答: Since $\left(1+\frac{1}x\right)^x$ converges to $e$, for any constant $\epsilon > 0$ there exists $M>0$ such that for all $x>M$, $$0 \leq |a_k| \leq \frac{e+\epsilon}{|x|}.$$ Then $a_k \to 0$ from sandwich lemma.

Title: Limit as $x\to \infty,$ of $\;\frac{(1+\frac{1}{x})^{x}}{x}$

Body:
I'm having some trouble evaluating the $\lim_{x\to\infty}\frac{\left(1+\frac{1}{x}\right)^{x}}{x}$ in the context of real analysis. Denote this sequence by $(a_k)$

If so, then I need to prove that for all $\varepsilon>0$

$$\left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}-0\right| <\varepsilon$$

We know that $\lim_{x\to\infty}1/x=0$, however $|1/x|\leq\left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}\right|$, so I'm not sure this information is useful. I considered the sequential characterization of continuity, so if the sequence above converges to 0, then $f((a_k))$ converges to $f(0)$, but there is also not an obvious choice of $f$. Any suggestions into how to evaluate the limit?


问:
Conditional Expectation: $E(X|Y) \geq Y$ and $E(Y|X) \geq X$ implies $X=Y$ a.e.
Suppose for random variables $X$ and $Y$ with finite expectation we he have $E(X|Y) \geq Y$ and $E(Y|X) \geq X$.  We want to show that $X=Y$ almost everywhere.

My attempt:

Let $A \in \sigma(X)$ and $B \in \sigma(Y)$.  Then we have

$\int_B X = \int_B E(X | Y) \geq \int_B Y$ and $\int_A Y = \int_B E(Y | X) \geq \int_A X$.

My idea now is to consider sets of the form $\{X>a>Y\}$ for $a$ rational but I'm having trouble with getting the other variable involved.  The issue seems to be that $\{X>a\}$ is in $\sigma(X)$ but not necessarily in $\sigma(Y)$ and it seems like we cannot get the other inequality working.  Any help would be appreciated.

答:
Rewriting what you have in the question body, we have:

$\int_B X - Y \geq 0$ and $\int_A X - Y \leq 0$ for $B \in \sigma(Y)$ and $A \in \sigma(X)$. We need to show that these are actually equalities, so let us just do one. Let $B$ be an arbitrary set, and note that $\int_{B^c} X - Y \geq 0$ since $B^c$ is in $\sigma(Y)$ as well. We then have (by some abuse of notation) $\int_{B^c} + \int_B = \int_\Omega$. So, we overall have

$$0 \leq \int_B X - Y = \int_\Omega X - Y - \int_{B^c} X - Y \leq 0$$

since the first term is negative ($\Omega \in \sigma(X)$) and the second one is positive. Thus, for any $B \in \sigma(Y)$, $\int_B X - Y = 0$. You can do the same argument for all $A$ in $\sigma(X)$ as well, so we have that $X = Y$ for all sets $A \in \sigma(X) \cup \sigma(Y)$. Now, if you want the full sigma algebra $\sigma(X, Y)$, let us note that $1_{A \cap B} = 1_{A} + 1_{B} - 1_{A \cup B}$, so you can show that $X = Y$ in an algebra generated by the two sigma algebras. You can then use monotone convergence to go to a monotone class containing these to show closure under countable unions and intersections. Then, monotone class theorem for sets should give you the overall result.

ContentLicense: CC BY-SA 4.0


问:
Find the length of the curve.
Let $x(t) = \sin(t)^4$, $y(t) = \cos(t)^2$, $t \in [0, \frac{\pi}{2}]$

We want to find this curve's length.

There lots of these type questions, but I have a problem with just counting an answer :

Here's what I know:
$l = \int_0^{\frac{\pi}{2}}\sqrt{x'(t)^2 + y'(t)^2}{\rm d}t$

$x'(t) = 4\sin^3(t)\cos(t), y'(t) = -2\cos(t)\sin(t)$

We substitute the resulting into the formula:
$l = \int_0^{\frac{\pi}{2}}\sqrt{16\sin^6(t)\cos^2(t) + 4\cos^2(t)\sin^2(t)}{\rm d}t = $

$= \int_0^{\frac{\pi}{2}}2\cos(t)\sin(t)\sqrt{4\sin^4(t) + 1}{\rm d}t$

But what can I do next to get an answer ? I don't know any technique for this. Can someone help me ?

答: HINT: Substitute $u = 4\sin^2(x) + 1$.

Title: Find the length of the curve.

Body: HINT: Substitute $u = 4\sin^2(x) + 1$.


答:
Put $s=4\sin^{2}t+1$. You get $\frac 1  4 \int_1^{5}\sqrt {s}ds$.

EDIT. It should have been $\sqrt {4\sin^{4}t+1}$ inside the integral instead of $\sqrt {4\sin^{2}t+1}$. So you have to make the substitution $s=4\sin^{4}t+1$ and use the fact that $\sin^{2}t= \sqrt {\frac {s-1} 4}$.

Title: Find the length of the curve.

Body:
Let $x(t) = \sin(t)^4$, $y(t) = \cos(t)^2$, $t \in [0, \frac{\pi}{2}]$

We want to find this curve's length.

There lots of these type questions, but I have a problem with just counting an answer :

Here's what I know:
$l = \int_0^{\frac{\pi}{2}}\sqrt{x'(t)^2 + y'(t)^2}{\rm d}t$

$x'(t) = 4\sin^3(t)\cos(t), y'(t) = -2\cos(t)\sin(t)$

We substitute the resulting into the formula:
$l = \int_0^{\frac{\pi}{2}}\sqrt{16\sin^6(t)\cos^2(t) + 4\cos^2(t)\sin^2(t)}{\rm d}t = $

$= \int_0^{\frac{\pi}{2}}2\cos(t)\sin(t)\sqrt{4\sin^4(t) + 1}{\rm d}t$

But what can I do next to get an answer ? I don't know any technique for this. Can someone help me ?


问:
Finding a specific polynomial function, with conditions.

Find a polynomial such that $f(x) = a + bx + cx^2 +dx^3 + ex^4$ such
that $f(1)=1, f(2) = -1, f(-1)=5, f(3) = -59, f(-2) = -29$

Any hints on how to approach this? I was thinking about plugging in the f value and then the result in order to build a matrix like this, but I am not sure if that is what I am supposed to do here.

$\displaystyle \begin{pmatrix}
a & b & c & d & e & 1\\
a & -b & -c & -d & -e & -1\\
a & -b & -c & -d & -e & 5\\
a & 3b & 3c & 3d & 3e & -59\\
a & -2b & -2c & -2d & -2e & -29
\end{pmatrix}$

does this make any sense?

答:
You need to calculate f(1), f(2), f(-1), f(3) and f(-2) to find the coefficients.

f(1) = a+b+c+d+e = 1

f(2) = a+2b+4c+8d+16e = -1

f(-1) = a-b+c-d+e = 5

f(3) = a+3b+9c+27d+81e= -59

f(-2)= a-2b+4c-8d+16e = -29

Next we take the coefficients and the result and we put them in a matrix:

A=

| 1  1  1  1  1   |

| 1  2  4  8  16  |

| 1 -1  1 -1  1   |

| 1  3  9 27  81  |

| 1 -2  4 -8  16  |

C will be the matrix with the outputs:

| 1|

|-1|

| 5|

|-59|

|-29|

B will be the matrix with the letters:

|a|

|b|

|c|

|d|

|e|

Now we need to solve the matrix equation A*B=C.

But because in matrix algebra there is no division operation, we need to multiply both sides with the inverse of the matrix A. We will call the inverse A'. So the equation becomes:

(A')(A)(B) = (A')*(C)

Since A' is the inverse of A, (A')*(A)= I and I is the identity matrix(a matrix that when its multiplied by another one doesnt change the result) So the equation is:

(B)=(A')*(C)

So now we need to calculate the inverse, and multiply it by the matrix C, and those thats all. To do that there are multiple ways and honestly its been too long since high school. But i leave you with this link: https://www.mathsisfun.com/algebra/systems-linear-equations-matrices.html


问:
Let X and Y be jointly distributed random variables having the joint probability density function
Let $X$ and $Y$ be jointly distributed random
variables having the joint probability
density function
[
f(x,y) = \begin{cases}
            \frac{1}{\pi}, &\text{if}\quad x^2 + y^2 \leq 1\\
             0, &\text{otherwise}\\
            \end{cases}

Then $P(Y > |X|)$ is

I tried writing integration for this but I am getting wrong. I tried writing cdf for this function as well but that did not help. How should I proceed further?

答:
A figure indicating the range of $(X, Y)$ often helps to determine the integral bounds and solve problems like this.

In this problem, the range of $(X, Y)$ is given by $0 \leq x^2 + y^2 \leq 1$, which is the unit circle (showing below).

Since the area of the unit circle is $\pi \times 1^2 = \pi$ and given that
$$f(x,y) = \frac{1}{\pi} \quad \text{if } x^2 + y^2 \leq 1,$$
and 0 otherwise, we conclude that $(X, Y)$ is uniformly distributed over the unit circle (think about why this is true).

Next, we want to find $P(Y > \left|X\right|)$. Since $(X, Y)$ is uniformly distributed, the problem reduces to finding the ratio between areas of the region defined by the event $\{Y > \left|X\right|\}$ (i.e. highlighted in light blue) and the unit circle. From the figure we can see that the highlighted region is in fact a quarter of the unit circle (the angle between $y = x$ and $y = -x$ is 90 degrees), we have
$$P(Y > \left|X\right|) = \frac{\frac{1}{4}\pi}{\pi} = \frac{1}{4}.$$

Remark: This approach makes use of the property of uniform distribution, which bypasses calculating the integral. If we do want to solve this problem using the integral approach, note that
\begin{align*}
P(Y > \left|X\right|) &= \iint_{\{y > \left|x\right|\}} f(x,y) dxdy\\
&= \iint_{\{y > \left|x\right|, \ x^2 + y^2 \leq 1\}} \frac{1}{\pi} dxdy\\
&= \frac{1}{\pi} \iint_{\{y > \left|x\right|, \ x^2 + y^2 \leq 1\}} dxdy\\
&= \frac{1}{\pi} \int_{\pi/4}^{3\pi/4} \int_0^1 rdrd\theta \quad (\text{polar coordinate})\\
&= \frac{1}{\pi} \int_{\pi/4}^{3\pi/4} \left[\frac{1}{2}r^2\right]\bigg|_0^1 d\theta\\
&= \frac{1}{\pi} \int_{\pi/4}^{3\pi/4} \frac{1}{2} d\theta\\
&= \frac{1}{2\pi} \left[\frac{3\pi}{4} - \frac{\pi}{4}\right]\\
&= \frac{1}{2\pi} \cdot \frac{\pi}{2}\\
&= \frac{1}{4}.
\end{align*}
More details on polar coordinate transformation can be found here.


问:
Roots of a integral having upper bound as variable

We are given above the graph of $f(x) = x \sin x$. Define $G(x) = \int\limits_0^x f(t) dt $. It is clear that since $G( \pi ) > 0$ and $G(2\pi) < 0$, then we have a root $c$ in $[\pi, 2 \pi ]$. Similarly, there is a root in the interval $[2 \pi, 3 \pi]$, same argument.

Here is the subtle part: Notice that $G(0) = \int_0^0 f = 0 $. Do we consider $x=0$ to be a root of $G(x)$? is it by convention or there any underlying issues to consider here?

Im uncomfortable about it since we do not know what happes as $x \to 0^-$. Any thoughts?

答:
As @heropup answered, you are looking for the zero's of function

$$F(x) = \sin (x)-x \cos (x) $$ which is an even function. For $x >0$, beside the trivial $x=0$, the roots are located closer and closet to $x_*^{(n)}=(2n+1)\frac \pi 2$.

We can approximate these roots building Taylor series around these points and using series reversion. This gives
$$x_n= t -\sum_{n=0}^\infty \frac {a_n}{t^{2n+1}} \qquad \text{where} \qquad t=(2n+1)\frac \pi 2$$ the first $a_n$'s making the sequence
$$\left\{1,\frac{2}{3},\frac{13}{15},\frac{146}{105},\frac{781}{315},\frac{16328}{346
   5},\frac{6316012}{675675},\frac{38759594}{2027025},\frac{9655714457}{241215975},\cdots\right\}$$

Transforming the series expansion into a Padé approximant gives with a good accuracy
$$x_n=t-\frac{5 \left(24 t^4-44 t^2+9\right)}{3 t \left(40 t^4-100 t^2+47\right)}$$


问:
necessary and sufficient conditions for weak convergence in $c_0$
First, some definitions:

A sequence $[x_n]$ in a normed linear space $X$ is said to converge weakly to an
element $x$ if for every $\phi \in X^*,\,\,  \phi(x_n) \rightarrow \phi (x),$ (where$X^*$ is the conjugate space of $X$.)

As for $c_0$, let $c_0$ denote the space of all real sequences that converge to zero, with the norm $\| x\| = \sup_n |x(n)|.$

Now, I need to show that in order for a sequence $[x_n]$ to converge weakly to an element $x$ in the space $c_0$, it is necessary and sufficient that the sequence be bounded and that (for each $i$) we have $\lim_{n \rightarrow \infty} x_n(i) =x(i).$

Here's my attempt at a proof:

First direction: assume that the sequence is weakly convergent, prove the boundedness and pointwise convergence:
We know that any weakly convergent sequence is bounded. Also, every continuous linear functional $\phi$ in $c_0$ can be written in the following form for some suitable point $\alpha$ in $l^1$. $$\phi(x) = \sum_{i=1}^{\infty } \alpha(i) x(i)$$
Similarly, $$\phi_n (x) = \sum_{i=1}^{\infty } \alpha^\prime (i) x_n(i).$$ Now, due to weak convergence, we have $lim_{n \rightarrow \infty} \phi_n(x) = \phi(x) $. We pick $\alpha(i) = \alpha ^ \prime (i) = e(i) $, where $e(i)$ is the $i$th basis for $c_0$. Thus, we can conclude $lim_{n \rightarrow \infty} x_n(i) = x(i).$

Am I right so far? And if so, I'd appreciate any hint/help on the inverse direction of the proof. Thanks.

