问:
Most basic equation: Heart rate is in units of bpm. What are the units of "b"?
This is the most simple question in the world, but it is surprisingly difficult to explain to a student.  The heart rate is measured in bpm, beats per minute.  Minutes are a true unit. Beats are not a true unit. However, we are using beats as if it where a unit.  I'm not sure how to explain this to a student, and I'm confusing myself.   Please help.</p>

答:
BPM means "beats per minute" but that doesn't mean that "beats" is a "unit."  Otherwise we could have an infinite number of "units" based on counting other things.

For instance:  How many red two-door cars with sagging taillights and broken headlights with expired license plates pass by a point on this road in a minute?  Oh, $7$, you say?  Well I suppose the rate is 7 RTDCSTBHELPPM."

So now "RTDCSTBHELP" is a "unit"?!?!

Insane.


问:
What is the definite integral of...
$$\int^L_{-L} x \sin(\frac{\pi nx}{L})$$

I've seen something like this in Fourier theory, but I'm still not sure how to approach this integral. Wolfram Alpha gives me the answer, but no method. Integrate by parts? Substitution?

答:
Whenever you have $p(x)f(x)$, where $p(x)$ is a polynomial, and $f(x)$ is a function like $\sin(x)$ or $e^x$, such that differentiating doesn't increase its "order," integration by parts is always a good technique.  Differentiate the polynomial so it becomes "less complicated," and integrate the other function.

$$\int_{-L}^Lx\sin\left(\frac{n\pi x}{L}\right)\,dx$$
Let $u = x, dv=\sin\left(\frac{n\pi x}{L}\right)dx$.  Then $du = dx$ and $v = \frac{-L}{n\pi}\cos\left(\frac{n\pi x}{L}\right)$.

Thus,

$$\int_{-L}^Lx\sin\left(\frac{n\pi x}{L}\right)\,dx = \frac{-xL}{n\pi}\cos\left(\frac{n\pi x}{L}\right)\Bigg|_{-L}^L - \int_{-L}^L\frac{-L}{n\pi}\cos\left(\frac{n\pi x}{L}\right)\,dx$$

ContentLicense: CC BY-SA 3.0

Title: What is the definite integral of...


问:
Problem's solving the integral $\int_0^\infty \frac{1} {x^{1/3}(x+a)} dx$ where a is a constant
I found this integral

$$ \int_{0}^{\infty} \frac{1} {x^{1/3}(x+a)} \, \mathrm{d}x $$

as part of an example of differentiation under the integral sign in the book Advanced Calculus Explored, by Hamza E. Asamraee. There is only the hint that you can use the substitution:

$$y= x^{1/3}$$

so in the end:

$$ \mathrm{d}x = 3y^{2}\mathrm{d}y$$

And after performing the substitution the integral becomes:

$$ \int_{0}^{\infty} \frac{3y^2} {y(y^3+a)} \, \mathrm{d}y = \int_{0}^{\infty} \frac{3y} {(y^3+a)} \, \mathrm{d}y $$

After this you can factorize the denominator in the form

$$y^3+a = (y+a^{1/3})(y^2-a^{1/3}y+a^{2/3})$$

and use partial fractions so that the integral becomes:

$$ -\frac{1}{a^{1/3}} \int_{0}^{\infty} \frac{\mathrm{d}y}{y+a^{1/3}} + \int_{0}^{\infty} \frac{y+a}{y^2-a^{1/3}y+a^{2/3}} \, \mathrm{d}y $$

And the problem is that these two integrals diverge but in the book or even in wolfram alpha the result is:

$$ \frac{2\pi}{a^{1/3}\sqrt{3}} $$

But I don't know if I made a mistake in the substitution or in which part, so I request for help :c

答:
Hint:

WLOG $a=1$ (you can rescale the variable).

$$\int\left(\frac{y+1}{y^2-y+1}-\frac{1}{y+1}\right)dy=\int\left(\frac{y-\frac12}{y^2-y+1}+\frac{\frac32}{y^2-y+1}-\frac{1}{y+1}\right)dy
\\=\ln\frac{\sqrt{y^2-y+1}}{y+1}+\int \frac{\frac32}{y^2-y+1}dy.$$

The first term will cancel out, and the remaining integral is solved by an arc tangent.

ContentLicense: CC BY-SA 4.0


问:
Eigenfunction of the Fourier transform
I want to show that $$ \frac{1}{ \sqrt{ 2 \pi}} \int_{-\infty}^{\infty} \frac{e^{-iwx}}{\cosh{ (x  \sqrt{\frac{\pi}{2}}} ) } = \frac{1}{\cosh{ (w  \sqrt{\frac{\pi}{2}}} ) }  .$$

My attempt is to first make a substitution $y = x \sqrt{ \frac{\pi}{2} }$, and write $\cosh$ as exponentials which yeilds

$$ \frac{1}{ \sqrt{ 2 \pi}} \int_{-\infty}^{\infty} \frac{e^{-iwx}}{\cosh{ (x  \sqrt{\frac{\pi}{2}}} ) }dx =  \frac{2}{  \pi} \int_{-\infty}^{\infty} \frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{e^{y} + e^{-y} }dy =  \frac{2}{  \pi} \int_{-\infty}^{\infty} e^{-y}\frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{1 - (-e^{-2y}) }dy$$
Now I can write it as a convergent geometric series

$$\frac{2}{  \pi} \int_{-\infty}^{\infty} e^{-y}\frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{1 - (-e^{-2y}) }dy = \frac{2}{  \pi} \int_{-\infty}^{\infty} \sum_{k = 0}^{\infty} -e^{-2yk}  e^{-y}e^{-iwy\sqrt{\frac{2}{\pi}}}dy$$ $$ = \frac{2}{  \pi} \sum_{k = 0}^{\infty}\int_{-\infty}^{\infty}  -e^{-y(2k+1+iw\sqrt{\frac{2}{\pi}})}dy$$

However this integral diverges. Does anyone have any hint on what to do next, or another strategy to show this?

Since this would mean that $\frac{1}{\cosh( x \sqrt{\frac{\pi}{2}})}$ would be an eigenfunctions of the Fourier transforms, should you be able to write it as a linear combination of gaussians?

Thanks!

$\textbf{Added:}$
So before writing it as a geometric series, I split the integration in two parts
$$\frac{2}{  \pi} \int_{-\infty}^{\infty} \frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{e^{y} + e^{-y} }dy =  \frac{2}{  \pi} \int_{0}^{\infty} e^{-y}\frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{1 - (-e^{-2y}) }dy +  \frac{2}{  \pi} \int_{-\infty}^{0} e^{y}\frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{1 - (-e^{2y}) }dy$$
$$= \frac{2}{  \pi} \sum_{k = 0}^{\infty}\int_{0}^{\infty} (-1)^k e^{-y(2k+1+iw\sqrt{\frac{2}{\pi}})}dy + \frac{2}{  \pi} \sum_{k = 0}^{\infty}\int_{-\infty}^{0} (-1)^k  e^{y(2k+1-iw\sqrt{\frac{2}{\pi}})}dy$$
$$ =  \frac{4}{\pi} \sum_{k=0}^{\infty} (-1)^k \frac{ 2k+1 }{ (2k+1)^2 + (w \sqrt{\frac{2}{\pi}} )^2} $$

Now in a previous exercise I proved this identity:
$$ \frac{\cosh(a( \pi -x))}{\sinh(a \pi)} = \frac{1}{a \pi} + \frac{2}{\pi} \sum_{n = 1}^{\infty}\frac{a}{a^2+n^2}\cos(nx). $$ for $ 0 \leq x \leq \pi$, which look very similar with $x = \pi$. However I'm not sure how to proceed.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{{1 \over \root{2\pi}}
\int_{-\infty}^{\infty}{\expo{-\ic\omega x} \over
\cosh\pars{x\root{\pi/2}}}\,\dd x =
{1 \over \cosh\pars{\omega\root{\pi/2}}}}:\ {\Large ?}}$.

Lets $\ds{\pars{~\root{\pi/2}x \mapsto x/4 \implies x \mapsto
\root{2/\pi}x/4~}}$ and $\ds{\nu = \root{2/\pi}\omega/4}$:
\begin{align}
&\bbox[5px,#ffd]{{1 \over \root{2\pi}}
\int_{-\infty}^{\infty}{\expo{-\ic\omega x} \over
\cosh\pars{x\root{\pi/2}}}\,\dd x} =
\root{2 \over \pi}\,\Re
\int_{0}^{\infty}{\expo{-\ic\omega x} \over
\cosh\pars{x\root{\pi/2}}}\,\dd x
\\[5mm] = &\
{1 \over 2\pi}\,\Re
\int_{0}^{\infty}{\expo{-\ic\nu x} \over
\cosh\pars{x/4}}\,\dd x =
{1 \over \pi}\,\Re\int_{0}^{\infty}{\sinh\pars{x/4}
\expo{-\ic\nu x} \over \sinh\pars{x/2}}\,\dd x
\\[5mm] = &\
{1 \over \pi}\,\Re\int_{0}^{\infty}{
\expo{-\pars{1/4 + \ic\nu}x} -
\expo{-\pars{3/4 + \ic\nu}x}\over 1 - \expo{-x}}\,\dd x
\\[5mm] = &\
{1 \over \pi}\,\Re\bracks{\int_{0}^{\infty}{%
\expo{-x} - \expo{-\pars{3/4 + \ic\nu}x}\over 1 - \expo{-x}}
\,\dd x -
\int_{0}^{\infty}{%
\expo{-x} - \expo{-\pars{1/4 + \ic\nu}x}\over 1 - \expo{-x}}
\,\dd x}
\\[5mm] = &\
{1 \over \pi}\,\Re\bracks{%
\Psi\pars{{3 \over 4} + \ic\nu} -
\Psi\pars{\bracks{{1 \over 4} + \ic\nu}}}
\end{align}

where $\ds{\Psi}$ is the Digamma Function. See $\ds{\color{black}{\bf 6.3.22}}$ in A & S Table.

Then,
\begin{align}
&\bbox[5px,#ffd]{{1 \over \root{2\pi}}
\int_{-\infty}^{\infty}{\expo{-\ic\omega x} \over
\cosh\pars{x\root{\pi/2}}}\,\dd x}
\\[5mm] = &\
{1 \over 2\pi}\bracks{\Psi\pars{{3 \over 4} + \ic\nu}
- \Psi\pars{{1 \over 4} - \ic\nu}}
\\[2mm] + &\
{1 \over 2\pi}\bracks{\Psi\pars{{3 \over 4} - \ic\nu} -
\Psi\pars{{1 \over 4} + \ic\nu}}
\\[5mm] = &\
{1 \over 2\pi}\bracks{%
\pi\cot\pars{\pi\bracks{{1 \over 4} - \ic\nu}} +
\pi\cot\pars{\pi\bracks{{1 \over 4} + \ic\nu}}}
\end{align}
In the last line I used the
Euler Reflection Formula.
See $\ds{\color{black}{\bf 6.3.7}}$ in A & S Table.

Therefore,
\begin{align}
&\bbox[5px,#ffd]{{1 \over \root{2\pi}}
\int_{-\infty}^{\infty}{\expo{-\ic\omega x} \over
\cosh\pars{x\root{\pi/2}}}\,\dd x} =
\Re\cot\pars{\pi\bracks{{1 \over 4} - \ic\nu}}
\\[5mm] = &\
\Re\bracks{%
-\cot\pars{\ic\pi\nu} - 1 \over -\cot\pars{\ic\pi\nu} + 1} =
\Re\bracks{%
\ic\coth\pars{\pi\nu} - 1 \over \ic\coth\pars{\pi\nu} + 1}
\\[5mm] = &\
-\,{1 - \coth^{2}\pars{\pi\nu} \over
\coth^{2}\pars{\pi\nu} + 1} = \on{sech}\pars{2\pi\nu} =
\on{sech}\pars{2\pi\root{2 \over \pi}{\omega \over 4}}
\\[5mm] = &\
\bbx{1 \over \cosh\pars{\omega\root{\pi/2}}} \\ &
\end{align}

Title: Eigenfunction of the Fourier transform


问:
If $n>1$, are there $f_1,\dots,f_k\in\mathbb{Z}[x]$ of degree $n$ such that, for any prime $p$, some $f_i$ is irreducible and of degree $n$, mod $p$?
In the comments section of this question, the original poster raised an interesting variation on the problem, which I will state as follows:

Let $n>1\in\mathbb{N}$. Do there exist $f_1,\dots,f_k\in\mathbb{Z}[x]$ of degree $n$ such that, for any $p$ prime, there exists $i(p)\leqslant k$ with $f_{i(p)}$ irreducible and of degree $n$, mod $p$?

I am nearly certain that, for any $n\in\mathbb{N}$, the answer to this question is no, and I suspect there is probably a solution to this using Dedekind's theorem or some similar tool. However, off the top of my head I am not able to see how such an argument might go in the general case. Can anyone give a proof?

To get the ball rolling, I have written up an elementary proof for the case $n=2$ below. Other proofs for specific values of $n$ are very welcome as well.

Theorem: Let $f_1,\dots,f_k\in\mathbb{Z}[x]$ be polynomials of degree at most $2$. Then there exists a prime $p$ such that each $f_i$ is either reducible or of degree $\leqslant 1$, mod $p$.

Proof: Suppose that $f_i=a_ix^2+b_ix+c_i$ for each $i$. By the quadratic formula, if $a_i$ does not vanish mod $p$, then $f_i$ will have a root mod $p$ ($p>2$) if and only if $b_i^2-4a_ic_i$ is a quadratic residue mod $p$. Thus let $\delta_i=b_i^2-4a_ic_i$ for each $i$. It will suffice to find a prime $p$ such that $\delta_i$ is a quadratic residue mod $p$, for every $i$.

