问:
Riemann integrability of a function

The function $f:[a, b] \rightarrow \mathbb{R}$ is defined to be $f(x) = |x|$ if $x \in \mathbb{Q}$, and $f(x)=0$ if $x \in \mathbb{R} \setminus \mathbb{Q}$. Is $f$ integrable? Why?

My attempt was:

$f$ has infinite points of discontinuity. And since $\mathbb{R} \setminus \mathbb{Q}$ is not countable, we cannot find a partition $P_n$ such that $U(P_n, f) - L(P_n, f) \rightarrow 0$ when $n \rightarrow \infty$. Hence $f$ is not integrable.

Is this correct? Would it be another way of proving it?

答:
It is indeed correct, but, unless you've proven it earlier, it is not obvious why an uncountable set of points of discontinuity implies the non-integrability of a function. The way I'd go about it without using the countability criterion would be as follows:

Lets first assume that $0\notin [a,b]$. Without loss of generality, we can say $0 < a$ because, if $b<0$ we apply the same argument on $-f$. This means that on $[a,b] f(x)=x$ for all rational points $x$. Let $P=\{x_0, x_1,...x_n\}$ be an arbitrary partition. Since for any $k\in\{1,2,...,n\}$ there exists an irrational number in the interval $I_k:=(x_{k-1},x_k)$ the infimum of $f$ on that set must be $0$ (since there are no values of $f$ less than $0$). Similarily, because the set $\mathbb{Q}$ is dense in $\mathbb{R}$ there must also be a rational number in the interval $I_k$. This implies that the supremum of $f$ on $I_k$ must be greater than or equal to that rational number, which is obviously greater than or equal to $a$. This means that $L(P,f)=0$ and $U(P,f)\geq a(b-a)$ for any partition $P$. Meaning that their difference is greater than or equal to $a$. This implies the sequence of differences from your question cant converge to zero (if it converges, then it converges to a number $\geq a > 0$).

To solve for the case $0\in[a,b]$ notice that it suffices to prove that the function isn't Riemann integrable on some subset of $[a,b]$. And a closed subinterval of $[a,b]$ that doesnt contain $0$ must exist.

Title: Riemann integrability of a function


问:
Proof that ellipse can't be represented by a parametric polynomial
I know Ellipses can be represented by rational parametric functions, ie. as the ratio of two polynomials.

For example, an arc of an ellipse with minor axis $1$ and major axis $2$ can be represented by the rational parametric function:

$$ C(u) = \left(\frac{1-u^2}{1+u^2}, \frac{4u}{1+u^2}\right) $$

with $u \in [0,1]$

I also know the same ellipse can be represented by:

$$ C'(u) = \left(sin(u), 2cos(u) \right) $$

with $u \in [0,\frac{\pi}{2}]$

But this second way isn't a rational parametrization.

I wanna know if there is a way to proof that an Ellipse can't be represented by a polynomial parametric function and where to find it (or how to build this proof).

My attempt until now (by contradiction):

Suppose $ x = P(u)$ and $y = Q(u)$ are polynomials such that

$$E(u) = \left(P(u), Q(u) \right) \text{ with } u \in [0,1] \text{ such that } E(0) = (a,0) \text{ and } E(1) = (0,b)$$

is an ellipse arch centered at the origin.

But I couldn't find any contradiction until now.

I would be very thankful with any help to this problem. :)

答:
The implicit equation of this ellipse is

$$x^2+\frac{y^2}4=1.$$

No polynomials of degree $>0$ are such that

$$p^2(t)+\frac{q^2(t)}4=1.$$

ContentLicense: CC BY-SA 4.0


问:
Number of points of intersections
Let $l_1$ and $l_2$ two parallel lines. We choose $8$ points on $l_1$ and $7$ on $l_2$. Then all the market points on $l_1$ are joined with the market points on $l_2$. How many points are obtained as an intersection (that are not on $l_1$ nor on $l_2$) of the traced segments?

I´ve solved the problem by doing all the lines and I got $588$ intersections, so, I´m looking for a solution without making all of them. Any hint?

If I take the first point on $l_1$ and then traced all the lines with the respective point on $l_2$ I got $0$ intersections. Then if I take the second point and joined to the first on $l_2$ I have $6$ intersections, I noticed that the next will add $5$... So, the first lines are not crossed.
Then from the second point, there are $6+5+4+3+2+1$ from the third point, $12+10+8+6+4+2$, from the fourth $18+15+12+9+6+3$, from the fifth $24+20+16+12+8+4$, from the sixth $30+25+20+15+10+5$, from the seventh $36+39+24+18+12+6$, finally from the eighth $42+35+28+21+14+7$.
Adding up all the numbers I get $588$.

答:
Pick two marked points on $l_1$ and two on $l_2$. There is exactly one way to draw two segments between those points that creates an intersection; since all possible segments are drawn in the original question, the number of ways to pick the pairs of points on $l_1$ and $l_2$ is the number of intersections.

On $l_1$ a pair of points can be picked in $\binom82$ ways; on $l_2$, $\binom72$ ways. The number of intersections is $\binom82\binom72=588$.

Title: Number of points of intersections


问:
Find the probability density function of the random variable y=y(x)=x^2 , x with known probability density function.
Find the probability density function of the random variable y=y(x)=x^2, x with known probability density function,

$f(x) =
     \begin{cases}
       0 &\quad\text{if} \; \infty \lt x \lt -1 \\
       \frac{1}{2}  &\quad\text{if} \; -1 \le x \lt 0 \\
       \frac{1}{2}e^{-x} &\quad\text{if} \; 0 \le x \lt \infty \\
     \end{cases}$

I want to use the theorem that says $g(y)=\lvert \frac{dx}{dy} \rvert f(x)$ but since y=x^2 is not bijective I don't know how to proceed. I know that I have to work with $x=-\sqrt{y}$ separately from $x=\sqrt{y}$ but don't know in wich way. Any other method to solve this is fine. Thanks

答:
We can write $f_X$ as a mixture, namely $f_X(x) = f_{X_1}(x) + f_{X_2}(x)$, where $f_{X_1}(x) = \frac{1}{2} \mathbb 1(-1 \le x < 0)$ and $f_{X_2}(x) = \frac{1}{2}e^{-x} \mathbb 1 (x \ge 0)$.  Then the transformed density is $$\begin{align}
f_Y(y) &= \frac{1}{2\sqrt{y}}\left( f_{X_1}(-\sqrt{y}) + f_{X_2}(\sqrt{y})\right) \\
&= \frac{1}{2\sqrt{y}} \left( \frac{1}{2} \mathbb 1 (0 < y \le 1) + \frac{1}{2} e^{-\sqrt{y}} \mathbb 1 (y \ge 0) \right) \\
&= \frac{1}{4 \sqrt{y}} \begin{cases} 1 + e^{-\sqrt{y}} , & 0 \le y \le 1 \\ e^{-\sqrt{y}}, & 1 < y < \infty. \end{cases}
\end{align}$$
This approach works because $f_X$ is a mixture of two densities for which the transformation $Y = X^2$ is monotone on their individual supports.


问:
How many possible ways can a rectangular space be filled with squares of 2 colors, so that the contiguous edges never have the same color?
How many possible ways can a rectangular space be filled with squares of 2 colors, so that the contiguous edges never have the same color?

答:
Assume we have rectangles of horizontal width $n$ and height $m$ that are tiled by quadratic tiles split diagonally, as in the figures of the OP.

We can place the top left tile in $4$ ways. Then we can add $n-1$ tiles in the first row with $2$ choices for each next tile, and we can add $m-1$ tiles in the first column with $2$ choices for each next tile. From then on everything is determined, because the fourth tile in an L-shape can be placed in exactly $1$ way, whatever the situation.

The total number $N$ of tilings therefore is given by
$$N=4\cdot 2^{n-1}\cdot 2^{m-1}=2^{m+n}\ .$$


问:
Finding the limit of a step function f(x)
I need to find and prove the possible existence of this limit :

The function $f(x)$ is defined as follows:
$$f(x) =
\left\{
\begin{array}{c}
1, x < -2 \\
0, x > -2 \\
\end{array}
\right.
$$

Does $f(x)$ have a limit as $x → −2$ ? If it does, state the value of the limit, briefly
justifying your answer. If there is no limit, briefly state why not.

I've tried using the $(\epsilon, δ)$-definition definition of the limit :

$\lim_{x\to-2} f(x) = a$ if for any $\epsilon > 0$ there exists $δ$ s.t. $|f(x) - a| < \epsilon$ for all $x$ s.t. $|x + 2| < δ$.

But I struggle because of two points :

I am not sure how to operate on a step function such as this one and use it as part of the $(\epsilon, δ)$-definition of a limit.

There are two limits, from both left and right which exist, but I am not sure if this answers the question "Does $f(x)$ have a limit as $x → −2 ?$"

Any help would be appreciated.

答:
Assume it has a limit
$$\lim_{x\to-2}f(x)=L$$

So, by sequential charactetisation of the limit, for any sequence $ (x_n) $ which goes to $ -2 $, the sequence of the images $ (f(x_n)) $ should go to $ L .$

But we have
$$\lim_{n\to +\infty}f(-2-\frac 1n)=1$$
and
$$\lim_{n\to+\infty}f(-2+\frac 1n)=0$$

Done.


问:
Solve for an Ellipse Tangent to 2 Lines
I'm trying to automate creation of a curve in PowerPoint.

Here's an image of what I'm working towards:

I'm trying to show a diagram of a rocket trajectory from a launch site on Earth to a circular orbit.  The math that I can't figure out is how to solve for an ellipse that connects the launch point (x1, y1) to the orbit insertion point (x2, y2) and is tangent to a vertical line at the launch point and tangent to the circular orbit.  My approach has been to use (x-h)^2/a^2 + y^2 / b^2 = 1 because while I can set the horizontal axis of the ellipse at y=0, I don't know where the x coordinate of the center might fall, so I have an x-offset, h.  I found an equation for the slope of a tangent line to an ellipse at a point as m = a^2*x/(b^2*y).  I'm not sure if this is the right equation because sometimes the desired ellipse is longer horizontally and sometimes it might be longer vertically.  If my insertion point is 45 degrees along the 1st quadrant of the orbit, then the tangent line has a slope of m = -1. Substituting the coordinates of the insertion point in to the slope equation, I can establish a relationship between a and b.  Then I solved the ellipse equation for h as a function of x, y, and a (substituting for b with the slope relationship).  Then I got 2 "h" equations by substituting the launch point, (x1, y1) and the insertion point, (x2, y2).  Then I iterated to find an "a" value that minimized the difference between the 2 "h" equations.  This successfully generated an ellipse that intersected the two points, but I think the slope at the insertion point is wrong (doesn't match the orbit), and it only appears to work for ellipses that are longer horizontally.

I apologize for this terribly formatted question.  I've spent all day at work trying to figure this out and I'm stumped.  Many of the solutions to similar problems online a just out of reach for me to understand how to apply it to my situation.

Thanks for any help anyone might can provide.

[edit]
Using a modification of alex.jordan's method I solved my problem.  Here's my work:
https://www.dropbox.com/s/61mmyuz4hmh2ry9/ellipses.pdf

