问:
All smooth functions have Taylor expansion?
The theories are:

$f(x)$ has Taylor expansion equals the remainder of Taylor polynomial converge to $0$.

a smooth function such as $f(x) = e^{-1/x^2}$, $x>0$, $f(x) = 0$, $x \leqslant 0$ do not has Taylor expansion near $0$.

However, I think about these propositions and find:

smooth - exist Taylor polynomial - select Peano remainder = $o(x^n$) - the remainder must converge to $0$ - smooth function must have Taylor expansion near $0$

I will be appreciated if someone can point out the error in this infer.

答:
For every smooth function $f$, you can consider the Taylor series around $0$
$$\sum_{n \geq 0} \frac{f^{(n)}(0)}{n!}x^n$$

However, there are two possible obstructions in general to link directly this series to the function :

It is possible that the radius of convergence is $0$. Actually, one can prove that for every sequence $(a_n)$, there exists a smooth function $f$ satisfying $f^{(n)}(0)=a_n$. Therefore the Taylor series can be every series, so it can be a series which converges nowhere.

It is possible that the Taylor series converges everywhere, but is not equal to $f$ in a neighbourhood of $0$. The classical example is the one you mention with $e^{-1/x^2}$, you get a Taylor series which is the null series, and the function is non constant equal to $0$ in any neighbourhood of $0$.

Finally, you have the following condition for a function to be analytic over an interval $I$ : a smooth function $f : I \rightarrow \mathbb{R}$ is analytic over $I$ iff $\forall [a,b]\subset I$, there exists $M \in \mathbb{R}$ and $\alpha > 0$ such that $\forall n \in \mathbb{N}$, $$||f^{(n)}||_{\infty|[a,b]} \leq Mn!\alpha^n$$


问:
Minimization problem of $e^x$ and straight line.

Given the curve $C$: $y=e^{x}$ and a point $P$: $(x_0,y_0)$ with $y_0 > e^{x_{0}}$, a straight line $L$ of slope $k$ goes through the point $P$ and intersects $C$ at two points $A$ and $B$, try to find out: $(1).$  $k$ that minimizes the distance of $A$ and $B$, $(2).$ k that minimizes the area of triangle OAB ($O$ is the origin $(0,0)$).

Edit:

I've tried the Lagrange multiplier method and only solved part of the problem(the minimal triangle when P on yAxis, in this case, $k=y_0$). However, the steps are quite complicated (for high school students). I'd like to know if there is some elementary methods to approach this problem. Thanks!

Edit 2:

(add the image)

I took the following Steps:

as shown in the chart, points P,Q,A,B are colinear. then we got the constraits of variables a,b and q, which are: $g(a, b) = 0, g(a,q) = 0, g(b,q) = 0$.

the distance between A and B is given by $f(a,b) = sqrt((b-a)^2+(e^b-e^a)^2)$

the area of the triangle OAB is given by $f(a,b,q)=1/2 * q * (b-a)$

${\mathcal {L}}(a,b,\lambda) = f(a,b)+\lambda*g(a,b)$

solve equations: $\nabla _{a,b,\lambda }{\mathcal {L}}(a,b,\lambda )=0$

The last step was the part that is very complicated.

答:
This is not an answer, just an illustration. The picture below represents the contour lines for $f(a,b) = (a-b)^2+(e^a-e^b)^2$ and, in red, the constraint $\dfrac{e^b-y_0}{b-x_0} = \dfrac{y_0-e^a}{x_0-a}$. This was obtained for $(x_0,y_0) = (\frac 12, 2)$

As you can see, the minimum will be attained at a point where the contour line of $f$ is tangent to the constraint. This is in fact the thinking behind Lagrange multipliers. I believe that Lagrange multipliers is probably the best way to go but, even in this case, we need handle the numerical solution of one or more equations.

ContentLicense: CC BY-SA 4.0


问:
Solve the equation $\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$
$$\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$$

$$x=?$$

I solved this but the equation $ (2x + 1) (3x ^ 4-x ^ 3 + 2x ^ 2-2x + 1) = 0 $ is formed I answer $ x =- \frac {1} {2} $ I know there is, but I couldn't do the next expression. I need help with that, or someone will solve it in a better way. I'd be happy with that.

答:
Your way is fine with a solution for $x=-\frac12$, now we have that

$$3x ^ 4-x ^ 3 + 2x ^ 2-2x + 1=x^2(3x ^ 2-x  + 1)+(x-1)^2 > 0$$

since $3x ^ 2-x  + 1>0$.

To find complex roots we can refer to Quartic equation.


问:
Taylor Series of analytic function
Why can the radius of convergence of the Taylor series of an analytic function around $z_0$ be larger than the largest disk centered at $z_0$ of the region of analyticity?

In such situation, could in the region outside of the domain of analyticity the Taylor series not converge to the function?

Thanks in advance.

答:
For example, consider $f =\exp : \mathbb{D} \rightarrow \mathbb{C}$ the restriction of the exponential map to $\mathbb{D} = \lbrace z \in \mathbb{C}, |z|<1 \rbrace$. The Taylor series of $f$ at $z=0$ has an infinite radius of convergence, but the largest disk centered around $0$ in $\mathbb{D}$ has radius $1$.

This situation where the domain is restricted (so is not "well-chosen") is the only possibility where the radius of Taylor series can be strictly larger than the one of the largest disk. Otherwise you will have an equality.

Title: Taylor Series of analytic function


问:
Let $a,b>0$ be real numbers satisfying $a^2+b^2=2$. Prove that $a+b \leq 2$.
I've tried messing around with some AM-GM stuff but I can't think of anything else for this. Can I get a proof for Real numbers? (I had a proof for rationals but can't work for reals).</p>

答:
Polar coordinates.

Let $a=r\sin t;$ $y=r \cos t,$ where $r=√2.$

Maximize $a+b= r(\sin t +\cos t);$

$√2(\sin t +\cos t)=$

$√2(√2(\sin (π/4+t)) \le 2 \cdot 1.$


问:
$G$ is Abelian if and only if $[e,e]$ is the only commutator of $G$
Wikipedia claims that for a given group $G$ with the identity element $e$ the commutator $[e,e]=e$ is the only commutator if and only if $G$ is Abelian.

I know that for a given $N \trianglelefteq G$ the quotient $G/N$ is Abelian if and only if $[G,G] \le N$ and this shows that the commutator subgroup of $G$ is the minimal normal subgroup for which $G/N$ is Abelian.

But does this help to prove that "$G$ is Abelian if and only if $[e,e]$ is the only commutator of $G$"?

If yes then how?

Also does there exist any name for $[e,e]$?

答:
Wikipedia claims that the identity element $e$ is the only commutator iff $G$ is Abelian.

$[e,e]=e$ always holds, and yes it has a name: the identity element.

The claim is immediate from the fact that $[a,b]=e\iff ab=ba$, which is clear by the definition of commutator: $[a,b]=a^{-1}b^{-1}ab$.


问:
Is it always possible to partition data in such a way that Simpson's paradox is achieved?
$A,B,C,D $ are natural numbers such that:
$$\frac AB ~>~\frac CD,~A\le B,~C\le D$$

Is it always possible to find natural numbers $a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2$ that satisfy:
$$\text{1) }~~ a_1+a_2 = A,~b_1+b_2 = B,c_1+c_2 = C,d_1+d_2 = D~; $$
$$\text{2) }~~ \frac{a_1}{b_1} <~\frac{c_1}{d_1}~,\frac{a_2}{b_2} <\frac{c_2}{d_2}$$

Trivial cases that I exclude are if: $$A=B, A=0, C=0,B=1,D=1$$

Intuitively I expect the answer to be no. Because in the examples of Simpson's paradox percentages A/B and C/D differ slightly.

答:
I assume (as antkam suggests in the comments) that you require $a_1\leq b_1$, etc, so that all of these fractions are probabilities.

Then there are a couple of potential barriers:

If $C\leq 1$ then one of the $\frac{c_i}{d_i}$ must be $0$, and so can't be larger than $\frac{a_i}{b_i}$.

If $B-A\leq 1$ then one of the $\frac{a_i}{b_i}$ must be $1$, since otherwise we would have $a_1+a_2\leq(b_1-1)+(b_2-1)<A$, and so it can't be smaller than $\frac{c_i}{d_i}$.

In all other cases we have the valid solution
$$\frac{0}{1}<\frac{C-1}{D-1}\quad\text{and}\quad\frac{A}{B-1}<\frac11.$$


问:
In which sense the convolution product is an average? [convolution in Fourier transform]
This is probably a silly question. But I'm studying the convolution for Fourier transforms
$$ (f*g)(x)=\frac{1}{T} \int_0^T f(y)g(x-y)dy $$
and my professor said that "the product $ f(y)g(x-y) $ is a sort of average of the two functions". Since it's not the first time I hear a product is a kind of average, but I can't convince myself of it, could anyone clarify this a bit?
Any other idea on the meaning of convolution will be highly appreciated!

答: I'm the OP: I found a beautiful explanation. This is actually an average: see here</p>


答:
Just to give a different approach, note that any solution to $\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$ must lie in $(-1,0)\cup(1,\infty)$ because the left hand side is negative on $(-\infty,-1)\cup(0,1)$ whereas the right hand side, being a square root, is always non-negative, and the expression is undefined at $x=0$ and $x=\pm1$. Now if $x\lt0$, then $3x^4-x^3+2x^2-2x+1$ is a sum of positive terms (i.e., $x^3$ and $2x$ are negative, so $-x^3$ and $-2x$ are positive), while if $x\gt1$, then

$$3x^4-x^3+2x^2-2x+1\gt3x^4-x^4+2x^2-2x^2+1=2x^4+1\gt0$$

so $3x^4-x^3+2x^2-2x+1$ has no roots in the feasible solution set for $x$. (In fact, as user's answer shows, it has no roots in $(0,1)$ either.)

Title: Solve the equation $\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$

Body:
$$\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$$

$$x=?$$

I solved this but the equation $ (2x + 1) (3x ^ 4-x ^ 3 + 2x ^ 2-2x + 1) = 0 $ is formed I answer $ x =- \frac {1} {2} $ I know there is, but I couldn't do the next expression. I need help with that, or someone will solve it in a better way. I'd be happy with that.


问:
Limit at non continuous point or arctangent
For an integral I need to calculate, I need to calculate the following limit:

$$\lim_{t \rightarrow \pi^-} \arctan \left(\frac{\tan(\frac{t}{2})}{3}\right).$$
I have used the formula:

$$\tan(\frac{t}{2})= \frac{\sin(t)}{\cos(t)-1}$$
and then applied L'Hopital's rule to show that the limit inside of the arctangent is $\infty$ but, although I know $\tan(\frac{\pi}{2})$ is also $\infty$, I need to justify that the limit is $\frac{\pi}{2}$ rigorously.

答:
There is no undeterminate form here :
$$\lim_{t \rightarrow \pi^-} \tan \left( \frac{t}{2}\right) = +\infty$$

so $$\lim_{t \rightarrow \pi^-} \frac{1}{3}\tan \left( \frac{t}{2}\right) = +\infty$$

so $$\lim_{t \rightarrow \pi^-} \arctan\left(\frac{1}{3}\tan \left( \frac{t}{2}\right) \right) = \lim_{x \rightarrow +\infty} \arctan\left(x\right) = \frac{\pi}{2}.$$


问:
How to show $E[X\mid X = x] =x$?
I read that $E[X\mid X = x] =x$ but I don't get that when I try to prove it:
\begin{align}
E[X\mid X = x] &= \sum x P(X=x|X=x) \\
&= \sum x \frac{P(X=x,X=x)}{P(X=x)} \\
&= \sum x \frac{P(X=x)}{P(X=x)} \\
&= \sum x  \\
\end{align}
So what am I doing wrong?

答:
You indexed wrong:

$$\mathbb E(X|X=x) = \sum_k k \mathbb P(X=k|X=x) = x \mathbb P(X=x|X=x) = x $$

Title: How to show $E[X\mid X = x] =x$?


答:
$$\mathbb{E}[X|X=x]=\mathbb{E}[x]=x$$

When you say $X=x$ it means that $x$ is a fixed value, not anymore a rv

Title: How to show $E[X\mid X = x] =x$?

Body:
I read that $E[X\mid X = x] =x$ but I don't get that when I try to prove it:
\begin{align}
E[X\mid X = x] &= \sum x P(X=x|X=x) \\
&= \sum x \frac{P(X=x,X=x)}{P(X=x)} \\
&= \sum x \frac{P(X=x)}{P(X=x)} \\
&= \sum x  \\
\end{align}
So what am I doing wrong?


答:
You are using $x$ for two different purposes: as a free variable and as a bound variable (index). The more correct solution (at least for a discrete random variable), using $x$ as a free variable and another variable ($y$) as a bound variable:

$$
\begin{align}
E[X|X = x] &= \sum_y y P(X=y|X=x) \\
&= \sum_{y\ne x} y \frac{P(X=y,X=x)}{P(X=x)} + x \frac{P(X=x,X=x)}{P(X=x)}\text{ (split out }y=x\text{ from }y\ne x\text{)} \\
&= x\frac{P(X=x)}{P(X=x)}\text{ (all terms in the sum above are zero)} \\
&= x  \\
\end{align}
$$

Title: How to show $E[X\mid X = x] =x$?