Let $p_1,\dots,p_l$ be all of the odd prime divisors of each $\delta_i$, and consider the system of congruences \begin{align}
p\equiv 1&\mod 4 \\
p\equiv 1&\mod p_1 \\
p\equiv 1&\mod p_2 \\
&\vdots \\
p\equiv 1&\mod p_l.
\end{align} The collection $\{4,p_1,\dots,p_l\}$ is pairwise coprime by construction, so Sun Zi's theorem tells us that the above system of congruences is equivalent to $p\equiv m\text{ }(\operatorname{mod} 4p_1\dots p_l)$ for some $m\in\mathbb{N}$. Now, $m$ is certainly coprime with $4p_1\dots p_l$, since $m$ is congruent to $1$ mod $4$ and $1$ mod $p_i$, for each $i$. Thus, by Dirichlet's theorem, there exists some $\lambda\in\mathbb{N}$ such that $m+\lambda\cdot 4p_1\dots p_l$ is prime; let $p=m+\lambda\cdot 4p_1\dots p_l$.

By construction, $p$ is congruent to $1$ mod $4$ and $1$ mod $p_i$, for each $i$. In particular, $p$ is a quadratic residue mod $2$ and mod each $p_i$, so, by quadratic reciprocity, $2$ and each $p_i$ are all quadratic residues mod $p$. Similarly, since $p$ is equivalent to $1$ mod $4$, we have that $-1$ is also a quadratic residue mod $p$. By construction, every $\delta_i$ is plus-or-minus a product of powers of $\{2,p_1,\dots,p_l\}$, and so – since products of quadratic residues are quadratic residues – every $\delta_i$ is a quadratic residue mod $p$. This means that each $\overline f_i$ is either linear or reducible, mod $p$, as desired.

答:
You're correct. A negative answer follows from the Frobenius density theorem (a slightly weaker and easier-to-prove version of the Chebotarev density theorem), which is a sort of converse to Dedekind's theorem and which says the following:

Theorem (Frobenius): Let $f(x) \in \mathbb{Z}[x]$ be a monic polynomial. The density of primes $p$ such that the factorization of $f(x) \bmod p$ into irreducible polynomials over $\mathbb{F}_p[x]$ has $c_i$ factors of degree $i$ is equal to the density of elements of the Galois group $\text{Gal}(f)$ acting with $c_i$ cycles of length $i$ on the roots of $f$ in a splitting field of $f$.

Corollary: If $f(x) \in \mathbb{Z}[x]$ is a monic polynomial with Galois group $G$, then the density of primes $p$ such that $f(x) \bmod p$ splits completely is $\frac{1}{|G|} \ge \frac{1}{(\deg f)!}$. In particular, there is at least one such prime.

Now apply this result to $f = \prod f_i$.

Edit: Several sources you can find by googling only state the Frobenius density theorem for the case that $f$ is irreducible, and for this application we really need the general case; it is stated without the irreducibility assumption, for example, in Lenstra and Stevenhagen's Chebotarev and his density theorem, which I believe is where I first learned about it. I don't think we really need the hypothesis that $f$ is monic either, since the result is really about the splitting field of $f$ anyway; we can just ignore the finitely many primes dividing the leading term since they have density $0$. Similarly here we're ignoring the finitely many primes dividing the discriminant to avoid dealing with the possibility of repeated roots, since they also have density $0$.

Edit #2: I have been very silly. There's an elementary proof of a weaker version of the corollary, totally avoiding any analytic number theory, which suffices for this purpose. I learned this argument from an MO answer by Bjorn Poonen and forgot about it until just now.

Proposition: If $f(x) \in \mathbb{Z}[x]$, then there are infinitely many primes $p$ such that $f(x) \bmod p$ splits completely.

Proof. Let $K$ be a splitting field for $f$ over $\mathbb{Q}$, let $\alpha \in \mathcal{O}_K$ be a primitive element for $K$ (so that $K = \mathbb{Q}(\alpha)$), and let $g \in \mathbb{Z}[x]$ be the minimal polynomial of $\alpha$. Then, ignoring finitely many primes dividing any denominators that show up, $f(x)$ splits completely $\bmod p$ iff $g(x)$ has a root $\bmod p$ (since by hypothesis every root of $f$ is a polynomial in $\alpha$) iff there exists an integer $k$ such that $p \mid g(k)$.

But the values of every non-constant polynomial in $\mathbb{Z}[x]$ are divisible by infinitely many primes; this can be proven using a Euclid-style argument but see also this blog post for a counting argument. (In one sentence: an eventually monotonically increasing sequence of integers divisible by only finitely many primes must grow faster than polynomially.) $\Box$


答:
Let $f_1, \dots, f_n \in \Bbb Z[x]$ be irreducible polynomials with no condition on the degrees. We claim that there exist infinitely many primes $p$ such that $f_i$ splits completely into linear factors mod $p$. By applying this to the irreducible factors of the polynomials $f_1, \dots, f_k$ in the question, we get that there are no such $f_1, \dots, f_k$.

Let $\alpha_i$ be a root of $f_i$ for all $i$. For each $i$, write $\alpha_i=\frac{\beta_i}{n_i}$, where $n_i \in \Bbb Z$ and $\beta_i$ is integral over $\Bbb Z$. Note that if $g_i$ is the minimal polynomial of $\beta_i$, then we have $g_i(x)=\pm f_i(n_ix)$. Now if $p$ is any prime not dividing any of the $n_i$, then we see that a splitting of $g_i$ into linear factors mod $p$ also gives a splitting of $f_i$ into linear factors mod $p$, so we replace $f_i$ with $g_i$ and assume wlog that all $f_i$ are monic and hence all the $\alpha_i$ are integral.

Consider the number field $K=\Bbb Q(\alpha_1, \dots, \alpha_n)$ and the order $\mathcal{O}:=\Bbb Z[\alpha_1, \dots, \alpha_n] \subset \mathcal{O}_K$. This order has a conductor $\mathfrak{c}$.
It's well-known that in every number field, there are infinitely many primes that split completely, this follows e.g. by applying the Cebotarev density theorem after passing to the normal closure. Of those infintely many primes $p$ that split completely in $K$, only finitely many are not coprime to the conductor $\mathfrak{c}$. For those that are coprime to the conductor, the factorization of $(p)$ in $\mathcal{O}_K$ and in $\mathcal O$ will have the same form. Thus for $p$ coprime to the conductor such that $(p)$ splits completely in $\mathcal{O}_K$, $p\mathcal{O}$ factors as $p\mathcal{O}=\mathfrak{p}_1 \cdot \ldots \cdot \mathfrak{p}_l$ for distinct prime ideals $\mathfrak{p}_1, \dots, \mathfrak{p}_l$. Thus the quotient is $\mathcal{O}/p\mathcal{O} \cong (\Bbb F_p)^l$ but we also have $\mathcal{O}/p\mathcal{O}=\Bbb Z[x_1, \dots, x_n]/(p,f_1(x_1), \dots, f_n(x_n))\cong \Bbb F_p[x_1, \dots, x_n]/(f_1(x_1), \dots, f_n(x_n)$, we see that this is only possible if all $f_i$ split completely in $\Bbb F_p$.

Title: If $n>1$, are there $f_1,\dots,f_k\in\mathbb{Z}[x]$ of degree $n$ such that, for any prime $p$, some $f_i$ is irreducible and of degree $n$, mod $p$?

Body:
In the comments section of this question, the original poster raised an interesting variation on the problem, which I will state as follows:

Let $n>1\in\mathbb{N}$. Do there exist $f_1,\dots,f_k\in\mathbb{Z}[x]$ of degree $n$ such that, for any $p$ prime, there exists $i(p)\leqslant k$ with $f_{i(p)}$ irreducible and of degree $n$, mod $p$?

I am nearly certain that, for any $n\in\mathbb{N}$, the answer to this question is no, and I suspect there is probably a solution to this using Dedekind's theorem or some similar tool. However, off the top of my head I am not able to see how such an argument might go in the general case. Can anyone give a proof?

To get the ball rolling, I have written up an elementary proof for the case $n=2$ below. Other proofs for specific values of $n$ are very welcome as well.

Theorem: Let $f_1,\dots,f_k\in\mathbb{Z}[x]$ be polynomials of degree at most $2$. Then there exists a prime $p$ such that each $f_i$ is either reducible or of degree $\leqslant 1$, mod $p$.

Proof: Suppose that $f_i=a_ix^2+b_ix+c_i$ for each $i$. By the quadratic formula, if $a_i$ does not vanish mod $p$, then $f_i$ will have a root mod $p$ ($p>2$) if and only if $b_i^2-4a_ic_i$ is a quadratic residue mod $p$. Thus let $\delta_i=b_i^2-4a_ic_i$ for each $i$. It will suffice to find a prime $p$ such that $\delta_i$ is a quadratic residue mod $p$, for every $i$.

Let $p_1,\dots,p_l$ be all of the odd prime divisors of each $\delta_i$, and consider the system of congruences \begin{align}
p\equiv 1&\mod 4 \\
p\equiv 1&\mod p_1 \\
p\equiv 1&\mod p_2 \\
&\vdots \\
p\equiv 1&\mod p_l.
\end{align} The collection $\{4,p_1,\dots,p_l\}$ is pairwise coprime by construction, so Sun Zi's theorem tells us that the above system of congruences is equivalent to $p\equiv m\text{ }(\operatorname{mod} 4p_1\dots p_l)$ for some $m\in\mathbb{N}$. Now, $m$ is certainly coprime with $4p_1\dots p_l$, since $m$ is congruent to $1$ mod $4$ and $1$ mod $p_i$, for each $i$. Thus, by Dirichlet's theorem, there exists some $\lambda\in\mathbb{N}$ such that $m+\lambda\cdot 4p_1\dots p_l$ is prime; let $p=m+\lambda\cdot 4p_1\dots p_l$.

By construction, $p$ is congruent to $1$ mod $4$ and $1$ mod $p_i$, for each $i$. In particular, $p$ is a quadratic residue mod $2$ and mod each $p_i$, so, by quadratic reciprocity, $2$ and each $p_i$ are all quadratic residues mod $p$. Similarly, since $p$ is equivalent to $1$ mod $4$, we have that $-1$ is also a quadratic residue mod $p$. By construction, every $\delta_i$ is plus-or-minus a product of powers of $\{2,p_1,\dots,p_l\}$, and so – since products of quadratic residues are quadratic residues – every $\delta_i$ is a quadratic residue mod $p$. This means that each $\overline f_i$ is either linear or reducible, mod $p$, as desired.


问:
Does $\prod_k (x - r_k)^{m_k} = \prod_k (1 - \frac{x}{r_k})^{m_k}$?
I am trying to understand the last line in the excerpt below from my course notes. The discussion concerns an algorithm for turning a (compact, i.e. finite) expression for a generating function into a formula for the coefficients of the generating function.

The "we can express" part of the last line is giving me trouble. I cannot see why the two products are equal. Is there something obvious I'm missing? I see that they are both equal at $x = 0$, and that we can put $r_k$ in the denominator when we know none of the $r_k$ equal $0$, but why are the two expressions equal? I don't get it.

I appreciate any help.

答: The products are not the same, but $g(x)=\lambda\prod_k (x-r_k)^{m_k}$ where $\lambda$ is a constant. $g(0)=1$ means that $\lambda\prod_k(-r_k)^{m_k}=1$ and thus $\lambda=\prod_k\frac{1}{(-r_k)^{m_k}}$. Therefore $$ g(x)=\prod_k\frac{(x-r_k)^{m_k}}{(-r_k)^{m_k}}=\prod_{k}\left(1-\frac{x}{r_k}\right)^{m_k} $$


问:
show that $\int_{0}^{\infty } \frac{\sin (ax)}{x(x^2+b^2)^2}dx=\frac{\pi}{2b^4}(1-\frac{e^{-ab}(ab+2)}{2})$
show that

$$\int_{0}^{\infty } \frac{\sin (ax)}{x(x^2+b^2)^2}dx=\frac{\pi}{2b^4}\left(1-\frac{e^{-ab}(ab+2)}{2}\right)$$

for $a,b> 0$

I would like someone solve it using contour but also I would to see different solution using different way to solve it

is there any help
thanks for all

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty }{\sin\pars{ax} \over x\pars{x^{2} + b^{2}}^{2}}\,\dd x} =
{1 \over 2}\on{sgn}\pars{a}\,
\Im\int_{-\infty}^{\infty }{\expo{\ic\verts{a}x} - 1 \over
x\pars{x^{2} + b^{2}}^{2}}\,\dd x
\\[5mm] = &\
{1 \over 2}\on{sgn}\pars{a}\,\Im\braces{%
2\pi\ic\,{1 \over \pars{2 - 1}!}\,
\lim_{x\ \to\ \ic\verts{b}}\,\,\,
\partiald{}{x}\bracks{\expo{\ic\verts{a}x} - 1 \over
x\pars{x + \ic \verts{b}}^{2}}}
\\[5mm] = &\
\bbx{\on{sgn}\pars{a}\,{\pi \over 2b^{4}}
\bracks{1 - {\expo{-\verts{ab}}\pars{\verts{ab} + 2} \over 2}}} \\ &
\end{align}


问:
Prove the definition of Riemann Integration for Step Functions
$f$ is RI on $[a,b]$ iff $\forall \epsilon > 0 \exists h_1,h_2$ (two step functions) s.t.

$h_1 \leq f \leq h_2$ and s.t.

$$\int_{a}^{b} (h_2(x) - h_1(x)) dx \leq \epsilon$$

I am thinking if I break the integral,

$\int_{a}^{b} h_2(x)dx - \int_{a}^{b}  h_1(x) dx \leq \epsilon$

And then somehow get both of these to correspond to $\frac{\epsilon}{2}$, I'd have the result. how to get there though, I'm not quite sure. I need to somehow make these step functions into regular functions to use the definitions I know. Does anyone have any advice on how to do this one?

