答:
“ I can't think of any way that will give us a negative power in the doninator ”

To put something in the denominator, you must divide by it as taught in elementary school.

$1\div x^{-3}=\frac 1 {x^{-3}}$ Then it is in the denominator.

To reiterate: $1 \div x^{-1}=\frac 1 {x^{-1}}$

Now to prove using this, $1 \div x^{-1}=1\times x$, you times the multiplicative inverse.

Title: Why is $\frac{1}{x^{-1}}=x$?


问:
Suppose $\{\vec{x}, \vec{y}\}$ is a basis for $R^{2}$ and $A$. Show $A$ is invertible.
Suppose $\{\vec{x}, \vec{y}\}$ is a basis for $R^{2}$ and $A = \begin{bmatrix}
a & b\\
c & d
\end{bmatrix}$. If $\{a\vec{x} + b\vec{y}, c\vec{x} + d\vec{y}\}$ is a basis for $R^2$ show $A$ is invertible.

My Attempt

For every $\vec{v} \in R^2$ we must have a unique $k_1,k_2 \in R$ such that $ \vec{v} = k_1(a\vec{x}+b\vec{y}) + k_2(c\vec{x}+d\vec{y})$. Hence we can construct the matrix and equation $Ak=\vec{0}$, from the fact that $ \vec{v} = (k_1a+k_2c)\vec{x}+(k_1b+k_2d)\vec{y}$

$$A \vec{k}= \begin{bmatrix}
a & b\\
c & d
\end{bmatrix} \vec{k} = \vec{0}$$

By the invertible matrix theorem, if this system $rankA=2$ (because we have a unique solution) then the matrix $A$ must be invertible. QED.

Can anyone offer suggestions to improve this proof / point out inconsistencies?

答: I assume that your $R$ is the real numbers $\mathbb R.$ Consider a basis $\{x, y\}$ of $\mathbb R^2$ and a real matrix $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ such that $\{ax + by, cx + dy \}$ is a basis of $\mathbb R^2.$ Observe that it suffices to prove that $\det A$ is nonzero. On the contrary, if it were the case that $\det A = ad - bc = 0,$ then we would have that $$d(ax + by) - b(cx + dy) = adx + bdy - bcx - bdy = (ad - bc)x + (bd - bd)y = 0.$$ By hypothesis that $\{ax + by, cx + dy\}$ is a basis (and hence, the vectors $ax + by$ and $cx + dy$ are linearly independent), we must have that $d = -b = 0.$ Consequently, we have that $\{ax, cx\}$ is a basis of $\mathbb R^2.$ But this is a contradiction. Explicitly, if $c = 0,$ then $cx = 0,$ and the zero vector cannot belong to a basis. On the other hand, if $c$ is nonzero, then $0 = ax - ax = ax - \frac a c (cx)$ shows that $ax$ and $cx$ are linearly dependent. We conclude therefore that $\det A$ is nonzero.

ContentLicense: CC BY-SA 4.0


问:
Simplifaction of Negative Fractional Exponent for Derivatives
I'm self learning calculus and I ran across this question during my practice.

The question asks to differentiate the following equation:

$$\sqrt{x^3 + \csc(x)}$$

Now, I was able to do the following...

$$(1/2(x^3 + \csc(x)))^{-1/2}\cdot\cot(x)\cdot-\csc(x) + 3x^2$$

Basically, I applied the chain rule

However, when I looked at the solution, I was only partially correct. My numerator was right, but the denominator was wrong. The solution provided was

$$\frac{3x^2-\cot(x)\csc(x)}  {2 (\csc(x) + x^3)^{1/2}}$$

I am confused as to how the "two" comes into the denominator. Doesn't $1/2$ as an exponent imply a regular square root? The negative in the exponent makes it a fraction and the $1/2$ just makes it a regular square root . Thus, I'm thinking there is no reason for the "2" to be there at all, becuase it's a regular square root.

Hoping someone can clear this up.

Thanks

答:
Let $u=x^3+\csc(x)$.

By the chain rule, you need
$$
\frac{d}{du}\sqrt{u}=\frac{1}{\color{red}{2}\sqrt{u}}
$$
and
$$
\frac{du}{dx}(x^3+\csc(x))=3x^2+\cot(x)\csc(x)
$$

Notes.

$$
\frac{d}{du}u^{1/2}=\frac12 u^{-1/2}=\frac{1}{\color{red}{2}\sqrt{u}}
$$


问:
If ABC is a right triangle with right angle at C, and if BD is any line cutting AC at point D, prove that $BD^2+AC^2=AB^2+DC^2$
I was considering using the Pythagorean theorem to start the proof but I have no idea where to take it or if that would even work. If anyone has any helpful advice that would be greatly appreciated.

Also does MathJax Latex work in the title? I thought all I needed for superscript was the corresponding symbol but that doesn't appear to be working.

答:
DBC is also a right triangle with a right angle at C so from the Pythagorean Theorem:

$$DB^{2} = CD^{2} + CB^{2}$$

$$AB^{2} = CA^{2} + CB^{2}$$

So,

$$ DB^{2} + CA^{2} = (CD^{2} + CB^{2}) + (AB^{2} - CB^{2}) = CD^{2} + AB^{2} $$


答:
Using Pythagoras in triangle  $ABC$,

$AC^2 + BC^2 = AB^2$

Now using Pythagoras in triangle $DBC$

$DC^2 + BC^2 = BD^2$

From the above equations, $AB^2 - AC^2 = BD^2 - DC^2$

or $\boxed{BD^2 + AC^2 = AB^2+DC^2}$

Title: If ABC is a right triangle with right angle at C, and if BD is any line cutting AC at point D, prove that $BD^2+AC^2=AB^2+DC^2$

Body:
I was considering using the Pythagorean theorem to start the proof but I have no idea where to take it or if that would even work. If anyone has any helpful advice that would be greatly appreciated.

Also does MathJax Latex work in the title? I thought all I needed for superscript was the corresponding symbol but that doesn't appear to be working.


问:
If $f\in C([a,b])$, $M=\max_{[a,b]}f$ and $ f\geq0$ then ${(\int_{a}^{b}f^n)}^{1/n} \rightarrow M$

This is exercise 18 section 7.2 of Bartle and Sherbert's real analysis.

It suffices to show that
$(\int_{a}^{b}(\frac{f(x)}{M})^ndx)^{\frac{1}{n}}\rightarrow 1$.

We know that
$0\leq \frac{f(x)}{M}< 1\Longrightarrow (\frac{f(x)}{M})^n\rightarrow 0 $ and we do have $0\leq \frac{f(x)}{M}< 1$ for every $x\in [a,b]\setminus \{c\}$ where $M:=f(c)$ ($c$ exists by the continuity of $f$ and the compactness of $[a,b]$). Therefore $\int_{a}^{b}(\frac{f(x)}{M})^ndx\rightarrow 0$....

答:
As you say, taking $\widetilde{f} := f/M$ if necessary - and noting that the case $M = 0$ holds beforehand - we may assume $M =1$. Note that since $(-)^n$ is monotone, the function $f^n$ attains a maximum precisely where $f$ does and hence it must be $M^n = 1$. Fix $c \in [a,b]$ such that $f(c) = 1$.

We know that

$$
M_n^n = \int_a^b f^n(t)dt \leq (b-a),
$$
so
$$
M_n = \left(\int_a^b f^n(t)dt\right)^{1/n} \leq \sqrt[n]{b-a}
$$
and taking limits we obtain $\lim_{n\to\infty} M_n \leq 1$.

Now fix $0 < \delta < 1$ and, by continuity, consider an interval $(c-\varepsilon/2,c+\varepsilon/2)$ where $f(z) \geq 1-\delta$, so that $f^n(z) \geq (1-\delta)^n$. Hence

$$
M_n^n = \int_a^b f^n(t)dt \geq \int_{c-\varepsilon}^{c+\varepsilon} f^n(t)dt \geq (1-\delta)^n\varepsilon
$$
and thus
$$
M_n \geq (1-\delta)\varepsilon^{1/n}.
$$

Taking limits, we see that $\lim_{n \to \infty}M_n \geq 1-\delta$ for all $\delta \in (0,1)$. This completes the proof.


Title: Simplifaction of Negative Fractional Exponent for Derivatives

Body:
I'm self learning calculus and I ran across this question during my practice.

The question asks to differentiate the following equation:

$$\sqrt{x^3 + \csc(x)}$$

Now, I was able to do the following...

$$(1/2(x^3 + \csc(x)))^{-1/2}\cdot\cot(x)\cdot-\csc(x) + 3x^2$$

Basically, I applied the chain rule

However, when I looked at the solution, I was only partially correct. My numerator was right, but the denominator was wrong. The solution provided was

$$\frac{3x^2-\cot(x)\csc(x)}  {2 (\csc(x) + x^3)^{1/2}}$$

I am confused as to how the "two" comes into the denominator. Doesn't $1/2$ as an exponent imply a regular square root? The negative in the exponent makes it a fraction and the $1/2$ just makes it a regular square root . Thus, I'm thinking there is no reason for the "2" to be there at all, becuase it's a regular square root.

Hoping someone can clear this up.

Thanks

Body:
Power rule (with a power of $\frac 12$): $(x^k)' = kx^{k-1}$.  So if $k =\frac 12$ then $(\sqrt x)' = (x^{\frac 12})' = \frac 12 x^{\frac 12 -1} =\frac 12 x^{-\frac 12} =\frac 1{2\sqrt x}$

So by chain rule:

$(\sqrt{x^3 + \csc(x)})'=$

$\frac 1{2\sqrt{x^3 + \csc(x)}}\cdot( x^3 + \csc(x))'=$

$\frac 1{2\sqrt{x^3 + \csc(x)}}\cdot(3x^2  - \csc(x)\cot (x))=$

$\frac {3x^2  - \csc(x)\cot (x)}{2\sqrt{x^3 + \csc(x)}}$


问:
An exercise about zerodivisors
If $A$ is a commutative ring with unity, $f\in A$ and $x\in SpecA$,  with the notation $f(x)$ I mean the coset $x+f\in A/x$. Now look at this exercise:

Prove that a nonzero element $f\in A$ is a zerodivisor if and only if there exists a decomposition $SpecA=X\cup X'$ where $X\subseteq$ $SpecA$ and $X'\subsetneq Spec A$ such that $f(x)=0$ for all $x\in X$.

One direction is easy: ($\Rightarrow$) If $f$ is a zerodivisor, it's not invertible, so $f$ is contained in a maximal ideal $\mathfrak m$. We have that $SpecA= V(f)\cup D(f)$ where $D(f)\neq Spec A$ because $\mathfrak m\notin D(f)$.

I have problems to show that the other direction. Any hint?

答:
Part of the standard definition of irreducibility is that the components $X,X'$ in the decomposition are closed - this, I think, is what Georges  was pointing out.

The direction you were looking for can be solved using the following idea from Hartshorne:

Given two closed sets $X \cup X' = \operatorname{Spec} A$, taking complements gives us two open sets $U,U'$ such that $U \cap U' = \varnothing$. Now we can glue together a zero function on one with a nonzero function on the other, which gives us zerodivisors in $A$.

ContentLicense: CC BY-SA 3.0

Title: An exercise about zerodivisors


问:
Integration with complex constant
This may be an overly simple question, but:

If I am integrating something like

$$\int -i\ dx\,,$$

is this the same as

$$-i \int\ dx = -i x + c, c \in \mathbb{R}\,$$

that is, the complex constant $i$ can just be treated as how one typically treats real constants in integration?

Thank you.

答:
Yes, this is true; but why this is true is bit more delicate question.

To keep it simple: one can define integrals of complex-valued functions along curves in the complex plane. The resulting integral-notion will be $\Bbb C$-linear (as one might expect) instead of only $\Bbb R$-linear (as the usual, say, Riemann-integral is). However, this more general integral notion still includes the usual one as special case and hence the more general linearity follows at once.

(Note: Usually, complex integrals are defined as line integrals and not as general expressions involving an integral sign. But, intuitevely, I do not think this should matter here.)

Title: Integration with complex constant


问:
$A \oplus B \cong A \oplus C$ implies $B \cong C$ where $A,B,C$ are finitely generated $R$ modules and $R$ is a PID
There are number of similar questions to this, but I have read through them all and the answers either rely on results that I don't have access to or I'm unsure how to translate them to this situation.

The exact problem is as follows. Suppose $R$ is a PID, $A,B$ and $C$ are finitely generated $R$ modules and that $A \oplus B \cong A \oplus C$, then $B \cong C$.

It seems like we should be able to apply the structure theorem for finitely generated modules over a PID. In the version I'm most familiar with, this result says that given a PID $R$ and an $R$ module $M$, we have that
$$
M \cong R^n \oplus R/(a_1) \oplus ...\oplus R/(a_m)
$$
where $n \in \mathbb{N}$, $a_i \in R$, and $a_1|a_2|...|a_n$. Further, if
$$
M \cong R^s \oplus R/(b_1) \oplus ...\oplus R/(b_t)
$$
and $b_1|...b|t$, then we have that $n=s$, $m=t$ and $a_i$ is associate to $b_i$ for all $a_i$.

Applying that to this problem at hand, we can write down decompositions for $A,B$, and $C$:
$$
A \cong R^a\oplus R/(a_1)\oplus ... \oplus R/(a_l)
$$
$$
B \cong R^b\oplus R/(b_1)\oplus ... \oplus R/(b_m)
$$
$$
C \cong R^c\oplus R/(c_1)\oplus ... \oplus R/(c_n)
$$
So we have that
$$
R^a\oplus R/(a_1)\oplus ... \oplus R/(a_l) \oplus R^b\oplus R/(b_1)\oplus ... \oplus R/(b_m) \cong R^a\oplus R/(a_1)\oplus ... \oplus R/(a_l) \oplus R^c\oplus R/(c_1)\oplus ... \oplus R/(c_n)
$$
$$
R^{a+b}\oplus R/(a_1)\oplus ... \oplus R/(a_l) \oplus R/(b_1)\oplus ... \oplus R/(b_m) \cong R^{a+c} \oplus R/(a_1)\oplus ... \oplus R/(a_l) \oplus R/(c_1)\oplus ... \oplus R/(c_n)
$$

What I would like to do here is to apply the uniqueness part of the structure theorem I referenced above, but it is not at all clear to me why the divisibility condition on the invariant factors would still hold, since now we have factors from from $A$ and $B$ on the left hand side and the same problem on the right. What am I missing here? Is this easier to understand with the elementary divisor form?

Also as I mentioned at the beginning, I know there are some similar questions already posed, but I have read them and I am still unsure, so if you give a link to another question please try to explain more precisely how that question applies.

