问:
Can we solve $u(x)=-1+x+\frac{x^2}{2}+2e^x-\int_0^xu(t)dt$ with using noise terms phenomenon?
question is can  $u(x)=-1+x+\frac{x^2}{2}+2e^x-\int_0^xu(t)dt$ be solved via noise terms phenomenon?

I made

$u_0(x)=-1+x+\frac{x^2}2+2e^x $

$u_{k+1}(x)=-\int_0^xu_k(t)dt$, $k\geq0$

$u_1(x)=-\int_0^xu_0(t)dt=-\int_0^x(-1+x+\frac{x^2}2+2e^x)dt=x-\frac{x^2}2-\frac{x^3}6-2(e^x-1)$

so I canceled $\pm \frac {x^2} {2}$ and $\pm2e^x$ from $u_0(x)$ and found the solution as $u(x)=-1+x$
But it doesnt justify the integral equation. Does this mean it cant be solved with noise terms p.? Is this a proper way to check?

答:
$u(x)=-1+x+\frac{x^2}{2}+2e^x-\int_0^xu(t)dt \Rightarrow \frac{du}{dx}=2x+2e^x+1-u$

$\frac{du}{dx}=2x+2e^x+1-u \Rightarrow e^x\frac{du}{dx}+e^xu=2e^xx+2e^{2x}+e^x$

$e^x\frac{du}{dx}+e^xu=2e^xx+2e^{2x}+e^x \Rightarrow e^xdu+e^xudx=d(e^xu)=(2e^xx+2e^{2x}+e^x)dx$

$d(e^xu)=(2e^xx+2e^{2x}+e^x)dx \Rightarrow e^xu+c=\int(2e^xx+2e{^2x}+e^x)dx$

$e^xu+c=\int(2e^xx+2e^{2x}+e^x)dx=e^x+e^{2x}+2(xe^x-e^x)$

$e^xu+c=e^x+e^{2x}+2(xe^x-e^x) \Rightarrow u=2x-1+e^x-\frac{c}{e^x}$


问:
How to find all third roots of $-i$?
How to find all third roots of $-i$?
I have no idea.
Could you give me some hint?

答:
If you know de Moivre's law, or representation of complex numbers as exponentials, this problem is simple.

If you don't, here is an algebraic approach.

There exists some complex number $a+bi$ such that $(a+bi)^3 = -i$

Multiplying it out and combining terms.

$a^3 + 3a^2bi + 3a(bi)^2 + (bi)^3 = -i\\
(a^3 - 3ab^2) + (3a^2b - b^3)i = -i$

Set the real parts and the imaginary parts equal to one annother.

$a^3 -3ab^2 = 0\\
a^2b - b^3 = -1$

Working with the first equation.
$a(a^2-3b^2) = 0$

$a = 0$ or $a^2 = 3b^2$

If $a = 0$ substituting into the second equation.

$-b^3 = -1\\
b = -1\\
a+bi = -i$

This is one of our answers.

If $a^2 = 3b^2$

$9b^3 - b^3 = -1\\
8b^3 = -1\\
b^3 = -\frac {1}{8}\\
b = -\frac {1}{2}$

Substituting back into $a^2 = 3b^2$

$a^2 = \frac {3}{4}\\
a = \pm \frac {\sqrt 3}{2}\\
\frac {\sqrt 3}{2} - \frac 12 i, -\frac {\sqrt 3}{2} - \frac 12 i$

are our other 2 solutions.

ContentLicense: CC BY-SA 4.0

Title: How to find all third roots of $-i$?


问:
Compute the Bias of the estimates
If the true expection of response is
\begin{align}
E(y)=\beta_0+\beta_1x+\beta_2x^2
\end{align}
but you assume $E(y)=\beta_0+\beta_1x$ insted. With $x=-2,-1,0,1,2$. How can i compute the Bias of the estimates of $\beta_0$ and $\beta_1$?. I did sth wrong because my solution is 0. Pls support me. Thx a lot

答:
If you simply ignore the $\beta_2$ term but kept the correct $\beta_0$ and $\beta_1$ terms then you would be $\beta_2\times 4$ too low for $x=-2$, $\beta_2\times 1$ too low for $x=-1$, $\beta_2\times 0$ too low for $x=0$, $\beta_2\times 1$ too low for $x=1$, and  $\beta_2\times 4$ too low for $x=2$.

Your linear regression will attempt to minimise these errors.  The best fit to $(-2,4\beta_2),  (-1,\beta_2),  (0,0),  (1,\beta_2), (2,4\beta_2)$ would be $\hat y=2\beta_2 +0x$

so the expected bias in the estimate of $\beta_0$ is $2\beta_2$, and the expected bias in the estimate of $\beta_1$ is $0$.

Title: Compute the Bias of the estimates


问:
Help with calculus of variations question Chiang
I am trying to solve the following problem from Chiang's Fundamental Methods of Dynamic Optimisation:
$$
V \left[y \right] = \int_{0}^{2} \left( 2ye^t + y^2 +y'^2\right) dt
$$

with boundary conditions, $y(0) = 2$ and $y(2) = 2e^2+e^{-2}$

I can get to the general solution, but the definitive solution escapes me. The general solution (I think) is as follows:

Using Euler's equation
$$
F_{y'y'} y''(t) + F_{yy'}y'(t)+F_{ty'}-F_y = 0
$$

We have

$$
2y''(t) -(2e^t+2y) =0 \\
y''(t) =e^t+y
$$

Integrating with respect to time twice gives the general solution

$$
y{*}(t) = e^t +\frac{1}{2}yt^2 +c_1t +c_2
$$

Solving for the constants $c1$ and $c2$ using the boundary conditions yeilds

$$
y(t=0) = e^{(0)} + \frac{1}{2}y(0)^2+c_1(0)+c_2 = 2 \\
c_2 = 2-1 - 0-0 \\
c_2 = 1
$$

And

$$
y(t=2) = e^{(2)} + \frac{1}{2}y(2)^2+c_1(2)+c_2 = 2e^2+e^{-2} \\
$$

Substituting $c_2 = 2$
$$
y(t=2) = e^{2} + 2y+2c_1+2 = 2e^2+e^{-2} \\
y(t=2) =  2c_1 = 2e^2-e^{2}+e^{-2}-2y-2 \\
y(t=2) =  2c_1 = e^2+e^{-2}-2y-2 \\
y(t=2) =  c_1 = \frac{1}{2}(e^2+e^{-2})-y-1 \\
$$

Substituting these back into the general solution

$$
y^{*}(t) = e^t +\frac{1}{2}yt^2 +(\frac{1}{2}(e^2+e^{-2})-y-1)t +2
$$

Which I don't think is correct. The answer given in the solutions appendix is which I am miles away from and can't see how to get there

$$
y*(t) = e^t +e^{-t}+\frac{1}{2}te^t
$$

答: Your Lagrangian is $$L(t,y,y')= 2y{\rm e}^t + y^2 + (y')^2,$$so the Euler-Lagrange equation $$\frac{\partial L}{\partial y} -\frac{{\rm d}}{{\rm d}t}\frac{\partial L}{\partial y'}=0$$becomes $$2{\rm e}^t + 2y -2y''=0,$$which can be rearranged to $$y''-y = {\rm e}^t.$$The general solution is $$y(t)=A{\rm e}^t+B{\rm e}^{-t}+\frac{t{\rm e}^t}{2}.$$Since $y(0)=2$, we have $A+B=2$. And $y(2)=2{\rm e}^2+{\rm e}^{-2}$ becomes $$A{\rm e}^2 +B{\rm e}^{-2} +{\rm e}^2 =2{\rm e}^2+{\rm e}^{-2},$$so $A=B=1$ fits the bill. The solution you're looking for is $$y(t)={\rm e}^t+{\rm e}^{-t}+\frac{t{\rm e}^t}{2}=2\cosh t + \frac{t{\rm e}^t}{2}.$$


问:
Correlation between multiplied numbers?
I do not have a strong math background, but I'm curious as to what this pattern is from a mathematical standpoint.

I was curious how many minutes there were in a day, so I said "24*6=144, add a 0, 1440."  Then it immediately struck me that 12*12=144, 6 is half of 12, and 12 is half of 24.  So, I checked to make sure it worked in other circumstances:

4*16=64
8*8=64

11*44=484
22*22=484

9*36=324
18*18=324

So what exactly is going on here from a logical standpoint to create that pattern?  Thanks in advance for satiating my curiosity!

答:
You are recombining the prime factors of the numbers.

64 => 2*2*2*2*2*2

So any split of these numbers will result in the total 64.

Same with the other numbers, you are just rearranging the prime factors of the numbers to form different combinations which will end up as the same product.

ContentLicense: CC BY-SA 3.0

Title: Correlation between multiplied numbers?


问:
Connection on direct sum of vector bundles written as a matrix
I am trying to solve the following problem:

I have to admit I am completly stuck and don't even know where to start. The notes I am reading introduce $E_{1}\oplus E_{1}$ rather briefly. I had a look at Lee's books and I found a short discussion about the direct sum of two vector bundles. I think I am fine with $E_{1}\oplus E_{1}$ being a vector bundle and that a connection on $E_{1}\oplus E_{1}$ can be defined by $\nabla:=\nabla_{1}\oplus\nabla_{2}$. What puzzles me is the matrix representation of a connection. We never talked about anything like a correspondence between connections and matrices. Furthermore forms are introduced in the next chapter. Yet two entries are forms. I guess, since they are 1-forms, they are "only" linear maps.

I would really appreciate some tips or source where I can look it up or that explains this "correspondence" so that I can solve the problem myself.

Many thanks in advance!

答:
If $\nabla$ and $\nabla'$ are connections on $E$, then their difference $\nabla - \nabla'$ is an "endomorphism-valued $1$-form" $B \in \Omega^1(M,\mathrm{End}(E))$. If $E=E_1 \oplus E_2$ then we can define the connection $\nabla^1$ by

$$ \nabla^1_X(s_1) = \mathrm{pr}_1 \nabla_X(s_1) $$

for a (local) section $s_1$ of $E_1$, where $\mathrm{pr}_1 \colon E \to E_1$ is the projection defined by the direct sum. You can do the same for $\nabla^2$. Then $\nabla-\nabla^1-\nabla^2$ is an endomorphism-valued $1$-form on $E$, and taking a look at the definition you see that

$$ \tilde{A}_X(s_1) = \mathrm{pr}_2 \nabla_X(s_1) .$$


问:
Second order differential equation and its related power series
Consider the DE $$y''(x)+(1-x)^{-1}y'(x)+y(x)=0$$ where it has an ordinary point at x=0.

a) There are two different power series solutions about x=0; what is their radius of convergence?

b) Let these solutions be $$y=\sum_{n=0}^\infty a_nx^n$$ Find the corresponding recurrence solution.

c) Obtain the solutions up to and including the term in $$x^4$$ first for $$a_0 = 1, a_1 = 0, \textrm{ then for } a_0 = 0, a_1 = 1$$

So my understanding is that I am supposed to plug in the respective series for y, y', y'' into the original equation then get a recurrence relationship from it. But I am lost on the concept of radius of convergence and on how to start on c).

Body:
hint

Write your equation as

$$(1-x)y''+y'×(1-x)y=0$$

with

$y=a_0+a_1x+a_2x^2+a_3x^3+$
$$...+a_nx^n+...$$

$y'=a_1+2a_2x+$
$$...+(n+1)a_{n+1}x^n+...$$

$y''=2a_2+2.3a_3x+$
$$...+(n+1)(n+2)x^n+...$$


问:
What's the word for being able to "pull something out of a function"?
If I have something like $(5x + 7x) = x(5 + 7),$ I've "factored" out the $x$.

With linear functions you can say $f(a \cdot x) = a \cdot f(x)$ "by linearity", you've (for lack of better phrasing) "pulled out the multiplication".

Or for preimages of sets, the preimage of union = union of preimage $f^{-1}(\cup_i A_i) = \cup_i f^{-1}(A_i)$.

Is there a general term for saying some operation inside a function works in a similar way? Something like

$$f(A \odot B) = f(A) \odot f(B)$$

where $\odot$ is some general operator. I guess we could say "linearity in $\odot$", is that standard?

答:
I see this as a special case of what category theory calls commuting.

In your first example, you start with the two numbers $5$ and $7$. If you first multiply them both by $x$, then add them together, you get the same result as you do if you first add them, then multiply the result with $x$. This isn't difficult to describe with a square-shaped commutative diagram. The same goes for your $\odot$ example.

If you have a linear map, applying that map commutes with multiplication by a constant in this category theoretical sense. This aligns with the matrix fact that the identity matrix and multiples thereof commute with other (square) matrices.

For your preimage example, anyone familiar with category theory would know exactly what you mean if you say that unions commute with taking preimages of a function.


问:
Bertrand's paradox: "... centre point lies in one of the two centre quarters of the diameter perpendicular to this direction ..."?
I am currently studying Photogrammetric Computer Vision – Statistics, Geometry, Orientation and Reconstruction by Förstner and Wrobel. Chapter 2 Probability Theory and Random Variables gives the following example:

In the case of alternatives which are not countable, e.g., when the event is to be represented by a real number, we have difficulties in defining equally probable events. This is impressively demonstrated by Bertrand’s paradox (Fig. 2.1), which answers the question: What is the probability of an arbitrarily chosen secant in a circle longer than the side of an inscribing equilateral triangle? We have three alternatives for specifying the experiment:

Choose an arbitrary point in the circle. If it lies within the concentric circle with half the radius, then the secant having this point as centre point is longer than the sides of the inscribing triangle. The probability is then $1/4$.

Choose an arbitrary point on the circle. The second point of the secant lies on one of the three segments inducing sectors of $60^\circ$. If the second point lies in the middle sector the secant through these points is longer than the side of the inscribing triangle. The probability is then $1/3$.

Choose an arbitrary direction for the secant. If its centre point lies in one of the two centre quarters of the diameter perpendicular to this direction the secant is longer than the side of the inscribing triangle. The probability is then $1/2$.

I don't quite understand the description given by 3.:

If its centre point lies in one of the two centre quarters of the diameter perpendicular to this direction the secant is longer than the side of the inscribing triangle. The probability is then $1/2$.

What is meant by "two centre quarters of the diameter perpendicular to this direction", and how is the probability $1/2$? A clearer explanation would be greatly appreciated.

