问:
In a formal axiomatic system with both the law of the excluded middle and the axiom of infinity does an uncountable set have to exist?
This was something I have been thinking around when examining writings of some “philosophers” who seemingly argue against the existence of a continuation but at the same time accept the law of the excluded middle and also the axiom of infinity.

I know from cantor’s proof of the unaccountability of the reals if you have something that obeys the real number axions then the set of all of them will be uncountable (if you have LEM).  However what I am interested in is if you just have the natural numbers and LEM does that entail the existence of some sort of uncountable set.

Clearly introducing the power set axiom would also entail an uncountable set so that isn’t the type of FAS I am interested in.

答:
Note that ACA0 (one of the common base theories in reverse mathematics) is conservative over PA for arithmetical sentences and yet powerful enough to handle essentially all applied real analysis. And ACA0 has a nice countable model where its sets are simply the arithmetical subsets of $ℕ$. There is no set-theoretic infinity axiom in ACA0, but any model of ACA0 obviously has an infinite set (i.e. the model of the first-order subtheory, which is a model of PA). Of course ACA0 has no uncountable sets; it has just two levels after all.

But I don't know whether you realize that even with no set-theoretic axioms, not to say power-set kind of axiom, Cantor's theorem still holds in the following form:

For every binary arithmetical predicate $Q$ there is some unary arithmetical predicate $R$ such that $∀k{∈}ℕ\ ∃m{∈}ℕ\ ¬( \ Q(k,m) ⇔ R(m) \ )$.

Why is this like Cantor's theorem? It says that for every arithmetically definable function $f$ from $ℕ$ to subsets of $ℕ$ (i.e. $f(k) = \{ m : m{∈}ℕ ∧ Q(k,m) \}$ for each $k∈ℕ$) there is some arithmetically definable subset $S$ of $ℕ$ (i.e. $S = \{ m : m{∈}ℕ ∧ R(m) \}$) that is not in the range of $f$. So, clearly, the arithmetically definable subsets of $ℕ$ are 'uncountable' in the sense of being not enumerable by any arithmetically definable function.

The proof mirrors that of the usual Cantor's theorem, which I shall leave as an easy exercise.


问:
Validity of this particular "Chain Rule"
For two multivariable function $y(m,n)$ and $x(m,n)$ does it make sense/is it mathematically correct to talk of the following chain rules -

$$\frac{dy}{dx}=\frac{\partial y}{\partial m}\frac{dm}{dx}+\frac{\partial y}{\partial n}\frac{dn}{dx}\tag{1}$$
and
$$\frac{dx}{dy}=\frac{\partial x}{\partial m}\frac{dm}{dy}+\frac{\partial x}{\partial n}\frac{dn}{dy}\tag{2}$$

simultaneously, where m, n can be rewritten as $m=m(x,y),\,n=n(x,y)$ and $d/dx,\,d/dy$ of m, n exists.

The context is this PhysicsSE question - if that is relevant.

How correct/wrong is this "Chain rule"?

Any intuitive/geometric argument together with a mathematical proof, if possible, is highly appreciated.

答:
The expression $\frac{dy}{dx}$ suggests that we have a rule which associates a value of $x$ to a single, well-defined value of $y$ - let's say $y=g(x)$ - and that we are differentiating it, so $\frac{dy}{dx} = g'(x)$.  In your example, you have one function of two variables $f:\mathbb R^2 \rightarrow \mathbb R$ and two functions of one variable $m,n : \mathbb R \rightarrow \mathbb R$; $g$ is then given by their composition,
$$y = g(x) = f\bigg(m(x),n(x)\bigg)$$
$$\implies \frac{dy}{dx}= g'(x) = (\partial_1f)\big(m(x),n(x)\big)\cdot m'(x) + (\partial_2f)\big(m(x),n(x)\big) \cdot n'(x)$$
where $\partial_i f$ is the partial derivative of $f$ with respect to its $i^{th}$ entry.  Standard convention is to suppress the arguments and associate each entry with the function we plug into it, which leads to the prettier expression
$$\frac{dy}{dx} = g'(x) = \frac{\partial f}{\partial m}  \frac{dm}{dx} + \frac{\partial f}{\partial n} \frac{dn}{dx}$$

It is entirely possible that $g$ can be inverted, at least locally.  This would imply the existence of a function $\tilde g$ such that $\tilde g(g(x)) = g(\tilde g(x)) = x$, which means that we can write $x = \tilde g(y)$.  It's not hard to show that $\tilde g'(y) = \frac{1}{g'\big(\tilde g(y)\big)}$ as long as $g'\big(\tilde g(y)\big)\neq 0$; in such cases, we would have that
$$\frac{dx}{dy} = \tilde g'(y) = \frac{1}{g'\big(\tilde g(y)\big)} = \left(\left.\frac{dy}{dx}\right|_{x=\tilde g(y)}\right)^{-1}$$

However, this $\tilde g$ will generally not be able to expressed in terms of the functions $f$, $m$, and $n$ in any meaningful way.  Take for example $f(m,n)=m^2n$, $m(x)=x$, $n(x)=x^4$.  Then $y=g(x) = f\big(m(x),n(x)\big) = x^6$, and $\frac{dy}{dx} = 6x^5$.

Inverting this, $x = \tilde g(y) = y^{1/6}$, so $\frac{dx}{dy} = \frac{1}{6 y^{5/6}} = \left(6 x^5\right)^{-1}$, and all is well.

Your expression (2) implies that $\tilde g(y)$ can be expressed as $F\big(m(y),n(y)\big)$ for yet another function $F:\mathbb R^2\rightarrow \mathbb R$.  In such a case, we would indeed have (skipping right to the pretty notation),
$$\frac{dx}{dy} = \tilde g'(y) = \frac{\partial F}{\partial m}\frac{dm}{dy} + \frac{\partial F}{\partial n}\frac{dn}{dy}$$

In my simple example above, note that $\tilde g(y) = y^{1/6}$, so we could let $F(m,n) = m^{1/6}$.  Of course, this $F$ is not unique; $F(m,n)=n^{1/12}$ and $F(m,n) = m^{1/12} n^{1/24}$ would also work, and would also give us the result we're looking for.

More generally, if $m$ is invertible with inverse $\tilde m$, we could let $F(m,n) = \tilde g\big(\tilde m (m)\big)$, which would yield the result we're looking for.

Of course, the above is what I would consider idle mathematical curiosity.  Given a relationship $y = g(x) = f\big(m(x),n(x)\big)$, it is broadly possible to find a function $F$ such that $x = \tilde g(y) = F\big(m(y),n(y)\big)$. This $F$ is highly non-unique in general, and I can think of no particularly good reason to do this, but you can bend over backward to do so if you really want to.  However, $F$ could not be determined from $f$ alone; it would inevitably involve the inverse function of $m$, $n$, or both.

ContentLicense: CC BY-SA 4.0


问:
A simple $n\times n$ system of linear equations
Can anyone provide the analytical solution of the following $n\times n$ system of linear equations

$\underbrace{\begin{pmatrix}
 a & -b & -b & ... & -b\\
 -b & a & -b & ... & -b\\
 -b & -b & a & ... & -b\\
  : &  : & : & ... & : \\
  : & : & : & ... &: & \\
 -b & -b & -b & ... & a \\
 \end{pmatrix}}_{A}\underbrace{\begin{pmatrix}x_1\\
x_2\\
x_3\\
:\\
:\\
x_n
\end{pmatrix}}_{X}=\underbrace{\begin{pmatrix}y_1\\
y_2\\
y_3\\
:\\
:\\
y_n
\end{pmatrix}}_{Y}
$
where $A_{(n\times n)}$ is a symmetric matrix and $a$, $b$, $y_i$ are positive real numbers.

Note: I would appreciate if someone could provide more details about the theory of eigenvalues-eigenvectors related to the solutuion of a linear system of eaquations.

答:
Hint

Use linear combinations in the main matrix:

Take every line (except the first) minus the first:

$$\begin{pmatrix}
 a & -b & -b & ... & -b\\
 -(a+b) & a+b & 0 & ... & 0\\
 -(a+b) & 0 & a+b & ... & 0\\
  : &  : & : & ... & : \\
  : & : & : & ... &: & \\
 -(a+b) & 0 & 0 & ... & a+b \\
 \end{pmatrix}
$$

Now sum at the first column, the sum of the other columns:

$$\begin{pmatrix}
 a-(n-1)b & -b & -b & ... & -b\\
 0 & a+b & 0 & ... & 0\\
 0 & 0 & a+b & ... & 0\\
  : &  : & : & ... & : \\
  : & : & : & ... &: & \\
 0 & 0 & 0 & ... & a+b \\
 \end{pmatrix}
$$

Can you finish?


问:
Prove that the amount {0.1} ∗ of all bit-strings are countable
Let B be the set of (final) bit strings starting with 1, in addition to the empty string. Show on B is countable. Tip: a bit string can be considered as a representation of a natural number. For example. the bit string 101 represents the number 20 ∗ 1 + 21 ∗ 0 + 22 ∗ 1 = 5. In general: bn ··· b1b0 represents 20 ∗ b0 +21 ∗ b1 + ··· + 2n ∗ bn.

And heres where I'm really struggeling. How do I prove that the amount {0.1} ∗ of all bit-strings are countable?

答:
By the given tip, every integer corresponds to a unique string and every string corresponds to a unique integer (if two strings differ in some bits, the integers differ in a sum/difference of distinct powers of $2$, which cannot be zero).

This proves countability.


问:
Cusps of the modular forms

According to the definition of the cusps attached above the set of cusps is infinite and to be more precise it is $Q \cup \infty$ since $G$ is a subgroup of $SL(2,Z)$ so the identity matrix (that is a parabolic element) belongs to $G$ so every point of $Q \cup \infty$ is cusp but they mentioned before that it is finite.

How could this be explained? (In addition, what is cusp?)

答:
$SL_2(\Bbb{Z}) i\infty = \Bbb{Q}\cup i\infty$

$SL_2(\Bbb{Z}) i\infty $ is only one point of the modular curve, the one that you need to add to get a compact Riemann surface. The chart is $q\to SL_2(\Bbb{Z}) \frac{\log(q)}{2i\pi}$ for $|q|<1/3$

It works the same way for $\Bbb{Q}\cup i\infty=\bigcup_{j=1}^J G \gamma_j i\infty$ a finite disjoint union, for some $\gamma_j \in SL_2(\Bbb{Z})$, and we need to add $J$ points to $G\backslash \Bbb{H}$ to get a compact Riemann surface. The charts are $q\to G \gamma_j k_j\frac{\log(q)}{2i\pi }$ for some $k_j$.

This is what we mean with the cusps of the modular curves.

Title: Cusps of the modular forms


问:
How to approximate a sharply peaked function with deltas?
I have the function

$$
f(x)=\cases{\frac{1}{\sqrt{\sinh^2(a)-\sinh^2(ax)}} & $-1 \leq x \leq1$ \\ 0 & otherwise}
$$

Where $a \in \mathbb{R}^+$. Here is a plot of $f$:

It would be very convenient if I could justify (in any way: maybe distributionally?) that

$$
f(x)=A\left[ \delta(x-1)+\delta(x+1) \right]+B+\mathcal{O}(?) \quad ; \quad a \to \infty \tag{1}
$$

For some $x$ independent constants $A$ and $B$. I am uncertain of how to do this, or how to estimate the error.

What I've tried

My thought was to look at the Fourier series of $f$, and associate terms with the Fourier series of (2). By expanding (1) around $x= \pm1$, I can find a Laurent series in $(x \pm 1)$, and perform the Fourier integral term by term. The result is messy, and worse: does not allow easy comparison to the Fourier series of (2). This may not be a fruitful approach.

By looking at the wiki page about mollifiers, it seems I need to claim something like: $a f(x) \to \delta (x \pm1), \ a \to \infty$. Superficially, this appears to work, but I'm not sure how to make it concrete, or if there should be any other factors next to $af(x)$ (other than a factor to ensure normalization to unity).

Questions

How can I estimate the error and find $A$ and $B$ in (1)?

If $A$ and $B$ cannot be found, how can I find the ratio $A/B$?

Background

$f(x)$ will eventually appear beneath an integral with a Green's function $G$. I'd like to formulate this for arbitrary $G$. We can take some liberties about the 'niceness' of $G$, as necessary. In particular, it has a well behaved Fourier transform.

答:
The first thing that must be done with $f(x)$ is to make it zero for $x=0$. Redefine:
$$
f^*(x)=\cases{1/\sqrt{\sinh^2(a)-\sinh^2(ax)}-1/\sinh(a) & $-1 \leq x \leq 1$ \\ 0 & otherwise}
$$
The next step is to divide it by the area underneath it.
With other words, $f^*(x)$ shall be normed, with a number called $N$:
$$
N(a) = \int_{-1}^{+1} f^*(x)\,dx = \int_{-1}^{+1} \left[\frac{1}{\sqrt{\sinh^2(a)-\sinh^2(ax)}}-\frac{1}{\sinh(a)}\right] dx
$$
Calculating the integral turns out to be a cumbersome job. It can hardly be done by hand.
Therefore we have invoked a computer algebra system, MAPLE to be precise:

```
> f(x,a) := 1/sqrt(sinh(a)^2-sinh(a*x)^2)-1/sinh(a);
> for k from 0 to 9 do evalf(int(subs(a=k+1,f(x,a)),x=-1..+1)) end do;
```

It is thus ensured that the area beneath the iterands $f^*(x)/N(a)$ is always unity.
The numbers $N(a)$ have been plugged into a computer program. Here is a plot of the convergents of $f^*(x)$ for these
parameter values $\,a=1\cdots 10\,$. Scaling is almost the same as in the OP's picture, but the $y$-axis has been elongated to $y \le 9$,
to demonstrate that the approximations do indeed "peak" near $x = \pm 1$.

The function is symmetric around $x=0$. It remains to be shown that:
$$ \begin{cases}
\lim_{a\to\infty} f^*(x = \pm 1)/N(a) = +\infty \\
\lim_{a\to\infty} f^*(x \ne \pm 1)/N(a) = 0
\end{cases} $$
Symbolically for the first case, because that limit does not really "exist".
But anyway the first case is trivial, because of the zero denominator.
The second case (a "true" limit) seems to be much more difficult.
Despite of the last technicality - to be repaired later eventually - we jump to conclusions:
$$
\lim_{a\to\infty} \frac{f^*(x)}{N(a)} = \frac{1}{2}\left[\,\delta(x+1)+\delta(x-1)\,\right]
$$
It is not possible to put this in a form as seems to be required in the question.

Later

In order to be able to proceed, we want to have an impression of the behaviour of the norming integral $N(a)$.
The problem is that $N(a)$ is a (too) quickly decreasing function, spoiling the party when trying to prove that
$\;\lim_{a\to\infty} f^*(x \ne \pm 1)/N(a) = 0\;$. Consider instead:
$$
N^*(a) = N(a)\,a\,e^a
$$
Our hope is that $\,N^*(a)\,$ is more or less a constant as a function of $a$.
Numerical experiments (with MAPLE) reveal that such indeed may be the case:
```
> f(x,a) := 1/sqrt(sinh(a)^2-sinh(a*x)^2)-1/sinh(a);
> for k from 0 to 19 do evalf(int(subs(a=k+1,f(x,a)),x=-1..+1)*(k+1)*exp(k+1),30) end do; evalf(ln(16),30);
```
Resulting in a sequence of numbers that seems to converge to $2.772588722239 \approx \ln(16)$.

