问:
Generalization of order of the product of elements in an abelian group as follows.
Let $(G,*)$ be a group. Let $a_i \in G, |a_i|=n_i, 1 \le i \le m$. Suppose $\gcd(n_i,n_j)=1$ and $a_ia_j=a_ja_i$, for all $i$ and $j$. Let $x=a_1*a_2*\ldots*a_m$. Show that
$|x| = n_1n_2\ldots n_m$.

Attempt:
Here I got so far. Please give me a correction if I make some mistakes.

Write $a*b=ab$.
Let $a_i \in G, |a_i| = n_i$ where $1\le i \le m$. Given that $\gcd(n_i,n_j) = 1$ and
$a_ia_j = a_ja_i$ for all $i$ and $j$.
Let $e_G$ be the identity element in $G$.

Let $P(m)$ be the statement:

The order of $a_1a_2\ldots a_m$ is $n_1n_2\cdots n_m$.

Base Case:
Consider $P(2)$. That is, the order of
$a_1a_2$ is $n_1n_2$. We'll show it now.
Note that $|a_1| = n_1$ and $|a_2|=n_2$.
Let $|a_1a_2| = s$.

To show that $s=n_1n_2$:

$s \mid n_1n_2$:
Notice that
\begin{align*}
(a_1a_2)^{n_1n_2} &= \overset{n_1n_2}{\overbrace{abab \ldots ab}} \\
&= \overset{n_1n_2}{\overbrace{a_1a_1\ldots a_1}}
\overset{n_1n_2}{\overbrace{a_2a_2\ldots a_2}} \\
&= a_1^{n_1n_2}a_2^{n_1n_2} \\
&= (a_1^{n_1})^{n_2}((a_2)^{n_2})^{n_1} \\
&= (e_G)(e_G) \\
&= e_G
\end{align*}
Hence, $s \mid n_1n_2$, as desired.

$n_1n_2 \mid s$:
Consider that
\begin{align*}
e_G = (a_1a_2)^s &= (a_1a_2)^{sn_1} \\
&= a_1^{sn_1}a_2^{sn_1} \\
&= (a_1^{n_1})^sa_2^{sn_1} \\
&= (e_G)a_2^{sn_1} \\
&= a_2^{sn_1}.
\end{align*}
Hence, $n_2 \mid sn_1$. Since
$\gcd(n_1,n_2)=1$, then we must have
$n_2 \mid s$. Similarly, consider
\begin{align*}
e_G = (a_1a_2)^s &= (a_1a_2)^{sn_2} \\
&= a_1^{sn_2}a_2^{sn_2} \\
&= a_1^{sn_2}(a_2^{n_2})^s \\
&= a_1^{sn_2}(e_G) \\
&= a_1^{sn_2}.
\end{align*}
Hence, $n_1 \mid sn_2$. Since
$\gcd(n_1,n_2) = 1$, then we must have
$n_1 \mid s$.
Thus, we have $n_1n_2 \mid s$, as desired.

Hence, $s = n_1n_2. \Box$.

Thus, the order of $a_1a_2$ is $n_1n_2$
i.e. $P(2)$ is true.

Inductive Step:
Let $x=a_1a_2\ldots a_k$ and $n=n_1n_2\cdots n_k$.
Assume that $P(k)$ is true. That is, the order of $x$ is $n$.

To show $P(k+1)$ is also true (that is, the order of $xa_{k+1}$ is $nn_{k+1}$.):
Notice that
\begin{align*}
a_1a_2\ldots a_{k+1} &= a_1a_2\ldots a_ka_{k+1} \\
&= (a_1a_2\ldots a_k)a_{k+1} \\
&= xa_{k+1}.
\end{align*}
Now, we apply the same method as in the base case above. Here we go:

Let $|xa_{k+1}| = t$. To show that
$t = nn_{k+1}$:

$t \mid nn_{k+1}$:
Consider that
\begin{align*}
(xa_{k+1})^{nn_{k+1}} &=
\overset{nn_{k+1}}{\overbrace{xa_{k+1}xa_{k+1}\ldots xa_{k+1}}} \\
&= \overset{nn_{k+1}}{\overbrace{xx\ldots x}}\overset{nn_{k+1}}{\overbrace{a_{k+1}a_{k+1}\ldots a_{k+1}}} \\
&= (x^n)^{n_{k+1}}(a_{k+1}^{n_{k+1}})^n
\\
&= (e_G)(e_G) \\
&= e_G.
\end{align*}
Hence, $t \mid nn_{k+1}$, as desired.

$nn_{k+1} \mid t$:
Consider that
\begin{align*}
e_G = (xa_{k+1})^t &= (xa_{k+1})^{tn} \\
&= x^{tn}a_{k+1}^{tn} \\
&= (x^n)^ta_{k+1}^{tn} \\
&= (e_G)a_{k+1}^{tn} \\
&= a_{k+1}^{tn}.
\end{align*}
Hence, $n_{k+1} \mid tn$.
But, since $\gcd(n,n_{k+1}) = 1$, then we must have $n_{k+1} \mid t$.
Similarly, consider that
\begin{align*}
e_G = (xa_{k+1})^t &= (xa_{k+1})^{tn_{k+1}} \\
&= x^{tn_{k+1}}a_{k+1}^{tn_{k+1}} \\
&= x^{tn_{k+1}}a_{k+1}^{tn_{k+1}} \\
&= x^{tn_{k+1}}(a_{k+1}^{n_{k+1}})^t \\
&= x^{tn_{k+1}}(e_G) \\
&= x^{tn_{k+1}}.
\end{align*}
Hence, $n \mid tn_{k+1}$. Since
$\gcd(n,n_{k+1}) = 1$, then we must have
$n \mid t$.
Thus, we have $nn_{k+1} \mid t$, as desired.

Hence, $nn_{k+1} = t. \Box$.

Thus, the order of $xa_{k+1}$ is
$nn_{k+1}$ i.e. $P(k+1)$ holds.

Therefore, the order of $a_1a_2\ldots a_m$ is $n_1n_2\cdots n_m$ and hence proved. $\Box$

答:
Putting aside whether or not this is an efficient proof of the result or not and a few minor typos this is a correct and carefully carried out proof.

The one part where it might be better to write things slightly differently occurs at the point where you prove $n_1n_2$ divides $s$ (and similarly for the general case).

Your (correct) argument here would be much clearer if you  wrote:-

Assume $ e_G=(a_1a_2)^s $, then $e_G=(a_1a_2)^{sn_1}$ etc.

In other words always make it clear to the reader when you are assuming something.

In terms of efficiency, you realised that you were repeating the same argument when you said "we apply the same method as in the base case above". What you could have done is the following:-

Assume $P(k)$ is true and consider  $a_1a_2\ldots a_{k+1}=(a_1a_2\ldots a_k)a_{k+1}$.

Now $a_1a_2\ldots a_k$  has order $n_1n_2\ldots n_k$ by assumption and $a_{k+1}$ has order $n_{k+1}$. Therefore, by the base case, $a_1a_2\ldots a_{k+1}$ has order $n_1n_2\ldots n_{k+1}$.

Look how much work that has saved!

P.S. I wish many of my students had written out proofs as carefully as this - so don't think my comments mean your work is not good!


问:
Probability that cubic of integer ends with 11
Let $x$ be an integer between $1$ and $10^{12}$. What is the probability that $x^3$ ends with 11?

I started with expressing $x^3=(a+10b)^3$ and applied binomial theorem to get $$x^3 = a^3+300b^2a+30a^2b+1000b^3 $$
How do I proceed from here?
Thanks.

答:
You've already noticed that only $a^3$ and $30a^2b$ affect the last two digits.

Notice further that only $a^3$ affects the final digit--namely, the final digit of $x^3$ is the final digit of $a^3.$ Thus, $a=1.$

Consequently, the second-to-last digit of $x^3$ is the last digit of $3b,$ and so we need the last digit of $b$ to be $7.$

Can you take it from there?

ContentLicense: CC BY-SA 4.0


问:
Volume of fundamental domain in lattice of $\mathbb{Z}^2$
I'm trying to prove exercise 6.1 on Algebraic number theory by Stewart and Tall and essentially it states:

If $L$ is a lattice in $\mathbb{R^2}$ with basis $u, v\in \mathbb{Z^2}$, $T$ its fundamental domain, i.e.
$$T=\{xu+yv: 0\leq x,y <1\}$$
and $n$ is the number of points of $\mathbb{Z}^2$ that are in $T$ then
$$\operatorname{vol}(T)=n$$

I know Pick's theorem but the exercise is stated without any knowledge on that topic and I don't know how to prove it just knowing the definitions I gave. I've tried to prove it inductively on $n$ but I can't even prove the base case and I don't know how to relate in any way the volume and $n$. Any ideas?

答: This exercise does not seem to be correct:  Take for example T = \Z^2, the standard integer lattice in \R^2.  By your definition of fundamental domain, the only lattice point in T would be the origin so n=1, but of course vol(T)=1 \not= n-2.</p>


问:
Radius of Convergence of $\sum_{n=1}^{\infty} c_n(x-3)^n$
I'm trying to figure out the following:
Say we know the series $\sum_{n=1}^{\infty} c_n(x-3)^n$ diverges for $x=8$ and converges at $x=1$, what would the bounds be of the radius of convergence $?\le R \le?$

I was thinking, well, since the center is at $x = 3$ and it still converges at $x=1$ we know for certain that $x=4$ still converges, but that $x=5$ could actually be an endpoint and thus not converge (since $x=1$ could be an endpoint that happens to converge). For some reason though, the answer to this question appears to be $1\le R \le5$ which I don't understand. How can we say from the above information that $5$ is definitely included in our radius?

答: The answer to the question seems to be incorrect. Rather, the radius of convergence cannot be less than $2,$ since it converges at $x=1,$ but cannot be greater than $5,$ since it diverges at $x=8.$ Hence, $2\le R\le 5.$ I suspect that the text simply has a typo.


问:
What is the current school of thought concerning precision of numeric conversions of measurements?
(EDIT: Changed title to say "precision" not "accuracy".)

(Now asked at https://matheducators.stackexchange.com/q/19373/8931)

I had a heated argument with someone online who claimed to be a school mathematics teacher of many years standing. The question which spurred this discussion was something along the lines of:

"A horseman was travelling from (location A) along a path through a forest to (location B) during the American War of Independence. The journey was of 22 miles. How far was it in kilometres?"

To my mind, the answer is trivially obtained by multiplying $22$ by $1.6$ to get $35.2$ km, which can be rounded appropriately to $35$ km.

I was roundly scolded by this ancient mathematics teacher for a) not using the official conversion factor of $1.60934$ km per mile and b) not reporting the correct value as $35.405598$ km.

Now I have serious difficulties with this analysis. My argument is: this is a man riding on horseback through a forest in a pre-industrial age. It would be impractical and impossible to measure such a distance to any greater precision than (at best) to the nearest $20$ metres or so, even in this day and age. Yet the answer demanded was accurate to the nearest millimetre.

But when I argued this, I was told that it was not my business to round the numbers. I was to perform the conversion task given the numbers I was quoted, and report the result for the person asking the question to decide how accurately the numbers are to be interpreted.

Is that the way of things in school? As a trained engineer, my attitude is that it is part of the purview of anybody studying mathematics to be able to estimate and report appropriate limits of accuracy, otherwise you get laughably ridiculous results like this one.

答:
"The current school of thought concerning accuracy of numeric conversions of measurements" is that the conversion is done with infinite precision. This allows to use the same formulas about falling stones independently of the used units.

Here we are in a high school environment with their special conventions. The standard rule is to use the given accuracy of the data, do exact numerical calculations with them, and perform a reasonable final rounding.

In the example of the horseman it is not expected that the pupil finds out the officially certified $60$ decimal places ISO factor between miles and kilometers. In his math book there is somewhere a line saying that $1$ mile $=1.609$ kilometers, or similar. The pupil is then expected to use this figure, compute with it, and make a final rounding to the precision of, e.g., meters, so that (s)he obtains $22\cdot1.609=35.398$ kilometers.


问:
If $A, B$ are Morita equivalent, then $\underline{\text{Nat}}(I_A, I_A)\cong\underline{\text{Nat}}(I_B, I_B)$.
Let $A, B$ two rings and $I_A: {}_A\text{Mod} \to{}_A\text{Mod}$ the identity functor. I am trying to show that if $A, B$ are Morita equivalent, then $\underline{\text{Nat}}(I_A, I_A)\cong\underline{\text{Nat}}(I_B, I_B)$ (where $\text{Nat}(I_A, I_A)$ is the set of all natural transformations from $I_A$ to $I_A$).

My idea is the following: If the $A, B$ are Morita equivalent, then ${}_A \text{Mod}$ is equivalent to ${}_B \text{Mod}$ and the equivalence is given by the adjunction
$$(P \otimes_A - , Q\otimes_B -)$$
where $P \in {}_B \text{Mod}_A$ and $Q \in {}_A \text{Mod}_B$. Consequently, $\lambda \in \underline{\text{Nat}}(I_A, I_A)$ can be send to $\overline{\lambda} \in \underline{\text{Nat}}(I_B, I_B)$ such that for each $N \in{}_B\text{Mod}$, there exists a unique (up to isomorphism) $M \in {}_A \text{Mod}$ with
$$\varphi: N \cong P\otimes_A M,$$
and we define
$$\overline{\lambda}_N: N \overset{\varphi}{\longrightarrow} P\otimes_A M \overset{\text{id}_P \otimes \lambda_M}{\longrightarrow} P\otimes_A M  \overset{\varphi^{-1}}{\longrightarrow}N.$$
With the same idea, we can send $\gamma \in \underline{\text{Nat}}(I_B, I_B)$ to $\overline{\gamma} \in \underline{\text{Nat}}(I_A, I_A)$. But here I am not able to show that these applications are mutual inverse (I even have some difficulties to show that $\overline{\lambda}$ is indeed in $\underline{\text{Nat}}(I_B, I_B)$). Is it the right way to show the statement ? Can someone help me?