答: Does this figure help you see?

Body: Does this figure help you see?


答:
In the hexagram picture on the right, the chosen direction of the secant is horizontal

so the centre of the secant is on the vertical line

and the secant will be too short if the centre is the top quarter of the vertical line or the bottom quarter (the thin parts)

but the secant will long enough if the centre is the middle two quarters (the thick parts of the vertical line)

Title: Bertrand's paradox: "... centre point lies in one of the two centre quarters of the diameter perpendicular to this direction ..."?

Body:
I am currently studying Photogrammetric Computer Vision – Statistics, Geometry, Orientation and Reconstruction by Förstner and Wrobel. Chapter 2 Probability Theory and Random Variables gives the following example:

In the case of alternatives which are not countable, e.g., when the event is to be represented by a real number, we have difficulties in defining equally probable events. This is impressively demonstrated by Bertrand’s paradox (Fig. 2.1), which answers the question: What is the probability of an arbitrarily chosen secant in a circle longer than the side of an inscribing equilateral triangle? We have three alternatives for specifying the experiment:

Choose an arbitrary point in the circle. If it lies within the concentric circle with half the radius, then the secant having this point as centre point is longer than the sides of the inscribing triangle. The probability is then $1/4$.

Choose an arbitrary point on the circle. The second point of the secant lies on one of the three segments inducing sectors of $60^\circ$. If the second point lies in the middle sector the secant through these points is longer than the side of the inscribing triangle. The probability is then $1/3$.

Choose an arbitrary direction for the secant. If its centre point lies in one of the two centre quarters of the diameter perpendicular to this direction the secant is longer than the side of the inscribing triangle. The probability is then $1/2$.

I don't quite understand the description given by 3.:

If its centre point lies in one of the two centre quarters of the diameter perpendicular to this direction the secant is longer than the side of the inscribing triangle. The probability is then $1/2$.

What is meant by "two centre quarters of the diameter perpendicular to this direction", and how is the probability $1/2$? A clearer explanation would be greatly appreciated.


问:
What is the algebraic way to solve inequalities like this: $|x+4|<2x$?
What is the algebraic way to solve inequalities like this: $|x+4|<2x$?

Can I rewrite this as $-2x<x+4<2x$ or [$x+4<2x$ and $-(x+4)<2x$]?

I found out that the answer was $x>4$, but how can I do it algebraically instead of by inspection? I also did some algebra that lead me to $x>{-4\over 3}$ when I considered $-(x+4)<2x$, this can't be right.

答:
Consider $$f(x)=2x-|x+4| = \begin{cases} 3x+ 4, & x < -4 \\ x-4, & x \ge 4\end{cases}$$
Obviously $f(x)$ is continuous, piecewise linear and strictly increasing, and $f(4)=0$. Therefore $$f(x) > 0 \iff x>4.$$


问:
symbolizations in FOL
I am doing a question of symbolizations in FOL
The problem is ：If Jupiter has a moon, it has a moon with an atmosphere.
Keys are:

M(x) x is a moon.

O(x, y)  x orbits y.

j    Jupiter

T(x) x has an atmosphere.

I wrote: Ax((M(x)/\O(x,j))->(M(x)/\T(x)/\O(x,j)))

Ax here stands for "for all x", and jupiter has a moon means there is a moon orbiting jupiter

I cannot figure out what I did wrong please help

答:
What you've written, translated back into natural language, is:

"For every $x$, if $x$ is a moon orbiting Jupiter then $x$ is a moon orbiting Jupiter with atmosphere."

There is a big problem with this, however: what if Jupiter has two (or more) moons?

The sentence you're trying to translate merely says that if Jupiter has a moon, then Jupiter has a moon with atmosphere; it does not say that all of Jupiter's moons (if it has any) have atmosphere.

Basically, the two clauses "has a moon" and "has a moon with atmosphere" are a bit more independent than you're treating them here. One correct implementation is the following:

 Try $$[\exists x(M(x)\wedge O(x,j))]\implies [\exists y(M(y)\wedge O(y,j)\wedge T(y))].$$ Using different variables here isn't necessary, but it does make things clearer to read.

Title: symbolizations in FOL


问:
Prove that the sequence $a_{n} = 1+\frac{1}{2n}$ converges to 1.
This is what I have so far. Let $\varepsilon >0$.

We must show that there exists an $N\in\mathbb{N}$ such that $n\geq N$ implies that $|(1+\frac{1}{2n} - 1| < \varepsilon$.

Since $a_{n} = 1+\frac{1}{2n}$, $∣1 + \frac{1}{2n} - 1∣ = \frac{1}{2n}$ and $\frac{1}{2n} < \varepsilon$ if $n > \frac{1}{2\varepsilon}$.

答:
Yes

You look for $ N\in \Bbb N $ satisfying

$$(\forall n\in \Bbb N) \;\;\Bigl(n\ge N\implies n>\frac{1}{2\epsilon}\Bigr)$$

So, you can take
$$N=\lfloor \frac{1}{2\epsilon}\rfloor +100$$
You will be sure that
$$n\ge N\implies n>>\frac{1}{2\epsilon}$$
$$\implies 1+\frac{1}{2n}-1<\epsilon$$


答:
Let $\varepsilon > 0$. According to the Archimedean  property, there exsts $n_{\varepsilon}\in\mathbb{N}$ s.t

\begin{align*}
\frac{1}{2\varepsilon} < n_{\varepsilon} \Rightarrow \frac{1}{2n_{\varepsilon}} < \varepsilon
\end{align*}

Consequently, for every $\varepsilon > 0$, there corresponds $n_{\varepsilon}\in\mathbb{N}$ s.t. for every $n\in\mathbb{N}$
\begin{align*}
n\geq n_{\varepsilon} \Rightarrow \frac{1}{n} \leq \frac{1}{n_{\varepsilon}} \Rightarrow \frac{1}{2n} \leq \frac{1}{2n_{\varepsilon}} < \varepsilon
\end{align*}
thence we conclude the proposed result holds.

Hopefully this helps!

Title: Prove that the sequence $a_{n} = 1+\frac{1}{2n}$ converges to 1.

Body:
This is what I have so far. Let $\varepsilon >0$.

We must show that there exists an $N\in\mathbb{N}$ such that $n\geq N$ implies that $|(1+\frac{1}{2n} - 1| < \varepsilon$.

Since $a_{n} = 1+\frac{1}{2n}$, $∣1 + \frac{1}{2n} - 1∣ = \frac{1}{2n}$ and $\frac{1}{2n} < \varepsilon$ if $n > \frac{1}{2\varepsilon}$.


答:
You are on right track. But you have yet to show what your $N$ is.

Depending on how the set of real numbers are introduced, you can either set $N$ to be $[\frac{1}{2\epsilon}]+1$ or apply the Archimedean principle to show that there exists a positive integer $N$ such that $N>\frac{1}{2\epsilon}$.


答: You got most of it. Now, you need to find an $N(\epsilon)\in \Bbb N$ which is bigger than the lower bound for $n$ (i.e. $\frac{1}{2\epsilon}$). In this case $N_\epsilon=\lceil\frac{1}{2\epsilon}\rceil$ satisfies.


问:
Can we omit surjectivity of homomorphism between structures?
I know the following fact (it can be proved using induction on the length of $\Phi$):

Let $\mathfrak{A,B}$ be algebraic structures (their domains are $A,B$), let $f:\mathfrak A\rightarrow\mathfrak B$ be their isomorphism and let $\Phi(x_1,\dots,x_n)$ be a formula, $a_1,\dots,a_n\in A$. Then $\mathfrak A\models\Phi(a_1,\dots,a_n)\Longleftrightarrow\mathfrak B\models\Phi(fa_1,\dots,fa_n)$.

I want to weaken this statement. My new version is (in the accepted notation):

Let $f:\mathfrak A\rightarrow\mathfrak B$ be a surjective homomorphism. Then $\mathfrak A\models\Phi(a_1,\dots,a_n)\Longrightarrow\mathfrak B\models\Phi(fa_1,\dots,fa_n)$.

The motivation for surjectivity is: let $\Phi=\forall x\Psi$.

$\mathfrak A\models\Phi(a_1,\dots,a_n)$ means that for every interpretation $\gamma:\{x\}\cup FV(\Phi)\rightarrow A$ such that $\gamma x_i=a_i, i\in\{1,\dots,n\},$ we have $\mathfrak A\models\Psi(x_1,\dots,x_n,x)[\gamma]$. Since $\Psi$ is shorter than $\Psi$, we can conclude that $\mathfrak B\models\Psi(fx_1,\dots,fx_n,fx)[f\circ\gamma]$.

The last thing we need to say that $\mathfrak B\models\Phi(fx_1,\dots,fx_n)$, as far as I understand, is surjectivity of $f$. Otherwise there may exist interpretation $\Gamma:\{x\}\cup FV(\Phi)\rightarrow B$ such that $\Gamma(x_i)=b_i, i\in\{1,\dots,n\}, \Gamma(x)=b$ and $f^{-1}(b_i)\ne\varnothing=f^{-1}(b), i\in\{1,\dots,n\},$ in which $\mathfrak B\nvDash\Psi(b_1,\dots,b_n,b)$.

Am I right? Or it is possible to omit surjectivity somehow?

答:
Let $f:\mathfrak A\rightarrow\mathfrak B$ be a surjective homomorphism. Then $\mathfrak A\models\Phi(a_1,\dots,a_n)\Longrightarrow\mathfrak B\models\Phi(fa_1,\dots,fa_n)$.

This is false. Your argument works for the inductive step of adding a universal quantifier, but not for the inductive step of adding a negation.

We say that a map $f:\mathfrak A\rightarrow\mathfrak B$ preserves a formula $\Phi(x_1,\dots,x_n)$ if $\mathfrak A\models\Phi(a_1,\dots,a_n)\Longrightarrow\mathfrak B\models\Phi(fa_1,\dots,fa_n)$.

If $f$ is not injective, it will fail to preserve the formula $\lnot(x_1=x_2)$. And if $f$ is an injective homomorphism which is not an embedding, then it will fail to preserve the formula $\lnot R(x_1,\dots,x_n)$ for some relation symbol $R$. So if a surjective homomorphism preserves all formulas, it must also be an injective embedding, so it must be an isomorphism!

Of course, there is a wider class of maps that preserve all formulas, namely the elementary embeddings. But for a general theory, there is not a much cleaner characterization of elementary embeddings than "those embeddings which preserve all formulas". A major reason to prove that a theory of interest has quantifier elimination (or is model-complete) is to obtain a more concrete understanding of the elementary embeddings between its models.

On the other hand, though surjective homomorphisms fail to preserve all formulas, we can characterize those formulas which are preserved by surjective homomorphisms. These are exactly the formulas which are equivalent to a positive formula: one built from atomic formulas using $\land$, $\lor$, $\forall$, and $\exists$ (but not $\lnot$).

And you ask about removing surjectivity. The formulas which are preserved by all homomorphisms are exactly those which are equivalent to a positive existential formula: one built from atomic formulas using $\land$, $\lor$, and $\exists$ (but not $\lnot$ or $\forall$).

These are two examples of a whole world of "preservation theorems" in model theory, which link up syntactic properties of formulas with semantic relationships between models. Good references are the model theory textbooks by Chang and Keisler and by Hodges.


答: You are doing $$(kx)\frac { x }{ k } $$ where $k=2$ </p>

Title: Correlation between multiplied numbers?

Body:
I do not have a strong math background, but I'm curious as to what this pattern is from a mathematical standpoint.

I was curious how many minutes there were in a day, so I said "24*6=144, add a 0, 1440."  Then it immediately struck me that 12*12=144, 6 is half of 12, and 12 is half of 24.  So, I checked to make sure it worked in other circumstances:

4*16=64
8*8=64

11*44=484
22*22=484

9*36=324
18*18=324

So what exactly is going on here from a logical standpoint to create that pattern?  Thanks in advance for satiating my curiosity!


问:
l'Hospitals rule for multivariate limit
I have the following bivariate function where $\alpha,\beta,\gamma,\delta$ are constants, $a,b,c,d$ are integers and $x,y\in [0,1]$
$$
f(x,y) = \frac{\alpha(1-x^ay^b) + \beta(1-x^cy^d)}{\gamma(1-x^2)+\delta(1-y^3)}
$$
which I would like to examine the limit of as $x=y=1$ from below. My instinct was to apply l'Hospital's rule by setting $x=y$ and converting to a univariate function.
$$
f(x,x) = \frac{\alpha(1-x^{(a+b)}) + \beta(1-x^{(c+d)})}{\gamma(1-x^2)+\delta(1-x^3)}
$$
from which we can gather the limit is
$$
\lim_{x\rightarrow 1}f(x,x) = \frac{\alpha(a+b)+\beta(c+d)}{2\gamma +3\delta}
$$
However, am also aware that l'Hospital's rule is strictly for univariate functions. In this particular case, how can I be certain that my limit is correct?

答: If $g,\,h$ are each $0$ at $x=y=1$ and can be expanded about it as Taylor series,$$g=g_x(1,\,1)(x-1)+g_y(1,\,1)(y-1)+o(x-1)+o(y-1)$$etc., but the limiting behaviour of $g/h$ is path-dependent unless$$\Delta:=g_x(1,\,1)h_y(1,\,1)-g_y(1,\,1)h_x(1,\,1)=0.$$With $g:=\alpha+\beta-\alpha x^ay^b-\beta x^cy^d,\,h=\gamma+\delta-\gamma x^2-\delta y^3$, we get$$\Delta=(b\alpha+d\beta)2\gamma-(a\alpha+c\beta)3\delta,$$so in general the limit is undefined.

Title: l'Hospitals rule for multivariate limit


问:
The ratio of two areas of triangles
On the picture below is a square. The question is to find $s/r$.

I tried to draw a line from the middle of one side of the square to the middle of the opposite one, and it appears that the area of $s$ is $1/4$ the area of our square. So I stopped to calculate the ratio of triangle's area $r$ to the area of whole square.

Any hints?

答:
I don't know if this is the easiest way.  It certainly isn't particularly elegant.

I made the assumption that the side length of the square is 1.
I then found the coordinates of the points on the corners of R.

