问:
The matrix diagonalizable coefficients in a field of characteristic different from $2$
I read a property that I did not understand:

Let $M\in GL(2,\mathbb{F}_3)$ the matrix such that $M= I_2$ or $M^2=I_2$. As it has coefficients in a field of characteristic different from $2$, it is diagonalizable.

Is it a theorem? Can you help me prove this result

答:
If not already diagonal, the matrix has trace zero.

When the characteristic is not two, we have $2=1+1 \neq 0$

$$
\frac{1}{2} \;
\left(
\begin{array}{cc}
2&b\\
0&1\\
\end{array}
\right)
\left(
\begin{array}{cc}
1&b\\
0&-1\\
\end{array}
\right)
\left(
\begin{array}{cc}
1&-b\\
0&2\\
\end{array}
\right)=
\left(
\begin{array}{cc}
1&0\\
0&-1\\
\end{array}
\right)
$$

For the next one, we need $a^2 + bc = 1 \; .$

$$
- \frac{1}{2c} \;
\left(
\begin{array}{cc}
c&-1-a\\
-c&a-1\\
\end{array}
\right)
\left(
\begin{array}{cc}
a&b\\
c&-a\\
\end{array}
\right)
\left(
\begin{array}{cc}
a-1&a+1\\
-c&c\\
\end{array}
\right)=
\left(
\begin{array}{cc}
-1&0\\
0&1\\
\end{array}
\right)
$$


问:
How to determine the leading-order asymptotic behaviour of this integral?
$\int^{1}_{0} \cos(xt^{3})\tan(t)\, dt$

as x → ∞

I am stuck on how to apply the stationary phase method when $\tan(t)$ vanishes at the stationary point. Should I expand tan in its Taylor series? Is the stationary point $0\,$? Thanks in advance

答:
Just to provide a bit different approach to the evaluation of the leading asymptotic term.
$$I(x)=\int^{1}_{0} \cos(xt^{3})\tan(t)\, dt=\frac{1}{3}\int^{1}_{0} \frac{\cos(xs)\tan(s^{\frac{1}{3}})}{s^{\frac{2}{3}}}\, ds=\Re\frac{1}{3}\int^{1}_{0} \frac{e^{ixs}\tan(s^{\frac{1}{3}})}{s^{\frac{2}{3}}}\, ds$$
Next, we change the variable $s\to{is}=se^{\frac{\pi{i}}{2}}$. Integral along a closed contour (from small $r$ to $1$ of axis $X$; along a quarter-circle (angle $\phi$ - from $0$ to $\frac{\pi}{2}$); from $i$ to $i{r}$ and along the quarter-cercle of small radius $r$ clockwise to the starting point) is zero - there is no singularity inside the contour. Integral along a small quarter-circle (around $s=0$) $\,\to0$ at $r\to0$ - due to the integrand asymptotics near zero.

Integral along the quarter-circle of radius $1$ counter clockwise is limited by
$$\int_{R=1}\frac{e^{ixs}\tan(s^{\frac{1}{3}})}{s^{\frac{2}{3}}}\, ds\leqslant{C}\int_0^{\frac{\pi}{2}}e^{-sin(\phi){x}}dx\leqslant{C}\frac{\pi}{2x}\,,$$ where C is some constant.

Representing $\tan((is)^{\frac{1}{3}})$ as a series and expanding integration to $\infty$  we get the leading asymptotic term:

$$I(x)\sim\frac{1}{3}\Re \,{e}^{\frac{\pi{i}}{6}}\int^{1}_{0} \frac{e^{-xs}\tan((is)^{\frac{1}{3}})}{s^{\frac{2}{3}}}\, ds+O\biggl(\frac{1}{x}\biggr)\sim\frac{1}{3}\Re \,{e}^{\frac{\pi{i}}{6}}\int^{\infty}_{0} e^{-xs}\frac{{e}^{\frac{\pi{i}}{6}}s^{\frac{1}{3}}}{s^{\frac{2}{3}}}\, ds+O\biggl(\frac{1}{x}\biggr)$$

$$I(x)\sim\frac{1}{3}\Re \,{e}^{\frac{\pi{i}}{6}} \biggl(\frac{{e}^{\frac{\pi{i}}{6}}\Gamma(\frac{2}{3})}{x^{\frac{2}{3}}}\biggr)+O\biggl(\frac{1}{x}\biggr)\sim\frac{\cos(\frac{\pi}{3})\Gamma(\frac{2}{3})}{3x^{\frac{2}{3}}}+O\biggl(\frac{1}{x}\biggr)$$
$$I(x)\sim\frac{\Gamma(\frac{2}{3})}{6x^{\frac{2}{3}}}+O\biggl(\frac{1}{x}\biggr)$$

ContentLicense: CC BY-SA 4.0


问:
Checking if an Extension is normal
I want to see if my reasoning is correct. I am given a list o field extensions and asked to determine which ones are normal extensions:

(a) $\mathbb{Q}(\alpha) : \mathbb{Q}$ where $\alpha$ is the real seventh root of 5

(b) $\mathbb{Q}(\sqrt{-5}) : \mathbb{Q}$

(c) $\mathbb{Q}(\sqrt{5}, \alpha) : \mathbb{Q}$ where $\alpha$ is as described in (a)

(d) $\mathbb{R}(\sqrt{-7}) : \mathbb{R}$

(e) $\mathbb{Q}(t) : \mathbb{Q}$

For extensions (b) and (d), the argument seems pretty straightforward: the splitting polynomial for (b) is $x^2 + 5$, and the extension $\mathbb{Q}(\sqrt{-5}) : \mathbb{Q}$ contains all the roots of the polynomial, so the extension is normal. For (d), the splitting polynomial is (I think) $x - \sqrt{-7}$, and the extension $\mathbb{R}(\sqrt{-7}) : \mathbb{R}$ contains all the roots of the polynomial, so the extension is normal (there is only one root, since square roots are in the field of real numbers).

For (a), I noted that since the splitting polynomial $(x^7 - 5)$ have multiple roots in the complex field, the extension cannot be normal. I'm not really sure where to start with (c) or (e), particularly (e) because it seems like I would need more information about $t$ to know anything about splitting polynomials, roots, etc.

答:
For c), the same argument that for $a)$ works: $\mathbb{Q}(\sqrt{5},\alpha)\subset \mathbb{R}$, so you won't get the non-real conjugates of $\alpha$ (which are $\alpha$ times some $7^{\text{th}}$ root of unity).

For e), the extension is not even algebraic, so automatically not normal (it's normal that you cannot find the minimal polynomial of $t$ because there is none, $t$ is transcendental).

Title: Checking if an Extension is normal


问:
Philosophically motivated risk metrics
During a random process over a sample space $\Omega$, an agent incurs a cost (say, in money) given by a random variable $Z:\Omega \rightarrow R$, which is determined by the agent's action. The agent chooses the action which minimizes its "risk", which it calculates as some function $\rho(Z)$. Examples:

A risk-neutral agent uses $\rho=E$, the expectation operator, to minimize expected cost.

A minimax agent uses $\rho(Z)=\inf_\omega Z(\omega)$ to minimize the maximum possible cost.

A maxipok agent uses $\rho(Z)=P(Z > z)$ to minimize the chance of a cost exceeding a threshold $z$.

The latter two agents are risk-averse: given two cost distributions with the same mean, they will associate more risk with whichever has higher variance. Risk-aversion is often quite sensible: would you rather certainly keep your house, or bet it against a second house on a fair coinflip? But expected utility theorists typically justify risk-aversion on the basis that an agent's true utility function is usually concave in their net worth--money has decreasing marginal utility. Equivalently, cost in utility is convex in monetary cost.

If this is the rationale for risk-aversion, then a risk-averse agent's risk metric should satisfy $\rho(Z)=E(f \circ Z)$, where $f:R \rightarrow R$ is a convex function that relates cost in money to cost in utility. It's not hard to show that the maxipok metric satisfies this relationship for $f(c) = I(c > z)$, but minimax does not for any $f$. Nor do many of the other most popular risk metrics in finance: none of the VaR$_\alpha$, CVaR$_\alpha$, and mean-variance metrics can be interpreted as minimizing the expectation of a latent disutility function. The only well-studied risk-averse metric besides maxipok with this property is entropic risk, which has $f(c) = e^{-ac}$.

Ok, here's the question: I'm curious whether there is a pleasant characterization of the set of risk metrics which compute the expectation of some disutility. In other words, when does $\rho:R^\Omega \rightarrow R \cup \{\pm\infty\}$ satisfy $\rho(Z) = E(f \circ Z)$ for some $f:R \rightarrow R$? Furthermore, if $f$ exists, when does it have nice properties, e.g. monotonicity, continuity, convexity?

The $|\Omega| < \infty$ case is probably the place to start, to make the measure theory less horrifying.

答:
I now have a condition which I believe is necessary and sufficient in the special case of a finite sample space. My attempted proof is below.

Claim

Let $\Omega = [n]$, equipped with probability $Q=(q_1,\ldots,q_n)$ in the probability simplex in $R^n$. For a given risk metric $\rho:R^n \rightarrow R$, a latent disutility function $f:R \rightarrow R$ exists satisfying $\rho(Z)=E(f \circ Z)$ if and only if for every $\{x_1, \ldots, x_n\} \subseteq R$, $${\scr Q{\bf f}=R},$$ where $\scr Q$ is an $\underbrace{n \times \cdots \times n}_\text{$n+1$ times}$ tensor, ${\bf f} \in R^n$, and $\scr R$ is an $\underbrace{n \times \cdots \times n}_\text{$n$ times}$ tensor, with the components
$${\scr q}_{\omega_1 \ldots \omega_n k}=\sum_j q_j I(\omega_j=k)$$
$${\bf f}_\omega = \rho(x_\omega, \ldots, x_\omega)$$
$${\scr r}_{\omega_1 \ldots \omega_n}=\rho(x_{\omega_1}, \ldots, x_{\omega_n}).$$.

Proof

$(\implies)$ Assume such an $f$ exists. A cost variable $Z=(z_1,\ldots,z_n)$ is an element of $R^n$, and $E(Z) =\sum_j q_j z_j$. If $Z$ is constant, then $z_j = c$ for each $j \in [n]$, and thus $$E(f \circ Z) =\sum_j q_jf(z_j)= \sum_j q_jf(c)=f(c).$$
Hence $f(c) = \rho(c,\ldots,c)$ for each $c \in R$.

For any $\{x_1, \ldots, x_n\} \subseteq R$, there are $n^n$ random variables supported on the $x_i$, each of which has the form $Z = (x_{\omega_1}, \ldots, x_{\omega_n})$. Each such $Z$ places a linear constraint on $\rho$: $$\rho(x_{\omega_1}, \ldots, x_{\omega_n})=\rho(Z)=E(f \circ Z)=\sum_j q_jf(x_{\omega_j})=\sum_j q_j\rho(x_{\omega_j}, \ldots, x_{\omega_j}).$$ The equation ${\scr Q{\bf f}=R}$ expresses all of these constraints in tensor form, so it must hold.

$(\impliedby)$ From above, $f:R \rightarrow R$ defined by $f(c) = \rho(c,\ldots,c)$ has the desired property. This completes the proof.

Discussion

This result allows one to easily reject risk functions which do not have associated latent disutilities, by evaluating $\rho$ for particular $Z$ and enforcing linear constraints. It also provides an explicit construction of the unique $f$ from $\rho$ in cases where such an $f$ exists. This makes it easy to check monotonicity/continuity/convexity.

Unfortunately, even though the condition I present is sufficient, it would be quite difficult to use this result to confirm risk functions which do have associated latent disutilities. I'd love to see if someone who knows more about crazy linear algebra could find a more easily verifiable sufficient condition. And of course, a generalization to infinite sample spaces would be incredible, although at this point I'm afraid I wouldn't understand it without a PhD in real analysis.

Title: Philosophically motivated risk metrics


问:
Consider a random straight line $L$ passing through the origin with equation $y=Mx$
Consider a random straight line $L$ passing through the origin with equation $y=Mx$. Suppose that the angle $\theta$ that it makes with the positive $x$-axis is uniformly distributed in $(0,\pi/2)$.

Find the density of $M$, the slope of the line $L$. (Hint: Express the slope $M$ in terms of the angle.)

Let $(X,Y)$ be the coordinates of the point where the line $L$ intersects the unit circle $\{(u,v):u^2+v^2=1\}$. Find the expected value of $Y$.

The line $L$ divides the unit square, with corners $(0,0),(1,0),(0,1),(1,1)$, in two pieces. Find the expectation of the area of the smaller piece.

My thinking:

For this one, is it saying that the density of $M$ is the same as the slope of the line $L$? If so, does this mean that the slope of the line is simply $M=\tan(\theta)$?

If I think of the unit circle as a function $f(x)=\sqrt{1-x^2}$ for $x>0$. Then the expectation of Y is $\mathbb{E}[Y]=\int_{-\infty}^\infty xf(x)dx$? I am not too sure about this one because I don't see a relation to line L.

I am not sure how to find this one, but my thinking so far is to find an expression for the area of, which represents probability. Then similarly to part (2), I can use $\mathbb{E}[X]=\int_{-\infty}^\infty xf(x)dx$.

答:
The random variable $M$ is indeed related to the random variable $\theta$ by $M = \tan \theta$.  But the density of $M$ requires a little work: Either apply a general result you may have seen, or work from scratch by calculating $F_M(m) = \mathbb{P}(M\le m) = \mathbb{P}(\tan \theta \le m) = \mathbb{P}(\theta \le \arctan m)$ and applying the uniform distribution $\mathbb{P}(\theta \le \theta_0) = \frac{2}{\pi} \theta_0$ for any $\theta_0 \in [0,\pi/2]$.  This gives you the cumulative distribution function $F_M$, from which you find the density $f_M$ by taking the derivative.

