问:
Ambiguous notation in Spivak's Calculus: $f \circ g = I$
There are a few concepts I have come across that are causing me a little confusion. This question is motivated by the repeated usage of the notation $f \circ g = I$ in Michael Spivak's Calculus

Firstly, consider the function composition $f \circ g$. Let $f: A \to B$ and $g: C \to D$.

When understanding the notion of a composition's domain $\text{dom}(f \circ g)$, we need to consider the fact that $g$'s range $g[C]$ may not necessarily fully intersect with the $\text{dom}(f)=A$. Therefore, a given $x \in \text{dom}(g)= C$ is only in $\text{dom}(f \circ g)$ if $g(x) \in A$. From this, it is apparent that $\text{dom}(f \circ g) \subseteqq C$. Call this $C^*$.

(The above description is consistent with Spivak's definition of a composition's domain)

Now that we have characterized the notion of a composition's domain as $C^*$, it is straightforward to understand what we mean by the range of the composition $\text{ran}(f \circ g)$...specifically, $\text{ran}(f \circ g)=\text{ran}(f \upharpoonright C^*)$.

Secondly, consider a relation inverse $R^{-1}$, which we define as $R^{-1}=\{\langle y,x \rangle | \langle x, y \rangle \in R\}$. For a function $f$, $f^{-1}$ will only be a function if $f$ is injective (also referred to as '$\text{1-1}$').

I want to take a look at what one means implicitly when the following is written: $f \circ g = I$, where $I$ is the identity function. Given our earlier comments on compositions, $I$ is forced to be defined as follows: $I = \{\langle x,x \rangle \  |\  x \in C^*\}$.

Let $f$ be neither injective nor surjective. Let there be no restrictions on how we construct $g$. Prove that there exists a $g$ such that $f \circ g = I$.

Because $f$ is not injective, there exists an $x_1 \neq x_2$ such that $f(x_1)=f(x_2)$. Let's create a $g$ that purposefully avoids mapping into such elements (we will denote these as "bad elements" of $f$). So as to avoid invoking the Axiom of Choice, we will specify that for all elements in $\text{dom}(f)$ that are mapped to a common element in $\text{ran}(f)$ (the so-called bad elements), $g$ will not take any of its elements from $\text{dom}(g)$ into those bad elements of $\text{dom}(f)$. As a second property, let $\text{dom}(g) \subseteqq \text{ran}(f)$. For the final property, let us impose that $g$ is injective; this condition may require the $\text{dom}(g)$ to be trimmed so that the adherence to the first condition doesn't make the $|\text{ran}(g)|$ too small.

Define $g$ to have the following mapping rule: $x \mapsto f^{-1}(x)$ where $f(y)=x$. Importantly, because $g$ avoids mapping into any of the bad elements of $f$, the notion of $f^{-1}$ is unambiguous. That is to say, $f\upharpoonright \text{dom}(g)$ is injective...by design.

It would seem that I have successfully created a $g$ that satisfies the statement: $f \circ g = I$

To me, the statement $f \circ g = I$ is incredibly ambiguous. There is a problem in Spivak's book that amounts to saying: Prove that $f$ must be surjective in order for there to exist a function $g$ such that $f \circ g = I$. Well, I just did this...but as far as I can tell, it did not require $f$ to be surjective.

The issue seems to be in how one defines the domain of the Identity Function $I$. With a notation like $f \circ g = I$, it appears there are degrees of freedom in my choice of $g$ that should not be there.

答:
Is this the problem? (Chapter 3, Problem 23, 3rd Ed.):

Suppose that $f\circ g = I$, where $I(x) = x$. Prove that

(a) if $x\neq y$, then $g(x) \neq g(y)$;

(b) every number $b$ can be written $b = f(a)$ for some number $a$.

I probably don't know enough about modern function theory to address much of this but there was a related discussion you might find useful here

If I have the correct problem, Spivak in this case seems to want us to assume that here, for every real number $x$, $f(g(x)) = x$. Were this not the case, (b) would not necessarily be true. By asking us to prove something is true he implies enough information to make it true, but it wouldn't be true without those assumptions.

The takeaway from this problem, that $f\circ g = I$ $\implies$

(a) $g$ must be one-one, and

(b) every real number $x = f(a)$ for some real number $a$

are perhaps not interesting to folks with previous experience dealing with function and set theory :)


ContentLicense: CC BY-SA 4.0

Title: Is there a formula for (0.5)²+(1)²+(1.5)²+(2)²+...+(25)²

Body:
(1/4)(1³) + (1/9)(1³+2³) + (16)(1³+2³+3³) + ... + (1/2704)(1³+2³+3³+...+51³) = ?

Initially, I factored it as 1³ (1/2² + 1/3² + ... + 1/52²) + 2³ (1/3² + ... + 1/52²) + 3³ (1/4² +...+ 1/52²) + ... + 51³ (1/52²), but got stuck to split them as separate series.

Alternatively, I observed this pattern

1³/4 + (1³/9 + 2³/9) + (1+8+27)/16 + 100/25 + ... + (1+2+3+...+51)/52

= (1/2)² + (3/3)² + (6/4)² + (10/5)² + ... + ((1+...+51)/52)²

= (1/2)² + (2/2)² + (3/2)² + (4/2)² + (5/2)² ... + (51/2)²

= (1/2)² (1+2²+3²+...+51²)

= (1/2)² (1/6)(51)(2(51)+1)(51+1)

= (1/2)² (1/6)(51)(103)(52)

= (1/2)² (17)(103)(26)

= 45526/4

= 11381.25

Does this seem correct?

Body:
The sum can be expressed as:
$$
\frac1{2^2}\cdot 1^3+\frac1{3^2}\cdot (1^3+2^3)+\frac1{4^2}\cdot (1^3+2^3+3^3)+\cdots +\frac1{52^2}\cdot (1^3+2^3+\cdots +51^3)=$$
$$\sum_{i=2}^{52} \frac1{i^2} \sum_{j=1}^{i-1} j^3\stackrel{(1)}{=}\\
\sum_{i=2}^{52} \frac1{i^2}\cdot (\frac{(i-1)i}{2})^2=\sum_{i=2}^{52} \frac{(i-1)^2}4=\frac14\sum_{i=1}^{51} i^2\stackrel{(2)}{=}\cdots$$
where:
$$\sum_{i=1}^{n} i^3=(\frac{n(n+1)}{2})^2   \quad  \quad  (1)$$
$$\sum_{i=1}^{n} i^2=\frac{n(n+1)(2n+1)}{6}   \quad  \quad  (2)$$
Can you finish?


问:
Binomial expansion on quadratic solution
Given the following equation:

$$ p_{\gamma}^2+(-2m_dc)p_{\gamma} +2m_dB_d=0 \tag 1$$

Solving it for $p_{\gamma}$, using the quadratic formula, gives me:

$$ p_{\gamma} = \frac{2m_dc \pm \sqrt{4m_d^2c^2-4(2m_dB_d)}}{2}$$

$$ p_{\gamma} = m_dc \pm \sqrt{m_d^2c^2-(2m_dB_d)}$$

$$ p_{\gamma} = m_dc \pm m_dc \sqrt{1-\frac{2B_d}{m_dc^2}}$$

$$ p_{\gamma} \approx m_dc \pm m_dc\left(1 -\frac{B_d}{m_dc^2} \right)$$

But Im supposed to get the following solution for $p_{\gamma}$:

$$ p_{\gamma} = \frac{B_d}{c} \left(  1+ \frac{B_d}{2m_dc^2} \right) \tag 2$$

and I am aware that I have to use the quadratic formula and Binomial expansion to get to equation (2) but I can get there. Any help?

答:
Let's simplify the notation by dropping subscripts, and turn this into an arithmetic problem by making everything dimensionless:

$$
\frac{p}{mc} = 1 \pm \sqrt{1 - \frac{2B}{mc^2}}
$$

We can treat the square root as a binomial and expand.
The binomial expansion is

$$
(1 + \epsilon)^n = 1 + n\epsilon + \frac{n(n-1)}{2!}\epsilon^2 + \cdots
+ {n\choose m} \epsilon^m + \cdots
$$

and the trick is to neglect the higher-order terms; the trick works better when $n\epsilon \ll 1$. Taking the negative discriminant in your solution, the first-order approximation gives

\begin{align}
\frac{p}{mc}
&\approx 1 - \left( 1 - \frac12 \frac{2B}{mc^2}
\right)
= \frac{B}{mc^2}
\end{align}

To get the correction you're looking for, retain the second-order term, $-\frac18\left(\frac{2B}{mc^2}\right)^2$.


问:
Quotient space of the solid cube via identification of faces
Consider the quotient space of a cube $I^3$, where $I=[-1,1]$, obtained by the following identification.
$$
(1,y,z)\sim (−1,y,z),\quad (x,1, z)\sim (−x,−1,z)\quad \text{and} \quad (x,y,1)\sim(x,y,−1).
$$
how to describe the quotient space $X \!/\!_{\sim}$ as a product space?

So we identify the cube ABFE with the cube DCGH, identify the cube BFGC with the cube DAEH and twist the cube ABCD with the cube EFGH.

it is not possible to draw such pictures as Escher's illustration

I'm confused about following the process of how such an object is made.

Any help is warmly appreciated.

答:
Let $K$ be a spce obtianed as $I^2/R$ where R is as follows:  $(-1,y)\sim (1,y), (x,1)\sim (-x,-1)$.

This space is just Klein bottle. (See diagram with squares in https://en.wikipedia.org/wiki/Klein_bottle)

Now your space is just $K\times S^1$ where $S^1$ is a circle.(In the first two equations $z$ remains unchanged). It follows from the following general fact:

Let $Y$ be some space and $T=(Y\times I)/R'$ where $R'$ is $(t,1)\sim (t,-1)$. Then $Y\cong Y\times S^1$


问:
How many words of length $n$ is it possible to create from $\{a,b,c\}$ without any “$c$”s after a “$b$”?
This post is a duplicate of this post.

The question is,

How many words of length $n$ is it possible to create from $\{a,b,c\}$ without any “$c$”s after a “$b$”?

For example, the string $\color{blue}{\text{aabac}}$ is allowed but $\color{blue}{\text{cc}\color{red}{\text{bc}}\text{ab}}$ is not allowed because it contains $\text{bc}$ in that order. It could, however include $\text{cb}$.

I worked it for small values of $n$, and searched the resulting sequence on OEIS.

I have been told of a method by forming a recurrence relation, which I have to still think upon, but the question is why is it related to even indexed fibonacci sequence? Is there any natural interpretation of this? I was amazed to see so many equivalent definitions of this problem in the OEIS page and this definition is one of them. I would be glad to see some different answers to this problem.

Here is another related post.

答:
Let $a_n$ be the number of such strings of length $n$, $b_n$ be the number of such strings which end with b, and $c_n$ the number of such strings which do not end with b.  Then

$a_n = b_n+c_n$, since these are the  total number of such strings with length $n$

$b_n = a_{n-1}$, since you can take a string of length $n-1$ and put a b on the end, and this gives $a_n = a_{n-1}+c_n$

$c_n = b_{n-1}+2c_{n-1}$, since you can take a string of length $n-1$ ending in b and put an a on the end or you can take a  string of length $n-1$ not ending in b and put an a or a c on the end, and this gives $c_n =  a_{n-1}+c_{n-1}$

$c_n =  a_{n-1}+c_{n-1}$ and $a_n = a_{n-1}+c_n$ look Fibonacci-like.  Suppose we define $d_n$ such that $d_{2n-1}=c_n$ and $d_{2n}=a_n$.  Then we have $d_{2n-1}=d_{2n-2}+d_{2n-3}$ and $d_{2n}=d_{2n-1}+d_{2n-2}$ so combined $$d_{n}=d_{n-1}+d_{n-2}$$ which is the Fibonacci recurrence.

You also have to check the initial terms: $d_0=a_0=1$, $d_1=c_1=2$, $d_1=a_1=3$, so $d_n$ is the Fibonacci sequence offset by two positions, and $a_n=d_{2n}$ is the even terms of the Fibonacci sequence again with an offset.


问:
Solving integral with substitution gives wrong result
I want to solve the integral:

$$ \int_{-\infty}^{\infty} \frac{x^2}{(1+|x|^3)^4} \ dx.$$

I thought I could solve that via the substitution $u=1+|x|^3$ since then one has the integral:

$$ \int_{-\infty}^{\infty} \frac{1}{3u^4} \ du.$$

But that seems to be wrong (probably the substitution does not work here since it is not differentiable at 0), and that's why this integral is 0, but it should be $\frac29$. So how do I solve this correctly?

答:
The function is even, so rewrite the integral as

$$\int_{-\infty}^\infty \frac{x^2}{(1+|x|^3)^4}dx = 2\int_0^\infty\frac{x^2}{(1+x^3)^4}dx$$

Now try your substitution.


问:
Closed form for $\int_0^\infty\left(\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy\right)^3\mathrm dx.$
I need to find a closed form for these nested definite integrals:
$$I=\int_0^\infty\left(\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy\right)^3\mathrm dx.$$
The inner integral can be represented using the hypergeometric function $_2F_1$ or the complete elliptic integral of the 1st kind $K$ with an imaginary argument:
$$\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy=\frac\pi2   {_2F_1}\left(\frac12,\frac12;1;-x^2\right)=K(x\,\sqrt{-1}).$$
My conjecture is the integral $I$ has a closed-form representation:$$I\stackrel{?}{=}\frac{3\,\Gamma (\frac14)^8}{1280\,\pi^2}=7.09022700484626946098980237...,$$
but I was neither able to find a proof of it, nor disprove the equality using numerical integration. Could you please help me with resolving this question?</p>

答:
Using
$$ K(ik) =
\frac{1}{\sqrt{1+k^2}}K\left(\sqrt{\frac{k^2}{k^2+1}}\right) $$
and a substitution $t^2 = \frac{k^2}{1+k^2}$, rewrite the integral as
$$ \int_0^\infty K(i k)^3\,dk = \int_0^1 K(t)^3\,dt. $$

There is a paper "Moments of elliptic integrals and critical L-values"
by Rogers, Wan and Zucker (http://arxiv.org/abs/1303.2259; also one of
the authors' earlier papers: http://arxiv.org/abs/1101.1132), and the
authors, by relating this integral to an L-series of a modular form (their theorems 1 and 2),
show that
$$ \int_0^1 K(k)^3\,dk = \frac{3}{5}K(1/\sqrt{2})^4 =
\frac{3\Gamma(\frac14)^8}{1280\pi^2}, $$
using $K(1/\sqrt{2}) = \frac14 \pi^{-1/2}\Gamma(\frac14)^2$.