答:
By uniqueness of the structure theorem, we have $b = c$ so we may well assume that all modules are of torsion. As you say, it is not clear from the way we've written things how the invariant factors relate to each other.

One may use a different formulation of the structure theorem, splitting each $a_i$ into irreducible factors, using the Chinese remainder theorem, and regrouping quotients based on the irreducible appearing. We end up with a decomposition

$$
M = M[p_1] \oplus \ldots \oplus  M[p_r]
$$

where $M[p_i]$ is a sum of quotients of the form $R/(p_i^{\alpha_i})$. Going through the proof of the structure theorem one sees that the $p_i$'s appearing in this decomposition are unique up to associates and so are the powers $\alpha_i$ appearing in each so called $p_i$-torsion $M[p_i]$. One can show that $(A\oplus B)[p_i] = A[p_i] \oplus B[p_i]$ (think of $M[p_i]$ as the elements of $M$ such that $p_i^k m = 0$ for some $k$).

Hence, we many assume that both $A, B$ and $C$ are of $p$-torsion,

$$
A = \bigoplus_{i=1}^n R/(p^{a_i}), \quad B = \bigoplus_{i=1}^k R/(p^{b_i}), \quad C = \bigoplus_{i=1}^l R/(p^{c_i}).
$$

If $A \oplus B \simeq A \oplus C$, by uniqueness of the structure theorem (in this version) we have that $(a_1,\ldots, a_n,b_1,\ldots,b_k)$ up to reordering coincides with $(a_1,\ldots,a_n,c_1,\ldots,c_l)$. Hence $(b_j)$ and $(c_j)$ are - up to reordering - the same. This shows that $B \simeq C$.


问:
Which knots can be embedded in $\mathbb{R}^3$ in a centrally symmetric manner?
For which knots $K$ is there an embedding of $K$ in $\mathbb{R}^3$ for which $x\in K\iff (-x)\in K$?

Obviously, the unknot can be: take the circle given by $(\sin(t),\cos(t),0)$ for $t\in[0,2\pi)$.

More generally, given a knot $K$ and its mirror image $K'$, the knot sum $K\# K'$ can be embedded in a centrally symmetric manner by placing $K$ far from the origin, placing its negated copy, and bridging the gap between them by two straight lines.

In the negative direction, any chiral knot can't be embedded symmetrically, because the inversion of $\mathbb{R}^3$ reverses chirality but fixes such an embedding. (I think this can be made more precise with notions of amphichirality, but I'm not sure I understand the definitions and consequences well enough to make confident statements here.)

What other things can we prove about such embeddings? At a minimum, does anyone have an example of a centrally symmetric drawing of a nontrivial prime knot?

答:
Any knot with an involution which reverses the orientation on $\mathbb{R}^3$, but preserves the orientation on the knot has an embedding of the type you describe. (Equivalently an involution which reverses the orientation on the meridian but preserves the orientation on the longitude.)

I believe such knots are usually called "strongly positive amphicheiral" in the literature. See for example Figure 1(c) in this paper: https://projecteuclid.org/download/pdf_1/euclid.rmjm/1250127863

For a specific hyperbolic knot I would use Snappy (https://snappy.math.uic.edu）to check if the symmetry group has such an element. For example, $4_1$ has exactly two symmetries which reverse the orientation on $S^3$ but preserve the orientation on $4_1$; however, both of these symmetries have order 4 (not 2) so that $4_1$ is not strongly positive amphicheiral.

You can see these order 4 symmetries in the following diagram as a $2\pi/4$ rotation (in either direction) followed by reflection across an $S^2$ intersecting the diagram in the dotted green circle.

https://i.stack.imgur.com/AJeSJ.png


问:
Follow-up Question about Cesaro mean proof
I am trying to understand the proof behind the Cesaro mean converging. I am using https://math.stackexchange.com/a/2342856/633922 (hopefully it is also correct) as a guide because it seems very direct. I will comment on the steps I understand and where I need help.

The statement: If $(x_n)$ converges to $x$, the sum of averages $y_n=\dfrac{x_1+x_2+\cdots+x_n}{n}$ also converges to the same limit.

Proof:

Since $(x_n)$ converges, given an arbitrary $\epsilon >0$, there exists an $N_1\in\mathbb{N}$ such that whenever $n\geq N_1$ we have $|x_n-x|<\epsilon$.
(Definition of convergent sequence)

Now,
$$\begin{align*}
\left|\frac{x_1+x_2+\cdots+x_n}{n}-x\right|=&\left|\frac{(x_1-x)+\cdots+(x_{N_1-1}-x)}{n}+\frac{(x_{N_1}-x)+\cdots+(x_{n}-x)}{n}\right|\\
\leq& \left|\frac{(x_1-x)+\cdots+(x_{N_1-1}-x)}{n}\right|+\left|\frac{(x_{N_1}-x)+\cdots+(x_{n}-x)}{n}\right|\text{ (Triangle inequality)}
\end{align*}
$$

Now we want to make a statement about the first $N_1-1$ terms, $\color{red}{why?}$ That is:

By the Archimedean principle we can find an $N_2$ such that whenever $n\geq N_2$ we have that

$$\left|\frac{x_1+x_2+\cdots+x_{N-1}}{n}\right|<\epsilon
$$
(Thought: is it because $x_1,\dots,x_{N_1-1}$ is finite?)

Now we can choose an $N_3=\max\{N_1,N2\}$ such that for all $n\geq N_3$ we have (My thought: Is this because choosing the max of both will always guarantee the final inequality to always work?)
$$
\left|\frac{x_1+x_2+\cdots+x_n}{n}-x\right|\leq \underbrace{\epsilon}_{N_1-1}+\underbrace{\color{red}{\frac{n-N_1}{n}}}_{\text{why and how?}}\epsilon< 2\epsilon
$$

And this finishes the proof. I always assumed the ending statement has to be (something)$<\epsilon$ or is this saying that each sum of the right side of the triangle inequality is less than $\epsilon/2$. I would really appreciate the help on the areas I am thoroughly confused about.

答:
The first part is controlled because $N_1$ is fixed and the denominator $n$ can be arbitrarily large.

For the second part you are using the estimate $|x_n-x|<\epsilon$ for large $n$. Note that there are totally $n-N_1$ terms in the numerator, the absolute value of each is less than $\epsilon$.

To summarize, you are first given $\epsilon>0$. Then you have $N_1$ such that $n\ge N_1$ implies $|x_n-x|<\epsilon$, in particular
$$
|x_{N_1+1}-x|<\epsilon,\quad |x_{N_1+2}-x|<\epsilon,\cdots,|x_{n}-x|<\epsilon\tag{1}
$$
which implies by the triangle inequality that
$$
\left|\frac{(x_{N_1+1}-x)+\cdots+(x_{n}-x)}{n}\right|<\frac{(n-N_1)\epsilon}{n}\tag{1'}
$$

Then, for this (fixed) $N_1$, since
$$
\lim_{n\to\infty}\frac{|(x_1-x)+\cdots+(x_{N_1}-x)|}{n}=0
$$
there exists $N_2$ such that $n\ge N_2$ implies
$$
\frac{|(x_1-x)+\cdots+(x_{N_1}-x)|}{n}<\epsilon\tag{2}
$$

So if you pick $N_3=\max(N_1,N_2)$, then $n\ge N_3$ implies both (1') and (2).

If you can show that for every $\epsilon>0$, there exists $N>0$ such that $n\ge N$ implies $|a_n|<2\epsilon$, it follows that
$$
\lim_{n\to\infty}a_n=0
$$
You don't have to have exactly $\epsilon$ in the estimate.


答:
first red text: You want to make a seperate statement about the first $N_1-1$ terms since you don‘t know how the sequence behaves in those terms. You know that $|x_n-x|<\varepsilon$ for all $n\geq N_1$ but you don‘t know what‘s going on earlier in the sequence.

second red text: this follows again from the triangle inequality:
$$
\left| \frac{x_{N_1}+x_{N_1+1}+\ldots+x_n}{n} - x \right| = \left| \frac{x_{N_1}-x+x_{N_1+1}-x+\ldots+x_n-x}{n}  \right|
\leq \frac{|x_{N_1}-x|+|x_{N_1+1}-x|+\ldots+|x_n-x|}{n}
$$

that are $N_1-n$ terms of value $<\varepsilon$ (that’s because of how we chose $N_1$) in the numerator giving you the term
$$
\frac{n-N_1}{n}\varepsilon
$$

edit: regarding the term $2\varepsilon$: it is only important that we can make this expression arbitrarily small. Although it is „convention“ to use $\varepsilon$ it is also fine to use $2\varepsilon$ since we can make this expression as small as we want if we just make $\varepsilon$ small enough.

Title: Follow-up Question about Cesaro mean proof

Body:
I am trying to understand the proof behind the Cesaro mean converging. I am using https://math.stackexchange.com/a/2342856/633922 (hopefully it is also correct) as a guide because it seems very direct. I will comment on the steps I understand and where I need help.

The statement: If $(x_n)$ converges to $x$, the sum of averages $y_n=\dfrac{x_1+x_2+\cdots+x_n}{n}$ also converges to the same limit.

Proof:

Since $(x_n)$ converges, given an arbitrary $\epsilon >0$, there exists an $N_1\in\mathbb{N}$ such that whenever $n\geq N_1$ we have $|x_n-x|<\epsilon$.
(Definition of convergent sequence)

Now,
$$\begin{align*}
\left|\frac{x_1+x_2+\cdots+x_n}{n}-x\right|=&\left|\frac{(x_1-x)+\cdots+(x_{N_1-1}-x)}{n}+\frac{(x_{N_1}-x)+\cdots+(x_{n}-x)}{n}\right|\\
\leq& \left|\frac{(x_1-x)+\cdots+(x_{N_1-1}-x)}{n}\right|+\left|\frac{(x_{N_1}-x)+\cdots+(x_{n}-x)}{n}\right|\text{ (Triangle inequality)}
\end{align*}
$$

Now we want to make a statement about the first $N_1-1$ terms, $\color{red}{why?}$ That is:

By the Archimedean principle we can find an $N_2$ such that whenever $n\geq N_2$ we have that

$$\left|\frac{x_1+x_2+\cdots+x_{N-1}}{n}\right|<\epsilon
$$
(Thought: is it because $x_1,\dots,x_{N_1-1}$ is finite?)

Now we can choose an $N_3=\max\{N_1,N2\}$ such that for all $n\geq N_3$ we have (My thought: Is this because choosing the max of both will always guarantee the final inequality to always work?)
$$
\left|\frac{x_1+x_2+\cdots+x_n}{n}-x\right|\leq \underbrace{\epsilon}_{N_1-1}+\underbrace{\color{red}{\frac{n-N_1}{n}}}_{\text{why and how?}}\epsilon< 2\epsilon
$$

And this finishes the proof. I always assumed the ending statement has to be (something)$<\epsilon$ or is this saying that each sum of the right side of the triangle inequality is less than $\epsilon/2$. I would really appreciate the help on the areas I am thoroughly confused about.


问:
Need help to express the Taylor series for $\coth(x)$
I am finding a way to find the general formula for the series expansion of $\dfrac{x}{2}\coth\left(\dfrac{x}{2}\right)$. I know that this is the generating function of even Bernoulli numbers. Its general formula is

$$\dfrac{x}{2}\coth\left(\dfrac{x}{2}\right)=\sum_{n=0}^{+\infty} \dfrac{B_{2n}x^{2n}}{(2n)!}$$

In order to prove this formula, I wish to first find the formula for $\coth(x)$. Looking on wikiproof, I find this site:

I am going to copy the proof here, and wish to ask the questions based on the numbered line of the proof

$$\coth(x)=\dfrac{e^x+e^{-x}}{e^{x}-e^{-x}}\tag{1}$$

Multiply denominator and numerator for $e^{x}$, we have

$$\frac{e^{2x}+1}{e^{2x}-1}=\frac{e^{2x}-1+2}{e^{2x}-1}=1+\frac{2}{e^{2x}-1}\tag{2}$$

Factor $\dfrac{1}{x}$ out, we have

$$1+\dfrac{1}{x}\left(\frac{2x}{e^{2x}-1}\right)\tag{3}$$

It is known and without proving here that $\dfrac{x}{e^{x}-1}=\sum_{n=0}^{+\infty}\dfrac{B_nx^{n}}{n!}$, substituting $2x$ and draw out the constant $2$, we have:

$$1+\dfrac{1}{x}\left(\sum_{n=0}^{+\infty} \dfrac{B_{n}(2x)^{n}}{(n!)}\right)\tag{4}$$

The next step is to "release" the first two terms of the infinite series from the sigma notation chunk, with $B_{0}=1$ and $B_{1}=-\dfrac{1}{2}$, the first two terms are $1$ and $-x$

$$1+\dfrac{1}{x}\left(1-x+\sum_{n=2}^{+\infty}\dfrac{B_{n}(2x)^{n}}{n!}\right)=1+\dfrac{1}{x}-1+\dfrac{1}{x}\left(1-x+\sum_{n=2}^{+\infty}\dfrac{B_{n}(2x)^{n}}{n!}\right)=\dfrac{1}{x}+\dfrac{1}{x}\left(\sum_{n=2}^{+\infty}\dfrac{B_{n}(2x)^{n}}{n!}\right)\tag{5}$$

Convert the index from $n=2$ to $n=1$ in the sigma notation and change the index $n$ to $2n$ in the $B_n$ of the above equation to include all even Bernoulli numbers, we have:

$$\dfrac{1}{x}+\dfrac{1}{x}\left(\sum_{n=1}^{+\infty}\dfrac{B_{2n}(2x)^{2n}}{2n!}\right)\tag{6}$$

Now, here is my question, the next step is to change the of the sigma notation from $n=1$ to $n=0$. So why wikiproof writes as followed:

$$\dfrac{1}{x}\sum_{n=0}^{+\infty} \dfrac{B_{2n}(2x)^{2n}}{(2n)!}\tag{7}$$

Is this the intermediate step that is left out by wikiproof?, we need to minus the $B_0x^{2\cdot0}=1$:

$$\dfrac{1}{x}+\dfrac{1}{x}\left(-1+\sum_{n=0}^{+\infty}\dfrac{B_{2n}(2x)^{2n}}{2n!}\right)\tag{8}$$

As with $(5)$, if you increase the summation index, you release more terms. So if you decrease the index, you should, and this is my guess, eliminate terms to avoid adding duplicated terms. Is this interpretation correct?