答:
For any partition $P:a < x_0< x_1 < \ldots < x_n = b$, define  $M_j = \sup \{f(x) \, | \, x \in [x_{j-1},x_j]\}$ and  $m_j = \inf \{f(x) \, | \, x \in [x_{j-1},x_j]\}$, and the step functions

$$h_1(x) = \begin{cases}\sum_{j=1}^n m_j \mathbf{1}_{[x_{j-1},x_j)}(x), & a \leqslant x < b\\f(b), &x = b\end{cases}\\ h_2(x) = \begin{cases}\sum_{j=1}^n M_j \mathbf{1}_{[x_{j-1},x_j)}(x), & a \leqslant x < b\\f(b), &x = b\end{cases}$$

We have $h_1(x) \leqslant f(x) \leqslant h_2(x)$ for all $x \in [a,b]$ and

$$\int_a^b(h_2(x) - h_1(x)) \, dx = \sum_{j=1}^nM_j(x_j- x_{j-1}) - \sum_{j=1}^nm_j(x_j- x_{j-1}) = U(P,f) - L(P,f)$$

If $f$ is Riemann integrable, then for any $\epsilon > 0$ there exists a partition $P$ such that the difference between the upper and lower Darboux sums is less than $\epsilon$, that is $U(P,f) - L(P,f)< \epsilon$, which proves the forward implication.

For the reverse implication, step functions are integrable (although this is already assumed in the statement).

Since, $h_1 \leqslant f \leqslant h_2$, we have

$$ \inf \{h_1(x) \, | \, x \in [x_{j-1},x_j]\}\leqslant m_j \leqslant M_j \leqslant  \sup \{h_2(x) \, | \, x \in [x_{j-1},x_j]\},$$

and

$$L(P, h_1) \leqslant L(P,f) \leqslant U(P,f) \leqslant U(P,h_2)$$

Thus,

$$\int_a^b h_1(x)\, dx =\sup_PL(P,h_1)\leqslant \sup_PL(P,f) = \underline{\int_a}^bf(x) \, dx \\\leqslant \overline{\int_a^b}f(x) \, dx = \inf_PU(P,f) \leqslant \inf U(P,h_2)=\int_a^bh_2(x) \, dx  $$

Hence, for any $\epsilon > 0$ there exist step functions $h_1$ and $h_2$ such that

$$0 \leqslant \overline{\int_a^b}f(x) \, dx - \underline{\int_a}^bf(x) \, dx\leqslant \int_a^b(h_2(x)-h_1(x)) \, dx \leqslant \epsilon,$$

and the upper an lower Darboux integrals of $f$ must be equal, proving the reverse implication that $f$ is Riemann integrable.


答:
$${\rm M} := \frac27 \, {\rm I}_5 + \frac17 \, {\Bbb 1}_5 {\Bbb 1}_5^\top$$

Since matrix ${\Bbb 1}_5 {\Bbb 1}_5^\top$ is rank-$1$, its nonzero eigenvalue is equal to its trace. Thus, its characteristic polynomial is

$$q (s) := s^4 (s - 5)$$

Hence, the characteristic polynomial of ${\rm M}$ is

$$\det \left( s {\rm I} - {\rm M} \right) = \frac{1}{7^5} q ( 7 s - 2 ) = \cdots = (s-1) \left( s - \frac27 \right)^4$$

Since matrix ${\rm M}$ is stochastic, it has the eigenpair $\left(1, \frac{1}{\sqrt{5}}{\Bbb 1}_5 \right)$. Using the spectral decomposition,

$${\rm M}^k = 1^k \left( \frac{1}{\sqrt{5}}{\Bbb 1}_5 \right) \left( \frac{1}{\sqrt{5}}{\Bbb 1}_5 \right)^\top + \left( \frac27 \right)^k {\rm R}$$

and, thus,

$$\lim {\rm M}^k = 1^{\infty} \left( \frac{1}{\sqrt{5}}{\Bbb 1}_5 \right) \left( \frac{1}{\sqrt{5}}{\Bbb 1}_5 \right)^\top + \left( \frac27 \right)^{\infty} {\rm R} = \color{blue}{\frac15 {\Bbb 1}_5 {\Bbb 1}_5^\top}$$

Title: Infinite power of doubly stochastic matrix without using diagonalization

Body:
Given
$$M := \frac17 \begin{bmatrix}
3 & 1 & 1 & 1 & 1 \\
1 & 3 & 1 & 1 & 1 \\
1 & 1 & 3 & 1 & 1 \\
1 & 1 & 1 & 3 & 1 \\
1 & 1 & 1 & 1 & 3
\end{bmatrix}$$ find $$\lim _{n \to \infty} M^n$$

We can express,

$$
M = \frac{1}{7}\left(\begin{matrix}
-1 & -1 & -1 & -1 & 1 \\
1 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & 1
\end{matrix}\right)\left(\begin{matrix}
2 & 0 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 2 & 0 \\
0 & 0 & 0 & 0 & 7
\end{matrix}\right)\left(\begin{matrix}
\frac{-1}{5} & \frac{4}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{-1}{5} & \frac{4}{5} & \frac{-1}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{4}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{4}{5} \\
\frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5}
\end{matrix}\right)
$$

And evaluate,
$$
\lim _{n \rightarrow \infty} M^{n} = \lim _{n \rightarrow \infty} P\left(\begin{matrix}
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1
\end{matrix}\right)P^{-1}
= \frac{1}{5}\left(\begin{matrix}
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1
\end{matrix}\right)
$$
where I used that,
$$
P = \left(\begin{matrix}
-1 & -1 & -1 & -1 & 1 \\
1 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & 1
\end{matrix}\right)
\,\,\,\,\,\&
\,\,\,\,\,
P^{-1} = \left(\begin{matrix}
\frac{-1}{5} & \frac{4}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{-1}{5} & \frac{4}{5} & \frac{-1}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{4}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{-1}{5} & \frac{4}{5} \\
\frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5}
\end{matrix}\right)
$$

This traditional way of diagonalization involves not just finding eigenvalues but also eigenvectors (in other words find P).

Finding eigenvalues for this is easy.
I consider,
$$
A=\left[\begin{array}{lllll}
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1
\end{array}\right]
$$ which clearly has eigenvalues $\lambda_1 = 0$ (with algebraic multiplicity = $4$) and $\lambda_2 = 5$ (with algebraic multiplicity = $1$).

The eigenvalues for our first Matrix $M = \frac{1}{7}\left(A + 2I\right)$ are therefore, $\lambda_1 = \frac{2}{7}$ (with algebraic multiplicity = $4$) and $\lambda_2 = 1$ (with algebraic multiplicity = $1$).

Is there any way we can efficiently compute, $\lim _{n \rightarrow \infty} M^{n}$, without calculating eigenvectors i.e., without computing matrices $P$ and $P^{-1}$?