答:
For the converse: Let $f \in c_0^*=\ell_1$. Then, there exists $z \in \ell_1$ such that
$$f(y)=\sum_{k =1}^\infty y(k) z(k), \ y \in c_0.$$
Let $\varepsilon>0$. Pick a sufficiently large integer $N$ such that
$$ \sum_{k>N} |z(k)| < ε/2M$$
where $M=\sup ||x_n||$. Since $x_n(k) \xrightarrow{n \to \infty} x(k)$ for all $k$, we get that
$$ \lim_{n\to \infty} \sum_{k=1}^N |x_n(k)-x(k)| \ |z(k)|=0.$$
Therefore,
\begin{align*}
|f(x_n)-f(x)| &= |f(x_n-x)| \\
&= | \sum_{k=1}^\infty (x_n(k)-x(k)) z(k) | \\
&\leq \sum_{k=1}^N |x_n(k)-x(k)| \ |z(k)| + \sum_{k>N} |x_n(k)-x(k) \ |z(k)|.
\end{align*}
Conclude that $ \limsup |f(x_n)-f(x) | \leq ε$.


答:
Yes, the first part is right.

Now suppose $(x_n)$ is bounded and $x_{ni} \to x_i$ for all $i$. Let $\phi$ be in the dual space so that we can write $\phi (x)=\sum \alpha_ix(i)$ with $\sum |\alpha_i| <\infty$. Let $\epsilon >0$ and choose $n$ such that $\sum\limits_{i=N+1}^{\infty}  |\alpha _i| <\epsilon$. Now $\sum\limits_{i=1}^{\infty}\alpha_i (x_{ni}-x_i)|=\sum\limits_{i=N}^{\infty} \alpha_i(x_{ni}-xi)|+\sum\limits_{i=N+1}^{\infty} \alpha_i(x_{ni}-x_i)|$. Show that the second term does not exceed $2M\epsilon$ where $M=\sup |x_{ni}|$. Can you finish?

Title: necessary and sufficient conditions for weak convergence in $c_0$

Body:
First, some definitions:

A sequence $[x_n]$ in a normed linear space $X$ is said to converge weakly to an
element $x$ if for every $\phi \in X^*,\,\,  \phi(x_n) \rightarrow \phi (x),$ (where$X^*$ is the conjugate space of $X$.)

As for $c_0$, let $c_0$ denote the space of all real sequences that converge to zero, with the norm $\| x\| = \sup_n |x(n)|.$

Now, I need to show that in order for a sequence $[x_n]$ to converge weakly to an element $x$ in the space $c_0$, it is necessary and sufficient that the sequence be bounded and that (for each $i$) we have $\lim_{n \rightarrow \infty} x_n(i) =x(i).$

Here's my attempt at a proof:

First direction: assume that the sequence is weakly convergent, prove the boundedness and pointwise convergence:
We know that any weakly convergent sequence is bounded. Also, every continuous linear functional $\phi$ in $c_0$ can be written in the following form for some suitable point $\alpha$ in $l^1$. $$\phi(x) = \sum_{i=1}^{\infty } \alpha(i) x(i)$$
Similarly, $$\phi_n (x) = \sum_{i=1}^{\infty } \alpha^\prime (i) x_n(i).$$ Now, due to weak convergence, we have $lim_{n \rightarrow \infty} \phi_n(x) = \phi(x) $. We pick $\alpha(i) = \alpha ^ \prime (i) = e(i) $, where $e(i)$ is the $i$th basis for $c_0$. Thus, we can conclude $lim_{n \rightarrow \infty} x_n(i) = x(i).$

Am I right so far? And if so, I'd appreciate any hint/help on the inverse direction of the proof. Thanks.


答:
Welcome to MSE!

$\left(\begin{array}{ccccc|c} 1&1&1&1&1&1\\ 1&2&4&8&16& -1\\ 1&-1&1&-1&1&5\\1&3&9&27&81&-59\\ 1&-2&4&-8&16&-29\end{array}\right)\simeq_{RREF} \left(\begin{array}{ccccc|c}1&0&0&0&0&1\\ 0&1&0&0&0& -5\\ 0&0&1&0&0&4\\0&0&0&1&0&3\\ 0&0&0&0&1&-2\end{array} \right)$

thus $a=1,b=-5,c=4,d=3,e=-2$, and I used an RREF calculator to find the latter matrix.

The first matrix is the matrix representation of the $5$ equations with $5$ unknowns $a,b,c,d,e$:

$$\begin{align}a+b+c+d+e &=1\\
\\
a+2b+4c+8d+16e & =-1\\
\\
a-b+c-d+e &=5\\
\\
a+3b+9c+27d+81e &=-59\\
\\
a-2b+4c-8d+16e &=-29\end{align}$$

Note that in the matrix representation we are only interested in manipulating the coefficients of these $5$ equations, without having to rewrite $a,b,c,d,e$ multiple times, each column is representative of $a,b,c,d,e$ thus you don’t need to plug in $\left(\begin{array}{ccccc|c} a&b&c&d&e&1\end{array}\right)$ for the first row, same for the other rows.

Check:

After plugging in the values of $a,b,c,d,e$
$$f(x)=1-5x+4x^2+3x^3-2x^4$$

and plugging in $x=1,2,-1,3,-2:$

$$f(1)=1-5+4+3-2= 1 \checkmark$$
$$f(2)=1-5(2) +4(2)^2 +3(2)^3-2(2)^4 = 1-10+16+24-32=-1\checkmark$$
$$f(-1)=1+5+4-3-2=10-5=5\checkmark$$
$$f(3)=1-5(3)+4(3)^2+3(3)^3-2(3)^4=1-15+36+81-162=22-81=-59\checkmark$$
$$f(-2)=1-5(-2)+4(-2)^2+3(-2)^3-2(-2)^4=1+10+16-24-32=-29\checkmark$$

Title: Finding a specific polynomial function, with conditions.

Body:
Find a polynomial such that $f(x) = a + bx + cx^2 +dx^3 + ex^4$ such
that $f(1)=1, f(2) = -1, f(-1)=5, f(3) = -59, f(-2) = -29$

Any hints on how to approach this? I was thinking about plugging in the f value and then the result in order to build a matrix like this, but I am not sure if that is what I am supposed to do here.

$\displaystyle \begin{pmatrix}
a & b & c & d & e & 1\\
a & -b & -c & -d & -e & -1\\
a & -b & -c & -d & -e & 5\\
a & 3b & 3c & 3d & 3e & -59\\
a & -2b & -2c & -2d & -2e & -29
\end{pmatrix}$

does this make any sense?


答:
$f(1)=1, f(2)=-1,f(-1)=5,f(3)=-59,f(-2)=-29$
Provided a linear $5\times 5$ sysytem of equation as $MV=U$
where $$M=\begin{pmatrix} 1 & 1 &1 &1 &1 \\ 1 & 2 & 4 & 8 & 16 \\ 1 & -1 & 1 &-1 & 1\\ 1 & 3 & 9 & 27 & 81 \\ 1 & -2 & 4 & -8 & 16 \end{pmatrix}, V=\begin{pmatrix} a \\ b\\ c\\ d\\ e \end{pmatrix}, U =\begin{pmatrix} 1 \\ 2 \\ -1 \\ 5 \\ -59 \\ -29 \end{pmatrix}.$$ Solving this equation by LinearSolve of Mathematica we get get $$V=\begin{pmatrix} 1 \\ -5 \\ 4 \\ 3 \\ -2 \end{pmatrix}$$
Finally, we get $f(x)=1-5x+4x^2+3x^3-2x^4.$

Mathe matica Command: LinearSolve[m,u]


问:
Have any notable mathematical constants been found which are known to be rational?
Has anyone else thought about this? I've browsed the Wikipedia for mathematical constants and only things I saw that were definitely rational AFAIK were 0, 1, 2, and 1/2, which don't really count.

I think 777480/8288641 counts also. Fermat's/ Euler's Diophantine problem.

Are there any questions which have been asked where a rational answer is expected to be the answer?

Seeking tags also because I'm unsure what area of math this relates to most.

答: From Wikipedia, Legendre's constant was first conjectured to be $1.08366...$ but later proved to be exactly $1$.


问:
Limit as $x\to \infty,$ of $\;\frac{(1+\frac{1}{x})^{x}}{x}$
I'm having some trouble evaluating the $\lim_{x\to\infty}\frac{\left(1+\frac{1}{x}\right)^{x}}{x}$ in the context of real analysis. Denote this sequence by $(a_k)$

If so, then I need to prove that for all $\varepsilon>0$

$$\left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}-0\right| <\varepsilon$$

We know that $\lim_{x\to\infty}1/x=0$, however $|1/x|\leq\left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}\right|$, so I'm not sure this information is useful. I considered the sequential characterization of continuity, so if the sequence above converges to 0, then $f((a_k))$ converges to $f(0)$, but there is also not an obvious choice of $f$. Any suggestions into how to evaluate the limit?

答:
$\lim_{x \rightarrow \infty} (1+1/x)^x=e.$

Hence  $x$ it is bounded above, say $M >0$ is a bound.

Then

$0<|(1/x)(1+1/x)^x| <|(1/x)M|;$

Take the limit $x \rightarrow \infty$ (Squeeze Theorem).


问:
Function to Check the Parity of an Integer
Is there any function, f(x), that outputs whether a given integer is odd or even? For example, if x were 1237, the function would output 1; and if it were 80, it would output 2.

I know this is quite easy to accomplish in programming but when it comes to paper and pen math, I am having difficulties.

答:
You can use $n-2\big\lfloor \frac{n}{2} \big\rfloor$, where $\lfloor n \rfloor$ is the floor function.

Also $\frac{1-(-1)^n}{2}$ and $\sin(\frac{\pi n}{2})$.


问:
$0^0 = 1$ or indeterminate?
Is $0^0 = 1$ or indeterminate?  I've heard a number of conflicting answers.
If it was indeterminate, how would

$$e^x = \sum_{n = 0}^\infty \frac{x^n}{n!}$$

make sense since $e^0 = 1$?

答:
The sum you've written is short hand for

$$e^x=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\ldots$$

You'll notice that I wrote $1$ and not $x^0$, so there is no problem with the equation $e^0=1$.

You are right, however, to say that $0^0$ is indeterminate.  In fact, for any $a\in\mathbb{R}$, notice the following:

$$\lim_{x\to 0}a(x^2)^{(x^2)}=a$$

ContentLicense: CC BY-SA 3.0

Title: $0^0 = 1$ or indeterminate?


答: For each $a\in\mathbb{Z}$ we define $f:\mathbb{Z}\to\{0,1\}$ by letting $f(a)$ be the remainder after dividing $a$ by $2$. This definition is possible due to the division algorithm. We have $f(odd)=1$ and $f(even)=0$.

Title: Function to Check the Parity of an Integer

Body:
Is there any function, f(x), that outputs whether a given integer is odd or even? For example, if x were 1237, the function would output 1; and if it were 80, it would output 2.

I know this is quite easy to accomplish in programming but when it comes to paper and pen math, I am having difficulties.


答:
There are whole sequences of important rational numbers. See these pages on "integer" sequences (although they may have some non-integer rational values, e.g. the earliest Bernoulli numbers, which are the basis for results such as the famous $-\tfrac{1}{12}$ result @jojobo mentioned) & this page for rational ones.

Then there are quantities of the form $\zeta(2n)/\pi^{2n}$ with $n$ a positive integer.

One could argue every rational $s$ is important for a suitably chosen complex analysis problem due to factors of $e^{2\pi is}$ or $(1-e^{2\pi is})^{\pm1}$.

Title: Have any notable mathematical constants been found which are known to be rational?

Body:
Has anyone else thought about this? I've browsed the Wikipedia for mathematical constants and only things I saw that were definitely rational AFAIK were 0, 1, 2, and 1/2, which don't really count.

I think 777480/8288641 counts also. Fermat's/ Euler's Diophantine problem.

Are there any questions which have been asked where a rational answer is expected to be the answer?

Seeking tags also because I'm unsure what area of math this relates to most.


问:
Composition of a harmonic function with a holomorphic function is still harmonic

If $f$ is a harmonic function in a domain $D \subset \mathbb{C}$, and $g$ is a conformal mapping of a domain $D_0$ onto $D$, is $f \circ g$ harmonic in $D_0$?

I noticed this question while reading several pdf of lecture notes, and I'm having trouble figuring it out. Can anyone help?

Thank you so much!

答:
I was wondering the same question and reached here.

Here is how I thought about it:

Let $u$ be harmonic. And let $v$ be its harmonic conjugate. Then $f(z) = u(z) + iv(z)$ is analytic.
If $g$ is another analytic function such that range of $g$ is contained in the domain of $f$ then
$$f(g(z)) = u(f(z)) + iv(g(z))$$
is also analytic with $Re (f(g(z))) = u(f(z))$.


答: It really depends on what the context is. There are very good reasons to define $0^0 = 1$ like the one you just mentioned. Refer to here for a list of arguments for and against.</p>

Title: $0^0 = 1$ or indeterminate?

Body:
Is $0^0 = 1$ or indeterminate?  I've heard a number of conflicting answers.
If it was indeterminate, how would

$$e^x = \sum_{n = 0}^\infty \frac{x^n}{n!}$$

make sense since $e^0 = 1$?


问:
Test for convergence of a series
How to test the convergence of the series $\sum_{n=1}^{\infty}(n!)^{\frac{1}{n}}$?

I think ratio test or comparison test is doable. But for ratio test, I cannot calculate the limit value.

答:
$$a_n=(n!)^{\frac{1}{n}}\implies \log(a_n)=\frac{1}{n}\log(n!)$$ Using Stirling approximation twice and continuing with Taylor series, you would find that

$$\log(a_{n+1})-\log(a_n)=\frac{1}{n}-\frac{\log (2 \pi  n)}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{a_{n+1}}{a_n}=e^{\log(a_{n+1})-\log(a_n)}=1+\frac{1}{n}+\frac{1-\log (2 \pi  n)}{2 n^2}+O\left(\frac{1}{n^3}\right)$$

Title: Test for convergence of a series


问:
If $n-1$ integers are so that no partial sums are divisible by $n$, then the integers are congruent mod $n$ to one number coprime to $n$.
Let $a_1,a_2,...,a_{n-1}$ be integers.  Suppose that there is no subset $S\ne \emptyset$ of $\{1,2,...,n-1\}$ such that $\sum_{k\in S}a_k\equiv 0\pmod{n}$.  Show that there exists an integer $a$, $0< a< n$ st $\gcd(a,n)=1$ and $a_1,a_2,...,a_n\equiv a\pmod{n}$.

Source : https://artofproblemsolving.com/community/c6h626708p3759215 (mavropnevma's comment)

My attempt : Let $s_k=a_1+a_2+...+a_k$.  Then $s_1,s_2,...,s_{n-1}$ form a permutation of $1,2,...,n-1$ (mod n).  Let $s_0=0$ and consider the subscripts mod $n$.  Then the condition that $\sum_{k\in S}a_k\ne 0\pmod{n}$ for all $S\subseteq\{1,2,...,n-1\}$ st $S\ne \emptyset$ implies $\sum_{k\in S}s_k \ne \sum_{k\in S}s_{k-1}\pmod{n}$.  This is where I'm stuck.