And here's some Matlab code showing an application of the method and plotting a demonstration of the solution:

```<code>figure('Name','Ellipse Tangent to Orbit', ...
    'Color',[1,1,1],'NumberTitle','off', ...
    'Units','Normalized', 'Position',[0.1,0.1,0.8,0.8], ...
    'Visible','on');
axes;

hold on;

earthX = 0;
earthY = 0;

plot(earthX,earthY,'bo');

r = 700;

theta = 45 * pi/180;

%Insertion Point
s = earthX + r * cos(theta);
t = earthY + r * sin(theta);

%orbit
theta_mat = linspace(0,pi/2,100);
x_orbit = earthX + r*cos(theta_mat);
y_orbit = earthY + r*sin(theta_mat);

plot(x_orbit, y_orbit,'b--')

%launch point
x0 = 50;
y0 = 100;

plot(x0,y0,'ro');

plot(s,t,'ro');

%Slope at tangency point
m = -s/t;

%Vertical offset of ellipse equals coordinate of launch point
k = y0;

%b polynomial

% b^2 term
a0 = 2*t^3*x0*s^3*(t-k)^3-2*t^3*s^4*(t-k)^3-t^4*s^2*(t-k)^4;
% b term
a1 = 0;
% constant term
a2 = t^2*x0^2*s^4*(t-k)^4-2*t^2*x0*s^5*(t-k)^4-2*t^3*x0*s^3*(t-k)^5 + ...
    t^2*s^6*(t-k)^4+2*t^3*s^4*(t-k)^5+t^4*s^2*(t-k)^6;

%b = max(roots([a0,a1,a2]));

%Solving for b (reduced quadratic formula)
b = abs(sqrt(-a0*a2)/a0);

%Solve for a
a = sqrt((b^4*t^2-b^2*t^2*(t-k)^2)/(s^2*(t-k)^2));

%Solve for h
h = s-a^2*s*(t-k)/(b^2*t);

% X values for ellipse
x = linspace(h-a,h+a,100);

%y values for ellipse
y = b/a*sqrt(a^2-(x-h).^2)+k;

plot(x,y,'b-')

axis square
</code>```

Finally here's the plot:

https://www.dropbox.com/s/71t4tuw3bqv0xr1/orbits_verified.png

答:
There is an equation for the circle $$x^2+y^2=1\tag{1}$$ (and if you desire a bigger radius, scale everything up once this is finished.) and at any point along it, $$2x+2yy'=0\tag{2}$$ Then there is an equation for the ellipse $$\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1\tag{3}$$ and at every point along it $$\frac{2(x-h)}{a^2}+\frac{2yy'}{b^2}=0\tag{4}$$

Your aim is to find $a,b$, and $h$ that provide the desired tangency. Let's call the point of tangency $(s,t)$. Then you have a system of three equations:
$$\begin{align}
s^2+t^2&=1\tag{1b}\\
a^2s&=b^2(s-h)\tag{2b}\\
\frac{(s-h)^2}{a^2}+\frac{t^2}{b^2}&=1\tag{3b}
\end{align}$$
where (2b) equation has come from eliminating $yy'$ in equations (2) and (4).
This is five unknowns in three equations, so the system is likely very underdetermined. You can throw in a fourth equation by specifying the coordinates of the launch point $$\frac{(x_0-h)^2}{a^2}=1\tag{4b}$$ but it makes sense that there will still remain one degree of freedom, when you think about elongating the ellipse vertically while contracting it horizontally and maintaining the tangency. So your task is to solve the system
$$\begin{align}
s^2+t^2&=1\tag{A}\\
a^2s&=b^2(s-h)\tag{B}\\
\frac{(s-h)^2}{a^2}+\frac{t^2}{b^2}&=1\tag{C}\\
\frac{(x_0-h)^2}{a^2}&=1\tag{D}
\end{align}$$

I would use (B) to isolate $h$, and then substitute that into (C) and (D) and scale by $b^4$, resp $a^2b^4$:
$$\begin{align}
s^2+t^2&=1\tag{A}\\
a^2s^2+b^2{t^2}&=b^4\tag{C'}\\
(b^2x_0+a^2s-b^2s)^2&=a^2b^4\tag{D'}
\end{align}$$
(A) can be used to isolate $t^2$ and insert into (C'):
$$\begin{align}
a^2s^2+b^2-b^2s^2&=b^4\tag{C''}\\
(b^2x_0+a^2s-b^2s)^2&=a^2b^4\tag{D'}
\end{align}$$
$a^2$ can be isolated in (C'') and substituted into (D'), after which we rescale to keep denominators away:
$$\begin{align}
(sx_0-1+b^2)^2&=-b^2+b^2s^2+b^4\tag{D''}
\end{align}$$ This appears quartic in $b$, but once expanded, the $b^4$ terms cancel:
$$\begin{align}
(sx_0-1)^2+2sx_0b^2&=b^2(1+s^2)\tag{D'''}
\end{align}$$

and you should be free to choose any $b$, solve for $s$ with the quadratic formula and work your way back to get the other values, or vice versa. You have specified an insertion point at 45 degrees, so that would make $s=\frac{1}{\sqrt{2}}$. If we put this in equation (D''') and solve for $b^2$ we have
$$\begin{align}
b^2&=\frac{(x_0-\sqrt{2})^2}{3-2\sqrt{2}x_0}\tag{E}
\end{align}$$

ContentLicense: CC BY-SA 3.0

Title: Solve for an Ellipse Tangent to 2 Lines


问:
proof two general dimension inequalities
I have just showed that $A+B$ is indeed a subspace by showing it is non empty and closed under multiplication and addition, however I am not sure how to do these types of problems with $\mathbf{dim}$.

Problem: Dimensions

$A+B=\left \{ a+b : a\in A, b\in B \right \}$ and $C$ is a vectorspace and $A \subset C, B \subset C$.

Let $a_1,..,a_n$ be elements of $A$ and let $b_1,..,b_k$ be elements of $B$ be two families of vectors in $C$. Let $A=\operatorname{Span}(a_1,..,a_n)$ and $b=\operatorname{Span}(b_1,..,b_k)$

Now, Show that $\mathbf{dim}(A+B)\leq n+k$ and show that $\mathbf{dim}(A+B)\leq \mathbf{dim}(C)$

....................................................................

I am new to the subject and I could use some help since I am really struggling with both span and dim. I get the definitions but no idea on how to approach such a problem.

答:
Let $C$ be vector space over field $K$. You have proved that $A+B$ is subspace of $C$, then it is itself linear space over $K$. $$A=\operatorname{Span}(a_1,...,a_n)=\bigg\{\sum\limits_{t=1}^{n}\alpha_ta_t\bigg|~\alpha_t \in K\bigg\}.$$
$$B=\operatorname{Span}(b_1,...,b_k)=\biggl\{\sum\limits_{t=1}^{k}\beta_tb_t\;\biggm|~\beta_t \in K\biggr\}.$$
$$A+B=\operatorname{Span}(a_1,...,a_n)+\operatorname{Span}(b_1,...,b_k)=\biggl\{\sum\limits_{t=1}^{n}\alpha_ta_t+\sum\limits_{i=1}^{k}\beta_ib_i\;\biggm|~\alpha_t,\beta_i \in K\biggr\}=\biggl\{\sum\limits_{t=1}^{n+k}\gamma_tc_t\:\biggm|~\gamma_t \in K\biggr\},$$ where $c_t=a_t$ if $t \leq n$ and $c_t=b_{t-n}$, if $n+1 \leq t \leq n+k$. So basis of $A+B$ contains no more than $n+k$ vectors. Therefore $\dim(A+B) \leq n+k$.

2)If $\dim(C)=\infty$, then inequlity $\dim(A+B) \leq \dim(C)$ is obious.
Let $\dim(C) < \infty$ and $\{e_1,...,e_s\}$ be basis of $A+B$. Hence, $\{e_1,...,e_s\}$ is linearly independent in $C$ over $K$. Basis of $C$ is maximal linearly independent system of vectors from $C$ and $\{e_1,...,e_s\}$ is some linearly independent system. Therefore $\dim(A+B)=s \leq \dim(C)$.


问:
How to prove $\int_{0}^{\infty}\frac{\tan^{a} (x) \;dx}{x^2+b^2}=\frac{\pi}{2b}\frac{\tanh^{a} (b)}{\cos\frac{a\pi}{2}}$
I'm having trouble to prove the following:
$$
\int_{0}^{\infty}\frac{\tan^{a}\left(x\right)}
{x^{2} + b^{2}}\,\mathrm{d}x =
\frac{\pi}{2b}\,\frac{\tanh^{a}\left(b\right)}{\cos\left(\pi a/2\right)}\quad
\mbox{where}\ \left\vert a\right\vert < 1\
\mbox{and}\ b > 0
$$
The main difficulty is that the sign of the $\tan\left(x\right)$  alternates between positive and negative and so we need $\left(-1\right)$ to raise to power of noninteger number $a$.

I have no idea how to proceed.

答:
The integrand has poles at $x=\pm ib$ and singularities on the real axis at $x=(2n+1)\frac{\pi}{2}$ with cuts where the tangent becomes negative. Nevertheless, the integral can be viewed as a complex contour in the upper half-plane with the segment $(-\infty,\infty)$ shifted slightly by $+i0$ to avoid the singularities and make the integrand holomorphic in the upper half-plane except at $x=ib$. Let
$$C_R=(-R,R) \cup \gamma_R$$
be the closed contour where $\gamma_R$ is the large semi-circle counter-clockwise in the upper half-plane of radius $R$ s.t. $R\neq (2n+1)\frac{\pi}{2}$. Since the tangent is bounded on $\gamma_R$, the integral over the arc $\gamma_R$ vanishes in the limit $R \rightarrow \infty$. Using the residue theorem for the limiting closed contour $C_\infty$, we arrive at
$$\frac{\pi}{b} \, \tan^a(ib) = \int_{C_\infty} \frac{\tan^a(x)}{x^2+b^2} \, {\rm d}x = \int_{-\infty+i0}^{\infty+i0} \frac{\tan^a(x)}{x^2+b^2} \, {\rm d}x = (e^{i\pi a}+1) \int_0^{\infty} \frac{|\tan(x)|^a}{x^2+b^2} \, {\rm d}x  \,. \tag{1}$$
Rearranging gives the desired result.

Technically the last step is a bit more intricate than it appears at first sight. When substituting $x\rightarrow -x$ for the segment $(-\infty+i0,i0)$, it becomes $(-i0,\infty-i0)$ which matters because it changes the phase of the tangent. When $\tan(x)$ becomes negative, it has phase $+\pi$ for $\Im(x)>0$ and $-\pi$ for $\Im(x)<0$. This can be seen for e.g. $x\in(-\pi/2+i0,i0)$ and the fact that for $x$ close to $0$, $\tan(x) \approx x$ and hence they share the same phase properties i.e. $$\arg(\tan(x))=\arg(x)=\arg(-1+i0)=\arg(e^{i\pi})=\pi$$ and by the periodicity this translates to $\mathbb{R}$.

Therefore the correct sequence of steps leading to the RHS of (1) is as follows:
$$\int_{i0}^{\infty+i0} \frac{\tan^a(x)}{x^2+b^2} \, {\rm d}x + \int_{-i0}^{\infty-i0} \frac{(-\tan(x))^a}{x^2+b^2} \, {\rm d}x \\
=\sum_{n=0}^\infty  \int_{n\pi + i0}^{n\pi +\pi/2+i0} \frac{\tan^a(x)}{x^2+b^2} \, {\rm d}x + \sum_{n=0}^\infty \int_{n\pi + \pi/2 + i0}^{n\pi + \pi+i0} \frac{\tan^a(x)}{x^2+b^2} \, {\rm d}x \\
+ \sum_{n=0}^\infty  \int_{n\pi - i0}^{n\pi +\pi/2 - i0} \frac{(-\tan(x))^a}{x^2+b^2} \, {\rm d}x + \sum_{n=0}^\infty \int_{n\pi + \pi/2 - i0}^{n\pi +\pi - i0} \frac{(-\tan(x))^a}{x^2+b^2} \, {\rm d}x \, .$$
The argument inside $(\dots)^a$ of the first and last term is positive (on this part of the decomposed domain), i.e. the integrand has phase $0$ and both terms can be combined to $\int_0^\infty \frac{|\tan(x)|^a}{x^2+b^2} \, {\rm d}x$. The second term has $\Im(x)>0$ and thus phase $+\pi$. Likewise in the third term $\Im(x)<0$ and hence $-\tan(x)$ has phase $+\pi$ also. Pulling out the phase and using absolute values instead, these two terms combine to $e^{i\pi a} \int_0^\infty \frac{|\tan(x)|^a}{x^2+b^2} \, {\rm d}x$
giving the RHS of (1).


问:
Definition of Arithmetic Mean
I am currently in school and we are learning about Arithmetic Sequences. In one of the sections in our learning material, the arithmetic mean is defined as the term of the arithmetic sequence that is between two given terms. For me, at least, I know this is wrong. In all the reliable sources I found about the definition of arithmetic mean, its definition is the sum of the numbers divided by the count of the numbers. Here, an example is given that the arithmetic mean(s) of 2, 4, 6, 8, 10 are 4, 6, and 8 (which should be 6). Our material also mentions about inserting arithmetic means between two numbers like "Insert 4 arithmetic means between 7 and 47," wherein the answer would be 15, 23, 31, and 39 because it will form an arithmetic sequence. As far as I know I don't know anywhere where this definition is taught. I think this is stupid. So my questions are,

Is this really accepted? or where is it accepted?

What should I do about it? We have been taught this 2 months ago but no one bothered to ask the teacher about it that time

答:
I don't personally love this terminology, but there is a reason for it.  In your arithmetic sequence $$2,4,6,8,10,$$ each term between $2$ and $10$ is the arithmetic mean of the two numbers adjacent to it, i.e.
\begin{align}
\frac{2 + 6}{2} &= 4\\
\frac{4+8}{2} &= 6\\
\frac{6+10}{2} &= 8,
\end{align}
and so in some sense it makes sense to call these the arithmetic means of the sequence between $2$ and $10$.  In general, given an arithmetic sequence $$a_{1},a_{2},\dotsc,a_{n},\dotsc$$ with common difference $a_{2} - a_{1} = d$, the arithmetic mean of any two terms of the form $a_{k}$ and $a_{k+2}$ will be the middle term, $a_{k+1}$:
$$\frac{a_{k} + a_{k+2}}{2} = \frac{a_{k} + a_{k} + 2d}{2} = \frac{2(a_{k} + d)}{2} = a_{k} + d = a_{k+1}.$$

Title: Definition of Arithmetic Mean


问:
Tutte's theorem - How does assuming a maximal graph prove the theorem true for non-maximal graphs?
From Introduction to Graph Theory by West:

A graph $G$ has a $1$-factor if and only if $o(G - S) \le |S|$ for every $S \subseteq V(G)$ where $o(G)$ means the number of odd components of $G$.

Tutte shows:

$(1.)$ If $G' = G + e$ and $G$ satisfies Tutte's condition, then $G'$ also satisfies Tutte's condition.

and

$(2.)$  If $G'$ has no $1$-factor, then $G$ has no $1$-factor.

He then proceeds to prove sufficiency by stating "the theorem holds unless there exists a simple graph $G$ such that $G$ satisfies Tutte's condition, $G$ has no $1$-factor, and adding any missing edge to $G$ yields a graph with a $1$-factor".  He then proceeds to obtain a contradiction by showing that $G$ actually contains a $1$-factor.

