问:
Does $\sigma(T)=\overline{\sigma_p(T)}$ imply $\sigma_r(T)=\emptyset$?
Let $X$ be a $\mathbb C$-Banach space and $T\in\mathfrak L(X)$ with $\sigma(T)=\overline{\sigma_p(T)}$, where $\sigma_p(T)$ denotes the point spectrum of $T$. Does it necessarily follow that the residual spectrum $\sigma_r(T)$ is empty?

The situation I've got in mind is $X=\ell^2$ and $(Tx)_n:=\lambda_n x_n$ for some bounded $(\lambda_n)\subseteq\mathbb C$. In that situation, $\sigma_p(T)=\{\lambda_n:n\in\mathbb N\}$ and it is easy to show that $\mathbb C\setminus\overline{\sigma_p(T)}\subseteq\rho(T)$. On the other hand, $\overline{\sigma_p(T)}\subseteq\overline{\sigma(T)}=\sigma(T)$ so that we can conclude $\sigma(T)=\overline{\sigma_p(T)}$. Now I'd like to conclude the continuous spectrum $\sigma_c(T)$ is equal to $=\overline{\sigma_p(T)}\setminus\sigma_p(A)$.

EDIT: According to the answer below, the claim is wrong in general, but how do we see it in the particular situation described above?

答: No, take any operator of the form $T=T_1\oplus T_2$, on $X=X_1\oplus X_2$, with $\overline{\sigma_p(T_1)}=\sigma_r(T_2)$.


问:
Finding example of sets that satisfy conditions
give examples of sets such that:

i)$A\in B$ and $A\subseteq B$

My answer : $B=\mathcal{P(A)}=\{\emptyset,\{1\},\{2\},\{1,2\}\}$ and $A=\{1,2\}$ then $A\in B$ and $A\subseteq B$

ii) $|(C\cup D)\setminus(C\cap D)|=1$

My answer is: $C=\{1,2,3\}$, $D=\{2,3\}$ then $C\cup D=\{1,2,3\}$ and $C\cap D=\{2,3\}$ so $(C\cup D)\setminus(C\cap D)=\{1\}$ and $|(C\cup D)\setminus(C\cap D)|=1$

Can we find sets A and B such that $A\in B$ and $B\subseteq A$?
My answer is no.

Are my answers correct?

答:
Your answer is incorrect. Because $1,2\in A$, but $1,2\notin B$. Your second answer is correct.

To the last question, the answer is again correct (assuming $\sf ZF$), because $A\in B$ and $B\subseteq A$ would imply that we have $A\in A$, which is impossible due to the axiom of regularity.

To correct the first answer, consider the empty set.

ContentLicense: CC BY-SA 3.0


问:
Geometry on Isosceles Triangles
As shown in the figure, $$\overline{BD}=\overline{DE}=\overline{AE}\\\overline{EH}=\overline{HF}=\overline{FI}=\overline{IG}=\overline{GC}$$
Where $\overline{EH}$ is the bisector of $\angle AEC$.
What is the ratio of $\angle B$ and $\angle C$ ($\frac{\angle B}{\angle C}$) in degrees?

I put all the statements on the figure, but it doesn't seem to give me much useful information.

I also tried to assume a specific angle (for example, $\angle B$) to be some unknown angle $x$, but even with a lot of isosceles triangles given, there are still a lot of unknown anges.

答:
Let $\angle B = x$ and $\angle C = y$.

We have $\angle DEB = x$, and thus $\angle EDA = 2x$. $\angle DAE$ is $2x$ as well.

On the right we have $\angle GIC = y$. Hence $\angle IGF = 2y$, and so is $\angle IFG$.

Now $\angle FIH = 180^\circ - \angle FIG - \angle GIC = 180^\circ - (180^\circ - 2y - 2y) - y = 3y = \angle FHI$.

Similarly, $\angle HFE = \angle HEF = \angle HEA = 4y$.

Considering the angles at $E$, we have:

$$180^\circ = \angle BED + \angle DEA + \angle AEH + \angle HEF = x + (180^\circ - 4x)+4y+4y$$

Therefore $3x = 8y \leadsto \dfrac {\angle B}{\angle C} = \dfrac 83$.

ContentLicense: CC BY-SA 4.0

Title: Geometry on Isosceles Triangles


答:
Assume $\angle B$ and $\angle C$ to be $y$ and $x$ respectively.

By angle chasing,  you can show that $\angle AEB=180-3y$ and $\angle  HEC=4x$. Since $EH$ bisects $\angle AEC$, $\angle AEC=8x$

$\Rightarrow   3y=8x$

$\Rightarrow   \frac{\angle B}{\angle C}=\frac{y}{x}=\boxed {\frac{8}{3}}$

Title: Geometry on Isosceles Triangles

Body:
As shown in the figure, $$\overline{BD}=\overline{DE}=\overline{AE}\\\overline{EH}=\overline{HF}=\overline{FI}=\overline{IG}=\overline{GC}$$
Where $\overline{EH}$ is the bisector of $\angle AEC$.
What is the ratio of $\angle B$ and $\angle C$ ($\frac{\angle B}{\angle C}$) in degrees?

I put all the statements on the figure, but it doesn't seem to give me much useful information.

I also tried to assume a specific angle (for example, $\angle B$) to be some unknown angle $x$, but even with a lot of isosceles triangles given, there are still a lot of unknown anges.


问:
Domain/range finding of composite trig functions
I am stuck on this problem in my book for finding the domain and range of composite functions

For problems 8a-e I used a developed method to solve for the implied domain of these functions which produced correct results

Steps were:

Use the domain of the first function and the range of the other trig function contained inside it

Find the intersection of these two values

substitute the contained trig function between the intersection value to find the domain

substitute the domain values into the function to find the range

However, once I got to "8f" I found that the method I was using for the previous questions didn't carry over into arctan problems.

I see now that I probably have a fundamental misunderstanding of the process needed to solve these forms of problems. I'm guessing that question 10 is also related to this type of question

Could you please help explain the solving process for these kind of problems to me?

Thanks

答:
It's not your fault that at least some of the questions are clearly incorrectly framed. In particular, 10 seems to be missing the inverse function on each of its outside $\sin$ or $\cos$ terms.

On 8f, you would want to note that $\cos x\in[-1,1]$ for all real numbers, and then identify the portion of the range of $\tan^{-1}$ corresponding to those values, i.e. find $\tan^{-1}([-1,1])$. Let me know if that gets you where you need to go!


问:
Unitary operator on a Hilbert space
Let $U$ be a unitary operator with $\langle Ux,x \rangle\ge 0\quad  \forall x$.

Show that $U$ is the identity operator.

To prove this statement i'm reasoning in this way:

since $U$ is an isometry, $\langle Ux,Uy \rangle = \langle x,y \rangle$, i.e. $U^*U=UU^*=I$, with $U^*$ the adjoint of $U$.

I know that $\langle Ux,Ux \rangle =\langle x,x \rangle \ge 0$, but I don't know how to go on.

Can someone help me?

Thanks.

答:
False for real scalars but for complex scalars here is a proof: $U$ positive implies $U^{*}=U$  in this case, so $U^{2}=I$. Thus $(U-I)(U+I)=0$. But $U+I$ is invertible so $U-I=0$.

[$\sigma (U) \subseteq [0,\infty)$ since $U$ is positive and $\sigma (U) \subseteq S^{1}$ since $U$ is unitary. Hence, $\sigma (U) \subseteq\{1\}$ which shows that $-1 \notin \sigma (U) $. Hence, $U+I$ is invertible].

Title: Unitary operator on a Hilbert space


问:
How do I integrate $\int\dfrac{e^x - e^{-x}}{e^x + e^{-x}}\,dx$?

$$\displaystyle\int\frac{e^x - e^{-x}}{e^x + e^{-x}}\,dx$$

In Mathematica:

Integrate[(E^x - E^-x)/(E^x + E^-x), x]

I'm using the substitution $u=e^x + e^{-x}$, so

$du = e^x + e^{-x}$

$e^{-x} = u - e^x$

$e^{x} = u - e^{-x}$

Substituting into the the equation gives:

$$\int\frac{2e^x-u}{u}du$$

But I'm not sure where to go from here to get the supposed answer of:

$$\log(1 + e^{2x}) - x$$

答:
Let the denominator $e^x+e^{−x}$ be $t$
then differentiating both sides
$$e^x+e^{−x}$$
$$e^x-e^{−x}=dt/dx \iff (e^x-e^{−x})dx=dt$$

substitute this in the question
$$∫1/t \cdot dt
\log t$$


答:
(i) This doesn’t quite work, unfortunately, because $A\nsubseteq B$, i.e., $\{1,2\}\nsubseteq\{\varnothing,\{1\},\{2\},\{1,2\}\}$. In order for $A$ to be a subset of $B$, each element of $A$ must be an element of $B$. The elements of $A$ are $1$ and $2$, and neither of them is an element of $B$. It’s true that $\{1\}\in B$ and $\{2\}\in B$, but that’s very different from having $1\in B$ and $2\in B$. Try the same idea with $A=\varnothing$.

(ii) This is fine.

(iii) Your answer is correct: if $A\in B\subseteq A$, then $A\in A$, which is ruled out by the axiom of regularity (also called foundation).

Title: Finding example of sets that satisfy conditions

Body:
give examples of sets such that:

i)$A\in B$ and $A\subseteq B$

My answer : $B=\mathcal{P(A)}=\{\emptyset,\{1\},\{2\},\{1,2\}\}$ and $A=\{1,2\}$ then $A\in B$ and $A\subseteq B$

ii) $|(C\cup D)\setminus(C\cap D)|=1$

My answer is: $C=\{1,2,3\}$, $D=\{2,3\}$ then $C\cup D=\{1,2,3\}$ and $C\cap D=\{2,3\}$ so $(C\cup D)\setminus(C\cap D)=\{1\}$ and $|(C\cup D)\setminus(C\cap D)|=1$

Can we find sets A and B such that $A\in B$ and $B\subseteq A$?
My answer is no.

Are my answers correct?


问:
Optimal Strategy in Dice Game
In a recent interview I received the following question (an optimisation/strategy game)...which left me a bit stumped. The rules of play, you start with 0 points, then:

Roll three fair six-sided dice;

Now you have the option:

Hold: i.e. accept the values shown on your dice as the score for your turn. There is a caveat, if two or more dice show the same values, then all of them are flipped upside down - e.g. 1 becomes 6

OR

reroll the dice: You may choose to hold any combination of the dice on the current value shown (so you can choose to keep 1 dice the same and then reroll the other two). Rerolling costs you 1 point – so during the game and perhaps even at the end your score may be negative.
You can roll an infinite number of times...

My thoughts:

So clearly the best possible score is 18 and is achieved by rolling three 1s on the first roll
The reroll penalty prevents rolling forever to get 18.
If the value of the dice is greater than the expected value of rerolling them (accounting for the penalty), then you should stick...
I guess what I am asking is how do I work out the expected value of rerolling them (accounting for the penalty) and how does this fit into the optimal strategy...

答:
Comments only:

A good programmer (which I'm not) can implement Hugo Manet's idea and get the optimal strategy.  But I believe there is no nice pattern for one to solve mathematically based on my calculation with only one step projection (thus it's not optimal). Still one can get a feeling as to why it's not so simple (I could be wrong). Here's a summary:

If you get $3$ identical numbers, stick if it's $111,222,333$; reroll all three if you have $444,555,666$;

If you get $3$ distinct numbers:

If total points $\ge 11$ then stick;

If total points $\le 9$ then keep the smallest, reroll the two bigger ones;

If total points $=10$: keep the bigger two and reroll $1$ if it's $\{1,3,6\}$ or $\{1,4,5\}$; stick if it's $\{2,3,5\}$.

You get a pair and a third one which is different. This is very complicated.

When the pair is $1, 2$ or $3$: if the third one is $4/5/6$, stick; otherwise reroll the third one. Notice that if we have $\{2,2,3\}$ we get a tie: either stick, or reroll $3$;

When the pair is $4$: if the third one is $5/6$, stick; otherwise keep the third one but re-roll both $4$'s;

When the pair is $5$: if the third one is $1/2$, keep it and  reroll both $5$'s; otherwise reroll only one $5$ and keep the other two;

When the pair is $6$:  if the third one is $1$, keep it and reroll both $6$'s; otherwise reroll only one $6$ and keep the other two.

Again, the above is not the optimal because it only looks at one step. The optimal strategy should be decided by going backward from step 18, for example, where you know you should stick because otherwise you'd end up with a negative score.