问:
Generate samples of sinusoid in matlab
I have to generate 100 samples of sinusoid in matlab with frequency 0.2Hz and sampling rate 2Hz. Until now i have pass into matlab the values:

```<code>>> t=[0:99];
>> f=0.2;
>> fs=2;
</code>```

The sin-wave function is this:

$y(t)=A*sin(2pft+φ)$

Does the Amplitude has some connection with the sampling rate? For example Amplitude values start from 0 and with step of 0.2 did i need to take 100 samples? And how i could find φ in order to plot this signal?

答:
http://en.wikipedia.org/wiki/Sampling_rate
Amplitude is generally independent of sampling rate,about  phase shift you can omit it now,because  actually finding this parameters or  method called frequency estimation are many,you can search it on internet</p>

```<code>>> f=0.2;
>> fs=2;
>> A=100;%suppose that amplitude is 100
>> q=20;
>> t=0:1/fs:49;
>> x=A*sin(2*pi*f*t+q);
total length is 101,1 more then you want
</code>```

ContentLicense: CC BY-SA 3.0

Title: Generate samples of sinusoid in matlab

Body:
http://en.wikipedia.org/wiki/Sampling_rate
Amplitude is generally independent of sampling rate,about  phase shift you can omit it now,because  actually finding this parameters or  method called frequency estimation are many,you can search it on internet

>> f=0.2;
>> fs=2;
>> A=100;%suppose that amplitude is 100
>> q=20;
>> t=0:1/fs:49;
>> x=A*sin(2*pi*f*t+q);
total length is 101,1 more then you want


问:
$x^3-6x^2+2x-2=0$ solve for x
I solved it by approximation but its very long way to solve, so i was wondering if there is a quicker way to solve this kind of questions.</p>

答:
Don't you prefer the nice
$$x=2 \left(1+\sqrt{\frac{10}{3}} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{21
   }{10}\sqrt{\frac{3}{10}}\right)\right)\right)$$ that you could easily obtain using the steps given in the Wikipeadia page.

Title: $x^3-6x^2+2x-2=0$ solve for x


问:
Calculus question with oblique asymptote involved
This question is let $$f(x) = \frac{ax^2+bx-c}{x-4}.$$ Determine the values of $a,b,$ and $c$ so that $f(x)$ has an oblique asymptote of $y=3x+1$ and has local minimum at $x=-3$.

I know this requires two equations. The first one requires taking derivative of $f(x)$ and subbing local minimum. But how do I create second equation involving the oblique asymptote.

答:
Be aware that, given a $f(x)$ :

one thing is to find its "asymptotic behaviour" as $x \to \infty$, which means to find a $g(x)$  such that
$$
\mathop {\lim }\limits_{x \to \infty } {{f(x)} \over {g(x)}} = 1
$$
and we say that $f$ and $g$ are asymptotically equivalent;

another thing is to find the function $g(x)$ which is asymptotically equal to $f(x)$, i.e. such that
$$
\mathop {\lim }\limits_{x \to \infty } \left( {f(x) - g(x)} \right) = 0
$$

Now,  when we talk to find a horizontal or oblique asymptote to a curve we mean to find a line
which is asyptotic to the curve in the second meaning, i.e. that the distance goes to $0$.

In the first way (J. Wang's answer) you can only determine the slope of the line, but not the intercept because
$$
\mathop {\lim }\limits_{x \to \infty } {{f(x)} \over {mx}}
 = 0\quad  \Rightarrow \quad \mathop {\lim }\limits_{x \to \infty } {{f(x)} \over {mx + q}} = 0
$$

Therefore in your case we have
$$
\eqalign{
  & f(x) = {{ax^2  + bx - c} \over {x - 4}} =   \cr
  &  = {{a\left( {x - 4} \right)^2  + b\left( {x - 4} \right) + 8ax - 16a + 4b - c} \over {x - 4}} =   \cr
  &  = {{a\left( {x - 4} \right)^2  + b\left( {x - 4} \right) + 8a\left( {x - 4} \right) + 16a + 4b - c}
 \over {x - 4}} =   \cr
  &  = a\left( {x - 4} \right) + b + 8a + {{16a + 4b - c} \over {x - 4}} \cr}
$$

So for the asymptote we shall have
$$
ax + b + 4a = 3x + 1\quad  \Rightarrow \quad a = 3,\quad b =  - 11
$$
as already indicated by A.Chin.

Then the function becomes
$$
f(x) = {{3x^2  - 11x - c} \over {x - 4}} = 3x + 1 - {{c - 4} \over {x - 4}}
$$

It is clear that the function has the required oblique asymptote, as well as a vertical asymptote at $x=4$, where we have:

$$
\mathop {\lim }\limits_{x \to 4 + } f(x) = - {\rm sgn}(c - 4) \cdot \infty \quad
 \mathop {\lim }\limits_{x \to 4 - } f(x) =  {\rm sgn}(c - 4) \cdot \infty
$$
This means that for $c < 4$ you will  have a minimum at $4 < x$ and a maximum at $x < 4$, while

for $4 < c$ you do not have any local extremum.

You can easily check this analytically.

Therefore a minimum at $x=-3$ is unattainable.


问:
Double integrals in polar coordinates with unexpected result
I have been working on an exercise where I need to convert to polar coordinates :

$$\iint_R (x+y)\,dA\,,$$

where $$R=\left\{(x,y)\mid 1\le x^2+y^2\le 4\,,\, x\le 0\right\}$$

To me, it makes perfect sense that $ (x + y)\,dx\,dy $ became : $(r\cos(\theta) + r\sin(\theta))r\,dr\,d\theta$

What I don't understand is that, I was expecting to work between $r = -1$ and $r = -2$ ; $\theta = - \pi/2$ and $\theta = \pi/2$

Instead, the solution puts $r = 1$ and $r = 2$ ;
$\theta = \pi/2$ and $\theta = 3\pi/2$

Could you tell me why ? (I am trying to grasp this new concept)

答: Keep in mind that $r = \sqrt{x^2+y^2}$ cannot be negative... It represents the distance to the origin. The condition $1\leq x^2+y^2 \leq 4$ puts our domain between the circles with radius 1 and 2, and so $r \in[1,2]$. The condition $x \leq 0$ puts our domain in the second and third quadrants, for which $\theta \in [\frac{\pi}{2}, \frac{3\pi}{2}]$


问:
Integration of $\int_0^1 \int_0^1 \int_0^1 \frac{\ln(1-xyz)}{xyz} dx dy dz = - \frac{\pi^4}{90}$
Using wolfram alpha,

$$-\int_0^1 \int_0^1 \int_0^1 \frac{\ln(1-xyz)}{xyz} dx dy dz = \frac{\pi^4}{90}$$

How do you solve this integral, and why is it equal to $\zeta(4)$? I can see that

$$-\frac{\ln(1-z)}{z} = 1 + \frac{z}{2} + \frac{z^2}{3} + \ldots$$

答:
If $f$ is a continuous function on $[0,1]$ then $\displaystyle \int_0^1\int_0^1 f(xy)dxdy=-\int_0^1 f(u)\ln u du$

(proof: integration by parts)

Application:

\begin{align}I&=\int_0^1 \int_0^1 \int_0^1 \frac{\ln(1-xyz)}{xyz} dx dy dz\\
&\overset{f(x,y)=\frac{\ln(1-xyz)}{xyz}}=-\int_0^1 \left(\int_0^1 \frac{\ln u\ln(1-uz)}{uz}du\right)dz\\
&=-\frac{1}{2}\int_0^1 \int_0^1\frac{\ln(uz)\ln(1-uz)}{uz}dudz\\
&\overset{f(u,z)=\frac{\ln(uz)\ln(1-uz)}{uz}}=\frac{1}{2}\int_0^1\frac{\ln^2 v\ln(1-v)}{v}dv\\
&\overset{\text{IBP}}=\frac{1}{6}\Big[\ln^3 v\ln(1-v)\Big]_0^1+\frac{1}{6}\int_0^1 \frac{\ln^3 v}{1-v}dv\\
&=\frac{1}{6}\int_0^1 \frac{\ln^3 v}{1-v}dv\\
&=\frac{1}{6}\times -6\zeta(4)\\
&=-\zeta(4)\\
&=\boxed{-\frac{\pi^4}{90}}
\end{align}

NB:

I assume that: $\displaystyle \int_0^1 \dfrac{\ln^3 v}{1-v}dv=-6\zeta(4)$ and $\displaystyle \zeta(4)=\dfrac{\pi^4}{90}$


问:
prove $(V1⊕...⊕Vk)∩(W1⊕...⊕Wk)=(V1∩W1)⊕...⊕(Vk∩Wk)$
I was thinking an hour over this question and I really didn't make any progress.
Though to ask here and maybe get a hint ;) .

Question : if $V=U_1⊕...⊕U_k$ and for each $i=1,...,k$ given $V_i,W_i ⊆ U_i$.

Prove :
$(V_1⊕...⊕V_k)∩(W_1⊕...⊕W_k)=(V_1∩W_1)⊕...⊕(V_k∩W_k)$

I though to make this proof as in Discreme math when you assume $v∈$ one side and prove $v∈$ the second side so you prove $⊆$ from both sides and it yelds $\rightarrow$ $=$ symbol but I failed to do so.

I got stuck in the middle of the way in the both side of the proof.

any hints ideas please ?
Thank you and have a nice day!

答:
I think I got the answer but I'm not sure.
First let's notice that because $Vi,Wi⊆Ui$.

Intersection between this two can only occur it their index is the same (because $V_1$ can't intersect with $W_2$ because $U1⊕...⊕Uk$, $\rightarrow$ $U_1∩...∩U_i={0}$.

Lets assume $v∈(V_1⊕...⊕V_k)∩(W_1⊕...⊕W_k)$

then $v∈(V_1⊕...⊕V_k)$ And $v∈(W_1⊕...⊕W_k)$
and because intersection can only occur between the same index $\Rightarrow$ $v∈(V_1∩W1)⊕...⊕(V_k∩W_k)$.

Second side is pretty much the same , is it ok to prove like that ?


答:
Perhaps this drawing can help you in understanding the solution

To integrate the purlple area you have to chose a radius in

$$1<r<2$$

and a corresponding angle in

$$\frac{\pi}{2}<\theta<\frac{\pi}{2}+\pi=\frac{3}{2}\pi$$

Thus rotating this positive radius in the angle's interval you get exactly the desired area to integrate

Title: Double integrals in polar coordinates with unexpected result

Body:
I have been working on an exercise where I need to convert to polar coordinates :

$$\iint_R (x+y)\,dA\,,$$

where $$R=\left\{(x,y)\mid 1\le x^2+y^2\le 4\,,\, x\le 0\right\}$$

To me, it makes perfect sense that $ (x + y)\,dx\,dy $ became : $(r\cos(\theta) + r\sin(\theta))r\,dr\,d\theta$

What I don't understand is that, I was expecting to work between $r = -1$ and $r = -2$ ; $\theta = - \pi/2$ and $\theta = \pi/2$

Instead, the solution puts $r = 1$ and $r = 2$ ;
$\theta = \pi/2$ and $\theta = 3\pi/2$

Could you tell me why ? (I am trying to grasp this new concept)


问:
Does it exist an inverse operation for the vertical lift of a vector field on $M$ to $TM$?
Let $M$ be a differentiable manifold and $TM$ its tangent bundle. Let $X$ be a vector field on $M$, locally given by
$$ X=X^i \frac{\partial  } {\partial q^i}.$$
Then its vertical lift is the vector field on $TM$ given by
$$X^v= X^i \frac{\partial  } {\partial \dot{q}^i},$$
where $q$ are the coordinates on $U\subset M$, and $(q,\dot{q})$ are the induced coordinates on $TU\subset TM$. I would like to know if there exists a certain operation over vertical vector fields, let me denote it $X^\star$, such that

$$(X^v)^\star=X.$$

答:
Notice that the vertical lift produces vector fields which are constant on each fiber of $VTM$. As such, there is a bit of freedom in choosing a left inverse; here's one way of doing so.

Let $\xi:M\to TM$ be a vector field, and let $\Xi\subseteq TM$ be its image. We can restrict the vertical tangent bundle $VTM$ to a bundle over $\Xi$, given by $VTM|_{\Xi}$. The vertical lift then gives a bundle isomorphism $\operatorname{vl}_\xi:TM\to VTM|_{\Xi}$. A left inverse for the vertical lift of vector fields, $\varphi:\Gamma(VTM)\to\Gamma(TM)$ can then written as the restriction to $VTM|_{\Xi}$ followed by applying the identification between $VTM|_{\Xi}$ and $TM$. That is, for $p\in M$, we have
$$
(\varphi(X))(p)=\operatorname{vl}_{\xi}^{-1}(X(\xi(p))
$$
Since $\operatorname{vl}_\xi(X(p))=X^v(\xi(p))$ (since both are vertical lifts of $X(p)$) we have $\varphi(X^v)=X$.


问:
What is a sequence of partition for integration?
I'm currently reading a chapter on integration in a real analysis textbook and there is a corollary of the cauchy criterion of integration. Now they define the cauchy criterion for integration as: there exist a partition where the difference of the upper and lower sum is less than epsilon.

Now for this corollary, it is stated that there exist a sequence of partitions Pn where the limit as n approaches infinity difference between the upper sum and lower sum is 0.

So my question is what exactly is a sequence of partitions ?

答:
One example of a sequence of partitions $(P_k)_{k \geq 0}$ for the interval $[0,1]$ is the following.

$$P_0 = \{[0,1]\}$$

$$P_1 = \{[0,1/2), [1/2,1]\}$$

$$P_2 = \{[0,1/4), [1/4,1/2), [1/2,3/4), [3/4,1]\}$$

$$\vdots$$

$$P_k = \{[i/2^k, i+1/2^k), [(2^k-1)/2^k,1] : 0 \leq i \leq 2^k-2\}$$

$$\vdots$$

and so each $P_k$ is a partition of $[0,1]$ and as $k$ increases, we (in this case) get a finer and finer partition.

Hopefully that helps.


问:
Is it possible to improve on the bound $D(q^k) < \varphi(q^k)$ if $k > 1$?
The problem is as is in the title:

Is it possible to improve on the bound $$D(q^k) < \varphi(q^k)$$ if $k > 1$?

Here, $q$ is a prime number and $k$ is a positive integer.  The (deficiency) function $D(x)$ is defined as follows:
$$D(x) = 2x - \sigma(x)$$
where $\sigma(x)$ is the classical sum of divisors of $x$.

Of course, $\varphi(x)$ is just the Euler totient function of $x$.  Finally, let $I(x)=\sigma(x)/x$ be the abundancy index of $x$.