答:
For a subset $S$ of $1\ldots n-1$ let $f(S)= \{ \sum_{j\in T} a_j, \emptyset\subsetneq T\subset S\}$.

Claim: if $i\not \in S$ then $$|f(S \cup i)|\ge |f(S)|+1$$
Proof: $a_i+\sum_{j\in S} a_j$ can't be in $f(S)$, as otherwise we'd have $a_i+\sum_{j\in S} a_j=\sum_{j\in T} a_j$ for some $T\subset S$ so $a_i+\sum_{j\in S,\not \in T} a_j = 0$.

If some $a_j\ne a_1$ then $|f(\{1,j\}|=3$ so that for $1,j\not \in S$:
$|f(\{1,j\}\cup S)|\ge 3+|S|$ which gives that $f(\{1,\ldots,n-1\})$ contains the whole of $0\ldots n-1$.


答: Let's assume $n$ is integer. Your parity function can be expressed as $p \equiv n $ mod $2$, with $p = 0$ for $n$ even and $p = 1$ for $n$ odd.


问:
Let $f : \mathbb{Z} \to \mathbb{R}, f(x)=\begin{cases}\frac1x, x >0 \\ x , x \le 0 \end{cases}$ show that $f$ is a continuous injection.

Let $f : \mathbb{Z} \to \mathbb{R}, f(x)=\begin{cases}\frac1x, x >0 \\ x , x \le 0 \end{cases}$ show that $f$ is a continuous injection.

Starting from continuity for the case $x  \le 0$. Let $\varepsilon >0$ and $\delta = \varepsilon$. Now $$|f(x)-f(0)| = |x-0|=|x|<\delta = \varepsilon$$

For $x >0 $. Let $\varepsilon >0$ and $\delta =\varepsilon$ I have $$|f(x)-f(0)| = |\frac1x-0| = \frac1x < \delta = \varepsilon$$

So $f$ is continuous. For injectivity let $a,b \in \mathbb{Z}$ now when $x >0$ I have $$f(a)=f(b) \\\frac1a = \frac1b \implies a = b$$

For $x \leq 0$ I have $$f(a)=f(b) \iff a=b$$ so $f$ is also injective. Is this a working solution?

答:
Let $X$ be a topological space such that each one point sets of $X$ is open and let $Y$ be any topological space. Then any function $f:X\to Y$ is continuous.

Note that any subset of $X$ is open. The mentioned topology is known as discrete topology on $X$.

Definition of continuity to be used: Inverse image of open set is open.

Given any open set $V$ of $Y$, $f^{-1}(V)=\{ v\in X: f(v)\in V\}$ which is open in $X$. Consequently, $f$ is continuous.

For your example: $\mathbb{Z}$ equipped with the usual topology is discrete, therefore, the given function is continuous.


问:
How can I prove a function is bijective and get its inverse
I have the following function $$f:\Bbb{R}\rightarrow\Bbb{R^+}$$

which is defined as $$f(x) = 2^x$$

How do I show that this function is a bijection and find its inverse?

答:
First we prove that the function is onto. Let us assume that it is into.

 Then there exists some real value $y\in\Bbb R^+$ such that there is no real value of x for the condition $$f(x)=2^x=y$$
Taking $log_2$ on both sides, $$x=\log_2 y$$
Since $y>0$, there is a real, defined value of $x$. This is a contradiction. Thus our assumption is incorrect and the function is onto.

Now we prove that the function is one-one. Let us assume that it is many-one.

 Then we can take two real values $x_1$, $x_2$ $\in\Bbb R$ such that $x_1\neq x_2$ and  $f(x_1)=f(x_2)$, i.e. $$2^{x_1}=2^{x_2}$$ $$2^{x_1-x_2}=1$$
$$x_1-x_2=0$$
$$x_1=x_2$$
This is a contradiction. Thus our assumption is incorrect and the function is one-one.

 Since the function is both one-one and onto, it is a bijective function.
thus we can find its inverse.
$$$$
To find the inverse, we take $y=f(x)$ and then interchange $x$ and $y$ to get
$$x=2^y$$
$$\log_2 x=y$$
Therefore the inverse function of f(x) is
$$f(x)=\log_2 x$$


问:
Angular distribution of lines passing through two squares.
Let's say I've got two squares with side length $d$ that are held parallel at a distance $m$ apart.

Suppose that particles are randomly falling from above in such a way that the polar angle $\varphi$ of the trajectory of the particles has a probability distribution proportional to $\operatorname{cos}^2(\varphi)$, while the azimuthal angle is uniformly distributed.

A particle is admissible if it passes through both squares.

Can we find a closed form expression for the probability distribution of the polar angle of admissible particles?  (or at least a good numerical solution?)

My attempt at a stronger formulation:

First, let's normalize the problem by letting $\alpha=m/d$ and assuming each square has unit side length.  We may assume that the point at which the particle passes through Panel $1$ (the upper panel) is uniformly distributed, that is, there are a uniformly distributed variables $X_1$ and $Y_1$, each between $0$ and $1$, and the point of contact on the plane of panel $1$ is $$\mathfrak{X}_1=\left(\begin{array}{c}X_1\\Y_1\\ \alpha\end{array}\right).$$  In order to be admissible, a particle must pass through Panel $1$, and we are only interested in admissible particles, so this is our universe of particles.

Let's find the point of contact $\mathfrak{X}_2$ on the plane of Panel $2$ (the lower panel).  We need $$\left(\begin{array}{c}X_1+t\operatorname{sin}(\varphi)\operatorname{cos}(\theta)\\ Y_1+t\operatorname{sin}(\varphi)\operatorname{sin}(\theta)\\ \alpha+t\operatorname{cos}(\varphi)\end{array}\right)=\left(\begin{array}{c}X_2\\Y_2\\0\end{array}\right)$$
so $t=-\alpha\operatorname{sec}(\varphi)$, from which we have $$\mathfrak{X}_2=\left(\begin{array}{c}X_1-\alpha\operatorname{tan}(\varphi)\operatorname{cos}(\theta)\\Y_1-\alpha\operatorname{tan}(\varphi)\operatorname{sin}(\theta)\\0\end{array}\right).$$
So, the question is, what is the probability distribution of $\varphi$ for the lines satisfying
$$\begin{array}{rcl}0 \leq& X_1-\alpha\operatorname{tan}(\varphi)\operatorname{cos}(\theta) &\leq  1 \\
0 \leq& Y_1-\alpha\operatorname{tan}(\varphi)\operatorname{sin}(\theta) &\leq 1 \end{array}
$$
given that $$\begin{array}{rl}X_1 \text{ and }Y_1& \text{are uniformly distributed from }0\text{ to }1 \\ \theta &\text{is uniformly distributed from }0\text{ to }2\pi \\ \varphi & \text{has probability distribution }\frac{4}{\pi}\operatorname{cos}^2\varphi\text{ from }\frac{\pi}{2}\text{ to }\pi\end{array}$$
?

EDIT 2:

So I ran a simulation, yielding the following histogram (displayed below with $3$ different bin widths).  I haven't fit a curve to this yet, but maybe it will give us a hint.  Here $\alpha$ is about $6$.

答:
Use Bayes' theorem:
$$ P(\phi\mid\text{admissible}) = \frac{P(\text{admissible}|\phi)P(\phi)}{P(\text{admissible})}. $$

Unless I am misreading your question (what do you do about rays that are parallel to an admissible ray? I am assuming here they are admissible, so only the direction or a ray matters), a ray parametrized by
$$ (t \sin\phi \cos\theta, t \sin\phi \sin\theta, t \cos\phi), \qquad t\in\mathbb{R}, \quad 0\leq\phi\leq\frac\pi2, \qquad 0\leq\theta\leq2\pi $$
is admissible if at the point $m = z(t) = t\cos\phi$, we have
$$ |t\sin\phi\cos\theta| \leq d, \quad\text{and}\quad |t\sin\phi\sin\theta| \leq d, $$
because if it passes through the two squares, the most it can move at most $d$ in either $x$- or $y$-direction as it moves a distance $m$ in $z$-direction. This condition is equivalent to
$$ |\tan\phi\cos\theta| \leq d/m, \qquad |\tan\phi\sin\theta| \leq d/m = \alpha, $$
which is equivalent to
$$ |\tan\phi|^2 \leq 2\alpha^2 = (\tan\phi_0)^2, \qquad \max(|\cos\theta|,|\sin\theta|) \leq \frac{\alpha}{|\tan\phi|}. $$

Finally, $ P(\phi) = \frac{4}{\pi}\cos^2\phi$, and, setting $\beta=\alpha/|\tan\phi|$, $P(\text{admissible}) = \text{const}$, and
$$ P(\text{admissible}|\phi) = [\phi\leq\phi_0] P\left( \max(|\cos\theta|,|\sin\theta|) \leq \beta \right)
\\ = [\phi\leq\phi_0] 4 P(\theta\geq\arccos\beta, \theta\leq\arcsin\beta\mid0\leq\theta\leq\pi/2)
\\ = [\phi\leq\phi_0]([\beta\geq1] + \frac{2}{\pi}(\arcsin\beta-\arccos\beta)[\beta\leq1]). $$

Putting things together,
$$ P(\phi\mid\text{admissible}) \propto \cos^2\phi \times [\phi\leq\phi_0]([\beta\geq1]+\frac{2}{\pi}[\beta\leq1](\arcsin\beta-\arccos\beta)). $$
The constant of proportionality is found by integrating the rhs over $0\leq\phi\leq\frac\pi2$ and setting it to $1$, but probably doesn't have a closed form.


问:
Why is square root by long division found so?
We were taught the long division method of finding square root in junior classes. The  logic behind the method used to be unclear, it remains so even now! However, we learnt and practiced the algorithm well. To me it appears that the identity $$(a+b)^2=a^2+2ab+b^2$$ lies behind the method. However this seems to be only the minimal idea. So extensions and generalizations are welcome here. See the picture below as I can't type this stuff. Some one else may edit it properly.

答:
Good explanation from geimetey. My original submission has been done by my friend Zaffar. That's because I am not familiar with formalities if stack exchange.
Few points I clarify in this context:
(1) The division method can be extended for any n-root. Sat for cube root :
Start with a number bB+s where B is the base; a and b are integers. (bB +a) ^3 =(bB)^3 + 3a(bB)^2 + 3(bB)a^2 + a^3.
Let us group three at a time from right. So, two groups. Left group has only (bB)^3 and right group has 3a(bB)^2 + 3(bB)a^2 + a^3. One has to find suitable b such that right group matches. Fir actual numbers, one has to search largest close number for a and b at every step.
Generalising the geometry approach, this can thenbe compared with funding suitable 3D visualisation where in stead of area, it willill be  volume.
(2) For any n-root, the method is similar. For visualisation, one can conceive n dimensional hypercubes.</p>


问:
$\log_x(y)=m^4$ and $\log_y(x)=5/(m^3)$, what is the value of $m$?
$\log_x(y)=m^4$ and $\log_y(x)=5/(m^3)$, what is the value of $m$? </p>

答: Hint: $$\log_a b = \frac{\log b}{\log a}$$


问:
nth term of $2,2+\frac{1}{2},2+\frac{1}{2+\frac{1}{2}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}...$
What is the nth term of the sequence: $$2,2+\frac{1}{2},2+\frac{1}{2+\frac{1}{2}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}...$$

in terms of $s_{n-1}$.

I have tried for some time to calculate this but to no avail. Hopefully somebody with a better understanding of the world of sequences could help/point me in the right direction!

答:
In my answer to this question, I detailed the steps for solving a first-order rational difference equation such as
$${ a_{n+1} = \frac{ma_n + x}{a_n + y} }=m+\frac{x-m y}{a_n+y}$$
For your case $m=2$, $x=1$ and $y=0$. So, using the initial condition,
$$a_n=\frac{\left(1+\sqrt{2}\right)^n-\left(1-\sqrt{2}\right)^n } { \left(1+\sqrt{2}\right) \left(1-\sqrt{2}\right)^n+\left(\sqrt{2}-1\right)\left(1+\sqrt{2}\right)^n}$$

Edit

In the documentation of sequence $A000129$ in $OEIS$, there is superb formula given by Peter Luschny in year $2018$. It write
$$a_n=\frac 1{\sqrt{2}}\, e^{\frac{i \pi  n}{2}}\,\sinh \left(n \cosh ^{-1}(-i)\right)$$


问:
Pseudocompact metacompact Tychonoff spaces are compact
I was working in some exercises of the famous book Extensions and absolutes of Hausdorff Spaces by Porter and Woods but I'm stuck in some parts of the exercise 6N. The full exercise:

For (1), is clear that $\mathcal{U}'$ is an open cover of $Y$. Suposse that $\mathcal{U}'$ is not point-finite. Take $p\in\beta X\setminus X$ and $\{U_n'\mid n<\omega \}\subseteq \mathcal{U}'$ an infinite collection such that $p\in\bigcap_{n<\omega}U_n'=U$. Then $U$ is a $G_\delta$ set in $\beta X=\upsilon X$ by the pseudocompactness of $X$ and therefore $U\cap X\neq\emptyset$. But then $\bigcap_{n\in\omega} U_n\cap X\neq\emptyset$ and this contradicts the fact that $\mathcal{U}$ is a point-finite famly of $X$.

For (2), is clear that $F_n$ is a cover of $Y$. Take $n\in\mathbb{N}$ and $z\in Y\setminus F_n$. Then $|(\mathcal{U}')_z|>n $ and therefore $(\mathcal{U}')_z=\{U_1',U_2',\dots,U_m' \}$ with $m>n$. Define $V=\bigcap_{i=1}^{m}U_i'$. $V$ is an open set of $Y$ and if $w\in V$ then $|(\mathcal{U}')_w|>n$. Then $z\in V\subseteq Y\setminus F_n$. Therefore $F_n$ is a closed set of $Y$.

For (3). Here, I followed the hint. As $Y$ is an open subset of $\beta X$ and $\beta X$ is compact then $Y$ is locally compact and clearly $T_2$. Then $Y$ is a Baire space and we can conclude that $\bigcup\{\operatorname{int}_{Y}(F_n)\mid n\in\mathbb{N} \}$ is dense in $Y$. How can I conclude from here that $W=\bigcup\{\operatorname{int}_{Y}(F_n)\setminus F_{n-1} \}$ is dense in $Y$? The first idea that came to my mind was to consider an open set $U$ of $Y$ and prove that for all $n\in\mathbb{N}$, $U\setminus F_n\neq\emptyset$. If it is true, $U\setminus F_n$ is an open subset of $Y$ and by the hint we are done. But I think that this is so hard to prove or, actually, is not true. Any hint?

