问:
Conjunction Elimination and Implication (Is a test for contradiction necessary?)
I am interested in whether it is required to test for contradictions in assumptions within proofs, together with the rules of working with conjunction elimination & implication (I believe these two questions to be related due to the following). Specifically, consider the following:

Assume P ∧ Q

Therefore, by conjunction elimination we have P is true and Q is true

Lets try to prove P ∧ Q ⇒ R

If we can prove P ⇒ R, if my understanding is correct, we have proven P ∧ Q ⇒ R since P ∧ Q ⇒ P ⇒ R. Please correct me if I'm wrong here.

But what if we also have Q ⇒ ¬ R? This would create a contradiction and therefore P ∧ Q is false?

As I understand, in proof by contradiction of P ⇒ Q , we would assume ¬ Q and try to prove a contradiction to show that if P is true, then Q must be true. However, here, we are not contradicting any of the original assumptions (P).

The question is what if the original assumption (which is assumed to be true) has a contradiction in it? Does that mean my above logic of using conjugation elimination and then implication is not formally correct and I also have to prove there is no contradiction within any of the assumptions that are combined via conjugation whenever I am trying to show (P ∧  Q) ⇒ R (i.e. I cannot rely on P ⇒ R to show (P ∧  Q) ⇒ R)).

I know I cannot rely on P ⇒ R to show P ∨ Q  ⇒ R. What are the ground rules for working with conjugation elimination & implication?

A good summary of the logical implication rules I have found in https://www2.seas.gwu.edu/~ayoussef/cs1311/Logic.pdf (page 5). However I cannot find a good resource on the internet for my query above on the mechanics of working conjunction elimination and then implication.

I believe this answer helps quite a bit in terms of removing the need to test for contradiction within the assumptions: In a proof by contradiction, how do we know the assumption is the cause of the contradiction?

However, I still feel uncomfortable not testing for contradiction when applying conjunction elimination and then implication. If my assumptions like in the above result in two contradicting statements, how do I know for certain which one to choose (and how do I get comfortable that there are no contradicting statements without testing for it). The reason I ask this is because conjunction elimination results in a less general condition (i.e. P is less general than P ∧ Q); I am much more comfortable if I derived my implications directly from the more general case P ∧ Q and accordingly questioning the validity of the approach of relying on P ⇒ R.

答:
Your point 5 is correct: given that $P \land Q$ eventually leads to a contradiction, you can conclude that  $P \land Q$ is not true.

However, that does not invalidate any of the inferences you made from the assumption that $P \land Q$. That is, from the assumption that $P \land Q$, you can still infer $P$ as well as $Q$, and given those, you can respectively infer $R$ as well as $\neg R$ is true. And hence, you can infer a contradiction.  All those inferences are still correct. And hence you can still say that "If $P \land Q$ is true, then $P$, $Q$, $R$, $\neg R$, and $\bot$ (logic symbol for contradiction) are all true as well".  But, of course, since a contradiction can never be true, it follows that $P \land Q$ cannot be true either.

OK, so far so good. You also said that you were trying to prove $(P \land Q) \to R$. And, once again, you assumed $P \land Q$, inferred $P$, and from $P$ you were able to infer $R$.  At that point, you can indeed infer $(P \land Q) \to R$. Once again, you have basically shown that "If $P \land Q$ is true, then $R$ is true as well", and by conditional proof we can wrap that up with $(P \land Q) \to R$.

Now (and I think this is your real/main question): does the fact that you can also infer $Q$, and from that $\neg R$, and thus a contradiction, take anything away from that? No. It is still true that "If $P \land Q$ is true, then $R$ is true as well". So, when you do a conditional proof, there is really no need to see if your assumption leads to a contradiction.

In fact, if some assumption does lead to a contradiction, then anything follows from your assumption, since anything follows from a contradiction. Hence the conditional that has the assumption as its antecedent ('if' part) will definitely be the case, no matter what you put down for the consequent ('then' part)


问:
A continuous and derivable decreasing function s.t. the derivative is $0$ for $t>t_0$?
The question may be silly, but I am stucked with this thought...

I ask for a nonconstant continuous and derivable nondecreasing function s.t. the derivative approaches $0$ for $t\to\infty$, but more than that I might have $f'(t_0)=0$ for some $t_0$ and also for all $t>t_0$.

I really cannot think this example. For instance, $f(t)=\text{log}(t)$ does not mind, because $f'$ approaches to $0$, but $f'(t)\neq0\forall t$. And so on.

But because this question?

Because I have an equation on a model of form $\dfrac{d f}{d t}=g(t)$ with $g(t)$ nonpositive and I might have $g(t)=0$ for all $t>t_0$ for some $t_0>0$, but I cannot realize this.

Thank you so much.

答: The constant function f(x)=2 (say) satisfies all your requirements.</p>

ContentLicense: CC BY-SA 4.0


问:
Particular integral for $a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=\ln x$
I would like to know how to find a particular integral for $$a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=\ln x$$
where $a,b,c$ are constants. So far, the only functions I've come across for $f(x)$ in
$$a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=f(x)$$
are $$f(x)=x^ne^{kx}$$ $$f(x)=A\sin\alpha x+B\cos\alpha x$$ $$f(x)=\text{polynomial in $x$ with positive integer exponents}$$
and the particular integrals I've been taught to use respectively  are $$x^ne^{kx}+x^{n-1}e^{kx}+\cdots+e^{kx}$$ $$y=C\sin\alpha x+D\cos\alpha x$$ $$y=\text{polynomial in $x$ with positive integer exponents of the same degree as $f(x)$}$$ unless complications arise when finding the complimentary function.

However, how do I find the particular integral for when $f(x)=\ln x$?

答:
Here's a trick

For a linear, second-order, constant-coefficient ODE, with arbitrary function $h(x)$,
\begin{align}
a\frac{\mathrm d^2y}{\mathrm dx^2}+b\frac{\mathrm dy}{\mathrm dx}+cy=h(x)
\end{align}
making the substitution
\begin{align}
z(x)=\delta y(x)+\frac{\mathrm dy}{\mathrm dx},
\end{align}
\begin{align}
\delta =\left(\frac{b\pm\sqrt{b^2-4ac}}{2a}\right)
\end{align}
reduces the order of your ODE:
\begin{align}
a\frac{\mathrm dz}{\mathrm dx}+\frac{c}{\delta}z(x)=h(x).
\end{align}
I think you can take it from there.


问:
Proving an identity for the quantum angular momentum operator
I am trying to prove the following identity for the quantum angular momentum operator:

$$\vec L^2=-\hbar^2r^2(\hat r \times \boldsymbol{\nabla})\bullet (\hat r \times \boldsymbol{\nabla})=-\hbar^2r^2\left[\nabla^2-\dfrac{1}{r^2} \dfrac{\partial}{\partial r}(r^2\dfrac{\partial}{\partial r})\right]$$

So basically I have to prove:

$$(\hat r \times \boldsymbol{\nabla})\bullet (\hat r \times \boldsymbol{\nabla})=\nabla^2-\dfrac{1}{r^2} \dfrac{\partial}{\partial r}(r^2\dfrac{\partial}{\partial r})$$

I was trying to use vector identities like $A \times (B \times C)=(A \cdot C)B-(A\cdot B)C$ , $A\cdot(B\times C)=C\cdot(A\times B)$ or $A\times B=-B\times A$, but I am unsure if they hold when $\boldsymbol{\nabla}$ is one or more of the vectors. Any help please?


问:
Show that f is uniformly continuous
I am having difficulty determining what value I should assign to $\delta$ for the following problem.  How do I determine what it should be?

Define $f:[3.4,5] \rightarrow \mathbb{R}$ by $f(x)=2/(x-3)$.  Show that $f$ is uniformly continuous on [3.4,5].

This is what I have so far.

Choose $\varepsilon >0,$ and $\delta=???$.  If $|x-y|< \delta$ and $x,y \in [3.4,5]$, then

$|f(x)-f(y)|=|2/(x-3)-2/(y-3)| = |2(y-x)/((x-3)(y-3))|<???$

答: Your function is continuous over the interval, with continuous derivative there. Weierstrass' theorem ensures that $f'$ attains its maximum, say $M$, over $[3.4\;,\; 5]$. Then, by the Mean Value Theorem, we have that $$|f(x)-f(y)|\leq M|x-y|$$ over $[3.4, \; 5]$. Thus, given $\epsilon>0$, take $\delta =\frac{\epsilon}{M}$ and conclude.</p>

ContentLicense: CC BY-SA 3.0

Title: Show that f is uniformly continuous


答:
$P\to R,Q\to\lnot R$ do entail $P\land Q\to R$.

These premises do also entail $P\land Q\to\lnot R$

This is not by itself a contradiction, however $P\land Q$ would entail a contradiction under those premises.

So these premises entail that $\lnot(P\land Q)$

Title: Conjunction Elimination and Implication (Is a test for contradiction necessary?)

Body:
I am interested in whether it is required to test for contradictions in assumptions within proofs, together with the rules of working with conjunction elimination & implication (I believe these two questions to be related due to the following). Specifically, consider the following:

Assume P ∧ Q

Therefore, by conjunction elimination we have P is true and Q is true

Lets try to prove P ∧ Q ⇒ R

If we can prove P ⇒ R, if my understanding is correct, we have proven P ∧ Q ⇒ R since P ∧ Q ⇒ P ⇒ R. Please correct me if I'm wrong here.

But what if we also have Q ⇒ ¬ R? This would create a contradiction and therefore P ∧ Q is false?

As I understand, in proof by contradiction of P ⇒ Q , we would assume ¬ Q and try to prove a contradiction to show that if P is true, then Q must be true. However, here, we are not contradicting any of the original assumptions (P).

The question is what if the original assumption (which is assumed to be true) has a contradiction in it? Does that mean my above logic of using conjugation elimination and then implication is not formally correct and I also have to prove there is no contradiction within any of the assumptions that are combined via conjugation whenever I am trying to show (P ∧  Q) ⇒ R (i.e. I cannot rely on P ⇒ R to show (P ∧  Q) ⇒ R)).

I know I cannot rely on P ⇒ R to show P ∨ Q  ⇒ R. What are the ground rules for working with conjugation elimination & implication?

A good summary of the logical implication rules I have found in https://www2.seas.gwu.edu/~ayoussef/cs1311/Logic.pdf (page 5). However I cannot find a good resource on the internet for my query above on the mechanics of working conjunction elimination and then implication.

I believe this answer helps quite a bit in terms of removing the need to test for contradiction within the assumptions: In a proof by contradiction, how do we know the assumption is the cause of the contradiction?

However, I still feel uncomfortable not testing for contradiction when applying conjunction elimination and then implication. If my assumptions like in the above result in two contradicting statements, how do I know for certain which one to choose (and how do I get comfortable that there are no contradicting statements without testing for it). The reason I ask this is because conjunction elimination results in a less general condition (i.e. P is less general than P ∧ Q); I am much more comfortable if I derived my implications directly from the more general case P ∧ Q and accordingly questioning the validity of the approach of relying on P ⇒ R.


问:
Standard path for integral with complex limits?
my question is quite simple and might be a duplicate though I couldn't find one. Is there an accepted meaning for the notation
$$\int_{z_1}^{z_2}$$
With $z_1,z_2\in\mathbb{C}$? Is the standard to just integrate over the line connecting the two points?

Thanks.

答:
Yes. When your function $f$ is "path-independent" in the domain, this notation is sometimes used for
$$
\int_\gamma f(z)dz=\int_{z_1}^{z_2} f(z)dz
$$
where $z_1$ is the initial point of $\gamma$ and $z_2$ the final point.

However, when $f$ is not path-independent, one usually use the left-hand side and give the explicit description of $\gamma$.


问:
Finding area of quadrilateral inscribed inside of a semicircle
I've been practicing problems recently to study for the upcoming AMC10, and came across one I could not figure out how to solve one of them. The diagram of it is attached below, and the problem goes a little like this.

Quadrilateral PTSU is inscribed in semicircle O, as shown, with PQ = 3 units, QR = 5 units, and RS = 4 units. What is the area of PSTU?

I haven't been able to find a way to get to the area. I'm not completely sure how to find the length of the other sides, and I'm not aware of any method I can use for this quadrilateral to easily find the area of it. Any ideas?

答:
The diameter is $3+5+4=12$ and the radius is thus $12/2=6$.

You can connect $OT$ and $OU$ as auxiliary lines. Then we have $|OR|=|OS|-|RS|=6-4=2$ and $$|RT|=\sqrt{|OT|^2-|OR|^2}=\sqrt{6^2-2^2}=4\sqrt{2}.$$ Similarly, we have $|OQ|=|OP|-|PQ|=6-3=3$ and $$|UQ|=\sqrt{|OU|^2-|OQ|^2}=\sqrt{6^2-3^2}=3\sqrt{3}.$$ Then you can calculate the area immediately.


问:
Matrix Equation, Solving for Variables.
I'm going through my exercises, and came across a problem that wasn't covered in our lectures. Here's the question:

$
\begin{align}
\begin{bmatrix}
a-b & b+c\\
3d+c & 2a-4d
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
8 & 1\\
7 & 6
\end{bmatrix}
\end{align}
$

What I have done so far is:

$
\begin{align}
\begin{bmatrix}
a-b-8 & b+c-1\\
3d+c-7 & 2a-4d-6
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix}
\end{align}
$

And the solving for variables,

$$
a-b-8 = 0
$$
$$
a-b = 8
$$
$$
a = \frac{8}{-b}
$$

$$
b+c-1=0
$$
$$
b+c=1
$$
$$
b=\frac{1}{c}
$$

$$
3d+c-7=0
$$
$$
3d+c=7
$$
$$
3d=\frac{7}{c}
$$
$$
d=\frac{7}{3c}
$$

$$
2a-4d-6=0
$$
$$
2a-4d=6
$$
$$
\frac{16}{-2b}-\frac{28}{12c}=6
$$

Am I going about this correctly? Or am I just doing this completely incorrect?