ContentLicense: CC BY-SA 3.0


答:
Since $x^2dx=1/3d(x^3)$ using $u=x^3$ we have

$$\int_0^\infty \dfrac{x^2}{(1+x^3)^4}=\int_0^\infty1/3\dfrac 1{(1+u)^4}du=1/3\left.\left( -1/3 \dfrac1{(1+u)^3}\right)\right|=1/9.$$

Now in your integral the function is even, so the answer is just $2\cdot 1/9=2/9$

Title: Solving integral with substitution gives wrong result

Body:
I want to solve the integral:

$$ \int_{-\infty}^{\infty} \frac{x^2}{(1+|x|^3)^4} \ dx.$$

I thought I could solve that via the substitution $u=1+|x|^3$ since then one has the integral:

$$ \int_{-\infty}^{\infty} \frac{1}{3u^4} \ du.$$

But that seems to be wrong (probably the substitution does not work here since it is not differentiable at 0), and that's why this integral is 0, but it should be $\frac29$. So how do I solve this correctly?


问:
Euler-Lagrange equations for functions with multiple dependent arguments
Consider the following functional equation:

$$ g[f] = \int_{t_1}^{t_2} f(x_1, x_2, x_1',x_2',u ) du.\tag{1}$$

Now, suppose there is some dependency between $x_2$ and $x_1$ that is there exists $x_2(x_1)$, without substituting directly the dependency, can I use euler-lagrange equation? I tried deriving it as follows by petrubing $x_1 \to x_1 + \epsilon \eta(u)$

$$ \delta g = \int_{t_1}^{t_2} \frac{\partial f}{\partial x_1} \eta + \frac{\partial f}{\partial x_2} \frac{ dx_2}{dx_1} \eta + \frac{\partial f}{\partial x_1' } \eta' + \frac{ \partial f}{\partial x_2' } \frac{ \partial x_2'}{\partial x_1' }  \eta' du.\tag{2}$$

Or by algebra and integration by parts

$$ \delta g = \int_{t_1}^{t_2} \eta \left[ \partial_{x_1} f  + (\partial_{x_2} f) \frac{dx_2}{dx_1} \right]  - \frac{d}{du} \left[ \partial_{x_1'} f  + (\partial_{x_2'} f )\frac{dx_2'}{dx_1'} \right] \eta du.\tag{3}$$

Hence, for the above expression to be true for all $ \eta$, it must be that:
$$ \partial_{x_1} f + (\partial_{x_2} f) \frac{dx_2}{dx_1}=\frac{d}{du} \left[ \partial_{x_1'} f  + (\partial_{x_2'} f) \frac{dx_2'}{dx_1'} \right].\tag{4}$$

Is the condition for optimum under optimizing with $x_1$?

Questions:

Is my result correct?

Suppose $x_2 \to x_1$ is not a bijective map, then there would be different forms of $f$ depending on which branch we substitute into the functional equation, so seems to me there would be multipile optimums but doing it without substituting lead me to only one answer. So, what's the buisness going on there? eg: say $x_1 = (x_2)^2 $ then $x_1 = \pm x_2$

I considered an analogy with implicit equations, say that of a circle

$$ y = \pm \sqrt{1-x^2}$$

If we isolate for $y$ and differentiate, it turns out that we get two branches for slope value for same $x$ but if we directly different $x^2+y^2=1$ then we would get a derivative which is function of $(x,y)$ , so the branch is directly accounted for , so is that a similar idea here?

Note:

$$ \partial_z f = \frac{\partial f}{\partial z}$$

答:
Comments:

The dependence between $x_1$ and $x_2$ can usually be implemented via a constraint $$\Phi(x_1,x_2)~=~0,$$
and a Lagrange multiplier $\lambda$ with extended Lagrangian
$$\widetilde{f}=f+\lambda \Phi .$$
This is e.g. the case with OP's circle example.

Often the constraint function $\Phi$ is of graph type $$\Phi(x_1,x_2)~=~x_2-\phi(x_1),$$
i.e. the constraint is a graph
$$x_2~=~\phi(x_1).$$
It is not difficult to see (by eliminating $\lambda$) that this is equivalent to OP's approach (4).


问:
If $\gcd(a,b,c) = 1$ and $c = {ab\over a-b}$, then prove that $a-b$ is a square.
If $\gcd(a,b,c) = 1$ and $c = {ab\over a-b}$, then prove that $a-b$ is a square. $\\$

Well I tried expressing $a=p_1^{a_1}.p_2^{a_2} \cdots p_k^{a_k}$ and $b = q_1^{b_1}.q_2^{b_2}\cdots q_k^{b_k}$ and $c=r_1^{c_1}.r_2^{c_2}\cdots r_k^{c_k}$ basically emphasizing on the fact that the primes which divide $a$ are different from those that divide $b$ and $c$, but I couldn't come up with anything fruitful. $\\$

Any help would be appreciated. $\\$

Thanks

EDIT:- $a,b,c$ are positive integers.

答:
Suppose $p\mid\gcd(a,b)$, where $p$ is prime. Let $p^r$ be the highest power dividing $a$, and $p^s$ be the highest power dividing $b$. Now since $p\not\mid c$ we must have $p^{r+s}\mid (a-b)$, and in fact it is the highest power dividing $a-b$ (since $(a-b)\mid ab$). If $r>s$, then this is impossible since $p^r\mid a$ but $p^r\not\mid b$, and similarly if $s>r$. So $s=r$.

If $p$ divides exactly one of $a$ and $b$ then it doesn't divide $a-b$. Finally, if $p$ divides neither then it can't divide $a-b$ since $(a-b)\mid ab$.

Thus every prime which divides $a-b$ does so an even number of times.


问:
Limits of convergent sequences, where one is strictly smaller
So I have been learning about sequences recently and I am struggling with the following problem:

Given two convergent sequences $(a_n)_{n \geq 0}$ and $(b_n)_{n \geq 0}$, where $a_n < b_n$ for all $n$, prove that $\lim a_n \leq \lim b_n$ follows.

It seems obvious that this would be true, but I can't just quite prove it. I thought maybe looking at the difference of the two sequences $b_n - a_n$ would be a good idea. Since $b_n > a_n$ for all $n$ this sequence would contain only positive numbers. If I could somehow show, that this always converges to some $d \geq 0$ than I thought this would prove it, however I haven't really been able to make some progress there so now I'm back to square one. I would really appreciate some help.

答: Let $d_n\geq 0$. If $d=\lim d_n<0$  then for $\varepsilon =-d/2>0$ there is $N$ such that for $n>N$ $|d_n-d|<-d/2$. But $|d_n-d|=d_n-d\geq -d> -d/2$. A contradiction.


答:
the best behaved indefinite ternary quadratic form is $y^2 - zx.$ It takes little to prove that, demanding $x>0$ and $\gcd(x,y,z) = 1,$

$$ x = u^2, \; \; y=uv, \; \; z = v^2   $$

But then all solutions to your $bc-ca+ab=0$   come from
$$ a=x+y, \; \; b = y, \; \; c = y+z . $$

You wanted $a-b.$
Here
$$ a-b = x+y-y = x = u^2  $$

Alright, let me work on the reverse direction a bit

Not bad: In $y^2 - zx = 0,$ take $x=a-b, y = a, z=a+c.$  So that, demanding $x>0,$  it is a square.   We have $\gcd(x,y,z) = \gcd(a,b,c)$  and a little effort tells us this is $1$

ADDED. There is already an answer that refers to unique factorization; here let me talk about $zx=y^2$ using just gcd. The extra demands are that $x >0$  and $\gcd(x,y,z) = 1.$  We begin with $g=gcd(z,y),$  so that $z=gs, y=gt,$ and $\gcd(s,t)=1.$

We reach $gsx = g^2 t^2,$  or $sx = g t^2.$  As  $\gcd(s,t)=1,$  we know that $s|g$  so that we may write $g = su.$  This leads to $sx= sut^2$ and $x=ut^2,$   so that $u>0.$

Next we combine to get $y=stu$ and $z = us^2$

We have reached $$ x=u t^2, \; \; y = stu, \; \; z = u s^2  $$
However, as $u$ divides all three and $\gcd(x,y,z) = 1,$  we know $u=1$  Finally
$$ x= t^2, \; \; y = st, \; \; z =  s^2  $$

Body:
If $\gcd(a,b,c) = 1$ and $c = {ab\over a-b}$, then prove that $a-b$ is a square. $\\$

Well I tried expressing $a=p_1^{a_1}.p_2^{a_2} \cdots p_k^{a_k}$ and $b = q_1^{b_1}.q_2^{b_2}\cdots q_k^{b_k}$ and $c=r_1^{c_1}.r_2^{c_2}\cdots r_k^{c_k}$ basically emphasizing on the fact that the primes which divide $a$ are different from those that divide $b$ and $c$, but I couldn't come up with anything fruitful. $\\$

Any help would be appreciated. $\\$

Thanks

EDIT:- $a,b,c$ are positive integers.


问:
Difference between $p\left(x\right)\cdot{q\left(x\right)}$ and $\left(p\cdot{q}\right)\left(x\right)$
Let $p\left(x\right),q\left(x\right)\in{R\left[x\right]}$ such that $p\left(x\right)=x$ and $q\left(x\right)=x+a$.

My lecture notes state that "$p\left(x\right)\cdot{q\left(x\right)}=x^2+xa$ and $\left(p\cdot{q}\right)\left(x\right)=x^2+ax$ are different if $R$ is not commutative."

I was wondering why $p\left(x\right)\cdot{q\left(x\right)}$ and $\left(p\cdot{q}\right)\left(x\right)$ are different (I see that $xa\neq{ax}$) because some books write that $\left(p\cdot{q}\right)\left(x\right)=p\left(x\right)q\left(x\right)$.

Which one must we use in the ring of polynoms?

答:
The point is that the variable in a polynomial ring commutes with everything. This is needed to simplify a product of polynomials down to a polynomial again. To use your example: If you multiply $p(x)$ and $q(x)$ and don’t assume commutativity, you get
$$
  p(x) \cdot q(x) = x (x + a) = x^2 + x a.
$$
To turn this into the standard form of a polynomial, you have to swap $a$ and $x$ at the end, i.e. $x$ should commute with $a$. We call the resulting polynomial (i.e. what you get after swapping $x$ and $a$ in the example) $p \cdot q$.

Now, an important feature of polynomials is that you can replace the variable $x$ by an element $z$ of the ring (or even an element of an $R$-algebra). If $z$ commutes with everything (as is would in the more familiar case of commutative rings), swapping the order is justified for $z$ as well and “evaluating at $z$” becomes a ring homomorphism $R[x] \to R$. If it does not, however, $(p \cdot q)(z)$ is different from $p(z) \cdot q(z)$. This means that the polynomial ring is a lot less nice in the non-commutative setting.

Having written this, is occurs to me that I find the statement in your lecture notes confusing as well, because $p(x) q(x)$ should be equal to $(pq)(x)$ if $x$ is the variable of the polynomial ring (which it appears to be) by definition of multiplication. What really should be stated is that $p(z) q(z)$ is not necessarily equal to $(p q)(z)$ if $z$ is some other value and $p(z)$ denotes evaluation of $p$ at $z$.


答:
One can quickly derive the recurrence relation for even-indexed Fibonacci numbers :

$$F_{2k+2}=F_{2k+1}+F_{2k}$$
$$F_{2k+2}=(F_{2k}+F_{2k-1})+F_{2k}$$
$$F_{2k+2}=(F_{2k}+(F_{2k}-F_{2k-2}))+F_{2k}$$
$$\Rightarrow \boxed{F_{2k+2}=3F_{2k}-F_{2k-2}} \tag{$k \ge 1$}$$

where $F_{0}=0$, $F_{2}=1$, $F_{4}=3$, $F_{6}=8$, $\ldots$

The problem satisfies the same recursive relation as given by @lulu :
$$\boxed{T_{n+1}=3T_{n}-T_{n-1}}$$

Also note, $T_1=3$, $T_2=3^2-1=8$.

The recurrence relation as well as the terms of sequence match. Hence the two sequences are identical.

Body:
This post is a duplicate of this post.

The question is,

How many words of length $n$ is it possible to create from $\{a,b,c\}$ without any “$c$”s after a “$b$”?

For example, the string $\color{blue}{\text{aabac}}$ is allowed but $\color{blue}{\text{cc}\color{red}{\text{bc}}\text{ab}}$ is not allowed because it contains $\text{bc}$ in that order. It could, however include $\text{cb}$.

I worked it for small values of $n$, and searched the resulting sequence on OEIS.

I have been told of a method by forming a recurrence relation, which I have to still think upon, but the question is why is it related to even indexed fibonacci sequence? Is there any natural interpretation of this? I was amazed to see so many equivalent definitions of this problem in the OEIS page and this definition is one of them. I would be glad to see some different answers to this problem.

Here is another related post.


