问:
Basis of Image in a GF(5) matrix with variables
consider $A_\alpha := \begin{pmatrix} \alpha & 3 & 4\\ 3 & 1 & 3 \\ 4 & 2 & \alpha \end{pmatrix} \in \mathbb{Z_5}^{3\times3}$. Determine, dependant on $\alpha$, a Basis and the dimension of the Image of the standardinterpretation $F_{A_\alpha}$. What is the dimension of the kernel of $F_{A_\alpha}$? The standardinterpretation is defined as $F_A : K^{n\times1} \rightarrow K^{m \times 1}, x \mapsto F_A(x):=Ax $.

Okay. To determine the Image of a Matrix we just row reduce it and then we will see which columns are in the Image of that matrix.

Let's row reduce the matrix
$\begin{pmatrix}
\alpha &  3 & 4\\
 3&  1& 3\\
 4&  2& a
\end{pmatrix} \underset{\overset{R1 \leftarrow R_1+2\cdot R_2}{\longrightarrow}}{\overset{R_3 \leftarrow R_3 + 3 \cdot R_2}{\longrightarrow}}
\begin{pmatrix}
a+1 &  0& 0\\
 3&  1& 3\\
 0&  0& a+4
\end{pmatrix}$.

This doesn't look too bad, let's just look at the cases we get:

Let $\alpha = 0$ then we get $\begin{pmatrix}
1 &  0& 0\\
 3&  1& 3\\
 0&  0& 0
\end{pmatrix} \overset{R_2 \leftarrow R_2+2\cdot R_1}{\longrightarrow}
\begin{pmatrix}
1 &  0& 0\\
 0&  1& 3\\
 0&  0& 0
\end{pmatrix}
$ then we can see that a basis for $Image(A_\alpha)$ is  $\{\begin{pmatrix} a \\ 3 \\ 4 \end{pmatrix}, \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix}\}$. Since we have 2 elements in our basis for the Image, we know that $dim(Image)=2$.

We also remember that $dim(V) = dim(kernel) + dim(Image)$ therefore $3 = dim(kernel) + 2$ so we get $dim(kernel)=1$.

Do I have to repeat this process for all other 4 cases ($\alpha = 1,2,3,4$) or am I completly wrong about my approach? Thanks for any hints on how to solve this problem.

答:
Your row reduction is indeed correct. Your steps from there are problematic.  We have the reduced matrix
$$
\pmatrix{
a+1 &  0& 0\\
 3&  1& 3\\
 0&  0& a+4}.
$$
It is easy to see that for $\alpha \notin \{1,4\}$, further row-reduction brings us to the identity so that $A$ is invertible. In the case that $\alpha = 1$, we get
$$
\pmatrix{
2 &  0& 0\\
 3&  1& 3\\
 0&  0& 0} \leadsto
\pmatrix{
2 &  0& 0\\
 0&  1& 3\\
 0&  0& 0}.
$$
We deduce that the first two columns of $A$ form a basis of the image. That is, the image has basis
$$
\{(\alpha,3,4),(3,1,2)\} = \{(1,3,4),(3,1,2)\}.
$$
So, the dimension of the image is $2$ and the dimension of the kernel is $3 - 2 = 1$.

Similarly, for $\alpha = 4$, we get
$$
\pmatrix{
0 &  0& 0\\
 3&  1& 3\\
 0&  0& 3} \leadsto
\pmatrix{
 3&  1& 0\\
 0&  0& 3\\
0&0&0}.
$$
We deduce that the first and third columns of $A$ form a basis of the image. So, the dimension of the image is $2$ and the dimension of the kernel is $3 - 2 = 1$.


问:
Solve $\displaystyle\lim_{x \to \frac{\pi}{2}}{\biggl[(1+\cot x)^{\tan x}\biggr]^3}$

Evaluate the limit: $$\displaystyle\lim_{x \to \frac{\pi}{2}}{\biggl[(1+\cot x)^{\tan x}\biggr]^3}$$

Could someone help me to solve this limit?

I found the answer in symbolab, however I could not understand how the steps are explained.

Please explain this to me. Thank you.

答:
Although the easier way has already been shown, we usually solve the limits of the form $1^{\infty}$ by taking the natural logarithm on both sides.

So, if we call the limit as $L$, then we have

$\begin{align}
\ln L&=\displaystyle \lim_{x \to \frac{\pi}{2}}\ln {\biggl[(1+\cot x)^{\tan x}\biggr]^3}\\
&=\lim_{x \to \frac{\pi}{2}} 3\tan x \ln {(1+\cot x)} \\
&=\lim_{x \to \frac{\pi}{2}} \frac{3\ln {(1+\cot x)}}{\cot x} \\
&=3 \hspace{100pt} \left(\text{Since} \color{red}{\lim_{x \to 0}\frac{\ln (1+x)}{x} =1}\right) \\
\end{align}$

Therefore $L=e^3$.

ContentLicense: CC BY-SA 4.0


问:
Question on limit of Riemann Sum
I'm trying to follow the following calculation of the definite integral using a limit of Riemann Sums.  I have a question on the reasoning for a step in derivation of the limit which I will detail.

The original post is here.

$$\int_0^3 x^3 dx = \lim \limits_{n \to \infty} \sum_{i=1}^n f(x^*_i) \Delta x$$

$$= \lim \limits_{n \to \infty} \sum_{i=1}^n f\left(\frac{3i}{n}\right)\left(\frac 3n\right)$$

$$= \lim \limits_{n \to \infty} \left(\frac 3 n\right) \sum_{i=1}^n \left(\frac{3i}{n}\right)^3$$

$$= \lim \limits_{n \to \infty} \left(\frac 3 n\right)^4 \sum_{i=1}^n i^3$$

My first question is how is $\left(\frac{n(n+1)}{2}\right)^2$ multiplied to both sides when the RHS is unknown. Is the RHS assumed to be 1?

$$= \lim \limits_{n \to \infty} \left(\frac 3 n\right)^4 \left(\frac{n(n+1)}{2}\right)^2 \sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2$$

My second question is how are the next two equations derived from the prior?

$$= \left(\frac {3^4} 4\right) \lim \limits_{n \to \infty} \left(\frac 1 {n^4}\right) \left(\frac{n(n+1)}{2}\right)^2$$

$$= \left(\frac {3^4} 4\right) \lim \limits_{n \to \infty} \left(1 + \frac 1 n\right)^2$$

$$= \frac {3^4} 4$$

答:
$= \lim \limits_{n \to \infty} (\frac 3 n)^4 (\frac{n(n+1)}{2})^2 \sum_{i=1}^n i^3 = (\frac{n(n+1)}{2})^2$ should be

$= \lim \limits_{n \to \infty} (\frac 3 n)^4 (\frac{n(n+1)}{2})^2$ makes use of $\sum_{i=1}^n i^3 = (\frac{n(n+1)}{2})^2$.

(Also, I saw Riemann spelled incorrectly there.)

Addendum to answer additional question posted as comment on this question:

It should be

$$= \lim \limits_{n \to \infty} \left(\frac 3 n\right)^4 \left(\frac{n(n+1)}{2}\right)^2 = \left(\frac {3^4} 4\right) \lim \limits_{n \to \infty} \left(\frac 1 {n^4}\right) \left(\frac{n(n+1)}{\color{red}1}\right)^2,$$

because $$\left(\dfrac3n\right)^4 \left(\dfrac{n(n+1)}{2}\right)^2=3^4\left(\dfrac1n\right)^4\dfrac1{2^2}(n(n+1))^2 .$$

Finally,

$$\left(\dfrac1n\right)^4(n(n+1))^2=\dfrac{n^2}{n^4}(n+1)^2=\dfrac1{n^2}(n+1)^2=\left(\dfrac{n+1}n\right)^2=\left(1+\dfrac1n\right)^2.$$

Title: Question on limit of Riemann Sum


问:
Prove that a ring of fractions of $\mathbb Q[x]$ is a PID
I'm trying to prove the following:

Let
$$
R = \left\{\frac{f(x)}{g(x)}\mid f(x),g(x)\in\mathbb{Q}[x], g(x)\not=0\right\}\subseteq\mathbb{Q}(x).
$$
Show that $R$ is a principal ideal domain.

My approach was to first prove that for any integral domain $S$ we have
$$
\begin{equation}
\text{Frac}(S[x])\cong\text{Frac}(S)(x)
\end{equation}
$$
and so $R\cong\mathbb{Q}(x)$, hence it's a field and so a PID. But I couldn't come up with a proof of this statement. Is this the right approach or is there a better way?

答: $\Bbb Q$ is a field $\implies \Bbb Q[X]$ is a PID. Then as $R=S^{-1}(\Bbb Q[X])$ where $S=\Bbb Q[X] \setminus \{0\}$ and localization of a PID is a PID, hence, $R$ is also a PID.


问:
If $X, Y \sim N(0,1)$ are iid rv, then $U=\frac{X Y}{\sqrt{X^{2}+Y^{2}}} \sim N(0,1 / 4)$
I need to prove that if $X$ and $Y$ are iid $N(0,1)$ random variables, then $U=\frac{X Y}{\sqrt{X^{2}+Y^{2}}} \sim N(0,1 / 4)$.

I know that:

$U=\frac{\left(\frac{X+Y}{2}\right)^2-\left(\frac{X-Y}{2}\right)^2}{\sqrt{X^2 + Y^2}}$;

$\left(\frac{X+Y}{2}\right)^2, \left(\frac{X-Y}{2}\right)^2$ follows a chi-squared distribution with parameter 1;

$X^2 + Y^2 \sim \chi_2^2$ follows a chi-squared distribution with parameter 2.
But I don't know how to move forward from this or even if this is the right path.

答:
We can express in polar coordinates:
$$
\begin{aligned}
X &= R \cos \Theta \\
Y &= R \sin \Theta
\end{aligned}
$$
Where $R,\Theta$ are independent and $\Theta \sim U(0, 2\pi)$, that is, uniformly distributed on the interval $(0, 2\pi)$. (We could define $\Theta$ differently but this will do.) Then by the double angle formula we have:
$$
U = \frac{1}{2} R \sin 2 \Theta
$$
Since $\sin$ is $2\pi$-periodic and $\Theta$ is uniformly distributed, $\sin 2\Theta$ is identically distributed to $\sin \Theta$. And since $R$ and $\Theta$ are independent $R \sin 2\Theta$ is identically distributed to $R \sin \Theta$. Hence $2U = R \sin 2\Theta \sim N(0,1)$ and $U \sim N(0,1/4)$.

Note we don't need to know much about the distribution of $R$ other than the above relationship to the normal distribution.


问:
Cubic parametric curve rotation
I am drawing a spline based parametric curve which is described with the following functions (U for horizontal axis and Y for vertical).

$$U = A_u+B_u*p +C_u*p^2+D_u*p^3\\V = A_v+B_v*p +C_v*p^2+D_v*p^3$$
$$$$
How to rotate such a function $v(u)$ by $\theta$ degrees?

$$$$
My output equation should be composed of powers of three at max. Many thanks in advance

PS. I have seen a video How to Rotate any Curve by any Angle, but couldn't come up with a proper solution to my more complicated equation.

答:
A plane rotation is represented as

$$
R(\theta) = \left(
\begin{array}{cc}
 \cos (\theta ) & -\sin (\theta ) \\
 \sin (\theta ) & \cos (\theta ) \\
\end{array}
\right)
$$

and the rotated curve

$$
\left(
\begin{array}{c}
U\\
V
\end{array}
\right)_{\theta} = R(\theta)\left(
\begin{array}{c}
U\\
V
\end{array}
\right)
$$

Title: Cubic parametric curve rotation


问:
General formula for nth element of the sequence 1, 1, 0, 1, 1, 0, 1, 1, 0, ...

Give a general formula for nth element of the sequence 1, 1, 0, 1, 1, 0, 1, 1, 0, ...

I have been trying any combination, but I can't find a solution.

Body: How about $a_n=\dfrac43\sin^2\left(\dfrac{n\pi}3\right)$?


问:
"Boundaries" of cells in a cell decomposition
In Lee's text, a cell decomposition is defined as follows:

I was wondering, based on this definition, for any $e\in \mathcal{E}$, how can one characterize $\bar{e}\setminus e$? We know that the pre-image can be written as follows (under $\Phi$):
$$\Phi^{-1}(\bar{e}\setminus e) = \Phi^{-1}(\bar{e})\setminus \Phi^{-1}(e) = D\setminus \mathrm{Int}(D) = \partial D$$
with the second equality coming from the continuity of $\Phi$. However, we cannot simply apply $\Phi$ to both sides and say that $\bar{e}\setminus e$ is contained in the union of all cells of dimensions strictly less than that of $e$. Is the above procedure correct and, if so, how can one properly characterize $\bar{e}\setminus e$? Thanks!

答:
Since $D$ is compact, the closed map lemma shows that $\Phi$ is a closed map. Because it's closed and continuous, it follows that $\Phi\big(\overline A\big) = \overline {\Phi(A)}$ for every subset $A\subseteq D$ (see Proposition 2.30 in my book). In particular,
$$
\Phi(D) = \Phi\big(\overline {\operatorname{Int}D}\big)=\overline {\Phi(\operatorname{Int} D)} = \overline e,
$$
and so
$$
\overline e\smallsetminus e = \Phi(D)\smallsetminus e = \Phi(\partial D),
$$
which is contained in the union of all cells of lower dimension.


问:
Is it independent?
Let there be two tests for high blood sugar performed by two different
hospitals, A and B. Blood samples are taken from people are first sent to Hospital A and then Hospital B. Hospital A does not reveal the outcome of its tests to Hospital B.

Suppose that the tests are based on similar biological mechanisms. Let
A+ = Event that A declares blood sample has high blood sugar and
B+ = Event that B declares blood sample has high blood sugar

Are A+ and B+ independent or dependent?

I think that it's independent because Hospital A does not reveal the outcome of its tests to Hospital B.

However, does the fact that blood samples are taken from people are first sent to Hospital A and then Hospital B affect its independence?

答:
The events are absolutely not independent, since if you know the first test came back positive, that significantly increases the chances that the second one will come back positive. Remember that "dependence" does not imply any kind of causal relationship, just a statistical one.

Strictly speaking, your question is impossible to answer because it's rather vaguely stated - you would need to specify a probability distribution etc etc in order to actually be able to answer the question mathematically. However, for a reasonable interpretation of your scenario, the answer is that the two events are dependent.

For example, let's say a blood test works like this:

If the person has high blood sugar, the test comes back positive with probability $0.9$.

If the person has normal blood sugar, the test never comes back positive.