MY PROOF FOR THE BOUND

Since in general we just have $k \geq 1$, we have
$$\frac{q+1}{q} = I(q) \leq I(q^k)$$
from which we get
$$\frac{D(q^k)}{q^k} = 2 - I(q^k) \leq 2 - \frac{q+1}{q} = \frac{q-1}{q} = \frac{\varphi(q^k)}{q^k},$$
whereby we finally obtain
$$D(q^k) \leq \varphi(q^k).$$

So if $k > 1$, then I just get
$$I(q) < I(q^k),$$
right?  Proceeding similarly as before, I obtain
$$D(q^k) < \varphi(q^k).$$

Here is my specific question:

Will it be possible to come up with a tighter upper bound than $\varphi(q^k)$ for $D(q^k)$ when $k > 1$?

Edited (in response to a comment on November 5, 2020)

The reason I need a bound for $D(q^k)$ when $k>1$ is because of this closely related question.

答:
Let
$$f(k,q):=\varphi(q^k)-D(q^k)=q^{k-1}(q-1)-\bigg(2q^k-\frac{q^{k+1}-1}{q-1}\bigg)=\frac{q^{k-1}-1}{q-1}$$

Then we have

$$\frac{\partial f(k,q)}{\partial q}=\frac{((q-1)(k-2)-1)q^k+q^2}{(q - 1)^2 q^2}$$
which is non-negative.

So, we have
$$f(k,q)\ge f(k,2),$$

i.e.
$$\color{red}{D(q^k)\le \varphi(q^k)-(2^{k-1}-1)}$$

Added :

If $k\equiv q\equiv 1\pmod 4$, then we have $$D(q^k)\le \varphi(q^k)-\frac{5^{k-1}-1}{4}$$ and $$D(q^k)\le\varphi(q^k)-\frac{q^4-1}{q-1}$$ So we have $$3−\frac{1}{q^{k+1}}\bigg(\varphi(q^k)-\frac{5^{k-1}-1}{4}\bigg)\lt I(q^k)+I(n^2)$$ and $$3−\frac{1}{q^{k+1}}\bigg(\varphi(q^k)-\frac{q^4-1}{q-1}\bigg)\lt I(q^k)+I(n^2)$$ But I think that these are not better lower bounds for $I(q^k)+I(n^2)$ than $3−\dfrac{q−2}{q(q−1)}$.


问:
Find the least integral value of $n$ for $f(n)={\left(\frac{9}{10}\right)}^n<\frac{1}{2}$
I was doing a probability question where it was required to find out the least integral value of n with $$f(n)=\left(\frac 9 {10}\right)^n<\frac{1}{2}$$

A little calculation(by calculator of course) yields $\min(n)=7$.

as:

$f(5)=0.59$

$f(6)=0.53$

$f(7)=0.43$

My question is there a good way to approximately tell which integer $n$ can be without computational software and log tables (even $\log 2,\log 3$ etc.)vCalculus seems to be an unlikely option

答:
Rewrite what you want as

$$2\approx\left(10\over9\right)^n=\left(\left(1+{1\over9}\right)^9\right)^{n/9}\approx e^{n/9}$$

If you allow yourself to know that $\ln2\approx0.7$ (which is the basis for the "Rule of $70$" described in Rivers McForge's answer), then you get $n\approx9\ln2\approx6.3$, which accords with what you found.


问:
What do I need to do to figure out bounds for integrating over triangle that does not touch both axes?
What do I need to do to figure out bounds for integrating over triangle that does not touch both axes?

Such as the one with edges $(0,0), (1,1), (4,0)$.

If this was $(0,0), (0,1), (4,0)$ I'd view this simpler.

I'm thinking I'd need to subtract some rectangle or then find a way to describe the sides.

答:
Plotting or sketching the region helps. In the example you gave with points $(0,0), (1,1), (4,0)$, when we plot and based on the equation of lines, we realize that if we integrate along $y$ axis, we can do it as one integral.

For example if we were finding area, $A = \displaystyle \int_0^1 ((4-3y) - y) \, dy $

If we integrate along $x$ axis, we have to break it into two regions $(0 \leq x \leq 1)$ and $(1 \leq x \leq 4)$ or we have to find a change of variable.


问:
Finding series convergence by using the Comparison test
I want to find whether the Series diverges using the Comparison test
$$ \sum_{n=1}^\infty  \left(\frac{n+1}{n^2 - n}\right)  $$
would it be correct to compare it to $\left(\frac{n}{n^2}\right)$ so I would get
$$ \left(\frac{n+1}{n^2 - n}\right) \geq \left(\frac{n}{n^2}\right)  $$
and since
$$  \left(\frac{n}{n^2}\right) = \left(\frac{1}{n}\right)  $$
and $\left(\frac{1}{n}\right)$ is a harmonic series and is divergent so the original series also diverges?

答:
Yes your way is fine, as an alternative we can use that

$$\frac{n+1}{n^2 - n}=\frac{n-1+2}{n^2 - n}=\frac{n-1}{n^2 - n}+\frac{2}{n^2 - n}=\frac1n+\frac{2}{n^2 - n}$$


答:
Your intuition is correct. For me, it's simpler to use some elementary asymptotic analysis:

A polynomial is asymptotically equivalent to its leading term, hence
$$\frac{n+1}{n^2-n}\sim_\infty \frac n{n^2}=\frac1n,$$
which diverges.

Title: Finding series convergence by using the Comparison test

Body:
I want to find whether the Series diverges using the Comparison test
$$ \sum_{n=1}^\infty  \left(\frac{n+1}{n^2 - n}\right)  $$
would it be correct to compare it to $\left(\frac{n}{n^2}\right)$ so I would get
$$ \left(\frac{n+1}{n^2 - n}\right) \geq \left(\frac{n}{n^2}\right)  $$
and since
$$  \left(\frac{n}{n^2}\right) = \left(\frac{1}{n}\right)  $$
and $\left(\frac{1}{n}\right)$ is a harmonic series and is divergent so the original series also diverges?


问:
Is there a theory of extensible definitions?
We can define $+$ as a function $\mathbb{N}^2 \rightarrow \mathbb{N}$, and then prove:

Theorem 1. The range of $+$ is $\mathbb{N}$.

If we later wish to extend $+$ to a function $\mathbb{Z}^2 \rightarrow \mathbb{Z}$, then we have a problem; according to our new definition, the theorem we proved isn't true any more. To be rigorous, we must therefore define a new operation $ +' : \mathbb{Z}^2 \rightarrow \mathbb{Z}$. But this is clumsy. In practice, what we usually do is refrain from talking about $+$ in certain ways, and this allows us to extend the definition. So for instance, the following sentence is fine:

Theorem 2. For all $a,b \in \mathbb{N}$ it holds that $a \leq a+b$.

By writing sentences like Theorem 2 only, and refraining from sentences like Theorem 1, we're able to extend $+$ without invalidating our previous theorems, as opposed without having to introduce a new operation $+'$.

So what I'm looking for is a rigorous approach to the idea that, by pairing a definition with some additional information about how that definition is allowed to be used, it becomes extensible. And the more stringent we are about how a definition can be used, the more extensible it becomes.

Is there a theory of extensible definitions?

P.S. I don't know which tags are appropriate for this question. Please feel free to retag it.

答:
I don't think there's any formal distinction between your (i) and (ii) that would have told you, knowing only about addition in $\mathbb N$, that one is going to keep holding for the extendion and the other isn't.

It would have been simple enough instead to define an extension of $+_{\mathbb N}$ to $\mathbb Z$ which instead preserves the validity of (ii) but doesn't preserve (i). For example, we could have defined $a+_{\mathbb Z}b=|a|+_{\mathbb N}|b|$ and the resulting $+_{\mathbb Z}$ would still have been a strict extension of $+_{\mathbb N}$, satisfying (ii) but not (i).

So it's not really that some statements are inherently extensible, but merely that in choosing which extensions to consider we implicitly choose which properties are going to continue holding. Conversely, our desire to keep some properties valid will inform our choice of which extensions we allow to keep the same name and symbol.

Of course, when we originally defined + we could have paired it with an explicit list of properties that we intend all our extensions of it to keep satisfying. But that's not a very satisfactory solution in a larger setting, because in general we'll want to be able to consider many different kinds of extensions, some that will preserve many properties and others that will leave most of them behind.

For example, most extensions of $+$ preserve the property $a+b=b+a$, and in most cases one would certainly want to consider that "extensible" if the word is to have any useful meaning. However, $a+b=b+a$ is not true about the usual extension of $+$ to addition of transfinite ordinals.


问:
Major and Minor axis of an ellipse obtained from applying a linear transformation to the unit circle
We have a linear transformation $T: \Bbb R^2 \to \Bbb R^2$ whose matrix is $$\begin{pmatrix}
2&3\\
0&2
\end{pmatrix}.
$$ Suppose we start with the unit circle in $\Bbb R^2$, we know that its image under the transformation is an ellipse. I am trying to find the major and minor axis of the ellipse.

I tried finding the eigenvalues and eigenvector but there is only one eigenvector $(1,0)^t$ corresponding to the eigenvalue $2$.

答:
Here is one way to do it.

A parametric representation of the ellipse is $$E(\theta)=(2\cos\theta+3\sin\theta,2\sin\theta),$$ where $0\leq\theta\leq 2\pi$. The major axis in the direction which maximises the length of $E(\theta)$, and the minor axis in the direction which minimises the length. The square of the length is
$$\|E(\theta)\|^2=4\sin^2\theta+(2\cos\theta+3\sin\theta)^2$$
Elementary calculation shows that derivative of this is $3(4\cos 2\theta+3\sin 2\theta)$, whose four zeros in $[0,2\pi]$ are $$\arctan(2),\arctan(2)+\pi/2,\arctan(2)+\pi,\arctan(2)+3\pi/2$$

The major axis is therefore in the direction of $\arctan(2)$, and its length is $8$, and the minor axis is in the direction of $\arctan(2)+\pi/2$, and its length is $2$.


问:
Open ball of the Discrete Metric
Let $(X,d)$ be the discrete metric space, such that for $x,y\in X$,

$$d(x,y)=\cases{0 & $x=y$\\ 1 & $x\ne y$}$$

The open ball is defined as

$$B_r(x)=\cases{\{x\} & $0<\varepsilon \le 1$\\
X & $\varepsilon > 1$}$$

I am having trouble on understanding how we come to this definition of the open ball of the discrete metric. Any explanation would be highly appreciated.

答:
No, the open ball is not defined that way. The open ball $B_r(x)$ is the set $\{y\in X\mid d(x,y)<r\}$, for any metric space. But, if $d$ is the discrete metric, then:

if $r\leqslant1$ then $d(x,y)<r\iff d(x,y)=0\iff y=x$, and therefore $B_r(x)=\{x\}$;

if $r>1$, then, for each $y\in X$, $d(x,y)\leqslant1<r$, and therefore $B_r(x)=X$.

Title: Open ball of the Discrete Metric


问:
For the map $f\left( x \right) = x\sin \pi x$, $x>0$, points at which $f'(x)$ vanishes
Let $f\left( x \right) = x\sin \pi x$, $x>0$. Then for all natural number n, $f'(x)$ vanishes at

(A) A unique point in the interval $(n,n+\frac{1}{2})$

(B) A unique point in the interval $(n+\frac{1}{2},n+1)$

(C) A unique point in the interval $(n,n+1)$

(D) two points in the interval $(n,n+1)$

The official answer is B&C

My approach is as follow

$f'\left( x \right) = \pi x\cos \pi x + \sin \pi x = 0$

$\Rightarrow  - \tan \pi x = \pi x$

From here onward how do I proceed further

答:
$$\sin\pi x$$ keeps a constant sign in all unit intervals $(n,n+1)$, and the derivative changes sign once. We can reject $D$ and accept $C$, but the choice between $A$ and $B$ is unsure. To decide, we can consider the sign of

$$f'(n)f'\left(n+\frac12\right)=\pi n\cos \pi n \sin \pi\left(n+\frac12\right)>0.$$

You can conclude.


问:
Riemann integrable function is bounded, without any type of choice
I know a proof that Riemann integrable positive function $f$ defined on $[a,b]$ is bounded.

Suppose $f\geq 0$ is unbounded. First, for each $n$, choose a point $a_n$ in $[a,b]$ such that $f(a_n) > n$. Let $a$ be an arbitrary positive real number. Choose $n$ with $a > \frac1n$, and we can have a partition $p$ with each interval length bigger than $\frac1n$ while mesh($p$) $< a$. Then $R(f, p) > f(a_{n^2}) \frac1n = n$  so the Riemann sums do not tend to a finite limit.

But I think while we choose $a_n$ in the first step, countable choice is used. Are there any other ways to prove it without using some kind of choice, since this is a very 'basic' theorem in 'elementary 'analysis so it seems weird that it requires some kind of choice axiom.

答:
Suppose $f:[a,b]\to\mathbb{R}$ is Riemann integrable but assume $f$ is unbounded. Let $I_R=\int_{a}^{b}f$, which is also defined as $\lim_{||P||\to 0}R(f,P,x^*)$, where $P=\{a=t_0<t_1<...<t_n=b\}$ is a partition of $[a,b]$ and $x^*$ is a compatible n-tuple with $P$, that is, $x_i^*\in[t_{i-1},t_i]$ for all $1\leq i\leq n$. Also, $||P||$ is the norm of $P$, which is the maximum of $t_i-t_{i-1}$.

Since $f$ is integrable, then for all $\varepsilon>0$, there exists a $\delta>0$ such that for all partitions $P$ with $||P||<\delta$, then $|R(f,P,x^*)-I_R|<\varepsilon$.

Consider the particular choice $\varepsilon=57$. Then there exists a $\delta>0$ such that for all partitions $P$ with $||P||<\delta$, we have $|R(f,P,x^*)-I_R|<57$.