For (4), I followed the hint and made the inductive construction that suggests but in the last part, where I need to prove that $\{V_n\mid n\in\mathbb{N} \}$ is an infinite and locally finite family is where I am stucked. Clearly every $V_n$ is an open subset of $X$ because is the intersection of two open sets of $\beta X$ with $X$. Any idea or suggestion? I tried to prove that if $[\bigcap(U')_{y(n)}]\cap [\bigcap(U')_{y(m)}]\neq\emptyset$ then $y(n)=y(m)$ but I dont know if it is true.

With the another parts of the exercise I'm done and I don't have any trouble. Thanks. I appreciate any help.

Edit. The inductive construction that is suggested in (4):

First, asumme that the union of every finite subfamily of $\mathcal{U}'$ is not dense en $Y$. Take $y(1)\in W$ and consider $\mathcal{U}'_{y(1)}$. This set is a finite subfamly of $\mathcal{U}'$ and therefore there exist $V\subseteq Y$ and open set of $Y$ such that $V\cap \bigcup \mathcal{U}'_{y(1)}=\emptyset$. As $\mathcal{U}'$ is an open cover of $Y$, take $U_2'\in\mathcal{U}'$ such that $U_2'\cap V\neq\emptyset$. Then $U_2'\cap V$ is a non empty open set. Since $W$ is dense (by (3)) then $U_2'\cap V\cap W\neq\emptyset$. Take $y(2)\in U_2'\cap V\cap W$. This $y(2)$ have the desired properties and this completes the basis of induction. Now, suposse that we have been constructed $y(1),\dots,y(n)$ with the desired property. Observe that $\{\bigcup (\mathcal{U}')_{y(k)}\mid k\leq n \}$ is a finite subfamly of $\mathcal{U}'$. Doing the same as in the basis of induction, we can construct a point $y(n+1)$ such that $y(n+1)\in W\setminus \bigcup\{\bigcup (\mathcal{U}')_{y(k)}\mid k\leq n \}$. Now our construction is complete. Let $n\in\mathbb{N}$. As $y(n)\in W$ then there exist $m(n)\in\mathbb{N}$ such that $y(n)\in \operatorname{int}_{Y}(F_{m(n)})\setminus F_{m(n)-1}$.

答:
I agree with $(1)$, where you use that $\beta X = \nu X$ for $X$ pseudocompact and so $X$ is $G_\delta$-dense in $\beta X$ (Thm. 5.11(b) is also used here). So indeed $\mathcal{U}'$ is a point-finite cover of $Y$.

$(2)$: That the $F_n$ are closed is indeed witnessed by the intersection of $\mathcal{U}'_y$ for $y \notin F_n$. Also, the $F_n$ are increasing in $n$ by definition and they cover $Y$ by point-finiteness of $\mathcal{U}'$ (it's good to be explicit).

$(3)$: That $\bigcup_n \operatorname{int}_Y(F_n)$ is dense in $Y$ follows more subtly: suppose not, then some non-empty $Y$-open $O$ misses that union, but then this contradicts the Baire theorem applied to $O$ (which is also locally compact Hausdorff, hence a Baire space), as the $F_n \cap O$ then are nowhere dense in $O$ but cover $O$, making $O$ first category in itself.

So given a non-empty $O$ open in $Y$, we know $O$ intersects $\bigcup_n \operatorname{int}_Y(F_n)$ so there is minimal $n_0$ so that $O \cap \operatorname{int}_Y(F_n) \neq \emptyset$ and then $O$ also intersects $\operatorname{int}_Y(F_{n_0}) \setminus F_{n_0-1}$ by that minimality, proving the claim that $W$ is dense. Openness of $W$ is obvious.

The recursive construction  of the $y(n)$ you gave is fine IMO, so the only question is local finiteness of the $(V_n)$, $n \in \Bbb N$.

I'm still thinking on $(4)$, but I made it a Community wiki so others (Brian maybe?) can add to and improve the answer.


问:
proving $|| \vec{x} \times \vec{y}||^{2} = ||\vec{x}||^{2}||\vec{y}||^{2} - (\vec{x} \cdot \vec{y})^{2}$ using index notation
R.T.P  $|| \vec{x} \times \vec{y}||^{2} = ||\vec{x}||^{2}||\vec{y}||^{2} - (\vec{x} \cdot \vec{y})^{2}$ using index notation.

where $\epsilon_{ijk} = \begin{cases} 1 : (i,j,k) \in \lbrace (1,2,3);(2,3,1);(3,1,2) \rbrace \\ -1: (i,j,k) \in \lbrace (1,3,2);(2,1,3);(3,2,1) \rbrace \\ 0: i=j;i=k;j=k  \end{cases}$

initially i used the fact that:

$\vec{x} \times \vec{y} = \epsilon_{ijk} \vec{e_{i}} x_{j} y_{k}$

and that

$|| \vec{x} \times \vec{y} || = \epsilon_{ijk} x_{j} y_{k}$

$|| \vec{x} \times \vec{y} ||^{2} = (\epsilon_{ijk} x_{j} y_{k})^{2}$

im unable to split it into 2 terms that is the main issue.

答:
Observe that
\begin{equation*}
\begin{split}||x||^2||y||^2 -(x\cdot y)^2 & = \left(\sum_{j}x_j^2\right)\left(\sum_{k}y_k^2\right) - \left(\sum_{l}x_ly_l\right)^2 \\
&= \sum_{j\neq k}\left(x^2_jy_k^2 + x_k^2y_j^2-2x_jy_jx_ky_k\right) \\
& = \sum_{j\neq k}(x_jy_k - x_jy_k)^2 \\
& = \sum_{i}\left(\sum_{j}\sum_{k} \epsilon_{ijk}x_jy_k\right)^2 =||x\times y||^2
\end{split}
\end{equation*}


答:
First, observe that the dot product $\vec{x}\cdot\vec{y}$ of two vectors can be written as
$$\vec{x}\cdot\vec{y}=\delta_{ij}x_iy_j=x_iy_i,$$
and that the square of the norm $|\vec{x}\rvert^2 = \vec{x}\cdot \vec{x}$. Now,
\begin{align}\lvert\vec{x}\times \vec{y}\rvert^2&=(\vec{x}\times \vec{y})\cdot (\vec{x}\times\vec{y})\\
&= (\varepsilon_{ijk}x_jy_k)\:(\varepsilon_{ilm}x_ly_m) \\
&= (\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl})\:x_jx_ly_ky_m \\
&= x_jx_jy_ky_k-x_jy_jx_ky_k \\
&=|\vec{x}|^2|\vec{y}|^2-(\vec{x}\cdot\vec{y})^2.
\end{align}
We have also used the following identity:
$$\varepsilon_{ijk}\varepsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}.$$

Title: proving $|| \vec{x} \times \vec{y}||^{2} = ||\vec{x}||^{2}||\vec{y}||^{2} - (\vec{x} \cdot \vec{y})^{2}$ using index notation

Body:
R.T.P  $|| \vec{x} \times \vec{y}||^{2} = ||\vec{x}||^{2}||\vec{y}||^{2} - (\vec{x} \cdot \vec{y})^{2}$ using index notation.

where $\epsilon_{ijk} = \begin{cases} 1 : (i,j,k) \in \lbrace (1,2,3);(2,3,1);(3,1,2) \rbrace \\ -1: (i,j,k) \in \lbrace (1,3,2);(2,1,3);(3,2,1) \rbrace \\ 0: i=j;i=k;j=k  \end{cases}$

initially i used the fact that:

$\vec{x} \times \vec{y} = \epsilon_{ijk} \vec{e_{i}} x_{j} y_{k}$

and that

$|| \vec{x} \times \vec{y} || = \epsilon_{ijk} x_{j} y_{k}$

$|| \vec{x} \times \vec{y} ||^{2} = (\epsilon_{ijk} x_{j} y_{k})^{2}$

im unable to split it into 2 terms that is the main issue.


问:
Two questions about Euler's number $e$
I am on derivatives at the moment and I just bumped into this number $e$, "Euler's number" . I am told that this number is special especially when I take the derivative of $e^x$ , because its slope of any point is 1. Also it is an irrational ($2.71828\ldots$) number that never ends, like $\pi$.

So I have two questions, I can't understand

What is so special about this fact that it's slope is always 1?

Where do we humans use this number that is so useful, how did Mr Euler come up with this number?

and how come this number is a constant? where can we find this number in nature?

答:
Just a slight correction, as Jon Claus notes about the derivative of $e^x$:

what you may be remembering is that "$e$ is the unique real number such that the value of the derivative (slope of the tangent line) of the function $f(x) = e^x$ at the point $x = 0$ is equal to $1$.

See the Wikipedia article on Euler's number $e$ for more fascinating information:

The number e is the unique positive real number such that

$$\frac{d}{dt}e^t = e^t.$$

The number e is the unique positive real number such that

$$\frac{d}{dt} \log_e t = \frac{1}{t}.$$

The following three characterizations can be proven equivalent:

The number e is the limit

$$e = \lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n$$

Similarly:

$$e = \lim_{x\to 0} \left( 1 + x \right)^{1/x}$$

The number e is the sum of the infinite series

$$e = \sum_{n = 0}^\infty \frac{1}{n!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots$$

where $n!$ is the factorial of n.

The number e is the unique positive real number such that

$$\int_1^e \frac{1}{t} \, dt = 1.$$

The number $e$ is of eminent importance in mathematics, alongside $0, 1, \pi, \;\text{and}\; i.$ All five of these numbers play important and recurring roles across mathematics, and are the five constants appearing in one formulation of Euler's identity: $$e^{i\pi} + 1 = 0$$ Like the constant $π, e$ is irrational: it is not a ratio of integers; and it is transcendental: it is not a root of any non-zero polynomial with rational coefficients.

Title: Two questions about Euler's number $e$


问:
Definite integral between variables
I am trying to understand the book, when it says that: "Integrating between time 0 and time T, we get.."

What I need to integrate is this: $\frac{d S}{S}=\mu d t$

And when this is integrated between 0 and T, the book states that we get: $S_{T}=S_{0} e^{\mu T}$

I have tried solving this by doing definite integration, but I have no idea how to get to this result.

If you find that some information is missing in the post, it is because i didn't know, and probably why i couldn't figure it out.

答:
Your integral should look equivalent to
$$
\int_{S_0}^{S_T} \frac{dS}{S}= \mu\int_{0}^{T}dt
$$
where $S$ has been labelled with subscripts corresponding to the time. The left-hand-side is a "standard" integral ( a log).

Title: Definite integral between variables


问:
Smallest subgroup of $A_{5}$ containing the permutation $(12)(13)$.
I need to find the smallest subgroup of $A_{5}$ that contains the permutation $(12)(13)$ and to also explain why.
I guess we can write this permutation as $(123)$ and I guess, according to a list of subgroups that I found, that the smallest would be $A_{3}$ in $A_{5}$.

If somebody could help me out with some reasoning or hints to understand this, I would appreciate it a lot.

It's important to say that this is my first ever algebra course, so it could be nice if the help would be kept simple and understandable for beginners.

Thanks in advance

答:
You can see that,  order of the element $(12)(13)=(132)\ne(123)$ is 3.
Therefore, there are only 3 elements in the smallest subgroup containing $(12)(13)$
The required group is the one that generated by (12)(13), consisting of $(12)(13),((12)(13))^2$ and the identity permutation.


问:
Limit of integral appears in the Fourier series
It’s a problem when I read the Fourier Series in one calculus textbook.

Suppose that $0<a<\pi$,$f\in C^0[0,a]$,show that $$
\underset{\lambda \rightarrow \infty}{\lim}\frac{1}{\lambda}\int_0^a{f\left( x \right) \frac{\sin ^2\lambda x}{\sin ^2x}}dx=\frac{\pi}{2}f\left( 0 \right) .
$$

I have made two attempts to solve this problem. One is the approximation of $f$ by trigonometric polynomials, and the other is the fourier expansion for $sin^2x$.However,I have failed.

答:
Hints: First verify that this is true when $f$ is a constant by making the change of variable $t=\lambda x$.  You will need the fact that $\int_0^{\infty} \frac {\sin ^{2}t} {t^{2}} dt=\frac {\pi} 2$.

Next reduce the proof to the case $f(0)=0$ by considering the function $f(x)-f(0)$.

Choose $r\in (0,a)$ such that $|\int_0^{r} f(x) \frac {\sin ^{2}(\lambda x)} {\sin^{2}x} dx| <\epsilon$. Now observe that  $|\int_r^{a} f(x) \frac {\sin ^{2}(\lambda x)} {\sin^{2}x} dx| \leq M |\int_r^{a} f(x) \frac {\sin ^{2}(\lambda x)} {\sin^{2}x} dx|$ where $M=sup |f(x)|$. Now use the inequlity $\sin t \geq \frac 2 {\pi}t$ for $t \in (0,\pi)$. Also $\int_r^{a} f(x)\sin^{2}(\lambda x) dx=\frac 1 2 \int_r^{a} f(x)[1-\cos (2\lambda x)d$. Use Riemann Lebesgue Lemma to show that $\int_r^{a} f(x) \frac {\sin ^{2}(\lambda x)} {\sin^{2}x} dx $ is bounded.


问:
How to prove a linear map to be the trace function,3x
Let $M_n(F)$ be the matrix space on the field $F$, $f$ be the linear map from $M_n(F)$ to $F$ such that $f(I)=n$, where $I$ is the identitiy matrix. Furthermore, $f(AB)=f(BA)$ for any $A,B\in M_n(F)$. Show that $f=tr$, where $tr(A)=\sum_{i=1}^n a_{ii}$ for $A=(a_{ij}).

I do not have any idea....

答:
It is probably already on this forum, this is well known.
Denote by $E_{i,j}$ the $n\times n$ matrix with a $1$ in position $(i,j)$ and zero elsewhere.

For $i,j$ in $\{1,\ldots,n\}$ we have $f(E_{i,i}) = f(E_{i,j}E_{j,i}) = f(E_{j,i}E_{i,j}) = f (E_{j,j}) = 1$.

If $i\neq j$, we have $f(E_{i,j}) = f (E_{i,j}E_{j,j}) = f (E_{j,j}E_{i,j}) = f (0) = 0$.

We conclude writing $A= \sum_{i,j=1}^n a_{i,j}E_{i,j}$.