问:
Can the del operator be "separated" like this?
In class, we were asked to prove the identity $\nabla\times(\Omega\vec{V})=\Omega(\nabla\times\vec{V})-\vec{V}\times\nabla\Omega$. One of the possible approaches involved separating the del operator into two operators,

\begin{equation}\nabla\equiv\nabla_{\Omega}+\nabla_{\vec{V}}\end{equation}

where $\nabla_{\Omega}$ differentiates $\Omega$ and leaves it operates on $\vec{V}$ zero, and vice versa. I know how to prove the identity, once I accept that $\nabla\equiv\nabla_{\Omega}+\nabla_{\vec{V}}$, but I find ti difficult to convince myself that it is true. Can del be separated like this?

答:
The Del operator $\nabla$ operates on the independent coordinate variables, $x$, $y$, and $z$,not the dependent variables $\vec V(x,y,z)$ and $\Omega(x,y,z)$.  So, we have

$$\begin{align}
\nabla \times (\Omega \vec V)&=\sum_{i=1}^3 \hat x_i \frac{\partial}{\partial x_i}\times \left(\Omega \sum_{j=1}^3 \hat x_jV_j\right)\\\\
&=\sum_{i=1}^3 \sum_{j=1}^3 (\hat x_i \times \hat x_j )\left(\Omega \frac{\partial V_j}{\partial x_i}+V_j \frac{\partial \Omega}{\partial x_i}\right)\\\\
&=\Omega \nabla \times \vec V+\nabla \Omega \times \vec V\\\\
&=\Omega \nabla \times \vec V-\vec V\times \nabla \Omega
\end{align}$$

as was to be shown!


问:
A sequence is defined by $x_1=1$, $x_{n+1}= \sqrt{8+x_n}$. Prove each term of the sequence after the first is greater than the preceding term.
I tried to use mathematical induction, but letting $n=k$ and $n=k+1$ is getting nowhere. I also tried squaring both sides to get rid of the square root but that doesn't seem to help either. Need some hints...

After reading some hints,
For $n=k$, assume $x_k > x_{k-1}$

$\sqrt{8+x_{k-1}} > \sqrt{8+x_{k-2}}$

Prove it is true for $k+1$,

$x_k = \sqrt{8+x_{k-1}}$

Squaring both sides,

$8+x_k > 8+x_{k-1}$

$x_k > x_{k-1}$

Since it is true for $n=k+1$, it is also true for $n=k$.

Seems right?

答:
Hint

You can try induction. If you suppose that $x_{n}>x_{n-1}$, then

$$8+x_n>8+x_{n-1}\to \sqrt{8+x_{n}}>\sqrt{8+x_{n-1}}.$$

Can you finish?


答:
Well $\frac {ab}{a-b}$ being an integer seems unlikely and specific.

In particular if $p$ is a prime divisor of $c$ then $p|\frac {ab}{a-b}$ so $p|ab$ so $p|a$ or $b|b$.  But $p$ can't divide both as $\gcd(a,b,c) =1$.

But we have that if $p|c$ then either $p|a$ or $p|b$ but not both but then $p\not \mid a-b$ yet for any $q|a-b$ we must have $q|a$ or $q|b$ so $q$ must divide both $a$ and $b$ so....

Okay.... Let $\gcd(a,b) = d$ and with $a=a'd; b= b'd$ and as $\gcd(a,b,c) = 1$ we have $\gcd(d,c) =1$. (Also we have $\gcd(a',b') = 1$.   so.....

$c = \frac {ab}{a-b} = \frac {a'b'd^2}{d(a'-b')} = \frac {a'b'}{a'-b'}d$.

But two things to note.  1) $d\not \mid c$ so we must have $d|(a'-b')$ and $\gcd(a',b') = 1$ so $a' - b'$ can't have any factors in common with $a'$ or $b'$ and so cant have any factors in common with $a'b'$[1].  So we must have $a'-b'|d$.

So we must have:

$c = a'b'\frac d{a'-b'} = a'b'$ and $a'-b' = d$ and $a-b = d(a'-b') = d^2 = \gcd(a,b)^2$.

Too find a case where this is possible we can let $a',b'$ but any two relatively prime integers.  Say $a' = 7$ and $b'=3$ so $a'-b' = 4$.  Multiply both by $4$ to get $a = 28$ and $b = 12$ and let $c = \frac {ab}{a-b} = a'b'$ or $c = \frac {28\cdot 12}{28-12}=\frac {28\cdot 12}{16} = 7\cdot 3=21$.

SO $\gcd(28,12, 21) = 1$ and $21 = \frac {28\cdot 12}{28-12}$.

Cute.

=====

[1] A cute, but distractingly tangential lemma:  If $\gcd(m,n) =1$ then $\gcd(mn, m- n)=1$.  Pf: If $p|mn$ then $p|m$ or $p|n$  but not both.  So $p\not \mid m-n$.

Title: If $\gcd(a,b,c) = 1$ and $c = {ab\over a-b}$, then prove that $a-b$ is a square.


问:
Topology of fundamental groupoid.
Given fundamental groupoid $\Pi_1(S^1)$ of the circle, how can one define a topology on it? The information on nlab did little help other than the fact that it can be done since $S^1$ is path-connected. Set theoretically we can make the association with $S^1 \times \mathbb{R}$ but how can we define a topology on $\Pi_1(S^1)$ so that we have a homeomorphism between them?

答:
Grupoid of $Y$ is the factor of the set of all map $[0,1]\to Y$ modulo homotpies. On the set of all maps between two toplogical spaces $Map(X,Y)$ there is so-called compact-open toplogy:https://en.wikipedia.org/wiki/Compact-open_topology

Then take quotient toplogy.

Definition of compact-open topology:

Let $K\subset X$ be a compact set and $U\subset Y$ be some open. Let $V(K,U)$ be the set of all maps $X\to Y$ such that $f(K)\subset U$. Then take the smallest topology such that all $V(K,U)$ are open.

Title: Topology of fundamental groupoid.


答:
In order to make the groupoid $\Pi_1(X)$ into a topological groupoid we need to put a topology on the set of objects of $\Pi_1(X)$ and a topology on the set of morphisms of $\Pi_1(X)$. This is an example of a category internal to $\text{Top}$.

The set of objects already has a topology since objects are just points in $X$.

We can define a topology on the set of homotopy classes of paths in $X$ by defining an open set $S_{[\gamma],U,V}$ around $[\gamma]$ to be the set of all $[\eta * \gamma * \phi]$ where $\phi$ is completely contained in some open $U \subset X$ and $\eta$ is completely contained in some open $V \subset X$.

Assuming that the composition of paths is defined, i.e that $\eta(0) = \gamma(1)$ etc.

The topology on the set of homotopy classes of paths in $X$ is the topology generated by all the $S_{[\gamma],U,V}$.

This is similar to how the set of homotopy classes of paths beginning at a base point $x_0 \in X$ is topologized in Hatcher to construct the universal cover.

Title: Topology of fundamental groupoid.

Body: Given fundamental groupoid $\Pi_1(S^1)$ of the circle, how can one define a topology on it? The information on nlab did little help other than the fact that it can be done since $S^1$ is path-connected. Set theoretically we can make the association with $S^1 \times \mathbb{R}$ but how can we define a topology on $\Pi_1(S^1)$ so that we have a homeomorphism between them?


答:
For CW complexes (like $S^1$), the topology on the fundamental groupoid will be discrete, i.e. it is just a usual groupoid. The idea for an arbitrary space is that the morphism sets in the fundamental groupoid is a quotient of the path space from point a to point b. Quotients have a natural topology on them, so this is a reasonable topology to put on the homotopy classes of paths from point a to point b. The issue is that composition of paths does not give a continuous map, so these topologies do not make the fundamental groupoid into a topological groupoid.

These issues can be fixed by hand, and this is done so by Jeremy Brazas (who is pretty active on these sites) Here, in his paper "The fundamental group as a topological group." Of course, this is not directly what you ask, but the techniques will work for the groupoid.

As an aside, a very reasonable way to have topological groupoids interact with CW complexes in an interesting way, is to not take things up to homotopy. The fundamental infinity groupoid of $X$ has its object space the points of $X$ and the morphism spaces the (Moore) path spaces of $X$. Then by taking path components of morphism spaces, we can recover the usual fundamental groupoid. This object is very important and studying it is part of the foundations of infinity category theory.

Title: Topology of fundamental groupoid.


问:
Prove that $\int^{1}_{0} \sqrt{x^2+x}\,\mathrm{d}x < 1$

Prove that $\int\limits^{1}_{0} \sqrt{x^2+x}\,\mathrm{d}x < 1$

I'm guessing it would not be too difficult to solve by just calculating the integral, but I'm wondering  if there is any other way to prove this, like comparing it with an easy-to-calculate integral. I tried comparing it with $\displaystyle\int\limits^{1}_{0} \sqrt{x^2+1}\,\mathrm{d}x$, but this greater than $1$, so I'm all out of ideas.

Body:
HINT:

$x^2+x<x^2+x+\frac14=\left(x+\frac12\right)^2$


问:
Clarification on Outer Measure (Folland)
On page 29 of Folland's Real Analysis for an outer measure $\mu^*$ a set $A \subset X$ is called $\mu^*$-measurable if

$$
\mu^*(E) = \mu^*(E \cap A) + \mu^*(E \cap A^C)
$$

for all $E \subset X$.

Folland goes on to say that the inequality

$$
\mu^*(E) \leq \mu^*(E \cap A) + \mu^*(E \cap A^C)
$$

is obvious and holds for any $A$ and $E$.  I don't understand exactly how this is obvious though, am I missing something about the outer measure/sets in general that makes this assertion trivial?

答:
Let $A$ be any set $A=(A\cap B)\cup (A\cap B^c),$

$$m^*(A)=m^*((A\cap B)\cup (A\cap B^c))$$

$$m^*(A)\leq m^*(A\cap B)+m^* (A\cap B^c)$${ By subadditive property of outer measure.}

I am leaving this proposition( subadditive property) to prove.
Proposition: If $\{E_n\}_{n=1}^\infty$ is any countable collection of sets, disjoint or not , then
$$m^*(\bigcup_{k=1}^\infty E_k)\leq \sum_{k=1}^\infty m^*(E_k)$$


问:
What is the total derivative of a total derivative?
Suppose a function $y(t)$. Its total derivative is:

$$
d y(t)=y'(t)dt
$$

Now I want to take the total derivative again:

$$
d(dy(t))=d(y'(t)dt)=(dy'(t))dt+y'(t)d(dt)=y''(t)dt+y'(t)d(dt)
$$

What is $d(dt)$ - is it zero?

答: You will understand an aswer if you understand for what argument do you want to take a derivative. Derivative depends on two numbers: $(d f)[x,t]=f'[x]t$. So \begin{equation}(d^2 f)[x,t]=d(f'[x]t)[x,t]=td(f'[x])[x,t]=tf''[x]t=f''[x]t^2,\end{equation} where in square brackets I wrote an argument of a function. Replacing $t$ with more convenient $dt$ you have that $(d^2 f)[x,dt]=f''[x](dt)^2$. The second derivative is a quadratic form by $dt$.


Title: Prove that $\int^{1}_{0} \sqrt{x^2+x}\,\mathrm{d}x < 1$

Body:
Prove that $\int\limits^{1}_{0} \sqrt{x^2+x}\,\mathrm{d}x < 1$

I'm guessing it would not be too difficult to solve by just calculating the integral, but I'm wondering  if there is any other way to prove this, like comparing it with an easy-to-calculate integral. I tried comparing it with $\displaystyle\int\limits^{1}_{0} \sqrt{x^2+1}\,\mathrm{d}x$, but this greater than $1$, so I'm all out of ideas.

Body:
Well, you could use the AM-GM inequality:
$$\sqrt{x^2 + x} = \sqrt{x(x+1)} < \frac{x + (x+1)}{2} = x + \frac{1}{2}$$
and then
$$\int_0^1 \sqrt{x^2 + x} \, dx < \int_0^1 x + \frac{1}{2} \, dx = \frac{1}{2} + \frac{1}{2} = 1.$$


问:
Integral of a modified softplus function
In a manuscript I am currently reading, the authors propose a modified softplus function

$$g(a)=\frac{\log\left(2^a +1 \right)}{\log(2)}$$

for some $a \in \mathbb{R}$. The authors then claim that if $a$ is a polynomial, e.g. $a(x)=c_0 + c_1x+c_2x^2$, then the following integral has a closed-form solution:

$$\int_0^tg\left( \partial a(t) \right)\,dt$$

where $\partial$ denotes the derivative. Naively plugging in $g$ and then $\partial a(t)$, I obtain:

$$\int_0^t\frac{\log\left(2^{\partial a(t)} +1 \right)}{\log(2)} \,dt =\int_0^t\frac{\log\left(2^{c_1+2c_2t} +1 \right)}{\log(2)} \,dt$$

WolframAlpha informs me that this still has a closed form solution, but if we move one order higher, the integral seems to be indefinite. I assume there is some simplification or trick which I don't see to solve the integral more easily. Would you have any idea what this might be?

答:
For the indefinite integral $$\int\frac{\log\left(2^{c_1+2c_2t} +1 \right)}{\log(2)}\, dt,$$ assuming that $\log$ is the natural logarithm, we can use the substitution
\begin{align*}
u &= -2^{c_{1} + 2c_{2}t}\\
du &= -2c_{2}\log(2)2^{c_{1} + 2c_{2}t} \,dt = 2c_{2}\log(2)u\,dt.
\end{align*}

This gives us
\begin{align*}
\int \frac{\log\left(2^{c_1+2c_2t} +1 \right)}{\log(2)}\, dt &= \int\frac{\log(1-u)}{2c_{2}\log^{2}(2)u}\,du\\[5pt]
&= \frac{1}{2c_{2}\log^{2}(2)}\int\frac{\log(1-u)}{u}\,du\\[5pt]
&=-\frac{1}{2c_{2}\log^{2}(2)}\operatorname{Li_{2}}(u)+C\\[5pt]
&=-\frac{\operatorname{Li_{2}}(-2^{c_{1} + 2c_{2}t})}{2c_{2}\log^{2}(2)}+C.
\end{align*}
Here $\operatorname{Li_2}$ is the dilogarithm given by $$\operatorname{Li_2}(z) = -\int_{0}^{z}\frac{\ln(1-u)}{u}\,du.$$
So, for the definite integral: $$\boxed{\int_0^x\frac{\log\left(2^{c_1+2c_2t} +1 \right)}{\log(2)}\, dt = \frac{\operatorname{Li_{2}}(-2^{c_{1}}) - \operatorname{Li_{2}}(-2^{c_{1} + 2c_{2}x})}{2c_{2}\log^{2}(2)}.}$$


问:
What is the name of this shape? (spacetime)

(source: relativitet.se)

After seemingly endless searching for terms such as curved cone, hyper-cone etc I am at a loss as to what this shape is called. I believe it is commonly used to depict the curvature of space time.