答:
Your approach is correct, and in fact it's a special case of a more abstract category-theoretic observation. Suppose that $\mathcal{C}$ and $\mathcal{D}$ are equivalent categories, say $F \colon \mathcal{C} \to \mathcal{D}$ provides us with an equivalence with inverse $G$. Then I claim that I can find an isomorphism $F^* \colon \operatorname{Nat}(I_{\mathcal{D}},I_{\mathcal{D}}) \to \operatorname{Nat}(I_{\mathcal{C}},I_{\mathcal{C}})$ with inverse $G^*$. Here's what $F^*$ does. Suppose we start with a natural transformation $\beta \colon I_D \to I_D$. Then I define $F^* \beta$ to be the natural transformation $I_C \to I_C$ associating to an object $c$ the morphism
$$c \to G(F(c)) \xrightarrow{G(\beta(F(c)))} G(F(c)) \to c$$
You need to check that this is well-defined. For instance, if you have a map $c \to c'$, then
$$\require{AMScd} \begin{CD}
c @>{}>> c'\\ @V{\beta(c)}VV @VV{\beta(c')}V\\
c @>>> c'
\end{CD}$$
must commute. This may sound daunting, but when you write it out you'll find that it boils down to the naturality of the transformations $I \leftrightarrow G\circ F$ that commute from the fact that they're mutual inverses.

Through a symmetrical process one can find what $G^*$ should be.

Now we need to check that $F^* \circ G^* = I$ and $G^* \circ F^* = I$. This is going to be quite a trainwreck when writing it all out, but what you should always keep in mind is that, no matter how technical it seems, proofs in category theory always proceed by just 'following your nose'. Without further ado, let me try to unravel $G^* F^* \beta$ for a natural transformation $\beta$ and hope I don't make any mistakes. It is the natural map sending an object $d$ to the morphism
$$d \to FG(d) \to FGFG(d) \xrightarrow{FG\beta FG(d)} FGFG(d) \to FG(d) \to d.$$
Proving that it's a natural transformation will ultimately rely simply on the naturality of the transformations $I \leftrightarrow FG$. Now to show that this natural transformation $G^* F^* \beta$ is just $\beta$ (up to isomorphism) note that the diagram
$$\require{AMScd} \begin{CD}
d @>{\beta(d)}>> d\\ @V{}VV @VV{}V\\
FG(d) @>>{\\beta (FG(d))}> FG(d)\\ @V{}VV @VV{}V\\
FGFG(d) @>>{FG\beta FG(d)}> FGFG(d)
\end{CD}$$
commutes by all kinds of naturalities.


问:
Basic question about linear transformations.
I was solving this question, and encountered a very "basic" problem that I want to make sure if what I did was right:

Let $P:V\rightarrow V$ be a linear transformation such that $P\circ P=P , dimV=n$, Prove that $(I-P)\circ (I-P)=I-P$.

At the ending stages I reached something that looked like this:

$-P(I(v)-P(v))$ , and I did this explaining that I used the linearity : $-P(I(v)-P(v))=-P(I(v))+P(P(v))$.

 And I checked everywhere for something about subtraction but everywhere I see the linearity property defined with $T(v+u)=T(v)+T(u)$, I wanted to ask if it applies for subtraction too just to make sure I am not missing something.
Thanks in advance.

答:
You need to apply linearity several times.

\begin{align}
&-P(I(v)-P(v))
\\
&= -[P(I(v)) + P(-P(v))] &\text{since }P(u+w)=P(u)+P(w)
\\
&= -[P(I(v)) - P(P(v))] & \text{since } P(-u) = -P(u)
\\
&= -P(I(v)) + P(P(v)) & \text{since } -(u-w) = -u+w
\end{align}


答: Sure. After all,\begin{align}T(u-v)+T(v)&=T\bigl((u-v)+v\bigr)\\&=T(u)\end{align}and therefore$$T(u-v)=T(u)-T(v).$$

Title: Basic question about linear transformations.

Body:
I was solving this question, and encountered a very "basic" problem that I want to make sure if what I did was right:

Let $P:V\rightarrow V$ be a linear transformation such that $P\circ P=P , dimV=n$, Prove that $(I-P)\circ (I-P)=I-P$.

At the ending stages I reached something that looked like this:

$-P(I(v)-P(v))$ , and I did this explaining that I used the linearity : $-P(I(v)-P(v))=-P(I(v))+P(P(v))$.

 And I checked everywhere for something about subtraction but everywhere I see the linearity property defined with $T(v+u)=T(v)+T(u)$, I wanted to ask if it applies for subtraction too just to make sure I am not missing something.
Thanks in advance.


问:
$T(x_1,x_2,x_3,x_4,\ldots)=\left(x_2,\frac{x_3}{2},\frac{x_4}{3},\frac{x_5}{4},\ldots\right)$, spectral radius and spectrum
Define a pair of operators $S\colon\ell^2 \rightarrow \ell^2$ and $L\colon\ell^2 \rightarrow \ell^2$ (of sequences of complex or real numbers) as follows:

$S(x_1,x_2,x_3,x_4,\ldots)=\left(x_1,\frac{x_2}{2},\frac{x_3}{3},\frac{x_4}{4},\ldots\right)$

$L(x_1,x_2,x_3,x_4,\ldots)= (x_2,x_3,x_4,\ldots)$

Let $T=S \circ L$, so $T(x_1,x_2,x_3,x_4,\ldots)=\left(x_2,\frac{x_3}{2},\frac{x_4}{3},\frac{x_5}{4},\ldots\right)$ and $T^n(x_i)_{i\in\Bbb N}=\left(\frac{(i-1)!}{(i+n-1)!}x_{i+n}\right)_{i\in\Bbb N}$ if I'm not wrong. I know that $S$ is compact and $L$ is continuous, therefore T is compact. I know the norm of $T$ is $1$. How do I calculate the norm of $T^n$, and the spectral radius and the spectrum of $T$? Thank you in advance.

My attempt for the norm of $T^n$: I've tried to find a constant $K$ such that $\|T^n(x)\| \leq K \|x\|\ $ and an $x_0\in\ell^p$ such that $\|x_0\|=1$, $\|T^n(x_0)\|=K$, what would prove that $\|T^n\|=K$. That worked for $n=1$ but I don't know how to do it for the general case. Easier alternatives are welcome.

答:
Note that
$$\|T^nx\|_2^2 = \sum_{i=1}^\infty \left(\frac{(i-1)!}{(i-1+n)!}\right)^2 x_n^2 \le \sum_{i=1}^\infty \left(\frac1{n!}\right)^2 x_n^2 = \frac{1}{n!^2}\|x\|_2^2$$
so $\|T^n\| \le \frac1{n!}$. Therefore the spetral radius is
$$r(T) = \lim_{n\to\infty} \|T^n\|^{\frac1n} \le \limsup_{n\to\infty} \frac{1}{\sqrt[n]{n!}} = 0$$
and hence $\sigma(T) = \{0\}$. For the norm of $T$ we got $\|T\| \le 1$ and considering $Te_2 = e_1$ gives $\|T\|=1$.


问:
Mathematical Systems Question Help
Alright, this is another question for my math for teachers course. This question is not actually in the homework, but there are problems similar to it. I'd really like to learn how to do problems like this one, so it would be much appreciated if I was given a solid explanation/demonstration. I don't really have a clue where to start, so I've defined the terms as best as I can under each question... The question should be answered using the data from the table.

A mathematical system is defined by the following table:

Is there an identity element in this system? If so, what is it?
Identity is: aXe=a and eXa=a
Is closure satisfied by this system? Explain.
Closure is: aXb
Is this system commutative? Explain.
Commutative is: aXb=bXa
Does every element have an inverse? Explain.
Inverse is: aXx=e and xXa=e
Is this system a group? Explain.
Not sure what they mean by this...
Are they referring to another property?.

答:
Scanning along the rows, we see that the second row is in alphabetical order. Similarly, scanning along the columns, we see that the second column is in alphabetical order. Thus, $b$ must be the identity element.

Since all elements in the grid are either $a,b,$ or $c$, we know that the system is closed under multiplication.

Since the grid is symmetrical with respect to the main diagonal (draw a line going from the top left to the bottom right), the system is commutative.

Since the identity element $b$ appears at least once in every row and column, every element has an inverse (note that inverses aren't necessarily unique, suggesting that it isn't a group).

A group is a set $G$ with a multiplication operation $\times$ that satisfies 4 properties: identity, closure, inverses, and associativity. Since the first 3 conditions are satisfied, it suffices to consider the last condition:

Definition: An operation $\times$ is considered associative iff for all $x,y,z\in G$: $$ (x \times y) \times z = x \times (y \times z)$$

Observe that this system is not associative (and thus not a group), since:
  $$
(a \times a) \times c=b \times c = c\\
a \times (a \times c)=a \times b = a
$$
  Yet $c \ne a$.

ContentLicense: CC BY-SA 3.0

Title: Mathematical Systems Question Help


问:
Which method to solve this differential equation? Laplace?
How can I solve this differential equation?
$\frac {dP(t)} {dt}=Ae^{t/\tau}P(t)-BP(t)+C$
where $A,B,C, \tau$ are constant.
Should I use Laplace transform or some other method?

答:
If we shuffle things around a bit this is the same as:
$$P'-(Ae^{t/\tau}-B)P=C$$
so we can try first of all to solve for $C=0$:
$$P'(t)=(Ae^{t/\tau}-B)P(t)$$
$$\frac{P'}{P}=Ae^{t/\tau}-B$$
$$\ln P=A\tau e^{t/\tau}-Bt+Const.$$
then substituting this back into your original equation you can try to find the particular solution


答:
Linearity also includes scalar multiplication: $T(\alpha v)=\alpha T(v).$ In particular, then, $$T(v-u)=T\bigl(v+(-1)u\bigr)=T(v)+T\bigl((-1)u)=T(v)+(-1)T(u)=T(v)-T(u).$$

As for your figuring, it seems something may have gone wrong: $$(I-P)\circ(I-P)=I(I-P)-P(I-P)=I-P-\bigl(P(I)-P(P)\bigr)=I-P-(P-P)=I-P.$$


问:
there exists a $c>0$ such that every $3$-regular bipartite graph is a $(2n,3,c)$-expander
Definition: A graph $G = (V, E)$ is called an $(n, d, c)$-expander if it has $n$ vertices, the maximum degree of a vertex is $d$, and, for every set of vertices $W \subset V$ of cardinality $|W| \le n∕2$, the inequality $|N(W)| \ge c|W|$ holds, where $N(W)$ denotes the set of all vertices in $V \backslash W$ adjacent to some vertex in $W$

By considering a random bipartite three-regular graph on $2n$ vertices obtained by picking three random permutations between the two color classes, prove that there is a $c > 0$ such that for every n there exists a $(2n, 3, c)$-expander.

This is question 9.6.1 from The Probabilistic Method and I am tempted to argue that for any $c\le 1$ the graph is trivially an expander.

In the question we are asked to consider the 3-regular bipartite graph obtained by taking the edge union of three permutations on $[n]$. A permutation is basically a perfect matching, therefore this construction contains a perfect matching and by Hall's theorem, every subset $W\subset[n]$ satistifes $|N(W)|\ge|W|\ge c|W|$, in particular for sets $|W|\le 1/2$.

(it is actually even simpler than that, no need for Hall's theorem: by 3-regularity $|N(W)|<|W|$ would imply $3|N(W)|<3|W|$ which in english means $\#\{\text{outedges of } |N(W)|\} < \#\{\text{outedges of } |W|\}$ which is obviously false).

But then these hints from this website point to a completely different approach. I would be interested in additional hints on how to compute the probability that $N(W)<(1+\epsilon)|W|$ for a given $W\subset[n]$ as they say in the hints.

And of course I'd like to know if taking $c\le 1$ is OK...

答:
You might have been confused by the definition of $N(W)$. Usually this just means "the set of vertices adjacent to $W$". In the definition you've quoted from Alon and Spencer, we don't include vertices that are in $W$, which is the standard thing to do for measuring vertex expansion.

As a result, many cubic bipartite graphs are not expanders for $c =1$. For an extreme example, take two copies of any cubic bipartite graph, and let $W$ be the set of all vertices in one copy. Then $|W| = \frac n2$ and $|N(W)| = 0$, so the vertex expansion of such a graph is $0$.

For a family of connected examples with bad expansion, consider the prism graphs, such as the one below:

These are $3$-regular with $n=2k$ vertices, and when $k$ is even, they are bipartite. However, we can cut the prism in half vertically (if it is drawn as in the picture above) and take $W$ to be the vertices in one half. Then $|W| = \frac n2$ and $|N(W)|=4$, so at best this graph has vertex expansion $\frac 8n$, which goes to $0$ as $n \to \infty$.

So, no: not all bipartite cubic graphs are expanders.

The reason why you're seeing hints for proving that $|N(W)| \ge (1+\epsilon)|W|$ is not because $c$ can't be less than $1$. It's because in general, if $W = W_1 \cup W_2$ where $W_1$ is on one side of the bipartition and $W_2$ on the other,
$$
   N(W) = (N(W_1) \setminus W_2) \cup (N(W_2) \setminus W_1)
$$
which leads us to
$$|N(W)| \ge (|N(W_1)| - |W_2|) + (|N(W_2)| - |W_1|) = |N(W_1)| + |N(W_2)| - |W|.$$
if we can show that $|N(W_1)| \ge (1+\epsilon)|W_1|$ and $|N(W_2)| \ge (1+\epsilon)|W_2|$, then we conclude that $|N(W)| \ge \epsilon|W|$. If this always holds, then we have an $\epsilon$-expander.

We have to be careful because $|W| \le \frac12|V|$ doesn't imply that $|W_1| \le \frac12n$ and $|W_2| \le \frac12n$. But we also have $|N(W)| \ge |N(W_1)| - |W_2| \ge |W_1| - |W_2|$, and similarly $|N(W)| \ge |W_2| - |W_1|$, so if the two sides of $W$ are too unbalanced, we get vertex expansion immediately. That tells us that $|W_1|, |W_2| \le (\frac12 + \frac12\epsilon)n$ in all hard cases.