答:
You are completely correct. The general principle is the following (as you have noted):
$$\sum_{n=1}^\infty a_n=(-a_0+a_0)+\sum_{n=1}^\infty a_n=-a_0+\sum_{n=0}^\infty a_n$$

You confused yourself. Line $(5)$ should look like this:

\begin{align*}
1+\frac1x\left(\sum_{n=0}^\infty\frac{B_n(2x)^n}{n!}\right)
&=1+\frac1x\left(\color{blue}{1-x}+\sum_{\color{red}{n=2}}^\infty\frac{B_n(2x)^n}{n!}\right)
\\
&=1+\color{blue}{\frac1x-1}+\frac1x\left(\sum_{\color{red}{n=2}}^\infty\frac{B_n(2x)^n}{n!}\right)
\end{align*}

This is perfectly in sync with what the ProofWiki writes as:

$$1+\frac1x\left(\sum_{n=0}^\infty\frac{B_n(2x)^n}{n!}\right)=\frac1x+\frac1x\left(\sum_{n=2}^\infty\frac{B_n(2x)^n}{n!}\right)$$

The later index shift relies on the same principle as you have noted correctly (and as I have mentioned above).


问:
Collecting coupons which arrive as a Poisson process
I'm trying to answer the following question:

"A person collects coupons one at a time, at jump times of a Poisson process $(N_t)_{t\geq 0}$ of rate $\lambda$. There are m types of coupons, and each time a coupon of type j is obtained with probability $p_j$, independently of the previously collected coupons and independently of the Poisson process. Let T be the time the complete set is obtained.

Show that $\mathbb{P}(T<t)=\prod_{j=1}^m(1-e^{-p_j\lambda t})$. let L be the number of coupons collected by time T. Show that $\mathbb{E}L=\lambda \mathbb{E}T$."

I'm able to find the distribution of T, by splitting the process into independent subprocesses of rate $\lambda p_i$, but am not sure where to go for the second part. It seems that explicit computation $\mathbb{E}L$ or $\mathbb{E}[L|T]$ is infeasible. Is it possible to use the fact that $T$ is a stopping time here and perhaps the strong Markov property?

Thank you.

答:
Dear Ben, the coupon collector is just running $m$ Poisson processes at the same moment. The coupons of the $j$th type are being collected at rate $p_j\lambda$ and they're totally independent of the other types of the coupons. It's not hard to see why. Imagine you have an infinitesimal period of time ${\rm d}t$. What's the probability that you collect $j$th coupon in it? Well, the probability is $\lambda\,{\rm d} t$ that you collect any coupon, and you still have to multiply it by $p_j$ to be sure that it's the right coupon.

It follows that the probability that he hasn't collected any coupon of the $j$th type by time $t$ is equal to $\exp(-p_j \lambda t)$ - it's exponentially dropping like it always does in the Poisson distribution.

The probability that the guy has collected the coupon of $j$th time by time $t$ is the complement, $1-\exp(-p_j \lambda t)$. Because the coupons of different types are independent throughout the calculation, the probability that he has collected the coupons of all types by time $t$ (or even earlier, at time $T<t$, of course) is simply the product of the last expression over $j$,
$${\rm Prob(All\,\,by\,\,}t{\rm)}=\prod_{j=1}^m [1-\exp(-p_j \lambda t)]$$
The expectation value of the total number of collected coupons is just $\lambda T$. One doesn't even have to divide the coupons into types here. One returns to the overall Poisson distribution and this is just what the rate means. In a period of time whose length is $T$, the number of events is $\lambda T$ where $\lambda$ is the rate - that's why it's called the rate of the Poisson distribution. Of course, if $T$ is given by a distribution itself, the expectation value of $L$ will be the expectation value of $\lambda T$ because the relationship is linear.


问:
Smith Normal Form for PIDs

Could someone provide a good reference to look up the existence and uniqueness of Smith Normal Form (SNF) for a PID?

I have seen it done for Euclidean domains but not for a PIDs. I know the difference is not much, but I would like a good reference. Thanks.

答: See Hungerford, Algebra, 1974, Proposition 2.11, page 339.

Title: Smith Normal Form for PIDs


问:
Probability of throwing darts when likelyhood of hitting one side is more than the other
Currently learning probability from MIT OCW 6.041. I want to understand the following regarding Problem #5 from the PSET1(read the question here at MIT OCW). I am unable to understand part c of the given problem.

We need to calculate the probability of John hitting 50 point mark on the dartboard. I understand that we are looking for values of $\alpha$ and $\beta$ which are the proportionality constants, I want to understand why the probability of throwing the dart in the right half of the board is $\frac{2}{3}$

The question being asked above(in bold) is being referenced from the answer given for the Problem #5 referenced from the solution of the PSET1-Solution from the MIT OCW site.

The  first  term  in  the  sum  is  proportional  to  the  area  of  the  right  half  of  the  inner  disk and  is  equal  to απR2/2 = απ/2, where α is  a  constant  to  be  determined. Similarly,  the  probability  of  him  throwing  in  the  left  half  of  the  board  and  scoring  50 points  is  equal  to βπ/2, where β is  a  constant (not  necessarily  equal  to α). In  order  to  determine α and β,  let  us  compute  the  probability  of  throwing  the  dart  in the  right  half  of  the  board.  This  probability  is  equal  to απR2/2 = απ102/2 = α50π.
Since  that  probability  is  equal  to  2/3, α = 1/(75π). In  a  similar  fashion, β can  be determined  to  be  1/(150π).   Consequently,  the  total  probability  is  equal  to  1/150  + 1/300  = 0.01

答:
I want to understand why the probability of throwing the dart in the
right half of the board is $\frac{2}{3}$

We are told that "John is right handed and is twice more likely to throw in the right half of the board than in the left half."

So, $P(\text{Right half of the board}) = 2P(\text{Left half of the board})$.

Since there are only two sides of the board, $$P(\text{Right half of the board}) + P(\text{Left half of the board}) = 1 = P(\text{hit on the board})$$ $$\text{(we're gonna assume missing the board is impossible)}$$

Substituting in our first equation,

$$2P(\text{Left half of the board}) + P(\text{Left half of the board}) = 1$$
$$3P(\text{Left half of the board}) = 1$$
$$P(\text{Left half of the board}) = \frac{1}{3}$$
$$P(\text{Right half of the board}) = 2P(\text{Left half of the board}) = \frac{2}{3}$$


问:
Proof for the fact that $(a,b)$ is not compact.
Here is the proof for the fact that $(a,b)$ is not compact.

Proof

$\bigg\{ \left(a+\dfrac{b-a}{2^n}, b \right) \Bigg| n \in \mathbb{N} \bigg\}$ is an open covering of $(a,b)$. Let $U_i:=\left(a+\dfrac{b-a}{2^i}, b\right)$.
Suppose we can cover $(a,b)$ with $U_1 \cup U_2 \cup \cdots \cup U_k $. Then, $a+\dfrac{b-a}{2^{k+1}} \in (a,b)$ and $a+\dfrac{b-a}{2^{k+1}} \notin U_1 \cup \cdots \cup U_k.$ This is contradictory.
Thus, $(a,b)$ is not compact.

I understood this proof, but why does this proof use
$\left( a+\dfrac{b-a}{2^n}, b\right)$ ?

I think $\left(a+\dfrac{1}{2^n},b \right)$ also works because $\lim_{n\to \infty} \left(a+\dfrac{1}{2^n} \right)=a.$

Body: What if $b = a + \frac{1}{2^{10000000000}}$?


答: It's very much a detail, but the interval $(a+\frac{1}{2^n}, b)$ is not always a subset of $(a, b)$. Take $a=0, b=0.1$ then $(a+0.5, b)$ is not a valid subset of $(0, 0.1)$.

Title: Proof for the fact that $(a,b)$ is not compact.

Body:
Here is the proof for the fact that $(a,b)$ is not compact.

Proof

$\bigg\{ \left(a+\dfrac{b-a}{2^n}, b \right) \Bigg| n \in \mathbb{N} \bigg\}$ is an open covering of $(a,b)$. Let $U_i:=\left(a+\dfrac{b-a}{2^i}, b\right)$.
Suppose we can cover $(a,b)$ with $U_1 \cup U_2 \cup \cdots \cup U_k $. Then, $a+\dfrac{b-a}{2^{k+1}} \in (a,b)$ and $a+\dfrac{b-a}{2^{k+1}} \notin U_1 \cup \cdots \cup U_k.$ This is contradictory.
Thus, $(a,b)$ is not compact.

I understood this proof, but why does this proof use
$\left( a+\dfrac{b-a}{2^n}, b\right)$ ?

I think $\left(a+\dfrac{1}{2^n},b \right)$ also works because $\lim_{n\to \infty} \left(a+\dfrac{1}{2^n} \right)=a.$


问:
Is there a way to show $\frac{1646-736\sqrt{5}}{2641-1181\sqrt{5}}=\frac{17+15\sqrt{5}}{7+15\sqrt{5}}$ without multiplying large numbers?
I need to show that the following equality holds $$\frac{1646-736\sqrt{5}} {2641-1181\sqrt{5}} =\frac{17+15\sqrt{5}} {7+15\sqrt{5}} $$ The only way I could prove it was via cross multiplication using a calculator.

Is there any simpler way to do this? Following prime factorization is given $$1646=2\times 823,736=2^5\times 23,2641=19\times 139$$ and $1181$ is prime. I strictly want to avoid multiplying large numbers. I don't know if that's possible here.

The above simplification is used to obtain a famous approximation to $\pi$ given by Ramanujan.

Body:
Note that
\begin{eqnarray*}
\frac{1646-736\sqrt{5}} {2641-1181\sqrt{5}} =1 -5\frac{199-89\sqrt{5}}{2641-1181\sqrt{5}}
\end{eqnarray*}
and
\begin{eqnarray*}
\frac{89\sqrt{5}-199}{2641-1181\sqrt{5}}=\frac{2}{7+15 \sqrt{5}}.
\end{eqnarray*}


问:
For real $2\times 2$ matrices $P$, $Q$, Is $P>Q$ given their eigenvalues?

Suppose $P,Q\in \mathbb R ^{2\times 2}$, where $P=P^*$ and $Q=Q^*$.

If $P$ has eigenvalues $2$ and $4$, and $Q$ has eigenvalues $1$ and $3$, is $P>Q$? (1)

If $P$ have eigenvalues $2$ and $3$, and $Q$ has eigenvalues $.5$ and $1$, is
$P>Q$? (2)

Attempts:

Recall that $P>Q$ if $P-Q>0$, i.e. is positive definite.

$\displaystyle \left(\begin{matrix}2&0\\0&4\end{matrix}\right)-\left(\begin{matrix}3&0\\0&1\end{matrix}\right)=\left(\begin{matrix}-1&0\\0&3\end{matrix}\right)$ has a negative eigenvalue, the answer to (1) is no.

$\displaystyle \left(\begin{matrix}2&0\\0&4\end{matrix}\right)-\left(\begin{matrix}1 \text{ or }.5&0\\0&\text { .5 or 1 } \end{matrix}\right)=\left(\begin{matrix}\text{ 1 or 1.5 }&0\\0&\text{
3.5 or 3 }\end{matrix}\right)>0$, where I simply took both possibly combination of eigenvalues in $Q$. Is this enough to conclude that $P>Q$ for all $P, Q$ - is the answer to (2) yes? If so, what property of positive definite matrices justifies this?

答: A matrix $A$ is positive definite if and only if $A = A^*$ and all eigenvalues of $A$ are positive.  Thus in case (2) if $1 < \mu < 2$, $P > \mu I > Q$.


问:
What are the binary strings of length 10 where the sum of the digits is $\leq 4$?
I tought that for binary strings we mean a string made just by $0$s and $1$s, and the problem is to find all the possible strings whose sum of digits  is $\leq 4$.
But i don't know if considering the positional notation is required (because the problem is meaning that the sum must be $\leq 4$ in base 10), or it means that the base 2 addition between the digits must be equal to a binary string that in base 10 is equal to 4.

I would really appreciate a hint to have a better understanding of this.

p.s: This is my first post

答: The "sum of digits" in a binary string would just count the number of ones in that string. So for instance $1011110001$ has $6$ ones. You would not be interested in this string since it has more than four ones.  On the other hand you would be interested in the binary string $0110010010$.


Title: What are the binary strings of length 10 where the sum of the digits is $\leq 4$?

Body:
I tought that for binary strings we mean a string made just by $0$s and $1$s, and the problem is to find all the possible strings whose sum of digits  is $\leq 4$.
But i don't know if considering the positional notation is required (because the problem is meaning that the sum must be $\leq 4$ in base 10), or it means that the base 2 addition between the digits must be equal to a binary string that in base 10 is equal to 4.

I would really appreciate a hint to have a better understanding of this.

p.s: This is my first post

Body: The number of such strings is $$\sum_{k=0}^4 \binom{10}{k} = \binom{10}{0} + \binom{10}{1} + \binom{10}{2} + \binom{10}{3} + \binom{10}{4} = 1 + 10 + 45 + 120 + 210 = 386.$$


问:
Fundamental solution to 2D Diffusion Equation
I was reading about the diffusion equation $$-\frac{1}{D}\frac{\partial p(x,y,t)}{\partial t}+\frac{\partial^2}{\partial x^2}p(x,y,t)+\frac{\partial ^2}{\partial y^2}p(x,y,t)=0$$ where D is a constant.
This has an analytical fundamental solution of
$$p(x,y,t)=\frac{1}{4\pi D t}\exp\frac{-(x+y)^2}{(4Dt)}$$
(Source: http://rpdata.caltech.edu/courses/aph162/2006/Protocols/diffusion.pdf)

I was thinking that the solution to the similar equation where we take $D=1$ and introduce a constant $A$,
$$-\frac{\partial p(x,y,t)}{\partial t}+A\frac{\partial^2}{\partial x^2}p(x,y,t)+A\frac{\partial ^2}{\partial y^2}p(x,y,t)=0$$
would be something like $$p(x,y,t)=\frac{1}{4\pi At}\exp\frac{-(x+y)^2}{(4At)}$$

But that doesn't seem to work out. I am hoping someone can help me find the actual solution. Thank you in advance for your help.

Body: It should be $-(x^2 + y^2)$ not $-(x + y)^2$.


答:
I agree with the answer given by mrtaurho.

I just want to add that $\mathbb{C}$-linearity still holds if you are interested in finding an antiderivative of a given complex valued function, which seems to be your case to me.