What I don't understand is:

How does this show that for any graph $G$ that satisfies Tutte's condition it must have a $1$-factor?  This just seems to show:  If $G$ is a graph that satisfies Tutte's conditions, and $G'$ is the maximal graph created from $G$, then $G'$ has a $1$-matching.

Anyone have any ideas?

答:
Assume $G$ satisfies the theorem's conditions but does not have a $1$-factor. We know that:

We can keep adding edges to $G$ to get a maximal graph $G'$ that doesn't have a $1$-factor, but is such that by adding any other edge anywhere it will have a $1$-factor.

$G'$ satisfies the theorem's conditions (since adding edges preserves them).

The proof shows that a graph with properties of $G'$ cannot exist.

So we reach a contradiction and we conclude that the original graph $G$ must have a $1$-factor.


问:
Probability of at least one boy and one girl
Assume that the probability of a new born child being a boy is 1-p and the probability of being a girl is p. Assume that the sex of a newborn child is independent of the sex of previous children (if any) with the same parents.

A couple with those probabilities of having a boy or girl decides to continue having children until they have at least one child of each sex. What is the distribution of the number of children,X, the couple must have for this to occur and its expected value.

I found that P(X)=(1-p)^(x-1) *p +(p)^(x-1) *(1-p) and that E(X)= 1/(1-p) + 1/p .

Can anyone confirm whether or not this is correct?

If it is incorrect, could you please explain why?

答:
The distribution is correct but the expected value is not.

I think it is simpler to approach this by noting that you expect it to take $\frac 1p$ tries for a girl and $\frac 1{1-p}$ for a boy so the expected number of times to get one of each is $$E=p\times \left(1+\frac 1{1-p}\right)+(1-p)\times \left(1+\frac 1p\right)=\boxed {\frac {p^2-p+1}{p\,(1-p)}}$$

We remark that, for $p=\frac 12$, this yields $E=\frac {3/4}{1/4}=3$, which is easily seen to be correct (after the first child, you expect it to take two tries to get the opposite gender).  The formula provided by the OP gives $2$, so presumably there was an error in summing the infinite series.  Indeed, the infinite series is a bit messy, which is a good reason for looking for alternate methods of solution.


问:
Simplify $\sqrt n+\frac {1}{\sqrt n}$ for $n=7+4\sqrt3$
If $n=7+4\sqrt3$,then what is the simplified value of
$$\sqrt n+\frac {1}{\sqrt n}$$
I was taking LCM but how to get rid of  $\sqrt n$ in denominator</p>

答: Hint:  We have $\left(\sqrt n+\dfrac {1}{\sqrt n}\right)^2=n+2+\frac 1n$, then multiply top and bottom of the last by the conjugate.</p>


问:
$\frac{n}{2} < 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2^n -1}$ is not true for every $n > 1, n \in \mathbb{N}$, right?
So I found an exercise where it is required to prove that for every $n > 1, n \in \mathbb{N}$ the inequality is true:

$$\frac{n}{2} < 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2^n -1} < n$$

I found that disturbing since I think that for n > 5 that is simply not true (the left hand side is wrong). Can somebody ensure me in my thinking or tell me if I'm wrong?

答:
This inequality arises from a familiar argument to show that the harmonic series is divergent.

Let $$a_n = \sum_{k = 0}^{2^n - 1} \frac{1}{2^n + k} = \frac{1}{2^n} + \frac{1}{2^n + 1} + \cdots + \frac{1}{2^{n+1} - 1}.$$  So for instance, $$a_2 = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}.$$  Each term in $a_n$ is bounded below by $1/2^{n+1}$, and above by $1/2^n$.  Since $a_n$ contains $2^n$ terms, it follows that $$\frac{1}{2} < a_n \le 1$$ for all $n \ge 0$, with equality on the RHS occurring only for $n = 0$; and $$\frac{n}{2} < \sum_{m=0}^{n-1} a_m < n$$ for $n \ge 1$.  But $$\sum_{m=0}^{n-1} a_m = \sum_{k=0}^{2^{n-1} - 1} \frac{1}{k} = 1 + \frac{1}{2} + \cdots + \frac{1}{2^n - 1}.$$

In fact, we can prove a much tighter bound with calculus, but this is beyond the scope of the question.


答:
Hint:

Supposing $C$ is the vector space everyone lives in, you  have $\dim A \le n$ and $\dim B\le k$. Concerning $A+B$, you have  the short exact sequence
\begin{alignat}{2}
0\longrightarrow A\cap B &{} \longrightarrow & A\oplus B &{}\longrightarrow A+B\longrightarrow 0 \\ c&\longmapsto& (c,-c)\\
&&(a,b)&\longmapsto a+b
\end{alignat}
which shows that $\:\dim A+\dim B=\dim(A\oplus B) =\dim(A+B)+\dim(A\cap B)$.

Title: proof two general dimension inequalities

Body:
I have just showed that $A+B$ is indeed a subspace by showing it is non empty and closed under multiplication and addition, however I am not sure how to do these types of problems with $\mathbf{dim}$.

Problem: Dimensions

$A+B=\left \{ a+b : a\in A, b\in B \right \}$ and $C$ is a vectorspace and $A \subset C, B \subset C$.

Let $a_1,..,a_n$ be elements of $A$ and let $b_1,..,b_k$ be elements of $B$ be two families of vectors in $C$. Let $A=\operatorname{Span}(a_1,..,a_n)$ and $b=\operatorname{Span}(b_1,..,b_k)$

Now, Show that $\mathbf{dim}(A+B)\leq n+k$ and show that $\mathbf{dim}(A+B)\leq \mathbf{dim}(C)$

....................................................................

I am new to the subject and I could use some help since I am really struggling with both span and dim. I get the definitions but no idea on how to approach such a problem.


答:
Let $X$ be the number of required children. Notice that, after the first child is born, the remaining number of children follows a Geometric distribution with a parameter depending on the first child's gender.

Let $G$ denote the gender of the first child. Then we have
\begin{align}
\mathbb E[X\,\vert\, G = M] &= 1 + \frac 1p\\
\mathbb E[X\,\vert\, G = F] &= 1 + \frac 1{1-p}\\
\end{align}
Therefore, we have
$$
\mathbb E[X] = (1-p)\mathbb E[X\,\vert\, G = M] + p\mathbb E[X\,\vert\, G = F] = 1 + \frac p{1-p} + \frac {1-p}p.$$

Title: Probability of at least one boy and one girl

Body:
Assume that the probability of a new born child being a boy is 1-p and the probability of being a girl is p. Assume that the sex of a newborn child is independent of the sex of previous children (if any) with the same parents.

A couple with those probabilities of having a boy or girl decides to continue having children until they have at least one child of each sex. What is the distribution of the number of children,X, the couple must have for this to occur and its expected value.

I found that P(X)=(1-p)^(x-1) *p +(p)^(x-1) *(1-p) and that E(X)= 1/(1-p) + 1/p .

Can anyone confirm whether or not this is correct?

If it is incorrect, could you please explain why?


问:
Probability of Not Being Caught of Cheating
Suppose that there is a website on Internet. There are n number of students in a university class and you are the 1 one of n students. Probability of each person who may enter this website is 1/2. If at least two students enter this website, they are all being caught of cheating. What is the probability that you won't be caught of cheating given that you already entered this website?</p>

答:
I found following probability but I am not sure whether it is correct or not:

1/2^(n-1) , if n=1

1/2^(n) , otherwise


问:
Is my method of solving equation correct?
The problem in question is
$$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$

using $$a+b=2$$ where $a=\sqrt[5]{16+\sqrt{x}}$ and $b=\sqrt[5]{16-\sqrt{x}}$

$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$
$$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
$$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$

Got $\frac{32}{(a+b)^2}$ from the fact that $(a+b)^2(a+b)^3=32$ and $a+b=2$

$$a^5+b^5+5ab\biggl(\frac{32}{(2)^2}\biggr)+10a^2b^2(2)=32$$
$$a^5+b^5+40ab+20a^2b^2=32$$
From when I defined a and b earlier, I substitute and get
$$\left(\sqrt[5]{16+\sqrt{x}}\right)^5+\left(\sqrt[5]{16-\sqrt{x}}\right)^5+40\sqrt[5]{\left(16-\sqrt{x}\right)\left(16+\sqrt{x}\right)}+20\sqrt[5]{\left(16-\sqrt{x}\right)^2\left(16+\sqrt{x}\right)^2}=32$$
$$\require{cancel}\cancel{16}\cancel{+\sqrt{x}}+\cancel{16}\cancel{-\sqrt{x}}+40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=\cancel{32} 0$$
$$40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=0$$
$$20\biggl(2\sqrt[5]{256-x}+\sqrt[5]{\left(256-x\right)\left(256-x\right)\biggr)}=0$$

Then let $u=\sqrt[5]{256+{x}}$,
$$20(2u+u^2)=0$$
$$u(u+2)=0$$
$$u=0,-2$$

Substituting u to get x from $u=\sqrt[5]{256+{x}}$, I get $$x=\cancel{-288},256$$

However, since the original equation has a $\sqrt{x}$, which can't be negative, I eliminate $x=-288$, leaving just $$x=256$$ as my answer.

So, this is how I arrived on my answer. Did I perform any mathematical errors or any illegal mathematical maneuvers? Please let me know.
Thank you!

答:
There is another solution.

Let $\sqrt[5]{16+\sqrt{x}}=1+t$, where $t>0$; therefore, $\sqrt[5]{16-\sqrt{x}}=1-t$. Hence
$$(1+t)^5+(1-t)^5=32,$$
or
$$10\,t^4+20\,t^2+2=32.$$
Immediately we have
$$t^4+2\,t^2-3=0,$$
and $t^2=1$. Since $t>0$, $t=1$ is the single solution. So we have
$$\sqrt[5]{16-\sqrt{x}}=0\Rightarrow \sqrt{x}=16\Rightarrow x=256.$$


问:
Spivak Rising Sun Lemma
Suppose that all points of $(a,b)$ are shadow points, that is $\forall x \in (a,b) \space \exists y\space(y>x \mbox{ and } f(y)>f(x))$.

For all $x\in(a,b)$, prove that $f(x)\leq f(b)$.

let $A=\{y: x\leq y \leq b \mbox{ and } f(x)\leq f(y)\}$. Then in the solutions manual he proceeds:

Notice that we must have $f(x)\leq f(\sup A)$ because $f$ is continuous at
$sup A$ and there are points $y$ arbitrary close to $\sup A$ with
$f(x)\leq f(y)$. (A simple $\varepsilon - \delta$ argument is being
suppressed).

How does $f(x)\leq f(\sup A)$ follow from the continuity of $f$? It would've been very helpful if he had not suppressed the argument.

答:
Edit: I think I better understand the book solution. It's a bit confusing because of Spivak's choice of $x$ as some arbitrary fixed point in $(a,b)$. Up to this point we've grown accustomed to looking at how things change as we change $x$, but here $x$ is a fixed point, at least for this stage  of the argument.

The part you're asking about boils down to showing that $\alpha = \sup A$ is in $A$, i.e. $f(\alpha) \geq f(x)$.

Suppose instead $f(\alpha) < f(x)$

Take $\varepsilon = f(x) - f(\alpha) > 0$.

Because $f$ is continuous, $f$ can be made arbitrarily close to $f(\alpha)$ for for all $y$ sufficiently close to $\alpha$.

$\lim_{y \to \alpha-}f(y) = f(\alpha)$

For our $\varepsilon$ there exists some $\delta > 0$ such that for all $y$ if
$$\alpha - \delta < y \leq \alpha \text{, then } |f(y) - f(\alpha)| < \varepsilon$$

or

$$f(y) - f(\alpha) < \varepsilon $$
$$f(y) - f(\alpha) < f(x) - f(\alpha)$$
$$f(y) < f(x)$$

If this were true then there would be some $y_0$ in $(\alpha - \delta, \alpha)$ such that:
$$f(y) < f(x) \text{ for all } y \text{ in } [y_0, \alpha]$$

But this contradicts $\alpha$ being the least upper bound of $A$. ($y_0$ would be an upper bound less than $\alpha$).

Therefore, $f(\alpha) \not\lt f(x)$

You could show the same result by building the continuous function $g(y) = f(y) - f(x)$ and using Theorem 6-3 to show $g(y) < 0$ near $\alpha$...

Original post
I think I found a proof that is a bit simpler. Unfortunately it doesn't directly use any of the supremum/infimum arguments which are the whole point of the chapter, so it may not be helpful.

I haven't been back to check, but it's likely this is identical to Spivak's solution, only phrased differently...

First, we know that $f(b) \geq f(x)$ for all $x > b$. (If this were not the case, $f(b)$ would be a shadow point.)

Next, let's look at the interval $[a,b]$.

Suppose there exists some $x_0$ in (a,b) with $f(x_0) > f(b)$.

In this case, take the interval $[x_0,b]$. $f$ is continuous, so it must take on its max value for some $x$ in this interval, i.e. there exists some $x_1$ in $[x_0,b]$ such that $f(x_1) \geq f(x_0) > f(b)$.

