$ is the polar of $Q$.

(b) $Q$ is the pole of $Q_1Q_1$.

(c) $X,Q$ are harmonic conjugates w.r.t. $A,D$.

(d) $Y,Q$ are harmonic conjugates w.r.t. $B,C$.

(e) $XY$ is the polar of $Q$.

(f) $Q_1,Q_2;Q^*;X,Y;P,R$ are colinear.

Proof: Some parts are simple, they are enumerated to have the steps of the result displayed clearly.

(a), (b) are standard.

(c) follows by applying the theorems of Ceva and Menelaus in $\Delta RAD$:
$$
\begin{aligned}
-1 &=
\frac{XA}{XD}\cdot
\frac{CD}{CR}\cdot
\frac{BR}{BA}\ ,
&&\text{ Ceva w.r.t. the point $P$, and cevians $RPX$, $APC$, $DPB$,}
\\
+1 &=
\frac{QA}{QD}\cdot
\frac{CD}{CR}\cdot
\frac{BR}{BA}\ ,
&&\text{ Menelaus w.r.t. the secant line $QBC$.}
\end{aligned}
$$
(d) is (c) for the situation obtained by exchanging
$A\leftrightarrow B$,
$C\leftrightarrow D$.

(e) From (c), (d) both $X,Y$ are on the polar of $Q$, so this polar line is $XY=XYPR=PR$.

$\square$

Proposition:
$QR$ is the polar of $P$.

Proof: $P$ lies on the polar of $Q$, so $Q$ lies on the polar of $P$, La Hire's theorem. Similarly, $R$ lies on the polar of $P$. So $QR$ is the polar of $P$.

$\square$

Note: The parts (a), (b) of the first proposition are not needed, but this is the way i constructed $Q^*$ in geogebra.

Title: Diagonal triangle is autopolar


问:
Question on Tom Apostol's Proof for Method of Exhaustion
Currently, I am going through Tom Apostol's "Calculus: Volume 1 Second Edition" as I heard this was a good book to get a solid understanding of calculus and I wasn't satisfied with James Stewart's 9th edition calculus book.

I am on page 6 of the book and I am having some trouble. On pages 5 and 6, Apostol is proving that the area under the curve of $y = x^2$ between $x = 0$ and $x = b$ is equal to $\frac{b^3}3$, or $\frac{1}3$ of a rectangle with a base of length $b$ and a height of $b^2$. After proving that $1^2 + 2^2 +...+(n-1)^2 = $ $\frac{n^3}3 +$  $\frac{n^2}2 + $ $\frac{n}6$ (although I didn't fully understand how Apostol proved that either, this video by blackpenredpen https://www.youtube.com/watch?v=OpA7oNmHobM&ab_channel=blackpenredpen cleared that up for me so if anyone has trouble with that, check the link out), Apostol goes on to say:

For our purposes, we do not need the exact expressions given in the right-hand members of (I.3) and (I.4), or they can be proved directly by induction. (A proof by induction is given in Section I 4.1.)

All we need are the two inequalities

$1^2 + 2^2 + ... +   (n-1)^2$ < $\frac{n^3}3$ < $1^2 +    2^2 + ... + n^2$

which are valid for every integer n $\geq$ 1. These inequalities can be deduced easily as consequences of (I.3) and (I.4), or they can be proved by induction. (A proof by induction is given in Section I 4.1.)

If we multiply both inequalities in (I.5.) by $\frac{b^3}{n^3}$ and make use of (I.1) and (I.2), we obtain

$s_n$ < $\frac{b^3}3$ < $S_n$

for every n. The inequalities in (1.6) tell us that $\frac{b^3}3$ is a number which lies between $s_n$ and $S_n$ for every n. We will now prove that $\frac{b^3}3$ is the only number which has this property. In other words, we assert that if A is any number which satisfies the inequalities

$s_n$ < A < $S_n$

for every positive integer n, then A = $\frac{b^3}3$. It is because of this fact that Archimedes concluded that the area of the parabolic segment is $\frac{b^3}3$.

My questions arises at the inequality: $1^2 + 2^2 + ... +   (n-1)^2$ < $\frac{n^3}3$ < $1^2 +    2^2 + ... + n^2$

I understand that the values less than $\frac{n^3}3$ are the sum of squares of the heights of rectangles under the parabolic segment where $y=x^2$ and the values greater than $\frac{n^3}3$ are the sum of squares of the heights of rectangles that extend beyond the parabolic segment where $y=x^2$.

My question is, how did Apostol come up with the idea to use $\frac{n^3}3$ in the inequality?

