问:
Easy Question about P-names
Notation(from Kunnen): $M$ is countable transitive model of $ZFC$, $P$ is a partial order that belongs to $M$, $G$ is a $P$-generic filter over $M$, $\hat{b}$ will be a $P$-name for $b\in M$, and $M[G]$ is the smallest model of $ZFC$ containing $G$ and $M$.

The exercise i'm trying to do is the following: If $f$ is a function with range contained in $M$ and $f$ belongs to $M[G]$ then there is a set $B$ that belongs to $M$ containing the range of $f$. The hint given is to consider

$B=\left \{ b:\exists p  \in P(p\Vdash \hat{b} \in \operatorname{Rng}(\tau ) ) \right \} $ where $\operatorname{val} (\tau,G) = f$

I understand that this describes the desired $B$ as a class in $M$ but i can't find a way to show that this is actually a set in $M$.

Thanks for the help and sorry for the trivial question.

答:
The class $B$ is the class of all elements that some condition forces them into $\rm{Rng}(f)$.

The question is, indeed, why is this a set in $M$.

Note that: $B_p = \{b\colon p\Vdash \hat b\in\operatorname{Rng}(\tau)\}$ is a definable class in $M$, since $p$ cannot force a proper class of elements into the range of $f$ we have that this is a set in $M$.

Suppose some $p$ would force $B_p$ to be a proper class of elements, then any generic filter concentrating on $p$ would force $f$ to be a proper class. Take $G$ to be a generic filter concentrating on $p$ and derive contradiction.

Now $\displaystyle B=\bigcup_{p\in P}B_p$ is also a set in $M$ by axioms of replacement and union.

Let's look at a concrete example. Consider a Cohen forcing that introduces one new real number, that is a new function $f\colon\omega\to 2$, or a new $A\subseteq\omega$.

Of course, for different $P$-generic filters, the resulting set is going to be different (that is, one filter might force $f(0)=1$ and another the opposite). However for every $p\in P$ we can define the set $B_p = \{n\colon p\Vdash n\in A\}$.

Note if we consider $P$ to be partial functions from some $n$ into $2$ (ordered by extension) then $B_p$ is exactly $\{n\colon n\in\operatorname{Dom}(p)\land p(n)=1\}$, i.e. all the numbers on which $p$ decides that are in the set.

Of course that in this rather trivial example $B = \omega$, and the $B_p$ was extremely simple to define, but I hope the idea got through.

Title: Easy Question about P-names


问:
Galois group of $\mathbb{Q}(\sqrt[3]{5}, e^{2\pi i/3})$ over $\mathbb{Q}$ and $\mathbb{Q}(e^{2\pi i/3})$
I want to find $Gal(\mathbb{Q}(\sqrt[3]{5}, e^{2\pi i/3}) | \mathbb{Q})$ and $Gal(\mathbb{Q}(\sqrt[3]{5}, e^{2\pi i/3})| \mathbb{Q}(e^{2\pi i/3}))$

Let $w = e^{2\pi i/3}, \alpha = \sqrt[3]{5}, L = \mathbb{Q}(\sqrt[3]{5}, w)$.

Since $L$ is the splitting field of $f(x) = x^3 - 5$, it is a finite, normal, seperable extension, and therefore $|Gal(L|\mathbb{Q})| = [L:\mathbb{Q}] = [\mathbb{Q}(w, \alpha):\mathbb{Q}(w)][\mathbb{Q}(w):\mathbb{Q}]$.

The minimal polynomial for $w$ over $\mathbb{Q}$ is $x^3 - 1$, so $[\mathbb{Q}(w):\mathbb{Q}] = 3$

I thought that since the minimum polynomial for $\alpha$ over $\mathbb{Q}(w)$ is $x^3 - 5$, we would have $ [\mathbb{Q}(w, \alpha):\mathbb{Q}(w)] = 3 \implies |Gal(L|\mathbb{Q})| = 9$, but I have been told that $|Gal(L|\mathbb{Q})| = 6$.

Where is the problem in my reasoning above? Also, using the same logic, I found $Gal(L: \mathbb{Q}(w))$ to be $\{id, \sigma_1, \sigma_2\}$, where $\sigma_1(\alpha) = w\alpha, \sigma_2(\alpha) = w^2\alpha$. Is this wrong as well?

答: Since $x^3-1 = (x-1)(x^2+x+1)$ and $e^{2\pi i/3} \neq 1$, the minimal polynomial of $e^{2\pi i/3}$ over $\mathbb Q$ is $x^2+x+1$.

ContentLicense: CC BY-SA 4.0


问:
Construct a TM to show that {<G,s,t,k> | G is a directed graph with path length $\le$ k from s to t} is in class P
I would like to construct a deterministic TM that decides $L=${$<G,s,t,k>$ | $G$ is a directed graph that has a path of length at most $k$ from vertex $s$ to vertex $t$} in polynomial time.

So far, my construction is as follows.

$M=$"On input $<G,s,t,k>:$

For each path $p$ from $s$ to a vertex $v \in G$ of $k$ length:

------1.0 Clear all marks from vertices in $G$.

------1.1 Mark the vertices in $p$.

------1.2 Scan input. If $t$ is marked, accept.

Reject, since $t$ is not within $k$ steps from $s$."

(1.0), (1.1), and (1.2) are each $O(n)$, and $(2)$ is $O(1)$.

However, I think (1) is $O(n!)$.

What is a better approach that would make this $O(n^m)$ (i.e., polynomial)?

Edit:

The TM does not have to be "rigorously made".

For example, to find determine whether PATH={<G,s,t>| G is a directed graph that has a path from s to t} is in $P$, the following TM is acceptable:

M="On input <G,s,t>"

Mark vertex s.

Repeat until no additional vertex is marked:

------Scan all edges of G. If there is an edge from a marked vertex u to an unmarked vertex v, mark v.

If t is marked, accept. Otherwise, reject."

((1) is $O(1)$, (2) is $O(\#$of edges $*\#$of vertices$)=O(n^2*n=n^3)$, and (3) is $O(1)$).

答: Given a sequence of vertices $(s, v_1, \ldots, v_m, t)$, where $0 \leq m \leq k-1$, how many operations does it take to decide whether the sequence defines a path in $G$? How many such sequences are there?


问:
Interesting integral involving $\Gamma (z)$.
Find the value of $$\int_0^\infty t^{x-1}e^{-\lambda t \cos(\theta)} \cos(\lambda t \sin (\theta)) dt$$ where $\lambda >0$, $x>0$, and ${-1\over 2}\pi < \theta < {1\over 2}\pi$ in terms of the Gamma function.

I'm assuming this requires at least two different substitutions in order to evaluate it.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}t^{x - 1}\,\,
\expo{-\lambda t\cos\pars{\theta}}\,\,
\cos\pars{\lambda t\sin\pars{\theta}}\,\dd t}
\\[5mm] = &\
\Re\int_{0}^{\infty}t^{x - 1}\,\,
\expo{-\lambda t\cos\pars{\theta}}\,\,
\expo{\ic\lambda t\sin\pars{\theta}}\,\,\,\dd t
\\[5mm] = &\
\left.\Re\int_{0}^{\infty}t^{\color{red}{x} - 1}\,\,
\expo{-\lambda zt}\,\dd t\,\right\vert_{\,z\ =\ \exp\pars{-\ic\theta}}
\end{align}
Note that
$\ds{\expo{-\lambda zt} =
\sum_{k = 0}^{\infty}{\pars{-\lambda zt}^{k} \over k!} =
\sum_{k = 0}^{\infty}\color{red}{\pars{\lambda z}^{k}}\,\,{\pars{-t}^{k} \over k!}}$.

Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}t^{x - 1}\,\,
\expo{-\lambda t\cos\pars{\theta}}\,\,
\cos\pars{\lambda t\sin\pars{\theta}}\,\dd t}
\\[5mm] = &\
\Re\bracks{\Gamma\pars{\color{red}{x}}
\pars{\lambda z}^{-\color{red}{x}}}\quad
\pars{\substack{\ds{Ramanujan's}\\[1mm] \ds{Master}
\\[1mm] \ds{Theorem}}}
\\[5mm] = &\
\Gamma\pars{x}\lambda^{-x}\,\,
\Re\bracks{\pars{\expo{-\ic\theta}}^{-x}} =
\bbx{\Gamma\pars{x}\lambda^{-x}\cos\pars{\theta x}} \\ &
\end{align}

ContentLicense: CC BY-SA 3.0


问:
If $M$ is a real symmetric matrix such that $M^{k}=0$ for $k \in N$ then $M=0$.
My attempt :
$M$ is a real symmetric matrix so $PMP^{-1}=D$ for some $D$ being a $n \times n$ diagonal matrix.Now, $PM^kP^{-1}=D^k$.So we see that $M^k$ is also similar to a diagonal matrix and $D^k=0$, (as $M^k=0$)so $D=0$.If $D$ is a zero matrix then $M=P^{-1}DP$ from here we can conclude that $M=0$.

Is my reasoning OK?

答:
Your solution is absolutely correct.
You can see;

$D^k=\begin{bmatrix}\lambda_1^k &0&0\\ 0&\lambda_2^k&0\\0&0&\lambda_3^k\end{bmatrix}$.

Therefore $D^k=0 \iff D=0$(provided $D$ is a diagonal matrix).


问:
How to prove that $50^{44}+30! \cdot44^{50}-24 $ is divisible by $31$?
I am struggling with this example. This is a new topic for me. I would really appreciate some hints to understand it. I need part a) and b) to conclude the excercise.

I was thinking about Fermat, Euler or Wilson theorems but I am not sure if it is correct, of course I can use properties of congruences in number theory.

Part a): show that $50^{44}+30! \cdot 44^{50}-24 $ is divisible by $31$

Part b) : Conclude that $ (50^{44}+30!\cdot 44^{50})^{\varphi(31)} \equiv 1 \pmod{ 31}$

Body:
Considering that

$\quad 2^5 \equiv 1 \pmod{31} \land 50 \equiv 19 \pmod{31} \land 44 \equiv 13 \pmod{31}$

we calculate (looking for 'presentation loop closure'),

$\quad 13^2 \equiv 2^1 \cdot 7^1 \pmod{31}$

$\quad 19^2 \equiv 2^2 \cdot 5^1 \pmod{31}$

$\quad 7^2 \equiv 2^1 \cdot 3^2 \pmod{31}$

$\quad 5^3 \equiv 1 \pmod{31}$

$\quad 3^5 \equiv 2^1 \cdot 13^1 \pmod{31}$

and write

\begin{align}
50^{44} + 30! \cdot 44^{50} &\equiv 19^{44} - 13^{50} \pmod{31}, \text{ By Wilson's Theorem } \\
&\equiv 2^{44} \cdot 5^{22} - 2^{25}\cdot 7^{25} \pmod{31} \\
&\equiv 2^{4} \cdot 5^{1} -  2^{12} \cdot 3^{24} \cdot 7 \pmod{31} \\
&\equiv 2^{4} \cdot 5^{1} -  2^{12} \cdot 2^{4} \cdot 13^{4} \cdot 3^4 \cdot 7 \pmod{31} \\
&\equiv 2^{4} \cdot 5^{1} -  2^{12} \cdot 2^{4} \cdot 2^2 \cdot 7^2 \cdot 3^4 \cdot 7 \pmod{31} \\
&\equiv 2^{4} \cdot 5^{1} -  2^{12} \cdot 2^{4} \cdot 2^2 \cdot 2 \cdot 3^2 \cdot 3^4 \cdot 7 \pmod{31} \\
&\equiv 2^{4} \cdot 5^{1} -  2^{4}  \cdot 3^6 \cdot 7 \pmod{31} \\
&\equiv 2^{4} \cdot 5^{1} -  2^{4}  \cdot 2^1 \cdot 13^1 \cdot 3 \cdot 7 \pmod{31} \\
&\equiv 2^{4} \cdot 5^{1} -  3 \cdot 7 \cdot 13 \pmod{31} \\
&\equiv 80 + 130 \pmod{31} \\
&\equiv 24 \pmod{31}
\end{align}


问:
Proof $L{\rm{[}}\frac{{x(t)}}{t}{\rm{] = }}\int_s^\infty {X(u)du} $
I see that we usually use the theorem to solve the Laplace transform, however i want to proof the theorem, who could give me some details!!!

$L{\rm{[}}\frac{{x(t)}}{t}{\rm{] = }}\int_s^\infty  {X(u)du} $

where:
$L{\rm{[x(t)] = X(s)}}$

答:
Note that $\dfrac{\mathrm e^{-st}}t=\displaystyle\int_s^\infty\mathrm e^{-ut}\mathrm du$ hence, using Fubini theorem,
$$\int_0^\infty\mathrm e^{-st}\frac{x(t)}t\mathrm dt=\int_0^\infty x(t)\left(\int_s^\infty\mathrm e^{-ut}\mathrm du\right)\mathrm dt=\int_s^\infty\left(\int_0^\infty\mathrm e^{-ut}x(t)\mathrm dt\right)\mathrm du,
$$
that is, considering $y(t)=\dfrac{x(t)}t$ and $X=Lx$,
$$
(Ly)(s)=\int_s^\infty X(u)\mathrm du.
$$</p>


问:
Why is $1,x,x^2,x^3,\cdots$ a bad basis?

For functions of $x$, the first basis I would think of contains the powers $1, x, x^2 , x^3 , ...$. Unfortunately this is a terrible basis. Those functions $x^n$ are just barely independent. $x^{10}$ is almost a combination of other basis vectors $1, x, ... , x^9$. It is virtually impossible to compute with this poor "ill-conditioned" basis.

This is said in Page 426, Chapter 8-Linear Transformations, Introduction to Linear Algebra, Gilbert Strang, 5th Edition.

Why do we say that $1,x,x^2,x^3,\cdots$ is a bad basis ?

Taking the Wronkian
$$
\begin{vmatrix}x&x^2\\1&2x\end{vmatrix}=2x^2-x^2=x^2\\
\begin{vmatrix}x^2&x^3\\2x&3x^2\end{vmatrix}=3x^4-2x^4=x^4\\
\begin{vmatrix}x^3&x^5\\3x^2&5x^4\end{vmatrix}=5x^7-3x^7=2x^7
$$

Looks like linear independence is almost fine. So what does it mean to say it an ill-conditioned basis ?

And how does one show that "those functions $x^n$ are just barely independent" ?

答:
Just to add to angryavian's answer, if you are working in the space of functions defined over say, $[-1, 1]$, then what Strang likely refers to is the fact that the projection $p(x)$ of $x^{10}$ onto the space spanned by $\{1, x, x^2, \cdots, x^9\}$ is quite close to $x^{10}$ itself in a least-squares sense, namely that
$$\|p(x) - x^{10}\|_{L^2} = \int_{-1}^1 \left|p(x) - x^{10}\right|^2 dx$$
is small. We don't even need to calculate $p(x)$ exactly to verify this, because we can observe that even $x^8$ is a decent approximation to $x^{10}$ on the interval $[-1,1]$, as
$$\|x^8 - x^{10}\|_{L^2} \approx 0.0023$$
and we know that the square distance between $p(x)$ and $x^{10}$ is upper bounded by this value. And indeed, the distance between $x^k$ and the least squares approximation of $x^k$ restricted to the subspace spanned by $\{1, x, x^2, \cdots, x^{k-1}\}$ will tend to zero as $k \to \infty$. Intuitively, this means that the higher and higher $k$ goes, the closer and closer you get to adding a linearly dependent vector to your set.

A remark: even though I assumed for the sake of example that the polynomials were defined over the range $[-1, 1]$, this can be extended to any finite interval. Furthermore, this concept is not restricted to the canonical inner product $\langle f, g \rangle = \int f \bar g dx$. I suspect that under many inner products Strang's statement may hold in the way I've described.


答:
A polynomial time algorithm is for example a slightly modified breadth first search:

Start with $s$ and consider all neighbors. Then consider all the neighbor‘s neighbors etc. up to level $k$. If you found $t$ along the way (you might as well stop the algorithm in this situation) the graph $G$ is in your class, otherwise not.

If $\Delta$ denotes the maximal outgoing degree of $G$ this algorithm has a runtime bounded by $\Delta + \Delta^2 + ... + \Delta^k$ hence is in $\mathcal{O}(\Delta^k)$. This is of cause polynomial in the number of vertices $n$, the number of edges $m$ and the maximal degree $\Delta$, but it is exponential in $k$.

Title: Construct a TM to show that {<G,s,t,k> | G is a directed graph with path length $\le$ k from s to t} is in class P

Body:
I would like to construct a deterministic TM that decides $L=${$<G,s,t,k>$ | $G$ is a directed graph that has a path of length at most $k$ from vertex $s$ to vertex $t$} in polynomial time.

So far, my construction is as follows.

$M=$"On input $<G,s,t,k>:$

For each path $p$ from $s$ to a vertex $v \in G$ of $k$ length:

------1.0 Clear all marks from vertices in $G$.

------1.1 Mark the vertices in $p$.

------1.2 Scan input. If $t$ is marked, accept.

Reject, since $t$ is not within $k$ steps from $s$."

(1.0), (1.1), and (1.2) are each $O(n)$, and $(2)$ is $O(1)$.

However, I think (1) is $O(n!)$.

What is a better approach that would make this $O(n^m)$ (i.e., polynomial)?

Edit:

The TM does not have to be "rigorously made".

For example, to find determine whether PATH={<G,s,t>| G is a directed graph that has a path from s to t} is in $P$, the following TM is acceptable:

M="On input <G,s,t>"

Mark vertex s.

Repeat until no additional vertex is marked:

------Scan all edges of G. If there is an edge from a marked vertex u to an unmarked vertex v, mark v.

If t is marked, accept. Otherwise, reject."

((1) is $O(1)$, (2) is $O(\#$of edges $*\#$of vertices$)=O(n^2*n=n^3)$, and (3) is $O(1)$).


问:
Solution of an integral equation $\phi(x)+\int^1_0 xt(x+t)\phi(t)\,dt=x $ , $0 \le x \le 1 $
Solve the following integral equation:

$\phi(x)+\displaystyle \int^1_0 xt(x+t)\phi(t)\,dt=x $ , $0 \le x \le 1 $

I need to solve the integral equation above. Can anyone help me please?

答:
Related problems: (I). Here is a detailed solution that maybe someone benefits from it. Rearranging the equation as
$$ \phi(x)= x- \displaystyle \int^1_0 xt(x+t)\phi(t)\,dt \longrightarrow (1). $$

$$  \phi(x) = x - x^2\int_{0}^{1}t\phi(t)\,dt - x \int_{0}^{1}t^2\phi(t) \,dt $$

$$ \implies  \phi(x) = x - x^2 c_1 - x c_2 \longrightarrow (2).$$

Now, just subs back in the integral equation $(1)$ and compare the coefficients of $x's$, you will get a system of 2 equations in $c_1$ and $c_2$

$$ \frac{5}{4}c_{1}+\frac{1}{3}c_{2}= \frac{1}{3} $$

$$ \frac{1}{5}c_{1}+\frac{5}{4}c_{2}= \frac{1}{4}. $$

Solving the above system gives

$$ \left\{ c_{{1}}={\frac {80}{359}},c_{{2}}={\frac {59}{359}} \right\}.  $$

Subs back in $(2)$ yields the solution

$$ \phi \left( x \right) ={\frac {300}{359}}\,x-{\frac {80}{359}}\,{x}^{2
},$$

which can be checked by plugging back in $(1)$.


问:
Prove the following sequence is convergent: $a(1) = 1$ and $a(n+1) = 1/2 \cdot (a(n) + 3/a(n))$
I know the limit is $\sqrt 3$. And I tried proving that the sequence is decreasing from $a(2)$, yet when I got to this point, I have been stuck:

$a(n+1) - a(n) = (3 - a(n)^2) / 2a(n)$ ?

Thank you very much!

Body:
A more direct proof.

$s_{n+1}
=\dfrac12(s_n+\dfrac{a}{s_n})
$.

$\begin{array}\\
s_{n+1}^2
&=\dfrac14(a_n^2+2a+\dfrac{a^2}{s_n^2})\\
&=a+\dfrac14(s_n^2-2a+\dfrac{a^2}{s_n^2})\\
&=a+\dfrac14(s_n-\dfrac{a}{s_n})^2\\
&\gt a\\
\end{array}
$

so
$s_n > \sqrt{a}
$.

$\begin{array}\\
s_{n+1}-\sqrt{a}
&=\dfrac12(s_n-2\sqrt{a}+\dfrac{a}{s_n})\\
&=\dfrac12(\sqrt{s_n}-\dfrac{\sqrt{a}}{\sqrt{s_n}})^2\\
&=\dfrac1{2s_n}(s_n-\sqrt{a})^2\\
\dfrac{s_{n+1}-\sqrt{a}}{s_n-\sqrt{a}}
&=\dfrac1{2s_n}(s_n-\sqrt{a})\\
&\lt \dfrac12\\
\end{array}
$

so $s_n$ converges to
$\sqrt{a}$.


答:
The polynomial $x^3-5$ actually gives you all of $\Bbb Q(\sqrt[3]5, e^{2\pi i/3})$ by itself. It is irreducible, by Eisenstein, so $[\Bbb Q(\sqrt[3]5),\Bbb Q]=3$. However, the polynomial factors over $\Bbb Q(\sqrt[3]5)$ as
$$
(x-\sqrt[3]5)(x^2+\sqrt[3]5x+\sqrt[3]5^2)
$$
And it is the second-degree factor on the right that adjoins $\sqrt[3]5e^{2\pi i/3}$ and thus $e^{2\pi i/3}$ to $\Bbb Q(\sqrt[3]5)$. Which means that $[Q(\sqrt[3]5, e^{2\pi i/3}), Q(\sqrt[3]5)]=2$ (or technically, by its own, it only means ${}\leq 2$, but the equality is easily proven as we move from a field contained in $\Bbb R$ to one that isn't).

This happens in general: given any irreducible degree-$n$ polynomial, if you extend your field by adjoining one root of that polynomial, the polynomial will now factor into one linear term (given by the root you just adjoined) and one term of degree $n-1$ (which might even be reducible over the extended field). So the next root you adjoin will have strictly smaller degree. The Galois extension given by a degree $n$ polynomial can therefore at most have degree $n!$.

Title: Galois group of $\mathbb{Q}(\sqrt[3]{5}, e^{2\pi i/3})$ over $\mathbb{Q}$ and $\mathbb{Q}(e^{2\pi i/3})$

Body:
I want to find $Gal(\mathbb{Q}(\sqrt[3]{5}, e^{2\pi i/3}) | \mathbb{Q})$ and $Gal(\mathbb{Q}(\sqrt[3]{5}, e^{2\pi i/3})| \mathbb{Q}(e^{2\pi i/3}))$

Let $w = e^{2\pi i/3}, \alpha = \sqrt[3]{5}, L = \mathbb{Q}(\sqrt[3]{5}, w)$.

Since $L$ is the splitting field of $f(x) = x^3 - 5$, it is a finite, normal, seperable extension, and therefore $|Gal(L|\mathbb{Q})| = [L:\mathbb{Q}] = [\mathbb{Q}(w, \alpha):\mathbb{Q}(w)][\mathbb{Q}(w):\mathbb{Q}]$.

The minimal polynomial for $w$ over $\mathbb{Q}$ is $x^3 - 1$, so $[\mathbb{Q}(w):\mathbb{Q}] = 3$

I thought that since the minimum polynomial for $\alpha$ over $\mathbb{Q}(w)$ is $x^3 - 5$, we would have $ [\mathbb{Q}(w, \alpha):\mathbb{Q}(w)] = 3 \implies |Gal(L|\mathbb{Q})| = 9$, but I have been told that $|Gal(L|\mathbb{Q})| = 6$.

Where is the problem in my reasoning above? Also, using the same logic, I found $Gal(L: \mathbb{Q}(w))$ to be $\{id, \sigma_1, \sigma_2\}$, where $\sigma_1(\alpha) = w\alpha, \sigma_2(\alpha) = w^2\alpha$. Is this wrong as well?


问:
Sum of inverse of imaginary part of non trivial zeroes of Riemann zeta function
I would like to know if $$\sum_{k \geq 1} \frac{1}{\gamma_k}$$ is convergent or not. ($\gamma_k$ is the imaginary part of the k-th non trivial zero of $\zeta$).

In fact, I think this series is actually divergent, because of the "Explicit Formula" : $$\psi(x) = x-\sum_\rho \frac{x^\rho}{\rho} - \ln(2 \pi) - \ln(1-x^{-2})/2$$Where $\psi$ is the Chebyshev function.
So we have $$\sum_\rho \frac{x^\rho}{\rho} = x - \psi(x) - \ln(2\pi) - \frac{1}{2} \ln (1-x^{-2})$$But at $x=1$, the function $\ln(1-x^{-2})$ diverges. So the series $\sum_\rho 1/\rho$ must too. However, this "proof" may not be one because the "$\rho$" are $1/2+i \gamma_k$ (iff the RH is true) and not just $\gamma_k$.

Now, what I would like to know, is the value (if it converges) of $$\sum_{k \geq 1} \frac{1}{\gamma_k^s}$$For any $s \in \mathbb{C}$. I'm aware that this series propably diverges for $\Re(s) \leq 1$ but perhaps there is a way to extend this function analytically as in the case of Riemann's zeta function. Even if I can have an asymptotic approximation.

答:
If you have a look at this question, you will see an approximation
$$\Im\left(\rho _{n}\right)\sim \frac{2 \pi  n}{W\left(\frac{n}{e}\right)}$$ where $W(.)$ is Lambert function. So
$$\sum_{n=1}^\infty \frac 1 {\big[\Im\left(\rho _{n}\right)\big]^s}\sim\sum_{n=1}^\infty\left(  \frac{W\left(\frac{n}{e}\right)}{2 \pi  n}\right)^s\sim \sum_{n=1}^\infty \left(\frac{\log (n)}{n}\right)^s$$ which converges only if $s>1$.


问:
What is the PMF and CDF of the difference between two discrete random variables?
I have found a lot of references for the continuous case but I can't seem to find much information about how to find the PMF and CDF of the difference for two discrete r.v. for which we know their CDF and PMF.

答:
If $X$ and $Y$ are two independent discrete random variables, the PMF of $Z=X-Y$ is the convolution of $X$ and $-Y$:

$$P(Z=z)=\displaystyle\sum_{k=-\infty}^{\infty}P(X=k)P(Y=z-k)$$

Source: https://stats.stackexchange.com/a/450958/40817


问:
How find this $\int_{1}^{\infty}\frac{1}{x^2\sqrt{x^3-1}}dx$
find The integral
$$\int_{1}^{\infty}\dfrac{1}{x^2\sqrt{x^3-1}}dx$$
My try:let
$$\sqrt{x^3-1}=t\Longrightarrow x^3=t^2-1$$
Thank I can't,I think this answer maybe use Gamma integral</p>

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{1}^{\infty}{\dd x \over
x^{2}\root{x^{3} - 1}}} =
{1 \over 3}\int_{1}^{\infty}{\dd x \over
x^{4/3}\root{x - 1}}
\\[5mm] = &\
{1 \over 3}\int_{0}^{\infty}x^{\color{red}{1/2} - 1}\,\,
\pars{1 + x}^{-4/3}\,\,\dd x
\end{align}

$\ds{\pars{1 + x}^{-4/3} =
\sum_{k = 0}^{\infty}{-4/3 \choose k}x^{k} =
\sum_{k = 0}^{\infty}\bracks{{k + 1/3 \choose k}
\pars{-1}^{k}}x^{k} =}$

$\ds{\sum_{k = 0}^{\infty}\color{red}
{\Gamma\pars{4/3 + k} \over \Gamma\pars{4/3}}{\pars{-x}^{k} \over k!}}$.

Ramanujan's Master Theorem:
\begin{align}
&\bbox[5px,#ffd]{\int_{1}^{\infty}{\dd x \over
x^{2}\root{x^{3} - 1}}} =
{1 \over 3}\,\Gamma\pars{\color{red}{1 \over 2}}\,{\Gamma\pars{4/3 - \color{red}{1/2}} \over \Gamma\pars{4/3}}
\\[5mm] = &\
{1 \over 3}\,\,{\Gamma\pars{1/2}\Gamma\pars{5/6} \over \Gamma\pars{4/3}} =
\bbx{{1 \over 3}\on{B}\pars{{1 \over 2},{5 \over 6}}}
\approx 0.7468 \\ &
\end{align}
$\ds{\on{B}}$: Beta Function.


问:
Differentiability of $f(x) = x^2 \sin{\frac{1}{x}}$ and $f'$

Let $f(x) = x^2 \sin{\frac{1}{x}}$ for $x\neq 0$ and $f(0) =0$.

(a) Use the basic properties of the derivative, and the Chain Rule to show that $f$ is differentiable at each $a\neq 0$ and calculate $f'(a)$.

You may use without proof that $\sin$ is differentiable and that
$\sin' =\cos$.

Not even sure what this is asking.

(b) Show that $f$ is differentiable at $0$ and that $f'(0) =0$.

$\frac {f(x)-f(0)}{x-0} \to \lim_{x \to 0} x \sin(1/x)$.

$x \sin(1/x) \leq |x|$ and $\lim_{x \to 0} |x|=0$.

Thus $f(x)$ is differentiable at $0$; moreover $f^{'}(0)=0$.

(c) Show that $f'$ is not continuous at $0$.

$f{'}(x)=x^{2} \cos(1/x) (-x^{-2}) + 2x \sin (1/x)$.

In pieces: $\lim_{x \to 0} \cos (1/x)$.

$f^{'}(0-)$ nor $f{'}(0+)$ exists as $x \to 0$ $f^{'}(x)$ oscillates infinity between $-1$ and $1$ with ever increase frequency as $x \rightarrow 0$ for any $p>0$ $[-p,0]$, $[-p,p]$ or $[0,p]$ $f$ is not continuous.

Question: How to show more rigorously?

答:
Part (b). The function $f$ is differentiable at $0$ and has $f'(0)$ equal to the limit if the following limit exists: \begin{align} \lim_{x \to 0} \dfrac{f(x) - f(0)}{x-0} & = \lim_{x \to 0} \dfrac{f(x) - 0}{x}  & \textrm{ as } f(0) = 0 \\ & = \lim_{x \to 0} \dfrac{x^2 \sin\left(\frac{1}{x}\right)}{x} & \\
& = \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) & \end{align}

Now we can use the Squeeze Theorem. As $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$, we have that $$0 = \lim_{x \to 0} x \cdot -1 \leq \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) \leq \lim_{x \to 0} x \cdot 1 = 0$$

Therefore, $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0$ and we have $f'(0)=0$.


问:
Compact sets are closed.
I am a beginner in studying functional analysis, so please pardon me if my argument seems absurd. I am having hard time, coming in terms with the proof of "Compact subsets of $\mathbb{R}$, are closed". Now lets say that I have a set $ A = [1,2) $, and follow the proof. If I take the fixed point $x$ as $ {2} \in \mathbb{R} \backslash A $, then for each point $ y \in A$, I $\mathbf{can}$ find small disjoint neighborhoods for  $x$ and $y$. I am not sure if these small neighborhoods of $y$ cover $A$, but it is given that $A$ is compact. Or is it that I have to prove that these neighborhoods of $y$ doesn't cover $A$? Thanks

答:
$A$ is not compact:

Let $\alpha=\{(-1,1.1),(1,2-\frac{1}{n}) \mid \forall n \in \mathbb{N}\}$ be an open cover of $A$.

There is not any finite subcover of $\alpha$.

Moreover, in metric spaces, any compact set is closed and bounded, and Heine–Borel theorem says that in $\mathbb{R}^n$, any bounded and closed set is compact.

Edit: I tried to comment this but it is too long:

@Prashanth If I'm right, the proof's claim you sent is: A compact set $K$ in a Hausdorff topological space $X$ is closed. Which means that the complement of any compact set $K$ in a $T_2$ topological space $X$ is open. To see that, you have to show that $\forall x \in K^c$, there is an open neighbourhood $V$ such that $x \in V \subseteq K^c$ , (The definition of open set). and to show that, try to use the definition of compactness; the fact that any finite intersection of open sets is an open set; and the fact that $X \in T_2$ ($X$ is Hausdorff).

Hint: given a point $x \in K^c$ , because $X$ is Hausdorff, try to look at the (maybe infinite) union of the open sets $U_k\subseteq K$ such that  $k\in U_k \subseteq K$ such that $U_k \cap V =\emptyset$, when $x\in V\subseteq K^c$, and $V$ is an open set. Do this for all $x\in K^c$, for all $x\in K^c$ use the fact that $K$ is compact and see what you got.

Title: Compact sets are closed.


问:
Show that $n^4+4$ is a composite number

Show that if $n>1$, then the number $z=n^4+4$ is a composite number.

I only tried to analyze the parity of the number $n$: if $n$ is even then $z$ is even; if $n$ is odd then $z$ is odd.

But the second part didn't help me much, how can I proceed?

More generally, for which values of $k$ is the number $n^k+k$ composite?

答:
Observe that
\begin{align}
x^4+4=(x^2-2x+2)(x^2+2x+2).
\end{align}

Title: Show that $n^4+4$ is a composite number


问:
Hypergeometric distribution involving defective electronic components

A high-tech company receives a shipment of 100 sensitive electronic components from an electronic manufacturing firm.

This shipment will not be acceptable if six or more of the components are faulty. The high-tech company will randomly select a sample of M $\geq$ 1 components from this shipment and if there are no faulty components in this sample of size M, the shipment will be accepted.

What is the smallest value of M that ensures that the probability that the high-tech company will accept an unacceptable shipment is less than 10 percent?

I am stuck on trying to figure out what is p and n since it is not given? If someone can help get me further on this problem that would be great. Thank you.
So far I have gotten this far below:

X is a Hypergeometric (100,6,K)

X = # of defective in sample size K

p = rate of defectives

$P(A|U)=∑n=6100−K\frac{\binom{n}{0}\binom{100−n}{K}}{(100K)}P(N=n∣N≥6)$

$P(N=n|N≥6)=\frac{\binom{100}{n}p^n(1−p)^{100−n}}{\sum_{j=6}^{100}\binom{100}{j}(1−p)^{100−j}}$

答:
Your notation is strange and your reasoning is not clear.  Why, if the question uses $M$ for the number of components sampled, do you use $K$?  What does $p$ have to do with anything?  What do you mean by $P(A \mid U)$ when you have not defined $U$ or $A$?  Where does $6100$ come from?

If $M$ is the number of components sampled, there can be anywhere from $0$ to $M$ defects.  Let the random number of defects be $X$; then $$X \sim \operatorname{Hypergeometric}(n = 100, d, M)$$ where $d$ is the true number of defective items in the shipment.   We can easily calculate $$\Pr[X = 0 \mid d, M] = \frac{\binom{d}{0} \binom{100-d}{M-0}}{\binom{100}{M}} = \frac{\binom{100-d}{M}}{\binom{100}{M}}. \tag{1}$$  This is the conditional probability that, given $d$ defects, a sample of size $M$ will be defect-free and will therefore be accepted.  If $d \ge 6$, then the shipment is defective.  Note that the denominator of $(1)$ does not depend on $d$, and that, for a fixed choice of $M$, the probability of acceptance decreases with increasing $d$--this is intuitively obvious, since the more defects in your shipment, the less likely you are to pull a sample without any defects.  Since we want to ensure that the chance of erroneously accepting a defective shipment is less than $0.1$, we may select $d = 6$ as an upper bound:  $$\Pr[X = 0 \mid (d \ge 6) \cap M] \le \Pr[X = 0 \mid (d = 6) \cap M] = \frac{\binom{94}{M}}{\binom{100}{M}} = \prod_{i=0}^5 \left( 1 - \frac{M}{100-i} \right). \tag{2}$$  We want to choose the smallest $M$ such that the product in $(2)$ is less than $0.1$.  To save computational effort in guessing $M$, we have via the AM-GM identity
$$\prod_{i=0}^5 \left( 1 - \frac{M}{100-i} \right) \le \left( \frac{1}{6} \sum_{i=0}^5 1 - \frac{M}{100-i} \right)^6 = \left(1 - M \sum_{i=95}^{100} \frac{1}{i} \right)^6 \approx (1 - 0.0102596 M)^6.$$  Equating this with $0.1$ gives $M \approx 31.0645$, which furnishes our initial guess.  Now if we compute the exact probability for values around this guess, we have the table
$$\begin{array}{c|c}
M & \Pr[X = 0 \mid (d = 6) \cap M] \\
\hline
31 & \frac{681122}{6773025} \approx 0.100564 \\
32 & \frac{29614}{322525} \approx 0.0918192
\end{array}$$
Therefore, $M = 32$ is the least such $M$.


问:
Check the following inequalities with the Chebyshev inequality
Let $X \sim Poi(\lambda).$ Use the Chebyshev inequality to check the next inequalities:

$\mathbb{P}[X\leq{\frac{\lambda}{2}}] $   $\leq{\frac{4}{\lambda}}.$

$\mathbb{P}[X\geq{\frac{2}{\lambda}}] $  $\leq{\frac{1}{\lambda}}.$

I assumed that $X$ is non-negative, then as $\mathbb{E}(X) = \mathbb{V}ar(X) = \lambda$, I got $\mathbb{E}(X^2)$ and substituted in Chebyshev inequality but my inequalities weren't met. Help!

答:
For any random variable $X$ with $\operatorname{E}[X^2]<\infty $ and $c< \operatorname{E}[X]$ we have that

$$
\begin{align*}
\Pr [X\leqslant c]&=\Pr [-X\geqslant -c]\\
&=\Pr [\operatorname{E}[X] -X\geqslant \operatorname{E}[X] -c]\\
&\leqslant \Pr [|X-\operatorname{E}[X]|\geqslant \operatorname{E}[X] -c]\\
&\leqslant \frac{\operatorname{Var[X]}}{(\operatorname{E}[X] -c)^2}
\end{align*}
$$

Then if $X\sim\operatorname{Poisson}(\lambda )$ and $c=\lambda /2$ we find that $\Pr [X\leqslant \lambda /2]\leqslant 4/\lambda $, as desired. For the second case it seems that there is a typo, it probably must be $\Pr [X\geqslant 2\lambda ]\leqslant 1/\lambda $ instead of the given inequality (by example taking $\lambda =2$ we can see that the given inequality doesn't hold as $\Pr [X\geqslant 1]>3/4$).


答:
For the $k=4$ case, one can simply note that

$$n^4 + 4 = (n^2 - 2n + 2)(n^2 + 2n + 2)$$

The motivation is realizing that you have something that looks like it could factor as two factors of $n^2 + ?n + 2$, since $(n^2)^2 = n^4$ and $2^2 = 4$, and trying to determine what that unknown quantity is. One observes that, letting that quantity be variables $b,c$ in two factors, that

$$(n^2 + bn + 2)(n^2 + cn + 2) = n^4 + n^3 (b+c) + n^2 (bc + 4) + n(2b+2c) + 4$$

So one requires that

$$\begin{cases}
b+c = 0 \\
bc + 4 = 0 \\
2b + 2c = 0
\end{cases}$$

though the latter equation is redundant. So you end up having two numbers whose sum is zero, and whose product is $-4$. Some trial and error readily let's one see that $b = -2$ and $c=2$. (Or I suppose you could try substitution or another method.)

That said, I'm not really sure about generalizations on the $n^k + k$ case.

Obviously $k=1$ is a dud. Take $n=6$. ($7$ is prime.)

In the case $k=2$, take $n=3$. ($11$ is prime.)

In the case $k=3$, take $n=2$. ($11$ is prime.)

In the case $k=5$, take $n=2$. ($37$ is prime.)

In the case $k=6$, not sure. (Nothing sticks out as one to me with small cases.)

In the case $k=7$, take $n=4$. ($16,391$ is prime.)

In the case $k=8$, take $n=3$. ($6,569$ is prime.)

In the case $k=9$, take $n=2$. ($521$ is prime.)

In the case $k=10$, not sure. (Nothing sticks out as one to me with small cases.)

So I know it's true for $k=4$, and maybe $6$ and $10$, but I'm not sure.

Not really up for trying more cases, but I'm not sure of a pattern offhand.

Title: Show that $n^4+4$ is a composite number

Body:
Show that if $n>1$, then the number $z=n^4+4$ is a composite number.

I only tried to analyze the parity of the number $n$: if $n$ is even then $z$ is even; if $n$ is odd then $z$ is odd.

But the second part didn't help me much, how can I proceed?

More generally, for which values of $k$ is the number $n^k+k$ composite?


问:
Prove that middle point of $PQ$ is a circle
Point $A$ is one of the points of intersection of two given intersecting circles. Any line is drawn through $A$ to cut the circles again in $P$ and $Q$. Prove that the locus of middle point of $PQ$ is a circle.

My Attempt:

I was able to obtain the result with difficulty through analytic geometry but is there a geometrical solution to the problem

答:
OK, let me give a quick analytic proof of my claim in the comments.

Choose polar coordinates with $A$ being the origin.  The general equation of circles through $A$ is $r=d\cos(\theta-\theta_0)$ where $d$ is the diameter and $\theta=\theta_0$ is the half-ray containing the diameter through $A$.

So let $r=d_i\cos(\theta-\theta_i)$, $i=1,2$ be the two circles.  For any $\lambda\in\mathbb{R}$, the locus of $P_\lambda$ is therefore $$r=(1-\lambda) d_1\cos(\theta-\theta_1)+\lambda d_2\cos(\theta-\theta_2)$$
which yields the Cartesian version $(1-\lambda)S_1+\lambda S_2=0$.


问:
$\int \frac{(e^z)}{z-\pi i} dz$, if C is the ellipse |z - 2| + |z+2| = 6
$$\oint_{C} \frac{e^z}{z-\pi i}\ \mathrm dz$$
If $C$ is the ellipse $|z - 2| + |z+2| = 6$, how is it $0$? I can find the singular point in $2$ and $-2$.

答:
First, consider the set $S=\{z\in\mathbb{C}:|z-2|+|z+2|<2\sqrt{13}\}$. This set can be shown to be simply connected.

Let $z\in C$

$\Rightarrow z\in\mathbb{C}$

Suppose $z-\pi i=0$

$\Rightarrow z=\pi i\Rightarrow |z-2|+|z+2|=|\pi i-2|+|\pi i+2|=2\sqrt{\pi^{2}+4}>2\sqrt{9+4}=2\sqrt{13}\Rightarrow z\notin C$

Thus by the contrapositive, $z\in C\Rightarrow z-\pi i\neq 0$

$\Rightarrow z-\pi i\neq0$

Thus, $\frac{e^z}{z-\pi i}$ is holomorphic on $S$.

Let $z$ be in the curve.

$\Rightarrow |z-2|+|z+2|=6=2\sqrt{9}<2\sqrt{13}$

$\Rightarrow z\in S$

Thus, the curve is inside $S$, and as $\frac{e^z}{z-\pi i}$ is holomorphic on $S$, so $\oint_C {\frac{e^z}{z-\pi i}}\, {dz}=0$


问:
Can conservative systems have unstable equilibrium points?
In general, can conservative systems have a equilibrium(fixed) point which is unstable?
If so, I don't know why so in spite of these systems have conserved quantity.</p>

答:
It depends on what quantity is conserved. Is the conserved quantity is a positive definite function of the state variables, then the answer is no. In general, yes, they can. Consider, for example, the system
$$
\dot x= y\qquad \dot y= x-x^3.
$$
It has a conserved quantity $E(x,y)=\frac12 y^2-\frac12 x^2+\frac14x^4$ and the origin is unstable:


问:
Proof for exponentiation in modular arithemetic
I have found out, that the following is true for modular arithmetic when $t$ is a natural number.

$$a^t \bmod\ n \equiv (a\bmod\ n)^t\bmod\ n$$

But I have been unable to find a proof for this, does anyone have a source that proves this conjecture?

答:
We want to prove $$a^t \bmod\ n \equiv (a\bmod\ n)^t\bmod\ n.$$
Let $ a \bmod n \equiv b $.

Thus by substitution we are proving $$a^t \bmod\ n \equiv b ^t\bmod\ n.$$
But this means we need to show $ n \mid a^t - b^t $ under the assumption that $ n \mid a - b $ .

Then using the fact that $a^t-b^t=(a-b)(a^{t-1}+a^{t-2}b+\cdots+ab^{t-2}+b^{t-1})$


问:
$I$ a maximal ideal in ring $R$ if and only if for all $a \in R\setminus I, \exists r\in R, b \in I$ such that $b+ar=1$.
Let $R$ be a commutative Ring with $1\neq 0$. Let $I$ be a proper ideal of $R$. Show that $I$ is maximal $\iff$ for all $a \in R\setminus I$ there are $r\in R$ and $ b \in I$ such that $b+ar=1$.

I'm not sure how to approach proving this.

答:
It is well-known that if

$I \subset R \tag 1$

is a maximal ideal, then $R/I$ is a field; and thus if

$a \in R \setminus I, \tag 2$

we have

$a \ne 0 \mod I \tag 3$

which implies $a + I$ has an inverse in $R/I$; that is, there is some

$r \in R \tag 4$

with

$ar = 1 \mod I, \tag 5$

or

$ar - 1 \in I; \tag 6$

this of course implies that

$1 - ar = -(ar - 1) \in I, \tag 7$

and thus

$\exists b \in I \; 1 - ar = b, \tag 8$

from which we readily infer that

$ar + b = 1. \tag 9$

Now suppose that for every $a$ as in (2) there are

$r \in R, \; b \in I \tag{10}$

such that

$ar + b = 1 \tag{11}$

or

$ar - 1 = -b \in I, \tag{12}$

whence

$ar = 1 \mod I; \tag{13}$

this implies that

$r \in R \setminus I \tag{14}$

lest

$r = 0 \mod I \tag{15}$

in contradiction to (13), which now implies $r + I$ is the inverse to $a + I$ in $R/I$, which is thus seen to be a field, implying $I$ is maximal in $R$. $OE\Delta$.


问:
Coordinates on 3d Circle
Problem:

Given is is the centerpoint $C$ of a circle and two extra reference points $R$ and $E$, which are laying on the circle.
While as the line $\overline{CE}$ is defining the zero point on the circle of a given degree $α$ (clockwise rotation). Point $F$ is also on the very same circle. Therefore all of the points are on one in space rotated plane, on the circle.
Line $\overline{CF}$ is $α$ degrees away from line $\overline{CE}$. How can I get the XYZ coordinates of Point $F$?
($R ≠ E$)
Fig. 1

Fig. 1: Illustration of described problem

Note that $\overline{CR}$ is NOT necessarily parallel to $\overline{CF}$

$\overline{CR}$ to $\overline{CE}$ not necessarily orthogonal

all Points lay in one 3d circular plane

in Fig. 1 a second circle is drawn orthogonal to the XY Plane just for better relation understanding (not a projection)

The points $E$, $C$ and $R$ are defining the circles rotation in space, while $F$ is the point to move around on this circle (with a specific degree $\alpha$)

答:
I am going to answer how such a problem can be solved in a different way:

All lines to the points must be converted to vectors. A vector axis perpendicular to the circles face must be created. Then the vector of $CE$ is taken as the vector to rotate around the created axis vector, by theta degrees(counter clockwise rotation). This rotation and the vector of the new position can be calculated with the Rodrigues' rotation formula!
See also: Quaternion rotation

Title: Coordinates on 3d Circle


问:
How to prove the existence of infinitely many $n$ in $\mathbb{N}$,such that $(n^2+k)|n!$

Show there exist  infinitely many $n$ $\in \mathbb{N}$,such  that
  $(n^2+k)|n!$ and  $k\in N$

I have a similar problem:

Show that there are infinitely many $n \in \mathbb{N}$,such  that
  $$(n^2+1)|n!$$

Solution: We consider this pell equation,$n^2+1=5y^2$,and this pell equation have
$(n,y)=(2,1)$,so this equation have infinite solution$(n,y)$,and
$2y=2\sqrt{\dfrac{n^2+1}{5}}<n$.

so $5,y,2y\in \{1,2,3,\cdots,n\}$,
so $5y^2<n!$
then $(n^2+1)|n!$
 but for $k$ I have consider pell equation,But I failed,Thank you everyone can help

答:
Similar to your solution of $k=1$.

Consider the pell's equation $n^2 + k = (k^2+k) y^2$. This has solution $(n,y) = (k,1)$, hence has infinitely many solutions. Note that $k^2 + k = k(k+1) $ is never a square for $k\geq 2$, hence is a Pell's Equation of the form $n^2 - (k^2+k) y^2 = -k$.

Then, $2y = 2\sqrt{ \frac{ n^2+k} { k^2 +k } } \leq n$ (for $k \geq 2$, $n\geq 2$) always.


问:
Ratio of obtuse triangles to acute triangles in the square.
I came across OEIS sequences A190020 and A190019, and I noticed that they seemed to grow at a similar rate.

A190020: Number of obtuse triangles on a (n X n)-grid (or geoboard)

A190019: Number of acute triangles on a (n X n)-grid (or geoboard)

In particular, I'm interested in their ratio, which gives a sense of the relative frequency of obtuse vs acute triangles when points are chosen in a square.

Empirically, it looks like $\displaystyle \lim_{n \rightarrow \infty}\frac{A190020(n)}{A190019(n)} \approx 2.6391$. (See graph of the ratio below.)

What is a heuristic for why the limit appears to be finite and greater than 1?

How do you prove a finite upper bound for this ratio?

How do you prove a nonzero lower bound for this ratio?

Does this limit converge? Is it known what it converges to?

If three points are selected uniformly at random to form a triangle $\Delta$ in the unit square $[0,1] \times [0,1]$, does $$\frac{P(\Delta \text{ is obtuse})}{P(\Delta \text{ is acute})} = \lim_{n \rightarrow \infty}\frac{A190020(n)}{A190019(n)}?$$

答:
I have an outline for the calculation.  It is a six-fold integral, for the six coordinates of the triangle.  I think the six-fold integral can be evaluated to a single integral, which has several dozen elementary terms.  I would need Maple or Mathematica to finish this.

Let $S$ be the unit square $[0,1]\times[0,1]$.

Let $A$ and $B$ have coordinates $A(x,y)$, and $B(x+r\cos\theta,y+r\sin\theta)$

Let the ray $AB$ meet the edge of $S$ at $F$.  Let the line through $A$, perpendicular to $AB$, meet the square's edges at $D$ and $E$.

The probability $P$ that the triangle is obtuse is three times the probability that angle $\angle BAC$ is obtuse.  In the following formula, $I_{BAC \text{ is obtuse}}$ is $1$ if the angle at $A$ is obtuse, and $0$ otherwise.

$$P = 3\iint_S da\iint_S db \iint_S dc\text{  } I_{BAC \text{ is obtuse}}$$

The integral over $c$ is the area bounded by $S$ and $DE$, on the opposite side from $B$.  This has a dozen formulas, depending on which edges $D$ and $E$ are on.  Let $T(x,y,\theta)$ be this area.  The specific formula for the picture is $\frac12(x+y\tan\theta)(y+x\cot\theta)$

$$T(x,y,\theta)=\iint_S dc \text{  }I_{BAC\text{ is obtuse}}\\
P = 3\iint_Sda\iint_Sdb\text{  } T(x,y,\theta)$$

One can integrate over $b$ using polar coordinates centered at $A(x,y)$.  Since $T$ is independent of $r$, this can be integrated out
$$\iint_S db T(x,y,\theta)=\int dr \int d\theta \text{  }rT(x,y,\theta)\\
= \int d\theta\frac12(AF)^2 T(x,y,\theta)\\
P = 3\iint_S da \int_0^{2\pi}d\theta \frac12(AF)^2 T(x,y,\theta)$$
As $\theta$ runs through $2\pi$, there are twelve different formulas for $(AF)^2T(x,y,\theta)$, depending on which sides $D$, $E$ and $F$ are on.  All twelve have an elementary integral.  We can thus reach a formula
$$U(x,y) = \int_0^{2\pi}d\theta\frac12(AF)^2 T(x,y,\theta)\\
P = 3\iint_S \text{ }dx\text{ } dy\text{  }U(x,y)$$

$U(x,y)$ has several dozen terms, involving $x^i(1-x)^jy^k(1-y)^l$ for exponents that run from $-2$ to $4$.  It also has terms involving $\log x,\log (1-x),\log y$ and $\log(1-y)$.  The log terms are multiplied by polynomials in $x$ and $y$; the exponents for those terms are not negative.

$U(x,y)$ has a different form if $A$ is in the yellow region, or orange region; and fourteen other formulas in the rest of the square.  By symmetry, we just need to integrate over the yellow and orange regions, and multiply by eight because of the square's symmetry.

Each formula of $U(x,y)$ must be integrated over the appropriate region.  Suppose $U_1(x,y)$ is the formula for the yellow region and $U_2$ for the orange region.

$$P = 24\int_0^{1/2}dx\int_{1/2-\sqrt{1/4-x^2}}^xdy \text{  }U_1(x,y) + \\
24\int_0^{1/2}dx\int_0^{1/2-\sqrt{1/4-x^2}}dy\text{  } U_2(x,y)  $$

I think the $y$ integrals have elementary form, but since the limits are on the circular arc, I haven't found an elementary form for the final $dx$ integral.


问:
Uncorrelated but not independent random variables.

Let $X$ and $Y$ be random variables with the following joint distribution in $[0, 1]$ × $[0, 1]$: $f (x, y) = 0$ in the white areas, $f (x, y) = 2$ in the black areas.  These two variables are uncorrelated but are not independent?

答:
They are indeed not independent as you stated. We know that two random variables $X, Y$ (for which the probability densities $f_X$, $f_Y$ and the joint probability density $f_{X,Y}$ exist) are independent iff:

$f_{X,Y}(x, y) = f_X(x)f_Y(y)$ for all $x, y$

An here it is easy to find a counterexample:

$f_{X,Y}(1/3, 1/3) = 0$

$f_X(1/3) = 1$, $f_Y(1/3) = 1$

So the condition for the variables to be independent does not hold.

However when we are talking about correlation we are usually referring to the Pearson Correlation Coefficient which measures $linear$ relationship between two random variables. Here there is no linear relationship between the two variables, so the correlation is $0$. You can see that in the bottom of the following picture some examples of obviously dependent random variables (like in your example), where the correlation is $0$.

In real-world examples the Pearson Coefficient works well in many cases. For when it doesn't, there are some other correlation measures such as Spearman's rank coefficient, Kendall rank coefficient and others that can be used (although there is no guarantee that they will always manage to point out the existence of a relationship between your variables).

Edit:

As user asked, I am explaining how the Pearson Coefficient can be calculated here. The formula is:

$ρ_{X, Y} = \frac{cov(X, Y)}{σ_Xσ_Y} = \frac{E[XY] - E[X]E[Y]}{σ_Xσ_Y}$

Here the most parts of the formula can be easily calculated:

$E[X] = E[Y] = 1/2$

$σ_X = σ_Y = \sqrt{1/12}$

The only difficult part is the calculation of $E[XY]$:

$E[XY] = \int_0^1 \int_0^1 xy f_{XY}(x, y) dx dy$

The problem here is that $f_{XY}$ isn't continuous in $[0, 1] \times [0, 1]$, so to calculate this integral we need to calculate the eight integrals for the eight areas where the probability density function is non-zero, and add them:

$\int_0^1 \int_0^1 xy f_{XY}(x, y) dx dy = \\
\int_{1/4}^{1/2} \int_0^{1/4} xy f_{XY}(x, y) dx dy +
\int_0^{1/4} \int_{1/4}^{1/2} xy f_{XY}(x, y) dx dy + \\
\int_{1/4}^{1/2} \int_{1/2}^{3/4} xy f_{XY}(x, y) dx dy +
\int_{1/2}^{3/4} \int_{1/4}^{1/2} xy f_{XY}(x, y) dx dy + \\
\int_{0}^{1/4} \int_{3/4}^{1} xy f_{XY}(x, y) dx dy +
\int_{3/4}^{1} \int_{0}^{1/4} xy f_{XY}(x, y) dx dy + \\
\int_{1/2}^{3/4} \int_{3/4}^{1} xy f_{XY}(x, y) dx dy +
\int_{3/4}^{1} \int_{1/2}^{3/4} xy f_{XY}(x, y) dx dy
$

Here you can notice that due to the symmetry in your PDF, the two integrals in each of the lines have the same value, so you only need to calculate four integrals instead of eight. Perhaps there are smarter things you can do to calculate the original integral with less sweat, but I can't think of anything else.


问:
Can you express this as two summations?
$$\sum_{i=0}^{\infty} \left(\frac{x}{1+r}\right)^{i} = \ ?$$

Can you write this as:

$$\sum_{i=0}^{\infty} (1+r)^{-i}\  \times\ \sum_{i=0}^{\infty} x^{i} \ $$

This is probably wrong, but in any case, how would I evaluate the summation?

答:
The Cauchy product of infinite series gives us that

$$\left( \sum_{i=0}^\infty a_i \right) \left( \sum_{j=0}^\infty b_j \right) = \sum_{k=0}^\infty c_k$$

where

$$c_k = \sum_{\ell=0}^k a_\ell b_{k-\ell} = \sum_{\substack{i+j = \ell \\ \ell = k}} a_i b_j$$

So essentially taking the product of two series gives you summation, with a finite summation inside. You can see why that might not work for your method of factorization.

That said, your series appears to be a geometric series with ratio $R = x/(1+r)$, provided $x,r$ are such that $|R|< 1$. If so, then note that

$$\sum_{i=0}^{\infty} \left(\frac{x}{1+r} \right)^{i} = \sum_{i=0}^\infty R^i = \frac{1}{1-R}$$

Of course, if $x,r$ are such that $|R| \ge 1$, then the summation diverges.

Title: Can you express this as two summations?


答:
Hint
$$z=n^4+4=n^4+4n^2+4-4n^2$$
is the difference of two squares.

Title: Show that $n^4+4$ is a composite number


问:
An integral problem?
How do you integrate $e^{e^x}$? I was able to get it down to du/(ln u) but I wasn't able to go further. Thanks!</p>

答:
No, not Calculus AB level.  This antiderivative is "not elementary" in the technical meaning of that term.  https://en.wikipedia.org/wiki/Elementary_function

added

Maple says
$$
\int \!{{\rm e}^{{{\rm e}^{x}}}}{dx}=-{\rm Ei}_1 \left(-{{\rm e}^{x}}
 \right)
$$
where this "exponential integral" function is
$$
\mathrm{Ei}_1(z) = \int_1^\infty\frac{e^{-tz}}{t}\;dt
$$

Title: An integral problem?


问:
Prove that a polynomial function is a linear transformation
I've only seen examples of proving linear transformation of matrices from our module but none of this kind so i'm lost.

Say given the function:
$$ L: R_{2}[t] \rightarrow R_{1}[t]$$
$$ at^{2} + bt + c \rightarrow at + 2c - b $$

We are asked to prove that L is a linear transformation but I don't really know where to start. Could someone point me in the right direction?

答:
To show a transformation $T:V \to V$ on a vector space $V$ over a field $F$ is linear, what you do is take $\alpha v + \beta w$, where $\alpha,\beta \in F$ and $v,w \in V$ (and thus $\alpha v + \beta w \in V$), and show that

$$T(\alpha v + \beta w) = \alpha T(v) + \beta T(w)$$

Thus, in your case, take two polynomials, $p_i(t) = a_i t^2 + b_i t + c_i \in \Bbb R_2[t]$ for $i = 1,2$. Then, for scalars $\alpha_i \in \Bbb R$, $i=1,2$, show that

$$L(\alpha_1 p_1(t) + \alpha_2 p_2(t) ) = \alpha_1 T(p_1(t)) + \alpha_2 T(p_2(t))$$


问:
What are the constant solutions for separation of variables used for?
In my calc $2$ textbook there is an initial-value problem that requires the use of separation of variables.

The differential equation is $\frac{dy}{dx} = \left(2x+3\right)\left(y^2-4\right)$ with initial value $y(0)=1$.

In the list of steps to take to solve this equation using separation of variables, the very first step is "Check for any values of $y$ that make $g(y) = 0$. These correspond to constant solutions". When I solve $g(y)=0$ I get $g(y) = \pm 2$, but I don't really understand how these values are used to solve this equation?

答:
You need to solve an initial value problem so you need:

Solve the ordinary differential equation.

Use the initial condition for to find the constants.

For example, in your problem you have the following initial value problem (an ordinary differential equation+ initial condition).

\begin{eqnarray*}
(\mathbf{IVP}): \left\{\begin{aligned} \frac{dy}{dx}=(2x+3)(y^{2}-4), \\ y(0)=1 .\end{aligned} \right.
\end{eqnarray*}

Now, first we solve the ordinary differential equation (ODE) $\displaystyle \frac{dy}{dx}=(2x+3)(y^{2}-4)$ indeed, this's a separate ODE so we have $$ \frac{dy}{dx}=(2x+3)(y^{2}-4) \implies \frac{1}{y^{2}-4}dy=(2x+3)dx \implies \int \frac{1}{y^{2}-4}dy=\int (2x+3)dx$$
note that  that
\begin{eqnarray*}
\int \frac{1}{y^{2}-4}dy&=&\int \frac{1}{(y-2)}\frac{1}{(y+2)}dy\\
&=&\int \frac{A}{y-2}dy+\int \frac{B}{y+2}dy\\
&=&\frac{1}{4}\int \frac{1}{y-2}dy-\frac{1}{4}\int \frac{1}{y+2}dy
\end{eqnarray*}
since that  $$1=A(y+2)+B(y-2) \implies \left\{\begin{aligned} y=-2 \implies 1=-4B \implies B=-\frac{1}{4}\\ y=2 \implies 1=4A \implies A=\frac{1}{4}\end{aligned} \right.$$
so, we have $$\int \frac{1}{y^{2}-4}dy=\frac{1}{4}\ln|y-2|-\frac{1}{4}\ln|y+2|+c$$
Therefore
$$\int \frac{1}{y^{2}-4}dy=\int (2x+3)dx \implies \boxed{\frac{1}{4}\ln|y-2|-\frac{1}{4}\ln|y+2|=x^{2}+3x+\ln|C|}$$
2. Using the initial condition, since $y(0)=1$, we have $$ \frac{1}{4}\cdot 0-\frac{1}{4}\cdot 0 =\ln|C| \iff \ln|C|=0 \implies |C|=e^{0}=1 \implies \boxed{C=\pm 1}$$
3. So, the solution for the initial value problem is $$\boxed{\frac{1}{4}\ln|y-2|-\frac{1}{4}\ln|y+2|=x^{2}+3x \pm 1}$$


问:
$ABCD$ is a trapezium where $AB$ || $CD.$ Let $AB = b$ , $CD = a$ where $a < b$. Let $S$ be the area of trapezium $ABCD$.

$ABCD$ is a trapezium where $AB$ || $CD.$ Let $AB = b$ , $CD = a$ where $a < b$. Let $S$ be the area of trapezium $ABCD$. It is given that $[\Delta BOC] = \frac{2S}{9}$. Find the value of $\frac{b}{a}$.

Related Question :- $ABCD$ is a trapezium where $AB$ $||$ $CD$. Suppose $[\Delta BOC] = \frac{25}{9}$, and let $AB = b$, $CD = a$ where $a < b$.

Note:- The real fact is that I copied the initial question wrong, saying that $[\Delta BOC] = \frac{25}{9}$ and there were no unique trapeziums of that kind, so here is the real question and sorry for the inconvenience.

What I Tried: Here is a picture :-

$ [\Delta BOC] = [\Delta AOD]$ follows from carpet strategy, so :-
$$ [\Delta DOC] + [\Delta AOB] = \frac{5S}{9}$$

We have that $\Delta DOC \sim \Delta BOA$ , which might help find $\frac{b}{a}$, but I couldn't understand how I can use it. One way is to put variables for $DO$ , $OC$ , $AO$ and $OB$ , but that will make it more complicated and I don't want to do  it.

Another thing I thought of was to draw the perpendicular from $O$ , touching $AB$ and $CD$ at $M$ and $N$ , suppose. And let $[\Delta DOC] = x$ . I will get :-

$$\rightarrow \frac{1}{2} * ON * a = x$$
$$\rightarrow \frac{1}{2} * OM * b = \frac{5S}{9} - x$$

But that still deals with a variable $x$ , how do I fix it?

Can anyone help me? Thank You.

答:
$$\frac{DO}{OB}=\frac{ar(\Delta DOC)}{2S/9}=\frac{a}{b}={\left(\frac{ar(\Delta DOC)}{ar(AOB)}\right)}^{1/2}$$

let $ar(DOC)=p,ar(AOB)=q$ then by above $$pq=\frac{4S^2}{81}$$ then use $$p+q+4S/9=S$$ from here $p=?,q=?\Rightarrow \frac{a}{b}=?$


答:
Take two polynomials $at^2+bt+c$ and $a't^2+b't+c'$ and take the image by $L$. You get:

\begin{align*}
L((at^2+bt+c)+(a't^2+b't+c'))
&=L((a+a')t^2+(b+b')t+(c+c'))\\
&=(a+a')t+2(c+c')-(b+b')\\
&=(at+2ct-b)+(a't+2c'-b')
\end{align*}

This is $L$ of the first polynomial + $L$ of the second polynomial. Same when multiplying a polynomial by a constant $\alpha$.

Title: Prove that a polynomial function is a linear transformation

Body:
I've only seen examples of proving linear transformation of matrices from our module but none of this kind so i'm lost.

Say given the function:
$$ L: R_{2}[t] \rightarrow R_{1}[t]$$
$$ at^{2} + bt + c \rightarrow at + 2c - b $$

We are asked to prove that L is a linear transformation but I don't really know where to start. Could someone point me in the right direction?


问:
Could a "fuzzy", or non-symmetric, relationship change the size of a power set?
Let's say we have a set of numbers called the Unnatural Numbers (UN). These numbers have the following properties.

They're equinumerous with the natural numbers

They have no defined position

They have no defined value

The well ordering principle doesn't apply

They have just enough "distinctness" that we can add n UNs to a set and have the set contain n members

I think of them like a countably infinite pile of absolutely identical grains of sand.

The power set of the UNs seems to be countably infinite, which is the same size as the set of UN. This seems to be because the relationships between the UNs are low. My questions are:

What am I getting wrong?

Can power sets be used as a measure of how many relationships there are or is there a better way to do it?

Can relationships be made "fuzzy", or non-symmetrical, and what does that do to power sets?

Regarding (3), there seems to be no issue numerically, but I can't really work out what it means. Does a "fuzzy" relationship only carry certain properties, would this mean that subsets we think of as distinct would now be the same, would the power set only change under certain conditions, etc.

I may be talking complete nonsense with this, but it's been fun thinking about what "fuzziness" could mean.

答:
I know this is a zombie response to a two-year-old question, but it seems to me that the OP here reinvented/rediscovered the concepts of cardinality and equivalence classes, restricted to the set of natural numbers $\Bbb{N}_0 := \{0, 1, 2, 3, ...\}$. From OP's comments:

as I see it a subset that contains 3 members is exactly the same as every other member that contains 3 members, which means there's only 1 subset for each number of members, and only 1 infinite subset.

In other words, OP is really suggesting that 2 subsets of the "Unnatural Numbers" are to be considered "the same" if they can be paired off in a 1-1 correspondence. This is functionally equivalent to taking the subsets of the regular natural numbers $\Bbb{N}_0$ and forgetting the (labels/values/numerical properties) on the elements of any subset.

If we do this, then the only way to tell two subsets of $\Bbb{N}_0$ apart is if they have different numbers of elements. Contrapositively, this is the same as saying two subsets of $\Bbb{N}_0$ are "the same" if they have the same number of elements (i.e. there's a bijection between them). "Indistinguishability", in this sense, turns out to define what is usually referred to as an equivalence relation $\sim$ on the subsets of $\Bbb{N}_0$. Although OP specifies "fuzziness" to mean non-symmetric in their post, their example of "fuzziness" is more like the "forgetfulness" one gets from an equivalence relation--we can indeed check that "indistinguishability" is an equivalence relation on subsets of $\Bbb{N}_0$, which actually means it's symmetric (as well as reflexive and transitive). And the names of these equivalence classes are exactly the finite and countable cardinalities, because "cardinal numbers" are exactly what one gets by "forgetting" the labels, or interpretations, or relationships, between the elements of a set, and just keeping its "size".

(The process of "forgetting" some or all irrelevant aspects of the structure on a highly structured set is very common in math and plays a role in foundations: see, for instance, forgetful functors. I suppose this could be considered another topic the OP 'discovered' for themselves, in a particular context.)


答:
I showed in my answer to a related post if $[COD]=P$, $[AOB]=Q$, then for trapezium $ABCD$, $$[BOD]=\sqrt{PQ}=[AOC] \quad , \quad [ABCD]=(\sqrt{P}+\sqrt{Q})^2$$

Given here is $$\dfrac{P}{Q}=\dfrac{a^2}{b^2} \quad , \quad \dfrac{\sqrt{PQ}}{(\sqrt{P}+\sqrt{Q})^2}=\dfrac{2}{9}$$

Dividing second one by $Q$, and letting $a/b=x$ we get $$\dfrac{x}{(1+x)^2}=\dfrac{2}{9}$$ which can be solved to obtain $$x= \dfrac{1}{2} \Rightarrow \dfrac{b}{a}=2$$

Body:
$ABCD$ is a trapezium where $AB$ || $CD.$ Let $AB = b$ , $CD = a$ where $a < b$. Let $S$ be the area of trapezium $ABCD$. It is given that $[\Delta BOC] = \frac{2S}{9}$. Find the value of $\frac{b}{a}$.

Related Question :- $ABCD$ is a trapezium where $AB$ $||$ $CD$. Suppose $[\Delta BOC] = \frac{25}{9}$, and let $AB = b$, $CD = a$ where $a < b$.

Note:- The real fact is that I copied the initial question wrong, saying that $[\Delta BOC] = \frac{25}{9}$ and there were no unique trapeziums of that kind, so here is the real question and sorry for the inconvenience.

What I Tried: Here is a picture :-

$ [\Delta BOC] = [\Delta AOD]$ follows from carpet strategy, so :-
$$ [\Delta DOC] + [\Delta AOB] = \frac{5S}{9}$$

We have that $\Delta DOC \sim \Delta BOA$ , which might help find $\frac{b}{a}$, but I couldn't understand how I can use it. One way is to put variables for $DO$ , $OC$ , $AO$ and $OB$ , but that will make it more complicated and I don't want to do  it.

Another thing I thought of was to draw the perpendicular from $O$ , touching $AB$ and $CD$ at $M$ and $N$ , suppose. And let $[\Delta DOC] = x$ . I will get :-

$$\rightarrow \frac{1}{2} * ON * a = x$$
$$\rightarrow \frac{1}{2} * OM * b = \frac{5S}{9} - x$$

But that still deals with a variable $x$ , how do I fix it?

Can anyone help me? Thank You.


问:
Is $V= W\oplus W^{\perp}$ always?

Is $V= W\oplus W^{\perp}$ always?

Here, $W$ is a subspace of the vector space $V$.

I think this holds if $V$ is finite-dimensional, but what if it is not? Could someone help me prove $V= W\oplus W^{\perp}$ in the infinite-dimensional case, or produce a counterexample? I'm really not sure where to start.

The thing is, I am not able to see how moving from finite dimensions to infinite dimensions changes anything. Shouldn't it be the case that a vector lies in $W$ or $W^\perp$? Where else can it lie?

P.S.

This isn't part of a homework exercise or something, it just came to my mind. I'm taking a class in Linear Algebra right now that deals mostly with finite-dimensional vector spaces only, so I don't know where to start in this case.

答:
In order to define orthogonal complements we need $V$ to be an inner product space, not just a vector space. Then it is generally false that $V = W \oplus W^{\perp}$ if $V$ is an infinite-dimensional inner product space. A standard counterexample is to take $V$ to be the $\ell^2$ space of sequences

$$\ell^2(\mathbb{N}) = \{ a_n \in \mathbb{R}^{\mathbb{N}} : \sum_{n=1}^{\infty} |a_n|^2 < \infty \}$$

equipped with the inner product $\langle a, b \rangle = \sum_{n=1}^{\infty} a_n b_n$, and to take $W$ to be the subspace consisting of sequences with finite support (sequences such that $a_n = 0$ for sufficiently large $n$). It's a nice exercise to check that $W^{\perp} = 0$, so $W \oplus W^{\perp} = W$ is strictly smaller than $V$, and does not contain e.g. the sequence $a_n = \frac{1}{n}$.

The standard fix is to assume that $V$ is a Hilbert space (which is true here) and that $W$ is a closed subspace (which is not true here); then the statement is true. $V$ being a Hilbert space and $W$ being closed allows you to define the orthogonal projection of an element of $V$ onto $W$ which is what makes the proof work; you can check in the above example that the sequence $a_n = \frac{1}{n}$ does not admit an orthogonal projection onto $W$.

In a bit more detail, one way to define the orthogonal projection which works in finite dimensions is to take the closest vector in $W$, and another is to use an orthonormal basis of $W$ and define the orthonormal projection as a certain sum of projections onto the basis; both approaches generally fail in infinite dimensions because a limit you need to exist doesn't always exist, unless $V$ is Hilbert and $W$ is closed.

Shouldn't it be the case that a vector lies in $W$ or $W^\perp$?

Be careful; this is not what a direct sum means. A direct sum means that a vector decomposes into a component in $W$ and a component in $W^{\perp}$, not that it lies in either $W$ or $W^{\perp}$.

Title: Is $V= W\oplus W^{\perp}$ always?


问:
$\sum_{k=1}^{n} \arcsin(\sin(k))= a_n+b_n \pi$, for $a_n,b_n\in\mathbb{Z}$; what can be said about the sequence $(a_n,b_n )$?
We have $\sum_{k=1}^{n} \arcsin(\sin(k))= a_n+b_n \pi$ for some integers $a_n,b_n$. I have several questions about the behavior of these numbers:

For which $n$ does $b_n=0$ and $a_n\ne 0$, i.e. the sum is a non-zero integer?

For which $n$ does $(a_n,b_n)=(0,0)$, i.e. the sum vanishes?

Aside from these subsequences, based on numerical evidence I conjecture that $\lim_{n\to\infty} a_n/b_n=-\pi$, but I don't know how to prove it.

Recall the convergents of $\pi$ are $3, 22/7, 333/106, 355/113, 103993/33102$, etc. I computed $(a_n,b_n)$ for $1\le n\le 120000$ and here's what I found:

As expected, I saw 'increased aberrations' for $n$ near the numerator of the numerators of the convergents. For example, $b_n=0$ and $a_n\ne 0$ when $$n=\{1,8,16,52,60,96,104,140,148,184,192,228,236,272,280,316,324,360,103632,103668,103676,103712,103720,103756,103764,103800,103808,103844,103852,103888,103896,103932,103940,103976,103984\}$$ and $(a_n,b_n)=(0,0)$ when $$n=\{24,44,68,88,112,132,156,176,200,220,244,264,288,308,332,352,103640,103660,103684,103704,10372
   8,103748,103772,103792,103816,103836,103860,103880,103904,103924,103948,103968,103992\}.$$ In particular, when $b_n=0$ and $a_n \ne0$, aside from $n=1$ I found that $a_n=2$.

Aside from these values, the ratio is quite close to $\pi$. We have $a_{120000}/b_{120000} =-\frac{248162}{78993}$, which differs from $-\pi$ by about $0.0000231474$. Here is a plot:

Any information or insight would be much appreciated. I've heard of several continued fractions for $\pi$ and was wondering if perhaps this has to do with that.

Update: I checked oeis.org for both the numerator and denominator and found nothing. Since $\arcsin(\sin(k))=x_k+y_k \pi$, $x_k,y_k\in\mathbb{Z}$, is essentially the signed distance between $k$ and its nearest multiple of $\pi$, it is is clear that $x_k/y_k\approx -\pi$. Then perhaps using the CLT or another probabilistic argument, one could examine the convergence of the random variable $a_n/b_n$.

答:
This is very interesting !

For the fun of it, I computed the ratio's
$$R_k=-\frac{a_{10^k}}{b_{10^k}}$$ and obtained the following sequence
$$\left\{1,\frac{17}{6},3,\frac{22}{7},\frac{22}{7},\frac{931}{296},\frac{10559}{3361},\frac{1093611}{348107},\cdots\right\}$$
For the last one
$$\frac{1093611}{348107}-\pi=1.74 \times 10^{-6}$$


问:
Dirichlet-Multinomial probability that a specific category has the lowest count
Say we have $X=(X_1,X_2,\dots,X_n)\overset{d}{=}\operatorname{Dirichlet-Multinomial}(\boldsymbol{\alpha},N)$. How can we find the probability that $X_m$ is the smallest of the $X_i$, $\operatorname{P}(X_m<X_k, \forall k\neq m)$, for each $m$?

Edit: The marginal distributions for $X_i$ are $\operatorname{Beta-Binomal(\alpha_i,\sum_{j\neq i}\alpha_j, N)}$. Hence:$$
\operatorname{P}(X_i=x_i)=\binom{N}{x_i}\frac{\operatorname{B}(x_i + \alpha_i, N-x_i + \sum_{j\neq i}\alpha_j)}{\operatorname{B}(\alpha_i,\sum_{j\neq i}\alpha_j)}.
$$

答:
I don't think there can be a closed-form solution for $\operatorname{P}(X_m<X_k, \forall k\neq m)$ because for any particular $x_m$, we would need to work out partitions with a minimum size for each part for the remaining $x_k$ in order to ensure that each was at least $x_m+1$.  If, for example, $x_m$ was $2$, we would need to partition the other $N-2$ samples into $i-1$ parts of at least $3$ each.  If, as contemplated in your comment, equally low values were included, the minimum would decrease to $2$, but there would still be some minimum for each non-$m$ $x_k$ to satisfy.

Each acceptable partition might, in turn, represent multiple combinations, depending on the exchangeability of the remaining $x_k$ and $\boldsymbol{\alpha}_k$.  If all of the remaining $x_k$ values were identical and all of the associated $\boldsymbol{\alpha}_k$ values were identical, then a single calculation would suffice to cover every instantiation of that partition.  At the other extreme, if no $x_k$ equalled another $x_k$ and no $\boldsymbol{\alpha}_k$ equalled another $\boldsymbol{\alpha}_k$, then each acceptable partition could be realized in $k!$ unique ways (mappings of partition values to $i$'s), each of which would require a separate calculation.  In any event, the impossibility of counting in closed form even the number of acceptable partitions for a given $x_m$ (let alone summing the probabilities associated with all the different realizations of that partition) dooms, it seems to me, the possibility of a closed-form solution.

Once we move to approximations, moreover, we are inevitably trading off accuracy for computational ease.  So, without knowing how you would weigh these, it's hard to know what kinds of approximations you would or would not find acceptable.


问:
How analytic continuation allows for proof of these 2 theorems in theory of Partitions
Consider these 2 theorems in textbook apsotol introduction to analytic number theory.

1st is generating functions for partitions

I have self studied text and need help in verifying the argument of use of analytic continuation in  last image( last line of proof) : Is it due to reason that we can differentiate the formula derived infinitely many times for all complex numbers in disk |x|<1?

Similarly, in case of proof of Euler Pentagonal Theorem here :

See 2nd line of First image : Author says about analytic continuation. My understanding is that is it due to the fact that infinitely times differentiable in |x|<1 .

Am i right or wrong. Do I need to add something else.

I ask my questions here because there is no one to whom I can ask as it is not taught in my university.

Kindly shed some light on this.

答:
The principle of analytic continuation is reliant on the identity theorem.

Suppose that $f,g$ are analytic on $\Omega$, a connected open set, and the set of points where $f = g$ has a limit point. Then, $f=g$ on $\Omega$.

Now, the question is this : we have the functions $F(x) = \frac{1}{\prod (1+x^k)}$ and the power series $G(x) = \sum_{k=0}^\infty p(k)x^k$. These are functions , for now defined on $[0,1)$, which agree on $[0,1)$.

Now, analytic continuation is basically this : suppose we find a holomorphic function $F_1$ on the unit disk which extends $F$, and a holomorphic function on the unit disk $G_1$ which extends $G$. Since $F=G$ on $[0,1)$, $F_1 = G_1$ on $[0,1)$, which contains the limit point $0$, and therefore by the identity theorem, $F_1=G_1$ on the entire unit disk.

But what are $F_1$ and $G_1$? The answer to this is fairly obvious : the expressions which define $F$ and $G$ also end up making sense on the entire unit disk.

That is, for $\{|z|<1\}$ define $F_1(z) = \frac 1{\prod (1+z^k)}$. Does this make sense? Indeed it does, for the exact same reason it did for $x$. We have the following theorem for infinite products.

Suppose that $a_n \neq -1$ is a sequence of complex numbers with $\sum_{n=1}^\infty |a_n| < \infty$. Then the infinite product $\prod_{n=1}^\infty (1+a_n)$ converges absolutely to a non-zero quantity.

With this in mind, the obvious check on $\prod_{k=1}^\infty{(1+z^k)}$ shows that $F_1$ exists on $\{|z|<1\}$. This extends $F$, obviously.

Now, for $G_1$ take the power series defining $G$ i.e. define $G_1(z) = \sum_{k=0}^{\infty} p(k)z^k$. A theorem concerning power series easily shows that $G_1$ is defined on $|z| < 1$ :

Suppose that $\sum a_kz^k$ is a power series centered around zero, and suppose that $\sum a_kr^k < \infty$ for some $r>0$. Then , for every $|z|<r$, we have $\sum a_kz^k < \infty$.

That is, finding  one number for which a power series converges, gives a lot of other numbers for which that power series converges.

Now, $G_1$ is easily defined on $|z|< 1$ : for any $|z|<1$, we know that $G_1(\frac{1+|z|}{2}) = G(\frac{1+|z|}{2}) < \infty$, so $G_1(z) < \infty$.

Thus, $G_1$ and $F_1$ are well defined, so by the usual theorem they agree on the unit disk. This is what is meant by "analytic continuation" of the identity $ F(x)= \sum_{k=0}^\infty p(k)x^k$.

Part $2$ is exactly the same. The side $\prod(1-x^m)$ extends by the result for infinite products on $|z|<1$, and the power series extends by the theorem for power series. Thus, the same argument yields equality on $|z|<1$.


问:
In how many ways can you choose $k$ out of $n$ people standing in line, So there's a space of at least 3 people between them
In how many ways can you choose $k$ out of $n$ people standing in line, So there's a space of at least $3$ people between them.

Actually, I don't even know how to start on this one.

答:
HINT: You can think of this as a stars-and-bars problem. The $k$ people chosen are the bars, the $n-k$ other people are the stars, and you have to distribute the stars so that there are at least $3$ between each pair of adjacent bars. Your problem is to count the ways to do this.

Equivalently, let $x_0$ be the number of people to the left of the first person chosen, $x_i$ the number between the $i$-th and $(i+1)$-st people chosen (reading from left to right) for $i=1,\dots,k-1$, and $x_k$ the number to the right of the $k$-th person chosen. Then you want the number of solutions to

$$x_0+x_1+\ldots+x_k=n-k$$

in non-negative integers such that $x_i\ge 3$ for $i=1,\dots,k-1$.

You’ve probably seen other stars-and-bars problems by now, but even if you haven’t, the linked article gives a reasonably good explanation.


问:
Limit $\lim_{x\rightarrow0}\frac{1}{x}\int_{0}^{\sin x}\sin \frac{1}{t} \cos t^{2}\mathrm{d}t$
Determine whether or not the limit below exists.
$$\lim_{x\rightarrow0}\frac{1}{x}\int_{0}^{\sin x}\sin \frac{1}{t} \cos t^{2}\mathrm{d}t$$
I tried to use the Mean value theorem integrals to prove the limit exists, but it does not exist for $\lim\limits_{x\rightarrow0}\sin \frac{1}{x}$. So I guessed the limit does not exist and used the Cauchy principle to prove it, but I failed.

Any idea will be helpful.

答:
Since $\cos t^2$ is monotone decreasing for $t$ sufficiently close to $0$, by the second mean value theorem for integrals there exists $\xi_x \in (0, \sin x)$ such that

$$\int_0^{\sin x} \sin \frac{1}{t} \cos t^2 \, dt = \cos(0) \int_0^{\xi_x}\sin \frac{1}{t}  \, dt = \int_0^{\xi_x}\sin \frac{1}{t}  \, dt$$

Taking $g(t) = t^2 \cos \frac{1}{t}$ for $t > 0$ and $g(0) = 0$,  we have $g’(0) =0$ and for $t>0$,

$$g'(t) = 2t \cos \frac{1}{t} + \sin \frac{1}{t},$$ and, using the FTC,

$$\int_0^{\xi_x}\sin \frac{1}{t}  \, dt = \int_0^{\xi_x} g'(t) \, dt-\int_0^{\xi_x}2t \cos \frac{1}{t} \, dt=\xi_x^2\cos \frac{1}{\xi_x} -  \int_0^{\xi_x}2t \cos \frac{1}{t} \, dt$$

Thus,

$$\frac{1}{x} \int_0^{\sin x} \sin \frac{1}{t} \cos t^2 \, dt = \xi_x \frac{\xi_x}{x}\cos \frac{1}{\xi_x} - \frac{1}{x} \int_0^{\xi_x}2t \cos \frac{1}{t} \, dt$$

We can apply the mean value theorem to the integral on the RHS (since the integrand is continuous) to find $\theta_x \in (0,\xi_x)$ such that

$$\frac{1}{x} \int_0^{\sin x} \sin \frac{1}{t} \cos t^2 \, dt = \xi_x \frac{\xi_x}{x}\cos \frac{1}{\xi_x} - \frac{\xi_x}{x} 2\theta_x \cos \frac{1}{\theta_x} $$

Since $\xi_x/x < 1$, we have

$$\left|\frac{1}{x} \int_0^{\sin x} \sin \frac{1}{t} \cos t^2 \, dt\right| \leqslant \xi_x +2 \theta_x$$

Since $\xi_x , \theta_x \to 0$ as $x  \to 0+$, we get

$$\lim_{x \to 0+}\frac{1}{x} \int_0^{\sin x} \sin \frac{1}{t} \cos t^2 \, dt = 0$$

Similarly we can show that the limit as $x \to 0-$ is $0$ as well.


问:
Solving for a differential equation Gompertz growth equation
What is the general solution of this differential equation?
$$ \frac{dy}{dt} = k \enspace  y \enspace \ln(\frac{a}{y})$$ where $a$ and $k$ are positive constants.

So far, my solution is:
$$ \frac{dy}{y \enspace \ln(\frac{a}{y})} = k \enspace dt$$
When I let $u=lny$,
$$ \int \frac{1}{y(\ln a-\ln y)} \,dy =\int k dt $$
$$ \int \frac{1}{\ln a-u} \,du= kt + C $$

How to continue this?

答:
$$\frac{dy}{dt} = k \,  y \, \log\left(\frac{a}{y}\right)$$ Let $y=e^z$ to make
$$z' + k z=k \log (a)$$ which looks to be simple.


问:
Quadratic formula and isolating $f(x)$

$f(x) - \frac{1}{f(x)} = g(x)$

I need to isolate $f(x)$. So far I got

$f(x)f(x) - g(x)f(x) - 1 = 0$

Now I've tried to use the quadratic root formula with no success..
What should I do?

答:
As the comments indicated, replace $f(x)$ by $X$ and $g(x)$ by $a$. So we have to solve
$$X-\frac{1}{X}=a$$
Multiplying both sides by $X$, we get
$$X^2-1=aX$$
Using the quadratic formula, we get
$$X=\frac{a\pm\sqrt{a^2+4}}{2}$$
So
$$f(x)=\frac{g(x)\pm\sqrt{g(x)^2+4}}{2}$$

Title: Quadratic formula and isolating $f(x)$


问:
Proving trigonometric Identity: $\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$
I would like to try and prove

$$\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$$

using $LHS=RHS$ methods, i.e. pick a side and rewrite it to make it identical to the other side.

I found a quick way by doing this:

$$LHS = \frac{1+\sin x}{\cos x} = \frac{1+\sin x}{\cos x} \cdot \frac{1 - \tan x + \sec x}{1 - \tan x + \sec x}= \frac{1+\sin x+\cos x}{1-\sin x+\cos x} = RHS$$

but I feel that this is not a good way because I am manipulating the denominator of the LHS somewhat artificially, because I know it must be, in the end, $1-\sin x+\cos x$.

Does anyone have a better way of doing this?

答:
We'll work on the RHS. We'll multiply $\frac{1+\sin x+\cos x}{1-\sin x+\cos x}$ with the conjugate of the denominator. We'll do this from right to left, i.e. we pick the RHS and walk through to LHS.

Solution 1:
$$\require{cancel}\begin{aligned}\frac{1+\sin x+\cos x}{1-\sin x+\cos x}&=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}\cdot\frac{1+\sin x-\cos x}{1+\sin x-\cos x}\\&=\frac{1+\sin x\cancel{-\cos x}+\sin x+\sin^{2}x\cancel{-\sin{x}\cos{x}}\cancel{+\cos x}\cancel{+\sin{x}\cos{x}}-\cos^{2}x}{1\cancel{+\sin x}\cancel{-\cos x}\cancel{-\sin x}-\sin^{2}x+\sin{x}\cos{x}\cancel{+\cos x}+\sin{x}\cos{x}-\cos^{2}x}\\&=\frac{1+2\sin x+\sin^{2}x-\cos^{2}x}{1-\sin^{2}x+2\sin{x}\cos{x}-\cos^{2}x}\\&=\frac{1+2\sin x+\sin^{2}x-\cos^{2}x}{\cancel{\cos^{2}x}+2\sin{x}\cos{x}\cancel{-\cos^{2}x}}\\&=\frac{1+2\sin x+\sin^{2}x-\cos^{2}x}{2\sin{x}\cos{x}}\\&=\frac{\sin^{2}x\cancel{+\cos^{2}x}+2\sin x+\sin^{2}x\cancel{-\cos^{2}x}}{2\sin{x}\cos{x}}\\&=\frac{\cancel2\sin^{2}x+\cancel2\sin x}{\cancel2\sin{x}\cos{x}}\\&=\frac{\sin^{2}x+\sin x}{\sin{x}\cos{x}}\\&=\frac{\cancel{\sin{x}}\left(\sin x+1\right)}{\cancel{\sin{x}}\cos{x}}\\&=\frac{1+\sin x}{\cos x}\end{aligned}$$
Solution 2:
$$\begin{aligned}\frac{1+\sin x+\cos x}{1-\sin x+\cos x}&=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}\cdot\frac{1-\sin x-\cos x}{1-\sin x-\cos x}\\&=\frac{1\cancel{-\sin x}\cancel{-\cos x}\cancel{+\sin x}-\sin^{2}x-\sin{x}\cos{x}\cancel{+\cos x}-\sin{x}\cos{x}-\cos^{2}x}{1-\sin x\cancel{-\cos x}-\sin x+\sin^{2}x\cancel{+\sin{x}\cos{x}}\cancel{+\cos x}\cancel{-\sin{x}\cos{x}}-\cos^{2}x}\\&=\frac{1-\sin^{2}x-2\sin{x}\cos{x}-\cos^{2}x}{1-2\sin x+\sin^{2}x-\cos^{2}x}\\&=\frac{\cancel{\sin^{2}}x+\cancel{\cos^{2}x}\cancel{-\sin^{2}x}-2\sin{x}\cos{x}\cancel{-\cos^{2}x}}{\sin^{2}x-2\sin x+\sin^{2}x}\\&=\frac{-\cancel2\sin x\cos x}{\cancel2\sin^{2}x-\cancel2\sin x}\\&=\frac{-\sin{x}\cos{x}}{\sin^{2}x-\sin x}\\&=\frac{\sin{x}\cos{x}}{\sin x-\sin^{2}x}\\&=\frac{\cancel{\sin{x}}\cos{x}}{\cancel{\sin x}(1-\sin x)}\\&=\frac{\cos{x}}{1-\sin{x}}\cdot\frac{1+\sin{x}}{1+\sin{x}}\\&=\frac{\cos{x}(1+\sin{x})}{1-\sin^2{x}}\\&=\frac{\cancelto{1}{\cos{x}}(1+\sin{x})}{\cancelto{\cos{x}}{\cos^2{x}}}\\&=\frac{1+\sin x}{\cos x}\end{aligned}$$
Solution 3:
$$\begin{aligned}\frac{1+\sin x+\cos x}{1-\sin x+\cos x}&=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}\cdot\frac{1-\sin x+\cos x}{1-\sin x+\cos x}\\&=\frac{1+\sin x\cancel{-\cos x}+\sin x+\sin^{2}x\cancel{-\sin{x}\cos{x}}\cancel{+\cos x}\cancel{+\sin{x}\cos{x}}-\cos^{2}x}{1+\cancel{\sin x}\cancel{-\cos x}\cancel{-\sin x}-\sin^{2}x+\sin{x}\cos{x}\cancel{+\cos x}+\sin{x}\cos{x}-\cos^{2}x}\\&=\frac{1+2\cos x-\sin^{2}x+\cos^{2}x}{1-2\sin x+2\cos x+\sin^{2}x-2\sin{x}\cos{x}+\cos^{2}x}\\&=\frac{\cancel{\sin^{2}x}+\cos^{2}x+2\cos x\cancel{-\sin^{2}x}+\cos^{2}x}{\sin^{2}x+\cos^{2}x-2\sin x+2\cos x+\sin^{2}x-2\sin{x}\cos{x}+\cos^{2}x}\\&=\frac{\cancel2\cos^{2}x+\cancel2\cos x}{\cancel2\sin^{2}x+\cancel2\cos^{2}x-\cancel2\sin x+\cancel2\cos x-\cancel2\sin{x}\cos{x}}\\&=\frac{\cos^{2}x+\cos x}{\sin^{2}x+\cos^{2}x-\sin x+\cos x-\sin{x}\cos{x}}\\&=\frac{\cos{x}\left(1+\cos x\right)}{1-\sin x-\sin{x}\cos{x}+\cos x}\\&=\frac{\cos{x}\left(1+\cos x\right)}{1-\sin x+\cos{x}\left(1-\sin x\right)}\\&=\frac{\cos{x}\cancel{\left(1+\cos x\right)}}{\cancel{\left(1+\cos x\right)}\left(1-\sin x\right)}\\&=\frac{\cos{x}}{1-\sin{x}}\cdot\frac{1+\sin{x}}{1+\sin{x}}\\&=\frac{\cos{x}(1+\sin{x})}{1-\sin^2{x}}\\&=\frac{\cancelto{1}{\cos{x}}(1+\sin{x})}{\cancelto{\cos{x}}{\cos^2{x}}}\\&=\frac{1+\sin x}{\cos x}\end{aligned}$$
Solution 4:
$$\begin{aligned}\frac{1+\sin x+\cos x}{1-\sin x+\cos x}&=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}\cdot\frac{1+\sin x+\cos x}{1+\sin x+\cos x}\\&=\frac{1+\sin x+\cos x+\sin x+\sin^{2}x+\sin{x}\cos{x}+\cos x+\sin{x}\cos{x}+\cos^{2}x}{1\cancel{+\sin x}+\cos x\cancel{-\sin x}-\sin^{2}x\cancel{-\sin{x}\cos{x}}+\cos x\cancel{+\sin{x}\cos{x}}+\cos^{2}x}\\&=\frac{1+2\sin x+2\cos x+\sin^{2}x+2\sin{x}\cos{x}+\cos^{2}x}{1+2\cos x-\sin^{2}x+\cos^{2}x}\\&=\frac{\cancelto{1}{2}+\cancel2\sin x+\cancel2\cos x+\cancel2\sin{x}\cos{x}}{\cancel2\cos x+\cancel2\cos^{2}x}\\&=\frac{1+\sin x+\cos x+\sin{x}\cos{x}}{\cos x+\cos^{2}x}\\&=\frac{1+\sin x+\cos{x}\left(1+\sin x\right)}{\cos{x}\left(1+\cos x\right)}\\&=\frac{\cancel{\left(1+\cos x\right)}\left(1+\sin x\right)}{\cos{x}\cancel{\left(1+\cos x\right)}}\\&=\frac{1+\sin x}{\cos x}\end{aligned}$$
I hope this helps.


答:
Using your results you can set up directly the following two equations using the notations:

$\triangle DOC$ with area $\Delta_u$ and height $h_u$ and

$\triangle ABO$ with area $\Delta_l$ and height $h_l$

$h= h_l+h_u$ - height of the trapezium

So, because of similarity we have

$$r := \frac{h_l}{h_u}=\frac{b}{a}$$

Now, using your result, you have

$$\frac{5}{9}S = \Delta_u + \Delta_l =\Delta_u(1+r^2)$$
and
$$S = \frac{a+b}{2}h = \frac{a+b}{2}(h_u+h_l)=\frac a2h_u(1+r)(1+r)=\Delta_u(1+r^2)$$

Divide the equations to get
$$\frac 95 = \frac{(1+r)^2}{1+r^2}$$

This has two solutions for $r$ where the one in question is $\boxed{r=2}$, since $b>a$.

Title: $ABCD$ is a trapezium where $AB$ || $CD.$ Let $AB = b$ , $CD = a$ where $a < b$. Let $S$ be the area of trapezium $ABCD$.


问:
Power of a point tangents
Two circles $A$ and $B$ with centers $P$ and $Q$, respectively, are externally tangent to each other. The power of point $P$ with respect to circle $B$ is $8$. The power of point $Q$ with respect to circle $A$ is $15$. What is the ratio of the areas of circle $A$ and circle $B$?

How do I go about solving this problem? Is  there perhaps a formula for ratios of areas of circles given different powers of points?

Title: Power of a point tangents

Body:
Power of a point $X$ wrt circle $(O,r)$ is $OX^2-r^2$

Given is $$(a+b)^2-b^2=8 \quad , \quad (a+b)^2-a^2=15$$

Define $x:= a/b$
$$\dfrac{(a+b)^2-a^2}{(a+b)^2-b^2}=\dfrac{15}{8}$$

$$\Rightarrow \dfrac{(x+1)^2-x^2}{(x+1)^2-1^2}=\dfrac{15}{8}$$

$$\Rightarrow x=\dfrac{2}{5}$$


问:
For $s\in\mathbb{N}$ and $s<t=$LCM($l_1,...,l_m)$, $\sigma^s \neq I$ where $\sigma = \sigma_1...\sigma_m$

Let $n\in\mathbb{N}$ and consider $\sigma\in S_n$. Then $\exists k\in \mathbb{N}$ such that $\sigma^k = I$. Show this, and also that if $\sigma = \sigma_1...\sigma_m$ where $\sigma_i$'s are disjoint cycles, then the least $k$ such that $\sigma^k = I$ is the LCM of $(l_1,l_2,...,l_m)$ where $l_i$ is the length of the cycle $\sigma_i$.

For the first part, here's what I did:

Let $\sigma=\sigma_1...\sigma_m$ be the disjoint cycle decomposition of $\sigma$ where the length of each cycle $\sigma_i$ is $l_i$. Note that if $\Gamma = (a_1 a_2 a_3...a_l) \in S_{n}$ is a cycle in $S_n$, $l$ is its length. It is easily seen that $\Gamma^l = I$. Now define $t = LCM(l_1,...,l_m)$. Since disjoint cycles commute, we have $$\sigma^t = \sigma_1^t\sigma_2^t....\sigma_m^t$$
As $l_i\vert t$, each $\sigma_i^t = I$ and so $\sigma^t = I$.

To show that $t$ is the least such number, I must show that if $s\in\mathbb{N}$ and $s<t$ then $\sigma^s \neq I$.

Since $t$ is the LCM, we conclude that there exists at least one $i$ such that $l_i$ does not divide $s$, so that $\sigma_i^s \neq I$. I think I should now consider $$\sigma^s = \sigma_1^s\sigma_2^s....\sigma_i^s...\sigma_m^s$$

but I'm stuck here, what do I do next?

P.S.

I have realized that my questions boils down to:
If cycles $\sigma_1, \sigma_2,....,\sigma_k$ are disjoint then $\sigma_1\sigma_2...\sigma_k = I$ iff $\sigma_i= I$ for all $1\leq i\leq k$. Any hints?

答:
Although the results we are going to look into can be formulated at a very general level, I shall limit the frame of the discussion to general permutation groups, since you only seem to be familiar with them. It is however very easy to see how the propositions I am about to formulate extend in the general case (of arbitrary groups).

Let us consider an arbitrary set $A$ and the general symmetric group $\Sigma(A)$, structure by which I am referring to the set $\Sigma(A)$ of all permutations of set $A$ with the implicit binary operation of composition of maps. In general, this operation takes the syntax $\sigma \circ \tau$ for two arbitrary permutations $\sigma, \tau \in \Sigma(A)$, however for simiplicity we shall employ the simplified notation $\sigma\tau$ (used for "abstract" multiplications).

To say that $\sigma$ is a permutation of finite order has a very specific meaning in the general sense of group theory, however in our given context we shall be content to understand it as there exists nonzero $n \in \mathbb{N}^{\times}$ such that $\sigma^n=\mathbf{1}_A$. We also introduce the order of the finite order permutation $\sigma$ -- denoted by $\mathrm{ord}(\sigma)$ -- to be the unique natural number $k$ rendering the following relation true:
$$\left\{n \in \mathbb{N}^{\times} \mid \sigma^n=\mathbf{1}_A\right\}=k\mathbb{N}^{\times}.$$
Once again, may I point out that this is not the actual definition of the notion of order, but it is an easy to obtain equivalent description of the order of a permutation of finite order (there are also permutations of infinite order; nevertheless, the order of a permutation is always an elements of the set $\mathbb{N}^{\times} \cup \{\aleph_0\}$).

For arbitrary natural numbers $r, s \in \mathbb{N}$ let us denote their least common multiple by $[r; s]$. In case of an arbitrary family $r \in \mathbb{N}^I$ we use the notation $[r_i;]_{i \in I}$ for the least common multiple of the entire family $r$.

With the above mentions in place, let us consider two permutations of finite order $\lambda, \mu \in \Sigma(A)$ which:

commute, meaning that $\lambda\mu=\mu\lambda$

are both of finite order

are such that the subsets of powers of $\lambda$ and $\mu$ intersect trivially, in other words $\left\{\lambda^n\right\}_{n \in \mathbb{Z}} \cap \left\{\mu^n\right\}_{n \in \mathbb{Z}}=\{\mathbf{1}_A\}$.

Then $\lambda\mu$ is also of finite order and the relation $\mathrm{ord}(\lambda\mu)=\left[\mathrm{ord}(\lambda); \mathrm{ord}(\mu)\right]$ holds.

For simplicity, let us write $m=\mathrm{ord}(\lambda) \neq 0$, $n=\mathrm{ord}(\mu) \neq 0$ and let us introduce $p\colon=[m; n] \neq 0$. This means that $m, n \mid p$ so there exist $k, l \in \mathbb{N}$ such that $p=km=ln$. Since $\lambda$ and $\mu$ commute we have by virtue of a very general rule of algebraic calculus that $(\lambda\mu)^r=\lambda^r\mu^r$ for any $r \in \mathbb{N}$ (even for any $r \in \mathbb{Z}$, but you might not be familiar with the notion of raising to negative powers). This in particular entails that $(\lambda\mu)^{p}=\lambda^{p}\mu^{p}=\left(\lambda^m\right)^k\left(\mu^n\right)^l=\left(\mathbf{1}_A\right)^k\left(\mathbf{1}_A\right)^l=\mathbf{1}_A$. Since $p \neq 0$, this establishes $\lambda\mu$ as a permutation of finite order. Let us set $q\colon=\mathrm{ord}(\lambda\mu)$ and $M=\left\{r \in \mathbb{N}^{\times} \mid (\lambda\mu)^r=\mathbf{1}_A\right\}$.

By virtue of the characterisation of the order (in the finite case), we have $M=q\mathbb{N}^{\times}$ and since -- as we have just seen above -- $p \in M$, we gather that $q \mid p$. We now argue that the converse relation $p \mid q$ also holds in order infer that $p=q$. Since in particular $q \in M$ we have $(\lambda\mu)^q=\lambda^q\mu^q=\mathbf{1}_A$ which entails $\nu\colon=\lambda^q=\left(\mu^q\right)^{-1}=\mu^{-q}$. It is obvious that $\nu \in \left\{\lambda^n\right\}_{n \in \mathbb{Z}} \cap \left\{\mu^n\right\}_{n \in \mathbb{Z}}=\{\mathbf{1}_A\}$, whence $\lambda^q=\mu^{-q}=\mathbf{1}_A$, which is equivalent to $\lambda^q=\mu^q=\mathbf{1}_A$. Since $q \in \mathbb{N}^{\times}$ we have by virtue of the characterisation of the orders $m$ respectively $n$ of $\lambda$ respectively $\mu$ that $m, n \mid q$. By elementary arithmetic (the definition of the least common multiple), this entails $p \mid q$ and establishes our claim.

Before stating the next result, let us define the support of an arbitrary permutation $\sigma \in \Sigma(A)$ as the subset of elements of $A$ it actually permutes:
$$\mathrm{Supp}(\sigma)\colon=\{x \in A \mid \sigma(x)\neq x\}.$$
It is easy to prove that two permutations $\lambda, \mu \in \Sigma(A)$ of disjoint supports necessarily commute (I will let you think about it).

The above proposition concerning finite orders generalises to finite families of permutations $\lambda \in \Sigma(A)$ such that:

$\lambda_i$ is of finite order for each $i \in I$

$\lambda_i\lambda_j=\lambda_j\lambda_i$ for each pair $(i, j) \in I$

$\mathrm{Supp}(\lambda_i) \cap \mathrm{Supp}(\lambda_j)=\varnothing$ for each pair of distinct indices $i, j \in I, i \neq j$.
Then the product $\mu\colon=\displaystyle\prod_{i \in I}\lambda_i$ is also of finite order and the relation $\mathrm{ord}(\mu)=\left[\mathrm{ord}(\lambda_i);\right]_{i \in I}$ between the orders holds.

The way to prove this (somewhat) generalised version is by induction on $|I|$ (the cardinality of $I$). You might not find it straightforward to supply a proof for this on your own, so you can let me know. For the time being I think you have sufficient new material to digest.

The manner in which the generalised version applies to your problem at least I hope will become clear: if you have a finite family $\kappa$ of disjoint cycles then their supports satisfy the condition of pairwise disjointness in the generalised version above and their individual orders are finite, given by their lengths. Thus, their product $\pi$ is also of finite order $r$, number which is equal to the least common multiple of the family of lengths. Bearing in mind the characterisation of the order $r$ of $\pi$, $\pi^n=\mathbf{1}_A$ can only hold for nonzero natural $n$ if $n$ is a multiple of the order $r$ and therefore never in the case $n<r$ (any strictly positive multiple of $r$ being at least equal to $r$).


问:
$\int_0^x f(t)dt = [f(x)]^2$ and f is never 0
Problem says:

Suppose that continuous $f(x)$ is such that:

$$\int_0^x f(t) dt= [f(x)]^2$$

and $f(x)$ is never zero. Then find $f(x)$.

I do the question thinking it say "$f(x)$ is not the zero function"  I find $f(x) = \frac12 x$ as solution.

But, I see from again reading again $f(x)$ can never be zero. Now my solution is not correct. $f(0) = 0$ for my solution.

From how I have came to solution I wonder:

Is this possible even to find?

In the way I have solved it seemed this was the only way possible solution.

答:
EDIT: Expanding on J. M.'s advice.

From $\int_0^x {f(t)dt}  = f^2 (x)$, we get $f(x) = \frac{d}{{dx}}f^2 (x)$ (using that $f$ is continuous). From this, one wants to write $f(x)=2f(x)f'(x)$. In general, however, differentiability of $f^2$ does not imply that of $f$ (consider, for example, the functions $|x-1|$ and $(x-1)^2$, $x \geq 0$; the former is not differentiable at $x=1$), and we are not given that $f$ is differentiable.
Nevertheless, in our case, for any $x > 0$ we have
$$
f(x) = \frac{d}{{dx}}f^2 (x) =
\mathop {\lim }\limits_{h \to 0} \frac{{f^2 (x + h) - f^2 (x)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{[f(x + h) + f(x)][f(x + h) - f(x)]}}{h},
$$
leading to
$$
1 = \mathop {\lim }\limits_{h \to 0} \bigg[\frac{{f(x + h) + f(x)}}{{f(x)}}\frac{{f(x + h) - f(x)}}{h}\bigg],
$$
where we have used the assumption $f(x)>0$.
Since $f$ is continuous at $x$, it thus follows that
$$
1 = 2\mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h},
$$
and hence the conclusion that
$$
f'(x) = \frac{1}{2}.
$$
(The rest is straightforward.)

Original answer (see first comment below).

If $f$ is continuous on $[0,\infty)$, then, on the one hand,
$$
\mathop {\lim }\limits_{x \to 0^ +  } \int_0^x {f(t)dt}  = 0,
$$
and, on the other hand,
$$
\mathop {\lim }\limits_{x \to 0^ +  } f^2 (x) = f^2 (0).
$$
So the condition $\int_0^x {f(t)dt}  = f^2 (x)$ implies that $f^2 (0) = 0$, hence also $f(0)=0$.


问:
Integral of $\int^1_0\frac{dx}{\sqrt{x+3}-1}$
I want to solve this integral and need some directions.
$$\int^1_0\frac{dx}{\sqrt{x+3}-1}$$
I decided to call $x+3 = t^2  \rightarrow 2tdt = dx$ then : $$\int^1_0 \frac {2tdt}{t^2-1}$$ Now what should I do? call $t^2-1 = u$ ? and do the same thing on u?

Thanks!

答:
Let  $\sqrt{x+3}-1=t\implies x=(t+1)^2-3=t^2+2t-2\implies dx=2(t+1)dt$

If $x=0\implies t=\sqrt{3}-1, x=1\implies t=\sqrt{1+3}-1=1$

$$\int^1_0\frac{dx}{\sqrt{x+3}-1}=\int_{\sqrt3-1}^1\frac{2(t+1)}t dt$$

$$=2\int_{\sqrt3-1}^1\left(1+\frac1t\right) dt$$

Can you take it from here?


问:
Product of Radon Nikodym derivatives
Q) Suppose that $\mu_1, \mu_2, \mu_3$ are $\sigma-$finite positive measures on X such that $\mu_1\ll\mu_2$ and $\mu_2\ll \mu_3$. Show that $\mu_1\ll\mu_3$ and its Radon-Nikodym derivative satisfies

$$\frac{d\mu_1}{d\mu_3}(x)=\frac{d\mu_1}{d\mu_2}(x)\frac{d\mu_2}{d\mu_3}(x) \text{ for } \mu_3-\text{a.e. }x\in X \tag{1}$$

Since $\mu_1\ll \mu_2$, there exists a unique $f$ upto $\mu_2-$a.e. and since $\mu_2\ll \mu_3$, there exists a unique $g$ upto $\mu_3-$a.e.. To show $(1)$, I need to show that $f(x)g(x)$ is unique upto $\mu_3-$a.e.. Since $\mu_2\ll \mu_3$, $\mu_3(A)=0\implies \mu_2(A)=0$ and so if $g$ is unique upto $\mu_3-$a.e., then $g$ is unique upto $\mu_2-$a.e.. So $f(x)g(x)$ is unique upto $\mu_2-$a.e. but how can I show that it is unique upto $\mu_3-$a.e.?

答:
As $\mu_3(A)=0\Rightarrow\mu_2(A)=0\Rightarrow\mu_1(A)=0$, we have that $\mu_1\ll\mu_3$ and the Radon-Nikodym theorem says that there exists a unique $\mu_3-$a.e $h$  such that $\mu_1(A)=\int_Ah\,d\mu_3.$ We want to show that $h=fg$. For that let $f_n=\sum_{i=1}^na_i^n\chi_{A_i^n}$ and $g_k=\sum_{j=1}^kb_j^k\chi_{B_j^k}$ be simple functions such that $f_n\nearrow f$ and $g_k\nearrow g$. By the Monotone Convergence Theorem, we have

\begin{align*}\mu_1(A)&=\int_A f\,d\mu_2=\lim_{n\rightarrow \infty}\int_A f_n\,d\mu_2=\lim_{n\rightarrow \infty}\sum_{i=1}^na_i^n\mu_2(A\cap A_i^n)=\lim_{n\rightarrow \infty}\sum_{i=1}^na_i^n\int_{A\cap A_i^n}g\,d\mu_3\\&=\lim_{k,n\rightarrow \infty}\sum_{i=1}^n \sum_{j=1}^k a_i^nb_j^k\mu_3(A\cap A_i^n\cap B_j^k)=\lim_{k,n\rightarrow \infty}\int_A f_ng_k\, d\mu_3=\int_Afg\,d\mu_3.\end{align*}

Title: Product of Radon Nikodym derivatives


答:
We connect midpoints of DC and AB, if passes O. We can easily see that $S_{EGBC}=S_{ADEG}= \frac S 2$ abd we have:

$S_{DOC}=S-4(\frac{2s}9)=\frac S 9$

$2 a \times h=\frac{4s}9+2 S_{DOC}=\frac{6S}9=\frac{2S}3$ ⇒ $a=\frac{2S}{3h}$

$S=(a+b)\frac h 2$  ⇒ $a\times \frac h 2 +b\frac h 2=S$  ⇒ $b=\frac{4 S}{3 h}$

⇒ $\frac b a = 2$


问:
$F$ is an increasing function and $F(0) > 0$. For $x \geq 0$, define $G(x) = \int_{0}^{x} F$. Prove that $G$ is one-to-one
I am given a practice question where I am supposed to show that the function of G is one-to-one. Where $G(x) = \int_{0}^{x} F$, where $F$ is an increasing function and $F(0)>0$.

I thought about thinking about the second part of Fundamental Theorem of Calculus, but I don't know if that is the right method to show that the function is one to one. Also, I don't know if I can just say

Suppose that $F(0) = 1$ since $F(0) > 0$, then $G(x) = F(x) - 1$.

Any help will be great.

Edit: It was clarified that $x \geq 0$

答:
I would think this is a proof by contradiction:

Let's assume $G(x)$ is not one-to-one. Therefore there exists two different $x$ values, $x_1 < x_2$ such that $G(x_1) = G(x_2)$:

\begin{align*}
G(x_1) =&\ \int\limits_0^{x_1}F(t)dt \\
G(x_2)=&\ \int\limits_0^{x_2}F(t)dt = \int\limits_0^{x_1}F(t)dt + \int\limits_{x_1}^{x_2}F(t)dt = G(x_1) + \int\limits_{x_1}^{x_2}F(t)dt
\end{align*}

If $G(x_1) = G(x_2)$ then $\int\limits_{x_1}^{x_2}F(t)dt = 0$. You need to show that this isn't possible if $x_2 > x_1$ and $F(t)$ is increasing.

Edit (clarification of my comment):

This question was edited (which is fine and is good), but I do want to explain for anyone curious why we need $x \geq 0$ for the given problem statement.

If we are only given that $F(t)$ is monotonically increasing and $F(0) > 0$, that's not enough to conclude that $G(x) = \int\limits_0^xF(t)dt$ is one-to-one.

Consider the following:

$$
F(t) = 4t^3 + 2t + 1
$$

Clearly $F(0) = 1 > 0$ and it's clearly monotonically increasing (not even just monotonically non-decreasing--$F'(t) = 12t^2 + 2 > 0$ for all $t$)

$$
G(x) = \int\limits_0^xF(t)dt = x^4 + x^2 + x
$$

A simple graph of this function will show that it is one-to-one when $x\geq0$, in fact it's one-to-one for all $x \geq \xi$ such that $4\xi^3 + 2\xi + 1 = 0$, but if we allow a domain such that $\xi - \delta < x < \infty$, where $\delta > 0$, then it's clearly not one-to-one.

Which brings me to my ultimate criticism (of the question): we don't need $F(0) > 0$, we only need $F(0) \geq 0$ and $x \geq 0$. With that constraint I can easily explain the requirement for $x \geq 0$:

$$
F(t) = 4t^3 + 2t \rightarrow G(x) = \int\limits_0^xF(t)dt = x^4 + x^2
$$

This is $F(t)$, an odd function (and is clearly monotonically increasing), results in a even function: cannot possibly be one-to-one unless we restrict to $x \geq 0$.

Please Note: I'm talking about a "counterexample" as to why we must have $x \geq 0$ (i.e. a function, $F(t)$, that is monotonically increasing that does not result in a one-to-one function if $x < 0$)! This does not mean $F(t)$ needs to be odd--only monotonically increasing on the interval $0 < t < \infty$, with $F(0) \geq 0$ and $x \geq 0$.

Second Note: $F(t)$ is monotonically increasing but $F(0) < 0$: (assuming $x \geq 0$)

Simple counterexample:

$$
F(t) = 4(t - 1)^3 \rightarrow G(x) = \int\limits_0^xF(t)dt = (x - 1)^4 - 1
$$

Here $G(x)$ is not one-to-one because you get $G(0) = G(2) = 0$


问:
What are the constant solutions for separation of variables used for?
In my calc $2$ textbook there is an initial-value problem that requires the use of separation of variables.

The differential equation is $\frac{dy}{dx} = \left(2x+3\right)\left(y^2-4\right)$ with initial value $y(0)=1$.

In the list of steps to take to solve this equation using separation of variables, the very first step is "Check for any values of $y$ that make $g(y) = 0$. These correspond to constant solutions". When I solve $g(y)=0$ I get $g(y) = \pm 2$, but I don't really understand how these values are used to solve this equation?

答:
In this case, when $y(0)=1$, those constant solutions don't play a role.

But if the initial condition were $y(0)=2$, then the solution would be $y(t)=2$ (for all $t$), and likewise with $y(0)=-2$ you would get $y(t)=-2$. And those two solutions can't be obtained by writing $dy/(y^2-4)=(2x+3)dx$, since when you're dividing by $y^2-4$ you have to assume that $y \neq \pm 2$.