R code below:

It was done in a rush so there's plenty of space for improvement.

```<code>pts = function(a)
{
  u = unique(a);
  if(length(u)==3) {
    p = sum(a);
  } else if(length(u)==1) {
    p = 3*(7-u);
  } else
  {
    M = sum(a)-sum(u)
    p = sum(u)-M + 2*(7-M);
  }
  p
}

onestep <- function(a)
{
  p = rep(-1,8);

  p[8]=pts(a);

  for(i in 1:6)
  {
    # 001
    p[1] = p[1] + pts(c(a[1], a[2], i))/6

    # 010
    p[2] = p[2] + pts(c(a[1], i, a[3]))/6

    # 100
    p[4] = p[4] + pts(c(i, a[2], a[3]))/6

    for(j in 1:6)
    {
        # 011
        p[3] = p[3] + pts(c(a[1], i, j))/36

        # 101
        p[5] = p[5] + pts(c(i, a[2], j))/36

        # 110
        p[6] = p[6] + pts(c(i, j, a[3]))/36

        for(k in 1:6)
        {
        # 111
        p[7] = p[7] + pts(c(i,j,k))/216;
        }
    }

  }
  indx = which(p==max(p));
  c(p, p[indx], indx)
}

sink("mseoutput.txt")

for(i in 1:6)
{
    for(j in i:6)
    {
        for(k in j:6)
        {
            out <- onestep(c(i,j,k));
            cat(i,j,k,out);
            cat('\n');
        }
    }
}

sink()
</code>```

Title: Optimal Strategy in Dice Game

Body:
Comments only:

A good programmer (which I'm not) can implement Hugo Manet's idea and get the optimal strategy.  But I believe there is no nice pattern for one to solve mathematically based on my calculation with only one step projection (thus it's not optimal). Still one can get a feeling as to why it's not so simple (I could be wrong). Here's a summary:

If you get $3$ identical numbers, stick if it's $111,222,333$; reroll all three if you have $444,555,666$;

If you get $3$ distinct numbers:

If total points $\ge 11$ then stick;

If total points $\le 9$ then keep the smallest, reroll the two bigger ones;

If total points $=10$: keep the bigger two and reroll $1$ if it's $\{1,3,6\}$ or $\{1,4,5\}$; stick if it's $\{2,3,5\}$.

You get a pair and a third one which is different. This is very complicated.

When the pair is $1, 2$ or $3$: if the third one is $4/5/6$, stick; otherwise reroll the third one. Notice that if we have $\{2,2,3\}$ we get a tie: either stick, or reroll $3$;

When the pair is $4$: if the third one is $5/6$, stick; otherwise keep the third one but re-roll both $4$'s;

When the pair is $5$: if the third one is $1/2$, keep it and  reroll both $5$'s; otherwise reroll only one $5$ and keep the other two;

When the pair is $6$:  if the third one is $1$, keep it and reroll both $6$'s; otherwise reroll only one $6$ and keep the other two.

Again, the above is not the optimal because it only looks at one step. The optimal strategy should be decided by going backward from step 18, for example, where you know you should stick because otherwise you'd end up with a negative score.

R code below:

It was done in a rush so there's plenty of space for improvement.

pts = function(a)
{
  u = unique(a);
  if(length(u)==3) {
    p = sum(a);
  } else if(length(u)==1) {
    p = 3*(7-u);
  } else
  {
    M = sum(a)-sum(u)
    p = sum(u)-M + 2*(7-M);
  }
  p
}

onestep <- function(a)
{
  p = rep(-1,8);

  p[8]=pts(a);

  for(i in 1:6)
  {
    # 001
    p[1] = p[1] + pts(c(a[1], a[2], i))/6

    # 010
    p[2] = p[2] + pts(c(a[1], i, a[3]))/6

    # 100
    p[4] = p[4] + pts(c(i, a[2], a[3]))/6

    for(j in 1:6)
    {
        # 011
        p[3] = p[3] + pts(c(a[1], i, j))/36

        # 101
        p[5] = p[5] + pts(c(i, a[2], j))/36

        # 110
        p[6] = p[6] + pts(c(i, j, a[3]))/36

        for(k in 1:6)
        {
        # 111
        p[7] = p[7] + pts(c(i,j,k))/216;
        }
    }

  }
  indx = which(p==max(p));
  c(p, p[indx], indx)
}

sink("mseoutput.txt")

for(i in 1:6)
{
    for(j in i:6)
    {
        for(k in j:6)
        {
            out <- onestep(c(i,j,k));
            cat(i,j,k,out);
            cat('\n');
        }
    }
}

sink()


问:
What would be the distribution of $S_1+S_2$?
Note : For details about the Wishart distribution $\mathcal W_p(\Sigma,n)$, see here.

Let $S_1\sim \mathcal{W}_p(\Sigma_1,n_1)$ and $S_2\sim \mathcal{W}_p(\Sigma_2,n_2)$

I know that if for example we have the case that $\Sigma_1=\Sigma=\Sigma_2$, then $S_1+S_2\sim \mathcal{W}_p(\Sigma,n_1+n_2)$

But what if for example $\Sigma_1=\frac{n_1-1}{n_1}\Sigma$ and $\Sigma_2=\frac{n_2-1}{n_2}\Sigma$

What would be the distribution of $S_1+S_2$?

答: If $S_1$ and $S_2$ are independent, then I think this paper can provide the answer you are looking for: https://arxiv.org/pdf/1406.6638.pdf. Specifically equations (10) and (11). The expression involves hypergeometric functions.


问:
Need help figuring out a pattern question
my cousin asked for help on this math problem and the answer is apparently 6 but i can't find a coherent pattern that fits all of the triangles. I included an image of the question.

答: The missing no is 6. Sum of numbers in each horizontal layer is 36.</p>


问:
How to self study real analysis
I'm a high school student, who is interested in learning real analysis, but it seems like every textbook I take doesn't really provide a way to intuitively think about the concepts. I understand it to some extent but I'm not satisfied with the level of intuition provided so far. Can someone help me recommend and textbook or a way to study it</p>

答:
It seems you need a book that guides you as a real lecturer, so any suggestion to the super terse Rudin's Principles of Mathematical Analysis is to avoid. I would recommend two books: the former is to fill any gap from calculus to real analysis, and the other one to have a light introduction to real analysis.

Calculus: Early Transcendentals, by Bernard Gillett, Lyle Cochran, and William L. Briggs

Introduction to Analysis, by Edward D. Gaughan

If you couple them together, you will be able to have a smooth transition from high school level math to (introductory) undergraduate math classes.

Title: How to self study real analysis


问:
On the proof of the Khintchine-Kolmogorov Convergence Theorem
Let $(X_n)_{n=1}^{\infty}$ be a sequence of independent, mean-zero random variables with $\sum_{n=1}^{\infty}\text{Var}(X_n) < \infty$.

This is about an intermediate step in the proof of the  Khintchine-Kolmogorov Convergence Theorem. Denote $S_j = X_1 + \cdots + X_j$. I have that for any $h > 0$, $$\mathbb{P}\left(\sup_{i \geq 1}|S_{k + i} - S_k| \geq h \right) \leq \dfrac{1}{h^2}\sum_{j=k+1}^{\infty}\sigma^2_j$$ where $\sigma^2_j$ denotes the variance of $X_j$. How do you make the leap from this to $$\mathbb{P}\left(\sup_{n, k \geq N}|S_{n} - S_k| \geq h \right) \leq \dfrac{2}{h^2}\sum_{j=N+1}^{\infty}\sigma^2_j$$
After sorting through nearly a dozen measure-theoretic probability books, I haven't seen this exact leap being made, but Durrett is very close.

Here's my attempt: let's write
$$\sup_{n, k \geq N}|S_n - S_k|  = \sup_{n, k \geq N}|S_n - S_{k +i} + S_{k+i} - S_k| \leq \sup_{n, k \geq N}|S_{k + i} - S_n| + \sup_{n, k \geq N}|S_{k + i} - S_k|\text{.}$$
I'm not exactly sure how the $\sup_{n, k \geq N}$ get translated to $\sup_{i \geq 1}$ here, assuming this is correct.

答:
You can prove  he inequality with a bigger constant on the right but that is good enough for proving the theorem.

$\sup_{n,k \geq N} |S_n-S_k| \geq h$ implies $\sup_{n \geq N} |S_n-S_N| \geq h/2$ or $\sup_{k \geq N} |S_k-S_N| \geq h/2$. Form this you get the required inequality with RHS repaced by $2\frac  1 {(h/2)^{2}} \sum\limits_{j=N}^{\infty} \sigma_j^{2}$ which means $\frac 2 {h^{2}}$ gets replaced by $\frac  {8} {h^{2}}$.

[$\sup_{n \geq N} |S_n -S_N| $ is same as $\sup_{i \geq 1}  |S_{N+i}-S_N|$].


问:
How to prove the following calculus theorem
Let $ f:A\subseteq \mathbb{R}^n\to\mathbb{R}$ be a function which is defined in some punctured neighborhood of $ x_0 $. Then $ \lim_{x\to x_{0}}f\left(x\right)=L $ if and only if, for any path $ \gamma:[0,1]\to\mathbb{R}^{n} $ such that $ \lim_{t\to1^{-}}\gamma\left(t\right)=x_{0} $ and  $ \forall t\in[0,1]\thinspace\thinspace:\thinspace\thinspace\thinspace\gamma\left(t\right)\neq x_{0} $ we have $ \lim_{t\to1^{-}}f\left(\gamma\left(t\right)\right)=L $.

If I know that $ \lim_{x\to x_{0}}f\left(x\right)=L $, then its not hard to prove that for any path $ \gamma $ we have $ \lim_{t\to1^{-}}f\left(\gamma\left(t\right)\right)=L $.

Im struggling with the other direction. Assume for any path $ \gamma(t) $ as described in the theorem we have $ \lim_{t\to1^{-}}f\left(\gamma\left(t\right)\right)=L $. How can I show that the limit of $ f $ is $ L $ ?

Also, does this mean that I could choose whatever path I want in order to check if the limit exists? for example, say I have the function $ f\left(x,y\right) $ and I want to check its limit at $(0,0)$. If I'll choose the paths:

$ \gamma_{1}\left(t\right)=\left(t^{2}-1,t\sin\left(t-1\right)\right),\thinspace\thinspace\gamma_{2}\left(t\right)=\left(\ln\left(t\right),\sqrt{\left(1-t\right)}\right) $

and I'll get that one of the limits $ \lim_{t\to1^{-}}f\left(\gamma_{1}\left(t\right)\right) $ or $ \lim_{t\to1^{-}}f\left(\gamma_{2}\left(t\right)\right) $ does not exist, or they both exist but $ \lim_{t\to1^{-}}f\left(\gamma_{1}\left(t\right)\right)\neq\lim_{t\to1^{-}}f\left(\gamma_{2}\left(t\right)\right) $.

Does this mean that $ \lim_{\left(x,y\right)\to\left(0,0\right)}f\left(x,y\right) $ does not exist?

答:
Given any sequence $x_i$ tending to $x_0$ you can construct a path $\gamma $ such that $\gamma(1-\frac 1 i)=x_i$ for all $i$ and $\gamma (1)=x_0$. Just join the points by line segments. Hence, $f(x_i)=f(\gamma (1-\frac 1  i)) \to f(\gamma (1))=f(x_0)$ as $ i \to \infty$.

If the limit along some path does not exist or it exists but it is not $L$ then we can assert that $\lim_{x \to x_0} f(x)=L$ is false. If yo want to prove that $\lim_{x \to x_0} f(x)=L$ is  true then you will have to prove that the limit is $L$ along every path ending at $x_0$.


问:
Extension of scalar in tensor
In the book ('introduction to commutative algebra' by Atiyah), from $A$-module $M$, we get an $A$-module $M_B=B\otimes_A M$ that has a $B$-module structure such that $b(b'\otimes x)=(bb'\otimes x)$ for any $b,b'\in B$ and all $x\in M$ provided that there is a ring homomorphism $f:A\to B$ (view $B$ as an $A$ module via $f$). How is this operation possible? $b$ is in $B$ not $A$. And also, once we get $M_B$, then is it both $A$-module and $B$-module? I'm a beginner in module theory so I want explanation in detail. any comment will be appreciated

Edit: $A,B$ are commutative rings with unity.

答:
Yes, it will be both an $A$-module and a $B$-module.  The $A$-action on $M_B$ is defined using your $f$, namely if $a \in A$ then $a \cdot (b \otimes x) = (f(a) \cdot b) \otimes x$.  This makes sense since $f(a) \in B$ by definition of $f$.  This will be an action because $f$ is a homomorphism, so all the linearity conditions will be satisfied.

A lot of this left/right/bi-module stuff can be annoying and easy to forget; don't be frustrated, it was annoying for everyone who ever learned it, including Atiyah, Serre and Grothendieck I'm sure haha  It's good that you are trying to nail it down, because it can be useful to know for some slick categorical proofs later on in the theory.

Title: Extension of scalar in tensor


答:
The number is $n=66$. The sum of all alternating sums of  the rows should equal zero. So we need
$$
36+(30-n)-8+8=0,
$$
which says that $n=66$.
For example, the last row has alternating sum $$4-7+5-4+5-3+8=8$$

Title: Need help figuring out a pattern question

Body: my cousin asked for help on this math problem and the answer is apparently 6 but i can't find a coherent pattern that fits all of the triangles. I included an image of the question.


问:
$\theta$ distribution
reading number theory texts in one part I found myself "assuming that the prime numbers have distribution $\theta > 1/2$..."

what does this refer to?
Does it have to do with the density given by the prime number theorem?