If $x^*$ and $z^*$ are compatible with $P$, then we have $$\begin{align*}|R(f,P,x^*)-R(f,P,z^*)|&= |R(f,P,x^*)-I_R+I_R-R(f,P,z^*)|\\&\leq|R(f,P,x^*)-I_R|+|I_R-R(f,P,z^*)|\\&<114\end{align*}$$

This is a set up for the following, where we will redefine $x^*$ and $z^*$.

By the Archimedean property, choose $n\in\mathbb{N}$ such that $P_n$ is a regular partition of $[a,b]$ so that if $P_n=\{a=t_0<t_1<...<t_n=b\}$, then $t_i-t_{i-1}=\frac{b-a}{n}$ and $||P_n||<\delta$.

Suppose $x^*$ is compatible with $P_n$. Since $f$ is unbounded, then there exists some $j\in\mathbb{N}$ with $1\leq j\leq n$ such that $f|_{[t_{j-1},t_j]}:[t_{j-1},t_j]\to\mathbb{R}$ is unbounded.

In particular, there exists a $z_j\in[t_{j-1},t_j]$ such that $|f(z_j)-f(x_j^*)|>\frac{570n}{b-a}$.

Now, define $z^*$ as $x_i^*$ if $i\neq j$ and $z_j$ if $i=j$. Then we have $$\begin{align*}|R(f,P_n,x*)-R(f,P_n,z^*)|&=\big|\sum_{i=1}^{n}f(x_i^*)(t_i-t_{i-1})-\sum_{i=1}^n f(z_i^*)(t_i-t_{i-1})\big|\\&=|[f(x_j^*)-f(z_j)](t_j-t_{j-1})|\\&>\frac{570n}{b-a}\cdot\frac{b-a}{n}\\&=570 \end{align*} $$

However, we have shown that this quantity should be less than $114$, therefore we have reached a contradiction. Thus $f$ must be bounded.


问:
Solve for $\lambda$
I have the following system of two equations:

\begin{align}
w &= \frac{1}{k}\Omega^{-1}(\mu + \lambda \mathbf{1})\\
w^T\mathbf{1}&=1
\end{align}

where $\Omega$ is a matrix of size $n \times n$, $\mu$, $w$ and $\mathbf{1}$ are vectors of size  $n$ and $k$, $\lambda$ and $1$ are magnitudes. As you can see by the second equation is that the sum of the values of $w$ should be $1$.

I want to solve for $\lambda$ respecting both equations. So I input the first one on the second one.

My naïve attempt have been the following:

\begin{equation}
w^T\mathbf{1} = \mathbf{1}^Tw =1
\end{equation}
\begin{equation}
\mathbf{1}^T\frac{1}{k}\Omega^{-1}(\mu + \lambda \mathbf{1})=1
\end{equation}
\begin{equation}
\frac{1}{k}\mathbf{1}^T\Omega^{-1}(\mu + \lambda \mathbf{1})=1
\end{equation}

Here I got somehow stuck because I cannot send the terms $\mathbf{1}^T\Omega^{-1}$ to the right hand side of the equation. I thought of multiplying from the left by $\Omega\mathbf{1}^{T+}$ both sides of the equation where $\mathbf{1}^{T+}$ is the pseudoinverse of $\mathbf{1}^T$, that would give me something like the following $\Omega\mathbf{1}^{T+}\mathbf{1}^T\Omega^{-1}$. However the term $\mathbf{1}^{T+}\mathbf{1}^T$ would give me a $n \times n$ which is not the identity matrix.

I'm out of ideas, I though I wasn't not going to spend a lot of time to solve for $\lambda$ but I've been hours and I see that I don't have the right background or that I'm just missing something.

答:
Based on Ian's comment,

$$\frac{1}{k}\mathbf{1}^T\Omega^{-1}(\mu+\lambda\mathbf{1})=1\to \mathbf{1}^T\Omega^{-1}\mu+\lambda\mathbf{1}^T\Omega^{-1}\mathbf{1}=k$$

$$\lambda\mathbf{1}^T\Omega^{-1}\mathbf{1}=k-\mathbf{1}^T\Omega^{-1}\mu$$

but see that $\mathbf{1}^T\Omega^{-1}\mathbf{1}\in \mathbb R$, which is the sum of all elements of $\Omega^{-1}$. So, if $\mathbf{1}^T\Omega^{-1}\mathbf{1}\ne 0$ then you can write,

$$\lambda=\frac{1}{\mathbf{1}^T\Omega^{-1}\mathbf{1}}(k-\mathbf{1}^T\Omega^{-1}\mu).$$

Ps.: $\mathbf{1}^T\Omega^{-1}\mathbf{1}$ may be equal to $0$. For instance, take
$$\Omega^{-1}=\begin{pmatrix}
2&-1\\
0&-1
\end{pmatrix}.$$

Title: Solve for $\lambda$


问:
What conditions ensure that checking if a shape covers a box can be done just by evaluating the shape pointwise?

Background

I have described an algorithm for generating random points inside an arbitrary shape (such as a circle, polygon, or an arbitrary closed curve) contained within a box. It involves checking whether the box is outside or partially or fully inside the shape, and then—

Generating a uniform random point inside the box if the box is inside the shape,

rejecting the box and starting over if the box is outside the shape, and

subdividing the box, choosing a random sub-box, and repeating this process for that sub-box otherwise.

This algorithm uses a function called InShape that determines whether a shape covers an axis-aligned bounding box. It takes such a bounding box as input and returns—

YES if the box is entirely inside the shape;

NO if the box is entirely outside the shape; and

MAYBE if the box is partly inside and partly outside the shape.

Now, take a particular implementation of InShape that has certain knowledge about a particular shape. Assume the following:

The shape is closed, has nonzero finite volume, and has a boundary of measure zero.

The InShape implementation can determine only pointwise whether a point is either outside the shape, or on or inside the shape.

The InShape implementation has access to arbitrary-precision arithmetic, as well as interval arithmetic using arbitrary-precision rational numbers. See my library, for example.

Other than this, it doesn't matter how the shape is described -- it could be described as a sequence of line and/or curve segments describing the shape's outline; as a signed distance function; as an inequality; as a union or intersection of multiple shapes; etc.

The InShape implementation is given an axis-aligned bounding box as input. The goal is to correctly classify the box just by evaluating the shape pointwise.

Under certain conditions, this is trivial to do. For example, if the shape is enclosed by a 1x1 rectangle, the point (0, 0) is on the shape, and every horizontal or vertical line crosses the shape (inside the rectangle) at most once (think of one quarter of a circle centered at the origin), then the box can be correctly classified just by checking the point's corners. The algorithm (Algorithm 1) is thus to return—

YES if all the box's vertices are on or inside the shape;

NO if none of the box's vertices are on or inside the shape; and

MAYBE in any other case.

More generally, I believe Algorithm 1 will work if—

the shape is enclosed by a hypercube $[0, 1]\times[0, 1]\times...\times[0, 1]$,

the point $(0, 0, ..., 0)$ is on or inside the shape, and

every open axis-aligned line segment that begins in one face of the hypercube and ends in another face crosses the shape at most once (an example is one quarter of a circular disk whose center is at $(0, 0)$),

Or if—

the shape is enclosed by a $2$-dimensional rectangle $[0, 1]\times[0, 1]$,

the line segment $((0, 0), (1, 1))$ is entirely on or inside the shape,

the shape is convex and symmetric about that line segment, and

the box being tested arose out of a recursive subdivision of the 2-dimensional rectangle into smaller boxes with half the size, $\frac{1}{4}$ the size, etc.

However, for more general convex shapes (which are the shapes that I care about most), this is not so easy. For example, if the shape is convex and the point $(0, 0)$ is on the shape, the correct algorithm to classify the shape (Algorithm 2) is to return—

YES if all the box's vertices are on or inside the shape;

NO if none of the box's vertices are on or inside the shape and if the shape's boundary does not intersect the box's boundary; and

MAYBE in any other case.

This is not so easy because checking whether a box intersects a shape might not be robust especially if the shape is described by an inequality (such as $x^2 + y^2 - 1 <= 0$). Under certain cases, the algorithm might miss an intersection even though it's present. But at least when the shape is convex and when InShape uses interval arithmetic and builds one interval for each dimension of the box (here, $[x, x+\epsilon]$ and $[y, y+\epsilon]$), and evaluates the inequality only once with the intervals, InShape can still get robust results. In this algorithm (Algorithm 3), InShape returns—

YES if the result's upper bound is less than 0;

NO if the result's lower bound is greater than 0; and

MAYBE in any other case.

Questions

Thus my questions are:

What are necessary and/or sufficient conditions (such as convexity or regularity conditions, or other requirements on the shape) that allow Algorithm 1 to work correctly? Are the sufficient conditions I gave above for this algorithm correct? If so, can they be relaxed?

What are necessary and/or sufficient conditions that allow Algorithm 2 to work correctly, if the InShape method can only evaluate the shape point-by-point? In particular, how can Algorithm 2 robustly check for intersections as required to determine whether to return NO or MAYBE?

What are other conditions that allow InShape to correctly classify whether a box is outside or on or inside a shape when InShape can only evaluate the shape point-by-point, or when InShape proceeds as in Algorithm 3?

Is it possible (or what additional conditions make it possible) to correctly classify a bounding box as NO or MAYBE, using only pointwise evaluation of the shape, if—

the shape is enclosed by a hypercube $[0, 1]\times[0, 1]\times...\times[0, 1]$,

the point $(0, 0, ..., 0)$ is on or inside the shape,

the shape is convex, and

the box being tested arose out of a recursive subdivision of the hypercube into smaller boxes with half the size, $\frac{1}{4}$ the size, etc.?

(Note that in this case, classifying a bounding box as YES is trivial; just check its four corners.  On the other hand, I know that it's not enough to classify the box as NO or MAYBE this way.)

Note

An answer suggested the "Separating Axis Theorem for Polygons". But I don't see how the separating axis theorem can be applied to my questions given above. (Notably, the algorithm would have to determine whether a separating axis exists in order to apply the theorem, which seems more complicated than checking the corners of the bounding box, especially if the shape is described as a signed distance function, an implicit equation, etc.) In addition, the theorem doesn't seem to address the issue of whether to return MAYBE or YES, given that the algorithm shouldn't return NO.

Thus, if the separating axis theorem really related to this question, then the answer should give further detail on how the separating axis theorem answers the question of finding an algorithm to classify whether a box is in a shape given only pointwise access to the shape.

Examples

Take the following shapes, all of which are convex and equal 0 at the origin:

$v^2 - (u/v)^{1.4-1}*exp(-(u/v)) \le 0$ - Ratio-of-uniforms shape for the gamma(1.4) distribution

$v^2 - exp(-(u/v)^2/2) \le 0$ - Ratio-of-uniforms shape for the normal distribution

$v^2 - (u/v)^2*exp(-(u/v)^2/2) \le 0$ - Ratio-of-uniforms shape for the Maxwell distribution

All three shapes don't work under Algorithm 1, but they appear to give correct results under Algorithm 3, even without the intersection checks required by Algorithm 2.

答:
You may want to check the Separating Axis Theorem for $n$-dimensions. There is an answer by @Jens Renders here which may be what you are looking for.

See: Proof of Separating Axis Theorem for polygons


问:
Analogies between Prime Ideals and Knots
While reading this question posted at this link:

Subjects studied in number theory

i interestingly landed up on this Wikipedia page, and was quite amazed to see the variety of branches opening up. That's how I came to know about the subject Arithmetic Topology. Wikipedia describes this subject as:

Arithmetic topology is an area of mathematics that is a combination of algebraic number theory and topology. In the 1960s topological interpretations of class field theory were given by John Tate based on Galois cohomology, and also by Michael Artin and Jean-Louis Verdier based on Étale cohomology. In subsequent years, Barry Mazur and Yuri Manin pointed out a series of analogies between prime ideals and knots.

Can anyone explain me as to how Prime Ideals and Knot's are analogous (elementary explanation would be helpful) and why are we interested in studying these analogies.

答: Here's a useful "conceptual analogy dictionary" between 3-manifolds and number rings, excerpted from Morishita: Analogies between knots and primes, 3-manifolds and number rings.


问:
what will be the procedure to prove the following relationship?
Let $U$ follows standard uniform distribution , that is, $U\sim U(0,1)$ and
$X$ follows Pareto distribution, that is, $X\sim Pa{(\alpha,a,h)}$

where , $a=$location parameter ; $-∞<a<∞$

$h=$scale parameter ; $h>0$

$\alpha=$shape parameter ; $\alpha>0$

then How can i prove the relationship that $X$ and $a+hU^{-\frac{1}{\alpha}}$ have same distribution

My procedure was :

i derived the pdf of $X$ when $X=a+hU^{-\frac{1}{\alpha}}$
then i found that the derived distribution of $X$ is nothing but a Pareto distribution $Pa(\alpha,a,h)$

so, i concluded that if $X$ follows Pareto distribution $Pa(\alpha,a,h)$
& $U$ follows standard Uniform distribution $U(0,1)$ then

$$ "X\quad and\quad a+hU^{-\frac{1}{\alpha}}\quad have\quad same\quad distribution\quad". $$

is that my procedure correct?

i am confused because i have been asked for prove the relationship that $$ "X\quad and\quad a+hU^{-\frac{1}{\alpha}}\quad have\quad same\quad distribution\quad". $$
not to derive the pdf of $X$ when $X=a+hU^{-\frac{1}{\alpha}}$

please tell me the procedure to prove the relationship that $$ "X\quad and\quad a+hU^{-\frac{1}{\alpha}}\quad have\quad same\quad distribution\quad". $$

i have thought of another process using moment generating function technique(mgf) but i couldn't compute the mgf of pareto distribution

if i generalized my problem "i actually want to know that"

How can i prove a relationship between two different distributions that they follow the same distribution after some transformation

答:
If you want to show that two random variables follow the same distribution, there's not just one procedure to do that. This is because there are several quantities that uniquely determine the distribution of a random variable.