问:
SVD based method for predicting missing values in future experiments
I would like to pose my question in terms of a hypothetical example, where I have a matrix:

```
11 12 13...1m
21 22 23 ..2m
.  .  .    3m
.  .  .  .  .
11 n2 n3 .  nm```

The matrix has no missing values. Each column corresponds to one experiment and each row corresponds to a particular measurement parameter ( eg: temperature @ point 1, temperature @ point 2... Pressure....)

next, I conduct a New experiment, and I measure most parameters but not all n parameters. I want to estimate the missing parameter(s) in my New experiment.

Is there a formula based on SVD/PCA to calculate the missing value in my New experiment based on my 'training' matrix?

Are there any better machine learning / statistical methods to perform this calculation ?

Thanks in advance!

答:
See this paper:
Hastie, Mazumder, Lee and Zadeh, Matrix completion and low-rank SVD via fast alternating least squares.</p>


问:
Find eigenvalues by inspection

I was given this matrix:

\begin{pmatrix}
4 & 0 &0  &0 \\
-1 &  -2& 7 &0 \\
1 &2  &3  & 0\\
 0&-1 & 2 & 6
\end{pmatrix}
The task was to find the eigenvalues of this matrix without using the characteristic polynomial. What I've managed to do so far is this:

One of the eigenvalues is 6, because, given a vector $$(0,0,0,t)^T; t\neq0$$
it only gets multiplied by 6. This can be seen from the matrix, because the transformation of any vector $$(x,y,z,a)^T$$ will result in a vector $$(4x,-x-2y+7z,x+2y+3z,-y+2z+6a)^T$$ We can see that the $a$ appears only in the last part of the vector, which helps us determine the eigenvalue.

I also know that if I manage to find a second eigenvalue, I will be able to determine the other two easily, knowing that the sum of the eigenvalues is equal to the sum of the elements on the diagonal of the original matrix and that the product of the eigenvalues is equal to the determinant of the matrix, which can be easily computed. Knowing this, I will be able to get the other two eigenvalues. I, however, have trouble finding a second eigenvalue in this matrix by pure intuition. I will appreciate any help.

答:
The "inspection with bare eyes method" may work here as follows. We restate the question as follows:

Which number should we subtract from (each entry in) the diagonal of the given matrix, so that after this we are left with a matrix with zero determinant?

Obviously, the subtraction of $4$, respectively $6$ does the job. (We get $0$ entries in a row, respectively column.) Thus we have "used the corners", so let us try to use the "middle block" in
$$
A=
\begin{bmatrix}
4  & 0  & 0  &0 \\
 * & \color{blue}{-2} & \color{blue}{7}  &0 \\
 * & \color{blue}{ 2} & \color{blue}{3 } &0\\
 0 &  * & * & 6
\end{bmatrix}
\ ,
$$
and only the blue entries are now interesting.
What can we subtract from the diagonal of this blue block?

One observation is that adding $4$ on the (blue) diagonal, we get a (blue) matrix
with same blue columns, so $-4$ is also an eigenvalue.

The other value is $5$. (This can be guessed or we quickly compute the trace or the determinant of the blue block, and "guess" it from the sum or the product with the known $-4$ eigenvalue.)

Title: Find eigenvalues by inspection


问:
Finding covariance of $f_{(x,y)} (x,y) = \frac{1}{4}(y-x)e^{-y}$ for $-y<x<y$ and $y>0$
Our joint density function is

$f_{(x,y)} (x,y) = \frac{1}{4}(y-x)e^{-y}$  for $-y<x<y$ and $y>0$ and 0 elsewhere.

I know $Cov(X,Y) = E(XY) -E(X)E(Y).$

But I get that E(X) is not finite:

$f_x(x) = \int_{0}^{\infty} .25*(y-x)e^{-y}dy = \frac{1}{4}(1-x)$

$E(X)=\int_{0}^{\infty}x*\frac{1}{4}(1-x)dx$ which I got to diverge.

Where did I mess up or does the covariance not exist?

答:
Your formula for marginal distribution should be $f_X(x) = \frac{1}{4}e^{-|x|}(|x|-x+1)$), but we don't need it. We can just use a double integral to compute all the moments from the the joint distribution:
$$
\begin{aligned}
E[X^m Y^n]
&= \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{e^{-y}}{4}x^m y^n (y-x) \left[|x| < y\right] \ dx \ dy \\
&= \int_{0}^\infty \int_{-y}^y \frac{e^{-y}}{4}x^m y^n (y-x)\ dx \ dy \\
&= \int_{0}^\infty \frac{y^n e^{-y}}{4} \int_{-y}^y (yx^m - x^{m+1}) dx \ dy \\
&= \int_{0}^\infty \frac{y^n e^{-y}}{4} \left. \left(\frac{yx^{m+1}}{m+1}-\frac{x^{m+2}}{m+2}\right) \right|_{-y}^y \ dy \\
&=  \left(\frac{1-(-1)^{m+1}}{m+1}-\frac{1-(-1)^{m+2}}{m+2}\right)  \int_{0}^\infty \frac{y^{n+m+2} e^{-y}}{4}  \ dy \\
&= \begin{cases}
\frac{(m+n+2)!}{2(m+1)}& m \text{ even}\\
-\frac{(m+n+2)!}{2(m+2)} & m \text{ odd}\end{cases}
\end{aligned}
$$
Where we use $\int_{0}^\infty y^{n} e^{-y} \ dy =\Gamma(n+1) = n!$.
Hence
$$
\begin{aligned}
E[X] &= -\frac{3!}{2(3)} = -1 \\
E[Y] &= \frac{3!}{2(1)} = 3\\
E[XY] &= -\frac{4!}{2(3)} -4 \\
\mathrm{Cov}(X,Y) &= -4 - (-1)(3)=-1
\end{aligned}
$$


问:
Is there a way to show R2(s + 1, t + 1) > s · t .
Is there a way to show R2(s + 1, t + 1) > s · t ?

The only way I know finding ramsey numbers to by graphing. But to prove the above inequality, what would be the best method?

答:
Given a set of size $s\cdot t$, partition it into $s$ blocks of $t$ elements each. Then color the 2-element subsets by making a pair red if it's included in one block and blue otherwise.

For this coloring, any red-homogeneous set is included in a single block and therefore has at most $t$ elements. Any blue-homogeneous set has at most one element per block and therefore has at most $s$ elements.


问:
Reference Requestion: what's the proof of elementary equivalence of hierarchies raised over sets of quine atoms using Ehrenfeucht-Fraisse games?
In a prior posting Noah Schweber had presented this claim:

$(*)\quad$ Suppose $A,B$ are infinite sets of Quine atoms. Then for each ordinal $\alpha$ we have $$L_{\alpha,A}\equiv L_{\alpha,B}.$$ $(*)$ can be proved via Ehrenfeucht-Fraisse games.

Where $L_{\alpha,A}$ means $L$ hierarchy on set $A$, built as:

$L_{0,A}=A$,

$L_{\lambda,A}=\bigcup_{\alpha<\lambda} L_{\alpha,A}$ for $\lambda$ limit, and

$L_{\alpha+1,A}=L_{\alpha,A} \cup\mathcal{P}_{def}(L_{\alpha,A})$.

The argument posed makes no mention of difference in cardinality between $A$ and $B$, so it's applicable when $A$ is countable and $B$ is not.

My question: what is the proof of that result? Or if it's long then can one point me to a reference of that result? preferably one that is reachable online openly.

答:
In retrospect, EF-games are overkill here; we can just use Lowenheim-Skolem + a bit of constructibility theory.

First, note that any bijection $b:A_0\rightarrow A_1$ extends to a (unique) isomorphism $\hat{b}: L_{A_0,\alpha}\cong L_{A_1,\alpha}$.

Next, let $B$ be an uncountable "starting set" and $\alpha$ any countable ordinal. By downward Lowenheim-Skolem, there is a countable elementary $\mathcal{X}\subseteq L_{B,\alpha}$ with $\alpha\subseteq \mathcal{X}$. Let $A=B\cap\mathcal{X}$; clearly $A$ is countable, and by the usual absoluteness-of-$L$-hierarchy argument we get $\mathcal{X}=L_{A,\alpha}$.

(Specifically, we argue by induction on $\beta<\alpha$ that the thing $\mathcal{X}$ thinks is $L_{A,\beta}$ actually is $L_{A,\beta}$. Since $\mathcal{X}$ thinks every set is in some $L_{A,\beta}$, this gives the desired result. Note that we're crucially using here the fact that $\mathcal{X}\cap Ord$ is downwards-closed.)

We can put these together as follows. Let $A,B$ be arbitrary infinite sets of Quine atoms and $\alpha$ a countable ordinal. By the previous paragraph, we get countably infinite $A_0\subseteq A,B_0\subseteq B$ such that $$L_{A_0,\alpha}\preccurlyeq L_{A,\alpha}\quad\mbox{and}\quad L_{B_0,\alpha}\preccurlyeq L_{B,\alpha}.$$ But since there is a bijection between $A_0$ and $B_0$ we get an isomorphism between the corresponding $L$-structures, and putting this all together we have $$L_{A,\alpha}\equiv L_{A_0,\alpha}\cong L_{B_0,\alpha}\equiv L_{B,\alpha}.$$


问:
Group theory: Cayley's theorem

For $n$ large, we see that $G$ is a tiny subgroup of $S_n$, which has $n!$ elements, as compared
to $n$ elements. In contrast, for $n$ small, the situation is quite different, as shown by the
following examples. The group $\Bbb Z_2$ is in fact the same as $S_2$. (Check this; it’s trivial.). But
the group $\Bbb Z_3$, with three elements, clearly cannot be the same as $S_3$, with $3!= 6$ elements,
but it is the same as $A_3$. (Why?)

I was reading the Zee's book about group theory, as a beginner I am stuck in this citation above. $G$ is any finite group with $n$ elements, and we are talking about Cayley's theorem.

答:
When the author says "the group $\Bbb Z_2$ is the same as $S_2$," he means that $\Bbb Z_2$ is isomorphic to $S_2$.

In particular, denote $S_2 = \{e,\tau\}$, where $e$ denotes the identity map and $\tau:\{1,2\} \to \{1,2\}$ denotes the transposition defined by $\tau(1) = 2, \tau(2) = 1$. The map $\phi:\Bbb Z_2 \to S_2$ defined by $\phi(e) = 1,$ $\phi(\tau) = -1$ is an isomorphism.

In the same vein: when the author says that $\Bbb Z_3$ is the same as $A_3$, he means that these groups are isomorphic.

Title: Group theory: Cayley's theorem


问:
Wave equation and falling bar
I want to understand the following: consider an (elastic) bar of one dimension (length only) with length $L=1$m floating at some height $h=2m$:

Vertically

Horizontally

I know that the equation describing the deformation $u(x,t)$ of its points is
$$
u_{tt} = u_{xx} + \gamma u_{txx}.
$$
For parameter $\gamma =0$ this is the 1-dimensional wave equation. The initial conditions for each point $x \in L = [0,1]$ must be

zero initial deformation $u(x,0)=x$

zero initial velocity $u_t(x,0)= 0$

Aren't these i.c.s sensible?

Then I let the bar fall either vertically or horizontally. I am curious, shouldn't the impact and thus the resulting solution for $u$ depend on gravity constant $g$ somehow?

Since the bar is 1d if it fall horizontally I guess there is no sulution to $u$ since the impact will be only a function of $y$ (height). But what would the deformation be if the bar fall vertically or at some angle?

Note: I normally would expect some solution proportioanl to $e^{-(x-t)+e^{-(x+t)}}$ but with the initial conditions as above I get a linear solution on Mathematica.

答:
Yes. This problem isn't really 1D. The bar is described by the position vector $$
{\bf y}({\bf x}, t) = {\bf x} + {\bf u}({\bf x}, t)
$$ where $\bf u$ denotes the displacement field. Of course the displacement of the bar's particles is oriented along its length (a unit vector $\bf n$ orientating the bar might be introduced here). The material is viscoelastic of Kelvin-Voigt type, and the equation of motion reads $$
\rho {\bf y}_{tt} = \operatorname{div}{\boldsymbol\sigma} + \rho {\bf g}
$$
where ${\bf g}$ is the gravity force, and the stress tensor reads $$
{\boldsymbol\sigma} = (E \partial_{\bf n} {\bf y} + \eta \partial_{\bf n} {\bf y}_{t})\cdot {\bf n} \, ({\bf n}\otimes {\bf n}) \, .
$$
The constants $\rho$, $E$, $\eta$ are respectively the mass density, the elastic modulus, and the viscosity. In the vertical case where ${\bf y} = y {\bf e}_3$, ${\bf n} = {\bf e}_3$ and ${\bf g} = -g {\bf e}_3$, we have $$
\rho y_{tt} = E y_{zz} + \eta y_{zzt} - \rho g
$$
along the vertical direction ${\bf e}_3$ with coordinate $z$.

Note: the corresponding Lagrangian density is
$$
L = \tfrac12\rho (y_t)^2 - \tfrac12 E (y_z)^2 +\rho g z
$$ if $\eta=0$. The dissipative case is less straightforward as non-conservative forces arise.

Title: Wave equation and falling bar


答:
Group theory is all about the underlying structure of the elements, not the way we write them. As you mentioned, we can view $\mathbb{Z}_2$ as the set $\{1,-1\}$. In order to be a group and not just a set, we also must decide on the operation. Here it would be multiplication, and we get the following multiplication table:

\begin{array}{r|rr} & 1 & -1\\ \hline 1 & 1 & -1\\ -1 & -1 & 1 \end{array}

The group $S_2$ is usually written as the set $\{(12),\ ()\}$, where I just wrote $()$ to represent the identity, or do-nothing permutation, along with the operation of composition:

\begin{array}{r|rr} & () & (12)\\ \hline () & () & (12)\\ (12) & (12) & () \end{array}

As you can see, these sets and their operations look eerily similar, and in fact we say that they are the same group (technically, they are isomorphic groups). This happens if two groups have the same multiplication table, and the only difference is how the elements are written.

I'll let you explore the groups $\mathbb{Z}_3$ and $S_3$. The book you're reading wants you to notice that there is actually a subgroup of $S_3$ that is isomorphic to $\mathbb{Z}_3$ called $A_3$. Can you write out the multiplications for $\mathbb{Z}_3$ and $A_3$ to show that?

Title: Group theory: Cayley's theorem

Body:
For $n$ large, we see that $G$ is a tiny subgroup of $S_n$, which has $n!$ elements, as compared
to $n$ elements. In contrast, for $n$ small, the situation is quite different, as shown by the
following examples. The group $\Bbb Z_2$ is in fact the same as $S_2$. (Check this; it’s trivial.). But
the group $\Bbb Z_3$, with three elements, clearly cannot be the same as $S_3$, with $3!= 6$ elements,
but it is the same as $A_3$. (Why?)

I was reading the Zee's book about group theory, as a beginner I am stuck in this citation above. $G$ is any finite group with $n$ elements, and we are talking about Cayley's theorem.


问:
Optimal probability measure
Let $A$ be a finite set and let $\Bbb P$ be a probability measure on $A^{\Bbb N_0}$. Further, let $x_i:A^{\Bbb N_0}\to A$ be projection maps, so that $(x_i)_{i=0}^\infty$ can be treated as a stochastic process. Suppose that for some integer number $m\geq 1$ the measure $\Bbb P$ satisfies the following condition:
$$
  (x_i,\dots,x_{i+m})_*\Bbb P = (x_0,\dots,x_m)_*\Bbb P\quad \forall i\in \Bbb N,\tag{1}
$$
that is any sequence of the length $m+1$ has equal probability of appearance anywhere along the trajectory of the process. For example, $\Bbb P$ is shift-invariant (stationary) if and only if it satisfies $(1)$ for any choice of $m$. In my case, the number $m$ is fixed though, so $\Bbb P$ does not have to be a stationary measure.

Let us denote $\Bbb P^m := (x_0,\dots,x_m)_*\Bbb P$ to be a measure on $A^{m+1}$, and let $F:A^{m+1}\to\Bbb R$ be some arbitrary function. Suppose, I need to maximize
$$
  \Bbb P^m[F]:=\sum_{s\in A^{m+1}}F(s)\cdot\Bbb P^m(\{s\}) \tag{2}.
$$
where the optimization is done over all possible measures $\Bbb P$ that satisfy $(1)$. Have such problem been already studied, perhaps the solution is already known?

答:
For a finite sequence $Y= (y_0,\dots,y_m)$ let $\left <Y\right>\subset A^{m+1}$ be the set of sequences
$$\left<Y\right> = \{ (y_i, y_{i+1 \mod (m+1)}, \dots y_{i+1 \mod (m+1)}) :  0 \leq i \leq m \}$$
be the set of shifts of $Y$.

Now let $$\mathscr F = \left\{\left<Y\right> : Y\in A^{m+1}\right\}$$
be the shift invariant $\sigma$-algebra on $A^{m+1}$.  As $\mathscr F$ is a partition $\sigma$-algebra we have
$$\mathbb P^m\left[F \left| \mathscr F\right.\right](Y) = \frac 1{m+1} \sum_{\hat Y\in\left<Y\right>} F(\hat Y).$$

Therefore if $\bar Y$ is such that
$\sum_{\hat Y\in\left<\bar Y\right>} F(\hat Y)\geq\sum_{\hat Y\in\left< Y\right>} F(\hat Y)$ for every $Y\in A^{m+1}$ we have

$$\begin{array}{rl}
\mathbb P^m\left[F \right] &= \int_{A^{m+1}}  \mathbb P^m\left[F \left| \mathscr F\right.\right](X) d \mathbb P^m(X) \\
&\leq \int_{A^{m+1}}  \mathbb P^m\left[F \left| \mathscr F\right.\right](\bar Y) d \mathbb P^m(X) \\
&=\frac 1{m+1}\sum_{\hat Y\in\left<\bar Y\right>} F(\hat Y)
\end{array}$$

So  set $\mathbb P$ to be the measure constructed by choosing $(x_0,\dots,x_m)$ uniformly from $\left<\bar Y\right>$ and setting $x_k = x_{k \mod (m+1)}$ for $k> m$ then $\mathbb P$ is shift invariant by construction and
$$\mathbb P^m\left[F \right] =\frac 1{m+1}\sum_{\hat Y\in\left<\bar Y\right>} F(\hat Y)$$
attains its maximal value.

Title: Optimal probability measure


问:
Show the inequality : $a^{(2(1-a))}+b^{(2(1-b))}+c^{(2(1-c))}+c\leq 1$

Claim :

Let $0.5\geq a \geq b \geq 0.25\geq c\geq 0$ such that $a+b+c=1$  then we have :

$$a^{(2(1-a))}+b^{(2(1-b))}+c^{(2(1-c))}+c\leq 1$$

To prove it I have tried Bernoulli's inequality .

For $0\leq x\leq 0.25$ we have :

$$x^{2(1-x)}\leq 2x^2$$

As in my previous posts we have the inequality $x\in[0,0.5]$ :

$$x^{2(1-x)}\leq 2^{2x+1}x^2(1-x)$$ applying this for each variables $a,b$ we want to show :

$$2^{2a+1}a^2(1-a)+2^{2b+1}b^2(1-b)+2c^2+c\leq 1$$

Now by Bernoulli's inequality we have:

$$2^{2x+1}\leq 2(1+2x)$$

Remains to show :

$$2(1+2a)a^2(1-a)+2(1+2b)b^2(1-b)+2c^2+c\leq 1\quad(1)$$

The function :

$$f(x)=2(1+2x)x^2(1-x)$$ is concave for $x\in [\frac{1}{8}+\frac{\sqrt{\frac{19}{3}}}{8},0.5]$

So we can use Jensen's inequality remains to show :

$$2\left(2(1+a+b)\left(\frac{a+b}{2}\right)^2\left(1-\left(\frac{a+b}{2}\right)\right)\right)+2c^2+c\leq 1$$

So it reduces to a one variable inequality and using derivatives it's not hard to show that :

$$g(c)=2f\left(\frac{1-c}{2}\right)+2c^2+c\leq 1$$

For $c\in[0,1-2\left(\frac{1}{8}+\frac{\sqrt{\frac{19}{3}}}{8}\right)]$

It shows the equality case $a=b=0.5$ and $c=0$ but inequality $(1)$ is false for the other equality case $a=0.5$ and $b=c=0.25$.

We have also the inequality for $x\in[0.25,0.5]$ (we can prove it using logarithm and then derivative)

$$x^{(2(1-x))}\leq x^22^{-5(x-0.25)(x-0.5)+1}$$

Using Bernoulli's inequality :

$$x^22^{-5(x-0.25)(x-0.5)+1}\leq 2(x^2+x^2(-5(x-0.25)(x-0.5)))$$

So Remains to show :

$$2(a^2+a^2(-5(a-0.25)(a-0.5)))+2(b^2+b^2(-5(b-0.25)(b-0.5)))+2c^2+c\leq 1\quad (2)$$

Question :

Have you a proof ? How to show $(2)$ ?

Thanks in advance !

答:
Some thoughts

We first give some auxiliary results (Facts 1 through 4). The proofs are not difficult and thus omitted.

Fact 1: If $\frac{1}{2} \ge x \ge \frac{1}{4}$, then
$x^{2(1-x)} \le \frac{528x^2+572x-93}{650}$.

Fact 2: If $0\le x \le \frac{1}{4}$, then $x^{2(1-x)} \le 3x^2 - 2x^3$.
(Hint: Use Bernoulli inequality.)

Fact 3: If $\frac{1}{2} \ge x \ge \frac{1}{4}$, then
$x^{2(1-x)} \le \frac{752x^2-24x-1}{320}$.

Fact 4: If $0\le x \le \frac{1}{4}$, then $x^{2(1-x)} \le 3x^2 - 4x^3$.
(Hint: Use Bernoulli inequality.)

Now, we split into two cases:

$c \le \frac{1}{5}$:

By Facts 1-2, it suffices to prove that
$$\frac{528a^2+572a-93}{650} + \frac{528b^2+572b-93}{650} + 3c^2 - 2c^3 + c \le 1$$
or
$$650c^3-264a^2-264b^2-975c^2-286a-286b-325c+418 \ge 0.$$
It is verified by Mathematica.

$\frac{1}{5} < c \le \frac{1}{4}$:

By Facts 1, 3, 4, it suffices to prove that
$$\frac{528a^2+572a-93}{650} + \frac{752b^2-24b-1}{320} + 3c^2 - 4c^3 + c \le 1$$
or
$$83200c^3-16896a^2-48880b^2-62400c^2-18304a+1560b-20800c+23841 \ge 0.$$
It is verified by Mathematica.


问:
Mean and Median Calculations
A group of 20 has a mean = 85 and median = 80. A different group of 30 has a mean = 75 and median = 72. Now, what is the mean of and median for the entire group of 50?  </p>

答:
This is an elaboration on the answer from response, in case you didn't see weighted averaging before.

So you have $(x_i)_{i=1}^{20}, (y_i)_{i=1}^{30}$ with $\frac{1}{20} \sum_{i=1}^{20} x_i = 85$ and $\frac{1}{30} \sum_{i=1}^{30} y_i = 75$.

Thus, $\sum x_i = 20 \cdot 85$ and $\sum y_i = 30 \cdot 75$ hence total average is
$$
\frac{\sum x_i + \sum y_i}{50} = \frac{20 \cdot 85 + 30 \cdot 75}{50} = 79.
$$

EDIT Not sure what to do with the weighted median. To tell exactly, you would need the entire numer range, since it's the middle value. The only thing you can guarantee is that at least half the numbers will be $\geq 72$ and at least half the numbers will be $\leq 85$, so the median $m$ is guaranteed to satisfy $72 \leq m \leq 85$.

Title: Mean and Median Calculations


问:
Find $u$ for the Householder reflection matrix $I-2\frac{uu^T}{u^Tu}$

Given that the reflection of a vector $v=(3,0,0,4)$ across a plane $P=\{x:x^Tu=0\}$ is parallel to $w=(2,0,1,3)$. Find $u$

The reflction across a plane perpendicular to $u$ is given by the House holder matrix
$$
H_u=I-2\frac{uu^T}{u^Tu}
$$
Here I think it says
$$
H_uv=\lambda w
$$
Since $||H_uv||=||v||=\lambda||w||\implies \lambda =\frac{||v||}{||w||}$
$$
H_uv=v-2\frac{uu^T}{u^Tu}v=\frac{||v||}{||w||}w\\
\begin{bmatrix}3\\0\\0\\4\end{bmatrix}-2\frac{uu^T}{u^Tu}\begin{bmatrix}3\\0\\0\\4\end{bmatrix}=\frac{5}{\sqrt{14}}\begin{bmatrix}2\\0\\1\\3\end{bmatrix}
$$
How do I find $u$ ?

Can I take $u=v-\frac{||v||}{||w||}w$ ?

答:
Yes. Your solution is correct. You can make the substitution to convince yourself of this.

Alternatively, you can exploit geometrical intuition. First, normalize $v$ and $w$:

$$
v' = \frac{v}{\|v\|} \\
w' = \frac{w}{\|w\|}.
$$

Now you can decompose $v'$ and $w'$ into the components parallel and orthogonal to the reflection plane. Note that the last component is parallel to $u$.

By geometry, the parallel components of $v'$ and $w'$ will be the same and the orthogonal components will have equal length and direction and opposite orientation. Therefore, we can recover the component orthogonal to the reflection plane, i.e. parallel to $u$, by subtracting $v'$ and $w'$ (since then the components parallel to the reflection plane cancel out).

We conclude that

$$
u = v' - w' = \frac{v}{\|v\|} - \frac{w}{\|w\|}
$$

which is a scalar multiple of your result:

$$
\|v\| \left(\frac{v}{\|v\|} - \frac{w}{\|w\|}\right) = v-\frac{\|v\|}{\|w\|}w.
$$


问:
$T(n) = nT(n/2) ,T(1)=1$ How can I solve this with backward subsitituon?
$n$ can be power of 2 or we can assume $Integerpart(n/2)$ ==> int casting.

I find the recurrence relation of function but How can I solve this with using backward subsitituon ?

My recurrence relation solution

答:
Notice that the expression $T(n)$ starts with $n$ so in general you have

$$T(n)=n\frac{n}{2}\frac{n}{4}...\frac{n}{2^k}...1$$

Obviously the last $1$ is in the same format so it must be

$$\frac{n}{2^M}=1$$

or

$$n=2^M$$

From there it is obvious that we have $M+1$ terms in form of

$$\frac{n}{2^k}=1$$

where $M$ is $\log_2(n)$

We can write then

$$T(n)=\frac{n^{M+1}}{ 2^{ \frac{M(M+1)} {2} } }$$

or precisely after the substitution of $M$

$$T(n)=n^{\log_4(2n)}$$


问:
Extension of trace preserving $*$-homomorphisms on dense subalgebras of von Neumann algebras.
Let $(\mathcal{M}, \tau)$ be a tracial von Neumann algebra, that is, $\tau$ is a faithful normal trace on $\mathcal{M}$. Let $S$ be a WOT dense unital $*$-subalgebra of $\mathcal{M}$.

Given another tracial von Neumann algebra $(\mathcal{N}, \nu)$ and an algebraic $*$-homomorphism $\varphi: S \to \mathcal{N}$ that preserves the trace ($\nu \circ \varphi =\tau$), can we always extend $\varphi$ to a a trace preserving $*$-homomrphism from $\mathcal{M}$ to $\mathcal{N}$?

This situation has arised many times when dealing with separable von Neumann algebras, where $\varphi$ is defined on some countable dense subalgebra, and one would like to extend to all of $\mathcal{M}$ (for example, when $\mathcal{M} = L(\Gamma)$ for some countable group $\Gamma$, and $S$ is the algebra generated by the canonical unitaries $\{u_\gamma \vert \gamma \in \Gamma\}$). Yet for some reason, in the literature I encountered, this is often ignored or considered as trivial.

My approach for explaining this was using the 2-norms induced by the traces $\Vert x \Vert_{2,\tau} = \tau (x^*x), \; \Vert \cdot \Vert_{2, \nu}$, which turn out to be complete on the unit balls of $\mathcal{M, N}$, and capture the WOT density of $S$. However, I hit a severe roadblock since I do not assume $\varphi$ respects the operator norm in any way, and I do not see how in general trace preservation should say something about this.

Any reference or help will be greatly appreciated.