答:
On the second page of https://annals.math.princeton.edu/wp-content/uploads/Maynard.pdf it says,

Given $\theta>0$, we say the primes have level of distribution $\theta$ if, for every $A>0$, we have $$\sum_{q\le x^{\theta}}\max_{(a,q)=1}\left|\pi(x;q,a)-{\pi(x)\over\phi(q)}\right|\ll_A{x\over(\log x)^A}$$

In a footnote it says, "different authors have given slightly different names or definitions to this concept."

Title: $\theta$ distribution


问:
(Proof) Equality of the distances of any point $P(x, y)$ on the isosceles hyperbola to the foci and center of the hyperbola
I searched but couldn't find the proof.

Isosceles hyperbola equation:
$${H:x^{2}-y^{2} = a^{2}}$$

And let's take any point $P(x, y)$ on this hyperbola. Now, the product of the distances of this point $P(x, y)$ to the foci of the  isosceles hyperbola is equal to the square of the distance from point $P$ to the center of the hyperbola.

Proof?

I took this question from my analytical geometry project assignment. I tried various ways (I found the foci $F(x,y)$ and $F^{'}(x,y)$ terms of $x$, $y$ and chose any point on the hyperbola...) but I couldn't prove.
I request your help.

Body:
$$PS_1.PS_2=\sqrt{[(x-ae)^2+y^2][(x+ae)^2+y^2]}$$
$$=\sqrt{(x^2-a^2e^2)^2+y^2(x^2+a^2e^2+2aex)+y^2(x^2+a^2e^2-2aex)+y^4} $$
$$=\sqrt{x^4+a^4e^4-2a^2e^2x^2+y^2x^2+y^2a^2e^2+2aexy^2+y^2x^2+y^2a^2e^2-2aexy^2}$$
$$=\sqrt{x^2+2x^2y^2+y^4+a^4e^4-2a^2e^2(x^2-y^2)}$$
$$=\sqrt{(x^2+y^2)^2+a^4e^4-2a^4e^2}$$
Since $e=\sqrt{2}$ for this hyperbola, so finally we prove that
$$PS_1.PS_2= x^2+y^2=OP^2.$$


问:
separable equation problem
how can i solve this as a separable equation?

$$p'-\frac1yp=-\frac1p$$

i tried to find integrating factor:

$$\eta  = e^{\int 1/ydy}$$
by this way which makes eta=y, but that didn't work

答:
$$p'-\frac py=-\frac1p$$
It's not a linear DE, integrating factor only works for linear DE.
$$p'p-\dfrac {p^2}y=-1$$
Substitute $p^2=u$ and $u'=2pp'$ then the DE becomes linear of first order. Then find an integrating factor.
$$yu'-2u=-2y$$
$$y^2u'-2yu=-2y^2$$
$$\dfrac {y^2u'-2yu}{y^4}=-\dfrac 2 {y^2}$$
$$\left(\dfrac {u}{y^2}\right)'=-\dfrac 2 {y^2}$$
Integrate and unsubstitute $u$.

Title: separable equation problem


问:
Function "evaluation" just means "composition"?
I am self-studying and have a basic or naive question that follows from a simple observation. I have also included tags for type theory, etc because "evaluation" probably has a different meaning in these constructive approaches, which I'm also curious about.

Any set $X$ with elements $x_i$ can be characterized by a set denoted $(X)$ of pairs $(\star, x_i)$, alternatively denoted $(\star \mapsto x_i)$ (for all $x_i \in X$). Each one of these latter pairs encodes a distinct function from common domain $\{ \star \}$ to (each different element of) $X$. If we name such functions after their image, i.e. $(x_i):\star \mapsto x_i$, then the symbol "$f(x_i)$", traditionally interpreted as function evaluation, is reinterpreted as function composition, i.e. $f\circ (x_i)$ and the $\circ$ is often notationally suppressed. And roughly, $f = \cup_xf(x)$, every function decomposes to a unique set of singleton functions (better word?).

Question: Is there anything more to evaluation, or can I exactly equate evaluation with composition and forever de-clutter my mind of the term "evaluation"? Is the above perspective unusual or fruitful beyond reducing terminological clutter?

The notion of evaluation above - perhaps "extensional" is the right term? - means essentially subsetting the function at the chosen argument $x$. Perhaps I will see that a full answer to my question requires considering an intentional (?) or constructive perspective on "evaluation" (as in lambda calculus, etc). Perhaps only there will composition and evaluation legitimately have distinct meanings.

Edit: The notation $(x)$ seemed the most natural way to denote any function with singleton domain because it coheres with the conventional way to write evaluation as $f(x)$. The notation seems unambiguous because there is no context I can think of, other than "evaluation", where a single symbol, say $x$, is enclosed in brackets "()"? It does seem useful to notationally distinguish functions on singleton domain in some way: I would like the notation to be a mnemonic for whether the resulting function composite is a singleton function (i.e. a "value" like $f(x_i) = f \circ (x_i) = (\star, y_j)$) versus a non-singleton function. Perhaps there is a conventional notation for functions from singleton domain, other than $(x)$ and a conventional name for such functions, other than "singleton functions"?

答: That’s right, evaluation is not in any serious way distinct from composition with a function with singleton domain—or to remove all talk of elements entirely, a function whose domain is a terminal object. This kind of function can entirely replace the role of elements, which is at the heart of the categorical axiomatization of set theory called ETCS.</p>


问:
The value of $\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$
What is the value of this expression :
$$\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$$
Using calculator and wolfram alpha, the answer is, $-\frac{1}{\sqrt{5}}$

But, by solving it myself the result comes out to be different. My solution is as follow:
$$\begin{aligned} Put,\, &\cot^{-1}\left(\frac{-3}{4}\right) = \theta
\\\implies &\cot(\theta) = \frac{-3}{4} =\frac{b}{p}
\\So,\,&\cos(\theta) = \frac{-3}{5}\end{aligned}$$
$$\begin{aligned}\\Then,
\\\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right) &= \sin\left(\frac{\theta}{2}\right) \\&= \sqrt{\frac{1-\cos{\theta}}{2}}
\\&= \sqrt{\frac{1+\frac{3}{5}}{2}}
\\&= \frac{2}{\sqrt{5}}
\end{aligned}$$
This solution is used by a lot of websites.

So, I got two different values of single expression but I am not sure which one is correct. Can you point out where I have done the mistake?

答:
Since $$0\lt \cot^{-1} (-\frac 34 ) \lt \pi$$

, $$0\lt \frac 12 \cot^{-1} (-\frac 34) \lt \frac{\pi}{2} $$ which means its sine must be positive, so Wolfram’s answer is wrong. $\frac{2}{\sqrt 5}$ is correct.


问:
Second Stiefel-Whitney class of $\mathbb{C}\text{P}^2\#\overline{\mathbb{C}\text{P}^2}$
I know that $w_2(\mathbb{C}\text{P}^2\#\overline{\mathbb{C}\text{P}^2})\neq 0$, where $\overline{\mathbb{C}\text{P}^2}$ is $\mathbb{C}\text{P}^2$ with opposite orientation. But how do you prove this? I can't figure it out and I haven't found a single reference justifying it (I suppose it should be quite elementary...).

答:
The second Stiefel Whitney class of $Y=\mathbb{CP}^2$ is non-trivial. Similarly for $Z=\mathbb{CP^2}\setminus \{pt\}$. Namely the inclusion $Z\rightarrow Y$ pulls back the tangent bundle from $Y$ to $Z$, and on cohomology it induces an isomorphism on $H^2(\cdot;\mathbb{Z}_2)$.

Consider now the inclusion $i:Z\rightarrow X$, where $X$ is the space of the OP. The tangent bundle is again pulled back to Z, and $i^*(w_2(X))=w_2(Z)\not=0$. Thus $w_2(X)\not=0$.

This is a more general phenomenon: The first stiefel whitney class of a vector bundle over a $CW$ complex: i.e. if the bundle orientable or not , only depends on the restriction of the bundle over the one skeleton. Something similar is true of the second Stiefel Whitney class over the two-skeleton.


问:
Prove that $\lim_{p\to \infty}\Big(\frac{1}{\mu(A)}\int_{A}|f|^p\Big)^{1/p}=\|f\|_{\infty}$
We assume that $\mu(A)<\infty$. Prove that $$\displaystyle \lim_{p\to \infty}\Big(\frac{1}{\mu(A)}\int_{A}|f|^p\Big)^{1/p}=\|f\|_{\infty}$$

答:
We assume that $f \neq 0$ (otherwise is trivial), so that $|| f||_{\infty} \neq 0 $.

One inequility is clear: using the mean value theorem you have that

$$ \left ( \frac{1}{\mu(A)} \int_A |f|^p \right )^ p  = (|f(x_0)|^p)^{1/p} = |f(x_0) | \le || f|| _{\infty}$$

For the other one, let us call $A_{\varepsilon} := \{ x \in A: |f(x)| \ge ||f||_{\infty} (1-\varepsilon) \} $. Note that

$$ \int_A |f|^p = \int_{A_{\varepsilon} } |f|^p + \int_{A \setminus A_{\varepsilon} } |f|^p  \ge \mu(A_{\varepsilon}) || f||_{\infty}^{p} (1-\varepsilon)^p$$
Extracting the $p$-th root and dividing by $\mu(A)^{1/p}$ we have
$$ \left ( \frac{1}{\mu(A)} \int_A |f|^p \right )^{1/p} \ge \left (\frac{\mu(A_{\varepsilon})}{\mu(A)} \right )^{1/p} ||f||_{\infty} (1-\varepsilon ) $$

By the very definition of $L^{\infty}$ norm, we have that $\mu(A_{\varepsilon}) > 0$: if it was zero, then $f(x) \le ||f||_{\infty} (1-\varepsilon)$ for almost all $x$, which implies that $||f||_{\infty} \le ||f||_{\infty} (1-\varepsilon)$, a contradiction. For any $0 < s \le 1$ we have that $\lim_{p \to \infty} s^{1/p} = 1$, so that taking the liminf of the above expression yields

$$ \liminf_{p \to \infty}  \left ( \frac{1}{\mu(A)} \int_A |f|^p \right )^{1/p} \ge || f||_{\infty} ( 1- \varepsilon) $$

for an arbitrary choice of $\varepsilon > 0$, so that

$$ \liminf_{p \to \infty}  \left ( \frac{1}{\mu(A)} \int_A |f|^p \right )^{1/p} \ge || f||_{\infty} $$
and we are done.


答:
Both are correct, but the answer you obtain depends on the definition $\cot^{-1}(x)$, in particular its range, since inverse trigonometric functions are multi-valued.

If you write $\cot^{-1}(x) = \tan^{-1}(1/x)$ [as a calculator seldom has $\cot^{-1}$] and use the range $-\pi/2 < \tan^{-1}x < \pi/2$, we see that $\theta = \cot^{-1}(-3/4)$ must be negative.

In $-\pi/2 < x < 0$, $\cos x > 0$ and $\sin x < 0$. Hence $\cos \theta = \dfrac35$, and $\sin x = -\sqrt{\dfrac {1-3/5}2} = -\dfrac {1}{\sqrt 5}$.

Title: The value of $\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$

Body:
What is the value of this expression :
$$\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$$
Using calculator and wolfram alpha, the answer is, $-\frac{1}{\sqrt{5}}$

But, by solving it myself the result comes out to be different. My solution is as follow:
$$\begin{aligned} Put,\, &\cot^{-1}\left(\frac{-3}{4}\right) = \theta
\\\implies &\cot(\theta) = \frac{-3}{4} =\frac{b}{p}
\\So,\,&\cos(\theta) = \frac{-3}{5}\end{aligned}$$
$$\begin{aligned}\\Then,
\\\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right) &= \sin\left(\frac{\theta}{2}\right) \\&= \sqrt{\frac{1-\cos{\theta}}{2}}
\\&= \sqrt{\frac{1+\frac{3}{5}}{2}}
\\&= \frac{2}{\sqrt{5}}
\end{aligned}$$
This solution is used by a lot of websites.

So, I got two different values of single expression but I am not sure which one is correct. Can you point out where I have done the mistake?


答:
For any point on the hyperbola, $x^2 - y^2 = a^2$

Foci of the hyperbola are $(\pm a\sqrt2,0)$ and the center is $(0, 0)$.

So product of distance of point $P(x,y)$ on the hyperbola to foci is

$\sqrt{(x-a\sqrt2)^2 + y^2} \times \sqrt{(x+a\sqrt2)^2 + y^2}$

$\sqrt{x^2 + 2a^2 + y^2 - 2 \sqrt2 a x} \times \sqrt{x^2 + 2a^2 + y^2 + 2 \sqrt2 a x}$

Using $y^2 = x^2 - a^2$,

$ = \sqrt{((2x^2+a^2) - 2\sqrt2 a x) ((2x^2+a^2) + 2\sqrt2 a x)}$

$ = \sqrt{ 4x^4 + a^4+4a^2x^2 - 8a^2x^2}$

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

Which is the square of distance of point $P$ to the center of the hyperbola $(0, 0)$

Title: (Proof) Equality of the distances of any point $P(x, y)$ on the isosceles hyperbola to the foci and center of the hyperbola

Body:
I searched but couldn't find the proof.

Isosceles hyperbola equation:
$${H:x^{2}-y^{2} = a^{2}}$$

And let's take any point $P(x, y)$ on this hyperbola. Now, the product of the distances of this point $P(x, y)$ to the foci of the  isosceles hyperbola is equal to the square of the distance from point $P$ to the center of the hyperbola.

Proof?

I took this question from my analytical geometry project assignment. I tried various ways (I found the foci $F(x,y)$ and $F^{'}(x,y)$ terms of $x$, $y$ and chose any point on the hyperbola...) but I couldn't prove.
I request your help.


问:
Regarding the notation $f: a \mapsto b$