Also, if you have any relevant information on the geometry of it, such as parametric equations, curved surface area, enclosed area at height z, volume etc that would be much appreciated.

Edit: Funnel.

Any idea how to get the surface area of a disk at height z and volume in a limited domain?

I'll give some background to this problem. I am designing a building modeled on the pillar shape that emerges when a water droplet falls into a body of water, the shape of which is close modeled by this. As part of the design brief it is required that the average floor area (so disks at several discrete heights in z) is adhered to. Surface area and volume are also required for inputs into Building Information Modeling system. So essentially I need to set the parameters to give the building envelope while meeting the average floor area requirements.

答:
The name of the shape depends on the exact curvature in the longitudinal direction and what happens at infinity. You have shown a finite segment, so one cannot say for sure.

If the surface has constant negative Gaussian curvature, it is known as a pseudosphere (the sphere itself has constant positive Gaussian curvature). It has an outer limit where the surface turns parallel to its cross-section, similar to the outermost (largest) circle in the illustration but further down the curve.

If the longitudinal curve is a segment of a parabola or hyperbola, then the shape is an example of a quadric surface. The full curve extends to infinity in both directions.


问:
LCM of two periodic signals,
Why is $$\text{lcm}\left(\frac{\pi}{5},\frac{\pi}{2}\right)=\pi$$
where the $10$ here represents the period of $2\cos(10t+1)-\sin(4t-1)$

where $\displaystyle\frac{\pi}{5}$ is the period of $2\cos(10t+1)$ and $\displaystyle\frac{\pi}{2}$ is the period of $-\sin(4t-1)$.

The question is not why the lcm but it's why does the lcm of these two give $\pi$.

答: Suppose $n\pi/5=m\pi/2$ where $n,m\in\Bbb Z_{>0}$. For least multiple we need to minimize $m$. We get $n/m=5/2$ and the minimum $m$ that gives $n$ integral is $m=2$.

Title: LCM of two periodic signals,


问:
derivative trace of matrix inverse square
Consider matrix M = $\sum_{j=1}^n p_jf(x_j)f^T(x_j)$, P,X - scalar vectors, $ f=\left(1,x_1,x_2,x_1x_2,x_1^2,\ x_2^2\right)^T, x1,x2 \in (-1, 1), p_j > 0$. I need find solution for $\frac{\partial tr(M^{-2})}{\partial p_j}$. Thanks!!!!

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}\def\d{\operatorname{diag}}\def\D{\operatorname{Diag}}\def\m#1{\left[\begin{array}{r}#1\end{array}\right]}$Define
the $f$-vectors a bit more concisely along with some related matrices
$$\eqalign{
f(x) &= \m{1&x_1&x_2&x_1x_2&x_1^2&x_2^2}^T \\
f_j &= f(x_j) \\
F &= \m{f_1&f_2&\ldots &f_n} \\
p &= \m{p_1&p_2&\ldots &p_n}^T \\
M &= F\D(p)\,F^T \;=\; M^T \\
}$$
And let's use a colon to denote the inner/trace product
$$A:B = {\rm Tr}(A^TB)$$
Write the objective function using the above variables and calculate its gradient.
$$\eqalign{
\phi &= M^{-1}:M^{-1} \\
d\phi &= 2M^{-1}:dM^{-1} \\
 &= -2M^{-1}:M^{-1}\,dM\,M^{-1} \\
 &= -2M^{-3}:dM \\
 &= -2M^{-3}:F\D(dp)\,F^T \\
 &= -2F^TM^{-3}F:\D(dp) \\
 &= -2\d(F^TM^{-3}F):dp \\
\p{\phi}{p} &= -2\d(F^TM^{-3}F) \\
}$$
So there's the gradient. To enforce the positivity constraint use a
projected gradient rather than a regular gradient descent method.


问:
Sequential discontinuity verification
I am trying to prove that the function below is not continuous by the sequential definition of continuity, I did a tentative approach but I have never done continuity by the sequential definition so I am not sure if it is correct.

Questions:

Consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by:

$f(x) = \sin(\frac{1}{x})$ if $x$ is different from $0$

$f(x) = a$ if $x = 0$

My proof:

Regardless of the value of $a$, the sequence $x_n = \frac{1}{n}$ converges to $0$ when $n$ goes to infinity, however $f(x_n) = \sin(n)$ and therefore the sequence $f(x_n)$ does not converge when $n$ goes to infinity, so we are done.

Is this correct? If so, can it be improved.  If it is not correct, please point out where I slipped up.

答: Your proof is correct but you should say that $\displaystyle \lim_{n \to \infty} \sin n$ does not exist hence $f(x_n)$ does not converge to $f(0) = a$. Thus $f$ is not continuous at $x = 0$. You could take $x_n = \dfrac{1}{(4n+1)\frac{\pi}{2}}$ if $a \neq 1$ and $x_n = \dfrac{1}{(4n-1)\frac{\pi}{2}}$ if $a \neq -1$ then it is not the case that $f(x_n) \to f(0) = a$ .

Title: Sequential discontinuity verification


问:
When to use exponent rule $\sqrt x^2$ and when to square both sides?
I am solving equations to see if the equation is a function or not.  I would like to know when to use exponent rules and when to square both sides?  The equation is $$x= \sqrt {1 - y^2}$$ I want to use an exponent rule but am unsure because the y squared term is under the radical. The exponent rule I want to use is $$ \sqrt {a^2}= a^\frac 22$$ If I apply that rule to the equation I get $$x=1^\frac 22 -y^\frac22$$ Then I can solve for y $$y=1-x$$. Yes y is a function of x.

The above is wrong because $ \sqrt { x^2}$ =
$ \lvert x \rvert$.  The $x^2$ means there will be a negative and positive number equal distance away.

Upon further research I’d like to find out if this following method is acceptable for the same problem.   $$ x = \sqrt {1-y^2} $$  $$x=1- \lvert y \rvert$$

$$ \lvert y \rvert= -x + 1$$

Not sure if this correct and still wondering if the absolute value sign would stop this equation from being a function.

答:
If $y = f(x)$ is a function then if you ever have an $(x_1, y_1)$  you can not have an $(x_1, y_2)$ unless $y_1 = y_2$.

so that should tell you that if you ever have $x = \sqrt{1 - y^2}$ then as $y^2 = (-y)^2$ you will also have $x = \sqrt{1 - (-y)^2}$ so if you ever have $(x,y)$ you will also have $(x,-y)$ and if $y \ne 0$ then $y \ne -y$ so it can't be a function.

Example:  If $y= \frac 12$ then $x = \sqrt{1-\frac 1{2^3}} = \frac {\sqrt 3}2$ and you have both $(\frac{\sqrt 3}2, \frac 12)$ and $(\frac {\sqrt 3}2, -\frac 12)$.  So it can't be a function.

There not really any rule that $\sqrt{a^2} \underbrace{=}_{true} (a^2)^{\frac 12}\underbrace{=}_{!!!FALSE!!!!} a^{\frac 22} = a$.

The is a rule that $\sqrt{a^2} = |a|$ and $|a| = \begin{cases} a\\-a\end{cases}$.

So $x = \sqrt{1 - y^2}$ does mean that $x^2 = 1-y^2$ but WE MUST PUT A PIN IN AND REMEMBER THAT $1-y^2 \ge 0$ and $x \ge 0$.

And so $y^2 = 1-x^2$  (and we MUST remember that $x\ge 0$)

And so it is true that $\sqrt{y^2}=\sqrt {1-x^2}$.

!!BUT!!!! $\sqrt{y^2} \ne y$.

$\sqrt {y^2} = |y| = \begin{cases}y\\-y\end{cases}$ so

$\begin{cases}y\\-y\end{cases} = \sqrt{1-x^2}$ and

$y = \begin{cases} \sqrt{1-x^2}\\-\sqrt{1-x^2}\end{cases}$.

So not a function.


答:
A method which is generally applicable to concave functions:

If $f:[a, b] \to \Bbb R$ is concave then its graph lies below any tangent line:
$$
 f(x) \le f(c) + f'(c)(x-c) \, .
$$
If one chooses $c =(a+b)/2$ then the integral of the second term on the right over the interval $[a, b]$ is zero, and therefore
$$
 \int_a^b f(x) \, dx \le (b-a) \cdot f\left( \frac{a+b}{2}\right) \, .
$$

In our case this gives
$$
 \int_0^1 f(x) \, dx \le f\left(\frac 12 \right) = \frac 12 \sqrt 3 < 1 \, .
$$
The estimate $\frac 12 \sqrt 3 \approx 0.86603$ comes pretty close to the exact value of the integral, which is $\approx 0.84032$.


Title: LCM of two periodic signals,

Body:
Why is $$\text{lcm}\left(\frac{\pi}{5},\frac{\pi}{2}\right)=\pi$$
where the $10$ here represents the period of $2\cos(10t+1)-\sin(4t-1)$

where $\displaystyle\frac{\pi}{5}$ is the period of $2\cos(10t+1)$ and $\displaystyle\frac{\pi}{2}$ is the period of $-\sin(4t-1)$.

The question is not why the lcm but it's why does the lcm of these two give $\pi$.

Body: $$LCM(\frac{a}{b},\frac{c}{d},\frac{e}{f})=\frac{LCM(a,c,e)}{HCF{(b,d,f)}}.$$


问:
Bernoulli Process Probability that m-th failure occurs before n-th successes.
We are given that the probability of success is $p$ for each trial.

The questions asks me to find the probability that $m$-th failure occurs before the n-th success.

The answer I ended up with is seemingly incorrect, I was told that instead I managed to find the probability that $m$ failures occur before the $n$-th success.

My Solution

Any help would be appreciated.

Edit:My second attempt.

答:
Your formula holds if $n\leq m$

Oddly, if $m = n$, it represents the simplest way to solve the question of finding the Pr of losing a two player tournament of best of $(2m -1)$ matches.

In order to lose a $5$ set tennis match, for example, you can allow your opponent to win a maximum of $2$ of the $5$ sets, thus

$P(lose) =\dbinom5 0 p^0q^5 +\dbinom5 1 p^1q^4 + \dbinom 5 2 p^2q^3$


问:
How is the area of a country calculated?
As countries' or states' borders are not straight lines but they are irregular in nature, I wonder how anyone can calculate the area of a country or state.

When do you think the area of a country or state was first calculated?  Was it before satellites provided us accurate picture of the earth?

Note: I am not asking about surface area of a country. They are assumed as flat while calculating the area.

答:
Geographers used to measure areas with a planimeter:

http://en.wikipedia.org/wiki/Planimeter

Nowadays I suppose they use something like ArcGIS, which has a built-in area calculator, but I don't know what algorithm is used.


问:
How to find the indefinite integral of $\arctan((x^2+1)^{1/3})$?
I have to find the indefinite integral of $\arctan((x^2+1)^{1/3})$. I have tried to do different variable substitutions, then integration by parts, but in the end I always got a another integral that I could not calculate. Please give a hint how to calculate this integral.

答: If you do $x=y^3-1$ and $\mathrm dx=3y^2\,\mathrm dy$, then that indefinite integral becomes$$\int3y^2\arctan(y)\,\mathrm dy.$$You can deal with it by integration by parts$$\int3y^2\arctan(y)\,\mathrm dy=y^3\arctan(y)-\int\frac{y^3}{y^2+1}\,\mathrm dy.$$Can you take it from here?


问:
Counting the number of 2-cycles, 3-cycles and 4-cycles in $S_4$
I know that in $S_4$ there are five different cycle types: double transpositions, 2-cycles, 3-cycles, 4-cycles and 1-cycle (i.e. identity).

My question is, does there exist a combinatorial formula or similar for counting the number of, for example, the number of 3-cycles in $S_4$?

答:
The set $\{1,2,3,4\}$ has $\binom{4}{3}=4$ subsets of cardinality 3, e.g. $\{1,2,3\}$.
Each of them gives $(3-1)!=2$ different 3-cycles: $(1,2,3)$ and $(1,3,2)$, once you fix the first entry there are $2!$ possibilities for the rest.

In total there are $4\cdot 2=8$ different 3-cycles in $S_4$.

In general there are
$\binom nk\cdot(k-1)!$ differen
$k$-cycles in $S_n$.


答:
Yet another way: If $0 < x < 1$, then $x^2 < x$. So

$$\int_0^1 \sqrt{x^2+x} \ dx < \int_0^1 \sqrt{2x}\ dx = \frac{2 \sqrt{2}}{3} < 1$$

(This bound is about 0.9428, so not as good as Martin R's answer)


答:
It depends on what notation you are using.  Using standard notation, where the second derivative if $\frac{d^2y}{dt^2}$, this isn't actually possible.  However, if you use a more flexible notation, $d(dt) = d^2t$.  In this notation, the second derivative is

$$ \frac{d^2y}{dt^2} - \frac{dy}{dt}\frac{d^2t}{dt^2} $$

This notation allows for the algebraic manipulation of higher-order differentials, which the standard notation does not allow for.

Historically, many have considered $d(dt) = 0$ iff you believe $t$ to be moving at a constant rate.  This is where the origin of the more simplified version of the second derivative came from.  Basically, pre-relativity, we imagined time to be flowing constantly, so $dt$ was a constant, so its differential was zero.