问:
Some questions about determinants
I'm studying some Linear Algebra and determinants are being a real struggle for me. I can't understand them at all, but my biggest problem is with all the equivalent definitions and why they are in fact equivalent.

Given $V, W$ vector spaces, we define the tensor product $\otimes$ of $f \colon V^k \to \mathbb{R}$ and $g \colon W^l \to \mathbb{R}$ as $(f \otimes g )(v_1, \dots, v_{k+l})$ $= f(v_1, \dots, v_k)g(v_{k+1}, \dots, v_{k+l})$. Also, we define $V \otimes W = \langle v \otimes w \colon v \in V, w \in W \rangle$ (note that the tensor product of vectors is defined, since $V = V^{**}$, which is a space of functions).

Given $V$ a vector space, we define $X = V^{\otimes k} = V \otimes V \otimes \dots \otimes V$ ($k$ times) and $Y =$ $\langle v_1 \otimes \dots \otimes v_k \colon \exists i,j \textrm{ } v_i = v_j \rangle$ (i.e. the set generated by the tensor product of $k$ vectors in which at least two are equal). We define $\wedge^k V = X/Y$ (call it exterior $k-$power of $V$) and denote $\overline{u_1 \otimes \dots \otimes u_k} = u_1 \wedge \dots \wedge u_k$.

Given those important definitions of tensor product and exterior power of a space, let's introduce the definitions of determinant.

Given $V$ a vector space with $\dim V = n$ and $T\colon V \to V$ a linear transformation, note that $\dim \wedge^n V = 1$. Hence, the linear transformation $\wedge^n T \colon \wedge^n V \to \wedge^n V$ must be of the form $\wedge^n T(v) = cv$ for some constant $c$. This constant is called the determinant of $T$ and denoted $c = \det T$.

The first thing that bothers me here is the assumption that $\wedge^n T(v) = cv$ for some constant $c$, why is that? It seems trivial but I'm not getting it.

The usual way we see determinants (in high school) is by beeing introduced to determinants of matrices, not of linear operators, but we can translate this definition to matrix just by saying that the determinant of $T$ is the determinant of it's matrix in some basis. My first question is: why is this definition independent of the chosen basis?

Given a matrix $M = [m_{ij}]_{n \times n}$ we define its determinant $$\det T = \sum_{\sigma \in S_n} \textrm{sgn}(\sigma) m_{i\sigma(1)}\cdots m_{i\sigma(n)}.$$

Now given both definitions, why are those equivalent?

Thanks in advance and if there is anything I can do to make this question better, let me know please!

答:
Given a vector space $V$ of dimension $1$, then we may find a vector $v\in V$ which generates $V$ (trivially it is linear independet, as we have a set with just one element!). That is to say you can write any $w\in V$ as $w=\lambda v$ for a $\lambda\in K$ the field over which you define the vector space $V$. Now, any linear application $T$ from $V$ to itself as to map $v$ somewhere: either to $0$, in which case we have the trivial map or to some $w\in V$, which means:
$$T(v)=w=\lambda v$$
as we understood. Now, to any linear map there corrisponds a matrix: it is a very basic result you see in any linear algebra course.

Why is the definition determinant of a matrix independent of the basis is a bit more tricky, but once you show that det$MN=$det$M$det$N$, that $SMS^{-1}=M$ where $S$ is the change of basis, and det$(S^{-1})=\text{det}(S)^{-1}$ then you are over, as this clearly this means the determinant doesn't change when changing a basis.

As for your last question, it is actually totally not trivial and is shown in any text of linear algebra.

Title: Some questions about determinants


问:
If $\mu(A)\ne0$, is there always a $f\in L^p(\mu)$ with $\{f\ne0\}\subseteq A$ and $\mu(\{f\ne0\})\ne0$?
Let $(\Omega,\mathcal A,\mu)$ be a measure space and $p\in[1,\infty]$.

Are we able to show that if $A\in\mathcal A$ with $\mu(A)\ne0$, then there is a $f\in\mathcal L^p(\mu)$ with $\{f\ne0\}\subseteq A$ and $\mu\left(\{f\ne0\}\right)\ne0$?

If not, are we able to give a suitable condition on $\mu$ to ensure that the desired conclusion holds?

答:
If the space is $\sigma$-finite, say $\bigcup_{i=1}^\infty X_i = \Omega$ with $\mu(X_i) < \infty$ and $X_i \subseteq X_{i+1}$, then
$$ \lim_{i\to \infty} \mu(X_i \cap A) = \mu(A) > 0.$$
That is, for sufficiently large $i$, $\mu(X_i \cap A)$ becomes positive and remains finite.


问:
Covering paired squares on a $10 \times 10$ board
In a game, the board is a $10 \times 10$  table (in the style of a chessboard). Two players make moves in turns. In a single step, a player covers two neighboring free cells of the table by $2\times 1$ tile. The player who cannot make a step loses the game. Who has a winning strategy - the player who starts the game or his opponent?

答:
Correction.

My answer is marked as correct though it is in fact inaccurate. The game the OP is describing is called Cram, see this Wikipedia page, in which an argument is discussed as to why the second player will have a winning strategy in the case of a $10 \times 10$ board.

Old answer.

I initially misrecognised the problem as a very similar game called Domineering, which also has its own Wikipedia page. Curiously, unlike Cram, Domineering seems to lack the symmetry to have straightforward strategies like Cram does, and finding an optimal strategy for Domineering seems to be hard. For a $10 \times 10$ board, the Wikipedia page claims that a perfect strategy was found in the Master thesis of Nathan Bullock.


问:
$ A=\begin{pmatrix} 0 & 2 & 0\\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{pmatrix}$, calculate $e^A$
I have encounter a question in my book , it was For $ A=\begin{pmatrix}
0 & 2 & 0\\
0 & 0 & 3 \\
0 & 0 & 0
\end{pmatrix}$, calculate $e^A$

My solution way : I tried to find its eigenvalues , so i found that the only eigenvalue is $0$ and the eigenspace is $(1,0,0)$ .Hence , it cannot be diagonalized.

Then , i tried to use taylor exponential and it gives me $ A=\begin{pmatrix}
1 & 2 & 0\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}$ .However the answer is $ A=\begin{pmatrix}
1 & 2 & 3\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}$ .

What am i missing ,can you help me?

答:
Note that $$A^2=\begin{pmatrix}
0 & 0 & 6\\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix},$$ which is where the $3$ in the upper-right corner comes from.

Since $A^3$ is just the $3\times 3$ matrix of zeroes, $e^A=I+A+\frac12 A^2.$


答:
The structure of $A$ tells you immediately that $A$ is not diagonalizable without any calculation.

The matrix $A$ is a Jordan block that is nilpotent; there is an integer $N$ such that $A^N=0$. ($N$ is small in your particular case.) So you have $A^{m}=0$ for all $m>N$.

By definition, to calculate $e^A$, you need to know all the powers of $A$: $e^A:=\sum_{k=0}^\infty\frac{1}{k!}A^k$. But now you only need to calculate finitely many.

Body:
I have encounter a question in my book , it was For $ A=\begin{pmatrix}
0 & 2 & 0\\
0 & 0 & 3 \\
0 & 0 & 0
\end{pmatrix}$, calculate $e^A$

My solution way : I tried to find its eigenvalues , so i found that the only eigenvalue is $0$ and the eigenspace is $(1,0,0)$ .Hence , it cannot be diagonalized.

Then , i tried to use taylor exponential and it gives me $ A=\begin{pmatrix}
1 & 2 & 0\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}$ .However the answer is $ A=\begin{pmatrix}
1 & 2 & 3\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}$ .

What am i missing ,can you help me?


问:
Showing the spectral radius is smaller than the induced norm

The spectral radius of $A \in \mathbb{R}^{n \times n}$ denoted by $\rho(A)$ is the largest absolute value $\lvert \lambda \rvert$ of an eigenvalue of $A$. Show that $\rho(A) \le \lvert \lvert A \rvert \rvert $ for all induced matrix norms $\lvert \lvert A \rvert \rvert $.

Attempt at solution

If $\lambda$ is an eigenvalue of $A$ then $Ax = \lambda x$ for $x$ an eigenvector. Then, for $\rho(A) := \lvert \bar{\lambda}\rvert$ the largest eigenvalue,
\begin{align}
\lvert \lvert A \rvert \rvert  \overset{def.}{=} \max_{\lvert \lvert x \rvert \rvert=1 } \lvert \lvert Ax \rvert \rvert  &= \max_{\lvert \lvert x \rvert \rvert =1} \lvert \lvert \lambda x \rvert \rvert  \\
&= \lvert \lambda \rvert \underbrace{\max_{\lvert \lvert x \rvert \rvert =1} \lvert \lvert x \rvert \rvert}_{\color{red}{= 1 ?}}  \\
&\le \lvert \bar{\lambda} \rvert\\
\therefore \lvert \lvert A \rvert \rvert  &\le \rho(A)
\end{align}

The inequality is the other way around, can someone help me find my mistake?

答:
Your mistake lies in $$\max_{\|x\|=1}\|Ax\|=\max_{\|x\|=1}\|\lambda x\|.$$
Did you fix the eigenvalue $\lambda$ beforehand ? Not all vectors are necessarily $\lambda$-eigenvectors.

What you want is fixing $\lambda$ with $\rho(A)=|\lambda|$, and notice that $$\{x\in E\;/\;\|x\|=1\text{ and }Ax=\lambda x\}\subset\{x\in E\;/\;\|x\|=1\},$$
so that : $$\max_{\|x\|=1}\|Ax\|\geqslant\max_{\|x\|=1\text{ and }Ax=\lambda x}\|Ax\|.$$
Now, the rest is easy :

$\begin{align*}\||A\||&=\max_{\|x\|=1}\|Ax\|&\\&\geqslant\max_{\|x\|=1\text{ and }Ax=\lambda x}\|Ax\|\\&=\max_{\|x\|=1\text{ and }Ax=\lambda x}\|\lambda x\|\\&=\max_{\|x\|=1\text{ and }Ax=\lambda x}|\lambda|\|x\|\\&=\rho(A)\max_{\|x\|=1\text{ and }Ax=\lambda x}1\\&=\rho(A).\end{align*}$


答:
Citing accuracy (precision is a better term) in calculations involving measurements can be a tricky thing. 22 miles is not the same as saying 22.0 or 22.00 etc... 22 miles has 2 sig figs whereas 22.0 has 3.

So 22 miles can be between 21.5 and 22.4 - even more tricky because 21.5 is not closer to 22 than 21 and what about 22.49? which has 4 sig digs. Most people have this assumption that when they state 22 miles they are meaning "exactly" 22 miles but in the case of measurement this can never be so. Better to say 22 miles +/- .5 but we all know  that's never going to happen. So the teacher admonishing you is incorrect because you are dealing with a measured quantity and using 1.6 which has 2 sig digs is fine by me (since 22 has 2 sig digs). 35 is better than 35.2 since 35 also has 2 sig digs whereas 35.2 has 3

There are a lot of complicated rules about how many sig digs your answer should have when multiplying (and you can look them up), based on the number of sig digs in the 2 numbers you are multiplying. This is further complicated here because the conversion factor is not a measured quantity.

The easiest rule of thumb (but not complete) is that your answer should have the same number of sig digs as the least significant measurement in the calculation. I know that this does not completely address the the problem of precision in an answer, but it is at least a start.

Another thought. The use of calculators further confuses the concept of precision when multiplying measured quantities as they give an answer which is "way" beyond the precision of the numbers being multiplied. In the dumbed-down curriculum of the 21st century the teaching of precision has pretty much been abandoned in high school as it has been classified as too difficult to learn and not worth the effort.

Title: What is the current school of thought concerning precision of numeric conversions of measurements?

Body:
(EDIT: Changed title to say "precision" not "accuracy".)

(Now asked at https://matheducators.stackexchange.com/q/19373/8931)

I had a heated argument with someone online who claimed to be a school mathematics teacher of many years standing. The question which spurred this discussion was something along the lines of:

"A horseman was travelling from (location A) along a path through a forest to (location B) during the American War of Independence. The journey was of 22 miles. How far was it in kilometres?"

To my mind, the answer is trivially obtained by multiplying $22$ by $1.6$ to get $35.2$ km, which can be rounded appropriately to $35$ km.

I was roundly scolded by this ancient mathematics teacher for a) not using the official conversion factor of $1.60934$ km per mile and b) not reporting the correct value as $35.405598$ km.

Now I have serious difficulties with this analysis. My argument is: this is a man riding on horseback through a forest in a pre-industrial age. It would be impractical and impossible to measure such a distance to any greater precision than (at best) to the nearest $20$ metres or so, even in this day and age. Yet the answer demanded was accurate to the nearest millimetre.

But when I argued this, I was told that it was not my business to round the numbers. I was to perform the conversion task given the numbers I was quoted, and report the result for the person asking the question to decide how accurately the numbers are to be interpreted.

Is that the way of things in school? As a trained engineer, my attitude is that it is part of the purview of anybody studying mathematics to be able to estimate and report appropriate limits of accuracy, otherwise you get laughably ridiculous results like this one.


问:
Proof whether or not it is Hausdorff (T_2) space $X = \mathbb{R} \cup \{\alpha\} $ with $\alpha \notin \mathbb{R}$
Let $X = \mathbb{R} \cup \{\alpha\} $ with $\alpha \notin \mathbb{R}$. In $X$ we have the topology $\tau$ with the basis $$B = \{(a,b): a, b \in \mathbb{R}, a  0 \}$$

I want to show if $(X, \tau)$ is Hausdorff ($T_2$) or not, I tried to show that the open sets do not collide but I am confused with the $\alpha$. Any help is appreciated!

答: This space is sometimes referred to as the line with two origins. $0$ and $\alpha$ cannot be separated by open neighborhoods.