For instance, if $X$ is a random variable, then the cumulative distribution function $$F(x)=P(X\leq x),\quad x\in\mathbb{R},$$ determines the distribution of $X$ in the sense that if $Y$ is another random variable with cumulative distribution function given by $F$, then $X\sim Y$. Other quantities that determines the distribution uniquely are the moment-generating function, the characteristic function, the Laplace transform, the density function (if $X$ is absolutely continuous) and the probability mass function (if $X$ is discrete).

So in general if you want to show that $X\sim Y$ for two random variables $X$ and $Y$ you may freely choose any of the above quantities. For example we can choose to show that $\varphi_X(t)=\varphi_Y(t)$ for all $t\in\mathbb{R}$, where $\varphi$ is the characteristic function. Then we may conclude that $X\sim Y$.

Note that not all such quantities determines the distribution uniquely. For example, the sequence of moments $\{{\rm E}[X^n]\mid n\geq 1\}$ does not completely determine the distribution in the sense that one can find two random variables $X$ and $Y$ such that $X\not\sim Y$ but
$$
{\rm E}[X^n]={\rm E}[Y^n],\quad \text{for all }n\geq 1.
$$


问:
Why does the integral domain "being trapped between a finite field extension" implies that it is a field?
The following is an exercise from Qing Liu's Algebraic Geometry and Arithmetic Curves.

Exercise 1.2.

Let $\varphi : A \to B$ be a homomorphism of finitely generated algebras over a field. Show that the image of a closed point under $\operatorname{Spec} \varphi$ is a closed point.

The following is the solution from Cihan Bahran. http://www-users.math.umn.edu/~bahra004/alg-geo/liu-soln.pdf.

Write $k$ for the underlying field. Let’s parse the statement. A closed point in $\operatorname{Spec} B$ means a maximal ideal $n$ of $B$. And $\operatorname{Spec}(\varphi)(n) = \varphi^{−1}(n)$. So we want to show that $p := \varphi{−1}(n)$ is a maximal ideal in $A$. First of all, $p$ is definitely a prime ideal of $A$ and $\varphi$ descends to an injective $k$-algebra homomorphism $ψ : A/p \to B/n$. But the map $k \to B/n$ defines a finite field extension of $k$ by Corollary 1.12. So the integral domain $A/p$ is trapped between a finite field extension. Such domains are necessarily fields, thus $p$ is maximal in $A$.

In the second last sentence, the writer says that the integral domain $A/p$ is trapped between a finite field extension. I don't exactly know what it means, but I think it means that there are two injective ring homomorphisms $f:k\to A/p$ and $g:A/p\to B/n$ such that $g\circ f$ makes $B/n$ a finite field extension of $k$. But why does it imply that $A/p$ is a field?

答: $A$ and $B$ be finitely generated algebras over $k$. Let $\mathfrak m $ be maximal ideal of $B$. We have an injective map $A/\phi ^{-1}(\mathfrak m) \rightarrow B/\mathfrak m $. Identify $A/\phi ^{-1}(\mathfrak m)$ to its image via this map. Let $T\in A/\phi ^{-1}(\mathfrak m) $, then $1/T \in B/ \mathfrak m $- which is algebraic extension of the field $k$. So $1/T $ is there is a monic polynomial over $k$ which $1/T$ satisfies, multiplying this by $T^{n-1}$ you get that $1/T \in A/\phi ^{-1}(\mathfrak m) $ and you are done.


问:
How can we relate this sequence of integrals?
Let:
$$T_n=\int_0^1\int_0^t \frac{1}{x(1-x)}\frac{x^{2^n}}{x^{2^n}+1}\,\mathrm{d}x\,\mathrm{d}t$$

Through experimentation it looks like $T_n=\log 2/2^n$, so we should have the relation $2T_n=T_{n+1}$ and $T_0=\log 2$. However I having trouble actually proving this, the substitution $u=2x$ doesn't seem to work. I'm not hugely familiar with double-integration so perhaps that is where I am going wrong. I would love some help!

答:
$\textbf{Hint}$: Since the integrand has no $t$ dependence, we can change the order of integration to simplify the problem like so

$$= \int_0^1\int_x^1 \frac{1}{x(1-x)}\frac{x^{2^n}}{x^{2^n}+1}\,\mathrm{d}t\,\mathrm{d}x = \int_0^1 \frac{x^{2^n-1}}{x^{2^n}+1}\mathrm{d}x$$


问:
Exercise on convergent series
I am stumped by the following exercise (3.24 in Biler--Witkowski's book "Problems in mathematical analysis"):

Let $f$ be a continuous, increasing function from $[0,+\infty]$ to itself. Show that $$\sum \frac{1}{f(n)}$$ converges iff
  $$\sum \frac{f^{-1}(n)}{n^2}$$ converges.

Initially, I thought one could apply the integral test and do a change of variables in the integral (assuming $f$ to be $C^1$, say, which does not change anything) to move from one series to the other. This does not quite work - one of the issues being that $x \mapsto \frac{f^{-1}(x)}{x^2}$ does not need to be decreasing.

I think I'm probably missing something obvious, as the exercise does not even deserve a hint in the book - would anyone have an idea?

Note: this is not homework! I have not used series in a while and need to brush up on them for a class I'll be teaching over the summer, so I thought I'd spend the weekend doing exercises - so far, so good, except for this particular exercise.

Thanks for your help!

答:
Suppose that $f$ is $C^1$ and that the series $$\sum_{n=1}^\infty \frac{1}{f(n)}$$ is convergent then by test integral, the integral
$$\int_1^\infty \frac{dt}{f(t)}\tag{1}$$ is also convergent, then we integrate by parts
$$\int_1^\infty \frac{dt}{f(t)}=\lim_{t\to\infty}\frac{t}{f(t)}-\frac{1}{f(1)}+\int_1^\infty \frac{tf'(t)}{f^2(t)}dt$$
and by variable change
$$\int_1^\infty \frac{dt}{f(t)}=\lim_{t\to\infty}\frac{t}{f(t)}-\frac{1}{f(1)}+\int_{f(1)}^\infty \frac{f^{-1}(t)}{t^2}dt$$
The convergence of the improper integral $(1)$ imply
$$\lim_{t\to\infty}\int_t^{2t}\frac{dt}{f(t)}=0 $$
hence we can deduce by monotonicity of $f$ that
$$\lim_{t\to\infty}\frac{t}{f(t)}=0$$
hence we can see that the integral
$$\int_{f(1)}^\infty \frac{f^{-1}(t)}{t^2}dt$$
is convergent. Moreover we have
$$\sum_{n=1}^\infty \frac{f^{-1}(n)}{(n+1)^2}\leq \sum_{n=1}^\infty\int_n^{n+1}\frac{f^{-1}(t)}{t^2}dt=\int_{f(1)}^\infty\frac{f^{-1}(t)}{t^2}dt<+\infty$$
hence the series
$$\sum_{n=1}^\infty \frac{f^{-1}(n)}{n^2}$$
is convergent.

By the same manner we can prove the other implication.

Title: Exercise on convergent series


问:
Closure inside interior under quotient map
I am trying to show the following: given $A \subset B \subset X$ topological space. Assume $Cl(A) \subset Int(B)$. Let the quotient map $p: X \rightarrow X/A$. Show that $Cl(A/A) \subset Int(B/A)$. I have been trying to use the fact that for any continuous function $f$, $Cl(f^{-1}(E)) \subset f^{-1}(Cl(E))$, but $Cl(p^{-1}(A/A)) \subset p^{-1}(Cl(A/A))$, but what I need is the other direction.

答:
You can view it this way: for any quotient map $q: X \to Y$ we may call $A ⊆ X$ saturated if it is the $q$-preimage of some set $B ⊆ Y$. So we have a $1:1$ correspondence between subsets of $Y$ and saturated subsets of $X$. Having the quotient topology on $Y$ means that open sets are exactly those whose corresponding subsets of $X$ are open.

Now your particular quotient if quite simple – you just collapse $A$ to a point (if its nonempty), so every subset of $X$ that contains $A$ or misses $A$ is saturated. Now can you use these observations to prove your proposition?


问:
Having problem to understand the expression $∆x = ∆x_1 + ∆x_2 + · · · + \Delta x_n$
I am studying this pdf file. in the file they say:

$2.1.1$ Errors in sum and difference. Let us consider the exact

numbers $X_1, X_2, . . . , X_n$ and their corresponding approximate

number be respectively $x_1, x_2, . . . , x_n$. Assumed that $\Delta
> x_1, \Delta x_2, . . . , \Delta x_n$ be the absolute errors in $x_1,
> x_2, . . . , x_n$. Therefore, $X_i = x_i ± \Delta x_i$ , $i = 1, 2, .
> . . , n$. Let $X = X_1 + X_2 + · · · + X_n$ and $x = x_1 + x_2
> + · ·     · + x_n$. The total absolute error is: $|X − x| = |(X_1 − x_1) + (X_2    − x_2) + · · · + (X_n − x_n)| ≤ |X_1 − x_1| + |X_2 −
> x_2| + · · · +      |X_n − x_n|$ This shows that the total absolute
error in the sum is      $∆x = ∆x_1 + ∆x_2 + · · · + \Delta x_n$.

I don't understand how they conclude $∆x = ∆x_1 + ∆x_2 + · · · + \Delta x_n$. I mean the inequality $|(X_1 − x_1) + (X_2 − x_2) + · · · + (X_n − x_n)|≤ |X_1 − x_1| + |X_2 − x_2| + · · · + |X_n − x_n|$, shows  $∆x \le ∆x_1 + ∆x_2 + · · · + \Delta x_n$. but why they put equal sign between those exression?

答:
The wording in the text you are reading is a bit sloppy.

In fact, the problem is this: if you can measure quantities $X_1,\ldots X_n$ with absolute errors $\Delta x_1,\ldots,\Delta x_n$, so your measurements $x_1,\ldots,x_n$ are guaranteed to satisfy $|x_1-X_1|\le\Delta x_1,\ldots |x_n-X_n|\le\Delta x_n$, then what can you say about the sum $x_1+\ldots+x_n$ as an approximation of $X_1+\ldots+X_n$? The answer is:

$$|(x_1+\ldots+x_n)-(X_1+\ldots+X_n)|=|(x_1-X_1)+\ldots+(x_n-X_n)|\le|x_1-X_1|+\ldots+|x_n-X_n|\le\Delta x_1+\ldots+\Delta x_n$$

In other words, the error you make is at most $\Delta x_1+\ldots+\Delta x_n$.

On the other hand, you can easily see that you cannot assume you will make a smaller error than that, because it may happen that, for example $x_1=X_1+\Delta x_1,\ldots, x_n=X_n+\Delta x_n$ - i.e. you have "overshot" all the values by precisely the maximum you could, and in that case you will also have the equality:

$$|(x_1+\ldots+x_n)-(X_1+\ldots+X_n)|=|(x_1-X_1)+\ldots+(x_n-X_n)|=|\Delta x_1+\ldots+\Delta x_n|=\Delta x_1+\ldots+\Delta x_n$$

The same would happen if you "undershot" all the values by as much as their corresponding absolute error.

Thus, $\Delta x_1+\ldots+\Delta x_n$ is the smallest guaranteed absolute error, so it makes sense to call that number $\Delta x$. Your sum $x_1+\ldots+x_n$ is certainly at most that far from the correct sum $X_1+\ldots+X_n$, and this error may actually happen in the worst case(s).


问:
VC dimension of union and intersection
Let $\mathbb{A}$ and $\mathbb{B}$ be sets of subsets of $X$ of finite VC dimension. Show that:

(a) If $\mathbb{C}=\{A\cap B : A \in \mathbb{A},B \in \mathbb{B}\}$, then $\prod_{\mathbb{C}}(n)\leq\prod_{\mathbb{A}}(n) \cdot \prod_{\mathbb{B}}(n) $

(b) If $\mathbb{C}= \mathbb{A}\cup \mathbb{B}$, then $\prod_{\mathbb{C}}(n)\leq\prod_{\mathbb{A}}(n) + \prod_{\mathbb{B}}(n) $

The definition is of Vapnik-Chernovenkis (VC) dimension of a set of classifiers $\mathbb{H}$ is $$\text{VC}(\mathbb{H})=\text{max}\{n\in\mathbb{N}:\prod_{\mathbb{H}}(n)=2^n\}$$

The growth function $\prod_{\mathbb{H}}$ is defined as:

$$\prod_{\mathbb{H}}(n)=\text{max}_{\mathbb{x}\in X^n}|\{h(\mathbb{x}):h\in \mathbb{H}\}|.$$

Attempt for (a):

Fix a set $X$ of $n$ points. Let $Y_1,...,Y_k$ be the set of intersections of $\mathbb{A}$ with $X$. By defintion of $\prod_{\mathbb{A}}(X),k\leq\prod_{\mathbb{A}}(X)\leq \prod_{\mathbb{A}}(n)$. By definition $\prod_{\mathbb{B}}(Y_i)$, the intersection of the $\mathbb{B}$ with $Y_i$ is at most $\prod_{\mathbb{B}}(Y_i)\leq \prod_{\mathbb{B}}(n)$. Thus, the number of sets intersections of $\mathbb{C}$ with $X$ is at most

$$k \prod_{\mathbb{B}}(Y_i)\leq \prod_{\mathbb{A}}(n) \prod_{\mathbb{B}}(n)$$

I'm not sure how to go further for (a)

答:
As to (a): your answer is correct.

As to (b): cuts/dichotomies in $\Pi_{\mathbb{C}}(n)$ can be generated from $\mathbb{A}$ or from $\mathbb{B}$. Therefore (b) follows straightforwardly.

Title: VC dimension of union and intersection


问:
Does every element of an integral domain have an inverse?
I am reading a first course in algebra 7th edition written by John B. Fraleigh. I have seen the following two definitions:

1) A field is a commutative ring in which every nonzero element has
  multiplicative inverse.

2) An integral domain is a commutative ring with unity 1 and
  containing no zero divisors.

Then i saw a picture in the book that shows fields as subsets of integral domains like in the following picture:

My question is, how do we understand from these two definitions that fields are subsets of integral domains? In the definition of integral domain, i do not see anything saying that every element in the ring should have an inverse, it just says that there must be a multiplicative identity. Am i missing something or is there something missing in the definitions?