问:
Rolling a D6 five times; will there be at least three adjacent numbers among the results?
The order in which a number appear does not matter.
How do I calculate the probability that this will happen?</p>

答:
There can be 3 adjacent numbers among the 5 outcomes. for example, the outcomes can be 1,2,3,4, and 6.
To find the probability, you have to find the number of such favorable outcomes over the total number of outcomes.</p>


答: $(123)$ has order $3$ and as a result generates a copy of $\mathbb Z_3$ in $A_5$.  You're right that this is $A_3$.

Title: Smallest subgroup of $A_{5}$ containing the permutation $(12)(13)$.

Body:
I need to find the smallest subgroup of $A_{5}$ that contains the permutation $(12)(13)$ and to also explain why.
I guess we can write this permutation as $(123)$ and I guess, according to a list of subgroups that I found, that the smallest would be $A_{3}$ in $A_{5}$.

If somebody could help me out with some reasoning or hints to understand this, I would appreciate it a lot.

It's important to say that this is my first ever algebra course, so it could be nice if the help would be kept simple and understandable for beginners.

Thanks in advance


问:
Combinatorial Minimization Problem for sets of rational numbers.
If we have the set

$\Bigg\{ \frac{1}{1},\frac{1}{2},\frac{1}{3},...,\frac{1}{k} \Bigg\}=:A$

for some $k\in \mathbb{N}$ (known). We pick some $a\in \mathbb{Q}$ such that we have

$\Bigg\{ \frac{a}{1},\frac{a}{2},...,\frac{a}{k} \Bigg\}=:B$

The problem is, how do we pick the number $a\in \mathbb{Q}$ such that $a \neq k$ optimally so that the non-duplicates in the first and the second set are minimized i.e. $B$ shares as many elements as possible with $A$. By duplicates I refer to instances such as $\frac{1}{2}=\frac{2}{4}=\frac{4}{8}$ for example.

答:
Hugo Manet has the right idea in his comment. I'll assume that the condition $$a \neq k$$ is actually $$a \neq 1,$$ as $a=1$ is the obvious solution with maximal intersection of $A$ and $B$, because then $A$=$B$.

First, $A$ is a $k$-element set, and the same is true for $B$ if $a \neq 0$. $a=0$ means $A \cap B =\emptyset$, so $a=0$ will never be the sole optimal solution, so we can forget about $a=0$. Similiarly, if $a<0$, we get $A \cap B =\emptyset$, so we can restrict ourselves to $a > 0$.

Let's consider an element $\frac1i \in A$, with $1 \le i \le k$. It can be equal to at most one $\frac{a}j \in B, 1 \le j \le k$. Similiarly, each $\frac{a}j \in B, 1 \le j \le k$ can b equal to at most one $\frac1i \in A$.

That means the cardinality of $A \cap B$ is the cardinality of

$$C_a=\{(i,j) \in \mathbb N^2, 1 \le i,j \le k \mid  \frac1i=\frac{a}j\} = \{(i,j) \in \mathbb N^2, 1 \le i,j \le k \mid  a =\frac{j}i\}.$$

Obviously, for each $i$ in the given range there can only be at most one $j$ in that range that fulfilles the qotient condition, and vice versa, so we have $|C_a| \le k$ with equality holding for the forbidden trivial case of $a=1$.

Also note that $C_a=C_{1/a}$, as this this just means transposing every pair $(i,j)$ into $(j,i)$. So we can concentrate on looking at $0 < a < 1$ cases, any solution here will have a corresponding solution $\frac1a$.

So let's assume

$$a=\frac{p}q,\, p,q \in \mathbb Z,\, p,q > 0,\, \gcd(p,q)=1,\, p < q$$

wich means we need to find $(i,j)$ such that

$$\frac{p}q=\frac{j}i \Longleftrightarrow pi=qj.$$

Since $q$ and $p$ have no common divisor greater than 1, that means $q|i$, so $i=fq$ for some integer $f \ge 1$. That means $j=\frac{pi}q=p\frac{i}q=fp \in \mathbb Z$. So if $q|i$, we can find a positive (since $f,p > 0$) integer $j=fp$ that makes $\frac{j}i=a$ come true.

In addition, if we assume $i \le k$ we get that $j = \frac{pi}q = i\frac{p}q < i \le k$, as we assumed $a=\frac{p}q < 1$.

So that means we know that for each $(i,j) \in C_a$, with $a=\frac{p}q < 1$ in reduced form, $q|i$ is a necessary condition, but also suffienct, as the corresponding $j=ai$ is an integer in the range $1 \le j \le n$.

That means

$$|C_a|=|\{i\in \mathbb N, 1\le i\le k : q|i \}|$$.

The cardinality if the second set above is easy to calculate, its simply $\lfloor\frac{k}q\rfloor$, which is maximized if $q$ is as low as possible, and for $a=\frac{p}q < 1$ in reduced form that happens when $q=2$, so $a=\frac12$.

It never gets better, though there are a few small $k$ cases where it's not the only optimum. For $k=1$, we get $|C_a|=0$ for all $a \neq 1$. For $k=2$ it's the only optimum. For $k=3$, $q=3$ also works, as $|C_{1/2}|=|C_{1/3}|=|C_{2/3}|=1$. For $k=4,5$ there are 2 even integers in the range $1\ldots k$, but only one divisble by $3$, so again $q=2$ is the only optimal value.

For $k=6$, both counts get upped by $1$ at the same time, so still $q=2$ is the only optimal value. In addition, as $6$ is divisible by both $2$ and $2$, the count for $q=3$ can never reach the count for $q=2$, as the count for $2$ always gets incremented earlier than the correpondig count for $3$.

To get the final answer:

For $k=1$, each $a\neq 1$ is qually bad, producing an empty intersection of $A$ and $B$.

For $k\ge 2$, $a=\frac12, a=2$ produce optimal solutions, with $|A \cap B|=\lfloor\frac{k}2\rfloor$.

Those are the only optimal solutions, except for $k=3$, where $a=\frac13, a=\frac23, a=\frac32, a=3$ also produce an optimal solution of $|A \cap B| =1$.


问:
Density of Primes in an interval
Is it possible to characterize all or at least some $k \in \mathbb{N}$ such that there are more than $k$ primes between $k^2$ and $\lfloor k^2/2\rfloor$? Or is it possible to characterize all or at least some $k \in \mathbb{N}$ such that there are at most $k$ primes between $k^2$ and $\lfloor k^2/2\rfloor$?

答:
Let $k \geq 16$. By Wikipedia bounds on the prime-counting function, $\pi(k^2/2) <1.25506\frac{k^2}{4\ln{k}-2\ln{2}} \leq 1.25506\frac{k^2}{3.5\ln{k}}$, and $\pi(k^2) \geq \frac{k^2}{2\ln{k}}$. Thus $\pi(k^2)-\pi(k^2/2) \geq \frac{k^2}{\ln{k}}(0.5-1.25506/3.5)$. So as soon as $\frac{k}{\ln{k}} (0.5-1.25506/3.5) > 1$, your statement holds.

Using Wolfram, we can thus see that the statement holds for all $k \geq 22$, and the smaller numbers can be tested manually.

Title: Density of Primes in an interval


问:
Can real non-symmetric matrices have real eigenvalues?
From the spectrum theorem, we know real symmetric matrices have real eigenvalues.

But can real non-symmetric matrix have real eigenvalues?

What are the necessary and sufficient conditions for a real matrix to have real eigenvalues?

答:
This is an old question, but a more complete answer than the ones given may still be of interest. Necessary and sufficient conditions are given below.

The basic story is that in a quantum mechanical (QM) system the energy levels are given by the eigenvalues of a matrix $\mathbf{H}$ called the Hamiltonian. In the usual formulation of QM, $\mathbf{H}$ is required to be Hermitian to ensure real energy eigenvalues but one can ask what wider class of matrices would be eligible. It turns out that if the physical system satisfies parity (spatial reflection) and time-reversal symmetry, $\mathbf{H}$ may have a real spectrum. This is formulated in mathematical language below.

Theorem 1 (necessary condition): Suppose a matrix $\mathbf{H}\in\mathbb{C}^{N \times N}$ has real or complex conjugate eigenvalues. Then there exist two matrices/operators $\mathbf{P}$ and $\hat{\mathbf{T}}$ s.t.:

$\mathbf{P}^2=\mathbf{I}$ and $\mathbf{P}$ is also a Hermitian and unitary matrix

$\hat{\mathbf{T}}^2=\mathbf{I}$ and $\hat{\mathbf{T}}=\hat{K}\mathbf{U}$ where $\hat{K}$ is the complex conjugation operator and $\mathbf{U}$ is a unitary matrix

$[\mathbf{P},\hat{\mathbf{T}}]=\mathbf{0}$

$[\mathbf{P}\hat{\mathbf{T}},\mathbf{H}]=\mathbf{0}$.

Theorem 2 (sufficient condition): Suppose that there exists in addition to the above operators/matrices a further matrix which has $\mathbf{C}^2=\mathbf{I}$ and $[\mathbf{C},\mathbf{H}]=\mathbf{0}$ but is itself not the unit matrix. Then $\mathbf{H}$ has real spectrum only if $\mathbf{C}$ commutes with $\mathbf{P}\hat{\mathbf{T}}$.

The proofs to the above two theorems are given in Bender, Carl M., and Philip D. Mannheim. "PT symmetry and necessary and sufficient conditions for the reality of energy eigenvalues." Physics Letters A 374.15-16 (2010): 1616-1620

Note 1: There is a more general version of Theorem 1 given in Bender, Carl M., M. V. Berry, and Aikaterini Mandilara. "Generalized PT symmetry and real spectra." Journal of Physics A: Mathematical and General 35.31 (2002): L467.

Definition: $\hat{\mathbf{A}}$ is antiunitary if $(\hat{\mathbf{A}}\mathbf{u})^\dagger(\hat{\mathbf{A}}\mathbf{v})=(\mathbf{u}^\dagger\mathbf{v})^*$. (Any antiunitary $\hat{\mathbf{A}}$ can be expressed as the complex conjugation operator times a unitary matrix.)

Theorem 1.1: A matrix $\mathbf{H}$ has real or complex conjugate eigenvalues if it commutes with an antiunitary $\hat{\mathbf{A}}$ that also satisfies $\hat{\mathbf{A}}^{2k}=\mathbf{I}$ for $k$ odd integer. ($\mathbf{P}\hat{\mathbf{T}}$ corresponds to $k=1$.)

Note 2: A nice sufficient condition is indeed given by the paper referenced by @MossMurderer, albeit it's less general (only works for real matrices): Bose, N. K. "On real eigenvalues of real nonsymmetric matrices." Proceedings of the IEEE 56.8 (1968): 1380-1380.

Note 3: A non-trivial non-Hermitian operator which has real and positive eigenvalues is e.g. $\hat{H}=\frac{\mathrm{d}^2}{\mathrm{d}x^2}+m^2x^2-(ix)^N$ with a real parameters $m$ and $N$. More examples are given in many works by Carl Bender et. al., but one needs a bit (or sometimes a lot) of QM to understand them. Check out e.g. Bender, Carl M., and Stefan Boettcher. "Real spectra in non-Hermitian Hamiltonians having P T symmetry." Physical Review Letters 80.24 (1998): 5243. .


问:
Tweak equation to "add" fixed points
I hope someone can at least point me in a right direction.

I managed to close in on the equation required for a project, but have no idea about how to "add" certain characteristics that I need it to have. The equation is: $$y=-1/a(x-5)$$
It has two asymptotes that form a right angle at $(5,0)$ and as $a$ is increased, the graph (at least seems like) is symmetric along the $y=-x+5$ axis. All of this fits the purpose perfectly.

What it's missing though is two points that stay fixed as $a$ is varied, $(0,0)$ and $(5,5)$.

Basically I need to start from a straight line $y=x$ and "bend it" towards $(5,0)$ symmetrically by varying $a$, as if someone is pulling on it.

答:
An equation for a hyperbola like yours but centred at the origin is $y=-\frac{1}{ax}$.

If we shift its centre to $(c,-b)$ we get the equation $y=-b-\frac{1}{a(x-c)}$.

We want this to go through the points $(0,0)$ and $(5,5)$.
Substituting those coordinates results in the two equations
$$abc=1\\a(c-5)(b+5)=1$$

From $abc=a(c-5)(b+5)$ we find that $c=b+5$, which makes sense since this puts the centre $(c,-b)$ on the line of symmetry $y=5-x$.

Substituting this value for $c$ we get $ab(b+5)=1$, and solving this quadratic for $b$ we find that $b = -2.5 + \sqrt{2.5^2 + 1/a}$.

So the equation has become $y=-b-\frac{1}{a(x-b-5)}$ where $b = -2.5 + \sqrt{2.5^2 + 1/a}$. Not very nice, but I think it does what you want it to.

Title: Tweak equation to "add" fixed points


问:
Finding $\operatorname{Im}T$ - Linear Algebra
Regarding :

$T: \mathbb{R}^{n+1} \to \mathbb{R}^{n}, T(x_1,...,x_{n+1})=(x_1+x_{n+1},x_2,...,x_n) $

I need to find $\ker T$ and $\operatorname{Im}T.$

To find $\operatorname{Im}T,$ I want to take a spanning set of $\mathbb{R}^{n+1}$, which will be $(e_1,e_2,...,e_n,e_{n+1}),$ and by a theorem. I can say that :

$\operatorname{Im}T=\operatorname{span}\{T(e_1),...,T(e_{n+1})\}=\operatorname{span}\{e_1+e_{n+1},e_2\ldots, e_{n}\}$

But now I have a problem, because those vectors doesn't belong to $\mathbb{R}^n.$

Any help ?

Thanks a lot!

答:
Your computation of $Te_1,Te_2,..$ is wrong. $Te_1=T(1,0,0..,0)=(1,0,0..,0)$; In fact $Te_i=e_i$ for $i \leq n$ and $Te_{n+1}=e_1$. Now it should be clear that $T$ is surjective.

If $Tx=0$ then $x_2=x_3,...,x_n=0$ and $x_1+x_{n+1}=0$. So the kernel consists of vectors $x$ such that $x_1+x_{n+1}=0$. (It is the one dimensional space  spanned by $(1,0,0,..,0,-1)$).


Title: Finding $\operatorname{Im}T$ - Linear Algebra

Body:
Regarding :

$T: \mathbb{R}^{n+1} \to \mathbb{R}^{n}, T(x_1,...,x_{n+1})=(x_1+x_{n+1},x_2,...,x_n) $

I need to find $\ker T$ and $\operatorname{Im}T.$

To find $\operatorname{Im}T,$ I want to take a spanning set of $\mathbb{R}^{n+1}$, which will be $(e_1,e_2,...,e_n,e_{n+1}),$ and by a theorem. I can say that :

$\operatorname{Im}T=\operatorname{span}\{T(e_1),...,T(e_{n+1})\}=\operatorname{span}\{e_1+e_{n+1},e_2\ldots, e_{n}\}$

But now I have a problem, because those vectors doesn't belong to $\mathbb{R}^n.$

Any help ?