I have looked to this link Method of Exhaustion applied to Parabolic Segment in Apostol's Calculus before I asked this question as this was the closest I found to someone asking a similar question but I could not find the answer.

For context, I have provided images of pages 1 through 8 so I can show you all that Apostol is doing for this proof.

Also, sorry for any formatting errors as this is my first time posting on this site.

Images:

Pages 1 and 2

Pages 3 and 4

Pages 5 and 6

Pages 7 and 8

答:
In the early 1700's, Bernoulli proved (without Calculus) that for any positive integer $k$, there exists a unique set of rational numbers $c_0, c_1, \cdots, c_k$ such that

for any positive integer $n$,
$$\sum_{i=1}^n i^k = c_0n^{(k+1)} + c_1n^k + \cdots + c_kn^1.$$

Actually, Bernoulli took it one step further, and originated the idea of Bernoulli numbers.  For the purposes of this discussion, that is irrelevant.

Bernoulli's demonstration is (partially) recreated here.

So, it was well known, for centuries that

$$\sum_{i = 1}^n i^3 = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n.$$

What Apostol is doing in these first few pages is giving you an informal (but valid) glimpse of estimating the area under the curve through a series of step functions (e.g. rectangles).  His algebra in the pages that you cited is valid.

Whoever originated the idea that Apostol is merely repeating had the vision that the area of the inner rectangles would always be less than $\frac{b^2}{3}$ but would get closer and closer to $\frac{b^2}{3}.$

The analyis is that the $k$-th rectangle has width $\frac{b}{n}$ and height $\left(\frac{kb}{n}\right)^2.$  Therefore, the $k$-th rectangle has area $\frac{b^3}{n^3}k^2.$

Therefore, the sum of the area of the inner rectangles is

$$\frac{b^3}{n^3} \times \sum_{k=0}^{(n-1)}k^2$$

and the sum of the area of the outer rectangles is

$$\frac{b^3}{n^3} \times \sum_{k=1}^{n}k^2.$$

From the above algebra, it is immediate that as $n \to \infty$, the sum of the area of the inner triangles will converge to $\frac{b^3}{3}$ from below, and the sum of the area of the outer triangles will converge to $\frac{b^3}{3}$ from above.

Therefore, Apostol is (through the informal use of the squeeze theorem of limits) repeating the idea that some Calculus pioneer envisioned: the exact area under the curve has to be $\frac{b^3}{3}.$

What may be confusing is that Apostol's (algebra-only, no Calculus) derivation of the formula for $\sum_{i=1}^n i^3$, on page 6, though valid, is somewhat non-standard.  The non-Calculus derivation normally uses induction, as in the mathSE link that I cited.


问:
curve with constant curvature
Let $\gamma$: $I\twoheadrightarrow C\subset \mathbb{R}^2$ be a plane curve with constant curvature $\kappa>0$. Show that $C$ is part of a circle with radius $\kappa^{-1}$.
How to start this question, I don't have any clue.

答:
Parametrizing the curve in terms of it's arclength, we can assert that $\gamma(s)=(x(s), y(s))$ with $||\gamma'(s)||=1$. Then the normal vector is defined by

$$\mathbf{N}=\frac{\frac{d\mathbf{T}}{ds}}{||\frac{d\mathbf{T}}{ds}||}=\frac{\gamma''}{||\gamma''||}$$
since of course $\mathbf{T}=\gamma'(s)$. Setting $v(s)=||\gamma''(s)||$ we write the only Frenet equation:

$$\frac{d\mathbf{N}}{ds}=\frac{\gamma'''v-\gamma'' v'}{v^2}=-\kappa\gamma'$$

This is a differential equation for $\gamma(s)$. It is pretty difficult to solve in general but in this case if we take a dot product with $\gamma'$ and taking into account that
$$\gamma'''\cdot\gamma'=(\gamma''\cdot\gamma')'-\gamma''^2=\frac{1}{2}(\gamma'^2)''-\gamma''^2=-v^2$$
we see that
$$-\kappa=-v-\frac{v'}{v^2}\gamma''\gamma'\Rightarrow ||\gamma''(s)||=\kappa$$
So we learned that the norm of the acceleration vector is a constant. Putting everything back in our equation we get the linear 3rd order ODE
$$\gamma'''(s)+\kappa^2\gamma'(s)=0$$ which has the general solution
$$\gamma(s)=\gamma_0+\frac{\mathbf{A}}{\kappa}\cos(\kappa s)+\frac{\mathbf{B}}{\kappa}\sin(\kappa s)$$
where $\mathbf{A,B}$ are arbitrary vectors. We need to fulfill both conditions on the norm of the acceleration and the velocity however which give
$$||\gamma''(s)||=\kappa||\mathbf{A}\cos(\kappa s)+\mathbf{B}\sin(\kappa s)||=\kappa$$
This condition makes it easy to see now that
$$||\gamma(s)-\gamma_0||=\frac{||\gamma''(s)||}{\kappa^2}=\frac{1}{\kappa}$$
which is evidently the equation for a circle in $\mathbb{R}^2$. This also offers an explicit parametrization of the circle:
The constraint requires for all values of the arclength that
$$\frac{\mathbf{A}^2+\mathbf{B}^2}{2}+\frac{\mathbf{A}^2-\mathbf{B}^2}{2}\cos(2\kappa s)+\mathbf{A}\cdot\mathbf{B}\sin(2\kappa s)=1$$
which means that
$$||\mathbf{A}||=||\mathbf{B}||=1~,~ \mathbf{A}\cdot\mathbf{B}=0$$
Note that the constraint $||\gamma'(s)||=1$ is automatically satisfied with these constraints on the vectors. This allows us to parametrize the vectors as
$$\mathbf{A}=(\cos\theta, \sin\theta)~,~\mathbf{B}=(-\sin\theta, \cos\theta)$$
which in turn give the explicit form
$$\gamma(s)=\gamma_0+\frac{1}{\kappa}(\cos(\kappa s+\theta),\sin(\kappa s+\theta))$$
where $\gamma_0, \theta$ are arbitrary.

Title: curve with constant curvature


问:
Prove that $(2,1+\sqrt{-5})^2 = (2)\,$ in $\,\Bbb Z[\sqrt{-5}]$

Let $p$ be the ideal $\{2a+(1+\sqrt{-5})b\mid a,b\in\mathbb{Z}[\sqrt{-5}]\}$ in $\mathbb{Z}[\sqrt{-5}]$. Prove that $p^2$ is the principal ideal $(2)$.

I tried multiplying the ideal with itself and tried to simplify to see if I could relate it to the principal ideal $(2)$ but I can't seem to get it. This is what I have so far.

$(2a+(1+\sqrt{-5})b)^2=4a^2+4(1+\sqrt{-5})ab+(1+\sqrt{-5})^2b^2$

$=4a^2+(4+4\sqrt{-5})ab+(2\sqrt{-5}-4)b^2$

$=2(2a^2+(2+2\sqrt{-5})ab+(\sqrt{-5}-2)b^2)$

If $x\in p^2$, then $x$ is a multiple of $2$, so $x\in(2)$. Thus $p^2\subseteq (2)$.

I am unsure of how to show that $(2)\subseteq p^2$.

答:
Here’s another approach:

To multiply two ideals, say $I=(a_1,a_2,\cdots,a_m)$ and $J=(b_1,\cdots,b_n)$, where what’s in each pair of parentheses is a list of generators of the ideal, all you need to do is write down the products $a_ib_j$, all of them, and see what ideal they generate.

In the present case, we do this:
\begin{align}
\left(2,1+\sqrt{-5}\,\right)\left(2,1+\sqrt{-5}\,\right)&=\left(4,2+2\sqrt{-5},-4+2\sqrt{-5}\,\right)\\
&=\left(4,2+2\sqrt{-5},2\sqrt{-5}\,\right)\\
&=\left(4,2,2\sqrt{-5}\,\right)\\
&=\left(2,2\sqrt{-5}\,\right)=(2)
\end{align}
Notice that each simplification is reversible: you can go from left to right, as written, but also from right to left.


问:
$\lim_{(x,y)\rightarrow(0,0)}\frac{x^2y^2}{x^2y^2+(x-y)^2}$
I think the answer to this limit is zero-- but I can't seem to prove it. I've thought of using the squeeze theorem, or using the $\epsilon-\delta$ definition, but they don't seem to work out-- neither does applying polar coordinates solve the question. If I try proving that it does not exist, i try evaluating the limit along $y=mx$, along $y=x^2$ and all of them lead to the answer $0$. Any help would be appreciated!

答:
Consider the curves $\Gamma_{1}(t) = (t,t)$ and $\Gamma_{2}(t) = (t,-t)$. Then we have that
\begin{align*}
\lim_{t\to 0}f(\Gamma_{1}(t)) = \lim_{t\to 0}\frac{t^{4}}{t^{4} + 0^{2}} = 1
\end{align*}

On the other hand, one has that
\begin{align*}
\lim_{t\to 0}f(\Gamma_{2}(t)) = \lim_{t\to 0}\frac{t^{4}}{t^{4} + 4t^{2}} = \lim_{t\to 0}\frac{t^{2}}{t^{2} + 4} = 0
\end{align*}

Consequently, the proposed limit does not exist.