Let's also say that 5% of the population has high blood sugar. Well then with no information about anything, the probability that even $B$ will occur (the second test comes back positive) is $0.05\cdot 0.9=0.045$, so a 4.5% chance. However, if you know the first test came back positive, that means the person must have high blood pressure, which means the probability that the second test comes back positive is suddenly 90% rather than just 4.5%.

Title: Is it independent?


问:
prove the following linear functional is not induced by measure
Let the linear functional $I$ below defined on $C^1_c(\mathbb{R})$. as follows:

$$I(f) = \frac{df}{dx}(0)$$

Prove it's not induced by a measure,or even signed measure.that is no such measure makes $I(f) = \int fd\mu = \frac{df}{dx}(0)$.

I try to use sequence of function $f_n$ such that makes $\frac{df_n}{dx}(0) \to \infty$,but the total mass of $\int f_nd\mu \to 0$ ?If exist such measure it can not be the Lebesgue measure ,since we can construct such function,can it holds for some other measures?

答:
Take $f \in C^1_c(\mathbb{R})$ with $f \geq 0$ but $f'(0)<0.$ Then $I(f)<0$ but $\int f\,d\mu \geq 0.$ Contradiction!

Such an $f$ can be constructed by smoothing
$$
f_0(x) = \begin{cases}
2-x & \text{when $-1<x<1$} \\
0 & \text{otherwise}
\end{cases}
$$
for example by convolution with a narrow enough nascent delta function.


问:
How to show that transformation L is lineair?
Define L : $\mathbb{R}$[X, Y]$_\leq$ $_2$ $\longrightarrow$ $\mathbb{R}$[X]$_\leq$ $_2$

By L : f(X, Y) $\longrightarrow$ f(X, 2)

How to show that transformation L is lineair?

Of course I first tried it myself, but I ended up with a non-linear transformation and that cannot be correct because L is linear. I am very curious if someone can elaborate this question so that I can see what I did wrong. Thanks in advance for your help!

答:
We want to show that for all $f,g \in \Bbb R_{\leq 2}[X,Y]$ and $k \in \Bbb R$, $L(f + k g) = L(f) + kL(g)$. To that end, note that
\begin{align}
L(f + kg) & = L(f(x,y) + kg(x,y)) = f(x,2) + kg(x,2)\\
L(f) + kL(g) &= [f(x,2)] + k[g(x,2)] = f(x,2) + kg(x,2).
\end{align}
The result is indeed the same, so $L$ is linear.


问:
$n$-linear alternating form with $\dim{V}<n$ $\overset{?}{\text{is}}$ the $0$-form
Prove that every $n$-linear alternating form on a vector space of dimension
less than $n$ is the zero form.</p>

答: Just use the definition of alternating, along with the fact that multilinearity says that the function is totally determined by what happens when you put basis vectors in each slot.</p>

ContentLicense: CC BY-SA 3.0


答: I think $a_n =1-\frac{1}{3}\left(1+e^{\frac{2i\pi}{3}n}+e^{\frac{4i\pi}{3}n}\right)$ works.

Title: General formula for nth element of the sequence 1, 1, 0, 1, 1, 0, 1, 1, 0, ...

Body:
Give a general formula for nth element of the sequence 1, 1, 0, 1, 1, 0, 1, 1, 0, ...

I have been trying any combination, but I can't find a solution.


答:
Hint: This transformation can be written as

$L: aX^2 + b XY + c Y^2 + d X + e Y + f \mapsto aX^2 + 2bX + 4c + dX+ 2e + f= aX^2 + (2b+d)X + (4c+2e+f).$

Body:
Define L : $\mathbb{R}$[X, Y]$_\leq$ $_2$ $\longrightarrow$ $\mathbb{R}$[X]$_\leq$ $_2$

By L : f(X, Y) $\longrightarrow$ f(X, 2)

How to show that transformation L is lineair?

Of course I first tried it myself, but I ended up with a non-linear transformation and that cannot be correct because L is linear. I am very curious if someone can elaborate this question so that I can see what I did wrong. Thanks in advance for your help!


问:
Show rigorously that the sum of integrals of $f$ and of its inverse is $bf(b)-af(a)$

Suppose $f$ is a continuous, strictly increasing function defined on a closed interval $[a,b]$ such that $f^{-1}$ is the inverse function of $f$. Prove that,
  $$\int_{a}^bf(x)dx+\int_{f(a)}^{f(b)}f^{-1}(x)dx=bf(b)-af(a)$$

A high school student or a Calculus first year student will simply, possibly, apply change of variable technique, then integration by parts and he/she will arrive at the answer without giving much thought into the process. A smarter student would probably compare the integrals with areas and conclude that the equality is immediate.

However, I am an undergrad student of Analysis and I would want to solve the problem "carefully". That is, I wouldn't want to forget my definitions, and the conditions of each technique. For example, while applying change of variables technique, I cannot apply it blindly; I must be prudent enough to realize that the criterion to apply it includes continuous differentiability of a function. Simply with $f$ continuous, I cannot apply change of variables technique.

Is there any method to solve this problem rigorously? One may apply the techniques of integration (by parts, change of variables, etc.) only after proper justification.

The reason I am not adding any work of mine is simply that I could not proceed even one line since I am not given $f$ is differentiable. However, this seems to hold for non-differentiable functions also.

I would really want some help. Pictorial proofs and/or area arguments are invalid.

答:
Let $P' = \{f(x_0=a), f(x_1), \cdots, f(x_n=b)\}$ be a partition of $f^{-1}$ on $[f(a), f(b)]$. Then we know $P = \{x_0, \cdots, x_n\}$ is a partition of $f$ on $[a,b]$. Observe that
\begin{align*}
    L(f,P) + U(f^{-1}, P') &= \sum_{k=0}^{n-1}f(x_k) (x_{k+1} - x_k) + \sum_{k=0}^{n-1}x_{k+1}(f(x_{k+1}) - f(x_k))\\
    &= \sum_{k=0}^{n-1} x_{k+1}f(x_{k+1}) - x_kf(x_k)\\
    &= bf(b) - af(a)
\end{align*}
Similarly we have
\begin{align*}
    U(f,P) + L(f^{-1}, P') = bf(b) - af(a)
\end{align*}
Thus, for all $P'$ we have
\begin{align*}
    bf(b) - af(a) - U(f^{-1}, P') = L(f,P) \le \int_a^b f(x)dx \le U(f,P) = bf(b) - af(a) - L(f^{-1}, P')
\end{align*}
Thus $\int_a^b f(x) dx = bf(b) - af(a) - \int_{f(a)}^{f(b)} f^{-1}(x)dx$.


问:
Value of $\left \lfloor{x}\right \rfloor+\left \lfloor{-x}\right \rfloor$?
While reading about greatest integer function from a book, I found a question as $\left \lfloor{x}\right \rfloor+\left \lfloor{-x}\right \rfloor$  ?

I attempted it as follows:

We know:

$x-1<\left \lfloor{x}\right \rfloor< x\tag1$

Also then: $-x-1 < \left \lfloor{-x}\right
\rfloor < -x\tag2$

Adding $(1)$ & $(2)$, we get

$-2< \left \lfloor{x}\right \rfloor+\left \lfloor{-x}\right \rfloor<0$.

This is the answer which I got, but the actual answer was  $\left \lfloor{x}\right \rfloor+\left \lfloor{-x}\right \rfloor= -1$. I am not getting this. Where my method has gone wrong? Please help me.

答:
$T=\left\lfloor x \right\rfloor  + \left\lfloor { - x} \right\rfloor $

When x=n, only integer

$\underbrace {\left\lfloor x \right\rfloor }_n + \underbrace {\left\lfloor { - x} \right\rfloor }_{ - n} \Rightarrow n - n = 0$

When x=n+a, $a\in(0,1)$

$\left\lfloor {n + a} \right\rfloor  + \left\lfloor { - n - 1 + 1 - a} \right\rfloor $

$a \in \left( {0,1} \right)$, therefore $a' = 1 - a \in \left( {0,1} \right)$

$\underbrace {\left\lfloor {n + a} \right\rfloor }_n + \underbrace {\left\lfloor { - n - 1 + a'} \right\rfloor }_{ - n - 1} = n - n - 1 =  - 1$.

Hence when x is an integer $T=0$, else $T=-1$


问:
Combinatorics, counting and number theory.
Find number of ordered quadruples (a,b,c,d) [of positive integers] such that $lcm[a,b,c]= lcm[a,b,d]= lcm[a,c,d]= lcm[b,c,d] = 2^r\cdot3^s$

So i approached it like two of a,b,c,d has max power of 2 = r. For rest two, they have r+1 choices each. So it would be $\binom{4}{2} \cdot (r+1)^2$. Same for the power of three.

So as per me answer should be $(\binom{4}{2} \cdot (r+1) \cdot (s+1))^2$ but the answer to this problem is not given in my book.

I would really appreciate if you could tell whether my answer is correct and whether this is the correct approach and if it is not correct approach, then why?

Thanks!

答:
It seems you have double counting.  If $2^r|a,b,c$ then you count that case three times more than it should. (When you choose $a,b$ from the ${4\choose 2}$ options and choose power of $2$ of $c = r$ as one of the $r+1$ cases, that is the same as choosing $a,c$ from the ${4\choose 2}$ options and choosing the power of $2$ of $b = r$ as one of the $r+1$ cases.

I'd do it by cases  If exactly $2$ of $a,b,c,d$ have power of $2=r$ and the other two powers of $2$ are not $r$ there are ${4\choose 2}r^2$.  If exactly $3$ have $r$ there are ${4\choose 3}r$ and if all $4$ have $r$ there are ${4\choose 4}$.

So there are $6r^2 + 4r + 1$ ways to allocate the powers of $2$.

And there are $6s^2 + 4s + 1$ ways to allocate the powers of $3$.

So $(6r^2 + 4r + 1)(6s^2 + 4s + 1)$.

Someone might want to check my reasoning.


问:
Implicit Differentiation Help
I have to use implicit differentiation to find $\frac{dy}{dx}$ given:

$$x^2 \cos(y) + \sin(2y) = xy$$

I don't even know where to begin, I missed the class where we went over implicit differentiation, and because of that, I am completely stuck.

Thank you everyone.

Edit: I don't know how to make the equation look all nice and whatnot, so sorry about that

答:
Differentiating w.r.t $x$ we have $$2x \cos{y} - x^2\sin{y}\frac{\text dy}{\text dx} + 2\cos{2y}\frac{\text dy}{\text dx} = y + x \frac{\text dy}{\text dx}$$

From the above you have
\begin{align*}
2x \cos{y} - y &= x \cdot\frac{dy}{dx} + \sin{y}\cdot\frac{dy}{dx} -2\cos{2y}\cdot\frac{dy}{dx} \\ &= \frac{dy}{dx} \cdot \Bigl[ x + \sin{y} - 2\cos{2y}\Bigr]
\end{align*}

Take all the $\frac{\text dy}{\text dx}$ to one side and then simplify. And look at this Wikipedia Link. You have some worked out examples.

Title: Implicit Differentiation Help


答:
Let $k$ be a field, and let $V$ be a vector space over $k$, with $\dim V <n$.  Let $\omega$ be an alternating $n$-form on $V$.  Then $\omega : V^n \to k$ is an alternating, multilinear map.  For general input $v_1,\ldots,v_n$, there exists a linear dependence $\sum c_i v_i=0$ among the $v_i$, by dimension of $V$.  If $c_n \neq 0$ (e.g.), we can scale $c_n=1$.  Then solving for $v_n$, we may write
$$\omega(v_1,\ldots,v_n)=\omega\left(v_1,v_2,\ldots,v_{n-1},-\sum_{i<n} c_iv_i\right)=-\sum_{i<n}c_i \omega\left(v_1,v_2,\ldots,v_{n-1},v_i\right),$$
in which the last step uses multilinearity.  Since $\omega$ is alternating, switching the $i$th and $n$th inputs negates the value of $\omega$.  Yet here $\omega(v_1,\ldots,v_{n-1},v_i)$ is clearly invariant under such a permutation, hence
$$\omega(v_1,\ldots,v_{n-1},v_i)=0$$
for all $i$.  Thus $\omega$ is the zero map, as desired.

In general, we can show (by constructing a basis) that the dimension of the space of alternating $n$-forms on a $d$-dimensional space is given by $\binom{d}{n}$, which vanishes for $n > d$.

Title: $n$-linear alternating form with $\dim{V}<n$ $\overset{?}{\text{is}}$ the $0$-form

Body:
Prove that every $n$-linear alternating form on a vector space of dimension
less than $n$ is the zero form.</p>


问:
Basic Series Question-Euler
I'm studying Calc 2 and I have a basic series question.

Leonhard Euler discovered that $\sum _{n=1}^{\infty }\:\frac{1}{n^2}=\frac{\pi ^2}{6}$.

Given $\sum \:_{n=1}^{\infty \:}\:\frac{1}{\left(2n\right)^2}$, what does the series equal?

My work:

Let k=2n

$\sum \:_{k=2}^{\infty \:}\:\frac{1}{k^2}=\sum \:_{n=1}^{\infty \:}\:\frac{1}{\left(2n\right)^2}$.

Now $\sum \:_{k=2}^{\infty \:}\:\frac{1}{k^2}=-1+\sum \:_{k=1}^{\infty \:}\:\frac{1}{\left(k\right)^2}$

Finally: $\frac{\pi ^2}{6}-1=-1+\sum \:\:_{k=1}^{\infty \:\:}\:\frac{1}{\left(k\right)^2}$

My textbook says that $\sum \:_{n=1}^{\infty \:}\:\frac{1}{\left(2n\right)^2}$=$\frac{\pi ^2}{24}$ though I cannot see how, based on my work above.

Does anyone see an error?

Thanks in advance!