We can find the equations of two lines that make the sides of R

$y - \frac 12 x\\
y + 2x = 1$

And solve for the point of intersection.

$(x,y) = (\frac 25, \frac 15)$

Now we have a line though this point and the point $(1,0)$

$3y + x = 1$

Which hits the side at $(0, \frac 13)$

Then there is a side with a line through this line at the point (1,1)

$3y - 2x = 1$ which intersects the side whose equation we already have $y+2x = 1$

$4y = 2$

$(x,y) = (\frac 14, \frac 12)$

The points on the corners of $R: (\frac 25,\frac 15),(0,\frac 13),(\frac 14,\frac 12)$

Using the shoelaces algorithm we get the Area of $R = \frac {1}{20}.$

S has base 1, height $\frac 12$

Area of $S = \frac 14$

$\frac {\text {Area of S}}{\text{Area of R}} = \frac {5}{1}$

Title: The ratio of two areas of triangles


问:
Value of $c$ that makes $c(\bar X (1 - \bar X))$ an unbiased estimator of $p(1 - p)$
I need to find the value of $c$ that makes $c(\bar X (1 - \bar X))$ an unbiased estimator of $p(1 - p)$.

I am confused about how to do this. I want to say that the value of $c$ must be $1$ because $E(\bar X) = \frac{\mu}{n} = \frac{np}{n} = p$, but that seems too easy.

Any advice? Thanks in advance.

答:
$E(\bar{X}^2)=\frac{1}{n}E(X^2)+\frac{n-1}{n}p^2=\frac{\sigma^2}{n}+p^2$  You need  to take into account the $\sigma^2$ term to get $c$, since $E(\bar{X}(1-\bar{X}))=p(1-p)-\frac{\sigma^2}{n}$.

Assuming a Bernoulli distribution $\sigma^2=p(1-p)$  so that $E(\bar{X}(1-\bar{X}))=\frac{n-1}{n}p(1-p)$ leading to $c=\frac{n}{n-1}$ with max $c=2$ at $n=2$.


问:
Divisibility bound on sums of numbers
Suppose $p$ and $q$ are two prime numbers. Let $N(p)$ be the set of natural numbers that are divisible only by primes $\leq p$. Define $N(q)$ similarly.

If $N(p,q)$ is the set of numbers that can be written as a sum of a number in $N(p)$ and a number in $N(q)$, what is the upper bound on the primes that can divide numbers in $N(p,q)$, if any?

答:
For every prime number $p$ we have $1,2^{2^n}\in N(p)$, and hence $2^{2^n}+1\in N(p,q)$ for any pair of prime numbers $(p,q)$. The number $F_n:=2^{2^n}+1$ is called the $n$-th Fermat number. It is currently unknown whether there are only finitely many Fermat primes or not, despite the fairly generous attention it receives.

It is known, however, that every prime factor $p$ of the Fermat number $F_n=2^{2^n}+1$ satisfies
$$p\equiv1\pmod{2^{n+2}}.$$
In particular every prime factor of $F_n$ satisfies $p>2^{n+2}$, and hence there exist arbitrarily large prime numbers dividing numbers of the form $2^{2^n}+1$. This shows that $N(p,q)$ is unbounded for all pairs of primes $(p,q)$.

Title: Divisibility bound on sums of numbers


问:
How to find derivative of $\sin\sqrt{x}$ using difference quotient?
The definition of derivative of a function $f(x)$ is $$\lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$

Using this definition, the derivative of $\sin\sqrt{x}$ will be:

$$\lim_{h\to0} \frac{\sin\sqrt{x+h}-\sin\sqrt{x}}{h}$$

$$\lim_{h\to 0} \frac{\cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$$

Now i got stuck. How to find the limit or simplify this expression?

I get intuition that we have to use $$\lim_{x\to0}\frac{\sin x}{x} = 1$$

but that too is leading no where. I am unable to remove h from denominator.

NOTE

I know the derivative of $\sin\sqrt{x}$ is $\frac{\cos\sqrt{x}}{2\sqrt{x}}$ using chain rule, but this exercise was given to us for practice using division quotient.

答:
You got stuck here

$$2 \cdot \lim_{h\to 0} \frac{\cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$$

You can break that one limit up into the product of two limits.

$$\lim_{h\to 0} \cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)
\cdot\lim_{h\to 0} \frac{\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$$

The first limit is easy.

$$\lim_{h\to 0} \cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)
  = \cos\left(\frac{\sqrt{x}+\sqrt{x}}{2}\right) = \cos \sqrt x$$

The second isn't that much harder.

\begin{align}
   \lim_{h\to 0} \frac{\sin\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}
   &= \lim_{h\to 0} \frac{\sin\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}
                         {\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)} \cdot
                    \frac{\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h} \\
   &= \lim_{h\to 0} \left(1 \cdot
                    \dfrac{\sqrt{x+h}-\sqrt{x}}{2h}\right) \\
   &= \lim_{h\to 0} \dfrac{(x+h)-x}{2h(\sqrt{x+h}+\sqrt{x})} \\
   &= \dfrac{1}{4\sqrt{x}} \\
\end{align}

Putting it all together, you get $\dfrac{\cos\sqrt{x}}{2\sqrt{x}}$


问:
For $x,y\in l_2, x^{(k)}\to x, y^{(k)}\to y$ then $\langle x^{(k)},y^{(k)}\rangle \to \langle x,y\rangle$

For $x,y\in l_2$ if $x^{(k)}\to x, y^{(k)}\to y$ then show that $\langle x^{(k)},y^{(k)}\rangle \to \langle x,y\rangle$. Here $x^{(k)} = (x_1,x_2,\ldots,x_k,0,0,\ldots,0)$ and $\langle x,y\rangle = \sum_{i=1}^\infty x_iy_i$.

Just want to check if my proof is alright:

$$x^{(k)}\to x \implies \forall \epsilon > 0 \exists N_1\in\mathbb N\ \forall k\ge N_1 (\|x^{(k)}- x\|_2 < \epsilon)$$
$$y^{(k)}\to y \implies \forall \epsilon > 0 \exists N_2\in\mathbb N\ \forall k\ge N_2 (\|y^{(k)}- y\|_2 < \epsilon)$$

Pick $N_1,N_2$ such that $\|x^{(k)}- x\|_2 < \sqrt\epsilon$ for all $k\ge N_1$ and $\|y^{(k)}- y\|_2 < \sqrt\epsilon$ for all $k\ge N_2$. Put $N=\max (N_1,N_2)$.

Now note that $$|\langle x^{(k)},y^{(k)}\rangle - \langle x,y\rangle| \le \|x^{(k)}- x\|_2\|y^{(k)}- y\|_2 \le \epsilon, \forall k\ge N$$

So, the proof is complete.

Sounds good?

答:
This is a corollary of the following fact:

Let $\mathcal{H}$ be an inner product space, with inner product $\langle\cdot,\cdot\rangle$.
Let $x,y,x_{n},y_{n}\in\mathcal{H}$. If $x_{n}\rightarrow x$ and
$y_{n}\rightarrow y$, then $\langle x_{n},y_{n}\rangle\rightarrow\langle x,y\rangle$.

Proof: Since $(x_{n})$ and $(y_{n})$ are convergent sequences, they
are bounded. Choose $M>0$ such that $||x_{n}||\leq M$ and $||y_{n}||\leq M$
for all $n$. Note that
\begin{eqnarray*}
||x|| & \leq & ||x-x_{n}||+||x_{n}||\\
 & \leq & ||x-x_{n}||+M.
\end{eqnarray*}
Letting $n\rightarrow\infty$ and noticing that $||x-x_{n}||\rightarrow0$,
we have that $||x||\leq M$. Similarly $||y||\leq M$. We also assume
Cauchy-Schwarz ineqality (i.e., $|\langle x,y\rangle|\leq||x||\cdot||y||$)
without proof.

Consider
\begin{eqnarray*}
 &  & \left|\langle x_{n},y_{n}\rangle-\langle x,y\rangle\right|\\
 & \leq & \left|\langle x_{n},y_{n}\rangle-\langle x,y_{n}\rangle\right|+\left|\langle x,y_{n}\rangle-\langle x,y\rangle\right|\\
 & = & \left|\langle x_{n}-x,y_{n}\rangle\right|+\left|\langle x,y_{n}-y\rangle\right|\\
 & \leq & ||x_{n}-x||\cdot||y_{n}||+||x||\cdot||y_{n}-y||\\
 & \leq & M\left(||x_{n}-x||+||y_{n}-y||\right)\\
 & \rightarrow & 0.
\end{eqnarray*}
This shows that $\langle x_{n},y_{n}\rangle\rightarrow\langle x,y\rangle$.


问:
Prove convergence of improper integral $\frac{4}{(4x \ln^2(y))} \ dx$
I wanna prove that the following improper integral converge but I dont't have any clear idea of how to do this:

$$\int_3^\infty \frac{4}{4x \ln^2 (y)} \ dx$$

From what I have learned, usually improper integral could be separated it into two parts, but I don't whether I can apply that method for the improper integral. Does someone have any thoughts?

答:
hint

Your integral has the same nature than

$$J=\int_e^\infty\frac{dx}{x\ln^2(x)}$$

now, make the substitution
$$t=\ln(x)$$
with

$$dt=\frac{dx}{x}$$

So, $ J $ has the same convergence than

$$\int_1^\infty \frac{dt}{t^2}$$
which clearly converges.


问:
Finding UMP test for $\text{Cauchy}(0, \theta)$ distribution

If $X$ follows $\text{Cauchy}(0, \theta)$ distribution, construct a UMP size $\alpha$ test for testing $H_0:\theta=\theta_0$ against $H_1:\theta>\theta_0$

My attempt:

Let us take any $\theta_1>\theta_0$. First we have to find a most powerful test of size $\alpha$ for testing $H_0:\theta=\theta_0$ against $H'_1:\theta=\theta_1$

Now, $f_\theta (x)=\frac{\theta}{\pi(x^2+\theta^2)}$, $x\in\mathbb R, \theta>0$

So, $\lambda(x)=\frac{f_{\theta_1}(x)}{f_{\theta_0}(x)}=\frac{\theta_1(x^2+\theta_0^2)}{\theta_0(x^2+\theta_1^2)}$

$\frac{d\lambda(x)}{dx^2}>0$, so $\lambda(x)$ is an increasing function of $x^2$ or $|x|$, i.e, $\lambda(x)>k \iff |x|>c$

Therefore, by N-P lemma, a most powerful test of size $\alpha$ for testing $H_0:\theta=\theta_0$ against $H'_1:\theta=\theta_1$ is given by,

$\phi(x)= 1$ if $|x|>c$

$\phi(x)= 0$ if $|x|<c$

where c is such that $E_{\theta_0}\phi(x)=\alpha$, i.e, $P_{\theta_0}(|X|>c)=\alpha$

I'm getting stuck here. I cannot understand how to find the value of c. If $X$ follows $\text{Cauchy}(0, \theta_0)$, then what does $|X|$ follow?

Any kind of hints and suggestions are appreciated. Thanks in advance.

答:
$P_{\theta_0}(|X|>c)=\alpha$

$\displaystyle \Rightarrow P_{\theta_0}(X>c)+P_{\theta_0}(X<-c)=\alpha$

$\displaystyle \Rightarrow \frac{\theta_0}{\pi}\int_c^{\infty}\frac{dx}{x^2+{\theta_0}^2}+\frac{\theta_0}{\pi}\int_{-\infty}^{-c}\frac{dx}{x^2+{\theta_0}^2}=\alpha$

$\displaystyle \Rightarrow 1-\frac{2}{\pi}\tan^{-1}\left(\frac{c}{\theta_0}\right)=\alpha$

$\displaystyle \Rightarrow c=\theta_0\tan\frac{\pi\left(1-\alpha\right)}{2}$


问:
Matrix representation of the operator $A$ with $\frac{d}{dt}U(t)=AU(t)$
Consider the following equations

\begin{eqnarray}
u_{tt}+(u_{xx}-\phi u_t)_{xx}&=&0, \quad (l_0,l_1)\times(0,\infty)\\
v_{tt}-v_{xx}+\psi v_t&=&0,\quad (l_1,l_2)\times(0,\infty)\\
u(x,0)=u_0,\, u_t(x,0)&=&u_1\quad (l_0,l_1)\\
v(x,0)=v_0,\, v_t(x,0)&=&v_1 \quad (l_1,l_2)
\end{eqnarray}

whit $\phi(x),\psi(x)>0$ in $(l_0,l_1)$ ,$(l_1,l_2)$ respectively.

I want to show that the previous problem is well defined, i.e, the problem has unique solution treating it as an abstract Cauchy problem. I have to represent the problem as $\frac{d}{dt}U(t)=AU(t)$.

If $U=(u,v,w,z)^T$, then I define the system operator as $AU=(w,z,-(u_{xx}-\phi w)_{xx},v_{xx}-\psi z)^T$. Now, if $U(t)=(u,v,u_t,v_t)^T$ it follows that

$$\frac{d}{dt}U(t)=AU(t)=(u_t,v_t,-(u_{xx}-\phi u_t)_{xx},v_{xx}-\psi v_t)^T$$

It is a fact that I have to find the domain of the operator $A$, but my question is: Can I write the operator $A$ in matrix form?

答:
Writing $$-(u_{xx}-\phi u_t)_{xx}=-\partial_x^4 u+\partial_{x}^2 (\phi u_t)$$
it would seem that

$$
AU(t)
=\begin{pmatrix} u_t \\ v_t\\ -\partial_{x}^4 u+\partial_{x}^2(\phi u_t) \\ \partial_x^2 v-\psi v_t\end{pmatrix}
 =
\begin{pmatrix}
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1\\
 -\partial_x^4 & 0 & \partial_x^2\phi & 0 \\
 0 & \partial_x^2 & 0 & -\psi
\end{pmatrix}
\begin{pmatrix} u \\ v \\ u_t \\ v_t\end{pmatrix}
$$
where $ \partial_x^2 \phi = \phi_{xx}+2\phi_x\partial_x+\phi \partial_x^2$ has been used to simplify the typesetting.