Next consider the series expansion of $f^*(x)$ around $x=0$:
```
> series(f(x,a),x=0);
```
Resulting in:
$$
f^*(x) = \frac{1}{2}\frac{a^2}{\sinh(a)^3}x^2
+ \left[\frac{1}{6}\frac{a^4}{\sinh(a)^3} + \frac{3}{8}\frac{a^4}{\sinh(a)^5}\right]x^4 + O(x^6)
$$
For large $a > 0$ we know that $\,\sinh(a)\approx e^a\,$. Thus giving:
$$
f^*(x) \approx \frac{1}{2}a^2 e^{-3a}x^2 + \left[\frac{1}{6} a^4 e^{-3a} + \frac{3}{8} a^4 e^{-5a}\right]x^4 + O(x^6)
$$
Time to plug in our conjecturial approximation for the norming integral:
$$
\frac{f^*(x)}{N(a)} = \frac{f^*(x)}{N^*(a)}a\,e^a \approx \frac{a\,e^a}{\ln(16)} f^*(x) \approx \\
\frac{a^3 e^{-2a}}{2\ln(16)}x^2 + \left[\frac{a^5 e^{-2a}}{6\ln(16)} + \frac{3 a^5 e^{-4a}}{8\ln(16)}\right]x^4 + O(x^6)
$$
Which shows that $f^*(x)/N(a)$ is quickly (exponentially) decreasing to zero for $a\to\infty$ and $|x|\lt 1$, the limit that was "to be repaired later" :-)

The finishing touch is provided with
contributions by DinosaurEgg and Claude Leibovici:
$\lim_{a\to\infty} N^*(a) = \ln(16)\;$ does indeed hold.

At last

This response is intended to provide a far more detailed heuristic account of the mathematical approach as employed by @DinosaurEgg in a previous
answer.

The function as prescribed in the question does not converge to a delta-function in the limit $a\to\infty$.
However it converges to one with appropriate modifications:
$$
f^*(x)=N(a)\left[\frac{1}{\sqrt{\sinh^2(a)-\sinh^2(ax)}}-\frac{1}{\sinh(a)}\right] \\
N(a) = \int_{-1}^{+1} \left[\frac{1}{\sqrt{1-\left[\,\sinh(ax)/\sinh(a)\,\right]^2}}-1\right]\frac{dx}{\sinh(a)}
$$
A good thing is performing the proposed change of variables:
$$
\sinh(ax)=\sinh(a)\tanh(t) \quad \Longrightarrow \\
\frac{1}{\sqrt{1-\left[\sinh(ax)/\sinh(a)\right]^2}} = \frac{1}{\sqrt{1-\tanh^2(t)}} \\
= \frac{1}{\sqrt{\left[\cosh^2(t)-\sinh^2(t)\right]/\cosh^2(t)}}=\cosh(t) \\
\cosh(ax)=\sqrt{1+\sinh^2(ax)}=\sqrt{1+\sinh^2(a)\tanh^2(t)}
$$
With implicit differentiation we get:
$$
h(x,t) = \frac{\sinh(ax)}{\sinh(a)}-\tanh(t) = 0 \quad \Longrightarrow \\
dh = \frac{\partial h}{\partial x}dx + \frac{\partial h}{\partial t}dt = a\frac{\cosh(ax)}{\sinh(a)}dx - \frac{1}{\cosh^2(t)}dt = 0
\quad \Longrightarrow \\ dx = \frac{dt}{a\cosh^2(t)\cosh(ax)/\sinh(a)}= \frac{\sinh(a)}{a}\frac{dt}{\cosh^2(t)\sqrt{1+\sinh^2(a)\tanh^2(t)}}
$$
Change of integration bounds:
$$
\tanh(-\infty)=-1 \quad ; \quad \tanh(+\infty)=+1
$$
It follows that:
$$
N(a) = \frac{1}{a}\int_{-\infty}^{+\infty} \frac{\left[\,\cosh(t)-1\,\right]dt}{\cosh^2(t)\sqrt{1+\sinh^2(a)\tanh^2(t)}} \\
= \frac{1}{a\sinh(a)}\int_{-\infty}^{+\infty} \frac{\left[\,\cosh(t)-1\,\right]dt}{\cosh^2(t)\sqrt{1/\sinh^2(a)+\tanh^2(t)}}
= \frac{1}{a\sinh(a)}N^*(a)
$$
By previous heuristics, it has been assumed that the following limit exists:
$$
\lim_{a\to\infty} N^*(a) = \int_{-\infty}^{+\infty} \lim_{a\to\infty} \frac{\left[\,\cosh(t)-1\,\right]dt}{\cosh^2(t)\sqrt{1/\sinh^2(a)+\tanh^2(t)}} \\
= \int_{-\infty}^{+\infty}\frac{\left[\,\cosh(t)-1\,\right]dt}{\cosh^2(t)\sqrt{\tanh^2(t)}} =
\int_{-\infty}^{+\infty}\frac{\left[\,\cosh(t)-1\,\right]dt}{\cosh^2(t)|\tanh(t)|}
$$
Take care of the absolute value $\,|\tanh(t)|\,$! This means that the integrand is symmetric in $t$, so we can write:
$$
\lim_{a\to\infty} N^*(a) = 2\int_0^\infty\frac{\cosh(t)-1}{\cosh(t)\sinh(t)}\,dt
$$
We could have asked MAPLE for some help:

```
> 2*int((cosh(t)-1)/(cosh(t)*sinh(t)),t=0..infinity);
```

And the outcome is quite in agreement wth our prediction (because of $\,\sinh(a) \to e^a/2\,$):
$$
\lim_{a\to\infty} N^*(a) = 2\ln(2) = \ln(16)/2
$$
But let us not be lazy and evalutate that last integral by hand.
Remember l'Hôpital's rule for the last part:
$$
\sinh(2t)=2\sinh(t)\cosh(t) \quad \quad \Longrightarrow \quad \sinh(t)=2\sinh(t/2)\cosh(t/2) \quad \Longrightarrow \\
\int_0^\infty\left[\frac{1}{\sinh(t)}-\frac{1}{\cosh^2(t)\tanh(t)}\right]dt = \\
\int_0^\infty\left[\frac{d(t/2)}{\tanh(t/2)\cosh^2(t/2)}-\frac{dt}{\tanh(t)\cosh^2(t)}\right]dt \\
= \left[\;\ln(\tanh(t/2))-\ln(\tanh(t))\;\right]_0^\infty = \left[\;\ln\left(\frac{\tanh(t/2)}{\tanh(t)}\right)\right]_0^\infty \\
= \ln\left(\lim_{t\to\infty}\frac{\tanh(t/2)}{\tanh(t)}\right) - \ln\left(\lim_{t\to 0}\frac{\tanh(t/2)}{\tanh(t)}\right) \\
= \ln(1)-\ln\left(\lim_{t\to 0}\frac{\tanh(t/2)}{\tanh(t)}\right)=-\ln\left(\lim_{t\to 0}\frac{(1/2)/\cosh^2(t/2)}{1/\cosh^2(t)}\right) \\
= -\ln(1/2)=\ln(2)
$$

Body:
The first thing that must be done with $f(x)$ is to make it zero for $x=0$. Redefine:
$$
f^*(x)=\cases{1/\sqrt{\sinh^2(a)-\sinh^2(ax)}-1/\sinh(a) & $-1 \leq x \leq 1$ \\ 0 & otherwise}
$$
The next step is to divide it by the area underneath it.
With other words, $f^*(x)$ shall be normed, with a number called $N$:
$$
N(a) = \int_{-1}^{+1} f^*(x)\,dx = \int_{-1}^{+1} \left[\frac{1}{\sqrt{\sinh^2(a)-\sinh^2(ax)}}-\frac{1}{\sinh(a)}\right] dx
$$
Calculating the integral turns out to be a cumbersome job. It can hardly be done by hand.
Therefore we have invoked a computer algebra system, MAPLE to be precise:

> f(x,a) := 1/sqrt(sinh(a)^2-sinh(a*x)^2)-1/sinh(a);
> for k from 0 to 9 do evalf(int(subs(a=k+1,f(x,a)),x=-1..+1)) end do;

It is thus ensured that the area beneath the iterands $f^*(x)/N(a)$ is always unity.
The numbers $N(a)$ have been plugged into a computer program. Here is a plot of the convergents of $f^*(x)$ for these
parameter values $\,a=1\cdots 10\,$. Scaling is almost the same as in the OP's picture, but the $y$-axis has been elongated to $y \le 9$,
to demonstrate that the approximations do indeed "peak" near $x = \pm 1$.

The function is symmetric around $x=0$. It remains to be shown that:
$$ \begin{cases}
\lim_{a\to\infty} f^*(x = \pm 1)/N(a) = +\infty \\
\lim_{a\to\infty} f^*(x \ne \pm 1)/N(a) = 0
\end{cases} $$
Symbolically for the first case, because that limit does not really "exist".
But anyway the first case is trivial, because of the zero denominator.
The second case (a "true" limit) seems to be much more difficult.
Despite of the last technicality - to be repaired later eventually - we jump to conclusions:
$$
\lim_{a\to\infty} \frac{f^*(x)}{N(a)} = \frac{1}{2}\left[\,\delta(x+1)+\delta(x-1)\,\right]
$$
It is not possible to put this in a form as seems to be required in the question.

Later

In order to be able to proceed, we want to have an impression of the behaviour of the norming integral $N(a)$.
The problem is that $N(a)$ is a (too) quickly decreasing function, spoiling the party when trying to prove that
$\;\lim_{a\to\infty} f^*(x \ne \pm 1)/N(a) = 0\;$. Consider instead:
$$
N^*(a) = N(a)\,a\,e^a
$$
Our hope is that $\,N^*(a)\,$ is more or less a constant as a function of $a$.
Numerical experiments (with MAPLE) reveal that such indeed may be the case:

> f(x,a) := 1/sqrt(sinh(a)^2-sinh(a*x)^2)-1/sinh(a);
> for k from 0 to 19 do evalf(int(subs(a=k+1,f(x,a)),x=-1..+1)*(k+1)*exp(k+1),30) end do; evalf(ln(16),30);

Resulting in a sequence of numbers that seems to converge to $2.772588722239 \approx \ln(16)$.

Next consider the series expansion of $f^*(x)$ around $x=0$:

> series(f(x,a),x=0);

Resulting in:
$$
f^*(x) = \frac{1}{2}\frac{a^2}{\sinh(a)^3}x^2
+ \left[\frac{1}{6}\frac{a^4}{\sinh(a)^3} + \frac{3}{8}\frac{a^4}{\sinh(a)^5}\right]x^4 + O(x^6)
$$
For large $a > 0$ we know that $\,\sinh(a)\approx e^a\,$. Thus giving:
$$
f^*(x) \approx \frac{1}{2}a^2 e^{-3a}x^2 + \left[\frac{1}{6} a^4 e^{-3a} + \frac{3}{8} a^4 e^{-5a}\right]x^4 + O(x^6)
$$
Time to plug in our conjecturial approximation for the norming integral:
$$
\frac{f^*(x)}{N(a)} = \frac{f^*(x)}{N^*(a)}a\,e^a \approx \frac{a\,e^a}{\ln(16)} f^*(x) \approx \\
\frac{a^3 e^{-2a}}{2\ln(16)}x^2 + \left[\frac{a^5 e^{-2a}}{6\ln(16)} + \frac{3 a^5 e^{-4a}}{8\ln(16)}\right]x^4 + O(x^6)
$$
Which shows that $f^*(x)/N(a)$ is quickly (exponentially) decreasing to zero for $a\to\infty$ and $|x|\lt 1$, the limit that was "to be repaired later" :-)

The finishing touch is provided with
contributions by DinosaurEgg and Claude Leibovici:
$\lim_{a\to\infty} N^*(a) = \ln(16)\;$ does indeed hold.

At last

This response is intended to provide a far more detailed heuristic account of the mathematical approach as employed by @DinosaurEgg in a previous
answer.

The function as prescribed in the question does not converge to a delta-function in the limit $a\to\infty$.
However it converges to one with appropriate modifications:
$$
f^*(x)=N(a)\left[\frac{1}{\sqrt{\sinh^2(a)-\sinh^2(ax)}}-\frac{1}{\sinh(a)}\right] \\
N(a) = \int_{-1}^{+1} \left[\frac{1}{\sqrt{1-\left[\,\sinh(ax)/\sinh(a)\,\right]^2}}-1\right]\frac{dx}{\sinh(a)}
$$
A good thing is performing the proposed change of variables:
$$
\sinh(ax)=\sinh(a)\tanh(t) \quad \Longrightarrow \\
\frac{1}{\sqrt{1-\left[\sinh(ax)/\sinh(a)\right]^2}} = \frac{1}{\sqrt{1-\tanh^2(t)}} \\
= \frac{1}{\sqrt{\left[\cosh^2(t)-\sinh^2(t)\right]/\cosh^2(t)}}=\cosh(t) \\
\cosh(ax)=\sqrt{1+\sinh^2(ax)}=\sqrt{1+\sinh^2(a)\tanh^2(t)}
$$
With implicit differentiation we get:
$$
h(x,t) = \frac{\sinh(ax)}{\sinh(a)}-\tanh(t) = 0 \quad \Longrightarrow \\
dh = \frac{\partial h}{\partial x}dx + \frac{\partial h}{\partial t}dt = a\frac{\cosh(ax)}{\sinh(a)}dx - \frac{1}{\cosh^2(t)}dt = 0
\quad \Longrightarrow \\ dx = \frac{dt}{a\cosh^2(t)\cosh(ax)/\sinh(a)}= \frac{\sinh(a)}{a}\frac{dt}{\cosh^2(t)\sqrt{1+\sinh^2(a)\tanh^2(t)}}
$$
Change of integration bounds:
$$
\tanh(-\infty)=-1 \quad ; \quad \tanh(+\infty)=+1
$$
It follows that:
$$
N(a) = \frac{1}{a}\int_{-\infty}^{+\infty} \frac{\left[\,\cosh(t)-1\,\right]dt}{\cosh^2(t)\sqrt{1+\sinh^2(a)\tanh^2(t)}} \\
= \frac{1}{a\sinh(a)}\int_{-\infty}^{+\infty} \frac{\left[\,\cosh(t)-1\,\right]dt}{\cosh^2(t)\sqrt{1/\sinh^2(a)+\tanh^2(t)}}
= \frac{1}{a\sinh(a)}N^*(a)
$$
By previous heuristics, it has been assumed that the following limit exists:
$$
\lim_{a\to\infty} N^*(a) = \int_{-\infty}^{+\infty} \lim_{a\to\infty} \frac{\left[\,\cosh(t)-1\,\right]dt}{\cosh^2(t)\sqrt{1/\sinh^2(a)+\tanh^2(t)}} \\
= \int_{-\infty}^{+\infty}\frac{\left[\,\cosh(t)-1\,\right]dt}{\cosh^2(t)\sqrt{\tanh^2(t)}} =
\int_{-\infty}^{+\infty}\frac{\left[\,\cosh(t)-1\,\right]dt}{\cosh^2(t)|\tanh(t)|}
$$
Take care of the absolute value $\,|\tanh(t)|\,$! This means that the integrand is symmetric in $t$, so we can write:
$$
\lim_{a\to\infty} N^*(a) = 2\int_0^\infty\frac{\cosh(t)-1}{\cosh(t)\sinh(t)}\,dt
$$
We could have asked MAPLE for some help:

> 2*int((cosh(t)-1)/(cosh(t)*sinh(t)),t=0..infinity);

And the outcome is quite in agreement wth our prediction (because of $\,\sinh(a) \to e^a/2\,$):
$$
\lim_{a\to\infty} N^*(a) = 2\ln(2) = \ln(16)/2
$$
But let us not be lazy and evalutate that last integral by hand.
Remember l'Hôpital's rule for the last part:
$$
\sinh(2t)=2\sinh(t)\cosh(t) \quad \quad \Longrightarrow \quad \sinh(t)=2\sinh(t/2)\cosh(t/2) \quad \Longrightarrow \\
\int_0^\infty\left[\frac{1}{\sinh(t)}-\frac{1}{\cosh^2(t)\tanh(t)}\right]dt = \\
\int_0^\infty\left[\frac{d(t/2)}{\tanh(t/2)\cosh^2(t/2)}-\frac{dt}{\tanh(t)\cosh^2(t)}\right]dt \\
= \left[\;\ln(\tanh(t/2))-\ln(\tanh(t))\;\right]_0^\infty = \left[\;\ln\left(\frac{\tanh(t/2)}{\tanh(t)}\right)\right]_0^\infty \\
= \ln\left(\lim_{t\to\infty}\frac{\tanh(t/2)}{\tanh(t)}\right) - \ln\left(\lim_{t\to 0}\frac{\tanh(t/2)}{\tanh(t)}\right) \\
= \ln(1)-\ln\left(\lim_{t\to 0}\frac{\tanh(t/2)}{\tanh(t)}\right)=-\ln\left(\lim_{t\to 0}\frac{(1/2)/\cosh^2(t/2)}{1/\cosh^2(t)}\right) \\
= -\ln(1/2)=\ln(2)
$$


问:
Bijection between $(0,1)$ and the set of infinite binary sequence
I have been searching this problem in the internet for a while but I am still unable to get around the infinite 0's and 1's problem.

So far I have gathered some information. Please correct me if I am wrong:

$\forall x\in (0,1)$, there exists decimal and binary expression.

$\forall x\in (0,1)$, the expansion are in 4 forms: repeating, non-repeating, infinite 0's and 1's

Schroder-Bernstein theorem which states two injections between two sets implies same cardinality between the two sets

However, I still can't establish an bijection between $(0,1)$ and the set of infinite binary strings. Is it possible to show how?

答:
Every number in $(0,1)\setminus \bigcup_{n\in \mathbb{N}}\left\{ \frac{k}{2^n}: \ k=1,2,\ldots 2^{n-1} \right\}$ has unique binary expansion. The set $A:= \bigcup_{n\in \mathbb{N}}\left\{ \frac{k}{2^n}: \ k=1,2,\ldots 2^{n-1} \right\}$ is countable as a countable union of finite sets.

Now every number of the form $\frac{k}{2^n}$, $n\in \mathbb{N}$, has two expansions one of them ends with infinite zros (for example $\frac{1}{2} = \frac{1}{2} + \sum_{i=2}^{\infty} \frac{0}{2^i} = 0 + \sum_{i=2}^{\infty} \frac{1}{2^i}$). Furthermore, if a binary expansion of an $x$ ends with infinite zeros then $x$ clearly is in $A$. ($\ast$)

Also if $(x_1, x_2, \ldots ) \in \{ 0,1 \}^{\mathbb{N}}$, then $x:=\sum_{i=1}^{\infty} \frac{x_i}{2^i} \in (0,1)$. ($\ast \ast$)

So define $h:\{ 0,1 \}^{\mathbb{N}} \to (0,1)$ by $h((x_1, x_2, \ldots ))=x:=\sum_{i=1}^{\infty} \frac{x_i}{2^i} \in (0,1)$

Then $h$ is onto and thus $|\{ 0,1 \}^{\mathbb{N}}| \geq |(0,1)|$. In particular $\{ 0,1 \}^{\mathbb{N}}$ is uncountable.

Now set $B\subset \{0,1\}^{\mathbb{N}}$, the set of the sequences that end with infinite zeros, i.e. the set the binary expansions of each $x\in A$ such that it ends with infinite zeros.

Observe that the restriction  $h: \{ 0,1 \}^{\mathbb{N}}\setminus B \to (0,1) $ is, by ($\ast$) and ($\ast \ast$), a bijection.
Thus $|\{ 0,1 \}^{\mathbb{N}}\setminus \{1^n0^{\infty}: \ n\in \mathbb{N}\}| = |(0,1)|$

Also $\{ 0,1 \}^{\mathbb{N}} = \{ 0,1 \}^{\mathbb{N}}\setminus B \ \dot\bigcup \ B$, disjoint union, and the last set is countable.
Hence, $|\{ 0,1 \}^{\mathbb{N}}| = |\{ 0,1 \}^{\mathbb{N}}\setminus B| = |(0,1)|$


问:
Mellin transform Gaussian
Is there a known formula for the Mellin transform
$$M(s) = \int_0^\infty f(x)x^{s-1}\, dx$$
of the Gaussian probability density function

$$f(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac 12\left(\frac{x-\mu}{\sigma}\right)^2}$$

???

If the answer is yes, I would like to see a proof ;-)

答:
For $\Re(s)>0$ with $y=x^2$
$$\int_0^\infty x^{s-1}e^{-x^2}dx=\frac12\int_0^\infty y^{s/2-1}e^{-y}dy=\frac12 \Gamma(s/2)$$
Then
$$\int_0^\infty x^{s-1}e^{-(x-a)^2}dx= e^{-a^2}\sum_{k\ge 0} \frac{(2a)^k}{k!} \frac12 \Gamma((s+k)/2)$$ is some kind of hypergeometric function

ContentLicense: CC BY-SA 3.0

Title: Mellin transform Gaussian


问:
Prove there exist infinite value of $n$ such that $p\mid n2^{n}+1$
Let $p$ be an odd prime number. Prove that there exist infinite values of $n\in \mathbb{N}$ such that: $p\mid n2^n +1$.

$\textbf{My attempt}$: I have proved that $p\mid (p-1)2^{p-1}+1$. Therefore, we first define $n=p-1$, and then $n\mid n2^n +1$ holds. Since by cungruences $p\mid (p-1)2^{p-1}+1$ means that $(p-1)2^{p-1}\equiv -1 \equiv p-1 \pmod{p}$. Now, since $(p,p-1)=1$ because they are consecutive, we can multiply the equation by $p-1$, without any changes in the modulo, and we get $(p-1)^{2}\cdot 2^{p-1}\equiv (p-1)^2 \pmod{p}$. Recall that $p-1$ is the invers of itself modulo $p$ if and only if $p$ is prime. Thus, $$(p-1)^{2}\cdot 2^{p-1}\equiv (p-1)^2 \equiv 1\pmod{p}$$
We multiply again by $p-1$, and therefore: $$(p-1)^3 \cdot 2^{p-1}\equiv -1 \pmod{p}$$
By little's Fermat theorem,
$$(2^{p-1})^{(p-1)^2}\equiv 1 \pmod{p}$$
We conclude that: $$(p-1)^3\cdot 2^{(p-1)^3}\equiv -1\pmod{p}$$
So again we can define $n=(p-1)^3\neq p-1$, and by that we get another different value of $n$, such that $p\mid n2^{n}+1$, and so we proceed til we get infinite different numbers of $n$. $\textbf{Q.E.D}$

Now, I'm asking if I got it right because I showed an algorithm, such that $p\mid n2^n+1$, for each odd power of $p-1$, but I haven't written that down. Thanks!

答: If $p\mid n2^n+1$, then $p\mid (n+kp(p-1))2^{n+kp(p-1)}+1$ for all $k\in\mathbb{N}$, since $2^{p-1}\equiv1$ mod $p$, so as soon as you verify $p\mid (p-1)2^{p-1}+1$, you have an infinite family.


问:
Half-line - open or closed set
Half-line is given by

$$A=\{x\in R^n \mid x=x_0 + t(a-x_0), \forall t\geq 0\}$$

Is it open or closed set?

答:
You should consider the different cases:

if $ n = 1 $ then $A=\mathbb{R} $ and $\mathbb{R} $ is your hole space and them open by definition, and closed also by definition since $\varnothing$ is closed.

if $n\geq2$ then clearly $A\neq\mathbb{R} $ and as someone says before, for each point at $A^{C}$ you can find a hole neighborhood which is contained in $A^{C}$ which means $A^{C}$ is open which means $A$ is closed. $A$ is not open since the ball $B_{\varepsilon}\left(x_{0}\right)$ is not contained at $A$ for each $\varepsilon>0$ so $A$ is not open by definition.

Title: Half-line - open or closed set


问:
How to prove the following using invariants?
Problem Link: https://cutt.ly/GzRnhPo , slide 38

Problem : Given numbers 1, 2, 3, ..., 9.

we place signs + or - to form an expression.

we know that if x is a possible result to such an expression then

x is odd

-45 <= x <= 45

We want to prove the converse using invariants and the following greedy algorithm:

Start
from right to left and place the signs greedily:
if the current sum is less than the goal, increase the
sum, otherwise decrease it

Now, my guess for the invariant is the following:

assume that S denote the current remaining sum after some iteration, ex: initially S = x. if we place '-' in front of 9, then S become x-9

invariant is that (-total <= S <= total), such that total = 0 + 1 + 2 + ... + rightmost number that did not receive a sign yet.

Initially, the assumed-invariant property holds.   -45 <= S <= 45

After any iteration, S will be constrained by the suggested invariant.. proving this is my problem!

This imply that after algorithm is done,   0 <= S <= 0, which mean S = 0, which mean we made an expression with result equal x..

Is my invariant correct ?

How to prove it ?

Any other ideas to prove same thing using invariants and the provided greedy

答:
Your invariant works until there are only two numbers left for which we need to choose signs. At that point we can easily work out the remaining cases by hand.

Let's call $S_n$ the sum after the $n$-th iteration of your algorithm with $S_0$ being $x$.

Let's prove by induction on $n$ that $|S_n| \le \frac{(9-n) (10-n)}{2}$. This is just the sum that you called total. It's the sum of an arithmetic series.

Base case: if $n = 0$, then $-45 \le x \le 45 \implies |S_0| \le \frac{9 \cdot 10}{2}$.

Now suppose that $|S_n| \le \frac{(9-n) (10-n)}{2}$ for some $0 \le n \le 6$ and let's prove that $|S_{n+1}| \le \frac{(8 - n)(9 - n)}{2}$.

We will split this into two cases:

First if $S_n \le 0$ then $|S_n| = -S_n$. As such, by the induction hypothesis,
$$
0 \le -S_n \le \frac{(9-n) (10-n)}{2} \\
-\frac{(9-n) (10-n)}{2} \le S_n \le 0 \\
-(1 + 2 + 3 + \ldots + (9-n)) \le S_n \le 0
$$
We put a $+$ sign next to $9 - n$: $S_{n+1} = S_n + 9 - n$, as such,

$$
-(1 + 2 + 3 + \ldots (8 - n)) \le S_{n+1} \le 9 - n \le 1 + 2 + \ldots + (8 - n) \\
\implies |S_{n+1}| \le \frac{(8 - n)(9 - n)}{2}
$$

The reasoning is the same for the case $S_n > 0$, we just put a $-$ sign next to $9-n$.

Now we get $|S_7| \le 3$.

$S_7$ is the sum of $8$ numbers $5$ of which are odd (including $x$) so it's odd.
As such, $S_7 = \pm 1$ or $S_7 = \pm 3$.

If $S_7 = 1$, then if we put $-$ next to 2 and $+$ next to 1 we get $S = S_7 - 2 + 1 =0$.

If $S_7 = 3$, then if we put $-$ signs next to both $1$ and $2$ we get $S = S_7 - 2 - 1 = 0$.

For $-1$ and $-3$ just invert the signs.

As such, for every odd integer $x$ in $[-45,45]$ we can get $x$ by placing signs next to the numbers $1,2,\ldots, 9$ and taking the sum.


问:
Convergence of $\sum_{i=0}^{\log_2(n)} (\frac{3}{2})^i$
I have to solve the recurrence $T(n) = 3T(\frac{n}{2}) + n$ by using recursion tree. By applying this method I came to

$$T(n) = \sum_{i=0}^{\log_2(n)} (\frac{3}{2})^i$$

Since $|\frac{3}{2}| > 1$, I can not use convergence of geometric series for this problem. What else can I do to solve this problem?

答: As Gary said, the closed-form for a geometric series easily solves the problem to give you $T(n)=\frac{\bigl(\frac{3}{2}\bigr)^{\log_2(n)+1}-1}{\frac{3}{2}-1}=2\bigl(\frac{3}{2}\bigr)^{\log_2(n)+1}-2.$


问:
Find the polynomial $f(x)$ which have the following property
Find the polynomial $p(x)=x^2+px+q$ for which $\max\{\:|p(x)|\::\:x\in[-1,1]\:\}$ is minimal.

This is the 2nd exercise from a test I gave, and I didn't know how to resolve it. Any good explanations will be appreciated. Thanks!

Body:
Note that $p(0)=q, p(1)=1+p+q$ and $p(-1)=1-p+q$.

Thus

$$p(-1)+p(1)-2p(0)=2 \,.$$

Thus

$$2= p(-1)+p(1)-2p(0) \leq 4 \max \{ p(-1), p(1), -p(0) \} \,.$$

It follows that $\max \{ p(-1), p(1), -p(0) \} \leq \frac{1}{2} (*)$

This proves that

$$\max\{\:|p(x)|\::\:x\in[-1,1]\:\} \geq \frac{1}{2} \,.$$

Moreover, we can only get equality in $(*)$ if and only if $ p(-1)=\frac{1}{2}, p(1)=\frac{1}{2}, p(0)=-\frac{1}{2}$, if and only if $q=-\frac{1}{2}, p=0$.

It is now easy to check that $f(x)=x^2-\frac{1}{2}$ satisfies

$$\max\{\:|p(x)|\::\:x\in[-1,1]\:\} = \frac{1}{2} \,.$$

This proves that

$$\max\{\:|p(x)|\::\:x\in[-1,1]\:\} \geq \frac{1}{2} \,,$$
with equality if and only if $p=0, q=-\frac{1}{2}$.