答:
If you don't require $\nu$ to be faithful, then answer is no. For instance take $M=L^\infty[0,1]$, with $\tau$ integration against Lebesgue measure. Take $N=L^\infty[0,1]\oplus L^\infty[1,2]$, with $\nu$ given by $\nu(f\oplus g)=\tau(f)$. Let $S$ be the polynomials, and $\varphi(p)=p\oplus p$. Then $\nu\circ\varphi=\tau$, but $\varphi$ is not contractive, so it cannot extend to a $*$-homomorphism of C$^*$-algebras.

If we require that $\nu$ is faithful and that $\varphi$ has dense image, then it can be extended. To see this, consider $M$ and $N$ represented in GNS for $\tau$ and $\nu$ respectively. Let $\Omega_\tau, \Omega_\nu$ denote the corresponding cyclic vectors, then:
\begin{align}
\frac{\|\varphi(x)\varphi(z)\Omega_\nu\|^2}{\|\varphi(z)\Omega_\nu\|^2}
&=\frac{\langle \varphi(xz)\Omega_\nu,\varphi(xz)\Omega_\nu\rangle}{\langle \varphi(z^*z)\Omega_\nu,\Omega_\nu\rangle}\\[0.3cm]
&=\frac{\nu(\varphi(z^*x^*xz)}{\nu(\varphi(z^*z))}=\frac{\tau(z^*x^*xz)}{\tau(z^*z)}\\[0.3cm]
&=\frac{\|xz\Omega_\tau\|^2}{\|z\Omega_\tau\|^2}.
\end{align}
Taking the supremum all nonzero $z$, we get
$$
\|\varphi(x)\|=\|x\|.
$$
Note that $\varphi(z)=0$ implies $\tau(z^*z)=\nu(\varphi(z^*z))=0$, and $z=0$, so $\varphi$ is faithful.


问:
Complex coefficients giving real roots?
Edit: I figured it out, I just miscalculated. No, there are no complex solutions to k.

For what values of $k$ does the equation $kx^2+3kx+k=-9-kx$ have real roots? Just to simplify it, I'll write the equation as $kx^2+4kx+k+9=0$. Now the obvious solution would probably be to plug it into the quadratic formula and try to solve such that the discriminant is greater than zero. In the quadratic formula, $a = k$, $b = 4k$, and $c = k + 9$. Plugging these values in, I get

$$\begin{align}
&\frac{-4k±\sqrt{(4k)^2-4(k)(k+9)}}{2k}\\
&=\frac{-4k±\sqrt{16k^2-4k^2-36k}}{2k}\\
&=\frac{-4k±\sqrt{12k^2-36k}}{2k}\\
&=-2±\frac{\sqrt{12k^2-36k}}{2k}\\
&=-2±\frac{\sqrt{3k^2-9k}}{k}
\end{align}$$

Since the only thing that matters is finding real values, $-2$ can just be ignored as well as the $\pm$ sign, since nothing really changes anyways. Now the logical thing to do here would be to take the discriminant, set it to be greater than or equal to zero, and get the inequality for all the real values of $k$ (in case you're wondering, $k ≤ 0$ and $k ≥ 3$). But the more interesting part of the question would be finding the complex solutions to $k$. While attempting this, I use assumptions:

First, the square root of a complex number with an imaginary part is always another complex number with an imaginary part. I don't see how this isn't true and I haven't found anything that says otherwise yet, so it seems good to me.

Second, if an equation with real and imaginary parts sums to $0$, then the real and imaginary parts alone will also sum to zero. I don't see how this is false either, since imaginary parts can't contribute to real parts, so both must just equal zero.

So in order to solve this problem, I set $k$ to equal $a+bi$ such that $a$ and $b$ are real numbers. Therefore,

$$\begin{align}
&-2±\frac{\sqrt{3k^2-9k}}{k}\\
&=-2±\frac{\sqrt{3(a+bi)^2-9(a+bi)}}{a+bi}\\
&=-2±\frac{\sqrt{3(a^2+2abi-b^2)-9a-9bi)}}{a+bi}\\
&=-2±\frac{\sqrt{3a^2+6abi-3b^2-9a-9bi}}{a+bi}\\
\end{align}$$

Now if the numerator is equal to the denominator, then the fraction would equal 1 so I set it to that.

$$\begin{align}
\sqrt{3a^2+6abi-3b^2-9a-9bi} &= a+bi\\
3a^2+6abi-3b^2-9a-9bi &= (a+bi)^2\\
3a^2+6abi-3b^2-9a-9bi &= a^2+2abi-b^2\\
2a^2+4abi-2b^2-9a-9bi &= 0
\end{align}$$

Taking imaginary parts only:

$$\begin{align}
4abi-9bi &= 0\\
4abi &= 9bi\\
4ab &= 9b
\end{align}$$

Assuming that $b ≠ 0$:

$$\begin{align}
4a &= 9\\
a &= \frac{9}{4}
\end{align}$$

Now taking real parts only:

$$\begin{align}2a^2-2b^2-9a &= 0\\
2\left(\frac{9}{4}\right)^2-2b^2-9\left(\frac{9}{4}\right) &= 0\\
2\left(\frac{9}{4}\right)^2-9\left(\frac{9}{4}\right) &= 2b^2\\
\frac{486}{16} &= 2b^2\\
\frac{243}{16} &= b^2\\
±\frac{9\sqrt{3}}{4} &= b
\end{align}$$

Therefore $k = \frac{9}{4} + i\frac{9\sqrt{3}}{4}$ should be a solution. Whether or not cases like $k = \frac{9}{2} + i\frac{9\sqrt{3}}{2}$ that are multiples of the original answer would work don't really matter, but this seems like it should work. Another case that should also work would be setting the numerator of the original fraction to equal to zero. This also produces a complex solution for $k$.

The question is, can complex coefficients create real roots for quadratic equations, or did I calculate/assume/do the question wrong? If it is possible, did I miss any solutions? Any help would be appreciated.

答:
You mention that we should be able to set the numerator of $\frac{\sqrt{3k^2 - 9k}}{k}$ to $0$. This would give $3k^2 = 9k$ and so $k = 3$, (discarding $k = 0$). This isn't complex so it isn't providing a solution to your question.

Other than that, your proposed solutions for $k$ do not seem to provide real solutions to the quadratic given.

Note that you can rewrite your quadratic as $k ( x^2 + 4x +1) = -9$. So we have
$$k = \frac{-9}{ x^2 + 4x +1}$$
So if we plug in real numbers for $x$ (which would be the proposed solutions to your quadratic), then $k$ will also be real.

Title: Complex coefficients giving real roots?


问:
$N$ independent geometric Random Variables and their matchings

Suppose we have $n$ independent geometric RVs $X_1, X_2, \dots, X_n$ all with same $p \in (0,1]$. We say a pair $(i,j)$, $1\leq i<j\leq n$ matches if $X_i = X_j.$ Let $N$ denote the number of matches.

Then, if $E_{(i,j)}$ is the event $(i,j)$ matches, what is $P(E_{(i,j)})$ and $E(N)$. Also find $\operatorname{Var}(N)$ if $p=.5$?

For finding $P(E_{(i,j)})$ I did

$$P(E_{(i,j)}) = \sum_{k=1}^{\infty}P(X_i=k)P(X_j=k) = \frac{p}{2-p}.$$

I believe this is right but correct me if I'm wrong.

Then for finding $E(N)$:

For each pair create an indicator variable $I_{(i,j)}$ that is $1$ if that pair matches, $0$ otherwise.

Then $N=\sum_{1\leq i<j\leq n} I_{(i,j)}$ and since each indicator has expectation $p/(2-p)$, I got that

$$ E(N) = \binom{n}{2} \frac{p}{2-p}.$$

Again please let me know if this is wrong.

What I'm having trouble with is calculating the variance.

I'm not sure if each $E_{(i,j)}$ is independent. If they are independent then I think the variance is $\binom{n}{2} p(1-p)$. Not sure how to solve this if they are not, so I think they are independent but how would I prove that they are independent?

答:
Define a set $S\subseteq [n]^2$ by $S=\Big\{(i,j)\in[n]^2\Big|1\leq i<j\leq n\Big\}$. We can express $V(N)$ by $$V(N)=\sum_{(i,j)\in S}V\big(I_{(i,j)}\big)+\sum_{(i_1,j_1),(i_2,j_2)\in S\\(i_1,j_1)\neq (i_2,j_2)}\text{cov}\big(I_{(i_1,j_1)},I_{(i_2,j_2)}\big)$$

The trick here is to recognize that if $i_1,i_2,j_1,j_2$ are all distinct then $E_{(i_1,j_1)},E_{(i_2,j_2)}$ are independent. This means only $n(n-1)(n-2)$ remain with a non$-$vanishing covariance so we can express $V(N)$ as  $$V(N)={n \choose 2}\cdot V\big(I_{(1,2)}\big)+n(n-1)(n-2)\cdot \text{cov}\big(I_{(1,2)},I_{(1,3)}\big)$$ Now we have $$V\big(I_{(1,2)}\big)=\mathbb{E}\big(I_{(1,2)}^2\big)-\Big(\mathbb{E}\big(I_{(1,2)}\big)\Big)^2=\frac{2p(1-p)}{(p-2)^2}$$ $$\text{cov}\big(I_{(1,2)},I_{(1,3)}\big)=P(X_1=X_2=X_3)-\mathbb{E}\big(I_{(1,2)}\big)\mathbb{E}\big(I_{(1,3)}\big)=\frac{p^2}{p^2-3p+3}-\frac{p^2}{(p-2)^2}$$ We got $V(N)$.


问:
$f(0)=0$ and $f'(x)=f(x)^2$
At first sight, this exercise seems to be already seen many times in this website but I could not find a Analysis-1 level proof of the following :
Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(0)=0$ and $\forall x \in \mathbb{R}$, we have $f'(x)=[f(x)]^2$. Show that $f(x)=0, \forall x \in \mathbb{R}$.

My attempt is : Since $f'(x) \geq 0$ $\forall x$, we have that $\forall x \in \mathbb R^+ f(x) \ge 0$. By contradiction, suppose that $\exists a > 0$ such that $f(a)>0$. Since $f'=f^2\geq0$, we get that $f(x)>0$  for all $x \in (a, +\infty)$.

Let $g: (a,+\infty) \to \mathbb{R}$ be given by  $g(x)=\frac{1}{f(x)}$. As $g'(x)=-1$ we have $g(x)=-x+c$, $c \in \mathbb{R}$. So, $f(x)=\frac1{-x+c}, \forall x \in (a,+\infty)$. From $f(a)>0$ we get that $c>a$. But then $f$ isn't continuous at $x=c$, a contradiction.

I think I showed that the function is constantly equal to $0$ on $(0,+\infty)$, but how do I show it on $(-\infty,0)$ ? Should I just do the same proof with the g function on $(-\infty,a)$ with $a < 0$?

答:
Maybe I'm looking at this wrong, but it seems like we could easily do this with a separable ODE. Let $y = f(x)$. Then,

$\frac{dy}{dx} = y^2$

$\frac{dy}{y^2} = dx$

$\frac{-1}{y} = x + C$

$\frac{-1}{x + C} = y$

However, upon using the initial condition, we find a problem.

$\frac{-1}{C} = 0$

$-1 = 0$ which is obviously not true.

Therefore, the only possible solution to the ODE is the trivial solution. (i.e. f(x) = 0 is the only solution). And this is exactly what you want.

You mentioned you are in an Analysis class, so this may not suffice, but it seems like a decent way to go about it using methods from ODE.

Title: $f(0)=0$ and $f'(x)=f(x)^2$


问:
Show that no matter how we colour all edges of K5,5 with red and blue you can find a copy of K2,2 which either has all edges red or all edges blue.

Show that no matter how we colour all edges of $K_{5,5}$ with red and blue you
can find a copy of $K_{2,2}$ which either has all edges red or all edges blue.

I have done this: $R_2(G_1,G_2)\le R_2(10,4)$($10$ and $4$ being the number of vertices of the graphs)

but it doesn't seem right as work after this step (calculating $R_2(10,4)$ is not a reasonable task to do). Any hint on this question?

答:
Here's an equivalent formulation that might be more familiar:

In a $ 5 \times 5 $ square grid, each square is colored red or blue. Then, there exists 2 rows and 2 columns who's intersection yields squares of the same color.

The equivalence follows from: the color of square $(i,j)$ is the color of the edge connecting top vertex $i$ to bottom vertex $j$.

The standard incidence matrix proof works with this problem.

WLOG, there is a color (say red) with $ \geq 13$ squares.

Let $r_i$ be the number of red squares in the $i$th column.

For each column, let's count the number of red-pairs of squares (formed by 2 rows).

There $ { r_i \choose 2 } $ column pairs in each column.

There are at least $ \sum { r_i \choose 2 } \geq 5 { \frac{ \sum r_i } { 5} \choose 2 } = 5 { 13/5 \choose 2 } = 10.4 $  such column pairs. Since this is an integer, there are at least 11 such column pairs.

Since there are $ \frac{ 5 \times 4 } { 2 } = 10 $ pairs of rows, by the pigeonhole principle, there must be a pair of rows that have at least 2 column pairs.

This gives us the 2 rows and 2 columns that are of the same color.

The good news is that this easily generalizes to finding (say) $K_{3,3}$ in a $K_{n, n }$.


问:
Algebraic to parametric form for a surface equation
The surfaces given by the equation

$$(x^2+y^2)\cos^2 \psi + z^2 \cot^2 \psi = \alpha^2; 0<\psi<\pi$$
for $\psi$ constant are equipotential surfaces. I'm asked to express $\psi$ in terms of $x,y,z$.
This seems basic yet I'm not seeing how to proceed. Can you provide guidance on the method/approach to use in solving problems of this type?

答:
Define

$$r^{2}:=x^{2}+y^{2}$$

and

$$C:=cos(\psi )$$

using

$$cot^{2}(\psi ) =\frac{cos^{2}(\psi ) }{sin^{2}(\psi ) }=\frac{cos^{2}(\psi ) }{1-cos^{2}(\psi ) }$$

we get

$$r^{2}C^{2}+z^{2}\frac{C^{2}}{1-C^{2}}=\alpha ^{2}$$

which is equivalent to

$$r^{2}(1-C^{2})C^{2}+z^{2}C^{2}=(1-C^{2})\alpha ^{2}$$

which is just a quadratic in $C^{2}$. Note that $\cos(\psi )$ is invertible in the domain that you gave.

Can you take it from there?


问:
SO(2,1) not connected
I am trying to show that $SO(2,1)$ is not connected but I have no idea where to start really, I know that it is connected if there is a path between any two points. My definition of $SO(2,1)$ is:

$SO(2,1)=\{X\in Mat_3(\mathbb{R}) \mid X^t\eta X=\eta, \ \det(X)=1\}$ where $\eta$ is the matrix defined as: $$\left ( \begin{array}{ccc} 1 &0&0\\0&1&0\\0&0&-1\end{array}\right )$$

Thanks for any help

答: Consider the orbit of the vector $(0,0,1)$ under $SO(2,1)$; you should find that it's disconnected (note that there are elements of $SO(2,1)$ which map $(0,0,1)$  to itself, or to $(0,0,-1)$, and then show that it can not be mapped to any vector $(a,b,0)$).  So this gives us a continuous map from $SO(2,1)$ to a disconnected space, which implies that $SO(2,1)$ is disconnected.

Title: SO(2,1) not connected


问:
Polynomial equation root
I need to show whether:

$$\frac{1}{x}\times A - \left(\frac{1}{(x+b)(x-b)}-\frac{1}{(x+c)(c-x)}\right)$$

has a single root between $b$ and $c$, where $c>b>0$ and $A>0$. My numerical plots suggest so, but not 100% sure.

答:
Well, we have:

$$P(x)=\frac Ax - \frac{1}{x^2-b^2}+\frac{1}{c^2-x^2}$$

To show that the root exists, you need only show that $$\lim_{\epsilon \to 0} P(b+\epsilon)=-\infty \text{ and } \lim_{\epsilon \to 0} P(c-\epsilon)=\infty$$

Since the function is continuous whenever $x$ is not $0, \pm b, \pm c$ it is continuous on the range $(b, c)$ as defined, and therefore the intermediate value theorem proves the root exists.

Title: Polynomial equation root


问:
Approximation of $\arcsin(\sqrt{1-x^2})$ by $2\arcsin(\sqrt{\frac{1-x}2})$
I stumbled upon an use of $2\arcsin(\sqrt{\frac{1-x}2})$ to approximate $\arcsin(\sqrt{1-x^2})$. The latter is equal to $\frac\pi 2-\arcsin(x)$ when $0\le x\le 1$ and is useful if you have a good approximation of $\arcsin$ around $0$ and want to use it for values near $1$. On the flip side the starting approximation needs to cover $[-\frac1{\sqrt2},\frac1{\sqrt2}]$ in order to use the previous fact to approximate $\arcsin$ on the entire domain.

However if we use the fact that $\arcsin(\sqrt{1-x^2})\approx2\arcsin(\sqrt{\frac{1-x}2})$, our starting approximation only needs to cover $[-0.5, 0.5]$. Empirically this approximation is extremely good, but I'm wondering how would one come up with this kind of approximation and are there any insights to explain why the two are so close?

答:
Check the following ilustration:

$$
\begin{align}
\theta&=\sin^{-1}{\sqrt{1-x^{2}}}\\
\\
\frac{\theta}{2}&=\sin^{-1}{\frac{\sqrt{1-x^{2}}}{\sqrt{2+2x}}}\\
&=\sin^{-1}{\sqrt{\frac{1-x}{2}}}
\end{align}
$$


问:
$\lim_{(x,y) \to (\frac{-k\pi}{2},\frac{k\pi}{2})} \frac{|\sin(x-y)|^a}{|x^2-y^2|}$ with $a>0, k \in \Bbb Z$
Let $a>0$, $k \in\Bbb Z$ and $k \neq 0$.

Evaluate: $\lim_{(x,y) \to (\frac{-k\pi}{2},\frac{k\pi}{2})} \frac{|\sin(x-y)|^a}{|x^2-y^2|}$

The solution indicated is $+\infty$, for every $a$, but I don't understand why.

For example, if $a=2$, moving on $(\frac{-k\pi}{2}+2t,\frac{k\pi}{2}+t)$, $t \to 0$, I obtain:

$$\frac{\sin(t)^2}{|3t^2-k\pi t|}\sim \frac{t^2}{k\pi |t|}\sim \frac{|t|}{k\pi} \to 0$$

Where am I wrong?

Thanks in advance.

答:
Let $a>0$, $\;k\in\Bbb Z$ and $\;k\neq 0\;.\;$
Evaluate: $\lim\limits_{(x,y)\to\left(-\frac{k\pi}{2},\frac{k\pi}{2}\right)}\dfrac{|\sin(x-y)|^a}{|x^2-y^2|}$

It results that

$\begin{align}
\dfrac{|\sin(x-y)|^a}{|x^2-y^2|}&=\dfrac{|\sin(x-y+k\pi)|^a}{|x-y+k\pi|^a}\cdot\dfrac1{|x-y|}\cdot\\
&\cdot\dfrac{|x-y+k\pi|^a}{|x+y|}\;\;,
\end{align}$

for any $\;(x,y)\in\mathbb{R}^2\land y\ne\pm x\land y\ne x+k\pi\;.$

$\lim\limits_{(x,y)\to\left(-\frac{k\pi}{2},\frac{k\pi}{2}\right)}\dfrac{|\sin(x-y+k\pi)|^a}{|x-y+k\pi|^a}=\lim\limits_{t\to0}\dfrac{|\sin t|^a}{|t|^a}=1\;,$

$\lim\limits_{(x,y)\to\left(-\frac{k\pi}{2},\frac{k\pi}{2}\right)}\dfrac1{|x-y|}=\dfrac1{|k|\pi}\;,$

but

$\lim\limits_{(x,y)\to\left(-\frac{k\pi}{2},\frac{k\pi}{2}\right)}\dfrac{|x-y+k\pi|^a}{|x+y|}\;$ does not exist, indeed,

by letting $\;X=x-y+k\pi\;$ and $\;Y=x+y\;,\;$ we get that

$\lim\limits_{(x,y)\to\left(-\frac{k\pi}{2},\frac{k\pi}{2}\right)}\dfrac{|x-y+k\pi|^a}{|x+y|}=\lim\limits_{(X,Y)\to(0,0)}\dfrac{|X|^a}{|Y|}\;.$

If $\;\lim\limits_{(X,Y)\to(0,0)}\dfrac{|X|^a}{|Y|}\;$ existed, then the value of the limit would be the same on every subset $\;S\;$ of the domain of the function $\;\dfrac{|X|^a}{|Y|}\;$ for which $\;(0,0)\;$ is a limit point of $\;S\;,\;$ but

$\;\lim\limits_{(X,Y)\to(0,0)\\Y=m|X|^a}\dfrac{|X|^a}{|Y|}=\dfrac1{|m|}\;$

is not the same in fact it depends on $\;m\;,\;$ hence

$\lim\limits_{(x,y)\to\left(-\frac{k\pi}{2},\frac{k\pi}{2}\right)}\dfrac{|x-y+k\pi|^a}{|x+y|}=\lim\limits_{(X,Y)\to(0,0)}\dfrac{|X|^a}{|Y|}\;$

does not exist, consequently

even $\;\lim\limits_{(x,y)\to\left(-\frac{k\pi}{2},\frac{k\pi}{2}\right)}\dfrac{|\sin(x-y)|^a}{|x^2-y^2|}\;$ does not exist.


答:
$$
T\left(2^{\log_2 n}\right) = n T\left(2^{\log_2 \frac n2}\right)
$$

now calling

$$
\cases{
\mathcal{T}\left(\cdot\right) = T\left(2^{(\cdot)}\right)\\
z = \log_2 n
}
$$

we have the equivalent recurrence

$$
\mathcal{T}\left(z\right)=2^z \mathcal{T}\left(z-1\right)
$$

This recurrence has as solution

$$
\mathcal{T}\left(z\right) = c_0 2^z\cdot 2^{z-1}\cdot 2^{z-2}\cdots 1 = c_0 2^{\frac{z(z+1)}{2}}
$$

now with the backwards substitution $z = \log_2 n$ we recover

$$
T(n) = c_02^{\frac{(\log_2 n)^2+\log_2 n}{2}} = c_02^{\frac{(\log_2 n)^2}{2}}\sqrt{n}
$$

and due to the initial condition

$$
T(n) = 2^{\frac{(\log_2 n)^2}{2}}\sqrt{n}=n^{\frac{1+\log_2 n}{2}}
$$

Title: $T(n) = nT(n/2) ,T(1)=1$ How can I solve this with backward subsitituon?

Body:
$n$ can be power of 2 or we can assume $Integerpart(n/2)$ ==> int casting.

I find the recurrence relation of function but How can I solve this with using backward subsitituon ?

My recurrence relation solution


问:
How does Cantor's diagonal argument work?
I'm having trouble understanding Cantor's diagonal argument.  Specifically, I do not understand how it proves that something is "uncountable".  My understanding of the argument is that it takes the following form (modified slightly from the wikipedia article, assuming base 2, where the numbers must be from the set $ \lbrace 0,1 \rbrace $):

\begin{align}
s_1 &= (\mathbf{0},1,0,\dots)\\
s_2 &= (1,\mathbf{1},0,\dots)\\
s_3 &= (0,0,\mathbf{1},\dots)\\
\vdots &= (s_n \text{ continues})
\end{align}

In this case, the diagonal number is the bold diagonal numbers $(0, 1, 1)$, which when "flipped" is $(1,0,0)$, neither of which is $s_1$, $s_2$, or $s_3$.

My question, or misunderstanding, is: When there exists the possibility that more $s_n$ exist, as is the case in the example above, how does this "prove" anything?  For example:

\begin{align}
s_0 &= (1,0,0,\mathbf{0},\dots)\ \ \textrm{ (...the wikipedia flipped diagonal)}\\
s_1 &= (\mathbf{0},1,0,\dots)\\
s_2 &= (1,\mathbf{1},0,\dots)\\
s_3 &= (0,0,\mathbf{1},\dots)\\
s_4 &= (0,1,1,\mathbf{1},\dots)\\
s_4 &= (1,0,0,\mathbf{1},\dots)\ \ \textrm{ (...alternate, flipped } s_4\textrm{)}\\
s_5 &= (1,0,0,0,\dots)\\
s_6 &= (1,0,0,1,\dots)\\
\vdots &= (s_n \text{ continues})
\end{align}

In other words, as long as there is a $\dots \text{ continues}$ at the end, the very next number could be the "impossible diagonal number", with the caveat that it's not strictly identical to the "impossible diagonal number" as the wikipedia article defines it:

For each $m$ and $n$ let $s_{n,m}$ be the $m^{th}$ element of the $n^{th}$ sequence on the list; so for each $n$,

$$s_n = (s_{n,1}, s_{n,2}, s_{n,3}, s_{n,4}, \dots).$$

...snip...

Otherwise, it would be possible by the above process to construct a sequence $s_0$ which would both be in $T$ (because it is a sequence of 0s and 1s which is by the definition of $T$ in $T$) and at the same time not in $T$ (because we can deliberately construct it not to be in the list). $T$, containing all such sequences, must contain $s_0$, which is just such a sequence. But since $s_0$ does not appear anywhere on the list, $T$ cannot contain $s_0$.

Therefore $T$ cannot be placed in one-to-one correspondence with the natural numbers. In other words, it is uncountable.

I'm not sure this definition is correct, because if we assume that $m = (1, \dots)$, then this definition says that "$s_n$ is equal to itself"&mdadsh;there is no "diagonalization" in this particular description of the argument, nor does it incorporate the "flipping" part of the argument, never mind the fact that we have very clearly constructed just such an impossible $T$ list above.  An attempt to correct the "diagonalization" and "flipping" problem:

$$s_n = (\lnot s_{m,m}, \lnot s_{m,m}, \dots) \quad \text{where $m$ is the element index and} \quad\begin{equation}\lnot s_{m,m} = \begin{cases}0 & \mathrm{if\ } s_{m,m} = 1\\1 & \mathrm{if\ } s_{m,m} = 0\end{cases}\end{equation}$$

This definition doesn't quite work either, as we immediately run in to problems with just $s_1 = (0),$ which is impossible because by definition $s_1$ must be $ = (1)$ if $s_1 = (0)$, which would also be impossible because... it's turtles all the way down!?  Or more generally, with the revised definition there is a contradiction whenever $n = m$, which would seem to invalidate the revised formulation of the argument / proof.

Nothing about this argument / proof makes any sense to me, nor why it only applies to real numbers and makes them "uncountable". As near as I can tell it would seem to apply equal well to natural numbers, which are "countable".

What am I missing?

答:
[This comment avoids formality. At the bottom, it draws a contrast with the @Arturo Magidin answer.]

A goal of this comment will be to help understand the following claim:

Among ALL the "decimal" numbers between 0 and 1 lies at least
      one number that is not rational.

[You said]… in other words, as long as there is a …(continues…) at the end, the very next number could be the "impossible diagonal number"

So you think there is a number not yet specified but which matches the irrational we are building?

OK..

Do you agree that any set of rational numbers (even the set of all of them) can be mapped to the set of positive integers? [At least one such way is usually shown in text books.]

Do you agree that mapping to the set of positive integers is the same as making a "list" in some precise order? The sequence order would match the integer number.

Now, let's consider the set of ALL decimal numbers between 0 and 1. For the moment, we'll informally accept "decimal number" to mean what we all expect: a real number written out in base 10 using decimal point notation.

Do you agree that if this set composed of ALL decimals between 0 and 1 (hereafter "decimals") included only rational numbers, then we could list our set, each decimal number, in a list in some order (ie, in some mapping to positive integers)? We already agreed that any set of rational numbers can be mapped, so if the decimals form such a set....

Do you agree this now implies that if we can't find at least one way to list each decimal number in order, then the set of decimal numbers was not composed solely of rational numbers? [Any such decimal that might not be rational can be referred to as an "irrational".]

So we have to assume that at least one listing exists (one such mapping to the positive integers). We'll consider such a list and do the Cantor diagonalization process on it.

Below, I cover the Cantor process a little more, but for the moment let's believe the result, that we found a decimal number that is not on an alleged ordered listing of ALL decimals.

This is a contradiction. We can't have a list of ALL decimals but at the same time find a decimal that is not on that list.

This means one or more of our assumptions were incorrect.

What were our assumptions?

Primarily we stated we had a set consisting exactly of all decimals, and we assumed these were all rational numbers.

Unless you point out another specific "gotcha" or assumption, I'll assume this is it.

So we conclude that