$f(x_1)$ is a max on $[x_1, b]$ and it's certainly greater than all values of $f(x)$ to the right of $b$, therefore $x_1$ must not be a shadow point.

Since this is not the case, we must have $f(x) \leq f(b)$ for all $x$ in $(a,b)$.

Given this, if $f(x) = f(b)$ for any $x$ in $(a,b)$, again, $x$ would not be a shadow point. Therefore, $f(x) < f(b)$ for all $x$ in $(a,b)$.

Finally we look at $a$. We must have $f(a) \geq f(b)$, otherwise $a$ would be a shadow point.

If $f(a) > f(b)$ then we have

$$f(a) > \frac{f(a) + f(b)}{2} > f(b)$$

and the IVT tells us there exists some $x$ with $a < x < b$ such that

$$f(x) = \frac{f(a) + f(b)}{2} > f(b)$$

but this is a contradiction because we know $f(x) < f(b)$ for all $x$ in $(a,b)$.

Therefore, we must have $f(a) = f(b)$.

Title: Spivak Rising Sun Lemma


问:
Find $G'(0)$ where $G(x)=\int_0^{x+\pi} \frac{\sin(x+t)}{1+(x+t)^2} dt$
I want to find $G'(0)$ where $$G(x)=\int_0^{x+\pi} \frac{\sin(x+t)}{1+(x+t)^2} dt,\hspace{2cm} x\in\mathbb{R}.$$

I notice that this decomposes nicely into being $G(x)=\int_0^{x+\pi} f(x+t) dt$ if $f(x)=\frac{\sin x}{1+x^2}$, but am not sure if that is meaningless. I don't know how to find $G'$. Please help.

Body:
hint

Put
$$u=t+x$$
it becomes

$$G(x)=\int_x^{2x+\pi}\frac{\sin(u)du}{1+u^2}$$
$$=\int_{-\pi}^{\color{red}{2}x}\frac{\sin(v+\pi)dv}{1+(v+\pi)^2}-\int_0^x\frac{\sin(u)du}{1+u^2}$$

thus
$$G'(x)=\color{red}{2}\frac{\sin(2x+\pi)}{1+(2x+\pi)^2}-\frac{\sin(x)}{1+x^2}$$


问:
Determinants: A Special Condition
Under what conditions is

$$ \det(A_1 + \cdots + A_n) = \det(A_1)+\cdots+\det(A_n), $$

just curious.

答:
Of course we're only interested in $n \geq 2$. Let $A_{jk}$ be the $k$th row of $A_j$. The determinant map $\det(A_j)$ can be instead interpreted as an alternating $n$-multilinear map $\delta: \mathbb{F}^n \times \ldots \times \mathbb{F}^n \rightarrow \mathbb{F}$ such that $\delta(e_1,\ldots,e_n) = 1$; then $\det(A_j) = \delta(A_{j1}, \ldots, A_{jn})$. Then the condition
$$ \det(A_1 + \ldots + A_n)  = \det(A_1) + \ldots + \det(A_n)$$
can be rewritten as
$$ \delta(A_{11} + \ldots + A_{n1}, \ldots, A_{1n} + \ldots + A_{nn}) = \delta(A_{11}, \ldots, A_{1n}) + \ldots + \delta(A_{n1}, \ldots, A_{nn})$$
The LHS can be expanded, by multilinearity, and subtracting from the RHS, we obtain
$$ \sum_{i \in \{1, \ldots, n\}^n, i(k) \text{ not constant}} \delta(A_{i(1),1}, \ldots, A_{i(n),n}) = 0$$
In other words, the sum of all determinants of matrices where you mix and match all possible arrangements of rows (of the correct index) from any of the matrices $A_1, \ldots, A_n$, such that not all the rows come from the same matrix, must equal zero. The above condition is, of course, necessary and sufficient, but not probably not very satisfying since it is merely a bunch of algebraic manipulations. But I think that is all you can get with such a general algebraic condition.

You could construct a variety of sufficient conditions as special cases of the above condition, though again I'm not sure which would be particularly interesting without being unnecessarily restrictive or obvious. For example, you could require that each summand is zero; this would be equivalent to saying that every matrix where you mix up rows between the matrices $A_1, \ldots, A_n$ is not invertible.

Title: Determinants: A Special Condition


问:
Algebra which becomes a field after a base change
Let $A$ be an associative algebra over a field $k$, and let $L/k$ be any field extension.

Assume that $A \otimes_k L$ is a field (not necessarily a finite extension of $L$ !).  Can we deduce that $A$ is a field?

I know that $A$ is a (commutative, unital) integral domain, since $A \cong A \otimes_k k$ embeds in $A \otimes_k L$. But how to go further?

Maybe if $L/k$ is Galois, we can use $A \cong (A \otimes_k L)^{Gal(L/k)}$, after checking that this is indeed a $k$-algebra isomorphism ?

答:
Yes, $A$ is a field. We don't need to assume that the field extension is Galois or finite.

Lemma: If $k \to L$ is a field extension, then the base change / extension of scalars functor $(-) \otimes_k L$ (which we'll write $(-)_L$ to keep things short) is faithful (induces injections $\text{Hom}_k(V, W) \to \text{Hom}_L(V_L, W_L)$) and conservative (reflects isomorphisms: if $T : V \to W$ is a linear map such that $T_L : V_L \to W_L$ is an isomorphism, then $T$ is an isomorphism).

Proof. Faithfulness can be proven pretty explicitly by choosing bases if necessary. As for conservativity, if $V$ and $W$ are finite-dimensional we can argue using determinants; in general we can use the fact that faithful functors reflect epimorphisms and monomorphisms, so if $T_L$ is an isomorphism then $T$ is both an epimorphism and a monomorphism and hence an isomorphism (note that we really need to be working in a nice category like an abelian category here; this is false in general categories). $\Box$

Now we can argue as follows. An element $a \in A$ is invertible iff left multiplication $A \xrightarrow{a} A$ is an isomorphism (of $k$-modules). If $a \in A$ is nonzero then $a_L \in A_L$ is nonzero by faithfulness, and hence invertible since $A_L$ is a field by hypothesis. By conservativity, $a$ is invertible. Hence $A$ is a field.

This argument applies more generally to any faithfully flat ring homomorphism $R \to S$ replacing $k \to L$: these are also faithful and conservative, and so also don't make anything invertible that wasn't invertible already.


问:
normal distribution candy problem
There are 35 kids who are given candy, the amount of candy each kid gets distributes $N(8,2^2)$.
different amount of candy for each kid.

first question is, what's the probability that there are more than 300 candies overall?

I defined rv $s$  of $N(35*8,35^2*2^2)$ and $P(s>300)=1-P(s<=300)=1-Φ((300-35*8)/(2*35))=1-0.6123$

I hope that it's correct.
Second part is where I'm stuck.
out of 35 kids 15 are older and the others are younger. An adult marks the wraper of different candies from 5 different kids. What is the probability that all of the marked candies will belong to older kids?

I thought about approaching the problem as a binomial distribution of $Bin(5,15/35)$ but it doesn't seem correct beause not all kids have the same amount of candy. Would love to get some help as to how to solve this problem.

thank you.

答:
The probability to mark the candies exactly of 5 older kids is the probability to chose 5 older kids among the 35 total, say

$$\frac{\binom{15}{5}\binom{20}{0}}{\binom{35}{5}}$$

(an hypergeometric...)

The first point is wrong. The total candies distribution is the sum of 35 independent gaussian, thus the total amount of candies is

$$X=X_1+X_2+\dots+X_{35}$$

Assuming independence, you get

$$X\sim N(35\cdot8;4\cdot 35)$$

Thus

$$P(X>300)=1-\Phi\Bigg(\frac{300-35\cdot8}{\sqrt{35\cdot4}}\Bigg)\approx 4.55\%$$

Title: normal distribution candy problem


问:
Exponential Models Help

The volume of a substance, A, measured in cubic centimeters increases according to the exponential growth model $\frac{\mathrm{d}A}{\mathrm{d}t} = 0.3A$, where $t$ is measured in hours.

The volume of another substance, B, also measured in cubic centimeters increases at a constant rate of
$1\ \mbox{cm}^{3}$ per hour according to the linear model $\frac{\mathrm{d}B}{\mathrm{d}t} = 1.$

At $t = 0$, substance A has a volume $A\left(0\right) = 3$ and substance B has size $B\left(0\right) = 5$.

I got $3.4531$ using the equations $\frac{1}{0.3}log|\frac{A}{3}|= t$ and $B-5=t,$ is that correct?

答:
For the Initial Value Problem, $\frac{dy}{dt} = ky$ with $y(0) = y_0$ we obtain the solution $$y = y_0e^{kt}$$
So for the two fluids we have (both in units of $cm^3$)
$$A = 3e^{0.3t}\space\text{ and }\space B = 5+t$$
So
\begin{align*}
A &= B\\
3e^{0.3t} &= 5+t\\
t &= -5+\frac{1}{-0.3}W\left(-0.3\cdot3e^{0.3\cdot(-5)}\right)
\end{align*}
where $W$ is the Lambert W function. However this gives the negative $t$ value which satisfies this equation, a numerical approximation gives that the positive $t$ values is $t \approx 3.45305018$.

Title: Exponential Models Help


问:
Polynomials Problem in Abstract Algebra
this is my very first question here and I have many others..
So I have this problem that I have no clue about, I hope to get a solution for it and I hope you guys tell me what is it that I need to learn in order to be able to solve such problems

The problem:

The polynomial f(x) ∈ Z[x] has the root (a), it's known that f(1) − f(15) = 42 and f(3) + f(4) = 181
find  the remainder of the division of (a) by 3.

答:
Lemma: Let $f\in\mathbb Z[X]$. Also let $b,x,y\in\mathbb Z$ such that $x=y\mod b$. Then $f(x)=f(y)\mod b$.

Proof: I leave this to you.

Now to the actual question. If $a\in\mathbb Z$ is a root of $f$, that means $f(a)=0$, which also means that $f(a)=0\mod3$. That is, if $f(x)\neq0\mod 3$, then $x\neq a\mod 3$. Since the only possible remainders are $0,1,2$, it is enough to discard two of these possibilities.

Let's look at the other information we have:

$$f(1)-f(15)=42,\\
f(3)+f(4)=181.$$

With the lemma, it follows that
$$f(1)-f(0)=0\mod3,\\
f(0)+f(1)=1\mod3.$$

This is a system of linear equations $\mod3$, which we can solve for $f(1)$ and $f(0)$. We get

$$f(0)=2\mod3,\\
f(1)=2\mod3.$$

So neither $f(0)$, nor $f(1)$ is $0\mod3$, so $a$ is neither $0$, nor $1\mod3$. Which only leaves us with $a=2\mod3$.

Title: Polynomials Problem in Abstract Algebra


问:
$\int _0^{2\pi}\Biggl(\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}\Biggr)^2 dx$ by changing integral and summation.

Let $f(x)=\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}$. Prove that $f$ converges uniformly on $[0,2\pi]$ and evaluate $\int _0^{2\pi}f(x)^2dx$.

Since $\left|\frac{\sin nx}{2^n}\right|\le\left|\frac{1}{2^n}\right|$ and $\sum \limits _{n=1}^\infty \frac{1}{2^n}$ converges, $f(x)$ converges uniformly on $[0,2\pi]$.

After reading some related post, Changing integral and summation, and Calculate $\int_{-\pi}^\pi\bigg(\sum_{n=1}^\infty\frac{\sin(nx)}{2^n}\bigg)^2dx$. I need to prove $$\Biggl(\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}\Biggr)^2 \text{converges uniformly on }[0,2\pi].$$

Write $\Biggl(\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}\Biggr)^2=\sum_\limits{n=1}^\infty\sum_\limits{m=1}^\infty \frac{\sin nx}{2^n}\frac{\sin mx}{2^m}=\sum_\limits{k=1}^\infty c_k$, where $c_k=\sum_\limits{j=1}^k\frac{\sin jx\sin (k-j)x}{2^k}$.

So $$|c_k|\le\left|\frac{k}{2^k} \right|$$
Is $\sum_\limits{n=1}^\infty \frac{k}{2^k}$ converges?

I have another question. In the linked post, they write $$\begin{align} \int _0^{2\pi}\Biggl(\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}\Biggr)^2 dx & = \int _0^{2\pi} \sum_\limits{n=1}^\infty\sum_\limits{m=1}^\infty \frac{\sin nx}{2^n}\frac{\sin mx}{2^m} dx \\&=\sum_\limits{n=1}^\infty\sum_\limits{m=1}^\infty \int _0^{2\pi} \frac{\sin nx}{2^n}\frac{\sin mx}{2^m}dx . \end{align}$$

But I think it is a Cauchy product and so $$\begin{align} \int _0^{2\pi}\Biggl(\sum_\limits{n=1}^\infty \frac{\sin nx}{2^n}\Biggr)^2 dx &=\int _0^{2\pi}\sum_\limits{k=1}^\infty \sum_\limits{j=1}^k\frac{\sin jx\sin (k-j)x}{2^k}dx .   \end{align}$$

What is the difference between these two integrals?

答:
(1). A useful lemma (which has a short elementary proof) is that if $g_n\to g$ uniformly on a domain $D\subseteq \Bbb R$ and if each $g_n$ is continuous on $D$ then $g$ is continuous on $D.$ From this, if, also, $D$ is a closed bounded interval, then $g_n^2\to g^2$ uniformly on $D,$ as follows:

Let $M=\sup_{x\in D}|g (x)|.$ Then $M\ne \infty$ because $D$ is a closed bounded interval and $g$ is continuous on $D.$

Let $n_0\in\Bbb N$ such that $\forall n\ge n_0 \,(\sup_{x\in D}|g_n(x)-g(x)|\le 1).$ Then $\forall n\ge n_0 \,(\sup_{x\in D} |g_n(x)|\le 1+M).$

So if $n\ge n_0$ and $x\in D$ then $$|g_n(x)^2-g(x)^2|=|g_n(x)-g(x)|\cdot |g_n(x)+g(x)|\le$$ $$\le |g_n(x)-g(x)|\cdot (|g_n(x)|+|g(x)|) \le$$ $$\le |g_n(x)-g(x)|\cdot (2M+1).$$ So for all but finitely many $n$ we have $$\sup_{x\in D} |g_n(x)^2-g(x)^2|\le (2M+1)\cdot \sup_{x\in D}|g_n(x)-g(x)|.$$

