问:
How to construct natural numbers by set theory?

Definition 1:  For any set $a$ , its successor $a^+=a\cup \{a\}$.

Informally , we want to construct natural numbers such that :

$0=\emptyset,1=\emptyset^+,2=\emptyset^{++},3=\emptyset^{+++}$,...

Definition 2: A natural number is a set that belongs to every inductive set.

Then we can construct a set $\omega$ whose members are exactly the natrual numbers . $\{x | x \text{ belongs to every inductive set } \}$

The discussion above was in Herbert B.Enderton's book . However , I did not see how to make a connection between $\{0,1,2,3,... \}$ and the natural number we define above .

My attempt :

$(1)$ $\omega$ is inductive , and is a subset of every other inductive set .

$(2)$ $\omega$ is the smallest inductive set . Every inductive subset of $\omega$ is concides with $\omega$ .

$(3)$ If we can prove $N=\{\emptyset,\emptyset^+,\emptyset^{++},\emptyset^{+++}, ... , ... \}$ is actually a set , then $N=\omega$ since $N$ is inductive and every member $x \in N$ also belongs to $\omega$ .

My question :

I want to define $N$ as $$\{x\in \omega | x \text{ is finite times successor of  }\emptyset \}$$
However , the question happens when we want to define "finite times" . Although we can define finite by members belong to $\omega$ , how can we define finite times while we did not define even one number in our usual sense ?

答:
There is a way to define the natural numbers directly. Unlike the standard definition, it is not self-referential in nature. Before getting to that, I will review the standard definition. The most well-known construction of the natural numbers, found in texts such as those by Enderton or Hrbacek and Jech, begins as follows:

An inductive set is any set $X$ such that $\emptyset \in X$, and for all sets $a$, if $a \in X$ then $S(a) \in X$.

Here, $S(a) = a \cup \{a\}$ denotes the successor of a. Then the Axiom of Infinity states:

There exists an inductive set.

This definition is self-referential in the sense that the property "if $a \in X$ then $S(a) \in X$" references $X$ itself. Therefore, rather than defining the extension (members) of $X$ explicitly, as the Axiom Schema of Specification does for example, this property defines the extension implicitly, making it less clear what $X$ actually is.

The final step is to define the natural numbers $\mathbb N$ as the intersection of all inductive sets. This does result in a unique set, but it does not fully resolve the self-referential nature of the definition, in the sense that it still doesn't immediately describe the members of $\mathbb N$ and their properties.

There is nothing necessarily wrong with this, nor is there any formal circularity or logical contradiction here. Rather the issue is more of a meta-circularity, and has more to do with bringing the self-evidence of the Axiom of Infinity into question. None of the other axioms of ZFC use these kind of self-referential definitions.

One could argue that rigorous (e.g. ZFC) set theory should attempt to model the intuition of naive set theory, at least as far as it can while avoiding contradictions like Russell's Paradox. In naive set theory, sets are thought of as obtained by collecting together objects using some criterion. This is formalized in the Axiom Schema of Unrestricted Comprehension. While this axiom schema leads to paradoxes, these can be avoided using its restricted form, the Axiom Schema of Specification. The idea is to specify exactly what elements are in the defined set.

This is a unifying and intuitively appealing conceptual approach, which the standard definition of $\mathbb N$ does not follow. On the other hand, defining $\mathbb N$ as the smallest inductive set does immediately indicate that proof by induction is applicable. So that is perhaps one motivation for that definition.

But in the case of your question, it is precisely this self-referential aspect of the the Axiom of Infinity which I believe is the issue. Your attempt to define $\mathbb N$ as

$\mathbb N = \{x \mid x = S^n(\emptyset) \, \text{for some finite} \, n\}$.

will not work, because we cannot write $S^n$, the $n$-fold composition of the successor function $S$, in our first-order language, only in the meta-language. In fact, we can't write $f^n$ in our language for any function $f$ (set or class). We can only express it indirectly as an element of a set guaranteed to exist by the Axiom of Infinity and the Axiom Schema of Replacement. But then, this already requires a definition of $\mathbb N$.

However, we can define what it means to be finite directly. As I mentioned above, there is a direct (i.e. comprehension-like) way to define the natural numbers. We do this using ordinals.

If we take the common von Neumann definition of ordinals, then an ordinal is a transitive set that is well-ordered by set membership ($\in$). Furthermore, we define an ordinal $\alpha$ to be finite if every non-empty subset of $\alpha$ has a maximal element (with respect to the order defined by $\in$). Note the similarity of this definition to the well-founded property of a well-ordering, namely that every non-empty subset has a minimal element. From these definitions, it follows that the first few finite ordinals are $\emptyset$, $\{\emptyset\} = S(\emptyset)$, $\{\emptyset, \{\emptyset\}\} = S^2(\emptyset)$, etc.

Then we can define the natural numbers $\omega$ to be

$\omega = \{\alpha \mid \alpha \, \text{is a finite ordinal}\}$.

Of course, without an axiom asserting that such a set exists, this is just a class. We could just replace the Axiom of Infinity with the assertion that the above class is a set.
But we may as well just use the Axiom of Infinity, because as it turns out, using the definitions above, $\omega = \mathbb N$. In the remainder of this answer, I will prove this.

Proof. First we prove $\mathbb N \subseteq \omega$. We will show that $\omega$ is an inductive set. First note that $\emptyset \in \omega$. Now let $\alpha \in \omega$. Then $\alpha$ is a transitive set, and $\alpha \subseteq \alpha \cup \{\alpha\} = S(\alpha)$, so $S(\alpha)$ is a transitive set as well. Furthermore, since $\alpha$ is an ordinal, the set membership relation $\in$ defines a trichotomous, transitive, well-founded order on $\alpha$. Since for all $\beta \in S(\alpha)$, we have either $\beta \in \alpha$ or $\beta = \alpha$ by the definition of $S(\alpha)$, this order extends to a trichotomous, transitive order on $S(\alpha)$. To see that this order is also well-founded on $S(\alpha)$, let $A \subseteq S(\alpha)$ be non-empty. If $\alpha \notin A$ then $A \subseteq \alpha$, hence $A$ has a minimal element because $\alpha$ is an ordinal. Otherwise, by trichotomy of the order, the minimal element of $A$ is either $\alpha$ or the minimal element of $A \cap \alpha$, the latter of which exists because $A \cap \alpha \subseteq \alpha$. Finally, note that $\alpha$ is a maximal element of $S(\alpha) = \alpha \cup \{\alpha\}$, since $\alpha \not \in \alpha$, and for all $\beta \in \alpha$, $\alpha \notin \beta$ by trichotomy of the order. Therefore, every subset $A \subseteq S(\alpha)$ has a maximal element, which is either guaranteed to exist by the finiteness of $\alpha$ if $A \subseteq \alpha$, or is equal to $\alpha$ otherwise. Thus $S(\alpha)$ is a finite ordinal, so $S(\alpha) \in \omega$. This shows that $\omega$ is indeed an inductive set. Since $\mathbb N$ is the intersection of all inductive sets, $\mathbb N \subseteq \omega$.

Now we prove that $\omega \subseteq \mathbb N$. Suppose, for a contradiction, that there is an $\alpha \in \omega$ for which $\alpha \notin \mathbb N$. Since $\emptyset \in \mathbb N$ by definition, we may assume $\alpha \neq \emptyset$. Since $\alpha$ is a non-empty finite ordinal, it has a maximal element, call it $\gamma_0$. Then by the trichotomy of the set membership order on $\alpha$, for all $\beta \in \alpha$ either $\beta = \gamma_0$ or $\beta \in \gamma_0$. Therefore $\alpha = S(\gamma_0)$. Since $\alpha \notin \mathbb N$ and $\mathbb N$ is inductive, we must have $\gamma_0 \notin \mathbb N$. So the set $B = \{\beta \in \alpha \mid \beta \notin \mathbb N\}$ is non-empty. Since $B \subseteq \alpha$ and $\alpha$ is an ordinal, $B$ has a minimal element, call it $\beta_0$. Then $\beta_0$ must be non-empty, for otherwise $\beta_0 = \emptyset \in \mathbb N$ by definition of $\mathbb N$, contradicting the definition of $B$. Since $\beta_0 \in \alpha$ and $\alpha$ is a transitive set, $\beta_0 \subseteq \alpha$. Therefore, being a non-empty subset of the finite ordinal $\alpha$, $\beta_0$ has a maximal element, call it $\gamma_1$. By the same reasoning as above, $\beta_0 = S(\gamma_1)$. But then, because $\beta_0 \notin \mathbb N$ and $\mathbb N$ is inductive, we must also have $\gamma_1 \notin \mathbb N$. Note that $\gamma_1 \in \beta_0 \in \alpha$, so by transitivity $\gamma_1 \in \alpha$. This, together with the fact that $\gamma_1 \notin \mathbb N$, means that $\gamma_1 \in B$, contradicting the minimality of $\beta_0$. Therefore $\alpha$ must be an element of $\mathbb N$, and thus $\omega \subseteq \mathbb N$.

We conclude that $\omega = \mathbb N$. $\blacksquare$


问:
Question on finding maximum of AM GM inequality
How can i solve for minimum value of function $f(x)=\frac{x^2+x+1}{x^2-x+1} $ using AM GM inequality.

I am aware about the AM GM inequality, but am not able to split the terms into the correct form in order to apply the inequality

答:
If $\frac{x^2+x+1}{x^2-x+1} > k$ for all $k \in \mathbb R$, since $x^2+x+1 = (x-0.5)^2+0.75 > 0$, $x^2+x+1 > kx^2-kx+k$ or $(k-1)x^2+(-k-1)x+k-1<0$ again for all real $k$.

For the maximum such $k$, the discriminant must be zero, or $(k+1)^2 - 4(k-1)^2 = 0$ which gives $k=\frac{1}{3}, 3$. This gives the maximum and the minimum, as the process is the same with the maximum, just with the inequality sign being flipped.

ContentLicense: CC BY-SA 4.0


问:
If we don't know the transition probability, then how to prove a markov chain is recurrent or irreducible?
For example, we only have the next state information: $q(n+1) = q(n) + a - c$, where $a$ can be regarded as some arrival and $c$ can be regarded as some action. Also, $a,c \in \{0,1,\cdots\}$, which follows independently and identically distributed.

Can any friends give some comments on how to prove it?

答:
Assuming  that $E|a-c|<\infty$, the proposed chain is recurrent if and only if
$E[a-c] =0$. See, e.g.

Spitzer, Frank. Principles of random walk. Vol. 34. Springer Science & Business Media, 2013.


问:
Which law of logical equivalence says $P\Leftrightarrow Q ≡ (P\lor Q) \Rightarrow(P\land Q)$
I'm going through the exercises in the book Discrete Mathematics with Applications. I'm asked to show that two circuits are equivalent by converting them to boolean expressions and using the laws in this table.

$$\begin{array}{lcc}
\hphantom{1}\mathsf{1.\; Commutative\; laws:}       &  p\land q \equiv q\land p                          &  p\lor q \equiv q\lor p\\
\hphantom{1}\mathsf{2.\; Associative\; laws:}       &  (p\land q)\land r \equiv p\land (q\land r)        &  (p\lor q)\lor r \equiv p\lor (q\lor r)\\
\hphantom{1}\mathsf{3.\; Distributive\; laws:}      &  p\land (q\lor r) \equiv (p\land q)\lor (p\land r) &  p\lor (q\land r) \equiv (p\lor q)\land (p\land r)\\
\hphantom{1}\mathsf{4.\; Identity\; laws:}          &  p\land t \equiv p                                 &  p\lor c \equiv p\\
\hphantom{1}\mathsf{5.\; Negation\; laws:}          &  p\lor \lnot p \equiv t                           &  p\land \lnot p  \equiv c\\
\hphantom{1}\mathsf{6.\; Double\; negative\; law:}  &  \lnot(\lnot p) \equiv p                           &\\
\hphantom{1}\mathsf{7.\; Idempotent\; laws:}        &  p\land p \equiv p                                 &  p\lor p \equiv p\\
\hphantom{1}\mathsf{8.\; Universal\; bound\; laws:} &  p\lor t \equiv t                                  &  p\land c\equiv c\\
\hphantom{1}\mathsf{9.\; De\; Morgan}\text{'}\mathsf{s\; laws:}     &  \lnot(p\land q) \equiv \lnot p\lor \lnot q        &  \lnot(p\lor q) \equiv \lnot p\land\lnot q\\
\mathsf{10.\; Absorption\; laws:}                   &  p\lor (p\land q) \equiv p                         &  p\land (p\lor q) \equiv p\\
\mathsf{11.\; Negations\; of\; t\; and\; c:}        &  \lnot t \equiv c                                  &  \lnot c \equiv t\\
\end{array}$$

so as which law/s of logical equivalence says $P\Leftrightarrow Q ≡ (P\lor Q) \Rightarrow(P\land Q)$

I can see their equivalence clearly with a truth table. But the book is asking me to show it using the equivalence laws in the above table, and I can't see how any of them apply here. So, do any of those laws apply here in a way I'm not understanding? Or is there some other known law that does apply here?