答:
In summations, one should avoid scaling of the summation index like $2k=p$.
Because then then you have to cosider the even terms in the summand.
The correct way would be to write $$S=\sum_{k=1}^{\infty } \frac{1}{(2k)^2}$$, then use $2k=p$, to write $$S=\sum_{p=2}^{\infty~~'}\frac{1}{p^2}$$ notice prime over sigma, this will remind you that you cannot take $p=2,3,4,5,6,...$ instead you have to take $p=2,4,6,8...$
Then, $$S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...$$
$$\implies S=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....\right)=\frac{\pi^2}{24}$$

However, it is better to avoid transformation which scales the summing index like $p=2k, p=2k+1, p=3k+1, p=3k+4..$ transformations which shift the summation index are OK like $p=n+2, p=n-2, p=n+4, p=n+7,....$

Title: Basic Series Question-Euler


问:
Question related to the following function $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$
Let $f:\left( {0,1} \right) \to R$ be defined by $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$
, where 'b' is a constant such that $0 < b < 1$. Then

(A) $f$ is not invertible on (0,1)

(B) $f\ne f^{-1}$ on (0,1)and $f'(b)=\frac{1}{f'(0)}$

(C) $f= f^{-1}$ on (0,1)and $f'(b)=\frac{1}{f'(0)}$

(D) $f^{-1}$ is differentiable on (0,1)

My approach is as follow

$f:\left( {0,1} \right) \to R$, $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$,$0 < b < 1$

$f'\left( x \right) = \frac{{\left( {1 - bx} \right)\frac{d}{{dx}}\left( {b - x} \right) - \left( {b - x} \right)\frac{d}{{dx}}\left( {1 - bx} \right)}}{{{{\left( {1 - bx} \right)}^2}}}$

$f'\left( x \right) = \frac{{\left( {1 - bx} \right) \times  - 1 - \left( {b - x} \right) \times \left( { - b} \right)}}{{{{\left( {1 - bx} \right)}^2}}} = \frac{{\left( {1 - bx} \right) \times  - 1 + \left( {b - x} \right) \times b}}{{{{\left( {1 - bx} \right)}^2}}} = \frac{{\left( { - 1 + bx} \right) + \left( {{b^2} - bx} \right)}}{{{{\left( {1 - bx} \right)}^2}}} = \frac{{{b^2} - 1}}{{{{\left( {1 - bx} \right)}^2}}}$

$x\in(0,1)$ and $b\in(0,1)$, then $bx\in(0,1)$, hence $f'(x)<0$, hence my assumption is that $f$ is invertible as it is decreasing function when $0<x<1$ but official answer is (A) that is $f$ is not invertible on (0,1)

答:
Just observe that there is no $x$ such that $f(x) =\frac{1}{b}$. If any such $x$ exists then it would imply that $b^2=1$ which is certainly not possible. This shows that $f$ is not surjective which implies that $f$ is not invertible.

Edit:
One more way to see that $f$ is not surjective is to observe that $-1 < f(x) < b$ by simply manipulating the inequalites $0<x<1$ and $0<b<1$.


问:
Computing Sum of an Interesting Series
Let $f_{n}=\displaystyle\prod_{k=1}^{n}(e^{kx}-1)$ with $x\in\mathbb{R}$ and $n\in\mathbb{N^{*}}$ and define :
$$
a_{n}=f_{n}^{(n)}(0)\qquad\text{and}\qquad b_{n}=f_{n}^{(n+1)}(0)
$$
How can we show that :
$$
\frac{1}{4}\sum_{n=1}^{\infty}\frac{a_{n}}{b_{n}}=2-\frac{\pi^{2}}{6}
$$

答:
$a_{n}=(n!)^{2}$

$\displaystyle b_{n}=\frac{n(n+1)^{2}(n!)^{2}}{4}$

This can be seen if you rewrite $f_{n}(x)$ as :
$$f_{n}(x) :=n!x^{n}\displaystyle\prod_{k=1}^{n}\frac{e^{kx}-1}{kx}$$ and then apply the general Leibniz rule.

Title: Computing Sum of an Interesting Series


答: You also may show that $f(f(x))=x$.

Title: Question related to the following function $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$

Body:
Let $f:\left( {0,1} \right) \to R$ be defined by $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$
, where 'b' is a constant such that $0 < b < 1$. Then

(A) $f$ is not invertible on (0,1)

(B) $f\ne f^{-1}$ on (0,1)and $f'(b)=\frac{1}{f'(0)}$

(C) $f= f^{-1}$ on (0,1)and $f'(b)=\frac{1}{f'(0)}$

(D) $f^{-1}$ is differentiable on (0,1)

My approach is as follow

$f:\left( {0,1} \right) \to R$, $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$,$0 < b < 1$

$f'\left( x \right) = \frac{{\left( {1 - bx} \right)\frac{d}{{dx}}\left( {b - x} \right) - \left( {b - x} \right)\frac{d}{{dx}}\left( {1 - bx} \right)}}{{{{\left( {1 - bx} \right)}^2}}}$

$f'\left( x \right) = \frac{{\left( {1 - bx} \right) \times  - 1 - \left( {b - x} \right) \times \left( { - b} \right)}}{{{{\left( {1 - bx} \right)}^2}}} = \frac{{\left( {1 - bx} \right) \times  - 1 + \left( {b - x} \right) \times b}}{{{{\left( {1 - bx} \right)}^2}}} = \frac{{\left( { - 1 + bx} \right) + \left( {{b^2} - bx} \right)}}{{{{\left( {1 - bx} \right)}^2}}} = \frac{{{b^2} - 1}}{{{{\left( {1 - bx} \right)}^2}}}$

$x\in(0,1)$ and $b\in(0,1)$, then $bx\in(0,1)$, hence $f'(x)<0$, hence my assumption is that $f$ is invertible as it is decreasing function when $0<x<1$ but official answer is (A) that is $f$ is not invertible on (0,1)

Body: You also may show that $f(f(x))=x$.


问:
Evaluate the Sum $\sum_{i=0}^\infty \frac {i^N} {4^i}$

$$\displaystyle\sum_{i=0}^\infty \frac {i^N} {4^i}$$

I'm supposed to evaluate this as I'm working through Data Structures and Algorithm Analysis in C++.  I've solved similar problems, and after trying all of the algebra-fu manipulations I could think of, I'm stumped.

My first efforts were, similar to how I solved the three previous problems in the series, to try doing things like setting it equal to S, dividing both sides by four, and subtracting the equations, in various ways in an attempt to cancel terms.  However, I'm missing some step to make it happen.

答:
Note that for $\vert x \vert < 1$, we have
$$\sum_{i=0}^{\infty} x^i = \dfrac1{1-x}$$
We then have
$$\sum_{i=0}^{\infty} ix^i = x \dfrac{d}{dx}\left(\sum_{i=0}^{\infty} x^i \right) = x \dfrac{d}{dx}\left(\dfrac1{1-x}\right)$$
If we denote the operator $x\dfrac{d}{dx}$ as $L$, we then have
$$\sum_{i=0}^{\infty} ix^i = L\left(\dfrac1{1-x}\right)$$
$$\sum_{i=0}^{\infty} i^2x^i = L\left(L\left(\dfrac1{1-x}\right) \right) = L^2\left(\dfrac1{1-x}\right)$$
And in general,
$$\sum_{i=0}^{\infty} i^Nx^i = L^N\left(\dfrac1{1-x}\right)$$
Hence,
$$\sum_{i=0}^{\infty} \dfrac{i^N}{4^i} = \left.L^N\left(\dfrac1{1-x}\right) \right \vert_{x=1/4}$$
Also, this is nothing but
$$\text{Li}_{-N}\left(\dfrac14\right),$$
where $\text{Li}_s(z) = \displaystyle \sum_{k=1}^{\infty} \dfrac{z^k}{k^s}$, the Polylogarithm function.</p>


答:
If $n$ is divisible by $3$, then $a_n=0$. Otherwise, $a_n=1$.

So, the formula $a_n=\min(n \bmod 3,1)$ should work.


问:
Does the limit product rule apply if one limit does not exist?
Assume we want to calculate $\lim_{x \rightarrow 0} g(x) f(x)$.

Moreover, $\lim_{x \rightarrow 0} g(x) = \infty$ and
$\lim_{x \rightarrow 0} f(x) = n$ where $n \neq 0$ is some natural number.

Is it possible to conclude:
$$ \lim_{x \rightarrow 0} g(x) f(x) = \lim_{x \rightarrow 0} g(x) \lim_{x \rightarrow 0} f(x).$$

I am kind of insecure since I read on some pages that it is only possible if both limits (i.e. the one of $f$ and $g$) exist. However, on other pages, I read that in the case stated above the claim does hold.

Thank you in advance for your help!

答:
The expression
$$
\lim_{x \rightarrow 0} g(x) \lim_{x \rightarrow 0} f(x)
$$
makes no sense, as $\infty$ isn't a number and you can't do arithmetic with it.

However, as a mnemonic device, saying that $\lim_{x \rightarrow 0} g(x) f(x)$ is of "type $\infty\cdot n$", and therefore diverges to infinity is fine, after you prove that this is indeed always the case for $n\neq0$.


问:
Confused: using Taylor series to find derivative
TL;DR: read bolded parts

Lets say I have f(x) = sin(x^2) and I want the f''''''(0) (6th derivative). Using taylor series, this is really simple. We plug in x^2 into the taylor polynomial of sin(x), and get this:

Then the 6th derivative is 1/3! * 6! = 120.

I am confused because taylor series seems really unrelated; there should be an equally easy way to do this just with derivatives and chain rule (no detour to taylor series). But when I bash it out, I don't get a simple solution.

(120 on the last line, typo)

Why does taylor series come up in finding derivatives?

答:
There is (a-priori) no reason to believe, that calculating the sixth derivative of a function
will be easy.
In fact, if you were to fully write down the 6th derivative, this would be a really complicated expression.

The reason that the trick with the Taylor polynomial works so well here
is because $\sin(x)$ has a nice and well-known Taylor polynomial at the point $x=0$.

And for many functions, calculating the Taylor polynomial is also some work,
because it requires calculating derivatives.

And overall, Taylor polynomials already provide a lot of useful information about derivatives,
which is one of the reasons why mathematicians like Taylor polynomials.


答: Let $\omega$ be an $n$-multilinear, alternating form on $V$, where $\dim(V)=m<n$. Let $\{e_1,\ldots,e_m\}$ be any basis of $V$. Because $\omega$ is multilinear, $\omega$ is determined entirely by its values on $n$-tuples of elements from $\{e_1,\ldots,e_m\}$ (apply $\omega$ to an $n$-tuple of arbitrary vectors; write each of them as a linear combination of the $e_i$'s, and use the linearity of $\omega$ in each entry). But $n>m$, so every such $n$-tuple has a repeat...</p>


问:
A formula for the exterior angles of an irregular polygon, given interior angles?
So I took a challenge from my Geometry teacher to create code that when the user gives the computer how many angles / sides a polygon has and the angle of each of the interior angles it could find each of the exterior angles whether it is regular or irregular. For example the user tells the computer they have a four-sided shape (quadrilateral), the interior angles are $70, 75, 110, 145$. The computer would then be able to output each corresponding angle in the proper order. My problem is that with coding you can't just type in the human logic so I need a formula to work from: what on earth would be a formula that could apply to any polygon with any number of sides (irregular or regular)?

答:
The other answer is not complete, because you can have a non convex polygon (i.e one interior angle can be 240) and to say that $180-$input_angle is not all correct.

Thr proper way to to this is:

read angles and save them as a list angles_list

initialize and empty array with size equal to angles_list with name exteriors.

loop over elements in angles_list, if element is greater than 180, then append to exteriors $360-$element, else append $180-$element

return exteriors

Also you can add validations to check if the input angles indeed correspond to a polygon (i.e, the sum of angles must be a multiple of 180)


问:
How many "ANGLES" does a pentagon have?
If asked "How many angles does a pentagon have?" would the exterior and interior angles count as 2 angles for each vertex?</p>

答: The "angles" in the question "How many angles does [some shape] has" almost always refers to the number of internal angles. So, a pentagon has $5$ angles.

Title: How many "ANGLES" does a pentagon have?


问:
Show that $\| y\| =1$ and $\| y-x_j\|\geq \| x_j\|$
I have an idea to solve this exercise but I am confused. The exercise says:

Let $X$ be an infinite dimensional normed vector space over $\mathbb{K}$, where $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$

Let $x_1, x_2, \dots ,x_n\in X$. Show that there exists $y\in X$ such that $\| y\| =1$ and $\| y-x_j\|\geq \| x_j\|$  for all $j=1,2, \dots ,n$.
We know that if $0≠x\in X$, then there exists $f\in X^*$ such that $\|f\|=1$ and $f(x)=\mid \mid x\mid \mid$.

I was thinking we can show that $f(y)=1$ for some $y\in X$ then we are done. Or else I am thinking that we can say $\|f(y)\|= \| \| y\| \|=\|y\|=\|f\|\|y\|\Leftrightarrow \|y\|=\frac{\|f(y)\|}{\|f\|}$, and probably try to show that $\frac{\|f(y)\|}{\|f\|}=1$, but I don't know how I would do that.

Thank you

答:
You are on the right track. For $1 \le j \le n$ let $f_j \in X^*$ be a bounded functional such that $f_j(x_j) = \|x_j\|$ and $\|f_j\| =1$. Then the intersection of kernels $$\bigcap_{j=1}^n \ker f_j$$
is a nontrivial subspace of $X$. Indeed, we can define a linear map $f : X \to \Bbb{K}^n$ by $f(x) = (f_1(x), \ldots, f_n(x))$. Then $\ker f = \bigcap_{j=1}^n \ker f_j$ so if $\bigcap_{j=1}^n \ker f_j = \{0\}$ we would have that $f$ is an injective linear map from an infinite-dimensional space $X$ to a finite-dimensional space $\Bbb{K}^n$. This is a contradiction so $\bigcap_{j=1}^n \ker f_j$ is nontrivial.

Pick $y \in \bigcap_{j=1}^n \ker f_j$ such that $\|y\|=1$ and notice
$$\|y-x_j\| = \|f_j\|\|y-x_j\| \ge |f_j(y-x_j)| = |f_j(y)-f_j(x_j)| = |0-\|x_j\||=\|x_j\|$$

which proves the claim.


问:
Two Intuition Questions: Conservative VF and Dot Product
All,

In doing multivariable calculus, I have encountered line integrals. I have no problem calculating these, but I do have a question about the intuition behind two ideas (one of which, dot product, is early on in MV calculus).

First, the role of gradients in a conservative vector field. I do understand how to determine if a vector field is conservative: it is the gradient of a potential function. And I can do the math from there. But what I don't understand is the intuition behind this. What does the gradient have to do with the fact that I don't have to calculate a curve in the conservative vector field? I'm having a hard time understanding the WHY even though I get the HOW.

Second, on dot products. These are easy to do, and I understand that you need to dot product the force and the dr before taking the integral. But, again, why? The dot product, I guess, shows the relationship between two vectors - if it is zero, the vectors are orthogonal; and the dot product increases the closer the vectors get to parallel. But what is the intuition behind why the dot product can be used to determine work, i.e., w = F * d. Again, I can solve this, but I don't understand the intuition.

Thanks.

答:
Gravity is a good example of a conservative force.

If you expend energy to go up a hill, you get that energy back when you come down the hill.  Regardless of the path you take from A to B, the only thing that matters is the net change in altitude between the two points.  If you travel back to your starting point there is no net change in altitude and no net-energy spent.