(2). In your Q, with $D=[0,2\pi],$ let $g_n(x)=\sum_{j=1}^{n}(\sin jx)/2^j.$ Then $g_n(x)$ converges to a value $g(x)$ for each $x$ by term-by-term comparison with the absolutely convergent series $\sum_{j=1}^{\infty}1/2^j.$ And $g_n$ converges uniformly because $\sup_{x\in D}|g_n(x)-g(x)|=$ $=\sup_{x\in D}|\sum_{j=n+1}^{\infty}(\sin jx)/2^j|\le$ $ \sum_{j=n+1}^{\infty}1/2^j=1/2^n.$ And each $g_n$ is continuous.

Therefore by (1), $g_n^2\to g^2$ uniformly.

Remark: In (1), to obtain uniform convergence of $g_n^2$ it is usually necessary that $D$ is closed and bounded. Examples: (i). $D=\Bbb R$ and $g_n(x)=x+1/n$ and $g(x)=x.$ (ii). $D=(0,1]$ and $g_n(x)=1/x +1/n$ and $g(x)=1/x.$ In both (i) and (ii), $g_n\to g$ uniformly but $\sup_{x\in D}|g_n(x)^2-g(x)^2|=\infty$ for every $n\in \Bbb Z^+.$


问:
Finding specific alternative form of $\frac{(x-y)x+{y\over(y-z)}}{(x+y)z}=ab$
How does one approach; $$\frac{(x-y)x+{y\over(y-z)}}{(x+y)z}=ab$$ to find the form:  $$-a b z (x+y) (y-z) = x^2 (-y)+x^2 z+x y^2-x y z-y$$</p>

Body:
$$\frac{(x-y)x+{y\over(y-z)}}{(x+y)z}=ab$$
take LCM on upper part
$$\frac{(x-y)(y-z)x+{y}}{(x+y)(y-z)z}=ab$$
$${(x-y)(y-z)x+{y}}=ab(x+y)(y-z)z$$
$$-({(x-y)(y-z)x+{y}})=-(abz(x+y)(y-z))$$
$$-({(xy-xz-y^2+yz)x+{y}})=-abz(x+y)(y-z)$$
$$-({x^2y-x^2z-xy^2+xyz+{y}})=-abz(x+y)(y-z)$$
$${-x^2y+x^2z+xy^2-xyz-y})=-abz(x+y)(y-z)$$
$${x^2(-y)+x^2z+xy^2-xyz-y})=-abz(x+y)(y-z)$$


问:
Integrating $\int{\frac{x\,dx}{x^2 + 3x -4}}$
I saw this "Beat the Integral" problem and wanted to be sure I was approaching it correctly.

The integral is $$\int{\frac{x}{x^2 + 3x -4}dx}$$

So I decide I want to decompose this, because it's a degree-2 polynomial in the denominator and a degree-1 in the numerator, so that'd be a go-to. Also, since the degree of th enumerator is one less than the denominator we can maybe treat this as $du/u$ which would imply a natural log, though we don't quite know that yet.

So we decompose: I know that $x^2+3x-4$ can be factored as $(x-1)(x+4)$.

That gets me

$$\int{\frac{x}{(x-1)(x+4)}dx}$$
so I can do this:
$$\frac{Ax}{x-1}+\frac{B}{x+4} = \frac{x}{(x-1)(x+4)}$$
Which implies
$$A(x+4)+B(x-1) = x$$
and since the roots are at $x=1$ and $x=-4$, I can set it up like this:
$$5A = 1;-5B=1 \text{  and } A = \frac{1}{5}, B=-\frac{1}{5}$$

Leading to:

$$\frac{x}{5(x-1)}-\frac{1}{5(x+4)}$$
Which I can set up the integral like so:
$$\int{\frac{x}{5(x-1)}-\frac{1}{5(x+4)}dx}=\int{\frac{x}{5(x-1)}dx-\int{\frac{1}{5(x+4)}dx}}$$
I can now integrate by addition here  $$\int{\frac{x}{x-1}}dx=\int{\frac{x-1+1}{x-1}}dx=x+\int{\frac{1}{x-1}}dx=x-\ln(x-1)$$
And doing the same thing for the second term and bringing back my $\frac{1}{5}$
$$\frac{1}{5}(x-\ln(x-1)+\ln(x+4))$$

I suspect there is a further simplification I could do. On a problem like this I also saw it integrated as an arctangent, but that seemed needlessly complex? In any case I was curious if I did this correctly.

答:
$$I=\int\frac{x}{x^2+3x+4}dx=\frac12\int\frac{2x+3}{x^2+3x+4}-\frac{3}{x^2+3x+4}$$
$$I=\frac12\ln|x^2+3x+4|-\frac32\int\frac{1}{(x+3/2)^2+7/4}dx$$
then you can let $u=x+3/2,\,du=dx$ so you have $u^2+(\sqrt{7}/2)^2=\frac74\left((2u/\sqrt{7})^2+1\right)$ now let $v=2u/\sqrt{7},\,dv=2/\sqrt{7}du$ now you have an integral of the form:
$$\int\frac{1}{v^2+1}dv$$ which can be easily solved using the substitution $v=\tan(t)$


问:
No polynomials of degree > 0 are such that $p(t)^2 + q(t)^2 =1$
I'm trying to prove here that ellipses can't be represented by a parametric polynomial and got stuck on this part of the problem:

No real polynomials of degree $> 0$ are such that $$p(t)^2 + q(t)^2 =1$$

I want to prove the above statement.

答:
Over $\mathbb C$ you would have $(p+iq)(p-iq)=1$, which is a factorisation of a constant polynomial $1$ in $\mathbb C[x]$. This implies that $p+iq$ and $p-iq$ are constant polynomial themselves, which implies that $p$ and $q$ are both constant.

(The last conclusion follows directly if $p$ and $q$ are real polynomials, however we can see it is valid even if $p$ and $q$ were complex polynomials to start with, using: $p=\frac{1}{2}((p+iq)+(p-iq))$ and $q=\frac{1}{2i}((p+iq)-(p-iq))$)


问:
Why does lhs of this integral seem to diverge while the rhs does not?
$$
\int_{0}^\infty \exp(x(c-1))dx = \int_{0}^\infty \exp(-x(-c+1))dx \\ \frac{\exp(x(c-1))}{c-1} \Big|_0^\infty=\frac{\exp(-x(c+1))}{c-1} \Big|_0^\infty \\
$$

答: The top row is correct, you have something of the form $e^{-ax}$ where you need $a>0$ for convergence. The bottom row is however wrong


问:
chess probability placing 6 rooks , 2 on black squares and the other 4 on the white
The question is as the title says , we need to place 2 rooks on the black squares and the rest on the white.

My attempt , I tried first picking 2 rooks out of 6 then picking 2 black squares out of 32 for them then picking 4 out of the 4 rooks left and multiplying by 4 out of 32 white circles and our $Ω$ is 6 out of 64 since we need 6 squares for the 6 rooks out of the total 64 squares so my attempt was  $\frac {{6 \choose 2}{32 \choose 2} {4\choose 4}{32\choose 4}} {64 \choose 6}$ which is wrong, according to the book it should be $\frac {{32 \choose 2} {32\choose 4}} {64 \choose 6}$ but it has no explanation on why it is like that.

and I am failing to understand it don't we need to pick the 2 rooks on the black squares first?

答:
"Pick two rooks out of 6"

The rooks are indistinguishable.  There is effectively only one way in which we can do this step, not $\binom{6}{2}$ ways.  The denominator treats the rooks as indistinguishable as well with the $\binom{64}{6}$.

It doesn't really matter whether we treated the rooks as distinguishable or indistinguishable so long as we are consistent in how we treat them.  If the denominator treats them as indistinguishable then the numerator should treat them the same way too.

Removing the erroneous $\binom{6}{2}$ (and unnecessary $\binom{4}{4}$) you will find that your answer then matches the book's.

If you were to treat the rooks as distinguishable, then in reality we have $64\frac{6}{~}=\binom{64}{6}\cdot 6! = 64\cdot 63\cdot 62\cdot 61\cdot 60\cdot 59$ ways in which we can place the rooks.

The numerator would have been to choose the black spaces occupied, choose the white spaces occupied, and then choosing which rook it was in which space in $\binom{32}{4}\binom{32}{2}\cdot 6!$ ways, or if you insisted on breaking them up first in $\binom{32}{4}\binom{32}{2}\binom{6}{4}4!\binom{2}{2}2!$ ways leading to the same final answer as if we treated them as indistinguishable


问:
I have to find a matrix$A_{4 \times 4}$ with eigenvalues $2$ and $4$ and that both $E(2, A)$ and $E(4, A)$ are $1$ dimensional
I know that matrix with eigenvalues $2$ and $4$ can be made from just putting $2$ and $4$ on the diagonal. I have tried many different matrices that have the eigenvalue of $2$ and $4$ but I can not find one that satisfies the rule that that both $E(2, A)$ and $E(4, A)$ are $1$ dimensional. I have come up with some that satisfy either $E2$ or $E4$ but I can not come up with one that satisfies both of them.

答:
Check $A=\begin{bmatrix}
2 & 0 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 0 & 4 & 0 \\
0 & 0 & 1 & 4
\end{bmatrix}$. It satisfies the conditions which are you wanted.

P.S.= If you wanted to know more about such constructions, study Jordan canonical forms. We already know that diagonalizable matrices can be written in the form of $MDM^{-1}$. Jordan canonical form say that every square matrix can be written in the form of $MHM^{-1}$ where $H$ is lower triangular matrix whose diagonal entries are eigenvalues and you may see 1s under the some diagonal entries.


问:
What is wrong with this proof that inaccessible cardinals are inconsistent
Disclaimer: I believe this proof is wrong, but I'm asking because I can't find what's wrong with it, which means I must have some basic misunderstanding of the concepts involved.

First, some definitions. Recursively, we define an ordinal $\alpha$ to be $\eta$-unboundedly elementary:

$\alpha$ is $0$-unboundedly elementary iff $\alpha$ is an ordinal

$\alpha$ is $\eta$-unboundedly elementary iff for all $\beta<\eta$, for all $\gamma$, there exists $\delta>\gamma$, such that $V_\alpha\prec V_\delta$ and $\delta$ is $\beta$-unboundedly elementary. That is, $V_\alpha$ is elementary in unboundedly many $V_\delta$ for $\delta$'s lower in this hierarchy.

Some examples: $1$-unboundedly elementary ordinals are those $\alpha$'s such that $V_\alpha\prec V_\beta$ for unboundedly many ordinals $\beta$. In this blogpost they are called totally otherworldly cardinals. $2$-unboundedly elementary ordinals are those $\alpha$'s such that $V_\alpha$ is elementary in unboundedly many totally otherworldly cardinals.

An interesting feature of these unboundedly elementaries is that they exhibit increasing degree of correctness (where $\kappa$ is $\Sigma_n$-correct iff $V_\kappa\prec_{\Sigma_n}V$). Let us first see that $1$-unboundedly elementary ordinals are $\Sigma_3$-correct:

That they are $\Sigma_2$ correct is proven in the blogpost linked above. To see that they are in fact $\Sigma_3$-correct, suppose $\exists x~\varphi(x)$ is a true $\Sigma_3$ statement with parameters in $V_\alpha$. Fix a witness $a$ such that $\varphi(a)$ is true and fix $\beta$ large enough with $a\in V_\beta$ and $V_\alpha\prec V_\beta$. Then since $\beta$ must be a $\beth$-fixed point (this follows from the fact that, for any ordinals $\alpha,\beta$, $V_\alpha\prec V_\beta$ implies $V_\beta\vDash \mathsf{ZFC}$), $V_\beta\prec_{\Sigma_1}V$. Then the truth of the $\Pi_2$ statement $\varphi(a)$ is preserveed downward to $V_\beta$. This means $V_\beta\vDash \exists x~\varphi(x)$, and by elementarity, so does $V_\alpha$. This shows that $\alpha$ is $\Sigma_3$-correct.

Essentially the same trick can give us that $2$-unboundedly elementary ordinals are $\Sigma_4$ correct: this is because there are unboundedly many $\Sigma_3$-correct ordinals that they are elementary in, and if $\exists x~\varphi(x)$ is a true $\Sigma_4$ statement, then we can find some target $V_\beta$ large enough to include a witness, then proceed the same way as in the last paragraph.

Now here's the problem: the recursive definition of $\eta$-unboundedly elementary ordinals is $\Pi_3$. This means ``there exists a $2$-unboundedly elementary ordinal" a $\Sigma_4$ statement. So the least $2$-unboundedly elementary ordinal must be below the least $\Sigma_4$-correct cardinal. But this contradicts the fact we've just shown, that $2$-unboundedly elementary ordinals are $\Sigma_4$-correct!

Nevertheless, assuming there is an inaccessible cardinal, then the usual Löwenheim–Skolem argument shows the consistency of essentially any degree of unboundedly elementary ordinals, and in particular the consistency of $2$-unboundedly elementary.

Taking stock, we have: $\mathsf{ZFC}\vdash ``\text{there exists an inaccessible cardinal}"\rightarrow$ Con($2$-unboundedly elementary), and also $\mathsf{ZFC}\vdash``2\text{-unboundedly elementary ordinals are inconsistent}"$. Putting things together, this shows that $\mathsf{ZFC}$ refutes the existence of inaccessible cardinals. What went wrong?

Some possibilities I can think of: 1) the proof of their increasing degree of correctness is wrong, 2) the complexity calculation of the recursive definition wrong (or maybe it's not recursive after all, that there is some metamathematical issues), 3) inaccessible cardinals doesn't show that these unboundedly elementary ordinals are consistent.

答:
Your claim that "is $n$-unboundedly elementary" is $\Pi_3$ for every finite $n$ (since we already run into issues when $n=2$, there's no point in going transfinite yet) is unjustified: the complexity of the definition as given increases as $n$ does. In general, writing "$E_n$" for "is $n$-unboundedly elementary," we have the suggestive relative complexity expression $$E_{n+1}=\Pi_2(\Pi_1\wedge E_n).$$ The outer $\Pi_2$ is "$\forall b<n\forall \gamma\exists \delta>\gamma\exists W$." This $W$ is implicitly invoked in your post - it's intended to be the pair $\langle V_\alpha,V_\beta\rangle$, and the "$\Pi_1$" in the parenthetical part takes care of saying this. Finally, we have the reference to $E_n$. In general, for $n>0$ the given definition of $E_n$ is $\Pi_{1+2n}$.

