问:
Functional not attaining its norm
Consider the space $(\ell^1,||\cdot||_{\ell^1}+||\cdot||_{\ell^p})$ for some $p\in (1,\infty)$. Can someone provide an example of any $F:\ell^1\to \mathbb{R}$ linear and continuous such that $F(x)\neq ||F||^*$ for any $x\in \ell^1$ with $||x||_{\ell^1}+||x||_{\ell^p}=1$?

Title: Functional not attaining its norm

Body: Let $\displaystyle{F(x)=\sum_{n=1}^\infty x_n}$. $|F(x)|\leq \|x\|_1<\|x\|_1+\|x\|_p$ so that $\|F\|^*\leq 1$. Let $u_k$ be defined by $u_{k,n}=1/k$ if $n\leq k$ and $u_{k,n}=0$ if $n> k$. Since $F(u_k)=1$ and $\|u_k\|_1+\|u_k\|_p= 1+k^{1/p-1}\rightarrow 1$ as $k\rightarrow\infty$, $\|F\|^*= 1$.


问:
Theorem 7.13 Rudin's Real and Complex Analysis. Regularity of measure
I'm having trouble with the proof of the following:

Theorem 7.13 Associate to each $x \in \mathbb{R}^k$ a sequence $\left\{ E_i(x) \right\}$ that shrinks to $x$ nicely. If $\mu$ is a complex Borel measure and $\mu \perp m$, then
$$
\lim_{i,+\infty} \frac{\mu(E_i(x))}{m(E_i(x))} = 0 \;\;\; a.e. [x] \;\;\; (1)
$$

Proof: The Jordan decomposition theorem shows that it suffices to prove under the additional assumptions that $\mu \geq 0$. In that case, arguing as in the proof of theorem 7.10
$$
\frac{\alpha(x)\mu(E_i(x))}{m(E_i(x))} \leq \frac{\mu(E_i(x))}{m(B(x,r_i))} \leq \frac{\mu(B(x,r_i))}{m(B(x,r_i))}
$$
Hence (1) is a consequence of the special case
$$
(D\mu)(x) = 0 \;\;\; a.e. \;\;\; [m],
$$
which will now be proved.
The upper derivative $\overline{D}\mu$, defined by
$$
(\overline{D}\mu)(x) = \lim_{i \to \infty} \left[ \sup_{0 < r < 1/n} (Q_r\mu)(x) \right] \;\;\; (x \in \mathbb{R}^k)
$$
is a Borel function, because the quantity in brackets decreases as $n$ increases and is, for each $n$, a lower semicontinuous function of $x$; the reasoning used in Sec. 7.2 proves this.

Here it comes the part I don't understand.

Choose $\lambda > 0$, $\epsilon > 0$. Since $ \mu \perp m$, $\mu$ is concentrated on a set of Lebesgue measure $0$. The regularity of $\mu$ (Theorem 2.18) shows therefore that there is a compact set $K$, with $m(K) = 0$, $\mu(K) > \left\lVert \mu \right\rVert - \epsilon$.

I don't understand where the existance of $K$ comes from (is maybe given by the property of the measure in theorem 2.18? Or some construction maybe used in the proof of such theorem?).

Where do $m(K) = 0$ and $\mu(K) > \left\lVert \mu \right\rVert - \epsilon$ come from?

Probably my two questions are actually connected, but I can't really figure.

答: We are assuming that $\mu $ is  a positive measure. There exists $A$ such that $m(A)=0$ and $\mu (A^{c})=0$. By regularity there exists  compact set $K \subseteq A$ such that $\mu (K) >\mu (A)-\epsilon$. Note that $\|\mu\|=\mu (A)+\mu (A^{c})=\mu (A)$.  Hence $\mu (K) >\|\mu\| -\epsilon$. Of course $m(K) \leq m(A)=0$.

ContentLicense: CC BY-SA 4.0


问:
Simultaneous diagonalization and normal matrices

Let $A,B \in M_n (\mathbb{C})$ be normal matrices such that $AB = BA$. Prove that these matrices can be simultaneously unitarily diagonalized.

A matrix $U$ is unitary if and only if $U^{*} = U^{-1}$. Simultaneous diagonalization by a unitary matrix therefore means there exists a unitary $U$ such that $U^{-1} AU = U^{*} AU$ is diagonal and $U^{-1} BU = U^{*} BU$ is diagonal. I believe I need to first show that $A$ and $B$ have the same eigenvalues; if $u$ is an eigenvector of $A$ with eigenvalue $\lambda$, then
$$
ABu = BAu = \lambda Bu,
$$
which tells me that $Bu$ is an eigenvector of $A$ with the same eigenvalue $\lambda$ provided that $Bu \neq 0$, but that doesn't tell me anything about the eigenvectors of $B$.

Any help would be appreciated.

答:
Recall that the eigenspaces of $A$ are the same as the eigenspaces of $A^*$ (by normality). Then show that these $A/A^*$-eigenspaces are both $B$ and $B^*$-invariant.

So you can decompose $V=V_1\oplus\cdots\oplus V_k$, where the $V_i$ are $A/A^*$-eigenspaces, and are all $B$ and $B^*$-invariant. It follows that the restrictions $B_i\colon V_i\to V_i$ (more precisely, the 'compression') satisfies $B_i^*=B^*|_{V_i}$, so they are also normal.

By the spectral theorem, each $V_i$ has an orthonormal eigenbasis $\mathcal{B}_i$ for $B_i$. These will also be orthonormal eigenvectors for $A$, since $A$ acts as some multiple of the identity on each $V_i$. The union $\mathcal{C}=\mathcal{B}_i$ is an orthonormal eigenbasis both $A$ and $B$ on all of $V$. The matrix $U$ whose columns are the elements of this eigenbasis diagonalizes both $A$ and $B$.


问:
Heading angle change traveling along circle arc in 3D
I have a hypothetical vehicle traveling along an arc in 3D defined by curvature ($1/r$) and the "tilt angle" - if I let my vehicle travel indefinitely, the trajectory would be a circle; by "tilt angle" I mean the angle between the plane in which lies the circle and the xy plane (I don't know if there actually is a proper name for that). I need to calculate the heading angle (projection of into xy plane) at any given point of the circle (Curvature, the distance traveled along the circle and the tilt angle are given)

In 2D, this is quite easy: $heading angle = curvature * distance$

In 3D, it gets more complicated. The angle depends obviously on curvature and distance and on tilt angle.

My first intuition was to take the formula from 2D and multiply it by $cos(tilt)$, but that does not work, because if I travel exactly quarter of the circle, the heading angle is exactly perpendicular to inital heading angle for any tilt.

My second intuition was, that for tilt angle $\pi/2$ (and $-\pi/2$) is the heading angle at $1/4$ ($3/4$) of the circle undefined as the heading angle completely disappereas if the when the vehicle travels straight up/down, which made me think maybe division by zero. For tilt angle $\pi/2$ ($-\pi/2$), the heading angle does not change, except for the two undefined points, which makes me again think multiplying by $cos(tilt)$.

Does the formula (if there is any) change significantly, if the vehicle enters the arc already at and angle? (heading angle in the xy plane, as well as up/down angle) My intuition would be that this does not matter much, because I could compute the change and add it to the initial angles, but I am not really sure.

答:
The trick here is to forget about the circle. In the plane of the circle, we have the following diagram:

In the 2-dimensional case, $\theta$ is the change in angle that you want, and we see that indeed $\theta = \frac{d}{r}$ (with everything measured in radians). In three dimensions, we just need to adjust that:

In this diagram $\psi$ is your "tilt angle".

Consider first the triangle $ABC$. The angle at $B$ is right ($AB$ is parallel to the $y$ axis, and $BD$ is parallel to the $x$ axis, by construction), and we know the hypotenuse and one other angle. We can therefore find $BC$ to be $r\sin\theta$ and $AB$ to be $r\cos\theta$.

Now consider triangle $BCD$. The angle at $D$ is right ($BD$ is parallel to the $x$ axis, $CD$ is parallel to the $z$ axis). We again know the hypotenuse and one other angle, so we can obtain $BD = r\sin\theta\cos\psi$.

Next consider triangle $ABD$, which is where the magic is going to happen. The angle at $B$ is right, and we know two sides. Let $\alpha$ be the angle at $A$. Then we have $\tan \alpha = \frac{\sin\theta\cos\psi}{\cos\theta} = \tan\theta\cos\psi$. Pythagoras also gives $AD$ to be $s := r\sqrt{\cos^2\theta+\sin^2\theta\cos^2\psi}$.

Now, look closely at triangle $ADE$. It lies in the $xy$ plane, and is just a copy of the first diagram, with $\theta$ replaced by $\alpha$ and $r$ replaced by $s$. Thus, our change in heading is precisely $\alpha$.

To get this into your desired form, we now need only note that $\theta = \frac{d}{r}$, and so our change in heading is precisely $\arctan(\tan\theta\cos\psi) = \arctan\left(\tan\left(\frac{d}{r}\right)\cos\psi\right)$.

Note, however, that "tilt angle" is ambiguous: I've assumed that you're pitching around an axis perpendicular to the original direction of travel, but there are other equally valid interpretations that will give entirely different answers.


问:
Multivariate Dirac delta function

I was told that the multivariate Dirac delta function $\delta(\mathbf x), \mathbf x \in \mathbb R^n$ with

\begin{align}
0 &= \delta(\mathbf x) \quad \forall \mathbf x \neq (0,0,...,0) \\
1 &= \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \delta(\mathbf x) \; \text d x_1 \cdots \text d x_n
\end{align}

is equivalent to the product of the it's marginals, so

\begin{align}
\delta(\mathbf x) &= \delta(x_1,x_2,...,x_n) = \delta(x_1) \cdot \delta(x_2) \cdots \delta(x_n) = \prod_{i=1}^n \delta(x_i)
\end{align}

Why is that the case?

Also, I was told that a Dirac delta function that not only non-zero for the origin

\begin{align}
0 &= \delta(x_1, x_2, ..., x_m, ... , x_n) = \prod_{i=1}^m \delta(x_i) \quad \forall x_1, x_2, ... , x_m \neq 0 \\
1 &= \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \prod_{i=1}^m \delta(x_i) \; \text d x_1 \cdots \text d x_n
\end{align}

for $m<n$. Why does the second equation (still) hold (while still integrating over the entire n-dimensional set/space)?

EDIT: For illustrative purposes I imagine

For illustration I image the full case in 1 to be equivalent to

\begin{align}
\delta(\mathbf x) = \begin{cases} \lim\limits_{a\rightarrow0} \quad \dfrac{1}{a^n} & \forall  x_i \in [-\frac a2,\frac a2], 1\le i\le n \\[6pt]
\quad 0 & \text{otherwise}
\end{cases}
\end{align}

Thais how it was introduced to me at university.

So, I was guessing I would need for the second case something like

\begin{align}
\delta'(\mathbf x) = \begin{cases} \lim\limits_{a\rightarrow0} \quad \dfrac{a^{n-m}}{a^m}
& \forall
x_i \in [-\frac a2,\frac a2], x_j, \in [-\frac2a,\frac2a],
1\le i\le m < j\le n,   \\[6pt]
\quad 0 & \text{otherwise}
\end{cases}
\end{align}

so that the "volume" / integral over $\delta'$ would be the
support multiplied by the "height" of the Dirac:

$$
\frac{a^m}{a^{n-m}} \cdot \frac{a^{n-m}}{a^m} = 1
$$

independent of $a$.

So, if we consider $\delta'$ to be measures for a measure space $(X, \Sigma, \delta')$, we take the $\delta'$-measure for an $S_1\subset X$ and then translate $S_1$ only in directions of $x_i, m<i\le n$ to $S_2$, then the $\delta'$-measure of $S_2$ should be the same as for $S_1$.

But I do not see how this relates to what I wrote in 2 above (before this edit). My main issue is, that $\delta(x_1) \cdot \delta(x_2) \cdots \delta(x_m)$ seems to "lack" the "normalization" so that the integral becomes 1: For example if we have $m=1, n=2$, then shouldn't the "height" of the Dirac delta function be 1? Because we have $a$ in one direction and $1/a$ in the other, so to get 1 for the integral, we need 1 as the height. But this does not seem to be incorporated into $\delta(x_1) \cdot \delta(x_2) \cdots \delta(x_m)$.

答:
Observe that
\begin{align}
\delta(x_1) \cdots \delta(x_n)=0 \Leftrightarrow x_1,...,x_n\neq0.
\end{align}
Just like it should be for $\delta(x_1,...,x_n)$.
Also:
\begin{align}
\int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \delta(x_1) \cdots \delta(x_n) dx_1\cdots dx_n= 1\cdots 1=1=\int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \delta(x_1,...,x_n) dx_1\cdots dx_n
\end{align}
(they are equivalent functions/distributions)

As the function $\Pi_{i=0}^m \delta(x_i)$ is not defined in the last n-m Variabels it is also integrated over, the integration is an integration over a Null Set. I think this is what is bothering you. But note that the function $\delta(x)$ itself only has a value other than zero at x=0, which also is a Null Set in the domain of the function (e.g. reel Numbers). One must conclude that the value of $\delta(0)$ is something like infinity (just for intuition), or else the integral would be zero. This is how you can think about why integrating over the dirac delta function is always going to evaluate to 1, regardless if you are integrating over Null Sets.

Title: Multivariate Dirac delta function


问:
There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$?
There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$, what I tried is if possible then define $g(z)=f(z)-{1\over 2^{1\over z}}$ Then $g({1\over n})=0$ and so $g(z)$ is entire and its $0$ set has limit point in it and so $f(z)={1\over 2^{1\over z}}$ which is not analytic at $0$?
Please help!

Edit: OOps! the way I defined $g(z)$ that is not entire! could any one give me hint?

答:
Hints:

$$(1)\;\;\;\;\;f\left(\frac1n\right)\xrightarrow[n\to\infty]{}0$$

$$(2)\;\;\;\;\;\text{The zeros of entire functions are isolated.}$$

ContentLicense: CC BY-SA 3.0


问:
Use of Borel Canteli Lemma for Almost sure convergence, proof verification
I attempted to prove that $X_{n}$ converges almost surely to $0$ with the use of Borel Cantelli and Chebyshev inequality.
\begin{align*}
\mathbb{P}(\left \{ \omega \in \Omega:\left | X_{n} \right | \leq \epsilon, \exists \ n_{0} \ s.t. \ \forall  \ n \geq n_{0} \right \})
&= \mathbb{P}\left(\bigcap_{n_{0}=1}^{\infty}\bigcup_{n\geq n_{0}}\left \{ \omega \in \Omega:\left | X_{n} \right | \leq \epsilon\right \}\right)\\
&= 1-\mathbb{P}\left(\bigcap_{n_{0}=1}^{\infty}\bigcup_{n\geq n_{0}}\left \{ \omega \in \Omega:\left | X_{n} \right | > \epsilon\right \}\right).
\end{align*}

I define as $A_{n}=\left \{ \omega \in \Omega:\left | X_{n} \right | > \epsilon\right \}.$ Now from Chebyshev inequality I get

$$\mathbb{P}(\left \{ \omega \in \Omega:\vert X_{n}\vert > \epsilon\right \})\leq \frac{\sigma^{2}}{\epsilon}\Rightarrow \sum_{n}\mathbb{P}(A_{n}) \leq \frac{\sigma^{2}}{\epsilon}.$$

Hence, from the Borel Cantelli lemma https://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma

and I get that $\mathbb{P}\left(\bigcap_{n_{0}=1}^{\infty}\bigcup_{n\geq n_{0}}\left \{ \omega \in \Omega:\left | X_{n} \right | > \epsilon\right \}\right)=0$. So,
$$\mathbb{P}(\left \{ \omega \in \Omega:\left | X_{n} \right | \leq \epsilon, \exists \ n_{0} \ s.t. \ \forall  \ n \geq n_{0} \right \})=1.$$

答:
There seems to be a typo the first equality.