Conservative force fields tend to be centripetal.  e.g $F(x,y,z) = -x\mathbf {\hat i}+-y\mathbf {\hat j}+ -z\mathbf {\hat k}$ is a conservative force with all of the force vectors pointing toward the origin.  It could be that all of the force vectors are parallel, and still be a conservative force field.  e.g. $F(x,y,z) = 1\mathbf {\hat i}+2\mathbf {\hat j}+ 3\mathbf {\hat k}$

What about the dot product?  Again using gravity as our model for our force field.  When you are traveling at constant altitude, you are spending no energy to defeat gravity.  You direction of travel is perpendicular to the force and $\mathbf {a\cdot b} = 0$ when $\mathbf a$ is perpendicular to $\mathbf b$.  As you climb, the dot product will give you the vertical component of your travel, and thus the energy required to climb.


答:
Well, one can argue and justify a variety of answer but it comes down to what we mean when we ask the question.

I figure if we ask how many angles a pentagon has we absolutely should mean exactly the same thing as when we ask how many angles does any type of polygon have.

And if we ask how many angles a triangle has the universally accepted answer is ... $3$.

So what do we mean by that?

Presumably we mean how many interior angles and we assume one interior angle at each vertex.

So if we mean the same thing then the answer to "how many angles does a pentagon have" should be $5$.

Of course, we could (and many of us will; we just can't help ourselves... we are mathematicians after all) argue that a triangle really has $6$ angles because we should count exterior angles. But such arguments are usually met with a smirk and walking away.

The answer is $5$.

Title: How many "ANGLES" does a pentagon have?

Body: If asked "How many angles does a pentagon have?" would the exterior and interior angles count as 2 angles for each vertex?</p>


问:
Does a function from the empty set to the empty set exist?
I know that every function from the empty set to any other set is the empty function.

I also know that there is no function to the empty set from any other set.

Now, what if both the domain and codomain are empty? Would such a function exist?

答:
Two key things to know about the empty set are: Every statement that says

$$\forall x \in \emptyset $$

is (vacuously) true, and every statement that says

$$\exists x \in \emptyset $$

is (trivially) false.

So if you express the conditions for "$f$ is a function from $A$ to $B$" in the formal language of $\forall$ and $\exists$ you can answer your question.

(I think I also need to add that you "work from the outside in", i.e.

$$ \forall x \in \emptyset \;\;\exists y \in \emptyset \;\;\;P(x,y)$$

is true, no matter what statement $P(x,y)$ is.)


问:
Game theory/probability puzzle
Part 1:

You play a game where there is a box with 100$ and there are two players, each of you should write a number 0-100 on paper, then you show your numbers, if the sum is higher than 100 then each of you get 0 dollars, else you get what you wrote. What is your strategy?

Part 2:

You play the same game but your opponent told you that he is putting 80 (he might change his mind) - what is your plan?

For part 1, I think the Nash equilibria are (my number, opponent's number) = (x, 100 - x) for x between 0 and 100? Is this right, and if so, how to choose between them to pick your move? I was thinking about saying 50 as my number for symmetry/aesthetic reasons but would be interested in how one is supposed to tackle this problem. I drew out a reward matrix and tried iterated removal of weakly dominated strategies to get (50, 50) as the only cell remaining - would be grateful for any thoughts on the validity of this.

For part 2, I thought, if you're allowed to say something to the opponent, you could say that you were going to put 50, to try to force your opponent down to 50? Again I have no idea how to approach this. Any help would be much appreciated.

答: For the second part, I think I'd say to her "I'll choose the same number as you do". This hints to the other player that she should choose $50$, because both larger and smaller numbers result in lower payments ($0$ if higher, $x<50$ if lower). Of course, this depends on the credibility of your "threat".

Title: Game theory/probability puzzle


问:
Hello, i need to test this property of Logic Statements, Connectives and Quantifiers
(∃xp(x) → ∃xq(x)) → ∃x(p(x) → q(x))

I have to explain the procedure, starting with (∃xp (x) → ∃xq (x)) until reaching → ∃x (p (x) → q (x)).
To solve this exercise I have taken advantage of this theory:
Recall that given any referential S and conditions p (x),
q (x) appropriate (relative to the elements of S), we have defined ∃xp (x) as
{x ∈ S | p (x)} = ∅ and ∀xp (x) as {x ∈ S | p (x)} = S.
so i take (∃xp(x) → ∃xq(x)) following the theory, I define (∃xp(x) → ∃xq(x)) as {x ∈ S | p(x)} = ∅ → {x ∈ S | q(x)} = ∅

答:
You used semantics in your proof sketch, which is fine. A formal proof however doesn't do this and goes for example along the following lines (in my favorite deduction system for first-order logic):

We assume $\exists xq(x)\lor\neg\exists xp(x)$. (Because $a\rightarrow b\equiv b\lor\neg a$, which can be proved in my favorite deduction system.)
This implies $\exists xq(x)\lor\exists x(\neg p(x))$. (Because $\forall x\phi\rightarrow\exists x\phi$, which can also be proved.)
So, as any of $q(x)$ and $\neg p(x)$ implies $p(x)\rightarrow q(x)$, you can use the introduction and elimination rule for $\exists$ to establish $\exists x(p(x)\rightarrow q(x))$.

Good luck with logic!


问:
Sufficient condition for irreducibility of polynomial $f(x,y)$
Suppose we have a non-constant polynomial $f(x,y)\in\mathbb{Q}[x,y]$ such that the following two conditions are satisfied:

1) For every $x_0\in\mathbb{Q}$, the polynomial $f(x_0, y)\in\mathbb{Q}[y]$ is irreducible.

2) For every $y_0\in\mathbb{Q}$, the polynomial $f(x,y_0)\in\mathbb{Q}[x]$ is irreducible.

My question is this:

Can we conclude from these two conditions that $f(x,y)$ is an
  irreducible polynomial in $\mathbb{Q}[x,y]$?

I am leaning on "yes" side. This was my unsuccessful attempt: Suppose $f(x,y)$ is reducible, then $f(x,y)=g(x,y)h(x,y)$ for some non-constant polynomials $g, h\in\mathbb{Q}$. Now, for each fixed $x_0\in\mathbb{Q}$, we have $f(x_0,y)=g(x_0, y)h(x_0,y)$. Since $f(x_0, y)$ is irreducible, it means $g(x_0,y)$ or $h(x_0,y)$ must be constant. I think from here we should be able to conclude either $g$ or $h$ does not depend on $y$. Is this approach correct? Or perhaps there is a counter-example?

Thanks!

Body: No. Let $f(x, y) = (x^2 + 1)(y^2 + 1)$.


问:
Explicit bijection between the set of binary sequences and $\mathbb{R}$?
I know there is a standard proof (Cantor’s diagonalization argument) to show that the set of infinite binary sequences, call it $\Omega := \{0,1\}^\mathbb{N}$, is uncountable.

However, I would like to describe an explicit bijection between $\mathbb{R}$ and $\Omega$ to show that they are equipotent. Is there a direct way to do this? Or is my best bet to find a bijection between $\Omega$ and $(0,1)$ (or something) and compose it with a bijection from $(0,1)$ to $\mathbb{R}$?

On that note, is there a "standard" bijection between $\Omega$ and $(0,1)$?

Many thanks.

答:
The Cantor-Bernstein Theorem (aka Cantor-Schröder-Bernstein Theorem and Schröder-Bernstein Theorem) provides a "sort of" constructive proof that given injections $f\colon A\to B$ and $g\colon B\to A$, there exists a bijection $h\colon A\to B$. It's "sort of" constructive because it requires you to know whether a given element can be traced back "infinitely", and that may not be immediately apparent.

So my answer would involve several steps.

First, there is a standard bijection from $(0,1)$ to $\mathbb{R}$; there are several ways of doing so: one is given by Dark Malthorp. I usually like to take the bijection between $(-\frac{\pi}{2},\frac{\pi}{2})$ and $\mathbb{R}$ given by $\arctan(x)$, and of course the interval $(0,1)$ can be bijection with $(-\frac{\pi}{2},\frac{\pi}{2})$ given by $f(t) = -\frac{\pi}{2}+t\pi$.

Now, we can embedd $(0,1)$ into the set of binary sequences using the base-2 expansion, with the convention that for numbers with two expansions, we always pick the one with a tail of $1$s (you can pick whichever you want). So $\frac{1}{2}$ is $0.011111\ldots$, rather than $0.1$. Then $g\colon (0,1)\to 2^{\omega}$ is given by letting $g(t)(n)$ be the $n$th digit in the expansion.

And we can embedd $f\colon 2^{\omega}\to (0,1)$ as follows: given a sequence $(a_n)$ of $0$s and $1$s, we let $f((a_n))$ be
$$\frac{1}{10} + \sum_{n=0}^{\infty}\frac{5+a_n}{10^{n+2}}.$$
That is: the decimal expansion of the number $f((a_n))$ has a $5$ in position $n+1$ after the decimal digit if $a_n$ is $0$, and has $6$ in that position if $a_n=1$. And it has a $1$ in the first position after the decimal point. The latter is to ensure we are in $(0,1)$ even if we take the sequence of zeros. And I use $5$ and $6$ to avoid the problem of multiple expansions for the same number.

Then we use Cantor-Bernstein to construct a bijection $h\colon 2^{\omega}\to(0,1)$ using $f$ and $g$. The construction runs as follows:

We say an element $(a_n)$ of $2^{\omega}$ has a "parent" if there exists $x\in(0,1)$ such that $g(x)=(a_n)$. Similarly, an element $t\in (0,1)$ has a "parent" if there exists $(b_n)\in 2^{\omega}$ such that $f((b_n))=t$.

Given an element $(a_n)\in 2^{\omega}$, we have two possibilities: either we can use the "parent" relation to bounce back and forth between $(0,1)$ and $2^{\omega}$ infinitely many times, or else there comes a point where the process stops (by reaching either a $t\in (0,1)$ that is not in the image of $f$, or a $(b_n)$ in $2^{\omega}$ that is not in the image of $g$).

We can partition $2^{\omega}$ into three sets: $A_1$, the elements where the process stops in an element of $2^{\omega}$; $A_2$, the elements where the process stops in an element of $(0,1)$; and $A_{\infty}$, the elements where the process continues indefinitely without stopping.

Similarly, we can partition $(0,1)$ into three sets: $B_1$, the elements where the process stops in an element of $2^{\omega}$; $B_2$, the elements where the process stops in an element of $(0,1)$; and $B_{\infty}$, the elements where the process continues indefinitely without stopping.

Now, if $(a_n)\in A_1$, then $f((a_n))\in B_1$; if $(a_n)\in A_2$, then there exists a unique $t\in B_2$ such that $g(t)=(a_n)$; if $(a_n)\in A_{\infty}$, then $f((a_n))\in B_{\infty}$. Moreover, every element in $B_1$ is the image of someone in $A_1$; every element in $B_{\infty}$ is the image of someone in $A_{\infty}$. Thus, we define $h\colon 2^{\omega}\to(0,1)$ by
$$h((a_n)) = \left\{\begin{array}{lll}
f(a_n)&\quad&\text{if }(a_n)\in A_1;\\
g^{-1}(a_n)&&\text{if }(a_n)\in A_2;\\
f(a_n)&&\text{if }(a_n)\in A_{\infty}.
\end{array}\right.$$

So, composing $h$ with the bijection of $(0,1)$ to $(-\frac{\pi}{2},\frac{\pi}{2})$, and then using $\arctan(x)$ to biject with $\mathbb{R}$ would give you an "explicit} bijection between $2^{\omega}$ and $\mathbb{R}$.

To be honest, one almost never actually does this. Instead, we rely on knowing that such bijections exist. So I would usually explicitly use $\arctan(x)$ to see $\mathbb{R}$ and $(-\frac{\pi}{2},\frac{\pi}{2})$ can be bijected. Then have a general result about how any two finite open intervals can be bijected (to biject $(a,b)$ and $(c,d)$, use $g(t) = c + \left(\frac{t-a}{b-a}\right)(d-c)$). Then define the two injections above, and invoke Cantor-Schröder-Bernstein.


问:
Setting up equations in a mixing problem
In an exercise sheet for solving 2x2 systems of ODE's, the following question came up:

"Consider two tanks, A and B, each holding 200 litres of water. A pipe pumps
water from tank A to tank B at a rate of 5 l/min. At the same time another
pipe pumps liquid from tank B to tank A at the same rate. At time t = 0, $x_0$
kg of a chemical X is dissolved into tank A, and tank B has $y_0$ kg of the same
chemical X dissolved into it.

Write down the system of differential equations satisfied by $x(t)$ and $y(t)$,
the quantity of the chemical X in tanks A and B respectively."

However I am really lost as to how to set up these systems of equations.

答:
In a small time interval, tank A is losing chemical in proportion to what is currently within it, but also gaining chemical in proportion to what is currently in the other tank (B).

The situation is exactly mirrored for tank B.

At all times, the volumes of liquid in each tank remains constant at $200$ litres. The proportion of the liquid of one tank transferred to the other (and vice-versa) in a minute is always $\frac 5{200} = \frac 1{40}$.

Denote the small time interval $\Delta t$. The change in $x$ in tank A is given by:

$\Delta x = (-\frac 1{40}x +  \frac 1{40}y) \Delta t$

And the change in $y$ in tank B is given by:

$\Delta y = (+\frac 1{40}x -  \frac 1{40}y) \Delta t$

Dividing throughout by $\Delta t$ and taking the limit $\Delta t \to 0$ allows us to arrive at:

$\displaystyle 40\frac {dx} {dt} = y-x$

and

$\displaystyle 40\frac {dy} {dt} = x - y$

The next steps in the problem will probably involve you solving this system (to do that, try subtracting the second from the first and introduce the variable $z = x-y$).


问:
How can I find these velocities without using the quadratic formula?

If a ball is thrown vertically upward with a velocity of $160 \text{ ft/s}$, then its height after t seconds is $s = 160t − 16t^2$.

a) What is the velocity of the ball when it is $384 \text{ ft}$ above the ground on its way up?

  b) What is the velocity of the ball when it is $384 \text{ ft}$ above the ground on its way down?

$$\begin{align*}
384 &= 160t - 16t^2\\
16t^2 - 160t + 384 &= 0\\
16 (t^2 - 10t + 24) &= 0
\end{align*}$$

I can't factor the above... and I'm not supposed to used the quadratic formula. Am I stupid or is this unsolvable without without the formula?

答:
$$16(t^2 - 10 t + 24) = 0 \iff 16(t-4)(t-6) = 0 \implies t = 4,\;\; t = 6$$

Heading up: $384$ feet at time $t = 4$,

Descendng down after reaching maximum height: $384$ ft. at time $t = 6$.