答:
The truth table involves "t" and "c" throughout.  So, you can rewrite the truth table using the negation laws.  That is, instead of using "t" and "c" in your truth table, use (x∨¬x), and (x $\land$ $\lnot$x) respectively.  Then, since you'll have (x∨¬x) throughout the last column of the truth table, you use the equivalence law (x∨¬x)=t in an additional column, and thus you've showed the formula true using the equivalence laws...  specifically your last two columns will look like this:</p>

```<code>(x∨¬x)  t (by the equivalence law (x∨¬x)==t)
(x∨¬x)  t (by the equivalence law (x∨¬x)==t)
(x∨¬x)  t (by the equivalence law (x∨¬x)==t)
(x∨¬x)  t (by the equivalence law (x∨¬x)==t)
</code>```

ContentLicense: CC BY-SA 3.0

Body:
The truth table involves "t" and "c" throughout.  So, you can rewrite the truth table using the negation laws.  That is, instead of using "t" and "c" in your truth table, use (x∨¬x), and (x $\land$ $\lnot$x) respectively.  Then, since you'll have (x∨¬x) throughout the last column of the truth table, you use the equivalence law (x∨¬x)=t in an additional column, and thus you've showed the formula true using the equivalence laws...  specifically your last two columns will look like this:

(x∨¬x)  t (by the equivalence law (x∨¬x)==t)
(x∨¬x)  t (by the equivalence law (x∨¬x)==t)
(x∨¬x)  t (by the equivalence law (x∨¬x)==t)
(x∨¬x)  t (by the equivalence law (x∨¬x)==t)


问:
Does the short exact sequence $0\mapsto \mathbb{Z}^k\mapsto \Gamma\mapsto \mathbb{Z}^\ell\mapsto 0$ split?
I've seen in Nonsplit extension of $\mathbb{Z}$ by itself that every short exact sequence of the form $0\mapsto \mathbb{Z}\mapsto G\mapsto \mathbb{Z}\mapsto 0$ splits.

I wonder if this is still valid if we change the exact sequence by $$0\mapsto \mathbb{Z}^k\mapsto \Gamma\mapsto \mathbb{Z}^\ell\mapsto0$$ where $\Gamma$ is a discrete group.

Thanks in advance

答:
For abelian groups, The proof is the same than for vector spaces.

Choose a $\bf Z$ base of the kernel $e_1,..,e_k$ and a family of vectors $f_1,...,f_l$ which projects onto a base $f'_1,...,f'_l$of $\bf Z ^l$. We wil prove that $(e_1,..e_k, f_1,...f_l)$ is a $\bf Z$ base of $\Gamma$, ie the obvious map $\bf Z^{k+l}$ $\to \Gamma$ is an isomorphism.

Let $x\in \Gamma$; then we write $\pi (x)= \sum y_i f'_i$, so that $x-\sum y_i f_i$ is in the kernel, and there exists $x_1,..x_k$ so that $x= \sum x_i e_i +\sum y_i f_i$.

To check that we have a $\bf Z$ base, assume $x=0$. By projecting and using that $f'_i$, we see that $y_i=0$, and using that $e_i$ is a base of the kernel, we see that $x_i=0$.

But for the Heisenberg group,(see https://en.wikipedia.org/wiki/Heisenberg_group), the sequence do not split. This groupe is the group of upper triangular $(3,3)$ matrices with 1 on the diagonal and coefficients in $\bf Z$.


问:
Finding the global maxima when 1st derivative has a singularity?
Let $f$ be real function which is differentiable everywhere except at a finite set of singularity point $S$.

Assume that $\lim_{x\to \alpha} f(x) = -\infty$ for every $\alpha \in S$. Is it possible to prove that the global maxima is a local maxima? (i.e it can be found with the first and second derivative test?)

I'm not very strong in analysis so any help would be appreciated, thanks.

答: No. After the last singularity point, the function could increase so $\lim_{x\to\infty}f(x)=\infty$. Here, there would be no global maximum. If a global maximum exists, it occurs at a local maximum since this is true of all differentiable functions.


问:
Evaluate the flux of $F=\langle\sin(xyz), x^2y, z^2e^{x/5}\rangle$ through surface $S$ ... $4y^2+z^2=4, \space x\in [-2,2]$
I am trying to find the flux of  $\vec F=\langle\sin(xyz), x^2y, z^2e^{x/5}\rangle$ through the surface $S$ where $S$ consists of the elliptical cylinder defined by

$S$ ... $4y^2+z^2=4, \space x\in [-2,2]$

My first instinct is to parametarize the surface to get $\vec F \cdot \vec n$,

where

$$\vec r = \left\langle x,y,\pm \sqrt{1-y^2} \right\rangle$$

I get $$\vec n = \left\langle 0, \pm \frac{2y}{\sqrt{1-y^2}},1\right\rangle$$

so $$\vec F \cdot \vec n = \pm \frac{2x^2y^2}{\sqrt{1-y^2}} + 4(1-y^2)e^{x/5}.$$

I know that $x \in [-2,2]$ and $y \in \left[-\sqrt{1-\left(\frac{z}{2}\right)^2}, \sqrt{1-\left(\frac{z}{2}\right)^2} \right]$, and using a calculator I would be able to evaluate it, but with certain techniques the answer simplifies to

$$\frac{16\pi + 160 \sinh(2/5)}{3}.$$

I am not as familiar with integration involving hyperbolic functions and that is probably the reason why I am not seeing this.

Is there anyone that can confirm this fact?

答:
While you can apply divergence theorem, this one is quite straightforward for direct surface integral as well. Given it is a cylinder, it is easier to set up the integral in cylindrical coordinates. An elliptic cylinder with form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ can be parametrized as $x = a \cos\theta, y = b \sin \theta, z$. Following the same parametrization but adjusting for the fact that axis of our cylinder is along $x-$axis instead of $z$.

We choose $r(t) = (x, \cos\theta, 2 \sin\theta)$, so you can find that $r'(t) = (0, 2\cos\theta, \sin\theta)$

So, $\vec{F} \cdot \vec{n} = (2x^2\cos^2\theta + 4\sin^3\theta)$. The integral becomes,

$\displaystyle \int_0^{2\pi} \int_{-2}^{2} (2x^2 \cos^2\theta + 4 \sin^3\theta e^{x/5}) \ dx \ d\theta$.

Please note that $\sin^3\theta$ is an odd function and over $(0,2\pi)$, its integral is zero. So all you are left with is to integrate $\displaystyle \int_0^{2\pi} \int_{-2}^{2} 2 x^2 \cos^2\theta \ dx \ d\theta$ which is a straightforward integral and the answer comes to $\frac{32\pi}{3}$.

Now coming to your working,

You should avoid parametrizing surface where you have $\pm$ sign and have to split the integral into two as we work with oriented surface and one will have to be very careful with orientation for both integral.

As far as the integral you have set up is not a hyperbolic integral. I am taking the plus sign part for example.

$\vec F \cdot \vec n = \frac{2x^2y^2}{\sqrt{1-y^2}} + 4(1-y^2)e^{x/5}$.

$x \in \mathbb{(-2,2)}$ is correct but $y \in \left[-\sqrt{1-\left(\frac{z}{2}\right)^2}, \sqrt{1-\left(\frac{z}{2}\right)^2} \right]$ is not correct as you have parametrized your surface in terms of $(x, y)$. So you have to just take limits of $y$ which is $y \in \mathbb{(-1,1)}$.

When we integrate wrt $x$, integral of $e^{x/5}$ is simply $5 e^{x/5}$ and between $-2, 2$, the integral comes to $5(e^{2/5} - e^{-2/5})$. Note that $\sinh({\frac{2}{5}}) = \frac{e^{2/5} - e^{-2/5}}{2}$ and hence the integral result can be written that way too.

$\int_{-1}^{1} \int_{-2}^{2} \frac{2x^2y^2}{\sqrt{1-y^2}} + 4(1-y^2)e^{x/5} \ dx \ dy = \frac{16\pi}{3} + \frac{160}{3} \sinh(\frac{2}{5})$

Similarly for $\vec{F} \cdot \vec{n} = - \frac{2x^2y^2}{\sqrt{1-y^2}} + 4(1-y^2)e^{x/5}$, you get the integral of $\frac{16\pi}{3} - \frac{160}{3} \sinh(\frac{2}{5})$

Adding both, you get $\frac{32\pi}{3}$.


问:
$\frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$
Prove that
$$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$$

saw some similar problems like
show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true

but didn't manage to get 115. I could get a weaker conclusion of $x>100$ though.

\begin{align}
x^2 &= \left(\frac 21 \times \frac 21\right) \times \left(\frac 43 \times \frac 43\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10000}{9999}\right) \\
&\ge \left(\frac 21 \times \frac 32\right) \times \left(\frac 43 \times \frac 54\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10001}{10000}\right) \\
&= 10001
\end{align}

so $x > 100$

答:
You can make the problem more general since you have

$$S_p=\frac{\prod_{n=1}^p (2n) }{\prod_{n=1}^p (2n-1) }=\frac{2^p \Gamma (p+1) } {\frac{2^p \Gamma \left(p+\frac{1}{2}\right)}{\sqrt{\pi }} }=\sqrt{\pi }\,\,\frac{ \Gamma (p+1)}{\Gamma \left(p+\frac{1}{2}\right)}$$ Take logarithms, use Stirling approximation and continue with Taylor series for large $p$
$$A=\log (\Gamma (p+1))-\log \left(\Gamma \left(p+\frac{1}{2}\right)\right)$$
$$A=\frac{1}{2}\log (p)+\frac{1}{8 p}-\frac{1}{192
   p^3}+O\left(\frac{1}{p^5}\right)$$
$$\frac{ \Gamma (p+1)}{\Gamma \left(p+\frac{1}{2}\right)}=e^A=\sqrt{p}\left(1+\frac{1}{8 p}+\frac{1}{128 p^2}+O\left(\frac{1}{p^3}\right) \right)$$

$$S_p >\sqrt{p\pi}\left(1+\frac{1}{8 p}\right) $$ Using it for $p=5000$ gives $125.334547017$ while the exact value is $125.334547056$


问:
If an arithmetic progression contains a perfect square, then it must contain a perfect square strictly less than ...
(Question): If an arithmetic progression of positive integers $a, a+d, a+2d, \dots$ contains a perfect square, then it must contain a perfect square strictly less than $a+2d\sqrt{a}+d^2$.

I noticed that
\begin{equation}
(m-d)^2=m^2-d(2m-d)
\end{equation}
But I have no clue on how to prove the "strictly less" part. Thank you in advance for your comment/answer!

答:
I assume the arithmetic progression is non-degenerate, i.e., $d \neq 0$. Also, since the progression is of positive integers, then $a \gt 0$ and $d \gt 0$.

Let $a + kd$, for some integer $k \ge 0$, be the smallest element of the progression which is a perfect square. Thus, there's a positive integer $e$ such that

$$a + kd = e^2 \tag{1}\label{eq1A}$$

You're asking to prove

$$e^2 \lt a + 2d\sqrt{a} + d^2 \tag{2}\label{eq2A}$$

As Joffan's question comment states, the right side is equivalent to $(\sqrt{a} + d)^2$. Thus, \eqref{eq2A} is equivalent to trying to prove

$$e \lt \sqrt{a} + d \tag{3}\label{eq3A}$$

Assume, instead, that

$$e \ge \sqrt{a} + d \implies e - d \ge \sqrt{a} \implies (e - d)^2 \ge a \tag{4}\label{eq4A}$$

Note, though, that

$$\begin{equation}\begin{aligned}
(e - d)^2 & = e^2 - 2ed + d^2 \\
& = (a + kd) - 2ed + d^2 \\
& = a + (k - 2e + d)d
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$

This is also a perfect square smaller than $e^2$ (due to $e \gt e - d \gt 0$ using \eqref{eq4A}), and of the form $a + jd$ for an integer $j = k - 2e + d$, so it's a member of the arithmetic progression if $j \ge 0$. However, since $e^2$ is the smallest perfect square which is a member of the progression, this means $(e - d)^2$ cannot be a member, so this requires $j \lt 0$ and

$$(e - d)^2 \lt a \tag{6}\label{eq6A}$$

Note this contradicts \eqref{eq4A}, so the assumption it is correct is false. This means \eqref{eq3A} and, thus, \eqref{eq2A}, must be true instead.


问:
Simplify integral using partial fractions
If $I = \int x^{3}\sin x\mathrm{d}x$, how do I simplify this using partial fractions? Simplify not calculate. I chose $v=x^{3}$ and got $\frac{1}{3} x^{4}\sin x-\frac{1}{3} \int x^{3} \sin x+ x^{4} \cos x\mathrm{d}x$

答:
$v=x^3$ and $du=\sin x\,dx \Rightarrow dv=3x^2\,dx$ and $u=-\cos x$

$\int v\, du=uv-\int u\, dv \Rightarrow \int x^{3}\sin x\, dx=-x^3\cos x +\int 3x^2\cos x\, dx$

$\int 3x^2\cos x dx=3x^2\sin x-\int 6 x \sin x\, dx$

$\int 6 x \sin x\, dx=6 x \cos x-\int 6 \cos x\, dx=6 x \cos x - 6 \sin x$

After this point, all you have to do is plug last integral to other equations.


问:
Find $\int^\infty_0 \frac{\sin x}{x(1+\cos ^2 x)}dx$
find the value:
$$\int^\infty_0 \frac{\sin x}{x(1+\cos ^2 x)}dx$$
I try to integrate by parts:
$$-\int^\infty_0 \frac{1}{x}d\arctan(\cos x)$$
But it's not run,help me,thank you.

答:
Too long for a comment

Introduce a parameter $a$:

$$\bbox[5px,#ffd]{I(a)=\int \frac{\sin ax}{x(1+\cos ^2 x)}dx}$$

$$I'(a)=\int \frac{\cos ax}{(1+\cos ^2 x)}dx$$

$$I''(a)=\int \frac{-x\sin ax}{(1+\cos ^2 x)}dx=-xI(a)+\int I(a)$$

$$I'''(a)=-xI'(a)+I(a)$$

The solution of this differential equation is this.

Note that $c=-x,y=I(a)$ and $x=a$


问:
If $B⊆E$ is compact and $B\setminus\{x_0\}$ has no accumulation point in $E\setminus\{x_0\}$, an infinite net in $B\setminus\{x_0\}$ tends to $x_0$?
This should be rather simple and hold in more generality: If $B\subseteq\mathbb C$ is compact and $B\setminus\{0\}$ has no accumulation point in $\mathbb C\setminus\{0\}$, why does it follow that every infinite sequence (and/or every infinite net) in $B\setminus\{0\}$ converges to $0$?

I guess this question can be more generally formulated as follows: If $(E,\tau)$ is a topological space, $x_0\in E$, $B\subseteq E$ is $\tau$-compact and $B\setminus\{x_0\}$ has no $\left.\tau\right|_{E\setminus\{x_0\}}$-accumulation point, then every infinite sequence (and/or every infinite net) in $B\setminus\{x_0\}$ converges to $x_0$. Maybe we need to assume Hausdorfness. Or that $\tau$ is metrizable.