Your application of Chebysev's inequality is not correct. What you can say is $P(|X_n| >\epsilon) \leq \frac {EX_n^{2}} {\epsilon^{2}}$. You provided no information about variance of $X_n$. If you know that $\sum EX_n^{2} <\infty$ then the rest of your proof would be correct.


问:
How to evaluate such integral with pole structure?
Let's have integral:
$$
I = \int \limits_{-\infty}^{\infty} \frac{e^{-\frac{x^{2}}{2}}}{x - a - i0}
$$
How to evaluate it?

I tried to do following:
$$
\frac{1}{x -a - i0} = \int \limits_{0}^{\infty}d\tau e^{-\tau (x - a - i 0)} \Rightarrow I = \sqrt{2 \pi}\int \limits_{0}^{\infty}d\tau e^{-\frac{\tau^{2}}{2} + ia\tau},
$$
which is expressed in terms of Erf. Is this right?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
I = &\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{\expo{-x^{2}/2} \over x - a - \ic 0^{+}}\,\dd x} =
\mrm{P.V.}\int_{-\infty}^{\infty}{\expo{-x^{2}/2} \over x - a}
\,\dd x +
\ic\pi\expo{-a^{2}\,/2}
\\[5mm] & =
\ic\pi\expo{-a^{2}\,/2}  +
\mrm{P.V.}\int_{-\infty}^{\infty}{%
\expo{-\pars{x + a}^{\,\,2}\,\,/2} \over x}\,\dd x
\\[5mm] & =
\ic\pi\expo{-a^{2}\,/2}  +
\int_{0}^{\infty}{%
\expo{-\pars{x + a}^{\,\,2}\,\,/2} -
\expo{-\pars{x - a}^{\,\,2}\,\,/2} \over x}\,\dd x
\\[5mm] & =
\ic\pi\expo{-a^{2}\,/2}  -
2\expo{-a^{2}\,/2}\
\underbrace{\int_{0}^{\infty}{%
\expo{-x^{2}\,/2}\sinh\pars{2ax}  \over x}\,\dd x}
_{\ds{{1 \over 2}\,\pi\on{erfi}\pars{\root{2}a}}}
\\[5mm] = &\
\bbx{-\pi\expo{-a^{2}\,/2}\on{erfi}\pars{\root{2}a} +
\ic\pi\expo{-a^{2}\,/2}} \\ &\
\end{align}


问:
Linear Algebra Proof on Eigenvectors
Let $A$ and $B$ be $n\times n$ matrices, with $B$ invertible. Assume that $\lambda$ is an eigenvalue of $A$ with eigenvector $v$.

Prove that $\lambda$ is still an eigenvalue of $B−1AB$ with eigenvector $B−1v$.

答:
If $Av=\lambda v$, then $(B^{-1}AB)(B^{-1}v)=B^{-1}A(BB^{-1})v=B^{-1}Av=B^{-1}\lambda v=\lambda (B^{-1}v)$,

so indeed $B^{-1}v$ is an eigenvector of $B^{-1}AB$ with eigenvalue $\lambda$.

Title: Linear Algebra Proof on Eigenvectors


问:
What’s wrong in the following reasoning that every idempotent matrix is the zero matrix or the identity matrix correct?
Suppose A is an nxn idempotent matrix.

Then this means that $AA=A$
$\iff AA-A= 0_{n\times n}$
$\iff A(A-I_{nxn})=0_{n \times n}$

$A=0_{n\times n}$ or $A=I_{n \times n}$

This would imply that any idempotent matrix is either the identity or the zero matrix...

答:
Any matrix that is a diagonal matrix and the diagonal entries are zeroes and ones only, is idempotent. For example,
$$A=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}.$$
Convince yourself that indeed $AA=A$. Also observe that $A(A-I)=0$, but neither $A$ nor $A-I$ is zero.


问:
The entries of a column of a stochastic, idempotent matrix all have the same value
Let $A \in [0,1]^{n\times n}$,  $A = A^2$ and $\sum _{j=1}^n a_{ij} = 1 \;\forall i \in \{1,..,n\}$ (sum of row values equals $1$ for every row). I want to show, that the column values of $A$ are the same for every row ($ a_{ij} = a_{kj} \;\forall i,j,k$).
I am pretty disconnected from linear algebra right now and a direct proof doesn't work for me right now. I would also be thankful for a hint.

EDIT: I forgot one condition; every entry of the matrix is $ > 0$.

答:
$A^2-A=0$ thus $A$ is diagonalizable in $\mathbb{R}$ and its eigenvalues are $0$ or $1$. If $A=(a_{i,j})_{1\leqslant i,j\leqslant n}$ where we suppose $a_{i,j}>0$ for all $i,j\in[\![1,n]\!]$, then $(a_{i,j}-\delta_{i,j})_{1\leqslant i,j\leqslant n-1}$ is invertible because for all $i\in[\![1,n-1]\!]$, we have
$$ \sum_{\substack{1\leqslant j\leqslant n-1 \\j\neq i}}|a_{i,j}-\delta_{i,j}|=\sum_{\substack{1\leqslant j\leqslant n-1 \\j\neq i}}a_{i,j}=1-a_{i,i}-a_{i,n}>|a_{i,i}-1| $$
and thus ${\rm rank}(A-I_n)\geqslant n-1$. In the general case where we only have $a_{i,j}\geqslant 0$, let $\varepsilon\in ]0,1[$ and $A_{\varepsilon}=(1-\varepsilon)A+\frac{\varepsilon}{n}U$ where the coefficients of $U$ are all $1$. Then $A_{\varepsilon}$ verifies the same hypothesis as $A$, and $(1-\varepsilon)a_{i,j}+\frac{\varepsilon}{n}>0$ for all $i,j\in[\![1,n]\!]$ and thus, because of what said above, ${\rm rank}(A_{\varepsilon}-I_n)\geqslant n-1$. But since $\lim\limits_{\varepsilon\rightarrow 0}(A_{\varepsilon}-I_n)=A-I_n$, we have ${\rm rank}(A-I_n)\geqslant {\rm rank}(A_{\varepsilon}-I_n)$ for all $\varepsilon$ small enough and thus ${\rm rank}(A-I_n)\geqslant n-1$. If we had ${\rm rank}(A-I_n)=n$, then $A-I_n$ would be invertible and thus all the eigenvalues of $A$ would be $0$ which is not because $A\neq 0$. Thus ${\rm rank}(A-I_n)=n-1$ and $1$ is an eigenvalue of $A$ of multiplicity $1$. Since all the other eigenvalues of $A$ are $0$, ${\rm rank}(A)=1$ and, if $i_0\in[\![1,n]\!]$ is such that $C_{i_0}\neq 0$ where $C_1,\ldots,C_n$ are the columns of $A$, then for all $j\in[\![1,n]\!]$, there exists $\lambda_j\in\mathbb{R}$ such that $C_j=\lambda_j C_{i_0}$. Summing all the coefficients of $C_j$ with $j$ fixed gives $\lambda_j=1$ and thus all the columns of $A$ are identical.

EDIT : Since $a_{i,j}$ is supposed to be $>0$ by the OP, we can get rid of the $A_{\varepsilon}$ trick.


问:
Every piecewise linear function can be written as a sum
Let $g$ be a real valued piecewise linear function on some interval $[a,b]$ such that if we divide $[a,b]$ into $k$ pieces, $g$ is linear on each piece. Then, for some $d, a_1,...,a_k, c_1,...,c_k$, $g(x)=d+\sum_{i}^{k} a_i|x-c_i|$. How can I justify that?</p>

答: Hint: Note that the function $$x\mapsto\sum_{i=1}^ka_i|x-c_i| $$ is continuous and (assuming $a=c_1<c_2<\ldots <c_k=b$) is linear in each interval $[c_i,c_{i+1}]$ with slope $\sum_{i\le k}a_i-\sum_{i>k}a_i$. Let $m_i$ be the slopes of the given function in $[c_i,c_{i+1}]$. Can you find suitable $a_i$ then? What is missing after that?


问:
Smallest axis-aligned bounding box of hyper-ellipsoid
Let $E$ be the $n$-dimensional ellipsoid defined by
$$E:=\{x \in \mathbb{R}^n: (x-c)^T A (x-c) \le 1\},$$
where $c \in \mathbb{R}^n$ is the center of the ellipsoid, and $A \in \mathbb{R}^{n \times n}$ is a symmetric positive definite matrix.

Question:
How can one efficiently compute the coordinate-axis-aligned bounding box that just barely contains the ellipsoid?

For a 2D example, see the following picture:

Note:
I ask this question, and answer it myself, because this question (in general form) is surprisingly absent from math.stackexchange even after 10+ years. Good answers to this question are hard to find on the internet more generally. After googling around, I eventually had to figure this out myself, and am posting here to spare future people the same trouble. Many websites discuss the question in the special case of $2D$ and $3D$, but the the format of the ellipse is given in terms of axes and angles rather than SPD matrices, and the formulas do not generalize to n-dimensions. The good answer is given by achilles hui in the comments to the following closed question:
Bounding box of Ellipsoid
but no proof is provided there, and the question is closed so I cannot provide the answer with proof there. Even if that question were reopened, it is focused on the 3D case with axes and angles rather than the n-dimensional case with SPD matrices.

答:
The bounding box, $B$, is given by
$$B = \prod_{i=1}^n\left[c_i - \sqrt{d_i}, c_i + \sqrt{d_i}\right],$$
where $d_i$ is the $i^\text{th}$ diagonal entry of $A^{-1}$.

Proof:

Let $e_i = (0,\dots,0,1,0,\dots,0)$ be the vector with $i^\text{th}$ entry equal to one, and all other entries equal to zero. The $i^\text{th}$ coordinate difference between a point $x$ and the point $c$ is given by $e_i^T (x-c)$. The points on the surface of the ellipse satisfy $x \in \mathbb{R}^n: (x-c)^T A (x-c) = 1$. Therefore, the distance from the center of the ellipse to the bounding box in direction $i$ is the solution to the following optimization problem:
$$
\begin{aligned}
\max_{x} &\quad e_i^T (x-c) \\
\text{such that}&\quad (x - c)^TA(x-c) = 1.
\end{aligned}
$$
Now let
$$A^{-1} = R^TR$$
be a factorization of $A^{-1}$, and let $r_i$ be the $i^\text{th}$ column of $R$. For example, $R$ could be the Cholesky factor, or $R$ could be $A^{-1/2}$, or $R$ could be the factor in any other factorization of this form.  Making the change of variables
$u := R^{-T}(x-c),$ performing simple algebraic manipulations,
and using the fact that $e_i^T R^T = r_i^T$, the optimization problem becomes
$$
\begin{aligned}
\max_{u} &\quad r_i^T u \\
\text{such that}&\quad \|u\| = 1.
\end{aligned}
$$
The solution to this optimization problem is given by $u = r^i/\|r_i\|$, and the optimal value is
$$r_i^T u = \frac{r_i^Tr_i}{\|r_i\|} = \sqrt{r_i^Tr_i} = \sqrt{\left(A^{-1}\right)_{ii}} = \sqrt{d_i}.$$

Therefore, in the $i^\text{th}$ direction, the bounding box for the ellipsoid extends from $c_i - \sqrt{d_i}$ to $c_i + \sqrt{d_i}$. This holds for all coordinate directions $i$, which implies the desired result. $\blacksquare$


问:
Show that a triangle delimited by parallel lines to the sides of another triangle is similar to said triangle.
I was looking at geometric constructions of similar triangles, and at one point I came across the statement in the question.

A triangle delimited by vertices which are intersections of lines parallel to the sides of some other triangle is similar to said triangle.

I drew an example of the statement as follows:

Where the small green triangle's vertices are given by the intersections of some lines which are parallel to the sides of the blue triangle. The statement seems intuitively obvious, but I can't seem to find a way to justify it. I tried using the intercept theorem which is what I normally see is used to prove statements about similar triangles, but since I'm not sharing a side in both triangles I didn't think I could use it. Does anyone know a simple argument to justify this? Thank you!

答:
An easy way to make the connection:
Given triangle $ABC$, and through external points $D$, $E$, $F$ parallels drawn to $AB$, $BC$, $CA$, respectively, forming triangle $GHJ$.

For explicit proof that $\triangle ABC\sim\triangle GHJ$, extend $AB$ to meet the parallels to $CA$ and $BC$ at $K$ and $L$. Since by parallels$$\angle CAB=\angle JKL$$and$$\angle JKL=\angle JGH$$therefore$$\angle CAB=\angle JGH$$For the same reason$$\angle CBA=\angle JLK=\angle JHG$$Therefore remaining angles$$\angle ACB=\angle GJH$$ and$$\triangle ABC\sim\triangle GHJ$$


问:
Determine what kind of stationary point you have in $(0,0)$ where $f(x,y)=(x^2+y^2+1)^2 - 2(x^2+y^2) +4\cos(xy)$

Given the function

$$f(x,y)=(x^2+y^2+1)^2 - 2(x^2+y^2) +4\cos(xy)$$

I know that  the taylor polynomial of degree $4$ of $f$ is equal to $5+x^4+y^4$. And I know that $x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$. Now I need to determine what kind of stationary point you have in $(0,0)$. Any hints ?

答:
Since you know the Taylor series you can rewrite it such as
$$\hat f(0,0)=5+(\Delta x)^4+(\Delta y)^4+O^5$$
It is clear that any variation in $x$ or $y$ increases the value of $\hat f$ and therefore $(0,0)$ is a minimum.</p>


问:
"Lebesgue" measurabillity on Riemannian manifolds
Let $X$ be a smooth oriented manifold of positive dimension; Let $g_1,g_2$  be two Riemannian metrics on $X$. Define $\Lambda_1,\Lambda_2:C_c(X) \to \mathbb{R}$ by

$$ \Lambda_i(f)=\int_X f \, Vol_{g_i},$$ where $Vol_{g_i}$ is the Riemannian volume form of $g_i$.