问:
Simple numerical methods for calculating the digits of $\pi$
Are there any simple methods for calculating the digits of $\pi$? Computers are able to calculate billions of digits, so there must be an algorithm for computing them. Is there a simple algorithm that can be computed by hand in order to compute the first few digits?</p>

答:
Since the AGM hasn't been mentioned yet (apart from my plug to read the Borwein and Borwein book), here's how to use the AGM to compute $\pi$:

```<code>a←(1+1/√(2))/2
c←1/(8a)
s←a^2
f←1
repeat
a←(a+√((a-c)(a+c)))/2
c←(c^2)/(4a)
f←2f
s←s-fc^2
until |c| is small enough
(2a^2)/(2s-1)
</code>```

which exploits the Legendre relation.

Body:
Since the AGM hasn't been mentioned yet (apart from my plug to read the Borwein and Borwein book), here's how to use the AGM to compute $\pi$:

a←(1+1/√(2))/2
c←1/(8a)
s←a^2
f←1
repeat
a←(a+√((a-c)(a+c)))/2
c←(c^2)/(4a)
f←2f
s←s-fc^2
until |c| is small enough
(2a^2)/(2s-1)

which exploits the Legendre relation.

ContentLicense: CC BY-SA 2.5


答:
You don't need induction, as mac points out, but if you want to complete the inductive argument you started, you could do something like this:

$10^{n+1}=10\cdot 10^n\equiv(-1)\cdot(-1)^n=(-1)^{n+1} \; (\mod 11)$

Title: Complete induction of $10^n \equiv (-1)^n \pmod{11}$

Body:
To prove $10^n \equiv (-1)^n\pmod{11}$, $n\geq 0$, I started an induction.

It's $$11|((-1)^n - 10^n) \Longrightarrow (-1)^n -10^n = k*11,\quad k \in \mathbb{Z}. $$

For $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$

$n\Rightarrow n+1$
$$\begin{align*}
 (-1) ^{n+1} - (10) ^{n+1} &= k*11\\
(-1)*(-1)^n - 10*(10)^n &= k*11
\end{align*}$$
But I don't get the next step.


问:
Learning differential/Riemannian geometry for PDEs
I know there have been threads on which books to learn DG/RG from but hopefully this is sufficiently different to avoid closure.

Can anyone recommend a book to learn DG/RG (whichever is appropriate) so that I can do PDEs on manifolds? At the moment I am reading through John M Lee's Introduction to Smooth Manifolds and I am wondering whether I really need to learn all the topics in it since it doesn't touch RG which I believe is more used in the theory of PDEs. Maybe there is a better text.

Also, any topics to particularly study or avoid would be useful.

Thanks

答:
If you're assuming the Riemannian manifold has a fixed metric, then the most introductory source I've found is Folland, Introduction to Partial Differential Equations, which discusses aspects of PDEs on hypersurfaces and the Laplace-Beltrami operator, for example.

If you're looking for something more advanced, but which avoids getting into curvatures,
then Grigor'yan's Heat kernel and analysis on Manifolds (AMS,2009) is excellent IMHO.


问:
Extracting function from sensor
and sorry if I am on the wrong stackexchange.

I have a problem. I have collected data from a distance sensor and extracted its function with respect to temperature. The data extracted proved that the sensor is completely linear (mx + b).

The problem is that at different distances, the m and b coefficiente of the lines change. Of course b will change, but my problem is with the m coefficient. The slope coefficient changes according to the real distance and as stated above the distance read changes according to temperature in a very linear manner.

How would I go about extracting this behaviour and putting it into a function.

EDIT:

Ok. I was not very clear, so I'll try to add more information.

I noticed that my sensor was behaving very strangely during hot days. This is normal for well ... everything, however this specific distance sensor I'm using is incredibly more sensitive to temperature.

With that problem I tried analyzing my sensor's behaviour over temperature and noticed that the output distance changes with temperature in a line (mx + b). This behaviour repeats itself for several distances. I noticed that m and b coefficients change according to its actual distance (the m coefficient has the Ae^(-Db)+c, where D is the real distance).

So, distance measured changes with temperature, slope (m) coefficient changes with real distance, as well as c.

Since I can't actually know the real distance and my code has to give ma that information, I was trying to regress a function that could give me this information.

However, if I try to regress it according to mx+b, it will work correctly over a small distance range, and I can't use real distance (I think) for the m and c coefficients because I don't know it. How would I go about measuring the distance using only the temperature and given distance by the sensor? Is it possible?

Below it's possible to see what I'm talking about. The x axis is temperature, the y axis is measured distance and the different lines are the lines at 5 different distances (5, 4, 3, 2 and 1 m)

答:
(im newbie here, feel free to delete this comment if offtopic

distance=(temperature*x + b)

but sometimes x and b changes - it seems its relationship is "almost linear", but not 100% linear

you can use differences
distance=distance[2:length(distance)]-distance[1:(length(distance)-1)]
similiar to temperature

maybe relationship between "changes of distance" is linear to "changes of temperature"

i consider error, as something to be forecasted
(but often, predictions are worse then "good constant mean"... sometimes is better to accept linear relationship you already have, even with errors

ahaaa you have like not 1 equation..but 5 equations
so its not    distance=(temperature*x + b)... .x and b increase with distance
i feel now its not about forecasting but picking right x and b.
but u said it " extracting this behaviour and putting it into a function."

(btw i dont know the answer just trying)

edit:
so you know distance, for example at distance 2, equation is distance=(temperaturex + b)
when you have distance 3.123456 there is equation(temperaturexother + bother)

you already know which values to assign to x=, b= ... according to "start distance" you just dont want to search for them manualy, but via mathematics :D
seems so clear now .. but sry i dont know how to calculate it

Title: Extracting function from sensor


问:
Prove that there are infinitely many primes using $Q=p_{1}\cdot p_{2}\cdot \ldots \cdot p_{m}$ and $R=p_{m+1}\cdot p_{m+2}\cdot \ldots \cdot p_{n}$
In an exercise I am supposed to prove the infinitude of the primes using the following:

Let $p_{1}, p_{2}, \cdots , p_{n}$ be the first $n$ primes and $m\in \mathbb{Z}$ with $1<m<n$. Let $Q$ be the product of all primes up to $m$. That is $$Q=p_{1}\cdot p_{2}\cdot \ldots \cdot p_{m}$$ and $R$ the product of the remaining primes in the list, meaning $$R=p_{m+1}\cdot p_{m+2}\cdot \ldots \cdot p_{n}.$$ Now, the exercise tells me to consider $Q+R$ and show that it is not divisible by any of the primes in the list $p_{1}, p_{2}, \cdots , p_{n}$ and hence must have a factor not included in the list. From this on may conclude that there are infinitely many primes (?). My attempt:

Consider $Q+R$. $$Q+R=(p_{1}\cdot p_{2}\cdot \ldots \cdot p_{m})+(p_{m+1}\cdot p_{m+2}\cdot \ldots \cdot p_{n}).$$ Now, suppose (by way of contradiction) that $\exists p_{k}$ in the list $p_{1}, p_{2}, \cdots , p_{n}$ such that $p_{k}\mid Q+R$. That is $$(p_{1}\cdot p_{2}\cdot \ldots \cdot p_{m})+(p_{m+1}\cdot p_{m+2}\cdot \ldots \cdot p_{n})=l\cdot p_{k}$$ for some $l\in \mathbb{Z}$. But this is impossible since all prime factors of $Q$ are different from the prime factors of $R$ which implies that either $p_{k}\mid Q$ or $p_{k}\mid R$, but not both. Thus $Q+R$ must be divisible by a prime number which does not appear in the list $p_{1}, p_{2}, \cdots , p_{n}$. Therefore there must be infinitely many primes.

答: Your proof is correct, and the classic proof (with no splitting of primes and just adding $1$) corresponds to the case where $R$ is the empty product, i.e. $1$.


问:
Proving or disproving: $a^b<b^a$?
So, I have two real numbers, and I need to prove or disprove the statement: $a^b<b^a$.

Note: The question contains real numbers and not parameters. I just don't want you to solve it for me, so I don't reveal the numbers... This is a calculus I question.

I have thinking about this question for a while, with no clue on what to do.

I thought of defining some functions, and use some theorems we learned...

Any hints on how to deal with those kind of questions will be much appreciated!

Thanks a lot!

答:
I will try to give a good answer:

First, as the question is stated, the answer is no. Because you are not ordering $a$ and $b$, so you can reverse the tuples, for example $(3,10)$ and $(10,3)$.

Even asking for $b<a$, the tuple (4,2) does not satisfy the inequality because $2^4=4^2$. You can even generate an infinite family of counterexamples with $a=t^{\frac{1}{t-1}}$ $b=t^{\frac{t}{t-1}}$.

However, as the comments above showed, with the condition $e<b<a$ then the inequality holds.

Title: Proving or disproving: $a^b<b^a$?


答:
If we want to show that $a^b<b^a$, then we can raise both powers to the $\frac{1}{ab}$ power (assuming neither value is $0$) to get that our problem is equivilant to showing that

$$\left(a^{b}\right)^{\frac{1}{ab}}=a^{\frac{1}{a}}<\left(b^{a}\right)^{\frac{1}{ab}}=b^{\frac{1}{b}}$$

We now observe that

$$\lim_{x\to\infty}\ln\left(x^{\frac{1}{x}}\right)=\lim_{x\to\infty}\frac{\ln(x)}{x}=0$$

and thus

$$\lim_{x\to\infty}x^{\frac{1}{x}}=1$$

and at $x=0$ we have that $0^{\infty}=0$, and so

$$\lim_{x\to0}x^{\frac{1}{x}}=0$$

Finding extrema of $x^{\frac{1}{x}}$ yields that

\begin{align*}
\frac{d}{dx}x^{\frac{1}{x}}&*=\frac{d}{dx}e^{\frac{\ln\left(x\right)}{x}}\\
&=e^{\frac{\ln\left(x\right)}{x}}\frac{d}{dx}\frac{\ln(x)}{x}
&=e^{\frac{\ln\left(x\right)}{x}}\left(\frac{1}{x^{2}}-\frac{\ln\left(x\right)}{x^{2}}\right)
\end{align*}

and thus our extreme point is when $\frac{d}{dx}x^{\frac{1}{x}}=0$, i.e $\ln(x)=1$, i.e $x=e$. Plugging in $x=e$ to the equation yields $e^{\frac{1}{e}}\approx 1.44>1$, and so this must be a maximum (comparing to endpoints). Thus, if both numbers $a,b$ are on the same side of $e$ then whichever is closest to $e$ has a larger value of $x^{\frac{1}{x}}$. If they are on opposite sides, then you are out of luck with this method.

Title: Proving or disproving: $a^b<b^a$?

Body:
So, I have two real numbers, and I need to prove or disprove the statement: $a^b<b^a$.

Note: The question contains real numbers and not parameters. I just don't want you to solve it for me, so I don't reveal the numbers... This is a calculus I question.

I have thinking about this question for a while, with no clue on what to do.

I thought of defining some functions, and use some theorems we learned...

Any hints on how to deal with those kind of questions will be much appreciated!

Thanks a lot!


问:
Proof of the hockey stick/Zhu Shijie identity $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$
After reading this question, the most popular answer use the identity
$$\sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1},$$
or, what is equivalent,
$$\sum_{t=k}^n \binom{t}{k} = \binom{n+1}{k+1}.$$

What's the name of this identity? Is it the identity of the Pascal's triangle modified.

How can we prove it? I tried by induction, but without success. Can we also prove it algebraically?

Thanks for your help.

EDIT 01 : This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself are highlighted, a hockey-stick shape is revealed.

答:
Recall that for $k\in\Bbb N$ we have the generating function

$$\sum_{n\ge 0}\binom{n+k}kx^n=\frac1{(1-x)^{k+1}}\;.$$

The identity in the question can therefore be rewritten as

$$\left(\sum_{n\ge 0}\binom{n+k}kx^n\right)\left(\sum_{n\ge 0}x^n\right)=\sum_{n\ge 0}\binom{n+k+1}{k+1}x^n\;.$$

The coefficient of $x^n$ in the product on the left is

$$\sum_{i=0}^n\binom{i+k}k\cdot1=\sum_{i=0}^n\binom{i+k}k\;,$$

and the $n$-th term of the discrete convolution of the sequences $\left\langle\binom{n+k}k:n\in\Bbb N\right\rangle$ and $\langle 1,1,1,\dots\rangle$. And at this point you’re practically done.


问:
What do the || symbols mean when working with sets?
I was doing research work for a paper that I need to write for my master studies and I came across the cosine similarity concept.

After a quick look in wikipedia there's this mathematical formula that I don't fully understand:

In my use-case we're working with sets of words (so called text strings) and the top of the formula is the dot product of the two sets.

What I've never seen is the || symbols for a set. I know that |A| stands for the cardinality of the set A but I've never seen the other symbol. Is it the same?

答:
The Wikipedia article on "Cosine similarity" you mentioned says explicitly that $A$ and $B$ are vectors:

For a vector $A=(A_1,\cdots,A_n)$ in the Euclidean space $\mathbb{R}^n$, its "norm" is given by
$$
\|A\|:=\sqrt{A_1^2+\cdots+A_n^2}
$$


问:
Show that $\frac{46!}{48}+1$ is not a power of 47
The original exercise was:

Prove that there is no non-negative integers such that $m!+48=48(m+1)^n$

I could type something, but I've stopped at the moment I had to prove that that $\frac{46!}{48}+1$ is not a power of 47.Could anyone help me? I'll post my solution below:

Assume that there is such $m,n$, This, we must have that $48 \mid m!$, where $m$ must be at least 6. Note that $\frac{6!+48}{48} = 46 \neq 7^n$ and so, $m\geq 7$. By the same way, we have that $m \neq 7$ and we find $m \geq 8$. Note that, in this case, $6 \mid \frac{m!}{48}$, i.e, $$1=6 \wedge \frac{m!}{48}+1 = 1 = 6 \wedge (m+1)^n$$ and, so, $6 \text { and } m+1$ are coprimes.
If we had $m+1$ compound, then, there is a prime $p >3$, since both $2 \text{ and } 3$ do not divide $m+1$, such that $p \mid m+1$. By the other hand, once $48=2^4.3$, we have that $p \mid m!$, but $p \nmid m!+48$, which is an absurd, since $p \mid m+1$. Then, $m+1$ is prime and, by Wilson's Theorem, we have that $m+1 \mid m!+1$. Using this last argument with $m+1 \mid m!+48$, we reach that $m+1 \mid 47 \Rightarrow m=46$.
Now, remains to be shown that $\frac{46!}{48}+1$ is not a power of $47$

答:
$$46!=48(47^k-1)$$

First since $47^{7}\equiv1\pmod{43}$ we know $7\vert k$. (Easy to see by brute force $7$ times)

Next since $47^{40}\equiv1\pmod{41}$ and $47^{2}\equiv35\neq1\pmod{41}$ and $47^{4}\equiv24\neq1\pmod{41}$ and $47^{5}\equiv26\neq1\pmod{41}$ we know at least one of $8$ or $10$ divides $k$.

Either case, $k\geq56$ which implies the right side is much larger than the left side, contradiction.


答:
Apparently $\dfrac {46!}{48}+1 = 47 \times 22115034632797133 \times 361673892624894911 \times 304948016756743701941$, so finding a small prime factor is out of the question.

Alternatively we can note that $\log\left(\dfrac {46!}{48}+1\right) \div \log(47)$ is not an integer.

There is also a sneaky way: note that $\dfrac {46!}{48}+1$ ends in $\dots 000001$. We can attempt to find the multiplicative order of $47$ modulo $10^k$ for a managable $k$, say $k=3$.

We have $47 \equiv -1 \pmod 8$; this part is easy.

We have $47 \equiv -3 \pmod {25}$. We also know that if a power of $47^n$ ends in $1$, $n$ must be divisible by $4$; the multiplicative order of $47$ must also divide $\phi(100)=40$, hence it could only be one of $4,8,20,40$. Noting that:

$$(-3)^4 = 81\equiv 6 \pmod {25}$$

$$6^2 = 36 \equiv 11 \pmod {25}$$

we see that it is at least $20$ (which it is indeed). Finally we show that

$$47^{20} < \dfrac {46!}{48}+1 < 47^{40}$$

and we are done. We write:

$$\dfrac {46!}{48}+1 = 3 \times 4 \times \dots \times 23 \times 25 \times \dots \times 46 + 1$$

and we have

$$3 \times 4 \times \dots \times 23 \times 25 \times \dots \times 46 + 1 > (3\times 25) (4 \times 26)\dots (23 \times 45) > 47^{21}$$

$$3 \times 4 \times \dots \times 23 \times 25 \times \dots \times 46 + 1 = (3 \times 10)(4 \times 9)(5\times 8)(6\times 7)\times 11 \times \dots \times 46 + 1 < 47^{39}$$

Title: Show that $\frac{46!}{48}+1$ is not a power of 47

Body:
The original exercise was:

Prove that there is no non-negative integers such that $m!+48=48(m+1)^n$

I could type something, but I've stopped at the moment I had to prove that that $\frac{46!}{48}+1$ is not a power of 47.Could anyone help me? I'll post my solution below:

Assume that there is such $m,n$, This, we must have that $48 \mid m!$, where $m$ must be at least 6. Note that $\frac{6!+48}{48} = 46 \neq 7^n$ and so, $m\geq 7$. By the same way, we have that $m \neq 7$ and we find $m \geq 8$. Note that, in this case, $6 \mid \frac{m!}{48}$, i.e, $$1=6 \wedge \frac{m!}{48}+1 = 1 = 6 \wedge (m+1)^n$$ and, so, $6 \text { and } m+1$ are coprimes.
If we had $m+1$ compound, then, there is a prime $p >3$, since both $2 \text{ and } 3$ do not divide $m+1$, such that $p \mid m+1$. By the other hand, once $48=2^4.3$, we have that $p \mid m!$, but $p \nmid m!+48$, which is an absurd, since $p \mid m+1$. Then, $m+1$ is prime and, by Wilson's Theorem, we have that $m+1 \mid m!+1$. Using this last argument with $m+1 \mid m!+48$, we reach that $m+1 \mid 47 \Rightarrow m=46$.
Now, remains to be shown that $\frac{46!}{48}+1$ is not a power of $47$


问:
Why this converges to $0$?
Let $f:\Bbb C \to \Bbb C$ be an entire function. I wanna show that if there is an $M >0$ and an $n\in \Bbb N$ s.t. $$ |f(z)|\leq M \cdot |z|^n$$
forall $z\in \Bbb C$ with $|z|\geq 1$. Then $f$ is a polynomial of degree at most $n$.

Suppose we have a $M$ and a $n$ s.t. that $ |f(z)|\leq M \cdot |z|^n$. Now because $f$ is an entire function, we can represent $f$ as a power serie
$$f(z)=\sum^{\infty}_{i=0}a_iz^i $$
We have that the coefficients are bound by the cauchy bound
$$|a_i| \leq r^{-i} \max\{|f(u)|:u \in B(0,r)\}, \; \forall r >0$$
but with our $M$ and $n$ we have that $|a_i| \leq r^{-i} \cdot M \cdot |z|^n$, $\forall r>0$.Now if $r \to \infty$ t follows that $a_i=0$ $\forall i >n$. Why is it true? I have some problems to see this statement.

答: The inequality$$|a_i|\leqslant r^{-i}\max\{|f(z)|\mid z\in B(0,r)\},$$together with $|f(z)|\leqslant M|z|^n$ allows you to deduce that$$|a_i|\leqslant r^{-i}Mr^n=Mr^{n-i}.$$So,$$i>n\implies|a_i|\leqslant\lim_{n\to\infty}Mr^{n-i}=0.$$

Title: Why this converges to $0$?


问:
Show that $\left\lceil x-\frac{1}{2} \right\rceil$ is the closest integer to the number $x$
Show that $\left\lceil x-\frac{1}{2} \right\rceil$ is the closest integer to the number $x$, except when $x$ is midway between two integers $n$ and $n+1$, when it is the smaller of these two integers.

We want to prove 2 cases here:

When $x$ is midway between the two integers, $\left\lceil x-\frac{1}{2} \right\rceil$ is equal to the smaller of the two integers.

When $x$ is not midway between the two integers, $\left\lceil x-\frac{1}{2} \right\rceil$ is equal to the closest integer to the number $x$.

If we let $x$ be midway between the two integers $n$ and $n+1$, then

$$x = n + \frac{(n+1)-n}{2} = n+ \frac{1}{2}$$

Now, substitute back in $x$ in $\left\lceil x-\frac{1}{2} \right\rceil$,

$$\left\lceil n+ \frac{1}{2} - \frac{1}{2}\right\rceil  = \left\lfloor n \right\rfloor$$

The problem is why the ceiling function was changed to the floor function?

答:
The problem is why the ceiling function was changed to floor function?

Because $n$ is an integer, $\lfloor n\rfloor= \lceil n \rceil = n.$


问:
Alternating functional Series Convergence SOS....
Does the following series converge?
$\sum_{k=0}^\infty \frac{(-1)^k x^k \sqrt{k}}{k!}$

what is the radius of convergence?!!

答:
Note that $\sqrt{k}<k.$ Then
$\left| \frac{(-1)^k x^k \sqrt{k}}{k!} \right|<\frac{|x|^k k}{k!}=\frac{|x|^k}{(k-1)!}.$

For the series $\sum\limits_{k=0}^\infty {\frac{|x|^k}{(k-1)!}}$ the radius of convergence $R=\lim\limits_{k\to\infty}\frac{k!}{(k-1)!}=\ldots$


Body:
Suppose that
$$
47^n=\frac{46!}{48}+1.
$$
Then, $7\mid 47^n-1$, so $47^n\equiv (-2)^n\equiv 1\pmod 7$. Hence, $6\mid n$. However, it means that
$$
47^6-1\mid 46!
$$
but $47^2-47+1$ divides $47^6-1$ and $47^2-47+1=2163=21\cdot 103$. Thus, $2163\mid 46!$, but $103$ is prime and $103>46$, contradiction.


答: This game is called Cram and the second player has a winning strategy. For every domino the first player places, the second player places a domino at the same spot rotated $180^\circ$ around the original board's centre, thus always leaving a $C_2$-symmetric board to the first player; this is always possible at every move because of the symmetry and because there is no way a domino can be placed "centrally" (either a centre square or edge) to ruin the strategy. Eventually the first player will be unable to move and lose, handing the win to the second player.

Title: Covering paired squares on a $10 \times 10$ board

Body: In a game, the board is a $10 \times 10$  table (in the style of a chessboard). Two players make moves in turns. In a single step, a player covers two neighboring free cells of the table by $2\times 1$ tile. The player who cannot make a step loses the game. Who has a winning strategy - the player who starts the game or his opponent?


问:
How Borel sets make $\sigma$-algebra in a topological space?
I am trying to wrap my head around random variables and can't prove the following questions:

How, in a topological space $(X, \mathcal{T})$, the collection of all Borel sets, say $\mathfrak{B}$, make a $\sigma$-algebra? And, Why is it the smallest $\sigma$-algebra containing $\mathcal{T}$?

In the introduction of the Wikipedia entry of Borel sets these two statements appear. So, I'm using the definition of Borel set given there:

A Borel set is any set in a topological space that can be formed from open sets through the operations of countable union, countable intersection, and relative complement.

I will appreciate any help. Thank you.

答:
A $\sigma$-algebra $\mathcal{A}$ on a set $X$ is a subset of the powerset of $X$, $2^X$ and is defined by the following axioms:

$\emptyset$ and $X$ are in $\mathcal{A}$

$\mathcal{A}$ is closed under complement

$\mathcal{A}$ is closed under finite union

Let's prove that each of these axioms holds for the Borel $\sigma$-algebra.

Let $(X, \tau)$ be a topological space, where $\tau$ is a topology on the set $X$. As a quick reminder, let us list the axioms that define a topological space, where $\tau$ is also a subset of the $2^X$:

$\emptyset$ and $X$ are in $\tau$

$\tau$ is closed under arbitrary union

$\tau$ is closed under finite intersection

Once again, for convenience, let us remind ourselves of how the Borel $\sigma$-algebra $B$ is defined: as all sets resulting from the iterated (as pointed out by Michael Greinecker in the comments) application of three operations, namely countable union, countable intersection and complement to the open sets (the elements of $\tau$).

Axiom 1

Right off the bat, we can see that $\emptyset$ and $X$ are in $B$, since they are in $X$ by the definition of a topological space, and you can build them into the Borel $\sigma$-algebra trivially (e.g. union and intersection of the two).

Axiom 2

To prove that the complement of any element of $B$ is also in $B$, notice the following. The complement of the complement of a set is the set itself. The complement of a countable union is a countable intersection of complements. The complement of a countable intersection is a countable union of complements. As such, $B$ is closed under complement.

Axiom 3

A finite union is a countable union, which is one of the allowed operations to build a Borel set, so $B$ is closed under finite union.


答: You could compute $46! \equiv 281 \mod 47^2$, so $46!/48+1$ is not divisible by $47^2$.


问:
Need help on this closest point optimisation problem
I am getting stuck on this optimisation problem and I would be grateful if anyone could help me solve it.

Submarines are located at $(2,0)$ and $(1,1)$. Choose the slope $m$ so the line $y = mx$ goes between the submarines but stays as far as possible from the nearest one.

Please do not give away the answer but point me in the right direction.

答:
For fixed $m$, you can find the point on $y = mx$ that is nearest to $(2,0)$  by solving
\begin{align}
y &= mx \\
y-0 &= -\frac{1}{m}(x-2)
\end{align}
for $(x,y)$, yielding $(x_1,y_1)$.
Similarly, you can find the point on $y = mx$ that is nearest to $(1,1)$ by solving
\begin{align}
y &= mx \\
y-1 &= -\frac{1}{m}(x-1)
\end{align}
for $(x,y)$, yielding $(x_2,y_2)$.
Let these two nearest distances be
\begin{align}
d_1(m) &= \sqrt{(2-x_1)^2+(0-y_1)^2} \\
d_2(m) &= \sqrt{(1-x_2)^2+(1-y_2)^2}
\end{align}
Now let $f(m)=\min(d_1(m),d_2(m))$ be the distance from the line $y=mx$ to the nearer submarine.  You want to maximize $f(m)$ subject to $0 \le m \le 1$.  Hint: this maximum will occur where $d_1(m)=d_2(m)$.

Drag the slider for $m$ here to see a graphical representation:
https://www.desmos.com/calculator/snnpapyfln


问:
If $\mu(A)\ne0$, is there always a $f\in L^p(\mu)$ with $\{f\ne0\}\subseteq A$ and $\mu(\{f\ne0\})\ne0$?
Let $(\Omega,\mathcal A,\mu)$ be a measure space and $p\in[1,\infty]$.

Are we able to show that if $A\in\mathcal A$ with $\mu(A)\ne0$, then there is a $f\in\mathcal L^p(\mu)$ with $\{f\ne0\}\subseteq A$ and $\mu\left(\{f\ne0\}\right)\ne0$?

If not, are we able to give a suitable condition on $\mu$ to ensure that the desired conclusion holds?

答: No. Consider the $\sigma-$algebra $\mathcal{A}=\{\emptyset, \mathbb{R}\}$ and the Lebesgue measure. Then for this space, $L^p = \{0\}$ because the only measurable functions on $\mathcal{A}$ are constants and $\int_\mathbb{R}|\alpha|^pdm <\infty$ iff $\alpha=0$. Thus $\{f\not=0\} = \emptyset$ for any $f\in L^p$.


问:
Evaluate the integral using the given data

Let $f:[0,1]\to [0,1]$ be a continuous function such that $x^2 +(f(x))^2\le 1$ for all $x\in [0,1]$ and $\int_0^1 f(x).dx=\frac{\pi}{4}$, then find $\int_{-1/2}^{1/\sqrt 2} \frac{f(x)}{1-x^2}.dx$

The part that’s throwing me off the inequality function. I have no idea how to solve such that problems. Can I get a hint?

答:
It is a badly written question. It says for all $x \in (0,1)$ and does not define $f(x)$ for $x \lt 0$ and then asks to integrate for negative values of $x$. Anyway I assume the given equation is true for all $x \in (-1, 1)$.

$x^2 + (f(x))^2 \leq 1$ for all $x \ $ and $\displaystyle \int_0^1 f(x) \ dx = \frac{\pi}{4}$.

And we need to find $\displaystyle \int_{-1/2}^{1/ \sqrt2} \frac{f(x)}{1-x^2} \ dx$

We know from the equation that $x^2 + (f(x))^2 \leq 1$ is a circle. The inequality ensures that you cover all the points inside the circle, in addition to covering points on the circle because the question is about area.

The next point which show the integral $I = \frac{\pi}{4} \ $ for $0 \leq x \leq 1$ only reiterates that the function is never negative otherwise the area would be $\frac{\pi}{2}$ for $0 \leq x \leq 1$.

So the integral that we need to find simply translates to $\displaystyle \int_{-1/2}^{1/ \sqrt2} \frac{1}{\sqrt{1-x^2}} \ dx$


问:
Poisson conditional probability question
So I have a question that asks the following:

The probability that $n$ electrons are emitted by a thermionic cathode in a time interval $t$ is given by a Poisson distribution with mean $\lambda$. What is the probability that $n$ electrons are emitted in a time interval $t$  in which at least 2 electrons are emitted?

I've tried to use conditional probability with the Poisson pmf given that $n$ is more than or equal to 2 but I can't seem to get anywhere and at this point any help would be really appreciated. Thanks in advance!

答:
Let $X$ be the number of electrons emitted in a time interval $t$. Note that $X\sim \text{Po}(\lambda)$. Then the probability is $$\mathbb P(X=n|X\ge 2)=\frac{\mathbb P(X=n\cap X\ge 2)}{\mathbb P(X\ge 2)}$$
It is easy to see that for $n=0,1;\mathbb P(X=n\cap X\ge 2)=0$. For $n\ge 2$, $$\frac{\mathbb P(X=n\cap X\ge 2)}{\mathbb P(X\ge 2)}=\frac{\mathbb P(X=n)}{1-\mathbb P(0\le X\le 1)}=\frac{e^{-\lambda}\cdot \frac{\lambda^n}{n!}}{1-\sum_{k=0}^1e^{-\lambda}\cdot \frac{\lambda^k}{k!}}$$


问:
Maximum number of non crossing line segments endpoints chosen in a random set of points in the plane
Given a set of n randomly scattered points for even
n = 2,4,6,...,50 . Find the maximum number of line segments between the pairs of nodes in
such a way these line segments do not cross each other.</p>

答:
As stated in my comment, if you use every vertex exactly once, then you obtain $\frac n2$ edges, which is boring.

Otherwise you might want to look at Euler's formula which tells you that in order to maximize the number of edges, you also want to maximize the number of faces. So how many faces can there be? Looking at OEIS, and using the right search terms, you may find A008486

Expansion of $\displaystyle\frac{1+x+x^2}{(1-x)^2}$:

1, 3, 6, 9, 12, 15, 18, 21, …

with the following comment:

Conjecture from Dmitry Kamenetsky, Jun 29 2008: This is also the maximum number of edges possible in a planar simple graph with $n+2$ vertices.

The conjecture is correct. Proof: For $n=0$ the theorem holds, the maximum planar graph has $n+2=2$ vertices and $1$ edge. Now suppose that we have a connected planar graph with at least $3$ vertices. If it contains a face that is not a triangle, we can add an edge that divides this face into two without breaking its planarity. Hence all maximum planar graphs are triangulations. Euler's formula for planar graphs states that in any planar simple graph with $V$ vertices, $E$ edges and $F$ faces we have $V+F-E=2$. If all faces are triangles, then $F=\frac{2E}3$, which gives us $E=3V-6$. Hence for $n>0$ each maximum planar simple graph with $n+2$ vertices has $3n$ edges. – Michal Forisek, Apr 23 2009

One problem here is the fact that your graph has to be not only planar, but straight-edged. The above argument is still easy to see for all faces except the exterior one. So in order to use this result, you'll still have to show that you can always make the outer face a triangle as well. The easiest approach to this effect is probably starting out with an outer triangle, and then successively subdividing it, each additional vertex adding three edges and dividing an existing triangle into three new triangles.

But taking a closer look at your problem, there is this statement about points being “randomly scattered”. Which means you are not free to choose the locations of the outer points. In which case the number of vertices of the outer face is predetermined: it is the number of points in the convex hull of your input. The interior you could always triangulate, so you can give the resulting number of edges as a function of both the total number of points and the number of points in the convex hull. I don't see how the number of points being even might be relevant to all of this.


问:
Can we show this family be equicontinous via Ascoli Arzelà?
Let $\{f_{a}(x)=x^{-a},\hspace{1mm} a\in\mathbb{R}^{+}\}$ for $x\in I=(1,+\infty)$. Then we have $f_{a}(x)\in (0,1)\subset[0,1]$ thus for any sequence $(x_{n})$ there exists $(x_{n_{k}})$ and $y\in[0,1]$ such that:
$$f_{a}(x_{n_{k}})\longrightarrow y$$
So that $f_{a}(x_{n})$ admits converging subsequences. We can not directly apply Ascoli Arzelà as $I$ is not compact. But on any compact $K\subset I$ this result holds and $f_{a}$ is equicontinous on $K$. From such information can we derivate the applicability of Ascoli-Arzelà on $I$ as numerable union of compact subsets?

答:
The family is not uniformly equicontinuous on $(1, +\infty)$. If it were, since each $f_a$ has a continuous extension to $[1, +\infty)$, the family of extensions would be uniformly equicontinuous on $[1, +\infty)$. (For every $\varepsilon > 0$ the same $\delta$ would work on $[1, +\infty)$ as on $(1, +\infty)$.) A fortiori we would obtain a uniformly equicontinuous family on $[1,2]$ by restriction, and the Ascoli-Arzelà theorem says every sequence in that family would have a uniformly convergent subsequence. But we have
$$\lim_{n \to \infty} f_n(x) = \begin{cases} 1 &\text{if } x = 1,\\ 0 &\text{if } x > 1, \end{cases}$$
thus a sequence whose pointwise limit is discontinuous. It follows that the sequence has no uniformly convergent subsequence, hence the family is not equicontinuous on $[1,2]$. (Note: on a compact domain the concepts of uniform equicontinuity and pointwise equicontinuity coincide.)

As for pointwise equicontinuity, we can show that the family is pointwise equicontinuous on $(1, +\infty)$ using the Ascoli-Arzelà theorem.

Consider a sequence $(f_{a_n})_{n \in \mathbb{N}}$ and the sequence $(a_n)$ of indices. Either $(a_n)$ has a convergent subsequence - in which case we may assume the whole sequence converges, say to $\alpha \in [0, +\infty)$ - or we have $a_n \to +\infty$.

Consider any compact interval $[u,v] \subset (1, +\infty)$. In the former case, $(f_{a_n})$ converges to $f_{\alpha}$ uniformly on $[u,v]$, in the latter case it converges uniformly to $0$ on $[u,v]$.

Thus by the Ascoli-Arzelà theorem the family of the restrictions $\{ f_a\rvert_{[u,v]} : a \in \mathbb{R}^+\}$ is equicontinuous, and that implies the original family is equicontinuous at all $x \in (u,v)$. Since every $x \in (1, +\infty)$ lies in some $(u,v)$ with $1 < u < v < +\infty$, the family is pointwise equicontinuous on $(1, +\infty)$.


问:
Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $-\frac{7}{3}<a+b<-2$

Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $$-\frac{7}{3}<a+b<-2$$

I have shown that $a+b<-2$. My approach: $-3=8-11=a^3+b^3-6ab+2^3=\frac{1}{2}(a+b+2)((a-b)^2+(a-2)^2+(b-2)^2)$. From this we must have that $a+b<-2$.

Please give some idea/hint for the other part.

答:
$$a^3-6 a b+b^3=-11$$
can be written as
$$(a+b)^3-3 a b (a+b)-6 a b=-11$$
setting $$a+b=s;\;ab=p$$
we get
$$ s^3-3ps-6p+11=0$$
The equation $$z^2-sz+p=0\tag{1}$$
Gives the values of $a,b$. In order to have  real roots, discriminant must be positive
$$s^2-4p\ge 0$$
Solve
$$\begin{cases}
s^3-3ps-6p+11=0\\
s^2-4p\ge 0\\
\end{cases}
$$

Solving the first equation wrt $p$ we get
$$p=\frac{s^3+11}{3 (s+2)}$$

From the second equation we get $p\le \frac{s^2}{4}$.

Substituting in the first we get the inequality

$$\frac{s^3+11}{3 (s+2)}\le  \frac{s^2}{4}$$

$$\frac{s^3+11}{3 (s+2)}=  \frac{s^2}{4}\to s^3-6 s^2+44=0\to s=-2.30214$$
The solution of the inequality is
$$-2.30214 \le s<-2$$
that is
$$-2.30214 \le a+b<-2$$

edit

which is a more accurate result than
$$-\frac73< a+b <-2$$
indeed if $s=-2.32>-\frac73$ then $p=1.54913$ and the equation $(1)$ becomes
$$z^2+2.32 z+1.54913=0$$

In other words the claim is wrong because values of $a+b$ from $-\frac73$ to $-2.30214$ the numbers $a$ and $b$ can't be real.
which has no real solutions because discriminant is negative.


问:
Why $\omega$ must be compactly contained in $\Omega$ for this proof?
I was reading the following proof from Brezis' book on Sobolev spaces.

Proposition : Let $G\in C^1(\mathbb{R})$ be such that $G(0)=0$ and $|G'(s)|\le M\forall s\in\mathbb{R}$ for some constant $M$. Let $u\in W^{1,p}(\Omega)$ with $1\le p \le \infty$. Then

$$G\circ u \in W^{1,p}(\Omega)\text{ and }\frac{\partial}{\partial x_i}G\circ u = (G'\circ u))\frac{\partial u}{\partial x_i}, i=1,\ldots,N.$$

Proof : We have $|G(s)|\le M|s| \forall s\in\mathbb{R}$ and thus $|G\circ u|\le M|u|$; as a consequence, $G\circ u \in L^p(\Omega)$ and also $(G'\circ u)\frac{\partial u}{\partial x_i}\in L^p(\Omega)$. It remains to verify that

$$\int_{\Omega}(G\circ u)\frac{\partial \phi}{\partial x_i}=-\int_{\Omega}(G'\circ u)\frac{\partial u}{\partial x_i}\phi, \forall \phi\in C^1_c(\Omega).$$

When $1\le p< \infty$, one chooses a sequence $(u_n)$ in $C^{\infty}_c(\mathbb{R}^N)$ such that $u_n\to u$ in $L^p(\Omega)$ and a.e. on $\Omega, \nabla u_n \to \nabla u$ in $L^p(\omega)^N \forall \omega\subset \subset \Omega$. We have

$$\int_{\Omega}(G\circ u_n)\frac{\partial \phi}{\partial x_i} = -\int_{\Omega}(G'\circ u_n)\frac{\partial u_n}{\partial x_i}\phi, \forall \phi\in C^1_c(\Omega).(6)$$
But $G\circ u_n\to G\circ u in L^p(\Omega)$ and $(G'\circ u_n)\frac{\partial u_n}{\partial x_i}\to (G'\circ u)\frac{\partial u}{\partial x_i}$ in $L^p(\omega)\forall \omega\subset\subset \Omega$ (by dominated convergence), so that (6) follows from the above.

I don't really understand why we need $\omega\subset\subset \Omega$ (meaning $\omega$ is compactly contained in $\Omega$). Since $|G'\circ u_n|\le M$ and $u_n, \phi$ are smooth functions with compact support we have the result even if $\omega$ was not compactly contained in $\Omega$ no?

答:
One wants to approximate $W^{1,p}$ functions by smooth functions. For arbitrary $\Omega$ it is not possible to approximate $u$ by functions from $C^\infty_c(\mathbb R^d)$ or $C^\infty_c(\Omega)$. What is possible however, is to approximate $u$ on compact subsets of $\Omega$. This can be proven by simple mollification and use of cut-off function.

In the proof in your post, one simply does not have the existence of a sequence of smooth functions $(u_n)$ such that $\nabla u_n\to \nabla u$ in $L^p(\Omega)$.


问:
why integrating only alternating forms?