Note: $$(-4)\cdot (-6) = + 24;\quad -4 + - 6 = -10$$

Noticing those facts allow you to deduce that the factors must be $$\;t^2 - 10 t + 24 = (t - 4)(t - 6)$$


问:
Calculate the determinant of the following $n \times n$ symmetric matrix:
Let $n$ be a positive integer and $A = (a_{ij})_{n\times n}$, where $a_{ij} = |i-j|$, for $i = 1, 2, \dots, n$ and $j = 1, 2, \dots, n$. Calculate $\det A$.

I noticed that $a_{ii} = 0$ and $a_{ij} = a_{ji}$, so A is a symmetric matrix. Also, I saw that, if we make the notation $A_n$ for the A with n elements, $A_n$ is constructed from $A_{n-1}$ with $n-1, n-2, \dots, 0$ as elements for the last line and last column. I tried to use Laplace expansion but with no result.

This is how $A_n$ looks like: $A_n=\begin{bmatrix}
0&1&2& .&.&. &n-1\\
1&0&1&2& .&.&n-2 \\
2&1&0&1&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&2 \\
.&.&.&.&.&.&1 \\
n-1&n-2&.&.&2&1&0
\end{bmatrix}$

I calculated for a few small numers: $\det A_1 = 0$, $\det A_2 = -1$, $\det A_3 = 4$, $\det A_4 = -12$, $\det A_5 = 32$, but I didn't figure out a rule such that I could find the determinant through induction. Can you help me on this one?

答:
Hint. Consider the second-order differences of the rows. When $n\ge3$,
$$
\pmatrix{1&-2&1\\ &1&-2&1\\ &&\ddots&\ddots&\ddots\\ &&&1&-2&1\\ &&&&1&0\\ &&&&&1}A_n
=\pmatrix{0&2\\ \vdots&0&\ddots\\ \vdots&\ddots&\ddots&\ddots\\ 0&\cdots&\cdots&0&2\\ n-2&n-3&\cdots&1&0&1\\ n-1&n-2&\cdots&\cdots&1&0}.
$$


问:
Prove that if $\tau$ and $\sigma$ are two dyadic tensors then $\tau\cdot \sigma=\text{tr}(T\cdot S^T)$.
So using orthonormal bases, I proved that if $\tau$ and $\sigma$ are dyadic tensors then its inner product is the trace of the product of the matrices $T$ and $S$ of its image in $\hom(V,V)$ under a certain isomorphism. So since the product of a matrices is referred to the composition of linear function and since the trace is an invariant I thought that the result is generally true so that it holds even if we does not use orthonormal bases but unfortunately I was not sure about the correctness of the this argument and so I proved to calculate the inner product of two dyadic tensors with not using an orthonormal basis and strangerly I did not able to obtain the same result so I ask it my above argumentations are correct and if not I ask to prove the result if it is true even with not using an orthonormal basis. To follow some explanations about the definition I use of inner product between tensors: naturally if you know it you can read immediately only the final calculation about the inner product of two tensors.

Theorem 1

If $V$ is a vector space equipped with an inner product then there exist a unique isomorphism
$$
\#:V\rightarrow V^*
$$
with is induced by the inner product in such a way that
$$
\big\langle\#(\vec v),\vec w\big\rangle=\vec v\cdot\vec w
$$
for any $v,w\in V$. Moreover under this isomorphism the image of the reciprocal basis $\text E:=\{\vec\epsilon_{_{1}},...,\vec\epsilon_{_{n}}\}$ of any basis $\mathscr E:=\{\vec e_{{_1}},...,\vec e_{_{n}}\}$ is the dual basis $\mathscr E^*:=\{\vec e^{\,_1}\,...,\vec e^{_{n}}\}$, that is
$$
\#(\vec\epsilon_{_i})=\vec e^{\,_{i}}
$$
for each $i=1,...,n$

If $\tau$ is a mixed simple tensor in $\mathscr T^p_q(V)$ say
$$
\tau:=\vec v_{_{1}}\otimes...\otimes\vec v_{_{p}}\otimes \vec v^{\,_{1}}\otimes...\otimes \vec v^{\,_{q}}
$$
for some $\vec v_{_{1}},...,\vec v{_{p}}\in V$ and for some $\vec v^{\,_{1}},...,\vec v^{\,_{q}}\in V^*$, then we define the pure covariant representation of $A$ to be the tensor
$$
\#^p_q(\tau):=\#(\vec v_{_{1}})\otimes...\otimes\#(\vec v_{_{p}})\otimes \vec
 v^{\,_{1}}\otimes...\otimes\vec v^{\,_{q}}
$$

Theorem 2

The above condition define an isomorphism between $\mathscr T^p_q(V)$ and $\mathscr T_{p+q}$ whose inverse is given by the equality
$$
(\#^p_q)^{-1}(\vec v^{\,_{1}}\otimes...\otimes\vec v^{\,_{p}}\otimes \vec v^{\,_{1}}\otimes...\otimes\vec v^{\,_{q}})=\#^{-1}(\vec v^{\,_{1}})\otimes...\otimes\#^{-1}(\vec v^{\,_{p}})\otimes \vec v^{\,_{1}}\otimes...\otimes\vec v^{\,_{q}})
$$
for each simple tensor.

Theorem 3

If $\tau\in\mathscr T^p_q(V)$ is an arbitrary simple tensor, say
$$
\tau:=\vec v_{_{1}}\otimes...\otimes\vec v_{_{p}}\otimes \vec v^{\,_{1}}\otimes...\otimes\vec v^{\,_{q}}
$$
for some $\vec v_{_{1}},...,\vec v{_{p}}\in V$ and for some $\vec v^{\,_{1}},...,\vec v^{\,_{q}}\in V^*$, then for eanch pair of integers $(i,j)$, wherere $1\le i\le p$ and $1\le j\le q$, we seek a unique linear transformation called simple contraction
$$
C^i_j:\mathscr T^p_q(V)\rightarrow\mathscr T^{p-1}_{q-1}(V)
$$
such that
$$
C^i_j(\tau):=\langle v^{\,_j},\vec v_{_i}\rangle v_1\otimes...\otimes v_{_{i-1}}\otimes v_{_{i+1}}\otimes...\otimes v_{_p}\otimes\vec v^{\,_1}\otimes...\otimes\vec v^{\,_{j-1}}\otimes\vec v^{\,_{j+1}}\otimes...\otimes\vec v^{\,_{q}}
$$
for each all simple tensor $\tau$.

So using the contraction operation, we can define a double product for tensors. If $\tau\in\mathscr T^p_q(V)$ and $\sigma\in\mathscr T^q_p(V)$ then the tensor product $\tau\otimes\sigma$ is a tensor in $\mathscr T^{p+q}_{p+q}(V)$ so that if we apply the contraction
$$
C:=\underset{\text{q times}}{\underbrace{C^1_1\circ...\circ\ C^1_1}}\circ \underset{\text{p times}}{\underbrace{C^1_{q+1}\circ...\circ\ C^1_{q+1}}}
$$
to $\tau\otimes\sigma$ then the result is a scalar and moreover we observe that it has the same property of the dual operator $\langle\,\cdot\,\rangle$ so that we can define a isomorphism between $\mathscr T^p_q(V)$ and the dual of $\mathscr T^q_p(V)$ and so we can identify the two spaces. So we can define the dual product between two tensor through the equation

$\langle\tau,\sigma\rangle:=C(\tau\otimes\sigma)$

for any $\tau\in\mathscr T^p_q(V)$ and for any $\sigma\in\mathscr T^q_p(V)$.

Theorem 4

The equality defined for any $\tau,\sigma\in\mathscr T^p_q(V)$

$\tau\cdot\sigma:=\big\langle\#^p_q(\tau),(\#^{p+q})^{-1})\big(\#^p_q(\sigma)\big)\rangle$

define an inner product on $\mathscr T^p_q(V)$.

Definition 1

If $\tau\in\mathscr T^p_q(V)$ and $\sigma\in\mathscr T^r_s(V)$ an external contraction of order $l$ is a composition of simple contractions applied to $\tau\otimes\sigma$ each of whose contract an idices of $\tau$ with an indices of $\sigma$ respectively.

Theorem 5

If $\overline p\le p$ and $\overline q\le q$ then for any external contracion of order $(\overline p+\overline q)$ the function $J_C$ from $\mathscr T^p_q(V)$ in $\hom\big[\mathscr T^{\overline q}_{\overline p},\mathscr T^{p-\overline p}_{q-\overline q}\big]$ defined as
$$
[J_C(\tau)](\sigma):=C(\tau\otimes\sigma)
$$
for each $\tau\in\mathscr T^p_q$ and for each $\sigma\in\mathscr T^{\overline q}_{\overline p}$ is an isomorphism.

Corollary 5.1

Let be $\mathscr E:=\{\vec e_{_1},...,\vec e_{_n}\}$ a basis of $V$. So the spaces $\mathscr T^2(V)$ and $\hom(V,V)$ are isomorphic via the composition of an isomorphism $h$ from $\hom(V^*,V^{**})$ to $\hom(V,V)$ and an isomorphim $J_C$ from $\mathscr T^2(V)$ to $\hom(V^*,V^{**})$ such that
$$
\big[h(J_C(\tau)\big](\vec v):=T^{i,j}(\vec v\cdot\vec e_{_j})\vec e_{_i}
$$
for each $v\in V$ where the $n^2$ scalars $T^{i,j}$ are the components of $\tau$ with respect the basis of $\mathscr T^2(V)$ induced from $\mathscr E$.

Definition 2

The transpose operator is the endomorphism $T_{pq}$ of $\mathscr T^{p+q}$ defined throught the equation

$T_{pq}(\vec v_{_1}\otimes...\otimes\vec v_{_p}\otimes\vec u_{_1}\otimes...\otimes\vec u_{_q}):=u_{_1}\otimes...\otimes\vec u_{_q}\otimes\vec v_{_1}\otimes...\otimes\vec v_{_p}$

for any simple tensor. Moreover the general transpose operator $T^{pq}_{pq}$ could be defined thorugh the equation

$T^{pq}_{pq}(\tau):=\big((\#^p_q)^{-1}\circ T_{pq}\circ\#^p_q\big)(\tau)$

for each $\tau\in\mathscr T^p_q$

Theorem 6

The image of the transpose tensor of a tensor $\tau$ under the isomorphism $[h\circ J_C]$ above defined is the transpose function of the the image of $\tau$.

Now if $\mathscr E:=\{\vec e_{_1},...,\vec e_{_n}\}$ is a basis for $V$ then the componet of the image of a tensor $\tau$ and $\sigma$ under the isomorphism $h\circ J_C$ are the scalars $t_{i,k}:=T^{i,j}(\vec e_{_j}\cdot\vec e_{_k})$ and $s_{h,i}:=S^{hk}(\vec e_{_k}\cdot\vec e_{_i})$. So we observe that

$$
\tau\cdot\sigma:=\big\langle\#^p_q(\tau),(\#^{p+q})^{-1}\big(\#^p_q(\sigma)\big)\big\rangle=...=T^{i,j}S^{h,k}(\vec e_{_i}\cdot\vec e_{_h})(\vec e_{_j}\cdot\vec e_{_k})=\big(T^{i,j}(\vec e_{_j}\cdot\vec e_{_k})\big)\big(S^{h,k}(\vec e_{_i}\cdot\vec e_{_h})\big)
$$

so that if $\mathscr E:=$ is an orthonormal basis then
$$
\tau\cdot\sigma=...=\big(T^{i,j}(\vec e_{_j}\cdot\vec e_{_k})\big)\big(S^{h,k}(\vec e_{_i}\cdot\vec e_{_h})\big)=\big(T^{i,j}\delta_{j,k}\big)\big(S^{h,k}\delta_{i,h}\big)=T^{i,j}S^{i,j}=\text{tr}(T\cdot S^T)
$$
where $T$ and $S$ are the matrix of the image of $\tau$ and $\sigma$. So if $\mathscr E$ is not an orthonormal basis then  the result seems that does not hold. So what can you say about?

答:
First of all for practical reasons we define the matrices $L$ and $M$ through the equalities

$$L_{i,j}:=(\vec e^{\,_i}\cdot\vec e^{\,_j})$$

$$M_{i,j}:=(\vec e_{_i}\cdot\vec e_{_j})$$

for each $i,j=1,...,n$ and then let be $S$ and $T$ the matrices whose elements are the component of two dyadic tensor $\sigma$ and $\tau$. Now using the theorem $5$ it is possible to prove that the dyadic pure covariant, mixed and pure contravariant tensors are isomorphic to the set of endomorphisms through the applications $f$, $g$ and $h$ and in particular if $\tau$ is a pure covariant tensor then
$$
\big[f(\tau)\big]\big(\vec v)=\big(\,e^{_{i,h}}T_{h,j}v^{_j}\,\big)\vec e_{_h}
$$
for each $\vec v\in V$ and if $\tau$ is a mixed tensor then
$$
\big[g(\tau)\big](\vec v)=\big(\,T^i_jv^{_j}\,\big)\vec e_{_i}
$$
for each $\vec v\in V$ and finally if $\tau$ is a pure contravariant tensor then
$$
\big[(h(\tau)\big](\vec v)=\big(\,T^{i,h}e_{_{h,j}}v^{_j}\,\big)\vec e_{_i}
$$
for each $\vec v\in V$ so that the matrices associated to the image of a dyadic tensor $\tau$ are respectively $LT$ for a pure covariant tensor, $T$ for a mixed tensor and $TM$ for a pure contravariant tensor so that generally the component of the imge of a dyadic tensor are different from the same tensor whether the basis used is not orthonormal because in this case the matrix $L$ and $M$ are the identity matrix. So if $\tau$ and $\sigma$ are pure covariant tensors then
$$
\tau\cdot\sigma\equiv C\Big(\big(T_{i,j}\vec e^{\,i}\otimes\vec e^{_j}\big)\otimes\big(S_{h,k}\vec e^{\,_h}\otimes\vec e^{\,_k}\big)\Big)=T_{i,j}S_{h,k}C\big(\vec e^{\,_i}\otimes\vec e^{\,_j}\otimes\vec e^{\,_h}\otimes\vec e^{\,_k}\big)=T_{i,j}S_{h,k}e^{_{i,h}}e^{_{j,k}}=T_{i,j}e^{_{j,k}}S^t_{k,h}e^{_{h,i}}=\text{tr}\big(TLS^tL)
$$
while if $\tau$ and $\sigma$ are mixed tensors then
$$
\tau\cdot\sigma:=\Big\langle G^p_q(\tau),(G^p_q)^{-1}\big(G^p_q(\sigma)\big)\Big\rangle=T^i_jS^h_k\Big\langle G^p_q(\vec e_{_i}\otimes\vec e^{\,_j}),(G^p_q)^{-1}\big(G^p_q(\vec e_{_h}\otimes\vec e^{\,_k})\big)\Big\rangle=T^i_jS^h_k\Big\langle G\vec e_{_i}\otimes\vec e^{\,_j},\vec e_{_h}\otimes G^{-1}\vec e^{\,_k}\Big\rangle=T^i_jS^h_hC\big(\vec e_{_h}\otimes G^{-1}\vec e^{\,_k}\otimes G\vec e_{_i}\otimes\vec e^{\,_j}\big)=T^i_jS^h_ke_{_{i,h}}e^{\,_{j,k}}=T^i_je^{\,_{j,k}}\tilde{S}^k_he_{_{h,i}}=\text{tr}\big(TLS^tM\big)
$$
where $\tilde{S}^k_h$ are the scalar of the transpose matrix of $S$ and finally if $\tau$ and $\sigma$ are pure contravariant tensor then
$$
\tau\cdot\sigma:=\Big\langle G^p_q(\tau),(G^p_q)^{-1}\big(G^p_q(\sigma)\big)\Big\rangle=...=T^{i,j}S^{h,k}e_{_{i,h}}e_{_{j,k}}=T^{i,j}e_{_{j,k}}S^{k,h}_te_{_{h,i}}=\text{tr}\big(TMS^tM\big)
$$
where $S^{k,h}_t$ are the scalar of the transpose matrix of $S$. So since if the basis adoped is not orthonormal then the matrix $L$ and $M$ are different from the identity so we conclude that the result is generally false using a not orthonormal basis.