Thank you

答:
No, it is not necessarily the case that every element in an integral domain has a multiplicative inverse.

Every field is an integral domain, but not every integral domain is a field. Hence we have that the set of all fields is a proper subset of the set of all integral domains.


问:
Root of $x = \tanh(a x)$
Consider the equation:

$$x = \tanh(a x)$$

Given $a\in\mathbb{R}$, let $x=f(a)$ denote the maximum non-negative solution of this equation.

Is there a name for the function $f(a)$? Or can it be expressed in terms of known functions?

This equation comes up in statistical physics, for instance in the Curie-Weiss model. See Eq. 2.12 in https://www.unige.ch/math/folks/velenik/smbook/Curie-Weiss_Model.pdf.

答:
I don't think that there is a suitable function. Lambert's W seems insufficient.

Anyway, notice that the inverse function can easily be expressed:

$$a=\frac{\text{artanh}(x)}{x}.$$

Title: Root of $x = \tanh(a x)$


问:
Primitive of differential form $\omega = a\arctan(xe^y)dx+e^{-y}\ln(1+x^2e^{2y})dy$

$\omega = a\arctan(xe^y)dx+e^{-y}\ln(1+x^2e^{2y})dy$

Find $a$ s.t. $\omega$ is exact. Find a primitive of $\omega$.

I'm very confused about differential forms, so I don't know how to approach this question. To find $a$, I tried using the lemma that we saw in class i.e. if
$ \frac{\partial \alpha_i}{\partial x_j}=\frac{\partial \alpha_j}{\partial x_i}$ then $\omega$ is exact. So I calculated $\frac{\partial}{\partial x_2}a\arctan(xe^y)$ and $\frac{\partial}{\partial x_1}e^{-y}\ln(1+x^2e^{2y}) $ and that got me to the conclusion that $a=2$.

Now for the primitive, we only saw one example in class and it wasn't a very helpful one. So far I've integrated $2\arctan(xe^y)$for x. But what about $e^{-y}\ln(1+x^2e^{2y})$? What do I do with that? The idea would be to have this equation and find C: $\frac{\partial f}{\partial x_2} + C'(y)=e^{-y}\ln(1+x^2e^{2y})$ but it gets weird... Any idea of what I'm doing wrong?

Thanks!

答:
Turns out I was doing everything right. So here are the details:

If $\frac{\partial f}{\partial x}=2\arctan(xe^y)$, then $f$ is the primitive of $2\arctan(xe^y)$ = $2x\arctan(xe^y)-e^{-y}\ln(x^2e^{2x}+1)+C(y)$

Then you figure out C by deriving f. So you would have $\frac{\partial f}{\partial y}(2x\arctan(xe^y)-e^{-y}\ln(x^2e^{2x}+1)+C(y))=e^{-y}\ln(x^2e^{2x}+1)+C'(y)$

So that brings us to the conclusion that $C'(y)=0$ i.e. it's a constant.

So the final answer is $f=2x\arctan(xe^y)-e^{-y}\ln(x^2e^{2x}+1)+C$


问:
Evaluate $\iiint_{R} (2x+y) \,dx \,dy \,dz$
Evaluate $$\iiint _{R} (2x+y) \,dx \,dy \,dz\,,$$ where $R$ is the region bounded by the cylinder $z = 4 - x^{2} $ and the planes $x = 0$, $y = 0$, $y = 2$ and $z = 0$.

How do I extract the limits from the given conditions?

答:
To visualize this shape. Focus on the $x$-$z$ plane first and draw the curve $z=4-x^2$ and then note that $0\le y \le 2$.

$x$ start from value $0$ and when $z=0$, $x=2$. Hence $0 \le x \le 2$.

$z$ is bounded below by $0$ and bounded above by $4-x^2$.

$$\int\int\int_R (2x+y) \, dxdydz =  \int_0^2 \int_0^2\int_0^{4-x^2} (2x+y) \, dzdxdy$$


问:
(Non) Complete, (Non) separable, (Non) metrizable spaces
After studying several kinds of topological spaces (Like $L_p, C[0,1]$) etc., I thought it would be useful to me (and to others) if I tabulated some of them under 3 categories: Completeness(C), Separability(S) and Metrizability(M). Now this yields a total of 8 possibilities from spaces that are neither to spaces that are all 3. So far I have the following:

C,S,M : $\mathbb{R}^n$, $C[0,1]$, $L_p(\mu), 1\leq p < \infty$ where $\mu$ is the Lebesgue measure.

C,~S,M : $L_{\infty}(\mu)$

~C,S,M : $D[0,1]$ space under the Skorokhod's metric. However, under Billingsley's Metric (ref. Richard Bass "Stochastic Processes"), it is C,S,M.

~C,S,~M  : The Lower Limit topology of $\mathbb{R}$.

Since completeness is a property of a metric space, you can never have (C,~M). That eliminates two cases.

So that leaves me with (~C,~S,M) and (~C,~S,~M) for which I have no examples as of yet. I require at least 1 example in each to complete my list.

Question: Kindly help me find the remaining examples. References are welcome.

My searches: I searched for non complete non separable non metrizable spaces, but that didn't add to the table. Google searches were usually putting the "non" in front of separable and not anywhere else.

Purpose: I offer you my word that this is not a homework. Although it is interesting in it's own merit.

Thanks to everyone who helps make the table.

答: The book Counterexamples in Topology by Steen and Seebach would interest you. In the back it has a big table of hundreds of spaces and their properties. From that table I found a number of examples to (~C,~S,~M), for example the countable complement topology (there is a proof on M.SE: Properties of the countable complement topology on $\mathbb{R}$. ). That table does not list completeness, so I couldn't at a glance find an example of (~C,~S,M), but if one looks up all the spaces which are (~S,M) in the book, one might find one.


问:
Matrix exponential with given Jordan form
Find the matrix exponential for the matrix
$$A=\begin{bmatrix}
-14 & -16 & -16 \\
21 & 24 & 23 \\
-10 & -12 & -12 \end{bmatrix}$$

I know that $A$ can be represented in the form $A=T^{-1}JT$, where $$T=\begin{bmatrix}
0 & 1 & 2 \\
1 & 2 & 3 \\
2 & 2 & 1 \end{bmatrix}$$ and $$J=\begin{bmatrix}
-2 & 1 & 0 \\
0 & -2 & 0 \\
0 & 0 & -2 \end{bmatrix}$$

Shouldn't the matrix exponential then simply be $T^{-1}e^{Jt}T$? I think this should be correct but Wolfram Alpha gave a different result. This has terms which involve $e^{4t}$ while the matrix $J$ has no such terms- all its terms should only have $e^{-2t}$ right? And then simple matrix multiplications shouldn't change the powers of the exponent. I am unable to figure out where am I going wrong with this.

Thanks

答:
I obtain
$$
J=\begin{pmatrix}
-2 & 1 & 0 \\
0 & -2 & 0 \\
0 & 0 & 2 \end{pmatrix},
$$
with

$$T=\begin{pmatrix}
0 & 1 & 2 \\
1 & 2 & 3 \\
1 & 1 & \frac{1}{2} \end{pmatrix}, \quad
T^{-1}=\begin{pmatrix}
-4 & 3 & -2 \\
5 & -4 & 4 \\
-2 & 2 & -2 \end{pmatrix}
$$
We should indeed have $\exp(T^{-1}JT)=T^{-1}\exp(J)T$.

Do you have the factor $2$ of the last row reflected in your inverse of $T$? You should check again all three matrices, $T^{-1},J,T$. Of course, $T$ is not unique.


答: Those terms with $e^{4t}$ vanish after simplifying the expression. For instance, the top left entry of the answer is $-e^{-2t}(4t+2e^{4t}-3)$, which is equal to $-4te^{-2t}-2e^{2t}+3e^{-2t}$.

Title: Matrix exponential with given Jordan form

Body:
Find the matrix exponential for the matrix
$$A=\begin{bmatrix}
-14 & -16 & -16 \\
21 & 24 & 23 \\
-10 & -12 & -12 \end{bmatrix}$$

I know that $A$ can be represented in the form $A=T^{-1}JT$, where $$T=\begin{bmatrix}
0 & 1 & 2 \\
1 & 2 & 3 \\
2 & 2 & 1 \end{bmatrix}$$ and $$J=\begin{bmatrix}
-2 & 1 & 0 \\
0 & -2 & 0 \\
0 & 0 & -2 \end{bmatrix}$$

Shouldn't the matrix exponential then simply be $T^{-1}e^{Jt}T$? I think this should be correct but Wolfram Alpha gave a different result. This has terms which involve $e^{4t}$ while the matrix $J$ has no such terms- all its terms should only have $e^{-2t}$ right? And then simple matrix multiplications shouldn't change the powers of the exponent. I am unable to figure out where am I going wrong with this.

Thanks


问:
How to prove the sum of the reciprocals of the product of natural numbers with their increments?
How can I prove the following expression?

$$ \sum_{n=x}^{m} \frac{1}{n(n+1)} = \frac{1}{x} - \frac{1}{m+1} $$

Thanks in advance if anyone is able to help me!

答:
Note that (by partial fractions) we have $$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$
and thus $$ \sum_{n=x}^{m} \frac{1}{n(n+1)} =\sum_{n=x}^{m}(\frac{1}{n}-\frac{1}{n+1})$$
$$=\big(\frac{1}{x}-\frac{1}{x+1}\big)+\big(\frac{1}{x+1}-\frac{1}{x+2}\big)+...+\big(\frac{1}{m-1}-\frac{1}{m}\big)+\big(\frac{1}{m}-\frac{1}{m+1}\big)$$
$$=\frac{1}{x}-\frac{1}{m+1}$$
since all the terms in the middle cancel each other.


问:
Is it possible to make the trace of this matrix vanish?
Given a diagonal  $N \times N$ matrix $\mathbf X$ whose diagonal elements are $1-i$, let

$$\mathbf Y := \mathbf X\cdot \mathbf p\cdot \mathbf p^H\cdot \mathbf X^H$$

Is it possible to set the vector $\mathbf p$ of size $N \times 1$ to have $\operatorname{tr}(\mathbf Y) = 0$,  where $\mathbf p \neq \mathbf 0$?

答:
Rephrasing and generalizing, given matrix ${\rm A} \in \Bbb C^{n \times n}$, we would like to find vector ${\rm x} \in \Bbb C^n$ such that

$$0 = \mbox{tr} \left( {\rm A} \, {\rm x} \, {\rm x}^* {\rm A}^* \right) = \mbox{tr} \left( {\rm x}^* {\rm A}^* {\rm A} \, {\rm x} \right) = {\rm x}^* {\rm A}^* {\rm A} \, {\rm x} = \| {\rm A} \, {\rm x} \|_2^2$$

where $(\cdot)^*$ denotes the Hermitian transpose. Can you take it from here?

OwnerDisplayName: Rodrigo de Azevedo


问:
Finding function from its solutions
Let's say I want to find a function of which I only know its solutions, i.e. the points where it goes to zero.

$x_0 \: | \: f(x_0) = 0$

$x_1 \: | \: f(x_1) = 0$

$...$

Is there a way to programmatically find the function?

Thanks a lot.

答:
Yes and no.

Yes, you can always find some function, but no, you cannot find  the function with this property.

Here's why. Suppose we know that there's only one root (it's called a "root", not a "solution"), e.g. $f(0) = 0$. Then
$$
f(x) = x, f(x) = xe^x, f(x) = x^{13}, f(x) = 13x, \ldots
$$
are all different functions with exactly that one root.

But if you want to find just one, there's a particularly simple one that's easy to write down: the Lagrange Interpolating polynomial. (It actually solves a more general problem). For this particular problem, that turns out to be
$$
L(x) = (x-x_0) (x - x_1) \cdots (x - x_n)
$$
where  $x_0, \ldots, x_n$ are your roots. If you have infinitely many roots, this may not work, in which case a different solution will do:

$$
H(x) = \begin{cases}
0 & x = x0, x1, \ldots...\\
1 & \text{otherwise}
\end{cases}
$$
$H$ has the disadvantage that there's no pretty algebraic expression, and it's not generally continuous. If you want a continuous or differentiable function with infinitely many specified roots, you sneak into the realm of analysis, where such questions get addressed one at a time.

For instance, if you say "I'd like a continuous function whose roots are exactly all the rational numbers", the answer is "there's no such function: if $f$ is continuous and zero on the rationals, then it's everywhere zero on the reals."

Title: Finding function from its solutions


问:
point on triangle
let we have a domain, for example square that partitioned by some triangles that are not necessarily similar, and we know the coordinates of all vertices and the maximum size of the sides of all triangles, now let  we have  a point in the square, my problem is that I need to know that this point lying on which triangle?  How can I know it?

pleas help me. Thanks

答:
One way would be to iterate through the triangles. For each, get the barycentric coordinates $(u,v,w)$ of the point with respect to the triangle, and if $u,v,w$ are all $>= 0$ then the point lies in the triangle.

I found this PDF resource helpful when implementing barycentric coordinates in an application recently.

Title: point on triangle


问:
Major and Minor axis of an ellipse obtained from applying a linear transformation to the unit circle
We have a linear transformation $T: \Bbb R^2 \to \Bbb R^2$ whose matrix is $$\begin{pmatrix}
2&3\\
0&2
\end{pmatrix}.
$$ Suppose we start with the unit circle in $\Bbb R^2$, we know that its image under the transformation is an ellipse. I am trying to find the major and minor axis of the ellipse.

I tried finding the eigenvalues and eigenvector but there is only one eigenvector $(1,0)^t$ corresponding to the eigenvalue $2$.

答:
If we define
$$
A=\pmatrix{2 & 3 \\ 0 & 2}
$$

then the cartesian equation of the ellipse is $x^T M x=1$, where:
$$
M=(A^{-1})^TA^{-1}=\pmatrix{1/4 & -3/8 \\ -3/8 & 13/16}.
$$

Its semi-axes are given by $1/\sqrt{\lambda_i}$, where $\lambda_1=1$ and $\lambda_2=1/4$ are the eigenvalues of $M$, while the eigenvectors are aligned with the axes.