This is quite obvious if you think about it: you can picture $\mathbb{C}$ simply as $\mathbb{R}^2$ endowed with a product operation that gives it a field structure. Then, functions $\mathbb{C}\rightarrow\mathbb{C}$ are just functions $\mathbb{R}^2\rightarrow\mathbb{R}^2$, and you have just traced back your problem to real numbers.

However, the problem of finding an antiderivative in several variables is not as straightforward as in one variable, and it is better understood in terms of closed and exact differential forms.

But to keep it simple and short: yes, complex constants behave in the exact same way.

Title: Integration with complex constant

Body:
This may be an overly simple question, but:

If I am integrating something like

$$\int -i\ dx\,,$$

is this the same as

$$-i \int\ dx = -i x + c, c \in \mathbb{R}\,$$

that is, the complex constant $i$ can just be treated as how one typically treats real constants in integration?

Thank you.


问:
Conditional probability "brain teaser"
I'm reading A Practical Guide to Quantitative Finance Interviews by Xinfeng Zhou.

The book has the following two part question.

Part A: A company is holding a dinner for working mothers with at
least one son.  Ms. Jackson, a mother with two children, is invited.
What is the probability that both children are boys?

Part B: Your new colleague, Ms. Parker is known to have two children.
If you see her walking with one of her children and that child is a
boy, what is the probability that both children are boys?

According to the book, in Part A, the answer is
\begin{equation*}
1/3 = \frac{1/4}{3/4} = \frac{\mathbb{P}\{\text{two boys}\}}{\mathbb{P}\{\text{at least one boy out of two kids}\}} = \mathbb{P}\{\text{two boys} \, \mid \, \text{at least one out of two is a boy}\},
\end{equation*}
while, in Part B, the answer is
\begin{equation*}
1/2 = \mathbb{P}\{\text{second kid is a boy}\}.
\end{equation*}

I would have thought the "right" answer --- in quotes since the wording is clearly (?) vague --- is $1/3$ in both cases.  Am I missing something?

(Assume a kid is equally likely to be a boy or a girl...)

The best thing I can come up with is if we let $F_{1}$ and $F_{2}$ be coin flips taking values in the symbolic space $\{H,T\}$, then the solution to Part A is finding $\mathbb{P}\{(F_{1},F_{2}) = (H,H) \, \mid \, F_{1} \, \, \text{or} \, \, F_{2} = H\}$, while Part B is finding $\mathbb{P}\{F_{1} = H \, \mid \, F_{2} = H\}$.  On the other hand, just because I "saw" the kid (or the coin flip), doesn't mean I know which kid (or which coin) it was so I don't see why "What is $\mathbb{P}\{F_{1} = H \, \mid \, F_{2} = H\}$?" should be the "right" interpretation of the question.

答:
"Just because I 'saw' the kid (or coin flip) doesn't mean I know which kid (or which coin)..."

But in fact that kid (or coin) does become distinguishable, as it is the one you saw.

Title: Conditional probability "brain teaser"


问:
Given Eigen vector find eigen values.
The vector $
  \begin{bmatrix}
    1\\
    2 \\
    -1 \\
  \end{bmatrix}
$ is an Eigen vector of $A=$$
  \begin{bmatrix}
    -2 & 2 &-3\\
    2&1 & -6\\
    -1&-2 & 0\\
  \end{bmatrix}
$. One of the Eigen value of $A$ is

$(A)\ \ 1$  $\ \ \ (B)\ \ 2$ $\ \ \ (C)\ \ 5$ $\ \ \ (D)\ \ -1$

The way I tried to solve is like this

$AX=\lambda X$

$    \begin{bmatrix}
    -2 & 2 &-3\\
    2&1 & -6\\
    -1&-2 & 0\\
  \end{bmatrix}\begin{bmatrix}
    1\\
    2 \\
    -1 \\
  \end{bmatrix}=\begin{bmatrix}
    5\\
    10 \\
    -5 \\
  \end{bmatrix}=5\begin{bmatrix}
    1\\
    2 \\
    -1 \\
  \end{bmatrix}$

So, $5$ is the correct option. I am looking for quick alternate ways to solve this problem.Thanks

答:
The eigenvector assumption is redundant for this problem. But using this extra information actually gives you a quick way to solve it, as you did.

I don't think there is any "quicker" way to do this problem (except for looking directly at the solution manual...).

Title: Given Eigen vector find eigen values.


问:
Spectral mapping theorem in a subset
Let $A$ be a self-adjoint operator (specifically it is allowed to be unbounded) and let $f$ be a bounded Borel-measurable function. Through the functional calculus we may make sense of $f(A)$.

Let $P$ be an orthogonal projection, which does not commute with $A$. But $PAP$ is of course also self-adjoint.

Is there any relationship between $\sigma(P f(A)P)$ and $\sigma(P A P)$? In particular any relation of inclusion? We may also consider the problem as the relation between $\sigma(\left.A \right|_{W})$ and $\sigma(\left.f(A) \right|_{W})$ where $W$ is the closed subspace which is the range of $P$.

答:
Do you mean to ask about a relationship between $\sigma\big (Pf(A)P\big )$ and $f\big (\sigma(PAP)\big )$?

In any case, there is very little one can say regarding the relationship between these sets.  Consider,  for instance, the following two examples:

Let $ f(x)=x^2$, and
$$
  P=\pmatrix{1 & 0 \cr 0 & 0},\quad\text{and} \quad A=\pmatrix{0 & 1 \cr 1 & 0}.
  $$
Then
$\sigma\big (Pf(A)P\big )=\{0, 1\}$ and $\sigma(PAP)=\{0\}$, whence $f\big (\sigma(PAP)\big ) = \{0\}$.

Let $ f(x)=x^2$, and
$$
  P=\pmatrix{1 & 0 \cr 0 & 0},\quad\text{and} \quad A=\pmatrix{1 & 2 \cr 2 & 1}.
  $$
Then
$\sigma\big (Pf(A)P\big )=\{0, 5\}$ and $\sigma(PAP)=\{0, 1\}$, whence $f\big (\sigma(PAP)\big ) = \{0, 1\}$.

I suspect the only thing one can  say for sure is that both
$\sigma\big (Pf(A)P\big )$ and $\sigma(PAP)$ contain zero!

EDIT.  Here is a more radical example showing that,  for any two compact  subsets $E$ and $F$ of $\mathbb R$, both
containing zero, one can find a bounded operator $A$, and a function $f$, such that
$$
  f\big (\sigma(PAP)\big )=E, \quad \text{and}\quad \sigma\big (Pf(A)P\big )=F.
  $$

Let $x$ and $y$ be any two  bounded,  self-adjoint operators on $H$,  and put
$$
  z=y-x^3-2x.
  $$
Consider the operator $A$ on $H\oplus H$ given by
$$
  A=\pmatrix {x & 1\cr 1 & z}.
  $$
With $f(t)=t^3$, one then has that
$$
  f(A)=  A^3=\pmatrix {x^3+2x+z & x^2+z^2+xz+1 \cr x^2+z^2+zx+1  & z^3+2z+x}.
  $$
Letting $P=\pmatrix{1 & 0 \cr 0 & 0}$,  it follows that
$$
  Pf(A)P=\pmatrix {x^3+2x+z & 0 \cr 0 & 0} = \pmatrix {y& 0 \cr 0 & 0}.
  $$
Recalling that we were given closed sets $E$ and $F$ above, choose $x$ and $y$ such that
$\sigma (x) = \root 3 \of E$, and
$\sigma (y)=F$.  So,
$$
  \sigma (PAP)= \sigma \pmatrix {x & 0 \cr 0 & 0} = \sigma (x)\cup \{0\} = \root 3 \of E \cup \{0\} = \root 3 \of E,
  $$
whence
$$
  f\big (\sigma (PAP)\big ) =  f\big (\root 3 \of E \big ) = E,
  $$
while
$$
  \sigma \big (Pf(A)P\big )= \sigma (y)\cup \{0\} = F,
  $$

Title: Spectral mapping theorem in a subset


问:
Bertrand's postulate in another point of view
I was just wondering why can't we use prime number's theorem to prove Bertrand's postulate.
We know that if we show that for all natural numbers $n>2, \pi(2n)-\pi(n)>0$ we are done.
Why can't it be proven by just showing (By using the prime number's theorem) that for every natural numbers $n>2, \frac{2n}{ln(2n)}-\frac{n}{ln(n)}>0$?

答:
The following long-standing conjecture, due to Legendre, has a flavour similar to Bertrand's Postulate. For every $n\gt 1$, there is always a prime between $n^2$ and $(n+1)^2$. </p>

<p>By using the Prime Number Theorem, we can see that "usually," for large $n$, there are many primes between $n^2$ and $(n+1)^2$. However, the PNT is consistent with substantial irregularities of distribution, and is not strong enough to settle Legendre's conjecture. </p>

Body:
The following long-standing conjecture, due to Legendre, has a flavour similar to Bertrand's Postulate. For every $n\gt 1$, there is always a prime between $n^2$ and $(n+1)^2$.

By using the Prime Number Theorem, we can see that "usually," for large $n$, there are many primes between $n^2$ and $(n+1)^2$. However, the PNT is consistent with substantial irregularities of distribution, and is not strong enough to settle Legendre's conjecture.


问:
Uniqueness and existence of solution $\frac{dy}{dx} = \sqrt{x-y}$ given $y(2)=2$
I want to determine if there exists a solution and the uniqueness of it for: $\frac{dy}{dx} = \sqrt{x-y}$ given $y(2)=2$

A solution said that we cannot determine if there exists a solution since it was discontinuous at (2,2). This solution seems to imply that it is possible for x to be a number for example 1.99 and y be 2.00 or x be 2.00 and y be 2.01. I'm confused as to why the textbook makes this sort of assumption - I thought we were just determining the continuity of the point.

If we were to look at the neighborhood then why then if we are given $y(2)=1$ as the condition that we know that the above has a solution? It seems arbitrary as to how far we can go left and right or above and down a certain point to say that it was not continuous?

答:
Since the hypothesis $\lvert \frac{\partial f}{\partial y}\rvert<M$ in the nbd. of $(x_0,y_0)$ of Picard's uniqueness theorem doesn't hold here, hence the theorem fails to conclude anything.

Rather you should look to find the solution by using the substitution $x-y=u$


答:
To explain the discrepancy between the two situations, in the first situation you can have either of the two children be boys.
So call the children $A$ and $B$ then there are three possible situations, only $A$ is a boy, only $B$ is a boy or both $A$ and $B$ are boys.

In the second scenario, there are only 2 possible situations. If you say the boy you saw is $A$ then either $A$ is the only boy or $B$ is also a boy. The situation where $A$ is not a boy but $B$ is, is not a part of this condition.

Title: Conditional probability "brain teaser"

Body:
I'm reading A Practical Guide to Quantitative Finance Interviews by Xinfeng Zhou.

The book has the following two part question.

Part A: A company is holding a dinner for working mothers with at
least one son.  Ms. Jackson, a mother with two children, is invited.
What is the probability that both children are boys?

Part B: Your new colleague, Ms. Parker is known to have two children.
If you see her walking with one of her children and that child is a
boy, what is the probability that both children are boys?

According to the book, in Part A, the answer is
\begin{equation*}
1/3 = \frac{1/4}{3/4} = \frac{\mathbb{P}\{\text{two boys}\}}{\mathbb{P}\{\text{at least one boy out of two kids}\}} = \mathbb{P}\{\text{two boys} \, \mid \, \text{at least one out of two is a boy}\},
\end{equation*}
while, in Part B, the answer is
\begin{equation*}
1/2 = \mathbb{P}\{\text{second kid is a boy}\}.
\end{equation*}

I would have thought the "right" answer --- in quotes since the wording is clearly (?) vague --- is $1/3$ in both cases.  Am I missing something?

(Assume a kid is equally likely to be a boy or a girl...)

The best thing I can come up with is if we let $F_{1}$ and $F_{2}$ be coin flips taking values in the symbolic space $\{H,T\}$, then the solution to Part A is finding $\mathbb{P}\{(F_{1},F_{2}) = (H,H) \, \mid \, F_{1} \, \, \text{or} \, \, F_{2} = H\}$, while Part B is finding $\mathbb{P}\{F_{1} = H \, \mid \, F_{2} = H\}$.  On the other hand, just because I "saw" the kid (or the coin flip), doesn't mean I know which kid (or which coin) it was so I don't see why "What is $\mathbb{P}\{F_{1} = H \, \mid \, F_{2} = H\}$?" should be the "right" interpretation of the question.


问:
"Advice to young mathematicians"
I have been suggested to read the Advice to a Young Mathematician section  of the Princeton Companion to Mathematics, the short paper Ten Lessons I wish I had been Taught by Gian-Carlo Rota, and the Career Advice section of Terence Tao's blog, and I am amazed by the intelligence of the pieces of advice given in these pages.

Now, I ask to the many accomplished mathematicians who are active on this website if they would mind adding some of their own contributions to these already rich set of advice to novice mathematicians.

I realize that this question may be seen as extremely opinion-based. However, I  hope that it will be well-received (and well-answered) because, as Timothy Gowers put it,

"The most important thing that a young mathematician needs to learn is
  of course mathematics. However, it can also be very valuable to learn
  from the experiences of other mathematicians. The ﬁve contributors to
  this article were asked to draw on their experiences of mathematical
  life and research, and to offer advice that they might have liked to
  receive when they were just setting out on their careers."

答:
Here is my answer in short

Use YouTube videos .Some of them are good and explain basic content.

The Math Sorcerer  channel is an excellent for math advice and

inspiration

George Polya’s Mathematics and Plausible Reasoning vol 1 &2  These are great books to build mathematical skills

Circle around a particular problem to try solving it,instead of a
attacking it head on.

If you can’t solve a problem, go to another one and when     you go to bed, think about it and forget it. Let “Deep mind”,
ie,subconscious mind ,find a way to solve it. You might be surprised it finds hints for your conscious mind to solve it

Use this site,it is wealth of info.