EDIT: The situation I've got in mind is $B=\sigma(T)$ for some compact linear operator $T$ on a complex Banach space $X$. It is known that $B\setminus\{0\}$ is closed and discrete in $\mathbb C\setminus\{0\}$ and the spectral theorem asserts that an enumeration of the eigenvalues is either finite or tends to $0$.

Let $(E,\tau)$ be a topological space, $\mathcal N_\tau(x):=\{N\subseteq E:N\text{ is a }\tau\text{-neighborhood of }B\}$ for $x\in E$ and $B\subseteq E$. Remember that $x\in E$ is called $\tau$-accumulation point of $B$ if $$\forall N\in\mathcal N_\tau(x):B\cap N\setminus\{x\}\ne\emptyset. \tag1$$ Moreover, $(E,\tau)$ is called accumulation point compact if every infinite subset of $E$ has a $\tau$-accumulation point. It can be shown that if $(E,\tau)$ is compact, it is accumulation point compact as well.

答:
You need to assume that $B$ is infinite in order to draw any interesting conclusions; with that added assumption you can get a result along the lines that you want.

The space $E$ here is really irrelevant. You have a compact space $B$ and a point $x_0\in B$ such that $B\setminus\{x_0\}$ has no accumulation point in itself. This means that $B\setminus\{x_0\}$ is a relatively discrete subset of $B$. If $B$ is infinite, compactness of $B$ implies that every open nbhd of $x_0$ is cofinite in $B$: if $x_0$ had an open nbhd $U$ such that $B\setminus U$ were infinite, $\{U\}\cup\big\{\{x\}:x\in B\setminus U\big\}$ would be an open cover of $B$ with no finite subcover.

Let $\langle y_n:n\in\Bbb N\rangle$ be any sequence of distinct points of $B\setminus\{x_0\}$. If $U$ is an open nbhd of $x_0$, then $\{y_n:n\in\Bbb N\}\setminus U$ is contained in the finite set $B\setminus U$, so there is an $n_0\in\Bbb N$ such that $y_n\in U$ for all $n\ge n_0$, and the sequence therefore converges to $x_0$. Essentially the same argument works as long as each point of the sequence occurs only finitely many times. Without that restriction you may have a sequence with a constant subsequence, even if the original sequence hits infinitely many distinct points of $B\setminus\{x_0\}$, and in that case the original sequence won’t converge to $x_0$. However, the argument does show that every sequence in $B\setminus\{x_0\}$ that hits infinitely many different points of $B\setminus\{x_0\}$ has a subsequence converging to $x_0$.


问:
Find the mean and variance of a random variable composed of random variables.
Let $X_1,X_2,...,X_n$ be random variables all with unknown distributions, whose mean is 0, their deviation is 1 and all are independent from one another. Let $Y=\frac{X_1+X_2+...+X_n}{n}$. Find $E[Y]$ and $\sigma(Y)$.

This problem may look easy, but I got stuck in how to approach it. What I thought is that if the mean of all the $X_i$'s is $0$, then the mean of $Y$ should also be 0. But I'm still pretty confused. Any help is appreciated

答:
Y can also be written $\frac {\sum X} n$.

The expectation of Y is $E(\frac {\sum X} n)=\frac 1 nE(\sum X)=\frac 1 n \sum E(X)=\frac 1 n0=0$, in accordance with your intuition.

The variance of Y is $var(Y)=var(\frac {\sum X} n)=\frac 1 {n^2}var(\sum X)=\frac 1 {n^2}\sum var(X)=\frac n {n^2}=\frac 1 n$. Standard deviation is the square root of this, $\sigma(Y)=\frac 1 {\sqrt n}$.

You can only do this if the variables are iid, which they are.


问:
Clockwise , Counter-clockwise convention
When we calculate the solid angle subtended by half a sphere we simply do it as follows:

$\Omega=\int_{0}^{2\pi}\,d\phi \int_{0}^{\pi/2}\,sin \theta \,d \theta  =2\pi$

Here,:

$\theta$ = co-latitude (clockwise taken positive)

$\phi$ = azimuthal angle (anti-clockwise taken positve)

Now this can also be calculated as:

$\Omega=\int_{0}^{\pi}\,d\phi \int_{0}^{\pi}\,sin\theta \,d\theta  =2\pi$

Here, the azimuthal ring corresponding to a co-latitude just doesn't take a full $\,2\pi$ circle at once but gets added to another equal half from the lower portion in the $\pi/2$ to $\pi$ part of the co-latitude to give precisely the same thing.

The first integral above calculates the solid angle subtended by the upper hemispherical cap surface area  while the second one calculates the angle subtended by the surface area of the $0$ to $\pi$ the right of the axis (speaking loosely). Now of course they should match due to the isotropy of the sphere and other obvious things.

The problem I face is this:
What part of the convention (of choosing the co-latitude, as I have done) makes, an attempt to calculate the same thing like this:

$$\Omega\int_{0}^{\pi}\,d\phi \int_{-\pi/2}^{\pi/2}\,sin\theta \,d\theta   $$ futile ?

Since this i integral is obviously zero.

Now this might sound like a stupid question and it probably is but it bugs me. It is obvious on one hand that since $\sin $ is an odd function about 0 so the area under its curve gets cancelled in the full-wave from $-\pi/2$ to $\pi/2$ thus giving zero.

But on the other hand it seems that nothing of the clockwise-anticlockwise convention seems to stop me from writing the wrong integral and getting the wrong answer.

Probably in the last integral the half azimuthal ring at a particular positive value of the co-latitude gets cancelled by the half azimuthal ring at the negative of that co-latitude value (this explanation doesn't seem ryt to me though).

To put my rambling into context,

This thing came to mind when I was trying to solve a problem where I had to calculate the electric flux through a part of a sphere.

For that I needed to find the solid angle subtended by the shaded surface area.
Did it as follows:
$\int_{-\pi/2}^{\pi/2}\,d\phi \int_{\pi/2-\alpha}^{\pi/2}\,sin\theta \,d\theta  $

which yields

$\pi\sin\alpha$

This was listed as the correct option even when at $\alpha=\pi$ it becomes zero instead of $2$

In think this is again related to the fact that co-latitude changes sign to the left of the axis so probably the integral needs to to be split up or something.

So to sum up,

How would you put the limits for that integral so that it gives $2\pi$ ? (but remember I want the upper part to  become the surface that subtends)

Do conventions and the clockwise
counter-clockwise messing up is whats confusing me or is it totally unrelated to all of the mumbo-jumbo I said?

答:
When you write an integral in the form
$$ \int_{\phi_1}^{\phi_2}\,\mathrm d\phi
 \int_{\theta_1}^{\theta_2} \sin \theta \,\mathrm d \theta,$$
you gain a tighter, unambiguous visual relationship between the variables of integration and the integral signs they apply to (that is, $\mathrm d\phi$ is sitting right next to the integral sign it applies to, rather than all the way on the other side of $\mathrm d\theta$), but you lose the representation of a vital concept of integration over a region of a surface, namely the area element.

When you integrate over the surface of a unit sphere in spherical coordinates, as long as you keep $\theta$ in the range $0 \leq \theta \leq \pi,$ the area element has the form of a "rectangle" of height $\mathrm d\theta$ and width
$\sin\theta\,\mathrm d\phi.$
The "area" of this "rectangle" is therefore
$\sin\theta\, \mathrm d\theta\, \mathrm d\phi.$

You can integrate a function $f$ over a suitable region on the sphere by integrating
$$
 \int_{\phi_1}^{\phi_2} \int_{\theta_1}^{\theta_2}
 f(\phi,\theta) \sin \theta \,\mathrm d \theta\,\mathrm d\phi.
$$
You can turn this integral into a simple area measurement by setting
$f(\phi,\theta) = 1$ everywhere.
The term $\sin\theta$ is not a function you integrate over the surface of the sphere;
it is an essential part of how you integrate anything else over the surface of a sphere.

Conventionally, when doing integrations like these people almost always stay within the bounds of $0 \leq \theta \leq \pi$; after all, there is no point on the sphere that you can't identify using spherical coordinates with $0 \leq \theta \leq \pi$.
So writing $\sin \theta \,\mathrm d \theta\,\mathrm d\phi$ works fine.

If you take $\theta$ outside the range $0 \leq \theta \leq \pi,$
you can find values of $\theta$ where $\sin\theta$ is negative.
But the area elements on the surface of the sphere don't suddenly have negative areas just because you decided to use some unusual values of $\theta.$
If you want to be able to use arbitrary $\theta$ coordinates,
the width of an area element actually is $\lvert\sin\theta\rvert\,\mathrm d\phi,$
and the area element is
$\lvert\sin\theta\rvert \,\mathrm d \theta\,\mathrm d\phi$.

In the language of coordinate transformations, where the coordinates
$(\theta,\phi,\rho)$ represent a transformation from the local coordinates $(u,v,w)$ over the orthonormal basis $e_\theta,e_\phi,e_\rho,$ the area element in spherical coordinates is derived from the orthonormal area element by the rule
$$ \mathrm du \,\mathrm dv = \lvert \det(J)\rvert \,\mathrm d \theta\,\mathrm d\phi$$
where $J$ is the Jacobian of the transformation.
Note that for ordinary area (no "wedge products"), we always take the absolute value of the determinant of the Jacobian.
For this particular transformation, the determinant of the Jacobian is $\sin\theta$,
so the correct area element is technically always
$$ \lvert\sin\theta\rvert \,\mathrm d \theta\,\mathrm d\phi. $$
The reason why people write $\sin\theta$ in the integral rather than
$\lvert\sin\theta\rvert$ is that their plan is that they will never ever in their lives use this expression in an integral where $\theta$ ventures outside the range
$0 \leq \theta \leq \pi.$

The solution to your dilemma is to write the integral in a form that is correct for all $\theta,$ not just for $0 \leq \theta \leq \pi.$ That is, in order to integrate over a region $\phi_1 \leq \phi \leq \phi_2,$ $\theta_1 \leq \theta \leq \theta_2,$
write the area-measuring integral in the form
$$
 \int_{\phi_1}^{\phi_2} \int_{\theta_1}^{\theta_2}
   \lvert\sin\theta\rvert \,\mathrm d \theta\,\mathrm d\phi,
$$
or if you like,
$$
 \int_{\phi_1}^{\phi_2} \,\mathrm d\phi \int_{\theta_1}^{\theta_2}
   \lvert\sin\theta\rvert \,\mathrm d \theta.
$$
Using this always-correct area element for the unit sphere, you will not see cancellation effects when you allow negative values of $\phi_1.$
For example,
$$
 \int_0^\pi \,\mathrm d\phi \int_{-\pi/2}^{\pi/2}
   \lvert\sin\theta\rvert \,\mathrm d \theta = 2\pi,
$$
which is the correct area of the region integrated.

The real fly in the ointment here is not what happens for negative $\theta.$
It's the unfortunate fact that if you take the dashed line in the figure in that electric flux problem as the axis of your spherical coordinates, that is,
if $\theta=0$ you are on the dashed line,
then the shaded region on the surface of the sphere is not a region with bounds of the form $\phi_1 \leq \phi \leq \phi_2,$ $\theta_1 \leq \theta \leq \theta_2.$
If you say that $-\frac\pi2 \leq \phi \leq \frac\pi2$ on that region, then
when $\phi = 0$ the minimum value of $\theta$ is $\frac\pi2 - \alpha,$ but when
$\lvert\phi\rvert$ is near $\frac\pi2$ the minimum $\theta$ value is much closer to $\frac\pi2.$
The area of the region was never going to be computed by
$$
 \int_0^\pi \,\mathrm d\phi \int_{\pi/2-\alpha}^{\pi/2}
   \lvert\sin\theta\rvert \,\mathrm d \theta,
$$
even for small positive values of $\alpha.$

This is not an easy region to integrate using the dashed line as the axis of your spherical coordinates.
The choice of axis of spherical coordinates is an arbitrary choice in a problem like this, so you're better off choosing an axis along the line where the two shaded half-discs meet.

By the way, clockwise vs. anticlockwise depends on the location from which you view the rotation. Conventionally, when thinking about rotation in the $x,y$ plane, we look at it from a point of view far up the positive $z$ axis in a right-hand coordinate system, so the direction of $\phi$ in your figure is anti-clockwise.
And this same fixed point of view on the $z$ axis gives the same anticlockwise result for the direction of increasing $\phi$ at every $\phi$ and every $\theta$.

But consider the point $Q$ at spherical coordinates
$(\theta, \phi, \rho) = \left(-\frac\pi6,\frac\pi2,1\right)$,
which also has coordinates
$\left(\frac\pi6,-\frac\pi2,1\right).$