问:
Need help understanding the motivation for parametric statistical models
First, let me present the definition of a statistical model and a parametric statistical model.

Definition: A statistical model consists of a set of measures $\mathcal{P}$ on a measurable space $(\mathcal{X},\mathbb{E})$.

Definition: A statistical model is said to be parametric if there exists some surjective function $v:\Theta\to\mathcal{P}$, where $\Theta\subseteq\mathbb{R}^{k}$ is called the parameter space. The parameterised model is then given by $\mathcal{P}=\{v(\theta)\mid \theta\in\Theta\}$.

In regards to the first definition, is it not true to say that the goal in statistics is to infer which one of the measures in $\mathcal{P}$ is the true measure of some underlying experiment? And likewise, in regards to the second definition, we would like to infer which of the parameters in $\Theta$ is the true parameter of some underlying experiment, right?

This made me wonder what the motivation for parametric models is; I mean why do we want to infer a parameter instead of a measure?

Additional question: How would one extend the two definitions such that they include random variables?

I hope my questions are somewhat clear, but let me know if they aren't.

答:
Do you know George Box' famous quote: "All models are wrong, but some are useful."? In this light, I would say that the goal of statistics is not necessarily to find the true underlying model, i.e. the true data generating process, but to find an approximation that is useful in some way. This could be to get a better understanding about the underlying data generating process, but also to make good predictions for future data. These two tasks are fundamentally different: for either task you would learn a different model.

There is always a trade-off between choosing a simple model, and a model that fits the data well. If the model is too complex, e.g. it contains too many parameters, it can fit the data too well: it overfits, i.e. it not only learns the signal, but also the noise. If the model is too simple, it underfits, it doesn't learn the signal well enough. I also would say that there is no dichotomy between measures and parametric models, a statistical model could for example be a set of Gaussian distributions with a real mean and positive variance, this is a parametric model.


问:
Exponential map and Lie groups
As I understand it, the exponential map allows us to locally compare a manifold $\mathcal{M}$ around a point $p\in\mathcal{M}$ and its tangent space $T_p\mathcal{M}$. This is done by associating to any tangent vector $X_p\in T_p\mathcal{M}$ to a point $q$ in $\mathcal{M}$ obtained by the following method : we know that there is a unique geodesic $\lambda(t)$ such that $\lambda(0)=p$ and $X_p$ is tangent to $\lambda$ in $t=0$. We can then take $q=\lambda(1)$, i.e. $q$ is then point obtained by going "one unit" along $\lambda$ starting from $p$. The parameter $p$ doesn't necessarily take values in all $\mathbb{R}$ so this only well-defined for some subset of tangent vectors around the origin in $T_p\mathcal{M}$.

The exponential map is a very important notion that we use non-stop when working with Lie groups (as they are manifolds and the tangent space to the identity is its corresponding Lie algebra). But the notion of geodesics depends on the connection that we take. In GR, it's clear that we take the Levi-Civita connection, defined from the metric. But in general, when we work with an arbitrary Lie group, wich connection do we take ?

答: A Lie group is a parallelizable manifold, i.e there are vectors fields $X_1,..,X_n$, such that for every $g\in G$, $(X_1(g),...,X_n(g))$ is a basis of the tangent space $T_gG$. Let ${\cal G}=T_eG$ be the Lie algebra of $G$ and $a_1,...,a_n$ its basis. $a_i$ generated the left invariant vector fiel $X_i(g)=dL_g(a_i)$. It is possible to define a connection on $G$ by setting $\nabla_{X_i}X_j=0$. The geodesics of this connection at the neutral point $e\in G$ are $exp(tu), u\in {\cal G}$.

Title: Exponential map and Lie groups


答:
Please note that $z = 4-x^2$ is not a cylinder... In the picture below it is represented the surface $z = 4-x^2$. The planes $x=0$, $y=0$ and $y=2$ are the lateral faces, while $z=0$ is the bottom face.

I think that with this in mind you can place the integration limits.

 The final result is $\frac{80}{3}$.

Title: Evaluate $\iiint_{R} (2x+y) \,dx \,dy \,dz$

Body:
Evaluate $$\iiint _{R} (2x+y) \,dx \,dy \,dz\,,$$ where $R$ is the region bounded by the cylinder $z = 4 - x^{2} $ and the planes $x = 0$, $y = 0$, $y = 2$ and $z = 0$.

How do I extract the limits from the given conditions?


问:
Characterization of the center of a polygon
Let $O$ be the circumcenter of the regular polygon $P_n$. Then for any $A\in P_n$ one has $d(A,O)\leq r$, where $d$ is the usual Euclidean distance and $r$ is the polygon's circumradius.

Prove that this property uniquely determines the point $O$.

答:
Let $X$ be a point in the plane. We say that $X$ is a central point of $P_n$ if $d(A,X)\leq r$ for all $A \in P_n$.

If $\sigma$ is a symmetry of $P_n$ and $X$ is a central point, then so is $\sigma(X)$.

Thus, set of central points is fixed by all symmetries of $P_n$. But the only point fixed by all symmetries of $P_n$ is its circumcenter $O$.


问:
G delta sets and denseness
Let $(X,d)$ be a metric space and is complete. As $G_\delta$ sets are the countable intersection of open sets. So if $G_\delta$ is a dense set in $X$ then does this implies that the open sets whose intersection was $G_\delta$ is also dense.

答:
As some users and I pointed out in the comments, any set that contains a dense set has to be dense. Let's see why:

Imagine $D$ is dense and $D\subseteq A$. Take any open set $U$. Since $D$ is dense, $D\cap U\neq\emptyset$, so there is some $x\in D\cap U$. Then $x\in D$ and $x\in U$. Since $D\subseteq A$, $x\in A$ too, so $x\in A\cap U$. All this is just saying that given $D\subseteq A$ we have $\emptyset\neq D\cap U\subseteq A\cap U$. Since $U$ was a non-specific open set, this works for every open set, so we conclude that $A$ is dense too.

As for your question, as I said, if a dense set $D$ is a $G_\delta$, then it is a countable intersection of open sets: $D=\bigcap\limits_{n=1}^\infty U_n$. Since $D=\bigcap\limits_{n=1}^\infty U_n\subseteq U_n$ for every $n$, we have that each $U_n$ is dense.

Title: G delta sets and denseness


问:
How many 5 letters/number combinations?
How many combinations that are 5 characters can be made with numbers 0-9 and letters A-Z and all caps (e.g. ASX51)?</p>

答:
There are $26$ characters in the alphabet, and there are $10$ in $0$-$9$. So we have $36$ characters overall.

Therefore, there are
$$
\binom{36}{5}=\frac{36!}{(36-5)!5!}=\frac{36!}{31!5!}=\frac{36\cdot 35\cdot 34\cdot 33\cdot 32}{5\cdot4\cdot3\cdot2}=3\cdot7\cdot 17\cdot 33\cdot 32=376,992
$$
different $5$ characters.

Title: How many 5 letters/number combinations?


问:
Suggestions for books on combinatorics
I am very weak in combinatorics and I have not got my understanding clear to think a problem in combinatorics and come up with a logic/analysis.I want to develop both the interest and the understanding so that I can proceed further and solve questions.Are there books on combinatorics that are both interesting and has good problems(problems of all difficulties)</p>

答:
Two suggestions:

Schaum's Outline of Theory and Problems of Combinatorics by V.K. Balakrishnan has lots of worked problems and extensive coverage.  Inexpensive, too.

A Walk Through Combinatorics: An Introduction to Enumeration and Graph Theory, Second Edition by Miklos Bona. "The strong points of this book are in particular a very inviting style of exposition, in which developments are always well motivated and well illustrated by numerous examples, and the long list of problems at the end of each chapter, with detailed solutions..." (Zentralblatt MATH, quoted on the back cover of the book). Of the combinatorics books on my shelf, I think this is the one most likely to get the reader hooked on combinatorics.

Title: Suggestions for books on combinatorics


问:
At least one page contains two misprints in a 250 page book

A book is published, after using a Poisson distribution you find that the probability a given page has at least two misprints is $\frac{7}{400}$. We know discover that the book is 250 pages long, what is the probability that at least one page in this book has at least two misprints?

I tried to tackle this problem using a binomial distribution, i.e.: $\binom{250}{0}(\frac{7}{400})$ to determine the probability of exactly zero pages having two misprints and subtracting that from 1 to get the probability that at least 1 has two misprints. Am I correct in my understanding?

答: Since the probability of one page having at least $2$ misprints is $\frac{7}{400}$, the probability of a page having less than $2$ misprints is $\frac{393}{400}$. Then, the probability of all $250$ pages having less than $2$ misprints is $\big(\frac{393}{400}\big)^{250}$, so the probability of at least one page having at least $2$ misprints is $1-\big(\frac{393}{400}\big)^{250}\approx\boxed{0.98789}$


问:
Propositional Logic Problems
Im struggling with the following problems. In each case determine whether the statement is true or false.

a)

$(A \implies B) \iff C \models A \lor \neg B \lor C$

b)

$(\neg A \implies \neg B) \land (A \land \neg B)$ is satisfiable.

c)

$(\neg A \iff \neg B) \land (A \land \neg B)$.

Part b) I believe is true as it will hold for certain combinations of true/false for A and B. From my understanding if the statement is satisfiable then it is also true, meaning c) is also true?

I have not been able to make much progress on part a).

答:
(A) In order to prove that $$(A \rightarrow B) \iff C \tag{1}$$ tautologically implicate $$A  \vee \neg B \vee C  \tag{2}$$

You will need to prove that for every interpretation where (1) is true - (2) is true as well. Meaning that if you find an interpretation that makes (1) true and (2) false you can say (a) is not true. One way of doing it is by constructing a truth table, the second option is to prove

$$ ((A \rightarrow B) \iff C) \rightarrow A \vee \neg B \vee C$$

Is a tautology. I would prefer this way, because it is very straight forward, as I said, if this tautological implication is not correct (false) then we can find a line where $ A \vee \neg B \vee C$ is false.

This proposition evaluates to "false" if and only if $A = \bot, B = \top , C = \bot$ because otherwise, it would be true (as we have a DNF)
Plugging those values to (1) we get it also evaluates to false, meaning the logical implication is evaluated to true ( $\bot \rightarrow \bot \equiv \top$)

Other interpretation of $A,B,C$ would make (2) true - and that means the logical implication would also be true, so there is no need to check these interpretations.

Title: Propositional Logic Problems


问:
If $f, g, h$ are continuous, then $f(g,h)$ is continuous
Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^k, g:\mathbb{R}^n\rightarrow \mathbb{R}, h:\mathbb{R}^n\rightarrow \mathbb{R}$ be three continuous functions. Then, is $f(g(x),h(x))$ a continuous function?

I know that if $g$ and $f$ are continuous functions, then $g\circ f$ is a continuous function. However, the function in the question isn't a composition like this, so I am a bit confused. I proved in an exercise from the book I am using that $f:\mathbb{R}^k \rightarrow \mathbb{R}^p$ is continuous if each component is continuous i.e. for every $i$ between $1$ and $p$, $f_i:\mathbb{R}^k \rightarrow \mathbb{R}$ is continuous where $f=(f_1, f_2, \dots, f_p)$. I'm not sure if that would be helpful.

答:
I proved in an exercise from the book I am using that $f:\mathbb{R}^k \rightarrow \mathbb{R}^p$ is continuous if each component is continuous i.e. for every $i$ between $1$ and $p$, $f_i:\mathbb{R}^k \rightarrow \mathbb{R}$ is continuous where $f=(f_1, f_2, \dots, f_p)$.

Then you have everything you need. Define the function $\phi: \Bbb R^n \to \Bbb R^2$ by
$$
\phi(x) = (g(x),h(x)).
$$
$\phi$ is continuous because its components are continuous. So, $f \circ \phi$ is continuous. We have
$$
f(\phi(x)) = f(g(x),h(x)),
$$
so we have reached the desired conclusion.


问:
Solving a differential equation via reduction of order
I want to solve the following differential equation

$$ u^{''}(t)+\frac{2(2-t)}{t(1-t)} u'(t)+\frac{2(1+t)}{t^2(1-t)} u(t)=0, \ t \in (0,1).$$

In order to find the general solution I have to find two linearly independent solutions. I tried to determine the first solution and then use reduction of order to find the second solution. For the first solution I used the ansatz  $u(t)=t^{\alpha}, \ \alpha \in R$. Inserting this into the equation leads to

$$ 0=\alpha (\alpha-1) t^{\alpha-2}+\frac{2(2-t)}{t(1-t)} \alpha \ t^{\alpha-1}+\frac{2(1+t)}{t^2(1-t)} t^{\alpha} \\
=\alpha (\alpha-1) t^{\alpha-2} + \frac{2(2-t)}{(1-t)} \alpha \ t^{\alpha-2}
+\frac{2(1+t)}{(1-t)} t^{\alpha-2}$$

Dividing by $t^{\alpha-2}$ leads to

$$ 0=\alpha (\alpha-1) + \frac{2(2-t)}{(1-t)} \alpha
+\frac{2(1+t)}{(1-t)} \\
=\alpha^2+\left(\frac{2(2-t)}{(1-t)}-1 \right) \alpha+\frac{2(1+t)}{(1-t)} \\
=\alpha^2+\frac{3-t}{1-t} \alpha+\frac{2(1+t)}{(1-t)}.$$

Now, in order to find the first solution I need to determine $\alpha \in R$. However, my attempts to solve the equation above have lead to solutions for $\alpha$ that depend on $t \in (0,1)$ which doesn't seem right. What am I missing here?