While I have come to understand that $f:a\mapsto b$ means that for input from the set $a$, the function will return a value from the set $b$, I am curious as to how far one may "drag" this notation. Say $f(x) = x$, would $f: 1 \mapsto 1$ be valid, or does one have to write $f: \mathbb{R} \mapsto \mathbb{R}$? (While I understand that my initial example would seldom prove useful, I am curious as to what is valid notation or not.)</p>

答: In my experience, it is much more common to indicate the domains and codomains of functions using \rightarrow, as in $f : A \rightarrow B.$ Likewise, the \mapsto operator is more commonly used to define the rules of functions, such as in $x \mapsto x^2$.

Title: Regarding the notation $f: a \mapsto b$


答:
You got the expression $\cot \theta= \frac{-3}{4}$

As you may know, the value of $\cot \theta$ is postive in the first and third quadrants, which means it is negative in the second and fourth quadrants.
When $\cot \theta=\frac{-3}{4}$, $\cos \theta$ can either equal $\frac{-3}{5}$ or $\frac{3}{5}$

$\cos\theta=\frac{3}{5}$ is true wrt the fourth quadrant.

If $\cos \theta=\frac{3}{5}$, $\sin \theta= \frac{1}{\sqrt{5}}$

But, we know that in the fourth quadrant $\sin \theta$ is negative.

Thus, $\sin\theta=\frac{-1}{\sqrt{5}}$

In short, both answers are correct, for their respective quadrants.


问:
Prove the median of a symmetric distribution is the point of symmetry.
I'm trying to prove that if $f(a+\epsilon) = f(a-\epsilon)$, i.e. $f(x)$ is symmetric about $a$, then $a$ is the median of a continuous random variable with pdf $f(x)$. Using the fact that $a$ being the median means that $$\int_{-\infty}^a f(x)dx = \int_{a}^{\infty}f(x)dx = 1/2$$ I thought I could do something like

$$1 = \int_{-\infty}^{\infty}f(x)dx = \int_{-\infty}^a f(x)dx + \int_{a}^{\infty}f(x)dx,$$

but couldn't see why I could claim that the two integrals had to then be equal/ both then must be $1/2$.

I saw a similar version of what I'm trying to prove with $a=0$ here, but couldn't figure out how to apply a transform to my integrals so that it worked out the same way. I tried using $x=a-y$, but still just couldn't get the negative signs to work out correctly.

Any advice is apricated, and thank you in advance.

答:
You start by knowing that  $\int_{-\infty}^{a} f(x) \,dx + \int_{a}^{\infty} f(x) \,dx =1 $
where $a$ is the point of symmetry.
Then use substitution $y=x-a$ and you will get

$\int_{-\infty}^{0} f(a+y) \,dy + \int_{0}^{\infty} f(a+y) \,dy =1 $

Now another substitution only for the first integral $y=-z$

$\int_{-\infty}^{0} f(a+y) \,dy = \int_{\infty}^{0}- f(a-z) \,dz = \int_{0}^{\infty} f(a-z) \,dz = \int_{0}^{\infty} f(a+z) \,dz$

So you can rewrite the equation using the first integral as above

$\int_{0}^{\infty} f(a+y) \,dy + \int_{0}^{\infty} f(a+y) \,dy =1 $

$2*\int_{0}^{\infty} f(a+y) \,dy =1 $

$\int_{0}^{\infty} f(a+y) \,dy =1/2 $

Now undo the substitution

$\int_{a}^{\infty} f(x) \,dx =1/2 $

and rewrite the equation

$\int_{-\infty}^{a} f(x) \,dx + 1/2 =1 $

$\int_{-\infty}^{a} f(x) \,dx = 1/2 $

that means that the two integrals are equal and both must be 1/2

$\int_{-\infty}^{a} f(x) \,dx = \int_{a}^{\infty} f(x) \,dx = 1/2 $


问:
Show that this ODE with boundary conditions has a unique solution
We have the ODE $$-\ddot{x}+q(t)x=g(t),\quad x(a)=x_a,\, x(b)=x_b$$
with $g\in C([a,b],\mathbb{R}$ and $q\in C([a,b],\mathbb{R}_0^+)$. Show that there is a unique solution.

First of all, I transformed this problem in the form
$$\dot{X}(t)=U(t)X(t)+b(t),
$$
where $X=(x,\dot{x})^T$, $U(t)=\begin{pmatrix}0 & 1 \\ q(t) & 0 \end{pmatrix}$, $b=-(0, g(t))^T$. Also we can write the boundary condition as
$$AX(a)+BX(b)=r(X(a),X(b)),\quad A=\begin{pmatrix}1 & 0 \\0 & 0\end{pmatrix},\, B=\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}.
$$
I know that there is a theorem that states that this problem has a unique solution if
$$E=A+BW(b,a)
$$
is invertible, where
$$ \frac{\partial}{\partial t}W(t,a)=f_x(t,X(t))W(t,a)=U(t)W(t,a).
$$
How would I show that? Or is there some other way to show it?

EDIT: Forgot the "x" in the equation.

答:
Here's an alternative argument to show that the solution of the boundary value problem is unique.  Let $\ x_1\  $ and $\ x_2\ $ be solutions and let $\ z=x_2-x_1\ $. Then $\ z\ $must satisfy the homogeneous differential equation
$$
\ddot{z}=q(t)z\ ,
$$
and the boundary conditions
$$
z(a)=z(b)=0\ .
$$
Therefore
\begin{align}
\frac{d^2z^2}{dt^2}&=2\big(z\ddot{z}+\dot{z}^2\big)\\
&\ge2q(t)z^2\\
&\ge0\ .
\end{align}
It follows that $\ \frac{dz^2}{dt}\ $ is nondecreasing, and since $\ \frac{dz^2}{dt}(a)=$$z(a)\dot{z}(a)=0\ $, it also follows also that $\ \frac{dz^2}{dt}\ge0\ $ for $\ t\ge a\ $, and hence that $\ z^2\ $ is nondecreasing for $\ t\ge a\ $.  Thus
$$
0=z(a)^2\le z(t)^2\le z(b)^2=0\,
$$
and hence $\ z(t)^2=0\ $, for $\ a\le t\le b\ $. That is , $\ x_1(t)=x_2(t)\ $, and the solution must be unique.

Existence of solutions

First replace the transformed version of the differential equation with an equivalent homogeneous version:
$$
\dot{Z}(t)=M(t)Z(t)\ ,
$$
where $\ Z=\big(x,\dot{x},v\big)^T\ $ and
$$
M(t)=\pmatrix{0&1&0\\q(t)&0&-g(t)\\0&0&0}\ .
$$
The third entry, $\ v\ $, of $\ Z\ $ must be constant, and if we impose the initial condition $\ v(a)=1\ $, then it must be identically $1$, and the second row of the differential equation becomes $\ \ddot{x}=q(t)x-g(t)v=q(t)x-g(t)\ $.

The differential equations $\ \dot{Z}(t)=M(t)Z(t)\ $ and $\ \dot{X}(t)=U(t)X(t)\ $ are then special cases of the equations
$$
\hspace{-1em}\text{(L)}\hspace{5em}x_i^\prime=\sum_{j=1}^na_{ij}(t)x_j\hspace{2em}(i=1,\dots,n)
$$
appearing on p.19 of Coddington and Levinson's Theory of Ordinary Differential Equations ($1955$ edition) for which the following theorem appears on p.20:

Theorem 5.1. For the linear system (L), where the functions $\ a_{ij}\in C\ $ on $\ [a,b]\ $, there exists one and only one solution $ \varphi\ $ of (L) on $\ [a,b]\ $ passing through any point $\ (\tau,\xi)\in D\ $ $\big[$here, $\ D=[a,b]\times\mathbb{R}^n\ \big]$, that is, $\ \varphi(\tau)=\xi\ $.

Applying this theorem to the differential equation $\ \dot{Z}(t)=M(t)Z(t)\ $ with initial condition
$$
Z(a)=\pmatrix{0\\x_a\\1}
$$
tells us that it has a solution
$$
Z(t)=\pmatrix{\dot{x_0}(t)\\x_0(t)\\1}\ ,
$$
which must satisfy the equations $\ \ddot{x_0}=q(t)x_0-g(t)\ $, $\ x_0(a)=x_a\ $ and $\ \dot{x}_0(a)=0\ $.  This is just exactly the same $\ x_0\ $ that Lutz Lehmann defines in his answer, which the above theorem tells us must exist, and is unique.

Now applying the theorem to the differential equation $\ \dot{X}(t)=U(t)X(t)\ $ with initial condition
$$
X(a)=\pmatrix{1\\0}
$$
tells us that it has a solution
$$
X(t)=\pmatrix{\dot{y}(t)\\y(t)}\ ,
$$
which must satisfy the equations $\ \ddot{y}=q(t)y\ $, $\ y(a)=0\ $ and $\ \dot{y}(a)=1\ $. Again, this is exactly the same $\ y\ $ that Lutz Lehmann defines in his answer, which the theorem again tells us must exist, and is unique.  Exactly the same argument used above to show that $ z^2\ $ is nondecreasing shows also that $\ y^2\ $ must be non decreasing, and since $\ y(a)=0\ $, $\ \dot{y}=1\ $, there must exist $\ \epsilon\in(a,b)\ $ such that $\ y(\epsilon)>0\ $.  Therefore $\ y(b)^2\ge y(\epsilon)^2>0\ $.  So if we now define
$$
x(t)= x_0(t)+\left(\frac{x_b-x_0(b)}{y(b)}\right)y(t)
$$
just as Lutz Lehmann does in his answer, then $\ x\ $ will satisfy the differential equation
$$
\ddot{x}=q(t)x-g(t)\ ,
$$
and the boundary conditions $\ x(a)=x_a\ $, $\ x(b)=x_b\ $.


问:
How to find out length of bisector?
I found this challenge online but can't solve it.

I have a triangle C is 90 degrees. The bisector of A cuts BC into 5 and 3, how do i find out how long the bisector is?

答:
Suppose $AD$ bisects $\angle BAC$. From the angle bisector theorem, we have
$$\frac{BD}{DC} = \frac{AB}{AC} = \frac{5}{3}$$

Letting $l(AB) = 5x$ and $l(AC) = 3x$, by Pythagoras thoerem,
$$25x^2 = 64 + 9x^2$$
$$\implies 16x^2 = 64$$
$$\implies x = 2$$

Therefore $l(AC) = 6$, $l(DC) = 3$, now we use Pythagoras theorem again to compute $l(AD)$,
$$l(AD) = \sqrt{36+9} = \sqrt{45} = 3\sqrt{5}$$

Title: How to find out length of bisector?


答:
From the given, $$\frac{\tan2\alpha}{\tan\alpha}=\frac{5+3}3,$$

and we are looking for the value of

$$\frac3{\sin\alpha}.$$

$$\tan\alpha=\frac12\to\sin\alpha=\frac1{\sqrt5},\\3\sqrt5.$$

Title: How to find out length of bisector?

Body:
I found this challenge online but can't solve it.

I have a triangle C is 90 degrees. The bisector of A cuts BC into 5 and 3, how do i find out how long the bisector is?


问:
Show that: $\varphi$ maps $\alpha$ to a $\beta_i$ and $\varphi (\alpha)=\varphi^´(\alpha)\rightarrow \varphi = \varphi^´$

Let $\varphi:K \rightarrow L$ be a homomorphism of fields and $N/K$ be a finite field extension. The continuation is definded as $\phi:N \rightarrow L$. Let $\alpha\in N$ and $f$ the minimal polynom of $\alpha$ over $K$. Now consider the zero points $\beta_1,...,\beta_k$ of $f$ in $L$ and show that:

$\varphi$ maps $\alpha$ to a $\beta_i$

$\varphi (\alpha)=\varphi^´(\alpha)\rightarrow \varphi = \varphi^´$

Unfortunately another exercise from an old algebra exam which I'm really struggling to solve. Maybe someone could give me a hint? Thanks in advance!

Further information:

Definition of a continuation:

A continuation of a function is a another function. The domain of the other function includes a subset which is equal to the original function.
So the official definition is:

Let $X, Y$ and $A$ sets. A function $f:X \rightarrow Y$ is called a continuation of the function $g: A \rightarrow Y$, if $A$ is a Subset of $X$ and $g(x)=f(x)$ for all $x \in A$.

Definition of a finite field extension

$(N/K)$ is a finite field extension of the fields $N$ and $K$

A field extension is a pair of fields $K \subseteq N$, such that the operations of $K$ are those of $N$ restricted to $K$. In this case, $N$ is an extension field of $K$ and $K$ is a subfield of $N$. For example the complex numbers are an extension field of the real numbers and the real numbers are a subfield of the complex numbers.

If written like above $N/K$ then $K$ is a subfield  of $N$ and $N$ is the extension field/extension of $K$. Finite meaning there are a finite amounts of elements

Edit: $N$ is given by $N=K(\alpha)$. For the first task you need to set and $f(\alpha)=0$, but I am unsure how to procced.

答:
Let $\phi: K \to L$ be a fixed field morphism, let $n =$ deg$(f)$ and let $\phi(f)$ be the corresponding polynomial in $L[x]$. If $\psi: N \to L$ is a field morphism that extends $\phi$, then $\psi(\alpha)$ is a root of $\phi(f)$ since if
$$ 0 \: = \: f(\alpha) \: = \: a_{n}\alpha^{n} \: + \: ... \: + \: a_{1}\alpha \: +  \: a_{0} $$
then an application of $\psi$ shows that
$$ \psi(0) \: = \: 0 \: = \: \phi(a_{n}) \cdot \psi(\alpha)^{n} \: + \: ... \: + \: \phi(a_{1}) \cdot \psi(\alpha) \: +  \: \phi(a_{0}) \: = \: \phi(f)(\psi(\alpha)) $$
because $\psi$ is a field morphism that equals $\phi$ on $K$ (and therefore on  the coefficients of $f$).

Next, since $N = k(\alpha)$ and $f$ is the minimal polynomial of $\alpha$ over $K$, the sequence $(\alpha_{i})_{i = 0}^{n - 1}$ forms a linear basis for $N$ over $K$. In other words, each element $b$ of $N$ may be written as $b_{n}\alpha^{0} + ... + b_{n}\alpha^{n}$ for coefficients $b_{i}$ in $K$. After applying $\psi$, we get
$$ \psi(b) \: = \: \phi(b_{0}) \cdot \psi(\alpha)^{0} \: + \: ... \: + \: \phi(a_{n}) \cdot \psi(\alpha)^{n} $$
Hence every field morphism extension of $\phi$ to $N$ is completely determined by where it maps $\alpha$.


问:
Generating a random point on the unit circle
I'm trying to figure out a way to generate a random point on the unit circle in an application I am developing (I'm a programmer).

So far I have the following (in pseudo-code), where Z is a random number between 0.0 and 1.0:

```<code>theta = (2.0 * PI) * Z

2DVector.x = cos(theta)
2DVector.y = sin(theta)

result: 2DVector
</code>```

I know that it's wrong, as I'm getting nothing but massive x values and tiny y values. But I'm not familiar enough with the unit circle mathematics to know where I'm going wrong!

答: The example I provided works fine, so long as it's implemented properly.</p>


问:
How to self study real analysis
I'm a high school student, who is interested in learning real analysis, but it seems like every textbook I take doesn't really provide a way to intuitively think about the concepts. I understand it to some extent but I'm not satisfied with the level of intuition provided so far. Can someone help me recommend and textbook or a way to study it</p>

答:
"Calculus" Volumes 1 and 2, 2nd Edition 1966 : Tom Apostol

Recommended to me as an alleged jewel.  It is.

Title: How to self study real analysis


问:
Finding a limit of a rational function when the function isnt moving towards 0
Consider this:

$$\lim_{x \to 2} \frac{x^2+3x-4}{x^2-7x+10}$$ Now I've solved this for the case when x is approaching 0,but when I try to use the same method for when it is approaching 2,I dont get very far.So first I put in 2 and got that in the numerator we get a 2 and denominator we get a 0,so division by 0 we want to avoid that.Now I need to "change" the look of my polynomials to get the limit,but I'm not sure how to do that? I am not sure if I can use L'Hospital here since it isn't 0/0 but rather 2 divided by zero? A bit of advice would be much appreciated

Many thanks!

答: Note that $x^2-7x+10=(x-2)(x-5)$. Therefore your limit is$$\lim_{x\to2}\frac{\frac{x^2+3x-4}{x-5}}{x-2}.\tag1$$Since $\lim_{x\to2}\frac{x^2+3x-4}{x-5}=-2$, the limit $(1)$ doesn't exist, because$$\lim_{x\to2^+}\frac{\frac{x^2+3x-4}{x-5}}{x-2}=-\infty\quad\text{and}\quad\lim_{x\to2^-}\frac{\frac{x^2+3x-4}{x-5}}{x-2}=\infty.$$


问:
Finding no solutions and many infinitely many solutions for $k$
$$\left(
    \begin{array}{crr|c}
    1 & k & -1 &   2\\
    2 & -1 & k &   5 \\
    1 & 10 & -6 &  1 \\
     \end{array}
\right)  \xrightarrow[\text{$R_3=R_3-R_1$}]{\text{$R_2=R_2-2R_1$}}$$ $$\left(
    \begin{array}{ccc|r}
    1 & k & -1 &  2\\
    0 & -1-2k & k+2 &  1 \\
    0 & 10-k & -5 & -1 \\
     \end{array}
\right) \xrightarrow[\text{$R_3=R_3-(\frac{10-k}{2k+1})R_2$}]{} (2k+1\neq0)$$
$$\left(
    \begin{array}{ccc|c}
    1 & k & -1 & 2\\
    0 & -1-2k & k+2 & 5 \\
    0 & 0 & \frac{-k^2-2k+15}{2k+1} & \frac{-3k+9}{2k+1} \\
     \end{array}
\right) $$
Which has a unique solution $\iff \frac{-3k+9}{2k+1} \neq 0 \iff -3k+9\neq 0 \iff k\neq3$

Now how can I show that it got no solutions and infinitely many solutions. Not sure how to do it when I got so difficult equation for $k$. Any help would be appreciated

答:
You didn't have to divide by $2k+1$! Once you have row reduced to an upper triangular form, you can just consider the equations they represent:

There are no solutions if you end up with a row like

$$(0\ 0\ \ldots\ 0\ |\ x)$$

for some nonzero $x$ (the corresponding equation is a contradiction!), and

$$(0\ 0\ \ldots\ 0\ |\ 0)$$

indicates freedom to choose the variables as you like, hence choice (with infinitely many choices).

One way to think about it is about singularness of the 3×3 matrix on the left:

If it is invertible (full rank), there is exactly one solution.

If it is non-invertible (i.e., does not have full rank), then it will have either infinitely many solutions, or none, depending on whether the target on the right gets hit or not.

As it so happens, your matrix is non-invertible for $k = 3$ or $-5$ (by determinant of the original or reduced matrix), so, that given, you only really need check these values.


问:
Modified Euler's Totient function for counting constellations in reduced residue systems
I am working on a modified totient function for counting constellations in reduced residue systems for the same range that Euler's totient function is defined over.  This post is separated into three parts: "background" describes how this function was derived and the logic involved, "question" defines the function and questions involved, and "terminology" explains some of the terminology I use, which may or may not be a bit obfuscated.

NOTE: this entire question has been edited in an attempt to improve understandability. If you have any suggestions, please let me know.

Background:

The primorial is defined as the product of all primes that are smaller than or equal to some positive integer; defined for all positive integers, it is written $n\#$, but the convention I have found to be most useful is $P_n\#$, where $P_n$ is the nth prime number, and the function is the product of all prime numbers up to and including $P_n$.

The primorial function can be modified to evaluate Euler's totient function for primorial inputs.  The idea is this: the primorial itself can be seen as the product of the number of possible equivalence classes for each prime number.  To eliminate those numbers that are divisible by a prime number, we subtract one from each set of equivalence classes modulo a prime number.  This is better expressed below:

$$\phi(p_n\#) = \prod_{p\mid p_n\#} (p - 1)$$

Euler's totient function, in one of its many forms, is written as follows:

$$\phi(p^k) = (p-1)p^{k-1}$$

The difference between the two functions is negligible because primorials, as the products of the primes, have no prime powers in their factors beyond the initial power of $1$; this effectively cancels out the $pk-1$ term, leaving only the $(p-1).$ As both functions are measuring the same thing, it is the first function that, while correct, cannot be extended for any positive integer input.  The two expressions are equivalent.

The OEIS sequence A059861 is the result of the following expression:

$$\prod_{p\mid p_n\#} (p - 2)$$

Please take the time to look this up in the OEIS for verification; it calculates the 2 and 4 differences in the differences between the numbers in reduced residue systems for primorials. If we multiply it by 2, we count all of the "twins", that is, each number in the reduced residue system for a primorial that is $+2$ or $-2$ away from another number in that same reduced residue system.

We can see these functions a little differently when only considering primorial inputs; there are two possible equivalence classes modulo $2$: $0$, and $1$. There are $3$ modulo $3$: $0$, $1$, and $2$. If a positive integer is equivalent to $0$ modulo $2$ or $3$ or any other prime, it is divisible by that prime, and must therefore be equal to the prime in question, ir it is not prime itself as it is divisible by that prime. Thus, to count the number of positive integers smaller than the product of $2$ and $3$, we can take the product of $(2-1)(3-1) = 1\times2=2$.  By subtracting one from each prime number, we have eliminated the equivalence classes modulo those prime numbers that cannot support other prime numbers.

This same process works for counting twins in reduced residue systems for primorials, but instead of subtracting one from each prime in the product, we subtract 2; however, when subtracting $2,$ we must define the function differently for the first prime number, because otherwise, we will have a zero in our product, and while we don't know if there are infinitely many twin primes, we do know there are a lot of them. To fix this, the term involving the first prime number must be defined to evaluate to $1$; we still need to remove all those numbers which are equivalent to $0$ mod $2.$

Now, what if we were to extend this function to have the capacity to subtract any positive integer k from each prime number in the product? I believe this will create a function that can count constellations in reduced residue systems for primorials.  This function is written as follows:

$$p_n\#(-k) = \prod_{a = i}^n (p - k), i = \min(\{z\mid p_z > k\})$$

Remember that in terms of the totient function, we are missing something. To generalize this formula, we need to add that "something" back in (please forgive the notation, as I am unsure of how to write this otherwise):

$$\phi(-k)(n) = \prod_{p|n} (p-k)p^{m-1},$$

where $m$ is the power of $p$ that is a factor of $n.$  There is, however, a problem with that last function; when $p < k$, we introduce negative numbers, and when $p = k$, we introduce zero in our product. This will not do; however, we must be careful in fixing this problem, as the new addition is essential for the generalization of this formula.

Question:

$$\phi(-k)(n) = \left( \prod_{p|n \wedge p \leq k} p^{m-1} \right) \left( \prod_{p|n \wedge p > k} (p - k)p^{m-1} \right),$$

where m is the power of the prime as it appears in the prime factorization of n (in other words, the number of times p shows up in the prime factorization of n)

Questions:

1: is this last formula a valid method of counting constellations in reduced residue systems?

2: Have you seen this in existing literature? If so, please provide a reference!

EDIT: This formula, as stated in my comments to the chosen answer, does not correctly count all elements in a given constellation; it merely counts residues. It may, however, be possible that this formula either does or can be modified to count only the first element in constellations of given forms. The result can then be multiplied by the number of elements in the target constellation to count all members of given constellations in reduced residue systems.

Terminology:

Primorial: the product of prime numbers up to a given limit.  The limit can be given as any positive integer n or as a prime number indexed by a subscript.

Phi function, Euler's totient function: The function that counts how many numbers smaller than its input are coprime with that input.

Reduced Residue System: the numbers that the totient function counts relative to some input $n.$

Twin: as used here, it is any integer $n$ that is a member of a pair of integers such that the other member of the pair is either $n+2$ or $n-2.$

Constellation: again, as used here, a sequence of consecutive elements of the reduced residue system for a given input that appear according to some pattern; the Hardy-Littlewood $k$-tuple conjecture states that, given any constellation in the prime numbers, identical patterns will appear in constellations in the primes infinitely many times.

答:
You note the need to add something back; that is correct. For each p, you must add back the number of times that p occurs as a prime factor of the pair-wise differences between elements of the constellation, counting p at most once for each smallest element in multiple pairs. For example, for twins (diff=2), you get p-k+1, where 1 represents the occurrence of p as a prime factor of the difference. You can, indeed, generalize the formula to all constellations in the same manner. For example,   For constellation (0,4,6,10), you would “add back” 3 for p=2, 2 for p=3, and 1 for p=5, which I denote fsubp. Constellations that fail the basic congruence condition, that is, where p-k+fsubp < 1 are inadmissible in sieve theory terms. All others are admissible. The product of p-k+fsubp over the primes answers this question: “For every psubn instances of a constellation that survived the n-1 sift, how many will survive the n-th sift?” You’ll note that the ratio of the infinite products for two constellations equals the ratio of the Hardy Littlewood constants for those same constellations.

I have searched extensively and unsuccessfully for any publication of this simplified and generalized Euler totient function; my understanding, therefore, is that it is previously unexplored (subject to the possibility that I've failed to find or recognize something that is published).


问:
Algebraic varieties being geometrically irreducible
I have trouble comprehending the so-called notion of being geometrically irreducible. More specific, I was doing an exercise in Hartshorne, Algebraic Geometry, First Properties of Schemes, exercise $3.15$.

Briefly, let $k$ be a field and $X$ a scheme over $k$. Then $X$ is said to be geometrically irreducible if $X_{k'} = X \times_{k} k'$ (by the abuse of notation, I write $k$ to mean $\mathrm{Spec}(k)$). The exercise asks readers to show that the following three conditions are equivalent provided $X$ is a scheme of finite type over $k$:

$X \times_k \overline{k}$ is irreducible, where $\overline{k}$ is the algebraic closure of $k$.

$X \times_k k^{sep}$ is irreducible, where $k^{sep}$ is the separable closure of $k$.

$X$ is geometrically irreducible.

Note that the third implies the first one trivially, the first one implies the second because the property of being surjective is preserved under base change.

The solutions which are available online seem to be so terse and vague, therefore I have had to search for several sources, including Vakil's note FOA (section $9.5$) and QingLiu, Algebraic Geometry and Arithmetic Curves (section $3.2.2$). The two core lemmas in QingLiu are

Let $X$ be an algebraic variety over $k$, $K/k$ an algebraic extension of fields. Then for every reduced closed subvariety $W$ of $X \times_k K$, there exists a finite subextension $K'/k$ of $K/k$, and a unique reduced closed subvariety $Z$ of $X \times_k K'$ such that $W = Z \times_k K$.

Let $X$ be an algebraic variety over $k$, $K/k$ an algebraic extension. If $K/k$ is purely inseparable, then the projection $X \times_k K \to X$ is a homeomorphism.

I understand their proofs, but can not deduce the general case (an arbitrary extension $K/k$ instead of just being purely inseparable or algebraic). I think my problem is I do not truly understand the roles of two extensions $\overline{k}/k,k^{sep}/k$ here. Any suggestion would be greatly appreciate.

答:
For the first part, it's an important fact in itself that $X_{\overline{k}} \rightarrow X_{k^{sep}}$ is a homeomorphism.

To do that, it's not an issue to reduce to the case of affine $X$, and thus you want to show that for any $k$-algebra $A$, we have a natural correspondance between the spectra of $A \otimes \overline{k}$ and $A \otimes k^{sep}$. To do that, the main ingredient is to note that for each $x \in A \otimes \overline{k}$, in prime characteristic $p > 0$, there is an integer $n \geq 0$ such that $x^{p^n} \in A \otimes k^{sep}$. (It's not entirely obvious how to go on after that, but it's not very difficult).