The discussion here implies that for each $i$, there is a unique $\sigma$-algebra $\mathfrak B_i$, and a unique measure $\mu_i$ such that $I(f)=\int_X f d \mu$ for all $f \in C_c(X)$, and the conditions (a)-(f)** are satisfied. (This is essentially Riesz Representation theorem, with the additional observation that $X$ is $\sigma$-compact $\Rightarrow$ uniqueness of the $\mathfrak B_i$).

Questions:

(1) Is it true that $\mathfrak B_1= \mathfrak B_2$? (independence of the metric chosen)

Assuming this is true, is there a way to define $\mathfrak B_i$ without passing through a Riemannian metric? (This is a natural expectation now since $\mathfrak B_i$ is an invariant of $X$ as a differentiable manifold, thus unrelated to the added Riemannian structure)

My guess is that $\mathfrak B_i$ will be the completion of the Borel $\sigma$-algebra w.r.t a suitable measure (which should be any one of the "Lebesgue" measures $\mu_i$ mentioned above. On a first glance, this does seem to go through a Riemannian metric, since the $\mu_i$ was induced by it.

However, the completion w.r.t a measure $\mu$ is dependent only on the  subsets that have $\mu$-measure zero, and this is independent of the Riemannian metric chosen, and can be defined invariantly (see Lee's book on smooth manifolds, chapter 6: A subset of a manifold has measure zero if its image under every coordinate chart has measure zero  in $\mathbb{R}^n$)

(2) A function $f:X \to \mathbb{R}$ is measurable w.tr.t $\mathfrak B_i$ if and only if it is measurable after (pre)composing with a coordinate chart diffeomorphism?

(3) Is this the standard way to define $L^p(X)$ spaces (from the perspective of measurability, are the elements of $L^p(X)$ exactly the measurable functions in the sense above, with the additional requirement of finiteness of the integral?)

**
(a) $\mathfrak{B_i}$ contains all Borel sets,

(b) $\mu(V)=\sup \{I(f): f \in C_c(X), 0\leq f \leq 1, \operatorname{supp} f \subset V\}$ for each open $V$,

(c) $\mu(K) < \infty$  for compact $K$,

(d) $\mu(E)=\inf \{\mu(V): E \subset V, \ V \mbox{ open}\}$ for each $E \in \mathfrak{B}$,

(e) $\mu(E)=\sup \{\mu(K): K \subset E, \ K \mbox{ compact} \}$ for each open $E$ and for each $E\in \mathfrak{B}$ such that $\mu(E)< \infty$,

(f) $\mu$ is a complete measure on $\mathfrak{B}$.

答:
To put it brief, as hkr noted in the comment, "you buy the charts and topology, you buy the measurable sets".

Here is a more detailed answer:

(1) It is possible to define the Lebesgue sigma-algebra on a general manifold (with or `without' boundary) without alluding to Riemannian metrics or specific choices of densities.

Details can be found in Chapter XII of Amann, H. & Escher, J. "Analysis III". (Birkhäuser, 2009).

(2) This is implied by (1), as otherwise the construction would not work. A function is integrable with respect to a given measure---an invariant notion. (1) only gives you the sigma-algebra on the manifold, it does not give you the measure, however. Whether a function is integrable may depend on the choice of volume form/density, but I do not know of any more detailed discussion of this topic.

(3) Again, if you take a measure-theoretic perspective, the notions of Lp spaces carry over. In the mentioned reference, they even treat the case that the functions take values in some Banach space. A more interesting question would be, if one can make sense of "Lp-spaces of forms".


问:
How do I show that a 5 degree reducible polynomial must be divisible by some second degree irreducible polynomial?
I need to show that if I have a reducible polynomial $f\in \mathbb{F}_{3}[X] $ of degree $5$ with no roots. Then there exists a monic irreducible polynomial of degree 2 dividing f.

My attempt:

$\mathbb{F}_{3}[X]$ is a unique factorization domain, so f can be factored uniquely by irreducible elements.

If f is factored by two polynomials $f=q_1 q_2$, then it needs to satisfy $\deg(f)=\deg(q_1)+\deg(q_2)=5$.

So if $p_n$ is an irreducible polynomial of degree n, then f can be factored in two ways:
$f=p_1p_4$ or $f=p_2p_3$

(But do we don't know anything about the multiplicity of the roots? So couldn't it in theory be factored as a multiple of the same degree polynomial? For example 1 one degree and 2 second degree polynomials?)

I'm somehow supposed to conclude that the only way it can be factored is by $f=p_2p_3$ and thus there must be a second-degree polynomial dividing f (since it is a factor of a second degree). But how do I draw this conclusion?

答:
If the polynomial is reducible, it can be factored as $f=pq$ with $0\leq \deg(p),\deg(q)\lt \deg(f)$. In this case, $\deg(f)=5$. Moreover, you know that $f$ does not have any degree $1$ factors, since it has no roots and roots correspond to degree $1$ factors.

Now, the factorization of $f$ into irreducibles will correspond to a partition of $5$, given by the degrees of the irreducibles. The partitions of $5$ are:
$$\begin{align*}
5 &= 5\\
5 &= 4+1\\
5 &= 3+2\\
5 &= 3+1+1\\
5 &= 2+2+1\\
5 &= 2+1+1+1+1\\
5 &= 1+1+1+1+1.
\end{align*}$$

Of these, only two of them correspond to factorizations with no factor of degree $1$, and one of those two corresponds to the case where $f$ is irreducible. That leaves only one possibility.


问:
Suppose $g:\mathbb{R}\rightarrow \mathbb{R}$ is infinitely differentiable
Suppose $g:\mathbb{R}\rightarrow \mathbb{R}$ is infinitely differentiable, and that for each $n\in \mathbb{N}$ there are positive numbers $c_{n}$ and $\delta_{n}$ such that $|g(x)| \leq c_{n}|x|^n$ if $|x|<\delta_{n}$. Prove that for each natural number $n$, $g^{(n)}(0) = 0$.

My attempt using induction is as follows:

Base case ($n = 1$): By supposition there are $c_{1} >0$ and $\delta_{1} >0$ such that $|g(x)| \leq
 c_{1}|x|$ if $|x|<\delta_{1}$. Setting $x = 0$ yields that $|g(0)| \leq c_{1}|0| = 0$ and therefore $g(0) = 0$

Fix $k\in \mathbb{N}$ and suppose that $g^{(k)}(0) = 0$. This is the point where I get stuck. I have tried applying the Lagrange Remainder Theorem at this point, but have been unsuccessful. I understand this is not much to work with, but any help would be greatly appreciated.

答: Hint: Show recursively that the coefficients of the Taylor serie   ( or the limited development)  are zero by dividing by $x^n$ and computing the limit.


问:
How many subgroups of orders $3$ and $6$ are there in a non-cyclic Abelian group of order $18?$
How many subgroups of orders $3$ and $6$ are there in a non-cyclic Abelian group of order $18$?

答:
There is one noncyclic abelian group of order $18$, isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_2$.

These are the subgroups of order $6$. They are cyclic.
$$ \{(0,0,0), (0,0,1), (0,1,0), (0,1,1), (0,2,0), (0,2,1)\}. $$
and
$$ \{(0,0,0), (0,0,1), (1,0,0), (1,0,1), (2,0,0), (2,0,1)\}. $$

As for subgroups of order $3$, well, these have to be generated by elements of order $3$. There are four elements of order $3$ in this group, $(0,1,0), (0,2,0), (1,0,0), (2,0,0)$. These generate two distinct subgroups of order $3$.


问:
Antiderivative of piecewise function
I want to calculate the anti-derivative of: $$ f(x) = \left\{
    \begin{array}{ll}
        -\sin(x) & \mbox{if } x \geq 0 \\
        1-x^2 & \mbox{if } x<0
    \end{array}
\right.
$$
such as $F(\pi/2)=0$

I calculated both antiderivatives, thus I did get $\cos(x)$ for the first one and $x-x^3/3$.
Question is to calculate $F(\pi)$ +$F(-1)$.
But If I calculate that with null constant, I get $-5/3$ which is wrong.

Then could you help me on which value should I give to the constants $C_1+C_2$ ?
Thank you very much

答:
As you calculated that antiderivative for $x \geqslant 0$ is $\cos x +c_1$ and for $x<0$ it is $x-\frac{x^3}{3}+c_2$. As antiderivative should be continuous we have $c_1+1=c_2$, so we obtain
$$F(x) = \left\{
    \begin{array}{ll}
        \cos(x)+c & \mbox{if } x \geq 0 \\
        x-\frac{x^3}{3}+c+1 & \mbox{if } x<0
    \end{array}
\right.$$
now you can get $c$ from condition $F(\pi/2)=0$

Title: Antiderivative of piecewise function


问:
Dirac delta integral on the bounded set
How to solve that an integral?
$$\int_{[0,10)}x^2 d\mu  $$ where $$\mu = \sum_{k=1}^{4} {{1}\over{k}} \delta_{k}$$
I know that $$\int_{X} {f(x) \delta_{x_0} (dx)}= f(x_{0}) $$ but in that example I don't know how to do it on the interval (0,10]. Does anyone know how to solve this?

答:
Most likely how you're intended to solve this is to know the following two rules, or at least for them to be "intuitively obvious" to you if you haven't covered them in class specifically:

$\int fd\delta_x=f(x)$

$\int fd(\sum\lambda_i\mu_i)=\sum\lambda_i\int fd\mu_i$

You can use these rules to work out your integral.

Title: Dirac delta integral on the bounded set


答:
The question, as you stated it, is ill-posed. The anti-derivative of a piecewise continuous function like $f$ is defined up to a constant on each "piece". Thus, the anti-derivatives of $f$ are
$$
F(x) = \left\{
    \begin{array}{ll}
        \cos(x)+C_1 & \mbox{if } x \geq 0 \\
        x-\frac{x^3}{3}+C_2 & \mbox{if } x<0
    \end{array}
\right.
$$
With the condition that $F(\pi/2)=0$, we can solve for $C_1$, but we do not have enough information left to solve for $C_2$. Presumably, the question is implicitly assuming something; if I had to guess, it's assuming that $F(x)$ is continuous. If so, then you can use the equation
$$
\lim_{x\to0^+}F(x) = \lim_{x\to0^-}F(x)
$$
to solve for $C_2$.

Title: Antiderivative of piecewise function

Body:
I want to calculate the anti-derivative of: $$ f(x) = \left\{
    \begin{array}{ll}
        -\sin(x) & \mbox{if } x \geq 0 \\
        1-x^2 & \mbox{if } x<0
    \end{array}
\right.
$$
such as $F(\pi/2)=0$

I calculated both antiderivatives, thus I did get $\cos(x)$ for the first one and $x-x^3/3$.
Question is to calculate $F(\pi)$ +$F(-1)$.
But If I calculate that with null constant, I get $-5/3$ which is wrong.

Then could you help me on which value should I give to the constants $C_1+C_2$ ?
Thank you very much


问:
Problems with coins related to the generating function
I'm solving a discrete math problem, but I have no idea how to solve the following problems. Regarding No. 1, should I just multiply the generating functions a (x) = 1 / 1-x and b (x) Σbn x ^ n? Would you please write what it would look like?
No. 2 is b (x) = 1 / (1-x) ^ 2 as a result of differentiating a (x). Is this way of thinking correct?
To be honest, I'm sorry, but I don't know what to do about numbers 3 and 4.

Suppose there are an infinite number of  1＄,5＄, and 10＄ coins. At this time, n (n = 0, 1, 2, ...) coins are allowed to be duplicated from these coins. Let fn be the total number of choices to choose.
Also, let f (x) be the population function of the sequence f0, f1, f2, ....

1: Let a (x) be the generating function of the sequence 1, 1, 1, ..., and b (x) be the generating function of the sequence b0, b1, b2 .... At this time, Express the coefficient of the nth order term of a(x)b(x) using b0, b1, b2, ....

2: Let b (x) be the generating function of the sequence 1, 2, 3, ... again. Express b (x) in a closed form.

3: Explain that it is given by f (x) = (1 + x + x2 + x3 + ...) ^ 3.

4: Express fn by the formula of n.

答:
For #1, note that
\begin{align}
a(x)b(x)
&= \sum_{n \ge 0} x^n \sum_{k \ge 0} b_k x^k \\
&= \frac{1}{1-x}\sum_{k \ge 0} b_k x^k \\
&= \sum_{k \ge 0} b_k \frac{x^k}{1-x} \\
&= \sum_{k \ge 0} b_k \sum_{n \ge k} x^n \\
&= \sum_{n \ge 0} \left(\sum_{k=0}^n b_k\right)x^n,
\end{align}
so $a(x)b(x)$ is the generating function of the partial sums of $(b_n)$.

For #2, you can differentiate $\frac{1}{1-x}$, but you can also use #1 by noting that $1,2,3,\dots$ is the sequence of partial sums of $1,1,1,\dots$, so just multiply $\frac{1}{1-x}$ by $\frac{1}{1-x}$.

For #3 and #4, please start a new question.


问:
meaning of "$\wedge$" in set theory
I have been reading Analysis I by Terence Tao and this symbol, $\wedge$, keeps popping in proof's in this forum as well as from other sources, for example, Power Set Axiom. Terence Tao' analysis 3.4.6. This symbol is not described in the book so I do not understand what this means, could anyone explain it please? Thank you.

答:
Since there's a bit of confusion on this (maybe a bit overused) symbol, I think it's worth mentioning at least some of the most common uses for $\wedge$ (which does, after all, descend from the caret ^, if I'm not mistaken).

In set theory and propositional logic, $\wedge$ is usually used for the logical conjunction AND. This is why this symbol is encoded both by \wedge and \land in LaTeX (hence, in MathJax). It's "opposite" is the "vel" operator $\vee$, which is the logical disjunction OR.

In lattice theory, $\vee$ and $\wedge$ are respectively the join and meet operators, which are used to denote supremum and infimum of a subset of a partially ordered set.

Another use of $\wedge$ is the exterior (or wedge) product in geometry, which is an antisymmetric binary operation. You can find it for example in differential forms (sections of the exterior power of the cotangent bundle, or locally/heuristically speaking, "wedge products of differentials").

Finally, in measure theory and analysis, $f\wedge g$ is often used to denote a function defined by $f\wedge g(x)=\inf \{f(x),g(x)\}$, as opposed to $f\vee g(x)=\sup\{f(x),g(x)\}$.

The caret ^ usually defines some kind of exponentiation, but it's not used as "proper" notation, but rather as encoding only (for example, $x^{a+b}$ is encoded as $x^{a+b}$).

Title: meaning of "$\wedge$" in set theory


问:
Defining the predicates using choosen symbols.
Let the range of variatons of variables be the set of natural numbers. Show that using the symbols 0, 1, + and | we can define the predicate "x * y = z" (the symbol | means divisibility without remainder).

I have also a hint:
Define the predicate (∃y) (x = y^2) first.
You may need the following identities
(x + y)^2 = x^2 + xy + xy + y^2 , GCD(x, x + 1) = 1 and x^2 + x = LCM(x, x + 1),
where GCD is greatest common divisor, LCM is least common multiple.

答:
Welcome to MSE!