Title: "Advice to young mathematicians"


问:
Multiple integration with Dirac delta
I am reading a paper and trying to solve collision integral. In Appendix A, there is an integral, $$ I^{(n)} = \int \frac{p^2_1\,dp_1 \,p^2_3\,dp_3\,p^2_4\,dp_4}
{2E_12E_32E_4} 2\pi \delta (p_1 − p_3 − p_4)e^{−E_1/T} I_\Omega,$$
where
\begin{align}
I_\Omega
&=\left( 4\pi \right)^3 \int d^3\lambda \, \frac{\sin p_1\lambda }{p_1\lambda}\frac{\sin p_3\lambda }{p_3\lambda}\frac{\sin p_4\lambda }{p_4\lambda} \\
&=\dfrac{32\pi^5}{p_1p_3p_4}\Bigg[\dfrac{\left( p_1-p_3+p_4\right) }{\left| p_1-p_3+p_4\right|}+\frac{(p_1+p_3-p_4) }{| p_1+p_3-p_4|}-\frac{(p_1-p_3-p_4) }{|p_1-p_3-p_4|}-\frac{(p_1+p_3+p_4) }{|p_1+p_3+p_4|}\Bigg].
\end{align}

They get a result $$I^{(n)}=\int \frac{16\pi^6 p_1^2}{E_1}e^{-E_1/T} \, dp_1$$ after integrating out dirac delta. I can't seem to understand how do they remove both $p_3$ and $p_4$ using one delta integral. Also, since all the terms in brackets of $I_\Omega$ are positive, it seems that term goes to zero. Can anyone help?

Note: $E_i$ and $p_i$ are energy and momentum of $i^{th}$ particle and here it is assumed that $m_3 =0$ and $m_4=0$ hence $E_3 = \sqrt{p_{3}^{2}+m_{3}^{2}}=p_{3}
$ & similarly $E_4 = p_4$.

答:
First, it should be noted that $p_1,p_3,p_4$ are limited to positive real values (they represent magnitudes of momenta). Since the delta function restricts to $p_1=p_3+p_4$, it follows that

$$p_1-p_3+p_4=2p_4\geq 0,\quad p_1+p_3-p_4=2p_3\geq 0,\quad p_1+p_3+p_4=2p_1\geq 0.$$

Therefore for three of the four terms in $I_\Omega$ we have $$\dfrac{p_1-p_3+p_4}{| p_1-p_3+p_4|}=\dfrac{p_1+p_3-p_4}{| p_1+p_3-p_4|}=\dfrac{p_1+p_3+p_4}{| p_1+p_3+p_4|}=1.$$

This leaves the third term in $I_\Omega$, which at first glance seems ill-defined since $p_1-p_3-p_4=0$. But this absolute value arises from the representation $\int\frac{d\lambda}{\lambda p}\sin(\lambda p)=|p|^{-1},$ and thus
$$\frac{p}{|p|}=\int_0^\infty \frac{d\lambda}{\lambda}\sin(\lambda p)\to \int_0^\infty \frac{d\lambda}{\lambda}0=0$$ in the limit $p\to 0^+$.

Combining the above, we see that $I_\Omega=\dfrac{32\pi^5}{p_1p_3p_4}$
on the restriction $p_1=p_3+p_4$. Hence

\begin{align}
I^{(n)}
&= \int \frac{p^2_1\,dp_1}{2E_1} \frac{p^2_3\,dp_3}{2E_3} \frac{p^2_4\,dp_4}{2E_4}
2\pi \delta (p_1 − p_3 − p_4)e^{−E_1/T} I_\Omega\\
&= \int \frac{p^2_1\,dp_1}{2E_1} \frac{p^2_3\,dp_3}{2E_3} \frac{p^2_4\,dp_4}{2E_4}
2\pi \delta (p_1 − p_3 − p_4) \dfrac{32\pi^5}{p_1p_3p_4}e^{−E_1/T}\\
&= \int \frac{8\pi^6 p_1}{E_1}e^{−E_1/T}\,dp_1 \,dp_3\,dp_4
 \delta (p_1 − p_3 − p_4)
\end{align}
where I've used $E_3=p_3$, $E_4=p_4$.
Eliminating the $p_4$-integral using the delta function yields

$$I^{(n)}=\int_0^\infty \frac{8\pi^6 p_1}{E_1}e^{−E_1/T}\,dp_1\int_0^{p_1}\,dp_3=\int_0^\infty \frac{8\pi^6 p_1^2}{E_1}e^{−E_1/T}\,dp_1.$$
This is almost in agreement with the stated result: there's a missing factor of two which I haven't tracked down yet. I'll see what I find.

I haven't succeeded in finding the error, and the more checks I do make me suspicious about the original source. In particular, the factor of two would be accounted for if one assumes that $$\frac{p_1-p_3-p_4}{|p_1-p_3-p_4|}\to -1$$ when $p_1=p_3+p_4$ rather than zero (as I have above). But as far as I can tell, this term should genuinely vanish given its integral representation in the Appendix. Beyond that I don't see an obvious explanation, and so I would tentatively propose a mathematical error in the source.

After having examined the source material (along with some awareness of the underlying physics) the current conjecture of the OP and myself is that the delta function should actually be $$\delta(E_1-E_3-E_4)=0.$$ This reflects the fact that the constraint should arise from conservation of 4-momentum, which includes energy conservation. With this reading in mind, the behavior of $I_\Omega(p_1,p_3,p_4)$ when subject to this constraint is a bit different and I'll address the terms in pairs. First, note that energy conservation implies $$p_3+p_4=E_3+E_4=E_1=\sqrt{p_1^2+m_1^2}\geq p_1 >0.$$
Since $p_1\geq 0 \geq -p_1$, we conclude that $p_1-p_3-p_4$ and $p_1+p_3+p_4$ have opposite signs. As such, the last two terms in $I_\Omega$ will vanish identically. The analogous question for the remaining two terms are the values of $p_1-p_3+p_4$ and $p_1+p_3-p_4$, with $I_\Omega$ vanishing unless the signs match. But $$(p_1-p_3+p_4)+(p_1+p_3-p_4)=2p_1\geq 0$$ and so cannot both be negative. For both to be positive, we require
\begin{align}
0
&<(p_1-p_3+p_4)(p_1+p_3-p_4)\\
&=p_1^2-(p_3-p_4)^2\\
&=(p_3+p_4)^2-m_1^2-(p_3-p_4)^2\\
&=4p_3 p_4-m_1^2
\end{align}
Thus the terms are both positive if $4p_3p_4>m_1^2$ and we conclude that energy conservation dictates $$I_\Omega(p_1,p_3,p_4)=\dfrac{64\pi^5}{p_1p_3p_4}\theta(4p_3p_4-m_1^2)$$ whenever energy conservation is fulfilled. As such, the complete integral is

$$I^{(n)}
= \int \frac{p^2_1\,dp_1}{2E_1} \frac{p^2_3\,dp_3}{2E_3} \frac{p^2_4\,dp_4}{2E_4}
2\pi \delta (E_1 − E_3 − E_4)e^{−E_1/T}\dfrac{64\pi^5}{p_1p_3p_4}\theta(4p_3p_4-m_1^2)$$

Since energy conservation dictates $$p_4=E_4=E_1-E_3=\sqrt{p_1^2+m_1^2}-p_3,$$ integrating out $p_4$ first yields

$$I^{(n)}
= \int \frac{16\pi^6 p_1\,dp_1}{E_1} dp_3 e^{−E_1/T}\theta\Bigg(4p_3\Big[\sqrt{p_1^2+m_1^2}-p_3\Big]-m_1^2\Bigg)$$

The limits for the $p_3$ integral are dictated by the condition

$$4p_3\Big[E_1-p_3\Big]-m_1^2\geq 0$$

which for $p_1,p_3\geq 0$ is equivalent to $$\frac12\left(E_1-p_1\right)\leq p_3\leq \frac12\left(E_1+p_1\right)$$
As such the $p_3$-integral is simply

$$\int_{(E_1-p_1)/2}^{(E_1+p_1)/2} dp_3 = \frac{E_1+p_1}{2}-\frac{E_1-p_1}{2}=p_1$$
and therefore
$$I^{(n)}
= \int \frac{16\pi^6 p_1^2\,dp_1}{E_1} e^{−E_1/T}$$
as claimed. As a check we consider the integral denoted $I^{(p)}$ in the source material. Then the presence of the factor $U=E_3=p_3$ renders the $p_3$ integral as

$$\int_{(E_1-p_1)/2}^{(E_1+p_1)/2} p_3\,dp_3 = \frac12\left(\frac{E_1+p_1}{2}\right)^2-\frac12\left(\frac{E_1-p_1}{2}\right)^2=\frac12 E_1 p_1$$
and therefore

$$I^{(p)}
= \int 8\pi^6 p_1^2\,dp_1 e^{−E_1/T}$$
which again matches the source material.

Title: Multiple integration with Dirac delta


问:
Cross Product in Linear Algebra
I was looking at the index of "Linear Algebra Done Right" by Sheldon Axler and I couldn't find the cross product mentioned at all while the dot product is mentioned in pretty much all the later chapters. Is the dot product actually the more important concept in linear algebra?</p>

答: Cross product is for $\mathbb{R}^3$. The general notion is called exterior/wedge product. You will see this notion in  multilinear algebra. Inner product, on the other hand, is a fundamental concept for inner product spaces. I would not say which one is "more important". It is just that you are learning different topics in algebra.

Title: Cross Product in Linear Algebra


问:
Completeness of a quantifier-free axiomatization of Boolean algebra using partial-ordering
Consider the following set of axioms:

$a \leq a$

$(a \leq b \mathrel{\&} b \leq a) \Rightarrow a = b$

$(a \leq b \mathrel{\&} b \leq c) \Rightarrow a \leq c$

$0 \leq a \mathrel{\&} a \leq 1$

$a \leq a \lor b \mathrel{\&} b \leq a \lor b \mathrel{\&} ((a \leq c \mathrel{\&} b \leq c) \Rightarrow a \lor b \leq c)$

$a \land b \leq a \mathrel{\&} a \land b \leq b \mathrel{\&} ((c \leq a \mathrel{\&} c \leq b) \Rightarrow c \leq a \land b)$

$((a \leq b \mathrel{\&} a \leq \neg b) \Rightarrow a \leq 0) \mathrel{\&} ((b \leq a \mathrel{\&} \neg b \leq a) \Rightarrow 1 \leq a)$

$a \nleq \neg b \Rightarrow \exists x (x \neq 0 \mathrel{\&} x \leq a \mathrel{\&} x \leq b)$

[1] says axioms 7 and 8 are independent from axioms 1-6, and that axioms 1-8 are complete for Boolean algebras, but not that axioms 7 and 8 are independent from one another in the presence of axioms 1-6, leaving open the possibility that axioms 1-7 are also complete for Boolean algebras. So my question is: are they? If not, is there a quantifier-free extension of axioms 1-7 that is?

[1]: Huntington, E. V.. "Sets of Independent Postulates for the Algebra of Logic", Transactions of the American Mathematical Society, vol. 5, no. 3, 1904, pp. 288-309.

答:
Here's an answer to the first question: $(8)$ cannot be proved from $(1)$-$(7)$. There is in fact a nice countermodel, namely the lattice $M_3$ equipped with a "negation" operation which yields a derangement of the intermediate elements.

More precisely, our countermodel will have underlying set $\{0,1,x,y,z\}$. The partial order is given by setting $0$ at the botom, $1$ at the top, and no other inequalities holding besides reflexivity; $\wedge$ and $\vee$ are the the usual join and meet operations. See here for a picture of this. We then let negation be the unary function sending $0$ to $1$, $1$ to $0$, and $$x\mapsto y\mapsto z\mapsto x.$$

Incidentally, the lattice $M_3$ itself (so, the countermodel without negation) is an important example in order theory; it's one of the two lattices characterizing non-distributivity, the other being $N_5$.

Of course, this leaves the second question open. EDIT: per Eric Wofsey's answer, the answer to the second question is yes.


问:
How to form a generator and parity-check matrix given the code
So, I am given that we have a $[n,k,d]=[7,4,2]$ code, and I am asked to find its parity-check matrix and generator matrix but my notes are so unclear that I can't understand it (also can't find any helpful information on google). Would somebody be able to help and tell me how to do it in this case or in the general case? I am really struggling to understand it.

The most I have got is constructing the code $C$, say:

$$C=\{0000000,1100000,0110000,0011000,\dots,1111000,0111100,0011110,\dots,1111111\}$$

I don't know what to do now. Any help would be greatly appreciated :) Thank you!

I know that I need to find $k=4$ linearly independent codewords to construct the generator matrix, say $G$, but I just don't know how to do this.

Okay, I now attempted the questions and got an answer different to the mark scheme. I know it's possible to have more than one generator matrix solution so would someone be able to check if mine is correct?

My generator matrix:

$$G =
    \begin{pmatrix}
    1 & 1 & 0 & 0 & 0 & 0 & 0 \\
    0 & 1 & 1 & 1 & 1 & 0 & 0 \\
    0 & 0 & 0 & 1 & 1 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 1 & 1 \\
    \end{pmatrix}
$$

which in standard form gives:

$$G =
    \begin{pmatrix}
    1 & 0 & 0 & 0 & 0 & 1 & 0 \\
    0 & 1 & 0 & 0 & 0 & 1 & 0 \\
    0 & 0 & 1 & 0 & 1 & 0 & 0 \\
    0 & 0 & 0 & 1 & 0 & 0 & 1 \\
    \end{pmatrix}
$$

and the parity-check matrix is thus:

$$H=
    \begin{pmatrix}
    0 & 0 & 1 & 0 & 1 & 0 & 0 \\
    1 & 1 & 0 & 0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 & 0 & 0 & 1 \\
    \end{pmatrix}
$$

答:
As you correctly pointed out $n$ tells you the length of the codewords and $k$ the rank of your generator - from this you can easily derive the dimensions of your parity check matrix $H$.

For linear codes, if $C$ has a distance of $d$, any $d-1$ columns of $H$ are linearly independent, while there exists at least one set of $d$ columns of $H$ that are linearly dependent (remember that $d_{\mathrm{min}}(C) = w_{\mathrm{min}}(C)$ - when is $w_{\mathrm{min}}(C)\leq d$ and when is $w_{\mathrm{min}}(C)\geq d$ ?)

In your case, you want to find a couple of linearly dependent columns in your $H$ which is rather trivial


问:
Find $\lim_{k\rightarrow \infty} A^k x$
Let $\displaystyle A=\left(\begin{matrix}.5&1\\0&.75\end{matrix}\right),x=\left(\begin{matrix}-2\\1\end{matrix}\right)$.

Find $\displaystyle \lim_{k\rightarrow \infty} A^k x$

Find $\displaystyle \lim_{k\rightarrow \infty} \frac 1 {\Vert A^k x\Vert}A^k x$

I found the eigenvalues and eigenvectors to be $\displaystyle \lambda_1=.5, v_1=\left(\begin{matrix}1\\0\end{matrix}\right),\lambda_2=.75,v_2=\left(\begin{matrix}4\\1\end{matrix}\right)$.

x can be expressed as $x=-6v_1+v_2$, then $\displaystyle Ax=-6(.5)\left(\begin{matrix}1\\0\end{matrix}\right)+.75\left(\begin{matrix}4\\1\end{matrix}\right)=\left(\begin{matrix}-6(.5)+.75(4)\\.75\end{matrix}\right)$

Then $\displaystyle \lim_{k\rightarrow \infty} A^k x=\lim_{k\rightarrow \infty} \left(\begin{matrix}-6(.5)^k+(.75)^k(4)\\(.75)^k\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$ ?