Hello I was reviewing some concepts of differential forms. I cannot recall why only multilinear alternating forms can be integrated on manyfolds and not general multilinear forms... Why is the hypothesis of being alternating so important? In an intuitive way I'd like to understand it... Is it for having a covariant concept? Or for making some riemann sums convergent?</p>

答:
$\def\d{\mathrm{d}}
\def\np{\star}
\newcommand\pder[2][]{\frac{\partial #1}{\partial #2}}$Here we motivate this fact with a small example.

Suppose we wish to integrate something of the form
$\d x^1\np \d x^2$, where $\np$ is some, not necessarily alternating, product.
(The superscripts here are simply labels.)
We assume that the product is distributive and acts nicely with respect to scalar multiplication.
Under coordinate transformation we find
\begin{align*}
\d x^1\np \d x^2
&= \left(\pder[x^1]{y^1}\d y^1+\pder[x^1]{y^2}\d y^2\right)
\np
\left(\pder[x^2]{y^1}\d y^1+\pder[x^2]{y^2}\d y^2\right) \\
&=\pder[x^1]{y^1}\pder[x^2]{y^1}\d y^1\np\d y^1
+ \pder[x^1]{y^1}\pder[x^2]{y^2}\d y^1\np\d y^2 \\
& \quad + \pder[x^1]{y^2}\pder[x^2]{y^1}\d y^2\np\d y^1
+ \pder[x^1]{y^2}\pder[x^2]{y^2}\d y^2\np\d y^2.
\end{align*}
To agree with our standard notion of integration, we must have
$$\d x^1\star\d x^2 = \det\pder[x^i]{y^j} \d y^1\np \d y^2.$$
We immediately find
\begin{align*}
\d y^1\np\d y^1 &= \d y^2\np\d y^2 = 0 \\
\d y^2\np\d y^1 &= -\d y^1\np\d y^2,
\end{align*}
that is, the product $\np$ is antisymmetric.

Title: why integrating only alternating forms?


问:
Show that 7 is a quadratic residue for any prime p of the form 28k + 1 and 28k + 3.
Interesting number theory question, which I feel should be reasonably straight forward, but I can't seem to crack it.

Show that 7 is a quadratic residue for any prime p of the form 28k + 1 and 28k + 3.

Now, I know that a quadratic residue $a$ (mod m) means that there exists $x$ such that $x^2\equiv a$(modm).
So if 7 is a quadratic residue mod 28k+1 or 28k+3 then $x^2\equiv 7$ (mod 28k+1 or 28k +3).  But I am unsure of how to tackle this problem.
Perhaps, I could go on to say that for the first case, $x^2\equiv 6$ (mod 28k) $\equiv$ 3 (mod 14k)??

Or perhaps I am going about this wrong.

I also know the laws of quadratic reciprocity, so I need to show $(\frac{7}{28k+1})$ =1 and likewise for 28k +3

Help would be greatly appreciated.  Thank you

答:
Using Quadratic Reciprocity Theorem,

$$\left(\frac p7\right)\left(\frac 7p\right)=(-1)^{\frac{3(p-1)}2}$$

Now, if $p\equiv1\pmod4,\left(\frac p7\right)\left(\frac 7p\right)=1\iff \left(\frac 7p\right)=\left(\frac p7\right)$

Now, $(\pm1)^2\equiv1\pmod 7,(\pm2)^2\equiv4,(\pm3)^2\equiv2\implies p$ needs to be $1,2,4\pmod 7$

$p\equiv1\pmod4$ and $p\equiv1\pmod7\implies p\equiv1\pmod {4\cdot7}$

Using CRT, we can show

$p\equiv1\pmod4$ and $p\equiv2\pmod7\implies p\equiv9\pmod {4\cdot7}$

$p\equiv1\pmod4$ and $p\equiv4\pmod7\implies p\equiv25\pmod {4\cdot7}\equiv-3$

Now, if $p\equiv-1\pmod4,\left(\frac p7\right)\left(\frac 7p\right)=-1\iff \left(\frac 7p\right)=-\left(\frac p7\right)$

and $p$ needs to be $3,5,6\pmod 7$

$p\equiv-1\pmod4\equiv3$ and $p\equiv3\pmod7\implies p\equiv3\pmod {4\cdot7}$

$p\equiv-1\pmod4$ and $p\equiv6\pmod7\equiv-1\implies p\equiv-1\pmod {4\cdot7}$

Using CRT, we can show

$p\equiv-1\pmod4$ and $p\equiv5\pmod7\implies p\equiv19\pmod {4\cdot7}\equiv-9$

So, $7$ is a quadratic residue for any prime $p\equiv\pm1,\pm3,\pm9\pmod{28}$


问:
Question about variational inequality
I have a question about the following proposition.

Let $H$ be a Hilbert space, $x^* \in H^*$ a bounded functional, $U\subseteq H$ a closed linear subspace and $\bar{u} \in U$. Then $\Re x^*(u-\bar{u})\leq\Re (\bar{u},u-\bar{u})$ for every $u \in U$ if and only if $x^*(u)=(u,\bar{u})$ for every $u\in U$.

$\Re, \Im$ are real and imaginary parts, and it is assumed that the inner product $(\cdot,\cdot)$ is linear in the first slot.

Context. This equivalence helps in proving properties of the metric projection onto closed subspaces, and also the Riesz representation theorem.

Try. The proof proceeds as follows:

show that $\Re x^*(u-\bar{u})\leq\Re (\bar{u},u-\bar{u})$ for every $u \in U$ is equivalent to $\Re x^*(u-\bar{u})\leq\Re (u-\bar{u},\bar{u})$ for every $u \in U$

use linearity of $U$ and $x^*$ to prove $\Re x^*(u-\bar{u})\leq\Re (u-\bar{u},\bar{u})$ for every $u \in U$ holds if and only if the equality holds

still using linearity of $U$ and $x^*$ we can easily prove that $\Re x^*(u-\bar{u})=\Re (u-\bar{u},\bar{u})$ for every $u \in U$ holds if and only if $\Im x^*(u-\bar{u})=\Im (u-\bar{u},\bar{u})$ for every $u \in U$

put together all the pieces and conclude

What I don't have clear is how to prove step $1$.It seems to me that we are shifting the linearity of the inner product to the wrong slot if we proved it, and I don't see a way of doing so, since we have the real parts in between. Do you have any suggestion?

答:
Recall that $(x,y) = (y,x)$ so $(\cdot,\cdot)$ is antilinear in the second slot. We have $(\overline{u},\overline{u}) \in \Bbb{R}$ so
$$(\overline{u},u-\overline{u}) = (\overline{u},u) - (\overline{u},\overline{u}) = \overline{( u,\overline{u})} -(\overline{u},\overline{u}) = \overline{(u,\overline{u})- (\overline{u},\overline{u})}  = \overline{(u-\overline{u},\overline{u})}$$
so $(\overline{u},u-\overline{u})$ and $(u-\overline{u},\overline{u})$ have equal real parts.

Title: Question about variational inequality


问:
Question on marginal probability density function
Consider $\zeta \sim U[-2; 2]$, $\eta \sim U[0; 1]$, $Z = \zeta + \eta\zeta$, $\zeta$ and $\eta$ are independent.

First of all, I need to find the conditional density of $Z \vert\zeta=x$. Let $\zeta = x$, then $Z = x(1+\eta) = $
\begin{cases}
U[x; 2x], \hspace{0.2cm} x \ge 0, \\ U[2x; x], \hspace{0.2cm}x <0
\end{cases}

Writing this in one line,
$$\rho_{Z\vert\zeta}(z\vert x) = \frac{1}{x}\mathbb{1}_{x\le z\le 2x}\mathbb{1}_{x\ge0}-\frac{1}{x}\mathbb{1}_{2x\le z\le x}\mathbb{1}_{x<0}.$$

My first concern is whether this density is correct? What happens at $x=0$? I mean, is $\rho$ finite at this point. I think should be, but I do not understand it properly

Secondly, I need to find conditional expectation $\mathbb{E}[Z\vert \eta =x]$

\begin{equation}
\mathbb{E}[Z\vert \eta =x] = \int_\mathbb R z\cdot\rho_{Z\vert\zeta}(z\vert x)dz = \int_\mathbb Rz(\frac{1}{x}\mathbb{1}_{x\le z\le 2x}\mathbb{1}_{x\ge0}-\frac{1}{x}\mathbb{1}_{2x\le z\le x}\mathbb{1}_{x<0})dz =\\= \int_x^{2x}\frac{z}{x}\mathbb{1}_{x\ge0}dz-\int_{2x}^x \frac{z}{x}\mathbb{1}_{x<0}dz = \frac{3x}{2}
\end{equation}

Finally, I need to compute marginal density $\rho_Z(z)$.

\begin{equation}
\rho_Z(z) = \int_\mathbb R \rho_{(\zeta, Z)}(x, z)dx = \int_\mathbb R \rho_\zeta(x) \cdot \rho_{Z\vert\zeta}(z\vert x)dx = \\ = \int_\mathbb R \left(\frac{1}{4x}\mathbb{1}_{-2\le x\le2}\mathbb{1}_{x\le z\le 2x}\mathbb{1}_{x\ge0} - \frac{1}{4x}\mathbb{1}_{-2\le x\le2}\mathbb{1}_{2x\le z\le 2}\mathbb{1}_{x<0}\right)dx = \\ =\int_0^2\frac{1}{4x}\mathbb 1_{x\le z \le 2x}dx - \int_{-2}^0\frac{1}{4x}\mathbb 1_{2x\le z \le x}dx
\end{equation}

In the last line I have troubles with integration of indicator function depending on the variable with respect to I have integrate.

Thanks for the help and for correcting me in case of mistakes.

答:
Using the definition of CDF and indicating ($X,Y$ are a simple notation rather than you Greek letters IMHO)

$X\sim U[-2;2]$

$Y\sim U[0;1]$

$Z=X+XY$

with $X\perp\!\!\!\!\!\!\perp Y$

To calculate Z law you can do

$$F_Z(z)=\mathbb{P}[Z\leq z]=\mathbb{P}[XY\leq z-X]$$

It is only a matter of evaluating the areas involved by the parametric function $XY\leq z-X$ in the rectangle of the two uniforms...The function is parametric in $z \in [-4;4]$ which is your Z-Support

The requested CDF is this area multiplied by the reciprocal of the rectangle area: $1/4$

Deriving F you get $f_Z(z)$


问:
$ A=\begin{pmatrix} 0 & 2 & 0\\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{pmatrix}$, calculate $e^A$
I have encounter a question in my book , it was For $ A=\begin{pmatrix}
0 & 2 & 0\\
0 & 0 & 3 \\
0 & 0 & 0
\end{pmatrix}$, calculate $e^A$

My solution way : I tried to find its eigenvalues , so i found that the only eigenvalue is $0$ and the eigenspace is $(1,0,0)$ .Hence , it cannot be diagonalized.

Then , i tried to use taylor exponential and it gives me $ A=\begin{pmatrix}
1 & 2 & 0\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}$ .However the answer is $ A=\begin{pmatrix}
1 & 2 & 3\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}$ .

What am i missing ,can you help me?

答:
The minimal polynomial of your matrix is $x^3$. To find $f(A)$ for a holomorphic function $f$ defined on some neighbourhood of $\sigma(A) = \{0\}$, you need to find a polynomial $p \in \Bbb{C}[x]$ of degree $< 3$ such that
$$p(0) = f(0), \quad p'(0) = f'(0), \quad p''(0)=f''(0)$$
and then $f(A)=p(A)$.
Trying out $p(x) = ax^2+bx+c$, we find that $p(x) = \frac12 x^2+x+1$ is the desired polynomial if $f$ is the exponential function.

Therefore
$$e^A = p(A) = \frac12A^2+A+I = \begin{pmatrix}
1 & 2 & 3\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}.$$


问:
Existence of Unit-speed parametrization.
It seems obvious to me that for all regular smooth functions, there exists a unit-speed parametrization.  Is there a proof of that statement?
Andrew Pressley's Elem.Diff.Geometry tells me that "A parametrized curve has a unit-speed reparametrization if and only if it is regular". Does the proof of his statement prove existence?  In a follow-up example, he shows the twisted cubic, $\gamma(t)=(t,t^2,t^3),-\infty\le t \le +\infty$ and says that since it results in an elliptic integral when trying to find the arc length that we cannot express the unit-speed parametrization with elementary functions. That doesn't seem to exclude its existence.

答: Consider the arc length function $L:t\mapsto s$ of the curve $C$. Its derivative is just the length element, which is non-zero since $C$ is regular. It follows by the inverse function theorem that $L^{-1}:s\mapsto t$ exists, and thus the arc length parametrisation $\alpha(s)=C(L^{-1}(s))$  exists.


问:
The convergence of an integral, dependant on a parameter
I have to say if the integral
$$\int_0^\infty \frac{e^{-xy}}{1+x}\,dx$$
that depends on the parameter $y$ is convergent (I don't know the right terminology, since I'm not studying in english, but I will provide the criterion for such convergence) when 1) $y\in (0,1)$ and 2) $y\in [1,\infty]$. The criterion we use to determine such convergence is as follows:
If
$$\lim_{b\rightarrow +\infty}\sup_{y\in D}\left | \int_b^\infty \frac{e^{-xy}}{1+x} \, dx \right |=0$$
where $D$ is the domain of parameter $y$ then we say that the integral converges properly when $y$ is in that domain. The problem is, I have no idea how to integrate that function. Any ideas? Thank you!

答:
I would adress the problem this way: if $y=0$ then the integrand behaves like $$g(x)=\frac{1}{1+x}$$ (you can try to integrate $g$ and use the definition of indefinite integral) and therefore the integral diverges. If $y>0$, since $$\lim_{x \to \infty} \frac{e^{-xy} x^2}{1+x} =0\text{,}$$ then there is $M$ such that for each $x > M$ then
$$\frac{e^{-xy}}{1+x}<\frac{1}{x^2}\text{.}$$
Thus the integral converge (you can try to apply the defintion to $x^{-2}$ and use a suitable squeeze theorem. Do you know about that?). Finally, if $y<0$ then you are integrating
$$\frac{e^{\vert y \vert x}}{1+x}$$ which surely diverges! (Think about that and feel free to ask if you are not sure). Hope this helped.


问:
Why $x^2+3x^4+\omicron(x^4)=\omicron(x^3)$is false?
Why $x^2+3x^4+\omicron(x^4)=\omicron(x^3)$is false for $x\to 0$?

I know that:

A. $-\omicron(f)=\omicron(f)$

B. $\omicron(f)+\omicron(f)=\omicron(f)$

C. If g is a function $g*\omicron(f)=\omicron(gf)$

C2. if g is costant$g*\omicron(f)=\omicron(gf)=\omicron(f)$

D. $\omicron(f)^n=\omicron(f^n)$

E. $\omicron(f+\omicron(f))=\omicron(f)$

From C I try to write $\omicron(x^4)=x^4\omicron(1)$ but I don't know what my next move is.

答:
If
$$x^2+3x^4+\omicron(x^4)=\omicron(x^3)$$ then

$$\lim\limits_{x \to 0} \frac{x^2+3x^4+\epsilon(x)x^4}{x^3} = 0$$

where $\lim\limits_{x \to 0} \epsilon(x) =0$.

And this is not the case as

$$\frac{x^2}{x^3}$$ is not bounded around zero.


问:
Why is the index of the stabilizer neccesarily greater than 1?