答:
I assume that you are establishing the equation holds for all $n$ of the form $n = (p-1)^{2k+1}$ for all integers $k$ i.e., all $n$ that is an odd power of $(p-1)$. You do need to put this in explicitly though.

Your idea looks like it could be correct, except  you should have instead
$$(p-1)^32^{p-1} \equiv -1 \pmod p.$$

The idea is that $(p-1)^{2k+1} \equiv_p -1$ because $(p-1)^{2k+1}$ is an odd power of $p-1$, whereas $2^{(p-1)^{2k+1}} \equiv_p 1$ because $(p-1)^{2k+1}$ is a multiple of $p-1$.

Title: Prove there exist infinite value of $n$ such that $p\mid n2^{n}+1$

Body:
Let $p$ be an odd prime number. Prove that there exist infinite values of $n\in \mathbb{N}$ such that: $p\mid n2^n +1$.

$\textbf{My attempt}$: I have proved that $p\mid (p-1)2^{p-1}+1$. Therefore, we first define $n=p-1$, and then $n\mid n2^n +1$ holds. Since by cungruences $p\mid (p-1)2^{p-1}+1$ means that $(p-1)2^{p-1}\equiv -1 \equiv p-1 \pmod{p}$. Now, since $(p,p-1)=1$ because they are consecutive, we can multiply the equation by $p-1$, without any changes in the modulo, and we get $(p-1)^{2}\cdot 2^{p-1}\equiv (p-1)^2 \pmod{p}$. Recall that $p-1$ is the invers of itself modulo $p$ if and only if $p$ is prime. Thus, $$(p-1)^{2}\cdot 2^{p-1}\equiv (p-1)^2 \equiv 1\pmod{p}$$
We multiply again by $p-1$, and therefore: $$(p-1)^3 \cdot 2^{p-1}\equiv -1 \pmod{p}$$
By little's Fermat theorem,
$$(2^{p-1})^{(p-1)^2}\equiv 1 \pmod{p}$$
We conclude that: $$(p-1)^3\cdot 2^{(p-1)^3}\equiv -1\pmod{p}$$
So again we can define $n=(p-1)^3\neq p-1$, and by that we get another different value of $n$, such that $p\mid n2^{n}+1$, and so we proceed til we get infinite different numbers of $n$. $\textbf{Q.E.D}$

Now, I'm asking if I got it right because I showed an algorithm, such that $p\mid n2^n+1$, for each odd power of $p-1$, but I haven't written that down. Thanks!


问:
Why is $||Ax||=||x||$ if and only if $A^TA$ is the identity matrix?
I am going through my lecture nots in linear algebra and during a proof we state that

$||{Ax}||_2^2=<Ax, Ax>=<x, A^TAx>=<x,x>=||x||_2^2$ if and only if $A^TA=E$ with $E$ being the identity matrix. Unfortunately, I do not understand why from $||Ax||_2^2=||x||_2^2$ it has to follow that $A^TA$ has to be the identity matrix. Would not any isometric matrix satisfy this condition?

答:
First of all, I assume that when you make a statement about $x$, you mean that this statement should hold for every $x \in \Bbb R^n$ (where $n$ is the size of the square matrix $A$).

Second, note that the question in your title is distinct from the question in the body of your post. It is true that if $\|Ax\| = \|x\|$ holds for all $x \in \Bbb R^n$, then $A^TA$ must be the identity matrix. It is not true that if $\|Ax\| = \|x\|$ holds for all $x \in \Bbb R^n$, then $A$ is the identity matrix.

Sticking to the correct statement, it suffices to show the following: if $\langle x,A^TA x\rangle = \langle x,x \rangle$ holds for all $x \in \Bbb R^n$, then it must hold that $A^TA = E$. To that end, first observe that $A^TA$ is symmetric. Indeed, the fact that $(PQ)^T = Q^TP^T$ holds in general means that we have
$$
(A^TA)^T = A^T A^{TT} = A^TA.
$$
We see that
$$
\langle x, A^TAx \rangle = \langle x, x \rangle  \quad \text{ for all } x\in \Bbb R^n \implies\\
\langle x, A^TAx \rangle - \langle x, Ex \rangle = 0 \quad \text{ for all } x\in \Bbb R^n\implies\\
\langle x,(A^TA - E)x \rangle = 0 \quad \text{ for all } x\in \Bbb R^n.
$$
Now, Let $M = A^TA - E$, and let $m_{jk}$ denote the $j,k$ entry of $M$. Because $E$ and $A^TA$ are symmetric, $M$ is symmetric. Let $e_j \in \Bbb R^n$ denote the vector with a $1$ as its $j$th entry and $0$s elsewhere. We note that
$$
0 = \langle e_j,M e_j \rangle = m_{jj}.
$$
Thus, the diagonal entries of $M$ are all equal to zero. Similarly, for $j \neq k$,
$$
\begin{align}
0 &= \langle e_j + e_k, M(e_j + e_k) \rangle = m_{jj} + m_{jk} + m_{kj} + m_{kk}
\\ & = 0 + m_{jk} + m_{kj} + 0 = 2 m_{jk}
\end{align}
$$
where $m_{kj} = m_{jk}$ holds because $M$ is symmetric. Thus, we see from the above that for all $j \neq k$, $m_{jk} = 0$. That is, the off-diagonal entries of $M$ are also equal to zero.

Thus, $M$ is the zero matrix, which is to say that
$$
A^TA - E = 0 \implies A^TA = E,
$$
which was what we wanted.


问:
Lagrangian multiplier problem
I have a function $f=x^3+xy^2$ and a constraint $xy=z^2$ where $z\in \mathbb R$ that I have to solve using Lagrangian multipliers. This is what I have done so far:

The Lagrangian function is as follows: $L(x,y,z)=x^3+xy^2+\lambda(xy-z^2)$

Calculating the first order derivatives and setting them equal to zero:

(1)$\quad\frac{\partial L}{\partial x}=3x^2+y\lambda+y^2=0$

(2)$\quad\frac{\partial L}{\partial y}=x(2y+\lambda)=0$

(3)$\quad\frac{\partial L}{\partial z}=-2z\lambda=0$

(4)$\quad\frac{\partial L}{\partial \lambda}=xy-z^2=0$

Solving the system of equations:

$\quad$Using (2) I get that $x=0$.

$\quad$Using (4) I get $z=\sqrt{xy}$, which substituting $x=0$ is $z=\sqrt{0\times y}=0$, so $z=0$.

$\quad$Substituting $x=0$ and $z=0$ back into (4) I get that $y=0$.

$\quad$Shifting (2) around, I get that $\lambda=-2y$, and substituting in $y=0$ gets me $\lambda=0$.

So I get that $x=y=z=\lambda=0$.
This feels like a strange result to me, could anyone verify that this is correct?

答:
Note that $z$ is a constant, so you don't care about the derivative with respect to $z$. Then:
$$3x^2+\lambda y+y^2=0\\x(2y+\lambda)=0\\xy=z^2$$
Let's discuss now the case $z=0$. From the 3rd equation it implies $y=0$ or $x=0$ (or both). If $y=0$, from the first equation you get $x=0$. If $x=0$ you get $y=-\lambda$. This makes sense, since $x=0$ implies $f(x,y)=0$, independent of $y$. So the minimum value of $f(x,y)$ is $0$ for $x=0$ and any $y$.

Now for $z\ne 0$, it means $x\ne 0$ and $y\ne 0$. Then, from the second equation $\lambda=-2y$. Plugging into the first equation you get $$3x^2-y^2=0$$Together with $xy=z^2$ you can now solve for $x$ and $y$.

Title: Lagrangian multiplier problem


问:
What about $\mathbb{Z}[\sqrt[n]{2}]$?
Most textbooks talk about $\mathbb{Z}[\sqrt{d}]$ but I haven't succeed in my search of direct references about $\mathbb{Z}[\sqrt[n]{d}]$.

I would like to ask about $\mathbb{Z}[\sqrt[n]{2}]$ for $n>2$. I know it inherits the norm from $\mathbb{Q}(\sqrt[n]{2})$, but what is it known about $\mathbb{Z}[\sqrt[n]{2}]$? Is it a euclidean domain, or a unique factorization domain, or at least a GCD domain? Can their prime or irreducible elements be clasiffied as nicely as in $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{-1}]$? I will be very grateful with answers or references for this questions, even for just $\mathbb{Z}[\sqrt[3]{2}]$.

Thank you.

答:
The discriminant of $\alpha=\sqrt[3]{2}$ is $\pm 27\cdot 4$, so only $2$ and $3$ may ramify, since $\pm 27\cdot 4=[\mathcal{O}_K:\mathbb{Z}[\alpha]]^2 d_K$, where $K=\Bbb Q\!\left(\sqrt[3]2\right)$.

Now $\mu_\alpha=X^3-2$ is $2$-Eisenstein, so $2\nmid  [\mathcal{O}_K:\mathbb{Z}[\alpha]]$ and $\mu_{\alpha}(X-1)$ is $3$-Eisenstein, so  $3\nmid  [\mathcal{O}_K:\mathbb{Z}[\alpha +1]]=  [\mathcal{O}_K:\mathbb{Z}[\alpha]]$. Therefore,  $\mathcal{O}_K=\mathbb{Z}[\alpha]$ and $d_K=\pm 27\cdot 4$. It follows that the factorisation of $(p)$ into prime ideals is reflected by the decomposition of $X^3-2$ modulo $p$.

The  Minkoswki bound is $\approx 2.94$, hence the class group is generated by the class of prime ideals lying above $2$. Now, since $(2)=(\alpha,2)^3=(\alpha)^3$, the unique prime ideal above $2$ is $(\alpha)$, which is principal, so $\mathbb{Z}[\alpha]$ is a PID. A conjecture (supported by lot of positive results) suggests that a ring of integers is a PID if and only it is Euclidean (but not necessarily norm Euclidean, so the Euclidean function may be very hard to find).

Since $\mathbb{Z}[\alpha]$ is a PID, irreducible elements are exactly the generators of prime ideals. Depending of the factorisation of $X^3-2$ mod $p$, you will have.

For $p\neq 2,3$:

If $2$ is not a cube modulo $p$, $p$ is irreducible

If $2$ is a cube modulo $p$ but $X^2+X+1$ is irreducible modulo $p$, we have $(p)=P_1P_2$, where $P_1=(\alpha-m,p), P_2=(\alpha^2+m\alpha+m^2,p)$n where $m^3\equiv 2 \ [p]$.

The generators $\pi_1,\pi_2$ of $P_1,P_2$ will give you non associate irreducible elements.

If $2$ is a cube modulo $p$ and $X^2+X+1$ is reducible modulo $p$, whe have $(p)=P_1P_2P_3$, where $P_k=(\alpha-j^km,p)$ with $j^2+j+1\equiv 0 \ [p]$ and $m^3\equiv 2 \ [p]$.

The generators $\pi'_1,\pi_2',\pi'_ 3$ of $P_1,P_2,P_3$ will give you non associate irreducible elements.

If $p=2$, we get that $\alpha$ is irreducible

If $p=3$, $(3)=P^3$, where $P=(\alpha-1,3)=(\alpha-1)$ , hence $\alpha-1$ is irreducible.

Hence, up to association, the irreducible elements are $\alpha,\alpha-1$ and the various $\pi_1,\pi_2,\pi'_1,\pi'_2,\pi'_3$. I doubt that you can find the last ones explicitely for general $p$...

Title: What about $\mathbb{Z}[\sqrt[n]{2}]$?


问:
A Euclidean algorithm problem
Suppose that, after running the Euclidean algorithm on two integers $a$ and $b$, we find that $r_n=3$, where $r_n$ is the last remainder in the Euclidean algorithm. Furthermore, we find that $r_{n-1}=6$ and $r_{n-2}=21$. What can we say about $a$ and $b$?

Obviously, by the definition of the Euclidean algorithm, $\gcd(a, b)=3$. However, that isn't sufficient (isn't strong enough for an iff condition). I tried for a long time to think of more conditionns, but only came up with a bunch of false conjectures. Can someone help me?

答:
In the Euclidean algorithm, performed on the starting couple of positive integers  $(n,m)$

the remainders follow the sequence
$$
\eqalign{
  & n = r_{\, - 2}  =
 \left\lfloor {{n \over m}} \right\rfloor m + r_{\,0}   \cr
  & m = r_{\, - 1}  =
 \left\lfloor {{m \over {r_{\,0} }}} \right\rfloor r_{\,0}  + r_{\,1}   \cr
  &  \vdots   \cr
  & r_{\,k - 2}  =
 \left\lfloor {{{r_{\,k - 2} } \over {r_{\,k - 1} }}} \right\rfloor r_{\,k - 1}  + r_{\,k}   \cr
  &  \vdots   \cr
  & r_{\,h - 4}  =
 \left\lfloor {{{r_{\,h - 4} } \over {r_{\,h - 3} }}} \right\rfloor r_{\,h - 3}  + r_{\,h - 2}   \cr
  & r_{\,h - 3}  =
 \left\lfloor {{{r_{\,h - 3} } \over {r_{\,h - 2} }}} \right\rfloor r_{\,h - 2}  + r_{\,h - 1}   \cr
  & r_{\,h - 2}  =
 \left\lfloor {{{r_{\,h - 2} } \over {r_{\,h - 1} }}} \right\rfloor r_{\,h - 1}  +
 \left( {0 = r_{\,h} } \right)
\quad \left| {\;r_{\,h - 1}  = \gcd (n,m) = g} \right. \cr}
$$

where the sequence shall be strictly decreasing down to $0$ at step $h$,
i.e.
$$
0 = r_{\,h}  <  \cdots  < \,r_{\,k} \, < r_{\,k - 1}  <  \cdots  < r_{\,0}
 \quad \left\{ {\matrix{
   { \le \left| {\,n\,} \right|} & {\left| {\;0 = m} \right.}  \cr
   { < \left| {\,m\,} \right|} & {\left| {\;0 \ne m} \right.}  \cr
 } } \right.
$$
That also implies that the $ r_k $ are all multiples of $g= \gcd(n,m)$

You are asking practically how the sequence can be reconstructed backwards.
Let's start from the last line
$$
\eqalign{
  & s_{ - 2}  = {{r_{\,h} } \over g} = 0  \cr
  & s_{ - 1}  = {{r_{\,h - 1} } \over g} = 1  \cr
  & s_0  = {{r_{\,h - 2} } \over g} =
 \left\lfloor {{{r_{\,h - 2} } \over {r_{\,h - 1} }}} \right\rfloor
  = k_{\,0}  = k_{\,0} s_{ - 1}  + s_{ - 2}   \cr
  & s_1  = {{r_{\,h - 3} } \over g} =
 \left\lfloor {{{r_{\,h - 3} } \over {r_{\,h - 2} }}} \right\rfloor {{r_{\,h - 2} } \over g}
 + {{r_{\,h - 1} } \over g} = k_{\,1} s_0  + s_{ - 1}   \cr
  & s_2  = {{r_{\,h - 4} } \over g} =
 \left\lfloor {{{r_{\,h - 4} } \over {r_{\,h - 3} }}} \right\rfloor {{r_{\,h - 3} } \over g}
 + {{r_{\,h - 2} } \over g} = k_{\,2} s_1  + s_0   \cr
  & \quad \quad  \vdots   \cr
  & s_j  = {{r_{\,h - j - 2} } \over g} =
 \left\lfloor {{{r_{\,h - j - 2} } \over {r_{\,h - j - 1} }}} \right\rfloor {{r_{\,h - j - 1} } \over g}
 + {{r_{\,h - j} } \over g} = k_{\,j} s_{j - 1}  + s_{j - 2}  \cr}
$$
where the  $ k_j$ are arbitrary integers greater than $1$.