However, if you are just dealing with equation manipulations like you are doing, then you need to use the expanded notation.

For more information, see my paper, "Extending the Algebraic Manipulability of Differentials".

Title: What is the total derivative of a total derivative?

Body:
Suppose a function $y(t)$. Its total derivative is:

$$
d y(t)=y'(t)dt
$$

Now I want to take the total derivative again:

$$
d(dy(t))=d(y'(t)dt)=(dy'(t))dt+y'(t)d(dt)=y''(t)dt+y'(t)d(dt)
$$

What is $d(dt)$ - is it zero?


问:
Let $G$ be a group and $H\unlhd G$ with $|H|=25$. Also $G/H \cong A_4$. How many subgroups of order $75$ in $G$? And order $100$? How many are normal?
Let $G$ be a group and $H$ a normal subgroup with $|H| = 25$. Also we know that $G/H \cong A_4$.

I want to find out answers for these 2 questions:

How many subgroups of order 75 in $G$?

How many subgroups of order 100 in $G$? And how many of them are normal?

Here's what I found about that:

By the Lagrange theorem we can say that $|G| = 300$ (Because $|G| = |H||G/H|)$.

By the same reason for every $H'$ such that $|H'| = 75$ we can say that $|G/H| = 4$ (May be we can say that there's not many groups or order 4 for answer the 1'st question).

Second fact I think we can use is that order of every $g \in G$ is divisor of 300. And may be exist some pretty easy proof based on number theory.

答:
You don't need Sylow's Theorem. If $H$ is a normal subgroup with order coprime to its index in $G$, then every subgroup of $G$ of order a multiple of $H$ must contain $H$.

Hint: $|NX:N|=|X:X\cap N|$.

So all the subgroups you are asked about must contain $H$, and you can use the correspondence theorem.


问:
Given area of parallelogram, find smallest distance from origin to furthest point (complex numbers)

On the complex plane, the parallelogram formed by the points 0, $z,$
$\frac{1}{z},$ and $z + \frac{1}{z}$ has area $\frac{35}{37}.$ If the
real part of $z$ is positive, let $d$ be the smallest possible value
of $\left| z + \frac{1}{z} \right|.$ Compute $d^2.$

I begin by letting $z=x+yi$ and $\frac{1}{z}=\frac{x-yi}{x^2+y^2}$. Then, I applied the cross product to get that the

$$\text{Area of Parallelogram} = |z| \left|\frac{1}{z}\right| \sin(\theta) . $$

This implies that

$$ \frac{35}{37}=\left(\sqrt{x^2+y^2}\right) \cdot \left(\frac{1}{\sqrt{x^2+y^2}}\right) \sin(\theta) \Longrightarrow \frac{35}{37} = \sin(\theta) , $$

where $\theta$ is the angle between the $\frac{1}{z}$ and $z$. Subsequently, $\sin\left(\frac{\theta}{2}\right)=\frac{5}{\sqrt{74}}$ meaning that $z$ is in the form $x+\frac{5x}{7}i$. The problem reduced to an optimization problem, minimizing

$$ \left(x+\frac{x}{x^2+y^2}\right)^2+\left(y-\frac{y}{x^2+y^2}\right)^2=\left(x+\frac{x}{\frac{x\sqrt{74}}{7}}\right)^2+\left(\frac{5x}{7}-\frac{\frac{5x}{7}}{\frac{x\sqrt{74}}{7}}\right)^2 . $$

I eventually got the answer to be $\left(\frac{35}{37}\right)^2=\frac{1225}{1369}$. However, this answer is wrong. I am not sure where I messed up. Where did I go wrong?

答:
Let $a$ and $b$ two $2$-dimensional vectors satisfying $||a||\cdot||b||=1$.  Then (via Gram) the square of area of the parallelogram spanned by these vectors is $(35/37)^2=||a||^2\cdot||b||^2-\langle a,b\rangle^2=1-\langle a,b\rangle^2$ from where $\langle a,b\rangle=\pm12/37$.

Now
$$\begin{align}
d^2=||a+b||^2&=||a||^2+||b||^2+2\langle a,b\rangle\\
&=||a||^2+\frac{1}{||a||^2}\pm\frac{12}{37}\\
&=\left(||a||+\frac{1}{||a||}\right)^2-2
\pm\frac{12}{37}\\
&\geq 4-2
\pm\frac{12}{37}\\
&=2
\pm\frac{12}{37},
\end{align}$$
as for all reals $x$ we know $-2<x+1/x<2$.


问:
If $\gcd(a,b,c) = 1$ and $c = {ab\over a-b}$, then prove that $a-b$ is a square.
If $\gcd(a,b,c) = 1$ and $c = {ab\over a-b}$, then prove that $a-b$ is a square. $\\$

Well I tried expressing $a=p_1^{a_1}.p_2^{a_2} \cdots p_k^{a_k}$ and $b = q_1^{b_1}.q_2^{b_2}\cdots q_k^{b_k}$ and $c=r_1^{c_1}.r_2^{c_2}\cdots r_k^{c_k}$ basically emphasizing on the fact that the primes which divide $a$ are different from those that divide $b$ and $c$, but I couldn't come up with anything fruitful. $\\$

Any help would be appreciated. $\\$

Thanks

EDIT:- $a,b,c$ are positive integers.

Body: Note $\,(a\!-\!b)(a\!+\!c)=a^2\,$ and $\, \overbrace{(\color{#c00}{a\!-\!b,a\!+\!c})=1}^{\text{see below}}\,$ hence $\,a\!-\!b\,$ is a square. $\ \small\rm QED$


问:
Counting possible meal question
Sara has 3 varieties of pasta, 4 different sauces, 5 kind of fruits, milk, juice and water. A meal is made from a variety of pasta and sauce, one fruit and a beverage, how many different meal can she prepare if she never drink juice?

I first begin with 3.4 for pasta and sauce but I don't know if I should add or multiply 5 kind of fruits and the 2 left kind of beverage.
Also there's 3 varieties of pasta but I don't think the question meant that different sauces can be mixed.

My answer is 3.4.5.2 = 120 but I'm not sure this is the right way of thinking or if there's a trick in the question.

答:
you are correct, you multiply the options for each meal component and take a factor of 2 for the beverages to exclude juice and only keep milk and water.

So $4\times 3 \times 5 \times 2 = 5!= 120$ different meals are possible if each consists of exactly one pasta, one sauce, 1 fruit and one of 2 beverages.