```<code>we can't have a set of all decimals AND
have these all be rational numbers.
</code>```

Having made the conclusion, let's backtrack to consider more carefully the Cantor diag process and a related issue of representation of a number.

For the table, we just list the digits after the decimal point (each decimal being in the range (0,1) ).

You claim that the number D we were constructing might match some other decimal s4 that is further down on the list.

Do you agree that s4 appears on that list at location n for some positive integer n? That n is taken to be the positive integer to which we mapped s4.

Do you agree, then, that, by construction, D differs from s4 at least in position n?

Question:

```<code>Can s4 and D be the same number even though
they differ in at least one digit?
</code>```

[You said]... with the caveat that it's not strictly identical to the "impossible diagonal number"

Two decimal numbers, by a definition that forms a part of what we decided it means to be a decimal number (ie, a real number), satisfy exactly one of the following:

```<code>x=y
x<y
x>y
</code>```

So the allegation you have made (your doubt) is that

```<code>s4=D (?)
</code>```

That is, you think that although s4 and D might be represented differently on the table, it's possible that (in the sense of equality under real numbers) s4 and D might be equal.

(to be continued)

... time passes...

I took a careful look at the answer provided by Arturo Magidin. Besides the obvious differences in degree of formality between this comment and that one (formality implies greater accuracy and correctness, perhaps at the cost of understandability to someone not comfortable with the material), there are a few other points of contrast I want to mention.

1 -- One point is that I considered mappings from "decimals" to the natural numbers (ie, to positive integers). Arturo's answer looks at mappings from natural numbers to the set in question (ie, flips the mapping's origin and destination). That flip is not a major issue as goes this comment (ie, for the particular focus of this comment), but keep it in mind if you read both replies. Surjective is the same as "onto", btw. It's perhaps easier and more natural to use that other mapping direction to be able to also label finite sets as "countable". Also, I have not checked, but that approach is very likely in agreement with accepted definitions (something always useful to consider if you want to entertain formality).

2 -- [From the other answer:] "...so do some numbers have two representations in binary (for example, 0.01 and 0.0011111… are equal). There is a way of fixing this problem, but it is a bit technical and may obscure the issue, so I would rather not get into it." So Arturo's more complete answer bypasses the "tricky" point I left off on right before taking a break (about multiple digit arrangements corresponding to values deemed to be =). However, although not via Cantor's argument directly on real numbers, that answer does ultimately go from making a statement on countability of certain sequences to extending that result to make a similar statement on the countability of the real numbers. This is covered in the last few paragraphs of the primary proof portion of that answer. Pay attention to the order of mappings and how surjective (as well as injective) are used.

3 -- This brings up distinction three. Arturo's answer focuses on binary sequences (similar to the concept of binary representations) before later extending to "decimals". I informally just looked at anything that looks like an ordinary base 10 decimal number we are all used to using when we deal with numbers and measurements. The distinction taken into account there is to (perhaps more easily) deal with formal issues (which were not a concern for this comment).

4 -- [From the other answer:] "argue by contradiction; I don't like to introduce that into these discussions because the argument is difficult enough to 'grok' without the added complication." In contrast, this comment argues by contradiction a fair amount. While it is easier to do this when you don't want to define precise definitions, you may have noted that I kept bringing up the issue of "assumptions". We can consider that open-endedness to be one type of "complication"; however, this open-endedness is an important concern. I was addressing, not a proof under an abstract system loaded with its formal definitions, but the fact that abstract models ultimately "have to" find a way to be useful to society and the real world. It is this feedback/reality-check that leads the abstract models to be refined over time. This idea that a model might be deficient or that we might have to go back and analyze our assumptions or our axioms more carefully in order to keep in sync with the real world is an issue that more easily steals a certain amount of focus when you are dealing with proofs by contradiction. [The recognition of physical models (based on sound mathematical foundation) as "guesses" of reality plays a big part in engineering disciplines and engineering mindset.] On the other hand, within an accepted/proven framework, there is certainly something very comforting about being able to do direct proofs rather than indirect proofs.

OK, the direct proof approach and greater degree of formality is perhaps an important reason Arturo's comment took the path it took and differed from this comment in the above respects.

With all of this out of the way, I have decided for the moment to refer you to Arturo's comment if you want a more satisfying (precise) argument, and I may not finish up this comment addressing this last question of multiple/unique representations. [I may come back and edit, but likely I won't.]

Body:
[This comment avoids formality. At the bottom, it draws a contrast with the @Arturo Magidin answer.]

A goal of this comment will be to help understand the following claim:

Among ALL the "decimal" numbers between 0 and 1 lies at least
      one number that is not rational.

[You said]… in other words, as long as there is a …(continues…) at the end, the very next number could be the "impossible diagonal number"

So you think there is a number not yet specified but which matches the irrational we are building?

OK..

Do you agree that any set of rational numbers (even the set of all of them) can be mapped to the set of positive integers? [At least one such way is usually shown in text books.]

Do you agree that mapping to the set of positive integers is the same as making a "list" in some precise order? The sequence order would match the integer number.

Now, let's consider the set of ALL decimal numbers between 0 and 1. For the moment, we'll informally accept "decimal number" to mean what we all expect: a real number written out in base 10 using decimal point notation.

Do you agree that if this set composed of ALL decimals between 0 and 1 (hereafter "decimals") included only rational numbers, then we could list our set, each decimal number, in a list in some order (ie, in some mapping to positive integers)? We already agreed that any set of rational numbers can be mapped, so if the decimals form such a set....

Do you agree this now implies that if we can't find at least one way to list each decimal number in order, then the set of decimal numbers was not composed solely of rational numbers? [Any such decimal that might not be rational can be referred to as an "irrational".]

So we have to assume that at least one listing exists (one such mapping to the positive integers). We'll consider such a list and do the Cantor diagonalization process on it.

Below, I cover the Cantor process a little more, but for the moment let's believe the result, that we found a decimal number that is not on an alleged ordered listing of ALL decimals.

This is a contradiction. We can't have a list of ALL decimals but at the same time find a decimal that is not on that list.

This means one or more of our assumptions were incorrect.

What were our assumptions?

Primarily we stated we had a set consisting exactly of all decimals, and we assumed these were all rational numbers.

Unless you point out another specific "gotcha" or assumption, I'll assume this is it.

So we conclude that

we can't have a set of all decimals AND
have these all be rational numbers.

Having made the conclusion, let's backtrack to consider more carefully the Cantor diag process and a related issue of representation of a number.

For the table, we just list the digits after the decimal point (each decimal being in the range (0,1) ).

You claim that the number D we were constructing might match some other decimal s4 that is further down on the list.

Do you agree that s4 appears on that list at location n for some positive integer n? That n is taken to be the positive integer to which we mapped s4.

Do you agree, then, that, by construction, D differs from s4 at least in position n?

Question:

Can s4 and D be the same number even though
they differ in at least one digit?

[You said]... with the caveat that it's not strictly identical to the "impossible diagonal number"

Two decimal numbers, by a definition that forms a part of what we decided it means to be a decimal number (ie, a real number), satisfy exactly one of the following:

x=y
x<y
x>y

So the allegation you have made (your doubt) is that

s4=D (?)

That is, you think that although s4 and D might be represented differently on the table, it's possible that (in the sense of equality under real numbers) s4 and D might be equal.

(to be continued)

... time passes...

I took a careful look at the answer provided by Arturo Magidin. Besides the obvious differences in degree of formality between this comment and that one (formality implies greater accuracy and correctness, perhaps at the cost of understandability to someone not comfortable with the material), there are a few other points of contrast I want to mention.

1 -- One point is that I considered mappings from "decimals" to the natural numbers (ie, to positive integers). Arturo's answer looks at mappings from natural numbers to the set in question (ie, flips the mapping's origin and destination). That flip is not a major issue as goes this comment (ie, for the particular focus of this comment), but keep it in mind if you read both replies. Surjective is the same as "onto", btw. It's perhaps easier and more natural to use that other mapping direction to be able to also label finite sets as "countable". Also, I have not checked, but that approach is very likely in agreement with accepted definitions (something always useful to consider if you want to entertain formality).

2 -- [From the other answer:] "...so do some numbers have two representations in binary (for example, 0.01 and 0.0011111… are equal). There is a way of fixing this problem, but it is a bit technical and may obscure the issue, so I would rather not get into it." So Arturo's more complete answer bypasses the "tricky" point I left off on right before taking a break (about multiple digit arrangements corresponding to values deemed to be =). However, although not via Cantor's argument directly on real numbers, that answer does ultimately go from making a statement on countability of certain sequences to extending that result to make a similar statement on the countability of the real numbers. This is covered in the last few paragraphs of the primary proof portion of that answer. Pay attention to the order of mappings and how surjective (as well as injective) are used.

3 -- This brings up distinction three. Arturo's answer focuses on binary sequences (similar to the concept of binary representations) before later extending to "decimals". I informally just looked at anything that looks like an ordinary base 10 decimal number we are all used to using when we deal with numbers and measurements. The distinction taken into account there is to (perhaps more easily) deal with formal issues (which were not a concern for this comment).

4 -- [From the other answer:] "argue by contradiction; I don't like to introduce that into these discussions because the argument is difficult enough to 'grok' without the added complication." In contrast, this comment argues by contradiction a fair amount. While it is easier to do this when you don't want to define precise definitions, you may have noted that I kept bringing up the issue of "assumptions". We can consider that open-endedness to be one type of "complication"; however, this open-endedness is an important concern. I was addressing, not a proof under an abstract system loaded with its formal definitions, but the fact that abstract models ultimately "have to" find a way to be useful to society and the real world. It is this feedback/reality-check that leads the abstract models to be refined over time. This idea that a model might be deficient or that we might have to go back and analyze our assumptions or our axioms more carefully in order to keep in sync with the real world is an issue that more easily steals a certain amount of focus when you are dealing with proofs by contradiction. [The recognition of physical models (based on sound mathematical foundation) as "guesses" of reality plays a big part in engineering disciplines and engineering mindset.] On the other hand, within an accepted/proven framework, there is certainly something very comforting about being able to do direct proofs rather than indirect proofs.

OK, the direct proof approach and greater degree of formality is perhaps an important reason Arturo's comment took the path it took and differed from this comment in the above respects.

With all of this out of the way, I have decided for the moment to refer you to Arturo's comment if you want a more satisfying (precise) argument, and I may not finish up this comment addressing this last question of multiple/unique representations. [I may come back and edit, but likely I won't.]


问:
Decomposing representation of $sl(3,\mathbb{C})$ into direct sum of irreducibles
In his book, Lie Groups, Lie Algebras, and Representations, Hall asks us to consider the nine-dimensional representation of $\mathfrak{sl}(3,\mathbb{C})$ obtained by taking the tensor product of the representations with highest weights (1,0) and (0,1) (that is, the standard and dual representations). We want to decompose this representation as a direct sum of irreducibles. I am hoping that there is a method for doing this similar to the Clebsch-Gordan procedure in the case of $\mathfrak{sl}(2,\mathbb{C})$. Is this the case? If not, how does one usually proceed here?

I compute the possible weights and the corresponding weight vectors for the tensor product representation to be:

(1,1) --- $e_1\otimes f_1$ (highest weight)

(-1,2) --- $e_2\otimes f_1$

(2,-1) --- $e_1\otimes f_2$

(1,-2) --- $e_3\otimes f_2$

(-1,-1) --- $e_3\otimes f_3$

(-2,1) --- $e_2\otimes f_3$

(0,0) --- $e_1\otimes f_3, e_2\otimes f_2, e_3\otimes f_1$

where $\{e_i\}$ is the standard basis for $\mathbb{C}^3$ and as in Hall's book, we define $f_1=e_3$, $f_2=-e_2$, and $f_3=e_1$ to be the basis for the second copy of $\mathbb{C}^3$ in our tensor product (recall the dual space is isomorphic to $\mathbb{C}^3$).

Note: There is an answer to a similar question here. If this is the procedure I'm looking for, my question becomes what does one mean by "Remove a copy of this irreducible representation from $\mathbb{C}^3\otimes V_{1,1}$" in step (iv)? I suspect that it has something to do with the decomposition of the tensor product representation into a direct sum of its weight spaces.

答: The statement "Remove a copy of this irreducible representation from $\mathbb{C}^3\otimes V_{1,1}$" just means remove the weights occurring in the irreducible representation from those appearing in your representation, and see what's left. In your case, the irreducible representation with highest weight (1,1) is 8-dimensional (and is actually isomorphic to the adjoint representation), and its weights are listed in Section 6.5 of Hall: they are (1,1), (-1,2), (2,-1), (0,0) (twice), (1,-2), (-2,1), and (-1,-1). Removing them from your list of weights leaves only (0,0), which is the weight of the 1-dimensional trivial representation (with highest weight (0,0)). So your tensor product representation decomposes into the 8-dimensional adjoint representation plus a 1-dimensional trivial representation. This is sometimes written as $3\otimes\overline{3} = 8\oplus 1$.


答:
To answer the more general question, "can complex coefficients lead to real-valued roots of polynomials," the answer is yes.  We can readily see this by working backwards:  construct a polynomial for which one root is real and the roots with nonzero imaginary part do not occur in complex conjugate pairs.  Then the coefficients are guaranteed to not be pure real.  For example, consider $$f(z) = (z - 3)(z - 2 - i) = z^2 - (5+i)z + (6+3i).$$  The coefficients are complex but by construction, there is a real root.

If we require all roots of a polynomial to be real-valued, then the answer is no unless we allow the polynomial to be a scalar multiple of some arbitrary complex constant.  This follows from the fundamental theorem of algebra.

Title: Complex coefficients giving real roots?

Body:
Edit: I figured it out, I just miscalculated. No, there are no complex solutions to k.

For what values of $k$ does the equation $kx^2+3kx+k=-9-kx$ have real roots? Just to simplify it, I'll write the equation as $kx^2+4kx+k+9=0$. Now the obvious solution would probably be to plug it into the quadratic formula and try to solve such that the discriminant is greater than zero. In the quadratic formula, $a = k$, $b = 4k$, and $c = k + 9$. Plugging these values in, I get

$$\begin{align}
&\frac{-4k±\sqrt{(4k)^2-4(k)(k+9)}}{2k}\\
&=\frac{-4k±\sqrt{16k^2-4k^2-36k}}{2k}\\
&=\frac{-4k±\sqrt{12k^2-36k}}{2k}\\
&=-2±\frac{\sqrt{12k^2-36k}}{2k}\\
&=-2±\frac{\sqrt{3k^2-9k}}{k}
\end{align}$$

Since the only thing that matters is finding real values, $-2$ can just be ignored as well as the $\pm$ sign, since nothing really changes anyways. Now the logical thing to do here would be to take the discriminant, set it to be greater than or equal to zero, and get the inequality for all the real values of $k$ (in case you're wondering, $k ≤ 0$ and $k ≥ 3$). But the more interesting part of the question would be finding the complex solutions to $k$. While attempting this, I use assumptions:

First, the square root of a complex number with an imaginary part is always another complex number with an imaginary part. I don't see how this isn't true and I haven't found anything that says otherwise yet, so it seems good to me.

Second, if an equation with real and imaginary parts sums to $0$, then the real and imaginary parts alone will also sum to zero. I don't see how this is false either, since imaginary parts can't contribute to real parts, so both must just equal zero.

So in order to solve this problem, I set $k$ to equal $a+bi$ such that $a$ and $b$ are real numbers. Therefore,

$$\begin{align}
&-2±\frac{\sqrt{3k^2-9k}}{k}\\
&=-2±\frac{\sqrt{3(a+bi)^2-9(a+bi)}}{a+bi}\\
&=-2±\frac{\sqrt{3(a^2+2abi-b^2)-9a-9bi)}}{a+bi}\\
&=-2±\frac{\sqrt{3a^2+6abi-3b^2-9a-9bi}}{a+bi}\\
\end{align}$$

Now if the numerator is equal to the denominator, then the fraction would equal 1 so I set it to that.

$$\begin{align}
\sqrt{3a^2+6abi-3b^2-9a-9bi} &= a+bi\\
3a^2+6abi-3b^2-9a-9bi &= (a+bi)^2\\
3a^2+6abi-3b^2-9a-9bi &= a^2+2abi-b^2\\
2a^2+4abi-2b^2-9a-9bi &= 0
\end{align}$$

Taking imaginary parts only:

$$\begin{align}
4abi-9bi &= 0\\
4abi &= 9bi\\
4ab &= 9b
\end{align}$$

Assuming that $b ≠ 0$:

$$\begin{align}
4a &= 9\\
a &= \frac{9}{4}
\end{align}$$

Now taking real parts only:

$$\begin{align}2a^2-2b^2-9a &= 0\\
2\left(\frac{9}{4}\right)^2-2b^2-9\left(\frac{9}{4}\right) &= 0\\
2\left(\frac{9}{4}\right)^2-9\left(\frac{9}{4}\right) &= 2b^2\\
\frac{486}{16} &= 2b^2\\
\frac{243}{16} &= b^2\\
±\frac{9\sqrt{3}}{4} &= b
\end{align}$$

Therefore $k = \frac{9}{4} + i\frac{9\sqrt{3}}{4}$ should be a solution. Whether or not cases like $k = \frac{9}{2} + i\frac{9\sqrt{3}}{2}$ that are multiples of the original answer would work don't really matter, but this seems like it should work. Another case that should also work would be setting the numerator of the original fraction to equal to zero. This also produces a complex solution for $k$.

The question is, can complex coefficients create real roots for quadratic equations, or did I calculate/assume/do the question wrong? If it is possible, did I miss any solutions? Any help would be appreciated.


答:
Let $\theta = \arcsin \sqrt{1-x^2}$. Since the argument of $\arcsin$ is non-negative, consider the range $0\le \theta\le \frac \pi2$.

$$\begin{align*}
\sin\theta &= \sqrt{1 - x^2}\\
\sin^2\theta &= 1 - x^2\\
x &= \cos \theta\\
\end{align*}$$

Then using half-angle formula,

$$\begin{align*}
1-2\sin^2\frac\theta2&= \cos\theta\\
\sin^2\frac\theta2 &= \frac{1-\cos \theta}2\\
\sin\frac\theta 2&= \sqrt{\frac{1-\cos \theta}2}\\
&= \sqrt{\frac{1-x}2}\\
\frac\theta2 &= \arcsin \sqrt{\frac{1-x}2}\\
\theta &= 2\arcsin \sqrt{\frac{1-x}2}
\end{align*}$$

Title: Approximation of $\arcsin(\sqrt{1-x^2})$ by $2\arcsin(\sqrt{\frac{1-x}2})$

Body:
I stumbled upon an use of $2\arcsin(\sqrt{\frac{1-x}2})$ to approximate $\arcsin(\sqrt{1-x^2})$. The latter is equal to $\frac\pi 2-\arcsin(x)$ when $0\le x\le 1$ and is useful if you have a good approximation of $\arcsin$ around $0$ and want to use it for values near $1$. On the flip side the starting approximation needs to cover $[-\frac1{\sqrt2},\frac1{\sqrt2}]$ in order to use the previous fact to approximate $\arcsin$ on the entire domain.

However if we use the fact that $\arcsin(\sqrt{1-x^2})\approx2\arcsin(\sqrt{\frac{1-x}2})$, our starting approximation only needs to cover $[-0.5, 0.5]$. Empirically this approximation is extremely good, but I'm wondering how would one come up with this kind of approximation and are there any insights to explain why the two are so close?


问:
How many different strings of 5 lower case letters are there?
How many different strings of 5 lower case letters are there? Assume that letters may repeat and the string must contain at least one letter a?
I tried  this:
5×26^5=5×26 ×26 ×26 ×26=2684880</p>

答:
$(26)^5 - (25)^5.$

First term is total # of 5 character strings, where each character has 26 choices.

Second term is total # of 5 character strings, where each character has only 25 choices, re no "a" is used.

The difference between the two is the # of 5 character strings with at least 1 "a".