问:
R has an uncountable number of distinct dense subsets.
Does someone know how to prove this?

Let S and T be non-empty subsets of a topological space (X, τ) with S ⊆ T.
(iii) Hence show that if S is dense in X, then T is dense in X.

(iv) Using (iii) show that R has an uncountable number of distinct dense subsets.

I just wanna prove the fourth item using the result of the third.

I tried to do it by contradiction, but i don't know what exactly i have to do!

答:
The answer was found in the comments. Here is a concluding proof.

We know that $|\mathbb{R}\setminus\mathbb{Q}|=|\mathbb{R}|$, so $|\mathcal{P}(\mathbb{R}\setminus\mathbb{Q})|=|\mathcal{P}\mathbb{R}|$.

By (iii) and because $\mathbb{Q}$ is dense in $\mathbb{R}$, for any $U\subseteq\mathbb{R}\setminus\mathbb{Q}$ the set $\mathbb{Q}\cup U$ is dense in $\mathbb{R}$.
The function from
$$
\mathcal{P}(\mathbb{R}\setminus\mathbb{Q})\rightarrow\{\text{dense subsets of $\mathbb{R}$}\}
$$
defined by
$$
U\mapsto\mathbb{Q}\cup U
$$
is well-defined by (iii), and it is clearly injective.
So there are at least $|\mathcal{P}\mathbb{R}|$ (usually denoted $2^{2^{\aleph_0}}$) dense subsets of $\mathbb{R}$.
There are, of course, at most $2^{2^{\aleph_0}}$ dense subsets of $\mathbb{R}$.
So there are exactly $2^{2^{\aleph_0}}$ dense subsets of $\mathbb{R}$.
The cardinal $2^{2^{\aleph_0}}>\aleph_0$ is uncountable.


问:
Tensor product of injective linear maps
Let $X,Y,U,V$ be vector spaces. Let $S:X\to U$ and $T:Y\to V$ be linear maps. Then by linearisation there exists a linear map
$$S\otimes T : X\otimes Y \to U\otimes V,\quad x\otimes y \mapsto (Sx) \otimes (Ty). $$
If $S$ and $T$ are injective, then their tensor product is, too. I think I managed to justify this as follows.

Note that $U = S(X) \oplus U'$. Define $S' : U\to X$ such that $S'(S(x_u) +u' ) = x_u$. One readily verifies linearity and that is a left inverse for $S$. Analogously, define $T':V\to Y$. Then by linearisation, there exists linear
$$S'\otimes T' : U\otimes V \to X\otimes Y,\quad (S(x_u)+u') \otimes (T(y_v) + v') \mapsto x_u\otimes y_v. $$
This would give a left inverse for $S\otimes T$ hence making it injective.

I seem to recall that tensoring injective morphisms between modules does not always preserve injectivity. In fact, a certain module $F$ for which $\mbox{id}_F \otimes \phi$ is injective for all injective morphisms $\phi$ is called flat.

What am I invoking here that allows this to work for vector spaces?

Is it the decomposition $U=S(X) \oplus U'$?

答:
I know, this is an old question, but I both hate it when questions are solely answered within the comments, and I think the crucial term has not been named yet:

For a field $\mathbb{K}$, vector spaces over $\mathbb{K}$ are the same as free $\mathbb{K}$-modules! Freeness means exactly that they admit a basis, and it implies the flatness as $\mathbb{K}$-modules. Of course, the proof for this goes exactly as it was explained in the comments. It is an important and useful property to keep in mind.

Title: Tensor product of injective linear maps


问:
Find a mistake in summing a series
I was doing one task and i can't find where I'm wrong. I have to find sum $S$ of given series. The answer is  $\frac{1}{8}$.
$$S = \sum_{n=1}^{\infty}\frac{n}{(2n-1)^2(2n+1)^2} $$
First $\frac{n}{(2n-1)^2(2n+1)^2} = \frac{1}{2}(\frac{1}{(n - \frac{1}{2})^2}-\frac{1}{(n + \frac{1}{2})^2}) $
From this $ S_N = \frac{1}{2}(4-\frac{1}{(N+\frac{1}{2})^2}) $ and $ S = \lim_{N\to\infty}S_N = 2$
Where did I make a mistake?

Title: Find a mistake in summing a series

Body: As @Gary notes, you've miscounted powers of $2$. Since$$\begin{align}\frac{1}{(n-1/2)^2}-\frac{1}{(n+1/2)^2}&=\frac{2n}{(n-1/2)^2(n+1/2)^2}\\&=\frac{2n(2^2)^2}{(2n-1)^2(2n+1)^2}\\&=\frac{32n}{(2n-1)^2(2n+1)^2},\end{align}$$you're wrong by the expected factor of $\frac{\frac12}{\frac{1}{32}}=\frac{1}{16}$.


问:
Euler characteristic of a sphere with $n$ holes?
If I understood correctly, a sphere with a single hole is a torus, and has genus 1. Its Euler characteristic is 0.

If it has two holes, the Euler characteristic is -2. Three holes, -4.

However, I don't know how to generalize for a sphere with $n$ holes. Although it seems somewhat intuitive given the initial results, I couldn't prove it formally... Can anybody help me on that?

答: To elaborate on the previous answer, you know that $\chi(M\#N)=\chi(M)+\chi(N)-\chi(S^2)$ where $M\#N$ is the connected sum. Recognizing that a genus $g$ surface is a genus $g-1$ surface together with a torus, the result follows by an easy induction.


问:
If two sets have empty interior closures, does the interior of the closures's union is contained in the union of the interior of the closures?
Is that true that $(\overline{X})^{\circ} = (\overline{Y})^{\circ}  = \emptyset \Rightarrow (\overline{X \cup Y})^{\circ} \subseteq (\overline{X})^{\circ} \cup (\overline{Y})^{\circ}$, where $^{\circ}$ stands for "interior of the set"?

答:
It's true.

Assume that $U \subseteq \left(\overline{X \cup Y}\right)^\circ$ is open and non-empty. Then $U \setminus \overline{Y}$ is open (as intersection of open $U$ and open complement to $\overline{Y}$) and non-empty (otherwise $U \subseteq \overline{Y}$ and $\overline Y$  has non-empty interior). But $U \setminus \overline{Y} \subseteq \overline{X}$, so $\overline X$ has non-empty interior.


问:
Working on a generalized Cantor set, with Lebesgue measure, and a certain inequality.
Let's consider the interval $[0,1]$ in the same way that we constructed the Cantor set, we can use the same idea, but instead of removing in the step $n$ middle open intervals of length $\frac{1}{3^n}$ we remove of length $\frac{1}{5^n}$. Call each step of this construction $C_n$ i.e

$C_0 = [0,1]$

$C_1= C_0-\left(\frac{1}{2}-\frac{1}{10},\frac{1}{2}+\frac{1}{10}\right)=\left[0,\frac{2}{5}\right]\cup \left[\frac{3}{5},1\right]$

And then to construct $C_2$ consider the middle point of $\left[0,\frac{2}{5}\right]$ which is $\frac{2}{10}$ , and remove from that interval the open interval $\left(\frac{3}{20}-\frac{1}{2\cdot 5^2},\frac{3}{20}+\frac{1}{2\cdot 5^2}\right)$ , the same with the interval $\left[\frac{7}{10},1\right]$ .

(It's the same as the usual Cantor construction)
So we define $ C = \bigcap C_n$.  Let's define the function $f_n$ as the characteristic function of the set $C_n$ , and $f$ as the characteristic function of the set $C$.

Prove that $$
\mathop {\sup }\limits_{m \geqslant n} \int\limits_0^1 {\left| {f_n \left( x \right) - f_m \left( x \right)} \right|\,\mathrm dm}  \leqslant \frac{1}
{3}\left( {\frac{2}
{5}} \right)^n
$$
and then prove that under any modification of the function $f$ on a set of measure zero, is not Riemann integrable.

I'm very scared of this problem, I don't know how to prove the inequality please help me, I'm a little dizzy .
Thanks for the EDIT

答:
Let $\{I_{k,j}:j=1,\ldots,2^k\}$ be the set of open intervals we are throwing away on $k$-th step of our construction. Note that $m(I_{k,j})=5^{-k-1}$ for all $k\in\mathbb{N}$ and $j=1,\ldots,2^k$.

Let
$$
C_0:=[0,1]\qquad\qquad
C_{k+1}:=C_k\setminus\bigcup\limits_{j=1}^{2^k}I_{k,j}
$$
Now we proceed to the proof of inequality. Define $f_n=\chi_{C_n}$, then for $m\geq n$ we have
$$
\int_0^1|f_m(x)-f_n(x)|dm=
\int_0^1|\chi_{C_m}(x)-\chi_{C_n}(x)|dm=
\int_0^1\chi_{C_n\setminus C_m}(x)dm=
m(C_n\setminus C_m)=
m\left(\bigcup\limits_{k=n}^{m-1}\bigcup\limits_{j=1}^{2^k} I_{k,j}\right)=
\sum\limits_{k=n}^{m-1}\sum\limits_{j=1}^{2^k}m(I_{k,j})=
\sum\limits_{k=n}^{m-1}\sum\limits_{j=1}^{2^k}5^{-k-1}=
\sum\limits_{k=n}^{m-1}2^k5^{-k-1}
$$
Hence
$$
\sup\limits_{m\geq n}\int_0^1|f_m(x)-f_n(x)|dm=
\sup\limits_{m\geq n}\sum\limits_{k=n}^{m-1}2^k5^{-k-1}=
\sum\limits_{k=n}^\infty 2^k5^{-k-1}=\frac{2^n 5^{-n-1}}{1-2\cdot 5^{-1}}=\frac{1}{3}\left(\frac{2}{5}\right)^n
$$

We proceed to the second part. Define
$$
C:=\bigcap\limits_{k=1}^\infty C_k
$$
Its measure equals
$$
m(C)=
1-m(C_0\setminus C)=
1-m\left(\bigcup\limits_{k=0}^\infty\bigcup\limits_{j=1}^{2^k}I_{k,j}\right)=
1-\sum\limits_{k=0}^\infty \sum\limits_{j=0}^{2^k} m(I_{k,j})=\\
1-\sum\limits_{k=0}^\infty \sum\limits_{j=0}^{2^k} 5^{-k-1}=
1-\sum\limits_{k=0}^\infty 2^k 5^{-k-1}=
1-\frac{5^{-1}}{1-2/5}=\frac{2}{3}
$$
Now note that $x\in C$ iff its representation in a fivefold number system have no digit $2$ at any place. Hence for a given $x\in C$ you can change some digit to $2$ to leave the set $C$. Moreover you can change digit with very big place number to get a number outside $C$ but very close to $x$. Thus for each $x\in C$ and each $\varepsilon>0$ there exist $\hat{x}\notin C$. This is equivalent to discontinuity of $f$ in point $x\in C$. So $f$ is discontinuous at any point of $C$.

Since $m(C)>0$ by Lebesgue critirion $f\notin\mathcal{R}([0,1])$, because set of discontinuities of $f$ have positive measure. Let $\hat{f}$ be a function equivalent to $f$. Assume $\hat{f}$ is Riemann interable, then it is bounded. Hence the function $\delta=\hat{f}-f$ is bounded. Since $\hat{f}$ is equivalent to $f$, then $\delta$ is non zero only on the set of zero measure. By Lebesgue criterion $\delta\in\mathcal{R}([0,1])$. Then $f=\hat{f}-\delta\in$$\mathcal{R}([0,1])$ as difference of two Riemann integrable functions. Contradiction, so $\hat{f}\notin\mathcal{R}([0,1])$


问:
Given $a^7=b^5=a^2b=1_G. $ show that $a=b=1_G.$
This is a problem from my algebra book:

Let G be a group and a,b $\in $ G. Suppose that
$$a^7=b^5=a^2b=1_G. $$

Show that $a=b=1_G$.

I know I have to manipulate it somehow and tried doing
$$a^7=a^2b \iff a^5a^2=a^2b \iff a^5=b, $$
and the same for
$$b^5=a^2b \iff bb^4=a^2b \iff b^4=a^2.$$

Then I use the above
$$bb^3=a^2 $$
$$a^5b^3=a^2 \iff a^3b^3=1_G $$
$$a^3b^3=1_G=a^2b.$$
$$ ab^2=1_G=a^2b.$$
Then
$$ab^2=a^2b=1_g \iff b=a=1_G.$$

However, I have no feeling for algebra yet so I'm not sure if this 'solution' is considered correct or if there is a easier way to solve it so I would appreciate some feedback, thanks in advance!

答:
$G$ is a group, so $\forall g\in G$ $\exists h\in G:gh=hg=1_G$

$$a^2b=1_G \underset{\exists b^{-1}\in G}{\implies} a^2=b^{-1}\implies (a^2)^7=(b^{-1})^7\implies$$ $$(a^7)^2=(b^7)^{-1}\implies 1_G^2=(b^2b^5)^{-1} \underset{b^5=1_G}{\implies}1_G=b^{-2}$$
From Hp we have $b^7=1_G$, so $o(b)|7$ and, since $\forall g\in G$ $o(g^{-1})=o(g)$, from the equivalence $b^{-2}=(b^{-1})^2=1_G$, we have that $o(b^{-1})=o(b)|2$.

So, $o(b)|2,7\iff o(b)=1\implies b=1_G$.