Your hint is a good one. I'll try to explain it in slightly more depth since this is a kind of tricky problem.

We want to express $xy = z$. But we only have access to the symbols $0,1,+,\mid,=$.

The trick is to leverage the identity $(x+y)^2 = x^2 + 2xy + y^2$. From this, we know that

$$z=xy \iff (x+y)^2 = x^2 + z + z + y^2.$$

So we've reduced our problem (defining all multiplication) to a special case (defining perfect squares).

OK, can we find a way to write $k = x^2$? Here is where the hint regarding gcd and lcm comes into play. Since $\text{lcm}(x,x+1) = x^2 + x$, we see

$$ k = x^2 \iff \text{lcm}(x,x+1) = k+x $$

So now we're almost done. Here is a step-by-step guide you might use to finish the problem:

Can you write $\mathtt{LCM}(x,y,z)$, an expression which is true if and only if $z = \text{lcm}(x,y)$?

You should be able to do this using the divisibility relation $\mid$.

Then define $\mathtt{Square}(x,y) = \exists l . \text{isLCM}(x,x+1,l) \land l = y+x$.

Do you see why $\mathtt{Square}(x,y)$ is true if and only if $y=x^2$?

Finally, define $\mathtt{Prod}(x,y,z) = \exists X . \exists Y . \exists S . \mathtt{Square}(x,X) \land \mathtt{Square}(y,Y) \land \mathtt{Square}(x+y,S) \land S = X + z + z + Y$

Do you see why this correctly defines $\mathtt{Prod}(x,y,z) \iff xy = z$?

I hope this helps ^_^


问:
Are all 2-colorable graphs bipartite?
I believe bipartite means you can split the vertices of a graph into 2 groups, where each vertex in the first is connected to all in the second (correct me if Im wrong) Does that mean every 2-colorable graph is bipartite? If not, what are some counterexamples, and an explanation why?</p>

答: Yes, in fact bipartite is equivalent to 2-colorable.</p>

Title: Are all 2-colorable graphs bipartite?


问:
If the cardinality of a subgroup is half the cardinality of the original group, is the subgroup normal?
Suppose $(G,\cdot)$ is a group and $H$ is a subgroup of $G$, with $|H|=|G|/2$. Is $H$ a normal subgroup of $G$?

It's a theorem that's asserted without proof in a book for my Abstract Algebra class, so it's probably true, I'm just having trouble convincing myself. Lagrange's theorem tells us that there are two unique (left) cosets of $H$, and that they form a partition for $G$. The same can be said for the right cosets, so we essentially have two partitions for $G$, that each split the number of elements evenly between the two cosets. In order for $H$ to be normal we need the left and right cosets to be equal, but:

(1) I fail to see how $xH \cup yH$ and $Hx \cup Hy$, with $x,y\in G$, must necessarily be the partitions of $G$. Why not $xH \cup yH$ and $Hw \cup Hz$, for $w,z\in G$?

(2) Even if $xH \cup yH$ and $Hx \cup Hy$ are the two partitions, what is preventing $xH=Hy$ and $yH=Hx$, for $x\neq y$?

答:
We have $xH=H$ or $xH=G\setminus H$, where the former holds iff $x\in H$.
Likewise $Hx=H$ or $Hx=G\setminus H$, where the former holds iff $x\in H$. At any rate $xH=Hx$.


问:
Prove that $243^{2020} \equiv 1 \pmod 5$
I was supposed to use that  $a^4 \equiv 1 \pmod 5$, but that's not helping me, specifically, I don't know how to prove that $243^{505}$ is not as $5k, k \in \mathbb{Z}$.

Body: $$243^{2020}=(243^4)^{505}\equiv (3^4)^{505}\equiv 1^{505}\equiv 1\pmod 5 $$


问:
Basic topology of a 4 point set
Let $X=\{1,2,3,4\}$ be given by the topology $\tau=\{\emptyset,X,\{1\},\{1,2\},\{1,3\},\{1,2,3\}\}$.

Show that any two disjoint closed sets have disjoint neighborhoods.

So I define my closed sets (as complements of the given open sets) as

$\mathcal{C}=\{X,\emptyset,\{2,3,4\},\{3,4\},\{2,4\},\{4\}\}$

So my question is, since $4$ is an element of every closed set except $\emptyset$, is there even anything to show because every closed set except for $\emptyset$ is contained in the neighborhood $X$ (because $X$ is the only member of $\tau$ who contains the element $4$). Is there even anything to show or?

答:
Since @Hossien Sahebjame already answered your question in the comments, let formulate the same answer in a slightly more general context, because why not.

Let $\bar{X} = X \cup \{a\}$ be a set and define the open sets of the excluded point topology on $\bar{X}$ to be all subsets, not containing $a$ and the whole set, so $\tau = \mathcal{P}(X) \cup \bar{X}$. In every such topological space every closed set contains the point $a$ and hence there are no non-trivial disjoint closed sets which implies that "any two disjoint closed sets have disjoint neighborhoods" by means of a vacuous truth (as pointed out by you and in the comments already).

Now, having a coarser topology $\sigma \subset \tau$ can only reduce the number of closed sets; therefore all topologies that are coarser than an excluded point topology have this property. Since the topology you described is coarser than the excluded point topology $\bar{X} = \{1,2,3\} \cup \{4\}$, this completes this unnecessarily overcomplicated "answer".

Title: Basic topology of a 4 point set


问:
Does the decimal representation of $\pi$ contain almost all numbers?
Let $\Pi \in \Sigma^\infty$ be the decimal representation of $\pi$ where $\Sigma=\{0..9\}$. It is not known whether $\Pi$ contains each natural number. On the other hand, no number is known not to occur in $\Pi$. (Hope I've got that right)

First of all: is this property unique to $\pi$ or do all irrational numbers share it?
UPDATE: As pointed out by Brian and MJD, irrationality is completely unrelated: 1. there are irrational numbers which are known not to contain all numbers in their representation (Brian) 2. there are irrational numbers whose representation contains all numbers (MJD).

At any event, I think it should be much easier to show that almost all numbers are present, i.e. $ \exists N\in\mathbb{N}\forall k>N\exists p\in\Sigma^\star\exists s\in\Sigma^\infty: \Pi=pR(k)s$ where $R:\mathbb{N} \to \Sigma^\star$ maps each number to its decimal representation. So "allmost all" here means all except finitely many.

Has something like this been proven or disproven?

In general I don't think it's possible to construct a non-repetitive stream of symbols $\sigma \in \Sigma^\infty $ that does not contain almost all $w\in\Sigma^\star$ ("almost all" with respect to some enumeration of $\Sigma^\star$). (EDIT: This is possible as Brian points out)

答: Every decimal that is not eventually repeating represents an irrational number, so it’s very easy to construct irrational numbers that do not have the property, e.g., $0.01001000100001\dots~$.</p>


问:
Surface area of spherical cap by summing up circumferences
I have seen calculations of the surface area of a spherical cap using a surface of revolution approach. Is it possible to instead cut the spherical cap into slices horizontally (imagine the cap is formed by a horizontal slice of the sphere) and then adding them up (integrating them)?</p>

答:
Yes - if you remember that the sides are sloping

Suppose $x$ is the height of a slice from the origin of a sphere radius $r$

Then the circumference of a circle at that height is $2\pi \sqrt{r^2-x^2}$ and a slice thickness $\delta x$ is sloping in with sloping distance $\frac{r}{ \sqrt{r^2-x^2}}\delta x +o(\delta x)$ using the almost similar triangles shown below,  so the surface area for that slice is $2\pi r\,\delta x + o(\delta x)$.

It is then a simple integration of $2\pi r$ with respect to $x$ to say that the curved surface area of a spherical cap of height $h$ for a sphere of radius $r$ is $2 \pi r h$


答:
Take the sphere $x^2+y^2+z^2=r^2$ as an example and calculate the area of the cap over the angle $\theta$. The area is  integrated over a stack of sliced rings with radius $r\sin t$ and width $rdt $, hence area $ds= 2\pi r^2\sin t dt$
$$A = \int_0^{\theta}ds=2\pi r^2 \int_0^{\theta}\sin t dt= 2\pi r^2(1-\cos\theta)
$$

Title: Surface area of spherical cap by summing up circumferences

Body: I have seen calculations of the surface area of a spherical cap using a surface of revolution approach. Is it possible to instead cut the spherical cap into slices horizontally (imagine the cap is formed by a horizontal slice of the sphere) and then adding them up (integrating them)?</p>


问:
Why can you check closedness on the fibers? (Hartshorne's proof of Bertini's theorem)
In Hartshorne's proof of Bertini's theorem, given a linear system $|H|$, he defines the locus of "bad" hyperplanes $B_x$ for each point $x\in X\subset \Bbb P^n$ a projective variety, shows that this is a proper linear subset of $\{x\}\times |H|$, defines $B$ to be the union of all pairs $(x,H)$ so that $H\in B_x$, and then claims that "clearly $B$ is the set of closed points of a closed subset of $X\times|H|$".

I don't understand this. Clearly the statement "if a subset has proper closed intersection with all fibers of a map over closed points, then it's a proper closed subset" is false - consider the inclusion of a copy of $(\Bbb P^1\setminus\{0\})\times\{p\} \to \Bbb P^1\times\Bbb P^1$ followed by a projection. How can I rigorously see Hartshorne's claim?

答:
I pieced this together after the discussion with Mindlack in the comments. Define $B$ to be the subset of $X\times\Bbb P^n$ cut out by the minors of size $n-\dim X+1$ of the matrix $$\begin{pmatrix}
 \frac{\partial f_1}{\partial x_0} & \cdots & \frac{\partial f_1}{x_n} \\
 \vdots & \ddots & \vdots \\
\frac{\partial f_m}{\partial x_0} & \cdots & \frac{\partial f_m}{x_n} \\
a_0 & \cdots & a_n
\end{pmatrix}$$

where $f_i$ are a generating set for $X$ and $a_j$ are coefficients for the hyperplane $V(\sum a_jx_j)$ where the $x_j$ are coordinates on $\Bbb P^n$. It's not so hard to see that the vanishing of all of these minors at a point $x\in X$ means that the dehomogenization of $\sum a_jx_j$ is in the square of the maximal ideal there via the Jacobian criterion. This exactly gives $B_x$ as the fiber of $B$ over $x$.


问:
find limit of $\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$ with squeeze theorem
I'm trying to prove with squeeze theorem that the limit of the following series equals 1:

$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$

For the left side of the inequality I did:

$$\frac{1+\sqrt{1}+\sqrt[3]{1}+...+\sqrt[n]{1}}{n} <  \frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$

For the right side, at first I did the following:

$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n} <  \frac{n\sqrt[n]{n}}{n}$$

But then I realized it wasn't true and that the direction of this inequality is the opposite.

Do you have any idea which series with limit 1 is bigger from the original series?

Thanks!

答:
First of all we have, for any $ n\in\mathbb{N}^{*} $, the following : $$ \sqrt[n]{n}=1+\frac{\ln{n}}{n}\int_{0}^{1}{n^{\frac{x}{n}}\,\mathrm{d}x} $$

Since : \begin{aligned}0\leq\int_{0}^{1}{n^{\frac{x}{n}}\,\mathrm{d}x}&\leq n^{\frac{1}{n}}\\ &\leq 2\end{aligned}

We have : \begin{aligned} 1\leq \sqrt[n]{n}\leq 1+\frac{2\ln{n}}{n}&=1+\frac{4\ln{\sqrt{n}}}{n}\\ &\leq 1+\frac{4\sqrt{n}}{n}= 1+\frac{8}{2\sqrt{n}}\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \leq1+\frac{8}{\sqrt{n}+\sqrt{n-1}} \end{aligned}

That remains true for any $ n\in\mathbb{N}^{*} $, which means given $ n\in\mathbb{N}^{*} $, we have : \begin{aligned} 1\leq\frac{1}{n}\sum_{k=1}^{n}{\sqrt[k]{k}}&\leq 1+\frac{8}{n}\sum_{k=1}^{n}{\frac{1}{\sqrt{k}+\sqrt{k-1}}} \\ &\leq 1+\frac{8}{n}\sum_{k=1}^{n}{\left(\sqrt{k}-\sqrt{k-1}\right)}\\ &\leq 1+\frac{8}{\sqrt{n}} \end{aligned}

Thus, using the squeezing theorem, the limit would be $ 1 \cdot$


问:
Solving $f'''+\frac{n+1}{2}ff''-nf'^2+n=0$ with $n=e^\pi$
How do I solve

$$f'''+\frac{n+1}{2}ff''-nf'^2+n=0$$

with $n=e^\pi$ or arbitrary $n$?

This equation occurs in my model for the time evolution of the value of Bitcoin.

答:
Similar to How do you solve this kind of third order ODE? (non-linear Blasius equation),

Let $u=(f')^2$ ,

Then $\pm\dfrac{\sqrt u}{2}\dfrac{d^2u}{df^2}+\dfrac{n+1}{4}f\dfrac{du}{df}-nu+n=0$


问:
Solve $3^{1-x}=2$
I've been trying to solve a problem:

$3^{1-x}=2$

I converted this to a log as:
$\log_{3}{2} = (1-x)$

But I couldn't see how to progress from there. Having put it into a solving team, it suggests that it can be translated to:
$\left(1-x\right)\ln \left(3\right)=\ln \left(2\right)$

...but I can't see why this makes sense. Could anyone point me in the right direction?

答: Use the property of logarithms that $\log_b x=\dfrac {\log_e x}{\log_e b}$.

Title: Solve $3^{1-x}=2$


答:
This is due to the change of base formula, $$\log_{b}x = \frac{\ln x}{\ln b}.$$

This is because if you have $\log_{b}x = a$, then you can rewrite it as $x = b^{a}$.  Taking the natural logarithm of each side gives you $\ln x = \ln (b^a)$, which can be rewritten as $\ln x = a\ln b$, which upon dividing both sides by $\ln b$ gives you $a = \frac{\ln x}{\ln b}$.  Hence, you have $$\log_{b}x = a = \frac{\ln x}{\ln b}.$$

Title: Solve $3^{1-x}=2$

Body:
I've been trying to solve a problem:

$3^{1-x}=2$

I converted this to a log as:
$\log_{3}{2} = (1-x)$

But I couldn't see how to progress from there. Having put it into a solving team, it suggests that it can be translated to:
$\left(1-x\right)\ln \left(3\right)=\ln \left(2\right)$

...but I can't see why this makes sense. Could anyone point me in the right direction?


答: If you mean to find the value of $x$ you almos have it. Another option is to write $3^{(1-x)}=2$ as $\frac{3}{3^x}=2$ which is equivalent to $\frac{3}{2}=3^x$ and using $log_3$ the value of $x$ is $log_3\frac{3}{2}$.