Thanks a lot!

Body:
Note that:

$T(e_1)=(1,0,0,\ldots,0)$;

$T(e_2)=(0,1,0,0,\ldots,0)$;

$\vdots$

$T(e_n)=(0,0,0,\ldots,0,1)$;

$T(e_{n+1})=(1,0,0,\ldots,0)$.

So $\operatorname{Im}T=\Bbb R^n$.


问:
A question about generic points.
Let $X$ be a topological space and let $x$ be a point in $X$.
In Hartshorne's book, I saw the the following definition of a generic point:

$x$ is a generic point of $X$ if $\{x\}$ is dense in $X$.

But, in Qing Liu's book, I saw another definition of a generic point:

$x$ is a generic point of $X$ if $x$ is the unique point of $X$ that specialize to $x$
  that is, for a point $y \in X$ if x is contained in the closure of $\{y\}$ then $x=y$.

I don't know why these definitions are equivalent....

答: The definitions are different. With Hartshorne's, only irreducible schemes have a (necessarily unique) generic point. With Liu's "generic point" means "Hartshorne generic point of an irreducible component." They coincide precisely for irreducible schemes.</p>

Title: A question about generic points.


问:
I found this CHALLENGING problem . Help me with this?
Find⠀the⠀average⠀distance⠀of two⠀randomly-selected⠀points⠀in⠀a⠀circle⠀of⠀Radius $=R$.

This⠀is⠀my⠀approach: Consider a function $I(R)$ that is given an input (the radius $R$ of the circle) and outputs the average distance between two randomly-selected points inside that circle. The size of the radius seems like it would only scale the answer, so we can say that $I(R)=R\cdot I(1)$. This reduces the problem to merely considering the case when $R=1$.

One’s first instinct might be to consider two random variables, $r\in [0,1]$ and $\theta \in [0,2\pi]$, and considering $x=r\cos\theta$ and $y=r\sin\theta$, but a closer inspection shows us that these won’t correspond to random points, and the points would cluster more around the center of the circle than they would closer to the edges (notice to that $dA=r\cdot dr\cdot d\theta$, not just "⠀$dr\cdot d\theta$ "). The correct transformation is $x=r\sqrt{\cos\theta}$ and $y=r\sqrt{\sin\theta}$ (when I first considered this problem myself, I understood that the first approach wouldn’t work, and the points would cluster around the center, but even now I couldn’t exactly tell you why the correct transformation requires taking precisely the $\sqrt{r}$ ; this is evidence for the question’s conceptual difficulty).

Considering this transformation and integrating each variable with respect to its PDF, taking (without loss of generality) the first point to be $(r_1,0)$ and the second to be $(r_2,θ)$, and noticing that $\theta$ only needs to range from $0$ to $\pi$ yields the integral⠀:

\begin{align}I(1)&=\frac{\int_{0}^{1}\int_{0}^{1}\int_{0}^{\pi}\sqrt{(\sqrt{r_1}-\sqrt{r_2}\cos\theta)^2+(\sqrt{r_2}\sin\theta)^2}⠀d\theta dr_1 dr_2}{\int_{0}^{1}\int_{0}^{1}\int_{0}^{\pi}d\theta dr_1 dr_2}\\&=\frac{1}{\pi}\int_{0}^{1}\int_{0}^{1}\int_{0}^{\pi}\sqrt{r_1+r_2+\sqrt{r_1r_2}\cos\theta}⠀d\theta dr_1 dr_2 \end{align}

It took me 2hr to type the question and 1hr to type this integral. This took me really Long, So without showing the steps to anti-differentiate, This equals to : $⠀I(R)=\frac{128}{45\pi}R $ Is my answer is correct?

Kindly provide⠀your⠀solution⠀too.

答:
The desired average is obtained by a quadruple integral in Cartesian coordinates.

$$\frac1{\pi^2}\int_{x^2+y^2\le1,\\u^2+v^2\le1}\sqrt{(x-u)^2+(y-v)^2}dx\,dy\,du\,dv$$

Switching to polar and using rotational symmetry,
$$\int_0^{2\pi}\int_0^1\int_0^{2\pi}\int_0^1\sqrt{(r\cos\theta-s\cos\phi)^2+(r\sin\theta-s\sin\phi)^2}\,rdr\,d\theta\,s ds \,d\phi$$
$$=2\pi\int_0^1\int_0^{2\pi}\int_0^1\sqrt{r^2-2rs\cos\theta+s^2}\,rdr\,d\theta\,s ds.$$

From here, computation seems nightmarish. Integration on $\theta$ yields a complete elliptic function. Integration on $r$ or $s$ is doable but leads to relatively complex expressions.


问:
Let x and y be real numbers such that $6x^2 + 2xy + 6y^2 = 9 $.Find the maximum value of $x^2+y^2$
I tried to re-arrange the terms

$6x^2 + 2xy + 6y^2 = 9 $

$6x^2 + 6y^2 = 9 - 2xy $

$6 (x^2 + y^2) = 9 - 2xy $

$x^2 + y^2 = \frac{9 - 2xy}{6} $

Using A.M $\geq$ G.M

$\frac{x^2 + y^2}{2} \geq xy $

Can ayone help me from here? Am I going correct?

答: Let $x= r\cos\theta , y= r\sin\theta$ , put it back into the given relation and get a function of $r^2 $in terms of $\theta$ we need to maximize this $r^ 2 $wrt to theta  ,use calculus or just bounds of trigo function


问:
A $T_4$ space satisfying that 'every continuous map from it to a Hausdorff space is a closed map' is compact.
Suppose a $T_4$ space $X$ satisfies a property that every continuous map $f$ from $X$ to any Hausdorff space $Y$ is a closed map. Show that $X$ is compact.

What I am thinking is to let $\mathcal{U}$ be an open cover of $X$ and find a way to choose finite open sets from $\mathcal{U}$ that can cover $X$ again. Do we need to define a suitable continuous map by ourselves and use the fact that it is a closed map to choose suitable open sets from $\mathcal{U}$? I have tried to define a continuous map $f \colon X \to Y$ as mapping different open sets from $\mathcal{U}$ to different points in $Y$. However, different open sets from $\mathcal{U}$ may intersect with each other and hence $f$ may not be well-defined. Apart form that, I have no idea about how to choose suitable $f$ and $Y$. Can someone help me please?

Edit 1: I am trying to apply the Urysohn Lemma or the Tietz Extension Theorem due to the normality of $X$. But still have no idea.

Edit 2: Finally found a way to prove without using compactification. I will post the proof when I’m free. Hint: Use Urysohn Lemma and Tychonoff Theorem.

答: Any such $X$ has a compactification $(\gamma X, e)$ in which $e: X \to \gamma X$ is an embedding with a dense image (this uses $T_{3\frac12}$). Then the assumption says that $e$ must be a closed map, so $\gamma X = e[X]$ and $X$ is compact.


答:
For AM-GM, $-2xy \le 2|xy| \le x^2+y^2$ so $2xy\ge -x^2-y^2$.

$$5(x^2+y^2)\le 6x^2+2xy+6y^2 =9.$$
Then $x^2+y^2\le 9/5$.  The maximum is reached when $x=-y=\pm\sqrt{9/10}$.

You can also get the min value of $x^2+y^2$.  Because $2xy\le 2|xy|\le x^2+y^2$
$$7(x^2+y^2)\ge 6x^2+2xy+6y^2=9.$$
Then $x^2+y^2\ge 9/7$.  The minimum is reached when $x=y=\pm\sqrt{9/14}$.

I think the shape $6x^2+2xy+6y^2=9$ is an ellipse with major/minor axes given by the lines $x=\pm y$. The axes have length $2\sqrt{9/5}$ and $2\sqrt{9/7}$.

Title: Let x and y be real numbers such that $6x^2 + 2xy + 6y^2 = 9 $.Find the maximum value of $x^2+y^2$

Body:
I tried to re-arrange the terms

$6x^2 + 2xy + 6y^2 = 9 $

$6x^2 + 6y^2 = 9 - 2xy $

$6 (x^2 + y^2) = 9 - 2xy $

$x^2 + y^2 = \frac{9 - 2xy}{6} $

Using A.M $\geq$ G.M

$\frac{x^2 + y^2}{2} \geq xy $

Can ayone help me from here? Am I going correct?


问:
AP Calculus multiple choice question
The position of a particle along a line is given by $z(t) = 2t^3 -24t^2 + 90t +7$ for $ t \geq 0$. For what values of $t$ is the speed of the particle increasing?

a) $3 < t < 4$ only

b) $t > 4$ only

c) $t > 5$ only

d) $0 < t < 3$ and $t > 5$

e) $3 < t < 4$ and $t > 5$

For this problem I said d) based on my calculations:

$z'(t) = 6t^2 - 48t + 90 = 6(t-5)(t-3)$. I then did a chart and found out that $z'(0) > 0$ ,$z'(4) < 0$ and $z'(6) > 0$ leading me to conclude d) however the textbook says e). I thought that I did everything right, what is wrong with my conclusion?

答: A particle is speeding up when velocity and acceleration have the same signs and likewise is slowing down when velocity and acceleration have opposite signs. So you must take two derivatives. Technically you need to test the intervals created by all of the critical values in both derivatives. However since this is a MC question, just test those intervals into the derivatives. Use the factored form of each derivative for easy checking of positives and negative intervals.</p>

Title: AP Calculus multiple choice question


答:
Let $G$ be a finite group and $x\in G$. Then the smallest subgroup $H$ of $G$ containing $x$ is the cyclic group $H=\langle x\rangle$ generated by $x$. It consists of all integral powers $x^n$ of $x$.

Here we have $G=S_3$ and
$$H=\langle (123) \rangle =\{(1),(123),(132)\}\cong C_3\cong A_3.$$


问:
A tree coding all of $L_\alpha$ within $L_\alpha\cap\mathcal{P}(\omega)$
This question is about a sentence in the paper Gaps in the Constructible Universe by Marek & Srebrny. A countable ordinal is called a gap ordinal iff $(L_{\alpha+1}\smallsetminus L_{\alpha})\cap \mathcal{P}(\omega)=\emptyset$; that is, no reals are added in going from $L_\alpha$ to $L_{\alpha+1}$. Lemma 2.4 in the paper states:

Let $\alpha$ be a gap ordinal. If $X\in \mathcal{P}(\omega)\cap L_{\alpha}$ and $X$ is a real well-ordering, then the type of $X$ is less than $\alpha$.

The authors proved this by contradiction: assume not. We can construct a tree coding all of $L_{\alpha}$ within $L_{\alpha}\cap \mathcal{P}(\omega)$. Now, we apply Cantor's diagonal procedure to obtain a new set of natural numbers.

My question is about the italicized sentence in the paragraph above: what does this mean? More specifically, how do we construct such a tree?

答:
The point is that $X$ is a well-order, thus we may use recursion along $X$, but we do not know that the result of the recursion has to be a member of $L_\alpha$. Fortunately, we know that $L_\alpha$ for a gap ordinal $\alpha$ satisfies the complete arithmetical
schema of choice. (Lemma 2.2 of Marek and Srebrny:)

Lemma. If $\alpha$ is a gap ordinal, and $\phi(x,y)$ is a set-theoretic formula, then $L_\alpha$ satisfies
$$\forall n\in\omega\exists A\subseteq \omega \phi(n,A)\to \exists A\subseteq\omega \forall n\in\omega \phi(n,A^{(n)}).$$

Let $X\in L_\alpha\cap\mathcal{P}(\omega)$ be a well-order. I assume that $X$ is a well-order over $\omega$. Moreover, we may assume that the ordertype of $X$ is $>\alpha$.
Now consider the following formula $\phi(n, A)$ given by the conjunction of

$A$ is a function whose domain is of the form $\{m\in\omega\mid m\le_X n\}$,

For any $k$,

if $k$ is the least element of $X$, then $A(k)=\langle(\varnothing,\varnothing)\rangle$,

if $k$ is a $X$-successor of some $m\in\omega$, then $A(k)=(M,E)$, where $(M,E)=\operatorname{Def}(A(m))$, and

if not, then $A(k)$ is the direct limit of all $A(m)$, $m<_Xk$.

By the lemma, there is $A$ such that $\phi(n,A^{(n)})$ for all $n\in\omega$. Especially, if $n$ is the $\alpha$th element of $X$, then $A^{(n)}$ codes an isomorphic copy of $\langle L_\xi \mid \xi\le\alpha\rangle$.


问:
Acceleration, deceleration, distance - calculating point of deceleration
I'm working on game in space and I encounter problem that took me few hours to solve and I reached point when I think it would be easier to ask someone more clever, that trying to learn from beginning physics formulas from internet :)

I need to find formula for calculating point at given distance the body will need to start decelerating. I know acceleration(a=1m/s2), deceleration(d=2m/s2) and distance(s=100m) the body needs to travel. Initial velocity is 0, and final velocity is also 0. Body start to accelerate and at some point it needs to start decelerating. Acceleration & deceleration are const.

After what distance the body needs to start decelerating. What is formula for this i can use to calculate it.

Like above but additionally it needs some time (for example 2s) before it starts decelerating (time needed to rotate the ship before it starts decelerating with main thruster). In that time its not accelerating.

Every help is will be really appreciated :)

Edit - i made some calculations based on what I've already found in other posts and ideas other showed me and was able to solve it on my own :), answer below

$\displaystyle V_{i}$ - initial speed at start position (known)

$\displaystyle V_{f}$ - final speed at end position (destination) (known)

$\displaystyle a$ - acceleration of body (known)

$\displaystyle d$ - deceleration of body (known)

s - distance to travel (known)

$\displaystyle V_{x}$ - speed after we reach we need to decelerate (unknown)

$\displaystyle s_{dec}$ - distance after which we need to start decelerating  (unknown)

$\displaystyle s_{acc}$ - distance after which we need to start decelerating  (unknown)

$\displaystyle t_{acc}$ - time for which we accelerate  (unknown)

$\displaystyle t_{dec}$ - time for which we decelerate  (unknown)

答:
Was able to calculate on my own :) Checked with few set of variables and it works for me :) Thanks Matti P. for showing direction.

