答:
According to the prime factorization $60=2^2\cdot 3\cdot 5$ so it suffices to show $abc\equiv 0\pmod 4$, $\pmod 3$, and $\pmod 5$.

$abc\equiv 0\pmod 4$:  If two or all three of the numbers are even, we are done. Hence assume at least two are odd. As the sum of odd numbers is even, this means that exactly two of $a,b,c$ are odd. Thess cannot be $a$ and $b$ as that would make $c^2\equiv 1+1\pmod 8$, which is absurd. Hence $c$ and wlog $a$ are odd. Then $b^2=c^2-a^2\equiv 0\pmod 8$, which means $4\mid b$ and we are done.

$abc\equiv 0\pmod 3$: If any of $a,b,c$ is a multiple of $3$, we are done. So assume otherwise. But then $a^2\equiv b^2\equiv c^2\equiv 1\pmod 3$, whereas $1+1\not\equiv 1\pmod 3$.

$abc\equiv 0\pmod 5$: If any of $a,b,c$ is a multiple of $5$, we are done. So assume otherwise. But then each of $a^2,b^2,c^2$ is $\equiv \pm1\pmod 5$, so $a^2+b^2\equiv$ one of $-2,0,2$, but $\not\equiv \pm1$.

Body:
If $a,b,c \in \mathbb{Z}$ and $a^2+b^2=c^2$ show that $abc \equiv 0 \pmod{60}$.

I once read, on a number theory textbook - forget the title, in one of the problems list that all Pythagorean triplets when multiplied are divisible by 60


问:
Isn't $\lim\limits_{x \to 0}1^{1/x}$ be indeterminate
Isn't $\lim\limits_{x \to 0}1^{1/x}$ be indeterminate

Since $1^{\infty}$ is indeterminate.

Thanks

答:
No. The limit is equal to $1$. This is because for every $x\in\mathbb R\setminus \{0\}$, you have the equality $1^{\frac1x} = 1$.

There is no such thing as "$1^\infty$", and you will do yourself a massive favor if you stop thinking in sloppy terms and start thinking in definitions.

The definition you need here is the following:

If $f$ is a function defined on $D_f$, then $$L = \lim_{x\to a} f(x)$$ if and only if for every $\epsilon > 0$, there exists some $\delta > 0$ such that for all $x\in D_f\cap((a-\delta, a)\cup(a, a+\delta))$, the equality $|f(x)-L|<\epsilon$ is true.

This definition is all you need. You don't need vague concepts like "$1^\infty$ or anything like that. In fact, you don't need to think about the concept of infinity at all to prove that the limit of your function is $1$.

You can prove that the limit is $1$ in simple steps:

Let $\epsilon > 0$ be arbitrary.

Define $\delta = 1$.

Take any $x\in (-1, 0)\cup (0, 1)$.

Evaluate $|f(x)-L|$, which is straightforward.

Conclude that $|f(x)-L|<\epsilon$ is true.

Because $\epsilon$ was arbitrary, the claim is true for all epsilon values, so the proof is complete.

ContentLicense: CC BY-SA 4.0


答:
No, here we have

$$(\forall x\ne 0)\;\; f(x)=1^{1/x}=1$$

thus $ f $ is constant at $ (-\infty,0)\cup(0,+\infty)$ and

$$\lim_{x\to 0}f(x)=1$$

Body:
Isn't $\lim\limits_{x \to 0}1^{1/x}$ be indeterminate

Since $1^{\infty}$ is indeterminate.

Thanks


问:
count increasing functions on natural numbers.
I want to count all functions with these two conditions:
it is from {1,2,....,7} to {1,2,....,5}.
it is increasing .

I have some idea to start with f(1) different values but do not know how to organize solution.

答:
Solution if fuctions are surjective.

You probably mean nondecreasing functions. You can list all such functions:

If $3$ consecutive elements take the same value then you have:

$f(1)=f(2)=f(3)=1$ ...

$f(1)=1$ and $f(2)=f(3)=f(4)=2$ ...

$f(1)=1$ and $f(2)=2$ and $f(3)=f(4)=f(5)=3$ ...

... 2 more possibilities.

If two times $2$ consecutive elements take the same value.

$f(1)=f(2)=1$ and $f(3)=f(4)=2$ ...

$f(1)=f(2)=1$ and $f(3)=2$ and $f(4)=f(5)=3$ ...

... 2 more possibilities.

$f(1)=1$ and $f(2)=f(3)=2$ and $f(4)=f(5)=3$ ...

$f(1)=1$ and $f(2)=f(3)=2$ and $f(4)=3$ and
$f(5)=f(6)=4$ and $f(7)=5$.

... 1 more possibilities.

$f(1)=1$ and $f(2)=2$ and $f(3)=f(4)=3$ and $f(5)=f(6)=4$ and $f(7)=5$.

$f(1)=1$ and $f(2)=2$ and $f(3)=f(4)=3$ and $f(5)=4$ and $f(6)=f(7)=5$.

$f(1)=1$ and $f(2)=2$ and $f(3)=3$ and $f(4)=f(5)=4$ and $f(6)=f(7)=5$.

This is it.


问:
Matrix repesentation of a linear operator with respect to its basis
Consider the vector space of,$\mathbb{R}^{2\times 2}$, and the linear operator,$L_{A,B}:\mathbb{R}^{2\times 2}\to\mathbb{R}^{2\times 2}$, given by,$L_{A,B}(X)=(AX)+(XB)$

Where,$A=\begin{bmatrix}
1&3\\
2&4\\
\end{bmatrix}$,and,$B=\begin{bmatrix}
6&5\\
7&8\\
\end{bmatrix}$

Find the matrix representation of,$L_{A,B}$, with respect to the basis:

$(\begin{bmatrix}
1&0\\
0&0\\
\end{bmatrix},\begin{bmatrix}
0&1\\
0&0\\
\end{bmatrix},\begin{bmatrix}
0&0\\
1&0\\
\end{bmatrix},\begin{bmatrix}
0&0\\
0&1\\
\end{bmatrix})$

I defined X as:,$X=\begin{bmatrix}
x_1&x_2\\
x_3&x_4\\
\end{bmatrix}$

Then I found:,$L_{A,B}(X)=x_1\begin{bmatrix}
7&5\\
2&0\\
\end{bmatrix}+x_2\begin{bmatrix}
7&9\\
0&2\\
\end{bmatrix}+x_3\begin{bmatrix}
3&0\\
10&5\\
\end{bmatrix}+x_4\begin{bmatrix}
0&3\\
7&12\\
\end{bmatrix}$

But I have no idea how to go further. I know that there are several examples about this topic, but i do not understand those problems, and I hope that when I solved this problem, I get it. Can someone help me with this problem?

答:
Let's call $\mathcal{B} = (e_1,e_2,e_3,e_4)$ your basis.

In order to find the matrix representation of $L_{A,B}$ all you have to do is calculating $L_{A,B}(e_i)$ for each vector $e_i$ of your basis and then express this in your basis.

Let's give it a try:

$L_{A,B}\left(\begin{bmatrix}
1&0\\
0&0\\
\end{bmatrix}\right) = \begin{bmatrix}
7&5\\
2&0\\
\end{bmatrix} = 7 \times \begin{bmatrix}
1&0\\
0&0\\
\end{bmatrix}+5\times \begin{bmatrix}
0&1\\
0&0\\
\end{bmatrix}+2\times \begin{bmatrix}
0&0\\
1&0\\
\end{bmatrix} + 0 \times \begin{bmatrix}
0&0\\
0&1\\
\end{bmatrix}$

So the first column of $\text{Mat}_{\mathcal{B}}(L_{A,B})$ will be

$$\begin{pmatrix}
  7 \\
  5 \\
  2 \\
  0 \\
\end{pmatrix}$$

Do you need more explanation?


答:
Write your basis as $e_1,e_2,e_3,e_4$. Then $X = x_1e_1+x_2e_2+x_3e_3+x_4e_4$. You've already written out an expression of $L_{A,B}(X)$--from here you just need rewrite the output in terms of the basis, and the coordinate representation of the output of the $i$th basis vector tells us what the $i$th column of the matrix will be. For example, $L_{A,B}(e_1) = 7e_1+5e_2+2e_3+0e_4$. So the first column of the matrix for $L_{A,B}$ with respect to this basis will be $(7,5,2,0)^T$. Applying similar reasoning for the other basis vectors, we get
$$M = \begin{bmatrix}
7 & 7 & 3 & 0 \\
5 & 9 & 0 & 3 \\
2 & 0 & 10 & 7 \\
0 & 2 & 5 & 12
\end{bmatrix}$$
as the matrix representation for $L_{A,B}$. This looks a bit wonky, since our inputs $X$ should be $2\times 2$ matrices, but keep in mind we have to translate $X$ into our basis before applying $M$, turning $X$ into a $4\times 1$ column vector. The dimensions make sense since $M_2(\Bbb R)\cong \Bbb R^4$, so the matrix representation $M$ for $L_{A,B}$ should be $4\times 4$.

Title: Matrix repesentation of a linear operator with respect to its basis

Body:
Consider the vector space of,$\mathbb{R}^{2\times 2}$, and the linear operator,$L_{A,B}:\mathbb{R}^{2\times 2}\to\mathbb{R}^{2\times 2}$, given by,$L_{A,B}(X)=(AX)+(XB)$

Where,$A=\begin{bmatrix}
1&3\\
2&4\\
\end{bmatrix}$,and,$B=\begin{bmatrix}
6&5\\
7&8\\
\end{bmatrix}$

Find the matrix representation of,$L_{A,B}$, with respect to the basis:

$(\begin{bmatrix}
1&0\\
0&0\\
\end{bmatrix},\begin{bmatrix}
0&1\\
0&0\\
\end{bmatrix},\begin{bmatrix}
0&0\\
1&0\\
\end{bmatrix},\begin{bmatrix}
0&0\\
0&1\\
\end{bmatrix})$

I defined X as:,$X=\begin{bmatrix}
x_1&x_2\\
x_3&x_4\\
\end{bmatrix}$

Then I found:,$L_{A,B}(X)=x_1\begin{bmatrix}
7&5\\
2&0\\
\end{bmatrix}+x_2\begin{bmatrix}
7&9\\
0&2\\
\end{bmatrix}+x_3\begin{bmatrix}
3&0\\
10&5\\
\end{bmatrix}+x_4\begin{bmatrix}
0&3\\
7&12\\
\end{bmatrix}$

But I have no idea how to go further. I know that there are several examples about this topic, but i do not understand those problems, and I hope that when I solved this problem, I get it. Can someone help me with this problem?


问:
Complete factorisation of $x^8-2x^4\cos (4\theta)+1$ with roots of unity

Factorize completely $x^8-2x^4\cos (4\theta)+1$ using complex numbers and $n$-th root of unity.

The answer given is
$$\prod_{r=0}^3 \left(x^2-2x\cos\left(\theta+\frac{r\pi}{2}\right)+1\right)\,.$$
Can anyone please help . I just want to know how to tackle this . I just want the conceptual trick to solve such sum.

答:
$$
\begin{aligned}
x^8-2x^4\cos4\theta+1
&=x^8-x^4(e^{4i\theta}+e^{-4i\theta})+1 \\
&=(x^4-e^{4i\theta})(x^4-e^{-4i\theta}) \\
&=(x^2+e^{2i\theta})(x^2-e^{2i\theta})(x^2+e^{-2i\theta})(x-e^{-2i\theta}) \\
&=(x+ie^{i\theta})(x-ie^{i\theta})(x+e^{i\theta})(x-e^{i\theta}) \\
&\cdot(x+ie^{-i\theta})(x-ie^{-i\theta})(x+e^{-i\theta})(x-e^{-i\theta}) \\
&=\prod_{r=0}^3(x-i^re^{i\theta})(x-i^{-r}e^{-i\theta})\quad{(*)} \\
&=\prod_{r=0}^3[x^2-x(i^re^{i\theta}+i^{-r}e^{-i\theta})+1] \\
\end{aligned}
$$

in which the asterisk step follows from interchanging each linear factor.

Now, using the fact that for every integer $i^r=e^{\pi ir/2}$ for all integer $r$, we obtain

$$
\begin{aligned}
x^8-2x^4\cos4\theta+1
&=\prod_{r=0}^3\left[x^2-2x\left(e^{i\theta+i\pi r/2}+e^{-(i\theta+i\pi r/2)}\over2\right)+1\right] \\
&=\prod_{r=0}^3\left[x^2-2x\cos\left(\theta+{\pi r\over2}\right)+1\right]
\end{aligned}
$$

Hope this answers your question!


答:
If $\color{red}{|{\rm range}(f)|=5}$

Some consecutive elements must take the same value. So we have the following possiblities:

If $3$ consecutive elements take the same value then you have such situation $$*,*, (*,*,*),*,*$$ (elements in the bracket take the same value, in this case $3$, $4$ and $5$ takes value $3$) and that is ${5!\over 4!1!}=5$ possibilities.

If two times $2$ consecutive elements take the same value, then you have such situation $$*,(*,*),*,(*,*),*$$ and that is ${5!\over 3!2!}=10$ possibilities.

So in this case there is $15$ such functions.

If $\color{red}{|{\rm range}(f)|=4}$
Then some we have to take from codomain 1 element, that we can do on $5$ ways. Some elements take the same value so we have $$5\times \Big({4!\over 1!3!} +{4!\over 2!2!}+{4!\over 3!1!}\Big)= 70$$

If $\color{red}{|{\rm range}(f)|=3}$
Then some we have to take from codomain 2 element, that we can do on ${5\choose 2}=10$ ways. Some elements take the same value so we have $$10\times \Big(2{3!\over 1!2!} +{3!\over 1!1!1!}\Big)= 120$$

If $\color{red}{|{\rm range}(f)|=2}$
Then some we have to take from codomain 3 element, that we can do on ${5\choose 3}=10$ ways. Some elements take the same value so we have $$10\times 12= 120$$

If $\color{red}{|{\rm range}(f)|=1}$
Then we have only constant functions and that is $5$

So we have in total $330$ such functions.

Title: count increasing functions on natural numbers.

Body:
I want to count all functions with these two conditions:
it is from {1,2,....,7} to {1,2,....,5}.
it is increasing .

I have some idea to start with f(1) different values but do not know how to organize solution.


问:
Order of operations (BODMAS)
$$40-20/2+15\times1.5\\\hspace{.1in}\\40-20/2+15\times1.5=\\ 40-10+22.5=7.5$$

I'm studying and this is from an example.

In BODMAS, aren't addition and subtraction have same level?

So, in the 3rd line, it should be from left to right, correct?

ContentLicense: CC BY-SA 3.0

Title: Order of operations (BODMAS)

Body:
$\boxed{40-20/2+15\times1.5}$

Using brackets we can remove ambiguity:

$40+(-20/2)+(15\times1.5)$

$=40+(-10)+(22.5)$

$=(40+22.5)+(-10)$

$=62.5-10$

$=52.5$


答:
You have done the correct computation, now it's just a matter of writing this answer in terms of a matrix.