Title: Solve $3^{1-x}=2$


问:
Does convergence of $\lim_{n \rightarrow \infty} \int h d\mu_n$ for all continuous, bounded $h$ imply weak convergence of $(\mu_n)$?
Let $(\mu_n)$ be a sequence of positive finite Borel measures on $\mathbb{R}$. Suppose $\lim_{n \rightarrow \infty} \int h d\mu_n$ converges for every bounded, continuous function $h$ on $\mathbb{R}$. Must there exist a finite Borel measure $\mu$ on $\mathbb{R}$ such that
$$
\lim_{n \rightarrow \infty} \int h d\mu_n = \int h d\mu
$$
for every bounded, continuous function $h$ on $\mathbb{R}$?

My initial thought was that the Lévy–Prokhorov metric for finite Borel measures might be useful. But then I realized that Cauchy-ness of the sequence $\lim_{n \rightarrow \infty} \int h d\mu_n$ may not be the same thing as Cauchy-ness of the sequence $(\mu_n)$ in the Lévy–Prokhorov metric.

答:
We prove that the assertion is true. This has already been proved in the literature, such as in [1, Corollary 8.6.3], and the following proof is also partially motivated by the proof therein.

1. Define $T : C_b(\mathbb{R}) \to \mathbb{R}$ by

$$ T(f) = \lim_{n\to\infty} \int_{\mathbb{R}} f \, \mathrm{d}\mu_n, \qquad \forall f \in C_b(\mathbb{R}). $$

This gives rise to a well-defined functional on $C_b(\mathbb{R})$ by the assumption. It is also clear that $T$ is a positive linear functional. Moreover, if $\| f \| := \sup \left| f \right| $ denotes the supremum norm, then

$$ \left| T (f) \right| \leq T(\left| f \right|) \leq \|f\| T(1), $$

and so, $T$ is continuous. Now by the Riesz-Markov-Kakutani representation theorem, there exists a unique Borel measure $\mu$ on $\mathbb{R}$ such that

$$ T(f) = \int_{\mathbb{R}} f \, \mathrm{d}\mu, \qquad \forall f \in C_c(\mathbb{R}). $$

In other words, we know that $\mu_n \to \mu$ in a vague topology. Moreover, it is clear that $\mu(\mathbb{R}) \leq T(1)$. Indeed, if $f \in C_c(\mathbb{R})$ and $0 \leq f \leq 1$, then

$$ \int_{\mathbb{R}} f \, \mathrm{d}\mu = T(f) \leq T(1), $$

and letting $ f \uparrow 1$ along a subsequence together with the monotone convergence theorem shows the desired inequality. This $\mu$ is certainly the very candidate for the weak limit of $\mu_n$.

2. To establish this convergence, however, we need the uniform tightness of $(\mu_n)_{n\geq 1}$. Considering that the tightness condition intuitively means "no mass escapes towards the infinity", this essentially amounts to showing that

$$\mu(\mathbb{R}) = T(1) = \lim_{n\to\infty} \mu_n(\mathbb{R}). $$

In order to show this, assume otherwise that $\mu(\mathbb{R}) < T(1)$. We will prove that this leads to a contradiction. Intuitively, we will construct a sequence of "bumps" $(\rho_j)_{j\geq 1}$ that moves along with the position of the mass escaping towards the infinity. This will then be used to derive a contradiction.

To this end, write $\epsilon = \frac{T(1) - \mu(\mathbb{R})}{2020} > 0$. Then we will recursively construct a seqeunce $(\rho_j)_{j\geq 1}$ in $C_c(\mathbb{R})$ and a subsequence $\mu_{n_j}$ as follows:

We will conveniently set $\chi_0 = 0$ and $n_0 = 0$.

Suppose that $j \in \{1, 2, \dots \}$, and $\chi_{j-1} \in C_c(\mathbb{R})$ with $0 \leq \chi_{j-1} \leq 1$ and $n_{j-1}$ has been chosen. Since
$$ \lim_{n\to\infty} \int_{\mathbb{R}} \chi_{j-1} \, \mathrm{d}\mu_n
= \int_{\mathbb{R}} \chi_{j-1} \, \mathrm{d}\mu
\leq \mu(\mathbb{R})
\qquad\text{and}\qquad
\lim_{n\to\infty} \mu_n(\mathbb{R}) = T(1), $$
we can find $n_j > n_{j-1}$ such that
$$ \int_{\mathbb{R}} \chi_{j-1} \, \mathrm{d}\mu_{n_j} < \mu(\mathbb{R}) + \epsilon \qquad \text{and} \qquad \mu_{n_j}(\mathbb{R}) > T(1) - \epsilon. $$
Also, choose $r > 0$ such that $\chi_{j-1}(x) = 0$ whenever $\left| x \right| > r$. Then we can find $\chi_j \in C_c(\mathbb{R})$ such that
$$ \mathbf{1}_{[-r-1,r+1]} \leq \chi_j \leq 1 \qquad \text{and} \qquad \int_{\mathbb{R}} \chi_{j} \, \mathrm{d}\mu_{n_j} > T(1) - 2\epsilon. $$

Finally, set $\rho_j = \chi_j - \chi_{j-1}$ for each $j \in \{ 1, 2, \dots \}$. Then

$$ \int_{\mathbb{R}} \rho_{j} \, \mathrm{d}\mu_{n_j} > T(1) - \mu(\mathbb{R}) - 3\epsilon, \qquad \forall j \geq 1. \tag{*} $$

Also, the construction ensures that $0 \leq \rho_j \leq 1$ for all $j $ and that for each given $x \in \mathbb{R}$ there is at most $2$ values of $j$ for which $\rho_j(x) > 0$. Altogether, for any bounded sequence $a = (a_j)_{j\geq 1} \in \ell^{\infty}$, the function

$$ f_a = \sum_{k \geq 1} a_k \rho_k $$

defines a bounded continuous function on $\mathbb{R}$. Then the assumption tells that

$$ \int_{\mathbb{R}} f_a \, \mathrm{d}\mu_{n_j}
= \sum_{k \geq 1} a_k \int_{\mathbb{R}} \rho_k \, \mathrm{d}\mu_{n_j} $$

converges as $j\to\infty$ for any $a \in \ell^{\infty}$. This shows that the sequence $c^{(j)} = (c^{(j)}_k)_{k\geq 1} \in \ell^1$ defined by $c^{(j)}_k = \int_{\mathbb{R}} \rho_k \, \mathrm{d}\mu_{n_j} $ converges weakly as $j\to\infty$. Now it is well-known that this implies that $c^{(j)}$ converges in $\ell^1$. (See this posting, for instance.) Moreover, since

$$ \lim_{j\to\infty} c^{(j)}_k = \int_{\mathbb{R}} \rho_k \, \mathrm{d}\mu, $$

the $\ell^1$-limit of $c^{(j)}$ is the sequence $c = (c_k)_{k\geq 1}$ defined by $c_k = \int_{\mathbb{R}} \rho_k \, \mathrm{d}\mu$. So we must have

$$ \lim_{j\to\infty} \left| c^{(j)}_j - c_j \right|
\leq \lim_{j\to\infty} \| c^{(j)} - c \|_{\ell^1}
= 0. $$

On the other hand, $\lim_{j\to\infty} c_j = 0$ and $\text{(*)}$ shows that

$$ \lim_{j\to\infty} \left| c^{(j)}_j - c_j \right|
\geq T(1) - \mu(\mathbb{R}) - 3\epsilon
> 0 $$

which is a a contradiction.

3. Now we are ready to prove the claim. Fix a sequence $\chi \in C_c(\mathbb{R})$ such that $0 \leq \chi(x) \leq 1$ for all $x \in \mathbb{R}$. Then for any $ f \in C_b(\mathbb{R})$ and $n$,

\begin{align*}
\left| \int f \, \mathrm{d}\mu_n - \int f \, \mathrm{d}\mu \right|
&\leq \left| \int f \chi \, \mathrm{d}\mu_n - \int f  \chi \, \mathrm{d}\mu \right| \\
&\quad + \left| \int f(1 - \chi) \, \mathrm{d}\mu_n \right| + \left| \int f(1 - \chi) \, \mathrm{d}\mu \right| \\
&\leq \left| \int f \chi \, \mathrm{d}\mu_n - \int f  \chi \, \mathrm{d}\mu \right| \\
&\quad + \|f\| \left( \mu_n(\mathbb{R}) - \int \chi \, \mathrm{d}\mu_n \right) + \|f\| \left( \mu(\mathbb{R}) - \int \chi \, \mathrm{d}\mu \right)
\end{align*}

Letting $n\to\infty$ and using the fact that $\mu(\mathbb{R}) = \lim_{n\to\infty} \mu_n(\mathbb{R})$,

\begin{align*}
\limsup_{n\to\infty} \left| \int f \, \mathrm{d}\mu_n - \int f \, \mathrm{d}\mu \right|
&\leq 2 \|f\| \left( \mu(\mathbb{R}) - \int \chi \, \mathrm{d}\mu \right)
\end{align*}

Since the left-hand side is independent of $\chi$, letting $\chi \uparrow 1$ along a subsequence proves the desired claim.

[1] Bogachev, Vladimir Igorevich. Measure Theory. Springer, 2007.


问:
Exponential stability of damped wave equation
I like to proof the exponential stability of a damped wave equation with the following form.

Let $G \subset \mathbb{R}^n$ be bounded with smooth margin. For $\gamma > 0$ we are given the damped wave equation:

$u_{tt} - \triangle  u + \gamma u_t = 0$

$u(0,\cdot) = u_0(\cdot)$

$u_t(0,\cdot) = u_1(\cdot)$

$u|_{\partial G} = 0$

We are given the energy of this equation as $E(t) = \Vert u_t(t,\cdot) \Vert_2^2 + \Vert \nabla u(t,\cdot) \Vert_2^2$. I want to proof that $\exists C,c > 0$ such that $E(t) \leq C e^{-ct} E(0)$.

My approach is to use a lyapunov functional of the form $L(t) = E(t) + f(t)$ such that $\exists \alpha , \beta > 0: \alpha E(t) \leq L(t) \leq \beta E(t)$ and $\dfrac{d}{dt} L(t) \leq -dE(t)$ for $d>0$. With that we could esaly get the desired solution.

I obtained the derivative of $E(t)$ by multiplying the pde with $u_t$ and then integraiting it which gives me $\dfrac{d}{dt} E(t) = -2\gamma \int \limits _{G} u_t^2(t,x)dx$. But I can't think of a fitting definition for $L(t)$. Any suggestion or hint would be appreciated.

答:
From the inequality you obtained it follows in particular that
$$
\frac{dE}{dt} \leq -2\gamma E(t).
$$
The desired estimate then follows from Gronwall's inequality. More concretely, note that if you move all the terms to the left and then multiply by $e^{2\gamma t}$, then the inequality is equivalent to
$$
\frac{d}{dt}(e^{2\gamma t}E(t)) \leq 0.
$$
Integrating, it follows that for $t>0$,
$$
e^{2\gamma t}E(t) = E(0) + \int_0^t \frac{d}{ds}(e^{2\gamma s}E(s))~ds \leq E(0).
$$
Therefore
$$
E(t) \leq e^{-2\gamma t}E(0).
$$
If you want to do it your way in terms of Lyapunov functionals, you could just take $L = E$.


问:
Finiteness in infinite collection of non-cofinite sets implies finiteness?
For any set $X$, let $W_X=\{S\subset X\mid X\setminus S\text{ is infinite}\}$.

Does there exist a set $X$ and an infinite set $A\subset X$ such that for all $Y\in W_X$, $A\cap Y$ is finite?

I've been trying to prove the negation of this statement by contradiction for a bit now, but I can't get a grasp on it. Is it true or false?

答:
I think it is false, here goes a improved proof.
As $A$ is infinite there exist a collection of non repeated elements $(a_n)_n\subset A$ with $n\in \mathbb{N}$. Then you can pick the subsequence given by $S=(a_{2n})_n$.

Now, $S$ is obiously an element of $W_X$ as $A\setminus S\subset X\setminus S$ and $A\subset S=(a_{2n-1})_n$ which is an infinite set.
Finally $A\cap S=S$ which is an infinite set and that proof tha the statement is false.


问:
Substitution that results in upper and lower bounds of the same value
I'm looking at this integral which has a logarithm, and it's completely obvious I should use substitution to make the integral easier... then the bounds become equivalent. I honestly have never seen this happen, but it feels like I'm doing something that is forbidden - maybe even illegal.

Here's the substitution and the original bounds: $u=3+2\cos{\theta}$ over the interval $[-\pi ,\pi]$. Then our new region of integration will be $[1,1]$. This certainly cannot be correct. How do I work around this little dilemma?

答: Yes you're right, it's illegal. To use the substitution you have to check that the new function is a bijection between the two regions os integration, and $f(\theta)=3+2\cos(\theta)$ is not injective in $[-\pi,\pi]$. You may have to consider to change the domain from $[-\pi,\pi]$ to another interval.


答:
Champernowne's constant is irrational and obviously contains every integer in its digits.

Not every irrational number contains every integer in its digits.  For example, consider $0.1010010001000010000001\ldots$ or $0.101011010101011010101\ldots$ where the $n$th block is $1$ if $n$ is a square, $01$ otherwise. Obviously both of these omit nearly all sequences of digits; they even omit nearly all sequences of zeroes and ones.

Title: Does the decimal representation of $\pi$ contain almost all numbers?

Body:
Let $\Pi \in \Sigma^\infty$ be the decimal representation of $\pi$ where $\Sigma=\{0..9\}$. It is not known whether $\Pi$ contains each natural number. On the other hand, no number is known not to occur in $\Pi$. (Hope I've got that right)

First of all: is this property unique to $\pi$ or do all irrational numbers share it?
UPDATE: As pointed out by Brian and MJD, irrationality is completely unrelated: 1. there are irrational numbers which are known not to contain all numbers in their representation (Brian) 2. there are irrational numbers whose representation contains all numbers (MJD).

At any event, I think it should be much easier to show that almost all numbers are present, i.e. $ \exists N\in\mathbb{N}\forall k>N\exists p\in\Sigma^\star\exists s\in\Sigma^\infty: \Pi=pR(k)s$ where $R:\mathbb{N} \to \Sigma^\star$ maps each number to its decimal representation. So "allmost all" here means all except finitely many.

Has something like this been proven or disproven?