答:
How familiar are you with basic linear algebra? As @response has posted, you have numerous algebra mistakes. However, as for the method used to solve the problem, I think it's worth noting that you can easily transform this system to one more familiar looking and then use standard technique of Gaussian elimination. That is, rewrite the system as follows:

$
\begin{align}
\begin{bmatrix}
1 & -1 & 0 &  0\\
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 3 \\
2 & 0 & 0 & -4 \\
\end{bmatrix}
\end{align}
$
$
\begin{align}
\begin{bmatrix}
a \\
b \\
c \\
d \\
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
8 \\
1\\
7 \\
6
\end{bmatrix}
\end{align}
$

And then solve.

Title: Matrix Equation, Solving for Variables.


问:
What values of $j$ makes the congruence solvable when $a \leq b$?
What values of $j$ makes the congruence solvable when $a \leq b.$?

$$0 \equiv xp^{s}j + yp^t\pmod{p^a} $$ where $x,y$ are not divisible by $p.$ and $s,a$ are any integers and $a \leq b.$

Since the given congruence above is equivalent to $xp^{a+s−b}j + yp^t = u p^b $ for some $u$; and then I have to divide both sides by $p^m$ where $m= \operatorname{min} \{s,t,a\}$, then my answer for the question what values of $j$ that makes this congruence solvable is as follows:

If $m=a$

Then the equation is satisfied for all $j$.

If $m\ne a, t = m$

Then $x p ^{s-t}j + y = u p^{a-t}$ and since $p$ is not a factor of $y$ then there are no solutions.

If $m\ne a, t \ne m$

Then $xj + yp^{t-s} = u p^{a-s}$ and since $p$ is not a factor of $x$ there are no solutions.

Is my solution correct?

I have followed exactly the same procedure given to me in the solution here:

what values of $j$ makes this congruence solvable?

答:
You seem to have the right idea and there are lots of correct deductions, but there are errors:-

The equation on line 4 should be $$xp^sj + yp^t = u p^a.$$

The $m=a$ case is handled correctly.

In the second case, there can be solutions when $s=t$.

In the final case, your equation $xj + yp^{t-s} = u p^{a-s}$ is correct but there may be solutions when $j$ is divisible by $p$. For example, when $a=t$ and $j=Jp^{a-s}$, the equation becomes $xJ+y=u$ and so there are infinitely many solutions for $x$ and $y$.

Let's deal with one such case which you didn't consider:-

If $m=s=t$, $m\ne a$

Then $$xj + y = u p^{a-m}.$$

For any $j$ not divisible by $p$ we can therefore choose $y=up^{a-m}-xj$.


问:
Correct arrow notation for function with natural domain
I saw a definition stating that given a function $f:X \to Y$ the natural domain of $f$ is the largest possible subset $D \subseteq X$ for which the rule defining $f$ is valid.

From what I understand, the $X$ in $f:X \to Y$ must be the domain of $X$. Let’s suppose $D$ is a subset of $X$ but is not $X$ itself, then that seems to imply that there is at least one element in $X$ not in $D$. But if $X$ contains an element not in the domain of $f$ then we can’t write $f:X \to Y$. If such a subset $D$ exists and is not $X$, then that implies $f$ is not a function in the first place.

My question is: is this definition flawed? Or, when written in arrow notation, is it acceptable for $X$ to be a set containing the domain; even if it is not the domain of the function itself?

答:
Mathematical notations are mostly a matter of conventions and ultimately the understanding of notations depends very much on the context.

Sometimes people write $f:X\to Y$ while $X$ means a set that contains the domain of $f$ and the author explicitly denotes the domain of $f$ with $D$, which is a proper subset of $X$. Sometimes people make a combination and write something like
$$
f:D(f)\subset \mathcal{H}\to\mathcal{H}
$$
where $D(f)$ denotes the domain of $f$.

Mostly when people write $f:X\to Y$, $X$ means the domain of $f$.

If you assume in the first place that $X$ is the domain of $f$, and write $f:X\to Y$, then it never happens that "$X$ contains an element not in the domain of $f$".

If $B$ is a proper subset of $X$, which is assumed to be the domain of $f$, then it is true that $X$ has an element that is not in $B$. This does not contradict anything about the definition of the domain of $f$: your domain of $f$ is $X$, not $B$.

Think about the following example.

Suppose $f:[0,1]\to\mathbb{R}$ is a function defined on $X=[0,1]$. Then

$B=[0,1)$ is a (proper) subset of $X$.

$X$ contains an element that is not in $B$, i.e., $x=1$.


问:
How does composition in the category Rel work if not all relations are transitive?
My understanding of categories is that composition is always possible: if you have an arrow f from A to B and an arrow g from B to C, then you also have a third arrow from A to C that represents the composition of f and g. However, I'm confused about what this means in the category Rel if some of the relations are not transitive. If f represents a non-transitive relation between A and B, and g represents a non-transitive relation between B and C, then the composition of these relations might include pairs of elements that aren't actually related. What am I misunderstanding here?

Edit: Is the composition just the null set?

Edit 2: I see where I went wrong here: I was treating the arrows of Rel as itself a relation on the set of objects in the category, in which case composition would make it into a transitive relation. But each arrow represents a different relation, and transitivity is a property of a single relation, not a collection of them. Whoops. (Also, now I realize why it makes no sense to speak of a heterogenous relation being transitive.)

答:
I don't know what you mean by a transitive relation between two sets. The composition is a new relation, which is different from the previous relations (in general); in particular, even if $A = B = C$, it may relate new pairs of elements that neither the original relations relate.

One way to connect this to transitivity is that a relation $R$ on a set $X$ is transitive iff the composition $R \circ R$ is contained in $R$ (in the sense of subset containment). Moreover we can calculate the transitive closure of $R$ (the smallest transitive relation containing $R$) by taking the union over all composites $\bigcup_{n \ge 1} R^{\circ n}$.


问:
Integrating the Gaussian Distribution, Error Function Confusion
I am tasked with the following problem for a homework assignment:

Evaluate the following integrals by hand and check your results on Mathematica.

Gaussian integral: $\int_{0}^\infty e^{-(x-\mu)^2 / (2\sigma^2)} \, dx$, assume $\mu$ and $\sigma^2$ are real scalars and $\sigma^2 > 0$.

My question pertains to the end result of this integration. Mathematica lists the result as
\begin{equation}
\sigma\sqrt{\dfrac{\pi}{2}} \left( 1 + Erf\left(\dfrac{\mu}{\sigma \sqrt{2}}\right) \right)
\end{equation}

I'm having some trouble getting this same result, and I believe it's because I don't fully understand the error function. Below I lay out my work and show my final answer, which is similar to the above, but there are some glaring differences.

The below text is my completed solution:

Firstly, take $a = 1/(2\sigma^2)$. Take
\begin{align}
I
&=
\int_{0}^\infty e^{-a(x-\mu)^2} \, dx
\nonumber
\\[.5em]
I^2
&=
\left[ \int_{0}^\infty e^{-a(x-\mu)^2 } \, dx  \right]^2
\nonumber
\\[.5em]
&=
\left( \int_{0}^\infty e^{-a(x-\mu)^2} \, dx \right) \left( \int_{0}^\infty e^{-a(y-\mu)^2} \, dy \right)
\nonumber
\end{align}
In the above, $y$ is simply a dummy variable for the sole purpose of easing the difficulty of integration. Now, perform a change of variables such that $\beta = x - \mu$, $\gamma = y - \mu$, $dx = d\beta$, $dy = d\gamma$. Note that with this substitution, our limits of integration change. When $x=0$, $\beta = \mu$ (similarly for $y$). So,
\begin{align}
I^2
&=
\int_{0}^\infty \int_{0}^\infty
    e^{-a(x-\mu)^2} e^{-a(y-\mu)^2 }
\, dx dy
\nonumber
\\[.5em]
&=
\int_{-\mu}^\infty \int_{-\mu}^\infty
    e^{-a(\beta^2 + \gamma^2)}
\, d\beta d\gamma
\nonumber
\end{align}
Changing to polar coordinates, we have $\beta = r\cos\theta$, $\gamma = r\sin\theta$, and $d\beta d \gamma = r dr d\theta$ such that,
\begin{align}
I^2
&=
\int_0^{2\pi} \int_{\mu}^\infty
r e^{-a r^2}
dr d\theta
\nonumber
\\[.5em]
& \text{Take } u = r^2 \text{ such that } du = 2r dr
\nonumber
\\[.5em]
&=
2\pi \int_{\mu^2}^\infty
\dfrac{e^{-a u}}{2}
du
\nonumber
\\[.5em]
&=
\pi \left[- \dfrac{e^{-u / (2\sigma^2)}}{a} \right]_{\mu^2}^\infty
\nonumber
\\[.5em]
&=
\dfrac{\pi}{a} \, e^{-a\mu^2}
\nonumber
\end{align}

Therefore,
\begin{equation}
\boxed{
I
=
\int_{0}^\infty e^{-(x-\mu)^2 / (2\sigma^2)} \, dx
=
\sigma \sqrt{2\pi} \, e^{-\mu^2 / (4\sigma^2)}
}
\end{equation}

I know how the error function is defined as an integral of an exponential, and the $\mu/(\sigma\sqrt{2})$ seen in the Mathematica solution seems to almost match the term in the exponential of my solution, but I'm still very confused on how to get the error function in the solution. Any help is appreciated, thank you.

答:
You wrote that you need to change the limits of integration when you change the variables, but you didn't do it.

....

Now, perform a change of variables such that $\beta = x - \mu$, $\gamma = y - \mu$, $dx = d\beta$, $dy = d\gamma$. Note that with this substitution, our limits of integration change. When $x=0$, $\beta = -\mu$ (similarly for $y$). So,
\begin{align}
I^2
&=
\int_{0}^\infty \int_{0}^\infty
    e^{-a(x-\mu)^2} e^{-a(y-\mu)^2 }
\, dx dy
\nonumber
\\[.5em]
&=
\int_{-\mu}^\infty \int_{-\mu}^\infty
    e^{-a(\beta^2 + \gamma^2)}
\, d\beta d\gamma
\nonumber
\end{align}
...

I don't think polar coordinates are going to help you with this one.


问:
Simplified form of a combinatoric sum : $\sum_{k=1}^n k^{n-k} \binom {n}{k}$
Let $E$ be a set, such that $|E|=n$. We consider the set $I_n=\{ f:E\to E \; | \; f\circ f = f\}$ and for all $k=1,\dots,n$, the set $A_k=\{f \in I_n$ : $f$ has exactly $k$ fixed points}.

First, I needed to determine the cardinality of $A_k$, so I considered the surjective mapping :
\begin{align}
\Psi :& A_k \to C^k(E) \\
      &f \mapsto\{x\in E  : f(x)=x\}
\end{align}
such that $C^k(E)$ is the set of the subsets of $E$ that have $k$ elements, note that $|C^k(E)|=\binom{n}{k}$.

For all $K=\{x_1,\dots,x_k\}$ in $C^k(E)$, we can construct a preimage of $K$ by creating a function that maps all the $x_i$ to themselves, and have it map all the other elements of $E$ to $K$ (we can show that $f(E)=\Psi(f)$) so in other words :
\begin{equation}      \Psi^{-1}(K)=\{f : f|_K=Id_K \; \text{ and }\; f|_{E-K}\in K^{E-K} \}
\end{equation}

where $K^{E-K}$ is the set of mappings from $E-K$ to $K$. Therefore, $|\Psi^{-1}(K)|=|K^{E-K}|=k^{n-k}$. Therefore :
\begin{equation}
|A_k|=\left|\bigcup\limits_{K\in C^k(E)}\Psi^{-1}(K)\right|=\sum_{K\in C^k(E)} |\Psi^{-1}(K)| =\sum_{K\in C^k(E)}k^{n-k}=k^{n-k} |C^k(E)|=k^{n-k} \binom{n}{k}
\end{equation}
And since the $A_k$'s form a partition of $I_n$, so :
\begin{equation}
|I_n|=\left|\bigcup_{k=1}^n A_k \right|=\sum_{k=1}^n |A_k|=\sum_{k=1}^n k^{n-k} \binom{n}{k}
\end{equation}
So, this is where this sum comes from. I tried using that $k\binom{n}{k}=n\binom{n-1}{k-1}$, but it seems to get complicated pretty quickly, so I wonder if this sum has a simplified form, or rather I made a mistake in my reasoning and got the wrong expression. Any help is welcome.

答:
We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write for instance
\begin{align*}
k![z^k]e^{qz}=k![z^k]\sum_{j=0}^\infty \frac{(qz)^j}{j!}=q^k\tag{1}
\end{align*}

We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^n\binom{n}{k}k^{n-k}}&=\sum_{k=1}^n\binom{n}{k}(n-k)![z^{n-k}]e^{kz}\tag{2}\\
&=[z^n]\sum_{k=1}^n\frac{n!}{k!}z^ke^{kz}\tag{3}\\
&\,\,\color{blue}{=n![z^n]\sum_{k=1}^n\frac{\left(ze^z\right)^k}{k!}}\tag{4}\\
\end{align*}

Comment:

In (2) we use the coefficient of operator according to (1).