I am currently self-studying Aluffi's Algebra book at chapter IV on on Group actions in which he gives the counting formula where $G$ is a group that acts on a finite set $S$ being:

$$\lvert S \rvert = \lvert Z \rvert +  \sum_{a \in A} [G: G_a],$$

where $Z$ is the set of fixed point under the action and $A \subseteq S$ contains exactly one element for each nontrivial orbit of the action and $G_a$ is the stabilizer of $a$.

He claimed that each $[G:G_a]$ is a divisor of $\mid G \mid$ and also each $[G:G_a] > 1$ in which I understand the divisor part since it is equal to the cardinality of the orbit of $a$ and orbits partion the set, but I just don't get why $[G:G_a] > 1$ since isn't $1$ is a divisor for every integers?

Since I am self-studying, any information would be helpful.

答: If $[G:G_a] = 1$, then $G = G_a$ which would mean that $a$ is fixed by the group action and hence $a \in Z$ and hence $a \notin A$ since it would have a trivial orbit.


答: This follows from the definition of the set $A$. If $a\in A$ then $a$ belongs to a nontrivial orbit. So there must be some element $g\in G$ such that $g.a\ne a$. Hence $G_a\ne G$.

Title: Why is the index of the stabilizer neccesarily greater than 1?

Body:
I am currently self-studying Aluffi's Algebra book at chapter IV on on Group actions in which he gives the counting formula where $G$ is a group that acts on a finite set $S$ being:

$$\lvert S \rvert = \lvert Z \rvert +  \sum_{a \in A} [G: G_a],$$

where $Z$ is the set of fixed point under the action and $A \subseteq S$ contains exactly one element for each nontrivial orbit of the action and $G_a$ is the stabilizer of $a$.

He claimed that each $[G:G_a]$ is a divisor of $\mid G \mid$ and also each $[G:G_a] > 1$ in which I understand the divisor part since it is equal to the cardinality of the orbit of $a$ and orbits partion the set, but I just don't get why $[G:G_a] > 1$ since isn't $1$ is a divisor for every integers?

Since I am self-studying, any information would be helpful.


问:
How to prove this fractional limit is $0$?
I guess that
$$\frac{\lambda^2-||y||_\infty^2}{||y||_\infty||x||_1-\langle y, x\rangle}
\xrightarrow{ x\longrightarrow x^*} 0
$$
under the assumptions:

$f: x\in \mathbb R^n \longmapsto  y\in \mathbb R^n$ is continuous, $x\longrightarrow x^*\neq 0$,

$||y||_\infty \xrightarrow{  x\longrightarrow x^*} \lambda$ and $||y||_\infty\leq \lambda$, i.e. the nominator is always non-negative and converges to zero.

$||y||_\infty||x||_1-\langle y, x\rangle \xrightarrow{  x\longrightarrow x^*} 0$, i.e. the denominator is always positive and converges to zero.

I have the intuition that the limit holds true but I have no clues. Could anyone give me some hints how to prove or disprove it?

答:
I have found a counter example as follows:

Choose $y=f(x)=x$

Choose $x_k \rightarrow \lambda_{\mathbb R^n}$ such that $x_{k,1}= \lambda-1/k$ and $x_{k,i}=\lambda-1/k-1/2^k$ for $i=2,\ldots, n$.

Then the limit tends to $\infty$ not $0$.


问:
continuous Fourier transform
For some reals $a$ and $b$, we have $h(t)=p\left(\cfrac{t-a}{b}\right)$. Also, $p(s)$ is real and even function of $s$.
By the transformation rules, how can I determine the continuous Fourier transform $H$ of $h$ using the Fourier transform $P$ to $p$.

I am thinking I need to use $\displaystyle p(f)=\int^{\infty}_{\infty} p(t)\cdot\exp(2\pi i f t)\ \text dt$    as well as the time shifting and scaling.

答:
Note that, as $p$ is even, we have for $b < 0$: $\def\abs#1{\left|#1\right|}$
$$ h(t) = p\left(\frac{t-a}b\right) = p\left(-\frac{t-a}{b}\right) = p\left(\frac{t-a}{\abs b}\right) $$
(for $b > 0$ we obviously also have $h(t) = p\bigl((t-a)/\abs b\bigr)$. Now
\begin{align*}
  H(f) &= \int_{-\infty}^\infty h(t) \exp(2\pi i ft)\,dt\\
       &= \int_{-\infty}^\infty p\left(\frac{t-a}{\abs b}\right)\exp(2\pi i ft)\,dt\\
       &= \int_{-\infty}^\infty p(\tau)\exp\bigl(2\pi i f(\abs b \tau + a)\bigr)\,\abs b \,d\tau & \text{we substituted $\tau = \frac{t-a}{\abs b}$)}\\
       &= \abs b \exp(2\pi i fa)\int_{-\infty}^\infty p(\tau)\exp(2\pi i f\abs b\tau)\, d\tau\\
       &= \abs b \exp(2\pi i fa)\cdot P(f\abs b)
\end{align*}</p>

Title: continuous Fourier transform


答:
The proof is given pretty much by just explicitly constructing the reparametrization:

Let $c:I \rightarrow \mathbb{R}^3$ be a regular parametrized curve. Let $t_0 \in I$ and define:

$$\psi(t) := \int_{t_0}^{t}||\dot{c}(\tau)||d\tau.$$

Set $J = \psi(I)$. Now $\psi$ is a diffeomorphism from $I$ to $J$: By the fundamental theorem of calculus it is differentiable, and since

$$\psi'(t) = ||\dot{c}(t)|| > 0,$$

it is monotonic increasing, hence bijective. Define $\phi(t) := \psi^{-1}(t)$. Now  $\phi(t)$ is the reparametrization of $c$ that makes it parametrized by arc-length. You can check this by computing the norm of the derivative of the composition.

Now this gives you the existence of the reparametrization, but as you may well know not all integrals are expressable in terms of elementary functions, so we have no way to know if we can actually 'do' the integral. Apparently, the twisted cubic you mentioned is such a case (which I haven't checked...).

Title: Existence of Unit-speed parametrization.

Body:
It seems obvious to me that for all regular smooth functions, there exists a unit-speed parametrization.  Is there a proof of that statement?
Andrew Pressley's Elem.Diff.Geometry tells me that "A parametrized curve has a unit-speed reparametrization if and only if it is regular". Does the proof of his statement prove existence?  In a follow-up example, he shows the twisted cubic, $\gamma(t)=(t,t^2,t^3),-\infty\le t \le +\infty$ and says that since it results in an elliptic integral when trying to find the arc length that we cannot express the unit-speed parametrization with elementary functions. That doesn't seem to exclude its existence.


问:
Misunderstanding geometric series of matrices
$A$ is a $3$ by $3$ real matrix, $rank(A) = 2$, $\lambda_{max}(I_3 - A) < 1$. I am interested in the series $\displaystyle \sum_{k=0}^{\infty}(I_3 - A)^k$.

Going through the usual proof, we have $\displaystyle S_n = \sum_{k=0}^{n}(I_3-A)^k$, multiply by $(I_3 - A)$ on the right to get $S_n(I_3 - A) = \sum_{k=1}^{n}(I_3-A)^{k+1}$, subtract them to get $S_n(I_3 - (I_3-A)) = S_nA=I_3 -(I_3-A)^{n+1}$.

Similarly if we would have multiplied on the left, we would have gotten $AS_n = I_3-(I_3-A)^{n+1}$

Looking at the limit, since $\lim_{n \to \infty}I_3-(I_3-A)^{n+1} = I_3$, then $A$ and $S_\infty$ are inverses of each other.

How does this sit with the fact $rank(A) = 2$ but $A$ is 3 by 3?

答:
If the rank of $A$ is $2$, then the nullspace of $A$ is non-trivial and there must be an eigenvector $v$ of $A$ with eigenvalue $0$.

Thus $(I-A)v=v$, so the largest eigenvalue of $I-A$ would not be less than $1$. The two statements are contradictory.


答: You need to hunderstand that the greater the power of x^n is, the closer you are from zero. Personnaly I prefer O(x) rather than o(x), because i find it easyer to hunderstand, and to control. However, forget about all the rules of o(x), and try to focus on what they mean. Here x^4 IS a o(x^3). And it's the same with o(x^4). So your expression is equivalent x^2+o(x^3)=o(x^3). Wich is equivalent to x^2=o(x3). It can't be true, because x^2 is very big compare with o(x^3).</p>

Title: Why $x^2+3x^4+\omicron(x^4)=\omicron(x^3)$is false?

Body:
Why $x^2+3x^4+\omicron(x^4)=\omicron(x^3)$is false for $x\to 0$?

I know that:

A. $-\omicron(f)=\omicron(f)$

B. $\omicron(f)+\omicron(f)=\omicron(f)$

C. If g is a function $g*\omicron(f)=\omicron(gf)$

C2. if g is costant$g*\omicron(f)=\omicron(gf)=\omicron(f)$

D. $\omicron(f)^n=\omicron(f^n)$

E. $\omicron(f+\omicron(f))=\omicron(f)$

From C I try to write $\omicron(x^4)=x^4\omicron(1)$ but I don't know what my next move is.


问:
How to use maximum likelihood method to find $\mu$ and $\sigma$ from erroneous observations.
Let $X$ be a random variable having a normal distribution with mean $\mu$ and standard deviation $\sigma$. The observations $(x_i; i=1,2,\cdots,n)$ have measurement errors that are normally distributed with zero mean and known standard deviation $\delta$. Use the method of maximum likelihood to estimate $\mu$ and $\sigma^2$ from erroneous observations.

答:
It seems what you're saying is that you have $X \sim N(\mu, \sigma^2)$, that you want to estimate $\mu$ and $\sigma$, and that you don't have samples of $X$ but rather of $Y = X + \varepsilon$ where $\varepsilon \sim N(0, \delta^2)$. Maybe you have a process generating $X$ but then there's measurement noise tacked on to that.

If that's the case you know $Y \sim N(\mu, \sigma^2 + \delta^2)$, so use the maximum likelihood approach using your measurements $y_1, \dots, y_n$ to estimate $\mu$ and $\sigma^2 + \delta^2$.


问:
Pseudo-Cauchy Sequences are bounded help.
I'm currently studying real analysis with the book of "Understanding Analysis" by Abott, I was making Ex. 2.6.5, which states:

"Consider the following (invented) definition: A sequence $(s_n)$
is pseudo-Cauchy if, for all $\epsilon$ > 0, there exists an $N$ such that if $n ≥ N$, then
$|s_{n+1} − s_n| < \epsilon$. Decide which one of the following two propositions is actually true. Supply
a proof for the valid statement and a counterexample for the other.

(i) Pseudo-Cauchy sequences are bounded.

(ii) If $(x_n)$ and $(y_n)$ are pseudo-Cauchy, then $(x_n + y_n)$ is pseudo-Cauchy as
well."
\

So it appears that (i) is false, but I "managed" to provided a proof for it, so I would like to know where I went wrong, here's what I did:

Let $(s_n)$ be pseudo-Cauchy, therefore there exist $N$ such that if $n≥N$, then $|s_{n+1}-s_{n}| <\frac{\epsilon}{k}$, for some $k\in \mathbb{N}$. Without loss of generality assume $m>n$ and let $m =n+k$. Suppose that $m,n≥N$, then $$ |s_m-s_n|=|s_{n+k}-s_n|=|s_{n+k}-s_{n+k-1}+s_{n+k-1}+\dots+s_{n+1}-s_n|\\\leq|s_{n+k}-s_{n+k-1}|+\dots+|s_{n+1}-s_n|<\frac{\epsilon}{k}+\dots+\frac{\epsilon}{k}=\epsilon$$That is $|s_m-s_n|<\epsilon$, therefore $(s_n)$ is Cauchy and is bounded.

Any insight would be appreciate it, thanks.

答: Yes, the first assertion is indeed false (take $s_n=\log(n)$, for instance). The problem with your proof lies in that $k$. What you actually proved was that, for each $k\in\Bbb N$ and for each $\varepsilon>0$, $|s_{n+k}-s_n|<\varepsilon$, if $n$ is large enough. So what?  It doesn't follow from that that $(s_n)_{n\in\Bbb N}$ is a Cauchy sequence. You did not prove that you have $|s_m-s_n|<\varepsilon$ if $m$ and $n$ are large enough.


答:
The terminology you are looking for is uniform convergence, and you have provided a necessary and sufficient condition for the uniform convergence of an improper integral depending on a parameter $y \in D$, viz.

$$\lim_{b \to \infty}\sup_{y \in D} \left|\int_b^\infty f(x,y)\, dx\right| = 0$$

In case (2), we have $\left| \frac{e^{-xy}}{1+x}\right| \leqslant e^{-x}$ for all $y \in [1,\infty)$. Since $\int_0^\infty e^{-x} \, dx = 1 < \infty$ we have uniform convergence of the improper intergral for $y \in [1,\infty)$ by the Weierstrass M-test.

In case (1), we have

$$\sup_{y \in (0,1)} \left|\int_b^\infty \frac{e^{-xy}}{1+x}\, dx\right| \geqslant \sup_{y \in (0,1)}\int_b^{2b} \frac{e^{-xy}}{1+x}\, dx \geqslant \sup_{y \in (0,1)} b \cdot \frac{e^{-2by}}{1+2b} \geqslant \underbrace{\frac{b}{1+2b}e^{-2}}_{y = 1/b}, $$

and, hence,

$$\lim_{b\to \infty}\sup_{y \in (0,1)} \left|\int_b^\infty \frac{e^{-xy}}{1+x}\,dx \right|\geqslant \lim_{b \to \infty}\frac{b}{1+2b}e^{-2} = \frac{e^{-2}}{2} \neq 0$$

Therefore, convergence is not uniform for $y \in (0,1)$.

Title: The convergence of an integral, dependant on a parameter

Body:
I have to say if the integral
$$\int_0^\infty \frac{e^{-xy}}{1+x}\,dx$$
that depends on the parameter $y$ is convergent (I don't know the right terminology, since I'm not studying in english, but I will provide the criterion for such convergence) when 1) $y\in (0,1)$ and 2) $y\in [1,\infty]$. The criterion we use to determine such convergence is as follows:
If
$$\lim_{b\rightarrow +\infty}\sup_{y\in D}\left | \int_b^\infty \frac{e^{-xy}}{1+x} \, dx \right |=0$$
where $D$ is the domain of parameter $y$ then we say that the integral converges properly when $y$ is in that domain. The problem is, I have no idea how to integrate that function. Any ideas? Thank you!