You could find $\mathbb{E}[Y]$ with an approach like the one you give, but $f(x) = \sqrt{1-x^2}$ is not the density of $Y$: You would have to find it using the relation $Y = g(\theta) = \sin \theta$ first, as in (1).  Instead, it's easier to remember $\mathbb{E}[g(X)] = \int g(x) f_X(x) \, dx$ and calculate $\mathbb{E}[Y] = \mathbb{E}[\sin \theta] = \int_0^{\pi/2} \sin \theta \cdot \frac{2}{\pi} \, d\theta$

Express the area $A$ as a function of $\theta$, which is easy if you consider the cases $\theta \le \frac{\pi}{4}$ and $\theta > \frac{\pi}{4}$ separately.  Then proceed as in (2).


问:
For $p$ prime and $\omega = {e^{\frac{{2\pi i}}{p}}}$, show that $\omega \left( {1 - {\omega ^{k - 1}}} \right)$ divides p
In Number fields, Marcus, problem 17 on page 9 reduces to the following:

Show that $\omega \left( {1 - {\omega ^{k - 1}}} \right)|p$ for $k \geqslant 2$ in $\mathbb{Z}\left[ \omega  \right]$, where $\omega  = {e^{\frac{{2\pi i}}{p}}}$ and $p$ is a prime integer. The hint says to use $\prod\limits_{k = 1}^{p - 1} {\left( {1 - {\omega ^k}} \right)}  = p$, which is useful to see that $1 - \omega |p$, but I can't see how to use this to get the divisibility above.

I have proven that $\prod\limits_{k = 1}^{p - 1} {\left( {x - {\omega ^k}} \right)}  = \sum\limits_{i = 0}^{p - 1} {{x^i}} $ and ${x^p} + 1 = \prod\limits_{k = 0}^{p - 1} {\left( {x + {\omega ^k}} \right)} $, substituting $x = 1$ in the first equality gives $\prod\limits_{k = 1}^{p - 1} {\left( {1 - {\omega ^k}} \right)}  = p$ and I'm not sure if the second equality is useful at all.

Unnecessary context added for completeness, here's the problem and what I've done thus far:

Suppose that $\mathbb{Z}\left[ \omega  \right]$ is a unique factorization domain and that $\pi $ is irreducible such that $\pi |x + \omega y$. Show that $\pi $ does not divide any of the other factors in the equation $\prod\limits_{k = 0}^{p - 1} {\left( {x + y{\omega ^k}} \right)}  = {z^p}$ by showing that if it did, it would divide both $z$ and $yp$.

My solution up to this point: If it divides at least one other factor then it divides their difference, so either $\pi |y\left( {1 - \omega } \right),1 - \omega |yp \Rightarrow \pi |yp$ or $\pi |y\omega \left( {1 - {\omega ^{k - 1}}} \right)$. If $\omega \left( {1 - {\omega ^{k - 1}}} \right)|p$, we're practically done.

答: $p = {\omega ^p}p = {\omega ^p}\prod\limits_{k = 1}^{p - 1} {(1 - {\omega ^k})}  = \omega\cdot\prod\limits_{k = 2}^p {\omega (1 - {\omega ^{k - 1}})} $.


问:
Proof of rational number
Prove that if $a$ and $b$ are rational numbers satisfying $a^5+b^5=2a^2b^2$, then $1-ab$ is the square of a rational number.

I am just a Year 2 student learning Abstract Algebra. This problem is a challenging one that my teacher gives us. However, I have no ideas about how to solve it. I tried to represent rational numbers by $p/q$ ($p$ and $q$ are both integers) but failed. I would appreciate it very much if anyone can help me.

答:
I don't know abstract algebra. I'm just going to use algebra precalculus.

If $b=0$, then the statement is correct.

Let  $\dfrac {a}{b}=x, b≠0$,  then we have

$$\begin{align}&a^5+b^5=2a^2b^2\\
\implies &a\times x^4+b=2x^2 \\
\implies &a \left(x^2\right)^2-2x^2+b=0\\
\implies &\Delta=1-ab=T^2, T\in\mathbb Q.\end{align}$$

I add additional information to the answer:

If $a,b$ is rational and $b≠0$, then $\dfrac {a}{b}$ is also rational.

Our equation is a quadratic equation with respect to $x^2=\left(\dfrac {a}{b}\right)^2$.

In order for the root of a polynomial equation whose coefficients are rational to be rational, the polynomial discriminant must also be a perfect square of the rational number.

Title: Proof of rational number


问:
Dot Products and Linear Progamming
Let $v_{1}, \ldots, v_{m}$ and $w$ be vectors in $\mathbb{R}^{n}$. Suppose that whenever $d \in \mathbb{R}^{n}$ satisfies $v_{i} \cdot d \leq 0$ for $i = 1, \ldots, m$, it also satisfies $w \cdot d \leq 0$. Prove that $w = \lambda_{1} v_1 + \cdots + \lambda_{m} v_{m}$ for some nonnegative real
numbers $\{\lambda_{i}\}_{i=1}^m$.

I have formulated the given constraint as a linear problem: $\max_{v_{i} \cdot d \leq 0 \text{ for } i=1,\ldots,m} w \cdot d = 0$. The dual will then be $\min_{\sum_{j=1}^{n}w_{j}d_{i} \geq d_{i} \text{ for } i=1,\ldots,m} 0$. How should I proceed and prove the proposition (all that I can gather is that $\sum_{j=1}^{n}w_{i}\geq1$), given that the dual is correct?

答:
If the primal LP is to maximize $\sum_{j=1}^n w_j d_j$ subject to
\begin{align}
\sum_{j=1}^n v_{i,j} d_j &\le 0 &&\text{for $i\in\{1,\dots,m\}$} \\
\end{align}
then the dual LP is to minimize $\sum_{i=1}^m 0 \lambda_i$ subject to
\begin{align}
\sum_{i=1}^m v_{i,j} \lambda_i &= w_j &&\text{for $j\in\{1,\dots,n\}$} \tag1\\
\lambda_i &\ge 0 &&\text{for $i\in\{1,\dots,m\}$}
\end{align}

We are given that $$\sum_{j=1}^n w_j d_j \le 0,$$  so the trivial feasible solution $d_j^*=0$ is optimal. By strong duality, the dual problem is also optimal, yielding the existence of the desired $\lambda_i$ via dual equality constraint $(1)$.

Title: Dot Products and Linear Progamming


问:
Does $\int_0^{+\infty}\cos(x^3)\,\mathrm{d}x$ converge absolutely or conditionally?
$$\int\limits_0^{+\infty}\cos(x^3)\,\mathrm{d}x$$

I think that the integral absolutely does not converge, since the function does not tend to zero, but conditionally it converges, since if you look at the graph of the function, the areas of the cosine humps decrease, and the signs alternate, and then the Leibniz integral converges, but I think this solution is not strict and I would like to get a normal one, but there are no ideas.

答:
$$I=\int\limits_0^{+\infty}\cos(x^3)dx=\frac{1}{3}\int\limits_0^{+\infty}\frac{\cos(t)}{t^{\frac{2}{3}}}dt=\frac{1}{3}\Re\int\limits_0^{+\infty}\frac{e^{it}}{t^{\frac{2}{3}}}dt$$

The variable change: $t\to{s}$ $ (t={is}=se^{\frac{\pi{i}}{2}})$.

The change is valid: due to Jordan lemma integral in the complex plane along a big quarter-circle $\to0$ as its radius $R\to\infty$. Integral along a small quarter-circle around $s=0$  also $\to0$ at $r\to0$ - due to the integrand asymptotics near zero. The change of the variable (and, therefore, the limits of integration) does not affect the integral value.

$$I=\frac{1}{3}\Re\,\,e^{\frac{\pi{i}}{6}}\int\limits_0^{+\infty}\frac{e^{-s}}{s^{\frac{2}{3}}}ds=\frac{1}{3}\cos(\frac{\pi}{6})\Gamma\Bigl(\frac{1}{3}\Bigr)=\frac{1}{2\sqrt3}\Gamma\Bigl(\frac{1}{3}\Bigr)$$


问:
Classic example of a non exact form
Let $\dfrac{xdy-ydx}{x^2+y^2}$ be a 1-form defined in $\mathbb{R}^2\backslash\{0\}$.
Where can I find a detailed proof that it is not exact? I would prefer a proof that doesn't use results about conservative vector fields (I know that way). I have searched Jefree Lee, Do Carmo & some others, but they all skip this part.  Thanks a lot.</p>

答:
Yes, everyone's doing the standard FTC proof we all know. Instead, let's try to integrate to find a potential function $f$. In the right half-plane you get $f(x,y) = \arctan(y/x)$. Now, to get a continuous continuation of this into the upper left half-plane, we need to use
$g(x,y) = \arctan(x/y)+\pi/2$. To continue this into the lower left half-plane, we need
$h(x,y) = \arctan(y/x)+\pi$, and, finally, into the lower right half-plane, we need $k(x,y)=\arctan(x,y)+3\pi/2$. So, piecing, these together, we have a continuous potential function defined on $\mathbb R^2 - \{(x,0): x>0\}$. But $f(1,0) = 0$ and $\lim_{(x,y)\to (1,0)} k(x,y) = 2\pi$, so there's no continuous potential function on $\mathbb R^2-\{(0,0)\}$.

Of course, this $2\pi$ is the same thing we got by integrating the closed $1$-form around the circle. It represents the fact that the polar angle has made one full revolution.

Now, how do I know my construction is essentially unique and that there is no cleverer way around it?  Potential functions on a connected set are unique up to constants. So, up to shifting by a constant, my attempt is the only possible function.

ContentLicense: CC BY-SA 3.0

Title: Classic example of a non exact form


问:
Combinations' Problem
A person has six friends and during a certain vacation, he met them during several dinners. He found that he dined:-

with all the six exactly on one day,

with every five of them on $2$ days,

with every four of them on $3$ days,

with every three of them on $4$ days, and,

with every two of them on $5$ days.

Further every friend was present at $7$ dinners and every friend was absent at seven dinners.
How many dinners did he have alone?

I got puzzled with so much of information and couldn't get the answer. Please help.

答:
If he ate with $\ell\ge k$ friends, that counts as eating with $k$ friends in $\binom{\ell}{k}$ different ways. You can then use the principle of inclusion-exclusion to determine the number of days he ate with exactly $k$ friends. The expression that results is
$$
\sum_{\ell=k}^6 (-1)^{\ell-k}d_\ell\binom{\ell}{k}\binom{6}{\ell},
$$
where $d_\ell$ is the number of days on which he ate with each group of $\ell$ friends, that is, $d_6=1$, $d_5=2$, $d_4=3$, $d_3=4$, $d_2=5$, and $d_1=7$.

More formally, let $S$ be a subset of the friends and let $S'$ be the complementary subset. Let $E_S$ be the set of days on which only the members of $S$ were present for dinner. For $F\subseteq S'$, let $D_{S,F}$ be the set of days on which the members of $S$, the members of $F$, and possibly other friends were present for dinner. Then
$$
\lvert E_S\rvert=\lvert D_{S,\emptyset}\rvert-\sum_{\lvert F\rvert=1}\lvert D_{S,F}\rvert+\sum_{\lvert F\rvert=2}\lvert D_{S,F}\rvert-\ldots+(-1)^{\lvert S'\rvert}D_{S,S'}.
$$
Summing over all $S$ of cardinality $k$ gives the expression of the previous paragraph since there are $\binom{6}{\lvert S\rvert+\lvert F\rvert}$ ways to choose the set $S\cup F$, $d_{\lvert S\rvert+\lvert F\rvert}$ days on which that set of friends was present for dinner, and $\binom{\lvert S\rvert+\lvert F\rvert}{\lvert S\rvert}$ ways to decompose that set as a set $S$ and a set $F$.

Evaluating the sum explicitly gives

# days with exactly $6$ friends:
$$
1\cdot1\cdot1 = 1,
$$

# days with exactly $5$ friends:
$$
2\cdot1\cdot6-1\cdot6\cdot1 = 6,
$$

# days with exactly $4$ friends:
$$
3\cdot1\cdot15-2\cdot5\cdot6+1\cdot15\cdot1 = 0,
$$

# days with exactly $3$ friends:
$$
4\cdot1\cdot20-3\cdot4\cdot15+2\cdot10\cdot6-1\cdot20\cdot1 = 0,
$$

# days with exactly $2$ friends:
$$
5\cdot1\cdot15-4\cdot3\cdot20+3\cdot6\cdot15-2\cdot10\cdot6+1\cdot15\cdot1 = 0,
$$

# days with exactly $1$ friend:
\begin{align}
&7\cdot1\cdot6-5\cdot2\cdot15+4\cdot3\cdot20-3\cdot4\cdot15+2\cdot5\cdot6\\
&\quad-1\cdot6\cdot1 = 6.
\end{align}

This accounts for $13$ of the $14$ days, so he ate alone on one day.

Note that there is a more compact expression for the sum of these expressions over all values of $k$ (from $1$ to $6$):
\begin{align}
\sum_{k=1}^6\sum_{\ell=k}^6(-1)^{\ell-k}d_\ell\binom{\ell}{k}\binom{6}{\ell}&=\sum_{\ell=1}^6 (-1)^\ell d_\ell\binom{6}{\ell}\sum_{k=1}^\ell(-1)^k\binom{\ell}{k}\\
&=-\sum_{\ell=1}^6 (-1)^\ell d_\ell\binom{6}{\ell}\\
&=-(-7\cdot6+5\cdot15-4\cdot20+3\cdot15-2\cdot6+1\cdot1)\\
&=13.
\end{align}

The lengthier calculation, however, has the advantage that it shows that all the  numbers fall between $0$ and $14$, which is necessary for the problem to make sense.

Title: Combinations' Problem


问:
Prove that the set of Turing-decidable languages are closed under reversal.
I.e., prove if $L$ is Turing-decidable, then $L^ℛ=\{x\in\Sigma^* | x^ℛ\in L, \text{ where }x^ℛ\text{ is the reverse of }x\}$ is Turing-decidable. I know that the statement is true, but struggling with how to prove it formally.