In matrix form this recursion reads
$$
\left( {\matrix{ {s_{n - 1} }  \cr {s_n }  \cr  } } \right) =
 \left( {\matrix{   0 & 1  \cr   1 & {k_n }  \cr  } } \right)
\left( {\matrix{   {s_{n - 2} }  \cr    {s_{n - 1} }  \cr  } } \right)
\quad \left| {\;\left( {\matrix{ {s_{ - 2} }  \cr {s_{ - 1} }  \cr  } } \right)
 = \left( {\matrix{0  \cr 1  \cr  } } \right)} \right.
$$

In the example you give
$$
\eqalign{
  & g = 3,  \cr
  & s_{ - 1}  = {{r_{\,h - 1} } \over g} = {g \over g} = 1,\quad   \cr
  & s_0  = {{r_{\,h - 2} } \over g} = {6 \over g} = 2 = k_0 ,\;  \cr
  & s_1  = {{21} \over g} = \left\lfloor {{{r_{\,h - 3} } \over {r_{\,h - 2} }}}
 \right\rfloor {{r_{\,h - 2} } \over g} + {{r_{\,h - 1} } \over g} =   \cr
  &  = \left\lfloor {{{21} \over 6}} \right\rfloor {6 \over g} + {g \over g} =
 3s_0  + s_{ - 1}  = 7 \cr}
$$
then
$$
\eqalign{
  & \left( {\matrix{   {s_{ - 1} }  \cr    {s_0 }  \cr  } } \right) =
 \left( {\matrix{   0 & 1  \cr    1 & {k_0 }  \cr  } } \right)\left( {\matrix{   0  \cr    1  \cr  } } \right) =
 \left( {\matrix{   0 & 1  \cr    1 & 2  \cr  } } \right)\left( {\matrix{   0  \cr    1  \cr  } } \right) =
 \left( {\matrix{   1  \cr    2  \cr  } } \right)  \cr   & \left( {\matrix{   {s_0 }  \cr    {s_1 }  \cr  } } \right) =
 \left( {\matrix{   0 & 1  \cr    1 & {k_1 }  \cr  } } \right)\left( {\matrix{   1  \cr    2  \cr  } } \right) =
 \left( {\matrix{   0 & 1  \cr    1 & 3  \cr  } } \right)\left( {\matrix{   1  \cr    2  \cr  } } \right) =
 \left( {\matrix{   2  \cr    7  \cr  } } \right) \cr}
$$
and from here we can proceed with multiplying  by the matrix, with any selected constant, e.g.:
$$
\eqalign{
  & \left( {\matrix{ 0 & 1  \cr  1 & 1  \cr } } \right)
\left( {\matrix{ 2  \cr  7  \cr  } } \right) =
 \left( {\matrix{ 7  \cr 9  \cr } } \right)  \cr
  & \left( {\matrix{ 0 & 1  \cr 1 & 2  \cr  } } \right)
\left( {\matrix{ 2  \cr  7  \cr  } } \right) =
 \left( {\matrix{ 7  \cr  {16}  \cr  } } \right)  \cr
  & \left( {\matrix{ 0 & 1  \cr  1 & 1  \cr  } } \right)^{\,2}
 \left( {\matrix{ 2  \cr 7  \cr } } \right) =
 \left( {\matrix{  9  \cr   {16}  \cr  } } \right) \cr}
$$
which correspond to the couples $(21,27), \,(21,48) , \, (27,48)$ already indicated by @robjohn.

In conclusion we have the general solution to the recursion as
$$
\left( {\matrix{ {s_{n - 1} }  \cr {s_n }  \cr } } \right) =
 \prod\limits_{j = 0}^n {\left( {\matrix{  0 & 1  \cr  1 & {k_j }  \cr } } \right)
\left( {\matrix{ 0  \cr  1  \cr  } } \right)}
$$
so that when the $k$'s are unitary we get the Fibonacci numbers, which are known to have the longest
chain in the algorithm.
But for general values of the constants, there is not a simple characterization of the numbers
that can be obtained..

Title: A Euclidean algorithm problem


问:
Relation between an equation coefficients and equations's solutions
Let consider a second degree
$$a(k)x^2+b(k)x+c(k)=0$$

Suppose I want to find conditions on $k$ so that, called $x_1$ and $x_2$ the two solutions of the equation, they satisfy the following

$$\frac{1}{x_1^2}
+\frac{1}{x_2^2}=\frac{1}{4}$$

Could you give hints?

Body:
Hint:

We have:

\begin{cases}
 x_1 +x_2 =  -\frac{b(k)}{a(k)} \\
x_1x_2 =\frac{c(k)}{a(k)}
\end{cases}

and
\begin{align}
\frac{1}{4} &= \frac{1}{x_1^2} +\frac{1}{x_2^2}\\ &= \frac{x_1^2 +x_2^2}{x_1^2x_2^2}\\ &=  \frac{(x_1 +x_2)^2-2x_1x_2}{x_1^2x_2^2}
\end{align}


问:
Classifying Vector Spaces without AC
Using the axiom of choice we can give a simple classification of all vector spaces over a given field $K$ up to isomorphism: Any $K$-vector-space $V$ is just isomorphic to $\bigoplus_{i\in B}K$ where $B$ is a basis for $V$. Given AC we even know that there is some aleph $\aleph_\alpha$ such that $\vert B\vert = \aleph_\alpha$ and thus any vector-space-structure over $K$ appears in the sequence $\lbrace \bigoplus_{i\in \aleph_\alpha}K:\alpha\in Ord\rbrace$.

Now I was wondering whether a similar (weaker) classification still holds without AC. In this case not every vector space has a basis and not every cardinality is equal to some aleph, so clearly  the classification above fails. However one could try something like this:

Edit* Thanks to Asaf's answer I realised that my previous attempt at "covering all vector spaces" was flawed, so here is a more general approach:

As far as I can tell "new" vector spaces can be created using the following  "Gödel operations for vector spaces":

taking the sum or direct product over a family of spaces,

taking the quotient of two spaces,

constructing the space of homomorphisms between two spaces

considering the subspace of any space resulting from the above operations

We can thus construct a hierarchy which should be closed under all these operations: Let $X_0:=\lbrace K\rbrace$ and for $\alpha>0$ let $X_\alpha:=$

$\lbrace W:$ there exists $\beta < \alpha$ and $W_0,W_1\in \overline{X}_\beta$ such that $W=W_1/W_0$ or $W=Hom_K(W_0,W_1)$ or there exists $I\in V_\alpha$ (in the Von-Neumann-hierarchy) and a family $\lbrace W_i:i\in I\rbrace \subseteq \overline{X}_\beta$ such that $W=\sum_{i\in I}W_i$ or $W=\prod_{i\in I}W_i\rbrace$

where $\overline{X}_\alpha:=\lbrace W'\leq W: W\in X_\alpha\rbrace$. I suspect that every vector space over $K$ should be isomorphic to some element of $\bigcup_{\alpha\in Ord}\overline{X}_\alpha$. Is that correct and is there maybe a more "practical" approach to characterizing all vector spaces over $K$ in ZF?

答:
To the edit, the answer is yes. For a simple reason.

Suppose that $V$ is a $K$-vector space, then $V$ is a quotient of $W=\bigoplus_{v\in V}K$ by mapping $(\alpha_v\mid v\in V)$ to $\sum_{v\in V}\alpha_v v$ in $V$ itself, this is well-defined since all but finitely many $\alpha_v$ must be $0$.

So by allowing sums and quotients, every vector space is reached in sum+quotient operations from the field $K$.

Unfortunately, this does not help us to give any practical structure to vector spaces. If anything, it highlights how badly we don't understand them when the axiom of choice is taken away from us.

No. You can have a vector space over any given field (and indeed, over every field) that is not finitely generated, but every proper subspace is finitely generated.

Not only this vector space doesn't have a basis, it is not of the form of $\mathrm{Hom}_K(V,W)$ for any two vector spaces. Now, suppose that $\alpha$ is the least rank of such vector space, then for every $V,W$ of lower rank, including those generating our space, $V$ has an infinite dimensional subspace. Fix such $V_0\subsetneq V$ and consider $\{T\mid V_0\subseteq\ker T\}$. Easily, this is an infinite dimensional subspace, so it has to be equal to everything.

So that means that for any infinite subspace $V_0$, all operators must be $0$ on $V_0$. But now, if $v\notin V_0$, then $V_0\oplus\langle v\rangle$ is another infinite dimensional subspace. Therefore one of two options must hold:

$V=V_0\oplus\langle v\rangle$, and every operator $T\colon V\to W$ is just a function mapping $v$ somewhere in $W$, but that means that $W$ is isomorphic to $\operatorname{Hom}_K(V,W)$, and $W$ has a lower rank, so it has infinite dimensional proper subspaces. That's impossible.

For every $v\in V$, $v$ is in some infinite dimensional proper subspace of $V$, and therefore $\operatorname{Hom}_K(V,W)=\{0\}$, which is very much not infinite dimensional like we assumed. That's also impossible.

This argument should in principle work for every type of "terrible enough indecomposable" vector space, and so $\sf ZF+DC_\kappa$ will not solve your problems either, for any $\kappa$, as per my masters thesis.

Of course, one could ask if your statement implies choice. But there's a lot of really good questions about vector spaces without choice, and we don't really know how to solve most of them... throw this one on the pile.

Title: Classifying Vector Spaces without AC


问:
I need help in a mapping problem
I have a problem that I'm having trouble understanding it. In the book it says to find the inverse Laplace transform of the function $$\mathcal L^{-1}\left(\frac{1}{s(1+e^{as})}\right)$$. In order to find this first I have to prove that the function $F(s)=\frac{1}{s(1+e^{as})}$ satisfies the condition $$|F(s)|\le \frac{M}{|s|^p}$$ for some $p \gt 0$. So to prove this they show that $|1+e^s| \ge c \gt 0$ like the picture below.

I can't seem to understand this mapping at all. I would really appreciate some help

答:
This will be an incomplete answer, but in case no one else comes along:

The idea of the map $H(s) = 1 + e^s$ is that you are mapping points $s$ on a circle of radius $R_n = 2n\pi$ to another complex number $\zeta = 1+ e^x e^{iy}$, or say $\zeta - 1 = e^x e^{iy}$. From this form, you can see that the $x$ component of the coordinate $s$, which lies on the circle, becomes the magnitude of this new number $\zeta -1$. The $y$ component, $iy$, becomes the new phase. In a way you're mapping a circle onto another circle, since the components of one circle $C_n$ become the magnitude and angle of another circle-looking curve.

They first ask you to consider the region $0 < x \le x_0$. If $R_n = 2n\pi$ is big, the circle will hardly vary at all from $x=0$ to $x = x_0$, so you can think of the region as a rectangle in the complex plane with top and bottom at $y = i \pm 2n\pi$. Then you can see as you traverse the rectangle from $x = 0$ to $x = x_0$, the number $\zeta - 1$ goes from $e^x e^{iy} = e^{iy} = e^{i2n\pi}$ to $e^{x_0} e^{i2n\pi}$. The key is that the radius of this circle happens to be a multiple of $2\pi$, so then $\zeta - 1$ goes from $1$ to $e^{x_0}$. And $e^{x_0}$ will always be greater than 1, which is where they get the first part $\left|{1 + e^s}\right| \ge 2$.

Then in the region $x < 0$, the number $ \zeta - 1$ will be $e^x e^{iy} = e^{R_n \cos{\phi}} e^{iR_n \sin{\phi}} = e^{2n\pi \cos{\phi}} e^{i 2n\pi \sin{\phi}}$. Then if $ y = 2n\pi \sin{\phi}$ goes from $ 2n\pi $ to $(2n-1)\pi$, the number $\zeta - 1$ will make a half revolution, since its phase $y$ went down by $\pi$. From there, you are essentially at the conclusion.

Title: I need help in a mapping problem


问:
QR Factorisation of congurent vectors
There are n linearly independent vectors $x_1,x_2.........x_n$ $\in R^{m}$, another set of vectors $y_1,y_2.........y_n$ $\in R^{m}$ is congurent to it in the sense that the all the length and  distances are equal i.e.
$\|x_i\|_2 = \|y_i\|_2$ and $\|x_i - x_j\|_2  = \|y_i - y_j\|_2$  defining matrices X and Y as $[x_1,x_2..........x_n]$ and $[y_1,y_2.............y_n]$ respectively. I need to show that in reduced QR factorisation $R$ is same for both these matrices.

I started with multiplying both X and Y by there transpose.

$X^T * X = R^T * R$ $\& $
$Y^T * Y = R_1^T * R_1$

now i am unable to find any relation between these 2 terms please suggest me approach to proceed from here

答:
Fact 1: $X$ and $Y$ have congruent columns if and only if then $X^TX=Y^TY$.

From $\|x_i\|_2=\|y_i\|_2$ we have that
$$\tag{1}
x_i^Tx_i=y_i^Ty_i, \quad i=1,\ldots, n.
$$
This shows that the diagonals of $X^TX$ and $Y^TY$ are the same. Next, from $\|x_i-x_j\|_2=\|y_i-y_j\|_2$, we have
$$
x_i^Tx_i+x_j^Tx_j-2x_i^Tx_j=y_i^Ty_i+y_j^Ty_j-2y_i^Ty_j.
$$
Using (1) we get
$$
x_i^Tx_j=y_i^Ty_j,\quad i,j=1,\ldots,n.
$$
This proves the equality of the off-diagonal entries. $\square$

Now we could take a shortcut and use the fact that $X^TX=Y^TY$ implies that $X=UY$ for some orthonormal $U$ and we would be done (see, e.g., here). If $X=QR$ is a QR factorization of $X$, then $Y=(U^TQ)R$ is that of $Y$.

We can also take a more elementary approach.

Fact 2: If $X^TX=Y^TY$ and $R_x$ and $R_y$ are the triangular factors of their QR factorizations, then $R_y=DR_x$, where $D$ is a diagonal matrix with $\pm 1$ entries on the diagonal.

Assume that $X=Q_xR_x$ and $Y=Q_yR_y$. Then from $X^TX=Y^TY$ we have
$$\tag{2}
\begin{split}
R_x^TR_x=R_y^TR_y&\iff I=R_x^{-T}R_y^TR_yR_x^{-1}=(R_yR_x^{-1})^T(R_yR_x^{-1})
\\&\iff (R_yR_x^{-1})^{-T}=R_yR_x^{-1}.
\end{split}
$$
Since the matrix on the left and right-hand side is lower and upper triangular, respectively, this implies that $R_yR_x^{-1}=D$ is a diagonal matrix and putting this back to (2) we get that $I=D^2$. That is, $D$ has $\pm 1$ elements on the diagonal. $\square$

This is a well-known fact about the QR factorization or Cholesky decomposition, that the triangular factors are not unique unless we fix the signs on their diagonals (e.g., positive). If we do that, we have necessarily that $D=I$ and hence $R_x=R_y$.

Title: QR Factorisation of congurent vectors


问:
Is there an image to the point at infinity through this map?
I encountered a conformal mapping on the complex plane:$$z\rightarrow e^{i\pi z}$$
and I am not sure about where it does send the point at infinity. If I could say something along the lines: $$\text{Im}(\infty) = \infty$$
Then it would map it to the origin but there is still a voice in my head saying that this equality is non-sense. And from the usual definition of the imaginary part:$$\text{Im}(z)=\frac{z+\bar{z}}{2i}$$ it makes even less sense.
I'd appreciate some enlightenment.</p>

答:
You are seeking

$$
\lim_{z \to \infty} e^{i\pi z}.
$$

Note that if you consider $z = x + iy$, then by De Moivre's Theorem, we have
$$
e^{i\pi z} = e^{i\pi (x+iy)} = \frac{\cos(\pi x) + i \sin(\pi x)}{e^y}.
$$

Now everything depends on along which path $z \to \infty$: if $x$ is fixed and $y \to \infty$, the limit converges to $0$. Alternatively, if $y$ is fixed and $x \to \infty$, the limit does not exist.

Since the value of the limit is path-dependent, the limit does not exist.


问:
Finding the coordinates of the farthest point from a point inside an ellipse without knowing a or b?
I was wondering if anyone could help me figure this out. I have an ellipse and the known info is that the major axis is horizontal the center is at (0,0), there is a point P at (0, $\dfrac{3}{2}$), the maximum distance from this point to anywhere on the ellipse is $\sqrt{7}$, and $\dfrac{c}{a}$ =  ${\dfrac{\sqrt{3}}{2}}$ (c is the foci length, a is the major axis length). Is there any way to find the equation of this ellipse and the coordinates of the furthest point on the ellipse from P? Any help would be greatly appreciated thank you!

答:
$$\frac{c}{a}=\frac{\sqrt{3}}{2}\to c=\frac{\sqrt{3} a}{2}$$
we know that $$b^2=a^2-c^2\to a^2=4b^2$$
The equation of the ellipse is
$$\frac{x^2}{4b^2}+\frac{y^2}{b^2}=1$$
Write the ellipse in parametric form
$$E:(x=2b\cos t,y=b\sin t)$$
The distance between the point $P(0,3/2)$  and a generic point of the ellipse is
$$EP^2=f(b)=(2b\cos t)^2+\left(b\sin t-\frac32\right)^2=b^2 \sin ^2 t+4 b^2 \cos ^2 t-3 b \sin t+\frac{9}{4}$$
derivative is zero in $[0,2\pi]$ if
$$f'(b)= -3 b \cos t (2 b \sin t+1)=0$$
$$ t=\pm\pi/2;\;t=2\pi-\arcsin\left(\frac{1}{2b}\right);\;t=\pi+\arcsin\left(\frac{1}{2b}\right)$$
Looking at the second derivative $f''(b)=3 b (\sin t-2 b \cos 2t)$, if $b>\frac{1}{2}$ the maximum is $t=\pi+\arcsin\left(\frac{1}{2b}\right)$
$$f\left(\pi+\arcsin\left(\frac{1}{2b}\right)\right)=7$$ if $$4 b^2+3=7\to b^2=1$$
The equation of the ellipse is
$$\frac{x^2}{4}+y^2=1$$
and the coordinates of the most distant point are
$$\left(\pm\sqrt{3},-\frac{1}{2}\right)$$

Edit

You can't have an ellipse with point $P$ internal and $\sqrt 7$ maximum distance. See the second picture below.

$$...$$


问:
Mean curvature flow vs. diffusion
Mean curvature flow (MCF) and diffusion-type flows both have smoothing effects on a curve.

I can tell there is a deep connection between the two, as seen in the Merriman-Bence-Osher (MBO) numerical scheme, which uses diffusion to approximate MCF: it takes a characteristic function of a region $\Omega$ and iterates the following

apply the heat kernel (i.e. allow the boundary to diffuse), and

threshold (to recreate a sharp boundary).

What is the relationship between MCF and diffusive flows and how can one explain this geometrically? Honestly I'm not even sure whether "diffusive flow" is a term, but by that I mean evolution of a curve as described by the heat equation/heat kernel.

Edit: I also read somewhere that Ricci flow on a surface effectively evolves the curvature of the surface with a heat equation. Now, how does Ricci flow come into the picture?

答:
So the best reference for this would probably be Merriman, Bence, and Osher's original paper. In it, they show that the heat equation is locally a mean curvature flow with rate $D\kappa$.

Specifically, consider a region $\Omega$ and its characteristic function $\chi_\Omega$ (let's just call it $\chi$). We can consider the result of a diffusion of $\chi$, and in particular the effect of the diffusion at a point $P$ on the boundary of $\chi$.

The diffusion equation applied to $\chi$ would be $$ \frac{\partial \chi}{\partial t} = D\Delta \chi.$$

Consider the tangent circle to $P$, i.e. if $P$ has local curvature $\kappa = \frac{1}{r}$, the circle with radius $r$.

What is the effect of the diffusion at this point? Rewrite the diffusion equation in the polar coordinates based on the tangent circle to $P$, and we get
$$ \frac{\partial \chi}{\partial t} = \frac{D}{r} \frac{\partial \chi}{\partial r} + D\frac{\partial^2 \chi}{\partial r^2} + \frac{D}{r^2}\frac{\partial^2 \chi}{\partial \theta^2}$$

(note this is the diffusion equation LOCAL to the point $P$).

Since $\chi$ is tangent to the circle at $P$, the curve $\chi$ has locally unchanging angle at $P$, i.e. $\frac{\partial^2 \chi}{\partial \theta^2} = 0$.

Thus the diffusion equation at $P$ becomes
$$ \frac{\partial \chi}{\partial t} = \frac{D}{r} \frac{\partial \chi}{\partial r} + D\frac{\partial^2 \chi}{\partial r^2} $$

This is an advection-diffusion equation: the term $D\frac{\partial^2 \chi}{\partial r^2}$  acts to diffuse the the boundary of $\chi$, i.e. smoothing the sharp step that the characteristic function has. The term $\frac{D}{r} \frac{\partial \chi}{\partial r}$is the velocity term, which moves the point $P$ (and all points around it) at speed $\frac{D}{r}$. Since the local curvature at $P$ is $\kappa = \frac{1}{r}$, we have that the local change at $P$ simultaneously

diffuses out/spreads

moves in the direction of its local curvature with rate $D\kappa$.

This is why the MBO scheme alternates diffusion term and thresholding term. The diffusion term takes a MCF step, but the boundary gets diffuse. The thresholding term resharpens the boundary by defining it to be the level set $\chi = \frac{1}{2}$ (which evolves via MCF since it moves exactly with the advective velocity $D\kappa$).

TLDR the diffusion equation applied to the characteristic function $\chi_\Omega$ of a set $\Omega$ causes the boundary $\partial \Omega$ to move according to a mean curvature flow in a small time step, with rate $D\kappa$ where $\kappa$ is the curvature at each point. It only works in a small time step because this analysis relies on local polar coordinates at each point on the boundary.

Title: Mean curvature flow vs. diffusion


问:
Limit of holomorphic and bounded functions that can be extended to a ball
Let $(f_n)$ be a sequence of bounded and holomorphic functions on the annulus $A=\{z\in\mathbb{C}\,|\, 1<|z|<2\}$ such that each $f_n$ can be holomorphically extended to $B_2(0) = \{ z\in\mathbb{C}\,|\, |z|<2\}$. Let $(f_n)$ converge uniformly to a holomorphic and bounded function $f$ on $A$. Show that $f$ can be holomorphically extended to $B_2(0)$.

My attempt: We find a sequence $(g_n)$ of holomorphic functions on $B_2(0)$ with $f_n = g_n|_A$ for all $n\in\mathbb{N}$. As each $f_n$ is bounded, we have $|g_n| = |f_n| \leq M$ on $A$. In addition, as holomorphic functions are continuous, we infer $m:=\max_{z\in \overline{B_{3/2}(0)}} |g_n(z)|<\infty$. Together, this yields $|g_n|\leq \max\{m, M\}$ on $B_2(0)$, i.e. each $g_n$ is bounded on $B_2(0)$. If I could now show that $(g_n)$ is Cauchy with respect to the supremum norm on $B_2(0)$ (I achieved to show compact convergence on $B_2(0)$ by using the maximum modulus principle, but according to this post, this is not equivalent to uniform convergence), there would be a holomorphic and bounded limit $g: B_2(0)\rightarrow \mathbb{C}$, as the set of holomorphic and bounded functions on a domain of $\mathbb{C}$ is Banach. But then $f_n$ converges uniformly to $f$ on $A$ and $f_n = g_n|_A$ converges uniformly to $g$ on $A$. By uniqueness of limits, $f=g$ on $A$. Thus, $g$ is the desired holomorphic extension of $f$ to $B_2(0)$. How can I show that $(g_n)$ is Cauchy?

答:
Based on the comments by Ruy, we have using the maximum modulus principle
\begin{align}
\sup_{|z|\leq 1} |g_n(z)-g_m(z)| &\leq \sup_{|z|\leq 3/2} |g_n(z)-g_m(z)| \leq \sup_{|z|= 3/2} |g_n(z)-g_m(z)|\\
&= \sup_{|z|=3/2} |f_n(z)-f_m(z)|\leq \sup_{z\in A} |f_n(z)-f_m(z)|<\epsilon,
\end{align}
as $f_n\to f$ uniformly on $A$ and convergent sequences are Cauchy. Then
\begin{equation}
\sup_{z\in B_2(0)} |g_n(z)-g_m(z)|\leq \max\Bigg\{ \sup_{|z|\leq 1} |g_n(z)-g_m(z)|,\,\, \sup_{z\in A} |g_n(z)-g_m(z)|\Bigg\}<\epsilon,
\end{equation}
so we indeed have that $(g_n)$ is Cauchy with respect to the supremum norm on $B_2(0)$.


问:
Can all points in the plane be represented like this?
Solving a task regaring affine geometry, I've come across a problem:

Is it  true that, for every point $(x,y)\in \mathbb{R}^2$, there  exist $t\in \mathbb{R}, \alpha\in[0,2\pi]$, such that

$$x = t\cos\alpha + \sin \alpha \\ y = t\sin\alpha + \cos \alpha?$$

I don't have any idea how to prove this or find a counterexample. All hints appreciated!

答:
Let's play around and see what happens.

I write $a$ for $\alpha$,
$c$ for $\cos \alpha$,
and $s$ for $\sin \alpha$
cause I'm lazy.

We have
$x = t c + s$
and
$y = t s + c$.

$x^2+y^2 = t^2 c^2+2tcs + s^2 + t^2 s^2 + 2tcs + c^2
= t^2+1+4tsc$.

$xy = t^2 cs + tc^2+t s^2 + sc
= cs(t^2+1)+t$.

We can solve for $cs$ in each of these.

$cs = (x^2+y^2-t^2-1)/(4t)$
and
$cs = (xy-t)/(t^2+1)$
so
$(x^2+y^2-t^2-1)/(4t) = (xy-t)/(t^2+1)$.

Now we can solve for $t$.

$4t(xy-t) = (t^2+1)(x^2+y^2-t^2-1)$
or
$4txy-4t^2 = (t^2+1)(x^2+y^2) - (t^2+1)^2
= t^2(x^2+y^2)+(x^2+y^2)-t^4-2t^2-1
$
so
$t^4-t^2(x^2+y^2+2)+4txy+1=0$.

This is a quartic, unfortunately,
and I don't see an obvious root.

We can also do this:

$xs-yc = (tcs + s^2)-(tsc+c^2)
= s^2-c^2
= 2s^2 - 1
$,
so
$yc = xs-2s^2+1$.

Squaring this,
and using $c^2 = 1-s^2$,
we get another quartic for $s$.

In other words,
we can get equations for
$t$
and $\sin \alpha$,
but they are quartics.


问:
What is the largest $k∈N$ with the property that there are non-trivial events $A_1,...,A_k ⊆ Ω$ such that $A_1,...,A_k$ are independent?
Let $\mathbb{P}$ be the uniform distribution on $Ω = {1,2, . . . ,10}$. An event $A⊆Ω$ is said to be non-trivial if $0<P(A)<1$. What is the largest $k∈N$ with the property that there are non-trivial events $A_1,...,A_k ⊆ Ω$ such that $A_1,...,A_k$ are independent?

Found this question in the textbook but really unsure of how to approach this, any advice is appreciated.

答:
Hint:

Subsets $A$ and $B$ are independent here iff: $$10|A\cap B|=|A|\times|B|$$

This (together with the fact that $A$ and $B$ are non-trivial) provides info about $(|A|,|B|)$.

Now find $2$ non-trivial independent subsets and wonder if you can find a third one.


问:
Compute integral $\int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx$
While solving a more general problem, I came across the following integral

$$I= \int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx$$

To start off, I defined $t:=x^2/2$ to get

\begin{align*}
I= \int_{-\infty}^\infty x^{4n} e^{-x^2/2} \, dx &= 2\int_0^\infty x^{4n} e^{-x^2/2} \, dx \\
&= 2 \int_0^\infty (\sqrt{2t})^{4n} e^{-t} \frac{dt}{\sqrt{2t}} \\
&= 2 (\sqrt{2})^{4n - 1} \int_0^\infty t^{2n - 1/2} e^{-t} \, dt
\end{align*}

This seems to be a gamma function integral. I would start integrating by parts

\begin{align*}
\int_0^{+\infty} t^{2n-\frac{1}{2}}e^{-t} \, dt&=[ -t^{2n-\frac{1}{2}}e^{-t}]_0^{+\infty} - (-)(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t} \, dt \\
    &=(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t} \, dt
\end{align*}

But how to proceed further?

I looked up in a table and the result should be

$$I=2 (\sqrt{2})^{4n - 1} \Gamma\left(2n+ \frac 1 2 \right)$$

PS: Alternative methods avoiding the gamma function are very welcomed! :)

EDIT: I think that what Yves suggested was the following

$$I= \int_{-\infty}^\infty x^{4n} e^{-x^2/2}= (2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2} \int_0^{+\infty}t^{-\frac{1}{2}}e^{-t}\,dt=(2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2}\ \ 2\int_0^{+\infty}e^{-t} \, d(\sqrt{t})$$

Alrigtht! I see that $(2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2} = \Gamma\left(2n+ \frac 1 2 \right)$
NAIVE question though: how to see that $\int_0^{+\infty}e^{-t} \, d(\sqrt{t})=2 (\sqrt{2})^{4n - 1}$?

Body:
Note that

$$I(a)= \int_{-\infty}^{\infty} e^{-a x^2} \, dx=\sqrt{\frac \pi a}
$$

Then

$$\int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx
= \frac{d^{2n} I(a)}{da^{2n}}\bigg|_{a=\frac12}= \sqrt{\pi}\cdot \frac{d^{2n} (a^{-\frac12})}{da^{2n}}\bigg|_{a=\frac12}
= \frac{(4n-1)! \sqrt{2\pi}}{2^{2n-1}(2n-1)!}
$$


问:
Does pointwise convergence against a continuous function imply uniform convergence?
Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of functions with $f_n:\mathbb{R}\rightarrow[0,1]$ for all $n\in\mathbb{N}$ and $f:\mathbb{R}\rightarrow[0,1]$ continuous such that $\lim_{n\rightarrow\infty}f_n(x)=f(x)$ for all $x\in\mathbb{R}$.
Is it now possible to show that $f_n$ converges uniformly to $f$?</p>

答:
Your example,
$$
f_n(x) = \begin{cases}
 e^{1-1/({1-((n+1) x-1)^2})} & 0<x<\frac{2}{n+1} \\
 0 & \text{otherwise,}
\end{cases}
$$
behaves something like so:

I think it's crystal clear that this is a continuous (even differentiable, I think) sequence of functions that converges pointwise to zero. The convergence clearly can't be uniform, since there's always a point where the value of $f_n$ is $1$.

The example I had in mind looks something like this:

Note that the graph consists of just straight line segments, so it's quite simple to write down a piecewise formula for this.  In fact:

$$
f_n(x) = \begin{cases}
 (n+1) x & 0<x<\frac{1}{n+1} \\
 2-(n+1) x & \frac{1}{n+1}<x<\frac{2}{n+1} \\
 0 & \text{otherwise}.
\end{cases}
$$

I emphasize, though, that (in my approach, at least) the picture comes first.


问:
Expectation in cumulative distribution functions
I have this CDF:

$$F_X(x) =
\begin{cases}
0,  & \text{if $x<-4.5$} \\
0.1,  & \text{if $-4.5\leq x<3$} \\
0.3,  & \text{if $3\leq x<6$} \\
0.6,  & \text{if $6\leq x<10$} \\
1,  & \text{if $x\geq 10$}
\end{cases}$$

The questions involved were:

Calculate P(X=0)

I just said it was 0, because 0 is in the middle of an interval

Calculate P(X=3)

I said it was 0.2, because 3 is the start of an interval, so I just subtracted all the intervals before it before it's own interval

Find the expected value of X

This is where I'm a little lost. I tried looking up how to do it online, and tried the integral method where I sum up the integrals of each range, but the answer I got for that was wrong. Any help?

答:
You are almost there...1 and 2 are correct. To calculate the expectation observe that your rv is discrete and it is defined in

$$\{-4.5;3;6;10\}$$

with probabilities that you are able to calculate...thus the expectation is simply

$$\mathbb{E}[X]=\Sigma_i x_i p(x_i)$$

$$\mathbb{E}[X]=-4.5\times0.1+3\times 0.2+6\times0.3+10\times0.4=5.95$$

No integral is needed in this case.


问:
Unbounded second derivative of continuous bounded function
Does a continuous function $f$ exist where:

$f$ is continuous,

$f$ is known to be bounded with a codomain of $\left[0,1\right]$

$f''$ (second derivative) is unbounded with a codomain of $(-\infty,+\infty)$

答:
Presumably the answer is yes: a function can be bounded but wiggle very vigorously.

So let's try to make one.  Maybe we should start with $\cos(x)$, that's between $-1$ and $1$, easily fixed by adding $1$ and dividing by $2$. Unfortunately all derivatives are bounded.  So give it a little more vigor by say looking at $f$ where
$$f(x)=\frac{\cos(x^2) +1}{2}$$

We have $f'(x)=-x\sin(x^2)$ and
$$f''(x)=-(2x^2\cos(x^2) +\sin(x^2))$$.

Now show that we can make $f''(x)$ arbitrarily large positive or negative.  Then the Intermediate Value Theorem will show that the range of $f''(x)$ is the one you were looking for.

To make $f''(x)$ very large positive, just choose $x$ large such that $\cos(x^2)=-1$.  To make  $f''(x)$ very large negative, choose $x$ large such that $\cos(x^2)=1$.  Each is easy to arrange.


问:
How to prove that $\frac{1}{2} \left( 1 - \sqrt{3} \pm i \sqrt{2} \sqrt[4]{3} \right)$ is not a root of unity?
I have a proof for the complex roots $\sqrt{2} - 1 \pm i \sqrt{2\sqrt{2}-2}$ which involve a contradiction using the minimal polynomial of the real part of the root, but the proof does not work on the requested number, as the minimal polynomial of $\frac{1 - \sqrt{3}}{2}$ is $t^2 - t - \frac{1}{2}$. (not integral, so a lemma I used does not work).

The lemma:

If $\alpha$ is a root of unity whose real part is an algebraic integer, then $\alpha^4 = 1$.

Perhaps such a proof is impossible?

If this unfortunately turns out to be the case, the most elementary proof possible will be accepted.

答:
Let $\alpha:=\frac12(1-\sqrt{3}+i\sqrt{2}\sqrt[4]3)$, then $|\alpha|=1$, whence $\overline{\alpha}=\frac1\alpha$. Therefore,
$$1-\sqrt{3}=2\operatorname{Re}(\alpha)=\alpha+\overline{\alpha}=\alpha+\frac1\alpha.$$
We conclude that
$$3\alpha^2=\left(\alpha^2-\alpha+1\right)^2=\alpha^4+\alpha^2+1+2(-\alpha^3+\alpha^2-\alpha),$$
which may be simplified to
$$\alpha^4-2\alpha^3-2\alpha+1=0.$$
Now, we use @JyrkiLahtonen's remark that the above polynomial is irreducible because $\alpha+\overline{\alpha}\not\in\mathbb{Q}$, and it has two real zeroes, at approximately $0.435$ and $2.297$. Those are clearly not roots of unity, and if $\alpha$ were a root of unity, all its conjugates would be too.

Alternatively, once you know the polynomial is (monic) irreducible, it can only have a root of unity as one of its roots if it is cyclotomic. Take a quick look at the first $20$ or so cyclotomic polynomials, and you'll see that's not the case.


问:
Derivatives of Volume Integral and the Fundamental Theorem of Calculus
My question is regarding the Fundamental Theorem of Calculus and if it applies to a volume integral. I will try to present the problem but I am a bit rusty on my calculus.

The given situation I have is a volume integral of a multivariable function $ F = F(x,y,z,t) $
which is equal to zero:

$$ \int [F(x,y,z,t) ]\ dV = 0$$

$ F(x,y,z,t) $ is a scalar function describing temperature and heat - I can go into more details if that is relevant but the function is not fully defined (it is unknown).

To solve for $ F $ I would like to take the derivatives. I am not sure if the following steps are valid!

First, write explicitly the volume integral:

$$ \iiint[F(x,y,z,t)]\ dxdydz = 0$$

Take the derivatives:

$$ \frac{\partial}{\partial x} \frac{\partial}{\partial y} \frac{\partial}{\partial z} \Big( \iiint[F(x,y,z,t)]\  dxdydz \Big) = 0 $$

Then, apply fundamental theorem one by one (beginning from the inside $ \frac{\partial}{\partial z} \int \big[F(x,y,z,t)]\ dz = F(x,y,z,t)\  $) until you get the result:

$$ F(x,y,z,t) = 0 $$

Is this allowed? In my research I read that the multivariable "analog" to the Fundamental Theorem is Stokes Theorem, but I do not see how that is relevant in this situation. Again my calculus is rusty and it was never good to begin with so I hope this isn't something appears in the first result of a google search (I googled!).

As a test to verify I took a couple random multivariable functions, integrated them over $x$, $y$ and $z$ then then took the the derivatives and I did indeed return the integrand. I did this on wolfram alpha. I would like to know if this is a general principle that I just don't know the name of (Multivariable FT of $C$?) or if there is a mistake in my work.

Thank you kindly!

答:
No, it does not work that way.

The fundamental theorem works for an integral with variable bounds, hence a function of the bounds values. You just have a constant $V$.

Now think of a function $F$ that is anything positive in half of the domain, and the opposite in the rest of the domain: $V=0$, but $F\ne 0$.


问:
Is $\mathbb{Q}^{2}\cup \mathbb{I}^{2} $ connected?
It is well-known that $\mathbb{Q}^{2}$ and $\mathbb{I}^{2}$ (where $\mathbb I = \mathbb R\setminus \mathbb Q$) are disconnected spaces with the usual topology from $\mathbb{R}^{2}$, but how about the union? I figure it is, but I couldn't solved it. I've tried to express the set as the union of connected sets with nonempty intersection, because I think that is the key; but I reached nowhere.

答: Another way of saying the same idea as Robert Israel. Take families of rationals $p_n,q_n$ ($n\in\mathbb Z$) such that: $p_n<p_{n+1}$; $q_n\ne q_{n+1}$; $p_n\to a$ and $q_n\to b$ as $n\to-\infty$; and  $p_n\to c$ and $q_n\to d$ as $n\to\infty$. Take segment of every $(p_n,q_n)$ to $(p_{n+1},q_{n+1})$. The union of all of these segments, together with the points $(a,b)$ and $(c,d)$, is the graph of a continuous function $f:[a,c]\to\mathbb R$ satisfying what Robert Israel asserts.


问:
Is there an unlimited number of sequences of consecutive numbers with $\mu(n)=0$ of any length?
EDIT

I have received interesting comments to my post. Especially the comment of @Martin Hopf showed that this problem is a "classic" and by no means new.

Here is Eric Weissten's article on "squareful (nice term!) numbers" https://mathworld.wolfram.com/Squareful.html from which you can find the topic exposed.

This article https://arxiv.org/pdf/1210.3829.pdf is an exhaustive study of the problem.

Original post

Consider the subset of the natural numbers $n\ge 1$ which have at least one square prime factor. These can be formally defined by $\mu (n) = 0$, where $\mu$ is the Möbius function.

The first 40 of these are

$$s_{0}=\{4,8,9,12,16,18,20,24,25,27,28,32,36,40,44,45,48,49,50,52,54,56,60,63,64,68,72,75,76,80,81,84,88,90,92,96,98,99,100,104\}$$

Closer inspection of $s_{0}$ shows that there are sequences of consecutive such numbers. We call them compact sequences. If sorted by the length $m$ of the compact sequence we find in $s_{0}$ the following

$$m=2: \{8,9\}, \{24,25\}, \{27,28\},\{44,45\},\{48,49\},\{49,50\},\{63,64\},\{75,76\},\{80,81\},\{98,99\},\{99,100\}$$
$$m=3:  \{48,49,50\}, \{98,99,100\}$$

We see that for a given length $m$ there is more than one sequence, and that for the given length 40 of $s_{0}$ there are no sequences with length $m\gt3$.

Prolonging the list $s_0$ we find also longer compact sequences, and, again, for a given length there is more than one sequence.

Quoting only the first member of the first appearance of the corresponding sequence
I found numerically in the format $\{m,\text{first term of first appearance}\}$

$$c=\left(
\begin{array}{cc}
 2 & 8 \\
 3 & 48 \\
 4 & 242 \\
 5 & 844 \\
 6 & 22'020 \\
 7 & 217'070 \\
 8 & 1'092'747 \\
 9 & 8'870'024 \\
10 & 221'167'422\\
\end{array}
\right)$$

Remark: this sequence is contained in OEIS (https://oeis.org/A045882). Thanks to @Martin Hopf for pointing this out in a comment.

Conjecture

(1)  there is a compact sequences for any given length and

(2)  there are infinitely many compact sequences for any given length

Unfortunately, I was not able to prove or disprove the conjecture. Can you do better?

Additional question

(3) can you devise a formula for the first appearance of the compact series for given $m$?

答:
Letting $p_1,\dots,p_m$ be distinct primes. Then, using Chinese remainder there, find $x$ such that:

$$x\equiv  -i \pmod{p_i^2}$$

Then $p_i^2\mid x+i$ for $i=1,\dots,m.$ and there are infinitely many such $x.$

(This assumes you don’t require $\mu(x)\neq 0$ and $\mu(x+m+1)\neq 0.)$

But it does mean we can get arbitrarily long consecutive sequences.

The smallest for a particular $m$ is probably tricky. This is similar to the case where $x+1,x+2,\dots,x+m$ are all non-primes. It is easy to show that such $x$ exists, but it is hard in general to find the smallest $x$ for a given $m.$


问:
Hahn-Banach theorem seems to permit "unpermitted" operations
I've been taught to be very careful with the types of operations that we perform on functions in Sobolev spaces. Yet, the Hahn-Banach theorem seems to completely do away with the necessity of restraint. It says that any functional in the dualspace of a linear subspace may be extended to a functional in the dualspace of the complete space. If the subspace is dense, then this extension is even unique. Two examples that bother me:

The trace of the gradient is an unbounded operator in $H^1$, so the surface integral thereof is not a bounded linear functional. But on $H^2$ it would be, and this is a dense linear subspace. So Hahn-Banach gives us a unique extension of this operation to $H^1$.

The Dirac delta distribution is in the dualspace of $H^1_0([0,1])$, a dense linear subspace of $L^2([0,1])$. Per Hahn-Banach we are thus permitted to take point evaluations in $L^2([0,1])$.

These are two operations that I know are not allowed, so I must be misunderstanding the theorem or drawing the wrong conclusion. If we focus on the second example: are we really allowed to use Hahn-Banach to extend $\delta[\cdot]$ to $L^2([0,1])$? And if so, then what kind of extension would this produce if not a point evaluation?

答: The misunderstanding here is that in these spaces your norms are changing. Remember that a set is dense in a topological space if the closure of that set in the topology equals the whole space, and this topology very much depends on the norm. So indeed the set $H^1_0$ is a dense subspace of $L^2$, when equipped with the $L^2$ norm (and the corresponding topology), but that does not mean that the usual $H_0^1$ equipped with the $H^1$ norm is a dense subspace of $L^2$ topologically speaking, because these are in fact two entirely different topological spaces, so that notion does not even make sense. The $H^1$ norm and $L^2$ norm are very different, so the dual of $H_0^1$ as a Banach space is different from the dual of the subspace of $L^2$ consisting of the vectors of $H_0^1$. These are the same vector spaces, but different normed spaces. The Hahn-Banach theorem allows us to extend linear functionals on the latter space to linear functionals on $L^2$, but not on the former.


问:
Latin word that Euler used for "illustrative example"
I'm getting old and forgetful.

Everyone here knows the difference between theorem, corollary, lemma, proposition, conjecture, axiom, and postulate.

I once heard another one, a Latin term, that IIRC meant something like "illustrative example." The only example I ever found of it being used was from Euler.

It may be rare, but I like having a complete list. Can someone help me out and remind me what this mathematical term is?