One approach is to build the matrix column-by-column, as I have done on this post. Alternatively, we can build the matrix one row at a time: we have
$$
T(X) = \pmatrix{y_1&y_2\\y_3&y_4},
$$
where for intance $y_1 = 7x_1 + 7x_2 + 3x_3 + 0x_4$. Our goal is to find a matrix $M$ for which
$$
M \pmatrix{x_1\\x_2\\x_3\\x_4} = \pmatrix{y_1\\y_2\\y_3\\y_4}.
$$
We can find the first row of $M$ based on $y_1$. In particular, the matrix of the transformation will have the form
$$
M = \pmatrix{7&7&3&0\\?&?&?&?\\?&?&?&?\\?&?&?&?}
$$


答:
You're free to consider it an indeterminate form, if you so wish.

There is no mathematical content in the phrase “indeterminate form”: it's just a way to say that the particular limit may need additional work to be computed and no “direct substitution” is possible. Which is mainly done for the benefit of students, because sometimes they're too fast in jumping to conclusions.

Whether it actually benefits students or not is debatable. It's probably a question about how the notion of “indeterminate form” is presented.

Bear in mind that “indeterminate form” doesn't mean “not computable”. So whether you deem $\lim_{x\to0}1^{1/x}$ to be in indeterminate form has no real consequence. The limit is $1$ nonetheless (because it's the limit of a constant function, by the way).

Title: Isn't $\lim\limits_{x \to 0}1^{1/x}$ be indeterminate


答:
A limit is never indeterminiate. In can be undefined or it may perhaps not exist.
The form $1^\infty$ is said to be indeterminate because from $a_n\to 1$ and $b_n\to+\infty$ alone, we cannot determine tha limit (if it exists at all) of $a_n^{b_n}$ as $n\to \infty$. This doesn't mean that the limit never exists.

Similarly, the form $0^0$ is said to be indeterminate because $a_n\to 0$ and $b_n\to 0$ allow various behaviours for $a_n^{b_n}$ -- even though the expression $0^0$ is evaluates to $1$.


答:
With the 4th root of unity $e^{\pm i\frac{a+2\pi r}4} $ for $x^4 - e^{\pm ia}=0$, factorize as follows
\begin{align}
x^8-2x^4\cos  (4\theta)+1
=&\> (x^4-e^{i 4\theta})(x^4-e^{-i 4\theta}) \\
= &\prod_{r=0}^{3}(x-e^{i\frac{4\theta+2\pi r}4})
 \prod_{r=0}^{3}(x-e^{-i\frac{4\theta+2\pi r}4})  \\
 = &\prod_{r=0}^{3}(x-e^{i(\theta+ \frac{\pi r}2)})
  (x-e^{-i(\theta+ \frac{i\pi r}2)}) \\
= &\prod_{r=0}^{3}(x²-2x\cos(\theta+\frac{\pi r}{2})+1)
\end{align}

Title: Complete factorisation of $x^8-2x^4\cos (4\theta)+1$ with roots of unity

Body:
Factorize completely $x^8-2x^4\cos (4\theta)+1$ using complex numbers and $n$-th root of unity.

The answer given is
$$\prod_{r=0}^3 \left(x^2-2x\cos\left(\theta+\frac{r\pi}{2}\right)+1\right)\,.$$
Can anyone please help . I just want to know how to tackle this . I just want the conceptual trick to solve such sum.


问:
Show that $\mathbb{Z}/5\mathbb{Z} = \langle a\rangle$
Show that $\mathbb{Z}/5\mathbb{Z} = \langle a\rangle$.

I have that $a$ is an element of the non zero complex numbers $\mathbb{C}^*$ and $a^{5}=1$ where $a \neq 1$. I have shown already that $|a| = 5$ and the function that maps $(\mathbb{Z},+) \to (\mathbb{C}^*,\times)$ given by $n \mapsto a^n$ is a homomorphism.

How do I now go on to show that $\mathbb{Z}/5\mathbb{Z} = \langle a\rangle$?

答: Denote your homomorphism by $\varphi$. Using the fact that the order of $a$ is $5$, one can show that Ker $\varphi$ = $5\mathbb{Z}$. Now, the fundamental homomorphism theorem for groups says that $\mathbb{Z}/$Ker $\varphi$ $\cong$ Im $\varphi$. Now we have Im $\varphi = ⟨a⟩$ such that $\mathbb{Z}$/ $5\mathbb{Z} \cong ⟨a⟩$.


答:
Actually $\mathbb{Z}/5\mathbb{Z}$ is a cyclic group of order $5$, and since $5$ is prime then all the non-identity elements of $\mathbb{Z}/5\mathbb{Z}$ are of order $5$ then all the non-identity elements are generators of this group. Now, $a$ is an element of $\mathbb{C}^*$ and a is not identity and $a^5=1$ then $|a|=5$ and $$ is a cyclic group of order $5$ and having the elements $\{a,a^2,a^3,a^4,1\}$ where all the elements are of order $5$. So, $\{a,a^2,a^3,a^4,1\}$ is isomorphic to $\mathbb{Z}/5\mathbb{Z}$, Hence , $\mathbb{Z}/5\mathbb{Z}=$.
Is my way correct or convincing?

Body:
Show that $\mathbb{Z}/5\mathbb{Z} = \langle a\rangle$.

I have that $a$ is an element of the non zero complex numbers $\mathbb{C}^*$ and $a^{5}=1$ where $a \neq 1$. I have shown already that $|a| = 5$ and the function that maps $(\mathbb{Z},+) \to (\mathbb{C}^*,\times)$ given by $n \mapsto a^n$ is a homomorphism.

How do I now go on to show that $\mathbb{Z}/5\mathbb{Z} = \langle a\rangle$?

Body:
Actually $\mathbb{Z}/5\mathbb{Z}$ is a cyclic group of order $5$, and since $5$ is prime then all the non-identity elements of $\mathbb{Z}/5\mathbb{Z}$ are of order $5$ then all the non-identity elements are generators of this group. Now, $a$ is an element of $\mathbb{C}^*$ and a is not identity and $a^5=1$ then $|a|=5$ and $<a>$ is a cyclic group of order $5$ and having the elements $\{a,a^2,a^3,a^4,1\}$ where all the elements are of order $5$. So, $\{a,a^2,a^3,a^4,1\}$ is isomorphic to $\mathbb{Z}/5\mathbb{Z}$, Hence , $\mathbb{Z}/5\mathbb{Z}=<a>$.
Is my way correct or convincing?


问:
Semantics and expressive power of generic element quantifiers
In real analysis (and other applications like the graph-theoretic structural approach to sparse matrices), it would sometimes be convenient to quantify over all generic elements. A generic element $\xi$ would basically satisfy $(\exists xf(x)\neq g(x))\to f(\xi)\neq g(\xi)$ for all functions $f$, $g$ relevant to the current context. So a generic element quantifier would be a kind of "last word" quantifier, i.e. it has to wait until all other context is laid down, before it can fix its meaning. But even as a "last word" quantifier, it seems to be more ambiguous and brittle then other "overly expressive" quantifiers like second order quantifiers or Henkin quantifiers.

Let's assume that the first order language (with equality) contains only function symbols, but no predicate symbols. Is it possible to give a semantics for generic element quantifiers in this case? If this shouldn't be possible in general, even in the weak sense of full semantics for second order quantifiers, then maybe it helps if we restrict the allowed models to $\mathbb R^n$?

If it is possible to give a semantics for generic element quantifiers, do they extend the expressive power of the corresponding logic significantly?

答:
By Ηerbrand theorems all one can expect for an universal theory Γ:

```<code>     Γ |- ∃xP(x)
</code>```

That we can find terms t1,..,tn such that:

```<code>     Γ |- P(t1) & ... & P(tn)
</code>```

Herbrands theorem is for when predicate symbols exist. But I guess
a function symbol can also act as a predicate symbol, if it is the
characteristic function of a relation, and we would use f(x1,..,xn)=1 to
express the predicate. So I guess banning predicates doesn't help.

On the other hand if you do something with the connectives and quantifiers in your theory Γ, you might get a so called existence property. For example intuitionistic logic has the existence property, but also Horn clauses classically.

Body:
By Ηerbrand theorems all one can expect for an universal theory Γ:

     Γ |- ∃xP(x)

That we can find terms t1,..,tn such that:

     Γ |- P(t1) & ... & P(tn)

Herbrands theorem is for when predicate symbols exist. But I guess
a function symbol can also act as a predicate symbol, if it is the
characteristic function of a relation, and we would use f(x1,..,xn)=1 to
express the predicate. So I guess banning predicates doesn't help.

On the other hand if you do something with the connectives and quantifiers in your theory Γ, you might get a so called existence property. For example intuitionistic logic has the existence property, but also Horn clauses classically.


问:
When $\int |f|=\left|\int f\right|$ holds?
I was just wondering when did the equality hold for the following inequality:
$$\left|\int_{R^d}f(x)\, d x\right|\leq\int_{R^d}|f(x)|\, d x$$
where $f:R^d\to R$ is Lebesgue integrable on $R^d$.

Obviously, if $f\geq0$ a.e. on $R^d$, the equality will hold trivially, but I don't know whether it is the only case?

答: Let $E_+$ be the set where $f\geq 0$ and let $E_-$ be the set where $f<0$. Then $\int f=\int_{E_+}f+\int_{E_-}f=\int_{E_+}|f|-\int_{E-}|f|$. This is equal to $\int|f|=\int_{E_+}|f|+\int_{E_-}|f|$ in absolute value if and only if either $\int_{E_+}f$ or $\int_{E_-}f$ is $0$. An integral $\int_{E_+}f$ of a non-negative function is positive if and only if $f>0$ almost everywhere. Thus you have equality if and only if $f\geq 0$ almost everywhere or $f\leq 0$ almost everywhere.</p>


答: The string of values of such a function can be written uniquely as $$1^a2^b3^c4^d 5^e$$ where the $5-$ tuple $(a,b,c,d,e)$ of non-negative integers sums to $7$.  A routine application of Stars and Bars tells us that there are $$\binom {5+7-1}{7}=330$$ of these.


问:
Find the first four non-zero terms of the Maclaurin series of $f(z)=e^\frac{1}{1-z}$
Consider the following complex function:

$$f(z)=e^\frac{1}{1-z}$$

Find the first four non-zero terms of the Maclaurin series for $f(z)$, so around $z_0=0$.

I've asked my lecturer what's the best way to solve this and she simply said "just differentiate the function 4 times and find its value at $z=0$". I'm refusing to accept this and I'm sure there's a smarter way to approach this.

Unfortunately I'm clueless about this one and have no idea how to even start. I tried thinking of a function $g(z)$, which I know its Maclaurin series already, such that $g(z)f(z)=1$ and then comparing coefficients but no luck of finding one.

Any help would be grealy appreciated!

答:
One can compose the Taylor series for the exponential function with the geometric series. The “trick” is to write
$$
 \frac{1}{1-z} = 1 + \frac{z}{1-z}
$$
where the second term on the right has a Taylor series without constant term, so that it can be substituted into the exponential series:
$$
f(z)=e^{1/(1-z)} = e \cdot e^{z/(1-z)} = e \cdot \sum_{k=0}^\infty \frac{1}{k!} \left( \frac{z}{1-z}\right)^k \\
= e \cdot \sum_{k=0}^\infty \frac{1}{k!} z^k(1 + z + z^2 + z^3 + \ldots )^k
$$
Now collect all terms with powers from $z^0$ up to $z^3$:
$$
f(z)= e \cdot \left( 1 + (z + z^2 + z^3 + \ldots) + \frac 12(z^2 + 2 z^3 + \ldots) + \frac 16(z^3 + \ldots) + \ldots \right) \\
= e \cdot \left(1 + z + \frac 32 z^2 + \frac{13}{6} z^3 + \ldots\right)
$$


答:
Let $x_n=f(n)$, then we want the number of tuples $(x_1,\dots,x_7)$ such that $$1\leq x_1\leq\dots\leq x_7\leq5.$$
We can regard this as the number of ways to split $7$ balls into $5$ bins, which is $\binom{7+5-1}{7}=330$.


问:
Exponential distribution?
I am stucking on this problem from days, I need help, if anyone could help me.
I have $Y=\log(1+X)$ with $X$ that follows an exponential distribution, I would understand which is the $Y$ distribution, is $Y$ also exponential?

Thank you in advance

答: No. If $X\sim\operatorname{Exp}(\lambda)$, the CDF of $Y$ is$$P(Y\le y)=P(X\le\exp y-1)=1-\exp(-\lambda(\exp y-1))$$for $y\ge0$. This is a Gompertz distribution with $\eta=\lambda,\,b=1$.

Title: Exponential distribution?


问:
A 4-Regular graph with 7 vertices is non planar
A Graph with 7 vertices each having degree 4 cannot be planar. Any hints on the proof?</p>

答:
Up to isomorphism, there are two $4$-regular graphs on $7$ vertices, which can be exhaustively enumerated using geng which comes with nauty.  They are these two following graphs:

In the first graph, I highlighted a $K_{3,3}$ subgraph in orange (and thus it cannot be planar since $K_{3,3}$ is not planar).

In the second graph, I highlighted a $K_{2,3}$ subgraph in orange.  We observe that by identifying the two blue vertices we obtain a vertex adjacent to all three red vertices, thereby giving a minor isomorphic to $K_{3,3}$ (we delete the unnecessary edges).  Thus Wagner's Theorem implies this is also non-planar.


问:
Free groups as free product of infinite cyclic groups
Let $S$ be an arbitrary set (countable or uncountable). It is clear that the free abelian group generated by $S$ is isomorphic to the direct sum

$$\bigoplus_{s\in S}\mathbb{Z}.$$

Is the free group generated by $S$ then isomorphic to

$$\ast_{s\in S}\mathbb{Z}$$
where $\ast$ denotes the free product of groups?

答:
Yes.

It's a general pattern based on the free-forgetful adjunction: the free functor preserves colimits, in particular coproducts, so since $S=\coprod_{s\in S}1$ where $1$ is a set with one element, then
$$F(S)=\coprod_{s\in S}F(1)\,,$$
and the coproduct in the category of groups is free product and $F(1)=\Bbb Z$.


问:
Writing statements in logic symbols and decide if it is True or False.
I have a statement that I want to express in logic symbols and decide if its true or false. The following is my attempt:

Every real number can be written as a difference of two positive real numbers.
$$(\forall x \in \mathbb R)(\exists a,b \in \mathbb R)[(a,b>0) \wedge(x=a-b)]$$
I know this one I definitely didn't get it right. But that was my best attempt.

This statement is True because for positive real numbers, take x to be that number and y to be 0. Then x-0=x. For negative numbers, take x = 0 and y to be that negative number. Then 0-y = -y. For 0, take x = y>0. Then x-y=0.

Thanks in advance.