No matter what coordinates you use, the point $Q$ is at the end of a great-circle arc of length $\frac\pi6$ that starts at the pole, whose coordinates can be written
$\left(0,\frac\pi2,1\right)$ or $\left(0,-\frac\pi2,1\right).$
When $\phi =\frac\pi2,$ we measure the angle of this arc "backwards" from the pole,
but when $\phi = -\frac\pi2$ we measure it "forwards". That is, the exact same arc from $P$ to $Q$ goes in a "negative" $\theta$ direction for one value of $\phi$ and a "positive" $\theta$ direction for another value of $\phi.$ There is no place where we can stand where the positive direction of $\theta$ will always be clockwise, just as there is no place where we can stand where it will always be anti-clockwise.

The usual convention is that $\theta$ is measured in a positive direction away from the positive $z$ axis and toward the negative $z$ axis, but that only works because we usually limit the values of $\theta$ to the range $0 \leq \theta \leq \pi.$
If you have to give a meaning of $\theta$ for values outside that range, you can use the conventional definition for the range $0 \leq \theta \leq \pi$ and simply extend it around the same great circle in both directions from the points at $\theta = 0$ and $\theta = \pi.$ I still would not call this either clockwise or counterclockwise unless you also have a continually moving point of view as $\phi$ changes, and then it depends on whether your point of view "leads" $\phi$ or "trails" $\phi.$
In the figure in the question showing an example of an angle $\phi,$
the point of view "trails" $\phi$, but that's an accident of that particular figure.


问:
Prove or Disprove: If a function is differentiable on $\mathbb{R}$ and $|f(x)|\leq |x|^\alpha$ with $\alpha>1$, then $f'(0)=0$
I believe it's true, but I'm stuck. I think it involves MVT somehow, but we can't say anything about the function in order to conclude that the derivative at a defined point $0$ is equal to zero. I would appreciate any guidance on this, as I have been struggling with it for a long time.

答: First, $|f(0)| \leq |0|^\alpha=0$, so $f(0)=0$. Next, if $x \neq 0$, $|f(x)| \leq |x|^\alpha$ implies that $$\bigg| \frac{f(x)}{x} \bigg| \leq |x|^{\alpha-1}.$$


问:
How many unique patterns can a $4\times 4$ grid make?
I have a grid of $4\times 4$ size and has values starting from $0$ to $15$. How many unique $4\times 4$ patterns can this grid make?  I have no idea which formula or mathematical concept is required here.

I found it on the internet that told if you take $16!$ (factorial) this is how many unique patterns you. can have with a 4x4 grid. and $16!$ is in trillions? Is this the right way or is there anything else?

UPDATE:

What qualifies as a unique pattern?

Lets say we have $0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$ values in our $4\times 4$ grid.

Some unique pattern would be:

$1,0,2,3,4,5,6,7,8,9,10,11,12,13,14,15$

$2,1,0,3,4,5,6,7,8,9,10,11,12,13,14,15$

and so on. The key is that every unique pattern. will have the same values $0-15$ but will be randomized. So how many unique random patterns of $4\times 4$ can we make?

答:
$16!$ is correct. It's easy to see from where it comes from.

The first square could contain any of the 16 integers from $0$ to $15$. So, you have $16$ possibilities.

The second square could contain any of the 15 remaining integers. So the number of combinations for 2 squares and 16 integers is $16 \times 15$.

For the i-th square, the number of combinations would be $16 \times 15 \times \cdots \times [16-(i-1)]$.

Think about it with 2 squares and 3 integers $(0,1,2)$, for example.

First square: $(0)$,$(1)$,$(2) \rightarrow 3 \text{ combinations}$.

Second square:  For each integer we can pick any of the other 2 integers remaining. $(0,1)$ and $(0,2)$, $(1,0)$ and $(1,2)$, $(2,0)$ and $(2,1)$. Clearly 3 pairs. $3\times2$

If you considered a third square, there would be only one option for each previous combination, so the number of combinations keeps the same. $3\times2\times1$


答: $!16$ is correct, you start with the first value having 16 possibilities here, for the 2nd value you have then only 15 possibilities and so on.

Title: How many unique patterns can a $4\times 4$ grid make?

Body:
I have a grid of $4\times 4$ size and has values starting from $0$ to $15$. How many unique $4\times 4$ patterns can this grid make?  I have no idea which formula or mathematical concept is required here.

I found it on the internet that told if you take $16!$ (factorial) this is how many unique patterns you. can have with a 4x4 grid. and $16!$ is in trillions? Is this the right way or is there anything else?

UPDATE:

What qualifies as a unique pattern?

Lets say we have $0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$ values in our $4\times 4$ grid.

Some unique pattern would be:

$1,0,2,3,4,5,6,7,8,9,10,11,12,13,14,15$

$2,1,0,3,4,5,6,7,8,9,10,11,12,13,14,15$

and so on. The key is that every unique pattern. will have the same values $0-15$ but will be randomized. So how many unique random patterns of $4\times 4$ can we make?


问:
Is the following set empty?
$$
sp\left \{
\begin{pmatrix}
1 \\
-1 \\
1 \\
-1
\end{pmatrix} , \begin{pmatrix}
4\\
-2 \\
4 \\
-2
\end{pmatrix} , \begin{pmatrix}
1\\
1\\
1\\
1
\end{pmatrix}
 \right \} \bigcap \left \{ \begin{pmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{pmatrix} | \begin{matrix}
x_1 + x_2 = 0\\
x_3 = 2x_4
\end{matrix} \right \}$$

I was asked to find the basis of the following set but the vectors of the LHS set never fit to both requirements on the RHS set , or maybe I'm totally wrong and the binary relation between the requirements is OR?

答: The set in your question consists of only the zero vector. This is a $0$-dimensional vector space, and the empty set is a basis of it.</p>

Title: Is the following set empty?


问:
The successor of a set
Definition : The successor of a set $x$ is the set $S(x) = x \bigcup \{x\}$

Prove that $x \subseteq S(x)$ and there is no $z$ such that $  x \subset z \subset S(x)$

I really battle with proofs :(

Here is what I have:

let $ y \in x $

then $y \in S(x)$

therefore $ x \subseteq S(x) $

I think I have over simplified something here and am not really proving anything :(

The second thing to prove is that there is no $z$ such that $ x \subset z \subset S(x) $

I really don't know how to approach this one.
Would it be by contradiction?
Assume that such $z$ exists, and then show that it can't exist?

I know that the statement $ x \subset z \subset S(x) $ means there is at least one element in $z$ that is not in $x$ and at least one element in $S(x)$ that is not in $z$ which implies that there are elements in $S(x)$ which are not in $x$.  Doesn't this contradict the definition of $S(x)$?

答:
Normally this construction is used to build the numbers "out of nothing", but for the purposes of clarity I will pretend that $x=\{a,b,c\}$.  Now, $S=x\cup \{x\}=\{1,2,3\}\cup\{\{1,2,3\}\}$, so $$S=\{a,b,c,\{a,b,c\}\}, \textrm{ a set with four elements}$$

To prove that $x\subseteq S$, you need to prove that each element of $x$ is an element of $S$ (or you can use that $A\subseteq A\cup B$ for any $B$).  Study the example above to convince yourself that $S$ differs from $x$ only by the inclusion of one extra element (possibly no extra elements if $\{x\}\in x$, but this is rare).  That should give you the intuition you need to complete the second part of the proof.

Title: The successor of a set


问:
To find $m$ of ranges of integrals
If $$\int_{4}^{m} \frac{4}{x^{2}+3x-18}\,dx =\frac{4}{9}\ln6$$ and $m>4$, then how I should find the range of $m$? I tried solving for integral first. But this is ranges of $m$, is not just find a single number.

Title: To find $m$ of ranges of integrals

Body:
$$\frac{4}{x^{2}+3x-18} =\frac{4}{(x-3)(x+6)}=\frac{4}{9 (x-3)}-\frac{4}{9 (x+6)}$$
$$\int_4^m\frac{4}{x^{2}+3x-18} =\frac{4}{9} (\log (m-3)-\log (m+6)+\log (10))$$
$$\implies \frac{4}{9} \log \left(\frac{10 (m-3)}{m+6}\right)=\frac{4}{9} \log(6)\implies m= ???$$


问:
Is the image of the intersection of preimage distributive and equal to the intersection of original set in the image?
If $x\in\alpha^{-1}(A)\cap\alpha^{-1}(B)$, then we apply $\alpha$ to both sides and get $\alpha(x)\in\alpha(\alpha^{-1}(A)\cap\alpha^{-1}(B))$. Ok now this is the part where I am a bit unsure of. Can we just distribute $\alpha$ over the intersection and get $\alpha(x)\in A\cap B$?

Because it is true that $f(A)\cap f(B)\subseteq f(A\cap B)$. The intersection of two image is just a subset of the image of intersection so they are not equal, so that means $\alpha(\alpha^{-1}(A))\cap\alpha(\alpha^{-1}(B))\subseteq\alpha(\alpha^{-1}(A)\cap\alpha^{-1}(B))$ right? So if I distribute the $\alpha$ over the intersection, wouldn't I be getting a smaller set?

Edit: Ok the above is not correct. It is the case that $f(A\cap B)\subseteq f(A)\cap f(B)$ if the function is not injective and equal if injective.

答:
If $x \in \alpha^{-1}[A] \cap \alpha^{-1}[B]$ then $\alpha(x) \in A$ from being in the first set and also $\alpha(x) \in B$ from being in the second set and so indeed $\alpha(x) \in A \cap B$. This is direct from the definitions.

This is also immediate if you know that inverse images distribute over intersections and unions so that

$$\alpha^{-1}[A] \cap \alpha^{-1}[B] = \alpha^{-1}[A \cap B]$$

This does not hold for forward images of a function in general, but you don't even need that for the conclusion that $\alpha(x) \in A \cap B$.


问:
Pigeonhole principle: Prove that any (2n+1)-element subset of{1,2, ...,3n}contains three consecutive integers.
Prove that any (2n+1)-element subset of{1,2, ...,3n}contains three consecutive integers.

My general thinking is that pigeonholes are sets of consecutive integers from {1,2,...,3n} which are {1,2,3}{4,5,6},...,{3n-2,3n-1,3n} and that each one is a pigeonhole filled 2/3 of the way, so the (2n+1)th element would have to fill a pigeonhole fully, but I don't know how to prove it.

答:
You've got $n$ many holes, $\{1,2,3\},\{4,5,6\}, \ldots,\{3n-2,3n-1,3n\}$.

If you've got $2n+1$ balls to put in them, you cannot have at most $2$ per hole because that only leaves room for at most $2n$ balls, so one hole must contain $3$ and you're done.

So that partition indeed does work.


问:
What is the projective bundle over $(\Bbb P^1\times\Bbb P^1)/(\Bbb Z/2)$ which isn't $\Bbb P(\mathcal{E})$ for some vector bundle $\mathcal{E}$?
I'm thinking about when a (Zariski-locally trivial) $\Bbb P^n$-bundle comes from the projectivization of a vector bundle, and I've found Sasha's answer here on MO stating that $(\Bbb P^1\times\Bbb P^1)/(\Bbb Z/2)$ has a $\Bbb P^1$-bundle which isn't the projectivization of a vector bundle. Unfortunately, it doesn't say what this bundle is. Does anyone know an explicit description of what the intended bundle is, and an easy way to see it's not the projectivization of a vector bundle?

答:
Let $X = (\mathbb{P}^1 \times \mathbb{P}^1)/(\mathbb{Z}/2)$. This is a toric surface with 4 ordinary double points. Let
$$
\pi \colon Y \to X
$$
be the blowup of one of these points and let $E \subset Y$ be the exceptional curve. Then $E^2 = -2$ and one has
$$
Ext^1(\mathcal{O}_Y, \mathcal{O}_Y(E)) =
H^1(Y, \mathcal{O}_Y(E)) \cong
H^1(Y, \mathcal{O}_E(E)) \cong \Bbbk.
$$
Let the vector bundle $F$ on $Y$ be defined as the non-trivial extension
$$
0 \to \mathcal{O}_Y(E) \to F \to \mathcal{O}_Y \to 0.
$$
Then it follows from the construction that
$$
F\vert_E \cong \mathcal{O}_E(-1)^{\oplus 2}.
$$
Consider the projective bundle $\mathbb{P}_Y(F)$, the surface $S = \mathbb{P}_E(F\vert_E) \subset \mathbb{P}_Y(F)$, and the composition
$$
\mathbb{P}_Y(F) \to Y \to X.
$$
One can check that this composition has another factorization
$$
\mathbb{P}_Y(F) \to Z \to X,
$$
where the first morphism contracts the surface $S$ and the second morphism is a $\mathbb{P}^1$-bundle, which is not isomorphic to the projectivization of a vector bundle.

For the proof see https://arxiv.org/abs/1809.10628, Section 4.3.

EDIT. Here is an explanation why $Z \to X$ is not a projectivization.