The crucial point is that when we invoke $E_n$ in defining $E_{n+1}$, we're doing so within the scope of some already-introduced quantifiers (the "$\Pi_2$" bit). So we don't get this for free.

Note that this is exactly the same reason that the recursive definition of truth doesn't give a contradiction to Tarski's undefinability theorem.


问:
Simplify $\sqrt n+\frac {1}{\sqrt n}$ for $n=7+4\sqrt3$
If $n=7+4\sqrt3$,then what is the simplified value of
$$\sqrt n+\frac {1}{\sqrt n}$$
I was taking LCM but how to get rid of  $\sqrt n$ in denominator</p>

答:
Ross Millikan's answer is much more elegant, but I am posting this answer as well because it is in a sense more general.  (By this I mean it will work for any expression involving $\sqrt{n}$, and not just this epxression in particular.)  First, we make a guess that
$$
\sqrt{n} = a + b\sqrt{3}
$$

Squaring both sides, we find that
$$
7 + 4\sqrt{3} = a^2 + 3b^2 + 2ab\sqrt{3}
$$

Can you think of $a$ and $b$ integers that will make $a^2 + 3b^2 = 7$ and $2ab = 4$?  If so, you can write $\sqrt{n}$ in a nicer form and the expression should be simple to calculate.

(Note that this method doesn't always work.  But in general if you ever have a square root of a number of the form $x + y\sqrt{z}$, it can't hurt to try making $\sqrt{x + y\sqrt{z}} = x' + y'\sqrt{z}$.)


问:
Calculation of angles in regular polyhedrons
For any regular polyhedron (tetrahedron, cube, octahedron, dodecahedron, icosahedron, ...), how do I calculate an angle from the center of the regular polyhedron to the center of each face? Obviously, the angle decreases with the increasing number of faces. What is the formula, when the number of faces n is the variable?

For cube, n= 6, the angle is 90 degrees:

答:
Théophile's answer is probably the more practical one, but here is a nice technique that works for all regular polytopes in all dimension:

If $P$ is a regular polytope, let $G^\circ$ be the edge-graph of its polar dual (e.g., if $P$ is the cube, then $G^\circ$ is the edge-graph of the octahedron).
Alternatively, $G^\circ$ is the graph in which the vertices are the facets of $P$, and two are adjacent in $G^\circ$ if they are incident.

Let $\theta_2$ be the second-largest eigenvalue of the adjacency matrix of $G^\circ$, then the desired angle $\alpha$ is

$$\alpha=\cos^{-1}\!\Big(\frac{\theta_2}{\mathrm{deg}(G^\circ)}\Big),$$

where $\mathrm{deg}(G^\circ)$ is the vertex-degree of $G^\circ$.


问:
$ 3 \sum^{n}_{k=1} \sqrt{k} \geq 2n \sqrt{n} + 1, n \in \mathbb{N} \setminus \{0\}$ Induction proof. How to separate multiplication.
I have a problem with induction proofs like this one:

$$ 3 \sum^{n}_{k=1} \sqrt{k} \geq 2n \sqrt{n} + 1, n \in \mathbb{N} \setminus \{0\}$$

In the base case it works fine (for $n = 1$):
$$3 * \sqrt{1} \geq 2 + 1 \implies 3 = 3$$

Thesis ~ as above.

Now for $n + 1$:

$$ 3 \sum^{n+1}_{k=1} \sqrt{k} \geq 2(n+1) \sqrt{n+1} + 1$$

$$ 3 \sum^{n}_{k=1} \sqrt{k} + 3\sqrt{n+1} \geq 2n\sqrt{n+1} +1 + 2\sqrt{n+1}$$

And here is my problem: I don't know how to separate $1$ form $n$ in $2n\sqrt{n+1}$ on the left hand side.

On the right hand side of the inequality I got a sum which is good, but I can not separate that multiplication on the right. I thought about:

using Bernoulis inequality, but then I "loose" the square root and can not use my induction thesis.

changing $1$ under the square root for $1$ outside, but then the inequality doesn't work - right hand side is too big

putting both sides of inequality in $e^{ln\{\}}$, but it doesn't work either

I don't know whot to do and it is not the first time I am having exactly the same problem in induction proofs. Can anybody help me?

Body:
For the induction step, we have:

\begin{align}
3\sum_{k=1}^{n+1}\sqrt k &= 3\sum_{k=1}^{n}\sqrt k + 3\sqrt{n+1}\\
&\ge2n \sqrt n + 1 + 3\sqrt {n+1}\\
&=2(n+1)\sqrt{n+1}+1-2n(\sqrt{n+1}-\sqrt n)+\sqrt {n+1}\\
&=2(n+1)\sqrt{n+1}+1-\frac {2n}{\sqrt{n+1}+\sqrt n}+\sqrt {n+1}\\
&\ge 2(n+1)\sqrt{n+1}+1-\frac {2n}{2\sqrt n}+\sqrt {n+1}\\
&= 2(n+1)\sqrt{n+1}+1-\sqrt n+\sqrt {n+1}\\
&> 2(n+1)\sqrt{n+1}+1
\end{align}


答:
Induction based proof:

Assume true for $n$. For $n + 1$, we have
\begin{align*}
\sqrt{n + 1} & \geq \sqrt{n} \\
\implies \sqrt{n + 1} & \geq \frac{2n}{\sqrt{n} + \sqrt{n+1}} \\
\implies \sqrt{n + 1} & \geq 2n(\sqrt{n+1} - \sqrt{n}) \\
\implies 2n \sqrt{n} + \sqrt{n + 1} & \geq 2n\sqrt{n+1}.
\end{align*}

Thus,
$ \displaystyle 2(n+1)\sqrt{n + 1} + 1 \leq 2n \sqrt{n} + 1 + 3\sqrt{n + 1} \leq 3\sum_{k = 1}^n \sqrt{k} + 3\sqrt{n + 1} \leq 3\sum_{k = 1}^{n+1} \sqrt{k}$, as required.

Title: $ 3 \sum^{n}_{k=1} \sqrt{k} \geq 2n \sqrt{n} + 1, n \in \mathbb{N} \setminus \{0\}$ Induction proof. How to separate multiplication.

Body:
I have a problem with induction proofs like this one:

$$ 3 \sum^{n}_{k=1} \sqrt{k} \geq 2n \sqrt{n} + 1, n \in \mathbb{N} \setminus \{0\}$$

In the base case it works fine (for $n = 1$):
$$3 * \sqrt{1} \geq 2 + 1 \implies 3 = 3$$

Thesis ~ as above.

Now for $n + 1$:

$$ 3 \sum^{n+1}_{k=1} \sqrt{k} \geq 2(n+1) \sqrt{n+1} + 1$$

$$ 3 \sum^{n}_{k=1} \sqrt{k} + 3\sqrt{n+1} \geq 2n\sqrt{n+1} +1 + 2\sqrt{n+1}$$

And here is my problem: I don't know how to separate $1$ form $n$ in $2n\sqrt{n+1}$ on the left hand side.

On the right hand side of the inequality I got a sum which is good, but I can not separate that multiplication on the right. I thought about:

using Bernoulis inequality, but then I "loose" the square root and can not use my induction thesis.

changing $1$ under the square root for $1$ outside, but then the inequality doesn't work - right hand side is too big

putting both sides of inequality in $e^{ln\{\}}$, but it doesn't work either

I don't know whot to do and it is not the first time I am having exactly the same problem in induction proofs. Can anybody help me?


问:
Showing uniqueness of Riemann's Integral
I am given the definition: Le $f$ be defined on $[a,b]$. we say that $f$ is Riemann Integrable on $[a,b]$ if there is a number $L$ with the following property: for every $\epsilon>0$, there is a $\delta > 0$ such that $\left\|P\right\|< \delta$ implies $| \sigma -L| < \epsilon$ where $\sigma$ is the Riemann Sum of $f$ over the partition $P$ of $[a,b]$. In this case, we say that $L$ is the Riemann Integral of $f$ over $[a,b]$, and write $\int_{a}^{b} f(x) dx=L$

I am then asked to show why $L$ is a unique limit. It does make sence if not unique but how to show is well above me.

答:
Assume that $L_1$ and $L_2$ are the Riemann inttegrals of $f$ over $[a,b]$. We want to show that $L_1=L_2$. Let $\epsilon >0$. Then for each $i=1,2$, there exists $\delta_i>0$ such that
$$\|P \|<\delta_i \quad \Rightarrow \quad |\sigma-L_i|<\frac{\epsilon}{2}$$
whenever $P$ is a partition of $[a,b].$ Take $\delta$=min $\{\delta_1,\delta_2\}$. Fix a partition $P$ of $[a,b]$ and suppose $\|P\|<\delta$. Note that $\delta\le \delta_i$ for $i=1,2.$ Hence
$$0\le|L_1-L_2|\le|\sigma-L_1|+|\sigma-L_2|<\epsilon.$$
Since $\epsilon>0$ was arbitrary,
$$0\le|L_1-L_2|<\epsilon$$
holds for all $\epsilon >0.$ This forces us to conclude that $|L_1-L_2|=0.$ Hence, $L_1=L_2.$</p>


问:
Cohomology of a complex of sheaf
Let $\mathscr{F}^{\bullet}$ be a complex of sheaves of abelian groups on a scheme $X$, e.g., $\Omega_X^0\to\Omega_X^1\to\Omega_X^2\to...$. In my mind there are three ways of definition the cohomology.

Just take the global section, it forms a chain complex of abelian groups. Then we can just compute the usual cohomology.

We can define the cohomology sheaves $\mathscr{H}^i:=\mathrm{ker}(d^i)/\mathrm{im}(d^{i-1})$ as a quotient sheaf. Then take its global section.

Find an injective resolution of $\mathscr{F}^{\bullet}\to I^\bullet$. Then take global section on $I^\bullet$ and compute the cohomology of these chain of abelian groups.

Are they actually the same? Why do we take the the last one to define the de Rham cohomology?

答:
No, these are very badly not the same. Consider a sheaf $\mathcal{F}$ and the complex which has $\mathcal{F}$ in degree zero and is trivial elsewhere. Then procedures 1 and 2 produce the same wrong answer: global sections of $\mathcal{F}$ in degree zero, always. But we can find $X$ and $\mathcal{F}$ so that a single sheaf has interesting cohomology by the third procedure: take $X=\Bbb P^1_\Bbb C$ and $\mathcal{F}=\mathcal{O}(-2)$, for instance.

The reason we want to take hypercohomology (your third way) is that it makes de Rham cohomology useful: for a smooth complex variety $X$, we get that $\Bbb H(X,\Omega_X^\bullet) \cong H^i(X^{an},\Bbb C)$, where the first term is the de Rham cohomology on a scheme and the second is the de Rham cohomology of the complex manifold $X^{an}$. This MO post lays out more reasons.

Title: Cohomology of a complex of sheaf


问:
Solving $x^2 + py^2 = 4z^3$ with a fixed $z$
Given an odd prime $p$ and a positive integer $z$, I'd like to find all positive integers $x$, $y$ such that
$$
x^2 + p y^2 = 4 z^3.
$$

Because the left is positive definite there is only a finite number of solutions.

I could solve by incrementally guessing values of $y$ and checking if $4z^3-py^2$ is a square, until $py^2> 4 z^3$. But this is slow, and many times the answer just turns out to be there are no solutions, or wasted time after finding the first solution which turns out to be unique.

I've learned the left hand side is called a binary quadratic form. Which has a class size $h(-4p)$. If it  somehow helps, assume I know this value.

Is there an easy way to determine that, if a solution exists, it is unique?

Is there an easy way to find all the solutions?

答:
I recommend Lehman as an introduction, simultaneous to binary quadratic forms and quadratic number fields.

I'm going to ignore the $4.$  It is there for a reason, but things begin more simply without it.

$$\left( x^2 + p y^2 \right)^3  = \left( x^3 -3 px y^2 \right)^2 + p  \left( 3x^2 y - p y^3 \right)^2 $$

When $p \equiv 3 \pmod 8$ ( and $p>3$) we have
class number  divisible by 3. Sometimes we can use the easiest description (available if we can solve $a^3 - t^2 = p$):

$$\left( ax^2 +2txy + a^2 y^2 \right)^3  = \left(tx^3 + 3a^2x^2 y + 3atxy^2 +(t^2 - p) y^3  \right)^2 +  p  \left( x^3 - 3axy^2 - 2t y^3 \right)^2 $$

The smallest primes for which this works are $11, 19, 67, 83, 107.$ When there is no such expression, for example primes $43, 59,$ there is a less formulaic way to write it out, but there are still infinitely many cubes where your $z$ has no expression as $x^2 + p y^2$

Example $p=11, a=3,t=4.$
$$\left( 3x^2 +8xy + 9 y^2 \right)^3  = \left(4x^3 + 27x^2 y + 36xy^2 +5 y^3  \right)^2 +  11  \left( x^3 - 9xy^2 - 8 y^3 \right)^2 $$

Example $p=19, a=7,t=18.$
$$\left( 7x^2 +36xy + 49 y^2 \right)^3  = \left(18x^3 + 147x^2 y + 378xy^2 +305 y^3  \right)^2 +  19  \left( x^3 - 21xy^2 - 36 y^3 \right)^2 $$


答:
For many problems of this kind, once you prove the base case you only need to compare the increment from $n$ to $n+1$ on both sides,

$$3\sqrt{n+1} \ge 2((n+1)\sqrt{n+1}-n\sqrt n) \\
\iff 2n\sqrt{n} \ge (2n-1)\sqrt{n+1}\\
\iff 4n^3 \ge (2n-1)^2(n+1)=4n^3-3n+1\\
\iff 3n\ge 1$$
which is true.

One added benefit is this can be easily converted to a non-induction, telescopic proof.


问:
Evaluating $\sum_{n=1}^{\infty}\frac{(n-1)!}{\prod_{r=1}^{n}(x+r)}$ for $x\in\mathbb{R}^{+}$
I am trying to evaluate the following sum. Here $x\in\mathbb{R^{+}}$.

$$f(x)=\sum_{n=1}^{\infty}\frac{(n-1)!}{\prod_{r=1}^{n}(x+r)}$$

At present I do not know how to proceed. Any hints are appreciated. Thanks.

Body:
$$\prod _{r=1}^n (r+x)=\frac{\Gamma (n+x+1)}{\Gamma (x+1)}$$