Now, we want to explain why, if $X_{\overline{k}}$ is irreducible, then for any extension $k'/k$, $X_{k'}$ is irreducible.

First, we can assume $k$ is algebraically closed. Indeed, let $K$ be an algebraic closure of $k'$, so that we have a map $\overline{k} \rightarrow K$ (respecting $k' \cap \overline{k}$). Now $K/\overline{k}$ is a field extension, and $X_{\overline{k}}$ is irreducible. If $X_K$ is indeed irreducible, then, as $X_K \rightarrow X_{k'}$ is surjective, it follows that $X_{k'}$ will be irreducible.

So assume that $k$ is an algebraically closed field and $X$ is an irreducible scheme over $k$. Let $K/k$ be a field extension, we want to show that $X_{K}$ is irreducible. It's enough to show it when $X$ is affine (because the affine subsets of a general irreducible $X$ are dense and irreducible, and their base changes will be irreducible, and will pairwise intersect, which forces $X_K$ to be irreducible by topology), and we can furthermore assume $X$ is reduced (so $X$ is the spectrum of an integral $k$-algebra $A$), and we will show that if $K/k$ is an extension, $A\otimes K$ is an integral domain.

Let $K/k$ be an extension and let $a,b \in A \otimes K$ have a null product. There is a finitely generated (and integral) $k$-subalgebra $B \subset K$ such that $a,b$ are in the image of the monomorphism (flatness) $A \otimes_k B \rightarrow A \otimes K$, so we want to show that, in $A \otimes B$, $a$ or $b$ must be zero when their product is zero.

Now, take a $k$-basis $e_i$ of $A$, and let $a_i,b_i$ be the coordinates of $a,b$ in the $B$-basis $e_i \otimes 1$. Then, $(a_i)_i,(b_i)_i$ are almost null sequences of elements of $B$. Assume that there are some nonzero $a_i,b_j$. Then $a_ib_j \in B$ is nonzero (and $B$ is a finitely generated $k$-algebra), hence is not contained in some maximal ideal $\mu$ by the Nullstellensatz. Consider the reduction map mod $\mu$: $A \otimes B \rightarrow A$ (as $k$ is algebraically closed): then the images of neither $a$ nor $b$ are zero, but their product is, so $A$ isn't an integral domain, so we have a contradiction, which shows that $A \otimes K$ is a domain and we are done.


问:
Compute $\int_{-\infty}^{\infty} t \tan^{-1}(t) \exp (-t^2)\,dt$
I would like to obtain
$$
\int_{-\infty}^{\infty} t \tan^{-1}(t) \exp (-t^2)\,dt
$$
My idea is to use Fourier transform and go with generalized Parseval. I choose $x_{1}(t)=\tan^{-1}(t)$ and $x_{2}^{*}(t)=t \exp(-t^2)$. I don't know how to proceed after.
My problem with my idea is that I don't know the Fourier transform of the reciprocal function of tan(t).
Any suggestion??

Or maybe another choice of functions for the Generalized Parseval theorem?

答:
I would do integration by parts first and then employ Fourier transform techniques.

$$\int \arctan(x)xe^{-x^2}\,dx = -\frac{1}{2}\arctan(x)e^{-x^2} + \frac{1}{2}\int \frac{1}{1+x^2}e^{-x^2}\,dx.$$

Evaluating at $\pm\infty$, the first term on the left hand side will go to $0$ since $\arctan$ is bounded and $e^{-x^2}$ converges to $0$ at infinity. The latter term is very amenable to Fourier transform techniques via Parseval. Can you take it from here?


问:
Finding the minimum $n$ such that there are $n$ reals $x_i$ with $\sum_{i=1}^n\sin x_i=0$ and $\sum_{i=1}^ni\sin x_i=2019$
How to solve this problem?

$\sin x_i$ satisfies that, $\left(1\leqslant i\leqslant n\right)$
\begin{equation}
\sum_{i=1}^{n}\sin x_i=0,\qquad \text{where }x_i\in\mathbb{R}
\end{equation}
Moreover, $\sin x_i$ also satisfies that
\begin{equation}
\sum_{i=1}^{n}i\cdot\sin x_i=2019
\end{equation}
Here is the question. How to find the minimum of $n$, where $n$ is a positive integer?

答:
Consider the following expression, which (assuming $n$ is even) represents $\sum i \cdot \sin(x_i)$ where the first $\frac{n}{2}$ entries satisfy $\sin(x_i) = -1$ and the rest satisfy $\sin(x_i) = 1$, which clearly maximizes your second sum.

The expression, which amounts to adding the first $n$ numbers, then subtracting the first $\frac{n}{2}$ numbers twice (since we don't just want to take them away, we actually want to change the sign on them) is:

$$f(n) = \frac{(n)(n+1)}{2} - 2 \cdot \frac{(\frac{n}{2})(\frac{n}{2}+1)}{2}$$

Suppose $f(n) = 2019$, which would be right-on-the-nose-optimal.  Then you get:

$$\frac{(n)(n+1)}{2} - 2 \cdot \frac{(\frac{n}{2})(\frac{n}{2}+1)}{2} = 2019 \implies \text{multiply by 4}$$

$$2(n)(n+1) - (n)(n+2) = 8076 \implies n^2 = 8076 \implies n = 89.86\text{ish}$$

Since this formula applies to even $n$, it means that $n$ must be greater than $88$.  So it's sufficient to find a solution with $n = 89$.  Note that $(89)(90)/2 - (44)(45) = 2025$ which represents selecting the first $44$ entries so that $\sin(x_i) = -1$ and the rest to $1$.  Zeroing out the middle entry gives us an insufficient sum again, and this is maximal among such sums with $89$ entries.

Thus we are looking for a solution with $n = 90$.  For such a solution, consider $(1/2)(90)(91) - (45)(46) = 2025$, which is setting the first 45 entries to negative and the last $45$ entries to positive.  To decrease this sum by $6$ and achieve our desired $2019$, just swap the signs on $x_{44}$ and $x_{47}$, giving us a solution with $n = 90$, as desired.


答:
The rational function can be written as
$$
q(x)=\frac{x^2+3x-4}{x^2-7x+10}=\frac{(x+4)(x-1)}{(x-2)(x-5)}=\frac{1}{x-2}\cdot \frac{(x+4)(x-1)}{x-5}=:\frac{1}{x-2}\cdot g(x)
$$

Note that
$$
\lim_{x\to 2}g(x)=-2
$$
But $\displaystyle \lim_{x\to 2}\frac{1}{x-2}$ does not exist.$\dagger$ So the whole limit $\lim_{x\to 2}q(x)$, does not exist since otherwise, by one of the limit laws:
$$
\lim_{x\to 2}\frac{1}{x-2}=\frac{\lim_{x\to 2}q(x)}{\lim_{x\to 2}g(x)}
$$

$\dagger$ Notes.

This is a case where you should go over the materials about two-sided limits.

The following two limits show that $\displaystyle \lim_{x\to 0}\frac{1}{x}$ does not exist:
$$
\lim_{x\to 0-}\frac{1}{x},\quad \lim_{x\to 0+}\frac{1}{x}
$$

In general, $\lim_{x\to a}f(x)$ exists if and only if
$$
\lim_{x\to a-}f(x)=\lim_{x\to a+}f(x)=L
$$
where $L$ is a real number. $L$ could also be $+\infty$ or $-\infty$ if you want to deal with generalized limits.

Title: Finding a limit of a rational function when the function isnt moving towards 0

Body:
Consider this:

$$\lim_{x \to 2} \frac{x^2+3x-4}{x^2-7x+10}$$ Now I've solved this for the case when x is approaching 0,but when I try to use the same method for when it is approaching 2,I dont get very far.So first I put in 2 and got that in the numerator we get a 2 and denominator we get a 0,so division by 0 we want to avoid that.Now I need to "change" the look of my polynomials to get the limit,but I'm not sure how to do that? I am not sure if I can use L'Hospital here since it isn't 0/0 but rather 2 divided by zero? A bit of advice would be much appreciated

Many thanks!


问:
Why is $\pi $ equal to $3.14159...$?
Wait before you dismiss this as a crank question :)

A friend of mine teaches school kids, and the book she uses states something to the following effect:

If you divide the circumference of any circle by its diameter, you get the same number, and this number is an irrational number which starts off as $3.14159... .$

One of the smarter kids in class now has the following doubt:

Why is this number equal to $3.14159....$? Why is it not some other irrational number?

My friend is in a fix as to how to answer this in a sensible manner. Could you help us with this?

I have the following idea about how to answer this: Show that the ratio must be greater than $3$. Now show that it must be less than $3.5$. Then show that it must be greater than $3.1$. And so on ... .

The trouble with this is that I don't know of any easy way of doing this, which would also be accessible to a school student.

Could you point me to some approximation argument of this form?

答:
"Why is this number equal to 3.14159....? Why is it not some other irrational number?"

One can answer the second question "pi can equal some other irrational number" and it will have little bearing on any sort of application of mathematics.  There exist an uncountable infinity of numbers close to pi by the absolute value metric.  So, any engineer, physicist, or applied mathematician seeking a certain precision could use a value within that level of precision instead of pi.

So, I don't think the approximations you've gotten here answer your first question.

Title: Why is $\pi $ equal to $3.14159...$?


问:
$7$ friends promise to each send a postcard to $3$ others. Is it possible that everybody receives postcards only from people they sent postcards to?
This is a task meant to be solved using graph theory. My intuition says that it is not possible, however I don't know how to prove it apart from just drawing graphs with seven nodes in various configurations of connections between those nodes. I will appreciate any help or nudge in the right direction. Thank you.</p>

答:
Imagine if you will a directed graph detailing the scenario.  A vertex for each of the friends.  Draw a directed arrow from one friend to another iff the first friend sent a postcard to the other friend.

Now... the condition "everybody receives postcards only from people they sent postcards to" would imply that in every case where you have a directed edge from one friend to another you would also necessarily have a directed edge in the other direction and there would be no other such edges.

So... let us imagine now a graph rather than a directed graph, just like before, except this time we only draw the one undirected edge between friends indicating that both had mailed each other.

Such a graph would have $7$ vertices and each vertex would be of degree $3$.

Why is this not possible?

 Because the handshaking lemma implies that we may not have an odd number of odd-degree vertices.


问:
Divergence theorem(calculating flux of vector field)
I need to find flux of vector field $F=(x,y,z)$ over area $T=\{|x|+|y|+|z|\le 4|(x,y,z)\in \mathbb R^3\}$  oriented in the direction of outside normal, using:a)divergence theorem; b)directly.
First, using divergence theorem i get integral $\iiint 3dV$. Then, for the bounds, $z$ should always be in $[0,4]$. In $OXY$ plane I get square which contains two symmetric triangles. Then I thought to find bounds for the triangle in first and fourth quadrant like $0\le x\le 4, x-4\le y\le 4-x$(and then multiply integral with $2$). And what is left is to integrate. Is this correct?
For the second part I should calculate integral $F\cdot n_0$ where $n_0$ is normal vector, but to find it, first I need to calculate gradient which I guess won't be that easy because of absolute values.

答:
Yes you have correctly applied the divergence theorem. Now it is about finding the volume of the region.

See the base of the pyramid in the below diagram, which is in $XY$ plane. Now there are two square pyramids, one above $XY$ plane and one below with vertex at $(0,0,4)$ and $(0,0,-4)$.

Side of the base $ b = 4\sqrt2$ and height is $h = 4$. You can simply use the formula for pyramid volume $V = \frac{1}{3}b^2h$. Multiply by $2$ as you have two pyramids.

If you are going integral route, take the base above $x-$axis and that part of the pyramid for positive $z$. Find volume of it and then multiply by $4$.

$V = 4 \displaystyle \int_0^4 \int_{0}^{4-z} \int_{y+z-4}^{4-y-z} \ dx \ dy \ dz$


问:
A question on proving an equation to be an $n$-linear system in linear algebra
While studying Determinants from text book Hoffman and Kunze, I have a in an argument in a theorem whose reasoning is not provided .

Questions:
1st question is in underlined part of theorem.

It's image :

How did authors deduced that $A_{ij} D_{ij}(A) $ is n-linear function of A?

Question 2:  How does author derived the last line which is "Therefore $ E_{j} (A) $ = $(-1)^{k+j} $ .... .

Can anyone please give some hints.

Any help would be really appreciated

答:
At here, $D_{ij}A_{ij}$ should be interpreted as a function $f_i:M_{n\times n}(F)\to F$, $f_i(A)=A_{ij}D_{ij}(A)$.
$$f_i
\begin{pmatrix}
r_1\\
r_2\\
\vdots\\
v+\lambda w \\
\vdots\\
r_n
\end{pmatrix}
=(v_i+\lambda w_i)
D
\begin{pmatrix}
r_1'\\
r_2'\\
\vdots\\
r_n'
\end{pmatrix}
=v_i
D\begin{pmatrix}
r_1'\\
r_2'\\
\vdots\\
r_n'
\end{pmatrix}
+\lambda w_i
D\begin{pmatrix}
r_1'\\
r_2'\\
\vdots\\
r_n'
\end{pmatrix}
=f_i
\begin{pmatrix}
r_1\\
r_2\\
\vdots\\
v\\
\vdots\\
r_n
\end{pmatrix}
+
\lambda f_i
\begin{pmatrix}
r_1\\
r_2\\
\vdots\\
w \\
\vdots\\
r_n
\end{pmatrix}
$$
That is, $f_i$ is $n$-linear.