答: If $L$ is decidable, then we can decide $x \in L^{\mathscr{R}}$ by reversing $x$ and checking if the result is in $L$.


答:
Consider
$$I=\int \cos( x^3)\,dx=-\frac{1}{6} x \left(E_{\frac{2}{3}}\left(-i x^3\right)+E_{\frac{2}{3}}\left(i
   x^3\right)\right)$$
$$J=\int_0^t \cos( x^3)\,dx=\frac{\Gamma \left(\frac{1}{3}\right)}{2 \sqrt{3}}-\frac{1}{6} t \left(E_{\frac{2}{3}}\left(-i t^3\right)+E_{\frac{2}{3}}\left(i t^3\right)\right)$$ For large values of $t$, the last term is
$$t \left(E_{\frac{2}{3}}\left(-i t^3\right)+E_{\frac{2}{3}}\left(i t^3\right)\right)\sim \frac{i e^{-i t^3}}{6 t^2}$$
$$K=\int_0^\infty \cos( x^3)\,dx=\frac{\Gamma \left(\frac{1}{3}\right)}{2 \sqrt{3}}$$

Title: Does $\int_0^{+\infty}\cos(x^3)\,\mathrm{d}x$ converge absolutely or conditionally?

Body:
$$\int\limits_0^{+\infty}\cos(x^3)\,\mathrm{d}x$$

I think that the integral absolutely does not converge, since the function does not tend to zero, but conditionally it converges, since if you look at the graph of the function, the areas of the cosine humps decrease, and the signs alternate, and then the Leibniz integral converges, but I think this solution is not strict and I would like to get a normal one, but there are no ideas.


问:
How to interpret a range from 0 to 600 to a range between 1 to 0
Sorry for the poor wording, I'm not sure how to describe this.

As the value $x$ increases from $0$ to $600$, how can I make the value $y$ to range from $1$ to $0$. So for example, when $x = 0$, $y = 1$, and when $x=600$, $y = 0$ (and if $x=300$ then $y = 0.5$, and so on.) Any pointers in the right direction would be helpful.

答:
There might be many ways to achieve this, here's one:

Define $y = 1 - \frac{x}{600}$


问:
Find all of the solutions of $4(x,y) = (1,2)$
Find all of the solutions of $4(x,y) = (1,2)$ in $\mathbb{Z}_3$ x $\mathbb{Z}_8$. What I have so far:

$4(x,y) = (1,2) \implies (4x,4y) = (1,2) \implies 4x = 1 \pmod 3$ and $4y = 2 \pmod 8$.

答: Remember that you are in $\mathbb{Z_3}$ for $x$. Thus $x \in \{0,1,2\}$ and similarly $y \in \{0,1,2,3,4,5,6,7\}$. Thus by checking case by case we have: $x = 1$. Observe that $4y = 2 \pmod 8 \implies 2y = 1\pmod 4$, and this equation has no solution in $y$. Thus there is no solution at all.


答:
Find the equation of the line through the points (0,1) and (600,0).

In this case, if a line has $x$-intercept $a$ and $y$-intercept $b$, then the equation of the line is

$$\frac xa + \frac yb = 1$$

In your case this would be

$$\frac{x}{600} + \frac y1 = 1$$

$$x + 600y = 600$$

$$y = 1 - \frac{1}{600}x $$

Title: How to interpret a range from 0 to 600 to a range between 1 to 0

Body:
Sorry for the poor wording, I'm not sure how to describe this.

As the value $x$ increases from $0$ to $600$, how can I make the value $y$ to range from $1$ to $0$. So for example, when $x = 0$, $y = 1$, and when $x=600$, $y = 0$ (and if $x=300$ then $y = 0.5$, and so on.) Any pointers in the right direction would be helpful.


问:
How to determine if a relation is an equivalence relation?
So I am trying to figure out which of these are equivalence relations. I think I understand the basic idea in the sense that I need to check and see if each one is reflexive, transitive, and symmetric, but I am unsure of how to proceed with these particular relations. Any ideas here?

1.) Let $A$ be a set and $B$ be a subset of $A$. Define $R$ on $\mathcal{P}(A)$ by $X ~R~ Y$ iff $X \cap B = Y \cap B$.

2.) Define $R$ on $\mathbb{N}$ by $m ~R~ n$ iff $2 | mn$.

答:
Definitions are not ornaments. They are the tools to be used. "Reflexive, symmetric, transitive" are merely the $names$ of the tools.

$R$ is reflexive iff $(\bullet)\;\forall x\, (xRx).$

$R$ is symmetric iff $(\bullet\bullet)\;\forall x,y\,(xRy\implies yRx).$

$R$ is transitive iff $(\bullet\bullet\bullet)\;\forall x,y,z\,(\,[xRy\land yRz]\implies xRz).$

So which, if any, of $(\bullet),(\bullet\bullet),(\bullet\bullet\bullet)$ are true?

Someone else on this site said "Definitions are your friends".

For 1). Reflexive: If $x\subseteq A,$ let $y=x.$ Then $xRx \iff xRy \iff x\cap B=y\cap B\iff x\cap B=x\cap B.$

Symmetric: If $x,y\subseteq A$ let $x'=y$ and $y'=x.$ Then $$xRy\implies x\cap B=y\cap B\implies$$ $$\implies x'\cap B=y\cap B=x\cap B=y'\cap B\implies x'Ry'\implies yRx.$$

Transitive: $[xRy\land yRz]\implies$ $ [x\cap B=y\cap B\,\land \, y\cap B=z\cap B]\implies$ $\implies [x\cap B=y\cap B=z\cap B]\implies$ $ [x\cap B=z\cap B]\implies xRz.$


问:
What is the fastest way to find the characteristic polynomial of a matrix?
Finding the characteristic polynomial of a matrix of order $n$ is a tedious and boring task for $n > 2$.

I know that:

the coefficient of $\lambda^n$ is $(-1)^n$,

the coefficient of $\lambda^{n-1}$ is $(-1)^{n-1}(a_{11} + a_{22} + \dots + a_{nn})$,

the constant term is $\det{A}$.

When finding the coefficient of the linear term $\lambda$ of the characteristic polynomial of a $3\times 3$ matrix, one has to calculate the determinant of the matrix $A - \lambda I_n$ anyway. (But you don't have to sum all the terms, only the linear terms.)

Does anybody know a faster way?

答:
Once upon a less enlightened time, when people were less knowledgeable in the intricacies of algorithmically computing eigenvalues, methods for generating the coefficients of a matrix's eigenpolynomial were quite widespread. One of the more prominent methods for computing the coefficients was a method ascribed to both the Frenchman Leverrier, and the Russian Faddeev (who was an (co-)author of one of the oldest references on the practice of numerical linear algebra).

The (Faddeev-)Leverrier method is a method that will require you to do a number of matrix multiplications to generate the coefficients of the characteristic polynomial. Letting the $n\times n$ matrix $\mathbf A$ have the monic characteristic polynomial $(-1)^n \det(\mathbf A-\lambda\mathbf I)=\lambda^n+c_{n-1}\lambda^{n-1}+\cdots+c_0$, the algorithm proceeds like so:

$\mathbf C=\mathbf A;$

$\text{for }k=1,\dots,n$

$\text{if }k>1$

$\qquad \mathbf C=\mathbf A\cdot(\mathbf C+c_{n-k+1}\mathbf I);$

$c_{n-k}=-\dfrac{\mathrm{tr}(\mathbf C)}{k};$

$\text{end for}$

If your computing environment can multiply matrices, or take their trace (sum of the diagonal elements, $\mathrm{tr}(\cdot)$), then you can easily program (Faddeev-)Leverrier. The method works nicely in exact arithmetic, or in hand calculation (assuming you have the stamina to repeatedly multiply matrices), but is piss-poor in inexact arithmetic, as the method tends to greatly magnify rounding errors in the matrix, ever yielding coefficients that become increasingly inaccurate as the iteration proceeds. But, for the simple $3\times 3$ case envisioned by the OP, this should work nicely.

People interested in this old, retired method might want to see this paper.


问:
Method for estimating number of fish in my Aquarium
Context: I have an aquarium (50 Liters) with lots of fishes from the same specie (Guppy) and i have noticed that recently their number as grown exponentially due to their reproductive cycle. I would like to estimate their number.

Problem: It's not possible for me to count them individually, and i cannot dry the aquarium due to the fact that the newborns are translucid and have sizes bellow 4 milimeters, it would be an impossible task to distinguish them from the grains of smalls rocks in my aquarium.

Pre-Condition: Due to the fact that newborns stay relatively close to the ground and as they grow they will occupy (swim) in places increasingly closer to the surface level, we can consider that the fishes are uniformly distributed in the aquarium. Fish Tank

Possible method: Take a cup of water (for example 150 milliliters) consistently from the same place of the aquarium, and count the number of fishes from inside the cup. By repeating this method multiple time i could in theory estimate the total number of fishes in my aquarium.

Question: Is this a good method? Are there any big flaws in my lines of thinking? Could you think of a better, more pratical method?

Thanks in advance for your help :)

答:
Given your assumption that the fish are evenly distributed this is a reasonable method.

You have assumed that each level of the tank has fish of a given age and that the fish are laterally evenly distributed.  Maybe the fish avoid the walls and you sample in the corner, getting a much lower than average density of fish.

Your assumption of equal distribution is not reasonable because you are sampling the old fish and if the population grows exponentially there are many more young fish.  If the fish are stratified by age the lower levels will be denser with fish than the surface.  If you have observations over a span of time you will be able to estimate the rate of exponential growth.  If you know how the age of the fish changes with level in the tank you can estimate the increase in density as you go lower and use that to estimate the total population.

It would be much better to sample the water with a random distribution in space.  That would eliminate some of the systemic bias in the last two paragraphs.  If you believe that the age distribution is tied to the level you sample at, you can use your observations to find the function between level and density.  That can improve your estimate of the total population.

Even distribution is an easy assumption to make.  It makes a number of things easier.  It is hard to defend because many things can make the distribution uneven.  If you take samples from various places you collect data that can support or deny the even distribution assumption.


问:
How to get $-4(S+\cos\theta)$ when it depends on the area of the triangle in the unit circle?
The problem is as follows:

Assume $S$ square units $(u^2)$ is the area of the region whose color is blueberry. Consider that in this unit circle the diameter of the circle is $1$ decameter. Given what it is known in the figure from below, find $-4(S+\cos\theta)$.

The choices given are as follows:

$\begin{array}{ll}
1.&\cos 2\theta\\
2.&\sin 2\theta\\
3.&-\sin\theta\cos\theta\\
4.&-\sin\theta\cos\theta\\
\end{array}$

I'd like to note that there might be some typo in the given choices as the original source it is not very clear in this aspect. Thus I'm okay with an answer as long it is well justified. My main question here is not exactly trying to get an answer matching the choices but rather a way to understand the meaning of the arrows indicating the direction of the angle in the unit circle with that being mentioned. Please read my question from below.

For this problem I'm also stuck, and the reason for that is that the angle $\theta$ I'm not sure if it comes from the $\textrm{x-axis}$ or from the vertex on the right triagle which is contiguous to the small square looking from the bottom.

When an angle goes as this, in the clockwise direction I assume it is negative right?. But it doesn't help me much on how should I properly put this information in the solution.

I mean I always get confused when that arrow goes in the clockwise direction.

Because of this weird looking figure (at least to my understanding) there is not proper way to find that area. Does it exist something that can I do?. Because of the way how I'm stuck with this it would really help a lot a solution which would solve this sort of problem.

It seems that a drawing might be required to solve this problem thus can someone help me?. Since the part which confuses me the most is the way how it is used the arrow in the clockwise direction I kindly appreciate and beg someone could explain this to me in the right way to avoid getting confused.

答:
You have to think how you can connect $S$ to $\theta$. I am assuming $\theta$ is the point $N$ for which its $cos$ value is negative. Let's say the diameter is $d$. I cannot finish it since it seems it goes nowhere. You might get some idea:

$$ S = \frac{1}{2} MH \times PN = \frac{1}{2} (OM sin\alpha) \times PN = \frac{1}{2} (\frac{d}{2}sin\alpha) \times d = \frac{d^{2}}{4}sin\alpha $$

Note that:

$$ sin(\alpha) = - cos\theta $$

Thus:

$$ -4 (S + cos\theta) = -4 (\frac{d^{2}}{4}(- cos\theta) + cos\theta) = (d^{2}-4)cos\theta  $$

Even if you change the reference point of $\theta$ from $A$ to $A'$, it seems it doesn't help simplifying this equation. Is the source of problem trustworthy?


答:
The angle $\theta$ is irrelevant. The related acute angle is relevant as well as that $cos(\theta)=-cos(RAA)$.

The area S can be expressed in terms of RAA or $\theta$:

$S=R^2\cdot cos(RAA)=-R^2\cdot cos(\theta)$

The problem is that neither of the provided answers works.

$4(S+cos(\theta))=4cos(\theta)\cdot(1-R^2)=3cos(\theta)$

Title: How to get $-4(S+\cos\theta)$ when it depends on the area of the triangle in the unit circle?

Body:
The problem is as follows:

Assume $S$ square units $(u^2)$ is the area of the region whose color is blueberry. Consider that in this unit circle the diameter of the circle is $1$ decameter. Given what it is known in the figure from below, find $-4(S+\cos\theta)$.

The choices given are as follows:

$\begin{array}{ll}
1.&\cos 2\theta\\
2.&\sin 2\theta\\
3.&-\sin\theta\cos\theta\\
4.&-\sin\theta\cos\theta\\
\end{array}$

I'd like to note that there might be some typo in the given choices as the original source it is not very clear in this aspect. Thus I'm okay with an answer as long it is well justified. My main question here is not exactly trying to get an answer matching the choices but rather a way to understand the meaning of the arrows indicating the direction of the angle in the unit circle with that being mentioned. Please read my question from below.