In (3) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

In (4) we see we have the first terms of the series expansion of $ze^z$ at $0$.

We find in OEIS the entry A000248 with synopsis Expansion of e.g.f. $\exp(x\exp(x))$ which corresponds to (4) . We do not expect any further simplification of this expression.


问:
Quadratic Formula problem?
There is a right triangle. The hypotenuse is 17 units. The sum of the other two sides is 23. Find the length of the two other sides.

Thanks for everyone's help in advance!

答:
Well, we know that
$$a+b=23 \qquad\text{and}\qquad a^2+b^2=17^2=289.$$
Substituting $b=23-a$ into the second equation will give you a quadratic equation in $a$. It's quite likely that of the two solutions to that quadratic equation, only one will be positive. There's your value of $a$, and then $b=23-a$ recovers $b$.</p>

Title: Quadratic Formula problem?


问:
Proving a function is operation preserving
I am given the task to prove that f is operation preserving. $f(e^{2πix/n})= [x]_n$ from $G$ $\to$ $\mathbb Z_n$ where $G$ is the group of nth roots of unity under complex multiplication and $\mathbb Z_n$ is the group of integers mod $n$ under modular addition.

I understand that to show a function is operation preserving, I need to show $f(g\cdot h)=f(g)*f(h)$ where "$\cdot$" is the first group's operation and $*$ is the second group's. In my case, "$\cdot$" is complex multiplication and "$*$" is modular addition. The issue I have is that the input of $f$ is not just $x$, but $e^{2πix/n}$, so I am not sure what I'm supposed to replace the $x$ from $[x]_n$ with

答:
At issue here is essentially whether $f$ is well-defined in the first place, that is, is independent of the choice $x$ that gives a particular value $e^{2 \pi i x / n}$; more explicitly, this means that for any $x, y$ such that $e^{2 \pi i x / n} = e^{2 \pi i y / n}$ we need to ensure that $[x]_n = [y]_n$. Once we know that $f$ is well-defined, then checking whether it preserves the operation is just a matter of using the definition of $f$, namely, checking that
$$f(e^{2 \pi i x / n} \cdot e^{2 \pi i y / n}) = f(e^{2 \pi i x / n}) +_n f(e^{2 \pi i y / n}) .$$

Remark Another way to think about this is as follows: We're checking that the map $q_n : (\Bbb Z, +) \to (\Bbb Z_n, +_n)$, $a \mapsto [a]_n$ is constant on the fibers of the map $\epsilon: (\Bbb Z, +) \to (G, \,\cdot\,)$, $a \mapsto e^{2 \pi i a / n}$. It is, so there is a map $F : G \to \Bbb Z_n$ satisfying $F \circ \epsilon = q_n$, and unwinding definitions we see that the latter statement is just that the induced map $F$ is just the given map $f$. In fact, if we already know that $q_n$ and $\epsilon$ preserve operations, then we can conclude that $f$ does, too.


问:
$C^0$-close of a homeomorphism
Let $f, g:X\to X$ be homeomorphism. We say that $f$ is $\delta$-close to $g$ if $d(f(x), g(x))<\delta$ for all $x\in X$.

Let $X=\{(x, y): x^2+y^2=1\}\bigcup \{(x, y): (x-2)^2+y^2=1\}$ i.e. $X= \mathbb{S}^1(0, 0)\cup \mathbb{S}^1(2, 0)$. Consider $f:X\to X$ such that $f(p)=p$ if $p\in\{(-1, 0), (1, 0), (3, 0)\}$ and the first coordinate of $f(x, y)$ is larger than
$x$. Indeed the homeomorphism $f$ has three fixed point and if $(x, y)\notin \{(-1, 0), (1, 0), (3, 0)\}$ and $f(x, y)=(a, b)$ then $a>x$.

Let $g:X\to X$ be close enough to $f$. Is it true that $g(\mathbb{S}^1(0, 0))= \mathbb{S}^1(0, 0)$ and $g(\mathbb{S}^1(2, 0))= \mathbb{S}^1(2, 0)$?

Would you please help me to know it.

答:
Each homeomorphism $h : X \to X$ keeps $p = (1,0)$ fixed. This is true because $p = (1,0)$ is the only point such that $X \setminus \{p\}$ is not connected.

Let $X_0 = S^1(0,0) \setminus \{p\}$ and $X_2 = S^1(2,0) \setminus \{p\}$. These are the components of $X \setminus \{p\}$. Thus each homeomorphism $h : X \to X$ either maps $X_0$ to $X_0$ and $X_2$ to $X_2$ or maps $X_0$ to $X_2$ and $X_2$ to $X_0$. Hence $h$ either maps $S^1(0,0)$ to $S^1(0,0)$ and $S^1(2,0)$ to $S^1(2,0)$ or maps $S^1(0,0)$ to $S^1(2,0)$ and $S^1(2,0)$ to $S^1(0,0)$.

Let $f : X \to X$ be a homeomorphism such that $f(-1,0) \in S^1(0,0)$ and let $d(f(-1,0),(1,0)) = r$. Note that $r > 0$ because $f(-1,0) \ne (1,0)$. Then if $d(f(x),g(x)) < r$ we have $g(S^1(0,0)) = S^1(0,0)$ and $g(S^1(2,0)) = S^1(2,0)$. In fact, we have $g(-1,0) \in S^1(0,0)$ since $f(-1,0) \in S^1(0,0)$ and $d(f(-1,0),g(-1,0)) < r$.

Title: $C^0$-close of a homeomorphism


问:
Find the polynomial that goes through points B and C and is tangent to $f(x) = (2/3)x+4$ at A, Interpolation
I have 3 points:

$$A = (0,4) \\
B=(-5,0) \\
C=(5,0)$$

I need to find a polynomial that goes through B and C, and is tangent to $f(x) = (2/3)x+4$ at A.

I know that tangent means it must be equal to the derivative of f(x) at that point.

This is probably wrong, but I did the interpolation using points B, C and $(0,2/3)$. I got

Help?

Body:
$P(x)=(x^2-25)(ax+b)$

$P(0)=-25b=4\iff b=\dfrac{-4}{25}$

$P'(x)=3ax^2-25a-\dfrac{-4x}{25}\Rightarrow P'(0)=-25a=\dfrac23\Rightarrow a=\dfrac{-2}{75}$


问:
Can you compute the average value of dependent probabilities?
I want to construct a formula to compute the expected value from a card game I am developing. Is this possible or will I have to use a Monte Carlo simulation?

The game has a concept of slots and objects to fill the slots, and your deck of shuffled, face down cards contains a mix of both types. Players turn over their cards one by one with slots added to the end of a row and objects placed on top of vacant slots.  Cards are continually revealed until an object is drawn with no slots free to place it.  To add further complexity the player can start with a number of open slots in addition to those cards in their hand.

So given I have o starting open slots, s slot cards, and n object cards in my hand what is the average number of objects each player could expect to reveal in any one turn?

答:
You can compute the expected number of slots covered recursively.  Let $p,q,$ and $r$ be respectively the number of open slots, the number of slot cards remaining in the deck, and the number of object cards remaining in the deck, and let $e_{p,q,r}$ be the expected number of slots covered in this round, exclusive of any slots that have already been covered in the round.  We have $$e_{p,q,r}=\begin{cases}
0,&r=0\\
r,&r\geq p\\
\frac r{q+r}(1+ e_{p-1,q,r-1})+\frac q{q+r}e_{p+1,q-1,r},&\text{otherwise}
\end{cases}$$

It just becomes a matter of arranging the calculations in a convenient order, and eventually computing $e_{0,s,n}/$.  If I were doing this, I would do it first for a small deck, and check the theoretical against a simulation.  I would convince myself that the program was correct before running it with the true values of $s$ and $n$.


问:
Connections between Mersenne Primes and Codes
I recently watched the video for the talk The Biggest Known Prime Number (slides located here) by Keith Conrad, which is on Mersenne primes $M_n = 2^n-1$.
It is well-known that such numbers can be prime only if $n$ is prime, but that this condition is not sufficient for primality of $M_n$.
The smallest counterexample is $M_{11} = 2^{11} -1 = 23\cdot 87$.
Conrad's talk includes the remark:

That $2^{11}−1$ has factor 23 is related to an important construction in coding theory called the perfect binary Golay code(a special subset of 23-dimensional space over integers mod 2)

My questions are:

What is the relation precisely?

Are there further relations between divisors of "Mersenne composites" and coding-theoretic constructions?

答:
Well, there is a direct connection between composite $M_n$'s and coding theory.
Such a composite Mersenne number exists iff there is a (non-degenerate) binary irreducible cyclic code of dimension n and length a non-trivial divisor of $M_n.$ Of course, not all irreducible cyclic codes are as exciting as the $[23,11],$ the dual of the perfect binary Golay!


问:
Is the distribution of a random variable determined by its partial integrals?
Let $(\Omega_i, \mathcal{F}_i, \mathbb{P}_i)$ be a probability space for $i \in \{1,2\}$, and let $X_i:\Omega_i\rightarrow[-\infty, \infty]$ be a $\mathbb{P}_i$-integrable random variable, respectively. If $\int_{\{X_1 \in A\}} X_1\ d\mathbb{P}_1 = \int_{\{X_2 \in A\}} X_2\ d\mathbb{P}_2$ for every Borel set $A$, does it follow that $X_1 \overset{d}{=} X_2$?

答:
Yes. Let $\mu_1$  and $\mu_2$ be the measures induced by $X_1$ and $X_2$. Then $\int_A xd\mu_1(x)=\int_A xd\mu_2(x)$ for every Borel set $A$. This implies that $xd\mu_1(x)=xd\mu_2(x)$ which implies $\mu_1$ and $\mu_2$ coincide on $\mathbb R \setminus \{0\}$. But $\mu_1$  and $\mu_2$ are probability measures so we necessarily have $\mu_1(\{0\})=\mu_2(\{0\})$ also.

Here is  a proof without using signed measures:

Lemma

Let $\mu$ and $\nu$ be two measures on the Borel sets of $\mathbb R$ such that $\int_A fd\mu=\int_Afd\nu$ for all Borel sets $E$ where $f$ is a non-negative meaurable function. Then $\mu (A)=\nu (A)$ for any Borel set contained in $\{x: f(x) >0\}$.

Proof: Taking linear combinations we get $\int fgd\mu=\int fg\nu$ for any simple function $g$. By Monotne Convergenec Theorem it follows that the equation holds for all non-negative measuarble functions $g$ (since we can express such a fucntion as an increasing limit of a sequence of simple functions). Let $E$ be any Borel set. Taking $g(x)=\chi_E\frac 1 {f(x)}$ of $f(x) >0$ and $0$ if $f(x)=0$ we see that $\mu(E\cap F)=\nu (E \cap F)$ where $F=\{x: f(x) >0\}$. Thsi proves te lemma.

Using this lemma we can complete the proof easily. Apply the lemma to restrictions of  $\mu_1$ and $\mu_2$ to $(0,\infty)$ to see that $\mu_1(E)=\mu_2(E)$ for any Borel set $E$ contained in $(0,\infty)$. Similarly, the same holds for any Borel set $E$ contained in $(-\infty,0)$. Finally, we get $\mu_1(\{0\})=\mu_2(\{0\})$ by the argument above.


答: From the Pythagorean theorem, you have $a^2+b^2=c^2$ with $c=17, a+b=23$.  This is three equations in three unknowns.</p>

Title: Quadratic Formula problem?

Body:
There is a right triangle. The hypotenuse is 17 units. The sum of the other two sides is 23. Find the length of the two other sides.

Thanks for everyone's help in advance!


问:
Simple Inverse Logic
I have the following statement: "Every number besides 0 has a unique multiplicative inverse."

I am curious as to why the following does not work:

$$\forall x \in \mathbb{R_{ \neq 0}} \exists y,z \in \mathbb{R_{\neq 0}} : ((xy = 1) \cap (xz =1)) \iff (y =z)$$

From my understanding, the following is correct, but am told that there needs to be universal quantifier for z instead. Why is this the case?

$$\forall x \in \mathbb{R_{ \neq 0}} \exists y,z \in \mathbb{R_{\neq 0}} :
(xy =1 ) \land (xz = 1) \implies (y = z) $$

I'm assuming the following doesn't hold because (y = z) can be false, while the other predicate is true:

$$\forall x \in \mathbb{R_{ \neq 0}} \exists y,z \in \mathbb{R_{\neq 0}} :
(y = z) \implies (xy =1 ) \land (xz = 1) $$

答:
$\exists y\exists z~(\varphi(y)\land\varphi(z)\leftrightarrow y=z)$ can be satisfied by the existence of two distinct entities where at least one does not satisfy the predicate $\varphi$.   If there is some $y,z$ where $y\neq z$ and at least $\lnot\varphi(y)$ or $\lnot\varphi(z)$, then these values do satisfy $\varphi(y)\land\varphi(z)\leftrightarrow y=z$.

In short, the fact that $(2\cdot 3=1\land 2\cdot 4=1)\leftrightarrow 3=4$ is valued as true does not suggest that either $3$ or $4$ are the multiplicative inverse of $2$.

You wish to state "For every (non-zero real) $x$, there is some $y$ that is its multiplicative inverse, and there is no other."

$$\forall x\in\Bbb R_{\neq 0}~\exists y\in\Bbb R_{\neq 0}~\forall z\in\Bbb R_{\neq 0}~\big(xy=1\land(xz=1\to z=y)\big)$$

Note: the parenthesis placement is important.

Title: Simple Inverse Logic


问:
Evaluate $\sum\limits_{r=1}^\infty(-1)^{r+1}\frac{\cos(2r-1)x}{2r-1}$
I would like to know how to evaluate $$\sum\limits_{r=1}^\infty(-1)^{r+1}\frac{\cos(2r-1)x}{2r-1}$$

There are a couple of issues I have with this. Firstly, depending on the value of $x$, it seems, at least numerically, that this value is always
$$\pm\sqrt{\frac{\sqrt5-1}{2}}$$
If this is true, how would I prove this?

Secondly, I tried using complex numbers to evalaute this, as shown briefly below:

$$C=\sum\limits_{r=1}^\infty(-1)^{r+1}\frac{\cos(2r-1)x}{2r-1}$$
$$S=\sum\limits_{r=1}^\infty(-1)^{r+1}\frac{\sin(2r-1)x}{2r-1}$$
$$\implies C+iS=\arctan e^{ix}$$
on using the power series of $\arctan x$. Differentiating yields a completely imaginary number:
$$\frac{i}{2}\sec x$$
which proves that the value of $C$ is a constant. However, this gives me no information whatsoever on the value of $C$.

Thank you for your help.

答:
As you noted, the real part of the sum, $C$, is constant no matter what the value of $x$ (so long as $x$ is real). Thus plug in $x = 0$

$$C = \operatorname{Re}(\arctan e^{i0}) = \frac{\pi}{4}$$


问:
Simple examples of undergraduate real analysis that cannot be written in the language of first-order logic
The answer to this question would of course depend on what one means by "undergraduate real analysis". So my question, asked by a layman in logic myself, may be vague. You may consider it as the real analysis in the usual undergraduate curriculum. (Maybe it is more appropriate to replace "real analysis" with just "undergraduate math" or something else.)

I do "know" a bit of ZFC set theory and basic first-order logic. But I have not taken a course in the foundation of mathematics before, neither an advanced course in mathematical logic. So I apologize in advance if my question sounds silly to experts.

Do we have a simple example (axioms/definitions/theorems) in real analysis that can not be written in the language of first-order logic?

答:
It is all depends on your context, where you are working in.

Let's say we are working in First order ZFC, our language includes all of the logical symbols, and in addition we have the one 2-ary relation symbol: ∈, and our axioms are, well the ZFC axioms.

In this settings, you can pretty much do everything there is in real analysis$^1$, both first and second order(and third an...), that is because ZFC is pretty strong, we can think about $ℝ$ as a special set we construct, then by looking at $\mathcal P(ℝ)$ and $\mathcal P\mathcal P(ℝ)$ we can talk about "high order real analysis".

Things starts to become more interesting when we are in the settings of real analysis, more specifically, let's look at the example of ordered fields:

Our language contains the logical symbols: $\cdot, +, <$(multiplication, addition, and order) and the axioms of ordered fields.

In this context we can find a lot of "natural" second order sentences, every sentence that quantify over subsets, over functions or over relations. For example: "for every subset of the field, the subset is either unbound(from above) or there exists a supremum to this subset", this is the "Least upper bound property", and it is a second order sentence.

It is a bit harder to show that this sentence cannot be written using first order sentence, but it is possible to prove that.

But wait, why in ZFC we are allowed to quantify over those subsets? Well, because we are not quantifying over subset of our universe, in the context of ordered fields, our universe is the field we are working in, but in ZFC our universe is some weird collection of a lot of sets, and just like in ordered fields, it is possible to find "natural" second order sentences that you cannot talk about without quantifying over arbitrary sub-collection of the universe, that is, you cannot talk about them in FOL in the context of ZFC. But even then, we can just take a stronger theory, such as NBG set theory, or TG set theory, which allow you to talk about second order ZFC using first order sentences.

$^1$by that I mean that we express most of the things we care about, it doesn't mean that there is nothing interesting to say about real analysis in a different context, other than ZFC.


问:
Evaluating $\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $
The question is $$\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $$ I know the answer is $\frac{1}{2}$ and I found it using this equality :

$$(\sqrt{x^2+1} - x)(x+1) = \frac{x+1}{\sqrt{x^2+1} + x}$$

But is there any other way to solve this? Any hints would be appreciated.

答:
Use binomial series:

$\sqrt{1+\dfrac1{x^2}}=1+\dfrac1{2x^2}\cdots$

Addendum with further explanation:

$\lim\limits_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $

$=\lim\limits_{x\to\infty}\left(\sqrt{1+\dfrac1{x^2}}-1\right)x(x+1)$

$=\lim\limits_{x\to\infty}\left(1+\dfrac1{2x^2}\cdots-1\right)(x^2+x)$

$=\lim\limits_{x\to\infty}\left(\dfrac1{2x^2}\cdots\right)(x^2+x).$

Can you take it from here?


问:
Find all integrable functions $f$ satisfying for every $I\subseteq \mathbb{R}$ , $\left|\int_I f(t)dt\right|\leq (m(I))^2$

Find all integrable functions $f:\mathbb{R}\to\mathbb{R}$ satisfying
that for every measurable set $I\subseteq\mathbb{R}$,

$\left|\int_I f(t)dt\right|\leq (m(I))^2$

where $m$ is the Lebesgue measure.

I have tried to do it by case, when $m(I)<1$ and when $m(I)\geq 1$; however, I have not progressed much. Any help is appreciate it.

答: Hint: Lebesgue differentiation theorem.</p>

Body: Hint: Lebesgue differentiation theorem.


问:
non-equivalence of $p$-norms in $\ell_1$

For $x\in \mathbb{R}^{\mathbb{N}}, $ define $\|x\|_\infty := \sup_{i\in\mathbb{N}}|x_i|$ and for $1\leq p < \infty,$ the $p$-norm $\|x\|_p$ is given by $\|x\|_p := (\sum_{i} |x_i|^p)^{1/p}.$ The $\ell_p$ space is the set of all sequences with real entries with finite $p$-norm. Two norms $\|\cdot \|_a$ and $\|\cdot \|_b$ are equivalent on a real vector space $V$ if there are $m, M > 0$ so that $m\|x\|_a \leq \|x\|_b \leq M\|x\|_a$ for all $x\in V.$

Show that for $1\leq p < r \leq \infty,$ the norms $\|\cdot \|_p$ and $\|\cdot \|_r$ on $\ell_1$ are not equivalent.

I know how to show that $\ell_1\subseteq \ell_p$ for each $1\leq p\leq \infty.$ I think I should deal with the case where $r=\infty$ first as that might be easier. Taking the negation of the definition of equivalence, I think I should show that $\forall m, M > 0,$ there exists $x \in \ell_1$ so that $m\|x\|_p > \|x\|_r$ or $\|x\|_r > M\|x\|_p.$ However, I'm not sure how to find this particular $x$ for each choice of $m$ and $M.$

答:
Hint: For $r <\infty$ consider $((\frac 1 1)^{1/p},(\frac 1 2)^{1/p},\cdots (\frac 1 N)^{1/p},0,0\cdots)$.
If $\|x\|_p \leq C\|x\|$ we get $(\sum\limits_{k=1}^{N} \frac 1 k)^{1/p} \leq C (\sum\limits_{k=1}^{N}\frac 1 {k^{q/p}})^{1/q}$ for all $N$,  a contradiction.
For $r=\infty$ consider $(1,1,...,0,0....)$. If $\|x\|_p \leq C \|x\|_{\infty}$ then we get $N ^{1/p} \leq C$ for all $N$, a contradiction.


问:
Equinumerous sets, power set

(ZF) Prove that for an arbitrary set A the following holds:

$ \bar{A} = \bar{A ∪  [{A}]} ⇒ \bar{P(A)} =\bar{(P(A) ∪ [P(A)])}$ ;

$ \bar{A} =  \bar{(A ∪ [A])} ⇒  \bar{P(P(A))} =  \bar{P(P(A) × P(P(A))} $.

What I have made so far is:

By def, A and B are equinumerous if there is a bijection f: A->B.

Let  $ x  \subset A =>x \subset \bar{A ∪ {A}} $.

Then $ Dom(f) \subseteq A$   and $Rng(f) \subseteq \bar{A ∪  [{A}]} $. Obviously this is injection since $x\subset A$ or $x \equiv sing A $.

But because $ x \in A \Rightarrow x\in P(A).$ But $f(x)\subseteq P(A)$ and $f(x)\subseteq A\Rightarrow f(x) \in P(A)\cup [A].$ Hence there is a function $g$ for which $g(f(x))$ is $P(A)\mapsto (P(A)∪[P(A)])¯  $;
And hence they are equinumerous.

We know that $ (X)^(B∪C)=X^B×X^C $. But afterwards I do not know what to do.

With [x] I denote the singleton of the set x.
I am sorry for the syntax, but I am a beginner in Math Text Editors.

I will be really thankful to anyone who share the ideas he/she has about the problem since I am stuck on it.
I tried to explain what I think that will be useful with the solution even though I am not quite sure what to do about 2.
Thanks in advance!

答:
I’m afraid that your argument for $(1)$ simply doesn’t make much sense. You start by saying that if $x\subseteq A$, then $x\subseteq\overline{\overline{A\cup\{A\}}}$. However, this makes no sense, because $\overline{\overline{A\cup\{A\}}}$ has not been given an independent meaning as a set: it’s just part of the notation that you’re using to say when there is a bijection between two sets. What is true, and what I suspect you meant, is that if $x\subseteq A$, then $x\subseteq A\cup\{A\}$.

You then conclude from this that $\operatorname{dom}f\subseteq A$ and $\operatorname{ran}f\subseteq\overline{\overline{A\cup\{A\}}}$. There are multiple problems with this, starting with the fact that you have not defined $f$. We can guess that you mean the embedding $f:A\to A\cup\{A\}$, though the fact that you were originally looking at subsets $x$ of $A$ rather than elements of $A$ actually suggests that $f$ is a map from $\wp(A)$ to $\wp(A\cup\{A\})$, in which case the domain of $f$ is not a subset of $A$ but rather of $\wp(A)$. And in any case you are once again treating $\overline{\overline{A\cup\{A\}}}$ as if it were a set.

Then you say that $x\in A$ implies that $x\in\wp(A)$; this is simply false. And I can make no sense of the last sentence (about $g$).

You want to show that if there is a bijection between $A$ and $A\cup\{A\}$, then there is a bijection between $\wp(A)$ and $\wp(A)\cup\{\wp(A)\}$. The natural way to start is:

Let $f:A\to A\cup\{A\}$ be a bijection.

The rest is completely revised and corrected.

To get a bijection between $\wp(A)$ and $\wp(A)\cup\{\wp(A)\}$ we can use the fact that there is an obvious bijection $s$ from $A$ to $\mathscr{S}=\big\{\{a\}:a\in A\big\}$: let

$$s:A\to\mathscr{S}:a\mapsto\{a\}\,.$$

The map $f$ is a bijection from $A$ to a set, $A\cup\{A\}$, that is $A$ together with one extra element, so we should be able to use it and $s$ to get a bijection from $\mathscr{S}$ to the set $\mathscr{S}\cup\{\wp(A)\}$, which is $\mathscr{S}$ together with one extra element. Specifically, let

$$\hat f=f\circ s^{-1}:\mathscr{S}\to A\cup\{A\}\,,$$

and let

$$g:A\cup\{A\}\to\mathscr{S}\cup\{\wp(A)\}:x\mapsto\begin{cases}
s(x),&\text{if }x\in A\\
\wp(A),&\text{if }x=A\,;
\end{cases}$$

$\hat f$, being a composition of bijections, is a bijection, $g$ is easily seen to be a bijection as well.

Now let

$$\begin{align*}
&h:\wp(A)\to\wp(A)\cup\{\wp(A)\}:\\
&\quad X\mapsto\begin{cases}
g\left(\hat f(X)\right),&\text{if }X\in\mathscr{S}\\
X,&\text{if }X\in\wp(A)\setminus\mathscr{S}\,;
\end{cases}
\end{align*}$$

it’s not hard to check that $h$ is also a bijection.

Title: Equinumerous sets, power set


问:
Correct arrow notation for function with natural domain
I saw a definition stating that given a function $f:X \to Y$ the natural domain of $f$ is the largest possible subset $D \subseteq X$ for which the rule defining $f$ is valid.

From what I understand, the $X$ in $f:X \to Y$ must be the domain of $X$. Let’s suppose $D$ is a subset of $X$ but is not $X$ itself, then that seems to imply that there is at least one element in $X$ not in $D$. But if $X$ contains an element not in the domain of $f$ then we can’t write $f:X \to Y$. If such a subset $D$ exists and is not $X$, then that implies $f$ is not a function in the first place.

My question is: is this definition flawed? Or, when written in arrow notation, is it acceptable for $X$ to be a set containing the domain; even if it is not the domain of the function itself?

答:
After thinking way too long about this I'd say scrub the definition as given as you are correct: the notation "$f:X\to Y$ is function" does imply $X$ is the domain and the definition doesn't make sense.

But replace it with this:

If $f$ is a rule that can map some elements of $X$ to $Y$, and then the natural domain is the largest subset $D\subseteq X$ so that $f:D\subseteq X \to Y$ is a valid function. (i.e. in that $D$ is an acceptable domain and all elements of $D$ can be mapped and all elements of $X\setminus D$ can't be.)

That way we avoid saying the incorrect "$f:X\to Y$ is a function" (it isn't... that fails the definition of function) and avoid any weird and awkward "$f:X\to Y$ but not as function maps... we just mean some of the $X$ can be mapped"

========

I guess what the book is getting at is if you take a naive idea of function being a "rule" and there is some universal set we are working in then saying things like

$f(x) = \frac {x^2 +7}{x-3}$ or $h(x) = \arcsin x$ or $j(x) = \sqrt x$

and we make the naive and blanket statement that all theses are real functions and map $\mathbb R$ to $\mathbb R$.

Then the "unnatural" super-general-not-much thought put into "domain" of all these functions are $\mathbb R$.

ANd so the natural domain is the one's where the function actually works.  "Natural" domain of $f(x)$ is $\mathbb R\setminus \{3\}$.  Naural domain of $h$ is $[-1,1]$. and natural domain $j(x)$ is $[0,\infty)$.

But the thing is a real "grown-up" mathematician would never say $\mathbb R$ is the domain.  They'd simply say the if $f(x)$ isn't valid then $x$ is simply not in the domain. Period.

Unless.... well if this is practical math text.

Suppose I told you "I have function and it maps reals to reals and the function is $f(x) = \sqrt{x^3 - 57x} + \frac 1{\sqrt{431- x^2}}$ and I need to know what the domain is" well, the technical answer is the domain can be what I want it to be.  If I want the domain to be $\{0,-1, -2,-3\}$ so that $f(x)= \begin{cases}\frac 1{\sqrt {431}}&x=0\\\sqrt{56}+\frac 1{\sqrt{430}}&x=-1\\\sqrt{106}+\frac 1{427}&x=-2\\12 +\frac 1{\sqrt{422}}&x=-3\end{cases}$

well, then I can have $f:\{0,1,2,3\}\to \mathbb R$ via $f(x) =\sqrt{x^3 - 57x} + \frac 1{\sqrt{431- x^2}}$.  But that's not what I meant.

So then I say "C'mon.... you know what I mean.... I want the biggest set of real numbers that can be the domain.... I want the....um, let's call it 'natural domain'" and then .. well that is a legitimate question.

To define the function $f: X \to \mathbb R$ I must specify what the domain $X$ is.

But do I really have to?  Do I really have to say... "well we need $x^3 - 57 \ge 0$ and we need $431 -x^2 > 0$ so the means $x$ must be between...."  Do I even care?

Why shouldn't I just say:  Let $f: NaturalDomain(\sqrt{x^3 - 57x} + \frac 1{\sqrt{431- x^2}})\subset R\to \mathbb R$. and not really worry about knowing what values of $x$ that $ \sqrt{x^3 - 57x} + \frac 1{\sqrt{431- x^2}}$ id defined on?


问:
Solving an equation that contain floor function
Let suppose I have three positive integers $a, b,c$ and one unknown $x$ ($x$ is also a positive integer). Solve for the smallest $x$ that satisfies this equation

$$\left \lfloor\frac{x}{a} \right\rfloor + \left \lfloor \frac{x}{b} \right\rfloor \geq c$$

where $\lfloor x\rfloor$ is the floor function

For instance let $a = 1$, $b=1$ and $c=5$ in this case

$$\left \lfloor x \right\rfloor + \left \lfloor x \right\rfloor \geq 5$$

here the smallest $x$ is $3$.

For another case, if $a=3$ and $b=5$ and $c=4$ we obtain

$$\left \lfloor\frac{x}{3} \right\rfloor + \left \lfloor \frac{x}{5} \right\rfloor \geq 4$$

here the smallest $x$ is $9$

My question is, can we find the smallest $x$ exactly ?

答:
Assume that $a,b,c$ are positive
.

Let $y=\frac{abc}{a+b}$. Then $$\frac{y}{a}+\frac{y}{b}=c.$$

Therefore $$\left \lfloor\frac{y}{a} \right\rfloor + \left \lfloor \frac{y}{b} \right\rfloor$$ will be either $c$ or $c-1$.

If it is $c$

Then $y$ is a solution. We will still have a solution when $y$ is increased, until $y$ becomes the next multiple of either $a$ or $b$.

If it is $c-1$

Then increase $y$ until it becomes the next multiple of either $a$ or $b$. This will be the solution $x$ unless it is a multiple of both $a$ and $b$ in which case there is no solution. Given that we have a solution $x$, then we can increase it as above until the next multiple of either $a$ or $b$ is reached.

Example

Solve $\left \lfloor\frac{x}{5} \right\rfloor + \left \lfloor \frac{x}{7} \right\rfloor=8.$

We obtain $y=70/3$ and then $\left \lfloor\frac{y}{5} \right\rfloor + \left \lfloor \frac{y}{7} \right\rfloor=7.$

As $y$ is increased, the next two multiples of $5$ or $7$ are $25$ and $28$. The solution is $$25\le x<28.$$

Is this the sort of practical answer you were looking for?

Summary

The solutions can be completely expressed in terms of how $\frac{abc}{a+b}$ fits between successive terms of the ordered sequence of elements of $aZ\bigcup bZ$.

A formula for the revised question

A formula for the minimal $x$ is obtained as in the above answer and can be expressed as follows.

If $\frac{abc}{a+b}$ is in $aZ\bigcup bZ$ then that is the minimal $x$.

Otherwise $x=$ min$ (a\left \lfloor\frac{bc}{a+b}\right\rfloor+a,b\left \lfloor\frac{ac}{a+b}\right\rfloor+b)$.


问:
How to calculate $\int_{C}{}\frac{\bar{z}}{z+i} dz$ if $C$ is a circle: $|z+i|=3$?
The task:
$$
\text{Calculate } \int_{C}{}f(z)dz\text{, where } f(z)=\frac{\bar{z}}{z+i}\text{, and } C \text{ is a circle } |z+i|=3\text{.}
$$
Finding the circle's center and radius:
$$
|z+i|=|x+yi+i|=|x+(y+1)i|=3
\\
x^2+(y+1)^2=3^2
$$
Parametrizing the circle:
$$
z(t)=i+3e^{-2\pi i t}
$$

Now I need to calculate this integral:
$$
\int_{0}^{1}{f(z(t))z'(t)dt}=
2\pi\int_{0}^{1}{
\frac{1-3ie^{-2\pi i t}}{e^{4 \pi i t}}dt
}
$$

Unfortunately I calculated this integral, and it's equal to $0$. Is this correct? I don't think so. Where did I go wrong? Maybe I made a mistake when calculating the integral - what would be the best way to calculate it?

答:
You cannot directly use the Residue Theorem in this case since $\frac{\overline{z}}{z+i}$ is not holomorphic, so your approach is entirely reasonable. Your calculation of the integral is correct since

$$\int_{0}^{1}e^{2k\pi ti}dt=\left.\frac{1}{2\pi ki}e^{2k\pi kti}\right|_0^{1}=\frac{1}{2\pi ki}(1-1)=0$$

for any integer $k$, and so

\begin{align*}
2\pi\int_{0}^{1}\frac{1-3ie^{-2\pi it}}{e^{4\pi i t}}dt&=2\pi\int_{0}^{1}e^{-4\pi i t}dt-6\pi i\int_{0}^{1}e^{-6\pi it}dt\\
&=2\pi(0)-6\pi i(0)\\
&=0
\end{align*}


问:
Does "$G$ contains no $K_n$" imply "$G$ is $n-$colorable"?
This question occurs to me while I was thinking about the Four Color Theorem and Five Color Theorem. If we can prove that "$G$ contains no $K_n$" (where $K_n$ is the complete graph with $n$ nodes) implies "$G$ is $n-$colorable," then we can show that any planar graph is $5-$colorable, as any planar graph contains no $K_5$.

I would assume that this question is already answered somewhere, but I couldn't find anything.

Thanks for any help!

答:
The clique number $\omega(G)$ of a graph is the largest size of a clique in it. Your question is equivalent to asking whether $\chi(G) \le \omega(G) + 1$ where $\chi(G)$ is the chromatic number. Unfortunately this is known to be false: in particular, it's known that there exist triangle-free graphs (hence $\omega(G) = 2$) with arbitrarily large chromatic number; see, for example, here.

A variant of this problem where we ask for $K_n$ minors rather than $K_n$ subgraphs is apparently an open problem, the Hadwiger conjecture.


答: By decomposing an interval $I$ into $n$ subintervals of equal length, and summing the given inequality on them, you obtain that $|\int_I f| \le m(I)^2/n \;$ so $\int_I f=0$. From this it follows (e.g. by differentiating the integral $\int_0^x f$) that $f=0$ a.e.

Title: Find all integrable functions $f$ satisfying for every $I\subseteq \mathbb{R}$ , $\left|\int_I f(t)dt\right|\leq (m(I))^2$

Body:
Find all integrable functions $f:\mathbb{R}\to\mathbb{R}$ satisfying
that for every measurable set $I\subseteq\mathbb{R}$,

$\left|\int_I f(t)dt\right|\leq (m(I))^2$

where $m$ is the Lebesgue measure.

I have tried to do it by case, when $m(I)<1$ and when $m(I)\geq 1$; however, I have not progressed much. Any help is appreciate it.


问:
Cardinality of intersection of sets
Consider the following problem: find $n(A \cap B)$ if $n(A)=10$, $n(B)=13$ and $n(A \cup B) = 15$.

I know if I want to find the union I use the Cardinal Number formula:

$$n(A\cup B) = n(A) + n(B) - n(A\cap B)$$

But how do I do it the other way: to find $n(A\cap B)$?

Would it be $n(A\cap B) = n(A) + n(B) - n(A\cup B)$ ?

答:
Yes... $n(A\cap B) = n(A) + n(B) - n(A\cup B)$ is the way to go.

$n(A\cap B) = 8$ according to this.

A little Venn to visualize the formula:

Title: Cardinality of intersection of sets


问:
Prove that $p+\frac{2p-1}{ \left(\frac{1-p}{p}\right)^n -1} \leq \frac{1}{2} - \frac{1}{2n}$, if $0<p<1, n\in \mathbb N$

Prove that $p+\frac{2p-1}{ \left(\frac{1-p}{p}\right)^n -1} \leq \frac{1}{2} - \frac{1}{2n}$, if $0<p<1, n\in \mathbb N$

This is from a recently closed question.

Notice that the fraction on the LHS is well defined. The denominator can be written as

$$\frac{1}{p^n} \left( (1-p)^n - p^n \right)=\frac{1-2p}{p^n} g(p)$$
where $g(p)$ is a polynomial of $p$ taking positive values. When $p=\frac 12$, the LHS becomes
$p-\frac{1}{2^n g(\frac 12)}$.

My effort was shown here with variable substitution and taking derivatives. I'd like to know if there's a simpler, elementary proof without calculus. Thanks.

答:
No calculus is required.

Using your substitution in your linked effort:

Let $x=\frac{1-p}{p}$ then $p=\frac{1}{1+x}, x>0$. The inequality becomes
$$\frac{1}{1+x}+\frac{1}{1+x}\cdot \frac{1-x}{x^n-1} = \frac{x-x^n}{(1+x)(1-x^n)} \le \frac1 2-\frac{1}{2n}$$
$$\iff g(x) = \frac{x-x^n}{(1+x)(1-x^n)} \le \frac 12 - \frac{1}{2n}\tag 1$$

From here, notice that

$$ \begin{array}{ll} g(x) & = &\frac{ x}{1+x} \times \frac{ 1 - x^{n-1 }} { 1 - x^n} \\
&=& \frac{x}{1+x} \times \frac{ 1 + x + \ldots + x^{n-2} } {1 + x + \ldots + x^{n-1} } \\
& =& \frac{ x + \ldots + x^{n-1}} { 1 + 2x + 2x^2 + \ldots + 2x^{n-2} + 2x^{n-1} + x^{n} }  \\
& =& \frac{1}{2} - \frac{1}{2} \times \frac{ 1+x^n} {1 + 2x + 2x^2 + \ldots + 2x^{n-2} + 2x^{n-1} + x^{n}} \end{array}$$

So, it remains to show that

$$  1 + 2x + 2x^2 + \ldots + 2x^{n-2} + 2x^{n-1} + x^{n} \leq n ( 1 + x^n). $$

This is obvious, since $ x^k + x^{n-k} \leq 1 + x^n$ for $ 0 \leq k \leq n$.

(EG by factoring, if you want to avoid convexity-calculus arguments ).

As you mentioned, equality holds iff $ x =1$ or $ n = 1$.


问:
Show that any subring $R\subseteq L$ is a field if $K\subseteq L$ is module-finite field extension and $K\subseteq R$..
This is Exercise 1.50b in Fulton's Algebraic Curves.
It's a follow up to showing that algebraic elements form a subfield.
I found similar questions MSE, but with stronger hypothesis.

By module-finite extension we mean that $L$ is finitely-generated as a $K$-module.

My attempt:
take $r\in R$ and let $\rho\in L$ be its inverse in the field $L$.
I tried to show that $\rho \in R$, since we know that will happen.

By $K\subset L$ a finitely generated extension we mean that there are $l_1,\dots l_n$ in $L$ such that $r\in R$ can written as $r = l_1k_1+\dots+l_n k_n$.

When $r = l_jk_j$ we have $l_j = rk_j^{-1}\in L$, so $\rho = k_j^{-1}l_j^{-1}\in R$.
This strategy blowed off really hard when the above equation for $r$ had two or more terms.

答: Proof: Note that since $L$ is a module finite extension of $K$, we have that $L$ is a finite field extension of $K$ which implies that $L$ is algebraic over $K$. Now, let $y \in R$. In particular, $y \in L$ which gives us that $y$ is algebraic over $K$. Let $P(x)=x^n+c_{n-1}x^{n-1}+ \dots c_{1}x+c_{0}$ be the minimum polynomial for $y$ over $K$. We then have $y^{n} +c_{n-1}y^{n-1} + \dots c_{1}y+c_{0}=0$, which gives $y^{n}+c_{n-1}y^{n-1} + \dots + c_{1}y=-c_{0}$. This implies, $y^{n-1}+c_{n-1}y^{n-2}+ \dots + c_{1} = y^{-1}c_{0}$. Clearly, $y^{-1}c_{0} \in R$ and since $c_{0}^{-1} \in R$, we get $y^{-1} \in R$.


问:
Abstract Algebra Finding All Orbits - Is my answers correct?
Hope you are well.

I'm posting the following as a favour on behalf of a friend who is not a native speaker, she just wants to confirm her answer is indeed correct, or if it could be answered in a more efficient manner.

I have uploaded an image of her question and her answer respectivly, sorry iam unable to write it out due to my lack of mathmatic symbolic knowledge.

Any input would be greatly appreciated, Thanks in advance!

答:
Her calculations are perfectly correct. To be honest, it could happen that there is precisely one orbit, in which case the action is called transitive.

Besides, if she has learned the orbit-stabilizer theorem, she can check that $$|{\rm orb}(1)|=\frac{|G|}{|{\rm Stab}(1)|}=\frac{6}{2}=3,$$ but the size of the underlying set $H$ is precisely $3$. That is, the orbit containing $1$ must be the whole set! This also indicates the uniqueness of orbits in this action.


问:
Show if $|z+w| = |z|+|w|$ then there exist $\lambda \geq 0$ such that $z = \lambda w$
Show if $|z+w| = |z|+|w|$ then there exist $\lambda \geq 0$ such that $z = \lambda w$.

I'm in my first complex variables course, cant use polar coordinates (haven't seen them) so how would you approach this?

答:
By definition, if $z=a+bi$ then $|z|=\sqrt{a^2+b^2}.$ So you can use this and just work through the algebra:

Let $z=a+bi$ and $w=c+di.$ Now assume $|z+w|=|z|+|w|,$ then

$$\sqrt{(a+c)^2 + (b+d)^2}=\sqrt{a^2+b^2}+\sqrt{b^2+d^2}$$
square both sides:
$${(a+c)^2 + (b+d)^2} = {a^2+b^2}+{c^2+d^2}+2\sqrt{(a^2+b^2)(c^2+d^2)}$$
continue to expand
$$\underbrace{a^2+c^2+b^2+d^2}+2ac+2bd=\underbrace{a^2+b^2+c^2+d^2}+2\sqrt{(a^2+b^2)(c^2+d^2)}$$
and cancel where one can:
$$ac+bd=\sqrt{(a^2+b^2)(c^2+d^2)}$$
square both sides again:
$$(ac)^2+(bd)^2+2abcd=(ac)^2+(bc)^2+(ad)^2+(bd)^2$$
and finally
$$2abcd=(bc)^2+(ad)^2 $$
and you should be able to draw the necessary conclusion from here.


问:
A Question about Doctoral Theses in Mathematics
This is most definitely a soft question, which I'm sure may get some negative attention, and perhaps even be voted closed. However, I genuinely would like to generate answers on this matter as it concerns anyone that decides to get a doctorate in mathematics. (Since I believe this is the best forum to do so, here goes...)

Question: Does a doctoral thesis in mathematics have to contain more than an abstract, a proposition and proof of a new and exciting result?

I strongly believe that brevity is beauty to mathematicians and see no problem with a thesis that contains a lengthy abstract (~350 words) to generate enthusiasm and then jumps right into the heart of a novel and far-reaching result. The proof provided is extremely condensed in the spirit of Zagier, leaving almost all details that can be found elsewhere to be found elsewhere, in the works cited. If the result has merit, I see no reason why it can't be submitted or published in such a manner. To support my stance, I offer the following condensed theses:

David Rector, "An Unstable Adams Spectral Sequence", MIT (1966), 9 pgs.

Burt Totaro, "Milnor K-Theory is the simplest part of algebraic K-theory", Berkeley (1989), 12 pgs.

Herman Buvik, A New Proof of Torelli's Theorem, NYU (1962), 12 pgs.

Eva Kallin "A non-local function algebra" Berkeley (1963), 13 pgs.

Edmund Landau "Neuer Beweis der Gleichung $\sum_{k = 1}^{\infty} \frac{\mu(k)}{k} = 0$", Berlin University (1899), 14 pgs.

Barry Mazur, "On Embeddings of Spheres", Princeton (1959), 26 pgs.

John F. Nash "Non-Cooperative Games", Princeton (1950), 27 pgs.

Kevin Walker, "An Extension of Casson's Invariant to Rational Homology Spheres", Berkeley (1989). 29 pgs.

Although I'd appreciate to hear from anyone worth giving his two cents, I'm specifically eager to hear from those senior members of the community.

Thanks!

答:
Eva Kallin is a colleague of mine from Brown, and she always expressed pride in the brevity of her thesis. In all my professional years, I never heard anyone disparage that dissertation. It seems to me that generally, mathematicians envy those people who have produced unusually short theses.

Remember, no mathematician is expected to show encyclopedic knowledge of mathematics, especially in the thesis that’s submitted.


问:
Find the Order of an Elliptic Curve
I have an Elliptic Curve represented by the following equation and values:

Elliptic Curve: y^2 = x^3 + A*x + B mod M

```<code>M = 93556643250795678718734474880013829509320385402690660619699653921022012489089
A = 66001598144012865876674115570268990806314506711104521036747533612798434904785
B = 25255205054024371783896605039267101837972419055969636393425590261926131199030
</code>```

What is the order of this Elliptic curve?

I thought it should be the same as the modulo, M?

In sage math, if I check the following, I get a different result:

```<code>F = FiniteField(M)
E = EllipticCurve(F, [A, B])

E.order() = 93556643250795678718734474880013829509196181230338248789325711173791286325820
</code>```

However, the value of M is:
93556643250795678718734474880013829509320385402690660619699653921022012489089

These values look similar but they are not the same.

So, I want to know the following:

How is the order of the Elliptic Curve calculated?
How is it related to modulus? Since there is a similarity between the two values?

Thanks.

答:
Here is a late answer to this rather sage-related triple question:

(1) How is the order of the Elliptic Curve calculated? (...in sage i suppose.)

(2) How is it (the order) related to modulus?

(3) (...) *a similarity between the two values?

Let us re-type the code, so that it comes with printed messages:

```<code>M = 93556643250795678718734474880013829509320385402690660619699653921022012489089
A = 66001598144012865876674115570268990806314506711104521036747533612798434904785
B = 25255205054024371783896605039267101837972419055969636393425590261926131199030
print(f"Is M prime, M = {M}...?\n{M.is_prime()}")

F = GF(M)
print(f"F is {F}")

E = EllipticCurve(F, [A, B])
# print(f"E is {E}")    # please decomment

ord = E.order()
print(f"The order of E is:\n{ord}\nFactorized:\n{ord.factor()}")
</code>```

The above delivers:

```<code>Is M prime, M = 93556643250795678718734474880013829509320385402690660619699653921022012489089...?
True
F is Finite Field of size 93556643250795678718734474880013829509320385402690660619699653921022012489089
The order of E is:
93556643250795678718734474880013829509196181230338248789325711173791286325820
Factorized:
2^2 * 3 * 5 * 7 * 137 * 593 * 24337 * 25589 * 3637793 * 5733569 * 106831998530025000830453 * 1975901744727669147699767
</code>```

So definitively, ord, the order of the curve, and the prime M, the characteristic of the field of definition for the curve $E$, differ.

To have an answer to (2) and (3) in the same time, let us go to:

Hasse's Theorem on Elliptic Curves (wiki page)

The notation $M$ is rather unusual for the prime, which is the characteristic of the base field for the curve. Usually, the notation is $p$ for such a prime, and in case the base field is finite, one denotes by $q$ the number of elements in this field, it is a power of $p$. So in our case, we would expect the relation
for the order $\# E(F)$ of the finite group $(E(F),+)$ of points of $E$, ( elliptic curve with affine equation $y^2 = x^3 + Ax + B$, $A,B\in F$), defined over the field $F=\Bbb F_p$ with $p=M$ elements:
$$
(p+1)-2\sqrt p\le \# E(F)\le  (p+1)-2\sqrt p\ .
$$
And indeed:

```<code>sage: p = M
sage: bool( (p+1) - 2*sqrt(p) <= ord )
True
sage: bool( (p+1) + 2*sqrt(p) >= ord )
True
</code>```

To see what happens in a "simple case", let us consider some small prime instead of $p$, for instance $p=11$ (in base ten). We build the list of all elliptic curves of the shape $y^2 = x^3+ax+b$, $a,b\in \Bbb F_{11}$, compute the order, and make a dictionary for which curves appear for which order. We expect cardinalities that differ from $(p+1)=12$ with "tolerance" $\pm 2\sqrt p=\pm2\sqrt{11}\approx 6.633\dots$.

Sage code (given, since it takes less space that the results):

```<code>p, F = 11, GF(11)
curves = [EllipticCurve(F, [a, b]) for a in F for b in F if (a/3)^3 + (b/2)^2]
orders = [E.order() for E in curves]
possible_orders = list(set(orders))
possible_orders.sort()
</code>```

And the possible_orders are:

```<code>sage: possible_orders
[6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]
</code>```

and all of them are between the bounds $12\pm6$. Note that any value occurs!
To have a quick statistics how the $110$ curves fit in this list of orders...

```<code>for ord in possible_orders:
    ord_ab = [(c.a4(), c.a6()) for c, o in zip(curves, orders) if o == ord]
    print(f"ord = {ord:>2} :: {len(ord_ab):>2} curves :: {ord_ab}")
</code>```

And we get:

```<code>ord =  6 ::  5 curves :: [(1, 8), (3, 10), (4, 2), (5, 6), (9, 7)]
ord =  7 ::  5 curves :: [(2, 7), (6, 6), (7, 2), (8, 10), (10, 8)]
ord =  8 :: 10 curves :: [(1, 9), (2, 10), (3, 3), (4, 5), (5, 4), (6, 7), (7, 6), (8, 8), (9, 1), (10, 2)]
ord =  9 :: 10 curves :: [(1, 4), (2, 2), (3, 5), (4, 1), (5, 3), (6, 8), (7, 10), (8, 6), (9, 9), (10, 7)]
ord = 10 :: 10 curves :: [(1, 10), (2, 5), (3, 7), (4, 8), (5, 2), (6, 9), (7, 3), (8, 4), (9, 6), (10, 1)]
ord = 11 ::  5 curves :: [(1, 5), (3, 9), (4, 4), (5, 1), (9, 3)]
ord = 12 :: 20 curves :: [(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9), (0, 10), (1, 0), (2, 0), (3, 0), (4, 0), (5, 0), (6, 0), (7, 0), (8, 0), (9, 0), (10, 0)]
ord = 13 ::  5 curves :: [(1, 6), (3, 2), (4, 7), (5, 10), (9, 8)]
ord = 14 :: 10 curves :: [(1, 1), (2, 6), (3, 4), (4, 3), (5, 9), (6, 2), (7, 8), (8, 7), (9, 5), (10, 10)]
ord = 15 :: 10 curves :: [(1, 7), (2, 9), (3, 6), (4, 10), (5, 8), (6, 3), (7, 1), (8, 5), (9, 2), (10, 4)]
ord = 16 :: 10 curves :: [(1, 2), (2, 1), (3, 8), (4, 6), (5, 7), (6, 4), (7, 5), (8, 3), (9, 10), (10, 9)]
ord = 17 ::  5 curves :: [(2, 4), (6, 5), (7, 9), (8, 1), (10, 3)]
ord = 18 ::  5 curves :: [(1, 3), (3, 1), (4, 9), (5, 5), (9, 4)]
</code>```

For short, the values of characteristic $p$ (in the OP $M$) of the base field $F=\Bbb F_p$ of the elliptic curve $E$, and the order of the group of its $F$-rational points $E(F)$, do not coincide, all we know in this generality is the double inequality (Hasse).

I can't to the same for $M$, but there are $M^2$ equations (and $M^2$ is huge) of the shape $y^2=x^3 +ax+b$ that can be written in characteristic $M$, some of them are leading to singular curves, (well, $\# F_M^\times =(M-1)$ is not a multiple of $3$, so $a\to a^3$ is injective, so for each $b$ there exists exactly one $a$ leading to a zero discriminant, so there are exactly $M$ singular curves...) but "a lot of them" are non-singular, thus elliptic, and they will fit somehow into the "small range" of possible orders between the two bounds $(M+1)\pm 2\sqrt M$.
The curve from the OP is one of them...

First question, (1) How does sage compute the order?

Well, we have to look into the code. This can be done for instance by having an instance of the class defining the method order, and asking for...

```<code>sage: E.order??
</code>```

we get a lot of information...

```<code>Signature: E.order(algorithm=None, extension_degree=1)
Docstring:
   Return the number of points on this elliptic curve.

   INPUT:

   * "algorithm" -- (optional) string:

     * "'pari'" -- use the PARI C-library function "ellcard".

     * "'bsgs'" -- use the baby-step giant-step method as
          implemented in Sage, with the Cremona-Sutherland version of
          Mestre's trick.

     * "'exhaustive'" -- naive point counting.

     * "'subfield'" -- reduce to a smaller field, provided that the
       j-invariant lies in a subfield.

     * "'all'" -- compute cardinality with both "'pari'" and "'bsgs'";
       return result if they agree or raise a "AssertionError" if they
       do not
</code>```

and so on, many, many doc string lines. We get then same information for the delegated work, this is the method E.cardinality, and we can immediately put the hands on the relevant piece of code:

```<code>try:
    return self._order
except AttributeError:
    pass

jpol = None
if algorithm is None:
    # Check for j in subfield
    jpol = self.j_invariant().minimal_polynomial()
    if jpol.degree() < self.base_field().degree():
        algorithm = "subfield"
    else:
        algorithm = "pari"

if algorithm == "pari":
    N = self.cardinality_pari()
elif algorithm == "subfield":
    if jpol is None:
        jpol = self.j_invariant().minimal_polynomial()
    N = self._cardinality_subfield(jpol)
elif algorithm == "bsgs":
    N = self.cardinality_bsgs()
elif algorithm == "exhaustive":
    N = self.cardinality_exhaustive()
elif algorithm == "all":
    N = self.cardinality_pari()
    N2 = self.cardinality_bsgs()
    if N != N2:
        raise AssertionError("cardinality with pari=%s but with bsgs=%s" % (N, N2))
else:
    raise ValueError("algorithm {!r} is not known".format(algorithm))

self._order = N
return N
</code>```

So the computation is...

if we have already computed the ordered (in the same sage session), then it is cached, i.e. stored in the attribute _order. If this attribute is set, we deliver it. Not the case in the OP.

then we check if the subfield algorithm should be used, this makes sense for "bigger fields" $\Bbb F_q$ with prime field $\Bbb F_p$ (so $q$ is a power of the characteristic $p$), and can be applied iff the $j$-invariant lives in a (strict) subfield of $\Bbb F_q$. Not the case in the OP. (We already work over the prime field.) So the code decides to use the algoritm = "pari".

and this delegates the work to the computation of the order N = self.cardinality_pari(). So the true work goes to this method, and the only line of code in it is

```<code>Is M prime, M = 93556643250795678718734474880013829509320385402690660619699653921022012489089...?
True
F is Finite Field of size 93556643250795678718734474880013829509320385402690660619699653921022012489089
The order of E is:
93556643250795678718734474880013829509196181230338248789325711173791286325820
Factorized:
2^2 * 3 * 5 * 7 * 137 * 593 * 24337 * 25589 * 3637793 * 5733569 * 106831998530025000830453 * 1975901744727669147699767
</code>```0

So sage uses the pari/gp associated object, from it the method ellcard.
It is as if we would require instead:

```<code>Is M prime, M = 93556643250795678718734474880013829509320385402690660619699653921022012489089...?
True
F is Finite Field of size 93556643250795678718734474880013829509320385402690660619699653921022012489089
The order of E is:
93556643250795678718734474880013829509196181230338248789325711173791286325820
Factorized:
2^2 * 3 * 5 * 7 * 137 * 593 * 24337 * 25589 * 3637793 * 5733569 * 106831998530025000830453 * 1975901744727669147699767
</code>```1

Note: Using the "other algorithms" (implemented in E.order) we get the same value. Code for them can be also inspected, this is the advantage of open source.

```<code>Is M prime, M = 93556643250795678718734474880013829509320385402690660619699653921022012489089...?
True
F is Finite Field of size 93556643250795678718734474880013829509320385402690660619699653921022012489089
The order of E is:
93556643250795678718734474880013829509196181230338248789325711173791286325820
Factorized:
2^2 * 3 * 5 * 7 * 137 * 593 * 24337 * 25589 * 3637793 * 5733569 * 106831998530025000830453 * 1975901744727669147699767
</code>```2

Title: Find the Order of an Elliptic Curve

Body:
Here is a late answer to this rather sage-related triple question:

(1) How is the order of the Elliptic Curve calculated? (...in sage i suppose.)

(2) How is it (the order) related to modulus?

(3) (...) *a similarity between the two values?

Let us re-type the code, so that it comes with printed messages:

M = 93556643250795678718734474880013829509320385402690660619699653921022012489089
A = 66001598144012865876674115570268990806314506711104521036747533612798434904785
B = 25255205054024371783896605039267101837972419055969636393425590261926131199030
print(f"Is M prime, M = {M}...?\n{M.is_prime()}")

F = GF(M)
print(f"F is {F}")

E = EllipticCurve(F, [A, B])
# print(f"E is {E}")    # please decomment

ord = E.order()
print(f"The order of E is:\n{ord}\nFactorized:\n{ord.factor()}")

The above delivers:

Is M prime, M = 93556643250795678718734474880013829509320385402690660619699653921022012489089...?
True
F is Finite Field of size 93556643250795678718734474880013829509320385402690660619699653921022012489089
The order of E is:
93556643250795678718734474880013829509196181230338248789325711173791286325820
Factorized:
2^2 * 3 * 5 * 7 * 137 * 593 * 24337 * 25589 * 3637793 * 5733569 * 106831998530025000830453 * 1975901744727669147699767

So definitively, ord, the order of the curve, and the prime M, the characteristic of the field of definition for the curve $E$, differ.

To have an answer to (2) and (3) in the same time, let us go to:

Hasse's Theorem on Elliptic Curves (wiki page)

The notation $M$ is rather unusual for the prime, which is the characteristic of the base field for the curve. Usually, the notation is $p$ for such a prime, and in case the base field is finite, one denotes by $q$ the number of elements in this field, it is a power of $p$. So in our case, we would expect the relation
for the order $\# E(F)$ of the finite group $(E(F),+)$ of points of $E$, ( elliptic curve with affine equation $y^2 = x^3 + Ax + B$, $A,B\in F$), defined over the field $F=\Bbb F_p$ with $p=M$ elements:
$$
(p+1)-2\sqrt p\le \# E(F)\le  (p+1)-2\sqrt p\ .
$$
And indeed:

sage: p = M
sage: bool( (p+1) - 2*sqrt(p) <= ord )
True
sage: bool( (p+1) + 2*sqrt(p) >= ord )
True

To see what happens in a "simple case", let us consider some small prime instead of $p$, for instance $p=11$ (in base ten). We build the list of all elliptic curves of the shape $y^2 = x^3+ax+b$, $a,b\in \Bbb F_{11}$, compute the order, and make a dictionary for which curves appear for which order. We expect cardinalities that differ from $(p+1)=12$ with "tolerance" $\pm 2\sqrt p=\pm2\sqrt{11}\approx 6.633\dots$.

Sage code (given, since it takes less space that the results):

p, F = 11, GF(11)
curves = [EllipticCurve(F, [a, b]) for a in F for b in F if (a/3)^3 + (b/2)^2]
orders = [E.order() for E in curves]
possible_orders = list(set(orders))
possible_orders.sort()

And the possible_orders are:

sage: possible_orders
[6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]

and all of them are between the bounds $12\pm6$. Note that any value occurs!
To have a quick statistics how the $110$ curves fit in this list of orders...

for ord in possible_orders:
    ord_ab = [(c.a4(), c.a6()) for c, o in zip(curves, orders) if o == ord]
    print(f"ord = {ord:>2} :: {len(ord_ab):>2} curves :: {ord_ab}")

And we get:

ord =  6 ::  5 curves :: [(1, 8), (3, 10), (4, 2), (5, 6), (9, 7)]
ord =  7 ::  5 curves :: [(2, 7), (6, 6), (7, 2), (8, 10), (10, 8)]
ord =  8 :: 10 curves :: [(1, 9), (2, 10), (3, 3), (4, 5), (5, 4), (6, 7), (7, 6), (8, 8), (9, 1), (10, 2)]
ord =  9 :: 10 curves :: [(1, 4), (2, 2), (3, 5), (4, 1), (5, 3), (6, 8), (7, 10), (8, 6), (9, 9), (10, 7)]
ord = 10 :: 10 curves :: [(1, 10), (2, 5), (3, 7), (4, 8), (5, 2), (6, 9), (7, 3), (8, 4), (9, 6), (10, 1)]
ord = 11 ::  5 curves :: [(1, 5), (3, 9), (4, 4), (5, 1), (9, 3)]
ord = 12 :: 20 curves :: [(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9), (0, 10), (1, 0), (2, 0), (3, 0), (4, 0), (5, 0), (6, 0), (7, 0), (8, 0), (9, 0), (10, 0)]
ord = 13 ::  5 curves :: [(1, 6), (3, 2), (4, 7), (5, 10), (9, 8)]
ord = 14 :: 10 curves :: [(1, 1), (2, 6), (3, 4), (4, 3), (5, 9), (6, 2), (7, 8), (8, 7), (9, 5), (10, 10)]
ord = 15 :: 10 curves :: [(1, 7), (2, 9), (3, 6), (4, 10), (5, 8), (6, 3), (7, 1), (8, 5), (9, 2), (10, 4)]
ord = 16 :: 10 curves :: [(1, 2), (2, 1), (3, 8), (4, 6), (5, 7), (6, 4), (7, 5), (8, 3), (9, 10), (10, 9)]
ord = 17 ::  5 curves :: [(2, 4), (6, 5), (7, 9), (8, 1), (10, 3)]
ord = 18 ::  5 curves :: [(1, 3), (3, 1), (4, 9), (5, 5), (9, 4)]

For short, the values of characteristic $p$ (in the OP $M$) of the base field $F=\Bbb F_p$ of the elliptic curve $E$, and the order of the group of its $F$-rational points $E(F)$, do not coincide, all we know in this generality is the double inequality (Hasse).

I can't to the same for $M$, but there are $M^2$ equations (and $M^2$ is huge) of the shape $y^2=x^3 +ax+b$ that can be written in characteristic $M$, some of them are leading to singular curves, (well, $\# F_M^\times =(M-1)$ is not a multiple of $3$, so $a\to a^3$ is injective, so for each $b$ there exists exactly one $a$ leading to a zero discriminant, so there are exactly $M$ singular curves...) but "a lot of them" are non-singular, thus elliptic, and they will fit somehow into the "small range" of possible orders between the two bounds $(M+1)\pm 2\sqrt M$.
The curve from the OP is one of them...

First question, (1) How does sage compute the order?

Well, we have to look into the code. This can be done for instance by having an instance of the class defining the method order, and asking for...

sage: E.order??

we get a lot of information...

Signature: E.order(algorithm=None, extension_degree=1)
Docstring:
   Return the number of points on this elliptic curve.

   INPUT:

   * "algorithm" -- (optional) string:

     * "'pari'" -- use the PARI C-library function "ellcard".

     * "'bsgs'" -- use the baby-step giant-step method as
          implemented in Sage, with the Cremona-Sutherland version of
          Mestre's trick.

     * "'exhaustive'" -- naive point counting.

     * "'subfield'" -- reduce to a smaller field, provided that the
       j-invariant lies in a subfield.

     * "'all'" -- compute cardinality with both "'pari'" and "'bsgs'";
       return result if they agree or raise a "AssertionError" if they
       do not

and so on, many, many doc string lines. We get then same information for the delegated work, this is the method E.cardinality, and we can immediately put the hands on the relevant piece of code:

try:
    return self._order
except AttributeError:
    pass

jpol = None
if algorithm is None:
    # Check for j in subfield
    jpol = self.j_invariant().minimal_polynomial()
    if jpol.degree() < self.base_field().degree():
        algorithm = "subfield"
    else:
        algorithm = "pari"

if algorithm == "pari":
    N = self.cardinality_pari()
elif algorithm == "subfield":
    if jpol is None:
        jpol = self.j_invariant().minimal_polynomial()
    N = self._cardinality_subfield(jpol)
elif algorithm == "bsgs":
    N = self.cardinality_bsgs()
elif algorithm == "exhaustive":
    N = self.cardinality_exhaustive()
elif algorithm == "all":
    N = self.cardinality_pari()
    N2 = self.cardinality_bsgs()
    if N != N2:
        raise AssertionError("cardinality with pari=%s but with bsgs=%s" % (N, N2))
else:
    raise ValueError("algorithm {!r} is not known".format(algorithm))

self._order = N
return N

So the computation is...

if we have already computed the ordered (in the same sage session), then it is cached, i.e. stored in the attribute _order. If this attribute is set, we deliver it. Not the case in the OP.

then we check if the subfield algorithm should be used, this makes sense for "bigger fields" $\Bbb F_q$ with prime field $\Bbb F_p$ (so $q$ is a power of the characteristic $p$), and can be applied iff the $j$-invariant lives in a (strict) subfield of $\Bbb F_q$. Not the case in the OP. (We already work over the prime field.) So the code decides to use the algoritm = "pari".

and this delegates the work to the computation of the order N = self.cardinality_pari(). So the true work goes to this method, and the only line of code in it is

  return Integer(self.__pari__().ellcard())

So sage uses the pari/gp associated object, from it the method ellcard.
It is as if we would require instead:

sage: Integer( E.__pari__().ellcard() )
93556643250795678718734474880013829509196181230338248789325711173791286325820

Note: Using the "other algorithms" (implemented in E.order) we get the same value. Code for them can be also inspected, this is the advantage of open source.

sage: E.order(algorithm='pari')
93556643250795678718734474880013829509196181230338248789325711173791286325820
sage: E.order(algorithm='bsgs')
93556643250795678718734474880013829509196181230338248789325711173791286325820
sage: E.order(algorithm='subfield')
93556643250795678718734474880013829509196181230338248789325711173791286325820
sage: # E.order(algorithm='exhaustive')    # well, please not in this case...
sage: E.order(algorithm='all')
93556643250795678718734474880013829509196181230338248789325711173791286325820


问:
I have not heard of this identity before
I am trying to increase my math abilities by working through the solutions of past Putnam exam problems. I am currently working on Putnam 1985-2a. I have been able to work through to the last part of this problem where I currently have:

1 - $\sum_{i=1}^n(a^2_i)$

and all that's left to find is the minimal value of this expression. (EDIT: Minimum value for $\sum_{i=1}^n(a^2_i)$ so that I can obtain the maximum value of the term possible given that $a_i>0$ and $a_{1} + \dots + a_{n} =1$).

I am currently working on using the C-S inequality to do this. In the answer to the problem they give a lot of examples of how to find this, but I am curious about the following identity that they don't name. This is the identity:

$n(a_{1}^2 + \dots + a_{n}^2) = (a_{1}+\dots+a_{n})^2 +\sum_{i<j}(a_{i}-a_{j})^2$

The summation looks a little like how we calculate variance, but I don't know where this identity comes from, or if it has a name. Can anyone help?

Thanks.

答:
This is a special case of Lagrange's identity:
$$ \Big(\sum_{k=1}^n a_k^2\Big)\Big(\sum_{k=1}^n b_k^2\Big) -
   \Big(\sum_{k=1}^n a_k b_k\Big)^2 = \sum_{k=1}^{n-1}\sum_{j=i+1}^n
  (a_ib_j-a_jb_i)^2 \tag{1} $$
where all of the $\,b_k=1\,$ which results in
$$ n \Big(\sum_{k=1}^n a_k^2\Big) -
   \Big(\sum_{k=1}^n a_k\Big)^2 = \sum_{1\le i<j\le n}  (a_i-a_j)^2. \tag{2} $$
Divide both sides of this equation by $\,n^2\,$ and change $\,a_k\,$
into $\,x_i\,$ to get
$$ \text{Var}(X) = \frac1n \Big(\sum_{i=1}^n x_i^2\Big) -
   \Big(\frac1n \sum_{i=1}^n x_i\Big)^2 = \frac1{n^2}\sum_{1\le i<j\le n}  (x_i-x_j)^2 \tag{3} $$
an equation which is almost exactly what appears in the Wikipedia
Variance article.


问:
Finite groups schemes over a field are killed by the order
I am trying to prove that a finite group scheme over a field is killed by the order. I think I can do the case where the field $k$ is perfect, here's a poorly written synopsis:

First, by definition a finite group scheme over a field is affine. For a finite group scheme $G = \text{Spec}A$ where $A$ is a $k$-algebra, which is finitely generated over $k$ as a module. It's clear to see that $G$ is just a disjoint union of points. Since $k$ is perfect, $G_\text{red}$ is actually a closed subgroup, and so we can assume WLOG that $G$ is reduced. Further, by base extension with $\overline{k}$, we can assume $k$ is algebraically closed. Now we can just use the fact that for any finite extension $K$ of $k$, $K\otimes_k \overline{k}$ is isomorphic to the sum of injections of $K$ into $\overline{k}$ to see that the orders of the group schemes are comparable, and so since $G$ is a finite group scheme over an algebraically closed field, it's a constant group scheme so the result follows from group theory.

Here we used twice that $k$ is perfect:

Using $G_{red}$ as a subgroup

In the base extension with $\overline{k}$

which is why I don't think this proof generalizes to the case where $k$ is not perfect.

I'm stuck trying to prove the case where $k$ is not perfect. I know that somehow I am meant to use Frobenius morphisms, but I am not quite sure how to do this. The only material that I can find online is regarding Deligne's theorem which as far as I can tell only applies to commutative group schemes. Any help or hints would be appreciated.

答:
[Posting my comment as an answer, following OP's request.]

On the webpage of Ben Moonen, there are drafts about a book on Abelian Varieties. It also covers some general theory of group schemes. I found your problem as a guided exercise, though it does not give a written solution. The exercise may be found in Chapter 4, at page 22 of the .pdf, exercise 4.4. In fact, in proves the result over any reduced irreducible base scheme, under the assumption that the group scheme is finite locally free. It's equivalent to it being finite flat, with the further assumption that the base is locally noetherian (which of course it is, in the case of a field).

Here is the link to Ben Moonen's webpage

There must be a published reference with a written proof somewhere else, but I wasn't able to find it. If somebody knows about such a paper, please let us know in the comments.


问:
$\sup$ of integrals with unknown function. (norm of linear operators)
So, lately, I'm struggling with functional analysis exercises like this one:

My linear operator is defined in $L^2 (\mathbb R)$:

$$T : T f = \cos xf(x)+\sin xf(\pi−x).$$

it asks me to find $\|L\|$.

One way to solve it is to consider the functions in there are trigonometric ones and this will make everything much more trivial but my question is about a method where you effectively try to maximize f(x) by building up some sort of Dirac's delta where the other functions naturally maximize themselves, (in this case $\sin^2x$ and $\cos^2x$) which seems I did not understand correctly how to apply.

My idea is to let $f_n := \begin{cases} n & \text{if } (n\pi-\frac{1}{n}) < x < (n\pi+\frac{1}{n}) \\ 0 & \text{if not.} \end{cases} $

then i can integrate over the nonnull domain $[n\pi-\frac{1}{n}, n\pi+\frac{1}{n}]$

and applying the def of $\|T\|_2$ i will get $\|T\|= \frac{g(n)}{\|f\|}$ where $\|f\|=1$ and some sort of function $g(n)$ which I'm asked to find the sup of. Sadly in this case what i get is $\sup \{2n^2/n\}.$

I got the $n^2$ from my choice of $f(x)$ being${}= n$ in the nonnull domain (it's $L^2$ so i need to integrate the square of $f(x)$) and the $1/n$ by choosing $1/n$ as my, let's say, "moving boundary" of my integration domain.

So here it comes the problem, this method is clearly misunderstood by me because the result is a function of my $f(x)$ and in this case, the $g(n)$ is not even limited, if i will chose $f_n=\sqrt n$ in the non null domain, for example, i should have found $\|T\|=1$ (wich is the right answer btw). So, my question is: how that method truly works and, if it does, when?

答:
With $f \in L^2(R)$, you have $(Tf)(x) := f(x)\cos x + f(\pi - x)\sin x$.
The squared $L^2$ norm of $Tf$ is
\begin{equation}
\vert Tf \vert^2 = \int_R (f(x)\cos x + f(\pi - x)\sin x)^2\, dx.
\end{equation}
Multiply out the integrand to get
\begin{equation*}
\vert Tf \vert^2 =
\int_R f^2(x)\cos^2 x\, dx
+ 2\int_R f(x) f(\pi - x) \cos x \sin x dx
+ \int_R f^2(\pi - x)\sin^2 x\, dx.
\end{equation*}
The first and third integrals can be combined if you make a change of variable in the third, yielding $\int_R f^2(x)(\cos^2 x + \sin^2 x) dx$, which is $\vert f\vert^2$. Now figure out the simple change of variable in the second integral that produces an integrand which is an odd function about the shifted origin, and you can do that integral in your head.

You end up with $\vert Tf \vert^2 = \vert f \vert^2$.


问:
Probability chosing -defective -series
even I have understand probabilities how it works I find it difficult to achieve to solve 10 exercises in 10 right.I am doing mistakes. I want to calculate those with a calculator online or with some other technique that it need it makes it easier to solve.I will write 3 different examples of exercises. I want to have an easy solution "same" way in all OR I want to use a calculator/software for all or for each one case . Is any calculator that can solve those exercises?that would be perfect I wouldn't do mistakes and I will be sure for the result.

Exercise 1(Boxes):I have a box 5 black balls,6 white balls.I pick 3 (I can't the balls back).

```<code>a)Find what is the possibility to be the third ball black
b)the last two balls white
</code>```

Exercise 2(defective) :I have two companies A and B with packages. We take 30% of A and 70% of B.The 5% from A is defective (packages).

```<code>a)If we take 1 package what is  the probability to be defective?
</code>```

Exercise 3:(with serie ) : 4 white,5 black,7 red.We chose 4. What is the probability to come with the right way?

```<code>a)W B B R W
b)B B W B R
</code>```

答:
Responding to the OP's reaction to my comment.

Find what is the possibility to be the third ball black

Thought experiment:

Instead of having the balls drawn, a moderator writes down the numbers 1 thru 11, each on a separate slip of paper, and then randomly assigns each slip of paper to one of the balls.  This random assignment will then represent the order that the balls are drawn in.

Therefore, the probability of the 3rd ball being black is equivalent to the probability that the piece of paper with the number 3 on it was assigned to a black ball.  Why would such an occurrence be more or less likely than the probability that the piece of paper with the number 1 on it was assigned to a black ball?

Answer: the probability of the two occurrences is the same.  That is, assigning (at random) one piece of paper to one of the black balls is just as likely (no more and no less) than assigning a different piece of paper to one of the black balls.

Title: Probability chosing -defective -series


答:
So first, let me say that it is quite easy to see geometrically/show using Lagrange multipliers, that the maximum of $\sum a_i^2$ on the simplex $\{\sum a_i = 1 : a_i \ge 0\}$ occurs when all but one of the variables are $0$. The Lagrange condition $\lambda \nabla f = \nabla g$ is $2\lambda a_i = 1$ so either $a_i = 0$ or $a_i = \frac{1}{2\lambda}$ (all the non-zero $a_i$'s are equal) now you can just check if there are $k$ variables which are non-zero then the value is

$$ k \left( \frac{1}{k} \right)^{2} = \frac{1}{k}.$$

Therefore the maximum occurs when $k = 1$. This avoids having to guess at some symmetric function identity that does exactly what you want.

Then for the symmetric function identity, it just boils down to expanding the right hand side.

$$ \left( \sum a_i \right)^2 = \sum a_i^2 + 2 \sum_{i < j} a_i a_j$$

and

$$
\sum_{i < j} (a_i - a_j)^2 = \sum_{i < j} a_i^2 - 2 \sum_{i < j} a_i a_j + \sum_{i < j} a_j^2 = (n - 1) \sum a_i^2
$$

since $a_k^2$ appears once for each $j > k$ and once for each $i < k$ (so for every index except $k$).

Title: I have not heard of this identity before

Body:
I am trying to increase my math abilities by working through the solutions of past Putnam exam problems. I am currently working on Putnam 1985-2a. I have been able to work through to the last part of this problem where I currently have:

1 - $\sum_{i=1}^n(a^2_i)$

and all that's left to find is the minimal value of this expression. (EDIT: Minimum value for $\sum_{i=1}^n(a^2_i)$ so that I can obtain the maximum value of the term possible given that $a_i>0$ and $a_{1} + \dots + a_{n} =1$).

I am currently working on using the C-S inequality to do this. In the answer to the problem they give a lot of examples of how to find this, but I am curious about the following identity that they don't name. This is the identity:

$n(a_{1}^2 + \dots + a_{n}^2) = (a_{1}+\dots+a_{n})^2 +\sum_{i<j}(a_{i}-a_{j})^2$

The summation looks a little like how we calculate variance, but I don't know where this identity comes from, or if it has a name. Can anyone help?

Thanks.


问:
Trig Identity Math Problem $\sqrt{3} \tan(3x)=1$ from $0\leq x<360$
Solve the equation from $0\leq x < 360$

$\sqrt{3}\tan(3x)=1,$

I have tried to solve this equation by dividing the square root into both sides, and then making $3x$ equal to inverse $\tan$ of 1/$\sqrt{3}$. I only get $3x=30$ and $x=10$. I do not believe that this is right.

答:
Let $y = 3x$, so $0^{\circ}\leq y < 1080^{\circ}$. Then, we must have:

$$\sqrt{3}\tan(y) = 1$$

$$\tan(y) = \frac{1}{\sqrt{3}}$$

Because $\tan(y) = \frac{1}{\sqrt{3}}$ whenever $y = 30^{\circ} + n\cdot 180^{\circ}$ for any integer $n$, we find that the solutions satisfying $0^{\circ}\leq y < 1080^{\circ}$ are $y= 30^{\circ},\ 210^{\circ},\ 390^{\circ},\ 570^{\circ},\ 750^{\circ},\ 930^{\circ}$. Then, we have:

$$x = \frac{y}{3} = \boxed{10^{\circ},\ 70^{\circ},\ 130^{\circ},\ 190^{\circ},\ 250^{\circ},\ 310^{\circ}}$$

Note that the solutions all differ by $60^{\circ}$, one third the period of $\tan$.


Title: Trig Identity Math Problem $\sqrt{3} \tan(3x)=1$ from $0\leq x<360$

Body:
Solve the equation from $0\leq x < 360$

$\sqrt{3}\tan(3x)=1,$

I have tried to solve this equation by dividing the square root into both sides, and then making $3x$ equal to inverse $\tan$ of 1/$\sqrt{3}$. I only get $3x=30$ and $x=10$. I do not believe that this is right.

Body: Just divide by $\sqrt{3}$ to get tan(3x) = 1/$\sqrt{3}$ then do $3x=tan^{-1}(1/\sqrt{3})+180^{o}k$ for k=0,1,2,3... So x = 10, 70, 130,..