答:
This integral is improper because of the infinite upper limit, not because of the lower limit. The integrand is defined and finite for all $x \geq 2$. Therefore, you need only replace the upper limit of integration with a finite limit (call it $b$), evaluate the resulting proper integral, then take the limit of the answer as $b$ approaches infinity:
$$\displaystyle\int_2^\infty \frac{1}{x \ln^2 x} \: dx = \lim_{b \to \infty} \int_2^b \frac{1}{x \ln^2 x} \: dx$$

Title: Prove convergence of improper integral $\frac{4}{(4x \ln^2(y))} \ dx$

Body:
I wanna prove that the following improper integral converge but I dont't have any clear idea of how to do this:

$$\int_3^\infty \frac{4}{4x \ln^2 (y)} \ dx$$

From what I have learned, usually improper integral could be separated it into two parts, but I don't whether I can apply that method for the improper integral. Does someone have any thoughts?


问:
Exercise 14, section 9 of Hammack's Book of Proof.
The exercise asks the reader to prove or disprove the following statement.

If $A$ and $B$ are sets, then $\mathscr{P}(A) \cap \mathscr{P}(B) = \mathscr{P}(A \cap B)$.

I know one way to prove such a statement is to simply produce a series of equalities that transform the LHS of the equation into the RHS. However, I am not sure if my translation of the powersets into set builder notation is valid. More specifically, the book never gives an example of a quantifier used in a set builder so I tried to translate from the book's definition of a subset written in English (as opposed to symbolic notation). Bellow is my proof.

Proof. Observe the following sequence of equalities.
\begin{align}
\mathscr{P}(A) \cap \mathscr{P}(B) &= \{ X: X \subseteq  A \land X \subseteq B\} &(\text{def. of powerset}) \\
&= \{ X: \forall x \in X, x \in A \land x \in B \} &(\text{def. of subset}) \\
&= \{ X:\forall x \in X, x \in A \cap B \} &(\text{def. of intersection}) \\
&= \{ X: X \subseteq A \cap B \} &(\text{def. of subset}) \\
&= \mathscr{P}(A \cap B) &(\text{def. of powerset})
\end{align}

The proof is complete. $$\tag*{$\blacksquare$}$$

Any feedback on the validity of this proof or on ways it could be better written is greatly appreciated.

答:
The proof is correct, and if you’re required to use that rather artificial format, it’s just fine. You probably are required to give that much detail, but if you’re not restricted to that format, it could be given in a form closer to normal mathematical writing:

$X\in\wp(A)\cap\wp(B)$ iff $X\in\wp(A)$ and $X\in\wp(B)$, which is the case iff $X\subseteq A$ and $X\subseteq B$. But that is true iff $X\subseteq A\cap B$, i.e., iff $X\in\wp(A\cap B)$, so $\wp(A)\cap\wp(B)=\wp(A\cap B)$.


问:
Second order differential equation and its related power series
Consider the DE $$y''(x)+(1-x)^{-1}y'(x)+y(x)=0$$ where it has an ordinary point at x=0.

a) There are two different power series solutions about x=0; what is their radius of convergence?

b) Let these solutions be $$y=\sum_{n=0}^\infty a_nx^n$$ Find the corresponding recurrence solution.

c) Obtain the solutions up to and including the term in $$x^4$$ first for $$a_0 = 1, a_1 = 0, \textrm{ then for } a_0 = 0, a_1 = 1$$

So my understanding is that I am supposed to plug in the respective series for y, y', y'' into the original equation then get a recurrence relationship from it. But I am lost on the concept of radius of convergence and on how to start on c).

答:
To find the radius of convergence, I would suggest using the Ratio Test. Remember that the Ratio Test asks you to calculate $$\displaystyle\lim_{n \to \infty} \left| \frac{a_{n + 1}}{a_n} \right|$$
You might get this value from the recurrence relation.


问:
(weak) homotopy equivalence
I have a question arising from chapter 3, page 41, in Switzer. He says "Note that every homotopy equivalence (in $\mathscr{T}$ [this is the category of topological spaces]) is a weak homotopy equivalence." This statement suggests that there are topological spaces which are weakly homotopy equivalent but not homotopy equivalent. Can anyone give an example of two such spaces?

Also, you could infer from his statement that in some other category there might be homotopy equivalences which are not weak homotopy equivalences. Is this true? That is, are there categories such that H.E. $\not\Rightarrow$ W.H.E.?

答:
First note that by Whitehead's Theorem, if you know that $X$ and $Y$ are CW complexes, then "weakly homotopy equivalent" and "homotopy equivalent" coincide.

On the other hand, for general topological spaces, there are counterexamples where the two notions don't coincide.

The Long Line is not contractible, and yet all its homotopy groups vanish.  So, the inclusion of a point into the long line induces a weak homotopy equivalence between them (trivially), yet they are not homotopy equivalent.

Title: (weak) homotopy equivalence


问:
Can this sum ever converge?
If I have a strictly increasing sequence of positive integers, $n_1<n_2<\cdots$, can the following sum converge?

$$ \sum_{i=1}^\infty \frac{1}{n_i} (n_{i+1}-n_{i}) $$

I suspect (and would like to prove) that it always diverges. Haven't made much progress so far, though.

On a related note, is there any characterization of which subsequences of $1/n$ have a convergent sum?

答:
To expand on my comment (oops, and even more on Achille Hui's comment which appeared when I was typing):
$$
\frac{n_{i+1}-n_i}{n_i}\geq \int_{n_i}^{n_{i+1}}\frac{1}{t}dt\;\Rightarrow \; \sum_{i=1}^K\frac{n_{i+1}-n_i}{n_i}\geq \int_{n_1}^{n_{K+1}}\frac{1}{t}dt=\log\left(\frac{n_{K+1}}{n_1}\right)\rightarrow +\infty.
$$
Of course, $n_K$ tends to $+\infty$ as $n_k\geq K$ for every $K$ by induction.

For your other question, a partial answer can be given with the help of the notion of natural/asymptotic density. If $\liminf \frac{k}{n_k}>0$, then there exists $m>0$ and $K$ such that $\frac{k}{n_k}\geq m$ for every $k\geq K$, whence $\sum_{k\geq K}\frac{1}{n_k}\geq m\sum_{k\geq K}\frac{1}{k}=+\infty$. That is, if the set $\{n_k\;;\;k\geq 1\}$ has positive asymptotic lower density, then the series $\sum_{k\geq 1}\frac{1}{n_k}$ diverges. But the converse is false. For instance, the set of prime numbers $\mathcal P$ has null asymptotic density and yet $\sum_{p\in\mathcal{P}}\frac{1}{p}=+\infty$.

Title: Can this sum ever converge?


问:
Proving $\lim_{n \rightarrow \infty} \frac{2\sqrt{2}\Gamma(\frac{n+3}{2})}{n^{3/2}\Gamma(\frac{n}{2})} =1$

Assume $n \in \mathbb{N}$. Using Wolfram Alpha I found that the following has a nice limit:
$$
\begin{align*}
\lim_{n \rightarrow \infty} \frac{2\sqrt{2}\cdot\Gamma(\frac{n+3}{2})}{n^{3/2}\cdot\Gamma(\frac{n}{2})} =1
\end{align*}
$$
However, I'm not really sure how to arrive to that result through half-integer gamma.

From what I'm gathering,

$$
\frac{\Gamma(\frac{n+3}{2})}{\Gamma(n/2)} = 2^{-3/2}\frac{(n+1)!!}{(n-2)!!}
$$
which, I can only guess, should evaluate at $2^{-3/2} n^{3/2}$ or something close, but cannot see why.

答:
If you're open to other methods, using Stirling's approximation,

$$\operatorname{\Gamma}(n+1)=n!\sim\sqrt{2\pi n}\left(\frac ne\right)^n$$

you have

$$\operatorname{\Gamma}\left(\frac{n+3}2\right)\sim\sqrt{\pi(n+3)}\left(\frac{n+3}{2e}\right)^{\frac{n+3}2}\\
\operatorname{\Gamma}\left(\frac n2\right)\sim\sqrt{\pi n}\left(\frac n{2e}\right)^{\frac n2}$$

and in the limit,

$$\begin{align}
\lim_{n\to\infty}\frac{2\sqrt2\operatorname{\Gamma}\left(\frac{n+3}2\right)}{n^{\frac32}\operatorname{\Gamma}\left(\frac n2\right)}&=\lim_{n\to\infty}\frac{2\sqrt2\sqrt{\pi(n+3)}\left(\frac{n+3}{2e}\right)^{\frac{n+3}2}}{n^{\frac32}\sqrt{\pi n}\left(\frac n{2e}\right)^{\frac n2}}\\[1ex]
&=\frac{2\sqrt2}{(2e)^{\frac32}}\lim_{n\to\infty}\left(\frac{n+3}n\right)^{\frac{n+3}2}\\[1ex]
\end{align}$$

and $\left(\frac{n+3}n\right)^{\frac{n+3}2}\to e^{\frac32}$, so the limit of $1$ follows.


问:
Implicitly Differentiating a System of Matrix Equations
Setup

Let $\mathbf a$ be an arbitrary $m\times 1$ vector, $\mathbf B$ be an arbitrary $n\times m$ matrix, with $m>n$, and $\mathbf C$ be a symmetric $m\times m$ matrix. The scalar $\lambda$ is also given.

The $n\times 1$ vector $\mathbf h$, the $m\times m$ matrix $\mathbf \Gamma$, and the scalars $w$, and $s$ satisfy the following system of equations:
\begin{align*}
\mathbf h =  & \;\left(\mathbf B \,{\mathbf \Gamma}\,\mathbf B'\right)^{-1}\,\mathbf B\,{\mathbf\Gamma}\,\mathbf{a}  ,\\[2ex]
\mathbf {\Gamma} =&\;\mathbf I_m   +w\,\mathbf C,\\[2ex]
w=&\;\frac{1}{1-\lambda s},\\[2ex]
s=&\ \left(\mathbf h'\,\mathbf B-\mathbf a' \right)\mathbf {C}\left(\mathbf B^{\prime}\,\mathbf h -\mathbf a \right).
\end{align*}

Question

I want to find

$$\frac{\partial w}{\partial \lambda}=\mathbf?$$

The problem is that changing $\lambda$ changes $w$ which changes $\mathbf \Gamma$, and so on. So, I have to somehow use the Implicit Function Theorem. For systems of scalar variables, I am used to totally differentiating the system, but I am not sure how to deal with the matrix equations in that case. Any help would be very welcome.

答:
$
\def\l{\lambda}
$
For ease of typing, define the variables
$$\eqalign{
G &= \Gamma = (I+wC) \quad\qquad F=(BGB^T)^{-1} \quad\qquad p=(B^Th-a)
}$$
and note that $F$ and $G$ are symmetric since $C^T=C$

Now calculate differentials in order of increasing complexity.
$$\eqalign{
G &= I+Cw \\
dG &= C\,dw \\
\\
F &= (BGB^T)^{-1} \\
dF &= -F(B\,dG\,B^T)F \\
 &= -(FBCB^TF)\,dw \\
\\
h &= FBGa \\
dh &= FB\,dG\,a + dF\,BGa \\
 &= (FBCa)\,dw - (FBCB^TFBGa)\,dw \\
 &= (FBCa)\,dw - (FBCB^Th)\,dw \\
 &= FBC(a - B^Th)\,dw \\
 &= -(FBCp)\,dw \\
\\
p &= B^Th - a \\
dp &= B^Tdh \\
 &= -(B^TFBCp)\,dw \\
\\
s &= p^TCp \\
ds &= 2p^TC\,dp \\
 &= -(2p^TCB^TFBCp)\,dw \\
\\
w &= (1-\lambda s)^{-1} \\
dw &= w^{-2}(s\,d\lambda + \lambda\,ds) \\
w^2\,dw &=  s\,d\lambda - (2\l p^TCB^TFBCp)\,dw \\
s\,d\lambda &= (w^2 + 2\lambda p^TCB^TFBCp)\,dw \\
}$$
Therefore
$$\boxed{\color{red}{
\quad\frac{dw}{d\lambda} \;=\; \frac{s}{w^2 + 2\lambda p^TCB^TFBCp}
}\quad}\qquad\qquad$$


问:
determine whether if a linear system has binary solutions.
Is there any general method that helps to determine a linear system have binary solutions?

E.g. For a linear system
$$
Ax=b
$$
Where $A \in \mathbb{N}^{m\times n}, b \in \mathbb{N}^{m\times 1}$. Is there any efficient way to determine if there is any solution fits this constraint $x \in \{0, 1\}^{n}$ ?

答:
No.  This problem is a well-known NP-Complete problem.  The best-known algorithms take time exponential in the size of the problem data.  Unless P=NP (something that most computer scientists consider unlikely), there won't be an efficient polynomial-time algorithm for this problem.

This is problem MP1 in Garey and Johnson's Computers and Intractability (page 245.)


问:
Fun calculus problem I can't seem to solve
I've recently picked up a math book I haven't read since college (highly recommended reading by the way!). I was reviewing multi-dimentional derivatives and such, and I stumbled upon a problem I've been trying to solve for two days, and I can't get it out of my head, so please help me out! = )

Problem (from memory):

There is a rabbit that runs in a perfect circle of radius $r$ with a constant speed $v$. A fox chases the rabbit, starting from the center of the circle and also moves with a constant speed $v$ such that it is always between the center of the circle and the rabbit. How long will it take for the fox to catch the rabbit?

I tried using the fact that $|x'(t)| = |y'(t)| = v$ where $x(t)$ is fox's position and $y(t)$ is rabbit's position, and that $x'(t)x''(t) = 0$ because of constant speed restriction, but I'm still failing to find a solution.

Anyone feel like attacking this one?

答:
For this problem it is advisable to introduce polar coordinates ($r$ and $\phi$). The rabbit runs at speed $v$ on a perfect circle with radius $R$. Therefore, $\phi =vt/R$.  As the fox stays between the center and the rabbit, it is at the same $\phi$. The fox's speed (which is constant) has two component (prime denotes the derivative with respect to time)
$$ v = \sqrt{r'^2+ r^2 \phi'^2}.$$
From the knowledge of $\phi'= v/R$ and the fact that $v$ is constant, we can deduce $$r' = v\sqrt{1 - \frac{r^2}{R^2}}.$$
This differential equation can be solve by separating the variables. The time $T$ it takes to reach the rabbit is given by
$$T=\int_0^R \frac{dr}{v \sqrt{1 - \frac{r^2}{R^2}}} = \frac{\pi R}{2v}.$$
(the last integral has been solve by substituting $r=R \sin\phi$)</p>


答:
In general, it is true that there are spaces which are weakly homotopy equivalent but not homotopy equivalent. One example would be the point and the Warsaw Circle where the unique map from the Warsaw circle to the point induces an isomorphism on all homotopy groups, and so they are weakly equivalent, but the Warsaw circle is not contractible and so not homotopy equivalent to a point.