$$f(x)=\sum_{n=1}^{\infty}\frac{\Gamma(n)}{\frac{\Gamma (n+x+1)}{\Gamma (x+1)}}=\sum_{n=1}^{\infty}\frac{\Gamma (n) \Gamma (x+1)}{\Gamma (n+x+1)}=\frac{\Gamma (x)}{\Gamma (x+1)}=\frac{1}{x}$$


问:
Knowing that $n= 3598057$ is a product of two different prime numbers and that 20779 a square root of $1$ mod $n$, find prime factorization of $n$.

Knowing that $n= 3598057$ is a product of two different prime numbers and that 20779 is a square root of $1$ mod $n$, find prime factorization of $n$.

What I have done so far:

$n = p \cdot q$

$x^2 \equiv 1\pmod{n}$

$x^2 -1 \equiv 0\pmod{n}$

$(x-1)(x+1) \equiv 0\pmod{n}$

$x-1 = 20779 \lor x + 1=20779$

I have also noticed that:

$(x-1)(x+1) \equiv 0\pmod{p \cdot q}$

$(x-1)(x+1) \equiv 0\pmod{p} \land (x-1)(x+1) \equiv 0\pmod{q}$

But I have no idea what to do next. Any hints?

答: From $(x-1)(x+1)\equiv 0 \pmod n$ you cannot conclude that $x-1 = 20779 \lor x + 1=20779$ when $n$ is not prime.  You can conclude that $x-1 \equiv 0 \pmod p$ and $x+1 \equiv 0 \pmod q$ (or the other way-we could swap $p,q$) when you know that $n$ has only two prime factors unless $pq$ divides one of $x+1, x-1$.  So factor $20778$ and $20780$ looking for factors that will multiply to make $3598057$.  The other factors are small enough to find by hand. </p>


问:
semi positiveness of characteristic function
Prove that a characteristic function $\varphi$ is semi positiveness in the following form :

Taken $n$ real numbers $t_{1},\cdots,t_{n}$ and $n$ complex ones $z_1,\cdots z_n$ we have

$$\sum\limits_{h,k=1}^n \varphi(t_{h} - t_{k})z_{h}\overline{z_k}$$

I only managed to rewrite the sum as follows :

$$\sum\limits_{h=k}\varphi(0)z_h\overline{z_h} + 2 \sum\limits_{h>k} \text{Re}(\varphi(t_h-t_k)z_h\overline{z_k}) \overset{?}{\geq 0}$$

But how to proceed from here ?

答:
Just use that $\lvert a\rvert^2 = a\overline{a} $

$$\sum\limits_{h,k = 1}^{n} y_k \overline{y_h}\varphi(t_k-t_h) = \sum\limits_{h,k=1}^n y_k\overline{y_h}\int e^{ix(t_k-t_h)} d\mu =$$

$$ = \int \sum\limits_{h,k=1}^n y_ke^{ixt_k}\overline{y_h e^{ix t_h}} d\mu = \int \lvert \sum\limits_{h,k=1}^ny_ke^{ixt_k}\rvert^2 d\mu \geq 0$$


问:
Type of singularity of $f(z) = \frac{1}{1-e^z}$ in $z_0 = 0$ with Laurent Series
I need to  classify the pole $z_0=0$  of the following function-
$$f(z) = \frac{1}{1-e^z}$$
The only method that I know is to expand the function in its Laurent Series. I cannot use Bernoulli Numbers since I have not viewed in class.
I began by expanding $e^z$ as its Taylor series, but I got a problem when I was constructing de funcion $f(z)$.

I got stucked in the following

$$ \frac{1}{1-e^z} = \frac{1}{1-\sum_{n=0}^{\infty} \frac{z^n}{n!}} $$

Also I tried to substitute de Taylor series of $e^z$ in the expansion os $Log(1+z)$ and the geometric series; nonetheless, I got a double series in each case.

This topic is very new form me. Any help is aprecciated. Thank you very much.

答: It is a simple pole. An easy way to check this is to see that $\lim_{z\to 0} \frac{z}{1-e^z}=\lim_{z\to 0}\frac{1}{-e^z}=-1$(which can be done by expanding the Taylor series of $e^z$ or you can just divide the numerator and denominator by z and use that $1-e^z$ is holomorphic at 0.)


问:
Symbol of self-adjoint pseudodifferential operator
It seems that the following result should hold, but I can't find it explicitly anywhere. If $A=A^*$ is a properly supported pseudodifferential operator, does this imply that $\sigma_A(x,\xi)=\sigma_{A^*}(x,\xi)$? More specifically, is the symbol of $A$ real?

I know that the asymptotic expansion of $\sigma_{A^*}$ is given from the asymptotic expansion of $\sigma_A$ as
$$
\sum_{\alpha}\partial^{\alpha}D^{\alpha}\overline{\sigma_{A}(x,\xi)}\cdot \frac{1}{\alpha}
$$

答: This is true. As noted in the comments, it is (of course) true that if two operators are equal, then their principal symbols are equal. Otherwise, it would be ill-defined. Then, you clarified that you wanted to know if the principal symbol is necessarily real, and this is also true. The principal symbol of the adjoint is the complex conjugate of the principal symbol, i.e. $\sigma_{A^*}=\overline{\sigma_A}.$ Since $A^*=A,$ it follows that $\sigma_A=\sigma_{A^*},$ i.e. $\sigma_A=\overline{\sigma_A}.$ Thus, $$\text{Im }\sigma_A=\frac{1}{2i}\left(\sigma_A-\overline{\sigma_A}\right)=0,$$ implying that $\sigma_A$ is real. The full symbol need not be real, as the principal symbol is only defined up to a symbol of a lower order (i.e. if $A\in OPS^m,$ then $\sigma_A\in S^m/S^{m-1}$).


答:
Here is a careful derivation that outlines the reasoning process in induction proofs.

First, it's important to distinguish what you know to be true from what you want to be true. If the proposition holds for $n$, you know that

$$3\sum_{k=1}^n\sqrt k\ge 2n\sqrt n + 1\tag1$$
To show the proposition holds for $n+1$ you want to show
$$3\sum_{k=1}^{n+1}\sqrt k\ge 2(n+1)\sqrt {n+1} + 1\tag2$$
You are allowed to use (1) to get to (2). In anticipation of using (1), you write
$$3\sum_{k=1}^{n+1}\sqrt k = 3\sum_{k=1}^n\sqrt k + 3\sqrt{n+1}
$$
and then plug in (1):
$$3\sum_{k=1}^{n+1}\sqrt k \ge 2n\sqrt n + 1 + 3\sqrt{n+1}
$$
So what you want to be true is:
$$2n\sqrt n + 1 + 3\sqrt{n+1}\ge 2(n+1)\sqrt {n+1} + 1\tag3
$$
To verify this, the first step is to do algebra to simplify (3) into an equivalent but simpler statement. So get rid of the $2\sqrt{n+1} + 1$ that's common to both sides. This leaves the equivalent assertion, which still requires proof:
$$ 2n\sqrt n +\sqrt{n+1}\ge 2n\sqrt{n+1}\tag4$$
The other answers offer several ways to arrive at (4). Another approach is to rearrange into yet another equivalent statement (still requires proof), with $\sqrt n$ on one side and $\sqrt{n+1}$ on the other:
$$
2n\sqrt n\ge (2n-1)\sqrt{n+1}\tag5
$$
Now square both sides:
$$4n^3\ge (2n-1)^2(n+1)\tag6$$
Is (6) true? If so, then run the argument backwards from (6) to (3). If the reverse chain of implications is still valid, you are done. (Note that $n$ is at least $1$, so the step (5) $\to$ (6) is indeed reversible.)


问:
Finding the powers for the variables of a polynomial, for it to be divisible by another polynomial
For what natural numbers k and n can the polynomial $x^k+x^n$ be divided by $x^2-x+1$

Please explain the solution, or tell me what is it that I should learn/look up, in order to know how to solve such problems.

答:
A separable polynomial (one with distinct roots) is a divisor of another polynomial if and only if every root of the first is also a root of the second. What are the roots of $x^2-x+1$? Well, this polynomial is part of the factorization of $x^3+1$, which is itself part of the factorization of $x^6-1$. Thus, the roots of $x^2-x+1$ are precisely the $6$th roots of unity.

(So, it would help to become familiar with cyclic groups, roots of unity, and cyclotomic polynomials.)

And $x^n+x^k=x^k(x^{n-k}+1)$ so you need to check of primitive $6$th roots of unity satisfy the equation $x^{n-k}=-1$. Well, what kind of powers can you take primitive $6$th roots to to get $-1$?


问:
$C^k_b$ with sup-norm not complete
Let $C^n_b=\{ f : I\rightarrow \mathbb{C}: f~n\textrm{-times continuously differentiable and } \|f\|_{n,\infty} < \infty\}$, where $\emptyset\neq I\subseteq\mathbb{R}$ denotes an open interval and $\|f\|_{n,\infty}=\Sigma_{k=0}^n\|f^{(k)}\|_\infty$ be the function space of our concern. Equipped with the supremum norm this should not be complete.

Does anyone have a (classic) example of a Cauchy sequence for proving the noncompletness?

答:
The idea is to consider sequence of continuous functions that converge to the function of class $C([a,b])\setminus C^1([a,b])$. Then one needs to integrate this functions $n$ times and prove that the result is Cauchy sequence without limits in $C^n([a,b])$. It will not have the limit because $n$-th derivatives will converge to non-$C^1([a,b])$ function.

For simplicity wee consider case $[a,b]=[-1,1]$. Define functions
$$
y_k(t)=\int_0^t\arctan(ks)ds
$$
One can check that
$$
\lim\limits_{k\to\infty}y_k(t)=\frac{\pi}{2}|t|\notin C^1([-1,1])
$$
One can check that $n$-th antideriveative of $y_k$ is
$$
x_k(t)=\int_0^t\arctan(ks)\frac{(t-s)^n}{n!}ds
$$
You can show that
$$
|x_k^{(r)}(t)-x_l^{(r)}(t)|\leq M\int_0^{|t|}|\arctan (ks)-\arctan(ls)|ds\tag{1}
$$
for any $r=0,\ldots,n$ $k,l\in \mathbb{N}$ and $t\in[-1,1]$. Here
$$
M=\max\limits_{r=0,\ldots,n}\max\limits_{(s,t)\in[-1,1]^2}\frac{d}{dt}\frac{(t-s)^m}{m!}
$$
Using $(1)$ you can show that $\{x_n:n\in\mathbb{N}\}\subset C^n([-1,1])$ is a Cauchy sequence. It is not convergent by arguments given above.

Title: $C^k_b$ with sup-norm not complete


答:
Hint: Let $L_n(x)$ be the remainder of $x^n$ on division by the polynomial $x^2 - x + 1$.

The sequence starts with
$$ \eqalign{L_0(x) &= 1\cr
           L_1(x) &= x\cr
           L_2(x) &= x-1\cr
           L_3(x) &= -1\cr} $$
Since $L_3(x) = -L_0(x)$, we will have $L_{n+3}(x) = -L_{n}(x)$ for all $n$.

Title: Finding the powers for the variables of a polynomial, for it to be divisible by another polynomial

Body:
For what natural numbers k and n can the polynomial $x^k+x^n$ be divided by $x^2-x+1$

Please explain the solution, or tell me what is it that I should learn/look up, in order to know how to solve such problems.


问:
Finding a polynomial with two remainders and dividends
I need to find an equation, given the information that when divided by (x-1) it has a remainder of 2, and when divided by (x+2) the remainder is -19. I've looked at many answers to this type of question but none of them have really explained what to do. Please help!</p>

答:
Let $n$ and $m$ be two natural numbers which respectively represent the dividends of the polynomial we're looking for, we can write two expressions for that polynomial:

$$p_1= n(x-1) + 2$$
$$p_2= m(x+2) - 19$$

Simplifying further yields:

$$p_1= nx + (2-n)$$
$$p_2= mx+(2m-19)$$

And since, by our definition, $p_1=p_2$:

$$
\left\{
\begin{array}{c}
n=m \\
2-n=2m-19
\end{array}
\right.
$$

The above is true, because 2 polynomials are equal to each other only if their corresponding coefficients are the same.

Solving the equation system above yields $n=m=7$ (2 natural numbers!), and therefore, the polynomial that satisfies the answer for this question would be:

$$p= p1 = 7(x-1)+2 = 7x-5$$


问:
Convergence of $\int_\mathbb{R^n} \frac1{x^p}dx$
1)$\int_\mathbb{R^n}$ $\frac{1}{x^p}dx$

For what values of $p$ would this converge? I mean should p be less than the dimension of $\mathbb{R^n}$ or what?

My second question is, if I have $\int_\mathbb{B(0,\alpha)} \frac1{|x|^{\beta}}dx$ s.t $\alpha>0$ and
$B(0,\alpha) \subset \mathbb{R^n}$ and I made a change of variable s.t $r=||x||_2:=|x|$, then can you put a slight hint for the integral structure pls?

$x=(x_1,x_2,....,x_n)$

My sol:
$r=|x|$ , then $dx=\frac{|x|^n}{x} dr$ but my Dr. continued to put $dx=|x|^{n-1}dr:=r^{n-1} dr$ , but how would $\frac{|x|^n}{x}$ give $|x|^{n-1}$?

答:
I think you mean $\int\limits_{\mathbb{R}^n}\frac{dx_1...dx_n}{|x|^p}$. You could try to make spherical change of variables (in both cases). Other ways - Kronrod-Federer theorem (private case - representation as integral over sphere) or distributional function $F_{|x|}(t)$ if you know what it is. After all: in the first case integral diverges for any $p$, in the second converges iff $p<n$.

Look, we have $x=(x_1,...,x_n) \in \mathbb{R}^n$. Let $r \in  \mathbb{R}$ and $\phi=(\phi_1,...,\phi_{n-1}) \in \mathbb{R}^{n-1}$. We have spherecal change $x=\Phi(r,\phi)$ given by: $$x_1=r\cos(\phi_{n-1})\cos(\phi_{n-2})\cdot...\cdot\cos(\phi_2)\cos(\phi_1),$$ $$x_2=r\cos(\phi_{n-1})\cos(\phi_{n-2})\cdot...\cdot\cos(\phi_2)\sin(\phi_1),$$
$$x_2=r\cos(\phi_{n-1})\cos(\phi_{n-2})\cdot...\cdot\sin(\phi_2),$$
$$.............$$
$$x_{n-1}=r\cos(\phi_{n-1})\sin(\phi_{n-2}),$$
$$x_{n}=r\sin(\phi_{n-1}).$$

$G=\{(r,\phi):~r\in(0,+\infty),~ \phi_1 \in (-\pi,\pi),~ \phi_i \in (-\pi/2,\pi/2)$ if $ i>1\}$. Then $\Phi(G)=\mathbb{R}^n \setminus \{x:x_1 \leq 0, x_2=0\}=Y$ and $\Phi:G \longrightarrow Y$ is diffeomorfism with $\det \Phi^{'}=r^{n-1}\cos^{n-2}(\phi_{n-1})\cdot...\cdot\cos^2(\phi_3)\cos(\phi_2)$.