In general I don't think it's possible to construct a non-repetitive stream of symbols $\sigma \in \Sigma^\infty $ that does not contain almost all $w\in\Sigma^\star$ ("almost all" with respect to some enumeration of $\Sigma^\star$). (EDIT: This is possible as Brian points out)


问:
Writing out triple integrals - do x, y, and z have to be in order?
I’m currently working on some problems in Multivariable Calculus and am wondering if there is any stipulation that when writing out triple integrals, they must be x - then y - then z or else z - then y - then x. I have never seen it any other way and I am wondering if perhaps this is a property of the Cartesian coordinate system, just like when drawing our x,y,z axes they must be in order (counterclockwise like angles)?</p>

答:
There is a key result called the Fubini-Tonelli Theorem (and a variety of related results) that roughly says that for any reasonable function, it does not matter what order you do the integrals in. Writing it down (actually two nested copies of it) in the $\mathbb{R}^3$ case, it becomes: For any measureable function $f$, we have

$$\int |f| dxdydz = \int |f| dxdzdy = \int |f| dydxdz = \int |f| dydzdx = \int |f| dzdxdy = \int |f| dzdydx,$$ and if $\int |f| dxdydz < \infty,$ then
$$\int f dxdydz = \int f dxdzdy = \int f dydxdz = \int f dydzdx = \int f dzdxdy = \int f dzdydx.$$

Note that the assumption that $\int |f|dxdydz < \infty$ is necessary for the latter part. In particular,
$$\int_0^1\int_0^1\int_0^1\frac{x^2 - y^2}{(x^2 + y^2)^2}dydxdz = \frac{\pi}{4},$$ but $$\int_0^1\int_0^1\int_0^1\frac{x^2 - y^2}{(x^2 + y^2)^2}dxdydz = \frac{-\pi}{4}.$$

The problem is that $$\int_0^1\int_0^1\int_0^1\left|\frac{x^2-y^2}{(x^2+y^2)^2}\right|dxdydz = \infty,$$ so the above does not apply.


问:
Why is this method right to solve some equations of degree 2?
Given this equation

$-x^2+x+6=4$

We can write it as

$(3-x)(x+2) = 4$

So weridly if I take
$(3-x) = 4$ and $(x+2) = 4$, I can get $x=-1$ and $x=2$ that are indeed the correct solutions.

Why does this happen? And in which cases can this occur?

答:
I think it happens because in both of the solutions one of the parts of the product become a "1". I mean, for example, when you look for the solution and you write $(3-x)=4$ you are asuming $x+2=1$ so $x=-1$ which is in fact a valid solution for the equation. An the same occurs in the other case.
I will have to give it a deeper thought but I think that with that it may be possible to findd out more examples.


问:
What exactly is a number?
We've just been learning about complex numbers in class, and I don't really see why they're called numbers.

Originally, a number used to be a means of counting (natural numbers).

Then we extend these numbers to instances of owing other people money for instance (integers).

After that, we consider fractions when we need to split things like 2 pizzas between three people.

We then (skipping the algebraic numbers for the purposes of practicality), we use the real numbers to describe any length; e.g. the length of a diagonal of a unit square.

But this is when our original definition of a number fails to make sense- when we consider complex numbers, which brings me to my main question: what is a rigorous definition of 'number'?

Wikipedia claims that "A number is a mathematical object used to count, label, and measure", but this definition fails to make sense after we extend $\mathbb{R}$.

Does anyone know of any definition of number that can be generalised to complex numbers as well (and even higher order number systems like the quaternions)?

答:
I just want to share my own experience with similar questions. I believe the correct answer to this question is really important in letting one see and enjoy the power of Mathematics without having curtains blocking its light.

More than anything, I myself may have asked such a question from myself when thinking about dimensions greater than 3. For instance, the $R^n$ space. It seemed very complicated. I knew what would an $R^3$ space look like(for a long time, we have believed we live in one), but I used to ask myself how would $R^n$ be useful if it's not imaginable in real world to see what things would look like in nD (e.g, in 5D). But after some while,  I learned some Algebra and realized that if I measured 5 things and put those 5 output numbers in a collection, that would be in 5D as well!

But wait, those were totally different "things"! The "geometrical" 1D 2D 3D spaces, and the dimensions in Algebra. So how would we be able to "ultimately" define what a dimension is?

The answer is, we will never be able to. A dimension can be many things depending on the context. But then how would we be able to know what we're talking about when we talk about "dimensions"? The answer is simple, we will pay attention to the context, and the context will "describe" the thing we're talking about.

So to sum it up, in Mathematics, we don't really care about all-encompassing definitions, neither do we care about imaginations, because both of these would strive for concreteness, while Maths is abstract. The things we truly care about in Maths are: "Descriptions". And those turn out to be very useful.

Title: What exactly is a number?


问:
Proof of normal space
Let $(X, \mathcal{T})$ a $T_1$ space. Suppose that, for all $F$ closed in $X$ and each $W \in \mathcal{T}$ such that $F \subset W$ exists a sequence of open sets $\{W_n\}_{n \in \mathbb{N}}$ such that $F \subset \bigcup_{n \in \mathbb{N}} W_n$ and $\overline{W_n} \subset W$. Prove that $X$ is normal.

My idea. Let $A,B \subset X$ closed sets and $A \cap B = \emptyset$ then $B \subset X-A$ so $X-A$ is open then exists a sequence of open sets $\{W_n\}_{n \in \mathbb{N}}$ such that $B \subset \bigcup_{n \in \mathbb{N}} W_n$ and $\overline{W_n} \subset X-A$

答:
SKETCH: You need to apply the hypothesis to both of the sets $A$ and $B$: there are open sets $V_n$ and $W_n$ for $n\in\Bbb N$ such that $A\subseteq\bigcup_{n\in\Bbb N}V_n$, $B\subseteq\bigcup_{n\in\Bbb N}W_n$, and $\operatorname{cl}V_n\subseteq X\setminus B$ and $\operatorname{cl}W_n\subseteq X\setminus A$ for each $n\in\Bbb N$.

Now let $G_0=V_0$ and $H_0=W_0\setminus\operatorname{cl}G_0$. If open sets $G_k$ and $H_k$ have been defined for $k=0,\ldots,n$, let

$$G_{n+1}=V_{n+1}\setminus\operatorname{cl}\bigcup_{k=0}^nH_k$$

and

$$H_{n+1}=W_{n+1}\setminus\operatorname{cl}\bigcup_{k=0}^{n+1}H_k\,.$$

Now let $G=\bigcup_{n\in\Bbb N}G_n$ and $H=\bigcup_{n\in\Bbb N}H_n$, and show that $G$ and $H$ are disjoint open nbhds of $A$ and $B$, respectively.

If you’ve seen a proof that $T_3$ Lindelöf spaces are normal, this should look quite familiar.

Title: Proof of normal space


问:
Suppose $g:\mathbb{R}\rightarrow \mathbb{R}$ is infinitely differentiable
Suppose $g:\mathbb{R}\rightarrow \mathbb{R}$ is infinitely differentiable, and that for each $n\in \mathbb{N}$ there are positive numbers $c_{n}$ and $\delta_{n}$ such that $|g(x)| \leq c_{n}|x|^n$ if $|x|<\delta_{n}$. Prove that for each natural number $n$, $g^{(n)}(0) = 0$.

My attempt using induction is as follows:

Base case ($n = 1$): By supposition there are $c_{1} >0$ and $\delta_{1} >0$ such that $|g(x)| \leq
 c_{1}|x|$ if $|x|<\delta_{1}$. Setting $x = 0$ yields that $|g(0)| \leq c_{1}|0| = 0$ and therefore $g(0) = 0$

Fix $k\in \mathbb{N}$ and suppose that $g^{(k)}(0) = 0$. This is the point where I get stuck. I have tried applying the Lagrange Remainder Theorem at this point, but have been unsuccessful. I understand this is not much to work with, but any help would be greatly appreciated.

答: If the claim is false, then let $k$ be the least natural number such that $g^{(k)}(0)\ne 0.$ Then in a suitable neighborhood of zero, one has $g(x)=x^k\left(a_k+a_{k+1}x^{k+1}+\cdots +\right)$ so $|g(x)|=|x^k||a_k+a_{k+1}x^{k+1}+\cdots +|\le c_k|x^k|.$ It follows that $\left|\frac{g(x)}{x^k}\right|\le c_k.$ Now, $g$ is bounded on $(-\delta_k,\delta_k)$, so letting $x\to 0$, we obtain a contradiction.


问:
Proving integrability of the martingale $\cosh(\lambda B_t)\textrm{e}^{-\frac12\lambda^2t}$
In trying to prove that $M_t:=\cosh(\lambda B_t)\textrm{e}^{-\frac12\lambda^2t}$, I have hit a few roadblocks. I know how to show the martingale property as has been done many times in literature; but I haven't been able to properly justify that $\mathbb{E}|M_t|<+\infty$ so that I can finally conclude that it is a true martingale. Clearly $\textrm{e}^{-\frac12\lambda^2t}$ is bounded but the $\cosh$ isn't. How do I deal with the $\cosh$ and prove the integrability? Thanks!

答:
The link here (Proof that the exponential martingale is a Brownian Motion) shows that for any $\delta\in\mathbb{R}$ that $$M_{\delta,t}= e^{\delta B_t - \frac{1}{2}\delta^2 t}$$
is a Martingale. Notice that for any $\lambda \in\mathbb{R}$ that
$$\frac{1}{2}(M_{\lambda,t} + M_{-\lambda,t}) = \frac{(e^{\lambda B_t}+e^{-\lambda B_t})}{2}e^{-\frac{1}{2}\lambda^2t} = \cosh(\lambda B_t)e^{-\frac{1}{2}\lambda^2 t}$$
Since linear combinations of Martingales are martingales, then $\cosh(\lambda B_t)e^{-\frac{1}{2}\lambda^2 t}$ is in fact a martingale.


问:
System of ODE without constant coefficients
I've got following system of ODE...

$$y_1'(x) = (3x-1)y_1(x)-(1-x)y_2(x)$$
$$y_2'(x)=-(x+2)y_1(x)+(x-2)y_2(x)$$

I'd like to have some tips on how to solve that. I've researched a lot on the Internet and on this site here but the message is always the same "It's hard to solve that but in this case, it's easy because you can use trick xy".

Well I don't see the trick here. Only thing I can think of, is to solve the first equation for $y_2$, so that...

$$y_2 = \frac{3x-1}{1-x}y_1(x) - \frac{1}{1-x}y_1'$$

and then differentiate so that I get $y_2' = ...$ and plug $y_2, y_2'$ into the second equation.

But that cant be a good idea, can it?

I'm hoping for some helpful tips. Thanks in advance =)

Body:
$$y_1'(x) = (3x-1)y_1(x)-(1-x)y_2(x)$$
$$y_2'(x)=-(x+2)y_1(x)+(x-2)y_2(x)$$
Sum both DE:
$$(y_1+y_2)'(x)=(2x-3)y_1(x)+(2x-3)y_2(x)$$
This is a  separable differential equation.
$$u'=(2x-3)u$$
Where $u=y_1+y_2$.
$$(\ln u)'=2x-3$$
$$\ln u=x^2-3x+C$$
$$y_1+y_2=C_1e^{x^2-3x}$$
$$y_1=-y_2+C_1e^{x^2-3x}$$
Plug this in the second DE and solve.


答:
Note that the given equation can be rewritten as

$$(x+1)(x-2)=-[(3-x)-4][(x+2)-4]=0$$

which leads to $3-x=4$ and $x+2=4$.

Title: Why is this method right to solve some equations of degree 2?

Body:
Given this equation

$-x^2+x+6=4$

We can write it as

$(3-x)(x+2) = 4$

So weridly if I take
$(3-x) = 4$ and $(x+2) = 4$, I can get $x=-1$ and $x=2$ that are indeed the correct solutions.

Why does this happen? And in which cases can this occur?


答:
Consider $(a-x)(x-b) = c$

With roots $r_1$ and $r_2$.

We need four equalities to be true

$$a-r_1 = c, \quad r_1-b = 1, \quad a-r_2 = 1, \quad r_2-b = c$$

We get

$$\text{$r_1=a-c=b+1$ and $r_2=a-1 = b+c$}$$

Which implies $c=a-b-1$

So for any $a$ and $b$

$$(a-x)(x-b) = a-b-1$$

will behave as you described.


问:
Quantifier elimination for $\text{ACF0}$ over signature $(+,·)$.
We know that the complex field has quantifier elimination. This is usually in reference to the signature $(+,·,-,0,1)$. My question is, does it also have quantifier elimination with reference to the signature $(+,·)$?

答:
Yes, it does.  To prove this, it suffices to check that every atomic quantifier-free formula in the signature $(+,\cdot,-,0,1)$ is equivalent to one in the signature $(+,\cdot)$.  Such an atomic quantifier-free formula is just a statement that two terms are equal, so by subtracting the terms we see that it is equivalent to the statement $$p(x_1,\dots,x_n)=0.$$ for some polynomial $p$ with coefficients with $\mathbb{Z}$, where $x_1,\dots,x_n$ are the free variables.

Putting all the terms with positive coefficients on one side and all the terms with negative coefficients on the other side, we can rewrite this in the form $$q(x_1,\dots,x_n)=r(x_1,\dots,x_n)$$ where $q$ and $r$ now have coefficients in $\mathbb{N}$.  As long as $q$ and $r$ are not $0$ and have no constant terms, this means we can write them using only addition and multiplication.  If $q$ is $0$, we can rewrite our equation as $r=r+r$, and similarly if $r$ is $0$.  If both $q$ and $r$ are $0$, our equation is equivalent to $\top$.

The only remaining issue is if $q$ or $r$ has a constant term.  To fix this, we can consider the equation $$x_1q(x_1,\dots,x_n)=x_1r(x_1,\dots,x_n)$$ which has no constant terms and is equivalent to our original equation as long as $x_1+x_1\neq x_1$ (i.e., $x_1\neq 0$).  So by taking Boolean combinations, it suffices to also be able to express the formula $$q(0,x_2,\dots,x_n)=r(0,x_2,\dots,x_n).$$  But this formula has one fewer variable than we started with, so we can handle it by induction on the number of variables.  (In the base case where there are no variables, every formula is equivalent to $\top$ or $\bot$.)


问:
How do I make spirograh diagrams using software?
There are these "string math artworks" I believe they are called "spirographs" where people usually take a circle or other geometric shape and put in nails on the shape and then use yarn and weave the yarn in a particular way such as to make these artworks.

I am wanting to make similar diagrams using software instead of physically making them with nails and yarn. I found this one link where someone is using Tikz with Latex to generate a circle with nodes. This might work, but I don't know exactly how to use Latex and this Tikz (plug-in) software.

If someone can point me in the right direction that would be great. Thanks.

Here are some links:

https://tex.stackexchange.com/questions/415330/make-graph-where-nodes-and-connecting-edges-are-circles?newreg=03724c430452487a8f756a4c311bbdc3

https://www.youtube.com/watch?v=0K2YbYTyEqY

答:
Horses for courses.

If I understand correctly this Wikipedia Spirograph article indicates that you are interested in creating various types of roulette curves.  This implies that you want to create a curve that corresponds to a mathematical equation [e.g. $r^2 = k,$ or $x^2 + y^2 = k,$ or $r^2 = 2\cos(2\theta)$].

Under the assumption that you will always be able to supply the mathematical equation that corresponds to the image that you are trying to display, your query can be split into two questions:

How do you create the image?

What do you do with the image?

For insertion of an image into a pdf file that can be printed or stored (e.g. electronic math notebook), I recommend starting with LaTeX.

From this foundation, you have two choices.  The first choice is to create an image using math software that does not dovetail with LaTeX.  Then, convert the image into (for example) a jpeg file (i.e. *.jpg) and then issue the LaTeX commands to insert the graphics file into your pdf file.