Let $x_0 = \pi(E)$ and let $C \subset Z$ be the fiber of $Z \to X$ over $x_0$. If $Z \to X$ is a projectivization, there is a line bundle $L$ on $Z$ such that $L\vert_C \cong \mathcal{O}(1)$. Then the pullback of $L$ to $\mathbb{P}_Y(F)$ restricts to $S \cong E \times \mathbb{P}^1$ as $\mathcal{O}(0,1)$. On the other hand, the Grothendieck bundle of $\mathbb{P}_Y(F)$ restricts to $S$ as $\mathcal{O}(1,1)$. From this we conclude that there is a line bundle on $Y$ that restricts to $E$ as $\mathcal{O}(1)$, but this is impossible, because $Pic(Y)$ is generated by $\pi^*Pic(X)$ (which restricts to $E$ trivially) and by $\mathcal{O}_Y(E)$ (which restrict to $E$ as $\mathcal{O}(-2)$).


问:
How to construct a one-one mapping from natural numbers to irrational numbers
I am stuck on the problem of finding a one-one mapping from natural numbers to irrational numbers. Can anyone help?</p>

答: $\sqrt{2}$ is irrational. The map $f(n) = (n+1)\sqrt{2}$ sends distinct natural numbers to distinct irrational ones.


问:
I couldn't find that what is $M_n$
$f(x)=\sum_{n=1}^\infty\frac{1}{n}\sin\left(\frac{x^2}{n}\right)$  and that $|sin(x)|\leq|x|$

a) Use the Weierstrass M-test and Term-by-term Continuity Theorem to show that $ f(x)=\sum_{n=1}^\infty\frac{1}{n}\sin\left(\frac{x^2}{n}\right)$ is continuous on any interval  $[-a,a]$

b) Use the Weierstrass M-test and Term-by-term Differentiability Theorem to show that $ f(x)=\sum_{n=1}^\infty\frac{1}{n}\sin\left(\frac{x^2}{n}\right)$ is differentiable on any interval $[-a,a]$

c) Is it true that $f ( x )$  is continuous and differentiable on all of $ \mathbb{R}$?

答:
Denoting $f_n(x) = \dfrac1n\sin\left(\dfrac{x^2}{n}\right),$ we have $$|f_n(x)| \le \left|\dfrac1n\sin\left(\dfrac{x^2}{n}\right)\right| \le \dfrac{x^2}{n^2} \le \dfrac{a^2}{n^2}.$$
Note that $\sum \frac{1}{n^2}$ converges. Can you complete it now? (Take $M_n = a^2/n^2$.)

Title: I couldn't find that what is $M_n$


问:
Why is the Nth derivate isolated in many textbook definitions?

My textbook keeps writing definitions in the above form, with the LHS of the equation being the highest order derivative, and the RHS being everything else. Why is this form stressed over a form like this

? Is there any mathematical reason, if it is just a convention then why?

Notice that the second equation has a function of t on the RHS and everything else on the LHS.

答:
Because in this explicit form, if the right side function $f$ is continuous (on some open domain) then you can directly say that this is an ODE. Then all the basic existence and uniqueness theorems can be applied without further restrictions.

In contrast, if the equation is in implicit form, and if $y$ is a vector, then it is not given that one can isolate the highest derivatives (or even that the highest order derivatives have the same order for all variables), one would have to demand the conditions of the implicit function theorem. Even then, there can be multiple solutions for the highest derivative. See Clairaut equations for simple examples.

Without that local solvability for the highest derivatives, an implicit differential equation could as well be a DAE, a differential-algebraic system of equations (or in exotic cases fail to fall into any classification).


问:
connected components of domain without preimage of critical values
Let $f:M \to N$ be a differentiable map between two connected compact manifolds both of dimension $n\ge 2$ (I think i know the answer to my question if $n=1$). Let us assume that both $M$ and $N$ are connected and the image of $f$ is dense in $N$. So we know that preimage of a regular value is finite. Consider the regular points in $M$ and take the connected components of it. Let $A\subset M$ be such a component and suppose furthermore that that there are two different points $x,y\in A$ such that $f(x)=f(y)$. I have a complicated example ($n=2$) where this setup is possible. Let $U$ be a largest neighbourhood of $x$ in $A$ such that $f|U$ is injective (so if there is any $U'\supset U$ such that $f|U'$ is diffeomorphic to $f(U')$ then $U=U'$).

My questions is:

Is it true that the boundary of this largest neighbourhood $U'$ of $x$ will necessarily contain a point $x'$ in $A$ such that $f(x')$ is a critical value?

Is it true that the preimage of the critical value "separates" two points with the same image, i.e. if I consider $M$ without the preimage of the critical values then $f$ restricted to a connected component of this set is injective?

I would appreciate a counterexample or rough proof.

Edit1: Apparently I asked a question some years ago that is related to this question and the answer to that question would also answer my question here (both 1. and 2. are not necessarily true).

Edit2: Actually the example that was given to me two years ago was when $M$ is not compact and in my question I assumed $M$ is compact. I wonder if someone can give me an example when $M$ is the torus (so $n=2$).

答:
I have to have more conditions for this work. Here is an example when $M$ is compact and $n=2$ such that neither 1. nor 2. is true:

Take $M$ to be the unit sphere $S^2$ and $N$ to be the projecitve plane. We now take $f$ to be the natural map sending antipodal points of $S^2$ to the same point in the projective plane (represented by one of these points). This map is smooth and yet $2:1$.


问:
Primality test based on initial conditions alone.
Let $m=252601$. Suppose we discover that
$$3^{126300} ≡ 67772 \pmod{252601}$$
$$3^{252600} ≡ 1\pmod{252601}$$
Is then $252601$ prime? composite? Or can we not decide for sure from the information given?</p>

答: Hint: $a^{\frac{p-1}{2}} \equiv \pm 1\pmod p$ for any prime $p$ and a positive integer $a$ smaller than $p$.</p>


问:
How do I bypass the limit for the tan function on the calculator?
So basically, my partner and I are creating problems for a project. She created one where we aren't sure what the correct answer is.

Problem:
$Find \ cos\frac{\theta}{2} \ if \ tan \ \theta = \frac{3}{4}; \pi < \theta<\frac{3\pi}{2}$

I'll show you what we did.

First, we used the pythagorean formula to find $cos \ \theta$ (we need it later). The half-angle formula for cosine is $cos \frac {\theta}{2} = \pm \sqrt{\frac{1 + cos \ \theta}{2}}$.

These are our steps.

$cos \frac {tan^{-1}\frac{3}{4}}{2} = \pm \sqrt{\frac{1 + cos \ (tan^{-1}\frac{3}{4})}{2}}$

$ = \pm \sqrt{\frac{1 + \frac{4}{5}}{2}}$

$ = \pm \sqrt{\frac{\frac{9}{5}}{2}}$

$ = \pm \sqrt{\frac{9}{10}}$

$ = \pm \frac{3\sqrt{10}}{10}$

To decide whether we have a negative sign or a positive, we have to look at the domain.

$ \pi < \theta<\frac{3\pi}{2} \ changes \ to  \ \frac{\pi}{2} < \frac{\theta}{2} < \frac{3\pi}{4}$

We now know that $cos \frac {\theta}{2}$ is located in the second quadrant, which makes our final answer   $cos \frac {tan^{-1}\frac{3}{4}}{2} = - \frac{3\sqrt{10}}{10}$
But then, I checked the calculator.
The values are equal except for the negative sign

My partner said that the calculator thinks tangent is in the first quadrant, and if it were, cosine would be positive. But since we are in the third quadrant, cosine is in the second quadrant. I know there are domain restrictions, but I don't entirely understand what she meant. Also, how do I get the calculator to display the same answer?

答:
You cannot do the inverse tangent because the range of inverse tangent is $(-\frac{\pi}{2},\frac{\pi}{2})$, and so your answer for step 2 (and also the answer from your calculator) is already incorrect.

I would suggest you do this:

Draw the triangle in the third quadrant (because it is in the range $(\pi, \frac{3\pi}{2})$, with adjacent side (on the x-axis) $x=-4$ and opposite side (on the y-axis) $y=-3$. The hypotenuse is $5$ by Pythagoras. Then, $\cos \theta = -\frac{4}{5}$. Using the cosine half-angle formula, $\cos \frac{\theta}{2}=\pm\sqrt{\frac{1+\cos \theta}{2}}=\pm\sqrt{\frac{1+(-\frac{4}{5})}{2}}=\pm\sqrt{\frac{1}{10}}=\pm \frac{\sqrt{10}}{10}$ Because the angle is in the range $(\pi, \frac{3\pi}{2})$, then half the angle is in the range $(\frac{\pi}{2}, \frac{3\pi}{4})$, so $\cos$ is negative. Therefore, your answer is $\boxed{-\frac{\sqrt{10}}{10}}$.


问:
Probability density confusion in $p(z|x)$ when $z = x + y$, is $p(z|x) = p_{y|x}(z-x) = p(y|x)$?
I am reading on conditional pdf and encountered this example below.

For random varaibles $x, y$, and $z$,

"If $z = x + y$, then $p(z|x) = p_{y|x}(z - x)$."

I tried to make sense of this with following reasons:

\begin{align}
p(z|x) &= \frac{p(z, x)}{p(x)} \\
       &= \frac{p(x+y, x)}{p(x)} \\
       &= \frac{p(y, x)}{p(x)} \qquad\qquad\qquad \text{since p(x+y, x) = p(y,x)?}\\
       &= p(y|x).
\end{align}

My first confusion is for the notation:

(1) From above, it seems that $p_{y|x}(z-x) = p(y|x)$. But I don't know what the argument $z-x$ is there for. For example, is $p_{y|x}(z-x) = p_{y|x}(y)$, and $p(y|x) = p(y|x)(y)$? I do not understand why $z-x$ is used instead of $y$ and feel like I am missing some important concept here.

Now, the book says if $x$ and $y$ are independent, it becomes:

$$
p(z|x) = p_y(z-x).
$$

(2) Again, I do not understand why the particular argument $z-x$ is used to denote this. Isn't that basically just $p(y)$? Are they different somehow?

Next, the example shows the same thing. When $x$ and $y$ are independent Gaussian random variables with mean $\mu_x$, $\mu_y$, and variances $\sigma_x^2$, $\sigma_y^2$,

$$
p(z|x) = p_y(z-x) = \mathcal{N}(z-x; \mu_y, \sigma_y^2).
$$
and it says that $p(z|x)$ is the pdf of $z-x$ with mean $\mu_y$ and vraince $\sigma_y^2$.
I don't understand why the author makes sure to distinguish between $p(y)$ and $p_y(z-x)$ except if there are some subtle differences I am not understanding at the moment.

Perhaps it's just a style? But since I am a novice, I cannot escape the feeling of missing something. Could anyone please clarify it or tell me what I am missing here?

答:
The author is abusing notation: using what looks like the same notation to mean two different things. Specifically, the author writes $p(z\mid x)$ to mean $p_{Z\mid X}(z\mid x)$ and writes $p(y\mid x)$ to mean $p_{Y\mid X}(y\mid x)$. (Note -- I'm using capital letters to denote the random variable and lower case to denote the value the random variable takes; I dunno if the author makes this distinction.) In essence, the author is being lazy and omitting subscripts when "the interpretation is clear".  This can lead to confusion when you are not aware that this convention is being adopted.

If you are aware that this is what is going on, the derivations should make more sense. So read the first argument this way: If $Z=X+Y$, then
$$\begin{aligned}
p_{Z\mid X}(z\mid x)&=p(Z=z\mid X=x)\\& = p(X+Y=z\mid X=x)\\&=p(Y=z-x\mid X=x)\\&=p_{Y\mid X}(z-x\mid x)\end{aligned}$$ If $X$ and $Y$ are independent, then $p_{Y\mid X}(y\mid x) = p_Y(y)$ for any $x$, which explains the second conclusion: $$p_{Y\mid X}(z-x\mid x) = p_Y(z-x) \qquad \text{when $X$ and $Y$ are independent.}$$


答:
$$\begin{align}
p(z|x) &= \frac{p(z, x)}{p(x)} \\
       &= \frac{p(x+y, x)}{p(x)} \\
       &= \frac{p(y, x)}{p(x)} \qquad\qquad\qquad \text{since p(x+y, x) = p(y,x)?}\\
       &= p(y|x).
\end{align}$$

Your discussion is about random variables $X,Y,Z$ , values $x,z$.  You introduced the symbol of $y$ without explaining it (eg $y=z-x$).  Always explain new symbols, and only introduce them if it makes things clearer.  Here, it does not.

You also confuse the joint distributions.  Indeed the notation is a source of confusion.  I recommend subscripting all such functions - don't leave it up to the value symbols.

$$\begin{align}
p_{\small Z\mid X}(z\mid x) &= \frac{p_{\small Z,X}(z, x)}{p_{\small X}(x)} &&\text{definition of conditional probability} \\
       &= \frac{p_{\small X+Y,X}(z, x)}{p_{\small X}(x)} &&\text{because }Z=X+Y \\
       &= \frac{p_{\small Y,X}(z-x, x)}{p_{\small X}(x)} &&\text{change of variable} \\
       &= p_{\small Y\mid X}(z-x\mid x).
\end{align}$$

Title: Probability density confusion in $p(z|x)$ when $z = x + y$, is $p(z|x) = p_{y|x}(z-x) = p(y|x)$?

Body:
I am reading on conditional pdf and encountered this example below.

For random varaibles $x, y$, and $z$,

"If $z = x + y$, then $p(z|x) = p_{y|x}(z - x)$."

I tried to make sense of this with following reasons:

\begin{align}
p(z|x) &= \frac{p(z, x)}{p(x)} \\
       &= \frac{p(x+y, x)}{p(x)} \\
       &= \frac{p(y, x)}{p(x)} \qquad\qquad\qquad \text{since p(x+y, x) = p(y,x)?}\\
       &= p(y|x).
\end{align}

My first confusion is for the notation:

(1) From above, it seems that $p_{y|x}(z-x) = p(y|x)$. But I don't know what the argument $z-x$ is there for. For example, is $p_{y|x}(z-x) = p_{y|x}(y)$, and $p(y|x) = p(y|x)(y)$? I do not understand why $z-x$ is used instead of $y$ and feel like I am missing some important concept here.

Now, the book says if $x$ and $y$ are independent, it becomes:

$$
p(z|x) = p_y(z-x).
$$

(2) Again, I do not understand why the particular argument $z-x$ is used to denote this. Isn't that basically just $p(y)$? Are they different somehow?

Next, the example shows the same thing. When $x$ and $y$ are independent Gaussian random variables with mean $\mu_x$, $\mu_y$, and variances $\sigma_x^2$, $\sigma_y^2$,

$$
p(z|x) = p_y(z-x) = \mathcal{N}(z-x; \mu_y, \sigma_y^2).
$$
and it says that $p(z|x)$ is the pdf of $z-x$ with mean $\mu_y$ and vraince $\sigma_y^2$.
I don't understand why the author makes sure to distinguish between $p(y)$ and $p_y(z-x)$ except if there are some subtle differences I am not understanding at the moment.

Perhaps it's just a style? But since I am a novice, I cannot escape the feeling of missing something. Could anyone please clarify it or tell me what I am missing here?


问:
Can someone show this strange equality?
The solution in the book says
$$
\frac{4}{5^{2/3}+5^{1/3}+5}
$$
and the solution I calculated is
$$
\frac{-25^{1/3}+5}{5}
$$

Apparently these describe the same number.

However, I tried to show the equality and came nowhere.

Any idea on how to prove the equality?

Title: Can someone show this strange equality?

Body:
Hint: A difference of cubes can be factored as $$x^3 - 1 = (x-1)(1+x+x^2).$$ Plugging in $x=5^{1/3}$, we have $$5-1=(5^{1/3}-1)(1+5^{1/3}+5^{2/3})=(1-5^{-1/3})(5^{1/3}+5^{2/3}+5).$$
Can you finish from here?


问:
How to find the least value of $\cot x + \frac{\csc^2x+\cot ^4x}{\csc^2x+\cot x}+\tan x + \frac{\sec^2x+\tan^4x}{\sec^2x+\tan x}$?
The problem is as follows:

First assume that $x$ is on:

$\frac{3\pi}{2}<x<2\pi$

Then find the least integer value of the expression from below:

$$\cot x + \frac{\csc^2x+\cot ^4x}{\csc^2x+\cot x}+\tan x + \frac{\sec^2x+\tan^4x}{\sec^2x+\tan x}$$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{5}\\
2.&\textrm{2}\\
3.&\textrm{4}\\
4.&\textrm{3}\\
\end{array}$

I am not sure how to tackle this problem. What I've attempted to do is to attempt to build the square.

This is seen in below:

$$\cot x +\frac{1+\cot^2 x + \cot ^4 x}{1+\cot^2x+\cot x}+\tan x + \frac{1+\tan^2 x + \tan^4x }{1+\tan^2x+\tan x}$$

This reduces to:

$$\cot x +\frac{\left(\frac{1}{2}+\cot^2 x\right)^2 +\frac{3}{4}}{\left(\cot x+\frac{1}{2}\right)^2+\frac{3}{4}}+\tan x + \frac{\left(\frac{1}{2}+\tan^2 x\right)^2+\frac{3}{4}}{\left(\frac{1}{2}+\tan x\right)^2+\frac{3}{4}}$$

But that's it this is the part where I got stuck. I don't know what else can be done here in order to simplify the expression?. You can notice that the expressions in the denominator are of a lesser power.

Does it exist any sort of algebraic manipulation which I'm overlooking? Can someone help me here please?

Body:
The key is to use the cubic identities $1-u^3 = (1-u)(1+u+u^2), 1+u^3 = (1+u)(1-u+u^2)$. For convenience, let $\cot x = c$, $\tan x = t$; substitute $c^2$ and $t^2$ into the first identity to get:

$$c + \frac{1+c^2+c^4}{1+c+c^2} + t + \frac{1+t^2+t^4}{1+t+t^2}$$
$$= c+t + \frac{1 - c^6}{1 - c^2} \frac{1 - c}{1 - c^3} + \frac{1 - t^6}{1 - t^2} \frac{1-t}{1 - t^3}$$
$$= c+t + \frac{(1-c^3)(1+c^3)}{(1-c)(1+c)} \frac{1 - c}{1 - c^3} + \frac{(1 - t^3)(1+t^3)}{(1-t)(1+t)} \frac{1-t}{1 - t^3}$$
$$= c+t + 1-c+c^2 + 1-t+t^2 = \tan^2 x + \cot^2 x + 2$$
$$= (\tan x + 1/\tan x)^2 - 2 + 2 ≥ 4$$

where in the last step, square both sides of $u+1/u < -2$ when $x < 0, x \in \mathbb R$ ($\tan x$ is negative in the given range).


答: You are correct about the first part, and your oversimplification is really a proof. For the second you are also correct when you proceed by contradiction. So there is an element in $z$ not in $x$ but in $S(x)$. That element can only be $\{x\}$. But then $z=S(x)$.</p>

Title: The successor of a set

Body:
Definition : The successor of a set $x$ is the set $S(x) = x \bigcup \{x\}$

Prove that $x \subseteq S(x)$ and there is no $z$ such that $  x \subset z \subset S(x)$

I really battle with proofs :(

Here is what I have:

let $ y \in x $

then $y \in S(x)$

therefore $ x \subseteq S(x) $

I think I have over simplified something here and am not really proving anything :(

The second thing to prove is that there is no $z$ such that $ x \subset z \subset S(x) $

I really don't know how to approach this one.
Would it be by contradiction?
Assume that such $z$ exists, and then show that it can't exist?

I know that the statement $ x \subset z \subset S(x) $ means there is at least one element in $z$ that is not in $x$ and at least one element in $S(x)$ that is not in $z$ which implies that there are elements in $S(x)$ which are not in $x$.  Doesn't this contradict the definition of $S(x)$?


问:
Condition of invertible diagonal operator on Hilbert space
I am reading C*-Algebras and Operator Theory - Gerald J. Murphy. Here is where I'm stack

$U $ is invertible if and only if $\inf_n|\lambda_n|>0$

1.4.3. Example. One should not be misled by Theorem 1.4.11—the spectral behaviour of compact operators is not typical of all operators. To illustrate this, let $H$ be a separable Hilbert space with an orthonormal basis $(e_n)_{n=1}^\infty$. If $(\lambda_n)$ is a bounded sequence of scalars, define $u\in B(H)$ by setting $u(x)=\sum_{n=1}^\infty\lambda_n\alpha_ne_n$ when $x=\sum_{n=1}^\infty\alpha_ne_n$. We call $u$ the diagonal operator with diagonal $(\lambda_n)$ with respect to the basis $(e_n)$. It is readily verified that $\|u\|=\sup_n|\lambda_n|$, and that $u$ is invertible if and only if $\inf_n|\lambda_n|>0$, and in this case $u^{-1}$ is the diagonal operator with respect to $(e_n)$ with diagonal $(\lambda_n^{-1})$. These observations imply that $\sigma(u)$ is the closure of the set $\{\lambda_n\mid n=1,2,\ldots\}$.

$u$ is diagonal operator on Hilbert space, then $u$  is invertible iff $\inf \lambda >0$.

答:
$u(e_n)=\lambda_ne_n$. So, if $u$ is invertible, then $\lambda_n \neq 0$,   $u^{-1}(e_n)=\frac 1  {\lambda_n}e_n$ and $\frac 1  {|\lambda_n|} \leq \|u^{-1}\|$ proving that $\inf_n |\lambda_n| >0$.

Conversely, if $\inf_n |\lambda_n| >0$ then $v(\sum \alpha_n e_n)=\sum \frac1 {\lambda_n} \alpha_n e_n$ defines a bounded operator and $u\circ v=v\circ u=I$.


答:
$$\sec^2 x=1+\tan^2 x \, , \quad \csc^2 x=1+\cot^2 x$$

Use these identities to rewrite the rational terms as

$$\frac{\sec^2x+\tan^4x}{\sec^2x+\tan x}=\frac{\tan^4x+\tan^2 x +1}{\tan^2x+\tan x+1}$$

Now use the identity : $$t^4+t^2+1=(t^2+t+1)(t^2-t+1)$$

The original expression simplifies
$$\cot x + (\cot^2 x -\cot x +1) + \tan x + (\tan^2 x - \tan x +1)$$
$$=\cot^2 x + \tan^2 x +2$$
$$=(\cot^2 x + \tan^2 x -2)+4$$
$$=(\cot x - \tan x)^2+4$$

This will be minimum when first term is zero. There exists a value of $x \in (3\pi/2, 2\pi)$ for which $\cot x =\tan x$ (namely $7\pi/4$). Thus the minimum attained is $4$.

Title: How to find the least value of $\cot x + \frac{\csc^2x+\cot ^4x}{\csc^2x+\cot x}+\tan x + \frac{\sec^2x+\tan^4x}{\sec^2x+\tan x}$?

Body:
The problem is as follows:

First assume that $x$ is on:

$\frac{3\pi}{2}<x<2\pi$

Then find the least integer value of the expression from below:

$$\cot x + \frac{\csc^2x+\cot ^4x}{\csc^2x+\cot x}+\tan x + \frac{\sec^2x+\tan^4x}{\sec^2x+\tan x}$$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{5}\\
2.&\textrm{2}\\
3.&\textrm{4}\\
4.&\textrm{3}\\
\end{array}$

I am not sure how to tackle this problem. What I've attempted to do is to attempt to build the square.

This is seen in below:

$$\cot x +\frac{1+\cot^2 x + \cot ^4 x}{1+\cot^2x+\cot x}+\tan x + \frac{1+\tan^2 x + \tan^4x }{1+\tan^2x+\tan x}$$

This reduces to:

$$\cot x +\frac{\left(\frac{1}{2}+\cot^2 x\right)^2 +\frac{3}{4}}{\left(\cot x+\frac{1}{2}\right)^2+\frac{3}{4}}+\tan x + \frac{\left(\frac{1}{2}+\tan^2 x\right)^2+\frac{3}{4}}{\left(\frac{1}{2}+\tan x\right)^2+\frac{3}{4}}$$

But that's it this is the part where I got stuck. I don't know what else can be done here in order to simplify the expression?. You can notice that the expressions in the denominator are of a lesser power.

Does it exist any sort of algebraic manipulation which I'm overlooking? Can someone help me here please?


问:
Lower Sum Inequality??
Let $f$ and $g$ be bounded function on $[a,b]$ and $h=f+g$.

How can I link
$$L_P(f) + L_P(g) \leq L_P(h)$$
this relationship to prove that
$$ \underline{\int^b_a}(f) + \underline{\int^b_a}(g) \leq \underline{\int^b_a}(h)$$

If there's any better way to prove the statement right above, tell me.
The integrals above represents lower integrals, it doesn't has to be Riemann integral.

答:
That's the standard proof.

By definition of supremum, if $\varepsilon > 0$, there exist partitions $P', P''$ of $[a,b]$ such that $$L_{P'}(f) > \underline {\int_a^b} (f)-\varepsilon$$ and $$L_{P''}(g) > \underline {\int_a^b} (g)-\varepsilon$$ Considering a common refinement $P$ of $P'$ and $P''$ and using the relationship in your post, you get $$\underline {\int_a^b} (f) + \underline {\int_a^b} (g) < \underline {\int_a^b} (h) + 2\varepsilon$$ So you are done because of the arbitrariness of $\varepsilon$.

Title: Lower Sum Inequality??


问:
Prove,that there exists bijection between the sets $\{0,1\}^*$ and $\{0,1,2\}^*$
Should I show, that both of the sets are countable by mapping initial words to natural numbers?</p>

答:
There are several ways to show that, with different "pre-requisites".

One first idea is to actually explicit a bijection between the two sets. This is somewhat tedious.

Another idea is to find another set - e.g. $\mathbb N$ - which is in bijection with your two sets. Depending on how you prove that $\mathbb N$ and $\{0, 1\}^\ast$ are in bijection, this might lead to an "un-constructive" proof, meaning that you will know that some bijection exists, but with no way of actually computing it explicitly.

Finally, you can also use various theorems and find:

An injection and a surjection from one set to the other.

Two injections (one in each sense)

Both of these will give the existence of a bijection between the two sets.

To be fair, the simplest way would be to show that they are both countable, and claim that this is enough to guarantee the existence of a bijection between the two sets.


问:
The successor of a set
Definition : The successor of a set $x$ is the set $S(x) = x \bigcup \{x\}$

Prove that $x \subseteq S(x)$ and there is no $z$ such that $  x \subset z \subset S(x)$

I really battle with proofs :(

Here is what I have:

let $ y \in x $

then $y \in S(x)$

therefore $ x \subseteq S(x) $

I think I have over simplified something here and am not really proving anything :(

The second thing to prove is that there is no $z$ such that $ x \subset z \subset S(x) $

I really don't know how to approach this one.
Would it be by contradiction?
Assume that such $z$ exists, and then show that it can't exist?

I know that the statement $ x \subset z \subset S(x) $ means there is at least one element in $z$ that is not in $x$ and at least one element in $S(x)$ that is not in $z$ which implies that there are elements in $S(x)$ which are not in $x$.  Doesn't this contradict the definition of $S(x)$?

答:
Your first step could be okay or not, depending on whether you understand why it’s justified. The point is that since $S(x)=x\cup\{x\}$, $x\subseteq S(x)$, simply because it’s always true that $a\subseteq a\cup b$, no matter what sets $a$ and $b$ are.

Now suppose that $x\subsetneqq z\subsetneqq S(x)$. Then there is some $u\in z\setminus x$. On the other hand, $z\subseteq S(x)$, so $u\in S(x)$. There is only one element of $S(x)$ that is not an element of $x$, so $z$ must be that element; what is it? Why does that contradict the assumption that $z\subsetneqq S(x)$?

Title: The successor of a set


问:
Some concrete examples of $M_q(2)$ points
Given $q \in \mathbb{C}$ invertible, Kassel says that an $M_q(2)$ point of an $R$ algebra is a $m=\left(\begin{array}{cc}
A & B\\
C & D
\end{array}\right)\in R^{4}$ such that $A,\,B,\,C,\,D \in R$ satisfy the following relations
$$CA=qAC,$$$$ DB=qBD, $$$$BA=qAB,$$$$DC=qCD,$$$$BC=CB,$$$$DA-qCB=AD-\left(q^{-1}\right)BC.$$
Can anybody give me a concrete example of $M_3(\mathbb{C})$ matrices that form such a point? If it's not possible in $M_3(\mathbb{C})$, every other concrete example is well accepted.
Thanks in advance

答:
Let me maybe give you a few answers that you do not want to hear. First, the zero matrix is, of course, an $M_q(2)$ point. That's not very interesting indeed. Then there are the following two points with entries of size one:
$$\pmatrix{1&0\cr0&1},\qquad\pmatrix{0&1\cr1&0}$$
I guess you are not interested in those either. But not really because some of the entries are equal to zero as somebody suggested, but since this example is kind of classical. In particular, the entries are one-dimensional. You could construct another example with 2-dimensional entries by somehow considering those two both "at once":
$$\pmatrix{
\pmatrix{1&0\cr0&0}&\pmatrix{0&0\cr0&1}\cr
\pmatrix{0&0\cr0&1}&\pmatrix{1&0\cr0&0}
}$$
But you should not be interested in this example either, because it still is classical although the entries are two-by-two matrices. The reason is that the entries mutually commute, so they are diagonal (or, in general, simultaneously diagonalizable) and you can restrict to the common eigenspaces and get exactly the two examples above.

So, you are probably looking for an example, where the entries generate a non-commutative algebra. These are not hard to construct from some elementary examples of $q$-commuting matrices. I guess you already got some examples of those in the answers to your another question Concrete cases where $YX=qXY$.

So, for instance, you can go for
$$\pmatrix{
\pmatrix{1&0\cr0&q}&\pmatrix{0&1\cr0&0}\cr
\pmatrix{0&1\cr0&0}&\pmatrix{q&0\cr0&1}
}.$$


问:
local minimum in 0
Consider the function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ given by $f(x,y)=(y-x^2)(y-3x^2)$ Prove that:

If $u\in\mathbb{R^2}\setminus\left \{0\right\}$ an arbitrary vector and $g:\mathbb{R}\rightarrow\mathbb{R}^2$ given by $g(t)=tu$, then the function $f\circ g$ has a local minimun at $0\in \mathbb{R}$

I have tried to do it through the Hessian matrix but I have gotten stuck in how it should derive, I also feel that in any case I would only get 0 in the Hessian matrix, can someone help me?

答: If $u=(a,b)$, then $g(tu)=3 a^4 t^4-4 a^2 b t^3+b^2 t^2$, whose second derivative at $0$ is $2b^2$. So, if $b\ne0$, $g(tu)$ has a local minimum at $0$. And if $b=a$, then $g(tu)=3a^4t^4$, which also has a local minimum at $0$.

Title: local minimum in 0


答: No. need to use the  Hessain matrix. $f(g(t))=(tu_2-t^{2}u_1^{2})(tu_2-3t^{2}u_1^{2})=t^{2}((u_2-tu_1^{2})(u_2-3tu_1^{2})$ which has the value $0$ when $t=0$. So we only have to check that $f(g(t)) \geq 0$ when $|t|$ is sufficiently small. But $((u_2-tu_1^{2})(u_2-3tu_1^{2}) \to u_2^{2}$  as $ t>0$ so it is clear that $((u_2-tu_1^{2})(u_2-3tu_1^{2}) >0$ when $|t|$ is sufficiently small provided $u_2 \neq 0$. The case $u_2=0$ is left to you.

Title: local minimum in 0

Body:
Consider the function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ given by $f(x,y)=(y-x^2)(y-3x^2)$ Prove that:

If $u\in\mathbb{R^2}\setminus\left \{0\right\}$ an arbitrary vector and $g:\mathbb{R}\rightarrow\mathbb{R}^2$ given by $g(t)=tu$, then the function $f\circ g$ has a local minimun at $0\in \mathbb{R}$

I have tried to do it through the Hessian matrix but I have gotten stuck in how it should derive, I also feel that in any case I would only get 0 in the Hessian matrix, can someone help me?


问:
A set $E$ is closed if and only if $E = E^-$
Let $E$ be a subset of a metric space $(S,d)$. I want to show that the set $E$ is closed if and only if $E = E^-$ where $E^-$ is the closure of a set $E$.

First I assumed $E = E^-$. Then since I know $E^-$ is the intersection of all closed sets containing $E$, $E^-$ must be closed because the intersection of any closed sets is a closed set. So it follows that $E$ is closed.

Now if $E$ is closed, then $S \setminus E$ is open. That also means $E^- \setminus E$ is open since $E \subseteq E^-$ by definition. Suppose $E \subsetneq E^-$. Let $x$ be on the boundary of $E^-$. Then there does not exist an $r>0$ such that $\{s \in S \ | \ d(s, x) < r \} \subseteq E^- \setminus E$. That means $E^- \setminus E$ is not open which gives a contradiction. So it must be the case that $E^- = E$.

Could someone give me feedback on my proof? And is there a better/shorter proof?

答: Well I personally have learned that $\overline{E}$ (closure of $E$) is defined as $\overline{E} = E \cup E'$ where $E'$ is the set of all limit points of $E$ and that $E$ is a closed set iff $\forall \; x \in E' x \in E$ (every limit point of $E$ is a point of $E$). From this definition it's easy to see that if $\overline{E} = E$ then $E \cup E' = E$ and it follows that $E' \subset E$ so $E$ must be closed. Now if we assume $E$ is closed then $\forall \; x \in E' x \in E$ or more simply $E' \subset E$ and again from this it's easy to observe that $E \cup E' = E$.</p>


问:
Find $\int^\infty_0 \frac{\sin x}{x(1+\cos ^2 x)}dx$
find the value:
$$\int^\infty_0 \frac{\sin x}{x(1+\cos ^2 x)}dx$$
I try to integrate by parts:
$$-\int^\infty_0 \frac{1}{x}d\arctan(\cos x)$$
But it's not run,help me,thank you.

答:
Observe that
\begin{align}
I=&\ \int^\infty_0 \frac{\sin x}{2x(1-\frac{1}{2}\sin^2x)}\ dx = \frac{1}{4}\int^\infty_{-\infty} \frac{\sin x}{x}\sum^\infty_{k=0}\frac{\sin^{2k}x}{2^k}\ dx\\
 =&\ \frac{1}{4}\sum^\infty_{k=0}\frac{1}{2^k}\int^\infty_{-\infty} \frac{\sin^{2k+1}x}{x}\ dx
\end{align}
If you believe that
\begin{align}
\int^\infty_{-\infty} \frac{\sin^{2k+1}x}{x}\ dx = \binom{2k}{k}\frac{\pi}{4^k}
\end{align}

then it follows that

\begin{align}
I=\frac{\pi}{4}\sum^\infty_{k=0}\binom{2k}{k}\frac{1}{8^k} = \frac{\sqrt{2}\pi}{4} = \frac{\pi}{2\sqrt{2}}.
\end{align}

Note that the last line follows from the series expansion
\begin{align}
\frac{1}{\sqrt{1-x}}=\sum^\infty_{k=0}\binom{2k}{k}\frac{x^k}{4^k}.
\end{align}


问:
Some proofs involving locally connected sets
Below are some proofs involving locally connected sets. I mainly need help on finishing iii, iv, and v.

"A topological space(X,τ)is said to be locally connected if it has a basis B consisting of connected (open) sets."

(i) Verify that Z is a locally connected space which is not connected
It has been previously shown that the integers are totally disconnected, so the only connected sets are the singletons which are also open. The singleton forms a basis for Z so Z is locally connected

(ii) Show that $R^{n}$ and $S^{n} $ are locally connected, for all n>1 ($S^{2}$ is the unit circle in $R^{2}$ and $S^{3}$ is the unit sphere, ect...)

$R^{n}$ is easily proven using the first definition of locally disconnected. Every interval, or open balls are connected and open and the form a basis which were previously proven.
Take this basis from $R^{n}$ and intersect them with $S^{n}$. This forms a basis of $S^{n}$ which are all connected and open.

iii) Let (X,τ) be the subspace of $R^{2} $consisting of the points in the line segments joining〈0,1〉to 〈0,0〉and to all the points〈$\frac{1}{n}$,$0$〉, n= 1,2,3,.... Show that(X,τ)is connected but not locally connected.

We probably use contradiction to show that X is connected. Assume it is not connected. Then X=C$\cup$D where C and D are disjoint nonempty open sets in the subspace topology. I am not sure how to proceed, but I think the reason why a separation is impossible has something to do with the fact that all the line segments share (0,1) in common which makes it hard to find the disjoint sets.

As for why X is not locally connected, this time I think the problem point is (0,0). Any open neighborhood around that point intersects parts of line segment from (0,1) to ($\frac{1}{n}$,$0$);near 0.

(iv) Prove that every open subset of a locally connected space is locally connected.
Let (X,T) be a locally connected space and A open in X with the subspace topology. Then X has a set of connected basis whose intersection with A forms a basis for A. However there is no lemma or theorem proving that intersection of open connected sets with open sets are connected, so I can't conclude those basis of A are connected. How to proceed?

v) Prove the finite product topology of locally connected space is locally connected if and only if each component if locally connected.

We prove this is true for just the product of two space X and Y. If X and Y are locally connected, they each have a set of open connected basis, $B_x$ and $B_y$ We claim $B_x \times B_y $forms a connected open basis for the product topology It is previously proven that the product of connected sets are connected in the product topology. To simplify the proof, we just show that each basic open sets in the product topology are unions of elements of $B_x \times B_y $ Let $O_x \times O_y $ be an open set in X $\times$ Y. Then $O_x$=($B_{x_1} \cup$ $B_{x_2} \cup B_{x_3}$...$B_{x_n}$) and $O_y$=($B_{y_1} \cup$ $B_{y_2}\cup B_{y_3}$....$B_{y_m}$). Then $O_x \times O_y$=($B_{x_1} \cup B_{x_2}$ $\cup B_{x_3}$...$B_{x_n}$)$\times$ ($B_{y_1} \cup$ $B_{y_2} \cup B_{y_3}$....$B_{y_m}$) = $B_{x_1} \times (B_{y_1 }$ $\cup B_{y_2} \cup B_{y_3}$....$B_{y_m}$) $\cup B_{x_2} \times (B_{y_1} \cup B_{y_2} \cup B_{y_3}$....$B_{y_m}$)....We just distribute using set theory operation with Cartesian product and get $O_x \times O_y$ is the union of elements of the form $B_x \times B_y$. For finite products greater than 2, we can just use induction.

Now for the other direction, assume X $\times$ Y is locally connected so there is a set of connected open basis for X $\times$ Y. open sets are either of the form $O_x$ $\times$ $O_y$ where $O_x$ and $O_y$ are open sets from X and Y respectively or unions of sets of this form.
Question: How do I justify the fact that connected open sets in X $\times $ Y are just unions of connected open sets of the form $O_x $ $\times$ $O_y$ where $O_x$ and $O_y$ are connected open sets?

答:
To show that the space in (iii) is connected, note that each of the line segments joining $\langle 0,1\rangle$ to a point on the $x$-axis is connected, and use the fact that if $\mathscr{C}$ is a family of connected sets, and there is a point $p\in\bigcap\mathscr{C}$, then $\bigcup\mathscr{C}$ is connected. If you’ve not yet proved this, now is a good time to do so. If $\bigcup\mathscr{C}$ is the union of disjoint relatively open subsets $U$ and $V$, assume without loss of generality that $p\in U$, and use the connectedness of the members of $\mathscr{C}$ to conclude that $U\supseteq C$ for each $C\in\mathscr{C}$.

You’re already on the right track to finish (iii): the space is indeed not locally connected at $\langle 0,0\rangle$. In fact, any open nbhd of $\langle 0,0\rangle$ that does not include $\langle 0,1\rangle$ fails to be connected.

For (iv) just observe that if $\mathscr{B}$ is a base for $X$, and $A\subseteq X$ is open, then $\{B\in\mathscr{B}:B\subseteq A\}$ is a base for $A$.

For the last part of (v), let $\pi_X:X\times Y\to X$ be the projection map. Let $x\in X$, and let $U$ be any open nbhd of $X$, so that $U\times Y$ is an open set in $X\times Y$. Let $y$ be any point of $Y$; $\langle x,y\rangle\in U\times Y$, so there is a connected open set $W$ in $X\times Y$ such that $\langle x,y\rangle\in W\subseteq U\times Y$. Then $\pi_X[W]$ is a connected open nbhd of $x$ contained in $U$. (Why?) It follows that $X$ has a base of connected sets and is therefore locally connected.


问:
Defining random variable which follows multiple distribution by case
It might be a stupid question but, I want to ask that it is possible to define a special random variable described as follow,

Let $Y$ to be a random variable which I want to define. Then,

$$
Y \sim \left\{
        \begin{array}{ll}
            N_{1}(\mu_{1}, \sigma_{1}) & \quad \text{with probability of 0.3} \\
            N_{2}(\mu_{2}, \sigma_{2}) & \quad \text{with probability of 0.5} \\
            N_{3}(\mu_{3}, \sigma_{3}) & \quad \text{with probability of 0.2} \\
        \end{array}
    \right.
$$

Is it possible to define $Y$ as a random variable following above probability distribution?

If possible, how can I get $E(Y)$ and $Var(Y)$? Can I calculate the $E(Y)$ in an intuitive way like the below or there is another right way to calculate the expectation and variation?

$$
E(Y) = 0.3E(N_{1}) + 0.5E(N_{2}) + 0.2E(N_{3})$$

If there is any theory or keyword about this, please let me know about it. I tried to google about my problem but I couldn't because I don't know what to search.