Title: Counting possible meal question


问:
How to convert to exact arithmetic?
I'm working through a numerical analysis exercise where I use Newton's method over five iterations. Below is my code.

```<code>x = [-5:.01:5];
x0 = 1;
f = 1+0.99*x-x;
fprime = -0.01;

% calculation: take five iterations
for i = 1:5
    h= x0 - f/fprime;
    xnew = h;
end

fprintf("The root after 5 iterations: %25.e", xnew);
</code>```

I have no issue with how this code works, but I am required to express the number of iterations to reach a root in exact arithmetic. My textbook doesn't contain information about exact arithmetic and Google wasn't too specific.

I know the root is found after one iteration. How do I express $1$ in exact arithmetic?

答:
That is exactly the answer. In exact arithmetic (for instance using a rational number data type over unlimited integers), Newton solves linear systems in one step.

(One could also directly use the linear solver that is used anyway to compute the Newton step.)

Title: How to convert to exact arithmetic?


问:
Proving an equation using Ricci / Bianchi identities
Consider the Ricci tensor given by $R_{\alpha \beta} = \nabla_\alpha \nabla_\beta \phi$ for scalar field $\phi$.  I've been trying to use the Ricci and contracted Bianchi identities to show that
$$\nabla_\alpha (\nabla_\beta \nabla^\beta \phi) = -2R_{\alpha \beta} \nabla^\beta \phi.$$

[In this case the Ricci identity is $\nabla_\alpha \nabla_\beta V^\gamma - \nabla_\beta \nabla_\alpha V^\gamma = R^\gamma_{\delta \alpha \beta} V^\delta$ and the contracted Bianchi identity is $\nabla^\beta R_{\alpha \beta} = \frac{1}{2}\nabla_\alpha R$.]

My 'attempt':

One can contract the definition with $g^{\alpha \beta}$ to get $R = \nabla_\beta \nabla^\beta \phi$, where $R$ is the Ricci scalar. Applying $\nabla_\alpha$ to both sides and using the contracted Bianchi identity, I got
$$2 \nabla^\beta \nabla_\alpha \nabla_\beta \phi = \nabla_\alpha (\nabla_\beta \nabla^\beta \phi ),$$
which is perhaps a non-zero amount of progress. I also played around with the two identities separately but couldn't produce anything that seemed useful. Any help in getting started is greatly appreciated!

答:
Consider

$$R_{\alpha \beta} = \nabla_\alpha \nabla_\beta \phi\tag{1}$$

Thus, applying $\nabla_\beta$ in the last equality and by Ricci identity and Bianchi identity, we have
\begin{align*}
\nabla_\beta R_\alpha^\beta&=\nabla_\beta(\nabla_\alpha\nabla_\beta\phi)\\
\frac{1}{2}\nabla_\alpha R&=\nabla_\alpha(\nabla_\beta\nabla^\beta\phi)+R^\beta_{\delta\beta\alpha}\nabla^\delta\phi.\tag{2}
\end{align*}
By other hand, tracing the equality (1), we obtain
$$R=R_\beta^\beta=\nabla_\beta\nabla^\beta\phi.$$
Applying $\nabla_\alpha$ in the last equality, we obtain
$$\nabla_\alpha R=\nabla_\alpha\nabla_\beta\nabla^\beta\phi\tag{3}$$

Putting, (3) in (2), we obtain
\begin{align*}
\frac{1}{2}\nabla_\alpha(\nabla_\beta\nabla^\beta\phi)&=\nabla_\alpha(\nabla_\beta\nabla^\beta\phi)+R_{\delta\alpha}\nabla^\delta\phi\\
-\frac{1}{2}\nabla_\alpha(\nabla_\beta\nabla^\beta\phi)&=R_{\delta\alpha}\nabla^\delta\phi\\
\nabla_\alpha(\nabla_\beta\nabla^\beta\phi)&=-2R_{\alpha\beta}\nabla^\beta\phi
\end{align*}

In the last equality, we change $\delta$ by $\beta$ (since is a summand) and the fact that $Ric$ tensor is symmetric.


问:
Show that $f(x)=n/(1+nx)$ does not uniformly converge on $(0, \infty)$
I have proved that this sequence of functions is uniform convergent on $[1,\infty)$ by choosing $N>1/\epsilon$ (correct me if i am wrong).

Now need to show that it is not uniform convergent on $(0,\infty)$. What should be the choice for $N$?

Body: Hint: If $x=\frac1{n^2}$, then$$\frac n{1+nx}=\frac n{1+1/n}=\frac{n^2}{n+1}.$$


问:
Prefix-Free vs Uniquely Decodable Codes
It has been proved on the Cover-Thomas book that for a set of uniquely decodable codes with lengths $\{l_i\}$ (finite or infinite) that satisfies the Kraft-McMillan inequality, $\sum_i 2^{-l_i}\le 1$,  we can construct an instantaneous code with the same code-word lengths, which can be expressed as nodes on a tree graph.

I'm a bit confused about a (counter)example: fixed-length code (for example, all binary codes with length 10, not more/less, of which there are $2^{10}=1024$ of them; they should be uniquely decodable but not prefix-free). It obviously satisfies the Kraft-McMillan inequality: the sum $\sum 2^{-10}=2^{10}\cdot2^{-10}=1$. Henceforth, we should be able to construct an instantaneous code with the same code-word length. However, the lengths are uniformly l=10; while for instantaneous codes, there is at most 1 code per length (which, on the tree graph, means that no code-word is an ancestor of any other code words). We cannot assign codes with the same length to nodes of different depths. There seems to be a contradiction.

I'm pretty sure that I misunderstood something. I would really appreciate it if you could let me know where I am wrong.

答:
fixed-length codes ... should be uniquely decodable but not prefix-free

This is false. A (non singular) fixed-length code is trivially prefix-free, because a codeword cannot be a prefix of another codeword.

while for instantaneous codes, there is at most 1 code per length

Again, false.

An instantaneous codes correspond to a rooted tree where the codewords correspond to leaves (i.e., non internal nodes). Both codes (source) here are instantaneous

Title: Prefix-Free vs Uniquely Decodable Codes


问:
Prove that $\int^{1}_{0} \sqrt{x^2+x}\,\mathrm{d}x < 1$

Prove that $\int\limits^{1}_{0} \sqrt{x^2+x}\,\mathrm{d}x < 1$

I'm guessing it would not be too difficult to solve by just calculating the integral, but I'm wondering  if there is any other way to prove this, like comparing it with an easy-to-calculate integral. I tried comparing it with $\displaystyle\int\limits^{1}_{0} \sqrt{x^2+1}\,\mathrm{d}x$, but this greater than $1$, so I'm all out of ideas.

答:
Still simpler.

For $0<x<1$ we have $x^2<x$.

Hence $\sqrt{x^2+x}\lt \sqrt{x+x}=\sqrt{2}\sqrt{x}$, and the integral can be estimated as $\int_0^1 \sqrt{x^2+x}<\sqrt{2}\int_0^1 \sqrt{x}=\frac{2}{3}\sqrt{2}=\sqrt{\frac{8}{9}}\lt 1$.
QED.


问:
Free homotopy of a closed curve must be a closed curve
I'm reading a textbook on differential geometry and the author proceeds to define a weaker version of homotopy.

Two continuous curves $c_0, c_1: [a,b] \to U \subseteq \mathbb{R}^n$ with the same endpoints are homotopic if there exists a continuous map $H: [a,b] \times [0,1] \to U$ such that $H(s,0) = c_0(s), H(s,1)=c_1(s)$ and $H(a,t) = c_0(a), H(b,t) = c_0(b).$ If the last condition is dropped, we say $H$ is a free homotopy.

A proof of line integral invariance under homotopy proceeds and at the end the author says

If $c_0, c_1$ are closed curves ($c_0(a) = c_0(b), c_1(a) = c_1(b)$) that are freely homotopic, then $H(a,t), H(b,t)$ are equal curves.

This is equivalent to every curve in the free homotopy being closed, but I don't see why this has to be true. In fact, why can't we slightly open a loop and then shrink it to a point? There has to be some continuity problem with what I'm imagining since in the case of a simple loop around the origin, this contradicts the fact that $\mathbb{R}^2 \setminus \{0\}$ is not simply connected. In fact, I didn't even think about this issue until reading through the book caused me to imagine it.

The continuity issue must be very subtle since the partial derivatives (with respect to $s, t$) exist and are continuous in the example of slightly opening a loop. I suspect that letting $(s,t)$ approach $(a, 0)$ from different directions will seal the deal. With that in mind, how do I go back and prove $c_t(s) = H(s,t)$ is a closed curve for all $t$?

Update: The book is Differential Forms and Applications by Do Carmo. In fact, the author defines simple-connectedness as every closed curve being freely homotopic to a point and says $\mathbb{R}^2 - \{0\}$ is not simply connected later, so my example better not work. But now I've changed my mind and am starting to think it works again.

Let $C_t$ be the unit circle with the top half dilated to have radius $1+t$ and center at $(t,0).$ Consider a free homotopy from $C_0$ to $C_1$ back to $C_0.$ We must already have a violation of continuity at $t=0$ else after getting to $C_{\epsilon},$ we can just shrink to a point. The exact equation for $H : [0, 2\pi] \times [0,1] \to \mathbb{R}^2$ is $H(s,t) = (t + (1+t)\cos s, (1+t)\sin s)$ when $s \le \pi$ and this is clearly a continuous function of $s, t.$ Thus, $H$ is continuous on $A = [0, \pi] \times [0,1], B = [\pi, 2\pi] \times [0,1], A \cap B = \{\pi\} \times [0,1],$ so is continuous on the whole domain $A \cup B.$

答:
do Carmo's definition of a free homotopy between curves does not really make sense because then any two curves $c_0, c_1: [a,b] \to U$ with the same endpoints are freely homotopic - and that would be fairly uninteresting. To see that take
$$H(s,t) = \begin{cases} c_0(a +(b-a)s(1-2t)) & t \le 1/2 \\ c_1(a +(b-a)s(2t-1)) & t \ge 1/2 \end{cases}$$
Note that for $t = 1/2$ both lines give  $H(s,1/2) = c_0(a) = c_1(a)$. Also observe that $H(a,t) = c_0(a) = c_1(a)$ for all $t$, but in general we do not have $H(b,t)$ constant.

I think what do Carmo really wants to define is free homotopy for closed curves. If $c_0,c_1$ are closed curves, not nessarily based at the same point of $U$, then a free homotopy of closed curves is a homotopy such $H_0 = c_0, H_1 = c_1$ and all $H_t$ closed curves. See for example Characterizing simply connected spaces .


问:
Question regarding the integral of random variables with same distribution.
Suppose $X,Y$ are independent random variables with the same distribution then is it the case that for all measurable sets $A$, $E[X1_A]=E[Y1_A]$. I think it is true but I can't prove why. Any hints or suggestions? Also if the statement isn't true, what conditions can be imposed on the sets $A$ such that the equality holds.

答:
This is not correct. Just flip a coin twice. Then $\Omega = \{0,1\}\times\{0,1\}$, $P(s,t)=1/4$ for all $(s,t)\in\Omega$, $X(s,t) = s$, $Y(s,t) = t$. These are definitely i.i.d. Now, choose $A = \{(0,1)\}\subset\Omega$. We have
$$
E(X1_A) = P(X1_A=1) = P(X=1,s=0,t=1) = 0,
$$
but
$$
E(Y1_A) = P(Y1_A=1) = P(Y=1,s=0,t=1) = P(s=0,t=1) = \frac 14.
$$


答:
From Roman's "Fundamentals of Group Theory: An Advanced Approach", page 200, we have:

Theorem 6.12:

The number of cycles of length $k$ in $S_n$ is $$\binom{n}{k}(k-1)!=\frac{n!}{k(n-k)!}.$$

The number of permutations in $S_n$ whose cycle structure consists of $r_i$ cycles of length $k_i$, for $i=1,\dots, m$ is $$\frac{n!}{r_1!\dots r_m!k_1^{r_1}\dots k_m^{r_m}}.$$

Title: Counting the number of 2-cycles, 3-cycles and 4-cycles in $S_4$

Body:
I know that in $S_4$ there are five different cycle types: double transpositions, 2-cycles, 3-cycles, 4-cycles and 1-cycle (i.e. identity).

My question is, does there exist a combinatorial formula or similar for counting the number of, for example, the number of 3-cycles in $S_4$?


问:
First-order logic: Understanding quantifiers
Suppose we have a signature $\sigma=\{E\}$ and a structure $\mathcal{G} = (V(G), E^\mathcal{G})$ where $V(G)$ is the vertex set of a graph $G$ and $E^\mathcal{G}$ is the binary relation symbol for the existence of an edge between two vertices.

Consider the formula $\varphi := \forall x \exists y(E(x, y)). $ The formula says for every vertex $x$, there exists a vertex $y$ such that $\{xy\} \in E(G)$.

Does this mean that $x$ and $y$ can be equal, resulting in a self-loop for every node? Or does the formula implicitly say that $x \neq y$? How should the formula be interpreted?

答:
There is no implicit assumption of inequality. One way in which $\varphi$ could be satisfied by $\mathcal{G}$ is if every node has a self-loop, although of course this isn't the only way $\varphi$ could be satisfied.

If you want to further demand inequality, you should write $$\forall x\exists y(x\not=y\wedge E(x,y))$$ instead.


答:
Pointwise, it converges to $x \mapsto \frac{1}{x}$. If it was uniformly convergent on $(0,\infty)$, we would have
$$
\mathop {\lim }\limits_{n \to  + \infty } \mathop {\sup }\limits_{x > 0} \left| {\frac{1}{x} - \frac{n}{{1 + nx}}} \right| =0.
$$
But
$$
\mathop {\sup }\limits_{x > 0} \left| {\frac{1}{x} - \frac{n}{{1 + nx}}} \right|\mathop  \ge \limits^{x = 1/n} \left| {n - \frac{n}{2}} \right| = \frac{n}{2}.
$$
A usual way showing that $f_n(x)\to f(x)$ is not uniform on $A$ is to give a lower bound on $\sup_{x\in A} |f(x)-f_n(x)|$ which does not tend to $0$ when $n\to +\infty$. This is generally done by choosing an appropriate $x$ in $A$ as a function of $n$ (in our example $x=\frac{1}{n} \in (0,\infty)$).

Title: Show that $f(x)=n/(1+nx)$ does not uniformly converge on $(0, \infty)$

Body:
I have proved that this sequence of functions is uniform convergent on $[1,\infty)$ by choosing $N>1/\epsilon$ (correct me if i am wrong).

Now need to show that it is not uniform convergent on $(0,\infty)$. What should be the choice for $N$?


问:
South Pole and Stereographic Projection
I'm trying to parameterize a sphere through the stereographic projection.

This projection fails on the south pole, once I'm not working with $\infty$ as an element of the ring $\mathbb{R}$.

How can I define a map that sends me to the south pole? I'm struggling with this because all the points on the plane have already been taken.

Thank you.

答: Stereographic projection gives a homeomorphism $S^n\setminus \{p\}\to \Bbb{R}^n$, where $p$ is the point you project from. So, if you project from the north pole $N$, you won’t be able to define the map for $N$ (and for good reason - the sphere and Euclidean space are not homeomorphic). Anyway, you can use the another stereographic projection from the south pole $S$ to find a chart on $S^n$ containing $N$ if you need to and vice versa.

Title: South Pole and Stereographic Projection


问:
Borel-Cantelli lemma, exponential distribution problem
Good evening,

I am currently solving an exercise :
Let Xn be a sequence of independent random variables, each with the exponential distribution with rate $\frac{1}{\lambda}$.
I want to prove the part : $\limsup_{n \to \infty}  (\frac{X_n}{\log(n)}) \leq \frac{1}{\lambda}$ a.s.

I found $\Pr\left(\frac{X_n}{\log(n)}\geq \alpha\right) = \frac{1}{n^{\alpha \lambda}},  \alpha >0$.

I used the Borel-Cantelli lemma with $\alpha > \frac{1}{\lambda}$ and the sum $\sum_{n \in   \mathbb{N}} \Pr(X_n
 \geq \alpha \ln(n)) < + \infty$

In the end, I found $\Pr\left(\frac{X_n}{\log(n)}\geq \alpha
\mbox{ i.o}\right) = 0
\Rightarrow \lim_{n \to \infty} \sup_n (\frac{X_n}{\log(n)}) \leq \alpha$ a.s.

How can I demonstrate that $\lim_{n \to \infty} \sup_n (\frac{X_n}{\log(n)}) \leq \frac{1}{\lambda}$ a.s. ?

答:
First, for any $\alpha>0$,
$$
\mathsf{P}(X_n>\alpha\ln n)=n^{-\alpha\lambda}.
$$
The RHS is summable iff $\alpha>\lambda^{-1}$. Therefore, by the B-C lemma, $X_n>\ln n/\lambda$ i.o., but $X_n<(1+\epsilon)\ln n/\lambda$ ev. for any $\epsilon >0$. Thus, $\limsup_{n\to\infty}X_n/\ln n=1/\lambda$ a.s.


问:
Statistics question in cosmology
The following are the approximate parameters for the Benchmark model in cosmology:
$H_0 = 73.8 km s^{-1} Mpc^{-1})$, $\Omega_{m,0} = 0.266$, and $\Omega_{\Lambda,0} = 0.734$.

Given that we know $H_0$ very precisely to this value in a flat universe (meaning neglecting the negligible contribution from radiation, $\Omega_{m,0} + \Omega_{\Lambda,0} = 1$) with $\Omega_{\Lambda,0}$ being an unknown value very close to the given number, if we want to make a measurement of $\Omega_{\Lambda,0}$ with an accuracy of 5%, what is the maximum percent error we can tolerate on our measurement of luminosity distance at z = 0.5, 1.0, and 1.5?

Extra note: luminosity distance can be approximated as
$$d_L \approx \frac{c}{H_0} z (1 + \frac{1 - q_0}{2} z)$$

Where
$$q_0 = \frac{1}{2} \Omega_{m,0} - \Omega_{\Lambda,0}$$

Due to covid, I've been unable to take the statistical physics course that was part of my plan, so it has been since the mid 2000's since I've had a stats course. There are a bunch of follow-up questions, but I'm hoping once I grasp the approach to this one I will be able to do them myself.

答:
Please see the comment, however it might be best to make $\Omega_{m,0}$ the subject of $\Omega_{m,0} + \Omega_{\Lambda,0} = 1$ and replace it and $q_0$ in your luminosity distance formula, so it only depends on the cosmological constant density parameter.  Redshifts are usually known really accurately, so you could then vary  the cosmological constant density parameter by 5% in each case and see what percentage variation occurs for the luminosity distance, for different redshifts.

There is probably a more mathematical way with differentials to do it, but it finds the answer.

It must be said that most of the assumptions being made in cosmology at the moment are likely to be wrong, theories such as this, which claims that the Hubble parameter should be halved and a different scale factor- redshift relation used, might be just as viable.

https://www.researchgate.net/publication/342040580_A_New_Solution_of_the_Friedman_Equations

P.S the speed of light should be in km/s

Title: Statistics question in cosmology


问:
What are 'contexts' actually called?
Consider the following argument by contradiction.

\begin{array}{|l}
\mbox{We wish to deduce A.} \\
{\begin{array}{|l}
\mbox{Suppose not A.} \\
\hline \\
\mbox{Then B. Thus C. Therefore, contradiction. }
\end{array} } \\
\mbox{Thus, A.}
\end{array}

So to actually perform the argument by contradiction, we had to open a new ‘context’ wherein additional assumptions hold. What are these ‘contexts’ actually called? And which deductive systems actually utilize them?

答:
The other answers have discussed subproofs in natural deduction systems for conditional proof.  There are also natural deduction systems that have subproofs for other types of contexts, such as necessity introduction in modal logic.  For instance, axiomatic modal logics typically have the inference rule $$\vdash \phi / \vdash \Box\phi$$ which states that if $\phi$ is a theorem then so is $\Box\phi$, and the axiom $$\Box(\phi \to \psi) \to (\Box\phi \to \Box\psi)$$ which says that “if $\phi$ necessarily implies $\psi$, then if $\phi$ is necessary, so is $\psi$.”  In natural deduction systems for modal logic, axioms can be captured by ‘modal contexts’ or modal subproofs.  For instance, in Introductory Modal Logic, Konyndyk uses, in addition to conditional subproofs of the type described in the question, modal subproofs of the form shown in the following:

\begin{array}{l}
{\begin{array}{|ll}
\Box P  \\
\hline
{\begin{array}{|ll}
\Box Q \\
\hline
{\Box\begin{array}{|ll}
P & \mbox{reit} \\
Q & \mbox{reit} \\
P \land Q & \mbox{$\land$ intro}
\end{array}} \\
\Box(P \land Q) & \Box\mbox{ intro} \\
\end{array}} \\
\Box Q \to \Box(P \land Q) & \to\mbox{ intro}
\end{array}} \\
\Box P \to (\Box Q \to \Box(P \land Q)) & \to\mbox{ intro}
\end{array}

The idea is that necessary things can be reiterated into a modal subproof, but they lose the leading $\Box$ when they move in.  Different modal logics generate different constraints on what can be reiterated in, and whether or not the modality is dropped upon reiteration.  These ‘contexts’ are still called (modal) subproofs.

Another point of interest comes up in first-order logic, where a subproof may introduce a context in which a something is taken as a term denoting an arbitrary individual (for universal introduction) or a witness individual with a particular property (for existential elimination).  These ‘contexts’ are also typically referred to as subproofs.

Title: What are 'contexts' actually called?


问:
Stokes theorem for a current
I am struggling to understand the Stokes theorem for currents(differential form with distributional coefficients). The statement is as follows:

Lest $S$ be a current of degree $N - 1$ with compact support in $\Omega$. Then $\int_{\Omega} dS = 0$. The proof is: Let $\chi$ be a smooth cut-off function in $\Omega$ such that $\chi \equiv  1$ in a neighborhood of $K$, a compact subset of $\Omega$ containing the support of $S$. Then $\int_{\Omega} dS = \int_{\Omega} \chi dS = \langle S, d\chi \rangle = 0$.

I am not really following the logic here. First of all, how would one define the integral of a current? I know that a current of degree $p$ is a continuous linear functional on the space of smooth (n-p) forms with compact support, and they are essentially forms with distributional coefficients. My guess is that the coefficients of a top degree form will always be locally integrable, then you just integrate the coefficients over the support of the current but I am not sure whether this is the correct interpretation or not.

答:
In my opinion the statement is a tautology but at the same time abuses the notation. $S$ is of degree $N-1$ (in view of the statement the dimension must be $N$) so it 'acts' upon 1 forms (with smooth and compact coefficients). The exterior derivative is defined by duality so $dS$ acts upon $0$-forms, i.e. functions. So the stated claim may be written:

$$ "\langle dS,1 \rangle" = \langle dS, \chi \rangle
= (-1)^N \langle S, d\chi \rangle = 0 $$
since $d\chi$ has support disjoint from that of $S$.
You may look e.g. in chapter 3 of
Currents for this type of calculations. I have put the first expression in quotation marks as it is also a slight abuse of notation  since the constant function 1 does not have compact support, but it still makes sense since $S$ is supposed to have compact support.

The above coincides with standard integrals when these are defined but the notation may be misleading.

Some more details: In $n$ dimensional Euclidean space and for (non-negative)
$p+q=n$ there is a
natural bi-linear map on smooth forms $(\alpha,\beta) \in \Omega^p({\Bbb R}^n) \times \Omega^q({\Bbb R}^n)) \rightarrow \int \alpha\wedge \beta\in {\Bbb R}$ defined when at least one of the forms (say $\beta$) has compact support.
The map, $\beta\in \Omega^q_c({\Bbb R}^n) \mapsto \langle \Lambda_\alpha, \beta\rangle =\int \alpha \wedge \beta$ then defines a continuous linear map on the space of $q$-forms with compact support (equipped with its natural Frechet space topology). The space of such linear maps is called the space of $p$-currents, which thus contains $\Omega^p({\Bbb R}^n)$ as a naturally embedded image under the map $\alpha \mapsto \Lambda_\alpha$ above.

For smooth forms $\alpha$ and $\beta$ of degree $p$ and $q$, respectively, one has $d(\alpha\wedge \beta)= d\alpha \wedge \beta + (-1)^p \alpha \wedge d\beta$. However, when $p+q=n-1$ then by Stokes,
$\int_D d(\alpha \wedge \beta) = \int_{\partial D} \alpha\wedge \beta = 0$ when the open set $D$ contains the support of $\beta$. Thus,
$$ \langle d\Lambda_\alpha, \beta \rangle = \int d\alpha \wedge \beta = -(-1)^{p} \int \alpha \wedge d\beta = (-1)^{p+1} \langle \Lambda , d\beta\rangle.$$
One then declares equality of extremities to define the exterior derivative of any $p$-current $\Lambda$ (when $p=n-q-1$). This now leads to the tautology as explained above.

Title: Stokes theorem for a current


问:
Convergence for improper integral $\int_0^\infty x^re^{-x} \,dx$
I'm trying to find for which values $r$ the following improper integral converges.
$$\int_0^\infty x^re^{-x}\, dx$$
What I have so far is that $x^r < e^{\frac{1}{2}x}$ for $x \geq a$, which splits the integral into
$$\int_0^a x^re^{-x} \,dx + \int_a^\infty e^{-\frac{1}{2}x} \, dx$$
We know the latter interval converges, but I don't know what to do with the first one. (For reference, graphing the functions reveals the answer to be $x > -1$.)

Edit: I would like a proof without the gamma function. Preferably one that uses the comparison test to compare limits.

答:
\begin{align}
& \text{For small enough }a>0, \\[8pt]
& x^r > x^r e^{-x} > 0.9 x^r \text{ when } 0<x<a.
\end{align}
Therefore $\displaystyle \int_0^a x^r e^{-x} \, dx$ converges for precisely those values of $r$ for which $\displaystyle\int_0^a x^r\,dx$ converges.


问:
What is the rational function associated to a dominant morphism to $\mathbb{P}^1$?
Let X be an integral scheme of finite type over an arbitrary field $k$. Let $f:X \rightarrow \mathbb{P}^1$ be a dominant morphism. What is meant by the rational function of $f$? Fulton uses the notation $\text{div}(f)$ but it is not at all clear to me how to obtain a rational function from a dominant morphism to the projective line.

答:
I'm turning my comment into an answer. By definition of $f$, $f$ induces a morphism $f^\sharp: K(\mathbb{P}^1) \rightarrow K(X)$ (the function fields are the rings of stalks at the generic points). The rational function of $f$ is $f^\sharp(t_0/t_1)$, where $t_0/t_1 \in K(X)$ generates that field. Depending on the conventions, it could be $t_1/t_0$ instead.

The basic idea is that $\mathbb{A}^1=\operatorname{Spec} k[t]$ is a dense open subset of $\mathbb{P}^1$ (in two different ways, hence the "convention" issue) and the map $f^{-1}(\mathbb{A}^1)\rightarrow \mathbb{A}^1$ is given by a regular function on $f^{-1}(\mathbb{A}^1)$ (the image of $t$), which is rational on $X$.


问:
About Generalized Hypergeometric Function
I have a definition of Generalized hypergeometric function as $$_1F_2(a_1;b_1,b_2;z) = \sum_{n=0}^\infty \frac{(a_1)_n}{(b_1)_n (b_2)_n} \frac{z^n}{n!}$$
My question is, what will be the result if $a_1$ equals to zero? I couldn't come across any definition defining this special case. </p>

<p>Thanks. </p>

答:
This has already been pointed out in the answers but I will put it here for the sake of completeness.

This question really comes down the the behavoir of the Pochhammer symbol. Begin with
$$
_1F_2(a_1;b_1,b_2;z) = 1+\sum_{n=1}^\infty \frac{(a_1)_n}{(b_1)_n (b_2)_n} \frac{z^n}{n!}.
$$
By definition, $(a_1)_n=\frac{\Gamma(a_1+n)}{\Gamma(a_1)}$ and as $a_1\to 0$ we find $1/\Gamma(a_1)\sim a_1+\mathcal O(a_1^2)$. Therefore, in the limit $\Gamma(a_1+n)/\Gamma(a_1)\to 0$. Hence, the series terms for $n\geq 1$ vanish and
$$
_1F_2(0;b_1,b_2;z) = 1.
$$


问:
An open and closed set
Provide an example of a non-empty subset $X \subseteq \mathbb{Q}$ that is both open and closed, and justify your example.

This is a question from a practice set I have. The first and foremost thing I would check is the empty set $\emptyset$ and the rationals $\mathbb{Q}$. But $X$ needs to be non-empty and $\mathbb{Q}$ is neither open nor closed. I then considered a slightly less trivial example: $(2, 3) \cup (4, 5)$ with the metric $d(a, b) = |a - b|$ (I think, $(2, 3)$ and $(4, 5)$ are both open and closed but I am not sure how to justify it).

Any assistance in finding and/or justifying such a set is much appreciated.

答: In every topological space $X$ the empty set $\emptyset$ and the whole $X$ are always clopen sets, i.e. sets that are simultaneously open and closed. Since your question asks about a non-empty example, we can use $\mathbb{Q}$. A more interesting clopen subset of $\mathbb{Q}$ (with the usual topology) is $(-\pi, \pi)$. It is immediate that the interval is open. It is also closed, because its complement $(-\infty, -\pi) \cup (\pi, +\infty)$ is open.

Title: An open and closed set


问:
six digit puzzle using five restrictions
I have been stumped on this puzzle for a while now. I know the answer, but of course this does not satisfy my need to understand why.

Puzzle:
Let $ABCDEF$ be a six digit number such that:

All digits are different,

The sum of the first two digits is the same as the sum of the last two digits, ($A+B = E+F$)

The sum of all the digits equals the last two digits of the number, ($A+B+C+D+E+F = 10E+F$)

The pairs $AB,CD$ and $EF$ are all prime, ($10A+B$, $10C+D$ and $10E+F$ all prime)

The sum of the last two digits is less than $10$ ($E+F<10$ and hence $A+B<10$ using $2$)

What is the number?

My attempt:

Since $10A+B$ is prime, we have $B$ cannot be divisible by $2$ or $5$, in other words, $B$ must be $1,3,7$ or $9$. This logic is the same for $D$ and $F$.

Doing $3) - 2)$ yields:

$$C+D+E+F=9E$$
$$C+D+F=8E$$

Therefore $C+D+F$ is divisible be $8$. If that is true then $C+D+F$ is of course even. Since $F$ and $D$ has to be odd $(1,3,7,9)$, $C$ must be even.

That's essentially as far as got. I feel as though I am missing something simple.

Note: (the answer is: $416723$)

答:
I can't see anything better than casework.  Conditions $2$ and $4$ tells us that $AB$ and $EF$ are two-digit primes with the same digit sum.  This should greatly restrict the possibilities.  (I'm assuming, as I think you did that we cannot have $E=0$.)  A little python script produced