And what is $\displaystyle \lim_{k\rightarrow \infty} \frac 1 {\Vert A^k x\Vert}A^k x=\lim_{k\rightarrow \infty} \left(\begin{matrix}\frac{-6(.5)^k+(.75)^k(4)}{\sqrt{(-6(.5)^k+(.75)^k(4))^2+((.75)^k)^2}}\\\frac{(.75)^k}{\sqrt{(-6(.5)^k+(.75)^k(4))^2+((.75)^k)^2}}\end{matrix}\right)=\left(\begin{matrix}\infty\\\infty\end{matrix}\right)$?

答:
Since you already did the first part, I will rewrite the second part in terms of the eigenvector decomposition

$$\lim_{k\to\infty} \frac{\left(\frac{1}{2}\right)^k\cdot(-6v_1)+\left(\frac{3}{4}\right)^k\cdot(v_2)}{\left|\left(\frac{1}{2}\right)^k\cdot(-6v_1)+\left(\frac{3}{4}\right)^k\cdot(v_2)\right|} \equiv \lim_{k\to\infty}\frac{av+bw}{|av+bw|}$$

The two vectors are not orthogonal, and the magnitude of the sum of non orthogonal vectors is given by

$$|av+bw| = \sqrt{(av+bw)\cdot(av+bw)} = \sqrt{a^2|v|^2+2abv\cdot w + b^2|w|^2}$$

Plugging in the vectors we have, this simplifies to

$$\sqrt{a^2+8ab+17b^2} = |b|\sqrt{\left(\frac{a}{b}\right)^2+ 8\frac{a}{b}+17}$$

Now why did we pull out $b$ instead of $a$ when the same formula could have applied either way? It's because $\frac{a}{b}\to 0$, thus we can take the limit without a L'Hopital situatuon occurring. Meaning the limit simplifies to

$$\lim_{k\to\infty} \frac{\frac{a}{b}v +  w}{\sqrt{\left(\frac{a}{b}\right)^2+ 8\frac{a}{b}+17}} \to \frac{w}{\sqrt{17}} = \frac{1}{\sqrt{17}}\begin{pmatrix} 4 \\ 1 \end{pmatrix}$$

because $b>0$. In general you can see that the second limit will approach the unit eigenvector with the largest (by absolute value) eigenvalue.

Title: Find $\lim_{k\rightarrow \infty} A^k x$


答: Axiom 8 can be replaced by the axiom $a\vee(b\wedge c)=(a\vee b)\wedge (a\vee c)$.  In the language of order theory, axioms 1-6 describe a (bounded) lattice, and adding axiom 7 gives you a complemented lattice (with a specific choice of complement operation).  Boolean algebras are the same thing as complemented distributive lattices, so all you need to add to axioms 1-7 is the distributive law.

Title: Completeness of a quantifier-free axiomatization of Boolean algebra using partial-ordering

Body:
Consider the following set of axioms:

$a \leq a$

$(a \leq b \mathrel{\&} b \leq a) \Rightarrow a = b$

$(a \leq b \mathrel{\&} b \leq c) \Rightarrow a \leq c$

$0 \leq a \mathrel{\&} a \leq 1$

$a \leq a \lor b \mathrel{\&} b \leq a \lor b \mathrel{\&} ((a \leq c \mathrel{\&} b \leq c) \Rightarrow a \lor b \leq c)$

$a \land b \leq a \mathrel{\&} a \land b \leq b \mathrel{\&} ((c \leq a \mathrel{\&} c \leq b) \Rightarrow c \leq a \land b)$

$((a \leq b \mathrel{\&} a \leq \neg b) \Rightarrow a \leq 0) \mathrel{\&} ((b \leq a \mathrel{\&} \neg b \leq a) \Rightarrow 1 \leq a)$

$a \nleq \neg b \Rightarrow \exists x (x \neq 0 \mathrel{\&} x \leq a \mathrel{\&} x \leq b)$

[1] says axioms 7 and 8 are independent from axioms 1-6, and that axioms 1-8 are complete for Boolean algebras, but not that axioms 7 and 8 are independent from one another in the presence of axioms 1-6, leaving open the possibility that axioms 1-7 are also complete for Boolean algebras. So my question is: are they? If not, is there a quantifier-free extension of axioms 1-7 that is?

[1]: Huntington, E. V.. "Sets of Independent Postulates for the Algebra of Logic", Transactions of the American Mathematical Society, vol. 5, no. 3, 1904, pp. 288-309.


问:
Fixed points of strong inversion of a knot
(Edit: added definition of strong inversion.)

(2nd edit: added motivation for question at the end.)

Definition: A knot $K$ in $S^3$ is strongly invertible if there is an involution of $(S^3,K)$ which preserves the orientation of $S^3$ and reverses the orientation of $K$.

A strong inversion is an element of Sym$(S^3,K)$ taken up to conjugacy in Sym$^+(S^3,K)$, where Sym$(S^3,K)$ is the the symmetry group of $K$, i.e. the group of diffeomorphisms of the pair $(S^3,K)$ modulo isotopies, and Sym$^+(S^3,K)$ is the group of diffeomorphisms preserving the orientation of $S^3$.

Equivalently, a strong inversion on $K$ is a $π$-rotation of $S^3$ which leaves $K$ invariant and has axis meeting $K$ at exactly two points.

My question:

Let $K$ be a knot (possibly non-alternating) that we know admits a strong inversion. A priori, we do not know anything about the strong inversion. The fixed points of the strong inversion consist of a circle in $S^3$ intersecting $K$ in two points (e.g. by the positive solution to the Smith conjecture). Is it possible to explicitly find these two fixed points of the strong inversion on some projection of $K$? As an additional question: is there any way to find these two fixed points on a minimal crossing projection?

For example, the trefoil is a strongly invertible knot. The two fixed points of $K$ under the strong inversion are described by the image below (taken from https://arxiv.org/pdf/1909.08556.pdf).

Another example: take the non-alternating knot $11$n$92$. From KnotInfo (https://knotinfo.math.indiana.edu/), we can see that this hyperbolic knot is reversible with symmetry group D1 ($ = \mathbb{Z}/2\mathbb{Z}$). Thus, it admits exactly one strong inversion. The diagram below includes minimal crossing diagrams of $11$n$92$ taken from KnotInfo. Can we find the two fixed points of the strong inversion in any of the diagrams below? Or, can we find the two fixed points in a a different projection of $11$n$92$?

Motivation for question:

Let $h$ be a strong inversion of a knot $K$. Let $p$ be the projection $S^3 \to S^3/h$. We define $G(K,h) := p(\text{Fix}(h)) \cup p(K)$, which is the $\theta$-curve induced by $h$ and consists of three knots.

In the paper "Identifying tunnel number one knots" by Morimoto, Sakuma, and Yokota (https://projecteuclid.org/euclid.jmsj/1226498920), the authors state the following fact:

Corollary 1.3. Let $K$ be a knot with tunnel number one. Then $K$ admits a a strong inversion such that the set of constituent knots of $G(K,h)$ consists of two trivial knots and a knot with a 2-bridge decomposition.

The authors also mentioned that they use Corollary 1.3 to obtain the tunnel number of almost all knots with $\leq$ 10 crossings (they were missing two), all of which have tunnel number one or two. Many of the knots with $\leq$ 10 crossings have minimal crossing diagrams which make it difficult to explicitly find the fixed points of the strong inversion. I was thus wondering, how can one implement Corollary 1.3?

答:
For alternating knots there is always a way to see a strong inversion from any alternating diagram as a sequence of flypes plus an elementary symmetry of the diagram, see https://arxiv.org/abs/2002.08191.

Even in this case, the symmetry is not always 'easily' visible in a minimal crossing diagram. For example, there is a strong inversion on $5_2$ for which there is no minimal crossing diagram in which the strong inversion is visible as rotation around an axis contained in the plane of the diagram, or perpendicular to the plane of the diagram (see the strong inversion on $5_2$ shown in the right column of the table at the end of https://arxiv.org/abs/1908.00082 where they show a 6-crossing diagram). This can be checked, for example, by enumerating all minimal crossing diagrams for $5_2$.

For non-alternating knots I don't know about any general visiblity results for strong inversions, although in practice they are also often visible on a minimal-crossing diagram by performing some flypes. In your example 11n92, you can see the strong inversion from your diagram by performing the $\alpha$ flype, then the $\beta$ flype, then $\pi$ rotation around the dotted black axis shown in the image below. In particular, the fixed points are the two intersection points of the black axis with the knot outside of the flypes. (The two intersection points inside the flypes are exchanged by the involution.)

https://i.stack.imgur.com/ArThF.png


问:
Let $f:[0,\infty) \to [0,\infty)$ be increasing and satisfy $f(0)=0$ and $f(x)>0 \forall x>0$.
Let $f:[0,\infty) \to [0,\infty)$ be increasing and satisfy $f(0)=0$ and $f(x)>0
\forall x>0$. If $f$ also satisfies $f(x+y) \leq f(x)+f(y) \forall x,y \geq 0$, then $f \circ d$ is a metric whenever $d$ is metric. Show each of the following conditions is sufficient to ensure that $f(x+y) \leq f(x)+f(y) \forall x,y \geq 0$:

a) $f$ has a second derivative satisfying $f'' \leq 0$;

b) $f$ has a decreasing first derivative.

c) $f(x)/x$ is decreasing for $x>0$.

My Work Thus Far:

Let $g(x)=f(x)x$, for $x>0$. According to the question we have that $f''<0$ then $f$ has a decreasing first derivative, meaning $f'$ is decreasing. Now, taking the derivative of $g$ yields: $g'(x)=xf'(x)-f(x)x^2=f'(x)x-f(x)x^2$, where $x>0$.

I know that $-f(x)x^2$ is decreasing, but how do I know for sure that $f'(x)x$ is decreasing so I can deduce that $g'(x)<0, \forall x>0$ and that $g(x)$ is a decreasing function?

答:
We assume that (a) or (b) holds. Let $x,y\in[0,\infty)$. If $x=0$
or $y=0$, we clearly have $f(x+y)=f(x)+f(y)$ because $f(0)=0$.
Consider the case that $x>0$ and $y>0$. Without loss of generality, we may assume that $x\leq y$. By mean-value theorem, there
exist $\xi\in(y,x+y)$ and $\eta\in(0,x)$ such that $f(x+y)-f(y)=f'(\xi)x$
and $f(x)-f(0)=f'(\eta)x$. Note that $\eta < \xi$, so by (a) or (b), we have $f'(\eta)\geq f'(\xi)$. Thererfore

\begin{eqnarray*}
 &  & f(x+y)-f(x)-f(y)\\
 & = & [f(x+y)-f(y)]-[f(x)-f(0)]\\
 & = & [f'(\xi)-f'(\eta)]x\\
 & \leq & 0.
\end{eqnarray*}


问:
How can I show two polynomials common roots with a given matix and resultant?
How can I show, that if $u:= ax + b$ and $v:= x^2 + cx + d$ ,  $(a, b, c, d ∈ \mathbb{R})$ polinoms have common root, if and only if $det(A[u, v]) = 0$, where
$A[u, v] =
\begin{bmatrix}
a & b & 0\\
0 & a & b\\
1 & c & d
\end{bmatrix}$ .

Should I do this with resultant? And how? I'm looking for guidance, not the exact solution.

答:
$\det A[u,v]=0$ if and only if there exists a nonzero vector $X$ such that
$$\begin{pmatrix}
a & b & 0\\
0 & a & b\\
1 & c & d
\end{pmatrix} X =0.$$
If $\alpha$ is a common root, then
$X=\begin{pmatrix} \alpha^2\\ \alpha \\ 1\\ \end{pmatrix}$ works. Now if such an $X$ exists can you prove that there is a common root?


问:
Proving inequality that (problably) using AM-GM
For $a, b$ and $c$ positive real number, prove following inequality $$\dfrac{a+b}{c^2}+\dfrac{b+c}{a^2}+\dfrac{a+c}{b^2}\geq 2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)$$

答:
Notice that the inequality is equivalent to the following nice form:

$$ ( a + b + c ) ( \frac{1}{a^2} + \frac{1}{b^2 } + \frac{1}{c^2} )  \geq 3 ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) $$

Recognize that this is Chebyshev (on oppositely arranged sequences).


问:
If $m \equiv m'$ mod $n$, then gcd$(m,n) = 1$ $\Longleftrightarrow$ gcd$(m',n) = 1$.
How to prove the following statement:

If $m \equiv m'$ mod $n$, then gcd$(m,n) = 1$ $\Longleftrightarrow$ gcd$(m',n) = 1$.

I have never learned number theory before, for me it's already hard to prove g.c.d of two integers is 1, not to mention the whole statement above.

These kinds of facts jump out frequently in my abstract algebra course. Is there any good introductory number theory book that has clear and illuminating proof for these "basic" facts? Then I can remember the proof technique and these "basic" results by rote.

答:
Actually the problem has a simple solution, if you are allowed to use the following result:

$$\ gcd(m,n)=1\ \mathrm{iff}\ \exists k_1,k_2\in\mathbb{Z}\ \mathrm{such\ that}\ k_1m+k_2n=1 $$
since $m\equiv m'(\mod n)$ then $m=m'+k_3n$ for some $k_3\in\mathbb{Z}$ replacing we have
$$1=k_1m+k_2n= k_1(m'+k_3n)+k_2n=k_1m'+(k_2+k_3)n $$
which shows that $$ gcd(m',n)=1$$

You can try the other way.


答:
$u$ and $v$ have a common root if and only if there are non-zero polynomials $\alpha$, with $\deg \alpha \leq \deg u -1$, and $\beta $, with $\deg \beta \leq \deg v -1$, such that

$$ \beta(x) u(x) + \alpha(x) v(x) = 0. \; \; (1)$$

(We can see this by denoting the common factor by $t$, non-constant polynomial, such that $u = tu_1$ and $v=tv_1$. We can then choose $\beta = v_1$ and $\alpha = -u_1$.)

In our example, $\alpha(x) = e$ and $\beta(x) = fx+g$, respectively.

