问:
Self-intersection of a cubic Bezier, interpretation of the solution
I am trying to understand the form of the determinants that I get when I try to calculate the self-intersection. Since I might have made some mistakes, I'm including the whole derivation:

Let $\vec{P}(t)$ be a cubic Bezier curve in two dimensions. In power basis:
$$
\vec{P}(t)=(−\vec{A}+3\vec{B}−3\vec{C}+\vec{D})t^3+(3\vec{A}−6\vec{B}+3\vec{C})t^2+(−3\vec{A}+3\vec{B})t+\vec{A}
$$
$$
=\vec{a}t^3+\vec{b}t^2+\vec{c}t+\vec{d}.
$$
For a loop intersection to occur, there must exist $(t,s), t\neq s$ :
$$
\vec{a}(t^3-s^3)+\vec{b}(t^2-s^2)+\vec{c}(t-s)=0 \quad\quad\quad/(t-s)
$$
$$
\vec{a}(t^2+ts+s^2)+\vec{b}(t+s)+\vec{c}=0
$$
If I make the substitution $ u :=t+s/2$, I get:
$$
a_x(u^2+\frac{3s^2}{4})+b_x(u+\frac{s}{2})+c_x=0
$$
$$
a_y(u^2+\frac{3s^2}{4})+b_y(u+\frac{s}{2})+c_y=0
$$
I can eliminate the quadratic term by doing $a_y(I)-a_x(II)$:
$$
(a_yb_x - a_xb_y)(u+\frac{s}{2})+(a_yc_x-a_xc_y)=0
$$
$$
M:=a_yb_x - a_xb_y\quad\quad\quad N:=-(a_yc_x-a_xc_y)
$$
$$
u=\frac{N}{M}-\frac{s}{2}
$$
Substituting $u$ back:
$$
\vec{a}\left[\left(\frac{N}{M}-\frac{s}{2}\right)^2+\frac{3s^2}{4}\right]+\vec{b}\frac{N}{M}+\vec{c}=0
$$
$$
\vec{a}\left[s^2 - \frac{N}{M}s+ \left(\frac{N}{M}\right)^2\right]+\vec{b}\frac{N}{M}+\vec{c}=0
$$
These quadratic equations have solutions:
$$
s_i=\frac{1}{2a_i}\left(
a_i\frac{N}{M}\pm\sqrt{D_i}\right)
$$
$$
D_i=a_i^2\left(\frac{N}{M}\right)^2-4a_i\left(a_i\left(\frac{N}{M}\right)^2+b_i\frac{N}{M}+c_i\right)
$$
And now I'm confused...

Why did I end up with two quadratic equations? Shouldn't there be just one? My understanding is that  $D = 0$ corresponds to the cusp shape, $D > 0$ to a loop intersection and $D < 0$ to no intersection.

Furtheremore, I would like to understand what $M=0$ (which blows up the determinant) corresponds to. The expression for the curve being quadratic explains it only partly, i.e. when $a_x = a_y = 0$.
What do the other conditions in:
$$
M=a_yb_x - a_xb_y=0
$$
correspond to? (E.g. what is 'wrong' with $(b_x = b_y = 0)$, $(a_yb_x = a_xb_y)$?)

答:
(1) $D_x$ and $D_y$ are almost identical:

```<code>from sympy import *
ax, ay, bx, by, cx, cy = symbols('ax ay bx by cx cy', real=True)
M = ay*bx - ax*by
N = ax*cy - ay*cx
Dx = ax**2 * (N/M)**2 - 4*ax*(ax*(N/M)**2 + bx*N/M + cx)
Dy = ay**2 * (N/M)**2 - 4*ay*(ay*(N/M)**2 + by*N/M + cy)
print(Dx.expand().simplify())
print(Dy.expand().simplify())

ax**2*(-3*ax**2*cy**2 + 6*ax*ay*cx*cy + 4*ax*bx*by*cy - 4*ax*by**2*cx - 3*ay**2*cx**2 - 4*ay*bx**2*cy + 4*ay*bx*by*cx)/(ax**2*by**2 - 2*ax*ay*bx*by + ay**2*bx**2)
ay**2*(-3*ax**2*cy**2 + 6*ax*ay*cx*cy + 4*ax*bx*by*cy - 4*ax*by**2*cx - 3*ay**2*cx**2 - 4*ay*bx**2*cy + 4*ay*bx*by*cx)/(ax**2*by**2 - 2*ax*ay*bx*by + ay**2*bx**2)
</code>```

So $D_x=a_x^2K$ and $D_y=a_y^2K$ where $K$ is common to both expressions. The $a_x^2$ or $a_y^2$ can be brought out of the square root, where it then cancels the other instances of $a_x$ or $a_y$:
$$s_i=\frac{N/M\pm\sqrt K}2$$
There is only one quadratic equation for $s$ – the possible values for $s_x$ and $s_y$ are identical. Your interpretation of $D$ (essentially equivalent to my $K$) is correct though.

(2) Let $\vec P(t)=(x(t),y(t))$. If $M=0$ there are real $\lambda$ and $\mu$, not both zero, for which $\lambda x(t)+\mu y(t)$ is a linear polynomial (cubic and quadratic terms both cancel). The rotated curve $\vec Q(t)=\begin{bmatrix}\lambda&\mu\\-\mu&\lambda\end{bmatrix}\vec P(t)$ is then the graph of a single-variable (generically) cubic function $y=f(x)$. In short, $M=0$ implies $\vec P(t)$ has no self-intersections and at most one inflection point.

(3) Your method is not the most elegant way of deriving the self-intersection parameters. See my answer here.

Body:
(1) $D_x$ and $D_y$ are almost identical:

from sympy import *
ax, ay, bx, by, cx, cy = symbols('ax ay bx by cx cy', real=True)
M = ay*bx - ax*by
N = ax*cy - ay*cx
Dx = ax**2 * (N/M)**2 - 4*ax*(ax*(N/M)**2 + bx*N/M + cx)
Dy = ay**2 * (N/M)**2 - 4*ay*(ay*(N/M)**2 + by*N/M + cy)
print(Dx.expand().simplify())
print(Dy.expand().simplify())

ax**2*(-3*ax**2*cy**2 + 6*ax*ay*cx*cy + 4*ax*bx*by*cy - 4*ax*by**2*cx - 3*ay**2*cx**2 - 4*ay*bx**2*cy + 4*ay*bx*by*cx)/(ax**2*by**2 - 2*ax*ay*bx*by + ay**2*bx**2)
ay**2*(-3*ax**2*cy**2 + 6*ax*ay*cx*cy + 4*ax*bx*by*cy - 4*ax*by**2*cx - 3*ay**2*cx**2 - 4*ay*bx**2*cy + 4*ay*bx*by*cx)/(ax**2*by**2 - 2*ax*ay*bx*by + ay**2*bx**2)

So $D_x=a_x^2K$ and $D_y=a_y^2K$ where $K$ is common to both expressions. The $a_x^2$ or $a_y^2$ can be brought out of the square root, where it then cancels the other instances of $a_x$ or $a_y$:
$$s_i=\frac{N/M\pm\sqrt K}2$$
There is only one quadratic equation for $s$ – the possible values for $s_x$ and $s_y$ are identical. Your interpretation of $D$ (essentially equivalent to my $K$) is correct though.

(2) Let $\vec P(t)=(x(t),y(t))$. If $M=0$ there are real $\lambda$ and $\mu$, not both zero, for which $\lambda x(t)+\mu y(t)$ is a linear polynomial (cubic and quadratic terms both cancel). The rotated curve $\vec Q(t)=\begin{bmatrix}\lambda&\mu\\-\mu&\lambda\end{bmatrix}\vec P(t)$ is then the graph of a single-variable (generically) cubic function $y=f(x)$. In short, $M=0$ implies $\vec P(t)$ has no self-intersections and at most one inflection point.

(3) Your method is not the most elegant way of deriving the self-intersection parameters. See my answer here.


问:
Evaluate sum of the series $\sum { 1\over n!(n^4+n^2+1)}$

Evaluate  sum of the series:  $\displaystyle \sum_{n=0}^\infty {1\over n!(n^4+n^2+1)}$

Comparison test $\left(\text {with } \sum {1\over n^2}\right)$ confirms the convergence of the series. But what's the sum of the series.
Using comparison tests I managed to show that ${19\over 14}\le$ sum $\le{e+121\over 91}$.

The series seems to converge to $\frac e2$. But how to show that ?

答:
${1\over n!(n^4+n^2+1)}$
$=\frac 12 \left[{n\over (n+1)!\{n(n+1)n+1\}}-{n-1\over n!\{n(n-1)+1\}}+{1\over (n+1)!}\right]$

$\displaystyle \sum_{n=0}^\infty \textstyle {1\over n!(n^4+n^2+1)}
=\displaystyle \frac 12 \left[ \sum_{n=1}^\infty \textstyle {n-1\over n! \{n(n-1)+1\}}-\displaystyle \sum_{n=0}^\infty \textstyle{n-1\over n!\{n(n-1)+1\}}+\displaystyle \sum_{n=0}^\infty \textstyle \frac{1}{n!}-1\right]$
$= \frac 12[-(-1)+e-1]=\frac e2$

ContentLicense: CC BY-SA 4.0


问:
Evaluate $ \int \frac{3x+1}{x^2+4x+4}dx $
$$
\int \frac{3x+1}{x^2+4x+4}dx
$$

How can I solve this integral? I can't find any way that lead to $$\frac{f^\prime(x)}{f(x)}$$

I tried with WolframAlpha but the explanation wasn't very intuitive this time!

Body:
Note that
$$3x+1=\frac32(2x+4)-5$$


问:
How many ways to arrange 5 different dogs, 1 cat and 1 rat such that the rat is always left to the cat (not necessarily near).

How many ways to arrange 5 different dogs, 1 cat and 1 rat such that the rat is always left to the cat (not necessarily near).

I started out by arranging the 5 different dogs which is $5!$, and from here basically I got stuck, I tried to do it by cases, like if I arranged the 5 dogs, I have 6 places to put the cat in,starting from the right, the rat also can be put in 6 places (can be near the cat from the left), and moving one to the left each case, so I got: $6*6+5*5+4*4+3*3+2*2+1*1=91$ and this multiplied by the number of ways to arrange the dogs gave me the answer: $91*5!=10920$, now this seems clearly wrong and I messed up since $7!=5040$, which is the number of ways to arrange all seven, I am trying to understand my mistakes and how to deal with this question.

Thanks in advance to any help.

答: As you say there are $7!=5040$ arrangements if we ignore the restriction on the cat and rat.  We can match up arrangements with the rat left of the cat with arrangements with the rat right of the cat, so the arrangements with the rat left of the cat are half the total.  The result is $$\frac 12 \cdot 7!=2520$$


问:
Calculate $\int_{\mu}^{2t} \frac{1}{E+2\mu -2\epsilon}\frac{1}{\sqrt{4t^2-\epsilon^2}} \mathrm dx$
I stumbled upon this integral while doing Solid-state physics homework:
$$\int_\mu^{2t} \frac{1}{E+2\mu -2\epsilon}\frac{1}{\sqrt{4t^2-\epsilon^2}} \mathrm d\epsilon$$
where $E<0$, $\mu \in (-2t,2t)$ and $t>0$.
I tried plugging it into Mathematica:

```<code>Integrate[1/(e + 2 \[Mu] - 2 x) 1/Sqrt[4 t^2 - x^2], {x, \[Mu], 2 t},
Assumptions -> -2 t < \[Mu] < 2 t && t > 0 && e < 0]
</code>```

and got the following answer:
$$-\frac{1}{2} \pi  \sqrt{\frac{1}{E^2+4 E \mu +4 \mu ^2-16 t^2}}-\frac{\arctan\left(\frac{\mu  (E+2 \mu )-8 t^2}{\sqrt{\left(4 t^2-\mu ^2\right) \left(E^2+4 E \mu +4 \left(\mu ^2-4 t^2\right)\right)}}\right)}{\sqrt{E^2+4 E \mu +4 \left(\mu ^2-4 t^2\right)}}$$

This result gives correct "physical" interpretation, so now I would like to know how this integral was
calculated.
My first guess was to use substitution $\epsilon = 2t\sin{x}$ and transform this integral to
$$\int_{\arcsin{\frac{\mu}{2t}}}^{\frac{\pi}{2}} \frac{\mathrm dx}{E+2\mu - 4t\sin{x}}
$$
Here I thought about trying to evaluate indefinite integral by doing the standard substitution $u = \tan{\frac{x}{2}}$, however I hesitated after reading this comment.
Is there any other way to obtain the result in closed form like the one obtained from Mathematica?

答:
Introduce
$$ f \colon (-1,1) \times (0,\infty) \to \mathbb{R} \, , \, f(m,w) = \int \limits_m^1 \frac{\mathrm{d} v}{(v-m+w)\sqrt{1-v^2}} \, ,$$
to simplify the notation. Your integral then equals $- \frac{1}{4t} f\left(\frac{\mu}{2t},-\frac{E}{4t}\right)$ (just let $v = \frac{\varepsilon}{2t}$).

In order to compute the integral we proceed as you suggested: we let $v = \cos(x)$ (slightly easier than $v = \sin(x)$) and then $\tan\left(\frac{x}{2}\right) = u$ to obtain
\begin{align}
f(m,w) &= \int \limits_m^1 \frac{\mathrm{d} v}{(v-m+w)\sqrt{1-v^2}} = \int \limits_0^{\operatorname{arccos(m)}} \frac{\mathrm{d} x}{\cos(x)-m+w} \\
&= 2 \int \limits_0^{\sqrt{\frac{1-m}{1+m}}} \frac{\mathrm{d} u}{1-m+w + [w-(1+m)] u^2}
\end{align}
for $(m,w) \in (-1,1) \times (0,\infty)$. The next step depends on the relation between $m$ and $w$ (or $\mu$ and $E$ in the original problem). More precisely, depending on the sign of $w - (1+m)$, after one more substitution we get the following results for $(m,w) \in (-1,1) \times (0,\infty)$:
$$ f(m,w) = \begin{cases}
\frac{2 \operatorname{artanh} \left(\sqrt{\frac{(1+m-w)(1-m)}{(w+1-m)(1+m)}}\right)}{\sqrt{1-(m-w)^2}} &, \, w < 1+m \\
\sqrt{\frac{1-m}{1+m}} &, \, w = 1+m \\
\frac{2 \arctan \left(\sqrt{\frac{(w-(1+m))(1-m)}{(w+1-m)(1+m)}}\right)}{\sqrt{(m-w)^2 - 1}} &, \, w > 1+m
\end{cases} \, .$$
Plugging the original variables into these results (and using the relation $\arctan(x) + \arctan\left(\frac{1}{x}\right) = \frac{\pi}{2}$ for $x > 0$), we see that your Mathematica expression corresponds to the third case, though it covers all cases if complex numbers and limits are taken into account properly.


问:
limits of sequences exponential and factorial: $a_n=e^{5\cos((\pi/6)^n)}$ and $a_n=\frac{n!}{n^n}$
Compute the limit $\lim\limits_{n\to\infty}a_n$ for the following sequences:

(a) $a_n=e^{5\cos((\pi/6)^n)}$

(b) $a_n=\frac{n!}{n^n}$

For part (a) do I just take the limit of the exponent part and then the answer would be $e$ raised to whatever the limit is?

And would the limit be $1$ or $-1$? because $\cos$ goes between those two.

For part $b$ it is in the form of infinity over infinity but how do you take the derivative of $n!$? Will it ever break out of infinity over infinity?

ContentLicense: CC BY-SA 3.0

Body:
For the second one
(I've done this before btw),
$n! = \prod_{i=1}^n i
=\prod_{i=1}^n (n+1-i)
$
so

$\begin{align}
n!^2
&= \prod_{i=1}^n i(n+1-i)\\
&= \prod_{i=1}^n (i(n+1)-i^2)\\
&= \prod_{i=1}^n ((n+1)/2)^2-(n+1)/2)^2+ i(n+1)-i^2)\\
&= \prod_{i=1}^n ((n+1)/2)^2-((n+1)/2-i)^2)\\
&\le \prod_{i=1}^n ((n+1)/2)^2)\\
&= ((n+1)/2)^{2n}
\end{align}
$

so $n! < (\frac{n+1}{2})^{n}
$
and
$\frac{n!}{n^n}
\le (\frac{n+1}{2n})^{n}
$
and this $\to 0$ as $n \to \infty$.


问:
Calculate fourier transfrom $f(x) = \frac{\cos(bx)}{a^2 +x^2} $ for $a, b >0$ how should I continue from here
$$f(x) = \frac{\cos(bx)}{a^2 +x^2} $$ for $a, b$ $>0$

$$\hat f(\omega) = \int\limits^{\infty}_{-\infty} \dfrac{1}{2{\pi}}\cdot\dfrac{\cos\left(bx\right)}{a^2+x^2}\mathrm{e}^{-\mathrm{i}wx}\,\mathrm{d}x = \int\limits^{\infty}_{-\infty} \dfrac{\mathrm{e}^{\mathrm{i}bx}+\mathrm{e}^{-\mathrm{i}bx}}{4\pi\left(a^2+x^2\right)}\mathrm{e}^{-\mathrm{i}wx}\,\mathrm{d}x=$$

$$ \int\limits^{\infty}_{-\infty} \dfrac{\mathrm{e}^{\mathrm{i}x\left(b-w\right)}}{2\left(a^2+x^2\right)}\,\mathrm{d}x  + \int\limits^{\infty}_{-\infty} \dfrac{\mathrm{e}^{-\mathrm{i}x\left(b+w\right)}}{2\left(a^2+x^2\right)}\,\mathrm{d}x$$

How should I continue from here?

I checked on wolfram alpha but the answer should not expressed with dirac function since we didn't learn it.

答:
Note that the factor $\cos(bx)$ is just a modulation, i.e., a translation of the spectrum. So we can try to determine the Fourier transform $G(\omega)$ of

$$g(x)=\frac{1}{a^2+x^2}\tag{1}$$

and from $G(\omega)$ determine the Fourier transform of $f(x)$:

$$F(\omega)=\frac12\big[G(\omega-b)+G(\omega+b)\big]\tag{2}$$

Rewrite $g(x)$ as

$$g(x)=\frac{1}{2a}\left[\frac{1}{a+jt}+\frac{1}{a-jt}\right]\tag{3}$$

It's straightforward to see that

$$\mathcal{F}\left\{\frac{1}{a+jt}\right\}=2\pi e^{a\omega}u(-\omega),\qquad a>0\tag{4}$$

and

$$\mathcal{F}\left\{\frac{1}{a-jt}\right\}=2\pi e^{-a\omega}u(\omega),\qquad a>0\tag{5}$$

where $u(\omega)$ is the Heaviside unit step function.

Using $(4)$ and $(5)$, the Fourier transform of $g(x)$ is obtained as

$$G(\omega)=\frac{\pi}{a}e^{-a|\omega|}\tag{6}$$

The Fourier transform of $f(x)$ can be obtained by plugging $(6)$ into $(2)$.


问:
(Relatively) Boundedness of self adjoint operator $A^q$.
Suppose $A$ is a self-adjoint operator on the Hilbertspace $\mathcal{H}$. Furthermore let $p,q \in \mathbb{R}$ with $0 \leq q < p$. I want to show that $A^q$ is relatively bounded w.r.t $A^p$ with $A^p$-bound zero. So basically I have to show the following:

$$\exists a,b\ge 0: \|A^qx\|^2 \leq a \|x\|^2 + b \|A^px\|^2$$
With the infimum of all $b´s$ satisfying this inequality being equal to zero. My thoughts so far: From functional calculus, I know that I can define $A^q$ as follows:
$A^q = \int t^q dE(t)$,  where $E$ is the spectal measure associated with $A$. Furthermore I know that I have:
$$\|A^q x\|^2 = \int t^{2q} \langle dE(t)x,x\rangle \leq \int t^{p+q} \langle dE(t)x,x\rangle $$

At this point I don´t see how to proceed. Can anyone help me out with this please? Is there any (maybe Hölder-like??) inqulity that I can use for this problem?

答:
Using functional calculus as indicated in the question, the problem boils down to showing that for every $b>0$ there exists $a>0$ such that $t^{2q}\leq a+bt^{2p}$ for all $t\geq 0$.

Since $\lim_{t\to\infty}(t^{2q}-bt^{2p})=-\infty$, there exists $T>0$ such that $t^{2q}-bt^{2p}\leq 0$ for $t\geq T$. On the compact interval $[0,T]$, the continuous function $t\mapsto t^{2q}-bt^{2p}$ is bounded above by a constant $a>0$. Altogether, $t^{2q}-bt^{2p}\leq a$ for all $t\geq 0$.


问:
Find least value of $9\tan\phi+4\cot\phi+3$ without derivatives
The problem is as follows:

In a pharmaceutical lab in Daegu, a group of scientists need to find
the power of a visible blue light lamp to get through a sucrose
crystal in order to measure its purity for a certain medication.

The power for such light is given by:

$$J=\frac{1}{H}$$

Assuming that $H$ has to be the minimum value of the equation from
below:

$$H=9\tan\phi+4\cot\phi+3$$

$\theta$ is an acute angle which lies in the first quadrant.

What would be the value of $J$ so that the scientists could set up
their lamp?

The alternatives given in my book are:

$\begin{array}{ll}
1.&\frac{1}{5}\\
2.&\textrm{1}\\
3.&\textrm{15}\\
4.&\frac{1}{15}\\
\end{array}$

This problem is making me feel confused. Should derivatives be used here?

I've attempted to do so and derivate the equation, by doing so this yields:

$H'=9\sec^2\phi-4\csc^2\phi$

But equating this to zero doesn't seem to be the intended approach.

Then I came up with the idea of adding the intervals from where $\phi$ in the first quadrant lies.

$0\leq \cot \phi < +\infty$

Tangent is also in the same interval:

$0\leq \tan \phi < +\infty$

Then multiplying by $9$ and by $4$ and adding $3$ to the second equation.

This results into:

$3\leq 9\tan\phi+4\cot\phi+3 < +\infty$

But without any further constrain there isn't any way to tell, which would be the least value?

If it would had been the integer value, wouldn't had been $4$?

Running out of idead, what it came to my mind was to equate both functions:

$\tan\phi=\cot\phi$

This corresponds to $\phi=\frac{\pi}{4}$

Assuming this: I'm guessing that the result would be

$9\tan\phi=1$

$4\cot\phi=1$

then $1+1+3=5$

But this is more like guessing and not a serious way to justify the answer. I cannot be sure if that's the answer. Can someone help me here and explain me why derivatives doesn't seem to work?

How to solve this without derivative?. Help me please, I'm confused.

答:
Set $\tan\phi=z$

$f(z)=9 z + \frac{4}{z} + 3$ is minimum when $f'(z)=0$ and $f''(z)>0$

$f'(z)=9 -\frac{4}{z^2}=0\to z=\pm\frac{2}{3}$.

$f''(z)=\frac{8}{z^3}$

as $f''\left(\frac{2}{3}\right)>0$ we have that $f\left(\frac{2}{3}\right)=15$ is the minimum.


问:
Given a triangle $ABC$ and its incenter I. If $P,Q$ are points on BI and CI so that $2PAQ=BAC$. Prove that $PDQ=90$
Given a triangle $ABC$ and its incenter I. If $P,Q$ are points on $BI$ and $CI$ so that $2\angle PAQ=\angle BAC$. If $D$ is the point where the inscribed circle of $ABC$ intersects $BC$, prove that $\angle PDQ=90$.

If $X\in BI$ we state $f(X)$ is the only point on $CI$ which is collinear with $A$ and is the points which becomes of X if we rotate $X$ around $A$ by angle $\frac{1}{2}\angle ABC$. Hence $f(P)=Q$.  We also have that $f(B)=I$ and $f(I)=C$. Also if $I_1$ and $I_2$ the incenter of $\triangle ABD, \triangle ACD$ respectively then $f(I_1)=f(I_2)$

This is as far as I got. I believe it should be able to finish like this, but I've gotten stuck. Could you please help me finish it off?

答:
To continue your idea: the map $f$ is a projective map from the set of points on $BI$ to the set of points on $CI$. We can consider another map $g$ that sends $X\in BI$ to the point $Y\in CI$ such that $\angle XDY=90^{\circ}$. This map $g$ is also projective, so to show that $f\equiv g$, it suffices to check that they are equal at three points.

But this is easy: $f(B)=I=g(B)$, $f(I)=C=g(I)$ and $f(I_1)=I_2=g(I_1)$.

An alternative approach: $\angle DIQ+\angle AIB=180^{\circ}$, so $I$ has an isogonal conjugate wrt quadrilateral $ABDQ$. But it's easy to see that the isogonal conjugate must in fact be $P$.

So $\angle PDQ=\angle IDB=90^{\circ}$.


答:
you see that $x^{2}+4x+4=(x+2)^{2}$. Now you can use substitution. Take $u=x+2$ then $du=dx$. this means that you can rewrite the integral as following:
\begin{align}
\int\frac{3x+1}{x^{2}+4x+4}dx&=\int\frac{3x+1}{(x+2)^{2}}dx\\
&=\int\frac{3(u-2)+1}{u^{2}}du\\
&=\int\frac{3u-5}{u^{2}}du\\
&=\int\frac{3u}{u^{2}}du-\int\frac{5}{u^{2}}du\\
&=\int\frac{3}{u}du-\int\frac{5}{u^{2}}du\\
&=3\ln(u)+\frac{5}{u}+C
\end{align}

I hope I used enough little steps to guide you trough your excercise. If I didn't calculate wrong this must be the solution.

Title: Evaluate $ \int \frac{3x+1}{x^2+4x+4}dx $

Body:
$$
\int \frac{3x+1}{x^2+4x+4}dx
$$

How can I solve this integral? I can't find any way that lead to $$\frac{f^\prime(x)}{f(x)}$$

I tried with WolframAlpha but the explanation wasn't very intuitive this time!


问:
Does continuity guarantee existence of double integral?
In single variable real analysis we know if $f: \mathbb{R} \to \mathbb{R}$ is continuous then its integral exists over any bounded domain. Does this generalize to the 2-variable case? Does the double integral of a function $f: \mathbb{R^2} \to \mathbb{R}$ exist over any bounded domain when the function is continuous, or is uniform continuity (or some other stronger condition) required?

答: Continuity over $R^2$ is good enough for existence of the double integral.


问:
Approximating partial series $\sum_{j=1}^{p} \sqrt{1-r^{2|k-j|}}/j$
I'm working with the following series:

$$S_k = \sum_{j=1}^{p} \frac{\sqrt{1-r^{2|k-j|}}}{j} $$
here $|r|<1$, $k \leq p$.

Naturally, for small $r$ the series approaches harmonic sums $H_k \approx \gamma + \log(k) + 1/2k - 1/12k^2$, however, I'm trying to find a more manageable expression to work with..

By numerical experiments (guesswork mostly) I find that the following works reasonably well for $|r|<0.9$:

$$ S_k \approx H_p + \log\left(\frac{\beta_{11}}{1 + \exp\{\beta_{12} + \beta_{13} \cdot r\}}\right)\frac{1}{k} + e^{\beta_{21} + \beta_{22} r} \frac{1}{k^2} $$
for some numerical values $\beta_{11},\beta_{12},\beta_{13},\beta_{21},\beta_{22}$. Further, accuracy for $r$ closer to 1 can be improved by additional terms $k^{-3}$ and, probably, deeper expansion into $\text{exp}(\beta_{21} + \beta_{22} r + \beta_{23} r^2 + \ldots )$ near the $k^{-2}$ term.

All this suggests a possible expansion through $k^{-1}, k^{-2}, \ldots$, similar to approximation of $H_k$, just the summation weights will probably depend from $r$.

Does there exist any similar approximation that I could use analytically?

答:
Just a few preliminary ideas (for the time being, I hope)

Using the expansion
$$\sqrt{1-r^{2|k-j|}}=\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n}r^{2n|k-j|}$$ Let
$$a_n=(-1)^n \binom{\frac{1}{2}}{n}r^{2n|k-j|}$$
$$\sum_{j=1}^k \frac{a_n}j=(-1)^{n+1} \binom{\frac{1}{2}}{n}  \left(r^{-2 n}\Phi \left(r^{-2 n},1,k+1\right)+r^{2
   k n} \log \left(1-r^{-2 n}\right)\right)$$ where appears  the Lerch transcendent function.

Difficult to give now more than for "small" values of $r$
$$r^{-2 n}\Phi \left(r^{-2 n},1,k+1\right)\sim -\frac 1 {k+1}$$

Now, the problem is to work

$$r^{2k n} \log \left(1-r^{-2 n}\right)$$ whish seems to be more problematic

To follow, I hope and wish


问:
What is an extension of an inner product space?
Chapter 5 of Alan Macdonald's "Linear and Geometric Algebra" begins by saying,

The geometric algebra $\mathbb{G}^n $ is an extension of the inner product space $\mathbb{R}^n.$

What is an extension of an inner product space?

答: I mean extension in the usual English sense: $\mathbb{R}$, its vectors and its operations (scalar product, vector addition, inner product) are contained in $\mathbb{G}$.


问:
Represent a vector with a scalar value
I have a system that returns vector representation of my input, but instead of a vector, I want to get a single number. Hence, my question is, is there any method that converts(represents, encodes) a vector to a single scalar value?

Using the norm of the vector might be a solution, but since I am not good at linear algebra(or maths in general), I thought it may be a good idea to ask at here...

Thank you.

答:
It kind of depends on the context.

There is no general way to encode a vector (which I assume to have more than one component, otherwise there is no problem) faithfully into a single value ie. in such a way that you can recover the original vector from this value. Intuitively the components correspond to degrees of freedom, so restricting from say 3 degrees of freedom to 1 degree of freedom is a irreversible loss of information/freedom.

Of cause there are instances, where a vector is uniquely determined by a single value:

Let’s say you only consider vectors in $\Bbb R^2$ and you know that all the vectors you are interested in, have length/magnitude/norm 1, then any such vector corresponds uniquely to its angle with the $x$-axis. Note that our assumption that the vectors have length 1 is a restriction making us loose a degree of freedom, so it is not surprising that only one degree of freedom remains...

If you are in a discrete setup, by which I mean vectors in $\Bbb N^k$, $\Bbb Z^k$ or $\Bbb F_p^k$ (strictly speaking only the last one has vectors, but ignore that for now), then you might have a chance to enumerate all vectors. In this case any vector corresponds to a unique number, but the usual operations on vectors (like addition, multiplication by a scalar etc.) will become weird, when trying to write them down with respect to that enumeration.

Of cause these examples are somewhat artificial and I doubt they are what you are after.

So there is one last option, what representing a vector of a scalar may be. Assuming the vector represents some kind of data, say a color, then you might be interested in some sort of information you can derive from the components of the vector, say brightness. The euclidean norm is just one of many useful values you can associate to a vector. Others may be the maximum of all components, the mean of the components, the variance of the components etc. etc. Which kind of value you use depends on the context of your data. But again, most of these values will not be faithful, in that you won’t be able to recover the vector, once you calculated the associated scalar.

Title: Represent a vector with a scalar value


问:
What's meant by radius vector of a curve?

Find the length of the minimum radius vector of the curve $\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$.

Let $r$ be the radius vector. So, $x^2+y^2=r^2$. Taking out the value of $x^2$ from the given curve and putting it here, we get value of $r$ in terms of $y$. Taking derivative and equating it to zero, we get the value of $y$ and $x$. Thus, $r$ comes out to be $a+b$.

But what's meant by radius vector of a curve? We are calculating its minima. So, can it have other values as well?

答:
Let $x=r \cos \theta, y=r \sin \theta$, then
$$\frac{r^2\cos^2 \theta}{a^2}+\frac{r^2 \sin^2 \theta}{b^2}=1$$
$$\implies r=\sqrt{\frac{1}{a^{-2} \cos^2 \theta + b^{-2} \sin^2 \theta}}>0,$$
is the radius vector for this ellipse.


问:
Find a minimal sufficient statistic for $U(\theta,\theta+c)$, where $(\theta,c)$ unknown.

Suppose $X_1,\cdots,X_n$ are $i.i.d$ from a distribution with p.d.f
$$\delta_{(\theta,c)}(x)=\frac{1}{c}\mathbb{1}_{(x\in[\theta,\theta+c])},$$
where $\theta\in\mathbb{R}$ and $c\in\mathbb{R}^+$ unknown.

Find a minimal sufficient statistic for $(\theta,c)$.

From the range of $x_i$, i.e. $x_i\in[\theta,\theta+c]$, we can determine that $\theta\leq x_i$ and $\theta+c\geq x_i$, $\forall i\in\{1,\cdots,n\}$. This implies $$\theta\leq x_{(1)}\text{ and }\theta+c\geq x_{(n)},$$ where $x_{(1)}=\underset{i\in\{1,\cdots,n\}}{\min}x_i$ and $x_{(n)}=\underset{i\in\{1,\cdots,n\}}{\max}x_i$.

The plots shows the area of $\theta\leq x_{(1)}\text{ and }\theta+c\geq x_{(n)}.$ It look like that $(x_{(1)},x_{(n)})$ is a minimal sufficient statistic for $(\theta,c)$ because $(x_{(1)},x_{(n)})$ uniquely determines the shape of the log-likelihood function. Is this correct?

答:
This is an attempt at an alternative proof that $(x_{(1)}, x_{(n)})$ is a minimal sufficient statistic for $(\theta, c)$ using a direct approach.

Consider the likelihood ratio
$$ R(x,y) = \frac{p(x | \theta, c)}{p(y | \theta, c)} $$
where $x = (x_{1}, x_{2}, \dots x_{n}), y = (y_{1}, y_{2}, \dots y_{n})$. The ''factorization'' theorem of minimal sufficient statistics states that if a statistic $T(x)$ has the following property P then it is a minimally sufficient statistic for $\theta$:

P: $R(x, y)$ does not depend on $\theta$ if and only if $T(x) = T(y)$.

I claim that $T(x) = (x_{(1)}, x_{(n)})$ has this property and therefore is minimally sufficient.

To see this, first consider what $p(x | \theta, c)$ is. I claim that
$$p(x | \theta, c) = c^{-n} I[x_{(1)} \geq \theta, x_{(n)} \leq \theta+c].$$
The indicator $I[x_{(1)} \geq \theta, x_{(n)} \leq \theta+c]$ just determines if $x$ is a valid value in the range; certainly $p(x | \theta, c) = 0$ if $x$ is not within $[\theta, \theta+c]$. Otherwise, since the distribution is uniform we have that $p(x | \theta, c)$ should just be $c^{-n}$ because the density function of each $x_{i}$ is just $c^{-1}$.

Therefore, we have that
$$
R(x, y) =  \frac{p(x | \theta, c)}{p(y | \theta, c)} = \frac{I[x_{(1)} \geq \theta, x_{(n)} \leq \theta+c]}{I[y_{(1)} \geq \theta, y_{(n)} \leq \theta+c]}
$$
since the $c^{-n}$ values cancel out.

At this point, I should admit that the factorization theorem is not technically be applicable here, but I'd guess that it is possible to extend that theorem to this situation. Intuitively, we can see that the ratio of $p(x | \theta, c)$ to $p(y | \theta, c)$ is independent of $(\theta, c)$ if and only if $T(x) = T(y)$, despite the technicality that $R(x, y) = \frac{0}{0}$ when both $x$ and $y$ are not in the interval. That is, if we pretend that $\frac{0}{0} = 1$, then $R(x, y) = 1$ whenever $T(x) = T(y)$, whereas it might either be 0 or $+\infty$ when $T(x) \not = T(y)$.


问:
number of injective and surjective functions? [verification]
Hello there's a question where you need to find the total possible injective and surjective possibilities. Could someone verify my  answer? I'm almost certain of answer a, but want to know if b is also correct?

If you know if its true can you reply, it would be very helpful. If it's wrong could you explain why?

Thanks in advance

question

assume |X|=2 and |Y|=2016

a) How many injective functions are there from X to Y?
My answer = $\frac{2016!}{2014!}$

b) How many surjective functions are there from Y to X?
My answer = $2^{2016}-2$

答:
Your answer for injections is correct.

Your answer for surjections is incorrect. In total we have $2^{2016}$ functions. However the only way a function is non-surjective in this case is if both of the elements in $X$ are mapped to the same element in $Y$, and there are $2016$ elements in $Y$.

So there are actually $2^{2016}-2016$ injections.


问:
Calculate an undefined size (n) determinant
Simply put, not quite sure where to even start with something like this..

Calculate the n-sized determinant
enter image description here

答:
The easiest way in my opinion, but requires more knowledge, is to recognize it as the product of the eigenvalues.  View the matrix as $\text{Ones}_{n\times n} + (n-1)I$ where $\text{Ones}_{n\times n}$ is the $n\times n$ matrix consisting of all $1$'s.

Note $\text{Ones}_{n\times n}$ has eigenvalue $0$ with geometric multiplicity $n-1$ (since it is clearly rank1) and eigenvalue $n$ (since the others are zero and the trace is $n$ which should have been the sum of the eigenvalues) with geometric multiplicity $1$.

$(n-1)I$ obviously has eigenvalue $n-1$ with geometric multiplicity $n$, no effort required, and can use the same eigenbasis as $\text{Ones}_{n\times n}$ since this is the identity matrix we are talking about.

It follows then that your matrix has eigenvalue $n-1$ with geometric multiplicity $n-1$ and eigenvalue $2n-1$ with geometric multiplicity $1$.  The determinant is then:

$$(2n-1)(n-1)^{n-1}$$


问:
Questions on differential equations of matrices
I have a differential equation $$N'_x(x)=G(x)N(x)$$ where $N, G$ are $2\times2$ matrices depending on $x$, and $G$ satisfies $\sigma G+G\sigma=0$, $\sigma$ is one half of the pauli matrix, i.e. $$\sigma=\begin{pmatrix}\frac{1}{2}&0\\
0&\frac{-1}{2}\end{pmatrix}$$ My question is:

Would $N^{\ast}\sigma N$ then be independent of $x$? Why or why not?

答:
Assuming you mean that $N^\prime(x) = G(x)N(x)$, it follows by linearity of the operator $A \mapsto A^\ast$ that  $$(N^\ast)^\prime(x) = (N^\prime(x))^\ast = (G(x)N(x))^\ast = N(x)^\ast G(x)^\ast.$$
Hence, by the noncommutative Leibniz rule,
$$
 (N(x)^\ast\sigma N(x))^\prime = (N(x))^\ast \sigma N(x) + N(x)^\ast \sigma^\prime N(x) + N(x)^\ast \sigma N(x)^\prime\\ = N(x)^\ast G(x)^\ast \sigma N(x) + N^\ast(x) \sigma G(x) N(x)\\ = N(x)^\ast\left(G(x)^\ast - G(x)\right)\sigma N(x).
$$
What can you conclude from this? In particular, what would be a convenient condition to impose on $G$ to guarantee that $(N(x)^\ast\sigma N(x))^\prime=0$?</p>


问:
Approximating $\int \log\big[1+\sin^2(t)\big]\,dt$
On yesterday, a now closed question asked for a while how to compute the antiderivative
$$I=\int \log\big[1+\sin^2(t)\big]\,dt$$ which, as given by a CAS, is just awful (see @Forester's comment and @Turing's answer). During the time the question was open, I tried on my side and gave up since being unable to find anything simpler.

Then, I tried to approximate the integrand since we just need a good representation of it for $0 \leq t \leq \frac \pi 2$ because of the symmetries and periodicity. What I am proposing is a polynomial function matching the function and its first and second derivative values at a minimum number of data points. For the time being, I chose $t=0$, $t=\frac \pi 4$ and $t=\frac \pi 2$. The final result is
$$f(t)=\log\big[1+\sin^2(t)\big]\sim t^2+\frac 19 \sum_{n=3}^8 \frac{a_n}{\pi^n} t^n=g(t)$$ the $a_n$'s being
$$\left(
\begin{array}{ccc}
n & a_n & \frac{a_n}{9\pi^n}\\
 3 & -768 \pi -217 \pi ^2-22608 \log (2)+18432 \log (3) &         +0.087806356 \\
 4 & 7680 \pi +1940 \pi ^2+286416 \log (2)-221184 \log (3) &      -1.360303642 \\
 5 & -27648 \pi -8500 \pi ^2-1501632 \log (2)+1105920 \log (3) &  +1.225383826 \\
 6 & 43008 \pi +19568 \pi ^2+3960576 \log (2)-2801664 \log (3) &  -0.512958585 \\
 7 & -24576 \pi -22720 \pi ^2-5170176 \log (2)+3538944 \log (3) & +0.102611702 \\
 8 & 10496 \pi ^2+2654208 \log (2)-1769472 \log (3) &             -0.007207813
\end{array}
\right)$$

About the norm
$$\Phi=\int_0^{\frac \pi 2} \big[f(t)-g(t)\big]^2 \, dt=1.79 \times 10^{-8}$$ while the beautiful approximation
$$\sin(t) \simeq \frac{16 (\pi -t) t}{5 \pi ^2-4 (\pi -t) t}\qquad \text{for} \qquad 0\leq t\leq\pi$$ would lead to  $\Phi=8.74 \times 10^{-7}$ (almost $50$ times larger).

From this
$$J(x)=\int_0^x \log\big[1+\sin^2(t)\big]\,dt\sim \frac 13 x^3+\frac 19 \sum_{n=3}^8 \frac{a_n}{(n+1)\,\pi^n} x^{n+1}$$

As a test
$$J\left(\frac{\pi }{2}\right)=\frac{\pi }{181440}\left(432 (128 \log (3)-87 \log (2))-55 \pi ^2 \right)\approx 0.591383$$ to be compared to the exact value
$$ \pi \log\left(\frac{1+\sqrt{2}}{2}\right)\approx 0.591331$$ recalled by @projectilemotion in comments.

My questions

Without adding more terms, could it be possible to improve the approximation ? Would changes of base points could have a significant impact (if yes, how to optimize them ?)

Would, by chance, be known (in a simple form) the exact values of the integrals
$$K_n=\int_{-\frac \pi n}^{\frac \pi n} \log\big[1+\sin^2(t)\big]\,dt$$ beside the case of $n=2$ ? If there were, I suppose that much better approximations could be done.

答:
A physicist's point of view. So i'm trying to keep things as simple as possible.
(I'm using $\ln x$ rather than $\log x$).

The first step is to choose an approximate, simple analytical expression for $\ln(1+y)$ in $y\in[0,1]:$

$$\ln(1+y)\approx\frac{\ln 2}{131}y(20y^2-75y+186) $$

Now, set $y=\sin^2t$

$$\ln(1+\sin^2t)\approx\frac{\ln 2}{131}(131+35\cos^2t+20\cos^4t)\sin^2t$$

Next, integrate this expression with respect to $t$ over the interval $[0, x]$ $x\in[0,\frac{\pi}{2}]$ and denote the result $I(x)$

A closed-form expression can be obtained for $I(x)$ but it is a little bit messy. But for $x=\frac{\pi}{2}$ we obtain:

$$I\left(\frac{\pi }{2}\right)=\frac{569}{2096}\pi\ln 2$$

Absolute error from exact value is about $0.0002$

It is about an order of magnitude more inaccurate than Claude's result $J\left(\frac{\pi }{2}\right)$ but this is the price we have to pay for simplicity.


问:
Chow ring of projetive space
The Chow ring of $\mathbb{P}^n$ is $\mathbb{Z}[H]/H^{n+1}$, where $H$ is the linear equivalence lass determined by the zero-set of a linear functional. I have a few (very basic) points of confusion:

Firstly, intersection of curves of various degrees in $\mathbb{Q}[H]/H^{n+1}$ should correspond to tensor product of line bundles in the Grothendieck ring. $aH$ is nilpotent in this ring, and therefore not invertible. Yet it corresponds to a line bundle in the grothendieck ring, which is invertible. Where do I go wrong here?

It seems like the intersection of a point and a curve could be either a point or $\emptyset$. Why is the intersection in $\mathbb{Z}[H]/H^{n+1}$ equal to $0$?

答:
The Chow ring tries to capture what happens in the generic case. Generically, a point and a curve do not intersect. Neither does a pair of curves if $n > 2$.

To your first point, line bundles correspond to divisors (codimension 1) not to curves. You're referring to the identity that looks something like

$$ D_1 \cdot D_2 \cdots D_n = \deg(\mathcal{O}(D_1) + \dots + \mathcal{O}(D_n)). $$

But as far as I know, the tensor product $\mathcal{O}(D_1) + \dots + \mathcal{O}(D_n)$ does not correspond to the intersection $D_1 \cap \cdots \cap D_n$ which is a finite set of points, it corresponds to the data $(D_1, \dots, D_n)$.

For example, in $\mathbb{P}^2$, given say $6$ points, we can factor $\mathcal{O}(6) = \mathcal{O}(1) + \mathcal{O}(6) = \mathcal{O}(2) + \mathcal{O}(3)$. Given two curves of degrees $(1,6)$ or $(2,3)$, their intersection is (generically) 6 points, but given 6 points, you can't just construct two curves whose intersection is those 6 points. If the 6 points are generic there are simply no lines or conics that pass through them in the first place.

So the symbol "$\mathcal{O}(2) + \mathcal{O}(3)$" contains more information than just the number "6". In particular, it contains more information than the intersection product $C_1 \cdot C_2$ of two curves of degrees 2 and 3. That's why we need to lose information by taking the degree for these to be equal.

Title: Chow ring of projetive space


问:
Show that a point inside an acute triangle is the orthocenter
Point $M$ lies inside an acute triangle $\triangle ABC$ such that the circumscribed circles of $\triangle ABM,\triangle BCM$ and $\triangle ABC$ have equal radii. Show that $M$ is the orthocenter.

I have been trying to solve the problem for an hour using the law of sines because we have just studied it, but I can't come up with anything. What about a solution using vectors?

答:
Let the centre of the circle opposite vertex $A$, $B$ and $C$ be $A_{1}$, $B_{1}$ and $C_{1}$ respectively. Draw $A_{1}B$, $A_{1}C$, $CB_{1}$, $B_{1}A$, $AC_{1}$ and $C_{1}B$.

Observe that, in quadrilateral $A_{1}BC_{1}M$, all of the sides are equal; Hence,it's a rhombus. Similarly, quadrilateral $AC_{1}MB_{1}$ and $CB_{1}MA_{1}$ are rhombi.

Now, let the circumcentre of $\triangle ABC$ be $O$. Draw $OA$, $OB$ and $OC$.

$\angle BA_{1}C=\angle BA_{1}M+\angle MA_{1}C=\angle BC_{1}M+\angle MB_{1}C=2\angle BAM+2\angle MAC=2\angle BAC$

Quadrilateral $BOCA_{1}$ is a kite since $OB=OC$ and $BA_{1}=A_{1}C$.

Also, in quadrilateral $BOCA_{1}$, $\angle BOC=2\angle BAC=\angle BA_{1}C$.
$\Rightarrow$   Quadrilateral $BOCA_{1}$ is a rhombus.

Similarly, it can be proven that quadrilateral $OBC_{1}A$ and $AOCB_{1}$ are rhombi as well.

$\Rightarrow  OB=A_{1}C$

Hence, the circumradii of $\triangle AMB$, $\triangle BMC$ and $\triangle CMA$ are all equal to the circumradius of $\triangle ABC$.

Notice that, in $\triangle AC_{1}B_{1}$ and $\triangle OBC$, $AC_{1}=OB$, $AB_{1}=OC$ and $\angle C_{1}AB_{1}=\angle BOC$ [Since $OB\parallel AC_{1}$ & $OC\parallel AB_{1}$]

Hence, $\triangle AC_{1}B_{1}\cong \triangle ABC$ by $S-A-S$ criterion of congruence.
$\Rightarrow  B_{1}C_{1}=BC$

Observe that, in quadrilateral $B_{1}C_{1}BC$, $BC_{1}=B_{1}C$ and $B_{1}C_{1}=BC$. Hence, it's a parallelogram.
$\Rightarrow  B_{1}C_{1}\parallel BC$

Now, since $AM\perp B_{1}C_{1}$$[$$AC_{1}MB_{1}$ is a rhombus$]$ and $B_{1}C_{1}\parallel BC$,  $AM\perp BC$.

Similarly,it can be proven that $BM\perp CA$. Therefore, $M$ is the orthocentre of $\triangle ABC$.


问:
Evaluating $(1+x)(1+x^2)\ldots (1+x^{p-1})$ at pth roots of unity for odd primes p
As part of a combinatorics question I was able to simplify it into evaluating $(1+x)(1+x^2)\ldots (1+x^{p-1})$ at pth roots of unity for odd primes p.

I have guessed and checked with a computer that this product is always $1$ except for when we use the root of unity $1$, in which case it's $2^{p-1}$ (the latter case is trivial)

I'm not sure how to prove that the product is $1$ for non $1$ pth roots of unity.

What I've tried:

Let $z$ be a pth root of unity that isn't equal to $1$.

Hence the product equals
$$ \prod_{k=1}^{\frac{p-1}2}(1+z^k)(1+z^{-k}) \quad\quad\text{(pairing opposite factors)}$$
$$ = \prod_{k=1}^{\frac{p-1}2}\left(2\cdot\cos\left(\frac{k\pi}{p}\right)\right)^2$$

However I see no way to go forward from here and prove the product equals $1$.

答: First note that since $p$ is prime for any p-th root of unity $x$ we know that $-x$ is a p-th root of $-1$. Furthermore we know $X^p-1=\prod_{i=0}^{p-1}(X-x^i)$, thus we have $X^p+1=\prod_{i=0}^{p-1}(X+x^i)$ as by the above mentioned we know all roots of $-1$ hence we know how the polynomial factors(alternatively multiply the equation with $-1$ to obtain $-X^p+1=-\prod_{i=0}^{p-1}(X-x^p)=-(-1)^p\prod_{i=0}^{p-1}(-X+x^i)=\prod_{i=0}^{p-1}(-X+x^i)$ as p is odd, and presume as below with the only difference evaluating at $-1$ and adjusting the sign of $X$). Now we observe that $x^0=1$ hence we have $X^p+1=(X+1)\prod_{i=1}^{p-1}(X+x^i)$. Hence evaluating at $X=1$ yields $2=1^p+1=2\prod_{i=1}^{p-1}(1+x^i)$ i.e. $1=\prod_{i=1}^{p-1}(1+x^i)$.


问:
how to dot product two vectors with different planes?
how to dot product two vectors with different planes?

I have vectors $A$,$B$ and $C$, vectors $A$ and $B$ is on $xy$ plane while vector $C$ is on $xz$ plane. I need to find the dot product of $A.C$ how should I do that? my book says that dot product of two vectors can be expressed in terms of their rectangular components. vector $B$ lies in $y$-axis. vector $A$ makes $60$ degrees to $B$, vector $C$ makes $37$ degrees to $x$-axis. $A=10$, $B=8$ and $C=5$.

please help me to solve this problem

答:
B lies in y- axis , then A makes 30 degrees with x-axis, so that:

$$A=10\cos30\, i+10\sin30\,j\qquad A=5\sqrt3\,i+5\,j$$

$$C=5\cos37\,i+5\sin37\,k\quad =4\,i+3\,k$$

$$A.C=20\sqrt3$$


答:
To your specific questions,

It's not true that there's a correspondence of curves to line bundles taking intersection to tensoring. There's a correspondence of divisors (co-dimension one subvarieties) to line bundles, but line bundle tensoring corresponds to taking a union of divisors.

A point intersecting a curve in a big-dimensional space is not a transverse intersection. For a heuristic picture of what intersection theory in algebraic geometry is trying to capture that explains why examples like this ought to be zero, you might consider reading Guillemin and Pollack's book on Differential Topology. Or, an algebro-geometric way to say it is that if the curve is passing thru a specified point, the curve is not "generic" in its equivalence class w.r.t. the point.

Title: Chow ring of projetive space

Body:
The Chow ring of $\mathbb{P}^n$ is $\mathbb{Z}[H]/H^{n+1}$, where $H$ is the linear equivalence lass determined by the zero-set of a linear functional. I have a few (very basic) points of confusion:

Firstly, intersection of curves of various degrees in $\mathbb{Q}[H]/H^{n+1}$ should correspond to tensor product of line bundles in the Grothendieck ring. $aH$ is nilpotent in this ring, and therefore not invertible. Yet it corresponds to a line bundle in the grothendieck ring, which is invertible. Where do I go wrong here?

It seems like the intersection of a point and a curve could be either a point or $\emptyset$. Why is the intersection in $\mathbb{Z}[H]/H^{n+1}$ equal to $0$?


问:
Find a function $f(x)$ on $(0,\infty)$ with arclength $(x+1)^2$

Give an analytic definition for a function $f(x)$ defined on $(0,\infty)$ for which the arclength from $0$ to $d$ is always $(d+1)^2$.

I’m generally pretty competent with calculus, and I think that my setup ought to be $$(d+1)^2=\lim_{a\to 0^+}\int_a^d \sqrt{1+f’(x)^2}dx$$ But I’m kind of lost as to how to actually evaluate this. For one thing, I know that the arclength integral usually has no “nice” antiderivative, but I don’t think that’s a problem because I’m just asking about a case in which it does. Next, I think the function has some sort of vertical asymptote at $0$ (because at $x=\varepsilon$ for $\varepsilon$ arbitrarily close to $0$, the arclength is already $(1+\varepsilon)^2>1$, and in particular $>\varepsilon$). Otherwise, I’m pretty much stumped. I could maybe take the derivative of both sides? But then I don’t know how to show that the derivative necessarily commutes inside the limit, and I’m not sure what to do with what would come out of that anyway.

I got this question by modifying a similar one asked on a math LinkedIn group that asked about arclength $d^2$, for which I noted that the arclength would by shorter than that of a straight line for $0<x<1$, so no such function could exist. But I’m pretty sure no similar argument would work for this function because $(x+1)^2>x$ (and in particular $(x+1)^2\not<x$) for all real $x$.

Any help or even hints would be appreciated. Thanks!

Body:
Pick $a>0$. Then for all $t>a$, we need
$$ (t+1)^2=(a+1)^2+\int_a^t \sqrt{1+f’(x)^2}dx.$$
Apply $\frac{\mathrm d}{\mathrm dt}$ to both sides:
$$ 2(t+1)=\sqrt{1+f'(t)^2}.$$
Solve for $f'(t)$:
$$ f'(t)=\sqrt{(2(t+1))^2-1}=\sqrt{4t^2+8t+3}.$$
Conclude that
$$ f(x)=f(0)+\int_0^x\sqrt{4t^2+8t+3}\,\mathrm dt.$$


问:
For every set $S \subseteq \mathbf{R}$, show that $int(S) \subseteq ext(ext(S))$.
I would like to ask, if the following constitutes a valid proof, or should I attempt it again?

Let $ext(S)$ denote exterior of set $S \subseteq \mathbb R$.

For every set $S \subseteq \mathbf{R}$, show that $int(S) \subseteq ext(ext(S))$.

My attempt.

Let $x$ be an arbitrary point in $int(S)$. By definition of the interior of a set, there exists an $\epsilon > 0$ such that the open interval $(x - \epsilon,  x + \epsilon)$ is contained in $S$. We are interested to show that, $(x - \epsilon, x + \epsilon) \in (ext(S))^C$.  Pick an arbitrary $y \in (x - \epsilon, x + \epsilon) \subseteq S$. We choose $\delta = \min \{\frac{y-(x-\epsilon)}{2},\frac{(x+\epsilon)-y}{2}\}$, so $(y - \delta, y+\delta) \subseteq (x - \epsilon, x + \epsilon) \not\subseteq S^C$. So, $y \notin ext(S)$. Therefore, $y \in (ext(S))^C$. This is true for all $y \in (x - \epsilon, x + \epsilon)$. So, $(x - \epsilon, x + \epsilon) \subseteq (ext(S))^C$. Consequently, $x \in (ext(ext(S))$.

Thus, $int(S) \subseteq ext(ext(S))$.

答: Easier : if $x \in O=\operatorname{int}(S)$ then the open $O$ witnesses also that $O$ is a subset of $\operatorname{ext}(S)^\complement$ so in its interior i.e. $\operatorname{ext}(\operatorname{ext}(S))$ and so you’re done showing the inclusion. The metric or order plays no role.


问:
Prove a relation is an ordering relation when the other is also one
Let $R$ be a relation on a set $A$, then define $R’$ on $A\times A$ by
$$(a_1,a_2)R'(b_1,b_2)\Leftrightarrow a_1R\,a_2\wedge b_1R\,b_2$$

Prove that $R’$ is an ordering relation when $R$ is an ordering relation.

the relation $R’$ is a linear ordering when $R=\mathrm{id}_A$ and the number of elements in $A$ is $1$

An ordering relation is reflexive, antisymmetric, and transitive.
Where as a linear one is connex , antisymmetric and transitive.
I have no clue how to use the definition of $R’$ compared to $R$.
For reflexivity, I said if $R$ is reflexive, it means that the set A must contain the $R=\mathrm{id}_A$ and therefor the set $A \times A$ would also contain $R=\mathrm{id}_A$, therefore $R’$ is reflexive exactly when $R$ is reflexive, but I still don’t know how to fill in the definition.
My idea so far is working on the right side of the equation where both $a_1$ and $a_2$ equals $x$, both $b_1$ and $b_2$ equals $y$ but I’m not sure how to proceed.

答:
Partial Answer:
$\newcommand{\op}[2]{ \left\langle #1 ,\, #2 \right\rangle }$
$\newcommand{\rel}[1]{ \mathcal{#1} }$

Let $a = \op{a_1}{a_2}$ and $b = \op{b_1}{b_2}$.
We can redefine $\rel{R'}$:
\begin{align}
\op{a}{b} \in \rel{R'} \iff \{ a ,\, b \} \subseteq \rel{R}
\end{align}

The order of elements in a set does not matter, i.e. $\{ a ,\, b \} = \{ b ,\, a \}$, so the relation $\rel{R'}$ is always symmetric.

Because $\rel{R}$ is reflexive, we have $\forall x \in A \left( \op{x}{x} \in \rel{R} \right)$.

Let $a = b = \op{x}{x}$, obviously $\{ a ,\, b \} = \{ a ,\, a \} = \{ a \} \subseteq \rel{R}$, therefore $\op{a}{a} \in \rel{R'}$, i.e. $\rel{R'}$ is reflexive.

Let $x = \op{x_1}{x_2},\; y = \op{y_1}{y_2},\; z = \op{z_1}{z_2}$, and $\{ x,\, y,\, z \} \subseteq \rel{R}$.

\begin{align}
\{ x ,\, y \} \subseteq \rel{R} &\implies \op{x}{y} \in \rel{R'} \\
\{ y ,\, z \} \subseteq \rel{R} &\implies \op{y}{z} \in \rel{R'} \\
\{ x ,\, z \} \subseteq \rel{R} &\implies \op{x}{z} \in \rel{R'}
\end{align}

Therefore, $\rel{R'}$ is always transitive.

To prove $\rel{R'}$ is antisymmetric, we have to prove $\rel{R'} \cap \rel{R'}^{-1} \subseteq \Delta x$, where $\Delta x = \{ \op{x}{x} :\: x \in A \}$.
Because $\rel{R'}$ is always symmetric, $\rel{R'} = \rel{R'}^{-1}$. Then we have to prove $\rel{R'} \cap \rel{R'}^{-1} = \rel{R'} \subseteq \Delta x$. Meanwhile, $\rel{R'}$ is reflexive, so $\Delta x \subseteq \rel{R'}$. Therefore, to prove $\rel{R'}$ is antisymmetric, we have to prove $\rel{R'} = \Delta x$, or $$\forall x,\,y \in A \left( \begin{array}{c} x = y \iff \op{x}{y} \in \rel{R'} \\ x \neq y \iff \op{x}{y} \notin \rel{R'} \end{array} \right)$$
Furthermore,
$$\forall x,\,y \in A \left( \begin{array}{c} x = y \iff \{ x,\,y \} \subseteq \rel{R} \\ x \neq y \iff \{ x,\,y \} \nsubseteq \rel{R} \end{array} \right)$$

For all $x_1,\,x_2 \in A$, let $x = \op{x_1}{x_2},\; y = \op{x_2}{x_1}$. Obviously, $x = y$ when $x_1 = x_2$, $x \neq y$ when $x_1 \neq x_2$.

Because $\rel{R}$ is antisymmetric, assumming $x_1 \neq x_2$ and $x = \op{x_1}{x_2} \in \rel{R}$, we can get $y = \op{x_2}{x_1} \notin \rel{R}$.
Now we have $x \neq y \implies \{ x,\, y \} \nsubseteq \rel{R}$.

Assume $\{ x,\, y \} \nsubseteq \rel{R}$, then $ x \notin \rel{R} \lor y \notin \rel{R}$, i.e. $ \op{x_1}{x_2} \notin \rel{R} \lor \op{x_2}{x_1} \notin \rel{R}$ holds. Because $\rel{R}$ is reflexive, $\op{x_1}{x_1} \in \rel{R}$ and $\op{x_2}{x_2} \in \rel{R}$. Now we get $x_1 \neq x_2$, i.e. $\{ x,\, y \} \nsubseteq \rel{R} \implies x \neq y$.

Therefore, $\rel{R'}$ is antisymmetric.

All in all, $\rel{R'}$ is an ordering relation when $\rel{R}$ is one.


问:
Why is the determinant of a rotation matrix equal to 1?
Why is the determinant of a rotation matrix equal to 1? I would like a geometric interpretation of this. Just curious.</p>

答:
The determinant of the standard rotation matrix is 1, this can be calculated very easy in 2D (see Answers above) and in 3D as well:

$\begin{bmatrix}
  1&0&0\\
  0&c\Phi&-s\Phi\\
  0&s\Phi&c\Phi\\
\end{bmatrix} = R \left( \Phi \right)$

$\begin{bmatrix}
  c\Theta&0&s\Theta\\
  0&1&0\\
  -s\Theta&0&c\Theta\\
\end{bmatrix} = R \left( \Theta \right)$

$\begin{bmatrix}
  c\Psi&-s\Psi&0\\
  s\Psi&c\Psi&0\\
  0&0&1\\
\end{bmatrix} =
R \left( \Psi \right)$

e.g.: $|| R \left( \Phi \right) || = 1 * cos^2(\phi) + sin^2(\phi) = 1$

and the others are identical.

Now using det(A*B) = det(A)*det(B) proof it is obvious to see, that any rotation matrix you builded from those three above will have a determinant = 1

edit: Thanks for link to the tutorial!


问:
Arithmetic question- invariance of objects involved
I have an exceedingly simple question, but something that has been
eating at me for some time now. Imagine that we wish to perform subtraction
of objects with the same units, say \$. I have a total of 10\$, composed
of 5\$ of my own money, and 5\$ of my brother's. My brother then requests
his money back (which I have identified by markings on the bills), and
I have left
$$
10\$-5\$=5\$
$$

Imagine, however, that I had not identified my brother's portion
of the money, and had jumbled it all up. My brother asks for his money
back, and again, he receives $5\$,$but not necessarily his own money.
I understand there is no difference in the quantity of money he receives-
what is the property of arithmetic/objects that renders these two
values the same? In which case is there an appreciable difference
between these two values? If not, why is there no difference, even though we intuitively understand there to be one?

Many
thanks!

答: There is no difference, because a dollar is a dollar, irrespective of where it comes from. If you want to make a difference you would have to add a label and you would have e.g. $(x,y)$ with $x$ the amount of dollars and $y$ and extra parameter, here $y=$ the money given by you or $y=$ the money given by your brother. If you call $V(x,y)$ the value of the money with parameters $(x,y)$ then we have generally accepted $V(x,y) = V(x,z)$ for all possible $x,y,z$. But you and your brother may decide that for your guys $V(x,y)\neq V(x,z)$ for whatever reason.


问:
Looking for a reference of integral involving product of four spherical harmonics
We know
$$\int d \Omega Y_{l_1m_1}(\theta,\phi) Y_{l_2 m_2}(\theta,\phi)  Y_{l_3 m_3 } (\theta,\phi) = \sqrt{ \frac{ (2l_1 + 1)(2 l_2+1)(2l_3+1)}{4\pi} } \pmatrix{ l_1 l_2 l_3 \\ m_1 m_2 m_3 } \pmatrix{ l_1 l_2 l_3 \\ 0 0 0 } $$

Is there any reference for integral involving four spherical harmonics? I could try to work it out myself, but a cross check with literature maybe useful

答:
We can always reduce the product of two spherical harmonics to a sum over one spherical harmonic (DLMF 34.3.20),
$$
Y_{l_1m_1} \, Y_{l_2m_2}
= \sum_{lm} \sqrt{\tfrac{(2l_1+1)(2l_2+1)(2l+1)}{4\pi}} \, \bigl(\begin{smallmatrix}l_1&l_2&l\\0&0&0\end{smallmatrix}\bigr) \, \bigl(\begin{smallmatrix}l_1&l_2&l\\m_1&m_2&m\end{smallmatrix}\bigr) \, \overline{Y_{lm}} \\
= \sum_{lm} (-1)^m \, \sqrt{\tfrac{(2l_1+1)(2l_2+1)(2l+1)}{4\pi}} \, \bigl(\begin{smallmatrix}l_1&l_2&l\\0&0&0\end{smallmatrix}\bigr) \, \bigl(\begin{smallmatrix}l_1&l_2&l\\m_1&m_2&-m\end{smallmatrix}\bigr) \, Y_{lm}
$$
Hence for the integral over $n+1$ spherical harmonics:
$$
\int Y_{l_1m_1} \cdots Y_{l_nm_n} \, Y_{l_{n+1}m_{n+1}} \\
= \sum_{lm} (-1)^m \, \sqrt{\tfrac{(2l_n+1)(2l_{n+1}+1)(2l+1)}{4\pi}} \, \bigl(\begin{smallmatrix}l_n&l_{n+1}&l\\0&0&0\end{smallmatrix}\bigr) \, \bigl(\begin{smallmatrix}l_n&l_{n+1}&l\\m_n&m_{n+1}&-m\end{smallmatrix}\bigr) \int Y_{l_1m_1} \cdots Y_{l_{n-1}m_{n-1}} \, Y_{lm}
$$
We can build expressions recursively from there. For example, the next formula after $n = 3$:
$$
\int Y_{l_1m_1} \, Y_{l_2m_2} \, Y_{l_3m_3} \, Y_{l_4m_4} \\
= \sum_{lm} (-1)^m \, \sqrt{\tfrac{(2l_3+1)(2l_4+1)(2l+1)}{4\pi}} \, \bigl(\begin{smallmatrix}l_3&l_4&l\\0&0&0\end{smallmatrix}\bigr) \, \bigl(\begin{smallmatrix}l_3&l_4&l\\m_3&m_4&-m\end{smallmatrix}\bigr) \int Y_{l_1m_1} \, Y_{l_2m_2} \, Y_{lm} \\
= \tfrac{\sqrt{(2l_1+1)(2l_2+1)(2l_3+1)(2l_4+1)}}{4\pi} \sum_{lm} (-1)^m \, (2l+1) \, \bigl(\begin{smallmatrix}l_1&l_2&l\\0&0&0\end{smallmatrix}\bigr) \, \bigl(\begin{smallmatrix}l_3&l_4&l\\0&0&0\end{smallmatrix}\bigr) \, \bigl(\begin{smallmatrix}l_1&l_2&l\\m_1&m_2&m\end{smallmatrix}\bigr) \, \bigl(\begin{smallmatrix}l_3&l_4&l\\m_3&m_4&-m\end{smallmatrix}\bigr)
$$
The sum over $m$ picks out the condition that $m_1 + m_2 + m_3 + m_4 = 0$, and the integral vanishes otherwise. Assuming the condition is fulfilled:
$$
\int Y_{l_1m_1} \, Y_{l_2m_2} \, Y_{l_3m_3} \, Y_{l_4m_4} \\
= (-1)^{m_1+m_2} \, \tfrac{\sqrt{(2l_1+1)(2l_2+1)(2l_3+1)(2l_4+1)}}{4\pi} \sum_{l} (2l+1) \, \bigl(\begin{smallmatrix}l_1&l_2&l\\0&0&0\end{smallmatrix}\bigr) \, \bigl(\begin{smallmatrix}l_3&l_4&l\\0&0&0\end{smallmatrix}\bigr) \, \bigl(\begin{smallmatrix}l_1&l_2&l\\m_1&m_2&m_3+m_4\end{smallmatrix}\bigr) \, \bigl(\begin{smallmatrix}l_3&l_4&l\\m_3&m_4&m_1+m_2\end{smallmatrix}\bigr)
$$
I don't immediately see if the remaining sum reduces to a known expression, but the $3j$-symbols with $000$ at least can be reduced further, so perhaps there is something more to be done. I am marking this answer community wiki in the hope that someone might be able to improve it.


问:
How do you deduce X given a set of truth value?
given some truth value, how can we indeed deduce what the form is like? For example,

P Q R X

T T T T

T T F F

T F T T

T F F T

F T T T

F T F F

F F T F

F F F T

Here, given the truth value of X, can you deduce the form of X in term of P,Q,R?? I think if time are given one can guess the formula but if the case become more complicated, it seem very inefficient to guess. Is there any more general technique or algorithm?

答:
For any truth table, there are many logical expressions that have that truth table. We give a simple mechanical method for producing such an expression. The main disadvantage is that in general, the expression so obtained is quite a bit longer than necessary.

Look at all the entries that give truth value $T$. The first one is $TTT$. Write down $P\land Q\land R$.

The next one is $TFT$. Write down $P\land \lnot Q\land R$.

The next one is $TFF$. Write down $P\land \lnot Q\land \lnot R$.

Continue, one for every combination of truth values that yields $T$. The remaining two are $FTT$ and $FFF$. Write down the appropriate conjunctions.

Put $\lor$'s between all the things you wrote down. We get the expression
$$(P\land Q\land R)\lor (P\land \lnot Q\land R)\lor (P\land \lnot Q\land \lnot R) \lor (\lnot P\land Q\land R)\lor (\lnot P\land \lnot Q\land \lnot R) .$$

A little thinking shows why this gives the right truth table. It certainly gives $T$ at the right places. And at the places where the truth table yielded $F$, everyone of the conjunctions that were $\lor$-ed is false.

The above procedure gives us what is called the full disjunctive normal form (DNF). This form may be quite inefficient. There are good algorithms for simplifying a DNF to use a much smaller number of connectives. This minimization process is an important part of circuit design.

We can also do some hand-optimization. For example, note that there are only $3$ combinations that give the value $F$, as against $5$ that give the value $T$. So write down the disjunctive normal form we would get if the $F$ and $T$ in the last column had been switched,  put an $\lnot$ in front. and use De Morgan's Laws to simplify. We will get a conjunctive normal form that is simpler than the DNF we got. There are many other tricks/methods.


问:
Prove a student participates in all 2009 available courses.
In a university there are available 2009 courses in total. Each course has 45 enrolled students. Between every 2 courses there is exactly one student who was enrolled in both. Prove that there is a student who has enrolled in all of the courses.

I tried building a hypergraph where the courses are nodes and the students edges but this did not bring much. Any help is appreciated.

答:
Assume not. Then, we claim that a student can be in at most $45$ courses.

Proof: Take a student $S$ and a course $C$ such that $S$ is not in the course $C$. Let $E$ be the set of courses which contain $S$. Now, for every $E_i$ in $E$, we have $E_i \cap C$ contains exactly one student $e_i$. Note that these students are all distinct. So, if $\lvert E \rvert > 45$, then we would have a contradiction since $C$ contains $45$ students.

Now, let $C$ be a course. Find $n$ courses with at least $1$ student from $C$ ($C$ is not among these). By claim, we know that $n \leq 45 \cdot 44 < 2008$. So, there is a course which does not intersect with $C$, which is a contradiction.


问:
How to graph thickness per volume added?
Say I have here a cube with side length $1$. If I wanted to double its volume, I would have to add a layer of material $0.16109$ units all around it. If I wanted to make that same cube have triple volume of the original, I would have to add a layer of $0.09469$ units thick of material on top of the first extra layer. I used these equations here:

$1=3(2x⋅1^2)+(2x)^3$ for the first layer,

$1=3(2x(1.32218^2)+(2x)^3$ for the second.

Now I want to graph volume to thickness of the added layer like $(1;0)$, $(2;0.16109)$, $(3;0.09469)$, etc.

答:
I will take different variables:

let $s$ be the sidelength of the cube

and $2a$ the sidelength increase

(coefficient $2$ is explained by the fact we want to have a "$a$" increase on each side)

Let us express the condition under the form:

$$(s+2a)^3-s^3=1$$

which is equivalent to:

$$6s^2a+12sa^2+8a^3-1=0. \tag{1}$$

Let us consider (1) as a quadratic in variable $s$ and parameter $a$. Its positive root is:

$$s=\dfrac{1}{6a}(-6a^2+\sqrt{6a(1-2a^3)})$$

with the following representation:

Remark: coefficients $6, 12, 8$ in (1) are resp. the number of faces, edges, and vertices of the cube. The following figure shows the interesting interpretation between the terms of (1):


问:
Prove that $T$ is invertible if and only if $p=m$ and $B$ is invertible.
Assume $p,n,m$ are positive integers, and $T: \mathbb M_{m \times n}(\mathbb F) \to \mathbb M_{p \times n}(\mathbb F)$ is a linear transformation such that $\forall A \in \mathbb M_{m \times n}(\mathbb F):A \mapsto BA$,where $B$ is a fixed $p \times m$ matrix.Prove that $T$ is invertible if and only if $p=m$ and $B$ is invertible.

If $T$ is invertible then it takes the a basis of  $\mathbb M_{m \times n}(\mathbb F)$ to a basis of  $\mathbb M_{p \times n}(\mathbb F)$,from here we have that $mn=pn \implies m=p$,but I don't know how to show that $B$ is invertible.

For the other direction the linear transformation $G:\mathbb M_{p \times n}(\mathbb F) \to \mathbb M_{m \times n}(\mathbb F)$ with $A \mapsto B^{-1}A$ can be seen as the inverse of $T$.

答:
Assume $T$ is invertible and $\ker B\ne0$. Since $n>0$ there is a $m\times n$ matrix $A$ with $0\ne\operatorname{im}A\subseteq\ker B$. But then $T(A)=BA=0$, so by the injectivity of $T$ we get $A=0$ which contradicts $\operatorname{im}A\ne0$.

Therefore $\ker B=0$ and since $B$ is a square-matrix this implies that $B$ is invertible.


问:
Remainder when divided by 75
The problem:

Let $x_1$, $x_2$, $x_3$, $\dots$ be the increasing sequence of positive numbers which are co-prime to $75$ ($x_1$ = 1, $x_2$ = 2, $x_3$ = 4, $x_4$ = 7, $\dots$). Find the remainder when $x_{2019}$ is divided by 75.

Can't get started. Haven't got the faintest idea on how to approach.

Hints?

答:
I assume you mean the positive integers coprime to 75 in order, not just any sequence of integers where they are all coprime to 75 (otherwise the answer can be any integer coprime to 75)

There are $\varphi(75) = 40$ residues mod 75 coprime to 75. Hence the sequence has period 40.

Hence $x_{2019} = x_{2019 (mod 40)} = x_{19}$, which you can easily find by hand.

Title: Remainder when divided by 75


问:
How to express "there exist a number of something such that ..." in the language of set theory
In multivariate real analysis, differentiability is defined as follows:

A function $f: \mathbb{R}^{n} \to \mathbb{R}$ is differentiable at a point $x_{0} \in \mathbb{R}^{n}$ if and only if $x_{0}$ is in the interior of the domain of $f$ and there exist $n$ constants $m_{1}$, $m_{2}$, $\dots$, $m_{n}$ such that
\begin{equation}
\lim_{x\to x_{0}}\frac{f\left(x\right)-f\left(x_{0}\right) - \sum_{i=1}^{n}m_{i}\left(x_{i}-{x_{0,i}}\right)}{\lVert x-x_{0} \rVert} = 0,
\end{equation}
where $x = \left(x_{1},x_{2},\dots,x_{n}\right)$ and $x_{0} = \left(x_{0,1},x_{0,2},\dots,x_{0,n}\right)$.

I have no problem with understanding this definition. However, this definition reminded me of one question: how to express "there exist a number of objects such that" in the language of set theory? It seems that the set of natural numbers has to play a role in the expression, but I am not sure how to construct such a proposition rigorously, especially when the universal quantification on $n$ is needed.

答:
If you want a way to talk about $n$-tuples without fixing a specific $n$, we can achieve this by shifting attention to sequences as individual objects. For example, in place of

For every $n$, there are points $x_1,...,x_n$ such that [stuff]

we can write

For every finite ordinal $x$ there is a function $T: x\rightarrow \mathbb{R}$ such that [stuff]

(think of $x$ as $\{0,1,...,n-1\}$ and of $T(i)$ as $x_i$). Keep in mind that "finite ordinal" is something we can actually express in the language of set theory (e.g. as "hereditarily transitive set with no non-surjective self-injection").

Note that this idea also shows up in the proof of Godel's incompleteness theorem, a key piece of which is showing that the language of arithmetic can talk about finite sequences of natural numbers even though a priori it's only able to refer to individual natural numbers.


问:
How can i show this inequality?
Let $n>1$ and $a_1,...,a_n \in \mathbb{R}^+$ be such that $\sum a_i=1$. For evey $i$, define $b_i=\sum_{j=1,j\neq i}a_j$. Show that

$\sum_{k=1}^n \dfrac{a_k}{1+b_k}\ge \dfrac{n}{2n-1}$

Thanks a lot for any suggestion.

Title: How can i show this inequality?

Body:
Note that $b_i=1-a_i$. Thus using AM-HM inequality gives

\begin{align}
\sum_{k=1}^{n}{\frac{a_k}{1+b_k}}=\sum_{k=1}^{n}{\frac{a_k}{2-a_k}}=\sum_{k=1}^{n}{(\frac{2}{2-a_k}-1)}& =2\left(\sum_{k=1}^{n}{\frac{1}{2-a_k}}\right)-n \\
& \geq \frac{2n^2}{\sum_{k=1}^{n}{(2-a_k)}}-n \\
&=\frac{2n^2}{2n-1}-n \\
& =\frac{n}{2n-1}
\end{align}


问:
Probability problem on Unif(0,1) random variables
Given three iid Unif(0,1) random variables a, b, and c, what is the probability that c is the maximum if a+b < 1?

Attempt:
The probability distribution of a+b for a+b < 1 is a+b and for 1 < a+b < 2 it is 2-(a+b). Therefore, the events a+b < 1 and a+b > 1 both have probability 1/2. Also, $P(c>a \cap c>b) = 1/3$. This means that

$$1/3 = P(c>a \cap c>b | a+b > 1)P(a+b > 1) + P(c>a \cap c>b | a+b < 1)P(a+b < 1) $$

$$2/3 = P(c>a \cap c>b | a+b > 1) + P(c>a \cap c>b | a+b < 1) $$

I am not sure how to continue. Can someone explain how to proceed with the solution?

答: Representing the possible outcomes of sampling $a,b,c$ as three coordinates into a unit cube, the volume where $c>a,c>b,a+b<1$ is a triangular dipyramid with vertices $(0,0,0),(0,1,1),(1,0,1),(0,0,1),(1/2,1/2,1/2)$, which has base area $\frac{\sqrt3}2$ and combined height $\frac{\sqrt3}2$. Thus $P(c>a,c>b,a+b<1)=\frac13\cdot\frac{\sqrt3}2\cdot\frac{\sqrt3}2=\frac14$. More clearly, $P(a+b<1)=\frac12$ (a right triangular prism). Hence the final answer is $\frac{1/4}{1/2}=\frac12$.


问:
Solution to IVP for ODE
ODE: $ 2tx'(t)-1=x^2(t) $

If I have the general solution:

$2 \arctan(x) = \ln(t) + C $

And I have the initial condition $ x(t=1) =1 $ then:

$2 \arctan(1) = \ln(1) + C $

$ \frac{\pi}{2} = C $

I just found C but I still need the solution to the IVP, right? How do I go from here?

Hope you can help!

答:
If you found that $C=\frac{\pi}{2}$, and that your ODE solution is
$$2 \arctan(x) = \ln(t) + C,$$
then the solution to your IVP is just the same but using the value of $C$ you just found:
$$\boxed{2 \arctan(x) = \ln(t) + \frac{\pi}{2}},$$
and this would be an $\underline{\text{implicit}}$ solution to your IVP. Now, if what you're asking for is the $\underline{\text{explicit}}$ solution, the you just have to transform it into the form you like, to be said:
$$\arctan(x)=\frac{\ln(t)}{2} + \frac{\pi}{4}$$
$$\Downarrow$$
$$\boxed{x=\arctan^{-1}\left(\frac{\ln(t)}{2} + \frac{\pi}{4}\right)}$$

Title: Solution to IVP for ODE


问:
Pointwise convergence of $\sum_{n=1}^{\infty} \chi_{\left[n, n+1 / n^{2}\right]}(x)$. Help in understanding solution from the book.
show that $$
\sum_{n=1}^{\infty} \chi_{\left[n, n+1 / n^{2}\right]}(x)
$$ is pointwise converge for $x \in \mathbb{R}$ while $\chi_{\left[n, n+1 / n^{2}\right]}(x)$ is the indicator function

The solution in the book :

$$
 \chi_{\left[n, n+1 / n^{2}\right]}(x)=\left\{\begin{array}{ll}
1 & n \leq x \leq n+\frac{1}{n^{2}} \space  \\
0 & \text { otherwise }
\end{array}\right.
$$
thus for every $x$: $$
\sum_{n=1}^{N} \chi_{\left[n, n+1 / n^{2}\right]}(x)=\sum_{n=1}^{N_{0}} \chi_{\left[n, n+1 / n^{2}\right]}(x)
$$ for $
N \geq N_{0} \geq x+1
$
and thus the series is pointwise.

I dont understand how:

$\sum_{n=1}^{N} \chi_{\left[n, n+1 / n^{2}\right]}(x)=\sum_{n=1}^{N_{0}} \chi_{\left[n, n+1 / n^{2}\right]}(x)$ imply that we have pointwise converge

what the inequality  $
N \geq N_{0} \geq x+1
$ mean to us in this problem, why is there $+1$ factor?

答:
To prove pointwise convergence of your series, I'd look at what happens at different points $x$. The "worst" case is when $x=2$ and your series is $1+1+0+0+0+0+0+\dots$, which converges to $2$. For all other values of $x$, the series is either all zeros (converging to $0$) or a single $1$ and all the rest zeros (converging to $1$).  So the series converges for every value of $x$, and that's what pointwise convergence means.

The solution you quoted uses the fact that, if the terms of a series are all zeros from some point on, then, no matter what happened before that point, the series will converge because the partial sums are equal from that point on. The quoted solution just writes that observation out in impressive notation.


问:
How can I prove that the $\lim _{n\to \infty }\frac{log_{a}n}{n}=0$ , $a>1$ with the simplest tools of analysis?
I tried solving this problem but I didn't know where to start so I searched it on the internet and the only solution that I found was by using L'Hopital's rule. The thing is we haven't studied that in class, but my professor uses this problem constantly in order to solve other problems and I don't understand why this works. I think I should use the squeeze theorem (since I am only familiar with the limits of sequences and that's the most helpful theorem I know), but I don't know which inequality to use. Could you please help me by giving a hint or some advice?

Thanks in advance.

答:
As $\log_a x = \frac{\ln x}{\ln a}$, it is enough to prove the result for $\frac{\ln n}{n}$.

We first prove that $\lim\limits_{n \to \infty} a_n = 0$ where $a_n = \frac{\ln 2^n}{2^n}$. As $\ln 2^n = n \ln 2$, it is enough to prove that $\lim\limits_{n \to \infty} b_n = 0$ where $b_n = \frac{ n}{2^n}$ which is a consequence of the inequality

$$2^n = (1+ 1)^n \ge 1 + n + \frac{n(n-1)}{2}$$ that you can get using binomial theorem.

Finally, you get $\lim\limits_{n \to \infty} \frac{\ln n}{n} =0$ by squeezing $n$ between $2^m$ and $2^{m+1}$

$$\frac{\ln 2^m}{2^{m+1}} = \frac{\ln 2}{2}\frac{m}{2^m} \le \frac{\ln n}{n }\le 2 \ln 2\frac{(m+1)}{2^{m+1}} = \frac{\ln 2^{m+1}}{2^{m}} $$


答:
You can use the Squeeze theorem. You probably know that (roughly speaking) when you apply an inverse (continuous) function to a inequality, it "reverts the inequality". So, if $e^x>x$, then $\log(x)<x$. This may be a given property in your notes (I am pretty sure because I saw this in my first real analysis course). Please try it first, and if you cannot, continue reading! :)

$0 \leq \dfrac{\log(n)}{n} = \dfrac{\log(n^{2/2})}{n} = \dfrac{2\log(n^{1/2})}{n} \leq
\dfrac{2\sqrt{n}}{n} = \dfrac{2}{\sqrt{n}} \to 0 $.

So $\dfrac{\log(n)}{n} \to 0$ by the Squeeze theorem.

The difference between $\log(\cdot)$ and $\log_a(\cdot)$ is just a constant, so it is the same proof. Hope this is usefl to you!

Title: How can I prove that the $\lim _{n\to \infty }\frac{log_{a}n}{n}=0$ , $a>1$ with the simplest tools of analysis?

Body:
I tried solving this problem but I didn't know where to start so I searched it on the internet and the only solution that I found was by using L'Hopital's rule. The thing is we haven't studied that in class, but my professor uses this problem constantly in order to solve other problems and I don't understand why this works. I think I should use the squeeze theorem (since I am only familiar with the limits of sequences and that's the most helpful theorem I know), but I don't know which inequality to use. Could you please help me by giving a hint or some advice?

Thanks in advance.


问:
Describing a bounded set in $\Bbb{R^n}$
Let $B\subset\Bbb{R^n}$ and bounded. May I claim that $B$ can be contained within a rectangle $D_{\delta}$, which is constructed by multiplying intervals of $\Bbb{R}$, with length $\delta>0$? So this rectangle would have a measure of $\delta^n$. Also, if this is true, is this then equivalent with the definition of a bounded set or a consequence of it?

答: This is equivalent, since we can always draw a box around a ball and a ball around a box in Euclidean space.</p>

Title: Describing a bounded set in $\Bbb{R^n}$


问:
How to find a derivative with respect to a matrix?
If I have $f: \mathbb{R}^{n \times m} \times \mathbb{R}^{n} \to \mathbb{R}^{m}$

And $f(K,t) = Kt + h$ where $h \in \mathbb{R}$

How would I find $\frac{\partial f}{\partial K}$ and $\frac{\partial f}{\partial t}$?

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}\def\e{\varepsilon}\def\R#1{\in{\mathbb R}^{#1}}$To expand on user7530's comment, coordinate-wise derivatives are a useful approach which avoids higher-order tensors or transformations (i.e. vectorization) which flatten those tensors into matrices.

Given the following variables
$$\eqalign{
A\R{m\times n}
\qquad&b&\R{n}
\qquad&c&\R{m}
}$$
their coordinate-wise derivatives are
$$\eqalign{
\p{A}{A_{ij}} &= e_i\,\e_j^T \qquad
\p{b}{b_{i}} &= e_i \qquad
\p{c}{c_{j}} &= \e_j \\
}$$
where {$\,e_i,\,\e_j\,$} are the cartesian basis vectors for their respective dimensions.

Applying this to the current problem yields
$$\eqalign{
f &= Kt +h \\
\p{f}{K_{ij}} &= \left(e_i\e_j^T\right)t \;&=\; e_i\,t_j \\
\p{f}{t_{i}} &= K\left(e_i\right) \;&= \big(i^{th}\,{\rm column\,of\,}K\big) \\
}$$


问:
Soving Recurrence Relation
I have this relation

$u_{n+1}=\frac{1}{3}u_{n} + 4$

and I need to express the general term $u_{n}$ in terms of $n$ and $u_{0}$.

With partial sums I found this relation

$u_{n}=\frac{1}{3^n}u_{0} + 4\sum_{n=1}^n\frac{1}{3^n-1}$

But I also need to prove by mathematical induction that my $u_{n}$ is ok, but I have no idea how to do this. Can anyone please help me?

Thanks in advance

答:
$$u_{n+1}=\frac{1}{3}u_{n} + 4\implies 3u_n=u_n+12$$

$$\implies 3u_{n+2}=u_{n+1}+12$$

On subtraction, $$3u_{n+2}- 3u_{n+1}=u_{n+1}-u_n\implies 3u_{n+2}- 4u_{n+1}+u_n=0 $$

We can use Characteristic equation of the Recurrence relation to solve it

Title: Soving Recurrence Relation


问:
Validity of challenge to common Euler product derivation strategy (infinitely long denominator prime factorisations)
Background

The following Euler product for the Riemann zeta function is well known.

$$ \sum_n \frac{1}{n^s} = \prod_p (1-\frac{1}{p^s})^{-1} $$

Here $n$ ranges over all integers, $p$ over a primes, and real $s>1$.

Common Proof Strategy

Many derivations / proofs, found in textbooks and papers, consider the following expression.

$$(1 - \frac{1}{p^s})^{-1} = 1 + \frac{1}{p^s} + \frac{1}{p^{2s}} + \frac{1}{p^{3s}} + \ldots$$

The LHS is finite for any given $p$ and the series expansion is valid because $\frac{1}{p} < 1$.

The following takes the product over all primes.

$$\prod_{p_i} (1-\frac{1}{p_i^s})^{-1} = 1 + \frac{1}{p_1^s} + \frac{1}{p_1^{2s}} + \ldots  + \frac{1}{p_1^sp_2^{s}} + \frac{1}{p_1^sp_3^{s}} + \ldots $$

The LHS is a product of finite and non-zero factors.

The RHS has terms of the form $\frac{1}{X}$ where $X$ contains all combinations of the primes, and all combinations of powers of the primes.

It is common to apply the Fundamental Theorem of Arithmetic to see that there is one term X for each integer $n$, and therefore the RHS is the desired $\sum\frac{1}{n^s}$.

Challenge

A challenge (for example here) that has been raised to this very common proof logic is that there are terms $X$ with an infinite number of factors in the denominator, for example:

$$ \frac{1}{(2^2\cdot3^2\cdot 5^2\cdot 7^2 \cdot\ldots)^s} $$

or another simpler example:

$$ \frac{1}{(2\cdot 2 \cdot 2\cdot 2\cdot\ldots)^s} $$

Question

Is the challenge valid?

I am not a trained mathematician, but in my opinion the proof strategy is valid because terms $X$ with denominators with an infinite number of prime factors are equivalent to zero. That is:

$$ \frac{1}{(2^2\cdot3^2\cdot 5^2\cdot 7^2 \cdot\ldots)^s} = 0$$

and

$$ \frac{1}{(2\cdot 2 \cdot 2\cdot 2\cdot\ldots)^s} = 0$$

My assertion is that the proof strategy remains valid because any finite integer $n$ has a single finite non-zero term $X$, and those $X$ with infinitely long denominators can be discarded because they are zero.

答:
By definition, where $p_i$ is the $i^{th}$ prime,
$$\prod_{p_i} \left(1-\frac{1}{p_i^s}\right)^{-1}:=\prod_{i=1}^\infty\left(1-\frac{1}{p_i^s}\right)^{-1}:= \lim_N \prod_{i=1}^N\left(1-\frac{1}{p_i^s}\right)^{-1}$$

Now $$\prod_{i=1}^N\left(1-\frac{1}{p_i^s}\right)^{-1} = \prod_{i=1}^N\left(\sum_{e_i=0}^\infty \frac 1{\,p_i^{se_i}\,}\right)$$
And
$$\begin{align}\prod_{i=1}^N\left(\sum_{e_i=0}^\infty \frac 1{\,p_i^{se_i}\,}\right) &= \prod_{i=1}^N \lim_{n_i} \sum_{e_i=0}^{n_i} \frac 1{\,p_i^{se_i}\,}\\
&= \lim_{n_1,\dots,n_N} \sum_{e_1=0}^{n_1}\dots\sum_{e_N=0}^{n_N}\frac 1{(p_1^{e_1}p_2^{e_2}\dots p_n^{e_N})^s}\end{align}$$
provided the RHS converges. But since the terms are all positive, the limitend is an increasing function of all its indices. And it is the sum of a finite subsequence of the known convergent sequence $\sum_k \frac 1{k^s}$, so it is bounded above by that value. Hence the RHS side must converge.

Further, it includes a term of the form $\frac 1{k^s}$ for every $k < p_{N+1}$. Thus
$$\sum_{k=1}^{p_{N+1}-1} \dfrac1{k^s} \le \lim_{n_1,\dots,n_N} \sum_{e_1=0}^{n_1}\dots\sum_{e_N=0}^{n_N}\frac 1{(p_1^{e_1}p_2^{e_2}\dots p_n^{e_N})^s} \le \sum_{k=1}^\infty \dfrac1{k^s}$$
$$\sum_{k=1}^{p_{N+1}-1} \dfrac1{k^s} \le \prod_{i=1}^N\left(1-\frac{1}{p_i^s}\right)^{-1}\le \sum_{k=1}^\infty \dfrac1{k^s}$$

Since $p_{N+1} \to \infty$ as $N \to \infty$, by the squeeze theorem

$$\sum_{k=1}^\infty \dfrac1{k^s} \le \prod_{i=1}^\infty\left(1-\frac{1}{p_i^s}\right)^{-1} \le \sum_{k=1}^\infty \dfrac1{k^s}$$

That is, $$\prod_{p_i} \left(1-\frac{1}{p_i^s}\right)^{-1} = \sum_{k=1}^\infty \dfrac1{k^s}$$

Note that at one stage in this proof, it required that $$\sum_{k=1}^\infty \dfrac1{k^s}$$ converge. I.e., it only works for $s > 1$.


问:
How can you prove n= $6k +2$ when the $x^2$ and $x^3$ coefficient of $(2k +x)^n$ is the same?
I've tried working out the coefficients for each

$$\binom{n}{2}(2k)^{n-2}=\binom{n}{3}(2k)^{n-3}$$

But I'm not sure if that's helpful in any way? And I don't know what to do next?

答:
Use the formula ${n\choose 2}={n(n-1)\over 2}$ and ${n\choose 3}={n(n-1)(n-2)\over 6}$.

Plug into your result you get $${n(n-1)\over 2}\times {(2k)}^{n-2}={n(n-1)(n-2)\over 6}\times{(2k)}^{n-3}$$

Cancel out both side you get $6k=n-2$


答:
Expand the binomials and cancel the $2k$'s:
$$\frac{n(n-1)}2(2k)=\frac{n(n-1)(n-2)}6$$
We can cancel $n(n-1)$ too, and multiply by $6$:
$$3(2k)=n-2$$
$$n=6k+2$$

Title: How can you prove n= $6k +2$ when the $x^2$ and $x^3$ coefficient of $(2k +x)^n$ is the same?

Body:
I've tried working out the coefficients for each

$$\binom{n}{2}(2k)^{n-2}=\binom{n}{3}(2k)^{n-3}$$

But I'm not sure if that's helpful in any way? And I don't know what to do next?


答:
Yes you're on the right track, we just need to simplify the equation.

First divide out by $(2k)^{n-3}$ and use the formula for binomial coefficients to get
$$ \frac12 n(n-1) \times 2k = \frac16 n(n-1)(n-2)$$
$$ k = \frac16 (n-2)$$
$$ n = 6k+2$$


答:
As mentioned in the other answer, the two are equivalent.

There is also a more "inherent" definition of bounded which says that $B \subset \Bbb R^n$ is bounded if there exists some $M > 0$ such that $\|x - y\| < M$ for all $x, y \in B$. You can show that this coincides with the ball definition (and hence, with the rectangle definition).

Title: Describing a bounded set in $\Bbb{R^n}$

Body: Let $B\subset\Bbb{R^n}$ and bounded. May I claim that $B$ can be contained within a rectangle $D_{\delta}$, which is constructed by multiplying intervals of $\Bbb{R}$, with length $\delta>0$? So this rectangle would have a measure of $\delta^n$. Also, if this is true, is this then equivalent with the definition of a bounded set or a consequence of it?


问:
Difference of induction and divide and conquer
I have seen that induction and divide and conquer are used as problem solving techniques but they are treated either as something different or the former as a way to support the latter.

To me it seems that they are both exactly the same thing i.e. defining the problem in a way that has (can be composed of) smaller instances of the same original problem, having a base case and a way to build a solution for instance $n  + 1$ having the solution to instance $n$.

So in my understanding they (induction and divide and conquer) are exactly the same thing/exactly the same problem solving approach or at least different terms for the same thing/topic.

Am I misunderstanding something? If yes what is the difference I am overlooking? If not why are they treated as different things in texts?

答:
They are somewhat related concepts, but not the same. Induction is a way of proving something by building up from smaller cases. "Divide and conquer" is an approach to solving a problem (which may be a proof).

Not all induction problems can be described as divide and conquer. For example, there are any number of arithmetic statements like $\sum_{i=1}^n i = \frac{n(n+1)}2$ that can be proved by induction, but these proofs wouldn't be considered divide and conquer since you're just incrementally adding one more element at a time. Usually, "divide and conquer" suggests splitting up the problem into chunks (e.g., halves or thirds).

On the other hand, not all divide and conquer approaches are instances of induction. Maybe you're trying to find a path visiting different delivery locations in a city (the Travelling Salesman Problem): you might break this down into neighbourhoods, then break it down again as needed, solve the smaller problems, then recombine them. But the result will not necessarily be optimal, and notably this wasn't part of a proof, just an approach to a problem.


问:
Prove $\forall p, 0<p<m$ the integral $\int_{\mathbb{R}^m}\frac{d\mu(x)}{||x-y||^p}<\infty$ for almost all $y\in\mathbb{R}^m$

$\mu $ is finite borel measure  in $\mathbb{R}^m$

Prove $\forall p, 0<p<m$ the integral $\int_{\mathbb{R}^m}\frac{d\mu(x)}{||x-y||^p}<\infty$ for almost all $y\in\mathbb{R}^m$

I know that this question can be solved with polar coordinates but I want to prove this with Tonelli thm. if possible. Any help will be welcome :)

答:
We may assume that $\mu$ is a probability measure. Let $B(0,r)$ denote the Euclidean ball of radius $r$ around $0$ in $\mathbb{R}^m$. First observe that for $0<p<m$ and $0<R<\infty$, the Lebesgue integral $$I_R(x):=\int_{B(0,R)}\frac{dy}{||x-y||^p}$$ can be bounded by some constant $C(R,m,p)<\infty$. For $x \notin B(0,2R)$ this is clear, and for $x \in B(0,2R)$ it can be inferred via polar coordinates or dividing space into shells: For any integer  $k$, the contribution of the shell $\{y: 2^{-k}<||y-x|| \le 2^{1-k}\}$ to $I_R(x)$ is
$O(2^{k(p-m)})$, and we can sum this over $-\log_2(R)-2<k<\infty$.

Denote $J(y)=\int_{\mathbb{R}^m}\frac{d\mu(x)}{||x-y||^p}$. Then   by Tonelli,
$$\int_{B(0,R)} J(y) \, dy =\int_{\mathbb{R}^m} I_R(x) \, d\mu(x) \le C(R,m,p) \,,$$
so $J(y)<\infty $ for a.e. $y \in B(0,R)$. Take countable union over integer $R<\infty$.


问:
GCSE: volume of cylinder( How I can find the same height of the container)
Looking please for any help.

In the figure below, 2 containers are given. Pour water from container A into container B until the height in both containers is the same. Find the same height in both container.

My attempt was the following:
6:4
6,5:x
6x= 26
x=4,3

I know I messed up!
All help/solutions appreciated.

Body: The total volume is $\pi (6)^2 (6.5)$. Suppose the common height is $h$. Then, $$\pi(6)^2 h + \pi(4)^2 h = \pi (6)^2 (6.5) \\ h = \frac{36\times 6.5}{52}=4.5 \ cm$$


问:
Finding attractors / fixed points for the circumference of the main bulb of the Mandelbrot Set
The Mandelbrot set is the set of all complex numbers $c$ that cause the function $z_{n+1} = z_n^2 + c$ to remain bounded when iterated from $z_0 = 0$. Plotted in the complex plane it includes a main cardioid, formed by rolling a circle of radius $0.25$ around another circle of the same radius centered on the origin with its cusp at $0.25 + 0i$; and a main bulb, also a circle of radius $0.25$, centered on the value $-1 + 0i$ and touching the main cardioid at the value $-0.75 + 0i$.

Taking values of $c$ on the perimeter of the main cardioid causes $z$ to converge on or orbit around fixed point values lying on a circle of radius $0.5$ centered on the origin. This circle is the path taken by the center of the rolling circle that generates the shape of the main cardioid. The fixed point value is specifically the location of the midpoint of the rolling circle where it generates the chosen value of $c$. It is easy to find the attracting value for any point on the cardioid given the angle of that point, or conversely, to find the point on the cardioid for any attracting value given the angle of that value.

Taking values of $c$ on the perimeter of the main bulb causes $z$ to converge on or orbit around pairs of fixed point values lying on a lemniscate of Bernoulli with foci at $0 + 0i$ and $-1 + 0i$. The fixed points are symmetrical about the value $-0.5 + 0i$. I have been unable to find an equation that gives the exact attracting values for any point on the circle, or conversely, to find the point on the circle for any attracting values.

What is the relationship between the point on the circle and the points on the lemniscate of Bernoulli?

I've gotten close by representing the lemniscate with the parametric equations
$$x = \frac{\sqrt{2}}{2} * \frac{-\sin(θ)}  {\cos^2(θ) + 1} - \frac{1}{2}$$
$$y = \frac{\sqrt{2}}{2} * \frac{\sin(θ) \cos(θ)}  {\cos^2(θ) + 1}$$
Where $θ$ is the angle around the main bulb, starting with 0° at the point where the bulb joins the main cardioid, but that only lines up exactly for angles 0° and 180° (and all further rotations mod 360°). For other angles it is either too far around, or not far enough.

答:
You are looking for solutions of $z = (z^2 + c)^2 + c$, which expanded gives:

$$ z^4 + 2cz^2 - z + c^2 + c = 0 = c^2 + (2 z^2 + 1) c + z^4 - z $$

Solving for $c$ given $z$ is easier than solving for $z$ given $c$, but setting $c = -1 + e^{2\pi i t}/4$ and doing so we find that
$$
z = \frac{1}{2} \left(-1\pm\sqrt{1-e^{2 i \pi  t}}\right).
$$
You can plot this to generate your lemniscate.


答:
They're related, but different. When you prove, by induction on $n$, the complexity (be it in time or memory) of a recursive algorithm for solving a size-$n$ problem, divide and conquer as a definition of that recursive algorithm proves a recurrence relation used in the inductive step.

For example, this algorithm for the tower of Hanoi takes $T_n$ movements with $n$ discs with $T_0=0$, and divide and conquer proves $T_{k+1}=2T_k+1$, allowing us to prove by induction $T_n=2^n-1$.

Note that divide and conquer can also be used to prove approximations, asymptotic behaviours, and bounds for the complexity; see here and here.

Title: Difference of induction and divide and conquer

Body:
I have seen that induction and divide and conquer are used as problem solving techniques but they are treated either as something different or the former as a way to support the latter.

To me it seems that they are both exactly the same thing i.e. defining the problem in a way that has (can be composed of) smaller instances of the same original problem, having a base case and a way to build a solution for instance $n  + 1$ having the solution to instance $n$.

So in my understanding they (induction and divide and conquer) are exactly the same thing/exactly the same problem solving approach or at least different terms for the same thing/topic.

Am I misunderstanding something? If yes what is the difference I am overlooking? If not why are they treated as different things in texts?


问:
How to compute and think intuitively about $\limsup$ and $\liminf$ of the below function?
$$x_n = \frac{n^2}{1+n^2} \cos \frac{n \pi}{16} $$

Could someone please let me know how to derive the limits in this case?

I can think — with using l'Hopital rule the limit of $\frac{n^2}{1+n^2} $ is $1$. So can I just say that, it will be multiplied with extremum values of $\cos$ function? How to prove this? and also how to think about $\limsup$ and $\liminf$ intuitively in this case?

答:
In this case you're right that the $\liminf$ and $\limsup$ are $-1,1$ respectively, and intuitively the reason you suggested is right. The sequence $\frac{n^2}{1+n^2}$ approaches $1$ (you don't even need L'Hopital's rule... personally I think L'Hopital's rule is over used even for simple limit calculations), while $\cos\left(\frac{n\pi}{16}\right)$ can oscillate between $\pm 1$.

More formally, we can prove things like this. Recall that one of the possible definitions of $\limsup$ is to take the supremum of all possible subsequential limits. In this case, we have that for all $n\in\Bbb{N}$, $-1\leq x_n\leq 1$, so definitely we have $-1\leq \limsup\limits_{n\to \infty}x_n \leq 1$. On the other hand, consider the subsequence
$\{x_{32n}\}_{n=1}^{\infty}$. We have
\begin{align}
x_{32n}&=\frac{(32n)^2}{1+(32n)^2}\cos(2n\pi) = \frac{(32n)^2}{1+(32n)^2},
\end{align}
and as $n\to \infty$, the limit is $1$. Thus, we have found a subsequence which has limit $1$, so because $\limsup\limits_{n\to\infty} x_n$ is by definition the supremum of subsequential limits, we must have $\limsup\limits_{n\to\infty}x_n \geq 1$. Combining with what we said above it follows that $\limsup\limits_{n\to\infty}x_n = 1$.

Another possible definition is $\limsup\limits_{n\to\infty} x_n:= \lim\limits_{n\to\infty}\sup\limits_{k\geq n}x_k$. Note that the sequence $x_k=\frac{k^2}{1+k^2}=1-\frac{1}{1+k^2}$ increases to $1$; also by taking $k$ to be even multiples of $16$, it follows that for every $n\in\Bbb{N}$, we have that $\sup\limits_{k\geq n}x_k = 1$. Hence, taking the limit as $n\to\infty$ still yields $1$; i.e
\begin{align}
\limsup_{n\to\infty}x_n&:= \lim_{n\to\infty}\sup_{k\geq n}x_k = \lim_{n\to\infty} 1 = 1.
\end{align}
I'll leave the case of $\liminf$ to you.


问:
3D software like GeoGebra
Does it exist a free interactive geometry software, like GeoGebra, which works for 3D geometry? I would be able to draw spheres, great circles, and so on.</p>

答: You can try pgftikz http://www.texample.net/tikz/examples/</p>

ContentLicense: CC BY-SA 2.5

Title: 3D software like GeoGebra

OwnerDisplayName: anonymous

LastEditorDisplayName: anonymous


问:
If $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, then $(\nu_t)_{t\in I}$ is tight
Let $E$ be a metric $\mathbb R$-vector space, $I$ be a nonempty set and $(\mu_t)_{t\in I},(\nu_t)_{t\in I}$ be families of nonnegative finite measures on $\mathcal B(E)$.

We can show that if $(\mu_t)_{t\in I}$ and $(\nu_t)_{t\in I}$ are tight$^1$ and bounded in total variation norm, then the family of convolutions $(\mu_t\ast\nu_t)_{t\in I}$ is tight and bounded in total variation norm as well.

Now I would like to show the converse: If $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, then $(\nu_t)_{t\in I}$ is tight. (Most probably we need to impose a boundedness assumption on $(\mu_t)_{t\in I}$ and on $(\mu_t\ast\nu_t)_{t\in I}$ or even a priori on $(\nu_t)_{t\in I}$.)

For the moment, I'm only able to prove the claim assuming that the $\mu_t,\nu_t$ are probability measures. Under this assumption, the proof goes as follows:

Let $\varepsilon>0$. Since $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, there is a compact $K\subseteq E$ with $$\sup\left(\sup_{t\in I}\mu_t(K^c),\sup_{t\in I}(\mu_t\ast\nu_t)(K^c)\right)<\varepsilon\tag1.$$ Now $K-K$ is compact and \begin{equation}\begin{split}1-\varepsilon<(\mu_t\ast\nu_t)(K)&=\underbrace{\int_K\mu_t({\rm d}x)\nu_t(\underbrace{K-x}_{\subseteq\:K-K})}_{\le\:\nu_t(K-K)}+\underbrace{\int_{K^c}\mu_t({\rm d}x)\underbrace{\nu_t(K-x)}_{\le\:1}}_{<\:\varepsilon}\\&<\nu_t(K-K)+\varepsilon\end{split}\tag2\end{equation} for all $t\in I$, which immediately yields $$\nu_t(K-K)>1-2\varepsilon\;\;\;\text{for all }t\in I\tag3.$$

Can we drop the probability measure assumption?

EDIT

I'm not able to obtain the claim, even with the help of the answer below. If $\varepsilon>0$, the goal is to show that $$\nu_t(E)-\nu_t(K-K)=\nu_t((K-K)^c)<\varepsilon\tag4.$$ If I'm choosing all the constants as generic as possible, we obtain $$\mu_t(E)\le c_1\tag5$$ $$\mu_t(K^c)\le k_1(\varepsilon)\tag6$$ $$(\mu_t\ast\nu_t)(K)<c_1\nu_t(K-K)+c_2k_1(\varepsilon)\tag6$$ $$\mu_t(E)\ge\alpha_1\tag7$$ $$(\mu_t\ast\nu_t)(E)\le c_3\tag8$$ $$\nu_t(E)\le\frac{c_3}{\alpha_1}=:c_2\tag9$$ $$(\mu_t\ast\nu_t)(E)\ge\alpha_3\tag{10}$$ $$\nu_t(E)\ge\frac{\alpha_3}{c_1}\tag{11}$$ $$(\mu_t\ast\nu_t)(K^c)<k_3(\varepsilon)$$

The assumptions that I'm imposing is that $(\mu_t)$ and $(\mu_t\ast\nu_t)$ are bounded above (by $c_1$ and $c_3$) and below (by $\alpha_1$ and $\alpha_3$) in total variation norm and are tight (so that we can find $K$ for arbitrary $k_1(\varepsilon)$ and $k_3(\varepsilon)$).

In order to obtain the claim we need to choose $k_1(\varepsilon)$ and $k_3(\varepsilon)$ such that $(4)$ holds. From everything above, we obtain $$\nu_t(E)-\nu_t(K-K)<\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)\tag{12}$$ Even if we simplify this by assuming $\alpha_1=\alpha_3$, $c_1=c_3$ and $k_1(\varepsilon)=k_3(\varepsilon)$, the equation $$\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)=\varepsilon\tag{13}$$ doesn't seem to admit a solution $(k_1(\varepsilon),k_3(\varepsilon))$ with positive $k_1(\varepsilon)$ and $k_3(\varepsilon)$; unless $\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}\ge0$; which clearly doesn't need to be the case (it is clearly the case for probability measures though ...)

$^1$ Remember that a family $\mathcal F$ of finite signed measures on $\mathcal B(E)$ is called tight if for all $\varepsilon>0$, there is a compact $K\subseteq E$ such that $$\sup_{\mu\in\mathcal F}|\mu|(K^c)<\varepsilon.$$

答:
My counter-example in the comments will still work. Let $\mu_n = n^{-1} \delta_0$ and let $\nu_n = \delta_n$. Then all of the hypotheses in the question are satisfied but $\nu_n$ is not a tight family.

If one adds the hypotheses that $\inf_t \mu_t(E) \wedge \nu_t(E) > 0$ then the result is true however. In this case, define the measures $m_t = \mu_t(E)^{-1}\mu_t, n_t = \nu_t(E)^{-1}\nu_t$. From the additional assumption, it is straightforward to see that $m_t, m_t \ast n_t$ are tight families of probability measures. Hence, from what you already deduced, $n_t$ is a tight family of probability measures. Then, if $\sup_t n_t(K^c) < \varepsilon$ one has that $\sup_t \nu_t(K^c) < (\sup_t \nu_t(E)) \varepsilon \leq c \varepsilon$ where $c = \sup_t \|\nu_t\|$.


Title: Soving Recurrence Relation

Body:
I have this relation

$u_{n+1}=\frac{1}{3}u_{n} + 4$

and I need to express the general term $u_{n}$ in terms of $n$ and $u_{0}$.

With partial sums I found this relation

$u_{n}=\frac{1}{3^n}u_{0} + 4\sum_{n=1}^n\frac{1}{3^n-1}$

But I also need to prove by mathematical induction that my $u_{n}$ is ok, but I have no idea how to do this. Can anyone please help me?

Thanks in advance

Body:
$$u_{ n }=\frac { 1 }{ { 3 }^{ n } } u_{ 0 }+4\sum _{ k=0 }^{ n-1 }{ \frac { 1 }{ { 3 }^{ k } }  } =\frac { 1 }{ { 3 }^{ n } } u_{ 0 }+4\left( \frac { { \left( 1/3 \right)  }^{ n }-1\quad  }{ 1/3-1 }  \right) =\frac { 1 }{ { 3 }^{ n } } u_{ 0 }-6\left( \frac { 1 }{ { 3 }^{ n } } -1 \right) $$ $$u_{ n }=\frac { 1 }{ { 3 }^{ n } } \left(u_{ 0 }-6\right)+6$$

See geometric series


问:
Proving a ring is commutative knowing that $(x+y)^2=x^2+y^2$
Let $(R, +, ·)$ be a ring and $(x+y)^2=x^2+y^2$. Prove that the ring is commutative. This is what I've tried:

$$(x+y)^2=x^2+y^2$$
$$x^2+xy+yx+y^2=x^2+y^2$$
$$xy+yx=0$$

And now I don't know what to do. I'm not sure if I'm on the right path or not so I need some  guidance.

EDIT: It is not specified that it is a ring with a unit.

答:
As a complement to the other answer: without unitality, the result is false.

Define a ring $R$ as follows: as a set, $R$ is the three-dimensional $\mathbb{R}$-vector space $\mathbb{R}^3$ spanned by the standard unit vectors $e_1, e_2 ,e _3$. Its addition comes from the vector space addition, and its multiplication is defined by  $$(a_1 e_1 + a_2 e_2 + a_3 e_3) \cdot (b_1 e_1 + b_2 e_2 + b_3 e_3) := (a_1 b_2 - a_2 b_1) e_3, \quad \forall a_i, b_i \in \mathbb{R},$$ in other words we impose the relations $e_1 e_2 = e_3 = - e_2 e_1$ and all other products of generators are zero.

You can check that this indeed defines a ring structure: the addition forms a group since it always does in vector spaces, the multiplication is associative (since any products of three elements are equal to zero), and the distributive laws are also fine.

By definition, this product fulfils $xy = - yx$, which is equivalent to $(x + y)^2 = x^2 + y^2$ by your calculation, but it does not fulfil $xy = yx$. Hence, there are nonabelian rings where your relation holds.

Note: This structure does not come out of thin air, this is closely related to the wedge product of multilinear forms! I do not know if you are familiar with these, but luckily you do not need to be to understand this counterexample.


问:
Finding a closed form expression for the recurrence relation $a(n,k+1)=(k+1)^n+\sum_{m=1}^k {k+1\choose m}a(n,m)$
I originally stumbled upon this beast whilst trying to find the number of surjective functions from $[n]=\{ 1,...,n\}$ to $[k]$.

Here is my idea, denote this number $a(n,k)$ hence $a(n,k) =k^n- \#\{{f:[n]\to [k]\; | \; f \; isn't \; surjective}\}$. For every $1\leq m\leq k-1$ we have ${m\choose k}$ ways of choosing a set of $m$ elements from $[k]$ and $a(n,m)$ functions in $\{{f:[n]\to [k]\; | \; f \; isn't \; surjective}\}$ with $|im(f)|=m$ and so:

$a(n,k+1)=(k+1)^n-\sum_{m=1}^k {k+1\choose m}a(n,m)$

From here I tried to use induction to get $a(n,k+1)$ to the form of a polynomial in $a(n,1)$ which is clearly $1$, but had no luck as to finding the polynomial itself.

Solutions to the original problem or to the recurrence relation would by much appreciated.

EDIT: from the inclusion-exclusion prinicple I got the follwing expression for $a(n,k)$ which is:

$a(n,k) = \sum_{m=0}^k (-1)^m{k\choose m}(k-m)^n$

and plugging it into $a(n,k+1)=(k+1)^n-\sum_{m=1}^k {k+1\choose m}a(n,m)$ yeilds:

$a(n,k+1)=(k+1)^n-\sum_{m=1}^k{\sum_{i=0}^m {k+1 \choose k+1-m,i,m-i}(m-i)^n}$

which, to be honest, looks terrible.

My original question still stands, what is the closed-form expression (if there even is one) for this monster double sum?

答:
After computing some values I have found OEIS A019538 and the formula:

$$k!{n \brace k}$$

You can find further information at the OEIS page.