For this problem I'm also stuck, and the reason for that is that the angle $\theta$ I'm not sure if it comes from the $\textrm{x-axis}$ or from the vertex on the right triagle which is contiguous to the small square looking from the bottom.

When an angle goes as this, in the clockwise direction I assume it is negative right?. But it doesn't help me much on how should I properly put this information in the solution.

I mean I always get confused when that arrow goes in the clockwise direction.

Because of this weird looking figure (at least to my understanding) there is not proper way to find that area. Does it exist something that can I do?. Because of the way how I'm stuck with this it would really help a lot a solution which would solve this sort of problem.

It seems that a drawing might be required to solve this problem thus can someone help me?. Since the part which confuses me the most is the way how it is used the arrow in the clockwise direction I kindly appreciate and beg someone could explain this to me in the right way to avoid getting confused.


问:
Let $E$ an $\mathbb{R}$-vectorial space and $f : E \rightarrow E $ an endomorphism. Prove that...
Let $E$ an $\mathbb{R}$-vectorial space and $f : E \rightarrow E $ an endomorphism. Prove that:

$(a)$ $dim (E)$ odd $\Rightarrow$ $f$ has at least one eigenvalue.

$(b)$ $dim (E)$ pair and $det(f) < 0 $ $\Rightarrow$ $f$ has at least two different eigenvalues.

$(c)$ It is also possible that $dim (E)$ pair and $f$ has no eigenvalue.

Any ideas to begin with?

答:
Here's a sketch/hints for you to think about:

(a) every odd degree polynomial has a real root.

(b) if $p(t)$ is the characteristic polynomial of $f$ and it has even degree, then $\lim_{t \to \pm \infty}p(t)=+\infty$, but $p(0)<0$, so the intermediate value theorem gives at least two distinct roots (one positive and one negative).

(c) reverse engineer some $f$, in the case $\dim E =2$, with $p(t)=t^2+1$.


问:
Proving an inference rule in sequent calculus

Is the following rule correct?

$\dfrac{\Phi, \varphi \Rightarrow \Delta, \psi}{\Phi, \varphi \rightarrow \psi \Rightarrow \Delta}$

I don't think it is. If I pick $\Phi = \{\top\}, \varphi = \bot, \psi = \bot$ and $\Delta = \emptyset$, then the upper sequence is an axiom (since $\varphi = \psi = \bot$) and is therefore valid, but the lower sequence isn't since there isn't anything on the right side to be satisfied ($\top \land \bot \rightarrow \bot \Rightarrow \emptyset$).

What do you think of my answer? Am I allowed to set $\varphi = \psi = \bot$ and call the upper sequence an axiom? Would really appreciate some feedback.

答:
Your argument that the inference rule is invalid is correct, but you can simplify it slightly by setting $\Phi$ to be $\emptyset$ rather than $\{\top\}$.

A sequent can have no premises or no conclusions. Having no premises is equivalent to having $\top$ as a premise and having no conclusions is equivalent to having $\bot$ as a conclusion. This can be justified by the fact that $\bigwedge \emptyset$ is $\top$ and $\bigvee \emptyset$ is $\bot$ and the fact that $\Gamma \vdash \Delta$ is equivalent to $\bigwedge \Gamma \vdash \bigvee \Delta$.

In particular $\vdash$ with no premises and no conclusions is false.

This gives us, using $F$ instead of $\bot$ so it looks a little nicer.

$$ \frac{F \vdash F}{F \to F \vdash} \;\;\text{is equivalent to}\;\; \frac{F \vdash F}{\vdash} \; \text{which is invalid} $$

I think that this inference rule is intentionally similar to the following inference rule, which is valid

$$ \frac{\Phi, \phi \vdash \psi, \Delta}{\Phi \vdash \phi \to \psi, \Delta} \;\;\text{is a valid rule that looks similar to the prompt} $$


问:
Spectrum of the sum and product of two commutative elements in a banach algebra
Let $A$ be a commutative unital Banach algebra. We know that for any two elements of $A$ $\sigma(a+b)\subset \sigma(a)+\sigma(b)$ and $\sigma(ab)\subset \sigma(a)\sigma(b)$.

I know that if we consider $2 \times 2$ matrices, then we get examples of a non-commutative unital banach algebra such that $\sigma(a)+\sigma(b)\not\subset \sigma(a+b) $ and $\sigma(a)\sigma(b)\not\subset \sigma(ab)$.

I wonder if there are examples of $a$ and $b$ in a non-unital commutative $A$ such that $\sigma(a)+\sigma(b)\not\subset \sigma(a+b) $ and $\sigma(a)\sigma(b)\not\subset \sigma(ab)$. (I can't think of any at the moment)

Thank you in advance.

答:
Let  $A=\mathbb C^2\oplus C_0(\mathbb R)$. For the sum, you could do
$$
a=\begin{bmatrix}1\\-1\end{bmatrix}\oplus0,\qquad b=\begin{bmatrix}-1\\1\end{bmatrix}\oplus0.
$$
Then
$$
\sigma(a)+\sigma(b)=\{-2,0,2\}\not\subset \{0\}=\sigma(a+b).
$$
For the product,
$$
a=\begin{bmatrix}1\\0\end{bmatrix}\oplus0,\qquad b=\begin{bmatrix}0\\1\end{bmatrix}\oplus0.
$$
Then
$$
\sigma(a)\sigma(b)=\{0,1\}\not\subset \{0\}=\sigma(ab).
$$


问:
How to deal with subtraction of sigma?
How to solve this?

$$\displaystyle\sum_{n=1}^\infty
 \frac{n}{3^{n-1}}-\displaystyle\sum_{n=1}^\infty \frac{n}{3^n}$$

My work

$$S=\displaystyle\sum_{n=1}^\infty \left(\frac{n}{3^{n-1}}-\frac{n}{3^n}\right)$$
$$S=\displaystyle\sum_{n=1}^\infty\frac{3n-n}{3^n}=\displaystyle\sum_{n=1}^\infty \frac{2n}{3^n}$$

which seems to yield nothing.

答:
You can do differentiation on series. So if you know (this is a basic geometric series)
$$\sum_{n=1}^{\infty}r^n = \frac{r}{1-r}$$
Then differentiate with respect to $r$.
$$\sum_{n=1}^{\infty}nr^{n-1} = \frac{1}{(1-r)^2}$$
or, by multiplying by $r$,
$$\sum_{n=1}^{\infty}nr^{n} = \frac{r}{(1-r)^2}$$
Do you see how your series is reflected in this formula?

Title: How to deal with subtraction of sigma?


问:
Article suggestion for Russian reading
I apologize if this question does not really fit here.

I appreciate if someone can suggest math research articles in Russian (to practice reading, learning some math related terminologies etc.)

I don't want to restrict any subject or level of difficulty, but I prefer an original work which will be interesting to read. Many thanks.

答:
A nice bite-sized place to start would be the Международная Математическая Oлимпиада: https://www.imo-official.org/?language=ru.

Click on задачи, then for most recent years you can download the problems in the language of your choice. Удачи!

Update: for articles, consider the publication Квант. It was a popular science magazine aimed at school students and teachers, and its archives are freely available. Here's the collection of its mathematics articles: http://kvant.mccme.ru/rub/19A.htm

Title: Article suggestion for Russian reading


问:
How to get the area of the triangle inscribed in the unit circle when a slope is tangent?
The problem is as follows:

The circle shown in the figure from below is a unit circle. $PT$ is a line tangent to that circle on point $T$. Angle alpha measured in radians is an acute angle. Using this information, find the area of the orange shaded region which is comprised $\triangle TQR$. Assume that the length of the radius of the unit circle is of 1 decameter.

The choices given in my book are as follows:

$\begin{array}{ll}
1.&200\cos\alpha\sin\left(\frac{\pi}{6}-\frac{\alpha}{2}\right)\\
2.&100\sqrt{2}\cos\frac{\alpha}{2}\sin\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\\
3.&50\sin\alpha\sin\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\\
4.&100\sqrt{2}\sin\frac{\alpha}{2}\sin(45^\circ+\alpha)\\
\end{array}$

I'm not sure exactly how to solve this question, it seems kind of difficult to my abilities.

I tried all sorts of manipulations on the image but I can't find a proper relation to get the base of the triangle. I also considered the oportunity to use the tangent secant theorem but this did not helped me much in the solution.

This half angle thing which is on the third quadrant it is making me feel confused. Needless to say that it is more than thirty minutes and I don't think I am getting anywhere with this problem.

Can someone help me here?.

I am totally lost here. Since I'm not good at spotting things it would help a lot that a drawing could accompany the solution as I guess that it is needed some labels to see where should additional or rearranging the angles would be needed to solve this problem.

答:
Coordinate of $Q$ is $(-1, 0)$.

Since $\angle AOR = \alpha$, coordinate of $R$ is $(\cos \alpha, -\sin \alpha)$.

Since $\angle TOA = \frac{\pi}{2}-\alpha$, coordinate of $T$ is $\left( \cos\left(\frac{\pi}{2}-\alpha\right), \sin \left( \frac{\pi}{2}-\alpha \right) \right)=(\sin \alpha, \cos \alpha)$.

The corresponding area is

\begin{align}
\frac12 \cdot 100 \cdot \begin{vmatrix} -1 & \cos \alpha  & \sin \alpha \\ 0 & -\sin \alpha  & \cos \alpha \\ 1 & 1  & 1\end{vmatrix} &= 50 \cdot (\sin \alpha + \cos \alpha  + \cos^2 \alpha +\sin^2 \alpha) \\
&= 50 \cdot \left[(1 + \sqrt2 \sin \left(\alpha + \frac\pi4 \right)\right] \\
&= 50 \sqrt{2}\left[\sin\left( \frac{\pi}{4} \right) + \sin \left( \alpha + \frac{\pi}4\right)\right] \\
&= 100 \sqrt2 \cos \left( \frac{\alpha}2\right)\sin \left( \frac{\alpha}2\color{red}+\frac{\pi}4\right)
\end{align}


问:
Prove that the open first quadrant in $\mathbb{R}^d$ is an open subset of $\mathbb{R}^d$
Let $Q$ be the “open first quadrant” in $\mathbb{R}^d,$ i.e., $Q=\{x= (x_1, . . . , x_d) \in \mathbb{R}^d:x_1, . . . , x_d>0\}.$ Prove that $Q$ is an open subset of $\mathbb{R}^d.$

A set is open if for every $x \in Q,$ there is an $r>0$ such that $B_r(x) \in Q.$ To start the proof, I think that I should let some variable, say $y$, be $y \in B_r(x),$ and try to show that $y \in Q.$ So by definition of an open ball of radius $r,$ with $r>0$, we have $|x-y|<r$. I am not sure how to use this to get to $y\in Q$ though. Is this the right path ?

答:
(To use this proof, you must apply some theory.) Let $S_k$ be the inverse image of $(0,\infty)$ by the $k$-th projection. (Projections are continuous.)
$$Q=S_1\cap...\cap S_k$$


答:
I suspect that you're supposed to shift the index of one of the series. What you've done is 100% valid, but as you said, it's not yielding something "useful". Instead, take the series
$$\sum_{n=1}^\infty \frac{n}{3^{n-1}},$$
and shift the index. Let $m = n - 1$. Then when $n = 1$, $m = 0$, so
$$\sum_{n=1}^\infty \frac{n}{3^{n-1}} = \sum_{m=0}^\infty \frac{m+1}{3^{m}} = \sum_{n=0}^\infty \frac{n + 1}{3^n} = 1 + \sum_{n=1}^\infty \frac{n + 1}{3^n}.$$
Thus,
$$\sum_{n=1}^\infty \frac{n}{3^{n-1}}-\sum_{n=1}^\infty \frac{n}{3^n} = 1 + \sum_{n=1}^\infty \frac{n + 1}{3^n} - \sum_{n=1}^\infty \frac{n}{3^n} = 1 + \sum_{n=1}^\infty \frac{1}{3^n},$$
which is a geometric series that you can calculate easily.

Title: How to deal with subtraction of sigma?

Body:
How to solve this?

$$\displaystyle\sum_{n=1}^\infty
 \frac{n}{3^{n-1}}-\displaystyle\sum_{n=1}^\infty \frac{n}{3^n}$$

My work

$$S=\displaystyle\sum_{n=1}^\infty \left(\frac{n}{3^{n-1}}-\frac{n}{3^n}\right)$$
$$S=\displaystyle\sum_{n=1}^\infty\frac{3n-n}{3^n}=\displaystyle\sum_{n=1}^\infty \frac{2n}{3^n}$$

which seems to yield nothing.


答:
You can prove this using the Green's function for the vector derivative.

There exists a Green's function $G$ for the vector derivative $\partial$:

$$\partial \cdot G = \delta \implies G(r) = \frac{r}{2\pi |r|^2}$$

The Hodge dual of this field is what you're looking for.

$$\partial \cdot G = \delta \implies \partial \wedge (Gi) = i \delta$$

All we have to do then is compute the Hodge dual $Gi$:

$$\omega = Gi = \frac{ri}{2\pi |r|^2} = \frac{x \sigma^y - y \sigma^x}{2 \pi |r|^2}$$

where $\sigma^x, \sigma^y$ are basis covectors.  Thus, $\partial \wedge \omega = i\delta$, the (dual of the) Dirac delta function, and $\omega$ is therefore not exact.

Title: Classic example of a non exact form

Body:
Let $\dfrac{xdy-ydx}{x^2+y^2}$ be a 1-form defined in $\mathbb{R}^2\backslash\{0\}$.
Where can I find a detailed proof that it is not exact? I would prefer a proof that doesn't use results about conservative vector fields (I know that way). I have searched Jefree Lee, Do Carmo & some others, but they all skip this part.  Thanks a lot.</p>


问:
is the math induction proof wrong?
Prove that for each positive integer n,

$$O_n+2=O_{n+1}$$Here $$O_n$$ is odd positive integer. That is$$O_1=1$$ $$O_2=3$$ etc.