However, in the category of CW-complexes, if $X$ and $Y$ are weakly homotopy equivalent, then they are homotopicy equivalent. This is a theorem known as Whitehead's Theorem.

Title: (weak) homotopy equivalence

Body:
I have a question arising from chapter 3, page 41, in Switzer. He says "Note that every homotopy equivalence (in $\mathscr{T}$ [this is the category of topological spaces]) is a weak homotopy equivalence." This statement suggests that there are topological spaces which are weakly homotopy equivalent but not homotopy equivalent. Can anyone give an example of two such spaces?

Also, you could infer from his statement that in some other category there might be homotopy equivalences which are not weak homotopy equivalences. Is this true? That is, are there categories such that H.E. $\not\Rightarrow$ W.H.E.?


问:
When sine is equal to one (depending on multiplier)?
I've made up the equation: $\sin\left(\frac{5n\pi}{2}\right)=1$. Through experimenting I figured that this equation holds for $n = 0.2 + 0.8\,m$. Why is that?

答:
Do you know the general answer to $\sin(x)=1$? It is $x=2k \pi + \pi / 2$. Thus:

$$2.5 n \pi = 2k \pi + \pi / 2 $$

Which yields:

$$ n = (4*k + 1) / 5 = 0.2 + 0.8 k$$


问:
Induction proof for a function being in $C^{\infty}(\mathbb{R})$
Let $$f(t) = \begin{cases}
     0 & \text{: } x \le 0\\
     e^{-\frac{1}{t^2}}& \text{: } x > 0
\end{cases}$$
Prove that $f(t)$ is in $C^{\infty}(\mathbb{R})$.

I know that for a function to be in $C^{\infty}(\mathbb{R})$ it has to be infinitely times differentiable and each derivative must be continuous. I'm a bit stuck on how to prove this. I'm assuming the answer will need to be proved inductively. Ie:

Suppose the $f^{(n)}(t)$ is continuous. I need to then prove that $f^{n+1}(t)$ exists and is continuous. I'm not quite sure how to do this, I'm new to proofs by induction. I can also image proving continuity at $t=0$ may be a challenge.

答: Find all derivatives of $f(x)$ everywhere on $\mathbb{R}$ except for $t = 0$. Then prove that all of them in fact are continuous at zero.


问:
Motivation behind Lambda Calculus?
The lambda calculus provides a formalism broad used in theoreretical cs to write functions without
giving them explicit names, it declares anonymous functions.
That is at first glance it's just an alternative
form of function notation to write $\lambda$.$f(x)$ instead of $f(x)=x$.
But I don't see the reason why this lambda calculus is so important and
provides a better way to deal with functions for functional programming
languages and from viewpoint of theoretical computer science instead of convential "mathematical"
notation $f(x)=x$?

The techniques like curring work fine also with later notation, so what what are
the main reasons why lambda calculus is prefered?

答:
There's no difference. Lambda Calculus can't tell if you write $\lambda x . M$, if you write $f(x) = M$ or if you write $\hat{x} . M$ (which apparently was church's original notation? I haven't fact checked that, though).

The reason people care about lambda calculus is because it has exactly 3 combinatorial rules from which all computation follows. The dream is that, by studying these (very simple!) combinatorial rules (and we as mathematicians, and arguably as humans, are very good at reasoning about combinatorics) we can understand something about computation itself.

If you compare even the length of the definition of lambda calculus to the length of the definition of turing machines you'll see why people like this framework.

As a last reason, it's very flexible. We have the 3 basic combinatorial rules, but if we want to express other properties, or make certain operations primitive, it's very easy to extend the lambda calculus to make reasoning about more complicated computation easy.

There's a deep connection between lambda calculus and category theory that's also interesting and worth studying in its own right, but that's a longer discussion.

I hope this helps ^_^

Title: Motivation behind Lambda Calculus?


问:
Grouping vertices of a graph
Suppose $G$ is a graph with $4n$ vertices. Prove that one can partition the vertices into two groups with $2n$ vertices, such that that there is an even amount of edges between the two groups.

How can one approach this problem? Induction doesn't seem to work and it is just so strange.

答:
For any graph $G(V,E)$ we have the handshaking lemma $\sum_{v\in V}d(v)=2|E|$, so the sum of vertex degrees is even. Therefore any graph must have an even number of vertices of odd degree.

Partition the vertices here into sets $A,B$ such that each set contains an even number of odd-degree vertices - for example, if there are $10$ odd-degree vertices, you could have $6$ in one part and $4$ in the other. If all vertices are of odd degree, you must have $2n$ odd-degree vertices in each part.

Then the sum of vertex degrees within each part will be even; in part $A$ (say) with degree sum $d_A$, the $f$ edges connecting within $A$ will contribute an even number $2f$ to that degree sum, and as a result there will be an even number of edges, $d_A{-}2f$, connecting outside the group.

Observations: We needed the graph to have $4n$ vertices so the partitions would each have an even number of vertices only for the case where every single vertex was of odd degree. If not every vertex has odd degree, the equal split will work for any graph with an even number of vertices. For a graph with an odd number of vertices ($2m{+}1$), there must be a vertex of even degree, so it will always be possible to partition the vertices into two groups that differ in number by one ($m,m{+}1$) and have an even number of edges connecting the two parts.

Title: Grouping vertices of a graph


问:
Can we solve $u(x)=-1+x+\frac{x^2}{2}+2e^x-\int_0^xu(t)dt$ with using noise terms phenomenon?
question is can  $u(x)=-1+x+\frac{x^2}{2}+2e^x-\int_0^xu(t)dt$ be solved via noise terms phenomenon?

I made

$u_0(x)=-1+x+\frac{x^2}2+2e^x $

$u_{k+1}(x)=-\int_0^xu_k(t)dt$, $k\geq0$

$u_1(x)=-\int_0^xu_0(t)dt=-\int_0^x(-1+x+\frac{x^2}2+2e^x)dt=x-\frac{x^2}2-\frac{x^3}6-2(e^x-1)$

so I canceled $\pm \frac {x^2} {2}$ and $\pm2e^x$ from $u_0(x)$ and found the solution as $u(x)=-1+x$
But it doesnt justify the integral equation. Does this mean it cant be solved with noise terms p.? Is this a proper way to check?

答:
Define:
$$U(x)=\int_0^xu(t)dt$$
Then you have:
$$U(x)+U'(x)=-1+x+\frac{x^2}2+2e^x$$
Which is just a first order ODE


问:
What is the optimal classification rule among all rules that are function of $\|x\|^2$?
I have a textbook problem which I'm not quite sure how to solve:

Suppose that you observe $(x_1,y_1),...(x_{100}, y_{100})$, which you assume to be i.i.d. copies of a random pair $(x, y)$ taking values in $\mathbb{R}^2 \times \{1,2\}$. Further, suppose that you observe $x$, and you would like to predict $y$. The data looks like the following:

Given that the data is rotationally symmetric and $\operatorname{Pr}(y=1)=\tfrac{1}{2}$,  and

$$
\big[\|x\|^2\mid y=1\big] \sim \text{Exp}(\tfrac{1}{2}) \qquad\text{and}\qquad
\big[\|x\|^2\mid y=2\big] \sim \text{Unif}([9,16])$$

What is the optimal classification rule among all rules that are functions of $\|x\|^2$?

Also, how do I show that the expected cost of this classification rule is equal to $\frac{1}{2}(e^{-9/2}-e^{-8})$? The misclassification costs are equal to $c_1=c_2=1$.

So, in the textbook i'm using, the expected cost of misclassification is defined as:

Suppose we use the classification rule $g:\mathbb{R}^p\rightarrow \{1,2\}$, that assigns to group $1$ when $x \in R_1$ adnto group $2$ when $x\in R_2$. The expected cost of misclassification associated to the rule $g$ is $$\mathbb{E}[\text{cost}(Y,g(X))]=c_2\mathbb{P}(x\in R_1 | Y=2)\pi_2+c_1\mathbb{P}(x\in R_2 | Y=1)\pi_1$$
Where $\pi_1=\mathbb{P}(Y=1|x)$ and $\pi_2=\mathbb{P}(Y=2|x)$

My attempt: We have that

$$\begin{aligned}
R_1:&=\{x: \int_{|X|^2|Y=1}(x|y=1)dx\gt \int_{|X|^2=2}(x|y=2)dx\}
\\  &=\{x: \tfrac{1}{7} \gt \tfrac{1}{2}e^{\frac{1}{2}x}\}
\\  &=\{x: \log(\tfrac{49}{4})\lt x\}
\\ R_2:&=\{x: \log(\tfrac{49}{4}) \gt x\}
\end{aligned}$$

So,

$$
\mathbb{E}[\text{cost}(y, g(x))]
=\tfrac{1}{2}\int_{R_1}\int_{|X|^2|Y=2}(x)dx+\tfrac{1}{2}\int^{R_2}\int_{|X|^2|Y=1}(x)dx
$$

The second integrand is equal to $0$, and the first integrand is equal to:

$$
\tfrac{1}{2}\int^{16}_9\frac{e^{-\frac{1}{2}x}}{2}dx
=\tfrac{1}{2}[e^{-\frac{-1}{2}x}]^{16}_9=\frac{e^{\frac{9}{2}}-e^8}{2}
$$

Would this be correct?

答:
Your $R_1$ and $R_2$ are definitely not correct, and there are overall a lot of problems with your attempt:

$\{x: \tfrac{1}{7} \gt \tfrac{1}{2}e^{\frac{1}{2}x}\}$ and $\{x: \log(\tfrac{49}{4})\lt x\}$ are nonsensical statements since $x\in\mathbb R^2$. You probably mean $\|x\|$ instead?

In this case, the statement $\{x: \tfrac{1}{7} \gt \tfrac{1}{2}e^{\frac{1}{2}\|x\|}\}=\{x: \log(\tfrac{49}{4})\lt \|x\|\}$ is also wrong, you miscalculated there. This should be obvious though, since $\log(49/4)\approx 2.5$, but the decision boundaries should quite obviously be at $\|x\|=3$ and $\|x\|=4$.

The statement $\mathbb{E}[\text{cost}(y, g(x))]
=\tfrac{1}{2}\int_{R_1}\int_{|X|^2|Y=2}(x)dx+\tfrac{1}{2}\int^{R_2}\int_{|X|^2|Y=1}(x)dx$ does not make any sense semantically. What are you integrating? What are you integrating over?

How do you get from $\tfrac{1}{2}\int_{R_1}\int_{|X|^2|Y=2}(x)dx$ to $\tfrac{1}{2}\int^{16}_9\frac{e^{-\frac{1}{2}x}}{2}dx$ ?? Where did the $\log(\tfrac{49}{4})$ go??

While in the very end, you somehow get the right numbers, it is not clear what your argument is for your choice of $R_1$ and $R_2$ being optimal, and that no cost below $\tfrac{1}{2}(e^{\frac{9}{2}}-e^8)$ is achievable.

I would recommend the following:

Try to write everything down in a clean fashion

The problem is rotationally symmetric $\leadsto$ express $x$ in polar coordinates


问:
Find the area using integrals

What is the area defined by the curves $y = \sin x$ and $y = \cos x$ for $0 < x < \frac{\pi}{4}$?

I think it's $$\int_{0}^{\frac{\pi}{4}} [\cos x - \sin x]\, \mathbb{d}x$$ but I'm not sure if my step is right or not.

答:
Notice that $\cos(0)=1,\sin(0)=0$ now find the points for which the two curves intersect:
$$\sin(x)=\cos(x)\Rightarrow \tan(x)=1$$
$$x=\arctan(1)=\pi/4$$
What this means is that for: $0\le x\le \pi/4,\,\cos(x)\ge\sin(x)$. And so if you want the area between two functions that is the same as the integral of the greater function minus the integral of the lesser function, in this case:
$$\int_0^{\pi/4}\cos(x)dx-\int_0^{\pi/4}\sin(x)dx$$
So yes you are correct. If you did it the other way around you would simply end up with a negative value and since we are considering area here it is clear that the sign is just wrong

Title: Find the area using integrals


问:
If $\lim_{x\to\infty}f'(x)=0$ and if $g\geq c>0$ on $(a,\infty)$, then $\lim_{x\to\infty}\left[f\left(x+g(x)\right)-f(x)\right]=0.$
Problem: Let $f:[0,\infty)\to\mathbb R$ be continuous on $[0,\infty)$ and differentiable on $(0,\infty)$. Assume that $\lim\limits_{x\to\infty}f'(x)=0$. Prove that if $g\geq c>0$ on $(a,\infty)$, then
$$\lim_{x\to\infty}\left[f\left(x+g(x)\right)-f(x)\right]=0.$$

My Thoughts: Let $\varepsilon>0$ be given. By hypothesis, there is some $N\in\mathbb N$ such that $\vert f'(x)\vert<\varepsilon$ for all $x>N$. Since $g\geq c>0$, it follows that $x+g(x)>N$ for all $x>N$. Fix $x>N$. Since $f$ is continuous on $[x,x+g(x)]$ and differentiable on $(x,x+g(x))$, the mean value threorem implies that there is some $u\in(x,x+g(x))$ such that
$$\left\vert\frac{f(x+g(x))-f(x)}{x+g(x)-x}\right\vert=\left\vert\frac{f(x+g(x))-f(x)}{g(x)}\right\vert=\vert f'(u)\vert<\varepsilon.$$
The above implies that
$$\lim_{x\to\infty}\frac{\vert f(x+g(x))-f(x)\vert}{g(x)}=0.$$
On the other hand, $1/g(x)\leq 1/c$, so that
$$\lim_{x\to\infty}\frac{\vert f(x+g(x))-f(x)\vert}{c}=0,$$
from which the result follows by the laws of limits.

Do you agree with my proof above?

Thank you for your help in advance.

Note: The last step is incorrect, ignore my attempted proof.

答:
The proposition is false. The following is a counter-example.

Let $f:[0,\infty)\rightarrow\mathbb{R}$ be defined by $f(x)=\sqrt{x}$.
Then $f$ is continuous on $[0,\infty)$, differentiable on $(0,\infty)$,
and $f'(x)\rightarrow0$ as $x\rightarrow\infty$. Let $g:(1,\infty)\rightarrow\mathbb{R}$
be defined by $g(x)=x^{2}.$ For sufficiently large $x$, we have
\begin{eqnarray*}
 &  & f(x+g(x))-f(x)\\
 & = & \frac{x^{2}}{\sqrt{x+x^{2}}+\sqrt{x}}\\
 & \rightarrow & \infty
\end{eqnarray*}
as $x\rightarrow\infty$.


答:
Hint.

A standard example of a smooth non-analytic function is
$$
g(x)=\left\{\begin{array}{ll}e^{-\frac{1}{x}} & \text { if } x>0 \\ 0 & \text { if } x \leq 0\end{array}\right.
$$

A detailed proof for smoothness is given in this Wikipedia article:

https://en.wikipedia.org/wiki/Non-analytic_smooth_function#The_function_is_smooth

You can adapt the techniques for your example. A key step is to note that for any positive integer $m$,
$$
\lim_{x\downarrow 0}\frac{e^{-\frac1{x}}}{x^{m}}=0
$$

Alternatively, observe that
$$
f(x)=g(h(x))
$$
where $h(x)=x^2$ for $x\ge 0$ and $h(x)=0$ for $x<0$. You can show that $h$ is smooth. The result then follows from the chain rule.

Title: Induction proof for a function being in $C^{\infty}(\mathbb{R})$

Body:
Let $$f(t) = \begin{cases}
     0 & \text{: } x \le 0\\
     e^{-\frac{1}{t^2}}& \text{: } x > 0
\end{cases}$$
Prove that $f(t)$ is in $C^{\infty}(\mathbb{R})$.

I know that for a function to be in $C^{\infty}(\mathbb{R})$ it has to be infinitely times differentiable and each derivative must be continuous. I'm a bit stuck on how to prove this. I'm assuming the answer will need to be proved inductively. Ie:

Suppose the $f^{(n)}(t)$ is continuous. I need to then prove that $f^{n+1}(t)$ exists and is continuous. I'm not quite sure how to do this, I'm new to proofs by induction. I can also image proving continuity at $t=0$ may be a challenge.


问:
Why does the logistic curve not grow the fastest when $y = \frac{L}{2}$?
I know that the y-value where a logistic model grows the fastest is the limit to infinity divided by 2.

$y = \frac{L}{2}, L = \lim_{t\to\infty} y$

However, this doesn't always seem to be the case. Sometimes, I am told that the point where the curve is the steepest is $\frac{L}{2}$, but sometimes, I am told that I need to find the points of inflection of the curve.

This problem is an example where the answer doesn't seem to fit the $\frac{L}{2}$ definition. The growth of a population is modeled by the logistic differential equation:

$\frac{dP}{dt} = 2P \left( 1-\frac{P}{5} \right)$

At what point $P$ does the population grow the fastest?

Since the model fits the template $\frac{dP}{dt} = kP \left( 1-\frac{P}{L} \right)$, finding the value of $L$ is straightforward: 5. The answer is $10$ and not $\frac{5}{2}$ for some reason. Why is that?

答:
Careful with your differential equation. You wrote $\displaystyle\frac{dP}{dt}$ on the LHS, then changed your variable to $y$ on the RHS. Here is your hint: The population grows the fastest when $\displaystyle\frac{dP}{dt}$ is maximized, so its derivative is equal to zero. What is the derivative of $\displaystyle\frac{dP}{dt}$?

Next, your value 10 does not make sense because it is above the carrying capacity $L =5$ for the population. If you want the point to be on the logistic curve, then it must be less than $L$. When $0 < P < L$, the rate of change of the population is maximized when $P = \displaystyle\frac{L}{2}$. To find the point of the rate of maximal change, set the derivative of the RHS (with respect to $y$) equal to zero, then solve.


答: A simple induction argument shows that for $x >0$ $f^{n}(x)$ can be written as $p_n(\frac 1 x) e^{-1/x^{2}}$ for some sequence of polynomials $(p_n)$ But $\lim_{y \to \infty} \frac {p_n(y)} {e^{y^{2}}} =0$ by L'Hopital's Rule. Put $y=\frac 1  x$ in this.


问:
Categorical semantics explained – what is an interpretation?
I’ve been really having a hard time trying to understand categorical semantics. In fact, I am confused to the point I am afraid I don't know how to ask this question!

I’ve been reading textbooks like Makkai and Reyes’ book, Jacobs’ Categorical logic and type theory (1992) or Categorical Logic by Andrew Pitts, but I simply cannot get the feeling of what are they trying to do. This entry and this one in the nlab do not seem to be very helpful for my understanding. In fact, it seems that the more I read the more confused I get.

Can those questions be answered crystal clearly?

How we define a categorical interpretation?

If it a functor, what is its domain? That is, in model theory, we interpret languages. What are we interpreting here, languages or theories .. or something else?

Let me take an example: I know, very informally, that intuitionistic propositional logic (IPL) correspond to bicartesian closed categories (BCC).

But how to make this statement mathematically precise?
How do we define such an interpretation?

If it is indeed a functor, what is its domain? The category of intuitionistic propositional logics or the category of intuitionistic propositional languages? Or even a particular intuitionistic propositional logic understood as a categry (possibly given by the category of proofs)?

I would really appreciate any answer that could shed some light not only into this example, but also into the big picture of it.

Background: I believe to have all the required background in category theory (UPs, natural transformations, equivalences, adjoints etc.) and type theory (polymorphism, dependent types etc.), having attended courses in both disciplines.

答:
What is an interpretation? A functor? I think that's a fair answer. So, the spotlight of your query focuses on the question: what is its domain?

Well, it is good that you say that you have the background, because I'm about to put that to the test - out of necessity so as to be able to make the answer (relatively) succinct.

Let's start from basics ... in a new and different setting.

The fundamental objects studied by logic are predicates, while the fundamental relation addressed is that of inference. We write $A ⊢ B$ to state that predicate $B$ may be inferred from predicate $A$. In a categorical formulation of logic, each predicate is an object, the inference relation is an arrow, so that $A ⊢ B$ could just as well be written $A → B$, while a morphism $f: A → B$ is a proof that witnesses that inference. For that relation, we can just as well write $f: A ⊢ B$, or equivalently as $B ⊣ A: f$; and for bi-directional inferences, we could write $f: A ⊣⊢ B: g$ to denote the relation pair $(f: A ⊢ B, A ⊣ B: g)$.

(Footnote: there actually are attempts to account for sequents within categorical logic, but I won't get into that here. So I restrict inferences so that they have one and only one predicate on the left and right.)

Suppose we call the category that contains the basics of logic $$ - that's our domain.

In logic, predicates are formed by connectives. We'll use this term very broadly to encompass a wide variety of operations, while yet providing uniform treatment for them all. A connective operates on predicates, and in this domain, each such operation will be represented by a functor from various combinations formed of $$ into $$, itself.

For the following, we'll use the following conventions that for a category $$, $||$ denotes its set of objects, while for each $A, B ∈ ||$, $(A,B)$ denotes the set of morphisms for $A → B$ (or $A ⊢ B$ in $$). The unit category is $1$, with $|1| = {0}$ and $1(0,0) = \{1_0\}$. The category $^†$ is the reverse of category $$; and $×$ is the direct product of categories $$ and $$. For a set $T$, we'll denote, by $^T$, the indexed product of $$, where $|^T| = ||^T$ and for each $A, B ∈ ||^T$, $^T(A,B) = ∏_{t∈T} (A(t),B(t))$.

Thus, if $f: A → B$ (or $f: A ⊢ B$) in $$, then $B ← A: f$ (or $B ⊣ A: f$) in $^†$.

Finally, we may also include the category $^⇔$, which is given by $|^⇔| = ||$ and for $A, B ∈ ||$, $^⇔(A,B) = (A,B)×(B,A)$. Distinguished subsets of $^⇔(A,B)$ include
$$\{ (f,g)∈^⇔(A,B) ¦ f∘g∘f = f, g∘f∘g = g \}$$
for generalized inverses and
$$\{ (f,g)∈^⇔(A,B) ¦ f∘g = 1_B, g∘f = 1_A \}$$
for inverses, which define respective subcategories, the last of which is the maximal groupoid contained in $$.

The corresponding relation in $^⇔$ is $f: A ⊣⊢ B: g$, for $(f,g) ∈ ^⇔(A,B)$.

(I'm a fan of symmetry - and also of Noether - so, I like it when things are made so as to respect duality as much as possible.)

The minimum set of connectives we'll want for $$, along with the functors we'll associate them with, are some or all of those drawn from the following set:

$⊤,⊥: 1 → $ respectively for True and False,

$∧,∨: × → $ respectively for And and Or,

$⊃: ^†× → $, $⊂: ×^† → $ respectively for OnlyIf and Unless,

$∀,∃: ^T → $, respectively for All and Some: quantifiers bound to a set $T$ of terms of a type, which we'll also call $T$.

With each connective, we associate a natural bijection between two families of inferences, one which contains the connective and the other which does not.

$(C,⊤) ⇔ 1(0,0) = \{1_0\}$, $(⊥,D) ⇔ 1(0,0) = \{1_0\}$, for $C, D ∈ ||$;

$(C,A∧B) ⇔ (C,A)×(C,B)$, $(A∨B,D) ⇔ (A,D)×(B,D)$, for $A, B, C, D ∈ ||$;

$(C,A⊃B) ⇔ (C∧A,B)$, $(A⊂B,D) ⇔ (A,B∨D)$, for $A, B, C, D ∈ ||$;

$(C,∀A) ⇔ ^T(C,A)$, $(∃B,D) ⇔ ^T(B,D)$, for $A, B ∈ |^T|$ and $C, D ∈ ||$.

And ... that's it!

As I said, I'll put your background to the test. So, as an exercise: describe all of what is entailed by requiring that these be natural bijections. Each of these natural bijections entails a bundle of identities (i.e. commutative diagrams) and corresponding proof rules, as do the corresponding functors.

Intuitionism includes all of these connectives, except the one for Unless; while bi-intuitionism includes them all. A Cartesian closed category is precisely characterized by one which includes the connectives for True, And and OnlyIf. A bi-Cartesian closed category also includes Or and False. (Edit: False is not in a Cartesian closed category, as I originally stated. I corrected that.) A bi-Cartesian bi-closed category also includes Unless ... and always reduces to a partial order (i.e. if $f, f': A ⊢ B$, then $f = f'$).

To this list, we can also add the defined connectives

$¬,⨽: ^† → $, respectively for strong Not (as "is false") and weak Not (as "not true"),

$≡,≢: ^⇔ × ^⇔ → $ respectively for Iff and Xor,

defining them in terms of the other connectives by

$¬A ⊣⊢ A⊃⊥$, $⨽B ⊣⊢ ⊤⊂B$,

$A≡B ⊣⊢ (A⊃B)∧(B⊃A)$, $A≢B ⊣⊢ (A⊂B)∨(B⊂A)$.

The corresponding natural bijections are

$(C,¬A) ⇔ (C∧A,⊥)$, $(⨽B,D) ⇔ (⊤,B∨D)$, for $A, B, C, D ∈ ||$;

$(C,A≡B) ⇔ (C∧A,B)×(C∧B,A)$, $(A≢B,D) ⇔ (A,B∨D)×(B,A∨D)$, for $A, B, C, D ∈ ||$;

The last two need further elaboration to make them consistent with the respective functors.

If $m:C'⊢C$, $n:D⊢D'$, $k:A⊣⊢A':k'$, $l:B⊣⊢B':l'$, then we have $k:A⊢A'$, $k':A'⊢A$, $l:B⊢B'$, $l':B'⊢B$. The morphisms in $(C,A≡B)$ and $(A≢B,D)$ are respectively $(m,(k,k')≡(l,l'))$ and $((k,k')≢(l,l'),n)$. The corresponding morphisms in $(C∧A,B)×(C∧B,A)$ should be $(m∧k',l)$ and $(m∧l',k)$, while those in $(A,B∨D)×(B,A∨D)$ should be $(k',l∧n)$ and $(l',k∧n)$.

Thus if $f: C∧A ⊢ B$ and $g: C∧B ⊢ A$ map to $h: C ⊢ A≡B$, then $(l'∘f∘m∧k,k'∘g∘m∧l)$ maps to $((k,k')≡(l,l'))∘h∘m$. Similarly, if $f: A ⊢ B∨D$ and $g: B ⊢ A∨D$ map to $h: A≢B ⊢ D$, then $(l'∨n∘f∘k,k'∨n∘g∘l)$ maps to $n∘h∘((k,k')≢(l,l'))$.

Finally, for the quantifiers, we can write

$(∀x:T)A(x) ⊣⊢ ∀[x:T ↦ A(x)]$,

$(∃x:T)A(x) ⊣⊢ ∃[x:T ↦ A(x)]$,

and adopt the convention, more broadly, of writing *any* operator that acts on functions equivalently as a binding operator. This treatment of quantifiers stays strictly first-order, avoiding the better-known higher-order type-theoretic formalisms making up the Lambda Cube and is suitable either for single-sorted term signatures, where there is just one type $T$ whose underlying set includes all terms, or multi-sorted term algebras, where there are multiple types $T$, each with its own pair of bounded quantifiers.

An interpretation is simply any category that $$ maps to by a functor - whichever way we choose to define $$. I've given you several alternatives.

As a side-note, it is worth pointing out the following:

Each of the natural bijections involving $C$ are "left-invariant": we can conjoin extra statements on the left to $C$ on both sides of an equivalence to come up with another instance of the equivalence.

Each of the natural bijections involving $D$ are "right-invariant": we can disjoin extra statements on the right to $D$ in a similar way.

It is not possible to create an "internal" lambda abstraction, unless all the rules are left-invariant. The corresponding dual (a "gamma" abstraction, we might perhaps call it) would similarly require all the rules to be right-invariant.

Right-invariance for the $C$-rules and left-invariance for the $D$-rules characterize "watershed" conditions seen in lattice-theory

For $⊤$, $⊥$: two-sided invariance is already assured.