答:
Here is one, which I like:

scholium NOUN (pl. scholia)

Pronunciation /ˈskəʊlɪəm/

historical

A marginal note or explanatory comment made by a scholiast.

‘They fall into two categories: the first, a group of ten plays which have been transmitted to us in our medieval manuscripts complete with the accumulation of ancient notes and comments that we call scholia.’

Origin

Mid 16th century modern Latin, from Greek skholion, from skholē ‘learned discussion’.

Lexico


问:
How do I show there exist $k \in [0,n]$ such that $\| f_k(y)-x\|< \epsilon$?
Let $X$ be a bounded and closed subset of $\mathbb{R}^4$. Let $f: X \rightarrow X $ be a homeomorphism. Write $f_n$ for the nth iterate of $f$ if $n>0$, for the $-nth$ iterate of $f^{-1}$ if $n<0$, and the identity map if $n=0$. Thus, $f_{n+1}(x)=f(f_n(x))$ for all $n \in \mathbb{Z}$.

Write $A(x)=\{f_n(x):n \in \mathbb{Z}\}$ for $x \in X$. Suppose that $A(x)$ is dense in $X$ for all $x \in X$.

Show that for each $x \in X$ and all $\epsilon >0$, there exist $n>0$ such that for all $y \in X$, there exists $k  \in [0,n]$ such that $\| f_k(y)-x\|< \epsilon$.

Anyone suggest some direction to solve this question?

答:
Too long for a comment, requested by OP.

Assume that $f$ is an isometry, and that $X$ is bounded (but not necessarily closed).

Let $\epsilon >0$. Then, by a simple compactness of closure argument, we can find $F=\{y_1,...,y_k \} \subseteq X$ so that for all $z \in X$ there exists some $1 \leq j \leq k$ with
$$
\|z-y_j \|_\infty < \frac{\epsilon}{2} \qquad (*)
$$

Since by assumption $A(x)$ is dense in $X$, for each $1 \leq j \leq k$ there exists some $n_j \in \mathbb Z$ so that
$$
\|f_{n_j}(x)-y_j\| \leq \frac{\epsilon}{2}
$$
Since $f$ is an isometry, this is equivalent to
$$
\|x-f_{-n_j}(y_j)\| \leq \frac{\epsilon}{2} \qquad (**)
$$

Now, pick $M,N \in \mathbb N$ so that $-M \leq n_j \leq N$ for all $1 \leq j \leq k$.
Set $n=N+M$. We claim that this $n$ works.

Let $y \in X$ be arbitrary. Since $f_{M}(y) \in X$, by (*) there exists some $1 \leq j \leq k$ so that
$$
\|f_{M}(y)-y_j \| \leq \frac{\epsilon}{2}
$$

Since $f$ is an isometry we get
$$
\|f_{M+n_j}(y)-f_{n_j}(y_j) \| \leq \frac{\epsilon}{2}
$$
and hence by (**)
$$
\|f_{M+n_j}(y)-x \| <\epsilon
$$
Since $0=M+(-M) \leq M+n_j \leq M+N=n$ this proves the claim.


问:
Using Green's formula for a vector field in a planar point set
I am trying to calculate the integral of the vector field $F = \begin{bmatrix}-y & x\end{bmatrix}^T$ in the point set $S = \{(x, y) \in \mathbb{R}^2\mid 1 \leq x^2 + y^2 \leq 4\}$ in two ways with Green's formula. Currently I am having both computational and conceptual issues.

The computational issue to which I ran into is that if I am not mistaken, $\int\int_S(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y})dA = \int_1^2\int_1^{\sqrt{4 - x^2}}2dydx = \int_1^2\sqrt{4 - x^2} - 1dx$. From this I don't really see any other way forward than to substitute $x = 2\sin(u)$ and recalculate the limits. Is this really the only way forward, or is there a better trick to this?

The conceptual issue I am having is that how do even start to compute the boundary integral? Sure, $\partial S = C_1 \cup C_4$, where $C_1 = \{(x, y) \in \mathbb{R}^2\mid x^2 + y^2 = 1\}$ and $C_4 = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 = 4\}$. How can I parametrize $\partial S$? I am bit rusty with path integrals, so does it hold that $\int_{\partial S}F_1 dx + F_2 dy = (\int_{C_1}F_1 dx + F_2 dy) + (\int_{C_4}F_1 dx + F_2 dy)$?

Edit: The question I am solving is that Calculate the integrals of Green's theorem w.r.t vectorfield $F$ and pointset $S$.

答:
Your bounds for the region is not correct. The region is the area between two circles $x^2+y2 = 1$ and $x^2+y^2 = 4$. If you do it in cartesian coordinates, you will have to split your integral. It is easier to do it in polar coordinates, the limits should be $1 \leq r \leq 2, 0 \leq \theta \leq 2 \pi$ and $dA = r \ dr \ d\theta$.

As far as the question, I think it could be written more clearly.

To apply Green's Theorem, the curve has to be positively oriented, piecewise smooth, simple and closed.

So if we define curve $C$ as going from point $A$ along $C_1$ in anti-clockwise direction to point $B$ and then to point $C$, along $C_2$ in clockwise direction (region continues to be on your left) back to point $C$, then to point $B$ and then along $C_1$ in anticlockwise direction back to point $A$, we have a region that meets all the conditions we spoke about earlier to apply Green's theorem.

When we are doing line integral, note that the line integral along $BC$ and $CB$ will cancel out.

So the line integral along $C$ is equivalent to line integral along $C_1 \cup C_2$ where $C_1$ is anti-clockwise along $r = 2$ and $C_2$ is clockwise along $r = 1$.

You can parametrize $C_1$ as $(2 \cos t, 2 \sin t), 0 \leq t \leq 2\pi$ and $C_2$ as $(\cos t, \sin t), 2\pi \leq t \leq 0$.


问:
Bell numbers and a card shuffle
A deck of n cards  may be 'shuffled' by moving the top card to any (random) position in the deck, and performing this operation n times. Martin Gardner asserts (Scientific American, May 1978) that the number of such shuffles that return the deck to its original state is equal to the nth Bell number. This appears to be true for the first few cases but I have not been able to explain why it should be so for all n.
A set partition that corresponds to the problem seems hard to find. Symmetric groups offer some insights but nothing conclusive, so far.

答:
It is slightly more convenient to consider the shuffles "played backwards", with each move taking an arbitrary card to the top of the deck. Then we're still counting shuffles (call them neutral) that leave the deck in its initial state when played.

Let's label the cards by (elements of) $[n]:=\{1,\ldots,n\}$, and let the initial state of the deck be $(1,\ldots,n)$. Then any (not only neutral) shuffle is uniquely determined by $\ell=(\ell_1,\ldots,\ell_n)$, where $\ell_k$ is (the label of) the card that is moved to the top at step $k$.

Now, to each neutral shuffle, we assign the partition of $[n]$ into the equivalence classes of the relation $j\sim k\iff\ell_j=\ell_k$. This is reversed as follows: let $[n]=N_1\cup\ldots\cup N_m$ be a partition of $[n]$, with $N_1,\ldots,N_m$ ordered so that $j\mapsto\max N_j$ is decreasing; then, we set $\ell_k=j$ iff $k\in N_j$.

The order of $N_j$ ensures that the latter gives a neutral shuffle (in the resulting deck, the card $1$ will be on top because it is moved last, and so on upto the card $m$; the remaining cards (if any) are not used, so these go to the bottom, with their order unchanged). It is also clear that different partitions map to different $\ell$'s.

Title: Bell numbers and a card shuffle


问:
Showing $\lim \sup_{n \to \infty} s_n = + \infty$ is in the set of subsequential limits of $\{s_n\} \in \mathbb R$

Let $E$ be the set of subsequential limits of $\{s_n\} \in \mathbb R$. If $\lim \sup_{n \to \infty} s_n = +\infty$, then $E$ is not bounded above in $\mathbb R$, so given any $N \in \mathbb R$, there's some $\{s_{n_k}\}$ that converges to $\color{green}{\text{something $\ge N.$}}$ So for any $\epsilon > 0$, there are infinitely many elements of $\{s_{n_k}\}$ which are $\ge N - \epsilon.$ Now fix some $\epsilon$ and let $M = N - \epsilon$. So, given any $M \in \mathbb R$, there are infinitely many elements $\{s_n\}$ that are $\ge M$ meaning there's some subsequence that diverges to $+\infty$ and so $+\infty \in E.$

If $E$ is not bounded above, then some $\{s_{n_j}\}$ diverges to $+\infty$ meaning $+\infty \in E$. Isn't that enough? Why do we need the part of the proof that follows the $\color{green}{\text{part in green}}$ of the quote above?

Given definition of subsequential limits

答:
that is perfecty right: Bounded above means
$$
\exists M>0 \forall n:  x_n \le M
$$
and so "not bounded above" means
$$
\forall M>0 \exists n: x_n > M
$$
Which means, that you can let $M=1$, find $n\ge 1$ such that $x_n > 1$.
Then you let $M= x_n+1$ and you know that there exists $k$ with $x_k> x_n+1$.
The point is: if there were no $k>n$ with this property, then all larger values were taken on indices $1..n$ - and the sequence hence bounded. That means that you can assume w.r.o.g. that $k>n$. Iterating you obtain an increasing subsequence that tents to $+\infty$.


问:
Find the normalized eigenvectors of a matrix
Let $T=\begin{pmatrix}5 & 0 & 0 \\ 0 & 2 & i\\ 0 & -i & 2 \end{pmatrix}$.

I found the eigenvalues and eigenvectors already and they are $1,3,5$ and $\begin{pmatrix}0\\-i\\1\end{pmatrix}$,$\begin{pmatrix}0\\i\\1\end{pmatrix}$, and $\begin{pmatrix}1\\0\\0\end{pmatrix}$.

According to mathematica, their orthoganalized form will be $\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
 0 & -\frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array}
\right)$

But I am getting wildly different answers when I try to find the normalized eigenvectors using the Gram-Schmidt process.

答: The matrix that Mathematica gives is not the matrix formed by the unit eigenvectors, but the conjugate transpose of it. That is, it is a unitary matrix $U$ such that $T = U^\ast\pmatrix{5\\ &1\\ &&3}U$. You can easily see that the columns of $U^\ast$ are scalar multiples of your three eigenvectors.


问:
Help with finding the basis of a polynomial set
let $U = \{p \in P_4(F) : p(2) = p(5)\}$. find a basis of $U$

this is what i tried:
\begin{align*}
p(x) &= ax^4 + bx^3 + cx^2 + dx + e\\
p(2) &= p(5)\\
a16 + b8 + c4 + d2 + e &= a625 + b125 + c25 + d4 + e\\
609a + 117b + 21c + 3d &= 0\\
d &= -609a/3 - 117b/3 - 21c/3\\
d &= -203a - 39b - 7c\\
p(x) &= ax^4 + bx^3 + cx^2 + (-203a - 39b - 7c)x + e\\
p(x) &= a(x^4 - 203^x) + b(x^3 -39x) + c(x^2 -7x) + e\\
\end{align*}

basis: $\{(x^3 - 203), (x^2 -39), (x - 7), (1)\}$

is that correct? it's my first time reading a linear algebra book (linear algebra done right), having a bit of trouble with the exercises, thanks for any help :), sorry for any mistakes in nomenclature.

答:
You're almost there, but you shouldn't have taken the factor of $x$ out of each term. This is because $x$ is not an element of the base field, thus you can't treat it as a coefficient, but rather, think of it as a vector.

As you correctly pointed out, $U$ consists of all polynomials of the form:
$$p(x) = a(x^4-203x)+b(x^3-39x)+c(x^2-7x)+e$$

This means that $U = \text{Span}(x^4-203x, x^3-39x,x^2-7, 1)$. Now you need to check wheter this vectors are linearly independent (Hint: use the fact that they have different degrees) and conclude the argument!


问:
Given $x \in {\rm int}(K)$ (covered by nondecreasing cones $\{K^r\}$), does there exist $r_0$ s.t. $x \in {\rm int}(K^{r_0})$?
Let $K\ (\subseteq \mathbb{R}^n)$ be a closed convex cone and $\{K^r\}_{r=0}^\infty$ be a family of closed convex cones satisfying $K^r \subseteq K^{r+1} \subseteq K\ (\forall r)$.

Assume that ${\rm int}(K) \subseteq \bigcup_{r=0}^\infty K^r$ and there exists $x \in {\rm int}(K)$.

Then, does there exist $r_0$ such that $x \in {\rm int}(K^{r_0})$?

Besides, can we remove the assumption that $\{K^r\}$ are closed?

Intuitively, this proposition is true, but I cannot prove it rigorously. Note that a similar problem appears in the proof of Theorem 6.1 in the following paper: https://www.jstor.org/stable/25151657?seq=1#metadata_info_tab_contents

答:
This is a bit of a round-about way of proving it, but it works. (There's probably a more straightforward way of doing this.)

First note that almost all of the cones $K_0,K_1,\dots$ must have non-empty interior. (Indeed, it is impossible for all of them to have empty interior, so there must exist an index $j\in\mathbb{N}$ such that $\operatorname{int}(K_r)\neq\emptyset$ for every $r\geq j$.)

Now let $x\in\operatorname{int}(K)$. We want to show that there exists an index $r_0\in\mathbb{N}$ for which $x\in\operatorname{int}(K_{r_0})$. By the assumption that $$\operatorname{int}(K)\subseteq\bigcup_{r\in\mathbb{N}} K_r,$$ there is an index $r_1\in\mathbb{N}$ for which $x\in K_{r_1}$. By the observation in the preceeding paragraph, we may suppose without loss of generality that $\operatorname{int}(K_{r_1})$ is non-empty. If $x\in\operatorname{int}(K_{r_1})$ then we are done, so suppose that $x\not\in\operatorname{int}(K_{r_1})$. (That is, $x$ is on the boundary of $K_{r_1}$.) There exists a vector $y\in\mathbb{R}^n$ such that $$x-y\in\operatorname{int}(K_{r_1}) \qquad\text{and}\qquad x+y\not\in K_{r_1}.$$ As $x\in\operatorname{int}(K)$, there exists a value $\epsilon>0$ small enough such that $x+\epsilon y\in\operatorname{int}(K)$. Define a new point $x_1=x+\epsilon y$. For this new point $x_1\in\operatorname{int}(K)$, there exists an index $r_0\in\mathbb{N}$ with $r_0>r_1$ such that $x_1\in K_{r_0}$. As it is the case that $$x-y\in\operatorname{int}(K_{r_1}) \qquad\text{and}\qquad K_{r_1}\subseteq K_{r_0},$$ it follows that $x-y\in\operatorname{int}(K_{r_0})$ as well. Every nontrivial convex combination of the points $x_1\in K_{r_0}$ and $x-y\in\operatorname{int}(K_{r_0})$ must be in $\operatorname{int}(K_{r_0})$. That is, for every $t\in(0,1)$, it holds that
$$tx_1 + (1-t)(x-y)\in\operatorname{int}(K_{r_0}).$$
Choosing $t = \frac{1}{1+\epsilon}$ gives us the desired result that $x\in\operatorname{int}(K_{r_0})$.