Answer for point 1

$\displaystyle V_{i}$ - initial speed at start position (known)

$\displaystyle V_{f}$ - final speed at end position (destination) (known)

$\displaystyle a$ - acceleration of body (known)

$\displaystyle d$ - deceleration of body (known)

s - distance to travel (known)

$\displaystyle V_{x}$ - speed after we reach we need to decelerate (unknown)

$\displaystyle s_{dec}$ - distance after which we need to start decelerating \ (unknown)

$\displaystyle s_{acc}$ - distance after which we need to start decelerating \ (unknown)

$\displaystyle t_{acc}$ - time for which we accelerate \ (unknown)

$\displaystyle t_{dec}$ - time for which we decelerate \ (unknown)
\begin{equation*}
\end{equation*}
$ $
\begin{equation*}
a\ =\ \frac{V_{x} \ -\ V_{i} \ }{t_{acc}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d\ =\ \frac{V_{x} \ -\ V_{f} \ }{t_{dec}} \ \ \
\end{equation*}

(1)
\begin{equation}
t_{acc} \ =\ \frac{V_{x} \ -\ V_{i} \ }{a} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t_{dec} \ =\ \frac{V_{x} \ -\ V_{f} \ }{d}
\end{equation}
(2)
\begin{equation}
s_{dec} \ =V_{f} \ *\ t_{dec} \ +\ \frac{1}{2} \ d\ *\ (t_{dec} )^{2\ } \ \ \ \ \ \ s_{acc} \ =V_{f} \ *\ t_{acc} +\ \frac{1}{2} \ a\ *\ (t_{acc} )^{2\ } \ \ \ \ \
\end{equation}

\begin{equation*}
s\ =\ \ s_{acc} \ +\ s_{dec} \ =V_{i} \ *\ t_{acc} +\ \frac{1}{2} \ a\ *\ (t_{acc} )^{2\ } \ +V_{f} \ *\ t_{dec} \ +\ \frac{1}{2} \ d\ *\ (t_{dec} )^{2\ } \ \
\end{equation*}

\begin{equation*}
s\ =V_{i} \ *\ \frac{V_{x} \ -\ V_{i} \ }{a} \ +\ \frac{1}{2} \ a\ *\ \left(\frac{V_{x} \ -\ V_{i} \ }{a}\right)^{2} \ +\ V_{f} \ *\ \frac{V_{x} \ -\ V_{f} \ }{d} \ +\ \ \frac{1}{2} \ d\ *\ \left(\frac{V_{x} \ -\ V_{f} \ }{d}\right)^{2}
\end{equation*}

\begin{equation*}
s\ =\frac{V_{x} V_{i} \ -\ V^{2}_{i} \ }{a} \ +\ \ \frac{V^{2}_{x} \ -\ 2V_{x} V_{i} \ +\ V^{2}_{i} \ }{2\ a} +\frac{V_{x} V_{f} \ -V^{2}_{f} \ \ }{d} +\ \ \ \frac{{V^{2}_{x} \ -\ 2V_{x} V_{f} \ +\ V_{f}}^{2}}{2\ d}
\end{equation*}

\begin{gather*}
{2\ ads\ =2dV_{x} V_{i} \ -2dV^{2}_{i} \ +\ \ dV^{2}_{x} \ -\ 2dV_{x} V_{i} \ +\ dV^{2}_{i} \ +\ 2aV_{x} V_{f} \ \ -\ 2aV^{2}_{f} \ +\ aV^{2}_{x} \ -\ 2aV_{x} V_{f} \ +\ aV_{f}}^{2}\\
\end{gather*}
\begin{equation*}
{2\ ads\ =\ dV^{2}_{x} \ -\ dV^{2}_{i} \ +\ aV^{2}_{x} \ -\ aV_{f}}^{2}
\end{equation*}
\begin{equation*}
{2\ ads\ +\ aV_{f}}^{2} \ +\ dV^{2}_{i} \ =\ V^{2}_{x} \ ( d\ +\ a)
\end{equation*}
\begin{equation*}
\frac{{2\ ads\ +\ aV_{f}}^{2} \ +\ dV^{2}_{i}}{( d\ +\ a)} \ =\ V^{2}_{x} \
\end{equation*}
(3)
\begin{equation}
\sqrt{\frac{{2\ ads\ +\ aV_{f}}^{2} \ +\ dV^{2}_{i}}{( d\ +\ a)}} \ =\ V_{x} \
\end{equation}

When I get result of Vx I can get t needed for acceleration to Vx and deceleration from Vx (formula (1)). Then I can get distance s for acceleration by using formula (3), which is the value I needed.

Answer for point 2

$\displaystyle t_{rot}$ - time for rotation of body \ (known)

$\displaystyle t$ - time to travel given distance (we can get this from previous formula)

At this point we want to know how long body need to accelerate before it can rotate and then decelerate.

\begin{equation*}
\ \ \ \ \ \ \ \ \ \ \ \ t\ =\ \ t_{acc} \ +\ t_{dec} \ +t_{rot} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t_{dec} \ =\ \ t_{acc}
\end{equation*}

\begin{equation*}
\ \ \ \ \ \ \ \ \ \ \ \ t\ =\ 2t_{acc} +t_{rot} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{equation*}

\begin{equation*}
\ \ \ \ \ \ \ \ \ \ \ \frac{\ t\ -t_{rot}}{2} \ \ \ =\ t_{acc} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\end{equation*}


答:
$T(e_i)$ belongs to $\mathbb R^n$, since that is the codomain of $T$, hence you get $\mathbb R^n$ for your image.  There are a number of ways to see this.  For instance, you could note that $T(e_i)=\hat {e_i}$, where $\hat{e_i}$ are the standard basis vectors for $\mathbb R^n$.

Alternatively, you could note that the kernel is one-dimensional, and use Rank-Nullity.


问:
A quadratic Approximation Function Expression
I have very clear question. I'm searching about polynomial approximation.

In this formula, it say's $$f = (z-z_2)(z-z_3).F_1 + (z-z_3)(z-z_1).F_2 + (z-z_1)(z-z_2)F_3$$

where $F_1 = \frac{f_1}{(z_1-z_2)(z_1-z_3)}$, etc.

I searched it and when we write quadratic approximation we can use taylor series and we can write f like this:

$$f(x) = f(a) + (x-a)\frac{f'(a)}{1!} + (x-a)^2\frac{f''(a)}{2!} +\ldots  $$

How it converts $f = (z-z_2)(z-z_3)\cdot F_1 + (z-z_3)(z-z_1)\cdot F_2 + (z-z_1)(z-z_2)\cdot F_3$. What is the source of this formula? It doesn't similar to taylor series. I'm searching it for $3$ days but I can't understand how can we write it? What's the relation between this f expression and taylor series? Can anybody help me please?

答:
That is the formula for Langrange polynomial:

Given a set of $k+1$ data points
:$(x_0, y_0),\ldots,(x_j, y_j),\ldots,(x_k, y_k)$
where no two $x_j$ are the same, the '''interpolation polynomial in the Lagrange form''' is a linear combination
$$L(x) := \sum_{j=0}^{k} y_j \ell_j(x)$$
of Lagrange basis polynomials
:$$\ell_j(x) := \prod_{\begin{smallmatrix}0\le m\le k\\ m\neq j\end{smallmatrix}} \frac{x-x_m}{x_j-x_m} = \frac{(x-x_0)}{(x_j-x_0)} \cdots \frac{(x-x_{j-1})}{(x_j-x_{j - 1})} \frac{(x-x_{j+1})}{(x_j-x_{j+1})} \cdots \frac{(x-x_k)}{(x_j-x_k)},$$
where $0\le j\le k$.

In your context, you have $3$ points, hence we get a quadratic interpolation polynomial. It is the quadratic polynomial that passes through the $3$ points.


问:
Find the area of an equilateral triangle inscribed in a unit square
If inside the square $ABCD$ there is an equilateral triangle $CMN$ inscribed, as shown in the figure. If the area of the square is 1, find the area of the triangle.

I attempted to say, I state that $x=MD$

then $MC^2=1+x^2$. $AM=1-x$ and I thought that something might come from Pythagoras, but it didn't. $\triangle AMN$ looks isosceles, but I can't prove that it is, although if it is proved to be isosceles, then the question can easily be solved by use of the pythagorean theorem. We also have that $\angle CNB+\angle CMD=150^o$. I got stuck here, could you please explain to me how to solve the question?

答:
Since the triangle is equilateral, the lengths of the sides must be equal; that is,
$$
2x^2=x^2-2x+2\tag1
$$
The solutions to $(1)$ are $x=-1\pm\sqrt3$. The only one that would apply to the case pictured above would be $-1+\sqrt3$. This would make the area of the triangle
$$
x-\tfrac12x^2=2\sqrt3-3\tag2
$$


问:
Why does $(g\circ f)'(x_0)=g'(f(x_0))\circ f'(x_0)$?
I had a Multivariable calculus class this morning and the teacher showed us a proposition that I'm not quite sure I understand. The proposition is the following:

Let:

$D\subset \mathbb R^n $

$D^* \subset  \mathbb R^m$

$x_0 \in D$

$f:D \to D^*$ be a differential function at $x_0$

$g: D^* \to \mathbb R^p$ be a differential function at $f(x_0)$

Then $g\circ f$ is differential at $x_0$ and: $$(g\circ f)'(x_0)=g'(f(x_0))\circ f'(x_0)$$

My problem with this proposition is that I'm not understanding two things:

I understand derivatives of functions $f: \mathbb R^n \to \mathbb R$ (Things like partial derivatives, directional derivatives, etc...) and I understand that the derivative of such function is a scalar, But what is the definition of the derivative of a function $f: \mathbb R^n \to \mathbb R^m$? I don't recall my teacher showing us the meaning of such derivative.

My second question is, in the Rhs of the equality, why do we have $g'(f(x_0))\circ f'(x_0) $ instead of $g'(f (x_0)). f'(x_0) $? I don't' understand why we are using composition of functions. Does that mean that $(g\circ f)'(x_0) $ is a function?

答:
Point 1

Regarding the definition of the derivative of a map $f : \mathbb R^n \to \mathbb R^m$, you should look at what the Fréchet derivative is. See Wikipedia entry. The Fréchet derivative at each point $a \in \mathbb R^n$ is a linear map $f^\prime(a) : \mathbb R^n \to \mathbb R^m$. See also what the Jacobian is.

In general, the derivative of a map $f : U \to V$ defined between two linear spaces $U,V$ is a map $f^\prime : U \to \mathcal L(U,V)$ where $\mathcal L(U,V)$ denotes the linear space of the linear maps between $U$ and $V$. The derivative $f^\prime$ of $f$ maps each point of the domain of $f$ to a linear map!

Point 2

When you understand that the derivative is a linear map, then using composition (of linear maps) for the derivative of the composition of maps makes perfect sense.

Comments

Your sentence ...I understand that the derivative of such function is a scalar... is not correct. The derivative of a map $g : \mathbb R^n \to \mathbb R$ at a point $ a \in \mathbb R^n$ is not a scalar. It is a linear form from $\mathbb R^n$ to $\mathbb R$. And the derivative $g^\prime$ itself is a map $ g^\prime : \mathbb R^n \to \left(\mathbb R^n\right)^\prime$ where $\left(\mathbb R^n\right)^\prime$ stands for the dual space of $\mathbb R^n$.

By understanding all this, you'll make a big jump in your comprehension of what the derivative really is!


问:
If $\sup_n P(\vert X_n -a_n\vert > b) \leq c$, then can I find a $d \in \mathbb{R}$ such that $\sup_n P(\vert X_n \vert > d) \leq c$?
Let $\{X_n\}_{n \in \mathbb{N}}$ denote a sequence of real random variables and $\{a_n\}_{n \in \mathbb{N}}$ a sequence of real numbers. Suppose $P(\vert X_n\vert>\epsilon) \to 0$ for $n \to \infty$ for all $\epsilon>0.$

If $\sup_n P(\vert X_n -a_n\vert > b) \leq c$, then can I find a $d \in \mathbb{R}$ such that $\sup_n
 P(\vert X_n \vert > d) \leq c$? And how to choose $d$?

So I know $X_n$ converges to $0$ in probabillity. Can I use that it is a deterministic limit to say something about convergence almost surely? Or can I use
$$
\sup_n P(\vert X_n \vert  > b+ \vert a_n\vert) = \sup_n P(\vert X_n \vert - \vert a_n\vert > b) \leq \sup_n P(\vert X_n -a_n\vert > b) \leq c
$$
somehow?

答: Actually, you do not need the assumption $\sup_n P(\vert X_n -a_n\vert > b) \leq c$ because convergence in probability is enough. Indeed, using $\varepsilon=1$ in the definition of convergence in probability, we can find $n_0$ such that for each $n\geqslant n_0$, $\mathbb P\left(\lvert X_n\rvert>1\right)\leqslant c$. For $1\leqslant n\leqslant n_0$, take $d_n>0$ such that $\mathbb P\left(\lvert X_n\rvert>d_n\right)\leqslant c$ [this follows from the fact that for any random variable $X$, $\mathbb P(\lvert X\rvert>R)\to 0$ as $R$ goes to infinity]. Finally, take $d=\max\{1,d_1,\dots,d_{n_0-1}\}$.