Do the same thing for $a$ with the Hp that $o(a)|5$ and from the equivalence $a^2b=1_G$, the order of $a$ is forced to be $1$ and we can conclude that $a=b=1_G$.


问:
Prove the equation implicitly represents a function
Unfortunately I got stuck solving the following exercise:

Prove that the equation $xe^z + ye^{-z} + z = 1$ implicitly defines z as a function of (x,y), in the set $A\times[0,\infty)$ where: $A = \{(x,y)\in[0,\infty)\times[0,\infty)|x+y\leq1\}$.

Now, from what I understand, I am to show that z has a single solution in $[0,\infty)$ for all
$(x,y)\in{A}$ (where A can be described as all the points inside a triangle the Cartesian plain that its vertices are the origin, (1,0) and (0,1)). Or, in other words, z is a function of (x,y) in the specific region of topic.
I tried solving the equation to no avail (that is extracting z). It occurred to me that it might be unnecessary to solve the equation
for the proof but I can't seem to find any other way.

Any hints or directions will be gratefully accepted.
Thanks :)

答:
Idea

Prove that $\exists!z\in[0,\infty)$ that satisfies the equation $\forall(x,y)\in{A}$ and to show that I first want to show that
$\frac{d(xe^z + ye^{-z} + z)}{dz} \geq 0$. Now show $xe^z + ye^{-z} + z \leq 1$ for $z = 0$ and due to the fact that
$xe^z + ye^{-z} + z$ is continuous as a function of $z$ and $\frac{d(xe^z + ye^{-z} + z)}{dz} \xrightarrow[z \to \infty]{} \infty$ the equation must be satisfied for a single $z$.

Nothing is mentioned about partial derivatives in my book up to this question so I am not sure it's even possible to claim what I've claimed here but I think this is the general idea. Don't really have anyway to confirm the validity of the proof and it is challenging for me to formalize it but I'll do my best and post an update.


问:
For $x\neq 1$ find $x$ such that ...
For $x\neq 1$, find $X$ such that:
$$x+1\mid 2$$
$$x+2\mid 3$$
$$x+3\mid 4$$
$$x+4\mid 5$$
Can someone help me with this ?

答:
Given that
$$x+1\mid 2$$
$$x+2\mid 3$$
$$x+3\mid 4$$
$$x+4\mid 5$$

We rewrite it in modular form.

$$x+1\equiv 0 (\mod 2)$$
$$x+2\equiv 0 (\mod 3)$$
$$x+3\equiv 0 (\mod 4)$$
$$x+4\equiv 0 (\mod 5)$$

Subtracting so only $x$ is on the left, you get, by rules of modular arithmetic,

$$x\equiv 1 (\mod 2)$$
$$x\equiv 1 (\mod 3)$$
$$x\equiv 1 (\mod 4)$$
$$x\equiv 1 (\mod 5)$$
Therefore, $x-1$ is divisible by $2, 3, 4, 5$. The least common multiple of $2, 3, 4, 5$ is $60$, and so one solution is $x-1=60$, and so $x=61$ is a solution.

A more general solution is all numbers one more than a multiple of 60, usually written as $x=60k+1$ for $k\in \mathbb {Z} $ (But $k \neq 0$ because $x \neq 1$).

Title: For $x\neq 1$ find $x$ such that ...


问:
Finding the probability that at most n events take place on any interval during a given time period
I am interested in the general case, but let us start with a smaller example. Suppose that cars arrive to a street with intensity $\lambda$ per minute. We would like to know the probability that at least two cars have arrived on the street during any five minute interval in the next hour. How can we find this?

My initial thought was to use complementary event: 1 - the probability that at most one car is on the street in any five minute period during the next hour. Hence my reasoning was something like $1 - \int_0^{55}\mathbb{P}(N(s + 5) - N(s) \leq 1)ds - \int_{55}^{60}\mathbb{P}(N(60) - N(55 + s) \leq 1)ds$.

But I quickly realized that finding this probability might not be that easy, since the intervals we are considering overlap, namely $N(5)$ and $N(5 + s)$ overlap except for the infinitesimal point $s$. So then, is the correct way to integrate just $\mathbb{P}(N(s) \leq 1$ over the region? Moreover, I think that my line of reasoning is missing something critical, since I do not see a reason, why the summation would not end up being negative.

答:
Conditional Probability Approach

Poisson says that in an hour, the probability that $n$ cars have entered is
$$
e^{-60\lambda}\frac{(60\lambda)^n}{n!}\tag1
$$
Given that $n$ cars have entered in the hour, the probability that none is within $5$ minutes of another is the ratio of the volumes of the simplices
$$
\frac{\left|\left\{x_k\ge0:\sum\limits_{k=1}^nx_k\le60-5(n-1)\right\}\right|}{\left|\left\{x_k\ge0:\sum\limits_{k=1}^nx_k\le60\right\}\right|}=\left(\frac{13-n}{12}\right)^n\tag2
$$
Each $x_k$ is the time from car $k-1$, or the beginning of the hour, to car $k$. With $n-1$ buffers of $5$ minutes removed, we get the numerator; without the buffers removed, we get the denominator.

Thus, Bayes' Theorem  gives
$$
\begin{align}
p(\lambda)
&=\sum_{n=0}^{12}e^{-60\lambda}\frac{(60\lambda)^n}{n!}\left(\frac{13-n}{12}\right)^n\tag{3a}\\
&=e^{-60\lambda}\sum_{n=0}^{12}\frac{(65-5n)^n}{n!}\lambda^n\tag{3b}\\
&=e^{-60\lambda}\left(\vphantom{\frac11}\right.
1+60\lambda+\frac{3025}2\lambda^2+\frac{62500}3\lambda^3+\frac{1366875}8\lambda^4\\[3pt]
&+\frac{2560000}3\lambda^5+\frac{367653125}{144}\lambda^6+\frac{30375000}7\lambda^7+\frac{30517578125}{8064}\lambda^8\\[3pt]
&+\frac{800000000}{567}\lambda^9+\frac{284765625}{1792}\lambda^{10}+\frac{15625000}{6237}\lambda^{11}+\frac{9765625}{19160064}\lambda^{12}
\left.\vphantom{\frac11}\right)\tag{3c}
\end{align}
$$
This matches the result from $(8)$ of my previous answer.


答:
$$\text{So, }v=\frac{ds}{dt}=160-32t$$

$$\text{Now, }160t-16t^2=384\implies t^2-10t+24=0\implies t=4,6$$

The smaller value of $t$ will occur during upward movement

Title: How can I find these velocities without using the quadratic formula?

Body:
If a ball is thrown vertically upward with a velocity of $160 \text{ ft/s}$, then its height after t seconds is $s = 160t − 16t^2$.

a) What is the velocity of the ball when it is $384 \text{ ft}$ above the ground on its way up?

  b) What is the velocity of the ball when it is $384 \text{ ft}$ above the ground on its way down?

$$\begin{align*}
384 &= 160t - 16t^2\\
16t^2 - 160t + 384 &= 0\\
16 (t^2 - 10t + 24) &= 0
\end{align*}$$

I can't factor the above... and I'm not supposed to used the quadratic formula. Am I stupid or is this unsolvable without without the formula?


问:
Counterexample to problem regarding Hall $\pi$-subgroups of a quotient group
In M. Isaacs' Finite Group Theory, there is a remark on a problem which the author didn't provide any more details. The problem is this: Given a surjective group homomorphism $\theta:G\rightarrow G/N$, where $G$ is a finite group. If $P/N$ is a Sylow $p$-subgroup of $G/N$, then $P/N=\theta (Q)$ for some Sylow $p$-subgroup $Q$ of $G$. The proof is easy, & it depends on the fact that a Sylow $p$-subgroup does exist.

Now, the author remarked that the problem would not remain true if 'Sylow $p$-subgroup' were replaced by 'Hall $\pi$-subgroup', where $\pi$ is a prime set. However, I don't know where to find a counterexample. If one attempts to imitate the proof as in the Sylow case, one would know that the problem would remain true provided $G$ is solvable. (Since a solvable group has a Hall $\pi$-subgroup for every prime set $\pi$). So if $G$ is a counterexample, $G$ must be nonsolvable.

I'm looking at building $G$ from nonabelian simple groups, which I'm awfully unfamiliar with. So far I haven't got any useful results. Any idea what the minimal counterexample is?

Edit: @SteveD mentioned a counterexample in the comment. But I think I want a counterexample in which whenever we say the Hall $π$-subgroups of a group, we actually require the members of $π$ to divide $|G|$. Perhaps it's only slightly more difficult to figure out such a counterexample.

答:
I think the remark only holds in that primes might vanish in the quotient, and the image thus is not any longer a $\pi$-subgroup:

Suppose that $H\le G$ is a $\pi$-Hall subgroup. Let $A=HN/N$ the image of $H$ in the factor. Then $A\le G/N$ is a subgroup such that $[G/N:A]$ is coprime to $\pi$, and $|A|=[HN:N]=[H:H\cap N]$ is a product of primes in $\pi$. It thus is a Hall subgroup, albeit might be so only for a subset of $\pi$, if certain primes do not divide the order of $G/N$.


问:
Alternative proof that the sum of two convergent series is convergent
For my Real Analysis course I'm being asked to prove this basic concept:

If $\sum_{n=1}^\infty |a_n|$ and $\sum_{n=1}^\infty |b_n|$ converge absolutely then prove that $\sum_{n=1}^\infty (a_n+b_n)$ also converges absolutely.

Now this question is very basic. It can be proved by adapting the proof of the sum law for limits to series instead. I've done this but my professors says the answer is wrong. Instead, he wants to me solve this question without using epsilon-proofs. The only hint he provides is that I must consider the rearrangement of each series and apply the triangle inequality to prove convergence. This logic is incomprehensible to me but he refuses to help any further.

Is there another method to prove this question without using limit laws and epsilon-proofs?

答:
Well, we only want to show that
$$
\sum_{n = 1}^\infty \lvert a_n + b_n \rvert < \infty.
$$
By the triangle inequality we have:
$$
\sum_{n = 1}^\infty \lvert a_n + b_n \rvert \leq \sum_{n = 1}^\infty (\lvert a_n \rvert + \lvert b_n \rvert)
$$
Since both
$$
\sum_{n = 1}^\infty \lvert a_n \rvert \quad \text{ and } \quad \sum_{n = 1}^\infty \lvert b_n \rvert
$$
converge, we can rewrite
$$
\sum_{n = 1}^\infty (\lvert a_n \rvert + \lvert b_n \rvert) = \sum_{n = 1}^\infty \lvert a_n \rvert + \sum_{n = 1}^\infty \lvert b_n \rvert < \infty.
$$
So
$$
\sum_{n = 1}^\infty \lvert a_n + b_n \rvert < \infty
$$
according to the direct comparison test.

In general: If we can prove that some series $\displaystyle \sum_{n = 1}^\infty \lvert c_n \rvert$ is bounded, we know (at least in $\mathbb{R}$) that it converges, since the partial sums
$$
S_m: m \mapsto \sum_{n = 1}^m \lvert c_n \rvert
$$
are bounded and increasing. To see the latter just note that ($m \in \mathbb{N})$:
$$
0 \leq \lvert a_{m+1} - b_{m+1} \rvert \implies S_m \leq S_{m} + \lvert a_{m+1} - b_{m+1} \rvert = S_{m+1}
$$


问:
Formulas for the definition of set pre/image
I refer to Wiki page below as I write down my question.

Let $f: X \to Y$ be a function with $A \subseteq X, B \subseteq Y$. On Wiki the definition of inverse image is given as $f^{-1}[B] = \{x \in X: f(x) \in B\}$ which in my book is also rendered as $x \in f^{-1}(B) \iff f(x) \in B.$

Wiki also gives the definition of image set as $f[A] = \{f(x): x \in A\}$. For this case my book doesn't have a pretty formula like the one above. The examples section of the Wiki page has $f(\{−2, 3\}) = \{4, 9\}$ where $f(x) = x^2$. But $f(2), f(-3) \in \{4, 9\}$, yet $2, -3 \not \in \{-2, 3\}$. Maybe that's why there's no pretty formula for this case.

Given what I wrote above, I'll try to justify why the formulas above (don't) work.

Statement: $x \in f^{-1}(B) \iff f(x) \in B.$

Proof: Let $x \in f^{-1}[B].$ Then by definition of $f^{-1}[B]$, we have $x \in X$ s.t. $f(x) \in B$. Now suppose $f(x) \in B.$ Since $f$ is a function, $f(x)$ must have a pre-image $x$ in the domain of $f$ which is $X$. But by definition of $f^{-1}[B]$ such an element $x$ must be in $f^{-1}[B]$.

Statement: it's not true that $x \in A \iff f(x) \in f(A)$.

Proof: The Wiki page already gave a counterexample, but I'd like to render it as a direct proof to hone my proof-writing skills. Suppose $x \in A$. Since $f$ is a function, $f(x)$ must be in the image of $f$ (distinct from the image of $A$) which is $f(X).$ The image of $A$ is $f(A)$ where $f(A) \subseteq f(X).$ But $f(x) \not \in f(X) \setminus f(A)$ as by definition of $f(A)$ we have that any $f(x) \in f(X) \subseteq Y$ with $x \in A$ must be in $f(A)$. Now let $f(x) \in f(A).$ Since $f$ is a function, $f(x)$ must have a pre-image $x$ in the domain of $f$ which is $X$. But $A \subseteq X$ and nothing precludes $x$ for being in $X \setminus A.$ That means we cannot claim $f(x) \in f(A) \implies x \in A.$

I have two questions:

(1) Do the proofs above work?

(2) Can I alter the definition of image set given on Wiki page to $f[A] = \{f(x) \in Y: x \in A\}$ to make it look a bit more symmetrical with  $f^{-1}[B] = \{x \in X: f(x) \in B\}$?

Wiki page on image set

答:
Your proof essentially comes down to the phrase "nothing precludes $x$ for being in $X∖A$." The only way to justify a statement such as this is to give a counterexample. Generally in math, if you wish to show that a statement is true, the way to do it is with a proof. If you wish to show that a statement is false, you must construct a counterexample. (Note that one direction of your statement is true, namely that $x\in A\implies f(x)\in f(A)$. The fact that the converse is false requires counterexample.)

For the second part of your question, you could define $f^{-1}(A)=\{y\in Y:\exists x\in A \mbox{ s.t. } f(x)=y\}$. I don't know if this is more to your liking or not.


问:
Extreme values of $2\cos^2 \frac{x}{2}\cdot\sin^2 x$ without calculus

To find the extreme values of
$$2\cos^2 \frac{x}{2}\cdot\sin^2 x$$

I Have not learnt calculus yet but this is a question within our syllabus .which made me to think if this can be done with it .