I always appreciate so much for your sincere help, thank you.

答:
The terminology is that $Y$ is a mixture distribution. More precisely, it is a finite mixture of the r.v.s $N_1,N_2,N_3$ with weights $0.3,0.5,0.2$. The probability density of $Y$ is then simply the mixture density
$$f_Y(y)=0.3f_{N_1}(y)+0.5f_{N_2}(y)+0.2f_{N_3}(y).$$
That immediately implies your conjecture for $E(Y)$, and indeed in general one has $$E[H(Y)]=0.3E[H(N_1)]+0.5E[H(N_2)]+0.2E[H(N_3)]$$
for any function $H(\cdot)$ with finite expectated value. (For proof and applications, see the Wikipedia link above.)


问:
Proving that a sphere has a minimal surface to volume ratio using Calculus of Variations
I know the problem is traditionally solved via the isoperimetric inequality, but I was hoping to solve it by minimizing a surface of revolution subject to a volume constraint.

The surface area of a surface of revolution is:

$$A=2\pi\int_{-1}^{1}y\sqrt{1+\dot{y}^2}~dx$$

and the volume will be:

$$V=\pi\int_{-1}^{1}y^2~dx$$

I'd like to show that the sphere (or in this case the function $x^2+y^2=1$) minimizes the surface area functional for any fixed volume.

The combined action will then be:

$$S=\int_{-1}^{1}2\pi y\sqrt{1+\dot{y}^2}+\lambda\pi y^2 dx$$

The Euler Lagrange Equation simplifies to:

$$-\frac{2 \pi  y'(x)^2}{\sqrt{y'(x)^2+1}}+2 \pi  \sqrt{y'(x)^2+1}+\frac{2 \pi
   y(x) y'(x)^2 y''(x)}{\left(y'(x)^2+1\right)^{3/2}}-\frac{2 \pi  y(x)
   y''(x)}{\sqrt{y'(x)^2+1}}+2 \pi  \lambda  y(x)=0$$

I have no clear idea how to proceed from here. I can solve for $\lambda$ using Mathematica, and get

$$\lambda=\frac{y(x) y''(x)-y'(x)^2-1}{y(x) \left(y'(x)^2+1\right)^{3/2}}$$

The ODE solution is extremely convoluted, involving about half a dozen elliptic integrals, and doesn't seem to be going anywhere. Is there something fundamentally wrong with my approach, or is there an obvious next step in the problem?

EDIT: Using the Beltrami identity instead, and reversing the problem so the volume is varied and a constant surface area is the constraint, the EL equation can be simplified to:

$$y(x)^2+\frac{2\lambda y(x)}{\sqrt{y'(x)^2+1}}=k$$

Which, while much simpler, doesn't get me closer to an answer.

If $k$ is equal to $0$ and $y(0)$ is likewise, then the ODE reduces nicely, and can be solved. One solutions is:

$$y(x)=\sqrt{4 \lambda ^2-x^2}$$

Which is the equation of a circle (which when rotated around the axis becomes a sphere). However, some assumptions are made there which are not necessarily true.

答:
In the original notation, I decided to maximize the volume keeping the surface area fixed, the action-integral being:
$$
I = \int_{-1}^{1} (πy^2 - \lambda 2πy\sqrt {1+y'^2} ) dx
$$
Then, solving straight the Euler-Lagrange DE, I could reduce it to a Bernoulli type DE. Solving it, I got the result:
$$
y'(x)= \frac {\sqrt{(2\lambda (y-C)-y^2)(2\lambda (y+C)+y^2)}}{y^2+2C\lambda}
$$
Where $C$ is the constant of integration obtained in the aforementioned Bernoulli DE and $\lambda$ the Lagrange multiplier. For $C=0$ this solves to a circle, i.e. a spherical surface of revolution. However, I'm stuck at the case $C≠0$ : either this means there exists some other surface of revolution that can maximise the volume, or there is some way to prove, by ab initio methods, that $C$ must be zero. I humbly invite some suggestions over this.

Edit

With little algebraic manipulation we can show that $C< \frac {\lambda}{2}$. With $\lambda=1$ let $C=1/2n$ for a natural number $n$. The integral is then:
$$
x+c=\int\frac{y^2+\frac{1}{n}}{\sqrt {4(1-\frac{1}{n})-(y^2-(2-\frac{1}{n}))^2}}
$$
As $n \to \inf $ we get the sphere solution. What is the general solution $y_n$?


问:
Probability of no matching pairs of shoes
There are 3 different pairs ( i.e. 6 units say aa, bb, cc) of shoes in a lot. Now three person come and pick the shoes randomly (each get 2 units). Let p be the probability that no one is able to wear shoes (i.e. no one gets a correct pair) , then the value of $\frac{13p}{4-p}$ is?

答:
We use a technique that is definitely overkill for this problem, the Principle of Inclusion/Exclusion. But the idea might be useful for a larger problem, say $5$ people and $5$ pairs of shoes.

Let our people be P, Q, and R. There are $\binom{6}{2}$ ways for P to pick two shoes. And there are $\binom{3}{1}$ ways to pick a matching pair. So the probability that P picks a matching pair is $\frac{\binom{3}{1}}{\binom{6}{2}}$.
This is $\frac{1}{5}$.

If we find the sum $\frac{1}{5}+\frac{1}{5}+\frac{1}{5}$, we will have counted twice the probability P and Q both choose a matching pair, also the same for Q and R, also for P and R. Given that P chose a matching pair, the probability Q does is $\frac{\binom{2}{1}}{\binom{4}{2}}=\frac{1}{3}$.

So the probability P and Q choose a matching pair is $\frac{15}{5}\cdot\frac{1}{3}$. We subtract $3$ times this from $\frac{3}{5}$.

We have subtracted too much. We must add back the probability they all choose a matching pair, which is $\frac{1}{5}\cdot\frac{1}{3}$.

So the probability there is at least one person with a matching pair of shoes is
$$3\cdot\frac{1}{5}-3\cdot \frac{1}{5}\cdot \frac{1}{3}+1\cdot\frac{1}{5}\cdot\frac{1}{3}.$$
This simplifies to $\frac{7}{15}$.

So the probability $p$ of no match is $1-\frac{7}{15}$.


答:
The "limit" you're trying to bypass is the fact that $\tan^{-1}$ only produces output values between $-\frac\pi2$ and $\frac\pi2.$
It will never directly give you an angle in the third quadrant.

The way you "get around" this is you remember that
$$ \tan\theta = \tan(\theta + \pi) = \tan(\theta + 2\pi) = \tan(\theta - \pi),$$
that is, no matter what whole multiple of $\pi$ you add or subtract from $\theta,$
the tangent of the resulting angle is always the same as $\tan\theta.$

Since you know $\pi < \theta<\frac{3\pi}{2}$, you know that if you subtract $\pi$ from $\theta$ you will get an angle $\alpha = \theta - \pi$ where
$0 < \alpha < \frac\pi2.$
Since $\tan\alpha = \tan\theta = \frac34,$
and $\alpha$ is an angle that you can get as an answer from $\tan^{-1},$
you know that $\alpha = \tan^{-1}\frac34.$
Therefore $\theta = \alpha + \pi = \pi + \tan^{-1}\frac34.$

To continue along the lines you were trying to follow, you will have to compute
$\cos\frac{\pi + \tan^{-1}(3/4)}{2}.$
One way to continue is to use the cosine half-angle formula,
just as you attempted, but with the correct angle, so you get
$$
\cos\frac\theta2
 = \cos\frac{\pi + \tan^{-1}\frac34}{2}
 = \pm\sqrt{\frac{1 + \cos\left(\pi + \tan^{-1}\frac34\right)}{2}}.
$$
Now use the fact that $\cos(\pi + \beta) = -\cos(\beta).$
So now you have
$$
\cos\frac\theta2
 = \pm\sqrt{\frac{1 - \cos\left(\tan^{-1}\frac34\right)}{2}}.
$$
You can get to the correct answer from there using similar methods to what you already used in your attempt for this question,
but you get a different final answer since you have a difference inside the square root rather than a sum.

Alternatively, write
$\cos\frac{\pi + \tan^{-1}(3/4)}{2}
  = \cos\left(\frac\pi2 + \frac{\tan^{-1}(3/4)}{2}\right),$
and use the fact that
$\cos\left(\frac\pi2 + \phi\right) = -\sin\phi.$
Therefore
$$
\cos\frac\theta2
 = -\sin\left(\frac{\tan^{-1}\frac34}{2}\right)
 = \pm\sqrt{\frac{1 - \cos\left(\tan^{-1}\frac34\right)}{2}}.
$$
The same result follows.

Having said all that, I think the easier way to solve this is not to "get around" the $\tan^{-1}$ function at all, but rather draw a triangle as recommended in the other answer. That answer uses straightforward geometric intuition that is much less likely to lead you astray.

Title: How do I bypass the limit for the tan function on the calculator?

Body:
So basically, my partner and I are creating problems for a project. She created one where we aren't sure what the correct answer is.

Problem:
$Find \ cos\frac{\theta}{2} \ if \ tan \ \theta = \frac{3}{4}; \pi < \theta<\frac{3\pi}{2}$

I'll show you what we did.

First, we used the pythagorean formula to find $cos \ \theta$ (we need it later). The half-angle formula for cosine is $cos \frac {\theta}{2} = \pm \sqrt{\frac{1 + cos \ \theta}{2}}$.

These are our steps.

$cos \frac {tan^{-1}\frac{3}{4}}{2} = \pm \sqrt{\frac{1 + cos \ (tan^{-1}\frac{3}{4})}{2}}$

$ = \pm \sqrt{\frac{1 + \frac{4}{5}}{2}}$

$ = \pm \sqrt{\frac{\frac{9}{5}}{2}}$

$ = \pm \sqrt{\frac{9}{10}}$

$ = \pm \frac{3\sqrt{10}}{10}$

To decide whether we have a negative sign or a positive, we have to look at the domain.

$ \pi < \theta<\frac{3\pi}{2} \ changes \ to  \ \frac{\pi}{2} < \frac{\theta}{2} < \frac{3\pi}{4}$

We now know that $cos \frac {\theta}{2}$ is located in the second quadrant, which makes our final answer   $cos \frac {tan^{-1}\frac{3}{4}}{2} = - \frac{3\sqrt{10}}{10}$
But then, I checked the calculator.
The values are equal except for the negative sign

My partner said that the calculator thinks tangent is in the first quadrant, and if it were, cosine would be positive. But since we are in the third quadrant, cosine is in the second quadrant. I know there are domain restrictions, but I don't entirely understand what she meant. Also, how do I get the calculator to display the same answer?


问:
Distribution of irrational or transcendental numbers
Lets say we select a random number $0.abc.......$ between $[0,1]$ where each digit after decimal is independently drawn from a fixed discrete probability distribution ($\sum P(X=n)=1$ for $n\in\{0,1,...9\}$). Obviously, $0=0.\bar{0}$ and $1=0.\bar{9}$ are included in this range. What is the probability that a random number is irrational/transcendental? Are there infinitely more transcendental numbers than irrational numbers? Can we get a closed-form solution for uniform distribution? Is it possible to select a distribution such that it always generates an irrational or rational number? Clearly, it is possible to generate only rational numbers by fixing the distribution to a fixed point mass.

答:
The rational numbers are merely countably infinite, while the real numbers are uncountably infinite, so the probability of a number in the interval [ 0 , 1 ] being irrational is "certainty".

The algebraic numbers are also countably infinite, so the probability of a number in that interval being transcendental is also "certainty".


问:
Why are trigonometric functions considered rational?
So I am reading spivaks calculus and in it he states that a rational function is a function of the form $\frac{p}{q}$ where $p$ and $q$ are polynomial functions and $q$ is not the function that is always equal to $0$ however he then states that the function $f(x)=\sin(x^2)$
Is rational .

How can this be?

How is $\sin(x^2)$ the ratio of 2 polynomials?

Thanks in advance.

答:
The source of this seems to be the bottom of page 42. But I think you're misinterpreting Spivak. The relevant passage is this:

"The rational functions are themselves quite special examples of an even larger class of functions, very thoroughly studied in calculus, which are simpler than many of the functions first mentioned in this chapter. The following are examples of this kind of function:"

Spivak then goes on to enumerate a few examples, including $\sin(x^2)$. However, Spivak is not claiming that these examples are rational but rather that they're examples of an "even larger class of functions". On the next page, he clarifies that he's talking about functions which can be produced by composition of simpler functions. In the present case, one has $f(x)=\sin(x^2)$ as the composition $f(x)=g(h(x))$ where $g(x)=\sin x$ and $h(x)=x^2$. Every rational function can be generated using such compositions, but Spivak's examples are certainly not rational.


答:
Suppose that $f(x)=\sin(x^2)$ is rational. Then there are polynomials $p,q$ as above such that $\sin (x^2)=\frac{p(x)}{q(x)}.$ By continuity we get

$$p(x)=q(x) \sin (x^2)$$

for all $x$. Now let $x_n:= \sqrt{n \pi}$  for $n \in \mathbb N.$ Then we have

$$p(x_n)=0$$

for all $n$. But  the set of zeros of $p$ is finite, a contradiction.

Title: Why are trigonometric functions considered rational?

Body:
So I am reading spivaks calculus and in it he states that a rational function is a function of the form $\frac{p}{q}$ where $p$ and $q$ are polynomial functions and $q$ is not the function that is always equal to $0$ however he then states that the function $f(x)=\sin(x^2)$
Is rational .

How can this be?

How is $\sin(x^2)$ the ratio of 2 polynomials?

Thanks in advance.