```<code>(4, [13, 31])
(5, [23, 41])
(7, [43, 61])
(8, [17, 53, 71])
(10, [19, 37, 73])
(11, [29, 47, 83])
(13, [67])
(14, [59])
(16, [79, 97])
(17, [89])
</code>```

On each line, the first number is the digit sum, and the list gives all two-digit primes $>11$ with that digit sum. Of course, it would be easy to do this by hand.

The list is really shorter than it may  appear.  We are told that the sum of the digits in $EF$ is less than $10$, so we only have to consider the first $4$ lines.  Also, we cannot have digit sum $4$, because it's not possible that one of $AB,EF$ is $13$ and the other is $31$.  For the same reason, if the digit sum is $8$, then one of the primes must be $53$.

I haven't carried it any further than this.

EDIT

Of course, $C+D+F=8E\implies E\neq0$, so there's no need for assumptions.

Body:
I can't see anything better than casework.  Conditions $2$ and $4$ tells us that $AB$ and $EF$ are two-digit primes with the same digit sum.  This should greatly restrict the possibilities.  (I'm assuming, as I think you did that we cannot have $E=0$.)  A little python script produced

(4, [13, 31])
(5, [23, 41])
(7, [43, 61])
(8, [17, 53, 71])
(10, [19, 37, 73])
(11, [29, 47, 83])
(13, [67])
(14, [59])
(16, [79, 97])
(17, [89])

On each line, the first number is the digit sum, and the list gives all two-digit primes $>11$ with that digit sum. Of course, it would be easy to do this by hand.

The list is really shorter than it may  appear.  We are told that the sum of the digits in $EF$ is less than $10$, so we only have to consider the first $4$ lines.  Also, we cannot have digit sum $4$, because it's not possible that one of $AB,EF$ is $13$ and the other is $31$.  For the same reason, if the digit sum is $8$, then one of the primes must be $53$.

I haven't carried it any further than this.

EDIT

Of course, $C+D+F=8E\implies E\neq0$, so there's no need for assumptions.


问:
Let $f(m,1) = f(1,n) = 1$ for $m \geq 1, n \geq 1,$ and let $f(m,n) = f(m-1,n) + f(m,n-1) +
Let $f(m,1) = f(1,n) = 1$ for $m \geq 1, n \geq 1,$ and let $f(m,n) = f(m-1,n) + f(m,n-1) + f(m-1,n-1)$ for $m > 1$ and $n > 1.$ Also, let

$$S(n) = \sum_{a+b=n} f(a,b), \text{ for } a \geq 1, b \geq 1.$$

Note: The summation notation means to sum over all positive integers $a,b$ such that $a+b=n.$

Given that

$$S(n+2) = pS(n+1) + qS(n) \text{ for all } n \geq 2,$$

for some constants $p$ and $q$, find $pq.$

答: I substituted in my values of $f(m,n)$ int the function, and found all the values of $$S(k) = \sum_{a+b=k} f(a,b), \text{ for } a \geq 1, b \geq 1$$ up to $k=5$, then wrote a system of equations, and solved for $p$ and $q$.


问:
Tiling an L-shape with "almost square"s
Define an "almost square" as a rectangles with aspect ratio between $1/2$ and $2$. What is the minimal number of interior-disjoint almost-squares required to tile the following L-shape (where $n$ is an integer)?

Background: It is known (e.g. Richard Kenyon, 1996) that tiling an $n \times (n-1)$ rectangle with squares requires $O(\log n)$ squares. This number grows to infinity as $n$ grows. However, if we relax our requirements and agree to tile the rectangle with almost squares, then one almost-square is sufficient. In practical scenarios, almost-squares can be almost as good as squares, and their number (in this case) is significantly smaller.

My question is: Does this fact extend to more general polygons? Specifically, is it possible to tile the L-shape in the above diagram (which is an $n \times n$ square, missing a single $1 \times 1$ square at the corner) with axis-parallel almost-squares whose number is constant (independent of $n$)?

Note: This question is related: https://math.stackexchange.com/questions/873586/size-of-minimal-covering-overlapping-and-disjoint

答:
Consider the problem as a tiling of an $n\times n$ square, where we have already fixed a $1\times 1$ in the corner. Given a partial placement of rectangles in the $L$-shape, let $a$ be the highest $x$-coordinate of any point on a rectangle and $b$ the highest $y$-coordinate of any point on a rectangle (where we parametrize the big square as $[0,n]\times [0,n]$).

Obviously, we start with $(a,b)=(1,1)$. Now, consider how $a$ and $b$ can change as we add rectangles. If we add the "wrong" rectangles, $a$ and $b$ will grow dramatically, but we can choose them in whatever order we like. Consider the rectangles occupying the points $(a+\epsilon,0)$ and $(0,b+\epsilon)$: the rectangles just on the "other side" of the current borders. They must be distinct from each other, and at most one of them can occupy the point $(a+\epsilon,b+\epsilon)$. Consider the other rectangle. If we add it, then one of $a$ and $b$ stays fixed, and the other increases by one of the dimensions of the rectangle, so we have either $(a,b)\mapsto (\le a+2b,b)$ or $(a,b) \mapsto (a,\le b+2a)$ (by the constraints on the ratio of our rectangles). Since we are trying to get $(a,b)$ as large as possible, we can forget about the $\le$ signs in these moves and just always pursue the maximum. Call these Move $1$ and Move $2$, respectively.

What is the fewest number of applications of Move $1$ and Move $2$ to get from $(1,1)$ to $(n,n)$? Note that if $a<b$, then the pair $(a+2b,b)$ is strictly larger than the pair $(a,b+2a)$ once we reverse the order of the latter (since our target is symmetric, we don't care about order). So the optimal strategy is to always alternate between Move 1 and Move 2, and our series of $(a,b)$ values goes $(1,1), (1,3), (7,3), (7,17), (41,17), \ldots$, with terms coming from $\text{A}001333$.

This puts a lower bound $k$ on the number of moves, which is asymptotically $\log_{1+\sqrt{2}}(n)$. But, conveniently, $k$ is also an upper bound!

To see this, note that we can easily add a rectangle at each step so that the rectangles placed so far completely tile the region $[0,a]\times[0,b]$, and then just clip off the final one or two rectangles when we exceed the bounds of our $n\times n$ square. So the only risk is that this clipping-off may bring the ratios outside of $[1/2,2]$.

Let $(a,b)$ be the furthest we get before either value exceeds $n$; WLOG, say $a<b$. With some thought, it is apparent that this only poses a problem for clipping if $b>n/2$; our solution is to shrink our $a\times b-2a$ rectangle so that its top border is at $y=n/2$, and proceed from there. Now our only concern is that this modification will have rendered the modified rectangle to have an unacceptable ratio, but with a bit of tedious algebra one can check that we're okay as long as $a\le 3b$, which is easily verified.

So the final answer is that the number of necessary rectangles is the smallest $k$ such that

$$\frac{(1-\sqrt{2})^k + (1+\sqrt2)^k}2\ge n$$

Title: Tiling an L-shape with "almost square"s


答:
There are only $20$ candidate primes out of which only $4$ pairs sum to less than $10$ and have distinct digits.

$$\{23,41\} , \quad \{17,53\} ,\quad \{53,71\}, \quad\{43,61\}$$

Now

$$1+2+3 \le C+D+F \le 7+8+9$$
$$\Rightarrow 6 \le 8E \le 24$$
$$\Rightarrow E \in \{1,2,3\}$$

From above we see that either $EF=17$, $AB=53$ $\Rightarrow CD=29$ due to

Since F and D has to be odd (1,3,7,9), C must be even.

which cannot satisfy the (third) sum condition or $EF=23$, $AB=41$.

Title: six digit puzzle using five restrictions

Body:
I have been stumped on this puzzle for a while now. I know the answer, but of course this does not satisfy my need to understand why.

Puzzle:
Let $ABCDEF$ be a six digit number such that:

All digits are different,

The sum of the first two digits is the same as the sum of the last two digits, ($A+B = E+F$)

The sum of all the digits equals the last two digits of the number, ($A+B+C+D+E+F = 10E+F$)

The pairs $AB,CD$ and $EF$ are all prime, ($10A+B$, $10C+D$ and $10E+F$ all prime)

The sum of the last two digits is less than $10$ ($E+F<10$ and hence $A+B<10$ using $2$)

What is the number?

My attempt:

Since $10A+B$ is prime, we have $B$ cannot be divisible by $2$ or $5$, in other words, $B$ must be $1,3,7$ or $9$. This logic is the same for $D$ and $F$.

Doing $3) - 2)$ yields:

$$C+D+E+F=9E$$
$$C+D+F=8E$$

Therefore $C+D+F$ is divisible be $8$. If that is true then $C+D+F$ is of course even. Since $F$ and $D$ has to be odd $(1,3,7,9)$, $C$ must be even.

That's essentially as far as got. I feel as though I am missing something simple.

Note: (the answer is: $416723$)


问:
approximate spectrum ${\sigma}_{ap}$ of a bounded linear transformation
if $\lambda\in{\sigma}_{ap}(T)$, where T is bounded linear transformation, how do i show that $|\lambda|≤\|T|\|$ ?

if $\lambda\in{\sigma}_{ap}(T)$, then for every $\varepsilon >0$ there exists $x\neq0$ such that $\|Tx−λx\|≤\varepsilon \|x\|$. I imagine that next it should be developed in some way to use the boundibility of T to conclude what is required, but I can't think of how to relate them, could someone give me a clue, or illustrate the path I should take?

Thanks in advance.

答:
For any $\varepsilon> 0$ there is $x\ne 0$ such that
\begin{align*}
\varepsilon\|x\|\ge\|Tx-\lambda x\|\ge\big|\|Tx\|-|\lambda|\|x\|\big|\ge |\lambda|\|x\| - \|Tx\|.
\end{align*}
Hence,
$$
|\lambda|\|x\|\le\|Tx\| + \varepsilon\|x\|\le(\|T\| + \varepsilon)\|x\|.
$$
Dividing by $\|x\|\neq0$ gives $|\lambda|\le\|T\|+\varepsilon$. Letting $\varepsilon\to 0$ yields the desired result.


问:
Bayes Classifier Needed Number of Parameters
I am trying to understand Bayesian Classifiers but I have the following question:

Let's assume we have a vector $X=(X_1,..X_n)$ and $Y$ where each $X_i$ and $Y$ have boolean values. We are trying to figure out $P(Y|X)=\frac{P(X|Y)P(Y)}{P(X)}$. The lecture notes that I am reading said that to estimate $P(X_1,...,X_n|Y)$, we need to estimate $2(2^n-1)$ parameters. What does it mean and how do we get this? For each $X_i$, there are $2^n$ possible values and there are 2 possible values for $Y$. How do we get $2(2^n-1)$? I mean for each $Y$ value, we need to consider $2^n$ parameters, which give us a total of $2^{n+1}$ parameters to consider, why do we subtract 2?

答:
Here we have $X_i, Y$ binary variables.

First consider $Y=0$, we need to have $P(X|Y=0)$ for $2^n$ different values of $X=X_1,X_2,\ldots,X_n$.

But also we know $\sum\limits_{X_i\in\{0,1\}}P(X_1,X_2,\ldots,X_n|Y=0)=1$, being a probability mass function.

This reduces the number of parameters to estimate by 1, as shown below.

If we estimate the probability values for all but one boolean value assignments for $X_i$s, e.g., let's consider estimating probabilities for all assignments except a specific one, namely, $V=\{X_1:b_1,\ldots,X_n:b_n\}$, i.e, estimating for a total of $2^n-1$ value assignments, and let's define $p=\sum\limits_{\{0,1\}^n - V}P(X_1,X_2,\ldots,X_n|Y=0)$, i.e., sum of all those $2^n-1$ probabilities estimated, then the value of $P(X=V|Y=0)$ is no longer needed to be estimated, it can be directly computed as $P(X=V|Y=0)=1-p$.

Hence, it's sufficient to estimate $2^n-1$ parameters for this case $Y=0$.

Similarly, we need to estimate $2^n-1$ parameter values corresponding to the case $Y=1$, i.e., for $P(X|Y=1)$.

Hence, total number of parameters to estimate = $2(2^n-1)=2^{n+1}-2$


问:
Algebraic problem for satisfying a given equation
I'm trying to solve the following exercise:

My backward equation looks like:
$P_{i,j}'(t) = i\lambda P_{i+1,j}(t) - i\lambda P_{i,j}(t) $

So i started with differentiating $P_{i,j}(t)$:

${j-1 \choose i-1}(e^{-i\lambda t}(1-e^{-\lambda t})^{j-i})'$
By using $(uv)' = u'v + uv'$

${j-1 \choose i-1}-i\lambda (e^{-i\lambda t}(1-e^{-\lambda t})^{j-i}) + {j-1 \choose i-1}(j-i)\lambda (e^{-(i+1)\lambda t}(1-e^{-\lambda t})^{j-i-1}) $

where the first term is cleary $-i\lambda P_{i,j}(t) $.

I don't see how i can turn $ {j-1 \choose i-1}(j-i)$ into  ${j-1 \choose i}i$  so the second term equals  $i\lambda P_{i+1,j}(t) $

答:
You seem to have the wrong definition of $P_{i,j}$ it should be ${j-1 \choose i-1}$, not ${j-i \choose i-1}$. If you do the same thing with this corrected version, then
$$
\begin{align*}
{j-1 \choose i-1}(j-i) &= \frac{(j-1)!}{(i-1)!(j-i)!}(j-i) \\
&= \frac{(j-1)!}{(i-1)!(j-i-1)!} = \frac{(j-1)!}{((i+1)-1)!(j-(i+1))!}i \\
&= {j-1 \choose (i+1)-1}i
\end{align*}
$$</p>


问:
How to find the indefinite integral of $\arctan((x^2+1)^{1/3})$?
I have to find the indefinite integral of $\arctan((x^2+1)^{1/3})$. I have tried to do different variable substitutions, then integration by parts, but in the end I always got a another integral that I could not calculate. Please give a hint how to calculate this integral.

答:
Wolfram Alpha says that the result 'cannot be expressed in terms of standard mathematical functions'.

The best you can do is say that
$$
\int \arctan(\sqrt[3]{x^2+1}) \, dx = \int_{0}^{x} \arctan(\sqrt[3]{t^2+1}) \, dt+C \, .
$$
This result follows directly from the fundamental theorem of calculus.


问:
Proof of $C_n = \frac{1}{n + 1} \binom{2n}{n}$
I want to comprehend the first proof of $C_n = \frac{1}{n + 1}\binom{2n}{n}$ on Wikipedia. Link: here.
I have problems with two steps:

$$\sum_{n=0}^{\infty} \binom{\frac{1}{2}}{n}y^n = \sum_{n=0}^{\infty} \frac{(-1)^{n + 1} }{4^n(2n - 1)} \binom{2n}{n}y^n$$ I calculated
$$\binom{\frac{1}{2}}{n} = \frac{(-1)^{n + 1}}{4^n} \frac{(n - 1.5) \cdot (n - 1.5 - 1) \cdot \ldots \cdot (n - 1.5 - (n - 3) + 1) }{n!}$$, but how to proceed?

One shall insert $$\sum_{n = 0}^{\infty} \frac{(-1)^{n+1}}{4^n(2n-1)}\binom{2n}{n} x^n$$ into $$\frac{1 - \sqrt{1 - 4x}}{2x} = \frac{1}{2} \frac{1}{x}(1 - \sqrt{1 - 4x})$$ by using standard power series operations. But how do we handle the division of $1$ by $x$ in terms of $x^n$? The wikipedia entry for power series says that the first coeffizient of the division is calculated as $d_0 = \frac{a_0}{b_0}$, but for $x = \sum b_n x^n$, $b_0$ is 0, such that we get a division by zero.

Please note that we consider formal power series, so I do not know if and how we can somehow detour via an other development point.

Background: The problem occured during computing the moments of the semicircular distribution in free probability theory.

EDIT: While elaborating the solution (I admit that I never had found it by myself) given by @BrianM.Scott, the specific second question was solved too: One does not calculate $\frac{1}{x}$ directly, but splits the sum up and lets the dangerous constant terms (for $n = 0$) eliminate one another before dividing the remaining sum terms (for $n > 0$) by $x$.

答: In this answer I wrote out a more detailed version of the argument in Wikipedia. It covers the evaluation of $\binom{\frac12}n$ in detail and also shows how to deal with the constant term, so it may answer your questions.


问:
Simplifying $f(\sqrt{7})$, where $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$

If $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$ ; then $f(\sqrt {7})=\; ?$

I tried solving this equation through many methods, I tried rationalizing, squaring, etc. But after each of them, the method became really lengthy and ugly.

I also noted that once we substitute $\sqrt {7}$ the inner part of the radical becomes imaginary. How to proceed with this piece of information?

Please help me with this problem. Any more innovative methods would be appreciated.

Answer: $4$.

Edit $[7^{th}$ March, $2021]$:

I was told yesterday that this question was wrong and that we were supposed to find $f(2\sqrt5)$. Although I can solve with $2\sqrt5$ the same way as given below in the answers.... What I don't understand is that why $\sqrt7$ doesn't work. Can someone please help?

答:
To answer your latest edit, if a complex number $x$ is not nonnegative real, there are two cases with regard to its square root:

$x<0$. In this case what I have learned is $\sqrt x=\sqrt{-x} i$. Maybe some math textbooks will consider this illegal but many authors comfortably regard $\sqrt{-1}$ as $i$.

$x=a+bi$ where $b\ne0$. In this case there are two numbers whose square equal to $x$ and one should avoid writing $\sqrt x$.

If $x=\sqrt 7$, it is ok to ask: find all complex numbers $z$ such that $z^2=\sqrt7 - 4\sqrt {\sqrt 7 - 4}$. It's also ok to ask you to find all complex numbers $z$ such that $z^2=\sqrt 7 + 4\sqrt {\sqrt 7 -4}$. Each will have two solutions (one being the opposite of the other). If you add the two (separate) solutions with positive real part you will get $4$. If you add the two solutions with negative real part you will get $-4$.