问:
An isomorphism between two finite fields
Suppose we have two fields $F_1$ and $F_2$ of order $9$ where both groups of units are cyclic, i.e. $$F_1=\{0\}\cup\{\alpha^i\,|\,0\leq i\leq 7\},\qquad F_2=\{0\}\cup\{\beta^i\,|\,0\leq i\leq 7\}$$ for some $\alpha\in F_1$ and $\beta\in F_2$. We know that those two fields have to be isomorphic as they have the same number of elements. A concrete isomorphism would be given by $\varphi: F_1\to F_2$ where $\varphi(\alpha):=\beta$. It is easy to see that $\varphi(xy)=\varphi(x)\varphi(y)$ for every $x,y\in F_1$ due to the representation of the fields as powers of the generators $\alpha$ and $\beta$; the fact that $\varphi$ is bijective follows immediately as well.

It is not clear to me, however, why $\varphi(x+y)=\varphi(x)+\varphi(y)$ for every $x,y\in F_1$. Is there any elementary way to see this?

Edit: As metamorphy noted in the answer below, $\varphi$ is not generally an isomorphism. Is there any way, then, to choose another suitable generator $\beta'\in F_2^\times$ such that $\varphi$ is actually an isomorphism?

答:
$\varphi(\alpha):=\beta$ (i.e. $\varphi(\alpha^i):=\beta^i$) does not necessarily define an isomorphism.

The automorphisms of $\mathbb{F}_q$, for $q=p^n$ and $p$ prime, are $x\mapsto x^{p^k}$ for $0\leqslant k<n$. In particular, for $q=9$, the only nontrivial automorphism is $x\mapsto x^3$. Hence, for $F_1=F_2=\mathbb{F}_9$ and $\beta=\alpha^5$ (say), the corresponding $\varphi$ is not an isomorphism.

As for the final subquestion, the choice of $\beta'$ is determined by the action of $+$ (in $F_1$ and $F_2$) or, as noted in the comments, by the action of $x\mapsto x+1$. For $\mathbb{F}_9$, the (only) two possibilities are:

\begin{array}{c|c|c|c|c|c|c|c|c|}
x & 0 & 1 & \alpha & \alpha^2 & \alpha^3 & \alpha^4 & \alpha^5 & \alpha^6 & \alpha^7 \\ \hline
(x+1)_1 & 1 & \alpha^4 & \alpha^7 & \alpha^3 & \alpha^5 & 0 & \alpha^2 & \alpha & \alpha^6 \\ \hline
(x+1)_2 & 1 & \alpha^4 & \alpha^2 & \alpha^7 & \alpha^6 & 0 & \alpha^3 & \alpha^5 & \alpha \\ \hline
\end{array}


问:
Solve $\sqrt{x+4}-\sqrt{x+1}=1$ for $x$
Can someone give me some hints on how to start solving $\sqrt{x+4}-\sqrt{x+1}=1$ for x?

Like I tried to factor it expand it, or even multiplying both sides by its conjugate but nothing comes up right.

答:
Multiplying by the conjugate as you originally suggested does work here. If you multiply both sides by $\sqrt{x+4} + \sqrt{x+1}$ you get
$$(x+4) - (x+1) = \sqrt{x+4} + \sqrt{x+1}$$
Which is the same as
$$\sqrt{x+4} + \sqrt{x+1} = 3$$
Add this to the original equation and divide by $2$ to obtain
$$\sqrt{x+4} = 2$$
Squaring you get
$$x +4 = 4$$
Therefore $x = 0$ is the only solution.

Also note the similarity to lab bhattacharjee's method.

Title: Solve $\sqrt{x+4}-\sqrt{x+1}=1$ for $x$


问:
The value of $\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$
What is the value of this expression :
$$\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$$
Using calculator and wolfram alpha, the answer is, $-\frac{1}{\sqrt{5}}$

But, by solving it myself the result comes out to be different. My solution is as follow:
$$\begin{aligned} Put,\, &\cot^{-1}\left(\frac{-3}{4}\right) = \theta
\\\implies &\cot(\theta) = \frac{-3}{4} =\frac{b}{p}
\\So,\,&\cos(\theta) = \frac{-3}{5}\end{aligned}$$
$$\begin{aligned}\\Then,
\\\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right) &= \sin\left(\frac{\theta}{2}\right) \\&= \sqrt{\frac{1-\cos{\theta}}{2}}
\\&= \sqrt{\frac{1+\frac{3}{5}}{2}}
\\&= \frac{2}{\sqrt{5}}
\end{aligned}$$
This solution is used by a lot of websites.

So, I got two different values of single expression but I am not sure which one is correct. Can you point out where I have done the mistake?

答:
In all four values. First, half angle formula

$$ \pm\sqrt{[1+\cos \left( \cot^{-1(...)})]/2\right)}$$

$$\pm \sqrt{\dfrac{1\pm\frac35 }{2}}$$

$$\pm \dfrac{2}{\sqrt 5}, \pm\dfrac{1}{\sqrt 5}.$$

So both answers correct.


问:
Function spaces- why are functions a vector space rather than an algebra?
I have Calculus + Analytic geometry credits and what I'm reading as an introduction to Functional analysis is to define $(f+g)(x) = f(x)+g(x)$ and call the functions a vector space and that; when do I also define $(f \times g)(x) = f(x) \times g(x)$? It's confusing me about what is a linear function and what's not a linear function.

答: Turning my comment into an answer.. The structure of the space of functions from $X$ to $Y$ depends heavily on the nature of the codomain $Y$. If $Y$ is a vector space, the space of functions will naturally inherit the vector space structure in the usual way. If $Y$ also happens to be a field, then the space of functions will even further be an algebra since you will be able to multiply elements in the codomain and track that back to the functions themselves naturally. It will however not in general be a field since you can define functions that are supported on opposite pieces to give a product of $0$. In order for the algebra of functions to be a field, you will need to add more structure by focusing on certain classes of functions.


问:
Spanning list reduction to basis & relation to Algorithm Correctness
From the book "Linear Algebra Done Right" I noticed the proof for reduction of a spanning list in vector space is algorithmic:

Now, I'm convinced of this algorithm. But If I wanted to prove it, would proof-of-correctness of the algorithm suffice? ( since we have a loop [ represented as steps ], we can prove correctness by finding a loop invariant ).  Is proof by simple induction here ( on $ n $ ) is any different?

答:
I'm not sure if I understand what you're asking, but if you show that the algorithm does what it is supposed to do, namely reduce a spanning list to a basis, then that gives you a proof of the theorem. In fact, it gives you more, because it also tells you how to reduce a spanning set into a basis if you were actually given one.

That said, if you would want to implement the procedure outlined in the proof of the book into say a computer, then you might want to add some details on checking whether $v_j$ is in the span of $v_{0},\ldots,v_{j-1}$. For the purpose of the theoretical algorithm, this is somewhat irrelevant, because we know that it either is in the span, or it's not, and the algorithm will proceed either way.

Edit. Regarding the steps in the comment:
1 and 3. is verified at the end of the proof: using (2.21) it is observed that the output of the algorithm is indeed a basis. 2. is true because at each step $j$, you turn your attention to vector $v_j$, and since there are finitely many vectors, you will surely at some point reach the last vector. I do not understand 4. As for 5.: The proof gives no indication on how to implement the algorithm in practice, or how long it would take for a computer to finish it. In that sense, it might not be considered a true algorithm. In fact, let me rephrase 'Step $j$' as follows:

Step $j$: Ask the Lord Almighty if $v_j$ is in the span of $v_1,\ldots,v_{j-1}$. If so, delete it; if not, leave it unchanged.

I claim that the proof is still valid when phrased in this rather crazy way. Indeed, all that matters is that there is an outcome to the step: either $v_j$ is in the span, or it's not. Whether a computer can actually figure that out is besides the point: there's an answer either way and you get to move on to step $j+1$ either way.


问:
If $A_n$ is a countable collection of sets in $F$ with $\mu(A_n) = 1 $ for $ n\geq 1$ then: $\mu\big(\bigcap_{n=1}^{\infty} A_n \big) = 1$
I just started learning Measure Theory in a Probability Course.
I am having trouble understanding the significance and the proof of the following theorem.

Let $(\Omega, F, \mu)$ be a measure space.

If $\mu$ is a probability measure and $A_n$ is a countable collection of events in $F$ with $\mu(A_n) = 1 $ for $ n\geq 1$ then:

$$\mu\big(\bigcap_{n=1}^{\infty} A_n \big) = 1$$

答:
Hint: The simple way to solve this is to show that

$$ \mu \Bigg( \Big( \bigcap_{n=1}^{\infty} A_n\Big)^c \Bigg)=0 .$$

Try using De-Morgan's law, and recall that $\mu\big( A_n^c \big)=0 $ for all $n$.

The significance of this statement, is that with respect to $\mu$, you only care what happens with probability $1$. Everything else is indistinguishable with respect to $\mu$.


问:
Dual operator on $C([0,2 \pi])$
Consider $T:C([0,2 \pi]) \rightarrow C([0,2 \pi]) $
$$
T(f(x)) =e^{ix} f(x).
$$
Find dual operator $T^*: M([0,2 \pi]) \rightarrow M([0,2 \pi]) $.

Using the Riesz representation for $\mu \in M([0,2\pi])$ and definition of dual operator we can write
$$
T^*\mu(f(x)) = \mu(T(f(x)) = \mu(e^{ix} f(x)) = \int_0^{2\pi} e^{ix} f(x) d \mu(x).
$$
How can I express $T^* (\mu)$?

Thanks.

答: You can use the change of variables formula for measures (see for instance this older StackExchange post:  Is there a change of variables formula for a measure theoretic integral that does not use the Lebesgue measure) to rewrite your last expression into the form $$\int_0^{2\pi}f(x) d\nu(x)$$ Obtaining something of the form $$\langle T^* \mu, f\rangle$$ should then allow you to define the operator $T^*$.

Title: Dual operator on $C([0,2 \pi])$


问:
An example of one-to-one function to two variables
I was searching for an one-to-one function $f:[0,1]\times[0,1] \rightarrow \mathbb{R}$. I failed to construct one. Neither can I prove that such a function does not exist. Do you have any answer for this?

答:
Hint: Cantor-Bernstein-Schroeder (=CBS) Theorem in set theory, or directly with arithmetic of cardinal numbers

$$|\,[0,1]\times[0,1]\,|=2^{\aleph_0}\cdot2^{\aleph_0}=2^{\aleph_0}=|\,\Bbb R\,|$$

Make sure you understand the above and can prove the different steps.


问:
Induced module and surjective morphism
I am trying to solve the following question:

Let $G$ be a finite group and $H$ a subgroup of $G$. Let $A$ be a $G$-module. Show that $\pi: I^{H}_{G}(A)\to A$ defined by $\pi(f)=\sum_{g\in G/H}g\cdot f(g^{-1})$ is surjective morphism.

Here, $I^{H}_{G}(A)$ is the induced module from $H$ to $G$. It is $Hom_{H}(\mathbb{Z}[G],A)$.

I tried to understand the image of $\pi$, but I could only conclude that it is invariant by $H$. I don't have any other idea what to do.

答:
Let's start with the following observation. If $H$ is a subgroup of $G$, then we have an isomorphism $f \colon \mathbb{Z}[G] \to \bigoplus_{G/H} \mathbb{Z}[H]$ of left $H$-modules. Here the direct sum is taken over the right cosets. For convenience, let's enumerate these cosets as $Hg_1,\ldots,Hg_r$, and let's write an element in the target as an $r$-tuple $(h_1,\ldots,h_r)$. The map $f$ is then the expected one: Start with an element $x \in G$, viewed as an element of $\mathbb{Z}[G]$. It lives in a unique $Hg_i$ for some $i$, and can thus be written as $hg_i$. Then $f(x) = (0,\ldots,h,\ldots,0)$, with the $h$ on the $i$-th index.

Now that we know this, we observe that
$$\begin{split}
\operatorname{Hom}_{\mathbb{Z}[H]}(\mathbb{Z}[G],A) &= \operatorname{Hom}_{\mathbb{Z}[H]}(\bigoplus_{G/H} \mathbb{Z}[H],A)\\
&= \bigoplus_{G/H} \operatorname{Hom}_{\mathbb{Z}[H]}(\mathbb{Z}[H],A)  \\
&= \bigoplus_{G/H} A\end{split}$$
Surely we should then expect a surjection onto $A$. The map $\pi$ you're interested in should, in this langauge, be precisely the 'expected' surjection $\bigoplus_{G/H} A \to A$, which is the identity on every component.

Edit. So I guess that if you meditate on these steps you should be able to 'see' an explicit pre-image of an element $a \in A$. Namely, split up $G$ into cosets $Hg_1,\ldots,Hg_r$, and assume WLOG that $g_1 = e$. Now consider the $H$-equivariant map $\alpha \colon \mathbb{Z}[G]$ sending $e$ to $a$ and hence $h$ to $ha$, while sending all the other cosets entirely to $0$. By a direct computation, $\pi(\alpha)$ should be $a$.

Title: Induced module and surjective morphism


问:
Function "evaluation" just means "composition"?
I am self-studying and have a basic or naive question that follows from a simple observation. I have also included tags for type theory, etc because "evaluation" probably has a different meaning in these constructive approaches, which I'm also curious about.

Any set $X$ with elements $x_i$ can be characterized by a set denoted $(X)$ of pairs $(\star, x_i)$, alternatively denoted $(\star \mapsto x_i)$ (for all $x_i \in X$). Each one of these latter pairs encodes a distinct function from common domain $\{ \star \}$ to (each different element of) $X$. If we name such functions after their image, i.e. $(x_i):\star \mapsto x_i$, then the symbol "$f(x_i)$", traditionally interpreted as function evaluation, is reinterpreted as function composition, i.e. $f\circ (x_i)$ and the $\circ$ is often notationally suppressed. And roughly, $f = \cup_xf(x)$, every function decomposes to a unique set of singleton functions (better word?).

Question: Is there anything more to evaluation, or can I exactly equate evaluation with composition and forever de-clutter my mind of the term "evaluation"? Is the above perspective unusual or fruitful beyond reducing terminological clutter?

The notion of evaluation above - perhaps "extensional" is the right term? - means essentially subsetting the function at the chosen argument $x$. Perhaps I will see that a full answer to my question requires considering an intentional (?) or constructive perspective on "evaluation" (as in lambda calculus, etc). Perhaps only there will composition and evaluation legitimately have distinct meanings.

Edit: The notation $(x)$ seemed the most natural way to denote any function with singleton domain because it coheres with the conventional way to write evaluation as $f(x)$. The notation seems unambiguous because there is no context I can think of, other than "evaluation", where a single symbol, say $x$, is enclosed in brackets "()"? It does seem useful to notationally distinguish functions on singleton domain in some way: I would like the notation to be a mnemonic for whether the resulting function composite is a singleton function (i.e. a "value" like $f(x_i) = f \circ (x_i) = (\star, y_j)$) versus a non-singleton function. Perhaps there is a conventional notation for functions from singleton domain, other than $(x)$ and a conventional name for such functions, other than "singleton functions"?

答:
Basically, yes, what you are saying is that in the category of sets and functions the action of composition of $f\colon A\to B$ with $(a)\colon \star \to A$ is precisely encoding function evaluation $f(a)$. So this can indeed neatly be written as $f(x)=f\circ (x)$. This paves the way to thinking about functions and elements in a dual way: a function $f\colon A\to B$ is a machine that converts elements of type $A$ into elements of type $B$; dually an element of type $A$ is a thing that any machine $f\colon A\to B$ can act on to turn into something of type $B$. This tells us that we can take elementhood as primitive and define functions in terms of elements (as is classically done) or that we can treat functions as primitive and define elementhood in terms of functions (as is categorically done). In some sense, for sets and functions (at least naively) the two approaches agree.

However, in many other situations the two options are very different. Suppose that instead of sets and functions your world consists of the non-negative reals $[0,\infty ]$ (with $\infty $ thrown in) and you keep track of magnitude by means of $x\to y$ iff $x\ge y$. The special element $0$, the additive neutral element, serves here as the 'singleton' $\star$ (the reason is that a singleton set is neutral (up to isomorphism) with respect to the cartesian product of sets; similarly $0$ is neutral with respect to addition of real numbers). Now, in the world of sets the arrows $\star\to A$ encode the elements of $A$. What is encoded in the world of non-negative reals? Well, $\star \to a$ is no simply saying that $a=0$. So, if $a=0$, then it has a unique 'element' while any $a>0$ has no elements at all.

This is probably not surprising since real numbers are quite different from sets. Places that behave more like sets than real numbers are called toposes. But there is really no need to abandon the "composition=evaluation" paradigm in non-toposes. Indeed, evaluation can further be understood in terms of a much simpler operation. Going back to sets, if $[A,B]$ is the set of all functions $f\colon [A\to B]$, then the evaluation mapping is $\mathrm {ev}\colon  [A,B]\times A\to B$ given by $\mathrm {ev}(f,a)=f(a)$. This evaluation function has a very special universal property. If $f\colon T\to [A,B]$ is any function from any set $T$ we can form the composition $\mathrm {ev}\circ (f\times \mathrm {Id}_A)$ to get a function $h\colon T\times A\to B$. But also conversely, any such function $h$ arises uniquely in this way from a function $f\colon T\to [A,B]$.

In other words, there is a natural bijection between functions $f\colon T\to [A,B]$ and functions $h\colon T\times A\to B$. The translation is performed universally by the evaluation function. So, we observe that the internal notion of the set of all functions $[A,B]$ together with the notion of evaluation is fully encoded by the much simpler notion of the cartesian product of sets. The encoding is done by stating the natural bijectiveness above. The technical way of saying it is that evaluation is the counit of an adjunction whose right adjoint is cross multiplying with a fixed set $A$.

Now we can replace the world of sets and functions with their cartesian product by other things. For instance, non-negative real numbers, magnitude comparison, and addition. It turns one (nice exercise) that there is here too an internal notion of "the real numbers of all magnitudes from $a$ to $b$", namely $[a,b]$, complete with an evaluation mapping with formally the same properties as for sets. Such places are called closed monoidal categories.


问:
Express a set in spherical coordinates

Given $B=\left\{(x, y, z) \in \mathbb{R}^{3}: x^{2}+y^{2}+z^{2} \leq 1\right\}$, $\alpha \in(0, \pi)$ find the volume of
$$
S_{\alpha}=\left\{(x, y, z) \in B: z \geq(\cos \alpha) \sqrt{x^{2}+y^{2}+z^{2}}\right\}
$$

How can I express $S_{\alpha}$ in spherical coordiates?

答:
In spherical coordinates, $x = \rho \cos\theta \sin\phi, y = \rho \sin\theta\sin\phi, z = \rho \cos\phi$

$S_{\alpha}=\left\{(x, y, z) \in B: z \geq(\cos \alpha) \sqrt{x^{2}+y^{2}+z^{2}}\right\}$

Given $z \geq(\cos \alpha) \sqrt{x^{2}+y^{2}+z^{2}} \ , \alpha \in (0,\pi)$

$\rho \cos\phi \geq \rho\cos\alpha \implies \phi \leq \alpha$

So to find volume of $S_{\alpha}$, you integrate the same way as you do for the whole sphere i.e. the bounds of $\rho$ and $\theta$ remain the same but instead of $0 \leq \phi \leq \pi$, you need to use the bounds $0 \leq \phi \leq \alpha$.

Title: Express a set in spherical coordinates


问:
Can finite induction be independent of mathematical induction?
Suppose we have a property $P(n)$ pertaining to natural numbers and $P(n) \to P(n+1)$.
We know by mathematical induction that $P(N)$ implies $P$ being true for all natural numbers greater than or equal to $N$.

It is an obvious consequence of this that for a finite number of natural numbers greater than or equal to $N$, $P$ is also true. I am wondering, however, that whether this "finite induction" can be applied even if we didn't know mathematical induction principle. This might seem trivial, just like $P(1) \to P(2)$, $P(2) \to P(3)$ and so on, but I don't know any logical rules approving this...

答:
We can prove the meta-theorem that in any logical system which lets you derive $Q$ from $P$ and $P \implies Q$, there is a proof of any given $P(n)$ from a base case less than $n$ and the induction step $P(n) \implies P(n+1)$. It is not necessary that the logical system have induction as an axiom.

In that logical system, such proofs will look like

Prove $P(1)$.

Prove $P(1) \implies P(2)$.

Prove $P(2) \implies P(3)$.

Prove $P(3) \implies P(4)$.

Prove $P(4) \implies P(5)$.

Conclude $P(2)$, then $P(3)$, then $P(4)$, then $P(5)$ by modus ponens.

It will be able to prove any given $P(n)$, and the meta-theorem tells us that it will be able to prove any given $P(n)$. Without the axiom of induction, it will not be able to prove $\forall n\ge 1\, P(n)$.


问:
Obtain the joint PDF of $\theta_{1} - \theta_{2}$ and $\theta_{1}$ where $\theta_{1}$ and $\theta_{2}$ are i.i.d $\text{Uniform}[0, 2\pi)$
I have 2 i.i.d Uniform random variables $\theta_{1}$ and $\theta_{2}$ on $[0, 2\pi)$. Here the problem pushes us to consider that all operations are made modulo $2\pi$.

My question is how to obtain the joint PDF of $\theta_{1} - \theta_{2}$ and $\theta_{1}$ and then obtain the conditional PDF of $\theta_{1}$ given $\theta_{1} - \theta_{2}$??.

For the second part It can be deduced from the first by writing it as the joint PDF divided by the PDF of $\theta_{1}$. But for the first part, A common idea that I saw is to derive the $\text{CDF}(x, y) = P(\theta_{1} - \theta_{2} \le x$, $\theta_{1} \le y$) and then perform derivations respect to $x$ and $y$. Is this a right approach? If yes, how to proceed with CDF especially with our uniform distribution restrictions?

答:
You have to take into consideration the following system

$$\begin{cases}
z=\theta_1-\theta_2\\
v=\theta_1
\end{cases}\rightarrow \begin{cases}
\theta_1=v\\
\theta_2=v-z
\end{cases}$$

The jacobian is obviously $|J|=1$ and thus

$$f_{ZV}(z,v)=\frac{1}{4\pi^2}\cdot\mathbb{1}_{(0;2\pi)}(v)\cdot\mathbb{1}_{(v-2\pi;v)}(z)$$

That is the joint distribution is defined over the following parallelogram

Knowing the joint distribution, the rest of the exercise follows as a consequence without difficulties

The fact that the uniform marginals are defined in $(0;2\pi)$ or $[0;2\pi)$ is irrelevant

EDIT: If you want to integrate in V it is very useful to define joint support in a different but equivalent manner (this is why I showed you its drawing)

$$\mathbb{1}_{(-2\pi;0)}(z)\cdot\mathbb{1}_{(0;z+2\pi)}(v)+\mathbb{1}_{(0;2\pi)}(z)\cdot\mathbb{1}_{(z;2\pi)}(v)$$


问:
Machine epsilon definition
Assume representation of floating point in decimal, in the form $0.q_1q_2...q_n\cdot 10^{exp}$. By definition $1+\epsilon_{\text{mach}} = 1$, but a textbook claims that with this definition, $\epsilon_{mach} = 10^{-n}$.

The question I have is, suppose I do $0+\epsilon_{\text{mach}} = 0.00....01 \cdot 10^0 \ne 0$. It is representable in floating point as non-zero number. Doesn't this contradict the definition of $\epsilon_{\text{mach}}$?

And if we were to say $\epsilon_{mach} = 10^{-(n+1)}$, then wouldn't another number, say $2\cdot 10^{-(n+1)} > \epsilon_{mach}$ also be ignored?

答:
What you are missing is that floating-point addition isn't quite the same as (theoretical) real addition.  A floating-point number has only finitely-many digits in its mantissa.  When you add two floating-point numbers of different exponents, the mantissa of the smaller has to be shifted rightward, to align the radix points correctly - e.g if you add 1 to 1e-5, you are actually adding 1.0000... and 0.00001000... together.  If you have to shift rightward more than the number of digits of the mantissa, when their sum is (truncated or rounded, depending) to the mantissa length, the extra digits past the end get dropped, so in effect the sum doesn't change the larger number.

This means the machine epsilon is the value 1 multiplied by the smallest (in magnitude) negative power of ten such that this dropping occurs.

Title: Machine epsilon definition


问:
Find necessary and sufficient conditions for $T$ to be compact
Let $b \in l^{1} (\mathbb{Z})$ and define an operator $T \in L(l^{2} (\mathbb{Z}))$ (space of sequences $\mathbb{Z}\to\mathbb{C}$ with the norm $\sum_{n = -\infty}^{\infty} |a_n|^{2}$) by $(Tx)_n = \sum_{n=-\infty}^{\infty} b_{n-m} x_m$. Find necessary and sufficient conditions on $b$ for $T$ to be compact.

I suspect that $T$ is compact only if $b\equiv 0$, since $Te_{n} = \sum_{m=-\infty}^{\infty} b_{m-n} e_m$, so the bounded sequence $e_{n}$ has an image without a convergent subsequence (I think? since $Te_n$ all have the same norm and $\langle Te_n | Te_k \rangle = \sum_{m=-\infty}^{\infty} b_{n-m}b_{k-m}^{*} $, and it seems to me that for sufficiently far away n and k this should be very small since $b \in l^{1} (\mathbb{Z})$ so the sequence cant converge ?)

Any help or hint would be very appreciated. Thanks!

Edit: Since $ \langle Te_n | Te_m \rangle $ can be rewritten az $\sum_{m=-\infty}^{\infty} b_{n-k+m}b_{m}^{*}$, it seems that it would be sufficient to show that if $S$ is the 'shift right' operator in $l_{2}$ then $S^{n} v$ weakly converges to $0$ for any $v \in l_{2}$, but im unsure if this is true.

答:
Let $\mathbb T$ be the unit circle and let $Z:\mathbb T\to {\mathbb C}$ be the function defined by $Z(z)=z$, for all $z\in \mathbb T$.

Consider the operator (Fourier transform)
$$
  \mathfrak F:\ell ^2({\mathbb Z})\to L^2(\mathbb T)
  $$
given on the canonical orthonormal basis $\{e_n\}_{n\in {\mathbb Z}}$ by
$$
  \mathfrak F(e_n) = Z^n, \quad\forall n\in {\mathbb Z}.
  $$
It is then easy to see that $\mathfrak F$ is a unitary operator and that
$$
  \hat T:=  \mathfrak F T\mathfrak F^*
  $$
is the so called multiplication operator given by
$$
  \hat T(f)|_z = \hat b(z) f(z), \quad\forall f\in L^2(\mathbb T),\quad\forall z\in  \mathbb T,
  $$
where
$\hat b$
is the function on $\mathbb T$ defined by
$$
  \hat b(z) = \sum_{n\in {\mathbb Z}}b_nz^n.
  $$

If $T$ is compact,  then so is $\hat T$,  but this cannot happen for a nonzero $b$,  since the spectrum of $\hat T$ coincides with
the range of $\hat b$, which is a closed interval rather than a  sequence converging to zero as it would be expected for a
compact operator.