The null polynomial $(1)$ can be rewritten as

$$ (fa+e)x^2 + (fb+ga +ec)x + (gb +ed) = 0, $$
which is equivalent to the system of equations in unknowns $ f, g, e$
$$ a\cdot f + 0\cdot g + 1\cdot e =0 $$
$$ b\cdot f + a \cdot g + c\cdot e = 0 $$
$$ 0\cdot f + b\cdot g + d \cdot e = 0 $$
Solution $ f, g, e$ is non-trivial if and only if $\det A = 0$.

Brute force solution: For $a\not= 0$, the common root can only be

$$-\frac{b}{a}. $$

In this case:

$$ v\left(-\frac{b}{a} \right) = 0,$$

which is equivalent to:

$$ b^2 -cba +da^2 =0.$$

One can then verify that:
$$\det A  = b^2 -cba +da^2. $$

Title: How can I show two polynomials common roots with a given matix and resultant?

Body:
How can I show, that if $u:= ax + b$ and $v:= x^2 + cx + d$ ,  $(a, b, c, d ∈ \mathbb{R})$ polinoms have common root, if and only if $det(A[u, v]) = 0$, where
$A[u, v] =
\begin{bmatrix}
a & b & 0\\
0 & a & b\\
1 & c & d
\end{bmatrix}$ .

Should I do this with resultant? And how? I'm looking for guidance, not the exact solution.


问:
Variational principle for the principal eigenvalue (Evans)
Evans tell us there exists an orthonormal basis $\{w_k\}_{k=1}^\infty$ of $L^2(U)$ in the following theorem 1. Whether  this is basis a maximal orthonormal set in $L^2(U)$ (and why?).
In the following theorem 2 , the author tells us $\left \{\frac{\omega _k}{ \lambda _k^{\frac{1}{2}}}\right\}_{k=1}^{\infty}$ is in fact an orthonormal basis of $H_0^1(U)$. Then he shows $$B[\omega _k,u]=\lambda _k(\omega_k,u)=0, (k=1,…)$$ force $u≡0$, as $\{\omega _k\}_{k=1}^{\infty}$ is basis of $L^2(U)$ .(Why this forces $u$≡0?).
Could someone give me some details, thank you !

Theorem is typed as follow.

THEOREM 1 (Eignevalues of symmetric elliptic operators).

$\quad$(i) Each eigenvalue of $L$ is real.

$\quad$(ii) Furthermore, if we  repeat each eigenvalue according to its (finite) multiplicity, we have $$\qquad \, \,\sum=\{\lambda_k\}_{k=1}^\infty$$

$\qquad \, \,$ where $0 < \lambda_1 \le \lambda_2 \le \lambda_3 \le \cdots$ and $\lambda_k \rightarrow \infty$ as $k \rightarrow \infty$.

$\quad \, \,$(iii) Finally, there exists an orthonormal basis $\{w_k\}_{k=1}^\infty$ of $L^2(U)$, where $w_k \in H_0^1(U)$ is an $\quad \, \,$eigenfunction corresponding to $\lambda_k$:

$$\quad \, \, \begin{cases}Lw_k = \lambda_k w_k & \text{in }U \\ w_k = 0 & \text{on } \partial U \end{cases}$$ for $k=1,2,\ldots$.

答:
The existence of an orthonormal eigenbasis follows from the spectral theorem for compact self-adjoint operators applied to the inverse $L^{-1}$ of the elliptic operator.

The $\lambda_k$ are all positive and therefore $\lambda_k (w_k, u) = 0$ implies that the inner product $(w_k, u)$ is zero for all $k$. Since $\{w_k\}$ are an orthonormal basis, $\{w_k\}$ is complete, which means exactly that if $(w_k, u) = 0$ for all $k$, then $u$ must be the zero element of the vector space $L^2(U)$. Thus $u = 0$ in $L^2(U)$; in particular $u = 0$ a.e. and thus $u = 0$ in $H^1_0(U)$.


问:
What is the length of $EF$?
In triangle $ABC$ , $AD$ is perpendicular to $BC$. It is given that $BC$=8 cm and $AD$=6 cm. If $E$ and $F$ are the mid points of $BD$ and $AC$ respectively, find $EF$.

I tried using the mid point theorem by constructing a line parallel to $AB$, but the equations led nowhere.

答: Let $M$ denote the midpoint of the segment $AB$. Then the midpoint theorem implies that $ME$ has length 3 (by comparison with the length of the segment $AD$), and another application of the midpoint theorem implies that $MF$ has length 4 (by comparison with the length of the segment $BC$). Because $ME$ is parallel to $AD$, $MF$ is parallel to $BC$, and $AD$ is perpendicular to $BC$, we have that $ME$ and $MF$ are perpendicular. Thus, $MEF$ is a right triangle with legs of length 3 and 4 so that the hypotenuse $EF$ has length 5.

Title: What is the length of $EF$?


问:
How can I measure variance efficiently?
I have bunch of values, for example {1,2,3,4}.
I need to measure variance in a very efficient way.
On wikipedia variance is defined as sum of squared differences
between the data examples and the mean, and then you normalize
that sum with 1/n, where n is the number of data examples.
One improvement is to skip the normalization. Any other ways
I can improve this and make it less expensive to compute?

Thanks

答:
I presume you're talking implementing the calculation in software (if it is to be done by hand, some of the remarks might be irrelevant).

The averaging/normalization at the end is usually the least expensive step of the calculations; after all, it's just one arithmetical operation. Depending on what your measure of efficiency is, you might find the following formula useful:
$$\mathrm{Var}=\mathrm{E}(X^2)-(\mathrm{E}(X))^2$$

If you are dealing with $n$ samples $x_1, x_2, \ldots, x_n$, it can be expressed as follows:
$$\mathrm{Var}=\frac{1}{n}\sum_{i=1}^n x_i^2 - \left(\frac{1}{n}\sum_{i=1}^n x_i\right)^2$$

Thus, in order to calculate the variance of the given samples, you only need to find their sum and sum of their squares, both divided by $n$ and perform one squaring and one subtraction at the end. The total number of arithmetic operations will then be $(n+1)$ squarings, $(2n-2)$ additions and one subtraction... which looks quite efficient to me. As an extra bonus, you will also get the (arithmetic) mean of the samples. Finally, this approach also allows you to process the samples as they are appearing without storing them (of course, unless you need them for other purposes); allowing you to process lots of samples without increasing the memory requirements. The only possible problem could arise if the variance is very small and thus the two quantities you're going to subtract are very close; you'd need to be well aware of the capabilities and limitations of your real-number processing tools.

Title: How can I measure variance efficiently?


问:
What is this polynomial?
I came across the "summation by parts" formula, and wondered how it could be useful in any way:
$$
\sum_{k=n}^m\ a_k \Delta b_k=\left[a_k b_k\right]_{n}^{m+1}-\sum_{k=n}^{m}\ b_{k+1} \Delta a_k
$$
Where $\Delta a_k$ is the Forward difference operator. I tried to do the following: with $|\alpha|<1$:
$$
\sum_{k=0}^m \ k \alpha^k=\frac{1}{\alpha-1}\sum_{k=0}^m \ k \alpha^k(\alpha-1)=\frac{1}{\alpha-1}\left(\left[ k\alpha^k\right]_0^{m+1}-\sum_{k=0}^m\ \alpha^{k+1}\right)
$$
And then I took the limit as  $m\to \infty$, getting:
$$
\sum_{k=0}^\infty\ k\alpha^k=\frac{\alpha}{(1-\alpha)^2}
$$
Which is a result I'm sure many of you know, but then I realized that, with this formula, i could do the same for $\sum_{k=0}^\infty\ k^2 \alpha^k$, being $\Delta k^2=2k+1$, arriving at:
$$
\sum_{k=0}^\infty\ k^2 \alpha^k=\frac{\alpha}{(1-\alpha)^2}\left(2\cdot\frac{\alpha}{1-\alpha}+1\right)
$$
But I could do the same for $\sum_{k=0}^{\infty}\ k^3\alpha^k$, and so on, obtaining this formula:
$$
\sum_{k=0}^\infty\ k^3\alpha^k=\frac{\alpha}{(1-\alpha)^2}\left(6\cdot \left(\frac{\alpha}{1-\alpha}\right)^2+6\cdot \left(\frac{\alpha}{1-\alpha}\right)+1\right)
$$
And so on with the fourth degree and the fifth:
$$
\sum_{k=0}^{\infty}\ k^4\alpha^k=\frac{\alpha}{(1-\alpha)^2}\left(24\cdot \left(\frac{\alpha}{1-\alpha}\right)^3+36\cdot\left(\frac{\alpha}{1-\alpha}\right)^2+14\cdot\left(\frac{\alpha}{1-\alpha}\right)+1\right)
$$

$$
\sum_{k=0}^{\infty}\ k^5\alpha^k=\frac{\alpha}{(1-\alpha)^2}\left(120\cdot\left(\frac{\alpha}{1-\alpha}\right)^4+240\cdot \left(\frac{\alpha}{1-\alpha}\right)^3+150\cdot\left(\frac{\alpha}{1-\alpha}\right)^2+30\cdot \left(\frac{\alpha}{1-\alpha}\right)+1\right)
$$
At this point I got bored and wanted to find a general formula, seeing a polynomial pattern, but all i could figure out was this: (I didn't prove any of the following, I just tried to figure out the patterns, which could be misleading)

Given $|\alpha|<1$, and $n\in \mathbb{N}$
$$
\sum_{k=0}^\infty k^n \alpha^k=\frac{\alpha}{(1-\alpha)^2}\left(\sum_{k=0}^{n-1} P_{n,k} \left(\frac{\alpha}{1-\alpha}\right)^k\right)
$$
Where $P_{n,k}$ is a coefficient defined for $n\in \mathbb{N},\ k\in \{0,1,2,...,n-1\}$, with these defining properties:
$$
\forall n \in \mathbb{N}, P_{n,0}=1\\
P_{n,k}=\sum_{h=1}^{n-k}\binom{n}{h}P_{n-h,k-1}
$$
I'm having some troubles on how to get to a closed form of theese coefficients, all i know is that the coefficient of the highest term is $n!$, the known term is $1$, and that the linear term is $2^n-2$, the formulas for the other coefficients are a mess beyond my comprehension, with nested sums of binomial coefficients. If you recognize these polynomials by any chance, or if you can find a closed form for them, or any kind of insight on this, please let me know.

Post Scriptum: I didn't prove any of the results i showed about $P_{n,k}$, so if you got a correction to make, please point that out.

答:
The results you obtained, say for $n=3$, are based on the fact that
$$k^3=k(k-1)(k-2)+3k(k-1)+k$$

So
$$\sum_{k=0}^\infty k^3 a^k=\sum_{k=0}^\infty k(k-1)(k-2) a^k+3\sum_{k=0}^\infty k(k-1)a^k+\sum_{k=0}^\infty ka^k$$ that is to say
$$a^3\sum_{k=0}^\infty k(k-1)(k-2) a^{k-3}+3a^2\sum_{k=0}^\infty k(k-1)a^{k-2}+k\sum_{k=0}^\infty ka^{k-1}$$ that is to say
$$\sum_{k=0}^\infty k^3 a^k=a^3\left(\sum_{k=0}^\infty a^{k} \right)'''+3a^2\left(\sum_{k=0}^\infty a^{k} \right)''+a\left(\sum_{k=0}^\infty a^{k} \right)'$$

Now, for the general case of
$$S_n=\sum_{k=0}^\infty k^n a^k=\Phi (a,-n,0)$$ where appears the Hurwitz-Lerch transcendent function.

This can write
$$S_n=\frac a {(1-a)^{n+1}}P_{n-2}(a)$$

The first polynomials are
$$\left(
\begin{array}{cc}
 n & P_n(a) \\
 1 & 1 \\
 2 & a+1 \\
 3 & a^2+4 a+1 \\
 4 & a^3+11 a^2+11 a+1 \\
 5 & a^4+26 a^3+66 a^2+26 a+1 \\
 6 & a^5+57 a^4+302 a^3+302 a^2+57 a+1 \\
 7 & a^6+120 a^5+1191 a^4+2416 a^3+1191 a^2+120 a+1 \\
 8 & a^7+247 a^6+4293 a^5+15619 a^4+15619 a^3+4293 a^2+247 a+1
\end{array}
\right)$$ where you can first notice that they are palindromic.

The  coefficientcorresponds to  Eulerian numbers (just seach for them in $OEIS$)

Title: What is this polynomial?


问:
Finding lengths related to an altitude of $\triangle PQR$ without Cosine Rule?

I wonder if there is another way to do this.
I did this question by using Cosine Rule.
I tried to use similar triangle, but it seems that there is not enough information to support the property. The best I could is Side and Angle + unknown..

Edited

COSINE RULE

$$102^{2} = 126^{2} + 60^{2} - [ 2*{60}*{126}*Cos{A}] $$
Where A = angle (RPQ)

After finding the angle, I then use trigonometry to find

$$Cos{A} = \frac{PS}{60}$$

答:
Use Heron's formula to find the area, where the semi-perimeter $s$ is $\frac{60+102+126}{2} = 144$.

Then $A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{144(144-60)(144-102)(144-126)} = 3024$, and $A = \frac{1}{2}bh \Rightarrow h = \frac{A}{(1/2)b} = \frac{3024}{1/2 \cdot 126} = 48$, hence $RS = 48$.

$PS$ and $QS$ can be found using Pythagoras: they are $\sqrt{60^2-48^2} = 36$ and $\sqrt{102^2-48^2} = 90$ respectively.

As a sanity check, $PS+QS = 36+90=126$ which is the length of $PQ$.


问:
Uniqueness and existence of solution $\frac{dy}{dx} = \sqrt{x-y}$ given $y(2)=2$
I want to determine if there exists a solution and the uniqueness of it for: $\frac{dy}{dx} = \sqrt{x-y}$ given $y(2)=2$

A solution said that we cannot determine if there exists a solution since it was discontinuous at (2,2). This solution seems to imply that it is possible for x to be a number for example 1.99 and y be 2.00 or x be 2.00 and y be 2.01. I'm confused as to why the textbook makes this sort of assumption - I thought we were just determining the continuity of the point.

If we were to look at the neighborhood then why then if we are given $y(2)=1$ as the condition that we know that the above has a solution? It seems arbitrary as to how far we can go left and right or above and down a certain point to say that it was not continuous?

答:
It turns out that (at least if you want real solutions) the solution is only defined for $x \ge 2$.

Strictly speaking the solution is not a solution at $x=2$, as writing $y' = \ldots$ implies that $y$ is differentiable, and a function differentiable at $x=2$ must be defined in $(2-\epsilon, 2+\epsilon)$ for some $\epsilon > 0$. If $y$ was a solution
in a neighbourhood of $x=2$ with $y(2)=2$, we'd have $y'(2)=0$, and then $x-y < 0$ for
$x < 2$ with $|x-2|$ sufficiently small.  But then $\sqrt{x-y}$ is not real.