I am not able approach this problem any help is appreciated.

答:
Via AM-GM,
$$2\cos^2\frac x2 \cdot \sin^2 x = (\cos x + 1)(1-\cos^2 x) = \frac 12 (1 + \cos x)(1+\cos x) (2- 2\cos x) \le \frac 12 \left( \frac{1+\cos x + 1 + \cos x + 2 - 2\cos x}{3}\right)^3=\frac 12 \left( \frac 43\right)^3 = \frac{32}{27}$$

where $"=" \iff 1+\cos x = 2- 2 \cos x \iff\cos x = \frac 13.$


问:
Advanced combinatorial probability
There are $5$ rooms $R_1,R_2,R_3, R_4, R_5$ in a hallway. $17$ people may enter any of those $5$ rooms upon their wish. What is the probability that $5$ of them will enter $R_1$, $2$ will enter $R_2$, $4$ will enter $R_3$, $4$ will enter $R_4$  and  $2$  will enter $R_5$?

答:
There is $17$ people but we should select $5 $people among them to place the first room.Then , we have $12$ people and we shoul select $2$ people among them.(I assume you know that combination i used for selecting objects in a set ) This process will continue until it is done.

Let come to the case of $5^{17}$ . It is said that each people can choose any of the rooms ,so the first person has $5 $ choice to get in a room and the second has also $5$ choice and so on ,i.e,  $5....5=5^{17}$

$\frac{C(17,5)C(12,2)C(10,4)C(6,4)C(2,2)}{5^{17}}$

Title: Advanced combinatorial probability


问:
Count number of maps from $2$ object set to $S_n$ up to equivalence
I need to find a way to count the number of maps from $\{0,1\}$ to $S_{n}$ the symmetric group on n symbols, where 2 maps $f,g$ are equivalent if $\exists \sigma \in S_{n}$, $\sigma f \sigma^{-1} = g$ (that is, $\sigma f(0) \sigma^{-1} = g(0)$ and $\sigma f(1) \sigma^{-1} = g(1)$. specifically im interested in the case $n=3$ but if there is some generalization for $n$ and maybe a set larger than $\{0,1\}$ I would love to hear it.

What I tried: I was thinking that it was enough to count the number of maps from $\{0,1\}$ to the set of conjugacy classes of $S_{n}$, but obviously the condition imposed is slightly stronger, since it isnt enough to ask that $g(0)$ is conjugate to $f(0)$ and same for $f(1),g(1)$. Perhaps there's a way to slightly modify this argument?

答:
Call $M_{n}$ the set of functions form $[2]=\{0,1\}$ to $S_n$ and consider the action by conjugation of $S_n$ there. You want to count the number of orbits. Burnside lemma tells us that
$$\text{# Orbits}=\frac{1}{|S_n|}\sum _{\sigma \in S_n}\left | M_n^{\sigma }\right |=\frac{1}{n!}\sum _{\sigma \in S_n}|\text{# functions fixed by $\sigma$}|.$$

$f$ is fixed by $\sigma$ iff $\sigma f \sigma ^{-1}=f$ so $f(0)\sigma ^{-1}f(0)^{-1}=\sigma ^{-1}$ that means that $f(0)$ fixes $\sigma ^{-1}$ under conjugation. Same for $f(1).$ so we have to get two permutations that fix $\sigma ^{-1}.$

To fix a permutation by conjugation we have to fix every cycle, and to fix a cycle of length $k$ we have $k$ choices. That means that if $\sigma ^{-1}$ has cycle type $\lambda _1,\cdots ,\lambda _s$ then we have $\prod _{i=1}^s\lambda _i$ ways to fix it, so if we want to choose two elements of this set we have to compute $$\left ( \prod _{i=1}^s\lambda _i\right ) ^2=\prod _{i=1}^s\lambda _i^2.$$ In that way we have that
$$\text{# Orbits}=\frac{1}{n!}\sum _{\lambda \vdash n}\sum _{\sigma \text{ has cycle type }\lambda} \prod _{i=1}^s{\lambda _i}^2=\frac{1}{n!}\sum _{\lambda \vdash n}\left ( \text{# permutations type $\lambda$}\right )\cdot \prod _{i=1}^s{\lambda _i}^2. $$
If you have a partition $\lambda =(\lambda_1,\cdots , \lambda _s)$ and you consider $a_i = \text{# times $i$ is a part of $\lambda$},$ then, because you can choose the elements in each cycle and then multiply by number of cycles, the sum above can be expressed as
$$\text{# Orbits}=\frac{1}{n!}\sum _{a_1+2\cdot a_2+\cdots +n\cdot a_n=n}\frac{n!}{\prod _{i=1}^ni^{a_i}}\cdot \prod _{i=1}^ni^{2\cdot a_i}=\sum _{a_1+2\cdot a_2+\cdots +n\cdot a_n=n}\prod _{i=1}^ni^{a_i},$$
from there is not hard to believe that if you have functions from $[m]$ to $S_n$ then the total number of functions is
$$\sum _{a_1+2\cdot a_2+\cdots +n\cdot a_n=n}\prod _{i=1}^ni^{(m-1)a_i}.$$
Which kind of makes sense for $m=1$ because you get the number of partitions of $n.$

This can be expressed as
$$[x^n]\prod _{k=1}^n\frac{1}{1-k^{m-1}\cdot x^k},$$
for $m=2$ you can find more info in OEIS. $m>2$ seems to not be there.


问:
Calculate the determinant of the following $n \times n$ symmetric matrix:
Let $n$ be a positive integer and $A = (a_{ij})_{n\times n}$, where $a_{ij} = |i-j|$, for $i = 1, 2, \dots, n$ and $j = 1, 2, \dots, n$. Calculate $\det A$.

I noticed that $a_{ii} = 0$ and $a_{ij} = a_{ji}$, so A is a symmetric matrix. Also, I saw that, if we make the notation $A_n$ for the A with n elements, $A_n$ is constructed from $A_{n-1}$ with $n-1, n-2, \dots, 0$ as elements for the last line and last column. I tried to use Laplace expansion but with no result.

This is how $A_n$ looks like: $A_n=\begin{bmatrix}
0&1&2& .&.&. &n-1\\
1&0&1&2& .&.&n-2 \\
2&1&0&1&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&2 \\
.&.&.&.&.&.&1 \\
n-1&n-2&.&.&2&1&0
\end{bmatrix}$

I calculated for a few small numers: $\det A_1 = 0$, $\det A_2 = -1$, $\det A_3 = 4$, $\det A_4 = -12$, $\det A_5 = 32$, but I didn't figure out a rule such that I could find the determinant through induction. Can you help me on this one?

Body:
You can take this approach:

$$A_n=\begin{bmatrix}
0&1&2& .&.&. &n-1\\
1&0&1&2& .&.&n-2 \\
2&1&0&1&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&2 \\
.&.&.&.&.&.&1 \\
n-1&n-2&.&.&2&1&0
\end{bmatrix} $$
Now, add the last column to the first one, notice it will always be equal to $n-1$. $(C_1 = C_1+C_n)$

$$\begin{bmatrix}
n-1&1&2& .&.&. &n-1\\
n-1&0&1&2& .&.&n-2 \\
n-1&1&0&1&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&2 \\
.&.&.&.&.&.&1 \\
n-1&n-2&.&.&2&1&0
\end{bmatrix} =(n-1)\begin{bmatrix}
1&1&2& .&.&. &n-1\\
1&0&1&2& .&.&n-2 \\
1&1&0&1&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&2 \\
.&.&.&.&.&.&1 \\
1&n-2&.&.&2&1&0
\end{bmatrix}$$
From here, we can do as follows:

go from the last row towards the first and decrease each row's value with the one above it (for any row but the first one).
($\forall i \neq 1, R_i=R_i-R_{i-1}$, Starting with $i=n$ then $i=n-1 ... i=2$)
$$ = (n-1)\begin{bmatrix}
1&1&2& .&.&. &n-1\\
0&-1&-1&-1& .&.&-1 \\
0&1&-1&-1&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&-1 \\
.&.&.&.&.&.&-1 \\
0&1&.&.&1&1&-1
\end{bmatrix}$$
Expand $C_1$:
$$ = (n-1)\begin{bmatrix}
-1&-1&-1& .&.&-1 \\
1&-1&-1&.&.&. \\
.&.&.&.&.&. \\
.&.&.&.&.&-1 \\
.&.&.&.&.&-1 \\
1&.&.&1&1&-1
\end{bmatrix}$$
Now Add the first row to all of the other rows ($\forall i \neq 1, R_i = R_i + R_1$)
$$ = (n-1)\begin{bmatrix}
-1&-1&-1& .&.&-1 \\
0&-2&-2&.&.&. \\
.&.&-2&.&.&. \\
.&.&.&.&.&-2 \\
.&.&.&.&.&-2 \\
0&.&.&0&0&-2
\end{bmatrix} = (n-1)[-1*(-2)^{n-2}]$$


问:
To show that $\sqrt{z^2-1}=\exp(\frac{1}{2} \log(z^2-1))$ is analytic in the plane minus [-1,1]
By the definition of logarithm branch, we set that $\sqrt{z^2-1}=\exp(\frac{1}{2} \log(z^2-1))$
However to show that the $\sqrt{z^2-1}$ is analytic in the entire plane minus the interval [-1,1], it is better to prove firstly $\sqrt{z^2-1}$ is continuous along the interval (-∞,1) and then show the above statement by applying Morera's theorem. (My course advisor said to me)

However, when I follow the instruction of my advisor, I just get stuck for solving it. He said that when Im z>0 the path of integration is in the upper half-plane, and vice versa. I understood that statement. Therefore, to prove that limit value of f(z) for z -> x for $x \in (-∞, -1)$, we must establish that the same limit exists also when we approach x to upper half-plane or the lower half-plane.

But I don't understand that the difference between these limits equals to the $\int_C \frac{2\zeta}{\zeta^2-1}$ (C is regular closed curve surrounding $\zeta=+1,-1$ . Why does that differnce is followed by above integral form, and why does the contour is defined as above? I barely grasp the concept..........

Also, assuming that I understand that thing, there is still problematic to complete the proof. By the argument principle, since the $z^2-1$ has its zero in C by +1 and -1, $\int_C \frac{2\zeta}{\zeta^2-1}d\zeta=2 \pi i \times 2 = 4 \pi i$ , Then, how we conclude that limit value exists? My professor just said me that $e^{w/2}=e^{(w+4\pi i)/2}$ but I cannot  understand why that thing satisfies. After that, how to applying Morera to complete this things?

Your suggestion will help me !!!!! thanks

答:
This is similar to Making a cube root function analytic on $\mathbb{C}\backslash [1,3]$

The plan is this:

Introduce $f(z)=\frac12\log (z^2-1)$ as a holomorphic function in the domain $\mathbb C\setminus (-\infty,1]$.

Show that $\exp(f)$ extends continuously to the interval $(-\infty,-1)$.

Use Morera's theorem to show that $\exp(f) $ is holomorphic in $\mathbb C\setminus [-1,1]$.

Step 1 uses the fact that $\mathbb C\setminus (-\infty,1]$ is simply-connected and does not contain the points $\pm 1$ where $f$ has singularities.

Step 2, on which you got stuck, is based on the fundamental theorem of calculus:  $f(b)-f(a)=\int_\gamma f'(\zeta)\,d\zeta$ where $\gamma$ can be any smooth curve that goes from $a$ to $b$. Notice that $f'(\zeta)=  \zeta/(\zeta^2-1)$. Integrating $f'(\zeta)$  along a simple   closed curve that passes through $x_0\in (-\infty,-1)$, we find that the integral is $2\pi i$. This  means that  $f$ takes on different values when approaching $x_0$ from above and from below. But since the exponential function is $2\pi i$ - periodic, it follows that $\exp(f)$ has the same limit  when approaching $x_0$ from above and from below. This is what we wanted.


问:
Constant function formal definition
A generic function $F : X \rightarrow Y$ can be defined as $\forall x \in X \exists !y \in Y (x, y) \in F$. How about the formal definition of a constant function? The formula $\exists !y \in Y \forall x \in X (x, y) \in F$ does not seem right to me, since it does not prevent that the same element in the set $X$ is mapped to multiple different elements in the set $Y$. Instead, it only prevents that two different constant lines can be mapped from the same set $X$.

答:
The definition $\exists !y \in Y \forall x \in X (x, y) \in F$ does not seem right to me, since it does not prevent that two different elements in the set $X$ are mapped to the same element in the set $Y$.

But you do not want to prevent that different elements from $X$ get mapped to the same element in $Y$.
In fact, you want exactly that to happen: different elements get mapped all to the same element in $Y$.

So your definition correctly describes constant functions, if you also require $F$ to be a function.
So you could use both axioms and combine them with a $\land$ symbol.

Title: Constant function formal definition


答:
How about just combining the two statements together?
$$(\forall x\in X\ \exists!y\in Y\ (x,y)\in F)\land(\exists y\in Y\ \forall x\in X\ (x,y)\in F)$$
Note that uniqueness of the $\exists$ in ths second statement is not required: suppose two such $y$'s existed, $y_1$ and $y_2$. Then for any element $x\in X$, $(x,y_1)$ and $(x,y_2)$ are in $F$, which is not allowed by the first statement. Thus only one such $y$ may exist.

Title: Constant function formal definition

Body: A generic function $F : X \rightarrow Y$ can be defined as $\forall x \in X \exists !y \in Y (x, y) \in F$. How about the formal definition of a constant function? The formula $\exists !y \in Y \forall x \in X (x, y) \in F$ does not seem right to me, since it does not prevent that the same element in the set $X$ is mapped to multiple different elements in the set $Y$. Instead, it only prevents that two different constant lines can be mapped from the same set $X$.


问:
Find the remainder R (x) if it holds that:
\begin{align*}
  P(x) &= q(x)(x^2+x+1)+x+1  \\
  P(x) &= Q(x)(x^2-x+1)+x-1  \\
P(x) &= M(x)(x^4+x^2+1)+ R(x)
\end{align*}

Body:
Algebraic useful identity$$x^4+x^2+1=x^4+2x^2+1-x^2=(x^2+1)^2-x^2=(x^2+1+x)(x^2+1-x)$$
to give remainder $(x+1)$ when divided by $(x^2+x+1)$ and remainder $(x-1)$ when divided by $(x^2-x+1)$ polynomial must be like
$$P(x)=\left(x^2+x+1\right) \left[\left(x^2-x+1\right) Q(x)+x-1\right]+x+1$$
$$P(x)=Q(x)(x^4+x^2+1)+[(x^2+x+1)(x-1)+(x+1)]$$
$$P(x)=\left(x^4+x^2+1\right) Q(x)+x^3+x$$
$$R(x)=x^3+x$$