Within LaTeX, the other choice is to find a graphics package that dovetails with LaTex - see CTAN.  For specialized graphics packages you can search for "graphics" from the opening screen at CTAN.  However, there are generally three all purpose Latex-plug-in graphics packages: PSTricks, Asymptote, and PGF-Tikz.  All three of these packages are capable of creating mathematical graphs directly into your LaTeX generated pdf file.

All three of these all purpose Latex graphics packages have significant learning curves, and are (perhaps) oriented towards LaTeX users with different skills.  What most LaTeX users (me included) do is to spend a brief amount of time exploring each one, including the software's gallery of samples, and documentation.  Then, the user will typically choose only one of the three, and focus on mastering that particular package.

This begs the question of how to produce various mathematical images totally outside of Latex.  I am largely ignorant in this area.  In fact, I stumbled on this query simply trying to educate myself.

I decided to start by searching mathSE for all queries with the [math-software] tag.  Once I complete that, I will then start with Wikipedia.  So far, my two ground-zero Wikipedia articles are Plotting Software and Math Software.

Once I gather links by exhausting mathSE and my two ground zero links then I'll explore the corresponding packages.  One key issue to consider is whether it is acceptable to use the application online through a browswer, or whether you wish to download an application for local use.  A peripheral issue is compatibility with your operating System (i.e. Windows, Linux derivative, or Apple-Mac).

Another issue is usage.  Do you wish to create animated graphics (e.g. *.png or *.gif files)?  Further, are you going to collect the graphical images into a folder on your hard drive?  Alternatively, do you simply want a method of quickly creating a temporary math image that you don't plan to store.  Examples of temporary usage include helping you to understand a specific math problem and illustrating a mathSE query or answer with an embedded image.


问:
Any combination of nilpotent operators is still nilpotent?

If $N$ and $N'$ are nilpotent operator on a vector space and $N$ commutes with $N'$ then $N+N'$ is nilpotent.

$$(N+N')^r=\sum_{j=0}^r \binom{r}{j}N^j(N')^{r-j}=0$$ for some r large enough. but it seems if $N$ does not commute with $N'$, $N+N'$ is still nilpotent. So, Can I say that any combination of nilpotent operators is still nilpotent?

答: No, you can't: take $A=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $B=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$. Then $A+B$ squares to the identity and isn't nilpotent.


问:
Logic and implication negation
I am new to predicate logic and there is something I can't seem to understand.

Let's have an implication:
If I have a sister, I have a sibling.

Now, it's negation should be:
I have a sister and I do not have a sibling.

But I can't understand why the following is not a valid negation:
If I have a sister, I do not have a sibling.

It should be a negation because the first part of the implication happened and the second did not.

答: The structure of your statement is $A\Rightarrow B$ where $A=\text{"I have a sister"}$ and $B=\text{"I have a sibling"}$. Then the negation is $\lnot B\Rightarrow \lnot A$ which is "If I don't have a sibling I don't have a sister".

Title: Logic and implication negation


问:
Does there getting from $1$ to $\sqrt[4]{2}$ using $\sqrt{\alpha^2+ 1}$
The aim is to get from $1$ to $\sqrt[4]{2}$ or prove it is impossible using only one of the following options:

Add or subtract  two previously constructed numbers.

Multiply two previously constructed numbers.

Using a previously constructed number $\alpha$ construct both solutions to $\alpha^2+1=\beta^2$.
I’ve managed to construct many numbers close to it, such as $\sqrt{4+2\sqrt{2}}$.

I’m pretty sure it’s impossible but haven’t managed to prove it.

Can any of you help?

Edit 1: reciprocals

We can construct all quadratic radicals and rational numbers.

In this case, we are seaching for an extension of $\mathbb Q$ that is closed under (3), The constructible numbers are closed under this operation, but i think there is a subfield of the constructible numbers closed under it and containing $\mathbb Q$ as a subfield.

答:
Alright, you should check your calculation again. If you really did construct $2^{1/4}$ you would immediately be able to construct $\sqrt{1 + \sqrt 2}.$  This is not possible: the quickest way to say it is that Hilbert's field is the set of totally real elements in the constructible field (closed under square roots of positive elements).

This is pages 145-148 in Geometry: Euclid and Beyond by Robin Hartshorne.

I repeated the first example search at
https://doc.sagemath.org/html/en/reference/number_fields/sage/rings/number_field/totallyreal_rel.html