问:
Rolling a D6 five times; will there be at least three adjacent numbers among the results?
The order in which a number appear does not matter.
How do I calculate the probability that this will happen?</p>

答:
The following Python returns 3480, so the answer is $3480/6^5 = 145/324 \approx 44.8\%$. Writing it was certainly faster and less error-prone than trying to get inclusion-exclusion or cases working.

```<code>from itertools import product

def three_adjacent(L):
    p = ['0']*(max(L)+1)
    for l in L:
        p[l] = '1'
    s = ''.join(p)
    return '111' in s

sum(1 for ds in product(range(1, 6+1), repeat=5) if three_adjacent(ds))
</code>```

Body:
The following Python returns 3480, so the answer is $3480/6^5 = 145/324 \approx 44.8\%$. Writing it was certainly faster and less error-prone than trying to get inclusion-exclusion or cases working.

from itertools import product

def three_adjacent(L):
    p = ['0']*(max(L)+1)
    for l in L:
        p[l] = '1'
    s = ''.join(p)
    return '111' in s

sum(1 for ds in product(range(1, 6+1), repeat=5) if three_adjacent(ds))


问:
Find the number of permutations of integers from $1$ to $10$ inclusive that do not start with $1$ and do not end with $10$
Find the number of permutations of integers from $1$ to $10$ inclusive that

$a)$ do not start with $1$ and do not end with $10$.

$b)$ no odd integer will be in its right place.

$c)$ numbers $1$, $4$ and $7$ are not in their right places.

For the question $a)$, I tried something like this:

Since the first place cannot be a $1$, there are $9$ ways to fill it. Similarly, the last position can't be a $10$, so there are only $8$ ways to fill it. The remaining digits can be filled in $8!$ ways.

So by rule of product, the answer would be $9 \cdot 8 \cdot 8!$.

But the answer given in the key seems different. I'm not sure if I have gone wrong somewhere in my approach.

答:
All three parts of this problem can be solved by using the Inclusion-Exclusion Principle.

Find the number of permutations of integers from $1$ to $10$ inclusive that do not start with $1$ and do not end with $10$.

We will subtract the number of permutations which start with $1$ or end with $10$ from the total number of permutations.

There are $10!$ permutations of the ten distinct integers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$.

If we place a $1$ in the first position, there are $9!$ ways to permute the remaining nine distinct integers in the remaining nine positions.

If we place a $10$ in the last position, there are $9!$ ways to permute the remaining nine distinct integers in the remaining nine positions.

Subtracting those permutations with a $1$ in the first position or a $10$ in the last position from the total number of permutations gives us a preliminary count of
$$10! - 9! - 9!$$
but we have subtracted too much.  We have counted those permutations which have both a $1$ in the first position and a $10$ in the last position twice, once when we subtracted those permutations which have a $1$ in the first position and once when we subtracted those permutations which have a $10$ in the last position.  We only want to subtract such cases once, so we must add them to the total.

If we place a $1$ in the first position and a $10$ in the last position, the remaining eight distinct integers can be permuted in the remaining eight positions in $8!$ ways.

Hence, the number of permutations of the integers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ which do not start with $1$ and do not end with $10$ is
$$10! - 2 \cdot 9! + 8!$$

I suggest that you try part (c) next.  Subtract the number of permutations which have a $1$ in the first position or a $4$ in the fourth position or a $7$ in the seventh position from the total number of permutations by using the Inclusion-Exclusion Principle.  Doing so is easier than doing part (b) since you have three excluded conditions rather than five.


问:
Probability of ending with 2 blue balls in a row after $n$ draws
Thank you for taking the time to read my question.

The scenario is, we're playing a game where we have in a box 9 balls: 4 red, 3 green, 2 blue.

We draw one ball at a time, and note down its color and then return it for the next draw, the game stops only when we have drawn the same color twice in a row.

What is the probability of the game ending after $n$ draws? $P(X=n)$?

What is the probability of the last two balls being blue?

I have tried to treat it as a geometric probability distribution, with: p: probability of having two balls in a row with the same color $P(X=n) = n-1$ fails and then a success but this ball number 2 and 3 could be the same color in this. I've tried to discuss it case by case, starting with if the first ball is Blue, Red, Green, in hopes of noticing some recursion, but no luck. Haven't had any other ideas.

答:
Here is an idea how to solve the problem using the Markov's chain. Possibly it is not the simplest solution but it is general enough so that it can be applied to more complicated problems as well.

Let the probabilities to draw the balls be $p_1,p_2,p_3$, respectively and let the state of the sequence be characterized by the color (type) of the last ball. Then the transition matrix  and the initial state read:
$$
T=\left(\begin{array}{ccc|c}
0&  p_1& p_1&\color{grey}0\\
p_2&0&p_2&\color{grey}0\\
p_3&p_3&0&\color{grey}0\\
\hline
\color{grey}{p_1}&\color{grey}{p_2}&\color{grey}{p_3}&\color{grey}{1}
\end{array}\right)\quad\text{and}\quad
P_1=\begin{pmatrix}
p_1\\
p_2\\
p_3\\
\color{grey}0\\
\end{pmatrix},\tag1
$$
respectively, and  the state of the sequence after $n$ draws is
$$
P_n=T^{n-1}P_1.\tag2
$$

To make the things simpler one can note that there is no back coupling from the final state, so that it can be omitted and only the action of the "active" $3\times3$ block of the matrix onto the $3\times1$ vector of "active" states [highlighted in $(1)$] be considered. In what follows these reduced versions of the transition matrix and the state vector are assumed.

The characteristic polynomial of the active block is:
$$
Q(x)=x^3-(p_1p_2+p_2p_3+p_3p_1)x-2p_1p_2p_3,\tag3
$$
so that the probability vector $P_{n}$ can be computed by the recurrence relation:
$$
P_{n}=(p_1p_2+p_2p_3+p_3p_1)P_{n-2}+2p_1p_2p_3P_{n-3}\tag4
$$
with the elements of the three initial state vectors being:
$$
P_{0,i}=\tfrac12,\quad P_{1,i}=p_i,\quad P_{2,i}=(1-p_i)p_i.\tag5
$$
Finally the probability of ending with two balls of $i$-th color after $n$-th draw is:
$$
P_{n-1,i}\cdot p_i.\tag6
$$


问:
Link between Graph of function & Graph theory
Definition for graph of function: For a function $ f:A \rightarrow B $ , $ \{ (x,f(x)) : x \in A \} $ is the graph of the function $ f $.

Graph [ with no loops, not directed, not weighted ] is a mathematical object of the format $ G = <V,E> $ , where $ V  $ is set of vertices, and $ E $ is a set of edges between vertices and is defined as $ E = \{
\{u,v\} : u,v \in V \} $

Question: What is the relation between 'Graph' as defined in graph theory and 'Graph of function' as defined above? It seems to me that the definition for graph of a function is some sort of set of edges ( and the vertices are $x \in A $? ), but I want to be sure.

答:
Briefly, "yes", the questioner is perfectly correct. In more detail:

A Directed Graph, in the sense of vertices and edges, provided we agree to allow loops is exactly the same as a subset of the cartesian product $V \times V$ where $V$ is the set of vertices. The same structure is also called a binary relation on $V$.

Some graphs in that sense are functions from $V$ to $V$. These are the special graphs where
for every $v \in V$ there is exactly one edge $(v,w)$ in the graph for some $w \in V$.

When the graph of a function $f: \mathbb{R} \to \mathbb{R}$ is drawn, the points in the drawing (on the curve of the function) are all coordinate points in the plane
$\mathbb{R} \times \mathbb{R}$.

A function $f: \mathbb{R} \to \mathbb{R}$ can always be thought of as a set of ordered pairs
$$\{(x,y) \in \mathbb{R} \times \mathbb{R} \mid y = f(x)\}.$$
From this viewpoint the "graph of a function" is just a special case of a "directed graph" as in graph theory where the vertices are all the elements of $\mathbb{R}$ and the edges are all the pairs $(x, f(x))$. So an edge of the graph is a point in the plane.

This story still works for functions which are multivalued,  partial, or are not from $\mathbb{R}$ to $\mathbb{R}$, etc.

If you think about the adjacency matrix of a graph on a finite set of vertices, you have exactly the same underlying idea as plotting the graph of a function. The places you put a $1$ (for the existence of an edge) in the matrix are points being selected as part of the plot. In the case of the adjacency matrix it's a discrete space rather than the dense real plane, but the graph in each case is determined by picking out points in a space with two axes along which the set of vertices is laid out.


问:
Is this integral calculated correctly?
Calculate the integral $\int \frac{dx}{\sqrt{a^2-x^2}}$.

The proof of my textbook says that: Let $t=\frac{x}{a},dt=d(\frac{x}{a})$,

Then we have $\int\frac{dx}{\sqrt{a^2-x^2}}=\int \frac{dx}{a\sqrt{1-(\frac{x}{a})^2}}=\int \frac{dt}{\sqrt{1-t^2}}=\arcsin t+C=\arcsin\frac{x}{a}+C$.

I think it's not right. It dosen't consider the condition that $a<0$, and if then I have the following process:

$\int\frac{dx}{\sqrt{a^2-x^2}}=\int \frac{dx}{-a\sqrt{1-(\frac{x}{a})^2}}=-\int \frac{dt}{\sqrt{1-t^2}}=-\arcsin t+C=-\arcsin\frac{x}{a}+C^{'}$.

It's different from the former result.

答:
Actually what happens, for $~a\ne0~,$
$$\int\frac{dx}{\sqrt{a^2-x^2}}=\int \frac{dx}{|a|\sqrt{1-\left(\frac{x}{a}\right)^2}}$$
Now substitute $~t=\frac{x}{|a|}\implies dt=d\left(\frac{x}{|a|}\right)=\frac{dx}{|a|}~,$ which gives$$\int\frac{dx}{\sqrt{a^2-x^2}}=\int \frac{dt}{\sqrt{1-t^2}}=\arcsin(t)+c=\arcsin\left(\frac{x}{|a|}\right)+c$$ where $~c~$ is arbitrary constant.

Title: Is this integral calculated correctly?


问:
Limit of $\int_0^1\left(\frac {2}{\sqrt {(1-t^2)(1-xt^2)}}-\frac{x}{1-xt}\right)\,dt$ as $x\to 1^{-}$
While going through this question I was reminded of one of my earlier questions and I found that there is some unfinished business which needs some further exploration.

Let $$F(x) =\int_0^1\left(\frac{2}{\sqrt{(1-t^2)(1-xt^2)}} -\frac{x}{1-xt}\right)\, dt=\int_0^1 f(x, t) \, dt\tag{1}$$ for $x\in[0,1]$.

Let's observe that $$\lim_{x\to 1^-}f(x,t)=\frac{1}{1+t}\tag{2}$$ and hence it is natural to expect that $\lim_{x\to 1^-}F(x)$ should equal $\int_0^1 dt/(1+t)=\log 2$ but numerical evidence as well as some amount of elliptic function theory tells (see one of my questions linked earlier for details) us that this particular limit is $4\log 2$.

This suggests that there is some weird behavior of integrand as $x\to 1^{-}$ (in particular the convergence is not uniform).

I would like to have this limit evaluated using some analysis related to convergence of integrand as $x\to 1^-$. Any help in this direction would be appreciated.

Note: I have asked a new question instead of bumping an old one. The old question is more about solution verification and is related to elliptic integrals. I wanted to have a different perspective which involves general issues of uniform convergence to handle the limit of this integral.

答:
The non-uniform convergence prevents the direct use of the given integrand.

However, the original problem is to compute $L=\lim\limits_{x\to1^-}\big(2K(\sqrt x)+\log(1-x)\big)$.

One may replace $x/(1-xt)$ with "something better" that approximates the original "elliptic" integrand uniformly (on $t\in(0,1)$ as $x\to1^-$) and is still elementarily integrable.

Let's replace $x$ by $x^2$ (to get rid of $\sqrt x$ everywhere), and consider $$F(x,t)=\frac1{\sqrt{(1-t^2)(1-x^2 t^2)}},\\G(x,t)=\frac1{a_x(1+t)\sqrt{(1-t)(1-xt)}},$$ where $a_x=\sqrt{(1+x)/2}$ is chosen to have $\lim\limits_{x\to1^-}\big(F(x,t)-G(x,t)\big)=0$ uniformly.

Now $\int_0^1 F(x,t)\,dt=K(x)$ and $\int_0^1 G(x,t)\,dt=a_x^{-2}\tanh^{-1}a_x$. Hence
\begin{align*}
L&=\lim_{x\to1^-}\big(2K(x)+\log(1-x^2)\big)
\\&=\lim_{x\to1^-}\big(\frac{1}{a_x^2}\log\frac{1+a_x}{1-a_x}+\log(1-x^2)\big)
\\&=\lim_{x\to1^-}\left[\log\left(\frac{1+a_x}{1-a_x}(1-x^2)\right)+\frac{1-x}{1+x}\log\frac{1+a_x}{1-a_x}\right]
\\&=\lim_{x\to1^-}\log\left(\frac{(1+a_x)^2}{1-a_x^2}(1-x^2)\right)
\\&=\lim_{x\to1^-}\log\big(2(1+x)(1+a_x)^2\big)=4\log 2.
\end{align*}


问:
$\log_x(y)=m^4$ and $\log_y(x)=5/(m^3)$, what is the value of $m$?
$\log_x(y)=m^4$ and $\log_y(x)=5/(m^3)$, what is the value of $m$? </p>

答:
Another Hint:

$$\log_x{y}=\frac{1}{\log_y{x}}$$

The above is equivalent to the hint provided by Stahl.


Body:
Without AM-GM inequality:

Let, $x=a+b$ and $y=a-b$, then you get

$$6(a+b)^2+2(a+b)(a-b)+6(a-b)^2=9$$

$$14a^2+10b^2=9$$

$$x^2+y^2=(a+b)^2+(a-b)^2=2a^2+2b^2$$

$$x^2+y^2=2a^2+2 \times \frac{9-14a^2}{10}$$

$$x^2+y^2=\frac{9-4a^2}{5}≤\frac95$$

$$\color {gold}{\boxed {\color{black} {\text{max}[x^2+y^2]=\frac 95.}}}$$

If you want to find the minimum value of the expression, you can write in the same way:

$$x^2+y^2=2a^2+2b^2$$

$$x^2+y^2=2b^2+2 \times \frac{9-10b^2}{14}$$

$$x^2+y^2=\frac{4b^2+9}{7}≥\frac 97$$

Then, you get

$$\color {gold}{\boxed {\color{black} {\text{min}[x^2+y^2]=\frac 97.}}}$$


问:
How to interpret / read the following mathematical notations?
In one of the paper I am reading, I came across two mathematical notations which I cannot understand. The author also did not give any explanation. In both, note that $N$ is a positive integer.

Term 1: $x \in \{ 0,1 \}^N$

Term 2: $h\in \mathbb{R}^N_{>0}$

Thanks

答:
In general, given two sets $A$ and $B, A^B$ represents set of all functions $f$ such that $f:B\to A$.

The first one represents the set of all functions $f$ such that $f:\mathbb N \to \{0,1\}$. Recall set of all sequences of $0$s and $1$s, you'll notice that both these sets are the same.

So if $x\in \{0,1\}^\mathbb N$, then $x$ is a function from $\mathbb N$ to $\{0,1\}$.

The second one represents set of all functions $f$ such that $f:\mathbb N\to \mathbb R^+$. Recall set of all sequences of positive numbers, you'll notice that both these sets are the same.

So if $h\in \mathbb R_{>0}^\mathbb N$, then $h$ is a function from $\mathbb N$ to $\mathbb R_{>0}$, where $\mathbb R_{>0}=\{x\in \mathbb R: x\gt 0\}$


问:
Simplest (completely) positive maps
So, this question may be very simple, but I have an element $a$ in a C*-algebra, we can assume $a\in M_n(\mathbb{C})$ and therefore
$$ a = b a b^{-1} $$
has the same spectrum. Therefore, $x\mapsto bxb^{-1}$ maps positive elements to positive elements. But on the other hand the simplest completely positive map is $x\mapsto cxc^*$ and I don't get now why this is a positive map or where is the mistake.

答:
If $x$ is positive, we may decompose it as $x= vv^*$ and then
$$cxc^* = cvv^*c^* = cv(cv)^*$$
is positive. Hence, $x \mapsto cxc^*$ is positive.

Title: Simplest (completely) positive maps