Note that $|x|=r$.

Now we can rewrite your integral as follows: $$\int\limits_{\mathbb{R}^n}\frac{dx_1...dx_n}{|x|^p}=\int\limits_{G}\frac{\det \Phi^{'}}{r^p}drd\phi_{n-1}...d\phi_1=\int\limits_{G}\frac{r^{n-1}\cos^{n-2}(\phi_{n-1})\cdot...\cdot\cos^2(\phi_3)\cos(\phi_2)}{r^p}drd\phi_{n-1}...d\phi_1=$$ $$=\underbrace{\Bigg(\prod\limits_{i=2}^{n-1}\int\limits_{-\pi/2}^{\pi/2}\cos^{i-1}(\phi_i)d\phi_i\Bigg)}_{V_n \in (0,+\infty)}\cdot\int\limits_{0}^{+\infty}r^{n-1-p}dr.$$ The last integral diverges for any p.

In the second case we have $$\int\limits_{\mathbb{R}^n}\frac{dx_1...dx_n}{|x|^p}=V_n\int\limits_{0}^{\alpha}r^{n-1-p}dr,$$ which converges iff $p<n$.

Addition: $V_n$ is Hausdorf measure of $(n-1)$-dimentional shpere with radius = 1.


问:
Extension of a group homomorphism
Let $G$ and $K$ be (possibly non-Abelian) groups and let $\phi:G\rightarrow K$ be a homomorphism. Let $\bar{G}$ be a group containing $G$ as a subgroup.

Is it always possible to extend $\phi$ to a homomorphism $\bar{\phi}:\bar{G}\rightarrow K$ (such that $\bar{\phi}$ restricted to $G$ is $\phi$ itself)? If not what conditions are required for this to be true?

I have seen somewhere that for the Abelian case if $K$ is divisible this is possible but I'm mostly interested in the non-Abelian case.

Is it always possible to extend $\phi$ to a homomorphism $\bar{\phi}:\bar{G}\rightarrow \bar{K}$ where $\bar{K}$ is some group containing $K$ as a subgroup? If not what conditions are required for this to be true?

答: In finitely generated torsion free nilpotent groups you can extend some homomorphisms to the Malcev completion and consider the restriction to the group which wants the extension. The problem is that this group can be no invariant, so the image is, if no more information, the Malcev completion. Lemma 2.7 in Baumslag's "Lecture notes on nilpotent groups" can be an alternative in f.g. t-f nilpotent.</p>

Title: Extension of a group homomorphism


问:
The truth value of $\exists\, y\geq 2\; \forall\, x\leq -1 \quad y\neq \dfrac{2x-1}{x+1}$
Determine the truth value of the following statement
$$P:"\exists\, y\geq 2\; \forall\, x\leq -1 \quad y\neq \dfrac{2x-1}{x+1}"$$
negation of $P$ is :
$$\neg P:"\forall\, y\geq 2\; \exists\, x\leq -1 \quad y= \dfrac{2x-1}{x+1}"$$
in order to know the truth of the $P$ I will investigate Its negation  which is,
$$\neg P:"\forall\, y\geq 2\; \exists\, x\leq -1 \quad y= \dfrac{2x-1}{x+1}"$$
let $y\in[2,+\infty[$, and I will look if x exists on $]-\infty,-1[$ such that $y=\dfrac{2x-1}{x+1}$

let $f(x)=\dfrac{2x-1}{x+1}$ and I will use the variation of the function $f(x)= \dfrac{2x-1}{x+1}$.

$f$ is continuous and strictly decreasing on the interval $]-\infty,-1[$ and $y\in [2,+\infty[$ then by mean value theorem, We conclude that there exists at least one solution on $]-\infty,-1[$

My questions:

Is my method correct?

Is there other ways?

答:
That is not correct. Note that the equation $\frac{2x-1}{x+1}=2$ is equivalent to $2x-1=2x+2$, which has no solutions. Therefore, $\neg P$ is false, and so $P$ is true.

If the restriction of $y$ was that $y\in(2,\infty)$, then your method would be correct.


问:
Proving $a_n = \sum_{k=1}^{n}\frac1{k}\sin(\frac{k\pi}{n+1})$ is increasing
Is the sequence $a_n = \sum_{k=1}^{n}\frac1{k}\sin(\frac{k\pi}{n+1})$ increasing?

Note that it converges to the integral $\int_0^\pi\frac{\sin x}{x} \,dx$. I was plotting this sequence on Desmos and apparently it is monotone. But I'm having an hard time proving this.

It seems that it's difficult to manipulate the sum: I don't think I can employ functions here and study some derivative; I used the sine sum-to-product formula while trying to show that $a_n < a_{n+1}$, but it makes things more cumbersome. What could be a way to start?

EDIT: for the sake of my problem, it suffices to show that the integral is un upper bound though. I'm not really interested whether it is increasing or not.

As suggested by a user, this question was asked to complete the proof given in the answer to another question, which I posted here Bound the absolute value of the partial sums of $\sum \frac{\sin(nx)}{n}$. Basically, what I'm trying to show is that the sequence of the maxima of the peaks of the partial sums of the Fourier series for the sawtooth wave is increasing to the integral given above (related to the Gibbs's phenomenon).

答:
Let $S_n(x) = \sum_{k=1}^n \frac{\sin kx}{k}$ denote the partial sum. We have

$$S_n'(x) =  \sum_{k=1}^n \cos kx = \frac{\sin \frac{nx}{2}\cos \frac{(n+1)x}{2}}{\sin \frac{x}{2}}, $$

and in $[0,\pi]$ the local maxima and minima of $S_n(x)$ are at
$ \frac{2m+1}{n+1} \pi$ for $m = 0,1, \ldots ,\lfloor\frac{n-1}{2} \rfloor$ and $\frac{2m}{n}\pi$ for  $m = 1, \ldots ,\lfloor\frac{n-1}{2} \rfloor$, respectively.

Note that $S_{n+1}(x) - S_n(x) = \frac{\sin (n+1)x}{n+1}$ and, thus, $S_{n+1}\left(\frac{2m+1}{n+1}\pi \right) = S_n\left(\frac{2m+1}{n+1}\pi  \right) $.

Note that $2m \leqslant n-1 < n$ and

$$\frac{2m+1}{n+2} \pi< \frac{2m+1}{n+1} \pi < \frac{2m+2}{n+2} \pi,$$

that is, the point $\frac{2m+1}{n+1} \pi$ lies between the adjacent relative maximum and minimum of $S_{n+1}$ so that

$$S_{n+1}\left(\frac{2m+1}{n+2}\pi \right) > S_{n+1}\left(\frac{2m+1}{n+1}\pi  \right) =  S_n\left(\frac{2m+1}{n+1}\pi  \right)$$.

In particular, for $m = 0$ we have

$$S_{n+1}\left(\frac{\pi}{n+2} \right) > S_{n}\left(\frac{\pi}{n+1}  \right)$$

as was to be shown.


问:
Find the linear transformation of a matrix knowing 4 linear transformations
I'm stuck with an exercise where they give me 4 linear transformations $T:M_{2\times2}\to\Bbb R$ for 4 matrices 2x2 and then they ask me for the linear transformation of a fifth matrix.

$$ T
  \begin{pmatrix}
    1 &0\\
    0 & 0\\
  \end{pmatrix}
=3,
T\begin{pmatrix}
    0 &1\\
    1 & 0\\
  \end{pmatrix}
=-1,
T
  \begin{pmatrix}
    1 &0\\
   1 & 0\\
  \end{pmatrix}
=0,
T
  \begin{pmatrix}
    0 &0\\
    0 & 1\\
  \end{pmatrix}
=0,
T
  \begin{pmatrix}
    a &b\\
    c & d\\
  \end{pmatrix}
=?$$

and I have worked with this situation for $T:\Bbb R^n \to\Bbb R^m$ but never with matrices so I have the doubt that how is supposed to find 4 constants that will multiply de 4 matrices I know to compose the fifth matrix to finally can find the linear transformation of this last one. (I have tried adding up the 4 matrices as follows)

$$
  \begin{pmatrix}
    1 &0\\
    0 & 0\\
  \end{pmatrix}
+
\begin{pmatrix}
    0 &1\\
    1 & 0\\
  \end{pmatrix}
+
  \begin{pmatrix}
    1 &0\\
   1 & 0\\
  \end{pmatrix}
+
  \begin{pmatrix}
    0 &0\\
    0 & 1\\
  \end{pmatrix}=
  \begin{pmatrix}
    a &b\\
    c & d\\
  \end{pmatrix}
$$

\begin{cases}
2=a  &  \\
1=b & \\
2=c\\
1=d\\
\end{cases}

and then use the coefficients
$$
  2\begin{pmatrix}
    1 &0\\
    0 & 0\\
  \end{pmatrix}
+
1\begin{pmatrix}
    0 &1\\
    1 & 0\\
  \end{pmatrix}
+
  2\begin{pmatrix}
    1 &0\\
   1 & 0\\
  \end{pmatrix}
+
  1\begin{pmatrix}
    0 &0\\
    0 & 1\\
  \end{pmatrix}=
  \begin{pmatrix}
    a &b\\
    c & d\\
  \end{pmatrix}
$$

and I just want to know if I continue in this way or I have committed any mistakes, thanks for your time <3

Body:
One standard approach is to find numbers $x_ij$ with $1 \leq i,j \leq 4$ for which
$$
x_{11} \pmatrix{1&0\\0&0} + x_{12} \pmatrix{0&1\\1&0} + x_{13} \pmatrix{1&0\\1&0} + x_{14} \pmatrix{0&0\\0&1} = \pmatrix{1&0\\0&0}\\
x_{21} \pmatrix{1&0\\0&0} + x_{22} \pmatrix{0&1\\1&0} + x_{23} \pmatrix{1&0\\1&0} + x_{24} \pmatrix{0&0\\0&1} = \pmatrix{0&1\\0&0}\\
x_{31} \pmatrix{1&0\\0&0} + x_{32} \pmatrix{0&1\\1&0} + x_{33} \pmatrix{1&0\\1&0} + x_{34} \pmatrix{0&0\\0&1} = \pmatrix{0&0\\1&0}\\
x_{41} \pmatrix{1&0\\0&0} + x_{42} \pmatrix{0&1\\1&0} + x_{43} \pmatrix{1&0\\1&0} + x_{44} \pmatrix{0&0\\0&1} = \pmatrix{0&0\\0&1}.\\
$$
Once you have done this, we find that
$$
T\pmatrix{a&b\\c&d} \\
= a\cdot T\left[ x_{11} \pmatrix{1&0\\0&0} + x_{12} \pmatrix{0&1\\1&0} + x_{13} \pmatrix{1&0\\1&0} + x_{14} \pmatrix{0&0\\0&1}\right]\\
+ b\cdot T\left[x_{21} \pmatrix{1&0\\0&0} + x_{22} \pmatrix{0&1\\1&0} + x_{23} \pmatrix{1&0\\1&0} + x_{24} \pmatrix{0&0\\0&1} \right]\\
+ c \cdot T\left[x_{31} \pmatrix{1&0\\0&0} + x_{32} \pmatrix{0&1\\1&0} + x_{33} \pmatrix{1&0\\1&0} + x_{34} \pmatrix{0&0\\0&1}\right]\\
+ d \cdot T \left[x_{41} \pmatrix{1&0\\0&0} + x_{42} \pmatrix{0&1\\1&0} + x_{43} \pmatrix{1&0\\1&0} + x_{44} \pmatrix{0&0\\0&1} \right]
\\ =
a\cdot  \left[x_{11} T\pmatrix{1&0\\0&0} + x_{12} T\pmatrix{0&1\\1&0} + x_{13} T\pmatrix{1&0\\1&0} + x_{14} T\pmatrix{0&0\\0&1}\right]\\
+ b\cdot \left[x_{21} \pmatrix{1&0\\0&0} + x_{22} \pmatrix{0&1\\1&0} + x_{23} \pmatrix{1&0\\1&0} + x_{24} \pmatrix{0&0\\0&1}\right] \\
+ c \left[x_{31} T\pmatrix{1&0\\0&0} + x_{32} T\pmatrix{0&1\\1&0} + x_{33} T\pmatrix{1&0\\1&0} + x_{34} T\pmatrix{0&0\\0&1}\right]\\
+ d \cdot \left[x_{41} T\pmatrix{1&0\\0&0} + x_{42} T\pmatrix{0&1\\1&0} + x_{43} T\pmatrix{1&0\\1&0} + x_{44} T\pmatrix{0&0\\0&1} \right].
$$
In other words, if $X$ is the matrix whose entries are $x_{ij}$, then we have
$$
T \pmatrix{a&b\\c&d} = \pmatrix{3&-1&0&0} X \pmatrix{a\\b\\c\\d}.
$$
So, we need to find the matrix $X$. If we write the full system of equations as a matrix, then we can see that the system of equations can be written as
$$
\pmatrix{1&0&1&0\\0&1&0&0\\0&1&1&0\\0&0&0&1}
\pmatrix{x_{11}&x_{12}&x_{13}& x_{14}\\
x_{21} & x_{22} & x_{23} & x_{24}\\
x_{31} & x_{32} & x_{33} & x_{34}\\
x_{41} & x_{42} & x_{43} & x_{44}} =
\pmatrix{1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1}.
$$
In other words,
$$
X = \pmatrix{1&0&1&0\\0&1&0&0\\0&1&1&0\\0&0&0&1}^{-1}.
$$


问:
Lattice isomorphism between $(\{A \subset \mathbb{N} : A$ is finite$\}, \cup, \cap)$ and $(\mathbb{N},lcm,gcd)$.
My problem is to know if there exists a lattice isomorphism between $B = (\{A \subset \mathbb{N} : A$ is finite$\}, \cup, \cap)$ and $C = (\mathbb{N},lcm,gcd)$. I think that is posible because have the same cardinality ($\mathbb{N}$), both have a minimum ($\emptyset$ and 1 respectively) and neither have maximum.

I tried to define a function (candidate to isomorphism) from B to A starting sending $\emptyset$ to the 1 and the sets of one element to the primes "in order" ($\{1\}$ to 2, $\{2\}$ to 3, $\{3\}$ to 5, ...) and then I need to define that sends the $\{1,2\}$ to 2*3=6 and so on. But that function isn't a surjection.

I also know that is enough to find a bijective homomorphism.

答:
Wikipedia notes the algebraic and order-theoretic definitions of lattices are equivalent, so we may as well discuss if there is an order isomorphism from $(\mathcal{P}_{\mathrm{fin}}(\mathbb{N}),\subseteq)$ and $(\mathbb{N},\mid )$.

The answer is no, for $(\mathbb{N},\mid )$ has elements that cover one atom but no other atoms (squares of primes) and $(\mathcal{P}_{\mathrm{fin}}(\mathbb{N}),\subseteq)$ doesn't (it's not possible for a set to strictly contain one singleton set but no other singleton sets).

It's kind of like comparing $2^{\mathbb{N}}$ to $\mathbb{N}^{\mathbb{N}}$ with lexicographic orderings, except your two lattices are sublattices of these comprised of functions with finite support.


问:
limit of a sequence relative to a norm
Assume we have some vector space $ V$ and a norm $ ||•|| $.

If I hve a sequence $ v_n $  that converge to a vector relative to the norm, that is, $ lim ||v_{n}-v || =0 $.

How can I show that $lim ||v_{n}||=||v|| $?

I tried to use the squeeze theorem and wrote:

$ ||v_n||=||v_n -v+v|| \leq ||v_n-v|| + ||v || $

This gives me the the limit is smaller or equals to ||v||.

What can I do in order to show the other direction?

Title: limit of a sequence relative to a norm

Body: From $\|v_n\| \leq \|v_n-v\|+\|v\|$ one has $\|v_n\|-\|v\| \leq \|v_n-v\|$, and from $$ \qquad \|v\| = \|\!-\!v\| = \|(v_n-v)-v_n\| \leq \|v_n-v\|+\|v_n\|$$ one has $-\|v_n-v\| \leq \|v_n\|-\|v\|$. Thus $$-\|v_n-v\| \leq \|v_n\|-\|v\| \leq \|v_n-v\|$$ and then taking the limit as $n \to \infty$ we conclude that $\lim_n (\|v_n\|-\|v\|) = 0$.