and got

```<code>jagy@phobeusjunior:~$ sage
┌────────────────────────────────────────────────────────────────────┐
│ SageMath Version 6.9, Release Date: 2015-10-10                     │
│ Type "notebook()" for the browser-based notebook interface.        │
│ Type "help()" for help.                                            │
└────────────────────────────────────────────────────────────────────┘
sage:  ZZx = ZZ['x']
sage:  F.<t> = NumberField(x^2-2)
sage: enumerate_totallyreal_fields_rel(F, 2, 10000)
[[1600, x^4 - 6*x^2 + 4, xF^2 + (t + 1)*xF + 3*t - 3],
 [2048, x^4 - 4*x^2 + 2, xF^2 + t - 2],
 [2304, x^4 - 4*x^2 + 1, xF^2 + t*xF - 1],
 [2624, x^4 - 2*x^3 - 3*x^2 + 2*x + 1, xF^2 + (t + 1)*xF + t - 1],
 [4352, x^4 - 6*x^2 - 4*x + 2, xF^2 + t*xF + t - 2],
 [7168, x^4 - 6*x^2 + 7, xF^2 + t - 3],
 [7232, x^4 - 2*x^3 - 5*x^2 + 4*x + 4, xF^2 + (t + 1)*xF + t - 2],
 [8768, x^4 - 2*x^3 - 5*x^2 + 6*x + 7, xF^2 + xF + t - 3],
 [9792, x^4 - 2*x^3 - 7*x^2 + 2*x + 7, xF^2 + (t + 1)*xF + 2*t - 3]]

sage: enumerate_totallyreal_fields_rel(F, 2, 100000)
[[1600, x^4 - 6*x^2 + 4, xF^2 + xF - 1],
 [2048, x^4 - 4*x^2 + 2, xF^2 + t - 10],
 [2304, x^4 - 4*x^2 + 1, xF^2 + t*xF - 1],
 [2624, x^4 - 2*x^3 - 3*x^2 + 2*x + 1, xF^2 + (t + 1)*xF + t - 1],
 [4352, x^4 - 6*x^2 - 4*x + 2, xF^2 + t*xF + t - 14],
 [7168, x^4 - 6*x^2 + 7, xF^2 + t - 3],
 [7232, x^4 - 2*x^3 - 5*x^2 + 4*x + 4, xF^2 + (t + 1)*xF + t - 2],
 [8768, x^4 - 2*x^3 - 5*x^2 + 6*x + 7, xF^2 + (t + 1)*xF + 4*t - 5],
 [9792, x^4 - 2*x^3 - 7*x^2 + 2*x + 7, xF^2 + (t + 1)*xF + 2*t - 3],
 [10304, x^4 - 2*x^3 - 7*x^2 + 8*x + 8, xF^2 + (t + 1)*xF + 3*t - 4],
 [10816, x^4 - 2*x^3 - 9*x^2 + 10*x - 1, xF^2 + (t + 1)*xF + 7*t - 9],
 [12544, x^4 - 8*x^2 + 9, xF^2 + t*xF - 3],
 [13888, x^4 - 2*x^3 - 7*x^2 + 6*x + 9, xF^2 + (t + 1)*xF + t - 3],
 [14336, x^4 - 8*x^2 + 14, xF^2 + t - 4],
 [16448, x^4 - 2*x^3 - 7*x^2 + 8*x + 14, xF^2 + (t + 1)*xF + 6*t - 8],
 [18432, x^4 - 12*x^2 + 18, xF^2 + 3*t - 6],
 [18496, x^4 - 2*x^3 - 11*x^2 + 12*x + 2, xF^2 + xF - 4],
 [18688, x^4 - 10*x^2 - 4*x + 14, xF^2 + t*xF + t - 4],
 [20032, x^4 - 2*x^3 - 9*x^2 + 10*x + 17, xF^2 + (t + 1)*xF + 5*t - 7],
 [21056, x^4 - 2*x^3 - 11*x^2 + 2*x + 17, xF^2 + (t + 1)*xF + 3*t - 5],
 [21568, x^4 - 2*x^3 - 11*x^2 + 12*x + 18, xF^2 + (t + 1)*xF + 4*t - 6],
 [22592, x^4 - 2*x^3 - 9*x^2 + 8*x + 16, xF^2 + (t + 1)*xF + t - 4],
 [22784, x^4 - 12*x^2 - 8*x + 17, xF^2 + t*xF + 2*t - 5],
 [23552, x^4 - 10*x^2 + 23, xF^2 + t - 5],
 [24832, x^4 - 14*x^2 - 12*x + 18, xF^2 + t*xF + 3*t - 6],
 [26176, x^4 - 2*x^3 - 9*x^2 + 10*x + 23, xF^2 + (t + 1)*xF + 8*t - 11],
 [28224, x^4 - 2*x^3 - 13*x^2 + 14*x + 7, xF^2 + xF - 5],
 [29248, x^4 - 2*x^3 - 11*x^2 + 6*x + 23, xF^2 + (t + 1)*xF + 2*t - 5],
 [30976, x^4 - 12*x^2 + 25, xF^2 + t*xF - 5],
 [31744, x^4 - 14*x^2 + 31, xF^2 + 3*t - 7],
 [31808, x^4 - 2*x^3 - 11*x^2 + 12*x + 28, xF^2 + (t + 1)*xF + 7*t - 10],
 [33344, x^4 - 2*x^3 - 11*x^2 + 10*x + 25, xF^2 + (t + 1)*xF + t - 5],
 [34816, x^4 - 12*x^2 + 34, xF^2 + t - 6],
 [35392, x^4 - 2*x^3 - 13*x^2 + 14*x + 31, xF^2 + (t + 1)*xF + 6*t - 9],
 [36416, x^4 - 2*x^3 - 15*x^2 + 2*x + 31, xF^2 + (t + 1)*xF + 4*t - 7],
 [36928, x^4 - 2*x^3 - 15*x^2 + 16*x + 32, xF^2 + (t + 1)*xF + 5*t - 8],
 [37952, x^4 - 2*x^3 - 11*x^2 + 12*x + 34, xF^2 + xF + t - 6],
 [41216, x^4 - 14*x^2 - 4*x + 34, xF^2 + t*xF + t - 6],
 [42048, x^4 - 2*x^3 - 13*x^2 + 8*x + 34, xF^2 + (t + 1)*xF + 2*t - 6],
 [45632, x^4 - 2*x^3 - 13*x^2 + 14*x + 41, xF^2 + xF + 2*t - 7],
 [46144, x^4 - 2*x^3 - 13*x^2 + 12*x + 36, xF^2 + (t + 1)*xF + t - 6],
 [47104, x^4 - 16*x^2 + 46, xF^2 + 3*t - 8],
 [48128, x^4 - 14*x^2 + 47, xF^2 + t - 7],
 [48704, x^4 - 2*x^3 - 15*x^2 + 6*x + 41, xF^2 + (t + 1)*xF + 3*t - 7],
 [49408, x^4 - 16*x^2 - 8*x + 41, xF^2 + t*xF + 2*t - 7],
 [51200, x^4 - 20*x^2 + 50, xF^2 + 5*t - 10],
 [51264, x^4 - 2*x^3 - 15*x^2 + 16*x + 46, xF^2 + (t + 1)*xF + 8*t - 12],
 [51776, x^4 - 2*x^3 - 13*x^2 + 14*x + 47, xF^2 + xF + t - 7],
 [53312, x^4 - 2*x^3 - 17*x^2 + 4*x + 46, xF^2 + (t + 1)*xF + 4*t - 8],
 [53824, x^4 - 2*x^3 - 17*x^2 + 18*x + 23, xF^2 + xF - 7],
 [54848, x^4 - 2*x^3 - 17*x^2 + 18*x + 49, xF^2 + (t + 1)*xF + 7*t - 11],
 [55552, x^4 - 18*x^2 - 12*x + 46, xF^2 + t*xF + 3*t - 8],
 [55872, x^4 - 2*x^3 - 19*x^2 + 2*x + 49, xF^2 + (t + 1)*xF + 5*t - 9],
 [56384, x^4 - 2*x^3 - 19*x^2 + 20*x + 50, xF^2 + (t + 1)*xF + 6*t - 10],
 [56896, x^4 - 2*x^3 - 15*x^2 + 10*x + 47, xF^2 + (t + 1)*xF + 2*t - 7],
 [57600, x^4 - 16*x^2 + 49, xF^2 + t*xF - 7],
 [59648, x^4 - 20*x^2 - 16*x + 49, xF^2 + t*xF + 4*t - 9],
 [60992, x^4 - 2*x^3 - 15*x^2 + 14*x + 49, xF^2 + (t + 1)*xF + t - 7],
 [61696, x^4 - 22*x^2 - 20*x + 50, xF^2 + t*xF + 5*t - 10],
 [63488, x^4 - 16*x^2 + 62, xF^2 + t - 8],
 [64512, x^4 - 18*x^2 + 63, xF^2 + 3*t - 9],
 [65600, x^4 - 2*x^3 - 17*x^2 + 8*x + 56, xF^2 + (t + 1)*xF + 3*t - 8],
 [67648, x^4 - 2*x^3 - 15*x^2 + 16*x + 62, xF^2 + xF + t - 8],
 [69184, x^4 - 2*x^3 - 17*x^2 + 18*x + 63, xF^2 + xF + 3*t - 9],
 [69696, x^4 - 2*x^3 - 19*x^2 + 20*x + 34, xF^2 + xF - 8],
 [71936, x^4 - 18*x^2 - 4*x + 62, xF^2 + t*xF + t - 8],
 [72256, x^4 - 2*x^3 - 19*x^2 + 6*x + 63, xF^2 + (t + 1)*xF + 4*t - 9],
 [72704, x^4 - 22*x^2 + 71, xF^2 + 5*t - 11],
 [73792, x^4 - 2*x^3 - 17*x^2 + 12*x + 62, xF^2 + (t + 1)*xF + 2*t - 8],
 [74816, x^4 - 2*x^3 - 19*x^2 + 20*x + 68, xF^2 + xF + 4*t - 10],
 [76864, x^4 - 2*x^3 - 21*x^2 + 4*x + 68, xF^2 + (t + 1)*xF + 5*t - 10],
 [77888, x^4 - 2*x^3 - 17*x^2 + 16*x + 64, xF^2 + (t + 1)*xF + t - 8],
 [79424, x^4 - 2*x^3 - 23*x^2 + 2*x + 71, xF^2 + (t + 1)*xF + 6*t - 11],
 [79424, x^4 - 2*x^3 - 17*x^2 + 18*x + 73, xF^2 + xF + 2*t - 9],
 [79936, x^4 - 2*x^3 - 23*x^2 + 24*x + 72, xF^2 + (t + 1)*xF + 7*t - 12],
 [80896, x^4 - 18*x^2 + 79, xF^2 + t - 9],
 [83968, x^4 - 20*x^2 + 82, xF^2 + 3*t - 10],
 [84224, x^4 - 20*x^2 - 8*x + 73, xF^2 + t*xF + 2*t - 9],
 [84544, x^4 - 2*x^3 - 19*x^2 + 10*x + 73, xF^2 + (t + 1)*xF + 3*t - 9],
 [85568, x^4 - 2*x^3 - 17*x^2 + 18*x + 79, xF^2 + xF + t - 9],
 [87616, x^4 - 2*x^3 - 21*x^2 + 22*x + 47, xF^2 + xF - 9],
 [89152, x^4 - 2*x^3 - 19*x^2 + 20*x + 82, xF^2 + xF + 3*t - 10],
 [92416, x^4 - 20*x^2 + 81, xF^2 + t*xF - 9],
 [92736, x^4 - 2*x^3 - 19*x^2 + 14*x + 79, xF^2 + (t + 1)*xF + 2*t - 9],
 [93248, x^4 - 2*x^3 - 21*x^2 + 8*x + 82, xF^2 + (t + 1)*xF + 4*t - 10],
 [94464, x^4 - 22*x^2 - 12*x + 82, xF^2 + t*xF + 3*t - 10],
 [96256, x^4 - 24*x^2 + 94, xF^2 + 5*t - 12],
 [96832, x^4 - 2*x^3 - 19*x^2 + 18*x + 81, xF^2 + (t + 1)*xF + t - 9],
 [96832, x^4 - 2*x^3 - 21*x^2 + 22*x + 89, xF^2 + xF + 4*t - 11],
 [99392, x^4 - 2*x^3 - 19*x^2 + 20*x + 92, xF^2 + xF + 2*t - 10],
 [99904, x^4 - 2*x^3 - 23*x^2 + 6*x + 89, xF^2 + (t + 1)*xF + 5*t - 11]]
sage:
</code>```

Body:
Alright, you should check your calculation again. If you really did construct $2^{1/4}$ you would immediately be able to construct $\sqrt{1 + \sqrt 2}.$  This is not possible: the quickest way to say it is that Hilbert's field is the set of totally real elements in the constructible field (closed under square roots of positive elements).

This is pages 145-148 in Geometry: Euclid and Beyond by Robin Hartshorne.

I repeated the first example search at
https://doc.sagemath.org/html/en/reference/number_fields/sage/rings/number_field/totallyreal_rel.html

and got

jagy@phobeusjunior:~$ sage
┌────────────────────────────────────────────────────────────────────┐
│ SageMath Version 6.9, Release Date: 2015-10-10                     │
│ Type "notebook()" for the browser-based notebook interface.        │
│ Type "help()" for help.                                            │
└────────────────────────────────────────────────────────────────────┘
sage:  ZZx = ZZ['x']
sage:  F.<t> = NumberField(x^2-2)
sage: enumerate_totallyreal_fields_rel(F, 2, 10000)
[[1600, x^4 - 6*x^2 + 4, xF^2 + (t + 1)*xF + 3*t - 3],
 [2048, x^4 - 4*x^2 + 2, xF^2 + t - 2],
 [2304, x^4 - 4*x^2 + 1, xF^2 + t*xF - 1],
 [2624, x^4 - 2*x^3 - 3*x^2 + 2*x + 1, xF^2 + (t + 1)*xF + t - 1],
 [4352, x^4 - 6*x^2 - 4*x + 2, xF^2 + t*xF + t - 2],
 [7168, x^4 - 6*x^2 + 7, xF^2 + t - 3],
 [7232, x^4 - 2*x^3 - 5*x^2 + 4*x + 4, xF^2 + (t + 1)*xF + t - 2],
 [8768, x^4 - 2*x^3 - 5*x^2 + 6*x + 7, xF^2 + xF + t - 3],
 [9792, x^4 - 2*x^3 - 7*x^2 + 2*x + 7, xF^2 + (t + 1)*xF + 2*t - 3]]

sage: enumerate_totallyreal_fields_rel(F, 2, 100000)
[[1600, x^4 - 6*x^2 + 4, xF^2 + xF - 1],
 [2048, x^4 - 4*x^2 + 2, xF^2 + t - 10],
 [2304, x^4 - 4*x^2 + 1, xF^2 + t*xF - 1],
 [2624, x^4 - 2*x^3 - 3*x^2 + 2*x + 1, xF^2 + (t + 1)*xF + t - 1],
 [4352, x^4 - 6*x^2 - 4*x + 2, xF^2 + t*xF + t - 14],
 [7168, x^4 - 6*x^2 + 7, xF^2 + t - 3],
 [7232, x^4 - 2*x^3 - 5*x^2 + 4*x + 4, xF^2 + (t + 1)*xF + t - 2],
 [8768, x^4 - 2*x^3 - 5*x^2 + 6*x + 7, xF^2 + (t + 1)*xF + 4*t - 5],
 [9792, x^4 - 2*x^3 - 7*x^2 + 2*x + 7, xF^2 + (t + 1)*xF + 2*t - 3],
 [10304, x^4 - 2*x^3 - 7*x^2 + 8*x + 8, xF^2 + (t + 1)*xF + 3*t - 4],
 [10816, x^4 - 2*x^3 - 9*x^2 + 10*x - 1, xF^2 + (t + 1)*xF + 7*t - 9],
 [12544, x^4 - 8*x^2 + 9, xF^2 + t*xF - 3],
 [13888, x^4 - 2*x^3 - 7*x^2 + 6*x + 9, xF^2 + (t + 1)*xF + t - 3],
 [14336, x^4 - 8*x^2 + 14, xF^2 + t - 4],
 [16448, x^4 - 2*x^3 - 7*x^2 + 8*x + 14, xF^2 + (t + 1)*xF + 6*t - 8],
 [18432, x^4 - 12*x^2 + 18, xF^2 + 3*t - 6],
 [18496, x^4 - 2*x^3 - 11*x^2 + 12*x + 2, xF^2 + xF - 4],
 [18688, x^4 - 10*x^2 - 4*x + 14, xF^2 + t*xF + t - 4],
 [20032, x^4 - 2*x^3 - 9*x^2 + 10*x + 17, xF^2 + (t + 1)*xF + 5*t - 7],
 [21056, x^4 - 2*x^3 - 11*x^2 + 2*x + 17, xF^2 + (t + 1)*xF + 3*t - 5],
 [21568, x^4 - 2*x^3 - 11*x^2 + 12*x + 18, xF^2 + (t + 1)*xF + 4*t - 6],
 [22592, x^4 - 2*x^3 - 9*x^2 + 8*x + 16, xF^2 + (t + 1)*xF + t - 4],
 [22784, x^4 - 12*x^2 - 8*x + 17, xF^2 + t*xF + 2*t - 5],
 [23552, x^4 - 10*x^2 + 23, xF^2 + t - 5],
 [24832, x^4 - 14*x^2 - 12*x + 18, xF^2 + t*xF + 3*t - 6],
 [26176, x^4 - 2*x^3 - 9*x^2 + 10*x + 23, xF^2 + (t + 1)*xF + 8*t - 11],
 [28224, x^4 - 2*x^3 - 13*x^2 + 14*x + 7, xF^2 + xF - 5],
 [29248, x^4 - 2*x^3 - 11*x^2 + 6*x + 23, xF^2 + (t + 1)*xF + 2*t - 5],
 [30976, x^4 - 12*x^2 + 25, xF^2 + t*xF - 5],
 [31744, x^4 - 14*x^2 + 31, xF^2 + 3*t - 7],
 [31808, x^4 - 2*x^3 - 11*x^2 + 12*x + 28, xF^2 + (t + 1)*xF + 7*t - 10],
 [33344, x^4 - 2*x^3 - 11*x^2 + 10*x + 25, xF^2 + (t + 1)*xF + t - 5],
 [34816, x^4 - 12*x^2 + 34, xF^2 + t - 6],
 [35392, x^4 - 2*x^3 - 13*x^2 + 14*x + 31, xF^2 + (t + 1)*xF + 6*t - 9],
 [36416, x^4 - 2*x^3 - 15*x^2 + 2*x + 31, xF^2 + (t + 1)*xF + 4*t - 7],
 [36928, x^4 - 2*x^3 - 15*x^2 + 16*x + 32, xF^2 + (t + 1)*xF + 5*t - 8],
 [37952, x^4 - 2*x^3 - 11*x^2 + 12*x + 34, xF^2 + xF + t - 6],
 [41216, x^4 - 14*x^2 - 4*x + 34, xF^2 + t*xF + t - 6],
 [42048, x^4 - 2*x^3 - 13*x^2 + 8*x + 34, xF^2 + (t + 1)*xF + 2*t - 6],
 [45632, x^4 - 2*x^3 - 13*x^2 + 14*x + 41, xF^2 + xF + 2*t - 7],
 [46144, x^4 - 2*x^3 - 13*x^2 + 12*x + 36, xF^2 + (t + 1)*xF + t - 6],
 [47104, x^4 - 16*x^2 + 46, xF^2 + 3*t - 8],
 [48128, x^4 - 14*x^2 + 47, xF^2 + t - 7],
 [48704, x^4 - 2*x^3 - 15*x^2 + 6*x + 41, xF^2 + (t + 1)*xF + 3*t - 7],
 [49408, x^4 - 16*x^2 - 8*x + 41, xF^2 + t*xF + 2*t - 7],
 [51200, x^4 - 20*x^2 + 50, xF^2 + 5*t - 10],
 [51264, x^4 - 2*x^3 - 15*x^2 + 16*x + 46, xF^2 + (t + 1)*xF + 8*t - 12],
 [51776, x^4 - 2*x^3 - 13*x^2 + 14*x + 47, xF^2 + xF + t - 7],
 [53312, x^4 - 2*x^3 - 17*x^2 + 4*x + 46, xF^2 + (t + 1)*xF + 4*t - 8],
 [53824, x^4 - 2*x^3 - 17*x^2 + 18*x + 23, xF^2 + xF - 7],
 [54848, x^4 - 2*x^3 - 17*x^2 + 18*x + 49, xF^2 + (t + 1)*xF + 7*t - 11],
 [55552, x^4 - 18*x^2 - 12*x + 46, xF^2 + t*xF + 3*t - 8],
 [55872, x^4 - 2*x^3 - 19*x^2 + 2*x + 49, xF^2 + (t + 1)*xF + 5*t - 9],
 [56384, x^4 - 2*x^3 - 19*x^2 + 20*x + 50, xF^2 + (t + 1)*xF + 6*t - 10],
 [56896, x^4 - 2*x^3 - 15*x^2 + 10*x + 47, xF^2 + (t + 1)*xF + 2*t - 7],
 [57600, x^4 - 16*x^2 + 49, xF^2 + t*xF - 7],
 [59648, x^4 - 20*x^2 - 16*x + 49, xF^2 + t*xF + 4*t - 9],
 [60992, x^4 - 2*x^3 - 15*x^2 + 14*x + 49, xF^2 + (t + 1)*xF + t - 7],
 [61696, x^4 - 22*x^2 - 20*x + 50, xF^2 + t*xF + 5*t - 10],
 [63488, x^4 - 16*x^2 + 62, xF^2 + t - 8],
 [64512, x^4 - 18*x^2 + 63, xF^2 + 3*t - 9],
 [65600, x^4 - 2*x^3 - 17*x^2 + 8*x + 56, xF^2 + (t + 1)*xF + 3*t - 8],
 [67648, x^4 - 2*x^3 - 15*x^2 + 16*x + 62, xF^2 + xF + t - 8],
 [69184, x^4 - 2*x^3 - 17*x^2 + 18*x + 63, xF^2 + xF + 3*t - 9],
 [69696, x^4 - 2*x^3 - 19*x^2 + 20*x + 34, xF^2 + xF - 8],
 [71936, x^4 - 18*x^2 - 4*x + 62, xF^2 + t*xF + t - 8],
 [72256, x^4 - 2*x^3 - 19*x^2 + 6*x + 63, xF^2 + (t + 1)*xF + 4*t - 9],
 [72704, x^4 - 22*x^2 + 71, xF^2 + 5*t - 11],
 [73792, x^4 - 2*x^3 - 17*x^2 + 12*x + 62, xF^2 + (t + 1)*xF + 2*t - 8],
 [74816, x^4 - 2*x^3 - 19*x^2 + 20*x + 68, xF^2 + xF + 4*t - 10],
 [76864, x^4 - 2*x^3 - 21*x^2 + 4*x + 68, xF^2 + (t + 1)*xF + 5*t - 10],
 [77888, x^4 - 2*x^3 - 17*x^2 + 16*x + 64, xF^2 + (t + 1)*xF + t - 8],
 [79424, x^4 - 2*x^3 - 23*x^2 + 2*x + 71, xF^2 + (t + 1)*xF + 6*t - 11],
 [79424, x^4 - 2*x^3 - 17*x^2 + 18*x + 73, xF^2 + xF + 2*t - 9],
 [79936, x^4 - 2*x^3 - 23*x^2 + 24*x + 72, xF^2 + (t + 1)*xF + 7*t - 12],
 [80896, x^4 - 18*x^2 + 79, xF^2 + t - 9],
 [83968, x^4 - 20*x^2 + 82, xF^2 + 3*t - 10],
 [84224, x^4 - 20*x^2 - 8*x + 73, xF^2 + t*xF + 2*t - 9],
 [84544, x^4 - 2*x^3 - 19*x^2 + 10*x + 73, xF^2 + (t + 1)*xF + 3*t - 9],
 [85568, x^4 - 2*x^3 - 17*x^2 + 18*x + 79, xF^2 + xF + t - 9],
 [87616, x^4 - 2*x^3 - 21*x^2 + 22*x + 47, xF^2 + xF - 9],
 [89152, x^4 - 2*x^3 - 19*x^2 + 20*x + 82, xF^2 + xF + 3*t - 10],
 [92416, x^4 - 20*x^2 + 81, xF^2 + t*xF - 9],
 [92736, x^4 - 2*x^3 - 19*x^2 + 14*x + 79, xF^2 + (t + 1)*xF + 2*t - 9],
 [93248, x^4 - 2*x^3 - 21*x^2 + 8*x + 82, xF^2 + (t + 1)*xF + 4*t - 10],
 [94464, x^4 - 22*x^2 - 12*x + 82, xF^2 + t*xF + 3*t - 10],
 [96256, x^4 - 24*x^2 + 94, xF^2 + 5*t - 12],
 [96832, x^4 - 2*x^3 - 19*x^2 + 18*x + 81, xF^2 + (t + 1)*xF + t - 9],
 [96832, x^4 - 2*x^3 - 21*x^2 + 22*x + 89, xF^2 + xF + 4*t - 11],
 [99392, x^4 - 2*x^3 - 19*x^2 + 20*x + 92, xF^2 + xF + 2*t - 10],
 [99904, x^4 - 2*x^3 - 23*x^2 + 6*x + 89, xF^2 + (t + 1)*xF + 5*t - 11]]
sage: