答:
Your approach to change the order of integration is plausible. Specifically, we have
$$
\int_0^1 \int_{\sqrt x}^1 \sin(\pi y^3) dy dx  = \int_0^1 \int_0^{y^2} \sin(\pi y^3) dx dy = \int_0^1 y^2 \sin(\pi y^3) dy.
$$
To see how to switch the limits, notice that the integration region is
$$ D = \lbrace (x,y) \in \mathbb{R}^2 \mid 0 \leq x \leq 1, \sqrt{x} \leq y \leq 1\rbrace = \lbrace (x,y) \in \mathbb{R}^2 \mid 0 \leq x \leq 1, x \leq y^2 \leq 1\rbrace $$
and the intersection point of the curve $ y = \sqrt{x} $ and $ x = 1 $ is (1,1). The result follows. You may convince yourself with the aid of other plotting software.

I'll leave the remaining integral to you, which can be solved by substitution of variable $ u = \pi y^3 $.

Title: Double integral $\int_{0}^{1}\int_{\sqrt{x}}^{1} \sin(\pi \ y^3)dy\ dx$

Body:
In this integral i have tried some substitution and have changed the order of integration too but i couldn't solve this

$$ \int_{0}^{1}\int_{\sqrt{x}}^{1} \sin(\pi \ y^3)dy\ dx$$
any suggestion would be great

Thanks


问:
Prove that $A$, a matrix of rank $3$, can't have characteristic polynomial of $p(x) = x^7 - x^5 + x^3$
Prove that $A$, a matrix of rank $3$, can't have characteristic polynomial of $p(x) = x^7 - x^5 + x^3$

My attempt to contradict:

Because of that characteristic polynomial, the matrix must be a $7 \times 7$ matrix. Also, $-\mathrm{tr}(A) = 0$, because $x^6 = 0$. $A$ is a matrix with rank of $3$ so the determinant is $0$ and therefore it isn't invertible.

I am unable to progress from this point onwards.

答:
Hint: The geometric multiplicity of an eigenvalue is always at most the algebraic multiplicity.

What is the eigenspace corresponding to the eigenvalue $0$?

ContentLicense: CC BY-SA 4.0


问:
Does there exist a topos with these $n+2$ truth values?
This question is based on the answers to this question.

The Question:

Let $n\in\Bbb N$. Let $N$ be a set with $n+2$ elements, labelled $0$ to $n$, and the $(n+2)$th element labelled $\infty$. Suppose we have a function

$$\begin{align}
t:N &\to N,\\
0 &\mapsto 0,\\
m &\mapsto m-1,\quad\text{(for } m\in \overline{1,n}\text{)}\\
\infty &\mapsto \infty.
\end{align}$$

Does there exist a topos whose internal logic corresponds to $\infty$ being "false", $0$ being "true", and each $m\in \overline{1,n}$ being "$m$ steps (through $t$) until truth"?

Motivation:

An answer to this question, hopefully, will kill two birds with one stone: objections to: (1) systems of logic with more than two truth-values and (2) whether infinity has a rigorous place in mathematics.

Thoughts:

It is my belief that such a topos can be created; however, I don't know how.

I'm aware that $(N, t)$ is a dynamical system. I don't have much experience with them.

For an idea of my abilities in topos theory, see this: The legitimacy of topos theory and intuitionism.

Please help :)

答:
I recommend you read SGL by MacLane and Moerdijk (available here) since I'm going to be quoting results from there and generally it's a great introduction to Topos theory.

In particular, in I.4 we are given the following characterisation for the subobject classifier of a presheaf topos ${\bf Sets}^{\mathcal{C}^{op}}$:
$$ \Omega\colon \mathcal{C}^{op} \to {\bf Sets}, \ \ \Omega(C) = \{\,S \mid S \text{ is a sieve on $C$}\,\}.$$
In the text, they offer one motivating description of $\Omega(C)$ as "the set of "paths to truth"".  The map ${\rm true} \colon 1 \to \Omega$ has as components $\ast \mapsto t_C$ (where $t_C$ is the maximal sieve of all arrows to $C$).

We might want to restrict to when $\mathcal{C}$ is a preorder category, in which case sieves correspond to down-sets in the order.  If $\alpha$ is an ordinal, the nonempty down-sets of $\alpha^{op}$ are in bijection with the elements of $\alpha$. In some sense these are the "truth values".  For a subfunctor $Q \rightarrowtail P$ in ${\bf Sets}^{\alpha^{op}}$, the characteristic map $\varphi^Q \colon P \to \Omega$ has components $\varphi^Q_\beta(x) = \gamma$, the least ordinal $\gamma \in \alpha$ such that $x\cdot f \in Q(\gamma)$ (where $f$ is the map induced by $\beta \leqslant \gamma$), or $\varphi^Q_\beta(x) = \bot$ should such a $\gamma$ not exist.

It's not hard to see that, taking $\alpha = n$, we get the topos you seek.


问:
Evaluation of a complex numbers partial sum
Let $w = e^{i\frac{2\pi}5}$. I would like to evaluate

$$w^0 + w^1 + w^2 + w^3 +...+ w^{49}$$

Can anyone please give me an idea how to evaluate the expression?

Thanks in advance

答: Note that $e^{2\pi i k/5}$, where $k$ ranges from $0$ to $4$, gves us all the roots of the equation $z^5-1=0$. The sum of these $5$ roots is the coefficient of $z^4$, which is $0$. </p>

ContentLicense: CC BY-SA 3.0


问:
Problem of relations between the coefficients of a quadratic equation.how to proceed?
I have tried several ways to solve this problem of using the root properties of a second degree equation but I can't come up with anything favorable.

If the roots of $ax^{2}+bx+c=0$  are of the form  $\dfrac{\lambda }{\lambda -1}$ and  $\dfrac{\lambda+1 }{\lambda}$, then the value $(a+b+c)^{2}$ in terms of $a$, $b$, $c$ is?

Options

$(a) \quad b^{2}-4ac$

$(b) \quad b^{2}-2ac$

$(c) \quad 2b^{2}-ac$

$(d) \quad 4b^{2}-2ac$

$(e) \quad b^{2}-4abc$

答:
Define $f(x)=ax^2+bx+c$ then
$$
f(x)=ax^2+bx+c=a\left(x-\frac{\lambda}{\lambda-1}\right)\left(x-\frac{\lambda+1}{\lambda}\right)
$$
Hence
$$
(a+b+c)^2=(f(1))^2=a^2 \left(1 - \frac{\lambda}{\lambda-1}\right)^2 \left(1 - \frac{\lambda+1}{\lambda}\right)^2\\
= a^2 \frac{1}{(\lambda-1)^2} \frac{1}{\lambda^2} = a^2 \left(\frac{\lambda}{\lambda-1} - \frac{\lambda+1}{\lambda}\right)^2 \\= a^2 (r-s)^2 = a^2 ((r+s)^2 - 4rs)\\=a^2(-b/a)^2 - 4 a^2 (c/a)=b^2-4ac
$$
where $r=\lambda/(\lambda-1)$ and $s=(\lambda+1)/\lambda$


问:
Is it possible to find the control points of a quadratic bezier curve, given the starting, ending and stationary point?
I know from this question on SO that it is possible to get the stationary point of a bezier curve given the control points, but I want to know wether the opposite is true:

If I have the start and end points of a Parabola, and I have the maximum point, is it possible to express this a quadratic bezier curve?

I have seen this answer explaining that you usually would need 6 points on a cubic to effectively describe a bezier curve, but looking at the question itself, it describes a few special cases where you can find the control points on a cubic, and I am curious on wether a quadratic bezier curve has such points, and wether a stationary point is one of them

The variable t of the stationary point is not avalible

I am coming from a computer science background, and although im not too weak in math, any answers/articles that don't use complicated maths and keep it simple would be appreciated

If this is not possible, would it help of the stationary point is always a maximum point and the x coordinate if the max point is the mean of the x coordinates of the start and end points?

By maximum point I mean the point with the maximum value for its y coordinate and by stationary I meant the point with the maximum or minimum value for y

答:
Let $P_0$ and $P_2$ be startpoint and endpoint of a quadratic Bézier curve. If by "maximum point" you mean that point $Q$ on the curve which is at the maximum distance from line $AB$ (this is not the same as the vertex of the parabola, in general), then finding the control point $P_1$ is easy, because $Q$ is the midpoint between $P_1$ and the midpoint $M$ of $AB$.

Note that the tangent to the curve at $Q$ is parallel to $AB$ and $Q$ is obtained by substituting $t=1/2$ in the equation of the curve:
$$
B(t)=(1-t)^{{2}}P_{0}+2t(1-t)P_{1}+t^{{2}}P_{2}.
$$

EDIT.

After the given clarifications by the OP, let's assume that the stationary point $P=(x,y)$ of the curve (having $dy/dt=0$) is also known, in addition to $P_0$ and $P_2$. We can simplify a bit the computation shifting all points with a translation so that $P_0=(0,0)$ (this can of course be undone at the end with the opposite translation). In this case we have then:
$$
(x,y)=2t(1-t)(x_1,y_1)+t^2(x_2,y_2).
$$
The condition $dy/dt=0$ gives $ty_2+(1-2t)y_1=0$, which can be solved for $y_1$:
$$
y_1={t\over2t-1}y_2.
$$
We can insert this into the equation for $y$, which becomes then
$t^2y_2=y(2t-1)$ and can be solved for $t$:
$$
t={y\pm\sqrt{y^2-y_2y}\over y_2}.
$$
Note that the sign must be chosen so that $0<t<1$, and this can happen only if $\min(0,y2)<y<\max(0,y2)$, as it was to be expected, and $y_2\ne0$ (but see below for this case).

We can substitute this into the equations for $y_1$ given above and into
$$
x_1={x-t^2x_2\over2t(1-t)}
$$
to obtain the coordinates of control point $P_1$.

Finally, if $y_2=0$ the above solution is singular, but only apparently, because that is exactly the case discussed in the first part of my answer. Hence we can find at once the solution if $y_2=0$:
$$
y_1=2y,\quad x_1=2x-{1\over2}x_2.
$$


问:
In how many ways can you select $4$ different numbers from the numbers $1,2,..,15$
In how many ways can you select $4$ different numbers from the numbers $1,2,..,15$ so that no pair of numbers among the selected ones are subsequent numbers.

I turned this into a minor problem. I assumed that we have $5$ numbers and we want to choose two different numbers so that no pair of numbers among the selected ones is subsequent. I came to the conclusion that it will be $5$ choose $2$ and then minus all subsequent pairs that I calculated in my head. But I wonder if there is a better method that can be applied to $15$ numbers.

Also, how would you make this problem more difficult.

答:
The question is equivalent to the following:

In how many ways may 4 people be seated in a row of 15 chairs given
that no two people may sit next to one another?

Consider some arrangement of the 4 people as specified, then take one chair out from between each pair of people. What you’re left with is a unique arrangement of $4$ people in 12 chairs without restrictions. Similarly, starting with an unrestricted arrangement of 4 people in 12 chairs, adding a chair between each pair of people gives a unique arrangement of 4 non-adjacent people in 15 chairs. (Convince yourself of these two assertions.) Thus there is a one to one correspondence between the restricted 15-chair arrangements of the problem and unrestricted 12-chair arrangements. The number of unrestricted 12-chair arrangements is the number of ways to choose 4 chairs out of 12
$$\binom{12}{4}$$


问:
Question about notation in Engelking's General Topology book
In Engelking's General topology book, (2nd, page 36) he states a theorem related to $T_0$ space.

For every $T_0$ space we have $|X| \leq \exp w(X)$.

What does $\exp$ mean? I know $|X|$ is the cardinality of $X$, $w(X)$ is the weight of $X$.

答:
For a cardinal number $\mathfrak m$ Engelking denotes by $\exp \mathfrak m$ the cardinal number $2^{\mathfrak m}$, i.e. the cardinality of the power set.

You can find this in the Introduction (Section I.2 "Cardinal numbers").


问:
Show the set is affine subspace.
About 2 days ago I was learning stuff about affine geometry and yesterday I got stuck with the following problem

Suppose that $S$ is a subset of affine space A. Show the set:
$$
\langle S \rangle \stackrel{\text{def}}{=} a + \text{span}\{\vec{ax}: x \in S\}, \text{ for some $a \in S$ }
$$

Does not depend on $a$ and also is the minimal affine subspace containing the set S. And also

$$
\langle S \rangle=\big\{ {\sum_i \lambda_i x_i}: \sum_i \lambda_i=1, \space x_i \in S \big\}
$$

My attempt

I know how to prove the second task:
$$
v \in \langle S \rangle \implies v=a+ \sum_i \lambda_i \vec{ax_i} \\
=a + \sum_i {\lambda_i (x_i-a)} \\
=(1-\sum_i \lambda_i)a+\sum \lambda_i x_i
$$

Assuming that we proved the set $\langle S \rangle$ does not depend on $a$ it is clear that $v$ does not depend on $a$. Hence $\sum_i \lambda_i=1$. Since $v$ is an arbitrary point from $S$ then it implies that the whole set $\langle S \rangle$ consists of linear combinations of points from $S$ with sum of coefficients equal to $1$.

But how to prove the first one? E.g. to show that the set

$$
\langle S \rangle = a + \text{span}\{\vec{ax}: x \in S\}, \text{ for some $a \in S$ }
$$

does not depend on $a$? It doesn't look straightforward to me.

Have you got any ideas?

答:
$\langle S\rangle$ does not depend on $a$ because $a=b+\overrightarrow{ba}$ for all $b\in S$. But $\overrightarrow{ba}$ is in $\operatorname{span}\{\overrightarrow{ax}~\vert~x\in S\}$ (which I will call $V_a$ from now on). So
$$a+V_a=b+\overrightarrow{ba}+V_a=b+V_a=b+V_b.$$
For the last equality you need to show that $V_a=V_b$ for all $a,b\in S$. Note that $V_a$ is spanned by vectors of the form $\overrightarrow{ax}$, while $V_b$ is spanned by vectors of the form $\overrightarrow{bx}=\overrightarrow{ax}-\overrightarrow{ab}$.

It is also the minimal affine subspace containing $S$ because:

It's obviously an affine subspace.

It clearly contains $S$.

If all points of $S$ are to be contained in this affine space, then its spanning vector space must contain vectors pointing between these points. So all the vectors which span $V_a$ must be contained in the vector space spanning any affine space containing $S$ anyway.

Also, disregard my comment that you can't multiply $a$ by a number since it's not a vector. You seem to be working with a definition of affine space where the underlying set is still the vector space, so it's fine.

Title: Show the set is affine subspace.


问:
What is a direct proof, formally?
In my logic class I took in college, I sometimes would get wrong marks because I proved a statement indirectly instead of directly. Informally, a direct proof of a conditional $P \implies Q$, is one where you assume $P$ and try to deduce $Q$ "directly", without using the contrapositive of that conditional. But no one has ever given me a formal, rigorous definition of direct proof. Can someone give me one, and with one, can someone give examples of statements which can't be proven directly, only indirectly?

答:
As Rob Arthan said in the comments, one way to distinguish between (one kind of) direct proof and indirect proof is simply to distinguish between proofs that use intuitionistically valid deductive rules and proofs that do not.

To show that there are some theorems (using classical logic) that have no intuitionistically valid proofs, it suffices to show that there is a way to interpret intuitionistic logic such that the intuitionistic deductive rules are sound for that interpretation, and also that there is a tautology of classical logic that is not a tautology under that interpretation. A popular one is LEM (i.e. $A∨¬A$), but it may not be convincing. Here is another:

$(A⇒B)∨(B⇒A)$.

You cannot prove this classical tautology in intuitionistic logic. In that precise sense there can be no direct proof of this tautology. This is more convincing than LEM because it is even slightly counter-intuitive and you really need to 'check cases' (relying on LEM) to prove it. Lest you think that 'checking cases' is direct, observe that every intuitionistic proof of ( $¬B⇒¬A$ ) can be translated to a proof that is intuitionistic except for a single LEM case-checking:

  If $A$:

    $B∨¬B$.

    If $B$:

      $B$.

    If $¬B$:

      [Insert proof of ( $¬B⇒¬A$ ) here.]

      $¬A$.

      Contradiction (i.e. $A,¬A$).

      $B$.  [by explosion]

    $B$.

  $A⇒B$.

There are many interpretations of intuitionistic logic under which this classical tautology does not hold. One is Kripke frames (see here for a sketch), and another is the BHK interpretation. Understanding either of them will give you a clear idea of how exactly an intuitionistic proof can be considered to be 'direct'.

Title: What is a direct proof, formally?


问:
Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$
I tried $$\int{x\sqrt{1-x^2}\arcsin{x}\ \mathrm{d}x}$$$$=\int{\arcsin{x}\ \mathrm{d}\left(\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\ \mathrm{d}\left(\arcsin{x}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3\sqrt{1-x^2}}\ \mathrm{d}x}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\frac{2}{3}\int{x^3\ \mathrm{d}x}+\frac{2}{3}\int{x^5\ \mathrm{d}x}$$$$=\cdots$$
This solution isn't true. Where am I wrong?

EDIT: Thank you all for the many different answers! If I could, I'd give you all the accepted answer.

答:
Let $x=\sin(\theta)$.* Then,
$$
\frac{dx}{d\theta}=\cos(\theta) \implies dx=\cos(\theta)d\theta \, .
$$
The integral becomes
$$
\int \sin(\theta) \left(\sqrt{1-\sin(\theta)^2}\right) \theta\cos(\theta)d\theta
$$

*To be more precise, we are making the substitution $\theta=\arcsin(x)$, so that $-\pi/2 \leq \theta \leq \pi/2$. This allows us to conclude that $\arcsin(\sin(\theta))=\sin(\theta)$.

Title: Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$


答:
Let $x=\sin(u)$, then $dx=\cos(u)du$. So
$$\begin{align}\int x\sqrt{1-x^2}\arcsin(x)dx&=\int u\sin(u)\cos^2(u)du\\
&=\int u\sin(u)du-\int u\sin^3(u)du\end{align}$$

Title: Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$

Body:
I tried $$\int{x\sqrt{1-x^2}\arcsin{x}\ \mathrm{d}x}$$$$=\int{\arcsin{x}\ \mathrm{d}\left(\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\ \mathrm{d}\left(\arcsin{x}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3\sqrt{1-x^2}}\ \mathrm{d}x}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\frac{2}{3}\int{x^3\ \mathrm{d}x}+\frac{2}{3}\int{x^5\ \mathrm{d}x}$$$$=\cdots$$
This solution isn't true. Where am I wrong?

EDIT: Thank you all for the many different answers! If I could, I'd give you all the accepted answer.


答: This is the sum of the fifth roots of unity, ten times over.  See here for a discussion of why each of these sums is zero.</p>

Title: Evaluation of a complex numbers partial sum

Body:
Let $w = e^{i\frac{2\pi}5}$. I would like to evaluate

$$w^0 + w^1 + w^2 + w^3 +...+ w^{49}$$

Can anyone please give me an idea how to evaluate the expression?

Thanks in advance


Title: Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$

Body:
Integration by parts
$$\int x \sqrt{1-x^2} \arcsin x\, dx$$
take
$$u'(x)=x \sqrt{1-x^2};\;v(x)=\arcsin x$$
so
$$u(x)=\int x \sqrt{1-x^2} \, dx=-\frac{1}{2}\int (-2x)(1-x^2)^{-1/2}\, dx=-\frac{1}{3} \left(1-x^2\right)^{3/2}+C$$
and
$$v'(x)=\frac{1}{\sqrt{1-x^2}}$$
apply IBP formula
$$\int u'(x)v(x)\,dx=u(x)v(x)-\int u(x)v'(x)\,dx$$

$$\int x \sqrt{1-x^2} \arcsin x\, dx=-\frac{1}{3} \left(1-x^2\right)^{3/2}\arcsin x-\int\left(-\frac{1}{3} \left(1-x^2\right)^{3/2}\right)\frac{1}{\sqrt{1-x^2}}\,dx=$$
$$=-\frac{1}{3} \left(1-x^2\right)^{3/2}\arcsin x+\int \frac{1}{3} \left(1-x^2\right)\,dx=$$
$$=-\frac{1}{3} \left(1-x^2\right)^{3/2} \arcsin x+\frac{1}{3} \left(x-\frac{x^3}{3}\right)+C$$
which can be simplified to
$$=\frac{1}{9} \left(3x-x^3-3 \left(1-x^2\right)^{3/2} \arcsin x\right)+C$$


问:
2D cubic Bezier curve. Point of self-intersection
I have a 2D cubic Bézier curve defined by a set of control points A, B, D and C.
How can I find a point of self-intersection P (two parameter values t)?

答:
I solved this problem years ago and the relevant code is still in my Kinross repository.

Represent the cubic Bézier curve in the power basis:
$$(x(t),y(t))=(x_3t^3+x_2t^2+x_1t+x_0,y_3t^3+y_2t^2+y_1t+y_0)$$
Let the self-intersection parameters be $\lambda$ and $\mu$, then obviously $x(\lambda)-x(\mu)=y(\lambda)-y(\mu)=0$. Expanding, then dividing by $\lambda-\mu$ (the trivial solution), gives
$$x_3(\lambda^2+\lambda\mu+\mu^2)+x_2(\lambda+\mu)=-x_1\\
y_3(\lambda^2+\lambda\mu+\mu^2)+y_2(\lambda+\mu)=-y_1$$
Now solve this linear system for $\lambda^2+\lambda\mu+\mu^2$ and $\lambda+\mu$ and obtain
$$\lambda\mu=(\lambda+\mu)^2-(\lambda^2+\lambda\mu+\mu^2)$$
Finally, by Viète's formulas, $\lambda$ and $\mu$ are the roots of $t^2-(\lambda+\mu)t+\lambda\mu$. Either may be outside $[0,1]$, indicating a self-intersection outside the curve proper, or both may be complex, indicating no self-intersection even in the extended curve.

My implementation of the above algorithm in Kinross first makes an affine transformation mapping $A,B,C$ to $(0,0),(0,1),(1,1)$ respectively. Because Bézier curves are equivariant under affine transformations of their control points, the parameters of self-intersection do not change. After transforming, $D$ is at $(x,y)$ and
$$\lambda+\mu=1-\frac y{x+y-3}\qquad\lambda\mu=(\lambda+\mu)^2+\frac{3(\lambda+\mu)}{x-3}$$


答: Your mistake is a sign error:$$\begin{align}\frac{d}{dx}x^3(1-x^2)^{3/2}&=x^2(1-x^2)^{1/2}[3(1-x^2)+x\tfrac32(-2x)]\\&=x^2(1-x^2)^{1/2}[3-3x^2-3x^2].\end{align}$$You seem to have mistaken that last $-$ sign for a $+$. Everything after that is error-carried-forward.

Title: Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$


问:
Factorization of Polynomial in an extension field
I'm trying to understand the proof of the following problem:

Let $L / K$ be a field extension, and assume that $K$ has $q=p^{r}$ elements. Let $P(X)$ be a irreducible polynomial in $K[X]$. Let $M$ be a subfield of $K(\alpha)$ for $\alpha$ a root of $P(X)$, and suppose that the degree of $K(\alpha)$ over $M$ is $r .$ Show that $P(X)$ factors, in $M[X],$ as a product of $\frac{d}{r}$ irreducible polynomials of degree $r$.

Proof: We have the previous results: (a) $P\left(\alpha^{q}\right)=P(\alpha)^{q}$ for all $\alpha \in L$, (b) If $P(X)$ is irreducible, and has a root $\alpha$ in $L$, then $P(X)$ factors in $L$ as $(X-\alpha)\left(X-\alpha^{q}\right)\left(X-\alpha^{q^{2}}\right) \ldots\left(X-\alpha^{q^{d-1}}\right),$ where $d$ is the degree of $P(X)$. Now, let $Q(X)$ be an irreducible element of $M[X]$ that divides $P(X) .$ Then, since $P(X)$ factors into linear factors in $K(\alpha), Q(X)$ has a root in $K(\alpha) ;$ by (b) this root is of the form $\alpha^{q^{s}}$ for some $s$. Applying (b) with $M$ in place of $K$ and $Q(X)$ in place of $P(X)$ (and noting that $M$ has $q^{\frac{d}{r}}$ elements) we see that the other roots are $\left(\alpha^{q^{s}}\right)^{q^{k} \frac{d}{r}},$ for $0 \leq k \leq r-1,$ so $Q(X)$ has degree $r$ as claimed.

This proof feels a bit too quick for me. Could anyone explain the line "Applying (b) with $M$ in place of $K$ and $Q(X)$ in place of $P(X)$ (and noting that $M$ has $q^{\frac{d}{r}}$ elements) we see that the other roots are $\left(\alpha^{q^{s}}\right)^{q^{k} \frac{d}{r}},$ for $0 \leq k \leq r-1,$ so $Q(X)$ has degree $r$ as claimed?" More precisely,

(1) Why is $0 \le k \le r-1$, and why are the $\left(\alpha^{q^{s}}\right)^{q^{k} \frac{d}{r}}$  distinct for distinct $k$?

(2) How do we conclude $Q(X)$ has degree $r$?

Partial answers (answering only some of the questions) are very welcome! (I will upvote them too)

By the way, this is a problem from a basic algebra course. I do not know Galois Theory.

答:
From the factorization of $P(X)$ we know that the elements $\alpha^{q^i}$, $i=0,1,\ldots,d-1$, are all distinct. It may be better to restate this in the form
$$
\alpha^{q^i}=\alpha^{q^j}
$$
if and only if $i\equiv j\pmod d$

If $M$ is an intermediate field, $K\subset M\subset K(\alpha)$, by the tower law of field extensions we know that $\ell:=[M:K]$ is a factor of $d=[K(\alpha):K]$.
Also, $M$ has $q^\ell$ elements.

As noted, the polynomial $Q(X)$ has a root, say, $\beta=\alpha^{q^s}$. By (b) the other roots of $Q(X)$ are
$$
\beta_j:=\beta^{q^{\ell j}}
$$
for $j=0,1,\ldots, t-1$, $t=\deg Q(X)$. For all $i$ we have
$\beta_i=\alpha^{q^{s+\ell i}}.$

Observe that by using (a) and induction on $i$, we see that the element $\beta_i$ is a zero of $Q(X)$ for all natural numbers $i$. That is, not only when $0\le i\le t-1$.

The earlier observation says that
$$\beta_j=\beta_i$$ if and only if $$s+\ell i\equiv s+\ell j\pmod d$$
if and only if
$$
i\equiv j\pmod {d/\ell}.
$$
So there are $d/\ell$ distinct elements $\beta_i$. We know that the zeros of $Q(X)$ are distinct, so we can conclude that $t=d/\ell$.

This answers both parts as we see that $\ell=d/r$.


问:
Is it possible to rewrite this equation using Einstein Summation?
Can we rewrite the below linked equation as: $$\overline{x} = \frac{x_i}{n} $$

Or is it necessary to have two variables and indices?

Here's the equation

答:
You could define $a:=(1,\dots,1)$ and then write $\bar x=\frac1n a_i x_i$. In general, Einstein is best used for expressions involving linear operations, that is, linear operators applied to vectors, or more generally, products of tensors. If you can rewrite an expression using linear operations, like the scalar product used above, you can use Einstein notation.

The reason why you need such operations is, as you said, that the notation only makes you take sums when an index comes up twice in a product. As for why the notation was chosen this way: exactly because physicists use it to describe linear operations. That's what it's made for.


问:
Eigen Vectors are not correct
I have this matrix
$\mathbf{A}=\left[ \begin{array}{rr}
-2& 2 \\
-0.4&0.2
\end{array} \right]$

I need to find its Eigen vectors. The book says its $\left[ \begin{array}{rr}
 2 \\
 0.4597
\end{array} \right]$

$\left[ \begin{array}{rr}
2 \\
1.7403
\end{array} \right]$
for $\lambda_1=-0.2597$ and $\lambda_2=-1.5403$. I used Matlab and it gave me the same result, but when I do the following, I'm not getting the same result

$(\mathbf{A}-\lambda \mathbf{I})\mathbf{x}$, for instance for $\lambda_2=-1.5403$ what I'm getting is $E_1=4.35E_2$ and $E_1=3.67E_2$. What am I doing wrong.

答:
Probably you're checking it incorrectly.
I don't know what's your $E_1$ and $E_2$, but if $\mathbf{x}_1$ and $\mathbf{x}_2$ are your two eigenvectors, then $(\mathbf{A}-\lambda_1\mathbf{I})\mathbf{x}_1=\mathbf{0}$ and $(\mathbf{A}-\lambda_2\mathbf{I})\mathbf{x}_2=\mathbf{0}$ indeed, so the eigenvalues and eigenvectors are correct.

Edit: the crucial additional information here is given in the comment by Jean-Claude Arbaut: The first eigenvector you give is associated with the second eigenvalue, and likewise for the second eigenvector and the first eigenvalue.

Title: Eigen Vectors are not correct


问:
Find convex closure of certain set
I'm having trouble solving this problem:

Given the set $X\subset\mathbb{R}^2$ defined as
$$X=\{(x,y)\in\mathbb{Z}\times\mathbb{R} : y\geq 0,x^2\leq b^2+y\},$$
being $b\in\mathbb{R}$. Is $X$ covex? Study the convex closure of $X$ in $\mathbb{R}^2$ in function of $b$. Define this closure in an analogous way to the set $X$, adding if necessary one new inequality. Note: consider absolute value.

I know how to prove $X$ is not convex, but I don't know how to answer the second question. The first thing I need to do is obviously change the condition $(x,y)\in\mathbb{Z}\times\mathbb{R}$ to $(x,y)\in\mathbb{R}^2$. But after this, I don't know how to express it. I clearly see the shape in my head, but I don't understand how to expess it with just one more inequality. I understand that
$$\{(x,y)\in\mathbb{R}^2 : y\geq 0,x^2\leq b^2+y\}$$
is NOT the convex closure of $X$, since you don't need the full parabola $x^2=b^2+y$ as the set's edge (instead, the edge of the convex closure will be an infinte number of segments joining each consecutive intersection points between $x\in\mathbb{Z}$ and $x^2=b^2+y$). I hope I have explained my point correctly

How can I solve this? Any help or hint will be appreciated, thanks in advance

答:
As we see clearly on the drawing provided by GNUSupporter :

The convex closure of $X$ is the set the one he provided you can see his answer as adding the inequality :

Because the fonction $ f: x \to x^2 -b^2 $ is convex it is under its cords.

You want $y$ to be upper the graph of $f$ but also upper the graph drawn by GNUSupporter :

That condition is the one given by GNU:

$$Y=\{(x,y)\in \mathbb{R}^2 \  y\geq 0, x^2\leq b^2 +y, \ y \geq \lfloor x \rfloor^2-b^2+(x-\lfloor x \rfloor)  (2 \lfloor x \rfloor+1) \} $$

but the second one is useless becausse it lies in  the third.

Which gives :

$$Y=\{(x,y) \in \mathbb{R}^2 \  y\geq 0, \ y \geq \lfloor x \rfloor^2-b^2+(x-\lfloor x \rfloor)  (2 \lfloor x \rfloor+1) \} $$

An advice, if a teacher ask you a question actually quite open saying if necessary. Remember with that kind of question and how it is ask you're rather free to propose a solution as given by GNU the importance is most of the time the methode and the answer more than follwing up too scrupulously. Being more scrupulossly can be a gain of time, energy and also , what is most beautiful, a liberty which allow you to solve an exercise rigourously but with a certain hindsight !

Title: Find convex closure of certain set


问:
Jacobian Transformations of Two Random Variables ( probability density function)
Let ${Y}_{1}$ and ${Y}_{2}$ be independent,each with density $\frac{1}{{y}^{2}}$ , $y>1$. Consider the transformation ${U}_{1}=\frac{{Y}_{2}}{{Y}_{1}+{Y}_{2}}$ and ${U}_{2}={Y}_{1}+{Y}_{2}$. Find probability density function of random variables $\left({U}_{1},{U}_{2}\right)$.

Here is my solution:

Since ${Y}_{1}$ and ${Y}_{2}$ are independent then $f({y}_{1},{y}_{2})=\frac{1}{{\left({y}_{1}{y}_{2}\right)}^{2}}$ , ${y}_{1}>1$ , ${y}_{2}>1$.

${U}_{1}=\frac{{Y}_{2}}{{Y}_{1}+{Y}_{2}}$ and ${U}_{2}={Y}_{1}+{Y}_{2}$ then ${Y}_{2}={U}_{1}{U}_{2}$ , ${Y}_{1}={U}_{2}-{U}_{1}{U}_{2}$

From here Jacobian is $J=\left|\begin{array}{cc}{u}_{2}& {u}_{1}\\ -{u}_{2}& 1-{u}_{1}\end{array}\right|={u}_{2}\Rightarrow \left|J\right|={u}_{2}$
Then I stuck here because I found the density function $f({u}_{1},{u}_{2})=\frac{{u}_{2}}{{\left({u}_{2}-{u}_{1}{u}_{2}\right)}^{2}{\left({u}_{1}{u}_{2}\right)}^{2}}$  but when I want to check it this function with integral definition I can not define the interval of integral. Any help will be appreciated.

答:
When defining joint support like this, you need to consider to specify the marginal support of which variable first. If you specify $u_1$ first, since

$$ u_1 = \frac {y_2} {y_1 + y_2} $$

then we see that

when $y_1$ fixed, $y_2 \to +\infty$, $u_1 \to 1$;

when $y_1 \to +\infty$, $y_2$ fixed, $u_1 \to 0$.

And obviously $0 < u_1 < 1$ so the support of $U_1$ is indeed $(0, 1)$

Let say we specify $u_1$ first. Then the "conditional" support of $u_2$ will be in terms of $u_1$. Since

$$ \begin{cases} u_2 - u_1u_2 &= y_1 > 1 \\ u_1u_2 &= y_2 > 1 \end{cases}
\Rightarrow \begin{cases} u_2 &> \displaystyle  \frac {1} {1 - u_1} \\  u_2 &> \displaystyle \frac {1} {u_1} \end{cases} $$

So the joint support can be specified as
$$ 0 < u_1 < 1, u_2 > \max\left\{\frac {1} {1-u_1}, \frac {1} {u_1} \right\}$$

Similarly if we want to specify $u_2$ first, obviously

$$u_2 = y_1 + y_2 > 1 + 1 = 2$$

so the marginal support of $U_2$ is $(2, +\infty)$. And

$$ \begin{cases} u_2 - u_1u_2 &= y_1 > 1 \\ u_1u_2 &= y_2 > 1 \end{cases}
\Rightarrow \begin{cases} u_1 &< \displaystyle 1 - \frac {1} {u_2} \\ u_1 &> \displaystyle \frac {1} {u_2} \end{cases} $$

So the joint support can be equivalently specified as

$$ u_2 > 2, \frac {1} {u_2} < u_1 < 1 - \frac {1} {u_2} $$

Verification:
Note
$$ \int \frac {1} {u^2(1-u)^2} du = \frac {1} {1-u} - \frac {1} {u}
- 2\ln (1-u) + 2\ln u + C$$

Therefore
$$ \begin{align}
&~ \int_{1/u_2}^{1-1/u_2} \frac {1} {u_1^2(1-u_1)^2u_2^3} du_1 \\
=&~ \frac {1} {1-1+1/u_2} - \frac {1} {1-1/u_2}
- 2\ln \left(1-1+\frac {1} {u_2}\right)
+ 2\ln\left(1- \frac {1} {u_2} \right) \\
&~ - \frac {1} {1-1/u_2} + \frac {1} {1/u_2}
+ 2\ln \left(1-\frac {1} {u_2}\right)
- 2\ln\left(\frac {1} {u_2} \right) \\
=&~ 2u_2 - \frac {2u_2} {u_2-1} + 4\ln(u_2-1)\end{align} $$

As a result,
$$ \begin{align}
&~ \int_2^{+\infty} \int_{1/u_2}^{1-1/u_2} \frac {1} {u_1^2(1-u_1)^2u_2^3} du_1du_2 \\
=&~ \int_2^{+\infty} \left[\frac {2} {u_2^2} - \frac {2} {u_2^2(u_2-1)} + \frac {4\ln(u_2-1)} {u_2^3} \right] du_2 \\
=&~ \left. - \frac {2\ln(u_2-1)} {u_2^2} - \frac {2} {u_2}  \right|_2^{+\infty} \\
=&~ 1 \end{align} $$


问:
Show that $\lim_{x \to 1} \sum_{n=1}^{\infty}\frac{n x^2}{n^3 + x^2}= \sum_{n=1}^{\infty} \frac{n}{n^3 +1}$.
What I understand from the question is that the author (N. L. Carothers) what me to show that the series $\sum_{n=1}^{\infty}\frac{n x^2}{n^3 + x^2}$ converges uniformly. Hence, the exchange between the limit and the summation is possible. However, It's been a week that I have been trying to find a summable upper bound so that the M-test will be applicable with no improvements. I tried AM-GM but did not get any results. Honestly, I do not publish the post here unless I try everything I know.

答:
Since you want to find the limit of the series on the left as $x$ goes to 1. You can prove the convergence in any neighborhood of 1.

Consider any neighborhood of 1, namely, consider the neigborhood $(1-\delta, 1+ \delta)$ where $\delta >0$. So

$\lVert \frac{n x^2}{n^3+x^2}\rVert_\infty = \sup_{x\in (1-\delta,1+\delta) } \big|\frac{n x^2}{n^3+x^2} \big| \le \big|\frac{x^2}{n^2}\big| \le \big| \frac{1+\delta}{n^2} \big|.$

Now, we found an upper bound $M_n = \frac{1+\delta}{n^2}$ and according to the M-test theorem $\sum_{n=1}^{\infty} \frac{1+\delta}{n^2}$ converges on the negoborhood $|x-1| <\delta,\,\,\, \forall \delta >0$.


问:
Shortest distance between a general point and a parabola.
Find the shortest distance between a point $(0, b)$ and the parabola $x^2=4y.$

My attempt: For a point $(x,y)$ on the parabola, the distance between a point $(x, y)$ on the parabola and given point $(0, b)$ is (by definition of distance formula)
 $$D^2=(x-0)^2+(y-b)^2.$$Since the point $(x, y)$ lies on the parabola, we have that $$D^2=4y+(y-b)^2.$$
Since $D$ is minimum implies that $D^2$  is also minimum. Therefore, I applied the first order necessary condition for minimum, i.e. $$\frac{dD^2}{dy}=0\implies y=b-2.$$Also, we have $\frac{d^2D^2}{dy^2}>0$ means that $D^2$ attains a minimum value at $y=b-2$ and the minimum value is  square root of $4b-4.$

But what if $b<1$? Please help why this method is not working or at what point I am not focusing on?

答:
The point you are missing is that for $b < 2$, we cannot have $y = b-2$, since $4y = x^2 \implies y\ge 0$.

Instead, notice that for $b<2$ (and $y\ge 0$):

$$\frac {d{D^2}}{dy} = 4+2(y-b)>4+2y-4\ge0$$

Hence $D^2$ is increasing as $y$ increases. This shows that the minimum is attained at $y=0$.


问:
Drawing dynamic circles based on input value
Is there a formula that will allow me to calculate the radius of a circle based on an input value?  The input value could be as small as zero or as large as $10^7$, or larger.  The circle is restricted to a minimum radius of $10$ and a maximum radius of $100$.

Does anyone know how to calculate something like this?

UPDATE

The input values correspond to state/country population.  I want to calculate the radius (how big the circle should be) of the circle based on the input value.

答: $$r=10+\frac { 90 }{ { 10 }^{ 7 } } P$$ where P is your input and r is the radius of the circle.</p>


问:
Why can we define an atlas on the unit circle $S^1$ via the logarithm?
I'm following Tu's Introduction to Manifolds to build an atlas on the unit circle $S^1$. In order to build charts, Tu defines the following maps:

$$ \phi_1(e^{it})=t, \; - \pi< t < \pi  , \\
\phi_2(e^{it})=t, \; 0< t < 2\pi. $$
He says that both maps are homeomorphisms onto their images. I don't understand this conclusion: an homeomorphism is supposed to be a continuous bijective map, and Tu himself says that $\phi_1$ and $\phi_2$ are branches of the complex logarithm. Isn't there a problem with the cut in the complex logarithm?

答:
Usually one defines a branch of the logarithm on a sliced plane $\mathbb C \setminus L(c)$, where $L(c) =\{\lambda c \mid \lambda \ge 0\}$ is a ray through some $c \ne 0$.

Of course we may assume $c \in S^1$. Writing $c = e^{i\gamma}$ with $\gamma \in \mathbb R$, for each $w \in \mathbb C \setminus L(c)$ there exists a unique $\varphi(w) = \varphi_\gamma(w) \in (\gamma,\gamma+2\pi)$ such that $e^{i\varphi(w)} = w/\lvert w \rvert$. Then a logarithm branch is given by

$$\ln = \ln_\gamma : \mathbb C \setminus L(c) \to \mathbb C, \ln w = \ln \lvert w \rvert + i\varphi(w)$$
The image of $\ln_\gamma$ is the open strip $\{x+iy \mid x > 0, \gamma < y < \gamma + 2\pi \} =(0,\infty) \times (\gamma,\gamma+2\pi)$.

If we pick another $\gamma'$ with $e^{i\gamma'} = c$, then we have $\gamma' - \gamma = 2k\pi$ for some $k \in \mathbb Z$ which gives another function $\varphi_{\gamma'} :  \mathbb C \setminus L(c) \to (\gamma+2k\pi,\gamma+ 2(k+1)\pi), \varphi_{\gamma'}(w) = \varphi_\gamma(w) + 2k\pi$, and another logarithm branch $\ln_{\gamma'}$ on $\mathbb C \setminus L(c)$.

Tu's maps are obtained as follows:

$c = 1$. Take $\gamma = 0$ and get $\phi_1(w) = \varphi_0(e^{it}) = t$ for $w = e^{it}$ with $t \in (0,2\pi)$.

$c = -1$. Takte $\gamma = -\pi$ and get $\phi_2(w) = \varphi_{-\pi}(e^{it}) = t$ for $w = e^{it}$ with $t \in (-\pi,\pi)$.

To see that these maps form an atlas on $S^1$, it is essential to know that $\varphi_\gamma \mid_{S^1 \setminus \{c\}} : S^1 \setminus \{c\} \to (\gamma,\gamma+2\pi)$ is a homeomorphism. Have a look at Why is $\sigma(u,v)=(\cos u,\sin u,v)$ a homeomorphism on $0< u <2\pi$? and Is $f(t)=(\cos(2\pi t),\sin(2\pi t))$ an open function?


问:
Let $B=\{f\in C[0,1] :\Vert f\Vert_{\infty} \leq 1\}$. Find all $f\in B$ such that $f=(g+h)/2$
Question says

For $(C[0,1], \Vert\cdot\Vert_{\infty})$, let $B=\{f\in C[0,1] :
 \Vert f\Vert_{\infty} \leq 1\}$. Find all $f\in B$ such that
there exist $g,h\in B$, $g\neq h$, with $f=\frac{g+h}{2}$

I studied some graphs and I feel that for $f \equiv 1$ and $f \equiv -1$, we can't find such $g,h, g\neq h$ which gives $f=\frac{g+h}{2}$ as only choice for $g$ and $h$ left is to be identically equal to  $1$ or $-1$, in respective case. For all other cases, we can increase and decrease function $f$ little bit(for those $x$ such that $\Vert f(x)\Vert \neq 1$) so as to get average equal to $f$. I can sketch that but not able to put it mathematically. Am I correct?

Thanks.

答: Hint for a rigorous proof: Suppose $|f(c)| <1$ for some $c$. Let $r=1-|f(c)|$. Let $g_n$ be a piece-wise linear function such that $g_n(c)=\frac r 2$ and $g_n(x)=0$ for $|x-c| \geq \frac 1 n$. [Draw a picture.] Then $f=\frac {(f+g_n) +(f-g_n)} 2$  and $|f(x)\pm g_n (x)| \leq 1$ for all $x$ provided $n$ is large enough.


问:
Conditional Probability: Bayesian Cause/Effect Question

The probability that a randomly chosen male has a circulation problem is 0.25.  Males who have a circulation problem are twice as likely to be smokers as those who do not have a circulation problem.  What is the conditional probability that a  male has a circulation problem, given that he is a smoker?

Let:

C = Circulation Problem

S = smoker.

The problem is asking you to solve for $\Pr(C \mid S)$.  I thought I could solve problem by using a table and I completed it below.  I interpreted the statements as such:

1:  "a randomly chosen male has a circulation problem is 0.25":  $\Pr(C)=0.25$

2:  "Males who have a circulation problem are twice as likely to be smokers as those who do not have a circulation problem.":  $\Pr(C \cap S) = 2x$ and $\Pr(C' \cap S) = x$

\begin{array}
{|c|c|c|c} \hline & C & C' &  \\ \hline S& 2x& x&\\ \hline S'& & &\\ \hline & 0.25 & &1\\
\hline . \end{array}

Which becomes:

\begin{array}
{|c|c|c|c} \hline & C & C' &  \\ \hline S& 2x& x & 3x\\ \hline S'& & &\\ \hline & 0.25 & 0.75 &1\\
\hline . \end{array}

So I did $\Pr(C\mid S) = \cfrac{\Pr(C \cap S)}{\Pr(S)}=\cfrac{2x}{3x}=2/3$ which is wrong.  The answer is 2/5!

Am I incorrectly interpreting statement 2 above as an intersection when instead it is a conditional probability?  In other words, should I be writing $\Pr(S\mid C)=2x$ and $\Pr(S\mid C')=x$?

Thank you in advance.

答:
I was able to answer/understand this question with Arthur's comments.  In this question, I had the problem of distinguishing between a conditional and nonconditional probability question.  This is also a good post to view for anyone who has problems:

How do we identify a probability problem as a conditional probability problem?

The proper way to mathematically represent the statements are as follows:

1:  $\Pr(C) = 0.25$:  The probability that a randomly chosen male has a circulation problem is 0.25.

2:  $\Pr(S \mid C) = 2 * \Pr(S \mid C')$:  Males who have a circulation problem are twice as likely to be smokers as those who do not have a circulation problem.  From this statement we can also infer:

3:  $\Pr(C \mid S):  $What is the conditional probability that a male has a circulation problem, given that he is a smoker?

In this question, we are asked for a specific group within a group which is hint to conditional probability.  This problem could be diagrammed as a tree where the first node would be whether the person has a circulation problem (C or C').  Then the C and C' nodes would each have two branches come from them:  S and S'.

In other words, whether you're a smoker or have a circulation problem are not equal status. One is higher than another if viewed as a hierarchy.

From definition of conditional probability:

$\Pr(C \mid S) = \cfrac{\Pr(C \cap S)}{\Pr(S)} = \cfrac{\Pr(C \cap S)}{Pr(S \cap C) + \Pr(S \cap C')}=\cfrac{0.25*2x}{0.25*2x + 0.75*x}=2/5$

Notice that in conditional probability problems, they can be both represented as a tree diagram or a venn box diagram.


答:
$$\int x\sqrt{1-x^2}sin^{-1}x dx$$
$$x = sin\theta$$
$$cos\theta= \frac{dx}{d\theta}$$
$$dx= cos\theta d\theta$$
$$\int x\sqrt{1-x^2}sin^{-1}x dx= \int sin\theta(\sqrt{cos^2\theta})\theta (cos\theta d\theta)$$
$$=\int \theta sin\theta cos^2\theta d\theta$$
$$=\int \theta sin\theta (1-sin^2\theta) d\theta$$
$$=\int \theta sin\theta -\int \theta sin^3\theta d\theta$$
$$=\int \theta sin\theta -\int \theta\Bigl(\frac{3sin\theta}{4}-\frac{sin3\theta}{4}\Bigl)  d\theta$$
$$=\int \theta sin\theta -\int\frac{3\theta sin\theta}{4}+ \int\frac{\theta sin3\theta}{4}d\theta$$
Applying product rule,
$$\int \theta sin\theta= \theta \int sin\theta - \int(\theta)' \int sin\theta d\theta= sin\theta - \theta cos\theta$$
$$-\int\frac{3\theta sin\theta}{4}= \frac{3\theta cos\theta}{4} -\frac{3 \sin\theta}{4}$$
$$\int\frac{\theta sin3\theta}{4}d\theta=\frac{1}{4}[\frac{sin3\theta}{9}-\frac{\theta cos3\theta}{3}]$$
$$\int x\sqrt{1-x^2}sin^{-1}x dx=sin\theta - \theta cos\theta\ + \frac{3\theta cos\theta}{4} -\frac{3 \sin\theta}{4}+\frac{sin3\theta}{36}-\frac{\theta cos3\theta}{12}$$
$$= \frac{1}{4}(sin \theta - \theta cos\theta)+ \frac{sin3\theta}{36}-\frac{\theta cos3\theta}{12}+ c$$
Substitute theta with corresponding expression in x

Title: Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$


问:
Understanding Complex Operators
I am not a follower of Schaum's Outline of Complex Variables but for seeking my professor lecture I read the third chapter called Complex Differentiation and the Cauchy-Riemann Equations. And I introduced some of the complex operators which isn't explained well manner in that chapter. Here are those:

If $A(x,y)=P(x,y)+iQ(x,y)$ is a complex
continuously differentiable function of $x$ and $y$ (vector? isn't those come from $\mathbb R$) then
$$\text{grad }A=\nabla A=\frac{\partial P}{\partial x}-\frac{\partial Q}{\partial y}+i\left(\frac{\partial P}{\partial y}+\frac{\partial Q}{\partial x}\right)$$

The $\nabla$ del symbol can be interpreted as a vector of partial derivative operators, and its three possible meanings—gradient, divergence, and curl—can be formally viewed as the product with a scalar, a dot product, and a cross product, respectively, of the del "operator" with the field. But I can't relate the above operation with any of those mentioned category (linked statement from Wikipedia). Rather it seems complex multiplication of two number.

The divergence of $A$ is,
$$\text{div }A=\nabla\cdot A=\text{Re}\{\bar\nabla A\}=\text{Re}\left\{\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)(P+iQ)\right \}$$

The curl of $A$ is,
$$|\text{curl }A|=|\nabla \times A|=|\text{Im}\{\bar \nabla A\}|=\left|\text{Im}\left\{\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y} \right)(P+iQ) \right\}  \right|=\left| \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right)\right|$$

There is no reference to how those formulas come from. I really want to know their derivation. Any reference or solution would be great helpful for me.

答:
In this context the nabla operator $\nabla$ is a vector differential operator defined for complex functions. Intuitively, if we have
$$ \nabla \equiv \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right) $$
in $ \mathbb{R}^2 $. Then the duple, viewed in the complex number sense, can be rewritten as
$$ \nabla \equiv \frac{\partial}{\partial x}+ i \frac{\partial}{\partial y}.  $$
By this virtue, we may also define the following:
$$ \bar\nabla \equiv \frac{\partial}{\partial x} - i \frac{\partial}{\partial y}.  $$
A few remarks can be made concerning the definitions above:

They are differential operators, which implies that they don't carry any meaning unless they are applied to a complex function.

The definitions of divergence and curl coincides with that for functions in $ \mathbb{R}^2 $ (you may check by direct computations).

The definition of gradient follows from multiplying $ \nabla $ and the function itself. However, it no longer carries the definition of gradient in $ \mathbb{R}^2 $.

Title: Understanding Complex Operators


问:
Proving that a Function does not have any extreme points
I need to find a continuous function, such that $$\lim_{x \to \infty}f(x)=L_{1}$$ $$\lim_{x \to -\infty}f(x)=L_{2}$$

And such that $f$ doesn't have any extreme points.

So I found this function here: $f(x)=\frac{x}{|x|+1}$ for every $x \in \mathbb{R}$.

I have proved that the function is continuous and also that  $\lim_{x \to \infty}f(x)=1$ and $\lim_{x \to -\infty}f(x)=-1$.

Now all I left to do is proving that this function has no extreme points at all.

I want to prove that the function is strictly increasing, and for that reason my theorem holds.

I have tried to choose $x_{1}, x_{2} \in \mathbb{R}$ such that $x_{1}<x_{1}$ and prove that $f(x_{1})<f(x_{2})$ but I find this difficult for me for some reason.

I will appreciate you help to finish my proof!

Thank you so much! :)

答: One we is to prove the function to be derivable for x $\in$ R. Then show derivative > 0.


答:
Suppose $f$ has a point $x$ such that $-1<f(x)<1$. Then, by continuity of $f$, there is a whole interval about $x$, say $(x-\delta,x+\delta)$ for which this is true. Let $s:=\max\{1-|f(t)|:x-\delta\le t\le x+\delta\}>0$.

Let $g(t):=f(t)+r(t-x)$ and $h(t):=f(t)-r(t-x)$ where $r(t)=s(1-\min(1,|t|))$ is the tent function centered at zero with height $s$. Then clearly $f=\frac{g+h}{2}$ and $|g(t)|\le|f(t)|+s\le1$; similarly for $h$. Hence $g,h\in B$ and $g\ne h$, with the required properties.

Title: Let $B=\{f\in C[0,1] :\Vert f\Vert_{\infty} \leq 1\}$. Find all $f\in B$ such that $f=(g+h)/2$

Body:
Question says

For $(C[0,1], \Vert\cdot\Vert_{\infty})$, let $B=\{f\in C[0,1] :
 \Vert f\Vert_{\infty} \leq 1\}$. Find all $f\in B$ such that
there exist $g,h\in B$, $g\neq h$, with $f=\frac{g+h}{2}$

I studied some graphs and I feel that for $f \equiv 1$ and $f \equiv -1$, we can't find such $g,h, g\neq h$ which gives $f=\frac{g+h}{2}$ as only choice for $g$ and $h$ left is to be identically equal to  $1$ or $-1$, in respective case. For all other cases, we can increase and decrease function $f$ little bit(for those $x$ such that $\Vert f(x)\Vert \neq 1$) so as to get average equal to $f$. I can sketch that but not able to put it mathematically. Am I correct?

Thanks.


问:
Determine the dimension of the subspace of all matrices that commutes with an all 1 matrix and find a basis

Let $\Gamma$ denote the $n$-by-$n$ matrix which has $1$ in all its entries. Denote
$$S := \{A \in \mathcal{M}_n(\mathbb{R}) : A\Gamma = \Gamma A\}$$
Determine the dimension of $S$, and find a basis for $S$.

Simple calculation tells that $S$ consists of all matrices with the sum of each row and column agreeing on a value. This is similar to the case of Magic matrices, so I found a literature for that and conclude that the dimension is $\,n^2 - 2n + 2\,$.

Problem is, how do I find a basis for it?

Exhausting all the possibilities by hand seems undesirable, and I can't think of a way to do it systematically.

答:
If $A \in  S$ and $s$ is the sum of each line or column of $A$, then $A-\frac{s}{n} {\Gamma} \in  {\mathcal{V}}_{n}$, the space of ${n}\times{n}$ matrices which sum of each line or column is $0$. We can find a basis of ${\mathcal{V}}_{n}$ and add ${\Gamma}$ to this basis to get a basis of $S$.

Let us now suppose that $A = \left({a}_{i , j}\right) \in  {\mathcal{V}}_{n}$ and let $m = n+1$. Let $\left({x}_{1} , \cdots  , {x}_{n}\right)$ and
$\left({y}_{1} , \cdots  , {y}_{n}\right)$ be $2 n$ arbitrary scalars. Let us build a matrix $B = {b}_{i , j} \in  {\mathcal{V}}_{m}$ by defining

\begin{equation}
\left\{\begin{array}{lcl}{b}_{i , j}&=&{a}_{i , j}+{x}_{j}+{y}_{i} \quad  \text{ if } i , j  \leqslant  n\\
{b}_{i , m}&=&{-{s}_{x}}-n {y}_{i} \quad  \text{ if } i  \leqslant  n\\
{b}_{m , j}&=&{-n} {x}_{j}-{s}_{y} \quad  \text{ if } j  \leqslant  n\\
{b}_{m , m}&=&n {s}_{x}+n {s}_{y}
\end{array}\right.
\end{equation}

where ${s}_{x} = \sum _{j} {x}_{j}$ and ${s}_{y} = \sum _{i} {y}_{i}$. Conversely, any matrix $B \in  {\mathcal{V}}_{m}$ can be written in this way
because the ${x}_{j}$ and ${y}_{i}$ must be defined by

\begin{equation}
\renewcommand{\arraystretch}{1.5}  \left\{\begin{array}{lcl}{x}_{j}&=&{-\frac{1}{n}} {b}_{m , j}-\frac{1}{n} {s}_{y}\\
{y}_{i}&=&{-\frac{1}{n}} {b}_{i , m}-\frac{1}{n} {s}_{x}
\end{array}\right.
\end{equation}

The sums of the last line and column of $B$ plus the constraint that
${b}_{m , m} = n {s}_{x}+n {s}_{y}$ then imply that ${s}_{x} = \sum _{j} {x}_{j}$ and ${s}_{y} = \sum _{i} {y}_{i}$. It follows that the matrix
${a}_{i , j} = {b}_{i , j}-{x}_{j}-{y}_{i}$ for $i , j  \leqslant  n$
belongs to ${\mathcal{V}}_{n}$. This decomposition is not unique because we have one degree of freedom in the choice of $s_x$ an $s_y$. We could get a unique decomposition by choosing for example $s_y=0$.

From this we deduce the induction relation $\dim  {\mathcal{V}}_{n+1} = (2 n-1)+\dim  {\mathcal{V}}_{n}$,
hence $\dim {\mathcal{V}}_{n} = (n-1)^2$ because $\dim {\mathcal{V}}_{1}=0$ and the construction gives us a way to build a basis of these vector spaces by taking the matrices in factor of one of the ${x}_{j}$'s or
${y}_{i}$'s in the process.


问:
In how many ways can you select $4$ different numbers from the numbers $1,2,..,15$
In how many ways can you select $4$ different numbers from the numbers $1,2,..,15$ so that no pair of numbers among the selected ones are subsequent numbers.

I turned this into a minor problem. I assumed that we have $5$ numbers and we want to choose two different numbers so that no pair of numbers among the selected ones is subsequent. I came to the conclusion that it will be $5$ choose $2$ and then minus all subsequent pairs that I calculated in my head. But I wonder if there is a better method that can be applied to $15$ numbers.

Also, how would you make this problem more difficult.

答:
Take out $4$ balls from a line of $15$ balls, so $11$ remain, then  insert the $4$ balls taken out in $\binom{12}4$ ways at the uparrows so no two can be adjacent

$\uparrow\bullet\uparrow\bullet\uparrow\bullet\uparrow\bullet\uparrow\bullet\uparrow\bullet\uparrow\bullet\uparrow\bullet\uparrow\bullet\uparrow\bullet\uparrow\bullet\uparrow{}$

Now number the balls $1\, thru\, 15$

The ones you inserted back represent the chosen numbers


问:
If a fair coin is tossed repeatedly, what is the probability that the $3rd$ head occurs on the $nth$ toss?
I have solved for the case in which I found the probability that $1st$ head occurs in $nth$ toss, but I don't know how to solve for this problem which is a little bit different. Can anyone guide or help??

$PROBLEM:$

If a fair coin is tossed repeatedly, what is the probability that the $3rd$
head occurs on the $nth$ toss?

答: This is equal to the probability of getting 2 heads in the first $n-1$ tosses, and the 3rd head in the $n$th toss. The probability is equal to $$\binom{n-1}{2}\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{n-3}\cdot \frac{1}{2}=\binom{n-1}{2}\left(\frac{1}{2}\right)^n$$


问:
Find the limit $ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$
For $a,b,c>0$, Find $$ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$$

how can I find the limit of sequence above?
Provide me a hint or full solution.
thanks ^^

答:
$a^{\frac 1n}=1 +\dfrac{\log a}{n}+O(\frac1{n^2})$

So,$\Big(\dfrac{a^{\frac 1 n}+b^{\frac 1 n}+c^{\frac 1 n}}3\Big)^n=(1+\frac{\log abc}{3n}+O(\frac1{n^2}))^n\to \sqrt[3]{abc}$ This generalizes easily to 4 or more terms also.


问:
Counter example of "True or False" statement that is not clear for me
Let $f$ be piecewise function $f: \mathbb{R} \rightarrow \mathbb{C}$ such that $ \int_{-\infty}^{\infty}|f(x)| d x<+\infty$

Statement:
There is necessary exist $g$ such that $ \int_{-\infty}^{\infty}|g(\omega)| d \omega<+\infty$ and $f(x)= \int_{-\infty}^{\infty}g(\omega)e^{i\omega x} d \omega$

Answer :
False

take  $f(x)$=the indicator $\mathbb{\chi}_{[-1,1]}$

I dont understand this counter example, how does it contradict the statement?

答: $\int e^{-iwx}f(x)dx=\frac {\sin w} w$. This implies that $f$ is the Fourier Transform of $g(w)=\frac {\sin w }w$ but $g$ is not inetgrable.


问:
Approximating Probabilities of Winning a Game with a 9% Winning Chance
A game has a winning probability of $9\%$. If you play $10$ times, what is the probability that you win $$(i)\text{ exactly once}$$ $$(ii) \text{at least once}$$ $$(iii) \text{less than 3 times or more than 5 times}$$

Since this is a binomial distribution $\frac{winning}{losing}$, I approached (i) the following way:

$$\frac{10!}{9!} \times {0.09}^1 \times {0.91}^9 = 38.5\%$$

Is this correct? Because it seems rather high to me that one has almost a $40%$ probability of winning EXACTLY once. But my intuition might be (as is usually is) flawed.

For the second subtask now I assume that one could simply calculate

$$1 - {0.91}^{10} = 61.1\%$$

Here I wanted to ask, I one were not to use this shortcut, but to actually add all binomial probabilities, if it would be admissible to approximate the result normally in the following way:

$$E(x) = np = 100.09 = 0.9~~ \text{and}~~ \sigma = \sqrt {({100.09}\times{0.91})} = 0.905$$

If we include all results from $1$ to $10$ wins, with continuity correction $(1-0.5 = 0.5)$ and standardisation we have

$$1 - P(X ≥ 1) = P\left(Z ≥ \frac{(0.5-0.9)}{0.905}\right) = P(Z ≥ -0.442) = 67.1\%$$

However, this approximation seems quite off form the actual solution of $61.1\%$, where did I do a mistake? Or can't you use a normal approximation for the task in question?

For the last subtask I again started by calculating the binomial probability:

$$1 - \text{P(three wins) - P(four wins) - P(five wins)} \\=1 - \left(\frac{10!}{3!7!} \times {0.09}^3 \times {0.91}^7 \right) - \left(\frac{10!}{4!6!} \times {0.09}^4 \times {0.91}^6) \right) - \left(\frac{10!}{5!5!} \times {0.09}^5 \times {0.91}^5 \right)= 94.6 \%$$ Would there be a faster way to calculate this?

Again, if you would approximate normally:

$$1 - P\left(\frac{(2.5-0.9)}{0.905} ≤ Z ≤ \frac{(5.5-0.9)}{0.905}\right) = 1 - P(1.768 ≤ Z ≤ 5.083) = 96.2\%$$

I seem to be doing something wrong with the normal approximation, but I can't figure out what is is. Can someone spot the mistake?

Body:
Comparison of Binomial and Poisson Distributions

Exactly one time:
$$
\binom{10}{1}.09^1.91^9=0.38513682
$$
Poisson approximation: (off by $5.0\%$)
$$
e^{-.9}\frac{.9^1}{1!}=0.36591269
$$
At least once:
$$
1-\binom{10}{0}.09^0.91^{10}=0.61058388
$$
Poisson approximation: (off by $2.8\%$)
$$
1-e^{-.9}\frac{.9^0}{0!}=0.59343034
$$
Less than three or more than five:
$$
1-\sum_{k=3}^5\binom{10}{k}.09^k.91^{10-k}=0.94604101
$$
Poisson Approximation: (off by $0.9\%$)
$$
1-\sum_{k=3}^5e^{-.9}\frac{.9^k}{k!}=0.93748656
$$


问:
If all edges in a network have distinct capacities, then the min-cut is unique
I'm stuck trying to prove that If all edges in a network have distinct capacities, then the min-cut is unique. I'd like to know if there is some proof that doesn't involve going through F&F algorithm, or designing one. Instead, I tried proving it by contradiction, assuming the existence of two different cuts with equal capacities (and both minimums) but I don't know how to arrive to a contradiction.

答: There is the following counterexample for the claim you trying to prove.


答:
There is no need for calculus. As you found,

$d^2=4y+(y-b)^2 = y^2 - (2b-4)y + b^2 = \big(y - (b-2)\big)^2 + 4b-4$

So you can clearly see that $d^2$ is minimum when we minimize $|y - (b-2)|$

But as we must have $y \geq 0$,

$y = b-2 \,$ for $b \geq 2$ otherwise $y = 0$.

Title: Shortest distance between a general point and a parabola.

Body:
Find the shortest distance between a point $(0, b)$ and the parabola $x^2=4y.$

My attempt: For a point $(x,y)$ on the parabola, the distance between a point $(x, y)$ on the parabola and given point $(0, b)$ is (by definition of distance formula)
 $$D^2=(x-0)^2+(y-b)^2.$$Since the point $(x, y)$ lies on the parabola, we have that $$D^2=4y+(y-b)^2.$$
Since $D$ is minimum implies that $D^2$  is also minimum. Therefore, I applied the first order necessary condition for minimum, i.e. $$\frac{dD^2}{dy}=0\implies y=b-2.$$Also, we have $\frac{d^2D^2}{dy^2}>0$ means that $D^2$ attains a minimum value at $y=b-2$ and the minimum value is  square root of $4b-4.$

But what if $b<1$? Please help why this method is not working or at what point I am not focusing on?


问:
compute$\iint_D x^2y+y\sin(x^9) \ dxdy$
Can someone please help me with this integral
$$\iint_D x^2y+y\sin(x^9) \ dxdy $$
$$D=\{(x,y):x^2+y^2\leq2,y>0\}$$

I have done like this:
$$\int_{x=-\sqrt2}^{\sqrt2}\int_{y=0}^{\sqrt{2-x^2}} x^2y+y\sin(x^9) \ dydx$$
but the answer is wrong why?

Any suggestion would be great
Thanks

答: The $y\sin x^9$ part, being odd in $x$, doesn't contribute. Since $D$ has polar specification $0\le r\le\sqrt{2},\,0<\theta<\pi$, the integral is $\int_0^\pi\cos^2\theta\sin\theta d\theta\int_0^\sqrt{2}r^4dr$, which I leave you to evaluate.

Title: compute$\iint_D x^2y+y\sin(x^9) \ dxdy$


问:
Number of way choose ordered distinct quadruples $(A,B,C,D)$ from $\{1,2,...,n\}(n\geq 2)$

Number of way choose ordered distinct quadruples $(A,B,C,D)$ from
$\{1,2,...,n\}(n\geq 2)$ such that  $A\subseteq B\cup C\cup D$.

I think in following manner:
$\forall x\in \{1,2,...,n\} $
if $A$ included it then it must be in  $B \vee C \vee D$ or if $x$ not in $A$ it can be at any $B \vee C \vee D$ such a way give us $7^n$ .Any one can verify my solution?

答:
Let's count the $(A,B,C,D)$ for a given choice of $S=B\cup C\cup D$ (where $k=|S|$).

Draw a Venn diagram for $B,C,D$. We can distribute the elements of $S$ among the $7$ regions, and that corresponds to a choice of $(B,C,D)$. Labelling the regions $1$-$7$, this is the same as a function from $S$ to the set $\{1,\cdots,7\}$, of which there are $7^k$. Also there are $2^k$ choices for $A\subseteq S$.

There are $\binom{n}{k}$ choices for $S\subseteq\{1,\cdots,n\}$, so summing over $k$ we get

$$ \sum_{k=0}^n \binom{n}{k}\cdot7^k\cdot2^k. $$

Do you see why this is $15^n$?


答:
We can choose $A$ on $2^n$ ways.

For each $a\in A$ there must be at least one from $\{B,C,D\}$ which contains $a$. That we can do on $2^3-1 =7$ ways. So if $|A|=k$ then we have $7^k$ options. The rest of $n-k$ elements we can arrange in sets $B,C, D$ arbitrary, so that is $8^{n-k}$ ways.

So we have $$\sum _{k=0}^n{n\choose k}\cdot 7^k\cdot 8^{n-k} = 15^n$$ 4-couples $(A,B,C,D)$.

Body:
Number of way choose ordered distinct quadruples $(A,B,C,D)$ from
$\{1,2,...,n\}(n\geq 2)$ such that  $A\subseteq B\cup C\cup D$.

I think in following manner:
$\forall x\in \{1,2,...,n\} $
if $A$ included it then it must be in  $B \vee C \vee D$ or if $x$ not in $A$ it can be at any $B \vee C \vee D$ such a way give us $7^n$ .Any one can verify my solution?


问:
Positive coefficients after the substitution?
Let $n>1$ be an integer and $\Phi_n(x)$ be the $n$ th cyclotomic polynomial.

If we subsititute $x$ by $x+1$ , do we always get a polynomial with positive coefficients ? If yes, how can this be proven ? I also want to prove that no
coeffcient upto the $\varphi(n)$ th is missing.

Examples :

$\Phi(6)=x^2-x+1$ transforms into $x^2+x+1$

$\Phi(24)=x^8-x^4+1$ transforms into $x^8+8x^7+28x^6+56x^5+69x^4+52x^3+22x^2+4x+1$

Upto $n=3\ 200$ , this is the case. Some special cases can be easily proven, for example that $n$ is a power of $2$ or that $n$ is a prime. But I have no idea for a general proof.

答:
Deal with the case $n=1,$ $ n = 2$ separately.

Notice that $ \Phi_1(x+1) = (x+1) - 1 = x$, which has a non-positive constant term.

This will be the only coefficient of $ \Phi_n (x+1)$ which is non-positive.

For $ n > 2$,

Let $A_n$ denote the set of integers $ 0 < k < n$ such that $ \gcd( n,k) = 1$.

Let $A_n^*$ denote the set of integers $ 0 < k < n/2$ such that $ \gcd( n,k) = 1$.

$$\Phi_n(x)  = \prod_{j \in A_n } (x - e^{2k\pi/n}) =  \prod_{j \in A_n^*} \left( x^2 - 2 \cos \frac{ 2\pi j}{n} \cdot x + 1 \right) .$$

Let $ f_m (x)  =  x^2 - 2mx + 1 $.

Since $f_m(x+1) = x^2 + 2(1-m)x + 2 (1-m)$, so the coefficients of $f_m(x+1)$ are positive for $ m < 1$.

Thus, for $n > 2$, since $ \cos \frac{2 \pi j } { n} < 1$ for $ j \in A_n^*$
$$\Phi_n(x+1) = \prod_{j \in A_n^*} f_{\cos \frac{2\pi j  }{n}}(x+1) ,$$
where the latter is a product of polynomials with positive coefficients, hence $\Phi_n(x+1)$ has positive coefficients for up to the degree $ \phi(n)$.


问:
How to show a derivative is a maximum, not a minimum
I am working through a maths text book. I study as a hobby. I have solved most of this problem except for the very last part.

The points A and B are on the same horizontal level and at a distance of $b$ apart. A particle P falls vertically at rest from B, so that at time $t$ its depth below B is $kt^2$, where $k$ is constant. At this time, the angle of depression of P from A is $\theta$.

Prove that $\frac{d\theta}{dt} = \frac{2bkt}{b^2 + k^2t^4}$. Show that $\frac{d\theta}{dt}$ is greatest (and not least) when $\theta = \frac{1}{6}\pi$.

I proved the first part and to then I differentiated $\frac{d\theta}{dt} = \frac{2bkt}{b^2 + k^2t^4}$ to get the second derivative:

$\frac{d^2\theta}{dt^2} = \{(b^2 + k^2t^4)(2bk) - (2bkt)(4k^3t^4)\}/(b^2 + k^2t^4)^2$

$= 0$ when $2b^3k +2bk^3t^4 - 8bk^3t^4 = 0$

$\implies b^2k = 3k^2t^4$

Now when $\theta = \frac{1}{6}\pi$, $\tan \theta = \frac{1}{\sqrt3}$ and

$\frac {kt^2}{b} = \frac{1}{\sqrt3}$, meaning $b^2 = 3k^2t^4$

So $\theta = \frac{1}{6}\pi$ is a maximum or a minimum.

But how do I show which? Taking a third derivative seems very complicated.

答:
I got that second derivative is zero when
$$b^2-3 k^2 t^4=0;\;t=\pm\sqrt[4]{\frac{b^2}{3 k^2}}$$
Compute the third derivative$^*$
$$\frac{d^3\theta}{dt^3}=-\frac{8 b k^3 t^3 \left(5 b^2-3 k^2 t^4\right)}{\left(b^2+k^2 t^4\right)^3}$$
which is negative for $t_0=\sqrt[4]{\frac{b^2}{3 k^2}}$ so that $t_0$ is a maximum.

This is called  [second derivative test][1].

Alternative method

Set second derivative positive
$$b^2-3 k^2 t^4>0\to -\sqrt[4]{\frac{b^2}{3 k^2}}<t<\sqrt[4]{\frac{b^2}{3 k^2}}$$
means that at $t_0$ the derivative goes from positive to negative therefore $t_0$ is a maximum

$(*)$

First derivative is
$$\frac{d\theta}{dt}=2 b k\left[ \frac{ t}{b^2+k^2 t^4}\right]$$
Second derivative is
$$\frac{d^2\theta}{dt^2}=2bk\left[\frac{1 \left(b^2+k^2 t^4\right)-t \left(4 k^2 t^3\right)}{\left(b^2+k^2 t^4\right)^2}\right]=\frac{2 b k \left(b^2-3 k^2 t^4\right)}{\left(b^2+k^2 t^4\right)^2}$$

Third derivative is
$$\frac{d^3\theta}{dt^3}=2bk\left[\frac{\left(-12 k^2 t^3\right) \left(b^2+k^2 t^4\right)^2-2 \left(4 k^2 t^3\right) \left(b^2-3 k^2 t^4\right) \left(b^2+k^2 t^4\right)}{\left(b^2+k^2 t^4\right)^4}\right]=$$
$$=-\frac{8 b k^3 t^3 \left(5 b^2-3 k^2 t^4\right)}{\left(b^2+k^2 t^4\right)^3}$$
[1]: https://en.wikipedia.org/wiki/Derivative_test#Second-derivative_test_(single_variable)


问:
computation and calculation of limits
We just completed limits in school, and I have quite a few doubts.

Limit x tends to 0 we found sin x=x and cos x=1-x.
Subsequently we could substitute x as 0 in cos x to give 1 but not sin x to give 0. Can someone explain why?

Secondly, can someone explain why substituing a polynomial partially is wrong? Consider 2x/x as limit x tends to 0. We could write this is (x+x)/x and then substitue x=0 for the first x on the top to get x/x.

Thirdly, how can a fraction which initially starts in some form 0/0 or something like that evaluate to a finite number? Geometrically I understand it, but not algebraicaly.

Lastly, Can someone refer some sort of course/book which deals in limits in details and rigour from scratch?

答:
Note that $\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x}$ makes sense, and so does $\displaystyle \lim_{x \to 0} \dfrac{\cos x}{1-x}$, and both are equal to $1$. What this means is, when $x$ gets closer and closer to $0$, the expressions approach the value $1$. You can read the proof for the first limit here.

But when you write $\sin x=x$, this means you want a solution to this equation, that is, a value of $x$ which will satisfy this equation. I assume that you have read the indeterminate forms wikipedia page which I mentioned in comments. So, if an expression on substituting $x=0$, becomes of the form $0/0$ or any other indeterminate form mentioned, you cannot evaluate it's value at that point, simply because it's not defined at that point. What then you can do is, try and see what value does the expression approaches when $x$ approaches $0$, because it is defined in it's neighbourhood, and that is what it means to take the limit at that point.

You cannot substitute $x=0$ partially in an expression. It's totally wrong. You have to replace every $x$ by a $0$ while substituting.

A good book for starting limits would be "Essential Calculus: Early Transcendentals by James Stewart".

Title: computation and calculation of limits


答:
$x$ should be integrated over first:
$$\int_0^{\sqrt2}y\int_{-\sqrt{2-y^2}}^{\sqrt{2-y^2}}(x^2+\sin x^9)\,dx\,dy$$
Since $\sin x^9$ is an odd function and we are integrating over a domain symmetric about its oddness point, its contribution to the integral is zero and we may ignore it:
$$=\int_0^{\sqrt2}y\int_{-\sqrt{2-y^2}}^{\sqrt{2-y^2}}x^2\,dx\,dy$$
$$=\int_0^{\sqrt2}y[x^3/3]_{-\sqrt{2-y^2}}^{\sqrt{2-y^2}}\,dy$$
$$=\frac23\int_0^{\sqrt2}y(2-y^2)^{3/2}\,dy$$
$$=-\frac13\int_0^{\sqrt2}-2y(2-y^2)^{3/2}\,dy$$
$$=-\frac13\int_2^0u^{3/2}\,du$$
$$=\frac13\cdot\frac25[u^{5/2}]_0^2$$
$$=\frac{2^{7/2}}{15}$$

Title: compute$\iint_D x^2y+y\sin(x^9) \ dxdy$

Body:
Can someone please help me with this integral
$$\iint_D x^2y+y\sin(x^9) \ dxdy $$
$$D=\{(x,y):x^2+y^2\leq2,y>0\}$$

I have done like this:
$$\int_{x=-\sqrt2}^{\sqrt2}\int_{y=0}^{\sqrt{2-x^2}} x^2y+y\sin(x^9) \ dydx$$
but the answer is wrong why?

Any suggestion would be great
Thanks


问:
Person X and Y throw a Fair die one after another. Whoever throws 6 first wins. What is the probability that X wins?
Two people take turns rolling a fair die. Person $X$ rolls first, then
person $Y$ , then $X$, and so on. The winner is the first to roll a $6$. What is the
probability that person $X$ wins?

Our teacher in class solved this question as follows:

Probability of winning in 1st round = Probability that 6 occurs = 1/6

So Probability of $Y$ losing in first round becomes $= 5/6$

Where it is assumed that X starts the game.

suppose if X throws and if 6 comes then the game will stop and X will win. but if X loses then Y will throw the die (here Y will have to lose because according to question we want X to win the game) and then X will throw . and so on.

it will be like
$$X+ \neg X\neg YX+ \neg X\neg Y\neg X\neg YX+\dots$$

HERE $X$ denotes X wins and $\neg X$ means X loses i.e. 6 does not come . similarly $\neg Y$ means Y loses.

This will form an infinite geometric progression series. Converting into probabilities we have
$$(1/6) + (5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(5/6)(1/6)+ \dots$$
The sum of this geometric series is the answer.

MY question is:
how the probability of $X$ in $2nd$ toss is $(5/6)(5/6)(1/6)$ and similarly in the following tosses? Why these terms are in multiplication not addition?

答:
Let $p$ be the probability that player $1$ (X) wins.

Then, the probability that player $2$ wins is $$\frac{5}{6}\cdot p$$

Both events are mutually excluded and one of them must happen, so we have

$$p+\frac{5}{6}p=1$$

This gives $$p=\frac{6}{11}$$


问:
Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$
I tried $$\int{x\sqrt{1-x^2}\arcsin{x}\ \mathrm{d}x}$$$$=\int{\arcsin{x}\ \mathrm{d}\left(\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\ \mathrm{d}\left(\arcsin{x}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3\sqrt{1-x^2}}\ \mathrm{d}x}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\frac{2}{3}\int{x^3\ \mathrm{d}x}+\frac{2}{3}\int{x^5\ \mathrm{d}x}$$$$=\cdots$$
This solution isn't true. Where am I wrong?

EDIT: Thank you all for the many different answers! If I could, I'd give you all the accepted answer.

答:
Let $x=\sin{u}$. Substitute $\arcsin{x}=u$ to change variables, notice that $d(\cos^3{u})=-3\sin{u}\cos^2{u}\,du$ and get the integral by parts
$$
\int x\sqrt{1-x^2}\arcsin{x}\,dx =
 \int{u}\,(\sin{u}\cos^2{u}\,du) =
 -\frac{1}{3}\int{u}\,d(\cos^3{u}) =
 \frac{1}{3}\Bigl(\int\cos^3{u}\,du - u\cos^3{u} \Bigr)
$$
After back substitution the last integral looks
$$
\frac{1}{3}\Bigl(\int(1-x^2)\,dx - (1-x^2)^\frac{3}{2}\,\arcsin{x} \Bigr) =
 \frac{1}{3}\Bigl( x - \frac{x^3}{3} - (1-x^2)^\frac{3}{2}\,\arcsin{x} \Bigr)
$$
Don't forget a constant and get the answer
$$
\int{x\sqrt{1-x^2}\arcsin{x}}\,dx =
 \frac{1}{3}\Bigl( x - \frac{x^3}{3} - (1-x^2)^\frac{3}{2}\,\arcsin{x} \Bigr) + C
$$
Check the math and good luck!

Title: Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$


问:
$\pi$-system, $\sigma$-field and Monotone class
I'm trying to solve exercise 1.9.42 in the Resnick book. This exercise involves the concept of monotonous class, $\pi$-system and $\sigma$-algebra.

Assume $\mathcal{P}$ is a $\pi-system$ and $\mathcal{M}$ is a monotone class. Show $\mathcal{P}\subset\mathcal{M}$ does not imply $\sigma(\mathcal{P})\subset\mathcal{M}$.

I started looking for examples in which I have a set that is a $\pi-system$ and a monotonous class, but that the $\sigma(\mathcal{P})$ is not a monotone class. But I doubt if this is really what the exercise calls for, as it would make no sense to find a $\sigma$-algebra that does not belong to the monotone class, since every $\sigma$-algebra is a monotone class.
Anyway I thought of the following example:

$\Omega$={1,2,3,4}

P={{1,2,3,4},{2,3,4},{3,4},{4}}

It's a $\pi$ system?

{1,2,3,4}$\bigcap${2,3,4}={2,3,4} ok

{2,3,4}$\bigcap${3,4} ok

{2,3,4}$\bigcap${4} ok

{3,4}$\bigcap{4}$ ok

It's a monotone classe?

{1,2,3,4}$\supset${2,3,4}$\supset${3,4}$\supset${4}

Now I would need to check the sigma algebra ...

That makes sense?

答:
I think you msiunderstood the question. You have to come up with two families $\mathcal P$ and $\mathcal M$.

Let $\Omega=\{1,2\}, \mathcal P=\{\{1\}\}$ and $\mathcal M=\{\{1\}, \{1,2\}\}$. Then $\mathcal P$ is a $\pi-$ system and  $\mathcal M$ is a monotone class. But $\{2\}=\Omega \setminus \{1\} \in \sigma (\mathcal P)$, $\{2\} \notin \mathcal M$.


答:
It is not that complicated.

$$\tan\theta\; = \; \frac{kt^2}{b}$$
$$\sec^2{\theta}\ d\theta\; = \; \frac{2ktdt}{b}$$
$$\frac{d\theta}{dt}\;= \frac{2kt \cos^2\theta}{b}$$
$$\frac{d^2\theta}{dt^2}\;= \frac{2k \cos^2\theta}{b}\; -\; \frac{4kt \cos{\theta}\sin{\theta}}{b}.\frac{d\theta}{dt}$$

While taking 3rd derivative put $\frac{d^2\theta}{dt^2}\: = \: 0$.

Note: Use $\sin2\theta$

Title: How to show a derivative is a maximum, not a minimum

Body:
I am working through a maths text book. I study as a hobby. I have solved most of this problem except for the very last part.

The points A and B are on the same horizontal level and at a distance of $b$ apart. A particle P falls vertically at rest from B, so that at time $t$ its depth below B is $kt^2$, where $k$ is constant. At this time, the angle of depression of P from A is $\theta$.

Prove that $\frac{d\theta}{dt} = \frac{2bkt}{b^2 + k^2t^4}$. Show that $\frac{d\theta}{dt}$ is greatest (and not least) when $\theta = \frac{1}{6}\pi$.

I proved the first part and to then I differentiated $\frac{d\theta}{dt} = \frac{2bkt}{b^2 + k^2t^4}$ to get the second derivative:

$\frac{d^2\theta}{dt^2} = \{(b^2 + k^2t^4)(2bk) - (2bkt)(4k^3t^4)\}/(b^2 + k^2t^4)^2$

$= 0$ when $2b^3k +2bk^3t^4 - 8bk^3t^4 = 0$

$\implies b^2k = 3k^2t^4$

Now when $\theta = \frac{1}{6}\pi$, $\tan \theta = \frac{1}{\sqrt3}$ and

$\frac {kt^2}{b} = \frac{1}{\sqrt3}$, meaning $b^2 = 3k^2t^4$

So $\theta = \frac{1}{6}\pi$ is a maximum or a minimum.

But how do I show which? Taking a third derivative seems very complicated.


问:
Area of square inside a triangle
Considering the attached image, please compute the area of the square.

$\\$To elaborate suppose we have a triangle $\Delta ABC$, that its angle $\angle ABC$ is equal to $45^o$. On side $\overline{AB}$ we have a point $E$ that divide the side to two segments, $\overline{BE}=4$ and $\overline{EA}=6$.

Find the area of the square $\square CDEF$, such that its $\overline{FC}$ side reside on $\overline{AC}$ side of the triangle $\Delta ABC$, its one vertex is on E point, and the alternate vertex of E is on vertex C.

答:
Hint: In $\triangle BCE$, $D$ is the circumcenter. Can you see why?

So $BD = CD = DE$ and $\angle BED = \angle A$.

So you can easily see that $\sin 2A = \frac{2}{3}$ and area of the square is $36 \sin^2A = 18(1-\cos 2A)$

Title: Area of square inside a triangle


问:
Which notation is better (more common) in math?
Which one is better?

$$\arccos^2x^2\quad\quad \text{or}\quad\quad(\arccos x^2)^2$$

Are both notations acceptable?

答:
Is $\arccos^2x$

the inverse function of $\cos^2x$?

the arccosine squared?

The first notation is therefore ambiguous. I always write $\cos^{-1}$ to mean $\arccos$, and so I would write the given expression as
$$(\cos^{-1}x^2)^2$$


问:
Basis of Ker f and Im f
Let $f : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be the linear application as following

$$
f\left(x,y,z\right)=\left(x-y,y-x,0\right)
$$

I found that a basis of Ker$f$ is $\left(1,1,0\right)$ and that a basis of Im$f$ is $\left(1,-1,0\right)$. But by rank theorem we should have
$$
\dim(\operatorname{Im} f)+\dim(\operatorname{Ker} f)=3
$$

but here I've found two spaces with dimension 1, what am I missing ?

答: You got the kernel wrong. $(0,0,1)$ is also in the kernel.

Title: Basis of Ker f and Im f


问:
Expected value and variance of estimator with a sum in the denominator
I have the following estimators:

$\hat{\theta_1} = \frac{n}{\sum_{i=1}^{n} x_i^{^{\delta}}} $ and

$\hat{\theta_2} = -\frac{n}{\sum_{i=1}^{n} \ln (x_i^{^{}})} $

I want to find the expected value and variance of those estimators to determine whether they are biased and/or consistent. Yet, I'm somehow stuck on how to deal with the sum in the denominator.

$\mathbf{E}\left( \frac{n}{\sum_{i=1}^{n} x_i^{^{\delta}}} \right)$ = n $\mathbf{E}\left( \frac{1}{\sum_{i=1}^{n} x_i^{^{\delta}}} \right)$

and

$\mathbf{V}\left( \frac{n}{\sum_{i=1}^{n} x_i^{^{\delta}}} \right) = {n}^2$ $\mathbf{V}\left( \frac{1}{\sum_{i=1}^{n} x_i^{^{\delta}}} \right)$

The same holds for $\hat{\theta_2}$, but at this point I don't know how to proceed.

答:
The first "expression" you wrote in the comment is

$$f_X(x)=\theta \delta x^{\delta-1}e^{-\delta \theta x}$$

please provide the support because i.e. if $X>0$ this is NOT a density.

Example: Set $\theta=2$ and $\delta=3$ and you get

$$\int_0^{\infty}6x^2e^{-6x}dx \ne 1$$

The second expression is a Beta.

$$f_X(x)=\theta x^{\theta-1}$$

$\theta>0$

$x \in [0;1 ]$

The ML estimator for $\theta$ you showed is wrong. The correct one is

$$\hat{\theta}=\frac{n}{\Sigma_i(-\log X_i)}$$

It is very easy to verify that

$$Y=-\log(X)\sim  Exp(\theta)$$

Thus $\Sigma_i Y_i\sim Gamma(n;\theta)$ with expectation $E(\Sigma_i Y_i)=n/\theta$

Thus, using Jensen's inequality you get

$$\mathbb{E}[\hat{\theta}]\ne \frac{n}{n/\theta}=\theta$$

so it is biased for $\theta$

On the other hand, as $\hat{\theta}$ is MLE, it is asymptotically unbiased and consistent for $\theta$ (these are basic properties of MLE)

If you want to calulate mean and variance of your MLE analithically it is not difficult... you have to solve some integrals connected with gamma distribution.

The first "modified " density is now clear: it's a Weibull. Now, assuming $\delta$ known,

$$\hat{\theta}=\frac{n}{\Sigma_x x^\delta}$$

is the MLE for $\theta$ thus it is asimptotically unbiased and consistent. To check unbiasedness let's take the likelihood

$$L(\theta)\propto \theta^n e^{-\theta \Sigma_x x^\delta}$$

$$l(\theta)=n\log\theta-\theta\Sigma_x x^{\delta}$$

The score is

$$l^*(\theta)=\frac{n}{\theta}-\Sigma_x x^{\delta}$$

Now remembering that $E(l^*)=0$ you get

$$E\Big(\frac{n}{\Sigma_x x^\delta}\Big)=\theta$$

Thus $\hat{\theta}$ is unbiased


问:
Is this set relatively open or closed?
Take the set  $\{(x,y) : (x-2)^2 + y^2 < 2\}$ .  Is this set relatively open or relatively closed in the subspace $B(2,0)$ radius $\sqrt2$.  (The open ball centered at $(2,0)$ with radius $\sqrt2$) ?

I want to say it is both relatively open and relatively closed although I've been told it is just relatively closed.  I don't understand how it is one and not the other.  You can get an open ball around every point in the set and it is still in the subspace.  Since the subspace is equal to the set, you can find any closed or open set in $\mathbb R^2$ s.t. the set is equal to the intersection.

Thanks!

答: Given a topological space, a subset $A \subset X$ becomes a topological space of its own by defining the open sets in $A$ to all sets of the form $U \cap A$ where $U \subset X$ is open (same for closed sets). So $A$ is open and closed in $A$.</p>

Title: Is this set relatively open or closed?


答:
For X to win on their second toss, three things need to happen:

X doesn't get a six on their first toss (else X wins earlier)

Y doesn't get a six on their first toss (else Y wins)

X does get a six on their second toss.

Now these things have probability $\frac56,\frac56,\frac16$ respectively. Also, they are independent events - what happens on one toss of the die doesn't affect what will happen on the next. (Here we can think of the players continuing to roll until X's second toss, whether or not the game is already over.) So the probability of all three happening is the product of the individual probabilities $\frac56\times\frac56\times\frac16$.

Title: Person X and Y throw a Fair die one after another. Whoever throws 6 first wins. What is the probability that X wins?

Body:
Two people take turns rolling a fair die. Person $X$ rolls first, then
person $Y$ , then $X$, and so on. The winner is the first to roll a $6$. What is the
probability that person $X$ wins?

Our teacher in class solved this question as follows:

Probability of winning in 1st round = Probability that 6 occurs = 1/6

So Probability of $Y$ losing in first round becomes $= 5/6$

Where it is assumed that X starts the game.

suppose if X throws and if 6 comes then the game will stop and X will win. but if X loses then Y will throw the die (here Y will have to lose because according to question we want X to win the game) and then X will throw . and so on.

it will be like
$$X+ \neg X\neg YX+ \neg X\neg Y\neg X\neg YX+\dots$$

HERE $X$ denotes X wins and $\neg X$ means X loses i.e. 6 does not come . similarly $\neg Y$ means Y loses.

This will form an infinite geometric progression series. Converting into probabilities we have
$$(1/6) + (5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(5/6)(1/6)+ \dots$$
The sum of this geometric series is the answer.

MY question is:
how the probability of $X$ in $2nd$ toss is $(5/6)(5/6)(1/6)$ and similarly in the following tosses? Why these terms are in multiplication not addition?


问:
retract by deformation
I'm learning about deformation retract and have this problem as an exercise:

let  :

$$V:=\left\{(x, y, z) \in \mathbb{R}^{3} / x^{2}+y^{2} \neq 0\right\}$$

Show that $W:=\left\{(x, y, 0) / x^{2}+y^{2}=1\right\}$ is a retract by deformation of V

If you can help me I will thankful to you.

答:
It might be helpful to visualize it: $V$ is the $\Bbb R^3$ without the $z$-axis and $W$ is the circle with radius $1$ and center $(0,0,0)$ inside the $x$-$y$-plane. Define the map $r:V\to W$ by first "flattening", i.e setting the $z$-coordinate $=0$, and then contracting everything to the circle along rays: $$r(x,y,z)=\frac{1}{\sqrt{x^2+y^2}}(x,y,0)$$
Show that this is a retraction. Now construct a homotopy between the identity on $V$ and $r$. You can do that in the simplest possible way, i.e. connecting two points $x$ and $r(x)$ via a straight line (why?).

Edit: A homotopy between $\operatorname{id}_V$ and $r$ (or more formally between $\operatorname{id}_V$ and $j\circ r$ where $j:W\hookrightarrow V$ is the inclusion) can be defined as follows:\begin{align*}
H:V\times[0,1]&\to V\\
(x,t)&\mapsto (1-t)x+t\cdot r(x)
\end{align*}
I.e. $H(x,t)$ is the point on the straight line between $x$ and $r(x)$ linearly parameterized by $t$. Now check what you get for $t=0,1$ and that the homotopy is indeed inside $V$.

Title: retract by deformation


答: Hint: See picture. Make sure you understand how each sides are calculated and make use of similar triangles.

Title: Area of square inside a triangle

Body:
Considering the attached image, please compute the area of the square.

$\\$To elaborate suppose we have a triangle $\Delta ABC$, that its angle $\angle ABC$ is equal to $45^o$. On side $\overline{AB}$ we have a point $E$ that divide the side to two segments, $\overline{BE}=4$ and $\overline{EA}=6$.

Find the area of the square $\square CDEF$, such that its $\overline{FC}$ side reside on $\overline{AC}$ side of the triangle $\Delta ABC$, its one vertex is on E point, and the alternate vertex of E is on vertex C.


问:
Prove that$ \frac{a+b}{\sqrt{ab(1-ab)}} + \frac{b+c}{\sqrt{bc(1-bc)}} + \frac{c+a}{\sqrt{ca(1-ca)}} \le \frac{\sqrt{2}}{abc}$
Prove the following inequality where positive reals $a$, $b$, $c$ satisfies $ab+bc+ca=1$.
$$
        \frac{a+b}{\sqrt{ab(1-ab)}} + \frac{b+c}{\sqrt{bc(1-bc)}} + \frac{c+a}{\sqrt{ca(1-ca)}} \le \frac{\sqrt{2}}{abc}
    $$
I got that $ \frac{a+b}{\sqrt{ab(1-ab)}}=\frac{a+b}{\sqrt{ab(bc+ca)}}=\frac{a+b}{\sqrt{abc(a+b)}}.$

So rationalising, we get  $\frac{a+b}{\sqrt{abc(a+b)}}=\frac{(a+b){\sqrt{abc(a+b)}}}{abc(a+b)}=\frac{\sqrt{abc(a+b)}}{{abc}}$

Then enough to show that $\sum_{cyc}\sqrt{abc(a+b)}\le \sqrt2.$

Any help?

答:
let $x=ab,y=bc,z=ca$  and $x+y+z=1$

after your work  we have to show
$$\sum_{cyc} \sqrt{2x(y+z)}\le 2$$ which is true by AM-GM $$\sum_{cyc} \sqrt{2x(y+z)}\le \sum_{cyc}\frac{2x+y+z}{2}=2$$


问:
bijective $C^{1}$-image
Let $F: \mathbb{R^{3}}\to \mathbb{ R}$ a $C^{1}$-image and assume that $(dF) (x, y, z)\neq 0$ as soon as $F (x, y, z) = 0$.

Define $O= \{(x, y, z) ∈ \mathbb{R^{3}}| F (x, y, z) = 0\}$.

Prove that every point of $O$ has an environment that looks like a piece of the xy plane in $\mathbb{R^{3}}$.

I think this means that I have to prove that for every point $(a, b, c) \in O$ there open parts $U, V \subset \mathbb{R^{3}}$ exist and a bijective $C^{1}$- image $\phi: U → V$ with $C^{1}$-inverse such that $(0, 0, 0) \in U, (a, b, c) \in V$ and $O \cap V = \phi (U \cap (\mathbb{R^{2}} \times {0})$

My attempt:

Let's take $(a,b,c) \in O$ random. Let's assume that (1x3)- matric $(dF)(x,y,z)$ is of rang $1$.
Because of the implicit function theorem there exist

an open part $V\subset \mathbb{R^{3}}$ so $(a,b,c)\in V$

an open part $U \subset \mathbb{R^{2}}$ with $(0,0) \in U$

a $C^{1}$-image $\phi:U\to \mathbb{R}$ with $\phi((0,0))=(a,b,c)$

and $ V \cap O=\phi(U)$ and the image of $(d\phi)(y)$ equals de kern of $(dF)(\phi(y))$ for all $y\in U$.

Because the condition that $(dF)(x,y,z) \neq 0$, $U$ needs to be a part of $\mathbb{R^{3}}$. This means that $(0,0,0) \in U$ and $O \cap V=\phi(U\cap (\mathbb{R^{2} \times {0} ))}$

I'm not that sure that the conclusions I made about the condition $(dF)(x,y,z) \neq 0$ are correct. Or maybe I started the question wrong. Can someone check my answer and see what I do wrong or right.

答: Here is an alternative, more straightforward solution: By the regular level set theorem, $O=F^{-1}(0)$ is an embedded submanifold of codimension 1. Hence it has a smooth structure which makes it into a manifold of dimension 2.

Title: bijective $C^{1}$-image


问:
Evaluation of a complex numbers partial sum
Let $w = e^{i\frac{2\pi}5}$. I would like to evaluate

$$w^0 + w^1 + w^2 + w^3 +...+ w^{49}$$

Can anyone please give me an idea how to evaluate the expression?

Thanks in advance

Body:
Hints:

$A=w^0 + w^1 + w^2 + w^3 +...+ w^{49}(w-1)=w^{50}-1=(w^5)^{10}-1=(w^5-1)( \dots$

$w=e^{i \frac{2 \pi k}{5}} \implies1+ w+w^2+w^3+w^4=0 \implies A=0$


问:
What does this pic mean?(guess it's a binary number)
This number was presented in yesterday class and explained by teacher that they were lined up according to a certain rule. I think we'll use nibbles to solve this. I also think that the circle means something like pi or a clock (but not sure).

答: This is in fact a de Bruijn sequence: every four-bit word appears exactly once in the circle.</p>


答:
Start a sequence with
$$ 1,1,1,1$$
and then repeatedly append the smallest choice $a_n\in\{0,1\}$ such that the $(a_{n-3},a_{n-2},a_{n-1},a_n)$ does not equal any previous $(a_{k-3},a_{k-2},a_{k-1},a_k)$. This will make you continue the sequence as follows
$$ 1,1,1,1,0,0,0,0,1,0,0,1,1,0,1,0,1,1,1$$
at which point you have to stop, but also have completed the circle (with overlap).
As a consequence, every quadruple $(x,y,z,u)\in\{0,1\}^4$ occurs exactly once in the circle.

Title: What does this pic mean?(guess it's a binary number)

Body: This number was presented in yesterday class and explained by teacher that they were lined up according to a certain rule. I think we'll use nibbles to solve this. I also think that the circle means something like pi or a clock (but not sure).


答:
$$\begin{align*}
\angle ACE &= 45^\circ\\
\triangle ABC &\sim \triangle ACE\\
\frac{AC}{AB} &= \frac{AE}{AC}\\
{AC}^2 &= 60
\end{align*}
$$

Consider $\triangle ACE$, taking sine law and considering only the case where $\angle AEC = \angle ACB$ are obtuse,

$$\begin{align*}
\frac{\sin\angle AEC}{AC} &= \frac{\sin \angle ACE}{AE}\\
\sin\angle AEC &= \frac{AC\sin\angle ACE}{AE}\\
&= \frac{\sqrt {60}\cdot \frac1{\sqrt2}}{6} = \frac{\sqrt{30}}{6}\\
\cos\angle AEC &= \color{red}-\sqrt{1-\sin^2\angle AEC}\\
&= \color{red}-\frac{\sqrt6}{6}
\end{align*}$$

The area of the square can be found from side $EF = AE\cos \angle AEF$:

$$\begin{align*}
(AE\cos \angle AEF)^2 &= {AE}^2 \cos^2(\angle AEC-45^\circ)\\
&= {AE}^2\cdot \frac{1+\cos 2(\angle AEC-45^\circ)}{2}\\
&= {AE}^2\cdot \frac{1+\cos (2\angle AEC-90^\circ)}{2}\\
&= {AE}^2\cdot \frac{1+\sin 2\angle AEC}{2}\\
&=\frac{6^2}2\left[1+2\cdot\frac{\sqrt{30}}{6}\left(\color{red}-\frac{\sqrt6}{6}\right)\right]\\
&= 18 \color{red}- 6\sqrt5
\end{align*}$$

If otherwise consider an acute $\angle ACB$ and flip the red $\color{red}-$ sign, this gives another solution.

Title: Area of square inside a triangle


问:
Can I get a light hint for this binomial proof?

Consider the geometric series

$$S=1+(1+x)+(1+x)^2+\dots+(1+x)^n$$

(a) Show that
$$S=\frac{(1+x)^{n+1}-1}{x}$$

(b) Hence show that

$$S={n+1\choose1}+{n+1\choose2}x+\dots+{n+1\choose r+1}x^r+\dots+{n+1\choose n+1}x^n$$

(c) Hence prove that

$${n\choose r}+{n-1\choose r}+\dots+{r\choose r}={n+1\choose r+1}$$

I have successfully solved parts a) and b). However, I am struggling with c). I can tell it's the hockey stick identity but the thing that's getting me is the hence. How can I use the previous parts to prove it? I just need a little push. Perhaps it has to be a specific binomials relationship. I tried using the definition of symmetry but that didn't work.

答:
Hint: What is the coefficient of $ x^r$ in $S$?

Answer that in 2 different ways using part b.


问:
Changing order of integration for a double integral
I just started dealing with double integrals, and I encountered the following question:

Change order of integration of the following:
$$\int^{2\pi}_{0}dx\int^{\sin(x)}_{0}f(x,y)dy.$$

So I drew the graph and tried integrating over $y$ first (meaning I take lines that are perpendicular to the $y$ axis), and got the following:

$0\leq y\leq 1$. If $0\leq x\leq \frac{\pi}{2}$ then $y=\sin(x)\leq x\leq \frac{\pi}{2}$, meaning $x=\arcsin(y)\leq x\leq \frac{\pi}{2}$, and if $\frac{\pi}{2}\leq x\leq \pi$ then $x=\arcsin(y)\geq x\geq \frac{\pi}{2}$. All in all I get:

$$\int^{1}_{0}dy\int^{\frac{\pi}{2}}_{\arcsin(y)}f(x,y)dx+\int^{1}_{0}dy\int^{\arcsin(y)}_{\frac{\pi}{2}}f(x,y)dx.$$

Is this the right way to approach this problem?

答:
Your formula leads to
$$\int^{1}_{0}dy\int^{\frac{\pi}{2}}_{\arcsin(y)}f(x,y)dx+\int^{1}_{0}dy\int^{\arcsin(y)}_{\frac{\pi}{2}}f(x,y)dx\\=
\int^{1}_{0}dy\left(\int^{\frac{\pi}{2}}_{\arcsin(y)}f(x,y)dx-
\int^{\frac{\pi}{2}}_{\arcsin(y)}f(x,y)dx\right)=0$$
which is not correct.

Recall that  the function $x\mapsto \sin(x)$ is not invertible on $[0,2\pi]$, and we have to pay attention when we try to solve the equation $y=\sin(x)$ with respect to $x$.

We have that
$$\int^{2\pi}_{0}dx\int^{\sin(x)}_{0}f(x,y)dy
=\int^{\pi}_{0}dx\int^{\sin(x)}_{0}f(x,y)dy-\int_{\pi}^{2\pi}dx\int_{\sin(x)}^{0}f(x,y)dy.$$
Therefore, when we change the order in the first integral we find,
$$\int^{\pi}_{0}dx\int^{\sin(x)}_{0}f(x,y)dy=
\int_{0}^{1}dy\int_{\arcsin(y)}^{\pi-\arcsin(y)}f(x,y)dx,
$$
whereas the second one is
$$\int_{\pi}^{2\pi}dx\int_{\sin(x)}^{0}f(x,y)dy=
\int_{-1}^{0}dy\int_{\pi-\arcsin(y)}^{2\pi+\arcsin(y)}f(x,y)dx.$$


问:
Derivation for finding vertex and turning point of parabola?
My teacher told me that form the equation of parabola .

$y + \frac{D}{4a} = a(x+\frac{b}{2a})^2$

So , to find vertex of turning point ,

Make $y=0$ and $x = 0$.

You get $-\frac{b}{2a}$ and $-\frac{D}{4a}$ but aren’t we getting these values in +?

I am not getting why would he say that and how to reach there.

Please do share if there is any other point you have.

答:
The "a" in the vertex form is the same "a" as
in $y = ax2 + bx + c$ (that is, both a's have exactly the same value). The sign on "a" tells you whether the quadratic opens up or opens down. Think of it this way: A positive "a" draws a smiley, and a negative "a" draws a frowny. (Yes, it's a silly picture to have in your head, but it makes is very easy to remember how the leading coefficient works.)

In the vertex form of the quadratic, the fact that (h, k) is the vertex makes sense if you think about it for a minute, and it's because the quantity "x – h" is squared, so its value is always zero or greater; being squared, it can never be negative.

Suppose that "a" is positive, so a(x – h)2 is zero or positive and, whatever x-value you choose, you're always taking k and adding a(x – h)2 to it. That is, the smallest value y can be is just k; otherwise y will equal k plus something positive. When does y equal only k? When x – h, the squared part, is zero; in other words, when x = h. So the lowest value that y can have, y = k, will only happen if x = h. And the lowest point on a positive quadratic is of course the vertex.

If, on the other hand, you suppose that "a" is negative, the exact same reasoning holds, except that you're always taking k and subtracting the squared part from it, so the highest value y can achieve is y = k at x = h. And the highest point on a negative quadratic is of course the vertex.


问:
Proving a ring is commutative knowing that $(x+y)^2=x^2+y^2$
Let $(R, +, ·)$ be a ring and $(x+y)^2=x^2+y^2$. Prove that the ring is commutative. This is what I've tried:

$$(x+y)^2=x^2+y^2$$
$$x^2+xy+yx+y^2=x^2+y^2$$
$$xy+yx=0$$

And now I don't know what to do. I'm not sure if I'm on the right path or not so I need some  guidance.

EDIT: It is not specified that it is a ring with a unit.

答:
EDIT: My apologies, I just saw your comment saying that you are not assuming the ring is unital. I'll leave the answer up but I will edit in a correct solution if I come up with one.

You're just about there. You've shown that $xy=-yx$ for all $x,y$ but you want to remove the negative sign to prove commutativity. I'll do this by proving that in such a ring, $1=-1$. That means that $xy=-yx=yx$ as desired. Indeed, let's consider $(1+1)^2$. Of course, this is $4$, but by assumption it also equals $1^2+1^2=2$. Hence, $4=2$ so we subtract $3$ from both sides and conclude that $1=-1$.

Just to note, when I write these numbers, I mean their associated object in the ring $R$. That is, when I write $n\in R$ I mean the $n-$fold sum of $1_R$.


答:
Similar to other answers. Identify $D$ as circumcenter of $\triangle BEC$. With more angle chasing, $AC$ is tangent to the circumcircle of $\triangle BEC$.

By secant-tangent theorem $AC^2=AE\cdot AB=60$

Denote side of square by $x$. Then by Pythagoras in $\triangle EAF$,

$$x^2 + (\sqrt{60}-x)^2=6^2$$

There'll be two values of $x$. Two squares are possible as explained by @peterwhy in the comment below the question. Obtain $x^2$.

Title: Area of square inside a triangle


答:
It is easier to understand it if we substitute variables.

Let:
$$p = y + \frac{D}{4a}$$
$$s = x+\frac{b}{2a}$$

Then the equation transforms to
$$p = as^2$$

Where is the vertex of this parabola? Obviously at $p = 0, s = 0$.

Vertex at $p = 0$ means it is at $y + \frac{D}{4a} = 0$, which means $y = -\frac{D}{4a}$.

Vertex at $s = 0$ means it is at $x+\frac{b}{2a} = 0$, which means $x = -\frac{b}{2a}$.

Title: Derivation for finding vertex and turning point of parabola?

Body:
My teacher told me that form the equation of parabola .

$y + \frac{D}{4a} = a(x+\frac{b}{2a})^2$

So , to find vertex of turning point ,

Make $y=0$ and $x = 0$.

You get $-\frac{b}{2a}$ and $-\frac{D}{4a}$ but aren’t we getting these values in +?

I am not getting why would he say that and how to reach there.

Please do share if there is any other point you have.


答:
To find the vertex of a parabola, equate the slope to 0.
that is $dy/dx=0$

$dy/dx= 2a(x+\frac{b}{2a})$

so
$2a(x+\frac{b}{2a})=0$

so $x=-b/2a$

putting the point in the equation, we get the y coordinates as $-\frac{D}{4a}$


问:
Solving an equation when the unknown is both a term and exponent
I'm taking a course (algorithms) and the instructor assigns us problems from the CLRS Introduction to Algorithms 3rd edition.  We aren't marked on the problems, they are just given as exercises.

Exercise 1.2-3 asks: at what value of n will an algorithm that runs at $100 n^2$ be faster than one that runs at $2^n$

Essentially I started out by arranging the question like so:

$100 n^2$ = $2^n$ and then I converted it to log format which gives: $n = lg\ 100 n^2$ as my limited math spidey senses told me if you are looking to find the value of an exponent, log is usually the way to do it.

But at this point I'm stuck, I'm not sure how to collect the n terms and get a meaningful answer to what n equals.  Would I divide by $lg\ n^2$ on both sides?  But if so, then what?

FYI: lg = $log_{2}$

答: This cannot be solved in closed form using elementary functions. Nor is that the point of the exercise. I suppose you should try some kind of binary serach starting from the observation that for instance $n=20$ suffices but $n=1$ does not.</p>


问:
Determine the dimension of the subspace of all matrices that commutes with an all 1 matrix and find a basis

Let $\Gamma$ denote the $n$-by-$n$ matrix which has $1$ in all its entries. Denote
$$S := \{A \in \mathcal{M}_n(\mathbb{R}) : A\Gamma = \Gamma A\}$$
Determine the dimension of $S$, and find a basis for $S$.

Simple calculation tells that $S$ consists of all matrices with the sum of each row and column agreeing on a value. This is similar to the case of Magic matrices, so I found a literature for that and conclude that the dimension is $\,n^2 - 2n + 2\,$.

Problem is, how do I find a basis for it?

Exhausting all the possibilities by hand seems undesirable, and I can't think of a way to do it systematically.

答:
First hint:
$\Gamma\,$ is symmetric and satisfies $\Gamma^2 =n\,\Gamma$ , so that $\,\frac1n\Gamma\,$ is an orthogonal projector. It is fairly straightforward to see that it projects onto a $1$-dimensional image. Hence $\,\frac1n\Gamma\,$ is similar to
$E_{nn}=\left(\begin{smallmatrix} 0 &\ldots &0 \\
\vdots &\ddots &\vdots \\
 0 &\dots &1\end{smallmatrix}\right)$,
note that $E_{nn}$ has just one nonzero entry. Similarity means there is some transformation matrix $T$, and $\,TE_{nn}T^{-1}=\frac1n\Gamma\,$ holds. In this context
you may consult this .

Consider the commutator map
$[\,\cdot\, ,E_{nn}]: \mathcal{M}_n(\mathbb{R})\to\mathcal{M}_n(\mathbb{R}), A\mapsto AE_{nn} -E_{nn}A$,
which is linear.

Second hint:
$\dim\ker\, [\,\cdot\, ,E_{nn}] = (n-1)^2+1$ may be read off from
$$[A,E_{nn}] \;=\;
\begin{pmatrix} 0 &\dots &0 &a_{1,n}\\
\vdots &\ddots &\vdots & \vdots\\
 0 &\dots &0 &a_{n-1,n}\\
 -a_{n,1} &\dots &-a_{n,n-1} &0\end{pmatrix}$$

First question:
How are $S$ and $\,\ker\, [\,\cdot\, ,E_{nn}]\,$ related to each other, in terms of the transformation matrix $T$ ?

Last question: How to systematically deduce from the preceding a basis of $S$ ?


问:
Composition of generic measurable functions.
Consider $(X,M_X),(Y,M_Y),(Z,M_Z)$ measurable spaces, namely $M_X,M_Y,M_Z$ are $\sigma$-algebras.

Consider $f\colon X \to Y$, $g\colon Y \to Z$.

Suppose that $f$ is $(M_X,M_Y)$-measurable (namely $\forall \,A \in M_Y \quad f^{-1}(A) \in M_X$), and that $g$ is $(M_Y,M_Z)$-measurable.

Now consider $h=g \circ f\colon X \to Z$.

Is $h$ $(M_X,M_Z)$-measurable?

$ $

My attempt is the following:

$\forall \,A \in M_Z \quad h^{-1}(A)=f^{-1}(g^{-1}(A)) \in M_X$, since $f,g$ are measurable.

So $h$ is $(M_X,M_Z)$-measurable.

Am I wrong? Thank you!

答:
Perhaps rewording it in the following way is better:

We want to show that $g\circ f$ is measurable. Let $B\in M_Z$. Then $(g \circ f)^{-1}(B)=f^{-1}(g^{-1}(B))$. As $g$ is measurable, $g^{-1}(B)\in M_Y$. As $f$ is measurable, $f^{1}(g^{-1}(B))\in M_X$. Hence the composition is measurable.