答: $(\forall x \in \mathbb{R})(\exists a,b \in \mathbb{R^{+}}) [x=a-b]$


答:
Let $A_+=f^{-1}((0,\infty))$, $A_-=f^{-1}((-\infty,0))$.
Then let $I_\pm=\int_{A_\pm}f(x)\,\mathrm dx$.
We have $I_\pm=0$ iff $\mu(A_\pm)=0$, $\left|\int_{\mathbb R^d}f(x)\,\mathrm dx\right|=|I_++I_-|$, $\int_{\mathbb R^d}|f(x)|\,\mathrm dx=I_+-I_-$, hence equality iff one of $I_\pm$ is zero, and that is iff one of $A_\pm$ has measure zero.</p>

Title: When $\int |f|=\left|\int f\right|$ holds?

Body:
I was just wondering when did the equality hold for the following inequality:
$$\left|\int_{R^d}f(x)\, d x\right|\leq\int_{R^d}|f(x)|\, d x$$
where $f:R^d\to R$ is Lebesgue integrable on $R^d$.

Obviously, if $f\geq0$ a.e. on $R^d$, the equality will hold trivially, but I don't know whether it is the only case?


问:
List the elements of the smallest set of pairs which needs to be added to R to form an equivalence relation, R∗

Suppose R = {(1, 1),(1, 4),(1, 5),(2, 2),(2, 7),(3, 3),(3, 6)} is a
relation on the set A = {1, 2, 3, 4, 5, 6, 7}

Q: List the elements of the smallest set of pairs which needs to be added to
R to form an equivalence relation, R∗

Attempt:

No clue how to answer this. What does equivalence mean? Would the smallest set be (1,1)? Would equivalence be (4,1), (5,1) (2,2) (7,2), (3,3), (6,3) etc etc? Not sure.

答:
An equivalence relation has to be reflexive wich means $(a,a) \in R$, for all $a \in R$, therefore you have to add $(4,4),(5,5),(6,6)$ to $R$.

It also has to be symmmetric wich means: for all $a,b \in R$, $(a,b) \in R \  \iff (b,a) \in R$. Therefore you have to add $(4,1),(5,1),(7,2),(6,3)$.

The last condition for $R$ to be a an equivalence relationship is that $R$ needs to be transitive, wich means for all $a,b,c \in R$, $(a,b)\in R$ and $(b,c) \in R$ then $(a,c) \in R$ therefore you don't need to add any elements to $R$ because $R$ is already transitive.


问:
Counter-intuitive result to an Integral calculus problem
Consider the equations $y=x$ and $y=x^2$ in the first quadrant,

My hypothesis is that the solids produced by rotating the area enclosed by the equations about x axis and y axis will have the same volume.

But when i use integration methods to find the volume of those solids

For rotation about x-axis:-

$ V_1=\pi\int_{0}^{1}( x^2-x^4 )\,dx $

For rotation about y-axis:-

$ V_2=\pi\int_{0}^{1}( y-y^2)\,dy $

$Clearly,  V_1\neq V_2$

But this is very counter intuitive for me since both volumes can be understood as infinite sum of the same area about the two different axes.
Is there something that i am missing?

Thanks for your help

Edit:
When i sweep the area slice by a small constant angle about both axes ,i will endup with a larger arclength as a result of larger outer radius in case of the rotation with respect to $y axis$ compared to $x axis$ giving more volume for the same angle no matter the angle.
This extends my intuition to considering areas as sum of arclengths,which i need to consider more.

答:
Rotating about the $y$ would be the same thing as using the shells method on $x$. Compare the two integrals

$$\pi\int_0^1 f_1^2(x)-f_2^2(x)dx$$

$$2\pi\int_0^1x(f_1(x)-f_2(x))dx$$

Assuming $f_1>f_2$ on $(0,1)$ et's see when the integrands are equivalent

$$\pi(f_1^2(x)-f_2^2(x)) = 2\pi x(f_1(x)-f_2(x)) \implies f_1(x)+f_2(x) = 2x$$

so if the sum of the functions is less than $2x$, than rotation about the $y$ will be larger.


问:
an exponent problem
I need the value of $x$, but I cannot figure out how to deal with ${\sqrt x-1}$ in the following equation.

$x^{\sqrt x-1} =3/2$

Could anyone please help me to figure this out.

答:
Consider that you are looking for the zero's of function
$$f(x)=x^{\sqrt x-1} -\frac 32$$
Its derivatives are
$$f'(x)=\frac{1}{2} x^{\sqrt{x}-2} \left(2 \sqrt{x}+\sqrt{x} \log (x)-2\right)$$
$$f''(x)=\frac{1}{4} x^{\sqrt{x}-3} \left(4 \left(x-2 \sqrt{x}+2\right)+\log (x)
   \left(4 x-5 \sqrt{x}+x \log (x)\right)\right)$$

The first derivative cancels at $x=1$ and $f(1)=-\frac{1}{2}$, $f''(1)=1$. So, this point is a minimum and there are two solutions
$$0 < x_1 < 1 \quad 1 < x_2$$

Build a Taylor series around $x=1$. It will be
$$-\frac{1}{2}+\frac{1}{2} (x-1)^2+O\left((x-1)^3\right)$$ This gives two solutions : $0$ and $2$ and $0$ must be discarded.

$$f(2)=2^{\sqrt{2}-1}-\frac{3}{2}\sim -0.167428 \implies x_2 >2$$ and you were already told that $x=\frac 94$ is the largest solution. Then, I shall not focus on this one.

Adding one more term to the expansion, we have
$$-\frac{1}{2}+\frac{1}{2} (x-1)^2-\frac{3}{8} (x-1)^3+O\left((x-1)^4\right)$$ which is a cubic in $(x-1)$. The discriminant being negative $(\Delta=-\frac{179}{256})$, there is only one real root. Using the hyperbolic method, we have
$$\frac{1}{9} \left(13-8 \cosh \left(\frac{1}{3} \cosh
   ^{-1}\left(\frac{211}{32}\right)\right)\right)\sim 0.207953$$

So, now we have our starting guess for Newton method. For the smallest root, the iterates are
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0.207953 \\
 1 & 0.291319 \\
 2 & 0.346314 \\
 3 & 0.360882 \\
 4 & 0.361632 \\
 5 & 0.361634
\end{array}
\right)$$

This number is not recognized by inverse symbolic calculators, but (given for the fun) it is close to
$$\frac{11+\sqrt{23}}{50}\, \sqrt[3]{\frac{3}{2}}  \sim   0.361634266$$ while the "exact" solution is $0.361634272$.

Even if this does not mean any thing, for this funny number $f(x)=1.56\times 10^{-8}$.

Title: an exponent problem


答:
Since the divisor polynomial is of degree $3$, the remainder will be of atmost degree $2$. Hence, instead of $Ax+B$ that you have taken, take $Ax^2+Bx+C$ as the remainder and you also have $3$ equations to plug the values.

Note that $f(1)=A+B+C=3$, $f(-3)=9A-3B+C=-1$ and $f(-2)=4A-2B+C=1$.

Title: Remainder when $\frac{f(x)}{x^3+4x^2+x-6}$ and given two other remainders

Body:
Determine the remainder $r$ when $$\frac{f(x)}{x^3+4x^2+x-6}\rightarrow remainder \ r$$.

The following is known $$\frac{f(x)}{x^2+2x-3}\rightarrow remainder\ (x+2)\\
\frac{f(x)}{x+2}\rightarrow remainder\ (1)\\$$

My work:

$$(x^3+4x^2+x-6)=(x^2+2x-3)(x+2)\\(x^2+2x-3)=(x-1)(x+3)\\
(x+2)\\
f(x)=(x^3+4x^2+x-6)g(x)+(Ax+B)\\
f(1)=A+B=x+2\\
f(-3)=-3A+B=x+2\\
f(-2)=-2A+B=1\\
\left\{\begin{matrix}
 A+B=x+2& x=-1\\
 -3A+B=x+2& B=1\\
 -2A+B=1& A=0
\end{matrix}\right.\\
r=(-1)(0)+1=1$$

Can the remainder really be $r=1$? My intuition says it's wrong.

EDIT:
$$f(x)=(x^3+4x^2+x-6)g(x)+(Ax^2+Bx+C)\\
f(1)=A+B+C=x+2\\
f(-3)=9A-3B+C=x+2\\
f(-2)=4A-2B+C=1\\
\left\{\begin{matrix}
 A+B+C=x+2& A=\frac{x+3}{6}\\
 9A-3B+C=x+2& B=\frac{3x+5}{6}\\
4A-2B+C=1 &C=\frac{4-2x}{6}
\end{matrix}\right.\\
r=\frac{x+3}{6}x^2+ \frac{3x+5}{6}x+\frac{4-2x}{6}=\frac{x^3+6x^2+3x+9}{6}$$


问:
MIT Statistics course: multi-dimensional matrix form for Regression
In the lecture here at time 31:13 the instructor is explaining regression in multi-dimonsinal matrix form but he starts with an example in 1 dimension ([screenshot of the illustration is below]) and assumes if $X\hat \beta$ is the line that minimizes the distance with observations then $\|Y\|^2 = \|X\hat \beta\|^2+\|Y- X\hat \beta\|^2$. From the illustration what he actually highlights as Y is a point rather than just just one value Y so I don't know how the equations actually maps to the illustration on the board.

Any help is very appreciated.

答:
you are confusing two different pictures that represent regression using matrices and least squares.
picture number 1: you have your data points on say, a 2 dimensional plane (x,y) and you are trying to draw a line that has the "least squared" errors and those errors are quite obviously the magnitude of the line coming out of those data points perpendicular to the line you want to draw.
I think you are mixing up that picture with what is drawn in here and what I'm going to elaborate on. notice, $X \hat \beta$ is a vector.
to fully understand this you need to remember how a system of equations. for each data point you have y=C+Dt (I'm writing that instead of y=mx+b cause I will be using x and b somewhere else) where t and y are known and you are trying to find C and D.
such C and D's that fit all the t and y's (the coordinates of your data points).
notice the system of linear equations? we have Ax=b each row of A is the coefficient of C and D for each data point, x (the unknown) is C an D. b is y for each data point.
in some cases this system is solvable, meaning a there is a line that goes through all the points, but what if there is no solution? this is where vector spaces come into play. when is Ax=b solvable? when b is in the column space of A.
what is the best fit if Ax=b isn't solvable, in other words when b isn't in the column space of A? the projection of b onto the column space of A (it has a formula which was mentioned in the lecture you have linked) and the error itself will be a vector. in this picture X beta hat is the projection of b onto the column space, Y is b itself and epsilon is b minus its projection (the error)
I highly recommend watching this this lecture of MIT 18.06: https://www.youtube.com/watch?v=osh80YCg_GM&t=378s


问:
HINT for summing digits of a large power
I recently started working through the Project Euler challenges, but I've got stuck on #16 (http://projecteuler.net/problem=16)

$2^{15} = 32768$ and the sum of its digits is $3 + 2 + 7 + 6 + 8 = 26$. What is the sum of the digits of the number $2^{1000}$?

(since I'm a big fan of generality, my interpretation is to find a solution to the sum of digits of $a^b$ in base $c$, and obviously I'm trying to solve it without resorting to "cheats" like arbitrary-precision numbers).

I guess this is simpler than I'm making it, but I've got no interest in being told the answer so I haven't been able to do a lot of internet searching (too many places just give these things away). So I'd appreciate a hint in the right direction.

I know that $2^{1000} = 2^{2*2*2*5*5*5} = (((((2^2)^2)^2)^5)^5)^5$, and that the repeated sum of digits of powers of 2 follows the pattern $2, 4, 8, 7, 5, 1$, and that the last digit can be determined by an efficient pow-mod algorithm (which I already have from an earlier challenge), but I haven't been able to get further than that… (and I'm not even sure that those are relevant).

答:
In this case, I'm afraid you just have to go ahead and calculate $2^{1000}$. There are various clever ways to do this, but for numbers this small a simple algorithm is fast enough.

The very simplest is to work in base 10 from the start. Faster is to work in binary, and convert to base 10 at the end, but this conversion is not completely trivial. A compromise (on a 32-bit machine) would be to work in base 1000000000, so that you can double a number without overflow. Then at the end the conversion to base 10 is much simpler.


问:
Find a function $f$ such that $\lim_{x\to{}0}{f(x^2)}$ exists, but $ \lim_{x\to{}0}{f(x)}$ does not.
Context:

I'm brushing up on some Analysis and currently going the exercises in M. Spivak's Calculus book, specifically chapter 5 on limits. Everything was going fine until I came across this question. I've been thinking about it for some time with no luck.

Question:
"Give an example where $\lim_{x\to{}0}{f(x^2)}$ exists, but  $\lim_{x\to{}0}{f(x)}$ does not."

My attempts:

A previous question showed that $\lim_{x\to{}0}{f(x^3)}=\lim_{x\to{}0}{f(x)}$, which I believe works because we can find the third root of any real number (which was useful in the epsilon - delta proof for it). Which makes me believe the above fails because we can't square root negative reals. This lead me to play around with functions involving $\sqrt{x}$ and utilising its 'undefinedness' on the negatives.

I started with $f(x)=\sqrt{x-1}$ which clearly has an undefined limit at $0$. But this is of course no different (considering the limit at $0$ that is) to $f(x^2)$.

Any hints? I feel as though I'm overlooking something so simple.

答:
I came up with another example, albeit after seeing Hagon von Eitzen's response.

We can choose $f(x)=\text{floor}(x)$.


答:
I performed an iterative solution with an Android phone simulator of an HP-$42$S. I started by taking the log of both sides.
$$x^{\sqrt{x}-1}=\frac{3}{2}$$
$$(\sqrt{x}-1)\cdot\log(x)=\log\left(\frac{3}{2}\right)$$
$$\sqrt{x}-1=\frac{\log\left(\frac{3}{2}\right)}{\log(x)}$$
$$\sqrt{x}=\frac{\log\left(\frac{3}{2}\right)}{\log(x)}+1$$
$$x=\left(\frac{\log\left(\frac{3}{2}\right)}{\log(x)}+1\right)^2$$
Start with an initial value of $x=5$ and watch a solution evolve.

After coding the below, plug in 5, and XEG "SME".
The HP-$42$S code follows:
LBL "SME"
STO 01
LBL 01
3
2
/
LOG
RCL 01
LOG
/
1
+
X^2
STO 01
VIEW ST X
PSE
GTO 01

This is at least one solution there may be others.

The simulator is so fast, if you take the PSE out, you will just see a solution. For example, incrementing X from 0 to 1,000,000 takes seconds.

Title: an exponent problem

Body:
I need the value of $x$, but I cannot figure out how to deal with ${\sqrt x-1}$ in the following equation.

$x^{\sqrt x-1} =3/2$

Could anyone please help me to figure this out.


问:
divergent or convergent as limits go to infinity?
My question concerns generalized integrals.

I can understand that $\int_{0}^\infty \frac{x}{1+x^2} \,dx$ is divergent and thus $\int_{-\infty}^\infty \frac{x}{1+x^2} \,dx$ is  divergent by definition (since an integral is convergent only if it is convergent on all its intervals). I can also understand that $\int_{-A}^A \frac{x}{1+x^2} \,dx \xrightarrow {} 0$ when $A \xrightarrow {}\infty$ since it is an odd function and corresponding values "cancel out".

What I can't wrap my head around is how it is possible for the 2 to be true at the same time. That is how $\int_{-\infty}^\infty \frac{x}{1+x^2} \,dx$ can be divergent and $\int_{-A}^A \frac{x}{1+x^2} \,dx$ can be convergent when A is close to $\infty$. I guess the answer lies in the limits, but I still find it hard to grasp.

Could you help me?

答:
Another way to think of divergence when a principal value exists is that we can make the integral equal to every real value we could want, thus the assignment of any particular value would be absurd. Take for example the limit

$$\lim_{A\to\infty}\int_{-A}^{2A}\frac{x}{1+x^2}dx = \lim_{A\to\infty}\frac{1}{2}\log\left(\frac{1+4A^2}{1+A^2}\right) = \log 2$$

In fact we can get any real number we want with the appropriate bounds

$$\lim_{A\to\infty}\int_{-A}^{e^tA}\frac{x}{1+x^2}dx = t$$

as a continuous version of the Riemann rearrangement theorem.


问:
Prove that the system $Ax = b, A \in \Bbb R^{m \times n}$ , has a unique solution if and only if rank $A$ = rank$[A \mid b] = n$.
So I am trying to prove the above statement and wanted to see if my line of thinking is in fact correct.

So first, because there exists a  unique solution then rank$(A) =$ rank $[A\mid b]$. I then said suppose that rank $(A) \neq n$ and instead that the rank is in fact $k < n$. This means there are $n-k$ linearly dependent columns which means we have infinitely many solutions to $Ax = b$ because we can have $n-k$ arbitrary elements of $x$ because we can 'eliminate' the linearly dependent columns. Is this line of thinking correct?

答:
You can simplify a bit the proof using a more general context:

Over any field $K$, the  linear  system $Ax=b$ has a solution if & only if $\operatorname{rank}(A)=\operatorname{rank}\bigl([A|b] \bigr)$ for the very reason you mentioned – that $b$ is in the span of the column vectors of $A$.

Using the rank-nullity formula,  you see this common rank is the codimension of the affine subspace of solutions. On another hand, the solution is unique if & only if the codimension is $n$ (which implies $m\ge n$).


问:
Proof: Find minimum moves to reach target on an infinite line algorithm
An infinite line is given, where we stand at $0$, and we can move by a distance of $i$ units to left or right in our $i$-th step: we need to find the minimum number of steps needed to reach a particular target.

The algorithm I found is the following one:

```<code>target = abs(target);
int sum = 0, step = 0;
while (sum < target || (sum - target) % 2 != 0) {
        step++;
        sum += step;
    }
return step;
</code>```

I do not know if the algorithm works mathematically. For example, I do know that for $i=11$ it gives the answer $5$, which is correct, but what is the mathematics behind it?

答:
You are given an integer $T$ (=target) and you are looking for the smallest number $n$ such that $\pm 1\pm2\pm\ldots\pm n=T$.

What the algorithm is doing is this: first it assumes $T$ is positive (because the solution is the same for $T$ and $-T$ - that is the purpose of target = abs(target)). Then it finds the smallest sum $S_n=1+2+\ldots+n$ that satisfies two conditions:

$S_n\ge T$

$S_n$ is of the same parity as $T$.

Then it outputs the value $n$.

Examples

Let's set $T=11$. The smallest sum $1+2+\ldots+n$ that is not smaller than $11$ is $1+2+3+4+5=15$. We have obviously "overshot" $11$, but by a suitable change in signs we will get $11$ as, say $1-2+3+4+5=11$.

Let's now set $T=12$. The smallest sum is still $1+2+3+4+5=15$ but it is of the wrong parity. Changing pluses into minuses will preserve the parity, so none of $\pm 1\pm 2\pm 3\pm 4\pm 5$ can be equal $12$. We can go further to $1+2+3+4+5+6=21$ but this sum is also odd. Only $1+2+3+4+5+6+7=28$ is of the correct parity (even). Now, we have "overshot" $12$, but we can now change some of the plus signs into minus, in hope that we can reach the sum of $12$. Indeed, this is possible as $-1+2+3+4+5+6-7=12$.

This is the precise motivation for the algorithm you've shown. First, "overshoot" the target, using a sum $1+2+\ldots+n$ of the right parity (the parity of $T$), and then prove that we can change some of the plus signs to minus to get the correct sum.

Proof: To prove that this algorithm is correct, one needs to prove that the above two conditions ($S_n\ge T$, $S_n$ of the same parity as $T$) are sufficient and necessary to be able to write $T$ as a sum of the form $\pm 1\pm 2\pm\ldots\pm n$. Necessity is easy to see:

If $S_n<T$ then, obviously, the biggest sum $1+2+\ldots+n<T$ and it cannot be made any bigger by converting some pluses into minuses.

If $S_n$ and $T$ are of different parity, then no way of putting pluses and minuses can convert the sum $S_n=1+2+\ldots+n$ into $T$. Namely, every conversion of $+i$ into $-i$ changes the sum by $2i$ - by an even number. Thus all the numbers of the form $\pm 1\pm 2\pm\ldots\pm n$ are of the same parity, and if one of them (the $S_n$ itself) is not of $T$'s parity - none of them is!

Now, for sufficiency: assume that $S_n=1+2+\ldots+n\ge T$ and is of the same parity as $T$. We need to prove that there is some ordering of $+$ and $-$ signs that convert this sum to exactly $T$. In other words, we need something like:

Lemma: If $S_n=1+2+\ldots+n$, then every number $T, 0\le T\le S_n$ of the same parity as $S_n$ can be represented in at least one way as a sum $\pm 1\pm 2\pm\ldots\pm n$.

Proof: Induction on $n$:

Trivial edge case: $n=0$, $S_0=0$ (empty sum). This is true because the only $T$ satisfying those conditions is $T=0$, but in that case $T=S_0$.

Inductive step: $n-1\to n$. Let $n\ge 1$ and let $0\le T\le S_n$ of the same parity as $S_n$. Then, either $T-n$ or $n-T$ will be between $0$ and $S_{n-1}$:

If $T\ge n$, then $0\le T-n\le S_n-n=S_{n-1}$.

If $0\lt T\le n$, then $0\le n-T\le n-1\le S_{n-1}$.

The only remaining case is $T=0$, in which case $n-T=n\le S_{n-1}$ for all $n\ge 3$. The cases $n=1,2$ are impossible because $T=0$ is not of the same parity as either $S_1=1$ or $S_2=3$.

Also, $T-n$ and $n-T$ are of the same parity, i.e. both of the same parity as $S_n-n=S_{n-1}$.

Now, we can apply the inductive hypothesis to either $T-n$ or $n-T$, as at least one of them satisfy the required conditions:

Either $T-n=\pm 1\pm 2\pm \ldots\pm (n-1)$, in which case $T=\pm 1\pm 2\ldots\pm (n-1)+n$

Or $n-T=\pm 1\pm 2\pm\ldots\pm(n-1)$, in which case $T=\mp 1\mp 2\mp\ldots\mp (n-1)+n$.


问:
A matrix w/integer eigenvalues and trigonometric identity
Any intuition and/or rigorous arguments on the proofs of the following statements would be appreciated:

Let $n$ be a natural number.
(a) Consider the following Toeplitz/circulant symmetric matrix:

$$
\Lambda_{kl} = \begin{cases}\frac{2n(n+1)}{3}, & k = l, \\[6pt]
\frac{-1}{1-\cos\frac{2\pi(k-l)}{2n+1}}, & k \ne l,\end{cases}
$$

where $k,l = 1,2, \ldots, 2n+1$.

Prove that its eigenvalues are natural numbers(!)

$$2n, 4n-2, 6n-6, \ldots, n(n+1)$$

with multiplicity $2$ and $0$ w/multiplicity $1$.

(b) Find a formula for the right signs in the following trigonometric identity

$$
\tan\left(\frac{l\pi}{2n+1}\right) = 2\sum_{k=1}^n (\pm)\sin\left(\frac{k\pi}{2n+1}\right),\qquad l = 1,2,\ldots, n.
$$

and show that the signs are uniquely determined in the case of prime $2n+1$.

The problems (that are formulated to be self-contained) arise from discrete approximation of continuous inverse boundary problem (see http://en.wikibooks.org/wiki/On_2D_Inverse_Problems for background) and the statements were formulated w/computer help.

答:
This is not a complete answer to (a), but it may be a start. It also explains the repeated eigenvalues.

It is known that a circulant matrix is diagonalized by the Fourier matrix and the eigenvalues of a circulant matrix are given by the discrete Fourier transform of its first row. See chapter 3 of here for a detailed explanation.

Let $c_j$ be the first row your matrix. Given the definition of $\Lambda$, we have

$$
c_j = \begin{cases} 2n(n+1)/3, & j = 0 \\[6pt]
\frac{-1}{1-\cos(\frac{2\pi j}{2n+1})}, & j = 1, \dots, 2n \\
\end{cases}
$$

Let $\lambda_m$ be the eigenvalues of $\Lambda$. Then

$$
\begin{align}
\lambda_m &= \sum_{j=0}^{2n}c_j \exp \left( -\frac{2\pi imj}{2n+1} \right) \\
&= \frac{2n(n+1)}{3} + \sum_{j=1}^{2n} c_j \exp \left( -\frac{2\pi imj}{2n+1} \right), \qquad m = 0, \dots, 2n
\end{align}
$$

The imaginary part of this sum will vanish, so we're left with

$$
\lambda_m = \frac{2n(n+1)}{3} + \sum_{j=1}^{2n} \frac{\cos(\frac{2\pi jm}{2n+1})}{\cos(\frac{2\pi j}{2n+1})-1}
$$

A closed form evaluation for this sum will solve the problem. I don't have any headway on this, but here are two observations.

Observation 1 $ \qquad \lambda_0 = 0$

Proof

$$
\begin{align}
\lambda_0 &= \frac{2n(n+1)}{3} + \sum_{j=1}^{2n} \frac{1}{\cos(\frac{2\pi j}{2n+1}) - 1}\\
&= \frac{2n(n+1)}{3} -\frac{1}{2} \sum_{j=1}^{2n} \csc^2(\frac{\pi j}{2n+1}) \\
&= \frac{2n(n+1)}{3} -\frac{1}{2} \frac{1}{3} ((2n+1)^2 -1) \\
&= \frac{2n(n+1)}{3} - \frac{1}{6} (4n^2 + 4n) \\
&= 0
\end{align}
$$

In the first equality above, I've used the double angle identity, $sin^2(\theta) = \frac{1}{2}(1-cos(2\theta))$, and in the second the identity $\sum_{k=1}^{n-1} csc^2(\frac{k\pi}{n}) = \frac{1}{3}(n^2 - 1)$, see proof below.

Observation 2 $\qquad \lambda_m = \lambda_{2n + 1 - m}$

Proof

$$
\lambda_m = \frac{2n(n+1)}{3} + \sum_{j=1}^{2n} \frac{\cos(\frac{2\pi jm}{2n+1})}{\cos(\frac{2\pi j}{2n+1})-1}
$$

$$
\lambda_{2n + 1 - m} = \frac{2n(n+1)}{3} + \sum_{j=1}^{2n} \frac{\cos(\frac{2\pi j(2n + 1 - m)}{2n+1})}{\cos(\frac{2\pi j}{2n+1})-1}
$$

The denominators in the fractions are the same. The result then follows by noticing that

$$
\begin{align}
\cos \left( \frac{2\pi j (2n + 1 -m )}{2n+1} \right) &= \cos \left( 2\pi j - \frac{2\pi j m}{2n+1} \right) \\
&= \cos \left( \frac{2\pi jm}{2n+1} \right)
\end{align}
$$

Proof of $\sum_{k=1}^{n-1} csc^2(\frac{k\pi}{n}) = \frac{1}{3}(n^2 - 1)$

This proof follows exercise 8 in chapter 5 of Stromberg, An Introduction to Classical Real Analysis.

Begin with De Moivre's formula:
$$
\cos(nx) + i\sin(nx) = (\cos(x) + i\sin(x))^n
$$

Expanding the RHS via the Binomial Theorem gives $$ \cos(nx) + i\sin(nx) = \sum_{k=0}^{n}\binom{n}{k}\cos ^{n-k}(x) (i\sin(x))^k $$

Equate imaginary parts, the RHS is imaginary for $k$ odd, so set $k = 2j+1$

$$ \sin(nx) = \sum_{j=0}^{\lfloor{\frac{n-1}{2}} \rfloor}(-1)^j\binom{n}{2j+1}\cos ^{n-2j-1}(x)\sin ^{2j+1}(x)$$

Setting $n = 2m+1$ and multiplying outside the sum by $\sin ^{2m}(x)$ and dividing on the inside gives

$$ \begin{align} \sin((2m+1)x)
=& \sin ^{2m+1}(x)\sum_{j=0}^{m} (-1)^j\binom{2m+1}{2j+1}\cos ^{2(m-j)}(x)\sin ^{2(j-m)}(x) \\
=& \sin ^{2m+1}(x)\sum_{j=0}^{m} (-1)^j\binom{2m+1}{2j+1}(\cot ^2(x))^{m-j}
\end{align}
$$

Now $\sin((2m+1)x)$ has roots $x = \frac{k\pi}{{2m+1}}, k = 1, \dots, m$. This implies that the polynomial

$$ \sum_{j=0}^{m} (-1)^j\binom{2m+1}{2j+1}t^{m-j} $$ has roots $t = \cot ^2(\frac{k\pi}{2m+1}), k = 1 \dots m$. The sum of the roots of a polynomial of degree $m$ is equal to the negative of the coefficient of the $(m-1)$ degree term divided by the coefficient of the leading term. The leading term has coefficient $2m+1$, the $m-1$ degree term has coefficient $-\binom{2m+1}{3}$, thus the sum of the roots is $m(2m-1)/3$ and we have the identity

$$ \sum_{k=1}^{m}\cot^2 \left( \frac{k\pi}{2m+1} \right) = \frac{m(2m-1)}{3} $$

Making the substitution $\csc^2 \theta = 1 + \cot ^2 \theta$ gives

$$ \sum_{k=1}^{m} \csc^2 \left(\frac{k\pi}{2m+1} \right) = \frac{2}{3}m(m+1) \tag{1}$$

The statement to prove is $$ \sum_{k=1}^{n-1} \csc^2 \left(\frac{k\pi}{n} \right) = \frac{1}{3}(n^2 - 1) \tag{2}$$

Making the substitution $n \rightarrow 2n + 1$, (2) is equivalent to $$ \sum_{k=1}^{2n} \csc^2 \left(\frac{k\pi}{2n + 1} \right) = \frac{4}{3}n(n+1) \tag{3}$$

Noticing that $\csc\left(\frac{k\pi}{2n+1}\right) = \csc\left(\frac{(2n+1-k)\pi}{2n+1}\right)$, we have that the sum over the first $n$ indices in $(3)$ equals the sum over last $n$ indices and thus that $(1)$ is equivalent to $(3)$. $\Box$


问:
Is any finite semigroup of this type a left monoid?
Let $(S, \cdot, e)$ be a semigroup $(S, \cdot)$ with binary operation $e$ in which the identities $e(x, y)\cdot x\approx x$ and $e(x, y)\approx e(y, x)$ hold.

In this question I asked if any such semigroup is necessarily a left monoid. Example given to me by J.-E. Pin shows that this isn't true. Clearly, $(\mathbb{Z}, \min, \max)$ is not a left monoid but satisfies those identities.

A left monoid is a semigroup with left identity.

Since I couldn't find a finite semigroup like this which wouldn't be a left monoid, and I've tried checking in GAP semigroups of order $\leq 4$, I suspect all finite semigroups of this form are left monoids for some combinatorial reasons.

Unfortunately, I'm not sure how to get all semigroups of orders, say, $\leq 7$, which would not be left monoids and be lwr semigroups, other than taking all the semigroups which aren't monogenic or monoids using GAP's Smallsemi package and checking if they are of this form by hand by creating a multiplication table. As you can imagine, this is very tedious.

Does there exist a finite semigroup of this form, not being a left monoid, and if yes, can you provide an example of least order?

答:
Any nonempty finite semigroup $S$ of this type has a left identity. First observe that for all $x, y \in S$,
$$
(1) \quad e(x,y)x = x \text{ and } e(x,y)y = e(y,x)y = y.
$$
Since $S$ is finite, it contains an idempotent $x_0$.
Let $S = \{x_0, x_1, \ldots, x_n\}$ and let $(a_i)_{0 \leqslant i \leqslant n}$ be the sequence of elements of $S$ defined by $a_0 = x_0$ and, for $1 \leqslant i \leqslant n$, $a_i = e(a_{i-1},x_i)$.

Claim: $a_n$ is a left identity of $S$.

First observe that, for $1 \leqslant i \leqslant n$,
\begin{align}
  &(2) \quad& a_ia_{i-1} &= e(a_{i-1},x_i)a_{i-1} = a_{i-1} \\
  &(3) \quad& a_ix_i &= e(a_{i-1},x_i)x_i = x_i
\end{align}
Let us now prove by induction on $k = i+j$ that, for $0 \leqslant i \leqslant j$,
$$
(4) \quad a_jx_i = x_i.
$$
If $k = 0$, then $i = j = 0$ and $a_0x_0 = x_0$ since $x_0$ is idempotent. Suppose that (4) holds for $i + j \leqslant k$ and suppose that $i + j = k+1$. If $i = j$, then (4) follows from (3). If now $i \leqslant j-1$, then $a_{j-1}x_i = x_i$ by the induction hypothesis. It follows by (2) that
$$
  a_jx_i = a_j(a_{j-1}x_i) = (a_ja_{j-1})x_i = a_{j-1}x_i = x_i.
$$
This proves the claim and concludes the proof.


问:
Numerical analysis objective question on bisection method.
Let $M$ be the length of the initial interval $[a_0,b_0]$ containing a solution of $f(x)=0$ . Let $\{x_0,x_1,\cdots \}$ represents the successive points generated by the bisection method. Then the minimum number of iterations required to guarantee an approximation to the solution with an accuracy of $\epsilon $ is given by

$1.$ $-2-\frac{log(\frac{\epsilon}{M})}{log(2)}$.

$2.$ $-2+\frac{log(\frac{\epsilon}{M})}{log(2)}$.

$3.$ $-2-\frac{log(\frac{\epsilon}{M})}{log^2(2)}$.

$4.$ $-2-\frac{log(M\epsilon)}{log(2)}$.

Using $$
n\ge \frac{\log{(b_0-a_0)}-\log{\epsilon}}{\log2}
$$ I only got that
$$n\ge \frac{\log{M}-\log{\epsilon}}{\log2}
$$
So
$$
n\ge \frac{\log(\frac{M}{\epsilon})}{\log2}
 =-\frac{\log(\frac{\epsilon}{M})}{\log2}$$

Now I don’t know from where $-2$ comes. Please help to find the correct option . Thank you .

答:
Ok so in my opinion it doesn't really matter, because you already have the answer to your question, which is more about the number of iterations when the ratio $\frac{\varepsilon}{M}$ is small and the $-2$ is not important, probably put in by the lecturer/course director/whoever.

Wikipedia states,

In mathematics, the bisection method is a root-finding method that
applies to any continuous functions for which one knows two values
with opposite signs.

So for your question, I think we do know (i.e. we are told) the sign of $f(a_0)$ and $f(b_0)$ because otherwise, $f(a_0), f(b_0), f\left(\frac{a_0+b_0}{2}\right)$, and almost every subsequent midpoint obtained by the bisection method could have the same sign, making it impossible to know which interval a root is in.

If $x_0 \in (a_0,b_0)$ or $x_1 \in (a_0,b_0)$, then assuming your bisection method will happen inside $[x_0,x_1],$ the roots could lie outside of $[x_0,x_1]$ but inside $(a_0,b_0),$ so the bisection method might not find the roots, in $(a_0,b_0)$, because it only searches in $[x_0,x_1].$

So I think the following is implicitly assumed by the question:

You know the sign of $f(a_0)$ and you know that the sign of $f(b_0)$ is different to the sign of $f(a_0)$: for example you know that $f(a_0)$ is negative and $f(b_0)$ is positive.

One of {$x_0, x_1$} equals $a_0$ and the other one of {$x_0, x_1$} equals $b_0.$

This way, the change of sign in the interval $[a_0,b_0] = [x_0,x_1]$ means that the first two points $x_0$ and $x_1$ told us that the root is in the interval $[a_0,b_0]$. So you only start obtaining "new information" that you didn't already know from the 3rd iteration, namely $x_2$, and onwards.


问:
Accessible natural numbers
A positive natural number is said to be accessible if it can be written as sum of poisitive integers whose sum of reciprocals adds up to $1$. It is easy to show that $3, 5 $ are not accessible while $4$ is. Is there a way to characterize all accessible positive integers, for example, an inductive process? It is clear that for any number $n$, the number of summands is at most $\lfloor n/2 \rfloor$ as all of these summands are greater than or equal to $2$. Is it possible to answer the problem in two cases: distinct and not necessarily distinct positive integers?

Please give some hint or reference.

Thank you.

答:
These are called Egyptian numbers in OEIS A125726.  The first $73$ are given
$$1, 4, 9, 10, 11, 16, 17, 18, 20, 22, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33,\\ 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53,\\ 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73,\\ 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86$$ There are a couple references.  All the squares are there, as $n^2$ can be represented as $n\ n$'s.  If $n$ is in the list, so is $2n+2$ as you can double all the numbers in an expansion and add a $2$.  If $n$ is in the list with a $2$ in the expansion, so is $n+6$ because you can replace the $2$ with $4,4$.  The last one my eye sees missing is $23$.  Mathworld states that every number $\ge 78$ can be expressed this way with distinct numbers in the expansion and calls those numbers strictly Egyptian.  There is a link to the paper in the OEIS entry.

Title: Accessible natural numbers


问:
positive recurrent, null recurrent, or transient.prove of 0-1 markov chain
Let $\left\{X_{n}\right\}_{n=0}^{\infty}$ be a Markov chain with state space $\{0,1,2, \ldots\}$. The transition probabilities are
$$
 p_{0,1}=1, \quad p_{i, i+1}+p_{i, i-1}=1, \quad p_{i, i+1}=\left(\frac{i+1}{i}\right)^{2} p_{i, i-1}, \quad i \geq 1
 $$

(1)Suppose that $X_{0}=0 .$ Find the probability that the chain never returns to state $0 .$

(2)Now suppose that the transition probabilities are
$$
  p_{0,1}=1, \quad p_{i, i+1}+p_{i, i-1}=1, \quad p_{i, i+1}=\left(\frac{i+1}{i}\right)^{\alpha} p_{i, i-1}, \quad i \geq 1
  $$
for some constant $\alpha .$ For every $\alpha \in(-\infty, \infty),$ indicate whether the chain is positive recurrent, null recurrent, or transient. Prove your conclusion.

I have a idear to get the probability that chain never returns to state 0 based on recurrence formula, but I have no idear about how to slove the complex relationship between $p_{i, i-1}$ and $p_{i, i+1}$, is there anyone who can help me?

答:
For (1), the answer is $\frac{6}{\pi^2}$. Let $h_n=\mathbb P_n(\text{hit }0)$, so $h_0=1$ and we want $1-h_1$. Then
$$h_n=p_{n,n-1}h_{n-1}+p_{n,n+1}h_{n+1}\implies \frac{h_{n+1}-h_n}{h_n-h_{n-1}}=\frac{n^2}{(n+1)^2}.$$
Telescoping, we get $h_{n+1}-h_n=\frac{1}{(n+1)^2}(h_1-h_0)$. So summing over $0\leq n\leq m-1$, we obtain $$h_m=1+(h_1-1)\sum_{k=1}^m\frac{1}{k^2}.$$
As $\sum_{k=1}^m\frac{1}{k^2}\to\frac{\pi^2}{6}$, the fact that $(h_m)_{m=0}^\infty$ is the smallest non-negative solution to this equation implies that $h_1=1-\frac{6}{\pi^2}$.

The chain is transient if $\alpha>1$, and recurrent if $\alpha\leq1$. Repeating the calculation above, we get
$$h_m=1+(h_1-1)\sum_{k=1}^m\frac{1}{k^\alpha}.$$
If $\alpha\leq1$, the series diverges, so as the $h_m$ are bounded, we need $h_1=1$. This implies the chain is recurrent.

If $\alpha>1$, the series converges to some $C>0$. As above, we get $h_1=1-\frac{1}{C}<1$, so the chain is transient.


问:
Integral of $\sqrt{1 + \sqrt{x}}$
My professor wants us to do this problem to refresh ourselves with substitution. We have to solve:

$$\int\sqrt{1 + \sqrt{x}}\,\mathrm dx$$

  $$\int\sqrt{1 + \sqrt{1 + \sqrt{x}}}\,\mathrm dx$$

  ...etc...

If someone could point me in the right direction with the first one, I think I can handle the others.

Thanks for the help guys!

答:
Hint:

With $u=\sqrt{1+\sqrt x}$, you have $x=(u^2-1)^2.$

With $u=\sqrt{1+\sqrt{1+\sqrt x}}$, you have $x=((u^2-1)^2-1)^2.$

In both cases, a polynomial and the integral is easy. This generalizes.

Title: Integral of $\sqrt{1 + \sqrt{x}}$


问:
Does continuity of joint density function imply continuity of marginal density functions?
We have two continuous random variables $X,Y$ (not neccessarily independent) and we know that their joint density function exists and is continuous. Does it follow that the probability density function of $X$ (and $Y$) also exists and is continous?

Equivalent (?) statement without mentioning r.v. :

we have continuous function $f:\mathbb{R}^2\to\mathbb{R}_{\geq0}$ such that $\iint_{\mathbb{R}^2} f(x,y)\mathrm{d}(x,y)=1$

is it true that function $g:\mathbb{R}\to\mathbb{R}_{\geq0},x\mapsto\int_{-\infty}^{+\infty}f(x,y)\mathrm{d}y$ is well-defined and is continuous?

My attempt:

I tried to make use of the "integral=1" condition so I tried to bound a $(1-\epsilon)$ fraction of $f$'s area-under-surface in some interval of $y$.

But it cannot guarantee that area-under-curve for every fixed $x$ is also at least $(1-\epsilon)*total\_area$ in this interval of $y$ - which is required in ordered to prove continuousness using definition.

答:
$g$ is well defined (a.e.) and is a probability density by Tonelli's theorem (and defined everywhere by Fatou's lemma).
I think that this should be a counterexample to continuity (if you eliminate the hypothesis that $X$ and $Y$ are continuous, by Daniel Adams' observation).

Let $\varphi:\mathbf{R}\to\mathbf{R}$ a continuous function such that $\varphi(0)=1$ and $\int_0^{\infty}\varphi=1$. Then
$$f(x,y)=
\begin{cases}
 \frac{1}{\pi}\frac{x}{1+x^2y^2}\phi(x)  & \text{if} & x>0\\
 0 & \text{if} & x\le 0
\end{cases}$$
should be a continuous probability density but $g$ is not continuous


问:
Why is it important to have a discrepancy between image and codomain?
A function $f:\mathbb{R}\rightarrow \mathbb{R}$ given by $f(x)=x^2$ has $\mathbb{R}_{\geq0}$ as its image and $\mathbb{R}$ as its codomain.

What's the need for this discrepancy? Why don't we just write $f:\mathbb{R}\rightarrow \mathbb{R}_{\geq0}$?

I believe the reason for this is that there may be some functions in which stating the image this way may be impractical - if this hypothesis is true, what is this class of functions? If yes or no, is there any other reason for this?

答:
The reason for this discrepancy is that for many functions it's not particularly important what its image is, while at the same time it would be tricky to figure this out.  The codomain however is easy, it describes the type of value one can get. Is it a real number, a complex number, a fraction?

For example, $$f(x)=e^{x\bar{x}}+\frac{7}{3^{x\bar{x}+4}}$$

It's true that $f:\mathbb{C}\rightarrow\mathbb{C}$, but in fact $f:\mathbb{C}\rightarrow \mathbb{R}$.  Hence the codomain gives useful information, but the image is difficult to determine.


答:
I think the general idea (for one of the directions) is correct, but it can be worded for explicitly. The following uses reasoning from first principles only.

Forward direction. Suppose that there exists uniquely $x_0$ such that $Ax_0=b.$ If $A$ has rank less than $n$, then there is some $z\neq 0$ such that $Az=0$. But then $x_0+z\neq x_0$ is another solution to $Ax_0=b$. So $A$ has rank $n$. This actually implies $[A|b]$ has rank $n$ as well: rank $[A|b]$ cannot be less than $n$ (because $[A|b]$ contains $A$) and cannot be $n+1$ (because $b$ belongs to the column space of $A$).

Backward direction. If both $A$ and $[A|b]$ have rank $n$, then $b$ must belong to the column space of $A$, so there is $x_0$ such that $Ax_0=b$. Moreover, this $x_0$ must be unique: if there is another $\tilde{x}$ such that $A\tilde{x}=b$, then $A(x-\tilde{x})=0$, contradicting $A$ having full column rank.

Title: Prove that the system $Ax = b, A \in \Bbb R^{m \times n}$ , has a unique solution if and only if rank $A$ = rank$[A \mid b] = n$.

Body:
So I am trying to prove the above statement and wanted to see if my line of thinking is in fact correct.

So first, because there exists a  unique solution then rank$(A) =$ rank $[A\mid b]$. I then said suppose that rank $(A) \neq n$ and instead that the rank is in fact $k < n$. This means there are $n-k$ linearly dependent columns which means we have infinitely many solutions to $Ax = b$ because we can have $n-k$ arbitrary elements of $x$ because we can 'eliminate' the linearly dependent columns. Is this line of thinking correct?


问:
Vectors subspace closure
Let W, U vector subspaces of V

and let $u,w \in V-{0}$ vectors such that

$u \in U \wedge u \notin W \wedge w \in W$

Can I say for a fact that $u + w \notin W$

答: Yes, a sum of a vector in $W$ and a vector outside of $W$ can't belong to $W$. If $u+w\in W$ then $u=(u+w)-w\in W$, a contradiction.

Title: Vectors subspace closure


答:
Yes you can.

Suppose for the purpose of contradiction that $u + w \in W$. By the closure properties of a subspace, it must hold that
$$
(u + w) + (-1) \cdot w = u + w - w = u \in W,
$$
contradicting our premise.

Title: Vectors subspace closure

Body:
Let W, U vector subspaces of V

and let $u,w \in V-{0}$ vectors such that

$u \in U \wedge u \notin W \wedge w \in W$

Can I say for a fact that $u + w \notin W$


问:
Bound power series with kind of random coefficients by it's finite versions?
Let*s define a power series:

$$
\begin{align}
p_n(x) = \sum\limits_{k=1}^n a_k x^k, \quad a_k \in \mathbb{N}
\end{align}
$$

Is it possible to give a certain bound $-1 \le x_b  \le x < 0 $ such that

$$
\begin{align}
p_n(x) &> p_{\infty}(x), \quad \text{if n is even} \\
p_n(x) &< p_{\infty}(x), \quad \text{if n is odd}
\end{align}
$$

although $a_n$ is kind of a random distributed sequence only bound between $a_n = n$ and $a_n = n^2$?

Edit: I would be happy also about a rough bound. F.e. let's say the best $x_b$ for a given $a_n$ sequence would be $-0.6$ but I would be able to say that $-0.25$ is a certain bound such that the conditions above are true.

Edit 2: With kind of randomly distributed I mean that I don't know the sequence. The values can jump up and down. I only know that this sequence is bounded. So, the "jumps" are bounded, too.

Clarification of the question: Is there any way to give a fixed $x_b$ only by the bounds such that the condition is true?

答:
I assume that $p_n(x) = \sum_{k=1}^n a_k x^k$.

As I understood the question, there exists a feasible sequence $(a_n)$ without the bound $x_b$, because the gap between the bounds for $a_n$ is too big. Namely, let $a_n=n$ when $n$ is odd and $a_n=n^2$, when $n$ is even. Given $-1<x<0$, pick an even $n$ such that $\tfrac 1n <-x$. It is easy to check that for each positive odd $k$ we have
$$ a_{n+k}+xa_{n+k+1}=(n+k)+x(n+k+1)^2<0.$$
Then
$$p_\infty(x)-p_n(x)= \sum_{k=n+1}^\infty a_k x^k=x^{n+1}((a_{n+1}+a_{n+2}x)+ x^2(a_{n+3}+a_{n+4}x)+\dots)>0.$$


问:
What is the opposite category of the category of Presheaves?
I know that the category of presheaves on $\textbf{C}$ is defined as $\textbf{Psh}(C):= \textbf{Func}(\textbf{C}^{op}, \textbf{Set})$.

I have been wondering what the category $(\textbf{Psh}(\textbf{C}))^{op}$ would be like.

I have a hunch that $(\textbf{Psh}(\textbf{C}))^{op}$ would be isomorphic to $\textbf{Func}(\textbf{C}, \textbf{Set})$.

Am I correct in this guess? I have tried to work on this but I do not even know whether the hunch is correct. Any help with this would be great.

Thank you.

答: This is not the case. For every category $\mathbf{Set}^{\mathbf{C}}$, its opposite category contains an object that is not the domain of any morphism except for its own identity (because the constantly $\varnothing$ functor is not the codomain of any morphism except its identity). But unless $\mathbf{C}$ is the empty category, $\mathbf{Set}^{\mathbf{C}}$ will never have such an object.


问:
Integration strategy for $\int_0^1(-\ln x)^2dx$
Calculus 2

Problem: The region bounded by $y=-lnx$ and the $x$-axis on the interval from $0$ to $1$ is rotated about the $x$-axis. Find the volume if possible

Thought Process:

So far, I have the problem setup for using washer/disk method to solve

$$\pi\int_0^1(-\ln x)^2dx$$

I'm not sure how exactly to proceed from here. A simple U-sub from here doesn't help because I have nothing swap for $du$. My next idea would be Integration by Parts setting $u=-lnx$ and $dv=-lnx$, but I don't know what $\int-\ln x$ would be.

I could try shells instead, but that would change my limits of integration, though that might not be a big deal. So...

$$y=-lnx$$

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

$$2\pi\int_0^\infty ye^{-y}dy$$

This looks like a decent candidate for parts, so I'll try $u=e^{-y}$ and $dv=y$, resulting in $du=-e^{-y}$ and $dv=ydy$

$$-\frac12e^{-y}y^2-\int_0^\infty \frac12y^2(-e^{-y})dy$$

Which doesn't look any easier than the first one, so maybe Integration by Parts isn't a great idea.

I cant think of how to proceed with either of these.

Question:

How can I proceed to find an antiderivative of either

Washer/Disk Method $\pi\int_0^1(-\ln x)^2dx$

Shells Method $2\pi\int_0^\infty ye^{-y}dy$

Note: I don't really need help with the Improper Integral portion of this, at least...I don't think I will. I only left the limits of integration on there to show that I know what the problem is asking. I know that I'll need to substitute the discontinuities with a variable and take the limit as the variable approaches the discontinuity and I don't think I'll have a problem with that. I'd really just like some advice on what strategy to use to find the antiderivative for either of these. I can't think of a method that seems like it will work.

Thanks in advance for any advice

Edit At the behest of users Babak and Triatticus, I swapped my $u$ and $dv$ choices for the Integration by Parts with the shells method and came to a much friendlier derivative

$u=y$  $dv=e^{-y}dy$

$du=dy$  $v=-e^{-y}$

$2\pi (-ye^{-y}-\int_0^\infty -e^{-y}dy)$

Since this is an Improper Integral, I must take the limit as the function approaches infinity

$$2\pi (-ye^{-y}-\lim_{t\rightarrow\infty}[e^{-y}]|_0^t)$$

And since it is integration by parts I need to apply the limits of integration to the $uv$ portion of $uv-\int vdu$ leading to

$$2\pi (-ye^{-y}|_0^\infty-\lim_{t\rightarrow\infty}[e^{-y}]|_0^t)$$

and again this is improper so I will replace the $\infty$ with another variable

$$2\pi (\lim_{m\rightarrow\infty}-ye^{-y}|_0^m-\lim_{t\rightarrow\infty}[e^{-y}]|_0^t)$$

For the sake of completion

$$-2\pi [\lim_{m\rightarrow\infty}(me^{-m}-0e^-0)+\lim_{t\rightarrow\infty}(e^{-t}-e^{-0})]$$

Skip to Edit 2, below is incorrect

$$-2\pi [\lim_{m\rightarrow\infty}(me^{-m})+\lim_{t\rightarrow\infty}(e^{-t})]$$

$$-2\pi [\lim_{m\rightarrow\infty}(m\frac1{e^{m}})+\lim_{t\rightarrow\infty}(\frac1{e^{t}})]$$

So, $\lim_{t\rightarrow\infty}(\frac1{e^{t}})$ will become $0$, and $\lim_{m\rightarrow\infty}(m\frac1{e^{m}})$ is an $\frac{\infty}{\infty}$ so we use L'Hospital's Rule to analyze by comparing the derivative of the numerator to the derivative of the denominator

Numerator:$\frac{d}{dm} m = 1$

Denominator: $\frac{d}{dm} e^m=e^m$

So ultimately, the denominator will approach infinity faster than the numerator, meaning that this is also equal to zero and the final equation becomes

$$-2\pi (0 + 0)$$

And now that I wrote this all out it seems wrong to me because that means that this function has a volume of zero...

The official question was answered long ago but I may need to make another thread to see if anyone would be willing to check my work because a volume of zero doesn't seem right to me

Edit 2 As I was going through my work again I noticed a problem, $e^{-0}$ does not equal $0$, will update for a last and final (hopefully) time after correcting my work on paper with what should be the correct answer

Final Edit This is the answer to the problem as a whole and not just my question, for anyone's future reference. The steps done above are correct except that $e^{-0}=1$ and not zero

$$-2\pi [\lim_{m\rightarrow\infty}(me^{-m})+\lim_{t\rightarrow\infty}(e^{-t})-1]$$

$$-2\pi [\lim_{m\rightarrow\infty}(\frac{m}{e^{m}})+\lim_{t\rightarrow\infty}(\frac{1}{e^{t}}-1)]$$

So, $\lim_{t\rightarrow\infty}(\frac1{e^{t}})$ will become $0$, and $\lim_{m\rightarrow\infty}(m\frac1{e^{m}})$ is an $\frac{\infty}{\infty}$ so we use L'Hospital's Rule to analyze by comparing the derivative of the numerator to the derivative of the denominator

Numerator:$\frac{d}{dm} m = 1$

Denominator: $\frac{d}{dm} e^m=e^m$

Resulting in $\lim_{m\rightarrow\infty}(m\frac1{e^{m}})$ ultimately equaling zero, leaving the following

$$-2\pi [0+0-1)]$$

$$-2\pi [-1]$$

$2\pi$ is the final answer for the volume

答:
Integrate $$\pi\int_{0}^{1}\left(-\ln x\right)^2dx=\pi \frac{d^2}{dt^2}\left(\int_{0}^{1} x^t dx\right)\bigg|_{t=0}
= \frac{d^2}{dt^2}\left(\frac\pi{t+1}\right) \bigg|_{t=0} =2\pi
$$


问:
Simple Number Theory Question
Lets say I have number in decimal form to the ten-thousands place $$p=0.abcd$$ and another number $$q=0.xyz$$ to the hundredth place. $$a.b,c,d,x,y,z$$ are all integers from 0 to 9, inclusive, except for z & d which can only range from 1 to 9, inclusive. If I want to know exactly where the decimal will lie in the multiplication of $$p*q$$ is there a formula I can employ?

Right now every time I multiply two decimals, I multiply $$abcd$$ and $$xyz$$ and then move the decimal point by the sum of number places d&z are away(so 7 moves to the eft). But this is all conditioned on the size of the number abcd*xyz yields. Is there a formula that can standardize this?

答:
$p = 0.abcd = abcd \times 10^{-4}$

$q = 0.xyz = xyz \times 10^{-3}$

$pq = abcd \times xyz \times 10^{-7}$

So you'd always want to move it by 7 placed, regardless of how many digits $abcd\times xyz$ has.

You can see how to 'standardize this', but using the $10^{-n}$ to understand the decimal, as compared to integers.

Title: Simple Number Theory Question


问:
$\sum_{n=0}^\infty \frac{1}{(2n)!}=\cosh(1)$
Is there any way to show that $$\sum_{n=0}^\infty \frac{1}{(2n)!}=\cosh(1) = \frac{1+e^2}{2e}$$ knowing that $$\sum_{n=0}^\infty \frac{1}{n!}=e$$
I guess that we can also show that $\sum_{n=0}^\infty \frac{1}{(2n-1)!}=\sinh(1) $ but I have no clue of how to get rid of the "2" in the factorial...

答:
No, the fact that $\sum_{n=0}^\infty \frac 1{n!} = e$ is insufficient (by any reasonable interpretation of what it means to "use a fact"). However, you could reach your result if you use the additional fact that
$$
\sum_{n=0}^\infty \frac{(-1)^n}{n!} = e^{-1}.
$$
In particular, we can note that
$$
\sum_{n=0}^\infty \frac{1}{(2n)!} = \frac 12 \left(
\sum_{n=0}^\infty \frac{1}{n!} +
\sum_{n=0}^\infty \frac{(-1)^n}{n!}
\right)
$$


答:
Write down the series expansions for $e^x$ and $e^{-x}$ around $0$. Add the two series and divide by $2$. Similarly, you can subtract them, to get $\sinh$.
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\\e^{-x}=\sum_{n=0}^\infty\frac{(-1)^nx^n}{n!}$$
When you add these two expressions, the odd terms will cancel. If you subtract them, the even terms will cancel out.

Title: $\sum_{n=0}^\infty \frac{1}{(2n)!}=\cosh(1)$

Body:
Is there any way to show that $$\sum_{n=0}^\infty \frac{1}{(2n)!}=\cosh(1) = \frac{1+e^2}{2e}$$ knowing that $$\sum_{n=0}^\infty \frac{1}{n!}=e$$
I guess that we can also show that $\sum_{n=0}^\infty \frac{1}{(2n-1)!}=\sinh(1) $ but I have no clue of how to get rid of the "2" in the factorial...


问:
Solvability by radicals of a separable field of characteristic $p$
Preliminary Definitions

A field $K$ is said to be solvable by proper radicals over $F$ if there exist fields $F_0,F_1,\dots,F_m$ such that $F=F_0\subseteq F_1\subseteq F_2\subseteq\dots\subseteq F_m$ and $K\subseteq F_m$ and for all $i$ we wave $F_i=F_{i-1}(a_i)$ with $a_i^{[F_i:F_{i-1}]}\in F_{i-1}$.

A field $N$ is said to be metacyclic over $F$ if $N$ is Galois over $F$ and its Galois group is solvable.

For every $n$ not multiple of the characteristic of the field $F$, let $C_n$ be the $n$-th cyclotomic field.

My Question

I know that a Galois field $N$ over $F$ of characteristic $0$ is solvable by proper radicals if and only if its Galois group is solvable.

I would like to know what would be an equivalent condition about the Galois group of a Galois field $N$ over $F$ of characteristic $p$ for $N$ to be solvable by proper radicals over $F$, not including Artin-Schreier polynomials.

I found only a partial answer to this, namely:

Theorem 1: Let $P$ be a set of primes different from $p$ such that for every $r\in P$ then all prime factors of $r-1$ are in $P$. If $N$ is metacyclic of degree $n$ over $F$ of characteristic $p$ and every prime divisor of $n$ is in $P$, then $N$ is solvable by proper radicals over $F$.

Theorem 2: If a Galois field $N$ of degree $n$ over $F$ of characteristic $p$ is solvable by proper radicals over $F$, then $N$ is metacyclic over $F$ and $p\nmid n$.

My Attempt

I tried to prove some facts and theorems, but I do not know how to improve them to complete the problem. For example, the set $P$ in Theorem 1 looks so ill-behaved or broken in pieces, an example of such a set being the set of all primes less than $p$.

Remark: In the definition of solvability by proper radicals, we can suppose without loss of generality that $[F_i:F_{i-1}]$ is prime.

Indeed, if $K=F(a)$ and $a^{mn}\in F$ and $[K:F]=mn$, then let $L=F(a^n)$, then $K=L(a)$ and $L=F(a^n)$ and $a^n\in L$ and $(a^n)^m=a^{mn}\in F$, so that $[K:L]\leq n$ and $[L:F]\leq m$, but $[K:L][L:F]=[K:F]=mn$, so $[K:L]=n$ and $[L:F]=m$, so $K=L(a)$ and $a^{[K:L]}\in L$ and $L=F(a^n)$ and $(a^n)^{[L:F]}\in F$.

Fact 1: Let $F$ be a field containing a primitive $n$-th root of $1$. Let $E=F(a)$ where $a^n\in F$ and no smaller power of $a$ is in $F$. Then $E$ is a Galois extension of $F$ with cyclic Galois group of order $n$. Conversely, if $n$ is prime and $E$ is a cyclic extension of $F$ of degree n, then $E=F(a)$ for some $a$ with $a^n\in F$.

a) Suppose that $F$ has a primitive $n$-root $\zeta$ of unity. Let $K=F(a)$ e suppose that $n$ is the least such that $a^n\in F$. Then $x^n-a^n\in F[x]$ and $x^n-a^n=(x-a)(x-\zeta a)\dots(x-\zeta^{n-1}a)$ is separable and all its roots are in $K$, so $K$ is Galois over $F$. Let $G=\mathrm{Aut}(K:F)$. For every $\sigma \in G$, then $\sigma(a^n)=a^n$, so $(\sigma(a)/a)^n=1$, so let $\zeta_\sigma=\sigma(a)/a$, then $\zeta_\sigma\in F$. For $\sigma,\theta\in G$, then $\sigma\tau(a)=\sigma(\zeta_\tau a)=\zeta_\tau\sigma(a)=\zeta_\tau\zeta_\sigma a$, so $\zeta_{\sigma\tau}=\zeta_\tau\zeta_\sigma$. Hence $f:\sigma\mapsto\zeta_\sigma$ is an homomorphism from $G$ to the group $H$ of the $n$-roots of unity. If $\zeta_\sigma=1$, then $\sigma(a)=a$, so $\sigma=I$; then $f$ is injective. If $d=|\mathrm{Im}(f)|$, then for all $\sigma\in G$ we have $\zeta_\sigma^d=1$, so $(\sigma(a)/a)^d=1$, so $\sigma(a^d)=a^d$, so $a^d\in F$, so $d=n$, therefore $[K:F]=|G|=n$.

b) If $n$ is prime and $E=F(\xi)$ is cyclic of degree $n$ over $F$, then designate the roots of the minimum function of $\xi$ by $\xi_i=\sigma^{i}(\xi)$. These elements are in $E$ and so are:

$$a_i=\xi_0+\zeta^i\xi_1+\dots+\zeta^{i(n-1)}\xi_{n-1},\quad\quad(i=0,\dots,n-1).$$

The determinant:

$$d=\mathrm{det}(\xi^{ij}),\quad\quad(i,j=0,\dots,n-1)$$

is the Vandermonde determinant $\prod_{i<j}(\xi^i-\xi^j)\neq 0$, so that if the $a_i$ are all in $F$ then $\xi_i$ would be all in $F$, contrary to the hypothesis $E=F(\xi_0)\neq F$. So some $a_i$ is not in $F$ and this is not $a_0$ which is $\xi_0+\dots+\xi_{n-1}\in F$. So $\sigma(a_i)=\zeta^{-i}a_i$ and then the conjugates $\sigma^k(a_i)$, $k=0,\dots,n-1$ are all distinct by primeness of $n$, so the minimal polynomial of $a_i$ over $F$ has degree at least $n$, so $E=F(a)$ and $a^n\in F$ because $\sigma(a_i^n)=\zeta^{-ni}a_i^n=a_i^n$.

Fact 2: Let $N$ be metacyclic over $F$ of characteristic $p$ of degree $n=p_1\dots p_r$ not multiple of $p$ over $F$, and let $F$ fave primitive $p_i$-roots of unity for primes $p_i$, $i=1,\dots,r$. Then there are fields $F=F_0\subseteq\dots\subseteq F_m=N$ such that for all $i$ we wave $F_i=F_{i-1}(a_i)$ with $a_i^{[F_i:F_{i-1}]}\in F_{i-1}$.

For $N$ has the form:

$$F=F_0\subseteq F_1\subseteq \dots\subseteq F_m=N$$

where $F_i$ is cyclic of prime degree $p_i\neq p$ over $F_{i-1}$, so by Fact 1 we have $F_i=F_{i-1}(a_i)$ with $a_i^{p_i}\in F_{i-1}$.

Fact 3: Let $p_1,\dots,p_r$ be primes different from $p$ and $C_{p_1},\dots,C_{p_r}$ the corresponding cyclotomic fields over $F$ over characteristic $p$. Then their composite $C$ is metacyclic over $F$.

The result is evident when $r=1$ since $C_{p_1}$ is cyclic over $F$. Assuming that the composite $D$ of $C_{p_1},\dots,C_{p_{r-1}}$ is metacyclic over $F$, then $D$ and $C_{p_r}$ are Galois over $F$, so their composite $C$ is Galois over $F$ and $\mathrm{Aut}(C:D)$ is isomorphic to a subgroup of $\mathrm{Aut}(C_{p_r}:F)$, which is isomorphic to a subgroup of the cyclic group $\mathbb{Z}_{p_r}^*$, so $\mathrm{Aut}(C:D)$ is cyclic. Then $C$ is metacyclic over $D$ and, since $D$ is metacyclic over $F$, the field is metacyclic over $F$.

Fact 4: Let $F$ be a field of characteristic $p$ and $P$ be a set of primes different from $p$ such that for every $r\in P$ then all prime factors of $r-1$ are in $P$. For every $r\in P$, if $C$ is the composite of the $C_q$ with $q\in P$ and $q\leq r$, then there are fields $F=F_0\subseteq \dots\subseteq F_m=C$ such that for all $i$ we have $F_i=F_{i-1}(a_i)$ and $a_i^{[F_i:F_{i-1}]}\in F_{i-1}$.

Let $r\in P$, $C$ be the composite of $C_q$ with $q\in P$ and $q<r$ and $D$ the composite of $C$ and $C_r$. As $C_r$ is Galois over $F$, then $D$ is Galois over $C$ and $\mathrm{Aut}(D:C)$ is isomorphic to a subgroup of $\mathrm{Aut}(C_r:F)$, which is isomorphic to a subgroup of the cyclic group $\mathbb{Z}_r^*$, so $\mathrm{Aut}(D:C)$ is cyclic of order $d\mid r-1$, but then for every $q$ prime divisor of $d$, then $q\in P$ and $q<r$, so that $C$ has a primitive $q$-root of unity; so by Fact 2 we have $C=K_0\subseteq\dots\subseteq K_n=D$ such that for all $j$ we wave $K_j=K_{j-1}(b_j)$ with $b_j^{[K_j:K_{j-1}]}\in K_{j-1}$. By induction hypothesis, there are $F=F_0\subseteq \dots\subseteq F_m=C$ such that for all $i$ we have $F_i=F_{i-1}(a_i)$ and $a_i^{[F_i:F_{i-1}]}\in F_{i-1}$.

Fact 5: If $K$ is a separable extenstion of $F$ of characteristic $p$ and is solvable by proper radicals, then there are fields $F=F_0\subseteq F_1\subseteq\dots\subseteq F_M$ such that $K\subseteq F_M$ and for all $i$ we have $F_i=F_{i-1}(a_i)$ and $a_i^{[F_i:F_{i-1}]}\in F_{i-1}$ and $[F_i:F_{i-1}]$ prime different from $p$. In particular, $p\nmid[K:F]$.

If $F$ is a field of characteristic $p$ and $K=F(a)$ with $a^p\in F$ and $[K:F]=p$ and $L=K(b)$ with $b^q\in K$ and $[L:K]=q$ prime with $q\neq p$, then the minimal polynomial of $b$ over $K$ is $x^q-b^q$, which is separable over $K$, so the minimal polynomial $f$ of $b$ over $K(b^p)$ is also separable, but it also divides $(x-b)^p=x^p-b^p\in K(b^p)[x]$, so $f=x-b$ and $b\in K(b^p)$, therefore $K(b^p)=K(b)$. For all $t\in K$, we have $t=\sum_kc_ka^k$ with $c_k\in F$, so $t^p=\sum_kc_k^pa^{pk}\in F$. So $(b^p)^q=(b^q)^p\in F$. As $L=K(b^p)=F(b^p,a)$ and $a^p\in F\subseteq F(b^p)$, then $[L:F(b^p)]\leq p$, and also $(b^p)^q\in F$ so $[F(b^p):F]\leq q$, but $[L:F(b^p)][F(b^p):F]=[L:F]=[L:K][K:F]=qp$, then $[L:F(b^p)]=p$ and $a^p\in F(b^p)$ and $[F(b^p):F]=q$ and $(b^p)^q\in F$.

Therefore, if $F$ is a field of characteristic $p$ and $K$ is separable over $F$ and solvable by proper radicals over $F$, then there are $F=F_0\subseteq F_1\subseteq\dots\subseteq F_m$ such that $K\subseteq F_m$ and for all $i$ we have $F_i=F_{i-1}(a_i)$ with $a_i\in{[F_i:F_{i-1}]}\in F_{i-1}$ and $[F_i:F_{i-1}]$ prime, so by the preceding paragraph we can suppose without loss of generality that there is $M$ such that for all $i\leq M$ we have $[F_i:F_{i-1}]\neq p$ and for all $i>M$ we have $[F_i:F_{i-1}]=p$, so for all $\xi\in K$ then $\xi$ is separable over $F$, so $\xi$ is separable over $F_M$, but as $\xi\in F_m$, we have $\xi^p\in F_{m-1}$ and by induction we have $\xi^{p^{m-M}}\in F_M$, so the minimal polynomial $f$ of $\xi$ over $F_M$ divides $(x-\xi)^{p^{m-M}}=x^{p^{m-M}}-\xi^{p^{m-M}}\in F_M[x]$, and $f$ is separable over $F_M$, so $f=x-\xi$ and $\xi\in F_M$. Therefore $K\subseteq F_M$.

Theorem 1: Let $P$ be a set of primes different from $p$ such that for every $r\in P$ then all prime factors of $r-1$ are in $P$. If $N$ is metacyclic of degree $n$ over $F$ of characteristic $p$ and every prime divisor of $n$ is in $P$, then $N$ is solvable by proper radicals over $F$.

For let $N$ be metacyclic of degree $n$ over $F$ of characteristic $p$ such that every prime divisor of $n$ is in $P$, and let $r$ be the greatest prime divisor of $n$, and $C$ be the composite of the $C_q$ with $q\in P$ and $q\leq r$. The composite $D$ of $N$ and $C$ is Galois over $C$ and its Galois group is isomorphic to a subgroup $H$ of the group $G$ of $N$ over $F$. The group $G$ is solvable and so is $H$. Therefore $D$ is metacyclic over $C$ and its degree $m$ is a divisor of $n$, so that for every $q$ prime divisor of $m$ then $C$ has a primitive $q$-root of unity. By Fact 2, there are fields $C=K_0\subseteq\dots\subseteq K_m=N$ such that for all $j$ we wave $K_j=K_{j-1}(b_j)$ with $b_j^{[K_j:K_{j-1}]}\in K_{j-1}$, and by Fact 4 there are fields $F=F_0\subseteq \dots\subseteq F_m=C$ such that for all $i$ we have $F_i=F_{i-1}(a_i)$ and $a_i^{[F_i:F_{i-1}]}\in F_{i-1}$. Therefore, $D$ is solvable by proper radicals over $F$.

Theorem 2: If a Galois field $N$ of degree $n$ over $F$ of characteristic $p$ is solvable by proper radicals over $F$, then $N$ is metacyclic over $F$ and $p\nmid n$.

Let $N$ be a Galois field over $F$ with automorphism group $G$ and assume that $N$ is solvable by proper radicals over $F$. Then, by Fact 5, there are $F=F_0\subseteq F_1\subseteq\dots\subseteq F_M$ such that $K\subseteq F_M$ and for all $i$ we have $F_i=F_{i-1}(a_i)$ and $a_i^{[F_i:F_{i-1}]}\in F_{i-1}$ and $[F_i:F_{i-1}]=p_i$ prime different from $p$ and let $C$ be the composite of the cyclotomic fields $C_{p_i}$. The group $G'=\mathrm{Aut}(C:F)$ is solvable by Fact 3. If $D=N\cap C$ and $H=\mathrm{Aut}(N:D)$ and $H'=\mathrm{Aut}(C:D)$, then $D$ is Galois over $F$ and $\mathrm{Aut}(D:F)$ is isomorphic to both $G'/H'$ and $G/H$. But $G'/H'$ is solvable, and so $G/H$ is solvable. Also, if $N_0$ is the composite of $N$ and $C$, then $H_0=\mathrm{Aut}(N_0,C)$ is isomorphic to $H$. We define $L_t=C(a_1,\dots,a_M)$, then $L_i=L_{i-1}(a_i)$ and $L_{i-1}$ contains a primitive $p_i$-root of unity and $a_i^{p_i}\in L_{i-1}$, so that by primeness of $p_i$ and Fact 1 either $L_i$ is cyclic of prime degree $p_i$ over $L_{i-1}$ or $L_i=L_{i-1}$. Is $N_1$ is the composite of $N_0$ and $L_1$, then $H_1=\mathrm{Aut}(N_1,L_1)$ is isomorphic either to $H_0$ or a normal subgroup of index $p_1$ of $H_0$. In fact, $[H_0:H_1]=[N_0\cap L_1:C]$ and, since $[L_1:C]$ is $p_1$ or $1$, then $[N_0\cap L_1:C]$ is $p_1$ or $1$. We define $N_i$ as the composite of $N_{i-1}$ and $L_i$ and obtain a sequence of groups:

$$H_0\supseteq H_1\supseteq\dots\supseteq H_M,$$

where $H_i=\mathrm{Aut}(N_i:L_i)$ and either $H_i$ is $H_{i-1}$ or a normal subgroup of prime index $p_i$ of $H_{i-1}$. But $N_M$ is the composite of $N_0$ and $L_M$ and since $N\subseteq F_M$ we have $N_0\subseteq L_M$ and $N_M=L_M$, $H_M$ must be the identity group. Each $H_i\subset H_{i-1}$ is evidently a maximal normal divisor of $H_{i-1}$ and if we delete the repeated groups from the sequence $H_0,H_1,\dots,H_M$ we obtain a composition series of $H_0$. Then $H_0$ is solvable, and so is $H$. Therefore the group $G$ is solvable.

答:
Radical means a tower of extensions of the form $M(a)/M$ with $a^d\in M$ for some $d$.

For a finite characteristic $p$ extension $F/K$, let $L$ be the normal closure, $F/K$ is contained in a radical extension iff $Aut(L/K)$ is solvable and with
$L_n=L^{p^{n!}}K(\zeta_{n!}), K_n=K(\zeta_{n!})$, for some thus all $n$ large enough

$$p\nmid [L_n:K_n]$$

This latter condition is because otherwise there is a subfield $E\subset L_m$ such that $L_m/E$ is cyclic of degree $p$ and $[L_n:EK_n]=p$ for all $n$,

$F/K$ contained in a radical extension implies that $L_n/EK_n$ is contained in a separable radical extension, for $n$ large enough we have the roots of unity to say it must be $EK_n(a)$ with $x^d-a^d\in EK_n[x]$ irreducible, which is inseparable since $p=[L_n:EK_n]$ must divide $d$, a contradiction.

Conversely if $Aut(L/K)$ is solvable and $p\nmid [L_n:K_n]$ then for $n$ large enough: $L_n/K_n$ is a tower of cyclic extensions of degree coprime with $p$ and we have the roots of unity to use the same argument as in characteristic $0$, that $\sum_j \zeta_d^j \sigma^j(a)$ is a $d$-th root.


问:
Showing that a function has a certain absolute minimum.
Suppose we have the function $$f(x) = \frac{x}{p} + \frac{b}{q} - x^{\frac{1}{p}}b^{\frac{1}{q}}$$ where $x,b \geq 0 \land p,q > 1 \land \frac{1}{p}+\frac{1}{q} = 1$

I am trying to show that $b$ is the absolute minimum of $f$.

I proceeded as follows:

$$\frac{df(x)}{dx} = \frac{1}{p} - \frac{x^{\frac{1}{p}-1}}{p} b^{\frac{1}{q}} = \frac{x - x^{\frac{1}{p}} b^{\frac{1}{q}}}{px}$$

Now I will look for critical points by searching for the zeros of this function.

$$\frac{x - x^{\frac{1}{p}} b^{\frac{1}{q}}}{px} = 0 \iff x - x^{\frac{1}{p}} b^{\frac{1}{q}} = 0 \iff x = x^{\frac{1}{p}} b^{\frac{1}{q}}$$.

Now I can see that $b$ is a critical point.

How ever when I continue my calculations to check whether there are any other critical points
$$x = x^{\frac{1}{p}} b^{\frac{1}{q}} \implies x^p = b^{\frac{p}{q}}x \implies x^{p-1} = b^{\frac{p}{q}} \implies x = b^{\frac{p}{(p-1)q}}$$

But this could not be equal to $b$, where did I go wrong?

答:
Let  $\dfrac{1}{p}=a$ and $\dfrac{1}{q}=k$

$ax+bk-x^a \cdot b^k=f(x)  $

$\dfrac{x + \dots x_{ath}+ b+ \dots b_{kth}}{a+k} \ge (x^ab^k)^{1/(a+k)}$ (By AM-GM inequality)

$x+ \dots x (a$ times)$=ax$ and $b+ \dots b (k$ times)$=bk$

$a+k=1 \implies ax+bk \ge x^a\cdot b^k \implies ax+bk-x^ab^k \ge 0$

Now you have $f(x) \ge 0$, the minimum is achieved when $x=b$ and $a=k$


问:
Topologicclad clasp property
If $\tau_{X}$ is the topology in $X$ then the subspace topology in any $Y\subset X$ is the induced topology:

$$\tau_{Y}=\{T\cap Y:T\in \tau_{X}\}$$

This follows that the closed in $Y$ are:

$$\overline{\tau_Y}=\{Y\backslash (T\cap Y):T\in \tau_{X}\}$$

So if $Z\subset Y\subset X$ then $Cl_{Y}(Z)=Y\cap Cl_{X}(Z)$.

I wonder why it's worth the equality $Cl_{Y}(Z)=Y\cap Cl_{X}(Z)$?

答:
Proposition. [2.2.1, Eng] Let $X$ be a topological space and $M$ a subspace of $X$. The set $A\subset M$
is closed in $M$ if and only if $A=M\cap F$, where $F$ is closed in $X$. The closure $\tilde{A}$ of a set
$A\subset M$ in the subspace $M$ and the closure $\overline{A}$ of $A$ in the space $X$ are related by the equality
$\tilde{A}=\overline{A}\cap M$.

Proof. If $A =M\cap F$, where $F =\overline{F}\subset X$, then $M\setminus A=M\cap (X \setminus F)$ and $A$ is
closed in $M$ as the complement of an open set. Conversely, if $A$ is a closed subset of $M$, then
$M\setminus A = M\cap U$, where $U$ is open in X. Thus
$$A = M\setminus(M\setminus A) = M\setminus (M\cap U) = M\cap(X\setminus U),$$
and $A = M\cap F$, where $F = X \setminus U$ is closed in $X$.

By definition of the closure operator, $\tilde{A}$ is equal to the intersection of all closed subsets
of $M$ that contain $A$, i.e., of all sets $M\cap F$, where $F = \overline{F}$ and $A\subset F$. This gives the equality
$\tilde{A} = M\cap\overline{A}$. $\square$

References

[Eng]  Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.

Title: Topologicclad clasp property


问:
How can I prove $\int_{0}^{+\infty }\frac{x-sinx}{\sqrt{x^{7}}}dx$ converges/diverges?
I need to determine if the following integral converges/diverges:

$$\int_{0}^{+\infty }\frac{x-\sin x}{\sqrt{x^{7}}}dx$$

We can write the integral as:

$$\int_{0}^{+\infty }\frac{x-\sin x}{\sqrt{x^{7}}}dx=\int_{0}^{a }\frac{x-\sin x}{\sqrt{x^{7}}}dx+\int_{a}^{+\infty }\frac{x-\sin x}{\sqrt{x^{7}}}dx=I_{1}+I_{2}$$

+For $I_{1}=\int_{0}^{a}\frac{x-\sin x}{\sqrt{x^{7}}}dx$.

Since $\sin x= x-\frac{x^{3}}{3!} +O(x^{3})$, I know that $I_{1}$ is convergent.

+For $I_{2}=\int_{a}^{+\infty }\frac{x-\sin x}{\sqrt{x^{7}}}dx$.

I don't know how to use $x-\sin x$ for other ideas such as the comparison test. The only thing I can exploit is $x - \sin x>0 , x>0$.

Can you give me some hints? Thank you very much!

答:
That's good work for the first part,

For $I_2$ you can see that $\sin(x) = o_{+\infty}(x)$ therefore $\dfrac{x-\sin(x)}{x^{7/2}} \sim_{+\infty} \dfrac{x}{x^{7/2}} = \dfrac{1}{x^{5/2}} $.

Moreover $\int_1^{+\infty} \dfrac{1}{x^{5/2}} \mathrm{d}x$ converges (Riemann's rule with $5/2>1$). And $\forall x\geq1, x-\sin(x)\geq 0$ so you can use the theorem of comparison for positive functions and it gives you the convergence of $I_2$.


问:
How can I prove $2^n+10n^2 \le 12 \cdot 2^n, n \ge 1$ is true by induction?
I'm having trouble with this problem.
$2^n+10n^2 \le 12 \cdot 2^n, n \ge 1$

Here is my process.

n = 1, it is true <-- basis step

Assume that arbitrary k is true. $2^k+10k^2 \le 12 \cdot 2^k$.
And I make  $2^{k+1}+20(k)^2 \le 12 \cdot 2^{k+1}$ by multiplying 2.

But if k become k+1,the equation should be $2^{k+1}+10(k+1)^2 \le 12 \cdot 2^{k+1}$..
I can't translate $20k^2$ into $10(k+1)^2$. How can I prove it by induction?
Or is this equation false?

Body:
$2^n+10n^2 \le 12 \cdot 2^n, n \ge 4^{**}$

Divide both sides by $2^n$
$$1+\frac{10 n^2}{2^n}\le 12;\;n\ge 4\tag{1}$$
For $n=4$ it is true
$$1+\frac{10\cdot 4^2}{2^4}=11\le 12$$
Now suppose $(1)$  is true for $n$ and let's prove it for $n+1$
$$1+\frac{10 (n+1)^2}{2^{n+1}}=1+\frac{1}{2}\left(\frac{10n^2}{2^n}\right)+\frac{1}{2}\left(\frac{20n+10^*}{2^n}\right) <1+\frac{1}{2}\left(\frac{10n^2}{2^n}\right)+\frac{1}{2}\left(\frac{10n^2}{2^n}\right)=\\=1+\frac{10n^2}{2^n}\le 12$$
$^*\quad 20n+10<10n^2$ for $n\ge 4$

$^{**}$ inequality is false for $n=3$

indeed
$$2^3+10\cdot 3^2 \le 12 \cdot 2^3\to 98\le 96$$
which is false


问:
Uniformly Integrable Second Moments and Weak Convergence.
Let $\mathcal{P}(\mathbb{R}^d)$ be the space of Borel probability measures, and let $\mathcal{P}_2(\mathbb{R}^d)$ be the space of Borel probability measures with finite second moment. Let $\{\mu_n\} \subset \mathcal{P}_2(\mathbb{R}^d)$ have uniformly bounded second moments, and assume $\mu_n\to \mu \in \mathcal{P}(\mathbb{R}^d)$ weakly.

Is it true that $\mu \in \mathcal{P}_2(\mathbb{R}^d)$ ?

答:
Yes.

By Skorohod representation theorem there exists random vectors $X_n$ and $X$ such that $\mu_n$ is the distribution of $X_n$,  $\mu$ is the distribution of $X$ and $X_n \to X$ a.s.

Denote by $X_n^{(1)}$ the sequence of first coordinates of vectors $X_n$. We know that $X_n^{(1)} \to X^{(1)}$ a.s., where $X^{(1)}$ is the first coorndite of vectors $X$.

As $X_n$ have uniformly bounded second moments $sup_{n \ge 1} E ||X_n||^2 < \infty$ then $sup_{n \ge 1} E |X_n^{(1)}|^2 < \infty$.

By Fatou's lemma $ E (X^{(1)})^2 = E \underline{lim} (X_n^{(1)})^2 \le \underline{lim} E(X_n^{(1)})^2 < \infty$.

Similarly $ E (X^{(i)})^2 < \infty$ for all $1 \le i \le d$. It follows that the limit of $P_n$ is in $\mathcal{P}_2(R^d)$, q.e.d.

Addition

By De la Vallée-Poussin criterion for Uniform Integrability  $\{X_n^{(1)} \}_{n\ge 1}^{\infty}$ is a uniformly integrable class of r.v. Hence $X^{(1)} \in L_1$ and $EX_n^{(1)} \to EX^{(1)}$. But $EX_n^2$ doesn't converges to $EX^2$ in general case.

Example. Consider $X_n \sim \sqrt{n} Bern(\frac{1}n)$, then $X_n \to 0$. Thus $P_n, P \in \mathcal{P}_2(R^d)$, $EX_n^2 \to 1 \ne 0 = EX^2 $.


问:
Find the highest natural number which is divisible by $30$ and has exactly $30$ different positive divisors.

Find the highest natural number which is divisible by $30$ and have exactly $30$ different positive divisors.

What I Tried: I am not sure about any specific approach to this problem. Of course as the number is divisible by $30$ , some of the factors of the number would be $1,2,3,5,6,10,20,30$ , but that only makes $8$ divisors and there will be $12$ more which I have no idea of .

Can anyone help me?

答:
30=2x3x5
So the exponents of 2,3, and 5 have to be 2,1,4 in no particular order. In order to maximize our product, we can do 5^4 x 3^2 x 2^1=11250


答:
You can write

$$
p=\frac{abcd}{10000},\qquad q=\frac{xyz}{1000}
$$

so

$$
pq=\frac{abcd\cdot xyz}{10000000}
$$

and the only cases where this eventually has less than seven decimal digits (after discarding trailing zeros) is when the product at the numerator ends with $0$, which is the case when either $d$ is $5$ and $z$ is even (or the other way around).

In fact a product of numbers not divisible by $10$ is divisible by $10$ if and only if one of them is divisible by $5$ and the other one by $2$. So take the maximum exponent $m$ by which one is divisible by $5$, the maximum exponent $n$ by which the other is divisible by $2$; set $k=\min\{m,n\}$; then exactly $k$ trailing zeros will appear in the product.

This phenomenon can never appear when the number base is prime, for example in binary arithmetic.

Title: Simple Number Theory Question

Body:
Lets say I have number in decimal form to the ten-thousands place $$p=0.abcd$$ and another number $$q=0.xyz$$ to the hundredth place. $$a.b,c,d,x,y,z$$ are all integers from 0 to 9, inclusive, except for z & d which can only range from 1 to 9, inclusive. If I want to know exactly where the decimal will lie in the multiplication of $$p*q$$ is there a formula I can employ?

Right now every time I multiply two decimals, I multiply $$abcd$$ and $$xyz$$ and then move the decimal point by the sum of number places d&z are away(so 7 moves to the eft). But this is all conditioned on the size of the number abcd*xyz yields. Is there a formula that can standardize this?


问:
General Topology: basic closing properties in induced topology
If $\tau_{X}$ is the topology in $X$ then the subspace topology in any $Y\subset X$ is the induced topology $\tau_{Y}=\{T\cap Y:T\in \tau_{X}\}$. This follows that the closed in $Y$ are $\overline{\tau_Y}=\{Y\backslash (T\cap Y):T\in \tau_{X}\}$. So if $Z\subset Y\subset X$ then $Cl_{Y}(Z)=Y\cap Cl_{X}(Z)$.
So if $Z=Cl_{Y}(Z)$ and $Y=Cl_{X}(Y)$ then:

$$Z=Cl_{Y}(Z)=Y\cap Cl_{X}(Z)=Cl_{X}(Y)\cap Cl_{X}(Z)=Cl_{X}(Z)$$.

I know that as $ Y $ is closed then $Cl_{X}(Y)=Y$. But I couldn't understand why this equality is worth here:
$$Cl_{X}(Y)\cap Cl_{X}(Z)=Cl_{X}(Z)$$

I would like to understand the intuitive notion of this equality.

答: Since $Z\subset Y$, $Cl_X(Z)\subset Cl_X(Y)$, so $Cl_X(Y)\cap Cl_X(Z)=Cl_X(Z)$.


问:
Integration of product of exponential function and natural logarithm
I am trying to calculate the following integral:

$$\int_{-1}^{1}\ln(1+\gamma x)\exp\left(-\frac{(z-x)^2}{2}\right)dx$$

with $\gamma\in[-1,1]$, $z\in\mathbb{R}$ and ln being the natural logarithm. I tried a lot of thinks like substitution, integration by parts, used the series expansion of the natural logarithm resp of the exponential function. I also searched for some help in Literatur but couldn't find anything helpful. My question is: Can we solve this integral?

答:
$$I=\int\log(1+\gamma x)\,\exp\left(-\frac{(z-x)^2}{2}\right)dx$$

Let $x=\sqrt{2} t+z$ to make
$$I=\sqrt{2}\int   \log \big[1+\gamma  \left(\sqrt{2} t+z\right)\big]\,e^{-t^2}\,dt$$ Using Taylor around $t=0$
$$ \log \big[1+\gamma  \left(\sqrt{2} t+z\right)\big]=\log (1+\gamma  z)-\sum_{n=1}^\infty \frac{2^{n/2} \left(-\frac{\gamma }{\gamma  z+1}\right)^n}{n} t^n$$ and
$$J_n=\int t^n\,e^{-t^2}\,dt=-\frac{1}{2} \Gamma \left(\frac{n+1}{2},t^2\right)$$

This would be a nightmare.