My attempt:
Base Case: $$O_1+2=3$$ = $$O_2$$
$$
Inductive step: Suppose $$P(k) =$$O_{k}+2=O_{k+1}$$ is true

Adding 2 to both sides we get,

$$(O_k +2)+2=O_{k+1}+2$$

this implies $$O_{k+1}+2=O_{k+2}$$
I am doubtful about the last step. I think it is wrong? Thanks in advance.

答: Your last step is correct, and you showed that $P(k+1)$ is true and by induction $P(n)$ is true for all $n \ge 1$ which means that $O_n +2 = O_{n+1}$ is true for all $n \ge 1$.

Title: is the math induction proof wrong?


问:
What is a suitable strict Lyapunov function for this non-linear system?
I have verified using the total energy function that around $(0,0)$ the nonlinear system is stable:

\begin{align}\dot x&=y \\ \dot y &= -x + \beta x^3 - \gamma y \end{align}

where $\beta, \gamma > 0$ are constants. However, I have been trying to find a suitable strict Lyapunov function $V$ but from the expressions I have come up with so far, I cannot definitively deduce that the derivative is less than zero (If I use linearization, then $(0,0)$ is an asymptotically stable sink. I have tried the classical $V = x^2 + y^2$ and tried changing up the coefficients and exponents so that I can have some cancellations of the odd terms. It has been several hours now and still no luck. Any hints will be much appreciated. Moreover, the total energy function that I use for this system is $H(x,y) = \frac{y^2}{2} + \frac{x^2}{2} - \frac{\beta x^4}{4}$.

答:
A Lyapunov function can be obtained from the linearized system.
The linearized system is
$$\dot x= y,$$
$$\dot y=-x - \gamma y$$
or
$$
\left(\begin{array}{c}\dot x\\\dot y\end{array}\right)=
A \left(\begin{array}{c}x\\y\end{array}\right),
$$
where
$$
A=\left(\begin{array}{rr}0&1\\-1&-\gamma
\end{array}\right).
$$
Solving the Lyapunov equation $A^TP+PA=-Q$, where $P$ is an unknown matrix,
$Q$ is any positive-definite matrix, one can obtain the solution $P$,
which can be used to construct the Lyapunov function
$$
V(x,y)=\left(\begin{array}{cc}x&y\end{array}\right)
P
\left(\begin{array}{r}x\\y
\end{array}\right).
$$
For example,
$$
Q=\left(\begin{array}{rr}2&0\\0&2
\end{array}\right)
$$
gives us
$$
P=\left(\begin{array}{cc}\frac2{\gamma}+\gamma&1\\1&\frac2{\gamma}
\end{array}\right).
$$
Thus, one of the possible Lyapunov functions is
$$
V(x,y)=\left( \frac2{\gamma}+\gamma\right)x^2+2xy+\frac2{\gamma}y^2.
$$


问:
Why is the formula for the area of a cardioid $ \int_a^b \frac{1}{2} r^2 d \theta$
I've seen this expression in many places :$\int_a^b \frac{1}{2} r^2 d \theta$ and was wondering if someone can explain where this came from? I've noticed that it's sometimes explained in conjunction with double integrals but I'm not very familiar with double integrals (as in what it means using a picture). </p>

答:
All area calculations are double integrals. However, when we are given a curve that bounds our area we can make simplify our integral and convert it into a single integral.

In rectangular coordinates we do this when we calculate an area under a curve. Technically, we start with $$\int_{a}^{b}\int_{0}^{f(x)}dydx$$  and then do the integral in the $y$ direction to get $$\int_{a}^{b}f(x)dx$$

We can do the same thing in polar coordinates. We start with $$\int_{a}^{b}\int_{0}^{r(\theta)}\rho d\rho d\theta$$ and then we integrate in the $\rho$ direction to get $$\int_{a}^{b}\frac{r^2(\theta)}{2}d\theta $$


问:
Exercise on Venn diagrams.
At an international conference on environmental pollution, they attend

90 specialists, of which 33 speak English, 52 Portuguese and 45 Spanish;

28 of them speak Portuguese and English; 22 English and Spanish; 15 Portuguese and

Spanish. How many attendees speak the three languages?

Who help me, with this. Please.

答: The 3 way intersection of 25 noted above in various responses is greater than 2 of the 2 way intersections of 22 and 15. That's a problem.</p>

Title: Exercise on Venn diagrams.


问:
Conflicting definitions of instanton number
In Nakahara's "Geometry, Topology and Physics" (and many other sources) the instanton number of an $SU(2)$ instanton $A$ with curvature $F^A$ is defined by

$$\int_{S^4}\text{ch}_2(E)=\frac{1}{2}\int_{S^4}\text{tr}(\frac{iF^A}{2\pi})^2$$
I came across https://esc.fnwi.uva.nl/thesis/centraal/files/f1909314163.pdf which is a thesis that discusses instantons, and on page 75 it performs a calculation that uses
$$\frac{1}{4\pi^2}\int_{S^4}|F^A|^2dV_g$$
as the definition for the instanton number. I do not understand how to reconcile these two definitions. By the linearity of the trace operator, Nakahara's definition can be written as
$$-\frac{1}{8\pi^2}\int_{S^4}\text{tr}(F^A)^2$$

Could anyone explain to me how these two definitions are equivalent? In particular, it isn't clear to me what $|F^A|^2$ precisely means. I think this is the main source of confusion.

答:
The curvature $F_A$ is locally a 2-form with values in $\mathfrak{su}(2)$. To define its norm we combine the Riemannian metric with the invariant inner product on $\mathfrak{su}(2)$ given by $\langle A, B\rangle = Tr(AB^*) = -Tr(AB)$. Explicitly, if locally $F_A = F_{ij}dx^i\wedge dx^j$ for $F_{ij}\in\mathfrak{su}(2)$, then
$$
|F_A|^2 = -Tr(F_{ij}F_{kl})(g^{ik}g^{jl} - g^{il}g^{jk})
$$
This is expressed in coordinate-free terms using the Hodge star:
$$
-Tr(F_A \wedge * F_A) = |F_A|^2 dV_g.
$$

On the other hand we can write $F_A = F_A^+ + F_A^-$ as the sum of self-dual and anti-self-dual parts. Since a self-dual form wedged with an anti-self-dual 2-form is zero, one finds
$$\begin{align}
-Tr(F_A\wedge F_A) &= -Tr(F_A^+\wedge F_A^+) - Tr(F_A^- \wedge F_A^-)\\
&=(|F_A^+|^2 - |F_A^-|^2)dV_g
\end{align}$$
by replacing $F_A^\pm$ by $\pm * F_A^\pm$  in one factor of each term in the first line.

By orthogonality of $F_A^+$ and $F_A^-$, we have $|F_A|^2 = |F_A^+|^2 + |F_A^-|^2$. Hence
$$\begin{align}
\frac{1}{8\pi^2}\int|F_A|^2 dV_g &= \frac{1}{8\pi^2}\int (|F_A^+|^2 + |F_A^-|^2)dV_g\\
&\geq \frac{1}{8\pi^2} \int(|F_A^+|^2 - |F_A^-|^2)dV_g\\
&=-\frac{1}{8\pi^2}\int Tr(F_A^2),
\end{align}$$
with equality if and only if $F_A^- = 0$, i.e. $F_A$ is self-dual.

The definition from Nakahara is the "right" one, in the sense that it is determined topologically and does not depend on the connection $A$. The above shows that it is equal to the definition in the thesis you cite (modulo a factor of 2) in the case of a connection with self-dual curvature, which seems to be what is under consideration in the thesis. Moreover, the above is the justification of the standard fact that the self-dual instantons are exactly the global minimizers of the Yang-Mills energy $\int |F_A|^2 dV_g$.

Note that the factor of 2 might be accounted for by a different convention on the norm in $\mathfrak{su}(2)$, where it is also natural to set $\langle A, B\rangle = -\frac{1}{2}Tr(AB)$.


答:
I consider the other answers as superior to the approach that I am going to document.  I offer it merely as an alternative perspective.

For $r \in (0,1),~~$ let $~~A = \sum_{k = 0}^\infty r^k = \frac{1}{1-r}~~$ and let $~~B = \sum_{k = 1}^\infty r^k = \frac{r}{1-r}.$

Let $T = \sum_{k=1}^\infty kr^k ~~\implies
~~T = r + 2r^2 + 3r^3 + 4r^4\cdots $

$= B + rB + r^2B + r^3B + \cdots = B(1 + r + r^2 + r^3 + \cdots) = BA.$

Title: How to deal with subtraction of sigma?


问:
For a set $E$ of finite Lebesgue measure, $\lim_{y \to 0} m((E + y) - E) = 0$
I am having some trouble proving the following:

Let $E \subseteq \mathbb{R}$ have finite measure. Then  $\lim_{y \to 0} m((E + y) - E) = 0$

Context: I am studying the Fourier transform, and I am trying to prove the following lemma:

Suppose that $f \in L^p(\mathbb{R})$. Then $\lim_{y \to 0}||f(x-y) - f(x)||_p = 0.$

My plan is to prove it for simple functions first, and then use their density in $L^p$ to prove it for all functions. But to prove it for simple functions, I have to prove that $\lim_{y \to 0} m((E + y) - E) = 0$, which I am having trouble with.

Thank you very much for any help!

答: Hint: Use the approximation thorem of measure theory: Theorem D, p. 56 of Halmos' Measure Theory,  to reduce this to the case when $E=(a,b]$.


问:
Express the series $\sum^\infty_{n=1} \frac{n}{(1)(3)(5)(7)...(2n+1)}$ as a telescoping series
The task now is to express $\sum^\infty_{n=1} \frac{n}{(1)(3)(5)(7)...(2n+1)}$ as a telescoping series. My attempt is to deduce a formula for simplifying $\frac{n}{(1)(3)(5)(7)...(2n+1)}$ by looking at the "shape" of the first few terms. I guess that $\frac{n}{(1)(3)(5)(7)...(2n+1)}= \frac{1}{N_n }(\frac{nC^n_0}{1}+\frac{-nC^n_1}{3}+...+\frac{(-1)^nnC^n_n}{2n+1})$, where $N_n = \frac{2^nn!}{n}$. This formula does look promising and true, though it seems difficult to prove the formula by induction and to actually use this for cancelling terms, in order to show the series is telescoping.
I think there should be(?) an even simpler way to get the formula. Any idea would be much appreciated!

答: Hint: $\dfrac{n}{1\cdot 3\cdot 5\cdots (2n+1)} = \dfrac{1}{2}\left(\dfrac{1}{1\cdot 3\cdots (2n-1)} - \dfrac{1}{1\cdot 3\cdots (2n+1)}\right)$


问:
Need help solving -
i was writing an paper on solutions of triangles when i encountered this sum -

In a $\Delta$ ABC , P is an interior point such that $\angle PAB = 10^\circ$ , $\angle PBA = 20^\circ$ , $\angle PCA = 30^\circ$ , $\angle PAC = 40^\circ$  and a , b , c are the sides of the triangle. Find the angles of the triangle and hence evaluate $cos(A-B) + cos(B-C) + cos(C-A)$

i used sine rule for the triangles but  am clueless when i proceed to calculate $\angle PBC $ or $\angle PCB $. I assumed $\angle PCB  = x $. My goal is to find  $x$.

答:
Here is what the extent of what I think we can say numerically for this set of triangles:

We are given that

$$m (\angle PAB) = 10º \ , \ m (\angle PAC) = 40º \ \Rightarrow \ m (\angle A) = 50º \ ; $$

$$m (\angle PBA) = 10º  \ \Rightarrow \ m (\angle APB) = 160º \ ; $$

$$m (\angle PCA) = 30º \  \Rightarrow \ m (\angle APC) = 110º \ ; $$

$$ \Rightarrow \ m (\angle BPC) = 90º \ . $$

If we designate  $ \ m (\angle PCB) = x \ $ , then we have $ \ m (\angle PBC) = 90º - x \ ; $ we can also conclude that $ \ m (\angle B) = 100º - x \ \text{and} \ m (\angle C) = 30º + x \ . $

$$ \\ $$

EDIT:  All right, I think Calvin Lin was right after all about using the "trig form" of Ceva's Theorem.  We still want the numerical information from above.  Applying the Theorem here gives us

$$\sin(\angle PCA) \cdot \sin(\angle PBC) \cdot \sin(\angle PAB) \ = \ \sin(\angle PCB) \cdot \sin(\angle PBA) \cdot \sin(\angle PAC)  $$

$$\Rightarrow \ \sin 30º \cdot \sin(90º - x) \cdot \sin 10º  \ = \ \sin x \cdot \sin 10º \cdot \sin 40º $$

$$ \Rightarrow \ \sin 30º \cdot \cos x   \ = \ \sin x  \cdot \sin 40º \ \Rightarrow \ \tan x   \ = \ \frac{\sin 30º}{\sin 40º} \ \approx \ 0.7779 $$

$$\Rightarrow \ m (\angle PCB) \approx 37.9º \  , \ \ m (\angle PBC) = 52.1º  $$

$$\Rightarrow \ m (\angle B) \approx 62.1º \  , \ \ m (\angle C) = 67.9º . $$

So, if I used this properly, the shape of the triangle is unique and the lengths of the sides don't enter into the answer for the question.  It looks like that Theorem is what it was intended that you apply here. (I now share CutTheKnot's lament that this theorem is not more widely taught.)

Title: Need help solving -


问:
A triangle is split into three parts as shown on the picture...
A triangle is split into three parts as shown in the picture (not drawn to scale). The areas of the three regions form an arithmetic progression (where the area of quadrilateral $A_2$ is the middle term). Using the information in the picture, find the length of the missing side of triangle $A_1$.

答:
$$A_{1}+A_{3}=2A_{2}$$

I used A and B to denote the areas of certain triangles in the figure below (I regret choosing A as a variable, it has nothing to do with the areas given in the problem statement). Let A be the leftmost triangle and let B be the triangle which is a part of $A_{2}$.

Split the areas according to the bases (since the triangles have the same height) and plug the expressions into the above equation:

$$\frac{1}{2} \left(A + \left(\frac{1}{2} (4 A + 3 B) + 2 B\right)\right) = 3 A + B, $$

$$B=2A$$

Also from area division note that:

$$\frac{3B}{3}=\frac{4A}{?}\ $$

$$2=\;?$$


问:
Number of solutions of $x+\frac{87}x=\lfloor x\rfloor+\frac{87}{\lfloor x\rfloor} $ where $x\in\mathbb{Q} , x\notin\mathbb{Z}$
How many solutions the equation $x+\frac{87}x=\lfloor x\rfloor+\frac{87}{\lfloor x\rfloor} $ has in rational non integer sets of numbers?

$1)\text{zero}\quad\quad\quad\quad\quad\quad\quad2)1\quad\quad\quad\quad\quad\quad\quad3
)2\quad\quad\quad\quad\quad\quad\quad4)87\quad\quad\quad\quad\quad\quad\quad5)\text{infinity}$

So the solution can't be an integer therefore $\lfloor x\rfloor\ne x$ and generally we can say $\lfloor x\rfloor< x$. also denominator of the fractions can't be zero so  $\lfloor x\rfloor \neq0$ hence $x\notin[0,1)$

I wrote the equation as:

$$x-\lfloor x\rfloor=\frac{87}{\lfloor x \rfloor}-\frac{87}{x}=\frac{87(x-\lfloor x\rfloor)}{x\lfloor x\rfloor}$$

But can't see a way to continue.

Body:
$$x+\frac{87}x=\lfloor x\rfloor+\frac{87}{\lfloor x\rfloor}$$

Let $x = n + \delta$ where $x \in \Bbb Z$ and $\delta \in (0,1)$

$$n+\frac{87}n=(n+\delta)+\frac{87}{n+\delta}$$

$$(n+\delta)(n^2 + 87) = n((n+\delta)^2 + 87) $$

$$n^3 + \delta n^2 + 87n + 87 \delta = n^3 + 2 \delta n^2 + \delta^2 n + 87n$$

$$\delta n^2 + \delta^2 n - 87\delta = 0$$

$$\delta n = 87-n^2$$

If $n > 0$

$$ 0 \lt 87 - n^2 < n$$

$$n \in \{1,2,\dots, 9\} \wedge n \in \{9, 10, \dots   \}$$
$$ n = 9$$

If $n < 0$

$$ n < 87-n^2 < 0$$

$$n \in \{-9, -8, \dots, -1\} \wedge n \in \{\dots -12,-11, -10\}$$

$$n \in \emptyset$$

For $n = 9$ we get $\delta = \dfrac{87-n^2}{n} = \dfrac 23$.

So the only solution is $x = 9\frac 23 = \frac{29}{3}$.


问:
Artin exercise 2.4.5

Prove that every subgroup of a cyclic group is cyclic. Do this by working with exponents, and use the description of the subgroups of $\mathbb{Z}^+$.

Having taken a look at a solution after having struggled with this problem, I'm very confused about the hint. There is no guarantee that $S$ be a subgroup of $\mathbb{Z}^+$, e.g., it could contain non-integers, so I don't know where I need to use the fact that all subgroups of the positive integers are of the form $b\mathbb{Z}$.

A hint on how to use this approach would be appreciated.

答:
You already know that if $H$ is a subgroup of $\mathbb{Z}$, then $H=a\mathbb{Z}$ for some unique $a\ge 0$.

Want to prove: If $G$ is a cyclic group and $H$ is a subgroup of $G$, then $H$ is cyclic too.

Let $G$ be a cyclic group and let $H\le G$ (notation for subgroup). Can you find a a surjective homomorphism $f\colon\mathbb{Z}\to G$?

Then $H=f(f^{-1}(H))$ and, since $f^{-1}(H)=a\mathbb{Z}$ is cyclic, $H$ is also cyclic, since it's the homomorphic image of a cyclic group (prove this).

Title: Artin exercise 2.4.5


问:
Check if nD point lies within nD Hypersphere defined by n+1 points
I am writing some code where I need to be able to determine if a specific n-D point lies within a n-D hypersphere defined by n+1 points. The 2D version of this is clear to me; to check if a point, p,  lies within a circle defined by a triangle abc, one can just compute the determinate of the following matrix:

[a_x, a_y, |a|, 1]

[b_x, b_y, |b|, 1]

[c_x, c_y, |c|, 1]

[p_x, p_y, |p|, 1]

(|n|= n_x^2+n_y^2)

if the determinate is greater than zero then the point lies outside of the circle, if its zero, its on the circle, else it is within the circle.

I am hoping that this can scale up to how ever many dimensions I need, but I am unsure if it will as I do not know what would go in the additional column would be, obviously the new row would represent the new point.

Any help or suggestion would be greatly appreciated! thanks!

答:
You can solve for the center of the circumsphere of the simplex defined by the $n+1$ points, as follows:

Suppose your circumcenter is $c.$ Then

$$\|c - v_i\| = R^2,$$ for all $i.$ You don't know $R^2,$ but it does follow that

$$\|c-v_i\| = \|c-v_j\|,$$ for all $i, j.$ Expanding the square of the norm as inner product, you see that:

$$\|v_i\|^2 - \|v_j\|^2 = 2 \langle c, v_i - v_j\rangle,$$ for every $i, j.$ Now, by a translation, put $v_0$ at the origin, so you have
$$\|v_i\|^2 = 2 \langle c, v_i\rangle$$ for every $i=1, \dotsc, n.$
You can solve this system of linear equations for $c$ and then compute $R$ from the very very first equation above. Now, for your new point, just check that the distance to $c$ is less than $R.$


问:
Counting numbers smaller than $N$ with exactly $k$ *distinct* prime factors
Using common notation, $\omega(n)$ is the number of distinct prime factors on $n$. Similiarly, $\Omega(n)$ is the number of prime factors of $n$, not necessarily distinct: $120=2^{3}\cdot 3 \cdot 5$ , therefore $\omega(120)=3$ , $\Omega(120) = 5$.

If we let $W_k(N)$ be the count of numbers not exceeding $N$ with exactly $k$ distinct prime factors {$i | i \leq N , \omega (i) = k$}, is there any way or formulation to calculate it? Or at least a relatively coincise algorithm for it, which does not include factoring each number and checking for $\omega$ individually?

For example, one can easily find that $W_3(100)=8$ , $W_3(1000000)=379720$ , $W_4(1000000)=208034$.

Please note here I am not looking for an asymptotic approximation but rather an exact count. And $N$ can be large, with order of magnitude of $10^{12}$ and over.

I have found that a solution exists for the $\Omega$ counterpart, known as k-Almost Primes, which admits closed-form formulas, but as for $\omega$, I have found non.

答:
$k=0$

$$N_0(n) = \begin{cases}
    0 & \text{if $n < 1$} \\
    1 & \text{if $n >= 1$}
    \end{cases}$$

Since $n=1$ is the only non-negative integer with $\omega(n) = 0$, this is simple.  Any $n >= 1$ returns 1, otherwise return 0.

$k=1$

$$ \begin{align}
N_1(n) & = \text{prime_power_count}(n) \\
    & = \sum_p^n \sum_{x=1}^{\lfloor\log_{p}(n)\rfloor} 1\\
    & = \sum_p^n \lfloor\log_{p}(n)\rfloor\\
    & = \pi\left(\lfloor n^{1/2} \rfloor\right) +
        \sum_{p}^{\lfloor\sqrt n\rfloor} \lfloor\log_{p}(n)\rfloor - 1\\
    & = \sum_{k=1}^{\lfloor\log_2(n)\rfloor} \pi\left(\lfloor n^{1/k}\rfloor\right)
 \end{align}$$

These are the prime powers.  We can do a sum over the primes, then the prime powers, or just do the obvious thing and turn the prime power loop into a single integer log.  The second to last equation is noting that after $\sqrt(n)$ all the logs are 1, so we can do a single prime count for all the $p^1$ cases, leaving us only having to examine the primes up to $\sqrt(n)$.  We can continue this logic to its end and get the last equation which is a series of prime counts.  This is what a number of software packages implement, and it is quite fast given a modern prime count method.

$k=2$

$$ \begin{align}
N_2(n) & = \sum_{p}^{\lfloor\sqrt n\rfloor}
             \sum_{i=1}^{\lfloor\log_{p}(n)\rfloor}
               \sum_{q = p+1}^{\lfloor n/{p^i} \rfloor}
                 \lfloor\log_{q}(\lfloor n/{p^i} \rfloor)\rfloor && \text{$p$ and $q$ over primes} \\
       & = \sum_{p}^{\lfloor\sqrt n\rfloor}
             \sum_{i=1}^{\lfloor\log_{p}(n)\rfloor}
               \bigl(
               \pi\left(\lfloor n/{p^i} \rfloor\right) -
               \sum_q^p
                 \lfloor\log_{q}(\lfloor n/{p^i} \rfloor)\rfloor
               \bigr) && \text{$p$ and $q$ over primes} \\
 \end{align}$$

The first two sums are for summing over the prime powers.  In an implementation it's still two loops, but a bit more efficient.  We'd also calculate $n/{p^i}$ once rather than doing it over and over inside the innermost loop.  The second version is much faster given a decent prime count method, because the final sum is over a small number of primes rather than many.

$k=3$

$$ \begin{align}
N_3(n) & = \sum_{p}^{\lfloor n^{1/3} \rfloor}
            \sum_{i=1}^{\lfloor\log_{p}(n)\rfloor}
             \sum_{q=p+1}^{\lfloor n^{1/2} \rfloor}
              \sum_{j=1}^{\lfloor\log_{q}(n/{p^i})\rfloor}
               \bigl(
               \pi\left(\lfloor n/{(p^i p^j)} \rfloor\right) -
               \sum_r^q
                 \lfloor\log_{q}(\lfloor n/{(p^i p^j} \rfloor)\rfloor
               \bigr) \\
 \end{align}$$

Just showing the more efficient innermost loop here.  $p$, $q$, and $r$ run over the primes.  This looks more complicated than it really is, as again we have two sums indicating the prime powers.  We are trying to count the $p^i q^j r^k <= n$.  We are going to count in order, so $p < q < r$.  So as we loop through the powers $p^i$ we then find $q^j r^k <= n/p^i$, and so on.

The solution for $k = 4$ should be clear at this point, as we just add another outer sum pair, and continue for higher $k$.

It is possible this can be written in a better style, especially turning the double sum for primes powers into a single one, though note we use the actual prime for the start of the next loop.  It is also not unlikely this can be further simplified.

In C or Perl code, this turns into a sub-20 line recursive function that handles arbitrary k.  Albeit this is with a library providing logint, rootint, prime_power_count, prime_count, and fast looping over primes.

As a timing example:

```<code>time say omega_prime_count(2,powint(10,9)):

5.9s  efficient ranged nfactor sieve
1.2s  first double summation method
0.03s second double summation method

time for n=10^10:  0.13 seconds.  (15s for the first, 61s for the sieve)
</code>```

For $k=8$, I get omega_prime_count(8,10^8) = 719 in 66 microseconds, vs. a naive loop in Pari/GP taking 38 seconds.  omega_prime_count(8,10^12) = 953640790 takes 2.8 seconds.

Some interesting papers:

Sums over primitive sets with a fixed number of prime factors, 2019

On the counting function of semiprimes, 2020, page 12

Body:
$k=0$

$$N_0(n) = \begin{cases}
    0 & \text{if $n < 1$} \\
    1 & \text{if $n >= 1$}
    \end{cases}$$

Since $n=1$ is the only non-negative integer with $\omega(n) = 0$, this is simple.  Any $n >= 1$ returns 1, otherwise return 0.

$k=1$

$$ \begin{align}
N_1(n) & = \text{prime_power_count}(n) \\
    & = \sum_p^n \sum_{x=1}^{\lfloor\log_{p}(n)\rfloor} 1\\
    & = \sum_p^n \lfloor\log_{p}(n)\rfloor\\
    & = \pi\left(\lfloor n^{1/2} \rfloor\right) +
        \sum_{p}^{\lfloor\sqrt n\rfloor} \lfloor\log_{p}(n)\rfloor - 1\\
    & = \sum_{k=1}^{\lfloor\log_2(n)\rfloor} \pi\left(\lfloor n^{1/k}\rfloor\right)
 \end{align}$$

These are the prime powers.  We can do a sum over the primes, then the prime powers, or just do the obvious thing and turn the prime power loop into a single integer log.  The second to last equation is noting that after $\sqrt(n)$ all the logs are 1, so we can do a single prime count for all the $p^1$ cases, leaving us only having to examine the primes up to $\sqrt(n)$.  We can continue this logic to its end and get the last equation which is a series of prime counts.  This is what a number of software packages implement, and it is quite fast given a modern prime count method.

$k=2$

$$ \begin{align}
N_2(n) & = \sum_{p}^{\lfloor\sqrt n\rfloor}
             \sum_{i=1}^{\lfloor\log_{p}(n)\rfloor}
               \sum_{q = p+1}^{\lfloor n/{p^i} \rfloor}
                 \lfloor\log_{q}(\lfloor n/{p^i} \rfloor)\rfloor && \text{$p$ and $q$ over primes} \\
       & = \sum_{p}^{\lfloor\sqrt n\rfloor}
             \sum_{i=1}^{\lfloor\log_{p}(n)\rfloor}
               \bigl(
               \pi\left(\lfloor n/{p^i} \rfloor\right) -
               \sum_q^p
                 \lfloor\log_{q}(\lfloor n/{p^i} \rfloor)\rfloor
               \bigr) && \text{$p$ and $q$ over primes} \\
 \end{align}$$

The first two sums are for summing over the prime powers.  In an implementation it's still two loops, but a bit more efficient.  We'd also calculate $n/{p^i}$ once rather than doing it over and over inside the innermost loop.  The second version is much faster given a decent prime count method, because the final sum is over a small number of primes rather than many.

$k=3$

$$ \begin{align}
N_3(n) & = \sum_{p}^{\lfloor n^{1/3} \rfloor}
            \sum_{i=1}^{\lfloor\log_{p}(n)\rfloor}
             \sum_{q=p+1}^{\lfloor n^{1/2} \rfloor}
              \sum_{j=1}^{\lfloor\log_{q}(n/{p^i})\rfloor}
               \bigl(
               \pi\left(\lfloor n/{(p^i p^j)} \rfloor\right) -
               \sum_r^q
                 \lfloor\log_{q}(\lfloor n/{(p^i p^j} \rfloor)\rfloor
               \bigr) \\
 \end{align}$$

Just showing the more efficient innermost loop here.  $p$, $q$, and $r$ run over the primes.  This looks more complicated than it really is, as again we have two sums indicating the prime powers.  We are trying to count the $p^i q^j r^k <= n$.  We are going to count in order, so $p < q < r$.  So as we loop through the powers $p^i$ we then find $q^j r^k <= n/p^i$, and so on.

The solution for $k = 4$ should be clear at this point, as we just add another outer sum pair, and continue for higher $k$.

It is possible this can be written in a better style, especially turning the double sum for primes powers into a single one, though note we use the actual prime for the start of the next loop.  It is also not unlikely this can be further simplified.

In C or Perl code, this turns into a sub-20 line recursive function that handles arbitrary k.  Albeit this is with a library providing logint, rootint, prime_power_count, prime_count, and fast looping over primes.

As a timing example:

time say omega_prime_count(2,powint(10,9)):

5.9s  efficient ranged nfactor sieve
1.2s  first double summation method
0.03s second double summation method

time for n=10^10:  0.13 seconds.  (15s for the first, 61s for the sieve)

For $k=8$, I get omega_prime_count(8,10^8) = 719 in 66 microseconds, vs. a naive loop in Pari/GP taking 38 seconds.  omega_prime_count(8,10^12) = 953640790 takes 2.8 seconds.

Some interesting papers:

Sums over primitive sets with a fixed number of prime factors, 2019

On the counting function of semiprimes, 2020, page 12


问:
Rolling four dice, what are the odds of having at least a pair of sixes?
Let's say I roll four dice. I want to know the odds of having at least two 6. There can be more too.

I did: 6C1 * 6C1, which gave me 36. I assumed the total possible combinations was 1296. I don't know if it's good, I doubt it is...

How do I know what the odds are of throwing 4 dice and at least getting two 6?

答:
Hint :

"At least 2" means you can get 2 sixes on 4 tries OR 3 sixes on four tries, OR all four sixes on four tries.

Can you calculate the probabilities of these individual events?

In the end, use the fundamental principle of counting, and give the answer as the sum of all these probabilites.


问:
Did I sketch this polar curve correctly?
The equation is:

$r^2=-4 \sin(2\theta)$

I first made a reference graph in cartesian coordinates using values $\displaystyle \frac{\pi}{4}$, $\displaystyle \frac{\pi}{2}$, $\displaystyle \frac{3 \pi}{4}$, $\displaystyle \pi$. Then from that I formed this:

Something seems off about that though. Should it be across the other axis instead?

答: Give the comment from Alon some consideration, though I wasn't able to immediately eliminate your potential graph based on that idea alone.  Let's think about the interval $\frac{\pi}{2}< \theta< \pi$.  On that interval, $\sin2\theta$ goes from $0$ to $-1$ to $0$, so $r^2=-4\sin2\theta$ goes from $0$ to $4$ to $0$, so $r$ goes from $0$ to $\pm 2$ to $0$ again.  No matter how I interpret your graph, it says that $r=\pm4$ at $\theta=\frac{\pi}{2}$ or $\theta=\frac{3\pi}{2}$ (or something along those lines).</p>


问:
Interesting Problems for NonMath Majors
Sometime in the upcoming future, I will be doing a presentation as a college alumni to a bunch of undergrads from an organization I was in college. I did a dual major in mathematics and computer science, however the audience that I am presenting to are not necessarily people who enjoy math. So, to get their attention, I was thinking about presenting an interesting problem in math, for example, the birthday problem, to get their attention and have them enjoy the field of math a little.

I feel a question in the field of probability would interest them the most (due to its instinctiveness) , although that's just a personal opinion. The audience studies a variety of majors, from sciences to engineers to literature and the arts.

So here's my question, besides the birthday problem, are there any other interesting problems that would be easy to understand for people who have limited knowledge of calculus and would, hopefully see math as an interesting subject and get their attention? (Doesn't have to be in the field of probability.)

答: Here is what I think is a very interesting problem. Little children know what "one half" means, and maybe even "one third," etc. When we study fractions, the basic idea is that there are quantities in between the counting quantities. After all, "one half" is really a quantity too. So how can we denote such a quantity? The ingenious scheme is to think of the quantity "one" broken into a counting number of equal parts like, for example, two equal parts. Then we can describe some other number of those parts by another counting number. These TWO counting numbers together then represent a quantity smaller than one, e.g., one out of two equal parts. Next comes the question of how to denote such a thing. So here is the problem: There are different ways we could denote the two counting numbers, e.g., 1|2 meaning "one our of two equal parts" or 1↓2 etc. So why do we denote a fraction by DIVISION of those two counting numbers, with the total number of equal parts as the denominator and the desired number of them as the numerator? In other words, does DIVISION really have anything to do with representing a quantity as a number of equal parts? Is the use of division ingenious? Also, division is the inverse of multiplication. Is the division notation for a fraction the inverse of anything? One purpose of this question is to point out that even though we all think we understand all about fractions, we actually may not. Another is to point out that mathematical definitions are not arbitrary. They are sensible. There are ALWAYS DERIVATIONAL reasons for them. A third is that although we are all comfortable with saying that mathematics is "abstract," quantity is perfectly concrete, as is the mathematical development of its aspects. </p>

Title: Interesting Problems for NonMath Majors


问:
On a conjectured upper bound for $k=\nu_q(N)$, if $N=q^k n^2$ is an odd perfect number with special prime $q$
Let $N=q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Denote the abundancy index of the positive integer $x$ as $I(x)=\sigma(x)/x$ where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$.  (For example, when $M$ is perfect, we have $\sigma(M)=2M$ and $I(M)=\sigma(M)/M=2$.  The smallest example of an even perfect number is $6$.  It is currently unknown if there are any odd perfect numbers.)

Fix a positive integer $a$.

We have
$$k \leq a \iff I(q^k) \leq I(q^a) \iff \frac{2}{I(q^a)} \leq \frac{2}{I(q^k)}=I(n^2)$$
$$\iff I(q^k) \leq I(q^a) < \frac{2}{I(q^a)} \leq I(n^2),$$
since
$$I(q^a) = \frac{q^{a+1} - 1}{q^a (q - 1)} < \frac{q}{q-1} \leq \frac{5}{4} < \sqrt{2}.$$
Therefore, we get
$$k \leq a \iff \Bigg[I(q^k) - I(q^a)\Bigg]\Bigg[I(n^2) - I(q^a)\Bigg] \leq 0$$
$$\iff 2 + (I(q^a))^2 \leq I(q^a)\Bigg[I(q^k) + I(n^2)\Bigg].$$
We thus obtain the following proposition:

PROPOSITION If $N=q^k n^2$ is an odd perfect number with special prime $q$, then $k \leq a$ holds if and only if
$$I(q^a) + \frac{2}{I(q^a)} \leq I(q^k) + I(n^2).$$

ALTERNATIVE PROOF OF THE PROPOSITION Note that the same proposition can be proved by using the fact that
$$f(k) = I(q^k) + I(n^2) = I(q^k) + \frac{2}{I(q^k)}$$
is a decreasing function of $k$.  (See mathlove's answer to a closely related question.)

But
$$I(q^a) + \frac{2}{I(q^a)} = \frac{q^{a+1} - 1}{q^a (q - 1)} + \frac{2q^a(q - 1)}{q^{a+1} - 1} = \frac{3q^{2a+2} - 4q^{2a+1} + 2q^{2a} - 2q^{a+1} + 1}{q^a (q - 1)(q^{a+1} - 1)},$$
whose partial fraction decomposition, per WolframAlpha, is
$$\frac{2(q - 1)}{q(q^{a+1} - 1)} - \frac{1}{q^a (q - 1)} + \frac{3q^2 - 4q + 2}{q(q - 1)},$$
where of course
$$\frac{3q^2 - 4q + 2}{q(q - 1)} < I(q^k) + I(n^2)$$
holds unconditionally (i.e. even if we do not have the condition $k \leq a$).

So here is my:

INITIAL QUESTION

Does the inequality
$$\frac{2(q - 1)}{q(q^{a+1} - 1)} > \frac{1}{q^a (q - 1)}$$
hold for all special primes $q$?

MY ATTEMPT

Suppose to the contrary that there exists a special prime $q$ such that
$$\frac{1}{q^a (q - 1)} \geq \frac{2(q - 1)}{q(q^{a+1} - 1)}.$$

This last inequality is equivalent to
$$q^{a+2} - q \geq 2{q^a}(q - 1)^2 = 2{q^a}(q^2 - 2q + 1) = 2q^{a+2} - 4q^{a+1} + 2q^a$$
$$\iff 0 \geq q^{a+2} - 4q^{a+1} + 2q^a + q = q^{a+1} (q - 4) + 2q^a + q > 0,$$
which is a contradiction.

FINAL QUESTION

Does the contradiction suffice to prove that $k \leq a$?

答:
I think that the answer to the final question is no.

If you want to prove $$A−B+C\le D$$
then $$A>B\qquad\text{and}\qquad C<D$$
are not sufficient where $$A=\frac{2(q−1)}{q(q^{a+1}−1)},\qquad B=\frac{1}{q^a(q−1)},
$$
$$C=\frac{3q^2−4q+2}{q(q−1)},\qquad D=I(q^k)+I(n^2)$$


问:
Understanding if statements in EWD931
EWD931 is: https://www.cs.utexas.edu/users/EWD/ewd09xx/EWD931.PDF, which proposes an interesting structure with continuous truth values. Note therefore that the symbols I use in this question use his notation not standard notation.

He remarks that $x \land y \equiv x$ (or $x \lor y \equiv y$) are equivalent to $x \implies y$. Why is this true? I can't seem to get it.

答:
Note that at the end of the second page he defined $x\Rightarrow y$ to be a shorthand for either expressions you mentioned ($x\vee y\equiv y, x\wedge y\equiv x$), and presumably you already know those two are equivalent.

If you really wonder why this definition is made, try the degenerate case of $x,y=0,1$. In fact the top of the third page already lists quite a few of interesting properties it has.

Title: Understanding if statements in EWD931


问:
Questions related to graphing of nested functions
Here is the problem from recently concluded AIME 2021.

Find the number of integers c such that the equation $||20|x|-x^2|-c|=21$ has 12 distinct real solutions.

I know how one can solve these algebraically. However I have always found (my personal preference) that plotting the graph of the following equation: $||20|x|-x^2|= c+21$ and $||20|x|-x^2|= c-21$ and then getting a pictorial understanding a much more elegant technique.

However under the exam situation I find it difficult to plot the above 2 graphs. My question to valuable members to guide me on the following points:

How do I quickly get critical points to sketch the graph? Is there any good resource which will help me in mastering the graphing of various special functions?

Can the algebraic way be simplified for newbies like me to ensure clear understanding?

答: You can go on unpacking the absolute value signs.  When you do, you should note what assumption implies each sign.  The innermost one is easy.  If $x \ge 0$ you have $|20x-x^2|$, which is $20x-x^2$ for $0 \le x \le 20$ and $x^2-20x$ for $x \gt 20$.  Similarly if $x \lt 0$ you have $|-20x-x^2|$ which is $-20x-x^2$ for $-20 \le x \lt 0$ and $x^2+20x$ for $x \lt -20$


问:
Trigonometric definite integral question
If $$I_n:=\int_0^{\pi/4} \tan^n x dx$$

what is

$$n(I_{n+1}-I_{n-1}) ?$$

I found

$$I_{n+1}-I_{n-1}$$

But it's not a value. It still has integration in it

Body:
Using $x=\tan ^{-1}(t)$
$$I_n=\int_0^{\frac \pi 4} \tan^n (x)\, dx=\int_0^1 \frac{t^n}{t^2+1}\,dt$$
$$I_{n+1}-I_{n-1}=\int_0^1 t^{n-1} \frac {t^2-1}{t^2+1}\,dt$$
Now, you could use
$$\frac {t^2-1}{t^2+1}=-1+2 \sum_{k=1}^\infty (-1)^{k+1}t^{2k}$$making
$$I_{n+1}-I_{n-1}=-\int_0^1 t^{n-1}\,dt+2 \sum_{k=1}^\infty (-1)^{k+1}\int _0^1 t^{2k+n-1}\,dt$$
$$I_{n+1}-I_{n-1}=-\frac 1 n+2 \sum_{k=1}^\infty (-1)^{k+1} \frac{1}{2 k+n}$$
$$I_{n+1}-I_{n-1}=-\frac 1 n+\frac{1}{2} \left(\psi \left(\frac{n}{4}+1\right)-\psi   \left(\frac{n}{4}+\frac{1}{2}\right)\right)$$ Now, if $n$ is large, using the asymptotics of the digamma function
$$n(I_{n+1}-I_{n-1})=-\frac{1}{n}+\frac{2}{n^3}-\frac{16}{n^5}+\frac{272}{n^7}+O\left(\frac{1}{n^9}\right)$$

For $n=10$ the exact result is
$$\frac{41}{6}-10 \log (2)\sim -0.09813847226611976083898788$$ while the above truncated series would give
$-\frac{61333}{625000}=-0.0981328$ corresponding to a relative error equal to $0.0058$%.


问:
Prove or disprove there does not exist any triangle $\triangle {ABC}$ in which the line $IG$ is perpendicular to the line $IC$
The problem, asked 14 Feb at 18.18 by user @r ne, (see figure $1$ below) would appear to have been abandoned because of some adverse comment about “How to ask a good question”.

The attached figure $2$ below shows the triangle with sides $14, 15$ and $16$ in which the incenter and the centroid have been graphed. This triangle would seem to offer an answer to the above problem but it does not, the slopes in play appear as perpendicular to the eye but are not so with geometric accuracy.

If the problem had a solution, then taking as unknowns the three sides $a, b, c$ the coordinates of the incenter $I$ and centroid $G$ can be easily calculated so, using the known perpendicularity condition with the slopes $ m_ {IC} \cdot m_{IG} = -1$, we arrive at a polynomial equation in $a,b,c$. No matter of “diophantine” with the sides but of lengths of geometric segments taking any positive real value, we have the following question:

Prove or disprove that there is no triangle in the plane in which the aforementioned perpendicularity is fulfilled.

I have easily proven this and if I have made a mistake then it would not be subtle at all (instead something trivial like putting $2+3=6$ for example).

What strikes me is that a certain geometric intuition suggests that a small variation in the position of vertex $A$ of the given example (Figure $2$ below) of "almost perpendicularity" could "fix" the point, although with non-integer real sides, which does not matter in the problem at hand.
In this case, it would be clear that my proof has an error that I cannot see and then the question would be devoid of any interest.

But if, as I believe, my proof is correct, then the apparent conflict between the geometric intuition in figure $2$ and the fact that such a triangle cannot exist must be explained in some way that I imagine not trivial in any respect. And I think this is due to $\pi$ which is linked to both incircle and perpendicularity.

答:
An old problem answered here, asks to prove that $CI \perp IG \Rightarrow$ the following holds true.

$$6ab=(a+b)(a+b+c) \tag{A}$$

The additional conditions imposed by @rne are that $a,b,c$ be integers and $(a,b)=2$. $(16,14,15)$ does not satisfy $(A)$.

One can easily find, for example, $(a,b,c)=(14,10,11)$ satisfies all the above. Here is a drawing of this triangle in Geogebra with $\angle CIG$ measured after construction.


问:
$ \log|x-4| -\log|3x-10| = \log\left|\frac{1}{x} \right| $ one formal way to say we won't choose all solutions in degree 2 equation

$$
\log|x-4| -\log|3x-10| = \log\left|\frac{1}{x} \right|
$$

After applying some properties we have:
$$
\log\left|\frac{x^2-4x}{3x-10}\right| = 0.
$$

After applying some operations, this equation becomes:
$$
x^2-7x+10 = 0\\
x=2 ;x=5.
$$

The problem is: if we substitute $x$ with "$5$" it's good, but if we substitute with "$2$" there's an obvious error, since $2$ is out of the logarithm's domain.

Are there some 'formal' ways to say why we won't choose number 2, that just say: "we won't choose number 2 because it doesn't make sense"?

EDIT:
Here is another exercise to explain that there are some solutions that doesn't work even with absolute values in the logarithms' arguments:

$$
\log(\sqrt{x+14})+\log(\sqrt{x+7})-\log(1.2)=1 \\
\cdots\\
x^2+21x-46=0\\
x_1=-23; x_2=2.$$

But again: what happened with $-23$ ? Does it just disappear ?

Note: I'm working in $\mathbb R.$

答:
No, $x=2$ is not out of domain. If $x=2$, you get

$$\ln 2-\ln 4=\ln \left(\frac 24\right)=\ln  \left(\frac 12\right)$$ which is correct.


问:
Proof of rational number
Prove that if $a$ and $b$ are rational numbers satisfying $a^5+b^5=2a^2b^2$, then $1-ab$ is the square of a rational number.

I am just a Year 2 student learning Abstract Algebra. This problem is a challenging one that my teacher gives us. However, I have no ideas about how to solve it. I tried to represent rational numbers by $p/q$ ($p$ and $q$ are both integers) but failed. I would appreciate it very much if anyone can help me.

答:
[1] Roots of quadratic equations

We'll need to recall a few facts about the roots of quadratic equations. For any quadratic equation $ax^2  + bx + c = 0$, the roots are $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

When will the roots of this equation be rational? Suppose $a$ and $b$ are rational. If the square root $\sqrt{b^2-4ac}$ of the discriminant is also rational, then note that the roots will be rational. The converse is also true: if the roots are rational, then $\sqrt{b^2-4ac}$ must be rational.

(You can prove these if you know that sums and products and ratios of rational numbers are rational.)

[2] The given equation

You can make the given equation $a^5 + b^5 = 2a^2b^2$ look like a quadratic equation. First rewrite it as $a(a^4) - 2a^2b^2 + b(b^4) = 0$ to expose even exponents. (This is the main trick to get the solution!) Divide by $b^4$ (this is always possible as long as $b\neq 0$) to get $a(a/b)^4 - 2(a/b)^2 + b = 0$.

Now this expression looks like the quadratic equation: $ax^2 - 2x + b = 0$, evaluated at $x=(a/b)^2$.  So our given equation "Assume that $a^5 + b^5 = 2a^2b^2$" is equivalent to:

Assume that the quadratic equation $ax^2 - 2x + b = 0$ has $x=(a/b)^2$ as a root.

[3] The discriminant

We have determined that, by assumption, the equation $ax^2 - 2x + b = 0$ has $x=(a/b)^2$ as a root. This root is a rational number. Therefore (by #1), the square root of the discriminant is rational. That is, $\sqrt{(-2)^2 - 4ab}$ is rational. Dividing by 2, we have that $\sqrt{1-ab}$ is rational— which was to be shown.

[4] The edge case $b=0$

In the above manipulation, we divided by $b$. This works in every case except when $b=0$. Let's consider that case now: When $b=0$, then we are asked to prove that $1-ab = 1-0 = 1$ is the square of a rational number, which is straightforwardly true.

Title: Proof of rational number


问:
If I have three points, is there an easy way to tell if they are collinear?
Points $(a,b)$, $(m,n)$, and $(x,y)$ are selected at random. What is the quickest/easiest way to tell if they are collinear? At first I thought it was a matter of comparing slopes but that doesn't appear to be enough.</p>

答:
"At first I thought it was a matter of comparing slopes"... and you were right!

If the line segments AB and BC have the same slope, then A, B, C are necessarily collinear. Note that there are some corner cases having to do with whether B is the "middle" point or not (in which case the slopes will still be equal), and one having to do with vertical lines (where some formula you use to compute slope might divide by 0).

Putting all this together, the points $(a, b)$, $(m, n)$ and $(x, y)$ are collinear if and only if $$(n-b)(x-m) = (y-n)(m-a)$$
(comes from $\frac{n-b}{m-a} = \frac{y-n}{x-m}$, but not writing it in fraction form to avoid division by $0$).


问:
What is the significance of power set of $\phi$?
The formula of subsets of any sets is : $2^n$

We have a subset of $\phi$ because $2^0 = 1$ and the subset is $\phi$
itself.

Therefore:

$p(\phi) = \{{ \phi}\}$

$p(p(\phi) = \{{ \phi, \{\phi\}}\}$

$p(p(p(\phi)) = \{{ \phi, \{\phi\}, \{\phi ,\{\phi\} \},  \{ \{ \phi \}\}}\}$

And so it goes on... but what is the mathematical significance of the above power sets?

答: $0$ is the empty set. The set whose only element is the empty set is $1.$ For more, google "von Neumann ordinals".


问:
Topology of a set of matrices
I am trying to find the topology of $M$ where

$$M := \left\{ A \in \Bbb R^{3 \times 3} \mid A = 2 \vec{u} \cdot (\vec{u})^T - (\mathbb{1}_{3 \times 3} - \vec{u} \cdot (\vec{u})^T) \right\},$$

with $\vec{u}$ three dimensional unit vectors with real coefficients and $\mathbb{1}_{3 \times 3}$ the $3 \times 3$ identity matrix.

What I have done so far is putting $\vec{u} = (a \ \ b \ \ c)^T$, with $a^2 + b^2 + c^2 = 1$, from which $A$ takes the general form

\begin{align*}
A = \begin{pmatrix} 3a^2-1 & 3ab & 3ac \\ 3ab & 3b^2 - 1 & 3bc \\ 3ac & 3bc & 3c^2 -1 \end{pmatrix}.
\end{align*}

From this general form I deduced that $A$ has determinant 2 (and is therefore invertible), $A$ is symmetric, and has a vanishing trace. At this point I do not quite see what the topology of $M$ would be, however, I should mention that I am not very familiar with topology yet.

Any help would be greatly appreciated.

答: The  map $f$ from the unit sphere in $\mathbb{R}^3$ to your space is continuous and $f(-\vec{u}) = f(u).$ So, your space is diffeomorphic to the projective plane $\mathbb{R}P^2.$

Title: Topology of a set of matrices


问:
Total number of multiplication of upper triangular matrices
Suppose we have to square upper triangular matrices that we want to multiply using the following code:

The question is to find the total number of multiplications. I noticed that every time the number of multiplications get decreased by 1 in each run. So the total number of multiplication is as follows:

$$[1+2+\cdots+n] + [1+2+\cdots+(n-1)] + \cdots + [1]$$

Question: In a solution, they say the total number of multiplications is $n(n + l)(n + 2)/6$, can you please explain how they got this number?

答:
You should know that $[1+2 + \dots + n] = \frac{n(n+1)}2.$ To prove that
$\sum_{k=1}^n \frac{k(k+1)}2 = \frac{n(n+1)(n+2)}6$ you need to know how to sum the squares of the first $n$ natural numbers.


答: You missed solutions. There are altogether four real solutions: $$2,5,\frac{1\pm\sqrt{41}}{2}$$ To find them, the easiest way is to consider the four intervals for $x$ respectively: $(-\infty,0),(0,\frac{10}{3}),(\frac{10}{3},4),(4,+\infty)$.

Title: $ \log|x-4| -\log|3x-10| = \log\left|\frac{1}{x} \right| $ one formal way to say we won't choose all solutions in degree 2 equation

Body:
$$
\log|x-4| -\log|3x-10| = \log\left|\frac{1}{x} \right|
$$

After applying some properties we have:
$$
\log\left|\frac{x^2-4x}{3x-10}\right| = 0.
$$

After applying some operations, this equation becomes:
$$
x^2-7x+10 = 0\\
x=2 ;x=5.
$$

The problem is: if we substitute $x$ with "$5$" it's good, but if we substitute with "$2$" there's an obvious error, since $2$ is out of the logarithm's domain.

Are there some 'formal' ways to say why we won't choose number 2, that just say: "we won't choose number 2 because it doesn't make sense"?

EDIT:
Here is another exercise to explain that there are some solutions that doesn't work even with absolute values in the logarithms' arguments:

$$
\log(\sqrt{x+14})+\log(\sqrt{x+7})-\log(1.2)=1 \\
\cdots\\
x^2+21x-46=0\\
x_1=-23; x_2=2.$$

But again: what happened with $-23$ ? Does it just disappear ?

Note: I'm working in $\mathbb R.$


问:
Consider the Sturm-Liouville problem $y^{\prime\prime}+ \lambda y=0,y(0)=y( \pi)=0.$
Consider the Sturm-Liouville problem $y^{\prime\prime}+ \lambda y=0,y(0)=y( \pi)=0.$
Which of the following statement(s) are true:

There exists only countably many characteristic values

There exists uncountably many characteristic values

Each characteristic function corresponding to the characteristic value $\lambda$ has exactly $\lfloor \sqrt\lambda \rfloor -1$ zeros in $(0, \pi)$

Each characteristic function corresponding to the characteristic value $\lambda$ has exactly $\lfloor \sqrt\lambda \rfloor $ zeros in $(0, \pi)$

My Approach: Solving for $\lambda >0$, $y=c_1 \cos \sqrt{\lambda}x+c_2 \sin \sqrt{\lambda}x$,

As $y(0)=0$, $c_1=0$; As $y(\pi)=0$, so either $c_2=0$ or $\sqrt{\lambda} \pi=n \pi$, $n \in \mathbb Z$, i.e. $\lambda_n=n^2$, $y_n=C_n \sin nx$

For $\lambda \le0$, $y=0$ is trivial solution. So, I guess option $1.$ is correct but I cannot understand about options $3.$ and $4.$ please help.

答:
So after solving the given Sturm-Liouville problem $($for $\lambda=m^2>0$, as for $\lambda\le0$ we get trivial solution$)$, we have

$$y(x)=B\sin(nx)~,$$ which is a characteristic function corresponding to the characteristic value $~m=n\in\mathbb Z~.$

For option $\bf {(1)}$ and $\bf {(2)}$ : Since $~\mathbb Z~$ is countable, so there exist only countably many characteristic value.
Hence option $(1)$ is correct but option $(2)$ is incorrect.

For option $\bf {(3)}$ and $\bf {(4)}$ : Now we have $$y_n=B_n\sin(nx)$$
For $~n=1~,~~~~y_1=B_1\sin(x)~$ which has no solution (i.e., no zero) in $~(0,\pi)~.$

i.e., characteristic function corresponding to the characteristic value $~\lambda=n^2=1~$ has exactly $~\lfloor \sqrt\lambda \rfloor -1=1-1=0~$ zero in $(0, \pi)$.

For $~n=2~,~~~~y_2=B_2\sin(2x)~$ which has only one solution in $~(0,\pi),\big[$which is $~x=\pi/2~\big].$

i.e., characteristic function corresponding to the characteristic value $~\lambda=n^2=4~$ has exactly $~\lfloor \sqrt\lambda \rfloor -1=2-1=1~$ zero in $(0, \pi)$.

Similarly, we can verify for the other values of $~n\in\mathbb Z~.$

Hence each characteristic function corresponding to the characteristic value $~\lambda~$ has exactly $~\lfloor \sqrt\lambda \rfloor -1~$ zeros in $(0, \pi)$.

Therefore option $(3)$ is correct but option $(4)$ is incorrect.