问:
Is there a link between two magic squares with the same constant?
For instance we consider the magic squares of order $3$ with the constant $15$.

We can find :
\begin{array}{ | l | c | r | }
     \hline
     8 & 3 & 4 \\ \hline
     1 & 5 & 9 \\ \hline
     6 & 7 & 2 \\
     \hline
    \end{array}
and
\begin{array}{ | l | c | r | }
     \hline
     -2 & 8 & 9 \\ \hline
     16 & 5 & -6 \\ \hline
     1 & 2 & 12 \\
     \hline
    \end{array}

Do we have a tool to transform the first matrix into the second ? I mean that does the invariant (same constant) is sufficient to classify all the magic squares of a given constant under a group's action ?

I know that of it was the same numbers for the two matrices, there exists an action from the group $\mathcal{D}_4$ on the set of magic squares of order $3$.

Thanks in advance !

答:
All magic squares of order 3 are based on the same structure. As you note in the question, actions from $\mathcal{D}_4$, the dihedral group of order 8, permute the numbers while preserving the row and column (and diagonal) sums, and hence preserve the magic property.

There are other things we can do which preserve the magic property. For example, we can add a constant to each cell value, or multiply each value by a constant. Thus we can apply any linear operation $x' = ax + b$ to each value.

To see what else preserves the magic property, we need to look at how magic squares can be produced.

Magic squares are closely related to Latin squares. "A Latin square is a $n\times n$ array filled with n different symbols, each occurring exactly once in each row and exactly once in each column". Eg,

$$\begin{array}{|r|r|r|}
\hline
A & B & C\\
\hline
C & A & B\\
\hline
B & C & A\\
\hline
\end{array}$$

We can create a magic square using a pair of
mutually orthogonal Latin squares, that is, two Latin squares of the same order that when superimposed produce distinct pairs of values in each cell. (An old name for this construction is the Graeco-Latin square, but as Wikipedia mentions, that name is rarely used these days, although you may encounter it in works on magic squares).

The order 3 magic square can be produced by combining

$$\begin{array}{|r|r|r|}
\hline
2 & 0 & 1\\
\hline
0 & 1 & 2\\
\hline
1 & 2 & 0\\
\hline
\end{array}$$

with

$$\begin{array}{|r|r|r|}
\hline
1 & 2 & 0\\
\hline
0 & 1 & 2\\
\hline
2 & 0 & 1\\
\hline
\end{array}$$

which is just the previous matrix reflected vertically. Note that this square is magic, with a constant sum of 3, although it's not a proper magic square because it has repeated values. When we combine the two squares we get

$$\begin{array}{|r|r|r|}
\hline
21 & 02 & 10\\
\hline
00 & 11 & 22\\
\hline
12 & 20 & 01\\
\hline
\end{array}$$

The 1st digit in each cell comes from the corresponding position in the 1st Latin square, the 2nd digit comes from the 2nd Latin square. Notice that all of these ordered pairs are distinct.

Now, we treat each of those values as a number written in base 3. This gives us

$$\begin{array}{|r|r|r|}
\hline
7 & 2 & 3\\
\hline
0 & 4 & 8\\
\hline
5 & 6 & 1\\
\hline
\end{array}$$

which is a valid magic square because there are no repeated values. If we add 1 to each cell, we get

$$\begin{array}{|r|r|r|}
\hline
8 & 3 & 4\\
\hline
1 & 5 & 9\\
\hline
6 & 7 & 2\\
\hline
\end{array}$$

which is your first matrix.

If we use any linear operation of the form $au + bv + c$ on the $(u,v)$ values in a cell, we'll also get a magic square. We just need to be a little careful when choosing $a$ and $b$ to avoid getting repeated values. For example, with $a=7, b=4, c=-6$ we get

$$\begin{array}{|r|r|r|}
\hline
7\cdot2+4\cdot1-6 & 7\cdot0+4\cdot2-6 & 7\cdot1+4\cdot0-6\\
\hline
7\cdot0+4\cdot0-6 & 7\cdot1+4\cdot1-6 & 7\cdot2+4\cdot2-6\\
\hline
7\cdot1+4\cdot2-6 & 7\cdot2+4\cdot0-6 & 7\cdot0+4\cdot1-6\\
\hline
\end{array}$$

or

$$\begin{array}{|r|r|r|}
\hline
12 & 2 & 1\\
\hline
-6 & 5 & 16\\
\hline
9 & 8 & -2\\
\hline
\end{array}$$

which is your 2nd matrix, rotated by 180°.

As I said earlier, all magic squares of order 3 are based on the same structure. Clearly, if you permute the rows or columns of a Latin square, the result is still a Latin square. So given any Latin square we can permute it so that its 1st row and column are in natural order, eg

$$\begin{array}{|r|r|r|}
\hline
0 & 1 & 2\\
\hline
1 & 2 & 0\\
\hline
2 & 0 & 1\\
\hline
\end{array}
$$

This is known as the reduced form.

The Wikipedia article on Small Latin squares and quasigroups states that there is only one reduced Latin square of order three. That's pretty easy to see. If we start with

\begin{array}{|r|r|r|}
\hline
0 & 1 & 2\\
\hline
1 &  & \\
\hline
2 &  & \\
\hline
\end{array}

there's only one possible way to complete the square.

For a magic square, we need the diagonals to sum to the row constant. So we need to permute the rows & columns of the standard Latin square so that the diagonals of our mutually orthogonal Latin squares are either $(0, 1, 2)$ or $(1,1,1)$. It's easy to see that both diagonals cannot be of the same form, so we need to start with an arrangement like

$$\begin{array}{|r|r|r|}
\hline
2 &  & 1\\
\hline
 & 1 & \\
\hline
1 &  & 0\\
\hline
\end{array}$$

and clearly there's only one way to complete that Latin square:

$$\begin{array}{|r|r|r|}
\hline
2 & 0 & 1\\
\hline
0 & 1 & 2\\
\hline
1 & 2 & 0\\
\hline
\end{array}$$

which is the first $(0,1,2)$ Latin square given earlier.

We need to choose $a$ and $b$ to ensure that all the cell values are distinct. Our cell value equation is $x = f(u, v) = au + bv + c$. Given two cells, $(u, v)$ and $(u', v')$, we want $f(u, v) \ne  f(u', v')$.

Let
$$au + bv + c = au' + bv' + c$$
$$a(u-u') = b(v'-v) \tag{1}$$
The possible values of $u-u', v'-v$ are in $\{0, \pm1, \pm2\}$.
We only want (1) to be true when $u-u' = v'-v = 0$. So we just need to choose non-zero $a$ and $b$ such that $a \ne \pm b$, $a \ne \pm 2b$, and $2a \ne \pm b$. We are free to choose any $c$.

Here is a Python script that generates order 3 magic squares using the above method, with MathJax output.

Body:
All magic squares of order 3 are based on the same structure. As you note in the question, actions from $\mathcal{D}_4$, the dihedral group of order 8, permute the numbers while preserving the row and column (and diagonal) sums, and hence preserve the magic property.

There are other things we can do which preserve the magic property. For example, we can add a constant to each cell value, or multiply each value by a constant. Thus we can apply any linear operation $x' = ax + b$ to each value.

To see what else preserves the magic property, we need to look at how magic squares can be produced.

Magic squares are closely related to Latin squares. "A Latin square is a $n\times n$ array filled with n different symbols, each occurring exactly once in each row and exactly once in each column". Eg,

$$\begin{array}{|r|r|r|}
\hline
A & B & C\\
\hline
C & A & B\\
\hline
B & C & A\\
\hline
\end{array}$$

We can create a magic square using a pair of
mutually orthogonal Latin squares, that is, two Latin squares of the same order that when superimposed produce distinct pairs of values in each cell. (An old name for this construction is the Graeco-Latin square, but as Wikipedia mentions, that name is rarely used these days, although you may encounter it in works on magic squares).

The order 3 magic square can be produced by combining

$$\begin{array}{|r|r|r|}
\hline
2 & 0 & 1\\
\hline
0 & 1 & 2\\
\hline
1 & 2 & 0\\
\hline
\end{array}$$

with

$$\begin{array}{|r|r|r|}
\hline
1 & 2 & 0\\
\hline
0 & 1 & 2\\
\hline
2 & 0 & 1\\
\hline
\end{array}$$

which is just the previous matrix reflected vertically. Note that this square is magic, with a constant sum of 3, although it's not a proper magic square because it has repeated values. When we combine the two squares we get

$$\begin{array}{|r|r|r|}
\hline
21 & 02 & 10\\
\hline
00 & 11 & 22\\
\hline
12 & 20 & 01\\
\hline
\end{array}$$

The 1st digit in each cell comes from the corresponding position in the 1st Latin square, the 2nd digit comes from the 2nd Latin square. Notice that all of these ordered pairs are distinct.

Now, we treat each of those values as a number written in base 3. This gives us

$$\begin{array}{|r|r|r|}
\hline
7 & 2 & 3\\
\hline
0 & 4 & 8\\
\hline
5 & 6 & 1\\
\hline
\end{array}$$

which is a valid magic square because there are no repeated values. If we add 1 to each cell, we get

$$\begin{array}{|r|r|r|}
\hline
8 & 3 & 4\\
\hline
1 & 5 & 9\\
\hline
6 & 7 & 2\\
\hline
\end{array}$$

which is your first matrix.

If we use any linear operation of the form $au + bv + c$ on the $(u,v)$ values in a cell, we'll also get a magic square. We just need to be a little careful when choosing $a$ and $b$ to avoid getting repeated values. For example, with $a=7, b=4, c=-6$ we get

$$\begin{array}{|r|r|r|}
\hline
7\cdot2+4\cdot1-6 & 7\cdot0+4\cdot2-6 & 7\cdot1+4\cdot0-6\\
\hline
7\cdot0+4\cdot0-6 & 7\cdot1+4\cdot1-6 & 7\cdot2+4\cdot2-6\\
\hline
7\cdot1+4\cdot2-6 & 7\cdot2+4\cdot0-6 & 7\cdot0+4\cdot1-6\\
\hline
\end{array}$$

or

$$\begin{array}{|r|r|r|}
\hline
12 & 2 & 1\\
\hline
-6 & 5 & 16\\
\hline
9 & 8 & -2\\
\hline
\end{array}$$

which is your 2nd matrix, rotated by 180°.

As I said earlier, all magic squares of order 3 are based on the same structure. Clearly, if you permute the rows or columns of a Latin square, the result is still a Latin square. So given any Latin square we can permute it so that its 1st row and column are in natural order, eg

$$\begin{array}{|r|r|r|}
\hline
0 & 1 & 2\\
\hline
1 & 2 & 0\\
\hline
2 & 0 & 1\\
\hline
\end{array}
$$

This is known as the reduced form.

The Wikipedia article on Small Latin squares and quasigroups states that there is only one reduced Latin square of order three. That's pretty easy to see. If we start with

\begin{array}{|r|r|r|}
\hline
0 & 1 & 2\\
\hline
1 &  & \\
\hline
2 &  & \\
\hline
\end{array}

there's only one possible way to complete the square.

For a magic square, we need the diagonals to sum to the row constant. So we need to permute the rows & columns of the standard Latin square so that the diagonals of our mutually orthogonal Latin squares are either $(0, 1, 2)$ or $(1,1,1)$. It's easy to see that both diagonals cannot be of the same form, so we need to start with an arrangement like

$$\begin{array}{|r|r|r|}
\hline
2 &  & 1\\
\hline
 & 1 & \\
\hline
1 &  & 0\\
\hline
\end{array}$$

and clearly there's only one way to complete that Latin square:

$$\begin{array}{|r|r|r|}
\hline
2 & 0 & 1\\
\hline
0 & 1 & 2\\
\hline
1 & 2 & 0\\
\hline
\end{array}$$

which is the first $(0,1,2)$ Latin square given earlier.

We need to choose $a$ and $b$ to ensure that all the cell values are distinct. Our cell value equation is $x = f(u, v) = au + bv + c$. Given two cells, $(u, v)$ and $(u', v')$, we want $f(u, v) \ne  f(u', v')$.

Let
$$au + bv + c = au' + bv' + c$$
$$a(u-u') = b(v'-v) \tag{1}$$
The possible values of $u-u', v'-v$ are in $\{0, \pm1, \pm2\}$.
We only want (1) to be true when $u-u' = v'-v = 0$. So we just need to choose non-zero $a$ and $b$ such that $a \ne \pm b$, $a \ne \pm 2b$, and $2a \ne \pm b$. We are free to choose any $c$.

<a href="https://sagecell.sagemath.org/?z=eJx9U9uK2zAQfTf4H6YKFCt1c_Nb2JRCoX1aWkqhD4kxsiPb2iayK8vJprv77x1ZvmVT6oCcOZpzNJ45mryZ15Wax0LOuTxBedF5IQPXIYTAFy65YppDofZcQQBHlokEqt81U7xyHcDnpxJacwnxBb7dw-q7kBmsFqvlbLGcBctGx3VcJ2YVhw1sLWm78mHhwzL029gEPqz62ASYYuLQ0Pc8hbLWUabE3svo2qalhQJVnEFIyFrIPKUSUnu4QS1mY_paJzoynT-wx0EvPwhpqlRk1_wlFo7rNDW1IxzzTMgnphS7vDw9q-b3QnzLDP9blVGZsbLkcu8ReAtk9lAI6W0rrbyaNqzacJAaUnhnitgR-m9-cxwdqutw5OCrra8jV1g72cn2PJN-1ZeqS8uLs5fr4wGRFnKdASR3efDha-uD-5EP0BS_uLqb4zah1NJc5yNKo3cSbXveGCfw2CbwId7gdBOzKJ4eeKKjx81ndqj4AFw6QCsmq7KoeAewWhdRXe7RlRbqZjeBT4XEXtaJvrKp3S1qbSY4mlAdYaN9OEXtqP6I0jMmxfpw3a7X75chHY3PpKECm9Y4m3h6wjWxQijSCYxFaTiQ8fh-RI0tu5p_mO9DlSMIbTGRDm1ZXyk0FizOtrSxy3AvvCFfbsi4Wm6f23f39qCDqOwVGh9kPnGKCTTsv6C_lAZubfX6ftkt15m0LqB_ATE7QuI=&lang=sage" rel="nofollow noreferrer">Here is a Python script that generates order 3 magic squares using the above method, with MathJax output.