Body:
Taking partial fraction of each term we have

$u''(t)+\left(\dfrac{4}{t}-\dfrac{2}{t-1}\right)u'(t)+\left(\dfrac{2}{t^2}+\dfrac{4}{t}-\dfrac{4}{t-1}\right)u(t)=0$

Let $u(t)=t^kv(t)$ ,

Then $u'(t)=t^kv'(t)+kt^{k-1}v(t)$

$u''(t)=t^kv''(t)+kt^{k-1}v'(t)+kt^{k-1}v'(t)+k(k-1)t^{k-2}v(t)=t^kv''(t)+2kt^{k-1}v'(t)+k(k-1)t^{k-2}v(t)$

$\therefore t^kv''(t)+2kt^{k-1}v'(t)+k(k-1)t^{k-2}v(t)+\left(\dfrac{4}{t}-\dfrac{2}{t-1}\right)(t^kv'(t)+kt^{k-1}v(t))+\left(\dfrac{2}{t^2}+\dfrac{4}{t}-\dfrac{4}{t-1}\right)t^kv(t)=0$

$v''(t)+\dfrac{2k}{t}v'(t)+\dfrac{k(k-1)}{t^2}v(t)+\left(\dfrac{4}{t}-\dfrac{2}{t-1}\right)v'(t)+\left(\dfrac{4k}{t^2}-\dfrac{2k}{t(t-1)}\right)v(t)+\left(\dfrac{2}{t^2}+\dfrac{4}{t}-\dfrac{4}{t-1}\right)v(t)=0$

$v''(t)+\left(\dfrac{2k+4}{t}-\dfrac{2}{t-1}\right)v'(t)+\left(\dfrac{k^2+3k+2}{t^2}+\dfrac{2k+4}{t}-\dfrac{2k+4}{t-1}\right)v(t)=0$

Choosing $k=-2$ , the ODE becomes

$v''(t)-\dfrac{2}{t-1}v'(t)=0$

$\dfrac{v''(t)}{v'(t)}=\dfrac{2}{t-1}$

$\ln v'(t)=2\ln(t-1)+c$

$v'(t)=C(t-1)^2$

$v(t)=C_1(t-1)^3+C_2$

$\therefore u(t)=\dfrac{C_1(t-1)^3}{t^2}+\dfrac{C_2}{t^2}$


问:
Write out the multiplication table for the field $\mathbb{Z}[\alpha] := \mathbb{Z}_2[x]/(x^3+x+1)$.

Write out the multiplication table for the field $\mathbb{Z}[\alpha] := \mathbb{Z}_2[x]/(x^3+x+1)$.

Q1. What is the inverse of $\alpha$?

Q2. Determine a generator of the multiplicative cyclic group $\mathbb{Z}[\alpha]^\times$

I've written down the table, but how do I get the inverse of $\alpha$? $\alpha$ can be any number. If $\alpha$ is not an integer, how do I take mod $2$?

And for Q2, I can see that, in my table, all $8$ elements of $\mathbb{Z}_2[x]/(x^3+x+1)$ generate all elements of $\mathbb{Z}_2[x]/(x^3+x+1)$. For example, $x+1$ generates all eight elements. Is this because $x^3 + x + 1$ is irreducible?

答:
For Q1: Here $\alpha$ is not 'any number' - $\alpha$ designates a root of $x^3+x+1$, which has been adjoined to make this field.  In effect, $\alpha==x$.  So ask: what polynomial over $\mathbb{Z}_2$, when multiplied by x, yields a polynomial in $\mathbb{Z}_2[x]$ with remainder 1 modulo $x^3+x+1$?

And for Q2: Yes it is because $x^3+x+1$ is irreducible over $\mathbb{Z}_2$.


问:
Calculate $\lim _{ n\to \infty } \int _{ |x|<n }{ \int _{ |y|<n }{\sin(x^2+y^2)\,dx\,dy } } $
$$\lim _{ n\to \infty } \int _{ |x|<n }{ \int _{ |y|<n }{ \sin(x^2+y^2)\,dx\,dy } }$$

Seeing that $n\to \infty$ I deduce that yes $|x|<n$ and $|y|<n$ then $-\infty<x<\infty$ and $-\infty<y<\infty$

But I don't know if I can use these limits.

Another observation I see is in the argument of the function $\sin(x^2+y^2)$ Which makes me think that I can use polar coordinates.

Any suggestions on how to calculate such a limit? How could you write the limits of the integral?

答:
The integral under the limit is equal to $2 C_n S_n$, where $$C_n=\int_{-n}^n\cos t^2\,dt,\qquad S_n=\int_{-n}^n\sin t^2\,dt$$ (which is easy to see using $\sin(a+b)=\sin a\cos b+\cos a\sin b$).

And it is well-known that $\lim\limits_{n\to\infty}C_n=\lim\limits_{n\to\infty}S_n=\sqrt{\pi/2}$. Hence the given limit is $\pi$.


问:
Solving $x^3\equiv2\bmod151$ using Lagrange's Theorem

I'm trying to show that the congruence
$$x^3\equiv2\bmod151$$
has no solution by using Lagrange's Theorem (group theory).

I know that $151$ is prime, but I don't know how can I use the theorem in this case.

Am I supposed to create cyclic groups s.t. one is subgroup of the other? How can I do that?

Thank you in advance!

答: Since $151$ is a prime, $\mathbb{Z}/151\mathbb{Z}$ is a field, and $(\mathbb{Z}/151\mathbb{Z})^{\times}$ is a multiplicative group of order $150$. Suppose $x^3\equiv 2\pmod{151}$. Notice that $2$ has multiplicative order $15$ modulo $151$ (its order has to divide $150$, and so you can check with a calculator that $151$ does not divide $2-1,2^3-1,2^5-1,2^{10}-1$, while it divides $2^{15}-1$). Since the (multiplicative) subgroup of $(\mathbb{Z}/151\mathbb{Z})^{\times}$ generated by $x$ contains the subgroup generated by $2$, we deduce that the order of the subgroup generated by $x$ is divisible by $15$ and divides $150$. So it can be $15,30,75,150$. But no one of these is admissible. For example, if $x$ has order $30$, then $$1\equiv x^{30}\equiv (x^3)^{10}\equiv 2^{10}\not\equiv 1\pmod{151},$$ contradiction.


答: Hint: $x^3 \equiv 2 \bmod 151 \implies 1 \equiv x^{150} \equiv 2^{50} \bmod 151$. But $2^{50} \not\equiv 1 \bmod 151$.

Title: Solving $x^3\equiv2\bmod151$ using Lagrange's Theorem

Body:
I'm trying to show that the congruence
$$x^3\equiv2\bmod151$$
has no solution by using Lagrange's Theorem (group theory).

I know that $151$ is prime, but I don't know how can I use the theorem in this case.

Am I supposed to create cyclic groups s.t. one is subgroup of the other? How can I do that?

Thank you in advance!


问:
Summation using binomial coefficient
I am trying to calculate the following using binomial coefficients and summation, but my memory is standing in the way:

$$
   \sum_{k=1}^n {k} * 2 ^ {k - 1}
$$

Thanks!

With great help I got to:

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

Can you please confirm this?

答:
Hint: switch to polynomial functions.
$$
 P(x)=\sum_{k=1}^n kx^{k-1}
$$</p>

Title: Summation using binomial coefficient


问:
Linear map and Linear Independence proof
Suppose that $V$ and $W$ are vector spaces and $T : V → W$ is linear. If $u_1, . . . , u_n ∈ V$ and
$Tu_1, . . . , Tu_n$ are linearly independent, then $u_1, . . . , u_n$ are linearly independent.

So here is my attempt at the proof:

Assume that $Tu_1, . . . , Tu_n$ is linear independent, then $a_1Tu_1+. . .+a_nTu_n=0$, and $a_1,...,a_n=0$

Then $T(a_1u_1+...+a_nu_n)=0$

Using that fact that $T(0)=0$, $\ a_1u_1+. . .+a_nu_n=0$

So $u_1, . . . , u_n$ are linearly independent.

I have seen two solutions that they started the proof with assuming $\ a_1u_1+. . .+a_nu_n=0$, and Take T both sides by $\ T(a_1u_1+. . .+a_nu_n)=T(0)$ and prove the statement from there. Is my way correct or this is?Thanks

答:
Assume $T(\mathbf u_1),\dots,T(\mathbf u_n)$ are linearly independent and let $\mathbf 0$, $\vec 0_{\mathcal V}$, $\vec 0_{\mathcal W}$ denote the additive identity of $\Bbb F$, $\mathcal V$, $\mathcal W$ respectively where $\Bbb F$ is the ground field of $\mathcal V$.

$$\forall c_1,\dots,c_n\in\Bbb F\;\;\;\sum_{i=1}^nc_i\mathbf u_i=\vec 0_{\mathcal V}\implies\sum_{i=1}^nc_iT(\mathbf u_i)=\vec 0_{\mathcal W}\implies c_1=\cdots=c_n=\mathbf 0$$


答:
The field is $GF(8)=\Bbb Z_2[x]/\langle x^3+x+1\rangle$ and contains a zero of $x^3+x+1$, namely the residue class $\alpha = \bar x = x +\langle x^3+x+1\rangle$. Then $\alpha^3+\alpha+1=0$ in $GF(8)$.

The elements of $GF(8)$ are $0, 1, \alpha,\alpha^2,\alpha^3=\alpha+1,\ldots,\alpha^6, \alpha^7=1$.
So the inverse of $\alpha$ is $\alpha^6=(\alpha^3)^2 = (\alpha+1)^2 = \alpha^2+1^2 = \alpha^2+1$.

The multiplicative group $GF(8)^*$ is cyclic and has $\alpha$ as a generator, see above.

Body:
Write out the multiplication table for the field $\mathbb{Z}[\alpha] := \mathbb{Z}_2[x]/(x^3+x+1)$.

Q1. What is the inverse of $\alpha$?

Q2. Determine a generator of the multiplicative cyclic group $\mathbb{Z}[\alpha]^\times$

I've written down the table, but how do I get the inverse of $\alpha$? $\alpha$ can be any number. If $\alpha$ is not an integer, how do I take mod $2$?

And for Q2, I can see that, in my table, all $8$ elements of $\mathbb{Z}_2[x]/(x^3+x+1)$ generate all elements of $\mathbb{Z}_2[x]/(x^3+x+1)$. For example, $x+1$ generates all eight elements. Is this because $x^3 + x + 1$ is irreducible?


问:
Are there any Combinatoric proofs of Bertrand's postulate?
I feel like there must exist a combinatoric proof of a theorem like: There is a prime between $n$ and $2n$, or $p$ and $p^2$ or anything similar to this stronger than there is a prime between $p$ and $(\prod_p p) + 1$ (Euclid's theorem).

I was trying to prove one by the Sieve on this grid

```<code>    1     2     3     4 ... p
  p+1   p+2   p+3   p+4 ... 2p
 2p+1  2p+2  2p+3  2p+4 ... 3p
...........................
...........................
........................... p^2
p^2+1 p^2+2 p^2+3 p^2+4 ...
</code>```

but it did not work.

Do any good arguments like this exist? I don't expect anything as strong as Prime Number Theorem or even Bertrand, but surely a direct combination proof can prove that there are lots of primes?

答:
Interesting idea to use a grid.  I doubt it will work for this question, since writing things in that grid format in a sense splits into arithmetic progressions modulo $p$, and usually saying things about the primes in a progression is harder.

However, I cannot help but mention that using a grid like that was applied brilliantly by Maier to prove a counter-intuitive result, (using the Prime Number Theorem for Arithmetic Progessions) and the idea is now called the Maier Matrix Method.

(A small digression, but it is interesting!)

The Maier Matrix Method

There are questions regarding primes in short intervals, and we can ask ourselves what does $$\pi(x+y)-\pi(x)$$ look like?  To say anything meaningful, $y$ cannot be too small, but here lets suppose $y=\log^B(x)$ for some $B>2$.  Selberg proved that under the Riemann Hypothesis, we have $$\pi(x+y)-\pi(x)\sim \frac{y}{\log x}$$ as $x\rightarrow \infty$ for almost all $x$. (A set with density $\rightarrow 1$)  It was then conjectured that this asymptotic must hold for all $x$ which are sufficiently large.  (This conjecture was made for several reasons, one of which is that it is true under Cramer's probabilistic model)

In a surprising turn of events, Maier showed it was false, and that there exists $\delta>0$ and arbitrarily large values of $x_1$ and $x_2$ such that both $$\pi(x_1 +\log^B (x_1))-\pi (x_1)> (1+\delta)\log^{B-1}(x_1)$$and $$\pi(x_2 +\log^B (x_2))-\pi (x_2)< (1-\delta)\log^{B-1}(x_2)$$hold, despite the fact that the asymptotic holds for a set of density $1$.

He proved this using a method which is now called the "Maier Matrix Method."  Essentially, it is just drawing a grid which is similar to the one above, and then applying a clever combinatorial argument.  The columns are arithmetic progressions, and by PNT4AP, we can easily say things about them to understand the number of primes in the grid.  There is a little trick with oscillation of the Dickman Function, but then basically by the pigeon hole principle the question is solved.

I definitely think you might find this expository article by Dr.Andrew Granville to be interesting.  (It is quite readable, and gives an a more in depth, and very clear explanation)


问:
Vector similarity for prediction
I have vectors of same length consisting of 1 and 0. I am trying to find out how similar they are. So far I am using hamming distance that I calculate sum of one vector then sum of second vector and the difference between this is the difference of the days. With 1 and 0 it works pretty well.

My problem is that it doesn't reflect in any way where is the difference in the vectors and what is the variance of the error. I have thought of counting of how many 1 been misplaced to 1 of the next vector and how many 0 have been misplaced. It gives little bit more of information but still doesn't tell anything about the variance of the error.

The vectors are used to represent occupancy of house in time, with every 1 indicating that house is occupied and 0 that it is not. From this I am trying to predict how next day will look.

I also found out that approach to compare vectors is not the best one but rather to find a trend across the vectors is the best way to create effective predictions.

Can you hint me which ways are there to tackle this?

答: You could try Cosine similarity or the Jaccard index.</p>

Title: Vector similarity for prediction