For $∧$, $∨$: two-sided invariance is equivalent to distributivity.

For $∀$, $∃$: two-sided invariance is equivalent to "indexed distributivity", of the respective forms:
$$(∀x)(A(x) ∨ D) ⊢ (∀x)A(x) ∨ D,$$
$$C ∧ (∃x)B(x) ⊢ (∃x)(C ∧ B(x)).$$

Bi-intutionism has both forms of indexed distributivity, intuitionism has only the one for $∃$, but it is entirely possible to modify intuitionism by adding the one for $∀$ as an extra rule.

For $⊃$, $⊂$ (or any of $¬$, $⨽$, $≡$, $≢$): two-sided invariance makes the logic a classical logic, and all connectives are defineable and all rules become two-sided invariant. It actually turns the category not just into a partial order, but a boolean lattice.

Ways to get around the "collapse" issue for bi-cartesian, bi-closed categories are an active area of research; but I personally don't see it as a problem. The relevant condition (that $f,f': A ⊢ B$ entails $f = f'$), in other contexts - like braided monoidal categories - is called "coherence" and is not only a desired property, but is specifically sought for - which is what the "hexagon diagrams" in braided monoidal categories are there for. We can get around collapse by defining different layers of equivalence by which of the commutative diagrams are to be permitted. For bi-intuitionist and classical logic, there would be a intuitionalist layer of weak equivalence, an indexed distributive intuitionist layer and a bi-intuitionist layer.

So, classical logic can be included in this framework by making all of the natural bijections invariant on both the left and right; conjoining a $C$ on the left of all the $D$-rules, and disjoining a $D$ on the right of all the $C$-rules. The roles played by $C$ and $D$ are entirely analogous to the role played by continuations in programming languages and may be deemed, respectively, the "left continuation" and "right continuation" of the rule. In sequent calculus, they are each replaced by finite sequences of statements.

The quantum logic of von Neumann cannot easily be fit into this framework, because it is non-distributive; although in more recent times - by virtue of the work of Isham et al. - von Neumann's logic framework has been superseded by a newer one rooted in bi-intuitionist logic.

The rule $A ⊢ ¬¬A$ is valid in intuitionist logic, while $⨽⨽B ⊢ B$ and $¬B ⊢ ⨽B$ are both valid in bi-intuitionist logic. The inclusion of any of the converses $¬¬A ⊢ A$, $B ⊢ ⨽⨽B$ or $⨽B ⊢ ¬B$ is also a watershed for classical logic. But instead of posing these rules separately - which breaks the simple scheme just laid out here - it's easier to stay within the paradigm by extending the natural bijections to allow continuations on both the left and right.


问:
$\lim_{n\to\infty}\sqrt[3]{3n}\sigma_n=1$
We put :
$S_n=\displaystyle\sum_{k=1}^{k=n}\sigma^2_k$
Where $(\sigma_n)_{n\geq 1}$  is a real sequence, its boundaries are positive.

_Assume that
$(\lim_{n\to\infty}\sigma_n S_n=1)$

Prove that :$\lim_{n\to\infty}\sqrt[3]{3n}\sigma_n=1$

I tried to prove this, but I did not reach any result.

If possible, find any method or idea that would help us in the solution, thank you in advance

答:
First results:

$\lim S_n=+\infty$  (otherwise: $\{S_n\}$ converges, $\lim \sigma_n^2=0$  , $\lim \sigma_n^2 S_n=0 $ , contradiction)

$\lim \sigma_n = 0 $ (because $\ \lim \sigma_nS_n=1\ $ and $\ \lim S_n=+\infty$)

$S_n \sim S_{n-1}$  (because $\ S_{n-1} = S_n-\sigma_n$)

The idea:

$S_n^3-S_{n-1}^3 = (S_n-S_{n-1})(S_n^2+S_nS_{n-1}+S_{n-1}^2)  $

$(S_n^3-S_{n-1}^3) \sim 3\sigma_n^2 S_n^2$

$\lim (S_n^3-S_{n-1}^3) = 3$

By Cesaro:  $\ \lim \dfrac{S_n^3-S_0^3}{n}=3 $

The conclusion:

$S_n\sim \root{3}\of{3n}\  $ , therefore $\ \sigma_n \sim \dfrac{1}{\root{3}\of3{n}} $


问:
How to calculate opposite direction angle
I searched but couldn't find on Google:

My question is, how do I find the opposite direction of an angle, for example 170 degree, how do I calculate the opposite direction in degrees?

Thanks in advance.

答:
Considering an angle normallized between -180° and 180°, the opposite angle is given by :</p>
```<code>if angle == 0 :
  opposite = 180 // or -180 as you wish
else :
  if angle > 0 :
    opposite = angle - 180
  else :
    opposite = angle + 180
</code>```

Body:
Considering an angle normallized between -180° and 180°, the opposite angle is given by :

if angle == 0 :
  opposite = 180 // or -180 as you wish
else :
  if angle > 0 :
    opposite = angle - 180
  else :
    opposite = angle + 180


问:
Values of Integral with Function
I have an integral $\int_{0}^{1}e^{2x}f(x)dx$. But $f(x)$ is another function ,not an integral. It is a function that is twice differentiable, and $f''(x)=kf(x)$, $k$ is constant. Then there is local max at $(0,1)$ and local min at $(1, -1)$. So what are possible values of the integral? How to use local max and local min to solve the problem? I have one value is $\frac{e^{2}+1}{k-2}$, is this right? What is other values?

答:
Ok so we know a few things to start with $f''(x)=kf(x),f(0)=1,f(1)=-1,f'(0)=f'(1)=0$. We can go ahead and solve this in a few ways:

ODE:

$$f''(x)-kf(x)=0$$
let $f(x)=Ae^{\lambda x}\therefore f''(x)=A\lambda^2e^{\lambda x}$ so:
$$Ae^{\lambda x}\left[\lambda^2-k\right]=0$$
This means that $\lambda=\pm\sqrt{k}$ so:
$$f(x)=Ae^{\sqrt{k}x}+Be^{-\sqrt{k}x}$$
now we can use our initial and boundary conditions:
$$f(0)=A+B=1$$
$$f(1)=Ae^{\sqrt{k}}+Be^{-\sqrt{k}}=-1$$
$$f'(0)=A\sqrt{k}-B\sqrt{k}=\sqrt{k}(A-B)=0$$
$$f'(1)=A\sqrt{k}e^{\sqrt{k}}-B\sqrt{k}e^{-\sqrt{k}}=\sqrt{k}e^{-\sqrt{k}}\left(Ae^{2\sqrt{k}}-B\right)=0$$
now using:
$$\begin{cases}A+B=1\\A-B=0\end{cases}$$
You can solve for $A$ and $B$ then you should be able to work out $k$ from there.

Title: Values of Integral with Function


问:
Show the following function is differentiable
$$f(x) = \int_{-\sqrt{x}}^{\sqrt{x}} e^{-\sqrt{1 + t^2}} dt$$

Explain why this function is differentiable at each $x \in [0,\infty)$ and compute $f'(x)$.

I know that $f'(x) = e^{-\sqrt{1 + \sqrt{x}^2}} - e^{-\sqrt{1 + (-\sqrt{x})}^2} = e^{-\sqrt{1 + x}} - e^{-\sqrt{1 + x}} = 0$. However, I am having a hard time explaining why it is differentiable at each $x$.

I tried to apply the limit definition $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$. However, it becomes pretty nasty since it has integrals within the limit. I'm not convinced this is the way to show it just because it doesn't really get me anywhere.

My next idea seemed golden to me.

I can state the Fundamenetal Theorem of Calculus as follows:

Let $f: [a,b] \to \mathbb{R}$ be integrable and define
$$F(x) = \int_{a}^{x} g(t) dt$$
$\forall x \in [a,b]$. Then, $F$ is continuous on $[a,b]$. If $f$ is continuous at some point $c \in [a,b]$, then $F$ is differeitable at $c$ and $F'(c) = f(c)$.

Our function $f$ has no discontinuities, so its set of discontinuities is of measure $0$. So it is integrable. $f$ is continuous $\forall c$ in the domain because it is an exponential, which is always continuous. So, this would mean $F$ is differentiable everywhere.

BUT, our function is $f: [0,\infty) \to \mathbb{R}$. not an interval of the form $[a,b]$. Is this option out of the picture as well then? It seems so close!

Any help is appreciated!

答: The question is wrong. The function is not differentiable at $0$. By definition $f'(0)=\lim_ {x \to 0+} \frac 1  x  \int_{-\sqrt x}^{\sqrt x} e^{-\sqrt {1+t^{2}}}dt=2\lim_ {x \to 0+} \frac 1  x  \int_{0}^{\sqrt x} e^{-\sqrt {1+t^{2}}}dt$. An applicatioin of L'Hopital's Rule shows that this limit is $\infty$.


问:
Confused about Bayes' theorem
One thing I don't understand for Bayes' theorem is that I end up in a circular position sometimes when I try to solve it.
$$P(B|C) = \frac{P(C\mid B) \times P(B)}{P(C)}$$
However, I don't know $P(C\mid B)$!

If I try to use Bayes' theorem to calculate it, then I end up right back where I started.

I know $P(B)$ and $P(C)$, but they are dependent on a variable $A$ whose probability I also know. How do I get out of the circle if I don't have a prior for $P(C\mid B)$?

答:
If you think of Bayes's theorem with the denominator cleared, i.e., $P(B|C)P(C)=P(C|B)P(B)$, you see that it's just two expressions for $P(B\cap C)$.  "Applying" it in the customary fraction form (as in the question) amounts to replacing one of these two expressions by the other.  So it's not surprising that, if you do it again, replacing the "other" expression by the "one", you just undo what you did the first time.

Furthermore, if you know only $P(B)$ and $P(C)$, you won't be able to "get out of the circle" and find $P(B|C)$ and/or $P(C|B)$, because these are not determined by the information you have.  Imagine, for example, flipping a fair coin and letting both $B$ and $C$ be "heads"; so $P(B)=P(C)=\frac12$ and $P(B|C)=P(C|B)=1$.  Now consider the same $B$ with a different $C$, namely "tails".  You still have $P(B)=P(C)=\frac12$ but now $P(B|C)=P(C|B)=0$.

Title: Confused about Bayes' theorem


问:
Value of $c$ that makes $c(\bar X (1 - \bar X))$ an unbiased estimator of $p(1 - p)$
I need to find the value of $c$ that makes $c(\bar X (1 - \bar X))$ an unbiased estimator of $p(1 - p)$.

I am confused about how to do this. I want to say that the value of $c$ must be $1$ because $E(\bar X) = \frac{\mu}{n} = \frac{np}{n} = p$, but that seems too easy.

Any advice? Thanks in advance.

答:
$$
\operatorname{var}(\overline X) = \frac{p(1-p)} n
$$
So
$$
\operatorname E(\overline X^2) = p^2 + \frac{p(1-p)} n
$$
$$
\operatorname E(\overline X(1-\overline X)) = p - p^2 - \frac p n + \frac {p^2} n = \left( 1 - \frac 1 n \right) p(1-p).
$$

So the value of $c$ that you need is the reciprocal of $1 - \dfrac 1 n.$


问:
Show that the lower limit topology with rationals is different than lower limit topology with reals
Consider the topology with basis $\{[a,b): a,b \in \mathbb{Q}\}$ and the topology $\{[a,b): a,b \in \mathbb{R}\}$.

I know that $[\sqrt{2},2)$ is open in the lower limit topology generated by the reals.

How do I prove that it is not open in the lower limit topology generated by rationals. Couldn't a union of elements in the rational lower limit topology approximate $[\sqrt{2},2)$?

答: If it were open in the lower limit topology generated by rationals, there would exist a basis element $[a, b)$ (with $a, b \in \mathbb{Q}$) contained in $[\sqrt2, 2)$ such that $\sqrt2 \in [a, b)$. Evidently, this is impossible, since for any $a \in \mathbb{Q}$ and $\mathbb{Q} \ni b > a$ such that $a > \sqrt2$, the interval $[a, b)$ cannot contain $\sqrt 2$.


问:
Let $2 = p_1 < p_2 < \dots$ be the increasing sequence of all prime numbers. For large $n$, which one is larger $2^{p_n}$ or $p_{2^n}$?
Let $2 = p_1 < p_2 < \dots$ be the increasing sequence of all prime numbers. For large $n$, which one is larger $2^{p_n}$ or $p_{2^n}$?

I need some help on this please.

I have tried using the Prime number theorem since $$\pi(x) \approx \frac{x}{\log x}$$

I used this expression for $x=2^{p_n}$ and $x=p_{2^n}$ but I couldn't make any progress.

$p_{2^n}$ is the $2^n$-th prime number.

答:
From the prime number theorem, for sufficiently large $n$,  the $n$th prime is less than $n^2$.

Furthermore, all primes except $2$ or $3$ are $1 \text{ or } 5 \pmod 6$ (which means intuitively, that at most $1/3$ of numbers are prime). So $p_n > 3n -c$ for some constant $c$.

$$p_{2^n} < (2^n)^2 = 2^{2n}, \text{ while } 2^{p_n} > 2^{3n-c} = 2^{3n} / 2^c.$$

As $n \rightarrow \infty$, $2^{3n} / 2^c$ will be bigger than $2^{2n}$.  Therefore, for sufficiently large $n$, $2^{p_n}$ is bigger


问:
Limit of Sequence??
The sequence: $-2, 0, -1, 0, -1/2, 0, -1/4, 0, -1/8, 0, -1/16, 0\ldots$

Heyy. So I was a little confused about this question. How would this have an existing limit if it keeps alternating between a value and $0$? I know if you have a sequence of $1, -1, 1, -1\ldots$ that would not have a limit.

So why is the sequence above have a limit?

Thank you in advance for any help!

答:
Just remember the definition of a sequence limit in term of $\varepsilon$-$\delta$: $(a_n)_{n} \subset \mathbb R$ converges to $a \in \mathbb R$ if and only if for all $\varepsilon > 0$, there exists $N_\varepsilon \in \mathbb N$ such that for all $n \ge N_\varepsilon$ we have
$$|a - a_n| < \varepsilon.$$
In this case, $a = 0$ and
$$
a_n =
\begin{cases}
-1/2^{n/2} & \text{if } n \text{ is even},\\
0 & \text{otherwise}.
\end{cases}
$$
Let $\varepsilon > 0$. For $N_\varepsilon = \lceil 2\log_2(1/\varepsilon)\rceil$, we have, $\forall n \ge N_\varepsilon$, that
$$\left|a - a_n\right| = \left|a_n\right| < \varepsilon.$$

Title: Limit of Sequence??


问:
Number of Sylow bases of a certain group of order 60
We have $G=\langle a,b \rangle$ where $a=(1 2 3)(4 5 6 7 8)$, $b=(2 3)(5 6 8 7)$.  $G$ is soluble of order 60 so has a Hall $\pi$ subgroup for all possible $\pi\subset\{2,3,5\}$. $\langle a\rangle$ is a normal subgroup.
A Hall $\{2,3\}$ subgroup is $\langle a^5,b \rangle$, a Hall $\{2,5\}$ subgroup is $$ and a Hall $\{3,5\}$ subgroup is $\langle a\rangle$.  This means a Sylow basis is $\langle a^3 \rangle$, $\langle a^5\rangle$ and $\langle b\rangle$.
I am trying to determine the number of Sylow bases.  Here is my attempt:

$\langle a^3\rangle$ and $\langle a^5 \rangle$ are normal as characteristic subgroups of the normal subgroup $\langle a\rangle$, and so as all Sylow bases are conjugate, their number is equal to the number of Sylow 2 subgroups, which is 1 or 5 by Sylow's theorem and $\langle b\rangle$ is not normal so it is $5$.

The answer we were given is 15, so I wondered if anyone else could spot where I'm going wrong.  Thanks.

答:
The number of Sylow bases is exactly equal to the number of Sylow complement bases. A Sylow basis has a complicated definition involving permuting subgroups, but a Sylow complement basis has a very simple definition: a choice of one Sylow $p$-complement for each prime $p$. Thus we just need to count the number of Sylow $p$-complements. This is just finding the index of their normalizer, and since the Sylow 2-complement is normal, but the other complements are not and have prime index, we can easily count them.

$p=2$: The Sylow $p$-complements are the Hall $\{3,5\}$-subgroups, the conjugates of the normal subgroup $\langle a \rangle$. So just one.

$p=3$: The Sylow $p$-complements are the Hall $\{2,5\}$-subgroups, $\langle a^3,b \rangle$ and its 3 conjugates (not normal, but has index 3)

$p=5$: The Sylow $p$-complements are the Hall $\{2,3\}$-subgroups, $\langle a^5, b \rangle$ and its 5 conjugates (nor normal, but has index 5)

All told that is $1 \times 3 \times 5 = 15$ possibilities.

Another method is to find a system normalizer. In this case it is a Sylow 2-subgroup, which can be seen from Carter's work on abnormal subgroups (given the large number of normal subgroups).


问:
Is there a general criterion in judging the rationality (or appropriateness) of the boundary conditions?
When we consider a partial differential equation, we need to give some boundary condition such that this question is reasonable. What criterion can we use? e.g., heat equation $\partial_t u(t,x) = \Delta_x u(t,x)$ we take Laplace transfrom, then we get $\tau u(\tau,x) - u(0,x) = \Delta u(\tau,x)$. It seems that a boundary condition in the form of initial condtion $u(0,x) = f(x)$ is more reasonable. But is there a general criterion in judging the rationality (or appropriteness) of the boundary conditions?

答: The paper of Kurt Friedrichs, Symmetric Positive Linear Differential Equations, Communications in Pure and Applied Mathematics, vol 11, p333-458 (1958), discusses this question.</p>


问:
Calculation for Standard Deviation Given a Gaussian Distribution
Quick background: I work as a Software Engineer in the Embedded Systems space, and my primary job function has me working with signal processing algorithms. I feel pretty good when it comes to statistics, but this one comment left in a codebase by a former employee has me stumped.

The calculation is supposed to be for the standard deviation of a signal to noise difference, and is described as such: sigma = pow(1-normdist(n), m).

I haven't been able to place the equation to anything standard deviation related, at least from my knowledge. The 1-normdist(n) part remind me of calculating tail probabilities, but even then, taking a power of a probability doesn't make much sense to me.

Any help and advice is welcomed!

答:
Building on Paul's comment about $(1-p)^m$

and assuming normidst is the standard normal CDF $\Phi$ in some programming language

you have $(1-\Phi(n))^m$ as the probability that all $m$ independent samples from a standard normal distribution are at least $n$


问:
Given a ergodic Markov chain find the limit $\lim_{n \to \infty} \frac{\log{P}(\omega_{i} \neq 1)}{n}$
Suppose we have an ergodic Markov chain on a finite state space $X=\{1,2, \ldots, r\}$ with transition matrix $A$ and stationary measure $\pi$ which we also take to be the initial distribution. I wanted the find the following limit:

$$\lim_{n \to \infty} \frac{\log(P\{\omega_{i} \neq 1: i=0,1,2 \ldots, n \}, )}{n}$$

I'm honestly quite lost on how to go about this, it seems like law of large numbers for Markov chains should be relevant.

答:
Let $M$ be the substochastic matrix obtained from the transition matrix $P$ by erasing the first row and column, and let $\lambda(M)<1$ denote its Perron eigenvalue [1]. The probability in the numerator of the question is $\pi M^n {\bf 1}$ where ${\bf 1}$ is the all ones vector. Thus
$$\lim_n \frac {\log(\pi M^n {\bf 1})}{n} =\log \lambda(M)<0 \,.
$$

[1] https://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem


问:
Confusion with the partial derivative of a natural logarithm
Why do we solve the following example using chain rule? I would assume that the answer would be $18/x^1$.

答:
Because you have to. The second equality is essentially this:

$$
\frac{d}{dx}\left(9(\ln x)^2\right)=9\cdot 2\ln x \cdot\frac{d}{dx}\ln(x)
=\frac{18\ln x}{x}
$$

[Edited:] From the context of your identities, $\ln^2x:=(\ln x)^2$.

Also, the superscripts on $x^1$ and $x^2$ mean different components, not exponents.


答:
Since $e^{-\sqrt{1+t^2}}$ is defined and continuous everywhere, it has an antiderivative $F(t)$ which is defined and differentiable everywhere, with $F'(t) = e^{-\sqrt{1+t^2}}$.  Since $F$ is defined and differentiable everywhere, so is the function $H(t) := F(t) - F(-t)$.  The fundamental theorem of calculus says that

$$\int\limits_{-\sqrt{x}}^{\sqrt{x}} e^{-\sqrt{1+t^2}}dt = F(\sqrt{x}) - F(-\sqrt{x}) = H(\sqrt{x})$$

which is differentiable for $x \in (0,\infty)$ (not on $[0,\infty)$), because a composition of two differentiable functions is differentiable.

The derivative of this function is given by the chain rule:

$$\frac{d}{dx} \bigg( F(\sqrt{x}) - F(-\sqrt{x}) \bigg) = F'(\sqrt{x}) \frac{1}{2 \sqrt{x}} - F'(-\sqrt{x}) \frac{-1}{2\sqrt{x}}$$ $$ = e^{-\sqrt{1+x}} \frac{1}{2\sqrt{x}} + e^{-\sqrt{1+x}}\frac{1}{2\sqrt{x}} = \frac{e^{-\sqrt{1+x}}}{\sqrt{x}}$$

Title: Show the following function is differentiable

Body:
$$f(x) = \int_{-\sqrt{x}}^{\sqrt{x}} e^{-\sqrt{1 + t^2}} dt$$

Explain why this function is differentiable at each $x \in [0,\infty)$ and compute $f'(x)$.

I know that $f'(x) = e^{-\sqrt{1 + \sqrt{x}^2}} - e^{-\sqrt{1 + (-\sqrt{x})}^2} = e^{-\sqrt{1 + x}} - e^{-\sqrt{1 + x}} = 0$. However, I am having a hard time explaining why it is differentiable at each $x$.

I tried to apply the limit definition $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$. However, it becomes pretty nasty since it has integrals within the limit. I'm not convinced this is the way to show it just because it doesn't really get me anywhere.

My next idea seemed golden to me.

I can state the Fundamenetal Theorem of Calculus as follows:

Let $f: [a,b] \to \mathbb{R}$ be integrable and define
$$F(x) = \int_{a}^{x} g(t) dt$$
$\forall x \in [a,b]$. Then, $F$ is continuous on $[a,b]$. If $f$ is continuous at some point $c \in [a,b]$, then $F$ is differeitable at $c$ and $F'(c) = f(c)$.

Our function $f$ has no discontinuities, so its set of discontinuities is of measure $0$. So it is integrable. $f$ is continuous $\forall c$ in the domain because it is an exponential, which is always continuous. So, this would mean $F$ is differentiable everywhere.

BUT, our function is $f: [0,\infty) \to \mathbb{R}$. not an interval of the form $[a,b]$. Is this option out of the picture as well then? It seems so close!

Any help is appreciated!


问:
Estimating probabilities using Bayes rule?
I am working on a past exam paper. I am given a data set as follows:

Hair: {brown, red} = $\{B,R\}$

Height: {tall, short} = $\{T,S\}$

Country: {UK, Italy} = $\{U,I\}$

Our sample is:

$$(B,T,U)\quad (B,T,U)\quad (B,T,I)\quad (R,T,U)$$
$$(R,T,U)\quad (B,T,I)\quad (R,T,U)\quad (R,T,U)$$
$$(B,T,I)\quad (R,S,U)\quad (R,S,U)\quad (R,S,I)$$

Question. Estimate the probabilities $P(B,T\mid U)$, $P(B\mid U)$, $P(T\mid U)$, $P(U)$ and $P(I)$.

As the question states estimate, I am guessing that I don't need to calculate any values. Is it just a case of adding up how many times $P(B,T\mid U)$ occurs over the whole data set e.g. $(2/12) = 16\%$.

Then would the probability of $P(U)$ be $0$?

答:
If I've read the question correctly, then I think your reasoning is sound, but your numbers are a little off.

If we assume that the probabilities associated with hair colour, height and country are all independent, then I think your approach of just counting works reasonably well. The numbers just need adjusting:
$$P(B,T \mid U) = \frac{\text{number of events of the form $(B,T,U)$}}{\text{number of events with country } U} = \frac{2}{8} = \frac{1}{4}.$$
Notice that the denominator is $8$, not $12$ as you stated. We're only looking at events where it's given that the country is $U$, so we don't count events where the country is $(\text{not } U)$.

Repeat similarly as appropriate for the other events.

Collected other thoughts:

We can easily see that $P(U)$ is probably non-zero, since we have people whose country is $U$. If $P(U)=0$, then this would clearly be ridiculous.

I'm not sure how much use Bayes' rule is here. I can see why you might want to use it (and maybe there's a way to apply it that I've missed), but you'll still need to calculate conditional probabilities from the sample somewhere, so you might as well go straight to it.

Finally, as Henning noted in a comment, the word "estimate" has a different meaning in statistics. Normally it's just a fancy word for a guess, but in statistics it really means an educated guess of some quantity (such as the probability distribution, which we're calculated here) based on some data we've collected.

For example, if I count the number of red cars and blue cars on a road (say, 10 and 20, respectively), then I can estimate the probability distribution as
$$P(\text{colour} = \text{red}) = \frac{10}{10+20} = \frac{1}{3}, \qquad P(\text{colour} = \text{red}) = \frac{20}{10+20} = \frac{2}{3}.$$
I've still done numerical calculations, but this is still called an estimate in statistics.


问:
Is a quotient set the same as a partition?
So I've just started trying to teach myself some topology and im really confused on what a partition is and how it is in any way related to an equivalence relation.

My main confusion however is that of the concept of a quotient set I've read that a quotient set is the set of all equivalence classes of a set namely if $S$ is some set with an equivalence relation $R$

Then is the quotient set $S/R=\{[a]  |  a \in S\} $

My question is how is this quotient set the same as a partition of a set?

答: When you have a quotient, each class is a subset of elements related by whatever your relation $R$ is. Those subsets are disjoint and each element of $S$ belongs to one of them. Therefore they form a partition of $S$. To see that they are disjoint, suppose $x\in [a]\cap[b]$ with $a$ and $b$ not related by $R$. Then you would have $xRa$ and $xRb$ and from transitivity $aRb$, a contradiction. It should be clear that each element of $S$ belongs to some class (which could be a singleton).


问:
Equivalence $\|A\|=\inf\left\lbrace c>0:\|Ax\|\leq c\|x\| \right\rbrace$
Prove that if $A$ is a matrix, and we defined $$\|A\|=\sup\left\lbrace\frac{\|Ax\|}{\|x\|}:x\neq 0\right\rbrace $$
then $$\|A\|=\inf\left\lbrace c>0:\|Ax\|\leq c\|x\| \right\rbrace $$

Attempt: $\|A\|$ is a lower bound, because $\|Ax\|\leq \|A\|\|x\|$, for all $x\in \mathbb{R}^n$ so $$ \|A\|\leq \inf\left\lbrace c>0:\|Ax\|\leq c\|x\| \right\rbrace$$ if I were able to prove that $\|A\|\geq \inf\left\lbrace c>0:\|Ax\|\leq c\|x\| \right\rbrace$ then this will be done! Or another criteria will be show that $$\forall \varepsilon>0, \exists \tilde{c}\in \left\lbrace c>0:\|Ax\|\leq c\|x\| \right\rbrace, \ \mathrm{such\ that}\ \|A\|+\varepsilon>\tilde{c} $$
I tried and the only thing that I achieved $$\forall \varepsilon>0,\exists\tilde{x}\in\mathbb{R}^n,\ \mathrm{such\ that},\ \|A\|-\varepsilon< \frac{\|A\tilde{x}\|}{\|\tilde{x}\|}$$ from the definition of $\sup$ of $\|A\|$, any help or hint?

答:
Let's call this infimum $I$. For all $x$ we have $||Ax||\leq ||A||\cdot ||x||$, and so $I\leq ||A||$.

Conversely, if $||Ax||\leq C||x||$ for all $x$, then for all $x\ne 0$ we have $\frac{||Ax||}{||x||}\leq C$. By taking the supremum on $x\ne 0$ we obtain $||A||\leq C$. Now we take the infimum on all such values of $C$ to get $||A||\leq I$.